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https://mathoverflow.net/questions/258954/mathbbz-linear-independence-of-arguments-of-units-in-non-cm-number-fields | # $\mathbb{Z}$-linear independence of arguments of units in non-CM number fields
Playing a little bit with Groessencharacters a stumbled on the following question:
Let $K$ be a non CM number field with $r_1$ real embeddings and $2r_2>0$ complex embeddings. Set $r=r_1+r_2-1$ and assume that $r\geq 1$. Let $\epsilon_1,\ldots,\epsilon_{r}$ be a system of $\mathbf{Z}$-linearly independent (in the multiplicative notation) units of $\mathcal{O}_K$. For $x\in K$ and $j\in\{1,2,\ldots,r_2\}$ let $x^{(j)}$ be the image of $x$ under the $j$-th complex embedding (where the $r_2$ choices of complex embeddings are made so that no complex embedding are related by complex conjugation).
Q: Is it possible to choose $K$ and $\{\epsilon_1,\ldots,\epsilon_{r}\}$ so that the above assumptions are satisfied and such that there exists an $(r+r_2)$-tuple $((m_1,\ldots,m_r);(n_1,\ldots,n_{r_2}))\in\mathbf{Z}^{r+r_2}\backslash\{0\}$ such that $$2\pi m_j-\sum_{k=1}^{r_2} n_k\cdot\arg\left(\frac{\epsilon_j^{(k)}}{|\epsilon_j^{(k)}|}\right)=0,$$ for all $j\in\{1,\ldots,r\}$ ?
This looks like a delicate question about the $\mathbf{Q}$-linear independence of the real numbers $\arg\left(\frac{\epsilon_j^{(k)}}{|\epsilon_j^{(k)}|}\right)$.
added The non-CM assumption is essential since for a CM unit $\epsilon_j$, we have $\arg\left(\frac{\epsilon_j^{(k)}}{|\epsilon_j^{(k)}|}\right)\in\pi\mathbf{Q}$. Also, if $\epsilon_j\in\mathcal{O}_K^{\times}$ is chosen to be totally real then $\arg\left(\frac{\epsilon_j^{(k)}}{|\epsilon_j^{(k)}|}\right)\in 2\pi\mathbf{Z}$. So may be one could come up with an example that would combine CM units and totally real units...
added following Kevin's comment. Let us look at the infinite part of Groesscharacter $\chi$ (so by character I mean continuous and unitary), which we denote by $\chi_{\infty}$. Then it is a continuous group homomorphism $\chi_{\infty}:(K\otimes_{\mathbf{Q}}\mathbf{R})^{\times}\rightarrow S^1$ which is invariant under a finite index subgroup of $U\leq\mathcal{O}_K^{\times}$. If we take $U$ to be the group generated by the $\epsilon_j$'s then such characters are "essentially" parametrized by
(1) an $(r+r_2)$-tuples $(m;n)\in\mathbf{Z}^{r+r_2}$,
(2) a "sign character", i.e., a character which is trivial on the connected component of the identity of $(K\otimes_{\mathbf{Q}}\mathbf{R})^{\times})$, and
(3) a real number $t$.
The question I raised is essentially equivalent to ask if the pair $(m;n)$ is uniquely determined by $\chi_{\infty}$ ? For example if $K$ is totally real then it is easy to see that (this is the calculation that was performed by Hecke in the first paper where he introduces such characters) two distinct $r$-tuples $m$ and $m'$ give rise to two distinct characters $\chi_{\infty}$ and $\chi_{\infty}'$. In general, this uniqueness already fails for quartic CM fields (and also for those of larger degrees). So it looked to me like a natural question to ask for what number fields does this "uniqueness" fail ?
• I think I'd rather see the underlying question about grossencharacters than this translation into a more concrete statement ;-) – Kevin Buzzard Jan 6 '17 at 16:46 | 2019-03-26 03:00:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 0.975361168384552, "perplexity": 150.98016475454864}, "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-2019-13/segments/1552912204768.52/warc/CC-MAIN-20190326014605-20190326040605-00531.warc.gz"} |
https://math.stackexchange.com/questions/914023/partial-differential-chain-rule-with-designation-of-some-variables-being-held-co | # Partial differential chain rule with designation of some variables being held constant
If a function $f$ can be expressed in two different coordinate systems say $(x,y)$ and $(\bar x,\bar y)$, how would one take the partial differential of $f$ with respect to $\bar x$ with $\bar y$ held constant? I've seen it written:
$$\left( \dfrac{\partial f}{\partial \bar x}\right)_{\bar y} = \left(\dfrac{\partial f}{\partial x}\right)_y \left(\dfrac{\partial x}{\partial \bar x}\right)_{\bar y} + \left(\dfrac{\partial f}{\partial y}\right)_x \left(\dfrac{\partial y}{\partial \bar x}\right)_{\bar y}$$
But I do not understand where this comes from, specifically which variables should be held constant in each derivative. For example why couldn't the $\dfrac{\partial f}{\partial x}$ derivative have $\bar y$ held constant instead of $y$?
why couldn't the $\frac{\partial f}{\partial x}$ derivative have $\bar y$ held constant instead of $y$?
In principle, you could consider such a derivative, provided that it is possible to change $x$ while holding $\bar y$ constant. E.g., if $\bar y =2x$ you would not be able to do that, and then $\left(\frac{\partial f}{\partial x}\right)_{\bar y}$ would not make sense. In any event, the derivative would likely be different, and then the equality would not hold.
The chain rule that you stated is the usual multivariable chain rule, although it's normally not written with those subscripts. When subscripts are not given, readers are expected to understand that $(x,y)$ and $(\bar x,\bar y)$ are the coordinate pairs of interest here, and that when we take partial derivatives, the other member of the same pair is held constant. With this convention understood, one does not need subscripts in the formula you gave.
The proof of multivariable chain rule is not hard to find. I'll give a sketch in terms of differentials: $$dx = \left(\dfrac{\partial x}{\partial \bar x}\right)_{\bar y} d\bar x + \left(\dfrac{\partial x}{\partial \bar y}\right)_{\bar x} d\bar y$$ $$dy = \left(\dfrac{\partial y}{\partial \bar x}\right)_{\bar y} d\bar x + \left(\dfrac{\partial y}{\partial \bar y}\right)_{\bar x} d\bar y$$ Plug the above into $$df = \left(\dfrac{\partial f}{\partial x}\right)_{ y} d x + \left(\dfrac{\partial f}{\partial y}\right)_{x} d y$$ and the result follows. | 2021-06-17 17:56:16 | {"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": 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": 0.9185829162597656, "perplexity": 148.43240318021674}, "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/1623487630518.38/warc/CC-MAIN-20210617162149-20210617192149-00080.warc.gz"} |
https://brilliant.org/discussions/thread/please-help-is-my-proof-correct-2/ | ×
Q) Diagonals AC and BD of quadrilateral ABCD intersects at O . If AB=CD , then show that 1) ar(DOC) = ar(AOB) 2) ar (DCB) = ar(ACB) [ question from NCERT class 9 textbook.
In triangle ABD , AO is the median therefore ar(AOB) = ar(AOD) [ since median of a triangle divides it into 2 triangles with equal areas ]...................................(1) and also in triangle BCD , CO is the median and ar(COB) = ar ( COD) ...................................(2) . Adding (1) and (2) we get ar(ABD)= ar(BCD) . Thus diagonal divides it into 2 congruent triangles. Now we consider equal lines AB and CD after joining their end points we would either get an isosceles trapezium if the lines are not parallel or a parallelogram if the lines are parallel . But the diagonal of a isos. trapezium does not divide it into 2 triangles with equal areas , therefore it must be a parallelogram . Now triangles ACB and BCD are on the same base BC and between same parallels AD and BC . therefore ar(DCB)=ar(ACB) [ proof 2] and subtracting ar(BOC) from both sides we get ar(DOC)=ar(AOB] [ proof 1 ] . Hence proved . [ The question can easily be done by drawing perpendiculars from B and D on AC.]
Note by Karandeep Singh Ludhar
2 years, 4 months ago
Sort by:
[Adding (1) and (2) we get ar(ABD)= ar(BCD) ].................................... this part of your answer goes wrong because this part can be only be prooved if bo would be the median if BO would be the median than TRIANGLE BOC AND BOA would have equal area. and then ar[cob]=ar[boa]. Then area of all the four triangles would be equal. then you can prove your first part. so, this prove is wrong. · 2 years, 4 months ago
But , I have not used BO . I have just used median AO and CO . · 2 years, 4 months ago
but by using ao and co u can not prove ar[abd] =ar [dcb]. i think you are not clear with my answer let me explain . let us take ar[abd]=x and ar[cbd]=y then since ao is the median of x ar [aod]=ar[aob]=1/2x. since co is the median ar[cod]=ar[cob]=1/2y. and 1/2y would be only equal to 1/2x if their areas are equal that is do would be the median.i hope your doubt would be clear by this for more doubt about this contact me at 0788(2290223). · 2 years, 4 months ago
@karandeep singh ludhar @abhishek anand .. Kaddu's way is correct.. I did it the same way. This question appeared in our exams too right?.. Kaddu apna e-mail id de · 2 years, 4 months ago
sir, tera answer sahi kiye kya · 2 years, 4 months ago
Hao.. Mere 2marks kate construction mei :'( · 2 years, 4 months ago
tu to median se kiya hai jo quetion me diya hua tha to tu construction kya kiya · 2 years, 4 months ago
DB pe perpendiculars draw karke (From A and C) dono ko congruent proove krde.. Easy and simple =D.. Ye shayad question 17 thaa na? · 2 years, 4 months ago
vahi to mai bhi kiya hoo lekin kara ka prove galat hai · 2 years, 4 months ago
kaise sahi hai bhai /////explain?????????? · 2 years, 4 months ago
Search for NCERT Model Solutions :) · 2 years, 4 months ago
solution me perpendicular bana ke kiye hai · 2 years, 4 months ago
Kaise galat hai /////explain?????????? · 2 years, 4 months ago
apna phone number de karta hoo explain\ · 2 years, 4 months ago
But the question appeared in our exams just one time and also not right but in paper . · 2 years, 4 months ago
Bhai english -_- .. Aur Kaddu Topper !!! IMO ka solve kr liya re :'( .. Rukk be dharta hoon tereko · 2 years, 4 months ago
Tereko ke baad bracket kon lagayega . · 2 years, 4 months ago
Bracket konchika :p :p .. Tu aara hai FIITJEE? · 2 years, 4 months ago
Why is a tomato red and round . bataa. · 2 years, 4 months ago
Bhai tu hi bta dey... Because Brinjal is long and purple :p _/_ · 2 years, 4 months ago
hybrid brinjal khata hai kya · 2 years, 4 months ago
Because if it would have been green and long it would be a cucumber . <english me isliye likha kyon ki question aur answer dono english me the . · 2 years, 4 months ago
Stud ^^ _/_ · 2 years, 4 months ago
tu apna number de na · 2 years, 4 months ago
Par maine kal hi to tujhe diya tha . · 2 years, 4 months ago
vapas se dede mere bhai · 2 years, 4 months ago
Kaddu ka 8959868806 hai.. I remember · 2 years, 4 months ago
karan apne nani ke ghar gaya hai vaha ka number mang raha hoo · 2 years, 4 months ago
The number is not correct .Haaaaaaaa · 2 years, 4 months ago
Bhai change ho gaya hoga.. Give me your new number and email id · 2 years, 4 months ago | 2017-07-27 14:54:20 | {"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": 0.9639887809753418, "perplexity": 12245.649185322241}, "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-2017-30/segments/1500549428300.23/warc/CC-MAIN-20170727142514-20170727162514-00191.warc.gz"} |
http://rlhxxb.sxicc.ac.cn/article/2022/8 | ## 2022年 第50卷 第8期
2022, 50(8): 1-8.
2022, 50(8): 937-953. doi: 10.1016/S1872-5813(22)60002-1
The co-combustion of the low-rank coal with coal derived semi-coke is of great significance to solve the urgent problem of excessively produced semi-coke in China. In this research, the oxy-fuel co-combustion characteristics of Zhundong sub-bituminous coal with bituminous coal derived semi-coke are systematically investigated using thermogravimetric analysis. Compared with air combustion, oxy-fuel atmosphere increased the ignition and burnout temperature by 10 and 40 °C, respectively. Increasing the oxygen concentration to 30% strongly compensated for the slight reduction of the combustion parameters under oxy-fuel condition and much better co-combustion performance was obtained. Three iso-conversional methods, namely, Flynn-Wall-Ozawa (FWO), Kissinger-Akahira-Sunose (KAS) and Starink, were applied to estimate the activation energy, which can be divided into two stages during the co-combustion process. The average activation energy of sub-bituminous coal, the blend and semi-coke were 49.31, 50.82 and 59.00 kg/mol, respectively. Further, the pre-exponential factor and thermodynamic parameters of the enthalpy change, Gibbs free energy change and entropy change were calculated. Interaction indices were innovatively used for both kinetic-thermodynamic parameters and DTG values. An obvious interaction can be observed during the co-combustion process. The kinetic and thermodynamic results demonstrated that the 30% semi-coke ratio was beneficial to co-combustion. Meanwhile, X-ray fluorescence (XRF) and ash fusion analyses proved that the slagging tendency of sub-bituminous coal ash reduced by blending of semi-coke.
2022, 50(8): 974-983. doi: 10.1016/S1872-5813(22)60001-X
2022, 50(8): 993-1003. doi: 10.19906/j.cnki.JFCT.2022014
2022, 50(8): 1023-1033. doi: 10.1016/S1872-5813(22)60006-9
Metal oxide-zeolite (OX-ZEO) bifunctional catalysts have been shown to have excellent aromatic selectivity and catalytic stability in syngas conversion; however, low CO conversion hinders their further development. In this paper, a series of In-ZrO2 bi-metallic oxides with In/Zr molar ratio ranging of 1/100−1/1 were prepared. After thoroughly investigated by X-ray diffraction, transmission electron microscopy, N2 sorption, pyridine-adsorbed infrared spectroscopy, X-ray photoelectron spectroscopy, electron paramagnetic resonance and temperature programmed desorption technologies, we found that introduction of indium has significantly influence on the catalytic performance due to the variation of sample’s physicochemical properties. Indium species was benefit to the dissociation of H2 that promotes CO activation. Nevertheless, it also induced the formation of more CH4. In-ZrO2 oxide with In/Zr ratio of 1/50 showed CO conversion of 18.2% with the selectivity of oxygenates of 86.4%. After combined with H-ZSM-5, In/Zr=1/50&H-ZSM-5 gave CO conversion of 46.5% with ${\rm{C}}_{5+}$ selectivity of 62.6% and the aromatic selectivity in ${\rm{C}}_{5+}$ reached 93.4%. However, the catalytic stability of this bifunctional catalyst was gradually decreased due to the aggregation of indium atoms.
2022, 50(8): 1075-1083. doi: 10.19906/j.cnki.JFCT.2022021 | 2023-01-31 07:09:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.5005824565887451, "perplexity": 10846.51304977684}, "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-2023-06/segments/1674764499845.10/warc/CC-MAIN-20230131055533-20230131085533-00363.warc.gz"} |
https://www.lesswrong.com/posts/zcPLNNw4wgBX5k8kQ/decision-theory?commentId=jpg96G4njLgRSMbM9 | ## LESSWRONGLW
I am talking about the surreal number ε, which is smaller than any positive real. Events of likelihood ε do not actually happen, we just keep them around so the counterfactual reasoning does not divide by 0.
Within the simulation, the AI might be able to conclude that it just made an ε-likelihood decision and must therefore be in a counterfactual simulation. It should of course carry on as it were, in order to help the simulating version of itself.
Why shouldn't the environment be learning?
To the Omega scenario I would say that since we have an Omega-proof random number generator, we get new strategic options that should be included in the available actions. The linear function then goes from the ε-adjoined open affine space generated by {Pick H with probability p | p real, non-negative and at most 1} to the ε-adjoined utilities, and we correctly solve Omega's problem by using p=1/2. | 2020-05-30 03:10:52 | {"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": 0.8317914605140686, "perplexity": 840.365065608927}, "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-2020-24/segments/1590347407001.36/warc/CC-MAIN-20200530005804-20200530035804-00358.warc.gz"} |
https://languagelog.ldc.upenn.edu/nll/?p=27691 | ## Annals of parsing
Two of the hardest problems in English-language parsing are prepositional phrase attachment and scope of conjunction. For PP attachment, the problem is to figure out how a phrase-final prepositional phrase relates to the rest of the sentence — the classic example is "I saw a man in the park with a telescope". For conjunction scope, the problem is to figure out just what phrases an instance of and is being used to combine.
The title of a recent article offers some lovely examples of the problems that these ambiguities can cause: Suresh Naidu and Noam Yuchtman, "Back to the future? Lessons on inequality, labour markets, and conflict from the Gilded Age, for the present", VOX 8/23/2016. The second phrase includes three ambiguous prepositions (on, from, and for) and one conjunction (and), and has more syntactically-valid interpretations than you're likely to be able to imagine unless you're familiar with the problems of automatic parsing.
On the basis of general linguistic and real-world knowledge, I interpret this title as promising lessons for the present time, asserting that these lessons are derived from a comparison with various phenomena in the period known as the Gilded Age, and specifying that those phenomena are inequality, labor markets, and conflict. Reading the article confirms this interpretation.
The relevant linguistic knowledge includes the fact that "lessons for X" and "lessons from Y" are both common idioms. It's reasonable to expect a modern stochastic parser to know this — but of course for and from are promiscuous modifiers, willing to hook up with almost any noun, verb, or adjective at all.
The relevant real-world knowledge includes the fact that socio-economic inequality is a frequently discussed feature of the Gilded Age and also of the present time, that such inequality is related to labour markets, and that it may lead to conflict. Modern parsers don't think about things at this level — current directions of progress, like distributional semantics, don't really help much in cases like this one.
To illustrate some of the ways that parsing can go wrong, I'll give the (variously wrong) results returned by three available on-line parsers. And let me note in passing that all of these parsers report their results in the general framework adopted by the Penn Treebank project, which flourished due to a sort of treaty that emerged from a summit meeting of computational linguists held 25 years ago (Ezra Black et al., "A Procedure for Quantitatively Comparing the Syntactic Coverage of English Grammars", HLT 1991). This approach has been so widely used in "treebank" projects, across many languages and types of text, that it makes sense in my opinion to teach it more widely, at least to linguistics students if not to a broader audience.
The (rather old-fashioned) Link Grammar parser messes up the worst:
(S (NP (NP Lessons)
(PP on
(NP inequality)))
,
(VP labour
(NP (NP markets)
, and
(NP conflict))
(PP from
(NP the Gilded Age ,
(PP for
(NP the present)
.)))))
You can visualize the induced structure more clearly in a tree diagram:
In other words, the article is about "Lessons on X, labour Y"; X is "inequality"; and Y is "markets and conflict from the Gilded Age", more specifically the "Gilded Age for the present".
The Berkeley parser is a bit better:
(ROOT
(NP
(NP (NNS Lessons))
(PP (IN on)
(NP
(NP (NN inequality))
(, ,)
(NP (JJ labour) (NNS markets))
(, ,)
(CC and)
(NP
(NP (NN conflict))
(PP (IN from)
(NP (DT the) (NNP Gilded) (NNP Age)))
(, ,)
(PP (IN for)
(NP (DT the) (NN present))))))
(. .)))
Note that this version adds the "part of speech" tags immediately above the individual words:
Now we're talking about lessons on X, where X="inequality, labour markets, and Y", and Y="conflict from the Gilded Age for the present". This is coherent but certainly not correct.
The Stanford parser almost gets it right:
(ROOT
(NP
(NP (NNS Lessons))
(PP (IN on)
(NP
(NP (NN inequality))
(, ,)
(NP (NN labour) (NNS markets))
(, ,)
(CC and)
(NP
(NP (NN conflict))
(PP (IN from)
(NP (DT the) (NNP Gilded) (NNP Age))))
(, ,)))
(PP (IN for)
(NP (DT the) (NN present)))
(. .)))
Here "from the Gilded Age" modifies only "conflict", and has no direct relationship to "lessons".
I think the correct analysis has the three PPs as all parallel dependents of "lessons"
• "on inequality, labour markets, and conflict"
• "from the Gilded Age"
• "for the present"
with the conjunction joining the three NPs
• "inequality"
• "labour markets"
• "conflict"
(ROOT
(NP
(NP (NNS Lessons))
(PP (IN on)
(NP
(NP (NN inequality))
(, ,)
(NP (NN labour) (NNS markets))
(, ,)
(CC and)
(NP (NN conflict))))
(PP (IN from)
(NP (DT the) (NNP Gilded) (NNP Age)))
(, ,)
(PP (IN for)
(NP (DT the) (NN present))))
(. .))
Some people might prefer to add additional structure, say by binding "on inequality, labour markets, and conflict" more closely to "lessons", or even stacking up the PP modifiers recursively:
I'll leave this question to the syntacticians.
But whatever exactly the right answer is, it's not what the three online parsers came up with. There may be some systems out there than can do better on this particular example — but PP attachment and conjunction scope in English remain hard problems for computational linguistics.
August 24, 2016 @ 8:42 am
Surely the editor of the journal could have made the title a lot more reader- (and presumably parser-) friendly before the piece was published, e.g. "Lessons for the present from the Gilded Age, on labour markets, inequality, and conflict". The presence of the comma before for the present is a real giveaway that the expression needs to be rearranged.
Also, even in the actually published version, what effect does the presence of the "Oxford comma" have on the parsers' behaviour? That is, would it have made a difference if the title had been "Lessons on labour markets, inequality and conflict from the Gilded Age, for the present"?
[(myl) You can try it at the links provided in the post. Omitting that comma makes the Stanford parser and the Berkeley parser do the same thing, and not the right thing.]
2. ### Daniel Barkalow said,
August 24, 2016 @ 11:45 am
It looks like the Link Grammar parser was misled by being given an NP when it thought it should get an S, and was pretty much doomed by the fact that "labour" is the only remotely plausible verb. You might get something more representative out of "There are lessons on…"
The Stanford parser's interpretation is probably the best guess without a bunch of pragmatics. (The present probably doesn't need lessons on [conflict from the Gilded Age], but could use lessons on [conflict from the Cold War Era].) I think [[…[NP PP]], PP] is more likely than [[… NP] PP, PP], all things being equal.
3. ### Gregory Kusnick said,
August 24, 2016 @ 12:03 pm
My first reading had "for the present" attached to the conjunction. That is, today we'll talk about lessons on inequality, labor markets, and conflict; additional lessons will be covered in our next installment.
4. ### D.O. said,
August 24, 2016 @ 8:50 pm
I would even rephrase it as "Labor markets, inequality, and conflict: lessons for the present from the Gilded Age", but maybe I was far too long around academic writing.
August 25, 2016 @ 12:47 am
@ D.O.: Yes, of course, much better (or at least much more academic)!
6. ### Idran said,
August 25, 2016 @ 9:30 am
Obviously there are ways that the title could have been made clearer, but I don't think that's the point of this article? Even unclear, it's still a valid English construction. The article isn't about improving the title, it's about automated efforts at parsing valid, if complicated, constructions.
7. ### Yuval said,
August 25, 2016 @ 2:05 pm
I read this at first like Stanford did.
Also, I think you misrepresented Link's parse: not markets and conflict are from the past, but labor. | 2021-09-20 16:27:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5748869776725769, "perplexity": 6780.401237938244}, "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/1631780057083.64/warc/CC-MAIN-20210920161518-20210920191518-00368.warc.gz"} |
https://www.physicsforums.com/threads/vector-analysis-resources.324911/ | # Vector Analysis Resources
1. Jul 13, 2009
### romistrub
So here I am, trying to learn vector analysis (vector calculus, call it what you want) and I'm all but befuddled by the number of ways to present the topic. I'm looking for a resource (preferably free) that expands on:
Lame coefficients / scale factors (the same thing, right?)
