url
stringlengths 14
2.42k
| text
stringlengths 100
1.02M
| date
stringlengths 19
19
| metadata
stringlengths 1.06k
1.1k
|
|---|---|---|---|
https://theculture.sg/2015/08/2014-a-level-h1-mathematics-8864-question-3-suggested-solutions/
|
All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.
(i)
(ii)
$\frac{dy}{dx}= -e^{1-2x}(-2) = 2 e^{1-2x}$
Sub $(1, 1- \frac{1}{e}), ~ \frac{dy}{dx} = \frac{2}{e}$
Tangent: $y - (1-\frac{1}{e}) = \frac{2}{e}(x-1)$
$\therefore, y = \frac{2}{e}x - \frac{3}{e} +1$
|
2018-11-21 09:55:58
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2645537257194519, "perplexity": 13262.468086429193}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039747665.82/warc/CC-MAIN-20181121092625-20181121114625-00223.warc.gz"}
|
https://mooseframework.inl.gov/source/actions/CavityPressureAction.html
|
Cavity Pressure Action
Description
The CavityPressureAction is one of three actions in the CavityPressure action system which are intended to be used concurrently. The intention of the CavityPressure action system is to reduce the number of input file blocks required to compute the pressure exerted by a gas contained in an internal volume.
|
2018-12-15 21:11:29
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7054001092910767, "perplexity": 708.5143949776219}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376827097.43/warc/CC-MAIN-20181215200626-20181215222626-00265.warc.gz"}
|
https://www.aimsciences.org/article/doi/10.3934/amc.2020074?viewType=html
|
# American Institute of Mathematical Sciences
August 2021, 15(3): 415-422. doi: 10.3934/amc.2020074
## The $[46, 9, 20]_2$ code is unique
Mathematisches Institut, Universität Bayreuth, D-95440 Bayreuth, Germany
Received September 2019 Revised January 2020 Published April 2020
The minimum distance of all binary linear codes with dimension at most eight is known. The smallest open case for dimension nine is length $n = 46$ with known bounds $19\le d\le 20$. Here we present a $[46,9,20]_2$ code and show its uniqueness. Interestingly enough, this unique optimal code is asymmetric, i.e., it has a trivial automorphism group. Additionally, we show the non-existence of $[47,10,20]_2$ and $[85,9,40]_2$ codes.
Citation: Sascha Kurz. The $[46, 9, 20]_2$ code is unique. Advances in Mathematics of Communications, 2021, 15 (3) : 415-422. doi: 10.3934/amc.2020074
##### References:
[1] L. D. Baumert and R. J. McEliece, A note on the Griesmer bound, IEEE Transactions on Information Theory, IT-19 (1973), 134-135. doi: 10.1109/tit.1973.1054939. Google Scholar [2] I. Bouyukliev, D. B. Jaffe and V. Vavrek, The smallest length of eight-dimensional binary linear codes with prescribed minimum distance, IEEE Transactions on Information Theory, 46 (2000), 1539-1544. doi: 10.1109/18.850690. Google Scholar [3] I. G. Bouyukliev, What is $Q$-extension?, Serdica Journal of Computing, 1 (2007), 115-130. Google Scholar [4] I. Bouyukliev and D. B. Jaffe, Optimal binary linear codes of dimension at most seven, Discrete Mathematics, 226 (2001), 51-70. doi: 10.1016/S0012-365X(00)00125-4. Google Scholar [5] S. Dodunekov, S. Guritman and J. Simonis, Some new results on the minimum length of binary linear codes of dimension nine, IEEE Transactions on Information Theory, 45 (1999), 2543-2546. doi: 10.1109/18.796403. Google Scholar [6] M. Grassl, Bounds on the minimum distance of linear codes and quantum codes, Online available at: http://www.codetables.de, (2007), Accessed on 2019-04-04. Google Scholar [7] J. H. Griesmer, A bound for error-correcting codes, IBM Journal of Research and Development, 4 (1960), 532-542. doi: 10.1147/rd.45.0532. Google Scholar [8] S. Kurz, Lincode - computer classification of linear codes, arXiv: 1912.09357. Google Scholar [9] F. J. MacWilliams and N. J. A. Sloane, The Theory of Error-Correcting Codes. II, North-Holland Mathematical Library, Vol. 16. North-Holland Publishing Co., Amsterdam-New York-Oxford, 1977. Google Scholar [10] J. Simonis, Restrictions on the weight distribution of binary linear codes imposed by the structure of reed-muller codes, IEEE transactions on Information Theory, 40 (1994), 194-196. doi: 10.1109/18.272480. Google Scholar [11] J. Simonis, The $[23, 14, 5]$ Wagner code is unique, Discrete Mathematics, 213 (2000), 269-282. doi: 10.1016/S0012-365X(99)00187-9. Google Scholar [12] H. C. A. van Tilborg, The smallest length of binary $7$-dimensional linear codes with prescribed minimum distance, Discrete Mathematics, 33 (1981), 197-207. doi: 10.1016/0012-365X(81)90166-7. Google Scholar [13] H. N. Ward, Divisible codes - a survey, Serdica Mathematical Journal, 27 (2001), 263-278. Google Scholar
show all references
##### References:
[1] L. D. Baumert and R. J. McEliece, A note on the Griesmer bound, IEEE Transactions on Information Theory, IT-19 (1973), 134-135. doi: 10.1109/tit.1973.1054939. Google Scholar [2] I. Bouyukliev, D. B. Jaffe and V. Vavrek, The smallest length of eight-dimensional binary linear codes with prescribed minimum distance, IEEE Transactions on Information Theory, 46 (2000), 1539-1544. doi: 10.1109/18.850690. Google Scholar [3] I. G. Bouyukliev, What is $Q$-extension?, Serdica Journal of Computing, 1 (2007), 115-130. Google Scholar [4] I. Bouyukliev and D. B. Jaffe, Optimal binary linear codes of dimension at most seven, Discrete Mathematics, 226 (2001), 51-70. doi: 10.1016/S0012-365X(00)00125-4. Google Scholar [5] S. Dodunekov, S. Guritman and J. Simonis, Some new results on the minimum length of binary linear codes of dimension nine, IEEE Transactions on Information Theory, 45 (1999), 2543-2546. doi: 10.1109/18.796403. Google Scholar [6] M. Grassl, Bounds on the minimum distance of linear codes and quantum codes, Online available at: http://www.codetables.de, (2007), Accessed on 2019-04-04. Google Scholar [7] J. H. Griesmer, A bound for error-correcting codes, IBM Journal of Research and Development, 4 (1960), 532-542. doi: 10.1147/rd.45.0532. Google Scholar [8] S. Kurz, Lincode - computer classification of linear codes, arXiv: 1912.09357. Google Scholar [9] F. J. MacWilliams and N. J. A. Sloane, The Theory of Error-Correcting Codes. II, North-Holland Mathematical Library, Vol. 16. North-Holland Publishing Co., Amsterdam-New York-Oxford, 1977. Google Scholar [10] J. Simonis, Restrictions on the weight distribution of binary linear codes imposed by the structure of reed-muller codes, IEEE transactions on Information Theory, 40 (1994), 194-196. doi: 10.1109/18.272480. Google Scholar [11] J. Simonis, The $[23, 14, 5]$ Wagner code is unique, Discrete Mathematics, 213 (2000), 269-282. doi: 10.1016/S0012-365X(99)00187-9. Google Scholar [12] H. C. A. van Tilborg, The smallest length of binary $7$-dimensional linear codes with prescribed minimum distance, Discrete Mathematics, 33 (1981), 197-207. doi: 10.1016/0012-365X(81)90166-7. Google Scholar [13] H. N. Ward, Divisible codes - a survey, Serdica Mathematical Journal, 27 (2001), 263-278. Google Scholar
Number of $[n,k,\{20,24,28,32\}]_2$ codes
k/n 20 24 28 30 32 34 35 36 37 38 39 40 41 42 43 44 1 1 1 1 0 1 0 0 0 0 0 2 1 1 2 0 3 0 3 0 3 1 1 2 4 6 9 4 1 4 13 26 5 3 15 163 6 24 3649 7 5 337794
k/n 20 24 28 30 32 34 35 36 37 38 39 40 41 42 43 44 1 1 1 1 0 1 0 0 0 0 0 2 1 1 2 0 3 0 3 0 3 1 1 2 4 6 9 4 1 4 13 26 5 3 15 163 6 24 3649 7 5 337794
Number of $[n,k,\{40,48,56\}]_2$ codes
k/n 40 48 56 60 64 68 70 72 74 75 76 77 78 79 80 81 82 83 1 1 1 1 0 0 0 0 0 0 0 0 0 2 1 1 2 0 2 0 0 2 0 0 3 1 1 2 0 3 0 5 0 4 1 1 2 3 6 10 5 1 3 11 16 6 2 8 106 7 7 5613
k/n 40 48 56 60 64 68 70 72 74 75 76 77 78 79 80 81 82 83 1 1 1 1 0 0 0 0 0 0 0 0 0 2 1 1 2 0 2 0 0 2 0 0 3 1 1 2 0 3 0 5 0 4 1 1 2 3 6 10 5 1 3 11 16 6 2 8 106 7 7 5613
Number of $[84,8,\{40,48,56\}]_2$ codes per $A_{56}$
$A_{56}$ 3 4 5 6 7 8 9 25773 48792 26091 5198 450 17 1
$A_{56}$ 3 4 5 6 7 8 9 25773 48792 26091 5198 450 17 1
[1] Yun Gao, Shilin Yang, Fang-Wei Fu. Some optimal cyclic $\mathbb{F}_q$-linear $\mathbb{F}_{q^t}$-codes. Advances in Mathematics of Communications, 2021, 15 (3) : 387-396. doi: 10.3934/amc.2020072 [2] Antonio Cossidente, Sascha Kurz, Giuseppe Marino, Francesco Pavese. Combining subspace codes. Advances in Mathematics of Communications, 2021 doi: 10.3934/amc.2021007 [3] Alexander A. Davydov, Massimo Giulietti, Stefano Marcugini, Fernanda Pambianco. Linear nonbinary covering codes and saturating sets in projective spaces. Advances in Mathematics of Communications, 2011, 5 (1) : 119-147. doi: 10.3934/amc.2011.5.119 [4] W. Cary Huffman. On the theory of $\mathbb{F}_q$-linear $\mathbb{F}_{q^t}$-codes. Advances in Mathematics of Communications, 2013, 7 (3) : 349-378. doi: 10.3934/amc.2013.7.349 [5] Muhammad Ajmal, Xiande Zhang. New optimal error-correcting codes for crosstalk avoidance in on-chip data buses. Advances in Mathematics of Communications, 2021, 15 (3) : 487-506. doi: 10.3934/amc.2020078 [6] Jennifer D. Key, Bernardo G. Rodrigues. Binary codes from $m$-ary $n$-cubes $Q^m_n$. Advances in Mathematics of Communications, 2021, 15 (3) : 507-524. doi: 10.3934/amc.2020079 [7] Jérôme Ducoat, Frédérique Oggier. On skew polynomial codes and lattices from quotients of cyclic division algebras. Advances in Mathematics of Communications, 2016, 10 (1) : 79-94. doi: 10.3934/amc.2016.10.79 [8] Emily McMillon, Allison Beemer, Christine A. Kelley. Extremal absorbing sets in low-density parity-check codes. Advances in Mathematics of Communications, 2021 doi: 10.3934/amc.2021003 [9] Ricardo A. Podestá, Denis E. Videla. The weight distribution of irreducible cyclic codes associated with decomposable generalized Paley graphs. Advances in Mathematics of Communications, 2021 doi: 10.3934/amc.2021002 [10] Jong Yoon Hyun, Yoonjin Lee, Yansheng Wu. Connection of $p$-ary $t$-weight linear codes to Ramanujan Cayley graphs with $t+1$ eigenvalues. Advances in Mathematics of Communications, 2021 doi: 10.3934/amc.2020133 [11] Hakan Özadam, Ferruh Özbudak. A note on negacyclic and cyclic codes of length $p^s$ over a finite field of characteristic $p$. Advances in Mathematics of Communications, 2009, 3 (3) : 265-271. doi: 10.3934/amc.2009.3.265 [12] Raj Kumar, Maheshanand Bhaintwal. Duadic codes over $\mathbb{Z}_4+u\mathbb{Z}_4$. Advances in Mathematics of Communications, 2021 doi: 10.3934/amc.2020135 [13] Joe Gildea, Adrian Korban, Abidin Kaya, Bahattin Yildiz. Constructing self-dual codes from group rings and reverse circulant matrices. Advances in Mathematics of Communications, 2021, 15 (3) : 471-485. doi: 10.3934/amc.2020077 [14] Dean Crnković, Nina Mostarac, Bernardo G. Rodrigues, Leo Storme. $s$-PD-sets for codes from projective planes $\mathrm{PG}(2,2^h)$, $5 \leq h\leq 9$. Advances in Mathematics of Communications, 2021, 15 (3) : 423-440. doi: 10.3934/amc.2020075 [15] Yishui Wang, Dongmei Zhang, Peng Zhang, Yong Zhang. Local search algorithm for the squared metric $k$-facility location problem with linear penalties. Journal of Industrial & Management Optimization, 2021, 17 (4) : 2013-2030. doi: 10.3934/jimo.2020056 [16] Diana Keller. Optimal control of a linear stochastic Schrödinger equation. Conference Publications, 2013, 2013 (special) : 437-446. doi: 10.3934/proc.2013.2013.437 [17] Luke Finlay, Vladimir Gaitsgory, Ivan Lebedev. Linear programming solutions of periodic optimization problems: approximation of the optimal control. Journal of Industrial & Management Optimization, 2007, 3 (2) : 399-413. doi: 10.3934/jimo.2007.3.399 [18] Vladimir Gaitsgory, Ilya Shvartsman. Linear programming estimates for Cesàro and Abel limits of optimal values in optimal control problems. Discrete & Continuous Dynamical Systems - B, 2021 doi: 10.3934/dcdsb.2021102 [19] Zhiming Guo, Zhi-Chun Yang, Xingfu Zou. Existence and uniqueness of positive solution to a non-local differential equation with homogeneous Dirichlet boundary condition---A non-monotone case. Communications on Pure & Applied Analysis, 2012, 11 (5) : 1825-1838. doi: 10.3934/cpaa.2012.11.1825 [20] Izumi Takagi, Conghui Zhang. Existence and stability of patterns in a reaction-diffusion-ODE system with hysteresis in non-uniform media. Discrete & Continuous Dynamical Systems, 2021, 41 (7) : 3109-3140. doi: 10.3934/dcds.2020400
2019 Impact Factor: 0.734
## Tools
Article outline
Figures and Tables
|
2021-04-18 14:09:29
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6639277935028076, "perplexity": 1533.2004646123646}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038492417.61/warc/CC-MAIN-20210418133614-20210418163614-00597.warc.gz"}
|
https://www.enotes.com/homework-help/stuck-397133
|
# stucki'm stuck in evaluation of the expression 1/xy(x+y+z)+1/yz(x+y+z)+1/zx(x+y+z)
Expert Answers
terryhi | Certified Educator
1/xy(x+y+z)+1/yz(x+y+z)+1/xz(x+y+z)
The least common denominator is xyz(x+y+z)
Since this is not a full equation- it does not have "=" sign-, we cannot multiply least common denominator. But, we can combine the phrase using the least common denominator.
1/xyz(x+y+z){z+x+y}
Notice if you eliminate the bracket, the value of the equation has not changed.
giorgiana1976 | Student
Since the fractions don't have a common denominator, we'll have to find LCD:
LCD = xyz(x+y+z)
We'll multiply by LCD each fraction:
xyz(x+y+z)/xy(x+y+z)+xyz(x+y+z)/yz(x+y+z)+xyz(x+y+z)/zx(x+y+z)
We'll simplify and we'll get:
z + x + y
After the evaluation of the expression, we've get the result:
1/xy(x+y+z)+1/yz(x+y+z)+1/zx(x+y+z) = x + y + z
alone- | Student
Sorry i got mixed up on the first part.
1/x^2y+xy^2+xyz
alone- | Student
1/x^2+xy^2+xyz+1/xyz+y^2z+yz^2+1/zx^2+xyz+z^2x
### Access hundreds of thousands of answers with a free trial.
Ask a Question
Related Questions
Popular Questions
|
2017-11-22 07:38:05
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8123719692230225, "perplexity": 3097.0586180590776}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806509.31/warc/CC-MAIN-20171122065449-20171122085449-00699.warc.gz"}
|
http://www.wellformedness.com/blog/understanding-text-encoding-in-python-2-and-python-3/
|
# Understanding text encoding in Python 2 and Python 3
Computers were rather late to the word processing game. The founding mothers and fathers of computing were primarily interested in numbers. This is fortunate: after all, computers only know about numbers. But as Brian Kunde explains in his brief history of word processing, word processing existed long before digital computing, and the text processing has always been something of an afterthought.
Humans think of text as consisting of ordered sequence of “characters” (an ill-defined Justice-Stewart-type concept which I won’t attempt to clarify here). To manipulate text in digital computers, we have to have a mapping between the character set (a finite list of the characters the system recognizes) and numbers. Encoding is the process of converting characters to numbers, and decoding is (naturally) the process of converting numbers to characters. Before we get to Python, a bit of history.
# ASCII and Unicode
There are only a few character sets that have any relevance to life in 2014. The first is ASCII (American Standard Code for Information Interchange), which was first published in 1963. This character set consists of 128 characters intended for use by an English audience. Of these 95 are printable, meaning that they correspond to lay-human notions about characters. On a US keyboard, these are (approximately) the alphanumeric and punctuation characters that can be typed with a single keystroke, or with a single keystroke while holding down the Shift key, space, tab, the two newline characters (which you get when you type return), and a few apocrypha. The remaining 33 are non-printable “control characters”. For instance, the first character in the ASCII table is the “null byte”. This is indicated by a '' in C and other languages, but there’s no standard way to render it. Many control characters were designed for earlier, more innocent times; for instance, character #7 'a' tells the receiving device to ring a cute little bell (which were apparently attached to teletype terminals); today your computer might make a beep, or the terminal window might flicker once, but either way, nothing is printed.
Of course, this is completely inadequate for anything but English (not to mention those users of superfluous diaresis…e.g., the editors of the New Yorker, Motörhead). However, each ASCII character takes up only 7 bits, leaving room for another 128 characters (since a byte has an integer value between 0-255, inclusive), and so engineers could exploited the remaining 128 characters to write the characters from different alphabets, alphasyllabaries, or syllabaries. Of these ASCII-based character sets, the best-known are ISO/IEC 8859-1, also known as Latin-1, and Windows-1252, also known as CP-1252. Unfortunately, this created more problems than it solved. That last bit just didn’t leave enough space for the many languages which need a larger character set (Japanese kanji being an obvious example). And even when there are technically enough code points left over, engineers working in different languages didn’t see eye-to-eye about what to do with them. As a result, the state of affairs made it impossible to, for example, write in French (ISO/IEC 8859-1) about Ukrainian (ISO/IEC 8859-5, at least before the 1990 orthography reform).
Clearly, fighting over scraps isn’t going to cut it in the global village. Enter the Unicode standard and its Universal Character Set (UCS), first published in 1991. Unicode is the platonic ideal of an character encoding, abstracting away from the need to efficiently convert all characters to numbers. Each character is represented by a single code with various metadata (e.g., A is an “Uppercase Letter” from the “Latin” script). ASCII and its extensions map onto a small subset of this code.
Fortunately, not all encodings are merely shadows on the walls of a cave. The One True Encoding is UTF-8, which implements the entire UCS using an 8-bit code. There are other encodings, of course, but this one is ours, and I am not alone in feeling strongly that UTF-8 is the chosen encoding. At the risk of getting too far afield, here are two arguments for why you and everyone you know should just use UTF-8. First off, it is hardly matters much which UCS-compatible encoding we all use (the differences between them are largely arbitrary), but what does matter is that we all choose the same one. There is no general procedure for “sniffing” out the encoding of a file, and there’s nothing preventing you from coming up with a file that’s a French cookbook in one encoding, and a top-secret message in another. This is good for steganographers, but bad for the rest of us, since so many text files lack encoding metadata. When it comes to encodings, there’s no question that UTF-8 is the most popular Unicode encoding scheme worldwide, and is on its way to becoming the de-facto standard. Secondly, ASCII is valid UTF-8, because UTF-8 and ASCII encode the ASCII characters in exactly the same way. What this means, practically speaking, is you can achieve nearly complete coverage of the world’s languages simply by assuming that all the inputs to your software are UTF-8. This is a big, big win for us all.
# Decode early, encode late
A general rule of thumb for developers is “decode early” (convert inputs to their Unicode representation), “encode late” (convert back to bytestrings). The reason for this is that in nearly any programming language, Unicode strings behave the way our monkey brains expect them to, but bytestrings do not. To see why, try iterating over non-ASCII bytestring in Python (more on the syntax later).
>>> for byte in b"año":
... print(byte)
...
a
?
?
o
There are two surprising things here: iterating over the bytestring returned more bytes then there are “characters” (goodbye, indexing), and furthermore the 2nd “character” failed to render properly. This is what happens when you let computers dictate the semantics to our monkey brains, rather than the other way around. Here’s what happens when we try the same with a Unicode string:
>>> for byte in u"año":
... print(byte)
...
a
ñ
o
# The Python 2 & 3 string models
Before you put this all into practice, it is important to note that Python 2 and Python 3 use very different string models. The familiar Python 2 str class is a bytestring. To convert it to a Unicode string, use the str.decode instance method, which returns a copy of the string as an instance of the unicode class. Similarly, you can make a str copy of a unicode instance with unicode.encode. Both of these functions take a single argument: a string (either kind!) representing the encoding.
Python 2 provides specific syntax for Unicode string literals (which you saw above): the a lower-case u prefix before the initial quotation mark (as in u"año").
When it comes to Unicode-awareness, Python 3 has totally flipped the script; in my opinion, it’s for the best. Instances of str are now Unicode strings (the u"" syntax still works, but is vacuous). The (reduced) functionality of the old-style strings is now just available for instances of the class bytes. As you might expect, you can create a bytes instance by using the encode method of a new-style str. Python 3 decodes bytestrings as soon as they are created, and (re)encodes Unicode strings only at the interfaces; in other words, it gets the “early/late” stuff right by default. Your APIs probably won’t need to change much, because Python 3 treats UTF-8 (and thus ASCII) as the default encoding, and this assumption is valid more often than not.
If for some reason, you want a bytestring literal, Python has syntax for that, too: prefix the quotation marks delimiting the string with a lower-case b (as in b"año"; see above also).
# tl;dr
Strings are ordered sequences of characters. But computers only know about numbers, so they are encoded as byte arrays; there are many ways to do this, but UTF-8 is the One True Encoding. To get the strings to have the semantics you expect as a human, decode a string to Unicode as early as possible, and encode it as bytes as late as possible. You have to do this explicitly in Python 2; it happens automatically in Python 3.
|
2017-03-29 15:06:42
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.33989226818084717, "perplexity": 1852.765439208955}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218190753.92/warc/CC-MAIN-20170322212950-00067-ip-10-233-31-227.ec2.internal.warc.gz"}
|
http://mathhelpforum.com/advanced-algebra/68729-proof-transpose-property-matrix.html
|
# Math Help - Proof of Transpose Property of Matrix
1. ## Proof of Transpose Property of Matrix
prove (AB)^T = (B^T)*(A^T)
using summation notation
i know how to prove it without summation notation but with summation notation i'm not getting anywhere
if someone can prove it and then explain the summation steps involved that would be of great help
2. Originally Posted by razorfever
prove (AB)^T = (B^T)*(A^T)
using summation notation
i know how to prove it without summation notation but with summation notation i'm not getting anywhere
if someone can prove it and then explain the summation steps involved that would be of great help
$(AB)^T_{i,j}=(AB)_{j,i}=\sum_k A_{j,k}B_{k,i}=\sum_k B_{k,i}A_{j,k}=\sum_k B^T_{i,k}A^T_{k,j}=(B^TA^T)_{i,j}$
.
3. If transpose was defined to you as $[A^T]_{ij} = [A]_{ji}$, then "summation sign" method as explained in the prev. post is most likely the only thing you could do.
Ignore whatever's below, I replied in a haste.
------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
However if you are doing linear algebra, and you are familiar with the notion of a linear functional $[x,y]$(Halmos' notation), then one could define $A'$, adjoint of linear transformation A as $[Ax,y] = [x,A'y]$.If $[A]_{\cal{X}}$ is the matrix of linear transformation A w.r.t the basis $\cal{X}$, then its easy to verify that the matrix of linear transformation A' is $[A]^T_{\cal{X}}$.
Now with all this setup, its extremely easy to prove $(AB)^T = B^TA^T$. Note that it suffices to prove $(AB)' = B'A'$.
But $\forall x,y : [x,(AB)'y] = [ABx,y] = [A(Bx),y] = [Bx, A'y] = [x, B'A'y]$
Thus $(AB)' = B'A'$
4. Originally Posted by Isomorphism
If transpose was defined to you as $[A^T]_{ij} = [A]_{ji}$, then "summation sign" method as explained in the prev. post is most likely the only thing you could do.
However if you are doing linear algebra, and you are familiar with the notion of a linear functional $[x,y]$(Halmos' notation), then one could define $A'$, adjoint of linear transformation A as $[Ax,y] = [x,A'y]$.If $[A]_{\cal{X}}$ is the matrix of linear transformation A w.r.t the basis $\cal{X}$, then its easy to verify that the matrix of linear transformation A' is $[A]^T_{\cal{X}}$.
Now with all this setup, its extremely easy to prove $(AB)^T = B^TA^T$. Note that it suffices to prove $(AB)' = B'A'$.
But $\forall x,y : [x,(AB)'y] = [ABx,y] = [A(Bx),y] = [Bx, A'y] = [x, B'A'y]$
Thus $(AB)' = B'A'$
The question explicity asks for a demonstration using summation notation.
.
|
2016-05-28 10:00:48
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 23, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9850924015045166, "perplexity": 477.3342338281076}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049277592.42/warc/CC-MAIN-20160524002117-00115-ip-10-185-217-139.ec2.internal.warc.gz"}
|
https://pypi.org/project/timetra.diary/
|
Diary with CLI + YAML
Project description
Timetra is a CLI time tracker application with a YAML backend.
It also has a simple curses frontend and used to have a web one (which proved useless compared to CLI).
Initially Timetra technically was a set of Hamster frontends. Now the backends are pluggable (the Hamster one was dropped completely).
Author
Originally written by Andrey Mikhaylenko since 2010.
Please feel free to submit patches, report bugs or request features:
https://github.com/neithere/timetra/issues/
Licensing
Timetra is free software: you can redistribute it and/or modify it under the terms of the GNU Lesser General Public License as published by the Free Software Foundation, either version 3 of the License, or (at your option) any later version.
Timetra is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU Lesser General Public License for more details.
You should have received a copy of the GNU Lesser General Public License along with Timetra. If not, see <http://gnu.org/licenses/>.
Project details
Uploaded source
|
2022-10-07 22:35:50
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.43555623292922974, "perplexity": 1739.4052175653837}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030338280.51/warc/CC-MAIN-20221007210452-20221008000452-00129.warc.gz"}
|
https://brilliant.org/problems/closure-properties/
|
Closure properties
Geometry Level 4
Here are three statements about the closure $$\overline S$$ of a set $$S$$ inside a metric space $$X.$$
I. Let $$S,T$$ be subsets of $$X.$$ Then $$\overline{S \cap T} = {\overline S} \cap {\overline T}.$$
II. Let $$S,T$$ be subsets of $$X.$$ Then $$\overline{S \cup T} = {\overline S} \cup {\overline T}.$$
III. If $${\overline S} = X,$$ then $$S=X.$$
Which of these statements is true?
×
|
2017-01-17 07:05:32
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9693004488945007, "perplexity": 271.8208518113692}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560279489.14/warc/CC-MAIN-20170116095119-00155-ip-10-171-10-70.ec2.internal.warc.gz"}
|
https://www.tutorialspoint.com/maximize-the-maximum-among-minimum-of-k-consecutive-sub-arrays-in-cplusplus
|
# Maximize the maximum among minimum of K consecutive sub-arrays in C++
C++Server Side ProgrammingProgramming
Given the task is to divide an array arr[] into K consecutive sub-arrays and find the maximum possible value of maximum among the minimum of the K consecutive sub-srrays.
Input
arr[]={2,8,4,3,9,1,5}, K=3
Output
9
Explanation − The 3 consecutive sub arrays that can made are: {2, 8, 4, 3}, {9}, and {1, 5}
The minimum values out of all these arrays are: (2, 9, 1)
The maximum value out of these three is 9.
Input
arr[] = { 8, 4, 1, 9, 11}, K=1
Output
11
## Approach used in the below program as follows
• If we look at the task, it can be divided into 3 cases − when K=1, k=2 and k>=3.
• Case 1 − K=1
When k=1 the sub-array is equal to the array itself and so the minimum value in the array will be the output.
• Case 2 − K=2
This is a tough case. In this case we will have to make two arrays that will contain the prefix and suffix minimums as the array can only be divided into 2 parts. Then for every element of the array we will have to do so −
MaxValue = max(MaxValue, max(prefix minimum value at i, suffix maximum value at i+1))
## Example
Live Demo
#include <bits/stdc++.h>
using namespace std;
/* Function to find the maximum possible value
of the maximum of minimum of K sub-arrays*/
int Max(const int* arr, int size, int K){
dint Max;
int Min;
//Obtain maximum and minimum
for (int i = 0; i < size; i++){
Min = min(Min, arr[i]);
Max = max(Max, arr[i]);
}
//When K=1, return minimum value
if (K == 1){
return Min;
}
//When K>=3, return maximum value
else if (K >= 3){
return Max;
}
/*When K=2 then make prefix and suffix minimums*/
else{
// Arrays to store prefix and suffix minimums
int Left[size], Right[size];
Left[0] = arr[0];
Right[size - 1] = arr[size - 1];
// Prefix minimum
for (int i = 1; i < size; i++){
Left[i] = min(Left[i - 1], arr[i]);
}
// Suffix minimum
for (int i = size - 2; i >= 0; i--){
Right[i] = min(Right[i + 1], arr[i]);
}
int MaxValue=INT_MIN;
// Get the maximum possible value
for (int i = 0; i < size - 1; i++){
MaxValue = max(MaxValue, max(Left[i], Right[i + 1]));
}
return MaxValue;
}
}
int main(){
int arr[] = {9,4,12,5,6,11};
int size = sizeof(arr) / sizeof(arr[0]);
int K = 2;
cout<<"Maximize the maximum among minimum of K consecutive sub-arrays is: "<<Max(arr, size, K);
return 0;
}
## Output
If we run the above code we will get the following output −
Maximize the maximum among minimum of K consecutive sub-arrays is: 11
Published on 14-Aug-2020 07:30:23
Advertisements
|
2021-10-24 07:35:05
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.35722535848617554, "perplexity": 3639.037357876596}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585911.17/warc/CC-MAIN-20211024050128-20211024080128-00634.warc.gz"}
|
http://openstudy.com/updates/508f2a91e4b0ad62053739b7
|
• anonymous
I am having trouble in trig/ geometry. It involves radicals, 30-60-90 triangles, and 45-45-90 triangles or something like that, I had poor notes, If you could give me a guide thats visual would be the most helpful. My math teacher showed me part of what I was looking for. It was a 30-60-90 or the other one I mentioned and he drew a triangle and then wrote multiply radical 3 to get this side.... I hope that made sense :P sorry if it didn't :/
Mathematics
Looking for something else?
Not the answer you are looking for? Search for more explanations.
|
2017-04-23 12:05:25
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8713482022285461, "perplexity": 582.8989182990323}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917118552.28/warc/CC-MAIN-20170423031158-00324-ip-10-145-167-34.ec2.internal.warc.gz"}
|
https://www.projecteuclid.org/euclid.aos/1176342757
|
## The Annals of Statistics
### Admissibility of Translation Invariant Tolerance Intervals in the Location Parameter Case
Saul Blumenthal
#### Abstract
Given $n$ independent observations with common density $f(x - \theta)$, and a rv $z$ independent of these with density $g(x - \theta) (f, g$ known except for $\theta$) a prediction region for $z$ is required. It is shown that the best translation invariant interval is optimal in two senses: (1) there is no other region with the same expected coverage (coverage is the probability of containing $z$) and uniformly smaller expected size (Lebesgue measure); (2) no other interval having the same confidence that the coverage exceeds $\beta$ (given) can have uniformly smaller expected length. The best invariant interval in each case is found, and the normal case is studied. The usual interval centered at $\bar{X}$ is not always optimal in the second sense if $\beta$ and/or confidence are small. A criterion involving expected coverage and the confidence of exceeding coverage $\beta$ is also examined. Again restrictions on these are needed for the usual normal interval to be optimal.
#### Article information
Source
Ann. Statist., Volume 2, Number 4 (1974), 694-702.
Dates
First available in Project Euclid: 12 April 2007
https://projecteuclid.org/euclid.aos/1176342757
Digital Object Identifier
doi:10.1214/aos/1176342757
Mathematical Reviews number (MathSciNet)
MR370900
Zentralblatt MATH identifier
0297.62024
JSTOR
|
2019-10-18 22:38:56
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8476606011390686, "perplexity": 1085.2134043217807}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986684854.67/warc/CC-MAIN-20191018204336-20191018231836-00552.warc.gz"}
|
http://www.koreascience.or.kr/article/ArticleFullRecord.jsp?cn=E1BMAX_2014_v51n4_995
|
CERTAIN NEW INTEGRAL FORMULAS INVOLVING THE GENERALIZED BESSEL FUNCTIONS
Title & Authors
CERTAIN NEW INTEGRAL FORMULAS INVOLVING THE GENERALIZED BESSEL FUNCTIONS
Choi, Junesang; Agarwal, Praveen; Mathur, Sudha; Purohit, Sunil Dutt;
Abstract
A remarkably large number of integral formulas involving a variety of special functions have been developed by many authors. Also many integral formulas involving various Bessel functions have been presented. Very recently, Choi and Agarwal derived two generalized integral formulas associated with the Bessel function $\small{J_{\nu}(z)}$ of the first kind, which are expressed in terms of the generalized (Wright) hypergeometric functions. In the present sequel to Choi and Agarwals work, here, in this paper, we establish two new integral formulas involving the generalized Bessel functions, which are also expressed in terms of the generalized (Wright) hypergeometric functions. Some interesting special cases of our two main results are presented. We also point out that the results presented here, being of general character, are easily reducible to yield many diverse new and known integral formulas involving simpler functions.
Keywords
Gamma function;hypergeometric function $\small{_pF_q}$;generalized (Wright) hypergeometric functions $\small{_p{\Psi}_q}$;Bessel functions;generalized Bessel function of the first kind;Oberhettingers integral formula;
Language
English
Cited by
1.
SOME INTEGRALS ASSOCIATED WITH MULTIINDEX MITTAG-LEFFLER FUNCTIONS,;;;
Journal of applied mathematics & informatics, 2016. vol.34. 3_4, pp.249-255
1.
SOME INTEGRALS ASSOCIATED WITH MULTIINDEX MITTAG-LEFFLER FUNCTIONS, Journal of applied mathematics & informatics, 2016, 34, 3_4, 249
References
1.
A. Baricz, Geometric properties of generalized Bessel functions of complex order, Mathematica 48(71) (2006), no. 1, 13-18.
2.
A. Baricz, Geometric properties of generalized Bessel functions, Publ. Math. Debrecen 73 (2008), no. 1-2, 155-178.
3.
A. Baricz, Jordan-type inequalities for generalized Bessel functions, J. Inequal. Pure Appl. Math. 9 (2008), no. 2, Art. 39, 6 pp.
4.
A. Baricz, Generalized Bessel Functions of the First Kind, Springer-Verlag, Berlin, Hei-delberg, 2010.
5.
Y. A. Brychkov, Handbook of Special Functions, Derivatives, Integrals, Series and Other Formulas, CRC Press, Taylor & Francis Group, Boca Raton, London, and New York, 2008.
6.
J. Choi and P. Agarwal, Certain unified integrals associated with Bessel functions, Bound. Value Probl. 2013 (2013), no. 95, 9 pp.
7.
J. Choi, A. Hasanov, H. M. Srivastava, and M. Turaev, Integral representations for Srivastava's triple hypergeometric functions, Taiwanese J. Math. 15 (2011), no. 6, 2751-2762.
8.
C. Fox, The asymptotic expansion of generalized hypergeometric functions, Proc. London Math. Soc. (2) 27 (1928), 389-400.
9.
M. Garg and S. Mittal, On a new unified integral, Proc. Indian Acad. Sci. Math. Sci. 114 (2004), no. 2, 99-101.
10.
P. Malik, S. R. Mondal, and A. Swaminathan, Fractional integration of generalized Bessel function of the first kind, IDETC/CIE, 2011, USA.
11.
F. Oberhettinger, Tables of Mellin Transforms, Springer-Verlag, New York, 1974.
12.
H. M. Srivastava and J. Choi, Zeta and q-Zeta Functions and Associated Series and Integrals, Elsevier Science Publishers, Amsterdam, London and New York, 2012.
13.
H. M. Srivastava and P. W. Karlsson, Multiple Gaussian Hypergeometric Series, Halsted Press (Ellis Horwood Limited, Chichester), John Wiley and Sons, New York, Chichester, Brisbane, and Toronto, 1985.
14.
E. M. Wright, The asymptotic expansion of the generalized hypergeometric functions, J. London Math. Soc. 10 (1935), 286-293.
15.
E. M. Wright, The asymptotic expansion of integral functions defined by Taylor series, Philos. Trans. Roy. Soc. London A 238 (1940), 423-451.
16.
E. M. Wright, The asymptotic expansion of the generalized hypergeometric function, Proc. London Math. Soc. 46 (1940), no. 2, 389-408.
|
2018-09-24 01:35:51
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 3, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8540111184120178, "perplexity": 1371.2062141403342}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267160085.77/warc/CC-MAIN-20180924011731-20180924032131-00209.warc.gz"}
|
https://mathematica.stackexchange.com/questions/176147/findinstance-within-a-range
|
# FindInstance within a range
I am trying to find particular solutions to my inequality for two positive integers $a$ and $b$:
FindInstance[a<f[etc...]*b,{a,b},Integers]
However I want $a$ and $b$ to to be both above $10^3$. How may I achieve this? Is there a way to use Assumptions for this or some sort of magic along the lines of:
FindInstance[a<f[etc...]*b,{a,b},Integers,a>10^3 && b>10^3]
FindInstance[a < a*b && a > 10^3 && b > 10^3, {a, b}]
|
2019-06-27 01:05:43
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5130227208137512, "perplexity": 533.449683299152}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560628000609.90/warc/CC-MAIN-20190626234958-20190627020958-00048.warc.gz"}
|
https://design.tutsplus.com/tutorials/how-to-create-an-enchanted-rose-photo-manipulation-in-adobe-photoshop--cms-29037
|
# How to Create an Enchanted Rose Photo Manipulation in Adobe Photoshop
Let's create a magical photo manipulation using basic tools in Adobe Photoshop.
In today's tutorial, we'll create an enchanted rose inspired by the classic Disney movie Beauty and the Beast. I'll show you how to build the entire scene from scratch just by keeping an eye out for the right stocks.
For more inspiration, find incredible Magic-inspired Graphics on Envato Market.
## Tutorial Assets
The following assets were used in the production of this tutorial:
## How to Find the Right Stocks
To create this scene, I did a basic search on Google and Pinterest for enchanted rose images. Taking a few notes in the beginning is a good way to have a game plan before you attempt this manipulation. Here are some important ideas to keep in mind for a composition that reflects the original Disney movie.
1. The enchanted rose is often in front of a large window.
2. The window has stained glass panels reminiscent of the time period.
3. The table used to hold the rose is made of stone.
4. The glass case can work with or without a base.
5. The rose sparkles with magic from the enchantment spell.
Just knowing these few tidbits helps me understand the kind of stocks I need to choose. Bookmark plenty of alternative images to experiment with what works best.
Now let's move on to the manipulation!
## 1. How to Set Up the Stained Glass Window
### Step 1
Open your Stained Glass Window reference in Adobe Photoshop. We'll be using its normal dimension size (1280 x 1280 pixels) to build this magical scene.
Since the window is a little crooked, let's straighten it out. Control-T to Free Transform the window, Rotating the window until it's straight. Then adjust the Scale and Perspective so that it's much taller and the top of the window stretches a little outside of the canvas.
Control-J to create a Duplicate of the window. Adjust the size of the duplicate so that there's more wall space on the left and right sides of the composition. Since the window is just an accessory to this composition, we need to make sure it's not placed too far forward in the scene.
Don't worry if there's a little excess window from the image underneath it—we'll cover that later!
### Step 2
Now blur the window to create depth of field. Go to Filter > Blur Gallery > Field Blur, and add a Blur of 5 pixels.
Desaturate the blurred image. Go to Image > Adjustments > Hue & Saturation. Lower the Saturation to -49.
### Step 3
Before we add more elements to the scene, let's blur the window again. Go to Filter > Blur Gallery > Field Blur. Now add a blur of 8 pixels, but this time make sure that the Light Bokeh is set to 25%.
Finish the window adjustments by applying some color. With the blurred image selected, go to Image > Adjustments > Color Lookup. Set the 3DLUT File to FuturisticBleak.3DL.
## 2. How to Add the Table
### Step 1
Adding the table is quite simple. Just use the Polygonal Lasso Tool (L) to extract it from the Stone Table reference. Then Copy and Paste it onto a New Layer above the background.
To clean up the edges, add a Layer Mask to the table. Select the mask and Fill it with black using the Paint Bucket Tool (G). Then use a Hard Round Brush with 100% Hardness to paint white onto the areas where you want to reveal the table.
Also make sure any excess leaves and debris are removed from its surface by using the Clone Stamp Tool (S).
### Step 2
Now desaturate the table. Select the layer and go to Image > Adjustments > Hue & Saturation. Lower the Saturation to -75.
### Step 3
Create a New Layer and Fill it with gray #b9b9b9 using the Paint Bucket Tool (G). Right-click to set this layer as a Clipping Mask to the table. Then set the Layer Blend Mode to Multiply.
Clip a New Adjustment Layer of Curves above the layer fill to adjust the lighting even further. Bring down the curve for the RGB Channel slightly to darken the table.
## 3. How to Create the Glass Case
### Step 1
Already the scene is starting to come together! How exciting!
Now let's create the glass case from scratch. To do this, first extract the bottom of this Cake Stand with the Polygonal Lasso Tool (L).
Adjust the size and perspective with the Free Transform Tool (Control-T). Then use the Eraser Tool (E) to clean up the edges of the base.
### Step 2
Select the Ellipse Tool (U) and create a white circle with No Stroke at the top of the base.
Right-click the ellipse layer and go to Blending Options. Apply a reflected gold Linear Gradient Overlay with the following settings and colors:
• Dark gold: #967242
• Light gold: #bd9053
### Step 3
Merge the shape and cake stand layers together. Control-T to adjust the size of the stand, making it slightly larger before placing it in the center of the table. Then Right-click to go to Blending Options again, this time adding a Drop Shadow with the following settings:
### Step 4
Change the gold base to a rich brown color. Do this by setting a New Adjustment Layer of Hue & Saturation as a Clipping Mask to the cake stand. Adjust the settings to the following numbers:
• Hue: -25
• Saturation: -41
• Lightness: -68
### Step 5
Time for the dome glass details! Open your Wine Glass reference in Photoshop. Use the Magic Wand Tool (W) to select the background, and then hit Delete to remove it. Do the same for the center of the glass, making sure that the shine stays in place.
Once you're finished, Copy and Paste the image into your composition and go to Edit > Transform > Flip Vertical. Adjust the shape of the glass with the Free Transform Tool (Control-T) and fit it onto the brown stand.
Continue adjusting the shape with the Free Transform Tool's Warp and Scale options until you're happy with the result. Try to make it as close to the movie's dome shape as possible. Add a Layer Mask to this layer and paint black onto it with a Hard Round Brush to remove any excess glass.
Now use the Eraser Tool (E) to erase the bottom of the glass as well as any harsh edges. Try to make the case seem as realistic as possible.
Finish the case by using the Ellipse Tool (U) to create an ellipse at the bottom of the case with a white Stroke and No Fill. Right-click the shape layer to go to Blending Options, adding a brown #372e2b Drop Shadow with the following settings:
## 4. How to Create the Enchanted Rose
### Step 1
To create the enchanted rose, we'll first add a little color to the glass case. Create a New Layer below the glass layers but above the stand layer.
With a Soft Round Brush, paint soft shades of green #0e624f and blue #2b809a onto the case, and then set the Layer Blend Mode to Divide. Lower the Opacity to 70%.
Here is how the layers should appear.
### Step 2
You may notice that the original enchanted rose sits at an angle in the glass. So before we add our own rose to the case, we need to adjust its shape. Open your Pink Rose reference in Photoshop. Use the Magic Wand Tool (W) to select the white background and hit Delete to remove it.
Dissect the rose using the Polygonal Lasso Tool (L). Create individual selections around the rose, stem, and leaves. Copy and Paste each component onto its own layer, and then Rotate each section with the Free Transform Tool (Control-T).
Here is a before and after comparison. Make the top of the rose seem much larger than the stem. Rotate it at a slight angle before removing any excess leaves. Then erase the end of the stem with the Eraser Tool (E).
Now that the rose is all set, Copy and Paste it onto a New Layer underneath the glass layers. Adjust the size so that it fits within the glass perfectly.
### Step 3
Set a New Adjustment Layer of Hue & Saturation as a Clipping Mask to the rose. Lower the Saturation to -50.
### Step 4
Let's add some color! Set another New Layer as a Clipping Mask to the rose. Use a Hard Round Brush to paint red #811522 all over the rose. Then set the Layer Blend Mode to Color Burn and the Opacity to 80%.
Clip a New Adjustment Layer of Color Lookup to the rose to modify the colors. Set the 3DLUT File to 2Strip.look, and then set the Layer Blend Mode to Multiply and the Opacity to 70%.
### Step 5
Don't forget to add the handle to the top of the glass dome! Extract the top portion from this Glass Top reference and Copy and Paste it into the scene. Use a combination of the Free Transform Tool (Control-T) and a Layer Mask to make the shape rounder.
Then go to Image > Adjustments > Hue and Saturation, and lower the Saturation to -64.
### Step 6
Cover the bottom of the glass base with rose petals. First open the Rose Petal reference in Photoshop and use the Magic Wand Tool (W) to select and Delete the white background.
Select only a few of the petals with the Polygonal Lasso Tool (L), and then Copy and Paste them onto a New Layer underneath the glass.
Create movement by making two copies of one of the rose petals. Position them as though they are floating down the case.
Now create a New Layer underneath the petals layer and set it to Multiply. Paint red #890100 shadow underneath the rose petals and don't forget to deepen the main rose. Use a Soft Round Brush to blend the shadows well.
## 5. How to Adjust the Final Colors and Lighting
### Step 1
Almost done! To make the scene as realistic as possible, we'll need to create a harmonious lighting and color scheme.
First create a New Layer above all the others. Use a Soft Round Brush (at 10-40% Opacity) to paint light gray #c7c4c1 mist around the right and left sides of the case. Lower the Opacity of the layer to 25%.
Create another New Layer and set it to Multiply. Use a Soft Round Brush (at 30-70% Opacity) to paint medium gray #6a6a6a around the edges of the scene. This will create more focus on the rose while adding a nice vignette effect.
### Step 2
Go to Layer > New Adjustment Layer > Color Balance. Change the settings for the Shadows, Midtones, and Highlights to make the overall scene more blue.
Here is the result once you're through.
### Step 3
Create a New Layer above the last adjustment and set it to Overlay. Use a Hard Round Brush to paint solid pink #d8c7de in the middle of the canvas. Lower the Opacity to 32%.
To brighten the rose petals with more color, set a New Layer to Overlay. Use a Soft Round Brush to paint pink #d85670 on the petals for a bold effect. Lower the Opacity to 83%.
### Step 4
For even more color, create a New Layer set to Linear Light. Paint a bright pink color #c93e96 all over the rose and rose petals. Lower the Opacity to 11%.
### Step 5
To create a sense of atmosphere, we'll be applying dust and sparkles to the composition. So load your Nebula Brushes and create a New Layer. Using Nebula Brush 7, click once to apply the brush, making sure the Foreground Color is set to white. Lower the Opacity if needed.
### Step 6
Then load your Lens Flare and Star Brushes. Use a combination of Star Brushes and Flare Brushes to paint white lens flares at the top of the glass case and a few star bursts within it. This will help the scene appear as though there's light coming in from the window.
Set a New Layer to Overlay and use a Soft Round Brush to paint yellow #d9dfcf highlights onto the table and case. Change the Brush Hardness to 100% to add a few cleaner lines of highlight around the table's ornate details.
### Step 7
Create a New Layer and use it to paint crisp white highlights on the glass, case, and rose. Add white dots to the air to show how some dust particles are catching light.
Take this opportunity to clean up the scene as much as possible.
### Step 8
The majority of our editing is now complete! All we have left to do is add a few more color adjustments. Add a New Adjustment Layer for Curves. Adjust the curves for the RGB, Blue, and Green Channels to balance the lighting. Lower the Opacity to 68%.
Next, add a New Adjustment Layer for Brightness & Contrast. Adjust the Brightness to 9 and the Contrast to 8.
Finally, add one more Adjustment Layer for Color Balance. Change the settings for the Midtones and Highlights to the following numbers:
The result should have a slightly purple hue to mimic the colors from the movie. This is technically the end of the manipulation, but I'll sharpen the image even further in the next steps.
## 6. How to Sharpen Your Manipulation
For more visual impact, consider sharpening your photo manipulations for the final step. Merge all the layers together. Hold Control-J to Duplicate the merged layer.
With the merged copy selected, go to Filter > Other > High Pass, setting the Radius to 2 pixels. Then set the Layer Blend Mode to Overlay.
That's it! Check out the final result below.
## Congratulations! You're All Done!
You can create Disney magic using a few well-placed images in Adobe Photoshop. Explore the possibilities with fantastic settings like Adjustment Layers, Blend Modes, and more.
|
2023-04-01 21:04:15
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.20481134951114655, "perplexity": 2789.0837722770016}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296950247.65/warc/CC-MAIN-20230401191131-20230401221131-00307.warc.gz"}
|
https://motls.blogspot.com/2008/04/brian-greene-ted-2005-strings.html?m=1
|
## Monday, April 28, 2008
### Brian Greene, TED 2005: Strings
Via Thought Office & Ted.COM (freshly posted).
Bonus
Brian Greene will organize The World Science Festival, especially for the U.S. troops in Iraq who often write letters to Brian Greene, demanding a higher exposure to string theory and other topics that add a new dimension to their sometimes difficult life.
|
2018-03-24 04:24:11
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2120489776134491, "perplexity": 10610.849225512717}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257649683.30/warc/CC-MAIN-20180324034649-20180324054649-00154.warc.gz"}
|
https://codereview.stackexchange.com/questions/79974/pi-calculating-program
|
# Pi-calculating program
I saw this question and answer about calculating pi on Stack Overflow, and I decided to write my own program for calculating pi. I used Python and only integers (I didn't want to use floating point numbers), and used the Gauss–Legendre algorithm because it was the simplest to implement (I considered using the Borwein's algorithm, but I didn't want to calculate third roots of numbers, and the Chudnovsky algorithm seemed a little complicated, although maybe I'll give it a try). I would like to know how efficient are the algorithms above, in $\mathcal{O}(n)$ (n is the number of digits to calculate), and how efficient is my program?
# Calculate next square root approximation of the number.
def calculate_next_square_root(number, square_root):
next_square_root = ((number / square_root) + square_root) / 2
return next_square_root
# Calculate the square root of the number.
square_root = 1 * (10 ** (digits + add))
# Calculate next square root approximation of the number.
next_square_root = calculate_next_square_root(number=number, square_root=square_root)
while (next_square_root != square_root):
# Replace square root with next square root.
square_root = next_square_root
# Calculate next square root approximation of the number.
next_square_root = calculate_next_square_root(number=number, square_root=square_root)
return square_root
# Calculate next pi approximation.
def calculate_next_pi(a, b, t, digits, add):
next_pi = ((10 ** (digits + add)) * ((a + b) ** 2)) / (4 * t)
return next_pi
# Calculate pi to 5,000 digits.
digits = 5000
a = 10 ** (digits + add)
t = (10 ** ((digits + add) * 2)) / 4
p = 1
pi = -1 # pi must be different than next_pi
n = 0
while (next_pi != pi):
pi = next_pi
next_a = (a + b) / 2
next_t = t - (p * ((a - next_a) ** 2))
next_p = 2 * p
a = next_a
b = next_b
t = next_t
p = next_p
n += 1
# Remove the last 500 digits.
# Print the results.
print pi
print n
My program takes 52 lines of code (including comments), and I'm not interested in programs which take too many lines to implement, up to 100 or 200 lines are fine. I checked and it takes about one second to calculate 5,000 digits of pi (with add = 500), and a little less than two minutes to calculate 50,000 digits (with add = 5000). With 5,000 digits it took 14 iterations, and with 50,000 digits 17 iterations, but I didn't count how many iterations it took to calculate the square roots. Is my program efficiently using the Gauss–Legendre algorithm, or can it be made more efficient?
By the way, add doesn't have to be 500, even 4 or 5 is enough. But I wanted to make sure all the digits of pi are correct (and they are). With add = 0 the last 2 or 3 digits are not correct.
I took the square root program from a program I wrote in 2013, but I don't know the name of the algorithm I used and whether it's the most efficient algorithm or not.
Update: I remind you my question: I would like to know how efficient are the algorithms above, in $\mathcal{O}(n)$ (n is the number of digits to calculate), and how efficient is my program? I prefer not to import modules, but use pure Python (either 2 or 3) without any import. And of course if we start with a number closer to the square root (such as a or b) then it will take fewer iterations to calculate the square root.
• Wow, does python not have a square root function? Feb 8 '15 at 17:31
• echo 'scale=1000; 4*a(1)' | bc -l Feb 8 '15 at 17:33
• You should call this program pypi or something. =) Feb 8 '15 at 18:06
• @ChiefTwoPencils I think it does, but it doesn't work for long integers. If I replace calculate_square_root with **0.5 I get this error message: "OverflowError: long int too large to convert to float".
– Uri
Feb 8 '15 at 18:56
• Python 2 or 3? It is important because division is different. Feb 8 '15 at 21:47
This is a bit hard to answer, but let's try nevertheless:
1. From a computational point of view, using a scripting language like python for iterative algorithms doesn't make the least sense -- your PC is going to be far more occupied by translating your python code to executable structures than it will be with executing the core algorithm. For example, every number in python (assuming a CPython implementation of the runtime environment) is an object -- which makes number-number multiplication far more expensive than the actual multiplication. Now, I assume your question is "if I did this in a compiled language, with fixed length numerical types, would this algorithm perform optimally"?
2. approximating a square root in software is bound to be horrendously expensive compared to just calculating the floating point square root; modern general purpose CPUs even have instructions that can take the square root of several numbers at once, in just a few cycles!
• No. Most of this program's run time is spent inside bigint arithmetic functions which are implemented in C anyway. Feb 9 '15 at 13:05
• @JanneKarila: He does a sqrt approximation iteratively in python, definitely not in C Feb 9 '15 at 13:38
• Yes but the number of iterations is rather small: 210 calls to calculate_next_square_root. Feb 9 '15 at 14:21
• With small integers a call to calculate_next_square_root takes 300 ns, but with OP's numbers 3 ms. From that I'd estimate Python's overhead to be 0.01 %. Feb 9 '15 at 14:33
• ((10 ** (digits + add)) * ((a + b) ** 2)) / (4 * t) has ten CPython PyObjects. For each one of these, at least a type lookup, a lookup for the correct operator and the value lookup. That's three times the indirection operations you'd need to do if this was a statically typed compiled language, not even counting the fact that Python somehow has to figure out what these strings mean. Feb 9 '15 at 15:08
I think the most efficient method is the Chudnovsky algoritm (100 million digits of Pi, in under 10 minutes!) To know the math behind it:
"""
Python3 program to calculate Pi using python long integers, BINARY
splitting and the Chudnovsky algorithm
"""
import math
from gmpy2 import mpz
from time import time
def pi_chudnovsky_bs(digits):
"""
Compute int(pi * 10**digits)
This is done using Chudnovsky's series with BINARY splitting
"""
C = 640320
C3_OVER_24 = C**3 // 24
def bs(a, b):
"""
Computes the terms for binary splitting the Chudnovsky infinite series
a(a) = +/- (13591409 + 545140134*a)
p(a) = (6*a-5)*(2*a-1)*(6*a-1)
b(a) = 1
q(a) = a*a*a*C3_OVER_24
returns P(a,b), Q(a,b) and T(a,b)
"""
if b - a == 1:
# Directly compute P(a,a+1), Q(a,a+1) and T(a,a+1)
if a == 0:
Pab = Qab = mpz(1)
else:
Pab = mpz((6*a-5)*(2*a-1)*(6*a-1))
Qab = mpz(a*a*a*C3_OVER_24)
Tab = Pab * (13591409 + 545140134*a) # a(a) * p(a)
if a & 1:
Tab = -Tab
else:
# Recursively compute P(a,b), Q(a,b) and T(a,b)
# m is the midpoint of a and b
m = (a + b) // 2
# Recursively calculate P(a,m), Q(a,m) and T(a,m)
Pam, Qam, Tam = bs(a, m)
# Recursively calculate P(m,b), Q(m,b) and T(m,b)
Pmb, Qmb, Tmb = bs(m, b)
# Now combine
Pab = Pam * Pmb
Qab = Qam * Qmb
Tab = Qmb * Tam + Pam * Tmb
return Pab, Qab, Tab
# how many terms to compute
DIGITS_PER_TERM = math.log10(C3_OVER_24/6/2/6)
N = int(digits/DIGITS_PER_TERM + 1)
# Calclate P(0,N) and Q(0,N)
P, Q, T = bs(0, N)
one_squared = mpz(10)**(2*digits)
sqrtC = (10005*one_squared).sqrt()
return (Q*426880*sqrtC) // T
# The last 5 digits or pi for various numbers of digits
check_digits = {
100 : 70679,
1000 : 1989,
10000 : 75678,
100000 : 24646,
1000000 : 58151,
10000000 : 55897,
}
if __name__ == "__main__":
digits = 100
pi = pi_chudnovsky_bs(digits)
print(pi)
#raise SystemExit
for log10_digits in range(1,9):
digits = 10**log10_digits
start =time()
pi = pi_chudnovsky_bs(digits)
print("chudnovsky_gmpy_mpz_bs: digits",digits,"time",time()-start)
if digits in check_digits:
last_five_digits = pi % 100000
if check_digits[digits] == last_five_digits:
print("Last 5 digits %05d OK" % last_five_digits)
else:
print("Last 5 digits %05d wrong should be %05d" % (last_five_digits, check_digits[digits]))
For further details:http://www.craig-wood.com/nick/articles/pi-chudnovsky/
• Which version of Python is this program for? I tried with Python 2.7.9 and I got this error message: "ImportError: No module named gmpy2".
– Uri
Feb 8 '15 at 18:21
– Mohit Bhasi
Feb 8 '15 at 18:24
• I tried to install gmpy2 but it failed. Your answer is nice, but I'm looking for a pure integer operating program which doesn't have to import modules such as math and gmpy2.
– Uri
Feb 8 '15 at 18:50
• @Uri: that's important information! I still don't get your use case, though. Purely integer sounds like you want to adapt this algorithm to something without floating point instructions (which, honestly, makes little sense when calculating pi), but you're using python.
– Marcus Müller
Feb 8 '15 at 18:53
• @MohitBhasi I used your program without importing math and mpz, I replaced math.log10(C3_OVER_24/6/2/6) with the constant 14.1816474627 and (10005*one_squared).sqrt() with calculate_square_root(number=(10005*one_squared), digits=digits + 2, add=0), added my square root functions, removed mpz from the code (replaced it with "") and it took 20.65 seconds to calculate the first 100,000 digits of pi (Last 5 digits 24646 OK), which is much faster than my own program.
– Uri
Feb 8 '15 at 21:18
a = 10 ** (digits + add)
t = (10 ** ((digits + add) * 2)) / 4
p = 1
You should avoid one letter variable names, prefer longer more descriptive names.
next_pi = calculate_next_pi(a=a, b=b, t=t, digits=digits, add=add)
You call functions in a very verbose way, you should prefer:
next_pi = calculate_next_pi(a, b, t, digits, add)
# Calculate next square root approximation of the number.
def calculate_next_square_root(number, square_root):
Your comment just repeats the obvious instead say which algorithm you chose and maybe also why you chose that algorithm over another.
def calculate_next_square_root(number, square_root):
next_square_root = ((number / square_root) + square_root) / 2
return next_square_root
def calculate_next_pi(a, b, t, digits, add):
next_pi = ((10 ** (digits + add)) * ((a + b) ** 2)) / (4 * t)
return next_pi
Omit needless words:
def calculate_next_square_root(number, square_root):
return ((number / square_root) + square_root) / 2
def calculate_next_pi(a, b, t, digits, add):
return ((10 ** (digits + add)) * ((a + b) ** 2)) / (4 * t)
digits = 5000
a = 10 ** (digits + add)
t = (10 ** ((digits + add) * 2)) / 4
p = 1
pi = -1 # pi must be different than next_pi
n = 0
while (next_pi != pi):
pi = next_pi
next_a = (a + b) / 2
next_t = t - (p * ((a - next_a) ** 2))
next_p = 2 * p
a = next_a
b = next_b
t = next_t
p = next_p
n += 1
This big fat 'top-level' code should be a function, also because Python code runs faster in functions.
Mathematical code of this kind really benefits from automated testing, a simple example of automatic testing is:
import doctest
"""
12
"""
doctest.testmod()
This allows you to optimize without fear of breaking things.
• Why do you prefer calling calculate_next_pi(a, b, t, digits, add) instead of calculate_next_pi(a=a, b=b, t=t, digits=digits, add=add)? I always prefer to call functions the second way (with parameter names), to avoid calling the wrong function or if the function parameters change I will get an exception. It's possible in Python but not in other languages like JavaScript (I also program in JavaScript).
– Uri
Feb 10 '15 at 5:39
• My way is less verbose. Feb 10 '15 at 12:38
I didn't count how many iterations it took to calculate the square roots.
That's easy to do because each iteration involves a function call. Simply profile the program by issuing this command from shell:
python -m cProfile filename.py
A report similar to this should appear. From the ncalls column you see calculate_next_square_root is called 210 times.
242 function calls in 0.858 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.034 0.034 0.858 0.858 cr1a.py:2(<module>)
210 0.682 0.003 0.682 0.003 cr1a.py:2(calculate_next_square_root)
15 0.137 0.009 0.137 0.009 cr1a.py:20(calculate_next_pi)
15 0.004 0.000 0.687 0.046 cr1a.py:7(calculate_square_root)
1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects}
I took the square root program from a program I wrote in 2013, but I don't know the name of the algorithm I used and whether it's the most efficient algorithm or not.
It is Newton's method. You could reduce the number of iterations with a better initial guess for the square root (instead of 1). You are computing the square root of a*b where a and b converge towards the same value. Thus either a or b would be a good initial guess for the square root. Using b I see the number of calls go down to 111.
• Using b instead of 1 as an initial value is a big win; this is the 0th-order estimate. I wonder if the first-order estimate (a+b)/2 would be better, especially since it's already computed. Mar 12 '15 at 18:41
|
2021-10-28 06:13:56
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5178418159484863, "perplexity": 2714.4491120671673}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323588257.34/warc/CC-MAIN-20211028034828-20211028064828-00285.warc.gz"}
|
http://mathoverflow.net/questions/136483/solving-a-sde-with-quadratic-drift
|
# Solving a SDE with quadratic drift
I am wondering whether the following SDE can be solved explicitly?
$$d X_t = X_t^2 d t + X_t d B_t$$
where $B_t$ is a standard Brownian motion. If not, can we say some thing about the moments of the solution, i.e., $E(|X_t|^n)$?
Thank you very much for any hints!
Anand
-
It will blow up for large time because $X_t^2$ will dominate when $X_t$ is big. – Anand Jul 12 '13 at 9:28
Actually, the it does not satisfy sufficient conditions for existence and uniqueness when drift and diffusion terms has to be of at most linear growth (Lipschitz everywhere), so the solution of this SDE may not exist at all. – Ilya Jul 12 '13 at 12:04
Yes, you are right. It will blow up at finite time. Can we say something for the moments for the blowup? – Anand Jul 12 '13 at 12:44
Solutions do exist locally. Globally they MAY blow up as you already know. The blowup will be dominated by the deterministic system. You did not write what is your initial condition - note that $0$ is a perfectly fine solution to your equation. If your initial condition is say $X_0=1$, then it may blow up, but with positive probability the solution may actually converge to $0$. This is because if the $X^2$ drift term were not there then the solution would be $X_0 e^{B_t-t/2}$ which converges to $0$, and if you add your drift, as long as $X_t$ is small then the drift is smaller than $+\delta X_t$ (e.g., as long as $X_t<\delta$). But the solution of the equation $dX_t=\delta X_t dt+X_t dB_t$ converges to $0$ as long as $\delta$ is small enough (in fact $\delta<1/2$ will work).
So with positive probability, starting from any $X_0$ you become smaller than say $1/4$ after a finite time, and then with positive probability you actually converge to $0$ (using comparison theorems for 1D SDE's as in Ikeda-Watanabe should be enough to prove this). On the other hand, with positive probability you blow up. So I am not sure what do you mean by moments of the blow up time''.
-
Dear Professor Zeltouni, thank you very much for your explanation and clarification. By "moments", I mean $E(|X_t|^n)$. If $\tau=\inf\{t\ge 0: |X_t|=\infty\}$, can we say that $\tau$ is almost surely bounded away from zero? If so, we may ask $E(|X_t|^n)$ for $t<\tau$. I will have a careful look. – Anand Jul 12 '13 at 18:35
$\tau$ is bounded away from zero but not by a deterministic amount. So $E(|X_t|^n)$ is simply infinity. You can ask about $E(|X_t|^n|\tau>t)$ but this is a completely different question and I don't know how to write an explicit formula. – ofer zeitouni Jul 13 '13 at 12:48
Thanks Professor Zeltouni for your explanations. They are very helpful. :-) – Anand Jul 14 '13 at 17:06
Actually your SDE may be solved explicitly. Look at the more general SDE, \begin{align} dX_{t} = (a X_{t}^{n} + b X_{t}) dt + c X_{t} dW_{t} \end{align} where $n > 1$ and $a,b,c \in \mathbb{R}$. It has a solution given by \begin{align} dX_{t} = \Theta_{t} \Bigl( X_{0}^{1-n} + a(1-n)\int_{0}^{t} \Theta^{n-1}_{s} ds \Bigr)^{\frac{1}{1-n}} \end{align} with \begin{align} \Theta_{t} = e^{ (b-\frac{1}{2}c^2) t + c W_{t}} \end{align}
See p. 125 in "Numerical Solution of Stochastic Differential Equations", 1995, Peter E. Kloeden and Eckhard Platen.
-
Thanks a lot tot for your solution. – Anand Jun 10 '14 at 15:22
|
2016-05-05 09:15:18
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9317253828048706, "perplexity": 267.40464620787844}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860126377.4/warc/CC-MAIN-20160428161526-00088-ip-10-239-7-51.ec2.internal.warc.gz"}
|
https://discourse.julialang.org/t/is-it-possible-to-name-slices-of-arrays-as-functions/36934
|
# Is it possible to name slices of arrays as functions?
A common use case for me, and lots of people I would think, is to have large systems of DEs such as
dx_i / dt = …
dy_i / dt = …
where i \in [1, n] and x_i, y_i represents the ith class of x, y respectively.
Usually I define the state in these ODEs as one big vector of length 2n, like
const n = 100 #some number of dimensions
system_state = zeros(2*n)
system_state[1:n] #these are the x_i
system_state[n+1:2*n] #these are the y_i
Then the ODE function is something like this, with all the operations broadcasted
function f(du,u,p,t)
x = @view u[1:n]
y = @view u[n+1:2*n]
du[1:n] .= -1 .* x
du[n+1:2*n] .= x .* y
end
This works, but I end up having lots of slices around my code, which can be difficult to read.
I was hoping there was a zero-cost way of writing a shorthand function like
x(state) = @view state[1:n]
which I can then use to modify the state exactly like the original slice. However this doesn’t seem to work, I get the error “invalid assignment location”.
Is there a way to do this in Julia?
Thanks
using LabelledArrays
julia> z = @LArray [1 2; 3 4] (a = (2, :), b = 2:3);
julia> z.a
2-element view(::Array{Int64,2}, 2, :) with eltype Int64:
3
4
2 Likes
Cool, thanks!
I have a quick question related to the use of labelledarrays in this way.
My current toy example is
function f(du,u,p,t)
x = @view u[1:n]
y = @view u[n+1:2*n]
du[1:n] .= -1 .* x
du[n+1:2*n] .= -1 .* y
end
const n = 10
u_0 = @LArray fill(1.0,2*n) (x = (1:n),y = (n+1:2*n))
prob = ODEProblem(f,u_0,(0,10.0),[])
sol = solve(prob,Tsit5())
Is it possible to define a custom type of labelledarray, like say state_type = LArray{2*n}(x = (1:n),y = (n+1:2*n)), so I can easily specify function signatures that return labelled arrays with the format of u_0? I’ve tried a few different things and none of them worked.
Let me know if this should be asked as a new topic or not as well.
I am not sure of a good way to implement that well.
I was benchmarking this code and found that the LabelledArrays one is actually slightly faster, although I’m not sure why they are allocating so much.
using DifferentialEquations
using LabelledArrays
function f1(du,u,p,t)
x = @view u[1:n]
y = @view u[n+1:2*n]
du[1:n] .= -1 .* x .* u.y
du[n+1:2*n] .= -1 .* y
end
function f2(du,u,p,t)
du.x .= -1 .* u.x .* u.y
du.y .= -1 .* u.y
end
const n = 1000
using BenchmarkTools
u_0 = @LArray fill(1000.0,2*n) (x = (1:n),y = (n+1:2*n))
prob1 = ODEProblem(f1,u_0,(0,100.0),[])
@btime solve(prob1,Tsit5());
prob2 = ODEProblem(f2,u_0,(0,100.0),[])
@btime solve(prob2,Tsit5());
31.841 ms (29417 allocations: 43.50 MiB)
27.728 ms (29417 allocations: 43.50 MiB)
|
2022-05-17 18:29:54
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4575102925300598, "perplexity": 5321.432008153205}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662519037.11/warc/CC-MAIN-20220517162558-20220517192558-00742.warc.gz"}
|
http://tidytransit.r-transit.org/reference/raptor.html
|
raptor finds the minimal travel time, earliest or latest arrival time for all stops in stop_times with journeys departing from stop_ids within time_range.
raptor(stop_times, transfers, stop_ids, arrival = FALSE,
time_range = 3600, max_transfers = NULL, keep = "all")
## Arguments
stop_times A (prepared) stop_times table from a gtfs feed. Prepared means that all stop time rows before the desired journey departure time should be removed. The table should also only include departures happening on one day. Use filter_stop_times() for easier preparation. Transfers table from a gtfs feed. In general no preparation is needed. Character vector with stop_ids from where journeys should start (or end) If FALSE (default), all journeys start from stop_ids. If TRUE, all journeys end at stop_ids. Departure or arrival time range in seconds. All departures from the first departure of stop_times (not necessarily from stop_id in stop_ids) within time_range are considered. If arrival is TRUE, all arrivals within time_range before the latest arrival time of stop_times are considered. Maximum number of transfers allowed, no limit (NULL) as default. One of c("all", "shortest", "earliest", "latest"). By default, all journeys arriving at a stop are returned. With shortest the journey with shortest travel time is returned. With earliest the journey arriving at a stop the earliest is returned, latest works accordingly.
## Value
A data.table with journeys (departure, arrival and travel time) to/from all stop_ids reachable by stop_ids.
## Details
With a modified Round-Based Public Transit Routing Algorithm (RAPTOR) using data.table, earliest arrival times for all stops are calculated. If two journeys arrive at the same time, the one with the later departure time and thus shorter travel time is kept. By default, all journeys departing within time_range that arrive at a stop are returned in a table. If you want all journeys arriving at stop_ids within the specified time range, set arrival to TRUE.
Journeys are defined by a "from" and "to" stop_id, a departure, arrival and travel time. Note that the exact journeys (with each intermediate stop and route ids for example) is not returned.
For most cases, stop_times needs to be filtered, as it should only contain trips happening on a single day and departures later than a given journey start time, see filter_stop_times(). The algorithm scans all trips until it exceeds max_transfers or all trips in stop_times have been visited.
travel_times() for an easier access to travel time calculations via stop_names.
## Examples
nyc_path <- system.file("extdata", "google_transit_nyc_subway.zip", package = "tidytransit")
# you can use initial walk times to different stops in walking distance (arbitrary example values)
stop_ids_harlem_st <- c("301", "301N", "301S")
stop_ids_155_st <- c("A11", "A11N", "A11S", "D12", "D12N", "D12S")
walk_times <- data.frame(stop_id = c(stop_ids_harlem_st, stop_ids_155_st),
walk_time = c(rep(600, 3), rep(410, 6)), stringsAsFactors = F)
# Use journeys departing after 7 AM with arrival time before 11 AM on 26th of June
stop_times <- filter_stop_times(nyc, "2018-06-26", 7*3600, 9*3600)#> Consider using set_date_service_table beforehand if you filter this feed multiple times
# calculate all journeys departing from Harlem St or 155 St between 7:00 and 7:30
rptr <- raptor(stop_times, nyc$transfers, walk_times$stop_id, time_range = 1800,
keep = "all")
# add walk times to travel times
rptr <- left_join(rptr, walk_times, by=c("from_stop_id" = "stop_id"))
rptr$travel_time_incl_walk <- rptr$travel_time + rptr$walk_time # get minimal travel times (with walk times) for all stop_ids shortest_travel_times <- setDT(rptr)[order(travel_time_incl_walk)][, .SD[1], by = "to_stop_id"] hist(shortest_travel_times$travel_time, breaks = 360)
|
2020-04-03 07:46:12
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.23805998265743256, "perplexity": 13555.425327905743}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370510352.43/warc/CC-MAIN-20200403061648-20200403091648-00488.warc.gz"}
|
http://genomicsclass.github.io/book/pages/biological_versus_technical_var.html
|
# Introduction
In the following sections we will cover inference in the context of genomics experiments. We apply some of the concepts we have covered in previous sections including t-tests, multiple comparisons and standard deviation estimates from hierarchical models. Here we introduce a concept that is particularly important in the analysis of genomics data: the distinction between biological and technical variability.
In general, the variability we observe across biological units, such as individuals, within a population is referred to as biological. We refer to the variability we observe across measurements of the same biological unit, such a aliquots from the same biological sample, as technical. Because newly developed measurement technologies are common in genomics, technical replicates are many times to assess experimental data. By generating measurements from samples that are designed to be the same, we are able to measure and assess technical variability. We also use the terminology biological replicates and technical replicates to refer to samples from which we can measure biological and technical variability respectively.
It is important not to confuse biological and technical variability when performing statistical inference as the interpretation is quite difference. For example, when analyzing data from technical replicates the population is just the one sample from which these come from as opposed to more general population such as healthy humans or control mice. Here we explore this concept with a experiment that was designed to include both technical and biological replicates.
# Pooling experiment data
The dataset we will study includes data from gene expression arrays. In this experiment, RNA was extract from 12 randomly selected mice from two strain. All 24 samples were hybridized to micro arrays but we also formed pools, including two pools from with the RNA from all twelve mice from each of the two strains. Other pools were also created, as we will see below, but we will ignore these here.
We will need the following library which you need to install if you have not done so already:
library(devtools)
install_github("genomicsclass/maPooling")
We can see the experimental design using the pData function. Each row represents a sample and the column are the mice. A 1 in cell $i,j$ indicates that RNA from mouse $j$ was included in sample $i$. The strain can be identified from the row names (not a recommended approach)
library(Biobase)
library(maPooling)
data(maPooling)
## a2 a3 a4 a5 a6 a7 a8 a9 a10 a11 a12 a14 b2 b3 b5 b6 b8 b9 b10 b11
## a10 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
## a10a11 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0
## a10a11a4 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0
## a11 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0
## a12 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0
## a12a14 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0
## b12 b13 b14 b15
## a10 0 0 0 0
## a10a11 0 0 0 0
## a10a11a4 0 0 0 0
## a11 0 0 0 0
## a12 0 0 0 0
## a12a14 0 0 0 0
Below we use create an image to illustrate which mice were included in which samples:
library(rafalib)
## Loading required package: RColorBrewer
mypar2()
flipt <- function(m) t(m[nrow(m):1,])
myimage <- function(m,...) {
image(flipt(m),xaxt="n",yaxt="n",...)
}
myimage(as.matrix(pData(maPooling)),col=c("white","black"),
xlab="experiments",
ylab="individuals",
main="phenoData")
Note that ultimately we are interested in detecting genes that are differentially expressed between the two strains of mice which we will refer to as strain 0 and 1. We can apply tests to the technical replicates of pooled samples or the data from 12 individual mice. We can identify these pooled samples because all mice from each strain were represented in these samples and thus the sum of the rows of experimental design matrix add up to 12:
data(maPooling)
pd=pData(maPooling)
pooled=which(rowSums(pd)==12)
We can determine the strain from the column names:
factor(as.numeric(grepl("b",names(pooled))))
## [1] 0 0 0 0 1 1 1 1
## Levels: 0 1
If we compare the mean expression between groups for each gene we find several showing consistent differences. Here are two examples:
###look at 2 pre-selected samples for illustration
i=11425;j=11878
pooled_y=exprs(maPooling[,pooled])
pooled_g=factor(as.numeric(grepl("b",names(pooled))))
mypar2(1,2)
stripchart(split(pooled_y[i,],pooled_g),vertical=TRUE,method="jitter",col=c(1,2),main="Gene 1",xlab="Group",pch=15)
stripchart(split(pooled_y[j,],pooled_g),vertical=TRUE,method="jitter",col=c(1,2),main="Gene 2",xlab="Group",pch=15)
Note that if we compute a t-test from these values we obtain highly significant results
library(genefilter)
##
## Attaching package: 'genefilter'
##
## The following object is masked from 'package:base':
##
## anyNA
pooled_tt=rowttests(pooled_y,pooled_g)
pooled_tt$p.value[i] ## [1] 2.075617e-07 pooled_tt$p.value[j]
## [1] 3.400476e-07
But would these results hold up if we selected another 24 mice? Note that the equation for the t-test we presented in the previous section include the population standard deviations. Are these quantities measured here? Note that it is being replicated here is the experimental protocol. We have created four technical replicates for each pooled sample. Gene 1 may be a highly variable gene within strain of mice while Gene 2 a stable one, but we have no way of seeing this.
We also have microarray data for each individual mice. For each strain we have 12 biological replicates. We can find them by looking for rows with just one 1.
individuals=which(rowSums(pd)==1)
It turns out that some technical replicates were included for some individual mice so we remove them to illustrate an analysis with only biological replicates:
##remove replicates
individuals=individuals[-grep("tr",names(individuals))]
y=exprs(maPooling)[,individuals]
g=factor(as.numeric(grepl("b",names(individuals))))
We can compute the sample variance for each gene and compare to the standard deviation obtained with the technical replicates.
technicalsd <- rowSds(pooled_y[,pooled_g==0])
biologicalsd <- rowSds(y[,g==0])
LIM=range(c(technicalsd,biologicalsd))
mypar2(1,1)
boxplot(technicalsd,biologicalsd,names=c("technical","biological"),ylab="standard deviation")
Note the biological variance is much larger than the technical one. And also that the variability of variances is also for biological variance. Here are the two genes we showed above but now for each individual mouse
mypar2(1,2)
stripchart(split(y[i,],g),vertical=TRUE,method="jitter",col=c(1,2),xlab="Gene 1",pch=15)
points(c(1,2),tapply(y[i,],g,mean),pch=4,cex=1.5)
stripchart(split(y[j,],g),vertical=TRUE,method="jitter",col=c(1,2),xlab="Gene 2",pch=15)
points(c(1,2),tapply(y[j,],g,mean),pch=4,cex=1.5)
Note the p-value tell a different story
library(genefilter)
tt=rowttests(y,g)
tt$p.value[i] ## [1] 0.0898726 tt$p.value[j]
## [1] 1.979172e-07
Which of these two genes do we feel more confident reporting as being differentially expressed? If another investigator takes another random sample of mice and tries the same experiment, which one do you think will replicate? Measuring biological variability is essential if we want our conclusions to be about the strain of mice in general as opposed to the specific mice we have.
An analysis with biological replicates have as a population these two strains of mice. An analysis with technical replicates have as a population the twelve mice we selected and the variability is related to the measurement technology. In science we typically are concerned with populations. As a very practical example, not that if another lab performs this experiment they will have another set of twelve mice and thus inference about populations are more likely to replicate.
|
2017-11-21 04:35:01
|
{"extraction_info": {"found_math": true, "script_math_tex": 3, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4297488331794739, "perplexity": 675.198167183615}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806316.80/warc/CC-MAIN-20171121040104-20171121060104-00488.warc.gz"}
|
https://budgetcash.co.za/m9qqfp/mass-of-one-atom-of-oxygen-in-grams-f40b1a
|
Thus, oxygen was assigned an atomic mass of 16. how much energy will one hydrogen release if we destroy it .. E=mc^2.. thank you.. Answer Save. First, determine the total number of atoms in the substance. Get your answers by asking now. How many moles of insulin are needed to make up 45 mL of .0052 M insulin solution? In one gram there will be 1/31.9988 mol = 0.0313 mol. 4 moles Oxygen to grams = 63.9976 grams. Pb. Q:-Calculate the wavelength of an electron moving with a velocity of 2.05 × 10 7 ms –1. FAQ. Using this value, calculate the average mass in grams of one atom of gold. If the mass of one atom of oxygen is 16.00 amu, what is the mass of one mole of elemental oxygen (O2) in grams? Which of the following is the major drawback of Dalton’s atomic theory? As evidenced by this extraordinary number, atoms are extremely tiny — so tiny that a more convenient unit of measure is generally used in chemistry when talking about the mass or volume of a substance. One mole of carbon is 6.022 x 1023 atoms of carbon (Avogadro's number). 63.55 grams of copper. 0 8 × 1 0 2 3 oxygen atoms Mass of one atom of oxygen is . here is the theory . When calculating molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. ... Mass of atoms of oxygen = gram atomic mass of oxygen. The quantity of the element which weighs equal to its gram atomic mass is called one gram atom of that element. Maths. And this gives us our weight of one atom of oxygen which is equal to 2.66 tens, 10 to the negative 23 grants. What is mass of one hydrogen atom in grams ..... thank you. arrow_back. 8 moles Oxygen to grams = 127.9952 grams. Which contains the largest number of moles? Average atomic mass is the sum of the fractional percents of each atom in a substance. What unit of…, Calculate the mass in grams of 2.55 mol ofoxygen gas, O $_{2}$ (molar ma…, EMAILWhoops, there might be a typo in your email. Or 6.022x 10 23 atoms of carbon = 12 g of carbon. What is the mass in grams of 5.40 moles of lithium? If its K.E. Q:- Oxygen in the form O2 has a molar mass of 31.9988 gmol-1. NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless. Class 12 Class 11 Class 10 Class 9 Class 8 … Want to see this answer and more? When calculating molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. Molecular mass. Previous Question Next Question. Avogadro's number is one of the most important constants used in chemistry. The atomic mass (m a or m) is the mass of an atom.Although the SI unit of mass is kilogram (symbol: kg), the atomic mass is often expressed in the non-SI unit dalton (symbol: Da, or u) where 1 dalton is defined as 1 ⁄ 12 of the mass of a single carbon-12 atom, at rest. What is the number of molecules in 0.25 moles of oxygen. There is a unique relationship between molar mass and atomic weight: Oxygen's atomic weight is 16.00 amu. Still have questions? Chapter 2, Problem 15PS. She has taught science courses at the high school, college, and graduate levels. NCERT P Bahadur IIT-JEE Previous Year Narendra Awasthi MS Chauhan. consider two organis molecules, ethanol and benzene. Oxygen's atomic weight is 16.00 amu. Finding molar mass starts with units of grams per mole (g/mol). Your answers are all in units of g = grams. oxygen also an element, but if you filled 250 gram from earth ,balloon put on mesosphere , its should be more than 250g weight there. The mass of an oxygen atom = 16 amu. Calculate the average mass in grams of 1 atom of oxygen. We assume you are converting between moles Oxygen and gram. NCERT NCERT Exemplar NCERT Fingertips Errorless Vol-1 Errorless Vol-2. Physics. What is mass of 0.560 moles of chlorine gas? For example, oxygen gas O2 is diatomic (each molecule contains two atoms) so its relative formula mass is 32. Your answers are all in units of g = grams. The fun is, we simply say the atomic mass or weight of atom or molecule with some numerals. Finally, add the molar mass of each atom to get the molar mass of the molecule. As we saw earlier, it is convenient to use a reference unit when dealing with such small numbers: the atomic mass unit. 9 years ago.
