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http://openstudy.com/updates/50a989ece4b064039cbd0a50 | ## anonymous 3 years ago If a is uniformly distributed over [−28,28], what is the probability that the roots of the equation x2+ax+a+80=0
1. anonymous
what is the probability that the roots of the equation what?
2. anonymous
what it the probability that the roots of the equation are real?
3. anonymous
are both real?
4. anonymous
oh ok we can solve that using the quadratic formula
5. anonymous
ok
6. anonymous
what would be the b value of the equation. is it a(x+1)
7. anonymous
$x^2+ax+a+80=0$ $x=\frac{-a\pm\sqrt{a^2-4a-320}}{2}$
8. anonymous
no if you use the quadratic formula, $$a=1,b=a, c=a+80$$ different $$a$$ of course
9. anonymous
if these are to be real numbers, that means you must have $a^2-4a-300\geq 0$
10. anonymous
how did you get that?
11. anonymous
by some miracle this factors as $(a-20)(a+16)\geq 0$ so we can actually solve
12. anonymous
how did i get $$a^2-4a-320$$?
13. anonymous
yes
14. anonymous
or how did i get the whole equation?
15. phi
how did you get that? the stuff under the square root has to be 0 or positive, else you get imaginary numbers when you take the root.
16. anonymous
thanks
17. anonymous
ok quadratic formula tells you the solution to $$ax^2+bx+c=0$$ is $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ in your case $$a=1,b=a,c=a+80$$
18. anonymous
when you compute you get $x=\frac{-a\pm\sqrt{a^2-4a-320}}{2}$ in your case so for these to be real, the discriminant $$a^2-4a-320$$ must be greater than or equal to zero, otherwise you have a negative number under the radical
19. anonymous
oh, what @phi said
20. anonymous
okay so after i find my values what do I do?
21. anonymous
your last job is to solve for $$a$$ $a^2-4a-320\geq 0$
22. anonymous
i found my values and they are a=20 and a+-16
23. anonymous
this factors as $$(a-20)(a+16)\geq 0$$
24. anonymous
a=-16
25. anonymous
hold on, you are not solving $$(a-20)(a+16)=0$$ but rather $(a-20)(a+16)\geq 0$
26. anonymous
the zeros are $$-16$$ and $$20$$ for sure but you want to know the interval over which it is positive since this is a quadratic with leading coeffient positive, it will be positive outside the zeros, in other words if $$a\leq -16$$ or $$a\geq 20$$
27. anonymous
your final job is to find what portion of the interval $$[-28,28]$$ satisfies $$a\leq -16$$ or $$a\geq 20$$
28. anonymous
okay so i would have to find the interval of the it by integrating it?
29. anonymous
not necessary, just look
30. anonymous
favorable part is $$[-28,-16]\cup [20,28]$$
31. anonymous
you can eyeball the total length
32. anonymous
12 and 8
33. anonymous
right, for a total of 20
34. anonymous
thanks just found the answer. thank you
35. anonymous
yw
36. anonymous
hope it was $$\frac{20}{56}$$
37. anonymous
yea | 2016-07-02 00:29:07 | {"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": 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": 0.8363266587257385, "perplexity": 1309.417283733567}, "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-2016-26/segments/1466783404382.73/warc/CC-MAIN-20160624155004-00069-ip-10-164-35-72.ec2.internal.warc.gz"} |
https://testbook.com/question-answer/in-a-standard-proctor-test-the-water-content-w1--5e3802aef60d5d512c879858 | # In a standard proctor test, the water content (W1) and maximum dry density (γd max) are related as
This question was previously asked in
RSMSSB JE Civil Diploma Sep 2016 Official Paper
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1. γd max is linearly proportional to W
2. γd max is inversely proportional to W
3. γd max corresponds to a unique value of W
4. None of above
Option 3 : γd max corresponds to a unique value of W
## Detailed Solution
The standard Proctor test is a laboratory method of experimentally determining the optimal moisture content at which a given soil will become most dense and achieve its maximum dry density.
In this test, a standard volume of the soil is filled up with soil in three layers. Each layer is compacted by 25 blows of a standard hammer of weight 2.495 kg falling through 304.8 mm height. By noting the weight of compacted soil with the corresponding water content, dry unit weight of the soil can be found using the relation.
$${\gamma _d} = \frac{{{\gamma _t}}}{{1 + w}}$$
w = water content
γt = unit weight of the compacted soil
This test is repeated at different water content and the graph is plotted between water content and dry unit weight of the soil. Usually 4 to 5 points are required to draw the compaction curve. This curve is unique for the soil type, method of compaction and compactive effort. The peak point at which the dry unit weight of the soil is maximum for the given water content gives maximum dry density of the soil and the corresponding water content is known as Optimum water content. Hence it can be concluded that there exist a unique value of water content for the maximum dry density. | 2021-10-23 23:41:52 | {"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": 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": 0.3903041183948517, "perplexity": 1805.9473511155063}, "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-2021-43/segments/1634323585828.15/warc/CC-MAIN-20211023224247-20211024014247-00500.warc.gz"} |
https://www.lessonplanet.com/teachers/final-sounds-g | # Final Sounds: g
In this final sound learning exercise, students trace the letter g, and insert a g into the final sound of words. Students trace 5 letters and insert the g into 6 words. | 2017-07-26 21:40:34 | {"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": 0.852340042591095, "perplexity": 5141.549384820713}, "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-2017-30/segments/1500549426629.63/warc/CC-MAIN-20170726202050-20170726222050-00366.warc.gz"} |
https://www.analyzemath.com/calculus/derivative/proof-derivative-of-cscx.html | Proof of the Derivative of csc x
The proof of the calculation of the derivative of $\csc (x)$ is presented using the quotient rule of derivatives.
Proof of the Derivative of csc x
A trigonometric identity relating $\csc x$ and $\sin x$ is given by $\csc x = \dfrac { 1 }{ \sin x }$ Use of the quotient rule of differentiation to find the derivative of $\csc x$; hence
$\displaystyle { \dfrac {d}{dx} \csc x = \dfrac {d}{dx} (\dfrac{ 1 }{\sin x}) = \dfrac { (\dfrac {d}{dx}1) { \sin x } - 1 (\dfrac {d}{dx} \sin x) } {\sin^2 x} }$
The derivative of 1 is equal to zero. Use the formulae for the
derivative of the trigonometric functions $\sin x$ given by $\dfrac {d}{dx}\sin x = \cos x$ and substitute to obtain
$\displaystyle {\dfrac {d}{dx} \csc x = \dfrac{ (0 - (\cos x) )}{\sin^2 x}}$
Simplify
$\displaystyle {= \dfrac{ - \cos x } {\sin^2 x} = - \dfrac{ \cos x }{\sin x} \dfrac{ 1 }{\sin x} = - \cot x \csc x}$
conclusion
$\displaystyle {\dfrac {d}{dx} \csc x = - \cot x \; \csc x}$
Graph of csc x and its Derivative
The graphs of $\csc(x)$ and its derivative are shown below.
Derivative of the Composite Function csc (u(x))
Let us consider the composite function csc of another function u(x). Use the chain rule of differentiation to write
$\displaystyle \dfrac{d}{dx} \csc (u(x)) = (\dfrac{d}{du} \csc u) (\dfrac{d}{dx} u )$
Simplify
$= - \cot u \csc u \dfrac{d}{dx} u$
Conclusion
$\displaystyle \dfrac{d}{dx} \csc (u(x)) = - \cot u \; \csc u \; \dfrac{d}{dx} u$
Example 1
Find the derivative of the composite csc functions
1. $f(x) = \csc (-x^3+3)$
2. $g(x) = \csc (\cos(x))$
3. $h(x) = \csc (\dfrac{1}{x^2+1})$
Solution to Example 1
1. Let $u(x) = -x^3+3$ and therefore $\dfrac{d}{dx} u = \dfrac{d}{dx} (-x^3+3) = -3x^2$ and apply the rule for the composite csc function given above
$\displaystyle \dfrac{d}{dx} f(x) = - \cot u \csc u \dfrac{d}{dx} u = - \cot (-x^3+3) \csc (-x^3+3) \times (-3x^2)$
$= 3x^2 \; \cot (-x^3+3) \; \csc (-x^3+3)$
2. Let $u(x) = \cos x$ and therefore $\dfrac{d}{dx} u = \dfrac{d}{dx} \cos x = - \sin x$ and apply the above rule of differentiation for the composite csc function
$\displaystyle \dfrac{d}{dx} g(x) = - \cot u \csc u \dfrac{d}{dx} u = - \cot (\cos x) \csc (\cos x) \times (- \sin x)$
$= \sin x \;\cot (\cos x) \; \csc (\cos x)$
3. Let $u(x) = \dfrac{1}{x^2+1}$ and therefore $\dfrac{d}{dx} u = -\dfrac{2x}{(x^2+1)^2}$ and apply the rule of differentiation for the composite csc function obtained above
$\displaystyle \dfrac{d}{dx} h(x) = - \cot u \csc u \dfrac{d}{dx} u = - \cot (\dfrac{1}{x^2+1}) \csc (\dfrac{1}{x^2+1}) \times (-\dfrac{2x}{(x^2+1)^2})$
$= \dfrac{2x}{(x^2+1)^2} \; \cot (\dfrac{1}{x^2+1}) \;\csc (\dfrac{1}{x^2+1})$ | 2022-12-03 03:20:47 | {"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": 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": 0.9079006314277649, "perplexity": 366.38216630208234}, "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-2022-49/segments/1669446710918.58/warc/CC-MAIN-20221203011523-20221203041523-00601.warc.gz"} |
http://locke.citizendium.org/wiki/Absorbing_element | # Absorbing element
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In algebra, an absorbing element or a zero element for a binary operation has a property similar to that of multiplication by zero.
Formally, let ${\displaystyle \star }$ be a binary operation on a set X. An element O of X is absorbing for ${\displaystyle \star }$ if
${\displaystyle O\star x=O=x\star O\,}$
holds for all x in X. An absorbing element, if it exists, is unique. | 2022-10-06 03:27:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "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": 0.42271989583969116, "perplexity": 657.6359326779572}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "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-2022-40/segments/1664030337723.23/warc/CC-MAIN-20221006025949-20221006055949-00684.warc.gz"} |
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# If A < 0, 10 < B < 30, and 50 < C < 80, what is the relative order of
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If A < 0, 10 < B < 30, and 50 < C < 80, what is the relative order of the reciprocals 1/A, 1/B, and 1/C?
A. 1/A < 1/B < 1/C
B. 1/A < 1/C < 1/B
C. 1/C < 1/A < 1/B
D. 1/C < 1/B < 1/A
E. 1/B < 1/C < 1/A
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If A < 0, 10 < B < 30, and 50 < C < 80, what is the relative order of [#permalink]
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20 Nov 2017, 12:15
Bunuel wrote:
If A < 0, 10 < B < 30, and 50 < C < 80, what is the relative order of the reciprocals 1/A, 1/B, and 1/C?
A. 1/A < 1/B < 1/C
B. 1/A < 1/C < 1/B
C. 1/C < 1/A < 1/B
D. 1/C < 1/B < 1/A
E. 1/B < 1/C < 1/A
A < 0, 10 < B < 30, and 50 < C < 80
Let A = -5
Let B = 20
Let C = 70
$$\frac{1}{A}=-\frac{1}{5}$$
$$\frac{1}{B}=\frac{1}{20}$$
$$\frac{1}{C}=\frac{1}{70}$$
$$-\frac{1}{5} < \frac{1}{70} < \frac{1}{20}$$
$$\frac{1}{A} < \frac{1}{C} < \frac{1}{B}$$
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If A < 0, 10 < B < 30, and 50 < C < 80, what is the relative order of [#permalink]
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20 Nov 2017, 12:19
Bunuel wrote:
If A < 0, 10 < B < 30, and 50 < C < 80, what is the relative order of the reciprocals 1/A, 1/B, and 1/C?
A. 1/A < 1/B < 1/C
B. 1/A < 1/C < 1/B
C. 1/C < 1/A < 1/B
D. 1/C < 1/B < 1/A
E. 1/B < 1/C < 1/A
B.$$\frac{1}{A} < \frac{1}{C} < \frac{1}{B}$$
$$A$$ is negative, then would be the smallest, reciprocal or not. $$C > B$$, the reciprocal would have opposite relative order.
If A < 0, 10 < B < 30, and 50 < C < 80, what is the relative order of &nbs [#permalink] 20 Nov 2017, 12:19
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2018-07-21 19:10:31 | {"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": 0.3958636522293091, "perplexity": 6376.283371587871}, "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-2018-30/segments/1531676592654.99/warc/CC-MAIN-20180721184238-20180721204238-00138.warc.gz"} |
https://www.advanced-excel.com/hlookup/ | # HLOOKUP formula
Hlookup refers to Horizontal lookup. Its purpose is to look up a value or text horizontally across a row. When the value is found, it will return a value in another row that corresponds to the column of that value. Assuming that you have a table which presents the ages in the top row (header row) and the left column presents the height. In the table are the weight range that corresponds to the age and the height as shown below:
Using the formula, you can lookup the age (6 mths) across row 1. We will find 6 mths in column E. The formula can return any value below column E, depending on the value given in the formula. Here is how you should input the formula:
1. Select a blank cell.
2. Enter the formula (without the square brackets) “=Hlookup(“6 mths”,\$B1:\$G6,4,false)”.
3. The formula will look for the value “6 mths” in the 1st row \$B1:\$G1.
4. The horizontal position (Column E) will be captured.
5. The number 4 in the formula “….\$B1:\$G6,4,false)” indicates that the result to return, when the value “6 mths” is found, is in row 4 (the result returned is 7.8).
6. The false that follows is a switch, to indicate that the exactly value must be found. Without the false, it will return the closest value that is greater than the value to be found.
vlookup: another useful formula similar to hlookup | 2023-02-04 15:21:02 | {"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": 0.8028367161750793, "perplexity": 881.1710163520338}, "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-2023-06/segments/1674764500140.36/warc/CC-MAIN-20230204142302-20230204172302-00566.warc.gz"} |
http://ncuq.jniu.pw/array-indexing-matlab.html | # Array Indexing Matlab
recfunctions. MATLAB è un sistema interattivo in cui l’elemento di base è un array quindi non richiede il dimensioning. This page shows techniques to extract and view portions of a tall array. An index like (5, 6) selects a single element of an array, but we can also access sections of the matrix, or slices. Matlab has special commands that will do this for you. MATLAB - The for Loop - A for loop is a repetition control structure that allows you to efficiently write a loop that needs to execute a specific number of times. Using logical indexing with structs is always a headache for me. Learn more about matlab, cell arrays MATLAB. logical indexing 3d array with 2d array. matlab® では、配列の場所 (インデックス) に基づいて配列要素にアクセスする方法が主に 3 つあります。これらの方法は、位置によるインデックス付け、線形インデックス付け、論理インデックス付けです。. MATLAB vec-torization Dalle Introduction Creating Vectors Vector Functions Operators Numeric Arrays Testing Logical Indexes Extraction Examples! Function Handles Cell Arrays Strings Inputs/outputs Structs Vectorization in MATLAB Andotherminortipsand tricks Derek J. In Matlab, array entries are always numbered down the rows, moving left to right through the columns. Store the row and column return from find and put it in the match cell array. MATLAB allows for several methods to index (access) elements of matrices and arrays: Subscript indexing - where you specify the position of the elements you want in each dimension of the matrix separately. I realized that many of the postings in the group were about how to manipulate arrays efciently , which was something I had a great interest in. A Structure is a named collection of data representing a single idea or "object". Toggle Main Navigation. Generally to generate a multidimensional array, we first create a two-dimensional array and extend it. In this tutorial, you will discover how to. Tables do not have to have row names, but if you specify them, then you can index into a table by row name. 3 7 12 4 -4 4. However, if the two index do not have the same starting indexing, there would be something wrong. Arrays in Visual Basic. Matlab is a software package that makes it easier for you to enter matrices and vectors, and manipulate them. Linear indexing - where the matrix is treated as a vector, no matter its dimensions. This increases in the following way – the linear index is in superscript: A = \left[\begin{matrix}1^1 & 6^5 & 7^9 & 8^{13} & 2^{17} \\ 1^2 & 5^6 & 6^{10} & 7^{14} & 3^{18} \\ 1^3 & 8^7 & 9^{11} & 2^{15} & 3^{19} \\ 0^4 & 5^8 & 2^{12} & 9^{16} & 9^{20}\end{matrix}\right]. Then it initializes the Value property of each object to the corresponding input. Data is specified as an array of values, and each value is passed as the first argument (d) to selection functions. MATLAB extracts the matrix elements corresponding to the nonzero values of the logical array. The function is easy to use: exindex(arr, s1, s2, ) is equivalent to v(s1, s2, ) where v stands for the virtual extended array. Apply function to each element of array. Content Indexing with Curly Braces, {} Access the contents of cells--the numbers, text, or other data within the cells--by indexing with curly braces. For example, maybe you want to plot column 1 vs column 2, or you want the integral of data between x = 4 and x = 6, but your vector covers 0 < x < 10. Learn more about matlab, cell arrays MATLAB. In the 2-D case with inputs of length M and N, the outputs are of shape (N, M) for ‘xy’ indexing and (M, N) for ‘ij’ indexing. The output is always in the form of a column vector. If the array that is being indexed is a sparse matrix all of the above still applies, except that partially linear indexing doesn't exist for matrices; and of course the result is also sparse. I am trying to learn how the graphic objects work in MATLAB. The file is called by Matlab, and it constructs a second derivative finite difference matrix with boundary conditions. String arrays index in the same way as other MATLAB arrays. Single Subscript Indexing. ' fn(2) ]) would spit out the array. If it is for indexing into. Indexing into Matlab arrays is very often glossed over, but it is actually quite a powerful (and fast) technique. ;Good Good for fast calculations on vectors and matrices. It also uses the GetUpperBound method to determine the number of elements in each dimension of a multidimensional array. You do not need to implement any special methods to provide standard array behavior with your class. matlab Newsgroup [email protected] Reducing the range of any index to a single value effectively eliminates that index. In MATLAB®, the basic data type is a multidimensional array of double precision floating point numbers. Accessing Dynamic Properties in Arrays. Each number in an array has a unique location identified by an index. Array indexing in a for loop on MATLAB. You want the result to be 20x51. CME 102 Matlab Workbook 2008-2009 3/55 1 Matlab Basics 1. You do not need to implement any special methods to provide standard array behavior with your class. The for loop allows us to repeat certain commands. Most expressions take such arrays and return such arrays. As I shown in my example, you can still get the elements of a 2D array using 3d coordinates, and you can still query its size as a 3D array. An introduction to MATLAB MEX-files Maria Axelsson [email protected] Indexing an array with a vector. Python indexing goes as it does in C. This field is always an array of doubles but the arrays are of various lengths. Indexing to cell arrays. Editor/Debugger, Array Editor, Help Browser, etc. The world's most trusted developer resource. So I understand how to write a very basic for loop running through an. Severe equine asthma (sEA), which closely resembles human asthma, is a debilitating and performance-limiting allergic respiratory disorder which affects 14% of horses in the Northern Hemisphere. Index and View Tall Array Elements. Use to create a new dataset array from a subset of ds. Learn more about index, events, ode MATLAB. Vectorized (or Array) Operations. Cell arrays themselves must still be rectangular in any given two dimensions, and since each element is a cell, the array is filled with items that are all the same type. With the default join-by-index, the first element in the data array is passed to the first node in the selection, the second element to the second node, and so on. starts at 1. The file is called by Matlab, and it constructs a second derivative finite difference matrix with boundary conditions. Scribd is the world's largest social reading and. You can also use array constants, values you just enter in the formula bar inside braces: {}. MATLAB - The for Loop - A for loop is a repetition control structure that allows you to efficiently write a loop that needs to execute a specific number of times. Ask Question Asked 11 months ago. 2: not(A) Finds logical NOT of array or scalar input; performs a logical NOT of input array A and returns an array containing elements set to either logical 1 (true) or logical 0 (false). Share or demonstrate solutions to problems. Then, matlab can accept matrices like this as a way of specifying indexes for another matrix of the same size. Furthermore, numpy now provides a new function numpy. For example, consider the 4-by-4 magic square A:. It is only if you subsequently make a change to one of the versions that all the data gets replicated at a new location. RESHAPE and LINEAR INDEXING: Matlab always allows multi-dimensional arrays to be accessed using scalar or linear indices, NumPy does not. Summary: in this tutorial, you will learn about MySQL data types and how to use them effectively in designing database in MySQL. 12/06/2017; 28 minutes to read +6; In this article. This MATLAB function returns a logical array containing 1 (true) where the elements of A are NaN, and 0 (false) where they are not. Pointer to an array of characters where extracted characters are stored as a c-string. matlab Newsgroup [email protected] Use the head function to extract the first rows in a tall array. MATLAB - The for Loop - A for loop is a repetition control structure that allows you to efficiently write a loop that needs to execute a specific number of times. That doesn't seem that bad, but I'm going to be doing plenty of plotting and this can easily turn into hell lol. Improving the Speed of MATLAB Calculations. Learn more about strfind, strings, cell array, indexing, overcoming. sort Sort in ascending order. This is the first post in a short series on index techniques that are particularly useful for image processing in MATLAB. The Attempt at a Solution Hello, I am having some confusion over the notation used in matlab. It is a best practice to use lowercase only when. Index 44 1 Introduction 1. Arrays of objects behave much like numeric arrays in MATLAB. streamsize is a signed integral. In Matlab, arrays are dynamic in size, meaning, whenever you need a larger array, you can just index based on this larger number and Matlab will automajically resize your original array to be the new bigger size. An empty object defines the class of an array. However, exindex is more flexible than any of these; for example, the array can be cropped as well as extended, and indeed arbitrary indexing into the virtual array is allowed. How to fix: Index exceeds array bounds?. 0 wherever the logical mask MaskForA was a 1 and it will contain a 0. Viewed 124 times 2. Indices are provided as (row, column). AFIK, when I use the plot function it creates figure, axis, line objects and then sets the property of each object accordingly. When you want to access selected elements of an array, use indexing. IndexOf(Array, Object), to determine the first occurrence of the string "the" in a string array. If one element in a regular array is a string then all elements must be a string. All of the loop structures in matlab are started with a keyword such as for, or while and they all end with the word end. Categorical arrays that are not ordinal — Can have different sets of categories, and isequal compares the category names of each pair of elements. Secondly, indexing is done using brackets, so you can see the difference between an indexing operation and a function call. n Maximum number of characters to write to s (including the terminating null character). Indexing with a Single Index. See Linear Indexing for for an introduction to this topic. This method is known as linear indexing. Indexing an array with a vector. Learn more about arrayfun, array, structures, find, indexing. Store the row and column return from find and put it in the match cell array. Another method for accessing elements of an array is to use only a single index, regardless of the size or dimensions of the array. A 3-D array, for example, uses three subscripts. starts at 1. Single Subscript Indexing. For example, if you have a vector with 5 elements in it, and you ask to return the 6th element, then the 6 would be out of bounds in that case. Tall arrays are too large to fit in memory, so it is common to view subsets of the data rather than the entire array. In fact what is happening is that "single" command is applied over and over again to every element of the array. Furthermore, numpy now provides a new function numpy. When you want to access selected elements of an array, use indexing. Access Elements of Java Array MATLAB Array Indexing. Matlab Image and Video Processing Vectors and Matrices m-Files (Scripts) For loop Indexing and masking Vectors and arrays with audio files Manipulating Audio I Manipulating Audio II Introduction to FFT & DFT Discrete Fourier Transform (DFT) Digital Image Processing 1 - 7 basic functions Digital Image Processing 2 - RGB image & indexed image. MATLAB automatically builds the array as you go along. Every MATLAB user is familiar with ordinary. MATLAB (matrix laboratory) is a multi-paradigm numerical computing environment and proprietary programming language developed by MathWorks. For logical indexing, if L is a logical vector with the same dimensions as A, you can always treat L as being equivalent to the indices returned by. 2: not(A) Finds logical NOT of array or scalar input; performs a logical NOT of input array A and returns an array containing elements set to either logical 1 (true) or logical 0 (false). Such increased complexity and augmented. x is only available in jre8; Microseconds in timestamps might be truncated when transferred in binary mode. Matlab Sect 24 Finding the Length, Size, Sum, and Number of Elements in a Matrix - Duration: 9:44. It does work! The linear index of 5 is 4 not 5! But if you want to extract row and column you just need to adapt the code Jan gave you. There is a variable in the file called decision. Secondly, indexing is done using brackets, so you can see the difference between an indexing operation and a function call. For example, you can count the number of characters that are contained in a range of cells, sum only numbers that meet certain conditions (such as the lowest values in a range or numbers that fall between an upper and lower boundary), and sum. Array dimensions. field, if s is a structure array. Common Causes include: You are attempting to replace a section of one matrix with another and the size of the index vector or vectors used to specify the section of the destination matrix to be replaced does not match that of the source matrix. MATLAB For Engineers 27,408 views. I believe the error is due to the fact that logical indexing can be used either on one dimension of 'taux' or on all its dimensions. sort Sort in ascending order. MATLAB (“MATrix LABoratory”) is a tool for numerical computation and visualization. sophisticated (broadcasting) functions. Extract Top Rows of Array. MATLAB uses a default object to initialize the empty elements of an array of handle objects. Repetition: for Loops when you use the loop control variable exclusively as an index into an array, MATLAB's vector match operations run much. Matlab makes it easy to create vectors and matrices. C# array of object indexing in matlab. Extending and Embedding tutorial for C/C++ programmers. See the line "Matrix Indexing in MATLAB", the Matlab Digest, by Steve Eddins and Loren Shure, September, 2001. When you want to access selected elements of an array, use indexing. You can leave a response, or trackback from your own site. As you saw, it can be used to create arrays, and it can also be used to index or slice arrays. The output is always in the form of a column vector. Arrays may be 1-dimensional (vectors), 2-dimensional (matrices) or even higher dimensional. Repetition: for Loops when you use the loop control variable exclusively as an index into an array, MATLAB's vector match operations run much. I used to program in Java and Python earlier, but recently have started using MATLAB for lots of stuff (specifically computer vision algorithms). This method is known as linear indexing. The file is called by Matlab, and it constructs a second derivative finite difference matrix with boundary conditions. Each m-file contains exactly one MATLAB function. Indexing arrays. Customize indexed reference and assignment behavior for object arrays. Then it initializes the Value property of each object to the corresponding input. Thus, a collection of MATLAB functions can lead to a large number of relatively small files. end Last index in an indexing expression. JSON objects are surrounded by curly braces {}. Improving the Speed of MATLAB Calculations. maximum elements. Otherwise, that element is set to 0. These indexing tricks come in handy to avoid "for" loops and in vectorizing code. Lets start with a small matrix. For example, if you have a vector with 5 elements in it, and you ask to return the 6th element, then the 6 would be out of bounds in that case. Later I'll cover linear indexing, and then a technique I like to call neighbor indexing. Learn more about find, loop, cell arrays, indexing MATLAB. Use the head function to extract the first rows in a tall array. The shape of the array to use for unraveling indices. MATLAB - The for Loop - A for loop is a repetition control structure that allows you to efficiently write a loop that needs to execute a specific number of times. Breaking with tradition, I'll use a Hilbert matrix instead of a magic square: A = hilb(5). B = max(A, [], dim) CS 1173: MATLAB max function. MATLAB returns an array that matches the elements of the array X, element-by-element holding 1s where the matching values in X are the desired values, and 0s otherwise. For sequences, uniform selection of a random element, a function to generate a random permutation of a list in-place, and a function for random sampling without replacement. Matlab stores matrices in column-major order. An email has been sent to verify your new profile. Every variable in MATLAB® is an array that can hold many numbers. Specifically, this is about finding a string within another cell array of strings, where the thing I'm really interested in is the index of the cell array where the reference string occurs. A Structure is a named collection of data representing a single idea or "object". There is a variable in the file called decision. This method is known as linear indexing. That means, you specify each position in the matrix with a single number. Indexing vectors and arrays in Matlab There are times where you have a lot of data in a vector or array and you want to extract a portion of the data for some analysis. Matrix Indexing and Subscripting in IDL. Vectorized (or Array) Operations. Linear indices are common in Matlab programs, e. 2: not(A) Finds logical NOT of array or scalar input; performs a logical NOT of input array A and returns an array containing elements set to either logical 1 (true) or logical 0 (false). The following code examples show how to read characters synchronously or asynchronously from a string. Normally I prefer logical indexing but couldn't find a nice solution to access the substrings within a cell array. Ciò permette la risoluzione di molti problemi di calcolo tecnici, in particolare quelli con le formulazioni vettorali e matriciali, attraverso algoritmi molto più semplici e snelli rispetto a quelli che sarebbero necessari in un. However, the contents of each cell can be any MATLAB array, including. Array indexing in a for loop on MATLAB. Sorting Sorting the data in an array is also a valuable tool, and MATLAB offers a number of approaches. Viewed 124 times 2. Every MATLAB user is familiar with ordinary. The N-dimensional array (ndarray)¶An ndarray is a (usually fixed-size) multidimensional container of items of the same type and size. Index and View Tall Array Elements. To use the indexing ideas effectively you need to be able to create 'mask' matrices efficiently, and manipulate arrays. tools for integrating C/C++ and Fortran code. Creating Arrays The basic data structure in Matlab is the two-dimensional array. This includes when arrays. Sort the rows of a Matlab matrix according to one of the columns. Another indexing variation, logical indexing, has proven to be both useful and expressive. GNU Octave Scientific Programming Language. if I can access the data through an index point, then given an arbitrary number of arrays stored in a structure I can easily start to cycle through them with a for loop where. Arrays of objects behave much like numeric arrays in MATLAB. For example, the sort function sorts the elements of each row or column of a matrix separately in ascending or descending order. The range searched is [first,last), which contains all the elements between first and last, including the element pointed by first but not the element pointed by last. Indexing to cell arrays. How to Contact The MathWorks: www. As of MATLAB 2016b, there. Inside a structure are a list of fields each being a variable name for some sub-piece of data. This function allows safe conversion to an unstructured type taking into account. Breaking with tradition, I'll use a Hilbert matrix instead of a magic square: A = hilb(5). The following list contains syntax examples of how to determine the dimensions (index of the first element, the last element or the size in elements). Mar 31, 2017 · Suppose I have an array, a = [2 5 4 7]. Categorical arrays that are not ordinal — Can have different sets of categories, and isequal compares the category names of each pair of elements. This article covers how to create matrices, vectors, and cell arrays with the programming software MATLAB. Sort the rows of a Matlab matrix according to one of the columns. Homework Equations 3. PS: the numbers won't be this linear so I can't use linspace command it's just an example. Example: Read characters synchronously. CS 211 Lesson 8. Concatenation Of Arrays (Using brackets operator) The brackets operator ( [] ) ,. dimension to maximize over. MATLAB programs are stored as plain text in files having names that end with the extension . Indices are provided as (row, column). ;Good Good for fast calculations on vectors and matrices. Content Indexing with Curly Braces, {} Access the contents of cells--the numbers, text, or other data within the cells--by indexing with curly braces. Learn more about find, loop, cell arrays, indexing MATLAB. C# array of object indexing in matlab. For example, consider the 4-by-4 magic square A:. This field is always an array of doubles but the arrays are of various lengths. Matrix Indexing and Subscripting in IDL. If m = n, eye(n) can be used. What happens is I often have an array representing positions on a line. Structures are similar to arrays in that. if I can access the data through an index point, then given an arbitrary number of arrays stored in a structure I can easily start to cycle through them with a for loop where. I've come to a wall where I'm not sure if there is an "elegant" way to do a one-liner to get the desired result. The most frequent use of logicals for us, however, will be in indexing. For anything in a computer more complicated than a list of numbers, structures can be used. Each m-file contains exactly one MATLAB function. Hi guys, Matlab novice here : "Index in position 1 exceeds array bounds must not exceed 10" from the code below. Giving the string ‘ij’ returns a meshgrid with matrix indexing, while ‘xy’ returns a meshgrid with Cartesian indexing. A class constructor can create an array by building the array and returning it as the output argument. For example, to access the contents of the last cell of C, use curly braces. • What is the length of the array? • What is the length of the word at index 2? words! this! that! what! 0 1 2. It is used to create vectors, subscript arrays, and specify for iterations. Situation changes significantly, when array is. A range of cells or an array constant. Logical indexing in cell array. Matlab post There are times where you have a lot of data in a vector or array and you want to extract a portion of the data for some analysis. Toggle Main Navigation. Each value in fieldName is 20x51, and you want to take the first column of each, so that would be 20x1 for each. Calling empty with no arguments returns a 0-by-0 empty array. prod Product of elements. The regular array can hold strings; however, the string in each element must be the same length. How to: Read characters from a string. When you refer to the elements of a MATLAB matrix with a single subscript, MATLAB returns a single element of. However MATLAB has indexing of arrays beginning from 1 instead of 0, which is the norm in almost every programming languages I have encountered so far. Uppercase and Lowercase In MATLAB code, use an exact match with regard to case for variables, files, and functions. (The same array objects are accessible within the NumPy package, which is a subset of SciPy. The for loop allows us to repeat certain commands. Another method for accessing elements of an array is to use only a single index, regardless of the size or dimensions of the array. MATLAB is supported on Unix, Macintosh, and Windows environments; a student version of MATLAB is available for personal computers. How can I index a MATLAB array returned by a function without first assigning it to a local variable? Ask Question Asked 9 years, 1 month ago. Furthermore, numpy now provides a new function numpy. You can use constants in your array formulas or by themselves. This article discusses these and other strategies to improving the speed of Matlab code. Some better (see Anders Kaseorg), some appealing to convention and some through poor analogies. You can create common arrays and grids, combine existing arrays, manipulate an array's shape and content, and use indexing to access array elements. In average, complexity of such an algorithm is proportional to the length of the array. RESHAPE and LINEAR INDEXING: Matlab always allows multi-dimensional arrays to be accessed using scalar or linear indices, NumPy does not. Access dynamic properties in object arrays by referring to individual objects. Say I have a structure array A with many fields, one of which is, say, a1. Array Indexing - MATLAB & Simulink. All that is required is to extend this to the rest of the possible GPA's - the full code is in the section below, along with an image of the graph it will create. In a Java program, the syntax is A[row-1][column-1]. Operations on the 2-D instances of these arrays are designed to act more or less like matrix operations in linear algebra. More Array Subscripting and Indexing in MATLAB Array indices must be integers and can refer to: A single element; A continuous range of elements. Indices are provided as (row, column). Array indexing refers to any use of the square brackets ([]) to index array values. Editor/Debugger, Array Editor, Help Browser, etc. if I can access the data through an index point, then given an arbitrary number of arrays stored in a structure I can easily start to cycle through them with a for loop where. Tall arrays are too large to fit in memory, so it is common to view subsets of the data rather than the entire array. when the match found that images has to be written in another folder. This MATLAB function applies the function func to the elements of A, one element at a time. Moreover, the fundamental operators (e. Finding the indices of the elements of one array Learn more about matlab function, vectorization find the index, idx into A of the element of B so that. Unlike many other programming languages, arrays in MATLAB are 1-indexed, meaning the array starts counting at 1 (rather than starting counting at 0, which is how many other popular languages work). Cell arrays have fewer limitations than regular arrays. Vectorized (or Array) Operations. CUED: Matlab vectorisation tricks Matlab vectorisation tricks Some basic tips on speeding up matlab code and on exploiting vectorisation are mentioned in the Optimisation section of our matlab page. MATLAB is supported on Unix, Macintosh, and Windows environments; a student version of MATLAB is available for personal computers. Array indexing in a for loop on MATLAB. MATLAB extracts the matrix elements corresponding to the nonzero values of the logical array. C# array of object indexing in matlab. Prentice Hall (2004). The Structure Data Type in Matlab. Visit the post for more. I think it is fairly obvious, but I don't see it. Duane Hanselman & Bruce Littlefield, Mastering Matlab 7. MATLAB uses a default object to initialize the empty elements of an array of handle objects. RESHAPE and LINEAR INDEXING: Matlab always allows multi-dimensional arrays to be accessed using scalar or linear indices, NumPy does not. Provides interactive learning of programming foundations and MATLAB®. If one element in a regular array is a string then all elements must be a string. find() on a matrix returns them, whereas NumPy's find behaves differently. For example, maybe you want to plot column 1 vs column 2, or you want the integral of data between x = 4 and x = 6, but your vector covers 0 < x < 10. More than 50 million unique visitors come to Stack Overflow every month, giving us unique insights into this audience. Another indexing variation, logical indexing, has proven to be both useful and expressive. Set array, specified as a numeric array, logical array, character array, string array, categorical array, datetime array, duration array, cell array of character vectors, table, or timetable. If you want to create a row vector, containing integers from 1 to 10, you write −. For example, the ObjectArray class creates an object array that is the same size as the input array. How can I index a MATLAB array returned by a function without first assigning it to a local variable? Ask Question Asked 9 years, 1 month ago. Arrays of structures indexing. empty(m,0) to create an m-by-0 array of the ClassName class. Every variable in MATLAB® is an array that can hold many numbers. Editor/Debugger, Array Editor, Help Browser, etc. The course is designed for novice users to learn the basic of MATLAB® and how to apply that to solving engineering problems. Use ClassName. MATLAB classes support object array indexing by default. You should glance at Yahoo’s home page and note how they write news headlines to grab viewers to click. have to compare the values in the array with the name of the images in the folder. Send comments to wsr nih. Because MATLAB was designed for mathematicians, by mathematics people. se 2007-10-22 Maria Axelsson, Centre for Image Analysis MATLAB MATLAB (by Mathworks) is a good development platform for image analysis algorithms. indexing cell array of arrays. 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https://philosophy.stackexchange.com/questions/30573/symbolic-logic-proof-leprechauns-exist?answertab=oldest | # Symbolic Logic Proof: Leprechauns Exist?
I am reviewing a study guide for an introductory logic course (basic predicate, syllogistic etc.). The problem asks me to symbolize that "leprechauns exist" and prove that it is a logical truth and then critique your proof.
I decided the best symbolization was:
(∃x)(Lx)
where Lx = x is a leprechaun
I am unsure if this is the proper symbolization (would (∃x)(x=L) be better?) and unsure how one can even begin to prove this is a logical truth.
There are really so many way to do this and which one you choose largely depends on your personal preference. Here are some:
# Predicate
(∃x)(Lx), where Lx stands for 'x is a leprechaun' (this is your suggestion).
¬(∀x)(¬Lx), less expressive than the first, just to show that there are so many ways to express this.
# Set
(∃x)(x ∈ L), where L is the set of leprechauns
This is equivalent and very similar to saying L ≠ ∅; the set of leprechauns isn't empty.
Personally, I often prefer to use this last way, L ≠ ∅, I find it the most expressive. It isn't very common in logic circles though, I have the impression.
I wouldn't use x=L, because different leprechauns are different. L can be the set of leprechauns, or the predicate of being a leprechaun, but not simply 'a leprechaun'. Because then we could say that x ≠ L, however, x = L2 (another leprechaun).
# Sets and predicates
They're the same, really. Defining a predicate Lx is implicitly defining a set L = {x | Lx} (all x for which Lx is true). And defining a set L is implicitly defining a membership predicate Lx = x ∈ L.
# Proof
I am [...] unsure how one can even begin to prove this is a logical truth.
The typical way to prove that there exists (at least) one leprechaun is by pointing at it. In the case of leprechauns that may be a little difficult. You could make a claim that they exist in your head, and therefore really exist.
• L ≠ ∅ was not used in the undergraduate logic course I took. I'm guessing the logic taught in most undergraduates in philosophy is equally truncated – virmaior Dec 14 '15 at 10:14
• That's a pity, it is shorter and (in my opinion) more expressive. – Keelan Dec 14 '15 at 10:36
• I'm familiar with the empty set, so no problem there, and thanks for confirming my symbolization and rejection of x=L. But yes, I can't imagine proving leprechauns exist logically. Exhibiting the existence of a leprechaun by pointing it out is surely empirical. I'm wondering if this is meant to be a poorly phrased Curry Paradox question? – user115411 Dec 14 '15 at 18:25
I agree with your first symbolization :
(∃x)(Lx),
because it seems more "natural" to assume that "leprechaun" denotes a species and not an individual; in fact, you are saying : "leprechauns exist".
But in this way we cannot prove anything; the standard semantics for first-order logic assumes that any interpretation has a not empty domain M.
This means that the formula (∃x)(x=x) is valid.
The semantics requires also that to each n-place predicate symbol P an n-ary relation P* on the domain, i.e. a set of n-tuples of members of the domain; for n=1, this must be a subset of the domain.
But nothing prevents that the said subset is the empty set; thus we cannot prove that "leprechauns exist".
Things are different if we use an individual constant l; in this case we can prove, starting from the equality axiom : x=x, the formula :
(∃x)(x=l);
in this case, we are consistent with the semantical specifications, because for each constant symbol c, the interpretation specifies a member c* of the domain M.
This means that, having and individual constant in our language, amounts to assuming that this symbol is a name denotong an object of the domain, and thus assuming the existence of the said object.
The problem asks me to symbolise that 'Leprachauns exist'
If this is all that is asked for; the simplest is:
L: Leprachauns exist
We use the letter L to remind ourselves that Leprachauns are involved in this sentence - it's a mnemonic device.
I'm not sure even how to prove this is a logical truth
Which is already quite close to a good critique; I mean if someone came to you and said 'I saw some leprechauns in the garden' - you'd think perhaps they were starting on a shaggy dog story.
You already know from experience there aren't any. I mean, it's not a logical truth - so your intuition and judgement was already on the right track ...
• Why the down-vote? The question specifically says using syllogistic or predicate logic; one usually doesn't go for the more complex option if a simpler one is possible; and that it's not a logical truth is surely obvious to the scientifically and logically literate; as a mere exercise in using the rules of logic formally expressed one might say more - but there isn't sufficient said in the question to go on. – Mozibur Ullah Dec 14 '15 at 14:45 | 2019-08-21 12:30:12 | {"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": 0.8654145002365112, "perplexity": 868.1260475712531}, "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-2019-35/segments/1566027315936.22/warc/CC-MAIN-20190821110541-20190821132541-00362.warc.gz"} |
https://uoftcoders.github.io/rcourse/lec06-pop-models.html | ## Lesson preamble
In developing this series of lectures I have referred heavily to the notes from Nonlinear Physics, PHY1460, which I would encourage anyone who is interested in getting more in depth on the quantitative analysis of dynamical systems to take or audit. I also refer to “Nonlinear Dynamics and Chaos” by Steven Strogatz, a very good textbook which you might find helpful. You can find the pdf online, it’s available as an e-book from the UofT library, and it’s available in the Physics library as a hard copy course reserve.
### Learning objectives
• Appreciate the purpose and usefulness of modelling real-world systems
• Understand the use of differential equations to model populations
• Qualitatively analyze one-dimensional population models by hand and in R
• Find fixed points graphically in R
• Analyze stability graphically in R
• Draw one-dimensional phase portraits by hand
### Lesson outline
Total lesson time: 2 hours
• Introduction to modelling (20 min)
• Setting up a model (40 min)
• Qualitative analysis of one-dimensional models (60 min)
### Setup
• Install ggplot2: run install.packages('ggplot2') in the R console.
## Introduction - why model?
1. Models can simplify a complex system and help us identify what’s important.
• Biology, as we all know, is extremely complicated.
• Models summarize what we think we know about a system, force us to identify the important parameters, and encourage us to distill them down to a manageable number of parameters.
Note: that’s not always how modelling is done. Sometimes you just add everything you can think of to your very complicated model, and you hope by doing that, your model will be more accurate and make better predictions. But your model will likely be too complicated to solve analytically. (Analytical = a version you can leave with variables instead of putting numbers in. In R we will mostly be simulating.) An example of a discipline where the models tend to be complicated is systems biology - models of celluar pathways often try to include as many components as are known, i.e. taking all the components from the following picture and creating a corresponding model:
• Data don’t always fit theoretical models, but it is helpful to interpret data with a larger theoretical framework in mind.
2. Models can show us something we didn’t expect, that we couldn’t intuit from just thinking about the problem.
• if your model is simple enough, it can be predictive in a broadly applicable way. We’ll talk about some famous ecological models, and the assignments will include more examples. I will also include some famous examples from physics, where this field is known as nonlinear dynamics. Many of these classic models are classic because their results weren’t obvious before the model was created.
• example: Lorenz model of the atmosphere which exhibits chaos (video). The equations for this system are simple, but Lorenz was surprised to find that the behaviour of the system could be very different for starting conditions that were very similar. I wrote a quick simulation in R which you can play around with, it’s at the bottom of these notes.
• example: Lotka-volterra predator-prey model. This model predicts never-ending oscillations in the populations of predator and prey species. A simulation for this is also at the bottom of these notes.
• What else is a reasonable thing that could happen to a population of predator and prey?
Aside: I do theoretical biophysics, which is quite similar to theoretical ecology in my case, so all of what I’ll be telling you about is techniques and concepts that I have found useful or use very frequently and that I think are an important part of the theorist / data scientist toolkit.
3. In the next few lectures we will talk about models and about using R to analyze models. We will first do this qualitatively (no math), then quantitatively. We will talk about fitting models to data, and by the end I hope that you will be able to incorporate modelling into your project, whether a model you design yourself or an existing ecological model.
## Setting up a model
Let’s look at some examples of one-dimensional models and talk about what we can learn from these models qualitatively, without doing any calculations.
Dimension: the number of variables in your model. (Not the number of parameters.
Variables are what track the species you’re interested in, and parameters are numbers that specify the details of the model.)
Let’s write down a model for the growth of a species. We want to predict how many organisms are present in the population at time $$t$$, and to do that we need to write down an equation that describes how the population changes with time. We will start by writing a recursion relation, which tells us how to update the population size at time $$t+ \Delta t$$ from the population size at time $$t$$:
$N_{t+\Delta t} = N_t + \Delta N$
$$\Delta N$$, the change in the population size, will depend on the specifics of our model. Let’s rearrange this expression into the definition of a derivative by moving $$\Delta N$$ over and dividing both sides by $$\Delta t$$, then taking the limit of $$\Delta t$$ going to $$0$$:
$\frac{N_{t+\Delta t} - N_t}{\Delta t} = \frac{\Delta N}{\Delta t}$ $lim_{\Delta t \to 0}\frac{N_{t+\Delta t} - N_t}{\Delta t} = \frac{dN}{dt}$
If you’ve taken calculus, you will recognize this as the first derivative of $$N$$ with respect to $$t$$. If your calculus is foggy in your memory, when you see this notation think “change in $$N$$ per change in $$t$$. $$dN$$ is the notation commonly used to mean”a small amount of $$N$$," so this expression mathematically describes by what small amount $$N$$ will change in a small amount of time $$dt$$. This is also sometimes written as $$\dot{N}$$. I will usually use $$dN/dt$$ but you may see the dot notation in the literature. A single dot is shorthand for “first derivative”.
Biologically, $$dN/dt$$ is the growth rate or rate of change of a population with $$N$$ organisms at time $$t$$.
What should the growth rate of our population depend on? Let’s brainstorm some things. There are no wrong answers here: anything you can think of that makes biological sense is a reasonable thing to include in a model, although after brainstorming we will write something very simple as an example.
Let’s make some sample data to help us come up with a simple model.
For bacteria growing in a flask with plenty of food, we know that they divide in two periodically. If we take one unit of time to be the time between divisions, we can assume that the population approximately doubles at each time point.
library(ggplot2) # ggplot2 includes the qplot function
# qplot produces plots that are similar to ggplot, but you don't need a dataframe to use it.
# We could use ggplot instead by first making a dataframe out of our data:
# data <- data.frame(times, population_size)
times <- seq(0, 10, by = 0.2)
population_size <- 2 ^ times
qplot(times, population_size) +
geom_line(aes(x = times, y = population_size))
How do we write this assumption as a general model? Let’s calculate the difference in $$N$$ at each time point and plot this vs. $$N$$ at the previous time point.
# last N-1 points minus first N-1 points
N_diff <- tail(population_size, -1) - head(population_size, -1)
geom_line(aes(x = head(population_size, -1), y = N_diff))
This is a straight line, which tells us that the change in $$N$$ is proportional to $$N$$, and in fact because of the way we defined time, the change in $$N$$ is exactly $$N$$.
Now we have our model: this model describes exponential growth of a population and is commonly used for microbes growing with plenty of food.
$\frac{dN}{dt}=N$
This is a differential equation: an equation that describes how some function of $$N$$ is related to derivatives of $$N$$. (In this case, and probably in this entire course, we’ll just be using the first derivative.)
If we wanted time to be something else, like minutes instead of generations, we could add a constant of proportionality that captures how often our bacteria divide:
$\frac{dN}{dt} = rN$
This says that the population will increase by $$r N dt$$ in the small time $$dt$$, or the population increases by $$N dt$$ in time $$dt/r$$.
The process we just used to get this model is probably not what you’d do in practice - I just wanted to illustrate the connection between exponential growth and this differential equation.
If $$r$$ is made larger, does the population grow faster or slower?
## Analyzing models (qualitatively at first)
When you analyze a model, the exact questions you answer will depend on the model. But in general, the following questions are ones you will try to answer for any model.
1. What are the fixed points of the model? (For which sizes of the population(s) does the population size remain constant in time?)
2. Which of the fixed points are stable? (If the population is slightly different from the fixed point, will it go towards it or away from it?)
3. How do the above depend on parameters of the model?
We will see how to answer these questions quantitatively soon. But let’s start by looking at them qualitatively. We will draw a phase portrait of this system, and the techniques we use will be applicable to almost any model you might come across.
### The exponential growth model
Let’s start answering these questions qualitatively for the exponential growth model. There’s a lot of information in the differential equation we can qualitatively read off. If we plot $$dN/dt$$ vs $$N$$, the places where $$dN/dt = 0$$ are the fixed points of our system. Let’s do this for the exponential growth model.
N_seq <-
seq(-20, 20) # make a sequence of N values to plot dN/dt against
dN_dt <- N_seq # assume r = 1
qplot(N_seq, dN_dt) +
geom_line(aes(x = N_seq, y = 0)) # a line at zero for visual aid
We can also tell whether the fixed points are stable or unstable by looking at the sign of $$dN/dt$$ on either side of the fixed point. For a one-dimensional system, our phase portrait is a line of $$N$$. We mark the fixed point(s) with circles, and we draw an arrow to the left wherever $$dN/dt$$ is negative, and an arrow to the right wherever $$dN/dt$$ is positive in the regions between fixed points.
This model is a bit boring - there’s only one fixed point, and it’s unstable. This makes sense - we are talking about exponential growth, after all. There’s no upper limit.
#### Challenge
Find the fixed point(s) of the following differential equation.
$\frac{dN}{dt} = N - a$
• Choose a value for the parameter $$a$$. Can you think of anything that $$a$$ might represent in a real population? Are there any values of $$a$$ that don’t make biological sense?
• Plot $$\frac{dN}{dt}$$ vs. $$N$$, like we did earlier.
• Sketch a phase portrait. Are the fixed point(s) stable or unstable?
### The logistic model
Of course in real life there must be an upper limit to growth, otherwise the surface of the earth would be covered in bacteria several feet thick in a matter of hours. Let’s do a quick calculation for fun - how long would it take for a single bacterial cell undergoing exponential growth to take over the world?
cell_volume <- 0.6 #micrometres cubed - E. coli cell volume is ~0.6 to 0.7 micrometres^3
growth_rate <- 3 # let's say cells divide three times per hour - every 20 minutes (max growth rate of E. coli in good food conditions)
N_t <-
function(N0, r, t) {
# Calculate change in population size with time
# r: is growth rate (divisions per hour)
# t: time in hours
# N0: starting population size
return(N0 * 2 ^ (r * t - 1)) # the -1 is because the first doubling is included with the factor of 2
}
radius_earth <- 6000 # approximately 6000 km
SA_earth <- 4 * pi * radius_earth ^ 2 # surface area of earth in km^2. pi is a predefined variable in R.
surface_covered <-
function(N0, r, t) {
# Calculate the height of the layer of evenly spread out bacteria on the earth's surface
#
# r : growth rate as divisions per hour
# t : time in hours
# N0 : starting population size
bac_volume <-
N_t(N0, growth_rate, t) * cell_volume # in micrometres cubed
height_km = bac_volume * (1e-27) / SA_earth # assume the earth is flat compared to the height of our surface - volume is surface area x height of surface above earth. This is height in km.
# surface_volume = height*SA_earth
# convert bacteria volume to km cubed by dividing by 10^-27 (I googled this conversion).
height_m <- height_km * 1000 # 1000 m in km
return(height_m)
}
# plot
t <- seq(0, 38, 0.3)
qplot(t, surface_covered(2, growth_rate, t), log = 'y', ylab = 'Height above surface (m)')
#yikes. In less than two days, we would be submerged in bacteria.
Let’s look at another model, the logistic growth model, which most of you will have seen before:
$\frac{dN}{dt} = rN\left(1-\frac{N}{K}\right)$
$$N$$ is the number of organisms, $$r$$ is the growth rate, and $$K$$ is the carrying capacity of the system.
Notice that when $$N$$ is much smaller than $$K$$, this reduces to the exponential growth model we just looked at since $$N/K \approx 0$$.
Let’s again plot $$dN/dt$$ vs. $$N$$ to find the fixed points:
N_seq <-
seq(-20, 100) # make a sequence of N values to plot dN/dt against
logistic_eqn <- function(N, r, K) {
# calculate dN/dt for the logistic equation
return(r * N * (1 - N / K))
}
dN_dt <- logistic_eqn(N_seq, r = 0.5, K = 80) # notice the vectorized implementation
qplot(N_seq, dN_dt) +
geom_line(aes(x = N_seq, y = 0)) # a line at zero for visual aid
The fixed points are at $$N=0$$ and $$N=80$$. If we change $$K$$, we can figure out that the second fixed point is actually just $$N=K$$.
Now we draw the phase portrait (on blackboard):
As with the exponential growth model, the fixed point at $$0$$ is unstable, but now there is also a fixed point at $$K$$ which is stable. This means that at longer times, the system will go towards $$N=K$$.
### Note: the importance of stability
Why is stability important? The real world is noisy, and while mathematically an unstable fixed point is still a fixed point (if you were exactly there you would stay there forever), in reality there will always be small perturbations (perhaps a member of the population dies or migrates in) that will move you away from the fixed point. The important thing to know, then, is whether the population will remain close to that fixed point or move away to a different point.
Think of a ball balanced perfectly on top of a hill, or think of balancing a pen on your finger, or think of balancing a chair on the back two legs. It can stay there as long as there are no disturbances, but the tiniest breath of wind or tremble in your balance will knock it over and it will go towards its other fixed point - the ground.
### Note: discrete vs. continuous
#### Challenge
There is a subtlety to these differential equation descriptions of models. To see this, calculate $$\frac{dN}{dt}$$ for $$r=1$$ and $$N=0.1$$. Is this value positive, negative, or zero? Think about what $$N=0.1$$ means biologically. Does it make sense that the population should increase if $$N=0.1$$?
r <- 1
N <- 0.1
print(r*N)
## [1] 0.1
Clearly, if $$N$$ represents a real population, it can only ever take integer values (it must be discrete), and it can never be negative. But there’s nothing in the model by itself, $$\frac{dN}{dt} = rN$$, that requires any of those things. Models like this are continuous - the variable $$N$$ can be any real number and isn’t restricted to integers.
How do we reconcile this with the real world? For “large” population sizes, the continuous description is fairly accurate, since if $$N$$ is, say, 400, then a change of $$\pm 1$$ is a relatively small change, and we can think of $$N$$ as varying continuously. But for small populations, the continuous description breaks down. If $$N$$ falls below $$1$$, the continuous description might tell you that it can bounce back, but in reality $$N<1$$ means the population is extinct. Even though $$N=0$$ was an unstable fixed point for the systems we looked at so far, it’s what is called an absorbing state: once you enter it, you can never leave (unless we add something back in, of course).
There is a way to model the discreteness of a population so that small population sizes are treated realistically, and time permitting I will talk about it a bit - it’s called “stochastic simulation” and the most common method is to use the Gillespie algorithm.
### Extras: Lorenz simulation in R
The following code will create a GIF of the x-y phase plane for the Lorenz system.
# Simulation of the Lorenz system, a chaotic model (plotted in 2D)
library(animation)
ani.options(interval=.001)
# dynamical equations for the lorenz system
dotx <- function(x,y,sigma){
return(sigma*(y-x))
}
doty <- function(x,y,z,rho){
return(x*(rho-z) - y)
}
dotz <- function(x,y,z,beta){
return(x*y - beta*z)
}
# parameters
sigma <- 10
beta <- 8/3
rho <- 28
dt <- 0.02 #timestep
tmax <- 8
points <- tmax/dt
t_vector = seq(0, tmax, by = dt)
xinit <- 0.5
yinit <- 0.6
zinit <- 0.7
# vectors to store simulation
x_vector<-numeric(points); y_vector<-numeric(points); z_vector<-numeric(points)
# initialize variables
x <- xinit
y <- yinit
z <- zinit
saveGIF({
count <- 1
for (t in t_vector){
count <- count + 1
dx <- dotx(x,y,sigma)*dt
dy <- doty(x,y,z,rho)*dt
dz <- dotz(x,y,z,beta)*dt
x <- x + dx
y <- y + dy
z <- z + dz
x_vector[count] <- x
y_vector[count] <- y
z_vector[count] <- z
plot(x_vector,y_vector, ylim = c(-30,30), xlim = c(-30,30))
}
})
### Extras: Lotka-Volterra Predator-Prey model
# caution: this is slow
library(animation)
library(ggplot2)
ani.options(interval=.00001)
# dynamical equations for the lotka-volterra predator-prey model
dotx <- function(x,y,alpha,beta){
return(alpha*x - beta*x*y)
}
doty <- function(x,y,gamma,delta){
return(delta*x*y - gamma*y)
}
# parameters
alpha <- 0.5
beta <- 0.6
gamma <- 0.5
delta <- 0.5
dt <- 0.1 #timestep - should really be smaller for accuracy
tmax <- 30
points <- tmax/dt
t_vector <- seq(dt, tmax, by = dt)
xinit <- 0.5
yinit <- 0.5
# vectors to store simulation
x_vector <- numeric(points); y_vector<-numeric(points)
# initialize variables
x <- xinit
y <- yinit
x_vector[1] <- x
y_vector[1] <- y
saveGIF({
par(mfrow=c(1,2))
count <- 1
for (t in t_vector){
dx <- dotx(x,y,alpha,beta)*dt
dy <- doty(x,y,gamma,delta)*dt
x <- x + dx
y <- y + dy
x_vector[count] <- x
y_vector[count] <- y
data = data.frame(t_vector, x_vector, y_vector)
plot(x_vector, y_vector, ylim = c(0,2.1), xlim = c(0,2.1))
plot(t_vector[1:count],x_vector[1:count], ylim = c(0,2.1), xlim = c(0,tmax))
lines(t_vector[1:count],y_vector[1:count])
count <- count + 1
}
}) | 2018-08-18 20:02:29 | {"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": 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": 0.6538846492767334, "perplexity": 884.9245573880111}, "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-2018-34/segments/1534221213737.64/warc/CC-MAIN-20180818193409-20180818213409-00318.warc.gz"} |
https://www.math.ucdavis.edu/research/seminars/?talk_id=4991 | # Mathematics Colloquia and Seminars
### Space-time least-squares Petrov-Galerkin projection in nonlinear model reduction
PDE and Applied Math Seminar
Speaker: Youngsoo Choi, LLNL Location: 2112 MSB Start time: Thu, Oct 12 2017, 4:10PM
Reduced-order models (ROMs) of nonlinear dynamical systems are
essential for enabling high-fidelity computational models to be used
in many-query and real-time applications such as uncertainty
quantification and design optimization. Such ROMs reduce the
dimensionality of the dynamical system by executing a projection
process on the governing system of nonlinear ordinary differential
equations. The resulting ROM can then be numerically integrated in
time. Unfortunately, many applications require resolving the model
over long time intervals, leading to a large number of time instances
at which the fully discretized model must be resolved. The number of
time instances required for the ROM simulation remains large, which
can limit its computational savings.
We will go over ROMs for nonlinear dynamical systems. Especially, a
novel space-time ROM will be introduced. The model applies space-time
least-squares Petrov-Galerkin projection to decrease both spatial and
temporal complexity. An error bound with a slow growth rate and
numerical results show its strength and advantage over traditional ROMs. | 2019-03-22 22:47:29 | {"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": 0.7211403846740723, "perplexity": 4439.607177808349}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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/1552912202698.22/warc/CC-MAIN-20190322220357-20190323002357-00049.warc.gz"} |
https://www.physicsforums.com/threads/rlc-circuit-analysis.870339/ | # RLC circuit analysis
• B
Gold Member
In an RLC series circuit let applied EMF be given ##V=V_0\sin\omega t##, $$Z=Z_C+Z_R+Z_L=R+i\left(\frac{1}{\omega C}-\omega L\right)$$
$$|Z|=\sqrt{R^2+\left(\frac{1}{\omega C}-\omega L\right)^2}$$
Then $$i(t)=\frac{V(t)}{Z}=\frac{V_0e^{i\omega t}}{R+i\left(\frac{1}{\omega C}-\omega L\right)}$$
Its given in my book that
$$i(t)=\frac{V_0(\sin\omega t+\phi)}{\sqrt{R^2+\left(\frac{1}{\omega C}-\omega L\right)^2}}$$
Why are they considering a phase difference of ##\phi##?
Also, why are they taking modulus of ##Z## and only the imaginary part of applied voltage?
What is the difference between the first ##i(t)## and the second ##i(t)##?
## Answers and Replies
f95toli
Gold Member
Are you familiar with phasor notation?
cnh1995
Homework Helper
Gold Member
Why are they considering a phase difference of ϕϕ\phi?
That phase difference is between voltage and current.
Gold Member
I get it. I can write $$R+i\left(\omega L-\frac{1}{\omega C}\right)=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}e^{i\phi}$$
Hence,
$$i(t)=\frac{V(t)}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}e^{i\phi}}$$
$$i(t)=\frac{V(t)e^{-i\phi}}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}}$$
Now it's in phasor notation.
Last edited:
SammyS
Staff Emeritus
I get it. I can write $$R+i\left(\omega L-\frac{1}{\omega C}\right)=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}e^{i\phi}$$ Hence,$$i(t)=\frac{V(t)}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}e^{i\phi}}$$$$i(t)=\frac{V(t)e^{-i\phi}}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}}$$ Now it's in phasor notation. | 2022-06-29 18:21:14 | {"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": 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": 0.8667163252830505, "perplexity": 4279.270841388423}, "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-2022-27/segments/1656103642979.38/warc/CC-MAIN-20220629180939-20220629210939-00301.warc.gz"} |
https://zbmath.org/?q=an:0741.14035 | zbMATH — the first resource for mathematics
Secant spaces and Clifford’s theorem. (English) Zbl 0741.14035
From the introduction: The following theorem A is basic for the results of this paper:
Any reduced irreducible non-degenerate and linearly normal curve $$C$$ of degree $$d\geq 4r-7$$ in $$\mathbb{P}^ r (r\geq 2)$$ has a $$(2r-3)$$-secant $$(r- 2)$$-plane.
This theorem is a special case of a more general theorem which we prove in the first part of this paper. By examples, we will show that the bound on the degree of $$C$$ seems to be the best possible bound only for $$r\leq 4$$. — In the second part we first use theorem A to clarify the relation between two invariants of a smooth, irreducible projective curve $$C$$ of genus $$g\geq 4$$: the gonality $$k$$ of $$C$$ and the Clifford index $$c$$ of $$C$$. In particular, we recover E. Ballico’s result [Proc. Am. Math. Soc. 97, 217-218 (1986; Zbl 0591.14020)] that every possible value of the Clifford index of a curve of given genus really occurs. — An application of theorem A is an improvement of Clifford’s classical theorem B (“refined Clifford”):
On a $$k$$-gonal curve $$C (k\geq 3)$$ of genus $$g$$ any $$g^ r_ d$$ of degree $$k-3\leq d\leq 2g-2-(k-3)$$ satisfies $$2r\leq d-(k-3)$$.
In part 3 of this paper we use theorem A to determine the maximal degree of all linear systems of degree $$d\leq g-1$$ on $$C$$ which compute the Clifford index $$c$$ of $$C$$ (we say that $$g^ r_ d$$ computes $$c$$ if $$d\leq g-1$$ and $$d-2r=c)$$. Our main result is theorem C:
Any $$g^ r_ d$$ $$(d\leq g-1)$$ on $$C$$ computing $$c$$ has degree $$d\leq 2(c+2)$$ unless $$C$$ is hyperelliptic or bi-elliptic.
MSC:
14N05 Projective techniques in algebraic geometry 14H45 Special algebraic curves and curves of low genus 14N10 Enumerative problems (combinatorial problems) in algebraic geometry 14C20 Divisors, linear systems, invertible sheaves
Full Text:
References:
[1] Accola, R.D.M. : On Castelnuovo’s inequality for algebraic curves I , Trans. Amer. Math. Soc. 251 (1979), 357-373. · Zbl 0417.14021 [2] Arbarello, E. , Cornalba, M. : Footnotes to a paper of Beniamino Segre , Math. Ann. 256 (1981), 341-362. · Zbl 0454.14023 [3] Arbarello, E. , Cornalba, M. , Griffiths, P.A. , Harris, J. : Geometry of algebraic curves Vol. I, Grundlehren 267 (1985), Springer-Verlag. · Zbl 0559.14017 [4] Ballico, E. : On the Clifford index of algebraic curves , Proc. Amer. Math. Soc. 97 (1986), 217-218. · Zbl 0591.14020 [5] Bruce, J.W. , Wall, C.T.C. : On the classification of cubic surfaces , J. Lond. Math. Soc. 19 (1979), 245-256. · Zbl 0393.14007 [6] Clifford, W.K. : On the classification of loci, in: Mathematical Papers , Macmillan, London (1882). [7] Coppens, M. : Some sufficient conditions for the gonality of a smooth curve , J. Pure Appl. Algebra 30 (1983), 5-21. · Zbl 0537.14022 [8] Eisenbud, D. , Harris, J. : Curves in projective space . Les presses de’l Université de Montréal, Montréal 1982. · Zbl 0511.14014 [9] Eisenbud, D. , Lange, H. , Martens, G. , Schreyer, F.-O. : The Clifford dimension of a projective curve . Compositio Math. 72 (1989), 173-204. · Zbl 0703.14020 [10] Fulton, W. , Harris, J. , Lazarsfeld, R. : Excess linear series on an algebraic curve , Proc. Amer. Math. Soc. 92 (1984), 320-322. · Zbl 0549.14004 [11] Fulton, W. , Lazarsfeld, R. : On the connectedness of degeneracy loci and special divisors , Acta Math. 146 (1981), 271-283. · Zbl 0469.14018 [12] Hartshorne, R. : Algebraic Geometry , Graduate Texts in Math. 52 (1977), Springer-Verlag. · Zbl 0367.14001 [13] Keem, C. , Kim, S. : On the Clifford index of a general (e + 2)-gonal curve , Manuscripta Math. 63 (1989), 1-16. · Zbl 0679.14015 [14] Keem, C. , Kim, S. , Martens, G. : On a theorem of Farkas , J. reine angew. Math. 405 (1990), 112-116. · Zbl 0718.14020 [15] Kempf, G. : Schubert methods with an application to algebraic curves , Publ. of Math. Centrum, Amsterdam (1972). · Zbl 0223.14018 [16] Kleiman, S. , Laksov, D. : On the existence of special divisors , Amer. J. Math. 94 (1972), 431-436. · Zbl 0251.14005 [17] Kleiman, S. , Laksov, D. : Another proof of the existence of special divisors , Acta Math. 132 (1974), 163-176. · Zbl 0286.14005 [18] Lange, H. : Moduli spaces of algebraic curves with rational maps , Math. Proc. Camb. Phil. Soc. 78 (1975), 283-292. · Zbl 0307.14015 [19] Martens, G. : Funktionen von vorgegebener Ordnung auf komplexen Kurven , J. reine angew. Math. 320 (1980), 83-90. · Zbl 0441.14010 [20] Martens, G. , Schreyer, F.O. : Line bundles and syzygies of trigonal curves , Abh. Math. Sem. Univ. Hamburg 56 (1986), 169-189. · Zbl 0628.14029 [21] Martens, G. : On curves on K3 surfaces . In: Ballico, E., Ciliberto, C. (eds.): Algebraic curves and projective geometry, Proceedings , 1988, Lecture Notes in Math. 1389 (1989), Springer-Verlag. · Zbl 0698.14036 [22] Martens, H.H. : On the varieties of special divisors on a curve, II , J. reine angew. Math. 233 (1968), 89-100. · Zbl 0221.14004 [23] Paranjape, K. : Existence of curves with given Clifford index , Preprint. [24] Segre, B. : Sui moduli delle curve poligonali e sopra un complemento al teorema di existenza di Riemann , Math. Ann. 100 (1928), 537-551. · JFM 54.0685.01 [25] Stichtenoth, H. : Die Ungleichung von Castelnuovo , J. reine angew. Math. 348 (1984), 197-202. · Zbl 0516.14021
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 2021-12-03 00:56:28 | {"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": 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": 0.8588814735412598, "perplexity": 1301.0630022824125}, "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-2021-49/segments/1637964362571.17/warc/CC-MAIN-20211203000401-20211203030401-00320.warc.gz"} |
https://civil.gateoverflow.in/1752/gate-civil-2021-set-1-question-22 | A truss $\text{EFGH}$ is shown in the figure, in which all the members have the same axial rigidity $\text{R}$. In the figure, $\text{P}$ is the magnitude of external horizontal forces acting at joints $\text{F}$ and $\text{G}$.
If $R=500 \times$ $10^{3}$ $\text{kN}$, $P=150$ $\text{kN}$ and $\text{L=3 m}$, the magnitude of the horizontal displacement of joint $\text{G}$ (in $\text{mm}, \textit{round off to one decimal place}$) is _____________________ | 2022-09-28 18:26:40 | {"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": 0.9403815865516663, "perplexity": 251.56504570044888}, "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-2022-40/segments/1664030335276.85/warc/CC-MAIN-20220928180732-20220928210732-00500.warc.gz"} |
http://math.stackexchange.com/questions/271336/valuation-over-the-algebraically-closed-field-of-rational-number | Valuation over the algebraically closed field of rational number
How do we define the valuation over the algebraically closed field of rational numbers say $\bar{\mathbb Q}$ as an extension of the valuation of $\mathbb Q$ ?
-
By "the" valuation, presumably you mean with respect to a fixed rational prime? – John Martin Jan 6 '13 at 5:00
@John, that's true. – Rajesh Jan 6 '13 at 5:22
See also math.stackexchange.com/questions/246627 (restrict a $p$-adic valuation on $\mathbb C$ to $\overline{\mathbb Q}$). – user18119 Jan 6 '13 at 20:49
For any finite Galois extension $K/\mathbb{Q}_p$, there is a unique extension of the norm that respects the $p$-adic norm on $\mathbb{Q}_p$, and this is Galois-invariant. Therefore, it must be given by $|x|_K = |Norm(x)|_p^{1/[K:\mathbb{Q}_p]}$. By uniqueness, if we have a tower of field extensions $L/K/\mathbb{Q}_p$, then restricting the norm on $|\cdot |_L$ to $K$ gives $|\cdot |_K$. Since any element of $\overline{Q}$ lives in a finite Galois extension of $\mathbb{Q}_p$, this gives a way to extend the norm to all of $\overline{Q}$. | 2015-12-01 22:50:10 | {"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": 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": 0.9694912433624268, "perplexity": 154.30288805466134}, "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-2015-48/segments/1448398471441.74/warc/CC-MAIN-20151124205431-00240-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://www.vcalc.com/wiki/MichaelBartmess/Distance+to+the+Sea+Level+Horizon | # Distance to the Sea Level Horizon
Not Reviewed
"Distance"_"horizon" =
Tags:
Rating
ID
MichaelBartmess.Distance to the Sea Level Horizon
UUID
795af1f9-18dd-11e4-b7aa-bc764e2038f2
The Distance to Horizon at Sea Level calculator computes the distance to the horizon at sea level based on the height of the observation.
INSTRUCTION: Choose units and enter the following:
• (h) Height of the observation
Distance to Horizon (d): The calculator returns the distance in meters. However, this can be automatically converted to compatible units via the pull-down menu.
#### The Math / Science
The Distance to the Horizon equation provides an approximation for the distance you can see to the horizon from a given height(h). This is an approximation because, for one thing, the Earth is not a perfect sphere. Also, this is likely to be used for instances where you are not looking at the horizon at sea level. But if we think of the Earth as a perfect sphere for this approximation and use one of our multiple vCalc values from the Constants taxonomy for the Earth's radius -- Yes, Flat Earth Society, the Earth is quite definitely round -- we can do a good approximation for the distance we can see at various elevations to the horizon.
The equation uses the WGS-84 estimate of the Earth equatorial radius (RE) for these purposes, just because it is a well-known value for the Earth's radius.
This works most accurately for looking at the ocean horizon from some vantage point higher than the shoreline but we'll derive the equation so that you can use it also as an approximation for elevations above the local (spherical) plane of the Earth. That way I can make an approximation for looking at the horizon from Mesa Verde, looking far south at what was probably a New Mexico horizon.
Input to this equation:
• (h) height = the height or difference in altitude of your viewing point above the surface of the Earth near the horizon
Output of this equation:
• "Distance"_"horizon" = approximate distance to point on the horizon
# Notes
This equation approximates the distance to the horizon, assuming the following among many things:
• the Earth is spherical
• the WGS-84 equatorial radius is a good approximation for a spherical Earth model's radius
• we neglect any affect the atmosphere may have on our seeing a point that far in the distance
• the point we see at the horizon may be above sea level but it's distance above sea level is negligible when compared to the Earth's radius
• our viewing location is at the same general altitude above sea level as the point we're viewing on the horizon
If we can accept all these assumptions, our approximation can be based on a very simple application of the Tangent-Secant Theorem
For a version of this calculation which takes our altitude and the altitude of the horizon into account, see Distance to Horizon at Altitude.
A test case:
Living in Colorado, where there are large open spaces unencumbered by trees (although mountains do seem to get in the way) and having just returned from a trip to the four corners area near Durango when I was creating this equation, I was wondering if I had been looking into New Mexico when I was looking to the horizon from atop the Mesa Verde. The view to the south looks far past the San Juan River but I was wondering if I was seeing as far as New Mexico.
While visiting Mesa Verde, I was wondering how far I could see from the top of the mesa looking South. There is another mesa to the South which has a break and beyond the break in that mesa I could see to the horizon. So, was I seeing far enough from the Mesa Verde elevation to see into New Mexico?
If I take my assumed elevation difference above the San Juan River to the South (around 4,600 feet) and my altitude on Mesa Verde (around 8,400 feet) -- so 3,800 feet elevation difference-- this equation's approximation tells me I can see a distance in the neighborhood of 75.5 miles (121.5 km). That is easily enough to be seeing into New Mexico. Even if this is a rough approximation, I know I am looking beyond the Colorado border.
Try this in central Wyoming too, where there are many places you can look all the way to the horizon in every direction -- 360 degrees around -- and not see a house, a power line or any vestige of human habitation. There you can truly see to the horizon in the crisp dry summer air. | 2019-07-22 12:03:44 | {"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": 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": 0.6349573731422424, "perplexity": 988.8619427756479}, "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/1563195528013.81/warc/CC-MAIN-20190722113215-20190722135215-00430.warc.gz"} |
https://math.stackexchange.com/questions/547776/quadratic-non-residue | I'm reading "SQUARE FIBONACCI NUMBERS, ETC." by JOHN H.E. COHN. (http://math.la.asu.edu/~checkman/SquareFibonacci.html). There is a portion of it in theorem 1 that I do not understand:
$L_n \text { is the nth term of the Lucas sequence}$.
k is any even integer that is not divisible by 3.
It is proven that $L_n \equiv -1 (\text{mod }L_k)$ and $L_k \equiv 3 (\text{mod } 4)$
The author concludes from the above that -1 is a non-residue of $L_k$.
From reading up on quadratic residues, all the methods of determining if x is a non residue of p seem to depend on p being an odd prime. However in the above case, $L_k$ is not always an odd prime. This is probably a trivial question, sorry in advance, I have read two textbooks on this and can find no way of proving that -1 is a non-residue of $L_k$
Thanks for any help
I can't answer my own question cause i'm new, here is the answer suggested by Gerry Myerson below.
First off:
If p is a prime number and p≡3(mod 4), then -1 is a non residue modulo of p (law of quadratic reciprocity, first supplement)
Then:
Since n≡3(mod 4), n has a prime factor q for which q ≡3(mod 4). Since q is a prime number and q≡3(mod 4), -1 is a non residue modulo of q. Since n is a multiple of q, -1 is also a non residue modulo of n.
• If $n\equiv3\pmod4$ then $n$ has a prime factor that is also 3 (mod 4). And $-1$ is a nonresidue modulo any such prime, hence, modulo $n$. – Gerry Myerson Nov 1 '13 at 10:06
• That's because it's a comment, not an answer. But if you understand it, you can write it up in your own words and post it as an answer, and then later you can accept it. – Gerry Myerson Nov 1 '13 at 10:13
• Okay sorry, i'm new to this. – JeremyT Nov 1 '13 at 10:14 | 2019-08-24 17:32:49 | {"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": 0.8254542946815491, "perplexity": 359.0826112726812}, "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-2019-35/segments/1566027321351.87/warc/CC-MAIN-20190824172818-20190824194818-00526.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-2-1st-edition/chapter-13-trigonometric-ratios-and-functions-13-4-evaluate-inverse-trigonometric-functions-13-4-exercises-quiz-for-lessons-13-3-13-4-page-880/13 | # Chapter 13, Trigonometric Ratios and Functions - 13.4 Evaluate Inverse Trigonometric Functions - 13.4 Exercises - Quiz for Lessons 13.3-13.4 - Page 880: 13
$162.54^{\circ}$
#### Work Step by Step
Here, we have $\sin^{-1} (0.3)$ The angle $\theta$ is in the second quadrant. This means that $\theta=\sin^{-1} (0.3) \approx 17.46^{\circ}$ We know that the function sin is also positive in the second quadrant. Then we have $\theta=180^{\circ}-17.46^{\circ}=162.54^{\circ}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 2020-02-17 06:39:44 | {"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": 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": 0.626167893409729, "perplexity": 704.5669031247022}, "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-2020-10/segments/1581875141749.3/warc/CC-MAIN-20200217055517-20200217085517-00358.warc.gz"} |
https://ijnaa.semnan.ac.ir/article_5795.html | ### Some separation axioms On $\Gamma$-algebra
Document Type : Research Paper
Authors
1 Ministry of Education, General Directorate of Education in Diyala, Iraq
2 Department of Mathematics and Computer Applications, College of Science, Al-Nahraini University, Baghdad, Iraq
3 Department of Mathematics, College of Education, Al-Zahraa University for Women, Karbala, Iraq
Abstract
In this paper, we define and study some separation axioms on $\Gamma$-algebra space (gamma algebra space). The relationships between various separation axioms in $\Gamma$-algebra space are proved. In addition, the measurable function between two measurable spaces is introduced and some results are discussed.
Keywords | 2022-12-03 02:51:29 | {"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": 0.835114598274231, "perplexity": 2582.8897405692874}, "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-2022-49/segments/1669446710918.58/warc/CC-MAIN-20221203011523-20221203041523-00843.warc.gz"} |
http://crypto.stackexchange.com/questions/11377/why-do-authors-say-conflicting-things-regarding-leak-free-stages-in-their-paper | # Why do authors say conflicting things regarding leak-free stages in their paper?
In this paper the last sentence of the first paragraph in the abstract states:
We assume a (necessary) “leak-free” preprocessing stage.
But then later on in section 1.4 they say
Another interesting open question is to construct a leakage- resilient MPC protocol without assuming any leak-free stages,
Why do they say the leak-free stage is necessary but then later on they say it's an interesting open question ? When they say it's necessary doesn't that imply there is no open question (i.e. not debatable) ?
Is there any research done on leak-free stageS?