Metric coefficients
Forms (maybe?)
Generalized Stokes' Theorem
Jacobian matrix
Basically, my problem is this: my professor introduced scale coefficients without explaining how to calculate them, and in trying to learn it by myself, I stumbled upon all sorts of confusing and/or revealing generalizations. Unfortunately, I don't understand how they fit together.
For example:
- I know how to calculate the Jacobian given a transformation
- I know how to derive div/grad/curl operators for orthogonal coordinate systems from reading the Appendix of Griffith's Electrodynamics. These forms depend on scale factors, which Griffiths, although including them, never explains how to find them or even what they really *are*.
- I understand, to a degree, where scale factors and metric coefficients come from, from reading "Geometrical Vectors" and trying to understand wikipedia.
- Wikipedia sucks.
Is there anything that just gives the briefest (and I mean briefest) summary of how to compute these factors given a general curvlinear coordinate system? I can handle a sentence or two explaining the steps, but I really just need some worked examples. I'm also short on time! :)
Thanks! | 2017-10-22 18:39:20 | {"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": 0.8243443965911865, "perplexity": 1107.9665229751515}, "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-2017-43/segments/1508187825399.73/warc/CC-MAIN-20171022165927-20171022185927-00727.warc.gz"} |
http://eprints3.math.sci.hokudai.ac.jp/1518/ | # High regularity of solutions of compressible Navier-Stokes equations
Preprint Series # 776
Cho, Yonggeun High regularity of solutions of compressible Navier-Stokes equations. (2006);
PDF - Requires a PDF viewer such as GSview, Xpdf or Adobe Acrobat Reader390Kb
## Abstract
We study the Navier-Stokes equations for compressible {\it barotropic} fluids in a bounded or unbounded domain $\Omega$ of $\mathbf{R}^3$. The initial density may vanish in an open subset of $\Omega$ or to be positive but vanish at space infinity. We first prove the local existence of solutions $(\rho^{(j)}, u^{(j)})$ in $C([0,T_* ]; H^{2(k-j)+3} \times D_0^1 \cap D^{2(k-j)+3} (\Omega ) )$, $0 \le j \le k, k \ge 1$ under the assumptions that the data satisfy compatibility conditions and that the initial density is sufficiently small. To control the nonnegativity or decay at infinity of density, we need to establish a boundary value problem of $(k+1)$-coupled elliptic system which may not be in general solvable. The smallness condition of initial density is necessary for the solvability, which is not necessary in case that the initial density has positive lower bound. Secondly, we prove the global existence of smooth radial solutions of {\it isentropic} compressible Navier-Stokes equations on a bounded annulus or a domain which is the exterior of a ball under a smallness condition of initial density.
Item Type: Preprint 5 viscous compressible fluids, compressible Navier-Stokes equations, vacuum 35-xx PARTIAL DIFFERENTIAL EQUATIONS 1518 | 2017-06-24 13:57:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.925723671913147, "perplexity": 449.0258873156568}, "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/1498128320263.78/warc/CC-MAIN-20170624133941-20170624153941-00094.warc.gz"} |
https://socratic.org/questions/how-do-you-graph-using-the-intercepts-for-4x-3y-5 | # How do you graph using the intercepts for 4x+3y=-5?
Mar 6, 2016
$\textcolor{b l u e}{{y}_{\text{intercept}} = - \frac{5}{3}}$
color(blue)(x_("intercept") =-5/4
#### Explanation:
Given:$\text{ } \textcolor{b r o w n}{4 x + 3 y = - 5}$......................(1)
$\textcolor{b l u e}{\text{To find the x-intercept}}$
The x-axis crosses the y-axis at $y = 0$ so all we need to do is substitute 0 for y and we are then able to determine x-intercept
Write equation (1) as $\text{ } 4 x + 3 \left(0\right) = - 5$
Divide both sides by 4 giving
$\frac{4}{4} x = - \frac{5}{4}$
But $\frac{4}{4} = 1$
color(blue)(x_("intercept") =-5/4
$\textcolor{b l u e}{\text{To find the y-intercept}}$
The y-axis crosses the x-axis at $x = 0$ so all we need to do is substitute 0 for x and we are then able to determine y-intercept
Write equation (1) as $\text{ } 4 \left(0\right) + 3 y = - 5$
Divide both sides by 3 giving
$\text{ } \frac{3}{3} y = - \frac{5}{3}$
But $\frac{3}{3} = 1$
" "color(blue)(y_("intercept")=-5/3)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Drawing the graph}}$
Put a mark on the x axis at $x = - \frac{5}{4}$
Put a mark on the y axis at $y = - \frac{5}{3}$
Put a ruler on the graph and draw a line that passes through both these points. Extend the line to the edges of the graph.
$\textcolor{red}{\text{////////////////////////////////////////////////////////////////////////////}}$
$\textcolor{g r e e n}{\text{If you so wish to, this is how you change the equation into the format of}}$
$\text{ } y = m x + c$
Subtract $\textcolor{b l u e}{4 x}$ from both sides
" "color(brown)(4xcolor(blue)(-4x)+3y=-5color(blue)(-4x)
$\text{ } 0 + 3 y = - 4 x - 5$
Divide both sides by (same as $\textcolor{b l u e}{\times \frac{1}{3}} \text{)}$
$\text{ } \textcolor{b r o w n}{3 y \textcolor{b l u e}{\times \frac{1}{3}} = \left(- 4 x - 5\right) \textcolor{b l u e}{\times \frac{1}{3}}}$
$\text{ } \frac{3}{3} \times y = - \frac{4}{3} x - \frac{5}{3}$
But $\frac{3}{3} = 1$ giving
$\text{ } y = - \frac{4}{3} x - \frac{5}{3}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
# | 2019-09-20 03:38:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 29, "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": 0.8132660984992981, "perplexity": 446.6669273851913}, "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/1568514573827.2/warc/CC-MAIN-20190920030357-20190920052357-00523.warc.gz"} |
https://ora.ox.ac.uk/objects/uuid:6b3c03c4-34f3-4187-a603-5cfa4df1a61e | Journal article
### First measurement of Bose-Einstein correlations in proton-proton collisions at √s=0.9 and 2.36 TeV at the LHC.
Abstract:
Bose-Einstein correlations have been measured using samples of proton-proton collisions at 0.9 and 2.36 TeV center-of-mass energies, recorded by the CMS experiment at the CERN Large Hadron Collider. The signal is observed in the form of an enhancement of pairs of same-sign charged particles with small relative four-momentum. The size of the correlated particle emission region is seen to increase significantly with the particle multiplicity of the event.
### Authors
Khachatryan, V More by this author
Sirunyan, AM More by this author
Tumasyan, A More by this author
Bergauer, T More by this author
Journal:
Physical review letters
Volume:
105
Issue:
3
Pages:
032001
Publication date:
2010-07-05
DOI:
EISSN:
1079-7114
ISSN:
0031-9007
URN:
uuid:6b3c03c4-34f3-4187-a603-5cfa4df1a61e
Source identifiers:
116066
Local pid:
pubs:116066
Language:
English
Keywords: | 2020-10-22 23:44:40 | {"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": 0.9019355177879333, "perplexity": 2234.272885474994}, "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/1603107880401.35/warc/CC-MAIN-20201022225046-20201023015046-00149.warc.gz"} |
https://support.gurobi.com/hc/en-us/community/posts/360056768152-Dense-solution-set | What do you mean by
|x' - x| \leq \delta ?
Are you comparing x and x' just component by component --meaning, you are looking for the largest L1-ball that can fit inside a polyhedron-- or are you comparing these with respect to Euclidean norm? The two problems are very different, and in the former case there is a very nice modelling approach.
Ah. Sorry, I meant L1 norm. Yes, I am trying to fit largest L1-ball that can fit inside a polyhedron. So yes, can you guide me towards that modelling approach?
With L1 norm in mind, I think one can give a nice compact formulation to this problem. But please take it with a grain of salt, so to speak, as I did not test it, and read and check things carefully. So, no guarantees, but I will give this a try:)
I will describe something a bit different first, from which move to your formulation.
1) Assume, in addition that 0\in \{x\in\Re^n: Ax \leq b\}.
Suppose you want to determine the largest volume of a centrally-symmetric rectangular box (with faces parallel to coordinate planes) that fits inside a polytope \{x: Ax \leq b\} (that contains the origin). Since the box has faces parallel to the coordinate planes, you can effectively describe it by a vector of numbers r_j --displacement from 0-- in each direction j, j = 1 through n.
Now, in principle you would have to check that every vertex of the box, with coordinates + or -r_j, is within the polytope, i.e., satisfies Ax \leq b; this is a lot of constraints (2^n, in fact).
Fortunately though, you only need to check "the worst case" scenario for each of the constraint, namely, noting that for each constraint i the worst case situation amounts to
(*) \sum_j abs( A_{i,j} ) r_j \leq b
as the vertex coordinates take on values plus or minus r_j as you go through all the vertices of the box.
S, to get the max-volume centrally symmetric box you need to maximize the product of r_j variables (equivalent to the sum of the logs, which is convex), subject to the set of constraints (*) above.
2) Note if you want just a max-volume cube in the above, all r_j are the same, i.e., you would use a single scalar variable r instead.
Now, lets drop the assumption that 0\in \{x\in\Re^n: Ax \leq b\}. We can "centre" the polytope around some feasible point c (assuming such point exists), i.e.,
(**) A c \leq b
and furthermore, drawing on the reasoning above, re-write the cube containment constraints (writing the box vertex coordinates x_j as c_j plus or minus r_j) as
(***) \sum_j abs( A_{i,j} ) r \leq b - A c
It remains to maximize r subject to (***).
I think this is it.
Hopefully, this helps (and I am not making major mistakes here :)).
ps: apparently I was thinking of L-\infintiy norm in
|x' - x| = max_j |x_j - x_j'| \leq \delta
in the above.
If you want the L1 norm, this should be easier, as you only need to check the containment of 2n vertices of the L1 ball of radius r (a "diamond", not a box), where each vertex has only one non-zero --either r or -r at some position j-- and zeros every place else. So, you should augment the sum in (***) accordingly.
Hope this helps.
Hey. Thanks for your answer. If I understand correctly, you convert the cube containment conditions to vertex satisfiability. But why only 2n vertices for n-dimension? Shouldn't it be 2^n, because 2 direction (+/-) for each dimensions.
As in if, upper and lower bounds of a box are [2, 2, 2] and [-2, -2, -2] respectively, then there are 2^3 vertices, not 2*3 vertices.
(-2, -2, -2),
(-2, -2, 2),
(-2, 2, -2),
(-2, 2, 2),
(2, -2, -2)
(2, -2, 2)
(2, 2, -2)
(2, 2, 2)
For L1 norm you have a unit ball defined by
|x| + |y| + |z| + ... <= 1
with respective vertices
(1;0;0;....), (-1;0;0;...), (0;1;0;0...) etc.giving you 2n vertices in dimension n,
whereas, the L-infinity ball is denied by
max{|x|,|y|,...} <= 1
so this is a standard unit "box" with 2^n vertices.
Depending on which norm you work with, (***) will be different.
Hope this helps.
This works for L-infinity norm:
maximize r
subject to
(***) \sum_j abs( A_{i,j} ) r \leq b - A c
Since it is could be a bit tricky to figure out how to avoid checking 2^n vertices, I made it explicit.
In case of L1 norm, you can substitute the vertices yourself as there are only 2n of those.
Hope this helps. | 2020-05-27 22:37:29 | {"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": 0.9401949048042297, "perplexity": 1562.87266851345}, "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/1590347396163.18/warc/CC-MAIN-20200527204212-20200527234212-00304.warc.gz"} |
https://eoinmackall.wordpress.com/2016/05/06/the-grothendieck-group-of-algebraic-vector-bundles-calculations-of-affine-and-projective-space/ | # The Grothendieck Group of Algebraic Vector Bundles; Calculations of Affine and Projective Space
In this post, we view the Grothendieck group of a scheme over a field $F$, $X/F$ as a functor $K_0(-):Sch/F\rightarrow Ab$ (from the category of schemes over $F$ to the category of abelian groups) applied to the scheme $X$. Our goal is to see calculations for the simplest examples of the Grothendieck group, and what machinery is used throughout. Citation for this machinery will be included, as well as precise statements for the results used throughout but, we often omit proof with our goal in mind.
Several motivations can be given as to why one would study the Grothendieck group. The original motivation, I think, being: it aids in studying extensions of vector bundles over a scheme $X$. Recall there is an equivalence of categories between vector bundles over $X$, and locally free sheaves of finite rank on $X$ (the equivalence is given as so: for every vector bundle $\pi:E\rightarrow X$ we can associate the sheaf $\mathscr{E}$ given by the presheaf where, for each open $U\subset X$, we define $\mathscr{E}(U)=\{s:U\rightarrow E: s\circ \pi=Id_U\}$ – it can be checked this is a locally free sheaf of finite rank – and conversely for every locally free sheaf $\mathscr{E}$ we construct the vector bundle $E$ as $\text{Spec}_X(Sym(\mathscr{E}))$ the relative Spec of the symmetric algebra of $\mathscr{E}$ – note that we could construct the vector bundle using $\mathscr{E}^\vee$ instead). We can define an extension of a sheaf $\mathscr{F}$ by a sheaf $\mathscr{G}$ to be a sheaf $\mathscr{E}$ which fits in a short exact sequence
$0\rightarrow \mathscr{F}\rightarrow \mathscr{E}\rightarrow \mathscr{G}\rightarrow 0$.
Any two extensions $\mathscr{E},\mathscr{E}'$ which admit a map from one to the other, commuting with the maps from $\mathscr{F}$ and to $\mathscr{G}$, give a commutative diagram which we may apply the 5-lemma or the Snake lemma to show $\mathscr{E}\cong \mathscr{E}'$. One can hope to classify all such extensions by classifying each distinct short exact sequence. Sounds like a big task, but a start is with the
Definition 1: the functor $K_0(-):Sch/F\rightarrow Ab$ applied to a scheme $X$ is called the Grothendieck group of $X$. For any scheme $X$, the group $K_0(X)$ is defined to be the free abelian group on isomorphism classes of locally free sheaves of finite rank on $X$ quotiented by the subgroup generated by the relations $[\mathscr{E}]=[\mathscr{F}]+[\mathscr{G}]$ for every short exact sequence $0\rightarrow \mathscr{F}\rightarrow \mathscr{E}\rightarrow \mathscr{G}\rightarrow 0$.
Immediate from this definition is that $[\mathscr{F}\oplus\mathscr{G}]=[\mathscr{F}]+[\mathscr{G}]$ and $[0]=0$. We remark the Grothendieck group satisfies a, slightly irrelevant for our purposes, equivalent characterization as the object satisfying the following
Universal Mapping Property 2: for any functor $F(-):Sch/F\rightarrow Ab$ which factors additively through the category of locally free sheaves of finite rank over a scheme, and for every scheme $X$, there is a unique morphism of abelian groups $K_0(X)\rightarrow F(X)$.
We could similarly define $K_0'$ by replacing the category of locally free sheaves of finite rank by coherent sheaves over a scheme $X$. Then for any scheme $X$, there is a homomorphism $K_0(X)\rightarrow K_0'(X)$ which maps the class of a locally free sheaf to itself $[\mathscr{E}]\mapsto [\mathscr{E}]$. For smooth, quasi-projective schemes this is in fact an isomorphism (the inverse given by $[\mathscr{F}]\mapsto\sum_i (-1)^i [\mathscr{E}_i]$ for a locally free finite rank resolution $\mathscr{E}_{\bullet}\rightarrow \mathscr{F}\rightarrow 0$). Since all of the schemes we will consider below are smooth and quasi-projective, we can work with either group.
Below we compute, for a field $F$, $K_0(\text{Spec}(F))$ in proposition 3, $K_0(\mathbb{A}_F^n)$ in proposition 6 proposition 10 and proposition 12, and $K_0(\mathbb{P}^n_F)$ in proposition 8.
Proposition 3: $K_0(\text{Spec}(F))\cong \mathbb{Z}$.
Proof. Since any locally free sheaf over $\text{Spec(F)}$ is a finite dimensional $F$ vector space, and any two finite dimensional vectorspaces are isomorphic if and only if they have the same dimension, we can define an isomorphism directly $\dim:K_0(\text{Spec}(F))\rightarrow \mathbb{Z}$ by $[V]\mapsto \dim(V)$.$\square$
Our next goal will be to compute the Grothendieck group of all higher dimensional affine spaces, which we do in three distinct ways throughout this post. The following lemma of Serre gives a useful characterization of vector bundles over an affine scheme which we will employ later.
Lemma 4 (Serre-Swan)Let $\mathscr{F}$ be a coherent sheaf on a connected affine variety $X=\text{Spec}(A)$. Then the following are equivalent:
i) $\Gamma(X,\mathscr{F})$ is a projective $A$-module
ii) $\mathscr{F}$ is locally free
iii) $\mathscr{F}$ is isomorphic to the sheaf of germs of sections of a vector bundle over $X$.
Reference. This result can be found in [Faisceaux Algébriques Cohérents – Serre, Chapter 1, Section 4]. There is an english translation available via Google at the moment.
Lemma 5 (Quillen-Suslin)Every finitely-generated projective $F[x_1,...,x_n]$-module with $F$ a field is free of finite rank.
Reference. Aside from the original papers, an exposition of this result aimed towards students is given in [Introduction to Commutative Algebra and Algebraic Geometry – Ernst Kunz, Chaper 4, Section 3, Theorem 3.15]. The necessary tools for the proof are also developed in this chapter.
Proposition 6: $K_0(\mathbb{A}_F^n)\cong \mathbb{Z}$ for every $n\geq 0$.
Proof. By Lemma 4, a locally free sheaf of finite rank $\mathscr{F}$ over $\mathbb{A}^n_k$ is canonically isomorphic with the associated sheaf, $\widetilde{\Gamma(\mathbb{A}_k^n, \mathscr{F})}$, of a finitely generated projective $F[x_1,...,x_n]$-module. Lemma 5 implies every class (or its additive inverse) in $K_0(\mathbb{A}_F^n)$ can be represented by $\widetilde{F[x_1,...,x_n]^{\oplus p}}$ for some $p\geq 0$. The function $\text{rank}:K_0(\mathbb{A}_F^n)\rightarrow \mathbb{Z}$ defined by $[\widetilde{F[x_1,...,x_n]^{\oplus p}}]\mapsto p$ is an isomorphism, which completes the proof.$\square$
Before proving proposition 6, again, we calculate the Grothendieck group of projective space. The way we’ll do it is by considering a theorem, which is very useful, about projective bundles over a scheme $X$. For any locally free sheaf of rank $n$, say $\mathscr{E}$, we can construct a vector bundle, as stated in the introduction, by constructing $E:=\text{Spec}_X(Sym(\mathscr{E}))$ . The related projective bundle associated to $\mathscr{E}$, $\text{Proj}_X(Sym(\mathscr{E}))$ is constructed as the relative Proj of the symmetric algebra of $\mathscr{E}$.
Lemma 7: Let $X$ be a connected, smooth, quasi-projective scheme and $\mathscr{E}$ a locally free sheaf of rank $n+1$ over $X$. Then $K_0(\text{Proj}_X(Sym(\mathscr{E})))\cong K_0(X)[T]/(\sum_{i=0}^{n+1}(-1)^i [\Lambda^{n+1-i}(\mathscr{E})]T^i)$ where $\Lambda^{k}(\mathscr{E})$ is the $k$th exterior power sheaf of $\mathscr{E}$. Further, $T=[\mathscr{O}_{Proj_X(Sym(\mathscr{E}))}(1)]$ in the Grothendieck group.
Reference(s): Two references are provided – [Lectures on the K-Functor in Algebraic Geometry – Yuri Manin, Theorem 4.5] and [Riemann Roch Algebra – Fulton and Lang, Chapter 5, Section 2, Theorem 2.3].
Proposition 8: $K_0(\mathbb{P}^n_F)\cong \mathbb{Z}[T]/(1-T)^{n+1}$.
Proof. Note for the scheme $X=\text{Spec}(F)$, a locally free sheaf $\mathscr{E}$ of rank $n+1$ is a $n+1$ dimensional vector space, and the associated projective bundle is isomorphic with $\mathbb{P}^n_F$. It was shown in Proposition 3 that $K_0(\text{Spec}(F))\cong \mathbb{Z}$ by the dimension map. Thus, the class $[\Lambda^{n+1-i}(\mathscr{E})]$ is associated with the dimension of this vector space, which is ${n+1 \choose n+1-i}$. By Lemma 7 this yields the isomorphism with $\mathbb{Z}[T]/(\sum_{i=0}^{n+1}{n+1 \choose n+1-i}T^i)$ but this latter polynomial factors as $(1-T)^{n+1}$ and the proposition follows. $\square$
There are other ways to calculate the Grothendieck group of affine space. We include an alternative calculation which uses the localization exact sequence of $K$-theory as its main tool. One interesting question raised by the appearance of this sequence was whether or not the localization exact sequence could naturally be extended (to the left) as a long exact sequence. An answer was given (yes) in the work of higher algebraic $K$-theory by Daniel Quillen. Using some of the results given by his generalization of this theory, we provide a third calculation of $K_0(\mathbb{A}_F^n)$. Along with some final remarks, this will conclude our post.
Lemma 9 (Localization Exact Sequence): For any closed subscheme $Z$ of a scheme $X$, there is an exact sequence of abelian groups $K_0'(Z)\rightarrow K_0'(X)\rightarrow K_0'(Z\setminus X)\rightarrow 0$ where the first map is given by $[\mathscr{F}]\mapsto [i_*\mathscr{F}]$ and the second by $[\mathscr{G}]\mapsto [j^*\mathscr{G}]$ for the canonical inclusions $i:Z\rightarrow X,j:X\setminus Z\rightarrow X$.
Proof. See [Le Théorème de Riemann–Roch – Borel and Serre, Proposition 7]. This is also shown in [Riemann Roch Algebra – Fulton and Lang, Chapter 6, Section 3, Theorem 3.2].
Proposition 10: $K_0(\mathbb{A}^n_F)\cong \mathbb{Z}$.
Proof. Since $\mathbb{P}^n_F=\mathbb{A}^n_F\cup \mathbb{P}^{n-1}_F$. Using Lemma 9 we obtain an exact sequence $K_0(\mathbb{P}^{n-1}_F)\rightarrow K_0(\mathbb{P}^n_F)\rightarrow K_0(\mathbb{A}^n_F)\rightarrow 0$. The result will follow if we show the generators $1, T,..., T^{n-1}$ of $K_0(\mathbb{P}^{n-1}_F)$ map to distinct elements in $K_0(\mathbb{P}^n_F)$. (Moreover since $\mathbb{Z}$ is on the right side of this sequence, it follows that $K_0(\mathbb{P}^{n-1}_F)$ surjects onto a summand of $K_0(\mathbb{P}^n_F)$. Currently, I can’t see where the basis elements go to but, I’m gonna mark it done anyways. If I can figure out how it splits on the left I’ll update it).$\square$
We end mentioning one result often titled the homotopy invariance property.
Lemma 11: For any map $p:E\rightarrow X$ which is flat and has fibres isomorphic with affine space, there is an isomorphism $K_0(E)\cong K_0(X)$.
ReferenceThis is [Higher Algebraic K-Theory: I – Quillen, Propsoition 4.1]. One should note the results Quillen derives are in a much more general framework than may be alluded to by this post.