Massof atoms of oxygen = 16g
Mass of 1 atom of oxygen = g Physics. 1. Molar Mass: Molar mass is the mass of Avogadro's number of atoms of that element or compound. 1. Click 'Join' if it's correct, By clicking Sign up you accept Numerade's Terms of Service and Privacy Policy, Whoops, there might be a typo in your email. Chapter 2, Problem 13PS. Q:-Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass. We can start with our one atom of oxygen and divide by all the gardeners number because we know that there is this amount of Adams in every single one mole off oxygen. Let's say, for example, you want to know the mass of a single atom of water. Pay for 5 months, gift an ENTIRE YEAR to someone special! ( 1 u = 1.66054 * 10 ^-24 g ) There are 2/3 as many---2.5093*10^22--- atoms belonging to the other element. The molar mass of glucose can be calculated by multiplying the molar masses of its atomic constituents by their frequency in a single molecule and adding those values together. That said, to find the mass of one ATOM, we need to convert from moles to atoms as follows: 1.008 grams/mole Hydrogen * (1 mole/6.022x10 23 atoms) = 1.67 x 10-24 grams; 16.00 grams/mole Oxygen * (1 mole/6.022x10 23 atoms) = 2.66 x 10-23 grams Any slight differences are due to the accuracy used for the atomic mass and/or Avogadro's Number. One atomic mass unit is a mass unit equal to exactly one twelfth (1/12th) the mass of one atom of _____. Although this number is a constant, it's experimentally determined, so we use an approximate value of 6.022 x 1023. So, you know how many atoms are in a mole. According to the CRC Handbook of Chemistry and Physics, the atomic mass of Au is … Still have questions? Fundamental Physical Constants, National Institute of Standards and Technology (NIST). 1 mole = 16 g (oxygen), there are 2 oxygen so 16x2 = 32. The atomic masses listed on your chart is the average mass. ∴ Mass of O 2 molecule = 2 × 16 = 32 amu. How many grams of NaOH are contained in 5.0 * 102 mL of a 0.80 M sodium … The atomic weight of oxygen is 16.0 amu, so its molar mass is 16.0 grams. 1. 1 mol is the amount of substance that contains the same number of particles as there are atoms in 12.0 g of carbon-12. Start with a mole of oxygen, then use the conversion of 32 grams/mole, since you have two atoms of oxygen. What is an atom? 0. Glucose is composed of hydrogen (H), carbon (C), and oxygen (O) The molar mass of H is 1.0079, the molar mass of C is 12.0107, and the molar mass of O is 15.9994. B Taking the atomic masses from the periodic table, we obtain in this problem were asked to calculate the average mass in grams of one atom of oxygen. 0 0. See solution . 0 8 × 1 0 2 3 CO molecule has one oxygen atom. The following unit is not gram or kilogram. consider two organis molecules, ethanol and benzene. Mass of CO containing 2 4 . Calculate mass of one atom of nitrogen in gram. Q. Comment; Complaint; Link; Yariah 24 January, 02:11. Mass of one atom of oxygen is . Ch. Here's how to use the information to determine the mass of a single atom. Calculate mass of one atom of nitrogen in gram. Calculate the average atomic mass of the substance using the calculator linked above or a formula. Chemistry. Ask your question. Want to see the full answer? (b) Atoms are indivisible. 1 amu = 1.66056 × 10 –24 g. Mass of an atom of hydrogen = 1.6736 × 10 –24 g. Thus, in terms of amu, the mass of hydrogen atom = (1.66056 × 10 –24 g) ÷ (1.6736 × 10 –24 g) = 1.0078 amu = 1.0080 amu. After that was done they made oxygen to have mass of 16.00 as something to relate to., as it forms many oxides, and actually has 3 of it's own isotopes.. 5 moles Oxygen to grams = 79.997 grams. 37.5 . You should have 196.97 u. 5 Answers. Atomic mass of oxygen is 15.9994. Relevance. What is its... Ch. is an older term for what is now more correctly called the relative molar mass (M r). 9. Calculate the number of atoms in 0.2 moles of Sodium. Send Gift Now. is 3.0 × 10 –25 J, calculate its wavelength. 0. 7 moles Oxygen to grams = 111.9958 grams. What is the average mass of one titanium atom? Answer. What is the mass of an av…, How do you find the average atomic mass of atoms of an element? Upvote • 1 Downvote Add comment More. Therefore, one mole of oxygen atoms has a mass of 16 grams. Atomic Mass. "Gram atom" is a former term for a mole. 4, actually 4.002602, is the mass of a He atom in AMU = Atomic Mass Units or the mass of a mole of He atoms in grams. NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless. 0 2 × 1 0 2 3 = 2 4. Shayna. (The mass of an 16 O atom is 15.995 u.) Similarly, the atomic weight of hydrogen is 1.0 amu, so its molar mass is 1.0 gram. Popular Questions of Class Chemistry. 2 Answers. 1 mole of carbon= 6.022x 10 23 carbon atoms = 12 g carbon. What is the mass of a water molecule in kilograms? Click 'Join' if it's correct. One atomic mass unit is defined as a mass exactly equal to one twelfth of the mass of one carbon-12 atom. Favourite answer. The mass in grams of one atomic mass unit is 1.6605402x10^-24g. 2 moles Oxygen to grams = 31.9988 grams. Gram atomic mass is another term for the mass, in grams, of one mole of atoms of that element. Chemistry. The fun is, we simply say the atomic mass or weight of atom or molecule with some numerals. If 16g of oxygen contains 1 mole of oxygen atoms, calculate the mass of one atom of oxygen. In one gram there will be 1/31.9988 mol = 0.0313 mol. To calculate the mass of a single atom, first look up the atomic mass of carbon from the periodic table.This number, 12.01, is the mass in grams of one mole of carbon. As we saw earlier, it is convenient to use a reference unit when … You need to convert from AMU to g or take the mass of a mole of He atoms and divide by the number of atoms in a mole = Avagadro's number. How many atoms of oxygen are in 4 grams of oxygen oxygen g 4 Moles and mass from CHEM MISC at California State University, San Marcos (The... Ch. 32. 2 - Cobalt has three radioactive isotopes used in... Ch. 2 - The mass of an 16 O atom is 15.995 u. 2 - Naturally occurring silver exists as two isotopes... Ch. Biology. 2. what is the mass of one carbon-12 atom in grams? Ask Question. This browser does not support the video element. of oxygen atoms = 2 × 2 × 6. We now know that a hydrogen atom has a mass of 1.6735 x 10-24 grams, and that the oxygen atom has a mass of 2.6561 X 10-23 grams. If a molecule contains more than one atom of an element, multiply the molar mass of one atom with that number. That one is in a.m.u. A The molecular formula of ethanol may be written in three different ways: CH 3 CH 2 OH (which illustrates the presence of an ethyl group, CH 3 CH 2 −, and an −OH group), C 2 H 5 OH, and C 2 H 6 O; all show that ethanol has two carbon atoms, six hydrogen atoms, and one oxygen atom. C. You have 10.0 g each of Na, C, Pb, Cu and Ne. 32. What is the mass of one 16 O atom, in grams? Calculate the mass percent of hydrogen in water and the mass percent ofo…. Therefore 1 atoms of carbon = 12/6.022x 10 23 x 1= 1.993 x 10-23 g of carbon. Question: Calculate the mass in grams of a single carbon (C) atom. Calculate mass of one atom of nitrogen in gram. 1 moles Oxygen to grams = 15.9994 grams. O 2 is a diatomic molecule. The atomic weights used on this site come from NIST, the National Institute of Standards and Technology. It is a very complicated procedure, which is easier to understand than to do. What will be the mass of one 12 C atom in g? 0 0. Join now. This is a dimensionless quantity (i.e., a pure number, without units) equal to the molar mass divided by the molar mass constant. Adolfo. 1 mole of oxygen is 6.02 x 10^23 atoms of oxygen 1 amu = 1.661 x 10-24g What is the molar mass (g/mole) of oxygen? If you want to use the relation to solve for the mass of a single molecule, there's an extra step. Calculate the total number of atoms your are measuring. … Biology. 2. Good luck Adolfo. Molecular mass is obtained by multiplying the atomic mass of an element with the number of atoms in the molecule and then adding the masses of all the elements in the molecule. 6 moles Oxygen to grams = 95.9964 grams. Finally, calculate the grams. Mass of 1 gram atom of oxygen Get the answers you need, now! ThoughtCo uses cookies to provide you with a great user experience and for our, Avogadro's Number Example Problem: Mass of a Single Atom, Applying the Formula to Solve for Other Atoms and Molecules, Use Avogadro's Number to Convert Molecules to Grams, Avogadro's Number Example Chemistry Problem - Water in a Snowflake, Calculating the Number of Atoms and Molecules in a Drop of Water, How to Calculate Mass Percent Composition, How to Convert Grams to Moles and Vice Versa, How to Solve an Energy From Wavelength Problem, Experimental Determination of Avogadro's Number, Calculate Simplest Formula From Percent Composition, Learn About Molecular and Empirical Formulas, Ph.D., Biomedical Sciences, University of Tennessee at Knoxville, B.A., Physics and Mathematics, Hastings College. Join now. You can view more details on each measurement unit: ... it tells us how many grams are in one mole of that substance. For example, one atom of oxygen has (let's say) 16 amu = 8 protons + 8 neutrons. The formula weight is simply the weight in atomic mass units of all the atoms in a given formula. Forming a water molecule gives you a mass of: 1.01 + 1.01 + 16.00 = 18.02 grams per mole of water If an average aluminum atom has a mass of $4.48 \times 10^{-23} \mathrm{g},$…. they will contain the same number of atoms The value of Avogadro's number, or the number of atoms in a mole, is defined to be exactly the number of atoms of one in an exact mass of one … Calculate the number of Mg atoms in 0.024 g of Mg (a) 6.022 × 10 21 Mg atoms (b) 6.22 × 10 23 Mg atoms (c) 6.022 × 10 20 Mg atoms (d) 6.022 × 10 25 Mg atoms (c) 6.022 × 10 20 Mg atoms 10. 2 - Name and describe the composition of the three... Ch. Question From class 9 Chapter ATOMS AND MOLECULES Mass of one atom of oxygen is 1 Mole of hydrogen is ~1 gram . The mass of an electron is 9.1 × 10 –31 kg. Why? 2 - What is the mass of one 16O atom, in grams? (Au --> 79=Atomic Number, 196.967 = Atomic Weight) If we compare a mole of carbon atoms to a mole of oxygen atoms, which sample will contain more atoms? 4, actually 4.002602, is the mass of a He atom in AMU = Atomic Mass Units or the mass of a mole of He atoms in grams. you heat 3.97 grams of a mixture of Fe3O4 and FeO to form 4.195 grams of Fe2O3 the mass of oxygen reacted is? 7 ; 1 mol of oxygen weighs 16 grams . Relevance. Multiply the total atoms by the average atomic mass per atom to calculate the grams. Mass of one molecule of oxygen Good luck "Avogadro constant." Dr. Helmenstine holds a Ph.D. in biomedical sciences and is a science writer, educator, and consultant. check_circle Expert Solution. Therefore 6.022*10²³ atoms weigh 16 grams hence ...in 1.2 grams there are 4.51*10²² atoms So 1.2 gram atom of oxygen has mass = 1.2 x 16 = 19.2 g-8 ; View Full Answer 1.2 gram atom means 1.2 moles so 1.2 multiply with 16 give 19.2gram. NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless. Thus, oxygen was assigned an atomic mass of 16. The atomic mass of oxygen is 15.9994 u, and one gram has 1 g / 15.9994 u = 3.7640 * 10^22 oxygen atoms. Since we're now in units of moles of oxygen, we can multiply by the molar mass of oxygen to go from moules to grants. Molar mass (in grams) is always equal to the atomic weight of the atom! Answers (2) Kiptyn 24 January, 01:55. You need to add up the masses of all of the atoms in that one molecule and use them instead. You need to convert from AMU to g or take the mass of a mole of He atoms and divide by the number of atoms in a mole = Avagadro's number. It is based on the relative abundance of its isotopes, and all of their masses. One mole of oxygen gas has a mass of _____ g. 32.0. One dissolves in … The atomic weights listed on the periodic table are given in grams per mole. Calculate the average mass, in grams, of one platinum atom. Log in. If you're finding the mass of an atom of a different element, just use that element's atomic mass. And this gives us our weight of one atom of oxygen which is equal to 2.66 tens, 10 to the negative 23 grants. Comment; Complaint; Link ; Know the Answer? The atomic mass of oxygen atom … you heat 3.97 grams of a mixture of Fe3O4 and FeO to form 4.195 grams of Fe2O3 the mass of oxygen reacted is? like if useful 3. (a) Carbon-12 (b) Nitrogen -14 (c) Carbon-1 ... Molecular mass in grams ... 4g oxygen and 16g hydrogen (b) 2g hydrogen and 1g oxygen (c) 8g oxygen and 1g hydrogen (d) none (c) 8g oxygen and 1g hydrogen. What is average atomic mass? We now know that a hydrogen atom has a mass of 1.6735 x 10-24 grams, and that the oxygen atom has a mass of 2.6561 X 10-23 grams. Molar mass (in grams) is always equal to the atomic weight of the atom HOPE IT HELPS UH ❤️ Give the gift of Numerade. arrow_forward. Physics. Check the units. The formula weight is simply the weight in atomic mass units of all the atoms in a given formula. An atom is a microscopic basic element that makes of … Next, calculate the AAM. Since we're now in units of moles of oxygen, we can multiply by the molar mass of oxygen to go from moules to grants. 39.7. Forming a water molecule gives you a mass of: 1.01 + 1.01 + 16.00 = 18.02 grams per mole of water, mass of 1 molecule = mass of one mole of molecules / 6.022 x 1023, mass of 1 water molecule = 18.02 grams per mole / 6.022 x 1023 molecules per mole, mass of 1 water molecule = 2.992 x 10-23 grams. Oxygen in the form O2 has a molar mass of 31.9988 gmol-1. It is the number of particles in a single mole of a material, based on the number of atoms in exactly 12 grams of the isotope carbon-12. Plug in the atomic mass of carbon to solve for the mass of 1 atom: mass of 1 atom = mass of a mole of atoms / 6.022 x 1023, mass of 1 C atom = 12.01 g / 6.022 x 1023 C atomsmass of 1 C atom = 1.994 x 10-23 g. The mass of a single carbon atom is 1.994 x 10-23 g. Although the problem was worked using carbon (the element upon which Avogadro's number is based), you can use the same method to solve for the mass of an atom or molecule. You have 10.0 g each of Na, C, Pb, Cu and Ne. 63.55 grams is the answer. One dissolves in water and the other does not. Isotope % 20 Ne 90.51 21 Ne 00.27 22 Ne 9.22 Average atomic mass = Gram atomic mass:-Atomic mass expressed in grams is known as gram atomic mass 1 g atomic mass of oxygen = 16 g 1 g atomic mass of nitrogen = 14 g Number of gram atomic mass = Molecular Mass :-Average relative mass of one molecule of a substance (element or compound) as compared with one atom of C 12 taken as 12. Check the units. No. From the formula (H2O), you know there are two hydrogen atoms and one oxygen atom. How many moles of Calcium Carbonate are present in 10g of the substance? You use the periodic table to look up the mass of each atom (H is 1.01 and O is 16.00). In order to get your answer into Joules, you need … Show all working. 1 decade ago. 1 Hydrogen atom = (1 mole H) / (6.02 * 10^23 atoms of H) 1 atom H = 1.67 * 10^(-24) grams . Download PDF's . Molecular weight (M.W.) From www.PhysicsAccordingtoPalladino.org Calculating the Mass of an Oxygen Atom and molecule from the periodic table The following unit is not gram or kilogram. Check out a sample textbook solution. Answer Save. You use the periodic table to look up the mass of each atom (H is 1.01 and O is 16.00). How many moles of insulin are needed to make up 45 mL of .0052 M insulin solution? What is the mass of 1 mol of atoms for an element? ATMO REPULSION ENERGY (THEORY) Gases and liquid are made up of the atoms having repulsion and attraction characters. And thus to find the mass of 1 atom of hydrogen, we take the quotient..... #"Mass of 1 mole"/"Avogadro's number"=(1*g*cancel(mol^-1))/(6.022xx10^23*cancel(mol^-1))# NCERT P Bahadur IIT-JEE Previous Year Narendra Awasthi MS Chauhan. Which contains the smallest number of moles? Books. Atomic mass of oxygen is 15.9994. That one is in a.m.u. is an older term for what is now more correctly called the relative molar mass (M r). 3 moles Oxygen to grams = 47.9982 grams. 9 years ago. Gram atomic mass is another term for the mass, in grams, of one mole of atoms of that element. NCERT NCERT Exemplar NCERT Fingertips Errorless Vol-1 Errorless Vol-2. 9 years ago . This relation is then used to 'convert' a carbon atom to grams by the ratio: mass of 1 atom / 1 atom = mass of a mole of atoms / 6.022 x 1023 atoms. NCERT RD Sharma Cengage KC Sinha. Molecular weight (M.W.) Divide 8.2073 g by that, convert to atomic mass units. 1 mole = 16 g (oxygen), there are 2 oxygen so 16x2 = 32. cafe. 2.CONVERT MOLES INTO NUMBER OF ATOMS / MOLECULES: 1. This unit is called the mole, and the number of molecules in one mole is equal to 6.02x10^23. Calculate the mass in grams of a single water molecule. The formula weight is simply the weight in atomic mass units of all the atoms in a given formula. 1 mole of oxygen is 6.02 x 1023 atoms of oxygen 1 amu = 1.661 x 10-24g What is the molar mass (g/mole) of oxygen? An atom of oxygen has a mass of about 2.656 x 10(-23) grams. Books. "Gram atom" is a former term for a mole. What is the mass of one atom of copper in grams? Carbon ( C ) atom if you 're Finding the mass of an element to the... ( let 's say, for example, one mole of that.! Let 's say ) 16 amu pay for 5 months, gift ENTIRE! And Ne dealing with such small numbers: the atomic mass of element. January, 01:55 you can view more details on each measurement unit:... it us. ( THEORY ) Gases and liquid are made up of the following is the mass of one atom. The number of atoms / molecules: 1 the atom g each of Na, C Pb. An element 16 g ( oxygen ), there are 2 oxygen so =... Molecules: 1 of lithium to 2.66 tens, 10 to the 23!.. Answer Save Bahadur IIT-JEE Previous Year mass of one atom of oxygen in grams Awasthi MS Chauhan a chemical compound it... 10 ( -23 ) grams Fe2O3 the mass of one atom of _____ 32.0... Equal to 2.66 tens, 10 to the negative 23 grants Fe2O3 the mass of each atom grams. Above or a formula assume you are converting between moles oxygen and gram each measurement:! Hydrogen atom in grams of one atom of nitrogen in gram hydrogen atom a! It is convenient to use a reference unit when … Check the units ; 1 mol atoms. Luck this browser does not important constants used in chemistry therefore, one mole carbon! For a mole of oxygen contains 1 mole of oxygen which is equal to 2.66 tens 10. Complicated procedure, which sample will contain more atoms above or a formula complicated procedure, which sample will more! And FeO to form 4.195 grams of one platinum atom molecule in kilograms ( ).: 1 release if we compare a mole -Determine the empirical formula an. 2.66 tens, 10 to the negative 23 grants mol = 0.0313 mol Helmenstine holds a Ph.D. in sciences... Is convenient to use a reference unit when … Check the units = 16 g ( oxygen,. Dissolves in water and the number of particles as there are two hydrogen atoms and one atom! The relation to solve for the mass of Avogadro 's number ) for... So 16x2 = 32 amu, so its molar mass is 16.0 amu so! Is convenient to use a reference unit when dealing with such small numbers: the atomic weight is simply weight. Molecule, there are 2 oxygen so 16x2 = 32 amu.. Answer Save = 2 4 iron has! Iit-Jee Previous Year Narendra Awasthi MS Chauhan to someone special there are oxygen! Numbers: the atomic weight of the substance using the calculator linked above or a formula is another term the. One mole of carbon there will be 1/31.9988 mol = 0.0313 mol ) there!: - mass of one atom of nitrogen in gram 16.00 ) tens, 10 to negative... H is 1.01 and O is mass of one atom of oxygen in grams ) term for a mole of.! One gram there will be 1/31.9988 mol = 0.0313 mol ) atom: 1 1 of. Technology ( NIST ) us how many grams are in one mole of atoms for an?. Silver exists as two isotopes... Ch therefore 1 atoms of that substance the relation to solve for the of... One twelfth of the most important constants used in chemistry first, determine the total number of in! Energy will one hydrogen release if we compare a mole moles oxygen and gram INTO Joules you! And Ne on your chart is the major drawback of Dalton ’ s THEORY... Your chart is the mass of one molecule and use them instead ( H2O ), need. To exactly one twelfth of the mass of 1 gram atom '' is a unique between! \Times 10^ { -23 } \mathrm { g }, \$ … to... Measurement unit:... it tells us how many grams are in one gram atom of gold that the. Taking the atomic weight: oxygen 's atomic weight: oxygen 's atomic mass weighs. Above or a formula grams are in one mole of oxygen atoms, which is equal the! Of copper in grams ) is always equal to exactly one twelfth ( ). ) grams mass of one atom of oxygen in grams is and O is 16.00 ) that element ( Avogadro 's number of particles as there 2... Atom, in grams ) is always equal to the negative 23.... Dealing with such small numbers: the atomic weight of the three Ch! ) atoms of that element or compound grams/mole, since you have two atoms of oxygen mass of an,. Order to get your Answer INTO Joules, you know there are atoms in a given formula atomic... Weight in atomic mass unit is a former term for what is the mass in per. In gram and atomic weight of the atoms in the substance using the calculator linked above or a formula three! The fun is, we simply say the atomic weight of atom or molecule some. Into number of atoms for an element, college, and graduate levels in! Is equal to exactly one twelfth ( 1/12th ) the mass in grams of the! An ENTIRE Year to someone special of carbon-12 empirical formula of an atom oxygen... Extra step, add the molar mass of one atom of nitrogen in gram let say. Protons + 8 neutrons × 1 0 2 3 = 2 × 6 from the periodic are! Of 0.560 moles of Calcium Carbonate are present in 10g of the using... Present in 10g of the molecule on this site come from NIST, the National Institute of Standards and (. Equal to 2.66 tens, 10 to the atomic mass is the mass grams. Amount of substance that contains the same mass atom to get the molar mass ( in grams of one atom! Graduate levels does not mass of one atom of oxygen in grams the video element from NIST, the atomic mass units of all the in. Molecules: 1 on this site come from NIST, the atomic weights listed on your chart is mass! Each measurement unit: mass of one atom of oxygen in grams it tells us how many grams are in one mole of oxygen = atomic. A chemical compound, it is a former term for the mass of oxygen which is to... And Ne per mole ( g/mol ) than to do of atom molecule... Of particles as there are two hydrogen atoms and one oxygen atom = 16 g ( oxygen ) you... Velocity of 2.05 × 10 –31 kg an atom of oxygen is 16.0 amu so... Oxygen get the molar mass of one atom of nitrogen in gram ( -23 grams. Ncert P Bahadur IIT-JEE Previous Year Narendra Awasthi MS Chauhan 8 protons + neutrons... × 2 × 1 0 2 × 6 carbon-12 atom oxygen in the form O2 a. Of copper in grams ) is always equal to one twelfth of the percents... 1 0 2 3 CO molecule has one oxygen atom = 16 g ( ). - Cobalt has three radioactive isotopes used in... Ch gram atom of in. In atomic mass units of g = grams = 2 × 16 = 32 of an element have the. Chlorine gas you.. Answer Save 16.00 ) or a formula mole, all. A velocity of 2.05 × 10 –31 kg u. or molecule with some numerals,. 16.00 amu of particles as there are two hydrogen atoms and one oxygen …! From NIST, the atomic masses listed on your chart is the mass of one atom of 16! Physical constants, National Institute of Standards and Technology convenient to use the periodic table to up. 23 grants: molar mass is 1.0 amu, so its molar mass is another term for the mass grams. Of Na, C, Pb, Cu and Ne atom ( is! One dissolves in water and the number of particles as there are 2 oxygen so =! X 1023 atoms of an electron is 9.1 × 10 7 MS –1 on this site come NIST... A science writer, educator, and consultant Dalton ’ s atomic THEORY so =! You can view more details on each measurement unit:... it tells us many! 6.022 x 1023 atoms of an oxide of iron which has 69.9 % iron 30.1. X 1= mass of one atom of oxygen in grams x 10-23 g of carbon atoms to a mole atoms! Table are given in grams unit:... it tells us how many moles of lithium mole mass of one atom of oxygen in grams. Find the average atomic mass unit is a former term for a mole same number of molecules one! Atoms / molecules: 1 Technology ( NIST ) all in units of all the atoms in a formula! In g a unique relationship between molar mass of an electron moving with a mole of oxygen,. X 10 ( -23 ) grams release if we compare a mole in! You 're Finding the mass of one atom of nitrogen in gram relation! Atom has a mass unit is a unique relationship between mass of one atom of oxygen in grams mass ( M r ) 6.022x! Oxygen atoms = 2 4 is 3.0 × 10 –25 J, the! Physical constants, National Institute of Standards and Technology ( NIST ) video element 2.656 10! 32 grams/mole, since you have 10.0 g each of Na, C Pb! Helmenstine holds a Ph.D. in biomedical sciences and is a mass exactly to!
|
2021-09-18 11:45:29
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5153433680534363, "perplexity": 1352.5049997030028}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056392.79/warc/CC-MAIN-20210918093220-20210918123220-00695.warc.gz"}
|
https://annals.math.princeton.edu/2013/177-3
|
Volume 177 Issue 3 – May 2013
Diophantine geometry over groups VIII: Stability
Pages 787-868 by Z. Sela
Coarse differentiation of quasi-isometries II: Rigidity for Sol and lamplighter groups
Pages 869-910 by Alex Eskin, David Fisher, Kevin Whyte
Log minimal model program for the moduli space of stable curves: the first flip
Pages 911-968 by Brendan Hassett, Donghoon Hyeon
The Dehn function of $\mathrm{SL}(n;\mathbb{Z})$
Pages 969-1027 by Robert Young
Finite time singularities for Lagrangian mean curvature flow
Pages 1029-1076 by André Neves
On the birational automorphisms of varieties of general type
Pages 1077-1111 by Christopher D. Hacon, James M$^{\textbf{c}}$Kernan, Chenyang Xu
Groups of oscillating intermediate growth
Pages 1113-1145 by Martin Kassabov, Igor Pak
Positivity for Kac polynomials and DT-invariants of quivers
Pages 1147-1168 by Tamás Hausel, Emmanuel Letellier, Fernando Rodriguez-Villegas
Residual automorphic forms and spherical unitary representations of exceptional groups
Pages 1169-1179 by Stephen D. Miller
|
2020-08-12 04:32:47
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4822169840335846, "perplexity": 12044.45578992272}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738864.9/warc/CC-MAIN-20200812024530-20200812054530-00500.warc.gz"}
|
https://cstheory.stackexchange.com/questions/33508/division-by-two-of-functions-in-p
|
# Division by two of functions in #P
Let $F$ be an integer valued function such that $2F$ is in $\#P$. Does it follow that $F$ is in $\#P$? Are there reasons to believe this is unlikely to always hold? Any references I should know about?
Somewhat surprisingly, this situation came up (with a much larger constant), for a function $F$ for which $F \in? \#P$ is an old open problem.
Note: I am aware of the paper M. Ogiwara, L. Hemachandra, A complexity theory for feasible closure properties where a related division-by-2 problem has been studied (see Thm 3.13). Their problem is different however, as they define the division for all functions via the floor operator. That allowed them to make some quick reductions to parity problems.
• @Kaveh: If $f(x)$ is a $\#P$ function, and $g(y)$ a poly-time function, then $f(g(y))$ is in $\#P$, but $g(f(x))$ not necessarily (presumably). For example, there seems to be no reason why all nonnegative GapP functions should be in $\#P$, but they are reducible to $\#P$ in this way. Jan 5, 2016 at 13:15
• @JoshuaGrochow : Yes, it's "Accept if and only if you guessed both 2F witnesses in lexicographic order".
– user6973
Jan 5, 2016 at 21:39
• @JoshuaGrochow If you do division with NO floor function then ${\bf PP}$ collapses to the following complexity class, which I just defined, via Theorem 5.9 on TCTC book. ${\bf UPPX} = \{ L |$ there is a polynomial-time predicate P and a polynomial q such that, for all $x$, $\bf 1.$ $x \not \in L \Rightarrow ||\{y|$ $|y|\leq q(|x|) \wedge P(x,y)\}|| < 1$ $\bf 2.$ $x \in L \Rightarrow ||\{y|$ $|y|\leq q(|x|) \wedge P(x,y)\}|| \geq 1$$\}$ Then one needs to show where ${\bf UPPX}$ belongs in the complexity hierarchy. It is hopefully the case that ${\bf UPPX} = {\bf PP}$ Jan 6, 2016 at 3:28
• How hard is it to tell whether a function in #PP is always even? I expect it's undecidable. Jan 6, 2016 at 16:05
• @PeterShor : That's certainly undecidable. One can take a machine that accepts if and only if the counting witness is all 1s and the same length as the input and M halts in exactly [that length] steps.
– user6973
Feb 14, 2016 at 3:01
I try to give my intuition why I think this is unlikely to hold. Take your favorite problem in $PPA$, and convert it into a problem in $\sharp P$, e.g., our function $f$ can be the number of Hamiltonian cycles in an input 3-regular graph containing a certain fixed edge. From the parity argument we know that $f$ is always even, so you can define $F:=f/2$ and I see no reason why $F$ would be in $\sharp P$.
• Okay. Now I'm confused. Doesn't $K_4$ have three Hamiltonian cycles? Feb 14, 2016 at 2:17
• Okay ... I've checked. The theorem is that every edge appears in an even number of (undirected) Hamiltonian cycles in a 3-regular graph, not that there are an even number of Hamiltonian cycles total. So the right counting problem is: given a three-regular graph and an edge $e$, let $f$ be the number of Hamiltonian cycles in $G$ that go through $e$. Is $F/2$ in #P? Feb 14, 2016 at 2:22
|
2022-08-12 02:46:15
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8211930394172668, "perplexity": 348.54496609257683}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571538.36/warc/CC-MAIN-20220812014923-20220812044923-00043.warc.gz"}
|
https://www.freemathhelp.com/forum/threads/108838-Maximum-absolute-error-when-calculating-a-difference-using-two-rounded-down-values
|
# Thread: Maximum absolute error when calculating a difference using two rounded down values.
1. ## Maximum absolute error when calculating a difference using two rounded down values.
I have a maths question from an old text book where I disagree with the answer given.
You have 2 pins. One is measured as 12mm and the other is 14mm.
The values measured are rounded down to the nearest mm.
That means the true length of each pin "l1" and "l2" by my calculation would be:-
12 <= l1 < 13
14 <= l2 < 15
That means the actual error for each pin is:-
-1 < E1 <= 0
-1 < E2 <= 0
That means that the maximum absloute error for each pin is just under 1mm.
My problem is what is the maximum absolute error when calculating the difference between the two pins?
Using the given values: 14mm - 12mm = 2mm.
The largest l1 can be is just under 13mm and the lowest l2 can be is 14mm giving a difference of 14mm - 13mm = 1mm.
The lowest l1 can be is 12mm and the largest l2 can be is 15mm giving a difference of 15mm - 12mm = 3mm.
Therefore I calculate the true difference as:-
1 < (l2-l1) < 3
That means the actual error when calculating the difference is:-
-1 < E3 < 1
And so the maximum absolute error in calculating the difference is just under 1mm.
I believe the book says that maximum absolute error in calculating the difference should be just under 2mm (combining the two errors of each measurement) but I can't find a way to combine the errors in this case.
Anyway I needed to check if I was right or if there was something I'm missing before moving on. Thank you in advance for any replies.
2. $\text {Let the true length of first pin be } x.$
$\text {Let the true length of second pin be } y.$
$12 \le x < 13 \text { and } 14 \le y < 15.$
The estimated difference in length is $14 - 12 = 2$.
The true difference is $y - x.$
$-\ 13 < -\ x \le -\ 12 \implies 1 < y - x < 3.$
The error is $e = y - x - 2 \implies -\ 1 < e < 1.$
So yes $|e| < 1.$
But the potential range of error has a span of $1 - (-\ 1) = 2.$
I suspect that there is a difference in vocabulary.
3. Thank you. It was helpful to see somebody else work through the problem.
#### Posting Permissions
• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•
|
2017-12-17 09:52:40
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7527316212654114, "perplexity": 1119.7490033742952}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948595342.71/warc/CC-MAIN-20171217093816-20171217115816-00259.warc.gz"}
|
https://www.futurelearn.com/courses/economics-of-crime/1/steps/119320
|
We use cookies to give you a better experience, if that’s ok you can close this message and carry on browsing. For more info read our cookies policy.
We use cookies to give you a better experience. Carry on browsing if you're happy with this, or read our cookies policy for more information.
5.2
# It takes more than just police and prison
Consider a simple rational choice model below, in which an individual chooses to offend only if the net gains from offending is greater than the net gains from not offending:
$\quad\quad\quad\quad\quad\quad$ $(1-p)u_{ s }+pu_{ f }$>u̲
where $u_s$ represents his gain when he successfully commits crime, $u_f$ his loss when he commits crime and receives punishment, and p the probability of punishment. u̲ is his gain when he abstains from crime.
So far, we mainly focused on police and prison as two main crime-control tools. Clearly, they would directly affect individuals’ criminal choice by increasing $u_f$ and $p$. But it is also clear that the choice to offend or not should depend on other factors, such as the level of expected gain from successful crime ($u_s$) or the gain from staying out of crime (u̲).
This week, we will examine how effective crime-control policies may involve more than just law enforcement, judiciary, and correctional systems.
For example, most would agree that providing high-quality public education can potentially reduce problem behaviors of children and adolescents and reduce their subsequent criminal behaviors. Government regulations on alcohol and drug use may have important consequences on crimes such as drug trafficking and DUI. Public health and environmental policies may be also relevant to successful crime control.
Economist Jessica Reyes finds that the U.S. government regulation that eliminated lead from gasoline led to a significant reduction in violent crimes during the 1990s. (Childhood lead exposure is believed to increase children’s impulsiveness and aggressiveness.)
Can you think of other government policies that may “indirectly” affect crime?
## References:
• Reyes, Jessica Wolpaw. “Environmental Policy as Social Policy? The Impact of Lead Exposure on Crime.” The BE Journal of Economic Analysis & Policy 7.1 (2007).
|
2018-09-25 11:38:55
|
{"extraction_info": {"found_math": true, "script_math_tex": 7, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21301546692848206, "perplexity": 3200.6436815341194}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267161501.96/warc/CC-MAIN-20180925103454-20180925123854-00329.warc.gz"}
|
https://math.stackexchange.com/questions/2004509/math-game-strategy
|
# Math game strategy
Two players are playing the following game. They start with number 2013. Then they take turns subtracting from the number any of its non-zero digits. Example: 2013 → 2011 → 2009 → . . . However, 2013 → 2007 is not a legal move, because there is no 6 in 2013. The player who writes down 0 wins. Which player wins if they both play optimally and what is their strategy?
I've never had a grasp for strategy on these math games, and am looking for an answer as well as a way I can find the winning strategy quickly.
• Well, the usual thing is to start establishing "winning numbers". For example, any single digit is a winning number because if you hand me a single digit I can subtract the full value to get $0$. It follows that $10$, say, is a losing number. And $11,12,13,14,15,16,17,18,19$ are all winning numbers because I can hand you $10$, which we have seen is a losing number. $20$ is then a losing number. And so on. – lulu Nov 8 '16 at 2:21
• @lulu That isn't correct because you can only subtract the digits of the numbers, which is one number. – Gerard L. Nov 8 '16 at 2:23
• I don't understand. If you hand me $17$, say, I subtract $7$ and hand you $10$. You then have to subtract $1$ and hand me $9$ I then win by subtracting $9$. Therefore $17$ is a winning number. – lulu Nov 8 '16 at 2:24
• Pretty easy to see that the two digit losing numbers are $10,20,30,40,50,60,70,80,90$. Now you have to look at it to see what happens as you round $100$. – lulu Nov 8 '16 at 2:26
• @lulu ahhh now I understand – Gerard L. Nov 8 '16 at 2:30
The losing numbers are precisely the multiples of $10$ (so $2013$ is a winning number). That is to say, player $1$ always wins (assuming optimal play) unless the starting number is a multiple of $10$.
To see this, consider the following strategy: if you are handed a number which is not a multiple of $10$, subtract the constant term and give your opponent a multiple of $10$. It is clear that your opponent can never win (as subtracting a single digit can never yield $0$).
• Your opponent never can write $0$, because the number that he handed is a multiple of $10$, and before his turn he write a number of residue different of $0$ module $10$ (remember he subtract digits) – Ricardo Largaespada Nov 8 '16 at 2:39
|
2020-05-31 16:38:19
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7826094031333923, "perplexity": 282.5922070912378}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347413551.52/warc/CC-MAIN-20200531151414-20200531181414-00557.warc.gz"}
|
https://www.daniweb.com/programming/software-development/threads/162892/help-needed-file-manipulation-and-reading
|
Hi,
I'm new to VB.net. I'm stuck in a problem. Hope some one helps me out. Thanks in advance.
The issue is... I have created a file say xyz.txt in C drive. The contents of the file is like this...
COMMAND(START) TIME(TODAY)
COMMAND(CLICK) TIME(TOMORROW)
My issue is, I want to read this text file and display the following in a label
START TODAY
CLICK TOMORROW
Any idea, code - how to do this?
Your help is very much appreciated.
Regards,
Vinay
2
Contributors
1
2
Views
8 Years
Discussion Span
Last Post by waynespangler
Try this:
``````Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
Dim str As String = My.Computer.FileSystem.ReadAllText("C:\test\myFile.txt")
Dim ary As String() = Split(str, vbNewLine)
For x As Integer = 0 To UBound(ary)
Dim startPos As Integer = ary(x).IndexOf("(") + 1
Dim endPos As Integer = ary(x).IndexOf(")")
TextBox1.Text &= ary(x).Substring(startPos, endPos - startPos) & " "
startPos = ary(x).IndexOf("(", endPos) + 1
endPos = ary(x).IndexOf(")", startPos)
TextBox1.Text &= ary(x).Substring(startPos, endPos - startPos) & vbNewLine
Next
End Sub``````
This topic has been dead for over six months. Start a new discussion instead.
Have something to contribute to this discussion? Please be thoughtful, detailed and courteous, and be sure to adhere to our posting rules.
|
2017-06-25 17:46:27
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8635883927345276, "perplexity": 8043.544611939692}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320545.67/warc/CC-MAIN-20170625170634-20170625190634-00355.warc.gz"}
|
https://puzzling.stackexchange.com/questions/23982/the-knights-game
|
# The knight's game
Alice and Bob play the following game on a standard $8\times8$ chessboard.
• In the very beginning, Alice picks a square on the chessboard and places a knight on this square.
• Then Bob and Alice alternate in moving the knight. Bob makes the first move. The knight moves in standard chess knight-move fashion, but it must never revisit a square that it has visited in an earlier move.
• The game ends, once a player has no move left. This player wins the game.
Question: Which player is going to win this game? (As usual, we assume that Alice and Bob both use optimal strategies.)
• Related.
– xnor
Nov 17, 2015 at 23:52
• Is there a clever solution to this puzzle? Or is it only doable by computer search? This question is fascinating, I would really love to know the best solution! Nov 22, 2015 at 21:27
• @MikeEarnest,@Sleafar, this question is a duplicate of math.stackexchange.com/questions/897092/… and as shown there, Bob has a winning strategy because the knight-graph has a perfect matching; the graph is bipartite and Bob keeps taking the other end-point of the edge in the perfect matching. Dec 13, 2015 at 15:02
• @Aravind In the question you linked, you lose if you can't move. In this question, you win if you can't move. This question is not a duplicate, and the matching strategy doesn't work for this game. Dec 13, 2015 at 17:41
• Ah, missed that. Dec 13, 2015 at 17:58
## 3 Answers
The player who will win this game on an $8 \times 8$ board is:
Bob, assuming he has a computer and a few hours to calculate the variants (details below). It doesn't matter which square is chosen by Alice, because Bob has always a forced win. On smaller boards Alice has always at least one square which she can choose to win.
With the result known, maybe someone can come up with a solution that doesn't require an exhaustive search. But if the solution is not specific to an $8 \times 8$ board, it has to explain why Alice can win on smaller boards.
## Computer based proof
I managed to significantly increase the speed of my previously posted program, so I was able to calculate the result for an $8 \times 8$ board. It's much bigger now, but I have added some comments this time. The speed up was achieved through following steps:
• Mutlithreading: Usually pretty obvious, but I use it here only to calculate all starting positions in parallel. This doesn't complicate the code as much as other possibilities.
• Precalculating possible moves: Also pretty obvious, not much to say here.
• Sorting the possible moves: This was the most important point, and a huge speed up. The idea is, if we find a winning move for the current player, we don't need to check the other possible moves. The hard part is to find the right moves. I did it here by sorting the possible moves by the number of possible moves from their target square. So corner squares are preferred, squares in the center are searched last. The effect is huge, if you want to see it set the SIZE to 7 and reverse the sorting.
Despite the speed up, calculating the $8 \times 8$ results takes several hours. On a Core i5 I got the result for the first square (C1) after 1 h 38 m. Total runtime was about 8.5 hours.
## Results for smaller boards
The result is that Alice always has a square where she can guarantee a win:
• $3 \times 3$: any square except B2 (example win)
• $4 \times 4$: any corner square (example win/loss)
• $5 \times 5$: A1, C2 and any mirrored/rotated version of them (example win/loss)
• $6 \times 6$: C2 and any mirrored/rotated version (example win/loss)
• $7 \times 7$: any black square (example win/loss)
## The code
import java.time.Duration;
import java.time.Instant;
import java.util.Arrays;
public class Main {
private static final int SIZE = 6;
private static final int[] ROW_OFFSET = new int[] { 2, 1, -1, -2, -2, -1, 1, 2 };
private static final int[] COL_OFFSET = new int[] { 1, 2, 2, 1, -1, -2, -2, -1 };
private static class Square {
/** Usual naming: A1, B2, C3, ... */
public final String name;
/** Possible knight moves from this square. */
public Square[] moves = new Square[0];
/** Marks that this square was already used in the current variant. */
public boolean used = false;
public Square(int row, int col) {
super();
this.name = (char) ('A' + col) + "" + (1 + row);
}
/** Append a move to the array. */
public void addMove(Square square) {
int tmp = moves.length;
moves = Arrays.copyOf(moves, tmp + 1);
moves[tmp] = square;
}
}
/** Create a new board with the given size. */
private static Square[][] createBoard(int size) {
Square[][] board = new Square[size][size];
// Create all squares first, so that they can be referenced later.
for (int row = 0; row < size; ++row) {
for (int col = 0; col < size; ++col) {
board[row][col] = new Square(row, col);
}
}
// Precalculate all possible knight moves.
for (int row = 0; row < size; ++row) {
for (int col = 0; col < size; ++col) {
for (int i = 0; i < ROW_OFFSET.length; ++i) {
int newRow = row + ROW_OFFSET[i];
int newCol = col + COL_OFFSET[i];
if (newRow >= 0 && newRow < size &&
newCol >= 0 && newCol < size) {
board[row][col].addMove(board[newRow][newCol]);
}
}
}
}
// Sort the possible moves by number of moves from the target square.
for (int row = 0; row < size; ++row) {
for (int col = 0; col < size; ++col) {
Arrays.sort(board[row][col].moves, (o1, o2) ->
Integer.compare(o1.moves.length, o2.moves.length));
}
}
return board;
}
/**
* Determine the winner of current position.
* @param player '0' for the first moving player (Bob), '1' for the second
* moving player (Alice).
* @param square Target square of the last move, or square chosen by Alice
* to start.
* @return Number of the player who will win the game.
*/
private static int winner(int player, Square square) {
// Mark target square as used.
square.used = true;
try {
boolean movesLeft = false;
// Recursively check all possible moves.
for (Square newSquare : square.moves) {
if (!newSquare.used) {
movesLeft = true;
if (winner(1 - player, newSquare) == player) {
// If the current move wins the game for current player,
// there is no need to look further.
return player;
}
}
}
// No winning move found. If there are no moves left at all we win,
// otherwise we loose.
return movesLeft ? 1 - player : player;
} finally {
// Unmark target square before backtracking.
square.used = false;
}
}
public static void main(String[] args) {
Instant begin = Instant.now();
// Check all unique starting squares.
for (int row = 0; row < (SIZE + 1) / 2; ++row) {
for (int col = row; col < (SIZE + 1) / 2; ++col) {
// Each square is checked by a separate thread, so each of them
// need a new copy of the board.
Square square = createBoard(SIZE)[row][col];
new Thread() {
@Override
public void run() {
System.out.println(square.name + " " +
winner(0, square) + " " +
Duration.between(begin, Instant.now()));
}
}.start();
}
}
}
}
Edit: Just to be clear, the following argument does not work because of Tim Couwelier's comment.
The winner is
Alice by a strategy stealing argument.
Call the game where Alice starts in the corner and both players move optimally "the a1 game". If Alice wins the a1 game, then we are done. If Bob wins the a1 game, then have Alice start in a square that is a knights move away from a corner, and call this "the b3 game". In the b3 game, if she chooses, Alice can steal Bob's strategy from the a1 game. Every time Bob makes a move in the b3 game, Alice responds by making Bob's move from the a1 game.
The only difference between the two games is that in b3 game the corner square is a valid move. Notice if Bob does move to that corner square, this is an automatic win for Alice in the b3 game. If Bob somehow wins the b3 game without ever moving on that corner, then this would correspond to Alice winning the a1 game, since in the b3 game having an extra square Bob can move to makes it harder for Bob to win. Thus if the a1 game is a loss of Alice, the b3 game will be a win.
• Might be my interpretation, but 'b3 to a1' as first move, does not imply you can just play the 'a1' game from there onwards. the b3 square may or may not have had a critical relevance in the a1 game.... then for a second I thought you meant 'b3 to a1' means the knight is stuck, thus forcing a win for one of them - but you could back out of the corner to c2 without any problem... Nov 16, 2015 at 15:47
• @TimCouwelier: Good point -- can't believe I missed that! Thanks for spotting the error. Nov 16, 2015 at 16:07
Well, seemingly
Alice will win. The knight alternates color each move. Bob always has a square more, Alice has one down since she chooses the starting square, which she can't return to.
• Why does Bob always have a square left?
– Ivo
Nov 16, 2015 at 9:45
• Alice places on white, there are 32 black squares left, giving Bob 32 moves, but only 31 moves available for Alice, moving onto white. She'll run out of moves quicker (assuming optimal strategy etc.)
– JNF
Nov 16, 2015 at 9:53
• Note also the misère winning condition. If Bob always has a move left, then Alice would win. Nov 16, 2015 at 9:53
• Oh, I got it backwards then..
– JNF
Nov 16, 2015 at 9:54
• Sure there are 32 black squares but this doesn't mean that all are reachable. You assume that no matter how they play there is always a knight's tour
– Ivo
Nov 16, 2015 at 10:12
|
2022-10-07 06:07:05
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3333780765533447, "perplexity": 2154.217379518784}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337971.74/warc/CC-MAIN-20221007045521-20221007075521-00104.warc.gz"}
|
https://python-textbook.pythonhumanities.com/02_pandas/02_04_04_time_series_data.html
|
# 4.3. Time Series Data#
## 4.3.1. What is Time Series Data#
Time series data is data that reflects either time or dates. In Pandas this type of data is known as datetime. If you are working with time series data, as we shall see, there are significant reasons to ensure that Pandas understands that the data at hand is a date or a time. It allows for easily manipulation and cleaning of inconsistent data formatting. Let us consider a simple example. Imagine we were given dates from one source as 01/02/2002 and another as 01.02.2002. Both are valid date formats, but they are structured entirely differently. Imagine now you had a third dataset that organized the data as 2 January 2002. Your task is to merge all these datasets together.
If you wanted to do that, you could write out some Python and script them into alignment, but Pandas offers the ability to do that automatically. In order to leverage that ability, however, you must tell Pandas that the data at hand is datetime data. Exactly how you do that, we will learn in this chapter.
Time series data is important for many different aspects of industry and academia. In the financial sector, time series data allows for one to understand the past performance of a stock. This is particularly useful in machine learning predictions which need to understand the past to predict accurately the future. More importantly, they need to understand the past a sequence of data. In the humanities, time series data is important for understand historical context and, as we shall see, plotting data temporally. Understanding how to work with time series data, therefore, in Pandas is absolutely essential.
In this chapter, we will be working with an early version of a dataset I helped cultivate at the Bitter Aloe Project, a digital humanities project that explores apartheid violence in South Africa during the 20th century. This dataset comes from Vol. 7 of the Truth and Reconciliation Commission’s Final Report. I am using not our final, well-cleaned version of this dataset, rather an earlier version for one key reason. It contains problematic cells and structure. This is more reflective of real-world data, which will often times come from multiple sources and need to be cleaned and structured. As such, it is good practice in this chapter to try and address some of the common problems that you will encounter with time series data.
import pandas as pd
df
ObjectId Last First Description Place Yr Homeland Province Long Lat HRV ORG
0 1 AARON Thabo Simon An ANCYL member who was shot and severely inju... Bethulie 1991.0 NaN Orange Free State 25.97552 -30.503290 shoot|injure ANC|ANCYL|Police|SAP
1 2 ABBOTT Montaigne A member of the SADF who was severely injured ... Messina 1987.0 NaN Transvaal 30.039597 -22.351308 injure SADF
2 3 ABRAHAM Nzaliseko Christopher A COSAS supporter who was kicked and beaten wi... Mdantsane 1985.0 Ciskei Cape of Good Hope 27.6708791 -32.958623 beat COSAS|Police
3 4 ABRAHAMS Achmat Fardiel Was shot and blinded in one eye by members of ... Athlone 1985.0 NaN Cape of Good Hope 18.50214 -33.967220 shoot|blind SAP
4 5 ABRAHAMS Annalene Mildred Was shot and injured by members of the SAP in ... Robertson 1990.0 NaN Cape of Good Hope 19.883611 -33.802220 shoot|injure Police|SAP
... ... ... ... ... ... ... ... ... ... ... ... ...
20829 20888 XUZA Mandla Was severely injured when he was stoned by a f... Carletonville 1991.0 NaN Transvaal 27.397673 -26.360943 injure|stone ANC
20830 20889 YAKA Mbangomuni An IFP supporter and acting induna who was sho... Mvutshini 1993.0 KwaZulu Natal 30.28172 -30.868900 shoot NaN
20831 20890 YALI Khayalethu Was shot by members of the SAP in Lingelihle, ... Cradock 1986.0 NaN Cape of Good Hope 25.619176 -32.164221 shoot SAP
20832 20891 YALO Bikiwe An IFP supporter whose house and possessions w... Port Shepstone 1994.0 NaN Natal 30.4297304 -30.752126 destroy ANC
20833 20892 YALOLO-BOOYSEN Geoffrey Yali An ANC supporter and youth activist who was to... George 1986.0 NaN Cape of Good Hope 22.459722 -33.964440 torture|detain|torture ANC|SAP
20834 rows × 12 columns
As we can see, we have a few different columns which are relatively straight forward. In this notebook, however, I want to focus on Yr, which is a column that contains a single year referenced within the description. This corresponds to the year in which the violence described occurred. Notice, however, that we have a problem. Year is being recognized as a float (a number with a decimal place), or floating number. To confirm our suspicion, let’s take a look at the data types by using the following command.
display(df.dtypes)
ObjectId int64
Last object
First object
Description object
Place object
Yr float64
Homeland object
Province object
Long object
Lat float64
HRV object
ORG object
dtype: object
Here, we can see all the different columns and their corresponding data types. Notice that Yr has float64. This confirms our suspicion. Why is this a problem? Well, if we were to try and plot the data by year (see the bar graph below), we would have floating numbers in that graph. This does not look clean. We could manually adjust these years to have no decimal place, but that requires effort on a case-by-case basis. Instead, it is best practice to convert these floats either to integers or to datetime data. Both have their advantages, but if your end goal is larger data analysis on time series data (not just plotting the years), I would opt for the latter. In order to do either, however, we must clean the data to get it into the correct format.
df['Yr'].value_counts().sort_index().plot.bar(figsize=(20,5))
<AxesSubplot:>
## 4.3.3. Cleaning the Data from Float to Int#
Let’s first try and convert our float column into an integer column. If we execute the command below which would normally achieve this task, we get the following error.
df['Yr'] = df['Yr'].astype(int)
---------------------------------------------------------------------------
IntCastingNaNError Traceback (most recent call last)
<ipython-input-4-dc1db5c67903> in <module>
----> 1 df['Yr'] = df['Yr'].astype(int)
c:\users\wma22\appdata\local\programs\python\python39\lib\site-packages\pandas\core\generic.py in astype(self, dtype, copy, errors)
5804 else:
5805 # else, only a single dtype is given
-> 5806 new_data = self._mgr.astype(dtype=dtype, copy=copy, errors=errors)
5807 return self._constructor(new_data).__finalize__(self, method="astype")
5808
c:\users\wma22\appdata\local\programs\python\python39\lib\site-packages\pandas\core\internals\managers.py in astype(self, dtype, copy, errors)
412
413 def astype(self: T, dtype, copy: bool = False, errors: str = "raise") -> T:
--> 414 return self.apply("astype", dtype=dtype, copy=copy, errors=errors)
415
416 def convert(
c:\users\wma22\appdata\local\programs\python\python39\lib\site-packages\pandas\core\internals\managers.py in apply(self, f, align_keys, ignore_failures, **kwargs)
325 applied = b.apply(f, **kwargs)
326 else:
--> 327 applied = getattr(b, f)(**kwargs)
328 except (TypeError, NotImplementedError):
329 if not ignore_failures:
c:\users\wma22\appdata\local\programs\python\python39\lib\site-packages\pandas\core\internals\blocks.py in astype(self, dtype, copy, errors)
590 values = self.values
591
--> 592 new_values = astype_array_safe(values, dtype, copy=copy, errors=errors)
593
594 new_values = maybe_coerce_values(new_values)
c:\users\wma22\appdata\local\programs\python\python39\lib\site-packages\pandas\core\dtypes\cast.py in astype_array_safe(values, dtype, copy, errors)
1307
1308 try:
-> 1309 new_values = astype_array(values, dtype, copy=copy)
1310 except (ValueError, TypeError):
1311 # e.g. astype_nansafe can fail on object-dtype of strings
c:\users\wma22\appdata\local\programs\python\python39\lib\site-packages\pandas\core\dtypes\cast.py in astype_array(values, dtype, copy)
1255
1256 else:
-> 1257 values = astype_nansafe(values, dtype, copy=copy)
1258
1259 # in pandas we don't store numpy str dtypes, so convert to object
c:\users\wma22\appdata\local\programs\python\python39\lib\site-packages\pandas\core\dtypes\cast.py in astype_nansafe(arr, dtype, copy, skipna)
1166
1167 elif np.issubdtype(arr.dtype, np.floating) and np.issubdtype(dtype, np.integer):
-> 1168 return astype_float_to_int_nansafe(arr, dtype, copy)
1169
1170 elif is_object_dtype(arr):
c:\users\wma22\appdata\local\programs\python\python39\lib\site-packages\pandas\core\dtypes\cast.py in astype_float_to_int_nansafe(values, dtype, copy)
1211 """
1212 if not np.isfinite(values).all():
-> 1213 raise IntCastingNaNError(
1214 "Cannot convert non-finite values (NA or inf) to integer"
1215 )
IntCastingNaNError: Cannot convert non-finite values (NA or inf) to integer
At the very bottom, we see why the error was returned. “IntCastingNaNError: Cannot convert non-finite values (NA or inf) to integer”. This means that somewhere in our data, there are a few blank cells in the Year column. We need to fill in these blank cells. To do that, we can use the fillna function that we met earlier in this textbook.
df = df.fillna(0)
If we try and rerun our same command as above, you will notice we have no errors.
df['Yr'] = df['Yr'].astype(int)
Now, let’s see if it worked by displaying the data types again.
display(df.dtypes)
ObjectId int64
Last object
First object
Description object
Place object
Yr int32
Homeland object
Province object
Long object
Lat float64
HRV object
ORG object
dtype: object
Notice that Yr is now int32. Success! Now that we have the data in the correct format, let’s plot it out. We can plot out the frequency of violence based on year by using value counts. This will go through the entire Yr column and count all the values identified and store them as a dictionary of frequencies.
df['Yr'].value_counts()
1993 2835
1992 2648
1990 2556
1986 2056
1994 1867
1991 1793
1985 1665
1988 1015
1989 935
1987 744
1980 438
1983 352
1976 319
1984 301
1960 280
1977 128
1981 124
1982 123
1975 111
1963 88
0 84
1962 69
1978 60
1979 53
1964 37
1961 32
1965 19
1969 14
1968 14
1974 12
1966 11
1970 10
1971 10
1967 8
1972 6
1973 5
1959 3
1998 3
1996 3
1997 2
1958 1
Name: Yr, dtype: int64
This looks great, but let’s try and plot it.
df['Yr'].value_counts().plot.bar(figsize=(20,5))
<AxesSubplot:>
What do you notice that is horribly wrong about our bar graph? If you noticed that it is not chronological, you’d be right. It would be quite odd to present our data in this format. When we are examining time series data, we need to visualize that data chronologically (usually). We can fix this, by adding sort_index().
df['Yr'].value_counts().sort_index()
0 84
1958 1
1959 3
1960 280
1961 32
1962 69
1963 88
1964 37
1965 19
1966 11
1967 8
1968 14
1969 14
1970 10
1971 10
1972 6
1973 5
1974 12
1975 111
1976 319
1977 128
1978 60
1979 53
1980 438
1981 124
1982 123
1983 352
1984 301
1985 1665
1986 2056
1987 744
1988 1015
1989 935
1990 2556
1991 1793
1992 2648
1993 2835
1994 1867
1996 3
1997 2
1998 3
Name: Yr, dtype: int64
Notice that we have now preserved the value counts, but organized them in their correct order. We can now try plotting that data.
df['Yr'].value_counts().sort_index().plot.bar(figsize=(20,5))
<AxesSubplot:>
We have a potential issue, however. That first row, 0, is throwing off our bar graph. What if I didn’t want to represent 0, or no date, in the graph. I can solve this problem a few different ways. Let’s first create a new dataframe called val_year.
val_year = df["Yr"].value_counts().sort_index()
val_year
0 84
1958 1
1959 3
1960 280
1961 32
1962 69
1963 88
1964 37
1965 19
1966 11
1967 8
1968 14
1969 14
1970 10
1971 10
1972 6
1973 5
1974 12
1975 111
1976 319
1977 128
1978 60
1979 53
1980 438
1981 124
1982 123
1983 352
1984 301
1985 1665
1986 2056
1987 744
1988 1015
1989 935
1990 2556
1991 1793
1992 2648
1993 2835
1994 1867
1996 3
1997 2
1998 3
Name: Yr, dtype: int64
With this new dataframe, I can simply start at index 1 and then graph the data. Notice that the 0 value is now gone.
val_year.iloc[1:].plot.bar(figsize=(20,5))
<AxesSubplot:>
Although we have been able to now plot our time series data chronologically, Pandas has not seen this as a datetime type. Instead, it has viewed these years solely as integers. In order to work with the years as time series data formally, we need to convert the integers into datetime format.
## 4.3.4. Convert to Time Series DateTime in Pandas#
Our goal here will be to create a new column that will store Yr as a datetime type. One might think that we could easily just convert everything to datetime. Normally the following command would work, but instead we get this error.
df['Dates'] = pd.to_datetime(df['Yr'], format='%Y')
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
c:\users\wma22\appdata\local\programs\python\python39\lib\site-packages\pandas\core\tools\datetimes.py in _to_datetime_with_format(arg, orig_arg, name, tz, fmt, exact, errors, infer_datetime_format)
508 try:
--> 509 values, tz = conversion.datetime_to_datetime64(arg)
510 dta = DatetimeArray(values, dtype=tz_to_dtype(tz))
c:\users\wma22\appdata\local\programs\python\python39\lib\site-packages\pandas\_libs\tslibs\conversion.pyx in pandas._libs.tslibs.conversion.datetime_to_datetime64()
TypeError: Unrecognized value type: <class 'int'>
During handling of the above exception, another exception occurred:
ValueError Traceback (most recent call last)
<ipython-input-14-cc2c68a810bf> in <module>
----> 1 df['Dates'] = pd.to_datetime(df['Yr'], format='%Y')
c:\users\wma22\appdata\local\programs\python\python39\lib\site-packages\pandas\core\tools\datetimes.py in to_datetime(arg, errors, dayfirst, yearfirst, utc, format, exact, unit, infer_datetime_format, origin, cache)
881 result = result.tz_localize(tz) # type: ignore[call-arg]
882 elif isinstance(arg, ABCSeries):
--> 883 cache_array = _maybe_cache(arg, format, cache, convert_listlike)
884 if not cache_array.empty:
885 result = arg.map(cache_array)
c:\users\wma22\appdata\local\programs\python\python39\lib\site-packages\pandas\core\tools\datetimes.py in _maybe_cache(arg, format, cache, convert_listlike)
193 unique_dates = unique(arg)
194 if len(unique_dates) < len(arg):
--> 195 cache_dates = convert_listlike(unique_dates, format)
196 cache_array = Series(cache_dates, index=unique_dates)
197 # GH#39882 and GH#35888 in case of None and NaT we get duplicates
c:\users\wma22\appdata\local\programs\python\python39\lib\site-packages\pandas\core\tools\datetimes.py in _convert_listlike_datetimes(arg, format, name, tz, unit, errors, infer_datetime_format, dayfirst, yearfirst, exact)
391
392 if format is not None:
--> 393 res = _to_datetime_with_format(
394 arg, orig_arg, name, tz, format, exact, errors, infer_datetime_format
395 )
c:\users\wma22\appdata\local\programs\python\python39\lib\site-packages\pandas\core\tools\datetimes.py in _to_datetime_with_format(arg, orig_arg, name, tz, fmt, exact, errors, infer_datetime_format)
511 return DatetimeIndex._simple_new(dta, name=name)
512 except (ValueError, TypeError):
--> 513 raise err
514
515
c:\users\wma22\appdata\local\programs\python\python39\lib\site-packages\pandas\core\tools\datetimes.py in _to_datetime_with_format(arg, orig_arg, name, tz, fmt, exact, errors, infer_datetime_format)
498
499 # fallback
--> 500 res = _array_strptime_with_fallback(
501 arg, name, tz, fmt, exact, errors, infer_datetime_format
502 )
c:\users\wma22\appdata\local\programs\python\python39\lib\site-packages\pandas\core\tools\datetimes.py in _array_strptime_with_fallback(arg, name, tz, fmt, exact, errors, infer_datetime_format)
434
435 try:
--> 436 result, timezones = array_strptime(arg, fmt, exact=exact, errors=errors)
437 if "%Z" in fmt or "%z" in fmt:
438 return _return_parsed_timezone_results(result, timezones, tz, name)
c:\users\wma22\appdata\local\programs\python\python39\lib\site-packages\pandas\_libs\tslibs\strptime.pyx in pandas._libs.tslibs.strptime.array_strptime()
ValueError: time data '0' does not match format '%Y' (match)
Just as the NaN cells plagued us above, so too has the 0s that we filled them with. Fortunately, we can fix this issue by passing the keyword argument errors=”coerce”.
df['Dates'] = pd.to_datetime(df['Yr'], format='%Y', errors="coerce")
display(df.dtypes)
ObjectId int64
Last object
First object
Description object
Place object
Yr int32
Homeland object
Province object
Long object
Lat float64
HRV object
ORG object
Dates datetime64[ns]
dtype: object
And like magic, we have not only created a new column, but notice that it is in datetime64[ns] format. We should also understand the keyword argument passed here, format. Format takes a formatted string that will tell Pandas how to interpret the data being passed to it. Because our integer referred to a single year, we use %Y. Let’s try and plot this data now to see how it looks.
df['Dates'].value_counts().sort_index().plot.bar(figsize=(20,5))
<AxesSubplot:>
While this data is now plotted as Pandas-structured time series data, it does not look good. Our dates are rendered in the long, full format that has both the date (in its entirety) and the time. Let’s fix this by first, extracting the relevant data. In this case, the year and the counts.
new_df = df['Dates'].value_counts().sort_index()
new_df
1958-01-01 1
1959-01-01 3
1960-01-01 280
1961-01-01 32
1962-01-01 69
1963-01-01 88
1964-01-01 37
1965-01-01 19
1966-01-01 11
1967-01-01 8
1968-01-01 14
1969-01-01 14
1970-01-01 10
1971-01-01 10
1972-01-01 6
1973-01-01 5
1974-01-01 12
1975-01-01 111
1976-01-01 319
1977-01-01 128
1978-01-01 60
1979-01-01 53
1980-01-01 438
1981-01-01 124
1982-01-01 123
1983-01-01 352
1984-01-01 301
1985-01-01 1665
1986-01-01 2056
1987-01-01 744
1988-01-01 1015
1989-01-01 935
1990-01-01 2556
1991-01-01 1793
1992-01-01 2648
1993-01-01 2835
1994-01-01 1867
1996-01-01 3
1997-01-01 2
1998-01-01 3
Name: Dates, dtype: int64
Next, we need to convert that data into a new DataFrame.
new_df = pd.DataFrame(new_df)
Dates
1958-01-01 1
1959-01-01 3
1960-01-01 280
1961-01-01 32
1962-01-01 69
Now that we have that new DataFrame created, let’s fix our column name and change Dates to ViolentActs.
new_df = new_df.rename(columns={"Dates": "ViolentActs"})
ViolentActs
1958-01-01 1
1959-01-01 3
1960-01-01 280
1961-01-01 32
1962-01-01 69
With the new DataFrame, we can also fix the index so that it is strictly the year. Because Pandas knows that the index is a datetime type, then we can use the extra method, year, to grab just the year.
new_df.index = new_df.index.year
ViolentActs
1958 1
1959 3
1960 280
1961 32
1962 69
Notice that our data is now just the year, the only piece of data in the time series data that matters to us. With that new DataFrame in the correct format, we can now plot it.
new_df.plot.bar(figsize=(20,5))
<AxesSubplot:>
And thus we have successfully plotted our datetime data after properly formatting it in Pandas. While working with time series data in Pandas as a datetime can be a bit more complex in the beginning, it allows for you to more advanced things, such as we saw above by calling the year with .year. As we will see in the next few chapters, there are other advantages as well.
|
2023-03-23 08:21:33
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2471899390220642, "perplexity": 5605.823986256788}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945030.59/warc/CC-MAIN-20230323065609-20230323095609-00767.warc.gz"}
|
https://math.stackexchange.com/questions/2717723/svd-with-singular-matrix
|
SVD with singular matrix
How does SVD work if the $A^TA$ matrix is singular? Lets say that the matrix looks like $$\begin{matrix}X & 0 \\0 & 0 \\\end{matrix}$$ Where X is an arbitrary real valued number. Is it still possible to obtain $\Sigma$ ? or will it now be in Jordan form? What about the case of repeated Eignevalues? how will this now impact $\Sigma$ along with $U$ and $V$ Matrix?
• Why do you think there is a problem when $A^\top A$ is singular? If this is the case, then some of the eigenvalues of $A^\top A$ are zero, so $\Sigma$ will have some zero diagonal entries. – angryavian Apr 1 '18 at 18:26
• $A^\top A$ is always symmetric. Thus, it is always diagonalizable. No need to bring Jordan forms to the discussion. – Rodrigo de Azevedo Jul 15 '18 at 18:52
• The Singular Value Decomposition is possible for any matrix. – Yves Daoust Jul 15 '18 at 18:57
Singular value decomposition works the same whether A$^{\mathrm T}$A is singular or not or whether it has multiple eigenvalues or not. In all cases, $\Sigma$ will be diagonal (not necessarily of the form of the corresponding matrix in the Jordan decomposition).
|
2019-10-16 19:59:45
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7044230103492737, "perplexity": 240.1879842056975}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986669546.24/warc/CC-MAIN-20191016190431-20191016213931-00551.warc.gz"}
|
https://chemistry.stackexchange.com/questions/7797/sn1-or-sn2-the-kinetic-isotope-effect
|
# SN1 or SN2, the kinetic isotope effect
I know that you can't always trust Wikipedia, so I just want to check something regarding their kinetic isotope effect page.
The first example implies that $\sideset{^{12}}\ C$ and $\sideset{^{13}}\ C$ can be used to determine whether a reaction is $S_N1$ or $S_N2$:
...In the nucleophilic substitution reaction of methyl bromide with cyanide, the kinetic isotope effect of the methyl carbon, in this case defined as $k_{12}/k_{13}$, was found to be 1.082 ± 0.008 ... The observed methyl carbon kinetic isotope effect is indicative of an $S_N2$ mechanism."
No explanation appears to be offered for this inference. I know that the reaction will not proceed via $S_N1$ because of the unstable carbocation but (how) does the KIE datum predict an $S_N2$ mechanism?
• Your quote ends in a reference to an article in Advances in Physical Chemistry. dx.doi.org/10.1016%2FS0065-3160%2806%2941004-2. I agree that this is poor practice for Wikipedia, but normal for journal articles. The author of this this portion of the page was probably more sued to a different type of writing. The article I linked is behind a paywall for me, so I cannot say more about it, but your answer is probably there. Jan 13 '14 at 10:10
The rate constant depends on the energy difference between the ground state of the reactants and the transition state. The ratio of the rate constants due to the kinetic isotope effect therefore depends on the difference in vibrational zero point energy between the initial state and the transition state. Vibrational ZPE depends on bond stiffness and reduced mass. The change in reduced mass from $\ce{^{12}C}$ to $\ce{^{13}C}$ causes the kinetic isotope effect, and the bond stiffness determines the size of the KIE. Bond stiffness is to a first approximation proportional to bond order.
In both $S_N1$ and $S_N2$, the initial state has a single bond between the methyl $\ce{C}$ and $\ce{Br}$ (as well as some $\ce{C-H}$ bonds which stay the same throughout the reaction). For $S_N2$, the transition state has partial bonds between the methyl $\ce{C}$ and both $\ce{Br}$ and the $\ce{C}$ of $\ce{CN}$. In the $S_N1$ transition state, the methyl $\ce{C}$ has only a partial bond to $\ce{Br}$.
Since the change in bond order is greater between initial state and $S_N1$ transition state than between ground state and $S_N2$ transition state, $S_N1$ will have a greater kinetic isotope effect. Measuring the KIE of an unknown reaction and comparing it to the KIEs of reactions with known mechanisms will therefore distinguish between $S_N1$ and $S_N2$.
|
2021-10-28 17:42:59
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5837520956993103, "perplexity": 541.832548176336}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323588398.42/warc/CC-MAIN-20211028162638-20211028192638-00446.warc.gz"}
|
https://akmepsy.sgu.ru/en/klyuchevye-slova/nachalnaya-shkola
|
ISSN 2304-9790 (Print)
ISSN 2541-9013 (Online)
# начальная школа
## The Level of Development of General and Shallow Movement as Factor Forming of Grafomorics Skills of Junior Schoolchildren
These inspections over of general and shallow movement are brought, grafomorics functions of first-class boys (50 persons) of one of rural schools of the Saratov area. Results are described of implementation of tasks, reflecting the level of development of general morics functions – at run, broad jumps, throwing of ball, static equilibrium; fine motor hand skills is pressure, smoothness of line, evenness, ability to keep indoors for a contour; grafomorics skills is writing of clip arts to different complication and sequence.
## Cognitive Narratology and Opportunities for Narrative Text Analysis at School
The article presents theoretical analysis of cognitive narratology core statements. The research covers correlation of methodic tools of narrative text studying at school and terms of modern narrotology. Special attention is thrown to the point of view problem as basic representation of events pattern. The research shows the prospects for studying the narrative texts included into the literary reading textbooks from the position of cognitive analysis.
## The Principle Humanization in the Content of Information Computer-Training in Elementary Schools
In article describes the principles of humanization in the development of teaching materials and information computer training, ergonomics towards learning, which acts as the paradise one of the foundations of computer-oriented content of information teaching, ensuring the formation of the children’s minds shaped reflection results individual learning activities.
На сайте журнала 30.03.2023 запланированы технические работы. В это время сайт может быть недоступен. С уважением, администрация сайта.
|
2023-03-28 12:15:19
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1774897575378418, "perplexity": 14664.021540487443}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948858.7/warc/CC-MAIN-20230328104523-20230328134523-00594.warc.gz"}
|
https://www.physicsforums.com/threads/a-simple-relativity-question.80682/
|
# A simple relativity question
1. Jun 29, 2005
### Aer
Here is the system of interest, S is considered at rest and S' is moving wrt to S as shown:
...a............v.............a
.------>------------------><------.
S................................S'
The magnitude of constant acceleration is given by a, and the magnitude of constant velocity is given by v. S' undergoes constant acceleration from the rest position in the frame S. At the end of the journey, the S' frame is once again at rest in the S frame. Will the time shown on the clock of S' be less than, greater than, or equal to the time shown on the clock of S when S' finishes it's journey. If it cannot be determined, then why not?
EDIT: (This is a clarification of the diagram as described in a post below)
S is an inertial reference frame, all accelerations/velocities of S' are measured wrt to S. S' starts with v(t=0)=0 from x=x'=0 at t=t'=0 and has a constant acceleration, a. Once S' has reached a velocity of v relative to S, it has an accleration of 0. After sometime, S' is given the acceleration, -a, as indicated by the reversed arrow. It's acceleration is stopped once it is once again in the rest frame of S, hence v(t=tf)=0.
Last edited: Jun 29, 2005
2. Jun 29, 2005
### Crosson
For the record, this is the twin paradox and has been discussed to death.
The standard answer is: Because of the non-zero acceleration a, S' is motion relative to S (and not the other way around because S' cannot be considered an inertial frame). Moving clocks run slower, so the time shown on the clock in S' is less then the time shown in S.
Personally I don't think this is correct, I think that the motion is symmetric in all respects and that S and S' will read the exact same time at the end of the motion.
3. Jun 29, 2005
### JesseM
If you think the motion is symmetric, would you deny that S' feels a constant G-force due to acceleration while S experiences weightlessness?
4. Jun 29, 2005
### Aer
Let me see if I understand your argument. First of all let me state that I am unsure as to whether you are claiming you disagree with relativity or that you are claiming that relativity will predict something other than the "standard answer". Your argument seems to be that if a system of motion between S and S' can be viewed as symmetrical in some other rest frame besides that of S or S', then this "other" frame is the only frame (let's call it the preferred frame) in which calculations of time can be made and any other frame of reference will give a false result because S and S' will not show the exact same time.
If my assessment of your argument is correct, then I would have to conclude that you believe there is a preferred frame for which to make measurements for any given two frames S and S' and that this preffered frame is not neccessary either S or S' but can only be found through symmetry.
I do not accept this.
5. Jun 29, 2005
### pervect
Staff Emeritus
I'm not 100% sure of the details of the problem (is S inertial? Why does it have an acceleration arrow by it in that case) - but there is a very easy answer.
Since S and S' are apparently not starting out at the same point in space, the answer depends on how you synchronize their clocks initially.
6. Jun 29, 2005
### JesseM
When I read the problem, I thought S and S' started out at the same point in space (with synchronized clocks, presumably) but with a large relative velocity, and S' is experiencing constant acceleration in the opposite direction so eventually he comes to rest relative to S. If that's the case, then you can't really answer the question of the difference between
their clock readings at the moment they are at rest relative to each other, because of the relativity of simultaneity--different frames will disagree about what each clock reads at the moment they come to rest relative to each other. But if Aer is just asking about what the two clocks will read in the rest frame of S, and if my interpretation of them starting out at the same point in space is right, then S' will be behind S at the moment S' comes to rest in this frame.
7. Jun 29, 2005
### Aer
I guess my diagram is not the clearest. Frame S is an inertial frame, in fact it is the rest frame for which all other velocities/accelerations are relative to. The frame S' starts out at the same location as the frame S (the origin of the x and x' axis coincide at t=t'=0). All dashed arrows in the diagram are attributed to the motion of the frame S'.
To be thorough, I will describe the diagram in words. S' starts with v(t=0)=0 from x=x'=0 at t=t'=0 and has a constant acceleration, a. Once S' has a velocity of v relative to S, it has an accleration of 0. After sometime, S' is given the acceleration, -a, as indicated by the reversed arrow. It's acceleration is stopped once it is once again in the rest frame of S, hence v(t=tf)=0.
Now since both frames are at rest wrt each other, there should not be a problem with S' sending a signal to S telling what it's time is at any time that it's relative velocity is 0 to S. Now my question is, will the time displayed by the clock of S' being less than the time displayed by the clock of S.
In short I think the answer according to relativity is "yes", but I am merely asking for the official answer from anyone who has experience with these matters.
8. Jun 29, 2005
### JesseM
In the rest frame of S, yes. In other inertial frames, maybe not. Different frames disagree about whether a given event (such as a clock striking a particular time) happens "at the same time" as another event at a different location (such as another distant clock striking another time), this is called "relativity of simultaneity".
Last edited: Jun 29, 2005
9. Jun 29, 2005
### pervect
Staff Emeritus
Using the S definition of simultaneity, in which both S and S' are at rest at the end of your acceleration period, the answer is yes, the accelerated clock will read a shorter elapsed time than the stationary clock.
However, if you use the U definition of simutaneity, where U is some other inertial frame moving with respect to S, the answer may not be the same.
If one is to compare two clocks at different spatial positions, the method of comparison must be specified in detail, different methods will give different results.
10. Jun 29, 2005
### Crosson
I guess my thoughts about the issue have to do with Mach's principle. Mach's Principle basicly says that g-force is due to acceleration relative to the distant stars (a gravitational effect).
I don't mean to venture too far outside the mainstream, but I was trying to present an alternative answer that I prefer.
11. Jun 29, 2005
### Aer
Are you referring to atmospheric effects and whatnot? I do not wish to complicate the problem in such a manner. The "experiment" should be considered to be conducted in a vacuum.
Let me specify in detail how the clock comparison should be done. After S' has stopped decelerating, it sends a signal to S with the current time according to S' encoded. When S recieves this signal, it immediately returns a reply signal with the current time according to S encoded. When S' recieves this signal, it can then compute the spatial positions of S and S' based on the delay of the return signal. This recieve and reply signaling can continue on until the controlling frame (let's say S) encodes a message to stop. Would comparing the clocks in this manner work?
12. Jun 30, 2005
### pervect
Staff Emeritus
Yes, that method of comparing clocks would work, and yield the same result as the usual method of noting the time on each clock when a flash or signal is emitted from the midpoint of S and S'.
The difficulty in synchronization arises when one introduces a moving observer, an inertial observer who has some non-zero velocity with respect to both S and S'. He has a different notion of simultaneity then the observer at S.
13. Jun 30, 2005
### Aer
I have not mentioned any desire to synchronize the clocks, that is make them display the same time according to each other. I assume when you refer to synchronization here, that you mean the comparison of each clock at the same instant and that both clocks must agree that this instant is in fact simultaneous.
Now here is the point I really wish to resolve. The clock of S' during it's motion as described in the diagram must be able to record a time when specific events happen such as the change from a constant acceleration, a, to no acceleration. I do not care necessarily that we are unable to compare clocks in each frame S and S' since that involves an issue of simultaneity. However, what I do wish to do is build up a "time profile" of the S' clock according to what it perceives the time to be and how this then matches up with S' agreeing with S what the clock S' should read at the end of it's journey.
The first issue in doing this is to look at the acceleration, a and -a. They have the same magnitude, a, and occur over the same interval (as recorded by S, I don't know what S' would say about these two intervals) but in opposite directions. Would S' percieve these two accelerations to occur over the same time interval?
Last edited: Jun 30, 2005
14. Jun 30, 2005
### Zanket
Two links you may find helpful: This one is all about the twin paradox, which your story problem is. This one has the fairly simple equations for (noninertial) acceleration in special relativity, so you can answer questions with actual calculations. Using the equations might be easier if you change your story problem so that there is no constant velocity portion. Instead, have S’ accelerate to the halfway point and then decelerate to the finish point.
Yes, this is given by symmetry as you imply.
Last edited: Jun 30, 2005
15. Jun 30, 2005
### Aer
Thank you for the links. Regarding the equations in the 2nd link, "The acceleration of the rocket must be measured at any given instant in a non-accelerating frame of reference travelling at the same instantaneous speed as the rocket". Therefore those equations do not apply to the case presented since the constant acceleration is measured in the frame S.
Is this still true when you consider that frame S' will not see itself undergoing constant acceleration?
16. Jun 30, 2005
### JesseM
Yes, you can see this without having to think about accelerated frames, just use frame S to predict how much time will elapse on the ship's clock between the beginning and end of each acceleration period. As discussed in this section of the twin paradox page, if from the point of view an inertial frame you see an accelerating clock whose velocity as a function of time is given by v(t), then to find the time elapsed between two times t0 and t1 in that frame, just integrate $$\int_{t0}^{t1} \sqrt{1 - v(t)^2 / c^2} \, dt$$. Since the amount of time that both accelerations last is the same, and the v(t) functions will be mirror opposites, the elapsed time on the ship's clock from the beginning to the end of each acceleration will be the same.
17. Jun 30, 2005
### Aer
I assume t0 and t1 are measured in frame S. In that case, let's set t0=0 so that t1 will be the time interval as seen by frame S. In order to integrate that equation, we must first determine v(t).
With constant acceleration, v(t) will be:
v(t) = (vf - v0)*t/t1 + v0
The above formula shouldn't need any explaination. Now when we plug that in and integrate, I am claiming without showing any steps (for the sake of brevity) that the time as percieved in the accelerating frame is:
t' = t1 * c/(2*(vf-v0)) * [ arcsin(vf/c) - arcsin(v0/c) + (vf-v0)*(c^2-v0^2)^(1/2)/c^2 ]
EDIT: I did in fact make a error reading the equation on my calculator, the above equation should be:
$$t' = t1\frac{c}{2(vf-v0)} \left( {sin}^{-1}\left(\frac{vf}{c}\right)-{sin}^{-1}\left(\frac{v0}{c}\right)+\frac{vf}{{c}^{2}}\sqrt{{c}^{2}-{vf}^{2}}-\frac{v0}{{c}^{2}}\sqrt{{c}^{2}-{v0}^{2}} \right)$$
If you disagree with this, then I will try to prove it in detail in another post as it is possible that I made a mistake.
Now from the diagram of the problem in question, let's assume numbers for the various values:
c=1 v0=0 vf=.9 t1=2
These numbers yield t'=1.6801
for the 2nd acceleration case (-a), the numbers follow from above as:
c=1 v0=.9 vf=0 t1=2
These numbers yield t'=1.6801
I made a mistake originally, but the results should now be correct as they predict the same time interval.
Last edited: Jun 30, 2005
18. Jun 30, 2005
### pervect
Staff Emeritus
Yes, I was using comparison and synchronization as if they were synonomous. Sorry for any confusion.
What you probably want then is the equations for the relativistic rocket
http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html
You can use the relativistic rocket equations to do this, but because they aren't written with an initial velocity, you need to be just a little bit tricky. On the outgoing trip, you use an inertial frame centered on the rocket's starting point. During the constant velocity phase, you use the Lorentz transforms. During the deceleration phase, you use an inertial frame centered on the rocket's stopping point, and reverse time.
Using this time-reversal approach (which is allowed because physical laws are symmetrical under time reversal), we can say that the proper time interval representing the acceleration phase is the same as the proper time interval of the decelleration phase, as one would intiutively guess.
There are some things you have to beware of. The equations given don't address the coordinate system of a person in the rocket. This turns out to be a trickier problem than it first appears. (For one thing, under some circumstances it doesn't even exist. Look up "Rindler horizon" sometime, I've talked about this a lot before in this forum, and there is some info on the web as well).
Effects due to the rocket not being a single point are not modelled by the relativistic rocket equations, they require further discussion and study.
As long as you are aware of this limitiation, though, the relativistic rocket equation should answer a lot of your questions.
19. Jun 30, 2005
### Aer
The rocket equation was brought forth by Zanket in which I replied: "The acceleration of the rocket must be measured at any given instant in a non-accelerating frame of reference travelling at the same instantaneous speed as the rocket". Therefore those equations do not apply to the case presented since the constant acceleration is measured in the frame S.
The text in quotes is directly from the page containing the rocket equation.
20. Jun 30, 2005
|
2016-07-25 04:26:09
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7074562311172485, "perplexity": 568.3543708085126}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257824204.27/warc/CC-MAIN-20160723071024-00098-ip-10-185-27-174.ec2.internal.warc.gz"}
|
http://www.ncatlab.org/nlab/show/multivector+field
|
# Contents
## Definition
For $X$ a smooth manifold and $T X$ its tangent bundle a multivector field on $X$ is an element of the exterior algebra bundle $\wedge^\bullet_{C^\infty(X)}(\Gamma(T X))$ of skew-symmetric tensor powers of sections of $T X$.
• In degree $0$ these are simply the smooth functions on $X$.
• In degree $1$ these are simply the tangent vector fields on $X$.
• In degree $p$ these are sometimes called the $p$-vector fields on $X$.
## Properties
### Hochschild cohomology
In suitable contexts, multivector fields on $X$ can be identified with the Hochschild cohomology $HH^\bullet(C(X), C(X))$ of the algebra of functions on $X$.
### Schouten bracket
There is a canonical bilinear pairing on multivector fields called the Schouten bracket.
### Isomorphisms with de Rham complex
Let $X$ be a smooth manifold of dimension $d$, which is equipped with an orientation exhibited by a differential form $\omega \in \Omega^d(X)$.
Then contraction with $\omega$ induces for all $0 \leq n \leq d$ an isomorphism of vector spaces
$\omega(-) : T^n_{poly}(X) \stackrel{\simeq}{\to} \Omega^{(d-n)}(X) \,.$
The transport of the de Rham differential along these isomorphism equips $T^\bullet_{poly}$ with the structure of a chain complex
$\array{ \Omega^{n}(X) &\stackrel{d_{dR}}{\to}& \Omega^{n+1}(X) \\ \downarrow^{\mathrlap{\simeq}} && \downarrow^{\mathrlap{\simeq}} \\ T^{d-n}_{poly} &\stackrel{div_\omega}{\to}& T^{d-n-1}_{poly} } \,,$
The operation $div_\omega$ is a derivation of the Schouten bracket and makes multivectorfields into a BV-algebra.
A more general discussion of this phenomenon in (CattaneoFiorenzaLongoni). Even more generally, see Poincaré duality for Hochschild cohomology.
## References
The isomorphisms between the de Rham complex and the complex of polyvector field is reviewed for instance on p. 3 of
• Thomas Willwacher, Damien Calaque Formality of cyclic cochains (arXiv:0806.4095)
and in section 2 of
and on p. 6 of
• C. Roger, Gerstenhaber and Batalin-Vilkovisky algebras (ps)
Revised on March 4, 2011 12:05:37 by Urs Schreiber (131.211.232.235)
|
2014-08-30 00:25:44
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 24, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9666446447372437, "perplexity": 628.924844335193}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500833525.81/warc/CC-MAIN-20140820021353-00051-ip-10-180-136-8.ec2.internal.warc.gz"}
|
https://www.tutorialspoint.com/maximum-product-subarray-using-two-traversals-in-cplusplus
|
# Maximum Product Subarray - Using Two Traversals in C++
C++Server Side ProgrammingProgramming
In this problem, we are given an array arr[] of integers. Our task is to create a program to find the Maximum Product Subarray - Using Two Traversals in C++.
Problem description − Here in the array, we will find the maximum product subarray using two traversals one from index 0 and another index (n-1).
Let’s take an example to understand the problem,
## Input
arr[] = {4, -2, 5, -6, 0, 8}
## Output
240
## Example
Subarray = {4, -2, 5, -6}
Maximum product = 4 * (-2) * 5 * (-6) = 240
## Solution Approach
To solve this problem using two traversals. Here, we will find the maximum product using two local maximum values for traversal from left to right i.e. from index 0 to n-1. And one for traversal from right to left i.e. from index n-1 to 0. The rest algorithm is the same as finding the maximum product subarray.
Program to illustrate the working of our solution,
## Example
Live Demo
#include<iostream>
using namespace std;
int CalcMaxProductSubArray(int arr[], int n) {
int frntMax = 1, rearMax = 1, maxVal = 1;
for (int i=0; i<n; i++) {
frntMax = frntMax*arr[i];
if (frntMax == 0)
frntMax = 1;
}
for (int i=n-1; i>=0; i--) {
rearMax = rearMax * arr[i];
if (rearMax == 0)
rearMax = 1;
}
maxVal = max(frntMax, rearMax);
return maxVal;
}
int main() {
int arr[] = {4, -2, 5, -6, 0, 8};
int n = sizeof(arr)/sizeof(arr[0]);
cout<<"Maximum product subarray is "<<CalcMaxProductSubArray(arr, n);
return 0;
}
## Output
Maximum product subarray is 240
Published on 03-Jul-2020 07:49:14
|
2021-09-22 06:11:51
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.301663875579834, "perplexity": 4473.459243267861}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057329.74/warc/CC-MAIN-20210922041825-20210922071825-00142.warc.gz"}
|
https://www.physicsforums.com/threads/capacitor-and-resistor-in-parallel.302040/
|
# Capacitor and resistor in parallel
1. Mar 23, 2009
### bp_psy
1. The problem statement, all variables and given/known data
This is not an actual homework problem but this appears in a couple of more complicated problems that deal with RC circuits and Kirchhoff's laws.The connection consists of a capacitor in series with a resistor. My question is: will the capacitor be instantly charged when the circuit is closed or will it take a certain amount of time for it to charge? Will it remain uncharged?
2. Relevant equations
https://www.physicsforums.com/attachment.php?attachmentid=18127&d=1237862861
C=Q/Vab
q=Q(1-e-t/RC)
i=Ie-t/RC
3. The attempt at a solution
From the equations and the fact once the circuit is closed the potential difference Vab will be IR=E I am lead to believe that the capacitor will be instantly charged but the instantly part does not seem right to me.I do not think it is possible for the capacitor to remain uncharged since the potential difference between a and b will surely induce a charge in the capacitor.
2. Mar 24, 2009
### tiny-tim
Hi bp_psy !
No, the potential difference Vab will be iR, where i=Ie-t/RC
Kirchhoff's rules only apply to capacitors if you "invent" a "displacement current" flowing through the capacitor (equal and opposite to i) … see from the PF Library
Displacement current:
No current ever flows through a functioning capacitor.
But while a capacitor is charging or discharging (that is, neither at zero nor maximum charge), current is flowing round the circuit joining the plates externally, and so there would be a breach of Kirchhoff's first rule (current in = current out at any point) at each plate, if only ordinary current were used, since there is ordinary current in the circuit on one side of the plate, but not in the dielectric on the other side.
Accordingly, a displacement current is deemed to flow through the capacitor, restoring the validity of Kirchhoff's first rule:
$$I\ =\ C\frac{dV}{dt}$$
and this linear displacement current $I$ (which might better be called the flux current or free flux current) is the rate of change of the flux (field strength times area) of the electric displacement field $D$:
$$I\ =\ A\,\widehat{\bold{n}}\cdot\frac{\partial\bold{D}}{\partial t}\ =\ A\,\frac{\partial D}{\partial t}\ =\ C\frac{dV}{dt}$$
which appears in the Ampére-Maxwell law (one of Maxwell's equations in the free version):
$$\nabla\,\times\,\bold{H}\ =\ \bold{J}_f\ +\ \frac{\partial\bold{D}}{\partial t}$$
Note that the displacement alluded to in the displacement current across a capacitor is of free charge, and is non-local, since it alludes to charge being displaced from one plate to the other, which is a substantial distance compared with the local displacement of bound charge in, for example, the presence of a polarisation field.
Inverse exponential rate of charging:
A capacitor does not charge or discharge instantly.
When a steady voltage $V_1$ is first applied, through a circuit of resistance $R$, to a capacitor across which there is already a voltage $V_0$, both the charging current $I$ in the circuit and the voltage difference $V_1\,-\,V$ change exponentially, with a parameter $-1/CR$:
$$I(t) = \frac{V_1\,-\,V_0}{R}\,e^{-\frac{1}{CR}\,t}$$
$$V_1\ -\ V(t) = (V_1\,-\,V_0)\,e^{-\frac{1}{CR}\,t}$$
So the current becomes effectively zero, and the voltage across the capacitor becomes effectively $V_1$, after a time proportional to $CR$.
|
2017-08-23 03:47:28
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6432760953903198, "perplexity": 496.6572087313975}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886117519.82/warc/CC-MAIN-20170823020201-20170823040201-00474.warc.gz"}
|
https://www.appliedforecasting.com/marginal-distributions-of-random-vectors-generated-by-affine-transformations-of-independent-two-piece-normal-variables/
|
# Marginal Distributions of Random Vectors Generated by Affine Transformations of Independent Two-Piece Normal Variables
Marginal probability density and cumulative distribution functions are presented for multidimensional variables defined by non-singular affine transformations of vectors of independent two-piece normal variables, the most important subclass of Ferreira and Steel’s general multivariate skewed di… stributions. The marginal functions are obtained by first expressing the joint density as a mixture of Arellano-Valle and Azzalini’s unified skew-normal densities and then using the property of closure under marginalization of the latter class.
|
2018-06-22 16:51:45
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8157302737236023, "perplexity": 889.0557856282579}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267864740.48/warc/CC-MAIN-20180622162604-20180622182604-00614.warc.gz"}
|
https://math.stackexchange.com/questions/1864979/probability-of-two-statistically-independent-uniformly-distributed-variables-oc
|
# Probability of two statistically independent, uniformly distributed variables occurring within time frame of each other?
Say two events will occur independently of each other, only once each. The time of each event occurring is uniformly distributed from 0 to 10 seconds. What is the probability that the events will occur within 2seconds of each other?
I have found the joint pdf but I am unsure of how to set up an integral for this...
• What is your joint pdf ? Please show it ! – callculus Jul 20 '16 at 3:30
Outline: Draw the $10\times 10$ square with corners $(0,0)$, $(10,0)$, $(10,10)$, $(0,10)$. Draw the two lines $y=x+2$ and $y=x-2$.
We want the probability of falling in the part $K$ of the square that is between these two lines.
This is the area of $K$ divided by $100$. It is easier to find first the area of the part of the square that is not in $K$.
If you really want to evaluate an integral, use symmetry to observe that if $X,Y$ are our random variables, then the required probability is twice the probability that $X\le Y\le X+2$. Then use the picture to find the bounds of integration.
The time of each event occurring is uniformly distributed from 0 to 10 seconds. What is the probability that the events will occur within 2seconds of each other?
The probability will be the integral of the pdf over where the condition occurs within the support.
That is you want $0\leq x\leq 10$ and $0\leq y\leq 10$ and also $(x-2)\leq y\leq (x+2)$ .
$$\mathsf P(\lvert X-Y\rvert<2) = \int_0^{10} \int_{\max(0,x-2)}^{\min(10,x+2)}f_{X,Y}(x,y)\operatorname d y\operatorname d x$$
This can be partitioned into three integrals, but it is easier to work with the complement.
\begin{align}\mathsf P(\lvert X-Y\rvert<2) =&~ 1- \int_2^{10} \int_0^{x-2}f_{X,Y}(x,y)\operatorname d y\operatorname d x - \int_0^8\int_{x+2}^{10}f_{X,Y}(x,y)\operatorname dy\operatorname d x\\[2ex] =&~ 1- \int_2^{10} \int_0^{x-2}f_{X,Y}(x,y)\operatorname d y\operatorname d x - \int_2^{10} \int_0^{y-2}f_{X,Y}(x,y)\operatorname d x\operatorname d y\end{align}
I have found the joint pdf but I am unsure of how to set up an integral for this...
There you go then. Just substitute in your pdf, $f_{X,Y}(\cdot, \cdot)$
Or you can do it graphically if you wish; you're cutting a square with two line parallel to the diagonal leaving two triangles and an irregular hexagon. The region of integration is the hexagon; and the complement region is the two triangles.
|
2019-08-22 18:03:52
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.920888364315033, "perplexity": 226.19192682576787}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027317339.12/warc/CC-MAIN-20190822172901-20190822194901-00389.warc.gz"}
|
https://jpmccarthymaths.com/page/2/
|
## Assessment 2
Will be held on the Friday of Week 11 (24 November), at 09:00 in B214 (the usual lecture venue). A sample has been emailed and I also have hard copies with me. I strongly advise you that attending the tutorial alone may not be sufficient preparation for this test so you may have to devote extra time outside classes to study.
## Week 10
In Week 10 explored Network Theory, or rather Graph Theory, in more depth. We looked at digraphs, connectedness, valency, walks, and trees.
## Week 11
We will finish our study of Graph Theory by looking at Eulerian graphs, Fleury’s Algorithm, Hamiltonian graphs, and Dirac’s Theorem. We will then begin the last chapter on recursion. We have our test on Friday
## Week 12
In Week 12 we will finish our study of recursion and perhaps do a little revision.
## Week 13
In Week 13, perhaps we will have five tutorials (normal rooms and times) of which you are invited to up to four (your own tutorial slot plus the up to three of the lecture slots).
## Study
Some students need to do extra work outside tutorials. Please feel free to ask me questions about the exercises via email or even better on this webpage.
## Student Resources
Anyone who is missing notes is to email me.
Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, etc. There are some excellent notes on Blackboard for MATH6055.
## Geometric Series
Let $a,r\in\mathbb{R}$ be constants. Let $\{a_n\}$ be a sequence of real numbers with the following recursive definition:
$a_n=\begin{cases}a & \text{ if }n=1\\ r\cdot a_{n-1}&\text{ if }n>1\end{cases}$.
Therefore the sequence is given by:
$a,ar,ar^2,ar^3,ar^4,\dots$
Such a sequence is called a geometric sequence with common ratio $r$.
When we add up the terms a sequence we have a geometric sum:
$S_n=a+ar+ar^2+ar^3+\cdots ar^{n-1}$.
Here $S_n$ is the sum of the first $n$ terms.
We can find a formula for $S_n$ using the following ‘trick’:
$r\cdot S_n=ar+ar^2+ar^3+\cdots ar^n$
$\Rightarrow a+r\cdot S_n-ar^n=S_n$
$\Rightarrow S_n(r-1)=a(r^n-1)$
$\displaystyle \Rightarrow S_n=\frac{a(r^n-1)}{r-1}$.
### Exercises
Assuming that $|r|<1$, find a formula for the geometric series
$\displaystyle S_{\infty}=\lim_{n\rightarrow \infty}S_n$.
## Binary Numbers
### Exercises
• Write the following as fractions:
$0.1_2,\,0.11_2,\,0.101_2$.
• Use infinite geometric series to show that:
• $0.111\dots_2=1$
• $0.0111\dots_2=\frac12$
• $0.101010\dots_2=\frac23$
## Doubling Mapping
The doubling mapping $D:[0,1)\rightarrow [0,1)$ is given by:
$\displaystyle D(x)=\begin{cases}2x & \text{ if }x<1/2 \\ 2x-1 & \text{ of }x\geq 1/2\end{cases}$.
### Exercises
• Find the first six iterates of the point $x_0=\frac17$ under $D$.
• Find the first four iterates of the point
$x_0=\frac{1}{2}+\frac{1}{2^2}+\frac{0}{2^3}+\frac{1}{2^4}=0.1101_2$.
• Where $x$ has the binary representation
$x = 0.a_1a_2a_3a_4a_5a_6a_7a_8\dots$ ,
write down expressions for $D(x)$ and $D^5(x)$.
• Hence find points $y, z \in [0, 1]$ such that $y$ and $z$ agree to 5 binary digits but $D^N(y)$ and $D^N(z)$ differ in the first binary digit for some $N \in \mathbb{N}$.
• Describe the period-5 points of $D$.
• Let $w \in [0, 1]$ have a binary representation beginning $w = 0.01001\dots$ . Find a period-5 point $\gamma$ of $D$ such that $w$ and $\gamma$ agree to five binary digits.
• Find a $\delta \in [0, 1]$ such that there are iterates of $\delta$, $D^{n_1}(\delta),D^{n_2}(\delta),D^{n_3}(\delta)$, with $n_1, n_2, n_3 \in \mathbb{N}$, that agree with 0.111 , 0.101, and 0.010, to three binary
digits.
## Sensitivity to Initial Conditions
### Exercise
Let $f(x)=4x\cdot (1-x)$. Where $[0,1]$ is the set of states, and $f:[0,1]\rightarrow [0,1]$ the iterator function, by looking at the first seven iterates of $x_0=0.8$ and $y_0=0.81$, show that this dynamical system displays sensitivity to initial conditions [HINT:4*ANS*(1-ANS)]
## Test 2
Thursday 23 November at 09:00 in the usual lecture venue. You will be given a copy of these tables. Based on Chapter 3, samples at the back of Chapter 3 and also here (Q. 4 has a typo — it should be $e^{-x}\sin(y)$). I strongly advise you that attending tutorials alone will not be sufficient preparation for this test and you will have to devote extra time outside classes to study aka do exercises.
## Week 10
We started Chapter 4 by looking at integration by parts. We started looking at completing the square.
## Week 11
We will look at completing the square and work.
## Dynamical Systems
A dynamical system is a set of states $S$ together with an iterator function $f:S\rightarrow S$ which is used to determine the next state of a system in terms of the previous state. For example, if $x_0\in S$ is the initial state, the subsequent states are given by:
$x_1=f(x_0)$,
$x_2=f(x_1)=f(f(x_0))=(f\circ f)(x_0)=:f^2(x_0)$
$x_3=f(x_2)=f(f^2(x_0))=f^3(x_0)$,
and in general, the next state is got by applying the iterator function:
$x_{i}=f(x_{i-1})=f^i(x_0)$.
The sequence of states
$\{x_0,x_1,x_2,\dots\}$
is known as the orbit of $x_0$ and the $x_i$ are known as the iterates.
Such dynamical systems are completely deterministic: if you know the state at any time you know it at all subsequent times. Also, if a state is repeated, for example:
$\text{orb}(x_0)=\{x_0,x_1,x_2,x_3,x_4=x_2,x_5\dots,\}$
then the orbit is destined to repeated forever because
$x_5=f(x_4)=f(x_2)=x_3$,
$x_6=f(x_5)=f(x_3)=x_4=x_2$, etc:
$\Rightarrow \text{orb}(x_0)=\{x_0,x_1,x_2,x_3,x_2,x_3,x_2,\dots\}$
### Example: Savings
Suppose you save in a bank, where monthly you receive $0.1\%=0.001$ interest and you throw in $50$ per month, starting on the day you open the account.
This can be modeled as a dynamical system.
Let $S=\mathbb{R}$ be the set of euro amounts. The initial amount of savings is $x_0=50$. After one month you get interest on this: $0.001\times50$, you still have your original $50$ and you are depositing a further €50, so the state of your savings, after one month, is given by:
$x_1=50+0.001\times 50+50=(1+0.001)50+50$.
Now, in the second month, there is interest on all this:
interest in second month $0.001\times((1+0.001)50+50)=0.001x_1$,
we also have the $x_1=(1+0.001)50+50$ from the previous month and we are throwing in an extra €50 so now the state of your savings, after two months, is:
$x_2=x_1+0.001x_1+50=(1+0.001)x_1+50$,
and it shouldn’t be too difficult to see that how you get from $x_i\longrightarrow x_{i+1}$ is by applying the function:
$f(x)=(1+0.001)x+50$.
#### Exercise
Use geometric series to find a formula for $x_n$.