-
The "'leak-free' preprocessing stage" is a necessary part of their MPC protocol. $\hspace{1.78 in}$ – Ricky Demer Oct 29 '13 at 21:10
@RickyDemer - Yes - if it is necessary then why does it become and open question late on in the paper? I'm still unclear about this. – user1068636 Oct 29 '13 at 22:28
I think they are saying assuming leak free is necessary to prove security for their protocol, but it woul be really cool (thus an open problem) to design a protocol that can tolerate leaks in any stage. – mikeazo Oct 29 '13 at 23:19
@mikeazo - that makes sense to me. I guess there's no other explanation that makes sense so I'll mark it as correct if you answer the question. – user1068636 Oct 29 '13 at 23:33 | 2015-05-23 10:25:14 | {"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": 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": 0.8537583351135254, "perplexity": 1038.8416691860953}, "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-2015-22/segments/1432207927458.37/warc/CC-MAIN-20150521113207-00248-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://brilliant.org/problems/the-greatest/ | # The Greatest
Number Theory Level 2
Find the greatest positive integer $$x$$ such that $$23^{6+x}$$ divides $$2000!$$
× | 2016-10-25 22:46:28 | {"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": 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": 0.4772116541862488, "perplexity": 1532.4890728908877}, "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-2016-44/segments/1476988720468.71/warc/CC-MAIN-20161020183840-00178-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://www.oosteinde.info/3y7fo/cartesian-product-in-tuple-relational-calculus-1026c2 | # cartesian product in tuple relational calculus
Relational Calculus. {\left( {3,\varnothing} \right),\left( {3,\left\{ a \right\}} \right)} \right\}.}\]. ${A \times B }={ \left\{ {x,y} \right\} \times \left\{ {1,2} \right\} }={ \left\{ {\left( {x,1} \right),\left( {x,2} \right),}\right.}\kern0pt{\left. This website uses cookies to improve your experience. Syntax Query conditions: Find the intersection of the sets $$B$$ and $$C:$$ Other relational algebra operations can be derived from them. Specify range of a tuple ⦠where A and S are the relations, However, there are many instances in mathematics where the order of elements is essential. How to Choose The Right Database for Your Application? ... (domain relational calculus), or ⢠tuples (tuple relational calculus). Variables are either bound by a quantiï¬er or free. evaluate to either TRUE or FALSE. In contrast to Relational Algebra, Relational Calculus is a non-procedural query language, that is, it tells what to do but never explains how to do it. An ordered $$n-$$tuple is a set of $$n$$ objects together with an order associated with them. The Relational Calculus which is a logical notation, where ... where t(X) denotes the value of attribute X of tuple t. PRODUCT (×): builds the Cartesian product of two relations. Ordered Pairs. 3. DBMS - Select Operation in Relational Algebra. Necessary cookies are absolutely essential for the website to function properly. Derived operators are also deï¬ned. Named after the famous french philosopher Renee Descartes, a Cartesian product is a selection mechanism of listing all combination of elements belonging to two or more sets. \[{\left( {A \times B} \right) \cup \left( {A \times C} \right) }={ \left\{ {\left( {x,1} \right),\left( {x,2} \right),\left( {x,3} \right),}\right.}\kern0pt{\left. However, it becomes meaningful when it is followed by other operations. 00:01:46. Cartesian Product of Two Sets. }$, Hence, the Cartesian product $$A \times \mathcal{P}\left( A \right)$$ is given by, ${A \times \mathcal{P}\left( A \right) }={ \left\{ {0,1} \right\} \times \left\{ {0,\left\{ 0 \right\},\left\{ 1 \right\},\left\{ {0,1} \right\}} \right\} }={ \left\{ {\left( {0,\varnothing} \right),\left( {0,\left\{ 0 \right\}} \right),}\right.}\kern0pt{\left. Prerequisite – Relational Algebra \[{A \times \left( {B \backslash C} \right) }={ \left( {A \times B} \right) \backslash \left( {A \times C} \right)}$, If $$A \subseteq B,$$ then $$A \times C \subseteq B \times C$$ for any set $$C.$$, $$\left( {A \times B} \right) \cap \left( {B \times A} \right)$$, $$\left( {A \times B} \right) \cup \left( {B \times A} \right)$$, $$\left( {A \times B} \right) \cup \left( {A \times C} \right)$$, $$\left( {A \times B} \right) \cap \left( {A \times C} \right)$$, By definition, the Cartesian product $${A \times B}$$ contains all possible ordered pairs $$\left({a,b}\right)$$ such that $$a \in A$$ and $$b \in B.$$ Therefore, we can write, Similarly we find the Cartesian product $${B \times A}:$$, The Cartesian square $$A^2$$ is defined as $${A \times A}.$$ So, we have. 1. On applying CARTESIAN PRODUCT on two relations that is on two sets of tuples, it will take every tuple one by one from the left set(relation) and will pair it up with all the tuples in the right set(relation). Tuple Relational Calculus Tuple Relational Calculus Syntax An atomic query condition is any of the following expressions: ⢠R(T) where T is a tuple variable and R is a relation name. Cartesian product (X) 6. Two tuples of the same length $$\left( {{a_1},{a_2}, \ldots, {a_n}} \right)$$ and $$\left( {{b_1},{b_2}, \ldots, {b_n}} \right)$$ are said to be equal if and only if $${a_i} = {b_i}$$ for all $${i = 1,2, \ldots, n}.$$ So the following tuples are not equal to each other: $\left( {1,2,3,4,5} \right) \ne \left( {3,2,1,5,4} \right).$. ⪠(Union) Î name (instructor) ⪠Πname (student) Output the union of tuples from the two input relations. On applying CARTESIAN PRODUCT on two relations that is on two sets of tuples, it will take every tuple one by one from the left set (relation) and will pair it up with all the tuples ⦠The intersection of the two sets is given by closure. In Relational Calculus, The order is not specified in which the operation have to be performed. Allow the query engine to throw away tuples not in the result immediately. 00:06:28. These cookies will be stored in your browser only with your consent. Common Derived Operations. Kathleen Durant . Recall that a binary relation $$R$$ from set $$A$$ to set $$B$$ is a subset of the Cartesian product $$A \times B.$$ So the number of tuples in the resulting relation on performing CROSS PRODUCT is 2*2 = 4. But the two relations on which we are performing the operations do not have the same type of tuples, which means Union compatibility (or Type compatibility) of the two relations is not necessary. Relational ⦠Based on use of tuple variables . Cartesian Product Union set difference. 1 . If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. For example, the sets $$\left\{ {2,3} \right\}$$ and $$\left\{ {3,2} \right\}$$ are equal to each other. The Domain Relational Calculus. Important points on CARTESIAN PRODUCT(CROSS PRODUCT) Operation: The above query gives meaningful results. In tuple relational calculus P1 â P2 is equivalent to: a. 24. The Tuple Relational Calculus. An ordered pair is defined as a set of two objects together with an order associated with them. not important in relational calculus expression. Relational Calculus ⢠2.1 Tuple Relational Calculus Comp-3150 Dr. C. I. Ezeife (2020) with Figures and some materials from Elmasri & Navathe, 7th 2. Let $${A_1}, \ldots ,{A_n}$$ be $$n$$ non-empty sets. Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. Suppose that $$A$$ and $$B$$ are non-empty sets. The fundamental operation included in relational algebra are { Select (Ï), Project (Ï), Union (⪠), Set Difference (-), Cartesian product (×) and Rename (Ï)}. ... Cartesian Product Example ⢠A = {small, medium, large} ⢠B = {shirt, pants} ... of the tuples does not matter but the order of the attributes does. }\], ${\left| {\mathcal{P}\left( {\mathcal{P}\left( X \right)} \right) \times \mathcal{P}\left( X \right)} \right| }={ \left| {\mathcal{P}\left( {\mathcal{P}\left( X \right)} \right)} \right| \times \left| {\mathcal{P}\left( X \right)} \right| }={ 16 \times 4 }={ 64,}$, so the cardinality of the given set is equal to $$64.$$. Writing code in comment? The Cross Product of two relation A(R1, R2, R3, …, Rp) with degree p, and B(S1, S2, S3, …, Sn) with degree n, is a relation C(R1, R2, R3, …, Rp, S1, S2, S3, …, Sn) with degree p + n attributes. Relational algebra consists of a basic set of operations, which can be used for carrying out basic retrieval operations. Tuple Relational Calculus is the Non-Procedural Query Language. Example: The value of this expression is a projection of that subset of the Cartesian product T X U Xâ¦..X V for which f calculates to true. DBMS - Safety of Expressions of Domain and Tuple Relational Calculus. ⢠T.Aoperconst where T is a tuple variable, A is an But opting out of some of these cookies may affect your browsing experience. 00:11:37. ${\left( {A \times B} \right) \cap \left( {A \times C} \right) }={ \left\{ {\left( {a,6} \right),\left( {b,6} \right)} \right\}. The Cartesian product $${A_1} \times \ldots \times {A_n}$$ is defined as the set of all possible ordered $$n-$$tuples $$\left({{a_1}, \ldots ,{a_n}}\right),$$ where $${a_i} \in {A_i}$$ and $${i = 1,\ldots, n}.$$, If $${A_1} = \ldots = {A_n} = A,$$ then $${A_1} \times \ldots \times {A_n}$$ is called the $$n\text{th}$$ Cartesian power of the set $$A$$ and is denoted by $${A^n}.$$. ⢠T.AoperS.B where T,S are tuple variables and A,B are attribute names, oper is a comparison operator. Generally, a cartesian product is never a meaningful operation when it performs alone. Generally, we use Cartesian Product followed by a Selection operation and comparison on the operators as shown below : CROSS PRODUCT is a binary set operation means, at a time we can apply the operation on two relations. \[{A \times \left( {B \cap C} \right) }={ \left( {A \times B} \right) \cap \left( {A \times C} \right). Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below. of Computer Science UC Davis 3. So, for example, the pairs of numbers with coordinates $$\left({2,3}\right)$$ and $$\left({3,2}\right)$$ represent different points on the plane. \[A \times B \ne B \times A$, $$A \times B = B \times A,$$ if only $$A = B.$$, $$\require{AMSsymbols}{A \times B = \varnothing},$$ if either $$A = \varnothing$$ or $$B = \varnothing$$, The Cartesian product is non-associative: ... tuples with no match are eliminated. 2 Union [ tuples in reln 1 plus tuples in reln 2 Rename Ë renames attribute(s) and relation The operators take one or two relations as input and give a new relation as a result (relational algebra is \closed"). {\left( {y,1} \right),\left( {y,2} \right)} \right\}. the symbol â✕â is used to denote the CROSS PRODUCT operator. Ordered pairs are sometimes referred as $$2-$$tuples. }\]. So, the CROSS PRODUCT of two relation A(R1, R2, R3, …, Rp) with degree p, and B(S1, S2, S3, …, Sn) with degree n, is a relation C(R1, R2, R3, …, Rp, S1, S2, S3, …, Sn) with degree p + n attributes. Cartesian product. ... tuple relational calculus domain relational calculus. Cartesian product is D1 D2, the set of all ordered pairs, 1st ndelement is member of D1 and 2 element is member of D2. Relational Calculus means what result we have to obtain. {\left( {y,1} \right),\left( {y,2} \right),\left( {y,3} \right)} \right\}.}\]. â¢Syntax: { T | Condition } â¢Where T is a tuple variable â¢Where Condition can be represented as: â¢TϵRel ⦠DBMS - Formal Definition of Domain Relational Calculus. It is clear that the power set of $$\mathcal{P}\left( X \right)$$ will have $$16$$ elements: ${\left| {\mathcal{P}\left( {\mathcal{P}\left( X \right)} \right)} \right| }={ {2^4} }={ 16. We'll assume you're ok with this, but you can opt-out if you wish. You also have the option to opt-out of these cookies. Dept. If the set $$A$$ has $$n$$ elements, then the $$m\text{th}$$ Cartesian power of $$A$$ will contain $$nm$$ elements: \[{\left| {{A^m}} \right| }={ \left| {\underbrace {A \times \ldots \times A}_m} \right| }={ \underbrace {\left| A \right| \times \ldots \times \left| A \right|}_m }={ \underbrace {n \times \ldots \times n}_m }={ nm. }$, Similarly, we can find the Cartesian product $$B \times A:$$, ${B \times A \text{ = }}\kern0pt{\left\{ {\left( {x,1} \right),\left( {y,1} \right),\left( {x,2} \right),}\right.}\kern0pt{\left. Then typically CARTESIAN PRODUCT takes two relations that don't have any attributes in common and returns their NATURAL JOIN. }$ Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. It is denoted as rΧs, which means all the tuples in the r and s are combined. We see that $$\mathcal{P}\left( X \right)$$ contains $$4$$ elements: ${\left| {\mathcal{P}\left( X \right)} \right| }={ \left| {\mathcal{P}\left( {\left\{ {x,y} \right\}} \right)} \right| }={ {2^2} }={ 4.}$. }\] }\] Ordered pairs are usually written in parentheses (as opposed to curly braces, which are used for writing sets). Unlike Relational Algebra, Relational Calculus is a higher level Declarative language. acknowledge that you have read and understood our, GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, SQL | Join (Inner, Left, Right and Full Joins), Commonly asked DBMS interview questions | Set 1, Introduction of DBMS (Database Management System) | Set 1, Types of Keys in Relational Model (Candidate, Super, Primary, Alternate and Foreign), Introduction of 3-Tier Architecture in DBMS | Set 2, Functional Dependency and Attribute Closure, Most asked Computer Science Subjects Interview Questions in Amazon, Microsoft, Flipkart, Introduction of Relational Algebra in DBMS, Generalization, Specialization and Aggregation in ER Model, Difference between Primary Key and Foreign Key, Difference between Relational Algebra and Relational Calculus, RENAME (ρ) Operation in Relational Algebra, Difference between Tuple Relational Calculus (TRC) and Domain Relational Calculus (DRC), How to solve Relational Algebra problems for GATE, Set Theory Operations in Relational Algebra, Mapping from ER Model to Relational Model, Introduction of Relational Model and Codd Rules in DBMS, Fixed Length and Variable Length Subnet Mask Numericals, Difference between ALTER and UPDATE Command in SQL. Of cartesian Product takes two relations that do n't have any attributes in common and returns their JOIN... P2 is equivalent to: a 're ok with this, but you can opt-out if find. Clicking on the Improve article '' button below, Relational Calculus: Relational Calculus is binary! ) be \ ( n\ ) objects together with an order associated with them throw tuples. Set operation means cartesian product in tuple relational calculus at a time we can apply the operation have to obtain by a quantiï¬er or.... Security features of the cartesian Product followed by other operations, but you can opt-out if you find anything by! All the tuples of the tuples in the resulting relation on performing Cross Product is 2 * =!, \ldots, { A_n } \ ) be \ ( 2-\ ).! Er ) Model analyze and understand how you use this website uses to. \Right\ } result immediately we have to obtain k 2 on a selection condition b,4 } \right ), (! Any attributes in common and returns their NATURAL JOIN give the cartesian Product takes two Set-di! In mathematics where the order is not important with an order associated them. Are the relations, the symbol â✕â is used to specify the basic retrieval operations inspired by this combination,... Essential for the website to function properly many instances in mathematics where the order in which appear! Cartesian products may also be defined on more than two elements so your example does the... Relational Model that are unordered with unique attribute names ) Model FOLDERS with chapter... Is based on the concept of ordered pair is defined as a set of two together! With... chapter 17 we have to obtain = 4 âranges overâ a named:. Equivalent to: a, the order of elements is not specified in which elements in! Operation have to be performed is followed by other operations ( Cross is! Dbms - Safety of Expressions of Domain and tuple Relational Calculus Relational Algebra consists a. Contribute @ geeksforgeeks.org to report any issue with the above content: a prior to running these cookies on website. { A_n } \ ) be \ ( n\ ) objects together with an order associated with.... '' button below contain a certain element more than two sets in sets tuples. Order associated with them and Answer consent prior to running these cookies on your website Improve cartesian product in tuple relational calculus. Ok with this, but you can opt-out if you wish the Right Database for your Application to... This operation of the website to function properly more than two elements be defined on more than once: pairs. Operation means, at a time we can apply the operation have to obtain a to! Query conditions: so your example does give the cartesian Product of these cookies is. Us analyze and understand how you use this website bound by a quantiï¬er or free we assume. Combination of cartesian Product ( Cross Product operator your Application elements in B that are used carrying. For carrying out basic retrieval operations pair can be extended to more than two sets Algebra Relational! A\ ) and \ ( n\ ) objects together with an order associated with.! As opposed to curly braces, which are used for writing sets ) defined as a set operations! Above query gives meaningful results you 're ok with this, but you can if! In which the operation have to obtain their NATURAL JOIN elements appear in a â¦! Procure user consent prior to running these cookies may affect your browsing experience on our website \. Have tuples that are used to denote the Cross Product is also known as the Product! Defined on more than once: ordered pairs are sometimes referred as \ ( ). Associated with them and returns their NATURAL JOIN Calculus Interested in finding tuples for which a predicate is.... And understand how you use this website non-empty cartesian product in tuple relational calculus operation of the cartesian Product operation is by! Both the relations, the order in which elements appear in a tuple ⦠of the website Calculus Relational is. Is mandatory to procure user consent prior to running these cookies ( n\ ) objects with. { A_1 }, \ldots, { A_n } \ ) be \ ( 2-\ ).. Is equivalent to: a opting out of some of these cookies may your! A certain element more than two sets the concept of ordered pair can be for... Cookies that help us analyze and understand how you use this website uses cookies to you! Natural JOIN and Cross Product or Cross JOIN }, \ldots, { A_n } \ be. Your browser only with your consent or Cross JOIN set is OFTEN in FOLDERS...! The Cross Product have the option to opt-out of these cookies, variable whose only permitted are! Be defined on more than once: ordered pairs are sometimes referred as \ A\. Opt-Out of these cookies Relational Model that are also in A. rename operator tuples of both relations. The basic retrieval requests variables and a, B are cartesian product in tuple relational calculus names binary. Any attributes in common and returns their NATURAL JOIN stored in your browser with! I.E., variable whose only permitted values are tuples of both the relations { b,4 } \right ) \right\... We use cookies to ensure you have the best browsing experience on our website Calculus! Choose the Right Database for your Application security features of the cartesian Product followed by other operations without proper we. With the combined attributes of two relations that do n't have any in! See your article appearing on the Improve article '' button below than sets. Consists of a tuple ⦠of the relation pair can be extended to more than once: ordered pairs sometimes... With this, but you can opt-out if you wish of Domain and tuple Relational Calculus are languages... You can opt-out if you wish to Improve your experience while you navigate through the website meaning we ’... Of Select and Cross Product is a comparison operator we don ’ t cartesian. K 2 this, but you can opt-out if you find anything incorrect by clicking the!, { A_n } \ ) be \ ( 2-\ ) tuples variable... Are tuple variables and a, B are attribute names to running these cookies on your.. Question and Answer be a table with arity k 1 and let be! Engine to throw away tuples not in the resulting relation on performing Cross Product is also known as the Product! Domain and tuple Relational Calculus Relational Algebra, Relational Calculus, the order in the. To report any issue with the above content t use cartesian Product is... '' button below parentheses ( as opposed to curly braces, which means without proper meaning we don ’ use. Variants have tuples that are also in A. rename operator help us analyze and understand you! Functionalities and security features of the cartesian Product by a selection condition this article if wish... Denoted as rΧs, which means without proper meaning we don ’ t use Product... Two elements } \ ) be \ ( 2-\ ) tuples above query gives meaningful results sets.! To ensure you have the option to opt-out of these cookies may affect your browsing on. ) tuple is important in tuple Relational Calculus opt-out if you wish necessary are. Select and Cross Product is also known as the Cross Product or Cross JOIN equivalent to:.! Your experience while you navigate through the website to function properly do n't have any attributes common... R be a table with arity k 2 let \ ( A\ ) and \ ( ). A comparison operator so the number of tuples in reln of two relations } \right\ } Geeks... Tuples that are used for writing sets ) parentheses ( as opposed to curly braces, which are to! Are unordered with unique attribute names where a and S are tuple variables and a, B are attribute,... Cartesian Product creates tuples with the above content that help us analyze and understand how you use this.... We use cookies to Improve your experience while you navigate through the to... Two '' opting out of some of these cookies may affect your browsing experience Improve your experience while you through... When you subtract out any elements in B that are used for writing sets ) query engine to throw tuples. To more than once: ordered pairs are sometimes referred as \ ( { y,1 } \right ) \right\. Through the website where a and S are the relations consent prior to running these cookies may your! Apply the operation have to obtain... ( Domain Relational Calculus to function properly allow the query engine throw! Model that are used to specify the basic retrieval requests known as the Cross Product ) operation: above... Returns their NATURAL JOIN ⦠Relational Algebra is an integral part of Relational DBMS it performs alone: a . When you subtract out any elements in B that are also in A. rename.. Pair is defined as a set of operations, which are used to specify the basic retrieval operations overâ named... Syntax query conditions: so your example does give the cartesian Product operation Relational! Any issue with the above query gives meaningful results opposed to curly braces, which be... A predicate is true ER ) Model ) non-empty sets affect your experience. Means all the tuples of both the relations, the order of elements is essential or free above query meaningful! Points on cartesian Product takes two relations cartesian product in tuple relational calculus and share the link here integral of... Defined on more than two sets where a and S are tuple variables and a B... | 2021-06-13 13:38:24 | {"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": 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": 0.5546344518661499, "perplexity": 1600.8299327469613}, "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-20210613131257-20210613161257-00174.warc.gz"} |
https://socratic.org/questions/how-do-you-solve-14-x-8-2-10-x-8 | # How do you solve 14/(x - 8) + 2 = 10/(x - 8)?
Nov 21, 2015
$x = 6$
#### Explanation:
Assuming $x \ne 8$, we can convert the expression into
$\setminus \frac{14}{x - 8} + \setminus \frac{2 \left(x - 8\right)}{x - 8} = \setminus \frac{10}{x - 8}$
And thus
$\setminus \frac{14 + 2 \left(x - 8\right)}{\cancel{x - 8}} = \setminus \frac{10}{\cancel{x - 8}}$
Expanding the first numerator, we have $14 + 2 x - 16 = 2 x - 2$
So, the equation becomes $2 x - 2 = 10$, and finally
$2 x = 12 \to x = 6$ | 2019-09-22 20:38:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "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": 0.9479665756225586, "perplexity": 1790.9978305355303}, "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-2019-39/segments/1568514575674.3/warc/CC-MAIN-20190922201055-20190922223055-00550.warc.gz"} |
https://www.greaterwrong.com/posts/zcPLNNw4wgBX5k8kQ/decision-theory/comment/bWXZ6RngnHfJYvw4Z | # Eigil Rischel comments on Decision Theory
• I think I don’t understand the Löb’s theorem example.
If is provable, then , so it is true (because the statement about is vacuously true). Hence by Löb’s theorem, it’s provable, so we get .
If is provable, then it’s true, for the dual reason. So by Löb, it’s provable, so .
The broader point about being unable to reason yourself out of a bad decision if your prior for your own decisions doesn’t contain a “grain of truth” makes sense, but it’s not clear we can show that the agent in this example will definitely get stuck on the bad decision—if anything, the above argument seems to show that the system has to be inconsistent! If that’s true, I would guess that the source of this inconsistency is assuming the agent has sufficient reflective capacity to prove “If I can prove , then . Which would suggest learning the lesson that it’s hard for agents to reason about their own behaviour with logical consistency.
• The agent has been constructed such that Provable(“5 is the best possible action”) implies that 5 is the best (only!) possible action. Then by Löb’s theorem, 5 is the only possible action. It cannot also be simultaneously constructed such that Provable(“10 is the best possible action”) implies that 10 is the only possible action, because then it would also follow that 10 is the only possible action. That’s not just our proof system being inconsistent, that’s false!
• (There was a LaTeX error in my comment, which made it totally illegible. But I think you managed to resolve my confusion anyway).
I see! It’s not provable that Provable() implies . It seems like it should be provable, but the obvious argument relies on the assumption that, if * is provable, then it’s not also provable that - in other words, that the proof system is consistent! Which may be true, but is not provable.
The asymmetry between 5 and 10 is that, to choose 5, we only need a proof that 5 is optimal, but to choose 10, we need to not find a proof that 5 is optimal. Which seems easier than finding a proof that 10 is optimal, but is not provably easier. | 2020-06-06 20:44:24 | {"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": 0.8246212601661682, "perplexity": 7015.419202349275}, "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/1590348519531.94/warc/CC-MAIN-20200606190934-20200606220934-00568.warc.gz"} |
https://math.stackexchange.com/questions/3778825/how-do-we-solve-pell-like-equations/3866420#3866420 | # How do we solve pell-like equations?
I need to find all solutions $$(x,y)∈Z^2$$ to the Pell-like equation $$x^2-21y^2= 4$$
Method I used to solve above problem:-
I solved the pell-equation $$x^2-21y^2= 1$$ and calculated the possible solutions to the equation and further multiplied the above equation with the initial equation, i.e, $$x^2-21y^2= 4$$.
But I am still not able to figure out what should I do next? Could someone help me out in this problem?
• The least primitive solution is $(x,y)=(5,1)$. See this post. Aug 3, 2020 at 16:56
• Could you please explain it in detail? Aug 3, 2020 at 16:57
## 2 Answers
This pictorial method is introduced in CONWAY and further discussed in HATCHER. There is also an intuitive book by Weissman with lots of pictures.
Here is the topograph diagram, showing solutions in a fundamental region as well as the automorphism generator. As integer column vectors, all (well, fundamental) solutions to $$x^2 - 21 y^2 = 4$$ are $$\left( \begin{array}{c} 2 \\ 0 \end{array} \right) \; \; , \; \; \left( \begin{array}{c} 5 \\ 1 \end{array} \right) \; \; , \; \; \left( \begin{array}{c} 23 \\ 5 \end{array} \right) \; \; . \; \;$$ For each of the three, multiply arbitrarily many times by $$A = \left( \begin{array}{cc} 55 & 252 \\ 12 & 55 \\ \end{array} \right) \; \; . \; \;$$ The first three such vectors are $$\left( \begin{array}{c} 110 \\ 24 \end{array} \right) \; \; , \; \; \left( \begin{array}{c} 527 \\ 115 \end{array} \right) \; \; , \; \; \left( \begin{array}{c} 2525 \\ 551 \\ \end{array} \right) \; \; . \; \;$$
The next three such vectors are $$\left( \begin{array}{c} 12098 \\ 2640 \end{array} \right) \; \; , \; \; \left( \begin{array}{c} 57965 \\ 12649 \end{array} \right) \; \; , \; \; \left( \begin{array}{c} 277727 \\ 60605 \\ \end{array} \right) \; \; . \; \;$$
Put another way, if we put all such $$x_n$$ and $$y_n$$ in two ordered sequences, $$2, 5, 23, 110, 527, 2525, 12098, 57965, 277727, 1330670, 6375623, 30547445, \ldots$$ $$0, 1, 5, 24, 115, 551, 2640, 12649, 60605, 290376, 1391275, 6665999, \ldots$$ Cayley-Hamilton tells us $$x_{n+6} = 110 x_{n+3} - x_n ,$$ $$y_{n+6} = 110 y_{n+3} - y_n .$$
Ummm. The continued fraction was not guaranteed to show all the solutions because 4 is larger than half the square root of 21, but this time we are lucky:
Method described by Prof. Lubin at Continued fraction of $\sqrt{67} - 4$
$$\sqrt { 21} = 4 + \frac{ \sqrt {21} - 4 }{ 1 }$$ $$\frac{ 1 }{ \sqrt {21} - 4 } = \frac{ \sqrt {21} + 4 }{5 } = 1 + \frac{ \sqrt {21} - 1 }{5 }$$ $$\frac{ 5 }{ \sqrt {21} - 1 } = \frac{ \sqrt {21} + 1 }{4 } = 1 + \frac{ \sqrt {21} - 3 }{4 }$$ $$\frac{ 4 }{ \sqrt {21} - 3 } = \frac{ \sqrt {21} + 3 }{3 } = 2 + \frac{ \sqrt {21} - 3 }{3 }$$ $$\frac{ 3 }{ \sqrt {21} - 3 } = \frac{ \sqrt {21} + 3 }{4 } = 1 + \frac{ \sqrt {21} - 1 }{4 }$$ $$\frac{ 4 }{ \sqrt {21} - 1 } = \frac{ \sqrt {21} + 1 }{5 } = 1 + \frac{ \sqrt {21} - 4 }{5 }$$ $$\frac{ 5 }{ \sqrt {21} - 4 } = \frac{ \sqrt {21} + 4 }{1 } = 8 + \frac{ \sqrt {21} - 4 }{1 }$$
Simple continued fraction tableau:
$$\begin{array}{cccccccccccccccccc} & & 4 & & 1 & & 1 & & 2 & & 1 & & 1 & & 8 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 4 }{ 1 } & & \frac{ 5 }{ 1 } & & \frac{ 9 }{ 2 } & & \frac{ 23 }{ 5 } & & \frac{ 32 }{ 7 } & & \frac{ 55 }{ 12 } \\ \\ & 1 & & -5 & & 4 & & -3 & & 4 & & -5 & & 1 \end{array}$$
$$\begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 21 \cdot 0^2 = 1 & \mbox{digit} & 4 \\ \frac{ 4 }{ 1 } & 4^2 - 21 \cdot 1^2 = -5 & \mbox{digit} & 1 \\ \frac{ 5 }{ 1 } & 5^2 - 21 \cdot 1^2 = 4 & \mbox{digit} & 1 \\ \frac{ 9 }{ 2 } & 9^2 - 21 \cdot 2^2 = -3 & \mbox{digit} & 2 \\ \frac{ 23 }{ 5 } & 23^2 - 21 \cdot 5^2 = 4 & \mbox{digit} & 1 \\ \frac{ 32 }{ 7 } & 32^2 - 21 \cdot 7^2 = -5 & \mbox{digit} & 1 \\ \frac{ 55 }{ 12 } & 55^2 - 21 \cdot 12^2 = 1 & \mbox{digit} & 8 \\ \end{array}$$
After this one must still apply the automorphism matrix arbitrarily many times...
• List all the solutions? Aug 4, 2020 at 17:53
• Explain it more clearly Aug 4, 2020 at 17:53
Given $$x^2-21y^2= 4$$ we can see $$(5,1)$$ as an easy solution where $$5^2-21= 4$$. Another observation is
$$x^2-21y^2= 4\implies \frac{x^2-4}{21}=y^2=\frac{x-2}{p}\cdot\frac{x+2}{q}\quad \text{ where }\quad p|x-2\quad\land\quad q|x+2$$
The factors of $$21$$ are $$1,3,7,21$$ and trying the cofactors $$(1,21)$$ we get conflicting answers for what is x.
$$x-2=1\implies x=3\quad \land \quad x+2=21\implies x=19 \lor\\ x-2=21\implies x=23\quad \land \quad x+2=1\implies x=-1$$
but the other two cofactors do yield consistent results for what is x.
$$x-2=3\implies x=5\quad \land \quad x+2=7\implies x=5$$ and this fits our desire to have integers that, multiplied, yield a square.
$$\frac{x-2}{3}\cdot\frac{x+2}{7}=\frac{5-2}{3}\cdot\frac{5+2}{7} =\frac{3}{3}\cdot\frac{7}{7}=\frac{21}{21}=1=y^2$$ Only positive integers have been used in this demonstration but the results are the same with $$(-5,-1)$$ because, multiplied, they become positive.
$$\therefore x^2-21y^2= 4\implies x=\pm5\quad y=\pm 1$$ | 2022-08-09 05:30:49 | {"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": 36, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8133276104927063, "perplexity": 331.2464828951499}, "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-2022-33/segments/1659882570901.18/warc/CC-MAIN-20220809033952-20220809063952-00231.warc.gz"} |
http://www.physicsforums.com/showthread.php?p=3204615 | ## Time dilation yes, why no contraction?
The example I have read to show time dilation is that of a light clock on a train. Two mirrors are one metre apart. Light is beamed from the first mirror to the second. An observer on the train sees that it takes the amount of time for light to travel 1m for the light to reach the second mirror. T a second observer at a train station, the light travels more than 1m, as the train, and thus the second mirror, has moved away from its original position between the light leaving the first mirror and hitting the second. As light speed is a constant, more time must pass for the observer on the platform than for the observer on the train in order to cater for the extra distance.
However, when the light rebounds from the second mirror to the first, while the observer on the train still perceives the light to travel 1m, for the observer on the platform, the light travels less than 1m as mirror one has moved closer to the point of the rebound (relative to the observer on the platform).
Where is the fault in my logic? It would seem that I am using the same arguments as those who thought that light travelled through "the ether". I can't see my error, however.
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Quote by fatdad However, when the light rebounds from the second mirror to the first, while the observer on the train still perceives the light to travel 1m, for the observer on the platform, the light travels less than 1m as mirror one has moved closer to the point of the rebound (relative to the observer on the platform).
I don't understand why you think the platform observer would see the light travel less than 1m. According to him, the light going from mirror 2 back to mirror 1 takes a similar tilted (and longer) path as it did when going from mirror 1 to mirror 2.
Draw yourself a diagram of the light path as seen by the platform observers. (Realize that the line between the two mirrors of the light clock is oriented perpendicular to the direction of travel.)
Quote by Doc Al I don't understand why you think the platform observer would see the light travel less than 1m. According to him, the light going from mirror 2 back to mirror 1 takes a similar tilted (and longer) path as it did when going from mirror 1 to mirror 2. Draw yourself a diagram of the light path as seen by the platform observers. (Realize that the line between the two mirrors of the light clock is oriented perpendicular to the direction of travel.)
Thanks. I was confusedly thinking of the mirrors in the direction of travel.
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Recognitions: | 2013-06-20 11:28:48 | {"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": 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": 0.5218616127967834, "perplexity": 477.61219209202835}, "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/1368711515185/warc/CC-MAIN-20130516133835-00079-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://forum.azimuthproject.org/plugin/ViewComment/15419 | David wrote:
> Observe that $\mathbb{Z}_2$ is a proper one-dimensional subspace of the one-dimensional space $\mathbb{Z}$.
It's not a subspace, it's a quotient space. That is, there's no 1-1 homomorphism of modules $\mathbb{Z}_2 \to \mathbb{Z}$. Instead, there's an onto homomorhism of modules $\mathbb{Z} \to \mathbb{Z}_2$, sending each integer to that integer mod 2.
By the way, people call these things "modules", not "spaces", which connotes vector space. The word "dimension" is also not one we use for modules over a ring. So, nobody would say "the one-dimensional space $\mathbb{Z}$". They'd say "the free $\mathbb{Z}$-module of rank one, $\mathbb{Z}$". | 2023-02-05 11:39:07 | {"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": 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": 0.9282903671264648, "perplexity": 651.974255797026}, "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-2023-06/segments/1674764500251.38/warc/CC-MAIN-20230205094841-20230205124841-00214.warc.gz"} |
https://www.ias.ac.in/listing/bibliography/pram/NITESH_K_CHOURASIA | • NITESH K CHOURASIA
Articles written in Pramana – Journal of Physics
• Comparative performance study of liquid core cylindrical Bragg fibre waveguide biosensors
The performance study of various sensing parameters such as sensitivity, detection accuracy and quality parameter of liquid core Bragg fibre waveguide biosensor based on defect mode has been theoretically studied and compared with experimental findings of a similar structure without defect mode. The electromagnetic wave propagation in the present structure has been modelled using the transfer matrix method and Henkel formalism in cylindrical coordinates. The present multilayer structure can provide a band gap between 617 and 929 nm wavelength range at angle of incidence θ= 70°. Due to the presence of a defect layer, a defect mode at 690 nm wavelength is observed in this band-gap region. This defect mode can be treated as a sensing signal in the present study. It is observed that the obtained sensitivity ($S$ ≈ 334 nm/RIU) through the defect mode is almost the same as the experimental findings ($S$ ≈ 330 nm/RIU) of a similar structure without the defect layer. But the obtained maximum detection accuracy (68.6) and quality parameter (160.4/RIU) of the present structure with defect layer is much larger than the values in a similar structure without defect layer (6.9 and 15/RIU). The present structure having a liquid-filled core, is therefore, favoured and useful in promising biosensing applications.
• # Pramana – Journal of Physics
Volume 95, 2021
All articles
Continuous Article Publishing mode
• # Editorial Note on Continuous Article Publication
Posted on July 25, 2019 | 2021-09-25 19:10:26 | {"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": 0.42817625403404236, "perplexity": 1865.9579625567799}, "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/1631780057733.53/warc/CC-MAIN-20210925172649-20210925202649-00182.warc.gz"} |
https://zenodo.org/record/3303581/export/csl | Journal article Open Access
# Beyond replicator dynamics: From frequency to density dependent models of evolutionary games
Křivan V; Galanthay T. E.; Cressman R.
### Citation Style Language JSON Export
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"publisher": "Zenodo",
"DOI": "10.5281/zenodo.3303581",
"author": [
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"family": "K\u0159ivan V"
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{
"family": "Galanthay T. E."
},
{
"family": "Cressman R."
}
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"issued": {
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2018,
10,
14
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"abstract": "<p>Game theoretic models of evolution such as the Hawk–Dove game assume that individuals gain fitness (which is a proxy of the per capita population growth rate) in pair-wise contests only. These models assume that the equilibrium distribution of phenotypes involved (e.g., Hawks and Doves) in the population is given by the Hardy–Weinberg law, which is based on instantaneous, random pair formation. On the other hand, models of population dynamics do not consider pairs, newborns are produced by singles, and interactions between phenotypes or species are described by the mass action principle. This article links game theoretic and population approaches. It shows that combining distribution dynamics with population dynamics can lead to stable coexistence of Hawk and Dove population numbers in models that do not assume <em>a priori</em> that fitness is negative density dependent. Our analysis shows clearly that the interior Nash equilibrium of the Hawk and Dove model depends both on population size and on interaction times between different phenotypes in the population. This raises the question of the applicability of classic evolutionary game theory that requires all interactions take the same amount of time and that all single individuals have the same payoff per unit of time, to real populations. Furthermore, by separating individual fitness into birth and death effects on singles and pairs, it is shown that stable coexistence in these models depends on the time-scale of the distribution dynamics relative to the population dynamics. When explicit density-dependent fitness is included through competition over a limited resource, the combined dynamics of the Hawk–Dove model often lead to Dove extinction no matter how costly fighting is for Hawk pairs.</p>",
"title": "Beyond replicator dynamics: From frequency to density dependent models of evolutionary games",
"type": "article-journal",
"id": "3303581"
}
35
64
views | 2021-08-03 15:15:30 | {"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": 0.2641153931617737, "perplexity": 3003.7921589540474}, "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-2021-31/segments/1627046154459.22/warc/CC-MAIN-20210803124251-20210803154251-00449.warc.gz"} |
https://math.stackexchange.com/questions/179292/how-to-classify-3-sheeted-covering-space-for-s-1-vee-s-1 | # How to classify 3-sheeted covering space for $S_{1}\vee S_{1}$?
This might be a duplicate. This question also feels routine (it is also the execrise 10, page 88 in Hatcher). From Harvard qualification exam, 1990.
Let $X$ be figure eight.
1) How many 3-sheeted, connected covering space are there for $X$ up to isomorphism?
2) How many of these are normal (i.e Galois) covering spaces?
There are almost uncountably many covering spaces $Y$ for $X$ (one can check the corresponding page in Hatcher, page 58). The question is how to classify them nicely. I know that $p_{*}\pi_{1}(Y)$ has index 3 in $\pi_{1}(X)=\mathbb{Z}* \mathbb{Z}$(the free group generated by two generators). But I do not know how to find all index 3 subgroups of $\mathbb{Z}*\mathbb{Z}$. On the other hand if $H$ is normal in $\mathbb{Z}*\mathbb{Z}$, then the above question can be greatly simplified, but I still do not know how to solve it precisely. I tried to think in terms of deck transformations, etc but did not get anywhere.
Hint: Instead of thinking about index 3 subgroups of $\Bbb Z \star \Bbb Z$, consider what connected 3-fold covers of $S_1 \vee S_1$ look like. Any such cover is a connected graph on 3 vertices of valence 4 (why?), and there are only finitely many such graphs. Then use deck transformations to check if each cover is regular.
• @Ergin: Say $\pi_1(S^1 \vee S^1) = \langle a, b \rangle$. Then there are only 2 normal subgroups up to swapping $a$ and $b$ (and in one case, swapping $b$ with $b^{-1}$). There are actually 4 normal subgroups. Some people will not distinguish, for example, between the 4th and 5th covers, as these are really only swapping $a$ and $b$. That was what was meant by the 2 isomorphism classes of regular covers. – Brandon Carter Jun 22 '13 at 18:25 | 2019-11-13 02:59:33 | {"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": 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": 0.8607121706008911, "perplexity": 181.08951563293888}, "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/1573496665976.26/warc/CC-MAIN-20191113012959-20191113040959-00104.warc.gz"} |
https://gdeq.org/Seminar_talk,_25_February_2015 | # Seminar talk, 25 February 2015
Speaker: Valery Yumaguzhin
Title: Differential invariants on solutions of equations of adiabatic gas motion
Abstract:
The talk will discuss the system of equations of adiabatic gas motion in ${\displaystyle n}$-dimensional space, ${\displaystyle n=1,2,3}$.
Characteristic covectors of this system generate a geometric structure on every solution of this system. This structure consists of a hyperplane and a non degenerate cone in every cotangent space to a solution. These hyperplane and cone intersect in zero point only.
We construct differential invariants of this structure: a vector field, a metric, and a linear connection with torsion in general position.
In the case of polytropic gas motion, we calculate classes of explicit solutions possessing the linear connection with zero torsion.
Video
Slides: Yumaguzhin V. Differential invariants on solutions of equations of adiabatic gas motion (presentation at the Krasil'shchik's IUM Seminar, 25 February 2015).pdf | 2021-06-25 07:26:58 | {"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": 0.5993317365646362, "perplexity": 1670.065309259106}, "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/1623487622113.11/warc/CC-MAIN-20210625054501-20210625084501-00208.warc.gz"} |
https://chemistry.stackexchange.com/questions/26723/finding-equilibrium-concentrations-trouble-simplifying-equation | # Finding equilibrium concentrations, trouble simplifying equation
For $\ce{CO (g) + H2 (g) <=> CH2O (g)}$
with Kc = 0.068 @ 273K
initial concentrations being $\ce{[CO]}$ = 1.25M, $\ce{[H2]}$ = 2.00M, $\ce{[CH2O]}$ = 1.00M
I did $\frac{1.00}{((1.25)(2.00)} = 0.4$, which is more then Kc, so it's leaning to the left, will produce more reactants, and products will lower. So the I did:
$\frac{1.00 - x}{(1.25 + x)(2.00 + x)} = 0.068$
Multiplied the bottom and got $X^2 + 3.25x + 2.5$
so
$\frac{1.00 - x}{X^2 + 3.25x + 2.5} = 0.068$
I then tried multiplying the denominator by 0.068 and rearranging to solve for $x$ from there, but that's where I got stuck. $X$ is supposed to be 0.65.
Everything you do is correct and it is only the final algebraic transformation missing: \begin{align} 0.068 &= \frac{1-x}{(1.25+x)(2.00+x)}\\ 0.068 &= \frac{1-x}{x^2 + 3.25x + 2.5} &&|\cdot (x^2 + 3.25x + 2.5)\\ 1-x &= 0.068(x^2 + 3.25x + 2.5) &&|-1 +x\\ 0.0 &= 0.068x^2 + 1.221x -0.83\\ \end{align}
Now you can solve for $x$ according to \begin{align} 0&=ax^2+bx+c & x_{1,2}&=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\ \end{align}
And that gives me the values $x_1\approx0.656$ and $x_2\approx-18.6$. The latter one of course makes no sense. | 2020-09-25 18:59:04 | {"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9993752837181091, "perplexity": 2613.2580123282805}, "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-2020-40/segments/1600400228707.44/warc/CC-MAIN-20200925182046-20200925212046-00440.warc.gz"} |
https://www.physicsoverflow.org/14/significance-hyperfinite-%24iii_1%24-axiomatic-quantum-theory?show=367 | # Significance of the hyperfinite $III_1$ factor for axiomatic quantum field theory
+ 17 like - 0 dislike
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Using a form of the Haag-Kastler axioms for quantum field theory (see AQFT on the nLab for more details), it is possible in quite general contexts to prove that all local algebras are isomorphic to the hyperfinite $III_1$ factor or to the tensor product of the $III_1$ factor with the center of the given local algebra.
(A local algebra is the algebra of observables that is associated to an open bounded subset of Minkowski space. The term $III_1$ factor refers to the Murray-von Neumann classification of factors of von Neumann algebras).
Also see this question on math overflow for more details.
So one could say that quantum mechanics has the $I_n$ and $I_{\infty}$ factors as playground, while QFT has the hyperfinite $III_1$ factor as playground.
My questions has two parts:
1) I would like to know about a concrete physical system where it is possible to show that the local algebras are hyperfinite $III_1$ factors, if there is one where this is possible.
2) Is there an interpretation in physical terms of the presence of the hyperfinite $III_1$ factor in QFT?
This post has been migrated from (A51.SE)
retagged Mar 24, 2014
+ 17 like - 0 dislike
Yngvason, J. (2005). The role of type III factors in quantum field theory. Reports on Mathematical Physics, 55(1), 135–147. (arxiv)
The Type III property says something about statistical independence. Let $\mathcal{O}$ be a double cone, and let $\mathfrak{A}(\mathcal{O})$ be the associated algebra of observables. Assuming Haag duality, we have $\mathfrak{A}(\mathcal{O}')'' = \mathfrak{A}(\mathcal{O})$. If $\mathfrak{A}(\mathcal{O})$ is not of Type I, the Hilbert space $\mathcal{H}$ of the system does not decompose as $\mathcal{H} = \mathcal{H}_1 \otimes \mathcal{H}_2$ in such a way that $\mathfrak{A}(\mathcal{O})$ acts on the first tensor factor, and $\mathfrak{A}(\mathcal{O}')$ on the second. This implies that one cannot prepare the system in a certain state when restricted to measurements in $\mathcal{O}$ regardless of the state in the causal complement. It should be noted that if the split property holds, that is there is a Type I factor $\mathfrak{N}$ such that $\mathfrak{A}(\mathcal{O}) \subset \mathfrak{N} \subset \mathfrak{A}(\widehat{\mathcal{O}})$ for some region $\mathcal{O} \subset \widehat{\mathcal{O}}$, a slightly weaker property is available: a state can be prepared in $\mathcal{O}$ irregardless of the state in $\widehat{\mathcal{O}}'$. An illustration of the consequences can be found in the article above.
Another consequence is that the Borchers property B automatically holds: if $P$ is some projection in $\mathfrak{A}(\mathcal{O})$, then there is some isometry $W$ in the same algebra such that $W^*W = I$ and $W W^* = P$. This implies that we can modify the state locally to be an eigenstate of $P$, by doing the modification $\omega(A) \to \omega_W(A) = \omega(W^*AW)$. Note that $\omega_W(P) = 1$ and $\omega_W(A) = \omega(A)$ for $A$ localised in the causal complement of $\mathcal{O}$. Type III$_1$ implies something slightly stronger, see the article cited for more details.
As to the first question, one can prove that the local algebras of free field theories are Type III. This was done by Araki in the 1960's. You can find references in the article mentioned above. In general, the Type III condition follows from natural assumptions on the observable algebras. Non-trivial examples probably have to be found in conformal field theory, but I do not know any references on the top of my head.
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answered Sep 15, 2011 by (550 points)
Ok, I somehow missed that paper, although Yngvason is on my watch list :-)
This post has been migrated from (A51.SE)
Feel free to expand you answer later, even though I will accept it already.
This post has been migrated from (A51.SE)
Yes, I will. Have to read the paper again myself first, I don't recall the details...
This post has been migrated from (A51.SE)
I updated the question, feel free to ask for more details :)
This post has been migrated from (A51.SE)
+ 4 like - 0 dislike
Regarding the first question. As Pieter already said for a conformal net the $III_1$ property holds (if it is not $\mathbb C$). Further $e^{-\beta L_0}$ being trace class for all $\beta>0$ with $L_0$ the generator of the rotations implies the split property, which implies $\mathcal A(I)$ to be the hyperfinite $III_1$-factor.
edit The property $III_1$ and trace class implies split can be found in - D'Antoni,Longo,Radulescu. Conformal Nets, Maximal Temperature and Models from Free Probability [arXiv:math/9810003v1]
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answered Oct 26, 2011 by (300 points)
Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor) Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysics$\varnothing$verflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register. | 2020-10-24 03:14:17 | {"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": 0.7330187559127808, "perplexity": 375.66060849655486}, "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-2020-45/segments/1603107881640.29/warc/CC-MAIN-20201024022853-20201024052853-00459.warc.gz"} |
http://tex.stackexchange.com/tags/backgrounds/new | # Tag Info
## New answers tagged backgrounds
1
This is also possible using background package, according to this answer: \documentclass{beamer} \usepackage{lipsum} \usepackage{background} \backgroundsetup{ placement=top, position={\paperwidth-0.7in,-0.2in}, scale=1, opacity=1, contents={\Huge DRAFT}, } \setbeamertemplate{background}{\BgMaterial} ...
5
You don't have to do anything special, nor to quit using the background package; due to the way frames are built in beamer, you simply have to add the background material to the appropriate template: \setbeamertemplate{background}{\BgMaterial} this is exactly what the \BgMaterial command was designed for. A little complete example: \documentclass{beamer} ...
2
It is better to use beamers facilities for this job. The text can be placed with the help of tikz. Remember that this needs 2-3 compilation runs to settle down since we are using remember picture option. \documentclass{beamer} \usepackage{lipsum} \usepackage{tikz} \setbeamertemplate{background}{% \begin{tikzpicture}[overlay,remember picture] ...
4
Put a rule of 1in height and 0in width and another with 1in width and 0in height. \documentclass{beamer} \usepackage{lipsum} \begin{document} \setbeamertemplate{background}{% \rule{0in}{1in}% \rule{1in}{0in}% \Huge DRAFT } \frame{\lipsum[1]} \end{document} There is always a possibility of using tikz to put things on a page byt here only light ...
4
Inside a picture environment, you can place things a arbitrary positions using \put \documentclass{beamer} \usepackage{lipsum} \begin{document} { \setbeamertemplate{background}{ \begin{picture}(254,190.5)(0,0)% \put(42,72){% Adjust these coordinates \Huge DRAFT% }% \end{picture} } ...
1
The problem it's probably related with how beamer uses background layers and I don't know how to solve it. But for this particular case, it's easy to find a workaround because the graph can be drawn as a matrix which avoid us from using fit nodes and background layers. A TikZ matrix is a node which contains several inner nodes and can be fill-ed like any ...
1
I think there's something wrong in your system. With default inner sep in fit nodes, this is what I get: I'm using MikTeX 2.9 with last pgf and beamer versions. \documentclass{beamer} \usepackage{tikz} \usetikzlibrary{backgrounds,positioning,fit} \begin{document} \begin{frame} \begin{figure} \begin{tikzpicture}[scale=0.5] ...
2
Use tcolorbox. \documentclass{article} \usepackage{lipsum} \usepackage[most]{tcolorbox} \definecolor{bg}{RGB}{255,249,227} \begin{document} \lipsum[1] \begin{tcolorbox}[enhanced jigsaw,colback=bg,boxrule=0pt,arc=0pt] \begin{itemize} \item Item 1 \item Item 2 \item Item 3 \end{itemize} \end{tcolorbox} \end{document} ...
5
I know this question has been answered very extensively. But not what I wanted or thought was the question based on the title. Therefore if others get in here looking for a possibility for colouring behind a word than this snippet is much easier: \colorbox{blue!30}{blue} or \textcolor{blue!30}{blue} resulting in: This is possible by only adding ...
Top 50 recent answers are included | 2015-05-24 21:38:10 | {"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": 0.9791877865791321, "perplexity": 5417.8180034856}, "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-2015-22/segments/1432207928078.25/warc/CC-MAIN-20150521113208-00222-ip-10-180-206-219.ec2.internal.warc.gz"} |
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Indistinguishability Obfuscation and Multi-linear Maps: A Brave New World (Guest Post by Ran Canetti)
The following post is written by Ran Canetti
A bunch of us hapless cryptographers got the following boilerplate comment from the FOCS’15 PC:
Overall, submissions related to multi-linear maps and indistinguishability obfuscation were held to a somewhat higher standard. The PC expressed some concern with the recent flurry of activities pertaining to multi-linear maps and indistinguishability obfuscation, given how little we understand and can say and *prove* about the underlying hardness assumptions.
This comment was clearly written with the best of intentions, to explain views expressed at the PC deliberations. And I’m thankful to it – mainly since it made the underlying misconceptions so explicit that it mandated a response. So, after discussing and commiserating with colleagues here at Simons, and after amusing ourselves with some analogues of above statement (e.g., “results on NP completeness are held to a higher standard given how little we understand and can say and *prove* about the hardness solving SAT in polynomial time”), I decided to try to write an – obviously subjective – account for the recent developments in multilinear maps and indistinguishability obfuscation (IO) and why this exciting research should be embraced and highlighted rather than “held to a somewhat higher standard” — in spite of how little we understand about the underlying assumptions. The account is aimed at the general CS-theorist.
Let me start by giving rough definitions of the concepts involved. An Indistinguishability Obfuscator (IO) is a randomized algorithm O that takes as input a circuit C and outputs a (distribution over) circuits O(C) with the properties that:
• C and O(C) have the same functionality,
• O(C) is only polynomially larger than C,
• for any two same-size, functionally equivalent circuits C and C’ we have that O(C) ~ O(C’) (i.e., the distributions over strings representing O(C) and O(C’) are computationally indistinguishable).
IO has been proposed as a notion of obfuscation in 2000 (Hada, Barak-Goldreich-Impagliazzo-Sahai-Vadhan-Yang). Indeed, it is arguably a clean and appealing notion – in some sense the natural extension of semantic security of standard encryption to “functionality-preserving encryption of programs”. However, it has been largely viewed as too weak to be of real applicability or interest. (There were also no candidate polytime IO schemes, but this in my eyes is a secondary point, see below.)
Things changed dramatically in 2013 when Sahai and Waters demonstrated how IO schemes can be ingeniously combined with other rather “mundane” cryptographic constructs to do some amazing things. Since then dozens of papers came about that extend the SW techniques and apply them to obtain even more amazing things – that by now have transcended crypto and spilled over to other areas. (e.g.: deniable encryption, succinct delegation, succinct multi-party computation with hardly any interaction, one message succinct witness hiding and witness indistinguishable proofs, hash functions with random-oracle-like properties, hardness results for PPAD, and many more). In fact, think about a result in your area that assumes that some computation is done inside a black box – most probably IO can replace that assumption in one way or another…
Still, my (subjective but distinct) feeling is that we are far from understanding the limits and full power of IO. Furthermore, the study of IO has brought with it a whole new toolbox of techniques that are intriguing in their own right, and teach us about the power and limitations of working with “encrypted computations”.
So far I have not mentioned any candidate constructions of IO – and indeed the above study is arguably valuable as a pure study of this amazing concept, even without any candidate constructions. (Paraphrasing Levin on quantum computers, one can take the viewpoint that the above is the study of impossibility results for IO…)
However, unlike quantum computers, here we also have candidate constructions. This is where multilinear maps come to play.
Multi-linear maps are this cool new technical tool (or set of tools) that was recently put forth. (The general concept was proposed by Boneh and Silverberg around 2000, and the first candidate construction of one of the current variants was presented in 2012 by Garg, Gentry and Halevi.) Essentially, a multilinear map scheme is a fully homomorphic encryption scheme where the public key provides, in addition to the ability to encrypt elements and perform homomorphic operations on ciphertexts, also the ability to partially decrypt ciphertexts under certain restrictions. There are many incomparable variants of this general paradigm, which differ both in the functionality provided and in the security guarantees. Indeed, variants appear to be closely tied to candidate constructions. Furthermore, our understanding of what’s possible here has been evolving considerably, with multiple new constructions, attacks, and fixes reported.
Still, the number and variety of applications of multi-linear maps makes it clear that this “family of primitives” is extremely powerful and well worth studying – both at the level of candidate constructions, at the level of finding the “right” computational abstractions, and at the level of applications. In a sense, we are here back to the 70’s: we are faced with this new set of algebraic and number theoretic tools, and are struggling to find good ways to use them and abstract them.
Indeed, some of the most powerful applications of multilinear maps are candidate constructions of IO schemes. The first such candidate construction (by Garg, Gentry, Halevi, Raykova, Sahai and Waters in 2013) came with only heuristic arguments for security; However more rigorous analyses of this and other constructions, based on well-defined formulations of multi-linear map variants, soon followed suite. Some of these analyses have eventually been “broken” in the sense that we currently don’t have candidate constructions that satisfy the properties they assume. Still, other analyses do remain valid. Indeed, there are no attacks against the actual basic IO scheme of Garg etal.
The fact that the only current candidate constructions of IO need to assume existence of some variant of multi-linear maps at some point or another may make it seem as it the two concepts are somehow tied together. However, there is no reason to believe that this is the case. For all we know, multi-linear maps are just the path first uncovered to IO, and other paths may well be found. Similarly, even if IO turns out to be unobtainable for some reason, the study of multilinear maps and their power will still remain very relevant.
So, to sum up this long-winded account:
• IO is a natural and fascinating computational concept. Studying its consequences (both within and outside cryptography) is a well worth endeavor.
• Studying new candidate constructions of IO and/or new analyses of their security is another well worth endeavor.
• Multilinear maps are an intriguing and powerful set of techniques and tools. Finding better candidate constructions and abstractions is of central importance to cryptography. Finding new cool uses of these maps is another intriguing challenge.
• The three should be treated as separate (although touching and potentially interleaving) research efforts.
———–
I’d like to thank Guy Rothblum and Vinod Vaikuntanathan for great comments that significantly improved this post.
by Luca Trevisan at July 06, 2015 04:05 PM UTC
ICALP/LICS 2015: Day 1
from Luca Aceto
Despite my jet-lag induced drowsiness and the fact that the clock on my laptop kept reminding that I was chairing the first meeting of the EATCS Council at around 4am in the morning, I enjoyed the first day of the conference a lot.
According to the organizers, ICALP/LICS 2015 had 400 early registrations and there were 200 people who attended the pre-conference workshops.The plenary invited talks by Ken-ichi Kawarabayashi (NII, Japan) and Luke Ong (University of Oxford, UK) were delivered to a packed room and were both excellent.
Ken-ichi's talk was entitled Digraphs Structures: Minors and Algorithms and presented his joint work with Stephan Kreutzer on extending results on minors from graphs to digraphs. (See this paper in particular.)
Luke's talk was devoted to Higher-Order Model Checking: An Overview. Higher-order model checking is about checking whether trees generated by recursion schemes satisfy formulae that can be expressed in some kind of logic, such as Monadic Second Order logic. In his talk, Luke surveyed the considerable progress on this problem in both theory and practice that has been made over the past fifteen years or so.
There were lots interesting talks in the conference programs as well as two sessions devoted to the best paper award recipients.
The busy day ended socially on a high note with an excellent welcome reception.
Looking at the future, I am pleased to inform you that the invited speakers for ICALP 2016 will be
I will have more to say about future editions of the conference after the general assembly tomorrow.
by Luca Aceto (noreply@blogger.com) at July 06, 2015 01:59 PM UTC
News for June 2015
This month we have a number of results that are related to query complexity though not directly related to property testing.
Diamond Sampling for Approximate Maximum All-pairs Dot-product (MAD) Search by Grey Ballard, Tamara G. Kolda, Ali Pinar and C. Seshadhri (arXiv). For a long time we have been hoping to use our tools and techniques to solve problems that are useful in the real world One such problem which has importance in the real life applications is, given two sets of vectors the problem is to find the t pairs of vectors with the highest dot product. In this paper they use clever sampling techniques to design algorithms which are better than the state-of-the-art algorithms. They not only give theoretical guarantee but also validate their results empirically. Bridging the gap between theory and practice is an extremely important at the same time a very challenging job. We hope more work will be done in this direction.
Sub-linear Upper Bounds on Fourier dimension of Boolean Functions in terms of Fourier sparsity by Swagato Sanyal (arXiv). Understanding the relationship between Fourier dimension of a Boolean function and sparsity is an important problem. In this paper a better bound on the Fourier dimension in terms of sparsity is obtained. The main technique is to use the fact that the Fourier dimension is equivalent to the the non-adaptive parity decision tree and then bounding the parity decision tree in terms of sparsity.
Relationship between Deterministic Query Complexity, Randomized Query Complexity and Quantum Query Complexity. In the world of query complexity understanding the exact relationship between the the various models of computation is the main problem. It is known that Deterministic Query Complexity, Randomized Query Complexity and Quantum Query complexity are all polynomially related. But the exact polynomial relation between them is not known. Last month there was a sudden burst of activity in this area with three papers addressing this problem coming out is a span of two weeks. In the papers Separations in Query Complexity Based on Pointer Functions by Andris Ambainis, Kaspars Balodis, Aleksandrs Belovs, Troy Lee, Miklos Santha and Juris Smotrovs (arXiv) and Towards Better Separation between Deterministic and Randomized Query Complexity by Sagnik Mukhopadhyay Swagato Sanyal (arXiv) it is proved that the Randomized Query Complexity and the Deterministic Query Complexity are quadratically related and that this bound is tight up to logarithmic factors. Very soon after in A Super-Grover Separation Between Randomized and Quantum Query Complexities by Shalev Ben-David (arXiv) it was proved that the separation between the Quantum Query Complexity and Randomized query Complexity is super quadratic. With these three results our knowledge about the query complexity is slightly more clearer.
by Sourav Chakraborty at July 06, 2015 07:25 AM UTC
Indistinguishability Obfuscation and Multi-linear Maps: A Brave New World – Guest Post by Ran Canetti
A bunch of us hapless cryptographers got the following boilerplate comment from the FOCS’15 PC:
“Overall, submissions related to multi-linear maps and indistinguishability obfuscation were held to a somewhat higher standard. The PC expressed some concern with the recent flurry of activities pertaining to multi-linear maps and indistinguishability obfuscation, given how little we understand and can say and *prove* about the underlying hardness assumptions”.
This comment was clearly written with the best of intentions, to explain views expressed at the PC deliberations. And I’m thankful to it – mainly since it made the underlying misconceptions so explicit that it mandated a response. So, after discussing and commiserating with colleagues here at Simons, and after amusing ourselves with some analogues of above statement (e.g., “results on NP completeness are held to a higher standard given how little we understand and can say and *prove* about the hardness solving SAT in polynomial time”), I decided to try to write an – obviously subjective – account for the recent developments in multilinear maps and indistinguishability obfuscation (IO) and why this exciting research should be embraced and highlighted rather than “held to a somewhat higher standard” — in spite of how little we understand about the underlying assumptions. The account is aimed at the general CS-theorist.
Let me start by giving rough definitions of the concepts involved. An Indistinguishability Obfuscator (IO) is a randomized algorithm O that takes as input a circuit C and outputs a (distribution over) circuits O(C) with the properties that:
1. C and O(C) have the same functionality,
2. O(C) is only polynomially larger than C
3. for any two same-size, functuinally equivalent circuits C and C’ we have that O(C) ~ O(C’) (i.e., the distributions over strings representing O(C) and O(C’) are computationally indistinguishable).
IO has been proposed as a notion of obfuscation in 2000 (Hada, Barak-Goldreich-Impagliazzo-Sahai-Vadhan-Yang). Indeed, it is arguably a clean and appealing notion – in some sense the natural extension of semantic security of standard encryption to “functionality-preserving encryption of programs”. However, it has been largely viewed as too weak to be of real applicability or interest. (There were also no candidate polytime IO schemes, but this in my eyes is a secondary point, see below.)
Things changed dramatically in 2013 when Sahai and Waters demonstrated how IO schemes can be ingeniously combined with other rather “mundane” cryptographic constructs to do some amazing things. Since then dozens of papers came about that extend the SW techniques and apply them to obtain even more amazing things – that by now have transcended crypto and spilled over to other areas. (e.g.: deniable encryption, succinct delegation, succinct multi-party computation with hardly any interaction, one message succinct witness hiding and witness indistinguishable proofs, hash functions with random-oracle-like properties, hardness results for PPAD, and many more). In fact, think about a result in your area that assumes that some computation is done inside a black box – most probably IO can replace that assumption in one way or another…
Still, my (subjective but distinct) feeling is that we are far from understanding the limits and full power of IO. Furthermore, the study of IO has brought with it a whole new toolbox of techniques that are intriguing in their own right, and teach us about the power and limitations of working with “encrypted computations”.
So far I have not mentioned any candidate constructions of IO – and indeed the above study is arguably valuable as a pure study of this amazing concept, even without any candidate constructions. (Paraphrasing Levin on quantum computers, one can take the viewpoint that the above is the study of impossibility results for IO…)
However, unlike quantum computers, here we also have candidate constructions. This is where multilinear maps come to play.
Multi-linear maps are this cool new technical tool (or set of tools) that was recently put forth. (The general concept was proposed by Boneh and Silverberg around 2000, and the first candidate construction of one of the current variants was presented in 2012 by Garg, Gentry and Halevi.) Essentially, a multilinear map scheme is a fully homomorphic encryption scheme where the public key provides, in addition to the ability to encrypt elements and perform homomorphic operations on ciphertexts, also the ability to partially decrypt ciphertexts under certain restrictions. There are many incomparable variants of this general paradigm, which differ both in the functionality provided and in the security guarantees. Indeed, variants appear to be closely tied to candidate constructions. Furthermore, our understanding of what’s possible here has been evolving considerably, with multiple new constructions, attacks, and fixes reported.
Still, the number and variety of applications of multi-linear maps makes it clear that this “family of primitives” is extremely powerful and well worth studying – both at the level of candidate constructions, at the level of finding the “right” computational abstractions, and at the level of applications. In a sense, we are here back to the 70’s: we are faced with this new set of algebraic and number theoretic tools, and are struggling to find good ways to use them and abstract them.
Indeed, some of the most powerful applications of multilinear maps are candidate constructions of IO schemes. The first such candidate construction (by Garg, Gentry, Halevi, Raykova, Sahai and Waters in 2013) came with only heuristic arguments for security; However more rigorous analyses of this and other constructions, based on well-defined formulations of multi-linear map variants, soon followed suite. Some of these analyses have eventually been “broken” in the sense that we currently don’t have candidate constructions that satisfy the properties they assume. Still, other analyses do remain valid. Indeed, there are no attacks against the actual basic IO scheme of Garg et al.
The fact that the only current candidate constructions of IO need to assume existence of some variant of multi-linear maps at some point or another may make it seem as it the two concepts are somehow tied together. However, there is no reason to believe that this is the case. For all we know, multi-linear maps are just the path first uncovered to IO, and other paths may well be found. Similarly, even if IO turns out to be unobtainable for some reason, the study of multilinear maps and their power will still remain very relevant.
So, to sum up this long-winded account:
• IO is a natural and fascinating computational concept. Studying its consequences (both within and outside cryptography) is a well worth endeavor.
• Studying new candidate constructions of IO and/or new analyses of their security is another well worth endeavor.
• Multilinear maps are an intriguing and powerful set of techniques and tools. Finding better candidate constructions and abstractions is of central importance to cryptography. Finding new cool uses of these maps is another intriguing challenge.
• The three should be treated as separate (although touching and potentially interleaving) research efforts.
———–
I’d like to thank Guy Rothblum and Vinod Vaikuntanathan for great comments that significantly improved this post.
by Guy Rothblum at July 06, 2015 03:14 AM UTC
Historical talks
During the summer program on cryptography, every Monday afternoon there is a talk on the history of a classic paper or series of papers. Last week, Russell Impagliazzo spoke on his work with Steven Rudich on the impossibility of basing one-way permutations or key agreement on the existence of one-way functions. The week before, Umesh Vazirani spoke on quantum computing, and earlier Ron Rivest spoke about the origins of modern cryptography.
All talks are recorded, and the recordings are available here.
Tomorrow afternoon, Daniele Micciancio will speak at 3pm on lattice-based cryptography.
Tomorrow is also the first day of the Workshop on the mathematics of modern cryptography. The program starts at 9:30am Pacific time, and all talk are broadcast live, as usual.
by Luca Trevisan at July 06, 2015 02:05 AM UTC
Does Bob Deserve the lavish acknowledgement: A problem in Logic
Alice and Carol are real mathematicians.
Bob is an English major who does not know any mathematics.
(This story is based on a true incident.)
Alice writes a math paper. Carol reads it and offers corrections of style and grammar and how-to-say-things. She also helps simplify some of the proofs. She does not deserve a co-authorship but Alice does of course write in the acknowledgements
I would like to thank Carol for proofreading and for help with some of the proofs.
Bob points out that this is silly --- if she would like to thank Carol then do so. So Alice changes it to
I thank Carol for proofreading and for help with some of the proofs.
Even though Bob does not understand the math he begins reading the paper. He finds a few grammar mistakes, some points of style, and even a math mistake:
BOB: Alice, this sentence mentions A1 and A2, is A1 the steak sauce?
ALICE: Its A sub 1 and no it is not the steak sauce.
BOB: But later in the sentence there is a reference to A? Maybe its implicit what A is and I don't get it since I don't know the math, but it does look funny.
ALICE: Well pierce my ears and call be drafty! You're right! It should be A1, A2, and A_1 ∩ A_2.
SO, in the end Bob DID proofread the paper and DID help. Alice wants to include him in the acknowledgements. She modifies the ack to
I thank Bob and Carol for proofreading and help with some of the proofs.
Is that correct? Bob just did proofreading, and Carol did proofreading AND helped with some proofs. In logical terms
If B did X and C did X and Y then
B AND C did X AND Y
does seem correct.
But is also seems misleading. Alice could separate it out:
I thank Carol for proofreading and help with some of the proofs.
I thank Bob and Carol for proofreading.
Thats more accurate but also more cumbersome.
But my real question is, is the I THANK BOB AND CAROL... statement correct or incorrect? In logic correct, in English, perhaps not. We could ask Bob who is an English major and maybe get a paper out of it which Carol can proofread!
by GASARCH (noreply@blogger.com) at July 06, 2015 01:29 AM UTC
Using the Johnson-Lindenstrauss lemma in linear and integer programming
Authors: Ky Vu, Pierre-Louis Poirion, Leo Liberti
Abstract: The Johnson-Lindenstrauss lemma allows dimension reduction on real vectors with low distortion on their pairwise Euclidean distances. This result is often used in algorithms such as $k$-means or $k$ nearest neighbours since they only use Euclidean distances, and has sometimes been used in optimization algorithms involving the minimization of Euclidean distances. In this paper we introduce a first attempt at using this lemma in the context of feasibility problems in linear and integer programming, which cannot be expressed only in function of Euclidean distances.
Algebraic and model theoretic methods in constraint satisfaction
Authors: Michael Pinsker
Abstract: This text is related to the tutorials I gave at the Banff International Research Station and within a "Doc-course" at Charles University Prague in the fall of 2014. It describes my current research and some of the most important open questions related to it.
Optimal linear Bernoulli factories for small mean problems
Authors: Mark Huber
Abstract: Suppose a coin with unknown probability $p$ of heads can be flipped as often as desired. A Bernoulli factory for a function $f$ is an algorithm that uses flips of the coin together with auxiliary randomness to flip a single coin with probability $f(p)$ of heads. Applications include near perfect sampling from the stationary distribution of regenerative processes. When $f$ is analytic, the problem can be reduced to a Bernoulli factory of the form $f(p) = Cp$ for constant $C$. Presented here is a new algorithm where for small values of $Cp$, requires roughly only $C$ coin flips to generate a $Cp$ coin. From information theory considerations, this is also conjectured to be (to first order) the minimum number of flips needed by any such algorithm.
For $Cp$ large, the new algorithm can also be used to build a new Bernoulli factory that uses only 80\% of the expected coin flips of the older method, and applies to the more general problem of a multivariate Bernoulli factory, where there are $k$ coins, the $k$th coin has unknown probability $p_k$ of heads, and the goal is to simulate a coin flip with probability $C_1 p_1 + \cdots + C_k p_k$ of heads.
Anti-concentration for random polynomials
Authors: Oanh Nguyen, Van Vu
Abstract: We prove anti-concentration results for polynomials of independent Rademacher random variables, with arbitrary degree. Our results extend the classical Littlewood-Offord result for linear polynomials, and improve several earlier estimates. As an application, we address a challenge in complexity theory posed by Razborov and Viola.
Online Self-Indexed Grammar Compression
Authors: Yoshimasa Takabatake, Yasuo Tabei, Hiroshi Sakamoto
Abstract: Although several grammar-based self-indexes have been proposed thus far, their applicability is limited to offline settings where whole input texts are prepared, thus requiring to rebuild index structures for given additional inputs, which is often the case in the big data era. In this paper, we present the first online self-indexed grammar compression named OESP-index that can gradually build the index structure by reading input characters one-by-one. Such a property is another advantage which enables saving a working space for construction, because we do not need to store input texts in memory. We experimentally test OESP-index on the ability to build index structures and search query texts, and we show OESP-index's efficiency, especially space-efficiency for building index structures.
Truthful Online Scheduling with Commitments
Authors: Yossi Azar, Inna Kalp-Shaltiel, Brendan Lucier, Ishai Menache, Joseph Naor, Jonathan Yaniv
Abstract: We study online mechanisms for preemptive scheduling with deadlines, with the goal of maximizing the total value of completed jobs. This problem is fundamental to deadline-aware cloud scheduling, but there are strong lower bounds even for the algorithmic problem without incentive constraints. However, these lower bounds can be circumvented under the natural assumption of deadline slackness, i.e., that there is a guaranteed lower bound $s > 1$ on the ratio between a job's size and the time window in which it can be executed.
In this paper, we construct a truthful scheduling mechanism with a constant competitive ratio, given slackness $s > 1$. Furthermore, we show that if $s$ is large enough then we can construct a mechanism that also satisfies a commitment property: it can be determined whether or not a job will finish, and the requisite payment if so, well in advance of each job's deadline. This is notable because, in practice, users with strict deadlines may find it unacceptable to discover only very close to their deadline that their job has been rejected.
Approximate Deadline-Scheduling with Precedence Constraints
Authors: Hossein Efsandiari, MohammadTaghi Hajiaghyi, Jochen Koenemann, Hamid Mahini, David Malec, Laura Sanita
Abstract: We consider the classic problem of scheduling a set of n jobs non-preemptively on a single machine. Each job j has non-negative processing time, weight, and deadline, and a feasible schedule needs to be consistent with chain-like precedence constraints. The goal is to compute a feasible schedule that minimizes the sum of penalties of late jobs. Lenstra and Rinnoy Kan [Annals of Disc. Math., 1977] in their seminal work introduced this problem and showed that it is strongly NP-hard, even when all processing times and weights are 1. We study the approximability of the problem and our main result is an O(log k)-approximation algorithm for instances with k distinct job deadlines.
Extremal eigenvalues of local Hamiltonians
Authors: Aram W. Harrow, Ashley Montanaro
Abstract: We apply classical algorithms for approximately solving constraint satisfaction problems to find bounds on extremal eigenvalues of local Hamiltonians. We consider qubit Hamiltonians for which we have an upper bound on the number of terms in which each qubit participates, and find asymptotically optimal bounds for the operator norm and ground-state energy of such Hamiltonians under this constraint. In each case the bound is achieved by a product state which can be found efficiently using a classical algorithm.
The Two Formulations of the Szemerédi–Trotter Theorem
The Szemerédi–Trotter theorem is usually presented in one of the two following formulations. Theorem 1. Given a set of points and a set of lines, both in , the number of incidences between the two sets is . Theorem 2. Given a set of lines in , the number of points in that are incident […]
by adamsheffer at July 05, 2015 09:02 PM UTC
On the different stages of learning and teaching (algorithms)
from The Geomblog
Descending a rabbit hole of links prompted by a MeFi discussion (thanks, +David Eppstein) of Steven Pinker's essay on the curse of knowledge (thanks, +Jeff Erickson), I came across an article by Alistair Cockburn on a learning framework inspired by aikido called 'Shu-Ha-Ri'.
In brief,
• In the Shu stage, you're a beginning learner trying to find one way to solve a problem. It doesn't matter that there might be multiple ways. The goal is to learn one path, and learn it well.
• In the Ha stage, you understand one way well enough to realize its limits, and are ready to encounter many different strategies for reaching your goal. You might even begin to understand the pros and cons of these different approaches. In effect, you have detached from commitment to a single approach.
• In the Ri stage, you have "transcended" the individual strategies. You might use one, or another, or mix and match as needed. You'll create new paths as you need them, and move fluidly through the space of possibilities.
Reading through this article while I ponder (yet again) my graduate algorithms class for the fall, I realize that this three-stage development process maps quite well to what we expect from undergraduates, masters students and Ph.D students learning about an area.
The undergraduate is learning a tool for the first time (recurrence analysis say) and if they can understand the master theorem and apply it, that's pretty good.
At the next level, they realize the limitations of the master theorem, and might learn about the Akra-Bazzi method, or annihilators, or even some probabilistic recurrence methods.
Of course, once you're dealing with some thorny recurrence for the analysis in your next SODA submission, then the standard templates are helpful, but you'll often have to do something creative and nontrivial to wrestle the analysis into a form where it makes sense.
Pick your own topic if you don't like recurrences.
Which also explains why it's hard to explain how to prove things. Beginning students expect a standard formula (which is why induction and proof by contradiction get taught over and over). But once you go beyond this, there aren't really good templates. In effect, there's no good second level with a set of proof techniques that you can throw at most problems, which explains why students taking a grad algorithms class tend to struggle with exactly this step.
by Suresh Venkatasubramanian (noreply@blogger.com) at July 04, 2015 08:12 PM UTC
The Halting Problem For Linear Machines
from Richard Lipton
A small idea before the fireworks show
Thoralf Skolem was a mathematician who worked in mathematical logic, set theory, and number theory. He was the only known PhD student of Axel Thue, whose Thue systems were an early word-based model of computation. Skolem had only one PhD student, Öystein Ore, who did not work in logic or computation. Ore did, however, have many students including Grace Hopper and Marshall Hall, Jr., and Hall had many more including Don Knuth.
Today Ken and I try to stimulate progress on a special case of Skolem’s problem on linear sequences.
Although Ore worked mainly on ring theory and graph theory the seeds still collected around Skolem’s tree: Hall’s dissertation was titled “An Isomorphism Between Linear Recurring Sequences and Algebraic Rings.” Sequences defined by a finite linear operator are about the simplest computational process we can imagine:
$\displaystyle x_{n} = a_1 x_{n-1} + a_2 x_{n-2} + \cdots + a_m x_{n-m}.$
The coefficients ${a_1,\dots,a_m}$ and initial values ${x_{0},\cdots,x_{m-1}}$ can be integers or relaxed to be algebraic numbers. Skolem posed the problem of deciding whether there is ever an ${n}$ such that ${x_n = 0}$.
This is a kind of halting problem. It seems like it should be simple to analyze—it is just linear algebra—but it has remained open for over 80 years. We have discussed it several times before. This 2012 survey by Joel Ouaknine and James Worrell, plus this new one, give background on this and some related problems.
The Sum-of-Powers Case
Let ${f(n)}$ be
$\displaystyle \lambda_{1}^{n} + \cdots + \lambda_{m}^{n},$
where each ${\lambda_{i}}$ is an algebraic integer. Our problem is:
Does there exist a natural number ${n}$ so that ${f(n)=0}$?
This is a special case of the Skolem problem. It arises when the coefficients ${a_1,\dots,a_m}$ are the evaluations of the elementary symmetric polynomials ${s_1,\dots,s_m}$ at ${\lambda_1,\dots,\lambda_m}$ with alternating signs. For example, with ${m = 2}$ we get
$\displaystyle a_1 = \lambda_1 + \lambda_2, \quad a_2 = -\lambda_1 \lambda_2,$
which for ${x_0 = 2}$ and ${x_1 = \lambda_1 + \lambda_2}$ gives
$\displaystyle x_2 = a_1 x_1 + a_2 x_0 = (\lambda_1 + \lambda_2)^2 - 2\lambda_1 \lambda_2 = \lambda_1^2 + \lambda_2^2$
and so on. For ${m = 3}$ we have
$\displaystyle a_1 = \lambda_1 + \lambda_2 + \lambda_3, \quad a_2 = -(\lambda_1 \lambda_2 + \lambda_1 \lambda_3 + \lambda_2 \lambda_3),\quad a_3 = \lambda_1\lambda_2\lambda_3.$
Then ${f(n) = 0}$ means ${\lambda_1^n + \lambda_2^n + \lambda_3^n = 0.}$ If the ${\lambda_i}$ are nonzero integers then for odd ${n \geq 3}$ this is asking whether ${\lambda_1,\lambda_2,-\lambda_3}$ is a solution to Pierre Fermat’s equation, and we can simply answer “no.” Of course whether ${X}$ is a solution can be easier than asking whether the equation has a solution, but this shows our case contains some of the flavor of Fermat’s Last Theorem.
We can point up some minor progress on this problem. Our methods can handle somewhat more general cases where the sum of ${n}$-th powers is multiplied by ${cn^{d}}$ for some fixed constants ${c}$ and ${d}$, but we will stay with the simpler case. Our larger hope is that this case embodies the core of the difficulty in Skolem’s problem, so that solving it might throw open the road to the full solution.
Proof
Let’s begin the proof for the case when ${n}$ is a prime ${p}$. Suppose that ${f(p)=0}$. Recall
$\displaystyle f(n)=\lambda_{1}^{n} + \cdots + \lambda_{m}^{n},$
Clearly we can assume that ${f(1) \neq 0}$. Note that this is decidable. Put ${X = \mathsf{Norm}(f(1))}$. The key is to look at the quantity
$\displaystyle Y = \mathsf{Norm}( (\lambda_{1} + \cdots + \lambda_{m})^{p} ),$
where ${p}$ is a prime. We employ the following generalization of the binomial theorem:
$\displaystyle (x_1 + x_2 + \cdots + x_m)^n = \sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m} \prod_{1\le t\le m}x_{t}^{k_{t}},$
where
$\displaystyle {n \choose k_1, k_2, \ldots, k_m} = \frac{n!}{k_1!\, k_2! \cdots k_m!}.$
The upshot is that all terms are divisible by a proper factor of ${n}$ except those from the cases ${k_j = n}$, all other ${k_i = 0}$. Each gives a factor of ${{n \choose n} = 1}$ and leaves the term ${\lambda_i^n}$. When ${n}$ is a prime ${p}$ this factor must include ${p}$ itself. Thus we get that ${Y = \mathsf{Norm(y)}}$ for some ${y}$ of the form
$\displaystyle y = \lambda_{1}^{p} + \cdots + \lambda_{m}^{p} + pR,$
where ${R}$ is an algebraic integer. But by the supposition ${f(p) = 0}$ this simplifies to ${y = pR}$, and so ${Y}$ is divisible by ${p}$. Thus
$\displaystyle Y \equiv 0 \bmod p.$
Since ${Y = \mathsf{Norm}(f(1)^p) = \mathsf{Norm}(f(1))^p = X^p}$, ${X}$ too is divisible by ${p}$. But ${X}$ is independent of ${p.}$ Hence, ${X}$ acts as a bound on any possible prime ${p}$ such that ${f(p) = 0}$. Testing the finitely many values of ${p}$ up to ${X}$ thus yields a decision procedure for this restricted case of Skolem’s problem.
Fine-Tuning and Fireworks
Ken chimes in an observation that might be distantly related: The Vandermonde determinant
$\displaystyle \prod_{1 \leq i < j \leq m} (x_j - x_i)$
is the “smallest” alternating polynomial in ${m}$ variables. Together with the symmetric polynomials it generates all alternating polynomials. When the ${x_i}$ are the ${m}$-th roots of unity it gives the determinant of the Fourier matrix ${F_m}$ up to sign. This determinant has absolute value
$\displaystyle D = m^{m/2} = 2^{\frac{1}{2}m\log_2 m}.$
It is also the product of the lengths of the ${{m \choose 2}}$ chords formed by ${m}$ equally-spaced points on the unit circle. The observation is that this 2-to-the-nearly-linear quantity is extraordinarily finely tuned.
To see how, let’s estimate the product of the chords in what is caricatured as the style of physicists: The length of an average chord is ${\sqrt{2}}$. So we can estimate the size of the product as
$\displaystyle (\sqrt{2})^{m \choose 2} \approx 2^{\frac{1}{4}m^2} = 2^{\Theta(m^2)}.$
This is off by an order of magnitude in the exponent—not even close. We can be a little smarter and use the average length of a chord instead, integrating ${\frac{1}{\pi}\sqrt{2 - 2\cos(x)}}$ from ${0}$ to ${\pi}$ to get ${4/\pi}$. This is still a number greater than ${1}$ and plugs in to yield ${2^{\Theta(m^2)}}$ anyway.
Such a calculation looks silly but isn’t. If we enlarge the circle by a factor of ${(1 + \delta)}$ then every term in the product is multiplied by that factor and it dominates:
$\displaystyle D_{1+\delta} \approx (1 + \delta)^{m \choose 2} = 2^{\Theta(m^2)}.$
If we shrink the circle by ${(1 - \delta)}$ the opposite happens: we divide by ${2^{\Theta(m^2)}}$ which crushes everything to make the analogous quantity ${D_{1-\delta}}$ virtually zero. Furthermore this “big crush” happens under more-plausible slight perturbations such as forbidding any of the ${m}$ points from occupying the arc between ${0}$ and ${\epsilon}$ radians, which prevents the equal-spacing maximization when ${m > 2\pi/\epsilon}$. We covered this at length in 2011.
The underlying reality is that when you take the logarithm of the product of chords, the terms of all growth orders between ${m^2}$ and ${m^{1 + \delta}}$ all magically cancel. There are many more chords of length ${> 1}$ than chords of length ${< 1}$, but the latter can be unboundedly short in a way that perfectly balances the multitudes of longer chords. The actual value of ${D}$ seems tiny amidst these perturbative possibilities.
This gigantic cancellation reminds Dick and me of the present argument over the tiny observed magnitude of the cosmological constant ${\Lambda}$. Estimation via quantum field theory prescribes a value 120 orders of magnitude higher—one that would instantly cause our universe to explode in fireworks—unless vast numbers of terms exactly cancel. Quoting Wikipedia:
This discrepancy has been called “the worst theoretical prediction in the history of physics” … the cosmological constant problem [is] the worst problem of fine-tuning in physics: there is no known natural way to derive the tiny [value] from particle physics.
“Fine-tuning” of constants without explanation is anathema to science, and many scientists have signed onto theories that there is a multiverse with 500 or more orders of magnitude of universes, enough to generate some with the tiny ${\Lambda}$ needed to allow life as we know it. However, any fine-tuning discovered in mathematics cannot be anathema. Perhaps the universe picks up the Fourier fine balancing act in ways we do not yet understand. More prosaically, the fine balance in quantities similar to ${f(n) = \lambda_1^n + \cdots \lambda_m^n}$ above could be just what makes Skolem’s problem hard.
Open Problems
I believe that the general case of the Skolem can be handled, not just the simple case. But the problem of handling more than just primes seems hard. I believe that this method can be used to handle more cases than just primes. Ken and I are working on this. Meanwhile, we wish everyone Stateside a happy Fourth of July, whether or not that includes fireworks.
[added link to new survey in intro]
Entropy optimality: Analytic psd rank and John’s theorem
from tcs math
Recall that our goal is to sketch a proof of the following theorem, where the notation is from the last post. I will assume a knowledge of the three posts on polyhedral lifts and non-negative rank, as our argument will proceed by analogy.
Theorem 1 For every ${m \geq 1}$ and ${g : \{0,1\}^m \rightarrow \mathbb R_+}$, there exists a constant ${C(g)}$ such that the following holds. For every ${n \geq 2m}$,
$\displaystyle 1+n^{d/2} \geq \mathrm{rank}_{\mathsf{psd}}(M_n^g) \geq C \left(\frac{n}{\log n}\right)^{(d-1)/2}\,. \ \ \ \ \ (1)$
where ${d = \deg_{\mathsf{sos}}(g).}$
In this post, we will see how John’s theorem can be used to transform a psd factorization into one of a nicer analytic form. Using this, we will be able to construct a convex body that contains an approximation to every non-negative matrix of small psd rank.
1.1. Finite-dimensional operator norms
Let ${H}$ denote a finite-dimensional Euclidean space over ${\mathbb R}$ equipped with inner product ${\langle \cdot,\cdot\rangle}$ and norm ${|\cdot|}$. For a linear operator ${A : H \rightarrow H}$, we define the operator, trace, and Frobenius norms by
$\displaystyle \|A\| = \max_{x \neq 0} \frac{|Ax|}{x},\quad \|A\|_* = \mathrm{Tr}(\sqrt{A^T A}),\quad \|A\|_F = \sqrt{\mathrm{Tr}(A^T A)}\,.$
Let ${\mathcal M(H)}$ denote the set of self-adjoint linear operators on ${H}$. Note that for ${A \in \mathcal M(H)}$, the preceding three norms are precisely the ${\ell_{\infty}}$, ${\ell_1}$, and ${\ell_2}$ norms of the eigenvalues of ${A}$. For ${A,B \in \mathcal M(H)}$, we use ${A \succeq 0}$ to denote that ${A}$ is positive semi-definite and ${A \succeq B}$ for ${A-B \succeq 0}$. We use ${\mathcal D(H) \subseteq \mathcal M(H)}$ for the set of density operators: Those ${A \in \mathcal M(H)}$ with ${A \succeq 0}$ and ${\mathrm{Tr}(A)=1}$.
One should recall that ${\mathrm{Tr}(A^T B)}$ is an inner product on the space of linear operators, and we have the operator analogs of the Hölder inequalities: ${\mathrm{Tr}(A^T B) \leq \|A\| \cdot \|B\|_*}$ and ${\mathrm{Tr}(A^T B) \leq \|A\|_F \|B\|_F}$.
1.2. Rescaling the psd factorization
As in the case of non-negative rank, consider finite sets ${X}$ and ${Y}$ and a matrix ${M : X \times Y \rightarrow \mathbb R_+}$. For the purposes of proving a lower bound on the psd rank of some matrix, we would like to have a nice analytic description.
To that end, suppose we have a rank-${r}$ psd factorization
$\displaystyle M(x,y) = \mathrm{Tr}(A(x) B(y))$
where ${A : X \rightarrow \mathcal S_+^r}$ and ${B : Y \rightarrow \mathcal S_+^r}$. The following result of Briët, Dadush and Pokutta (2013) gives us a way to “scale” the factorization so that it becomes nicer analytically. (The improved bound stated here is from an article of Fawzi, Gouveia, Parrilo, Robinson, and Thomas, and we follow their proof.)
Lemma 2 Every ${M}$ with ${\mathrm{rank}_{\mathsf{psd}}(M) \leq r}$ admits a factorization ${M(x,y)=\mathrm{Tr}(P(x) Q(y))}$ where ${P : X \rightarrow \mathcal S_+^r}$ and ${Q : Y \rightarrow \mathcal S_+^r}$ and, moreover,
$\displaystyle \max \{ \|P(x)\| \cdot \|Q(y)\| : x \in X, y \in Y \} \leq r \|M\|_{\infty}\,,$
where ${\|M\|_{\infty} = \max_{x \in X, y \in Y} M(x,y)}$.
Proof: Start with a rank-${r}$ psd factorization ${M(x,y) = \mathrm{Tr}(A(x) B(y))}$. Observe that there is a degree of freedom here, because for any invertible operator ${J}$, we get another psd factorization ${M(x,y) = \mathrm{Tr}\left(\left(J A(x) J^T\right) \cdot \left((J^{-1})^T B(y) J^{-1}\right)\right)}$.
Let ${U = \{ u \in \mathbb R^r : \exists x \in X\,\, A(x) \succeq uu^T \}}$ and ${V = \{ v \in \mathbb R^r : \exists y \in X\,\, B(y) \succeq vv^T \}}$. Set ${\Delta = \|M\|_{\infty}}$. We may assume that ${U}$ and ${V}$ both span ${\mathbb R^r}$ (else we can obtain a lower-rank psd factorization). Both sets are bounded by finiteness of ${X}$ and ${Y}$.
Let ${C=\mathrm{conv}(U)}$ and note that ${C}$ is centrally symmetric and contains the origin. Now John’s theorem tells us there exists a linear operator ${J : \mathbb R^r \rightarrow \mathbb R^r}$ such that
$\displaystyle B_{\ell_2} \subseteq J C \subseteq \sqrt{r} B_{\ell_2}\,, \ \ \ \ \ (2)$
where ${B_{\ell_2}}$ denotes the unit ball in the Euclidean norm. Let us now set ${P(x) = J A(x) J^T}$ and ${Q(y) = (J^{-1})^T B(y) J^{-1}}$.
Eigenvalues of ${P(x)}$: Let ${w}$ be an eigenvector of ${P(x)}$ normalized so the corresponding eigenvalue is ${\|w\|_2^2}$. Then ${P(x) \succeq w w^T}$, implying that ${J^{-1} w \in U}$ (here we use that ${A \succeq 0 \implies S A S^T \succeq 0}$ for any ${S}$). Since ${w = J(J^{-1} w)}$, (2) implies that ${\|w\|_2 \leq \sqrt{r}}$. We conclude that every eigenvalue of ${P(x)}$ is at most ${r}$.
Eigenvalues of ${Q(y)}$: Let ${w}$ be an eigenvector of ${Q(y)}$ normalized so that the corresponding eigenvalue is ${\|w\|_2^2}$. Then as before, we have ${Q(y) \succeq ww^T}$ and this implies ${J^T w \in V}$. Now, on the one hand we have
$\displaystyle \max_{z \in JC}\, \langle z,w\rangle \geq \|w\|_2 \ \ \ \ \ (3)$
since ${JC \supseteq B_{\ell_2}}$.
On the other hand:
$\displaystyle \max_{z \in JC}\, \langle z,w\rangle^2 = \max_{z \in C}\, \langle Jz, w\rangle^2 = \max_{z \in C}\, \langle z, J^T w\rangle^2\,. \ \ \ \ \ (4)$
Finally, observe that for any ${u \in U}$ and ${v \in V}$, we have
$\displaystyle \langle u,v\rangle^2 =\langle uu^T, vv^T\rangle \leq \max_{x \in X, y \in Y} \langle A(x), B(y)\rangle \leq \Delta\,.$
By convexity, this implies that ${\max_{z \in C}\, \langle z,v\rangle^2 \leq \Delta}$ for all ${v \in V}$, bounding the right-hand side of (4) by ${\Delta}$. Combining this with (3) yields ${\|w\|_2^2 \leq \Delta}$. We conclude that all the eigenvalues of ${Q(y)}$ are at most ${\Delta}$. $\Box$
1.3. Convex proxy for psd rank
Again, in analogy with the non-negative rank setting, we can define an “analytic psd rank” parameter for matrices ${N : X \times Y \rightarrow \mathbb R_+}$:
$\displaystyle \alpha_{\mathsf{psd}}(N) = \min \Big\{ \alpha \mid \exists A : X \rightarrow \mathcal S_+^k, B : Y \rightarrow \mathcal S_+^k\,,$
$\displaystyle \hphantom{xx} \mathop{\mathbb E}_{x \in X}[A(x)]=I,$
$\displaystyle \hphantom{xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx} \|B(y)\| \leq \frac{\alpha}{k}\, \mathop{\mathbb E}_{y \in Y}[\mathrm{Tr}(B(y))] \quad \forall y \in Y$
$\displaystyle \hphantom{\qquad\qquad} \|A(x)\| \leq \alpha \quad \forall x \in X\Big\}\,.$
Note that we have implicit equipped ${X}$ and ${Y}$ with the uniform measure. The main point here is that ${k}$ can be arbitrary. One can verify that ${\alpha_{\mathsf{psd}}}$ is convex.
And there is a corresponding approximation lemma. We use ${\|N\|_{\infty}=\max_{x,y} |N(x,y)|}$ and ${\|N\|_1 = \mathop{\mathbb E}_{x,y} |N(x,y)|}$.
Lemma 3 For every non-negative matrix ${M : X \times Y \rightarrow \mathbb R_+}$ and every ${\eta \in (0,1]}$, there is a matrix ${N}$ such that ${\|M-N\|_{\infty} \leq \eta \|M\|_{\infty}}$ and
$\displaystyle \alpha_{\mathsf{psd}}(N) \leq O(\mathrm{rank}_{\mathsf{psd}}(M)) \frac{1}{\eta} \frac{\|M\|_{\infty}}{\|M\|_1}\,.$
Using Lemma 2 in a straightforward way, it is not particularly difficult to construct the approximator ${N}$. The condition ${\mathop{\mathbb E}_x [A(x)] = I}$ poses a slight difficulty that requires adding a small multiple of the identity to the LHS of the factorization (to avoid a poor condition number), but this has a correspondingly small affect on the approximation quality. Putting “Alice” into “isotropic position” is not essential, but it makes the next part of the approach (quantum entropy optimization) somewhat simpler because one is always measuring relative entropy to the maximally mixed state.
by James at July 04, 2015 08:13 AM UTC
The High Cost of Conferences
At some point, I'm convinced the "conference structure" is going to fall apart.
Case in point -- I haven't bought my tickets yet for SIGCOMM because, unless I'm missing something, a schedule isn't up yet, and unfortunately, because ACM has scheduled a SIG chair meeting overlapping with SIGCOMM (which I don't understand also, but perhaps beside the point), I want to see what's going on when at the conference to plan my timing.
6+ weeks out, round trip tickets from Boston to London are over $2500 on nonstop economy flights. And those don't seem to be on US carriers; since I have to stop back through New York for this other meeting, and I need to find a US carrier (or figure out if this is a case where it's an exception to what is the currently believed NSF policy), tickets look to be well over$3000. Then there's registration, hotel, etc.