Proposition 12: $K_0(\mathbb{A}_F^n)\cong \mathbb{Z}$
Proof. Apply lemma 11 to the projection $\mathbb{A}^n_F\rightarrow \text{Spec}(F)$ and the result is then immediate from proposition 3.$\square$
Much more is true using the results provided above. For instance, lemma 11 implies the $K$-theory of a vector bundle is the same as the $K$-theory of its base space. There are also many other results one can find along the way. However, my main goal for writing this was to see if I could calculate some simple examples of the Grothendieck group (which I would say is pretty damned difficult from the definitions). | 2018-01-22 09:50:20 | {"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": 0, "img_math": 130, "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": 0.9625777006149292, "perplexity": 140.21254832946582}, "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-2018-05/segments/1516084891277.94/warc/CC-MAIN-20180122093724-20180122113724-00259.warc.gz"} |
https://math.stackexchange.com/questions/566848/representing-a-function-as-a-real-part-of-a-complex-variable | # Representing a function as a real part of a complex variable?
To represent a simple sinusoidally varying function $V(t)$ let us use $V(t)=Re(\hat V e^{i\omega t})$ where $\hat V$ can be a complex constant.Let $\hat V =-iV_o$ Therefore, $V(t)=V_o \sin(\omega t)$. Similarly, how can I represent the following function in terms of a complex function whose real value yields the function mentioned below:- $$Y(t)=|Y_0\sin(\omega t)|$$ To account for the the fact that both the half of the sinusoidal cycle will be positive, I first tried to represent $Y(t)=A\sin^2 (Ct)$ for some $A$ and $C$ and then use $\sin^2x=1-\frac{1+\cos(2x)}{2}$ to transform it into the required complex function.
But since the mathematical characteristics of the squared sine function is different from the absolute value of sine function (though the graphs have a similar appearance), the method doesn't yield the required function. How then, if possible to convert $Y(t)$ to a complex function?
Can we use Fourier series here to get the function as a sum of sines and cosines? I know:- $$|\sin(\omega t)|=-\sin(\omega t)\text{ for }\sin(\omega t)<0\text{ and } =\sin(\omega t)\text{ for }\sin(\omega t)>0$$
[I have used the physics tag because in AC circuit analysis, we characterize the varying impedances, currents and the voltages as complex numbers (exponentials). But at all times, the real situation is depicted by only the real part of complex currents. The exponentials are reduced by using $e^{i\theta}=\cos{\theta}+i\sin{\theta}$ into the real part which varies sinusoidally. Using a full wave recitifier we get $V_{in}(t)=V_0\sin\omega t$ but $V_{out}(t)=V_0|\sin\omega t|$]
$$Y(t) = |\sin (\omega t) | \\$$ can be extended to the complex plain using the formula $Im(z) = \dfrac{z - z^*}{2i}$. From this prove that $\sin : \Bbb{C} \to \Bbb{C}$, defined by
$$\sin(z) = \frac{e^{iz} - e^{-iz}}{2i}$$
is an extension of $\sin : \Bbb{R} \to \Bbb{R}$. Then $Y : \Bbb{C} \to \Bbb{C}$ using this definition is $Y(t) = \left |\dfrac{e^{i\omega t} - e^{-i \omega t}}{2i} \right |$, where $t \in \Bbb{C}$. But $|z| = \sqrt{z z^*}$ in $\Bbb{C}$. So you could expand that further if you wanted.
• can we get an expression for $Y(t)$ without using the absolute value function(so that it can be simply analysed)? – Satwik Pasani Nov 15 '13 at 2:22
• Like I said $|z| = \sqrt{zz^*}$ for all $z \in \Bbb{C}$. – Shine On You Crazy Diamond Nov 18 '13 at 15:55
• So $z= \left |\dfrac{e^{i\omega t} - e^{-i \omega t}}{2i} \right |$ and $z*= \left |\dfrac{e^{i\omega t} + e^{-i \omega t}}{2i} \right |$ ???? – Nick Dec 13 '15 at 0:20 | 2019-10-23 11:24:17 | {"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": 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": 0.8506309986114502, "perplexity": 233.9194652217481}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987833089.90/warc/CC-MAIN-20191023094558-20191023122058-00206.warc.gz"} |
https://www.anthonysmith.me.uk/2010/06/03/generating-latex-authors-lists-for-mnras-and-aa/ | Okay, a bit boring, but as promised, here are some Python functions to generate LaTeX source for long authors lists for MNRAS and A&A.
First, a small utility:
then the function for MNRAS:
and the function for A&A: | 2019-03-21 10:46:17 | {"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": 0.9394789338111877, "perplexity": 4326.899868212995}, "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-2019-13/segments/1552912202510.47/warc/CC-MAIN-20190321092320-20190321114320-00077.warc.gz"} |
https://directknowledge.com/single-variable-calculus/limits-and-continuity.html | In mathematics, both limits and continuity play a very important role in proofs and calculations. In this book, you will learn all about limits: what they are, how to find them, and how to use them in mathematical proofs. You will also learn about the different types of discontinuities and how to deal with them. With this knowledge, you will be able to perform complex calculations with ease and confidence.
Limits are the bread and butter of calculus. Without limits, we wouldn’t be able to find derivatives or integrals. In short, limits are essential for understanding calculus. So what are limits? Put simply, limits are a way of describing what happens when a function gets closer and closer to a certain value.
For example, if we have a function that is always increasing but never reaches a specific number, we say that the function approaches that number as its limit. This may not seem like a big deal, but limits are actually incredibly important. They allow us to define things like continuity and differentiability, which are essential concepts in calculus. So next time you’re struggling with a calculus problem, remember that it all comes down to limits!
Intuitively, one can think of limits as boundaries. Just as there are physical limits to what a person can do, there are limits to what a function can do. By understanding these limits, we can better understand the nature of continuity and how to take advantage of it in our calculations.
In short, limits are important in calculus because they allow us to understand how functions behave at specific points within their domains.
Limits are important because they help us understand continuity, derivatives, and integrals. And continuity is important because, well, the world is continuous (so it seems)! So limits are pretty important.
There are a few techniques that can be helpful for finding limits. One is to think about what the function is doing approaching the point in question. Is it getting closer and closer to a certain value? If so, that value is probably the limit. Another technique is to plug in values that are close to the point in question and see what happens. This can be helpful for getting an intuition for how the function behaves near the point in question. And finally, sometimes it can be helpful to use some algebraic manipulation to simplify the expression before taking the limit.
As with all things mathematical, practice makes perfect, so don’t be discouraged if limits give you some trouble at first. Keep at it and you’ll get the hang of it in no time!
Limits are important in calculus because they allow us to determine whether a function is continuous or not. Intuitively, a function is continuous if given any two points within its domain, there exists a smooth curve that connects those points. Limits enable us to test for continuity by seeing if the function’s value approaches the same number as we approach a specific point within the domain.
If the function’s value does approach the same number, then we say that the function is continuous at that point. If the function’s value does not approach the same number, then we say that the function is discontinuous at that point. Continuity is important because it allows us to determine how a function will behave as we approach certain values within its domain.
Continuity describes how a function behaves over a range of values. A function is continuous if it is smooth and uninterrupted over that range. In other words, there are no sudden jumps or discontinuities. Most functions you encounter in everyday life are continuous. For example, the position of a car on a road is continuous - there are no gaps or discontinuities in the car’s position. On the other hand, a function like 1/x is not continuous, because it has a discontinuity at x=0.
Together, limits and continuity allow us to understand and predict the behavior of functions at both the micro and macro levels. At the micro level, limits allow us to understand what happens to a function as x approaches a certain value. And at the macro level, continuity allows us to understand how a function behaves over a range of values. These concepts are essential for understanding and manipulating functions in calculus.
The existence of limits is one of the fundamental properties distinguishing calculus from ordinary algebra and geometry. Limits are basic to differential and integral calculus and play an important role in many other branches of mathematics. They also occur in physics and other sciences in connection with limiting cases and approximations. Informally, limits describe the behavior of a function as its arguments “approach” certain defined values; they provide a precise foundation for calculus, enabling earlier work with limits to be rigorously justified.
In most cases limits can be found by algebraic manipulations; more sophisticated means may sometimes be required, but even in difficult cases it is almost always possible to find some numerical indication of the desired limit. Many functions have limits at infinity that can be expressed in simple algebraic form, while others grow so rapidly that no finite limit exists.
Conversely, some functions have limits that do not exist when approached from either direction (the right or left), while others that have left and right limits may not have a limit as x approaches any real number other than these two specific values. The concept of a limit is thus seen to be closely related both to that of continuity (which deals with single-valued functions only) and to that of an infinite series, which enables discontinuities and multivalued functions also to be treated rigorously by means of limits.
One might wonder, “Why do we need a precise definition of limits? Can’t we just say that a limit is when a function gets close to a certain value?” While this informal definition suffices in many cases, there are situations where a more precise definition is required. For example, consider the function f(x) = 1/x. As x approaches 0, this function gets closer and closer to infinity. However, it never actually reaches infinity.
So if we were to use the informal definition of limits, we would say that the limit of f(x) as x approaches 0 is infinity. However, this is not very useful, since it doesn’t tell us anything about what happens to the function at x = 0. To get around this problem, mathematicians have devised a more precise definition of limits.
This definition allows us to say that the limit of f(x) as x approaches 0 is actually undefined. While this might not seem like a very satisfying answer, it does give us important information about the behavior of the function at x = 0. So in some cases, a more precise definition of limits can be quite helpful.
However, the definition of limits is actually quite technical, and it turns out to be very useful in calculus and other areas of mathematics. Without a precise definition of limits, many important theorems would be impossible to prove. So next time you’re feeling frustrated with your math homework, remember that your struggles could be helping to further the field of mathematics!
Have you ever been driving down the highway and suddenly had to brake for a stop sign or red light? If so, then you’ve experienced limits firsthand. Whenever you’re driving, you’re constantly changing speeds, and your car’s speedometer is measuring the rate of change of your car’s velocity.
But what happens when your car comes to a stop? The speedometer still measures a nonzero rate of change, but it’s obvious that your car’s velocity has changed dramatically. This is because the speedometer is measuring the instantaneous rate of change, or the limit of the average rate of change as the time interval approaches zero. This may seem like a lot of math jargon, but limits are actually quite intuitive.
In essence, they tell us how things are changing at a given moment. And limits are not just useful for calculus; they’re also essential for understanding rates of change in everyday life. So next time you’re stuck in traffic, just think of it as an opportunity to learn about limits!
One of the most important concepts in calculus is continuity. limits are a fundamental tool in calculus that allow us to determine whether a function is continuous at a point. Intuitively, continuity means that a function is “smooth” and doesn’t have any sharp jumps or abrupt changes.
Continuity is important because many of the most useful functions in calculus, such as the derivative and integral, are only defined for continuous functions. As a result, being able to identify and understand limits is essential for anyone who wants to study calculus. Luckily, limits are not as difficult as they may seem at first. With a little practice, anyone can learn to find them. And once you’ve mastered limits, you’ll be well on your way to understanding one of the most important branches of mathematics.
Limits and continuity may seem like dry, theoretical concepts, but they actually have a lot of real-world applications. For example, limits are used in calculus to determine the rate of change of a function, and they also play a role in physics when studying things like motion and acceleration. Similarly, continuity is important in many fields, including economics and computer programming. In fact, virtually any time you’re dealing with change or movement, limits and continuity are likely to be involved.
For example, limits can be used to understand how a car accelerates or how a projectile moves through the air. Continuity can be used to determine whether a path is smooth or whether a financial market is stable. In each of these cases, limits and continuity provide valuable insight that can help us to make better decisions.
So next time you’re stuck in traffic or trying to figure out why your computer keeps crashing, remember that you’re actually dealing with some pretty complex math!
Limits, the cornerstone of calculus, are a measure of how close a function gets to a certain value as it approaches that value from either direction. In other words, limits tell us how a function behaves near a point, such as whether it approaches the point from above or below, or oscillates around the point.
Continuity is a related concept: it deals with how well a function can be defined at a point, and whether it behaves predictably near that point. A function is continuous at a point if its limit exists at that point and if the function’s value at that point equals the limit. Intuitively, this means that a continuous function can be drawn without lifting one’s pencil from the paper. Many of the most important results in calculus depend on continuity; in fact, one could argue that calculus is really about continuity.
The idea of limits is used to define all sorts of important geometric objects such as tangent lines and curves, and these in turn are used to study functions and their behavior. Continuity is also used extensively in physics: for example, when solving problems involving fluids or electricity, one often assumes that the underlying functions are continuous. In short, continuity is a fundamental concept in calculus with far-reaching implications.
In the world of mathematics, limits are important because they help to define continuity. In other words, limits tell us when a function is “continuous” or “smooth.” Without limits, it would be difficult to calculate things like derivatives and integrals. So, in a sense, limits are the foundation of calculus. Without them, many of the important ideas in calculus would simply not be possible. Of course, limits can be a bit tricky to understand at first. But once you get the hang of it, you’ll see that they’re not so bad after all. And who knows? You might even come to enjoy working with them.
## 1.1 An Intuitive Introduction to Limits
This lecture illustrates when a limit does not exist by giving three case examples:
• A limit does not exist because the one-sided limits do not agree in value.
• A limit does not exist because of an oscillating behavior of a function, and
• a limit does not exist (no finite value) because of an unbounded behavior of a function.
## 1.2 Limits Using Tables
A limit is used to describe the behavior of a function near a point but not at the point. The function need not even be defined at the point. If it is defined there, the value of the function at the point does not affect the limit. Intuitively, $$$\lim _{x\to c}f(x)=L$$$ means we can make$$f(x)$$ as close to $$L$$ as we wish by taking any $$x$$ sufficiently close to, but different from $$c.$$
Example 1.1 Find the limit of $$f(x)$$ as $$x$$ approaches $$c$$ using a table of functional values for $$$f(x)=\frac{4x-9}{x^2-4}$$$ and $$c=3.$$
Solution. We compute, $\begin{array}{l|l} x & f(x) \\ \hline 2.997 & 0.599758 \\ 2.998 & 0.599839 \\ 2.999 & 0.59992 \end{array}$ $\begin{array}{l|l} x & f(x) \\ \hline 3.003 & 0.600238 \\ 3.002 & 0.600159 \\ 3.001 & 0.600079 \end{array}$ Thus as $$x$$ approaches $$3$$ from the left we estimate that $$f(x)$$ approaches $$3/5$$; and as $$x$$ approaches $$3$$ from the right we estimate that $$f(x)$$ approaches $$3/5.$$ Therefore, we estimate $$$\lim _{x\to 3}\frac{4x-9}{x^2-4}=\frac{3}{5}.$$$
Example 1.2 We will use a guessing method to show why the formal definition of a limit is a necessity.
Use a table to guess the values of $$$L=\lim _{x\to 0}\frac{2\sqrt{x+1}-x-2}{x^2}.$$$
Solution. From the table $\begin{equation*} \begin{array}{cccccccc} x & -0.5 & -0.1 & -0.01 & -0.001 & 0 & 0.001 & 0.005 \\ f(x) & -0.3431 & -0.2633 & -0.2513 & -0.2501 & \text{-} & -0.2499 & -0.2494 \end{array} \end{equation*}$ The number $$L$$ is suggested to be $$-0.25.$$ Interestingly, if you try $$$x=0.0000001$$$ just to make sure you have taken numbers close enough to 0, you may find that the calculator gives the value 0. Does this mean that the limit is 0? No, the calculator may give you a false answer because when $$x$$ is small enough (like $$0.0000001$$ ) then $$2\sqrt{x+1}-x-2$$ seems like 0. But in fact $$f(0.0000001)$$ is not equal to 0.
The point is, using technology to verify a computation can lead to misunderstanding; and in fact, a formal definition of a limit is needed. Using the formal definition of a limit, we can prove what the value of the limit is without any doubt. This type of proof is usually called an epsilon-delta proof since the formal definition is usually stated with the greek letters $$(epsilon) and$$ (delta).
Example 1.3 We will use a guessing method to show why the formal definition of a limit is a necessity.
Use tables of values to find the limit $$$\lim _{x\to 0}\left(x^3+\frac{\cos 5x}{10,000}\right).$$$
Solution. We construct a table of values. $\begin{equation*} \begin{array}{cccccc} x & 1 & 0.5 & 0.1 & 0.05 & 0.01 \\ f(x) & 1.000028 & 0.124920 & 0.001088 & 0.000222 & 0.000101 \end{array} \end{equation*}$ From the table it appears that $$$\lim _{x\to 0}\left(x^3+\frac{\cos 5x}{10,000}\right)=0.$$$
However, if we persevere with smaller values of $$x,$$ the next table suggests $$$\lim _{x\to 0}\left(x^3+\frac{\cos 5x}{10,000}\right) =0.0001000=\frac{1}{10,0000}$$$ $$$\begin{array}{ccc} x & 0.005 & 0.001 \\ f(x) & 1.00010009 & 0.00010000 \end{array}$$$ In fact, $$$\lim _{x\to 0}\left(x^3+\frac{\cos 5x}{10,000}\right)=\frac{1}{10,0000}$$$ which is easily proven once the formal limit definition is used to prove some interesting limit rules and continuity is discussed.
In summary, a three-pronged approach to solving limits is often:
• numerical approach by constructing tables of values,
• graphical approach by sketching a graph by hand or using technology,
• analytic approach by using algebra or calculus.
## 1.3 One-Sided and Two-Sided Limits
Consider, for example, a piecewise function with a jump where the function is pieced together (defined or not). Even though the one-sided limits might exist, they must agree in value for the two-sided limit to exist. In short, if the one-sided limits do not agree then the two-sided limit does not exist.
Theorem 1.1 The two-sided limit $$\lim _{x\to c}f(x)$$ exists if and only if the one-sided limits $$\lim _{x\to c^+}f(x)$$ and $$\lim _{x\to c^-}f(x)$$ both exist and $$\lim _{x\to c^-}f(x)=\lim _{x\to c^+}f(x).$$ In which case, $$\lim _{x\to c}f(x)=\lim _{x\to c^+}f(x).$$
Example 1.4 Sketch the graph of the piecewise function $$f$$ defined by $$$f(x)= \begin{cases} 2 x^2+1 & x<2 \\ 4 & x=2 \\ 3 x & x>2. \end{cases}$$$ Evaluate the following limits,
• $$\lim_{x\to 2^-}f(x)$$.
• $$\lim_{x\to 2^+}f(x)$$.
• $$\lim_{x\to 2}f(x)$$.
• $$\lim_{x\to 1^-}f(x)$$.
• $$\lim_{x\to 1^+}f(x)$$.
• $$\lim_{x\to 1}f(x)$$.
• $$\lim_{x\to 4^-}f(x)$$.
• $$\lim_{x\to 4^+}f(x)$$.
• $$\lim_{x\to 4}f(x)$$.
Example 1.5 Sketch the graph of the piecewise function $$g(x)$$ where $$g$$ is defined by $$$g(x)= \begin{cases} 3x-2 & x<-1 \\ 4 & x=-1 \\ x+5 & -1 < x < 3 \\ 4 & x=3 \\ 2-x & x>3. \end{cases}$$$ Evaluate the following limits,
• $$\lim_{x\to -1^-}g(x)$$
• $$\lim_{x\to -1^+}g(x)$$
• $$\lim_{x\to -1}g(x)$$
• $$\lim_{x\to 3^-}g(x)$$
• $$\lim_{x\to 3^+}g(x)$$
• $$\lim_{x\to 3}g(x)$$
• $$\lim_{x\to 4}g(x)$$
• $$\lim_{x\to 0}g(x)$$
• $$\lim_{x\to 1}g(x)$$
## 1.4 Oscillating Behavior
Also, consider another case where a function has an oscillating behavior . On one hand, the trigonometric functions all have an oscillating ( periodic ) behavior. However, imagine a function where the oscillation becomes much more pronounced as the variable approaches a fixed point; this type of oscillating behavior is where the function may not have a limit.
Example 1.6 Find $$\displaystyle\lim_{x\rightarrow 0} \cos \left(\frac{1}{x}\right).$$
Solution. The limit does not exist because $$\cos \frac{1}{x}=1$$ and $$\cos \frac{1}{x}=-1$$ for $\begin{equation*} x=\frac{1}{2\pi },\frac{1}{4\pi },\frac{1}{6\pi },\ldots \quad \text{and} \quad x=\frac{1}{\pi },\frac{1}{3\pi },\frac{1}{5\pi },\ldots \end{equation*}$ respectively.
The graph of $$y=\cos \left(\frac{1}{x}\right)$$ is oscillating around $$x=0,$$ so we infer that the limit does not exist because$$f(x)=\cos \frac{1}{x}$$ does not approach a number, but rather oscillates, as $$x$$ approaches 0.
## 1.5 Unbounded Behavior
Finally, we illustrate the case where a function becomes unbounded as the variable approaches a fixed point; for example, a function with a vertical asymptote. Without a finite number to assign the limit, we sometimes say that the limit does not exist.
Example 1.7 Determine $$\displaystyle \lim _{x\to 0}\frac{1}{x}.$$
Solution. Since $$f$$ decreases without bound as $$x\to 0^-$$ and $$f$$ increases without bound as $$x\to 0^+,$$ we say that $$\lim _{x\to0}f(x)$$ does not exist.
Example 1.8 Determine $$\displaystyle\lim _{x\to 0}\frac{1}{x^2}.$$
Solution. Since $$f$$ increases without bound as $$x\to 0^-$$ and $$f$$ increases without bound as $$x\to 0^+,$$ we say that $$\lim _{x\to0}f(x)=+\infty .$$
## 1.6 Exercises
Exercise 1.1 Estimate the limits, if they exist, by using a table of values to two decimal places.
• $$\lim _{x\to 2^+ }\frac{x^2-4}{x-4}$$
• $$\lim _{x\to 4^+ }\frac{\frac{1}{\sqrt{x}}-\frac{1}{2}}{x-4}$$
• $$\lim _{x\to 0 }\frac{\sin 2x}{x}$$
• $$\lim_{x\to 0}|x|\sin \frac{1}{x}$$
Exercise 1.2 Sketch the graph of $$f$$ and $$g.$$ Then identify the values of $$c$$ for which $$\lim _{x\to c}f(x)$$ and $$\lim _{x\to c}g(x)$$ exist given
• $$f(x)=\begin{cases} x^2 & x\leq 2 \\ 8-2x & 2<x<4 \\ 4 & x \geq 4 \end{cases}$$
• $$g(x)= \begin{cases}\sin x & x<0 \\ 1-\cos x & 0\leq x\leq \pi \\ \cos x & x>\pi \end{cases}$$
Exercise 1.3 Find $$a$$ so that the function $$f(x)=\begin{cases}a x+3 & x\leq 2 \\ 3-x & x>2\end{cases}$$ satisfies $$\lim _{x\to 2}f(x)=1.$$
Exercise 1.4 Estimate the limits by using tables of values for
• $$\lim _{x\to 13}\frac{x^3-9x^2-45x-91}{x-13}$$
• $$\lim _{x\to 13}\frac{x^3-9x^2-39x-86}{x-13}$$
Then using long division (or if you know it) explain why one of the limits exists and the other does not.
Exercise 1.5 Consider the function $$f(x) =\frac{|x+1|-|x-1|}{x}.$$ Estimate $$\lim _{x\to 0}f(x)$$ by evaluating $$f$$ at $$x$$-values near 0. Sketch the graph of $$f.$$
Exercise 1.6 Explain why $$\lim _{x\to 2}\frac{|x-2|}{x-2}$$ does not exist.