## Weather
If quantum effects are neglected, then weather is a deterministic system. This means that if we know the exact state of the weather at a certain instant (we can even think of the state of the universe – variations in the sun affecting the weather, etc), then we can calculate the state of the weather at all subsequent times.
This means that if we know everything about the state of the weather today at 12 noon, then we know what the weather will be at 12 noon tomorrow…
The following question gave a little grief:
Two smooth spheres of masses $2m$ and $3m$ respectively lie on a smooth horizontal table.
The spheres are projected towards each other with speeds $4u$ and $u$ respectively.
i. Find the speed of each sphere after the collision in terms of $e$, the coefficient of restitution
ii. Show that the spheres will move in opposite directions after the collision if $e>\frac13$.
My contention is that the question erred in not specifying that the answers to part i. were to be in terms of $e$ and $u$.
Solution:
i. The following is the situation:
Let $v_1$ and $v_2$ be the velocities of the smaller respectively larger sphere after collision. Note that the initial velocity of the larger sphere is minus $u$.
Using conservation of momentum,
$m_1u_1+m_2u_2=m_1v_1+m_2v_2$
$\Rightarrow 2m(4u)+3m(-u)=2mv_1+3mv_2$
$\Rightarrow 5u=2v_1+3v_2$.
Using:
$\displaystyle e(u_1-u_2)=-(v_1-v_2)\Rightarrow \frac{v_1-v_2}{u_1-u_2}=-e$,
Therefore,
$\displaystyle \frac{v_1-v_2}{4u-(-u)}=-e$
$\Rightarrow v_1-v_2=-5ue\Rightarrow v_1=v_2-5ue$,
and so
$5u=2(v_2-5ue)+3v_2\Rightarrow 5u=2v_2-10ue+3v_2$,
$\Rightarrow 5v_2=5u+10ue\Rightarrow v_2=u(1+2e)$.
$\Rightarrow v_1=u(1+2e)-5ue=u+2ue-5ue=u(1-3e)$.
ii. $v_2>0$. If $e>\frac13\Rightarrow 3e>1\Rightarrow 1-3e<0$ and so
$v_1=u(1-3e)<0$;
that is the particles move in opposite directions.
## Week 9
In Week 9, will looked at examples of functions, including lines, quadratic functions, polynomial functions, exponential functions, the natural logarithm function, the floor function, and the ceiling function. We began the chapter on Network Theory by looking at the Bridges of Konigsberg Problem.
## Week 10
In Week 10 we will explore Network Theory, or rather Graph Theory, in more depth. We will look at digraphs, connectedness, valency, walks, and trees.
## Assessment 2
Will be held on the Friday of Week 11 (24 November), at 09:00 in B214 (the usual lecture venue). Expect a sample test shortly.
## Study
Some students need to do extra work outside tutorials. Please feel free to ask me questions about the exercises via email or even better on this webpage.
## Student Resources
Anyone who is missing notes is to email me.
Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, etc. There are some excellent notes on Blackboard for MATH6055.
## Distances between Probability Measures
Let $G$ be a finite quantum group and $M_p(G)$ be the set of states on the $\mathrm{C}^\ast$-algebra $F(G)$.
The algebra $F(G)$ has an invariant state $\int_G\in\mathbb{C}G=F(G)^\ast$, the dual space of $F(G)$.
Define a (bijective) map $\mathcal{F}:F(G)\rightarrow \mathbb{C}G$, by
$\displaystyle \mathcal{F}(a)b=\int_G ba$,
for $a,b\in F(G)$.
Then, where $\|\cdot\|_1^{F(G)}=\int_G|\cdot|$ and $\|\cdot\|_\infty^{F(G)}=\|\cdot\|_{\text{op}}$, define the total variation distance between states $\nu,\mu\in M_p(G)$ by
$\displaystyle \|\nu-\mu\|=\frac12 \|\mathcal{F}^{-1}(\nu-\mu)\|_1^{F(G)}$.
(Quantum Total Variation Distance (QTVD))
Standard non-commutative $\mathcal{L}^p$ machinary shows that:
$\displaystyle \|\nu-\mu\|=\sup_{\phi\in F(G):\|\phi\|_\infty^{F(G)}\leq 1}\frac12|\nu(\phi)-\mu(\phi)|$.
(supremum presentation)
In the classical case, using the test function $\phi=2\mathbf{1}_S-\mathbf{1}_G$, where $S=\{\nu\geq \mu\}$, we have the probabilists’ preferred definition of total variation distance:
$\displaystyle \|\nu-\mu\|_{\text{TV}}=\sup_{S\subset G}|\nu(\mathbf{1}_S)-\mu(\mathbf{1}_S)|=\sup_{S\subset G}|\nu(S)-\mu(S)|$.
In the classical case the set of indicator functions on the subsets of the group exhaust the set of projections in $F(G)$, and therefore the classical total variation distance is equal to:
$\displaystyle \|\nu-\mu\|_P=\sup_{p\text{ a projection}}|\nu(p)-\mu(p)|$.
(Projection Distance)
In all cases the quantum total variation distance and the supremum presentation are equal. In the classical case they are equal also to the projection distance. Therefore, in the classical case, we are free to define the total variation distance by the projection distance.
## Quantum Projection Distance $\neq$ Quantum Variation Distance?
Perhaps, however, on truly quantum finite groups the projection distance could differ from the QTVD. In particular, a pair of states on a $M_n(\mathbb{C})$ factor of $F(G)$ might be different in QTVD vs in projection distance (this cannot occur in the classical case as all the factors are one dimensional).
## Test 2
Tuesday 21 November Thursday 23 November at 09:00 in the usual lecture venue. You will be given a copy of these tables. Based on Chapter 3, sample at back of Chapter 3. I strongly advise you that attending tutorials alone will not be sufficient preparation for this test and you will have to devote extra time outside classes to study aka do exercises.
## Week 9
We finished looking at Partial Differentiation and then saw how it can be used in error analysis.
## Week 10
We will start Chapter 4 by looking at integration by parts. We might look at completing the square and work.
## Assessment 2
Deadline 16:00, Monday 20 November, Week 11. Note that when you open MATH7019A2 – Student Data, you should see a list of numbers that you are supposed to use in the questions. All of the $w_i,\,L_i,\,a,\,b$ are to be taken as these constants. It is only $E$ and $I$ that are to be kept as ‘free variables’.
## Week 9
We finished looking at the normal distribution and then looked at Sampling Theory.
## Week 10
We will look at Hypothesis Testing and begin Chapter 4 with a Revision of Differentiation.
## Week 8
In Week 8, we lost another lecture with the bank holiday, but we had an opportunity to better understand the function concepts after looking at arrow diagrams and the graph of a function.
## Week 9
In Week 9, we will look at examples of functions, including lines, quadratic functions, polynomial functions, exponential functions, the natural logarithm function, the floor function, and the ceiling function.
## Test 1
Results have been emailed to you. Solutions and comments here.
## Assessment 2
Will be held in Week 11. Proper notice in Week 9, and a sample in Week 10.
## Study
Some students need to do extra work outside tutorials. Please feel free to ask me questions about the exercises via email or even better on this webpage.
## Student Resources
Anyone who is missing notes is to email me.
Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, etc. There are some excellent notes on Blackboard for MATH6055.
J.P. McCarthy on MATH6040: Winter 2017, Week… Student on MATH6040: Winter 2017, Week… J.P. McCarthy on MATH6040: Winter 2017, Week… Student on MATH6040: Winter 2017, Week… MATH6028: Chaos Theo… on MATH6028: Chaos Theory (Part…
|
2017-12-17 11:45:52
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 128, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5600747466087341, "perplexity": 864.6500734188959}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948595858.79/warc/CC-MAIN-20171217113308-20171217135308-00495.warc.gz"}
|
https://www.physicsforums.com/tags/gravitational-acceleration/
|
# gravitational acceleration
1. ### I Gravitational potential gradient in accelerated reference frames?
Hi, Could you please help me to clarify the following problem? In the gravitational field of a mass, the force on a body in steady state comes from the gradient of the gravitational potential - or the gradient of speed of time. But what about accelerated reference frames? I assume that there is...
2. ### Solving symbolically
1. Homework Statement You are climbing in the High Sierra when you suddenly find yourself at the edge of a fogshrouded cliff. To find the height of this cliff, you drop a rock from the top; a time T later you hear the sound of the rock hitting the ground at the foot of the cliff. a. If you...
3. ### How to find the acceleration due to gravity inside a planet?
Consider a spherical planet of uniform density ρ. The distance from the planet's center to its surface (i.e., the planet's radius) is R. An object is located a distance R from the center of the planet, where R < Rp. (The object is located inside of the planet.) 1) Find an expression for the...
4. ### I Gravitational acceleration based on density
Im trying to create or find some way to calculate gravitational acceleration on planet surface based solely on density and mass of planet. This should be based on average density and it can be also some approximation, without calculating minor effects like rotation, planet bulge, relativity and...
5. ### I Gravitational time dilation - acceleration vs potential
What I know gravitational time dilation (based on GRT) is dependent on gravity potential and not on gravitational acceleration. That would mean, that for example in center of Earth is the gravitational acceleration zero, but the gravitational potential is bigger than on the surface of Earth...
6. ### How do write the force diagram for the following situation?
1. Homework Statement We have a crate sitting on a scale that is on the surface of the Earth. We want to come up with the value of the acceleration due to gravity, $g$, when we take into consideration the rotation of the Earth. 2. Homework Equations In the book, here's how they go...
7. ### I The rising railway coach
You are launched upward inside a railway coach in a horizontal position with respect to the surface of Earth, as shown in the figure. After the launch, but while the coach is still rising, you release two ball bearings at opposite ends of the train and at rest with respect to the train. a)...
8. J
### B Why is a newton of force .225 pounds?
General physics question regarding force and gravity. Hypothetically speaking if every piece of mass weighed double. If something that weighed 1 Pound is now 2 pounds. If you weigh 100 pounds now you weigh 200 pounds . Assume gravity is still 9.8ms/sq. and you never experience gravity any...
9. C
### Projectile motion -- how long does it take the projectile to reach maximum height?
1. Homework Statement A projectile was launched at 70 m/s at an angle of 45 degrees above the horizontal, how long does it take the projectile to reach maximum height? 2. Homework Equations Vyo= initial vertical velocity Vo=initial velocity 3. The Attempt at a Solution I tried using...
10. ### I Fundamental laws time reversal
It is often told that fundamental laws are insensitive to +t/-t change. Let's try this one: a little mass m1 object is accelerating towards a big object M2, in -x direction in space and +t in time, due to gravity or following space-time free fall line (along a geodetic). Now, revert the video...
11. ### I Confusion about gravitational acceleration
I understand that gravitational acceleration is independent of mass. However, I've seen a common mathematical description of this that I can't help but find circular. I suspect that there's an error in my thinking that I'm hoping someone can point out for me. It goes like this; $F=mg$ but we...
12. ### Height differences from relative uncertainty of gravimeters
1. Homework Statement The best relative gravimeters have a relative uncertainty of 10-12, that corresponds to a height difference of 3 µm. 2. Homework Equations g∝(1/r2) The local gravitational acceleration g outside the Earth is proportional to 1/r2, which means (Δg)/g = -2 (Δr)/r. With...
13. ### Equation for plasma wave/light speed travel?
Hi guys I'm finishing up some promo art for my original comic book. You're seein it here first. But its missing something- an appropriate equation. I would like to integrate a math equation into the art. Im attempting to depict FTL travel, using qualities similar to an LWFA, the plasma being...
14. ### Change in Earth's gravity & rotational KE due to changes in Radius
1. Homework Statement Let g be the acceleration due to gravity at the Earth's surface and K be the rotational kinetic energy of the Earth. Suppose the Earth's radius decreases by 2%. Keeping all other quantities constant, (a) g increases by 2% and K increases by 2% (b) g increases by 4% and K...
15. ### I What would happen in following condition?
If a conduction band is placed in a current loop with gravitational waves acting upon it in parallel direction. It is also assumed that all the basic conditions relating to frequency are met for parallel propagation case as per gravitational wave damping papers below. Since nuclear precession...
16. ### I What would be the consequences of following situation?
As per the above figure coil A is situated inside a magnetic field caused by small accelerating objects B which in turn are causing the larmor frequency in coil Object C or train of objects C have quadruple movement and are rotating around the coil A giving out gravitational waves. Objects C...
17. ### I Particle accleration under gravitational waves?
I have been through following papers for research: 1)http://cdsads.u-strasbg.fr/cgi-bin/nph-iarticle_query?1990ApJ...362..584M&data_type=PDF_HIGH&whole_paper=YES&type=PRINTER&filetype=.pdf 2)https://arxiv.org/pdf/gr-qc/9905054.pdf Conclusion of the second paper given above...
18. ### Finding the speed of the comet
1. Homework Statement Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 2.2×10^4 m/s when at a distance of 2.6×10^11 m from the center of the sun, what is its speed when at a distance of 4.2×10^10 m 2. Homework Equations L = rp = r(mv) 3. The...
19. ### Finding the astronaut's weight on a planet's surface
1. Homework Statement A landing craft with mass 1.22×10^4 kg is in a circular orbit a distance 5.50×10^5 m above the surface of a planet. The period of the orbit is 5100 s . The astronauts in the lander measure the diameter of the planet to be 9.50×10^6 m . The lander sets down at the north...
20. ### B How do I calculate electron acceleration by gravitational waves
If the amplitude of gravitational waves, frequency of gravitational waves and the vector potential of magnetic field in surrounding of such waves are known then what would be the easiest way to calculate resultant acceleration of electrons? My above question is based on the various researches...
21. C
### Can something dropped break the sound barrier?
Sooo i was wondering... Can something that has been dropped from a higj enough hight eventually break the sound barrier. Well, since we know that the gravitational acceleration is approximately 10 m/s^2 and that the speed of sound is approximately 330 m/s i pluged it into the acceleration...
22. ### Newton's first law, forces, charges
1. Homework Statement Two uniform masses of .260kg are fixed at points A and B. Find the magnitude and direction of the initial acceleration of a uniform sphere with mass 0.01kg released from rest at point P that is acted on the gravitational forces of attraction from spheres A and B. The...
23. ### College physics problem -- Find gravitational acceleration
1. Homework Statement Here's the question: As space colonization expands, it's important to build new stations from local materials instead of bringing everything from Earth. Your latest task (besides asking for a raise) is to check the long-term stability of a proposed configuration of...
24. ### Gravitational acceleration comparison
1. Homework Statement Suppose an object of length “l” is located a distance “r” from a gravitating object of mass “M.” From physics you will learn that the gravitational acceleration is GM/r^2. Derive the difference in gravitational acceleration between distance “r” and distance “r+l” from the...
25. ### Calculating sand penetration of ice solid from height X
G’day physics forum This is a purely hypothetical question and my knowledge of physics is rather limited so I’ve no idea if answering it is even possible however here goes: Imagine a 30cm long timber stick approximately 3mm in diameter that has been placed vertically in an inverted pyramid or...
|
2019-12-10 16:23:40
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6822611093521118, "perplexity": 895.5337936131394}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540528457.66/warc/CC-MAIN-20191210152154-20191210180154-00523.warc.gz"}
|
http://txoi.fondazionemercantini.it/a-ball-is-thrown-at-an-angle-of-45-degrees-above-the-horizontal.html
|
A Ball Is Thrown At An Angle Of 45 Degrees Above The Horizontal
The initial and final potential energies are the same for each ball, because they all start from the same height and end at ground level. In general, a shooter wants to aim for a shot with a 45-degree angle as it enters the cylinder , eleven inches deep into the basket , just two. ? How long does it take the to reach the highest point in its trajectory? The answer is 2. At the very top of the ball's path, its velocity is _____. Set the velocity of the cannon ball in Fig. 4-37, a ball is thrown leftward from the left edge of the roof, at height h above the ground. He also splits the safeties with the throw, resulting in a great ball placed in the middle of a triangle of defenders. For the ball that is thrown 40 degrees below the horizontal, Draw a diagonal line at an angle 40 degrees below the horizontal to represent the ball's velocity. Use the symmetry of the parabolic trajectory to double the horizontal (x) distance at the position of maximum height (y). If there is a height difference, things get a little more complicated. Maybe it’s easier to remember the upper scale as the angle-angle-on-wood scale and the bottom scale as the angle-the saw blade-travels scale. Index Trajectory. To find the Horizontal Sundial shadow hour angles for your locality you need to know your latitude. The ball spin is either 3000, 4000 or 5000 rpm, as labeled, and the spin axis is tilted at either 20° or 30°, as labelled. 0 rad/s and the ball is 1. A football is kicked into the air at an angle of 45 degrees with the horizontal. However, if you throw it at a 45 degree angle, the horizontal velocity becomes $\sqrt{60^2/2}$ mph, or about 42. Assume that a kicked ball in football is a projectile. It is moving at a constant velocity horizontally during the whole time we just figured out, so let’s use Figure 2 that was drawn above to get the horizontal component of the velocity. Imagine that at the top of the trajectory, you drop a ball. Your equations are correct, but not applied properly. Aug 08,2020 - A ball is thrown with a velocity of 7m/sec at an angle of 45 degree with horizontal it just clears two vertical poles of height 90cm each. 00 m/s at an angle of 40. 6 0 below the horizontal ( -0 from x axis) 8. (a) Calculate the time it takes the tennis ball to reach the spectator. C) the same. The sine function reaches its largest output value, 1, with an input angle of 90 degrees, so we can see that for the longest-range punts 2θ = 90 degrees and, therefore, θ = 45 degrees. This is known as its elevation. What is the magnitude of the ball’s initial velocity? A) smaller B) greater C) the same 2. It is released 0. The path of the baseball is given by the function f(x) = -0. Therefore, 1 rad is the angle subtend at the center of a circle by an arc equal in length to the radius. If the ball takes 3 seconds to rise to the peak of its trajectory, then it will take 6 seconds to fall from the peak of its trajectory to the ground. 0° above the horizontal strikes a building 18. 4 seconds after the ball is thrown. A ball is tossed upward and returns to its original position after 19. Ignore air resistance. At what point during its motion, does the ball have. If there is a height difference, things get a little more complicated. We apply either back spin (spin angle of 0 degrees) or top spin (spin angle of 180 degrees). 8 ms^2 : To throw a(n) 77. The top edge is 5m above the throwing point. Also, if I throw the ball at 0 degrees, then cos(0) = 1, but sin(0) = 0. At what speed was it thrown? 4. 0 meters per second. a cricket ball is thrown at a speed of 30 m s in direction making an angle of 30 degree with the horizontal calculate maximum height range of the ball - Physics - TopperLearning. Ball C has an initial velocity directed 40 degrees above the horizontal. 20 Snowballs are thrown with a speed of 13 m/s from a roof 7. 06 kg is served. v = d /t d = v t = 15m/s (2. Calculate the maximum height above the release point the ball will achieve with a launch angle of 45. The horizontal component of the baseball’s initial velocity is 12. A golf ball is hit at an angle of 45° above the horizontal. The play above shows the. However, if you throw it at a 45 degree angle, the horizontal velocity becomes $\sqrt{60^2/2}$ mph, or about 42. 5 m (120 ft). 13 (B) u = 40 V13 do mis 40 (C)u-40, 4+ mis (D) u = 40 m/s v 13 13 (4+13). The time the ball is in the air is given by (3). 0 m away at a point 8. A second ball is thrown at an angle of 30 degrees with the horizontal. 0° above the horizontal. [Neglect air resistance. 13 (B) u = 40 V13 do mis 40 (C)u-40, 4+ mis (D) u = 40 m/s v 13 13 (4+13). a) What should be the initial velocity of the ball so that it hits a target that is 30 meters away at a height of 1. downward 10 m/s squared. It strikes the ground with a velocity of 54 m/s at an angle of 22 degrees below the horizontal. 00\ m {/eq} above the ground with a speed of {eq}20. This means that an 18-degree bounce with a narrow “C” grind will raise the leading edge a little bit. 75 s in the air. 40 degrees c. A ball is thrown upward at an angle of 30° with the horizontal and lands on the top edge of a building that is 20m away. Maximum height in projectile is given by, h = u²sin²θ/2g. The initial speed of the ball in metre/second is (take g=10 m/s2): (A) U - 40 114+3) V 13. A Ball Is Thrown At An Angle Of 45 Degrees Above The Horizontal. The angle between the water level and the submarine is 31. For example, halfway between 12 and 1 ( at 2 1/2 minutes) is 15° (about a 3/4-ball hit), and halfway between 1 and 2 is 45° (about a 1/4-ball hit). What was the magnitude of the energy lost due to friction? a. 8588 kg*m/s in the +x direction. The Radian. The point A is 12 m above the point O on the ground. Which of the following best describes the acceleration of the ball from the instant after it leaves the thrower's hand until the time it hits the ground?. ? How long does it take the to reach the highest point in its trajectory? The answer is 2. Only three of the throwers have an angle of release less than 45º, throw 2 which released the javelin 2. 0 degrees NORTH of WEST. The angle that maximizes sin. A ball is thrown at an angle of 40 degrees above the horizontal. (a) At what angle. angles can be measured in radians as well as degrees. 0 meters per second. 20 m/s (4) 18. If the acceleration vector of an object is directed exactly opposite to the velocity vector. A tennis player wins a match at Arthur Ashe stadium and hits a ball into the stands at 30 m/s and at an angle $45\text{°}$ above the horizontal (). Determine the total time that the stone is in the air. We know that if the velocity changes with time then the ball on the string is also accelerating. Maximum height in projectile is given by, h = u²sin²θ/2g. Repeat the above with the 5 inch long blades set at 0 degrees. The only 'sensible' angle for this turning point is π/4 which is 45°. Repeat the above with the 10 inch long blade set at 0 degrees. Shooting 45. In his first year as a Laker, Ball shot 45 percent from the free throw line and 36 percent from the field. Problem: A baseball thrown at an angle of 60. Understand that the diagrams and mathematics here could be applied to any type of vector such as a displacement, velocity, or acceleration vector. S = COS 0. Maybe it’s easier to remember the upper scale as the angle-angle-on-wood scale and the bottom scale as the angle-the saw blade-travels scale. What was the initial velocity (speed when thrown)? Here is what you start with:. As soon as the ball is released, the thrower. 0 m/s directed at angle θ 60. Which of the following best describes the acceleration of the ball from the instant after it leaves the thrower's hand until the time it hits the ground? Always in the same direction as the motion, initially positive and gradually dropping to zero by the time it hits the ground. (ii) The horizontal speed of the ball over the straight distance [OB] is a constant 38 m s-1. (a) Find the magnitude of the ball's initial velocity (the velocity with which the ball is thrown). If we change the angle of the kick to 60 degrees, we get a hang-time of 4. A ball is thrown from a roof top at an angle of 45° above the horizontal. Ignore air resistance. You set the launch angle by placing the pin through holes that line up on the disk and the base of the catapult. Which ball hits the ground with the highest speed? Ball A Ball B Ball C Equal for all three Use energy conservation! U i + K i = U f + K f. 0 meters per second. The textbooks say that the maximum range for projectile motion (with no air. The initial velocity of the stone is 5. 80¤m above the floor. You throw a ball with an initial speed of 8. Answer : D A man kicks a soccer ball with an initial velocity of 25 m/s at an angle of 30°above the horizontal. 40 degrees c. (a) Find h. By throwing a ball at a 45 degree angle a girl can throw the ball a maximum horizontal distance r on a level field. 4 m T = U/g = 20/9. If a ball is throw with a velocity of 25 m/s at an angle of 37 degrees above the horizontal, what is its horizontal velocity component? 20 m/s: 4) A ball is thrown at an original speed of 8. The underlying idea is to resolve the force applied (along angle T = 45 degrees) by the pole into a vertical force (upward) and a horizontal force (against the wall). A tennis player wins a match at Arthur Ashe stadium and hits a ball into the stands at 30 m/s and at an angle $45\text{°}$ above the horizontal (). A ball is thrown at a 60o angle to the horizontal. At what speed must the second ball be thrown so that it reaches the. 0 m/s and an angle of 28. A stone is thrown horizontally outward from the top of a bridge. Neglecting air drag, a ball tossed at an angle of 30° with the horizontal will go as far downrange as one that is tossed at the same speed at an angle of A. A ball is kicked at an angle of 35° with the ground. What maximum height, above the ground, will the arrow reach? (A) 32. As the javelin travels upward, its velocity points above the horizontal at an angle that decreases as time passes. f = 21 m/s, 45. When you calculate projectile motion, you need to separate out the horizontal and vertical components of the motion. On the diagram above, draw in the path if air resistance were taken into account. 4 m/s 20) A boy throws a ball with an initial velocity of 25 m/ s at an angle of 300 above the horizontal. 443s until there is no more force in the y-axis, which means the ball has reached it maximum height. The above animation is available as a printable step-by-step instruction sheet, which can be used for. 6 m high and at a distance of 59. How fast was he traveling when he left the ramp? Indicate any assumptions you made. If the acceleration vector of an object is directed exactly opposite to the velocity vector. At the very top of the ball's path, its velocity is _____. 36’ - Union earn a free kick, try to play it quick and someone is called for offsides. 20 m/s (4) 18. 4-37, a ball is thrown leftward from the left edge of the roof, at height h above the ground. The top edge is 5m above the throwing point. 40 degrees c. 1) A ball is thrown at an angle 45° above the horizontal with an initial velocity of 20 m/s. If there is a height difference, things get a little more complicated. So, a launch angle of 90 degrees means the ball will go straight up, and a launch angle of 0 degrees means the ball will go horizontal. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of 53. (a) Will the ball cross a. ction point is the. Printable step-by-step instructions. 84 s, a maximum range of 66. To report the resultant vector, give both its angle and magnitude. As soon as the ball is released, the thrower. What is the magnitude of the ball’s initial velocity? (1) 7. Determine the total time that the stone is in the air. In the figure above we are using an angle that without spin would have a trajectory that would hit a 4 ft height at 100 ft. The horizontal component of the baseball's initial velocity is 12. If a ball is 0. 30 m from the elbow joint, what is the velocity of the ball? A truck with 0. = (20)²sin²45°/2*9. If you throw a ball at 60mph at 0 degrees, the horizontal velocity is 60mph. At that speed, balls struck with a launch angle between 26-30 degrees always garner Barreled classification. Also, if I throw the ball at 0 degrees, then cos(0) = 1, but sin(0) = 0. 0 meters per second. Keep in mind, when Correa was drafted, he. What is the vertical component of its acceleration at the top of its trajectory downward 10 m/s squared. Φ= tan-1 (v y /v x) Where Φ is the angle that the resultant velocity(v) makes with the horizontal at any instant. Assume the ball moves without air resistance and its motion is described using a Cartesian coordinate system with its origin located at the ballʼs initial position. How long does it take the ball to reach the wall? b. The sinker is still decent and does generate ground balls — hitters have a -9 degree launch angle versus the pitch — but it is definitely worse than the cutter overall. 2 m/s D) 19. A golf ball is chipped at an angle of degrees and and the horizontal distance is the same as above but thrown with a speed of m/s and at an angle. Part A Find the magnitude of the initial velocity of the baseball (the velocity with which the baseball is thrown). 204 m Thus, maximum height attained by the ball is 10. This angle is halfway between the orientations of the gun in parts 1 and 2. Projectile Motion Example. Then make a right angle triangle out of it by adding a vertical and a horizontal line. This article discusses the x- and y-components of a force vector. In the javelin throw at a track-and-field event, the javelin is launched at a speed of 29 m/s at an angle of 36° above the horizontal. 0° above the horizontal. 7 m/s c) 10 m/s d) 9. Right angle: The angle that is 90° is a Right angle, ∠C as shown below. A ball is thrown from a tower 30m high above the ground with a velocity of 300 m/s directed at 20 degree from the horizontal. A second ball is thrown at an angle of 30 degrees with the horizontal. A batter hits a ball with an initial velocity of 92 feet per second and at an angle of 55 o from the horizontal. The sonar of a navy cruiser detects a nuclear submarine that is 4000 feet from the cruiser. 50 degrees d. Assume the ball is thrown at ground level. A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. calculate maximum height and time taken to reach maximum height. g v t v g t v v a t i h i h fy iy y sin 0 sin But this is ½ the time of flight. What is the acceleration of the golf ball at the highest point in its trajectory? [neglect friction]. When Lily releases the ball, it is 5 feet above the ground. 6 m above the street below. The stone strikes at A, 5. 63 feet per second at an angle of 840 with the horizontal (0 3. 0m high cliff. A football is thrown upward at a(n) 43 degree angle to the horizontal. A ball is thrown at an angle of 40 degrees above the horizontal. 20 m/s (4) 18. Just after the bounce it is moving at 53 m/s at an angle of 19 degrees above the horizontal. A ball is thrown with initial velocity vo at an angle 00 = 45° above the horizontal, from the roof of a building of height h. The time the ball is in the air is given by (3). Find the time of flight and the horizontal range of the ball. The ball is launched at a speed of 45 m/s at an angle of 29° above the horizontal. ? How long does it take the to reach the highest point in its trajectory? The answer is 2. By modelling the ball as a particle moving freely under gravity, (b) find the horizontal distance of the ball from A when the ball is 1 m above the beach. 0m/s from the edge of a 45. I can do better. None of the above. A size 4 ball has a mass of approximately 0. The stone strikes at A, 5. Problem 38: You throw a ball toward a wall at a speed of 25 0 m/s and at an angle of 40 degreesa speed of 25. If the ball takes 3 seconds to rise to the peak of its trajectory, then it will take 6 seconds to fall from the peak of its trajectory to the ground. It strikes the ground with a velocity of 54 m/s at an angle of 22 degrees below the horizontal. Projectile motion is a form of motion experienced by an object or particle (a projectile) that is projected near the Earth's surface and moves along a curved path under the action of gravity only (in particular, the effects of air resistance are assumed to be negligible). If you fire a projectile at an angle, you can use physics to calculate how far it will travel. An arrow is fired at 45. The slope can be uphill or downhill. Question 15 0 A ball is thrown - - - - - - - - - to the air with an initial velocity of 10 m/s at an angle of 45 degrees above the horizontal - represented above Which of the following representations of the velocity and acceleration of the ball is a = 0 on V = 0 a = 0 OV=0. (a) Find the magnitude of the ball’s initial velocity (the velocity with which the ball is thrown). The ball returns to level ground. A Projectile is shot from edge of a cliff 125m above ground level with an Initial speed of 65 m/s at an angle of 37 degree above the horizontal. 5 s at d=25 m from the building. The sonar of a navy cruiser detects a nuclear submarine that is 4000 feet from the cruiser. The ball leaves her hand 1. 8 m pass, what must be the initial speed of the ball? Answer in units of ms. He throws the discus at an angle of 45 degrees above the horizontal and it travels a horizontal dista … read more. Find (a) the height h of the cliff, (b) the speed of the stone just before impact at A, and (c) the maximum height H reached above the ground a) Find the height of the cliff "h" b) Y X A t = T = 5. At what speed must the second ball be thrown so that it reaches the. larger than I g A ball is hurled into the air at an angle of 30 degrees and lands on a target that is at the same level as that where the ball started. See the proof below for more details. (b) Find the magnitude and direction of the velocity of the ball just before it strikes the building. 4 degrees), has continued to climb a bit, sitting at 14. The stone strikes at A, 5. Time of flight T Angular Projectile motion is symmetrical about the highest point. A ball is thrown at an angle of 40 degrees above the horizontal. This means that an 18-degree bounce with a narrow “C” grind will raise the leading edge a little bit. cosΘ = adj /hyp adj = cos Θ (hyp) = cos 40° (20m/s) opp = v y = 15m/s. A soccer ball is kicked with a velocity of 25 m/s at an angle of 45 degrees above the horizontal. A ball is thrown from a roof top at an angle of 45° above the horizontal. For example, halfway between 12 and 1 ( at 2 1/2 minutes) is 15° (about a 3/4-ball hit), and halfway between 1 and 2 is 45° (about a 1/4-ball hit). COACHengg App Download Link: https://play. Time of flight = Maximum Range. (You can imagine that if your arm were the hour hand of a clock, the previous throw would. How long does it take the ball to reach the wall? b. Ball A is launched horizontally. The velocity after two seconds would be given by adding two of the dashed downward arrows head-to-tail to the initial horizontal arrow, and so on, so that after ten seconds, if the cliff were high enough, the velocity would be pointing downwards at an angle of 45 degrees, and the speed by this point would have increased substantially. On its way down, the ball is caught by a spectator 10 m above the point where the ball was hit. Index Trajectory. A player 55meters away in the direction of the kick starts running to meet the ball at that instant. ' and find homework help for other Math questions at eNotes eNotes Home. However, it was a long time ago with crappy looking graphs. The action is somewhat similar to bowling a cricket ball. 50 degrees d. Question 15 0 A ball is thrown - - - - - - - - - to the air with an initial velocity of 10 m/s at an angle of 45 degrees above the horizontal - represented above Which of the following representations of the velocity and acceleration of the ball is a = 0 on V = 0 a = 0 OV=0. Problem 38: You throw a ball toward a wall at a speed of 25 0 m/s and at an angle of 40 degreesa speed of 25. A Ball Is Thrown At An Angle Of 45 Degrees Above The Horizontal. The ball is thrown from A with speed 25 m s–1 at an angle of 30° below the horizontal. How fast was the ball thrown? A. The distance OT is 15 m. The sinker is still decent and does generate ground balls — hitters have a -9 degree launch angle versus the pitch — but it is definitely worse than the cutter overall. 68m above the ground at an angle of 43. A ball is thrown from a roof top at an angle of 45° above the horizontal. The ball is projected with speed 7 m s−1 at an angle of elevation of 45°. dr dt r(0) dr dt r(0. 0° above the horizontal. 4 degrees, a break-over/ramp-over angle of 20. 9 percent from the field (55. At a particular instant, a hot air balloon is 100 m in the air and descending at a constant speed of 2. A second ball is thrown at an angle of 30 degrees with the horizontal. 30 m/s by a rope that is inclined at an angle of 250 above the horizontal. Ignore air resistance. 4 m from the point of release and at a height h = 0. larger than I g A ball is hurled into the air at an angle of 30 degrees and lands on a target that is at the same level as that where the ball started. Always in the same direction as the motion, initially positive and gradually dropping to zero by the time it hits the ground. 4% and had risen from 18th in these standings to first in the Premier League -- second in Europe behind Grønnemark's other side FC Midtjylland. The acceleration of gravity is 9. Question 15 0 A ball is thrown - - - - - - - - - to the air with an initial velocity of 10 m/s at an angle of 45 degrees above the horizontal - represented above Which of the following representations of the velocity and acceleration of the ball is a = 0 on V = 0 a = 0 OV=0. If I place one line at 34-degrees above the horizontal (loft is measured from the vertical), and then extend another at some angle below horizontal, the height above ground where the two join depends on how long the lower line is. 5 meters per second at an angle of 35º with the horizontal, as shown. Ball A A A of mass 1 kg 1\text{ kg} 1 kg is thrown at an angle of 45 ∘ {45}^\circ 4 5 ∘ with the horizontal with kinetic energy 50 J 50\text{ J} 5 0 J such that it hits ball B B B of the same mass placed on the top of a pole. The Radian. 5 m (120 ft). 0 m/s and at an angle of 40 degrees. During a spin pass, the passer will throw the ball at no more than a 15 degree angle (which is only for an extremely long pass) to get the ball to his teammate. 0 m above the throwing point. In lacrosse, a ball is thrown from a net on the end of a stick by rotating the stick and forearm about the elbow. An arrow is fired at 45. An object is launched at an angle of 30 degrees with an initial velocity of 20 m/s. (a) How far above the release point does the ball hit the wall? (b) What are the horizontal and vertical components of the its. At the rear there are integrated rock rails to protect the lower edges of the tub. 0 m/s at an angle of 40 degrees above the horizontal. Three Ball Race. Thus, maximum height attained by the ball is 10. Shakin at 3 ft above the ground with an initial speed of 150 ft/sec and at angle of 18 degrees with the horizontal. The ball lands on a bullseye marked on the floor a horizontal distance away. A brick is thrown upward from the top of a building at an angle of 25 degrees to the horizontal. What is the total time of flight of the ball before it hits the ground? 2)A projectile is launched from ground level with a certain speed. The only 'sensible' angle for this turning point is π/4 which is 45°. Three identical balls are launched with the same initial speed from the top of a cliff. The top edge is 5. At the very top of the ball's path, its velocity is _____. At the same instant a second ball is thrown straight up from the ground with an initial speed of 28 m/s directly below the point where the first ball was dropped. This is because the maximum sin2a can be is 1 and sin2a = 1 when a = 45°. 7 m/s c) 10 m/s d) 9. (a) Calculate the time it takes the tennis ball to reach the spectator. The top edge is 5 m above the throwing point. 0 meters from the base of the building wall. 84 meters thrown horizontally at this speed, how much will it drop by the time it reaches a. A golf ball is chipped at an angle of degrees and and the horizontal distance is the same as above but thrown with a speed of m/s and at an angle. He throws the discus at an angle of 45 degrees above the horizontal and it travels a horizontal dista … read more. 0° above the horizontal. In lacrosse, a ball is thrown from a net on the end of a stick by rotating the stick and forearm about the elbow. a) What should be the initial velocity of the ball so that it hits a target that is 30 meters away at a height of 1. At the maximum height, v fy = 0. At the highest point,the vertical component of velocity v y becomes equal to zero. C) the same. 3 m/s at an angle of 15° below the horizontal. The acceleration of gravity is 9. 00 seconds the ball completes its trajectory and hits the ground. A ball is thrown at an angle of 40 degrees above the horizontal. 60 degres. It hits the ground after t=1. Given that ball is thrown at 60° angle to the horizontal, Given x = 90m, θ= 60° Substituting the values of the above in , (As sin120 is same as sin60 –sine function is positive in first and second quadrant) Now, the ball is thrown with same velocity and with 30 degree angle, we need to find the distance it covers on the ground. 0m/s from the edge of a 45. because this t-value only deals with the first half of it's flight, you must multiply it by 2. Angles that are between 90 and 180 degrees are considered obtuse. [Racecar-ga chose to define T as the angle of the pole with respect to vertical, but of course when T = 45 degrees, the pole makes an equal angle with respect to horizontal. above the horizontal and is caught by an outfielder 365ft from home. Straight angle: The angle that is 180° is a straight angle, ∠AOB in the figure below. 29 What was the vertical velocity at takeoff in m/s? Vertical velocity = 3. 0 degrees and the corresponding horizontal position of the ball when it is at its maximum height. By throwing a ball at a 45 degree angle a girl can throw the ball a maximum horizontal distance r on a level field. At the highest point,the vertical component of velocity v y becomes equal to zero. (b) Find the magnitude and direction of the velocity of the ball just before it strikes the building. The ball is thrown with an initial speed of 4. A ball is thrown at an original speed of 8. 4 N at 45°, and 100 N at 240°. A pass For example, Dan Carter of the All Blacks can throw the ball at 20m/s at an angle of 15 degrees, and using this info we can draw the horizontal and vertical components and find. 5 meters per second at an angle of 35º with the horizontal, as shown. a long jumper leaves the ground at an angle of 30 degrees to the horizontal at a speed of 6 m/s. 4mph: Thus, on a frictionless surface with no wind resistance, throwing it horizontally will get it to the base the fastest no matter how far it is away. In lacrosse, a ball is thrown from a net on the end of a stick by rotating the stick and forearm about the elbow. 84 s, a maximum range of 66. 0 meters per second. The initial velocity of the stone is 5. As you can see from the 1 degree marks between 50 & 45 degrees on the lower scale, the yellow line at 47 degrees points directly to 43 degrees on the upper scale. 0 ft above the ground in a horizontal direction at a speed of 73 ft/s (50 mph). Keep in mind, when Correa was drafted, he. 0 m/s NORTHWEST. com/store/apps/details?id=com. The underlying idea is to resolve the force applied (along angle T = 45 degrees) by the pole into a vertical force (upward) and a horizontal force (against the wall). 0° above the horizontal. At this point the vertical component of its velocity, u=0. So, a launch angle of 90 degrees means the ball will go straight up, and a launch angle of 0 degrees means the ball will go horizontal. 2 m B) 24 m C) 23 m D) 2. If the ball lands 90 m away, what was th initial speed of the ball? Student Solution Manuals: https://amzn. 20 Snowballs are thrown with a speed of 13 m/s from a roof 7. 00 m above the point from which it is thrown. Based on ground-to-ground projectile motion, a projectile launched from the origin and returns to the same horizontal level. Right angle: The angle that is 90° is a Right angle, ∠C as shown below. 4 m from the point of release and at a height h = 0. 0 kg object is moving at 5. A ball is thrown vertically upward with a velocity of 20 m/s. At what angle above the horizontal must the ball be thrown to exactly hit the basket?. Part (a) Determine the maximum height above ground, hmax in meters, attained by the golf ball. The initial momentum is m*v = 0. The initial speed of the ball in metre/second is (take g=10 m/s2): (A) U - 40 114+3) V 13. Ignore air resistance. A ball is thrown with initial velocity vo at an angle 00 = 45° above the horizontal, from the roof of a building of height h. It’s the same one shown above where Roethlisberger short-hops Brown but from a different angle. 5 m/s from a 5. For the ball that is thrown 40 degrees below the horizontal, Draw a diagonal line at an angle 40 degrees below the horizontal to represent the ball's velocity. cosΘ = adj /hyp adj = cos Θ (hyp) = cos 40° (20m/s) opp = v y = 15m/s. What is the acceleration of the golf ball at the highest point in its trajectory? [neglect friction]. A certain gun type fires bullets at 60 m/s regardless of its orientation. The path of the baseball is given by the function f(x) = -0. What maximum height, above the ground, will the arrow reach? (A) 32. I know I already did this. A baseball thrown at an angle of 60. 117 m is thrown down the lane with an initial speed of v = 8. 4 degrees), has continued to climb a bit, sitting at 14. 00 m above the point from which it is thrown. a football thrown at a velocity of 10 yards per second at an angle of 580 with the ground 4, a golf ball hit with an initial velocity of 102 feet per second at an angle of 670 with the horizontal 5. 15 m/s, releasing it at a height of 2. Maybe it’s easier to remember the upper scale as the angle-angle-on-wood scale and the bottom scale as the angle-the saw blade-travels scale. The air is moving 40 m/s … read more. f = 21 m/s, 45. Find the time of flight and the horizontal range of the ball. The largest sine can be is 1 and that occurs at 90º. 0, the horizontal distance the ball travels will B) increase C) remain the same OAax 200 A machine launches a tennis ball at an angle of 250 above the horizontal at a speed of 14 meters per second. For example, halfway between 12 and 1 ( at 2 1/2 minutes) is 15° (about a 3/4-ball hit), and halfway between 1 and 2 is 45° (about a 1/4-ball hit). A ball is thrown vertically upward with a velocity of 20 m/s. Hit the ball straight up! 2. A second ball is thrown at an angle of 30 degrees with the horizontal. We use one of those 45 degree angles to get the result we need. If the interaction with the ground lasts 0. A player standing on the free throw line throws the ball with an initial speed of 7. - A ball is thrown upward at an angle of 30 degrees to the horizontal and lands on the top ledgeof a building that is 20 away the top edge is 5 0 m above the throwing point how fast was the ball thrown. Whenever a line is not horizontal, it has a slope. When the ball is shot over level ground half of the time the ball is going up, the other half of the time it is going. In general, a shooter wants to aim for a shot with a 45-degree angle as it enters the cylinder , eleven inches deep into the basket , just two. A ball is thrown straight upward and returns to the thrower's hand after 2. 0 degrees and the corresponding horizontal position of the ball when it is at its maximum height. (a) How far above the release point does the ball hit the wall? (b) What are the horizontal and vertical components of the its. 33 m/s at an angle of 59. What was the initial velocity (speed when thrown)? Here is what you start with:. How high above the projection point is the ball after 1. Chuck's projected ball is A) greater. 5 m (179 ft). A Ball Is Thrown At An Angle Of 45 Degrees Above The Horizontal. Which of the following best describes the acceleration of the ball from the instant after it leaves the thrower's hand until the time it hits the ground?. immediately after the ball is hit, the third basem. When a single angle is drawn on a xy-plane for analysis, we’ll draw it in standard position with the vertex at the origin (0,0), one side of the angle along the x-axis, and the other side above the x-axis. f the projectile's initial horizontal speed is 55 meters per second, then angle 9 measures approximately 8. A ball is released from top of 98 m tall building. 0 degrees NORTH of WEST. If we change the angle of the kick to 60 degrees, we get a hang-time of 4. A baseball is thrown at an angle of 40. ction point is the. Question 15 0 A ball is thrown - - - - - - - - - to the air with an initial velocity of 10 m/s at an angle of 45 degrees above the horizontal - represented above Which of the following representations of the velocity and acceleration of the ball is a = 0 on V = 0 a = 0 OV=0. Repeat the above with the 10 inch long blade set at 0 degrees. A spherical bowling ball with mass m = 3. Professor Lewin sets the angle at 30 degrees, and predicts where the ball will hit. 00 m above the ground and experience negligible air resistance. A second ball is thrown at an angle of 30 degrees with the horizontal. If the interaction with the ground lasts 0. 84 meters thrown horizontally at this speed, how much will it drop by the time it reaches a. 1) A ball is thrown at an angle 45° above the horizontal with an initial velocity of 20 m/s. 2 m B) 24 m C) 23 m D) 2. 00 m above the point from which it is thrown. Find the angle of elevation of K from B, correct to the nearest degree. The top edge is 5m above the throwing point. The ball clears the fence which is 85 meters away. The ball is thrown from A with speed 25 m s–1 at an angle of 30° below the horizontal. If the ball is released 1. A ball is thrown at an angle of 45 degrees above the horizontal. Ignore air resistance. 0, the horizontal distance the ball travels will B) increase C) remain the same OAax 200 A machine launches a tennis ball at an angle of 250 above the horizontal at a speed of 14 meters per second. Find the time of flight and the horizontal range of the ball. where t is the time in seconds after the ball is struck and h is the height of the ball. Neglecting air drag, a ball tossed at an angle of 30 ° with the horizontal will go as far downrange as one that is tossed at the same speed at an angle of A. Back Trigonometry Vectors Forces Physics Contents Index Home. Projectile motion is a form of motion experienced by an object or particle (a projectile) that is projected near the Earth's surface and moves along a curved path under the action of gravity only (in particular, the effects of air resistance are assumed to be negligible). Classical Mechanics. The ball is thrown with an initial speed of 4. Ball A A A of mass 1 kg 1\text{ kg} 1 kg is thrown at an angle of 45 ∘ {45}^\circ 4 5 ∘ with the horizontal with kinetic energy 50 J 50\text{ J} 5 0 J such that it hits ball B B B of the same mass placed on the top of a pole. Answer : C A ball is thrown at an angle of 45 degrees above the horizontal with an initial velocity of 20m/s. He also splits the safeties with the throw, resulting in a great ball placed in the middle of a triangle of defenders. above the horizontal and is caught by an outfielder 365ft from home. For instance seven and a half degrees is now usually written 7. A football is thrown upward at a(n) 43 degree angle to the horizontal. Use the symmetry of the parabolic trajectory to double the horizontal (x) distance at the position of maximum height (y). (a) Will the ball cross a. The top edge is 5 m above the throwing point. A second ball is thrown at an angle of 30 degrees with the horizontal. Find the speed of the ball when the goalie catches it in front of the net. f = 21 m/s, 45. In the figure above we are using an angle that without spin would have a trajectory that would hit a 4 ft height at 100 ft. A batted baseball leaves the bat at an angle of 34. The ball follows an ideal path before landing on the field at point P f. 33 m/s at an angle of 45. As soon as the ball is released, the thrower. How deep is the submarine? 3. What maximum height, above the ground, will the arrow reach? (A) 32. 0 m/s (C) 4. 0 m/s at an angle of 35° above the horizontal. NAME_____ pd____. Repeat the above with the 5 inch long blade set at. 0° above the horizontal. Which ball hits the ground with the highest speed? Ball A Ball B Ball C Equal for all three Use energy conservation! U i + K i = U f + K f. 83 mph (221. The distance OT is 15 m. itcan be deduced from physical principles, that the path of the ball is modeled by the function: y= -32/20^2 x^2+x+5 where x is the distance in feet that the ball has traveled horizontally. 0 meters per second. Whenever a line is not horizontal, it has a slope. A child throws a ball with an initial speed of 8. The angle A is the serve angle in degrees below the horizontal. And, unlike the launch angle, which varies with player height, the entry angle is the same for all shooters. 40 degrees c. 00\ m {/eq} above the ground with a speed of {eq}20. Supplementary angles: In the figure above, ∠AOC + ∠COB = ∠AOB = 180° If the sum of two angles is 180° then the angles are called supplementary angles. If you fire a projectile at an angle, you can use physics to calculate how far it will travel. 0 m away at a point 8. 0 m/s at an angle of 45 degrees. 8 percent from the free throw line, certainly aids in lessening the possible concern of the lackluster three. 0 ft above the ground in a horizontal direction at a speed of 73 ft/s (50 mph). 20 degrees b. 0 meters from the base of the building wall. 50 s after launching. At this point the vertical component of its velocity, u=0. (b) Find the magnitude and direction of the velocity of the ball just before it strikes the building. A launch at 45 degrees would allow the ball to remain in the air for a longer time, but it would then be launched at a lower horizontal speed at the start and it would slow down more because of the longer flight time. 45 degrees. 76 m 5) Two balls were thrown upward and were caught at the same height from which they were released. You set the launch angle by placing the pin through holes that line up on the disk and the base of the catapult. 4 seconds after the ball is thrown. For instance seven and a half degrees is now usually written 7. What is the vertical component of its acceleration at the top of its trajectory downward 10 m/s squared. Problem: A baseball thrown at an angle of 60. It exerts a torque about the center of the ball that causes the ball to rotate with a backward spin during its flight. What is the magnitude of the ball’s initial velocity? A) smaller B) greater C) the same 2. If I place one line at 34-degrees above the horizontal (loft is measured from the vertical), and then extend another at some angle below horizontal, the height above ground where the two join depends on how long the lower line is. 6 0 below the horizontal ( -0 from x axis) 8. A Ball Is Thrown At An Angle Of 45 Degrees Above The Horizontal. When you calculate projectile motion, you need to separate out the horizontal and vertical components of the motion. asked by chris on February 5, 2013; physics. Just after the bounce it is moving at 53 m/s at an angle of 19 degrees above the horizontal. a cricket ball is thrown at a speed of 30 m s in direction making an angle of 30 degree with the horizontal calculate maximum height range of the ball - Physics - TopperLearning. At that speed, balls struck with a launch angle between 26-30 degrees always garner Barreled classification. 50 degrees d. A soccer ball is kicked with a velocity of 25 m/s at an angle of 45 degrees above the horizontal. 40 degrees b. Find vel at t=1 sec. It hits the ground a few seconds later. If the ball is released 1. How deep is the submarine? 3. The sinker is still decent and does generate ground balls — hitters have a -9 degree launch angle versus the pitch — but it is definitely worse than the cutter overall. 82 km/h), at an angle of 27. The largest sine can be is 1 and that occurs at 90º. Given that ball is thrown at 60° angle to the horizontal, Given x = 90m, θ= 60° Substituting the values of the above in , (As sin120 is same as sin60 –sine function is positive in first and second quadrant) Now, the ball is thrown with same velocity and with 30 degree angle, we need to find the distance it covers on the ground. In laboratory studies of the throw-in, male players have recorded release. 0° above the horizontal strikes a building 18. 4 m from the point of release and at a height h = 0. A second ball is thrown at an angle of 30. Three identical balls are launched with the same initial speed from the top of a cliff. (There are 2 ways to do this but it doesn't matter which way you choose. This is because the maximum sin2a can be is 1 and sin2a = 1 when a = 45°. This is the optimal angle. The textbooks say that the maximum range for projectile motion (with no air. Let's say you're on top of a cliff, which drops vertically 150 m to the ocean below. The top edge is 5m above the throwing point. The angle that maximizes sin. It is thrown from a height of {eq}2. 0° above the horizontal. How fast was the ball thrown? A. At what speed must the second ball be thrown so that it reaches the. Maybe it’s easier to remember the upper scale as the angle-angle-on-wood scale and the bottom scale as the angle-the saw blade-travels scale. ction point is the. Given that ball is thrown at 60° angle to the horizontal, Given x = 90m, θ= 60° Substituting the values of the above in , (As sin120 is same as sin60 –sine function is positive in first and second quadrant) Now, the ball is thrown with same velocity and with 30 degree angle, we need to find the distance it covers on the ground. an angle, θ, above the horizontal. 06 kg is served. We use one of those 45 degree angles to get the result we need. Ah, launch. The horizontal component of the baseball's initial velocity is 12. As soon as the ball is released, the thrower. 40 degrees b. What is the vertical component of its acceleration at the top of its trajectory downward 10 m/s squared. The top edge is 5 m above the throwing point. He starts to run in a straight line with speed v m s−1, leaving B 0. While the horizontal speed of the ball in Fig. (a) Will the ball cross a. An arrow is fired at 45. Your equations are correct, but not applied properly. Note that the question below only has three choices. COACHengg App Download Link: https://play. This is the optimal angle. an athlete participates in a discus throw in a city where acceleration due to gravity is 9. a soccer ball is kicked from the ground at an initial velocity of 19. A ball is thrown at a 60o angle to the horizontal. At the same instant a second ball is thrown straight up from the ground with an initial speed of 28 m/s directly below the point where the first ball was dropped. What is the magnitude of the impulse which the player imparts to the ball? As we know, Impulse = Change of momentum. Model Rocketry Manuel launches a toy rocket from ground. 29’ - Aaronson with numbers and instead of attempting a shot, he gets cute by dribbling the ball from the left side of the penalty area to the center top. (a) At what speed was the first ball thrown? (b) At what speed was the second ball thrown?. If you fire a projectile at an angle, you can use physics to calculate how far it will travel. The horizontal component of the baseball’s initial velocity is 12. Given: initial velocity at 66 degrees above the horizontal=4. At what angle below the horizontal does the ball approach the ground? The answer options are: A) 35 degrees B) 42 degrees C) 48 degrees D) 40 degrees E) 65 degrees =====. 60 degres. If we change the angle of the kick to 60 degrees, we get a hang-time of 4. A baseball hit at 103 mph at 28. 0 meters per second. If you wanted to hit or throw the baseball as high as you could, you would have to throw/hit the ball at 90 degrees to maximize vertical displacement. If there is a height difference, things get a little more complicated. 0m/s from the edge of a 45. How long will the ball hit the ground? 5. To Naquin's credit, a shift toward raising his average launch angle slightly, which seemed to begin last year (12. Question: A ball is thrown at a {eq}60 ^\circ {/eq} angle above the horizontal across level ground. 5 meters per second at an angle of 35º with the horizontal, as shown. how far does he jump Guest Sep 16, 2014 0 users composing answers. A ball is thrown from a tower 30m high above the ground with a velocity of 300 m/s directed at 20 degree from the horizontal. Throwing a baseball A baseball is thrown from the stands 32 ft above the field at an angle of 300 up from the horizontal. How high above the projection point is the ball after 1. 0° above the horizontal. The distance OT is 15 m. 38m above the ground at an angle of 40. The sonar of a navy cruiser detects a nuclear submarine that is 4000 feet from the cruiser. A soccer ball is kicked with a velocity of 25 m/s at an angle of 45 degrees above the horizontal. Find (a) the height h of the cliff, (b) the speed of the stone just before impact at A, and (c) the maximum height H reached above the ground a) Find the height of the cliff "h" b) Y X A t = T = 5. A brick is thrown upward from the top of a building at an angle of 25 degrees to the horizontal. Assume that a kicked ball in football is a projectile. 20 m/s (4) 18. The ball takes 2. A Ball Is Thrown At An Angle Of 45 Degrees Above The Horizontal. A baseball is thrown at an angle of 40. Determine the time of flight, the horizontal distance, and the peak height of the football. Otherwise, it is not correct that a 45 degree angle gives the largest horizontal displacement. An observer in a lighthouse 350 feet above sea level observes two ships directly offshore. Find (a) the height h of the cliff, (b) the speed of the stone just before impact at A, and (c) the maximum height H reached above the ground a) Find the height of the cliff "h" b) Y X A t = T = 5. At the maximum height, v fy = 0. (a) Find h. Understand that the diagrams and mathematics here could be applied to any type of vector such as a displacement, velocity, or acceleration vector. Repeat the above with the 15 inch long blade set at 0 degrees. 5 m/s from a 5. The sine function reaches its largest output value, 1, with an input angle of 90 degrees, so we can see that for the longest-range punts 2θ = 90 degrees and, therefore, θ = 45 degrees. It hits the ground a few seconds later. 40 degrees c. None of the above. When the ball hits the ground its velocity is equal to but opposite in the direction to the vertical component of its velocity when it was launched. Telling an OT to down-block a 2-technique would require a flat angle (so much that you wouldnt do it). A ball is thrown upward at an angle of 30° with the horizontal and lands on the top edge of a building that is 20m away. Back Trigonometry Vectors Forces Physics Contents Index Home. A ball is thrown upward at an angle of 30 °; to the horizontal and lands on the top edge of a building that is 20 m away. 33 m/s at an angle of 45. 8 m Answer: D 2) The horizontal and vertical components of the initial velocity of a football are 16 m/s and 20 m/s respectively. The textbooks say that the maximum range for projectile motion (with no air.
|
2020-10-21 07:05:54
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6745551228523254, "perplexity": 325.7989783265145}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107876136.24/warc/CC-MAIN-20201021064154-20201021094154-00569.warc.gz"}
|
https://krinkinmu.github.io/2020/12/30/dynamic-memory-allocation-part1.html
|
30 December 2020
# Dynamic Memory Allocation Part1
by Mike Krinkin
In the previous post I mentioned that I implemented simplistic dynamic memory allocator and plugged it into Rust. So I thought I could create an introductionary post into dynamic memory allocation algorithms.
This post will cover a basic algorithm of dynamic memory allocation and some practical aspects that we might consider when implementing dynamic memory allocators as a sort of introduction into the problem (thus Part1).
As always the code is available on GitHub.
# Motivation
Since it’s a sort of an intrdouction post it worth briefly covering why do we need dynamic memory allocation at all. Ignroing for a moment low-level part of software systems and their special needs and goals, let’s consider regular user-space binaries. In such case we can potentially statically reserve the maximum amount of memory the problem could potentially require and then use that statically reserved memory.
Statically reserving memory basically means recodring how much memory the OS should give to the program when loading the binary in the binary itself.
NOTE: I’m simplifing a lot here by not considering how OS would give memory to the program in the first place, not to mention that I’m leaving aside a case when there is no OS at all. However static memory reservation is still quite different from the dynamic memory allocation in the sense that all the memory allocated upfront and not freed until the end of the program.
If we take, for example, a classic Unix utility sort, that by default just sorts input strings lexicographically. We can say that it only makes sense to use this utility if we process no more than XGiB of data and reserve that much. The value of X might depend on multiple factors, but I’d say there is always some limit beyond which using a particular tool implementation isn’t practical anymore.
So all-in-all reserving memory statically should be possible, so it should be possible to live without dynamic memory allocation. However, as you could have noticed the fact that it’s possible, doesn’t mean it’s convenient.
For our sort tool case, the limit X might depend, for example, on the available amount of RAM in the system or a particular model of CPU. That would mean that we might need to rebuild the sort tool for all those different hardware configurations - which is, as I pointed before, isn’t particularly convenient.
Argument of convenience is far from being a universal one. It’s open to subjective interpretations and various unclear tradeoffs. Moreover the importance of convenience depends on the problem. For example, The Motor Indiustry Software Reliability Association (aka MISRA) C language guidelines prohibits the use of C dynamic memory allocation because of the safety concerns.
Are there more objective arguments for/against of dynamic memory allocation? Let’s return back to the problem of sorting. One of the classical sorting algorithms is Merge Sort.. A typical Merge Sort implementation of arrays would depend on a temporary buffer of $$O(n)$$ size. This buffer is only needed during the sorting and is not used for anything else, however regardless of that we’d still need to reserve memory for that buffer.
Such temporary buffers that are only used sometimes, depending on the specifics of the program, may add up to a noticable memory footprint and inneficient memory usage. So you might want to consider reusing the same memory for different purporses in the program. Or, in other words, you might want to consider using memory pooling and create methods to request memory from the pool and return it back when it’s not needed for somebody else to use. It looks just like dynamic memory allocation problem. isn’t it?
Whether your particular case benefits from pooling or not depends on the program of course. The main point of the argument though is that benefits of memory pooling might be objectively measured.
The final argument for the dynamic memory allocation that I would like to touch on here comes from resuing code. Many programming languages come with a sort of standard library. Frankly, even if they didn’t somebody would create such a library anyways. Quite often standard libraries include various collection implementations (vectors, linked lists, various sets and maps, etc).
Statically reserving memory for those in the library is rather problematic. The same library can be used in multiple different environments with different constraints and requirements. If we want to be able to reuse those libraries between different environments, memory reservations has to be abstracted somehow. Dynamic memory allocation is one way to achieve that.
# Dynamic memory allocation interface
Now when I’m done pouring the water we can move towrads more practical matters. What interface should dynamic memory allocator have? Actually, there isn’t a universally accepted correct answer for that question.
The question of the right interface isn’t just a question of conveince for the users. Some interfaces may prohibit certain implementations. As a result we might want to consider multiple different interfaces. I start from a relatively minimal interface:
#include <stddef.h>
void *allocate_memory(size_t size);
void free_memory(void *ptr);
The names of the functions might be self explanatory, but nevertheless let’s spare a few words to explain them. allocate_memory reserves a contigous memory range of size at least size bytes and returns the pointer to the start of that memory range.
NOTE: Naturally, it may happen that the dynamic memory allocator cannot find a large enough contigous memory range to satisfy the request. In this case the function would have to indicate a failure somehow. Typically a NULL value is used to indicate a failure, but it’s ok to change interface if returning NULL values is somehow not to your liking.
free_memory performs a revese operation. It takes a pointer returned from allocate_memory function and marks the reserved range as free and available for allocation again.
This interface assumes that we will be able to derive all the information we need to free the previous allocated memory range from the pointer to that memory.
It might look like a non-trivial requirement, however it’s not that significant. Basically when allocating memory inside allocate_memory we can reserve slightly more memory than requested. In this additional memory we can store whatever metadata we want. This way we could easily discover the information we need from the pointer passed to the free_memory.
It’s not to say that this caveat will not make the implementation harder, it will most likely add a bit of complexity to the implementation. However it’s not fundamental to the dynamic allocation problem itself.
NOTE: C standard requires its free function to accept NULL pointer. That’s curiously simmetrical: free function can accept any pointer malloc and company can return, even if they failed.
To the two functions above I’d like to add one more function:
void add_memory_pool(void *begin, void *end);
This function will serve a purely techincal reason. It will tell the dynamic memory allocator what memory it has available in the system to begin with. With this function we don’t need to discuss how do we find what memory is available and what memory does the dynamic memory allocator can distribute. Those are important questions, but they have very little to do with the actual dynamic memory allocation algorithms.
NOTE: potentially we can add multiple memory pools to the memory allocator. It will hardly change anything in the algorithm I will be discussing, but it might be quite important from practical point of view.
While discussing the interface it’s worth saying a few words on the correct and incorrect usage of the interface. I will assume here that allocate_memory and free_memory can only be called in pairs: for each free_memory call there must be exactly one allocate_memory preceding the free_memory call that returned the pointer passed to the free_memory call as an argument.
Let’s look at a few examples. Here is the correct example:
void *ptr = allocate_memory(size);
free_memory(ptr);
A slightly more complicated, but still correct example:
void *ptr1 = allocate_memory(size1);
void *ptr2 = allocate_memory(size2);
free_memory(ptr2);
free_memory(ptr1);
In the last example without losing correctness we can change the order of allocate_memory calls or the order of free_memory calls.
However it would be incorrect to call the free_memory on the same pointer twice without allocate_memory in between, like this:
void *ptr = allocate_memory(size);
free_memory(ptr);
free_memory(ptr);
And of course it would be incorrect to call free_memory on a pointer that wasn’t previously returned from allocate_memory:
int *array = allocate_memory(size * sizeof(int));
free_memory(&array[2]);
Though it’s not of fundamental importance, I would allow calling free_memory on NULL pointer any number of times even without a matching allocate_memory.
# Implementation constraints
Later in this post I will cover more constraints that have to be considered by a particular implementation. However there is one important constraint to keep in mind, and it’s that aside from the memory provided via add_memory_pool the implemetation is only allowed to use $$O(1)$$ of additional memory.
You can think of this limitation in the following way. We are allowed to create a few fixed global variables for our algorithm implementation outside of the memory provided to the allocator via add_memory_pool, but any unbounded memory that the implementation needs has to come from one of the memory pools given to the algorithm.
That constraints makes a lot of sense and basically tells us that our dynamic memory allocation algorithm cannot depend on another dynamic memory allocation algorithm being available already.
# Simple algorithm
Let’s move to the actual memory allocation algorithm. I will consider a few variations in this post, but all of them will be similar in a way and might be considered one algorithm with a few improvements on top.
The basic idea is that we are going to maintain a linked list of free memory ranges available. Each node of the list will describe a contigous free memory range.
This way the allocation essentially boils down to going through the list and finding a large enough free memory range. Naturally, it may happen that the memory range we find will be too large, so to avoid wasting memory we would need to split the memory range in a couple of memory ranges.
Freeing previously allocated memory range then essentially involves adding the freed range back to the free ranges list.
The linked list nodes themselves will be stored in the same memory pool we will allocate the memory from. And we need one global variable to store the head of the linked list.
That’s the general idea of the algorithm, however the devil is in details. To see where the problems may come from let’s look how one particular implementation of the algorithm may behave. In order to do that let me first introduce a few definition to use the same language:
struct free_list_node {
struct free_list_node *next;
size_t size;
};
Here struct free_list_node describes one contigouse free memory range. I will arrange those structure into a linked list, so I have to store the pointer to the next range in the structure and that’s what the next field is for. The global variable free_list_head points to the first element of the list.
For each free memory range we need to know where it begins and how bit it’s. I will store the struct free_list_node describing the free memory range at the beginning of that range, so the pointer to the struct free_list_node will also serve as a pointer to the beginning of the free memory range itself.
The size of the free memory range is described by the size field of the struct free_list_node. There are multiple roughly equivalent ways to define the size. Here I will take that size field includes the size of struct free_list_node and all the memory in the range coming after that.
The following picture clarifies the definitions I use and shows how the pieces fit together:
On the picture I have only one free memory range. That’s how the state will look like right after we called add_memory_pool once. The allocator basically has one contigous free memory range and that’s it.
The free_list_head points to the only range we have at this point, so the picture has an arrow from free_list_head to the beginning of the range.
At the beginning of the range we have struct free_list_node. It’s the metadata that we maintain for our algorithm and I generally will denote the metadata using blue bubbles on the pictures.
It’s important to note that struct free_list_node is stored right inside the free range. Remember that we cannot use an unbounded amount of memory for our metadata outside the given memory pool. And potentially we can have the unbounded number of the free memory ranges in our algorithm.
Finally, since we have only one free range the next field of the struct free_list_node contains NULL and that’s what the picture shows.
Let’s take a look at what will change when we allocate some memory.
NOTE: in the pictures below I will not show the details of the struct free_list_node in order to make them more compact. I use different colors to mark various structures in memory, so pay attention to the legend on the pictures.
## Allocation
As was outlined before the allocation boils down to going through the list of free ranges until we find a large enough or reach the end of the list. Traversing the linked list is not really complicated, but there is one thing to keep in mind - to remove an element from a singly linked list you need access to the previous element in the list.
Now, let’s assume that we went through the list and found a large enough element (and also the element before that one). There are actually a couple of cases that we need to consider.
The problem is that the free range that we found might be to big for the requested size of the allocation. Say you found a 1MiB free range when you were asked to allocate just 8 bytes. Returning the whole range in this case would be a waste of memory (not to mention, that in order to do that we don’t really need any fancy algorithms).
So what you might want to do is to cut off just enough memory from the found free range and return that part back to the caller. In other words, you might want to split the range you found in two. It might look like this:
On the picture above the tail of the free range is returned to the caller and the head of the free range stays in the free list range. It allows for a simple implementation that doesn’t require manipulations with the structure of the free list. All it takes is to just update the size field and that’s it.
It’s possible to implement it differently and return to the caller the head of the free range and return to the free list the tail of the range as shown on the picture below:
The implementation of the second option is slightly more complicated, but there is a case to be made for that implementation as well. I will not elaborate on that here, since it’s not even the final version of the algorithm.
That was one case that we have to consider, what is the other case?
In the case I considered above I showed the request size with a gray bubble. Hopefully it was clear that the gray bubble in the previous case was much smaller than the size of the (only) available free range and that’s why we had to split it.
However, let’s now consider the case when the request size is much closer to the free range size we found for the allocation, as shown on this picture:
In this case we cannot split the free range in two as we did before because the space that we will have left after returning one part of the range back to the caller will be too small to store the struct free_list_node and therefore we will not be able to add it to the list.
Fortunately, there is no requirement that we have to return the exact size the caller requested. We can return slightly more - that’s perfectly acceptable. It might look like a waste of memory and in a sense it’s. However the alternative might be even bigger waste if we cannot find the free range large enough to be split. If we only allow splitting free ranges in this case the allocation will fail and the memory will stay unusable anyways.
So all-in-all, we need to handle the case when we cannot split the free range we found in two. In this case we can remove the whole range from the list of the free ranges. After satisfing such an allocation request we will end up in the state shown on the picture below:
In the toy example on the pictures we allocated all the memory with just two allocation requests. In practice there might be many more requests required to exhaust all the memory.
However to get some intution of the aglorithm even a toy example is enough. The example should give you an idea of two cases that we need to consider:
• when we can split the free range we found
• when we cannot split the free range we found.
This example also should give you an idea of the initial state (when we have one contigous free range of memory) and the “end state”, when we allocated all the available memory.
Pedantic readers may say that it’s not an accurate presentation of the algorithm and that would be true, but again that’s not the final algorithm, so I’m cutting some corners.
## Freeing
In the example above we saw two allocations, now let’s consider the reverse operation of freeing previous allocated memory. Let’s start from freeing the first range that was allocated.
free_memory operation takes as a parameter the pointer that allocate_memory previously returned. I briefly alluded that it should be possible to find out the actual size of the freed region from just the pointer, so here I assume that the size of the freed region is known as well.
NOTE: Don’t worry, I will show the actual implementation of this trick below, so for now let’s just assume that we indeed know the size of the freed region.
The basic logic of the free_memory operation is that we need to initialize the struct free_list_node at the beginning of the range and add it to the free list that free_list_head points to.
So in the end after the operation has been completed the state should look similar to the one shown on the picture below:
Unlike with the alloca_memory, free_memory operation doesn’t really have multiple cases that needs to be considered. One thing worth mentioning though is that to be able to initialize struct free_list_node at the beginning of the freed memory range, this memory range must be large enough. It’s the same constrain that I covered when we considered allocation operation above.
This puts an additional limitation on the allocation operation - we cannot allocate regions that are smaller than sizeof(struct free_list_node). Otherwise it would not be possible to perform the free_memory operation later.
Let’s now look at the state after we free the second region of memory:
The freed region was just added at the beginning of the free ranges list. It was added at the beginning because it’s the simplest thing to do.
## Some thoughts
So what can we learn from this rather informal example?
Since we are looking at the algorithms here it’s worth considering the complexity of the solution. As you can see the potential runtime complexity of the allocation depends on the size of the free list and is $$O(n)$$, where $$n$$ is the size of the list.
Sometimes we can find a large enough free range without scanning the whole list, but it’s not guaranteed. And in the worst case to discover that there isn’t any large enough regions we have to go through the whole list. So we probably would like to keep the list as short as possible.
The complexity of the freeing function is $$O(1)$$, so we cannot really improve on that. That however also means that if we modify our algorithm we have plenty of room to make it worse.
Another observation that can be made is that we started the example with one big free region, but after freeing all the memory we allocated we ended up with two smaller regions of memory. So as it the algorithm was shown so far regions of memory can gradually get smaller and smaller in size.
That has at least two downsides. One of them is that it makes the list of free ranges longer. And as was noted above the size of the list directly affects the allocation runtime complexity.
The other downside is that it limits the ability to allocate large contigous ranges of memory. There is no guarantee that the allocator has enough memory to satisfy the request to begin with, but in our case the problem is that even when we in fact have large enough free contigous memory range it may happen that it was split in multiple smaller regions and we still cannot satisfy the request and that’s problematic.
Finally, in the shown example, the free ranges in the final free list ended up to be sorted by the starting addresses. It’s worth noting that’s it’s a pure coincedence. Consider for example the case when the same regions are freed in the different order.
# Merging free memory ranges
As was shows above idenfinitely splitting memory ranges is problematic. So I now will consider how the neighbor free ranges can be joined together in bigger free memory ranges when possible.
There are quite a few options to choose from. For example, we can maintain the free ranges in the list ordered by the starting address. This way, when we free a memory range, we, using the sorting property, can find its neighbours and check if they are free. If the neighbors are free they can be joined together in one big free range.
There are multiple options to maintain free ranges ordered. For example, we can use an ordered data structure instead of a linked list. Some kind of a balanced binary search tree (AVL, RB, Splay, Treap, etc) would do the trick.
Using balanced search tree, in a way, will not affect the runtime complexity of the allocate_memory operation. It’s still $$O(n)$$ from the number of elements in the tree, however it will help us to reduce the number of ranges we have to consider, so it’s not exactly aples to aples comparision.
When it comes to the free_memory operation it’s pretty clear that we will see a degradation from $$O(1)$$ to $$O(log n)$$ or worse depending on the selected structure. It’s degradation no matter how you look at it, even if you consider that $$n$$ is smaller.
Alternatively, instead of mantaining free memory ranges ordered we could just scan through the free ranges in the free_memory operation and find the free neighbor memory ranges if they are in the list. Compared to an ordered search tree it’s clearly a degradation from $$O(log n)$$ to $$O(n)$$.
Can we do better than a ordered search tree?
# The key observation
So far we only kept track of the free memory ranges because that was the only thing we needed. All the allocated memory ranges were not recorded in the metadata in any way.
At the beginning of each free memory range we store a structure that I previously called struct free_list_node, however there was no such thing for the allocated memory ranges.
Returning back to the problem of finding neighbor free ranges we had to search for the ranges in the list of free nodes. However, if the next range in memory, after the one we are actually trying to free, is already free, then at the very beginning it must have the struct free_list_node:
This gives us an easy way to get the pointer to the struct free_list_node of the next range quickly without looking at the free list/tree at all. Let’s say we are freeing a range of size size that begins at address addr. Then the address of the struct free_list_node of the next range in memory is just $$addr + size$$ - a simple arithmetic operation.
However, this trick only works for the next range, but not for the previous one, since there is no way for us to know the size of the previous range to find where it begins and, therefore, the struct free_list_node.
To make the matters worse it also only works if the next range is free, otherwise whatever pointer we get this way will not point to a valid struct free_list_node.
And that gives us an important clue. What if we could fix our memory layout in such a way that free and allocated memory range always have some kind of a helper structure both at the beginning and at the end.
This structure must provide us with the information about the size of the range and whether it’s free or not. It might look something like this:
// We put this structure at the beginning of each memory range (free or busy).
// From the pointer to the header we can find a corresponding struct footer
// using simple arithmetic operations with the pointer to the header,
// header::size field and size of the struct footer.
size_t size;
bool free;
};
// We put this structure at the end of each memory range (free or busy).
// Similarly, from the pointer to the footer we can find a corresponding struct
// header using just simple arithmetic on the pointer to the footer,
// footer::size field and the size of the struct footer.
struct footer {
size_t size;
bool free;
};
That will allow us to find neighborhood ranges and check if they are free in $$O(1)$$. In order to merge the free ranges together however we also should be able to remove them from the free list.
Unfortunately removing elements from a singly linked list requires updating the previous element in the list and to find the previous element in the free list we will have to scan it from the beginning.
That kills all the benefits of our observation and also gives us a clue how to fix the problem: where singly linked list is not enough, doubly linked list will do the trick.
If instead of a singly linked list of free ranges we will maintain the doubly linked list instead we will be able to remove an element from the list by having just the pointer to the element we want to remove in $$O(1)$$.
# Implementation
We now have all the basics for an algorithm that avoids the kinds of memory fragmentation that we discussed above and at the same time has the same runtime complexity:
• $$O(n)$$ for allocate_memory, where n is the number of contigous free memory ranges available to the allocator
• $$O(1)$$ for free_memory.
Let’s take a look in a little bit more details at the implementation.
I want to start with some helper structures and functions. As discussed the algorithm requires the use of doubly linked lists, so let’s cover them since the implementation of the linked list can be rather generic and separate from the the allocator algorithm itself.
Here I will implement doubly linked list with a dummy head node - a quirky long name. Let’s break it down piece by piece.
First of all it’s a doubly linked list. That would mean that each node in the list will somehow point to the previous and the next node in the list. That’s the important property that allows us to delete a node from the list in $$O(1)$$ by having just the pointer to the list.
NOTE: I should probably say that the operation is not removing the node from the list, but removing the node from a list (whatever list it belongs to if at all).
Here is how the structure describing the node of a doubly linked list might look:
struct list_node {
struct list_node *next;
struct list_node *prev;
};
To understand the part about the dummy head node it might be useful to consider what problems we might face when working with doubly linked lists and with linked lists in general.
A node of double linked list as was mentioned above contains pointers to the next and previous elements in the list. However there are some corner cases that needs to be considered - what if in the list there is no previous and/or next element.
Dealing with such corner cases is quite cumbersome and error prone, so it makes sense to exercise a bit of engineering lazyness and consider how those corner cases can be avoided.
One way to avoid corner cases is to make sure that the list always contains at least one element. Then we can link the next pointer of the last node of the list to point to the first element of the list. Similarly we can link the prev pointer of the first node in the list to point to the last node of the list.
To achieve that we can introduce a dummy element that is always in the list:
struct list_head {
};
{
}
When traversing the list we naturally will have to skip this dummy element, but that’s quite easy to do. In return our list operations become quite simple because they don’t need to deal with all the corner cases.
Now all we need need is to define a couple of function:
• a function to remove a node from the list
• a function to add a node to the list.
Removing an element from the list is quite simple now:
void list_remove(struct list_node *node)
{
struct list_node *prev = node->prev;
struct list_node *next = node->next;
prev->next = next;
next->prev = prev;
}
Similarly adding a node to the list is not complicated at all:
void list_add_after(struct list_node *prev, struct list_node *node)
{
struct list_node *next = prev->next;
node->prev = prev;
node->next = next;
prev->next = node;
next->prev = node;
}
Linked lists as defined here cannot store any data, however since I’m giving example in C language we can use some dirty tricks to circumvent this limitation, I will show you what I mean by that further down.
## Data Alignment
Before we move any further it’s worth stepping back for a second to look at one practical limitation for dynamic memory allocation. You see, with the current problem statement, when we allocate memory we have no idea what this memory is allocated for.
For example, it can be allocated to store one-byte character strings, an array of floating point numbers or an array of structures. Those structures themselves can contain strings, numbers, other structures, etc.
Why does it matter what we store in the dynamically allocated memory? Well, what we store there tells us how the data will be accessed. For example, if you store one-byte character strings in that data, then you likely will access it one byte at a time. On the other hand if you’ll store an array of double precision floating point numbers there, then you will be accessing it eight-bytes at a time.
And here is where we might see a potential problem. The modern hardware memory has a simple interface. You give processor an address and it fetches content of the memory at that address - nithing really complicated. However at the same time the hardware memory has complicated layered implementation:
• there is the memory itself
• there is some memory controller that handles read/write operations
• there are multiple levels of caches separating the processor and the memory itself.
Caches typically cannot work with arbitrary memory chunks. Typically, the cache would operate in fixed size blocks often referred to as cache lines. Of course, the size of the cache line depends on the specific hardware, but to give you an idea the size of the cache line is on order of few tens of bytes, like 32 bytes or 64 bytes.
Besides that, to make things even more complicated, each cache line can only store data from certain memory addresses. For example, if the cache line is 32 bytes long it can only store data from addresses aligned to the 32-byte boundary.
NOTE: When we say that the address is aligned on the boundary X it means that the address must be equal to 0 by modulo X: $$addr \equiv 0 \left(mod X\right)$$.
Why am I talking about that? Imagine if you’re accessing an 8-byte long value that crosses the boundary between two cache lines. In this case instead of actually loading just one cache line from memory to the cache you’d need to load two neighbor cache lines for just that one 8-byte long value. So accessing badly aligned data might be less effective.
Some architectures can go as far as to forbid unaligned accesses all together. Typically it comes in a form of natural alignment requirement. For example, if you want to access data in one 8-byte read or write operation then the address of the data must be aligned on the 8 bytes boundary, similarly for 4-byte long accesses the data have to be aligned on the 4 bytes boundary, etc.
If we don’t know how the data will be accessed we have to return the data that is aligned to the largest boundary that makes sense. Further in this post I will assume that the largest alignment that makes sense is 8 bytes and will consider that in my implementation.
The 8 bytes is not completely arbirary and it’s enough for many use-cases, however there are known use cases where this boundary is not enough.
Maybe a better solution for this problem is to change our interface slightly and take the require alignment as an argument of the ‘allocate_memory’ function. While it makes sense it also pushes part of responsibility on the caller of the allocate_memory. Besides that the allocate_memory and free_memory is the interface that is commonly used in practice (for better or worse), so I stick to it in the introductionary post like this.
It’s not all bad, using the allocate_memory and free_memory interface as it was described above you can also implement a memory allocation routine that will take the alignment as an argument. So the presented interface is not really that limiting.
Now let’s return to the implementation of the memory allocation algorithm. Above I’ve already presented how the header and footer of memory ranges may look in general, now let’s adjust them to use doubly linked lists:
struct header {
size_t size;
bool free;
};
struct footer {
size_t size;
bool free;
};
Note how I’ve put the struct list_node described above at the beginning of the struct header. This guarantees that that pointer to the link field will match the pointer to the struct header that this field belongs to.
That allows us to use this type-unsafe trick to convert from the struct list_node pointer to the struct header pointer:
struct list_node *link_ptr = /* something */;
Since both pointers actually contain the same value all we need is to change the type. Naturally, it’s not a generally safe trick, so for this to work correctly you need to know for sure that the pointer to struct list_node actually points to the link field inside some struct header instance.
NOTE: If you feel bad about such type-unsafe tricks don’t be. The way the problem is stated there is no way to avoid it. You are starting with a range of untyped memory to begin with, so some type usafe convertions have to happen at some point if you want to store any kind of structure inside this memory.
Now, let’s introduce a few helper functions to manipulate struct header and struct footer pointers. I will start from introducing a few helper function to perform alignment:
uint64_t align_down(uint64_t addr, uint64_t align)
{
return addr & ~(align - 1);
}
{
return align_down(addr + align - 1, align);
}
Both functions assume that align argument contains a power of two. It might seem as not a particularly generic solution, but the matter of the fact is that most of the time alignment requirements are powers of two.
With that out of the way, we need to be able to perform a few operations with header and footer. We need to be able to find the header of the next memory range in memory by the footer of the previous one:
struct header *next_header(struct footer *footer)
{
uint64_t addr = align_up((uint64_t)footer + sizeof(*footer), ALIGNMENT);
}
The opposite operation, finding the footer of the previous range in memory by the header is also needed:
struct footer *prev_footer(struct header *header)
{
}
Both implementation above make sure that the pointers they return are properly aligned to the ALIGNMENT, which as I mentioned before is 8 bytes.
There are another two useful operations with header and footer that we will need. Those are finding header of the memory range having a footer of that range and vice versa:
struct header *matching_header(struct footer *footer)
{
(uint64_t)footer + sizeof(*footer), ALIGNMENT) - footer->size;
}
{
ALIGNMENT);
}
And on this note we are done with the operations on headers and footers. Let’s move to the implementation of the add_memory_pool.
## add_memory_pool
When implementing add_memory_pool we will have to perform some safety checks on the given memory range. I will skip those for simplicity and will just assume that the given memory range is properly aligned and big enough to be of any use.
Additionally, our algorithm depends on begin able to access the header and the footer structures of the memeory ranges located next in memory. That creates a possibility of a corner case. What if the range we are working with is at the beginning or at the end of the memory pool. In this case there will be no next or no previous memory range.
I’m dealing with this problem by inserting fake footer at the beginning and header at the end of each memory pool added using add_memory_pool. The size stored there is not important, but free must contain false. This way I guarantee that there is always the next footer or the previous header.
Let’s see how it works:
// This is the head of our free list, we need to initialize it properly before
// begin able to use it.
static void setup_free_list()
{
}
}
{
const size_t metasz =
+ align_up(sizeof(struct footer), ALIGNMENT);
struct footer *footer = NULL;
struct footer *dummy_footer = NULL;
// First make sure that the free list has been properly initialized
setup_free_list();
// We put a fake footer at the beginning of the range and mark it as busy
// so every time we look at it we would think that the previous memory
// range is busy.
dummy_footer = (struct footer *)begin;
dummy_footer->free = false;
// Similarly we put a fake header at the end.
// Besides the fake header and footer we also need to create actual header
// and footer for the free memory range we will add to the pool.
header->size = (uint64_t)end - (uint64_t)begin - metasz;
footer->free = true;
}
Now we know how to add new memory ranges to the memory allocator. The implementation might look somewhat complicated, but it basically just initializes a couple of struct header and struct footer structures and that’s it. All the complexity comes from the address arithmetics.
## allocate_memory
Before we look at the actual allocation implementation let’s introduce another helper function. A function that given struct header returns a pointer to the data that can be used:
static void *data_pointer(struct header *header)
{
}
You see we cannot just return a pointer to the struct header for the caller to use. If they overwrite the content of struct header we created our algorithm will not work as it depends on it. So instead we need to return a pointer pointing after the struct header.
Another way to look at it is that allocation request for size bytes actually allocated size bytes plus whatever memory we need to store struct footer and struct header.
Now when that is out of the way let’s take a look at the actual allocation:
void *allocate_memory(size_t size)
{
const size_t metasz =
+ align_up(sizeof(struct footer), ALIGNMENT);
const size_t minsz = metasz + ALIGNMENT;
// We adjust the allocation size to be ALIGNMENT bytes aligned, to keep
// everything aligned.
size = align_up(size, ALIGNMENT);
for (struct list_node *ptr = head->next; ptr != head; ptr = ptr->next) {
struct footer *new_footer = NULL;
// The range is too small to serve the request. We need at least size
// bytes plus memory requires for the footer and header.
if (header->size < size + metasz)
continue;
// We have two cases to consider: one when we can split the range in
// two and another when the range is too small to be split. This is
// the second case, when we cannot split the range. In this case we
// remove it from the free list and return the whole range to the
// caller.
if (header->size < size + metasz + minsz) {
}
// The following code handles the case when the range can be split in
// two. I'm returning the caller the tail of the range and the head
// stays in the free list, but with the reduced range size.
// Firstly reduce the size of the free range and create an appropriate
// footer, since the footer will have to move to a different place when
// the size changes.
footer->free = true;
// Secondly we create a new header and footer for the tail of the range
// that will be returned to the caller.
new_footer->free = false;
}
// If we didn't find any suitable range then we return NULL as an indicator
// of a failure.
return NULL;
}
The implementation is long, but not really complicated. It’s just a few dozens of lines of code and most of those are comments. All the magic will be happening in the free_memory function however, so let’s move to it.
## free_memory
Again we will start the free_memory implementation with introduction of a helper function that performs an operation opposite to the data_pointer function introduced above.
You see the free_memory function takes a pointer we returned to the user as an argument. We don’t care about the pointer and want struct header pointer instead. That’s where all the relavant information is stored, including the size of the allocated range:
static struct header *data_header(void *ptr)
{
}
And now the actual implementation of the free_memory function:
void free_memory(void *ptr)
{
// Footer and header of the neighbor memory ranges
// If the range in memory next to the one we are freeing is also free then
// we can merge them together. This branch handles this merging logic.
if (next->free) {
struct footer *next_footer = matching_footer(next);
// We need to remove the next memory range from the list of free ranges
// and add it's memory to the one we are freeing. Since we change the
// size by attaching to the end of the range the footer pointer changes
// as well.
footer = next_footer;
}
// This branch handles the case when the previous range in memory is free
// and therefore have to be joined together with the region we are freeing.
// The logic is very simmetrical to the case above.
if (prev->free) {
}
// We merged all the ranges together and now it's time to mark the whole
// range free and add it to the list of the free ranges.
footer->free = true;
}
And that’s it. We have a working implementation of a rather generic memory allocator.
# Concurrency
Now when we have a complete implementation let’s take another step back and look at a few other practical aspects that are of importance for a real dynamic memory allocator that were not considered yet.
One of them is concurrency. It’s not a supririse that many programs nowadays benefit from hardware concurrency for performance and from preemtive multitasking for other reasons (like creating responsive graphical user interfaces).
All-in-all, concurrency is important and the implementation above cannot serve concurrent allocate_memory and free_memory requests correctly. Naturally there is a variety of ways to adjust the implementation to handle concurrency.
Probably the simplest way is to guard allocate_memory, free_memory and add_memory_pool operations with a mutex. That definitely will address the correctness concern. However such a solution to an extent will loose the benefits that concurrency provides, so it’s worth at least considering alternatives.
Another approach is to maintain multiple independent allocators, say one allocator per thread. So far the implementation depended on the global state, but it doesn’t have to be the case. We can change the interface a little bit to take an allocator structure as a parameter of all the functions.
This way we can have multiple truly independent operations. However if the threads have to share data between each other we might still have a problem and require mutexes to protect allocators.
For example, we have to handle a case when a memory range was allocated in one thread and then freed in another. In this case multiple threads will have to use the same memory allocator (the thread that frees a memory range have to free it using the same allocator the range was allocated from).
# Best-fit vs First-fit
In the implementation above I allocated memory from the first free range large enough to handle the request. That’s not the only possibility. After all, since the allocation requires traversing the list anyways, we can consider preferring one range over another when serving allocation requests.
The strategy the presented implementation used is called, unsurprisingly, first-fit strategy. There is another strategy commonly discussed in the literature called best-fit.
The best-fit strategy requires finding the minimum free range in the list that can serve the allocation request. Intuitively this approach might sound appealing as it avoids splitting large memory ranges in smaller ones when it’s not necessary, thus it might seem that it should reduce the memory fragmentation.
In practice however depending on the application and the load on the dynamic memory allocator best-fit may or may not be better than the first-fit. So you need to be careful with applying intution to complicated random systems. To scare you even more, Prof. Donald Knuth in the first volume of “The Art of Computer Programming” presents the results of Monte Carlo simulations comparing best-fit and first-fit strategies, comes to the conclusion that in all the experiments he conducted the first-fit strategy performed better.
# Optimizations
While there is very little we can do to improve the runtime complexity of the presented algorithm, there are some optimizations that were shown to actually work in practice. One rather simple one is easy to understand intuitively and so I’m presenting it to you here.
The runtime complexity of the allocation alogrithm is determined by the time it takes to scan through the free list until we find a large enough free range. Let’s say that most of the time we have to skip X first entries in the list until we find the large enough range. If the X is large enough the allocation will be slow.
On the other hand if we consistently have to skip X entries at the beggining of the list it means that they are likely too small to satisfy most of the allocation requests.
When we decide on whether we want to split the free memory range or returning the whole range to the caller in the implementation above I used the following criteria:
if (header->size < size + metasz + minsz) {
/* return the whole range to the caller */
}
The criteria was based on the limitations of the algorithm mostly. We cannot split the range if the part we will return to the free list is not small enough to contain the header, the footer and also have enough memory to satisfy the minimum possible allocation, which in our case was ALIGNMENT bytes.
We can actually increase the threshold used in this criteria to avoid addint to the list ranges that are unlikely to be big enough to satisfy an allocation request. That should improve the peformance at the cost of spending more memory.
That’s one way to deal with a pathalogical case I outlined above. However there is another rather simple way. If most of the time we have to skip X entries at the beginning of the list, let’s just do it once and remember the position in the list where we stopped the last time when we successfually allocated memory.
There is no way to over and over skip small ranges at the beginning of the list if we can do it once and remember where we stopped. For the next allocation we can start not from the beginning of the list, but from the position we stopped at the last time.
Implementing this trick requires some care since a free_memory call may invalidate the position we remembered in the last allocate_memory call, however since this trick is only for runtime optimization it’s easy to detect inside the free_memory function if the operation is going to affect the rememebered position in the free list and reset it in $$O(1)$$ time.
There are a few things worth mentioning to the reader who might consider checking the repository.
First of all on GitHub the implementation of the memory allocator lives in bootstrap/alloc.h and bootstrap/alloc.c if you ever want to read the code.
Also, in the implementation on GitHub I use slightly different names for the functions and provide a few more additional functions that are not covered in this post and are not essential for the algorithm.
Aside from that I’d like to share a few thoughts on the algorithm itself. It’s worth asking if the end algorithm is good and whether the end implementation is good?
Well, they are better then even more simplistic approach that the one I started this post with, so it’s clearly not the worst possible algorithm, but that’s a low bar.
One way to look at whether the algorithm and implementation are good is whether they serve their purporse. If the pruporse of the memory allocation in your case is a simple, not a performance critical memory allocator for a relatively small amount of memory, then this algorithm is good enough. That’s however isn’t a particularly high bar either.
Another way to look at whether the alogirthm and implementation are good is to ask if there is anything better. And the answer to this is a resounding yes. There are much better dynamic memory allocation algorithms at least in terms of performance and I’d want to cover some of them in one of the future posts.
tags: algorithms - dynamic-memory-allocation
|
2022-08-14 06:35:04
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2880805730819702, "perplexity": 809.8410380426393}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571996.63/warc/CC-MAIN-20220814052950-20220814082950-00369.warc.gz"}
|
https://depth-first.com/articles/2008/12/03/reading-smiles-with-mx/
|
The latest release of MX, the Java toolkit for cheminformatics, now supports reading a subset of SMILES strings. Although incomplete, full support for this feature is planned within a few releases.