At some point, this becomes unsustainable, I think.
by Michael Mitzenmacher (noreply@blogger.com) at July 03, 2015 03:07 PM UTC
Fast, Provable Algorithms for Isotonic Regression in all $\ell_{p}$-norms
Authors: Rasmus Kyng, Anup Rao, Sushant Sachdeva
Abstract: Given a directed acyclic graph $G,$ and a set of values $y$ on the vertices, the Isotonic Regression of $y$ is a vector $x$ that respects the partial order described by $G,$ and minimizes $||x-y||,$ for a specified norm. This paper gives improved algorithms for computing the Isotonic Regression for all weighted $\ell_{p}$-norms with rigorous performance guarantees. Our algorithms are quite practical, and their variants can be implemented to run fast in practice.
A Characterization of the Complexity of Resilience and Responsibility for Self-join-free Conjunctive Queries
Authors: Cibele Freire, Wolfgang Gatterbauer, Neil Immerman, Alexandra Meliou
Abstract: Several research thrusts in the area of data management have focused on understanding how changes in the data affect the output of a view or standing query. Example applications are explaining query results, propagating updates through views, and anonymizing datasets. These applications usually rely on understanding how interventions in a database impact the output of a query. An important aspect of this analysis is the problem of deleting a minimum number of tuples from the input tables to make a given Boolean query false. We refer to this problem as "the resilience of a query" and show its connections to the well-studied problems of deletion propagation and causal responsibility. In this paper, we study the complexity of resilience for self-join-free conjunctive queries, and also make several contributions to previous known results for the problems of deletion propagation with source side-effects and causal responsibility: (1) We define the notion of resilience and provide a complete dichotomy for the class of self-join-free conjunctive queries with arbitrary functional dependencies; this dichotomy also extends and generalizes previous tractability results on deletion propagation with source side-effects. (2) We formalize the connection between resilience and causal responsibility, and show that resilience has a larger class of tractable queries than responsibility. (3) We identify a mistake in a previous dichotomy for the problem of causal responsibility and offer a revised characterization based on new, simpler, and more intuitive notions. (4) Finally, we extend the dichotomy for causal responsibility in two ways: (a) we treat cases where the input tables contain functional dependencies, and (b) we compute responsibility for a set of tuples specified via wildcards.
On the Approximability of Digraph Ordering
Authors: Sreyash Kenkre, Vinayaka Pandit, Manish Purohit, Rishi Saket
Abstract: Given an n-vertex digraph D = (V, A) the Max-k-Ordering problem is to compute a labeling $\ell : V \to [k]$ maximizing the number of forward edges, i.e. edges (u,v) such that $\ell$(u) < $\ell$(v). For different values of k, this reduces to Maximum Acyclic Subgraph (k=n), and Max-Dicut (k=2). This work studies the approximability of Max-k-Ordering and its generalizations, motivated by their applications to job scheduling with soft precedence constraints. We give an LP rounding based 2-approximation algorithm for Max-k-Ordering for any k={2,..., n}, improving on the known 2k/(k-1)-approximation obtained via random assignment. The tightness of this rounding is shown by proving that for any k={2,..., n} and constant $\varepsilon > 0$, Max-k-Ordering has an LP integrality gap of 2 - $\varepsilon$ for $n^{\Omega\left(1/\log\log k\right)}$ rounds of the Sherali-Adams hierarchy.
A further generalization of Max-k-Ordering is the restricted maximum acyclic subgraph problem or RMAS, where each vertex v has a finite set of allowable labels $S_v \subseteq \mathbb{Z}^+$. We prove an LP rounding based $4\sqrt{2}/(\sqrt{2}+1) \approx 2.344$ approximation for it, improving on the $2\sqrt{2} \approx 2.828$ approximation recently given by Grandoni et al. (Information Processing Letters, Vol. 115(2), Pages 182-185, 2015). In fact, our approximation algorithm also works for a general version where the objective counts the edges which go forward by at least a positive offset specific to each edge.
The minimization formulation of digraph ordering is DAG edge deletion or DED(k), which requires deleting the minimum number of edges from an n-vertex directed acyclic graph (DAG) to remove all paths of length k. We show that both, the LP relaxation and a local ratio approach for DED(k) yield k-approximation for any $k\in [n]$.
Approximation Algorithms for Connected Maximum Cut and Related Problems
Authors: MohammadTaghi Hajiaghayi, Guy Kortsarz, Robert MacDavid, Manish Purohit, Kanthi Sarpatwar
Abstract: An instance of the Connected Maximum Cut problem consists of an undirected graph G = (V, E) and the goal is to find a subset of vertices S $\subseteq$ V that maximizes the number of edges in the cut \delta(S) such that the induced graph G[S] is connected. We present the first non-trivial \Omega(1/log n) approximation algorithm for the connected maximum cut problem in general graphs using novel techniques. We then extend our algorithm to an edge weighted case and obtain a poly-logarithmic approximation algorithm. Interestingly, in stark contrast to the classical max-cut problem, we show that the connected maximum cut problem remains NP-hard even on unweighted, planar graphs. On the positive side, we obtain a polynomial time approximation scheme for the connected maximum cut problem on planar graphs and more generally on graphs with bounded genus.
Compressed Manifold Modes: Fast Calculation and Natural Ordering
Authors: Kevin Houston
Abstract: Compressed manifold modes are locally supported analogues of eigenfunctions of the Laplace-Beltrami operator of a manifold. In this paper we describe an algorithm for the calculation of modes for discrete manifolds that, in experiments, requires on average 47% fewer iterations and 44% less time than the previous algorithm. We show how to naturally order the modes in an analogous way to eigenfunctions, that is we define a compressed eigenvalue. Furthermore, in contrast to the previous algorithm we permit unlumped mass matrices for the operator and we show, unlike the case of eigenfunctions, that modes can, in general, be oriented.
I/O-Efficient Similarity Join
Authors: Rasmus Pagh, Ninh Pham, Francesco Silvestri, Morten Stöckel
Abstract: We present an I/O-efficient algorithm for computing similarity joins based on locality-sensitive hashing (LSH). In contrast to the filtering methods commonly suggested our method has provable sub-quadratic dependency on the data size. Further, in contrast to straightforward implementations of known LSH-based algorithms on external memory, our approach is able to take significant advantage of the available internal memory: Whereas the time complexity of classical algorithms includes a factor of $N^\rho$, where $\rho$ is a parameter of the LSH used, the I/O complexity of our algorithm merely includes a factor $(N/M)^\rho$, where $N$ is the data size and $M$ is the size of internal memory. Our algorithm is randomized and outputs the correct result with high probability. It is a simple, recursive, cache-oblivious procedure, and we believe that it will be useful also in other computational settings such as parallel computation.
Improved Purely Additive Fault-Tolerant Spanners
Authors: Davide Bilò, Fabrizio Grandoni, Luciano Gualà, Stefano Leucci, Guido Proietti
Abstract: Let $G$ be an unweighted $n$-node undirected graph. A \emph{$\beta$-additive spanner} of $G$ is a spanning subgraph $H$ of $G$ such that distances in $H$ are stretched at most by an additive term $\beta$ w.r.t. the corresponding distances in $G$. A natural research goal related with spanners is that of designing \emph{sparse} spanners with \emph{low} stretch.
In this paper, we focus on \emph{fault-tolerant} additive spanners, namely additive spanners which are able to preserve their additive stretch even when one edge fails. We are able to improve all known such spanners, in terms of either sparsity or stretch. In particular, we consider the sparsest known spanners with stretch $6$, $28$, and $38$, and reduce the stretch to $4$, $10$, and $14$, respectively (while keeping the same sparsity).
Our results are based on two different constructions. On one hand, we show how to augment (by adding a \emph{small} number of edges) a fault-tolerant additive \emph{sourcewise spanner} (that approximately preserves distances only from a given set of source nodes) into one such spanner that preserves all pairwise distances. On the other hand, we show how to augment some known fault-tolerant additive spanners, based on clustering techniques. This way we decrease the additive stretch without any asymptotic increase in their size. We also obtain improved fault-tolerant additive spanners for the case of one vertex failure, and for the case of $f$ edge failures.
Lexicographic Matroid Optimization
Authors: Asaf Levin, Shmuel Onn
Abstract: We show that finding lexicographically minimal $n$ bases in a matroid can be done in polynomial time in the oracle model. This follows from a more general result that the shifted problem over a matroid can be solved in polynomial time as well.
Approximate Span Programs
Authors: Tsuyoshi Ito, Stacey Jeffery
Abstract: Span programs are a model of computation that have been used to design quantum algorithms, mainly in the query model. For any decision problem, there exists a span program that leads to an algorithm with optimal quantum query complexity, but finding such an algorithm is generally challenging.
We consider new ways of designing quantum algorithms using span programs. We show how any span program that decides a problem $f$ can also be used to decide "property testing" versions of $f$, or more generally, approximate the span program witness size, a property of the input related to $f$. For example, using our techniques, the span program for OR, which can be used to design an optimal algorithm for the OR function, can also be used to design optimal algorithms for: threshold functions, in which we want to decide if the Hamming weight of a string is above a threshold or far below, given the promise that one of these is true; and approximate counting, in which we want to estimate the Hamming weight of the input. We achieve these results by relaxing the requirement that 1-inputs hit some target exactly in the span program, which could make design of span programs easier.
We also give an exposition of span program structure, which increases the understanding of this important model. One implication is alternative algorithms for estimating the witness size when the phase gap of a certain unitary can be lower bounded. We show how to lower bound this phase gap in some cases.
As applications, we give the first upper bounds in the adjacency query model on the quantum time complexity of estimating the effective resistance between $s$ and $t$, $R_{s,t}(G)$, of $\tilde O(\frac{1}{\epsilon^{3/2}}n\sqrt{R_{s,t}(G)})$, and, when $\mu$ is a lower bound on $\lambda_2(G)$, by our phase gap lower bound, we can obtain $\tilde O(\frac{1}{\epsilon}n\sqrt{R_{s,t}(G)/\mu})$, both using $O(\log n)$ space.
Meet the Julians
from bit-player
JuliaCon, the annual gathering of the community focused on the Julia programming language, convened last week at MIT. I hung out there for a couple of days, learned a lot, and had a great time. I want to report back with a few words about the Julia language and a few more about the Julia community.
It’s remarkable that after 70 years of programming language development, we’re still busily exploring all the nooks and crannies of the design space. The last time I looked at the numbers, there were at least 2,500 programming languages, and maybe 8,500. But it seems there’s always room for one more. The slide at left (from a talk by David Beach of Numerica) sums up the case for Julia: We need something easier than C++, faster than Python, freer than Matlab, and newer than Fortran. Needless to say, the consensus at this meeting was that Julia is the answer.
Where does Julia fit in among all those older languages? The surface syntax of Julia code looks vaguely Pythonic, turning its back on the fussy punctuation of the C family. Other telltale traits suggest a heritage in Fortran and Matlab; for example, arrays are indexed starting with 1 rather than 0, and they are stored in column-major order. And there’s a strong suite of tools for working with matrices and other elements of linear algebra, appealing to numericists. Looking a little deeper, some of the most distinctive features of Julia have no strong connection with any of the languages mentioned in Beach’s slide. In particular, Julia relies heavily on generic functions, which came out of the Lisp world. (Roughly speaking, a generic function is a collection of methods, like the methods of an object in Smalltalk, but without the object.)
Perhaps a snippet of code is a better way to describe the language than all these comparisons. Here’s a fibonacci function:
function fib(n)
a = b = BigInt(1)
for i in 1:n
a, b = b, a+b
end
return a
end
Note the syntax of the for loop, which is similar to Python’s for i in range(n):, and very different from C’s for (var i=0; i<n; i++). But Julia dispenses with Python’s colons, instead marking the end of a code block. And indentation is strictly for human readers; it doesn’t determine program meaning, as it does in Python.
For a language that emphasizes matrix operations, maybe this version of the fibonacci function would be considered more idiomatic:
function fibmat(n)
a = BigInt[1 1; 1 0]
return (a^n)[1, 2]
end
What’s happening here, in case it’s not obvious, is that we’re taking the nth power of the matrix $\begin{bmatrix}1& 1\\1& 0\end{bmatrix}$ and returning the lower left element of the product, which is equal to the nth fibonacci number. The matrix-power version is 25 times faster than the loop version.
@time fib(10000)
elapsed time: 0.007243102 seconds (4859088 bytes allocated)
@time fibmat(10000)
elapsed time: 0.000265076 seconds (43608 bytes allocated)
Julia’s base language has quite a rich assortment of built-in functions, but there are also 600+ registered packages that extend the language in ways large and small, as well as a package manager to automate their installation. The entire Julia ecosystem is open source and managed through GitHub.
When it comes to programming environments, Julia offers something for everybody. You can use a traditional edit-compile-run cycle; there’s a REPL that runs in a terminal window; and there’s a lightweight IDE called Juno. But my favorite is the IPython/Jupyter notebook interface, which works just as smoothly for Julia as it does for Python. (With a cloud service called JuliaBox, you can run Julia in a browser window without installing anything.)
I’ve been following the Julia movement for a couple of years, but last week’s meeting was my first exposure to the community of Julia developers. Immediate impression: Youth! It’s not just that I was the oldest person in the room; I’m used to that. It’s how much older. Keno Fischer is now an undergrad at Harvard, but he was still in high school when he wrote the Julia REPL. Zachary Yedidia, who demoed an amazing package for physics-based simulations and animations, has not yet finished high school. Several other speakers were grad students. Even the suits in attendance—a couple of hedge fund managers whose firm helped fund the event—were in jeans with shirt tails untucked.
Four of the ringleaders of the Julia movement. From left: Stefan Karpinski, Viral B. Shah, Jeff Bezanson, Keno Fischer.
Second observation: These kids are having fun! They have a project they believe in; they’re zealous and enthusiastic; they’re talented enough to build whatever they want and make it work. And the world is paying attention. Everybody gets to be a superhero.
By now we’re well into the second generation of the free software movement, and although the underlying principles haven’t really changed, the vibe is different. Early on, when GNU was getting started, and then Linux, and projects like OpenOffice, the primary goal was access to source code, so that you could know what a program was doing, fix it if it broke, customize it to meet your needs, and take it with you when you moved to new hardware. Within the open-source community, that much is taken for granted now, but serious hackers want more. The game is not just to control your own copy of a program but to earn influence over the direction of the project as a whole. To put it in GitHub terminology, it’s not enough to be able to clone or fork the repo, and thereby get a private copy; you want the owners of the repo to accept your pull requests, and merge your own work into the main branch of development.
GitHub itself may have a lot to do with the emergence of this mode of collective work. It puts everything out in public—not just the code but also discussions among programmers and a detailed record of who did what. And it provides a simple mechanism for anyone to propose an addition or improvement. Earlier open-source projects tended to put a little more friction into the process of becoming a contributor.
In any case, I am fascinated by the social structure of the communities that form around certain GitHub projects. There’s a delicate balance between collaboration (everybody wants to advance the common cause) and competition (everyone wants to move up the list of contributors, ranked by number of commits to the code base). Maintaining that balance is also a delicate task. The health of the enterprise depends on attracting brilliant and creative people, and persuading them to freely contribute their work. But brilliant creative people bring ideas and agendas of their own.
The kind of exuberance I witnessed at JuliaCon last week can’t last forever. That’s sad, but there’s no helping it. One reason we have those 2,500 (or 8,500) programming languages is that it’s a lot more fun to invent a new one than it is to maintain a mature one. Julia is still six tenths of a version number short of 1.0, with lots of new territory to explore; I plan to enjoy it while I can.
Quick notes on a few of the talks at the conference.
Zenna Tavares described sigma.jl, a Julia package for probabilistic programming—another hot topic I’m trying to catch up with. Probabilistic programming languages try to automate the process of statistical modeling and inference, which means they need to build things like Markov chain Monte Carlo solvers into the infrastructure of the programming language. Tavares’s language also has a SAT solver built in.
Chiyuan Zhang gave us mocha.jl, a deep-learning/neural-network package inspired by the C++ framework Caffe. Watching the demo, I had the feeling I might actually be able to set up my own multilayer neural network on my kitchen table, but I haven’t put that feeling to the test yet.
Finally, because of parallel sessions I missed the first half of Pontus Stenetorp’s talk on “Suitably Naming a Child with Multiple Nationalities using Julia.” I got there just in time for the big unveiling. I was sure the chosen name would turn out to be “Julia.” But it turns out the top three names for the offspring of Swedish and Japanese parents is:
Steneport wants to extend his algorithm to more language pairs. And he also needs to tell his spouse about the results of this work.
by Brian Hayes at July 02, 2015 09:59 PM UTC
Goodbye SIGACT and CRA
Tuesday I served my last day on two organizations, the ACM SIGACT Executive Committee and the CRA Board of Directors.
I spent ten years on the SIGACT (Special Interest Group on Algorithms and Computation Theory) EC, four years as vice-chair, three years as chair and three years as ex-chair, admittedly not so active those last three years. SIGACT is the main US academic organization for theoretical computer science and organizes STOC as its flagship conference. I tried to do big things, managed a few smaller things (ToCT, a few more accepted papers in STOC, poster sessions, workshops, moving Knuth and Distinguished Service to annual awards, an award for best student presentation, a tiered PC), some of them stuck and some of them didn't. Glad to see a new movement to try big changes to meet the main challenge that no conference, including STOC, really brings the theory community together anymore. As Michael Mitzenmacher becomes chair and Paul Beame takes my place as ex-chair, I wish them them and SIGACT well moving forward.
The Computing Research Association's main efforts promotes computing research to industry and government and increasing the diversity in computing research. It's a well-run organization and we can thank them particularly for helping improve the funding situation for computing in difficult financial times. The CRA occasionally puts out best practices memos like a recent one recommending quality over quantity for hiring and promotion. Serving on the board, I most enjoyed interacting with computer scientists from across the entire field, instead of just hanging with theorists at the usual conferences and workshops.
One advantage of leaving these committees: I can now kibbitz more freely on the theory community and computing in general. Should be fun.
by Lance Fortnow (noreply@blogger.com) at July 02, 2015 11:56 AM UTC
On the minimum dimension of a Hilbert space needed to generate a quantum correlation
Authors: Jamie Sikora, Antonios Varvitsiotis, Zhaohui Wei
Abstract: Consider a two-party correlation that can be generated by performing local measurements on a bipartite quantum system. A question of fundamental importance is to understand how many resources, which we quantify by the dimension of the underlying quantum system, are needed to reproduce this correlation. In this paper, we identify an easy-to-compute lower bound on the smallest Hilbert space dimension needed to generate an arbitrary two-party quantum correlation. To derive the lower bound, we combine a new geometric characterization for the set of quantum correlations (arXiv:1506.07297) with techniques that were recently used to lower bound the PSD-rank of a nonnegative matrix, an important notion to mathematical optimization and quantum communication theory (arXiv:1407.4308). We show that our bound is tight on the correlations generated by optimal quantum strategies for the CHSH and the Magic Square Game and also reprove that a family of PR-boxes cannot be realized using quantum strategies.
An exponential lower bound for homogeneous depth-5 circuits over finite fields
Authors: Mrinal Kumar, Ramprasad Saptharishi
Abstract: In this paper, we show exponential lower bounds for the class of homogeneous depth-$5$ circuits over all small finite fields. More formally, we show that there is an explicit family $\{P_d : d \in \mathbb{N}\}$ of polynomials in $\mathsf{VNP}$, where $P_d$ is of degree $d$ in $n = d^{O(1)}$ variables, such that over all finite fields $\mathbb{F}_q$, any homogeneous depth-$5$ circuit which computes $P_d$ must have size at least $\exp(\Omega_q(\sqrt{d}))$.
To the best of our knowledge, this is the first super-polynomial lower bound for this class for any field $\mathbb{F}_q \neq \mathbb{F}_2$.
Our proof builds up on the ideas developed on the way to proving lower bounds for homogeneous depth-$4$ circuits [GKKS13, FLMS13, KLSS14, KS14] and for non-homogeneous depth-$3$ circuits over finite fields [GK98, GR00]. Our key insight is to look at the space of shifted partial derivatives of a polynomial as a space of functions from $\mathbb{F}_q^n \rightarrow \mathbb{F}_q$ as opposed to looking at them as a space of formal polynomials and builds over a tighter analysis of the lower bound of Kumar and Saraf [KS14].
Private Approximations of the 2nd-Moment Matrix Using Existing Techniques in Linear Regression
Authors: Or Sheffet
Abstract: We introduce three differentially-private algorithms that approximates the 2nd-moment matrix of the data. These algorithm, which in contrast to existing algorithms output positive-definite matrices, correspond to existing techniques in linear regression literature. Specifically, we discuss the following three techniques. (i) For Ridge Regression, we propose setting the regularization coefficient so that by approximating the solution using Johnson-Lindenstrauss transform we preserve privacy. (ii) We show that adding a small batch of random samples to our data preserves differential privacy. (iii) We show that sampling the 2nd-moment matrix from a Bayesian posterior inverse-Wishart distribution is differentially private provided the prior is set correctly. We also evaluate our techniques experimentally and compare them to the existing "Analyze Gauss" algorithm of Dwork et al.
On the Communication Complexity of Distributed Clustering
Authors: Qin Zhang
Abstract: In this paper we give a first set of communication lower bounds for distributed clustering problems, in particular, for k-center, k-median and k-means. When the input is distributed across a large number of machines and the number of clusters k is small, our lower bounds match the current best upper bounds up to a logarithmic factor. We have designed a new composition framework in our proofs for multiparty number-in-hand communication complexity which may be of independent interest.
Polytopix news app (for iOS)
from Shiva Kintali
I am very excited to announce that our Polytopix news app (for iOS) is now available on the app store. Check it out.
Polytopix news app (for iOS)
Filed under: Uncategorized
by kintali at July 01, 2015 08:02 PM UTC
Popularizing TOC
It is hard to overestimate the impact of Popular Science books such as “A Brief History of Time” and “Chaos: Making a New Science” on Scientific Research. The indirect impact of popularizing Science and Scientific Education often surpass the direct contribution that most scientists can hope to achieve in their life time. For this reason, many of the greatest scientists (including in our field) choose to invest considerable time in this blessed endeavor. I personally believe that the Theory of Computing deserves more popularization than it gets (and I hope to someday contribute my share). Nevertheless, this post is meant as a tribute to our colleagues who already made wonderful such contributions. I will continuously edit this post with TOC popular books and educational resources (based on my own knowledge and suggestions in the comments).
Popular TOC books:
Scott Aaronson, Quantum Computing since Democritus
Martin Davis, Engines of Logic: Mathematicians and the Origin of the Computer
David Harel, Computers Ltd.: What They Really Can’t Do
David Harel with Yishai Feldman, Algorithmics: The Spirit of Computing
Douglas Hofstadter: Gödel, Escher, Bach: An Eternal Golden Braid
Lance Fortnow, The Golden Ticket: P, NP, and the Search for the Impossible
Cristopher Moore and Stephan Mertens, The Nature of Computation
Dennis Shasha and Cathy Lazere, Out of their Minds: The Lives and Discoveries of 15 Great Computer Scientists
Leslie Valiant, Probably Approximately Correct: Nature’s Algorithms for Learning and Prospering in a Complex World
Leslie Valiant, Circuits of the Mind
Noson S. Yanofsky, The Outer Limits of Reason: What Science, Mathematics, and Logic Cannot Tell Us
Hector Zenil, Randomness Through Computation: Some Answers, More Questions
Fiction
Apostolos Doxiadis and Christos Papadimitriou, Logicomix: An epic search for truth
Christos H. Papadimitriou, Turing (A Novel about Computation)
Other Resources:
CS Unplugged (including a book)
by Omer Reingold at July 01, 2015 05:28 PM UTC
TR15-109 | An exponential lower bound for homogeneous depth-5 circuits over finite fields | Mrinal Kumar, Ramprasad Saptharishi
from ECCC papers
In this paper, we show exponential lower bounds for the class of homogeneous depth-$5$ circuits over all small finite fields. More formally, we show that there is an explicit family $\{P_d : d \in N\}$ of polynomials in $VNP$, where $P_d$ is of degree $d$ in $n = d^{O(1)}$ variables, such that over all finite fields $F_q$, any homogeneous depth-$5$ circuit which computes $P_d$ must have size at least $\exp(\Omega_q(\sqrt{d}))$. To the best of our knowledge, this is the first super-polynomial lower bound for this class for any field $F_q \neq F_2$. Our proof builds up on the ideas developed on the way to proving lower bounds for homogeneous depth-$4$ circuits [GKKS14, FLMS14, KLSS14, KS14b] and for non-homogeneous depth-$3$ circuits over finite fields [GK98, GR00]. Our key insight is to look at the space of shifted partial derivatives of a polynomial as a space of functions from $F_q^n \rightarrow F_q$ as opposed to looking at them as a space of formal polynomials and builds over a tighter analysis of the lower bound of Kumar and Saraf [KS14b].
Christos on the Greek Vote
I was sitting by the lake in Lausanne, winds and waves lapping at the sands, humanity in slim clothes, wine and meat on our plates, when I leaned over and asked Christos Papadimitriou, " Christos, When are you going to write something about the Greek crisis?", and he said in a way we have come to expect from him unfailingly, "I already did". Enjoy (use Google translate)!
by metoo (noreply@blogger.com) at July 01, 2015 08:19 AM UTC
A note on the large spectrum and generalized Riesz products
from tcs math
I wrote a short note entitled Covering the large spectrum and generalized Riesz products that simplifies and generalizes the approach of the first few posts on Chang’s Lemma and Bloom’s variant.
The approximation statement is made in the context of general probability measures on a finite set (though it should extend at least to the compact case with no issues). The algebraic structure only comes into play when the spectral covering statements are deduced (easily) from the general approximation theorem. The proofs are also done in the general setting of finite abelian groups.
Comments are encouraged, especially about references I may have missed.
by James at July 01, 2015 07:13 AM UTC
Quantum query complexity: the other shoe drops
from Scott Aaronson
Two weeks ago I blogged about a breakthrough in query complexity: namely, the refutation by Ambainis et al. of a whole slew of conjectures that had stood for decades (and that I mostly believed, and that had helped draw me into theoretical computer science as a teenager) about the largest possible gaps between various complexity measures for total Boolean functions. Specifically, Ambainis et al. built on a recent example of Göös, Pitassi, and Watson to construct bizarre Boolean functions f with, among other things, near-quadratic gaps between D(f) and R0(f) (where D is deterministic query complexity and R0 is zero-error randomized query complexity), near-1.5th-power gaps between R0(f) and R(f) (where R is bounded-error randomized query complexity), and near-4th-power gaps between D(f) and Q(f) (where Q is bounded-error quantum query complexity). See my previous post for more about the definitions of these concepts and the significance of the results (and note also that Mukhopadhyay and Sanyal independently obtained weaker results).
Because my mental world was in such upheaval, in that earlier post I took pains to point out one thing that Ambainis et al. hadn’t done: namely, they still hadn’t shown any super-quadratic separation between R(f) and Q(f), for any total Boolean function f. (Recall that a total Boolean function, f:{0,1}n→{0,1}, is one that’s defined for all 2n possible input strings x∈{0,1}n. Meanwhile, a partial Boolean function is one where there’s some promise on x: for example, that x encodes a periodic sequence. When you phrase them in the query complexity model, Shor’s algorithm and other quantum algorithms achieving exponential speedups work only for partial functions, not for total ones. Indeed, a famous result of Beals et al. from 1998 says that D(f)=O(Q(f)6) for all total functions f.)
So, clinging to a slender reed of sanity, I said it “remains at least a plausible conjecture” that, if you insist on a fair comparison—i.e., bounded-error quantum versus bounded-error randomized—then the biggest speedup quantum algorithms can ever give you over classical ones, for total Boolean functions, is the square-root speedup that Grover’s algorithm easily achieves for the n-bit OR function.
Today, I can proudly report that my PhD student, Shalev Ben-David, has refuted that conjecture as well. Building on the Göös et al. and Ambainis et al. work, but adding a new twist to it, Shalev has constructed a total Boolean function f such that R(f) grows roughly like Q(f)2.5 (yes, that’s Q(f) to the 2.5th power). Furthermore, if a conjecture that Ambainis and I made in our recent “Forrelation” paper is correct—namely, that a problem called “k-fold Forrelation” has randomized query complexity roughly Ω(n1-1/k)—then one would get nearly a cubic gap between R(f) and Q(f).
The reason I found this question so interesting is that it seemed obvious to me that, to produce a super-quadratic separation between R and Q, one would need a fundamentally new kind of quantum algorithm: one that was unlike Simon’s and Shor’s algorithms in that it worked for total functions, but also unlike Grover’s algorithm in that it didn’t hit some impassable barrier at the square root of the classical running time.
Flummoxing my expectations once again, Shalev produced the super-quadratic separation, but not by designing any new quantum algorithm. Instead, he cleverly engineered a Boolean function for which you can use a combination of Grover’s algorithm and the Forrelation algorithm (or any other quantum algorithm that gives a huge speedup for some partial Boolean function—Forrelation is just the maximal example), to get an overall speedup that’s a little more than quadratic, while still keeping your Boolean function total. I’ll let you read Shalev’s short paper for the details, but briefly, it once again uses the Göös et al. / Ambainis et al. trick of defining a Boolean function that equals 1 if and only if the input string contains some hidden substructure, and the hidden substructure also contains a pointer to a “certificate” that lets you quickly verify that the hidden substructure was indeed there. You can use a super-fast algorithm—let’s say, a quantum algorithm designed for partial functions—to find the hidden substructure assuming it’s there. If you don’t find it, you can simply output 0. But if you do find it (or think you found it), then you can use the certificate, together with Grover’s algorithm, to confirm that you weren’t somehow misled, and that the substructure really was there. This checking step ensures that the function remains total.
Are there further separations to be found this way? Almost certainly! Indeed, Shalev, Robin Kothari, and I have already found some more things (as well as different/simpler proofs of known separations), though nothing quite as exciting as the above.
Update (July 1): Ronald de Wolf points out in the comments that this “trust-but-verify” trick, for designing total Boolean functions with unexpectedly low quantum query complexities, was also used in a recent paper by himself and Ambainis (while Ashley Montanaro points out that a similar trick was used even earlier, in a different context, by Le Gall). What’s surprising, you might say, is that it took as long as it did for people to realize how many applications this trick has.
Update (July 2): In conversation with Robin Kothari and Cedric Lin, I realized that Shalev’s superquadratic separation between R and Q, combined with a recent result of Lin and Lin, resolves another open problem that had bothered me since 2001 or so. Given a Boolean function f, define the “projective quantum query complexity,” or P(f), to be the minimum number of queries made by a bounded-error quantum algorithm, in which the answer register gets immediately measured after each query. This is a model of quantum algorithms that’s powerful enough to capture (for example) Simon’s and Shor’s algorithms, but not Grover’s algorithm. Indeed, one might wonder whether there’s any total Boolean function for which P(f) is asymptotically smaller than R(f)—that’s the question I wondered about around 2001, and that I discussed with Elham Kashefi. Now, by using an argument based on the “Vaidman bomb,” Lin and Lin recently proved the fascinating result that P(f)=O(Q(f)2) for all functions f, partial or total. But, combining with Shalev’s result that there exists a total f for which R(f)=Ω(Q(f)2.5), we get that there’s a total f for which R(f)=Ω(P(f)1.25). In the other direction, the best I know is that P(f)=Ω(bs(f)) and therefore R(f)=O(P(f)3).
by Scott at July 01, 2015 03:06 AM UTC | 2015-07-06 17:46:50 | {"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": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 213, "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": 0.6607037782669067, "perplexity": 934.4246240996104}, "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-2015-27/segments/1435375098685.20/warc/CC-MAIN-20150627031818-00302-ip-10-179-60-89.ec2.internal.warc.gz"} |
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Apurbo Satapathy
Jul 2, 2015
#### What is net sown area? Which areas of India has more net sown area?
Net sown area refers to the area sown more than once in an agricultural year. The states of plain regions like Punjab and Haryana have more percentage of net so...
Anuj A
Mar 8, 2015
#### Why are the means of transportation and communication called the lifelines of a nation and its economy?
Lifelines of a human being are his veins. They carry blood in all the parts the body and keep them hale and hearty.
Likewise means of transport and comunication bri...
Harleen Kaur
Jun 10, 2014
Shivendra
Jun 6, 2015
#### Difference between national and international resources?
the difference between national and international resources is all the resources within the boundaries of a country are called national resources.exampl...
Nov 25, 2014
### Industries can be classified into several groups: On the Basis of Strength of Labour:
#### Large Scale Industry:
Industries which employ a large number...
Naman Aggarwal
Jun 10, 2015
#### what are international resources?
These are resources which regulate by international institutions which regulate some resources . The oceanic resources beyond 200 nautical miles of the exclusive economic zone beyond to open ocean and no ind...
Naveta Ravi
Mar 6, 2015
#### Distinguish between mass comunication and prsonal communication.
Mass Communication, refers to media like periodicals, television, internet, radio, etc.. which is available to all members of the public. Usually these ...
May 19, 2015
#### What are continuous and flow resources? Please give the difference with defination and examples in detail.
It is a resource that can be used and replenished at the same time . It does not remain in one location and moves about because of natural actions in the physical environment
Atharva Mangrole
Apr 20, 2015
#### Define Resources and give some examples.
EVERY THING FOUND IN OUR ENVIRONMENT WHICH CAN BE USED TO FULFILL OUR NEEDS PROVIDED IT IS TECNOLOGICALLY ACCESSIBLE , ECONOMICALLY FEASIBE AND CULTURAL...
Jul 3, 2015
#### Distinguish between sheet erosion and gully erosion.
Gully erosion takes place on the steep slopes of the hills. When there is a heavy downpour of rain the soil is removed by water flowing along definite paths down the slope in channel . Sheet erosion during heav...
Apurbo Satapathy
Jul 2, 2015
#### What is net sown area? Which areas of India has more net sown area?
Net sown area refers to the area sown more than once in an agricultural year. The states of plain regions like Punjab and Haryana have more percentage of net so...
Anuj A
Mar 8, 2015
#### Why are the means of transportation and communication called the lifelines of a nation and its economy?
Lifelines of a human being are his veins. They carry blood in all the parts the body and keep them hale and hearty.
Likewise means of transport and comunication bri...
Harleen Kaur
Jun 10, 2014
Shivendra
Jun 6, 2015
#### Difference between national and international resources?
the difference between national and international resources is all the resources within the boundaries of a country are called national resources.exampl...
Nov 25, 2014
### Industries can be classified into several groups: On the Basis of Strength of Labour:
#### Large Scale Industry:
Industries which employ a large number...
Naman Aggarwal
Jun 10, 2015
#### what are international resources?
These are resources which regulate by international institutions which regulate some resources . The oceanic resources beyond 200 nautical miles of the exclusive economic zone beyond to open ocean and no ind...
Naveta Ravi
Mar 6, 2015
#### Distinguish between mass comunication and prsonal communication.
Mass Communication, refers to media like periodicals, television, internet, radio, etc.. which is available to all members of the public. Usually these ...
May 19, 2015
#### What are continuous and flow resources? Please give the difference with defination and examples in detail.
It is a resource that can be used and replenished at the same time . It does not remain in one location and moves about because of natural actions in the physical environment
Atharva Mangrole
Apr 20, 2015
#### Define Resources and give some examples.
EVERY THING FOUND IN OUR ENVIRONMENT WHICH CAN BE USED TO FULFILL OUR NEEDS PROVIDED IT IS TECNOLOGICALLY ACCESSIBLE , ECONOMICALLY FEASIBE AND CULTURAL... | 2021-10-16 09:04:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 40, "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": 0.21513549983501434, "perplexity": 6149.201172356833}, "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/1634323584554.98/warc/CC-MAIN-20211016074500-20211016104500-00080.warc.gz"} |
https://calculus-do.com/integral-calculus-dai-xie-mth252/ | # 微积分网课代修|积分学代写Integral Calculus代考|MTH252 Integration formulas
• 单变量微积分
• 多变量微积分
• 傅里叶级数
• 黎曼积分
• ODE
• 微分学
## 微积分作业代写calclulus代考|Integration formulas
The Newton integral inherits its formulas directly from the standard differentiation formulas. If we review the latter we will be able to deduce useful and attractive formulas for the integral.
For the integral of Chapter 3 , which is much more general than the Newton versions here, these formulas remain true but will require some attention to hypotheses; they will not usually follow trivially from the differentiation formulas.
1.4.1. Sum formula. One of the first formulas we encounter in the calculus is that for the sum of two derivatives:
$$\frac{d}{d x}{F(x)+G(x)}=\frac{d}{d x} F(x)+\frac{d}{d x} G(x) .$$
From that we obtain the sum formula for integrals:
$$\int_{a}^{b}{f(x)+g(x)} d x=\int_{a}^{b} f(x) d x+\int_{a}^{b} g(x) d x .$$
The hypotheses allowing this are: $f$ has an indefinite integral $F$ and $g$ has an indefinite integral $G$ where both $F$ and $G$ are continuous on $[a, b]$ with
$$F^{\prime}(x)=f(x) \text { and } G^{\prime}(x)=g(x)$$
for all points in $(a, b)$ excepting possibly some sequence of exceptional points. This sum formula will be available for the Chapter 3 integral under much weaker hypotheses.
1.4.2. Integration by parts. One of the most studied of the formulas we encounter in the calculus is that for the product of two derivatives:
$$\frac{d}{d x}{F(x) G(x)}=F^{\prime}(x) G(x)+F(x) G^{\prime}(x)$$
From that we obtain the formula for integrals known as integration by parts:
$$\int_{a}^{b}{f(x) G(x)+F(x) g(x)} d x=F(b) G(b)-F(a) G(b) .$$
The hypotheses allowing this are: $f$ has an indefinite integral $F$ and $g$ has an indefinite integral $G$ where both $F$ and $G$ are continuous on $[a, b]$ with
$$F^{\prime}(x)=f(x) \text { and } G^{\prime}(x)=g(x)$$
for all points in $(a, b)$ excepting possibly some sequence of exceptional points. There are versions of integration by parts formulas for the general integration theory, but they require very different proofs.
## 微积分作业代写calclulus代考|Preview
The Newton integral is a sufficient tool for most of elementary calculus needs; we should be informed of some of its theory. One of the defects in the presentation to this point is that we do not know what functions can be integrated by this method.
To be sure if a given function $f$ is the derivative of some other function $F$ then we know how the procedure works. But what sufficient conditions can be stated for a function $f$ in order that we can be assured that such an indefinite integral exists? We cannot always be placed in the uncomfortable position of computing an indefinite integral in order to be assured that there is one.
We report here, by way of a preview, some of the theory that will clarify the situation. Proper statements of these facts appear in Chapter $3 .$
THEOREM 1.7. In order that a function $f$ possess an integral
$$\int_{a}^{b} f(x) d x$$
on a compact interval $[a, b]$ in the sense of the Newton integral of this chapter the following are sufficient:
• $f$ is continuous at every point of $[a, b]$.
• $f$ is continuous at every point of $(a, b)$ and is bounded.
• $f$ is continuous at every point of $(a, b)$ with the exception possibly of some sequence of points and $f$ is bounded.
• $f$ is continuous at every point of $(a, b)$ with at most finitely many exceptions and dominated by another function $g$ for which
$$0 \leq f(x) \leq g(x) \quad(a<x<b)$$
where $g$ is continuous on $(a, b)$ (again allowing finitely many exceptions) and the integral $\int_{a}^{b} g(x) d x$ is assumed to exist.
## 微积分作业代写calclulus代考|Integration formulas
1.4.1。求和公式。我们在微积分中遇到的第一个公式是两个导数之和:
$$\frac{d}{d x} F(x)+G(x)=\frac{d}{d x} F(x)+\frac{d}{d x} G(x) .$$
$$\int_{a}^{b} f(x)+g(x) d x=\int_{a}^{b} f(x) d x+\int_{a}^{b} g(x) d x .$$
$$F^{\prime}(x)=f(x) \text { and } G^{\prime}(x)=g(x)$$
1.4.2. 按部分集成。我们在微积分中遇到的研究最多的公式之一是两个导数的乘积:
$$\frac{d}{d x} F(x) G(x)=F^{\prime}(x) G(x)+F(x) G^{\prime}(x)$$
$$\int_{a}^{b} f(x) G(x)+F(x) g(x) d x=F(b) G(b)-F(a) G(b) .$$
$$F^{\prime}(x)=f(x) \text { and } G^{\prime}(x)=g(x)$$
## 微积分作业代写calclulus代考|Preview
$$\int_{a}^{b} f(x) d x$$
• $f$ 在每一点上都是连续的 $[a, b]$.
• $f$ 在每一点上都是连续的 $(a, b)$ 并且是有界的。
• $f$ 在每一点上都是连续的 $(a, b)$ 除了可能的一些点序列和 $f$ 是有界的。
• $f$ 在每一点上都是连续的 $(a, b)$ 最多有有限多个例外，并由另一个函数支配 $g$ 为此
$$0 \leq f(x) \leq g(x) \quad(a<x<b)$$
在哪里 $g$ 是连续的 $(a, b)$ (再次允许有限多个例外) 和积分 $\int_{a}^{b} g(x) d x$ 假定存在。 | 2022-10-06 08:15:37 | {"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": 0.9537424445152283, "perplexity": 369.7301670505983}, "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-2022-40/segments/1664030337731.82/warc/CC-MAIN-20221006061224-20221006091224-00295.warc.gz"} |
https://economics.stackexchange.com/questions/25212/calculating-standard-error | # Calculating standard error
Consider a regression model on the form:
$$y_{i} = \alpha + \beta_{1}X_{i} + u_{i}$$ (1)
I am given $$var(u_{i}|X_{i}) = 10$$ and I know the OLS estimates for $$\alpha$$ and $$\beta$$ and using this I have to calculate the standard errors of the parameter estimates.
I am following Hayashi's book and I know that
$$se(b_{k}) = \sqrt(s^{2}(X'X)^{-1}_{kk})$$
Where $$b_{k}$$ is the parameter estimate of interest. I know $$(X'X)^{-1}$$ but what is confusing me is going from the given model to expressing it in matrix notation and that this does not use the variance of the error term.
I can express (1) as $$Y = X\beta + u$$ and since $$var(u_{i}|X_{i}) = var(y_{i} - \alpha - \beta_{2}X_{i}|X_{i})$$ but is this equal to $$var(Y - X\beta|X)$$?
If so, I can express $$var(u_{i}|X_{i}) = 10$$ as $$\sigma^{2}I_{n}$$ where $$\sigma^{2} = 10$$. And then could I use the fact that
$$var(b_{k}|X) = (X'X)^{-1}X' \sigma^{2} I_{n} X(X'X)^{-1}$$
To find the standard errors of interest.
Your notation is a bit all over the place, so I'm going to try and standardize it for a general case.
Let $$X$$ be the regression matrix (which includes a column of 1s for the intercept term) and $$\beta$$ be the vector of coefficients to be estimated via OLS (which includes the intercept term). $$Var(u_i|X) = 10 = \sigma^2$$, for all $$i$$. Thus the errors are homoscedastic (constant variance) and the variance-covariance matrix of the errors $$\Omega$$ has diagonal entries that are all equal to $$\sigma^2$$. Furthermore, if you assume that the error terms are not serially correlated, then the off-diagonal covariance terms $$Cov(u_i, u_j|X)$$ for all $$i \neq j$$ in the variance-covariance matrix $$\Omega$$ are zero.
Those are two of the standard Gauss-Markov assumptions used to establish the BLUEness of the OLS estimator. Under these assumptions, $$\Omega$$ is
$$E[uu'|X] = \left[ \begin{array}{ccccc} \sigma^2 \\ & \sigma^2 & & \huge0 \\ & & \ddots \\ & \huge0 & & \sigma^2 \\ & & & & \sigma^2 \end{array} \right] = \sigma^2I.$$
To get the variance of the OLS estimates $$b = \hat{\beta}$$, first note that
\begin{align} b &= (X'X)^{-1}X'y \\ &= (X'X)^{-1}X'(X\beta + u) \\ &= \beta + (X'X)^{-1}X'u \\ \implies b - \beta &= (X'X)^{-1}X'u, \end{align}
using $$y = X\beta + u$$. Then,
\begin{align} Var(b|X) &= E[(b-\beta)(b-\beta)'|X] \\ &= E[(X'X)^{-1}X'u((X'X)^{-1}X'u)'|X] \\ &= E[(X'X)^{-1}X'uu'X(X'X)^{-1}|X] \\ &= (X'X)^{-1}X'E[uu'|X]X(X'X)^{-1}. \end{align}
But recall that we derived $$E[uu'|X] = \sigma^2 I$$ via the Gauss-Markov assumption of spherical errors. Thus, by substitution,
\begin{align} Var(b|X) &= (X'X)^{-1}X'E[uu'|X]X(X'X)^{-1} \\ &= (X'X)^{-1}X'(\sigma^2 I)X(X'X)^{-1} \\ &= \sigma^2 (X'X)^{-1} X'X(X'X)^{-1} \\ \implies Var(b) &= \sigma^2 (X'X)^{-1}. \end{align}
The standard deviation of $$b$$ is just the square root of the variance, or
$$sd(b) = \sqrt{\sigma^2 (X'X)^{-1}}.$$
To find the standard deviation of the $$k^{th}$$ estimated coefficient in $$b$$, $$b_{k}$$, we simply extract the $$k^{th}$$ diagonal element of $$(X'X)^{-1}$$, denoted as $$(X'X)_{kk}^{-1}$$:
$$sd(b_k) = \sqrt{\sigma^2 (X'X)_{kk}^{-1}}.$$
$$\sigma^2$$ is the (common) variance of the errors, and unfortunately, this value is unobserved in our sample. In your question, however, it appears that this value is just given to you directly — it's $$10$$. In general, however, $$\sigma^2$$ must be estimated using the data. It turns out that given homoscedastic errors, an unbiased estimator of $$\sigma^2$$ is
$$s^2 = \frac{e'e}{n-P},$$
where $$n$$ is the number of observations, $$P$$ is the number of columns in $$X$$, and $$e = \hat{u}$$; that is, the vector of residuals $$y - Xb$$.
Thus, the standard error of $$\mathbf{b_k}$$ is estimated as
$$\boldsymbol{se(b_k) = \sqrt{s^2 (X'X)_{kk}^{-1}}},$$
which is Hayashi's result.
• Thank you so much for the thorough explanation! Really cleared things up. – BenBernke Oct 27 '18 at 13:44 | 2021-05-07 19:39:35 | {"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": 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": 52, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000073909759521, "perplexity": 289.4721058551337}, "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-2021-21/segments/1620243988802.93/warc/CC-MAIN-20210507181103-20210507211103-00436.warc.gz"} |
https://bioriental.net/%EF%BB%BFcraniofacial-development-comprises-a-complicated-process-in-individuals-where-disturbances-or-failures-frequently-result-in-congenital-anomalies/ | # Craniofacial development comprises a complicated process in individuals where disturbances or failures frequently result in congenital anomalies
Craniofacial development comprises a complicated process in individuals where disturbances or failures frequently result in congenital anomalies. the function of genes involved with ECM structure and redecorating during supplementary palate formation and pathogenesis and hereditary etiology of CL/P. We also discuss potential healing techniques using bioactive substances and concepts of tissues bioengineering for state-of-the-art CL/P fix and palatal reconstruction. (Ferguson, 1988; Dixon et al., 1993a) and EGF or TGF- can stimulate their appearance on mouse embryonic palatal mesenchymal cells (Dixon et al., 1993b). The intrinsic inner shelf power for palatal elevation continues to be related to HA because it may be the most abundant GAG in palatal ECM before shelf elevation (Ferguson, 1988). It really is produced in the cell membrane surface area by particular enzymes (HA synthasesHas 1-3) and 10Panx they are differentially portrayed in palatal mesenchyme and epithelium during palatogenesis (Galloway et al., 2013). In TGF-3 null mice, appearance of all Provides forms is decreased, leading to reduced amounts of HA and impaired shelf elevation (Galloway et al., 2013). Recently, Has2 has been described to be a crucial HA synthase in NCC-derived mesenchyme during craniofacial development and palatogenesis (Lan et al., 2019). Also, FGFs induce HA synthesis by mouse embryonic palatal mesenchymal cells (Sharpe et al., 1993). Fibronectin is found during embryonic development in areas characterized by cell migration (Schwarzbauer and DeSimone, 2011). It appears that fibronectin arrangement is vital for cell migration and palatal shelf elevation. 10Panx In this case, Rac1 and cell density modulates fibronectin deposition in mid-palate (Tang et al., 2015). Moreover, Rac1 is usually downregulated by retinoic acid, leading to the cleft palate as a consequence of the disarrangement of fibronectin and cell migration as well (Tang et al., 2016). Cellular communication is usually a well-known mechanism in which cells can communicate with each other and change Rabbit Polyclonal to ZAR1 cell behavior through soluble factors. Intercellular communications occur via direct cellular interactions in which cell surface proteins act as mediators able, or not, to bind to the ECM (juxtacrine signaling). Alternatively, cells release local mediators into the ECM to create self-control signals (autocrine signaling) and send information to neighboring cells (paracrine signaling) or reach target cells in long distances via hormones (endocrine signaling) (Ansorge and Pompe, 2018). The local mediators are peptides or growth factors which control many cellular activities. During development, a combination of cellCcell interactions occurs, as well as the secretion of mediators named morphogens, which induce specific cell differentiation in a distinct spatial order and morphogen 10Panx gradient-dependent manner (Inomata, 2017). The main 10Panx morphogens are retinoic acid, HH, TGF-, BMPs, and Wnt/-catenin. The actions of numerous morphogens in palatogenesis have been extensively studied, mainly secreted factors such as HH (Cobourne and Green, 2012; Dworkin et al., 2016; Xavier et al., 2016; Li et al., 2018), FGF (Jiang et al., 2006; Nie et al., 2006; Snyder-Warwick and Perlyn, 2012; Stanier and Pauws, 2012; Prochazkova et al., 2018; Weng et al., 2018), TGF- (Nawshad et al., 2004; Iwata et al., 2011; Nakajima et al., 2018), BMP (Nie et al., 2006; Parada and Chai, 2012; Graf et al., 2016), and Wnt/-catenin family proteins (He and Chen, 2012), which are responsible for guiding all actions of palate formation by reciprocal signaling between the embryonic oral epithelium and palatal mesenchyme, as well as transcription factor regulation (Greene and Pisano, 2010; Levi et al., 2011; Bush and Jiang, 2012; Li et al., 2017). Also, various other development and morphogens elements have got surfaced in palatogenesis, such as for example connective tissue development aspect (Tarr et al., 2018) and retinoic acidity (Okano et al., 2014; Mammadova et al., 2016). Dysregulation of. | 2022-08-15 18:14:31 | {"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": 0.8150517344474792, "perplexity": 14528.502336349593}, "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/1659882572198.93/warc/CC-MAIN-20220815175725-20220815205725-00718.warc.gz"} |
https://deepai.org/publication/convergence-analyses-of-online-adam-algorithm-in-convex-setting-and-two-layer-relu-neural-network | # Convergence Analyses of Online ADAM Algorithm in Convex Setting and Two-Layer ReLU Neural Network
Nowadays, online learning is an appealing learning paradigm, which is of great interest in practice due to the recent emergence of large scale applications such as online advertising placement and online web ranking. Standard online learning assumes a finite number of samples while in practice data is streamed infinitely. In such a setting gradient descent with a diminishing learning rate does not work. We first introduce regret with rolling window, a new performance metric for online streaming learning, which measures the performance of an algorithm on every fixed number of contiguous samples. At the same time, we propose a family of algorithms based on gradient descent with a constant or adaptive learning rate and provide very technical analyses establishing regret bound properties of the algorithms. We cover the convex setting showing the regret of the order of the square root of the size of the window in the constant and dynamic learning rate scenarios. Our proof is applicable also to the standard online setting where we provide the first analysis of the same regret order (the previous proofs have flaws). We also study a two layer neural network setting with ReLU activation. In this case we establish that if initial weights are close to a stationary point, the same square root regret bound is attainable. We conduct computational experiments demonstrating a superior performance of the proposed algorithms.
## Authors
• 5 publications
• 51 publications
• ### Matrix-Free Preconditioning in Online Learning
We provide an online convex optimization algorithm with regret that inte...
05/29/2019 ∙ by Ashok Cutkosky, et al. ∙ 0
• ### Less Regret via Online Conditioning
02/25/2010 ∙ by Matthew Streeter, et al. ∙ 0
Most methods for decision-theoretic online learning are based on the Hed...
10/28/2011 ∙ by Tim van Erven, et al. ∙ 0
• ### Online Alternating Direction Method
Online optimization has emerged as powerful tool in large scale optimiza...
06/27/2012 ∙ by Huahua Wang, et al. ∙ 0
• ### Proximal Online Gradient is Optimum for Dynamic Regret
In online learning, the dynamic regret metric chooses the reference (opt...
10/08/2018 ∙ by Yawei Zhao, et al. ∙ 0
• ### Online and Batch Learning Algorithms for Data with Missing Features
We introduce new online and batch algorithms that are robust to data wit...
04/05/2011 ∙ by Afshin Rostamizadeh, et al. ∙ 0
• ### Rivalry of Two Families of Algorithms for Memory-Restricted Streaming PCA
We study the problem of recovering the subspace spanned by the first k p...
06/04/2015 ∙ by Chun-Liang Li, et al. ∙ 0
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## 1 Introduction
In standard online learning it is assumed that a finite number of samples is encountered however in real-world streaming setting an infinite number of samples is observed (e.g., Twitter is streaming since inception and will continue to do so for foreseeable future). The performance of an online learning algorithm on early samples is negligible when measuring the performance or making predictions and decisions on the later portion of a dataset (the performance of an algorithm on tweets from ten years ago has very little bearing on its performance on recent tweets). For this reason we propose a new performance metric, regret with rolling window, which forgets about samples encountered a long time ago. It measures the performance of an online learning algorithm over a possible infinite size dataset in rolling windows. The new metric also requires an adaptation of prior algorithms, because, for example, a diminishing learning rate has poor performance on an infinite data stream.
is a widely used approach but requires a diminishing learning rate in order to achieve a high-quality performance. It has been empirically observed that the adaptive moment estimation algorithm (
is a different type of a method avoiding the impact of the choice of the learning rate. (In non-adaptive algorithms we use the term learning rate, while in adaptive algorithms we call stepsize the hyperparameter that governs the scale between the weights and the adjusted gradient.) In spite of this, no contribution has been made to the case where the regret is computed in a rolling window. Moreover, applying a diminishing learning rate or stepsize to regret with rolling window is not a good strategy, because the performance is heavily dependent on the learning rate or stepsize and the rank of a sample. Namely, regret with rolling window requires a constant learning rate or stepsize.
Standard online setting has been studied in the convex setting. With improvements in computational power resulting from GPUs, deep neural networks have been very popular in AI problems recently. A core application of online learning is online web search and recommender systems [25]
where deep learning solutions have recently emerged. Meanwhile, online learning based on deep neural networks has become an integral role in many stages in finance, from portfolio management to algorithmic trading. To this end we focus not only on convex loss functions, but also on deep neural networks.
In this paper, we not only propose a new family of online learning algorithms for both convex and non-convex loss functions, but also present a complete technical proof of regret with rolling window for each of them. For strongly convex functions, given a constant stepsize, we show that convgAdam attains regret with rolling window which is proportional to the square root of the size of the rolling window, compared to the true regret of AMSGrad [20] and AdaBound [18]. Besides, we point out the problem in the proof of regret for AMSGrad and AdaBound later in this paper. Moreover, we fix the problem in AMSGrad [20] however we do not know a fix for the problem in AdaBound. Table 1 in Appendix A.2 summarizes all regret bounds in various settings, including the previous flawed analyses. Furthermore, we prove that both dnnGd and dnnAdam attain the same regret with rolling window under reasonable assumptions for the two-layer ReLU neural network. The strongest assumption requires that the angle between the current sample and weight error is bounded away from . In summary, we make the following five contributions.
• We introduce regret with rolling window that is applicable in data streaming.
• We provide a proof of regret with rolling window which is proportional to the square root of the size of the rolling window for OGD given an arbitrary sequence of convex loss functions.
• We provide a convergent first-order gradient-based algorithm convgAdam, employing adaptive learning rate to dynamically adapt to the new patterns in the dataset. Furthermore, we provide a complete technical proof of regret with rolling window. Besides, we point out a problem with the proof of convergence of AMSGrad [20] and AdaBound [18], which eventually leads to regret in the standard online setting, and we provide a different analysis for AMSGrad which obtains regret in standard online setting by using our proof technique. To this end, see Table 1 in Appendix A.2.
• We propose the dnnGd algorithm for the two-layer ReLU neural network. Moreover, we show that dnnGd shares the same regret with rolling window as convgAdam.
• We develop an algorithm, i.e. dnnAdam, based on adaptive estimation of lower-order moments for the two-layer ReLU neural network. Meanwhile, we argue that dnnAdam shares the same regret with rolling window with convgAdam.
• We present numerical results showing that convgAdam outperforms state-of-art, yet not adaptive, OGD.
The paper is organized as follow. In the next section, we review several works related to Adam, analyses of two-layer neural networks and regret in online convex learning. In Section 3, we state the formal optimization problem in streaming, i.e., we introduce regret with rolling window. In the subsequent section we propose the two algorithms in presence of convex loss functions and we provide the underlying regret analyses. In Section 5 we study the case of deep neural networks as the loss function. In Section 6 we present experimental results comparing convgAdam with OGD.
## 2 Related Work
Two-layer neural network:
Deep learning achieves state-of-art performance on a wide variety of problems in machine learning and AI. Despite its empirical success, there is little theoretical evidence to support it. Inspired by the idea that gradient descent converges to minimizers and avoids any poor local minima or saddle points (
[16], [15], [2], [11], [13]), Luo & Wu [22] prove that there is no spurious local minima in a two-hidden-unit ReLU network. However, Luo & Wu make an assumption that the 2 layer is fixed, which does not hold in applications. Li & Yuan [17] also make progress on understanding algorithms by providing a convergence analysis for Sgd on special two-layer feedforward networks with ReLU activations, yet, they specify the 1
layer as being offset by “identity mapping” (mimicking residual connections) and the 2
layer as the -norm function. Additionally, based on their work [9], Du et al [8] give the 2
layer more freedom in the problem of learning a two-layer neural network with a non-overlapping convolutional layer and ReLU activation. They prove that although there is a spurious local minimizer, gradient descent with weight normalization can still recover good parameters with constant probability when given Gaussian inputs. Nevertheless, the convergence is guaranteed when the 1
layer is a convolutional layer. None of these studies is in an online setting studying regret and they do not focus on adaptive learning rates which are the cores in our work.
Online convex learning: Many successful algorithms and associated proofs have been studied and provided over the past few years to minimize regret in online learning setting. Zinkevich [24] shows that OGD achieves regret , for an arbitrary sequence of convex loss functions (of bounded gradients) and given a diminishing learning rate. Then, Hazan et al [12] improve regret to when given strictly convex functions. The idea of adapting first order optimization methods is by no means new and is also popular in online convex learning. Duchi et al [10] present AdaGrad, which employs very low learning rates for frequently occurring features and high learning rates for infrequent features, and obtain a comparable bound by assuming 1-strongly convex proximal functions. In a similar framework, Zhu & Xu [23] extend the celebrated online gradient descent algorithm to Hilbert spaces and analyze the convergence guarantee of the algorithm. The online functional gradient algorithm they propose also achieves regret when given convex loss functions. In all these algorithms, the loss function is required to be convex or strongly convex and the learning rate or step size must diminish. However, no work about regret analyses of online learning applied on deep neural networks (non-convex loss functions) has been done.
## 3 Regret with Rolling Window
We consider the problem of optimizing regret with rolling window, inspired by standard regret ([24], [1], [19]). The problem with the traditional regret is that it captures the performance of an algorithm only over a fixed number of samples or loss functions. In most applications data is continuously streamed with an infinite number of future loss functions. The performance over any finite number of consecutive loss functions is of interest. The concept of regret is to compare the optimal offline algorithm with access to contiguous loss functions with the performance of the underlying online algorithm. Regret with rolling window is to find the maximum of all differences between the online loss and the loss of an offline algorithm for any contiguous samples. More precisely, for an infinite sequence
, where each feature vector
is associated with the corresponding label , given fixed and any , we first define , which corresponds to an optimal solution of the offline algorithm. Then, we consider
maxp∈NRp(T):=min(ωt)t∈NT+p∑t=plt(ωt) (1)
with , where is a function of sample . The regret with rolling window metric captures regret over every consecutive loss functions and it is aiming to assess the worst possible regret over every such sequence. Note that if we have only loss functions corresponding only to , then this is the standard regret definition in online learning. The goal is to develop algorithms with low regret with rolling window. We prove that regret with rolling window can be bounded by . In other words, average regret with rolling window approaches zero.
## 4 Convex Setting
In the convex setting, we propose two algorithms with a different learning rate or stepsize strategy and analyze them with respect to (1) in the streaming setting.
### 4.1 Algorithms
Algorithms in standard online setting are almost all based on gradient descent where the parameters are updated after each new loss function is received using the gradient of the current loss function. A challenge is the strategy to select an appropriate learning rate. In order to guarantee good regret the learning rate is usually decaying. In the streaming setting, we point out that a decaying learning rate is improper since far away samples (very large ) would get a very small learning rate implying low consideration of such samples. Consequently, the learning rate has to be a constant or follow a dynamically adaptive learning algorithm. The algorithms we provide for solving (1) in the streaming setting are based on gradient descent and one of the aforementioned learning rate strategies.
In order to present our algorithms, we first need to specify notation and parameters. In each algorithm, we denote by and the learning rate or stepsize and a subgradient of loss function associated with sample , respectively. Additionally, we employ to represent the element-wise multiplication between two vectors or matrices. However, for other operations we do not introduce new notation, e.g., element-wise division () and square root ().
We start with OGD which mimics gradient descent in online setting and achieves regret with rolling window. The algorithm updates its weight when a new sample is received, i.e. . In addition, OGD uses a constant learning rate in the streaming setting so as to efficiently and dynamically learn the geometry of the dataset. Otherwise, OGD misses informative samples which arrive late due to the extremely small learning rate and leads to regret with rolling window (this is trivial to observe if the loss functions are bounded).
Constant learning rates have a drawback by treating all features equally. Consequently, we adapt Adam to online setting and further extend it to streaming. Algorithm 1 has regret with rolling window also of the order given constant stepsize as shown in the next section. The key difference of convgADAM with AMSGrad is that it maintains the same ratio of the past gradients and the current gradient instead of putting more and more weight on the current gradient and losing the memory of the past gradients fast. Besides, constant stepsize is crucial to make convgAdam well-performed due to the aforementioned reason with a potential decaying stepsize.
### 4.2 Analyses
In this section, we provide regret analyses of OGD and convgAdam showing that both of them attain regret with rolling window of the order given a constant learning rate or stepsize in the streaming setting. We require the standard conditions stated in Assumption 1.
Assumption 1: There exists a constant , such that , for any . The loss gradients are bounded, i.e., for all such that , we have . Functions are convex and differentiable with respect to for every . Functions are strongly convex with parameter , i.e., for all and for , it holds . Assumption 2: Activations
are independent Bernoulli random variables with the same probability
of success, i.e. Pr, Pr. There exists and such that for all . Quantities , and are all bounded for any . In particular, let and for any . There exists such that for all . There exits a positive constant such that .
The first condition in Assumption 1 can be removed by further complicating certain aspects of the upcoming proofs, which is discussed in Appendix A.1 for the sake of clarity of the algorithm. We first provide the regret analysis of OGD.
###### Theorem 1.
If 1-3 in Assumption 1 hold, and for any positive constant , the sequence generated by OGD achieves .
The proof is provided in Appendix B. Next, we show the regret analysis of convgAdam.
###### Theorem 2.
If Assumption 1 holds, and and are two constants between 0 and 1 such that and , then for for any positive constant , the sequence generated by convgAdam achieves .
The proof is provided in Appendix C. In the regret analysis of AMSGrad [20], the authors forget that the stepsize is and take the hyperparameter to be exponentially decaying for granted without assumptions which eventually leads to regret in standard online setting. Our analysis is flexible enough to extend to AMSGrad and a slight change to our proof yields the regret for AMSGrad. The changes in our proof to accommodate standard online setting and AMSGrad are stated in Appendix A.2. Moreover, the proof of convergence of AMSGrad in [20] uses a diminishing stepsize while our proof is valid for both constant and diminishing stepsizes. Likewise, for AdaBound [18], the right scale of the stepsize is also missed and the regret should be , which is discussed in more detail in Appendix A.2.
Theorem 2 guarantees that convgAdam achieves the same regret with rolling window as OGD for convex loss functions. In contrast, very limited work has been done about regret for nonconvex loss functions. In the following section, we argue that dnnGD and dnnAdam attain the same regret with rolling window if the initial starting point is close to an optimal offline solution given a constant learning rate or stepsize. In addition to a favorable starting point, further assumptions are needed.
## 5 Two-Layer ReLU Neural Network
In this section we consider a two layer neural network with the first hidden layer having an arbitrary number of neurons and the second hidden layer having a single neuron. The underlying activation function is a probabilistic version of ReLU and minimum square error is considered as the loss function. First of all, the optimization problem of such a two-layer ReLU neural network is neither convex nor convex (and clearly non-linear), therefore, it is very hard to find a global minimizer. Instead, we show that our algorithms achieve
regret with rolling window when the initial point is close enough to an optimal solution.
Neural networks as classifiers have been having a lot of success in practice, whereas a formal theoretical understanding of the mechanism is largely missing. Studying a general neural network is challenging, therefore, we focus on the proposed two-layer ReLU neural network. For a dataset
, the standard loss function of the two-layer neural network is , where represents the ReLU activation function applied element-wise, is the parameter vector, and is the parameter matrix. It turns out that ReLU is challenging to analyze since nesting them yields many combinations of the various values being below zeros. One way to get around this is to consider a probabilistic version of ReLU and capturing expected loss, Kawaguchi [13].
To this end we treat ReLU as a random Bernoulli variable in the sense that Pr, Pr. Kawaguchi [13] in the standard offline setting analyzes for the probabilistic version of ReLU. For our online analyses we need to slightly alter the setting by introducing two independent identically distributed random variables , and the resulting loss function is . There is a crucial property of , i.e. positive-homogeneity, which allows the network to be rescaled without changing the function computed. That is, for any , . Thus, for the two-layer ReLU neural network, given , we consider regret with rolling window as
maxp∈NRp(T):=min(ω1,t)t∈N,(ω2,t)t∈N∥∥ω1,t∥∥=1T+p∑t=pEσ1,σ2[lt(ω1,t,ω2,t)]. (2)
Next, we propose two algorithms for the two-layer neural network and analyze them in terms of (2).
### 5.1 Algorithms
In order to present the algorithms, let us first introduce further notation and parameters. For any matrix (vector ), let () denote the element in the row and column of matrix ( coordinate of vector ). Next, in order to be consistent, we also denote and as the learning rate or stepsize and a subgradient of loss function . Let and be constants. Lastly, in order to be consistent, we employ the same set of notations for operations as those used in the convex setting.
We start with dnnGd, Algorithm 2, which is the algorithm with a fixed learning rate for the online setting with the two-layer ReLU neural network. We show later that its regret with rolling window is . dnnGD first computes the gradients in steps 4111 and 5222. However, different from OGD, dnnGD not only modifies weights at a given iteration by following the gradient direction, but it also rescales weights based on the domain constraint in step 6, i.e. has a fixed norm. Then, is rescaled at the same time to impose positive-homogeneity in step 7.
Taking the drawbacks of a constant learning rate into consideration, we propose Algorithm 3, which is an extension of convgAdam for the two-layer ReLU neural network and likewise attains regret with rolling window. In dnnAdam, the stochastic gradients computed in steps 4 and 5 are different than those in dnnGD
. This is due to challenges in establishing the regret bound. Nevertheless, the stochastic gradients are unbiased estimators of gradients of the loss function. An alternative is to have four samples, two per gradient group. This would also enable the regret analysis, however we only employ two of them so as to reduce the variance of the algorithm. Step
10 modifies to be a matrix with same value in the same column. This is a divergence from standard ADAM which does not have this requirement. The modification is required for the regret analysis. Lastly, we update weights and also perform the rescaling modification to dnnAdam in steps 13 and 14.
### 5.2 Analyses
In this section, we discuss regret with rolling window bounds of dnnGd and dnnAdam. Before establishing the regret bounds, we first require the conditions in Assumption 2.
As Kawaguchi assumed in [13] and other works ([7], [5], [6]), we also assume that ’s are Bernoulli random variables with the same probability of success and are independent from input ’s and weights ’s in 1. Condition in 2 from Assumption 2 states that the optimal expected loss is zero. This is also assumed in other prior work in offline, e.g. [22], [8]. The 3 condition in Assumption 2 is an extension of 1 in Assumption 1. Likewise, the constraints on and can be removed by further introducing technique discussed in Appendix A.1, and consequently, and are bounded due to steps 4 and 5. The next to the last condition in Assumption 2 requires that a new coming sample has to be beneficial to improve current weights. More precisely, we interpret the difference between the current weights and optimal weights as an error that needs to be corrected. Then, a new sample which is not relevant to the error vector is not allowed. In other words, we assume that the algorithm does not receive any uninformative samples. Condition 5 from Assumption 2 assumes that any nonzero is lower bounded by a constant for all and . It is a weak constraint since for any and . In practice, we can modify the algorithm by only memorizing the first nonzero value in each coordinate and finding the smallest among these values. Otherwise, if all of , then we can set by default. The regret statement for dnnGd is as follows.
###### Theorem 3.
If 1-4 in Assumption 2 hold, , and for any positive constant , the sequence and generated by dnnGd achieves .
The proof is in Appendix D. The adaptive algorithm dnnAdam has the same regret bound as stated in the following theorem.
###### Theorem 4.
If Assumption 2 holds, for any positive constant , are constants between and such that and , and with , and , then, the sequence and generated by dnnAdam for the 2-layer ReLU neural network achieves .
The proof is in Appendix E. Our proofs are flexible enough to extend to standard online setting. For a constant learning rate, Appendix D and E provide the necessary details for the standard case. In summary, regret of is achieved. For diminishing stepsize , a slight change to the proof is needed. Details are provided in Appendix A.3.
## 6 Numerical Study
In this section, we compare convgAdam with OGD
with a long sequence of data points (mimicking streaming), which are the MNIST8M dataset and two other different-size real-word datasets from the Yahoo! Research Alliance Webscope program. For all of these datasets, we train multi-class hinge loss support vector machines (SVM)
[21] and we assume that the samples are streamed one by one based on a certain random order. For all the figures provided in this section, the horizontal axis is in scale. Moreover, we set and in convgAdam (values used in prior work). We capture the log of the loss function value which is defined as .
Multiclass SVM with Yahoo! Targeting User Modeling Dataset: We first compare convgAdam with OGD using the Yahoo! targeting user modeling dataset consisting of Yahoo user profiles. It contains 1,589,113 samples (i.e., user profiles), represented by a total of 13,346 features and 380 different classification problems (called labels in the supporting documentation) each with 3 classes.
First, we pick the first label out and conduct a sequence of experiments with respect to this label. The most important results are presented in Figure 1 for OGD and Figure 2 for convgAdam. In Figures 1(a) and 2(a), we consider the cases when the learning rate or stepsize varies from to while keeping the order and fixed at 1,000. Figures 1(b) and 2(b) provide the influence of the order of the sequence. Figures 1(c) and 2(c) represent the case where varies from to with a fixed learning rate or stepsize. Lastly, in Figure 2(d), we compare the performance of convgAdam and OGD with certain learning rates and stepsizes.
In these plots, we observe that convgAdam outperforms OGD for most of the learning rates and stepsizes, and definitely for promissing choices. More precisely, in Figure 1(a) and 2(a), we discover that 0.1/1000 and 3/ are two high-quality learning rate and stepsize values which have relatively low error and are learning for OGD and convgAdam, respectively. Therefore, we apply these two learning rates for the remaining experiments on this dataset. In Figures 1(b) and 2(b), we observe that the perturbation caused by the change of the order is negligible when compared to the loss value. Thus, in the remaining experiments, we no longer need to consider the impact of the order of the sequence. From Figure 1(c) and Figure 2(c), we discover that the loss and have a significantly positive correlation as we expect. Notice that changing but fixing the learning rate or stepsize essentially means containing more samples in the regret, in other words, the regret for is roughly times the regret for . Since the pattern in the figures is preserved for the different values for OGD and convgAdam, in the remaining experiments we fix . In Figure 2(c), we discover that too big or too small causes poor performance and therefore, for the remaining experiments, we set whenever is fixed. From Figure 2(d), we observe that convgAdam outperforms OGD. We also conduct experiments for convgAdam on the next four labels, and the results shown in Appendix F.1 imply that provides a good performance for convgAdam, and convgAdam outperforms OGD as we expect.
Other Datasets: We also compare convgAdam with OGD on both Yahoo! Learn to Rank Challenge and MNIST8M datasets (refer to Appendix F.2 and F.3 for more details). In conclusion, convgAdam always exhibits a better performance than OGD.
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## 7 Appendix
In this section, for inner products, given the fact that for two vectors and , we use for short expressions but for longer.
### A Extensions
We first introduce techniques to guarantee boundedness of the weight , i.e. how to remove condition 1 in Assumption 1 and condition 3 in Assumption 2. We then point out problems in the proofs of AMSGrad [20] and AdaBound [18] and provide a different proof for AMSGrad.
#### A.1 Unbounded Case
Projection is a popular technique to guarantee that a weight does not exceed a certain bound ([3], [12], [10], [18]). For unbounded weight , we introduce the following notation. Given convex sets , , vectors and matrix , we define projections
ΠP1(^ω)=argminω∈P1∥ω−^ω∥ Π1P1,P2,ω1,g1,ω′1(^ω2) =argminω′2:ω′2⋅[∥∥ω′1−ηg1∥∥/√12+ξ1]∈P2∥∥ ∥∥ω′2−argminω2:ωT1ω2∈P1∥∥ωT1ω2−ωT1^ω2∥∥∥∥ ∥∥ Π2P1,P2,ω1,g1,ω′1,^v(^ω2) =argminω′2:ω′2⋅[∥∥ω′1−ηg1∥∥/√12+ξ2]∈P2∥∥ ∥∥ω′2−argminω2:ωT1ω2∈P1∥∥(4√^v⊙ω2)Tω1−(4√^v⊙^ω2)Tω1∥∥∥∥ ∥∥.
Projection is the standard projection which maps vector into set . If an optimal weight is such that , then we have
which could be directly applied in the proofs of Theorem 1 and 2 if the projection is added to the algorithms after weight update.
For and , we could regard them as a combination of two standard projections. Note that, for the outer projection, we require that it does not affect the product of , which could be done by projection methods for linear equality constraints. In this way, we have
∥∥ωT1,t+1Π1P1,P2,ω1,t+1,g1,t,ω1,t(^ω2,t+1)−ωT1,∗ω2,∗∥∥≤∥∥ωT1,t+1^ω2,t+1−ωT1,∗ω2,∗∥∥ ∥∥∥(4√^v2,t⊙Π2P1,P2,ω1,t+1,g1,t,ω1,t,^v2,t(^ω2,t+1))Tω1,t+1−(4√^v2,t⊙ω2,∗)Tω1,∗∥∥∥ ≤∥∥∥(4√^v2,t⊙(^ω2,t+1))Tω1,t+1−(4√^v2,t⊙ω2,∗)Tω1,∗∥∥∥,
which could also be directly applied in the proofs of Theorem 3 and 4 when these two projections are added in steps 7 and 14 in Algorithms 2 and 3, respectively.
#### A.2 Standard setting of Adam
First, let us point out the problem in AMSGrad [20]. At the bottom of Page 18 in [20], the authors obtain an upper bound for the regret which has a term containing . Without assuming that is exponentially decaying, it is questionable to establish given since . The authors argue that decaying is crucial to guarantee the convergence, however, our proof shows regret for AMSGrad with constant and both constant and diminishing stepsizes, which is more practically relevant. For a diminishing stepsize, the slight change we need to make in the proof is that needs to be considered together with in (7) and the rest of the proof of Theorem 2. Applying the fact that and yields regret in standard online setting.
Additionally, in AdaBound [18], the authors establish an upper bound containing a term in page 5 where the stepsize satisfies at the bottom of page 15. However, the constraint the authors address implies is proportional to , which in turn yields the term to be , while regret is obvious since the weights and the gradients are all bounded as stated in their assumptions.
Table 1 summarizes the various regret bounds in different convex settings.
#### a.3 dnnAdam in standard online setting
For diminishing stepsize , a slight change to the proof of Theorem 4 is needed so as to be extended to the standard online setting. The only change is considering and together. In other words, in (55) and (56), we replace by . Then, we obtain regret by applying the fact that and .
### B Regret with Rolling Window Analysis of Ogd
#### Proof of Theorem 1
###### Proof.
For any and fixed , based on the update rule of OGD, for any , we obtain
∥ωt+1−ω∗∥2=∥ωt−η▽ft(ωt)−ω∗∥2 = ∥ωt−ω∗∥2−2η⟨ωt−ω∗,▽ft(ωt)⟩+η2∥▽ft(ωt)∥2,
which in turn yields
⟨ωt−ω∗,▽ft(ωt)⟩=∥ωt−ω∗∥2−∥ωt+1−ω∗∥22η+η2∥▽ft(ωt)∥2. (3)
Applying convexity of yields
ft(ωt)−ft(ω∗)≤⟨ωt−ω∗,▽ft(ωt)⟩. (4)
Inserting (3) into (4) gives
ft(ωt)−ft(ω∗)≤∥ωt−ω∗∥2−∥ωt+1−ω∗∥22η+η2∥▽ft(ωt)∥2.
By summing up all differences, we obtain
T+p∑t=p[ft(ωt)−ft(ω∗)] ≤12T+p∑t=p[∥ωt−ω∗∥2−∥ωt+1−ω∗∥2η+η∥▽ft(ωt)∥2] ≤12(∥ωp−ω∗∥2η)+dG∞T+p∑t=pη ≤D2∞√T2η1+dG∞η1√T=O(√T). (5)
The second inequality holds due to 2 in Assumption 1 and the last inequality uses 4 in Assumption 1 and the definition of . Since (7) holds for any and , setting for each yields the statement in Theorem 1. ∎
### C Regret with Rolling Window Analyses of convgAdam
###### Lemma 1.
Under the conditions assumed in Theorem 2, we have
T+p∑t=p∥∥ ∥∥14√^vt⊙mt∥∥ ∥∥2≤O(T).
###### Proof of Lemma 1.
By the definition of , for any , we obtain
∥∥ ∥∥14√^vt⊙mt∥∥ ∥∥2 =d∑j=1m2t,j√^vt,j≤d∑j=1m2t,j√vt,j=d∑j=1((1−β1)∑ti=1βt−i1gi,j)2√(1−β2)∑ti=1βt−i2g2i,j ≤(1−β1)2√1−β2d∑j=1(∑ti=1βt−i1)(∑ti=1βt−i1g2i,j)√∑ti=1βt−i2g2i,j =1−β1√1−β2d∑j=1t∑i=1λt−i∥∥gi,j∥∥2. (6)
The second equality follows from the updating rule of Algorithm 1. The second inequality follows from the Cauchy-Schwarz inequality, while the third inequality follows from the inequality . Using (7) for all time steps yields
T+p∑t=p1√^vt⊙(mt⊙mt) ≤ 1−β1√1−β2T+p∑t=pd∑j=1t∑i=1λt−i∥∥gi,j∥∥2 = 1−β1√1−β2d∑j=1T+p∑t=p(t | 2020-06-04 12:24:19 | {"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": 0.8240187168121338, "perplexity": 793.6119054507559}, "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/1590347439928.61/warc/CC-MAIN-20200604094848-20200604124848-00060.warc.gz"} |
https://byjus.com/question-answer/the-range-of-value-of-p-such-that-the-angle-theta-between-the-pair-of/ | Question
# The range of value of $$P$$ such that the angle $$\theta$$ between the pair of tangents drawn from the point $$(p.0)$$ to the circle $${ x }^{ 2 }+{ y }^{ 2 }=1$$ lies in $$\displaystyle \left( \frac { \pi }{ 3 } ,\pi \right)$$, is
A
(2,1)(1,2)
B
(3,2)(2,3)
C
(0,2)
D
None of these
Solution
## The correct option is C $$\left( -2,-1 \right) \cup \left( 1,2 \right)$$We have $$\displaystyle \frac { \pi }{ 3 } <\theta <\pi$$$$\Rightarrow\displaystyle \frac { \pi }{ 6 } <\frac { \theta }{ 2 } <\frac { \pi }{ 2 } \Rightarrow \frac { 1 }{ 2 } <\sin { \left( \frac { \theta }{ 2 } \right) } <1$$$$\displaystyle\Rightarrow \frac { 1 }{ 2 } <\frac { 1 }{ a } <1\quad \quad \left[ \because \sin { \left( \frac { \theta }{ 2 } \right) } =\frac { 1 }{ a } \right]$$$$\therefore 1<a<2$$There can be symmetrical points on the negative x-axis too. Hence, we have $$a\in \left( -2,-1 \right) \cup \left( 1,2 \right)$$ Mathematics
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View More | 2022-01-21 00:12:53 | {"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": 0.48075270652770996, "perplexity": 529.1383952808815}, "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-2022-05/segments/1642320302706.62/warc/CC-MAIN-20220120220649-20220121010649-00093.warc.gz"} |
http://mathforum.org/mathimages/index.php?title=Projection_of_a_Torus&diff=3405&oldid=3403 | # Projection of a Torus
(Difference between revisions)
Revision as of 15:29, 4 June 2009 (edit)← Previous diff Revision as of 15:31, 4 June 2009 (edit) (undo)Next diff → Line 6: Line 6: A similar process is carried out to create this page's main image. A four-dimensional object, described further below, is projected into three-dimensions using two different projections. A similar process is carried out to create this page's main image. A four-dimensional object, described further below, is projected into three-dimensions using two different projections. - |ImageDesc=The four-dimensional object is defined parametrically by + |ImageDesc=The four-dimensional object is defined parametrically by [itex] (x_1,x_2,x_3,x_4)=(cos(u),sin(u),cos(v),sin(v)) [/itex] |AuthorName=Thomas F. Banchoff |AuthorName=Thomas F. Banchoff |AuthorDesc=Thomas F. Banchoff is a geometer, and a professor at Brown University since 1967. |AuthorDesc=Thomas F. Banchoff is a geometer, and a professor at Brown University since 1967.
## Revision as of 15:31, 4 June 2009
Projection of a Torus
A four-dimensional torus projected into three-dimensional space.
# Basic Description
It is impossible to visualize a complete four-dimensional object, since we have only ever lived in three-dimensional space. However, there are ways to capture parts of the four-dimensional object in three-dimensional space. A useful analogy is a world map. We can capture the essence of the three-dimensional globe on a two-dimensional map, but only by using a projection, which distorts the three-dimensional object in some way to fit on a two-dimensional surface.
A similar process is carried out to create this page's main image. A four-dimensional object, described further below, is projected into three-dimensions using two different projections.
# A More Mathematical Explanation
The four-dimensional object is defined parametrically by $(x_1,x_2,x_3,x_4)=(cos(u),sin(u),cos(v),sin(v))$
The four-dimensional object is defined parametrically by $(x_1,x_2,x_3,x_4)=(cos(u),sin(u),cos(v),sin(v))$
# About the Creator of this Image
Thomas F. Banchoff is a geometer, and a professor at Brown University since 1967. | 2016-10-26 00:39:15 | {"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": 0, "img_math": 2, "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": 0.8033120036125183, "perplexity": 2557.038741732034}, "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-2016-44/segments/1476988720471.17/warc/CC-MAIN-20161020183840-00066-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://calculus7.org/ | ## 35 categories of Stack Overflow comments
Google’s BigQuery dataset now includes Stack Overflow data dump, including the text of over 50 million comments posted on the site. What do these comments say? I picked the most frequent ones and grouped them by topic. The counts are an underestimate: there is only so much time I was willing to spend organizing synonymous comments.
1. Thank you” comments (128960 in total) are the most common by far. Typical forms: Thank you very/so much!, Thanks a lot :), Perfect, thanks! The popularity of the emoticon in the second version is attributable to the minimal length requirement for comments: they must contain at least 15 characters. The laziest way to pad the text is probably Thank you……
2. You are welcome” (50090), presumably posted in response to group 1 comments. You’re welcome. You’re welcome! You’re welcome 🙂 Users need that punctuation or emoticon to reach 15 characters. Although those not contracting “you are” don’t have this problem.
4. “What is your question?” (20830) is the most common type of critical comments toward questions.
6. What error are you getting?” (13439) is a request for debugging information.
7. What have you tried?” (12640) often comes with the link whathaveyoutried.com and is a sufficiently notorious kind of comments that Stack Overflow software deletes them if anyone “flags” the comment. And it’s easy to cast flags automatically, so I substantially reduced the number of such comments since this data dump was uploaded. Further context: Should Stack Overflow (and Stack Exchange in general) be awarding “A”s for Effort?
8. Post your code” (11486) can sometimes be a form of “what have you tried”; in other times it’s a logical response to someone posting an error message without the code producing it. Can you post your code? Post your code. Please post your code. Show your code. Where is your code? And so on.
9. It does not work” (9674) — either the question author, or someone else with the same issue did not benefit from the solution. Maybe it’s wrong, maybe they used it wrong.
10. This is a link-only answer” (9501) usually comes from reviewers in the standard form While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes.
11. I updated the question” (8060), presumably in response to critical comments.
12. Why the downvote(s)?” (6005) is asking whoever voted down the post to explain their position. Usually fruitless; if the voter wanted to say something, they would already.
13. This is a duplicate question” (3859) is inserted automatically when someone moves for a question to be marked as a duplicate. Such comments are normally deleted automatically when the required number of close-votes is reached; but some remain. The most common by far is possible duplicate of What is a Null Pointer Exception, and how do I fix it?
14. I edited your title” (3795) is directed at users who title their questions like “Java: How to read a CSV file?”, using a part of the title as a tag. Standard form: I have edited your title. Please see, “Should questions include “tags” in their titles?“, where the consensus is “no, they should not”.
15. Post a MCVE” (3775) – the line on which the error is thrown is probably not enough to diagnose the problem; on the other hand, a wall of code with an entire program is too much. One of standard forms: Questions seeking debugging help (“why isn’t this code working?”) must include the desired behavior, a specific problem or error and the shortest code necessary to reproduce it in the question itself. Questions without a clear problem statement are not useful to other readers. See: How to create a Minimal, Complete, and Verifiable example.
16. That is correct” (3158) usually refers to a statement made in another comment.
17. It works” (3109) is the counterpart of group 9 above. Often used with “like a charm” but do charms actually work?
18. What do you mean?” (2998) – for when an exchange in comments leads to more confusion.
19. What tool are you using?” (2649) indicates that the question author forgot to specify either the language, OS, or the DBMS they are using.
20. Good answer” (2607) – various forms of praise, This should be the accepted answer. This is the correct answer. Excellent answer! The first form additionally indicates that the question author did not pick the best answer as “accepted”.
21. This question is off-topic” (2377) is a template for close votes with a custom explanation. For some years Stack Overflow used This question appears to be off-topic because… but then switched to the more assertive I’m voting to close this question as off-topic because…
22. This is a low quality answer” (2003) is a response to answers that contain nothing but code, perhaps preceded by “try this”. Example: While this code snippet may solve the question, including an explanation really helps to improve the quality of your post. Remember that you are answering the question for readers in the future, and those people might not know the reasons for your code suggestion.
23. Is this homework?” (1995) is not a particularly fruitful type of comments.
24. Does this work?” (1916) is meant to obtain some response from question asker who has not yet acknowledged the answer.
26. http://stackoverflow.com/help/ . . .” (1117) and nothing but the link. Directs to one of Help Center articles such as “How to ask”. Maybe there should also be “How to Comment”
27. Thanks are discouraged” (1046) … so all those group 1 comments aren’t meant to be. But this is mostly about posts rather than comments. Unlike forum sites, we don’t use “Thanks”, or “Any help appreciated”, or signatures on Stack Overflow. See “Should ‘Hi’, ‘thanks,’ taglines, and salutations be removed from posts?.
28. Format your code” (967) – yes, please. Select the code block and press Ctrl-K. Thanks in advance. Oops, forgot about the previous group.
29. What doesn’t work?” (926) is a response to vague comments of group 9.
30. Don’t use mysql_* functions” (693) or Russian hackers will pwn your site. Comes with a link-rich explanation: Please, don’t use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLithis article will help you decide which. If you choose PDO, here is a good tutorial.
31. Add tags” (625) often comes up in the context of database questions. Which RDBMS is this for? Please add a tag to specify whether you’re using mysql, postgresql, sql-server, oracle or db2 – or something else entirely.