Exercise 1.7 Evaluate the function $$f(x)=x^2-\frac{2^x}{1000}$$ for $$x=1,$$ $$0.8,$$ $$0.6,$$ $$0.4,$$ $$0.2,$$ $$0.1,$$ and $$0.05.$$ Guess the value of $$\lim _{x\to 0}f(x).$$ Evaluate the function $$f(x)=x^2-\frac{2^x}{1000}$$ for $$x=0.04,$$ $$0.02,$$ $$0.01,$$ $$0.005,$$ $$0.003,$$ and $$0.001.$$ Guess again.
Exercise 1.8 The tabular approach is a convenient device for discussing limits informally, but if it is not used carefully, it can be misleading. For example, for $$x>0,$$ let $$f(x)=\sin \left(\frac{\pi }{\sqrt{x}}\right)$$
• Construct a table showing the value of $$x$$ and $$f(x)$$ for $$x=4,$$ $$4/25,$$ $$4/81,$$ $$4/169,$$ and $$4/289.$$ Based on this table what would you say about $$\lim _{x\to 0^+}f(x)?$$
• Construct a table showing the value of $$x$$ and $$f(x)$$ for $$x=4,$$ $$4/49,$$ $$4/121,$$ $$4/225,$$ $$4/361.$$ Based on this table what would you say about $$\lim _{x\to 0^+}f(x)?$$
• Based on your results in (a) and (b) what do you conclude about $$\lim _{x\to 0^+}f(x)?$$
Exercise 1.9 Sketch the graph of the function $f(x)= \begin{cases}-(x+1)^2+1 & -1\leq x\leq 0 \\ x^2 & 0 < x < 1 \\ 1 & 1 < x < 2 \\ 2 & x=2 \\ 1 & 2 < x\leq 3. \end{cases}$ and then use the graph to determine which the following statements about the function $$y=f(x)$$ are true and which are false?
• $$\lim _{\to -1^+}f(x)=1$$
• $$\lim _{x\to 2}f(x)$$ does not exist
• $$\lim _{x\to 2}f(x)=2$$
• $$\lim _{x\to 1^-}f(x)=2$$(e) $$\lim _{x\to 1^+}f(x)=1$$
• $$\lim _{x\to 1}f(x)$$ does not exist
• $$\lim _{x\to 0^+}f(x)=\lim _{x\to 0^-}f(x)$$
• $$\lim _{x\to c}f(x)$$ exists at every $$c$$ in the open interval $$(-1,1).$$
• $$\lim _{x\to c}f(x)$$ exists at every $$c$$ in the open interval $$(1,3).$$
• $$\lim _{x\to -1^-}f(x)=0$$
• $$\lim _{x\to 3^+}f(x)$$ does not exist
Exercise 1.10 Sketch the graph of the function $f(x)= \begin{cases} 3-x & x<2 \\ 2 & x=2 \\ \frac{x}{2} & x>2. \end{cases}$ and then use the graph to determine the following?
• Find $$\lim _{x\to 2^+}f(x),$$ $$\lim _{x\to 2^-}f(x),$$ and $$f(2).$$
• Does $$\lim _{x\to 2}f(x)$$ exist? If so, what is it? If not, why not?
• Find $$\lim _{x\to -1^-}f(x)$$ and $$\lim _{x\to -1^+}f(x).$$
• Does $$\lim _{x\to -1}f(x)$$ exist? If so, what is it? If not, why not?
Exercise 1.11 Let $$g(x)=\sqrt{x}\sin \left(\frac{1}{x}\right).$$ Use the graph of $$g$$ to determine the following.
• Does $$\lim _{x\to 0^+}g(x)$$ exist? If so, what is it? If not, why not?
• Does $$\lim _{x\to 0^-}g(x)$$ exist? If so, what is it? If not, why not?
• Does $$\lim _{x\to 0}g(x)$$ exist? If so, what is it? If not, why not?
Exercise 1.12 Graph $$f(x)=\begin{cases} x^3 & x\neq 1 \\ 0 & x=1. \end{cases}$$ Find $$\lim _{x\to 1^-}f(x)$$ and $$\lim _{x\to 1^+}f(x).$$ Does $$\lim _{x\to 1}f(x)$$ exist? If so, what is it? If not, why not?
Exercise 1.13 Graph $$f(x)=\begin{cases} 1-x^2 & x\neq 1 \\ 2 & x=1. \end{cases}$$ Find $$\lim _{x\to 1^-}f(x)$$ and $$\lim _{x\to 1^+}f(x).$$ Does $$\lim _{x\to 1}f(x)$$ exist? If so, what is it? If not, why not?
Exercise 1.14 Graph $$f(x)=\begin{cases} x & -1\leq x<0 \quad \text{or} \quad 0<x\leq 1 \\ 1 & x=0 \\ 0 & x<-1 \quad \text{or} \quad x>1.\end{cases}$$
• What is the domain and range of $$f?$$
• At what points $$c,$$ if any does $$\lim _{x\to c}f(x)$$ exist?
• At what points does only the left-hand limit exist?
• At what points does only the right-hand limit exist?
Exercise 1.15 Find the following limits.
• $$\lim _{x\to -7}(2x+5).$$
• $$\lim _{x\to -2}\left(x^3-2x^2+4x+8\right).$$
• $$\lim _{x\to 2}\left(\frac{x+3}{x+6}\right).$$
• $$\lim _{y\to 2}\left(\frac{y+2}{y^2+5y+6}\right).$$
• $$\lim _{y\to -3}(5-y)^{4/3}.$$
• $$\lim _{h\to 0}\left(\frac{5}{\sqrt{5h+4}+2}\right).$$
• $$\lim _{x\to 5}\left(\frac{x-5}{x^2-25}\right).$$
• $$\lim _{x\to 2}\left(\frac{x^2-7x+10}{x-2}\right).$$
• $$\lim _{x\to -2}\left(\frac{-2x-4}{x^3+2x^2}\right).$$
• $$\lim _{x\to 1}\left(\frac{x-1}{\sqrt{x+3}-2}\right).$$
• $$\lim _{x\to -2}\left(\frac{x+2}{\sqrt{x^2+5}-3}\right).$$
Exercise 1.16 Suppose $$\lim _{x\to 4}f(x)=0$$ and $$\lim _{x\to 4}g(x)=3.$$ Using limit laws find the limits $$\lim _{x\to 4}(g(x)+3),$$ $$\lim _{x\to 4}xf(x),$$ $$\lim _{x\to 4}(g(x))^2,$$ and $$\lim _{x\to 4}\frac{g(x)}{f(x)+1}.$$
Exercise 1.17 Suppose that $$\lim _{x\to -2}p(x)=4,$$ $$\lim _{x\to -2}r(x)=0,$$ and $$\lim _{x\to -2}s(x)=-3.$$ Using limit laws find the limits, $$\lim_{x\to -2}(p(x)+r(x)+s(x)),$$ $$\lim _{x\to -2}p(x)r(x)s(x),$$ and $$\lim _{x\to -2}\left(\frac{-4p(x)+5 r(x)}{s(x)}\right).$$
Exercise 1.18 Using limit laws evaluate the limit, $$\lim _{h\to 0}\frac{f(x+h)-f(x)}{h}$$ where $$f(x)=x^2$$ and $$x=1.$$
Exercise 1.19 Using limit laws evaluate the limit, $$\lim _{h\to 0}\frac{f(x+h)-f(x)}{h}$$ where $$f(x)=3x-4$$ and $$x=2.$$
Exercise 1.20 If $$\lim _{x\to -2}\frac{f(x)}{x^2}=1,$$ find $$\lim _{x\to -2}f(x)$$ and $$\lim _{x\to -2}\frac{f(x)}{x}.$$
Exercise 1.21 If $$\lim _{x\to 2}\frac{f(x)-5}{x-2}=3,$$ find $$\lim _{x\to 2}f(x).$$ Also if $$\lim _{x\to 2}\frac{f(x)-5}{x-2}=4$$ find $$\lim _{x\to 2}f(x).$$
## 1.8 Calculating Limits Using Limit Theorems
In this topic we concentrate not on the formal definition of a limit of a function of one variable but rather give several examples which emphasis algebra and trigonometry techniques to evaluate limits of functions using basic limit theorems.
::: {#thm- } [Limit Theorems] For any real number $$c,$$ suppose the functions $$f$$ and $$g$$ both have finite limits at $$x=c.$$ Then
• (Constant) $$\lim _{x\to c} k=k$$ for any constant $$k$$
• (Limit of $$x$$ ) $$\lim _{x\to c} x=c$$
• (Multiple) $$\lim _{x\to c}k f(x)=k \lim _{x\to c}f(x)$$
• (Sum) $$\lim _{x\to c}[f(x)+g(x)]=\lim _{x\to c}f(x)+\lim _{x\to c}g(x)$$
• (Difference) $$\lim _{x\to c}[f(x)-g(x)]=\lim _{x\to c}f(x)-\lim _{x\to c}g(x)$$
• (Product) $$\lim _{x\to c}[f(x)g(x)]=\left( \lim _{x\to c}f(x) \right)\left( \lim _{x\to c}g(x) \right)$$
• (Quotient) $$\lim _{x\to c}[f(x)/g(x)]=\left( \lim _{x\to c}f(x) \right)/\left( \lim _{x\to c}g(x) \right)$$
• (Power) $$\lim _{x\to c}[f(x)]^n=\left( \lim _{x\to c}f(x) \right){}^n$$ where $$n$$ is a rational number and whenever the limit on the right exists
• (Polynomial) $$\lim _{x\to c}P(x)=P(c)$$ for any polynomial $$P$$
• (Rational) $$\lim _{x\to c}R(x)=R(c)$$ for any rational function $$R$$ where $$c$$ is in the domain of $$R.$$ :::
Example 1.9 Find the limit of $$f(x)=\frac{2x^3-5x+8}{x^2-3}$$ at $$x=3.$$
Solution. By using several limit rules, we have \begin{align*} & \lim _{x\to 3}\frac{2x^3-5x+8}{x^2-3} =\frac{\lim _{x\to 3}\left(2x^3-5x+8\right)}{\lim _{x\to 3}\left(x^2-3\right)} \\ & \qquad =\frac{\lim _{x\to 3}\left(2x^3\right)-\lim _{x\to 3}(5x)+\lim _{x\to 3}(8)}{\lim _{x\to 3}\left(x^2\right)-\lim _{x\to 3}(3)} \\ & \qquad =\frac{2 \lim _{x\to 3}\left(x^3\right)- 5\lim _{x\to 3}(x)+8}{\lim _{x\to 3}\left(x^2\right)-3} \\ & \qquad =\frac{2 \left(\lim _{x\to 3} x\right){}^3- 5(3)+8}{\left(\lim _{x\to 3}x\right){}^2-3} =\frac{2 (3)^3- 5(3)+8}{(3)^2-3} =\frac{47}{6} \end{align*} Notice this is the same as evaluating the rational function $$f(x)$$ at $$x=3$$.
Example 1.10 Compute the limit of $$f(x)=\frac{2x^3+x^2-16x+12}{x^2-4}$$ at $$x=2.$$
Solution. By using several limit rules, we have \begin{align*} & \lim _{x\to 2}\frac{2x^3+x^2-16x+12}{x^2-4} =\lim _{x\to 2}\frac{(x-2)\left(2x^2+5x-6\right)}{(x-2)(x+2)} \\ & \qquad =\lim _{x\to 2}\frac{\left(2x^2+5x-6\right)}{(x+2)} =\frac{\left(2(2)^2+5(2)-6\right)}{((2)+2)} =3. \end{align*} In the previous example notice that we used $$$\label{limeq} \lim _{x\to 2}\frac{(x-2)\left(2x^2+5x-6\right)}{(x-2)(x+2)}=\lim _{x\to 2}\frac{\left(2x^2+5x-6\right)}{(x+2)}.$$$ Now it is not true that the functions $f(x)=\frac{\left(2x^2+5x-6\right)}{(x+2)} \qquad \text{and} \qquad g(x)=\frac{\left(2x^2+5x-6\right)}{(x+2)}$ are the same function because they have different domains. But the above equality is true because $$x$$ is approaching 2, and not equal to 2. So the point is,because $$f(x)=g(x)$$ when $$x\neq 2$$ we can indeed say $$\eqref{limeq}$$ holds. This is an important part of understanding limits.
Example 1.11 Suppose $$\lim _{x\to -2^-}f(x)=2,$$ $$\lim _{x\to -2^+}f(x)=4,$$ $$\lim _{x\to -2^-}g(x)=0,$$ and $$\lim _{x\to -2^+}g(x)=0,$$ find $\begin{equation*} \lim _{x\to -2}[f(x)+g(x)] \quad \text{and} \quad \lim _{x\to -2}[f(x)g(x)]. \end{equation*}$
Solution. Since $\lim _{x\to -2^-}f(x)\neq \lim _{x\to -2^+}f(x)$ we know $$\lim _{x\to -2}f(x)$$ does not exist, but this does not imply anything about $$\lim _{x\to -2}[f(x)+g(x)]$$ nor $$\lim _{x\to -2}[f(x)g(x)].$$
To find these limits we first find the two one-sided limits, $\lim _{x\to -2^-}[f(x)+g(x)]=\lim _{x\to -2^-}f(x)+\lim _{x\to -2^-}g(x)=2+0=2$ $\lim _{x\to -2^+}[f(x)+g(x)]=\lim _{x\to -2^+}f(x)+\lim _{x\to -2^+}g(x)=4+0=4$ and since $\lim _{x\to -2^-}[f(x)+g(x)]\neq \lim _{x\to -2^+}[f(x)+g(x)]$ we can now say the two-sided limit $$\lim _{x\to -2}[f(x)+g(x)]$$ does not exist. Similarly, $\lim _{x\to -2^-}[f(x)g(x)]=\left( \lim _{x\to -2^-}f(x) \right)\left( \lim _{x\to -2^-}g(x)\right)=2(0)=0$ $\lim _{x\to -2^+}[f(x)g(x)]=\left( \lim _{x\to -2^+}f(x) \right)\left( \lim _{x\to -2^+}g(x) \right)=4(0)=0$ and therefore, $$\displaystyle\lim _{x\to -2}[f(x)g(x)]=0.$$
## 1.9 Special Trigonometric Limits
Theorem 1.2 The following trigonometric limits hold: $\lim _{x\to 0}\frac{\sin x}{x}=1\text\qquad \text{and} \qquad \lim _{x\to 0} \frac{\cos x-1}{x}=0$
Example 1.12 Find $$\lim _{x\to 0} \frac{\sin ^2x}{2 x}.$$
Solution. We have $$\lim _{x\to 0}\frac{\sin ^2 x}{2x}=\lim _{x\to 0}\frac{\sin x}{2}\frac{\sin x}{x}=\frac{0}{2}(1)=0.$$
Example 1.13 Find $$\lim _{x\to 0}\frac{\sin x(1-\cos x)}{2 x^2}.$$
Solution. We have \begin{align*} & \lim _{x\to 0}\frac{\sin x(1-\cos x)}{2 x^2} =\lim _{x\to 0} \frac{1}{2}\frac{\sin x}{x}\frac{1-\cos x}{x} \\ & \qquad =\frac{1}{2}\lim _{x\to 0} \left(\frac{\sin x}{x}\right)\lim _{x\to 0}\left(\frac{1-\cos x}{x}\right) =\frac{1}{2}(1)(0) =0. \end{align*}
Example 1.14 Find $$\lim _{x\to 0} \frac{\sin 5x}{\sin 4x}.$$
Solution. We have \begin{align*} & \lim _{x\to 0} \frac{\sin 5x}{\sin 4x} =\lim _{x\to 0} \left(\frac{\sin 5x}{x}\right)\left(\frac{x}{\sin 4x}\right) \\ & \qquad =\lim _{x\to 0} \left(\frac{ 5\sin 5x}{5x}\right)\left(\frac{4x}{4 \sin 4x}\right) \\ & \qquad =\lim _{x\to 0} \left(\frac{ 5\sin 5x}{5x}\right)\lim _{x\to 0}\left(\frac{4x}{4 \sin 4x}\right) =\frac{5}{4}. \end{align*}
Example 1.15 Given $$g(x)=-27-9 x-6 x^2+x^3+x^4$$ and $$h(x)=21-16 x-3 x^2+2 x^3,$$ compute the limit $$g(x)/h(x)$$ at $$x=3.$$
Solution. Try a factor of $$(x-3)$$ from $$g(x)$$ obtaining $g(x)=(x-3)\left(9+6 x+4 x^2+x^3\right)$ and a graph of $$h$$ is which also inspires to try to factor of $$(x-3)$$ from $$h(x)$$ obtaining $$h(x)=(x-3)\left(2x^2+3x-7\right).$$
Therefore, \begin{align*} & \lim _{x\to 3}\frac{-27-9 x-6 x^2+x^3+x^4}{21-16 x-3 x^2+2 x^3} =\lim _{x\to 3}\frac{(x-3)\left(9+6 x+4 x^2+x^3\right)}{(x-3)\left(2x^2+3x-7\right)} \\ & \qquad =\lim _{x\to 3}\frac{9+6 x+4 x^2+x^3}{2x^2+3x-7} =\frac{9+6 (3)+4 (3)^2+(3)^3}{2(3)^2+3(3)-7} =\frac{9}{2}. \end{align*}
Theorem 1.3 If $$c$$ is any real number in the domain of the given function, then $\begin{array}{ccccc} \lim _{x\to c} \cos x=\cos c & & \lim _{x\to c} \sin x=\sin c & & \lim _{x\to c} \tan x=\tan c \\ \lim _{x\to c} \sec x=\sec c & & \lim _{x\to c} \csc x=\csc c & & \lim _{x\to c} \cot x=\cot c. \end{array}$
Example 1.16 Compute the limit of $$f(x)=\frac{x}{\sin x-2 \cos x}$$ at $$x=\pi .$$
Solution. By using several limit rules, we have $\lim _{x\to \pi }\frac{x}{\sin x-2 \cos x}=\frac{\pi }{\sin \pi -2\cos \pi }=\frac{\pi }{0-2(-1)}=\frac{\pi }{2}.$
Example 1.17 Compute $$\lim _{x\to \pi /4}\frac{1-\tan x}{\sin x- \cos x}.$$
Solution. We have \begin{align*} & \lim _{x\to \pi /4}\frac{1-\tan x}{\sin x- \cos x} =\lim _{x\to \pi /4}\frac{1-\frac{\sin x}{\cos x}}{\sin x- \cos x} \\ & \qquad =\lim _{x\to \pi /4}\frac{\frac{\cos x-\sin x}{\cos x}}{\sin x- \cos x} =\lim _{x\to \pi /4}\frac{\cos x-\sin x}{\cos x}\frac{1}{\sin x- \cos x} \\ & \qquad =\lim _{x\to \pi /4}\frac{-1}{\cos x} =\lim _{x\to \pi /4}\frac{-1}{\frac{1}{\sqrt{2}}} =-\sqrt{2} \end{align*}
Example 1.18 Compute the limit of $$f(x)=\frac{\tan x}{\sin x}$$ as $$x\to0.$$
Solution. We have $$\lim _{x\to 0}\frac{\tan x}{\sin x}$$ $$=\lim _{x\to 0} \left(\frac{\sin x}{\cos x}\right)\left(\frac{1}{\sin x}\right)$$ $$=\lim _{x\to 0} \frac{1}{\cos x}$$ $$=1.$$
## 1.10 Using Rationalization
Example 1.19 Compute: $$\lim _{x\to 1}\frac{\sqrt{x}-1}{x-1}.$$
Solution. We have \begin{align*} \lim _{x\to 1}\frac{\sqrt{x}-1}{x-1}=\lim _{x\to 1}\frac{\sqrt{x}-1}{x-1}\frac{\sqrt{x}+1}{\sqrt{x}+1}=\lim _{x\to 1}\frac{1}{\sqrt{x}+1}=\frac{1}{\sqrt{1}+1}=\frac{1}{2}. \end{align*}
Example 1.20 Compute the limit of $$f(x)=\frac{x-\pi }{\sqrt{x}-\sqrt{\pi }}$$ as $$x\to\pi .$$
Solution. We have \begin{align*} & \lim _{x\to \pi }\frac{x-\pi }{\sqrt{x}-\sqrt{\pi }} =\lim _{x\to \pi }\left(\frac{x-\pi }{\sqrt{x}-\sqrt{\pi }}\right)\left(\frac{\sqrt{x}+\sqrt{\pi }}{\sqrt{x}+\sqrt{\pi }}\right) \\ & \qquad =\lim _{x\to \pi }\frac{(x-\pi )\left(\sqrt{x}+\sqrt{\pi }\right)}{x-\pi } =2\sqrt{\pi } \end{align*}
Example 1.21 Compute the limit of $$f(x)=\frac{x-1}{\sqrt[3]{x}-1}$$ as $$x\to 1.$$
Solution. We have \begin{align*} & \lim _{x\to 1}\frac{x-1}{\sqrt[3]{x}-1} =\lim _{x\to 1}\left(\frac{x-1}{\sqrt[3]{x}-1}\right)\left(\frac{\sqrt[3]{x^2}+\sqrt[3]{x}+1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}\right) \\ & \quad =\lim _{x\to 1}\frac{(x-1)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}{x-1} =\lim _{x\to 1}\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right) =3. \end{align*}
## 1.11 Limits of Piecewise Functions
Example 1.22 Find $$$\lim _{x\to 0} \begin{cases} 2(x+1) & x<3 \\ 4 & x=3 \\ x^2-1 & x>3 \end{cases}$$$
Solution. Since $$x\to 0$$ we know that $$x<3$$ and so we use $$2(x+1)$$ to evaluate the limit of the piecewise function, as follows, $$$\lim _{x\to 0} \begin{cases} 2(x+1) & x<3 \\ 4 & x=3 \\ x^2-1 & x>3. \end{cases} =\lim _{x\to 0} 2(x+1)=2$$$
Example 1.23 At $$x=7,$$ compute the limit of
$$$f(x)= \begin{cases} 2\left(x-x^2\right) & x<7 \\ -83 & x=7 \\ (x-7)^2-84 & x>7 \end{cases}$$$
Solution. Since the function is pieced together at $$x=7$$ we will evaluate two one sided limit. First the limit from the left, we have $\lim_{x\to 7^-}f(x)=-84$ and for the limit from the right we have $\lim _{x\to 7^+}f(x)=-84.$
Since $\lim _{x\to 7^-} f(x)=\lim _{x\to 7^+}f(x)$ we know the two-sided limit must exist and we have $\lim _{x\to 7}f(x)=-84$ even though $$f(7)=-83.$$
## 1.12 Squeeze Theorem
Next we state the squeeze theorem and through an example show how to use it. Basically, the idea is to bound a function on both sides by functions whose limits can be more easily computed; and thus in the process squeeze the value of the limit of the original function out.
::: {#thm- } Squeeze Theorem Let $$f, g$$ and $$h$$ be functions of $$x$$ . If both of the following conditions hold
• $$g(x)\leq f(x)\leq h(x)$$ on an open interval containing $$c$$ and
• $$\lim _{x\to c}g(x) =\lim _{x\to c}h(x) =L$$
then $$\lim _{x\to c}f(x)=L.$$
:::
Example 1.24 Use the squeeze rule to find the limit of $$f(x)=x \cos \left(\frac{1}{x}\right)$$ as $$x\to 0^+.$$
Solution. We are interested in this function around $$x>0$$ . Knowing that the cosine function is always less than or equal to one, we see that when $$-1\leq x\leq 1$$ we have $-x\leq f(x)=x \cos \left(\frac{1}{x}\right)\leq x.$ Since $$\lim _{x\to 0} -x=\lim _{x\to 0} x=0$$ we have $$\lim _{x\to 0}x \cos \left(\frac{1}{x}\right)=0$$ by the squeeze theorem.