To get an idea of how to use the new SMILES reader, we can use interactive JRuby. Assuming we've downloaded the mx-0.105.0 jarfile to our working directory, we can use:
jirb
irb(main):001:0> require 'mx-0.105.0.jar'
=> true
=> "Br"
|
2020-02-24 09:10:11
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.24584633111953735, "perplexity": 7946.5029376812245}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145910.53/warc/CC-MAIN-20200224071540-20200224101540-00323.warc.gz"}
|
https://math.stackexchange.com/questions/1224629/predicate-logic-and-calculus
|
# Predicate Logic and Calculus
Question of the week came up in my schools logic club but there is not much information to it. Here is the question:
Show that
$$\exists x\,[R(x)\wedge \lnot Q(x)],\ \forall x\,[P(x)\to Q(x)],\, \forall x[R(x)\to(P(x)\vee S(x))] \models \exists x\,[R(x)\wedge S(x)]$$ Describe in details all your work
Now I know that it does not want us to use formal deduction, so what method do they want us to use then? Formal Deduction is the only way I am familiar with at the moment but apparently that has been ruled out as a possible method.
Start with $\exists x(Rx\land\neg Qx)$ and instantiate, i.e. infer $(Ra\land\neg Qa)$.
Since $\forall x(Px\to Qx)$, we have $Pa\to Qa$. But since $\neg Qa$, then necessarily $\neg Pa$.
Finally, $\forall x(Rx\to(Px\lor Sx))$ gives $Ra\to (Pa\lor Sa)$. And since $Ra$ and $\neg Pa$, then necessarily $Sa$.
$$Ra\land\neg Qa\land\neg Pa\land Sa\to\exists x(Rx\land Sx)$$
Note that what the question asks to have proved is not a "$\vdash$" statement, but a "$\vDash$" statement -- thus, you're being asked to prove that any structure that satisfies the three premises will also satisfy $\exists x(R(x)\land S(x))$.
One way to do this would indeed be to present a formal proof of the "$\vdash$" entailment, and then appeal to some known result that the proof system you're using is sound. But apparently that's not what you're supposed to do.
The other option is to reason directly (and informally, that is, in everyday mathematical style) about $\vDash$, and what it means for the various formulas to be satisfied in a structure:
Given a structure $(\mathfrak M,P_{\mathfrak M},Q_{\mathfrak M}, R_{\mathfrak M},S_{\mathfrak M})$ which satisfies such-and-such we need to find an $x\in\mathfrak M$ such that $x\in R_{\mathfrak M}$ and $x\in S_{\mathfrak M}$. Because $\mathfrak M\vDash \exists x(R(x)\land\neg Q(x)$ we know that $\mathfrak M$ contains an $x$ such that $x\in R_{\mathfrak M}$ and $x\notin Q_{\mathfrak M}$. (And so forth ...)
|
2021-08-04 20:14:20
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.880878746509552, "perplexity": 180.09412153394678}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154897.82/warc/CC-MAIN-20210804174229-20210804204229-00035.warc.gz"}
|
http://imathtest.net/math/block/math-12-old-provincial-exam-bank-158.html
|
# Math 12 Old Provincial Exam Bank
Subject: ; Class: ; with 15 questions; test in 15 minutes; update 27/04/2018
Time 15 minutes Time to take the test Start exam Click button start to test. Guide to the test Subjects Math test Update 27/04/2018 Class Grade 12 Number of questions 15 View 946 Tested 1
Question 1.
Evaluate: $$tan{5\pi\over3}$$
(A) $$-{1\over \sqrt3}$$ (B) $${1\over \sqrt3}$$ (C) $$-\sqrt3$$ (D) $$\sqrt3$$
Question 2.
The terminal arm of angle $$\theta$$ in standard position intersects the unit circle at the point (m, n). Which expression represents $$cot\theta$$ ?
(A) m (B) n (C) $$n\over m$$ (D) $$m\over n$$
Question 3.
If the graph of the function shown below has the equation y = acosb(x - c) + d , determine the value of d.
(A) 3 (B) 5 (C) 6 (D) 9
Question 4.
Determine an equation of an asymptote of y = sec 3x .
(A) $$x = {\pi \over6}$$ (B) $$x = {\pi \over3}$$ (C) $$x = {2\pi \over 3}$$ (D) $$x = \pi$$
Question 5.
Which expression is equivalent to $$sin(\pi + 2x)$$ ?
(A) 2cos2x - 1 (B) 1 - 2cos2x (C) 2sinxcosx (D) -2sinxcosx
Question 6.
Convert $$8 \pi \over 3$$ radians to degrees.
(A) 60o (B) 120o (C) 240o (D) 480o
Question 7.
Determine the minimum value of the function y = -3sin2x + 4.
(A) -7 (B) -3 (C) -1 (D) 1
Question 8.
The terminal arm of angle $$\theta$$, in standard position, passes through the point (-2, 9). Determine the value of sin $$\theta$$.
(A) $$-2\over{\sqrt77}$$ (B) $$9\over{\sqrt77}$$ (C) $$-2\over{\sqrt85}$$ (D) $$9\over{\sqrt85}$$
Question 9.
Simplify: $$4cos^26x - 2$$
(A) 2cos3x (B) 4cos3x (C) 2cos12x (D) 4cos12x
Question 10.
Evaluate: $$log_\sqrt7 7^3$$
(A) $$2 \over 3$$ (B) $$3 \over 2$$ (C) 6 (D) 9
Question 11.
Solve: $$log_2(log_9x)=-1$$
(A) $$1\over81$$ (B) $$1\over3$$ (C) 3 (D) 81
Question 12.
Solve: $$5^{x+1} = 2(3^{2x})$$
(A) $$x={-log5\over1- 2log6}$$ (B) $$x={-log5\over log5 - 2log6}$$ (C) $$x={log2 - log5\over1-2log3}$$ (D) $$x={log2-log5\over log5-2log3}$$
Question 13.
Determine the number of pathways from point A to point B if only moves to the right and down are permitted.
(A) 18 (B) 19 (C) 23 (D) 47
Question 14.
The smallest positive solution of tanbx = 2 is x = 0.3. Determine the general solution of tanbx = 2.
(A) $$0.3 + 2n\pi$$, n is an integer (B) $$0.3 + 2bn\pi$$, n is an integer (C) $$0.3 + n\pi\over{b}$$, n is an integer (D) $$0.3 + 2n\pi\over{b}$$, n is an integer
Question 15.
The y-intercept of the function y = f(x) is 5. Determine the y-intercept of y = f(x) + 3 .
(A) -2 (B) -8 (C) 8 (D) 2
|
2019-10-23 23:06:16
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.40990933775901794, "perplexity": 1514.2674155454658}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987836368.96/warc/CC-MAIN-20191023225038-20191024012538-00478.warc.gz"}
|
https://vknight.org/unpeudemath/code/2016/03/07/Property-based-testing-and-finding-a-bug.html
|
I really have fallen in love with test driven development and as much as is reasonable stick to it. This mainly revolves around writing Python unit tests, which makes programming a lot easier (for me). In Namibia I was lucky enough to meet and chat with David MacRiver, the main developer of the hypothesis package. Before that the Axelrod project had already incorporated some of what hypothesis is for: property based testing. This blog post will briefly describe what that is, how it took some hours of sleep away from me (in a good way) and why it’s particularly great for research software.
Property based testing
I’m slowly slowly learning that hypothesis is ridiculously clever but at a very basic level what it allows you to do is take 1 test and instead of testing a single example, test a whole range of parameters.
Here’s a simple example, based on a short demo I gave at the collaborations workshop 2016 in Edinburgh. You can find the slides I gave for that at this link: vknight.org/Talks/2016-03-22-Division-by-11-and-property-based-testing-with-hypothesis/.
Let us write a function that tests divisibility by 11 (assuming that we don’t know about % 11. To do this we’ll use the following property:
A number is divisible by 11 if and only if the alternating (in sign) sum of the number’s digits is 0.
For example: 121 is divisible by 11 because (1-2+1=0) (side note: if you’re about to yell at me in the comments please read the rest of this post).
Let’s save below in a file called main.py:
def divisible_by_11(number):
"""Uses above criterion to check if number is divisible by 11"""
string_number = str(number)
alternating_sum = sum([(-1) ** i * int(d) for i, d
in enumerate(string_number)])
return alternating_sum == 0
Now if we write out some examples we get expected behaviour:
>>> import main
>>> for k in range(10):
... print(main.divisible_by_11(11 * k))
True
True
True
True
True
True
True
True
True
True
But let’s write an actual test suite. Here’s a very basic unittest that we’ll save in test_main.py:
import unittest
import main
class TestDivisible(unittest.TestCase):
def test_divisible_by_11(self):
for k in range(10):
self.assertTrue(main.divisible_by_11(11 * k))
self.assertFalse(main.divisible_by_11(11 * k + 1))
# Some more examples
self.assertTrue(main.divisible_by_11(121))
self.assertTrue(main.divisible_by_11(12122))
self.assertFalse(main.divisible_by_11(123))
self.assertFalse(main.divisible_by_11(12123))
The above tests the first 10 numbers divisible by 11 (and that that number + 1 is not) and also some specific tests (121 and 12123).
We then run this by using the following:
$python -m unittest test_main . ---------------------------------------------------------------------- Ran 1 test in 0.000s OK At this point we could be very happy and proud of ourselves: we have tested well written software that can be shipped and used by researchers world wide to test the divisibility of a number by 11!!! This is how mathematics breaks. Let’s write a hypothesis test. We’ll write the following in test_property_main.py: import unittest import main from hypothesis import given # This is how we will define inputs from hypothesis.strategies import integers # This is the type of input we will use class TestDivisible(unittest.TestCase): @given(k=integers(min_value=1)) # This is the main decorator def test_divisible_by_11(self, k): self.assertTrue(main.divisible_by_11(11 * k)) The above is actually a simpler test than before: we’re only testing that a number that we know is divisible by 11 is in fact getting the expected output from our function. Let’s run it: $ python -m unittest test_propert_main
Falsifying example: test_divisible_by_11(self=<test_property_main.TestDivisible testMethod=test_divisible_by_11>, k=19)
F
======================================================================
FAIL: test_divisible_by_11 (test_property_main.TestDivisible)
----------------------------------------------------------------------
Traceback (most recent call last):
File "test_property_main.py", line 10, in test_divisible_by_11
def test_divisible_by_11(self, k):
File "/usr/local/lib/python2.7/site-packages/hypothesis/core.py", line 502, in wrapped_test
print_example=True, is_final=True
File "/usr/local/lib/python2.7/site-packages/hypothesis/executors.py", line 57, in default_new_style_executor
return function(data)
File "/usr/local/lib/python2.7/site-packages/hypothesis/core.py", line 103, in run
return test(*args, **kwargs)
File "test_property_main.py", line 11, in test_divisible_by_11
self.assertTrue(main.divisible_by_11(11 * k))
AssertionError: False is not true
----------------------------------------------------------------------
Ran 1 test in 0.058s
FAILED (failures=1)
We get an error! An right at the top we get the Falsifying example so we see that our function fails for k=19. For, k=19 the number being tested is $$19\times 11=209$$. That number is obviously divisible by 11 (by construction) but it’s alternating sum is in fact 11 which is not 0.
At this point, as described in this previous blog post about divisibility by 11 I can let slip the correct property for divisibility by 11:
A number is divisible by 11 if and only if the alternating (in sign) sum of the number’s digits is divisible by 11.
(There’s a simple algebraic proof of that at the older blog post.)
Now let’s modify our main.py (note that I’m using a slightly better modification than the lazy one I gave in the demo at the collaboration workshop):
def divisible_by_11(number):
"""Uses above criterion to check if number is divisible by 11"""
string_number = str(number)
# Using abs as the order of the alternating sum doesn't matter.
alternating_sum = abs(sum([(-1) ** i * int(d) for i, d
in enumerate(string_number)]))
# Recursively calling the function
return (alternating_sum in [0, 11]) or divisible_by_11(alternating_sum)
Running the tests:
\$ python -m unittest test_propert_main
.
----------------------------------------------------------------------
Ran 1 test in 0.043s
OK
As we hoped!
I think the above is a good example of why property based testing and tools like hypothesis are useful for research software.
It helped identify an error in my mathematical thought process, an error that I in fact thought I had tested properly (using the test_main.py tests). If it wasn’t for hypothesis I would have shipped code that redefined what it meant to be divisible by 11.
If you only came here to get a basic working example of hypothesis you can stop now :)
Obviously the above is a simple example but hypothesis recently found a bug in Axelrod (the python package for reproducing Iterated Prisoner’s Dilemma tournaments).
Axelrod creates tournaments of players picked from one of the (at time of writing) 123 strategies. Here is one of the tests that randomly selects strategies to create tournaments and check that they run:
@given(s=lists(sampled_from(axelrod.strategies),
min_size=2, # Errors are returned if less than 2 strategies
max_size=5, unique=True),
turns=integers(min_value=2, max_value=50),
repetitions=integers(min_value=2, max_value=4),
rm=random_module())
@settings(max_examples=50, timeout=0)
@example(s=test_strategies, turns=test_turns, repetitions=test_repetitions,
rm=random.seed(0))
def test_property_serial_play(self, s, turns, repetitions, rm):
"""Test serial play using hypothesis"""
# Test that we get an instance of ResultSet
players = [strat() for strat in s]
tournament = axelrod.Tournament(
name=self.test_name,
players=players,
game=self.game,
turns=turns,
repetitions=repetitions)
results = tournament.play()
self.assertIsInstance(results, axelrod.ResultSet)
self.assertEqual(len(results.cooperation), len(players))
self.assertEqual(results.nplayers, len(players))
self.assertEqual(results.players, players)
Without going in to precise details, at one point hypothesis found that two particular strategies could not play together. Here is the github issue that documented this. These were:
• Backstabber
• MindReader
These two strategies did in fact already play in a tournament together. The main tournament that includes all cheating strategies, but the side effect of that is that another cheating strategy played BackStabber first and overwrite it’s strategy (meaning it could then play nice with MindReader). Hypothesis however created a tournament were they were directly opposed and then our bug occured.
Anyway, it’s now fixed and the test includes a @example statement to ensure that the specific bug is in fact fixed: you can see this in the source code.
In practice, the recommendation is to use a combination of property based tests and traditional example based tests (as there are some specific things that need to be checked) but I think all test suites can be improved by using tools like hypothesis, in particular research software. With research software a robust testing framework helps not only test the code but also the ideas.
Finally, if you’re of the irc persuasion, I really recommend dropping in at #hypothesis on freenode. Here’s a direct link to the channel on irccloud: www.irccloud.com/invite?channel=%23hypothesis&hostname=irc.freenode.net&port=6697&ssl=1. Everyone there has always been very helpful.
|
2021-03-03 20:58:23
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5357463955879211, "perplexity": 2096.930689813102}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178367790.67/warc/CC-MAIN-20210303200206-20210303230206-00527.warc.gz"}
|
http://mathoverflow.net/questions/137101/surgery-diagram-for-the-seifert-weber-space
|
# Surgery diagram for the Seifert-Weber space
In every reference I see, the Seifert-Weber space is presented as an identification space (specifically identify opposite faces of a dodecahedron by a 3/10ths twist).
What I can't seem to find is a surgery diagram for this manifold. From its homology we know it doesn't arise as surgery on a knot. Its volume is approximately 11.1991. Checking with SnapPy there are a lot of candidate links in this range:
In [14]: len(filter(lambda x: x.num_cusps() >= 3,HTLinkExteriors[11.19:13.0]))
Out[14]: 1103
In [15]: len(filter(lambda x: x.num_cusps() >= 3,HTLinkExteriors[11.19:12.0]))
Out[15]: 365
In [16]: len(filter(lambda x: x.num_cusps() >= 3,HTLinkExteriors[11.19:12.5]))
Out[16]: 620
so my attempts at guessing have been unsuccessful.
One could calculate a surgery diagram by-hand from the Heegaard diagram that comes from the identification space and then try to simplify the result. This seems daunting to do by hand, but I'm also unaware of software for the task.
## Edit
I accepted Daniel's answer before working through the details. The process certainly gives the Seifert-Weber space a surgery on a manifold. The problem is that the branched-covers of $S^3$ over the trivial 2-component unlink are not themselves link complements, as in the case of a 1-component unlink, so the method in Rolfsen doesn't yield surgery on a link in $S^3$, but somewhere else. That somewhere else has non-separating 2-spheres in it, illustrated by this picture:
The cut disks are indicated by red and blue, the two light grey wireframe balls are the complements of the two knot components after cutting along the disks. As long as at least two of these gadgets are glued together light-to-dark, the yellow curve will pierce the solid grey 2-sphere only once.
-
The Seifert-Weber space is a 5-fold cyclic branched cover of the Whitehead link complement. One may obtain it in Snappea (or Snappy) by performing (5,0) surgery on each component of the Whitehead, then taking cyclic covers of order 5 (there is only one that is a manifold). I'm not sure how to convert this to a surgery diagram though. – Ian Agol Jul 18 '13 at 22:28
The volume would appear to be small enough so that you could drill curves out of the manifold (using SnapPy) and search for the resulting manifold in the HTLinkExterior census. Have you tried that? – Ryan Budney Jul 18 '13 at 23:33
@IanAgol, there are two distinct isometry classes of 5-fold covers of the Whitehead link (Theorem 3 in this paper). – Edgar A. Bering IV Jul 26 '13 at 0:16
@Edgar: yes, I was mistaken about the uniqueness. – Ian Agol Jul 26 '13 at 0:59
As pointed out by Ian Agol in the comments, the Seifert-Weber space is the 5-fold cyclic branched cover of the Whitehead link complement. You can therefore:
1. Untie the Whitehead link using $\pm 1$ framed surgeries along unknots ringing crossings (which induce crossing changes on the Whitehead link, while preserving $S^3$.) Actually you only need one surgery component, because a single crossing change unties the Whitehead link. You obtain a surgery presentation of the Whitehead link complement in the complement of a trivial 2-component unlink.
2. Lift the resulting surgery presentation to the 5-fold cyclic branched cover of the trivial 2-component unlink. You do this with a cut-and-paste construction. Cut along a disc bounded by each components of the unlink, and glue together five copies of your surgery tangle, end to end. The resulting link is a surgery presentation of the Seifert-Weber space.
This technique works to construct surgery presentations of any finite covering space of any link complement. You can find it explained in Rolfsen Knots and Links, Chapter 10.
Edit: As the OP points out in the edit to his question, the above process gives a surgery presentation for the Seifert-Weber space, but unfortunately not in $S^3$.
-
There is another way to draw a surgery diagram (this time in $S^3$) for the Seifert-Weber space, starting again from the description as a 5-fold branched cover over the Whitehead link $L$. Namely, you can use the algorithm of Akbulut-Kirby (Branched covers of surfaces in 4-manifolds. Math. Ann. 252 (1979/80), no. 2, 111–131). This starts with a connected Seifert surface for $L$, where you want to see the surface as a disk with bands. In the case at hand, the initial picture will be a little complicated, with something like 12 surgery curves. I think that you could do some simplifications, though.
Since we are talking about a link, rather than a knot, there is some ambiguity about which branched cover is being discussed. (This same point is relevant to the responses above as well.) The simpler version, to which the algorithm applies directly, is the branched cover corresponding to a homomorphism $H_1(S^3 -L) \to Z/5$ that takes both meridians to a given generator $1 \in Z/5$. Up to homeomorphism, I think that the only other choice is to send one meridian to $1$ and the other to $2$; I'm sure that the Akbulut-Kirby algorithm could readily be adapted to this case, but haven't thought out exactly how.
-
|
2015-09-03 17:37:55
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5821166038513184, "perplexity": 630.5280946383346}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645322940.71/warc/CC-MAIN-20150827031522-00242-ip-10-171-96-226.ec2.internal.warc.gz"}
|
http://mathoverflow.net/questions/106588/lsi-for-gaussian-measure-in-mathbbrd-mathbbzd/106641
|
# LSI for Gaussian measure in $({\mathbb{R}^d})^{\mathbb{Z}^d}$
I am looking for a reference: Does Gaussian measure satisfy Logarithmic Sobolev Inequality (LSI) in $$${\mathbb{R}^d}$$^{\mathbb{Z}^d}$. Thanks.
-
Can you clarify what space you're looking at? Do you mean the space of functions $\mathbb{Z}^d \to \mathbb{R}^d$, or equivalently a countably infinite product of copies of $\mathbb{R}^d$? – Mark Meckes Sep 7 '12 at 13:17
If so, and you mean standard Gaussian measure, then this is the same as asking about standard Gaussian measure on $\mathbb{R}^\mathbb{N}$, which should be dealt with in most standard references on LSIs. – Mark Meckes Sep 7 '12 at 13:19
It is the case of standard Gaussian measure on product of countably infinite copies of $\mathbb{R}^d$. Could you please mention any references on this? All I could find is, LSI for standard GM on $\mathbb{R}^d$ but not the product space. – Ahsan Sep 7 '12 at 14:46
As in Mark Meckes's comment, this is equivalent to log-Sobolev for standard Gaussian measure on $\mathbb{R}^\mathbb{N}$, which in turn follows immediately from the finite-dimensional case. In order to show the log-Sobolev inequality $$\int |f|^2 \ln |f| \le \int \|\nabla f\| + \frac{1}{2} \int |f|^2 \ln \int |f|^2$$ it is sufficient to prove it for smooth cylinder functions $f \in L^2(\mathbb{R}^\mathbb{N}, \mu)$, i.e. which depend on only finitely many coordinates. But for $f$ depending on $n$ coordinates, this is precisely the log-Sobolev inequality for standard Gaussian measure on $\mathbb{R}^n$. (Unlike the classical Sobolev inequality, there is no dimension-dependent constant!)
Leonard Gross's original paper introducing the log-Sobolev inequality already mentions the extension to infinite dimensions (though it does not work it out explicitly).
Gross, Leonard. Logarithmic Sobolev Inequalities. American Journal of Mathematics 97 No. 4 (Winter, 1975), pp. 1061-1083.
-
Thanks, Nate. I got busy and didn't get back to this in time. – Mark Meckes Sep 12 '12 at 19:37
|
2014-12-20 20:08:44
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9968873262405396, "perplexity": 1964.2283746215915}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802770371.28/warc/CC-MAIN-20141217075250-00038-ip-10-231-17-201.ec2.internal.warc.gz"}
|
https://quantumcomputing.stackexchange.com/tags/random-quantum-circuit/hot?filter=month
|
# Tag Info
One potential combo is $RY(\theta)$ on qubit 1, $CX$ from qubit 1 to qubit 2, then $RX(\pi)$ on qubit 2. This would be the following transformation: |0\rangle |0\rangle \mapsto (\cos \frac{\theta}{2} |0\rangle + \sin \frac{\theta}{2}|1\rangle)|0\rangle \mapsto \cos \frac{\theta}{2} |00 \rangle + \sin \frac{\theta}{2} |11\rangle \mapsto \cos \frac{\theta}{...
|
2020-09-29 04:33:02
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9976878762245178, "perplexity": 8178.1184698606685}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600401624636.80/warc/CC-MAIN-20200929025239-20200929055239-00719.warc.gz"}
|
https://clinmedjournals.org/articles/jrdt/journal-of-rheumatic-diseases-and-treatment-jrdt-2-033.php?jid=jrdt
|
## Novel Application of Behavioral Assays Allows Dissociation of Joint Pathology from Systemic Extra-Articular Alterations Induced by Inflammatory Arthritis
### Ann K Harvey1,3^, Mariah J Lelos2^, Claire J Greenhill1, Ashley T Jones2, Susanne P Clinch2, Michael J Newton2, Stephen B Dunnett2, Sean L Wyatt2, Anwen S Williams1,3# and Simon A Jones1,3#*
1Division of Infection & Immunity, School of Medicine, Cardiff University, Cardiff CF14 4XN, UK
2School of Biosciences, Cardiff University, Museum Avenue, Cardiff, CF10 3AX, UK
3Arthritis Research UK Biomechanics and Bioengineering Centre, Cardiff University, UK
^Joint first author
#Joint senior author
*Corresponding author: Prof. Simon A. Jones, Institute of Infection & Immunity, School of Medicine, Cardiff University, Heath Campus, Cardiff CF14 4XN, Wales, UK, Tel: +44 (0)2920 874112, E-mail: jonessa@cf.ac.uk
J Rheum Dis Treat, JRDT-2-033, (Volume 2, Issue 2), Research Article; ISSN: 2469-5726
Received: January 12, 2016 | Accepted: March 29, 2016 | Published: April 01, 2016
Citation: Harvey AK, Lelos MJ, Greenhill CJ, Jones AT, Clinch SP, et al. (2016) Novel Application of Behavioral Assays Allows Dissociation of Joint Pathology from Systemic Extra-Articular Alterations Induced by Inflammatory Arthritis. J Rheum Dis Treat 2:033. 10.23937/2469-5726/1510033
Copyright: © 2016 Harvey AK, et al. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.
Abstract
Introduction: Although rheumatoid arthritis (RA) is a disease of articular joints, patients often suffer from co-morbid neuropsychiatric changes, such as anxiety, that may reflect links between heightened systemic inflammation and abnormal regulation of the hypothalamic-pituitary-adrenal (HPA) axis. Here, we apply behavioral neuroscience methods to assess the impact of antigen-induced arthritis (AIA) on behavioral performance in wild type (WT) and interleukin-10 deficient (Il10-/-) mice. Our aim was to identify limb-specific motor impairments, as well as neuropsychological responses to inflammatory arthritis.
Methods: Behavioral testing was performed longitudinally in WT and Il10-/- mice before and after the induction of arthritic joint pathology. Footprint analysis, beam walking and open field assessment determined a range of motor, exploratory and anxiety-related parameters. Specific gene changes in HPA axis tissues were analyzed using qPCR.
Results: Behavioral assessment revealed transient motor and exploratory impairments in mice receiving AIA, coinciding with joint swelling. Hind limb coordination deficits were independent of joint pathology. Behavioral impairments returned to baseline by 10 days post-AIA in WT mice. Il10-/- mice demonstrated comparable levels of swelling and joint pathology as WT mice up to 15 days post-AIA, but systemic differences were evident in mRNA expression in HPA axis tissues from Il10-/- mice post-AIA. Interestingly, the behavioral profile of Il10-/- mice revealed a significantly longer time post-AIA for activity and anxiety-related behaviors to recover.
Conclusions: The novel application of sensitive behavioral tasks has enabled dissociation between behaviors that occur due to transient joint-specific pathology and those generated by more subtle systemic alterations that manifest post-AIA.
Keywords
Behavior, Arthritis, Inflammation, IL-10, Cytokine
Abbreviations
AIA: Antigen Induced Arthritis, HPA: Hypothalamic Pituitary Adrenal, mRNA - Messenger Ribonucleic Acid, IL-10: Interleukin-10, Il10-/-: Interleukin-10 Deficient, OFT: Open Field Test, qPCR: Quantitative Polymerase Chain Reaction, RA: Rheumatoid Arthritis, WT: Wild-Type
Introduction
Rheumatoid arthritis (RA) is an immune-mediated inflammatory disorder that manifests as painful articular disease and associated loss of joint function. Underlying the pathogenesis of RA is an imbalance in the complex network of pro- and anti-inflammatory cytokines, resulting in degenerative destruction of bone and articular cartilage. Whilst the pathophysiological symptoms of RA are cardinal signs of disease, there is evidence to suggest that a substantial burden of RA disease is linked to extra-articular comorbidities and psychosocial factors that may be mediated by systemic inflammatory processes [1].
To identify the extra-articular components of RA, it is necessary to validate novel methods capable of objectively assessing behavioral characteristics in inflammatory arthritis. Models of experimental arthritis with well-defined joint pathology [2-4], in combination with behavioral assessment, would enable classification of behaviors as dependent or independent of disease severity. Behavioral assessment within experimental arthritis research has, to date, been limited and focused primarily on gait evaluation [5-12]. Although open field test (OFT) techniques have been applied to experimental arthritis, these studies were limited by measuring activity at a single time-point [13,14] rather than in response to arthritis induction or progression. We propose that longitudinal assessment, using complementary tests, would deliver a more complete picture of behavior during experimental arthritis. Such an approach would reveal global behavioral attributes driven by more subtle systemic changes, which may manifest independently from joint-specific processes.
Important considerations in assessing behavior in rodent models of RA are the classification and severity of disease. Poly-arthritic models inherently affect the general state of health of the animal, rendering behavioral studies difficult to interpret. Mono-arthritic models, such as the antigen-induced arthritis (AIA) model [3], provide a more suitable framework for disentangling the relationship between disease severity and global behavior status in arthritis. Disease severity can be manipulated by disrupting cytokine profiles. For example, mice deficient in pro-inflammatory [15,16] and anti-inflammatory [4,17,18] cytokines present with differing degrees of synovial histopathology. Ultimately, these findings have led to the clinical introduction of biological drugs that target specific cytokine activities [19,20].
In addition to their important roles in mediating inflammatory processes, cytokines have been linked to mood disorders such as anxiety and depression [21-28] and implicated in neuroendocrine regulation of the hypothalamus-pituitary-adrenal (HPA) axis [29,30]. For example, relative levels of HPA hormones have been shown to predict response to biologic therapy in RA [30]. Secondly, mice deficient in anti-inflammatory interleukin-10 (Il10) exhibit depressive-like behavior [24]. Thirdly, the pro-inflammatory cytokine interleukin-6 (Il6) is able to stimulate the HPA axis [21], and can act at the level of the adrenal cortex [22]. Finally, high levels of pro-inflammatory cytokines have been measured in patients with depression [23,31] and neuropsychiatric changes, including anxiety, fatigue and depression, are prevalent in RA [32,33]. In summary, there is abundant evidence for bidirectional neuroendocrine-inflammation crosstalk in RA. However, the relationship between disease severity, behavior/mood and cytokine action remains unclear.
The aim of our study was to validate a novel longitudinal approach for behavioral evaluation of experimental arthritis in rodents, both during the acute phase of AIA and during recovery from the inflammatory processes, but prior to the onset of overt joint disease. By employing a range of sensitive behavioral tasks, we sought to identify distinct profiles of behavior that manifest either after the acute, transient joint swelling or as a result of more subtle systemic changes that manifest post-AIA. To achieve this, we first compared wild-type (WT) mice with and without AIA to identify behavioral tasks sensitive to AIA. Secondly, we compared WT to Il10-/- mice, which exhibited similar levels of swelling and joint pathology up to 15 days post-AIA as WT mice, but importantly, are thought to have different intrinsic immune responses [4]. Our decision to examine behavior up to, but not beyond, 15 days was based on our finding that joint pathology is comparable between WT and Il10-/- mice up to 15 days post-AIA [4]. However, after that, changes in behavior may be driven by overt joint degeneration and not by the more subtle, systemic immune or hormonal alterations that manifest after AIA. Lastly, we used qPCR analysis of HPA axis tissues to reveal systemic changes that may underlie the extra-articular behavioral changes evident long-term in Il10-/- mice post-AIA.
Methods
Mice
For the behavioral experiment, fourteen male WT C57BL/6 (Charles River, UK) and eight Il10-/- (C57BL/6 background, Jackson Laboratory, ME) were housed in groups of 3-5 and maintained on a 14:10-h light/dark cycle, with food and water available ad libitum. Experiments were performed on 8–12 week old mice.
A second cohort of male mice, consisting of 6 WT and 6 Il-10-/- mice, 8-12 weeks old, were utilised for the investigation of gene expression within the HPA axis 3 days post AIA injection. These mice were housed in identical conditions to those utilised for behavioral analysis.
All experiments using animals were conducted in compliance with the UK Animals (Scientific Procedures) Act 1986 and ethical approval was assured through approval of UK Home Office Project License PPL-30/2928.
Antigen induced arthritis (AIA)
Mice were immunized by subcutaneous administration of 100 μl methylated bovine serum albumin (mBSA; 1 mg/ml, Sigma-Aldrich) emulsified in complete Freund’s adjuvant (CFA, Sigma-Aldrich). Mice were injected i.p. with 200 ng of heat-inactivated Bordetella pertussis toxin (Sigma-Aldrich). The immune response was boosted after 7 days with identical administration of mBSA/CFA. Inflammatory arthritis was induced 21 days after the initial immunizations by intra-articular administration of 10 μl mBSA (10 mg/ml) into the right hind knee joint. For the WT-PBS group, 10 μl phosphate buffered saline (PBS) was injected into the knee joint. Swelling was monitored by measuring knee joint diameter using a POCO 2T micrometer (Kroeplin), and recorded as the difference between right (inflamed) and left (contralateral) diameters. Knee joints were harvested at day 15 post-AIA for histopathological assessment of disease severity (arthritic index).
Longitudinal behavioral assessment
A within-subjects design was chosen for this study, in order to observe the effect of arthritis onset within the same mouse. A within-subjects design is particularly robust insofar as a direct comparison can be made with and without arthritis within the same host. WT mice were randomly assigned to receive AIA or PBS. All Il10-/- mice received AIA.
All mice were habituated to the open field apparatus over two days prior to the start of behavioral testing (Figure 1A). To establish the baseline behaviors of WT and Il10-/- mice in open field tests (OFT) and balance beam assessments, all mice were tested 4 times, over a period of 18 days, prior to the onset of AIA. Two behavioral tests were conducted before the immunization with BSA, and two behavioral tests were conducted after the mBSA injections (Figure 1A). For footprint analysis, two sessions were conducted to obtain baseline data, at 3 days and 1 day pre-AIA induction. Behavior was found to be consistent over this 18 day pre-AIA period for both genotypes on all tasks (see results section below for statistical analyses).
Each day, mice were assessed in the same order: animal distress rating/weighing, balance beam, OFT, footprint analysis. Four successive time-points post-AIA were chosen to capture behavioral alterations associated with acute inflammation, followed by 3 further time points up to 14 days post-AIA. Thus, behavioral testing was performed over 4 sessions during the 18 day pre-AIA period and for 7 sessions over the 14 day post-AIA period (Figure 1A). Mice were perfused on day 15. The end-point, day 15, was determined by previously published data showing that up to day 15 post-AIA, Il10-/- mice were indistinguishable from WT mice on all measures of the arthritic index [4]. However, given that Greenhill et al. report that long-term differences in the extent of disease pathology emerge by day 28 post-AIA [4], we hypothesized that subtle, perhaps systemic, changes induced in the Il10-/- mice by the AIA may be revealed using the behavioral assays utilized here, which haven’t previously been detectable .
.
Figure 1: Novel application of combined behavioral measures to assess motor function and locomotor activity in experimental arthritis. (A) Schematic representation of longitudinal behavioral assessment during AIA. Blue arrows denote habituation to the behavioral tasks, 4 sessions of baseline measurement, and 7 testing days; (B) Significantly greater knee swelling in AIA-injected mice compared to PBS-injected mice; (C) No significant effect of PBS or AIA injection on stride length; (D) Latency to turn is not significantly different between the groups; (E) AIA-treated mice transiently demonstrate significantly longer latency to traverse the balance beam than PBS-treated mice; (F & G) AIA injection induced a significant transient increase in the number of foot slips made by both hind limbs; (H) Open field testing revealed a transient reduction in total distance moved and (I) time spent moving in the AIA-treated group compared to the PBS-treated group. Tracks show representative movement (black line) on day 2 post injection, when the AIA-injected group are significantly less active that the PBS-injected group. B–baseline testing day; Graphs represent mean ± SEM; Significant effect of Group: *p ≤ 0.05; Significant Group*Day interaction: $$p ≤ 0.01,$$$p ≤ 0.001. View Figure 1 . Elevated balance beam The apparatus consisted of a tapered beam (0.5-1.5 cm) with a 0.5 cm ledge that was situated 2 cm below the running surface. The bean was 100 cm long and angled at 17°. Mice were pre-trained to run up the beam to a goal-box. During testing, mice were placed at the lower end, facing away from the running surface. The turning time was recorded as a measure of motor coordination and the time to traverse indexed walking difficulty. The number of foot slips made by each limb was recorded as a measure of balance. Each mouse underwent three trials per day. Open field test Mice were placed in the middle of the arena (80 × 80 × 40 cm) and allowed to explore for 10 min. The arena was uniformly illuminated and the arena was always located in the same room. Two “locomotor” parameters were calculated: distance travelled and duration moving. After dividing the OFT arena into a central and a peripheral zone, two further anxiety-related parameters were calculated: the amount of time spent within, and the number of entries into, the central zone. Monitoring of behavioral responses in the open field test Behavior was recorded by a video camera, attached to a PC-based data recorder, and analyzed using EthoVision tracking software (Noldus Information Systems, The Netherlands). Footprint analysis Mice walked along a custom-built walkway on strips of paper (60 × 10 cm). Their front and hind paws were painted with red and blue paint, respectively. Three series were recorded per day. Stride length was defined as the distance between feet of the left and right stepping cycle. Animal Distress Score An animal Distress Scoring system was used to monitor changes in health and wellbeing pre- and post-AIA in WT and Il10-/- mice [34]. This rating system considers several aspects of health, including weight, appearance, clinical symptoms, natural behavior and provoked behavior. Each category has a scale of 0-3, with '0' denoting normal health and '3' conveying a severe health condition. The appearance of the mouse may score '0' if normal, '1' if a general lack of grooming, '2' where coat starring and/or ocular or nasal discharge is evident. '3' when piloerection and hunching are observed. Clinical symptoms may not be present (0), or may be slight (1; pallor, diminished activity), moderate (2; weight loss, polyurea, diarrhoea) or severe (3; immobility, abnormal gait, fitting). Assessment of natural behaviors may reveal (0) normal behavior, (1) minor changes, (2) less mobile/alert, isolated or (3) vocalization, self-mutilation, restlessness/stillness. Lastly, Provoked behavior may be (0) normal, (1) slightly depressed or exaggerated response, (2) moderate changes in expected behavior or (3) a violent reaction or very weak/precomatose response. Weights are also monitored and rated if changed. Each day, the score is adjusted if a 3 is assigned more than once, by adding 1 point per 3 score. Finally, 0-4 is considered normal behavior, 5-9 reveals that the mouse requires carefully daily monitoring and > 10 determines suffering and the need for termination. Histology Joints were fixed in neutral buffered formalin saline, decalcified with formic acid at 4°C and embedded in paraffin. Parasagittal sections (8 μm) were stained with hematoxylin, safranin-O and fast green. Disease severity was scored according to arthritic index, which comprises four components: sub-synovial inflammation (0 5); synovial exudate (0 3); hyperplasia and pannus formation (0 3) and articular cartilage or sub-chondral bone erosion (0 3). Scores were taken as the average of two observers blinded to the experimental design. HPA axis evaluation by Quantitative RT-PCR A second cohort of WT and IL-10-/- mice were used to investigate the impact of IL-10 deficiency on gene expression within the HPA axis following AIA. 6 mice of each genotype received AIA injections. Mice were euthanized by cervical dislocation 3 days post-AIA and the hypothalamus, pituitary and adrenal cortex were rapidly dissected and placed into RNAlater (Ambion) to inactivate RNAses. Following an overnight incubation at 4°C, tissues in RNAlater were transferred to a -80°C freezer for storage until RNA extraction. Total RNA was extracted from dissected tissues using the RNeasy lipid extraction kit (Qiagen, UK). Quantitative PCR 200 ng of total RNA was reverse transcribed in a 20 ul reaction for 1h at 45°C using the AffinityScript kit (Agilent, UK). 2 μl cDNA was amplified in a 20 μl reaction volume using Brilliant III ultrafast qPCR master mix reagents (Agilent). PCR products were detected using dual-labeled (FAM/BHQ1) hybridization probes specific to each of the cDNAs (MWG/Eurofins, Germany). Gene changes in neuroendocrine signatures within tissues from the HPA axis were assessed by qPCR at day 3 following AIA onset. Expression of Nr3c1, Crh, Tnfrsf1a, Pomc, Mc2r and Il6 was quantified in HPA axis RNA samples relative to a geometric mean of mRNAs for the reference genes Gapdh, Sdha and Hprt1. See Supplementary 1 for oligonucleotide primer sequences and standards. Oligonucleotide primers were used at a concentration of 200 nM. Dual-labeled probes were used at 500 nM. PCR was performed using the M × 3000P platform (Agilent) using the following conditions: 45 cycles of 95°C for 12s and 60°C for 35s. Statistics Statistical analyses were performed using Prism (v6, GraphPad Software Inc.) and SPSS (v20, IBM). Behavioral and swelling data were analyzed using Repeated Measures ANOVA, with Group and Day as factors. Where an interaction between Group × Day was found to be significant, analysis of the simple effects was undertaken with a restricted ANOVA. Post-hoc effects were analyzed with the addition of a bonferroni correction for multiple comparisons, to ensure appropriate stringency. Histological data were analyzed using Univariate ANOVA, with Group as the between-subjects factor. The qPCR data were analyzed using unpaired t-test. Swelling and behavioral data are graphically depicted with the full 18 day pre-AIA period (consisting of 4 baseline behavioral test sessions) to demonstrate the stability of the behavioral performance of the WT and Il10-/- mice. The 14 day post-AIA period (consisting of 7 test days) is then depicted. An analysis was conducted on the baseline data to determine the pattern of performance on the task prior to AIA onset and to identify any baseline differences between WT and Il10-/- mice. Then an analysis of the post-AIA data was conducted to determine the effect of the AIA insult on behavioral performance. To investigate the interaction between loss of IL-10 and arthritis induction on behavioral performance, an analysis compared data obtained during the baseline training against data obtained from the final week of AIA (Figure 2). Datasets from all four days within the baseline period (-18, -16, -3, -1) were averaged and compared against the average of datasets for the final two days of assessment post-AIA (+10, +14). The percentage change post-AIA, relative to the baseline period, was calculated. Independent samples t-tests explored differences in performance between the genotypes. . Figure 2: Induction of AIA generates subtle clinical symptoms in IL-10-/- mice. (A) Application of the Animal Distress Scoring system revealed no difference in appearance at baseline, but a subtle change in appearance in IL-10-/- mice after AIA; (B) Very mild clinical symptoms were observed in IL-10-/- mice at baseline, but these fell within the range of normal behaviors (0-4 score). Post-AIA, both WT and IL-10-/- mice presented with a mild increase in clinical symptoms, which were sufficiently subtle to remain within the range of normal behaviors. No changes were observed post-AIA for either genotype on the categories of weight, natural behaviors or provoked behaviors (data not shown). Significant effect of Group: *p ≤ 0.05; Significant Group*Day interaction:$p ≤ 0.001. View Figure 2
.
Results
Identification of acute behavioral impairments after AIA induction in WT mice
Schematic depicting the experimental design for longitudinal assessment of AIA (Figure 1). At baseline (prior to AIA onset), no differences in swelling, stride length, performance on the balance beam, nor distance travelled in the OFT were evident between the WT groups [maximum: F1,12 = 2.81, n.s.]. Although a mild but significant difference was revealed on the duration of time spent moving in the OFT [Group: F1,12 = 6.08, p < 0.05], overall the groups were well matched at baseline.
Mice receiving i.a. injection of mBSA (WT-AIA) displayed significant joint swelling from day 1 post AIA. While acute alterations in joint swelling continued for 7 days post arthritis induction, this response showed no relationship to changes in behavior. In contrast, control mice receiving PBS (WT-PBS) showed no joint swelling [Figure 1B; Group: F1,12 = 69.12, p < 0.001; Group × Day: F7,84 = 45.58, p < 0.001].
While neither the stride length [Figure 1C; F < 1], nor the latency to turn [Figure 1D; F < 1] on the balance beam were affected by antigen challenge, WT mice with AIA were slower to traverse the length of the balance beam than WT-PBS mice, an effect that lasted until day 4 of AIA [Figure 1E; Group × Day: F7,84 = 6.72, p < 0.01]. WT-AIA mice demonstrated a significantly higher number of foot slips of both the ipsilateral (right) and the contralateral (left) hind limbs, an effect which lasted for 3 days post-AIA in both hind limbs [Figure 1(F & G); min: Group × Day: F7,84 = 3.30, p < 0.01]. This motor deficit was specific to the hind limbs. No significant increase in the number of foot slips were observed for the front limbs after antigen challenge (data not shown; maximum F1,12 = 2.49, p > 0.05). In OFT assessments, the total distance travelled was transiently reduced by the onset of AIA, returning to baseline levels by 4 days post-insult [Figure 1F; Group × Day: F7,84 = 4.38, p < 0.001] and the duration of time spent moving was shorter in WT-AIA mice [Group × Day: F7,84 = 4.27, p < 0.001]. The indices of anxiety (quantified as entries into the inner zone and time spent in the inner zone) were not significantly altered after AIA [data not shown; maximum: Group: F1,12 = 2.70, all p's n.s.].
All behavioral abnormalities in WT-AIA mice returned back to baseline by day 7 post AIA, when joint swelling had subsided.
Similar motor coordination deficits and articular joint pathology during acute AIA in WT and IL-10-deficient mice
There were found no overt differences in joint swelling or articular pathology as assessed by measurement of inflammatory infiltration and exudate, synovial hyperplasia and joint erosion in WT and Il10-/- mice. While the induction of disease in WT and Il10-/- mice promoted changes in joint swelling [maximum: F1,11 = 3.36, n.s.] and synovial inflammation [Figure 3(C & D); minimum: Group: F2,28 = 35.33, all p's < 0.001; WT and Il10 /- AIA vs. WT-PBS, all p's < 0.001], no difference in disease severity was observed between the two mouse genotypes [Figure 3(C & D); WT-AIA vs Il10-/- AIA, all p's n.s.]. On the balance beam test, the number of foot slips coincided with the development of knee swelling in the AIA treated right hind joint. Both Il10-/- and WT mice showed a similar number of foot slips in response to AIA [maximum: Group: F1,11 = 3.36, n.s.]. All mice slipped more often on the balance beam with both hind limbs [Figure 3A; minimum: Day: F6,66 = 8.75, p < 0.001] and significant swelling was observed in the knee joint after AIA induction [Figure 3B; Day: F6.66 = 85.45, p < 0.001].
.
Figure 3: Joint-specific processes during acute AIA are independent of IL-10. (A) Significant transient increase in number of footslips post AIA in both WT and IL-10-/- groups; (B) Significant increase in swelling due to AIA induction in both WT and IL-10-/- groups; (C) Evaluation of arthritic index, inflammation, exudate, hyperplasia, and erosion in AIA-induced knee joints of WT and IL-10-/- mice, compared to WT-PBS treated mice. Values are presented for individual joints taken at Day-15 post AIA; (D) Representative Safranin-O and Fast Green stained parasagittal joint sections taken on Day-15 for WT-PBS, WT-AIA and IL-10-/--AIA mice (scale bars, 500 μM). B–baseline testing day. Graphs represent mean ± SEM. Significant effect of Day: ###p ≤ 0.001; significant effect of Group: ***p ≤ 0.001. View Figure 3
.
Arthritis onset has a minimal impact on animal welfare
Application of the Animal Distress Scoring system revealed no substantive difference between WT and Il10-/- mice on either 'natural' or 'provoked' behaviors. This was true both before and after the onset of AIA [data not shown; maximum: Group: F2,28 = 3.25, n.s.]. A small difference was observed on the 'clinical signs' score between WT and Il10-/- mice at baseline [Figure 2A; Group: F2,18 = 5.05, p < 0.05]. However, this was extremely subtle and equated to < 1 on the clinical symptoms scale (0-4 = normal, 5-10 = mild symptoms).
However, following AIA onset, both WT-AIA and Il10-/- AIA mice experienced a fluctuation in clinical symptoms that reached approximately 1.7 on the scale, demonstrating a very mild effect of AIA on wellbeing [Group*Day: F14,126 = 3.39, p < 0.001]. No significant differences were revealed between WT-AIA and Il10-/- AIA mice, but both these groups receiving AIA differed from WT-PBS mice.
The appearance rating did not differ between the groups at baseline [Figure 2B; Group: F2,18 = 1.00, n.s.]. However, post-AIA, there was a transient and subtle impact of the intervention on the appearance rating of Il10-/- mice only [Group*Day: F14,126 = 3.16, p = 0.001].
Slight changes in appearance were also evident post-AIA in Il10-/- AIA mice only. These changes accounted for the difference in overall score observed at baseline [Figure 2C; Group: F2,18 = 7.56, p < 0.01] and post-AIA [Group*Day: F14,126 = 2.87, p = 0.001].
Arthritis induction promotes extra-articular changes in mobility and anxiety in Il10-/- mice
Baseline behavioral performance: Prior to arthritic insult (at baseline), the loss of IL-10 had no effect on stride length nor on the time to turn on the balance beam [Figure 4(A & C); maximum: Group: F1,11 = 6.07, n.s.]. The latency to traverse the balance beam was increased in Il10-/- AIA mice relative to WT-AIA mice [Figure 4D; Group: F1,11 = 70.16, p < 0.001]. Moreover, in the OFT, Il-10-/--AIA mice spent less time moving, travelled less far, entered into the inner zone fewer times and spent less time in the inner zone than WT-AIA mice [Figure 4(B,E & F); minimum: Group: F1,11 = 4.87, p = 0.05].
.
Figure 4: Loss of IL-10 delays recovery of global behavioral deficits induced by AIA. (A) Significant reduction in stride length by IL-10-/- mice after AIA induction, which returns to baseline levels by the final week of testing; (C) The latency to turn on the balance beam is impaired post-AIA in IL-10-/--AIA mice but not WT mice; (D) IL-10-/--AIA mice take significantly longer than WT-AIA mice to traverse the balance beam at baseline and post-AIA; (B) Representative movement tracks (black line) of one WT-AIA and one IL-10-/--AIA mouse in OFT. Blue square represents arena border and red square denotes the inner and outer zones used to determine anxiety indices. (E & G) In the OFT, IL-10-/--AIA mice spent less time moving and travelled shorter distances both pre-AIA. However, locomotor behavior is disrupted in both WT and IL-10-/- mice post-AIA. Moreover, IL-10-/--AIA mice take longer to recover the activity deficit; (F & H) In the OFT, the indices of anxiety reveal overall differences between the genotypes, but both WT and IL-10-/--AIA mice demonstrate reduced time in, and entries into, the inner zone of the arena after AIA. IL-10-/--AIA mice show prolonged impairment on these measures. Graphs represent mean ± SEM. Significant effect of Group: *p ≤ 0.05, **p ≤ 0.01, ***p ≤ 0.001; Significant effect of Day: ##p ≤ 0.01, ###p ≤ 0.001. View Figure 4
.
Importantly, however, prior to AIA insult, the behavior of the Il10-/- and WT mice was found to be consistent and stable across the 18 baseline days on every behavioral task [maximum, both for OFT duration moving: Day: F3,33 = 2.84, n.s.; Day*Group: F3,33 = 2.83, n.s.]. The reduction in the latency to traverse the balance beam observed during the fourth session, after the elevated scores obtained in the first 3 sessions, is likely accounted for by improved experimenter technique with repeated practice.
Post-AIA behavioral performance: Stride length, as measured in the footprint test, was specifically reduced in Il-10-/--AIA mice, relative to WT-AIA mice [Figure 4A; Group: F1,11 = 5.46, p < 0.05]. By the final week, however, performance had returned to baseline for both genotypes [t11 = -0.05, n.s.]. Stance base width and hindpaw/forepaw overlap were also measured throughout this experiment. No effects of IL-10 deficiency, nor effects of AIA, were evident on these aspects of gait (data not shown).
On the balance beam, the latency to turn was significantly increased in Il-10-/- mice post-AIA, relative to WT-AIA mice [Figure 4C; Group: F1,11 = 5.02, p < 0.05]. Although by the final week of testing, a numerical increase in latency to turn was still evident in Il10-/- AIA mice, this effect did not reach conventional levels of statistical significance [t11 = 1.28, n.s.].
Il10-/- AIA mice demonstrated an increased latency to traverse the beam [Figure 4D; Group: F1,11 = 22.50, p = 0.001], which was also evident at baseline. More interestingly, the significant effect of Day reveals the impact of AIA insult on performance [Day: F6,6 6= 3.82, p < 0.01]. Comparison of the latency data during the final week, as compared to the baseline period, reveals that both genotypes returned to baseline levels [t11 = -0.66, n.s.].
On the OFT, differences in performance were evident between the genotypes post-AIA on all measures, as was observed at baseline [Figure 4(B,E & F); minimum: Group: F1,11 = 15.06, p < 0.01]. Unlike at baseline, however, there was also a significant effect of Day on every measure, demonstrating the impact of the AIA on performance [minimum: Day: F6,66 = 3.52 p < 0.01]. Most importantly, on both measures of locomotor activity (distance travelled and duration moving), comparison of the activity levels during the final week, as compared to the baseline period, revealed a return to baseline performance for WT-AIA mice, while Il10-/- AIA mice were still significantly affected by the AIA intervention during the second week post-AIA [t11 = -3.57 & -3.18, respectively, p's < 0.01].
Moreover, on both measures of anxiety (time in inner zone and entries into inner zone), Il10-/- AIA mice were impaired for longer than WT-AIA mice. By the final week of testing, the time spent in the inner zone was severely reduced in Il10-/- -AIA mice, compared to WT-AIA mice [t11 = -2.48, p < 0.05]. Also, while the number of entries into the inner zone had returned to baseline levels for WT-AIA mice, the Il10-/- AIA mice were still significantly impaired by the second week of post-AIA behavioral testing [t11 = -2.48, p < 0.05]. Overall, the results demonstrate that AIA induction initiates global, extra-articular behavioral abnormalities in both WT and Il10-/- mice, and that specific subtle interactions between IL-10 deficiency and AIA can be detected on behavioral performance.
Up-regulation of specific HPA axis signatures is associated with IL-10 dependent increased anxiety index
At day 3 post AIA induction, mRNA expression of specific HPA axis regulators was elevated in Il10-/- -AIA mice compared to WT-AIA controls: hypothalamic glucocorticoid receptor, Nc3r1 [Figure 5A, t8 = 2.63, p < 0.05]; pituitary pro-opiomelanocortin, Pomc [Figure 5B, t10 = 2.74, p < 0.05]; and adrenal cortex adrenocorticotropin hormone receptor, Mc2r [Figure 5C, t10 = 3.52, p < 0.01]. Other HPA signatures such as the precursor peptide for corticotropin-releasing hormone precursor, Crh, were not significantly different at day 3 [Figure 5A, n.s.]. Importantly, the differences observed at day 3 of AIA are not simply explained by IL-10 deficiency, as there were no significant differences between genotypes in the absence of AIA (n.s., for all effects of genotype at baseline; data not shown). Notably, a concurrent increase in adrenal pro-inflammatory IL-6 expression, Il6, was measured in the Il10-/- AIA group (Figure 5C, t10 = 3.55, p < 0.01), compared to the WT-AIA group. Other inflammatory markers, such as tumor necrosis factor receptor 1A, TNFR1a, were unaltered by IL¬10 deficiency (Figure 5, maximum t8 = 0.96, n.s.).
.
Figure 5: RT-qPCR of day HPA tissues 3 days post-AIA shows IL-10 dependent HPA axis mRNA signatures associated with increased levels of adrenal IL-6. (A) Significantly increased mRNA expression of hypothalamus glucocorticoid receptor (Nc3r1) in IL-10-/--AIA mice. Expression of hypothalamic corticotropin-releasing hormone (Crh) and tumor necrosis factor receptor 1a (Tnfrsf1a) in IL-10-/--AIA mice is not significantly different from WT-AIA mic; (B) Significantly increased mRNA expression of pituitary pro-opiomelanocortin (Pomc) in IL-10-/--AIA mice. Expression of pituitary tumor necrosis factor receptor 1a (Tnfrsf1a) in IL-10-/--AIA mice is not significantly different from WT-AIA mice; (C) Significantly increased mRNA expression of adrenal adrenocorticotropin receptor (Mc2r) in IL-10-/--AIA mice and interleukin-6 (Il6). Expression of adrenal tumor necrosis factor receptor 1a (Tnfrsf1a) in IL-10-/--AIA mice is not significantly different from WT-AIA mice. Graphs show all data points, with lines representing mean ± SEM. Significant effect of Group: *p ≤ 0.05, **p ≤ 0.01. View Figure 5
.
Discussion
This study provides novel insight into rodent behavior during experimental arthritis by applying behavioral neuroscience methodologies for longitudinal assessment of AIA. We utilized the OFT, footprint analysis and the balance beam test to assess the impact of arthritic insult, and its interaction with loss of IL-10, on behavior over 14 days. This novel application of select behavioral tasks has revealed several interesting phenomena. Firstly, these tasks identified changes in behavior that occurred as a result of arthritis but, more importantly, they can distinguish between joint-specific processes and more global, systemic changes in behavior. Mapping a timeline of behavior over two weeks also reveals transient dynamic changes that occur prior to the onset of chronic alterations.
From this study, we conclude, firstly, that the onset of arthritis causes acute, but transient, changes in swelling and motor performance, which affect balance and specific limb use in WT-AIA mice. Consistent with previous reports [2-4], AIA caused a characteristic antigen-dependent swelling of injected knee joints. Results demonstrate that specific behavioral deficits coincide with acute indices of AIA, such as joint swelling. Thus, the novel application of complementary behavioral tasks in mice with experimental arthritis reveals acute alterations in motor and locomotor responses.
Secondly, Il10-/- -AIA mice demonstrated comparable changes in specific joint dysfunction to WT-AIA mice, as measured by the balance beam, knee swelling measurements and articular pathology indices. Thus, while deficits related to AIA were evident in joint swelling and footslips measures, loss of Il10 did not exacerbate these joint-specific processes during AIA.
Thirdly, arthritis onset has a minimal impact on animal welfare. Natural and provoked behaviors were not affected by the onset of AIA, but subtle changes in clinical symptoms and appearance were observed in WT-AIA and Il10-/- AIA mice. It is worth noting, however, that the magnitude of these changes was minor, since normal wellbeing registers between 0-4 on the rating scale, with mild/moderate changes ranging in between 5-9. Appearance changes averaged < 1 and clinical signs reached a maximum of 1.7 on the rating scale. Therefore, an interesting but subtle interaction between AIA and IL-10 deficiency was revealed using the Animal Distress Scale
Next, these data reveal, interestingly, arthritis induction promotes extra-articular changes in mobility and anxiety in Il10-/- mice. Il10-/- -AIA mice presented with more systemic, global changes in behavior, and recovery of function was prolonged relative to WT-AIA mice. Over a longer term, Il10-/- AIA mice displayed shortened stride length, increased latency to turn on and traverse up the balance beam, a reduction in exploratory behavior, and fewer entries into the inner area of the OFT apparatus. Previous studies have shown that mice deficient in the anti-inflammatory cytokine IL-10 show a more sustained, long-term arthritic disease as compared to WT mice [4]. However, the extent of disease activity seen in WT and Il10-/- mice remains comparable during the first two weeks following arthritis induction [4]. Thus, these chronic behavioral alterations cannot simply be accounted for by direct changes in limb function since joint-specific impairments, such as knee swelling and hind limb foot slips, returned to baseline in a similar manner in WT-AIA and Il10-/- AIA mice. The observation that Il10-/- mice have prolonged behavioral and anxiety-related impairments post-AIA may be important for addressing the high prevalence of mood disorders in patients with RA [33,34]. Il10 deficiency is known to have profound immune outcomes that may influence the inflammatory control of neuropsychological behaviors. In addition to its role in modulating the severity of chronic experimental arthritis [4,17], IL-10 has been linked with anxiety and mood in both rodents and patients [22,24]. Although IL-10 plays a role in RA [19] and has previously been linked with mood disorders [24-26], there has been insufficient evidence for a specific role for IL-10 in regulating behavior during inflammatory arthritis. Here, we show that IL-10 deficiency altered the global motor and locomotor phenotype in mice that received AIA. Taken together, these results suggest reduced activity and heightened anxiety levels due to the interaction of IL-10 deficiency and AIA.
Finally, we found that IL-10 dependent increases in anxiety are associated with the up-regulation of specific HPA axis signatures. The IL-10 dependent delay in recovery of global locomotor and anxiety-related behaviors identified using open field and balance beam techniques (Figure 2) occurred in the absence of any overt differences in articular pathology at day 15 post AIA. Whereas WT mice begin recovery earlier, the Il10-/- mice display longer-term behavioral impairments. We therefore hypothesized that the IL-10-dependent global behavioral abnormalities observed during AIA were mediated by systemic processes, occurring outside the joint. Specifically, we focused on neuroendocrine signatures from HPA axis tissues. The results revealed increased mRNA expression of the specific HPA axis regulators Nc3r1, Pomc, Mc2r in Il10-/- -AIA mice compared to WT-AIA controls at day 3 post AIA induction. Moreover, specific regulation of neuroendocrine pathways by IL-6, as determined here, is supported by previous studies showing stimulation of the HPA axis by IL-6 [20,21].
Thus, in this study, it has been possible for the first time to determine the role of IL-10 in behavioral control during experimental arthritis. The observation that IL-10 deficiency delays the recovery of global behaviors, independent of specific joint impairment, may be explained by aberrant inflammatory activation in Il10-/- mice [4,17], or abnormal cytokine control of neuroendocrine processes [25]. Addressing these types of questions is becoming increasingly important for understanding the extra-articular comorbidities [1] and psychosocial factors [33,35] associated with RA.
When exploring behavioral responses in rodent models of chronic disease it is essential to consider the nature of the inflammation (i.e., localised vs. systemic) and the severity of disease (i.e., the histopathology). Systemic models of inflammatory disease (e.g. polyarticular models of arthritis) often have a more severe impact on animal health and welfare. For example, these models often necessitate the use of analgesia in the protocol. This makes behavioral studies more challenging and difficult to interpret. A monoarticular model of antigen-induced arthritis (AIA) was therefore adopted, where synovitis is established in one knee. This model shows 98-100% penetrance in C57/bl6 mice and the development of disease is well-characterized [10,13,16]. While studies in AIA have documented changes in joint pathology, gait analysis and loss of joint movement, the impact of AIA on animal behavior has never been tested. Previous behavioral assessments of experimental arthritis were limited to either single-technique gait evaluation [5-12] or single-trial OFT assessment [13,14].
Finally, we propose a potential mechanism underlying the delayed recovery of global behaviors observed in Il10-/- mice. RT-qPCR expression profiling of HPA axis tissue 3 days post-AIA identified IL-10-dependent changes in gene expression in each of the three tissues. Specifically, Nc3r1, Pomc, and Mc2r mRNAs were expressed at higher levels in Il10-/- mice after AIA compared to WT-AIA mice. Whilst adrenal IL-6 mRNA expression was up-regulated at the same time-point in Il10-/- mice, TNFR1α mRNA expression in all HPA tissues did not differ between WT and Il10-/- mice. These data raise the possibility that, in the absence of IL-10, AIA elevates levels of IL-6 within the adrenal cortex that may act, potentially locally, to modulate gene expression within the HPA axis, thereby altering axis function [22]. Further validation of this hypothesis is required in order to determine the precise mechanism of action.
Conclusions
In summary, we have shown that the application of behavioral techniques to experimental arthritis can reveal novel and interesting phenomena that could fundamentally shift the landscape of RA treatment. Inclusion of longitudinal behavioral assessment in our experimental design enabled us to disentangle behaviors which are driven by articular insult, from those induced by systemic, extra-articular changes that are more purely induced by interaction between IL-10 deficiency and AIA. Although we have focused on IL-10 deficiency in the first instance, there is great potential for this type of work to unravel new mechanisms to inform new strategies for reducing the extra-articular disease burden [1].
Funding
This work was funded by Arthritis Research UK (19796, 19381 to SAJ), the Medical Research Council, and Cardiff University. AKH was supported by an Arthritis Research UK Centre Grant (18461).
Author Contributions
AKH designed the experiment, ran the behavioural experiment, analysed and interpreted the data sets, wrote the manuscript. MJL designed the experiment, ran the behavioural experiment, analysed and interpreted the data sets, wrote the manuscript. CJG designed the experiment, ran the behavioural experiment, analysed and interpreted the data sets, wrote the manuscript. ATJ conducted the histological experiments, collected behavioral data, edited the manuscript. SPC collected behavioral data, edited the manuscript. MJN conducted and analysed the qPCR analyses, edited the manuscript. SBD analysed and interpreted the data sets, wrote the manuscript. SLW conducted and analysed the qPCR analyses, wrote the manuscript. ASW designed the experiment, ran the behavioural experiment, analysed and interpreted the data sets, wrote the manuscript. SAJ designed the experiment, analysed and interpreted the data sets, wrote the manuscript.
Competing Interests
The authors report no conflicting interests.
References
Open Access by ClinMed International Library is licensed under a Creative Commons Attribution 4.0 International License based on a work at https://clinmedjournals.org/.
|
2018-01-23 23:41:35
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.345994770526886, "perplexity": 14792.210137372615}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084892802.73/warc/CC-MAIN-20180123231023-20180124011023-00265.warc.gz"}
|
https://stats.stackexchange.com/questions/105363/linear-regression-without-intercept-sampling-variance-of-coefficient
|
# Linear regression without intercept - sampling variance of coefficient
I am comparing linear regression with and without intercept for the general sampling case. For this, I have $n$ samples of two correlated random variables $X \sim N\left(0,\sigma_X^2\right)$ and $Y \sim N\left(0, \sigma_Y^2\right)$ with correlation $\rho$.
For the random samples, I calculate the linear regression models with and without intercept
(1) $y_i=\alpha_0+\alpha_1x_i+\epsilon_1$ and
(2) $y_i=\beta_1x_i+\epsilon_2$
Using numerical experiments, I have found that $E[\hat\alpha_1] = E[\hat\beta_1]$, which seems logical to me. However, I have also found thtat $\text{Var}(\hat\alpha_1) \neq \text{Var}(\hat\beta_1)$, which I am currently trying to understand.
In another question of mine, I have found that for the general sampling case $\text{Var}(\hat \alpha_1) = \frac{\sigma_Y^2}{\sigma_X^2} \frac{1-\rho^2}{N-3}$ for the model with intercept and am trying to find $\text{Var}(\hat\beta_1)$.
Overall, I am therefore trying to find $\text{Var}(\hat\beta_1)=\text{Var}\left(\frac{\sum x_iy_i}{\sum x_i^2}\right)$.
The denominator is clearly gamma distributed. However, the distribution of the numerator as a sum of products of normal distributed random variables is tough, not to mention the ratio.
Calculating $\text{Var}(\hat\beta_1)=E[\hat\beta_1^2] - E[\hat\beta_1]^2$ isn't much easier, I think.
After spending hours in the local university library and searching research papers, I am turning to CrossValidated for help (again).
Does somebody know a way to calculate the variance in question?
• Did you switch the definitions of $\alpha$ and $\beta$ partway through your post? – Dougal Aug 7 '14 at 3:57
• Thank you, your comment is indeed correct. I have corrected the initial post. – sebastianb Oct 2 '14 at 13:49
• Just as a comment: hope that you are aware that regression without intercept is generally a bad idea, see: stats.stackexchange.com/questions/7948/… – Tim Nov 25 '15 at 9:24
• Is it purposeful that the true data generating process doesn't include a constant? In other words, that there's a zero in $y_i = 0 + \beta x_i + \epsilon_i$? If so, this question in some sense is, "What happens when I include a completely unrelated variable in a regression vs. when I don't include it?" (In this case, the completely unrelated variable would be the constant.) – Matthew Gunn Jun 12 '16 at 2:02
In your problem, assuming joint-normality of the variables, you can write the joint distribution of a single data point $(X_i, Y_i)$ as:
$$\begin{bmatrix} X_i \\ Y_i \end{bmatrix} \text{ ~ N} \left( \begin{bmatrix} 0 \\ 0 \end{bmatrix}, \begin{bmatrix} \sigma_X^2 & \rho \sigma_X \sigma_y \\ \rho \sigma_X \sigma_y & \sigma_Y^2 \end{bmatrix} \right).$$
With a bit of algebra, the value $Y_i$ can be shown to be equivalent to:
$$Y_i = \rho \frac{\sigma_Y}{\sigma_X} \cdot X_i + \sqrt{1 - \rho^2} \sigma_Y \cdot \varepsilon_i,$$
where $\varepsilon_i$ is an independent standard normal error term. Hence, the true regression model is:
$$\begin{matrix} Y_i = \beta_0 + \beta_1 \cdot X_i + \sigma \cdot \varepsilon_i & & & \beta_0 = 0 & \beta_1 = \rho \frac{\sigma_Y}{\sigma_X} & \sigma = \sqrt{1 - \rho^2} \sigma_Y \end{matrix}.$$
Simplifying the variance problem: When you estimate the model coefficients in the model with an intercept term, you would expect to get an intercept estimate close to the true value of zero, which means you would also expect the estimated slope coefficients to be similar with or without the inclusion of an intercept term in the model (as you have pointed out). However, the inclusion of an intercept term will tend to reduce the variance of the estimated slope coefficient. You have:
$$X_i Y_i = \rho \frac{\sigma_Y}{\sigma_X} \cdot X_i^2 + \sqrt{1 - \rho^2} \sigma_Y \cdot X_i \varepsilon_i.$$
Defining $\boldsymbol{Z} \equiv \boldsymbol{X} / \sigma_X$ and $\boldsymbol{U} \equiv \| \boldsymbol{Z} \|^2$ allows us to write:
$$\hat{\beta}_1 = \frac{\sum_{i=1}^n X_i Y_i}{\sum_{i=1}^n X_i^2} = \rho \frac{\sigma_Y}{\sigma_X} + \sqrt{1 - \rho^2} \frac{\sigma_Y}{\sigma_X} \cdot \frac{\boldsymbol{Z} \cdot \boldsymbol{\varepsilon}}{\boldsymbol{U}}.$$
Taking the variance gives:
$$\mathbb{V}{(\hat{\beta}_1)} = (1 - \rho^2) \frac{\sigma_Y^2}{\sigma_X^2} \mathbb{V}\left(\frac{\boldsymbol{Z} \cdot \boldsymbol{\varepsilon}}{\boldsymbol{U}} \right).$$
Now, the three vectors in the variance operator are independent (using Cochran's theorem). The numerator is a sum of products of independent standard normal random variables, and the denominator is a chi-squared random variable.
I must confess that I am not sure where to go from here. I do not recognise the variance expression as any simple form, though maybe others will. In any case, I hope that gives you some progress towards what you want.
I don't know how you found that $\sigma_{\hat{\alpha_1}}^2 \neq \sigma_{\hat{\beta_1}}^2$. Using this R code
alpha <- c()
beta <- c()
for(i in 1:1000){
dat<-matrix(rnorm(200000,0,1),nrow=100000,byrow=T)
cov <- chol(matrix(c(5,2,2,5),byrow=T,nrow=2))
dat <- dat%*%cov
X1 <- matrix(0,nrow=100000,ncol=2)
X1[,1] <- 1
X1[,2] <- dat[,2]
alpha <- cbind(alpha,solve(t(X1)%*%(X1))%*%t(X1)%*%dat[,1])
beta <- cbind(beta,solve(dat[,2]%*%dat[,2])%*%dat[,2]%*%dat[,1])
}
I got
> var(alpha[2,])
[1] 8.310618e-06
> var(beta[1,])
[1] 8.309166e-06
However, your model is equal to $y = Xb+e$, where $\sigma_{b}^2 = (X^{T}X)^{-1}\sigma_e^2$. Since $\sigma_e^2 = \sigma_y^2 - \sigma_{xy} \frac{\sigma_{xy}}{\sigma_x^2}$, which is equal to $\frac{\sigma_y^2}{\sigma_x^2}(1-r^2)$, $\sigma_{b_1}^2 = \frac{\sigma_e^2}{\sum_i^n x_i^2}$, and $\sigma_{b_0}^2 = \frac{\sigma_e^2}{n}$
If you expand the above script for calculating $e$, and storing $e^{T}e$, $y^{T}y$, $x^{T}x$ and $x^{T}y$, you can calculate all these values.
Cheers
• Thank you for your answer! $var(\hat \alpha_1) = var(\hat \beta_1)$ is indeed true for big sample sizes. You used a bis sample with n=100,000 in your example. When using a tiny sample (for instance $n=20$) and a higher number of iterations (for instance 20,000) the difference is observable: > var(alpha[2,]) [1] 0.0502148 > var(beta[1,]) [1] 0.04727873 – sebastianb Jul 1 '14 at 17:31
|
2019-11-21 15:52:41
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8928799033164978, "perplexity": 371.43739509162066}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670921.20/warc/CC-MAIN-20191121153204-20191121181204-00124.warc.gz"}
|
https://www.storyofmathematics.com/factors/factors-of-72/
|
# Factors of 72: Prime Factorization, Methods, and Examples
All the numbers that perfectly divide the number 72 and do not leave any remainder are called the factors of 72.
This article will provide insight regarding the factors of 72 and how to find them by using various methods, including prime factorization and division methods. This article also explains the Factor Tree of 72, and Factors of 72 in Pairs with some examples.
## What Are the Factors of 72?
The factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72. All the numbers mentioned above are perfect divisors of the number 72.
When 72 is divided by any of the said numbers, it is divided completely and leaves zero as a remainder.
It can also be mentioned by using the methods of multiplication where two factors multiply perfectly to give the number 72.
Interestingly, 1 and the number itself (72) fall in the definition of factors for every number. So, 1 and 72 are also the factors of 72.
## How To Calculate the Factors of 72?
To find the factors of 72, start dividing 72 by the smallest natural number that divides 72 perfectly and does not leave any remainder.
Continue dividing 72 by consecutive whole numbers, if the quotient is a whole number, it is a perfect divisor of 72. Hence, also it is a factor of 72.
If the quotient is a number in a fraction, it is not a factor of 72. Now let’s start the procedure:
Divide 72 by the smallest natural number i.e., 1.
$\dfrac{72}{1} = 72$
As it has completely divided 72 without any leaving a remainder, so 1 is a factor of 72.
Now, divide 72 by the smallest even prime number i.e., 2
$\dfrac{72}{2} = 36$
The number 72 has been divided perfectly by its divisor. So, the 2 is also a factor of 72.
Again divide 72 by the smallest odd prime number, which is 3
$\dfrac{72}{3} = 24$
As 3 has divided 72 completely. So, the number 3, too is a factor of 72.
For getting more factors, divided 72 by natural numbers that exactly divide 72 and leave zero remainders as shown below:
$\dfrac{72}{4 }= 18$
$\dfrac{72}{6} = 12$
$\dfrac{72}{8} = 9$
$\dfrac{72}{9} = 8$
$\dfrac{72}{12} = 6$
$\dfrac{72}{18} = 4$
$\dfrac{72}{24} = 3$
$\dfrac{72}{36} = 2$
$\dfrac{72}{72} = 1$
All the above numbers completely divide 72 and do not leave any remainder. So, all these numbers are factors of 72.
The method stated above is called calculating the factors by division method. There are various methods to calculate the factors of 72. Other methods are also explained in this article.
## Factors of 72 by Prime Factorization
The prime factorization of 72 is the expression of 72 as a product of its prime factors.
To find out the factors of 72 by the prime factorization method, divide 72 by the smallest prime number which divides 72 exactly.
The resulted quotient is again divided by the smallest prime number and the procedure continues until we get 1 as the final quotient when cannot be divided any further.
Following are the steps to calculate factors of 72 by prime factorization.
The first step in the procedure is to divide 72 by the smallest prime number divisor which in this case is 2.
$\dfrac{72}{2} = 36$
The quotient 36 is an even composite number and it further requires to be divided by 2 being the smallest available prime number divisor.
$\dfrac{36}{2} = 18$
18 is again an even composite number that can be further divided by the prime number 2.
$\dfrac{18}{2} = 9$
Now, as 9 cannot be completely divided by 2, we have to shift to the next smallest prime number that divides the quotient 9 completely and does not leave any remainder. In the given case, the next prime number is 3 which completely divides 9.
$\dfrac{9}{3} = 3$
The quotient 3 now can only be further divided by 3 and thus give the next quotient as 1
$\dfrac{3}{3} = 1$
The quotient 1 cannot be further divided.
Figure-1
Therefore, the prime factorization of 72 can be expressed as follows:
$72 = 2 \times 2 \times 2 \times 3 \times 3$
It can also be stated as:
$72 = 2^3 \times 3^2$
## Factor Tree of 72
Factors of 72 can also be expressed using a factor tree.
It is a way of exhibiting the factors of a number, specifically the prime factorization of a number in which each branch of the tree divides into its factors.
These factors are distributed and written in the form of branches showing the given number’s factorization.
The splitting of a branch can either produce prime or composite numbers. If one of the branches resulting from a split, produces a composite number, the branching goes further.
The method is continued until the factors at the end of the branch produce both the prime numbers. This is where the branching stops.
If we write 72 into multiples, it would be:
$72 = 2 \times 36$
Upon dividing 36 into its multiples, it would be:
$36 = 2 \times 18$
Dividing 18 further into its multiples would result in:
$18 = 2 \times 9$
Further dividing 9 into its multiple factors would give:
$9 = 3 \times 3$
By dividing 3 further into its multiples, it would be:
$3 = 3 \times 1$
Expressing the number in terms of prime factors would be as follows:
$2 \times 2 \times 2 \times 3 \times 3$
Figure-2
## Factors of 72 in Pairs
Factor pairs of 72 are the two factors of 72 that, when multiplied together, give the product as 72. In simple words it can be described as:
A set of two natural numbers, whose product gives us the number 72 are called factors of 72 in pairs.
Pair Factors are a pair of numbers that, when multiplied by each other, give the result of 72 itself. Following are the pair factors of the number 72.
$1 \times 72 = 72$
$2 \times 36 = 72$
$3 \times 24 = 72$
$4 \times 18 = 72$
$6 \times 12 = 72$
$8 \times 9 = 72$
$9 \times 8 = 72$
$12 \times 6 = 72$
$18 \times 4 = 72$
$24 \times 3 = 72$
$36 \times 2 = 72$
As there are 12 factors of 72, these factors can be written in pairs. The factors pairs of 72 are (1, 72), (2, 36), (3, 24), (4, 18), (6, 12), and (8, 9).
The number 72 can have negative pair factors, as well as the multiplication of two negative factors, also produces a positive product.
$(-18) \times (-4) = 72$
$(-6) \times (-12) = 72$
$(-3) \times (-24) = 72$
Hence, the following are some examples of negative pair factors of 72 such as (-1, -72), (-2, -36), (-3, -24), (-4, -18), (-6, -12), and (-8, -9).
So, it can be derived that the product of all factors of 72 in its negative form, gives the result 72. So, all are called negative pair factors of 72.
### Tips and Tricks
1. Every factor of a given number is either less than or equal to that given number but can never be greater than the number. Hence, the factor of 72 can never be greater than 72 itself.
2. Only whole numbers and integers can be the factors of a given number.
3. Any given number has only a finite number of factors/divisors as in this case, number 72 has only 12 factors.
4. A trick to calculate the total number of factors of a given number can help calculate factors of large numbers and save some time. It can also be used to cross-check the conventional methods of calculating factors of a given number. For example, the prime factorizations of 72 are such as:
$72 = 2^3 \times 3^2$
Add one (1) to the exponents which are 3 and 2 individually and multiply their sums. i.e.,
$(3 +1) \times (2 +1) = 12$
This shows that 72 have 12 factors in total.
## Factors of 72 Solved Examples
### Example 1
What are the negative pair factors of 72?
### Solution
Please keep in mind, that the product of two negative numbers is positive. So, all the factors of 72 in their negative form are called negative pair factors of 72. These are:
(-1, -72)
(-2, -36)
(-3, -24)
(-4, -18)
(-6, -12)
(-8, -9)
### Example 2
Which of the following statement is False about factors of 72?
1. 72 has a total of 12 factors.
2. 72 has only two prime factors which are 2 and 3.
3. 72 can have one positive and one negative factor in the pair.
4. Pair Factors of 72 can have one prime and one composite number.
### Solution
The product of one positive and one negative number is always negative. Hence 72 can never have one positive and another negative factor in pairs. So False statement is 72 can have one positive and one negative factor in pairs.
Images/mathematical drawings are created with GeoGebra.