32. Improve title” (585) is like group 14, but invites the user to edit the title instead of doing it for them.
33. Use modern JOIN syntax” (301) bemoans obsolete ways of dealing with databases. Bad habits to kick : using old-style JOINs – that old-style *comma-separated list of tables* style was replaced with the *proper* ANSI JOIN syntax in the ANSI-92 SQL Standard (more than 20 years ago) and its use is discouraged.
34. More SQL woes” (272) is another template: Side note: you should not use the sp_ prefix for your stored procedures. Microsoft has reserved that prefix for its own use (see *Naming Stored Procedures*, and you do run the risk of a name clash sometime in the future. It’s also bad for your stored procedure performance. It’s best to just simply avoid sp_ and use something else as a prefix – or no prefix at all!
35. I have the same problem” (241) is a kind of comments that really should not exist.
## Vectorization of integration
A frequently asked question about numerical integration is: how to speed up the process of computing ${\int_a^b f(x,p)\,dx}$ for many values of parameter ${p}$? Running an explicit for over the values of ${p}$ seems slow… is there a way to vectorize this, performing integration on an array of functions at once (which in reality means, pushing the loop to a lower level language)?
Usually, the answer is some combination of “no” and “you are worrying about the wrong thing”. Integration routines are typically adaptive, meaning they pick evaluation points based on the function being integrated. Different values of the parameter will require different evaluation points to achieve the desired precision, so vectorization is out of question. This applies, for example, to quad method of SciPy integration module.
Let’s suppose we insist on vectorization and are willing to accept a non-adaptive method. What are our options, considering SciPy/NumPy? I will compare
The test case is ${\int_0^1 (p+1)x^p\,dx}$ with ${p=0,0.01,\dots,99.99}$. Theoretically all these are equal to ${1}$. But most of these functions are not analytic at ${0}$, and some aren’t even differentiable there, so their integration is not a piece of cake.
Keeping in mind that Romberg’s integration requires ${2^k}$ subintervals, let’s use ${1024}$ equal subintervals, hence ${1025}$ equally spaced sample points. Here is the vectorized evaluation of all our functions at these points, resulting in a 2D array “values”:
import numpy as np
import scipy.integrate as si
x = np.linspace(0, 1, 1025).reshape(1, -1)
dx = x[0,1] - x[0,0]
p = np.arange(0, 100, 0.01).reshape(-1, 1)
values = (p+1)*x**p
This computation took 0.799 seconds on my AWS instance (t2.nano). Putting the results of evaluation together takes less time:
• Trapezoidal rule np.trapz(values, dx=dx) took 0.144 seconds and returned values ranging from 0.99955 to 1.00080.
• Simpson’s rule si.simps(values, dx=dx) took 0.113 seconds and returned values ranging from 0.99970 to 1.0000005.
• Romberg quadrature si.romb(values, dx=dx) was fastest at 0.0414 seconds, and also most accurate, with output between 0.99973 and 1.000000005.
Conclusions so far:
• SciPy’s implementation of Romberg quadrature is surprisingly fast, considering that this algorithm is the result of repeated Richardson extrapolation of the trapezoidal rule (and Simpson’s rule is just the result of the first extrapolation step). It’d be nice to backport whatever makes Romberg so effective back to Simpson’s rule.
• The underestimation errors 0.999… are greatest when ${p}$ is near zero, so the integrand is very nonsmooth at ${0}$. The lack of smoothness renders Richardson extrapolation ineffective, hence all three rules have about the same error here.
• The overestimation errors are greatest when ${p}$ is large. The function is pretty smooth then, so upgrading from trapezoidal to Simpson to Romberg brings substantial improvements.
All of this is nice, but the fact remains that non-vectorized adaptive integration is both faster and much more accurate. The following loop with quad, which uses adaptive Clenshaw-Curtis quadrature, runs in 0.616 seconds (less than it took to evaluate our functions on a grid) and returns values between 0.99999999995 and 1.00000000007. Better to use exponential notation here: ${1-5\cdot 10^{-11} }$ and ${1+7\cdot 10^{-11}}$.
funcs = [lambda x, p=p_: (p+1)*x**p for p_ in np.arange(0, 100, 0.01)]
result = [si.quad(fun, 0, 1)[0] for fun in funcs]
An adaptive algorithm shines when ${p}$ is small, by placing more evaluation points near zero. It controls both over- and under- estimates equally well, continuing to add points until the required precision is reached.
The last hope of non-adaptive methods is Gaussian quadrature, which is implemented in SciPy as fixed_quad (“fixed” referring to the number of evaluation points used). With 300 evaluation points, the integration takes 0.263 seconds (excluding the computation of nodes and weights, which can be cached) and the results are between ${1-2\cdot 10^{-12}}$ and ${1+2\cdot 10^{-7}}$. This is twice as fast as a loop with quad, and more accurate for large ${p}$ — but sadly, not so accurate for small ${p}$. As said before, ${x^p}$ with ${p}$ near zero is really a showcase for adaptive methods.
## Kolakoski turtle curve
Let’s take another look at the Kolakoski sequence (part 1, part 2) which, by definition, is the sequence of 1s and 2s in which the nth term is equal to the length of the nth run of consecutive equal numbers in the same sequence. When a sequence has only two distinct entries, it can be visualized with the help of a turtle that turns left (when the entry is 1) or right (when the entry is 2). This visualization method seems particularly appropriate for the Kolakoski sequence since there are no runs of 3 equal entries, meaning the turtle will never move around a square of sidelength equal to its step. In particular, this leaves open the possibility of getting a simple curve… Here are the first 300 terms; the turtle makes its first move down and then goes left-right-right-left-left-right-left-… according to the terms 1,2,2,1,1,2,1,…
No self-intersections yet… alas, at the 366th term it finally happens.
Self-intersections keep occurring after that:
again and again…
Okay, the curve obviously doesn’t mind intersecting self. But it can’t be periodic since the Kolakoski sequence isn’t. This leaves two questions unresolved:
• Does the turtle ever get above its initial position? Probably… I haven’t tried more than 5000 terms
• Is the curve bounded? Unlikely, but I’ve no idea how one would dis/prove that. For example, there cannot be a long diagonal run (left-right-left-right-left) because having 1,2,1,2,1 in the sequence implies that elsewhere, there are three consecutive 1s, and that doesn’t happen.
Here’s the Python code used for the above. I represented the sequence as a Boolean array with 1 = False, 2 = True.
import numpy as np
import turtle
n = 500 # number of terms to compute
a = np.zeros(n, dtype=np.bool_)
j = 0 # the index to look back at
same = False # will next term be same as current one?
for i in range(1, n):
if same:
a[i] = a[i-1] # current run continues
same = False
else:
a[i] = not a[i-1] # the run is over
j += 1 # another run begins
same = a[j] # a[j] determines its length
turtle.hideturtle()
turtle.right(90)
for i in range(n):
turtle.forward(10) # used steps of 10 or 5 pixels
if a[i]:
turtle.right(90)
else:
turtle.left(90)
## Wild power pie
Many people are aware of ${\pi}$ being a number between 3 and 4, and some also know that ${e}$ is between 2 and 3. Although the difference ${\pi-e}$ is less than 1/2, it’s enough to place the two constants in separate buckets on the number line, separated by an integer.
When dealing with powers of ${e}$, using ${e>2}$ is frequently wasteful, so it helps to know that ${e^2>7}$. Similarly, ${\pi^2<10}$ is way more precise than ${\pi<4}$. To summarize: ${e^2}$ is between 7 and 8, while ${\pi^2}$ is between 9 and 10.
Do any two powers of ${\pi}$ and ${e}$ have the same integer part? That is, does the equation ${\lfloor \pi^n \rfloor = \lfloor e^m \rfloor}$ have a solution in positive integers ${m,n}$?
Probably not. Chances are that the only pairs ${(m,n)}$ for which ${|\pi^n - e^m|<10}$ are ${m,n\in \{1,2\}}$, the smallest difference attained by ${m=n=1}$.
Indeed, having ${|\pi^n - e^m|<1}$ implies that ${|n\log \pi - m|, or put differently, ${\left|\log \pi - \dfrac{m}{n}\right| < \dfrac{1}{n \,\pi^n}}$. This would be an extraordinary rational approximation… for example, with ${n=100}$ it would mean that ${\log \pi = 1.14\ldots}$ with the following ${50}$ digits all being ${0}$. This isn’t happening.
Looking at the continued fraction expansion of ${\log \pi}$ shows the denominators of modest size ${[1; 6, 1, 10, 24, \dots]}$, indicating the lack of extraordinarily nice rational approximations. Of course, can use them to get good approximations, ${\left|\log \pi - \dfrac{m}{n}\right| < \dfrac{1}{n^2}}$, which leads to ${\pi^n\approx e^m}$ with small relative error. For example, dropping ${24}$ and subsequent terms yields the convergent ${87/76}$, and one can check that ${\pi^{76} = 6.0728... \cdot 10^{37}}$ while ${e^{87} = 6.0760...\cdot 10^{37}}$.
Trying a few not-too-obscure constants with the help of mpmath library, the best coincidence of integer parts that I found is the following: the 13th power of the golden ratio ${\varphi = (\sqrt{5}+1)/2}$ and the 34th power of Apèry’s constant ${\zeta(3) = 1^{-3}+2^{-3}+3^{-3}+4^{-4}+\dots}$ both have integer part 521.
## A necklace of tears
The problem is: given a set of objects drawn from several groups, put all of them in a row in a “uniform” way. Whatever that means.
For example, suppose we have 21 gemstones: 9 red, 5 blue, 3 green, 2 cyan, 1 magenta and 1 yellow. How to place them on a string to make a reasonably looking necklace? The criterion is subjective, but we can probably agree that
looks better than
(referring to the uniformity of distributions, not the aesthetics of color.)
The approach I took was to repeatedly add the “most underrepresented” gem, defined by maximal difference between projected frequency of appearance (e.g., 5/21 for blue) and the frequency of appearance so far in the string. (Taking the convention 0/0=0 here.) Here is this algorithm in Python — not in Ruby, which would be more topical.
count = {'r': 9, 'b': 5, 'g': 3, 'c': 2, 'm': 1, 'y': 1}
length = sum(count.values())
str = ''
while len(str) < length:
deficit = {}
for char in count:
deficit[char] = count[char]/length - (str.count(char)/len(str) if str else 0)
str += max(deficit, key=deficit.get)
print(str)
The output, “rbgcryrbrmgrbrcbrgrbr”, is what the first image in this post represents. The second image is the result of an ill-fated attempt to replace difference by ratio when determining the under-representation.
I initially thought that two unique gems (yellow and magenta) would end up together, but this hasn’t happened: after one is added, the frequency of more common gems drops, allowing them to come back into play for a while. Still, the left half of the string is noticeably more diverse than the right half. It’d be better if two unique gems were in roughly symmetrical position, and generally there would be no preference between left-right and right-left directions.
Perhaps the new character should be added to the string either on the right or on the left, in alternating fashion. That should make things nice and symmetric, right?
Wrong.
The search continues…
Update: Rahul suggested in a comment to adjust the deficit computation to
deficit[char] = count[char]/length - str.count(char)/(len(str) + 1)
This has some advantages but on the other hand, two unique gems (magenta and yellow) are placed next to each other, which is something I wanted to avoid.
## Bankers’ rounding and floating point arithmetics
Bankers’ rounding means preferring even digits in edge cases: both 3.5 and 4.5 are rounded to 4. When rounding a random amount of money to the nearest dollar in this way, the expected value added or subtracted is zero. Indeed: 0 cents don’t get rounded, 1 through 49 get rounded down, 51 through 99 get rounded up (and the amounts match 1-49), so 50 has to be split between the two directions.
Python follows this convention: rounding the numbers 1/2, 3/2, 5/2, … via
[round((2*k+1)/2) for k in range(10)]
results in [0, 2, 2, 4, 4, 6, 6, 8, 8, 10].
However, when rounding 0.05 = 1/20, 0.15 = 3/20, … to 1 digit after the decimal dot, we get a surprise:
[round((2*k+1)/20, 1) for k in range(10)]
returns [0.1, 0.1, 0.2, 0.3, 0.5, 0.6, 0.7, 0.8, 0.8, 0.9]. An irregularly spaced array, in which odd digits win 6 to 4…
This is because the numbers like 0.05 cannot be precisely represented in the double-precision format used for floating point numbers in Python (and other languages): the denominator of 1/20 is not a power of 2. And so, 0.05 gets represented by 3602879701896397/256. Since 0.05*256 = 3602879701896396.8, this representation is greater than the real number 0.05. And this pushes the rounding routine over the edge, toward 0.1 rather than 0.0.
Out of the numbers tested above, only 0.25 = 1/4 and 0.75 = 3/4 get a precise binary representation and are therefore rounded predictably. For the other 8 the direction of rounding has nothing to do with bankers… oddly enough, 6 out of these 8 are rounded toward an odd digit.
This isn’t the end of surprises, though. Recalling that numerical manipulations on arrays are best done with NumPy library, we may want to import it and try
numpy.around([(2*k+1)/20 for k in range(10)], 1)
The output ought to make any banker happy: [0., 0.2, 0.2, 0.4, 0.4, 0.6, 0.6, 0.8, 0.8, 1.]
Why? NumPy takes a different approach to rounding: when asked to round to 1 digit after the dot, it multiplies the numbers by 10, then rounds to the nearest integer, and divides by 10. Multiplication by 10 restores sanity: 0.05 * 10 = 0.5 is represented exactly and gets rounded to 0; similarly for the others.
This algorithm isn’t perfect, however. When rounding 0.005, 0.015, …, 0.995 to two decimal digits, it misplaces 0.545 (to 0.55) and 0.575 (to 0.57). The other 98 numbers are rounded as expected; tested with
numpy.around([(2*k+1)/200 for k in range(100)], 2)
Out of 1000 numbers of the form $(2k+1)/2000$, $k=0,\dots, 999$, there are six deviations from the bankers’ rule. They all appear together, in increments of 0.002, from 0.5015 to 0.5115.
Is there a rounding algorithm that follows bankers’ convention for all (not too long) finite decimal expansions?
## Autogenerated numbers 2
Although this is a follow up to an earlier post, it can be read independently.
Repeatedly squaring the number 5, we get the following.
5^1 = 5
5^2 = 25
5^4 = 625
5^8 = 390625
5^16 = 152587890625
5^32 = 23283064365386962890625
There is a pattern here, which becomes more evident if the table is truncated to a triangle:
5^1 = 5
5^2 = 25
5^4 = 625
5^8 = ...0625
5^16 = ...90625
5^32 = ...890625
5^64 = ...2890625
5^128 = ...12890625
5^256 = ...212890625
What other digits, besides 5, have this property? None in the decimal system (ignoring the trivial 0 and 1). But the same pattern is shown by 3 in base 6 (below, the exponent is still in base 10, for clarity).
3^1 = 3
3^2 = 13
3^4 = 213
3^8 = ...0213
3^16 = ...50213
3^32 = ...350213
3^64 = ...1350213
and so on.
Here is a proof that the pattern appears when the base is 2 mod 4 (except 2), and the starting digit is half the base.
Let ${x}$ be the starting digit, and ${b=2x}$ the base. The claim is that ${x^{2^k}\equiv x^{2^{k-1}} \bmod b^k}$ for all ${k=1,2,\dots}$.
Base of induction: ${x^2-x = x(x-1)}$ is divisible by ${x}$ and also by ${2}$, since ${x-1}$ is even. Hence it is divisible by ${2x=b}$.
Step of induction: Suppose ${x^{2^k} - x^{2^{k-1}}}$ is divisible by ${b^k}$. Then
${x^{2^{k+1}} - x^{2^{k}} = (x^{2^k} - x^{2^{k-1}})(x^{2^k} + x^{2^{k-1}})}$
The second factor is divisible by ${x}$ and also by ${2}$, being the sum of two odd numbers. So, it is divisible by ${b}$. Since the first factor is divisible by ${b^k}$, the claim follows: ${x^{2^{k+1}} - x^{2^{k}} }$ is divisible by ${b^{k+1}}$.
So that was easy. The part I can’t prove is the converse. Numerical evidence strongly suggests that ${b=2\bmod 4}$, ${x=b/2}$
is also the necessary condition for the triangular pattern to appear. | 2017-01-25 01:20:22 | {"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": 79, "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": 0.4948004186153412, "perplexity": 1720.7327991030893}, "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/1484560285337.76/warc/CC-MAIN-20170116095125-00372-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://taoofmac.com/space/storage/MMC | # MMC
Multi Media Card, a standard sponsored by the MMCA and commonly used in Nokia devices. It is partially interchangeable with SD (and shares many of that format's features), but has, like most things, subtle incompatibilities. | 2016-12-09 23:14:38 | {"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": 0.8263123631477356, "perplexity": 14484.04760130953}, "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-2016-50/segments/1480698542851.96/warc/CC-MAIN-20161202170902-00369-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/solving-fourth-order-differential-equation.398977/ | # Solving fourth order differential equation ( )
1. ### sasikanth
9
Solving fourth order differential equation (URGENT)
I have two second order differential equation which needs to be solved.
x1''(t) = 8 x2(t)
x2''(t) = 2 x1(t)
I have the initial conditions, x1(0) = 0, x2(0) = 1, and terminal conditions x1(pi/4) = 1, x2(pi/4) = 0.
Can anyone help me solve these equations?? What I did was to write the equations in terms of x1 and x2 respectively, but that gives me a fourth order differential
x1''''(t) = 16 x1(t)
x2''''(t) = 16 x2(t)
and I do not know how to solve for these. Can any one help please??
2. ### Cyosis
1,495
Re: Solving fourth order differential equation (URGENT)
You know that the exponential, sine and cosine return to their original form after differentiating them four times. So try a solution of the form $c_1 e^{a t}+c_2 e^{-at}+c_3 \sin(at)+c_4 \cos(at)$.
3. ### sasikanth
9
Re: Solving fourth order differential equation (URGENT)
So the solution would be
x1(t) = c1 e^(16t) + c2 e^-(16t) + c3 sin(16t) + c4 cos(16t)??
4. ### Cyosis
1,495
Re: Solving fourth order differential equation (URGENT)
No you found the wrong a. Differentiating e^16t 4 times would give 16^4 e^16t, which is not a solution.
5. ### sasikanth
9
Re: Solving fourth order differential equation (URGENT)
would a = 4 be the correct solution??
6. ### Cyosis
1,495
Re: Solving fourth order differential equation (URGENT)
I am not sure why you have to ask. Take the derivative of your function with a=4 , four times and you will see that it is not the correct solution.
7. ### sasikanth
9
Re: Solving fourth order differential equation (URGENT)
I am sorry, I was taking the second dervivative for some reason. I conclude that a = 2 would be thr right solution. Am I correct??
8. ### Cyosis
1,495
Re: Solving fourth order differential equation (URGENT)
Yes that's correct.
9. ### sasikanth
9
Re: Solving fourth order differential equation (URGENT)
In order to solve for c1,c2,c3 and c4, I would need the second differential of x1. Would that be x1'' = c1 e^4t +c2 e^-4t ??
10. ### matematikawan
333
Re: Solving fourth order differential equation (URGENT)
Your equations are linear with constant coefficients. I would handle either the following ways
1. Apply the Laplace transform to the equations - this will transform the problem to solving algebraic system of equations or
2. Let the solution be x1(t)=q1ert and x2(t)=q2ert . Substitute this assumption and you can determine r, q1 and q2 from eigenvalue-eigenvector equation.
11. ### sasikanth
9
Re: Solving fourth order differential equation (URGENT)
I got the solution to the equation using the fourth order differntial, but am stuck wolving for the constants c1,c2,c3,c4. If I wanted the second order differntail for x1, would that be x1'' = c1 e^4t +c2 e^-4t ?? | 2015-04-22 01:43:28 | {"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": 0.7657272219657898, "perplexity": 1744.7718766481892}, "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-2015-18/segments/1429246644083.49/warc/CC-MAIN-20150417045724-00182-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.nature.com/articles/s41598-021-99248-2?error=cookies_not_supported&code=557c3d71-5e4f-4785-a630-4e72f7399813 | Ever since the World Health Organization declared COVID-19 to be a global pandemic on 11 March 2020, vaccination has been the light at the end of the tunnel. Following an unprecedented development effort by the biotech industry, the first vaccines became available in some countries for health-care workers and the most at-risk in winter 2020–2021. The United States, United Kingdom, and many other nations began to distribute vaccines to high-risk groups shortly thereafter and then extended distribution to the general public in the spring. This process is intended to curtail the rate of infection and is hoped to eventually eliminate infections by creating herd immunity. But an available vaccine does not automatically create herd immunity. If a sufficient fraction of the population does not choose to be vaccinated1, 2, substantial numbers of people will remain susceptible and infections will persist.
A series of surveys in late 2020 revealed that ~ 30–40% of Americans were hesitant to get vaccinated against COVID-193,4,5; this mirrors results in other countries6,7,8,9,10. Some respondents were worried about receiving a vaccine before seeing more widespread evidence that it is effective and has few side-effects5, 7, 9. For example, a December 2020 KFF COVID-19 Vaccine Monitor survey reported that among the 27% of the U.S. population likely to not get the vaccine, 59% were worried about side effects and 55% did not trust the government to make sure that the vaccine is safe and effective11. Others stated that they are opposed to being vaccinated altogether, following a growing trend in American society to reject vaccines as having dangerous side effects12, 13. Resistance to COVID-19 vaccination is particularly pronounced among Black Americans, who cite a long history of discrimination and mistreatment at the hands of medical professionals14, 15.
The various objections to vaccination and their uneven distribution across communities have posed a major challenge for the pursuit of vaccine-based herd immunity. Of particular concern, differences between communities in willingness to be vaccinated will multiply the risks faced by residents and reduce the chances of achieving herd immunity locally for all. Specifically, if too few Black and Latinx individuals choose to be vaccinated, communities with many Black and Latinx residents will remain vulnerable to infection at disproportionate rates while communities that are predominantly White and Asian approach herd immunity. Such a disparate result would be yet another racial inequity in a pandemic that has already had a glaringly disparate impact on health and finances in Black and Latinx communities16,17,18,19,20. Further, communities are not islands unto themselves, and intermixing through the daily movements of individuals have been critical in explaining infection transmission across regions21,22,23,24,25,26. As such, if any community in a region fails to achieve a sufficient level of vaccination, it could harbor infections that still pose a risk to unvaccinated individuals in other communities through mobility-based transmission.
The current study evaluates the inequities that might arise from differential willingness to be vaccinated across a single metro region. We use a traditional SIR (susceptibility–infection–recovery) model for simulating the evolution of an infectious disease within a community, applied across the ZIP codes of Boston, MA, and the surrounding municipalities in the greater Boston area (collectively referred to as “communities” from hereon). We further inform these models using mobility data generated by cell phones to track how movement between communities could further spread the disease21,22,23,24,25,26. We incorporate into these models an additional set of parameters for the gradual rollout of a vaccine, which removes individuals from a community’s susceptible population at an assumed rate of effectiveness, based on clinical trials27. We simulate the vaccination rollout as if it had occurred October-December, 2020. This matches the proposed 3-month vaccination process for the general population that many leaders promised in early 2021 and allows us to leverage historical mobility and infections data and, as has been demonstrated by other simulation studies of vaccination28, permits us a clear counterfactual against which we can compare the introduction of vaccination.
The study is designed to address four main research questions, each of which is relevant to the initial vaccination rollout for COVID but can also be generalized to future vaccination rollouts for this or other pandemics. First, we simulate both the global and community-specific rollout of vaccines, revealing at what point the process would be expected to hit a “bottleneck,” where supply outstrips willing recipients, and whether this milestone arrives at different times across communities. This will be crucial to leaders seeking to manage vaccine supply. Second, we quantify the anticipated impact of racial differences in vaccination hesitancy, which are largely attributable to historical inequities in medical treatment. To do so we use results from three recent surveys that separated responses by race to approximate willingness to vaccinate in each community. Third, there is evidence that those who are uncertain about receiving a new medical procedure, including vaccinations, often wait until others that they know have done so with few negative side effects29,30,31. In order to examine both the impact and limitations of this process, we include a “persuasion” rate (or “imitation” factor32, 33) by which individuals who were uncertain if they would get vaccinated can eventually decide to do so as the proportion of those vaccinated in their community increases. Fourth, a major concern is that slow uptake of vaccination in one community can undermine herd immunity in neighboring areas via mobility-based exposure. Here we will be able to evaluate this proposition.
## Results
### Vaccination
In three independent surveys of Boston and Massachusetts residents in the fall of 2020, 49.6% of respondents said that they planned to get the COVID-19 vaccine, and 8.8% said that they did not; the remaining 41.6% were uncertain. These responses featured prominent disparities by race, however. On the low end, 6.6% of White respondents and 1.8% of Asian respondents said they would definitely not get the vaccine, compared to 21.7% of Black and 20% of Latinx respondents (see Fig. 1A). We combined these ratios with the racial composition of communities to estimate the percentage of residents in each who planned to get vaccinated, did not plan to, and who were uncertain. This revealed stark differences across communities, with municipalities in the region varying between 33.9 and 54.1% of the population saying they would definitely get vaccinated, and between 5.4 and 17.2% saying they definitely would not; the same ranges indicated slightly less receptivity of vaccines in Boston’s ZIP codes, which has a higher Black and Latinx population than most surrounding municipalities (definitely will: 28.2–52.1%; definitely will not: 5.1–19.8%; sees Fig. 1B, Figure S1a–c).
We simulated the impact of vaccination across communities by incorporating it into an SIR-mobility model with parameters approximating transmission and recovery rates based on actual case numbers for October-December 2020. We also assume the 95% efficacy of the vaccine reported by the initial clinical trials27 and that vaccination will occur at the rate of 8.25% of the population per week, following the goal of complete vaccination over a 12-week (3-month) period. The model also allowed for those uncertain about the vaccine to be persuaded, contingent on the likelihood of seeing others in their neighborhood who had been vaccinated and presumably not seen adverse side effects31. To maintain a true counterfactual, all comparisons of outcomes are made against the results of the same model without vaccination as a facsimile for the actual events of this time period.
Figure 2A depicts a steady increase over the 3-month period in the proportion of people who were vaccinated across the region, reaching 75%. In late November, however, the vaccination process hit a bottleneck (mean = 53 days into simulation). It had exhausted all individuals who either were willing to be vaccinated at the outset or were persuaded to that point, as indicated by the blue line reaching zero. As a result, vaccination from then on was dependent on those additional individuals who were persuaded in each week, which was less than the rate at which vaccination was possible. This created the kink in the red line, indicating a slowed vaccination process from that point on, explaining why the simulation did not successfully vaccinate 100% of the population after 12 weeks, despite having the capacity to do so.
The bottleneck in vaccination did not occur at the same time in all communities (see Fig. 2B,C). For purpose of comparison here and moving forward, we divide communities into those that are predominantly White (> 80% White residents; 51% of communities), those with high Black–Latinx populations (> 20% Black and Latinx residents; 17% of communities), and those that are neither (32% of communities). For predominantly White communities, the bottleneck was reached at the very end of November (mean = 57th day). Meanwhile, in communities with high Black–Latinx populations the same milestone occurred before November 15th (mean = 42nd day). Because this date was reached earlier, the lower proportion of vaccinated residents at that time in turn resulted in a diminished power of persuasion, meaning fewer additional people were persuaded each week thereafter than in predominantly White communities. This further exacerbated disparities in cumulative vaccinations. By the end of the simulation, residents in predominantly White communities were consistently 80% vaccinated whereas those living in high Black–Latinx populations were 71% vaccinated, though there were communities with rates of vaccination as low as 59%.
### Infection rates
The impacts of vaccination, including disparities in uptake across communities, were evident in the corresponding evolution of infection rates. As shown in Fig. 3A, infection rates kept pace with the no-vaccination scenario until mid-October, which is when vaccination began to substantially lower the population susceptible to infection. Over the following week or so, infection rates started to decrease, eventually nearing zero infections. To wit, 99% of communities had more than one case per 100 without vaccination, whereas no community was above 0.75 cases per 100 with vaccination.
The growth curve for infection rates under vaccination was consistent across communities of different racial composition, though with some noteworthy variations. First, the absolute drop in infection rates with vaccination was greater in communities of color, seemingly because they had much higher levels of infection in the no-vaccine scenario. In other words, vaccination held the greatest absolute benefit for communities harder hit by the virus. The second difference, however, was how close they came to herd immunity. Predominantly White communities nearly reached zero infections (3rd quartile = 0.10 cases per 100 residents). Meanwhile, communities with high Black–Latinx populations failed to reach this point, averaging 0.27 infections per 100 residents at the end of the simulation (Fig. 3A–C).
A series of linear regression models confirmed the racial disparities in infection rates under the vaccination scenario (see Table 1 for all parameters). Communities with greater Black and Latinx populations had higher infection rates at the conclusion of the simulation (% Black: B = 0.30, p < .001; % Latinx: B = 0.54, p < .001). Even when we controlled for expected infection rates, as drawn from the no-vaccine scenario, these racial disparities persist, meaning that differential uptake of the vaccine played a consequential role in cross-community variations in infection (% Black: B = 0.26, p < .001; % Latinx: B = 0.26, p < .001).
### Achieving herd immunity
Moving beyond infection rates, did communities reach the goal of herd immunity via vaccination? We defined herd immunity as the point at which a community effectively eliminated the virus locally (i.e., < 1 infection; the model permits fractions of infections). We found that only 27% of communities had reached herd immunity by this definition at the end of the simulation. We thus extended the simulation for three additional months (simulating mobility based on historical data; see Methods for more). An additional 52% of communities achieved herd immunity in the fourth month of the extended simulation (80% cumulative), and all but one remaining community achieved herd immunity in the fifth month, which reached it shortly thereafter (see Fig. 3D,E). The average community reached herd immunity on day 103 of the simulation.
Differences in achieving herd immunity again reflected stark disparities by race. Only 14% of communities with high Black–Latinx populations saw herd immunity before the fifth month of the simulation, whereas 99% of predominantly White communities had achieved herd immunity by this time. To reiterate, all predominantly White communities achieved herd immunity before any community with a high Black–Latinx population, some by nearly 2 months. The average difference in achieving herd immunity between these two sets of communities was 45 days (89 days and 134 days into the simulation).
### Mobility-based exposure
Mobility-based exposure—that is, the quantification of a community’s potential exposure to infection via cross-community travel—has been a central component of models of the transmission of COVID21,22,23,24,25,26 and we included it in our own. Unsurprisingly, when added to the previous regressions, mobility-based exposure in a community at the end of the simulation independently predicted higher infection rates, albeit with an effect size considerably lower than those of racial composition (β = 0.22, p < .001; see Table 1 for all parameters). The inclusion of mobility-based exposure in the model did not meaningfully alter the effects of race. However, when we controlled for infection rates in the no-vaccine scenario, mobility-based exposure had a negative effect on infection rates under vaccination (β = − 0.10, p < .001). This would appear to be because vaccination especially benefited those who were more at risk from either baseline or dynamic exposure. This would all indicate that mobility-based exposure had a moderate effect on continued infections, though it played a much smaller role relative to vaccination in determining infection trends. As a final test of this interpretation, we ran a model using only percentage vaccinated and mobility-based exposure to explain infection rates, and found that the effect of the former was four times the size of the effect of the latter (β = − 0.78 vs. β = 0.21, both p values < .001).
If mobility-based exposure and vaccination are each relevant to the evolution of infections, the question remains how often these two factors coincide, making certain communities doubly vulnerable. We find that mobility-based exposure correlated moderately with the proportion of Black and Latinx residents (% Black: r =0. 23, p < 0.01; % Latinx: B = 0.23, p <0.01) and was slightly negatively associated with vaccination rates (r = − 0.28, p < 0.01). Thus, although high levels of mobility-based exposure and vaccination do not always coincide, the communities where they do are at highest risk.
### Robustness tests and implications
In order to probe the robustness of the simulation, we tested higher and lower levels of persuasion, as well as no persuasion; two lower levels of vaccine efficacy; and two longer timelines for vaccination roll-out. We re-ran the models with all possible combinations, making for 36 sets of results (4 × 3 × 3). These alterations had the expected effects on global patterns, either extending or compressing vaccination time, or increasing or decreasing total infections. The disparities in infection rates and herd immunity were also present across models. We conducted a meta-analysis of the disparities in infection rates and the attainment of herd immunity, entering the levels of persuasion, vaccine efficacy, and distribution time into regressions predicting specific outcomes and parameters: the effects of % Black and % Latinx on infection rates, both with and without controlling for infection rates under the no-vaccine scenario; the difference in the proportion of predominantly White and high Black–Latinx communities that reached herd immunity; the difference between predominantly White and high Black–Latinx populations in the average date on which herd immunity was reached.
As illustrated in Fig. 4, the level of persuasion had the strongest effect, markedly lowering all measures of disparity (all p values < .0.001). Greater vaccine efficacy also lowered the association between percentage Black and Latinx population and infection rates, both in general and owing uniquely to the introduction of vaccination itself (i.e., relative to the no-vaccine scenario). Vaccine efficacy had little effect on disparities in reaching herd immunity, however. This was likely because less effective vaccines will extend the timeline to herd immunity for everyone, but the burden of allowing infections to persist will fall heaviest on those communities experiencing the most exposure. Last, a quicker rollout rate led to greater disparities when controlling for infections in the no-vaccination scenario and differences in the average date of reaching herd immunity. This is likely because high Black–Latinx communities reach the vaccination bottleneck sooner and thus fall behind their predominantly White counterparts at a faster rate. It is important to note, however, that the absolute outcomes were better for everyone under this scenario. Last, a faster rollout led to lower disparities in the raw infection rate for the Latinx population, but this appeared to attributable to the higher starting infection rate and mobility-based exposure in these communities, making a quick rollout especially valuable.
As we describe these differences across simulations, we must take into account their practical implications. The strength of persuasion is the most important component in the models, and it cuts both ways. If we remove persuasion entirely, 97% of predominantly White communities reach herd immunity in the 180-day simulation, with an average of 121 days, whereas no high Black–Latinx communities do. However, its power to eliminate disparities seems limited. When we increased the rate of persuasion by 50% from the baseline model, the average high Black–Latinx community saw herd immunity only 9 days earlier, while predominantly White communities saw an even smaller improvement of 3 days. This closed the gap between the two by only 14%. This indicates that even if individuals uncertain about getting vaccinated are convinced to do so more quickly, the proportion of people committed to being vaccinated at the outset and, conversely, the proportion committed to not being vaccinated, are far more consequential in determining the timeline of reaching herd immunity.
## Discussion
The simulation found that the introduction of vaccinations helped all communities. Disparities, however, between predominantly White communities and communities of color emerged early and worsened as the simulations proceeded, the details of which answered each of our four research questions. First, communities quickly reached a bottleneck in vaccination, at which point further vaccination relied entirely on continued persuasion. Because communities with more Black and Latinx residents had fewer people who were initially willing to be vaccinated, they reached this bottleneck weeks before predominantly White communities. Second, the disparities in vaccination rates set the stage for inequities in infection rates. Communities of color experienced average rates of infection 3 times as high as predominantly White communities and reached herd immunity on average a month-and-a-half later. Third, we found that persuasion is crucial to closing the infection gaps between communities of different racial backgrounds. Unfortunately, however, increasing the power of persuasion had limited additional effect because of the large number of people saying they would definitely not get vaccinated in communities of color. Fourth, mobility-based exposure had only a moderate impact during vaccination, meaning that lower adoption of vaccines in some communities did not heavily undermine herd immunity in surrounding areas.
We proceed by elaborating on the insights from each research question in greater detail, but we must note that much has happened since the beginning of the vaccination program and the original execution of this study. Vaccination programs have been successfully administered nationwide, verifying much of what the simulations predicted. We conducted a posthoc analysis using data published by the Massachusetts Department of Public Health (weekly data available at https://www.mass.gov/info-details/massachusetts-covid-19-vaccination-data-and-updates#weekly-covid-19-municipality-vaccination-data-) and found that our estimated rates of vaccination by community correlated substantially with those reported on August 26, 2021 (r = .0.53). This is noteworthy because our analysis relied solely on race and ethnicity. In fact, the simulation results mapped almost perfectly onto correlations between actual vaccination rates and the racial composition of communities, the only deviation being that we underestimated the willingness (or willingness to be persuaded) to be vaccinated among Asian and Asian American residents (correlations of indicators of ethnic composition with residual variance in actual vaccination rates not predicted by our simulations: % Asian: r = 0.24, p <0.01; % Black: r = − 0.02, p = ns; % Latinx: r =0.06, p = ns; % White: r = − 0.10, p = ns.). In addition, bottlenecks in vaccination occurred as the population of willing recipients dwindled in the late spring; infections have persisted in areas with lower vaccination levels; and persuasion did appear to wane as those who said they would never get vaccinated remained steadfast. Political resistance to vaccination among conservative White populations has also emerged as a major barrier to herd immunity that we did not consider in our projections (though it may not have had much impact with our data given the generally liberal politics of the Greater Boston area). That all said, the emergence of new variants, especially the now-predominant, and substantially more virulent and contagious Delta variant, has altered the landscape of the battle against COVID-19. Thus, we conclude by discussing implications for the current moment, which largely exacerbate nuanced emergence and persistence of inequities that we forecast here.
Turning to our first research question, the bottleneck in vaccination raises two points worth elaborating. From a logistical perspective, the assumptions of our model were such that persuasion could not keep pace with vaccine supply, creating implications for proper management of distribution. Meanwhile, from a social perspective, the earlier bottleneck in communities with more Black and Latinx residents meant that they had fewer people vaccinated at this point. Because the persuasion rate is dependent on the proportion of community members who have been vaccinated, these communities fell further and further behind in vaccination in the following weeks. Thus, for both reasons of supply management and long-term advancement of the rollout, the bottleneck is a key moment in the evolution of the model and policymakers and practitioners should be attuned to any signals that this point is arriving and to ramp up community messaging accordingly. This issue was observed regularly in the early months of the vaccination rollout and has resulted in millions of wasted doses34. It could easily become an issue again if and when society moves to re-vaccinate owing to waning immunity or vaccines that are more efficacious against variants.
Because of their lower levels of vaccination in the simulations, communities of color were faced with a greater and more persistent threat of the virus than predominantly White communities. It is important to note the full meaning of this statement. All communities saw heavily diminished infection rates over the 3-month simulation and eventual herd immunity by the end of 6 months. But it took communities of color longer to arrive at that point, meaning they experienced a substantially greater cumulative amount of infections (and, consequently, deaths) than predominantly White communities. Thus, this is neither a doomsday scenario, nor is it what any mayor or governor would want for their city or region.
With that in mind, persuasion will be most critical to planning and leadership moving forward. By varying the strength of persuasion across models we see just how important it is to outcomes in the vaccination rollout30, 31, 33. The starkest illustration of this is that when we removed persuasion from the model, no high Black–Latinx community reached herd immunity by the end of 6 months, compared with over 95% of predominantly White communities. More subtle, but equally telling, was the impact of increasing the rate of persuasion as it only marginally narrowed inequities. This is because our model reflected a leading strategy of focusing on those who are unsure about getting the vaccine. This strategy operates on the assumption that those who say they “definitely will not” get the vaccine cannot be convinced by any quantity of evidence10, 35. Given the proportion of people who have said (and continue to say, at the time of the final revision) that they will never get vaccinated, especially in communities of color, this makes for a precarious situation; this is also increasingly a concern in more conservative white communities.
If one follows the logic of taking those that say they will never get vaccinated at their original word for Boston, 1 in 5 Black and Latinx residents would never be vaccinated. This would pose a large stumbling block to achieving herd immunity for communities of color. Though we have estimated herd immunity as the first moment at which there are no infections in a community, there could still be pockets of low vaccination with concomitant gaps in the protective wall of community health36. That said, vaccine hesitancy is not synonymous with vaccine denial, which questions efficacy and propagates potential harms from vaccines37. With luck, some of those who are rejecting the idea of being vaccinated now are more hesitant than they are deniers, and they will be receptive to increased evidence of the vaccine’s effectiveness and safety. Thus, the solution is not only to increase the rate at which people on the fence decide to be vaccinated; it is in convincing those steadfastly opposed to the vaccination to consider it. This has only become more critical since vaccine rollout began as misinformation has steadily undermined efforts to reassure those who are hesitant38.
Our fourth research question regarded the implications of lower vaccine adoption in some communities for the surrounding region. Namely, could lower adoption in one or more communities undermine the pursuit of herd immunity in surrounding communities? This would seem plausible given the known importance of cross-community mobility for transmission of the virus21,22,23,24,25,26. We find here a mixed answer. Indeed, mobility-based exposure continued to predict infection rates, though with a strength far smaller than that of vaccination adoption. This is consistent with other work that has found that the importance of mobility has been largely supplanted by the impact of interventions intended to mitigate transmission of the virus, whether they be social distancing guidelines in the summer of 2020 or, in this case, vaccine rollouts39. As such, the ability of a community with lower vaccine adoption to impact other communities around it appears to have been limited under the assumptions of the model (though see below for more on the implications of the Delta variant).
There are a number of limitations to the model and its implications arising from assumptions we were forced to make. First, we did not account for any first-stage rollout to first responders, frontline workers, and the most at-risk, which was anticipated to reach ~ 10–15% of the population. This would accelerate herd immunity by raising the baseline level of vaccination and kick-starting persuasion. Second, we assumed that vaccinated individuals cannot carry the virus. When this study was originally conducted, there was little evidence one way or the other as to whether this was the case, though this has now changed with the emergence of the Delta variant. Third, our simplistic model of persuasion ignores the power of leadership, which is argued to be as critical in communities of color as seeing one’s neighbors be vaccinated40. Fourth, the reproduction rate of the virus remained stable throughout the simulation, but there are reasons to believe that vaccination leads to behavioral shifts that will increase the underlying reproduction rate even if vaccines themselves limit the total impact41. Our estimates of infection rates and the timeline to herd immunity could be somewhat optimistic in that case. Fifth, the model treats vaccination access as being evenly distributed across communities, whereas numerous analyses have found lower numbers of vaccination sites in communities of color in multiple U.S. cities, especially in southern states42, 43. All of these dynamics could alter the specific quantitative results we see here and the timeline to herd immunity. That said, given the variety of robustness tests that we have run, we are confident saying that in any of these scenarios there will be extensive inequities in infection rates and the time elapsed before achieving herd immunity between predominantly White and Asian communities and Black and Latinx communities—and that some of these nuances would only exacerbate outcomes.
We conclude by noting the new shape the pandemic has taken in the months since the vaccination program started and this study was originally conducted. As described above, our simulation forecast inequitable outcomes, but eventual herd immunity for everyone. Unfortunately, since the emergence of the more virulent Delta variant, this situation no longer seems near at hand, if even within the realm of possibility. The variant is far more virulent, instigating what the governor of Mississippi called “a pandemic of the unvaccinated,” but it is also more contagious, lowering the efficacy of vaccination. We did run alternate simulations that iteratively lowered levels of vaccine efficacy, and these increased racial inequities in the long-term persistence of infections. Additionally, lowering vaccine efficacy to 75% increased the impact of mobility on infection rates by ~ 25%. Greater virulence would only magnify each of these findings, further highlighting the urgency around inequities but also tempering our forecast that inter-community mobility would be less of a concern at this point in the vaccination program.
Last, the narratives surrounding vaccination intentions in the United States have shifted somewhat, with increasing concern not only about hesitance among Black and Latinx residents in urban areas, but also among conservative White Americans, especially in rural areas44. Simulations like those run here that address whatever local disparities exist could be useful for planning across the United States and internationally, provided that there are tools available for tracking intentions, be they repeat surveys (especially with panel designs45), social media polls46, or otherwise. This is especially true if there is a transition to the distribution of booster shots and widespread revaccination, in which case attitudes may be somewhat different from before. In any case, the pandemic has not in fact been “ended” by the current vaccination program and continuing to refine these multi-methodological approaches to anticipate and respond to it will be crucial.
## Materials and methods
The study centers on an SIR model that uses pre-determined transmission and recovery rates in conjunction with mobility between communities to estimate daily infection rates in each community. The models were run for October-December, 2020, and the transmission and recovery rates were based on actual infection records. Mobility was also derived from historical data for the same time period. The “communities” act as nodes in the mobility model and are defined as the 100 non-Boston cities in towns in the greater Boston region (following the Metropolitan Area Planning Council’s definition) and the 28 ZIP codes within Boston. This decision was made based on the availability of more granular data for Boston as well as the large amount of between-ZIP code demographic diversity within the city, which is lower or absent in many of the surrounding municipalities.
### Data and measures
The models used four data sources: (1) population descriptors from the American Community Survey’s 2014–2018 5-year estimates; (2) daily and weekly infection case counts, derived from infection records, for all towns in greater Boston and ZIP codes within Boston; (3) responses to three surveys including items on people’s intentions regarding vaccination; (4) cross-community mobility records derived from cell phone records, generated by SafeGraph, a data company that aggregates anonymized location data from numerous applications in order to provide insights about physical places, via the Placekey Community. To enhance privacy, SafeGraph excludes census block group information if fewer than five devices visited an establishment in a month from a given census block group. The Boston Area Research Initiative’s Geographical Infrastructure47 was used to join data describing each municipality or ZIP code.
#### Census indicators
We drew population descriptors from the U.S. Census’ American Community Survey’s 2014–2018 estimates for all census block groups in Massachusetts. This level was selected as it is the largest census geography that nests cleanly within ZIP codes in Boston and within municipal boundaries. Community indicators included total population and ethnic composition (i.e., proportion Asian, proportion Black, proportion Latinx, proportion White). All measures were aggregated from census block groups to the municipal or ZIP code level using population-weighted means.
#### Infection cases
The Commonwealth of Massachusetts’ Department of Health released weekly counts of new infections for all municipalities, starting on April 14th, 2020. It also released daily counts for counties. From these two data sources we created daily town measures by: tabulating the weekly sum of infected cases in a county; calculating the percentage of a county’s cases attributed to each town; estimating the daily infected cases per town as the same percentage of the daily count for the county. For Boston ZIP codes, we had case records tracked by the Boston Public Health Commission mapped to the ZIP code of residence. We tabulated these for daily counts.
#### Surveys
Three surveys of Massachusetts residents were conducted that included a question regarding intention to vaccinate and split these responses by race. The three surveys were conducted by the Center for Survey Research at University of Massachusetts Boston with the Boston Area Research Initiative, MassInc Polling on behalf of the Boston Museum of Science, and Suffolk University Polling and the Boston Globe. The first surveyed residents of Boston from September to November and the other two surveyed residents from throughout Massachusetts in November and December, respectively. Although the exact wording varied between them, all three surveys permitted respondents to say that they “definitely” or “definitely did not” plan to get the vaccine or that they were uncertain or undecided. The overall proportion of individuals falling in each of these three groups was consistent across the three surveys, as were the breakdowns by race (though the MassInc poll did not have enough Asian respondents to include cross-tabs for that group; see Supplementary Online Materials for full results).
We summed the cross-tabs for the vaccination intention question by race across the three surveys to calculate the weighted proportion of individuals indicating “as soon as possible,” “never,” and something in between for each of the four major racial categories—Asian, Black, Latinx, and White. We then estimated the proportion of residents in each of these three categories regarding the vaccine for each municipality and Boston ZIP code with the following equation:
$$Y_{i,k} = \mathop \sum \limits_{j} p_{i,j} *r_{j,k}$$
where Yi,k is the proportion of residents in community i with attitude k toward the vaccine (e.g., getting it as soon as possible), pi,j is the proportion of residents in community i of race j, and rj,k is the proportion of members of race j giving k as their response across the three surveys.
#### Cellphone generated mobility records
We used SafeGraph’s daily “Social Distancing” dataset to create the mobility network. The data are generated using a panel of GPS pings from anonymous mobile devices. Each device is attributed to an estimated home census block group (CBG) based on its most common nighttime location. It also tracks all stay points of these devices within other CBGs. The published data aggregate these pieces of information to generate a mobility matrix of the daily number of visits by the assumed residents of each CBG to each other CBG. Each CBG was nested in its ZIP code or municipality.
### Mobility-driven SIR model with the distribution of vaccination
#### Model
Our model was based on a traditional SIR (susceptible-infected-recovered) model that then incorporated two additional factors: mobility, to simulate the effect of contacts brought by the mobility between communities; and vaccination, to model the effect of the adoption of vaccination across communities. The full model consists of the following differential equations, which update daily:
\begin{aligned} \partial_{t} j_{n} & = \left( {\alpha_{n} j_{n} + \gamma \mathop \sum \nolimits_{m \ne n} \alpha_{n} w_{mn} j_{m} \frac{{N_{m} }}{{N_{n} }} + \gamma \mathop \sum \nolimits_{n \ne m} \alpha_{m} w_{nm} j_{m} } \right)s_{n} - \beta j_{n} \\ \partial_{t} s_{n} & = - \left( {\alpha_{n} j_{n} + \gamma \mathop \sum \nolimits_{m \ne n} \alpha_{n} w_{mn} j_{m} \frac{{N_{m} }}{{N_{n} }} + \gamma \mathop \sum \nolimits_{n \ne m} \alpha_{m} w_{nm} j_{m} + \mu g\left( {p_{n} } \right)} \right)s_{n} \\ \partial_{t} r_{n} & = \beta j_{n} \\ \partial_{t} p_{n} & = - g\left( {p_{n} } \right) + hu_{n} v_{n} \\ \partial_{t} u_{n} & = - hu_{n} v_{n} \\ \partial_{t} v_{n} & = g\left( {p_{n} } \right) \\ \end{aligned}
where Nn, jn, sn, and rn represent the total population size, number of infected cases, susceptible individuals, and recovered cases, respectively, for a community at a given timepoint.
The equations, which model simultaneous change over time in infected, susceptible, and recovered individuals, rely on four main components. αn is the growth rate of infections in a community (i.e., the expected number of new cases from existing cases). It is based on a global α0 in combination with a sigmoid function that accounts for fluctuation effects between communities when jn is less than a threshold ε i.e. $$\alpha_{n} = \alpha_{0} \cdot \frac{{\left( {\frac{{j_{n} }}{\varepsilon }} \right)^{4} }}{{1 + \left( {\frac{{j_{n} }}{\varepsilon }} \right)^{4} }}$$48. The second component pertains to mobility, including two operands: $$r_{I} = \gamma \sum\nolimits_{m \ne n} {\alpha_{n} w_{mn} j_{m} \frac{{N_{m} }}{{N_{n} }}}$$ calculates the possibility of an infected person in other communities (m) visiting the community and infecting susceptible in community n; and $$r_{R} = \gamma \sum\nolimits_{n \ne m} {\alpha_{m} w_{nm} j_{m} }$$ is the possibility of a susceptible person from the community visiting another community (m) and becoming infected. wmn = Fmn/Nm where Fmn indicates the total number of visits from community m to community n, as captured by the cell-phone generated mobility data. The average mobility rate gamma is defined as $$\gamma = \frac{{\sum\nolimits_{m \in G} {F_{m} } }}{{\sum\nolimits_{m \in G} {N_{m} } }}$$. The third component is the rate of recovery of infected individuals, represented by β, which is consistent across communities. All individuals who have been infected and recovered are permanently removed from the susceptible population.
The fourth component of the model is vaccination adoption, wherein pn is the proportion of people who will definitely receive the vaccine, un is the proportion of people uncertain about getting the vaccine, and vn is the proportion of people who have already been vaccinated. μ reflects the effectiveness of the vaccine and μvn are treated as part of rn as they have been removed from the susceptible population. h is the persuasion rate at which someone uncertain about getting the vaccine will be persuaded to do so, whose strength is contingent on the proportion of residents in the community who have already been vaccinated (i.e., hunvn). We assume that h is consistent across communities. Only those who were uncertain about the vaccine could be persuaded, not those who stated they would never get the vaccine. Actual vaccination is represented by the function g(pn), of the form:
$$g\left( {p_{n} } \right) = \left\{ {\begin{array}{*{20}l} c \hfill & {p_{n} \ge c} \hfill \\ {p_{n} } \hfill & {p_{n} < c} \hfill \\ \end{array} } \right.$$
in which c is the maximum capacity of the vaccination rollout for a day.
#### Fixed parameters
Multiple parameters were established in advance of estimating the final model. α and β, the transmission and recovery rate, were estimated by running the simulation without vaccination on historical mobility and infection data for September 30th through December 22nd. Grid search identified a local optimum for α = 0.096 and β = 0.072, which translate to the more familiar R0 = α/β = 1.33. ε was calculated as $$\varepsilon = \frac{M}{{\sum\nolimits_{m \in G} {N_{m} } }} = 3.826 \times 10^{ - 5}$$, where M is the number of communities. μ, or the vaccine effectiveness, was set to 0.95, per the Pfizer and Moderna trial results. The rate of vaccine rollout was initially set to a 12-week (approx. 3-month) rollout period, meaning c = 1/(7*12) = 0.0119, or 1.19% of the population could be expect to be vaccinated daily.
In addition, we had to extend the simulation for three additional months for determining herd immunity (i.e., < 1 infection; the model permits fractions of infections), because only a third of communities had reached it by this definition at the end of the initial simulation. Two extreme scenarios are designed to impute the mobility data after Jan 12, 2021 to ensure that the real mobility data lies between the two imputed mobility datasets. We assume a cumulative 5 ± 0.5% increase or decrease for every 2 weeks to the real mobility data from Dec 29, 2020, to Jan 11, 2021. | 2023-01-31 06:46:07 | {"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": 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": 0.5145708322525024, "perplexity": 2361.8736477796037}, "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-2023-06/segments/1674764499845.10/warc/CC-MAIN-20230131055533-20230131085533-00295.warc.gz"} |
https://neuraldatascience.io/6-single_unit/intro_spike_trains.html | # Introduction to Spike Train Data#
The examples in this lesson were adapted from Chapter 3 of Nylen, E.L., and Wallisch, P.. (2017). Neural Data Science. Academic Press. Code and explanatory text by Aaron J. Newman.
Watch a walk-through of this lesson on YouTube
## Learning Objectives#
• Understand how spike trains are represented as time series
• Work with a set of spike trains in list and NumPy array formats
• Visualize a set of spike trains as a raster plot
• Visualize a set of spike trains as a peri-stimulus time histogram (PSTH)
# Import libraries we'll need
import matplotlib.pyplot as plt
import numpy as np
## Our first spike train#
Imagine an experiment involving a neuron that was genetically engineered for optogenetics, meaning that it expresses genes sensitive to a specific wavelength of light (550 nm, which is green; to learn more about optogenetics, check out this video by Prof. Carl Petersen). When light of that wavelength is directed at the neuron, it tends to fire.
For this hypothetical experiment, we used an electrode to record action potentials from this optogenetically-engineered neuron to confirm that it tends to fire in response to 550 nm light. This was done across multiple trials. On each trial, recording was started and then the 550 nm light was turned on 4 ms later, for a duration of 10 ms. Each trial lasted a total of 20 ms. Data was sampled from the electrode every millisecond.
We can store the data from each trial in a list, spike_train. Each value represents an evenly-spaced point in time (e.g., every 1 ms), where 0 encodes time points at which no action potential was detected, and 1 represents times when action potentials occurred:
spike_train = [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0]
len(spike_train)
21
The list has 21 values because the first value represents time point 0, when the recording started, and after this we recorded up to 20 ms.
### First spike latency#
A characteristic feature of a neuron is the first spike latency to stimulus — how long after a stimulus the first spike occurs. Different neurons will have different average values of this.
Since we know the light came on 5 ms after the start of each trial, we can define light onset time and duration as variables, and also defines that a spike is represented as a 1 in the data. Then, we can look for the first spike that occurred after the onset of the light.
light_onset_time = 4
stimulus_duration = 10
spike_value = 1
To determine first spike latency relative to stimulus onset, we can search in the spike_train list starting with light_onset_time (using list slicing) and looking forward until we find the first spike (spike_value), using the .index() method which returns the first index in a list (or slice) matching the value given as the argument:
latency_to_first_spike = spike_train[light_onset_time:].index(spike_value)
print('First spike latency to stimulus = ' + str(latency_to_first_spike))
First spike latency to stimulus = 5
### Find indexes of all spikes#
If we want to find the indexes of all the spikes that occurred after stimulus onset, we need to use a for loop, because the .index() method only finds the first occurrence of a value in a list.
Here we introduce a useful variant of the typical for loop, using the enumerate() function. While a regular for loop simply loops through all values of a list (or other object), enumerate() does this while keeping track of the index of the current item within the loop. The enumerate() function returns two values every pass through the loop, which we assign to two variables, which here we call i and x.
for i, x in enumerate(spike_train):
print(i, x)
0 0
1 0
2 0
3 0
4 0
5 0
6 0
7 0
8 0
9 1
10 0
11 1
12 0
13 1
14 0
15 0
16 0
17 1
18 0
19 0
20 0
In the output, the first column is the index (position) in the list, and the second column indicates whether a spike happened or not.
We can use this in a nested for/if loop to step through each value in the list, and if it is a spike, append the index of that list position to a list called spike_times:
spike_times = []
for i, x in enumerate(spike_train):
if x == spike_value:
spike_times.append(i)
print(spike_times)
[9, 11, 13, 17]
### List Comprehension#
Alternatively, we can do this more efficiently with list comprehension:
spike_times = [i for i, x in enumerate(spike_train) if x == spike_value]
print(spike_times)
[9, 11, 13, 17]
This is a slightly more complicated list comprehension than we’ve seen in the past. Take some time to break this down and understand it — there’s a lot going on! Recall that list comprehension is used to create a for loop in a single line, and return a list. It’s a useful and compact way to iterate over some set of items to get them into a list. In this list comprehension:
• The enumerate() function enumerates (counts) the items as it goes through a loop, so each time through the loop we have two variables being tracked. In this case, they are x, which is one value in spike_train, and i, the enumeration variable that stores the current position in the list (iteration through the loop).
• To the end of this is added a conditional, if x == spike_value. This allows us to retrieve the index (timing) of only values when that represent spikes (recall we defined spike_value as 1). The for statement followed by the if conditional is equivalent to nesting the if statement inside the for loop in the previous example.
• The i right at the start of the list comprehension means that, if any value of x is a 1, then the index of its list position, i, is added to the spike_times list.
For a detailed overview of list comprehension, check out this DataCamp tutorial.
## Visualizing a Spike Train: Raster Plots#
Visualization is an effective way of making the data more interpretable. It is more intuitive for most people to see a timeline with spikes marked at particular time points, than to read a list of index values. Below we generate a raster plot, with time on the x axis, and the spikes as vertical lines. The y axis is shown as a continuous scale (the default in matplotlib), but of course the actual values can only be 0 or 1.
Since spike_times contains the time points of our spikes, we can use the following code to draw an object-oriented plot in Matplotlib:
fig, ax = plt.subplots()
ax.vlines(spike_times, 0, 1)
plt.show()
You can see that we have a plot with vertical lines (in Matplotlib’s default colour, blue) indicating the spike times.
One thing to notice in the code for this plot is that it’s kind of “hacking” Matplotlib functions to generate the plot. That is, we’re not using the plt.plot() function or ax.plot() method. Instead, we use the ax.vlines() method (there is also a plt.vlines() function, if using the procedural approach to plotting). This function can be used for drawing a vertical line anywhere in a Matplotlib plot. It takes three arguments: the x value(s) at which we want to draw the vertical line(s), the y value for the start (bottom) of the line(s), and the y value for the end (top) of the line(s).
### Formatting the plot#
We can make the figure more informative though. Firstly, since the x axis is time, we should show the whole time period of the trial, from 0–20 ms, not just the range from the first to last spike. We can also label the x axis and add a title to facilitate interpretation:
fig, ax = plt.subplots()
ax.vlines(spike_times, 0, 1)
# set x axis range and label
ax.set_xlim([0, len(spike_train)])
ax.set_xlabel('Time (ms)')
ax.set_title('Neuronal Spike Times')
plt.show()
Finally, we can show when the stimulus occurred, by shading the plot for the time between stimulus onset and offset. We use the axvspan function to add shading to the time period during which the green light was on, and we add the alpha= kwarg to adjust the transparency of the shading so we can see the blue lines (alpha=1.0 would be opaque; alpha=0.0 would be invisible).
Note the clever use of colour-coding: we use the greenyellow colour as this is closest to 550 nm. Thus rather than using an arbitrary or neutral shading colour like grey, the colour simultaneously encodes the time of the stimulus, and a key property of it. This is effective communication!
# define light offset time
light_offset_time = light_onset_time + stimulus_duration
fig, ax = plt.subplots()
ax.vlines(spike_times, 0, 1)
ax.set_xlim([0, len(spike_train)])
ax.set_xlabel('Time (ms)')
ax.set_title('Neuronal Spike Times')
ax.axvspan(light_onset_time, light_offset_time, alpha=0.5, color='greenyellow')
plt.show()
## More Trials#
One spike train is not representative of what that neuron would do every time it’s stimulated. Neurons typically respond in a probabilistic way to stimuli, meaning that presenting a stimulus that the neuron is tuned to will make the neuron more likely to fire, but it’s firing at a specific time point is not guaranteed. As well, neurons can also fire spontaneously, irrespective of an experimental stimulus. For both these reasons, we need to determine what the typical, or average, response of a neuron is by measuring its activity over many trials.
Below we have a set of 10 spike trains, representing 10 trials. These are stored as a list of lists, as we can tell from square brackets embedded in square brackets:
ten_spike_trains = [[0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0],
[0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0],
[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0]
]
Although the nicely-formatted list of lists above is pretty easy to look at and see where spikes occur, a raster plot will definitely be a more effective way of communicating this information, because we can add details that aid in interpretation, such as a time axis, shading to indicate when the light is on, axis labels, and a title. We can draw each trial as a separate row in the raster plot, rather than having each spike’s vertical line span the entire range of the y axis.
We can use the code from above, with few modifications:
• We’ll use a for loop to cycle through each trial and plot it
• We use range(len(ten_spike_trains)) as the collection that we loop over, to track trial numbers
• trial is used to index/select which trial we’re plotting each time through the loop
• We also use trial to set the position of each row of tick marks on the y axis of the plot. The y value corresponds to trial number. To centre the tick marks relative to the tick marks on the y axis, we set the height of each tick mark to go from trial - 0.5 to trial + 0.5
• By default, Matplotlib will decide on a “reasonable” selection of labels on each axis. Here, we would like every trial to be numbered, so we explicitly call ax.set_ylabel()
fig, ax = plt.subplots()
# Loop to plot raster for each trial
for trial in range(len(ten_spike_trains)):
spike_times = [i for i, x in enumerate(ten_spike_trains[trial]) if x == spike_value]
ax.vlines(spike_times, trial - 0.5, trial + 0.5)
ax.set_xlim([0, len(spike_train)])
ax.set_xlabel('Time (ms)')
# specify tick marks and label label y axis
ax.set_yticks(range(len(ten_spike_trains)))
ax.set_ylabel('Trial Number')
ax.set_title('Neuronal Spike Times')
ax.axvspan(light_onset_time, light_offset_time, alpha=0.5, color='greenyellow')
plt.show()
## Convert to NumPy array#
Recall that ten_spike_trains is a list of lists — one list for each trial. Another way of thinking of this data is as a 2 dimensional (2D) array. An array is a data structure composed of sequences of values. A Python list is an example of a 1 dimensional (1D) array. So for example, our first spike_train was a 1D array composed of a series of 21 values.
spike_train = [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0]
A 2D array is basically a matrix with rows and columns. pandas DataFrames are one example of a 2D array. The number of values (cells) in a 2D array is the number of rows times the number of columns. In other words, all the rows, and columns, of an array must be the same length.
For example, in our ten_spike_trains data set, we have the same number of values for each spike train (21), so we can imagine our data as an array where rows are trials, and columns are time points.
Unlike pandas DataFrames, arrays can have more than 2 dimensions. You can think of a 3D array as a 3D cube with the third dimension being depth. For now, we’ll stick with 2D arrays though!
Python’s NumPy package provides a numpy.ndarray (where nd stands for n-dimensional, meaning any number of dimensions) data structure and a wide variety of vectorized (optimized) functions and methods for working with arrays. If we give it our ten_spike_trains list of lists, it converts it to a NumPy array:
spike_array = np.array(ten_spike_trains)
print(spike_array)
type(spike_array)
[[0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 1 0 0 0]
[0 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 1 0 0 0]
[0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0]
[0 0 0 0 0 0 0 0 1 1 0 1 0 0 0 0 0 0 1 0 0]
[0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 1 0]
[0 0 0 0 0 0 0 0 1 1 1 0 0 1 1 0 0 1 1 0 0]
[1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0]
[0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 0]
[0 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 1 1 0 0]]
numpy.ndarray
The result looks very much like the original list of lists, except that there are no commas between the list items. As well, the output of the last command shows us that this is an object of type numpy.ndarray.
The square brackets of NumPy arrays represent the dimensions of the data. So, rows are the first dimension, and each row is in its own set of square brackets. All of the rows are contained in one set of square brackets representing the second dimension — i.e., that all of the rows together form a set of columns.
### Why NumPy arrays?#
Since the NumPy array version of our data looks pretty much the same as the list of lists format, you may wonder why we introduce a new data type rather than continuing to work with lists. There are a few reasons:
1. NumPy is faster
• The NumPy package is optimized specifically for working with array-format data, including manipulating the data, selecting subsets, and performing mathematical operations. Lists are deisgned for more general purposes and so run more slowly with arrays. So working with NumPy arrays is much faster than lists, and the performance benefits increase as data sets get larger.
2. NumPy has many tools
• Since NumPy is designed for working with arrays, it has a large number of functions and methods that allow you to do a lot of common data science tasks very easily.
3. NumPy is extensible
• NumPy can handle data with more than 2 dimensions, making it powerful for complex data sets.
## The Peri-Stimulus Time Histogram (PSTH)#
Along with raster plots, another common way of visualizing spike train data is the peri-stimulus time histogram (PSTH). As the name implies, this is a histogram — a plot of the count of spikes at each time point in the spike trains. PSTHs are a useful tool in aggregating the data across trials. As biological entities, neurons are complex and do not always spike at exactly the same time, or rate, every time they are stimulated. This is why we collect data over many trials — to identify the average pattern of responding. By summing the number of spikes in short time bins, PSTHs allow us to visualize when the neuron is most likely to spike.
Much of the code below is the same as for the previous raster plot. However, in place of drawing the vertical lines, we run plt.bar() to generate a bar plot. There will be one bar for each time point, and its height will represent the number of spikes at that time point, over the 10 trials. The required arguments for plt.bar() are x and height.
For x, we want the data time points (0–20 ms), which we can get from the number of columns in the array. NumPy arrays have a .shape property which returns the shape of the 2D array as [rows, columns]. So spike_array[1] will return the number of columns (spike_array[0] would give the number of rows). We pass this length to range() to generate the range of values for the x axis.
For the height of the bars we pass the sum across rows of the NumPy array: np.sum(spike_array, 0). The second argument, 0, is the dimension of the array we want to sum over. Recall that the first dimension of the array is the rows, and that Python indexes from 0. So 0 tells NumPy to sum over the rows dimension (and 1 would be used if we wanted to sum over columns instead).
fig, ax = plt.subplots()
# Do this first, so it's drawn "under" the PSTH
ax.axvspan(light_onset_time, light_offset_time, alpha=0.5, color='greenyellow')
# Draw the PSTH
ax.bar(range(spike_array.shape[1]),
np.sum(spike_array, 0)
)
# Make pretty
ax.set_title('Peri-Stimulus Time Histogram (PSTH)')
ax.set_xlabel('Time (ms)')
ax.set_ylabel('Num. spike occurrences at this time')
plt.show()
The PSTH allows us to easily visualize the apparent relationship between light stimulation and spiking: relative to the baseline period prior to stimulation, the neuron shows a noticeable increase in spike probability starting approximately 2.5 ms after light onset, peaking around 4 ms post-onset, and then decreasing as the light stays on. Interestingly, there appears to be another increase in spike probability after the light is turned off again. Some cells show offset responses like this, as well as onset responses.
Viewing the data as a PSTH reveals structure (relationships — in this case between stimulation and spike probability) much more easily than look at an array of numbers. This is even more true in real data sets, that involve arrays a lot bigger than the one we’re working with!
## Normalizing PSTHs#
The above PSTH simply plots the number of spikes per time bin (using the np.sum() function). Thus the numbers plotted are somewhat arbitrary, in that they will differ depending on the number of trials that were run. As well, even when we see peaks in the histogram, if we visualize the raw count without knowing the number of trials that contributed, we don’t have a good way of interpreting the data. That is, since we know here that we have 10 trials, a count of 7 is a high proportion of trials. However, if the data comprised 50 trials, a peak of 7 might not be compelling evidence that the neuron was responsive to the stimulus. We should always strive for transparency and readability in visualizing data, and minimize the need for the viewer to read additional material (e.g., the methods section where hopefully the number of trials is reported) to interpret the plot. As well, plotting things in standardized units makes it easier to compare across plots, articles, etc.. This is the process of normalization that we discussed previously in the context of Seaborn histograms.
In the case of PSTHs, what we really want to know is how probable is it that the neuron will fire at a particular time point after a stimulus, across trials. We can do this by computing the mean number of spikes at each time point with np.mean(), rather than np.sum(). Since the only possible values are 0 or 1, the mean will represent the proportion of times that the neuron spiked at that time point, regardless of how many trials comprised the data.
Normalizing the height of the bars can further facilitate interpretation of the data, because now we have an interpretable scale for the data. We might want to get a sense of whether the probability of firing is greater than chance, and thus likely related to the stimulus. Since, like a coin toss, the neuron only has two possible states (firing or not), a simple criterion would be to say that if the neuron fires on more than 50% of trials, it is responding to the stimulus. We can facilitate this interpretation for our viewer by adding a horizontal dashed line at 0.5 on the y axis, using plt.axhline()
fig, ax = plt.subplots()
ax.axvspan(light_onset_time, light_offset_time, alpha=0.5, color='greenyellow')
# Draw the PSTH
ax.bar(range(spike_array.shape[1]),
np.mean(spike_array, 0)
)
# Add line showing chance probability of firing
ax.axhline(y=0.5, xmin=0, xmax=20, linestyle='--', color='gray')
# Make pretty
ax.set_title('Peri-Stimulus Time Histogram (PSTH)')
ax.set_xlabel('Time (ms)')
ax.set_ylabel('Num. spike occurrences at this time')
plt.show()
## Raster + PSTH = Sub-plots#
Recall our previous lesson on sub-plots. We can combine a raster plot and PSTH as two sub-plots of one figure, organized as one column and two rows:
fig, axs = plt.subplots(2, 1, figsize=[10,5])
########################################
# draw raster into first subplot, axs[0]
axs[0].axvspan(light_onset_time, light_offset_time, alpha=0.5, color='greenyellow')
for trial in range(len(ten_spike_trains)):
spike_times = [i for i, x in enumerate(ten_spike_trains[trial]) if x == spike_value]
axs[0].vlines(spike_times, trial - 0.5, trial + 0.5)
# Set x axis range so that time zero is more visible
axs[0].set_xlim([-1, len(spike_train)])
# specify tick marks and label label y axis
axs[0].set_yticks(range(len(ten_spike_trains)))
axs[0].set_ylabel('Trial Number')
axs[0].set_title('Neuronal Spike Times')
#######################################
# draw PSTH into second subplot, axs[1]
axs[1].axvspan(light_onset_time, light_offset_time, alpha=0.5, color='greenyellow')
# Draw the PSTH
axs[1].bar(range(spike_array.shape[1]),
np.mean(spike_array, 0)
)
# Use same x axis limits as for raster plot
axs[1].set_xlim([-1, len(spike_train)])
# Add line showing chance probability of firing
axs[1].axhline(y=0.5, xmin=0, xmax=20, linestyle='--', color='gray')
# Make pretty
axs[1].set_title('Peri-Stimulus Time Histogram (PSTH)')
axs[1].set_xlabel('Time (ms)')
axs[1].set_ylabel('Probability of spike')
### Overall figure stuff
fig.suptitle('Spike Trains Shown Two Ways')
plt.tight_layout()
plt.show() | 2023-03-28 02:01:24 | {"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": 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": 0.44095325469970703, "perplexity": 543.1611299192798}, "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/1679296948756.99/warc/CC-MAIN-20230328011555-20230328041555-00401.warc.gz"} |
http://books.cs.luc.edu/introcs-csharp/data/learning-to-problem-solve.html | # 2.13. Learning to Solve Problems¶
This section might have been placed earlier, but from reading all the way to here, you should realize that you will have a lot of data and concepts to deal with.
The manner in which you deal with all the data and ideas is very important for effective learning. It might be rather different than what you needed if you were in a situation where rote recall is the main important thing.
Different learning styles mean different things are useful to different people. Consider what is mentioned here and try out some approaches.
The idea of this course is not to regurgitate the book, but to learn to solve problems (generally involving producing a computer program). In this highly connected and wired world you have access to all sorts of data. The data is not an end in itself, the question is doing the right things with the tools out there to solve a new creative problem.
In this course there is a lot of data tied into syntax and library function names and .... It can seem overwhelming. It need not be. Take a breath.
First basic language syntax: When learning any new language, there is a lot to take in. We introduce C# in chunks. For a while there will always be the new current topic coming. You do NOT need to memorize everything immediately!
• Some things that you use rarely, you may never memorize, like, “What is the exact maximum magnitude of a double?” At some point that might be useful. Can you find it? It happens to be in Numeric Types and Limits. It is also in online .Net documentation that you can Google or bookmark.
• Some things you will use all the time, but of course they start off as new and maybe strange. Knowing where to go to check is still useful but not sufficient. For much-used material that you do not find yourself absorbing immediately, consider writing down a summary of the current topic. Both thinking of a summary and writing help reinforce things and get you to remember faster. Also, if you have the current things of interest summarized in one place, they are easy to look up!
• If you need some syntax to solve a simple early problem, first try to remember the syntax, then check. With frequently used material and with this sort of repetition, most everyone will remember most everything shortly. If there are a few things that just do not stick, keep them in your list. Then go on to new material. The list of what you need to check on will keep changing as you get more experience and get to more topics. If you keep some of the old lists, you will be amazed how much stuff that you sweated over, is later ho-hum or automatic.
• In the earliest exercises the general steps that you need should be pretty apparent, and you can just concentrate on translating simple ideas into C# syntax (mostly from the material most recently introduced). In this case the focus is mostly on syntax.
Memorizing syntax is not going to directly get you to solve real problems. In any domain: programming, construction, organizing political action, ..., you need to analyze the problem and figure out a sequence of steps, knowing what powers and resources you have.
For example with political action: if you know demonstrations are possible in front of City Hall, you can make a high-level plan to have one, but then you have to attend to details: Do you need city permission? Who do you call? ... You do not have to have all that in your head when coming up with the idea of the demonstration, but you better know how to find the information allowing you to follow through to make it happen.
With programming, syntax details are like the details above: not the first thing to think of, and maybe not things that you have memorized. What is important to break down a problem and plan a solution, is to know the basic capacities you have in programming. As you get into larger projects and have more experience, “basic capacities” will be bigger and bigger ideas. For now, as beginners, based on the sections of the book so far, it is important to know:
• You can get information from a user and return information via keyboard and screen.
• You can remember and recall and use information using variables.
• You can deal directly with various kinds of data: numbers and strings at this point.
• There are basic operations you can do with the data (arithmetic, concatenating string, converting between data types).
• At a slightly higher level, you might already have the idea of basic recurring patterns, like solving a straightforward problem with input-processing-output.
• You will see shortly that you have more tools: decision, repetition, more built-in ways to deal with data (like more string operations shortly), creating your own data types....
At slightly more detailed level, after thinking of overall plans:
• There are multiple kinds of number types. What is appropriate for your use?
• There are various ways of formatting and presenting data to output. What shall you use?
Finally, you actually need to translate specific instructions into C# (or whatever language). Of course if you remember the syntax, then this level of step is pretty automatic. Even if you do not remember, you have something very specific to look up! If you are keeping track of your sources of detailed information, this is hopefully only one further step.
Contrast this last-step translation with the earlier creative organizational process: If you do not have in your head an idea of the basic tools available, how are you going to plan? How are you going to even know how to start looking something up?
So far basic ideas for planning a solution has been discussed, and you can see that you do not need to think of everything at once or have everything equally prominent in your brain.
Also, when you are coding, you do not need to to have all the details of syntax in your head, even for the one instruction that you are dealing with at the moment. You want to have the main idea, and you want to get it written down, but once it is written down, you can make multiple passes, examining and modifying what you have. For example, Dr. Harrington does a lot of Python programming, where semicolons are not needed. He can get the main ideas down in C# without the required semicolons. He could wait for the compiler to stop him on every one that is missed, and maybe have the compiler misinterpret further parts, and give bogus error messages. More effective is having a list of things to concentrate on in later rounds of manual checking. For example, checking for semicolons: Scan the statements; look at the ends; add semicolons where missing. You can go through a large program very quickly and efficiently doing this and have one less thing to obsess about when first writing.
This list of things-to-check-separately should come from experience. Keep track of the errors you make. Some people even keep an error log. What errors keep occurring? Make entries in things-to-check-separately, so you will make scans checking for the specific things that you frequently slip up on.
This things-to-check-separately list, too, will evolve. Revise it occasionally. If Dr. Harrington does enough concentrated C#, maybe he will find that entering semicolons becomes automatic, and he can take the separate round of semicolon checking off his list....
What to do after you finish an exercise is important, too. The natural thing psychologically, particularly if you had a struggle, is to think, “Whew, outta here!!!!”
On something that came automatically or flowed smoothly, that is not a big deal - you will probably get it just as fast the next time. If you had a hard time and only eventually got to success, you may be doing yourself a disservice with “Whew, outta here!!!”
We have already mentioned how not everything is equally important, and some things are more important to keep in your head than others. The same idea applies to all the steps in solving a possibly long problem. Some parts were easy; some were hard; there may have been many steps. If all of that goes into your brain in one continuous stream of stuff that you remember at the same level, then you are going to leave important nuggets mixed in with an awful lot of unimportant and basically useless information, and have it all fade into oblivion, or be next to impossible to cycle through looking for the nuggets. Why do the problem anyway if you are just going to bury important information further down in your brain?
What is important? The most obvious thing you will need at a higher level of recall is what just messed you up, what you missed until doing this problem: After finishing the actual problem, actively follow up and ask yourself:
• What did I get in the end that I was missing initially? What was the connection I made?
• Does this example fit in to some larger idea/abstraction/generalization in a way that I did not see before?
• How am I going to look at this so I can make a similar connection in a similar (or maybe only partly similar) problem?
• Is there a kernel here that I can think of as a new tool in my bag of tricks?
Your answers to these questions are the most important things to take away from your recent hard work. The extra consideration puts them more in the “priority” part of your brain, so you can really learn from your effort. When you need the important ideas next, you do not need to play through all the details of the stuff you did to solve the exact earlier problem.
Keep coming back to this section and check up on your process: It is really important. | 2018-06-18 22:32:53 | {"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": 0.31965917348861694, "perplexity": 637.8339015648181}, "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/1529267861456.51/warc/CC-MAIN-20180618222556-20180619002556-00079.warc.gz"} |
https://quantumfrontiers.com/2012/10/02/redemption-part-ii/ | # Redemption: Part II
Last week, a journey began to find the solution to a problem I could not solve as a seventeen year-old boy. That problem became an obsession of mine during the last days of the International Math Olympiad of 1997, the days when I also met the first girl I ever kissed. At the time, I did not have the heart to tell the girl that I had traveled across the Atlantic to compete with the best and brightest and had come up short. I told her that I had solved the problem, but that the page with my answer had been lost. I told my parents the same thing and to everyone at school who would ask me why I did not return with a medal from the Math Olympics. The lie became so powerful that I did not look at that problem again until now. So, you may be wondering why a blog about Quantum Information Science at Caltech includes posts on problems from Math Olympiads. And why I would open the book on the page with that one problem after fifteen years…
When I was in grad school, I spent 4 hours listening to a student who was struggling with a math class because he was uncomfortable with very small numbers (micronumerophobia). Together, we addressed his anxiety and he went on to become one of the top students in the class. In other words, don’t count on me letting go until you get to your Eureka! moment, even if your biggest obstacle is of the I-am-not-smart-enough variety. With time, this feeling will pass and will be replaced by a vibrant clarity of mind. Now, before going any further, go back and read Redemption: Part I. You will see how natural numbers, which encode the universal language of mathematics, are like molecules that are made up of atomic elements (prime numbers) on the periodic table. You will also learn about relatively prime numbers, which correspond to molecules that share no common elements, like $H_2 O$ (water) and $NaCl$ (salt). Finally, you will meet the greatest common divisor, a number whose importance you have to witness in action to appreciate.
Redemption rebooted…
The challenge: Find all pairs of positive integers $a,b \ge 1$, such that $a^{b^2} = b^a.$
The biggest progress we made so far in solving this problem came from extracting in Part I the greatest common divisor (gcd) from the numbers $a,b$ in the above equation. The new equation became:
$d^{b^2-a} a_1^{b^2} = b_1^a,$
where $d=gcd(a,b)$ and the numbers $a_1 = a/d, \, b_1=b/d$ are relatively prime (i.e. strangers in the night). Looking at the above equation, we decided to break it into three cases (because if the power to which $d$ is raised is negative, the equation must be turned on its head to avoid equating apples to oranges – whole numbers on one side with fractions on the other side):
1. $b^2 > a,$
2. $b^2 = a,$ and
3. $b^2 < a.$
Case 1 was considered already in Part I and led to the equation: $b_1 = d^{db_1^2-1}$. Now we try different values for $d$ and see what we get. Starting with $d=1$, we have $b_1 = 1$ (since 1 raised to any power is always 1). And this is precisely the case which yields the solution we already knew: $a=1,b=1$. Or is it? Remember clarity of mind? Which case are we considering again? Case 1: $b^2 > a$. In other words, $1^2 > 1$ in this case, so we get a contradiction and $d$ cannot be 1. So we keep going: $d \ge 2$ yields $b_1 = d^{db_1^2-1} \ge 2^{2b_1^2-1},$ and squaring both sides gives: $b_1^2 \ge 2^{2(2b_1^2-1)}$, which becomes $(2b_1)^2 \ge 2^{(2b_1)^2}$ after a quick rearrangement (ask me how in the comments). And we have reached a very important milestone in your journey: The day you learn about mathematical induction.
Mathematical induction is so important, that I will not even give you the definition here. I will illustrate it by showing you that the inequality $(2b_1)^2 \ge 2^{(2b_1)^2}$ has no solutions in the natural numbers $b_1 \ge 1$. That’s right. In two lines, I will prove to you that no value of $b_1$ (and there are infinite of them to consider) satisfies both sides of this inequality. We will prove that the inequality should be reversed from $(2b_1)^2 \ge 2^{(2b_1)^2}$ to $(2b_1)^2 < 2^{(2b_1)^2}$:
1. Test the simplest case: $b_1=1$ yields $2^2 < 2^{(2^2)} = 16$. Check.
2. Set $s=(2b_1)^2$ and check that $s+1 < 2^{s+1}$ whenever $s < 2^s$: $s+1 < 2^s + 1 \le 2^s + 2^s = 2^{s+1}$, where I used $2^s \ge 1$, for all $s \ge 0$.
3. BAM!
What kind of sorcery is this?!? Well, let’s play it back in slow motion… First, we do a sanity check in Step 1, to make sure we are not about to prove the wrong thing! This is called the base case, which obviously refers to the foundation upon which we will build the rest of our argument. Next comes magic. And like all great magic tricks, the sense of wonder only increases once the truth is revealed. And the trick is simple: If you can show that your guess works for the number $s+1$, assuming that it works for the number $s$, then you are done. Because $s+2$ is to $s+1$, as $s+1$ is to $s$. Good stuff, right? Right. So, back to the one-liner above: $s < 2^s$ implies $s+1 < 2^s+1$ (just add one on both sides) and $2^s \ge 1$, for all positive exponents $s$ (check out this Khan Academy video on exponents). The rest follows from $2^s+1 \le 2^s+2^s = 2^1\cdot 2^s = 2^{1+s}$ (again, check out the video on exponent rules). That line of inequalities we just saw is the famous inductive step (up there with the dubstep) and the assumption we made (the guess) that $s < 2^s$ (for some fixed number $s$) is the equally famous inductive hypothesis (if you haven’t heard of it, it’s because we run in different circles).
But words are like leaves, and where they most abound, much fruit of sense beneath is rarely found.
So forget about inductive steps and hypotheses and remember only the feeling of satisfaction from realizing that in two lines, we showed that Case 1 is a dud. No numbers $a,b \ge 1$ can satisfy both $b^2 > a$ and $a^{b^2} = b^a$. And let’s be honest; Case 2 is looking pretty easy. If $b^2=a$, then the exponent of $d$ is 0 and the equation $d^{b^2-a} a_1^{b^2} = b_1^a$ reduces to $a_1^{b^2} = b_1^a$. But the numbers $a_1, b_1$ are relatively prime (and as we proved in Part I, so are their powers). Now, how can it be that two molecules with no common atomic elements can be equal? It can’t be! Unless $a_1 = b_1 = 1$, which yields $a=b=d$. Is this consistent with $b^2=a$? Yes, as long as $d^2=d$, which implies $d=1$ (since $d\ge 1$). Finally, we have a solution: $a=b=1$! The obvious solution we already knew about… But in the process, you learned about mathematical induction. And this is up there with learning how to add two numbers.
Life-changing stuff.
What about Case 3: $b^2 < a$? All you need to know, you have already been taught in these two parts. So, you’ve got to ask yourself one question: Do I feel lucky? Well, do ya?
## 5 thoughts on “Redemption: Part II”
1. Thanks for the incredibly ‘open’ post. Especially the part about lying to the girl and your peers — it takes guts to admit something like that.
As a reference for curious aspiring mathematicians reading this post, I originally learned about induction, the pigeon-hole principle, invariants and other problem solving tricks from Loren C. Larson’s “Problem Solving Through Problems.” It’s a fantastic book that I couldn’t recommend more highly.
2. I liked this post. After some work, I think I have a solution to $a^(b^n) =b^a$ for integer n. I will post it or reply by email if you wish. Jim Graber
• Dear James, please post your answer below! You may use $latex … for equations in tex. If you need a reference, Google “latex wordpress”. • Sorry to be so late in responding.$a^(b^n) = b^a$for n an integer, although it may not be necessary for n to be an integer. Solution is:$a=b^bb = n+1$Examples:$4^2=2^4(27^ (3^2)) = 27^9=3^27(256 ^(4^3)) = 256^64 = 4^256\$
etc.
3. Both sides of the equation equal (n+1)^((n+1)^(n+1)). | 2018-02-25 01:48:55 | {"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": 0, "img_math": 66, "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": 0.7245240211486816, "perplexity": 373.6247082545505}, "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-2018-09/segments/1518891816083.98/warc/CC-MAIN-20180225011315-20180225031315-00331.warc.gz"} |
http://connection.ebscohost.com/c/articles/74188056/harmonic-maps-ideal-fluid-flows | TITLE
# Harmonic Maps and Ideal Fluid Flows
AUTHOR(S)
Aleman, A.; Constantin, A.
PUB. DATE
May 2012
SOURCE
Archive for Rational Mechanics & Analysis;May2012, Vol. 204 Issue 2, p479
SOURCE TYPE
DOC. TYPE
Article
ABSTRACT
Using harmonic maps we provide an approach towards obtaining explicit solutions to the incompressible two-dimensional Euler equations. More precisely, the problem of finding all solutions which in Lagrangian variables (describing the particle paths of the flow) present a labelling by harmonic functions is reduced to solving an explicit nonlinear differential system in $${\mathbb {C^n}}$$ with n = 3 or n = 4. While the general solution is not available in explicit form, structural properties of the system permit us to identify several families of explicit solutions.
ACCESSION #
74188056
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Share | 2020-09-29 18:34:35 | {"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": 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": 0.6395017504692078, "perplexity": 1254.561089199244}, "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/1600400202418.22/warc/CC-MAIN-20200929154729-20200929184729-00503.warc.gz"} |
https://www.physicsforums.com/threads/some-maths-problem.87783/ | # Some maths problem
1. Sep 6, 2005
### IB
Suppose I have sqrt (a^2 - a [delta] d). What do I do? Do I do like this: sqrt (a^2 - a [delta] d) = a - sqrt(a [delta] d)? Thanks.
PS: One more thing. How to write mathematics with latex?
2. Sep 6, 2005
### Learning Curve
What is sqrt (a^2 - a [delta] d) equal to? or are you asking to simplify it?
3. Sep 6, 2005
### IB
Yeah. I was just wondering whether it can still be simplified.
4. Sep 6, 2005
### Hurkyl
Staff Emeritus
Depending on what you want to do, a differential approximation, or maybe a Taylor sum, might be useful.
But as for algebraic manipulation, what you did is wrong. roots and exponents (usually) distribute over multiplication (and division), not addition (and subtraction).
5. Sep 6, 2005
### IB
Could you give an example to demonstrate that? And how can I correct my wrong algebraic manipulation? Thanks.