## 1.13 Exercises
Exercise 1.22 Find the following limits.
• $$\lim _{x\to -7}(2x+5)$$
• $$\lim _{x\to -2}\left(x^3-2x^2+4x+8\right)$$
• $$\lim _{x\to 2}\left(\frac{x+3}{x+6}\right)$$
• $$\lim _{y\to 2}\left(\frac{y+2}{y^2+5y+6}\right)$$
• $$\lim _{y\to -3}(5-y)^{4/3}$$
• $$\lim _{h\to 0}\left(\frac{5}{\sqrt{5h+4}+2}\right)$$
• $$\lim _{x\to 5}\left(\frac{x-5}{x^2-25}\right)$$
• $$\lim _{x\to 2}\left(\frac{x^2-7x+10}{x-2}\right)$$
• $$\lim _{x\to -2}\left(\frac{-2x-4}{x^3+2x^2}\right)$$
• $$\lim _{x\to 1}\left(\frac{x-1}{\sqrt{x+3}-2}\right)$$
• $$\lim _{x\to -2}\left(\frac{x+2}{\sqrt{x^2+5}-3}\right)$$
Exercise 1.23 Suppose $$\lim _{x\to 4}f(x)=0$$ and $$\lim _{x\to 4}g(x)=3.$$ Using limit laws find the limits $$\lim _{x\to 4}(g(x)+3),$$ $$\lim _{x\to 4}xf(x),$$ $$\lim _{x\to 4}(g(x))^2,$$ and $$\lim _{x\to 4}\frac{g(x)}{f(x)+1}.$$
Exercise 1.24 Suppose that $$\lim _{x\to -2}p(x)=4,$$ $$\lim _{x\to -2}r(x)=0,$$ and $$\lim _{x\to -2}s(x)=-3.$$ Using limit laws find the limits, $$\lim_{x\to -2}(p(x)+r(x)+s(x)),$$ $$\lim _{x\to -2}p(x)r(x)s(x),$$ and $$\lim _{x\to -2}\left(\frac{-4p(x)+5 r(x)}{s(x)}\right).$$
Exercise 1.25 Using limit laws evaluate the limit, $$\lim _{h\to 0}\frac{f(x+h)-f(x)}{h}$$ where $$f(x)=x^2$$ and $$x=1.$$
Exercise 1.26 Using limit laws evaluate the limit, $$\lim _{h\to 0}\frac{f(x+h)-f(x)}{h}$$ where $$f(x)=3x-4$$ and $$x=2.$$
Exercise 1.27 If $$\lim _{x\to -2}\frac{f(x)}{x^2}=1,$$ find $$\lim _{x\to -2}f(x)$$ and $$\lim _{x\to -2}\frac{f(x)}{x}.$$
Exercise 1.28 If $$\lim _{x\to 2}\frac{f(x)-5}{x-2}=3,$$ find $$\lim _{x\to 2}f(x).$$ Also if $$\lim _{x\to 2}\frac{f(x)-5}{x-2}=4$$ find $$\lim _{x\to 2}f(x).$$
## 1.15 Continuity at a Point
Definition 1.1 A function is continuous at a point $$c$$ means
• $$f(c)$$ is defined,
• $$\lim _{x\to c}f(x)$$ exists, and
• $$\lim _{x\to c}f(x)=f(c).$$
## 1.16 Discontinuity
Example 1.25 Find three examples of how a discontinuity might arise.
Solution. First, the function $\displaystyle f(x)=\frac{x^2-2x+1}{x-1}$ is discontinuous at $$x=1$$ because $$f(1)$$ is not defined. So one type of discontinuity is a hole in the function.
Secondly, the function $f(x)= \begin{cases} x^2+1 & x\geq 0 \\ -x^2-2 & x<0 \end{cases}$ is discontinuous at $$x=0$$ because $$\lim _{x\to 0^-}f(x)=-2$$ and $$\lim _{x\to 0^+}f(x)=1;$$ thus $$\lim _{x\to 0}f(x)$$ does not exist and so $$f$$ has a discontinuity at $$x=0.$$ This type of discontinuity is called a jump.
Thirdly, the function $f(x)=\frac{x-1}{x-2}$ is discontinuous at $$x=2$$ because $$f(2)$$ is not defined and this type of discontinuity is called a pole because $f(x)+$ as $$x\to 2^+$$.
## 1.17 Continuous Functions
Theorem 1.4 If $$f$$ and $$g$$ are functions that are continuous at $$x=c$$ then $$f\pm g,$$ $$f g,$$ $$f/g,$$and $$f\circ g$$ are continuous at $$x=c,$$ provided that $$c$$ is in the domain of the function.
Example 1.26 Give some examples of continuous functions.
Solution. For example, the functions $$2x^2-2x+5,$$ (polynomial), $$\displaystyle\frac{x-1}{x}$$ (rational), $$\csc x$$ (trigonometric), and $$\sec ^{-1}x$$ (inverse trigonometric) are continuous on their domains. Also the functions $\displaystyle 2x^2-2x+5+\frac{x-1}{x}(\csc x),$ $$\displaystyle\sec ^{-1}\left(\frac{x-1}{x}\right),$$ and $$\displaystyle\left(2x^2-2x+5\right)\left(\frac{x-1}{x}\right)$$ are continuous functions on their domains.
Theorem 1.5 If $$f$$ is a
• polynomial function,
• rational function,
• trigonometric function, or
• inverse trigonometric function,
then $$f$$ is continuous where it is defined.
::: {#thm- } [Composition Limit Theorem] If $$\lim _{x\to c}g(x)=L$$ and $$f$$ is a continuous function at $$L,$$ then $\lim _{x\to c}(f\circ g)(x)=f(L).$ :::
Example 1.27 Use the composition limit theorem to evaluate the following limits. $\lim _{x\to 3} \left(x^2+3\right)^2 \quad\text{and}\quad \lim _{x\to \pi /4} \sin ^4 x.$
Solution. By the composition limit theorem, we have $\lim _{x\to 3}\left(x^2+3\right)^2 =\left(\lim _{x\to 3}\left(x^2+3\right)\right)^2 =12^2 =144.$ By the composition limit theorem, we have
$\lim _{x\to \pi /4}\sin ^4x =\left(\lim _{x\to \pi /4}\sin x\right){}^4 =\left(\frac{\sqrt{2}}{2}\right)^4 =\left(\frac{1}{4}\right).$
## 1.18 One-Sided Continuity
Definition 1.2 The function $$f$$ is continuous from the right at $$c$$ if and only if $\lim _{x\to c^+}f(x)=f(c)$ and it is continuous from the left at $$c$$ if and only if $\lim _{x\to c^-}f(x)=f(c).$
Example 1.28 Give an example of a function that is continuous from the right (or right continuous ) at $$x=0.$$
Solution. The function $$f(x)=\sqrt{x}$$ is continuous from the right at $$x=0$$ because $$\lim _{x\to 0^+}\sqrt{x}=0.$$
## 1.19 Determining Parameters for Continuity
Example 1.29 Find constants $$a$$ and $$b$$ so that $\begin{cases} a x^2+b & x>2 \\ 4 & x=2 \\ x^2-a x+b & x<2 \end{cases}$ is continuous on $$\mathbb{R}$$.
Solution. Since $$f$$ is defined on $$\mathbb{R},$$ and $$f$$ is continuous for all $$x\neq 2$$ for any $$a$$ and $$b$$ that we choose, itis left to find an $$a$$ and $$b$$ such that $$\lim _{x\to 2^+}a x^2+b=4$$ and $$\lim _{x\to 2^-}\left(x^2-a x+b\right)=4.$$ Thus we have the system $$4 a+b=4$$ and $$4-2 a+b=4.$$ Solving this system we have, $$a=2/3$$ and $$b=4/3.$$ :::
Example 1.30 Find constants $$a$$ and $$b$$ such that $$f$$ is continuous at $$x=1$$ where $\begin{cases} a x+b & x>1 \\ 3 & x=1 \\ x^2-4x+b+3 & x<1 \end{cases}$
Solution. To have continuity at $$x=1$$ we must have $$\lim _{x\to 1^-}f(x)=3$$ and $$\lim _{x\to 1^+}f(x)=3,$$ thus $$a (1)+b=3$$ and $$(1)^2-4(1)+b+3=3.$$ Therefore, $$a+b=3$$ and $$b=3.$$ So $$a+3=3$$ and $$a=0.$$
## Removable Continuity
Example 1.31 Determine the value for which $$f(2)$$ should be assigned, if any, to have $\displaystyle f(x)=\sqrt{\frac{x^2-4}{x-2}}$ continuous at $$x=2.$$
Solution. Since $$\lim _{x\to 2^-}f(x)=2$$ and $$\lim _{x\to 2^+}f(x)=2$$ we have $$\lim _{x\to 2}f(x)=2.$$ Therefore, if we define $$f(2)=2$$ the function $$f$$ will be continuous at $$x=2.$$
## Intermediate Value Theorem
::: {#thm- } Intermediate Value Theorem If $$f$$ is a continuous function on the closed interval $$[a,b]$$ and $$L$$ is some number strictly between$$f(a)$$ and $$f(b)$$, then there exists at least one number $$c$$ on the open interval $$(a,b)$$ such that $$f(c)=L.$$
Example 1.32 The population (in thousands) of a colony of bacteria $$t$$ minutes after the application of a toxin is given by the function $P(t)= \begin{cases} t^2+1 & \text{if } 0\leq t<5 \\ -8t+66 & \text{if } t\geq 5 \end{cases}$ - When does the colony die out? - Show that at some time between $$t=2$$ and $$t=7,$$ the population is $$9,000.$$
Solution. Since $\lim _{t\to 5^+}P(t)=\lim _{t\to 5^+}(-8t+66)=26$ and $\lim _{t\to 5^-}P(t)=\lim _{t\to 5^-}\left(t^2+1\right)=26$ we know that $$P$$ is continuous at $$t=5$$ and thus, also for all $$t\geq 0.$$ (a) The colony dies out when $$-8t+66=0$$ which means $$t=33/4\approx 8.25.$$ Therefore, the colony dies out in 8 minutes and 15 seconds.
1. Since $$P(2)=5,$$ $$P(7)=10,$$ and $$P$$ is continuous on $$(2,7),$$ the intermediate value theorem yields at least one number $$c$$ between 2 and 7 such $$P(c)=9.$$ Therefore, there is some time $$t=c$$ between $$t=2$$ and $$t=7$$ such that the population is 9,000.
## 1.20 Exercises
Exercise 1.29 Sketch the graph of the function $f(x)= \begin{cases} x^2-1 & -1\leq x<0 \\ 2x & 0<x<1 \\ 1 & x=1 \\ -2x+4 & 1<x<2 \\ 0 & 2<x<3. \end{cases}$ Use the graph of the function $$f$$ to answer the following. - Does $$f(-1)$$ exist? - Does the limit, $$\lim _{x\to -1^+}f(x)$$ exist? - Does the limit, $$\lim _{x\to -1^+}f(x)=f(-1)$$? - Does $$f(1)$$ exist? - Does the limit, $$\lim _{x\to 1}f(x)$$ exist? - Does the limit, $$\lim _{x\to 1}f(x)=f(1)$$ exist? - Is $$f$$ defined at $$x=2?$$ - Is $$f$$ continuous at $$x=2?$$ - At what values is $$f$$ continuous? - What value should be assigned to $$f(2)$$ , to make the extended function continuous at $$x=2?$$ - To what new value should $$f(1)$$ be assigned to remove the discontinuity?
Exercise 1.30 For each of the following functions determine the largest set on which the function will be continuous.
• $$g(x)=\frac{x+1}{x^2-4x+3}$$
• $$g(x)=\frac{1}{|x|+1}-\frac{x^2}{2}$$
• $$g(x)=\frac{\sqrt{x^4+1}}{1+\sin ^2x}$$
• $$g(x)=\sqrt[4]{3x-1}$$
Exercise 1.31 For each of the following functions find constants $$a$$ and $$b$$ so that the function will be continuous for all $$x$$ in the domain.
• $$f(x)=\begin{cases} \frac{\tan a x}{\tan b x} & x<0 \\ 4 & x=0 \\ a x+b & x>0 \end{cases}$$
• $$f(x)=\begin{cases}\frac{\sqrt{a x+b}-1}{x} & x\neq 0 \\ 1 & x=0\end{cases}$$
• $$f(x)=\begin{cases}a x+3 & x<1 \\ 5 & x=1 \\ x^2+b & x>1\end{cases}$$
Exercise 1.32 Let $$u(x)=x$$ and $$f(x)=\begin{cases}0 & x\neq 0 \\ 1 & x=0\end{cases}$$. Show, for this given $$u$$ and $$f$$ , that $\lim _{x\to 0}f[u(x)]\neq f\left(\lim _{x\to 0}u(x)\right).$
Exercise 1.33 If a function $$f$$ is not continuous at $$x=c$$ , but can be made continuous at $$x=c$$ by being assigned a new value of that point, it is said to have a at $$x=c.$$ Which of the following functions have a removable discontinuity at $$x=c$$ ?
• $$f(x)=\frac{2x^2+x-15}{x+3}$$ at $$c=-3$$
• $$f(x)=\frac{x-2}{|x-2|}$$ at $$c=2$$
• $$f(x)=\frac{2-\sqrt{x}}{4-x}$$ at $$c=4$$
• $$f(x)=\frac{2-x}{4-\sqrt{x}}$$ at $$c=16$$
Exercise 1.34 Prove that the function $$f(x)=x^3-x^2+x+1$$ must have at least one real root.
Exercise 1.35 Prove that the function $$(x)=\sqrt{x+3}-e^x$$ must have at least one real root.
Exercise 1.36 Find a function(s) with the following properties. (a) Find functions $$f$$ and $$g$$ such that $$f$$ is discontinuous at $$x=1$$ but $$f g$$ is continuous there. (b) Give an example of a function defined for all real numbers that is continuous at only one point.
## 1.22 Average Rate of Change
Definition 1.3 Suppose $$y$$ is a function of $$x,$$ say $$y=f(x).$$
When a change in the variable is made from $$x$$ to $$x+\Delta x,$$ there is a corresponding change to the $$y,$$ namely $$\Delta y=f(x+\Delta x)-f(x)$$. The average rate of change of $$y$$ with respect to $$x$$ is $\frac{\Delta y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}$ and is also known as the difference quotient .
Example 1.33 Let $$f(x)=\sqrt{x^2-9}.$$ Find the average rate of change from $$x=3$$ to $$x=6.$$
Solution. The average rate of change of $$f$$ from $$x=3$$ to $$x=6$$ is given by, $\frac{f(x+\Delta x)-f(x)}{\Delta x}=\frac{f(6)-f(3)}{6-3}=\frac{\sqrt{6^2-9}-\sqrt{3^2-9}}{3}= \sqrt{3}$ which is also the slope of the secant line through $$(3,0)$$ and $$\left(6,3 \sqrt{3}\right).$$
In general, suppose an object moves along a straight line according to an equation of motion $$s=f(t),$$ where $$s$$ is the ( displacement} (\index{directed distance ) of the object from the origin at time $$t.$$
The function $$f$$ that describes the motion is called the position function of the object. In the time interval from $$t=a$$ to $$t=a+h$$ the change in position is $$f(a+h)-f(a)$$ and the average velocity over this time interval is $\frac{f(a+h)-f(a)}{h}$ which is the same as the slope of the secant line through these two points.
Example 1.34 If a billiard is dropped from a height of 500 feet, its height $$s$$ at time $$t$$ is given by the position function $$s=-16t^2+500$$ where $$s$$ is measured in feet and $$t$$ is measured in seconds. Find the average velocity over the intervals $$[2,2.5]$$ and $$[2,2.6]$$.
Solution. For the interval $$[2,2.5],$$ the object falls from a height of $$s(2)=-16(2)^2+500=436$$ feet to a height of $$s(2.5)=-16(2.5)^2+500=400.$$
The average velocity is $\frac{\Delta s}{\Delta t}=\frac{s(2.5)-s(2)}{2.5-2}=\frac{400-436}{2.5-2}=-72.$ For the interval $$[2,2.6],$$ the object falls from a height of $$s(2)=436$$ feet to a height of $$s(2.6)=391.84.$$
The average velocity is $\frac{\Delta s}{\Delta t}=\frac{s(2.6)-s(2)}{2.6-2}=\frac{391.84-436}{2.6-2}=-73.6.$ Note that the average velocities are negative indicating that the object is moving downward.
## 1.23 Instantaneous Rate of Change
The difference quotient $\frac{\Delta y}{\Delta x}=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}$ is the average rate of change of $$y$$ with respect to $$x$$ over the interval $$\left[x_1,x_2\right]$$ and can be interpreted as the slope of the secant line. Its limit as $$\Delta x\to 0$$ is the derivative at $$x=x_1$$ and is denoted by $$f'\left(x_1\right).$$
We interpret the limit of the average rate of change as the interval becomes smaller and smaller to be the instantaneous rate of change. Often, different branches of science have specific interpretations of the derivative.
As $$\Delta x\to 0$$ the average rate of change approaches the instantaneous rate for change ; that is, $\lim _{\Delta x\to 0} \frac{\Delta y}{\Delta x} = f'(x)$ and is also known as the derivative of $$f$$ at $$x.$$
Example 1.35 Let $$\displaystyle f(x)=\frac{x^2-x+5}{3-x}.$$
Find the instantaneous rate of change at $$x=2.$$
Solution. Since $\displaystyle f'(x)=\frac{-x^2+6 x+2}{(x-3)^2},$ the instantaneous rate for change of $$f$$ at $$x=2$$ is given by, $f'(2)=\frac{-2^2+6 (2)+2}{(2-3)^2}=10.$
## 1.24 What is a Tangent Line?
The tangent line problem is widely considered to be the instigating idea behind the derivative. Computing the slope of a tangent line was a problem that the French mathematician Pierre de Fermat developed. Picking up on these ideas were Isaac Newton and Gottfried Liebniz , who then developed differential calculus.
For a general curve it is not easy to define what is meant by a tangent line; for example a tangent line might mean a line touches the curve only once, but this does not work in all cases. Figure $$\ref{fig:tangentlines}$$ shows examples of tangent lines.
The important idea to remember is that a tangent line is a local concept, we say the tangent line at a point.
In our first example we will calculate a series of functions whose graphs are secant lines to the graph of a given function $$f$$ and use them to infer an equation of the tangent line at a point.
Example 1.36 Let $$f(x)=2x^3-2x+2$$. Find an equation of the line that passes through the points $$(1/2,5/4)$$ and $$(1,2)$$ and sketch the graph of both $$f$$ and the secant line. The equation of the secant line is $$y= \frac{3}{2} x+\frac{1}{2}$$.
Do the same for the points $$(1/2,5/4)$$ and $$(3/4,43/32)$$. We find an equation of the secant line is $$y=\frac{3}{8}x+\frac{17}{16}$$. We repeat this process several times and display the information in the following table. $\begin{array}{l|l|l|l} \small \Delta x & (x,f(x)) & (x+\Delta x,f(x+\Delta x)) & \text{Equation of secant line} \\ \hline 0.5 & (0.5,1.25) & (1,2) & y=1.5 x+0.5 \\ 0.25 & (0.5,1.25) & (0.75,1.34375) & y=0.375 x+1.0625 \\ 0.125 & (0.5,1.25) & (0.625,1.23828) & y=-0.09375 x+1.29688 \\ 0.0625 & (0.5,1.25) & (0.5625,1.23096) & y=-0.304688 x+1.40234 \\ 0.03125 & (0.5,1.25) & (0.53125,1.23737) & y=-0.404297 x+1.45215 \\ 0.015625 & (0.5,1.25) & (0.515625,1.24293) & y=-0.452637 x+1.47632 \end{array}$
From this table what would you say is the equation for the tangent line of the function $$f(x)=2x^3-2x+2$$ at $$(1/2,1/4)$$? Explain your conclusion. We infer the tangent line is $$y=-\frac{1}{2}x+\frac{1}{2}$$.
The difference quotient $\frac{\Delta y}{\Delta x}=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}$ is the average rate of change of $$y$$ with respect to $$x$$ over the interval $$\left[x_1,x_2\right]$$ $$_{}$$ and can be interpreted as the slope of the secant line. Its limit as $$\Delta x\to 0$$ is the derivative at $$x=x_1$$ and is denoted by $$f'\left(x_1\right).$$
We interpret the limit of the average rate of change as the interval becomes smaller and smaller to be the instantaneous rate of change. Often,different branches of science have specific interpretations of the derivative.
As $$\Delta x\to 0$$ the average rate of change approaches the instantaneous rate for change; that is, $\lim _{\Delta x\to 0} \frac{\Delta y}{\Delta x} = f'(x)$ and is also known as the derivative of $$f$$ at $$x.$$
## 1.25 Equation of Tangent Line
Theorem 1.6 If $$f'(a)$$ exists then an equation of the tangent line to the curve $$y=f(x)$$ at the point $$(a,f(a))$$ is $$y-f(a)=f'(a)(x-a).$$
Example 1.37 Find equations of the tangent lines to the curve $$\displaystyle y=\frac{x-1}{x+1}$$ that are parallel to the line $$x-2y=1$$.
Solution. The line $$x-2y=1$$ has slope $$m=1/2$$ and we use this with the derivative of $$y=(x-1)/(x+1)$$ to find the $$x$$. Since $$y'=2/(x+1)^2$$ we have $$1/2=(2/((x+1)^2))$$. Solving $$(x+1)^2=4$$ for $$x$$ we get $$x=1$$ and $$x=-3$$. Therefore, the points of tangency are at $$(1,0)$$ and $$(-3,2)$$.
The tangent lines are found by using $$y=1/2 x+b$$ where $$b=y-1/2 x$$ with $$(1,0)$$ and $$(-3,2)$$. We find $$b=-1/2$$ and $$b=7/2$$ respectively. Therefore, equations of the tangent lines are $$y=\frac{1}{2}x-\frac{1}{2}$$ and $$y=\frac{1}{2} x+\frac{7}{2}$$.
Example 1.38 How many tangent lines to the curve $$y=x/(x+1)$$ pass through the point $$(1,2)$$? At which points do these tangent lines touch the curve?
Solution. All tangent lines through $$(1,2)$$ have the form $$y-2=m (x-1)$$ where $$m=y'(x)=\frac{1}{(x+1)^2}.$$
Since we our looking for the intersection (point of tangency) we eliminate $$y$$ as follows: $y=\frac{x}{x+1}=\frac{1}{(x+1)^2}(x-1)+2.$ Solving for $$x$$ we obtain, $$x=-2\pm \sqrt{3}.$$ Thus there are two tangent lines and they are tangent at the point $\left(-2\pm \sqrt{3},\frac{-2\pm \sqrt{3}}{-2\pm \sqrt{3}+1}\right).$
Example 1.39 Find equations of both tangent lines through the point $$(2,-3)$$ that are tangent to the parabola $$y=x^2+x.$$
Solution. All tangent lines through $$(2,-3)$$ have the form $$y+3=m (x-2)$$ where $$m=y'(x)=2x+1.$$
Since we our looking for the intersection (point of tangency) we eliminate $$y$$ as follows: $y=x^2+x=(2x+1)(x-2)-3.$ Solving for $$x$$ we obtain, $$x=-1$$ and $$x=5.$$
Thus there are two tangent lines and they are tangent at the points $$(-1,0)$$ and $$(5,30).$$
The tangent lines are $$y=-x-1$$ and $$y=11 x-25.$$
## 1.26 Horizontal Tangent Lines
Theorem 1.7 If $$f'(a)=0$$ then the equation of the tangent line to the curve $$y=f(x)$$ at the point $$(a,0)$$ is $$y=f(a)$$ and $$f$$ is said to have a horizontal tangent line at $$x=a.$$
Example 1.40 For what values of $$x$$ does the graph of $$f(x)=2x^3-3x^2-6x+87$$ have a horizontal tangent?