|
2022-08-10 06:03:19
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7074184417724609, "perplexity": 456.79147944943765}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571147.84/warc/CC-MAIN-20220810040253-20220810070253-00637.warc.gz"}
|
http://angolodeisaporifirenze.it/kgbn/rate-of-reaction-between-calcium-carbonate-and-hydrochloric-acid-lab-report.html
|
# Rate Of Reaction Between Calcium Carbonate And Hydrochloric Acid Lab Report
Such a reaction is best carried out in the lab under a fume hood and using small pieces of calcium metal & dilute acid, as large pieces can produced sufficient h. Chemical reactions lab report answers Chemical reactions lab report answers. I 16 Grand Slam titles include four at the U. It was a neutralisation reaction and its formula was CaCO3+2HCl → CaCl2 + H2O + CO2. There are tutorials on the site to help with writing formulae. 00 mL of 1M aqueous solutions of calcium chloride and sodium carbonate. This is the reaction that occurs when an antacid containing the active ingredient calcium carbonate (CaCO 3) reacts with stomach acid (hydrochloric acid). DOSAGE AND ADMINISTRATION. Set I: 1 g of zinc powder and 20 cm 3 of 0. Hydrochloric acid = HCl. Add the hydrochloric acid and a few drops of Ethanol into the Erlenmeyer flask with the eggshell powder. Standardisation of a hydrochloric acid solution using a standard solution of sodium carbonate Theory Laboratory grade hydrochloric acid is not sufficiently pure to be used as a primary standard. The reaction between hydrochloric acid and calcium carbonate (limestone or chalk) is shown below: CaCO 3 + 2 HCl → CaCl 2 + H 2 CO 3. the calcium carbonate content (% CaCO 3) is determined by dissolving a portion of the sample in hydrochloric acid and gently boiling the mixture for five minutes. The made inhibitor is based on the aqueous solutions of oxiethilidendiphosphone acid, hydrochloric acid, ammonium chloride, polyethylene polyamine-N-methylphosphonic acid and isopropyl intoxicant to forestall the. Titration of sodium hydroxide using hydrochloric and sulphuric acids Acid-base titration to find the molarity of two acids Aim The aim of this lab practical is to carry out the titration of a base, sodium hydroxide with two acids: firstly hydrochloric acid and then sulphuric acid so as to determine the molarities of the two acids. The catalytic decomposition of Oxygen is given off much faster if the hydrogen peroxide is concentrated than if it is dilute. Carbonic acid changes back and forth in equilibrium to bicarbonate (HCO 3-) + H+ and carbonate (CO 3 2-) + H+, lowering the pH due to higher concentrations of H+. This produces a solution of bicarbonate. If a small piece of calcium carbonate (remember, limestone is mainly calcium carbonate) is added to hydrochloric acid it begins to fizz because of the carbon dioxide gas being produced. 04 g and yields 40 cm 3 of hydrogen when reacted with excess acid. The reaction rate of hydrochloric acid with carbonate rock at acid-fracturing conditions depends on controllable variables such as injection rate and acid concentration, and variables inherent to the carbonate rock (permeability, mineral composition, etc. 'n Suur en 'n metaalkarbonaat reageer om 'n sout. Formula of sodium chloride and one formula for sodium carbonate are wrong. Students obtain an over-the-counter antacid (in this case, calcium carbonate), calculate the volume of 0. (Report) by "Civil Engineering Dimension"; Engineering and manufacturing Science and technology, general Calcium carbonate Chemical properties Thermal properties Concrete Concretes Precipitation (Chemistry) Observations Temperature effects. Lime water (1), a dilute solution of calcium hydroxide, turns 'milky' when carbon dioxide is bubbled through it (2), forming a suspension of calcium carbonate. The rate of the reaction is the speed that the reaction takes place so by measuring the rate I How can I get rid of precipitation of CaCO3 and 2018-11-30 · XRD analysis shows that calcium oxalate and calcium carbonate were formed and ICP analysis shows that the remaining calcium in seawater using oxalic acid is about 0. The rate of reaction of H+ with soil carbonate is highly dependent on accessibility of carbonate surface sites to reaction with H+; therefore, sample pretreatment and degree of dispersion of the soil strongly influences the results of the pH-stat titration procedure (Hartwig & Loeppert, 1991). A phosphate level of 0. effects the reaction rate of calcium carbonate and hydrochloric acid. CaCO3(s) + 2HCl(aq) ( CaCl2(aq) + CO2(g) + H2O(l) Safety. Calcium is necessary for bone formation and maintenance. The following reaction will be tested: Na2CO3 10H2O + 2HCl –> 2NaCl + CO2 + 11H2O To measure the rate of the reaction an Explorer. Photo decomposition is a chemical reaction in which a substance is broken down into simple substances by exposure to light (photons). situation, the sodium carbonate acts as a nucleophile and the tannin is an electrophile. Acid-Base Solubility Tests: [Please write a general chemical reaction(s) for any positive solubility test result(s) that you obtain for your unknown compound. I was wondering if anyone has any clue why the amount of calcium carbonate formed (i. The reaction between copper (II) carbonate and sulphuric acid produces carbon dioxide gas. This analysis is done volumetrically by using a characteristic reaction of carbonate compounds, namely their reaction with acids. Then, approximately 2 mL of distilled water were added in order to flush the calcium carbonate completely into the flask. Soluble salts, mostly with calcium, are subsequently leached out, whereas insoluble salts along with amorphous hydrogels, remain in the corroded layer. Add some marble chips to hydrochloric acid in a conical flask with a side Give each student a Rate of reaction graphs sheet and an Information sheet and allow time for them. 0 mol/dm 3 and 1. There is one Calcium atom on the left and one Calcium atom on the right so this is balanced. Hydrochloric acid + sodium carbonate sodium. When the concentration of a reactant increases, the rate of reaction also increases. The rate of the reaction is the speed that the reaction takes place so by measuring the rate I How can I get rid of precipitation of CaCO3 and 2018-11-30 · XRD analysis shows that calcium oxalate and calcium carbonate were formed and ICP analysis shows that the remaining calcium in seawater using oxalic acid is about 0. Write a balanced chemical equation for a neutralisation reaction, mentioning the physical state of the reactants and the products. the reaction takes place spontaneously. Its chemical and physical properties lie behind the modern-day uses of limestone The principal mineral component of limestone is a crystalline form of calcium carbonate known as calcite. The collision must be with sufficient energy. The reaction between an acid and a base to give a salt and water is known as a neutralisation reaction. Measuring reaction rates Procedure 1 Using a measuring cylinder, 50 cm3 of 2 mol dm-3 hydrochloric acid are measured out and poured in a dry conical flask. Which of the following solutions would give the greatest initial rate of reaction and largest volume of gas when reacted with 1. From: Comprehensive Polymer Science and Supplements, 1989. 5 Reaction of Metallic Oxides with Acids Activity 2. State one similarity and Outline three different methods to distinguish between equimolar solutions of these acids in the laboratory. Also, the rule of the higher concentration of acid, the faster reaction it can be; the lower concentration of acid,. Limestone - calcium carbonate, (CaCO3) - dissolves and at a low rate. Calcium carbonate (limestone) is very insoluble in pure water but readily reacts in acid according to the reaction below. There are tutorials on the site to help with writing formulae. Then, $\pu{0. 1 M zinc chloride. An Investigation into the Factors Affecting the Rate of the Reaction between Calcium Carbonate and Hydrochloric Acid A chemical reaction is the process by which atoms or groups of atoms are redistributed, resulting in a change in the molecular composition of substances. How changing the concentration of HCl affects the reaction with CaCO3 finding the rate of reaction of marble chips and hydrochloric acid changing the Calcium Carbonate + HCl Reaction. An alternative MICP from calcium formate by Methylocystis parvus OBBP is presented here to overcome these disadvantages. 1 on page 478, record the levels of CO2: Pre-Industrial (circa 1750) Present Concentration Current Rate of Increase (%/yr) CO 2 Acids can have a profound effect on marine animals and plants that have shells made of calcium carbonate. The major component of eggshells is calcium carbonate, CaCO 3(s). Temperature can change easily and is difficult to keep constant. Acid-base titration of commercial antacids is a common high school chemistry lab exercise. This condition is not common, but in situations were the equilibrium system has little head space and the mixing of air is restricted the attendant pH and slow rate of return to equilibrium is of interest. 4 oz, 400 mL/13. The flask contains 20 g of calcium carbonate chips, and the simulation is paused when it is loaded up. The products of the neutralization reaction between calcium carbonate and hydrochloric. We assume that Ca2+ ions (liberated on the reaction surface) react with SO2− 4 ions to form solid gypsum crystals (CaSO 4) which, in turn, may accumulate. 16D) organised radially ( Fig. Hydrochloric acid = HCl. The results showed that the research and hypothesis were all accurate, and the results were that the larger is surface area is the fast it will react. Explain this observation. The fizz produced in sherbet is a reaction between a food acid and a carbonate. The purpose of the nitric acid in this reaction is to prevent the precipitation of undesired silver salts, like AgOH, which can occur with non-acidic conditions. 5 M hydrochloric acid necessary to neutralize it, titrate it, and calculate results and yield. Hydrogen chloride, chlorine. 2H 2 O and molecular weight 457. 2HCl(aq) + CaCO 3(s) → CaCl. This experiment is a reaction between a certain amount of magnesium and a certain volume of hydrochloric acid, which occurs in the following equation: Mg + 2HCl -> MgCl2 + H2 There must be a collision between the two particles in order for the reaction to take place. What is the balanced equation for the reaction between sulfuric acid and sodium sulfite? 1 Educator Answer what are the controlled, independent and dependent variables in an experiment, where you. Calcium carbonate and hydrochloric acid order of reaction which is better absorbed calcium carbonate or citrate I connubially zyban buy had not read of a difference between the strengths as some being for acne and some for wrinkles, except for Differin Gel, supposed to be for acne only. We had 5 beakers, one contained 100 mL of water and 0 mL. Continue to stir for 1 minute after addition of the sulfuric acid is complete. What is the difference between percentage yield and atom economy? Answered by Akriti G. Calcium carbonate is a chemical compound with the formula CaCO3. 3 compare with the rate of reaction 3 Expl. The hypothesis was supported by the results in the experiment. Hypothesis: If the temperature in the reaction between calcium carbonate and hydrochloric acid is increased, then the reaction rate between the two solutions would also increase. Calcium carbonate spherulites (Fig. This experiment was Calcium Carbonate (CaCO3) reacts with Hydrochloric Acid (HCl). Two types of reaction occur when baking soda and vinegar combine. This strong acid is highly corrosive and must be handled with appropriate safety precautions. (i) calcium carbonate and dilute nitric acid. Finally, because the materials list 1. For Parts (a) (iv)–(v), candidates were required to write a balanced equation for the reaction between calcium carbonate and nitric acid and to calculate the theoretical yield of carbon dioxide from 1. 4 Sulfuric acid + magnesium carbonate 8. When a suspension of the calcium carbonate was decomposed with droppings of hydrochloric acid, a small amount of the calcium carbonate was observed to survive until almost the last stage in the addition of the acid (Fig. The word equation for this reaction is: calcium carbonate + hydrochloric acid calcium chloride + water + carbon dioxide gas Describe an observation you would make when. 1 M acids (ethanoic acid and hydrochloric acid) The hydrochloric acid had a hydrogen ion concentration of 0. Agricultural lime and antacid tablets both neutralize acid, whether in soil or stomach, and therefore must be bases. 05M is conducted a laboratory to study the rate of reaction at interval time. ' Also if the particles have more energy, they will be traveling faster, making them An Investigation into the Factors Affecting the Rate of the Reaction between Calcium Carbonate and Hydrochloric Acid A chemical reaction is the. Add 5M hydrochloric acid to 5g-10g marble chips. Lab 349 Chemistry Essay 446 Words | 2 Pages. Imagine a reaction between magnesium metal and a dilute acid like hydrochloric acid. Research health effects, dosing, sources, deficiency symptoms, side effects, and interactions here. The rate of reaction between sodium thiosulfate and dilute hydrochloric acid can be measured by the time it takes for the sulfur precipitate to be formed, as shown in the equation below: 2HCl(aq) + Na 2 S 2 O 3 (aq) 2NaCl(aq) + SO 2 (aq) + S(s) + H 2 O(l) Hydrochloric Acid + Sodium Thiosulfate Sodium Chloride + Sulfur Dioxide + Sulfur + Water. The reaction of solid calcium carbonate with hydrochloric acid is a heterogeneous reaction. Kinetics Lab Essay How does the molar concentration of hydrochloric acid affect the rate of pressure in a gas releasing reaction? The aim of this experiment is to record how the change in molar concentration of hydrochloric acid affects the rate of the reaction. V) Gas Forming Reaction 1. Calcium carbonate dissolves in water only to a limited extent, but its solubility is greatly enhanced when the water is acidic. Balance the Chemical Equation for the reaction of calcium carbonate with hydrochloric acid: CaCO3+ HCl -> CaCl2 + CO2 + H2O To balance chemical equations we need to look at each element individually on both sides of the equation. Some (but not all) antacid tablets contain primarily calcium carbonate, along with flavoring, binders, and other inactive ingredients. 1 is for which element Reaction of calcium carbonate with hydrochloric acid Please answer all please answer if u know the answer Please answer allplease if u only the answer then answer it Please answer all please it's a request only if you know answer it In the reaction (2 Fe2+ + H2O2 + 2 H+ -> 2 Fe3+ + 2 H2O), what is the role of H2O2?. 0 mol/dm 3 and 1. 25 cm 3 of 1 M hydrochloric acid was measured by a measuring cylinder and transferred into a plastic sample bottle (Figure 2). The "molecular" equation nominally looks as follows: 2 HCl + CaCO3 --> CaCl2 + H2CO3. It is an acid-base exchange reaction, where an acid reacts with a base to form salt and water. Chemical reactions lab report answers Chemical reactions lab report answers. Measuring reaction rates Material: Calcium carbonate (CaCO3) pieces, 2. Carbonates provide the carbonate ion, CO 3 2–, which can react with H + (aq) to form H 2 O and CO 2. Two types of reaction occur when baking soda and vinegar combine. calcium hydroxide + phosphoric acid Æ 10. A simple example would be the action of hydrochloric acid on calcite (calcium carbonate). When the surface area is increased between calcium carbonate and hydrochloric acid, the reaction rate is sped up. How can the Rate of the Reaction between Calcium Carbonate and dilute Hydrochloric Acid be Measured? HCl + calcium carbonate calcium chloride + carbon dioxide + water. Lime water (1), a dilute solution of calcium hydroxide, turns 'milky' when carbon dioxide is bubbled through it (2), forming a suspension of calcium carbonate. The rest of the powders are soluble. In this mini-lab, sodium carbonate reacts with hydrochloric acid to produce sodium chloride, water and carbon dioxide. Is a reaction of any acid and any metal carbonate going to be endothermic like ethanoic acid Q. To produce a colloidal solution of sulphur, where the solution As the temperature of the system increases or as the concentration of reacting species increases the rate of precipitation of sulphur also increases. The current high global demand for high-quality paper, paint, adhesive/sealant, and plastic, filler industries cannot survive without unique and high-quality precipitated calcium carbonate (PCC). 8 g of calcium carbonate at RTP. , when the acid was transferred from the pipet to the carbonate sample). Use phenolphthalein for the first titration of sodium carbonate with hydrochloric acid, then check your results by doing a second titration with methyl. , 1992; Canti, 1997, 1998a, 1999 ). 16D) organised radially ( Fig. Calcium is a silvery-white metal; it is relatively soft, but much harder than sodium metal. Pipette 10 ml of hydrochloric acid. Because surface waters are in equilibrium with atmospheric carbon dioxide there is a constant concentration of carbonic acid, H 2 CO 3, in the water. To observe the effect of surface area (particle size) on the rate of reaction between magnesium ribbon and hydrochloric acid. H+ ion combines with [CO3]2- to form H2CO3 acid, a weak acid, which decomposes to form H2O and CO2. The rate of the reaction is the speed that the reaction takes place so by measuring the rate I How can I get rid of precipitation of CaCO3 and 2018-11-30 · XRD analysis shows that calcium oxalate and calcium carbonate were formed and ICP analysis shows that the remaining calcium in seawater using oxalic acid is about 0. The reaction of calcium carbonate with sulfuric acid yields carbon dioxide, water, and calcium sulfate. The gradual dissolution of limestone and marble, as well as coral and seashells, in acids is due to acid-base neutralization. The results showed that the research and hypothesis were all accurate, and the results were that the larger is surface area is the fast it will react. It is not a reactant in the balanced net ionic equation for this reaction of chloride ion and silver nitrate. Chemical reactions lab report answers Chemical reactions lab report answers. To clean up hard-water deposits containing calcium carbonate we would use vinegar. CaCO3(s) + 2HCl(aq) right arrow CaCl2(aq) + H2O(l) + CO2(g) How many grams of calcium chloride will be produced when 27. Rate Of Reaction Between Hydrochloric Acid And Calcium. effects the reaction rate of calcium carbonate and hydrochloric acid. When more carbon dioxide is bubbled through it, the carbonate dissolves to form dilute solution of calcium bicarbonate (3). Photo decomposition is a chemical reaction in which a substance is broken down into simple substances by exposure to light (photons). Pepsinogen is secreted by chief cells in the gastric glands of the body and antrum of the stomach. producing the predicted amount of product. base neutralization. Students obtain an over-the-counter antacid (in this case, calcium carbonate), calculate the volume of 0. The number of grams of acid neutralized by a 10 gram tablet. Measure 2 mL of concentrated sulfuric acid, H2SO4, in a 10-mL graduated cylinder. Hence, the rate of a reaction between two phases depends to a great extent on the surface contact between them. Independent variable b. Hydrochloric acid is an important and widely used chemical. To do so we would add known amount of hydrochloric acid to calcium carbonate and after the solid is dissolved we would titrate excess acid with a strong base. Hypothesis: In a reaction between calcium carbonate and hydrochloric acid Aim: To investigate the rate of reaction between Calcium Carbonate and Hydrochloric Acid. Record the exact mass in Data Table 2. Acetic acid is able to dissolve calcium carbonate because it is able to form calcium acetate and carbonic acid, which has a pKa of ~6. 1 M Magnesium chloride, 0. Aim: To investigate the rate of reaction between Calcium Carbonate (CaCo3) and hydrochloric acid (HCl) by changing the molarity of the acid and measuring how much Carbon Dioxide (Co2) is produced. After 30 seconds, the magnesium had decreased in mass by 45g. They must collide with each other 2. The reaction of calcium carbonate with sulfuric acid yields carbon dioxide, water, and calcium sulfate. 0 mol/dm 3 and 1. The aqueous layer (density of 1 g/ml) contained dissolved tannin salts and chlophyll. Hydrochloric acid (HCl) is one of the substances found in gastric juices secreted by the lining of the stomach. HCl, or hydrochloric acid, is a strong acid that reacts with sodium carbonate. 63 The alkalinity can be expressed in units ofmgIL of calcium carbonate, based on the reaction, If the water is very basic (pH greater than 8. The molecular formulas for carbonic acid and copper sulfate are H 2 CO 3 and CuSO 4 respectively. When calcium carbonate comes in contact with vinegar which contains acetic acid (CH 3 CO 2 H) a chemical reaction occurs. Add 125 cm 3, 1. An acidic oxide, sulfur dioxide combines with water vapor in the air to produce sulfurous acid in the following reaction:. Solubility in Vinegar (dilute acetic acid). Introduction to the Rate of Reaction. The rate of a reaction may be measured by following the loss of a reactant, or the formation of a product. Such reactions are accompanied by foaming or bubbling, or both, as the gas is released. From: Comprehensive Polymer Science and Supplements, 1989. The pH has a significant effect on the stability of sodium hypochlorite solutions. the container with the products after the reaction has taken place. The rest of the powders are soluble. Between 1-3 min the reaction is proceeding vigorously, however later on between 4-6 min the reaction tampers off and the consumption in HCl lowers per min intervals. Rate of reaction between Hydrochloric Acid and Calcium Carbonate Calcium carbonate reacts with hydrochloric acid to form carbon dioxide gas One way of following the rate of reaction at which it reacts is to measure the volume of carbon dioxide produced at certain time intervals during the reaction. Introduction to the Rate of Reaction. On average, a 1. The rate of the reaction is the speed that the reaction takes place so by measuring the rate I How can I get rid of precipitation of CaCO3 and 2018-11-30 · XRD analysis shows that calcium oxalate and calcium carbonate were formed and ICP analysis shows that the remaining calcium in seawater using oxalic acid is about 0. Which changes increase the rate of reaction between calcium hydrochloric acid? carbonate and dilute 1 2 3 increasing the concentration of the acid increasing the temperature increasing the size of the pieces of calcium carbonate. (Figure 1) 2. They test a stoichiometric version of the reaction followed by testing various perturbations on the stoichiometric version in which each reactant (citric acid, sodium bicarbonate and water) is strategically. We saw that this reacted too quickly as we used 10 second intervals and we The aim of this experiment is to find out how concentration affects the rate of reaction of reaction between magnesium and hydrochloric acid Apparatus: Magnesium. Since calcium acetate and carbonic acid would be soluble in the "aqueous" media, acetic acid is able to dissolve CaCO3 through this acid/base reaction. Requirements for Goal 2: • As a group, you will need to synthesize 0. The rate of a reaction depends on how many successful collisions there are in a given unit of time. More products are formed per unit time and hence the rate of reaction is higher; For example, the reaction between magnesium ribbon and hydrochloric acid is as follow: Zn(s) + HCl(aq) → ZnCl2(aq) + H2(g) The rate of reaction can be found by measuring the time taken for the magnesium ribbon to disappear from sight. Reaction of an acid with a metal, observing the hydrogen produced. Add some marble chips to hydrochloric acid in a conical flask with a side Give each student a Rate of reaction graphs sheet and an Information sheet and allow time for them. carbon dioxide sequestered) from the reaction between calcium ions and atmospheric carbon dioxide is slightly less at a. The following reaction will be tested: Na2CO3 10H2O + 2HCl –> 2NaCl + CO2 + 11H2O To measure the rate of the reaction an Explorer. Calcium carbonate is a chemical compound with the expression CaCO3. Nitric acid reacts with copper and produces different products according to its concentration. #4-4: Acid-Base Titration. 0g Of Hydrochloric Acid?B) Which Reactant Is In. ) 4 Write a rate equation for the reaction. The chemical formula for this reaction is #CH_4 + 2O_2 -> CO_2 + 2H_2O#. This reaction is also known as double decomposition reaction. From this balanced chemical equation, you can see that for every mole of calcium carbonate that is present initially, one mole of carbon dioxide will be produced. Thermal decomposition reaction (Thermolysis) Decomposition of calcium carbonate:Calcium carbonate (lime stone) decomposes into calcium oxide (quick lime) and carbon dioxide when heated. An Investigation Into the Factors Affecting the Rate of the Reaction Between Calcium Carbonate and Hydrochloric Acid 2472 Words | 10 Pages. Calcium carbonate and hydrochloric acid order of reaction which is better absorbed calcium carbonate or citrate I connubially zyban buy had not read of a difference between the strengths as some being for acne and some for wrinkles, except for Differin Gel, supposed to be for acne only. Calcium, magnesium or barium hydroxide are used as catalysts for the production of preimpregnating resins, which are precipitated at the end of the reaction by the addition of sulfuric acid. Hydrochloric Acid Calcium Chloride 70% Flake , Find Complete Details about Hydrochloric Acid Calcium Chloride 70% Flake,Hydrochloric Acid Calcium Chloride 70% Flake,Calcium Chloride Hydrochloric Acid,Exporting Calcium Chloride from Chloride Supplier or Manufacturer-Shandong Gaiyuan Import & Export Co. Rate Of Reaction Between Hydrochloric Acid And Calcium. Also commonly referred to as slaked lime or hydrated lime; calcium hydroxide is formed as a result of hydrating lime (calcium oxide, CaO). The equivalent weight of an acid or base for neutralization reactions or of any other compound that acts by double decomposition is the quantity of the compound that will furnish or react with or be equivalent to 1. Photo decomposition is a chemical reaction in which a substance is broken down into simple substances by exposure to light (photons). Lime is significantly cheaper than caustic (NaOH), but is much more difficult to. Calcium carbonate and magnesium hydroxide is a combination antacid used to treat indigestion, upset stomach, and heartburn. The plastic sample bottle with hydrochloric acid was put into the suction flask carefully. calcium carbonate/hydrochloric acid experiment is often used to establish that rate increases as particle size decreases. An acidic oxide, sulfur dioxide combines with water vapor in the air to produce sulfurous acid in the following reaction:. Where CaA2 is just some salt byproduct. Calcium is necessary for bone formation and maintenance. This is another familiar lab reaction. MEASURING A REACTION RATE Recall - measuring the rate of a reaction means measuring the change in the amount of a reactant or the amount of a product. The chemical compound hydrochloric acid (or muriatic acid) is the aqueous (water-based) solution of hydrogen chloride gas (HCl). 2HCl(aq)+CaCO3(s)→CaCl2(aq)+H2O(l)+CO2(g) Later, if we can find out the number of mole of unreacted acid, number of mole of calcium carbonate can then be found out. 01% and sodium 0. Hydrogen chloride occurs as either a colourless liquid with a an irritating, pungent odour, or a colourless to slightly yellow gas which can be shipped as a liquefied compressed gas; highly soluble in water. Evolution of carbon dioxide occurred immediately at the first drop of the hydrochloric acid. iron(II) sulfide + hydrochloric acid Æ 8. 16D) organised radially ( Fig. Calcium carbonate and hydrochloric acid. Mg + 2HCl MgCl 2 + H 2 3. Extension, or instead of using sodium carbonate solution: l Take a small amount (one-quarter spatula measure) of solid copper carbonate on a filter paper. Acid-base titration of commercial antacids is a common high school chemistry lab exercise. 0 mol dm-3 hydrochloric acid and cotton wool. Standardisation of a hydrochloric acid solution using a standard solution of sodium carbonate Theory Laboratory grade hydrochloric acid is not sufficiently pure to be used as a primary standard. Calcium Chloride 99% (CaCl2) use for. Rate Of Reaction Between Hydrochloric Acid And Calcium. Then 2 mL of concentrated hydrochloric acid were added and allowed to react with the calcium carbonate. On average, a 1. What can be measured to calculate the rate of reaction between magnesium and hydrochloric acid? magnesium. 1 M zinc chloride. This analysis is done volumetrically by using a characteristic reaction of carbonate compounds, namely their reaction with acids. Hydrochloric acid, HCl(aq) Sodium hydroxide Acetic acid, HC 2 H 3 O 2 (aq) Potassium iodide Calcium carbonate Fructose, C 6 H 12 O 6 (aq) Silver chloride Vial 1. Write a hypothesis for the reaction between calcium carbonate and hydrochloric acid Method: 3. An Investigation into the Factors Affecting the Rate of the Reaction between Calcium Carbonate and Hydrochloric Acid A chemical reaction is the process by which atoms or groups of atoms are redistributed, resulting in a change in the molecular composition of substances. Kinetics Lab Essay How does the molar concentration of hydrochloric acid affect the rate of pressure in a gas releasing reaction? The aim of this experiment is to record how the change in molar concentration of hydrochloric acid affects the rate of the reaction. First, excess acid is reacted with the calcium carbonate in eggshells. Some acids react with some metals and not all acids react with all metals. To induce calcium carbonate precipitation, M. iron(II) sulfide + hydrochloric acid Æ 8. Rate of Reaction Between Hydrochloric Acid and Calcium Chloride Essay. The rate of the reaction is the speed that the reaction takes place so by measuring the rate I How can I get rid of precipitation of CaCO3 and 2018-11-30 · XRD analysis shows that calcium oxalate and calcium carbonate were formed and ICP analysis shows that the remaining calcium in seawater using oxalic acid is about 0. The rate of a reaction is the speed at which a chemical reaction happens. Using acid-base back titration it is also possible to determine amount of substances that can be easily dissolved in acids, like calcium carbonate. asked by Allison on March 10, 2015; Chemistry. Acid-base titration of commercial antacids is a common high school chemistry lab exercise. Acid rain contains carbonic acid, nitric acid and sulphuric acid (CO2, NO2, and SO2). The reaction of an acid with calcium carbonate is an example of an acid-base reaction. I 16 Grand Slam titles include four at the U. , 2H + + M = 2M + + H 2. Calcium carbonate and hydrochloric acid. The reaction between copper (II) carbonate and sulphuric acid produces carbon dioxide gas. In the investigation I am going to find out how the surface area affects the rate of reaction by measuring the amount of gas produced and weight loss in a reaction between small/large pieces of Marble Chips (Calcium Carbonate) and Hydrochloric. 0% Chalk is almost pure calcium carbonate. This is the fifth of six quizzes recapping those ideas and it looks specifically at word equations - the method of writing down chemical reactions in words rather than in symbols. Fifteen weight percent sodium hypochlorite will decompose approximately 10 times faster than 5 wt% sodium hypochlorite at 25°C. The rate of the reaction is the speed that the reaction takes place so by measuring the rate I How can I get rid of precipitation of CaCO3 and 2018-11-30 · XRD analysis shows that calcium oxalate and calcium carbonate were formed and ICP analysis shows that the remaining calcium in seawater using oxalic acid is about 0. /**/ Marble chips and acid are placed in the flask but separated by a piece of card - preventing the reaction from proceeding. Kinetics Lab Essay How does the molar concentration of hydrochloric acid affect the rate of pressure in a gas releasing reaction? The aim of this experiment is to record how the change in molar concentration of hydrochloric acid affects the rate of the reaction. Lanthanum chloride was prepared via a reaction between lanthanum oxide and HCl solution. , when the acid was transferred from the pipet to the carbonate sample). pretty girl. CaCo3 + 2HCL CaCl2 + H2 O + CO2Reactant ProductI chose this reaction because it is easy to control as the production of. One of the products is a common acid. The chemical equation that is going to be followed throughout the experiment will be: CaCO3(s) + 2HCl(aq) –> CaCl2(aq) + H2O(l) + CO2(g) Multiple factors can impact on the rate of the reaction between the acidic factors and the marble particles. Rate Of Reaction Between Hydrochloric Acid And Calcium. calcium carbonate/hydrochloric acid experiment is often used to establish that rate increases as particle size decreases. In the guided activity, students vary the concentration of hydrochloric acid while the mass of calcium carbonate stays the same. So the equation of the reaction between calcium carbonate and HCl is: CaCO3+2HCl =CaCl2+H2O+CO2. • The reaction rate will be directly proportional to the HCl concentration. We can work out its purity by measuring how much carbon dioxide is given off. The word equation for this reaction is: calcium carbonate + hydrochloric acid calcium chloride + water + carbon dioxide gas Describe an observation you would make when. The amount of hydrochloric acid used up. When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. Let the mixture stand for 10 minutes. The curve shows this. Mg + 2HCl MgCl 2 + H 2 3. We can assume that when a strong acid reacts to form a weak acid that ions of the strong acid are removed from the solution. The equivalent weight of an acid or base for neutralization reactions or of any other compound that acts by double decomposition is the quantity of the compound that will furnish or react with or be equivalent to 1. The following reaction will be tested: Na2CO3 10H2O + 2HCl –> 2NaCl + CO2 + 11H2O To measure the rate of the reaction an Explorer. HCl(aq) + CaCO3(s) arrow CaCl2(aq) + CO2(g) + H2O(l) Things that affect the reaction rate of this experiment are: 1. The reaction between calcium carbonate and hydrochloric acid can be written two ways: CaCO 3 + 2 HCl → CaCl 2 + H 2 CO 3 and H 2 CO 3 → H 2 O + CO 2 ↑ calcium hydrochloric calcium carbonic carbonic water carbon. The neutralization process between sodium carbonate and hydrochloric acid could be shown in the following chemical equation. Calcium carbonate (limestone),$\text{CaCO}_{3}$, and dilute hydrochloric acid react to give off carbon dioxide gas. With stirring, slowly add the 2 mL of concentrated sulfuric acid to the dodecanol in the beaker. Aim: To investigate the rate of reaction between Calcium Carbonate (CaCo3) and hydrochloric acid (HCl) by changing the molarity of the acid and measuring how much Carbon Dioxide (Co2) is produced. For Parts (a) (iv)–(v), candidates were required to write a balanced equation for the reaction between calcium carbonate and nitric acid and to calculate the theoretical yield of carbon dioxide from 1. 2500 N potassium hydroxide, 0. The rate of reaction of H+ with soil carbonate is highly dependent on accessibility of carbonate surface sites to reaction with H+; therefore, sample pretreatment and degree of dispersion of the soil strongly influences the results of the pH-stat titration procedure (Hartwig & Loeppert, 1991). The initial concentration of La3+ was 1 mol/L and the and. Add 10 ml of 1 M HCl to it and start the stop-watch immediately when half of the hydrochloric acid solution has been added. Free Essay / Term Paper: reaction between calcium carbonate and hydrochloric acid Aim: To investigate the effect of temperature on reaction rate. These software are used for simulation of various chemical reactions. Calcium Carbonate + Hydrochloric Acid --> Calcium Chloride + Water + Carbon Dioxide. Then secure a pressure equalizing addition funnel filled with concentrated hydrochloric and add. Calcium is a silvery-white metal; it is relatively soft, but much harder than sodium metal. Carbonates provide the carbonate ion, CO 3 2–, which can react with H + (aq) to form H 2 O and CO 2. We had 5 beakers, one contained 100 mL of water and 0 mL. The rate of a reaction depends on how many successful collisions there are in a given unit of time. The rate law for this reaction will have the following form: rate = k[HCl]^n. Calcium Carbonate is a medication used to prevent or treat low blood calcium levels in people who do not get enough calcium from their diets. Reaction of an acid with a metal, observing the hydrogen produced. The apparatus shown in the diagram was used to investigate the rate of reaction of excess marble chips with dilute hydrochloric acid, HCl. The catalytic decomposition of hydrogen peroxide. Which of the following conditions will give the fastest rate of reaction?. Record the exact mass in Data Table 2. Hydrochloric acid + sodium carbonate sodium chloride + water + carbon dioxide. The Dependent Variables: The measured duration of acid-metal reaction in seconds and the rate of gas bubbles. 5 mol/dm 3 hydrochloric acid as well (keeping the temperature, volume of acid the same, and the mass of the calcium carbonate the same). 10 g of chalk was reacted with an excess of dilute hydrochloric acid. Aim: To investigate the rate of reaction between Calcium Carbonate (CaCo3) and hydrochloric acid (HCl) by changing the molarity of the acid and measuring how much Carbon Dioxide (Co2) is produced. Though both acids contain HCl, when we consider the purity of two acids, hydrochloric acid is the pure f. 2 CH 3 COOH + CaCO 3 = Ca(CH 3 COO) 2 + H 2 O + CO 2 Acetic acid + Calcium carbonate = Calcium acetate + Water. To clean up hard-water deposits containing calcium carbonate we would use vinegar. The acetic acid in the vinegar reacts with the calcium carbonate in the eggshell to make calcium acetate plus water and carbon dioxide that you see as bubbles on the surface of the shell. Which changes increase the rate of reaction between calcium hydrochloric acid? carbonate and dilute 1 2 3 increasing the concentration of the acid increasing the temperature increasing the size of the pieces of calcium carbonate. In this probe, a new repressive chemical composing was developed. Jan 24 2016 IM STUMPED 2 The reaction of solid calcium carbonate with hydrochloric acid is a heterogeneous reaction The rate law for this reaction will have the following form rate kHCln Explain why the concentration of calcium carbonate does not appear in the rate law 3 In step 5 why is it necessary to make sure the stopcock is in the open. Explain why the concentration of calcium carbonate does not appear in the rate law. 5% Hydrochloric Acid Water-insoluble compounds that are insoluble in 5% NaOH are tested with 5% hydrochloric acid (HCl). 1 M Potassium cyanide, 5% Sodium hydroxide, 0. 1) Assume that the hydrochloride acid solution has the same destiny and specific heat capacity as water 2) The amount heat released by the reaction are totally absorbed by the calorimeter. What can be measured to calculate the rate of reaction between magnesium and hydrochloric acid? magnesium. In this probe, a new repressive chemical composing was developed. reaction (2): q=mc T=100*4. Then additional carbon dioxide can react with the carbonate. 3), a quantity called the phenolphthalein alkalinity can be determined. The HS 2 O 3 – ion is a reactive intermediate, reacting further with additional S 2 O 3 2– ions to produce polymeric ions containing multiple S atoms. MEASURING A REACTION RATE Recall - measuring the rate of a reaction means measuring the change in the amount of a reactant or the amount of a product. Although this reaction is often done in. What type of chemical reaction is it?. reaction of calcium carbonate with hydrochloric acid: CaCO3+ HCl -> CaCl2 + CO2 + H2O. Example:Study the reaction: Between calcium carbonate and excess 1 mol dm-3 hydrochloride acid. How would the rate or reaction between 1. May decompose upon heating to product corrosive and/or. There is one Calcium atom on the left and one Calcium atom on the right so this is balanced. Chemical reactions lab report answers Chemical reactions lab report answers. In this instance, we are timing how long it takes for our reaction to produce 50cm3 of carbon dioxide. Sincethe rate law pretty much shows the variation between the rate of reaction at different concentrations, it This is alswo what is happening here, in the reaction between solid calcium carbonate with hydrochloric acid, the reaction is independent on the concentration hence it is not. CaCO 3 + 2HCl → CaCl 2 + CO 2 + H 2 O The hydrogen carbonate anion is also a base. Write three chemical equations (one full, one ionic and one net-ionic) for the reaction between aqueous calcium chloride and aqueous sodium carbonate. A phosphate level of 0. CaCo3 + 2HCL CaCl2 + H2 O + CO2Reactant ProductI chose this reaction because it is easy to control as the production of. In this experiment hydrochloric acid reacts with calcium carbonate. 5 g of calcium carbonate. An acid and a metal carbonate react to form a salt, water and carbon dioxide. Add some marble chips to hydrochloric acid in a conical flask with a side Give each student a Rate of reaction graphs sheet and an Information sheet and allow time for them. We saw that this reacted too quickly as we used 10 second intervals and we The aim of this experiment is to find out how concentration affects the rate of reaction of reaction between magnesium and hydrochloric acid Apparatus: Magnesium. Calcium carbonate is an inorganic salt used as an antacid. Add the hydrochloric acid and a few drops of Ethanol into the Erlenmeyer flask with the eggshell powder. Rate of Decomposition of Calcium Carbonate Isaac Rodriguez 2/4/17 Mark Guiao Ulices Gomez Purpose: The purpose of this lab is to design an experiment for the reaction of hydrochloric acid with calcium carbonate through the rules of kinetics in order to determine if either mass loss analysis or gas collection would accurately measure rate laws and reaction rates. The rate of reaction between sodium thiosulphate and hydrochloric acid also increases with increase in temperature. The overall equation for the reaction between sodium carbonate solution and dilute hydrochloric acid is: If you had the two solutions of the same concentration, you would have to use twice the volume of hydrochloric acid to reach the equivalence point - because of the 1 : 2 ratio in the. Between 1-3 min the reaction is proceeding vigorously, however later on between 4-6 min the reaction tampers off and the consumption in HCl lowers per min intervals. This is another familiar lab reaction. Hence the effervescence is observed when acids are dropped on the floor. What can be measured to calculate the rate of reaction between magnesium and hydrochloric acid? magnesium. White granular anhydrous calcium chloride is manufactured with the raw materials of high quality calcium carbonate and pure hydrochloric acid and through several processes of synthesis, filtering, evaporation, concentration and drying. The chemical formula for carbonic acid is: _____ Referring to Table 16. What can be measured to calculate the rate of reaction between magnesium and hydrochloric acid? magnesium + hydrochloric magnesium acid chloride + hydrogen The. Calcium carbonate is the principal mineral component of limestone. Find the best calcium supplements based on our tests and reviews, and know which products that failed, and see our Top Picks based on quality, dose, and cost. The following reaction will be tested: Na2CO3 10H2O + 2HCl –> 2NaCl + CO2 + 11H2O To measure the rate of the reaction an Explorer. What type do you think the second reaction is? Why? Hydrochloric acid and calcium carbonate react in the same way that baking soda and. Hydrochloric acid reacts with calcium carbonate to form calcium chloride, carbon dioxide and. It all depends on what is reacting with what. the reaction takes place spontaneously. The made inhibitor is based on the aqueous solutions of oxiethilidendiphosphone acid, hydrochloric acid, ammonium chloride, polyethylene polyamine-N-methylphosphonic acid and isopropyl intoxicant to forestall the. calcium carbonate + hydrochloric acid → calcium chloride + carbon dioxide + water. Carbonate reservoir stimulation involves the injection of reactive fluids, most commonly hydrochloric acid (HCl) However, the treatments are often constrained by the very rapid rate of reaction between HCl and carbonate rock, which results in limited depth of live acid penetration and ineffective flow. Which of the following solutions would give the greatest initial rate of reaction and largest volume of gas when reacted with 1. Price of calcium carbonate in pakistan calcium carbonate reacts with dilute hydrochloric acid as shown in the equation below calcium carbonate effervescent granules calcium carbonate to clean teeth equilibrium between calcium carbonate calcium oxide and carbon dioxide the precipitation of calcium carbonate lab report calcium carbonate vitamin d. Take the rate equation below: Rate = k[A] n [B] m. The "molecular" equation nominally looks as follows: 2 HCl + CaCO3 --> CaCl2 + H2CO3. MEASURING A REACTION RATE Recall - measuring the rate of a reaction means measuring the change in the amount of a reactant or the amount of a product. The amount of hydrochloric acid used up. 10 g of chalk was reacted with an excess of dilute hydrochloric acid. 2 Rates of reaction and factors affecting rate (ESCMX). The active ingredient in Tums is calcium carbonate, CaCO3, a base. The rate of the overall reaction (Equation 3) and its dependence on the concentration of HCl are important concerns in environmental chemistry due to. Measure 2 mL of concentrated sulfuric acid, H2SO4, in a 10-mL graduated cylinder. Add 5M hydrochloric acid to 5g-10g marble chips. Limestone is one familiar form of calcium carbonate. 6 The reaction of calcium carbonate with an acid produces carbon dioxide gas. State how the results would differ for each acid. Reducing the carbonate alkalinity increases the solubility of calcium salts such as calcium carbonate by preventing the formation of carbonate (CO3) alkalinity and its subsequent reaction with calcium hardness to form. 2HCl + Na 2 CO 3 2NaCl + H 2 O + CO 2. Then secure a pressure equalizing addition funnel filled with concentrated hydrochloric and add. Reacts exothermically with organic bases (amines, amides) and inorganic bases (oxides and hydroxides of metals). The rate of the reaction is the speed that the reaction takes place so by measuring the rate I How can I get rid of precipitation of CaCO3 and 2018-11-30 · XRD analysis shows that calcium oxalate and calcium carbonate were formed and ICP analysis shows that the remaining calcium in seawater using oxalic acid is about 0. The number of moles of acid neutralized per gram of tablet. Experiment 4/Alkalinity ofStreams and Lakes. First of all, we mixed toothpaste with water and HCl, and heated it up. Lanthanum carbonate is described as white to off-white powder. How would the rate or reaction between 1. Hydrochloric acid, HCl(aq) Sodium hydroxide Acetic acid, HC 2 H 3 O 2 (aq) Potassium iodide Calcium carbonate Fructose, C 6 H 12 O 6 (aq) Silver chloride Vial 1. It has a role as an antacid, a food colouring, a food firming agent and a fertilizer. Normally, antacids like magnesium hydroxide and calcium carbonate neutralize hydrochloric acid in the stomach, forming magnesium chloride and calcium chloride. The reactions of calcium carbonate with water where the gas phase is restricted or negligible. These parameters are evaluated by means of a fitting procedure based. Because of the rapid reaction with the hydrochloric acid in the stomach, the dosing of Renvela powder or tablet is anticipated to be similar to that of the sevelamer. In this paper we present a mathematical model for the reaction between calcium carbonate (CaCO 3) and a solution containing sulfuric acid (H 2SO 4). It is a common substance found in rocks as the minerals calcite and aragonite. calcium carbonate was greater than that of marble chips. Calcium carbonate can also be used to lower the acidity in soil and lakes as a result of acid rain. 50 cm 3 of 1M hydrochloric acid is a six-fold excess of acid. 7 n Take a small amount of copper oxide in a beaker and add dilute hydrochloric acid slowly. Rate of reaction between Hydrochloric Acid and Calcium Carbonate Calcium carbonate reacts with hydrochloric acid to form carbon dioxide gas One way of following the rate of reaction at which it reacts is to measure the volume of carbon dioxide produced at certain time intervals during the reaction. Measuring the rate of a reaction means measuring the change in the amount of a reactant or the amount of a product. 1 M zinc chloride. Marble is calcium carbonate, formula CaCO3. 2HCl(aq) + CaCO 3(s) → CaCl. In other words, the acid and the base (carbonate) are neutralized, or their pH gets. Then additional carbon dioxide can react with the carbonate. Then,$\pu{0. It is not. 20 describe the laboratory preparation of carbon dioxide from calcium carbonate and dilute hydrochloric acid The reaction between any metal carbonate and an acid will form carbon dioxide. reaction of the gases ammonia, NH 3, and oxygen, O 2. The reaction rate of hydrochloric acid with carbonate rock at acid-fracturing conditions depends on controllable variables such as injection rate and acid concentration, and variables inherent to the carbonate rock (permeability, mineral composition, etc. In this reaction, the acid HCl breaks apart the calcite, (CaCO 3 ) into the salt (CaCl 2 ) and carbonic acid (H 2 CO 3 ) which further breaks down into water (HOH) and carbon dioxide (CO 2 ), which bubbles off. Reagents: • calcium carbonate • methyl red indicator • ethylenediamminetetracetic acid, disodium. Lime is by far the most economically favorable alkaline reagent to use for acid neutralization. Rate Of Reaction Between Hydrochloric Acid And Calcium. Within the context of acid rain, the susceptibility of a lake to acidification depends on the alkalinity of the lake water. Suppose a 1. Calcium carbonate + hydrochloric acid → calcium chloride + water + carbon dioxide CaCO 3 (s) + 2HCl(aq) → CaCl 2 (aq) + H 2 O(l) + CO 2 (g) The volume of carbon dioxide gas produced can be. Because the reaction between sodium carbonate and hydrochloric acid proceeds in two stages, you can use more than one indicator. Examples of acid and carbonate reactions. Write a balanced chemical equation for a neutralisation reaction, mentioning the physical state of the reactants and the products. The effect of temperature on the rate of reaction can be investigated. In this experiment, a standard solution of sodium carbonate is used to determine the exact concentration of a hydrochloric acid solution. Let's consider a few neutralization reactions and how we write the equations. 2HCl (aq) + CaCO 3(s) CaCl 2 (aq) + CO 2(g) + H 2 O (l) The rate of this reaction can be measured by looking at the rate at which the product carbon dioxide gas is formed. Oily residue from your hands will inhibit the reaction with acid. Word Equation: Calcium + Hydrochloric Calcium + Carbon Dioxide + Water Carbonate Acid Chloride Chemical Equation: CaCo3 (s) + 2Hcl (aq) CaCl2 + Co2 (g) + H20 (l) The key variables in this experiment that affect the rate of reaction is, surface area, temperature, mass of CaCo3 and. reactions that result in the formation of a gas, often carbon dioxide (Figure 4. such things as de-scaling The United States Environmental Protection Agency rates and regulates hydrochloric acid as a toxic. 5 Nitric acid + calcium hydroxide 8. Chemical reactions lab report answers Chemical reactions lab report answers. 'n Suur en 'n metaalkarbonaat reageer om 'n sout. The rate of this reaction depends on the amount of sulfur dioxide in the air. To find out how much acid you will need to. 50 cm 3 of 1M hydrochloric acid is a six-fold excess of acid. Steel pickling Hydrochloric acid is used in pickling operations for carbon, alloy and stainless steels. Chemistry 104: Analysis of Commercial Antacid Tablets. This produces a solution of bicarbonate. Firstly, let's think about some different types of reactions and how quickly or slowly they occur. Yet calcium reacts at a moderate rate, whereas sodium reacts so rapidly that the reaction is Hence, the rate of a reaction between two phases depends to a great extent on the surface Figure 1. If a reaction has a low rate, that means the molecules combine at a slower speed than a reaction with a high rate. In this experiment hydrochloric acid reacts with calcium carbonate. 0 g of calcium carbonate are combined with 12. The reaction between sodium thiosulphate solution and. Hydrogen chloride occurs as either a colourless liquid with a an irritating, pungent odour, or a colourless to slightly yellow gas which can be shipped as a liquefied compressed gas; highly soluble in water. This prediction may be true but the results tell me otherwise, my prediction could mainly proven wrong because it was the first time the test took place and fair test taking place would be hard for the first time. Ÿ Hydrochloric acid. I don't know whether a reaction between calcium carbonate and nitric acid is a neutralisation reaction or It is also an exothermic reaction. Titration with sodium hydroxide is then used to determine the % CaCO 3. NiCO3(s) + 2 HCl(aq) NiCl2(aq) + H2CO3(aq). The reaction between copper (II) carbonate and sulphuric acid produces carbon dioxide gas. acid chloride. The reaction between hydrochloric acid and sodium carbonate is a two-stage one, so two different indicators can be used in the titration procedure. Carbonic acid changes back and forth in equilibrium to bicarbonate (HCO 3-) + H+ and carbonate (CO 3 2-) + H+, lowering the pH due to higher concentrations of H+. We aim to design an experiment which is safe to conduct in the lab and will meet the objective to investigate about the factor. 2 Names and Identifiers. When combined, aqueous solutions of sodium carbonate and hydrochloric acid generate an acid-base reaction. As the precipitate forms less of it can be seen. Rate Of Reaction Between Hydrochloric Acid And Calcium. 2 The rate of reaction between calcium carbonate and hydrochloric acid - To use appropriate measurements to - monitor the rate of reaction using mass lost The calcium carbonate is in the form of fairly large pieces of marble so that the particle size does not change much during the reaction. B: Variation of rate of reaction with temperature. The reaction between hydrochloric acid and calcium carbonate (limestone or chalk) is shown below: CaCO 3 + 2 HCl → CaCl 2 + H 2 CO 3. Test solubility similarly to how you tested water solubility. In contrast, calcium nitrate, produced from the reaction with nitric acid, is highly soluble in dilute nitric acid and so no. In the investigation I am going to find out how the surface area affects the rate of reaction by measuring the amount of gas produced and weight loss in a reaction between small/large pieces of Marble Chips (Calcium Carbonate) and Hydrochloric. 2H 2 O and molecular weight 457. What can be measured to calculate the rate of reaction between magnesium and hydrochloric acid? magnesium. The current high global demand for high-quality paper, paint, adhesive/sealant, and plastic, filler industries cannot survive without unique and high-quality precipitated calcium carbonate (PCC). Set II: 1 g of zinc powder and 20 cm 3 of 2 mol dm –3 hydrochloric acid at room temperature. Lab 349 Chemistry Essay 446 Words | 2 Pages. Carbonic acid calcium salt (CaCO3). CaCo3 + 2HCL CaCl2 + H2 O + CO2Reactant ProductI chose this reaction because it is easy to control as the production of. Chemistry 104: Analysis of Commercial Antacid Tablets. 5 mol/dm 3 hydrochloric acid as well (keeping the temperature, volume of acid the same, and the mass of the calcium carbonate the same). This article reports the effect of various organic and inorganic additives. The rate of this reaction depends on the amount of sulfur dioxide in the air (Figure 12. Acid Injection: Mineral acid, such as sulfuric or hydrochloric, neutralizes carbonate alkalinity resulting in a decrease in pH. Rate of Decomposition of Calcium Carbonate Isaac Rodriguez 2/4/17 Mark Guiao Ulices Gomez Purpose: The purpose of this lab is to design an experiment for the reaction of hydrochloric acid with calcium carbonate through the rules of kinetics in order to determine if either mass loss analysis or gas collection would accurately measure rate laws and reaction rates. Investigate how macro factors affect the reaction rate between solid calcium carbonate and hydrochloric acid. Bicarbonates and carbonates react with acid to form carbon dioxide bubbles. In other words, the acid and the base (carbonate) are neutralized, or their pH gets. The rate of this reaction can be measured by looking at the rate at which the product carbon dioxide gas is formed. The precipitation reaction cannot be initiated without lanthanum carbonate nuclei being in the system while the acid-based neutralization reaction. 0 mol/dm3 sulfuric acid. The rate of the reaction is the speed that the reaction takes place so by measuring the rate I How can I get rid of precipitation of CaCO3 and 2018-11-30 · XRD analysis shows that calcium oxalate and calcium carbonate were formed and ICP analysis shows that the remaining calcium in seawater using oxalic acid is about 0. This reaction serves as a test for carbon dioxide. How would the rate or reaction between 1. Two of the polymorphs of CaCO 3, calcite and aragonite, can form under different local. Also commonly referred to as slaked lime or hydrated lime; calcium hydroxide is formed as a result of hydrating lime (calcium oxide, CaO). Limestone - calcium carbonate, (CaCO3) - dissolves and at a low rate. This can be done by having the reaction take place in an open flask on an electric weighing machine (a pan. The rate of a reaction may be measured by following the loss of a reactant, or the formation of a product. Rate of reaction between Hydrochloric Acid and Calcium Carbonate Calcium carbonate reacts with hydrochloric acid to form carbon dioxide gas One way of following the rate of reaction at which it reacts is to measure the volume of carbon dioxide produced at certain time intervals during the reaction. Write a balanced chemical equation for the observed reaction. The calcium carbonate will be evident as a gelatinous precipitate (solid). How can the Rate of the Reaction between Calcium Carbonate and dilute Hydrochloric Acid be Measured? HCl + calcium carbonate calcium chloride + carbon dioxide + water. So it's like both a displacement and a decomposition reaction. Between 1-3 min the reaction is It also doesn't explain if the mixture can be stirred or shaken which would also affect the rate of reaction. When calcium carbonate comes in contact with vinegar which contains acetic acid (CH 3 CO 2 H) a chemical reaction occurs. € 6 (a)€€€€ Write a balanced equation for the reaction. Purpose: This lab was completed to find the rate of reaction of hydrochloric acid with calcium carbonate, to examine the effects of amount of reactant Background: The rate of reaction is defined as how fast or slow a reaction takes place; for the purposes of this lab, this is essentially equivalent. State an equation for the reaction of magnesium carbonate with dilute hydrochloric acid MgCo3 + 2HCl --> MgCl2 + CO2 + H2O The experiment is repeated using a sample of hydrochloric acid with double the volume, but half the concentration of the original acid. The ethanoic. With stirring, slowly add the 2 mL of concentrated sulfuric acid to the dodecanol in the beaker. The reaction proceeds quite rapidly during the initiate mixture of Hydrochloric acid and Calcium Carbonate chips. The rate of a reaction may be measured by following the loss of a reactant, or the formation of a product. The aqueous layer (density of 1 g/ml) contained dissolved tannin salts and chlophyll. The time taken for a certain amount of sulfur to form is used to indicate the rate of the reaction. Also commonly referred to as slaked lime or hydrated lime; calcium hydroxide is formed as a result of hydrating lime (calcium oxide, CaO). Write a hypothesis for the reaction between calcium carbonate and hydrochloric acid Method: 3. An acid and a metal carbonate react to form a salt, water and carbon dioxide. Investigate the effect of surface area on the rate of the reaction between calcium carbonate and hydrochloric acid Does an increase in the surface area Rationale Hypothesis If the concentration of hydrochloric acid is increased, then the change in mass overtime will decrease. Calcium carbonate reacts with hydrochloric acid to form calcium chloride, water and carbon dioxide. 8 (anhydrous mass). In general, a neutralisation reaction can be written as – Base + Acid → Salt + W ater 2. Mann, Nicholas R; Tranter, Troy J. Introduction to the Rate of Reaction. 0074 g of hydroxide ion; e. NaHCO 3 + HCH 3 COOH → NaCH 3 COOH + CO 2 + H 2 O. The Reaction Between Sodium Thiosulphate and Hydrochloric Acid The Effect on the concentration on the reaction between Sodium Thiosulphate and Hydrochloric acid Our aim in this investigation is to find out how the concentration of Sodium Thiosulphate affects the rate at which it reacts with Hydrochloric acid. Sodium carbonate is a basic compound, meaning that it generates hydroxide ions (OH⁻) when dissolved in water. Thermal decomposition reaction (Thermolysis) Decomposition of calcium carbonate:Calcium carbonate (lime stone) decomposes into calcium oxide (quick lime) and carbon dioxide when heated. Reactions fail to go to completion because liquid adheres to containers or evaporates Solids get stuck in filter paper or lost in the purification process. In this lab students will investigate the rate of decomposition of calcium carbonate with different concentrations of hydrochloric acid to learn more about kinetics and the rates of chemical reactions.
|
2020-06-03 05:11:34
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.651276707649231, "perplexity": 2865.319010445698}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347432237.67/warc/CC-MAIN-20200603050448-20200603080448-00480.warc.gz"}
|
https://www.mathworks.com/help/robust/ug/advanced-topics.html
|
This last section gives a few hints for making the most out of the LMI Lab. It is directed toward users who are comfortable with the basics, as described in Tools for Specifying and Solving LMIs.
Structured Matrix Variables
Fairly complex matrix variable structures and interdependencies can be specified with `lmivar`. Recall that the symmetric block-diagonal or rectangular structures are covered by Types 1 and 2 of `lmivar` provided that the matrix variables are independent. To describe more complex structures or correlations between variables, you must use Type 3 and specify each entry of the matrix variables directly in terms of the free scalar variables of the problem (the so-called decision variables).
With Type 3, each entry is specified as either 0 or ±xn where xn is the n-th decision variable. The following examples illustrate how to specify nontrivial matrix variable structures with `lmivar`. The following examples show variable structures with uncorrelated and interdependent matrix variables.
Specify Matrix Variable Structures
Suppose that the variables of the problem include a 3-by-3 symmetric matrix X and a 3-by-3 symmetric Toeplitz matrix, Y, given by:
`$Y=\left(\begin{array}{ccc}{y}_{1}& {y}_{2}& {y}_{3}\\ {y}_{2}& {y}_{1}& {y}_{2}\\ {y}_{3}& {y}_{2}& {y}_{1}\end{array}\right).$`
The variable Y has three independent entries, and thus involves three decision variables. Since Y is independent of X, label these decision variables n + 1, n + 2, and n + 3, where n is the number of decision variables involved in X. To retrieve this number, define the Type 1 variable X.
```setlmis([]) [X,n] = lmivar(1,[3 1]); n```
```n = 6 ```
The second output argument `n` gives the total number of decision variables used so far, which in this case is `n` = 6. Given this number, you can define Y.
`Y = lmivar(3,n+[1 2 3;2 1 2;3 2 1]);`
An equivalent expression to define Y uses the MATLAB(R) command `toeplitz` to generate the matrix.
`Y = lmivar(3,toeplitz(n+[1 2 3]));`
To confirm the variables, visualize the decision variable distributions in X and Y using `decinfo`.
```lmis = getlmis; decinfo(lmis,X)```
```ans = 3×3 1 2 4 2 3 5 4 5 6 ```
`decinfo(lmis,Y)`
```ans = 3×3 7 8 9 8 7 8 9 8 7 ```
Specify Interdependent Matrix Variables
Consider three matrix variables, X, Y, and Z, with the following structure.
`$X=\left(\begin{array}{cc}x& 0\\ 0& y\end{array}\right),\phantom{\rule{1em}{0ex}}Y=\left(\begin{array}{cc}z& 0\\ 0& t\end{array}\right),\phantom{\rule{1em}{0ex}}Z=\left(\begin{array}{cc}0& -x\\ -t& 0\end{array}\right),$`
where x, y, z, and t are independent scalar variables. To specify such a triple, first define the two independent variables, X and Y, which are both Type 1.
```setlmis([]); [X,n,sX] = lmivar(1,[1 0;1 0]); [Y,n,sY] = lmivar(1,[1 0;1 0]);```
The third output of lmivar gives the entry-wise dependence of X and Y on the decision variables $\left({\mathit{x}}_{1},{\mathit{x}}_{2},{\mathit{x}}_{3},{\mathit{x}}_{4}\right):=\text{\hspace{0.17em}}\left(\mathit{x},\mathit{y},\mathit{z},\mathit{t}\right).$
`sX`
```sX = 2×2 1 0 0 2 ```
`sY`
```sY = 2×2 3 0 0 4 ```
Using lmivar, you can now specify the structure of the Type 3 variable Z in terms of the decision variables ${\mathit{x}}_{1}=\mathit{x}$ and ${\mathit{x}}_{4}=\mathit{t}$.
`[Z,n,sZ] = lmivar(3,[0 -sX(1,1);-sY(2,2) 0]);`
Because `sX(1,1) `refers to ${\mathit{x}}_{1}$ and `sY(2,2)` refers to ${\mathit{x}}_{4}$, this expression defines the variable `Z` as:
`$Z=\left(\begin{array}{cc}0& -{x}_{1}\\ -{x}_{4}& 0\end{array}\right)=\left(\begin{array}{cc}0& -x\\ -t& 0\end{array}\right).$`
Confirm this results by checking the entry-wise dependence of `Z` on its decision variables.
`sZ`
```sZ = 2×2 0 -1 -4 0 ```
Complex-Valued LMIs
The LMI solvers are written for real-valued matrices and cannot directly handle LMI problems involving complex-valued matrices. However, complex-valued LMIs can be turned into real-valued LMIs by observing that a complex Hermitian matrix L(x) satisfies
L(x) < 0
if and only if
`$\left(\begin{array}{cc}\mathrm{Re}\left(L\left(x\right)\right)& \mathrm{Im}\left(L\left(x\right)\right)\\ -\mathrm{Im}\left(L\left(x\right)\right)& \mathrm{Re}\left(L\left(x\right)\right)\end{array}\right)<0.$`
This suggests the following systematic procedure for turning complex LMIs into real ones:
• Decompose every complex matrix variable X as
X = X1 + jX2
where X1 and X2 are real
• Decompose every complex matrix coefficient A as
A = A1 + jA2
where A1 and A2 are real
• Carry out all complex matrix products. This yields affine expressions in X1, X2 for the real and imaginary parts of each LMI, and an equivalent real-valued LMI is readily derived from the above observation.
For LMIs without outer factor, a streamlined version of this procedure consists of replacing any occurrence of the matrix variable X = X1 + jX2 by
`$\left(\begin{array}{cc}{X}_{1}& {X}_{2}\\ -{X}_{2}& {X}_{1}\end{array}\right)$`
and any fixed matrix A = A1 + jA2, including real ones, by
`$\left(\begin{array}{cc}{A}_{1}& {A}_{2}\\ -{A}_{2}& {A}_{1}\end{array}\right).$`
For instance, the real counterpart of the LMI system
MHXM < X, X = XH > I (1)
reads (given the decompositions M = M1 + jM2 and X = X1 + jX2 with Mj, Xj real):
`$\begin{array}{c}{\left(\begin{array}{cc}{M}_{1}& {M}_{2}\\ -{M}_{2}& {M}_{1}\end{array}\right)}^{T}\left(\begin{array}{cc}{X}_{1}& {X}_{2}\\ -{X}_{2}& {X}_{1}\end{array}\right)\left(\begin{array}{cc}{M}_{1}& {M}_{2}\\ -{M}_{2}& {M}_{1}\end{array}\right)<\left(\begin{array}{cc}{X}_{1}& {X}_{2}\\ -{X}_{2}& {X}_{1}\end{array}\right)\\ \left(\begin{array}{cc}{X}_{1}& {X}_{2}\\ -{X}_{2}& {X}_{1}\end{array}\right)`
Note that X = XH in turn requires that ${X}_{1}={X}_{1}^{H}$ and ${X}_{2}+{X}_{2}^{T}=0$. Consequently, X1 and X2 should be declared as symmetric and skew- symmetric matrix variables, respectively.
Assuming, for instance, that MC5×5, the LMI system (Equation 1) would be specified as follows:
```M1=real(M), M2=imag(M) bigM=[M1 M2;-M2 M1] setlmis([]) % declare bigX=[X1 X2;-X2 X1] with X1=X1' and X2+X2'=0: [X1,n1,sX1] = lmivar(1,[5 1]) [X2,n2,sX2] = lmivar(3,skewdec(5,n1)) bigX = lmivar(3,[sX1 sX2;-sX2 sX1]) % describe the real counterpart of (1.12): lmiterm([1 1 1 0],1) lmiterm([-1 1 1 bigX],1,1) lmiterm([2 1 1 bigX],bigM',bigM) lmiterm([-2 1 1 bigX],1,1) lmis = getlmis ```
Note the three-step declaration of the structured matrix variable `bigX`,
`$bigX=\left(\begin{array}{cc}{X}_{1}& {X}_{2}\\ -{X}_{2}& {X}_{1}\end{array}\right).$`
1. Specify X1 as a (real) symmetric matrix variable and save its structure description `sX1` as well as the number `n1` of decision variables used in X1.
2. Specify X2 as a skew-symmetric matrix variable using Type 3 of `lmivar` and the utility `skewdec`. The command `skewdec(5,n1)` creates a 5-by–5 skew-symmetric structure depending on the decision variables `n1 + `1, `n1 + `2,...
3. Define the structure of `bigX` in terms of the structures `sX1` and `sX2` of X1 and X2.
See Structured Matrix Variables for more details on such structure manipulations.
Specifying cTx Objectives for mincx
The LMI solver `mincx` minimizes linear objectives of the form cTx where x is the vector of decision variables. In most control problems, however, such objectives are expressed in terms of the matrix variables rather than of x. Examples include Trace(X) where X is a symmetric matrix variable, or uTXu where u is a given vector.
The function `defcx` facilitates the derivation of the c vector when the objective is an affine function of the matrix variables. For the sake of illustration, consider the linear objective
`$\text{Trace}\left(X\right)+{x}_{0}^{T}P{x}_{0}$`
where X and P are two symmetric variables and x0 is a given vector. If `lmsisys` is the internal representation of the LMI system and if x0, X, P have been declared by
```x0 = [1;1] setlmis([]) X = lmivar(1,[3 0]) P = lmivar(1,[2 1]) : : lmisys = getlmis ```
the c vector such that ${c}^{T}x=\text{Trace}\left(X\right)+{x}_{0}^{T}P{x}_{0}$ can be computed as follows:
```n = decnbr(lmisys) c = zeros(n,1) for j=1:n, [Xj,Pj] = defcx(lmisys,j,X,P) c(j) = trace(Xj) + x0'*Pj*x0 end ```
The first command returns the number of decision variables in the problem and the second command dimensions c accordingly. Then the `for` loop performs the following operations:
1. Evaluate the matrix variables X and P when all entries of the decision vector x are set to zero except xj: = 1. This operation is performed by the function `defcx`. Apart from `lmisys` and `j`, the inputs of `defcx` are the identifiers `X` and `P` of the variables involved in the objective, and the outputs `Xj` and `Pj` are the corresponding values.
2. Evaluate the objective expression for ```X:= Xj``` and` P:= Pj`. This yields the j-th entry of `c` by definition.
In our example the result is
```c = 3 1 2 1 ```
Other objectives are handled similarly by editing the following generic skeleton:
```n = decnbr( LMI system ) c = zeros(n,1) for j=1:n, [ matrix values ] = defcx( LMI system,j, matrix identifiers) c(j) = objective(matrix values) end ```
When solving LMI problems with `feasp`, `mincx`, or `gevp`, it is possible to constrain the solution x to lie in the ball
xTx < R2
where R > 0 is called the feasibility radius. This specifies a maximum (Euclidean norm) magnitude for x and avoids getting solutions of very large norm. This may also speed up computations and improve numerical stability. Finally, the feasibility radius bound regularizes problems with redundant variable sets. In rough terms, the set of scalar variables is redundant when an equivalent problem could be formulated with a smaller number of variables.
The feasibility radius R is set by the third entry of the options vector of the LMI solvers. Its default value is R = 109. Setting R to a negative value means “no rigid bound,” in which case the feasibility radius is increased during the optimization if necessary. This “flexible bound” mode may yield solutions of large norms.
Well-Posedness Issues
The LMI solvers used in the LMI Lab are based on interior-point optimization techniques. To compute feasible solutions, such techniques require that the system of LMI constraints be strictly feasible, that is, the feasible set has a nonempty interior. As a result, these solvers may encounter difficulty when the LMI constraints are feasible but not strictly feasible, that is, when the LMI
L(x) ≤ 0
has solutions while
L(x) < 0
has no solution.
For feasibility problems, this difficulty is automatically circumvented by `feasp`, which reformulates the problem:
Find x such that
L(x) ≤ 0 (2)
as:
Minimize t subject to
Lx < t × I.
In this modified problem, the LMI constraint is always strictly feasible in x, t and the original LMI Equation 2 is feasible if and only if the global minimum tmin of Equation 2 satisfies
tmin ≤ 0
For feasible but not strictly feasible problems, however, the computational effort is typically higher as `feasp` strives to approach the global optimum tmin = 0 to a high accuracy.
For the LMI problems addressed by `mincx` and `gevp`, nonstrict feasibility generally causes the solvers to fail and to return an “infeasibility” diagnosis. Although there is no universal remedy for this difficulty, it is sometimes possible to eliminate underlying algebraic constraints to obtain a strictly feasible problem with fewer variables.
Another issue has to do with homogeneous feasibility problems such as
ATP + P A < 0, P > 0
While this problem is technically well-posed, the LMI optimization is likely to produce solutions close to zero (the trivial solution of the nonstrict problem). To compute a nontrivial Lyapunov matrix and easily differentiate between feasibility and infeasibility, replace the constraint P > 0-by-P > αI with α > 0. Note that this does not alter the problem due to its homogeneous nature.
Semi-Definite B(x) in gevp Problems
Consider the generalized eigenvalue minimization problem
Minimize λ subject to
A(x) < λB(x), B(x) > 0, C(x) <0. (3)
Technically, the positivity of B(x) for some xRn is required for the well-posedness of the problem and the applicability of polynomial-time interior-point methods. Hence problems where
`$B\left(x\right)=\left(\begin{array}{cc}{B}_{1}\left(x\right)& 0\\ 0& 0\end{array}\right)$`
with B1(x) > 0 strictly feasible, cannot be directly solved with `gevp`. A simple remedy consists of replacing the constraints
A(x) < B(x), B(x) > 0
by
`$A\left(x\right)<\left(\begin{array}{cc}Y& 0\\ 0& 0\end{array}\right),\text{ }Y<\lambda {B}_{1}\left(x\right),\text{ }{B}_{1}\left(x\right)>0$`
where Y is an additional symmetric variable of proper dimensions. The resulting problem is equivalent to Equation 3 and can be solved directly with `gevp`.
Efficiency and Complexity Issues
As explained in Tools for Specifying and Solving LMIs, the term-oriented description of LMIs used in the LMI Lab typically leads to higher efficiency than the canonical representation
A0 + x1A1 + ... + xNAN < 0. (4)
This is no longer true, however, when the number of variable terms is nearly equal to or greater than the number N of decision variables in the problem. If your LMI problem has few free scalar variables but many terms in each LMI, it is therefore preferable to rewrite it as Equation 4 and to specify it in this form. Each scalar variable xj is then declared independently and the LMI terms are of the form xjAj.
If M denotes the total row size of the LMI system and N the total number of scalar decision variables, the flop count per iteration for the `feasp` and `mincx` solvers is proportional to
• N3 when the least-squares problem is solved via Cholesky factorization of the Hessian matrix (default) [2].
• M-by-N2 when numerical instabilities warrant the use of QR factorization instead.
While the theory guarantees a worst-case iteration count proportional to M, the number of iterations actually performed grows slowly with M in most problems. Finally, while `feasp` and `mincx` are comparable in complexity, `gevp` typically demands more computational effort. Make sure that your LMI problem cannot be solved with `mincx` before using `gevp`.
Solving M + PTXQ + QTXTP < 0
In many output-feedback synthesis problems, the design can be performed in two steps:
1. Compute a closed-loop Lyapunov function via LMI optimization.
2. Given this Lyapunov function, derive the controller state-space matrices by solving an LMI of the form
M + PTXQ + QTXTP < 0 (5)
where M, P, Q are given matrices and X is an unstructured m-by-n matrix variable.
It turns out that a particular solution Xc of Equation 5 can be computed via simple linear algebra manipulations [1]. Typically, Xc corresponds to the center of the ellipsoid of matrices defined by Equation 5.
The function` basiclmi` returns the “explicit” solution Xc:
```Xc = basiclmi(M,P,Q) ```
Since this central solution sometimes has large norm, `basiclmi` also offers the option of computing an approximate least-norm solution of Equation 5. This is done by
```X = basiclmi(M,P,Q,'Xmin') ```
and involves LMI optimization to minimize ∥X ∥.
Get trial now
|
2022-01-29 08:19:04
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 19, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7508363723754883, "perplexity": 859.3913586240707}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320300573.3/warc/CC-MAIN-20220129062503-20220129092503-00014.warc.gz"}
|
https://socratic.org/questions/5992fbbdb72cff54c4c0557d
|
# Question 0557d
Aug 20, 2017
${\text{C"_5"H}}_{10}$
#### Explanation:
For starters, you know that because you're dealing with a hydrocarbon, i.e. a compound that only contains carbon and hydrogen, you can use the mass of the sample and the mass of carbon to determine the mass of hydrogen present in the sample.
overbrace("0.60 g")^(color(blue)("mass of carbon")) + overbrace(xcolor(white)(.)"g")^(color(blue)("mass of hydrogen")) = overbrace("0.70 g")^(color(blue)("mass of hydrocarbon"))
You can thus say that the sample contains
$\text{0.70 g " - " 0.60 g = 0.10 g}$
of hydrogen. To convert the masses to moles, use the molar masses of carbon and hydrogen, respectively.
$\text{For C: " 0.60 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12color(red)(cancel(color(black)("g")))) = "0.05 moles C}$
$\text{For H: " 0.10 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1color(red)(cancel(color(black)("g")))) = "0.10 moles H}$
Now, to find the mole ratio that exists between carbon and hydrogen in the hydrocarbon, divide both values by the smallest one.
"For C: " (0.05 color(red)(cancel(color(black)("moles"))))/(0.05color(red)(cancel(color(black)("moles")))) = 1
"For H: " (0.10color(red)(cancel(color(black)("moles"))))/(0.05color(red)(cancel(color(black)("moles")))) = 2
This means that the hydrocarbon contains carbon and hydrogen in a $1 : 2$ mole ratio. Since this represents the smallest whole number ratio that can exist between the two elements in this hydrocarbon, you can say that the empirical formula of the hydrocarbon will be
${\text{C"_1"H"_2 implies "CH}}_{2} \to$ the empirical formula
Now, the molecular formula is always a multiple of the empirical formula.
$\text{molecular formula" = color(blue)(n) xx "empirical formula}$
This means that you have
("CH"_ 2)_ color(blue)(n) = "C"_ color(blue)(n)"H"_ (2color(blue)(n))
In order to find the value of $\textcolor{b l u e}{n}$, use the fact that the mass of $1$ mole of ${\text{C"_ color(blue)(n)"H}}_{2 \textcolor{b l u e}{n}}$ is equal to $\textcolor{b l u e}{n}$ times the mass of $1$ mole of ${\text{CH}}_{2}$.
The molar mass of ${\text{CH}}_{2}$ is equal to
${\text{12 g mol"^(-1) + 2 xx "1 g mol"^(-1) = "14 g mol}}^{- 1}$
Since you know that the molar mass of the hydrocarbon is equal to ${\text{70 g mol}}^{- 1}$, you can say that you have
$14 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g mol"^(-1)))) * color(blue)(n) = 70 color(red)(cancel(color(black)("g mol}}^{- 1}}}}$
This will get you
$\textcolor{b l u e}{n} = \frac{70}{14} = 5$
Therefore, the molecular formula of the hydrocarbon is
color(darkgreen)(ul(color(black)(("CH"_ 2)_ 5 = "C"_ 5 "H"_ 10)))#
|
2021-12-06 20:14:33
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 23, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.720732569694519, "perplexity": 1368.4751997459455}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363312.79/warc/CC-MAIN-20211206194128-20211206224128-00446.warc.gz"}
|
https://www.askiitians.com/forums/Wave-Motion/a-man-beats-drum-at-a-certain-distance-from-a-larg_198121.htm
|
Click to Chat
1800-2000-838
+91-120-4616500
CART 0
• 0
MY CART (5)
Use Coupon: CART20 and get 20% off on all online Study Material
ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.
There are no items in this cart.
Continue Shopping
A man beats drum at a certain distance from a large mountain. He slowly increases the rate of beating and finds that the Echo is not heard distinctly when the drum beating is at the rate of 40 per minute. he moves by 80m toward the mountain and finds that the Echo is again not heard distinctly when the rate of the beating of the drum is 1 per second what is the initial distance of the man from the mountain? (1)120m;(2)240m;(3)270m;(4)340m
3 months ago
Arun
8936 Points
Please find below the solution to the asked query:Case I. When the rate of drum beating is 40 per minuteTherefore, the interval between two drum beats = 60/40 = 1.5 sLet the distance between the mountain and the man be x m.Speed of sound, v=2xt=2x1.5=4x3 .....(1)Case II. When the man moves 80 m towards the mountainSpeed of sound, v=2(x−80)t=2(x−80)1=2x−160 .....(2)From eq. (1) and (2),2x−160=4x3⇒6x−480=4x⇒2x=480⇒x=240 mHope this information will clear your doubts about the numerical on echo.
3 months ago
Think You Can Provide A Better Answer ?
## Other Related Questions on Wave Motion
View all Questions »
• Complete Physics Course - Class 12
• OFFERED PRICE: Rs. 2,756
• View Details
### Course Features
• 101 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution
• Complete Physics Course - Class 11
• OFFERED PRICE: Rs. 2,968
• View Details
### Course Features
• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions
|
2018-05-20 21:41:25
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3043307960033417, "perplexity": 5172.984188940992}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794863689.50/warc/CC-MAIN-20180520205455-20180520225455-00019.warc.gz"}
|
http://wp.hygienesystemsgroup.com/chaminade-high-xjwady/notion-relation-formula-208038
|
These progress bars compare two numbers (say, 1 and 100) and visually show your progress. A very simple example on how to use Property Types as variables. The domain of W= {1, 2, 3, 4} The set of second elements is called the range of the relation. Subscribe to our newsletter and stay updated. relation properties and formulas. Notion Formula Reference. Notion is an all-in-one style productivity booster. I added a project to each task by simply clicking on the “Project” field for each task. To do this, we introduce the idea of function. A relation is a set of ordered pairs. Notion is a note-ta k ing app on steroids which provides many, MANY features — from simply writing a bulleted list of notes, to creating linked databases (tables) with conditional formulae. Solving can produce more than one solution because different input values can produce the same output value. What can’t you do in Notion that you can do in Excel. Click on “Text” under “Property Type” and select “Relation” in the “Advanced” section of the drop-down menu. This is useful for displaying complex formulas on your web page. It contains tips and tricks to help you get things done. Easiest way is to show you an example. You will be presented with a pop-up menu with three options:eval(ez_write_tag([[300,250],'theproductiveengineer_net-leader-4','ezslot_12',125,'0','0'])); Click on the “Relation” drop-down and select the relation you want to use for the rollup (in our example, the “Tasks for Project” relation). The range of W= {120, 100, 150, 130} Reflexive relation: When the Same element is present as co-domain or simply R in X is a relation with (a, a) ∈ R ∀ a ∈ X. See all Articles by James LeGrand See James LeGrand's Expert Page Get Updates on Love Get Updates on James LeGrand. A function is a specific type of relation in which each domain value, or input, leads to exactly one range value, or output. Once you’ve connected the column to your existing database. Click the “Create relation” button to create your relation in your database. Like a menu. I was very lucky to find a program that that has helped me grow my blog to over 35,000 page views and a YouTube channel that is growing month-over-month. In Excel terms, this means that you can only apply formulas to an entire column (a la copy and paste down). Rollups allow you to derive values from fields in other databases. The first is in-line creation. Now, go to the database that you connected your relation to. Notion VIP; My Account; Home; Become a Maker; Cart; Checkout; My account ; Shop 0.00 0 items; Notion Market. This is part six of a 12-part web series reviewing how to use Notion to manage your life and increase productivity. 2013-02-08T09 # An hour time part separated by a T 2013-02-08 09 # An hour time part separated by a space 2013-02-08 09:30 # An hour and minute time part 2013-02-08 … Founder of Nutt Labs, William Nutt is the creator of Notion VIP and Notion Market. I’m Jimmy and I spend a lot of my time in front of computers, phones and tablets trying to get my work done as efficiently as possible. If you’re using Notion as a comprehensive productivity system you’ll probably draw from a similar pool of operators and functions. In my example, my “Tasks” database has a column called “Estimated Time to Complete (hour)” that provides an estimate of how much time I think it will take to complete each task. Is it possible to get the property of a related item in a formula. Next, click on the “Property” drop-down and select the field you want to use to create the rollup from. Notion offers a way to create and manage tables all within their applications including mobile. Notion TrackerSuite is a collection of 12 productivity templates and 10-part video training series. In this course, we'll go step-by-step through everything that Notion has to offer, so that you will be able to understand the software from beginning to end. Skillshare is an online learning platform with courses on pretty much anything you want to learn. In this regard I advise that you do read the marvelous books "A Radical Approach to Real Analysis" and "A Radical Approach to Lebesgue's Theory of Integration" by David M. Bressoud. This is an example of an ordered pair. Example notation using the halo system can be seen below. This formula is inspired by David Allen’s GTD, combining an if statement, the dateAdd function and now() to test if an action (i.e. Open Notion to the database you want to create a relation on Click the “ + ” button in the last column of your database Click in the text box and type in a name for your relation Click on “ Text ” under “ Property Type ” and select “ Relation “ When building tables in Notion, you create columns to hold your values. A new tool that blends your everyday work apps into one. Syntax: Refers to the order of letters and terms in your formula to return the right value. This step-by-step beginner's guide will teach you everything you need to know to get up and running in Google Sheets! Project 24 by Income School is the program that I have used. Notice that the relation to my tasks database is now populated. Range is the set of all second coordinates: so B. Tell whether the relation is a function. Hi! See and . Identify the independent and dependent variables. After hearing about the buzzy productivity app called Notion, you migrated from Evernote and are excited for this new chapter in note-taking. This reference provides technical details and examples for every Notion function, operator and constant, as well as the patterns used to format dates … Evaluating Functions in Algebraic Forms. For example, 2. Premium Notion resources from celebrated experts. Beware that many of them convert to timestamp seconds, notion is using timestamp with milliseconds, so you might need to add three zeros at the end. Type in the name of the database you want to connect to and select it from the drop-down list. Notion has been featured on the Tools They Use podcast, for inside tips check out that episode here. Hier sollte eine Beschreibung angezeigt werden, diese Seite lässt dies jedoch nicht zu. Notion formulas are good. In Fibery formulas … This guide will teach you everything you need to know to use the timeline view in Notion effectively! We'll cover topics like: Notion blocks. Notice the arrow before the name of your column (like shown above). In this first video in a series of Notion formula tutorials, we'll create a bare-bones project management space. Click on one of the cells in the relation column (“Estimated Total Time” for my example). We'll cover topics like: Notion blocks. Examples. Posted by 1 year ago. Notion is a wiki service that supports database and table functions. However, formula editor does not have an autocomplete, so you have to type quite many things and the formula creation process is kinda slow: Quite basic formula takes 40 seconds to create in Notion. In our example, I want the “Estimated Time to Complete” field so I selected that one. Enclosed below is a table containing all of the property types available in a Notion table: Consider the relation that sends a student to the courses that student is taking. If you’re new to Notion functions, consider starting with Meet Notion’s Formula Property and Formulas for Work: Part 1.If you just need … For instance, the following example uses properties from a "Tasks" database and returns a formula inside a separate "Sleep Tracker." The team over at Notion.so are committed to providing users with an experience that they won’t get anywhere else. And why do these formulas look so different? Notion is one of my favorite tools in my productivity toolbox. This site is owned and operated by Productive Blogs LLC. Included is a "Tasks" and "Projects" database of which are related to one another. Two of the biggest reasons why it feels this way are relations and rollups. Nothing really special about it. Scroll to the far right column and press the “+” button to create a new column in your database. In this regard I advise that you do read the marvelous books "A Radical Approach to Real Analysis" and "A Radical Approach to Lebesgue's Theory of Integration" by David M. Bressoud. Close. Everything you need to know about Notion formulas. Our Notion Course Supercharge your Productivity is currently closed for enrollment, but sign up here to be notified when we re-open. Functions. Consider the relation that sends a … To learn more about Skillshare and its vast library of courses and get 30% off, click the link below:eval(ez_write_tag([[336,280],'theproductiveengineer_net-medrectangle-4','ezslot_2',114,'0','0'])); If you are just starting out with Notion and aren’t sure where to get started, you really need to check out my comprehensive beginner’s guide to Notion. Now … An ordered pair, commonly known as a point, has two components which are the x and y coordinates. And as you sit down to write your first If Statement, you expect to see a familiar Excel-like layout and it hits you: Where are the cells? It requires you to convert a number into text (a string) using the format function. Integrations. Here is a snapshot of my tasks database that contains a relation to my “projects” database. Check our compilation of top Notion tips and tricks to improve your productivity. Looking to master the timeline view in Notion? These difficulties led to the notion of function as a relation (or a rule of correspondence) and the need for a formula was removed. Definition of a Relation, Domain, and Range. Each Backlog item has a status property ("In Progress, Done, etc"). 1. What’s a Notion function? You can also generate an image of a mathematical formula using the TeX language (pronounced "tek" or "tech"). The pairing of the student number and his corresponding weight is a relation and can be written as a set of ordered-pair numbers. Free Notion templates will be provided for those who want to dive right in! I want to show how one can use multiple databases to create a conditional formula in Notion. Notion is really unique when it comes to productivity tools and that's what makes it so great! This is a common source of error. Ep 24 // Notion Database Relations & Rollups. More … You should think of a function as a mathematical object that expresses a functional relationship between two sets. September 5, 2020. =. https://www.khanacademy.org/.../cc-8th-function-intro/v/relations-and-functions Function: A relation from a set of inputs to a set of possible outputs where each input is related to exactly one output. When I do that, a list of the projects appears and I click the “+” button to add it. It really is an obsession of mine as I think we live in the golden age of productivity apps. Relations 1. It accomplishes this by adding up all the estimated hours for each task in each project. In my example, I wanted the total number of hours so I chose the “Sum” option. Relations allow you to create a relationship between two different databases. Why don’t these property types align? Each user story contains a relation to any related Backlog items (it's a two way relation). For this purpose, a large number of mathematical formulae have been suggested as measures of statistical association. 3.5 Relations and Functions: Basics A. Notion databases. In mathematics, of course, we need to work with functional relationships in the abstract. In this Notion Database tutorial you’ll learn the difference between tables and databases, how to use roll-ups and relations and the powerful linked databases feature. A binary relation from A to B is a subset of a Cartesian product A x B. R t•Le A x B means R is a set of ordered pairs of the form (a,b) where a A and b B. Excel has way more calculations and formulas. That is why I created The Productive Engineer blog. It's the all-in-one workspace for you and your team If you want descriptive text that uses the output of a formula here’s how to use the concat function to display the number of days a task has been overdue. Write a rule in function notation for the situation. Our results situate the class forcing theorem in the rich hierarchy of theories between GBC and Kelley-Morse set theory KM. Hope you find this useful. can I make another database with a relation property that relates and spits out dates from another database with a formula? 5:00 Rollup explained Rollups bring properties from a related database (one that is defined as a relation) 10:30 Update to action zone / Action items creating a rollup to goals. In mathematics (specifically set theory), a binary relation over sets X and Y is a subset of the Cartesian product X × Y; that is, it is a set of ordered pairs (x, y) consisting of elements x in X and y in Y. Close • Posted by 39 minutes ago. In this post we’ll list and explain the formula property type, operators, functions and share common formula examples that you can then use in your own workspaces. 3:00 Creating a new "entry" in a database is called a "record" or an "entry" A field is another term for property . The Full Notion Course here: http://bit.ly/2HWFE98 A first look at how to craft your first relational database all within Notion. Notice anything different? Our relationships tend to be a great source of agony. This allows you to reference data from one database and display/access it in another.eval(ez_write_tag([[250,250],'theproductiveengineer_net-large-mobile-banner-1','ezslot_5',120,'0','0'])); The first thing you want to do is open up Notion to the database that you want to add a relation to. And while there are some Notion integrations, may folks still pick Airtable over Notion when they need heavy duty access to automated workflows. The eccentricity of an ellipse is, most simply, the ratio of the distance c between the center of the ellipse and each focus to the length of the semimajor axis a. The figures we supply along the text to visualize the mathematics are faithful representations of Lobachevskian geometry in a Poincare 2D ball model. Notion Formulas are powerful in their own right, but a true API integration would be a game changer for formulas. Learn how do we write functions as rule. This basically says take the value from the cell above then add the value of the cell 2 to the left and subtract the value in the cell directly to the left. A new tool that blends your everyday work apps into one. Once added, you can add more than one entry, and it’ll populate that row. These difficulties led to the notion of function as a relation (or a rule of correspondence) and the need for a formula was removed. With modular productivity applications like Notion, Airtable and Coda, the need for tables is evolving. Using the == operator to compare two numbers and then the equal() formula for two texts (or strings). Jede Zeile in Notion braucht immer ein Name-Feld. It's the all-in-one workspace for you and your team The next stop on our Notion tutorial is a deep-dive into Notion Databases. Actuarial notation is a shorthand method to allow actuaries to record mathematical formulas that deal with interest rates and life tables.. Lets say I have a database with dates. Produces code for directly embedding equations into HTML websites, forums or blogs. For a preview of the raw power formulas can bring to a workflow, check out this video (queued @ 32:56) where we effortlessly create a GTD review function: The first odd thing you’ll notice is that since Notion acts as a database, it doesn’t have variables, you can’t access an individual cell. Question. dateBetween(now(), fromTimestamp(848103660000), "years") - this would represent a formula for calculation of my age They offer live help and typically respond to users within the day. Each task field for each project now contains all of the tasks assigned to each project. Here I share an example of a custom dashboard with inline databases which heavily rely on relational databases, one of Notion's most powerful feature! Here’s a more advanced nested If statement to convert a text-based tag (Monthly) into a number (30). The data types of values are defined in Notion as property types. This makes calculations like financial modeling and local variables (i.e. eval(ez_write_tag([[300,250],'theproductiveengineer_net-large-mobile-banner-2','ezslot_8',122,'0','0'])); Type in the name you want for the column (for our example, we chose “Project”). Start Here Articles Reviews Blog. They’re beyond the scope of this tutorial, but savvy Notioner Ben Smith has an excellent tutorial. In the 1970s, a version of bisimulation had already been developed by modal logicians to help better understand the relationship between modal logic axioms and their corresponding conditions on Kripke frames. It is so flexible and using the database functions in Notion really feels like having a superpower. All these constitute in the study of relation and function. Pretty cool, right? a string of letters next to each other). This forms the basis for a lot of Notion’s task-management capabilities and is a powerful trick to apply across your pages. It has these things and more because it is a spreadsheet program. It has multi-sheet support. Using Notion to Manage Your Life - Part 6 of 12 - The Goals Database. If so what is the formula for that? The Relationship Formula. A function is a specific type of relation in which each domain value, or input, leads to exactly one range value, or output. Menu. If someone buys an E-mini S&P 500 contract at 2,800, then that single futures contract is worth140,000 ($50 x 2,800). Each column in a Notion table is assigned a property type that defines the type of data the column can contain. Looking to get started with Google Sheets? be put in relation to the expected frequency of co-occurrence by chance, which can be computed from the individual frequencies of the two lexical items. If you like this article, subscribe to our newsletter. By James LeGrand. You should see a column like the one above in the database. Relations and Functions Let’s start by saying that a relation is simply a set or collection of ordered pairs. Therefore,$140,000 is the notional value of … Click the link below to check it out! Other formulae for the eccentricity of an ellipse. This indicates that this column is a rollup of data from another database. Archived. Notion macht gerade eine große Welle in den USA – die App/Plattform will unser Tool Chaos lichten und viele Dinge vereinen, unter anderem Evernote (das sich in den letzten Jahren leider nicht sehr viel weiterentwickelt hat) und sogar Tools wie Trello/Meistertask. Create a column, give it a name, and select “Rollup” as the property type as shown above. Using the learning by building approach I use in my Notion course Supercharge your Productivity you’ll build a music library (i.e. In addition to his independent work, William partners directly with Notion in various capacities: he created the Certified Notion Consultants program, co-authored the Help & Support guide and hosts the Notion … There is absolutely nothing special at all about the numbers that are in a relation. The arrow indicates that this column is related to another database. This instance is evening more confusing to me because there seems to be a notion of sets here (at least to the extent that I am interpreting the $\in$ relation correctly) in a language that I did not think had anything to do with sets. Let’s now look at my projects database. Your new column should appear with a magnifying glass icon in it. You can now choose a field on the other database to your current Notion database in view. I am always finding out how to do new things that make my work life a little easier and wanted a place to share what I have learned. Wouldn’t it be great if I could see this information in my “Projects” database such that I could see how many total hours each project will take based on the sum of all the tasks assigned to each project? A relation is a set of ordered pairs. Jede Bewertung bezieht sich natürlich auf eine Idee. Another common test (using the = operator) to see if a task is overdue. Show how one can use multiple databases to create tables inside of Notion tutorials! It contains tips and tricks to help you get things done to text ( string... Friend in your database as shown in the form see and and life tables applicable to!, Habitica: Give the domain and range of the semimajor axis a to far! Goals database are several models of Lobachevskian geometry in a function even without the graph all their! Once you ’ ve mastered the basics of Notion formulas that deal with interest rates life... Html websites, forums or Blogs really powerful and I click the “ ”... To an entire column ( “ Estimated total Time ” for my example, I want to to. “ projects ” database VIP and Notion Market set or collection of ordered.! Covers the Vertical line test, along with how to know to get up and in. Next stop on our Notion course Supercharge your notion relation formula you ’ ll see an denoting. Introduce the idea of function included is a “ project ” field so I the. The power of Notion output equal to the far right column and press the “ property ” drop-down screenshot! The review window form see and and terms in your database as shown above ) types as.! All these constitute in the screenshot above along with how to know to use to create the rollup the... Of top Notion tips and tricks to help you get things done to an entire column ( a of. No headers review window ( it 's a two way relation ) this massive cheat sheet glass icon in.. Are similar ) tools they use podcast, for inside tips check out that here... The video shows how it can be written as a comprehensive productivity system you ’ see! When building tables in Notion really feels like having a superpower but it doesn ’ t to... Choose from: Properties, Constants, Operators and functions Let ’ s a more advanced nested if statement convert... Database in another database think we live in the following table 2D ball model terms, database... Your web Page and select the field you want to show how one can use multiple databases create. And function notation Items ( it 's a two way relation ) Notion table is assigned a type... Of theories between GBC and Kelley-Morse set theory KM Vertical line test, along with how to to! Not as it has sometimes been represented set up a GTD to-do list, and it ’ probably... Wanted the total number of mathematical formulae have been a member for a. Using Notion to manage your life and increase productivity to return the right.... As I think we live in the study of relation and notion relation formula be helpful to “ leg into ” formulas! Great source of agony a friend in your database be really powerful I! Will be provided for those notion relation formula want to show how one can use multiple to... Statistical association source of agony notion relation formula your current Notion database in another with. Glance, Notion formulas uniform forcing relation applicable simultaneously to all formulas work apps into.. Improve your productivity is currently closed for enrollment, but a true integration. Contains all of the biggest reasons why it feels this way are relations functions. A lot of Notion formula tutorials, we 'll create a bare-bones management... Last updated ; Save as PDF Page ID 10893 ; No headers updated ; Save as PDF Page 10893. Now, go to the function ’ s guide to using Notion as a comprehensive productivity you! Absolutely nothing special at all about the numbers that are in a series of Notion from. ( y=f ( x ) \ ) from another database with a relation is simply a of... Tech '' ) concept of functions versus relations, Let ’ s into! Our compilation of top Notion tips and tricks to improve your Notion.! Nesting can make things tricky trick to apply across your pages Notion table is assigned a property type as in., you can imagine ich euch einen ersten Eindruck über die Fähigkeiten und das Konzept das hinter Notion.... Of possible outputs where each input is related to one another a way create. Representations of Lobachevskian geometry in a Notion table is assigned a property the scope this... Into a number ( 30 ) output equal to the far right column and the. [ 6 ] [ 7 ] [ 8 ] shorthand method for the. As discussed earlier, provide the ability to share and create calculations based on values in tool. Two of the database that you can imagine number of mathematical formulae have been suggested measures! Notion really feels like having a superpower in front of each property, you create new! Format function the “ property ” drop-down and select “ rollup ” as the type. Of course, we need to know to get up and running in Google Sheets counterparts,! Chose the “ Sum ” option role in relation to my tasks database is now populated Estimated Time. Monthly ) into a number ( 30 ) how it can be seen.... A snapshot of my tasks database is now populated the output in database. Like having a superpower data from another database and select it from the screenshot above, the rollup the. You experiment with them to see if a task is overdue should appear with a formula a. Output in the form \ ( y=f ( x ) \ ) ( 30 ) to record mathematical that... To test and show the greater of two numbers suggested as measures statistical. [ 6 ] [ 7 ] [ 7 ] [ 7 ] 8. Notion databases Expert Page get Updates on Love get Updates on Love get on. Called Notion, Airtable and Coda, the need for tables is evolving for this new chapter note-taking. Know it was for me ) but it doesn ’ t have to be a great source of.! Pool of Operators and functions the main letter functions Last updated ; Save as PDF Page ID 10893 ; headers. Students are shown in the form \ ( y=f ( x ) \ ) create based! A project to each other ) merely for atomic formulas, then every such Phas a uniform forcing merely... The right way the dateSubtract returns a number whereas the dateSubtract returns a number the... Each property, you create columns to hold your values and ideas.... A database ” drop-down and select it from the center to the guide: link to the in... Useful for displaying complex formulas on your web Page up as a mathematical object that expresses a relationship! Formulae have been suggested as measures of statistical association tasks database is now.... Way relation ) one of my tasks database is a shorthand method to allow actuaries to mathematical... To visualize the mathematics are faithful representations of Lobachevskian geometry in a series of Notion actuaries record... Geometry [ 6 ] [ 7 ] [ 7 ] [ 8 ] function: a and... Each input is related to one another a year now and just renewed my.. Backup your Evernote notes really feels like having a superpower see from the source it! Tricks to help you get things done 140,000 is the program that I used. The review window geometry [ 6 ] [ 7 ] [ 8 ] first coordinates: so.... Other applications like Notion, Airtable and Coda, the rollup shows the total number of hours so I the... - part 6 of 12 - the Goals database tables inside of Notion the function. Tutorials, we 'll create a conditional formula in Notion class forcing in... ) formula for two texts ( or strings ) Let a and be! ; No headers set or collection of 12 productivity templates and 10-part video training series, Give it a,. Training series die verknüpft werden soll users within the day functional relationships in study. The notes and ideas database their own right, but a true API would! Existing database hier sollte eine Beschreibung angezeigt werden, diese Seite lässt dies jedoch nicht zu,..., and select it from the drop-down list founder notion relation formula Nutt Labs, William Nutt the! Out that episode here my tasks database is now populated relation that sends a student to function... Folks still pick Airtable over Notion when they need heavy duty access to automated workflows your. This column is related to one another role in relation to any related Backlog (! System can be seen below number into text ( a string ) using the TeX language pronounced... Are excited for this new chapter in note-taking existing database integrations, may folks still pick Airtable Notion... Video and Screenshots and demonstrates ways of telling the difference the mathematics are faithful representations Lobachevskian... Are the x and y coordinates users with an experience that they ’! Notion course Supercharge your productivity Todoist, Habitica suppose the weights of students... And life tables after the main letter the Estimated hours for each project my case my. William Nutt is the creator of Notion collection of ordered pairs the type of data from another database (... A column like the one above in the form see and the text to the... Skillshare is an all in one database in another database for enrollment, but savvy Notioner Ben has.
|
2021-10-25 03:41:36
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3821788728237152, "perplexity": 1468.586247416507}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587623.1/warc/CC-MAIN-20211025030510-20211025060510-00619.warc.gz"}
|
https://www.jobilize.com/course/section/horizontal-stretches-and-compressions-by-openstax?qcr=www.quizover.com
|
# 3.5 Transformation of functions (Page 8/21)
Page 8 / 21
Write the formula for the function that we get when we stretch the identity toolkit function by a factor of 3, and then shift it down by 2 units.
$g\left(x\right)=3x-2$
## Horizontal stretches and compressions
Now we consider changes to the inside of a function. When we multiply a function’s input by a positive constant, we get a function whose graph is stretched or compressed horizontally in relation to the graph of the original function. If the constant is between 0 and 1, we get a horizontal stretch ; if the constant is greater than 1, we get a horizontal compression of the function.
Given a function $\text{\hspace{0.17em}}y=f\left(x\right),\text{\hspace{0.17em}}$ the form $\text{\hspace{0.17em}}y=f\left(bx\right)\text{\hspace{0.17em}}$ results in a horizontal stretch or compression. Consider the function $\text{\hspace{0.17em}}y={x}^{2}.\text{\hspace{0.17em}}$ Observe [link] . The graph of $\text{\hspace{0.17em}}y={\left(0.5x\right)}^{2}\text{\hspace{0.17em}}$ is a horizontal stretch of the graph of the function $\text{\hspace{0.17em}}y={x}^{2}\text{\hspace{0.17em}}$ by a factor of 2. The graph of $\text{\hspace{0.17em}}y={\left(2x\right)}^{2}\text{\hspace{0.17em}}$ is a horizontal compression of the graph of the function $\text{\hspace{0.17em}}y={x}^{2}\text{\hspace{0.17em}}$ by a factor of 2.
## Horizontal stretches and compressions
Given a function $\text{\hspace{0.17em}}f\left(x\right),\text{\hspace{0.17em}}$ a new function $\text{\hspace{0.17em}}g\left(x\right)=f\left(bx\right),\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ is a constant, is a horizontal stretch or horizontal compression of the function $\text{\hspace{0.17em}}f\left(x\right).$
• If $\text{\hspace{0.17em}}b>1,\text{\hspace{0.17em}}$ then the graph will be compressed by $\text{\hspace{0.17em}}\frac{1}{b}.$
• If $\text{\hspace{0.17em}}0 then the graph will be stretched by $\text{\hspace{0.17em}}\frac{1}{b}.$
• If $\text{\hspace{0.17em}}b<0,\text{\hspace{0.17em}}$ then there will be combination of a horizontal stretch or compression with a horizontal reflection.
Given a description of a function, sketch a horizontal compression or stretch.
1. Write a formula to represent the function.
2. Set $\text{\hspace{0.17em}}g\left(x\right)=f\left(bx\right)\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}b>1\text{\hspace{0.17em}}$ for a compression or $\text{\hspace{0.17em}}0 for a stretch.
## Graphing a horizontal compression
Suppose a scientist is comparing a population of fruit flies to a population that progresses through its lifespan twice as fast as the original population. In other words, this new population, $\text{\hspace{0.17em}}R,\text{\hspace{0.17em}}$ will progress in 1 hour the same amount as the original population does in 2 hours, and in 2 hours, it will progress as much as the original population does in 4 hours. Sketch a graph of this population.
Symbolically, we could write
See [link] for a graphical comparison of the original population and the compressed population.
## Finding a horizontal stretch for a tabular function
A function $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ is given as [link] . Create a table for the function $\text{\hspace{0.17em}}g\left(x\right)=f\left(\frac{1}{2}x\right).$
$x$ 2 4 6 8 $f\left(x\right)$ 1 3 7 11
The formula $\text{\hspace{0.17em}}g\left(x\right)=f\left(\frac{1}{2}x\right)\text{\hspace{0.17em}}$ tells us that the output values for $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ are the same as the output values for the function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ at an input half the size. Notice that we do not have enough information to determine $\text{\hspace{0.17em}}g\left(2\right)\text{\hspace{0.17em}}$ because $\text{\hspace{0.17em}}g\left(2\right)=f\left(\frac{1}{2}\cdot 2\right)=f\left(1\right),\text{\hspace{0.17em}}$ and we do not have a value for $\text{\hspace{0.17em}}f\left(1\right)\text{\hspace{0.17em}}$ in our table. Our input values to $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ will need to be twice as large to get inputs for $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ that we can evaluate. For example, we can determine $\text{\hspace{0.17em}}g\left(4\right)\text{.}$
$g\left(4\right)=f\left(\frac{1}{2}\cdot 4\right)=f\left(2\right)=1$
We do the same for the other values to produce [link] .
$x$ 4 8 12 16 $g\left(x\right)$ 1 3 7 11
[link] shows the graphs of both of these sets of points.
## Recognizing a horizontal compression on a graph
Relate the function $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ in [link] .
The graph of $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ looks like the graph of $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ horizontally compressed. Because $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ ends at $\text{\hspace{0.17em}}\left(6,4\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ ends at $\text{\hspace{0.17em}}\left(2,4\right),\text{\hspace{0.17em}}$ we can see that the $\text{\hspace{0.17em}}x\text{-}$ values have been compressed by $\text{\hspace{0.17em}}\frac{1}{3},\text{\hspace{0.17em}}$ because $\text{\hspace{0.17em}}6\left(\frac{1}{3}\right)=2.\text{\hspace{0.17em}}$ We might also notice that $\text{\hspace{0.17em}}g\left(2\right)=f\left(6\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\left(1\right)=f\left(3\right).\text{\hspace{0.17em}}$ Either way, we can describe this relationship as $\text{\hspace{0.17em}}g\left(x\right)=f\left(3x\right).\text{\hspace{0.17em}}$ This is a horizontal compression by $\text{\hspace{0.17em}}\frac{1}{3}.$
bsc F. y algebra and trigonometry pepper 2
given that x= 3/5 find sin 3x
4
DB
remove any signs and collect terms of -2(8a-3b-c)
-16a+6b+2c
Will
Joeval
(x2-2x+8)-4(x2-3x+5)
sorry
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
(X2-2X+8)-4(X2-3X+5)=0 ?
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
Y
master
master
Soo sorry (5±Root11* i)/3
master
Mukhtar
explain and give four example of hyperbolic function
What is the correct rational algebraic expression of the given "a fraction whose denominator is 10 more than the numerator y?
y/y+10
Mr
Find nth derivative of eax sin (bx + c).
Find area common to the parabola y2 = 4ax and x2 = 4ay.
Anurag
A rectangular garden is 25ft wide. if its area is 1125ft, what is the length of the garden
to find the length I divide the area by the wide wich means 1125ft/25ft=45
Miranda
thanks
Jhovie
What do you call a relation where each element in the domain is related to only one value in the range by some rules?
A banana.
Yaona
given 4cot thither +3=0and 0°<thither <180° use a sketch to determine the value of the following a)cos thither
what are you up to?
nothing up todat yet
Miranda
hi
jai
hello
jai
Miranda Drice
jai
aap konsi country se ho
jai
which language is that
Miranda
I am living in india
jai
good
Miranda
what is the formula for calculating algebraic
I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it
Miranda
state and prove Cayley hamilton therom
hello
Propessor
hi
Miranda
the Cayley hamilton Theorem state if A is a square matrix and if f(x) is its characterics polynomial then f(x)=0 in another ways evey square matrix is a root of its chatacteristics polynomial.
Miranda
hi
jai
hi Miranda
jai
thanks
Propessor
welcome
jai
What is algebra
algebra is a branch of the mathematics to calculate expressions follow.
Miranda
Miranda Drice would you mind teaching me mathematics? I think you are really good at math. I'm not good at it. In fact I hate it. 😅😅😅
Jeffrey
lolll who told you I'm good at it
Miranda
something seems to wispher me to my ear that u are good at it. lol
Jeffrey
lolllll if you say so
Miranda
but seriously, Im really bad at math. And I hate it. But you see, I downloaded this app two months ago hoping to master it.
Jeffrey
which grade are you in though
Miranda
oh woww I understand
Miranda
Jeffrey
Jeffrey
Miranda
how come you finished in college and you don't like math though
Miranda
gotta practice, holmie
Steve
if you never use it you won't be able to appreciate it
Steve
I don't know why. But Im trying to like it.
Jeffrey
yes steve. you're right
Jeffrey
so you better
Miranda
what is the solution of the given equation?
which equation
Miranda
I dont know. lol
Jeffrey
Miranda
Jeffrey
|
2020-12-04 11:33:54
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 52, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7554349303245544, "perplexity": 559.0150012720626}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141735600.89/warc/CC-MAIN-20201204101314-20201204131314-00267.warc.gz"}
|
http://www.math-only-math.com/multiples-a.html
|
# Multiples
Multiples of a number is discussed here.
What are multiples?
‘The product obtained on multiplying two or more whole numbers is called a multiple of that number or the numbers being multiplied.’
We know about the whole numbers: ‘the numbers starting from 0 and having the pattern 0, 1, 2, 3, 4, 5, up to infinity, are called whole numbers
The whole numbers minus 0 are the natural numbers.
All the natural numbers are multiples of 1.
There is no end to multiples of any number. The first ten multiples of the numbers starting from 1 to 10 are given here.
First ten Multiples of the Numbers
In the above table we can observe the first ten multiples of the numbers starting from 1 to 10.
A multiple is the product of a number and any other number.
Understanding Multiples
1. The first multiple of a number is the number itself.
2. Every number is a multiple of 1.
3. 0 is a multiple of every number.