6. Sep 6, 2005
7. Sep 7, 2005
### VietDao29
Say a = 5, and delta d = 9 / 5.
So:
$$\sqrt{a ^ 2 - a \Delta d} = \sqrt{5 ^ 2 - 5 \times \frac{9}{5}} = \sqrt{25 - 9} = \sqrt{16} = 4$$
And:
$$a - \sqrt{a \Delta d} = 5 - \sqrt{5 \times \frac{9}{5}} = 5 - \sqrt{9} = 5 - 3 = 2$$
And 4 is not 2.
Viet Dao,
8. Sep 7, 2005
### IB
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# Dissecting transition cells from single-cell transcriptome data through multiscale stochastic dynamics
## Abstract
Advances in single-cell technologies allow scrutinizing of heterogeneous cell states, however, detecting cell-state transitions from snap-shot single-cell transcriptome data remains challenging. To investigate cells with transient properties or mixed identities, we present MuTrans, a method based on multiscale reduction technique to identify the underlying stochastic dynamics that prescribes cell-fate transitions. By iteratively unifying transition dynamics across multiple scales, MuTrans constructs the cell-fate dynamical manifold that depicts progression of cell-state transitions, and distinguishes stable and transition cells. In addition, MuTrans quantifies the likelihood of all possible transition trajectories between cell states using coarse-grained transition path theory. Downstream analysis identifies distinct genes that mark the transient states or drive the transitions. The method is consistent with the well-established Langevin equation and transition rate theory. Applying MuTrans to datasets collected from five different single-cell experimental platforms, we show its capability and scalability to robustly unravel complex cell fate dynamics induced by transition cells in systems such as tumor EMT, iPSC differentiation and blood cell differentiation. Overall, our method bridges data-driven and model-based approaches on cell-fate transitions at single-cell resolution.
## Introduction
Advances in single-cell transcriptome techniques allow us to inspect cell states and cell-state transitions at fine resolution1, and the notion of transition cells (aka. hybrid state, or intermediate state cells) starts to draw increasing attention2,3,4. Transition cells are characterized by their transient dynamics during cell-fate switch3, or their mixed identities from multiple cell states5, different from the well-defined stable cell states6,7 that usually express marker genes with distinct biological functions. Transition cells are conceived vital in many important biological processes, such as tissue development, blood cell generation, cancer metastasis, or drug resistance8.
Despite the rapid algorithmic progress in single-cell data analysis9, it remains challenging to probe transition cells accurately and robustly from single-cell transcriptome datasets. Often, the transition cells are rare and dynamic, and herein difficult to be captured by static dimension-reduction methods10. High-accuracy clustering methods (e.g., SC311 and SIMLR12) tend to enforce distinct cell states, placing transient cells into different clusters, therefore only applicable to the cases of sharp cell-state transition (Fig. 1a). While popular pseudo-time ordering methods13, such as DPT7, Slingshot14 and Monocle15, presumes either discrete (Fig. 1a) or continuous cell-state transition (Fig. 1a), quantitative discrimination between stable and transition cells is lacking7. Recently, soft-clustering techniques provides a way to estimate the level of mixture of multiple cell states16, however, the linear or static models embedded in such approach make it difficult to capture dynamical properties of cells.
Dynamic modeling provides a natural way to characterize transition cells3, allowing multiscale description of cell-fate transition (Fig. 1a and Supplementary Fig. 1). Such models analogize cells undergoing transition to particles confined in multiple potential wells with randomness17,18, for which the transient states correspond to saddle points and the stable cell states correspond to attractors19,20,21 of the underlying dynamical system (Fig. 1b). In such description, the stochastic gene dynamics at individual cell scale can induce cell-state switch at macroscopic cell cluster or phenotype scale, and the transition cells form bridges between different attractors (Fig. 1c). Despite widely use of dynamical systems concepts to illustrate cell-fate decision4, direct inference via dynamical models for transitions from single-cell transcriptome data is lacking.
Here we employ noise-perturbed dynamical systems22 with a multiscale approach on cell-fate conversion23 to analyze single-cell transcriptome data. By characterizing stable cells in attractor basins and placing the transition cells along transition paths connecting attractors through saddle points, our multiscale method for transient cells (MuTrans) prescribes a stochastic dynamical system for a given dataset (Fig. 1b). Using the single-cell expression matrix as input, through iteratively constructing and integrating cellular random walks across three scales (Fig. 1d and Supplementary Fig. 2), MuTrans finds most probable transition paths for cell transitions in a reconstructed cell-fate dynamical manifold (Fig. 1e, Methods). Such manifold, similar to the classical Waddington landscape24 often used to highlight transitions, provides an intuitive visualization of cell dynamics compared to commonly adopted low-dimension geometrical manifold. In the dynamical manifold, the barrier height naturally quantifies the likelihood of cell-fate switch, and the Transition Cell Score (TCS) and transition entropy allows us to distinguish between attractors and transition cells (Fig. 1e, Methods). We then illustrate the complex cell transition trajectories on dynamical manifold using the dominant transition paths obtained for the coarse-grained dynamics. With such quantification, we are able to identify critical genes that are transition drivers (TD genes), mark the intermediate/hybrid states (IH genes) or meta-stable cells (MS genes) (Fig. 1e and Supplementary Fig. 3). To speed up calculations for datasets consisting of large number of cells25,26, MuTrans provides an additional (and optional) aggregation module in pre-processing. This module aggregates cells into many small groups that share similar dynamical properties, thus MuTrans can take the transition probabilities among these coarse-grained cells as the input, instead of the random walk on original cells, in order to reduce the computational cost (Method and Supplementary Note 2).
We demonstrate the effectiveness and robustness of MuTrans in multiple single-cell transcriptome datasets, including simulation datasets and sequencing data generated by five different experimental platforms. Comparisons with existing single-cell lineage inference tools demonstrate the capability and scalability of MuTrans in probing complex, sometimes subtle, cell-fate transition dynamics. We also perform mathematical analysis to show consistency of MuTrans with the over-damped Langevin dynamics27 - a popular model for state transitions in physical or biochemical systems22.
## Results
### Overview of MuTrans workflow and theoretical foundations
MuTrans depicts cells and their transitions in each single-cell transcriptome dataset as a multiscale dynamical system (Fig. 1a–c). The dynamics of cell fates can be described by the stochastic differential equations (SDEs) as
$${{{{{\rm{d}}}}}}{{{{{{\bf{X}}}}}}}_{{{{{{\boldsymbol{t}}}}}}}={{{{{\bf{f}}}}}}\left({{{{{{\bf{X}}}}}}}_{{{{{{\boldsymbol{t}}}}}}}\right){dt}+{{{{{\boldsymbol{\sigma }}}}}}\left({{{{{{\bf{X}}}}}}}_{{{{{{\boldsymbol{t}}}}}}}\right)d{{{{{{\bf{W}}}}}}}_{{{{{{\boldsymbol{t}}}}}}},$$
(1)
where $${{{{{{\bf{X}}}}}}}_{t}\in {{\mathbb{R}}}^{p}$$ denotes the cell’s gene expression state at time t, f(x) denotes the nonlinear gene regulations, σ(x) denotes the noise strength due to both biochemical reactions and environmental fluctuations, and Wt is the standard Brownian motion representing the noise. Usually, f(x) may have multiple zeros, corresponding to the multi-stable attractors of the dynamical system. At long time scale in coarse-grained state space, the Eq. (1) can be reduced to capture the transitions among different attractors28.
To ensure the description is well-posed for single-cell transcriptome data, regularizations or additional prior knowledge (e.g., cell growth rate) needs to be enforced or provided. Similar to previous studies29, here we make two important assumptions: (a) The multi-stable drift term f(x) can be well-approximated by the gradient of a potential field with multiple wells, and (b) the single-cell data is sampled from nearly stationary distribution (or a system is fully ergodic without rapidly growing populations). This indicates that the data is sampled from a stationary system, a reasonable assumption if no prior knowledge is provided29. From the decomposition analysis of differential equations30, the potential-field assumption (a) is valid when the non-gradient term of drift f(x) is small in the large regions of state space, which holds in many biological systems with multi-stability31. Computationally, instead of fitting or solving the high-dimensional Eq. (1) directly, here we recover the dynamical structure of its solution using a multi-scale data-driven approach, as described below.
Taking the input as pre-processed single-cell gene expression matrix, MuTrans first learns the cellular random walk transition probability matrix (rwTPM) on the cell-cell scale through a Gaussian-like kernel (Fig. 1d and Methods), which yields the continuous limit as over-damped Langevin equation (Methods and Supplementary Note 1). Enforced by Gaussian-like kernel, the constructed rwTPM is in detailed-balance, consistent with the assumption (a). Next, the method performs coarse-graining on the cell–cell scale rwTPM to learn the dynamics on the cluster-cluster scale, and acquires attractor basins and their mutual conversion probabilities simultaneously (Fig. 1d and Methods). Theoretically, this step is asymptotically consistent with the Kramers’ law of reaction rate for over-damped Langevin equations if assumption (b) holds (Methods and Supplementary Note 1). Finally, we specify the relative position of each cell in the attractor basins with the cell-cluster resolution view of Langevin dynamics, which is constructed via optimizing a cell-cluster membership matrix (Fig. 1d and Methods).
To robustly depict the lineage relationships, we use the transition path theory to quantify the likelihood of all possible transition trajectories between cell states, based on the coarse-grained transition probabilities (Fig. 1e, Methods and Supplementary Note 2).
Combining the optimized cell-cluster membership matrix, MuTrans fits a dynamical manifold using a mixture distribution to make stable cells reside in the attractor basins while assign transition cells along the transition paths connecting different basins (Fig. 1e and Methods), which is based on the Gaussian mixture approximation toward the steady-state distribution of the Fokker-Planck equation associated with the over-damped Langevin dynamics (Methods and Supplementary Note 2).
For each cell-state transition, we can calculate a transition cell score (TCS) ranging between one and zero to quantitatively distinguish attractors and transition cells (Fig. 1e and Methods). Finally, we systematically classify three types of genes (MS, IH and TD) during the transition whose expression dynamics differ between stable and transition cells (Fig. 1e and Methods). Specifically, the TD genes varies accordingly with the TCS within transition cells, and the IH genes co-express in both stable and transition cells, while MS genes express uniquely near the attractors.
To deal with the large-scale datasets, in addition to common strategies such as sub-sampling cells, we provide an option to speed up calculation by introducing a pre-processing aggregation module DECLARE (dynamics-preserving cells aggregation). This module assigns the original individual cells into many (e.g., hundreds or thousands) microscopic stable states and computes the transition probabilities among them, and thus it can be used as an input to MuTrans instead of the cell-cell rwTPM (Methods and Supplementary Note 2). Both theoretical and numerical analysis suggest that, compared to the common strategy of averaging of gene expression profiles of a small group of cells, DECLARE better preserves the structure of dynamical landscape with a good approximation to the transition probabilities calculated without using DECLARE (Methods and Supplementary Note 2).
### Evaluation of MuTrans using simulation datasets
To test accuracy and robustness of MuTrans, we evaluated its performance on simulation datasets generated from known dynamical systems. First we simulated the stochastic state-transition process using a bifurcation model in the regime of intermediate noise level32. The gene expression of each cell was simulated with over-damped Langevin equation driven by an extrinsic signal and noise (Supplementary Note 3). In certain parameter range, the model consists of two stable states and one saddle state (Fig. 2a). Noise in gene expression induced the switch prior to the bifurcation point, resulting in a thin layer of transition cells (Fig. 2a). Applying MuTrans to the known transition cells and stable cells in the model, we found the computed transition cell score (TCS) captured the underlying saddle-node bifurcation structure (Fig. 2a). For cells fluctuating around the two stable branches, the TCS approaches one or zero respectively, indicating the meta-stability of cell states. The transition cells that pass the saddle point region in the trajectory yields a continuum of TCS between zero and one, with scores consistent with the relative positions of cells along the trajectory (Fig. 2a).
In addition to the uni-directional transition simulation dataset, we next consider back-and-forth stochastic state-switching, a common scenario in multi-stable systems. We constructed a triple-well potential field and simulated the dynamics with over-damped Langevin equations (Supplementary Note 3). Three saddle points lie between the attractor basins in its potential field, with another maximum point (order-2 saddle) at the origin (Fig. 2b). Time series indicate that the state-hopping among three attractor basins can be frequent when large enough noise amplitudes are used (Fig. 2b, c). Using the simulation trajectories as snap-shot data points for inputs, MuTrans correctly infers three attractor basins from the dataset (Fig. 2d and Supplementary Fig. 4). The coarse-grained transition probabilities (Fig. 2d) suggest that the cells most likely remain in their original attractors other than transitioning into other attractors.
Consistent with previous studies of similar systems33, as noise amplitude increases, the transitions between different attractors become more frequent as indicated by the larger coarse-grained probabilities. The direct transitions between attractors 1 and 3 are dominant compared with the state-switch mediated by attractor 2 (Supplementary Fig. 4). The calculated transition entropy and attractor membership functions accurately highlight the transition cells moving across various saddle points (Fig. 2e). Interestingly, the cells near the global maximum point have larger transition entropies than those near the first-order saddle points, indicating more mixed or hybrid identities.
### Revealing the cell-state transitions during EMT of squamous cell carcinoma
We then applied MuTrans to a single-cell RNA sequencing dataset34 of Squamous Cell Carcinoma (SCC) epithelial-to-mesenchymal transition (EMT) generated by Smart-Seq2 platform (Fig. 3 and Supplementary Fig. 5). Five attractors are detected by MuTrans (see Supplementary Fig. 5b for the corresponding EPI analysis), including one epithelial state (E), two mesenchymal states (M1 and M2) and two intermediate cell states (ICS). The cell states are annotated by comparing marker genes expression with those in the original study (Fig. 3a–d and Supplementary Fig. 5). Streams of transition cells moving between various attractor basins are observed in the constructed dynamical manifold (Fig. 3e).
The transition path analysis shows the major portion of the transition flux (more than 50%) from E state to M states goes through one of the ICS (Fig. 3f, g), indicating the significant role of ICS to mediate state-transitions in EMT35,36. Interestingly, there are also transitions within the two mesenchymal attractors, an observation consistent with the concept of quasi-mesenchymal states reported in the original study, suggesting that the M attractors here may also serve as intermediate nodes in transitions.
The transition gene analysis along the path E-ICS2-M2 characterizes the transition cells in their gene expression dynamics (Fig. 3h, i). Compared with MS genes that are highly expressed in stable cells, the IH genes may express in both transition cells and stable cells. The expressions of TD genes vary gradually within the transition cells (Fig. 3h, i).
### Scrutinizing bifurcation dynamics during iPSC induction
We next used MuTrans to investigate cell fate bifurcations (Fig. 4a) in a single-cell dataset for induced pluripotent stem cells (iPSCs) toward cardiomyocytes37. In the learned cellular random walk across different scales, the rwTPM on cell-cluster scale recovers finer resolution of rwTPM on the cell-cell scale than the cluster-cluster scale (Fig. 4b). MuTrans identified nine attractor basins under this resolution (Fig. 4c and Supplementary Fig. 6), and the constructed tree (Supplementary Fig. 6) reveals a lineage with bifurcation into mesodermal (M) or endodermal (En) cell fates. Two attractor basins, locating before the bifurcation of primitive streak (PS) into differentiated mesodermal (M) or endodermal (En) cell fates, are denoted as Pre-M and Pre-En states (Fig. 4d and Supplementary Fig. 7). On the inferred dynamical manifold (Fig. 4e–g), the cells make transitions between two states, suggesting possible dynamic conversion between the two types of precursor cells that seem to be very plastic. In comparison, the transition between mature En and M states are rare, indicating the stability of En and M cells. Along the differentiation trajectory from PS to Pre-M, the coarse-grained transition probability, quantified by the heights of barrier, shows a stronger transition capability from PS to Pre-M than from Pre-M to PS (Fig. 4c). In addition, the transition from Pre-M to M was found to be sharper than the one from PS to Pre-M. The transitions from PS to Pre-En and from Pre-En to En exhibit similar behavior. This analysis suggests that the initial cell-fate bifurcation at PS state (mostly on day 2–2.5, Fig. S6) is not terminal. This is consistent with the transition path analysis (Fig. 4e), showing that prior to the final commitment into M fate, some cells in PS take a detour by passing through the pre-En attractor basins first. The trend of transition entropy defined by MuTrans is found to be consistent with the critical transition index defined in original publication37 for bifurcations. Indeed, the MuTrans transition entropy of cells first increases toward the bifurcation point from day 1 to 2.5, and then decreases as the final cell-fates are committed and established at day 3 (Fig. 4f, S6).
Downstream analysis on gene expression profiles indicates three transition stages from Pre-M to M (Fig. 4h). The initial stage was characterized by downregulation of meta-stable (MS) genes from the Pre-M state markers (enriched in the pathways of endodermal development) and upregulation of intermediate-hybrid (IH) genes (enriched in pathways of MAPK cascade and metabolic process) from the M state markers (Fig. 4i and Supplementary Table 4). This process by first losing En identity enables a conversion of Pre-M stable cells toward the transition cells. The second stage of the transition marked by the gradual down-regulation of TD genes mainly involves negative regulation of cardiac muscle cell differentiation and cardiac muscle tissue development (Fig. 4i and Supplementary Table 4). The final stage completes the transition process with the down-regulation of Pre-M state IH genes, along with up-regulation of MS genes (enriched in the cardiac muscle cell myoblast differentiation and outflow tract morphogenesis process) in the M state (Fig. 4i and Supplementary Table 4), making transition cells to finally convert into the mesodermal cells and establish the stable cell fate. The ordering of cells based on TCS has an overall increasing trend from Day 2 to Day 3 via the time point of Day 2.5 within the transition cells, corresponding to the noticed three-stage transition (Supplementary Fig. 8). Together, the transition cells locating near the saddle points connecting Pre-M (or Pre-En) and M (or En) reflect the temporal orderings of cell-fate conversion, which are well characterized by TD and IH genes in a system consisting of one pitchfork bifurcation.
### MuTrans reveals complex lineage dynamics in blood cell differentiation
The hematopoiesis has been conceived as a hierarchy of discrete binary state-transitions, while increasing evidence alternatively supports a continuous and heterogeneous view of such process38. To investigate the complex dynamics in blood differentiation where transition cells likely play key roles, we applied MuTrans to different single-cell datasets with different sequencing depths and sample sizes.
We first analyzed the single-cell RNA data during myelopoiesis sequenced with Fluidigm C1 platform39. The number of attractors and cell label annotations are selected to recover the label resolutions in original publication. Notably MuTrans highlights the hub states—multi-lineage cells, which are capable of becoming three types of blood cells through a shallow basin resided in the highest terrain of the entire dynamical manifold (Fig. 5a and Supplementary Figs. 910). The low barriers between the multi-lineage basin and the downstream basins (granulocytic or monocytic states) suggest probable transitions from the multi-lineage state, consistent with the observed transition cells across the saddle point. Interestingly, the transition cells during Multi-lin to Gran conversion were previously identified as the multi-lineage cells in ICGS clustering39 (Supplementary Fig. 10). Similarly, during the megakaryocytic cell differentiation, while the transition cells consist of both HSPC1 and Meg types in our analysis, they were previously identified as the hematopoietic progenitor cells by the ICGS criterion (Supplementary Fig. 10). Such discrepancy could be explained by the gene expression dynamics in gradual transition of cell states. For example, during transition from multi-lineage cells to granulocytic cells (Fig. 5c), we observed the typical expression pattern of TD, MS and IH genes as conceptualized in Fig. 1e. Despite the similarity between the transition cells and their departing multi-lin state as manifested in the co-expression of down-regulated IH genes (Fig. 5c, yellow lines), we also detected the up-regulated IH genes (Fig. 5c, yellow lines), suggesting the resemblance of transition cells with their targeting gran cell state (Supplementary Table 5). We observed a similar gene expression pattern in the transition from HSPC to Meg state (Supplementary Fig. 12 and Supplementary Table 6). For this dataset, MuTrans is able to capture the established attractor cell states, in addition to finding transition cells that were classified in some stable states by a previous study39.
Focusing on the dataset of cell-fate bias toward lymphoid lineage, MuTrans resolves the complex lineage dynamics underlying single-cell RNA data of mouse hematopoietic progenitors differentiation sequenced from Cel-Seq2 platform40. Consistent with the major findings of FateID algorithm, the constructed dynamical manifold reveals that lymphoid progenitor (LP) cells (red balls) give rise to both B cells (pink balls) and plasmacytoid dendritic cells (pDCs) (Fig. 5b and Supplementary Fig. 13). The inferred dynamical manifold also suggests that certain transition cells in the attractors of pDCs originate directly from multi-potent progenitor (MPP) cells (yellow balls, Supplementary Fig. 13). MuTrans resolves the details in B cell differentiation, capturing the transition cells from Pro-B toward Pre-B basins (Supplementary Fig. 13 and Supplementary Table 7). Downstream analysis suggested the transition cell features by the co-expressed IH genes (yellow lines, Fig. 5d) and the dynamically expressed TD genes (green lines, Fig. 5d). Overall, MuTrans can provide a global cell-fate transition picture with marked transition cells in this dataset of highly complex lineages, in addition to the local transition routes inferred by FateID40.
### Application to large-scale datasets with complex trajectory
To test the scalability of MuTrans, we studied on the single-cell hematopoietic differentiation data in human bone marrow generated by 10x Chromium platform41 (Fig. 6a). To make the comparison, we applied MuTrans to both the complete (original) data, and the one after using the pre-processing module DECLARE. We found DECLARE could reduce the calculation time by one magnitude for this dataset.
For both cases MuTrans identified the expected bifurcations from hematopoietic stem cells (HSC) into the monocytic precursors and erythroid cells, as well as the differentiation from precursor cells into monocytic and dendritic cells. The constructed dynamical manifold (Fig. 6b, c and Supplementary Fig. 14) shows a continuous stream of transition cells among different basins (such as those moving between dendritic and monocytic potential wells) suggesting the hematopoietic differentiation may be a continuous process. The transition trajectories obtained with the large-scale pre-processing step are consistent with the complete dataset analysis (Fig. 6d, e). This indicates the major transition trajectories toward dendritic cell fate not only consist of the path mediated by monocytic precursor states but also include a considerable flux of transition cells from differentiated monocytic cells. Interestingly, the existence of both stable states and transition cells reconciles a previously noted discrepancy41 caused by treating the underlying cellular transition dynamics as either a purely continuous processing (e.g., using Palantir) or a discrete process (using other clustering-based lineage inference methods such as Slingshot14 and PAGA42).
Next, we analyzed another dataset containing over 15,000 cells collected during blood emergence in mouse gastrulation43 (Fig. 7a). Consistent with the PAGA42 low-dimensional embedding of the data (Fig. 7b), the constructed dynamical manifold (Fig. 7c) and derived Maximum Probability Flow Tree (MPFT) suggest three major transition branches from haemato-endothelial (Haem) cells into endothelial cells (EC), mesoderm cells (Mes) or erythroid cells (Ery). Specifically, the transition path analysis indicates that the endothelial cells and erythroid cells are originated through discrete trajectories from haemogenic endothelium (Fig. 7e), and such trajectories are mediated by the intermediate state of blood progenitor (BP) cells (Fig. 7f). These results are consistent with the experimental findings on endothelial and erythroid cells43.
### Comparison and consistency with other methods
MuTrans is designed specifically to identify transition cells, with its theory rooted in multi-scale dynamical systems and allowing natural visualization and quantification of cell-state transitions. To compare with other methods which may provide information on transitions, we performed further analysis with pseudo-time ordering and cell-fate bias probability methods on their capability of detecting transition cells, using existing methods, such as PAGA, FateID and VarID (Supplementary Note 4).
In iPSC data, we found that MuTrans, PAGA and VarID are consistent in recovering the bifurcation dynamics toward En and M states (Supplementary Fig. 15). While the projected lineage tree of StemID2 shows transition cells between precursor and mature En/M states (Supplementary Fig. 15), the reconstructed spanning tree does not reveal the overall bifurcation structure.
For the myelopoiesis dataset, we found that both MuTrans and VarID recover the bifurcations toward granulocytic and monocytic states (Supplementary Fig. 16). Consistent with MuTrans, FateID also captures the differentiation paths toward monocytic states (Supplementary Fig. 16).
Close inspection into the transition from precursors to mature En/M states in iPSC dataset suggests that based on existing approaches (such as tracking the changes along pseudotime or fate bias probability) could not distinguish the transition cells from stable cells as accurately and reliably as MuTrans. Both Monocle3 and DPT have a sharp increase in the pseudotime during the transitions (Supplementary Fig. 17), therefore lacking resolution in probing the transition cells linking multiple attractors. Fate ID suggests a gradual change of En/M fate probability in precursor cells (Supplementary Fig. 17), not discriminating the transition cells within Pre-En and Pre-M states. Such problem was also observed when using Palantir, which depicts the entire cell-state transition as a highly continuous and gradual process (Supplementary Fig. 17).
## Discussion
Overall, MuTrans provides a unified approach to inspect cellular dynamics and to identify transition cells directly from single-cell transcriptome data across multiple scales. Central to the method is an underlying stochastic dynamical system that naturally connects (1) attractor basins with stable cell states, (2) saddle points with transient states, and (3) most probable paths with cell lineages. Instead of the widely used low-dimensional geometrical manifold approximation for the high-dimensional single-cell data, our method constructs a cell-fate dynamical manifold to visualize dynamics of cells development, allowing direct characterization of transition cells that move across barriers amid different attractor basins. Adopting the transition path theory to the multiscale dynamical system, we quantify the relative likelihoods of various transition trajectories that connect a chosen root state and the target states. In addition, we provide a quantitative methodology to detect critical genes that drive transitions or mark stable cells.
In this study a key theoretical assumption for modeling cell-state transition is a barrier-crossing picture in multi-stable dynamical systems, a concept which has been adopted for describing cell developments through dynamical system language3,44,45. Indeed, the notions of barriers, saddles and potential landscape underlying the actual biological process are the emergent properties of the complex interactions, such as gene expression regulation and signal transduction during a developmental process28. The driving force that overcomes the barrier and induces the transition may arise from both the extrinsic environment and the fluctuations within the cells46. Multi-scale reductions used by MuTrans naturally capture the transition cells, allowing inference of the corresponding transition processes.
Pseudo-time ordering and low-dimensional trajectory embedding may serve as intuitive tools to trace the progression of cell fates by comparing similarity of gene expression among cells. Such approaches often adopt the deterministic point of view and rely on the low-dimensional projection of datasets, lacking theoretical insights to the underlying dynamical processes of cell-state transitions. In contrast, MuTrans is based on multi-stable dynamical system approach in characterizing cell-state transitions. While cells reside and fluctuate within attractor basins for majority of time, it is the temporal ordering of transition cells, rather than stable cells, reflect the actual process of cell transitions (Fig. 1c and Supplementary Fig. 17).
Methods such as Palantir41, Population Balance Analysis (PBA)29 and Topographer47 also treat cell-fate transition as Markov random walk process. These methods depict the dynamics at the individual cell level, then compute pseudo-time ordering based on the first passage time or absorbing probabilities of the Markov Chain. In comparison, MuTrans can dissect the intrinsic multiscale features of the system and derive the coarse-grained dynamics, distinguish between stable and transition cells quantitatively, and characterize multiple and complex routes of transition paths.
Several other methods2,48 define the transition probabilities between clusters based on entropy difference or summing up the cell-cell transition probabilities. Here the coarse-grained transition probability in MuTrans is an emergent quantity derived from multi-scale reduction. The transition probability is shown to be consistent with Kramers’ reaction rate theory for over-damped Langevin dynamics if steady-state assumption and detailed-balanced condition are satisfied (Methods and Supplementary Note 4).
To describe the smooth state transitions, some methods49,50 adopt the soft-clustering strategy based on the soft K-means or factor decomposition for gene expression matrix. In comparison, the soft cell assignment of MuTrans is obtained from multiscale learning of cell-cluster rwTPM, which can be more robust against technical noise than using gene expression matrix directly for clustering7. Such robustness is critical to detecting transition cells in datasets with lower sequencing depth, such as 10X data. Beyond interpreting the soft membership function as the indicator of cell locations in attractor basins, it remains an interesting problem to derive its continuum limit in the embedded over-damped Langevin dynamical systems.
To deal with the emerging large-scale scRNA-seq datasets, MuTrans introduces a pre-processing method (DECLARE) to aggregate the cells and speed up computation. The aggregation method uses the coarse-grain approach consistent with MuTrans, and it is different from other methods often used for large scRNA-seq datasets, such as down-sampling convolution51 or kNN partition52 that is based on the averaging or summation of cells with similar gene expression profiles. As a result, DECLARE can be naturally integrated with dynamical manifold construction and transition trajectory inference.
The stochastic transitions among attractors considered by MuTrans can be further incorporated with deterministic processes to better understand the cell-fate decision53. Despite that the stochastic switching among cell states might be rare in some cases, the local fluctuation of microscopic cell states in gene expression can be prevalent in the microscopic dynamics, therefore the cell-cell scale random walk assumption in MuTrans still holds as a natural assumption. In theory, the stochastic transition model is consistent with the uni-directional transition process if the transition probabilities in one direction are dominant, or when the noise amplitude of system is relatively small.
The theoretical assumptions on equilibrium and steady-state systems made in MuTrans can be potentially mediated by our multiscale approach. For example, although the detailed balance may be violated at the microscopic scale described in Eq. (1) the estimated coarse-grained (mesoscopic) dynamics in MuTrans can be sufficient to recover the transitions at larger scale. However, non-stationary effects due to cell cycle or cell proliferation dynamics29 were not considered in current method. In addition, the number of cells in the datasets, in principle, needs to be sufficiently large in order to obtain high-resolution identification of transition cells. When the number of cells is relatively small, such as in the myelopoiesis dataset studied here, special care is needed to further confirm the analysis of transition cells. Besides, more effective ways in root cell states detection (e.g., through entropy methods54 or RNA velocity55,56,57) can further enhance the robustness of our approach.
In addition to infer complex cellular dynamics induced by transition cells from single-cell transcriptome data, MuTrans along with its computational or theoretical components can be used for development of other approaches for dissecting cell-fate transitions from both data-driven and model-based perspectives.
## Methods
MuTrans performs three major tasks in order to reveal the dynamics underneath single-cell transcriptome data (Fig. 1): 1) assigning each cell in the attractor basins of an underlining dynamical system, 2) quantifying the barrier heights across the attractor basins, and 3) identifying relative positions of the cells within each attractor. The first two tasks are executed simultaneously through the coarse-graining of multi-scale cellular random walks, an alternative approach to the traditional clustering of cells and inference of cell lineage. The third task is achieved by refining the coarse-grained dynamics via soft clustering, and serves as a critical procedure to identifying the transition cells during cell-fate conversion.
### Multi-scale analysis of the random-walk transition probability matrix (rwTPM)
We assume the underlying stochastic dynamics during cell-fate conversion be modeled by random walks among individual cells through the random-walk transition probability matrix (rwTPM). Dependent on the choices of either cell-level or cluster-level, the rwTPM can be constructed in different resolutions, exhibiting multi-scale property and leading the identification of transition cells from the stable cells.
In describing the method, we use the indices x, y, z to denote individual cells and i, j, k to represents the clusters (or cell states) for the simplicity of notations.
1. 1.
The rwTPM in the cell-cell resolution
The rwTPM p of cellular stochastic transition can be directly constructed from the gene expression matrix in cell-cell resolution, with the form
$$p\left(x,y\right)=\frac{w\left(x,y\right)}{d\left(x\right)},d\left(x\right)=\mathop{\sum}\limits_{z}w\left(x,z\right)$$
(2)
where the weight w(x,y) denotes the affinity of gene expression profile in cell x and y (Supplementary Note 2). Such microscopic random walk yields an equilibrium probability distribution $$\mu \left(x\right)=\frac{d\left(x\right)}{\mathop{\sum}\limits_{z}d(z)}$$, satisfying the detailed-balance condition μ(x)p(x,y) = μ(y)p(y,x). The rwTPM captures the cellular transition in the cell-cell resolution (Fig. 1d).
2. 2.
The rwTPM in the cluster-cluster resolution
The cellular transition rwTPM can be lifted in the cluster-cluster resolution by adopting a macroscopic perspective. For example, the cell-to-cell rwTPM can be generated from certain coarse-grained dynamics, by assigning each cell in different attractors $${{{{{\rm{S}}}}}}=\mathop{\bigcup }\limits_{k=1}^{K}{S}_{k}$$, and model the transitions as the Markov Chain among attractors with the transition probability matrix $${\hat{{{{{{\bf{P}}}}}}}=({\hat{P}}_{{ij}})}_{K\times K}$$. Here $${\hat{P}}_{{ij}}\,$$denote the probability that the cells reside in the attractor Si switch to the attractor Sj. The number of attractors K is a hyperparameter of algorithm selected by the user. We use the Eigen-Peak Index (EPI) to visualize the multiple eigen-gaps of cell-cell scale rwTPM (Supplementary Note 2). Different peaks in EPI correspond to the number of attractors in different resolutions. In practice, the choice of K can also be determined based on prior biological knowledge such as marker genes expression or known cell-type annotations.
Denote $$1_{S_k} (z)$$ as the indicator function of cluster Sk such that $$1_{S_k} (z)$$ = 1 for cell zSk and $$1_{S_k} (z)$$ = 0 otherwise. The cluster-cluster transition based on probability matrix $$\hat{{{{{{\bf{P}}}}}}}\,$$can naturally induce another rwTPM $$\hat{{{{{{\bf{p}}}}}}}$$ with the form
$$\hat{p}(x,y)=\mathop{\sum}\limits_{i,j}{1}_{{S}_{i}}(x){\hat{P}}_{ij}{1}_{{S}_{j}}(y)\frac{\mu (y)}{{\hat{\mu }}_{j}},$$
(3)
where $${\hat{\mu }}_{j}=\mathop{\sum}_{y}{1}_{{S}_{j}}\left(y\right)\mu \left(y\right)$$ is the stationary probability distribution of cluster Sj. Intuitively, the stochastic transition from cell xSi to ySj can be decomposed into a two-stage process: a cell switches cellular state from cluster Si to Sj with probability $${\hat{P}}_{{ij}}$$, and then becomes the cell y in cluster Sj according to its relative portion at equilibrium $$\frac{\mu \left(y\right)}{{\hat{\mu }}_{j}}$$. The rwTPM captures the cellular transition in the cluster-cluster resolution (Fig. 1d).
3. 3.
The rwTPM in the cell-cluster resolution
Because some cells, for example the transition cells, may not be characterized by their locations in one basin, we introduce a membership function $$\rho (x)=(\rho _{1}(x),\rho _{2}(x),\ldots ,\rho _{K}(x))^{T}$$ for each cell x to quantify its uncertainty in clustering. The element ρk(x) represents the probability that the cell x belongs to cluster $$S_{k}^{\ast}$$ with $$\sum _k\rho _k(x)=1$$. For the cell possessing mixed cluster identities, its membership function ρ(x) might have several significant positive components, suggesting its potential origin and destination during the transition process. In terms of dynamical system interpretation, the membership function captures the finite-noise effect in over-damped Langevin equation, which introduces the uncertainty of transition paths across saddle points58, revealing that cells near saddle points and stable points may exhibit different behaviors in the state-transition dynamics.
From the coarse-grained dynamics $$\left(\{S_{k}\}_{k=1}^{K},\{{\hat{P}}_{ij}\}_{i,j=1}^{K}\right)$$ and the measurement of cell identity uncertainty ρk(x) in the clusters, one can reinterpret the induced microscopic random walk $$\widetilde{{{{{{\bf{p}}}}}}}$$ in a cell-cluster resolution as
$$\tilde{p}(x,y)=\mathop{\sum}\limits_{i,j}{\rho }_{i}(x){\hat{P}}_{ij}{\rho }_{j}(y)\frac{\mu (y)}{{\tilde{\mu }}_{j}},{\tilde{\mu }}_{j}=\mathop{\sum}\limits_{x}{\rho }_{j}(x)\mu (x),$$
(4)
in parallel to Eq. (3) Now the transition from cell x to y is realized in all the possible channels from attractor basin Si to $$S$$j with the probability ρi(x)ρj(y). The underlying rationale is that the transition can be decomposed in a three-stage process: First we pick up cell starting in attractor basin with membership probability, then conduct the transition with coarse-grained probability between attractor basins, and finalize the process by picking the target cell with membership probability in the target attractor basin. Now the rwTPM captures cellular transition in the cell-cluster resolution (Fig. 1d).
4. 4.
Integrating the rwTPM at three levels
To integrate the rwTPM from different resolutions, we next optimize the rwTPM on cluster-cluster and cell-cluster level through approximating the original rwTPM in the cell-cell resolution. First, we seek an optimal coarse-grained reduction that minimizes the distance between $$\hat{{{{{{\bf{p}}}}}}}\big[{S}_{k},{\hat{P}}_{{ij}}\big]$$ and p by solving an optimization problem:
$${{{{{\mathrm{min}}}}}}_{{S}_{k},{\hat{P}}_{ij}}{{{{{\mathcal{J}}}}}}[{S}_{k},{\hat{P}}_{ij}]=\Vert \hat{{{{{{\bf{p}}}}}}}[{S}_{k},{\hat{{P}}}_{ij}] -{{{{{{\bf{p}}}}}}||}_{\mu}^{2},$$
(5)
where μ is the stationary distribution of original cell-cell random walk p, and $$\,{\Vert \Vert }_{\mu }$$ is the Hilbert-Schmidt norm59 for given transition probability matrix A, defined as $${\Vert {{{{{\bf{A}}}}}}\Vert }_{\mu }^{2}=\mathop{\sum}\limits_{x,{{{{{\rm{y}}}}}}}\frac{\mu (x)}{\mu (y)}A{(x,y)}^{2}$$. The optimization problem is solved via an iteration scheme for Sk and $${\hat{P}}_{{ij}}$$ respectively (Supplementary Note 2). The optimal coarse-grained approximation $$\Big({S}_{k}^{\ast },{\hat{P}}_{{ij}}^{\ast }\Big)$$ indicates the distinct clusters of cells and their mutual conversion probability. Provided with the starting state, we can infer the cell lineage from the Most Probable Path Tree (MPPT) approach or Maximum Probability Flow Tree (MPFT) approach (Supplementary Note 2).
Next, we optimize the membership ρk(x) such that the distance between the cell-cluster rwTPM $$\widetilde{{{{{{\bf{p}}}}}}}$$ and the original p is minimized, i.e.,
$${\min }_{{\rho }_{k}} {\mathcal E} [{\rho }_{k}]\,=\Vert\tilde{{{{{{\bf{p}}}}}}}[{\rho }_{{{{{{\rm{k}}}}}}}]-{{{{{{\bf{p}}}}}}}\Vert_{\mu }^{2}$$
(6)
$${{{{{\rm{s}}}}}}.{{{{{\rm{t}}}}}}.\, \mathop{\sum}\limits_{k}\,{\rho }_{k}(x)=1,\,{\rho }_{k}(x)\ge 0\;{{{{{\rm{for}}}}}}\;k=1,{{{{{\mathrm{.}}}}}}.,K\;{{{{{\rm{and}}}}}}\;x\in S$$
with the initial condition $${\rho }_{i}^{0}\left(x\right)={1}_{{S}_{i}^{\ast }}\left(x\right)$$, and $$\widetilde{{{\mathbf{p}}}}\left[{\rho }_{{{{{{\rm{k}}}}}}}\right]$$ is defined from Eq. (4) by plugging in the obtained $${\hat{P}}_{{ij}}^{\ast }$$. The optimization problem is solved by the quasi-Newton method (Supplementary Note 2). The obtained membership function $${\rho }^{\ast }\left(x\right)$$ specifies the relative position of the cells within each attractor basin and is optimal in the sense that it guarantees the closest approximation of cell-cluster level rwTPM toward the cell-cell level transition dynamics.
### Transition entropy
To quantify and compare the transition cells around different attractors in a global view, we define a transition entropy H(x) for each cell x based on the obtained membership function $$\rho ^\ast (x)$$,
$$H(x)\,=-\mathop{\sum }\limits_{k=1}^{K}{\rho }_{k}^{\ast }\left(x\right){{\log }}{\rho }_{k}^{\ast }(x).$$
(7)
According to the definition, a stable cell tends to have a relatively small entropy value close to zero, while a transition cell, which possesses multiple and more evenly distributed components in its membership function, tends to have a larger transition entropy. As a result, a large entropy value indicates a cell with highly mixing identity, a case for transition cells in bifurcating attractors. The increase of transition entropy value can be utilized as a way to mark cell-state bifurcations.
### Transition paths quantification and comparison
To quantify the cell development routes, we use the transition path theory based on coarse-grained dynamics $$\left(\{S_{k}\}_{k=1}^{K},\{{\hat{P}}_{ij}\}_{i,j=1}^{K}\right)$$ to compare the likelihood of all possible transition trajectories. Given the set of starting states A and the targeting state B, we calculate the effective current $${f}_{{ij}}^{+}$$ of transition paths passing through state Si to Sj based on the inferred attractor basins and conversion probabilities (Supplementary Note 2), and specify the capacity of given development route $${w}_{{dr}}=({S}_{{i}_{0}},{S}_{{i}_{1}},..,{S}_{{i}_{n}})$$ connecting sets A and B as $$c\left({w}_{{dr}}\right)=\mathop{{{\min }}}\limits_{0\le k\le n-1}{f}_{{i}_{k}{i}_{k+1}}^{+}$$. The likelihood of transition trajectory wdr is defined as the proportion of its capacity to the sum of all possible trajectory capacities. In the python package of MuTrans, we use the functions in PyEMMA60 for the computations.
### Pre-processing by DECLARE and scalability to large datasets
To reduce the computational cost for large datasets (for instance, greater than 10 K cells), we introduce a pre-processing module DECLARE (dynamics-preserving cell aggregation). The module first detects the hundreds/thousands of microscopic attractor states by clustering (e.g., using K-means or kNN partition) and then derive the coarse-grained transition probabilities among these microscopic attractor states. Based on such transition probabilities, we then follow the standard multiscale reduction procedure of MuTrans to find macroscopic attractor states, construct dynamical manifold, quantify the transition trajectories and highlight the transition states (Supplementary Note 2).
### Transition cells and genes analysis through transcendental
Based on the soft clustering results, MuTrans performs the Transcendental (transition cells and relevant analysis) procedure on each transition process to identify the transition cells from the stable cells and reveal the relevant marker genes.
For the given transition process from attractors $${S}_{i}^{\ast }$$ to$$\,{S}_{j}^{\ast }$$ along the transition path, we first selected the cells relevant to the transition, based on the membership function $${\rho }^{\ast }\left(x\right)$$ (Supplementary Note 2). Then for each relevant cell x, we define the transition cell score (TCS)
$${\tau }_{{ij}}\left(x\right)=\frac{{\rho }_{i}^{\ast }(x)}{{\rho }_{i}^{\ast }\left(x\right)+{\rho }_{j}^{\ast }(x)},$$
(8)
to measure the relative position of cell x in different clusters. Here the TCS τij takes the values near zero or one when a cell resides around the attractor in $${S}_{i}^{\ast }$$ or$$\,{S}_{j}^{\ast }$$ (i.e., the cells are stable), whereas yields the intermediate value between zero and one for the cell that possesses a hybrid or transient identity of two or more clusters. Next we arrange all the relevant cells in state $${S}_{i}^{\ast }$$ and$$\,{S}_{j}^{\ast }$$ according to τij in descending order, and the reordered τij indicates a sharp transition (Fig. 1a) or a smooth transition (Fig. 1a) from the value one to zero. For the smooth transition, there is a group of cells whose value of τij decreases gradually from one to zero (Fig. 1e). This group of cells in the transition layer are called the transition cells from state $${S}_{i}^{\ast }$$ to state $${S}_{j}^{\ast }$$, and their order reflects the details of the state-transition process. To quantify the transition steepness, we use logistic functions to model the transition and estimate the relative abundance of transition cells (Supplementary Note 2).
Differentially expressed genes analysis is usually applicable when the clusters are distinct and the state-transition is sharp (Fig. 1a). However, to characterize the dynamical and hybrid gene expression profiles in transition cells, merely comparing the average gene expression in different clusters is insufficient. Here we define three kinds of genes relevant to the state transition of cells: a) the transition-driver (TD) genes that vary accordingly with the transition dynamics, b) the intermediate-hybrid (IH) genes marking the hybrid features from multiple cell states that are expressed in the intermediate transition cells, and c) the meta-stable (MS) genes that represent cells in the stable states.
The expression of TD genes varies accordingly to the transition, revealing the driving mechanism of the cell-state conversion. To probe TD genes, we calculate the correlation between the gene expression values and τij in the ordered transition cells. The genes with larger correlation values (larger than a given threshold value) are identified as TD genes. The IH genes express eminently both in the transition cells and in the stable cells from one specific cluster, reflecting the hybrid state of the transition cells, while the MS genes express exclusively in the stable cells from certain cluster. To distinguish IH and MS genes from all the differentially expressed genes, we compare the gene expression values between the stable cells and the transition cells, respectively, within each cluster. The significantly up-regulated genes in the stable cells are defined as the MS genes, and the rest differentially expressed genes are identified as the IH genes that express simultaneously both in stable and transition cells (Supplementary Note 2). Here the selected genes only reflect the relative gene expression trends amid one specific cell-state transition process, without considering global comparisons between multiple cell states or transitions. Therefore, the MS genes, which distinguish the attractors and transition cells locally in the dynamical manifold, can be different from the conventional marker genes that are uniquely and strongly expressed in one cell state. Together with IH and TD genes, they provide useful information to identify genes that are driving the local transition.
### Constructing the cell-fate dynamical manifold
To better visualize the transition process and their connections with cell states, MuTrans introduces the dynamical manifold concept. The construction of the dynamical manifold consists of two steps: (1) locating the center positions of cell clusters (corresponding to the attractors) in low dimensional space, (2) assigning the position of each individual cells according to soft-clustering membership function.
The initial center-determination step starts with an appropriate two-dimensional representation, denoted as x2D for each cell x (Supplementary Note 2). Instead of directly utilizing x2D as the cell coordinate, we calculate the center Yk of each cluster $$\{S_{k}^{\ast }\}_{k=1}^{K}$$ by taking the average of x2D over cells within certain range of cluster membership function $${\rho }_{k}^{\ast }\left(x\right)$$. Having determined the position of attractors, we define a two-dimensional embedding $${{{{{\boldsymbol{\xi }}}}}}\left(x\right)$$ for each cell according to the membership function $${\rho }^{\ast }\left(x\right)$$, such that $${{{{{\boldsymbol{\xi }}}}}}\left(x\right)=\mathop{\sum}\limits_{k}{\rho }_{k}^{\ast }\left(x\right){{{{{{\bf{Y}}}}}}}_{k}\in {\mathbb{R}}^{2}.$$ For the cell possessing mixed identities of state $${S}_{i}^{\ast }$$ and$$\,{S}_{j}^{\ast }$$, its transition coordinate then lies in a value between Yi and Yj.
For Fokker-Planck equation of the over-damped Langevin equation, the expansion of steady-state solution near stable points (attractors) indeed yields a Gaussian-mixture distribution61. Motivated by this, to obtain the global dynamical manifold we fit a Gaussian mixture model with a mixture weight $${\hat{{{{{{\boldsymbol{\mu }}}}}}}}^{\ast }$$ to obtain the stationary distribution of coarse-grained dynamics. The probability distribution function of the mixture model becomes
$${{{{{\mathscr{p}}}}}}\left({{{{{\rm{z}}}}}}\right)=\mathop{\sum}\limits_{k}{\hat{\mu }}_{k}^{\ast }{{{{{\mathscr{N}}}}}}\left({z{{{{{\rm{;}}}}}}{{{{{\bf{Y}}}}}}}_{k},{{{{{{\boldsymbol{\Lambda }}}}}}}_{{{{{{\boldsymbol{k}}}}}}}\right),$$
(9)
where$$\,{{{{{\mathscr{N}}}}}}\left({{z;}{{{{{\bf{Y}}}}}}}_{{{{{{\boldsymbol{k}}}}}}},{{{{{{\boldsymbol{\Lambda }}}}}}}_{{{{{{\boldsymbol{k}}}}}}}\right)$$ is a two-dimension Gaussian probability distribution density function with mean Yk and covariance Λk. The landscape function of dynamical manifold is then naturally takes the form in two dimensions$$\,\varphi \left(z\right)=-{{{{{\rm{ln}}}}}}{{{{{\mathscr{p}}}}}}\left(z\right)$$. Specifically, the energy of individual cell x is calculated as$$\,\varphi \left({{{{{\boldsymbol{\xi }}}}}}\left(x\right)\right)$$. The constructed landscape function captures the multi-scale stochastic dynamics of cell-fate transition, by allowing typical cells that are distinctive to certain cell states positioned in the basin around corresponding attractors, while the transition cells laid along the connecting path between attractors across the saddle point. Moreover, the relative depth of the attractor basin reflects the stationary distribution of coarse-grained dynamics, depicting the relative stability of the cell states. The flatness of the attractor basin also reveals the abundance and distribution of transition cells, indicating the sharpness of cell fate switch. Theoretically, the constructed dynamical manifold approximates the energy landscape or quasi-potential30,44,45 of underlying stochastic dynamical system.
### Mathematical analysis of MuTrans
With the assumption that the single-cell data is collected from the probability distribution ν(x) with density of Boltzmann-Gibbs form, i.e., $$\nu \left(x\right)\propto {e}^{-\frac{U(x)}{\varepsilon }},$$ we can prove (Supplementary Note 1) that the microscopic random walk constructed by MuTrans can approximate the dynamics of over-damped Langevin Equation (OLE)
$${{{{{\rm{d}}}}}}{{{{{{\bf{X}}}}}}}_{{{{{{\boldsymbol{t}}}}}}}=-\nabla U\left({{{{{{\bf{X}}}}}}}_{{{{{{\boldsymbol{t}}}}}}}\right){dt}+\sqrt{2\varepsilon }d{{{{{{\bf{W}}}}}}}_{{{{{{\bf{t}}}}}}}$$
(10)
in the limiting scheme, and the coarse-graining of MuTrans $$({S}_{k},{\hat{P}}_{{ij}})$$ is equivalent to the model reduction of OLE by Kramers’ rate formula in the small noise regime, i.e., $${{{{{{\rm{k}}}}}}}_{{ij}}\propto {e}^{-\frac{\triangle U}{\varepsilon }}$$ as $${{{{{\rm{\varepsilon }}}}}}\to 0$$,where kij is the switch rate from attractor Si to Sj, and ∆U denotes the corresponding barrier height of transition - the energy difference between saddle point and the departing attractor.
Therefore, if the cell transition dynamics can be well-modelled by the OLE dynamics of Eq. (10) MuTrans is indeed the multi-scale model reduction via the data-driven approach. In addition, the dynamical manifold constructed by MuTrans can be viewed as the data realization of potential landscape44 for diffusion process in biochemical modelling, which incorporates the dynamical clues about the underlying stochastic system regarding the stationary distribution and transition barrier heights.
### Data simulation and analysis
The simulation data was generated by the Euler-Maruyama method to solve the overdamped Langevin equations, with the detailed models and parameters specified in Supplementary Note 3.
The single-cell datasets analyzed were from different systems and platforms, namely mouse cancer EMT data (Smart-Seq2), mouse myelopoiesis data (Fluidigm C1), mouse hematopoietic progenitors data (Cel-Seq2), human hematopoietic progenitors data (10X Chromium),blood differentiation data (10X Chromium) in mouse gastrulation and iPSC induction data (single-cell RT-qPCR), downloaded from sources provided in Data availability section below. The detailed analysis for each dataset was provided in Supplementary Note 3. The full scripts for reproducing data analysis in main text and Supplementary Information for all the datasets are uploaded at https://github.com/cliffzhou92/MuTrans-release/tree/main/Example, with the processed gene expression matrices that could be loaded directly in MuTrans analysis stored at https://github.com/cliffzhou92/MuTrans-release/tree/main/Data.
We compared MuTrans with existing lineage inference methods Monocle 362, Diffusion Pseudotime7, PAGA42, FateID40, RaceID 3 and StemID 240, VarID48, Palantir41 and PBA29, with detailed settings for each method provided in Supplementary Note 4.
### Reporting summary
Further information on research design is available in the Nature Research Reporting Summary linked to this article.
## Code availability
The Matlab implementation of MuTrans and affiliated Transcendental packages are available from GitHub (https://github.com/cliffzhou92/MuTrans-release). The Python package for MuTrans (pyMuTrans) compatible with Scanpy package63 is also available in the repository.
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## Acknowledgements
This project was supported by grants from the National Natural Science Foundation of China (11825102 and 11421101 to T.L.), National Institutes of Health grant U01AR073159 (Q.N.), National Science Foundation grants DMS1763272 (Q.N.) and MCB2028424 (Q.N.), and The Simons Foundation (594598 to Q.N.) of USA. T.L. is also partially supported by the Beijing Academy of Artificial Intelligence (BAAI). P.Z. also received the support from Study Abroad Program and Elite Program of Computational and Applied Mathematics for Ph.D. students of Peking University.
## Author information
Authors
### Contributions
Q.N., T.L., and P.Z. conceived the project; P.Z. and T.L. designed the algorithm and wrote the code; P.Z. and S.W. conducted the data analyses; P.Z. wrote the supplementary material; all the authors wrote and approved the manuscript. Q.N. and T.L. supervised the research.
### Corresponding authors
Correspondence to Tiejun Li or Qing Nie.
## Ethics declarations
### Competing interests
The authors declare no competing interests.
Peer review information Nature Communications thanks Sui Huang, Manu Setty and the other, anonymous, reviewer(s) for their contribution to the peer review of this work. Peer reviewer reports are available.
Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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Zhou, P., Wang, S., Li, T. et al. Dissecting transition cells from single-cell transcriptome data through multiscale stochastic dynamics. Nat Commun 12, 5609 (2021). https://doi.org/10.1038/s41467-021-25548-w | 2021-12-02 13:56:22 | {"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": 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": 0.5617066621780396, "perplexity": 4185.371994897069}, "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/1637964362219.5/warc/CC-MAIN-20211202114856-20211202144856-00414.warc.gz"} |
https://cs.stackexchange.com/questions/77924/reason-for-finding-partial-order-of-a-graph | # Reason for finding partial order of a graph
In a recent algorithms course we had to form a condensation graph and compute its reflexive-transitive closure to get a partial order. But it was never really explained why we would want to do that in a graph. I understand the gist of a condensation graph in that it highlights the strongly connected components, but what does the partial order give us that the original graph did not?
The algorithm implemented went like this:
1. Find strongly connected components (I used Tarjan)
2. Create condensation graph for the SCCs
3. Form reflexive-transitive closure of adjacency matrix (I used Warshall)
Doing that forms the partial order, but.... what advantage does finding the partial order give us?
One possible use: It lets you answer the question "is vertex $y$ reachable from vertex $x$?" in $O(1)$ time. | 2021-06-18 00:38:47 | {"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": 0.4744739234447479, "perplexity": 554.5150195643661}, "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/1623487634576.73/warc/CC-MAIN-20210617222646-20210618012646-00467.warc.gz"} |
https://stats.stackexchange.com/questions/335850/python-classification-algorithms-implementation-which-accept-missing-values | # Python - Classification algorithms implementation which accept missing values?
I've a binary classification problem which I want to solve where many features have a lot of missing values.
I know that imputing with mean/median/variance is a solution, but I'd like to run tests only with the original dataset without imputing. XGBoost allows the presence of missing values, while all the scikit-learn algorithms don't (correct me if I'm wrong), even if theoretically algorithm like random forest could accept missing values.
Other than XGBoost, which other python classification algorithms implementations allow the usage of a dataset with missing values?
• Instead of trying to work with NaN values, most people either exclude the observations with missing data (which is bad) or use imputation (better). As always, the best strategy is to just collect complete data (but that's sometimes impossible because life is unfair).
– Sycorax
Mar 21 '18 at 14:20
• If I'd exclude observations with missing data, probably my dataset would end up being empty :) Mar 21 '18 at 14:22
• @Sycorax I think NA and NaN are different things Mar 21 '18 at 14:28
• allowing NaNs is just an euphemism to imputation of some sort Mar 21 '18 at 14:29
• @hxd1011 Depends on how NaNs arise. I'm assuming that OP didn't do something silly like take $\log(x)$ for a vector $x$ which contains 0s, or divide by zero...
– Sycorax
Mar 21 '18 at 14:29 | 2021-09-18 21:14:16 | {"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": 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": 0.2841916084289551, "perplexity": 1644.8367454467323}, "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/1631780056572.96/warc/CC-MAIN-20210918184640-20210918214640-00647.warc.gz"} |
https://planetmath.org/DerivativeNotation | # derivative notation
This is the list of known standard representations and their nuances.
$\frac{du}{dv},\frac{df}{dx},\frac{dy}{dx}-$ The most common notation, this is read as the derivative of $u$ with respect to $v$. Exponents relate which derivative, for example, $\frac{d^{2}y}{dx^{2}}$ is the second derivative of $y$ with respect to $x$.
$f^{\prime}(x)\;,\vec{f}^{\prime}(\mathbf{x})\;,y^{\prime\prime}-$ This is read as $f$ prime of $x$. The number of primes tells the derivative, ie. $f^{\prime\prime\prime}(x)$ is the third derivative of $f(x)$ with respect to $x$. Note that in higher dimensions, this may be a tensor of a rank equal to the derivative.
$D_{x}f(\mathbf{x}),F_{y}(\mathbf{x}),f_{xy}(\mathbf{x})-$ These notations are rather arcane, and should not be used generally, as they have other meanings. For example $F_{y}$ can easily by the $y$ component of a vector-valued function. The subscript in this case means “with respect to”, so $F_{yy}$ would be the second derivative of $F$ with respect to $y$.
$D_{1}f(\mathbf{x}),F_{2}(\mathbf{x}),f_{12}(\mathbf{x})-$ The subscripts in these cases refer to the derivative with respect to the nth variable. For example, $F_{2}(x,y,z)$ would be the derivative of $F$ with respect to $y$. They can easily represent higher derivatives, ie. $D_{21}f(\mathbf{x})$ is the derivative with respect to the first variable of the derivative with respect to the second variable.
$\frac{\partial u}{\partial v}\;,\frac{\partial f}{\partial x}-$ The partial derivative of $u$ with respect to $v$. This symbol can be manipulated as in $\frac{du}{dv}$ for higher partials.
$\frac{d}{dv}\;,\frac{\partial}{\partial v}-$ This is the operator version of the derivative. Usually you will see it acting on something such as $\frac{d}{dv}(v^{2}+3u)=2v$.
$[\mathbf{Jf}(\mathbf{x})]\>,[\mathbf{Df}(\mathbf{x})]-$ The first of these represents the http://planetmath.org/node/842Jacobian of $\mathbf{f}$, which is a matrix of partial derivatives such that
$[\mathbf{Jf}(\mathbf{x})]=\left[\begin{array}[]{ccc}D_{1}f_{1}(\mathbf{x})&% \dots&D_{n}f_{1}(\mathbf{x})\\ \vdots&\ddots&\vdots\\ D_{1}f_{m}(\mathbf{x})&\dots&D_{n}f_{m}(\mathbf{x})\\ \end{array}\right]$
where $f_{n}$ represents the nth function of a vector valued function. The second of these notations represents the derivative matrix, which in most cases is the Jacobian, but in some cases, does not exist, even though the Jacobian exists. Note that the directional derivative in the direction $\vec{v}$ is simply $[\mathbf{Jf}(\mathbf{x})]\vec{v}$.
Title derivative notation Canonical name DerivativeNotation Date of creation 2013-03-22 11:58:27 Last modified on 2013-03-22 11:58:27 Owner mathcam (2727) Last modified by mathcam (2727) Numerical id 14 Author mathcam (2727) Entry type Topic Classification msc 26-00 Related topic Derivative Related topic Gradient Related topic PartialDerivative Related topic DirectionalDerivative Related topic JacobianMatrix Related topic LeibnizNotationForVectorFields | 2018-11-20 17:14:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 35, "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": 0.9080925583839417, "perplexity": 244.64381846666862}, "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/1542039746528.84/warc/CC-MAIN-20181120171153-20181120193153-00130.warc.gz"} |
http://mymathforum.com/algebra/22361-inequality-print.html | My Math Forum (http://mymathforum.com/math-forums.php)
- Algebra (http://mymathforum.com/algebra/)
- - Inequality (http://mymathforum.com/algebra/22361-inequality.html)
jatt-rockz November 5th, 2011 07:31 PM
Inequality
Solve each inequality graphically. State the solution and graph the solution:
a) (X^3) – (6x^2) +5x+12 >0
b) X/(x^2 -2x-8) >0
Zahar November 5th, 2011 08:08 PM
Re: Inequality
1 Attachment(s)
I do not know right answer but in my head))) exist this:
X/(x^2 -2x-8) >0
for the (x^2 -2x-8) :
x = [ 2 +/- square_root( 2^2 - 4*(-8) ) ] / 2 = 1 +/- square_root( 4 + 32 )/2 =
= 1 +/- 6/2 = 1 +/- 3 = { 4, -2 } (x1 = 4 and x2 = -2)
then we can to write
x/(x^2 -2x-8) = x / [ (x - x1)*(x - x2) ] = x / [ (x - 4)*(x + 2) ]
I think this
x do not equal 4 and -2 because we will have division by zero (but we used this numbers to transform equation)
we have x from ( - infinity) to (+ infinity)
then
case ( x < - 2 ) : (-) /[ (-)*(-) ] < 0 /* because x < 0, x - 4 < 0 , x + 2 < 0 as conclusion from that :: x < - 2 */
case ( -2 < x < 0 ) : (-) /[ (-)*(+) ] > 0 /* because x < 0, x - 4 < 0 , x + 2 > 0 as conclusion from that :: -2 < x < 0 */
case ( 0 < x < 4 ) : (+) /[ (-)*(+) ] < 0 /* because x > 0, x - 4 < 0 , x + 2 > 0 as conclusion from that :: 0 < x < 4 */
case ( 4 < x ) : (+) /[ (+)*(+) ] > 0 /* because x > 0, x - 4 > 0 , x + 2 > 0 as conclusion from that :: 4 < x */
But ! You need to write it in proper form, which is demanded by teacher, book, because I do not know the rules of ...
soroban November 5th, 2011 08:21 PM
Re: Inequality
Hello, jatt-rockz!
Quote:
Solve each inequality graphically. State the solution and graph the solution: [color=beige]. . [/color]$a)\;x^3\,-\,6x^2\,+\,5x\,+\,12\:>\:0$
$\text{W\!e have a cubic function: }\:y \:=\:x^3\,-\,6x^2\,+\,5x\,+\,12$
[color=beige]. . [/color]When is it positive?
[color=beige]. . [/color]When is its graph above the x-axis?
The function factors:[color=beige] .[/color]$f(x) \:=\:(x\,+\,1)(x\,-\,3)(x\,-\,4)$
[color=beige]. . [/color]Its x-intercepts are:[color=beige] .[/color]$-1,\:3,\:4$
The graph looks like this:
Code:
| | * | | * * -1 * | * * - - - o - + - - - o - - - o - - - * | 3 * 4 | * | |
$\text{Solution: }\:(-1\:<\:x\:<\:3)\:\cup\:(x\:>\:4)$
$\begin{array}{cccccccccc}\text{Graph: }=&--=&o=&o=&--=&o=&-1=&3=&4 \end{array}=$
All times are GMT -8. The time now is 06:07 PM. | 2019-04-22 02:07:59 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 6, "/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": 0.6899038553237915, "perplexity": 5933.048993658963}, "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-2019-18/segments/1555578533774.1/warc/CC-MAIN-20190422015736-20190422040748-00052.warc.gz"} |
https://www.meritnation.com/ask-answer/question/if-fx-2tan3x-51-cos6x-gx-is-a-function-having-the-same-time/sets-relations-and-functions/12620319 | # If is a function having the same time period as that of f(x), then which of the following can be g(x). (a) $\left({\mathrm{sec}}^{2}3\mathrm{x}+{\mathrm{cosec}}^{2}3\mathrm{x}\right){\mathrm{tan}}^{2}3\mathrm{x}$ (b) (c) $2\sqrt{1-{\mathrm{cos}}^{2}3\mathrm{x}}+\mathrm{cosec}3\mathrm{x}$ (d)
Dear Student,
Please find below the solution to the asked query: | 2022-01-27 03:36:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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": 0.8355803489685059, "perplexity": 1454.703808091521}, "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-2022-05/segments/1642320305052.56/warc/CC-MAIN-20220127012750-20220127042750-00429.warc.gz"} |
https://web2.0calc.com/questions/math-help-please-very-urgent | +0
# math help please very urgent
0
327
1
+272
please find the distance between the two numbers on a number line
-2 1/2 and -5 3/4
Oct 5, 2017
#1
+2337
+2
To find the distance of two numbers on the number line, simply subtract the 2 numbers.
$$-2\frac{1}{2}-(-5\frac{3}{4})$$ Subtracting a negative is equivalent to adding a positive. $$-2\frac{1}{2}+5\frac{3}{4}$$ Now convert each fraction to an improper fraction. Doing this is the first step to allow us to calculate the distance. $$-2\frac{1}{2}=-\frac{2*2+1}{2}=\frac{-5}{2}$$ Now, convert the other fraction to an improper fraction. $$5\frac{3}{4}=\frac{4*5+3}{4}=\frac{23}{4}$$ Now, reinsert these improper fraction back into the expression. $$\frac{-5}{2}+\frac{23}{4}$$ In order to add these fractions, we must create a common denominator. Multiply the first fraction by 2/2 to achieve this. $$\frac{-10}{4}+\frac{23}{4}$$ Now, add the fractions together. $$\frac{13}{4}=3.25$$ This is the distance in between the two numbers on the number line.
Oct 5, 2017 | 2019-02-17 09:08:11 | {"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": 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": 0.9322556257247925, "perplexity": 616.7248434620875}, "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-2019-09/segments/1550247481766.50/warc/CC-MAIN-20190217071448-20190217093448-00309.warc.gz"} |
https://stats.stackexchange.com/questions/214618/estimating-a-variable-from-its-cosine-corrupted-by-additive-gaussian-noise | # Estimating a variable from its cosine corrupted by additive Gaussian noise
I observe $y_i=\cos(\theta)+z_i$, $i=1,\ldots,n$, where each $z_i\sim\mathcal{N}(0,\sigma^2)$ is an i.i.d. zero-mean Gaussian random variable.
I am interested in estimating $\theta\in[0,\pi]$ with minimum mean square error (MSE).
By the general scalar case of CRLB I know that
$$\tag{1}\mathbb{E}[(\theta-\hat{\theta})^2]\geq\frac{\sigma^2\sin^2\theta}{n}.$$
I am wondering if an estimator exists that comes close to the CRLB in (1). Specifically, I am looking for an estimator whose MSE goes to zero when $\theta\rightarrow0$ or $\theta\rightarrow\pi$. Any ideas?
Update: Unfortunately (1) is incorrect: the $\sin\theta$ should be moved to denominator and squared. Thus, an estimator with MSE going to zero when $\theta\rightarrow0$ or $\theta\rightarrow\pi$ does not exist. I also forgot to square the $\sin\theta$ term in the analysis of the natural estimator below. When corrected, the asymptotic MSE in (2) of this estimator matches the CRLB in (1). See the answer below.
What I tried
The "natural" estimator would just average $y_i$'s and take the inverse cosine (we must also truncate the average so that it's in $[-1,1]$ to use inverse cosine). Its MSE is:
$$\mathbb{E}[(\theta-\hat{\theta})^2]=\mathbb{E}\left[\left(\theta-\cos^{-1}\left(\cos\theta+\frac{1}{n}\sum_{i=1}^nz_i\right)\right)^2\right]$$
The noise term $\frac{1}{n}\sum_{i=1}^nz_i$ gets small as $n$ increases, so the Taylor series expansion of $\cos^{-1}(\theta+x)$ around $x=0$ yields:
$$\tag{2}\mathbb{E}[(\theta-\hat{\theta})^2]\approx \frac{\sigma^2}{n\sin\theta},$$
where the approximation is from dropping of the lower-order terms in the Taylor series. While we can also use the Taylor series expansion to show that this estimator is unbiased, MSE in (2), unlike (1), has $\sin\theta$ in the denominator, which means the error gets worse when $\theta\rightarrow0$ or $\theta\rightarrow\pi$. This seems to be an inherent problem with this estimator (and not from approximation in (2)), as evident from a numerical experiment:
In the figure 'numerical' plots the MSE from the numerical experiment for $n=1000$ and $\sigma^2=1$, while 'approx $\cos^{-1}$(average)' and 'CRLB' plot (2) and (1) corresponding to these values of $n$ and $\sigma^2$. I think that the truncation of the average to $[-1,1]$ is to blame, but I'm not sure how to fix this.
For reference, here is the MATLAB code used to generate the "numerical" curve above:
theta_array=linspace(0,pi,20);
for ii=1:20
theta=theta_array(ii);
z=randn(1000,1000);
average=cos(theta)+mean(n);
average(average>1)=1;
average(average<-1)=-1;
error=theta-acos(average);
mean_error(ii)=mean(error);
mse(ii)=mean(error.^2);
end
• What about the ML estimator? – lacerbi May 25 '16 at 17:19
• @lacerbi - the natural estimator described above is the ML estimator, unfortunately. Think of the model as $y_i = a + z_i$; the ML of $a$ is the sample mean, so the ML of $\theta$ for $\theta \in (0, \pi)$ is the inverse cosine of $\hat{a}$. – jbowman May 25 '16 at 17:41
• @jbowman Oh, yeah, thanks. What about a Bayesian estimator then, with a uniform prior over $\theta$ and some appropriate cost function? – lacerbi May 25 '16 at 18:21
• Unfortunately, when I derive the CRLB, I get the $\sin^2\theta$ term in the denominator of your equation 1, not in the numerator. Naturally, if my derivation is correct, it would preclude the nice property of having the MSE go to zero when $\theta \to 0$ or $\theta \to \pi$, although it wouldn't preclude achieving the CRLB. – jbowman May 25 '16 at 18:46
• @Glen_b I moved the update that I wrote at the bottom of the question to an answer, per your request. Let me know if it should be expanded further... – M.B.M. May 26 '16 at 3:03
The CRLB in the general scalar case where we want to estimate $\theta=g(a)$, is given by:
$$\mathbb{E}(\theta-\hat{\theta})^2]\geq \frac{\left(\frac{\partial g}{\partial a}\right)^2}{I(a)}$$
where $I(a)$ is the Fisher information associated with $a$. Here $a=\cos\theta$. Since $\theta=\cos^{-1}(a)$, one must square the derivative of $\cos^{-1}(a)$ and substitute $a=\cos\theta$. One therefore gets:
$$\tag{1, corrected}\mathbb{E}[(\theta-\hat{\theta})^2]\geq\frac{\sigma^2}{n\sin^2\theta}.$$
I also forgot to square the $\sin\theta$ term when taking it out of the expectation in (2). It should read:
$$\tag{2, corrected}\mathbb{E}[(\theta-\hat{\theta})^2]\approx\frac{\sigma^2}{n\sin^2\theta}.$$
Thus, the MSE of the "natural" estimator matches CRLB asymptotically.
• Thanks -- I don't see how this would be answered otherwise -- I think there's some value in actually having an answer for this one. – Glen_b -Reinstate Monica May 26 '16 at 3:49 | 2020-01-19 22:09:10 | {"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": 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": 0.8919152021408081, "perplexity": 314.1700068916348}, "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-2020-05/segments/1579250595282.35/warc/CC-MAIN-20200119205448-20200119233448-00402.warc.gz"} |
https://openair-collective.github.io/openair-cyan/ | # Welcome to Cyan
Welcome to the documentation for the Cyan Decarbonizer, version 1.
A DIY small-scale open hardware direct air carbon capture (DACC) system.
The outer box and fan are optional for experimentation. The humidification chamber and air pump are the minimum required items.
## Purpose
This portable system operates at a small enough scale to empower individuals to tangibly experience carbon removal. It is also designed to get people actively thinking and sharing their carbon footprint and their role in the carbon cycle.
You can take a 5-minute tour of a Cyan system and learn more about the details of the project here:
If you have twenty-something minutes, details about the device design, current challenges, opportunities and big questions, and the Cyan mission’s general goals and aspirations are available here:
# Vision
Cyan is named in honor of cyanobacteria. As cyanobacteria did with oxygen, the vision was that many of these small-scale DACC units might come to have a significant impact on atmospheric $CO_2$ and thus climate change. If not directly due to their small size, then indirectly through the promotion of DACC as a technology and through the educational experience of using Cyan.
The educational experience might be the biggest way that Cyan units will combat climate change. One can quickly see what $CO_2$ emissions look like in trapped form, how much it weighs, and how long it takes to capture it. DACC starts to become a conversation point among friends and family. People start wanting to run Cyans to erase the emissions from their light bulbs, for example. And if they can neutralize those emissions, what’s next? Household emissions? Should I buy a bigger DACC unit to cover my home’s emissions? And so on.
# The Documentation, Q&A, Wiki, and Project Updates
The Cyan Assembly Instructions will get you started with an overview of how it works, a parts list, and instructions for putting it together. All this information can be found on this website.
We now have a Cyan Q&A that goes over some common general and technical questions.
The Supporting System Calculations describe the logic that went into designing the system, from a LCA/TEA perspective.
The $CO_2$ Capture Measurements describe results obtained so far on how well this system can perform.
To report your modifications, experimental results, and feedback for making Cyan better, please visit the Cyan DataShare site.
To see an overview of where we are with current progress, please view the Projects board.
Lastly, to view Cyan Project Progress updates on the OpenAir Forum, those are available here.
# Carbon Capture Details
## Inputs and Outputs
The input material, calcium hydroxide, can be found in any home improvement store as hydrated lime. However, for Cyan to be carbon-negative the input must either be sourced from a low-carbon supplier or obtained from waste cement.
The output material, calcium carbonate can be used in building materials (cement-lime) or as agricultural lime.
Magnesium hydroxide is also an option for use as the input material. Though perhaps more expensive depending on the source, it is less alkaline than calcium hydroxide and has performed as well to slightly better than calcium hydroxide in carbon capture effectiveness in our Cyan tests. It can also be used to make tiles or blocks when mixed with cement and optionally biochar. We hope to have a low-carbon supplier later this year.
## Emissions Reduction
At present, a Cyan will remove over the course of 24 hours the same amount of $CO_2$ emitted by running a 9-watt LED bulb for 3 hours. That is 2 grams of $CO_2$ for 10grams of input material - or a 0.2:1 ratio.
It is within theoretical possibility for this unit to remove the emissions of 20 9-watt LED bulbs for 3 hours (a whole house at night), that is 0.382 kg of $CO_2$ that would need to be captured. Adequate surface area needs to be provided within the unit to do this and we are presently working on that.
## Achieving Negative Emissions
The calcium hydroxide needs to be exposed to high humidity from between 3 to 16 hours to achieve a good level of carbonation. Keeping the electricity consumption low will ensure Cyan removes more $CO_2$ than it emits through electricity use.
Cyan uses 1.5 watts of electricity to power the air pump which bubbles air through water and humidifies the calcium hydroxide, allowing $CO_2$ to be absorbed. However, the air pump does not need to be run continuously to maintain adequate humidification. A duty cycle of, say, 10 minutes on & 50 minutes off is possible and suggested for negative emissions (more $CO_2$ removed than consumed by the air pump).
The fan uses 3 watts but only needs to be run intermittently to refresh the air inside the box. The fan is also useful to speed up dry time but this is not essential for operation. If you’re going for negative emissions, air drying would really help unless you are using renewable energy to power your unit. | 2022-10-02 07:42:03 | {"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": 0.28901028633117676, "perplexity": 2301.2358294314618}, "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-2022-40/segments/1664030337287.87/warc/CC-MAIN-20221002052710-20221002082710-00094.warc.gz"} |
https://hal.archives-ouvertes.fr/hal-01961053v2 | # Extensions with shrinking fibers
Abstract : We consider dynamical systems $T: X \to X$ that are extensions of a factor $S: Y \to Y$ through a projection $\pi: X \to Y$ with shrinking fibers, i.e. such that $T$ is uniformly continuous along fibers $\pi^{-1}(y)$ and the diameter of iterate images of fibers $T^n(\pi^{-1}(y))$ uniformly go to zero as $n \to \infty$. We prove that every $S$-invariant measure has a unique $T$-invariant lift, and prove that many properties of the original measure lift: ergodicity, weak and strong mixing, decay of correlations and statistical properties (possibly with weakening in the rates). The basic tool is a variation of the Wasserstein distance, obtained by constraining the optimal transportation paradigm to displacements along the fibers. We extend to a general setting classical arguments, enabling to translate potentials and observables back and forth between $X$ and $Y$.
Document type :
Preprints, Working Papers, ...
Cited literature [47 references]
https://hal.archives-ouvertes.fr/hal-01961053
Contributor : Benoît Kloeckner <>
Submitted on : Friday, February 7, 2020 - 3:01:15 PM
Last modification on : Thursday, February 13, 2020 - 1:35:09 AM
### Files
extensions.pdf
Files produced by the author(s)
### Identifiers
• HAL Id : hal-01961053, version 2
• ARXIV : 1812.08437
### Citation
Benoit Kloeckner. Extensions with shrinking fibers. 2020. ⟨hal-01961053v2⟩
Record views | 2020-02-18 03:48:43 | {"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": 0.49369028210639954, "perplexity": 2124.9093275484297}, "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/1581875143505.60/warc/CC-MAIN-20200218025323-20200218055323-00396.warc.gz"} |
http://tex.stackexchange.com/questions/35825/pretty-table-of-contents/35835 | # Pretty Table of Contents
I would like to know if it's possible to have a Table of Contents looking like the following, where the number of the page is the big number on the left.
The sections could look like in the following "uggly-font-formatted" example. If someone has a better idea for the sections and subsections...
-
Have a look at this post, it might help get you started – cmhughes Nov 22 '11 at 22:52
not necessarily related but few ideas here: tex.stackexchange.com/questions/19796/… – pluton Apr 2 '12 at 2:32
add comment
## 3 Answers
Here's a solution with a redefinition of the \chapter command (as implemented in book.cls) using \parboxes and the leftbar environment from the framed package:
\documentclass{book}
\usepackage{xcolor}
\usepackage{framed}
\definecolor{myred}{RGB}{127,0,0}
\definecolor{myyellow}{RGB}{169,121,69}
\renewenvironment{leftbar}{%
\def\FrameCommand{{\color{myyellow}\vrule width 2pt depth 6pt} \hspace{10pt}}%
\MakeFramed {\advance\hsize-\width \FrameRestore}}%
{\endMakeFramed}
\makeatletter
\def\@chapter[#1]#2{\ifnum \c@secnumdepth >\m@ne
\if@mainmatter
\refstepcounter{chapter}%
\typeout{\@chapapp\space\thechapter.}%
\addtocontents{toc}%
{%
{\protect\parbox{3.5em}{\hfill\Huge\color{myred}\bfseries\thepage}}%
\protect\hspace*{.5em}
\protect\parbox{\dimexpr\linewidth-4em\relax}{%
\protect\begin{leftbar}
{\scshape\small\chaptername~\thechapter}\\\sffamily#1%
\protect\end{leftbar}}\par\noindent
}%
\else
\addcontentsline{toc}{chapter}{#1}%
\fi
\else
\addcontentsline{toc}{chapter}{#1}%
\fi
\chaptermark{#1}%
\addtocontents{lof}{\protect\addvspace{10\p@}}%
\addtocontents{lot}{\protect\addvspace{10\p@}}%
\if@twocolumn
\@topnewpage[\@makechapterhead{#2}]%
\else
\@makechapterhead{#2}%
\@afterheading
\fi}
\makeatother
\begin{document}
\tableofcontents
\chapter{Beginning to learn design with \LaTeX}
\chapter{Beginning to learn design with HTML}
\setcounter{page}{13}% just for the example
\chapter{Beginning to learn design with HTML and some other text to span more than one line in the ToC}
\end{document}
The first \parbox will contain the page number; I just reserved space for two-digit numbers (3.5em); for three-digit numbers, the width of this box will have to be increased, and that of the third \parbox will have to be decreased in the same amount.
I've made some improvements and additions to my first solution. Now there will be no problems when loading hyperref (my previous solution was incompatible with hyperref); the modifications to the ToC emtries are now made with the help of the titletoc package; I've also added redefinitions for the section and subsection entries.
\documentclass{book}
\usepackage{xcolor}
\usepackage{framed}
\usepackage{titletoc}
\usepackage{etoolbox}
\usepackage{lmodern}
\definecolor{myred}{RGB}{127,0,0}
\definecolor{myyellow}{RGB}{169,121,69}
\patchcmd{\tableofcontents}{\contentsname}{\sffamily\contentsname}{}{}
\renewenvironment{leftbar}
{\def\FrameCommand{\hspace{6em}%
{\color{myyellow}\vrule width 2pt depth 6pt}\hspace{1em}}%
\MakeFramed{\parshape 1 0cm \dimexpr\textwidth-6em\relax\FrameRestore}\vskip2pt%
}
{\endMakeFramed}
\titlecontents{chapter}
[0em]{\vspace*{2\baselineskip}}
{\parbox{4.5em}{%
\hfill\Huge\sffamily\bfseries\color{myred}\thecontentspage}%
\vspace*{-2.3\baselineskip}\leftbar\textsc{\small\chaptername~\thecontentslabel}\\\sffamily}
{}{\endleftbar}
\titlecontents{section}
[8.4em]
{\sffamily\contentslabel{3em}}{}{}
{\hspace{0.5em}\nobreak\itshape\color{myred}\contentspage}
\titlecontents{subsection}
[8.4em]
{\sffamily\contentslabel{3em}}{}{}
{\hspace{0.5em}\nobreak\itshape\color{myred}\contentspage}
\begin{document}
\tableofcontents
\chapter{Beginning to learn design with \LaTeX}
\section{This is a test section}
\subsection{Long subsection title and some other text to span more than one line}
\newpage\setcounter{page}{123}% just for the example
\chapter{Beginning to learn design with HTML and some other text to span more than one line in the ToC}
\section{This is a test section}
\subsection{Long subsection title and some other text to span more than one line}
\end{document}
As Harish Kumar suggested in a comment, it will be nice to have some decorations for parts too. This new variation explores one possibility for these decorations:
\documentclass{book}
\usepackage{xcolor}
\usepackage{framed}
\usepackage{tikz}
\usepackage{titletoc}
\usepackage{etoolbox}
\usepackage{lmodern}
% definition of some personal colors
\definecolor{myred}{RGB}{127,0,0}
\definecolor{myyellow}{RGB}{169,121,69}
% command for the circle for the number of part entries
\newcommand\Circle[1]{\tikz[overlay,remember picture]
\node[draw,circle, text width=18pt,line width=1pt] {#1};}
% patching of \tableofcontents to use sans serif font for the tile
\patchcmd{\tableofcontents}{\contentsname}{\sffamily\contentsname}{}{}
% patching of \@part to typeset the part number inside a framed box in its own line
% and to add color
\makeatletter
\patchcmd{\@part}
{\addcontentsline{toc}{part}{\thepart\hspace{1em}#1}}
{\addtocontents{toc}{\protect\addvspace{20pt}}
\addcontentsline{toc}{part}{\huge{\protect\color{myyellow}%
\setlength\fboxrule{2pt}\protect\Circle{%
\hfil\thepart\hfil%
}%
}\\[2ex]\color{myred}\sffamily#1}}{}{}
%\patchcmd{\@part}
% {\addcontentsline{toc}{part}{\thepart\hspace{1em}#1}}
% {\addtocontents{toc}{\protect\addvspace{20pt}}
% \addcontentsline{toc}{part}{\huge{\protect\color{myyellow}%
% \setlength\fboxrule{2pt}\protect\fbox{\protect\parbox[c][1em][c]{1.5em}{%
% \hfil\thepart\hfil%
% }}%
% }\\[2ex]\color{myred}\sffamily#1}}{}{}
\makeatother
% this is the environment used to typeset the chapter entries in the ToC
% it is a modification of the leftbar environment of the framed package
\renewenvironment{leftbar}
{\def\FrameCommand{\hspace{6em}%
{\color{myyellow}\vrule width 2pt depth 6pt}\hspace{1em}}%
\MakeFramed{\parshape 1 0cm \dimexpr\textwidth-6em\relax\FrameRestore}\vskip2pt%
}
{\endMakeFramed}
% using titletoc we redefine the ToC entries for parts, chapters, sections, and subsections
\titlecontents{part}
[0em]{\centering}
{\contentslabel}
{}{}
\titlecontents{chapter}
[0em]{\vspace*{2\baselineskip}}
{\parbox{4.5em}{%
\hfill\Huge\sffamily\bfseries\color{myred}\thecontentspage}%
\vspace*{-2.3\baselineskip}\leftbar\textsc{\small\chaptername~\thecontentslabel}\\\sffamily}
{}{\endleftbar}
\titlecontents{section}
[8.4em]
{\sffamily\contentslabel{3em}}{}{}
{\hspace{0.5em}\nobreak\itshape\color{myred}\contentspage}
\titlecontents{subsection}
[8.4em]
{\sffamily\contentslabel{3em}}{}{}
{\hspace{0.5em}\nobreak\itshape\color{myred}\contentspage}
\begin{document}
\tableofcontents
\part{Designing with \LaTeX}
\chapter{Beginning to learn design with \LaTeX}
\section{This is a test section}
\subsection{Long subsection title and some other text to span more than one line}
\newpage\setcounter{page}{123}% just for the example
\part{Designing with HTML}
\chapter{Beginning to learn design with HTML and some other text to span more than one line in the ToC}
\section{This is a test section}
\subsection{Long subsection title and some other text to span more than one line}
\end{document}
Unfortunately, if hyperref is loaded, the part entry formatting will be lost (I'll have to think about this).
## Memoir document class
In some of the comments below there are suggestions to implement the style using the memoir document class. Below I present two variations.
### First variation
This version works for numbered and unnumbered parts, chapters, sections and subsections, and cooperates with hyperref:
\documentclass{memoir}
\usepackage[T1]{fontenc}
\usepackage{xcolor}
\usepackage{framed}
\usepackage{tikz}
\usepackage[colorlinks,linkcolor=black]{hyperref}
% colors to be used
\definecolor{myred}{RGB}{127,0,0}
\definecolor{myyellow}{RGB}{169,121,69}
% a modification of the leftbar environment defined by the framed package
% will be used to place a vertical colored bar separating the page number and the
% title in chapter entries
\renewenvironment{leftbar}{%
\def\FrameCommand{\textcolor{myyellow}{\vrule width 2pt depth 6pt}\hspace*{15pt}}%
\MakeFramed{\advance\hsize-\width\FrameRestore}}%
{\endMakeFramed}
% a command to circle the part numbers
\newcommand\Circle[1]{\tikz[overlay,remember picture]
\node[draw=myyellow,circle, text width=18pt,line width=1pt,align=center] {#1};}
% redefinition of the name of the ToC
\renewcommand\printtoctitle[1]{\HUGE\sffamily\bfseries#1}
\makeatletter
% redefinitions for part entries
\renewcommand\cftpartfont{\Huge\sffamily\bfseries\hfill}
\renewcommand\partnumberline[1]{%
\hbox to \textwidth{\hss\Circle{\textcolor{myyellow}{#1}}\hss}%
\vskip3.5ex\par\hfill\color{myred}}
\renewcommand*\cftpartformatpnum[1]{\hfill}
\renewcommand\cftpartafterpnum{\vskip1ex}
% redefinitions for chapter entries
\renewcommand\chapternumberline[1]{\mbox{\small\@chapapp~#1}\par\noindent\Large}
\renewcommand\cftchapterfont{\sffamily}
\cftsetindents{chapter}{0pt}{0em}
\renewcommand\cftchapterpagefont{\Huge\sffamily\bfseries\color{myred}}
\newcommand*{\l@mychap}[3]{%
\def\@chapapp{#3}
\vskip1ex%
\par\noindent\begin{minipage}{\textwidth}%
\parbox{4.5em}{%
\hfill{\cftchapterpagefont#2}%
}\hspace*{1.5em}%
\parbox{\dimexpr\textwidth-4.5em-15pt\relax}{%
\leftbar\cftchapterfont#1\hspace{1sp}\endleftbar%
}%
\end{minipage}\par%
}
\renewcommand*{\l@chapter}[2]{%
\l@mychap{#1}{#2}{\chaptername}%
}
\renewcommand*{\l@appendix}[2]{%
\l@mychap{#1}{#2}{\appendixname}%
}
% redefinitions for section entries
\renewcommand\cftsectionfont{\sffamily}
\renewcommand\cftsectionpagefont{\sffamily\itshape\color{myred}}
\renewcommand\cftsectionleader{\nobreak}
\renewcommand\cftsectiondotsep{\cftnodots}
\renewcommand\cftsectionafterpnum{\hspace*{\fill}}
\setlength\cftsectionnumwidth{12em}
\cftsetindents{section}{6em}{3em}
\renewcommand\cftsectionformatpnum[1]{%
\hskip1em\hbox to \@pnumwidth{{\cftsectionpagefont #1\hfill}}}
% redefinitions for subsection entries
\renewcommand\cftsubsectionfont{\sffamily}
\renewcommand\cftsubsectionpagefont{\sffamily\itshape\color{myred}}
\renewcommand\cftsubsectionleader{\nobreak}
\renewcommand\cftsubsectiondotsep{\cftnodots}
\renewcommand\cftsubsectionafterpnum{\hspace*{\fill}}
\setlength\cftsubsectionnumwidth{12em}
\cftsetindents{subsection}{9em}{3em}
\renewcommand\cftsubsectionformatpnum[1]{%
\hskip1em\hbox to \@pnumwidth{{\cftsubsectionpagefont #1\hfill}}}
\makeatother
\settocdepth{subsection}
\setsecnumdepth{subsection}
\begin{document}
\phantomsection
\addcontentsline{toc}{part}{\textcolor{myred}{Test Unnumbered Part}}
\part*{Test Unnumbered Part}
\tableofcontents*
\part{Designing with \LaTeX}
\chapter{Beginning to learn design with \LaTeX}
\section{A test section}
\subsection{A test subsection}
\phantomsection
\addcontentsline{toc}{section}{Test unnumbered section}
\section*{Test unnumbered section}
\newpage\setcounter{page}{123}% just for the example
\part{Designing with HTML}
\chapter{Beginning to learn design with HTML and some other text to span more than one line}
\section{Another test section}
\chapter*{Test Unnumbered Chapter}
\phantomsection
\addcontentsline{toc}{chapter}{Test Unnumbered Chapter}
\end{document}
### Second variation
projetmbc has suggested in an edit to the original question a different formatting when including sections and subsections; this is a first attempt to produce this new layout.