Solution. To find the horizontal tangent lines we find where the derivative is 0. We compute, $$y'(x)=6 x^2-6 x-6.$$
So we need to solve $$6 x^2-6 x-6=0.$$
We find, $$6 x^2-6 x-6=0$$ and so $$x^2-x-1=0$$. And using the quadratic formula we have $$x=\frac{1}{2} \left(1\pm \sqrt{5}\right).$$
Thus, the values of $$x$$ where the tangent lines are horizontal are $$\frac{1}{2} \left(1\pm \sqrt{5}\right).$$
Example 1.41 Find the points on the curve $$y=x^3-x^2-x+1$$ where the tangent line is horizontal.
Solution. To find the horizontal tangent lines we find where the derivative is 0. We compute, $$y'(x)=3 x^2-2 x-1$$. So we need to solve $$3 x^2-2 x-1=0.$$
Using the quadratic formula we have $$x=-\frac{1}{3}$$ and $$x=1.$$
Thus, the values of $$x$$ where the tangents lines are horizontal are $$1$$ and $$-1/3.$$
## 1.27 Relative Rate of Change
Next we illustrate the importance of the relative rate of change, as compared to the difference between the absolute rate of change and the average rate of change. The absolute change is not the same as the average rate of change. Namely, the absolute change is just the differences in the values of $$f$$ at the boundary of the interval $$[x,\Delta x],$$ namely $$f(x+\Delta x)-f(x);$$ whereas the average rate of change is the absolute change divided by the size of the interval: $\frac{f(x+\Delta x)-f(x)}{\Delta x}.$ The average rate of change is sometimes more useful; for example, suppose you want to know how long it takes to make some money and not just the size of the money made (absolute change). Knowing the rate at which the money is being made, (the average rate of change over a given time interval) is often useful.
Example 1.42 Temperature readings $$T$$ (in degrees Celsius) were recorded every hour starting at midnight on a day in April. The time $$x$$ is measured in hours from midnight. $\begin{array}{c|ccccccccccccc} x (h) & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline T (C) & 6.5 & 6.1 & 5.6 & 4.9 & 4.2 & 4.0 & 4.0 & 4.8 & 6.1 & 8.3 & 10.0 & 12.1 & 14.3 \end{array}$ $\begin{array}{c|cccccccccccc} x (h) & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 \\ \hline T (C) & 16.0 & 17.3 & 18.2 & 18.8 & 17.6 & 16.0 & 14.1 & 11.5 & 10.2 & 9.0 & 7.9 & 7.0 \end{array}$ - Find the average rates of change of temperatures with respect to time from noon to 3:00 p.m., 2:00 p.m. and 1:00 p.m. - Estimate the instantaneous rate of change at noon.
Solution. The average rates of change are, respectively, $\frac{\Delta T}{\Delta x}=\frac{T(15)-T(12)}{3}=\frac{18.2-14.3}{3}=1.3 \, C/h$ $\frac{\Delta T}{\Delta x}=\frac{T(14)-T(12)}{2}=\frac{17.3-14.3}{2}=1.5 \, C/H$ $\frac{\Delta T}{\Delta x}=\frac{T(13)-T(12)}{1}=\frac{16.0-14.3}{1}=1.7 \, C/h$ We plot the given data and use them to sketch a smooth curve that approximates the graph of the temperature function. Then we draw that tangent line at the point $$P$$ where $$x=12$$ and after measuring the sides of the triangle we estimate that the slope of the tangent line is $\frac{18.3-8}{14-8.5}=\frac{10.3}{5.5}\approx 1.9$ and so the instantaneous rate of change of temperature with respect to time at noon is about $$1.9 C/h.$$
Sometimes we are not interested in the instantaneous rate of change and instead we may want a relative rate of change (percentage). For example suppose a student makes a 39 on a test, this would be a very good grade if the score is out of 40 points. However if the score was out of a total of 100 points then the grade is not so good.
Definition 1.4 Let $$y=f(x),$$ then the relative rate of change at $$x=x_0$$ is the ratio $\frac{f'\left(x_0\right)}{f\left(x_0\right)}.$
Example 1.43 Let $$f(x)=\sqrt{x}.$$
Find the relative rate of change at $$x=5$$ and $$x=75.$$
Solution. Since $$f'(x)=\frac{1}{2 \sqrt{x}}.$$
The relative rate of change of $$f$$ at $$x=5$$ is $\frac{f'(5)}{f(5)}=\frac{\frac{1}{2 \sqrt{5}}}{\sqrt{5}}=\frac{1}{10} \approx 0.1 \text{ or } 10\%.$ The relative rate of change of $$f$$ at $$x=75$$ is $\frac{f'(75)}{f(75)}=\frac{\frac{1}{2 \sqrt{75}}}{\sqrt{75}} =\frac{1}{150} \approx 0.00666667 \text{ or } 0.6\%.$
Often we are more interested in the relative rate of change of a quantity instead of the instantaneous rate of change. If instance, if you are earning $$25,000/\text{yr}$$ and receive a 5,000 raise, you would probably be very please. However, if you were making $$100,000/\text{yr}$$ you may not be as please since the relative change is not as much. With the $$100,000/\text{yr}$$ pay you only have a $$5,000/100,000=0.05=5\%$$ increase whereas with the $$25,000/\text{yr}$$ pay you have a better percentage increase of $$5,000/25,000=0.20=20\%.$$
Example 1.44 Let $$f(x)=2x^2-3x+5.$$
- Find the average rate of change from $$x=2$$ to $$x=4.$$
- Find the instantaneous rate of change at $$x=2.$$
- Find the relative rate of change of $$f$$ at $$x=2$$
Solution. The average rate of change of $$f$$ from $$x=2$$ to $$x=4$$ is given by, $\frac{f(x+\Delta x)-f(x)}{\Delta x} =\frac{\left(2(4)^2-3(4)+5\right)-\left(2(2)^2-3(2)+5\right)}{2} =9$ Since $$f'(x)=4x-3,$$ the instantaneous rate for change of $$f$$ at $$x=2$$ is given by, $$f'(2)=4(2)-3$$ $$=5.$$ The relative rate of change of $$f$$ at $$x=2$$ is $\frac{f'(2)}{f(2)}=\frac{5}{2(2)^2-3(2)+5} =\frac{5}{7} \approx 0.714286 \text{ or } 71.$
## 1.28 Exercises
Exercise 1.37 Find a function(s) with the following properties. (a) Find functions $$f$$ and $$g$$ such that $$f$$ is discontinuous at $$x=1$$ but $$f g$$ is continuous there. (b) Give an example of a function defined for all real numbers that is continuous at only one point.
Exercise 1.38
1. Find an equation of the secant line to the graph of $$y=x^2-2x$$ through the points $$(1,-1)$$ and $$(-1,3)$$ .
2. Find an equation of the tangent line to the graph of $$y=x^2-2x$$ at the point $$(1,-1)$$ .
Exercise 1.39
1. Find the average rate of change of $$f(x)=x^2+3x$$ over the interval $$[0,1]$$ .
2. Find the instantaneous rate of change of $$f(x)=x^2+3x$$ at $$x=0$$ .
Exercise 1.40 A ball is thrown straight down from the top of a 220-foot building with an initial velocity of 120 meters per second. What is the velocity after 5 seconds? After 10 seconds?
Exercise 1.41 To estimate the height of a building, a stone is dropped from the top of the building into a pool of water at ground level. How high is the building if the splash is seen $$6.8$$ seconds after the stone is dropped?
Exercise 1.42 Let $$s(t)=-t^3+3t^2-3t$$ be the position function of a body moving on a coordinate line, with $$s$$ in meters and $$t$$ in seconds. (a) Find the body’s displacement and average velocity for the time interval $$0\leq t\leq 3.$$ (b) Find the body’s speed and acceleration at the endpoints of the time interval $$0\leq t\leq 3.$$ (c) When, if ever, during the time interval $$0\leq t\leq 3$$ does the body change direction?
Exercise 1.43
1. Let $$s(t)=\frac{25}{t^2}-\frac{5}{t}$$ be the position function of a body moving on a coordinate line, with $$s$$ in meters and $$t$$ in seconds. (b) Find the body’s displacement and average velocity for the time interval $$1\leq t\leq 5.$$ (c) Find the body’s speed and acceleration at the endpoints of the time interval $$1\leq t\leq 5.$$ When, if ever, during the time interval $$1\leq t\leq 5$$ does the body change direction?
Exercise 1.44
1. Let $$s(t)=\frac{25}{t+5}$$ be the position function of a body moving on a coordinate line, with $$s$$ in meters and $$t$$ in seconds. (b) Find the body’s displacement and average velocity for the time interval $$-4\leq t\leq 0.$$ (c) Find the body’s speed and acceleration at the endpoints of the time interval $$-4\leq t\leq 0.$$ When, if ever, during the time interval $$-4\leq t\leq 0$$ does the body change direction?
Exercise 1.45 If an arrow is shot upward on the moon with a velocity of $$58 m/s,$$ its height in meters after $$t$$ seconds is given by $$h=58t-0.83t^2.$$ (a) Find the velocity of the arrow after $$1 s.$$ (b) Find the velocity of the arrow when $$t=a.$$ (c) When will the arrow hit the moon? (d) With what velocity will the arrow hit the moon?
Exercise 1.46 If a cylindrical tank holds $$100,000$$ gallons of water, which takes $$1 h$$ to drain from the bottom of the tank, then Torricelli’s Law given the volume $$V$$ of water remaining in the tank after $$t$$ minutes as $V(t)=100,000\left(1-\frac{t}{60}\right)^2, \qquad 0\leq t\leq 60.$
Find the rate at which the water is flowing out of the tank after 20 minutes.
Exercise 1.47 Verify that the average over the time interval $$\left[t_0-\Delta t,t_0+\Delta t\right]$$ is the same as the instantaneous velocity at $$t=t_0$$ for the position function $$s(t)=\frac{-1}{2}a t^2+c.$$
Exercise 1.48 Let $$s(t)=t^2-3t+2$$ be the position function of a body moving on a coordinate line, with $$s$$ in meters and $$t$$ in seconds. (a) Find the body’s displacement and average velocity for the time interval $$0\leq t\leq 2.$$ (b) Find the body’s speed and acceleration at the endpoints of the time interval $$0\leq t\leq 2.$$ (c) When, if ever, during the time interval $$0\leq t\leq 2$$ does the body change direction?
Exercise 1.49 At time $$t$$ the position of a body moving along the $$s$$ -axis is $$s=t^3-6t^2-9t m.$$ (a) Find the body’s acceleration each time the velocity is zero. (b) Find the body’s speed each time the acceleration is zero. (c) Find the total distance traveled by the body from $$t=0$$ to $$t=2.$$
Exercise 1.50 A rock is thrown vertically upward from the surface of the moon at a velocity of 24 m/se (about 86 km/h) reaches a height of $$s=24t-0.8t^2$$ meters in $$t$$ sec. (a) Find the rock’s velocity and acceleration at time $$t.$$ (b) How long does it take the rock to reach its highest point? (c) How high does the rock go? (d) How long does it take the rock to reach half its maximum height?
Exercise 1.51 Had Galileo dropped a cannonball from the Tower of Pisa, 179 ft above the ground, the ball’s height above the ground $$t$$ sec into the fall would have been $$s=179-16t^2.$$ (a) What would have been the ball’s velocity, speed, and acceleration at time $$t?$$ (b) About how long would it have taken the ball to hit the ground? (c) What would have been the balls velocity at the moment of impact? | 2023-02-07 23:56:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "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": 16, "x-ck12": 0, "texerror": 0, "math_score": 0.9983775019645691, "perplexity": 2014.9350248754931}, "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/1674764500664.85/warc/CC-MAIN-20230207233330-20230208023330-00240.warc.gz"} |
https://electronics.stackexchange.com/questions/365554/bitwise-operators-whats-the-difference-vs-smthg | # Bitwise operators: Whats the difference ^= vs &= ~(smthg)?
I'm coding for microcontrollers and I've come across two styles of bitwise operations that quite confuse me a lot because they are used interchangeably. First let's assume our REGISTER has 4 bits; bit 0,1,2,3. The default value of REGISTER on reset is 1100.
1.what the difference between:
REGESTER ^= (1<<3) | (1<<2);
&
REGESTER &= ~( (1<<3) | (1<<2) );
2.Is the first one correct? ( I'm not certain that it's correct )
3.If the first one's correct, then why use the second statement as the resulting value is same as the second one?
Final Result: 0000
• google DeMorgan’s theorem – jsotola Mar 30 '18 at 18:33
These are not interchangeable in general. You get the same results in this particular case, because the first one toggles bits 3 and 2, the second one clears them. Which one is correct depends on what you want to accomplish
• Well, I'm writing a HAL for STM32 and from code examples I've seen, they use &= ~x to set the GPIO Register MODE. So, it kind of confused me, why not use ^x instead of &= ~x – Killerbee89 Mar 30 '18 at 18:46
• @Killerbee89 Because they cannot assume that the registers will be in their reset state. – duskwuff Mar 30 '18 at 18:57
1. ^= x is XOR. It flips the state of bits.
&= ~x is AND NOT. It clears bits.
2. ^= x is only correct if toggling the bits is what you want to do every time this code runs. This might be correct if you wanted to flash an LED, for example, but it shouldn't be used if you want to ensure that the register ends up in a specific state.
• and REGESTER |= (1<<2) set bit 2 – G36 Mar 30 '18 at 18:41
In addition to the already posted answers and understanding what these operators do, it is equally important to understand how dangerous bitwise operators are.
You are using integer literals like 1, a literal which is of type (signed) int. Signed operators are dangerous to use together with bitwise operators.
Many operators like |, ^ and ~ don't care about the type of the operands, they work on the raw binary level. But the shift operators can invoke various nasty forms of undefined behavior bugs, should you shift a signed operand. For example, the code 1 << 31 invokes undefined behavior, since it shifts data into the sign bits. Similarly, right shifts of signed variables have implementation-defined behavior if the operand is negative.
To make things worse, there are implicit integer promotions. Code such as this also invokes undefined behavior:
unsigned char byte = 1;
byte = ~byte << 1;
Because it is the same thing as 0xFFFF FFFE << 1. Anything can happen here, including a program crash.
Similarly, code such as this will behave unexpectedly:
uint8_t mask = 0xF0;
uint8_t data = 0x01; | 2019-10-17 03:36:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.6801235675811768, "perplexity": 1932.064632569022}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986672548.33/warc/CC-MAIN-20191017022259-20191017045759-00418.warc.gz"} |
https://www.scienceopen.com/document?vid=8c463150-91d7-42a5-96a0-4a0bbc28a1f1 | 6
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# Emergence of curved light-cones in a class of inhomogeneous Luttinger liquids
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### Abstract
The light-cone spreading of entanglement and correlation is a fundamental and ubiquitous feature of homogeneous extended quantum systems. Here we point out that a class of inhomogenous Luttinger liquids (those with a uniform Luttinger parameter $$K$$) at low energy display the universal phenomenon of curved light cones: gapless excitations propagate along the geodesics of the metric $$ds^2=dx^2+v(x)^2 d\tau^2$$, with $$v(x)$$ being the calculable spatial dependent velocity induced by the inhomogeneity. We confirm our findings with explicit analytic and numerical calculations both in- and out-of-equilibrium for a Tonks-Girardeau gas in a harmonic potential and in lattice systems with artificially tuned hamiltonian density.
### Most cited references40
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### The Evaluation of the Collision Matrix
(1950)
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### Quantum thermalization through entanglement in an isolated many-body system
, , (2016)
The concept of entropy is fundamental to thermalization, yet appears at odds with basic principles in quantum mechanics. While statistical mechanics relies on the maximization of entropy for a system at thermal equilibrium, an isolated many-body system undergoing Schr\"odinger dynamics has zero entropy because, at any given time, it is described by a single quantum state. The underlying role of quantum mechanics in many-body physics is then seemingly antithetical to the success of statistical mechanics in a large variety of systems. Here we observe experimentally how this conflict is resolved: we perform microscopy on an evolving quantum state, and we see thermalization occur on a local scale, while we measure that the full quantum state remains pure. We directly measure entanglement entropy and observe how it assumes the role of the thermal entropy in thermalization. Although the full state has zero entropy, entanglement creates local entropy that validates the use of statistical physics for local observables. In combination with number-resolved, single-site imaging, we demonstrate how our measurements of a pure quantum state agree with the Eigenstate Thermalization Hypothesis and thermal ensembles in the presence of a near-volume law in the entanglement entropy.
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### Time-dependence of correlation functions following a quantum quench
(2006)
We show that the time-dependence of correlation functions in an extended quantum system in d dimensions, which is prepared in the ground state of some hamiltonian and then evolves without dissipation according to some other hamiltonian, may be extracted using methods of boundary critical phenomena in d+1 dimensions. For d=1 particularly powerful results are available using conformal field theory. These are checked against those available from solvable models. They may be explained in terms of a picture, valid more generally, whereby quasiparticles, entangled over regions of the order of the correlation length in the initial state, then propagate classically through the system.
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### Author and article information
###### Journal
2017-05-01
1705.00679 | 2019-10-24 04:28:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6818113327026367, "perplexity": 1341.6371576792385}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987841291.79/warc/CC-MAIN-20191024040131-20191024063631-00333.warc.gz"} |
http://nrich.maths.org/6593/note | ### Shades of Fermat's Last Theorem
The familiar Pythagorean 3-4-5 triple gives one solution to (x-1)^n + x^n = (x+1)^n so what about other solutions for x an integer and n= 2, 3, 4 or 5?
### Exhaustion
Find the positive integer solutions of the equation (1+1/a)(1+1/b)(1+1/c) = 2
### Code to Zero
Find all 3 digit numbers such that by adding the first digit, the square of the second and the cube of the third you get the original number, for example 1 + 3^2 + 5^3 = 135.
# Classical Means
##### Stage: 5 Short Challenge Level:
Why do this problem? Doing this problem you need to visualise the relationship between the formulae for the arithmetic, geometric and harmonic means in terms of the radius of a circle and the geometry of similar triangles and Pythagoras Theorem. This process involves a beautiful blend of algebra and geometry. It provides a good exercise for learners in mathematical reasoning and proof. The mathematical concepts are simple and the interplay of ideas to give the necessary proofs is good experience for learners.
Possible approach
Ask the learners to work out the radius of the semicircle in terms of $a$ and $b$. Then ask them to identify all the triangles and to discuss with a partner, and write down, everything they observe about the triangles. They may then be able to do the problem for themselves. If not the following questions should provide some assistance.
Key questions
Can you find the radius of the semicircle?
Can you see a right-angled triangle with $G$ as the length of one side and use this to find $G$ in terms of $a$ and $b$?
Can you use similar triangles to find a relationship between $A$, $G$ and $H$?
Can you use similar triangles to find a formula for $H$ in terms of $a$ and $b$?
Can you you use the geometry of triangles to compare the lengths of $A$, $G$ and $H$?
Can you see a right-angled triangle with $Q$ as the length of one side and use this to find $Q$ in terms of $a$ and $b$?
Possible extension
Try the problem Pythagorean Golden Means and read the article About Pythagorean Golden Means
Possible support
Try the problem Harmonic Triangle. | 2016-08-30 09:05:49 | {"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": 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": 0.521917462348938, "perplexity": 243.5639267513153}, "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-2016-36/segments/1471982974985.86/warc/CC-MAIN-20160823200934-00261-ip-10-153-172-175.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/67512/level-structures-and-moduli-of-abelian-varieties | # level structures and moduli of abelian varieties
Hello,
In the definition of level structure of level $n$ for an elliptic curve $A$, there are two versions:
1. an isomorphism of group schemes $(\mathbf Z/n\mathbf Z)^2 \to A[n]$.
2. an isomorhpism of group schemes $(\mathbf Z/n\mathbf Z)^2 \to A[n]$ after passing to some etale cover of the base.
There is a natural map from the moduli problem of 1 to the moduli problem of 2. This map appears to be etale.
Question: for more general moduli problems (when the level is given by subgroups $H\subset GL(2,n)$), normally 2. is used. What is the reason for this?
Thanks
-
I think you're a bit muddled. A full level $n$ structure is 1. I am not sure what you mean by 2 -- is the cover part of the data for example? -- but my guess is that there's been a misunderstanding here. The reason that one has to be a bit more careful when doing the general $H$ case is because you want to allow the situation that the curve have a level $H$ structure but have no full level $n$ structures at all -- e.g. if $H=\Gamma_0(n)$ then you want an $H$-structure to mean a subgroup of order $n$, but a curve can have such a subgroup without having a full level $n$ structure. – Kevin Buzzard Jun 11 '11 at 16:17
So one way of doing a level $H$ structure (assuming $n$ invertible on the base) is to look at the isom scheme and then quotient out, and then observe that even if the isom scheme doesn't have points, the quotient still might. If you want to make this "explicit" you might start allowing base changes so you can see the points moving around under $H$. – Kevin Buzzard Jun 11 '11 at 16:19
[s/subgroup/cyclic subgroup/ in the above] – Kevin Buzzard Jun 11 '11 at 16:20 | 2013-12-20 17:26:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8976538181304932, "perplexity": 284.9467661199807}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1387345772826/warc/CC-MAIN-20131218054932-00041-ip-10-33-133-15.ec2.internal.warc.gz"} |
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## Alpern genericity theorem
Frédéric Le Roux has written a very lucid exposition of the Alpern genericity theorem. This theorem states that the same ergodic properties are generic in the set $Auto(T^d)$ of volume-preserving measurable transformations and in the set $Homeo(T^d)$ of volume-preserving homeomorphisms.
More precisely, endow $Auto(T^d)$ with the weak topology, i.e., the coarsest generated by $\mu\mapsto\mu(A)$, for all measurable subsets $A$ and equip $Homeo(T^d)$ with the uniform distance.
The theorem considers any $P\subset Auto(T^d)$ invariant under volume-preserving measurable isomorphisms. It states that $P$ is a dense $G_\delta$ subset of $Auto(T^d)$ if and only if $P\cap Homeo(T^d)$ is itself a dense $G_\delta$ subset of $Homeo(T^d)$.
## Orders of accumulation of entropy structures
Kevin McGoff gave a talk at Orsay on his work on the theory of entropy srtuctures and symbolic extensions. This theory was founded by Mike Boyle and Tomasz Downarowicz, among others.
This theory relates the continuity properties of the measure-theoretic entropy function with the existence of symbolic topological extensions with measure-theoretic entropies as close as possible to those of the initial system. This theory ascribes to any topological dynamical system an order of accumulation. M. Boyle and T. Downarowicz have shown that this is a countable ordinal.
David Burguet and K. McGoff have shown that any countable ordinal can be achieved by some topological dynamics. The proof relies on a realization theorem of T. Downarowicz and S. Serafin.
K. McGoff explained how he was able, by a more precise and direct construction to achieve the same on any prescribed compact manifold. The transformation can be chosen to be homeomorphic if the dimension is 2 or more.
## C^2 surface diffeomorphisms always have a symbolic extension
Most of topological dynamics studies systems of the form $T:X\to X$ where $T$ is a continuous self-map and $X$ is a compact metric space. One approach is to “reduce” such systems to symbolic dynamical system, i.e., $\sigma:S\to X$ where $S$ is a closed subset of $\{1,\dots,d\}^{\mathbb Z}$ and $\sigma((x_n)_{n\in\mathbb Z})=(x_{n+1})_{n\in\mathbb Z}$ such that $\sigma(S)=S$.