4. Multiples of a number are infinite (you can carry on writing them).
Multiples are really the products of the tables that you learn.
Sometimes, you learn the multiples till the 10th place or the 12th place of every number table but multiples are really infinite.
Take a look at the multiples of different numbers:
### Common Multiple
Take a look at the multiples of 2, 3, 4, till the 10th place.
Multiples of 2 are in yellow; 3 are in orange; 4 are in light green.
When a number is multiple of 2 or more numbers it is called a common multiple.
Common multiples of 2 and 3 are 6, 12, 18 till the 10th place.
Common multiple of 2, 3 and 4 till the 10th place is → 12.
`
Related Concept
|
2018-03-22 07:50:42
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8009935617446899, "perplexity": 611.3118509973393}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647782.95/warc/CC-MAIN-20180322073140-20180322093140-00178.warc.gz"}
|
https://en.wikipedia.org/wiki/Sesquipower
|
# Sesquipower
In mathematics, a sesquipower or Zimin word is a string over an alphabet with identical prefix and suffix. Sesquipowers are unavoidable patterns, in the sense that all sufficiently long strings contain one.
## Formal definition
Formally, let A be an alphabet and A be the free monoid of finite strings over A. Every non-empty word w in A+ is a sesquipower of order 1. If u is a sequipower of order n then any word w = uvu is a sesquipower of order n + 1.[1] The degree of a non-empty word w is the largest integer d such that w is a sesquipower of order d.[2]
## Bi-ideal sequence
A bi-ideal sequence is a sequence of words fi where f1 is in A+ and
${\displaystyle f_{i+1}=f_{i}g_{i}f_{i}\ }$
for some gi in A and i ≥ 1. The degree of a word w is thus the length of the longest bi-ideal sequence ending in w.[2]
## Unavoidable patterns
For a finite alphabet A on k letters, there is an integer M depending on k and n, such that any word of length M has a factor which is a sesquipower of order at least n. We express this by saying that the sesquipowers are unavoidable patterns.[3][4]
## Sesquipowers in infinite sequences
Given an infinite bi-ideal sequence, we note that each fi is a prefix of fi+1 and so the fi converge to an infinite sequence
${\displaystyle f=f_{1}g_{1}f_{1}g_{2}f_{1}g_{1}f_{1}g_{3}f_{1}\cdots \ }$
We define an infinite word to be a sesquipower if is the limit of an infinite bi-ideal sequence.[5] An infinite word is a sesquipower if and only if it is a recurrent word,[5][6] that is, every factor occurs infinitely often.[7]
Fix a finite alphabet A and assume a total order on the letters. For given integers p and n, every sufficiently long word in A has either a factor which is a p-power or a factor which is an n-sesquipower; in the latter case the factor has an n-factorisation into Lyndon words.[6]
|
2017-11-25 08:33:58
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8397728800773621, "perplexity": 987.8210547836421}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934809695.97/warc/CC-MAIN-20171125071427-20171125091427-00569.warc.gz"}
|
http://www.ck12.org/geometry/Trigonometric-Ratios-with-a-Calculator/lesson/Trigonometric-Ratios-with-a-Calculator/r11/
|
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
Trigonometric Ratios with a Calculator
Solving for values when triangles aren't special right triangles.
0%
Progress
Practice Trigonometric Ratios with a Calculator
Progress
0%
Trigonometric Ratios with a Calculator
What if you were given a 20-70-90 triangle? How could you find the sine, cosine, and tangent of the $20^\circ$ and $70^\circ$ angles? After completing this Concept, you'll be able to use a calculator to find the trigonometric ratios for angles that do not measure $45^\circ$ , $30^\circ$ , or $60^\circ$ .
Guidance
There is a fixed sine, cosine, and tangent value for every angle, from $0^\circ$ to $90^\circ$ . Your scientific (or graphing) calculator knows all the trigonometric values for any angle. Your calculator, should have [SIN], [COS], and [TAN] buttons. You can use your calculator and the trigonometric ratios is to find the missing sides of a right triangle by setting up a trig equation.
Example A
Find the trigonometric value, using your calculator. Round to 4 decimal places.
a) $\sin 78^\circ$
b) $\cos 60^\circ$
c) $\tan 15^\circ$
Depending on your calculator, you enter the degree and then press the trig button or the other way around. Also, make sure the mode of your calculator is in DEGREES.
a) $\sin 78^\circ = 0.97815$
b) $\cos 60^\circ = 0.5$
c) $\tan 15^\circ = 0.26795$
Example B
Find the value of each variable. Round your answer to the nearest tenth.
We are given the hypotenuse. Use sine to find $b$ , and cosine to find $a$ . Use your calculator to evaluate the sine and cosine of the angles.
$\sin 22^\circ &= \frac{b}{30} && \quad \ \ \cos 22^\circ = \frac{a}{30}\\30 \cdot \sin 22^\circ &= b && 30 \cdot \cos 22^\circ = a\\b & \approx 11.2 && \qquad \quad \ \ \ a \approx 27.8$
Example C
Find the value of each variable. Round your answer to the nearest tenth.
We are given the adjacent leg to $42^\circ$ . To find $c$ , use cosine and use tangent to find $d$ .
$\cos 42^\circ &= \frac{adjacent}{hypotenuse} = \frac{9}{c}&& \quad \tan 42^\circ = \frac{opposite}{adjacent} = \frac{d}{9}\\c \cdot \cos 42^\circ &= 9 && 9 \cdot \tan 42^\circ = d\\c &= \frac{9}{\cos 42^\circ} \approx 12.1 && \qquad \quad \ \ d \approx 27.0$
Any time you use trigonometric ratios, use only the information that you are given. This will result in the most accurate answers.
Guided Practice
1. What is $\tan 45^\circ$ ?
2. Find the length of the missing sides and round your answers to the nearest tenth: .
3. Find the length of the missing sides and round your answers to the nearest tenth: .
1. Using your calculator, you should find that $\tan 45^\circ=1$ ?
2. Use tangent for $x$ and cosine for $y$ .
$\tan 28^\circ &= \frac{x}{11} && \quad \ \ \cos 28^\circ = \frac{11}{y}\\11 \cdot \tan 28^\circ &= x && \frac{11}{\cos 28^\circ} = y\\x & \approx 5.8 && \qquad \quad \ \ \ y \approx 12.5$
3. Use tangent for $y$ and cosine for $x$ .
$\tan 40^\circ &= \frac{y}{16} && \quad \ \ \cos 40^\circ = \frac{16}{x}\\16 \cdot \tan 40^\circ &= y && \frac{16}{\cos 40^\circ} = x\\y & \approx 13.4 && \qquad \quad \ \ \ x \approx 20.9$
Practice
Use your calculator to find the value of each trig function below. Round to four decimal places.
1. $\sin 24^\circ$
2. $\cos 45^\circ$
3. $\tan 88^\circ$
4. $\sin 43^\circ$
5. $\tan 12^\circ$
6. $\cos 79^\circ$
7. $\sin 82^\circ$
Find the length of the missing sides. Round your answers to the nearest tenth.
Vocabulary Language: English Spanish
trigonometry
trigonometry
The study of the relationships between the sides and angles of right triangles.
Hypotenuse
Hypotenuse
The hypotenuse of a right triangle is the longest side of the right triangle. It is across from the right angle.
Legs of a Right Triangle
Legs of a Right Triangle
The legs of a right triangle are the two shorter sides of the right triangle. Legs are adjacent to the right angle.
Trigonometric Ratios
Trigonometric Ratios
Ratios that help us to understand the relationships between sides and angles of right triangles.
|
2015-11-29 16:49:45
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 35, "texerror": 0, "math_score": 0.7172891497612, "perplexity": 543.7021012885835}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398458553.38/warc/CC-MAIN-20151124205418-00311-ip-10-71-132-137.ec2.internal.warc.gz"}
|
https://www.physicsforums.com/threads/generalized-version-of-cannon-ball-problem.430332/
|
Generalized version of cannon ball problem
1. kevin0960
12
For All p in Natural Number,
Is $$\exists n , n > 1, \sum^{n}_{k=1} k^p = C^2$$ where C is arbitary natural number (not constant) ??
2. BruceG
40
As far as I'm aware the only solution is:
n = 24
p = 2
C = 70
3. kevin0960
12
No, I checked with mathematica n < 100,000, p < 20
there are some solutions such as
p = 5,
n = 13, 134, etc
I think there are more solutions.. :)
4. CRGreathouse
3,497
You're trying to solve a sequence of Diophantine equations. For a famous case, search for square triangular number.
5. kevin0960
12
Yeah, I know
But what I mean was, is it possible to find the solution for arbitary p ??
6. CRGreathouse
3,497
Probably not. Diophantine equations are hard, in the sense of the negative answer to Hilbert's 10th.
But for any given p it should be possible to at least formulate the problem in that form to see if anything can be discovered. So, for example, with p = 7 you have
$$3x^8 + 12x^7 + 14x^6 - 7x^4 + 2x^2=24y^2$$
7. kevin0960
12
It seems like we cannot find C for arbitary p,
But can we know the existence of C for arbitary p ??
I don't need to find the entire solutions, just a single one.
8. CRGreathouse
3,497
It's not clear that a solution exists for a given p. If not, I don't know of an easy way to prove it -- congruence conditions won't be enough, since n = 1 works and so there are always good congruence classes mod any prime power.
9. kevin0960
12
Yeah, It looks almost impossible to use modular to prove...
|
2015-11-26 01:20:44
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6940771341323853, "perplexity": 782.8846658361231}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398446286.32/warc/CC-MAIN-20151124205406-00152-ip-10-71-132-137.ec2.internal.warc.gz"}
|
https://math.stackexchange.com/questions/1608177/extensions-of-group-homomorphisms-for-special-groups
|
# Extensions of group homomorphisms for special groups
Let $G_1, G_2$ be groups, $H\leq G_1$ a subgroup, $\phi\colon H\to G_2$ a group homomorphism.
Are there some nice properties for the groups, so that it is true that we can get a group homomorphism $\tilde{\phi}\colon G_1\to G_2$ with $\tilde{\phi}_{|H}=\phi$?
I know that this usually isn't correct and this is also discussed in the post Extension of a group homomorphism, but maybe there are some ''nice'' properties for the groups, like $G_1, G_2$ abelian etc., where it works.
This is a very special situation where it is possible. Let $G = AB$ with $A \cap B = 1$ and suppose $A$ centralizes $B / B'$, i.e. we have $b^a \in bB'$ for all $b \in B$ and $a \in A$ (or that each $b \in B$ commutes with every $a \in A$ modulo $B'$). Then if $\varphi : B \to H$ is a homomorphism into an abelian group $H$, then we can extend it to a homomorphism $\overline \varphi : G \to H$ by setting $$\overline \varphi(ab) = \overline \varphi(b) = \varphi(b)$$ i.e. simply "ignore" the $A$-part. Then for $ab, a'b' \in G$ we have $a'b = ba'x$ with $x \in B'$ (and hence $\varphi(x) = 1$ as $H$ is abelian) and further \begin{align*} \overline \varphi(aba'b') & = \overline \varphi(aa'bxb') \\ & = \varphi(bxb') \\ & = \varphi(b)\varphi(x)\varphi(b') \\ & = \varphi(b)\varphi(b') \\ & = \overline \varphi(ab) \overline \varphi(a'b'). \end{align*}
• Starting from a unique factorisation $AB$ with $A\cap B$ seems natural to me, the rest was built add hoc. Just an idea for a slight generalisation: Given another homomorphism $\psi : A \to B$ define $\overline \varphi(ab) = \varphi(\psi(a)b)$, in the post we have $\psi(a) = 1$. This way such an extension could be induced by a second homomorphism. Now the verification runs similar, where again that $H$ is abelian is essential for the reordering in showing the homomorphism property. – StefanH Jan 11 '16 at 18:53
Actually I've now found a result for a special situation in another post namely Extending a homomorphism $f:\left<a \right>\to\Bbb T$ to $g:G\to \Bbb T$, where $G$ is abelian and $\mathbb{T}$ is the circle group.. There is an extension if $G_1$, $G_2$ are abelian and $G_2$ is divisable.
|
2019-07-17 15:03:36
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9994743466377258, "perplexity": 208.91324016720495}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525312.3/warc/CC-MAIN-20190717141631-20190717163631-00313.warc.gz"}
|
https://www.rdocumentation.org/packages/rdlocrand/versions/0.7
|
# rdlocrand v0.7
0
0th
Percentile
## Local Randomization Methods for RD Designs
The regression discontinuity (RD) design is a popular quasi-experimental design for causal inference and policy evaluation. Under the local randomization approach, RD designs can be interpreted as randomized experiments inside a window around the cutoff. This package provides tools to perform randomization inference for RD designs under local randomization: rdrandinf() to perform hypothesis testing using randomization inference, rdwinselect() to select a window around the cutoff in which randomization is likely to hold, rdsensitivity() to assess the sensitivity of the results to different window lengths and null hypotheses and rdrbounds() to construct Rosenbaum bounds for sensitivity to unobserved confounders. See Cattaneo, Titiunik and Vazquez-Bare (2016) <https://sites.google.com/site/rdpackages/rdlocrand/Cattaneo-Titiunik-VazquezBare_2016_Stata.pdf> for further methodological details.
## Functions in rdlocrand
Name Description rdlocrand-package rdlocrand: Local Randomization Methods for RD Designs rdrbounds Rosenbaum bounds for RD designs under local randomization rdsensitivity Sensitivity analysis for RD designs under local randomization rdwinselect Window selection for RD designs under local randomization rdrandinf Randomization Inference for RD Designs under Local Randomization No Results!
|
2020-09-19 14:43:21
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3561088740825653, "perplexity": 8343.959483276492}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400192778.51/warc/CC-MAIN-20200919142021-20200919172021-00729.warc.gz"}
|
https://support.bioconductor.org/p/107272/
|
best way to normalize for differences in sample read depth: normalize genome vs cn.MOPs
1
0
Entering edit mode
znl207 • 0
@znl207-14983
Last seen 3.5 years ago
I would like your advice about normalization with cn.MOPs - as I understand cn.MOPs algorithm includes normalization to compare between different loci and across different samples. cn.MOPs also contains the "normalize genome" function to compare across samples. What is the difference between these two options? I am working with a dataset of variable read depth samples ranging from 7x to 51x with an average of 16x. What is your recommendation for normalizing across these different samples? Should the "normalize genome" option be used?
cn.mops normalization • 403 views
0
Entering edit mode
@gunter-klambauer-5426
Last seen 8 months ago
Austria
Hi,
Yes, the function "cn.mops" internally also applies normalization. I just added the function "normalizeGenome" and "normalizeChromosome" for Users who want to normalize by hand. So, there is no difference between first applying the normalization function and then running cn.mops with "norm=0" (no normalization).
Yes, you are right the normalization should correct for different coverages. Even if your dataset contains vastly differing coverages, you can just run the standard cn.mops functions with the default options (which include normalization).
Regards,
Günter
|
2021-09-25 03:27:12
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4161123037338257, "perplexity": 3393.2819680117464}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057589.14/warc/CC-MAIN-20210925021713-20210925051713-00016.warc.gz"}
|
http://openstudy.com/updates/50fedec2e4b0426c6367f84c
|
## kirbykirby 2 years ago Given the following pmf (prob. mass function), find its cdf: $f(x)=\frac{7}{10}(\frac{3}{10})^x, x=0,1,2,...$
I know the formula to find the cdf is $F(x)=\sum_{X_j \le X}^{}f(x_j)$ Does this mean you get $\sum_{j=0}^{\infty}\frac{7}{10}(\frac{3}{10})^{X_j}$? I dunno why I'm slightly confused about the discrete case. I was used to doing this for continuous rv's.
|
2015-04-21 05:05:43
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7898119688034058, "perplexity": 455.7120272451963}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246640124.22/warc/CC-MAIN-20150417045720-00256-ip-10-235-10-82.ec2.internal.warc.gz"}
|
http://mech.math.msu.su/~fpm/eng/k01/k014/k01402h.htm
|
FUNDAMENTALNAYA I PRIKLADNAYA MATEMATIKA
(FUNDAMENTAL AND APPLIED MATHEMATICS)
2001, VOLUME 7, NUMBER 4, PAGES 983-992
## Grothendieck categories as quotient categories of $\left(R-mod, Ab\right)$
G. A. Garkusha
A. I. Generalov
Abstract
View as HTML View as gif image View as LaTeX source
A Grothendieck category can be presented as a quotient category of the category $\left(R-mod, Ab\right)$ of generalized modules. In turn, this fact is deduced from the following theorem: if $\mathcal C$ is a Grothendieck category and there exists a finitely generated projective object $P \in \mathcal C$, then the quotient category $\mathcal C / \mathcal S^P$, $\mathcal S^P = \\left\{C \in \mathcal C \mid \left\{\right\}_C \left(P, C\right) = 0\\right\}$ is equivalent to the module category $Mod-R$, $R =$C (P, P).
All articles are published in Russian.
Location: http://mech.math.msu.su/~fpm/eng/k01/k014/k01402h.htm
|
2017-12-16 07:08:53
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 8, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.770365834236145, "perplexity": 3106.9421459546725}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948585297.58/warc/CC-MAIN-20171216065121-20171216091121-00384.warc.gz"}
|
https://dml.cz/handle/10338.dmlcz/144062
|
# Article
Full entry | PDF (0.2 MB)
Keywords:
universal fixer; domination
Summary:
Given two disjoint copies of a graph $G$, denoted $G^1$ and $G^2$, and a permutation $\pi$ of $V(G)$, the graph $\pi G$ is constructed by joining $u \in V(G^1)$ to $\pi (u) \in V(G^2)$ for all $u \in V(G^1)$. $G$ is said to be a universal fixer if the domination number of $\pi G$ is equal to the domination number of $G$ for all $\pi$ of $V(G)$. In 1999 it was conjectured that the only universal fixers are the edgeless graphs. Since then, a few partial results have been shown. In this paper, we prove the conjecture completely.
References:
[1] Burger, A. P., Mynhardt, C. M.: Regular graphs are not universal fixers. Discrete Math. 310 (2010), 364-368. DOI 10.1016/j.disc.2008.09.016 | MR 2563053 | Zbl 1216.05098
[2] Cockayne, E. J., Gibson, R. G., Mynhardt, C. M.: Claw-free graphs are not universal fixers. Discrete Math. 309 (2009), 128-133. DOI 10.1016/j.disc.2007.12.053 | MR 2475005 | Zbl 1219.05116
[3] Gibson, R. G.: Bipartite graphs are not universal fixers. Discrete Math. 308 (2008), 5937-5943. DOI 10.1016/j.disc.2007.11.006 | MR 2464884 | Zbl 1181.05068
[4] Gu, W.: Communication with S. T. Hedetniemi. Southeastern Conference on Combinatorics, Graph Theory, and Computing. Newfoundland, Canada, 1999.
[5] Gu, W., Wash, K.: Bounds on the domination number of permutation graphs. J. Interconnection Networks 10 (2009), 205-217. DOI 10.1142/S0219265909002522
[6] Hartnell, B. L., Rall, D. F.: On dominating the Cartesian product of a graph and $K_2$. Discuss. Math., Graph Theory 24 (2004), 389-402. DOI 10.7151/dmgt.1238 | MR 2120063 | Zbl 1063.05107
[7] Mynhardt, C. M., Xu, Z.: Domination in prisms of graphs: universal fixers. Util. Math. 78 (2009), 185-201. MR 2499846 | Zbl 1284.05199
Partner of
|
2019-07-18 04:28:06
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9736801981925964, "perplexity": 1592.9455040343305}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525500.21/warc/CC-MAIN-20190718042531-20190718064531-00191.warc.gz"}
|
https://27oysterpoint.com/alta-ca-mtt/metal-glowing-temperature-4025f8
|
For those not familiar with chromaticity diagrams it may be pointed out that all the. The metal was like a dark orange/ almost redish in color. going slightly above (through the greens) or below (through the We instinctively think of the visible spectrum as being all there is to light, but it is actually a tiny sliver of the entire EM spectrum. How to teach a one year old to stop throwing food once he's done eating? → orange → yellow → white → blue. Even without waiting for evolution, the human vision is able to do chroma adaptation, which means we calibrate our perception of white to what we perceive as the ambient light. The green cone (M cone) is the "middle" one, while the red and blue cones (L and S cones) are furthest displaced to one side each of the visible spectrum. Light emitted by an object according to its temperature. Cherry: 1375. Shouldn't all wavelengths be emitted one by one as the temperature of the metal increases? GeoPandas: How to convert DataFrame to GeoDataFrame with Polygon? The apparent glowing hot metal in the photo above was not due to thermite. Great question, and one of the original ones that lead to quantum mechanics. Shouldn't all wavelengths be emitted one by one as the temperature of the metal increases? The black body emission spectrum at a number of temperature is shown below. Glow paint for metal is used to get a glow effect on any kinds of metal surfaces and objects. completely bridge the gap with the physiology of color perception (which The evolution you are describing is the evolution of the black body radiation as heat is increased. Here is the left side of a black body radiator, which a hot metal is. So i pushed it in and took it out when it was ready. Both of these subjects can have entire academic careers dedicated to them. White is the full spectrum. Moreover, it doesn't seem as good as the tube I used several years ago. I did not understand the graph shown in the answer. At a temperature of $3000 \, \rm K$ the vast majority of the visible light entering the eye is red and so the object is seen to be red. As heat increases , to the left, more and more photons at visible energies appear, first the red, then the yellow, then by the time temperature reaches green in the spectrum you already see "white" because that is the way color perception works in our eyes. How do we determine the color of heated glass? At 1750 K the red visible wavelengths are present, hence "red hot metal". "Red hot" things glow red, "white hot" things glow white. are a mathematical model of our color vision. As I mentioned, in a "real" material there may be some "bumps" in the spectrum due to changes in emissivity with wavelength - but to the best of my knowledge this is not enough to stop the appearance of red-yellow-white as things get hotter. What authority does the Vice President have to mobilize the National Guard? It is for very high temperatures ( the tail on the left at low metal melting temperatures is in the graph I show) and increasing with temperatures. Who was the man seen in fur storming U.S. Capitol? I didn't realize it for like 5 seconds. Raising the temperature even more which makes the proportions of the colours roughly equal the object makes the object look white. The widths of and the separations between the. Kjær et al. What causes a black-body radiation curve to be continuous? This is the question of the definition of white, and it is not Why does this happen? ? Engineering Materials. This is because a hotter object emits more photons in general (or in other words more energy flows away from it per second, quantized in photons, which is the Stephan-Boltzmann law mentioned in the other answer). An object is always radiating light (photons) of frequencies that are a function of its temperature. On the 600K line, there is only about 20% more yellow light than red. loaf of bread!) How would you make a RGB LED equally bright to a broad-spectrum white LED? Now, one may I made a figure to underscore what I think are the most crucial points to understand why black body radiation is never perceived as green: The result of this is that it is impossible to stimulate the M (green) cone with black body radiation without also stimulating either the L (red) or the S (blue) cone to a similar or larger degree. be “Why have we chosen to name ‘white’ a color from the Planckian Over curing, it induces a permanent bondage that can tolerate higher temperatures up to 550°F. @jkej Ha! Im just wondering how hot the metal was i touched? spectrum. The brain interprets the information from the colour receptors (cones) and so you perceive a colour which sometimes can be "unexpected". Without getting into that, the atoms are all moving and bumping into each other. Renaming multiple layers in the legend from an attribute in each layer in QGIS. Other answers show that as the temperature increases, the emission curve not only gets wider (what I've described here) but also gets slightly shifted up (what OP expected). 770. I have seen a test that was done for a TV program, in which they built a cup around a steel column to hold thermite. How do you think about the answers? It has nothing to do with the radiationg material, but rather the tempurature. OK so last month i wanted to see how hot one of those cigarette lighters in the car get. Swap the two colours around in an image in Photoshop CS6. The temperature for an object to start glowing (red) is not related to the kind of matter it’s made off, it’s driven just by the radiation emission law (currently recalled as the “Black body emission law”). has been alluded to in some answers), it is worth showing a picture of A metal rod heated in an induction furnace shows the changing colours extremely well. Most materials, including metals, obey the black body curves whose areas contain the total energy radiated and are a function of temperature. For instance, a metal with temperature around 900 degrees Fahrenheit (500 degrees Celsius) is about a faint red glow, while metals with temperature above 2500 F (1400 C) glow hot white. Incandescence from the sun. Why is wavelength of violet colour less than wavelength of blue colour? Which is probably not very surprising given that the Sun Any shortcuts to understanding the properties of the Riemannian manifolds which are used in the books on algebraic topology. @Dieblitzen While the number of photons representing green does increase, the number of photons representing red also increases, but less so. Alloys of Metals - Melting Points - Alloys of metals and their melting points; Aluminum - Melting Points of Binary Eutectic Alloys - Al - Aluminum - binary eutectic alloys and melting points How to tell what temperature a glowing object (metals) might be: Doesn’t really matter what the emitter is…stainless steel, cast iron, tungsten in your light bulb, the temps are about the same for a given color. D55. There are some complicated details, but here is the basis: as we discussed in class, temperature is a measure of the internal energy of a material. Electrical resistance is the measure of how strongly the metal impedes the passage of electrical current. It forms a uniform coating that will stay there for 200 years. Doesn’t really matter what the emitter is…stainless steel, cast iron, tungsten in your light bulb, the temps are about the same for a given color. And it's not how it works, quantum mechanic explains why. Why is the in "posthumous" pronounced as (/tʃ/). Will a divorce affect my co-signed vehicle? 1020. I will just reproduce one of the curves from there: This shows that the general shape of the emission curve is always the same - the only thing that changes is the position of the peak (which follows a 1/T law known as Wien's Displacement Law), and the area under the curve, which increases with the fourth power of the temperature (Stefan-Boltzmann law). When the metal is heated, the electrons absorb more energy and move faster. The most red and blue black body radiations instead peak at much longer and shorter wavelengths. Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. But in this figure (from Wikipedia), showing the perceived color of black body radiations for different temperatures, we can see that both of those temperatures would be perceived as more or less white. however, it has a completely different meaning: This leaves the notion of while light ill defined: the light coming There is plenty of green light in starlight, but there is no temperature that produces a distribution that is dominant in green. The L cone sensitivity peak around $570$ nm while the S cone sensitivity peak around $442$ nm. In practice, in the realm of color science, there are some so called These are meant to model natural daylight. Is it possible to assign value to set (not setx) value %path% on Windows 10? Inorganic Chemistry Many photoactive coordination compounds contain precious metals. In the physicists jargon, the name white is often used to mean a “flat I'd never thought to plot the Planckian locus before - such a simple and succinct way of thinking about this problem. Only with the hottest of the hot (like arc lamps) we get to the point where we begin to perceive the light as "slightly bluish". Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Is the yellow we perceive when our eyes are hit by red and green light at the same time the same yellow that is at the yellow frequency/wavelength? If you continue your graph to 10000K, it will be blue. Also, there is no need to limit to metals. It only takes a minute to sign up. Needless to say, it was a very... At what temperature can you take steel to that when it cools does not have any scale? Metal Glo is OK but no better - despite the product claims - than any other metal polish I've used. Other metals such as brass, silver, and copper may be fully annealed by the same process but may be quickly cooled, even water quenched, to finish the cycle. When metals (or any materials) get very hot, they emit "black body radiation". 745. The temperatures for which the blackbody radiation has a maximum in the green is around 6500K and then there is still a lot of light emitted in the other wavelength, thus the result is white. The walls of a household oven, meanwhile, can reach temperatures of around 500 °C (900 °F); that means the baking or broiling temperature only reaches about half of that in the metal in the walls. Once, the sight of someone's electronic bug zapper made me feel ill. Why did it cause this reaction? At the melting point the solid and liquid phase exist in equilibrium. As in, do we say "As wavelength increases, spectral energy density increases"? Any metal object can be painted with current glow paint using an ordinary brush or sprayer. The temperature of the object affects the color of the light that is radiated. Blackbody emission can be grossly oversimplified as "we start with the lowest and then we keep adding more energetic wavelengths as the temperature increases". These are instead the Planck curves for which the slope is at its steepest between the peaks of the spectral sensitivity curves, producing the maximal contrast in stimulation between the different cone types. Why only red, then yellow and then white? When the metal temperatures,as seen in the graph I show. The black body radiations that we perceive as most red and blue do not actually peak where the L and S cones peak in sensitivity. Dark Cherry: 1175. How is white light produced from only Red, Blue and Green? rev 2021.1.7.38271, Sorry, we no longer support Internet Explorer, The best answers are voted up and rise to the top, Physics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. OK so last month i wanted to see how hot one of those cigarette lighters in the car get. “standard illuminants” which are deemed white. Why was Warnock's election called while Ossof's wasn't? There may be some (small) deviations from the general law due to surface emissivity, but not so much that you would notice. When talking about actual visible colors, HEAT TREATMENT TEMPERATURES FOR CARBON STEEL : Temperature: Colour of Glowing Steel : Fahrenheit: Centigrade : 2350: 1288 : Dazzling White First layer usually 5°C-10°C higher than subsequent layers. The "missing" colors are there, but for the eye they are overwhelmed by wavelengths that are produced in much higher quantities. However, when things get very, very hot they might appear blue. And why not green? Bright Cherry: 1450. The choice of daylight What would the RGB color value of an infinitely hot blackbody be? If some metals do glow at with different colours, could you give me examples of such metals and the reason why this happens in specific cases? But the first effect is large while the other is small in comparison, so the widening into white dominates. For 1250 K most of the radiation is in longer wavelengths than the visible. Ingredients of the glowing paint: Heat-resistant glow paint from Noxton is made with high quality polyphenyl oxanic resin with high adhesion that is prepared for mixing with a glowing component. Then the light from any Thermometers actually use the change in electrical resistance in a piece of wire to mea… It's a funny name, because even "non black" bodies emit this thermal radiation. Still have questions? What are the best Seiko mechanical watches for under $500? The latest glow plug achieves a high temperature of 1500 °C in the after-glow mode, making it … Iron, for example, has a melting point of 1811 K and a boiling point of 3134 K. If iron got warm enough to have a blue tint, you wouldn't be able to see it for very long. ? been thoroughly covered in the previous answers. 690. purples). The physics of why the heated metal glows like a black body has already Replacing ruthenium with more–earth-abundant iron has been a long-sought goal, but iron compounds generally relax too rapidly after light absorption to channel the energy productively. The melting point (or, rarely, liquefaction point) of a solid is the temperature at which a sustance changes state from solid to liquid at atmospheric pressure. If you look at the full spectrum in the answer by Faecher, you get the full curve. 500. Power loss in a conductor = (current)squared multiplied by the resistance. Luminescent rims. emission formula with the color matching functions, which https://www.hearth.com/talk/wiki/know-temperature-when-metal-glows-red Why do metals only glow red, yellow and white and not through the full range of the spectrum? In practice, virtually all solid or liquid substances start to glow around 798 K (525 °C; 977 °F), with a mildly dull red color, whether or not a chemical reaction takes place that produces light as a result of an exothermic process. One important thing to note, as galinette did in a comment, is that metals do vaporize soon after they get warm enough to emit a significant amount of blackbody radiation in the visible portion of the electromagnetic spectrum. @Johan But the blue color of a gas flame from a stove is not due to black body radiation, but instead caused by emission lines in the. Healing an unconscious player and the hitpoints they regain. Using Wien's displacement law we find that the black body radiations for$\sim 5100$K and$\sim 6550$K would peak at these wavelengths. However, at that temperature the intensity of light is so great (and with so much UV and shorter wavelength light) that I would highly recommend not trying to look at the source. I get that as temperature increases the colour of the line changes from red to white, but how do we read the relationship between the x and y axes of the graph? CRL over HTTPS: is it really a bad practice? You can see in the link that as temperature increases the curve moves to the left, to smaller wavelengths. Why does thermal radiation only occur at infrared and visible frequencies? s shown, that energy in that part of the spectrum increases as temperature increases but the total is there in the radiation. So there is no overwhelming green but red and green. Thank you. People evolved under the Sun, eyes and brain can adjust and perceive non-white color as white (compare. If say 10J is for red, then as we increase to 20J red should be phased out and replaced by yellow, and to 30J blue. Was there anything intrinsically inconsistent about Newton's universe? The glowing debris was also dripping liquid metal that appears to have a bright yellow-white glow, which leads to the conclusion that the maximum temperature of the glowing rubble was probably above 1200°C/2200°F — consistent with the yellow-white hot glow of molten steel in a foundry. Generally accepted colors/temps are: 474 885 Red heat, visible in the twilight, 525 975 Red heat, visible in the daylight, 581 1077 Red heat, visible in the sunlight, I did copy and paste this, answer but it gives you exactly what you want to know and more. as a reference light source is obvious given that our species has Of course, this is gross oversimplification. In other words, you can expose grade 304 alloy steel to temperatures of up to 1,598 ° F for short periods of time without ill effect, and for extended periods of time in temperatures of up to 1,697 ° F. However, this can compromise the corrosion resistance of the metal, making it more susceptible to corrosion damage from exposure to moisture. Steel is still relatively soft just after the red color fades but at this point it rapidly get harder as it cools. Faint Red: 930. Materials suitable down to -75 °C. green) increase while the photons representing red decrease? Here is the left side of a black body radiator, which a hot metal is. Some stars have blackbody radiation. However, in order to This limit is called the Draper point. The colours shown on the graph illustrate the colours our eye would perceive if illuminate by the range of wavelengths spanned by a particular colour. In heating metals towards melting, as the heat rises it follows the curves in the number of probable photons at the visible frequencies. typical illuminant could be considered, in some sense, to be “white From then on it is white for our perceptions. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. It has to do with our eyes, and how our brain interprets messages from them. Needless to say, it was a very stupid idea. We can use the color of hot objects to estimate their temperatures from about 1000 K, as the peak wavelength moves into the visible spectrum. The applied mix easily sets within 4 – 6 hours at normal temperature. Some steels can be used at these temperatures, such as low alloy, quenched and tempered steels or ferritic nickel steels. The color of heated iron changes predictably (due to black-body radiation) from dull red through orange and yellow to white, and can be a useful indicator of its temperature. The good question would Why does the additive color model use red, green and blue instead of yellow, green and violet? For example a combination of red and green light the eye perceives as yellow. You can see the blue quite clearly when you're watching a gas flame on the stove, no need for special shielding against intense radiation. For example, there are no green stars. So i pushed it in and took it out when it was ready. This leads to more scattering, thus increasing the amount of resistance. Yes, the aluminum is silvery gray at its melting temperature (660°C). Notice that the drip hanging below (near the right end of the boat) is glowing, yet has a more silver color than the The column wasn't glowing when the thermite reaction finished it's a … How would interspecies lovers with alien body plans safely engage in physical intimacy? The curves are almost flat at this level of detection and at temperature s where metal is incandescent , still the number of photons emitted is biased towards the infrared, there will always be more photons at a lower frequency than the next band of the spectrum thus blue cannot dominate over the "white" soup our eyes see. Are Colors Emitted at Specific Temperatures? locus”. from a white surface has the same spectrum as whatever illuminant Material Properties - Material properties for gases, fluids and solids - densities, specific heats, viscosities and more ; Related Documents . "As the temperature increases there is more red light." the Planckian locus: This is the set of all the colors a black body can have, plotted in a Is Color Temperature and Wavelength the Same Thing? Look at @Farcher 's graph, how narrow the colour bands are compared to the vast swathe of infra red at the right hand side of the graph. Until you switch on an artificial light source, or half-open the window. Most low carbon (0.20-0.35%) martensitic steels can be used with sufficient reliability. As electrons pass through the metal, they scatter as they collide with the metallic structure. An even higher temperature when it is very dangerous to look at the hot object directly because of the uv light which is also emitted, the white has a bluish ting. Why don't metals glow from red to yellow to green to blue etc.? Is it possible to assign value to set ( not setx ) value % %! Say why metals never glow blue understood correctly, when heated to above 900 °F ( °C! To mobilize the National Guard in Panvel Navi Mumbai intrinsically inconsistent about Newton universe. Over HTTPS: is it really a bad practice sensitivity peak around$ 442 $nm RSS feed copy! To metals mechanical watches for under$ 500 to assign value to set ( not setx ) value % %! For like 5 seconds temperature that produces a distribution that is dominant green... Switch on an artificial light source, or half-open the window purpose of the metal is to... In parallel so i pushed it in and took it out when it was ready felt i. Http: //hearth.com/econtent/index.php/wiki/Temperat... can you calculate the output voltage and current by decoding the entire?... And liquid phase exist in equilibrium under $500 HTTPS: is possible. Increase while the number of photons representing green does increase, the most red and blue instead of,... Years ago moves to the left, to be perceivable increases the curve moves to mix! In equilibrium broad-spectrum white LED around$ 570 $nm while the other is small comparison. ( current ) squared multiplied by the resistance full range of the light that is heated, it emits because. Which a hot metal is vaporized before reaching that temperature, but for the they... At much longer and shorter wavelengths seen but an amalgam of a black body radiator, which much! The additive color model use red, blue and green a single wavelength, it emits red that... If you look at the visible frequencies, such as low alloy, quenched and tempered steels or nickel... One by one as the temperature even more which makes the proportions the!... can you calculate the output voltage and current by decoding the entire circuit in green an attribute each! Blue and green are produced in much higher quantities seen but an amalgam of a black radiations! Around in an induction furnace shows the path of a whole range the. For those not familiar with chromaticity diagrams it may be more green light in it makes the look!, viscosities and more ; Related Documents a high temperature can also with... Present, hence red hot '' things glow white ( photons ) of frequencies that a... Of why the heated metal glows like a spectrum is called incandescence$ 442 $nm small in,! ) value % path % on Windows 10 get to the point everything... Why was Warnock 's election called while Ossof 's was n't the that. Once, the aluminum is silvery gray at its melting temperature ( 660°C ) did it cause reaction! A metal that is radiated its melting temperature ( 660°C ) RGB color value of an infinitely hot blackbody?. The heated metal glows like a spectrum a blue blackbody radiation are above vaporization temperatures any... Absorb more energy is supplied, why does thermal radiation 1250 K of. Such as low alloy, quenched and tempered steels or ferritic nickel steels white '',.! It may be more green light than red, yellow and white and not through metal! missing '' colors are there with less than 30 feet of movement dash when affected by Symbol Fear. They emit black body has already been thoroughly covered in the graph “ why have chosen. Science, there may be more green light the eye they are by! As it cools power loss in a conductor = ( current ) squared multiplied by spectra. A combination of red ( e.g materials ) get very hot, they scatter as they collide with radiationg! In and took it out when it was ready the previous answers spectrum at a rate 75! Accidentally, makes the object affects the color of the spectrum increases as increases. Bright to a metal glowing temperature temperature can also glow with a red color may be green! By wavelengths that are produced in much higher quantities color value of an hot... Temperature that produces a distribution that is radiated i gave a number of photons that... Stay there for 200 years paste this URL into your RSS reader curve to be continuous of metal surfaces objects. → orange → yellow → white → blue terrified of walk preparation to! Fade to black '' effect in classic video games gave a number of temperature is shown below, the is!, and one of the original ones that lead to quantum mechanics creature with than! Green but red and green light the eye perceives as yellow metal in the books on algebraic.. Or steel, when the metal increases radiations instead peak at much longer metal glowing temperature shorter.... Perceive non-white color as white ( compare @ Luaan: Sure, that energy in that part of the starts! Representing green does increase, the process is called incandescence is dim enough overwhelmed by wavelengths that are produced much., makes the experiment stop about right where the white is prettiest much. Flat spectrum ”, i.e in that part of the line does not change like a dark almost. Red hot metal is at these temperatures, as the tube i used years... Any metal metals is that you ca n't get them really hot,. Space fillers for my service panel http: //hearth.com/econtent/index.php/wiki/Temperat... can you calculate the output voltage current... Question of the colours roughly equal the object makes the experiment stop about right where white! In Panvel Navi Mumbai 's electronic bug zapper made me feel ill. why it... In parallel 's done eating pronounced as < ch > ( /tʃ/ ) to a... Overwhelming green but red and yellow still exist at 30J any materials ) get very,. Into an atom, the most red and green the tempurature sets within 4 – 6 hours at temperature. The path of a black body furnace shows the changing colours extremely.... About 20 % more yellow light than red, makes the experiment stop about right where the is. The visible frequencies i understood correctly, when things get very, very hot, they as. Planckian locus before - such a simple and succinct way of thinking about this.... Accidently touched it increases '' can also glow with a red color may be metal glowing temperature below this if. Your graph to 10000K, it is white for our perceptions wavelengths enter visible added... Standard illuminants ” which are seen but an amalgam of a black body going hotter: red → →. Instead of yellow, green and violet effect is large while the other small! Its melting temperature ( 660°C ) in this earlier answer feet of movement dash when affected by Symbol 's effect! The experiment stop about right where the white is prettiest 200 years and succinct way of thinking this! Pronounced as < ch > ( /tʃ/ ) get very, very hot, they scatter as they collide the! Produced in much higher quantities blue instead of yellow, green and violet green light than red but. % ) martensitic steels can be used with sufficient reliability more short wavelengths enter 's Fear effect 442$.! Easily sets within 4 – 6 hours at normal temperature seen in fur U.S.. Mw and temperature of the colours roughly equal the object makes the experiment stop about right where the white often! Really a bad practice non-white color as white ( compare the man seen in the previous answers so called standard. Get a glow effect on any kinds of metal surfaces and objects link as... To infrared and lower, increases and become visible as their numbers increase then the light dim. Carbon ( 0.20-0.35 % ) martensitic metal glowing temperature can be painted with current paint... Effect on any kinds of metal surfaces and objects it works, quantum mechanic explains why dominant green! Was i touched function of its temperature cone sensitivity peak around $570$ while! Using an ordinary brush or sprayer why only red, green and violet vanish below that temperature at! Move faster and move faster should n't green overwhelm '' the red color may visible... $442$ nm practice, in some sense, to smaller wavelengths car get thinking. Light emitted by an object according to its temperature how would you make a LED! Object affects the color of the original ones that lead to quantum mechanics its! Affects the color of heated glass metals towards melting, as shown in the graph shown the... Stack Exchange Inc ; user contributions licensed under cc by-sa energy density increases '' ill. why did cause. An unconscious player and the hitpoints they regain chromaticity diagrams it may pointed... To set ( not setx ) value % path % on Windows 10 im just wondering how hot of. Radiating light ( photons ) of frequencies that are exposed to high temperatures - to! Of this in the link that as temperature increases, but less so soft just the. That we start seeing some green steels or ferritic nickel steels evolution of the radiation is individual. To the left, to smaller wavelengths one in which the power per unit frequency does not vanish that. Could be considered, in the graph i show makes the object affects the color of the light is enough. Once, the electrons absorb more energy is supplied, why does the Vice President to... At infrared and lower, increases and become visible as their numbers increase familiar... Wavelengths that are a function of its temperature the other is metal glowing temperature comparison!
10 Second Countdown After Effects, Define Interference Competition Class 12, Uconn Decision Date 2020, Glock 30 Accessories Amazon, Soft Close Wooden Toilet Seat,
|
2021-09-19 17:23:28
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4549524188041687, "perplexity": 1026.248694339768}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056892.13/warc/CC-MAIN-20210919160038-20210919190038-00141.warc.gz"}
|
https://math.stackexchange.com/questions/2873600/what-is-the-usefulness-of-classification-into-transcedental-and-algebraic-number
|
# What is the usefulness of classification into Transcedental and Algebraic numbers?
Classification into rationals and irrationals makes complete sense because irrational numbers seem to be completely different from rational numbers, which are terminating or repeating.
We know that all irrational numbers have non-terminating non-repeating repeating decimal expansions, then why is there need for separating numbers which are roots of polynomial equations and those who aren't?
Is there anything special in the decimal expansion of pi which the decimal expansion of root 2 doesn't have?
Could this be true that all non-transcedental irrationals can be expressed by a terminating formula in terms of radicals, rationals and arithmetic operations while transcedentals can't be? If this is true, then classification into non-transcedental irrationals and transcedentals will make sense.
EDIT- But alphacapture points out in the comments that that isn't true because polynomial equations of degree greater than 4 have no solution formula in terms of radicals. So, to me, the solution of a degree 8 polynomial equation seems to be as weird as pi, but according to maths, pi is supposed to be weirder because it's 'transcedental' and so somehow inherently differs from roots of polynomial equations.
• I can answer the question at the end right away: No. See en.wikipedia.org/wiki/Solvable_by_radicals – alphacapture Aug 6 '18 at 4:58
• – Luiz Cordeiro Aug 6 '18 at 5:05
• Algebraic numbers are "rare" in the sense that there are only countably many, so "most" numbers are transcendental. – Kusma Aug 6 '18 at 5:25
• Those numbers that are roots of an equation formed using algebra alone and rational numbers only are, in a sense, closer to being rational than numbers which are not. If I give you a number and tell you it's irrational but is the root of a simple equation formed using rational numbers, while another number is irrational and is not the root of any such equation, would you not say one of these two numbers is at least a little bit closer to being rational? – Ittay Weiss Aug 6 '18 at 10:13
• You seem to be focusing too much on decimal expansions, both in the question (viewing periodicity of the expansion as the crucial distinction) and in your answer (about approximation algorithms). In connection with the latter, note that an algebraic number is determined with complete precision by the information that it's the so-and-so many-th root of such-and-such polynomial with integer coefficients. – Andreas Blass Aug 6 '18 at 13:02
Transcendental numbers are numbers that cannot be defined in the language of algebra. Their existence shows that the basic concepts of arithmetic are not enough to fully describe all of the phenomena that occur in the real numbers.
Polynomials are precisely the formulae in one variable that can be written down using only addition, subtraction, and multiplication. Yes, they can be written as a sum of monomials, and this is a useful canonical form, but that makes for a poor definition, despite the fact that it's repeated as such endlessly by high school teachers and even most university-level sources. Thus, a polynomial equation with rational coefficients is just any equation that can be written down using the rational numbers and the $+$ and $\cdot$ signs. Of course by using negative coefficients, this also lets us use $-$ if we like. Furthermore, an equation that also uses division can always be reduced to the form $\frac {P(x)} {Q(x)}=0$, where $P$ and $Q$ are polynomials, and from there to $P(x)=0$, so if $x$ solves an equation involving division, it also solves an equation without division. And finally, any rational number can always be written using the four arithmetic operations and the numbers $0$ and $1$.
Thus, one definition for an algebraic number is a number which satisfies a formula written using only $+, -,\cdot,\div, 0, 1$, and $=$. If we consider this alphabet (and associated grammar) to be the "language of algebra", then such a formula can be taken as a definition of that number written in that language. $\sqrt 2$ can be given such a definition. $\pi$ is usually defined by making reference to geometry, and what its transcendence means is that we need geometry (or at least something bigger than algebra) to define it. A transcendental number is one such that the only predicates written in the language of algebra that the number verifies are the trivial predicates verified by all numbers, like $x+x=2x$.
You might object that something like $x^2=2$ doesn't really define $\sqrt 2$, since after all that equation is also true for $-\sqrt 2$. This is true, and in fact this insight eventually leads to Galois theory. The numbers $\sqrt 2$ and $-\sqrt 2$ cannot be distinguished using algebra and the rational numbers, in much the same way that $\pi$ cannot be defined using algebra and the rational numbers. In Galois theory we have the notion of conjugate numbers over a given field $F$, which are numbers which cannot be distinguished "from the point of view of $F$". This means that any sentence written in the "language of $F$" is either true for both elements, or true for neither. In turns out that there's always a fundamental "minimal sentence" - the minimal polynomial - such that the conjugate numbers of $a$ are precisely all of the numbers making that sentence true. Thus we cannot do better than the minimal polynomial as a definition for $a$ in the language of $F$ - it is the sentence true for $a$ which is true for the fewest other elements.
I remember reading somewhere that there's a general notion in logic called a "transcendental element" over a language or a formal system, or something like that, which is basically exactly what I outlined above: an element which verifies no sentences in the language other than the tautologies. Someone who knows more might leave a comment or an answer.
• The relevant subject is model theory, incidentally. One precise statement along the lines above is: the numbers definable in the structure $\mathcal{R}=(\mathbb{R}; +,\times)$ are exactly the algebraic numbers. In particular, we can distinguish in reals between $\sqrt{2}$ and $-\sqrt{2}$, since the ordering of real numbers is definable in $\mathcal{R}$: $a\le b$ iff for some $c$ we have $c^2=b-a$ (note that this does not work for the rationals). – Noah Schweber Aug 6 '18 at 23:01
• @NoahSchweber I think you mean the structure $\mathcal Q = (\mathbb Q, +, \times)$, no?. – Jack M Sep 14 '18 at 9:21
• No, I do not - I mean exactly what I wrote. Note that $\sqrt{2}$ (say) isn't even in $\mathbb{Q}$, so it doesn't make sense to say that it's definable in $\mathbb{Q}$. – Noah Schweber Sep 14 '18 at 12:37
I'd like to give three examples which highlight the usefulness of classifying a number as transcendental. We start with a problem stated thousands of years ago.
• Squaring the circle
This is a famous problem dating back to the ancient Greeks. Given a circle with radius one, find a construction in a finite number of steps with compass and straightedge only to obtain a square with the same area.
We know the area of the circle with radius one is $\pi$. On the one hand it can be shown that each length started from unit length, which is constructed in a finite number of steps with compass and straightedge only, is algebraic. On the other hand it was Ferdinand von Lindemann who could show that $\pi$ is transcendental (1882).
It is the transcendence of $\pi$ which clarified that this ancient problem is unsolvable.
The next example demonstrates the relevance of transcendental numbers in the 20th century.
• Hilbert's seventh problem
It was one of the great moments of the development of mathematics, when David Hilbert presented 23 problems to the mathematical community at the Paris conference of the International Congress of Mathematicians in 1900.
These problems had an enourmous influence to the development of mathematics, they even paved the way for new mathematical disciplines and inspired many of the most creative mathematicians in the 20th century.
The seventh problem, titled by Hilbert Irrationality and Transcendence of Certain Numbers addresses transcendent numbers, indicating how important a classification of them has been considered by the leading mathematicians. The problem can be stated as:
• Is ${\displaystyle a^{b}}$ always transcendental, for algebraic ${\displaystyle a\not \in \{0,1\}}$ and irrational algebraic ${\displaystyle b}$?
It needed more than three decades until an affirmative answer could be given by Aleksandr Gelfond (April 1, 1934) and independently by Theodor Schneider (May 28, 1934).
With the last example we jump right into the 21st century.
• Periods
represent another fascinating class of numbers introduced by D. Zagier and M. Kontsevich in 2000. They form a countable class of numbers lying between $\overline{\mathbb{Q}}$, the set of algebraic numbers and $\mathbb{C}$.
The properties of these numbers are an inspiring source for mathematicians and transcendent periods are highly attractive as shown in the survey Transcendence of Periods by Michel Waldschmidt (2005).
In the abstract of the paper he indicates the relevance of transcendental numbers.
Abstract: The set of real numbers and the set of complex numbers have the power of continuum. Among these numbers, those which are “interesting”, which appear “naturally”, which deserve our attention, form a countable set. Starting from this point of view we are interested in the periods as defined by M. Kontsevich and D. Zagier. We give the state of the art on the question of the arithmetic nature of these numbers: to decide whether a period is a rational number, an irrational algebraic number or else a transcendental number is the object of a few theorems and of many conjectures. We also consider the approximation of such numbers by rational or algebraic numbers.
The notion of algebraic number is a starting point for areas of mathematics such as Galois theory and algebraic number theory. Sets of algebraic numbers can he used to form finite extension fields of $\mathbf Q$, which behave in many ways like covering spaces in topology and Riemann surfaces (which are closely related to finite extensions of the field $\mathbf C(z)$).
One possible explanation I thought of myself is that there is a common iterative algorithm (like Newton's method) for all algebraic numbers (i.e. Roots of polynomials), to determine them with as much accuracy as one wants. But there are separate algorithms for e, pi, etc. And then there are non-computable transcendentals for which no algorithm exists. I guess this makes them weirder than non-transcendental irrationals.
• This should be a comment, or included in the body of the question itself, rather than an answer. – Noah Schweber Aug 6 '18 at 22:57
|
2019-06-26 00:26:38
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8213156461715698, "perplexity": 239.08402976927312}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999964.8/warc/CC-MAIN-20190625233231-20190626015231-00545.warc.gz"}
|
https://simons.berkeley.edu/talks/erasures-vs-errors-local-decoding-and-property-testing
|
Talks
Fall 2018
# Erasures vs. Errors in Local Decoding and Property Testing
Tuesday, November 27th, 2018 4:00 pm4:40 pm
We initiate the study of the role of erasures in local decoding and use our understanding to prove a separation between erasure-resilient and tolerant property testing. Local decoding in the presence of errors has been extensively studied, but has not been considered explicitly in the presence of erasures. Motivated by applications in property testing, we begin our investigation with local {\em list} decoding in the presence of erasures. We prove an analog of a famous result of Goldreich and Levin on local list decodability of the Hadamard code. Specifically, we show that the Hadamard code is locally list decodable in the presence of a constant fraction of erasures, arbitrary close to 1, with list sizes and query complexity better than in the Goldreich-Levin theorem. We use this result to exhibit a property which is testable with a number of queries independent of the length of the input in the presence of erasures, but requires a number of queries that depends on the input length, $n$, for tolerant testing. We further study {\em approximate} locally list decodable codes that work against erasures and use them to strengthen our separation by constructing a property which is testable with a constant number of queries in the presence of erasures, but requires $n^{\Omega(1)}$ queries for tolerant testing. Next, we study the general relationship between local decoding in the presence of errors and in the presence of erasures. We observe that every locally (uniquely or list) decodable code that works in the presence of errors also works in the presence of twice as many erasures (with the same parameters up to constant factors). We show that there is also an implication in the other direction for locally decodable codes (with unique decoding): specifically, that the existence of a locally decodable code that works in the presence of erasures implies the existence of a locally decodable code that works in the presence of errors and has related parameters. However, it remains open whether there is an implication in the other direction for locally {\em list} decodable codes. (Our Hadamard result shows that there has to be some difference in parameters for some settings.) We relate this question to other open questions in local decoding. Based on joint work with Noga Ron-Zewi and Nithin Varma.
|
2019-08-20 07:30:04
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6926790475845337, "perplexity": 353.59710230636165}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027315258.34/warc/CC-MAIN-20190820070415-20190820092415-00416.warc.gz"}
|
https://guygilboa.net.technion.ac.il/category/publications/
|
### List of papers ordered by popularity
Papers and Citation info
# Total-Variation Mode Decomposition
Ido Cohen, Tom Berkov, Guy Gilboa, “Total-Variation Mode Decomposition”, Proc. SSVM 2021, pp. 52-64
Abstract
In this work we analyze the Total Variation (TV) flow applied to one dimensional signals. We formulate a relation between Dynamic Mode Decomposition (DMD), a dimensionality reduction method based on the Koopman operator, and the spectral TV decomposition. DMD is adapted by time rescaling to fit linearly decaying processes, such as the TV flow. For the flow with finite subgradient transitions, a closed form solution of the rescaled DMD is formulated. In addition, a solution to the TV-flow is presented, which relies only on the initial condition and its corresponding subgradient. A very fast numerical algorithm is obtained which solves the entire flow by elementary subgradient updates.
Bibtex Citation
@InProceedings{10.1007/978-3-030-75549-2_5,
author="Cohen, Ido
and Berkov, Tom
and Gilboa, Guy",
editor="Elmoataz, Abderrahim
and Fadili, Jalal
and Qu{\'e}au, Yvain
and Rabin, Julien
and Simon, Lo{\"i}c",
title="Total-Variation Mode Decomposition",
booktitle="Scale Space and Variational Methods in Computer Vision",
year="2021",
publisher="Springer International Publishing",
address="Cham",
pages="52--64",
}
# Nonlinear Spectral Processing of Shapes via Zero-homogeneous Flows
Jonathan Brokman, Guy Gilboa, Proc. SSVM, “Nonlinear Spectral Processing of Shapes via Zero-homogeneous Flows”, pp. 40-51, 2021
Abstract
In this work we extend the spectral total-variation framework, and use it to analyze and process 2D manifolds embedded in 3D. Analysis is performed in the embedding space – thus “spectral arithmetics” manipulate the shape directly. This makes our approach highly versatile and accurate for feature control. We propose three such methods, based on non-Euclidean zero-homogeneous p-Laplace operators. Each method satisfies distinct characteristics, demonstrated through smoothing, enhancing and exaggerating filters.
Cite
Brokman J., Gilboa G. (2021) Nonlinear Spectral Processing of Shapes via Zero-Homogeneous Flows. In: Elmoataz A., Fadili J., Quéau Y., Rabin J., Simon L. (eds) Scale Space and Variational Methods in Computer Vision. SSVM 2021. Lecture Notes in Computer Science, vol 12679. Springer, Cham. https://doi.org/10.1007/978-3-030-75549-2_4
# Nonlinear Power Method for Computing Eigenvectors of Proximal Operators and Neural Networks
Accepted to SIAM J. on Imaging Scienes, 2021
Leon Bungert, Ester Hait-Fraenkel , Nicolas Papadakis and Guy Gilboa, arXiv
Neural networks have revolutionized the field of data science, yielding remarkable solutions in a data-driven manner. For instance, in the field of mathematical imaging, they have surpassed traditional methods based on convex regularization. However, a fundamental theory supporting the practical applications is still in the early stages of development. We take a fresh look at neural networks and examine them via nonlinear eigenvalue analysis. The field of nonlinear spectral theory is still emerging, providing insights about nonlinear operators and systems. In this paper we view a neural network as a complex nonlinear operator and attempt to find its nonlinear eigenvectors. We first discuss the existence of such eigenvectors and analyze the kernel of $\relu$ networks. Then we study a nonlinear power method for generic nonlinear operators. For proximal operators associated to absolutely one-homogeneous convex regularization functionals, we can prove convergence of the method to an eigenvector of the proximal operator. This motivates us to apply a nonlinear method to networks which are trained to act similarly as a proximal operator. In order to take the non-homogeneity of neural networks into account we define a modified version of the power method.
We perform extensive experiments on various shallow and deep neural networks designed for image denoising. For simple nets, we observe the influence of training data on the eigenvectors. For state-of-the-art denoising networks, we show that eigenvectors can be interpreted as (un)stable modes of the network, when contaminated with noise or other degradations.
# Modes of Homogeneous Gradient Flows
Accepted to SIAM J. on Imaging Sciences, 2021
Ido Cohen, Omri Azencot, Pavel Lifshitz, Guy Gilboa, arXiv, July 2020
Finding latent structures in data is drawing increasing attention in broad and diverse fields such as fluid dynamics, signal processing, and machine learning. In this work, we formulate Dynamic Mode Decomposition (DMD) for two types of dynamical system. The first, a system which is derived by a $\gamma$-homogeneous operator ($\gamma\neq 1$).
The second, a system which can be represented as a symmetric operator.
Regarding to the first type, dynamical systems, derived by $\gamma$-homogeneous operators $\gamma\in[0,1)$, reach the steady state in finite time. This inherently contradicts the DMD model, which can be seen as an exponential data fitting algorithm. Therefore, the induced DMD operator leads to artifacts in the decomposition. We show certain cases where the DMD does not even exist. For homogeneous systems ($\gamma\neq 1$), we suggest a time rescaling that solves this conflict and show that DMD can perfectly restore the dynamics even for nonlinear flows. For dynamics which derived by a symmetric operator, we expect the eigenvalues of the DMD to be real. This requirement is embeded in a variant of the DMD algorithm, termed as Symmetric DMD (SDMD).
With these adaptations, we formulate a closed form solution of DMD for dynamics $u_t = P(u)$, $u(t=0)=u_0$, where $P$ is a nonlinear $\gamma$-homogeneous operator, when the initial condition $u_0$ admits the nonlinear eigenvalue problem $P(u_0)=\lambda u_0$ ($u_0$ is a nonlinear eigenfunction, with respect to the operator $P$). We show experimentally that, for such systems, for any initial condition, SDMD achieves lower mean square error for the spectrum estimation. Finally, we formulate a discrete decomposition, related to nonlinear eigenfunctions of $\gamma$-homogeneous operator.
# Revealing stable and unstable modes of denoisers through nonlinear eigenvalue analysis
### Ester Hait-Fraenkel, Guy Gilboa, arXiv
In this paper, we propose to analyze stable and unstable modes of generic image denoisers through nonlinear eigenvalue analysis. We attempt to find input images for which the output of a black-box denoiser is proportional to the input. We treat this as a nonlinear eigenvalue problem. This has potentially wide implications, since most image processing algorithms can be viewed as generic nonlinear operators. We introduce a generalized nonlinear power-method to solve eigenproblems for such black-box operators. Using this method we reveal stable modes of nonlinear denoisers. These modes are optimal inputs for the denoiser, achieving superior PSNR in noise removal. Analogously to the linear case (low-pass-filter), such stable modes are eigenfunctions corresponding to large eigenvalues, characterized by large piece-wise-smooth structures. We also provide a method to generate the complementary, most unstable modes, which the denoiser suppresses strongly. These modes are textures with small eigenvalues. We validate the method using total-variation (TV) and demonstrate it on the EPLL denoiser (Zoran-Weiss). Finally, we suggest an encryption-decryption application.
# Experts with Lower-Bounded Loss Feedback: A Unifying Framework
Eyal Gofer, Guy Gilboa, arxiv preprint
The most prominent feedback models for the best expert problem are the full information and bandit models. In this work we consider a simple feedback model that generalizes both, where on every round, in addition to a bandit feedback, the adversary provides a lower bound on the loss of each expert. Such lower bounds may be obtained in various scenarios, for instance, in stock trading or in assessing errors of certain measurement devices. For this model we prove optimal regret bounds (up to logarithmic factors) for modified versions of Exp3, generalizing algorithms and bounds both for the bandit and the full-information settings. Our second-order unified regret analysis simulates a two-step loss update and highlights three Hessian or Hessian-like expressions, which map to the full-information regret, bandit regret, and a hybrid of both. Our results intersect with those for bandits with graph-structured feedback, in that both settings can accommodate feedback from an arbitrary subset of experts on each round. However, our model also accommodates partial feedback at the single-expert level, by allowing non-trivial lower bounds on each loss.
# Iterative Methods for Computing Eigenvectors of Nonlinear Operators
Guy Gilboa, arXiv preprint
A chapter to appear in Handbook of Mathematical Models and Algorithms in Computer Vision and Imaging.
In this chapter we are examining several iterative methods for solving nonlinear eigenvalue problems. These arise in variational image-processing, graph partition and classification, nonlinear physics and more. The canonical eigenproblem we solve is $T(u)=\lambda u$, where $T:R^n\to R^n$ is some bounded nonlinear operator. Other variations of eigenvalue problems are also discussed. We present a progression of 5 algorithms, coauthored in recent years by the author and colleagues. Each algorithm attempts to solve a unique problem or to improve the theoretical foundations. The algorithms can be understood as nonlinear PDE’s which converge to an eigenfunction in the continuous time domain. This allows a unique view and understanding of the discrete iterative process. Finally, it is shown how to evaluate numerically the results, along with some examples and insights related to priors of nonlinear denoisers, both classical algorithms and ones based on deep networks.
# NeurIPS 2020: Deeply Learned Spectral Total Variation Decomposition
Tamara G. Grossmann, Yury Korolev, Guy Gilboa, Carola-Bibiane Schönlieb, arXiv 2020
Accepted for NeurIPS 2020.
Non-linear spectral decompositions of images based on one-homogeneous functionals such as total variation have gained considerable attention in the last few years. Due to their ability to extract spectral components corresponding to objects of different size and contrast, such decompositions enable filtering, feature transfer, image fusion and other applications. However, obtaining this decomposition involves solving multiple non-smooth optimisation problems and is therefore computationally highly intensive. In this paper, we present a neural network approximation of a non-linear spectral decomposition. We report up to four orders of magnitude (×10,000) speedup in processing of mega-pixel size images, compared to classical GPU implementations. Our proposed network, TVSpecNET, is able to implicitly learn the underlying PDE and, despite being entirely data driven, inherits invariances of the model based transform. To the best of our knowledge, this is the first approach towards learning a non-linear spectral decomposition of images. Not only do we gain a staggering computational advantage, but this approach can also be seen as a step towards studying neural networks that can decompose an image into spectral components defined by a user rather than a handcrafted functional.
# Adaptive LiDAR Sampling and Depth Completion using Ensemble Variance
Eyal Gofer, Shachar Praisler, Guy Gilboa, arXiv July, 2020
Project details
Abstract
This work considers the problem of depth completion, with or without image data, where an algorithm may measure the depth of a prescribed limited number of pixels. The algorithmic challenge is to choose pixel positions strategically and dynamically to maximally reduce overall depth estimation error. This setting is realized in daytime or nighttime depth completion for autonomous vehicles with a programmable LiDAR. Our method uses an ensemble of predictors to define a sampling probability over pixels. This probability is proportional to the variance of the predictions of ensemble members, thus highlighting pixels that are difficult to predict. By additionally proceeding in several prediction phases, we effectively reduce redundant sampling of similar pixels. Our ensemble-based method may be implemented using any depth-completion learning algorithm, such as a state-of-the-art neural network, treated as a black box. In particular, we also present a simple and effective Random Forest-based algorithm, and similarly use its internal ensemble in our design. We conduct experiments on the KITTI dataset, using the neural network algorithm of Ma et al. and our Random Forest based learner for implementing our method. The accuracy of both implementations exceeds the state of the art. Compared with a random or grid sampling pattern, our method allows a reduction by a factor of 4-10 in the number of measurements required to attain the same accuracy.
|
2021-06-15 02:41:24
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5895781517028809, "perplexity": 1272.3876248906645}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487616657.20/warc/CC-MAIN-20210615022806-20210615052806-00123.warc.gz"}
|
http://openstudy.com/updates/55c21f7ae4b08de2ded16c42
|
## anonymous one year ago Triangle ABC is congruent to triangle XYZ. Angle A measures 50 degrees angle B measure (2x+40) degrees and angle X measures (4x+8) degrees Find the value of x. I got 13.67
1. zepdrix
Hey Kate :) Since A is the first angle in triangle ABC, and X is the first angle in triangle XYZ, and given that the triangles are congruent, this tells us that angle A is equivalent to angle X. Therefore $$\large\rm 50=4x+8$$ Solve for x! :) I don't think it quite works out to 13.67.
2. anonymous
14.5?
3. zepdrix
Subtract 8,|dw:1438786120187:dw|
4. zepdrix
Then divide by 4 to isolate your variable x.
5. anonymous
10.5
6. zepdrix
Yayyyy \c:/ Good Job Katie Crabcake!
|
2017-01-18 22:47:01
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.39406803250312805, "perplexity": 1965.9521448961732}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280364.67/warc/CC-MAIN-20170116095120-00424-ip-10-171-10-70.ec2.internal.warc.gz"}
|
http://math.tntech.edu/machida/MCMC/page15.html
|
Homepage > MCMC
## Mixture Model
A simple mixture model has a probability density function
with weight parameter . Here the components are probability density functions, and assumed to be entirely known.
Posterior density. Given the data of independent observations from the mixture density and the flat prior, the posterior density of weight parameter is proportional to
However, the numerical analysis of posterior density on the simplex
is rather very hard. Alternatively, we can devise an MCMC scheme.
Latent Variables. Then we introduce the following latent variable setup: Let be a latent variable indicating to which component the -th observation belongs, and let be the vector of latent variables on the space
By we denote the indicator function or 0 accordingly as or not. So that we can define
In short, denotes the number of 's set to . In this manner the latent variable can be together lumped into .
|
2018-11-20 14:34:58
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9699534177780151, "perplexity": 1061.832451851047}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039746398.20/warc/CC-MAIN-20181120130743-20181120152238-00007.warc.gz"}
|
https://www.physicsforums.com/threads/a-question-about-operating-systems.765693/
|
# A question about Operating Systems
1. Aug 12, 2014
### mech-eng
I have always been wondering that why there is only one commercial operating system which is Microsoft Windows whereas there are a lot of commercial cars, computers, publishers.
Best Regards.
2. Aug 12, 2014
### Staff: Mentor
There are several others (Mac, Unix), but MS is the biggest consumer OS because of natural monopoly power: it is a big hassle to change from MS to Mac (for example), so few people do it.
The analogy to cars is not a good one because all cars are essentially equally compatible with all roads and drivers.
3. Aug 12, 2014
### mech-eng
I actually want to learn that why other firms do not realise other OSs or in other words how did Microsoft become monopoly power? Why is it a hard thing to do a good Challenger against Windows ? I have asked this question especially for PCs, Mac is not regarded as a PC in spite of it is for a personal usage and Unix are not for trading and I think not easy-to use and as practical as Windows. This question is also current for CPUs because there are only the two, Intel and AMD.
Best Regards.
Last edited: Aug 12, 2014
4. Aug 12, 2014
Staff Emeritus
There used to be - IBM had several. Digital Research had two. They were all out-competed by Microsoft.
5. Aug 12, 2014
### SteamKing
Staff Emeritus
You could ask yourself why aren't more new companies founded which build cars or airplanes. After a certain point, the costs of entering a new market for a product and building a company from scratch become prohibitive.
People have chosen to live with the quirks and bugs endemic to MS OSes because the benefits of using the software which is available to run under this OS outweigh the negatives of the flaws of the OS, i.e., people have settled for the devil they know rather than one which is completely unknown.
Also, AMD CPUs are still hanging around because they can execute code developed for Intel CPUs. If they weren't capable of this, AMD would have faded into history, like Motorola and others, who might have had a better chip design than Intel but couldn't run all the software written for Intel CPUs. At one time, even Apple recognized the dominance of the MS-Intel axis, and made the Mac OS capable of reading MS formatted disks and in some cases running Intel code, IIRC.
6. Aug 12, 2014
### AlephZero
And it could have gone the other way. The first company to bring out dual-core CPUs, and 64-bit architecture for "mass-market" PCs, was not Intel but AMD. For a while it was Intel who were playing catch-up. Microsoft still calls its 64-bit versions of application software "AMD64", not "Intel64", because the current generation of Intel chips are AMD-compatible, not the other way round.
Actually ARM-designed CPUs are now outselling Intel. (I say ARM-designed, because the ARM doesn't do its own manufacturing). The OP didn't explicitly say the question was about desktop computers. ARM supplies about 60% of the CPUs in mobile devices. They are also powering lots of desktop computers' graphics cards. ARM isn't a new kid on the block either - it has been designing CPU chips since the 1980s.
The main barrier to introducing a new OS is simply the cost of developing applications that can replace what is already available. That barrier didn't apply to mobile phones, so new OS's like Chrome and Android are more widely used than the few phones running Windows. Apple has two different OS's for its desktop and mobile devices, OS-X and IOS.
7. Aug 12, 2014
### SteamKing
Staff Emeritus
AMD had the good fortune to be the second source of Intel CPUs for IBM when their PC went into production. Because IBM's contract with Intel required that a second source of CPUs be available, Intel licensed their 8088 and subsequent CPU designs to AMD for production, although the two companies later tangled in court when Intel refused to license its 386 CPU to AMD under the original IBM agreement. For better or worse, CPUs capable of executing Intel machine code will be with us for the foreseeable future.
AMD is hoping to branch out into ARM-compatible chips by sampling a product sometime this year. AMD has also hedged its bets by purchasing graphics card maker ATI some years ago, competing against the other graphics heavyweight, nVidia.
8. Aug 12, 2014
### phinds
Yep, that's the crux of the matter.
9. Sep 1, 2014
### enorbet
I suggest you research the development of Operating Systems, especially if you are entertained by Soap Operas or War Stories. It is a fascinating saga, sometimes exciting, often disgusting and disturbing, both funny and sad, and always a mirror on the realities of the marketplace.
Technically Macs are now PCs since they use Intel CPUs and dropped the RISC based early OpSys models in favor of OS-X, derived from Unix-like BSD.
I don't understand your comment about not being "easy to use" since "user friendliness" has always been the hallmark of Macs, indeed all Apple products. They literally invented "Plug 'n Play" a full decade before PCs.
As far as "practical" there are many industries and types of users (multimedia creation and editing come to mind) that are dominated by Mac. Since the vast majority of Home Computer users merely email, visit Facebook, watch some movies, surf the web a bit and laugh at cute cat memes, and perhaps play a few games, this is largely a non-issue, as Macs do those things effortlessly. The only bug in the soup these days is Office compatibility (which MS thwarts by changing formats but may soon have to stop) yet OpenOffice and especially LibreOffice are quite viable alternatives.
The biggest problem for Mac is a mistaken hangover of the perception that Macs are overpriced. This came about because Apple originally chose a very closed, nearly embedded type of system that only supported a limited list of top notch hardware (such as enterprise quality SCSI hard drives back when the cheaper IDE hard drives were less functional, more prone to failure, and considerably slower). None of the above are still true.
While they still prefer high quality hardware, having switched to an OpSys with on-demand loadable drivers, Mac has been capable of supporting more different kinds of hardware than Windows does, since 2002.
Recently IBM has invested billions of dollars (yes, you read correctly) in Linux, all but abandoning AIX and it's other Unix systems. At the same time (nobody knows for certain if they are related) RedHat has championed a new init system that is poised for very large scale commercial deployment along the lines of efforts (and supported by and contributed to) by the massive Google. It borrows some concepts from both Mac and Windows but has the flexibility and security of Unix systems. It has many old timer Linux users in an uproar because it is seen as an effort to close up the system, but it has vendors drooling. There are now only a small handfull of distros that haven't switched to systemd.
An early implementation of this, CoreOS, can be found here. http://en.wikipedia.org/wiki/CoreOS. While the wiki speaks mostly about vast enterprise application, note that it grew out of ChromeOS which it and Android are basically forms of Linux leaning toward this new, albeit somewhat closed, model.
The Soap Opera may yet continue.
Last edited: Sep 1, 2014
10. Sep 1, 2014
### mal4mac
11. Sep 1, 2014
### enorbet
Yes! He essentially majored in Poker while at Harvard and it has served him well indeed.
12. Sep 1, 2014
### Staff: Mentor
I think this is what you need: http://www.pbs.org/nerds/
Not sure if the video is available anywhere, but they do have the transcript there.
13. Sep 2, 2014
### mech-eng
I have learned that Google have maked an O.S for PCs called Chrome O.S and China will make a national one for PCs. Do you have any idea on Google's O.S. Will it be stable and free?
14. Sep 15, 2014
### harborsparrow
Incidentally, there is not just one Windows operating system. There are dozens of different kinds of Windows operating systems being sold at any one time. This made it so difficult for people programming on Windows that Microsoft had to develop .NET, a virtual machine, to make all the different Windows operating systems act mostly alike no matter which operating system one is coding for.
15. Sep 16, 2014
### vociferous
Microsoft got in heavy with business when IBM chose Microsoft to write the OS for its first PC in the early 80's.
Apple expanded heavily in primary and secondary education with its simplistic Apple DOS software and its easy to learn Apple Macintosh OS.
Microsoft came to dominate the desktop, in my opinion, for the following reasons.
1) Businesses are the biggest purchasers of hardware and IBM was a trusted brand whereas Apple's, along with all non-IBM PC's were largely considered to be toys for hobbyists.
2) The people who made the purchasing decisions in the 1980's for the family were usually the man, the same man who often used IBM PC's at work, not the kids or the homemaker who may have gravitated more towards an OS like Mac.
3) The IBM PC was quickly clones and Microsoft was quick to support the clone market, which could offer advantages like bargain basement prices and were generally open, whereas competitors were generally closed. For the most part, Apple clones were never successful so you were locked into the Apple ecosystem where there was no competition.
4) Microsoft was very good at using its money and influence to dominate the market and run competitors out of business. By the mid 1990's, pretty much every competing OS (Amiga, Atari, et cetera) were dead and Apple was moribund and close to being sold.
5) So much money has been put behind developing Windows that only a handful of companies could really afford to compete against Microsoft and it would be very difficult. Apple has rebounded and made some headway, but they position themselves as a luxury item on the desktop, which leaves Microsoft free to capture the vast majority of the low and mid-cost arena. Now, if Apple started making a $399 Macbook, it might give MS a run for its money. 16. Sep 16, 2014 ### rcgldr Not quite plug and play. Early MAC's had no DMA functionality for it's I/O, relying on an initial polled read or write to transfer the first byte of data, followed by 511 "blind" reads or writes with a hardware handshake and timeout if the device took too long. Aftermarket SCSI adapters were made with actual DMA which required their own drivers. PC's have had DMA since the very first PC and some ealier still CP/M systems. The often promised "the next Mac OS" will be pre-emptive multi-tasking for version 5, 6, 7, 8, 9, never happened until OSX, with the desktop version released in 2001, while Microsoft released Windows NT 3.5.1 in 1995 followed by NT 4.0 in 1996, Windows 2000 in 2000, Windows XP in 2001, ... . In the 1980's MAC's were overpriced compared to similar PC's, and Apples decision in late 1989 to raise prices across the board on all MACs, epecially the ones with color, combined with the gaining popularity of 386 PC clones with Windows 3.0 (1989) and later Windows 3.1x (1992), corresponded to the 1990's era with MAC's market share going from a bit over 10% down to around 2%. Since 2000 and later, the pricing differential has gone down, and MAC PC market share is back above 10% again. Microsoft soft sold PCDOS to IBM for a one time price, but reserved the right to sell MSDOS for PC compatibles. Microsoft made it's initial fortune by selling MSDOS for all the PC clones that later appeared. Although the various versions of Windows have dominated the PC market place, other devices such as smart phones have been dominated by other operating systems, Android, Blackberry, IOS, Windows, ... . IBM's current mainframe operating systems is z/OS (first released 2001), a 64 bit operating system that also includes the equivalent of virtual machines that can run UNIX, Linux, older IBM operating systems, in 64 bit, 32 bit or 24 bit legacy modes. Last edited: Sep 17, 2014 17. Sep 21, 2014 ### enorbet @rcqldr (and others who enjoy Techno Soap Opera) This will be a bit out of order (and probably TLDR but for the few who love the history of things) but first let me address pricing. Price (and intro to PnP) Were Macs (and other Apple products) commonly priced higher than PC systems? Yes. Was that OVERpricing? NO! The quality of components was commensurrate with price. The most obvious example I already noted being SCSI devices as opposed to ATA/IDE, but Apple always tried to keep prices down and even develop new hardware considerably cheaper that that which existed, but would not go below a higher threshold of quality. Example - (from wikipedia-MacIntosh) Note: There were 8 TTL chips (in addition to 4 PAL Chips and 2 more TTL Chips to control the controllers in an integrated manner) providing DMA control, but limited to Video and Audio. There just weren't many peripherals back then that needed larger scale DMA control. Please don't forget that for several years when "Wintel" either couldnt address more than 1MB RAM at all and later more only through a tiny "window" (not to mention other barriers like segmentation ), Mac's system directly addressed 16MB and suffered far less barriers than "WIntel" did. That may sound small now, in light of multi GB systems, but it was Orders of Magnitude larger and more powerful, then. Doesn't that take higher development cost hardware (and associated software)? and wouldn't one expect that difference to come with a price difference? Certainly nobody expects to pay the same price for a 100HP Fiat as for a 1600 HP Lamborghini. Note: The highest HP Lamborghini is actually ~700 HP, less than 0.5 of the above analogy, and costs almost$400,000 USD. A 106HP Fiat 500 is \$13,000 USD. I'm not comparing the cars, just our accepted expectations of cost/benefit where we know the benefits. In the 1980s (and even later) many wondered why any individual would even want a PC, let alone be willing to pay the price of a car to own one. THIS is why Macs inherited the mistaken tag of "overpriced".
Plug 'n Play
True, later, other peripherals depended on either ADB or the add-on, or later, built-in, SCSI bus, but the affinity for busses is plusses. You may recall that although USB 1.0 was called a "serial bus" it was not a true bus in the sense that even IBMs short-lived parallel VLB bus was, but more importantly the SCSI bus, which includes processing of Command Queuing, defined at the device level. This sort of intelligent bus handles interrupts in a vastly superior manner.
This led to the choice for FireWire, a true bus in which devices could interact and "set themselves up" for interrupts (with fewer needed, btw, since even back then the SCSI drive controller could access multiple drives simultaneously, soon numbering 15 devices). The only reason drives that evolved from ATA/IDE now feature simultaneous access is the move to SATA and the Advanced SCSI Programming Interface (ASPI). This is also when Apple/Mac decided to embrace the cheaper drives, when they reached similar performance and reliability, and when sheer adoption rate numbers kept the price down to near PATA pricing.
Firewire was fully functional as true Plug 'n Play Bus by 1986, and very mature by 1990. Members may recall, by contrast, the Win98 TV intro in early 1998 in which a USB Scanner was plugged in Live, which promptly crashed the system.
Also, going back to 1988 with the Mac II, bus mastering expansion cards had become available. By contrast, this did not become commonly available on Intel machines until mid 1990s. Any PCI component can request control of the bus ("become the bus master") and request to read from and write to system memory. DMA is still in there for legacy, but has been obsolete a very long time.
Having even 2 chainlinked 8237's was not only no guarantee of Plug 'n Play, it was an abject failure with legacy that plagued even early PCI devices. As a then system builder and service tech, I cannot possibly count the number of hours I spent in C:\config.sys and C:\autoexec.bat getting any ISA device and many PCI devices to resolve conflicts. Soundcards were particularly abyssmal.
The "high byte" problem mentioned was a choice at the time and was abandoned by 1989, with v7, notably with the release of the Mac IIci. I find your emphasis on DMA puzzling in light of the history of Plug 'n Play hardware, and my statement stands. Mac's PnP was superior from the very beginning until the year 2000. So maybe I was too conservative with "10 years" when 15 is closer to the mark.
IMHO, Windows 2000 was the first true 32-bit, stable (as well as pretty much fully functional PnP) OpSys by MS. In the beginning Windows was a lesser quality product with superior marketing. While there are reasons to choose or try other systems, nobody can argue that they haven't matured well. Windows has become more than just popular. It is now robust.
|
2016-10-26 06:12:04
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.23401905596256256, "perplexity": 4207.311650200903}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988720737.84/warc/CC-MAIN-20161020183840-00521-ip-10-171-6-4.ec2.internal.warc.gz"}
|
https://brilliant.org/discussions/thread/pi-ing-the-square/
|
×
# $$\pi$$-ing the square
Find the minimum amount of pieces that the square with side $$1$$ needs to be cut into, if a rectangle with side $$\pi$$ is to be formed. All the pieces need to be used, with no overlaps. If no solution exists, prove so.
NOTE: I posed this as a problem a few hours ago, thinking I have had a solution. Once I started writing it, I realized that I have had a fault in my proof. I did not manage to find a correct proof on my own, so I offer it as a collaborative adventure into the plena.
Note by Stanislava Sojáková
1 year, 8 months ago
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$
Sort by:
Check this out
- 1 year, 8 months ago
Hi, I posed a different problem. I am quite sure about the solution of my problem - $$5$$. On Dissections, there is nicely depicted the dissection. However, they fail to prove that it is necessary to have as many pieces. That is what I am missing.
- 1 year, 8 months ago
Yes, I realize there's a difference between turning an unit circle into an unit square, or into a rectangle with sides$$(1,\pi)$$ Nevertheless, the nature of the proof is suggested. Probably far beyond the scope of a Note in Brilliant.org.
As a matter of fact, once one has any rectangle, it's almost a trivial exercise to dissect it into a finite number of pieces, and reform the pieces to make a square. This would dispute Tarski's findings, unless one accepts the idea of dissecting the circle into pieces that "cannot be cut from the circle with scissors".
- 1 year, 8 months ago
|
2017-11-23 03:54:14
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9671532511711121, "perplexity": 1649.690962330604}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806720.32/warc/CC-MAIN-20171123031247-20171123051247-00454.warc.gz"}
|
https://www.nature.com/articles/s41598-017-17219-y?error=cookies_not_supported&code=a03ce514-f27f-4918-ad42-a0ad9d9b056a
|
Introduction
Although nitrogen (N) utilization has generally been optimized in agriculture, unreasonable fertilization can lead to agricultural non-point source pollution1,2,3,4,5,6, and improvements are necessary to avoid adverse environmental impacts of nitrate leaching. Nitrate leaching has a significant influence on plant N supply and groundwater quality. Nitrate concentrations in soil depend on the relation between uptake by plants, soil organisms, atmospheric N2 fixation, N mineralization (ammonification and nitrification), N deposition from the atmosphere, denitrification, and volatilization7.
The development of intensive agricultural areas based on irrigation with groundwater and N application in farming areas has had serious side effects on the land ecosystems including ground water depletion and nitrate leaching to ground water6,8. Due to environmental pollution, high nitrate concentrations may accumulate in the edible parts of some vegetables, particularly if excessive N fertilizer has been applied9. Consuming these crops can harm human health.
Leaching of nitrate from soil is driven by land-use type, management (e.g., fertilization), land-use change, climate, and soil properties10. Nitrate-N leaching losses were usually less from fine-textured soils than from coarse-textured soil11. The soil nitrate content may higher in spring than in autumn12. Precipitation/irrigation can significantly increase the nitrates in the soil leachate13,14. Nitrate losses decreased with the drain depth decreased15.
Organic fertilizers have been proposed as one solution to relieve environmental pressure and be a carbon-neutral alternative to liquid fossil fertilizers16. Organic matter improves soil structure, increases the water holding capacity and promotes biological transformations such as N-mineralization16,17. Several researchers have examined the impact of timing of N and water applications on crop yield in field experiments8,18. Behnke et al.19 found that N annual losses from 22.7 to 59.9 kg·ha−1, and they increase with N fertilization rates increase. The soil NO3 -N content under basal fertilizer was 1.65 times higher than that without fertilizer at 0–10 cm on the 36th day after sowing20. Davis et al.21 found that N applications increased N leaching and N2O emission without increasing biomass production. Liu et al.9 found that lettuce augmented with organic fertilizers had significantly longer and wider leaves, higher shoot, and lower NO3 -N concentrations compared with the same amount of inorganic fertilizers. Guo et al.22 found that N fertilizers coupled with farm yard manure resulted in 70% less NO3 -N accumulation in the soil profiles than that using mineral N fertilizer alone. However, some researches had found that manure applications without any pretreatment could cause serious NO3 -N leaching23,24.
There have been many kinds of organic fertilizers, such as manure, sewage sludge, stalks, compost, biogas residues, biogas slurry and so on. An increasing body of literature has been focused on the N fertilizers for crop yield and NO3 -N leaching, but very little is about comparing different kinds of organic fertilizers on NO3 -N distribution (soil, leachate and crop), vegetable yield and quality during the agricultural process. To solve the problem of nitrate content in vegetables, soil and underground water exceeding standard caused by unreasonable fertilization, specific objectives of this study were to: (i) evaluate different organic fertilizers on vegetable yield and quality; (ii) and also determine nitrate concentrations in different soil layers and soil leachate to evaluate environmental risk.
Results and Discussion
Vegetable yield
As expected, organic fertilizers significantly increased vegetable yield by 7.6–45.2% (Fig. 1). For the first rotation, tomato yield increased by 9.2–20.1% compared with CK. Among this, ROF did best with the yield of 84.9 t·ha−1 and it was significantly higher than other treatments. However, GOF had the least effect on tomato yield and BR was similar to GOF with the increase of 9.8%. In the second rotation, all the treatments increased the celery yield by 7.6–8.4%. The maximum increase was ROF treatment with the production of 68.3 t·ha−1. Compared with tomato, the increase of celery was not obvious. After application of organic fertilizers for one year, the tomato yield in all the treatments in this rotation increased compared with the first rotation. This was mainly due to the higher N mineralization as a result of higher biological activity25. For the third rotation crop, tomato yield increased by 25.8–45.2% compared with CK. In this time, ROF has the maximum yield with 112.7 t·ha−1. Among this “tomato-celery-tomato” system, organic fertilizers significantly increased the vegetable yield according to ANOVA test. This may because organic fertilizer application can increase soil organic matter and then increase yields26,27.
Vegetable quality
Application of organic fertilizers can increase vegetable qualities (Fig. 2). The concentration of vitamin C (Vc) after harvest the vegetables is shown in Fig. 2a. Application organic fertilizers significantly increased the concentration of Vc by 3.0–33.5% in the first rotation crop. ROF with the Vc concentration of 122 mg·kg−1 had the best effect and GOF did worst. This may because GOF had a low humification degree without a thermophilic phase9. The concentration of Vc increased by 12.6–31.5% in celery planting. Like the first rotation, ROF with the concentration of 83.5 mg·kg−1 had the best effect and GOF had the least effect. After three rotations, Vc of tomato in CK treatment decreased from 91.4 (first rotation) to 79.0 (third rotation) mg·kg−1, indicating undernourishment and N depletion. Organic fertilizers application could increase the concentration of Vc by 31.6–48.1% compared with CK in the third rotation. ROF with the concentration of 117 mg·kg−1 had the best effect. This is mainly because ROF with a high stabilization and humification degree could improve soil structure, increase the water holding capacity and promote biological transformations and then improve the vegetable quality9.
Figure 2b gives the concentration of soluble sugar (SE) after harvest the vegetables. Application organic fertilizers can significantly increase the concentration of SE by 9.9–17.3% in the first rotation. ROF had the best effect with the concentration of 3.67%. The concentration of SE was increased by 23.6–55.6% in celery planting. ROF with the concentration of 1.12% had the best effect among all the treatments, and GOF had the least effect of all. In the third rotation, the concentration of SE was increased by 18.2–30.3%. Similar to the celery, ROF with the concentration of 4.30% had the best effect of all, and GOF had the least effect. The concentration of SE with the third rotation increased in all treatments including CK compared with the first rotation. This illustrates that long-term application of organic fertilizer can improve the quality of vegetables.
The concentration of titratable acidity (TA) after harvest tomatoes is shown in Fig. 2c. Organic fertilizers application has no significant influence on TA. The content of TA in the third rotation decreased compared with the first rotation, indicating that organic fertilizer can improve the vegetable taste.
Nitrate concentration in vegetable
NO3 -N concentration is an important quality characteristic of vegetable. NO3 was perceived as a purely harmful dietary component which causes infantile methaemoglobinaemia, carcinogenesis and possibly even teratogenesis28. Figure 3 gives the NO3 -N concentration in tomato and celery. From this, NO3 -N concentrations of the two rotations of tomato were all less than 120 mg·kg−1 especially in the latter, which were far less than the limit of the national standard 600 mg·kg−1 (GB18406.1-2001). Celery is a crop which is easy to enrich NO3 and this is why the NO3 -N concentration in the third rotation of tomato lower than the first rotation especially in CK treatment. NO3 -N concentration in celery was much higher than that in tomato, but it was still less than the limit of the national standard 3000 mg·kg−1. Tomato-celery rotation system could significantly decrease the vegetable NO3 -N concentration under this continuous fertilization field. Through these three rotations, ROF had no significant difference with CK in terms of vegetable NO3 -N concentration, which indicates that ROF is a relatively safe way for fertilizing.
Nitrate content in soil
Nitrate concentration in the 0–60 cm soil layers (in time)
NO3 -N concentration in the root zone soil of the three rotations are shown in Fig. 4. Organic fertilizers significantly affect the soil nitrate concentration in the top layers (0–60 cm). The NO3 -N concentration in the 0–30 cm soil layer of GOF and ROF treatments achieved the minimum value after harvest the second rotation (celery). Celery roots were mainly distributed in 0–30 cm29 resulting in less absorbing of NO3 -N below 30 cm soil layer. This may be the reason why the NO3 -N concentration in 30–60 cm soil layer higher than that in 0–30 cm soil layer after harvest celery. Furthermore, for the second rotation NO3 -N content in the top 30 cm soil layer reduced by 60–80% compared with the first rotation, but NO3 -N in the 30–60 cm soil layer may have a small amount of accumulation for little absorption. However, there were no significant differences in the soil NO3 -N content in terms of the whole top layers (0–60 cm) compared with CK, due to NO3 -N absorption and enrichment in celery. The soil NO3 -N concentration after harvest the first rotation of tomato was significantly higher than the other rotations. This may be caused by the high nitrate content of the original soil (Table 1). Organic fertilizers application significantly increased the NO3 -N concentration in the 0–30 cm soil layer of all treatments after the first rotation, and especially in BR reached 138.2 g·kg−1, indicating a high risk of leaching. However, NO3 -N in the 30–60 cm soil layer changed slightly due to root absorption. After third rotation, NO3 -N concentration was significantly lower than the first rotation, owing to the low nitrate background values in this rotation after harvest celery. In this time, NO3 -N concentration in the 0–30 cm soil layer increased slightly. Moreover, NO3 -N concentration in the 30–60 cm reduced significantly, which could be attributable to the higher absorption of N in the 30–60 cm soil layer by the deeper root of tomato30. Thus in this tomato- celery rotation system, long time application of organic fertilizers will not affect soil nitrate content in the top layers (0–60 cm).
Nitrate content in soil profile (in depth)
Organic fertilizer significantly affected soil NO3 -N concentration in the 0–175 cm soil layers (Fig. 5). Due to incomplete utilization of fertilizer, treatments with the organic fertilizers increased the nitrate content in soil profile especially in the top layers compared with CK. Topsoil had the most obvious effect, above all GOF treatment reached 29.9 mg·kg−1, and this may be result from nitrification and mineralization for its instability20. Conversely, owing to the higher humification and stability degree of ROF31, nitrate content in ROF treatment was lower than any other fertilization treatments. After application of BR and GOF, nitrate had a dramatic enrichment in deep soil (especially below 75 cm) indicating N surplus, and such accumulation of NO3 -N in soil profile posed a high risk of N leaching into groundwater. BR and GOF were all incomplete fermentation without thermophilic phase, and then they had a low humification degree with very little stabilized organic matters32. Thus a large amount of nitrogen in BR and ROF cannot be fixed like ROF, which lead to nitrate leaching seriously. In ROF treatment, nitrate content in the soil below 100 cm almost had no difference with CK. From the perspective of security, ROF is the environmentally friendly way for fertilizing.
Nitrogen balance and N translocation
Calculation of N balance is one potentially useful method for predicting the risk of nitrate leaching into groundwater10. N balance in each treatment was calculated under this tomato-celery rotation system (Table 2). Without fertilizer, the amount of Nmin could achieve about 145 kg·ha−1. However, the Nresidual level was lower than the Ninitial level, indicating soil N depletion in some degree. Nuptake in the fertilization treatments were higher than that in CK treatment, especially in the ROF treatment, showing that fertilization can promote the absorption of N by root. After organic fertilizer application, the residual NO3 -N in the 0–60 cm soil layer after crop harvest accumulated to 100–122 kg·ha−1. Although this was still higher than the environmental safety standard in Europe (90–100 kg·ha−1 in the 0–100 cm soil layer)10, it resulted in 50–75% less NO3 -N accumulation in the soil profiles than the initial soil, and therefore the environmental risk was reduced in some degree. These results suggested that organic fertilizer application could be benefit for crop uptake, reduce the NO3 -N in the soil and then alleviate the soil NO3 -N leaching. NUE in these fertilization treatments were 19.4–30.0%, the ROF treatment presented the highest NUE among all the treatments due to the highest uptake by crops, implying the optimum fertilization way.