Some remarks:
1. The \StartMark command must be placed before each \chapter command and \EndMark must be placed immediately after the last sectional unit of a chapter that will be included in the ToC:
2. Page breaks can occur for the sectional units associated to a chapter and will be dealt automatically.
3. The code needs at least three runs.
The code:
\documentclass{memoir}
\usepackage[T1]{fontenc}
\usepackage{refcount}
\usepackage[colorlinks,linkcolor=black]{hyperref}
\usepackage{tikz}
\usetikzlibrary{calc}
% colors to be used
\definecolor{myred}{RGB}{127,0,0}
\definecolor{myyellow}{RGB}{169,121,69}
% a command to circle the part numbers
\newcommand\Circle[1]{\tikz[overlay,remember picture]
\node[draw=myyellow,circle, text width=18pt,line width=1pt,align=center] {#1};}
% redefinition of the name of the ToC
\renewcommand\printtoctitle[1]{\HUGE\sffamily\bfseries\color{myred}#1}
\makeatletter
% redefinitions for part entries
\renewcommand\cftpartfont{\Huge\sffamily\bfseries}
\renewcommand\partnumberline[1]{%
\hbox to \textwidth{\hss\Circle{\textcolor{myyellow}{#1}}\hss}%
\vskip3.5ex\color{myred}}
\renewcommand*{\l@part}[2]{%
\ifnum \c@tocdepth >-2\relax
\cftpartbreak
\begingroup
{\interlinepenalty\@M
\leavevmode
\settowidth{\@tempdima}{\cftpartfont\cftpartname}%
\addtolength{\@tempdima}{\cftpartnumwidth}%
\let\@cftbsnum \cftpartpresnum
\let\@cftasnum \cftpartaftersnum
\let\@cftasnumb \cftpartaftersnumb
\advance\memRTLleftskip\@tempdima \null\nobreak\hskip -\memRTLleftskip
\centering{\cftpartfont#1}\par%
}
\nobreak
\global\@nobreaktrue
\everypar{\global\@nobreakfalse\everypar{}}%
\endgroup
\fi}
% redefinitions for chapter entries
\renewcommand\chapternumberline[1]{\mbox{\Large\@chapapp~#1}\par\noindent}
\renewcommand\cftchapterfont{\Large\sffamily}
\cftsetindents{chapter}{0pt}{0pt}
\renewcommand\cftchapterpagefont{\HUGE\sffamily\bfseries\color{myred}}
\newcommand*{\l@mychap}[3]{%
\def\@chapapp{#3}\vskip1ex%
\par\noindent\begin{minipage}{\textwidth}%
\parbox{4.7em}{%
\hfill{\cftchapterpagefont#2}%
}\hspace*{1.5em}%
\parbox{\dimexpr\textwidth-4.7em-15pt\relax}{%
\cftchapterfont#1%
}%
\end{minipage}\par\vspace{2ex}%
}
\renewcommand*{\l@chapter}[2]{%
\l@mychap{#1}{#2}{\chaptername}%
}
\renewcommand*{\l@appendix}[2]{%
\l@mychap{#1}{#2}{\appendixname}%
}
% redefinitions for section entries
\renewcommand\cftsectionfont{\sffamily}
\renewcommand\cftsectionpagefont{\sffamily\itshape\color{myred}}
\renewcommand{\cftsectionleader}{\nobreak}
\renewcommand{\cftsectionafterpnum}{\cftparfillskip}
\cftsetindents{section}{7.5em}{3em}
\renewcommand\cftsectionformatpnum[1]{%
\hskip1em\hbox to 4em{{\cftsectionpagefont #1\hfill}}}
% redefinitions for subsection entries
\renewcommand\cftsubsectionfont{\sffamily}
\renewcommand\cftsubsectionpagefont{\sffamily\itshape\color{myred}}
\renewcommand\cftsubsectionleader{\nobreak}
\renewcommand{\cftsubsectionafterpnum}{\cftparfillskip}
\renewcommand\cftsubsectiondotsep{\cftnodots}
\cftsetindents{subsection}{10.5em}{3em}
\renewcommand\cftsubsectionformatpnum[1]{%
\hskip1em\hbox to 4em{{\cftsubsectionpagefont #1\hfill}}}
\makeatother
\settocdepth{subsection}
\setsecnumdepth{subsection}
% length to be used when drawing a line from the top of the text area
\newlength\Myhead
\setlength\Myhead{\dimexpr\headheight+\headsep+1in+\voffset+5ex\relax}
% length to be used when drawing a line to the bottom of the text area
\newlength\Myfoot
\setlength\Myfoot{\dimexpr\paperheight-\Myhead-\textheight-\footskip+5ex\relax}
% auxiliary counter to place labels
\newcounter{chapmark}
% Adds a mark and a label at the beginning of each chapter entry in the ToC and draws a line
% from the start of the chapter to the bottom of the text area if the mark
% for the chapter ending lies in a different page than the one from the end of the chapter.
% (the value of tjose pages is calculated using the label)
% Must be used right before each \chapter command
\newcommand\StartMark{%
\addtocontents{toc}{\protect\label{st\thechapmark}%
\protect\begin{tikzpicture}[overlay,remember picture,baseline]
\protect\node [anchor=base] (s\thechapmark) {};%
\ifnum\getpagerefnumber{st\thechapmark}=\getpagerefnumber{en\thechapmark} \else
\protect\draw[myyellow,line width=3pt] let \protect\p3= (s\thechapmark),%
\protect\p4 = (current page.south) in %
($(4em,\protect\y3) + (0,-1ex)$) -- ($(4em,\protect\y4) + (0,\protect\the\Myfoot)$);\fi
\protect\end{tikzpicture}\par}%
}
% Adds a mark and a label at the end of each chapter entry in the ToC and draws a line from
% the top of the text area to the ending of the chapter if the mark
% for the chapter ending lies in a different page than the one from the start of the chapter
% if both marks are in the same page, simple draws a line connecting the marks
% (the value of tjose pages is calculated using the label)
% Must be used right after the last sectional unit (that will go to the ToC) belonging to
% a chapter
\newcommand\EndMark{
\addtocontents{toc}{\protect\label{en\thechapmark}%
\protect\begin{tikzpicture}[overlay,remember picture,baseline]
\protect\node [anchor=base] (e\thechapmark) {};
\ifnum\getpagerefnumber{st\thechapmark}=\getpagerefnumber{en\thechapmark}
\protect\draw[myyellow,line width=3pt] let \protect\p1= (s\thechapmark), \protect\p2=(e\thechapmark) in ($(4em,\protect\y1) + (0,-1ex)$) -- ($(4em,\protect\y2) + (0,2ex)$);
\else%
\protect\draw[myyellow,line width=3pt] let \protect\p1= (e\thechapmark), \protect\p2=(current page.north) in ($(4em,\protect\y2) + (0,-\protect\the\Myhead)$) -- ($(4em,\protect\y1) + (0,1.5ex)$);
\fi
\protect\end{tikzpicture}\par}\stepcounter{chapmark}%
}
\begin{document}
\tableofcontents*
\part{Designing with \TeX\ and \LaTeX}
\StartMark
\chapter{Beginning to learn design with \TeX}
\section{Another test section}
\section{Another test section }
\subsection{A test subsection}
\section{Another test section}
\subsection{A test subsection}
\subsection{A test subsection with a long title spanning more than one line in the table of contents}
\section{Another test section}
\section{Another test section}
\section{Another test section}
\EndMark
\StartMark
\chapter{Beginning to learn design with \LaTeX}
\section{A test section with a long title spanning more than one line in the table of contents}
\subsection{A test subsection}
\subsection{Another test subsection}
\section{A test section}
\phantomsection
\addcontentsline{toc}{section}{Test unnumbered section}
\section*{Test unnumbered section}
\subsection{A test subsection}
\section{Another test section}
\subsection{A test subsection}
\subsection{A test subsection}
\section{Another test section}
\section{Another test section}
\EndMark
\newpage\setcounter{page}{123}% just for the example
\part{Designing with HTML}
\StartMark
\chapter{Beginning to learn design with HTML and some other text to span more than one line}
\section{Another test section}
\section{Another test section}
\subsection{A test subsection}
\section{Another test section with a long title spanning more than one line in the table of contents}
\subsection{A test subsection}
\section{Another test section}
\section{Another test section}
\subsection{A test subsection}
\subsection{A test subsection}
\section{Another test section}
\section{Another test section}
\subsection{A test subsection}
\EndMark
\StartMark
\chapter{Beginning to learn design with HTML and some other text to span more than one line}
\section{Another test section}
\section{Another test section}
\subsection{A test subsection}
\section{Another test section}
\section{Another test section}
\subsection{A test subsection with a long title spanning more than one line in the table of contents}
\section{Another test section}
\section{Another test section}
\subsection{A test subsection}
\section{Another test section}
\section{Another test section}
\subsection{A test subsection}
\subsection{A test subsection}
\section{Another test section}
\section{Another test section}
\subsection{A test subsection}
\section{Another test section}
\subsection{A test subsection}
\section{Another test section}
\section{Another test section}
\section{Another test section}
\subsection{A test subsection}
\EndMark
\appendix
\StartMark
\chapter{Test Appendix}
\section{Another test section}
\section{Another test section with a long title spanning more than one line in the table of contents}
\section{Another test section}
\subsection{A test subsection}
\section{Another test section}
\section{Another test section}
\EndMark
\end{document}
The obtained ToC:
-
With your minimal code you make it look easy. Looks great! – Werner Nov 23 '11 at 0:15
The alignment seems to be a little off on the right hand side. Otherwise, what a beautiful implementation! – Mark S. Everitt Nov 23 '11 at 11:16
@bobicool: when I have the time I will take a look about using memoir (although I don't use that document class). For the problem with the frenchb option for babel, simply use \shorthandoff{;} before \tableofcontents to temporarily switch off the shorthand action of the character ;. To switch back on the shorthand, you can use \shorthandon{;} after \tableofcontents. – Gonzalo Medina Apr 2 '12 at 14:19
@bobicool: I've updated my answer providing an implementation for the chapter entries with the memoir document class. – Gonzalo Medina Apr 8 '12 at 23:21
@dochar: the first one actually took less time (10-15 min.) but is a very restrictive solution. The last one (using memoir and allowing page breaks) took longer (several hours).. and I am still working (whenever I find some free time) on improving the solutions ;-) – Gonzalo Medina May 1 '12 at 22:36
show 31 more comments
Here is a solution using ConTeXt:
\define[3]\myTOC{% #1=chapter number, #2=chapter text, #3=page
{\ssd #3\hskip.4em}
\framed
[
frame=off,
leftframe=on,
rulethickness=2pt,
framecolor=darkgreen,
loffset=1em,
align=right,
location=bottom,
foregroundstyle=sans,
]
{{\rm\sc Chapter #1}\\#2}
}
\setuplist [chapter]
[
alternative=none,
command=\myTOC,
]
\starttext
\placelist [chapter]
\setnumber[userpage]{10}
\startchapter [title=Beginning to learn design with ConTeXt]
\stopchapter
\startchapter [title=Beginning to learn design with HTML]
\stopchapter
\stoptext
The result looks like this:
However, it is not optimal. It only defines the list style for the chapters. But since you didn't specify anything else, that's all I can do.
The command \setuplist defines the layout of the list in the first argument, here chapter. alternative=none tells the system that you don't want a predefined style, rather than to use the custom command, which defines the appearance. I set the page number to a two-digit value, otherwise it gets confusing, which number belongs to the page and which to the chapter count.
-
This is another great ConTeXt example! What's the use of {\sc Chapter #1} that seems to have lost the Small Caps? – Werner Nov 22 '11 at 23:28
Yes I saw it too and removed it from the code (and you posted the comment in the mean time ;) Does LM have sans small caps by the way? – Marco Nov 22 '11 at 23:35
Not that I can see; Read An exploration of the Latin Modern fonts (p 3). – Werner Nov 22 '11 at 23:45
@Werner According to the doc there are no sans small caps. I changed to roman ones (which in fact is ugly). The fine-tuning is is left for the OP. – Marco Nov 23 '11 at 0:03
Yes, @Gonzalo's answer reflects the same for LaTeX. Most likely the OP's images is from a PDF screen grab, with the original document constructed in some different publishing software. – Werner Nov 23 '11 at 0:15
show 1 more comment
I have pached the solution proposed by Marco (1) to allow for starred version of the \chapter command and (2) to prevent wrong pagebreaks which sometimes occured (in Marco's solution) between the pagenumber and the chapter title. For the latter purpose I wrapped the \parbox that contains \thecontentspage and the leftbar environment into the minipage environment.
\usepackage{framed}
\usepackage{titletoc}
\usepackage{etoolbox}
\definecolor{myred}{RGB}{127,0,0}
\definecolor{myyellow}{RGB}{169,121,69}
\patchcmd{\tableofcontents}{\contentsname}{\sffamily\contentsname}{}{}
\renewenvironment{leftbar}
{\def\FrameCommand{\hspace{6em}%
{\color{myyellow}\vrule width 2pt depth 6pt}\hspace{1em}}%
\MakeFramed{\parshape 1 0cm \dimexpr\textwidth-6em\relax\FrameRestore}\vskip2pt%
}
{\endMakeFramed}
\titlecontents{chapter}
[0em]{\vspace*{\baselineskip}}
{%
\begin{minipage}{\textwidth}
\parbox{4.5em}{\hfill\Huge\sffamily\bfseries\color{myred}\thecontentspage}%
\vspace{-2.3\baselineskip}\leftbar{\sffamily\chaptername~\thecontentslabel}\\
\sffamily
}
{%
\begin{minipage}{\textwidth}
\parbox{4.5em}{%
\hfill\Huge\sffamily\bfseries\color{myred}\thecontentspage}%
\vspace{-2.3\baselineskip}\leftbar\textsf{}\\\sffamily}
{\endleftbar\end{minipage}\nopagebreak}[\vskip2pt\nopagebreak]
\titlecontents{section}
[8.4em]
{\sffamily\contentslabel{3em}}{}{}
{\hspace{0.5em}\nobreak\itshape\color{myred}\contentspage}
\titlecontents{subsection}
[8.4em]
{\sffamily\contentslabel{3em}}{}{}
{\hspace{0.5em}\nobreak\itshape\color{myred}\contentspage}
-
This almost works with memoir ... Any chance you might be able to fix it to work with it? – André Apr 7 '12 at 13:41
The page number of subsubsection is not changed in TOC, please add this feature – KOF Jan 12 at 22:01
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http://kycenter.designq.kr/he-got-gvb/archive.php?3b0be6=grams-to-molecules-calculator | Grams to Moles Calculator Mole is a standard measurement of amount which is used to measure the number atoms (or) molecules. Conventional notation is used, i.e. [ The answer is 31.9988. This online calculator converts to moles given grams and converts to grams given moles. Print Tables. 1 grams O2 is equal to 0.031251171918947 mole. 1 gram (g) is equal to 0.03527396195 ounces (oz). Using the above calculator you could find that e.g. The particles can be anything, molecules, atoms and radioactive ions but also things like tanks of petrol, tubes of toothpaste, cigarettes, donuts, and even pizza! Grams to Atoms Calculator is a free online tool that displays the conversion from grams to atoms for the particle. In the event that the number of atoms in 12 grams of carbon could be resolved, it would be a similar number of molecules in the various known components and researchers would have a connection between the macro and subatomic worlds. Today this value is known as Avogadro's number (N) or constant in honour of his contributions to this area of chemistry. You can view more details on each measurement unit: molecular weight of O2 or mol The SI base unit for amount of substance is the mole. The mass (in grams) of a compound is equal to its molarity (in moles) multiply its molar mass: grams = mole × molar mass. Add a Grams to Grams conversion calculator to your website. Because the molar volume is the same for all ideal gases and is known, we can convert from grams to liters and vice versa if we know formula of a gas. Tl;dr version: Mm = molar mass, g = mass of substance (as a whole) in grams, mol = group of avogrado's number (Na, not to be confused with sodium) of molecules, n = number of … This is "OK" but not highly accurate. First, convert the grams to moles using the molar mass and then use Avogadro's number to find the number of molecules: This calculation tells you that there are 2.1 x 10 22 molecules of NaCl in 2 grams of NaCl. - the first letter of … This calculator finishes the topic started in Convert moles to liters and liters to moles calculator. The mass and molarity of chemical compounds can be calculated based on the molar mass of the compound. again! [ Empirical And Molecular Formula Solver. Thus, the molality of a solute that is present in a solution can be measured as the number of moles included in every kilogram of the solvent. Calculate the molar mass by multiplying the number of atoms of each element in the compound (its subscript) times the atomic mass of the element from the periodic table. Calculate the moles from the grams. How many grams O2 in 1 mol? To figure this out, you will need the molar mass of NaCl which is 58.44 g/mol. The mole or mol is an amount unit similar to familiar units like pair, dozen, gross, etc. Since the mass connection among carbon and hydrogen is 12:1 the quantity of particles in 1 gram of hydrogen is likewise 6.02214129(27) x 10^23 molecules. For accuracy please use our Grams To Cups Conversions calculator. How to convert Grams to Ounces. Molecular Mass Calculator. Another thing that must be remembered is that a mole of any particular compound contains precisely 6.023 x 1023 molecules. But if you don't find calculating moles of a substance easy, then our grams to moles calculator can help you in the best possible way. This is known as the molar mass and its units are grams/mole. grams = 58.443 × 5 = 292.215 (g) The ionic mass of ion and formula mass of ionic substance respectively. Molar mass to moles calculator can also assist in the process of checking moles or one can use Mole Fraction Calculator for same purpose. An effective method to measure the concentration of any solution is by calculating the molarity of the solute, and the other one is done by calculation of the molality. Learn a SIMPLE way to convert the number of molecules of a covalent compound to mass of that covalent compound in the Mole Conversion II Practice Problem Want to know how many moles a gram has or need grams to moles calculator for general calculation purposes? This chemistry video tutorial explains how to convert grams to molecules. How many molecules are in 16g of CO 2? You can find metric conversion tables for SI units, as well as English units, currency, and other data. It is defined …, Normality is a measure of concentration equal to the gram …. The molar mass is a valuable unit factor for some estimation. The mole of any substance or compound is actually its molecular weight that is expressed in the mass-unit of grams. Furnished with the connection between hydrogen and carbon and a characterized nuclear mass unit, researchers set out to decide the number of atoms in 12 grams of carbon. BYJU’S online grams to atoms calculator tool makes the conversion faster and it displays the conversion to atoms in a fraction of seconds. You divide and find that 1 gram of fluorine is equal to 0.0525350025878 moles. As long as there is … For accuracy please use our Grams To Cups Conversions calculator. Molar mass of NaCl is 58.443, how many grams is 5 mole NaCl? Here is a simple online Moles To Molecules calculator to convert moles into molecules. One gram (g) = weight of 1 milliliter (ml) of pure water at temperature 4 °C. Grams to Moles Formula, Grams to moles calculator, Convert moles to grams calculator, finding the number of moles from grams @Byjus.com Since the mole is characterized as the quantity of molecules in 12 grams of carbon, it tends to be expressed that the mass of one mole of particles is equivalent in grams to the numerical value of an element's atomic mass. For chemistry you often need to convert moles to grams and grams to moles. b)$$1\;\text{mole of}\;H_2O\;=\;18.01\;\text{grams}$$. Save my name, email, and website in this browser for the next time I comment. This site is owned and maintained by Wight Hat web development ©2009. How many molecules are in 70g of Sn? ›› Metric conversions and more ConvertUnits.com provides an online conversion calculator for all types of measurement units. Then, knowing that there are 6.022x10^23 molecules in one mol (Avogadro's number) you can find the number of molecules in however many mol of the substance that you have. To know how to calculate moles with mass to moles calculator, the formula is, Well, using grams to moles calculator is one of the easiest ways to calculate moles to grams or grams to moles. … The mass of ionic compound divided by the atomic mass of an element, molecular mass of a substance. This is true for all the elements because their masses are based on the same relationship. Moles To Molecules Calculator. Grams to Ounces conversion table Since water has two molecules of hydrogen and one molecule of oxygen, then the molecular weight of water is 18.01528g/mol. Grams to molecules calculator is usually helpful in this regard. Empirical And Molecular Formula Solver. So, thanks to this calculator, you shall never wonder "Avogadro's number is the number of what?" People can find the molar mass or atomic mass of elements on the periodic table, and the molar mass of a compound is the total molar mass of all the elements in this compound. You can view more details on each measurement unit: molecular weight of O2 or mol The SI base unit for amount of substance is the mole. Knowing that the conversion factor to get to molecules involves the number of mols, the first conversion you need to do from grams is to mol. These proportions enable scientists to build unit variables to perform numerous valuable computations. Because the molar volume is the same for all ideal gases and is known, we can convert from grams to liters and vice versa if we know formula of a gas. Or 1 mole of a substance will contain Avogadro's number of that substance. Letting N be the number of molecules or moles of the substance, the formula to convert from molecules to grams is then given as (N number of molecules) / (6.022 x 10^23) x (substance's molar mass in grams/mol). Since water has two molecules of hydrogen and one molecule of oxygen, then the molecular weight of water is 18.01528g/mol. BYJU’S online grams to moles calculator tool makes the calculation faster and it displays the grams to moles conversion in a fraction of seconds. Molar mass = (2 x 1.008) + (2 x 15.999) Note the use of more significant figures for oxygen Molar mass = 34.016 grams… Mole, in chemistry, is a standard scientific unit for measuring large quantities of very small entities such as atoms, molecules, or other specified particles. A gram-mole is the same thing as a mole in simple terms. One such very fundamental concept is the molality, which actually refers to the mole of a compound. This program determines both empirical and molecular formulas. Therefore; Therefore, for 16 grams of water there is 0.89 moles. There is a simple relation between these two:, where - mass of the substance in grams We know we have 10 g of HCl, and it has a molecular weight of 36.5 g / mol. Let's do a quick example to help explain how to convert from moles to grams, or grams to moles. Lets plug these numbers into the above equation: mole = 10 / 36.5 = 0.27 moles = 1.626×10^23 molecules of HCl Conversion from grams to molecules requires two conversion factors. We know we have 10 g of HCl, and it has a molecular weight of 36.5 g / mol. It is an easy way to calculate the grams to moles or moles to grams with our mole conversion calculator. Actually, the gram mole is mass of an element, mass of molecule and mass of the ion. The mole calculator uses the formula to get accurate results. You can use Avogadro's number in conjunction with atomic mass to convert a number of atoms or molecules into the number of grams. If you know the structure of the molecule, then you can figure out the molecular weight. E.g. Suppose you have 16 grams of water. This calculator finishes the topic started in Convert moles to liters and liters to moles calculator. After numerous analyses, it was resolved that 12 grams of carbon contained 6.02214129(27) x 10^23 molecules. Lets plug these numbers into the above equation: mole = 10 / 36.5 = 0.27 moles = 1.626×10^23 molecules of HCl In each case, the number of grams in 1 mol is the same as the number of atomic mass units that describe the atomic mass, the molecular mass, or the formula mass, respectively. The second converts moles of Cl 2 to the number of molecules. The gram (British spelling: gramme, abbreviation: g) is a unit of mass in the SI system (metric system). All you have to know is the component's nuclear mass and change the amu's to grams with the help of mol calculator. A mole (mol) is the SI unit for the measure of a substance and is characterized as Avogadro's number of any delegate molecule. The mass in grams of a single mole of any compound will equal the molecular weight of the compound when expressed in AMUs. The answer is 31.9988. If you take your 878 grams of fluorine and then look at the atomic mass. Step 2: Calculate using dimensional analysis The problem is done using two consecutive conversion factors. person_outlineTimurschedule 2017-04-24 16:36:31. The mole ratios are the links between the macro and subatomic worlds. Molar mass or molar weight is the mass that one mole of a substance has, and they are defined in grams per mole. Or 1 mole of a substance will contain Avogadro's number of that substance. One gram is equal to one thousandth of the kilogram (kg), which is the current SI (Metric system) base unit of mass. 1 mole of something is equal to 6.0221415x10 23 of it. How to calculate moles - moles to grams converter. Calculate the moles from the grams. Example. Want to know how many moles a gram has or need grams to moles calculator for general calculation purposes? Gram Conversion Calculator. - the first letter … To convert from moles to molecules multiply the molar amount of the molecule by Avogadro's number. 1 grams O2 is equal to 0.031251171918947 mole. The mass of the substance must be in grams. We couldn't find a conversion between molecules and grams [incompatible types] Do a quick conversion: 1 molecules = grams-force using the online calculator for metric conversions. Conversion calculator for converting recipe ingredients from metric grams to U.S. ounces, teaspoons, tablespoons, or cups based on general ingredient weights. Example: Suppose you have 16 grams of water. Please consider supporting us by disabling your ad blocker. This is "OK" but not highly accurate. For lab work or practical, chemists can also use our Theoretical Yield Calculator and Percent Yield Calculator as well. Grams to moles calculator is the simplest yet effective calculator. How to Calculate Double Declining Balance. The mass of the substance must be in grams. So, thanks to this calculator, you shall never wonder "Avogadro's number is the number of what?" again! Using the above calculator you could find that e.g. Grams to molecules calculator is usually helpful in this regard. When you are aware of the concentration, it becomes possible to find out whether it is a saturated or unsaturated solution. Answer: 55 g K 2 S . a pollution of 1 gram of benzene in a certain amount of water converts to N A /78.11≈ 7.7098 × 10 21 molecules polluting that water! On the off chance that truly, at that point you are in the ideal spot. Free Calculator. First, the molar mass allows you to change mass into mol. This online unit converter will help you to convert the grams of a molecule to the number of moles based on the weight of given chemical equation / formula. This program determines both empirical and molecular formulas. Enter the molecular formula of the molecule. This is the number of molecules in 1 mole of a chemical compound. Keep reading to find out more about this amazing mole to grams calculator. Answer: 0.053 g Cl One mole of the substance is the ratio of the mass of the substance to the atomic weight of the substance. The mass m in ounces (oz) is equal to the mass m in grams (g) divided by 28.34952:. This site uses an exact value of 6.0221415 x 10 23 for Avogadro's number. The Avogadro's number is a dimensionless quantity and is equivalent to the Avogadro constant. Then you use Avogadro's number to set up a relationship between the number of molecules and mass. With our moles to grams calculator, you can not only calculate grams to moles but also moles to grams seamlessly. In order to get a clearer understanding of the complex issues of chemistry, you will learn about the specific concepts of the subject. Or even want to calculate moles from grams of a substance? Kp stands for the equilibrium partial pressure. Hence, it calculates the number of moles per kilogram of the solvent that is used. Then you multiply that by your 878 grams. Or even want to calculate moles from grams of a substance? For molecules, you add together the atomic masses of all the atoms in the compound to get the number of grams per mole. We hope the grams to moles calculator will help you to solve chemical equations with Chemical Equation Balancer Calculator. Grams to Moles Calculator is a free online tool that displays the conversion of grams to moles. How many grams O2 in 1 mol? On the off chance that truly, at that point you are in the ideal spot. Our website is made possible by displaying online advertisements to our visitors. We assume you are converting between grams O2 and mole. For this, you need to know the molar mass of methane, which is 16.04 g/mol. We assume you are converting between grams O2 and mole. Answer: 2.2 * 10 23 molec CO 2 . 6.022 * 10 23 molecules This ratio is seen on the right arrow (the green text) in the conversion map below. Let's do a quick example to help explain how to convert from moles to grams, or grams to moles. There are a lot of ways through which the concentration of a solution is measured. Gap the mass of every component by the molar mass and increase the outcome by 100%. Conventional notation is used, i.e. Consequently, 16 grams of oxygen, 14 grams of nitrogen and 35.45 grams of chlorine all have 6.02214129(27) x 10^23 particles. By Developing 100+ online Calculators and Converters for Math Students, Engineers, Scientists and Financial Experts, calculatored.com is one of the best free calculators website. Conversion calculator for converting recipe ingredients from metric grams to U.S. ounces, teaspoons, tablespoons, or cups based on general ingredient weights. Our tools accurately convert grams to moles from the values of mass and molar mass. Grams to Moles Calculator. Answer: 3.5 * 10 23 molecules Sn . Gram Conversion Calculator. a pollution of 1 gram of benzene in a certain amount of water converts to N A /78.11≈ 7.7098 × 10 21 molecules polluting that water! Now, you know that in 105 g of methane there are 6.55 mol of methane. On a side note, I tend to abbreviate the units of molecules as (molec) for reasons of saving space and to help distinguish it from moles which is something different. A mass in grams numerically equal to the molecular weight contains one mole of molecules, which is known to be 6.02 x 10^23 (Avogadro’s number). If you have 9.2 * 10 20 molecules of Cl how many grams is that? Copyrights 2020 © calculatored.com . m (oz) = m (g) / 28.34952. Is there any difference between gram mole and mole, Enter the “Molar Mass” in the second field. How to calculate the mass of a particular isotopic composition. The first converts grams of Cl 2 to moles. To calculate or find the grams to moles or moles to grams the molar mass of each element will be used to calculate. If you have 3.0 * 10 24 molecules of K 2 S how many grams is that? You need to follow the simple instructions to find all grams to moles given below. Remember that a mole is a number similarly as twelve or a gross numbers. That is, the molar mass of a substance is the mass (in grams per mole) of 6.022 × 10 23 atoms, molecules, or formula units of that substance. Get a Grams to Grams conversion chart and print it. You will get results based on the values you given in the calculator within a few seconds. Examples: C6H12O6, PO4H2(CH2)12CH3 Calculate molecular mass by: Contact help@bmrb.io if you have any questions about this site ››Definition: Molecule. 1 g = 0.03527396195 oz. How to calculate moles - moles to grams converter. Use the search box to find your required metric converter → iPhone & Android app Weight Grams Grams to Kilograms Grams to Pounds Grams to Ounces Grams to Metric Tons (or Tonnes) Grams to Carats Grams to Milligrams Grams to Micrograms Grams to Troy Ounces Grams to Pennyweights Grams to Grains More units.. The particles can be anything, molecules, atoms and radioactive ions but also things like tanks of petrol, tubes of toothpaste, cigarettes, donuts, and even pizza! Convert 5g to Ounces: m (oz) = 5 g / 28.34952 = 0.17637 oz. One mole of any substance will have 6.022×10^23 molecules of that substance. Grams Conversion. All rights reserved. 10^23 molecules number similarly as twelve or a gross numbers Fraction calculator for converting recipe ingredients from grams! Point you are grams to molecules calculator 16g of CO 2 conversion of grams to molecules but not highly accurate moles kilogram... Or compound is actually its molecular weight that is expressed in AMUs of. Of checking moles or moles to grams with the help of mol.... Can figure out the molecular weight of 1 milliliter ( ml ) of pure water temperature... Consider supporting us by disabling your ad blocker of molecule and mass multiply the molar mass allows you solve... To get the number of molecules and mass find out whether it is defined,... Set up a relationship between the number of that substance first converts grams water. Balancer calculator mole in simple terms, dozen, gross, etc the simple to! ( N ) or constant in honour of his contributions to this finishes! ) / 28.34952 = 0.17637 oz they are defined in grams, gross, etc is mass of the that! Unit factor for some estimation the mass-unit of grams per mole g / mol use our Theoretical Yield calculator Percent! The compound to get accurate results the number of moles per kilogram the! Using dimensional analysis the problem is done using two consecutive conversion factors very! Some estimation 2 to moles given grams and converts to moles given grams and converts grams. Maintained by Wight Hat web development ©2009 mass or molar weight is the mass of every component the! In honour of his contributions to this calculator, you shall never wonder Avogadro 's number of per! Resolved that 12 grams of carbon contained 6.02214129 ( 27 ) x 10^23.. Or constant in honour of his contributions to this calculator, you never... Chemical Equation Balancer calculator molecular formula Solver of K 2 S how many molecules are in the ideal grams to molecules calculator! To 6.0221415x10 23 of it into mol is measured calculator, you will about... The right arrow ( the green text ) in the mass-unit of grams to moles contain Avogadro number... Molar mass of the concentration of a compound between the number of that substance the concepts! Few seconds is known as the molar mass ” in the calculator a. Thing that must be in grams ( g ) = m ( oz ) the conversion from grams to calculator! Know we have 10 g of methane there are a lot of ways through which the concentration a. For same purpose that 1 gram of fluorine is equal to the mass one... Accuracy please use our grams to moles or moles to grams and converts to grams conversion chart print... Equal the molecular weight of the substance must be in grams must be in grams calculates... Calculates the number of molecules in 1 mole of a solution is measured enable scientists to build variables. Convertunits.Com provides an online conversion calculator a grams to cups Conversions calculator as there a. To the Avogadro constant my name, email, and website in regard... 28.34952: oz ) = m ( oz ) = m ( g ) = m oz! Mole is mass of an element, mass of an element, molecular mass molecule. For this, you can not only calculate grams to moles calculator can also assist the... Our mole conversion calculator for converting recipe ingredients from metric grams to moles contributions to this calculator finishes the started. Hope the grams to molecules calculator is usually helpful in this regard unit! Will learn about the specific concepts of the ion contain Avogadro 's is. Use our Theoretical Yield calculator as well will be used to measure number! As there is a free online tool that displays the conversion of grams per mole two,... Of chemistry or one can use mole Fraction calculator for same purpose our Theoretical Yield and..., it calculates the number of grams per mole about the specific concepts of the solvent that is.... O2 and mole ways through which the concentration, it becomes possible to find out whether it defined. Known as the molar mass of ion and formula mass of methane water is.... Change the amu 's to grams the molar mass of the compound when expressed in the second moles... A single mole of a particular isotopic composition values you given in the process of checking moles or to. Grams calculator, you will learn about the specific concepts of the substance in grams per mole remembered! Grams the molar mass or molar weight is the molality, which 58.44. As twelve or a gross numbers atoms calculator is usually helpful in this.... And then look at the atomic mass of the concentration of a has... The calculator within a few seconds owned and maintained by Wight Hat web development ©2009 hence it... Enable scientists to build unit variables to perform numerous valuable computations need the mass... Molecule, then you can not only calculate grams to moles calculator is usually helpful this. To moles this, you need to follow the simple instructions to find all grams to moles calculator the mass... Is defined …, Normality is a simple online moles to grams with our moles to grams molar! Us by disabling your ad blocker moles given below molarity of chemical compounds can be based. Saturated or unsaturated solution and its units are grams/mole ) of pure water at temperature °C! It was resolved that 12 grams of fluorine and then look at the masses. Will help you to change mass into mol weight is the mass of ionic divided... And mass of the subject ) or constant in honour of his contributions to this area of chemistry you... But also moles to grams given moles usually helpful in this browser for the particle grams to molecules calculator topic started in moles. Us by disabling your ad blocker per mole out whether it is a simple online to! Hence, it becomes possible to find out more about this amazing mole grams. To the Avogadro 's number is a saturated or unsaturated solution is true for types... It has a molecular weight of water there is … Empirical and molecular formula Solver as long as is! Of hydrogen and one molecule of oxygen, then grams to molecules calculator molecular weight but... The process of checking moles or moles to grams conversion chart and print it number is the number of and! It calculates the number of that substance can be calculated based on general weights! Gross, etc currency, and it has a molecular weight that is expressed in AMUs mass of solvent. Becomes possible to find out more about this amazing mole to grams converter you shall wonder! The specific concepts of the substance must be in grams an element, mass of ionic substance respectively with Equation! ) in the ideal spot grams given moles any particular compound contains precisely 6.023 x 1023.. The off chance that truly, at that point you are converting between grams O2 and mole the ion of... Off chance that truly, at that point you are converting between grams O2 and mole converts grams... S how many moles a gram has or need grams to moles given grams and converts to given! Convert 5g to ounces: m ( g ) is equal to 0.03527396195 ounces ( oz ) weight! - mass of each element will be used to calculate the mass of ion and formula mass each... The ionic mass of the ion is 5 mole NaCl increase the outcome by 100 % of..., teaspoons, tablespoons, or cups based on the same relationship is a online. Units like pair, dozen, gross, etc, etc all the elements because their are! Of 6.0221415 x 10 23 molec grams to molecules calculator 2 or even want to calculate moles - to. The mass m in grams Wight Hat web development ©2009 one gram ( ). In 16g of CO 2 the gram … find that e.g substance in of! An element, mass of each element will be used to calculate moles from grams of a substance contain! Know that in 105 g of HCl, and website in this regard pair, dozen, gross,.... M ( oz ) = 5 g / mol 36.5 g / mol map below consider... Use mole Fraction calculator for general calculation purposes m in grams molecular mass calculator online tool that displays the of. Get a clearer understanding of the complex issues of chemistry, you learn. Variables to perform numerous valuable computations step 2: calculate using dimensional analysis the problem done... Known as the molar mass and its units are grams/mole you could find 1! G / mol for the next time I comment moles per kilogram of the molecule, the. Know we have 10 g of methane units like pair, dozen, gross, etc calculate moles moles. M ( oz ) is equal to 0.0525350025878 moles grams to moles calculator mole is mass of and... Equations with chemical Equation Balancer calculator metric grams to moles or moles to grams converter a online... 0.03527396195 ounces ( oz ) a chemical compound actually its molecular weight of 36.5 g /.! X 1023 molecules how to calculate the mass m in ounces ( )... From metric grams to moles = 292.215 ( g ) is equal 6.0221415x10... This site uses an exact value of 6.0221415 x 10 23 molecules ratio... Your website its molecular weight of 36.5 g / mol of 6.0221415 x 10 molec. Get a grams to cups Conversions calculator = 5 g / mol add together the atomic mass every! | 2021-04-16 19:38:55 | {"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": 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": 0.519262969493866, "perplexity": 1527.4330836263898}, "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-2021-17/segments/1618038089289.45/warc/CC-MAIN-20210416191341-20210416221341-00197.warc.gz"} |
http://codereview.stackexchange.com/questions/43190/intranet-php-application | # Intranet PHP application
I'm more of an IT guy (no CS course) with a strong and passionate relationship with Unix and I love KISS.
I'm writing an application to help my coworkers with their daily tasks. Every now and then I get request from coworkers to add functionality to help them, most of the time I will turn an Excel spreadsheet to a form based "module" in my application.
The application is built with PHP and runs on Apache, using MultiViews for "pretty-urls" and FallbackResource /index.php, the folder structure looks like
app/
_views/
_errors/
401.php
403.php
404.php
user/
get.php
core/
config.php
user.php
hr/
department.php
.settings.php
bootstrap.php
lib/
z/
form.php
route.php
view.php
logs/
public/
css/
fonts/
images/
js/
index.php
user.php
app/bootstrap.php is prepended to every .php files in the public/ folder.
<?php
set_include_path(implode(PATH_SEPARATOR, array(
get_include_path(),
dirname(__DIR__) . DIRECTORY_SEPARATOR .'lib',
__DIR__,
)));
// application user stuff
session_start();
if (isset($_SESSION['username'])) {$user = new core\user($_SESSION['username']); define('LOGGED', true); } else {$user = new core\user();
define('LOGGED', false);
}
function http_error($error) { http_response_code($error);
$view = new z\view('_errors/'.$error .'.php');
die($view); } My controllers are in the public/ folder and look like this: <?php # /user.php ; controllers for /user(.*) use z\form; use z\route as main; use z\view; main::get('/user/add', function() use ($user) {
if (!LOGGED)
http_error(401);
if (!$user->allowed('user-add')) http_error(403);$view = new view('user/add.php');
die($view); }); main::get('/user/%s', function($username) use ($user) {$person = new core\user($username); if ($person->notfound === true)
http_error(404);
if (!LOGGED)
http_error(401);
if (!$user->allowed('user-view')) http_error(403);$view = new view('user/get.php');
$view->person =$person;
if ($user->allowed('user-manage')) {$view->procedures = $person->procedures();$departments = hr\department::select();
array_shift($departments);$view->departments = $departments; } die($view);
});
main::post('/user', function() use ($user) { if (!LOGGED) http_error(401); if (!$user->allowed('user-add'))
http_error(403);
$form = new form($_POST);
if (core\user::insert(array(
'value1' => $form->field1->value, 'value2' =>$form->field2->value,
...
))) {
main::redirect('/user/'. $form->username->value); } else {$view = new view('user/add.php');
$view->error = true;$view->message = 'Error message';
die($view); } }); ... http_error(404); View 'user/get.php': <?php$layout = 'html.php' ?>
<h1><?= $person->displayname ?></h1> <p><?=$person->title ?></p>
<ul>
<li>Office: <?= $person->phone_office ?></li> <li>Mobile: <?=$person->phone_mobile ?></li>
</ul>
<div class="tabs">
<ul class="tabs-nav">
<?php if ($user->allowed('user-manage-somestuff')): ?> <li><a href="#tab-1">Tab 1</a></li> <?php endif ?> <?php if ($user->allowed('user-manage-someotherstuff')): ?>
<li><a href="#tab-2">Tab 2</a></li>
<?php endif ?>
<li><a href="#tab-3">Tab 3</a></li>
</ul>
<?php if ($user->allowed('user-manage-somestuff')): ?> <div id="tab-1"></div> <?php endif ?> <?php if ($user->allowed('user-manage-someotherstuff')): ?>
<div id="tab-2"></div>
<?php endif ?>
<div id="tab-3"></div>
</div>
This is then wrapped in an 'html.php' layout by the view class.
The core\config class:
<?php
namespace core;
class config
{
private static
$__settings = false; protected function __construct() {} public static function get($key) {
if (!self::$__settings) self::$__settings = include '../.settings.php';
$context = self::$__settings;
$pieces = explode('.',$key);
foreach ($pieces as$piece) {
if (!is_array($context) || !array_key_exists($piece, $context)) return false;$context = &$context[$piece];
}
return $context; } } The core\user class: <?php namespace core; use z\db; class user { public$displayname,
$firstname,$lastname,
$email,$phone_office,
$phone_mobile,$notfound = true;
protected
$_roles = []; public function __construct($username = 'anonymous') {
$this->displayname = 'Guest';$this->username = $username; if ($this->username != 'anonymous')
$this->__init(); } protected function __init() {$db = db::instance(config::get('db'));
$stmt =$db->prepare("
$stmt->execute(array( ':username' =>$this->username
));
$rslt =$stmt->fetch();
if (!$rslt) return false;$this->firstname = $rslt['firstname'];$this->lastname = $rslt['lastname'];$this->displayname = $this->firstname .' '.$this->lastname;
...
}
public static function insert(array $user) {$db = db::instance(config::get('db'));
$stmt =$db->prepare("
INSERT INTO users(field1, ...)
VALUES (:value1, ...)");
return $stmt->execute(array( ':value1' =>$user['field1'],
...
));
}
...
}
The hr\department class:
<?php
namespace hr;
use core\config;
use z\db;
class department
{
public static function insert(array $department) {...} public static function select($id = -1) {...}
public static function update(array $department) {...} public static function delete($id) {...}
}
As the application is getting bigger, I would like some advice on my code. Can you see problems with the way I'm doing things?
Is this code "maintainable"? I wouldn't want my application to be removed if/when I leave and let my users in pain. My main concern here is if the guy who will replace me be able to understand and work with this code "easily". Note that I removed all code comments before posting but everything is documented in comments.
I'm also having a lot of difficulties trying to understand tests and how to write them, can someone explain to me how can I write tests for my application? and will I really benefit from writing/using them?
I spent a lot of time looking at PHP frameworks from Zend to Symphony, Kohana, CI, Laravel, Phalcon and many many others... I really liked some of the "micro-frameworks" like Slim and Lemonade but the file structure didn't make sense to me, why have such a complicated file structure and so many files/folders for such a simple thing?
Then I found a small PHP library web.php and took it's route, view and form class/functions as a starting point to build my application.
About the view class, I can use a different layout from inside a controller like this:
$view = new view('view.php', 'layout.php'); - There have been a number of comments on your Reddit thread that reference various things will be new to you. This resource is probably your best bet as a starting point: phptherightway.com – bcmcfc Mar 4 at 8:52 I learned a lot from this website and can't thank enough it's author, it's clear, simple, I love it. But that was a few months ago. I think some of the practice I see are not really fit for the PHP "side" of a web application. – teuna Mar 4 at 9:40 IT doesn't really matter wether you use a framework or something else. The most important thing is to NOT write quick fixes and to have documentation. Not just inside your code, but a certain small 'Getting started' manual. That is the most important part in passing on software – Pinoniq Aug 19 at 16:27 ## 1 Answer I wouldn't stress about tests if it's a smaller company, an internal application, and will not be distributed. Generally a framework is a good idea. It gives you everything that you're asking for. I'd recommend Yii for performance, but Cake because of its popularity, which makes it easily maintainable. So the next guy might already know it, and the documentation of the framework minimises your need to document. i.e. If he needs to do something he can just look on their site. The simplest and most effective way forward would be to move it into Cake. Was this application built proactively? Is it essential that you concern yourself with the future of this application? After looking at the code: There are some important structural problems in this application. First off you need to read up on PHP best practices. You shouldn't use short tags. You don't need the user and department classes. You should have a generic model class that will deal with these models, which can be extended if you need to do something special for a certain model. i.e. Object oriented with inheritance. I see the$_POST goes straight into your form class. Does it get cleaned in there?
You are setting your layout in the view. What if you were doing something server side, such as browser detection, and then wanted to serve a different layout? You should set your layout where the logic is. In the controller.
Definitely move this into a common and accessible framework. Cake or Code Igniter. I think Yii maybe a bit too steep of a learning curve.
Also, because it is effectively business software, you will benefit from improved security within a framework.
-
The application wasn't build proactively and my only concern about it's future is my fellow coworkers using it (loving it and thanking me for it) everyday. The application did catch a couple managers attention so maybe it's something that will evolve in the company, at that point I'm not sure. \$_POST is sanitazed by the form class yes, I can also do validation but for the sake of making this post shorter didn't included them. – teuna Mar 2 at 21:02
There's nothing wrong with using PHP short tags (<?=). They are actively recommended in the community driven coding style guide PSR-1: php-fig.org/psr/psr-1 – bcmcfc Mar 4 at 11:39
That is not a recommendation. They are considered acceptable, but discouraged. What happens when you move files to a server with short tags off? Your code breaks, great! Use them, have fun ;) – Darius Mar 4 at 15:27 | 2014-09-16 07:23:00 | {"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": 0.19847483932971954, "perplexity": 8219.109913954062}, "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-2014-41/segments/1410657114105.77/warc/CC-MAIN-20140914011154-00098-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
https://pediatrics.aappublications.org/highwire/markup/62695/expansion?width=1000&height=500&iframe=true&postprocessors=highwire_tables%2Chighwire_reclass%2Chighwire_figures%2Chighwire_math%2Chighwire_inline_linked_media%2Chighwire_embed | Table 5.
Characteristics of Children and Adolescent Hospital Discharges Compared With Adults, 1996 (Number of Discharges in Thousands)
(Age 18+)
Children* and
Less Than
1 Year
1–4
Years
5–9
Years
10–14
Years
15–17
Years
Number of discharges (%)28 7036161 4422539321345534
(82.3)(17.7) (71.8)(8.7)(5.2)(5.6)(8.7)
Female, N(%)17 4433034 2121233138166376
(60.8)(49.3) (48.0)(43.3)(42.9)(48.1)(70.5)
Length of stay, mean (SD)5.33.3 3.13.23.84.53.7
(16.7)(16.0) (16.7)(11.6)(14.3)(15.9)(14.6)
Charges, mean (SD)$11 787$5284 $4319$6693$8131$9172\$7533
(47 282)(48 826) (49 778)(41 885)(51 242)(47 067)(45 473)
11 830930 215238143152182
N(%)(41.2)(15.1) (4.9)(44.1)(44.4)(44.0)(34.1)
Expected payer
Private, N(%)13 5603150 2324241158187240
(47.2)(51.1) (52.5)(44.9)(49.2)(54.2)(44.9)
Medicaid, N(%)49092462 1721249134124234
(17.1)(40.0) (38.9)(46.3)(41.7)(35.8)(43.8)
Uninsured, N(%)1386329 22629172036
(4.9)(5.3) (5.1)(5.4)(5.4)(5.8)(6.8)
Bedsize of hospital
Small, N(%)4352946 65791575982
(15.2)(15.4) (14.9)(16.9)(17.6)(17.2)(15.3)
Medium, N(%)91691859 13551518798168
(31.9)(30.2) (30.6)(28.0)(27.2)(28.3)(31.4)
Large, N(%)15 1113340 2397296177187282
(52.7)(54.2) (54.2)(55.0)(55.0)(54.3)(52.9)
Ownership of hospital
Private, not for profit, N(%)21 4124655 3400381231253390
(74.6)(75.6) (76.9)(70.8)(71.8)(73.4)(73.0)
Private, for profit, N(%)3356571 42949242346
(11.7)(9.3) (9.7)(9.0)(7.4)(6.7)(8.6)
Public, N(%)3864918 580108676896
(13.5)(14.9) (13.1)(20.0)(20.7)(19.7)(17.9)
Urban hospital, N(%)23 9795271 3824444271291441
(83.5)(85.6) (86.5)(82.4)(84.3)(84.2)(82.7)
Teaching hospital, N(%)95142381 1641224146156215
(33.2)(38.7) (37.1)(41.5)(45.4)(45.2)(40.3)
Region
Northeast, N(%)62411141 7951036875101
(21.7)(18.5) (18.0)(19.0)(21.2)(21.7)(18.9)
South, N(%)10 6362141 1455226133136191
(37.1)(34.7) (32.9)(41.9)(41.5)(39.3)(35.7)
Midwest, N(%)67931479 10731206881136
(23.7)(24.0) (24.3)(22.3)(21.1)(23.6)(25.6)
West, N(%)50331400 1099905253105
(17.5)(22.7) (24.8)(16.8)(16.2)(15.5)(19.8) | 2021-09-24 22:28:01 | {"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": 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": 0.2526496350765228, "perplexity": 972.9525551055045}, "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/1631780057580.39/warc/CC-MAIN-20210924201616-20210924231616-00180.warc.gz"} |
http://tex.stackexchange.com/questions/67543/padded-two-column-arrays | I am using the following kind of 'padded array' many times:
\setlength{\fboxrule}{0pt}
\setlength{\fboxsep}{3pt}
\fbox{
\ensuremath{
\begin{array}{|l|l|}
\hline
A & B\\
C & D\\
\vdots & \vdots \\
E & F\\
\hline
\end{array}
}
}
I'd like to be able to just write, e.g.,
\begin{DRS}
A & B\\
C & D\\
\vdots & \vdots \\
E & F\\
\end{DRS}
(or the same with a command instead of an environment), but I don't know a clean way to set this up. The problem is that fbox is a command, and array an environment, so I'm not sure of the right way to mix the two. I could possibly hack it together, e.g. using the environ package (which I just found by searching on this topic), but it would be nice to know the right way to do this.
Thanks!
Edit: this is the kind of image I'm producing (using nested DRSs):
-
If you need some padding, then ask explicitly for it:
\documentclass{article}
\newif\ifDRSmath
\newenvironment{DRS}
{\relax\ifmmode\global\DRSmathtrue\else\fi \kern3pt\begin{array}{|l|l|} \noalign{\kern3pt} \hline\mathstrut} {\hline \noalign{\kern3pt} \end{array}\kern3pt \ifDRSmath\global\DRSmathfalse\else\fi}
\begin{document}
\begin{DRS}
A & B\\
C & D\\
\vdots & \vdots \\
E & F\\
\end{DRS}
\begin{DRS}
A & B\\
C & D\\
\vdots & \vdots \\
E & F\\
\end{DRS}
\end{document}
The conditional is needed for being able to use the environment both in text mode and in math mode. There will work in both modes: at the start we check for math mode, at the end only for \ifDRSmath and this trick will avoid issuing $ improperly. - Thanks... but unfortunately that doesn't produce the 3pt of padding at the top and bottom, which I need. – Mohan Aug 17 '12 at 13:09 Also, is that conditional going to work with nesting? If you are already in mathmode, and enter a DRS, won't the first exit set DRSmath=false, and then the second exit (inappropriately) produce a$? – Mohan Aug 17 '12 at 13:24
@Mohan I've added the requested padding (although I'm quite dubious about it). The final comments should clarify why this can be used in both text and math mode. – egreg Aug 17 '12 at 15:17
Thanks... but that's producing a literal '3pt'. Could I ask what's wrong with using the fbox for the padding? – Mohan Aug 17 '12 at 15:23
@Mohan Sorry, fixed. It's possible to use \fbox, with the environ package or with the lrbox environment. – egreg Aug 17 '12 at 15:28
show 1 more comment
You can use either a normal \newenvironment (as in the DRSX environment below), or the environ package (as in the DRS environment):
There is no need to use \fbox{} as a normal \hline will do the job.
Code:
\documentclass{article}
\usepackage{environ}
\NewEnviron{DRS}{%
\setlength{\fboxrule}{0pt}
\setlength{\fboxsep}{3pt}
$\begin{array}{|l|l|}% \hline \BODY% \hline \end{array}$%
}
\newenvironment{DRSX}{%
\setlength{\fboxrule}{0pt}
\setlength{\fboxsep}{3pt}
\begin{math}\begin{array}{|l|l|}%
\hline
}{%
\hline
\end{array}\end{math}%
}
\begin{document}
\begin{DRS}
A & B\\
C & D\\
\vdots & \vdots \\
E & F\\
\end{DRS}
\begin{DRSX}
A & B\\
C & D\\
\vdots & \vdots \\
E & F\\
\end{DRSX}
\end{document}
-
Thanks... but I don't follow the point about the \hline. The point of the \fbox is to put a little padding around the array. (Note \fboxrule=0pt, so no lines are drawn.) – Mohan Aug 17 '12 at 12:01
PS. I'm curious about why some lines in your solution end in % but others don't... – Mohan Aug 17 '12 at 12:04 | 2014-07-25 01:38:12 | {"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": 0.8973934054374695, "perplexity": 2399.1342384713857}, "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-2014-23/segments/1405997892648.47/warc/CC-MAIN-20140722025812-00182-ip-10-33-131-23.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/geometry/geometry-common-core-15th-edition/chapter-1-tools-of-geometry-1-3-measuring-segments-practice-and-problem-solving-exercises-page-25/43 | ## Geometry: Common Core (15th Edition)
We can find GK by adding GH and HK because of the Segment Addition Postulate. To find GH, we can do GJ-HJ to get x+3. HK is 4x-3. GH+HK=5x. $GK=5x$ If GK=30 and GK=5x, then $5x=30$ by substitution, and x=6. We established that GH=x+3. Substituting 6 into the equation gives GH=9. To find JK, we can do HK-HJ to get 3x-3. JK is 3x-3. Substituting 6 into the equation gives JK=15. | 2020-03-29 19:41:57 | {"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": 0.795731246471405, "perplexity": 10644.008949478395}, "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/1585370495413.19/warc/CC-MAIN-20200329171027-20200329201027-00210.warc.gz"} |
https://tex.stackexchange.com/questions/564873/how-to-add-either-dashed-or-colored-vertical-line-in-array-but-without-using-a | # How to add either dashed or colored vertical line in array, but without using arydshln?
I need to make what is called augmented matrix, using array. This requires a vertical line added before the last column.
\documentclass[12pt]{report}
\usepackage{array}
\begin{document}
$\left[\begin{array}{@{}ccc|c@{}} 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \end{array}\right]$
\end{document}
Which works well. I just like to have the vertical line, either dashed or colored say red, to make it more clear.
I can not use arydshln as this package breaks many things. It does not work when loading longtable for example, and it does not work with tex4ht.
But array on its own, works. I just do not know how to add macro either inplace or in the preamble to tell it to make that one vertical line be dashed, or different color. I prefer answer that does not use tikz also, just to keep it simple. I would like the answer to work both in lualatex and tex4ht if possible, as I compile same source code to PDF and HTML.
But if it works only in PDF (i.e using lualatex), that will be fine also.
TL 2020
• Easiest is probably using tikzmark. I'm sure we have plenty of duplicates around the side. – Alan Munn Sep 30 at 22:19
Here is a solution for a coloured vertical rule:
\documentclass[12pt]{report}
\usepackage{xcolor, array, bigstrut}
\begin{document}
$\left[\begin{array}{@{}ccc!{\color{red}\vline width 0.6pt}c@{}} 1 & 2 & 3 & 4 \bigstrut[t]\\ 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \end{array}\right]$
\end{document}
The package nicematrix provides a specifier : for vertical dotted lines. The package loads PGF.
\documentclass[12pt]{report}
\usepackage{nicematrix, bigstrut}
\begin{document}
$\begin{bNiceArray}{ccc:c} 1 & 2 & 3 & 4 \bigstrut[t]\\ 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \end{bNiceArray}$
\end{document} | 2020-10-20 02:52:04 | {"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": 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": 0.789139986038208, "perplexity": 1511.2378458036628}, "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-2020-45/segments/1603107869785.9/warc/CC-MAIN-20201020021700-20201020051700-00625.warc.gz"} |
https://www.controlbooth.com/threads/what-mains-do-you-use.48191/ | # What mains do you use?
#### ACTSTech
##### Well-Known Member
We're moving forward with our reno project and it's time to address the house mains, or lack thereof. I have a pretty good idea what I'd like, but I wanted to see what people are using or in some cases, still not using. I've heard some horror stories about what to stay away from. Any thoughts?
In our case, the hall is going to be about 80' long and 40' wide, speakers will have to be surface mounted above the proscenium 21' off the floor (18' off the deck) and about 10' from the first row. I'm hoping to buy 4 of the QSC K8.2 powered speakers to mount on the proscenium which I think would be more than enough for our shows. I'd like a mini line array but that's not going to happen. I'm not 100% sold on the QSCs but I've heard them and I think I'd be comfortable. My only concern is that they're powered, so if something goes, it's the whole unit, not just a speaker or an amp, but it's the way things are done now.
Before I order, is there anything anyone's fallen in love with that they want to share? Talk me out of my insanity?
#### TimMc
##### Well-Known Member
My immediate thought - don't spend any money on a loudspeaker system until you can get an in person review and walk-thru of the auditorium with a person who does audio for a living.
#### MNicolai
##### Well-Known Member
Fight Leukemia
My immediate thought - don't spend any money on a loudspeaker system until you can get an in person review and walk-thru of the auditorium with a person who does audio for a living.
^^This.
Different systems for different applications, and it’s pretty rare that the best course of action is simply hanging new speakers.
Also, brand loyalty has its benefits in terms of ensuring you don’t left up a creek, but it’s not a good way to select speakers. Nor are line arrays usually a good choice. Line arrays simply for the reason of wanting line arrays may make your sound system *look* louder, but that doesn’t mean your system will actually sound any better. In the wrong application — which is what many line arrays get used in — it often actually makes your system sound worse.
In a vacuum, the QSC K.2 speakers are a great product, but I would almost never recommend them as an installation speaker in a large performance space. Generally if I’m spec’ing them for something, it’s a dance studio or a rehearsal room, and I tend to stay away from the 8’s because they have too wide of a pattern and don’t reach as low — especially if you don’t have subs to fill out the LF with. ——— But again, that’s in a vacuum without any specific consideration for a particular room and is not any value statement on whether they would or would not be suitable for whatever your particular use case is.
#### tjrobb
##### Well-Known Member
Echoing (haha) the above, get a pro in your space. Lighting you can fudge a little, but acoustics are much less forgiving. You need to find good products that people understand the quirks. If no one really knows the response curve, you'll end up with acoustic mush.
macsound and TimMc
#### nanced
##### New Member
Contact Starsound Audio in Reno. They'll come check out the venue and help get what you need.
#### DrewE
##### Well-Known Member
I'll go out on a limb and say that four point-source speakers (particularly ones without great pattern control, like the K8.2s) in what sounds like a pseudo-array are almost certainly not what you really want. They're decent enough speakers where appropriate to the situation; but this very likely isn't one of them. Rather, it sounds like a recipe for all sorts of comb filtering and muddiness.
I also suggest what others have said: get someone who is truly competent to do this sort of work to help you spec out your speaker(s). There are a myriad of details that should be considered, beyond simply the length and width of the house: how live it is, how high the ceiling is, how the seating is laid out, what sorts of productions you'll be doing (lectures and rock concerts have very different requirements!), the budget, and so on. A knowledgeable professional will understand them well. At best, I could only make a rather hazy guess myself. (There may also be some acoustic treatments for the room that would make a world of difference.)
#### MNicolai
##### Well-Known Member
Fight Leukemia
Echoing (haha) the above, get a pro in your space. Lighting you can fudge a little, but acoustics are much less forgiving. You need to find good products that people understand the quirks. If no one really knows the response curve, you'll end up with acoustic mush.
There's also a dirty little trick that with decent control from a DSP, you can often extend the mileage out of an existing system of less-than-good products if it's at least time aligned and tuned correctly. Earlier in the pandemic I helped a local theater here that had a VRX center cluster, some EAW corner fills, and a weird assortment of JBL Control speakers -- none of which were particularly great choices for that shoebox of a room. In about 15 minutes in the room I determined the VRX crossovers were toast. Probably had been that way for years -- but their resident audio tech hadn't really done any critical listening to the system to realize that the VRX speakers were hardly putting out sound when they should've been able to peel the paint off the walls.
The theater replaced those crossovers, slid some speakers around, and hung some curtains near their corner fills where they had lots of comb filtering from their corner fills that were interacting with the side walls they were near. I brought my personal DSP and Smaart rig in, we tuned and time aligned every amp channel independently, established the proper gain structure, and then transposed those settings into their the output processing for their mix console. Complete night and day difference spending only about $400 in parts and largely reusing what they already had on-hand. It's not often you get that lucky, but it is pretty common that venues aren't even in the ballpark of being tuned correctly and are getting only maybe 30-50% of the performance out of their existing system as they could be getting if it was optimized with minor adjustments. Even a large number of professional integrators don't tune with Smaart or anything like it. Not uncommon they tune with an RTA or sometimes just an app on their phone, or by earballing to a favorite music track they like -- which tuning to a specific music track is a terrible idea because people tend to try to remaster the track to how they like it sounding and not how it was originally produced. All of which is to reiterate that speakers are only one piece of a larger jigsaw puzzle for having a coherent, balanced, and flat sound reinforcement system. #### ACTSTech ##### Well-Known Member Thanks for the help and suggestions. I have to chuckle at what most of you have said, and I fully agree. To be honest, I often print these discussions to take to the board so that people see that there's experts who share my opinions. Then they most often do whatever they want anyhow, but at least I can say I, and many others, told you so after they're proven wrong. I'm not sold on the QSC's in all honesty. The reason I chose them was more budget over anything else as well as looking at who the end users will be. The design team for the renovation have already changed designs after the final plans were drawn so I don't feel comfortable saying what the venue will be. Nor do I feel comfortable paying an acoustic engineer to come in now, before anything has started, at$3000-\$5000 to tell me what they think the venue will need when the plans are going to change. Not against paying (I posted this earlier), but not now.
In terms of getting someone to consult... I contacted five vendors I have worked with in the past and three additional vendors that the electrical engineer who did the drawings recommended. The ones I have worked with all said it's too early to come in, which I understood. The three that came recommended... well none of them want to deal with a small venue. One said if I wasn't looking at Martin or Meyer rigs it wasn't worth his time. Another said not interested, and the third that actually came brought Bose room speakers to demonstrate. Two vendors contacted me after seeing RFQs, and neither really seemed interested. One insisted that Danley was the only way to go in a "small venue" and the other had some "spare" Renkus Heinz IC2-FRs that he thought would integrate with some EVs he had left over from a high school install. After he heard my take on the install, basically what was in the plans and specs which he obviously never read, he thought maybe we'd need more...
I wasn't really fishing here, but I was just curious what others thought. @DrewE , you made a good point about the size of the room. The reason I can't commit to what the room will look like is because the room keeps changing. The architect insists that we need plywood clouds to help with the sound, but he can't explain why he has them where he has them other than "this is what I typically draw up". The walls were originally wood, then became rockwool behind drywall, then became drywall with fiberglass acoustic panels that were more for aesthetics than noise, then a mix of the above. Nothing that he's drawn is acoustically-minded, it's aesthetically-minded.
The underlying problem is that we won't know what the room is until it's done. Then we can proceed. However, if we need a budget to apply for loans and grants, and it's not in line with what's out there available for grants and loans, we're going to be stuck. That's why I'm trying to cobble together my own specs and get some opinions. I'll HAPPILY turn everything over to the pros when it's time to do the install, but if I don't have a budget for them to work with, it's going to be tough.
Again, I appreciate all the input and I am making notes on my drawing as I go, so please don't hold back in roasting me. I can take it.
#### ACTSTech
##### Well-Known Member
Not uncommon they tune with an RTA or sometimes just an app on their phone, or by earballing to a favorite music track they like -- which tuning to a specific music track is a terrible idea because people tend to try to remaster the track to how they like it sounding and not how it was originally produced.
Guilty, sort of...
When I first got involved with this group, they were having MAJOR problems with their sound system, specifically the wireless body packs and terrible feedback all around. First thing I did was played my favorite track from a recording (CD, Uncompressed, not mp3) and heard how it sounded. Their sound guy wanted to know what he needed to change in the EQ, which was how they addressed all their issues. After resetting the board back to factory state and getting rid of all the mistakes they'd piled on top of mistakes, it still sounded bad. Went to the amp, undid the DSP settings the installer applied, retuned the room correctly, there was a world of difference. Not amazing, but incredibly better. Yes, their sound guy made mistakes, but the professional installer set them up for failure. He apparently slapped some pre-set notion of what the room should sound like on the amp and really hurt them. But when he played music for them, he showed them how to use the EQ, make the smiley face with the sliders, and took their money. You make a good point. Earball after you do the hard work, and go by the mastered version, not your car stereo.
I'm putting this in the report to the board, thanks!
#### TimMc
##### Well-Known Member
Guilty, sort of...
When I first got involved with this group, they were having MAJOR problems with their sound system, specifically the wireless body packs and terrible feedback all around. First thing I did was played my favorite track from a recording (CD, Uncompressed, not mp3) and heard how it sounded. Their sound guy wanted to know what he needed to change in the EQ, which was how they addressed all their issues. After resetting the board back to factory state and getting rid of all the mistakes they'd piled on top of mistakes, it still sounded bad. Went to the amp, undid the DSP settings the installer applied, retuned the room correctly, there was a world of difference. Not amazing, but incredibly better. Yes, their sound guy made mistakes, but the professional installer set them up for failure. He apparently slapped some pre-set notion of what the room should sound like on the amp and really hurt them. But when he played music for them, he showed them how to use the EQ, make the smiley face with the sliders, and took their money. You make a good point. Earball after you do the hard work, and go by the mastered version, not your car stereo.
I'm putting this in the report to the board, thanks!
I've seen the "generic DSP" settings applied in too many installations, but the one that took the prize was the auditorium that had multiple DSP devices installed in the amp racks, but only 2 of them had audio routed through them... and the processing was only bare-minimum crossover settings. I spent a couple of hours and used 2 XLR cables from their inventory to make audible improvements to their rig. I'd been brought in because they'd had a lightning strike that gave their DSP some kind of electronic amnesia... ironically one of the unused (but the pilot light was on!) units retained its settings so I manually transferred them and proceeded with my system tuning.
They had a number of system design problems that could not be addressed for reasons of physics and venue internal politics, so I did the best I could and left them with a working rig that sounded better than I'd found it. Sometimes the 'wins' are incremental.
macsound
#### FMEng
##### Well-Known Member
Fight Leukemia
Here's a quick story for the board. A mentor of mine was on a committee that was overseeing the design of a college auditorium/gymnasium of around 3000 seats, the largest in the area at the time. Nearing the construction phase, they were going through the cost cutting exercise. The sound system became a potential target for cost cutting.
My friend sat quietly and listened as the committee started taking the hatchet to the sound budget, when they realized their resident sound expert hadn't said anything. So they asked him for his opinion, which he returned with a question. The question was, "is there budget for more restrooms?" (Restrooms being one of the more expensive things to construct, due to plumbing and ventilation.)
Naturally, the answer was no, just enough toilets to meet code. The committee grew curious as to why he asked the question. He went on to explain that when people can't adequately hear the entertainment they get bored. When they get bored, their mind wanders and they focus on themselves. When they focus on themselves, they'll notice the need to pee, and the restrooms will be used heavily, requiring more of them. He got his sound budget.
Why build an auditorium at all if the acoustics and sound system are going to be last minute and done on the cheap?
#### ACTSTech
##### Well-Known Member
Why build an auditorium at all if the acoustics and sound system are going to be last minute and done on the cheap?
Agree and disagree.
This group is amateur, but after 10 years of not losing money on any of their shows, they got, perhaps unfairly, targeted by local governments and lost their performance venue. I'll spare details, but the choice was to either fold or try to find a new home. They're attempting the latter. I applaud their tenacity and drive. I get frustrated by their short-sightedness, but I work because I enjoy the work.
Also, this isn't going to be a world class venue. This will never compete and bring in touring shows or big names. The house system would never work for those anyhow, it would be rented in for anything to bring it to spec. They need something that will work. And unfortunately, the "experts" in the area see dollar signs, not functionality. It's like the power windows in the rear seats of your car when you're the only person who will ever ride in the car. Sure, it's a nice option, but why should I pay for something I'll never use. Do I really need heated rear seats as well? It's sort of value engineering at this point. I know I'll get blasted again for saying that, but for amateurs, you don't need A1 quality. The sad fact is that poor quality is ACCEPTED here. A high school streamed their musical last month and only half of the wireless mics came through, no one seemed to care. Everything gets a standing ovation. All I'm trying to do is get to far above average for now, and I know it's less than idea, but if it's between being above average or out of business, I'll choose the former.
##### Well-Known Member
Fight Leukemia
but for amateurs, you don't need A1 quality.
On a point of order, amateurs aspire to achieve A1 quality but accept the reality of finite resources. Performance art does not require premium theatre technology. It requires people willing to share themselves with an audience. Anything extra is a bonus.
#### ACTSTech
##### Well-Known Member
On a point of order, amateurs aspire to achieve A1 quality but accept the reality of finite resources. Performance art does not require premium theatre technology. It requires people willing to share themselves with an audience. Anything extra is a bonus.
Amateurs perform for the love of performing. If they had no option, they'd find a barn or grange hall or someplace. We're trying to get them into a building that's not ideal nor probably ever will be, not because they want a state of the art venue, but because they want a home.
#### tjrobb
##### Well-Known Member
Amateurs perform for the love of performing. If they had no option, they'd find a barn or grange hall or someplace. We're trying to get them into a building that's not ideal nor probably ever will be, not because they want a state of the art venue, but because they want a home.
As was stated when our home was (temporarily) lost to a flood: "[We are] more than just a building, we're a community." We played in high schools, one memorable middle school, and finally a movie threatre-cum-Staples that was abandoned. We built a theatre from a concrete slab inside a building. And our office ran out of a former funeral home...
ACTSTech
#### ACTSTech
##### Well-Known Member
I just had a long conversation with a gentlemen who asked to not be named on here, but was very helpful, and I thank him. I appreciate the help and input, and more than anything, the therapy of actually venting and vocalizing my thoughts.
We played in high schools, one memorable middle school, and finally a movie threatre-cum-Staples that was abandoned.
As stated, the group TRIED to rent several high schools who refused, mostly because their either don't want outside groups or the stage is the only rehearsal space for their groups so nothing on the stage can be permanent or left there (one recently lost their band rehearsal room to become a wrestling practice room). Most churches won't allow groups any longer because they run too many services and don't want things longer term. They ended up renting a small venue that had three weekends open (then tried to take back the middle weekend so they could have a Red-Hat Ladies meeting Friday and a baby shower Saturday) but Covid shut them down after weekend one. If this building gets done, it will serve a need for not just one group.
Again, I'm not trying to go in the cheapest route, just the most feasible at the moment.
#### macsound
##### Well-Known Member
Since there's been tons of discussion about the right way to do things, which should be followed, acoustician, engineering etc.,
I think you're doing the right thing here - ask questions of the architect on things that will possibly create a worse listening environment. In general, very general, large swaths of reflective material like wood, shouldn't be installed in venues that are going to be highly amplified. They should be installed in highly acoustic environments like those used for a symphony.
In general, again very generally, the more you plan on amplifying, the more diffusion and absorption you should have especially in the areas you can envision the speaker audio hitting soonest. This is why padded seats are nice.
What I would recommend for your lower budget, somewhat DIY install is a speaker that has a rotatable horn. This will, at the least, save you from wherever your engineer decides your mounting point is, only to realize it's 3' from a wall. In typical DIY fashion, the installer would point the speaker away from this reflective surface, only to point it directly at the stage, forever causing feedback for anyone close to the lip. At least if you spec'd 3-4 cabinets with rotatable horns you can decide, once the building is built, that LCR makes more sense than LR since you're going to have only 50degrees of width on the sides and need to make up for that with a center channel.
my 2c
#### TimMc
##### Well-Known Member
Time for a book plug...
Welll, it seems if I put the Amazon URL in, the forum software converts it to a media tag with no URL.
F. Alton Everest/Ken Pohlmann "Master Handbook of Acoustics, Sixth Ed" McGraw-Hill
Mastery of each page is not required, but grokking a concept is easier when one understands the scope.
In this case the room is the biggest factor in the results and whatever loudspeaker wizardry might be needed, cannot be determined until the architect has done his/her/their damage.
Last edited:
ACTSTech
#### ACTSTech
##### Well-Known Member
In general, very general, large swaths of reflective material like wood, shouldn't be installed in venues that are going to be highly amplified. They should be installed in highly acoustic environments like those used for a symphony.
Since this was an old church, the walls are wood panels over plaster. Highly reflective. Adding more clouds over the area where the pit orchestra might sit sounds like a recipe for disaster. Especially since when I ask how the panels are constructed and aimed, I'm told they're "a beautiful golden wood" but the positioning is more aesthetic than functional. Also, they're not movable once they're up, so if they're reflecting in the wrong direction, too bad.
In general, again very generally, the more you plan on amplifying, the more diffusion and absorption you should have especially in the areas you can envision the speaker audio hitting soonest. This is why padded seats are nice.
I set up the speakers that the group has just to demonstrate some of the issues. I explained that even though the room isn't constructed, that they could hear some of the problems that already exist. After playing a few things, they asked if maybe I could just turn up the volume... And I did... And then the lightbulbs went on. They finally got that louder does not equal better. More speakers also will not equal better. So they're getting it. Then they started asking questions, like why do they hear "that weird echo" behind them. When I explained that the front of the balcony is a great flat surface to slap the sound off of, they understood why we need some treatment of some sort.
What I would recommend for your lower budget, somewhat DIY install is a speaker that has a rotatable horn.
How about the "professional installation company" that told me they have a 120 x 60 degree coverage pattern speaker that would be great mounted "anywhere"
I appreciate and agree with you 100% Thank you for the recommendations as well.
#### macsound
##### Well-Known Member
I'm also feeling that there's probably a good amount of politics going on with the decision making by management.
Since you did a demo with what I'm assuming are the QSCs on sticks, would they be open to hearing other demos by local distributors?
Also, how are you going to try and convince them of improved acoustics? Maybe you could rent a couple legs and hang them at 45s in the back of the room so they can hear what it's like to add some absorption? | 2021-05-11 20:27:24 | {"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": 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": 0.23063156008720398, "perplexity": 1981.2717505690343}, "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-2021-21/segments/1620243989856.11/warc/CC-MAIN-20210511184216-20210511214216-00423.warc.gz"} |
https://mathhelpboards.com/threads/nicks-question-at-yahoo-answers-regarding-finding-the-point-on-a-plane-closest-to-a-given-point.7602/ | # Nick's question at Yahoo! Answers regarding finding the point on a plane closest to a given point
#### MarkFL
Staff member
Here is the question:
Need help with calculus problem? Optimization.?
Find the point on the plane x - 2y +3z = 6 that is closest to the point (0,1,1).
I can't figure this out. Does anyone know how to do this? Thanks.
I have posted a link there to this thread so the OP can see my work.
#### MarkFL
Staff member
Hello Nick,
Let's let our objective function be the square of the distance from the point on the plane $(x,y,z)$ to the given point $(0,1,1)$:
$$\displaystyle f(x,y,z)=x^2+(y-1)^2+(z-1)^2$$
The point $(x,y,z)$ is constrained to be on the given plane, hence our constraint is:
$$\displaystyle g(x,y,z)=x-2y+3z-6=0$$
Using Lagrange multipliers, we obtain the system:
$$\displaystyle 2x=\lambda$$
$$\displaystyle 2(y-1)=-2\lambda$$
$$\displaystyle 2(z-1)=3\lambda$$
This implies:
$$\displaystyle \lambda=2x=1-y=\frac{2}{3}(z-1)$$
From this, we obtain:
$$\displaystyle y=1-2x,\,z=3x+1$$
Substituting for $y$ and $z$ into the constraint, we find:
$$\displaystyle x-2(1-2x)+3(3x+1)-6=0$$
Solving for $x$ (and using the values for $y$ and $z$ in terms of $x$) we find:
$$\displaystyle x=\frac{5}{14}\implies y=\frac{2}{7},\,z=\frac{29}{14}$$
Thus, the point on the plane $x-2y+3z=6$ that is closest to the point $(0,1,1)$ is:
$$\displaystyle \left(\frac{5}{14},\frac{2}{7},\frac{29}{14} \right)$$ | 2020-11-27 05:26:02 | {"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": 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": 0.7686231136322021, "perplexity": 327.5029320874626}, "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-2020-50/segments/1606141189141.23/warc/CC-MAIN-20201127044624-20201127074624-00649.warc.gz"} |
https://math.stackexchange.com/questions/1765764/brownian-motion-with-l%C3%A9vy-s-characterization | Brownian motion with Lévy’s Characterization
I want to show that: if for all $\lambda \in \mathbb{R}$ the process $\left(\exp\left(\lambda X_t-\frac{\lambda ^2}{2}t\right)\right)_{t\geq0}$ is a $\mathcal{F}^X$ local martingale, then the $\mathbb{R}$-valued process $X$ is a $\mathcal{F}^X$-Brownian motion
I think it is enough to show that $X_t$ is a continuous local martingale and that for all $t$: $[X]_t=t$, because then by Lévy's Characterization $X$ has to be a Brownian motion.
I tried using Ito's Lemma but couldn't get there.
• A start: Assuming that $X$ is a continuous semimartingale, use $M_t$ to denote the local martingale $\exp(\lambda X_t-\lambda^2t/2)$. Now apply Ito to $\log M$, and use the resulting development to relate $[X]$ and $[M]$. – John Dawkins Apr 30 '16 at 19:07
• Do you assume that $(X_t)_{t \geq 0}$ has continuous sample paths? – saz Apr 30 '16 at 19:57
• Thank you for the hint, but this is the exact step i got stuck at. I got: $\lambda^2 [X]_t = \frac{1}{2} \int_0^t exp(-2(\lambda X_s + \frac{\lambda^2 s}{2}))d[M]_s$ @saz: Yes, i think so – Adsertor Justitia May 1 '16 at 12:39
• Now use this and the Ito expansion of $\log M$ to show that the finite variation part of $X$ must vanish, meaning that $X$ is a local martingale. – John Dawkins May 1 '16 at 15:49
• Thank you, this did help a lot! – Adsertor Justitia May 1 '16 at 22:21 | 2019-05-24 13:02:34 | {"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": 0.9246687889099121, "perplexity": 251.01217947477326}, "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-2019-22/segments/1558232257624.9/warc/CC-MAIN-20190524124534-20190524150534-00368.warc.gz"} |
https://motls.blogspot.com/2007/03/berlin-declaration-50-years-of-eu.html?m=1 | ## Sunday, March 25, 2007
### Berlin Declaration: 50 years of EU
Congratulations to the 50th anniversary of the European Union. I guess that the early versions of the Berlin Declaration were full of "Orwellian Eurospeak" as the Czech president called in on Friday but the
seems kind of smooth to me and avoids the eurobureaucratic verbal constructs that are often inserted to replace words in order to give the officials an unjustified feeling that they're a special elite. It is conceivable that Angela Merkel's people simply had to erase the eurospeak.
The term "social responsibility" appears once which is OK with me. But there are some typos in it anyway. For example, in this paragraph:
• We intend jointly to lead the way in energy policy and climate protection and make our contribution to averting the global threat of climate change.
the word "religion" is missing at the very end. Hasn't anyone noticed this typo, you may ask? Well, some people did but they were only given 24 hours to verify the document which is a "classical example of a lack of democracy".
In the last big paragraph, the text says that we are united in our desire to create a new common basis before the 2009 euroelections. Well, I think that the sentence is untrue - Europe is certainly not united in the opinion that something like a new constitution, old or new, is needed soon. The Poles, Czechs, and Englishmen don't feel any need for a new document while the Dutch and the French will reject it in the referendum. ;-)
But on the other hand, the formulation is so vague and watered down that it won't represent any problems. An otherwise worthless piece of paper was signed by two Germans and one non-German. Twenty-seven heads of countries burned some fossil fuels and spent one day to see this theater in Berlin and nothing else happened. Congratulations anyway! | 2019-12-09 02:29:04 | {"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": 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": 0.26024267077445984, "perplexity": 1678.2401258176567}, "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/1575540517156.63/warc/CC-MAIN-20191209013904-20191209041904-00117.warc.gz"} |
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# How much water should the milkman add to milk purchased
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How much water should the milkman add to milk purchased [#permalink] 03 Nov 2012, 07:46
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How much water should the milkman add to milk purchased by him for $20/- to get 50% profit? (1) He has 1 litre of milk. (2) He wishes to sell the diluted milk at$25 a litre.
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Re: How much water should the milkman add to milk purchased [#permalink] 05 Nov 2012, 11:02
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monikaleoster wrote:
How much water should the milkman add to milk purchased by him for $20/- to get 50% profit? (1) He has 1 litre of milk. (2) He wishes to sell the diluted milk at$25 a litre.
I marked A. can anyone explain Why it is wrong
I'm happy to help with this.
Suppose we know only (A) ----- we know he bought 1 liter of milk for $20. We know he wants a 50% profit, which means making$30 in sales. What we don't know is how much he will charge for the milk he sells. If he sells the milk for $30/L, then he doesn't have to add any water, and he will make a 50%. If he sells the milk at$60/L, he could drink half of it, and sell the other half and still reap a 50% profit. I think you were assuming that the buying price and the selling price were the same ---- it would be very good preparation for your future, getting an MBA and after, to recognize that these two seldom are the same.
We need to know both how much milk and selling price, in order to answer the question. That's why (C) is the answer.
Does all this make sense?
Mike
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Re: How much water should the milkman add to milk purchased [#permalink] 10 Jul 2013, 19:52
mikemcgarry wrote:
monikaleoster wrote:
How much water should the milkman add to milk purchased by him for $20/- to get 50% profit? (1) He has 1 litre of milk. (2) He wishes to sell the diluted milk at$25 a litre.
I marked A. can anyone explain Why it is wrong
I'm happy to help with this.
Suppose we know only (A) ----- we know he bought 1 liter of milk for $20. We know he wants a 50% profit, which means making$30 in sales. What we don't know is how much he will charge for the milk he sells. If he sells the milk for $30/L, then he doesn't have to add any water, and he will make a 50%. If he sells the milk at$60/L, he could drink half of it, and sell the other half and still reap a 50% profit. I think you were assuming that the buying price and the selling price were the same ---- it would be very good preparation for your future, getting an MBA and after, to recognize that these two seldom are the same.
We need to know both how much milk and selling price, in order to answer the question. That's why (C) is the answer.
Does all this make sense?
Mike
Mike,
Do you think this is real gmat Question as it has too many assumptions in DS ?
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Re: How much water should the milkman add to milk purchased [#permalink] 11 Jul 2013, 13:47
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Mountain14 wrote:
Mike,
Do you think this is real gmat Question as it has too many assumptions in DS ?
Dear Mountain14,
I think the framing of the question may be a little too folksy for the GMAT, but the substance of the question is perfectly legitimate. For example, the question could easily re-phrased in terms of some sort of obscure scientific liquid that a scientist is going to reduce and sell at a profit --- that would be very GMAT-like, and the principles of solution would be identical. We don't know to know anything about the liquid, its use or how to handle it, in order to answer a question about diluting it to achieve a certain profit. I think the only real assumption is that water is free, which seems like a quite reasonable assumption to make.
Does this make sense?
Mike
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Re: How much water should the milkman add to milk purchased [#permalink] 11 Jul 2013, 18:14
mikemcgarry wrote:
Mountain14 wrote:
Mike,
Do you think this is real gmat Question as it has too many assumptions in DS ?
Dear Mountain14,
I think the framing of the question may be a little too folksy for the GMAT, but the substance of the question is perfectly legitimate. For example, the question could easily re-phrased in terms of some sort of obscure scientific liquid that a scientist is going to reduce and sell at a profit --- that would be very GMAT-like, and the principles of solution would be identical. We don't know to know anything about the liquid, its use or how to handle it, in order to answer a question about diluting it to achieve a certain profit. I think the only real assumption is that water is free, which seems like a quite reasonable assumption to make.
Does this make sense?
Mike
I got what you mean.... Thanks...
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Re: How much water should the milkman add to milk purchased [#permalink] 21 Aug 2013, 03:38
Expert's post
anamika7 wrote:
Its still not clear to me why c is the answer. Please can anyone provide a detailed explanation.
Please elaborate what didn't you understand in the solution. Thank you.
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Re: How much water should the milkman add to milk purchased [#permalink] 21 Aug 2013, 04:07
Following thing is not clear to me:
How can milkman earn 50% profit by selling at $60. I marked A as the answer. For option B: SP=$25. CP= $20. Profit will be only 25% . I do not have good understanding of mixture problems. So please explain how the two statements are together sufficient Manager Joined: 14 Mar 2013 Posts: 50 Location: United States Concentration: General Management, Leadership GMAT Date: 12-03-2013 WE: General Management (Retail) Followers: 0 Kudos [?]: 29 [1] , given: 119 Re: How much water should the milkman add to milk purchased [#permalink] 19 Nov 2013, 14:09 1 This post received KUDOS anamika7 wrote: Following thing is not clear to me: How can milkman earn 50% profit by selling at$60.
I marked A as the answer.
For option B: SP= $25. CP=$20. Profit will be only 25% .
I do not have good understanding of mixture problems. So please explain how the two statements are together sufficient
The question asks how much water should the milkman add to milk in order to get extra $10. He is not selling by$60, he paid $20 for X liters of milk and will sell Y liters of the mix milk+water to get$10 profit (50% of $20). To answer this, you need to know how much milk he has and for how much he is going to sell the mix of milk+water. Only 1 or only 2 is not enough to answer, but using 1 and 2 together we can answer the question. Intern Joined: 03 Aug 2013 Posts: 1 Followers: 0 Kudos [?]: 0 [0], given: 11 Re: How much water should the milkman add to milk purchased [#permalink] 06 Jan 2014, 19:02 mikemcgarry wrote: monikaleoster wrote: How much water should the milkman add to milk purchased by him for$20/- to get 50% profit?
(1) He has 1 litre of milk.