J. Auslander asked about the obstructions for a topological system $T:X\to X$ to have a symbolic extension, i.e., a symbolic system $\sigma:S\to S$ and a continuous surjection $\pi:S\to X$ commuting with the dynamics: $\pi\circ\sigma =T\circ\pi$. There is an obvious one: a symbolic system (and therefore its topological factors) has finite topological entropy. Is there any other?
M. Boyle showed that this was indeed the case. With D. Fiebig and U. Fiebig, he showed that asymptotically h-expansive systems (including $C^\infty$ self-maps of compact manifolds by a result of mine based on Yomdin’s theory) always have a “nice” symbolic extension. T. Downarowicz and S. Newhouse showed that generic $C^1$ map have no symbolic extension whatever, leaving open the question of diffeomorphisms with finite smoothness.
T. Downarowicz and A. Maas showed that $C^r$ interval maps also always have symbolic extensions for $1.
David BURGUET has finally proved the same for arbitrary $C^2$ surface diffeomorphisms, see his preprint here.
Behind these works there is a rich and beautiful topological/ergodic/functional-analytical theory of entropy (called the entropy structure by T. Downarowicz) which does yet have the audience it deserves, in my opinion.
## Sylvain CROVISIER’s HDR
On November 25th, 2009, Sylvain CROVISIER defended his habilitation à diriger des recherches titled Perturbation de la dynamique de difféomorphismes en petite régularité. He first explained basic perturbation techniques:
• the Anosov-Katok procedure: you use more and more distorted conjugacies such that the limiting dynamics has new properties;
• the closing lemma of Pugh and the subsequent connecting lemmas of Hayashi and Bonatti-Crovisier: you try to glue pieces of orbits while avoiding intermediate visits to the support of the pertubation. This may force you to drop pieces of the orbits. Crovisier nevertheless managed to prove a generalized shadowing lemma constructing orbits guaranteed to visit some neighborhoods.
He then explained how the above was put to work. Together with François Béguin and Frédéric Le Roux, he used the first tool to realize almost arbitrary measurable dynamics as minimal, uniquely ergodic homeomorphisms on arbitray compact manifolds.
With Christian Bonatti, Enrique Pujals and other co-workers, he used the second set of tools (together with constructions of dominated splittings by Wen, Liao-Wen and his own central models) to get deep new results on the dynamics of C^1 generic diffeomorphisms.
Some of the most important ones state that, after a small C^1 perturbation, any diffeomorphism either has a strong global structure (i.e., a phenomenon in the sense of Pujals) or presents a simple obstruction (a mechanism, which one would like to be robust). Namely, up to C^1 perturbations, any diffeomorphism of a compact manifold is:
• Morse-Smale unless it has a transverse homoclinic intersection (this provides a description of an open and dense subset of the C^1 diffeomorphisms called the weak Palis conjecture);
• partially hyperbolic with a central bundle which is one-dimensional or a sum of two one-dimensional sub-bundles, unless it has a heterodimensional cycle or a homoclinic tangency;
• essentially hyperbolic (hyperbolic from the point of view of its attractors and repellers), unless it has a heterodimensional cycle or a homoclinic tangency.
• It is true that Mañé already characterized the non-hyperbolic diffeomorphisms as those having periodic points that can be made to bifurcate but the goal here is to get robust obstructions;
• It is not clear that dynamics with infinitely many chain recurrent classes can exist robustly: there is no known mechanism for that;
• It seems very difficult to get beyond C^1 with anything like the current techniques – even C^1+1/log seems out of reach;
• It is a reasonable to question to try and develop more precise description of the dynamics especially for the situations that hold on C^1 open sets and therefore occur on C^2 open sets, a usual requisite of ergodic theory techniques;
• The techniques have yielded results very analoguous to Zeeman Tolerance Stability conjecture. However such kinds of stability seem to have more philosophical appeal than a true rôle in the mathematical theory, as opposed to structural stability.
• It is not clear that the above decomposition of the space of C^1 diffeomorphisms extend to higher smoothness (even disregarding the enormous technical difficulties pertaining to the closing lemmas). For instance Bonatti and Diaz hope to show that any diffeomorphism can be C^1 approximated by a hyperbolic one or by one with a homoclinic tangency but this is not the case in the C^2 topology, because of Newhouse phenomenon.
• The techniques seem insufficient to study prevalent or Kolmogorov-typical dynamics. | 2017-05-27 21:20:59 | {"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": 0, "img_math": 30, "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": 0.8361559510231018, "perplexity": 1038.9899283542893}, "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-2017-22/segments/1495463609061.61/warc/CC-MAIN-20170527210417-20170527230417-00403.warc.gz"} |
http://castleage.wikia.com/wiki/Territories | Territories
1,055pages on
this wiki
Territories are a new system for purchasing land, which replaces the old Land interface for new players. In this feature, players can invade other players' armies in order to expand their plots, and purchase buildings in order to increase their territory level and their daily income.
At the moment, territories appear to be open to new players who have started since March 2014, coinciding with the release of version 1.300 of the Castle Age HD app. These players will have Territories as their gold system instead of Lands, while older players do not have territories at this time.
Overview
Territories provide a daily collection of gold income for players. Players start off with 100 plots to build on, and can increase their available plots by up to 20 by using 3 Stamina (one invasion) to invade other players' armies, and gain Territory XP for a territory. Upon reaching certain thresholds of XP, players will unlock plots, where they can spend gold to start construction of a building on the territory, with the end result being gain daily income.
Players can have as many buildings under construction at a time as they can afford and support.
Players with territories must manually collect any income every day, otherwise it is lost. The collection timer is not 24 hours from the last collection, but rather a time period that ends at Midnight US Pacific time (the same time that Conquest resources and Hero Crystals reset). Any buildings under construction at this time will be completed, and will have their income as part of the next collection total.
Types of Territory
Similar to lands, there are 7 types of Territory at this time, corresponding to the former first 7 lands. Each land has a daily income equal to 24 times the hourly income of the corresponding land, meaning the total costs and benefits are the same. Values and prices of territories have been changed, and are still in flux.
When invading other players, each territory has a specific amount of stamina needed per invasion. Upon winning, the player receives a set amount of Territory XP for that land, 6 player XP and 7 Hero XP per invasion, along with Gold and losing some Health like in any normal PvP battles.
Image Type Initial Level Requirement Stamina used per invade Territory XP per invade Income per unit
per day
Initial cost Incremental cost
per purchase
Small Tower 0 3 5 $3,600$4,500 +$450 Temple 5 3 5$12,000 $18,000 +$1,800
Marketplace 10 5 10 $52,800$90,000 +$9,000 Mountain Keep 15 5 10$180,000 $300,000 +$30,000
Castle 20 10 20 $1,800,000$2,250,000 +$225,000 Lake Fortress 25 10 20$6,000,000 $11,000,000 +$1,100,000
Sky Sanctuary 30 15 30 $8,400,000$22,000,000 +\$2,200,000
Territory Level/curve
When leveling up a territory, the total experience needed for a particular level follows a formula, dependent on the amount of energy needed per invasion. This formula is:
$\text{XP Needed} = \frac{XP per invade \cdot 5 \cdot level^2}{16}$
The total experience needed for each level is included in the table below. As of now, the maximum level for each territory is level 20.
Level Experience needed
Small Tower
Temple
Marketplace
Mountain Keep
Castle
Lake Fortress
Sky Sanctuary
124710
27132538
315295785
42550100150
54079157235
657113225338
777154307460
8100200400600
9127254507760
10157313625938
111903797571135
122254509001350
1326552910571585
1430761312251838
1535270414072110
1640080016002400
1745290418072710
18507101320253038
19565112922573385
20625125025003750
Territory Level and Player experience boost
When a player constructs buildings on their territories, their overall Territory level will increase. They will receive Territory XP equal to the experience needed to construct each of their buildings up to its maximum level. For example, if a player has 3 Small Towers, 2 Temples, and 1 Marketplace finished, they will have (15 + 7 + 4) = 26 total experience. The overall territory level formula is (rounded up):
$\text{XP Needed} = 1.25 \cdot level^2$
The current maximum territory level is 100, which can be achieved if all seven lands are at level 20.
If a player's total territory level exceeds their player level, they will receive a bonus of 5% per level exceeded to all sources of experience. This is not displayed in the experience text in most cases, but is taken into account in the actual experience granted. The total boost is capped at 35%, even if the player's territory level is more than 7 levels above their player level. This experience boost will only take effect until level 100, at which point it will disappear, even if a player's territory level is still greater than their player level. | 2017-03-25 07:49:43 | {"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": 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": 0.3824622631072998, "perplexity": 2729.3402925366527}, "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-2017-13/segments/1490218188891.62/warc/CC-MAIN-20170322212948-00475-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://www.biostars.org/p/340052/ | Kozak sequence strength calculation
0
3
Entering edit mode
4.0 years ago
Hello everyone,
I want to assess the probability of a translation initiation site to be actually translated (Interesting in human genome). Based on literature I know that one of the most important requirements is a good Kozak sequence. Is anyone aware of a bioinformatics tool that could calculate Kozak concensus strength based on a given sequence?
Alex
sequencing Kozak translation ribosome binding • 3.1k views
4
Entering edit mode
I won't post this as an answer since it doesn't answer the question as posed, but is perhaps some food for thought:
To the best of my knowledge, recent studies suggest that pretty much the whole chromosome is transcribed to some degree at any given time, but the levels are obviously modulated. If Shine-Dalgarno sequences in bacteria (closer to my area of expertise) are any indication, there is a relationship between sequence 'identity' and transcriptional/translational activity - however it's very complicated, as the sequence of the regulatory sequences is not the be-all and end-all.
There may well be existing literature which has benchmarked the transcriptional activity of different sequences, but the problem is that the data will be essentially incomparable between different experiments due to batch effects.
A quick google for kozak sequence effects on transcription turns up articles such as: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5563945/
and
https://journals.plos.org/plosone/article?id=10.1371/journal.pone.0108475
I'm certainly not aware of any tools that do this already. In principle I can see that it would work as a predicable phenomenon. Something to the effect of creating a standard curve of sequence distance from the canonical sequence (perhaps), versus transcriptional activity, but it would need to be based on a carefully curated reference set of data. I'm sort of thinking along the lines of a transcription initiation equivalent of the "N-end rule", or a similar kind of benchmarking to the study which showed translational response to different start codons (spoiler alert, ATG doesn't mean sh*t!)
The only thing that springs to mind otherwise, would be the Softberry site, which has several tools for promotor sequence analysis etc (though this usually only means predicting their locations in sequences etc), but maybe theres something there you can make work:
0
Entering edit mode
Hello Mr. Healey,
The two articles you referred me to are indeed some good extra knowledge and I have already gone through both.
Thanks a lot for sharing your thoughts regarding this issue. I will also have a look on the site you mentioned because I was not aware of it ( and whatever has to do with bioinformatic tools in general I am a newbie in this field)
Thanks a lot !
0
Entering edit mode
Hello,
A little late, but this website may be what you are looking for:
https://www.tispredictor.com/tis
It calculates Kozak sequence strength (referenced as Kozak Similarity Score) for each predicted initiation codon in a given sequence.
Here is the associated paper:
https://www.biorxiv.org/content/10.1101/2021.08.17.456657v1
I would recommend reading “Kozak similarity score algorithm” and “KSS as a reference for likelihood of translation initiation” in the Results section to understand the scoring metric.
The paper is still a preprint, so I would also keep that in mind. | 2022-10-03 13:55:13 | {"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": 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": 0.4825679659843445, "perplexity": 1344.120709531176}, "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/1664030337421.33/warc/CC-MAIN-20221003133425-20221003163425-00633.warc.gz"} |
http://mathematica.stackexchange.com/questions/10801/monospaced-fonts-are-not-monospaced/14014 | # Monospaced fonts are not monospaced
I'm using Linux-x86-64 as operating system. Can someone explain this behaviour and maybe give a hint how to resolve it? Here are two lines of exact the same number of letters of a monospaced font:
Style["ZZZZZZZZZZ\n__________", 20, FontFamily -> "Courier"]
As output I get (here with 30 Z and 30 _)
But what I expect from a monospaced font like Courier is to have the same size for every letter (and why is the line dashed anyway?). @Oleksandr was so kind to give me mental support and tried the things on his Windows machine. There the output seems ok:
## More detailed question
What I really want is an image of arbitrary resolution of every letter. I thought when I use a combination of Style and Rasterize I can get equally sized images of them in exact the same form as they appear in text.
But there is (at least here) a clear difference between an assembled image from rasterized letters and a rasterized version of the string
ImageAssemble[
Rasterize[Style[#, 30, FontFamily -> "Courier"], "Image"] & /@
CharacterRange["A", "Z"]]
Rasterize[
Style[StringJoin[CharacterRange["A", "Z"]], 30,
FontFamily -> "Courier"], "Image"]
What I noticed is that the height of the letters is always the same, only the width varies. Looking at how many letters have which width gives here
Sort@Tally@First@
Transpose[
ImageDimensions[
Rasterize[Style[#, 30, FontFamily -> "Courier"], "Image"]] & /@
CharacterRange["A", "z"]]
(*
Out[138]= {{18, 25}, {19, 16}, {20, 8}, {21, 8}, {24, 1}}
*)
-
I see the same issue, but to a much lesser degree (also on Linux-x86-64). The first example is almost not noticeable to the eye and for the last I get: {{18, 53}, {19, 3}, {20, 1}, {21, 1}} which is somewhat better. I have no glue where this comes from, though. – user21 Sep 19 '12 at 4:52
Well I don't have glue, but no clue either ;-) – user21 Sep 19 '12 at 5:39
@ruebenko maybe glue is really necessary here to put the letters together. – halirutan Sep 19 '12 at 16:32
@halirutan I get a similar output to yours, however changing the second line Style["ZZZZZZZZZZ\nAAAAAAAAAA", 20, FontFamily -> "Courier"] or Style["ZZZZZZZZZZ\n0123456789", 20, FontFamily -> "Courier"] seem to be good. Does it make sense ? – b.gatessucks Sep 19 '12 at 18:55
@b.gatessucks Yes, this makes sense in this scenario. Usually, in Courier all letters should have the same width. In Linux, some letters are wider than others. While the difference between "Z" and "_" is very large, the width of "A" and "Z" seem to be equal. Therefore, your examples look ok. – halirutan Sep 19 '12 at 21:44
While this is not an answer, I did not want to add it as a comment. The issue is not specific to Linux, happens under Windows as well (Win7 in particular). Also, it is not specific to font: any monospaced font seems to produce overhangs to the default frame of $n$ characters. While rasterized characters have different image dimensions, as pointed out by halirutan, this is not direclty related to the problem, as most of the operator characters comply with the expected dimensions (though they will cause a lot of problems) while those characters that have larger-than-unit dimensions (like A, W, M) might fit in nicely without an overhang. So I think that the different image dimensions and the kerning problem of e.g. "_" are different issues.
## 1. Rasterized characters
For the image assembly, I would suggest using a uniform image dimension applied via ImageCrop:
(* get standard character dimensions - find a suitable char, like "I" *)
unitDim = ImageDimensions@
Rasterize[Style["I", 30, Bold, FontFamily -> "Courier"], "Image"]
{18, 34}
"non cropped individual images"
ImageAssemble[
Rasterize[Style[#, 30, Bold, FontFamily -> "Courier"], "Image"] & /@
CharacterRange["A", "Z"]]
"cropped individual images"
ImageAssemble[
ImageCrop[
Rasterize[Style[#, 30, Bold, FontFamily -> "Courier"], "Image"],
unitDim] & /@ CharacterRange["A", "Z"]]
"rasterize as a whole"
Rasterize[
Style[StringJoin[CharacterRange["A", "Z"]], 30, Bold,
FontFamily -> "Courier"], "Image"]
While cropping individual images is a possible workaround, sadly I have no idea why rasterized characters end up being larger than the monospaced frame. The following test indicates, that this effect depends on FontSize and Magnification but not on ImageResolution:
chars = CharacterRange["A", "Z"]; charPos = Thread[chars -> Range@Length@chars];
resRange = {10, 100, 1000}; sizeRange = {10, 15, 20, 30, 50, 100}; magRange = {.5, 1, 2, 10};
data = Table[
style = {FontSize -> size, Magnification -> mag, FontWeight -> Bold, FontFamily -> "Courier"};
img = Rasterize[Style[#, Sequence @@ style], "Image", ImageResolution -> res] & /@ chars;
imgPos = Thread[img -> Range@Length@chars]; unitDim = ImageDimensions@img[["I" /. charPos]]; imgDim = ImageDimensions /@ img;
Cases[img, x_?((d = ImageDimensions@#) =!= unitDim &) :> chars[[x /. imgPos]]],
{res, resRange}, {size, sizeRange}, {mag, magRange}];
Column[MapThread[Labeled[TableForm[#1, TableHeadings -> {sizeRange, magRange}, TableDepth -> 2], "ImageResolution \[Rule] " <> ToString@#2, Top] &, {data, resRange}], Spacings -> 2]
Figure lists those rasterized characters that do not match the ImageDimension of the standard monospaced dimension (taken from "I"). Horizontal dimension: Magnification; Vertical dimension: FontSize. I would consider this a bug.
## 2. Monospaced characters are not monospaced
Further investigating the first issue of monospaced characters not being monospaced when printed:
n = 30;
font = "Courier"; (* alternatively: "Lucida Console" *)
text = StringJoin[
RandomChoice[{"=", "*", ":"}, n], "\n",
RandomChoice[{"M"}, n], "\n",
RandomChoice[{"@"}, n], "\n",
RandomChoice[{":"}, n], "\n",
RandomChoice[{"M", "A"}, n], "\n",
RandomChoice[{"_"}, n], "\n",
RandomChoice[{"@", "M"}, n], "\n",
RandomChoice[{":", "M"}, n], "\n",
RandomChoice[{":", "A"}, n], "\n",
RandomChoice[{":", "J"}, n]
];
Grid[{{
Style[StringJoin["Normal, 100% magnification, font size 10\n",
text], FontFamily -> font, FontSize -> 10],
Style[StringJoin["Bold, 100% magnification, font size 10\n", text],
FontFamily -> font, FontSize -> 10, Bold]
}, {
Style[StringJoin["Normal, 120% magnification, font size 10\n",
text], FontFamily -> font, FontSize -> 10, Magnification -> 1.2],
Style[StringJoin["Normal, 120% magnification, font size 12\n",
text], FontFamily -> font, FontSize -> 12, Magnification -> 1.2]
}}, Alignment -> {Left, Top}, Frame -> All]
Note the followings:
• Non-bold monospaced font with 100% magnification and 10 points height produces nicely fit lines - any divergence causes overhangs.
• Operators with certain letters (while independently are ok) together form an overhang. This seems to be true for all the operators I've tested.
• "_" is a renegate character as it does not even fit by itself into the frame, not even when not Bold.
• Overhang depends on magnification, font family and font size.
Again, I have no idea what causes these problems, but is definitely operator-character-related. Here, the problem is that no setup that is found on one machine is guaranteed to work on another, as a lot of these parameters depend on the OS.
-
Although we could not resolve the problem properly, you did a great job digging further into this. I'll accept this answer. It's a pity, I was the only one who upvoted your work. – halirutan Nov 19 '12 at 1:05
@halirutan Thanks for the accept, let's hope this will be cured in v9! – István Zachar Nov 19 '12 at 8:31
Just to update for v9: no, it is not cured yet, symptoms are the same for both issues. – István Zachar Oct 14 '13 at 10:03 | 2015-07-29 02:51:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.4641438126564026, "perplexity": 5912.236075606017}, "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-2015-32/segments/1438042985647.51/warc/CC-MAIN-20150728002305-00276-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://zxi.mytechroad.com/blog/tag/sum/page/5/ | # Posts tagged as “sum”
https://leetcode.com/problems/range-sum-query-2d-immutable/description/
Problem:
Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).
The above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (row2, col2) = (4, 3), which contains sum = 8.
Example:
Note:
1. You may assume that the matrix does not change.
2. There are many calls to sumRegion function.
3. You may assume that row1 ≤ row2 and col1 ≤ col2.
Idea:
Dynamic programming
Time complexity:
O(n^2)
sumRegion: O(1)
Solution:
Problem:
Given a 2D binary matrix filled with 0’s and 1’s, find the largest square containing only 1’s and return its area.
For example, given the following matrix:
1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0
Return 4.
Idea:
Dynamic programming
Solution 0: O(n^5)
Solution 1: O(n^3)
Solution 2: O(n^2)
Solution 1:
Solution 2:
https://leetcode.com/problems/map-sum-pairs/description/
Problem:
Implement a MapSum class with insert, and sum methods.
For the method insert, you’ll be given a pair of (string, integer). The string represents the key and the integer represents the value. If the key already existed, then the original key-value pair will be overridden to the new one.
For the method sum, you’ll be given a string representing the prefix, and you need to return the sum of all the pairs’ value whose key starts with the prefix.
Example 1:
Idea:
Prefix tree
Solution 1
Solution 2:
with std::unique_ptr
Problem:
Given an array containing n distinct numbers taken from 0, 1, 2, …, n, find the one that is missing from the array.
For example,
Given nums = [0, 1, 3] return 2.
Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?
Idea:
sum / xor
Solution: | 2021-04-15 04:44:20 | {"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": 0.4949771761894226, "perplexity": 1321.0972401703718}, "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-2021-17/segments/1618038083007.51/warc/CC-MAIN-20210415035637-20210415065637-00251.warc.gz"} |
https://dimag.ibs.re.kr/event/2020-08-18/ | • This event has passed.
# Tuan Tran, Anti-concentration phenomena
## Tuesday, August 18, 2020 @ 4:30 PM - 5:30 PM KST
Room B232, IBS (기초과학연구원)
### Speaker
Tuan Tran
IBS Discrete Mathematics Group
https://tuaentran.wixsite.com/homepage
Let $X$ be a real random variable; a typical anti-concentration inequality asserts that (under certain assumptions) if an interval $I$ has small length, then $\mathbb{P}(X\in I)$ is small, regardless the location of $I$. Inequalities of this type have found powerful applications in many branches of mathematics. In this talk we will discuss several recent applications of anti-concentration inequalities in extremal combinatorics, as well as random matrix theory. The talk is partially based on joint work with Matthew Kwan and Benny Sudakov.