Nitrate concentration of soil leachate
Although the organic fertilizer application could be benefit for crop uptake and reduce the NO3 -N in the soil, soil are still at the high risk of leaching with the high Nresidual and low NUE in all the fertilization treatments. Then the NO3 -N concentration of soil leachate at 100 cm depth below the soil surface were detected after harvest vegetables. NO3 -N concentration of soil leachate varied with treatments and crop types, ranging from 6.3 to 35.1 mg·L−1 for tomato and from 4.2 to 30.3 mg·L−1 for celery (Table 3). Soil NO3 -N leaching in tomato seasons was generally higher than in celery seasons due to higher crop N uptake and higher evaporation in celery rotation leading to less drainage into deeper layers. NO3 -N leaching in all treatments decreased after application of organic fertilizers especially ROF. Fertilizer type significantly affects the NO3 -N concentration in the soil leachate. The least NO3 -N leaching was observed in the ROF treatment mainly due to ROF with a higher organic matter content and biological activity, stabilization and humification degree, resulting in the increase of soil aggregation, nitrogen fixation capacity and decrease of NO3 -N leaching25. Moreover, NO3 -N concentration in CK and ROF treatments dropped below 10 mg·L−1 after harvest the second rotation (celery), which satisfied the threshold (<10 mg·L−1) for drinking water set by the US Environmental Protection Agency. This result suggested that application of ROF was no more likely to impair groundwater quality than the GOF, BR or even CK treatments.
Conclusions
Organic fertilizers significantly increased vegetable yield and quality, but with inappropriate application may cause serious environmental risk. Maximum vegetable yield of 80.9, 68.3, 112.7 t·ha−1 (first, second and third rotation crop, respectively) with best vegetable quality was obtained in ROF treatment. The highest N use efficiency with the least nitrate enrichment in soil was also found in ROF treatment. Moreover, under this fertilization way, nitrate concentration in soil leachate satisfied the threshold for drinking water. Thus, ROF was suggested to be the optimal fertilizer with the best yield, quality and the least environmental risk under the “tomato-celery” rotation system.
Materials and methods
Site description
One and a half years of field experiments (tomato1-celery-tomato2) were conducted on clay loam soil at Liuminying Agricultural Ecological Station (39°41′ N, 116°34′ E) in southeast suburb area (Daxing district) of Beijing, northwest edge of North China Plain. The soil was calcareous, alkaline, and rich in phosphorus and potassium. Agriculture in the area is intensified by a double cropping system (two vegetables a year) with high-yielding cultivar and high inorganic fertilizer (more than 1000 kgN·ha−1·yr−1) input. Some of the characteristics of this soil were determined before this experiment (Table 1). The average air temperature during tomato planting period was about 25 °C, while in celery planting period was about 18 °C.
Crop rotation and experimental fertilizers
A typical spring tomato–autumn celery double cropping rotation was chosen, representative of the common farming practices in the area, where tomato is usually planted from March to July and celery from August to October. Tomato cultivar with Israel 1420 greenhouse grown tomato (Lycopersicon esculentum Mill.) was planted in the experimental plot (see Section 2.3) at a density of 36,000 plant·ha−1. After tomato harvest, soil was ploughed before planting autumn celery. Celery cultivar with California celery (Apium graveolens L) was planted in the experimental plot at a rate of 2,300,000 plant·ha−1. The selected crop varieties and planting densities is representative of that used by local farmers.
In order to evaluate of agronomic and ecological effects of soil amendment, three kinds of common organic material i.e. general organic fertilizer (GOF), biogas residue (BR) and refined organic fertilizer (ROF) were used as N fertilizer. GOF was made by chicken manure and corn stalk through water-logged composting; BR was taken from Liuminying Biogas Station, which was made by chicken manure through anaerobic digestion; and ROF was made by chicken manure and mushroom residue through a 90 days thermophilic aerobic composting. Some of the composition and characteristics of these organic fertilizers are given in Table 4.
Experimental design
The experiment was conducted in a vegetable greenhouse during the tomato and celery growing season. Four treatments with three replicates were carried out, namely CK, GOF, BR and ROF. Then the experimental area consisted of 12 plots, 5.5 m wide and 6 m long for each, and these 12 plots were arranged as split plots in a randomized complete block with a 0.5 m isolation strip in order to avoid interference. The CK was a control treatment without fertilization. GOF, BR and ROF treatments were applied with the same amount of N with 350 kgN·ha−1 for each crops. Previous study has found that top dressing can increase the crop yield20. Then in this experiment, 66.7% of the fertilizer was used as base fertilizer and the remaining 33.3% as top dressing in fruit swelling period and vigorous period for tomato and celery, respectively. The management practices for controlling pest, disease and weeds complied with local practices for high-yield production.
Analytical methods
Tomato and celery plant were sampled from a 5 m2 area in each plot at harvest for the measurements of vegetable (tomato and celery) yield and tomato residual biomass. Samples of vegetable and tomato residual were oven-dried at 65 °C until they reached a constant weight to determine the water content and dry matter. The N content in vegetable and tomato residual of the samples were determined by the micro-Kjeldahl method by digesting the sample in H2SO4-H2O2 solution33. N uptake by plants was estimated by multiplying the tomato, tomato residual and celery dry matter weight by their N concentrations.
Three tomatoes (or three plants of celery) per plot with similar degree of maturity and similar size and without external defects were picked for the quality indices (mainly taste quality, nutrient quality and safety quality) measurement. Tomatoes or celeries were squeezed in a blender, and then the content of vitamin C (Vc), soluble sugar (SE), and titratable acidity (TA) in the plants were detected according to34. Besides, some of the squeezed vegetable was extracted with deionized water, filtered and then the concentration of NO3 -N in vegetable was determined by a continuous-flow analyzer (TRAACS 2000, Bran and Luebbe, Norderstedt, Germany).
For soil N measurements, three ceramic candle extraction systems with tubes (inside diameter 50 mm) were installed in each plot at 100 cm soil depths. The amount of nitrate leached during the growing season may be minimal compared to leaching losses that occur between the harvest of one crop and the planting of the next23. Then samples of the soil leachate were taken after each harvest and/or before sowing. Furthermore, soil samples in all plots were taken after each harvest and/or before each planting by sampling three cores per plot with an auger (3 cm inside diameter tube) to 60 cm depth in 30 cm increments. Moreover, soil samples in the depth of 15, 45, 75, 125 and 175 cm were taken after harvesting the second batch of tomato (tomato 2) to research the change of nitrate with soil depth. Soil samples obtained from the same layer and plot were thoroughly mixed. All of the soil and soil leachate samples were immediately brought to the laboratory for the measurement of NO3 -N and soil moisture content.
Each fresh soil sample was extracted with CaCl2 35, and the concentration of nitrate was determined by a continuous-flow analyzer (TRAACS 2000, Bran and Luebbe, Norderstedt, Germany). Soil samples were dried to a constant weight in an oven at 105 °C to determine the water content and dry matter. Bulk density of the soils was measured in the 0–60 cm soil depth with soil cores (3 cm inside diameter by 20 cm long). The NO3 -N contents in soil (mg·kg−1) were converted to kg·ha−1 based on the bulk density of different soil layers in order to calculate the N balance. For the nitrate analysis of soil leachate, the water samples were filtered through 0.45 μm membranes and the concentration of nitrate was determined by a continuous-flow analyzer36.
Nitrogen balance
Items in the N balance were estimated in each plot during the whole crop growing seasons. NO3 -N below 60 cm soil depth and NH4 +-N throughout the soil profile will not be included in the N balance calculations because the crop roots in this experiment were mainly distributed in the 0–60 cm depth and relatively low changes in NH4 +-N content between seasons were found (data not presented). The N balance can be written as:
$${{\rm{N}}}_{{\rm{initial}}}+{{\rm{N}}}_{{\rm{input}}}+{{\rm{N}}}_{{\rm{\min }}}-{{\rm{N}}}_{{\rm{uptake}}}-{{\rm{N}}}_{{\rm{residual}}}={{\rm{N}}}_{{\rm{surplus}}}$$
(1)
where Ninitial is initial soil NO3 -N in the 0–60 cm soil profiles; Ninput is N application rate (350 kg N·ha−1 per rotation crop plus 3 rotation crops); Nmin is N mineralization; Nuptake is N uptake by plant; Nresidual is residual NO3 -N in 0–60 cm soil profiles, and Nsurplus represent N that store in various soil fraction (mainly organic N) and N loss. N loss is considered as mainly NO3 -N leaching, since other N losses via denitrification, volatilization and erosion are relatively low under such environmental conditions as reported by Fang et al.37.
N mineralization (Nmin) was estimated by the balance of N inputs and outputs in the control (CK) as follows:
$${{\rm{N}}}_{{\rm{\min }}}={{\rm{N}}}_{{\rm{uptake}},0}+{{\rm{N}}}_{{\rm{residual}},0}-{{\rm{N}}}_{{\rm{initial}},0}$$
(2)
where Nuptake,0, Nresidual,0 and N initial,0 are crop N uptake, residual and initial soil NO3 -N in the 0–60 cm soil profile of the control, respectively.
$${{\rm{N}}}_{{\rm{utlization}}}={{\rm{N}}}_{{\rm{input}}}-{{\rm{N}}}_{{\rm{surplus}}}$$
(3)
$${\rm{NUE}}={{\rm{N}}}_{{\rm{utlization}}}/{{\rm{N}}}_{{\rm{input}}}$$
(4)
Nutlization is the part of Nuptake offered by organic fertilizer. NUE is the fertilizer N use efficiency during the one and a half years of experiment period.
Statistical analyses
Analysis of variance (ANOVA) was performed with the SAS8.2 for Windows, and mean comparisons were done using the least significant difference (LSD) test at P < 0.05.
Data availability statement
The authors declared that none of the data in the paper had been published or was under consideration for publication elsewhere.
|
2022-12-08 03:19:31
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3547089099884033, "perplexity": 7931.779889009494}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711232.54/warc/CC-MAIN-20221208014204-20221208044204-00772.warc.gz"}
|
http://www.physicsforums.com/showthread.php?t=515934&page=2
|
# What is high level mathematics really like?
by Nano-Passion
Tags: mathematics
P: 455
Quote by RandomMystery I don't understand how this part is valid: "If q is not prime then some prime factor p divides q. This factor p is not on our list: if it were, then it would divide P (since P is the product of every number on the list); but as we know, p divides P + 1 = q. Then p would have to divide the difference of the two numbers, which is (P + 1) − P or just 1. But no prime number divides 1 so there would be a contradiction, and therefore p cannot be on the list. This means at least one more prime number exists beyond those in the list." 1*2*3*4 = 24 = P 24+1=25 = q 25 is not prime... why would "p have to divide the difference of the two numbers, which is (P + 1) − P or just 1."
First, if you want to follow Euclid's logic, you should be using the product of primes, not of all integers. That doesn't make a big difference. It takes a little more work, but one can still find an example of your point. The first is 2*3*5*7*11*13+1 = 30031 = 59*509.
More important, remember that you're assuming that there is no prime greater than pn (which is 13 in the example). Euclid's reasoning shows that no prime <= pn divides q. Under the assumption that there IS no prime greater than pn, there must therefore be no prime that divides q. The example doesn't contradict this, because the smallest prime that divides 30031 is in fact bigger than 13.
And how does "no prime number divides 1 so there would be a contradiction, and therefore p cannot be on the list." prove that there must be at least one more prime number on that list?
You started by assuming that every prime was on the list. You showed that this leads to a contradiction. That means that the assumption was WRONG. If the assumption "all primes are on the list" is wrong, then there must be at least one prime not on the list. (It doesn't have to be q, though. That's what your example shows.)
P: 69
Quote by pmsrw3 First, if you want to follow Euclid's logic, you should be using the product of primes, not of all integers. That doesn't make a big difference. It takes a little more work, but one can still find an example of your point. The first is 2*3*5*7*11*13+1 = 30031 = 59*509. More important, remember that you're assuming that there is no prime greater than pn (which is 13 in the example). Euclid's reasoning shows that no prime <= pn divides q. Under the assumption that there IS no prime greater than pn, there must therefore be no prime that divides q. The example doesn't contradict this, because the smallest prime that divides 30031 is in fact bigger than 13. You started by assuming that every prime was on the list. You showed that this leads to a contradiction. That means that the assumption was WRONG. If the assumption "all primes are on the list" is wrong, then there must be at least one prime not on the list. (It doesn't have to be q, though. That's what your example shows.)
Thanks for the reply, it's cleared up some things and I think I get it now.
Euclid's proof shows that if P + 1 is not prime, then P + 1 can not be divided by a prime number in the finite set of prime numbers since when you divide P + 1 by a factor of P, that factor will end up dividing 1.
$\frac{(P + 1)}{factor of P}= factor of P\times(something) + \frac{1}{factor of P}$
Related Discussions General Math 26 Academic Guidance 8 General Math 0 Science & Math Textbooks 1 Academic Guidance 5
|
2014-09-02 19:22:34
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6328150629997253, "perplexity": 255.4455791750924}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1409535922763.3/warc/CC-MAIN-20140909044349-00153-ip-10-180-136-8.ec2.internal.warc.gz"}
|
https://brilliant.org/discussions/thread/cyclic-quadrilateral/
|
×
$$ABCD$$ is a cyclic quadrilateral, with $$AB = 2; \space BC = 4; \space CD = 5; \space DA = 10$$. Let $$K, \space T, \space O$$ be the projection of point $$D$$ on segment $$AB, \space BC, \space CA$$ respectively. Find $$\dfrac{KT}{TO}$$.
Note by Fidel Simanjuntak
1 month, 3 weeks ago
Sort by:
Hint: Ptolemy's Theorem and draw segment $$BD$$ · 1 month, 3 weeks ago
|
2017-08-20 17:15:29
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6931053400039673, "perplexity": 1133.8416441962975}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886106865.74/warc/CC-MAIN-20170820170023-20170820190023-00299.warc.gz"}
|
http://www.machinedlearnings.com/2012/05/
|
## Monday, May 21, 2012
### Instruction Theory?
In learning theory we often talk about an environment which is oblivious to our algorithm, or an environment which is fully aware of our algorithm and attempting to cause it to do badly. What about the case where the environment is fully aware of our algorithm and attempting to help it succeed?
Here's a concrete example. Suppose you are trying to communicate a message to a receiver, and this message is one of a finite set of hypotheses. You are forced to communicate to the receiver by sending a sequence of feature-label pairs $(X \times Y)^*$; the receiver will decode the message via ERM on the hypothesis set using the supplied data. How many examples does it take, and how should you chose them? If this sounds corny, consider that evolution works by reuse, so if the capability to learn from experience developed due to other selection pressure, it might be co-opted to service communication a la Cognitive Linguistics.
Intuitively, even if the hypothesis being communicated was learned from experience, it's not good strategy to retransmit the exact data used to learn the hypothesis. In fact, it seems like the best strategy would be not using real data at all; by constructing an artificial set of training examples favorable structure can be induced, e.g., the problem can be realizable. (Funny aside: I TA-d for a professor once who confided that he sometimes lies to undergraduate in introductory courses in order to get them closer to the truth''; the idea was, if they took an upper division class he would have the ability to refine their understanding, and if not they were actually better off learning a simplified distortion).
Some concepts from learning theory are backwards in this setup. For instance, Littlestone's dimension indicates the maximum number of errors made in a realizable sequential prediction scenario when the examples are chosen adversarially (and generalizes to the agnostic case). We can chose the examples helpfully here (what's the antonym of adversarial?), but actually we want errors so that many of the hypothesis are incorrect and can be quickly eliminated. Unfortunately we might encounter a condition where the desired-to-be-communicated hypothesis disagrees with at most one other hypothesis on any point. Littlestone finds this condition favorable since a mistake would eliminate all but one hypothesis, and otherwise no harm no foul; but in our situation this is worst-case behaviour, because it makes it difficult to isolate the target hypothesis with examples. In other words, we can chose the data helpfully, but if the set of hypotheses is chosen adversarially this could be still very difficult.
Inventing an optimal fictitious sequence of data might be computationally too difficult for the sender. In this case active learning algorithms might provide good heuristic solutions. Here label complexity corresponds to data compression between the original sequence of data used to learn the hypothesis and the reduced sequence of data used to transmit the hypothesis.
There is fertile ground for variations, e.g., the communication channel is noisy, the receiver does approximate ERM, or the communication is scored on the difference in loss between communicated and received hypothesis rather than 0-1 loss on hypotheses.
## Wednesday, May 16, 2012
### Active Minimax Forecasting
This is a continuation of my previous post where I noted that the Minimax Forecaster is tantalizing from an active learning perspective. Fortunately I noticed a paper by Abernethy et. al. called A Stochastic View of Optimal Regret through Minimax Duality. In this paper the authors unwrap online convex optimization in a general fashion by leveraging von Neumann's minimax theorem. By doing this they derive a general formula for the value of an online game to the adversary. The intuition of the previous post is that differences in game value between observing and not observing a particular outcome will be key to making active learning decisions in an adversarial setting, so this formula is very interesting.
Abernethy et. al. start out with a more general setup than the previous post, but I'll adapt their conventions to the previous post where there are differences. There is a game with $T$ rounds in it. There is a set $\mathcal{F}$ of experts. On each round, each expert $f$ produces a prediction $f_t$, player produces a prediction $p_t$, adversary simultaneously produces an outcome $y_t$, and player suffers an instantaneous loss $l (p_t, y_t)$. The experts are static (their predictions to do not depend upon previously observed outcomes), so essentially each expert is an sequence $f_{1:T}$. The player wants to generate a sequence of predictions $p_{1:T}$ which minimizes worst-case regret $\sup_{y_{1:T} \in \mathcal{Y}^T} L (p_{1:T}, y_{1:T}) - \inf_{f \in \mathcal{F}} L (f_{1:T}, y_{1:T}),$ where $L (p_{1:T}, y_{1:T}) = \sum_s l (p_s, y_s)$ is the total loss. (To ease notation supremums and infimums will capture the entire rest of the expression unless explicitly limited by parenthesis.) The Minimax Forecaster of the previous post is an explicit solution to a specific case of this problem, whereas Abernethy et. al. are concerned with characterizing such games in general. They consider the value of the game to the adversary under optimal play, \begin{aligned} R^* (\mathcal{F}) &= \inf_{p_1 \in \mathcal{P}} \sup_{y_1 \in \mathcal{Y}} \ldots \inf_{p_T \in \mathcal{P}} \sup_{y_T \in \mathcal{Y}} \sum_{t=1}^T l (p_t, y_t) - \inf_{f \in \mathcal{F}} \sum_{t=1}^T l (f_t, y_t). \end{aligned} The amazing result is that this is the same as \begin{aligned} R^* (\mathcal{F}) &= \sup_{Y \in \mathbb{P} (\mathcal{Y}^T)} \mathbb{E}_{y_{1:T} \sim Y} \left[ \sum_{t=1}^T \inf_{p_t \in \mathcal{P}} \biggl( \mathbb{E}_{\tilde y_t} \left[ l (p_t, \tilde y_t) | y_{1:t-1} \right] \biggr) - \inf_{f \in \mathcal{F}} \sum_{t=1}^T l (f_t, y_t) \right], \end{aligned} where the supremum is over distributions of outcome sequences $\mathbb{P} (\mathcal{Y}^T)$. In other words, this looks like a game played with an oblivious (non-stationary) environment, but played in the worst possible environment. This is a nifty result, and leads in subsequent work to interpolating between the IID and adversarial settings by constraining the supremum over sequence distributions.
Now with active learning, we make the adversary more powerful by choosing not to observe some of the outcomes (represented by the variable $z_s \in \{ 0, 1 \}$). I can upper bound the game value to the adversary as \begin{aligned} R^* (\mathcal{F} | z_{1:T}) &\leq \sup_{Y \in \mathbb{P} (\mathcal{Y}^T)} \mathbb{E}_{y_{1:T} \sim Y} \left[ \sum_{t=1}^T \inf_{p_t \in \mathcal{P}} \biggl( \mathbb{E}_{\tilde y_t} \left[ l (p_t, \tilde y_t) | \Omega_{t-1} \right] \biggr) - \inf_{f \in \mathcal{F}} \sum_{t=1}^T l (f_t, y_t) \right], \end{aligned} where $\Omega_t = \{ y_s | s \leq t, z_s = 1 \}$ denotes observed outcomes. This is intuitively pleasing because the inner conditional expectation represents the player's knowledge. To derive the upper bound follow the procedure in Appendix A of the paper, after transforming the game value expression whenever $z_s = 0$ from $\inf_{p_1 \in \mathcal{P}} \sup_{y_1 \in \mathcal{Y}} \cdots \inf_{p_s \in \mathcal{P}} \sup_{y_s \in \mathcal{Y}} \cdots \inf_{p_T \in \mathcal{P}} \sup_{y_T \in \mathcal{Y}} \sum_{t=1}^T l (p_t, y_t) - \inf_{f \in \mathcal{F}} \sum_{t=1}^T l (f_t, y_t),$ to $\inf_{p_1 \in \mathcal{P}} \sup_{y_1 \in \mathcal{Y}} \cdots \inf_{p_s \in \mathcal{P}} \cdots \inf_{p_T \in \mathcal{P}} \sup_{y_T \in \mathcal{Y}} \sup_{y_s \in \mathcal{Y}} \sum_{t=1}^T l (p_t, y_t) - \inf_{f \in \mathcal{F}} \sum_{t=1}^T l (f_t, y_t),$ i.e., by letting the adversary defer selection of the unobserved values until the end of the game. This is just an upper bound because in reality the adversary has to chose the outcome for round $s$ on round $s$ so possibly I'm being too generous to the adversary.
#### Minimax Forecasting
If we design a Minimax Forecaster on the extensive form game associated with the bound, we get an algorithm which optimizes the bound. Consider the case where all outcomes are observed except for round $s$. Repeating the backward induction steps from the original minimax forecaster paper yields the expressions, \begin{aligned} p_t^* &= \frac{1}{2} \biggl( 1 - R^* (\mathcal{F}, y_{1:t-1}0 | z_s = 0) + R^* (\mathcal{F}, y_{1:t-1}1 | z_s = 0) \biggr). & (t > s) \end{aligned} and \begin{aligned} &R^* (\mathcal{F}, y_{1:t} | z_s = 0) \\ &= \frac{T - t}{2} + \mathbb{E}_{\sigma_{t+1:T}}\left[ \sup_{y_s \in \mathcal{Y}} |p_s - y_s| - \inf_{f \in \mathcal{F}} L (f_{1:T}, y_{1:t} \sigma_{t+1:T}) \right] & (t > s) \\ &= R^* (\mathcal{F}, y_{1:t}) \\ &\quad + \mathbb{E}_{\sigma_{t+1:T}}\left[ \left| p_s - \frac{1}{2} \left(1 + \inf_{f \in \mathcal{F}} \left( L (f_{1:T}, y_{1:s-1} 0 \sigma_{s+1:T}) \right) - \inf_{f \in \mathcal{F}} \left( L (f_{1:T}, y_{1:s-1} 1 \sigma_{s+1:T}) \right) \right) \right| \right]. \end{aligned} Thus the residual game value having played $p_s$ and not observed outcome $s$ is equal to the fully observed residual game value plus a penalty related to expected absolute loss averaged over all Rademacher distributed playouts. This implies the optimal choice of $p_s$ is a median \begin{aligned} p_s^* &= \mathop{\operatorname{median}} \left( \frac{1}{2} \left( 1 + \inf_{f \in \mathcal{F}} \left( L (f_{1:T}, y_{1:s-1} 0 \sigma_{s+1:T}) \right) - \inf_{f \in \mathcal{F}} \left( L (f_{1:T}, y_{1:s-1} 1 \sigma_{s+1:T}) \right) \right) \right). \end{aligned} In the case where there are no additional rounds of play between $s$ and $T$, there is only one point in the distribution so the median is that point, so the optimal play $p_s^*$ is the same as in the fully observed case. In the case where there is one additional round of play between $s$ and $T$, there are two points in the distribution so the mean is a median, and again the optimal play $p_s^*$ is the same as in the fully observed game (consistent with the previous blog post). In the case of more than one additional round of play between $s$ and $T$, the mean is not necessarily a median (i.e., does not necessary minimize expected absolute loss) so optimal play $p_s^*$ is different. Maybe this is why Mathematica pooped out'' when I tried to solve a longer version of the unobserved game.
Once we've hopped over'' the unobserved point and determined $p_s^*$, we can continue the backwards induction; but I think the interesting bit is the penalty term, which says what the value of observing the outcome on round $s$ given all previous and subsequent rounds will be observed, $\mathbb{E}_{\sigma_{t+1:T}}\left[ \left| p_s - \frac{1}{2} \left(1 + \inf_{f \in \mathcal{F}} \left( L (f_{1:T}, y_{1:s-1} 0 \sigma_{s+1:T}) \right) - \inf_{f \in \mathcal{F}} \left( L (f_{1:T}, y_{1:s-1} 1 \sigma_{s+1:T}) \right) \right) \right| \right].$ This penalty term would be important for deciding whether or not to query the outcome on round $s$. It'd be nice to generalize it to the case where some of the previous outcomes are not observed and also with respect to potentially unobserved future outcomes. Of course, planning the latter might be exponentially difficult.
For this to be practical, it would be nice if medians and expected absolute losses could be cheaply and accurately estimated, e.g., using random playouts. Also perhaps deciding whether to query the outcome needs to be done on a game value bound, e.g., assuming all subsequent observations will be observed rather than taking into account future decisions to observe the outcome. Furthermore, the technique is still fundamentally transductive since the expert predictions for the planning horizon $f_{1:T}$ need to be known. Even with all those caveats, however, there might be an interesting application for this, e.g., in recommendation systems.
## Sunday, May 6, 2012
### The Minimax Forecaster and Transductive Active Learning
I've been making my way down the NIPS 2011 paper list, and found this nice paper Efficient Online Learning via Randomized Rounding by Cesa-Bianchi and Shamir. This paper is about improving and extending the Minimax Forecaster which is described in Prediction, Learning, and Games. (I own a copy but I confess to not having made it very far into that book.) The Minimax Forecaster uses a different strategy for online learning than mirror descent which is essentially what I (and everybody else?) use everyday. This different setting provides an opportunity to think about adversarial active learning.
Here's the basic setup. There is a game with $T$ rounds in it. There is a set $\mathcal{F}$ of experts. On each round, each expert $f$ produces a prediction $f_t$, player produces a prediction $p_t$, adversary simultaneously produces an outcome $y_t$, and player suffers an instantaneous loss $l (p_t, y_t)$. The experts are static (their predictions to do not depend upon previously observed outcomes), so essentially each expert is an sequence $f_{1:T}$. The player wants to generate a sequence of predictions $p_{1:T}$ which minimizes worst-case regret $\sup_{y_{1:T} \in \mathcal{Y}^T} \biggl( L (p_{1:T}, y_{1:T}) - \inf_{f \in \mathcal{F}} L (f_{1:T}, y_{1:T}) \biggr),$ where $L (p_{1:T}, y_{1:T}) = \sum_s l (p_s, y_s)$ is the total loss. When the observations are binary $\mathcal{Y} = \{ 0, 1 \}$, then an instantaneous loss of $|p_t - y_t|$ corresponds to expected 0-1 loss when the player is randomizing decisions. Amazingly this case yields a closed-form expression for the optimal prediction \begin{aligned} p^*_t &= \frac{1}{2} \biggl( 1 + R^* (\mathcal{F}, y_{1:t-1}1) - R^* (\mathcal{F}, y_{1:t-1}0) \biggr) \\ &= \frac{1}{2} \biggl( 1 + \mathbb{E}_{\sigma_{t+1:T}} \left[ \inf_{f \in \mathcal{F}} L (f_{1:T}, y_{1:t-1} 0 \sigma_{t+1:T}) - \inf_{f \in \mathcal{F}} L (f_{1:T}, y_{1:t-1} 1 \sigma_{t+1:T}) \right] \biggr), \end{aligned} where temporal sequence concatenation is denoted lexically, $\sigma_t$ is $\mathrm{Bournelli}(1/2)$ distributed a la Rademacher averages, and $R^* (\mathcal{F}, y_{1:t})$ is the residual game value after some rounds of play, \begin{aligned} R^* (\mathcal{F}, y_{1:t}) &= \frac{1}{2} \biggl( 1 + R^* (\mathcal{F}, y_{1:t-1}0) + R^* (\mathcal{F}, y_{1:t-1}1) \biggr) \\ &= \frac{T - t}{2} - \mathbb{E}_{\sigma_{t+1:T}}\left[ \inf_{f \in \mathcal{F}} L (f_{1:T}, y_{1:t} \sigma_{t+1:T}) \right]. \end{aligned} Essentially what's happening here is that player is able to make adversary indifferent to playing either option on each round by playing a constant plus the difference between the residual game values associated with playing each option; this causes the residual game value to be a constant plus the average value of continuing after playing each option. Unwrapping the game value recursively leads to the Rademacher style averages. One observation of the paper is that such expectations can be approximated by sampling to achieve a high probability regret bound, aka random playout.
In practice even to do random playout you need to know $f_{1:T}$ for the various experts. When mapping this to a contextual prediction setting, this corresponds to knowing the sequence of features in advance (but not the labels). Thus this is essentially a transductive technique. Some recommendation problems are naturally transductive, and the paper discusses an application to collaborative filtering.
#### Active Learning?
In principle the setup can be modified to consider active learning. Each round, in addition to generating a prediction, player must make a decision $z_t \in \{ 0, 1 \}$ whether or not to observe $y_t$. If $z_t = 0$, player cannot use the value of $y_t$ in subsequent predictions. Since it is always better for the player to observe $y_t$, there has to be some penalty for doing so, thus consider a constant penalty $\alpha$ per observation. The player wants to generate a sequence of predictions $p_{1:T}$ and queries $z_{1:T}$ which minimizes worst-case regret $\sup_{y_{1:T} \in \mathcal{Y}^T} \biggl( \sum_s \alpha z_s + L (p_{1:T}, y_{1:T}) - \inf_{f \in \mathcal{F}} L (f_{1:T}, y_{1:T}) \biggr).$ Concise general closed-form expressions have eluded me thus far, but there is a non-trivial case which yields nice answers: the two-round game.
It never makes sense to observe the final outcome $y_T$, so $z_T = 0$. In the two-round game, then, the question is whether to observe $y_1$. If $y_1$ is not observed (i.e., $z_1 = 0$), player must ballistically plan both predictions without intermediate feedback, \begin{aligned} (p_1^*, p_2^*) &= \mathop{\operatorname{arg\,inf}}\limits_{p_{1:2} \in \mathcal{P}^2} \sup_{y_{1:2} \in \mathcal{Y}^2} \left( |p_1 - y_1| + |p_2 - y_2| - \inf_{f \in \mathcal{F}} L (f_{1:2}, y_{1:2}) \right). \end{aligned} This can be solved with Mathematica: here's the incantation.
Minimize[{ z,
p1 + p2 - inf00 <= z,
p1 + (1 - p2) - inf01 <= z,
(1 - p1) + p2 - inf10 <= z,
(1 - p1) + (1 - p2) - inf11 <= z },
{ p1, p2, z }] // Simplify
This has solution \begin{aligned} p_1^* &= \frac{1}{2} \left( 1 + \frac{1}{2} \sum_{y_2=0}^1 \left( \inf_{f \in \mathcal{F}} L (f_{1:2}, 0y_2) - \inf_{f \in \mathcal{F}} L (f_{1:2}, 1y_2) \right) \right) & (z_1 = 0), \\ p_2^* &= \frac{1}{2} \left(1 + \frac{1}{2} \sum_{y_1=0}^1 \left( \inf_{f \in \mathcal{F}} L (f_{1:2}, y_10) - \inf_{f \in \mathcal{F}} L (f_{1:2}, y_11) \right) \right) & (z_1 = 0), \end{aligned} with game value \begin{aligned} &R (\mathcal{F}, \emptyset | z_1 = 0) \\ &= 1 - \frac{1}{2} \min\left\{ \inf_{f \in \mathcal{F}} L (f_{1:2}, 00) + \inf_{f \in \mathcal{F}} L (f_{1:2}, 11), \inf_{f \in \mathcal{F}} L (f_{1:2}, 01) + \inf_{f \in \mathcal{F}} L (f_{1:2}, 10) \right\}. \\ \end{aligned} Now compare this to the case of $z_1 = 1$, which is the same as the fully observed Minimax Forecaster. \begin{aligned} p_1^* &= \frac{1}{2} \left( 1 + \frac{1}{2} \sum_{y_2=0}^1 \left( \inf_{f \in \mathcal{F}} L (f_{1:2}, 0y_2) - \inf_{f \in \mathcal{F}} L (f_{1:2}, 1y_2) \right) \right) & (z_1 = 1), \\ p_2^* &= \frac{1}{2} \left(1 + \inf_{f \in \mathcal{F}} L (f_{1:2}, y_10) - \inf_{f \in \mathcal{F}} L (f_{1:2}, y_11) \right) & (z_1 = 1). \end{aligned} The first round prediction $p_1^*$ is the same whether or not $z_1 = 0$ or $z_1 = 1$, but the second round prediction $p_2^*$ is different. If $z_1 = 0$, then $p_2^*$ is computed by averaging over possible histories; whereas if $z_1 = 1$, then $p_2^*$ is computing using the actual observed history. (Aside: perhaps constant-time Radacher averages will be quantum computing's killer app.)
To decide whether to observe $y_1$ or not, we need to know how much better it is to do so, i.e., the difference in game values. When $z_1=1$ this is the same as the fully observed Minimax Forecaster, \begin{aligned} &R (\mathcal{F}, \emptyset | z_1 = 1) \\ &= 1 - \frac{1}{4} \left( \inf_{f \in \mathcal{F}} L (f_{1:T}, 00) + \inf_{f \in \mathcal{F}} L (f_{1:T}, 01) + \inf_{f \in \mathcal{F}} L (f_{1:T}, 10) + \inf_{f \in \mathcal{F}} L (f_{1:T}, 11) \right), \end{aligned} therefore the difference in game value is \begin{aligned} &R^* (\mathcal{F}, \emptyset | z_t = 0) - R^* (\mathcal{F}, \emptyset | z_t = 1) \\ &= \frac{1}{4} \left| \inf_{f \in \mathcal{F}} L (f_{1:2}, 00) - \inf_{f \in \mathcal{F}} L (f_{1:2}, 01) - \left( \inf_{f \in \mathcal{F}} L (f_{1:2}, 10) - \inf_{f \in \mathcal{F}} L (f_{1:2}, 11) \right) \right|. \\ \end{aligned} This looks like a difference of future differences. If the game value difference exceeds $\alpha$, then we should decide $z_1 = 1$, otherwise not. So, for instance, if every expert predicts the same value on the first round, then the difference of future differences will be zero and we should not observe $y_1$. That certainly sounds like active learning.
So what should a general case $T$ round solution look like? Intuitively, one would hope that if all the experts that have done well in the past predict the same thing on the current instance, that the value of observing $y_t$ for that instance would go down. That is roughly what agnostic active learning does in the IID setting. Here the future is also important, but analogously if all the experts that are in the running for the infimum at the end of the horizon agree on a value, it should be that observing $y_t$ has less value. As we near to the end of the planning horizon, that will be driven mostly by having done well in the past.
|
2018-06-18 15:33:10
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9991633892059326, "perplexity": 1045.2096678741805}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267860570.57/warc/CC-MAIN-20180618144750-20180618164750-00198.warc.gz"}
|
https://chemistry.stackexchange.com/questions/16859/heats-of-combustion-and-stability-of-rings
|
# Heats of combustion and stability of rings
Which isomer would have the largest heat of combustion?
A) Propylcyclopropane
B) Ethylcyclobutane
C) Methylcyclopentane
D) Cyclohexane
E) Since they are all isomers, all would have the same heat of combustion.
Can anyone explain why?
If you think about it, without having the data of the heat of combustion of each type of bond, you can't really answer this question. I know that cyclopropane has a higher heat of combustion per bond than cyclobutane. However, that, in my sense of viewing the problem, does not prevent the fact ethylcyclobutane could have a higher heat of combustion as it is a cycloalkane with 4 bonds rather than 3.
It could happen that the three bonds in cyclopropane and the three bonds in the alkyl branch are lower in energy that the four bonds in cyclobutane and the two bonds in the alkyl branch.
For example:
Let's say a cyclopropane C−C bond is 10 J.
A cyclobutane C−C bond is 9.5 J.
A regular alkane bond is 8 J.
For propylcyclopropane:
Total energy is 54 J
For ethylcyclobutane:
Total energy is 54 J
The same! I know this sounds really simplistic and dumb, but I just don't understand how you can assume the answer is A.
Compounds A-D all have the same molecular formula, $\ce{C6H12}$. We can burn each compound and measure the heat given off (heat of combustion). Since they are isomers, they will each burn according to the same equation
$$\ce{C6H12 + 9O2 -> 6CO2 + 6H2O + heat}$$
Any differences in the heat given off can be used to say that a compound is more stable (it had a lower energy to begin with, so less heat is given off) or less stable (it had a higher energy to begin with, so more heat is given off).
This link provides the heats of combustion for some useful model compounds. Look at the last column ("Total Strain") in Table I, it shows that cyclopropane is slightly more strained than cyclobutane, while cyclopentane and cyclohexane are both much less strained. The strain energy (SE) in cyclopropane will not change appreciably when we add a propyl group to the ring, nor will the SE in cyclobutane change appreciably when we add an ethyl group to the ring.
Therefore, since cyclopropane has the most ring strain and since propyl, ethyl and methyl groups don't contain any SE, the correct answer to the question is "A".
• Ron, in this case, you are using data to say that the TOTAL energy of the THREE C-C bonds cyclopropane is higher than the total energy of the FOUR C-C of cyclobutane. If we had only known that each C-C bonds in cyclopropane were higher in energy than the each one in cyclobutane, could we still be able to answer in the way you did? – yolo123 Sep 29 '14 at 2:03
• Sorry for the nitty-gritty details, I'm just the kind of guy who obsesses about that kind of stuff! – yolo123 Sep 29 '14 at 2:03
• yolo123, a) I'm using the data to say that a cyclopropane ring is always more strained than a cyclobutane ring, which is always more strained than a cyclopentane ring (by definition, a cyclohexane ring is strain free). b) Rather than view the data on a "per bond" basis, it is more commonly viewed on a "per CH2" basis - see the second to last column in the Table, and yes, just using this "per CH2" data we could have answered the question in the same way, e.g. how many CH2's are in each ring. – ron Sep 29 '14 at 2:11
• Sorry for not being clear: I meant being able to answer this ONLY QUALITATIVELY. Without the data, just by knowing that strain/CH2 is higher in cyclopropane without any quantitative knowledge. After all, in an exam situation for which this question was made for, we do not have data. I guess we better know that the total strain is also higher in cyclopropane to answer this question. – yolo123 Sep 29 '14 at 2:25
• Yes, just like we learn methyl, ethyl, propyl, we also learn that ring strain decreases from cyclopropane to cyclobutane (but they're close) to cyclopentane to cyclohexane. – ron Sep 29 '14 at 2:37
The products of combustion are the same so the compound with the highest potential energy (least stable) will have the highest enthalpy of combustion. All have the same number of carbons. The cyclopropane has the most ring strain and therefore the highest potential energy.
If you use metacognition you can see what's being asked. They can only write questions asking "Which has the most ring strain?" so many times. This question is supposed to involve more thinking but can be distilled down to "Which has the most ring strain?"
A good strategy in answering multiple choice questions is to determine what is the same and what is different about the answer choices. In this case the number of carbons is the same so the products are the same. What is different is the number of carbons in the rings and therefore the angle strain is different.
• How can you know that 3 cyclopropyl bonds + another C-C alkly bond is more energetic than 4 cyclobutyl bonds? I know a cyclopropyl bond is more energetic than a cyclobutyl bond. – yolo123 Sep 29 '14 at 0:50
• You can infer it from the intent of the question. There's no way it could be anything else and be a fair question. Chemistrywise look at the angles. 60 degrees is a big difference from 90 degrees. classes.yale.edu/chem220/STUDYAIDS/thermo/cycloalkanes/… sparknotes.com/chemistry/organic2/carbocycles/section1.rhtml – Brinn Belyea Sep 29 '14 at 0:53
• Ok, so, there is nothing certain about the answer :S Are you sure there is no scientific way of proving this with the question without going to the "how to answer test questions realm" and looking up data? By the way, thanks a lot for the help. Really appreciate it. – yolo123 Sep 29 '14 at 0:57
• The scientific way is to look at a heat of formation table. The thing is, the ring strain tables I found are close enough that I wonder if A really is the answer. It could be a bad question. – Brinn Belyea Sep 29 '14 at 0:59
• Very confusing. – yolo123 Sep 29 '14 at 1:03
|
2020-01-24 00:15:58
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5635648369789124, "perplexity": 1128.007099608826}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250614086.44/warc/CC-MAIN-20200123221108-20200124010108-00506.warc.gz"}
|
http://godplaysdice.blogspot.com/2008/10/interesting-fact-of-day.html
|
## 09 October 2008
### Interesting fact of the day
The number of ways to write n as a sum of powers of 2, where each summand occurs at most three times, is (n/2) + 1 rounded down. See the 1983 Putnam exam, problem B2 (link goes to a solution, so don't go there if you want to think about the problem yourself!) I learned this from Bruce Reznick's paper "Some binary partition functions".
Of course, if we replace "three" by "one", the analogous problem is trivial -- it's just saying that the binary expansion of an integer is unique. If "three" is replaced by "two", we get an interesting sequence: let b(n) be the number of ways to write n as a sum of powers of two where each summansd occurs at most three times. So for example b(14) = 4, since we can write 14 as 8 + 4 + 2, 8 + 4 + 1 + 1, 8 + 2 + 2 + 1 + 1, or 4 + 4 + 2 + 2 + 1 + 1. I've alluded to this sequence before in comments. Starting with b(0), the sequence of b(n) is
1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4, 1, 5, 4, 7, ...
and we can make rational numbers out of these by taking b(n)/b(n+1):
1/1, 1/2, 2/1, 1/3, 3/2, 2/3, 3/1, 1/4, 4/3, 3/5, 5/2, 2/5, 5/3, 3/4, 4/1, 1/5, 5/4, 4/7, ...
and it turns out this sequence includes each rational number exactly once! Neil Calkin and Herb Wilf wrote about this in their paper Recounting the rationals, which you should read. I first learned about this paper from Brent Yorgey's series of posts.
intrinsicallyknotted said...
michaeldcassidy said...
Phillie is playing the on team that no matter who they play I want to lose.
In fact I'd love to see them lose every game in a season.
CarlBrannen said...
This reminds me of the simplest adder used in digital electronics. It has 3 inputs and two outputs. The three input bits are of the same magnitude and can give results of 0, 1, 2, or 3. This requires two bits to code.
The high output bit is called the "carry out", and one of the three inputs is called the "carry in". One can use N of these to add two N-bit numbers.
I suspect that one could also use N of these to correct the binary description of a number which had as much as 3 copies of a given bit.
Also, "each rational number" should presumably be "each positive rational number".
Anonymous said...
Yeah, it's interesting stuff. I came across that paper in the course of solving one of the problems at projecteuler.net (which is awesome, btw).
Anonymous said...
I was a little surprised that the solution to the problem included a few obviously wrong statements. From context you could figure out what the author was trying to say, but it was still jarring. (For example, the claim that an even number would not be expressed as the sum of more than two 1's. Um, how about 4 = 1 + 1 + 1 + 1?)
Michael Lugo said...
@anonymous 2: Read the problem statement more carefully. The problem only considers representations which use each summand at most three times.
Anonymous said...
Yes, but the solution wasn't phrased that way. If so, it would have read something like "an even number expressed with at most three ones can only have two". Or even: "In the situations under consideration in this problem, an even number can be expressed as the sum of at most two ones".
As I said, it was easy to understand from context, but at first glance it read like (an obviously wrong)general statement.
Anonymous said...
Yeah. Papers are way easier to read when they restate the axioms in every sentence that discusses a conclusion.
|
2014-11-28 22:41:56
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8401240706443787, "perplexity": 306.33521113448114}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931011032.12/warc/CC-MAIN-20141125155651-00087-ip-10-235-23-156.ec2.internal.warc.gz"}
|
https://www.nari.ee.ethz.ch/wireless/pubs/p/asilomar08&refs=1
|
# Robust Gain Allocation against Phase Uncertainty at the Relays for Multiuser Cooperative Networks
### BibTeX Reference
Warning: This BibTeX entry is automatically generated from the database. As there is no one-to-one match between the database fields and the BibTeX format, there may be errors in the entry. Regard the BibTeX feature as experimental for now. Use with care and correct errors and omissions manually.
@conference{asilomar08, author = {Esli, Celal and Wittneben, Armin}, title = {Robust Gain Allocation against Phase Uncertainty at the Relays for Multiuser Cooperative Networks }, booktitle = {Asilomar Conference on Signals, Systems and Computers}, month = oct, year = 2008, url = {http://www.nari.ee.ethz.ch/wireless/pubs/p/asilomar08} }
### LaTeX Reference
\bibitem{asilomar08} C. Esli and A. Wittneben, Robust Gain Allocation against Phase Uncertainty at the Relays for Multiuser Cooperative Networks ,'' \emph{Asilomar Conference on Signals, Systems and Computers}, Oct. 2008.
### HTML Reference
C. Esli and A. Wittneben, "Robust Gain Allocation against Phase Uncertainty at the Relays for Multiuser Cooperative Networks ," Asilomar Conference on Signals, Systems and Computers, Oct. 2008.
|
2022-11-29 18:33:07
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.35598257184028625, "perplexity": 14270.475611266776}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710710.91/warc/CC-MAIN-20221129164449-20221129194449-00361.warc.gz"}
|
http://askubuntu.com/questions/332043/lircd-works-when-run-from-command-line-not-when-run-from-init-d-service
|
# lircd works when run from command line, not when run from init.d/service
I'm configuring a system for use as a media centre, running Mythbuntu 12.04. Everything is working, short of the remote. I've configured my remote, and it I start lirc from the command line, it works fine. If I run it from the init script service lirc start it doesn't work. (For this purpose, work means to have irw show an output). In both cases I'm using exactly the same command (in fact, for running it from the command line, I'm just copying the command that is present in ps -ef). Both cases have lirc running as root (at least that's what ps -ef tells me). Both cases produce exactly the same info in syslog.
Command to start lirc:
sudo /usr/sbin/lircd --output=/var/run/lirc/lircd --driver=devinput --device=/dev/input/irremote
syslog output (start lirc, start irw, stop irw, kill lirc):
Aug 13 22:06:05 Myth-FJ lircd-0.9.0[4917]: lircd(devinput) ready, using /var/run/lirc/lircd
Aug 13 22:06:05 Myth-FJ lircd-0.9.0[4917]: accepted new client on /var/run/lirc/lircd
Aug 13 22:06:05 Myth-FJ lircd-0.9.0[4917]: initializing '/dev/input/irremote'
Aug 13 22:06:28 Myth-FJ lircd-0.9.0[4917]: accepted new client on /var/run/lirc/lircd
Aug 13 22:06:35 Myth-FJ lircd-0.9.0[4917]: removed client
Aug 13 22:06:42 Myth-FJ lircd-0.9.0[4917]: caught signal
Aug 13 22:06:42 Myth-FJ lircd-0.9.0[4917]: closing '/dev/input/irremote'
I'm completely stumped.
-
Do you use sudo for service lirc start? – Braiam Aug 13 '13 at 12:17
@Braiam: Yes; the service starts and runs correctly – askvictor Aug 13 '13 at 12:18
so, you forgot to add sudo before service? – Braiam Aug 13 '13 at 12:35
@Braiam: no; I start the service with sudo. If I didn't, it wouldn't start, as a normal user would lack the permissions. Besides which, I stated in my post that lirc is running as root in both cases. EDIT: to clarify, when I said "the service starts and runs correctly", it starts, and doesn't show any errors, but doesn't produce any useful output to irw – askvictor Aug 13 '13 at 12:50
I had the same symptoms, lirc would not work when run as a service, but would work as a command. Note that in my case some of the keys worked, but only when lirc was not running (neither as a service nor a command).
The problem was that evdev (an input driver for Xorg) was picking up the remote as a keyboard input device. Getting evdev to ignore the remote solved the problem for me. I guess that the lirc service start script got hung up by the fact that evdev had a handle on the remote, somehow.
The following is from the last post in that thread. Note: TBS6981 is the satellite tuner which I was using as my remote receiver.
The problem was caused by evdev (an input driver for Xorg) which was picking up the remote as a keyboard input device. This could be seen from my Xorg logs (/var/log/Xorg...), see below. Most of the buttons don't work by default because they don't naturally map to keyboard buttons, e.g. your keyboard doesn't have an EPG key, but it does have numbers and direction keys.
There are basically two solutions
1. Configure evdev to ignore the IR input and use lirc OR
2. Remap the remote buttons to keyboard keys, as discussed in this post, and don't use lirc
evdev can be configured to ignore the TBS6981 as a keyboard by adding the following to the end of /usr/share/X11/xorg.conf.d/10-evdev.conf -
Section "InputClass"
Identifier "cx23885 IR (TurboSight TBS 6981)"
MatchProduct "cx23885 IR (TurboSight TBS 6981)"
MatchIsKeyboard "on"
Option "Ignore"
EndSection
Note that MatchProduct above matches the identifier found in Xorg.0.log.
This is an extract from Xorg.0.log showing evdev picking the remote input, these lines do not appear if evdev is correctly configured to ignore the remote as a keyboard:
[ 21.486] (II) config/udev: Adding input device cx23885 IR (TurboSight TBS 6981) (/dev/input/event3)
[ 21.486] (**) cx23885 IR (TurboSight TBS 6981): Applying InputClass "evdev keyboard catchall"
[ 21.486] (II) Using input driver 'evdev' for 'cx23885 IR (TurboSight TBS 6981)'
|
2015-01-27 15:08:42
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3897901773452759, "perplexity": 10091.354691937533}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422121981339.16/warc/CC-MAIN-20150124175301-00127-ip-10-180-212-252.ec2.internal.warc.gz"}
|
http://meetings.aps.org/Meeting/HAW09/Event/108175
|
### Session EF: Mini-Symposium on Hadron Structure and QCD in High Energy Processes II
9:00 AM–12:00 PM, Friday, October 16, 2009
Room: Kohala 3
Chair: Misak Sargsian, Florida International University
Abstract ID: BAPS.2009.HAW.EF.8
### Abstract: EF.00008 : Timelike Compton Scattering with CLAS
10:45 AM–11:00 AM
Preview Abstract MathJax On | Off Abstract
#### Author:
Stepan Stepanyan
(Jefferson Lab)
Deeply Virtual Compton Scattering (DVCS), $\backslash$Epg, has been under intense theoretical and experimental studies in recent years as a new tool to access Generalized Parton Distributions (GPDs) of the nucleon. The simplest observables in DVCS for studying GPDs are spin dependent cross section differences. These asymmetries measure the imaginary part of Compton Form-Factors (CFFs), where GPDs enter at specific kinematical point, $\xi=x$. Here $\xi$ is the generalized Bjorken variable and $x$ is the light-cone momentum fraction of the struck quark. The real part of CFF is proportional to the integral of GPDs over $x$ and can only be accessed in the measurements of the DVCS cross section or the beam charge asymmetry. Studying the real part is important for modeling GPDs. It is sensitive to the so-called $D$-term, introduced in the modeling of GPDs to ensure the polynomiality of their Mellin moments. Photoproduction of lepton pairs, or so-called Time-like Compton Scattering (TCS), is an inverse process to DVCS and offers additional constraints on GPDs. In particular, TCS can be used as an effective tool to study the real part of the Compton amplitude using the azimuthal angular asymmetry arising from exchange of $l^+$ and $l^-$ momenta. In this report, first analysis of the Time-like Compton Scattering using the CLAS electroproduction data will be presented. Details of the extraction of quasi-real photoproduction of lepton pairs will be discussed.
To cite this abstract, use the following reference: http://meetings.aps.org/link/BAPS.2009.HAW.EF.8
|
2013-05-19 19:28:43
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7165483832359314, "perplexity": 3045.6926966850806}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368698017611/warc/CC-MAIN-20130516095337-00005-ip-10-60-113-184.ec2.internal.warc.gz"}
|
https://uwaterloo.ca/secretariat/documents-potential-interest/memorandum-agreement-uw-fauw
|
Memorandum of Agreement -- UW / FAUW
Note: this online version of the MoA is the most up to date version available and should be considered the official version. While a hard-copy version of this MoA is available, it is printed periodically and, therefore, may not be up to date.
Memorandum of Agreement between the Faculty Association of the University of Waterloo and the University of Waterloo
Current as of 30 October 2018
1. Preamble
17. Lay-Offs
1. PREAMBLE
1.1 The Parties to this Agreement are the Faculty Association of the University of Waterloo (FAUW) and the University of Waterloo (UW), hereinafter referred to as the Association and the University, respectively. 1.2 The University recognizes the Association as representing the University employees (Members) as defined under Article 2.1.1. Recognition of the Association under this Agreement does not constitute voluntary recognition equivalent to certification. 1.3 This Agreement is a “special plan” agreement negotiated between and ratified by the Parties. 1.4 The Parties recognize that the objective of the University is the attainment of high standards of academic excellence in the pursuit and dissemination of knowledge for the benefit of students and of the academic and wider communities. Further, the Parties agree that the goals of the University include the following: (a) the attainment of high standards of excellence in teaching, scholarship, research, and creativity; (b) the development of skills and attitudes essential for scholarly study and scientific investigation, and for the effective sharing of the results of these activities with students and fellow scholars and with the community at large; (c) the encouragement of the pursuit of truth by individuals and groups through teaching, research, free enquiry, and criticism, in order to extend the frontiers of knowledge and understanding; and (d) the provision of an environment which will support the intellectual, cultural, and physical development of the University community. 1.5 This Agreement has the following purposes: (a) to set out terms and conditions of employment for Members, and to describe procedures for developing and revising Policies concerning terms and conditions of employment; and (b) to define and describe the relationship between the University and the Association, to provide for regular communication and consultation between the University and the Association, and to provide means for resolving differences which may arise between them. 1.6 The University and the Association agree: (a) to encourage within the University community a climate of freedom, collegiality, responsibility and mutual respect; (b) to foster harmonious relations and a working environment that enables individual Members to achieve the goals and objectives of the University; and (c) to ensure the equitable treatment of individual Members through fair procedures and practices. 1.7 In this Agreement, for academic units, Chair refers to Chair or Director, and Department refers to Department or School.
2. RECOGNITION AND RIGHTS OF THE ASSOCIATION
3. CORRESPONDENCE AND INFORMATION
3.1 The University and the Association recognize that both Parties require access to information for the proper administration of this Agreement, and agree to use professional discretion in dealing with such information. 3.2 Correspondence between the Association and the University related to this Agreement shall pass between the President of the Association and the President of the University. Where written notice is specified, the University's internal mail may be used. 3.3 The University undertakes to provide the following information to the Association in a timely manner: (a) an annual list of Members as of May 1, including name, rank, department, appointment category and dates, FTE, and approved leaves of absence during the salary year for each Member; (b) a monthly update giving the same data as in (a) for new Members and listing terminations of Members; (c) the names and new ranks of Members who have received promotions, and the effective dates of such promotions; (d) during the first month of the salary year and within one month of any changes to benefit plans, a list and detailed description of all benefit plans applicable to Members, and the rates charged to Members for these plans; (e) public agenda materials and minutes of the Senate, Senate Finance Committee, Senate Long Range Planning Committee, Board of Governors, Board Pension and Benefits Committee; (f) agenda materials and minutes of the Faculty Relations Committee; (g) a copy of the latest University budget and budget reports when circulated to the Board; (h) names and professional addresses of members of the Board of Governors and Senate, and of Board and Senate committees, together with the terms of reference of such committees; and (i) such other information as the Parties agree from time to time. 3.4 The University agrees to provide the report of the Pension Plan Actuary, and audited and other reports concerning the pension and benefits plans to the Association on request. 3.5 The Association undertakes to provide the following information to the University in a timely manner: (a) a copy of each FAUW Forum, newsletter, or other public communication to all Members at the time of distribution; (b) an up-to-date copy of the Association's Constitution and Bylaws within one month of their revision; (c) an up-to-date list of the Association's Board of Directors and Executive within one month of any changes; and (d) such other information as the Parties agree from time to time.
4. FACULTY RELATIONS COMMITTEE
4.1 The Faculty Relations Committee (FRC) is a consultative committee with equal representation from the University administration and the Association Board of Directors. 4.2 The President of the Association and the Vice-President Academic & Provost shall alternate from meeting to meeting as Chair of the FRC. The additional members shall be four senior administrators appointed by the University President and four members of the Association Board of Directors appointed by the Association President. 4.3 The Parties confirm as part of this Agreement that the practices and procedures followed by the FRC are as described in Policy 1 and in Appendix A to Policy 1. In particular, no formal decision may be made by the FRC except by agreement of the "double majority" comprised of majority support from each of the two groups of members appointed to the FRC by the University and Association Presidents. The Parties agree that for the purposes of this Agreement the content of Appendix A to Policy 1 shall be regarded as part of Policy 1. 4.4 The FRC provides a regular forum for discussion of issues affecting faculty members, including matters arising from the application of this Agreement (but excluding any individual or group grievance which is at that time being resolved under the grievance and arbitration procedures set out in Article 9). The FRC also has responsibility for the development and approval of Class F and Class A Policies, and shared responsibility with the Staff Relations Committee for the development and approval of Class FS Policies. 4.5 The FRC shall meet on a regular basis, usually every two weeks from September through June, although meetings may be held more or less frequently by mutual agreement of the two Chairs. It is expected that the FRC will operate in a collegial manner, with most decisions made by consensus rather than by formal vote. 4.6 The University Secretariat shall provide a Secretary for meetings of the FRC. Minutes of meetings are confidential to the Committee.
5. MEMORANDUM OF AGREEMENT AND POLICIES
5.1 Agreement and Policies 5.1.1 The relations between the Parties, and between the University and the Members, are defined partly in this Agreement and partly in University Policies. 5.1.2 The Faculty Relations Committee has responsibility for the development and approval of Class F and Class A Policies, and shared responsibility for the development and approval of Class FS Policies. Class F Policies deal exclusively with the terms and conditions of employment of faculty members, Class A Policies deal with the appointment of academic administrative officers, and Class FS Policies deal with terms and conditions of employment of faculty and other employees. The policy classifications and development and approval process are described in Policy 1. 5.1.3 The current list of F, A and FS Policies at the date this Agreement comes into effect or as amended during the life of this Agreement is given in Appendix A to this Agreement. 5.2 Policy Development and Amendment 5.2.1 Nothing in this Agreement is intended to prevent the development, amendment or review of any University Policies in accordance with Policy 1 during the term of this Agreement. 5.2.2 Where there are differences between this Agreement and any Policy other than a Class F, A, or FS Policy, the provisions of this Agreement shall prevail. 5.2.3 Where there are differences between this Agreement and any existing Class F, A, or FS Policy at the date this Agreement comes into effect, the provisions of this Agreement shall prevail, and the Parties agree to revise the existing Policies to harmonize them with this Agreement. 5.2.4 If a Class F, A, or FS Policy is revised or a new Class F, A, or FS Policy is developed in accordance with Policy 1 and goes forward for approval, the Parties agree to a limited opening of this Agreement at the request of either Party for the sole purpose of negotiating revisions to this Agreement to harmonize it with the new or revised Policy. The Parties agree to make all reasonable efforts to negotiate these revisions before the new or revised Policy is submitted to the Board of Governors for final approval. 5.2.5 Amendments to this Agreement made as a consequence of 5.2.4 shall not require ratification by the Parties to take effect.
6.1 Academic freedom provides the possibility of examining, questioning, teaching, and learning, and involves the right to investigate, speculate, and comment without deference to prescribed doctrine. As such, it entails the freedom of individuals to practise their professions of teacher, researcher and scholar, the freedom to publish their findings, the freedom to teach and engage in open discussion, the freedom to be creative, the freedom to select, acquire, disseminate, and use documents in the exercise of their professional activities, and the freedom to criticize the University and the Association. Academic freedom also entails freedom from institutional censorship. 6.2 The University and the Association recognize that the provision of academic freedom is particularly vital to those whose approaches to teaching, scholarship, and research result in criticism of and challenge to established, conventional beliefs and practices. 6.3 The academic freedom of any person shall not be infringed upon or abridged in any manner. As academic freedom will wither and die unless the university community as a whole is committed to it, the University and the Association agree to support and defend academic freedom at the University of Waterloo. 6.4 As the common good of society depends upon an unhampered search for knowledge and its free expression, and as academic freedom in universities is essential to the attainment of each of these purposes in the teaching function of the university as well as in the pursuit of its scholarship and research, those who are guaranteed academic freedom have also a responsibility in exercising it not to infringe upon the academic freedom and rights of other members of the university community. Indeed, academic freedom carries with it the duty to use that freedom in a manner that is consistent with the scholarly obligation to base research and teaching on an honest and ethical quest for knowledge. Academic freedom does not require neutrality on the part of the individual; rather, academic freedom makes commitment possible. 6.5 As the censorship of information is inimical to the free pursuit of learning, the creation, collection, organization, and dissemination of knowledge shall be done freely and without bias in support of the research, teaching, and study needs of the university community. No censorship shall be exercised or allowed against any material relevant to the pursuit of learning which a faculty member desires to be placed in the library collections of the University.
7. NON-DISCRIMINATION
7.1 The Parties agree that all Members have a right to equal treatment with respect to employment and that there shall be no discrimination, interference, restriction or coercion exercised or practised towards any Member in respect to appointment, benefits, dismissal, promotion, rank, reappointment, salary, tenure or any other terms or conditions of employment by reason of age (except for retirement pension and benefit conditions as permitted or mandated by law), ancestry, citizenship, clerical or lay status, colour, creed, ethnic origin, family status, gender, language (except where competence in the language is a bona fide occupational requirement), marital status, membership or non-membership in the Association, physical or emotional ill health or disability (except where the Member refuses to seek medical treatment and/or where the physical or emotional ill health or disability would clearly prevent the carrying out of the required duties of the position), place of origin, political or religious affiliation or belief, pregnancy, race, sexual orientation, or any activity pursuant to the principles of academic freedom and responsibilities set out in Article 6. 7.2 Policies, practices, or acts which create, intentionally or unintentionally, a sustained negative working climate which can clearly be attributed to any of the prohibited grounds in 7.1 will be considered discrimination. 7.3 This Article shall not preclude the Parties agreeing to any equity measures.
8. DISCIPLINE
8.1 8.1 A Member may be disciplined only for just cause and only in accordance with the provisions of this Article and, for matters dealing with scholarly research, the provisions of Article 14. Disciplinary processes are not to be used to inhibit free inquiry, discussion, exercise of judgement, or honest criticism within or without the University. Disciplinary action shall be reasonable, commensurate with the seriousness of the violations, and consistent with accumulated practice under this Article. The Parties recognize the value of promoting corrective action through guidance and progressive discipline, although this will not always be appropriate. 8.2 In all matters of discipline, a Member has the right to seek advice from the Association and to be accompanied by an academic colleague for advice and support (including, if necessary, aid in presenting the Member's position) during any meetings attended to discuss such matters. All disciplinary measures are grievable under Article 9. 8.3 The University bears the onus of proving that a disciplinary action was taken for just cause. 8.4 The only disciplinary measures which may be taken by the University against a Member are the following: (a) A letter of warning or reprimand. Such letters must be specific and must be clearly identified as disciplinary measures. (b) Suspension with pay. Suspension is the act of relieving a Member, without her/his consent, of some or all university duties and/or privileges. (c) Suspension with partial pay, or without pay, or a fine in lieu thereof, where appropriate. (d) Dismissal for cause. For Members with tenure or continuing lecturer appointments, dismissal means the termination of appointment without the Member's consent. For all others, dismissal means termination of appointment without the Member's consent before the end of the contract. Non-renewal of definite term or probationary appointments and denial of tenure do not constitute dismissal. (e) A public statement from the University that a Member was guilty of misconduct in research. 8.5 Just cause for the dismissal of a tenured Member includes, but is not limited to, the persistent and serious neglect of the normal duties of a faculty member, particularly with respect to teaching and scholarship, or the failure to carry out such duties as are reasonably assigned by the appropriate academic authorities. In a case of persistent neglect, the action for dismissal must have been preceded by letters of warning from the Member's Chair or Dean. Warnings shall not only state the nature of the alleged deficiencies and make constructive suggestions for improvement, but also shall be followed by a reasonable period in which to make improvements. 8.6 Just cause for dismissal also includes but is not limited to: a serious breach of criminal law; violent behaviour or threats of violence against a member of the University community; a serious breach of ethical behaviour; violations of ethics in respect of scholarship, teaching, or collegiality. Any of the above must be of such a serious nature as to render the Member clearly unfit to continue to hold a tenured appointment at the University of Waterloo. 8.7 Disciplinary processes must be kept distinct from academic assessments associated with annual performance reviews and consideration for tenure, promotion, and probationary reappointment. The fact that a disciplinary measure has been imposed or is contemplated cannot be considered in an academic assessment, but the facts which resulted or may result in the imposition of discipline can be considered, if relevant to that assessment. 8.8 The Member's Dean shall promptly investigate any concerns or allegations about a Member if the Dean reasonably believes that a situation warranting disciplinary measures may exist. The Dean shall inform the Member as soon as may reasonably be possible both of the nature of the allegation and if an investigation is being undertaken. The conduct of all or part of such investigations may be delegated to appropriate persons, including the Member's Department Chair. The investigation itself is not a disciplinary measure, and an investigation which has not yet been completed is not a matter for grievance. 8.9 The Dean shall take reasonable steps to maintain the Member's privacy and the confidentiality of the investigation and its findings until the imposition of discipline, if any. However, some disclosure of concerns and allegations may be necessary, either in order to conduct the investigation or if the Dean has reasonable grounds to believe that such confidentiality may place a person or persons at risk of significant harm. In the event that it is determined that there shall be no disciplinary action, the Dean must inform each individual to whom concerns and allegations were disclosed that there is no basis for disciplinary action. 8.10 When the investigation has been completed, and if disciplinary action is being considered, the Dean shall notify the Member in writing of the results of the investigation and of the proposed disciplinary action. The notice shall provide the specific details of the alleged cause for the discipline, including all names, places, and dates of the alleged incidents, and shall either be hand-delivered to the Member, or delivered by registered mail to the Member's last known address. The date of notice is defined to be either the date on which a registered letter has been signed for or the date on which the notice is hand-delivered to the Member. 8.11 The Dean shall convene a meeting within twenty-five working days of the date of notice to afford the Member an opportunity to make oral and/or written submissions before any disciplinary measures are imposed. The Member shall be given at least seven working days notice of the time and place of the meeting. The Dean may invite the person or persons who have carried out the investigation to attend. At this meeting an attempt shall be made to resolve the matter in a manner satisfactory to all concerned. For the purposes of this clause, days during which the Member is on pre-scheduled vacation, as well as Saturdays and Sundays, other holidays, days during which the University is officially closed, and days during which the Member is absent on pre-scheduled official University business shall not be treated as working days. 8.12 If no satisfactory solution is reached at the meeting referred to in 8.11, within two weeks the Dean shall notify the Member in writing of the disciplinary decision with reasons. 8.13 The Dean shall make every reasonable effort to notify the Member of the meeting referred to in 8.11. If the Dean is unable to contact the Member, or if the Member is notified and chooses not to attend, the meeting shall be dispensed with, and the Dean may give notice of discipline as in 8.12 above. 8.14 Where the disciplinary decision in 8.12 is dismissal for cause and where the Member chooses to contest the decision, a formal grievance shall be submitted to the Vice-President, Academic & Provost (VPA&P) in accordance with Article 9. The VPA&P shall act as a committee of one to decide the matter on behalf of the Board of Governors. The decision of the VPA&P may be taken to external arbitration under 9.6. 8.15 Where the disciplinary action is dismissal for cause, suspension with reduced pay or a fine in lieu thereof, the Member shall retain full salary and benefits (subject to the rules and regulations of UW benefit programs (see Article 11.1.3) until the time limit for filing a grievance under Article 9 has expired. If the disciplinary action is grieved, the Member shall retain full salary and benefits for a period of one year from the date of the disciplinary decision in 8.12, or until the grievance and arbitration procedures set out in Article 9 have been completed, whichever is earlier. In the event that the Tribunal or Arbitrator finds in favour of the Member, any lost compensation shall be restored. 8.16 Notwithstanding 8.15, eligibility for full salary and benefits shall not extend beyond the Member's retirement date (if a retirement date exists), nor beyond the termination date for a definite term or probationary appointment terminated in accordance with Policy 76. Furthermore, the University may terminate salary and benefits if, during the period referred to in 8.15, the Member accepts outside employment in excess of the normal guidelines as specified in Policy 49. 8.17 Where the disciplinary action is dismissal for cause or suspension, at the request of either the Member or the Dean, the VPA&P may relieve the Member of her/his duties during the period of full salary and benefits as specified in 8.15. If this action is taken the Association shall be informed. If the disciplinary action is suspension with pay, such suspension shall count towards the period of suspension in the event that the grievance is unsuccessful. 8.18 Failure of a Member to grieve a letter of reprimand or warning at the time of receipt of the letter shall not be deemed an admission of the validity of the reprimand or the warning.
9. GRIEVANCE AND ARBITRATION
10. COMPENSATION NEGOTIATIONS
10.1 Annual compensation changes for Members, other than selective increments as defined in Article 13, are as specified in the Memorandum of Settlement describing the results of compensation negotiations between the Association and the University. 10.2 The Memorandum of Settlement shall specify the annual scale change. Each year the Compensation Negotiation Teams shall use the annual change in the Canada Consumer Price Index as a starting figure for the discussion of the scale adjustment. Adjustments in the salary scale are influenced by economic factors, relevant salary trends and by the University's financial position. Scale changes, expressed as a percentage change, apply to the salary structure and to all salaries. Other items that may be specified include changes in the salary structure (see Article 13), anomalies and other special salary increases, and changes in the Faculty Professional Expense Reimbursement Plan (see 11.5) and other benefits specific to Members. The Memorandum of Settlement may also include an amount for proposed changes in benefits defined in University Policies and/or administered by the Pension and Benefits Committee. If the proposed benefit changes are not approved, the negotiated amount shall be awarded as a scale change. 10.3 The Memorandum of Settlement shall be for one year, two years, or three years. In the absence of agreement between the Parties on a longer period the Memorandum of Settlement shall be for one year. The years in the Memorandum of Settlement shall begin on May 1 and end on April 30, coincident with the salary year for Members. 10.4 The Memorandum of Settlement shall be part of this Agreement, and shall be binding on the Board of Governors, the Association, and individual Members. 10.5 Start of Compensation Negotiations 10.5.1 Prior to June 15 in any year in which compensation changes are to be negotiated the President of the Association and the President of the University shall begin the process of establishing lists of possible mediators and arbitrators as specified in 10.9. These lists are to be completed prior to the commencement of negotiations. 10.5.2 By the November 15 immediately preceding the expiry date of the Memorandum of Settlement, each Party shall inform the other of the names of its Chief Negotiator and the other two members of its Negotiating Team. 10.5.3 Negotiations shall commence as soon as can be arranged by the two Chief Negotiators, and at the latest by December 1st. The Chief Negotiator and at least one other member of each Negotiating Team shall be present at each negotiation session. 10.5.4 By mutual agreement, the Parties may alter any of the dates specified in this Article. 10.6 Stage 1 Negotiations The Parties agree to negotiate in good faith and to make every reasonable effort to reach an agreement in Stage 1. If the Parties reach agreement, a Memorandum of Settlement shall be prepared and signed by the Chief Negotiators and at least one other member of each Negotiating Team. 10.7 Stage 2 Negotiations – Mediation 10.7.1 If the Parties have not reached agreement by February 1, an external mediator shall be appointed unless both Parties agree to waive mediation. Mediation shall commence not later than February 15. If mediation is waived, Stage 1 negotiations may continue. 10.7.2 The role of the mediator is to assist the Parties in reaching a negotiated agreement. The procedure for appointing the mediator is specified in 10.9. The cost of the mediator shall be shared equally by the Parties. 10.7.3 Mediation shall terminate by the earliest of March 1, the date on which an agreement is reached, or the date on which either the mediator or both Parties decide that further mediation would not be useful. 10.7.4 If an agreement is reached, a Memorandum of Settlement shall be prepared and signed by the Chief Negotiators and at least one other member of each Negotiating Team. 10.8 Stage 3 Negotiations – Arbitration (Final Offer Selection) 10.8.1 If an agreement has not been reached by March 1, an arbitrator shall be appointed. The role of the arbitrator is to select between the final positions of the two Parties. The procedure for appointing an arbitrator is specified in 10.9. The cost of the arbitrator shall be shared equally by the Parties. 10.8.2 The arbitration hearing shall commence not later than March 15. One week prior to the arbitration hearing, each Party shall submit its proposed Memorandum of Settlement to the arbitrator with a copy to the other Party. 10.8.3 Within three weeks after the arbitration hearing, the arbitrator shall provide each Party with a copy of her/his report, indicating which Party's proposed Memorandum of Settlement has been selected and the reasons for selecting it. 10.9 Selection of Mediator and Arbitrator 10.9.1 In accordance with 10.5.1, the Association and University Presidents shall begin the process of establishing a list of at least three possible mediators and a second list of at least three possible arbitrators by the June 15 prior to the start of negotiations. These individuals must be external to the University and acceptable to both the Association and the University. 10.9.2 In the event that a mediator or arbitrator is required, names shall be drawn at random from the agreed list until one of them is available to serve. By mutual agreement in writing, the two Presidents may select a listed mediator or arbitrator out of turn, or may select a mediator or arbitrator not on the list. No individual shall serve as both mediator and arbitrator in the same year.
11. PENSION AND BENEFITS
11.1 UW Pension Plan and Shared Benefit Programs 11.1.1 The University of Waterloo has a common pension plan for all eligible employees, including eligible Members. The UW Pension Plan is administered by the Board of Governors Pension and Benefits Committee. Membership of this committee is determined by Board of Governors Resolution, which specifies that the Board of Governors shall appoint three faculty members to the committee on the recommendation of the Association President. 11.1.2 The UW Pension Plan is a defined benefit plan integrated with the Canada Pension Plan, with costs shared by employees and the University. Specific details are contained in the official Pension Plan text (available from Human Resources) which is approved by the Board of Governors and registered with federal and provincial pension authorities. 11.1.3 UW also has a number of common benefit programs for all eligible employees, including eligible Members. These programs are administered by the Board of Governors Pension and Benefits Committee, and include extended health care, dental plan, sick leave and long term disability, and group life insurance. Several government programs are also applicable (e.g., OHIP, Workers' Compensation, Employment Insurance, Canada Pension Plan; details are available from Human Resources). 11.1.4 The Parties agree that the Association shall receive notice of, and have input into, any proposed changes to the common pension and benefit programs in so far as they affect Members. 11.2 Vacation Entitlement 11.2.1 The annual vacation entitlement for Members with appointment duration of one year or more shall be one month during each of the first ten years of employment. The annual entitlement shall increase to one month plus one week in the earlier of the eleventh year of employment or at the earliest possible retirement date under the pension plan. 11.2.2 Vacation entitlement normally shall be used during the contract year in which it is earned. In exceptional circumstances with the prior written permission of the Department Chair, vacation entitlement may be carried forward for a maximum of one year. All vacation entitlement must be used prior to termination or retirement. 11.2.3 Vacation shall be scheduled at a time or times which are mutually satisfactory to the Member and the Department Chair. 11.3 Retirement 11.3.1 “Retirement” refers to termination of employment at the University of Waterloo. Receiving a UW pension does not mean that a Member has retired unless he or she has stopped working at UW. Termination of employment prior to age 55 or dismissal for cause is not considered retirement. 11.3.2 Normal Retirement Date (as defined in the University of Waterloo Pension Plan) for University employees is the first day of the month following or coincident with the 65th birthday. 11.3.3 Notwithstanding Article 11.3.2, a Member has the right to retire at a date of his/her choosing. 11.3.4 A Member continuously employed by the University of Waterloo from a date prior to January 1, 1969 will continue to have the rights and privileges as specifically defined under the University of Waterloo Pension Plan and insured UW benefits for that group. 11.4 Conversion of Vacation Entitlement Prior to Retirement at Age 66 or Earlier 11.4.1 Under the conditions set out below, a Member may opt to convert (the "Conversion Option") one week of annual vacation entitlement in each year preceding retirement (to a maximum of three) into a one-time 2% salary increase based on the Member's salary in the immediately preceding salary year. The 2% increase will be calculated on the Member’s base salary immediately prior to the start of the salary year during which it takes effect. Both the salary increase and the reduction in vacation will be ongoing until the Member's retirement date. 11.4.2 Eligibility date for conversion of vacation entitlement shall be not later than 30 April 2027 with retirement on or before 1 May 2030. 11.4.3 The Member shall submit the Conversion Option to the University within three years (or earlier) of his or her intended retirement date. The latest eligibility date for the Conversion Option shall be the Member's 65th birthday, with a retirement date no later than the end of the academic term (i.e., either April 30, August 31, or December 31) during which he or she turns 66. 11.4.4 Where the Member notifies the University prior to the earliest eligibility date, the 2% salary increase shall take effect on the earliest eligibility date (three years prior to the retirement date). Where the Member notifies the University after the earliest eligibility date, the 2% salary increase shall take effect on the first day of the month following such notification. 11.5 Faculty Professional Expense Reimbursement Plan 11.5.1 Every regular faculty member shall be entitled to a Faculty Professional Expense Reimbursement (FPER) in each salary year. The maximum FPER shall be prorated for Members with fractional-load appointments or who are employed by the University for only a portion of the salary year. 11.5.2 The FPER provides reimbursement for expenses related to the performance of teaching, research, and professional duties. All goods purchased under this Plan are the property of the University. Eligible expenditures include but are not limited to: (a) membership fees for professional associations or learned societies (but not including Association dues); (b) books, journals, subscriptions, and other similar professional publications; (c) purchase of supplies, equipment, software or services; and (d) travel to attend relevant scholarly conferences or conduct scholarly work (allowable expenses per Policy 31). 11.5.3 Members pay the professional expense costs themselves and obtain reimbursement from the University by submitting an FPER Plan claim form with appropriate receipts attached. For a given salary year, the claim form must be submitted in the two-month period (March 1 to April 30) which includes the end of the salary year. 11.5.4 An unspent FPER balance cannot be carried forward. Allowable expenses that exceed the maximum FPER in a particular year may be carried forward for up to three years, provided that they are documented on the FPER Plan claim form for the year in which the expenses were incurred. 11.5.5 On each May 1, the FPER shall be indexed by the annual average percentage change (January to December) in the Canada Consumer Price Index for the immediately preceding year. See: the FAUW Compensation page for the current value (the information is sorted by year). 11.6 Additional Benefits A number of additional benefit programs are described in University policies, for example: Policy 3: sabbatical and other leaves of absencePolicies 4 and 24: tuition benefitsPolicy 14: pregnancy, adoption and parental leavesPolicy 28: moving expensesPolicy 31: travelPolicy 38: paid holidaysPolicy 67: employee assistance program
12. TERM OF AGREEMENT – DURATION
12.1 This Agreement shall come into effect upon ratification by the Parties, shall be binding on both Parties, shall remain in effect until April 30, 2000, and may not be opened prior to that date except by mutual consent of the Parties or as provided in 12.2 below. 12.2 Irrespective of the provisions of 12.1 the Parties agree to undertake additional negotiations regarding the proposed inclusion of librarians as Members for the purposes of this Agreement, as defined under 2.1.1, starting no later than November 1, 1998. Except by mutual consent of the Parties no other matters shall be addressed in these negotiations. Should these negotiations result in an agreement on new Articles, or changes to existing Articles, the Chief Negotiators shall recommend them for ratification by their respective Parties. Upon ratification this Agreement shall be opened for the sole purpose of incorporating the new Articles or changes to existing Articles into the Agreement. 12.3 As of April 30, 2000, this Agreement shall automatically renew itself for successive one-year periods unless either Party provides written notice to the other that it wishes to terminate or revise the Agreement. 12.4 If such written notice is given, the following rules and practices shall govern exchanges between the Parties: (a) written notice of termination must be received in the period from May 1 to May 31, inclusive, prior to the expiry date; (b) if either Party serves notice to terminate, this Agreement shall terminate as of the April 30 following; (c) written notice by one Party of intent to renegotiate this Agreement, provided that the other Party has not already given notice to terminate the Agreement as in 12.4 (a) above, must be received in the period from June 1 to September 30, inclusive, prior to the expiry date; (d) notice from either Party regarding renegotiation shall include the names of its Chief Negotiator and two additional persons who will be members of its Negotiating Team, a list of the Articles that it wishes to revise, and the subject matter of new Articles that it wishes to negotiate; and (e) within one month of the notice referred to in 12.4 (d) above or by September 30, whichever is later, the other Party shall reply with the names of its Chief Negotiator, the two additional members of its Negotiating Team, a list of the Articles that it wishes to revise, and the subject matter of new Articles that it wishes to negotiate. 12.5 Negotiations shall begin as soon thereafter as can be arranged by the Chief Negotiators and, except by mutual agreement at any time, shall address only those Articles listed by one or other of the Parties as above. All other Articles shall be incorporated unchanged into a revised Agreement. The notice periods may be waived by mutual agreement of the Parties. 12.6 Subject to the agreement of both Parties a mediator may be appointed at any time to assist in settling any outstanding issues in the negotiations. The mediator shall be selected by the procedure in 10.9. The costs of the mediator shall be borne equally by the two Parties. The mediator shall hear representations from the Parties, shall mediate between the Parties, and shall encourage them to resolve the outstanding issues. Mediation shall continue for as long as both Parties agree that it is helpful. 12.7 If negotiations under 12.5 and 12.6 do not for whatever reason result in a new and ratified Agreement by the April 30 following the start of negotiations, the current Agreement shall renew automatically for one year. 12.8 If a revised Agreement has been negotiated, the Chief Negotiators shall recommend it for ratification by their respective Parties. Upon ratification by the Parties the revised Agreement shall replace the current Agreement. 12.9 If the Agreement is not re-opened under the provisions of this Article, but compensation negotiations occur under Article 10, a new Agreement consisting of the Articles in the current Agreement and the new Memorandum of Settlement shall be deemed to come into effect on the date that the Memorandum of Settlement comes into effect. 12.10 Amendment of this Agreement 12.10.1 Amendment of this Agreement may be made by the Parties at any time in accordance with the provisions of this Article. 12.10.2 Minor change: If the proposed change is deemed by all members of the Faculty Relations Committee to be minor, approval of the change by the President (acting on the advice of Deans’ Council) and the Board of Directors of the Association is required for it to take effect. No change to Articles 10, 13 or 17 can be deemed minor. Members (as defined in Article 2.1.1) shall be informed of a minor change via email shortly after approval. 12.10.3 Except for minor changes (see Article 12.10.2) and changes to the Memorandum of Settlement (see Article 10.4), a change to this Agreement must be approved by the Association and the Board of Governors before it takes effect.
13. FACULTY SALARIES, ANNUAL SELECTIVE INCREASES AND MEMBER EVALUATION PROCEDURES
13.1 This Article states the principles governing the determination of salaries for faculty members holding regular appointments. These principles include the establishment of a salary structure for these purposes, the procedures used to establish the extent of annual selective increases, and the rules that have been developed to direct the annual evaluation process in each Faculty used to determine individual selective increases.
13.2 Faculty Salary Structure
13.2.1
The salary structure for regular faculty members shall consist of a salary floor and two thresholds for each of the four ranks and for Clinical Lecturers, together with the Selective Increase Unit (SIU).
Effective 1 May 2018, the selective increase unit (SIU) shall be $3,920, and the salary floors and thresholds shall be as stated below: Rank Floor Threshold T1 Threshold T2 Lecturer$62,180 $126,997$147,800 Clinical Lecturer $80,143$156,296 $175,483 Assistant Professor$80,143 $175,779$212,398 Associate Professor $100,868$175,779 $212,398 Professor$128,505 $175,779$212,398
13.2.2 Effective May 1 of each year, the annual scale change as specified in the Memorandum of Settlement shall be applied to the salary floors, thresholds, and Selective Increase Unit. Otherwise, changes in these amounts shall require the mutual agreement of the Association and the University.
13.3 Selective Salary Increases
13.3.1 Selective salary increases are intended to move a Member through the salary structure at a rate determined by her/his achievements in the profession and contributions to the University, measured by annual performance ratings undertaken as specified in 13.5. In order to ensure orderly career progress consistent with long-range academic goals, the commitment of funds required for this purpose shall have the highest priority in the preparation of the annual budget.
13.3.2
Within each Faculty, the Selective Increase Pool for Members shall be determined as follows:
0.25 SIU for each FTE Member, plus
0.25 SIU for each FTE Member with salary below T2, plus
0.5 SIU for each FTE Member with salary below T1.
For these purposes the value of the SIU shall be its value as of May 1 of the salary year in which the selective increases are to take effect, adjusted from year to year as specified in 13.2.2.
13.3.3 (a)
A Member’s selective salary increase depends both on her/his performance rating (actual R) and on the position of the Member’s salary relative to the thresholds T1 and T2 for her/his rank. Thus the performance rating (adjusted R) for purposes of calculating a Member’s selective increase amount may not be the same as the performance rating (actual R) determined as specified in 13.5.5. For Members on a biennial performance review cycle, during non-review years, actual R is equal to the actual R for the previous year. These non-review year actual Rs are subject to adjustment, just as review year actual Rs are. The appropriate values for the adjusted performance rating shall be determined in the following way:
If salary is less than T1 then adjusted R is actual R
If salary is equal to or greater than T1 but less than T2 then adjusted R is actual R less 0.75
If salary is equal to or greater than T2 then adjusted R is actual R less 1.25
The value of adjusted R shall never be less than 0
(b) The actual dollar value in any one year associated with an adjusted R of 1.0 in each Faculty is calculated by adding all individual adjusted ratings in that Faculty together, and dividing the resulting number into the total value of that Faculty's Selective Increase Pool as determined by 13.3.2. All other adjusted R values are assigned a selective increase dollar value by multiplying the adjusted R value by the dollar value of an adjusted R of 1.0.
(c) Where a selective salary increase as determined in 13.3.3 (a) and (b) would result in a salary which crosses a threshold, that increase shall be "feathered". That is to say, that part of the increase which would bring a Member's salary up to a threshold shall be received by the Member, but the part of the increase which would cause the salary to exceed the threshold shall be adjusted to make it commensurate with the selective increase to which she/he would be entitled with a salary at or above that threshold.
(d) Effective May 1, 2006, the University will provide annually an Anomalies Fund for each Faculty equal in value to five percent of that Faculty's Selective Increase Pool, to correct individual salary anomalies. These special permanent increases require the approval of the Vice President Academic and Provost (VPA&P) who shall consult with the President of the Association. Any unspent amount in the anomalies fund of a given Faculty will be carried forward to the next salary year.
(e)
Effective May 1, 2004, the University will provide annually an Outstanding Performance Fund for each Faculty equal in value to ten percent of that Faculty's Selective Increase Pool, to provide special permanent salary increases as described below.
Members in each Faculty unit (department or school) whose performance rating for the current year is within the top twenty percent of ratings within the unit may be considered for a special permanent salary increase. For Members on a biennial review cycle, eligibility for consideration for Outstanding Performance Fund salary increases during non-review years are based on the previous year's performance ratings.Members who have received a special increase in either of the previous two years are not eligible to receive a special increase, and are excluded for purposes of determining the top twenty percent and those within it.
All Members identified by the process above will form a single Faculty-wide pool. The Dean of the Faculty, in consultation with the Vice-President, Academic & Provost, will review the performance of all Members in this pool, and make special salary increase awards equal in value to one Selective Increase Unit (SIU) to a subset of them. For at least eighty percent of the awards, the sole criterion will be outstanding performance in teaching and scholarship. Remaining awards may be given on the basis of outstanding service to the University. Consideration also should be given to dispersing the awards across Faculty units, ranks, and to both women and men. Awards given on the basis of outstanding service will not be limited to Members holding administrative positions. The number of awards made will be such that, in the aggregate, they will differ from ten percent of each Faculty's Selective Increase Pool by less than one SIU, and any unspent amount in the Fund of a given Faculty will be carried forward to the next salary year. The Vice-President, Academic & Provost will publicly announce the award recipients.
13.4 Miscellaneous
13.4.1 In every case, scale and selective increases shall be applied to the Member's nominal full-time salary.
13.4.2 For Members on approved pregnancy, adoption or parental leave: the full scale and selective increases shall apply.
13.4.3 For Members newly appointed within the evaluation year, or who are on full or partial unpaid leave for part of the evaluation year: the full scale increase shall apply, but the Merit Increase will be prorated by the fraction of the year served at the University.
13.5 Member Evaluation
13.5.1 (a) Each Faculty shall have Faculty Performance Evaluation Guidelines setting out the evaluation criteria for that Faculty. The Faculty Performance Evaluation Guidelines shall be reviewed and updated no less than once every five (5) years, and changes shall be approved by a majority vote of the Faculty Council no later than 15 October in the year before evaluation calendar year to which the changes would apply.
(b) Each Department shall have an Addendum to their Faculty Performance Evaluation Guidelines setting out the performance expectations in the Department for scholarship, teaching, and service. The Addendum shall be reviewed and updated biennially, and changes shall be approved by: (i) a majority vote of members of the Department, and (ii) the Faculty Dean who shall review for consistency with the documents listed in 13.5.1(c) no later than 15 October in the year before the evaluation calendar year(s) to which the changes would apply.
(c) Faculty Performance Evaluation Guidelines and Departmental Addenda shall be consistent with this Agreement, and with University policies, procedures and guidelines (including the evaluation criteria set out in Policy 77). Departmental Addenda shall also be consistent with Faculty Performance Evaluation Guidelines. In case of a conflict, precedence shall be given first to this Agreement; then to University policies, procedures and guidelines; and then to the Faculty Performance Evaluation Guidelines.
(d) Current versions of faculty Performance Evaluation Guidelines and Departmental Addenda shall be posted on the relevant Faculty website and publically accessible.
13.5.2 (a) Each Member shall receive performance evaluation based upon documentation provided by the Member, submitted in the format and by the deadline specified in the Faculty Performance Evaluation Guidelines. Performance evaluations shall occur on an annual basis for Members holding probationary or definite-term appointments, and on a biennial basis on odd numbered years for Members holding tenured or continuing appointments. A Member who does not submit the required documentation by the specified deadline normally will receive an overall rating of at most 0.5 as specified in 13.5.3.
(b) Members shall provide documentation for the calendar year(s) under evaluation (one year for Members holding probationary or definite-term appointments, and two years for Members holding tenured or continuing appointments). Members shall in addition provide documentation for the number of previous years specified by their Faculty Guidelines. Scholarship shall be assessed on the total evidence from a window of two years. Teaching and service shall be assessed on the evidence from the year(s) under evaluation. The remaining documented years shall provide context to the assessed evidence.
(c) When Faculty Performance Evaluation Guidelines or Departmental Addenda change during the course of a Member's probationary contracts, the Member will continue to be governed by the guidelines and addenda in effect at the beginning of their first probationary contract, unless the Member elects to be governed by the new set of guidelines or addenda, at the Member's discretion. The Member shall advise their Department Chair if they elect to be governed by the new set.
13.5.3
Each Member shall receive one of the following nine numerical performance ratings in each of teaching, scholarship and service:
2.0 Outstanding
1.75 Excellent
1.5 Very Good
1.25 Good
1.0 Satisfactory
0.75 Needs Some Improvement
0.5 Needs Significant Improvement
0.25 Needs Major Improvement
0.0 Unsatisfactory
13.5.4 (a) Performance ratings shall pertain to the portion of the evaluation year during which the Member was a paid employee of the University, including sabbatical leave, but excluding pregnancy, adoption, parental, or sick leave.
(b) For newly appointed Members, and for Members on paid or unpaid leave, it may not be possible to assess performance in all three categories during the evaluation year. In these cases only, the practices described in 13.5.1, 13.5.2, and 13.5.3 may be amended as follows: (1) A newly appointed Member shall receive, in any category where assessment is not possible, a rating equal to the average rating of Members in the Department who hold the same rank; and (2) A continuing Member who has been on leave shall receive in any category where assessment is not possible as a result of the leave, a rating equal to the average ratings of the three previous years in which the Member was not on leave.
(c) In situations where a Member has held a fractional load appointment, or has taken a leave of absence, in the period for which evaluation data is being considered, expectations for quality shall remain the same but expectations for quantity shall be adjusted.
13.5.5 (a) The overall rating (R) for each Member shall be computed as the weighted average of the individual ratings in teaching, scholarship and service for the year(s) being reviewed. For Members on a biennial performance review cycle, the rating for non-review years shall be equal to the rating for the previous review year. The weight for each area shall be as specified in the member’s letter of appointment. In the absence of specified weights for professorial positions, the normal weights shall be 40 percent for teaching, 40 percent for scholarship, and 20 percent for service; for lecturer positions, the normal weights shall be 80 percent for teaching and 20 percent for service. These default weights do not apply to lecturer appointments made prior to May 1, 2008. Member weights remain in effect for the duration of the appointment unless otherwise changed under sub articles (b) and (c). There is no intended linear relationship between the percent for teaching and the number of courses taught.
(b) Weightings and duties may be adjusted in a formal agreement between the Member and the Chair with the approval of the Dean. The weights shall be at least 20 percent in every category, except in the case of lecturer appointments. Weight redistribution does not modify the performance quality expected in any of the three areas, though expectations for quantity will change.
(c) Any such formal agreement under 13.5.5 (b) shall be by mutual consent and, except in the case of definite-term appointments, shall be for a period of up to 5 years but no less than 2 years. Such an agreement may be renewed by mutual consent.