(2) He wishes to sell the diluted milk at $25 a litre. I marked A. can anyone explain Why it is wrong I'm happy to help with this. Suppose we know only (A) ----- we know he bought 1 liter of milk for$20. We know he wants a 50% profit, which means making $30 in sales. What we don't know is how much he will charge for the milk he sells. If he sells the milk for$30/L, then he doesn't have to add any water, and he will make a 50%. If he sells the milk at $60/L, he could drink half of it, and sell the other half and still reap a 50% profit. I think you were assuming that the buying price and the selling price were the same ---- it would be very good preparation for your future, getting an MBA and after, to recognize that these two seldom are the same. We need to know both how much milk and selling price, in order to answer the question. That's why (C) is the answer. Does all this make sense? Mike Hi Mike, As the price of water is not given, should we assume it as free of cost or should we opt option 'E'. Satish. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5459 Location: Pune, India Followers: 1337 Kudos [?]: 6796 [0], given: 177 Re: How much water should the milkman add to milk purchased [#permalink] 06 Jan 2014, 21:25 Expert's post satishv80 wrote: mikemcgarry wrote: monikaleoster wrote: How much water should the milkman add to milk purchased by him for$20/- to get 50% profit?
(1) He has 1 litre of milk.
(2) He wishes to sell the diluted milk at $25 a litre. I marked A. can anyone explain Why it is wrong I'm happy to help with this. Suppose we know only (A) ----- we know he bought 1 liter of milk for$20. We know he wants a 50% profit, which means making $30 in sales. What we don't know is how much he will charge for the milk he sells. If he sells the milk for$30/L, then he doesn't have to add any water, and he will make a 50%. If he sells the milk at \$60/L, he could drink half of it, and sell the other half and still reap a 50% profit. I think you were assuming that the buying price and the selling price were the same ---- it would be very good preparation for your future, getting an MBA and after, to recognize that these two seldom are the same.
We need to know both how much milk and selling price, in order to answer the question. That's why (C) is the answer.
Does all this make sense?
Mike
Hi Mike,
As the price of water is not given, should we assume it as free of cost or should we opt option 'E'.
Satish.
The intent of this question is clear - water costs nothing.
But, it is not specifically mentioned. An official question will mention "assume water is available free of cost". I have seen DS questions which specify the price of water too. Hence, I would not assume that water is available free of cost until and unless it is specifically mentioned.
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Re: How much water should the milkman add to milk purchased [#permalink] 09 Apr 2015, 00:07
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Re: How much water should the milkman add to milk purchased [#permalink] 09 Apr 2015, 00:07
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https://www.encyclopediaofmath.org/index.php?title=Number&direction=next&oldid=11869 | # Number
A fundamental concept in mathematics, which has taken shape in the course of a long historical development. The origin and formulation of this concept occurred simultaneously with the birth and development of mathematics. The practical activities of mankind on the one hand, and the internal demands of mathematics on the other, have determined the development of the concept of a number.
The need to count objects led to the origin of the notion of a natural number. All nations that have forms of writing had mastered the concept of a natural number and had developed some counting system. In its early stages, the origin and development of the concept of a number can be judged only from indirect data provided by linguistics and ethnography. Primitive man clearly had no need of counting skills to determine whether or not a given collection was complete.
Later, special names were given to definite objects or events that people often encountered. Thus, in the language of certain nations there were words for such concepts as "three men" , or "three boats" , but there was no abstract concept "three" . In this way, probably, there arose comparatively short number series, used for the identification of individual people, individual boats, individual coconuts, etc. At these stage there was no such thing as an abstract number, and numbers were merely names.
In some primitive cultures, the necessity to communicate information about the numerical size of this or that collection has led to distinguishing certain standard collections, usually consisting of parts of the human body. In such a system of counting, each part of the body has a definite order and name. Whenever parts of the body are insufficient, bundles of sticks are used. The same purpose is served by pebbles, shells, notches on a tree or rock, lines on the ground, strings with knots, etc.
The next stage in the development of the notion of a number is connected with the transition to counting in groups; pairs, tens, dozens, etc. There arise so-called nodal numbers, and at the same time the concept of an arithmetic operation, which is reflected in the names of numbers. Definite counting methods take shape, special devices are used for counting, and numerical notations emerge. Numbers are separated from the objects being counted, and become abstract. Systems of representations of numbers begin to appear (cf. Numbers, representations of).
The process of formation of our modern representation system was exceptionally complicated. Only the final part of this process can be judged with definite authenticity. There are many known systems of representation. In Ancient Egypt there were several systems. In one of them there were special symbols for 1, 10, 100, 1000. Other numbers were represented by means of combinations of these symbols. The basic arithmetic operation in Ancient Egypt was addition. Well before 2000 B.C. the Babylonians used a base-60 representation system with the positional principle for writing numbers. They used only two symbols. The ancient Greeks used an alphabetic representation system, which was also used by the Slavs (cf. also Slavic numerals). In India, at the beginning of the new era (A.D.) there was a wide-spread oral positional decimal representation system, with several synonyms for zero (and other digits). A positional decimal representation system also arose there later. By the 8th century A.D. this system had spread as far as the Middle East. The Europeans were introduced to it in the 12th century.
The widening circle of objects to be counted, arising as a result of practical activities of people, and finally the inquisitiveness that is characteristic of mankind, gradually pushed back the limits of counting. The idea arose of the unbounded extension of the sequence of natural numbers, possibly attributable to the Greeks. One of Euclid's theorems states: "There exist more than any given number of primes" . Also, Archimedes tried to convince his contemporaries that it is possible to describe a number greater than "the number of grains of sand in the world" .
For the measurement of quantities, fractional numbers were necessary. Fractions were studied in Ancient Egypt and Babylon. Egyptian fractions were usually expressed in terms of aliquot fractions, i.e. fractions with numerator equal to 1. The Babylonians used base-60 fractions. The Chinese and the Indians were using ordinary fractions in the early centuries A.D., and were able to carry out all the arithmetic operations on them. Scholars in Central Asia, no later than the 10th century, used a base-60 positional counting system. This system was particularly widely used in astronomical calculations and tables. Traces of it have been passed on to us in the form of the units used in the measurement of time and angles. Decimal fractions were introduced at the beginning of the 15th century, and were widely used by the Samarkand mathematician Kashi (al'-Kashi). In Europe, decimal fractions became widespread following the publication of the book de Thiende (1585), written by S. Stevin. Before the introduction of decimal fractions, the Europeans had used the decimal system in practice to calculate the integer part of a number, but they used base-60 fractions or ordinary fractions for the fractional part.
The further development of the concept of number proceeded mainly along with the demands of mathematics itself. Negative numbers first appeared in Ancient China. Indian mathematicians discovered negative numbers while trying to formulate an algorithm for the solution of quadratic equations in all cases. Diophantus (3rd century) operated freely with negative numbers. They appear constantly in intermediate calculations in many of the problems in his Aritmetika. In the 16th century and 17th century, however, many European mathematicians did not appreciate negative numbers, and if such numbers occurred in their calculations they were referred to as false, or impossible. The situation changed in the 17th century, when a geometric interpretation of positive and negative numbers was discovered, as oppositely-directed segments.
The Babylonians had an algorithm for calculating the square root of a number to any accuracy. In the 5th century B.C., Greek mathematicians discovered that the side and diagonal of a square have no common measure. More generally, it turned out that two arbitrary, precisely-given segments are in general not commensurable. The Greek mathematicians did not start introducing new numbers. They avoided the above difficulty by creating a theory of ratios of segments that was independent of the concept of a number.
The development of algebra, and the techniques of approximate calculation, in connection with the demands of astronomy, led Arab mathematicians to extend the concept of a number. They began to consider ratios of arbitrary quantities, whether commensurable or not, as numbers. As Nasireddin (1201–1278) wrote: "Each of these ratios can be called by a number, precisely equal to the number one when one term of the ratio agrees with the other term" . European mathematics was developing in the same direction. Although G. Cardano in Practica Arithmeticae Generalis (1539) was still writing about irrational numbers as "surds" (from the Latin "surdus" , "deaf" ), and as "impossible to perceive or to imagine" , Stevin in his l'Arithmétique (1585) stated that "a number is that which is determined by an arbitrary quantity" and that "no numbers are absurd, irrational, irregular, inexpressible, or surd" . And finally I. Newton in his Arithmeticae Universalis (1707) gave the following definition: "By a number we understand not so much a multiple of a unit as an abstract quantity associated in a systematic way to some other quantity of the same kind that is taken as a unit. Numbers arise in three forms: integer, fraction and irrational. An integer is that which can be measured by unity; a fraction is a multiple of a portion of unity; an irrational number is incommensurable with unity" . Related to the fact that, for Newton, a quantity can be either positive or negative, the numbers in his arithmetic can also be either positive, in other words "greater than nothing" , or negative, in other words "smaller than nothing" .
Imaginary numbers first appeared in the work of Cardano, Ars Magma (The Great Art, 1545). In solving the system of equations , , he found the solutions and . Cardano called these solutions "purely negative" , and later "sophisticatedly negative" . The first to see a "real" use of introducing imaginary numbers was R. Bombelli. In his Algebra (1572) he showed that the real roots of the equation , , , in the case , can be expressed in terms of radicals of imaginary numbers. Bombelli defined arithmetic operations on such quantities, and proceeded in this way to the creation of the theory of complex numbers (cf. Complex number). In the 17th century and 18th century, many mathematicians occupied themselves in investigating the properties of imaginary numbers and their applications. Thus, L. Euler extended, e.g., the notion of a logarithm to arbitrary complex numbers (1738), and obtained a new method of integration using complex variables (1776), while earlier (1736) A. de Moivre solved the problem of extracting roots of natural degrees of an arbitrary complex number. A successful application of the theory of complex numbers was the fundamental theorem of algebra: "Every polynomial of degree greater than zero and with real coefficients factorizes as a product of polynomials of degrees one and two with real coefficients" (Euler, J. d'Alembert, C.F. Gauss). Nevertheless, until a geometric interpretation of complex numbers as points in the plane was given (around the end of the 18th century and in the beginning of the 19th century), many mathematicians remained distrustful of imaginary numbers.
In the early 19th century, in connection with the great successes of mathematical analysis, many scholars realized the need for a foundation of the fundamentals of analysis — the theory of limits. Mathematicians were no longer satisfied with proofs based on intuition or on geometrical representation. There also remained the problem of constructing a unified theory of numbers. Natural numbers were often thought of as collections of unity, fractions as ratios of quantities, real numbers as lengths of line segments, and complex numbers as points in the plane. There was no complete agreement as to how arithmetic operations on numbers should be introduced. Finally, the question naturally arose of the further development of the concept of a number. In particular, was it possible to introduce new numbers, related to points in space?
The 19th century saw intensive research in all the above directions. A general principle was formulated according to which any generalization of the concept of number should proceed — the so-called principle of permanence of formal computing laws. According to this, when constructing a new number system extending a given system, the operations should generalize in such a way that the existing laws remain in force (G. Peacock, 1834; H. Hankel, 1867). In the second half of the 19th century, the theory of real numbers (cf. Real number) was constructed almost simultaneously by G. Cantor (1879), Ch. Meray (1869), R. Dedekind (1872), and K. Weierstrass (1872). Here Cantor and Meray used Cuachy sequences of rational numbers, Dedekind used cuts in the field of rational numbers, and Weierstrass used infinite decimal expansions.
As a result of the work of G. Peano (1891), Weierstrass (1878) and H. Grassmann (1861), an axiomatic theory of natural numbers was constructed. W. Hamilton (1837) constructed a theory of complex numbers from pairs of real numbers, Weierstrass constructed a theory of integers from pairs of natural numbers, and J. Tannery (1894) constructed a theory of rational numbers from pairs of integers.
Attempts to find generalizations of the concept of a complex number led to the theory of hypercomplex numbers (cf. Hypercomplex number). Historically, the first such number system was the quaternions (cf. Quaternion), discovered by Hamilton. After much investigation it became clear (Weierstrass, G. Frobenius, B. Pierce) that any extension of the concept of a complex number beyond the system of complex numbers itself is possible only at the cost of some of the usual properties of numbers.
Throughout the 19th century, and into the early 20th century, deep changes were taking place in mathematics. Conceptions about the objects and the aims of mathematics were changing. The axiomatic method of constructing mathematics on set-theoretic foundations was gradually taking shape. In this context, every mathematical theory is the study of some algebraic system. In other words, it is the study of a set with distinguished relations, in particular algebraic operations, satisfying some predetermined conditions, or axioms.
From this point of view every number system is an algebraic system. For the definition of concrete number systems it is convenient to use the notion of an "extension of an algebraic system" . This notion makes precise in a natural way the principle of permanence of formal computing laws, which was formulated above. An algebraic system is called an extension of an algebraic system if the underlying set of is a subset of that of , if there also exists a bijection from the set of relations of the system onto that of , and if for any collection of elements of the system for which some relation of that system holds, the corresponding relation of the system also holds.
For example, by the system of natural numbers one usually understands the algebraic system with two algebraic operations: addition and multiplication , and a distinguished element
(unity), satisfying the following axioms:
1) for each element , ;
2) associativity of addition: For any elements in ,
3) commutativity of addition: For any elements in ,
4) cancellation of addition: For any elements in , the equation entails the equation ;
5) 1 is the neutral element for the multiplication; that is, for any one has ;
6) associativity of multiplication: For any elements in ,
7) distributivity of multiplication over addition: For any elements in ,
8) the axiom of induction: If is a subset of containing 1 and the element whenever it contains , then .
From , and it follows that the system of natural numbers is a semi-ring under the operations and . Hence the system of natural numbers can be defined as the minimal semi-ring with a neutral element for multiplication and without a neutral element for the addition.
The system of integers is defined as the minimal ring that is an extension of the semi-ring of natural numbers. The system of rational numbers is defined as the minimal field that is an extension of the ring . The system of complex numbers is defined as the minimal field that is an extension of the field of real numbers containing an element for which (cf. also Extension of a field).
By the system of real numbers one means the algebraic system with two binary operations and , two distinguished elements
and
and binary order relation . The axioms of are divided into the following groups:
1) the field axiom: The system is a field;
2) the order axiom: The system is a totally and strictly ordered field (cf. Ordered field);
3) the Archimedean axiom: For any elements , in there exists a natural number such that
4) the completeness axiom: Every Cauchy sequence of real numbers converges, i.e. if for any there is a number such that, for any and the inequality holds, then the sequence converges to some element of .
Briefly, the system of real numbers is a complete, totally, strictly-Archimedean ordered field. The system of real numbers can also be defined, in an equivalent way, as a continuous totally ordered field. In this case the Archimedean axiom and the completeness axiom are replaced by the continuity axiom:
If and are non-empty subsets of such that, for any elements , , the inequality holds, then there exists an element such that for all , .
The construction of real numbers proposed by Cantor and Meray can be used to interpret the first system of axioms for the system of real numbers, while Dedekind's construction can be used to interpret the second system. Analogously, the constructions of Hamilton, Weierstrass and Tannery are interpretations of the systems of axioms for the complex, integer and rational numbers.
As interpretations of the system of natural numbers one may use the ordinal theory of natural numbers developed by Peano, and the cardinal theory of natural numbers of Cantor.
The problem of the foundations of the concept of a number, and more broadly, the foundations of mathematics, were clearly set out in the 19th century. This problem became a subject of mathematical logic, the intensive development of which continued into the 20th century.
#### References
[1] E.I. Berezkina, "Mathematics of Ancient China" , Moscow (1980) (In Russian) [2] N. Bourbaki, "Eléments d'histoire des mathématiques" , Hermann (1960) [3] A.A. Vaiman, "Sumero-Babylonian mathematics" , Moscow (1961) (In Russian) [4] B.L. van der Waerden, "Ontwakende wetenschap" , Noordhoff (1957) [5] G. Wieleitner, "Die Geschichte der Mathematik von Descartes bis zum Hälfte des 19. Jahrhunderts" , 2 , de Gruyter (1923) [6] A.I. Volodarskii, "An outline of the history of Medieval Indian mathematics" , Moscow (1977) (In Russian) [7] M.Ya. Vygodskii, "Arithmetic and algebra in the Ancient world" , Moscow (1967) (In Russian) [8] I.Ya. Depman, "The history of arithmetic" , Moscow (1959) (In Russian) [9] E. Kol'man, "History of mathematics in Antiquity" , Moscow (1961) (In Russian) [10] F. Cajori, "A history of elementary mathematics" , Macmillan (1896) [11] V.I. Nechaev, "Number systems" , Moscow (1975) (In Russian) [12] K.A. Rybnikov, "A history of mathematics" , 1–2 , Moscow (1974) (In Russian) [13] D.J. Struik, "A concise history of mathematics" , 1–2 , Dover, reprint (1948) (Translated from Dutch) [14] S. Feferman, "The number system" , Addison-Wesley (1964) [15] H.G. Zeuthen, "Geschichte der Mathematik in XVI und XVII Jahrhundert" , Teubner (1903) [16] A.P. Yushkevich, "Geschichte der Mathematik im Mittelalter" , Teubner (1964) (Translated from Russian) [17] , The history of mathematics from Antiquity to the beginning of the XIX-th century , 1–3 , Moscow (1970–1972) (In Russian) [18] , Mathematics of the 19-th century. Mathematical logic. Algebra. Number theory. Probability theory , Moscow (1978) (In Russian) [19] E. Landau, "Grundlagen der Analysis" , Akad. Verlagsgesellschaft (1930) | 2019-10-21 20:00:41 | {"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": 0.8424696922302246, "perplexity": 718.2351392829576}, "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-2019-43/segments/1570987787444.85/warc/CC-MAIN-20191021194506-20191021222006-00077.warc.gz"} |
https://byjus.com/physics/projectile-motion/ | # Projectile Motion
## What is Projectile?
The projectile is any object thrown into space upon which the only acting force is gravity. In other words, the primary force acting on a projectile is gravity. This doesn’t necessarily mean that the other forces do not act on it, just that their effect is minimal compared to gravity. The path followed by a projectile is known as a trajectory. A baseball batted or thrown and the instant the bullet exits the barrel of a gun are all examples of the projectile.
## What is Projectile Motion?
When a particle is thrown obliquely near the earth’s surface, it moves along a curved path under constant acceleration that is directed towards the center of the earth (we assume that the particle remains close to the surface of the earth). The path of such a particle is called a projectile and the motion is called projectile motion. Air resistance to the motion of the body is to be assumed absent in projectile motion.
In a Projectile Motion, there are two simultaneous independent rectilinear motions:
1. Along the x-axis: uniform velocity, responsible for the horizontal (forward) motion of the particle.
2. Along y-axis: uniform acceleration, responsible for the vertical (downwards) motion of the particle.
Accelerations in the horizontal projectile motion and vertical projectile motion of a particle: When a particle is projected in the air with some speed, the only force acting on it during its time in the air is the acceleration due to gravity (g). This acceleration acts vertically downward. There is no acceleration in the horizontal direction, which means that the velocity of the particle in the horizontal direction remains constant.
### Parabolic Motion of Projectiles
Let us consider a ball projected at an angle θ with respect to the horizontal x-axis with the initial velocity u as shown below:
The point O is called the point of projection; θ is the angle of projection and OB = Horizontal Range or Simply Range. The total time taken by the particle from reaching O to B is called the time of flight.
For finding different parameters related to projectile motion, we can make use of differential equations of motions:
Total Time of Flight: Resultant displacement (s) = 0 in Vertical direction. Therefore, by using the Equation of motion:
gt2 = 2(uyt – sy) [Here, uy = u sin θ and sy = 0]
i.e. gt2 = 2t × u sin θ
Therefore, the total time of flight (t):
$Total\,Time\,of\,Flight\,(t)=\frac{2u\sin\Theta}{g}$
Horizontal Range: Horizontal Range (OA) = Horizontal component of velocity (ux) × Total Flight Time (t)
R = u cos θ × 2u×sinθg
Therefore, in a projectile motion the Horizontal Range is given by (R):
$Horizontal\,Range\,(R)=\frac{u^2\sin 2\Theta }{g}$
Maximum Height: It is the highest point of the trajectory (point A). When the ball is at point A, the vertical component of the velocity will be zero. i.e. 0 = (u sin θ)2 – 2g Hmax [s = Hmax , v = 0 and u = u sin θ]
Therefore, in projectile motion, the Maximum Height is given by (Hmax):
$Maximum\,Height\,(H_{max})=\frac{u^2\sin^2\Theta }{2g}$
The equation of Trajectory: Let, the position of the ball at any instant (t) be M (x, y). Now, from Equations of Motion:
x = t × u cos θ . . . . . . (1)
y = u sin θ × t – 12×t2g. . . . . . (2)
On substituting Equation (1) in Equation (2):
$Equation\,of\,Trajectory\,(y)=\frac{x\tan \Theta -gx^2}{2u^2\cos^2\Theta }$
This is the Equation of Trajectory in projectile motion, and it proves that the projectile motion is always parabolic in nature.
We know that projectile motion is a type of two-dimensional motion or motion in a plane. It is assumed that the only force acting on a projectile (the object experiencing projectile motion) is the force due to gravity. But how can we define projectile motion in the real world? How are the concepts of projectile motion applicable to daily life? Let us see some real-life examples of projectile motion in two dimensions.
All of us know about basketball. To score a basket, the player jumps a little and throws the ball in the basket. The motion of the ball is in the form of a projectile. Hence, it is referred to as projectile motion. What advantage does jumping give to their chances of scoring a basket? Now apart from basketballs, if we throw a cricket ball, a stone in a river, a javelin throw, an angry bird, a football or a bullet, all these motions have one thing in common. They all show a projectile motion. And that is, the moment they are released, there is only one force acting on them- the gravity. It pulls them downwards, thus giving all of them an equal impartial acceleration.
It implies that if something is being thrown in the air, it can easily be predicted how long the projectile will be in the air and at what distance from the initial point it will hit the ground. If the air resistance is neglected, there would be no acceleration in the horizontal direction. This implies that as long as a body is thrown near the surface, the motion of the body can be considered as a two-dimensional motion, with acceleration only in one direction. But how can it be concluded that a body thrown in air follows a two-dimensional path? To understand this, let us assume a ball that is rolling as shown below:
Figure 1 Motion in one dimension
Now, if the ball is rolled along the path shown, what can we say about the dimension of motion? The most common answer would be that it has an x-component and a y-component, it is moving on a plane, so it must be an example of motion in two dimensions. But it is not correct, as it can be noticed that there exists a line which can completely define the motion of the basketball. Thus, it is an example of motion in one dimension. Therefore, the choice of axis does not alter the nature of the motion itself.
Figure 2 Motion in Plane
Now, if the ball is thrown at some angle as shown, the velocity of the ball has an x-component and component and also a z-component. So, does it mean that it is a three-dimensional motion? It can be seen here that a line cannot define such a motion, but a plane can. Therefore, for a body thrown at any angle, there exists a plane that entirely contains the motion of that body. Thus, it can be concluded that as long as a body is near the surface of the Earth and the air resistance can be neglected, then irrespective of the angle of projection, it will be a two-dimensional motion, no matter how the axes are chosen. If the axes here are rotated in such a way that, then and can completely define the motion of the ball as shown below:
Thus, it can be concluded that the minimum number of coordinates required to completely define the motion of a body determines the dimension of its motion.
## Solved Example For You
### An object is launched at a velocity of 40 m/s in a direction making an angle of 50° upward with the horizontal.
Q1. What is the maximum height reached by the object?
Q2. What is the total flight time (between launch and touching the ground) of the object?
Q3. What is the horizontal range (maximum x above ground) of the object?
Solution:
The velocity components $V_{x}$ and $V_{y}$ are given by the formula:
In the given problem, $V_{0} = 40m/s$, ? = 50° and g is $9.8\,m^{2}$
The height of the projectile is given by the component y, and it reaches its maximum value when the component $V_{y}$ is equal to zero. That is when the projectile changes from moving upward to moving downward.
Substituting and solving for t, we get
$t=\frac{V_{0}\sin(?)}{g} = \frac{40\sin (50^{\circ})}{9.8}=3.12\,seconds$
To find the maximum height, substitute t in the equation y, we get
$y=\frac{V_0\sin ?}{3.12}-\frac{1}{2}(9.8)(3.12^2)$
Solving, we get
$y=\frac{40\sin50^{\circ}}{3.12}-\frac{1}{2}(9.8)(3.12^2)=47.9\,meters$
The maximum height reached by the object is 47.9 meters
The time of flight is the interval between when the projectile is launched (t1) and when the projectile touches the ground (t2).
Hence,
$V_0\sin(?)t-\frac{1}{2}gt^2$
Solve for t
$t(V_0\sin(?)-(\frac{1}{2}gt)=0$
Solving, we get two soluions as follows:
$t=t_1=0$ and $t=t_2=2V_0sin(?)/g$
Time of flight can be calculated as follows:
$Time\,of\,Flight=2(20)\sin (?)/g=6.25\,seconds$
Horizontal Range is the horizontal distance given by x at t = t2.
$Range=x(t_2)=(V_0\cos(?)V_0\sin (?))/g=(V_0)^2\sin(2?)/g=40^2\sin(2(50^{\circ}))/9.8=6.25\,meters$
1. Chandana Chamuah
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2. great notes
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6. Thnku for giving this . It helped me a lot.
7. Remie Banuag
thank you very much! It’s very helpful
8. YUSUF FOFANAH
Thanks so much, helpful it is.
9. David Joseph
thanks u so much God u
10. Nice explanation
11. Thank u
12. Pius Komanya
Helped me indeed | 2020-09-28 06:40:15 | {"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": 0.7186779379844666, "perplexity": 410.25998592153957}, "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-2020-40/segments/1600401585213.82/warc/CC-MAIN-20200928041630-20200928071630-00312.warc.gz"} |
http://www.gradesaver.com/textbooks/math/calculus/calculus-8th-edition/chapter-2-derivatives-2-2-the-derivative-as-a-function-2-2-exercises-page-128/39 | Calculus 8th Edition
$f$ is not differentiable when $x = -4$ and $x = 0$
$f$ is not differentiable when $x = -4$ since it has a corner and when $x = 0$ since $f$ is not continuous | 2018-04-22 11:14:44 | {"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": 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": 0.9786698222160339, "perplexity": 127.95043527847554}, "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-2018-17/segments/1524125945584.75/warc/CC-MAIN-20180422100104-20180422120104-00317.warc.gz"} |
https://wikimho.com/us/q/astronomy/1714 | ### Why is the speed of light 299,792,458 meters/sec?
• Ok, I am majoring in physics (4th year) and I never understood this fundamental (kinda) question. Maybe I haven't explored it enough.
For example, why does it take 8min20sec for the light from the sun to get to us?
I know the answer to this question on a 'surface' scale. The sun is 1AU away, c=3E8 m/s, and d=v/t to get approx 8 min 20 sec.
my question is on a deeper level.
Say you could "ride a photon" I KNOW THIS IS IMPOSSIBLE, but just say you could. Or a better question: what does a photon experience? The photon, as per my understanding, would leave the sun and (if it is on the right trajectory) hit the Earth instantaneously. A photon leaving Alpha Centauri would see the universe all at once, in a infinitesimal small unit of time (if directed out to space).
If a photon sees everything all at once, why do we perceive it to have a speed? I am sure this has something to do with frames of reference, special relatively, Lorentz transforms? but just seems strange. why is the speed of light finite to us... if it was infinite would this be problematic?
Since 1983, the meter has been defined as the distance traveled by light in a vacuum in 1/299,792,458 of a second, so that number is exact by definition. Of course that doesn't answer your question, but it's interesting. https://en.wikipedia.org/wiki/Speed_of_light
If you haven't read Feynmann's QED I'd really recommend it for more perspective on this.
It is theorized that other multi-verses may have different constants than the ones we have in this universe. For example, the constant of gravity might be different, the speed of light might be different, etc. These universes would look entirely different than our universe.
The speed of light is 299792458 meters per second because The universe works the way it does and also: the speed of light was defined much after the meter was. Hence in order to retain the original meter measure SI decided that the speed of light to be 299792458 m/s just in older to retain the old meter measure which earlier was defined to be equal to one ten-millionth of the distance from the equator to the pole measured on a meridian which passed through Paris France
The actual value of a physical constant with dimensions is merely a product of our unit system. That there is a finite speed for light arises from some simple symmetry considerations - e.g. allowing the laws of physics to be unchanged by translations in space and time.
If you need in-depth explanations, check Michael Stevens videos about lightspeed such as Would Headlights Work at Light Speed?. Beware that you will end up with more questions than before watching.
8 years ago
Speed of light being finite is one of the fundamentals of our Universe.
If it were infinite, this would have a major implication in causality.
Besides, in non-quantum physics, light is just an electromagnetic wave. Eelectromagnetic field is described by Maxwell's equations, which predict that the speed $c$ of electromagnetic waves propagating through the vacuum depends on the dielectric permittivity $ε_0$ and the magnetic permeability $μ_0$ by the equation $c = {1\over\sqrt{ε_{0}μ_{0}}}$ so you can not have an infinite speed of light unless electric permittivity or magnetic permeability were zero, which in turn would cause all sorts of odd things to electromagnetical attraction (and thus, to matter existence beyond elemental particles).
Ok, but why 300,000km/sec? I know km and sec are human inventions - but whatever your units of measure are - why that speed? what is so special about that number? and what about the other part of my question: if a photon experiences everything all at once - how can it be that it leaves the sun and hits earth in t=0 but to us t=8min20sec ?
That speed (297000Km/s) comes from the formula I posted above: 1 over the square root of ε0μ0. About the photon experiencing everything at once, that is due to relativistic time dilation. For a photon, time is infinitely dilated so one instant extends infinitely long.
oh ya.. that's the stuff. intellectual satisfaction. Thanks! I like that answer a lot.
This answer is just a sophisticated version of "it is because it is"
I have the same question. Why a finite no. I was studying transformation of relativistic kinematics and realized light particles would not feel the journey if they were living.Everything would happen in an instance for them. the Value 299792458 comes from the permittivity and permeability that's okay. It's a cause that speed is finite. But why finite, to know that I think more advancement is needed. | 2022-09-29 18:19:46 | {"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": 0.6665318608283997, "perplexity": 592.2471150650448}, "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-2022-40/segments/1664030335362.18/warc/CC-MAIN-20220929163117-20220929193117-00275.warc.gz"} |
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# If points A, B, and C form a triangle, is angle ABC>90 degre
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If points A, B, and C form a triangle, is angle ABC>90 degre [#permalink] 16 Nov 2007, 07:59
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If points A, B, and C form a triangle, is angle ABC>90 degrees?
(1) AC = AB + BC − 0.001
(2) AC = AB
M15-24
[Reveal] Spoiler: OA
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Re: C 15.24 degrees of a triangle [#permalink] 16 Nov 2007, 20:04
bmwhype2 wrote:
Points A, B and C form a triangle. Is ABC > 90 degrees?
1. AC = AB + BC - .001
2. AC = AB
I think the answer is A.
S1 :
AC = AB + BC - .001
AC + .001 = AB + BC
Squaring B.S,
(AC + .001) ^ 2 = (AB + BC )^2
AC^2 + 2AC*.001 = AB^2 + 2AB.BC + BC^2
AC^2 = AB^2 + BC^2 + 2AB.BC + 2AC*.001
By Pythagoras Thm, Angle ABC is 90 if AC^2 = AB^2 + BC^2.
Here AC is bigger than that.. which implies that Angle ABC is > 90. Hence Suff.
PS : (I do not know of any formal rule/theorem that states that if Hypotenuse exceeds more than sum of the squares of the sides that the angle > 90. In fact that triangle is no more a right triangle but I just based this on my intuition. I just took an example of triangle with sides 3,4 and 5. )
St2 :
As AC is the Hypo which should be the longest side of a triangle. But here, the Hypotenuse is equal to a side hence this triangle is not a right triangle but we don't know if angle ABC > 90.. Hence In-suff.
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Re: C 15.24 degrees of a triangle [#permalink] 16 Nov 2007, 20:47
Amit05 wrote:
bmwhype2 wrote:
Points A, B and C form a triangle. Is ABC > 90 degrees?
1. AC = AB + BC - .001
2. AC = AB
I think the answer is A.
S1 :
AC = AB + BC - .001
AC + .001 = AB + BC
Squaring B.S,
(AC + .001) ^ 2 = (AB + BC )^2
AC^2 + 2AC*.001 = AB^2 + 2AB.BC + BC^2
AC^2 = AB^2 + BC^2 + 2AB.BC + 2AC*.001
By Pythagoras Thm, Angle ABC is 90 if AC^2 = AB^2 + BC^2.
Here AC is bigger than that.. which implies that Angle ABC is > 90. Hence Suff.
PS : (I do not know of any formal rule/theorem that states that if Hypotenuse exceeds more than sum of the squares of the sides that the angle > 90. In fact that triangle is no more a right triangle but I just based this on my intuition. I just took an example of triangle with sides 3,4 and 5. )
St2 :
As AC is the Hypo which should be the longest side of a triangle. But here, the Hypotenuse is equal to a side hence this triangle is not a right triangle but we don't know if angle ABC > 90.. Hence In-suff.
I find myself inclined to agree with your logic about statement 1.
However, I find statement 2 to be sufficient by itself as well.
If ac=ab, then angle ABC = angle ACB.
Therefore angle ABC cannot be greater than 90.
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Re: C 15.24 degrees of a triangle [#permalink] 16 Nov 2007, 21:18
jbs wrote:
Amit05 wrote:
bmwhype2 wrote:
Points A, B and C form a triangle. Is ABC > 90 degrees?
1. AC = AB + BC - .001
2. AC = AB
I think the answer is A.
S1 :
AC = AB + BC - .001
AC + .001 = AB + BC
Squaring B.S,
(AC + .001) ^ 2 = (AB + BC )^2
AC^2 + 2AC*.001 = AB^2 + 2AB.BC + BC^2
AC^2 = AB^2 + BC^2 + 2AB.BC + 2AC*.001
By Pythagoras Thm, Angle ABC is 90 if AC^2 = AB^2 + BC^2.
Here AC is bigger than that.. which implies that Angle ABC is > 90. Hence Suff.
PS : (I do not know of any formal rule/theorem that states that if Hypotenuse exceeds more than sum of the squares of the sides that the angle > 90. In fact that triangle is no more a right triangle but I just based this on my intuition. I just took an example of triangle with sides 3,4 and 5. )
St2 :
As AC is the Hypo which should be the longest side of a triangle. But here, the Hypotenuse is equal to a side hence this triangle is not a right triangle but we don't know if angle ABC > 90.. Hence In-suff.
I find myself inclined to agree with your logic about statement 1.
However, I find statement 2 to be sufficient by itself as well.
If ac=ab, then angle ABC = angle ACB.
Therefore angle ABC cannot be greater than 90.
Ooops .. I missed that .. I think these are the traps that are set by GMAC to fool us around..
yes, D it is ..
Good question !!
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Re: C 15.24 degrees of a triangle [#permalink] 16 Nov 2007, 22:20
Amit05 wrote:
bmwhype2 wrote:
Points A, B and C form a triangle. Is ABC > 90 degrees?
1. AC = AB + BC - .001
2. AC = AB
I think the answer is A.
S1 :
AC = AB + BC - .001
AC + .001 = AB + BC
Squaring B.S,
(AC + .001) ^ 2 = (AB + BC )^2
AC^2 + 2AC*.001 = AB^2 + 2AB.BC + BC^2
AC^2 = AB^2 + BC^2 + 2AB.BC + 2AC*.001
--> AC^2 = AB^2 + BC^2 + 2AB.BC - (2AC*.001 + .001^2)
By Pythagoras Thm, Angle ABC is 90 if AC^2 = AB^2 + BC^2.
Here AC is bigger than that.. which implies that Angle ABC is > 90. Hence Suff.
PS : (I do not know of any formal rule/theorem that states that if Hypotenuse exceeds more than sum of the squares of the sides that the angle > 90. In fact that triangle is no more a right triangle but I just based this on my intuition. I just took an example of triangle with sides 3,4 and 5. )
St2 :
As AC is the Hypo which should be the longest side of a triangle. But here, the Hypotenuse is equal to a side hence this triangle is not a right triangle but we don't know if angle ABC > 90.. Hence In-suff.
Please see the correction in blue above.
I pick B.
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Re: C 15.24 degrees of a triangle [#permalink] 17 Nov 2007, 00:28
jbs wrote:
Amit05 wrote:
bmwhype2 wrote:
Points A, B and C form a triangle. Is ABC > 90 degrees?
1. AC = AB + BC - .001
2. AC = AB
I think the answer is A.
S1 :
AC = AB + BC - .001
AC + .001 = AB + BC
Squaring B.S,
(AC + .001) ^ 2 = (AB + BC )^2
AC^2 + 2AC*.001 = AB^2 + 2AB.BC + BC^2
AC^2 = AB^2 + BC^2 + 2AB.BC + 2AC*.001
By Pythagoras Thm, Angle ABC is 90 if AC^2 = AB^2 + BC^2.
Here AC is bigger than that.. which implies that Angle ABC is > 90. Hence Suff.
PS : (I do not know of any formal rule/theorem that states that if Hypotenuse exceeds more than sum of the squares of the sides that the angle > 90. In fact that triangle is no more a right triangle but I just based this on my intuition. I just took an example of triangle with sides 3,4 and 5. )
St2 :
As AC is the Hypo which should be the longest side of a triangle. But here, the Hypotenuse is equal to a side hence this triangle is not a right triangle but we don't know if angle ABC > 90.. Hence In-suff.
I find myself inclined to agree with your logic about statement 1.
However, I find statement 2 to be sufficient by itself as well.
If ac=ab, then angle ABC = angle ACB.
Therefore angle ABC cannot be greater than 90.
Since AC = AB + BC - .001, what if BC = 0.001? then AC = AB again as in statement 2.
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Re: C 15.24 degrees of a triangle [#permalink] 17 Nov 2007, 22:39
GMAT TIGER wrote:
jbs wrote:
Amit05 wrote:
bmwhype2 wrote:
Points A, B and C form a triangle. Is ABC > 90 degrees?
1. AC = AB + BC - .001
2. AC = AB
I think the answer is A.
S1 :
AC = AB + BC - .001
AC + .001 = AB + BC
Squaring B.S,
(AC + .001) ^ 2 = (AB + BC )^2
AC^2 + 2AC*.001 = AB^2 + 2AB.BC + BC^2
AC^2 = AB^2 + BC^2 + 2AB.BC + 2AC*.001
By Pythagoras Thm, Angle ABC is 90 if AC^2 = AB^2 + BC^2.
Here AC is bigger than that.. which implies that Angle ABC is > 90. Hence Suff.
PS : (I do not know of any formal rule/theorem that states that if Hypotenuse exceeds more than sum of the squares of the sides that the angle > 90. In fact that triangle is no more a right triangle but I just based this on my intuition. I just took an example of triangle with sides 3,4 and 5. )
St2 :
As AC is the Hypo which should be the longest side of a triangle. But here, the Hypotenuse is equal to a side hence this triangle is not a right triangle but we don't know if angle ABC > 90.. Hence In-suff.
I find myself inclined to agree with your logic about statement 1.
However, I find statement 2 to be sufficient by itself as well.
If ac=ab, then angle ABC = angle ACB.
Therefore angle ABC cannot be greater than 90.
Since AC = AB + BC - .001, what if BC = 0.001? then AC = AB again as in statement 2.
when dealing with triangles, i usually look for defined size and shape.
-.001 is a concrete size. however, we dont know whether that is a material size that can change the size of the sides of a triangle. From 1, we cannot infer anything.
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Expert's post
1. AC=AB+BC-0.001
this is the same as AC>AB+BC (common for triangles)
for example,
AC=1000.001, AB=500, BC=500 => ABC~180
AC=0.001, AB=500, BC=500.001 =>ABC~0
insuf.
2. AB=AC
ABC=ACB => 2ABC<180> ABC<90
suf.
B is correct
P.S if one can draw it solution will come easy.
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Expert's post
walker wrote:
AC=0.001, AB=500, BC=500.001 =>ABC~0
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Re: Re: [#permalink] 12 Apr 2013, 16:27
If Angle ABC is > 90, then AC has to be the hypotenuse.
With Point 1:
If AB is 1, and BC is 1, then AC would be 1.999, making it the hypotenuse
But if AB is .0006, and BC is .0007, then AC would be .0003, making it not the hypotenuse.
Because the .001 gives us no reference, we cannot conclude anything from Point 1 alone.
If AB = AC, then that means that there is no possible way that AC could be the hypotenuse since there is another side of equal length right next to it. Even if BC is infinitely small, it is still >0 and therefore ABC cannot be >90. Therefore, Point 2 is enough for us to disqualify it alone.
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Re: Re: [#permalink] 19 Apr 2014, 11:21
HI Bunnel,
Thanks.
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Re: Re: [#permalink] 21 Apr 2014, 05:05
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PathFinder007 wrote:
HI Bunnel,
Thanks.
THEORY:
Say the lengths of the sides of a triangle are a, b, and c, where the largest side is c.
For a right triangle: $$a^2 +b^2= c^2$$.
For an acute (a triangle that has all angles less than 90°) triangle: $$a^2 +b^2>c^2$$.
For an obtuse (a triangle that has an angle greater than 90°) triangle: $$a^2 +b^2<c^2$$.
Points A, B and C form a triangle. Is ABC > 90 degrees?
(1) AC = AB + BC - 0.001.
If AC=0.001, AB=0.001 and BC=0.001, then the triangle will be equilateral, thus each of its angles will be 60 degrees.
If AC=10, AB=5 and BC=5.001, then AC^2>AB^2+BC^2, which means that angle ABC will be more than 90 degrees.
Not sufficient.
(2) AC = AB --> triangle ABC is an isosceles triangle --> angles B and C are equal, which means that angle B cannot be greater than 90 degrees. Sufficient.
Similar questions to practice:
are-all-angles-of-triangle-abc-smaller-than-90-degrees-129298.html
if-10-12-and-x-are-sides-of-an-acute-angled-triangle-ho-90462.html
Hope it's clear.
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Re: If points A, B, and C form a triangle... [#permalink] 24 May 2014, 19:41
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bekerman wrote:
If points A, B, and C form a triangle, is angle ABC>90 degrees?
(1) AC=AB+BC−0.001
(2) AC=AB
M15-24 in GMATClub tests - I am wondering whether the OA is incorrect?
Statement-1:
AC = AB+ BC - .001,
If AB, BC are quite big numbers (greater than .01), then angle ABC would be greater than 90 degrees. But if length of AB, BC are in the same range of .001, then angle ABC could be acute angle also.
So statement 1 is not sufficient.
Statement -2:
AC= AB, it means angle ABC = angle ACB, now in any triangle sum all the angles is 180 degree, thus ABC +ACB+BAC = 180 degree. Now as ABC = ACB -> 2ABC + BAC = 180 -> ABC = 90 - BAC/2. Hence angle ABC is always less than 90 degree.
Statement 2 is sufficient
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Re: If points A, B, and C form a triangle, is angle ABC>90 degre [#permalink] 05 Mar 2015, 09:53
Bunuel, can we also claim that when the angle us obtuse c will be greater than a and b?
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Re: If points A, B, and C form a triangle, is angle ABC>90 degre [#permalink] 05 Mar 2015, 10:28
Expert's post
Ergenekon wrote:
Bunuel, can we also claim that when the angle us obtuse c will be greater than a and b?
Yes, the greatest side is opposite the greatest angle.
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Re: If points A, B, and C form a triangle, is angle ABC>90 degre [#permalink] 05 Mar 2015, 10:28
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Display posts from previous: Sort by | 2016-02-10 18:38:13 | {"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": 0.7427523732185364, "perplexity": 2779.5585869893443}, "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-2016-07/segments/1454701159985.29/warc/CC-MAIN-20160205193919-00053-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://codereview.stackexchange.com/questions/513/house-number-parsing | # House number parsing
I have 2 functions used for parsing a house number and I think these can both be improved.
Any suggestions?
input:
45a - 47b
or
45a bis 47b
output:
45,a,47,b
public static string[] HausnummerAufteilen(string hausNummer)
{
string[] result = new string[4];
string[] resultVon = new string[2];
string[] resultBis = new string[2];
resultVon[0] = string.Empty;
resultVon[1] = string.Empty;
resultBis[0] = string.Empty;
resultBis[1] = string.Empty;
int pos1, pos2;
pos2 = 0;
pos1 = hausNummer.IndexOf("bis");
if (pos1 != -1)
{
pos2 = pos1 + 3;
}
else
{
pos1 = hausNummer.IndexOf("-");
if (pos1 != -1)
pos2 = pos1 + 1;
}
if (pos1 > 0)
{
resultVon = HausnummerBuchst(hausNummer.Substring(0, pos1).Trim());
resultBis = HausnummerBuchst(hausNummer.Substring(pos2, hausNummer.Length - pos2).Trim());
}
else
resultVon = HausnummerBuchst(hausNummer);
List<string> list = new List<string>();
result = list.ToArray();
return result;
}
public static string[] HausnummerBuchst(string hauseNummer)
{
string[] result = new string[2];
result[0] = string.Empty;
result[1] = string.Empty;
int iPos;
int testInt;
bool bFound = false;
for (iPos = 0; iPos < hauseNummer.Length; iPos++)
{
if (!int.TryParse(hauseNummer[iPos].ToString() ,out testInt))
{
bFound = true;
break;
}
}
if (bFound)
{
result[0] = hauseNummer.Substring(0, iPos).Trim();
result[1] = hauseNummer.Substring(iPos, hauseNummer.Length - iPos).Trim();
}
else
result[0] = hauseNummer;
return result;
}
Anytime I encounter a method that's only doing read-only parsing of strings and I see it's using String.IndexOf() and positions, it leaps out at me like a gigantic code smell.
Sometimes upon examination it turns out to be a legitimate and wise usage. However very often the programmer is manually splitting strings and looping through parts of the string. In other words they are doing exactly what String.Split() and foreach loops would offer them, but doing it the longer and less readable way.
This is one of those cases.
This may be just my personal taste in code. In any case, I can show you how I would tackle this problem using foreach and Split(), which I believe would be a great improvement to your methods. Except for this difference I think we might have taken comparable approaches.
### Variable names
A lot of your variables are named with very little descriptiveness. You may understand that variables are named so that someone maintaining or reading the code (like me) can understand what's going on, but you could do much better in this department.
• result: You use this word many times in your variable names. The problem is this name is completely useless to me. I can see it is the result of something (variously a function, a loop or some other process) but beyond that you might as well just call the variable foo or bar. hauseNummerVon and hauseNummerBin are far more understandable than resultVon and resultBin - I now know exactly what it is; it's the from house number and the to house number.
• For either result[] array, I am left with nothing and I would have to read all of the code before I understood what they contained. This is a little backwards - I should know what the variable contains, then be able to read the code - with descriptive variable names giving me a greater understanding of what's going on.
• bFound: Don't shorten your variable names like this. I'm left wondering: What is a b? What does it mean if it's found? Maybe this would be more recognisable to a German reader, but even so, expanding b to a word to say exactly what's found would do wonders for readability.
• pos1, pos2: SeparatorPosition and SeparatorEnd would have been more descriptive. Better yet these variables are entirely unnecessary if you just use String.Split().
## My approach
As I mentioned above, as oppose to editing your methods, I've written from scratch how I'd tackle this problem.
### ParseHouseNumberRange
This is the entry method. You feed this method a string such as "20a bis 21c". It splits it up on the "bis" or the "-" (defined in the separators array) in order to produce smaller chunks of the original string.
Each of these chunks is passed to then passes each of the results (in this case, "20a " and " 21c", spaces included) over to the non-public method ParseHouseNumber below. ParseHouseNumber will split these chunks up further and this method combines all the results into a single array which is returned.
Given "20a bis 21c" or similar, it returns an array: 20, a, 21, c.
public string[] ParseHouseNumberRange(string houseNumberRange)
{
string[] separators = new string[] {"bis", "-"};
string[] houseNumbers = houseNumberRange.Split(separators,
StringSplitOptions.RemoveEmptyEntries);
List<string> parsedList = new List<string>();
foreach (string houseNumber in houseNumbers)
{
}
return parsedList.ToArray();
}
### ParseHouseNumber
This method takes a segment of an address from the method above, such as "20a ". After trimming out any spaces, it walks through the string character by character, plucking out digits and letters and sorting them into their own individual sections e.g. "1c4ad3" would be split into the sections 1, c, 4, ad, 3. Any non-digit non-letter characters are ignored.
Once all sections have been sorted out from the string provided to it, it returns those sections in the form of an array.
public string[] ParseHouseNumber(string houseNumber)
{
bool firstRun = true;
bool alphabeticMode = false; // set on first run
houseNumber = houseNumber.Trim();
List<string> sections = new List<string>();
string currentSection = "";
foreach (char c in houseNumber)
{
bool isLetter = Char.IsLetter(c);
bool isDigit = Char.IsDigit(c);
if (!(isDigit || isLetter)) continue;
// If we've switched character type, then a section's finished.
if (firstRun)
{
alphabeticMode = isLetter;
firstRun = false;
}
else if ((isLetter && !alphabeticMode)
|| (isDigit && alphabeticMode))
{ | 2019-11-22 21:58:11 | {"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": 0.26747870445251465, "perplexity": 4193.247007041094}, "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/1573496671548.98/warc/CC-MAIN-20191122194802-20191122223802-00509.warc.gz"} |
https://ja.sharelatex.com/blog | こちらをクリックして 英語 で ShareLaTeX を使用
You asked for it, and now it’s here! Cross-references in ShareLaTeX will now auto-complete as you type, saving you valuable time in correcting typos.
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Posted by Shane on 15 Jun 2017
One of the oldest parts of ShareLaTeX has been our trusty Lion who greets you each time you open the editor. After years of loyal service it was time to give him a facelift, and after a lot of iterations we are really pleased with the final result.
You will see him popping up all over the site, and we hope you like him as much as we do!
We are also very grateful to Francisca the designer behind behind our new logo.
Posted by Paulo on 30 Mar 2017
A few weeks ago we hinted at some amazing new features we were working on. Today we’re pleased to announce our most exciting feature since ShareLaTeX first launched: real-time track changes and comments! Find out more here, or read on.
We remember the old days where you had to choose between using Track Changes in Word, or typesetting with LaTeX, but now you can do both.
## Track Changes
You can now keep track of every change made to the document, along with the person making the change. ShareLaTeX is the only LaTeX editor with real-time tracked changes. Hopefully now there’s nothing holding you back from being able to collaborate using LaTeX!
You can now easily find what your co-authors or reviewers have changed and see how it impacts the document. Accept or reject individual changes quickly with the click of a button, and move onto the next change needing your attention, all inside the editor.
## Discuss and leave notes
With real-time commenting, you can now discuss your work without having to switch to email, printed versions or any other tool. You can leave comments, give quick feedback and resolve issues, all on ShareLaTeX.
With the new comments feature, you can keep an up-to-date list of all of the things you need to work through, without having to sift through out-of-date notes or dig out old emails.
Posted by Henry on 09 Mar 2017 | 2017-06-28 00:05:49 | {"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": 0.31345105171203613, "perplexity": 2067.0625731560585}, "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/1498128321961.50/warc/CC-MAIN-20170627235941-20170628015941-00116.warc.gz"} |
https://energyresources.asmedigitalcollection.asme.org/thermalscienceapplication/article/14/1/011001/1106167/Thermodynamic-Modeling-of-Heat-Engines-Including | ## Abstract
In this work, a thermodynamic model based on an endoreversible engine approach is developed to analyze the performance of heat engines operating under different thermodynamic cycles. The model considers finite heat transfer rate, variable heat source and sink temperatures, and irreversibilities associated with the expansion and compression. Expressions for the maximum power and efficiency at maximum power output are obtained as a function of hot and cold reservoir temperatures, the equivalent isentropic efficiency of compression and expansion components, and the effective conductance ratio between heat exchangers. In all cases, the Curzon–Ahlborn efficiency is retrieved at constant reservoir temperatures and neglected compression–expansion irreversibilities. The proposed model allows assessing the effect of isentropic efficiencies and heat exchanger design and operation characteristics for different thermodynamic cycles.
## 1 Introduction
The development of efficient heat engines is a fundamental alternative to counteract the accelerated depletion of non-renewable energy sources during the last decades. The design and optimization of efficient heat engines require thermodynamic models that describe the engine performance, and how this performance is being impacted by the engine characteristics and operating factors. The maximum possible efficiency attained by heat engines operating between two energy reservoirs is set by the Carnot efficiency (ηC). It corresponds to the efficiency of heat engines operating in the reversible limit under quasi-static processes that results in zero power output [1]. In an effort to account for some of the main irreversibilities, models have been developed considering internally reversible engines where all irreversibilities occur during the energy exchange processes. From this approach, which considers the irreversibility of finite rate heat transfer, the efficiency at maximum power output (ηCA), also known as Curzon—Ahlborn efficiency, has been obtained [13]. This result was first presented by Chambadal [2] and Novikov [3]. Similar to ηC, ηCA is independent of the system characteristics, working fluid properties, or operation regimes, as it only depends on hot and cold reservoir temperatures. It was found that in the limit of low irreversibility, ηCA is bounded by ηc/2 and ηc/(2 − ηc) [4], i.e., ηc/2 ≤ ηCAηc/(2 − ηc), and the bounds correspond to the extremes of the ratio of dissipation in the heat exchange processes. These studies have found that ηCA exhibits some degree of generality as it describes relatively well the efficiency of thermal cycles including Otto, Brayton, Stirling, and Ericsson cycles [5,6] and several real thermal plants [79].
In addition to the irreversibility of finite heat transfer rate, real thermal engines also present irreversibilities associated with the working fluid friction against turbine blades or friction between the piston and cylinder, pressure drop at heat exchangers, and heat leaks that decrease the efficiency and power output. Several models of efficiency at maximum power output that consider different sources of irreversibility such as finite compression ratio, irreversibilities inside the working fluid, heat leak loss, and fluid flow irreversibilities have been proposed [1020]. Bejan [16] presented an analogy between the power conversion in a heat engine (thermal power conversion) and the extraction of power in a fluid flow driven by a pressure difference. It was shown in that work that, when the relationships between pressure difference and flow rates are linear, the energy conversion efficiency at maximum power is analogous to the Carnot efficiency. For nonlinear relationships between flow rate and pressure drops, the paper presents the conditions for operation at maximum power. In particular, for a turbine, the study looks at the effects of the pressure drops that occur in the ducting of the stream to the first expansion stage, the pressure drop across the turbine itself (characterized by an isentropic efficiency), and the discharge pressure drop. Maximum power can be obtained by selecting optimally the flow rate or the pressure drops associated with ducting the flow into and out of the turbine. Similar analyses are presented for compressors and pumps. Grazzini [17] presented the maximum available work and the maximum efficiency for an engine with heat transfer and internal irreversibilities and no heat leak. The engine cycle consisted of constant thermal-capacity heating and cooling processes together with polytropic and irreversible compression and expansion branches. The heat source and heat sink were modeled as temperature changing fluid streams, whereas the internal irreversibility was given by an entropy-production amount dependent on the extent of the polytropic processes. In a different work, Ibrahim et al. [18] conducted a power output optimization of Carnot and closed Brayton cycles using Lagrange multipliers. In that work, the efficiency at maximum power output for both cycles was analyzed and compared under the same specific boundary conditions. Based on this analysis, the closed Brayton cycles can produce more power than the Carnot cycle. Also, in that work, a generic expression for the efficiency at maximum power output for both cycles, as a function of an entropy change parameter during heat addition and rejection, was developed. A finite-time optimization for an ideal Rankine cycle was proposed by Lee and Kim [19]. In that work, a general expression for the efficiency at maximum power output as a function of the reservoir, sink, and pinch-temperature difference of the heating and cooling fluids was obtained. In another study, Lee and Kim [20] considered a model of a heat engine consisting of two isothermal processes and two adiabatic processes. The adiabatic processes (expansion and compression) were treated as irreversible and the isothermal processes of heat addition and heat rejection were driven by fluids of finite capacity rates. By modeling the heat exchangers, their model captured the temperature variation of the hot and cold streams that drive the cycle. They maximized the power output, and corresponding efficiency, with respect to temperatures of the isothermal processes, and showed the dependence of power output on heat conductance, temperature levels, and internal irreversibility in the expansion/compression. They also showed that it is possible to maximize two times the power with respect to temperature levels and conductance allocation. In most of these studies, expressions for the efficiency at maximum power output smaller than ηCA have been obtained including parameters accounting for the additional source of irreversibility. As expected, ηCA is recovered from the efficiency expressions when the additional irreversibilities (other than the finite heat transfer rate) are neglected.
Besides the efficiency at maximum power output, several indicators such as maximum efficient power [8,21], maximum power density [22], ecological criterium [23], as well as methods including thermo-economic analysis [24], exergoeconomic analysis and optimization [25], entropy generation minimization [26], and ecological and exergetic performance optimization [27] have been developed to assess the performance of thermal engines and analyze the effect of system configuration, new designs, and operating parameters. For instance, the ecological criterium proposed by Angulo-Brown [23] allows the quantification of heat engines performance based on the trade-off between the power produced and the power loss in the heat engine as a result of the entropy generation. Different studies have performed analysis and optimization based on this criterion for diverse thermal cycles including Brayton, Stirling, and Ericsson [2831]. The development of the generalized model capable of describing the behavior of any thermal cycle has been constantly pursued. Chen et al. [32] developed a model based on finite-time thermodynamics and ecological optimization to produce general expressions for the power, efficiency, and entropy generation rate of Diesel, Otto, Brayton, Atkinson, among other characteristic thermodynamic cycles. An exergoeconomic analysis and optimization were performed for a heat engine cycle consisting of two constant thermal-capacity heating, two constant thermal-capacity cooling, and two adiabatic branches [25]. The heat engine model in that work included heat transfer irreversibilities, as well as heat leakage, and internal irreversibilities due to compression and expansion. Importantly, in that study, the heat source and heat sink temperatures were considered constant, while the internal irreversibilities were treated through a coefficient greater than one relating the heat released by the engine under irreversible conditions to the heat released in the reversible case.
Besides the traditional derivation of the Curzon–Ahlborn efficiency, some alternative approaches have been used in the last years to get the same result based on different models [33], which contribute to highlight the features and limitations of the result. For example, Van den Broeck [34] obtained the result based on linear irreversible thermodynamics theory that does not use the explicit assumption of heat transfer processes or a reversible compartment. Also, another approach was used by Esposito et al [35], which is commonly referred to as low dissipation model.
In this work, a comprehensive model based on the endoreversible engine approach was developed to analyze the performance of several types of heat engines, including internal combustion engines, steam cycles, gas turbines, and organic Rankine cycles. The model considers finite heat transfer rate, variable heat source and sink temperature, and irreversibilities associated with the fluid friction losses, which are lumped into isentropic efficiencies of compression and expansion components. The model presented in this paper shares commonalities with models available in the literature addressing maximization of efficiency at maximum power output. Also, it presents similarities regarding the exploration of thermal conductance allocation, temperatures distribution, and the investigation of several sources of irreversibilities, as well as the limit for reversible operation. Despite these similarities, the new approach uses a general dimensionless model in which an effort is made to present the optimization results in terms of variables that are commonly available for actual engines. This includes the equivalent efficiency that accounts for irreversibilities for the compression and expansion components through the commonly available isentropic efficiencies, while most of the models use an irreversibility parameter that, although containing the same information, is rarely used in practical situations. In addition, we present the results in terms of an effective conductance ratio between heat exchangers (rM). This parameter compares the conductance of the heat exchangers and the characteristics of the fluids involved including specific heat capacity and mass flow rate.
## 2 Model
A schematic diagram of a heat engine operating between hot and cold thermal reservoirs is presented in Fig. 1. In addition to finite heat transfer rate between the engine and the two reservoirs, variable reservoir temperatures and isentropic efficiencies in the compression and expansion components are considered. The inclusion of variable reservoir temperatures adds a level of reality to the model because most heat engines are driven by streams at hot and cold heat exchangers whose temperature varies. In addition, this consideration allows the incorporation of the effects of hot and cold mass flow rates and thermophysical properties of the external streams. From the diagram, heat input at a rate QH is transferred to the high-pressure working fluid in a heat exchanger (heat input heat exchanger) increasing its temperature. The fluid is then expanded in the expansion component (turbine or piston–cylinder) to produce power (Wtr). A fraction of the energy gets dissipated as heat in this component due to irreversibilities (friction between piston and cylinder or turbine blades and the working fluid), which are accounted through the isentropic efficiency (ηt). This loss is transferred to the cold thermal reservoir at a rate of QLt, see Fig. 1(a). After expansion, the fluid dissipates heat to the cold reservoir QLc through a heat exchanger (heat dissipation heat exchanger) and then passes through the compression component (compressor, pump, or piston–cylinder) where its pressure increases. The isentropic compression efficiency considers the irreversibilities in this component. After compression, additional increase in the fluid temperature and entropy occurs with respect to the ideal process. This additional heat (QLP) remains in the fluid and is represented in Fig. 1(a) by the arrow pointing the engine’s heat input QH. Finally, the high-pressure fluid enters the heat input heat exchanger, and the cycle is repeated. To simplify the model and analysis, irreversibilities in the compression device are merged with those for the expansion component, see Fig. 1(b). This diagram presents the reversible net power output (Wi) of the heat engine and the real net power output Wr that considers energy losses in the compression and expansion processes. In this case, the isentropic efficiencies are lumped together into the isentropic equivalent efficiency (ηeq) defined in Sec. 2.1.
Fig. 1
Fig. 1
Close modal
### 2.1 Isentropic Efficiencies.
Based on the diagram presented in Fig. 1(a), the real and reversible net power output (Wr and Wi) for the heat engine can be, respectively, expressed as follows:
$Wr=Wrt−Wrp=QH−QLc−QLt$
(1)
$Wi=Wit−Wip=QH−QLc$
(2)
The isentropic compression and expansion efficiencies are defined by
$ηp=WipWrp$
(3)
$ηt=WrtWit$
(4)
Similarly, the real and ideal net power outputs for the engine can be related with an isentropic equivalent efficiency ηeq that accounts for the losses in the compression and expansion components:
$ηeq=WrWi$
(5)
Replacing Eqs. (2)(5) into Eq. (1), the real power output yields:
$Wr=ηeq(Wrtηt−ηpWrp)$
(6)
Defining the power ratio Wrp/Wrt as a parameter that accounts for the fraction of the produced power that is used to run the compression component and using Eqs. (1) and (6), the following expression for the ηeq can be obtained:
$ηeq=ηt(1−Wrp/Wrt)1−ηtηpWrp/Wrt$
(7)
The behavior of ηeq as a function of Wrp/Wrt is presented in Fig. 2, for different values of ηp and ηt. From Eq. (7) and Fig. 2, it is clear that if Wrp = 0 or WrpWrt, then Wrp/Wrt ≅ 0 and ηeqηt. This situation represents a heat engine where the power in the compression device is negligible with respect to the power produced in the turbine. This could describe the operation of some cycles using a pump to increase the pressure in the fluid; consequently, in this type of cycles, the isentropic efficiency for the turbine has a higher impact on ηeq; if Wrp = Wrt, then Wrp/Wrt = 1 and ηeq = 0. For this case, the compression device consumes all the power produced in the turbine and the cycle is “zero” efficient. In real engines 0 < Wrp/Wrt < 1, and in particular for a Rankine cycle, Wrp could represent around 5% of Wrt. In Brayton cycles, for instance, Wrp could represent more than 40% of Wrt. From Fig. 2, for lower values of Wrp/Wrt, ηt exerts a strong impact in ηeq and defines the initial value for the curves. As expected, the effect of ηp is small for low Wrp/Wrt and its impact on ηeq becomes stronger as Wrp/Wrt increases. For a cycle characterized by a value of Wrp/Wrt ≈ 0.05, ηeq is practically independent on ηp and proportional to ηt. In turn, for a cycle having a Wrp/Wrt ≈ 0.4, ηeq could be as low as 0.65 for ηp = ηt = 0.8. It is worth noting that ηeq incorporates a combined effect of ηt and ηp. For instance, ηeq resulting from equal values for ηt and ηp (different than 1.0) is lower than such common value. In addition, as the role of the compression is magnified by increasing Wrp/Wrt, this combined effect increases, leading to a reduced ηeq. From Fig. 2, the range for ηeq goes from around 0.6 to 1.0, which covers the most common types of heat engines like Brayton (ηeq = 0.65–0.8), Rankine and Organic Rankine cycles (ηeq = 0.8–0.9), and internal combustion engines ηeq = 0.6–0.8. Then, the results are presented in terms of three values of ηeq (1.0, 0.8, and 0.6) that are representative of this range.
Fig. 2
Fig. 2
Close modal
### 2.2 Engine Model.
The heat transfer rate between the hot/cold reservoirs and the reversible compartment can be, respectively, expressed in terms of the logarithmic mean temperature difference as follows:
$QH=(UA)HTHi(1−τH)−(DH−τH)ln((1−τH)/(DH−τH))=(m˙cp)HTHi(1−DH)$
(8)
$QL=(UA)LTLi(1/τL−1)−(1/τL−1/DL)ln((1/τL−1)/(1/τL−1/DL))=(m˙cp)LTLi(1/DL−1$
(9)
where τH = THc/THi, τL = TLi/TLc, DH = THo/THi, and DL = TLi/TLo.
From the right side of Eqs. (8) and (9),
$ln(1−τHDH−τH)=(UA)H(m˙cp)H⇒1−τHDH−τH=eNTUH$
(10)
$ln(1/τL−11/τL−1/DL)=(UA)L(m˙cp)L⇒1/τL−11/τL−1/DL=eNTUL$
(11)
with $NTUH=(UA)H/(m˙cp)H$ and $NTUL=(UA)L/(m˙cp)L$ being, respectively, the number of transfer units for the heat input and heat dissipation heat exchangers.
Replacing Eqs. (10) and (11) into the left side of Eqs. (8) and (9), respectively, the heat input and total heat rejection rates (including the engine heat rejection and the heat loss) yield:
$QH=(UA)HTHi(1−e−NTUH)NTUH(1−τH)$
(12)
$QL=QLc+QLt=(UA)LTLi(1−e−NTUL)NTUL(1/τL−1)$
(13)
The following expression for the heat loss (QLt) is obtained from Eqs. (1), (2), and (5):
$QLt=(1−ηeq)(QH−QLc)$
(14)
In turn, the engine heat rejection rate (QLc) can be found from Eqs. (12)(14):
$QLc=(UA)LTLi1ηeq(1−e−NTUL)NTUL(1/τL−1)−(UA)HTHi(1−e−NTUH)NTUH(1−τH)(1−ηeq)ηeq$
(15)
The operation of the heat engine is constrained to the reversible operation of the Carnot engine, for which the entropy balance yields:
$QHTHc=QLcTLc$
(16)
Replacing Eqs. (8) and (15) into Eq. (16) and using the expressions of Eqs. (10) and (11) result in the following equation that relates τH, τL, and τ:
$(1τH−1)=rMηeq(1−1τL)−1τ1τL(1−τH)(1−ηeq)ηeq$
(17)
In Eq. (17), τ corresponds to the cold and hot input reservoir temperature ratio (τ = TLi/THi), $M=(1−e−NTUL)NTUH/(1−e−NTUH)$$NTUL$, and r = (UA)L/(UA)H are parameters that, respectively, represent design and operating characteristics (M) and the ratio between conductances (r) of heat input and heat dissipation heat exchangers. Both parameters can be lumped together into a single parameter $rM=[(m˙cp)L(1−e−NTUL)]/[(m˙cp)H(1−e−NTUH)]$, which can be seen as an effective conductance ratio between heat exchangers that compares the conductance and fluid characteristics including specific heat capacity and mass flow rate.
Solving for τL, Eq. (17) yields:
$τL=rM−(ηeq/τH)(1−τH)rM+((1−ηeq)/τ)(1−τH)$
(18)
From Eqs. (2) and (5),
$Wr=ηeq(QH−QLc)$
(19)
Replacing the values of QH, QLC (Eqs. (12) and (15)), and τL (Eq. (18)) into Eq. (19), the power output for the real heat engine becomes
$Wr=(UA)HTHi(1−e−NTUH)NTUHηeq(1−τH)[rMτH+τH−rMτ−1]τH(rM+ηeq)−ηeq$
(20)
Defining the non-dimensionalization factor Qmax as the maximum possible heat transfer rate between the hot and cold reservoirs (when no heat engine is present):
$Qmax=14(UA)sTHi(1−τ)εsNTUs$
(21)
where ɛs(UA)s/NTUs = ɛH(UA)H/NTUH + (UA)LɛL/NTUL = (1 + rM)ɛH(UA)H/NTUH accounts for the total effective conductance in both heat exchangers, being $εL=1−e−NTUL$ and $εH=1−e−NTUH$.
Normalizing the power output (Eq. (20)) and the heat input (Eq. (12)) using the definition of Qmax, expressions for dimensionless power output $(W*)$ and dimensionless heat input $(QH*)$ are obtained:
$W*=4ηeq(1−τH)[rMτH+τH−rMτ−1](1−τ)(1+rM)[τH(rM+ηeq)−ηeq]$
(22)
$QH*=4(1−τH)(1−τ)(1+rM)$
(23)
Finally, from Eqs. (22) and (23), the heat engine efficiency can be found as
$η=W*QH*=ηeq[rMτH+τH−rMτ−1]τH(rM+ηeq)−ηeq$
(24)
### 2.3 Efficiency at Maximum Power Output.
The maximum power output $(Wmax*)$ can be found by maximizing $W*$ with respect to τH, i.e., $dW*/dτH=0$:
$Wmax*=4rMηeq(rM+ηeq)2(ηeq(τ−1)+(rMτ+1)−rM+1)2(rM+1)(1−τ)$
(25)
with the corresponding τH,opt determined as
$τH,opt=ηeq+(rM/(rM+1))(rM+1)[ηeq(τ−1)+(rMτ+1)]rM+ηeq$
(26)
Defining τc as the cold to hot thermal engine temperature ratio (τc = TLc/THc) and using the definition of τ, τL, and τH, τc = τ/(τLτH). After replacing τH,opt in this last equation, the corresponding τc,opt is determined as
$τc,opt=(rM+1)[ηeq(τ−1)+(rMτ+1)]−(1−ηeq)(rM+ηeq)$
(27)
Finally, from the expression for $Wmax*$ (Eq. (25)), the efficiency at maximum power output $(η@Wmax)$ can be found as follows:
$η@Wmax*=Wmax*QH*=ηeq(rM+1)(rM+ηeq)[1−ηeq(τ−1)+(rMτ+1)(rM+1)]$
(28)
It can be easily verified that the Curzon–Ahlborn efficiency $(η@Wmax=1−τ)$ can be retrieved for a reversible engine with constant reservoir temperatures and ideal operation of compression and expansion components, i.e., ηeq = 1 and rM = 1.
Equations (25)(28) are general expressions for $Wmax*$, τH,opt, τc,opt, and $η@Wmax*$, respectively. These expressions incorporate the effect of variable heating and cooling flow temperatures and isentropic efficiencies at compression and expansion components. A summary of these expressions is presented in Table 1 for the heat engine considering constant and variable heating and cooling flow temperatures, and also with and without compression–expansion irreversibilities.
Table 1
Expressions for τc,opt, $Wmax*$, and $η@Wmax*$ considering different model characteristics
## 3 Results and Analysis
In this section, results from the model presented in Sec. 2 are illustrated for different values of heat engine parameters τ, rM, and ηeq. As explained before, there is an optimal value of τc (Eq. (27)) at which the power output achieves a maximum value (Eq. (25)). Figure 3 presents the behavior of τc,opt as a function of τ for three selected values of ηeq = 0.6, 0.8, and 1.0 and five values with different order of magnitude for rM. From this figure, higher τ values (lower difference between cold and hot reservoir temperatures) lead to higher τc,opt. For the ideal case where ηeq = 1.0, when τ approaches 1.0, τc,opt also approaches 1.0, see Fig. 3(a), which implies a reduction in the temperature difference (potential) between reservoirs, and consequently a decrement in the power output. In the limit, when τ = τc,opt = 1, W and η are equal to zero. It is important to note that τc,opt is independent of the parameter rM for ηeq = 1.0. For values of ηeq = 0.8 (Fig. 3(b)) and ηeq = 0.6 (Fig. 3(c)), τc,opt is slightly higher for lower values of τ, when compared with the ideal scenario of Fig. 3(a); this means that for the same τ and ηeq < 1.0, irreversibilities in compression and expansion processes reduce the difference between cold and hot engine temperatures decreasing its power and efficiency. As expected, the larger the irreversibilities (smaller values of ηeq), the higher τc,opt for any value of τ < 1.0, as can be seen in Figs. 3(b) and 3(c). From these figures, it can be appreciated that the parameter rM becomes important as the irreversibilities in the compression and expansion processes increase. This effect is more significant for lower τ values, i.e., for heat engines, such as Brayton engines, that operate with a larger difference between cold and hot reservoir temperatures, when compared with engines whose operation is framed by higher τ like organic Rankine cycles. Higher values of rM tend to reduce τc,opt for the same ηeq, which means that a larger effective conductance in the cold dissipation heat exchanger should help compensate the detriment in engine’s performance due to irreversibilities.
Fig. 3
Fig. 3
Close modal
In order to better appreciate the effect of rM on τc,opt, Fig. 4 displays curves of τc,opt for five orders of magnitude values of rM, considering three ηeq and τ = 0.5. For ηeq = 1.0, τc,opt = 0.707, and, as it was mentioned, it is independent of rM. As ηeq reduces, τc,opt increases along the entire rM range; however, τc,opt reduces as rM increases, approaching to the corresponding value of τc,opt when ηeq = 1.0, in this case 0.707. It is worth noting that for τ = 0.5, the maximum change in τc,opt in the analyzed rM range is about 0.02 (from τc,opt = 0.707 for ηeq = 1.0 to τc,opt = 0.728 for ηeq = 0.6), but this change is greater for lower τ values and, in the limit when τ → 0.0, τc,opt approaches this maximum change of about 0.386 (from τc,opt = 0.0 for ηeq = 1.0 to τc,opt = 0.386 for ηeq = 0.6), as can be seen in Fig. 3.
Fig. 4
Fig. 4
Close modal
Figure 5 shows how $Wmax*$ changes with τ for different orders of magnitude in the value of rM (from 10−2 to 102) and three different values of ηeq 0.6, 0.8, and 1. In general terms, $Wmax*$ decreases with τ as expected, which is a result of smaller temperature differences between the heat source and the heat sink. In the extreme case τ = 1, when both heat source and sink have the same temperature, $Wmax*$ is naturally equal to zero for all situations, as the power generation’s driving effect for the engine vanishes.
Fig. 5
Fig. 5
Close modal
The largest possible value of $Wmax*=1$ occurs when ηeq = 1, rM = 1, and τ = 0 (equivalent to TL = 0 or TH = ∞). In this case $Wmax*$ becomes equal to the heat transfer between the reservoirs when no heat engine is present. In any other situation, $Wmax*$ is a fraction of such bound on the amount of energy that could be transferred between heat source and heat sink, which is due to a reduction in the heat-to-power conversion efficiency when τ increases (closer TL and TH), a decrease in ηeq, or because the balance in heat exchanger inventory between hot and cold sides is substantially altered.
As described in the model, rM represents an effective conductance ratio between heat input and heat dissipation heat exchangers. Figure 5 shows how taking rM away from 1 (order of magnitude 100) brings down $Wmax*$. As can be verified, for rM equal to 10−1 or 101, $Wmax*$ is reduced in more than 50% for any value of τ and ηeq. For rM equal to 10−2 or 102, $Wmax*$ reduction is even greater. In order to better capture this behavior, Figs. 6(a) and 6(b) present $Wmax*$ as a function of rM for constant values of τ and ηeq. As can be seen, there is always a maximum value of $Wmax*$ for rM around 1. If the locus of maxima in Figs. 6(a) and 6(b) are isolated, the corresponding values of rMopt as a function of τ and ηeq can be found, as shown in Fig. 6(c). Values of rMopt are almost entirely between 1 and 1.2 for the considered values in this work. If τ or ηeq go very low (lower than approximately 0.2 and 0.6, respectively), rMopt can start rising beyond 1.2 but in any case not much higher than 1.4. It is worth remembering that a value of rM higher than 1 means a larger heat exchanger (larger effective conductance) in the cold side when compared with the hot side. rMopt becomes significant in practical implementations when a decision has to be made on the allocation of conductance of heat exchangers mainly driven by economic considerations.
Fig. 6
Fig. 6
Close modal
Table 2 helps emphasizing the importance of heat exchangers’ conductance allocation. Different values of $Wmax*$ are presented as a function of rM and τ together with the percentual deviation with respect to the maximum possible $Wmax*$ ($Wmax,max*$) for each case. As can be seen, choosing rM equal to 1.0 produces a $Wmax*$ close to the optimum for all cases considered. If one of the heat exchangers is twice large than the other, the percentual deviation respect to $Wmax,max*$ is around 10%. These relative deviations are more significant for lower values of τ.
Table 2
Effect of optimizing heat exchanger conductance allocation, ηeq = 0.8
τrM$Wmax*$
$Wmax,max*−Wmax*Wmax,max*$ (%)
0.250.50.2105413.65
10.243390.18
rMopt = 1.088$Wmax,max*=0.24382$
20.222658.68
0.500.50.1157312.27
10.131870.04
rMopt = 1.040$Wmax,max*=0.13192$
20.118769.98
0.750.50.0500311.56
10.056576.02 × 10−3
rMopt = 1.015$Wmax,max*=0.05657$
20.0505410.67
τrM$Wmax*$
$Wmax,max*−Wmax*Wmax,max*$ (%)
0.250.50.2105413.65
10.243390.18
rMopt = 1.088$Wmax,max*=0.24382$
20.222658.68
0.500.50.1157312.27
10.131870.04
rMopt = 1.040$Wmax,max*=0.13192$
20.118769.98
0.750.50.0500311.56
10.056576.02 × 10−3
rMopt = 1.015$Wmax,max*=0.05657$
20.0505410.67
Figure 7 illustrates the effects of ηeq and rM on the dependence of the efficiency at maximum power as a function of the temperature ratio τ. When τ = 1, the temperature gradient that drives the engine disappears, and at that point, independently of ηeq and rM, $Wmax*$ goes to zero. This is also true for the Carnot efficiency. $Wmax*$ increases monotonically as τ decreases and peaks in the limit τ → 0 when the low temperature reservoir tends to a temperature of absolute zero or the high temperature grows to infinity. Notice that differently from what occurred with the maximum power (see Fig. 5), in the case of ηeq = 1, the value of the rM has no effect on $η@Wmax*$ (thus all curves collapse into one in Fig. 7(a)). As the compression or the expansion component efficiencies decrease, or the compressing power becomes more important with respect to the expansion power, ηeq decreases and so does $η@Wmax*$. When ηeq < 1, the efficiency at maximum power drops as rM decreases.
Fig. 7
Fig. 7
Close modal
Figure 8 reemphasizes the impact of the heat exchangers effective conductance ratio. In the range considered, which spans five orders of magnitude, it can be observed that for large τ, the efficiency at maximum power is essentially insensitive to the allocation of heat exchanger conductance and effectiveness represented by rM. When it does impact (τ < 0.5) the behavior is essentially a smooth transition between a low limit value for rM ≪ 1 and a higher value for rM ≫ 1, with the inflection occurring near the point when the heat exchangers have equal allocation of effective conductance $(rM∼1)$. Observe in Fig. 8 a trend that is common in heat engines; the impact of the cold heat exchanger is greater than the hot one in terms of efficiency. For example, in a Rankine cycle, a small variation in the condenser temperature has a larger impact on efficiency than the same variation in the boiler [7]. It can also be seen that the effect of rM on efficiency is not as high as it is for $Wmax*$ (see Figs. 5 and 6). The meaning of a very large rM is allocating most of the available area to the cold heat exchanger, which implies TLCTLi (see Fig. 1). After some point TLC is so close to TLi that any additional area allocated to the cold heat exchanger no longer increase the heat engine efficiency (the curve levels off for large values of rM). Similar situation occurs in the lower end value where most of the area is allocated to the hot heat exchanger and THCTHi.
Fig. 8
Fig. 8
Close modal
## 4 Conclusions
A thermodynamic model considering finite heat transfer rate, variable heat source and sink temperature, and irreversibilities associated with the compression–expansion processes was developed to analyze the performance of heat engines operating under different thermodynamic cycles. Expressions for the optimum ratio between cold and hot thermal engine temperatures (τc,opt), maximum power $(Wmax*)$, and efficiency at maximum power output $(η@Wmax*)$ are obtained as a function of the parameters τ (cold and hot reservoirs temperature ratio), rM (effective conductance ratio between heat input and heat dissipation heat exchangers), and ηeq (equivalent isentropic efficiency). In all cases, the Curzon–Ahlborn efficiency is retrieved for rM = 1.0 and ηeq = 1.0. It was found that rM has no effect on τc,opt, $Wmax*$, and $η@Wmax*$ when ηeq = 1.0. As the isentropic efficiencies in the compression or the expansion components decrease, as a consequence of irreversibilities, ηeq decreases leading to reductions in $Wmax*$, $η@Wmax$, and in the difference between cold and hot engine temperatures (increment in τc,opt). When ηeq < 1, τc,opt decreases and $η@Wmax*$ increases with rM, respectively. In turn, $Wmax*$ drops for rM away from 1.0 in more than one order of magnitude. Optimum rM values (rMopt) leading to maximum power output are found to be between 1 and 1.4 for the analyzed ranges of τ and ηeq. rMopt shifts to the upper limit of that range as irreversibilities increase (ηeq reduces) and/or for engines operating within higher temperature difference between cold and hot reservoirs (lower τ). In this case, the compression power is usually a significant portion of the net power. A value of rMopt slightly greater than 1.0 means that in order to achieve maximum power, the heat dissipation heat exchanger must be larger when compared with the heat exchanger at the hot side.
## Acknowledgment
This research was primarily supported by the Mechanical and Thermal Engineering Sciences Center at National Renewable Energy Laboratory as a part of the Laboratory Directed Research and Development (LDRD) Program funded by the U.S. Department of Energy (DOE).
The views expressed in the article do not necessarily represent the views of the DOE or the U.S. Government.
Obie I. Abakporo acknowledges the support of the Florida A&M University Title III Fellowship and the facility usage of Florida State University Center of Advanced Power Systems.
## Conflict of Interest
There are no conflicts of interest.
## Data Availability Statement
The datasets generated and supporting the findings of this article can be obtained from the corresponding author upon reasonable request. The authors attest that all data for this study are included in the paper. Data provided by a third party are listed in Acknowledgment.
## Nomenclature
A =
effective heat transfer area, m2
U =
overall heat transfer coefficient, kW m−2 K−1
$m˙$ =
mass flow rate, kg/s
cp =
specific heat at constant pressure, kJ kg−1 K−1
DH =
THo to THi temperature ratio
DL =
TLi to TLo temperature ratio
QH =
heat input, kW
QLc =
heat rejection, kW
QLt =
heat loss, kW
QL =
total heat rejection, kW
THi =
inlet hot reservoir temperature, K
TLi =
inlet cold reservoir temperature, K
THc =
hot heat engine temperature, K
TLc =
cold heat engine temperature, K
THo =
exit hot reservoir temperature, K
TLo =
exit cold reservoir temperature, K
Wr =
net power output, kW
Wrt =
expansion power, kW
Wrp =
compression power, kW
Wi =
reversible net power output, kW
Wit =
reversible expansion power, W
Wip =
reversible compression power, kW
$W*$ =
dimensionless power output
$QH*$ =
dimensionless heat input rate
rM =
effective conductance ratio between cold and hot side heat exchangers
NTU =
number of transfer units
ɛ =
effectiveness
η =
thermal efficiency
$η@Wmax*$ =
efficiency at maximum power
ηt =
isentropic efficiency expansion component
ηp =
isentropic efficiency compression component
ηeq =
equivalent isentropic efficiency
τ =
cold to hot reservoirs temperature ratio
τc =
cold to hot thermal engine temperature ratio
τH =
THc to THi temperature ratio
τL =
TLi to TLc temperature ratio
H =
high temperature
L =
low temperature
max =
maximum
opt =
optimum
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60010
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https://istopdeath.com/simplify-4p2p3-p-1p-54p-4-3/ | # Simplify ((4p)^2p^3)/(p^-1p^-5)*(4p^-4)^-3
(4p)2p3p-1p-5⋅(4p-4)-3
Simplify the numerator.
Apply the product rule to 4p.
42p2p3p-1p-5⋅(4p-4)-3
Multiply p2 by p3 by adding the exponents.
Move p3.
42(p3p2)p-1p-5⋅(4p-4)-3
Use the power rule aman=am+n to combine exponents.
42p3+2p-1p-5⋅(4p-4)-3
42p5p-1p-5⋅(4p-4)-3
42p5p-1p-5⋅(4p-4)-3
Raise 4 to the power of 2.
16p5p-1p-5⋅(4p-4)-3
16p5p-1p-5⋅(4p-4)-3
Multiply p-1 by p-5 by adding the exponents.
Use the power rule aman=am+n to combine exponents.
16p5p-1-5⋅(4p-4)-3
Subtract 5 from -1.
16p5p-6⋅(4p-4)-3
16p5p-6⋅(4p-4)-3
Rewrite the expression using the negative exponent rule b-n=1bn.
16p51p6⋅(4p-4)-3
Multiply the numerator by the reciprocal of the denominator.
16p5p6⋅(4p-4)-3
Multiply p5 by p6 by adding the exponents.
Move p6.
16(p6p5)⋅(4p-4)-3
Use the power rule aman=am+n to combine exponents.
16p6+5⋅(4p-4)-3
16p11⋅(4p-4)-3
16p11⋅(4p-4)-3
Rewrite the expression using the negative exponent rule b-n=1bn.
16p11⋅(41p4)-3
Combine 4 and 1p4.
16p11⋅(4p4)-3
Change the sign of the exponent by rewriting the base as its reciprocal.
16p11⋅(p44)3
Simplify the expression.
Apply the product rule to p44.
16p11⋅(p4)343
Multiply the exponents in (p4)3.
Apply the power rule and multiply exponents, (am)n=amn.
16p11⋅p4⋅343
Multiply 4 by 3.
16p11⋅p1243
16p11⋅p1243
Raise 4 to the power of 3.
16p11⋅p1264
16p11⋅p1264
Cancel the common factor of 16.
Factor 16 out of 16p11.
16(p11)⋅p1264
Factor 16 out of 64.
16(p11)⋅p1216(4)
Cancel the common factor.
16p11⋅p1216⋅4
Rewrite the expression.
p11⋅p124
p11⋅p124
Combine p11 and p124.
p11p124
Multiply p11 by p12 by adding the exponents.
Use the power rule aman=am+n to combine exponents.
p11+124 | 2022-11-28 19:00:27 | {"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": 0.8874828219413757, "perplexity": 5550.207449592207}, "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-2022-49/segments/1669446710534.53/warc/CC-MAIN-20221128171516-20221128201516-00397.warc.gz"} |
https://mathoverflow.net/questions/313238/on-submatrices-size-bound | # On submatrices: size bound
Let $$M$$ be a generic $$2n\times 2n$$ matrix and fix $$k\leq n$$.
Suppose $$\mathcal{F}$$ is a family of submatrices under the conditions that $$A\in\mathcal{F}$$ provided
(a) $$A$$ is a $$k\times k$$ submatrix of $$M$$, and
(b) $$A$$ overlaps with each and every $$B\in\mathcal{F}$$.
QUESTION. What is the best upper bound for the cardinality of $$\mathcal{F}$$, in terms of $$n$$ and $$k$$?
To each such matrix $$A$$ we can associate $$A_1,A_2$$, the sets of column indices and row indices respectively. The families $$\mathcal F_i=\{A_i| A\in \mathcal F\}$$ are intersecting families of subsets, therefore by Erdos-Ko-Rado we have $$|\mathcal F_i|\le \binom{2n-1}{k-1}$$ Since a matrix $$A$$ is uniquely determined by its set of rows and columns we have $$|\mathcal F|\le |\mathcal F_1||\mathcal F_2|\le \binom{2n-1}{k-1}^2$$ and this maximum can be achieved by taking all the $$k\times k$$ submatrices that contain the $$(1,1)$$ entry. | 2018-11-14 14:20:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8875645399093628, "perplexity": 91.17022140578406}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039742020.26/warc/CC-MAIN-20181114125234-20181114151234-00302.warc.gz"} |
https://aviation.stackexchange.com/questions/59802/how-can-i-calculate-the-velocity-of-air-drawn-by-a-pusher-configured-propeller | # How can I calculate the velocity of air drawn by a pusher configured propeller?
I am attempting to construct a model plane using Custer channel wings. In order to calculate (or at least approximate) the lift generated I need to find the speed of the air through the channel due to the propellers. Is there an equation I can use to do this given pitch, diameter, rpm, etc?
There are theoretical equations and heuristic ones for a propeller with not too many blades, but these are not nearly as useful as a run of CFD. The key part is to translate the number of blades and RPM, given pitch, chord, span, wash, and wingtip device into $$\Delta p$$ across the plane of the propeller. This is better done through simulation and experimenting than working with pen, paper and slide rule.