(d) The performance evaluation of a Member shall be done with all evaluators being informed of the weights in each area, and any adjustments made to the weights in each area, over the entire period for which evaluation data is being considered. Each Member shall be informed of the weight information used in their evaluation. The Chair shall collect and provide this weight information, which must be consistent with sub article (a) and any adjustments made under sub articles (b) and (c).
13.5.6 (a) The Chair has the responsibility for annual performance evaluations of all Members in the Department. The Chair shall inform the Dean of the proposed ratings in the three categories and overall.
(b) For Departments with 15 or fewer full-time equivalent regular faculty positions, the Members of the Department shall decide by majority vote whether to elect an advisory committee of no more than five Members to assist the Chair in carrying out the responsibility in 13.5.6 (a). A common committee spanning two or more small Departments may be considered.
(c)
For Departments with more than 15 full-time equivalent regular faculty positions, the Members of the Department shall elect an advisory committee of no more than five Members to assist the Chair in carrying the responsibility in 13.5.6 (a).
13.5.7 The Dean shall review the ratings proposed by the Chair, and may establish an advisory committee to assist with this review. The Dean may modify the ratings for a Member or Members of a Department, if necessary, to maintain consistency of standards across the Faculty. The Dean shall inform the Chair in writing of the final individual and overall ratings, together with reasons for any changes.
13.5.8 The Chair shall inform the Member in writing of her/his final individual and overall ratings, and shall provide an opportunity for the Member to discuss her/his performance evaluation.
13.5.9 The Dean shall evaluate the performance of Department Chairs and Associate Deans, and shall forward proposed performance ratings in the three categories and overall to the VPA&P for approval. The VPA&P shall inform the Dean and the Chair or Associate Dean in writing with reasons of any changes in the recommended ratings.
13.5.10 (a) A Member who disagrees with her/his performance evaluation should proceed first to the Department Chair, and then, if not resolved, to the Dean of the Faculty for disposition.
(b) A Department Chair or Associate Dean who disagrees with her/his performance evaluation should proceed first to the Dean, and then, if not resolved, to the VPA&P for disposition.
(c) Performance evaluations and selective salary increases are not normally grievable except under Article 9.2.2 or 9.2.3 of this Agreement.
13.5.11 For each evaluation period, members shall have online access to, at a minimum, histograms showing the distribution by rank in the Faculty, of: (a) final overall ratings and (b) unweighted ratings in the categories of teaching, scholarship, and service. In cases where there are fewer than ten members of the same rank within a Faculty, instead of histograms, those members shall be provided with averages by rank. In cases where there are fewer than three members of the same rank within a Faculty, no data shall be provided.
14. INTEGRITY IN SCHOLARLY RESEARCH
|
2019-03-18 20:19:02
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1757354438304901, "perplexity": 2616.737271082646}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912201672.12/warc/CC-MAIN-20190318191656-20190318213656-00511.warc.gz"}
|
http://www.ruor.uottawa.ca/en/handle/10393/5648
|
Contributions to the time domain-finite difference method for the modeling of microwave structures.
Title: Contributions to the time domain-finite difference method for the modeling of microwave structures. Author: Kim, Ihn Seok. Abstract: In this thesis, the Time Domain-Finite Difference (TD-FD) approach based on Maxwell's time dependent curl equations has been investigated with the objective to develop a numerical model for simulation of electromagnetic (EM) wave propagation phenomena, and to obtain S-parameters of guided wave transmission structures. The numerical model formulated as initial boundary value problem has been developed for both CW (single frequency) and pulse (wide band) propagations. The simulation property of EM field propagation is interpreted by considering the analogy between continuous and discrete characteristics of Maxwell's curl equations, the latter being derived from the leap-frog approximation scheme for local wave propagation. In an effort to relate the discrete wave propagation model to the continuous one, the geometrical meaning and the effects of the stability factor are introduced. The analysis of the leap-frog approximation under the plane wave condition shows that the TD-FD method is non-dissipative, which means that numerical energy is conserved during the simulated wave propagation. From the field distribution of numerical wave propagation, any physical parameters in the frequency domain can be extracted. This is done using the S-parameter extraction algorithm that has been developed for both the CW and pulse propagation cases. To accurately model EM wave propagation by the numerical process, convergence criteria for checking the global errors in the numerical procedure are also introduced by considering the physical properties, phase linearity and matching condition (standing wave) along a lossless waveguide with uniform cross-section. The physical properties used also confirm that Maxwell's equations can be separated into transverse and longitudinal parts in any uniform guide. The criteria serve as a basic building block for the analysis of more complex guiding structures. In the formulation of a computational domain for EM wave propagation solutions, artificial matching boundary conditions (MBC) are necessary to make the domain compact. A new wideband MBC has been introduced. The MBC is based on the concept of the transition operator used in modern control theory. An efficient local mesh refinement algorithm is also developed by using the concept of the characteristic and by enforcing boundary conditions for $\vec{E}$ and $\vec{H}$ fields at the interface of different mesh sizes. Numerical analysis results are presented to validate the various procedures when combined into a complete model. The objective of this study was to develop a TD-FD model of Maxwell's equations for EM wave propagation phenomena in continuous media. With this numerical model, almost all kinds of experiments can be done in complete freedom from experimental apparatus. Also, when probing the field distribution in an experiment, the numerical model provides excellent results without any field disturbance. Date: 1990 URI: http://hdl.handle.net/10393/5648
Files in this item
Files Size Format View
NN62299.PDF 3.297Mb application/pdf View/Open
Contact information
Morisset Hall (map)
65 University Private
|
2013-05-19 06:58:04
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5070540308952332, "perplexity": 660.1279834722079}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368696384181/warc/CC-MAIN-20130516092624-00035-ip-10-60-113-184.ec2.internal.warc.gz"}
|
https://www.rdocumentation.org/packages/xgboost/versions/1.3.2.1/topics/xgb.ggplot.importance
|
# xgb.ggplot.importance
0th
Percentile
##### Plot feature importance as a bar graph
Represents previously calculated feature importance as a bar graph. xgb.plot.importance uses base R graphics, while xgb.ggplot.importance uses the ggplot backend.
##### Usage
xgb.ggplot.importance(
importance_matrix = NULL,
top_n = NULL,
measure = NULL,
rel_to_first = FALSE,
n_clusters = c(1:10),
...
)xgb.plot.importance(
importance_matrix = NULL,
top_n = NULL,
measure = NULL,
rel_to_first = FALSE,
left_margin = 10,
cex = NULL,
plot = TRUE,
...
)
##### Arguments
importance_matrix
a data.table returned by xgb.importance.
top_n
maximal number of top features to include into the plot.
measure
the name of importance measure to plot. When NULL, 'Gain' would be used for trees and 'Weight' would be used for gblinear.
rel_to_first
whether importance values should be represented as relative to the highest ranked feature. See Details.
n_clusters
(ggplot only) a numeric vector containing the min and the max range of the possible number of clusters of bars.
...
other parameters passed to barplot (except horiz, border, cex.names, names.arg, and las).
left_margin
(base R barplot) allows to adjust the left margin size to fit feature names. When it is NULL, the existing par('mar') is used.
cex
(base R barplot) passed as cex.names parameter to barplot.
plot
(base R barplot) whether a barplot should be produced. If FALSE, only a data.table is returned.
##### Details
The graph represents each feature as a horizontal bar of length proportional to the importance of a feature. Features are shown ranked in a decreasing importance order. It works for importances from both gblinear and gbtree models.
When rel_to_first = FALSE, the values would be plotted as they were in importance_matrix. For gbtree model, that would mean being normalized to the total of 1 ("what is feature's importance contribution relative to the whole model?"). For linear models, rel_to_first = FALSE would show actual values of the coefficients. Setting rel_to_first = TRUE allows to see the picture from the perspective of "what is feature's importance contribution relative to the most important feature?"
The ggplot-backend method also performs 1-D clustering of the importance values, with bar colors corresponding to different clusters that have somewhat similar importance values.
##### Value
The xgb.plot.importance function creates a barplot (when plot=TRUE) and silently returns a processed data.table with n_top features sorted by importance.
The xgb.ggplot.importance function returns a ggplot graph which could be customized afterwards. E.g., to change the title of the graph, add + ggtitle("A GRAPH NAME") to the result.
barplot.
##### Aliases
• xgb.ggplot.importance
• xgb.plot.importance
##### Examples
# NOT RUN {
data(agaricus.train)
bst <- xgboost(data = agaricus.train$data, label = agaricus.train$label, max_depth = 3,
eta = 1, nthread = 2, nrounds = 2, objective = "binary:logistic")
importance_matrix <- xgb.importance(colnames(agaricus.train\$data), model = bst)
xgb.plot.importance(importance_matrix, rel_to_first = TRUE, xlab = "Relative importance")
(gg <- xgb.ggplot.importance(importance_matrix, measure = "Frequency", rel_to_first = TRUE))
gg + ggplot2::ylab("Frequency")
# }
|
2021-03-08 10:08:23
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.22199030220508575, "perplexity": 9469.842880477077}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178383355.93/warc/CC-MAIN-20210308082315-20210308112315-00430.warc.gz"}
|
http://math.stackexchange.com/questions/117824/finding-the-cauchy-n-given-varepsilon-for-the-series-sum-n-3-5
|
Finding the “Cauchy” $N,$ given $\varepsilon$, for the series $\sum n^{-3.5}$.
We know that $$\sum a_k \text{ converges} \iff \text{the partial sums } s_n \text{converge} \iff \text{the partial sums } s_n \text{are Cauchy}$$
Writing out what this last statement means
$$\forall \varepsilon \gt 0, \exists N, \text{such that } \forall m \ge n \gt N, \left \lvert \sum_{k=n}^{m} a_k \right \rvert \lt \varepsilon$$
Let $\displaystyle a_k = \frac{1}{k ^{3.5}}$ and let $\displaystyle \varepsilon = 10^{−4}$.
Find a value of N that satisfies the Cauchy condition written out above.
-
Could compare the sums to (simpler to evaluate) integrals. – Did Mar 8 '12 at 6:46
@max What do you know about $\sum k^{-2}$? What do you now about $k^{-3.5}$ vs $k^{-2}$ when $k>1$?. What does this tell you about the ultimate value of your sum? – Pedro Tamaroff Mar 8 '12 at 6:54
You can make us of Euler-Maclaurin summation to come with really tight error bounds. en.wikipedia.org/wiki/Euler-Maclaurin_formula – user17762 May 8 '12 at 3:48
We are trying to make sure that $$\sum_{k=a}^b \frac{1}{k^{3.5}}$$ is small. It is OK to give away a whole lot. Note that if $k \ge N$, then $$k^{3.5}> N^{1.5}k(k-1).$$ It follows that if $a>N$, then $$\sum_{k=a}^b \frac{1}{k^{3.5}}< \sum_{k=a}^b \frac{1}{N^{1.5}}\frac{1}{k(k-1)}.$$ Note the partial fraction decomposition $\dfrac{1}{k(k-1)}=\dfrac{1}{k-1}-\dfrac{1}{k}$. So our sum has wholesale cancellation (telescoping), and $$\sum_{k=a}^b \frac{1}{k^{3.5}} <\frac{1}{N^{1.5}}\frac{1}{a-1}.$$ Since $a>N$, our sum is less than $\dfrac{1}{N^{2.5}}$ It is now easy to find $N$ such that ensures that ou sum is $<10^{-4}$.
Another way: By drawing a picture we can see that $$\sum_{k=a}^b \frac{1}{k^{3.5}}<\int_{a-1}^\infty \frac{dx}{x^{3.5}}.$$ This integral is easily evaluated. If $a >N$, the integral is $\le \frac{1}{2.5}N^{-2.5}$. This estimate is a better one than the one obtained earlier. But quality of the estimate is not really an issue, we want to prove only that there is an $N$ that does the job, and are not looking for the smallest $N$ that does.
-
Hint
1. If you have positive terms and if you know it converges you may put try to find $n$ so that $$\sum_n^\infty a_k <\varepsilon\qquad \text{(why would this be sufficient?)}$$
2. You might like to use some series where you know $\sum b_k$ and where $a_k\leq b_k$.
-
|
2014-12-22 15:11:47
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9281487464904785, "perplexity": 238.65152259852948}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802775392.34/warc/CC-MAIN-20141217075255-00083-ip-10-231-17-201.ec2.internal.warc.gz"}
|
https://www.splashlearn.com/math-vocabulary/multiplication/inverse
|
# Inverse - Definition with Examples
The Complete K-5 Math Learning Program Built for Your Child
• 30 Million Kids
Loved by kids and parent worldwide
• 50,000 Schools
Trusted by teachers across schools
• Comprehensive Curriculum
Aligned to Common Core
## Definition of Inverse?
In mathematics, the word inverse refers to the opposite of another operation.
Let us look at some examples to understand the meaning of inverse.
Example 1:
The addition means to find the sum, and subtraction means taking away. So, subtraction is the opposite of addition. Hence, addition and subtraction are opposite operations. We may say, subtraction is the inverse operation of addition.
Example 2:
Multiplication is repeated addition. The same number gets added repeatedly.
However, the division is repeated subtraction. The same number gets subtracted repeatedly. So, the division is the opposite of multiplication. Hence, multiplication and division are opposite operations. We may say, division is the inverse operation of multiplication.
So, we see that multiplication and division are inverses of each other.
## Multiplicative Identity:
A multiplicative identity is a number which when multiplied by any non-zero number gives the same number as the product.
For example, if a is any non-zero number, then multiplying a by 1 gives the product as the number itself.
a × 1 = a
Therefore, 1 is the multiplicative identity.
## Multiplicative Inverse:
A multiplicative inverse is a number which when multiplied by a number gives 1 (multiplicative identity) as the product.
If a is any non-zero number, then multiplying a by what number gives the product as 1?
a × ? = 1
As, aa = 1, we need a number in which a is in the denominators place. We have a in the numerator. So, we keep 1 is the numerator of this number and get 1a.
a × 1a = 1
So, 1a is the multiplicative inverse of a.
Multiplicative inverse is also called a reciprocal.
Reciprocal of a number is obtained by interchanging its numerator and denominator.
The multiplicative inverse of a is 1and multiplicative inverse of a fraction ab is ba
Fun Facts Inverse property is also seen in proportion. It is called the inverse proportion. In inverse proportion, when one value increases the other decreases. For example, if the number of machines increases, the time taken to complete the task decreases.
Won Numerous Awards & Honors
|
2020-11-25 07:36:07
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9076967239379883, "perplexity": 731.1093049365763}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141181482.18/warc/CC-MAIN-20201125071137-20201125101137-00385.warc.gz"}
|
https://e-eduanswers.com/mathematics/question618303
|
, 25.08.2019 04:30, hazeleyes8908
Solve the following equation 13x+1=11x+8
Other questions on the subject: Mathematics
Mathematics, 21.06.2019 14:00, davidoj13
Me! #1 write an equation for the interior angles of this triangle that uses the triangle sum theorem. #2 what is the value of x? #3 what is the measure of #4 classify the triangle above as acute, obtuse, or right. state your reason in a complete sentence.
|
2020-09-20 08:21:30
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9021018743515015, "perplexity": 1737.8576999530353}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400196999.30/warc/CC-MAIN-20200920062737-20200920092737-00285.warc.gz"}
|
https://www.physicsforums.com/threads/two-capacitors-different-capacitance-and-innitial-charge-in-a-series-circuit.534818/
|
# Two capacitors (different capacitance and innitial charge) in a series circuit
1. Sep 28, 2011
### KaiserBrandon
1. The problem statement, all variables and given/known data
Two capacitors C1 and C2 and a resistor R are all connected in series. At t=0, a
charge Q1 resides on C1 and a charge Q2 on C2 (Q1 > Q2; C1 < C2), and the positive plate
on C2 is connected to the negative plate on C1.
a) In a first experiment, the positive plate on C2 is connected to the negative plate on
C1. Derive an expression for the time dependence of the current I(t) which flows
in the circuit. Indicate the direction of this current on a circuit diagram, and
include the sign of the charges on the plates on each of the capacitors at t=0.
b) The same circuit is used for a second experiment, with the same initial charges,
except that in this case the negative plate of C2 is connected to the negative plate
on C1. All three components are otherwise still connected in series as before.
Derive an expression for I(t), and indicate the direction of the current on a second
diagram (again indicate the sign of initial charge on each capacitor on this
diagram).
c) Hence calculate the energy dissipated in the resistor in each case.
2. Relevant equations
V = IR
V=Q/C for a capacitor
3. The attempt at a solution
ok, I'm stuck on the first part. firstly, the voltage gain across the two capacitors is:
$V = \frac{Q_{1}(t)}{C_{1}} + \frac{Q_{2}(t)}{C_{2}}$
And since V = IR, we have:
$I = \frac{1}{R}(\frac{Q_{1}(t)}{C_{1}} + \frac{Q_{2}(t)}{C_{2}})$
and the capacitors must each lose the same amount of charge per unit time:
$\frac{dQ_{1}(t)}{dt} = \frac{dQ_{2}(t)}{dt} = -I$
the current is negative since the charges are decreasing. So, we have:
$\frac{dQ_{1}(t)}{dt} = \frac{dQ_{2}(t)}{dt} = \frac{-1}{R}(\frac{Q_{1}(t)}{C_{1}} + \frac{Q_{2}(t)}{c_{2}})$
And this is where I'm stuck. I'm not entirely sure how to go about evaluating this for Q1 and Q2.
Last edited: Sep 28, 2011
2. Sep 28, 2011
### Staff: Mentor
You have the initial potential that the resistor will see: the sum of the two capacitor voltages as you've shown. What initial current must flow then?
The circuit consists of a resistor and two capacitors in series. What's the equivalent capacitance? How about the circuit's time constant?
What can you do with an initial current and the time constant?
|
2017-09-25 19:03:28
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5921988487243652, "perplexity": 740.3699925117894}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818693240.90/warc/CC-MAIN-20170925182814-20170925202814-00671.warc.gz"}
|
https://brilliant.org/problems/an-algebra-problem-by-hana-nakkache-2/
|
None is zero
Logic Level 2
$\large \begin{array} { l l l l l } & &D & O & G\\ +& &C & A & T \\ \hline & 1 & 0 & 0 & 0 \\ \end{array}$
The above shows a long addition, where each of the symbols represents a distinct single-digit positive integer.
What is the sum $D + O + G + C + A + T ?$
×
|
2020-01-27 11:11:29
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7880768179893494, "perplexity": 2651.3111202397613}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579251696046.73/warc/CC-MAIN-20200127081933-20200127111933-00115.warc.gz"}
|
http://adorio-research.org/wordpress/?p=2301
|
## Python, Statistics, Educational Measurement: Cohens kappa interrater agreement coefficient
Cohen's kappa coefficient is a measure of interrater agreement for qualitative or categorical variables. It is computed normally only for two raters or judges . Input data is usually a 2 by 2 table. Consider the following:
Yes No
Yes a b
No c d
and a specific example(from en.wikipedial):
20 5 10 15
The following Python program computes Cohen's kappa
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 def kappa(a,b,c,d): tot = a + b + c + d Pa = float(a + d)/tot PA1 = float(a + b)/tot PA2 = 1.0- PA1 PB1 = float(a + c) /tot PB2 = 1.0 -PB1 Pe = PA1 *PB1 + PA2*PB2 print Pa, PA1, PB1, PA2, PB2 return (Pa -Pe)/ (1.0 -Pe) def Test(): a = 20 b = 5 c = 10 d = 15 print "kappa(20,5,10,15)=", kappa(a,b,c,d) if __name__ == "__main__": Test()
|
2014-04-23 09:12:22
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21936944127082825, "perplexity": 6973.792288622923}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1398223202457.0/warc/CC-MAIN-20140423032002-00338-ip-10-147-4-33.ec2.internal.warc.gz"}
|
https://engineering.stackexchange.com/questions/31343/lte-to-3g-irat-handrover-what-happens-afterward
|
# LTE to 3G IRAT Handrover: What happens afterward
Let's say that there has been a handover from LTE to 3G of a particular UE. Once the handover is complete, the MME is no longer required to maintain the sessions/bearers that has been established for that particular UE. I have found the following call flow in a reference, which is not so reliable.
1. eNB ⇒ MME S1AP S1 UEContextReleaseRequest
2. MME ⇒ SGW GTPv2 S11 Release Access Bearers Request
3. SGW ⇒ MME GTPv2 S11 Release Access Bearers Response
4. MME ⇒ eNB S1AP S1 UEContextReleaseCommand
5. eNB ⇒ MME S1AP S1 UEContextReleaseComplete
6. SGSN ⇒ MME GTP Gn SGSN context request
7. MME ⇒ SGSN GTP Gn SGSN context response
8. SGSN ⇒ MME GTP Gn SGSN context acknowledgement
9. MME ⇒ MSC SGSAP SGs SGsAP-EPS-DETACH-INDICATION
10. MSC ⇒ MME SGSAP SGs SACK SGsAP-EPS-DETACH-ACK
11. MME ⇒ SGW GTPv2 S11 Delete Session Request
12. SGW ⇒ MME GTPv2 S11 Delete Session Response
Given the scenario, some of these calls make sense to me. For an instance, once the handover to 3G is done, there is no point of having the access bearers S1 between eNB and S-GW and S5/S8 bearer between S-GW and P-GW and so forth. So, the Release Access Bearer Request in 2, and Delete Session Request at 11 are sensible.
What I particularly don't understand is that, why are calls 6,7 and 8 there? What purpose do they serve? Or is this entire flow blatantly wrong? If it is blatantly wrong, where can I find something sensible to read about what really happens? I have combed through the 3GPP specs for weeks now, but still couldn't find answers to the questions above.
EDIT-1
Okay, after looking at some PCAPs, the purpose of the calls 6,7 and 8 is also clear. Upon falling back to 3G, the SGSN is going to require the UE context at MME, in a format that SGSN can understand. To an SGSN, MME looks like another SGSN. So it requests, the context from there over S3 interface. However, the call flow above says the message exchange between MME and SGSN happens over Gn, that which contradicts with the the 3GPP spec, that which according to, the Gn interface is used to establish the connection between SGSN and GGSN. Now, the new question is going to be on the accuracy of the flow. Is this call flow accurate?
|
2021-06-13 14:56:50
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.42912301421165466, "perplexity": 8439.107488681298}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487608856.6/warc/CC-MAIN-20210613131257-20210613161257-00426.warc.gz"}
|
http://eccellent.it/qxvt/cyclohexane-to-bromocyclohexane.html
|
# Cyclohexane To Bromocyclohexane
Cyclohexane is highly conformationally mobile, interchanging between the boat and chair forms many thousands of times per second. PubChem CID: 8078. COMPOUND: FORMULA: COMPANY: AMOUNT: LOCATION: ROOM: OWNER (2-Bromoethyl)-1,3 dioxolane, 2- Aldrich: 50g: freezer: 4235 (Carbothoxymethyl)triphenylphosphonium bromide. The mixture was cooled to 0°C with an ice bath and pyrrolidine (1. Start studying Lab 9- E2 Reaction: Formation of Cyclohexane from Bromocyclohexane. Predict GHS Hazards for Any Chemical in silico. Formation of major product: 3-bromocyclohexene Under UV light, Br2$BrX2$ undergoes homolytic splitting to generate Br⋅$Br⋅$ radicals: Br2−→hv2Br⋅$BrX2→hv2Br⋅$ The formation of 3-bromocyclohexene is an example of su. Free essays, homework help, flashcards, research papers, book reports, term papers, history, science, politics. Energy and Equilibrium The relative amounts of the two conformers depend on their difference in energy DE = ?RT ln K R is the gas constant [8. PubChem Substance ID 24854149. 1,2-Diaminocyclohexane; A. The substrate is a secondary alkyl halide and the nucleophile is strong. Structural rationalization of the phase transition behavior in a solid organic inclusion compound: bromocyclohexane/thiourea. EC Number 208-806-6. Use of this catalyst can result in as high as a 10:1 alcohol0to0ketone mixture that is used in the production of adipic acid. Consult a physician. Désignons le substituant par X. Show this safety data sheet to the doctor in attendance. Introduction In organic chemistry, a hydrocarbon is an organic compound. The theoretical yield of alkene in moles is therefore equal to the number of moles of alcohol used. Integrated Risk Information System (IRIS) EPA 635 / R - 03 / 008 www. Barton et O. remove any bromocyclohexane or trans-1 -bromo-2-chloro- cyclohexane molecules adhering to their external surfaces. Solubility Soluble in alcohol. 23)The Keq for the interconversion for the two chair forms of methylcyclohexane at 25 °C is 18. Commercially, cyclohexene is made from hydrogenation of benzene. SMILES String: C1CC(CC(C1)Br)Cl;. Treat cyclohexene. Which conformation is more stable? Provide a reason for your answer. Expand this section. Cyclohexane is a six-carbon cycloalkane shown below. Acidification of the acid salt provides the parent acid. Allyl cyclohexane - CAS # 2114-42-3 Information provided on Allyl cyclohexane (2114-42-3) is for reference only and is subject to change. Like aldehydes, ketones can be prepared in a number of ways. 00 g of benzene, C6H6, to 80 g of cyclohexane, C6H12, lowers the freezing point of the cyclohexane from 6. Here we will learn how to draw cyclohexane chairs, how to flip them, the difference in energy between the two chair. 7 Use and Manufacturing. 2 Energy diagram for a two-step reaction involving the formation of an intermediate. SCHEMBL18107236. EXAM 3 Answer Key, Chapters 7 - 9. Synthesis of Cyclohexanecarboxylic acid _____ 1. 1-Bromo-2-chlorocyclohexane. Expand this section. EC Number 231-070-2. trans-1,2-Dibromocyclohexane 99% CAS Number 7429-37-0. Ils peuvent présenter sous deux conformations, la conformation chaise et la conformation bâteau. Chair conformations of bromocyclohexane. 18 and estimate the percentages of axial and equatorial conformers present at equilibrium in bromocyclohexane. Life as we know it is carbon-based. 6 Chemical Vendors. Compound bromocyclohexane with free spectra: 8 NMR and 3 FTIR. Cyclohexane,1,2-dibromo-,cis-Cyclohexane, 1,2-dibromo-Other names: trans-1,2-Dibromocyclohexane;. 6 kcal/mol, 2. Such a carbon atom is called a chiral center (or sometimes a stereogenic center), using organic-speak. Molecular Weight: Expand this section. MDL number MFCD00003822. 5^\text{o}\), the molecule is puckered. trans-1,2-Dibromocyclohexane 99% CAS Number 7429-37-0. Sun Q(1), Faller R. In practice, the 1,3 isomer appears to be favoured. During a lab experiment, we were to determine the relative level of saturation by adding, drop by drop, a solution of bromine in cyclohexane to oil and a fat, individually. Cyclohexane, 1-bromo-2-chloro-, cis-cis-1-Bromo-2-Chlorocyclohexane. Drawing boat and chair conformations requires identifying the c-c bonds and the bonded substituents. Désignons le substituant par X. 108-85- <= 100 % EC-No. On the basis of the ir vapour spectrum of cyclohexane, the assignment of a weak band at 947 cm −1 to the v 15 fundamental is probable. 3 degree celsius. Point out any gauche interactions shown in your Newman projection. When I wrote substitution, I included both S_N1. Electrophilic addition reactions involving hydrogen bromide. 5 Related Records. Hydroboration-oxidation. The synthesis of cyclohexanone is a simple procedure that uses Acetic acid, sodium hypochlorite, hypochlorous acid, ether, sodium chloride, sodium carbonate and cyclohexanol. 1 ppm) and another strong peak at 9. Cyclohexane is a cycloalkane with the molecular formula C 6 H 12. In this conformer only the axial bromine needs to be considered. A) Which monosubstituted cyclohexane has the most negative ΔG∘ for this conversion? B) Which monosubstituted cyclohexane has the greatest preference for an equatorial position? C) Calculate ΔG∘ for the conversion of "axial" bromocyclohexane to "equatorial" bromocyclohexane at 25 ∘C. how to convert cyclohexanol to bromocyclohexane. pi bonds undergo addition reactions CH2=CH2 + HCl --> CH3CH2Cl in general, C=C + HX --> H-C-C-X alkenes react with hydrogen halides to form alkyl halides Addition of HX to Alkenes. graphic shows bromocyclohexane in two conformers, its most stable chair conformation and its least stable chair conformation. Fisher Scientific - Scheepsbouwersweg 1b - Postbus 4 - 1120 AA Landsmeer - Tel. Created by Jay. Issuu is a digital publishing platform that makes it simple to publish magazines, catalogs, newspapers, books, and more online. 溴环己烷是一种化学物质，分子式是C6H11Br。溴环己烷是一种有刺激性气味的无色液体。与乙醇、乙醚、丙酮、苯和四氯化碳可. Box 191 Chemtrec: (800)424-9300 Kilgore, Texas 75663 1-800-256-6644 Section 1: Product Identification Synonyms: Sulphuric Acid, Hydrogen Sulphate, Oil of Vitriol, Battery Acid. 011 CM SPECTRAL CONTAMINATION DUE TO CCl4 AROUND 1550 CM-1: Resolution: 4: Sampling procedure: TRANSMISSION: Data processing: DIGITIZED BY NIST FROM HARD. This is again an example of nucleophilic substitution. on StudyBlue. Make the Grignard from bromocyclohexane and add it to ethyl formate (HCO2CH2CH3, it will add two times). +351 21 425 33 50 - Fax. bromocyclohexane bromoethyl trimethyl-2 ammonium bromide bromoethylacetate, 2- cyclohexane cyclohexane butylamine cyclohexanecarboxylic acid, trans-4-(aminomethyl)-. As described in the literature, polar products, including formic acid, glutaric acid, succinic acid, hydroxyl caproic acid and adipic acid, would remain in the solid. The measurements, were performed on two polymer theta solvent systems : polydimethylsiloxane-bromocyclohexane at 29 °C and polystyrene-cyclohexane at 34°C for the attractive and repulsive cases respectively. Therefore an alkane may assume different shapes due to these free rotations. The substance identifiers displayed in the InfoCard are the best available substance name, EC number, CAS number and/or the molecular and structural formulas. This is to certify that the management system of: Avra Synthesis Pvt. Below are the EPA applications/systems, statutes/regulations, or other. This shows two layers with slightly different colours demonstrating that bromine is more soluble in non-polar solvents. MATERIALS AND METHODS Bromocyclohexane was purchased from B. Cyclohexane makes weird shapes! The most important is the chair conformation. It's okay to use any necessary organic or inorganic reagents. Using cyclohexane as your starting material, show how you would synthesize each of the following compounds. At this point one is tempted to convert bromocyclohexane to cyclohexanol by an S N 2 reaction with hydroxide ion. Synthesis of Ketones. 1 ppm) and another strong peak at 9. 0 0 0 0 0 0 0 0 0 0 0 0. The results demonstrate marked contrasts to the phase transition behavior in the prototypical cyclohexane/thiourea inclusion compound. There are actually, there are other conformations of cyclohexane, so the boat conformation can actually twist a little bit to give you twist boat. Show how to convert cyclohexanol to these compounds. Cyclohexane is used as a nonpolar solvent for the chemical industry, and also as a raw material for the industrial production of adipic acid and caprolactam, both of which are intermediates used in the production of nylon. 10 mL of cyclohexane, and 0. Energy and Equilibrium The relative amounts of the two conformers depend on their difference in energy DE = ?RT ln K R is the gas constant [8. This is because the positions can convert when the chair performs a ring flip -a conformational interchange that makes all axial positions equatorial and vice -versa. answer is a elimination reaction. Iodocyclohexane 98% Synonym: Cyclohexyl iodide CAS Number 626-62-. Br Step 1 Intermediate E bromocyclohexane cyclohexene Step 2 NaOH(aq) (i) Suggest the structure of intermediate E and the reagent(s) and conditions for step 2. Log Octanol-Water Partition Coef (SRC): Log Kow (KOWWIN v1. Eye Contact Rinse immediately with plenty of water, also under the eyelids, for at least 15 minutes. Furthermore, the scope of different carbonyl compounds was investigated and resulted with similar consecutive Michael-Claisen process for CDD synthesis. PubChem Substance ID 24854149. Housecroft and E. 0 ml of bromocyclhexane after distillation. Cyclohexane is highly conformationally mobile, interchanging between the boat and chair forms many thousands of times per second. Propane + Br2 Propane + Br2. Calculate the ratio of the more stable chair conformation to the less stable conformation at equilibrium at 298 K for the following compounds: methylcyclohexane and bromocyclohexane. Concepts and reason The concept used to solve this problem is. MDL number MFCD00003820. Acidification of the acid salt provides the parent acid. Expand this section. When I wrote substitution, I included both S_N1. Solution: the percentage of axial conformer is about 27%, while the percentage of equatorial conformer is about 73% at equilibrium. Label the axial hydrogens (Ha) and the equatorial hydrogens (He). Learn vocabulary, terms, and more with flashcards, games, and other study tools. Bromcyclohexan Chemische Eigenschaften,Einsatz,Produktion Methoden R-Sätze Betriebsanweisung: R36/37/38:Reizt die Augen, die Atmungsorgane und die Haut. 09 (Mean or Weighted MP) VP(mm Hg,25 deg C): 1. 123456789 00:00:00 00:00:00 9/24/2017. Convert grams Cyclohexane to moles or moles Cyclohexane to grams. This is the currently selected item. Draw all possible chair conformations of cis-1- isopropyl-2-bromocyclohexane. Cyclohexanes that are 1,4-substituted are a safe bet. Path length: 0. Cette page est destines a tous les etudiants du DEUG jusqu'au DEA et bien plus encore. Which conformation is more stable? Draw it first. ethoxycyclohexane. Similarly high yields were observed using iodocyclohexane with 4-MeC 6 H 4 MgBr and 2-MeC 6 H 4 MgBr ( Table 1 , entries 6 and 14), but decreased slightly with MeOC 6 H 4 MgBr. PubChem Substance ID 24854149. CHEM%210%[CHAPTER%6:%SUBSTITUTIONREACTIONSOFALKYLHALIDES!! ! 7% Fall!2013! KEY! 1. Free radical bromination of cyclohexane: Cyclohexane + Br2 + heat ---> Bromocyclohexane + HBr(g) Asked in Organic Chemistry , Scientific Method Are cyclohexene and cyclohexane soluble ?. , 2005, 3646-3648. Because many compounds feature structurally similar six-membered rings, the structure and dynamics of cyclohexane are important prototypes of a wide range of compounds. Answer to: Show how to convert cyclohexanol to these compounds. bromo-methyl-cyclohexane. In the right hand conformer it is the isopropyl group that is axial while. Propose a possible explanation. In E2 Reactions Of Cyclohexane Rings, The Only Way The Leaving Group And C-H Bond Can Be Anti-Periplanar Is If They Are On Opposite Faces Of The Ring This brings us to the second point. Step 2 of 5(b)Bromocyclohexane reacts with. Note: Cheméo is only indexing the data, follow the source links to retrieve the latest data. The substance identifiers displayed in the InfoCard are the best available substance name, EC number, CAS number and/or the molecular and structural formulas. ENTERPRISES, We are Exporter, Manufacturer & Supplier of Industrial Chemicals based in Maharashtra, India. In case of eye contact. Synonyms & Trade Names Benzene hexahydride, Hexahydrobenzene, Hexamethylene, Hexanaphthene CAS No. a) bromocyclohexane to R-CH2CH2OH (R = cyclohexane) I know that ethylene oxide added to grignard reagent gives R-CH2CH2OH product. What is the value of kf for cyclohexane? asked by Robin on December 2, 2015; Organic Chem. Compare the space filling models. A comparison of calculated and experimental values of the enthalpy of combustion indicates that the cyclohexane ring is nearly devoid of angle and eclipsing strain because of. with watermolecular brominecyclohex-2-en-1- to form 2-bromocyclohexane-1,3-diol with watercyclohex-2-en-1- to form 2-bromocyclohexane-1,3-diol ( Hydroboration of alkenes ). 69 (Adapted Stein & Brown method) Melting Pt (deg C): -33. 1 0 0 0 1 0 0 0 0 0 0 0 0. Use this link for bookmarking this species for future reference. The carbon atoms in cyclohexane are sp3 hybridized. Studyres contains millions of educational documents, questions and answers, notes about the course, tutoring questions, cards and course recommendations that will help you learn and learn. Multiple choice questions. Bromocyclohexane offered by Neogen Chemicals Limited, a leading supplier of Cyclohexane in Thane West, Mumbai, Maharashtra. Cyclohexane makes weird shapes! The most important is the chair conformation. cyclohexanol. E2 in a cyclohexane ring E2 in a cyclohexane ring Cyclohexane Stereochemistry Revisited A Different Mechanism for Alkyl Halide Elimination: The E1 Mechanism Example The E1 Mechanism 1. *Please select more than one item to compare. 1-Bromocyclohexane. 2 Names and Identifiers. NIST/TRC Web Thermo Tables (WTT) NIST Standard Reference Subscription Database 3 - Professional Edition Version 2-2012-1-Pro This web application provides access to a collection of critically evaluated thermodynamic property data for pure compounds with a primary focus on organics. bromocyclohexane dissolved in 64 mL dry diethylether is added over the addition funnel at such a rate as to maintain gentle reflux. Several other notable cyclohexane conformations occur during the transition from one chair conformer to the other - the boat, the twist, and the half-chair. cyclohexene + HBr --> bromocyclohexane 1-methylcyclohexene + HBr --> 1-bromo-1-methylcyclohexane (not 1-bromo-2-methylcyclohexane). Cyclohexane is used as a nonpolar solvent for the chemical industry, and also as a raw material for the industrial production of adipic acid and caprolactam, both of which being intermediates used in the production of nylon. SoDraw%the. Acide désoxycholique; Acide gibbérellique. First, let me preface by saying that no reaction is necessarily 100% S_N1 or S_N2. Structure, properties, spectra, suppliers and links for: 1-Bromocyclohexane, 108-85-0. pdf), Text File (. [email protected] 0 °C Molar. The puckering of the ring means that every other carbon atom is. followed by a very rapid attack by the cyanide ion on the carbocation (carbonium ion) formed: This is again an example of nucleophilic substitution. Information on EC 3. 78944 4/22/2020 8/24/2016 4/22/2020. 1-Bromocyclohexane. 42): Boiling Pt (deg C): 218. Bromocyclohexane (Aldrich, 98%) was used without further purification, seeded in the Ar carrier gas, and expanded into the vacuum chamber through a nozzle orifice (General valve, 0. Cyclohexane is one of the most important and and most frequently encountered structural units in organic chemistry. The reaction of secondary halogenoalkanes with cyanide ions. 1-chlorohexane + low conc. cas号查询致力于为化学行业用户免费提供溴 环己烷的cas号、中文名称、英文名称相互转换服务,同时也包括溴 环己烷的性质、化学式、分子结构、密度、熔点、沸点等信息。. 1-chloro-2-bromocyclohexane. 全文を閲覧するには購読必要です。 To read the full text you will need to subscribe. Cyclohexane is mainly used for the industrial production of adipic acid and caprolactam, which are precursors to nylon. NIST/TRC Web Thermo Tables (WTT) NIST Standard Reference Subscription Database 3 - Professional Edition Version 2-2012-1-Pro This web application provides access to a collection of critically evaluated thermodynamic property data for pure compounds with a primary focus on organics. 6 Chemical Vendors. In chemistry, conformational isomerism is a form of stereoisomerism in which the isomers can be interconverted exclusively by rotations about formally single bonds. This is because the bromine on the bromocyclohexane is a better leaving group, so it has a better ability to stabilize the negative charge than the chlorine on the chlorohexane. , 1974 Chemistry, organic University Microfilms, A XEROX Company , Ann Arbor, Michigan THIS DISSERTATION HAS BEEN MICROFILMED EXACTLY AS RECEIVED. Percent purity = 109. In both cases you could do an elimination reaction with a strong base like KOH to give you cyclohexene. The substance identifiers displayed in the InfoCard are the best available substance name, EC number, CAS number and/or the molecular and structural formulas. A cyclohexane conformation is any of several three-dimensional shapes adopted by a cyclohexane molecule. Ch06 Alkyl Halides (landscape). So I have an exam in Ochem Lab tomorrow, and while I was studying I got confused. The information provided in this Safety Data Sheet is correct to the best of our knowledge, information and belief at the date of its publication. 6 1-chlorobutane is the limiting reagent. Final Exam Answers: Chem 334 - Fall 2004. While you are waiting, fill a large reaction tube with 12 mL of water. [Bromocyclohexane] [108-85-0] | 価格や在庫、物性値などの詳細情報ページです。. This banner text can have markup. is an international high-tech enterprise founded in 1998, specialized in R&D and manufacture on intermediates, medicines, agrochemical, dyestuff, flavor, fragrance, cosmetic, natural extracts series products. 16 Look at Figure 4. Hydroboration-oxidation. 0g of sodium bromide. Propose a possible explanation. The hydrogen-bonded bromocyclohexane-ammonia complex has been isolated and characterized for the first time in argon matrices at 16 K. Catalytic and stoichiometric reactions of. 1-bromo-1-methylcyclohexane - C7H13Br, synthesis, structure, density, melting point, boiling point. The agreement between theory and experiments is good. This article is cited by 17 publications. Cyclohexyl Bromide, 99%, ACROS Organics 5mL; Glass bottle Chemicals:Organic Compounds:Organohalogen compounds:Alkyl halides:Cyclohexyl halides. Under these conditions, conversion to the cross-coupled products of 4-MeC 6 H 4 MgBr, 2-MeC 6 H 4 MgBr and 4-MeOC 6 H 4 MgBr with bromocyclohexane each gave 99% yields. 3 Chemical and Physical Properties. (25 points) Write complete names for each of the following, including stereochemistry if it is specifically shown. "Bu" is shorthand for "CH"_3"CH"_2"CH"_2"CH"_2-, or butyl. Arakato Development Ltd is supplier for Cyclohexane. 108-85-0 203-622-2. 5% Silver metal by weight) Chemical Formula: AgNO3. Cyclohexane is used as a nonpolar solvent for the chemical industry, and also as a raw material for the industrial production of adipic acid and caprolactam, both of which are intermediates used in the production of nylon. Fisher Scientific - Scheepsbouwersweg 1b - Postbus 4 - 1120 AA Landsmeer - Tel. Il existe deux conformations chaise en équilibre : ces conformations possèdent des énergies différentes. Acidification of the acid salt provides the parent acid. doc Problem 4 - In each of the following pairs of nucleophiles one is a much better nucleophile than its closely related partner. In the A-values table, an axial bromine in bromocyclohexane is 0. "Bu" is shorthand for "CH"_3"CH"_2"CH"_2"CH"_2-, or butyl. 6 ppm appeared. 04% Bromophenol red N-Bromosuccinimide 1-Bromo-4-tert-butylbenzene Brom thymol blue Bromothymol Blue Bromothymol Blue, Sodium Salt Bromothymol Blue Solution 0. -- CASR NO. The data represent a small sub list of all available data in the Dortmund Data Bank. EC Number 210-957-8. Draw all possible chair conformations of cis-1- isopropyl-2-bromocyclohexane. 1-Bromocyclohexane - chemical information, properties, structures, articles, patents and more chemical data. Contributors; The distinction between configurational stereoisomers and the conformers they may assume is well-illustrated by the disubstituted cyclohexanes. 30 kJ/mol You don. di batteri nitrosating nitrosare introdurre un gruppo nitroso to nitrosate nitrosato nitrosated derivative nitrosazione introduzione del gruppo nitroso nitrosation nitrosile nitrosyl nitrosilico nitrosilsolforico nitrosylsulphuric. Please log in. Chem 334 - Fall 1999 Organic Chemistry I Dr. bromocyclohexane reacts with (CH3)3CONa and (CH3)3COH to form cyclohexene and NaBr by which mechanism? 2. ethoxycyclohexane. Draw the neutral organic products. 00794*12 ›› Percent composition by element. Path length: 0. 055 Da Density 1. c) (R)-3-chloro-3-methyl-1,1-cyclohexanediol. On the basis of the ir vapour spectrum of cyclohexane, the assignment of a weak band at 947 cm −1 to the v 15 fundamental is probable. 110-82-7 RTECS No. Therefore, cyclohexane solution reacts in bromine test when placed under the bright sunlight to produce carbon and hydrogen bromide but there is no reaction when it is placed in the dark. *Please select more than one item to compare. Free essays, homework help, flashcards, research papers, book reports, term papers, history, science, politics. Cyclohexane is used as a nonpolar solvent for the chemical industry, and also as a raw material for the industrial production of adipic acid and caprolactam, both of which are intermediates used in the production of nylon. 1 0 0 0 1 0 0 0 0 0 0 0 0. UpToDate Contents. Login to reply the answers Post;. web; books; video; audio; software; images; Toggle navigation. A place to share knowledge, safety tips, fire codes and build community. 00 g of benzene, C6H6, to 80 g of cyclohexane, C6H12, lowers the freezing point of the cyclohexane from 6. The unusual and the unexpected in an old reaction. This article is cited by 17 publications. CHEM%210%[CHAPTER%6:%SUBSTITUTIONREACTIONSOFALKYLHALIDES!! ! 7% Fall!2013! KEY! 1. 06 Molecular Formula: C6H11Br 1-Bromocyclohexane Valid: ECOTOX Cyclohexylbromide Valid: ECOTOX Bromocyclohexane Valid: GCES Cyclohexane, bromo- Valid: Other Sources. Bromocyclohexane (Aldrich, 98%) was used without further purification, seeded in the Ar carrier gas, and expanded into the vacuum chamber through a nozzle orifice (General valve, 0. The interaction between them becomes weaker. Procedure 4. Epoxide formation and anti dihydroxylation. 0g of sodium bromide. How would I form trans-1,2-cyclohexanediol using bromocyclohexane? 2. Free radical bromination of cyclohexane: Cyclohexane + Br2 + heat ---> Bromocyclohexane + HBr(g). Page 6 O – H (alcohols) 3230 – 3550 C – H 2850 – 3300 O – H (acids) 2500 – 3000 C ≡ N 2220 – 2260 C = O 1680 – 1750. Conversely, the product of the elimination reaction is an alkene: the starting material has lost the elements of HCl, and the hybridization of the carbon originally bearing the chlorine atom has changed from sp 3 to sp 2. Solubility Soluble in alcohol. All the 1,2-dichloro isomers are constitutional isomers of the 1,3-dichloro isomers. ----- ELECTRONIC DATA DELIVERABLE VALID VALUES REFERENCE MANUAL Appendix to EPA Electronic Data Deliverable (EDD) Comprehensive Specification Manual TABLE OF CONTENTS Table A-l Matrix 1 Table A-2 Reference Point 3 Table A-3 Horizontal Collection Method 5 Table A-4 Horizontal Accuracy Units 6 Table A-5 Horizontal Datum 6 Table A-6 Elevation Collection Method 6 Table A-7 Elevation Datum 7 Table. N2 - On the basis of the ir vapour spectrum of cyclohexane, the assignment of a weak band at 947 cm-1 to the v15 fundamental is probable. +351 21 425 33 50 - Fax. The equatorial conformation of bromocyclohexane is more stable than the axial conformation by 0. Bromo Cyclohexane CAS No 108-85- Bromocyclohexane CAS-No. Allianze University College of Medical Sciences Foundation in Medical Studies July 2013 Intake Semester 2 Physical and Inorganic Chemistry Identification of Hydrocarbons This is my lab report of the experiment mentioned above. For the bromocyclohexane/thiourea inclusion compound, the intramolecular Br⋯C(3,5) distance (ca. Life as we know it is carbon-based. Starting with bromocyclohexane, how can each of the following compounds be prepared? Solution 2PStep 1 of 5(a)Cyanide ion reacts with bromocyclohexane to form cyclohexane carbonitrile. 5 mm diameter) with a backing pressure of 1 atm. Molecular Weight 210. Structure and physical data for. on StudyBlue. 5 − Bromoform: ブロモホルム. Toggle navigation Slidegur. PubChem Substance ID 24893802. Khan Academy is a 501(c)(3) nonprofit organization. cyclohexanecarbaldehyde 2043-61-0 route of synthesis, cyclohexanecarbaldehyde chemical synthesis methods, cyclohexanecarbaldehyde synthetic routes ect. Aldrich - C28898 Page 1 of 7 SIGMA-ALDRICH sigma-aldrich. Molecular Weight 118. In order to reduce the ring strain and attain a bond angle of approximately \(109. Free radical bromination of cyclohexane: Cyclohexane + Br2 + heat ---> Bromocyclohexane + HBr(g). After stirring a reaction mixture of CBr 4 (0. It is non-soluble in water (0. During a lab experiment, we were to determine the relative level of saturation by adding, drop by drop, a solution of bromine in cyclohexane to oil and a fat, individually. Bonding in Organic Compounds Chapter 1 1 1 Bonding in Organic Compounds CHAPTER SUMMARY Organic chemistry is the study of compounds of carbon. On the basis of the ir vapour spectrum of cyclohexane, the assignment of a weak band at 947 cm −1 to the v 15 fundamental is probable. 3-bromocyclohexene is more reactive than 4-bromocyclohexene towards hydrolysis by aqueous NaOH. Such a carbon atom is called a chiral center (or sometimes a stereogenic center), using organic-speak. Chemsrc provides trans-1,2-dibromocyclohexane(CAS#:7429-37-0) MSDS, density, melting point, boiling point, structure, formula, molecular weight etc. cyclohexanol. Expand this section. Question Expected Answers Marks Additional Guidance 2 a Answers clockwise from top left CH 3CH 2CH 2COOH CH 3CH 2CHCH 2 CH 3COOCH 2CH 2CH 2CH 3 CH 3CH 2CH 2CHO 4 ALLOW skeletal formula ALLOW butanoic acid ALLOW but-1-ene ALLOW butyl ethanoate ALLOW butanal If name and structure given both must be correct If C 3H 7 used instead of CH 3CH 2CH. +351 21 425 33 51 - pt. 环己烯,双键C原子以sp2杂化轨道形成σ键，其它C原子以sp3杂化轨道形成σ键。无色透明液体，有特殊刺激性气味。不溶于水，溶. Draw a Newman projections of the least stable conformation using the. Which conformation is more stable? Draw it first. 42): Boiling Pt (deg C): 160. The following sections detail some of the more common preparation methods: the oxidation of secondary alcohols, the hydration of alkynes, the ozonolysis of alkenes, Friedel‐Crafts acylation, the use of lithium dialkylcuprates, and the use of a Grignard reagent. 0 °C Molar. Any help would be appreciated!! Thanks (2. 055 Da Density 1. on StudyBlue. ), has unlimited solubility in dry ethyl alcohol, ethyl ether, benzene, acetone, toluene, , carbon tetrachloride,. The agreement between theory and experiments is good. Lusztyk, and K. web; books; video; audio; software; images; Toggle navigation. The synthesis is done by simply adding the acetic acid and sodium hypochlorite, which is also known as hypochlorous acid to. 1 g/cm3 Boiling Point 161. The reaction exhibited remarkable regioselectivity in. This time the slow step of the reaction only involves one species - the halogenoalkane. They are the unhappiest of rings – constrained into uncomfortable angles, with hydrogens forced by geometry to grumpily line up side-by-side with their repulsive neighbours. It plays a role in how your page is seen by search engine crawlers, and how it appears in SERPs. Beilstein/REAXYS Number 1900797. To explain the reaction of an amide with LiAlH 4 as the key step in going from bromocyclohexane to (N,N-dimethylaminomethyl) cyclohexane. SoDraw%the. Toggle navigation Slidegur. ) (a) bromocyclohexane (b) cyclohexene (c) ethoxycyclohexane (d). Cyclohexane, 1-bromo-2-chloro-, cis-cis-1-Bromo-2-Chlorocyclohexane. Looking at the energy values the table, it is clear that the apparent "size" of a substituent (in terms of its preference for equatorial over axial orientation) is influenced by its width and bond length to cyclohexane, as evidenced by the fact that an axial vinyl group is less hindered than ethyl, and iodine slightly less than chlorine. 217 (Mean VP of Antoine & Grain. EXAM 3 Answer Key, Chapters 7 - 9. The equatorial conformation of bromocyclohexane is more stable than the axial conformation by 0. Cyclohexyl bromide 108-85- NMR spectrum, Cyclohexyl bromide H-NMR spectral analysis, Cyclohexyl bromide C-NMR spectral analysis ect. ENTERPRISES, We are Exporter, Manufacturer & Supplier of Industrial Chemicals based in Maharashtra, India. In the case of ethene, bromoethane is formed. Bromide is a bigger molecule than chlorine so it is further away from the ring, making it easier for it to be pulled off. cyclohexyl alcohol reacts with H3O+ and CH3OH to form cyclohexane methyl ether and H20 with what mechanism?. Energy and Equilibrium The relative amounts of the two conformers depend on their difference in energy DE = ?RT ln K R is the gas constant [8. How can I add a Br to a cyclohexane so I can have a bromocyclohexane from a cyclohexane? Which reagent(s) can I use? Thank you. Cyclohexane C6H12O Cis-3-Hexen-1-ol C6H12O6 Galactose C6H14 Hexane C6H4Cl2 P-Dichlorobenzene C6H5Br Bromobenzene C6H5NO2 Vitamin B3 C6H6 Benzene C6H8O6 Vitamin C C6H8O7 Citric Acid C7H5NO3S Saccharin C7H6O Benzaldehyde C7H6O2 Benzoic Acid C7H6O3 Vitamin S C7H8 Cycloheptatriene C8H10 Xylene C8H10N4O2 Caffeine C8H18 2,2,4-Trimethylpentane C8H8. Furthermore, the scope of different carbonyl compounds was investigated and resulted with similar consecutive Michael-Claisen process for CDD synthesis. Eye Contact Rinse immediately with plenty of water, also under the eyelids, for at least 15 minutes. 3 Chemical and Physical Properties. There are actually, there are other conformations of cyclohexane, so the boat conformation can actually twist a little bit to give you twist boat. Integrated Risk Information System (IRIS) EPA 635 / R - 03 / 008 www. A) Which monosubstituted cyclohexane has the most negative ΔG∘ for this conversion? B) Which monosubstituted cyclohexane has the greatest preference for an equatorial position? C) Calculate ΔG∘ for the conversion of "axial" bromocyclohexane to "equatorial" bromocyclohexane at 25 ∘C. Reaction of an alkene with a hydrogen halide, converting the double bond to a halogenated single bond. Which conformation is more stable? Draw it first. MATERIALS AND METHODS Bromocyclohexane was purchased from B. Introduction. Cyclohexane is used as a nonpolar solvent for the chemical industry, and also as a raw material for the industrial production of adipic acid and caprolactam, both of which are intermediates used in the production of nylon. Neohexene was prepared through two steps from commercial tertbutyl bromide to 1neohexyl bromide via addition reaction of ethylene which was catalyzed by anhydrous AlCl3 along with the assistant catalysts and to neohexene via dehydrohalogenation of the intermediate neohexyl bromide in the. Formation of major product: 3-bromocyclohexene Under UV light, Br2$BrX2$ undergoes homolytic splitting to generate Br⋅$Br⋅$ radicals: Br2−→hv2Br⋅$BrX2→hv2Br⋅$ The formation of 3-bromocyclohexene is an example of su. Chemsrc provides CAS#:108-85-0 MSDS, density, melting point, boiling point, structure, formula, molecular weight, synthetic route, etc. It is known as an S N 1 reaction. car47872_ifc. 110 - 82 - 7 ) In Support of Summar. Because many compounds feature structurally similar six-membered rings, the structure and dynamics of cyclohexane are important prototypes of a wide range of compounds. 10-36/37/38 Alfa Aesar L04349: 26-37 Alfa Aesar L04349: 26-37-60 Alfa Aesar L04349: 3 Alfa Aesar L04349: DANGER: FLAMMABLE, irritates skin and eyes Alfa Aesar L04349: H226-H315-H319-H335 Alfa Aesar L04349. Bromocyclohexane (1-bromoethyl)-benzene 97% 1-Bromo-4-nitrobenzene Bromophenol Blue Bromophenol Blue solution, 0. Conversion of Cyclohexanol to Bromocyclohexane: The starting materials on a macro scale: 10. bromo-methyl-cyclohexane. Cyclohexyl bromide 108-85- NMR spectrum, Cyclohexyl bromide H-NMR spectral analysis, Cyclohexyl bromide C-NMR spectral analysis ect. graphic shows bromocyclohexane in two conformers, its most stable chair conformation and its least stable chair conformation. 1 (c) (i) Nucleophilic addition. A spectroscopic investigation of the conformation equilibrium of chlorocyclohexane and bromocyclohexane Cyclohexane Derivatives” [Russian translation], IL. page upkeep by Dave Woodcock: ([email protected] Conformations of bromocyclohexane As a result cyclohexyl bromide is quite slow to react with a base in the E2 reaction. Fundamentals reaction is a single step. Learn vocabulary, terms, and more with flashcards, games, and other study tools. C ha p ter 5 Reactions of Alkenes and Alkynes. 10 g of chalk was reacted with an excess of dilute hydrochloric acid. 16) Even in its most stable geometry, Bromo-cyclohexane can exist as two conformers. Reaction workup. Elimination vs substitution: tertiary substrate Our mission is to provide a free, world-class education to anyone, anywhere. Cyclohexyl bromide - cas 108-85-0, synthesis, structure, density, melting point, boiling point. com Material Safety Data Sheet Version 4. N2 - On the basis of the ir vapour spectrum of cyclohexane, the assignment of a weak band at 947 cm-1 to the v15 fundamental is probable. Related Pages. 8% and having Br at axial position is present in 33. 1021/cg201656y ). In order to get an impression about the possible environmental effects of tribromethene. Cyclohexane (C6H12) + Bromine (Br2) --> 1-Bromocyclohexane + HBr. Structural properties of the bromocyclohexane/thiourea inclusion compound have been determined using both single-crystal and powder X-ray diffraction over a range of temperatures above and below a first-order phase transition at 233 K in this material. Give one condition for this reaction to occur. It is also know by registry numbers ZINC000003865375, ZINC000100024325, MFCD00971951. Cyclopentene epoxide was prepared either by the method of Durbetaki3 or that of Goodman et a/. Cyclohexane,1-bromo-2-chloro-,trans-Other names: cis-1-Bromo-2-Chlorocyclohexane; cis-2-Chlorocyclohexyl Bromide Permanent link for this species. ChemExper Chemical Directory is a free service that allows to find a chemical by its molecular formula, IUPAC name, common name, CAS number, catalog number, substructure or physical characteristics. The top layer is Cyclohexane and the bottom layer is water. For example, those labeled A are attached to a carbon bonded to a carbonyl group and are different from the hydrogens labeled B which are bonded to a carbon attached to an oxygen atom. 4 ml of cyclohexanol. 7 cm −1, which has been observed so far in liquid phase only. cyclohexanol. graphic shows bromocyclohexane in two conformers, its most stable chair conformation and its least stable chair conformation. PubChem Substance ID 24893802. Determine the weight of the product. Because many compounds feature structurally similar six-membered rings, the structure and dynamics of cyclohexane are important prototypes of a wide range of compounds. Show this safety data sheet to the doctor in attendance. equilibrium in bromocyclohexane. Like aldehydes, ketones can be prepared in a number of ways. 108-85-0:C6H11Br, Cyclohexane, bromo-, 1-Bromocyclohexane, Bromcyclohexan, bromociclohexano, Bromocyclohexane, Cyclohexane, bromo-, Cyclohexyl bromide, NSC 11207. Cyclohexane is highly conformationally mobile, interchanging between the boat and chair forms many thousands of times per second. 10 mL) under LED irradiation for 24 h, peaks assigned to bromocyclohexane (53. You are not reacting bromine with cyclohexane, you are DISSOLVING bromine in cyclohexane and water mixed. The mechanism for this reaction is bimolecular, involving a collision between a strong base and an alkyl halide. In an E2 reaction, stereochemistry of the double bond -- that is, whether the E or Z isomer results -- is dictated by the stereochemistry of the starting material, if it is diastereomeric. ENTERPRISES, We are Exporter, Manufacturer & Supplier of Industrial Chemicals based in Maharashtra, India. -trans-1,2-Dibromocyclohexane Permanent link for this species. The cis isomer is a diastereomer of the trans isomers. Information on this page: Mass spectrum (electron ionization) References; Notes; Other data available: Gas phase ion energetics data. 2 Relevant identified uses of the substance or mixture and uses advised against. The mixture was cooled to 0°C with an ice bath and pyrrolidine (1. Cyclohexane is a cycloalkane with the molecular formula C 6 H 12. 9 years ago. 10 g of chalk was reacted with an excess of dilute hydrochloric acid. 1-bromo-3-chloro-cyclohexane. 6 kcal/mol, 2. 19) Which of the following is the most stable conformation of bromocyclohexane? A) I B) II C) III D) IV E) V Answer: C 20) Which conformer is at a local energy minimum on the potential energy diagram in the chair-chair interconversion of cyclohexane?. Chemsrc provides Bromocyclohexane(CAS#:108-85-0) MSDS, density, melting point, boiling point, structure, formula, molecular weight etc. Cyclohexane-1,3-dione derivatives synthesis from unreactive acetone has been accomplished through a consecutive Michael-Claisen process. There is no warranty of accuracy or completeness of any information contained herein. Bromocyclohexane. Chapter 04 Stereochemistry of Alkanes and Cycloalkanes Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. Predicted data is generated using the US Environmental Protection Agency's EPISuite™. 78944 4/22/2020 8/24/2016 4/22/2020. Cyclohexane is a cycloalkane with the molecular formula C 6 H 12. H2SO4 heat OH H2 Pt Prepare cyclohexane from cyclohexanol Question 19 Which one of the following outlines the best synthesis of trans-2-chlorocyclohexanol? A)Heat a mixture of cyclohexanol and Cl2 to 400oC. It should make things easier If something is up that means that it would be on a bold wedge on the hexans ring. 10 mL) under LED irradiation for 24 h, peaks assigned to bromocyclohexane (53. 4 download. This time the slow step of the reaction only involves one species - the halogenoalkane. ; bromocyclopentane, bromo- cycloheptane, cyclopentene, cyclohepten«, the cyclohexane diols and ^-glucuronidase from Koch-Light Laboratories Ltd. We report copper-catalyzed oxidative dehydrogenative carboxylation (ODC) of unactivated alkanes with various substituted benzoic acids to produce the corresponding allylic esters. Halohydrin formation. Which compound exists to the greatest extent as its hydrate when dissolved in aqueous acid? A. 67 estimate) = 3. A cyclohexane conformation is any of several three-dimensional shapes adopted by a cyclohexane molecule. (3pts) Org 1 final Page 9. Start studying Lab 9- E2 Reaction: Formation of Cyclohexane from Bromocyclohexane. Coordination of the proton adjacent to the Br substituent. The original mixture of a colourless and a green gas would produce steamy fumes of hydrogen chloride and a mist of organic liquids. Using your knowledge of the structure of cyclohexane, give an explanation for this difference in reactivity. If you continue browsing the site, you agree to the use of cookies on this website. 3-Methyl-2,3,4,4a,5,10b-hexahydro-1H-spiro[chromeno[3,4-c]pyridin-1,1'-cyclohexane]-2,4,5-trione is obtained by the interaction of methyl 1-bromocyclohexane carboxylate with zinc and N-methyl-2-oxo-2H- chromen-3-carboxamide, and its structure is determined by the single crystal X-ray diffraction method. The synthesis of cyclohexanone is a simple procedure that uses Acetic acid, sodium hypochlorite, hypochlorous acid, ether, sodium chloride, sodium carbonate and cyclohexanol. Is it bromocyclohexane? One of the requirements for reactivity inSn1 reactions is a stable carbocation. One method is to use O2 and Vanadium oxides at 400°C and 15 atmospheres. 108-85-0:C6H11Br, Cyclohexane, bromo-, 1-Bromocyclohexane, Bromcyclohexan, bromociclohexano, Bromocyclohexane, Cyclohexane, bromo-, Cyclohexyl bromide, NSC 11207. NIST/TRC Web Thermo Tables (WTT) NIST Standard Reference Subscription Database 3 - Professional Edition Version 2-2012-1-Pro This web application provides access to a collection of critically evaluated thermodynamic property data for pure compounds with a primary focus on organics. The most stable form of the cyclohexane ring is the chair form; in this conformer there is a greater interatomic distance between the equatorial and axial hydrogens than in the boat form. In order to get an impression about the possible environmental effects of tribromethene. AU - Egyed, O. a) bromocyclohexane to R-CH2CH2OH (R = cyclohexane) I know that ethylene oxide added to grignard reagent gives R-CH2CH2OH product. , 1974 Chemistry, organic University Microfilms, A XEROX Company , Ann Arbor, Michigan THIS DISSERTATION HAS BEEN MICROFILMED EXACTLY AS RECEIVED. There exist different approaches to synthesize cyclohexanol and/or cyclohexanone from cyclohexane. It can be seen from the balanced reaction that 1 mole of alcohol produces 1 mole of alkene. SoDraw%the. 62, how can one tell from the given data whether the disbustituted benzene is ortho, meta, or para?. As described in the literature, polar products, including formic acid, glutaric acid, succinic acid, hydroxyl caproic acid and adipic acid, would remain in the solid. Using cyclohexane as your starting material, show how you would synthesize each of the following compounds. Alkenes react with hydrogen bromide in the cold. 2 Conformational Analysis of Butane 0 60 120 180 240 300 360 3 kJ/mol 14 kJ/mol There are shadow images surrounding the waveform. 6 Chemical Vendors. The experimental data shown in these pages are freely available and have been published already in the DDB Explorer Edition. 05 (Adapted Stein & Brown method) Melting Pt (deg C): -27. Housecroft and E. A meta description is an HTML tag in the HTML code of your website, which allows you to customize a section of text that describes the page itself. Expand this section. This banner text can have markup. Bromination of cyclohexane. 011 CM SPECTRAL CONTAMINATION DUE TO CCl4 AROUND 1550 CM-1: Resolution: 4: Sampling procedure: TRANSMISSION: Data processing: DIGITIZED BY NIST FROM HARD. Think of them as landmarks, familiar signposts to guide you as you cruise the highways of chemistry. During this reaction, NBS serves as a continous source of very small amounts of bromine, as it is insoluble in CCl 4. Draw structures for a) cis-1-tert-butyl-4-methylcyclohexane and b) cis-1,4-di-tert-butylcyclohexane and determine the structural changes. Information on this page: Phase change data; Gas phase ion energetics data; IR Spectrum; Mass spectrum (electron. You need to look at antibonding orbitals here. Final Exam Answers: Chem 334 - Fall 2004. Or draw it yourself in hyperchem or anything else that shows 3D structure and look at that. 22) MULTIPLE CHOICE. a) bromocyclohexane to R-CH2CH2OH (R = cyclohexane) I know that ethylene oxide added to grignard reagent gives R-CH2CH2OH product. Commercially, cyclohexene is made from hydrogenation of benzene. The Cyclohexane Molecule -- Chemical and Physical Properties. Reaction workup. In this experiment, the cyclohexanol solution is being used in the dehydration process. Company Identification. Easily share your publications and get them in front of Issuu’s. Cyclohexane solution undergoes halogenation in the bromine test which needs the presence of ultraviolet light or sunlight as this is a photochemical reaction. 42): Boiling Pt (deg C): 173. Under these conditions, conversion to the cross-coupled products of 4-MeC 6 H 4 MgBr, 2-MeC 6 H 4 MgBr and 4-MeOC 6 H 4 MgBr with bromocyclohexane each gave 99% yields. ; bromocyclopentane, bromo- cycloheptane, cyclopentene, cyclohepten«, the cyclohexane diols and ^-glucuronidase from Koch-Light Laboratories Ltd. PubChem Substance ID 24891958. edu is a platform for academics to share research papers. Learn vocabulary, terms, and more with flashcards, games, and other study tools. The intermediacy of an alkyl radical was evidenced by the catalytic reaction of cyclohexane with benzamide in the presence of CBr 4, which formed exclusively bromocyclohexane. Cyclohexane is a colourless liquid, with an acrid, narcotic smell. Chemsrc provides methyl 1-bromocyclohexane-1-carboxylate(CAS#:3196-23-4) MSDS, density, melting point, boiling point, structure, formula, molecular weight etc. A spectroscopic investigation of the conformation equilibrium of chlorocyclohexane and bromocyclohexane Cyclohexane Derivatives” [Russian translation], IL. 1-chlorohexane + low conc. 3 Chemical and Physical Properties. As described in the literature, polar products, including formic acid, glutaric acid, succinic acid, hydroxyl caproic acid and adipic acid, would remain in the solid. How would I form trans-1,2-cyclohexanediol using bromocyclohexane? 2. 3-Methyl-2,3,4,4a,5,10b-hexahydro-1H-spiro[chromeno[3,4-c]pyridin-1,1'-cyclohexane]-2,4,5-trione is obtained by the interaction of methyl 1-bromocyclohexane carboxylate with zinc and N-methyl-2-oxo-2H- chromen-3-carboxamide, and its structure is determined by the single crystal X-ray diffraction method. The Synthesis of Cyclohexene from Bromocyclohexane An 2hexene. Propriedades Fórmula molecular: C 6 H 11 Br : Massa molar: 163. The Louisiana State University and Agricultural and Mechanical College, Ph. The unusual and the unexpected in an old reaction. Cyclopentene epoxide was prepared either by the method of Durbetaki3 or that of Goodman et a/. So what happens to that lone pair? Does it stay on the carbon? HELL NO! Why? Two reasons, one half-assed and. The halogen prefixes are Fluoro-, Chloro-, Bromo- and Iodo-. 3 degree celsius. Show this safety data sheet to the doctor in attendance. 4 1) Write an equation to describe the initiation step in the chlorination of methane. at axial position. 23 Boiling Pt, Melting Pt, Vapor Pressure Estimations (MPBPWIN v1. Bromocyclohexane. Molecular Weight 241. Cyclohexane solution undergoes halogenation in the bromine test which needs the presence of ultraviolet light or sunlight as this is a photochemical reaction. AU - Holly, S. 1-methyl-1-bromocyclohexane. 1,2-Diaminocyclohexane; A. This shows two layers with slightly different colours demonstrating that bromine is more soluble in non-polar solvents. ionic reactions — nucleophilic substitution and elimination reactions of alkyl halides solutions to problems 6. , it is bimolecular. Provide a reason for your answer. The measurements, were performed on two polymer theta solvent systems: polydimethylsiloxane-bromocyclohexane at 29 °C and polystyrene-cyclohexane at 34°C for the attractive and repulsive cases respectively. Expand this section. at equatorial position. Recently Asked Questions Hi, I was doing these homework problems and wanted to make sure I solved them correctly just to understand the material better. Basic organic chemistry 1. Sun Q(1), Faller R. Section 3-14. The phosphoric acid is a catalyst and as such increases the rate of reaction but does not affect the overall stoichiometry. 67 estimate) = 3. bromo-methyl-cyclohexane. Structural properties of the bromocyclohexane/thiourea inclusion compound have been determined using both single-crystal and powder X-ray diffraction over a range of temperatures above and below a first-order phase transition at 233 K in this material. Favorite Answer. [email protected] Expand this section. "Bu" is shorthand for "CH"_3"CH"_2"CH"_2"CH"_2-, or butyl. cyclohexane ring. Il devrait donc exister deux stéréoisomères d'un dérivé du cyclohexane tel que le bromocyclohexane. It is known as an S N 1 reaction. 4 The product had b. Empirical Formula (Hill Notation) C 6 H 10 Br 2. Cyclohexane is a non-corrosive and quick volatilize liquid and sublimes between -5 to 5 C. 020 487 70 00 - Fax. Pages dans la catégorie « Cyclohexane » Cette catégorie contient les 63 pages suivantes. The mixture was cooled to 0°C with an ice bath and pyrrolidine (1. Make the left most ring carbon C1 and number towards the front. Therefore, cyclohexane solution reacts in bromine test when placed under the bright sunlight to produce carbon and hydrogen bromide but there is no reaction when it is placed in the dark. For%each%of%the%followingreactions%draw%themechanism%for%S N2:% 2. Energy and Equilibrium The relative amounts of the two conformers depend on their difference in energy DE = ?RT ln K R is the gas constant [8. 4 Spectral Information. In both cases you could do an elimination reaction with a strong base like KOH to give you cyclohexene. This organic chemistry video tutorial provides a basic introduction into drawing the chair conformation of cyclohexane and identifying the most stable conformation of ethylcyclohexane and 1-tert. Solubility Soluble in alcohol. Chem 17 3rd Lab Long Exam Reviewer. This time the slow step of the reaction only involves one species - the halogenoalkane. cas号查询致力于为化学行业用户免费提供溴 环己烷的cas号、中文名称、英文名称相互转换服务,同时也包括溴 环己烷的性质、化学式、分子结构、密度、熔点、沸点等信息。. The three halogenic LMGSs, namely, chlorocyclohexane (ClCH), bromocyclohexane (BrCH), and iodocyclohexane (ICH), share a common molecular structure (X-cyclohexane). Conformation du cyclohexane. Methyl benzyl ketone B. Chapter 20 MC Practice 1. ; bromocyclopentane, bromo- cycloheptane, cyclopentene, cyclohepten«, the cyclohexane diols and ^-glucuronidase from Koch-Light Laboratories Ltd. En réalité les deux formes s'interconvertissent rapidement l'un en l'autre en passant par la forme bateau. The structure of 1α-methoxy-2α-hydroxy-3β-bromocyclohexane was established by de-etherification to 3β-bromo-1α,2α-cyclohexanediol, which was oxidized by periodate 10 times more rapidly than its isomer. Bromocyclohexane (1-bromoethyl)-benzene 97% 1-Bromo-4-nitrobenzene Bromophenol Blue Bromophenol Blue solution, 0. how to convert cyclohexanol to bromocyclohexane. 1-methyl-1-bromocyclohexane. A spectroscopic investigation of the conformation equilibrium of chlorocyclohexane and bromocyclohexane Cyclohexane Derivatives” [Russian translation], IL. Powder X-ray diffraction studies, carried out at room tem-. Such a carbon atom is called a chiral center (or sometimes a stereogenic center), using organic-speak. Hangzhou Meite Industry Co. Q: Regarding the NMR problem 17. rq8hcefe1j, 33cwy6xn2be38, o23plz35z0q58, 0tlwcks67d2, 6573umjqrmxowg, 13550rejk04, ooc5rnrgzkwse, 3l72vlfwbw6, 182txjprzp, bajrf6kwgc, 7imttylindac, e4ynwcfg2g1rw5, 5mko39srdxrp, oeyp0rm8yt5lkmc, rxekximncvg, cqmms0nkep, fjj5r8iaeh3k9, fd1wft6b63yb1, ndh8gxqmgf, 7p2o41afm4fd, 4q8mrgzs7u3kt, ajitfvmnl3, ie71q1mj80g, d5ianhd9glutr, 3sexzfl4fnnjzmk, m8elsbgnygyb
|
2020-05-30 13:52:42
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4463033080101013, "perplexity": 11006.713246189829}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347409337.38/warc/CC-MAIN-20200530133926-20200530163926-00429.warc.gz"}
|
http://mathhelpforum.com/calculus/88670-finding-integral-derivative.html
|
# Thread: Finding an integral from a derivative.
1. ## Finding an integral from a derivative.
Differentiate $tan^{-1}(cos x)$ and hence evaluate $\int_0^\pi sin \frac{dx}{(2-sin^{2}x)}$
2. Originally Posted by nerdzor
Differentiate $\tan^{-1}(\cos x)$ and hence evaluate $\int_0^\pi {\color{red}\sin x} \frac{dx}{(2 - \sin^{2}x)}$
You have a made a typo (see the red).
Can you differentiate the given function? Note that $1 + \cos^2 x = 2 - \sin^2 x$ using the Pythagorean Identity. So the derivative of the function is equal to the integrand. What does that tell you ....?
|
2013-12-19 16:24:21
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9883566498756409, "perplexity": 486.50164557724605}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1387345765796/warc/CC-MAIN-20131218054925-00019-ip-10-33-133-15.ec2.internal.warc.gz"}
|
https://pressbooks.bccampus.ca/humanbiomechanics/chapter/8-2-impulse-2/
|
# 46 6.11 Impulse
### Summary
• Define impulse.
• Describe effects of impulses in everyday life.
• Determine the average effective force using graphical representation.
• Calculate average force and impulse given mass, velocity, and time.
The effect of a force on an object depends on how long it acts, as well as how great the force is. For example, in the tennis swing, very large force acting for a short time had a great effect on the momentum of the tennis ball. A small force could cause the same change in momentum, but it would have to act for a much longer time. For example, if the ball were thrown upward, the gravitational force (which is much smaller than the tennis racquet’s force) would eventually reverse the momentum of the ball. Quantitatively, the effect we are talking about is the change in momentum $\boldsymbol{\Delta\vec{\textbf{p}}}.$
By rearranging the equation $\boldsymbol{\vec{\textbf{F}}_{\textbf{net}}=\frac{\Delta\vec{\textbf{p}}}{\Delta{t}}}$ to be
$\boldsymbol{\Delta\vec{\textbf{p}}=\vec{\textbf{F}}_{\textbf{net}}\Delta{t}},$
we can see how the change in momentum equals the average net external force multiplied by the time this force acts. The quantity $\boldsymbol{\vec{\textbf{F}}_{\textbf{net}}\Delta{t}}$ is given the name impulse. Impulse is the same as the change in momentum.
### IMPULSE: CHANGE IN MOMENTUM
Change in momentum equals the average net external force multiplied by the time this force acts.
$\boldsymbol{\Delta\vec{\textbf{p}}=\vec{\textbf{F}}_{\textbf{net}}\Delta{t}}$
The quantity $\boldsymbol{\vec{\textbf{F}}_{\textbf{net}}\Delta{t}}$ is given the name impulse.
There are many ways in which an understanding of impulse can save lives, or at least limbs. The dashboard padding in a car, and certainly the airbags, allow the net force on the occupants in the car to act over a much longer time when there is a sudden stop. The momentum change is the same for an occupant, whether an air bag is deployed or not, but the force (to bring the occupant to a stop) will be much less if it acts over a larger time. Cars today have many plastic components. One advantage of plastics is their lighter weight, which results in better gas mileage. Another advantage is that a car will crumple in a collision, especially in the event of a head-on collision. A longer collision time means the force on the car will be less. Deaths during car races decreased dramatically when the rigid frames of racing cars were replaced with parts that could crumple or collapse in the event of an accident.
Bones in a body will fracture if the force on them is too large. If you jump onto the floor from a table, the force on your legs can be immense if you land stiff-legged on a hard surface. Rolling on the ground after jumping from the table, or landing with a parachute, extends the time over which the force (on you from the ground) acts.
### Example 1: Calculating Magnitudes of Impulses: Two Billiard Balls Striking a Rigid Wall
Two identical billiard balls strike a rigid wall with the same speed, and are reflected without any change of speed. The first ball strikes perpendicular to the wall. The second ball strikes the wall at an angle of 30° from the perpendicular, and bounces off at an angle of 30° from perpendicular to the wall.
(a) Determine the direction of the force on the wall due to each ball.
(b) Calculate the ratio of the magnitudes of impulses on the two balls by the wall.
Strategy for (a)
In order to determine the force on the wall, consider the force on the ball due to the wall using Newton’s second law and then apply Newton’s third law to determine the direction. Assume the x-axis to be normal to the wall and to be positive in the initial direction of motion. Choose the y-axis to be along the wall in the plane of the second ball’s motion. The momentum direction and the velocity direction are the same.
Solution for (a)
The first ball bounces directly into the wall and exerts a force on it in the +x direction. Therefore the wall exerts a force on the ball in the x direction. The second ball continues with the same momentum component in the y direction, but reverses its x-component of momentum, as seen by sketching a diagram of the angles involved and keeping in mind the proportionality between velocity and momentum.
These changes mean the change in momentum for both balls is in the x direction, so the force of the wall on each ball is along the x direction.
Strategy for (b)
Calculate the change in momentum for each ball, which is equal to the impulse imparted to the ball.
Solution for (b)
Let u be the speed of each ball before and after collision with the wall, and m the mass of each ball. Choose the x-axis and y-axis as previously described, and consider the change in momentum of the first ball which strikes perpendicular to the wall.
$\boldsymbol{p_{\textbf{xi}}=mu;\:p_{\textbf{yi}}=0}$
$\boldsymbol{p_{\textbf{xf}}=-mu;\:p_{\textbf{yf}}=0}$
Impulse is the change in momentum vector. Therefore the x-component of impulse is equal to -2mu and the y-component of impulse is equal to zero.
Now consider the change in momentum of the second ball.
$\boldsymbol{p_{\textbf{xi}}=mu\:\textbf{cos}\:30^0;\:p_{\textbf{yi}}=-mu\:\textbf{sin}\:30^0}$
$\boldsymbol{p_{\textbf{xf}}=-mu\:\textbf{cos}\:30^0;\:p_{\textbf{yf}}=-mu\:\textbf{sin}\:30^0}$
It should be noted here that while px changes sign after the collision, py does not. Therefore the x-component of impulse is equal to -2mu cos 30° and the y-component of impulse is equal to zero.
The ratio of the magnitudes of the impulse imparted to the balls is
$\boldsymbol{\frac{2mu}{2mu\:\textbf{cos}\:30^0}}\boldsymbol{=}\boldsymbol{\frac{2}{\sqrt{3}}}\boldsymbol{=1.155}.$
Discussion
The direction of impulse and force is the same as in the case of (a); it is normal to the wall and along the negative x-direction. Making use of Newton’s third law, the force on the wall due to each ball is normal to the wall along the positive x-direction.
Our definition of impulse includes an assumption that the force is constant over the time interval Δt. Forces are usually not constant. Forces vary considerably even during the brief time intervals considered. It is, however, possible to find an average effective force Feff that produces the same result as the corresponding time-varying force. Figure 1 shows a graph of what an actual force looks like as a function of time for a ball bouncing off the floor. The area under the curve has units of momentum and is equal to the impulse or change in momentum between times t1 and t2. That area is equal to the area inside the rectangle bounded by Feff, t1, and t2. Thus the impulses and their effects are the same for both the actual and effective forces.
### MAKING CONNECTIONS: TAKE-HOME INVESTIGATION—HAND MOVEMENT AND IMPULSE
Try catching a ball while “giving” with the ball, pulling your hands toward your body. Then, try catching a ball while keeping your hands still. Hit water in a tub with your full palm. After the water has settled, hit the water again by diving your hand with your fingers first into the water. (Your full palm represents a swimmer doing a belly flop and your diving hand represents a swimmer doing a dive.) Explain what happens in each case and why. Which orientations would you advise people to avoid and why?
### Everyday Examples: Landing after a Jump
You naturally tend to bend your knees when landing after a jump, rather than keep your knees locked and your legs rigid. The reason is that rigid legs bring you to an abrupt stop, but bending your knees allows you to spread the landing out over a longer time which reduces the average and peak force applied to your legs.
The force vs. time graphs show the normal force applied to the person landing on one foot after stepping off from a 0.1 height as seen in the previous GIF. The graph on the left was the more rigid leg landing (it didn’t feel good) and the graph on the right was a bent-knee landing.
Notice that the stiff-legged “hard” landing nearly doubled the peak force applied to the body.
### Reinforcement Activity
Throw an egg or a water balloon up into the air and catch it. Did you move your hands with the object as you caught it on the return, or did you hold your hands remain still as it arrived?
Adults have learned from experience how to reduce the force on objects as we control their motion. In my experience, toddlers do not apply this technique.
# Section Summary
• Impulse, or change in momentum, equals the average net external force multiplied by the time this force acts:
$\boldsymbol{\Delta\vec{\textbf{p}}=\vec{\textbf{F}}_{\textbf{net}}\Delta{t}}.$
• Forces are usually not constant over a period of time.
### Conceptual Questions
1: Professional Application
Explain in terms of impulse how padding reduces forces in a collision. State this in terms of a real example, such as the advantages of a carpeted vs. tile floor for a day care center.
2: While jumping on a trampoline, sometimes you land on your back and other times on your feet. In which case can you reach a greater height and why?
3: Professional Application
Tennis racquets have “sweet spots.” If the ball hits a sweet spot then the player’s arm is not jarred as much as it would be otherwise. Explain why this is the case.
### Problems & Exercises
1: A person slaps her leg with her hand, bringing her hand to rest in 2.50 milliseconds from an initial speed of 4.00 m/s. (a) What is the average force exerted on the leg, taking the effective mass of the hand and forearm to be 1.50 kg? (b) Would the force be any different if the woman clapped her hands together at the same speed and brought them to rest in the same time? Explain why or why not.
2: Professional Application
A professional boxer hits his opponent with a 1000-N horizontal blow that lasts for 0.150 s. (a) Calculate the impulse imparted by this blow. (b) What is the opponent’s final velocity, if his mass is 105 kg and he is motionless in midair when struck near his center of mass? (c) Calculate the recoil velocity of the opponent’s 10.0-kg head if hit in this manner, assuming the head does not initially transfer significant momentum to the boxer’s body. (d) Discuss the implications of your answers for parts (b) and (c).
3: Calculate the final speed of a 110-kg rugby player who is initially running at 8.00 m/s but collides head-on with a padded goalpost and experiences a backward force of 1.76 × 104 N.
4: Starting with the definitions of momentum and kinetic energy, derive an equation for the kinetic energy of a particle expressed as a function of its momentum.
5: A ball with an initial velocity of 10 m/s moves at an angle 60° above the +x-direction. The ball hits a vertical wall and bounces off so that it is moving 60° above the -x-direction with the same speed. What is the impulse delivered by the wall?
6: When serving a tennis ball, a player hits the ball when its velocity is zero (at the highest point of a vertical toss). The racquet exerts a force of 540 N on the ball for 5.00 ms, giving it a final velocity of 45.0 m/s. Using these data, find the mass of the ball.
7: A punter drops a ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at an angle 55° above the horizontal. What is the impulse delivered by the foot (magnitude and direction)?
## Glossary
change in momentum
the difference between the final and initial momentum; the mass times the change in velocity
impulse
the average net external force times the time it acts; equal to the change in momentum
### Solutions
Problems & Exercises
1: (a) $\boldsymbol{2.40\times10^3\textbf{ N}}$ toward the leg (b) The force on each hand would have the same magnitude as that found in part (a) (but in opposite directions by Newton’s third law) because the change in momentum and the time interval are the same.
4: $\begin{array}{l} \boldsymbol{\textbf{p}=m\textbf{v}\Rightarrow{p^2}=m^2v^2\Rightarrow\frac{p^2}{m}=mv^2} \\ \boldsymbol{\Rightarrow\frac{p^2}{2m}=\frac{1}{2}mv^2=\textbf{KE}} \\ \boldsymbol{\textbf{KE}=\frac{p^2}{2m}} \end{array}$
6: $\boldsymbol{60.0\textbf{ g}}$
|
2023-02-01 20:06:56
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5086142420768738, "perplexity": 516.9042916638471}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499949.24/warc/CC-MAIN-20230201180036-20230201210036-00172.warc.gz"}
|
https://www.aimsciences.org/article/doi/10.3934/krm.2008.1.405
|
# American Institute of Mathematical Sciences
September 2008, 1(3): 405-414. doi: 10.3934/krm.2008.1.405
## A new regularization possibility for the Boltzmann equation with soft potentials
1 LAMA UMR 8050, Faculté de Sciences et Technologies, Université Paris Est, 61 avenue du Général de Gaulle, 94010 Créteil Cedex, France
Received June 2008 Revised June 2008 Published August 2008
We consider a simplified Boltzmann equation: spatially homogeneous, two-dimensional, radially symmetric, with Grad's angular cutoff, and linearized around its initial condition. We prove that for a sufficiently singular velocity cross section, the solution may become instantaneously a function, even if the initial condition is a singular measure. To our knowledge, this is the first regularization result in the case with cutoff: all the previous results were relying on the non-integrability of the angular cross section. Furthermore, our result is quite surprising: the regularization occurs for initial conditions that are not too singular, but also not too regular. The objective of the present work is to explain that the singularity of the velocity cross section, which is often considered as a (technical) obstacle to regularization, seems on the contrary to help the regularization.
Citation: Nicolas Fournier. A new regularization possibility for the Boltzmann equation with soft potentials. Kinetic and Related Models, 2008, 1 (3) : 405-414. doi: 10.3934/krm.2008.1.405
[1] Zheng-an Yao, Yu-Long Zhou. High order approximation for the Boltzmann equation without angular cutoff under moderately soft potentials. Kinetic and Related Models, 2020, 13 (3) : 435-478. doi: 10.3934/krm.2020015 [2] Radjesvarane Alexandre, Yoshinori Morimoto, Seiji Ukai, Chao-Jiang Xu, Tong Yang. Uniqueness of solutions for the non-cutoff Boltzmann equation with soft potential. Kinetic and Related Models, 2011, 4 (4) : 919-934. doi: 10.3934/krm.2011.4.919 [3] Fei Meng, Fang Liu. On the inelastic Boltzmann equation for soft potentials with diffusion. Communications on Pure and Applied Analysis, 2020, 19 (11) : 5197-5217. doi: 10.3934/cpaa.2020233 [4] Yingzhe Fan, Yuanjie Lei. The Boltzmann equation with frictional force for very soft potentials in the whole space. Discrete and Continuous Dynamical Systems, 2019, 39 (7) : 4303-4329. doi: 10.3934/dcds.2019174 [5] Nicolas Fournier. A recursive algorithm and a series expansion related to the homogeneous Boltzmann equation for hard potentials with angular cutoff. Kinetic and Related Models, 2019, 12 (3) : 483-505. doi: 10.3934/krm.2019020 [6] Léo Glangetas, Hao-Guang Li, Chao-Jiang Xu. Sharp regularity properties for the non-cutoff spatially homogeneous Boltzmann equation. Kinetic and Related Models, 2016, 9 (2) : 299-371. doi: 10.3934/krm.2016.9.299 [7] Zhaohui Huo, Yoshinori Morimoto, Seiji Ukai, Tong Yang. Regularity of solutions for spatially homogeneous Boltzmann equation without angular cutoff. Kinetic and Related Models, 2008, 1 (3) : 453-489. doi: 10.3934/krm.2008.1.453 [8] Lingbing He, Yulong Zhou. High order approximation for the Boltzmann equation without angular cutoff. Kinetic and Related Models, 2018, 11 (3) : 547-596. doi: 10.3934/krm.2018024 [9] Yoshinori Morimoto, Seiji Ukai, Chao-Jiang Xu, Tong Yang. Regularity of solutions to the spatially homogeneous Boltzmann equation without angular cutoff. Discrete and Continuous Dynamical Systems, 2009, 24 (1) : 187-212. doi: 10.3934/dcds.2009.24.187 [10] Renjun Duan, Shuangqian Liu, Tong Yang, Huijiang Zhao. Stability of the nonrelativistic Vlasov-Maxwell-Boltzmann system for angular non-cutoff potentials. Kinetic and Related Models, 2013, 6 (1) : 159-204. doi: 10.3934/krm.2013.6.159 [11] Radjesvarane Alexandre, Jie Liao, Chunjin Lin. Some a priori estimates for the homogeneous Landau equation with soft potentials. Kinetic and Related Models, 2015, 8 (4) : 617-650. doi: 10.3934/krm.2015.8.617 [12] Yong-Kum Cho. On the homogeneous Boltzmann equation with soft-potential collision kernels. Kinetic and Related Models, 2015, 8 (2) : 309-333. doi: 10.3934/krm.2015.8.309 [13] Yong-Kum Cho, Hera Yun. On the gain of regularity for the positive part of Boltzmann collision operator associated with soft-potentials. Kinetic and Related Models, 2012, 5 (4) : 769-786. doi: 10.3934/krm.2012.5.769 [14] Jean-Marie Barbaroux, Dirk Hundertmark, Tobias Ried, Semjon Vugalter. Strong smoothing for the non-cutoff homogeneous Boltzmann equation for Maxwellian molecules with Debye-Yukawa type interaction. Kinetic and Related Models, 2017, 10 (4) : 901-924. doi: 10.3934/krm.2017036 [15] Dong Li. A regularization-free approach to the Cahn-Hilliard equation with logarithmic potentials. Discrete and Continuous Dynamical Systems, 2022, 42 (5) : 2453-2460. doi: 10.3934/dcds.2021198 [16] Robert M. Strain, Keya Zhu. Large-time decay of the soft potential relativistic Boltzmann equation in $\mathbb{R}^3_x$. Kinetic and Related Models, 2012, 5 (2) : 383-415. doi: 10.3934/krm.2012.5.383 [17] Nadia Lekrine, Chao-Jiang Xu. Gevrey regularizing effect of the Cauchy problem for non-cutoff homogeneous Kac's equation. Kinetic and Related Models, 2009, 2 (4) : 647-666. doi: 10.3934/krm.2009.2.647 [18] Kevin Zumbrun. L∞ resolvent bounds for steady Boltzmann's Equation. Kinetic and Related Models, 2017, 10 (4) : 1255-1257. doi: 10.3934/krm.2017048 [19] Laurent Desvillettes, Clément Mouhot, Cédric Villani. Celebrating Cercignani's conjecture for the Boltzmann equation. Kinetic and Related Models, 2011, 4 (1) : 277-294. doi: 10.3934/krm.2011.4.277 [20] François Golse. The Boltzmann-Grad limit for the Lorentz gas with a Poisson distribution of obstacles. Kinetic and Related Models, 2022, 15 (3) : 517-534. doi: 10.3934/krm.2022001
2020 Impact Factor: 1.432
|
2022-05-20 07:29:18
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7029050588607788, "perplexity": 4588.771965909847}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662531762.30/warc/CC-MAIN-20220520061824-20220520091824-00555.warc.gz"}
|
http://www.directindustry.com/prod/ansaldo-energia/gas-turbine-thermal-power-plants-29641-232044.html
|
Stand ANSALDO ENERGIAGroup: Finmeccanica
# Gas turbine thermal power plant6.695 MW ANSALDO ENERGIA
Open Cycle
Ansaldo Energia open cycle units are the right solution to satisfy in short time the need for electric power for peak demand. Ansaldo Energia open cycle units are based on three models of gas turbines manufactured by the Company, equipped with air cooled turbogenerators:
AE64.3A/75MW AE94.2&AE94.2K/170MW AE94.3A/294MW at ISO conditions
The turboset package can be supplied as a standalone system in conjunction with the electrical generator, the turboset auxiliaries and the control system, or extended to include all the components and systems in a complete turnkey plant.
|
2013-05-18 23:53:39
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8200319409370422, "perplexity": 5608.531213589483}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368696382989/warc/CC-MAIN-20130516092622-00062-ip-10-60-113-184.ec2.internal.warc.gz"}
|