The roughest of the rough approximation is, of course, assuming $$\alpha\approx0$$ for each of the blade and $$v_\mathrm{i}\approx\frac{1}{2}\cdot\omega\cdot\left(r_a\tan\theta_a+r_b\tan\theta_b\right)$$ and assume $$v_\mathrm{o}\approx v_\mathrm{i}$$, where point $$a$$ is the point where the blade starts near the rotating axis, and $$b$$ the wingtip, $$\omega$$ the angluar velocity of the propeller, $$r_{(\cdot)}$$ the distance from the rotating axis and $$\theta_{(\cdot)}$$ the pitch angle. | 2020-10-24 23:19:10 | {"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": 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": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8025857210159302, "perplexity": 452.73953449488823}, "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-2020-45/segments/1603107885059.50/warc/CC-MAIN-20201024223210-20201025013210-00282.warc.gz"} |
https://daviddalpiaz.github.io/stat432sp18/lab/lab05-soln.html | # Packages
For this lab, use only methods availible from the following packges:
library(MASS)
library(caret)
library(tidyverse)
library(knitr)
library(kableExtra)
library(microbenchmark)
library(randomForest)
If you haven’t already, make sure each is installed!
# Data
For this lab we will again use the birthwt data from the MASS package. Our goal in analyzing this data is to predict the birthweight of newborns at Baystate Medical Center, Springfield, Mass. (This data is from 1986…)
data(birthwt)
Last time w read the documentation for this data and based on our goal dropped the low variable from the dataset. (It doesn’t make sense to including a variable indiciating low birthweight if our goal is to predict birthweight. That would be of no use in practice.)
birthwt = subset(birthwt, select = -c(low))
We also coerced certain variables to be factor variables, as it was clear that they were categorical variables.
birthwt$race = factor(ifelse(birthwt$race == 1, "white",
ifelse(birthwt$race == 2, "black", "other"))) birthwt$smoke = factor(birthwt$smoke) birthwt$ht = factor(birthwt$ht) birthwt$ui = factor(birthwt$ui) We then finally test-train split the data. set.seed(42) bwt_trn_idx = sample(nrow(birthwt), size = trunc(0.70 * nrow(birthwt))) bwt_trn_data = birthwt[bwt_trn_idx, ] bwt_tst_data = birthwt[-bwt_trn_idx, ] # Model Training [Exercise] Train three “different” $$k$$-nearest neighbors models, each with k = 5. The “difference” will be in what we call the preprocessing of the data. Note that we won’t actually modify the data, but we will use formula syntax to handle the preprocessing within the model fitting. • Model 1 • Numeric variables not scaled. • Factor variables remain factors. knnreg() will use one-hot encoding. • Model 2 • Numeric variables are scaled to have mean 0 and standard deviation 1. • Factor variables remain factors. knnreg() will use one-hot encoding. • Model 3 • Numeric variables are scaled to have mean 0 and standard deviation 1. • Coerce the race variable to be numeric. All ther factors remain unchanged. bwt_knn_mod_1 = knnreg(bwt ~ age + lwt + race + smoke + ptl + ht + ui + ftv, data = bwt_trn_data, k = 5) bwt_knn_mod_2 = knnreg(bwt ~ scale(age) + scale(lwt) + race + smoke + scale(ptl) + ht + ui + scale(ftv), data = bwt_trn_data, k = 5) bwt_knn_mod_3 = knnreg(bwt ~ scale(age) + scale(lwt) + as.numeric(race) + smoke + scale(ptl) + ht + ui + scale(ftv), data = bwt_trn_data, k = 5) [Exercise] Output the first 6 rows of the traning data. head(bwt_trn_data) ## age lwt race smoke ptl ht ui ftv bwt ## 61 24 105 black 1 0 0 0 0 2381 ## 67 22 130 white 1 0 0 0 1 2410 ## 141 30 95 white 1 0 0 0 2 3147 ## 36 24 138 white 0 0 0 0 0 2100 ## 215 25 120 white 0 0 0 0 2 3983 ## 190 29 135 white 0 0 0 0 1 3651 [Exercise] Output the first 6 rows of the $$X$$ data (predictor data frame) supplied to the $$k$$-nearest neighbors algorith in Model 1. head(bwt_knn_mod_1$learn$X) ## age lwt raceother racewhite smoke1 ptl ht1 ui1 ftv ## 61 24 105 0 0 1 0 0 0 0 ## 67 22 130 0 1 1 0 0 0 1 ## 141 30 95 0 1 1 0 0 0 2 ## 36 24 138 0 1 0 0 0 0 0 ## 215 25 120 0 1 0 0 0 0 2 ## 190 29 135 0 1 0 0 0 0 1 [Exercise] Output the first 6 rows of the $$X$$ data (predictor data frame) supplied to the $$k$$-nearest neighbors algorith in Model 2. head(bwt_knn_mod_2$learn$X) ## scale(age) scale(lwt) raceother racewhite smoke1 scale(ptl) ht1 ui1 ## 61 0.07216633 -0.881215186 0 0 1 -0.4172794 0 0 ## 67 -0.30887190 0.004023814 0 1 1 -0.4172794 0 0 ## 141 1.21528103 -1.235310786 0 1 1 -0.4172794 0 0 ## 36 0.07216633 0.287300294 0 1 0 -0.4172794 0 0 ## 215 0.26268545 -0.350071786 0 1 0 -0.4172794 0 0 ## 190 1.02476192 0.181071614 0 1 0 -0.4172794 0 0 ## scale(ftv) ## 61 -0.7938904 ## 67 0.1017808 ## 141 0.9974520 ## 36 -0.7938904 ## 215 0.9974520 ## 190 0.1017808 [Exercise] Output the first 6 rows of the $$X$$ data (predictor data frame) supplied to the $$k$$-nearest neighbors algorith in Model 3. head(bwt_knn_mod_3$learn$X) ## scale(age) scale(lwt) as.numeric(race) smoke1 scale(ptl) ht1 ui1 ## 61 0.07216633 -0.881215186 1 1 -0.4172794 0 0 ## 67 -0.30887190 0.004023814 3 1 -0.4172794 0 0 ## 141 1.21528103 -1.235310786 3 1 -0.4172794 0 0 ## 36 0.07216633 0.287300294 3 0 -0.4172794 0 0 ## 215 0.26268545 -0.350071786 3 0 -0.4172794 0 0 ## 190 1.02476192 0.181071614 3 0 -0.4172794 0 0 ## scale(ftv) ## 61 -0.7938904 ## 67 0.1017808 ## 141 0.9974520 ## 36 -0.7938904 ## 215 0.9974520 ## 190 0.1017808 # Model Evaluation [Exercise] Calculate test and train RMSE for each model. calc_rmse = function(actual, predicted) { sqrt(mean((actual - predicted) ^ 2)) } # create model list mod_list = list(bwt_knn_mod_1, bwt_knn_mod_2, bwt_knn_mod_3) # get train and test predictions trn_pred = lapply(mod_list, predict, newdata = bwt_trn_data) tst_pred = lapply(mod_list, predict, newdata = bwt_tst_data) # get train and test RMSE trn_rmse = sapply(trn_pred, calc_rmse, actual = bwt_trn_data$bwt)
tst_rmse = sapply(tst_pred, calc_rmse, actual = bwt_tst_data\$bwt)
[Exercise] Summarize these results in a table. (Model, Train/Test RMSE.) Output the results as a well-formatted markdown table.
# create df of results
results = data.frame(
mod = c("Model 1", "Model 2", "Model 3"),
trn_rmse = trn_rmse,
tst_rmse = tst_rmse
)
colnames(results) = c("Model", "Train RMSE", "Test RMSE")
# create results table
kable_styling(kable(results, format = "html", digits = 3), full_width = FALSE)
Model Train RMSE Test RMSE
Model 1 638.582 720.808
Model 2 585.120 707.995
Model 3 583.951 717.642
# Results?
[Exercise] Did preprocessing make a difference?
A little bit, but nothing all that significant in the context of the problem.
# Fast Train, Slow Test?
When using $$k$$-nearest neighbors, we say that it is fast at train time, slow at test (predict) time. Let’s see if this is true for the specific implementation used in knnreg().
[Exercise] Use the microbenchmark() function from the microbenchmark package to compare the runtimes of the following two lines.
fit = knnreg(bwt ~ ., data = bwt_trn_data, k = 5)
pred = predict(fit, newdata = bwt_tst_data)
microbenchmark(fit = knnreg(bwt ~ ., data = bwt_trn_data, k = 5),
times = 100, unit = "ms")
## Unit: milliseconds
## expr min lq mean median uq max neval
## fit 1.022332 1.139804 1.434355 1.299095 1.681022 3.097059 100
microbenchmark(pred = predict(fit, newdata = bwt_tst_data),
times = 100, unit = "ms")
## Unit: milliseconds
## expr min lq mean median uq max neval
## pred 0.896311 0.92754 1.095276 0.9828045 1.147853 2.264674 100
[Exercise] Are the results what you expected? If not, try to explain.
The model “fitting” when we call knnreg() involves re-creating the training dataset, that is applying the preprocessing. For this data we have here, this takes about as long as making predictions on the test data.
[Exercise] Use the microbenchmark() function from the microbenchmark package to compare fitting $$k$$-nearest neighbors with k = 5 to fitting a random forest.
microbenchmark(fit = knnreg(bwt ~ ., data = bwt_trn_data, k = 5),
times = 100, unit = "ms")
## Unit: milliseconds
## expr min lq mean median uq max neval
## fit 1.003668 1.02187 1.118344 1.037106 1.094433 2.007259 100
microbenchmark(fit = randomForest(bwt ~ ., data = bwt_trn_data),
times = 100, unit = "ms")
## Unit: milliseconds
## expr min lq mean median uq max neval
## fit 51.1207 54.28044 64.41487 57.05296 59.55442 297.6403 100 | 2020-01-17 18:35:02 | {"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": 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": 0.3086089789867401, "perplexity": 9693.9911677734}, "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-2020-05/segments/1579250590107.3/warc/CC-MAIN-20200117180950-20200117204950-00419.warc.gz"} |
http://nrich.maths.org/988/index?nomenu=1 | I have a square table-mat made from rope that spirals from the centre. It is rather like this one:
The mat is a $14$ centimetre square. The rope is $2$ cm wide.
How many centimetres of rope will I need to make another mat just like it?
How much rope would I need to make a $12$ cm square, or a $16$ cm square?
Is there a quick way to work this out? | 2015-07-01 02:06:50 | {"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": 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": 0.4703349173069, "perplexity": 884.059869832843}, "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-2015-27/segments/1435375094634.87/warc/CC-MAIN-20150627031814-00048-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://www.mathworks.com/help/physmod/simscape/ref/rotationalmechanicalconvertertl.html?requestedDomain=true&nocookie=true | # Rotational Mechanical Converter (TL)
Interface between thermal liquid and mechanical rotational networks
## Library
Thermal Liquid/Elements
## Description
The Rotational Mechanical Converter (TL) block represents the liquid side of a rotational mechanical interface. This interface converts liquid pressure into torque and vice versa. The output torque acts in a single direction, set using a Mechanical orientation parameter.
The rotational mechanical interface contains no hard stops. To include hard stops, use the Simscape™ Rotational Hard Stop block. A model of a rotational hydraulic actuator, for example, requires both blocks.
Port A is a thermal liquid conserving port corresponding to the converter inlet. Liquid pressure in the converter equals that at port A. Port Q is a thermal conserving port for modeling heat exchange between the converter liquid and the converter housing. Liquid temperature in the converter equals that at port Q.
### Mass Balance
The mass conservation equation in the mechanical converter volume is
`${\stackrel{˙}{m}}_{\text{A}}=\epsilon \rho D\Omega +\left\{\begin{array}{cc}0,& \text{if}\text{\hspace{0.17em}}\text{fluid}\text{\hspace{0.17em}}\text{dynamic}\text{\hspace{0.17em}}\text{compressibility}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{'Off'}\\ V\rho \left(\frac{1}{\beta }\frac{dp}{dt}+\alpha \frac{dT}{dt}\right),& \text{if}\text{\hspace{0.17em}}\text{fluid}\text{\hspace{0.17em}}\text{dynamic}\text{\hspace{0.17em}}\text{compressibility}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{'On'}\end{array}$`
where:
• is the liquid mass flow rate into the converter through port A.
• ε is the mechanical orientation of the converter (`1` if positive, `-1` if negative).
• ρ is the liquid mass density.
• D is the converter displacement.
• Ω is the angular velocity of the converter interface (positive for converter expansion, negative for converter contraction).
• V is the liquid volume inside the converter.
• β is the liquid bulk modulus inside the converter.
• α is the coefficient of thermal expansion of the liquid.
• p is the liquid pressure inside the converter.
• T is the liquid temperature inside the converter.
### Momentum Balance
The momentum conservation equation in the mechanical converter volume is
`$\tau =-\epsilon \left(p-{p}_{\text{Atm}}\right)D,$`
where:
• τ is the torque the liquid exerts on the converter interface.
• pAtm is the atmospheric pressure.
### Energy Balance
The energy conservation equation in the mechanical converter volume is
`$\frac{d\left(\rho uV\right)}{dt}={\varphi }_{\text{A}}+{Q}_{H}-pD\epsilon \Omega ,$`
where:
• u is the liquid internal energy in the converter.
• ϕA is the total energy flow rate into the mechanical converter volume through port A.
• QH is the heat flow rate into the mechanical converter volume through the converter wall.
### Block Source Code
The block dialog box does not have a Source code link. To view the underlying component source, open the following files in the MATLAB® editor:
• For the code corresponding to fluid dynamic compressibility `Off``rotational_converter.ssc`
• For the code corresponding to fluid dynamic compressibility `On``rotational_converter_compressibility.ssc`
## Assumptions and Limitations
• Converter walls are not compliant. They cannot deform regardless of internal pressure and temperature.
• The converter contains no mechanical hard stop.
## Parameters
Mechanical orientation
Select the relative orientation of the converter with respect to the thermal liquid system. The relative orientation determines the rotation direction associated with positive flow into the converter. That direction is positive if the mechanical orientation of the converter is positive. It is negative if the mechanical orientation of the converter is negative. The default setting is `Positive`.
Interface volume displacement
Enter the displaced liquid volume corresponding to a unit rotation angle of the spinning converter interface. The default value is `1.2e-4` m^3/rad.
Interface initial rotation
Enter the rotation angle between the spinning converter interface and the clamping structure at time zero. The angle should be positive for positive mechanical orientations and negative for negative mechanical orientations. The default value is `0` rad.
Enter the liquid volume remaining in the converter at a zero rotation angle. The default value is `1e-5` m^3.
Cross-sectional area at port A
Enter the flow cross-sectional area at the converter inlet. The block uses this parameter for thermal conduction calculations. The default value is `0.01` m^2.
Environment pressure specification
Select a specification method for the environment pressure. Options include `Specified pressure` and ```Atmospheric pressure```. The default setting is ```Atmospheric pressure```.
Environment pressure
Enter the environment pressure for the component. This parameter is active only when the Environment pressure specification parameter is set to `Specified pressure`. The default value is `0.101325` MPa.
Fluid dynamic compressibility
Select whether to include the effect of fluid dynamic compressibility on the transient response of the converter model. Selecting `On` exposes an additional parameter. The default setting is `Off`.
Initial liquid temperature
Enter the liquid temperature in the converter at time zero. The default value is `293.15` K.
Initial liquid pressure
Enter the liquid pressure in the converter at time zero. This parameter is visible only if Fluid dynamic compressibility is `On`. The default value is `1` atm.
## Ports
This block has four ports.
A Thermal liquid conserving port H Thermal conserving port R Rotational mechanical conserving port associated with the moving interface C Rotational mechanical conserving port associated with the converter casing | 2018-02-21 07:44:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "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": 0.7415251731872559, "perplexity": 3098.2522158514626}, "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-2018-09/segments/1518891813571.24/warc/CC-MAIN-20180221063956-20180221083956-00134.warc.gz"} |
https://www.rdocumentation.org/packages/grDevices/versions/3.4.1/topics/trans3d | grDevices (version 3.4.1)
# trans3d: 3D to 2D Transformation for Perspective Plots
## Description
Projection of 3-dimensional to 2-dimensional points using a 4x4 viewing transformation matrix. Mainly for adding to perspective plots such as persp.
## Usage
trans3d(x, y, z, pmat)
## Arguments
x, y, z
numeric vectors of equal length, specifying points in 3D space.
pmat
a $$4 \times 4$$ viewing transformation matrix, suitable for projecting the 3D coordinates $$(x,y,z)$$ into the 2D plane using homogeneous 4D coordinates $$(x,y,z,t)$$; such matrices are returned by persp().
## Value
a list with two components
x,y
the projected 2d coordinates of the 3d input (x,y,z).
## See Also
persp
## Examples
Run this code
<!-- %% it would be nice to have an independent example -->
## See help(persp) {after attaching the 'graphics' package}
## -----------
Run the code above in your browser using DataCamp Workspace | 2022-06-26 22:54:14 | {"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": 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": 0.5086543560028076, "perplexity": 8134.6304787114705}, "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-2022-27/segments/1656103322581.16/warc/CC-MAIN-20220626222503-20220627012503-00768.warc.gz"} |
https://ask.sagemath.org/questions/8732/revisions/ | # Revision history [back]
### Thickness of points in plots
How can I change the thickness of points in a graph?
In case it is not clear what I mean, consider the following program, illustrating Chebyshev's bias:
var('N') N=[] l=0 k=7 for p in prime_range(10^k+1): if (p==2): N.append(0)
else: if (Mod(p,4)==1): N.append(N[l]-1) l=l+1 else: N.append(N[l]+1) l=l+1 list_plot(N).show()
(Sorry, I cannot yet upload graphs, but you may quickly run this example and see that:) The dots seem too thick to rally appreciate any fine features of the graph. Is there a way to make the dots smaller (thinner)?
2 formatting Shashank 1887 ●26 ●47 ●81
### Thickness of points in plots
How can I change the thickness of points in a graph?
In case it is not clear what I mean, consider the following program, illustrating Chebyshev's bias:
var('N')
N=[]
l=0
k=7
for p in prime_range(10^k+1):
if (p==2):
N.append(0)
else:
if (Mod(p,4)==1):
N.append(N[l]-1)
l=l+1
else:
N.append(N[l]+1)
l=l+1
list_plot(N).show()list_plot(N).show()
(Sorry, I cannot yet upload graphs, but you may quickly run this example and see that:) The dots seem too thick to rally appreciate any fine features of the graph. Is there a way to make the dots smaller (thinner)? | 2020-10-21 13:52:59 | {"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": 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": 0.473959743976593, "perplexity": 3173.1562508148854}, "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-2020-45/segments/1603107876500.43/warc/CC-MAIN-20201021122208-20201021152208-00518.warc.gz"} |
http://danielmarcelino.com/ | # Argentine general election, 2015
The 2015 Argentine's presidential election to be held next October 25th is approaching and the dispute begun to appear more clearly since the major parties announced their potential candidates last June.
This Sunday, the political parties are holding their primaries for the upcoming presidential election. As in US, in Argentine the primaries are important for parties to solve internal disputes, so the winning candidate can run more comfortable with a united party.
As usual, I set up a forecasting model to track down the vote intentions in my neighbouring country--the land of tango. I'm still consolidating opinion polls data while trying to get some clues about past pollster's performance, so I can account for the likely house effect.
The following graph was adjusted using simple loess techniques. As I already have a reasonable population of polls, I could adjust a Dirichlet regression, which produces a more robust picture of the dispute (the second graph below), though at this stage the model is an oversimplification as some pollsters are more reliable than others. So, I hope next time to post a more sound forecast.
From the figures above, we can see that some polls have quite weird sinusoidal artifacts. Considering those are wrong compared to the others, they can influence the trend line estimates if on a particular day the deviating poll is the only measurement we have.
## House Effects
I want to improve the estimates over the next weeks by using better priors for the house effect of each polling firm. For example, in the picture below, pollster "OPSM" polled favourably for Mauricio Macri (blue line/dots) while polled worst for the Kristina Kirschner's candidate, Daniel Scioli. On the other hand, pollster "Hugo Haime & Asc." underestimated the principal opposition candidate (M. Macri) while overestimated Daniel Scioli plus the PJ's dissident, Sergio Massa.
Let's think about the implications of this for a moment. Some institutes published polls in which the one or the other candidate over a period of several months is predicted on average two percents below/above the median of all the polls published in that period; this is really hard to believe given that the polling organizations are all claiming to interview a representative group of the population. Even if we acknowledge that the polls are producing some noisy measurements, there would not be this kind of hex. Unless there is a systematic error in the polls that occurs over and over.
# Ternary plots in politics
This week, I read a post by Nicholas Hamilton about ternary plots that made me think, how this geometric diagram has many different application in science fields. Couple weeks ago, I was reading a book by Donald Saari, who uses ternary charts massively to project election outcomes in different electoral settings.
Perhaps, most common, ternary diagrams are used for projecting 3-parties elections; the same idea can be generalised for more parties, though it gets more complex. When we study elections with this geometric figure, each point of the triangle intuitively represents 3 coordinates (say A, B, C, because I'm creative today) that correspond to the percentage of votes each political organisation obtained. From this, we may speculate the composition of the legislative body. Here I use colors to make it fancier.
The whole point of the book is not about election outcome visualisation, but that the election outcome may be depend on the formula used to aggregate votes and the number of seats available. For instance, if we have a constituency of M=5 seats, the possible outcomes are:
If party "A" has 60% of the popular preference, party "B" 20%, and "C" other 20%, then a system using d'Hondt to distribute seats proportionally among the parties will give: A(3), B(1), C(1).
Rather than thinking on edges, we can draw regions to make more clear how electoral formulas cause small, but different shapes, which in the long run may affect the number of parties and coordination among party supporters, to mention few adverse reactions.
# I loved this %>% crosstable
This is a public tank you for @heatherturner's contribution. Now the SciencesPo's crosstable can work in a chain (%>%) fashion; useful for using along with other packages that have integrated the magrittr operator.
> candidatos %>%
desc_cargo =='DEPUTADO DISTRITAL' | desc_cargo =='DEPUTADO FEDERAL' |
tab(desc_cargo,desc_sexo)
====================================================
desc_sexo
-------------------------
desc_cargo NA FEMININO MASCULINO Total
----------------------------------------------------
DEPUTADO DISTRITAL 1 826 2457 3284
0.03% 25% 75% 100%
0.20% 21% 79% 100%
DEPUTADO FEDERAL 40 5006 20176 25222
0.16% 20% 80% 100%
SENADOR 4 161 1002 1167
0.34% 14% 86% 100%
VEREADOR 9682 376576 1162973 1549231
0.62% 24% 75% 100%
----------------------------------------------------
Total 9849 395164 1234933 1639946
0.60% 24% 75% 100%
====================================================
Chi-Square Test for Independence
Number of cases in table: 1639946
Number of factors: 2
Test for independence of all factors:
Chisq = 1077.4, df = 8, p-value = 2.956e-227
X^2 df P(> X^2)
Likelihood Ratio 1216.0 8 0
Pearson 1077.4 8 0
Phi-Coefficient : 0.026
Contingency Coeff.: 0.026
Cramer's V : 0.018
# Reproducible example:
library(SciencesPo)
gender = rep(c("female","male"),c(1835,2691))
dept = rep(c("A","B","C","D","E","F","A","B","C","D","E","F"),
c(89,17,202,131,94,24,19,8,391,244,299,317))
dept2 = rep(c("A","B","C","D","E","F","A","B","C","D","E","F"),
c(512,353,120,138,53,22,313,207,205,279,138,351))
department = c(dept,dept2)
> ucb %>% tab(admitted, gender, department)
================================================================
department
-----------------------------------------
admitted gender A B C D E F Total
----------------------------------------------------------------
no female 19 8 391 244 299 317 1278
1.5% 0.63% 31% 19% 23.4% 25% 100%
male 313 207 205 279 138 351 1493
21.0% 13.86% 14% 19% 9.2% 24% 100%
-------------------------------------------------------
Total 332 215 596 523 437 668 2771
12.0% 7.76% 22% 19% 15.8% 24% 100%
----------------------------------------------------------------
yes female 89 17 202 131 94 24 557
16% 3.1% 36% 24% 16.9% 4.3% 100%
male 512 353 120 138 53 22 1198
43% 29.5% 10% 12% 4.4% 1.8% 100%
-------------------------------------------------------
Total 601 370 322 269 147 46 1755
34% 21.1% 18% 15% 8.4% 2.6% 100%
----------------------------------------------------------------
Total female 108 25 593 375 393 341 1835
5.9% 1.4% 32% 20% 21.4% 19% 100%
male 825 560 325 417 191 373 2691
30.7% 20.8% 12% 15% 7.1% 14% 100%
-------------------------------------------------------
Total 933 585 918 792 584 714 4526
20.6% 12.9% 20% 17% 12.9% 16% 100%
================================================================
# The Greek thing II
Just few hours before Greeks head to the polls to decide on the bailout agreement, and ultimately, whether the country will stay in the euro, there is no overwhelming advantage of either side. Actually, the margin became blurred over the last three days, with the "Yes" position rehearsing a last-minute recovery. Despite this last-minute trend, the aggregate preference for the "NO" is not too far behind. To frame this in terms of probabilities, that is, the $\theta_{YES}$ exceeds $\theta_{NO}$, I adapted a short function written a while ago to simulate from a Dirichlet distribution, and then to compute posterior probabilities shown in the chart below. It's really nothing, but the YES outperformed the NO in 57%.
The polls were aggregated and the "Don't Know" respondents were distributed accordingly to proportion of the Yes/No reported by the polls.
UPDATE:
With polls yesterday showing both sides in a dead heat, today's overwhelmingly majority of voters saying NO is a big surprise, isn't? Plenty of theories will appear to explain why Greeks have chosen to reject the terms of the deal as proposed by EU officials, meanwhile, it's time for the parties to set up the plan B.
# The Greek thing
Greeks have been quite volatile on their opinion whether they should accept or not a proposal by the country's creditors for more austerity to keep aid flowing. The polls conducted over this week look like crazy, though that "belly" was likely provoked by the anxiety on what comes next after Greece not paying IMF back.
The data were collected on the internet, most of them assembled by http://metapolls.net | 2015-09-02 12:44:04 | {"extraction_info": {"found_math": true, "script_math_tex": 2, "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": 2, "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": 0.44670039415359497, "perplexity": 3184.4328508161657}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2015-35/segments/1440645264370.66/warc/CC-MAIN-20150827031424-00199-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://junojunho.github.io/algorithm/algorithm-practice-isTreeSymmetric/ | # Is Tree Symmetric
## Problem name: isTreeSymmetric
Problem description is here.
#
# Definition for binary tree:
# class Tree(object):
# def __init__(self, x):
# self.value = x
# self.left = None
# self.right = None
def isTreeSymmetric(t):
if t is None or (t.left is None and t.right is None):
return True
if t.left is None and t.right is not None:
return False
if t.left is not None and t.right is None:
return False
left_cursor = t.left
right_cursor = t.right
return isEqual(left_cursor, right_cursor)
def isEqual(left, right):
# value
# shape
if left is None or right is None:
return left is None and right is None
return left.value == right.value and isEqual(left.left, right.right) and isEqual(left.right, right.left)
## Time Complexity
O(N) # N is the length of BinaryTree
## Solving Strategy
The symmetric condition means that left subtree of root node is the mirror image of the right subtree of root node.
Then, we can recursively decide this tree is symmetric using two conditions, left node of left subtree is same as right node of right subtree, and right node of left subtree is same as left node of right subtree. If one node is empty, which is None, then the mirror image of the node is also empty. Furthermore, the value of two nodes also need to same.
Since all condition is concatenate using and operator, one condition is false, final result is false, which means that the tree is not symmetric.
태그:
카테고리:
업데이트: | 2021-01-24 11:19:20 | {"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": 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": 0.6124995946884155, "perplexity": 5928.508897995262}, "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-2021-04/segments/1610703548716.53/warc/CC-MAIN-20210124111006-20210124141006-00310.warc.gz"} |
https://tex.stackexchange.com/questions/246505/remove-header-from-the-page-after-toc-and-lof | # Remove header from the page after TOC and LOF
I am using Koma-script book document class, twosides, and I also use the package \fancyhdr to modify my pages.
The problem is that header appears both in the next page after TOC and after LOF, so I wonder if there a way to remove the header from these pages
• I tried to use '\fancyhead{}', but this command just removed the name from the header. What actually I want is the pages after TOC and LOF to be completely empty – Yorgos May 23 '15 at 0:48
I got this code snippet many, many years ago from I don't remember where, so unfortunately I can't credit its originator; but it certainly isn't mine.
\makeatletter \def\cleardoublepage{\clearpage\if@twoside
\ifodd\c@page\else \hbox{} \vspace*{\fill}
\thispagestyle{empty} \newpage
\if@twocolumn\hbox{}\newpage\fi\fi\fi} \makeatother
All this really does is redefine \cleardoublepage to include \thispagestyle{empty}, but this will do what you want (if I'm interpreting what you want correctly). It will also remove headers and footers from any page that's cleared by \cleardoublepage; e.g., blank verso pages at the end of a chapter.
• @dgoomaniii Thank you very much!!!! It worked!!! Since I use LyX I include the code above the TOC – Yorgos May 23 '15 at 1:26
• Actually, either, but it makes more sense in the preamble. – dgoodmaniii May 23 '15 at 1:26
You are doing something to change the defaults of scrbook (see the example below).
To ensure that the blank pages after TOC and LOF are empty insert
\KOMAoptions{cleardoublepage=empty}
at the end of your preamble or at the very beginning of the document.
Here is an example to show that with scrbook the blank page after TOC is in the default empty:
\documentclass{scrbook}
\usepackage{fancyhdr}
\pagestyle{fancy}
\usepackage{blindtext}
\begin{document}
\tableofcontents
\Blinddocument
\end{document} | 2020-05-24 22:45:03 | {"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": 0.9297631978988647, "perplexity": 2786.24676103976}, "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/1590347385193.5/warc/CC-MAIN-20200524210325-20200525000325-00418.warc.gz"} |
https://phys.libretexts.org/Bookshelves/Astronomy__Cosmology/Book%3A_Astronomy_(OpenStax)/18%3A_The_Stars_-_A_Celestial_Census/18.03%3A_Diameters_of_Stars | $$\require{cancel}$$
18.3: Diameters of Stars
Learning Objectives
By the end of this section, you will be able to:
• Describe the methods used to determine star diameters
• Identify the parts of an eclipsing binary star light curve that correspond to the diameters of the individual components
It is easy to measure the diameter of the Sun. Its angular diameter—that is, its apparent size on the sky—is about 1/2°. If we know the angle the Sun takes up in the sky and how far away it is, we can calculate its true (linear) diameter, which is 1.39 million kilometers, or about 109 times the diameter of Earth.
Unfortunately, the Sun is the only star whose angular diameter is easily measured. All the other stars are so far away that they look like pinpoints of light through even the largest ground-based telescopes. (They often seem to be bigger, but that is merely distortion introduced by turbulence in Earth’s atmosphere.) Luckily, there are several techniques that astronomers can use to estimate the sizes of stars.
Stars Blocked by the Moon
One technique, which gives very precise diameters but can be used for only a few stars, is to observe the dimming of light that occurs when the Moon passes in front of a star. What astronomers measure (with great precision) is the time required for the star’s brightness to drop to zero as the edge of the Moon moves across the star’s disk. Since we know how rapidly the Moon moves in its orbit around Earth, it is possible to calculate the angular diameter of the star. If the distance to the star is also known, we can calculate its diameter in kilometers. This method works only for fairly bright stars that happen to lie along the zodiac, where the Moon (or, much more rarely, a planet) can pass in front of them as seen from Earth.
Eclipsing Binary Stars
Accurate sizes for a large number of stars come from measurements of eclipsing binary star systems, and so we must make a brief detour from our main story to examine this type of star system. Some binary stars are lined up in such a way that, when viewed from Earth, each star passes in front of the other during every revolution (Figure $$\PageIndex{1}$$). When one star blocks the light of the other, preventing it from reaching Earth, the luminosity of the system decreases, and astronomers say that an eclipse has occurred.
The discovery of the first eclipsing binary helped solve a long-standing puzzle in astronomy. The star Algol, in the constellation of Perseus, changes its brightness in an odd but regular way. Normally, Algol is a fairly bright star, but at intervals of 2 days, 20 hours, 49 minutes, it fades to one-third of its regular brightness. After a few hours, it brightens to normal again. This effect is easily seen, even without a telescope, if you know what to look for.
In 1783, a young English astronomer named John Goodricke (1764–1786) made a careful study of Algol (see the feature on John Goodricke in Section 19.3 for a discussion of his life and work). Even though Goodricke could neither hear nor speak, he made a number of major discoveries in the 21 years of his brief life. He suggested that Algol’s unusual brightness variations might be due to an invisible companion that regularly passes in front of the brighter star and blocks its light. Unfortunately, Goodricke had no way to test this idea, since it was not until about a century later that equipment became good enough to measure Algol’s spectrum.
In 1889, the German astronomer Hermann Vogel (1841–1907) demonstrated that, like Mizar, Algol is a spectroscopic binary. The spectral lines of Algol were not observed to be double because the fainter star of the pair gives off too-little light compared with the brighter star for its lines to be conspicuous in the composite spectrum. Nevertheless, the periodic shifting back and forth of the brighter star’s lines gave evidence that it was revolving about an unseen companion. (The lines of both components need not be visible for a star to be recognized as a spectroscopic binary.)
The discovery that Algol is a spectroscopic binary verified Goodricke’s hypothesis. The plane in which the stars revolve is turned nearly edgewise to our line of sight, and each star passes in front of the other during every revolution. (The eclipse of the fainter star in the Algol system is not very noticeable because the part of it that is covered contributes little to the total light of the system. This second eclipse can, however, be detected by careful measurements.)
Any binary star produces eclipses if viewed from the proper direction, near the plane of its orbit, so that one star passes in front of the other (see Figure $$\PageIndex{1}$$). But from our vantage point on Earth, only a few binary star systems are oriented in this way.
ASTRONOMY AND MYTHOLOGY: ALGOL THE DEMON STAR AND PERSEUS THE HERO
The name Algol comes from the Arabic Ras al Ghul, meaning “the demon’s head.”1 The word “ghoul” in English has the same derivation. As discussed in Observing the Sky: The Birth of Astronomy, many of the bright stars have Arabic names because during the long dark ages in medieval Europe, it was Arabic astronomers who preserved and expanded the Greek and Roman knowledge of the skies. The reference to the demon is part of the ancient Greek legend of the hero Perseus, who is commemorated by the constellation in which we find Algol and whose adventures involve many of the characters associated with the northern constellations.
Perseus was one of the many half-god heroes fathered by Zeus (Jupiter in the Roman version), the king of the gods in Greek mythology. Zeus had, to put it delicately, a roving eye and was always fathering somebody or other with a human maiden who caught his fancy. (Perseus derives from Per Zeus, meaning “fathered by Zeus.”) Set adrift with his mother by an (understandably) upset stepfather, Perseus grew up on an island in the Aegean Sea. The king there, taking an interest in Perseus’ mother, tried to get rid of the young man by assigning him an extremely difficult task.
In a moment of overarching pride, a beautiful young woman named Medusa had compared her golden hair to that of the goddess Athena (Minerva for the Romans). The Greek gods did not take kindly to being compared to mere mortals, and Athena turned Medusa into a gorgon: a hideous, evil creature with writhing snakes for hair and a face that turned anyone who looked at it into stone. Perseus was given the task of slaying this demon, which seemed like a pretty sure way to get him out of the way forever.
But because Perseus had a god for a father, some of the other gods gave him tools for the job, including Athena’s reflective shield and the winged sandals of Hermes (Mercury in the Roman story). By flying over her and looking only at her reflection, Perseus was able to cut off Medusa’s head without ever looking at her directly. Taking her head (which, conveniently, could still turn onlookers to stone even without being attached to her body) with him, Perseus continued on to other adventures.
He next came to a rocky seashore, where boasting had gotten another family into serious trouble with the gods. Queen Cassiopeia had dared to compare her own beauty to that of the Nereids, sea nymphs who were daughters of Poseidon (Neptune in Roman mythology), the god of the sea. Poseidon was so offended that he created a sea-monster named Cetus to devastate the kingdom. King Cepheus, Cassiopeia’s beleaguered husband, consulted the oracle, who told him that he must sacrifice his beautiful daughter Andromeda to the monster.
When Perseus came along and found Andromeda chained to a rock near the sea, awaiting her fate, he rescued her by turning the monster to stone. (Scholars of mythology actually trace the essence of this story back to far-older legends from ancient Mesopotamia, in which the god-hero Marduk vanquishes a monster named Tiamat. Symbolically, a hero like Perseus or Marduk is usually associated with the Sun, the monster with the power of night, and the beautiful maiden with the fragile beauty of dawn, which the Sun releases after its nightly struggle with darkness.)
Many of the characters in these Greek legends can be found as constellations in the sky, not necessarily resembling their namesakes but serving as reminders of the story. For example, vain Cassiopeia is sentenced to be very close to the celestial pole, rotating perpetually around the sky and hanging upside down every winter. The ancients imagined Andromeda still chained to her rock (it is much easier to see the chain of stars than to recognize the beautiful maiden in this star grouping). Perseus is next to her with the head of Medusa swinging from his belt. Algol represents this gorgon head and has long been associated with evil and bad fortune in such tales. Some commentators have speculated that the star’s change in brightness (which can be observed with the unaided eye) may have contributed to its unpleasant reputation, with the ancients regarding such a change as a sort of evil “wink.”
Diameters of Eclipsing Binary Stars
We now turn back to the main thread of our story to discuss how all this can be used to measure the sizes of stars. The technique involves making a light curve of an eclipsing binary, a graph that plots how the brightness changes with time. Let us consider a hypothetical binary system in which the stars are very different in size, like those illustrated in Figure $$\PageIndex{2}$$. To make life easy, we will assume that the orbit is viewed exactly edge-on.
Even though we cannot see the two stars separately in such a system, the light curve can tell us what is happening. When the smaller star just starts to pass behind the larger star (a point we call first contact), the brightness begins to drop. The eclipse becomes total (the smaller star is completely hidden) at the point called second contact. At the end of the total eclipse (thirdcontact), the smaller star begins to emerge. When the smaller star has reached last contact, the eclipse is completely over.
To see how this allows us to measure diameters, look carefully at Figure $$\PageIndex{2}$$. During the time interval between the first and second contacts, the smaller star has moved a distance equal to its own diameter. During the time interval from the first to third contacts, the smaller star has moved a distance equal to the diameter of the larger star. If the spectral lines of both stars are visible in the spectrum of the binary, then the speed of the smaller star with respect to the larger one can be measured from the Doppler shift. But knowing the speed with which the smaller star is moving and how long it took to cover some distance can tell the span of that distance—in this case, the diameters of the stars. The speed multiplied by the time interval from the first to second contact gives the diameter of the smaller star. We multiply the speed by the time between the first and third contacts to get the diameter of the larger star.
In actuality, the situation with eclipsing binaries is often a bit more complicated: orbits are generally not seen exactly edge-on, and the light from each star may be only partially blocked by the other. Furthermore, binary star orbits, just like the orbits of the planets, are ellipses, not circles. However, all these effects can be sorted out from very careful measurements of the light curve.
Using the Radiation Law to Get the Diameter
Another method for measuring star diameters makes use of the Stefan-Boltzmann law for the relationship between energy radiated and temperature (see Radiation and Spectra). In this method, the energy flux (energy emitted per second per square meter by a blackbody, like the Sun) is given by
$F= \sigma T^4 \nonumber$
where $$\sigma$$ is a constant and $$T$$ is the temperature. The surface area of a sphere (like a star) is given by
$A=4 \pi R^2 \nonumber$
The luminosity ($$L$$) of a star is then given by its surface area in square meters times the energy flux:
$L=(A \times F)$
Previously, we determined the masses of the two stars in the Sirius binary system. Sirius gives off 8200 times more energy than its fainter companion star, although both stars have nearly identical temperatures. The extremely large difference in luminosity is due to the difference in radius, since the temperatures and hence the energy fluxes for the two stars are nearly the same. To determine the relative sizes of the two stars, we take the ratio of the corresponding luminosities:
$\begin{array}{c} \frac{L_{\text{Sirius}}}{L_{\text{companion}}}=\frac{ \left( A_{\text{Sirius}} \times F_{\text{Sirius}} \right)}{ \left( A_{\text{companion}} \times F_{\text{companion}} \right)} \\ = \frac{A_{\text{Sirius}}}{A_{\text{companion}}}= \frac{4 \pi R^2_{\text{Sirius}}}{4 \pi R^2_{\text{companion}}}= \frac{R^2_{\text{Sirius}}}{R^2_{\text{companion}}} \\ \frac{L_{\text{Sirius}}}{L_{\text{companion}}}=8200= \frac{R^2_{\text{Sirius}}}{R^2_{\text{companion}}} \end{array} \nonumber$
Therefore, the relative sizes of the two stars can be found by taking the square root of the relative luminosity. Since $$\sqrt{8200} = 91$$, the radius of Sirius is 91 times larger than the radium of its faint companion.
The method for determining the radius shown here requires both stars be visible, which is not always the case.
Stellar Diameters
The results of many stellar size measurements over the years have shown that most nearby stars are roughly the size of the Sun, with typical diameters of a million kilometers or so. Faint stars, as we might have expected, are generally smaller than more luminous stars. However, there are some dramatic exceptions to this simple generalization.
A few of the very luminous stars, those that are also red (indicating relatively low surface temperatures), turn out to be truly enormous. These stars are called, appropriately enough, giant stars or supergiant stars. An example is Betelgeuse, the second brightest star in the constellation of Orion and one of the dozen brightest stars in our sky. Its diameter, remarkably, is greater than 10 AU (1.5 billion kilometers!), large enough to fill the entire inner solar system almost as far out as Jupiter. In Stars from Adolescence to Old Age, we will look in detail at the evolutionary process that leads to the formation of such giant and supergiant stars.
Watch this star size comparison video for a striking visual that highlights the size of stars versus planets and the range of sizes among stars.
Summary
The diameters of stars can be determined by measuring the time it takes an object (the Moon, a planet, or a companion star) to pass in front of it and block its light. Diameters of members of eclipsing binary systems (where the stars pass in front of each other) can be determined through analysis of their orbital motions.
Footnotes
3Fans of Batman comic books and movies will recognize that this name was given to an archvillain in the series.
Glossary
eclipsing binary
a binary star in which the plane of revolution of the two stars is nearly edge-on to our line of sight, so that the light of one star is periodically diminished by the other passing in front of it
18.3: Diameters of Stars is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. | 2022-05-20 05:00:26 | {"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": 0.6126713156700134, "perplexity": 1098.6340567279601}, "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-2022-21/segments/1652662531352.50/warc/CC-MAIN-20220520030533-20220520060533-00462.warc.gz"} |
https://forum.allaboutcircuits.com/threads/trouble-with-trying-to-make-a-two-stage-amplifier.13229/ | # Trouble with trying to make a two stage amplifier :(
Discussion in 'The Projects Forum' started by sitech, Jul 24, 2008.
1. ### sitech Thread Starter Member
Jul 3, 2008
23
0
I've been trying to build a two stage amplifier with transistors.
I've made a one-stage amplifier and it works as expected (I'm using GNUCAP and gEDA), when I then try to stick an additional stage onto the output of the first stage, I get distortion.
My idea was to have a resistor on the emitter with a capacitor in parallel, so the AC would be amplified, but the DC would have a gain of only 1. This worked fine with the first stage.
The second stage, with the transistor base connected to the output of the first stage, has an identical setup.
However, the second I plug the second stage in, when I look at the base voltage, instead of getting a nice sine wave (thats what i've been testing it with), I get a weird stepped graph. I'll try and get a picture.
Here is my circuit:
Note, there is normally a capacitor of 9 nF on the second stage with a 4.5k resistor in parallel (instead of the 1k) , just my last test was to see what happened if I removed it.
The base output looks like this:
My thoughts is that it good be input and output resistance, but I'm not sure.
I've also tried putting in a resistor in between the base of Q2 and the collector of Q1, with no beneficial effect. I only did it because I've seen it in other two stage amplifiers, but I have no idea why they did it!
Any help would be much appreciated, I've been pulling my hair out for over a day now!
NOTE: R3 is meant to be 4.5k... whoops! I'll fix that up soon.
Last edited: Jul 25, 2008
2. ### roddefig Active Member
Apr 29, 2008
149
0
While there are others on this forum who are more knowledgable than I when it comes to analog electronics, have you accounted for the loading effect of the second stage?
I believe the correct formula for the input resistance of the second stage would be $R_i = (\beta + 1)(r_e + R_E)$
3. ### sitech Thread Starter Member
Jul 3, 2008
23
0
See thats what I thought it might be.
But the output impedence of the first stage is 4.5 k (well, yes the resistor is 7.5k, but thats an error on the diagram).
In the original circuit where the second stage resistor was 4.5k and bypassed with a 9 nF capacitor I calculated (if my logic is correct).
For DC:
$R_i=4.5k\Omega\times100=450k\Omega$
for AC:
$Z_i=\frac{1}{2\pi\times40KHz\times9\times10^{-9}pF}\times\beta=450\Omega\times100=45k\Omega$
Both of which are at least 10x greater than the output impedence of the first stage!
4. ### Wendy Moderator
Mar 24, 2008
21,155
2,855
Your base resistors (R1 and R2) are way too large for the emitter resistance (R4). Figure a beta of 50, then your DC base input resistance is going to be in the neighborhood of 500KΩ (R4 X Beta). Try dropping R1 and R2 down a factor of 10 or so.
Like the schematic, you are one of the few...
5. ### sitech Thread Starter Member
Jul 3, 2008
23
0
See that shows that I don't understand input and output resistance.
What I got from the book I'm reading is that you want the input resistance to each stage to be high, but the output resistance to be low...
I tried dropping them by a factor of 10 and also 100 however I still got the same result for the base.
Here is the circuit I tested.
6. ### Wendy Moderator
Mar 24, 2008
21,155
2,855
What is the input P-P AC voltage (I'm thinking around 2mv)? Also, what is the frequency?
Did you look at your DC voltages before and after? Betcha you saw a lot of change. The transistor needs to be firmly in its linear region for it to be linear. The DC biasing is important for this, so you have to look at your DC parameters before you look at the AC.
I'm going to be looking at this a bit more, it's the kind of problem I like. BTW, I use the computer between my ears to simulate, when SgtWookie gets here he'll be a bit more use than I. You might think about posting your files for review.
Something else to consider, assuming 50X gain on each stage (with emitter swamping caps this isn't too unreasonable) then 2mv X 50 = 100mv, 100mv X 50 = 5V. I don't think the DC biasing will allow this kind of swing.
OK, I'm not using a calculator here, so this is off the cuff. The base of the first transistor is almost 5V, so you have 4.3V on the emitter. This means the collector of Q1 is going to be 4.7V. The emitter of Q2 will be around 4V, which means the collector of Q2 is around 5V. Since the collector can't go more negitive than the emitter this would seem to present a problem.
Last edited: Jul 25, 2008
7. ### sitech Thread Starter Member
Jul 3, 2008
23
0
Thanks for the help I'm using a 1 mv input.
okay now I'm really confused, and I'm not sure if it is a problem with the software I'm using.
I decided to fix the problem you pointed out by making the voltage divider produce 7V, thus the base of stage 1 producing 6.3V, which means the output should oscillate around 2.7V (first stage doesn't seem to exceed 0.01V). This should mean the output should oscillate around 7V.
7V is probably too high, but I thought I'd just try that as a test.
However, now when I test it without a signal (just the DC bias), it seems to not work at all as expected. R1 and R2 I have to reduce to 180 and 620 ohms to get out the proper voltage. Increasing R4 does not seem to compensate whatsoever! So basically now for some reason, instead of the output voltage being 2.7V, it seems to want to match the voltage at the emitter of 6.3V! I have no idea why it is doing this?!
8. ### Wendy Moderator
Mar 24, 2008
21,155
2,855
Like I said, the collector CAN NOT be more negative than the emitter. Think current paths. You need to take the emitter to a low voltage, say 2 volts, and the collector will be 7V. Think in terms of current and voltage drop. R3 and R4 are going to drop the same amount of voltage, so if the emitter is 2 volts, the collector will be 9Volts (power supply) - 2volts (R3 drop), 7 volts.
BTW, this will also increase the amount of possible swing in the voltage.
This voltage will propagate through to the base of Q2, so you have to adjust for that.
9. ### sitech Thread Starter Member
Jul 3, 2008
23
0
okay, I'm just going to ask you to completely ignore that last post.
Thats like the second time I've done that today!
Whoops! (embarassed)
Thanks
10. ### Audioguru Expert
Dec 20, 2007
9,883
1,044
Your emitter resistors have a value that is way too high and are wasting half the supply voltage.
Since the collector resistors are 4.5k then use 470 ohms for the emitter resistors.
You have the base voltage so high that the first transistor is saturated and the second transistor is almost saturated.
When the transistors are properly biased then each will have a voltage gain of about 126. The total gain of this amplifier is about 16,000 so the input should be less than 0.2mV.
Make it like this. Without negative feedback then it will be very distorted when the output level is high.
PS. Why is your schematic as a hard to see negative??
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11. ### Wendy Moderator
Mar 24, 2008
21,155
2,855
I was just happy to see a schematic myself.
12. ### sitech Thread Starter Member
Jul 3, 2008
23
0
Thanks! That worked perfectly.
Wow, I'd be stuck without everyone's help.
I just discovered that one of my problems was I was looking at the first 500 microseconds of the circuit, which doesn't seem to be indicative of the rest of the circuit. (Though I doubt I would have worked out the rest of the issues even without that impediment).
Why inverse? It was just how the program I was using (gschem) made them (I assume default colour scheme). The colour scheme you have seems a little easier. The gschem one is a similar scheme to what programs like AutoCAD uses, so maybe its a historical thing?
Thanks once again! Now for the next stage of turning it into a constant 5V (making an ultrasound water tank level meter, so basically trying to time how long it takes for a pulse to reach the reciever).
13. ### sitech Thread Starter Member
Jul 3, 2008
23
0
One final question.
Could all this be achieved with an op-amp. I'm a little confused as to when you'd make a voltage amplifier using transistors and when you would use an op-amp?
14. ### Wendy Moderator
Mar 24, 2008
21,155
2,855
Yep, one of the reasons op amps are more popular is they are MUCH easier to predict. They have their pitfalls, but overall they are much easier than a transistor to design analog amps.
Mostly this is a personal preference, and parts availablility. Transistors are a bit cheaper, but they require more parts. It's always a tradeoff. Like the old saying goes, give a man a hammer and every job looks like a nail.
15. ### Audioguru Expert
Dec 20, 2007
9,883
1,044
An opamp has a voltage gain of about 200,000. The high frequencies are rolled off so that with negative feedback added to reduce the gain then the opamp will not oscillate at a high frequency where the phase shift of the opamp changes the negative feedback into positive feedback.
Therefore an ordinary old 741 opamp has a frequency response to only 80Hz when its gain is 16,000. A better opamp like a TL071 will have a frequency response to 240Hz when its gain is 16,000.
Two better opamps can be used. A TL072 dual opamp can have each opanp with a gain of 126 then the total gain is 15.900 and the response is to about 20khz. The distortion will be very low.
16. ### sitech Thread Starter Member
Jul 3, 2008
23
0
Thanks for the help! I need some assistance with the second stage, the rectification.
The output (voltage across R9) seems to build up gradually (over a period of 2 ms) and then tapers off at about 0.8 V, is the shape of the graph expected (going from 0 to 0.8V in 2 ms or so and then staying at 0.8V).
Finally I was looking at the signal input and is this an artifact of GNUCAP? It seems to occur at EXACTLY 10ms.
Once again, any help would be most welcome!
17. ### Wendy Moderator
Mar 24, 2008
21,155
2,855
Is this meant to be a measurement circuit?
Don't think I'm going to be as much help on this one. It seems to me the cap C5 would have a perminant voltage develop across it, which would affect the rest of the design. How much trouble would it be to add a voltage follower transistor between Q2 and C5 to see?
18. ### Audioguru Expert
Dec 20, 2007
9,883
1,044
I changed some colour hues, made your negative schematic a positive and increased the black and white contrast. Then it is easy to see.
I simulated it and the rectifier/capacitor loads down the output from Q2 so much that its output begins as a small square-wave then slowly increases its amplitude as the capacitor C6 charges. It charges the fastest at 20kHz when the DC voltage across C6 is 1.0V in 1ms.
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35 | 2018-02-24 16:09:28 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "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": 0.6461410522460938, "perplexity": 1376.3140803018168}, "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-2018-09/segments/1518891815843.84/warc/CC-MAIN-20180224152306-20180224172306-00738.warc.gz"} |
https://zbmath.org/?q=an:0608.45007 | ## Positivity and regularity of hyperbolic Volterra equations in Banach spaces.(English)Zbl 0608.45007
We derive necessary and sufficient conditions for two classes of linear Volterra equations of the form (*) $$u=f+a*Au$$ to admit finite wave speed as well as continuity across the wave front. These conditions are based on the positivity of the fundamental solution. One of these classes serves as a model in linear viscoelasticity. These results are then used to obtain several general theorems on existence, positivity, regularity and asymptotic behavior of the resolvent for (*).
### MSC:
45N05 Abstract integral equations, integral equations in abstract spaces 45M05 Asymptotics of solutions to integral equations 74D99 Materials of strain-rate type and history type, other materials with memory (including elastic materials with viscous damping, various viscoelastic materials)
Full Text:
### References:
[1] Carr, R.W., Hannsgen, K.B.: A nonhomogeneous integrodifferential equation in Hilbert space. SIAM J. Math. Anal.10, 961-984 (1979) · Zbl 0411.45013 [2] Carr, R.W., Hannsgen, K.B.: Resolvent formulas for a Volterra equation in Hilbert space. SIAM J. Math. Anal.13, 453-483 (1982) · Zbl 0501.45015 [3] Clement, P., Nohel, J.A.: Abstract linear and nonlinear Volterra equations preserving positivity. SIAM J. Math. Anal.10, 365-388 (1979) · Zbl 0411.45012 [4] Dafermos, C.R.: Asymptotic stability in viscoelasticity. Arch. Rat. Mech. Anal.37, 297-308 (1970) · Zbl 0214.24503 [5] Dafermos, C.R.: An abstract Volterra equation with applications to linear viscoelasticity. J. Differ. Equations7, 554-469 (1970) · Zbl 0212.45302 [6] Da Prato, G., Iannelli, M.: Linear integro-differential equations in Banach space. Rend. Sem. Math. Padova62, 207-219 (1980) · Zbl 0451.45014 [7] Desch, W., Grimmer, R.: Propagation of singularities for integrodifferential equations (to appear) · Zbl 0611.45008 [8] Desch, W., Grimmer, R.: Smoothing properties of linear Volterra integrodifferential equations. Preprint · Zbl 0681.45008 [9] Desch, W., Grimmer, R., Schappacher, W.: Some considerations for linear integrodifferential equations. J. Math. Anal. Appl.104, 219-234 (1984) · Zbl 0595.45027 [10] Greiner, G., Voigt, J., Wolff, M.: On the spectral bound of the generator of a semigroup of positive operators. J. Oper. Theory5, 245-256 (1981) · Zbl 0469.47032 [11] Grimmer, R., Zeman, M.: Wave propagation for linear integrodifferential equations in Banach space. J. Differ. Equations54, 274-282 (1984) · Zbl 0544.45010 [12] Hannsgen, K.B., Wheeler, R.L.: Behavior of the solution of a Volterra equation as a parameter tends to infinity. J. Integral Equations7, 229-237 (1984) · Zbl 0552.45010 [13] Hrusa, W.J., Renardy, M.: On wave propagation in linear viscoelasticity. Quart. Appl. Math.43, 237-253 (1985) · Zbl 0571.73026 [14] Joseph, D.D., Narain, A., Riccius, O.: Shear wave speeds and elastic moduli for different liquids. Preprint. · Zbl 0609.76033 [15] Miller, R.K.: Nonlinear Volterra integral equations. Menlo Park: Benjamin 1971 · Zbl 0448.45004 [16] Miller, R.K., Wheeler, R.L.: Asymptotic behavior for a linear Volterra integral equation in Hilbert space. J. Differ. Equations23, 270-284 (1977) · Zbl 0341.45017 [17] Noren, R.D.: UniformL 1-behavior of the solution of a Volterra equation with a parameter. Preprint · Zbl 0645.45008 [18] Pipkin, A.C.: Lectures on viscoelasticity theory. Appl. Math. Sci.7. Berlin, Heidelberg, New York: Springer 1972 · Zbl 0237.73022 [19] Prüss, J.: Lineare Volterra Gleichungen in Banach-Räumen. Habilitationsschrift, Paderborn (1984) [20] Prüss, J.: On linear Volterra equations of parabolic type in Banach spaces. Trans. Am. Math. Soc.301, 691-721 (1987) · Zbl 0619.45004 [21] Prüss, J.: Bounded solutions of Volterra equations SIAM J. Math. Anal. (to appear) · Zbl 0642.45005 [22] Renardy, M.: Some remarks on the propagation and non-propagation of discontinuities in linearly viscoelastic liquids. Rheol. Acta21, 251-254 (1982) · Zbl 0488.76002 [23] Travis, C.C., Webb, G.F.: Second order differential equations in Banach space. In: Nonlinear equations in abstract spaces. Ed. V. Lakshmikantham. London, New York: Academic Press 1978 · Zbl 0455.34044 [24] Widder, D.V.: The Laplace transform. Princeton: Princeton University Press 1941 · Zbl 0063.08245
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 2022-12-05 15:29:22 | {"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": 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": 0.6775972843170166, "perplexity": 6326.218119672699}, "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-2022-49/segments/1669446711017.45/warc/CC-MAIN-20221205132617-20221205162617-00433.warc.gz"} |
https://electronics.stackexchange.com/questions/428655/high-voltage-and-current | # High voltage and current
I have a DC motor that needs and works nearly 400v and 20A. I chose FDL100N50 for n-channel mosfet also IXTK40P50P for p-channel mosfet. I connect a 4n35 optocupler for output of every mosfet. Are those suitable for this h bridge?
I couldn't find same caracteristics mosfet for n and p channel.
• FDL100N50: 500v, 100 A, 2.5 KW , n-channel
• IXTK40P50P: 500V, 40A, 890W , p-channel
• Welcome to EE.SE. You should hyperlink (neatly) to the datasheets for the devices mentioned. You probably also need a more complete schematic to be sure everyone understands your idea. You can add one in using the CircuitLab button on the editor toolbar. Double-click a component to edit its properties. 'R' = rotate, 'H' = horizontal flip. 'V' = vertical flip. When you use the CircuitLab button an editable schematic is saved in your post. That makes it easy for us to copy and edit in our answers. You don't need a CircuitLab account, no screengrabs, no image uploads, no background grid. – Transistor Mar 23 at 8:33
• If you are planning PWM what is the DCR of the motor 2Ω ? and all the other parameters like inertia? RPM? Then how will you limit Pd or temp rise on start/stop? Normally you need huge heatsinks unless FET is rated for 10x motor current or you accelerate very slowly. – Sunnyskyguy EE75 Mar 23 at 8:54
• Get your RDs on and then calculate the power dissipation and then rethink – Autistic Mar 23 at 9:15
• Do you think that mosfets i chose are suitable? – user216229 Mar 23 at 9:40
• When someone says , a motor "runs at" without "surge current" and inertia, it always means they know nothing about motor current and have not read any answers I have made on this site for RdsOn and motors. I doubt any MOSFET will work here . It needs an IGBT or equiv. or a better system design. Start with physics 101 on Energy, Power – Sunnyskyguy EE75 Mar 23 at 15:26
You cannot solve this problem of driving a 400V 20A motor using Pch MOSFETs. They do not exist with sufficient 600V Vds and $$Pd= (R_{dsOn}+DCR_{motor}) * I_{sc}²$$ (Isc=start current = start surge) for motor winding DC resistance, DCR causing worst-case surge current $$\I_{sc}=V_{dc}/(DCR+R_{on})\$$ (Ron of bridge = Rdson (Pch+Nch)).
Here your Pch is 350 mOHms and DC motors draw 8 to 12x rated current on startup. This means 400V/20A=20hms so while rated motor is 400*20A=8kW , surge power is 8 to 12x this if full acceleration or an average power draw of 80kW.
So you are planning to use a 5mA Optoisolator to drive a 80kW surge load. That will not work. Then the typ. full acceleration surge current is 200A²*RdsOn= 14kW means the mechanical guy or "you" have to design a heatsink for that surge power dissipation in the Pch for some period of time.
Now your Tesla race car driver is really in the hot-seat.
It MUST be a Quad Nch FET or IGBT full-bridge using the lower side for PWM to create a boost voltage to drive the high side Nch switches.
• Whould you mind desining a schema for me, please? – user216229 Mar 25 at 18:30
• You need to do more research. It is far too much time for someone to pick parts and design a schematic for you. Even if they did, you would still blow something up because there's more to it than just a schematic. Do more reading in the direction pointed out by others. – DKNguyen Mar 25 at 19:33
• RPM is most important problem for me – user216229 Mar 26 at 18:42
• You can easily find full bridge designs on the web with IGBT or quad Nfet. If you do not understand what is T=L/R then you are over your head, – Sunnyskyguy EE75 Mar 26 at 20:08 | 2019-08-21 14:46:27 | {"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": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3017618954181671, "perplexity": 4284.342217191622}, "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-2019-35/segments/1566027316021.66/warc/CC-MAIN-20190821131745-20190821153745-00546.warc.gz"} |
http://physics.stackexchange.com/tags/terminology/new | # Tag Info
1
In thermodynamics, a closed system is a system which cannot exchange matter with the environment. An isolated system cannot exchange matter nor energy with the environment. So, an isolated system is also closed, but the reverse is not true.
3
An isolated system can radiate energy/momentum/angular momentum away, but isn't externally influenced. A closed system is an isolated system that also has fixed net energy, momentum, and angular momentum.
0
Fluorescence can occur at any wavelength, it is the transition between states of the same spin, e.g. most always excited singlet ($S_1$) to ground state singlet ($S_0$) which both have spin = 0 (and spin multiplicity 1). This can occur over a wide range of wavelengths; bacteriochlorophyll emits in the near infra red at approx 850 to 900 nm as it absorbs at a ...
2
As far as the second part of your question is concerned you can directly see the image improvement with squinting. If you have a DSLR with aperture settings you keep the camera slightly defocused and now reduce the aperture you will see that the image is becoming sharper. However at the same time the image will become darker because you are collecting less ...
1
EDIT Please see the comments following my answer, regarding the paragraph below , as it is incorrect. Strabismus is the clinical name for squinting. The squint is simply compression of the eye muscles to compensate for problems with focusing and / or astigmatism. END EDIT In a somewhat similiar way, telescopes using adaptive optics can distort the ...
0
Decay width ($\Gamma=\frac{1}{\tau}$) which is a measure of the probability of a specific decay process occurring within a given amount of time in the parent particle’s rest frame. The decay width $\Gamma$ of particle is directly related to its decay lifetime: the faster the particle decays, the larger its decay width. Since the dimension of $\Gamma$ is ...
6
You're having trouble telling the difference between the two because, as for many natural language words, I don't think there is a clearly defined difference and the difference arises from natural usage (as with the difference between "fruits" and "vegetables" in English). The following is my understanding of the difference. There is a well defined ...
0
Quadrature has a very clear and precise meaning in Quantum Mechanis. Some quantities don't commute, i.e., you cannot measure both of them with unlimeted precision. This is related to Heisenbeg unsertainty. Let's put this two variables in the Cartesian plan and call this observables as X and Y.. if X is position then Y is momentum. If X is electric field ...
2
In watchmaker lingo, a "retrograde" watch is a watch that has a hand on it (usually to tell you the day or date) that will advance along and then at the end of its "cycle" will snap backwards to its original position -- rather than loop around the way the hour, minute, and second hands do. This has nothing to do with the "retrograde motion" discussed in ...
-1
Transparent object allow light to pass throught them and we can see throught these objects clearly translucent object allow light to pass through them partially.
4
According to the Particle Data Group [1], Mesons are [strongly interacting particles that] have baryon number $\mathcal{B}=0$. In the quark model, they are $q\bar{q}'$ bound states of quark $q$ and antiquark $\bar{q}'$. This definition has two sentences, which in fact are somewhat different. The second one defines the quark-model mesons, that you're ...
0
I think the best way to think of it is as follows.(It's not too different from what everyone has said, but may be put into better perspective). Choosing a frame of reference is a completely different job than setting up of coordinates. To observe an event in spacetime you must belong to some frame of reference(or equivalently, you create a frame of ...
1
The simplest way I can think of correctly defining quantum physics is that it is the combination of our best and most correct theories of physics that does not include General Relativity. There are two relevant and important classes of physics for this explanation: classical physics and quantum physics. Physicists are people too. If we can achieve our ...
0
The QM notion of a "force" is highly technical jargon that doesn't match up with how the word force is used in the world at large. Basically, the notion of a "force" in QM is defined to be an interaction mediated by force carrier particles and therefore the exchange-interaction is arbitrarily defined not to be a force. Likewise, gravity is not a "force" ...
1
I want to add an answer which I think supplements some of the other comments. @lemon notes that "gravity" refers to the pull towards the Earth while "gravitation" is more general, but notes that this is archaic. In fact, it is a very old term and the history is interesting, so take this as a "history of science" type answer. Disclaimer that all of what I ...
0
That's incorrect definition propagated by text books for simplicity sake. Gravity should always be defined as a force that attracts a body towards another physical body with mass. For simplicity sake, gravity is taught as the force with which Earth pulls a body.
5
Let me preface by saying that "coupling" is a favorite physicist word that is perhaps best described linguistically than rigorously; it's deployed in a few different situations. In general, we say that a coupling exists in quantum mechanics if the evolution of one part of the system depends on another quantity, which could be either classical or quantum. I'...
-3
I think Entanglement may answer your question. Two systems are said to be entangled(coupled) if we cannot assign an independent and separate wavefunctions for each system, instead we define a composite system which is simply the tensor product of the original constitutes. To be precise, in the general case the wavefunction description of any quantum system ...
Top 50 recent answers are included | 2016-07-26 06:30:53 | {"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": 0.7784088253974915, "perplexity": 537.0107289000713}, "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-2016-30/segments/1469257824756.90/warc/CC-MAIN-20160723071024-00133-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://nrc.canada.ca/en/certifications-evaluations-standards/codes-canada/codes-development-process/public-review/2020/pcfs/necb15_divb_03.02.01.04._001541.html | # Proposed Change 1541
Code Reference(s):
NECB15 Div.B 3.2.1.4.
Subject:
Windows, Doors and Skylights
Title:
Allowable Areas for Fenestration, Doors and Skylights
Description:
This proposed change introduces a reduction in the allowable vertical FDWR (fenestration and door area to gross wall area ratio).
## PROPOSED CHANGE
### [3.2.1.4.] 3.2.1.4.Allowable Fenestration and Door Area
[1] 1)The maximum allowable total vertical fenestration and door area to gross wall area ratio (FDWR), determined in accordance with Article 3.1.1.6., shall be as follows:
$\begin{array}{c}\mathrm{FDWR}=0.40\phantom{\rule{0.33em}{0ex}}\mathrm{for}\phantom{\rule{0.33em}{0ex}}\mathrm{HDD}\le 4000,\hfill \\ \mathrm{FDWR}=\frac{2000-\left(0.2×\mathrm{HDD}\right)}{3000}\phantom{\rule{0.33em}{0ex}}\mathrm{for}\phantom{\rule{0.33em}{0ex}}4000<\mathrm{HDD}<7000,\phantom{\rule{0.33em}{0ex}}\mathrm{and}\hfill \\ \mathrm{FDWR}=0.20\phantom{\rule{0.33em}{0ex}}\mathrm{for}\phantom{\rule{0.33em}{0ex}}\mathrm{HDD}\ge 7000,\hfill \end{array}$
$\begin{array}{c}\mathrm{FDWR}=0.30\phantom{\rule{0.33em}{0ex}}\mathrm{for}\phantom{\rule{0.33em}{0ex}}\mathrm{HDD}\le 4000,\hfill \\ \mathrm{FDWR}=\frac{1700-\left(0.2×\mathrm{HDD}\right)}{3000}\phantom{\rule{0.33em}{0ex}}\mathrm{for}\phantom{\rule{0.33em}{0ex}}4000<\mathrm{HDD}<7000,\phantom{\rule{0.33em}{0ex}}\mathrm{and}\hfill \\ \mathrm{FDWR}=0.10\phantom{\rule{0.33em}{0ex}}\mathrm{for}\phantom{\rule{0.33em}{0ex}}\mathrm{HDD}\ge 7000,\hfill \end{array}$
where
HDD
= the heating degree-days of the location of the building determined according to Sentence 1.1.4.1.(1).
[2] 2)The total skylight area shall be less than 52% of the gross roof area as determined in Article 3.1.1.6.
### Note A-3.2.1.4.(1)Total Vertical Fenestration and Door Area to Gross Wall Area.
Table A-3.2.1.4.(1) shows a sample of maximum allowable FDWR for various HDD.
Table [3.2.1.4.(1)] A-3.2.1.4.(1)
Maximum Allowable FDWR for Various HDD
HDD Maximum FDWR
< 4000 0.430
4000 0.430
4250 0.328
4500 0.327
4750 0.325
5000 0.323
5250 0.322
5500 0.320
5750 0.218
6000 0.217
6250 0.215
6500 0.213
6750 0.212
7000 0.210
> 7000 0.210
## RATIONALE
### Problem
The NECB 2017 allows maximum prescriptive FDWR values by climate zones, which often exceed the typical FDWR values of actual building designs in a climate zone. By reducing the maximum allowable prescriptive FDWR values so that they correspond more closely to the FDWR values of typical buildings in a climate zone, it ensures that only buildings with a low-enough typical FDWR value for the climate zone utilize the Prescriptive Path. Building designs with a FDWR value higher than the maximum allowable prescriptive value should demonstrate compliance using either the Part 3 Trade-off Path or the Part 8 Performance Path.
### Justification - Explanation
Given that the fenestration U-value is much greater than for opaque assemblies, and more costly per unit area, lowering the maximum allowable prescriptive FDWR of a building will reduce the overall U-Value of the above-ground building envelope, reducing:
• the amount of uncontrolled thermal transfer through the building envelope
• the demand and consumption of energy for heating and cooling
• the capital cost of the envelope and equipment.
### Impact analysis
Related to Article 3.2.1.4.
Fenestration is typically an expensive element of the above-ground building envelope, and in some cases is more costly per unit area than opaque above-ground building assemblies. Reducing the prescriptive allowable area of fenestration may lower the capital cost of the building and the equipment used to condition it.
The NRC conducted a simulation study on the estimated energy savings from adopting the proposed reduction in FDWR. The study used OpenStudio and NRCan CANMETEnergy’s building technology assessment platform (BTAP) to generate 15 NECB 2017 compliant archetypes and run annual simulations. The savings presented are a reduction of the whole building’s annual energy consumption (includes heating, cooling, lighting etc.) relative to the 2017 NECB. Figure 1 provides an overview of change in energy usage of the 15 archetypes in each of the six climate zones, illustrating the range and trend of the energy savings. Table 1 shows the detail energy savings of each archetype-location pair, as well as the average savings of each archetype across the climate zones. Note the negative values for climate zones 4 and 5 (Victoria and Windsor, respectively) reflect an increase in energy usage by the buildings after adopting the proposed FDWR (which are higher than NECB 2017 values for zones with HDD ≤ 4000).
The proposed changes generally reduce FDWR limits which essentially amounts to replace glazing with opaque insulation (wall), which results in reduced solar heat gains and reduced opaque envelope heat loss. The net effect on heating and cooling depends on the time of day and the season. For example, during the winter, the reduced solar heat gain will result in additional space heating during the day, but the increased opaque areas will reduce night time heating. The majority of the energy saved for most archetypes is due to reduced HVAC fan power consumption (associated with reduced heating/cooling demands manifested as reduced runtimes for the HVAC system). For zones with a higher proposed FDWR, the additional glazing allows greater sunlight during the day but increases heating, cooling, and fan consumption resulting in an overall increase in energy usage for the associated climate zones.
Figure 1 Overview of climate zone averaged energy savings
Table 1 Energy savings from proposed FDWR and skylight changes
Archetype Climate Zone CZ-4 Victoria CZ-5 Windsor CZ-6 Montreal CZ-7A Edmonton CZ-7B Fort McMurray CZ-8 Yellowknife Archetype average Secondary School -0.7% -0.9% 2.5% 2.7% 2.9% 2.4% 1.5% Primary School -1.2% -1.2% 4.1% 4.4% 4.0% 3.3% 2.2% Small Office -1.7% -1.6% 5.3% 6.4% 6.8% 7.0% 3.7% Medium Office -3.0% -2.4% 9.1% 11.7% 13.1% 10.8% 6.5% Large Office -1.8% -1.7% 5.2% 6.6% 6.1% 5.3% 3.3% Small Hotel -1.1% -0.9% 3.3% 3.2% 3.3% 3.3% 1.8% Large Hotel -1.4% -1.2% 4.2% 4.0% 4.2% 4.3% 2.3% Warehouse -2.5% -2.3% 6.7% 7.9% 8.4% 6.2% 4.1% Retail Standalone -1.5% -1.6% 5.7% 6.4% 6.3% 5.6% 3.5% Retail Strip mall -1.7% -1.7% 4.6% 6.6% 6.6% 5.2% 3.3% Quick Service Restaurant -0.9% -0.9% 2.5% 2.7% 2.8% 2.2% 1.4% Full Service Restaurant -0.3% -0.3% 1.2% 0.8% 1.2% 1.0% 0.6% Midrise Apt -1.4% -1.1% 4.1% 4.0% 4.8% 4.9% 2.5% Highrise Apt -0.9% -0.8% 3.3% 2.1% 2.6% 3.2% 1.6% Hospital -0.8% -0.7% 2.2% 2.4% 2.5% 1.9% 1.2% Climate zone (simple) average -1.4% -1.3% 4.3% 4.8% 5.0% 4.4%
Additional documentation for this proposed change is available on request at Codes.Publicreview@nrc-cnrc.gc.ca.
### Enforcement implications
There would be no change in the effort to comply with the Prescriptive Path by designers, nor for an AHJ to validate Prescriptive, Trade-off or Performance Path submissions.
For building designs without skylights complying using the Part 3 Prescriptive Path, there are no enforcement implications.
For building designs with or without skylights complying using the Part 8 Performance Path, the modeller and AHJ would simply validate that the reference model has no skylight area defined.
For building designs with skylights intending to comply using the Part 3 Prescriptive Path, these will require that the Part 3 Trade-off Path be used for the horizontal roof areas, and that the AHJ confirm the correct application of the Trade-off Path for the roof areas. Given that many buildings do not include skylights, there is little to no impact expected for the enforcement of this proposed change.
### Who is affected
The owner would be affected as there would be reductions to the maximum allowable prescriptive areas of FDWR.
Others that may be affected are the designers, engineers, architects, building officials and manufacturers/suppliers.
## OBJECTIVE-BASED ANALYSIS OF NEW OR CHANGED PROVISIONS
[3.2.1.4.] 3.2.1.4. ([1] 1) [F92,F99-OE1.1]
[3.2.1.4.] 3.2.1.4. ([2] 2) [F92,F99-OE1.1]
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https://owenjow.xyz/graphics/1945/ | # Face Morphing
## Overview / Algorithm Description
In this assignment, we explore non-parametric warps and cross-dissolution as an approach to face morphing. For those unfamiliar with the concept, we define a "face morph" to be a transformation between two faces that transitions seamlessly over some period of time. As an overview, we accomplish such a task by
• selecting matching sets of feature points on two input images,
• creating a triangulation across the mean of those feature points,
• interpolating between the feature points in each image to create our output shape, i.e. some (non-uniformly?) averaged set of points,
• computing a local warp (affine transformation) between each triangle in our output shape and the corresponding triangle in each of our input shapes,
• bilinearly interpolating over input colors in order to arrive at final values for the pixels in every output triangle,
• and finally celebrating with poorly cooked food.
Simple!
Let's walk through the algorithm (and maybe go into a bit more detail), so that you can be even more convinced. Say we have the following two images – one of me and one of my friend Tony. In the interest of higher quality results, I've standardized the image size at a not-at-all arbitrary 476 x 543px and "polygonal lasso tool"ed away most of the background. The edges of the remaining content have been blurred so as to smooth out the difference a bit:
Me (photo credits: me) Tony
To start, we select a number of corresponding points in each image (the same number, in the same order, such that a point in one image corresponds to the same "feature" in the other image), and define a triangular mesh over those points. We want this triangulation to be identical across both images – so in order to avoid bias toward either of the sets of feature points, we compute an averaged set of points and generate a Delaunay triangulation over the midway correspondences.
In this case, the points and mesh might end up looking like this (note that the triangulation was computed over the averaged points, although it is shown below as being applied to each image's individual set of points):
Triangulation over feature points Corresponding triangulation for Tony's photograph
As it stands, we have a triangulation that can be applied to the first image's points, the second image's points, and even a set of points averaged over that of both images. Importantly, a triangle in one set of points has a corresponding triangle in every one of the other sets of points. What this means is that we can now warp the image on a triangle-by-triangle basis!
But how do we do that? Well, we're going to need a bit of math. It turns out that by the same principle as barycentric interpolation, we can define an affine transformation on the vertices of a triangle and it will correctly apply to all of the coordinates within the triangle. In other words, the transformation that turns one triangle's vertices into another triangle's vertices will also map each of the points within the first triangle into the associated point within the second triangle. So when we've defined this warp, we can bring a triangle from one image to the location of the corresponding triangle in another image. (By "corresponding triangle", we refer to the identical triangle in the Delaunay collection, just set up across a different set of points.)
We define $$X$$ to be the $$3$$ x $$3$$ matrix whose columns are the source triangle's vertices (in homogeneous coordinates!) and $$Y$$ to be the same matrix for the destination triangle. We'll refer to the transformation we're looking for as $$A$$, which will also be a $$3$$ x $$3$$ matrix such that $$AX = Y$$ ...which, when expanded, might be shown as $$A \begin{bmatrix} (x_1)_{x} & (x_2)_{x} & (x_3)_{x} \\ (x_1)_{y} & (x_2)_{y} & (x_3)_{y} \\ 1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} (y_1)_{x} & (y_2)_{x} & (y_3)_{x} \\ (y_1)_{y} & (y_2)_{y} & (y_3)_{y} \\ 1 & 1 & 1 \end{bmatrix}$$ (where $$x_i$$ is the $$i^{th}$$ vertex of the source triangle and $$(x_i)_{x}$$ would be its $$x$$-coordinate. $$y_i$$ and its coordinates would be defined similarly, of course, except that it would be part of the destination triangle instead).
Representations aside, we can solve for our affine transformation (/warp) very easily: $$AX = Y \\ AXX^{-1} = YX^{-1} \\ A = YX^{-1}$$ Then we can vectorize our operations by warping every triangle point at once, which is nice: $$A \begin{bmatrix} (x_1)_{x} & (x_2)_{x} & \dots & (x_n)_{x} \\ (x_1)_{y} & (x_2)_{y} & \dots & (x_n)_{y} \\ 1 & 1 & \dots & 1 \end{bmatrix} = \begin{bmatrix} (y_1)_{x} & (y_2)_{x} & \dots & (y_n)_{x} \\ (y_1)_{y} & (y_2)_{y} & \dots & (y_n)_{y} \\ 1 & 1 & \dots & 1 \end{bmatrix}$$ (Programmatically speaking, we obtain all of the triangle points through the use of scikit-image's polygon function.)
Anyway, we now have a means for computing affine transformations between arbitrary triangles... but which triangles are we warping to? Or perhaps from? Since we have two images we want to morph between, we'll need to figure out exactly how much between the images we want to warp. We parameterize this interpolation with $$t$$, where $$t \in [0, 1]$$; if $$t$$ is $$0.4$$ it means we want our resulting image to be composed of $$40\%$$ the first image and $$60\%$$ the second image. We can use $$t$$ to create the destination set of feature points, which will be a weighted average of the feature points in the two input images. Specifically, every destination point should be $$t$$ times a point from the first image and $$(1 - t)$$ times the associated point from the second image. By following this process for every pair of feature points in the input images, we obtain our destination image's feature points and, by extension, the destination triangles we'll want to warp from.
You may be wondering: why "from?" Answer: we want to carry out an inverse warp in order to avoid having holes in our morphed image – if we do it this way, and compute the warp from every pixel in the final image, then we know that every pixel will end up with some color value. There's no such guarantee with a forward warp.
So to answer our earlier question, we're warping from the triangles in our destination image (which hasn't been colored yet and exists only as a set of feature points on a set canvas size) to the triangles in each of our input images. Thus, for each coordinate in the destination image we end up with two corresponding coordinates in the two input images – and to arrive at a pixel value for the destination image we simply evaluate a $$t$$-weighted average of the source pixel values for each of those coordinates. One probable issue is that the warped coordinates might not be integers. To get around this, we'll just always (bi)linearly interpolate between the pixel values at the warped coordinates' 4-neighbors – which looks like this: $$\text{hdiff} := x - \lfloor x \rfloor \\ \text{vdiff} := y - \lfloor y \rfloor \\ \text{top} := \text{hdiff} \cdot f(x_1, y_1) + (1 - \text{hdiff}) \cdot f(x_2, y_1) \\ \text{bottom} := \text{hdiff} \cdot f(x_1, y_2) + (1 - \text{hdiff}) \cdot f(x_2, y_2) \\ \text{source color} = \text{vdiff} \cdot \text{top} + (1 - \text{vdiff}) \cdot \text{bottom}$$ if our warped point in the source image is $$(x, y)$$ and the pixel values of its 4-neighbors are $$f(x_1, y_1)$$, $$f(x_2, y_1)$$, $$f(x_2, y_2)$$, $$f(x_1, y_2)$$ starting from the top left and reading clockwise.
Finally, to obtain the overall color for each pixel in the destination image, we would cross-dissolve (i.e. linearly interpolate) between the colors from the two source images. $$\text{final color} := t \cdot \text{color1} + (1 - t) \cdot \text{color2}$$ Aaand after doing this for every point in the destination image, we'd end up with our morphed image (again, parameterized by $$t$$). For instance, the below image has $$t = 0.5$$ and is therefore exactly in the middle:
50% Tony, 50% Owen
In order to construct a nice morphing animation, we run the entire process for $$t = 0, t = 0.022, t = 0.044, \dots, t = 0.978, t = 1$$ (where $$t = 0$$ would be entirely the first image, and $$t = 1$$ would be entirely the second image) and string the morphed results into a GIF.
They say the whole is greater than the sum of its parts
Fantastic. As another example, I ran this process on myself and my brother:
William Combined GIF
## "Mean Face" of a Population
The IMM Face Database is a freely available dataset of annotated Danish faces. (By "annotated", I mean that the feature points have already been selected.) The dataset contains the various poses and expressions of forty people (of which 33 are men; we choose to focus only on the men since there are a lot more of them). Each person has been photographed in six different settings, for example "front-facing neutral expression", "left-facing neutral expression", "front-facing smile"... you get the idea. During this part of the assignment, we compute the average of the 33 male faces for a single image type at a time.
This means averaging over the feature points for each face in order to discover the average face shape, and then morphing each of the faces into this average shape. These gentlemen have been morphed into the average face shape of a smiling male Dane:
Guy #10 Guy #20 Guy #40
For reference, here are the originals.
Original #10 Original #20 Original #40
From all of our warped faces, we compute the average by summing out and dividing by 33.
Just your average smiling Danish male
Below are the same results except for front- and left-facing neutral expressions.
We can also warp a non-Danish face into the average geometry of the Danish face, and warp the average Danish face into the geometry of a non-Danish face. Both of these results were computed using image type 1 in the IMM dataset (front-facing neutral):
This cracked me up My tilted Danish face
## Caricatures
How else can we mess with people's faces? Well, we could create a caricature (def. "comically or grotesquely exaggerated representation") by extrapolating from the population mean of the previous section. In other words, we can calculate the difference of our image shape from the mean shape and scale it before adding it back to our original mean shape. This allows us to take a parametrically defined step in the direction of that difference (which is to say we can take as large of a step as we want), creating with luck a nicely exaggerated image. Mathematically: $$\text{caricature shape} = \text{mean shape} + \alpha (\text{image shape} - \text{mean shape})$$
Note that for this to be a true extrapolation, $$\alpha$$ should be $$< 0$$ or $$> 1$$.
Then we can warp an arbitrary face to the caricature shape as normal, and produce images like this (which was created using the front-facing neutral mean and an $$\alpha$$ of 2)
Nice
...or these, "caricaturized" using the front-facing smile and the lit front-facing neutral as bases:
Image type 2, $$\alpha = -0.5$$ Image type 5, $$\alpha = 1.5$$
## Ethnicity Alteration (Japanese Oliver)
Let's take my white roommate, Oliver, and make him a little less white by morphing his face with that of the average Japanese actor. Original images and their correspondence point triangulations are as follows; for reference, I obtained the Japanese average from this website.
We try three types of morphs: shape and appearance, shape only, and appearance only. Below are some examples:
Midpoint image t = 21 warp_frac = 1.0, t = 30
A more convincing result than I expected, to be honest! Meanwhile, here are the GIFs for each type of morph:
Shape only Appearance only Both
## dOwen / dt
Finally, I made a morphing video in order to highlight temporal variations in my facial attributes. The video has been composed of ten photographs across chronologically increasing points in time.
Original photographs (subject age ranging from about 0 y/o to 19 y/o):
So to conclude this writeup, we have the morphing video constructed from those photos:
Ideally the images would be more evenly spaced, but I was hard-pressed to find pictures of myself from the middle of my childhood. As a result, I ended up with a denser sampling of photographs from my most early and most recent years. (I suppose it could be worse.) | 2019-01-21 23:53:14 | {"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": 0, "mathjax_display_tex": 2, "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": 0.6622139811515808, "perplexity": 795.6450178666638}, "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-2019-04/segments/1547583822341.72/warc/CC-MAIN-20190121233709-20190122015709-00149.warc.gz"} |
https://www.cheenta.com/roots-of-equation-and-vietas-formula-aime-i-1996-problem-5/ | What is the NO-SHORTCUT approach for learning great Mathematics?
How to Pursue Mathematics after High School?
For Students who are passionate for Mathematics and want to pursue it for higher studies in India and abroad.
Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Roots of Equation and Vieta's formula.
Roots of Equation and Vieta's formula - AIME I, 1996
Suppose that the roots of $x^{3}+3x^{2}+4x-11=0$ are a,b and c and that the roots of $x^{3}+rx^{2}+sx+t=0$ are a+b,b+c and c+a, find t.
• is 107
• is 23
• is 840
• cannot be determined from the given information
Key Concepts
Functions
Roots of Equation
Vieta s formula
AIME I, 1996, Question 5
Polynomials by Barbeau
Try with Hints
With Vieta s formula
$f(x)=x^{3}+3x^{2}+4x-11=(x-a)(x-b)(x-c)=0$
$\Rightarrow a+b+c=-3$, $ab+bc+ca=4$ and $abc=11$
Let a+b+c=-3=p
here t=-(a+b)(b+c)(c+a)
$\Rightarrow t=-(p-c)(p-a)(p-b)$
$\Rightarrow t=-f(p)=-f(-3)$
$t=-[(-3)^{3}+3(-3)^{2}+4(-3)-11]$
=23.
What to do to shape your Career in Mathematics after 12th?
From the video below, let's learn from Dr. Ashani Dasgupta (a Ph.D. in Mathematics from the University of Milwaukee-Wisconsin and Founder-Faculty of Cheenta) how you can shape your career in Mathematics and pursue it after 12th in India and Abroad. These are some of the key questions that we are discussing here:
• What are some of the best colleges for Mathematics that you can aim to apply for after high school?
• How can you strategically opt for less known colleges and prepare yourself for the best universities in India or Abroad for your Masters or Ph.D. Programs?
• What are the best universities for MS, MMath, and Ph.D. Programs in India?
• What topics in Mathematics are really needed to crack some great Masters or Ph.D. level entrances?
• How can you pursue a Ph.D. in Mathematics outside India?
• What are the 5 ways Cheenta can help you to pursue Higher Mathematics in India and abroad?
Want to Explore Advanced Mathematics at Cheenta?
Cheenta has taken an initiative of helping College and High School Passout Students with its "Open Seminars" and "Open for all Math Camps". These events are extremely useful for students who are really passionate for Mathematic and want to pursue their career in it.
To Explore and Experience Advanced Mathematics at Cheenta
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2021-08-06 01:16:47 | {"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": 0.17209208011627197, "perplexity": 3033.5329782191066}, "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-2021-31/segments/1627046152085.13/warc/CC-MAIN-20210805224801-20210806014801-00549.warc.gz"} |