## Details
Date:
Tuesday, August 18, 2020
Time:
4:30 PM - 5:30 PM KST
Event Category:
Event Tags:
Room B232
IBS (기초과학연구원)
## Organizer
Sang-il Oum (엄상일)
View Organizer Website
기초과학연구원 수리및계산과학연구단 이산수학그룹
대전 유성구 엑스포로 55 (우) 34126
IBS Discrete Mathematics Group (DIMAG)
Institute for Basic Science (IBS)
55 Expo-ro Yuseong-gu Daejeon 34126 South Korea
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https://tiergartenapo-app.de/logon-script-to-map-network-drive-server-2008-r2.html | logon script to map network drive server 2008 r2. The complete process has two sections. First navigate to one of the download links below, then: Scroll down to the "MICROSOFT SQL SERVER CONNECTIVITY FEATURE PACK COMPONENTS" section (it's far down) Under the "Microsoft SQL Server …. Simply select Advisor from the Application menu and follow the straightforward prompts. With a little help from my colleague Oddvar Håland Moe I have created a “one-liner” that will delete all the network printers on your client. But darn it, this person really needed to connect multiple times with different user ids. Expand the name of your domain and then expand the Group Policy Objects folder. I want to add another mapped drive as a test. View Details Intel® Network Adapter Driver for PCIe* Intel® Gigabit Ethernet Network Connections under Linux* Intel® Network Adapter Driver for Windows Server 2008 R2…. You could use this from a login script to pre-populate some profiles for end option. In the file explorer, click on the “Computer” option on the top navigation bar and then select “Add a Network Location” option. Once you login in the system run following statement for your username and it should fix your problem. Start Installer To connect this machine to a client computer and use …. As the title suggests, this does not happen for local accounts or for console connections, ONLY domain …. FTK Imager only sees internal drive serial number when connected to a Tableau device on Windows 10. the date and time when a user logged on to the Windows network in a hassle-free manner. You can add additional mapped drives by adding more If-Then statements. Computer Configuration –> Administrative Templates –> Windows Components –> Remote Desktop Services –> Remote Desktop Session Host -> Printer Redirection > Do not allow client printer redirection. If they don't currently have mapped network drives, then create a Group policy that will map whatever drives you want them to have access to. All seems ok but the drive is still being mapped …. Use one drive and sync the document library. Fill in the Folder field with the IP address or hostname of the server …. Otherwise have you considered using Group Policy Preferences instead of a script at logon. Powershell Library for Hyper-V Server 2008 R2…. On-premises licenses SQL Server 2008 and 2008 R2 Extended Security Updates will end on July 12, 2022. Right-click the new GPO and select Edit. Click on the SQL Server Services node and in the right panel right-click SQL Server (EXPRESS) and select restart to restart the service. We are in the process of moving our SIMS server from server A to server B. Many times the home folder (H) is not mapped when the user logs in. I need to be able to install Office 2016 Pro Plus on a Windows Server 2008 R2 Terminal Server, well, 2 actually. 2) Right-Click on Default Domain Policy. Following is an overview of a typical installation of the network operating system (NOS) Windows Server 2008 and configuration of your server. The NFS client is now an installable feature in Windows 8 and Windows Server 2012 and higher. I hope this gives you an overview about how you can do in-place upgrades from Windows Server 2008 R2 to Windows Server …. Open Group Policy Management Console (GPMC) and create a new policy. What Is a Server in Computer Networking?. Limit the size of overall profile cache. Now you could map network drives, set printers, or run just about any command Is a batch file running on Server 2008 R2, the commands: @REM == Now test the result. scr as the basis for all screensavers which is a blank screen. 1x Authentication Not Supported with Microsoft Server 2008 R2-based RADIUS Server. have same contents) We recently made changes to our File Servers. Check your TCP/IP Protocols in right side pane. We have Windows 2008 R2 standard Server and each user has a home directory but when users logon to the server with Windows 7 machine, their home drive and name does not appear. I am looking for the location for the windows logon script file. Choose Launch instances and create a …. Other computers on the network …. This has worked well since all our GPO's will be processed as we would expect before logon. In this example the backup file for the Northwind database will be placed on the "F$" share of server "server25". Customer will need to upgrade their Windows Server 2008 and Windows Server 2008 R2 to a newer version of Windows Server or migrate these servers to Microsoft Azure and get three years of extended security support for free, which will give you more time to upgrade your servers, redeploy your. วิธีการ แก้ปัญหาการลืม รหัสผ่าน Windows Server 2008 R2 11/05/2012. A Windows server 2008 R2 installation CD/DVD could save your life in an emergency. I'll cover the following topics …. I tried re-mapping the drive with a new name and it didn't work any different. Let's use a driver and his vehicle as an analogy. 2 New Remote Session Host features on server 2008. So, if my server was named DCServer1, and my domain was intelliadmin. Starting with Windows Server 2008…. Whenever we had Explorer open with one of the mapped drives and the policy refreshed (manual or automatic) then Explorer would crash. We have a bunch of drives mapped to various server - NONE of the rest are to a Windows 2003 server. Our fast, free, self-hosted Artificial Intelligence Server …. If the Sysvol share is missing, joining a Windows Server 2012 or 2012 R2 DC fails. Mar 06, 2020 · This fix is specific to users with Windows Server 2016 black screen after login to RDP sessions. In the opened up window click the "Connect to a Web site that you can use to store your documents and pictures" link. It was designed to run in scenario's where using the Onedrive for Business client is not feasible, such as legacy. Windows Server 2012 R2 Hardening Checklist. You disconnect the mapped drive. Applies to : Windows Server 2008 R2. The above command would map the share to the first available drive starting from Z: We can also specify the drive …. Right click on “Network” located on Windows Explorer and choose “Map network drive…”. cmd script that prepares the minimum WinPE operating system. Your WebDAV server must be using port 80, the default port. “Net use” is a command line method of mapping network drives to your local computer. Ensure that the drive is reconnected on the login. 2830140 Cumulative update package 6 (CU6) for SQL Server 2008 R2 Service Pack 2 (updated) 2013-06-13; 10. NOTE: The printers and network drives still install so i know the script is running. Site24x7 offers both free & paid monitoring services for your entire IT environment. User-1126563351 posted I am running Windows Web Server 2008 with all updates. The steps to remove a network drive from Windows 11 using File Explorer are similar to the ones in Windows 10. Copy two files into this folder: your logon script (for example, Logon. If a mapped network drive wasn't reconnected, any program that tries to access resources on the mapped network drive fails. Mapping network drive is convenient way to access shared folder on a network. The setup will complete and automatically reboot. The staff and students map to two different shares on the same (virtual) server …. If you encounter errors during backup initialization of Microsoft Server 2008 or 2008 R2, or Small Business Server (SBS) 2011, you may have misconfigured or disabled Volume Shadow Copy Services (VSS) drivers. The network drives map to shares on servers running the Server 2012 R2 operating system and those drives do not reconnect, they show disconnected after logon. The configured drive letter is the G drive. 1\DRIVENAME PASSWORD You can find out more about this command and its arguments via: net use ? Edit 2: As Tomalak mentioned in comments, you can later un-map it via net use Z: \delete Share. Step 1: Open the group policy management console. Windows Server 2019 Basic Video Tutorials By MSFTWebcast:In this step by step video guide, I will show you how to map network drives using bat file login …. Mapping network paths to drive letters has been a familiar task in corporate life. Right click New > Mapped Drive and set it as follows: Action = Update. Hi, I've migrated our main AD Server from 2003 to 2012R2 over the half term holidays and everything appears to be working just fine other then a intermittent issue with network scripts. To see what drives are mapped/available either look under "Computer" (or "My Computer"), or via command line enter Net Use. You may want to use the bypass option. The first step to do the task, open the My Computer then click the computer tab and click Map Network Drive. Hello there, I am about to consolidate multiple Windows 2000 and Windows Server 2003 File Servers into a new Windows Server 2008 box (these servers are all member servers of the same AD domain). Perform the steps to map a drive as listed above in the “ Map a Drive ” section. From here, I click Add, and click Browse. This chapter of Windows Server 2008 R2 Essentials is intended to provide an overview of the different commands currently provided in the command-prompt. If you have legacy scripts and PowerShell in the same GPO, be sure to configure the priority (see Fig. Home of Samba, the SMB file server. The issue that I am having is that one laptop is not automatically mapping network drives even though the user is in the correct active directory group. Blocks more than 7 million malicious domains and IPs while delivering high performance. To setup the OneDrive sync client as a Windows Service …. Then, you can right click this PC and choose "map network drive…. For this post, our servers runs Windows 2012 R2 with latest security patches. Step 2: Expand the Tree: node …. net use z: \\server\sharename If you're in a workground environment, if you create the same user name and password …. application will always use a service account to run services on the server). Since Server 2008, you are only allowed to keep one backup on remote shared folder. Mapping the Drives to the Users. To connect your PC to a server, follow these steps. For example, net use z: \\ Also, I added /p:no so that the login script will be the one that will assign the drive letter where it will not register to persist). Windows Server 2012 R2, Windows 8, Windows 10, and newer Windows Server versions now apply these settings in the background while a user is logged on, meaning if a change is made to a drive map …. Tick on Item-level targeting option, then click on Targeting button. Here’s the problem: You’ve got a small business with an Active Directory domain, but some of your computers have Windows XP/VISTA Home, so they can’t be joined to the domain. On the other hand, if possible, try on another supported desktop under a different network environment and see if the client can be installed. (I have tried the database name with and without brackets. Run WinServ on the source server and on the target server. A logon script runs automatically whenever a user logs on to a computer running a member of the Windows Server 2003 family of operating systems. To set a drive letter and path for any user in Active Directory, we need only use the Set-ADUser cmdlet with three parameters: HomeDrive, HomeDirectory, and the Alias of the user in Active Directory. I executed the xp_fixeddrives extended stored procedure and the above script on the same machine. How to Map Network Drives (Shared Folders…. In Windows Server 2008 and 2008 R2, if a server was installed as Server Core, it cannot be changed to the full GUI version and vice versa. Windows Server 2012 R2 Inside Out: Networking with TCP/IP. Wonderware InTouch 2012 (v10. You can open the network share in explorer with the drive letter. Map network drive on Windows 10 with Command Prompt. Click Browse to open the logon script directory, then select your logon script file and click OK. Windows User State Virtualization Using Logon Scripts in Pure and Mixed Active Directory Environments 9 Dec. Save 20% on Developer Reference …. It is not difficult to set up PowerShell logon script. The user must open file explorer and primary click each disconnected network drive in order to reconnect the drives. Featuring the latest innovations in simplified operations, performance, and security, HPE iLO allows you to manage your entire server environment with ease. This can be accessed either from the Start -> All Programs -> Remote Desktop Services -> RemoteApp Manager or by running remoteprograms. Choose a Drive letter for the mapped SharePoint site. 3) Click on " Map network drive " in the ribbon at the top. Alternatively, from Windows Explorer, map to a network drive to share a folder on the target server, and log in accordingly. Expand “Quota Management” -> “Quota Templates”. We use Scriptlogic as our login script processor and this can do pretty much anything we want in terms of mapping drives…. So you can save yourself the user mapping if you trust all users of the Linux machine, by granting rights to Anonymous Logon. On the next screen, click the bubble next to “Turn on Network …. The forest functional level is set to Windows Sever 2008. On May 17, 2016, Microsoft released a Convenience Rollup for Windows 2008 R2 and Windows 7. This fast, highly configurable IPv4/IPv6 scanner can streamline many of your network support procedures. [2] To make administration of a Server Core machine easier, a Windows script …. It syncs right away and if u select nothing it just stays empty. The NVIDIA vSphere VIBs for vSGA and the NVIDIA Windows 7 driver for vD. Action > Create Task General configuration. Send message to all users logged into Windows Server 2…. Windows Components\BitLocker Drive Encryption\Operating System Drives. Part 3 – Adding Session Hosts and Load Balancing session collections. Azure ad device registration error codes. In Windows Server 2012 R2, RD Connection Broker receives all incoming connection requests and determines what session host server will host the connection. But at least with Windows PowerShell in Windows Server 2012 R2, I can run a script in the background to keep looping and displaying the live network adapters —or at least catch the live ones and export their names to a CSV. My G drive mapping would map to other servers, and sometimes additional z mapped drive appeared that was a ghost. When you install a loopback adapter, the loopback adapter assigns a local IP address for your computer. calibre_ﾔ涜_ﾔ涜BOOKMOBI ｿ ・7PZ aﾒ h& oN u・ }､ ・ 蛍 燃 廁 ｣ ｪH ｱｯ ｸﾐ ｿﾟ ﾆ ﾌv"ﾒY$ﾘk&ﾟ8(・*・,. Go to Start > Administrative tools > Active directory users and computers. e can these map drives/shortcuts to the desktop or elsewhere). Mapping network drives is one of the most common uses of the net use command. Zoom untrusted server certificate your connection is not private. In the past we used to use a script, but there is a group policy setting that can save us the scripting effort. Windows 7: Choose Start→Computer. If your application is running with admin rights, you may not be able to see it on your mapped drives…. Engage in conversations about Citrix technologies — share your expertise, build your professional network…. Mapping network drives is one of the most common jobs for a network administrator. Note that this method is sessions specific, and the mapping will be lost when you close the Powershell session. System Platform 2014 R2 SP1 — archestra. You are still maintaining the UNC paths with this script so you know. I have also checked the event log and there are no errors. Jul 23, 2014 · In new deployment scenarios you will need the TFS 2013 or 2012 on an windows 2012 R2 server, that will never support SharePoint 2010, so we need an SharePoint 2013 SP1, that support windows 2012 R2 for now. On a Microsoft Windows Server with the Active Directory role installed, open the Group Policy Management. Based on SQL Server database technology and built on Microsoft’s Windows Azure cloud …. If it's through a login script or group policy then whatever drives they would normally have access to will be the drives they see remoting in to the server. This command can be used from windows command prompt also to open network connection window. Open the properties for Red teams drive mapping, and click on Common tab. If that user is in the Finance group, then the script will map the R: drive to the server\shared\finance share. IIS 6 Management Compatibility. Script to map sharepoint drive. Open “Windows Explorer” from your server. Create a GPO in Group Policy Management, named something related to network drive then right click and click edit. Create a new GPO and give it a name. View Details Intel® Network Adapter Driver for 82575/6 and 82580-Based Gigabit Network Connections under FreeBSD* 8/6/2021 Intel® Network Adapter Driver for Windows Server 2008 R2…. Blog: SBS 2008 Setup Guide V1. Since your script is checking for group membership it seems clear enough that active directory is working. It is the value named: DigitalProductId. For years, IT administrators have been relying on logon scripts for mapping users' network drives in a Windows domain environment. The mapping is now done by Active Directory itself. Logging out and back in also is no guarantee. Right-click and open as “Run as administrator”. It's wasn't in the batch script. Microsoft Windows is usually a presence in most computing environments, and UNIX administrators likely will be forced to use resources in Windows networks from time to time. Copy two files into this folder: your logon script (for example, . In this tutorial we will look how to create, mount, remove and manage network shares with net use and net share command. Fixing “Alert! TPM device is not detected” - Press F2 to access the BIOS and check if it comes with TPM. This will bring you to the Select Instance page. Access them from any PC, Mac or phone. Schedule sync: Continuously sync files to OneDrive …. Network shares are accessed using their UNC paths (begin with \\). Driver = {SQL Server Native Client 11. without password - connection is successful. AWS supports everything you need to build and run Windows applications including Active Directory,. For UTF-8 to be the native encoding, you need R at least 4. In Windows Server 2008, it is possible to allow connections only if Network Level Authentication is enabled. Tune up Windows Server for a secure and performant repository for your critical data. Go to Server Manager >> Features >> Add Features. I can also open the library in explorer. SQL SERVER – 2008 – 2008 R2 – Create Script to Copy Database Schema and All The Objects – Data, Schema, Stored Procedure, Functions, Triggers, Tables, She primarily focuses on the database domain, helping clients build short and long term multi-channel campaigns to drive …. The "Citrix Print Manager Service" log on account is set to "Network Service" which does not have permission to the printing virtual channel. SUSE, now with Rancher, offers the industry's most adaptable Linux operating system and the only open Kubernetes management platform. Windows Server 2008 Group Policy Drive Maps; Assign logon scripts the NT 4. "The network path was not found" error message when yo…. Creates CSV list of affected files and locations. In the Map Network Drive window, choose an available drive letter. If you Must delete network drives before the actual mappings using shares on Server 2012R2, use net use g: /del (ONLY)this will work on 2012R2 Server. Click the Start button, search for View network …. 1x Authentication integration does not succeed when Microsoft Server 2008 R2- based RADIUS service is being used for 802. 1\DRIVENAME Edit 1: Also, I believe the password should be simply appended: net use Z: \\10. 0 in Windows 7 SP1, Windows Server 2008 R2 SP1, or Windows Server 2008 SP2 2802886 - You cannot register an SPN from. Click Disconnect network drive…. How do you image a RAID? Microsoft SQL Server 2008 R2 Install Guide. logon scripts take a long time to run in windows vista, 2647452 paged pool leaks when you map a network drive and then disconnect it frequently in windows 7 or in windows server 2008 r2 …. Of course, you can always attach a VHD that you created in Microsoft Virtual PC or Virtual Server but you can also create your own VHDs inside Windows 7 or Win 2008 R2…. The security group is also applied in “Security” tab on the printsrv with PRINT rights and “everyone” is removed. Enable TCP/IP by right-clicking and selecting Enable, then OK. Zabbix team makes the product as good as it is,improving it day by day. Together with our Edge solutions, you can innovate fast and transform according to your own priorities. It supports zip compression with no size limit, incremental backup, network …. The reverse statement is true: all network drives mapped …. Intel® Network Adapter Driver for 82575/6, 82580 Intel® EMA Cloud Start Tool Terraform Scripts. How to Create users and security groups in active directory; 3. ps1 -windowstyle hidden -noprofile -executionpolicy bypass. (Note: For users looking to map drives on Windows, please see Mapping Drives (Shared Folders) on Windows) Getting Access to a Drive. To configure the EnableLinkedConnections registry value, follow these steps: Click Start, type regedit in the Start Search box, and then press ENTER. Network object; we then call the EnumNetworkDrives method to return a collection of all the mapped network drives on a computer. Find the permissions tab and select who you want to share the folder with. Customers trust Windows Server to run their business-critical and mission-critical workloads. 5 star reviews at Capterra, CNET, G2 and more. Forgetting Windows 2008 R2 Admin password is not the end of the world. These fixes improve overall performance and system reliability of both operating systems. Do a Google search like “proxy servers” and you’ll find dozens of PHP proxy scripts on the Internet that will help you create your own proxy servers in minutes for free. Right Click Drive Mappings, Select New – > Mapped Drive. Migrate apps from on-premises, multiple data centers, or multiple clouds in a fast, flexible, and safe way. RemovePrinterConnection “Your_print_serverprinter_name_being_removed” objNetwork. Searches all drives on system excluding mapped drives. Login to SharePoint using Internet Explorer. How to Map a Folder to a Drive Letter in Windows: 11 Steps. If you perform an in-place upgrade of Windows Server 2008 R2 and encounter problems related to the Windows Setup process (setup. After installing all his applications I was trying to decide on how to go about mapping his network drives after he logged onto the VPN. Right-click in the right empty space and select New -> Mapped Drive. The previous chapter covered the basic installation and configuration of Remote Desktop Services on Windows Server 2008 R2. CentOS, Ubuntu) Navigate Places → Connect to Server → Windows Share. Windows PowerShell Logon and Logoff scripts were first introduced with Windows Server 2008 R2 however many organization couldn’t take advantage of them because they were not supported on Windows XP. I have an Active Directory domain with a server outside of the domain in it's own workgroup. Tech Editor: Toby Phipps – MVP, Remote Desktop Services. To map a network drive using T as the drive letter and without username password-protected, use the following guide: Connect to a command prompt by clicking on start and search for cmd. On the Action drop down menu, there are four types of action: Create: This action will create a new mapped drive for user Replace: This action will delete the specified mapped drive, if it exist, and then re-create a new mapped drive. If your SQL Server listens on a non-default port you can specify that using the servername,xxxx syntax (note the comma, it's not a colon). Turn off smart multi-homed name resolution. We do not actually want to install them. To map a network drive using T as the drive letter and without username password-protected, use the following guide: Connect to a command prompt …. You can fix the mapped network drives problems by creating and running two scripts either using the Startup folder or running a task when the user signs into their account. To map a drive on the local computer. 5140: A network share object was accessed. The scheduled task was created as follows. In this post, we will learn the steps to configure Group Policy folder redirection policy by changing the path of users "My Document" from local computer to the network drive within a domain. You cannot log in because you have reached the maximum allowed number of concurrent sessions. Start by opening File Explorer (Windows + E) and, on the navigation section from the left of the window, select This PC. We have deployed NAC at one of our client sites. All you need is insert the Server 2008 …. Cosmetic Fix for Image Categories Icons missing on Investigator Login. Select it to change it if needed. Map network drive in Windows Using Command Line. The computing model for thin-client networking means that the horsepower is concentrated on the server end, not the client end. x of the script and it worked great for couple of years with both Windows 2012 R2 and 2019 RDS. Etsi töitä, jotka liittyvät hakusanaan How to install and configure exchange server 2016 on windows server 2016 step by step tai palkkaa maailman suurimmalta makkinapaikalta, jossa on yli 21 miljoonaa työtä. Once the GPO has refreshed on the Exchange server, the pesky printer mapping …. Next, press Confirm to start upgrading Windows. You can access the PowerShell ISE by typing "PowerShell_ISE" in the Run box. I need to have 2 drives mapped when ever a new user is added onto the server or when ever a user switches computers the drives should be . After restarting computer, the script should be executed (I checked gpresult and the script is there), but the drive is not mounted. Enabling Windows Server 2008 R2 File Sharing. ; Save the file below your Apache web server…. Before creating the GPO, create the share folder according to the above steps. Freely Move Pictures in Word Cut and How to Map Network Drives …. Before we configure Windows folder redirection GPO, login with the domain user, right click on "Documents", select Properties. If you installed Fiddler in the default location, this is. AutoRun will work with network drives (the DRIVE_REMOTE drive type) that are mapped to a drive …. You should see a newly created drive letter X:, with which you can access your mapped folder directly. SQL Server 2008 R2 Service Pack 3 contains Cumulative Update 1 to 13 from SQL Server 2008 R2 …. When the restore finishes, detach the database in development, and the production server can attach the database while it sits on the network share. Recently, the script stopped running on Windows 2012 R2 but still runs on 2019. If you're prompted for a confirmation, click Connect. Use the Net use command with the /DELETE switch to remove the mapped drive. Quick note: If you run the command as an administrator, the drive may not mount correctly and won’t appear in File. Welcome to the homepage of FileZilla®, the free FTP solution. The open source implementation of OpenVPN protocol, whose original code was authored by our co-founder, is licensed under GNU GPL. If you want to map a network drive on your Windows 10 PC, you’ll need to enable network discovery first. Supports SNMP, ICMP, DNS and TCP monitoring for devices that support it. exe), Google might only be able to provide limited assistance. Users without administrator privileges will not be allowed access to the Active Directory regardless of the server tools or network …. Like the This PC window, it lists local storage and network locations. The one book you absolutely need to get up and running with Windows Server 2008 R2. In Active Directory, we can create login scripts by using either a standard “DOS” shell scripts, or we can employ a Windows Script Host (WSH) compatible scripting …. It sure seems like there should be. Free Award-Winning File Manager. Step 1: Open server manager dashboard and click Group Policy Management. It originally recommended to disable IPv6 to overcome Exchagne 2007 installed on Windows 2008 (not 2008 R2) that have DSAccess NSPI to …. "D · User-1126563351 posted Thanks a lot Martin, I gave. Using Apache HTTP Server on Microsoft Windows. After searching the internet for a fix, nothing …. First published on MSDN on Jan 29, 2016. In the Group Policy Object (GPO) where drive maps are defined, edit User Configuration > Preferences > Windows Settings > Drive Maps. Plz watch and subscribe : [email protected] Use VBoxManage to Pass Through USB …. 2-From sharing tab select permissions. Script to download file from Sharepoint. These topics describe how to install operating systems, firmware, and hardware for the Sun Fire X4800 server. Proxy settings wont always work with logon script using server 2008. In my case I copied it to my server, and called it Wireless. , a user profile) are stored as a separate VHDX disk on dedicated file on the network …. Just found out that we can backup a database to a network address, like "\\network_path\db. DNS security services have delivered 100% uptime since 2006. This article will help you enable TLS security in Windows Server 2008 R2 or later versions by editing registry. 0) is a very well designed premium Windows password recovery tool. Here are a few additional tips for mapping a network drive: Assigning a drive letter to a network drive is called mapping the drive, or linking the drive…. While using the Windows Pro machine using Windows Explorer check for the Map Network drive in the Tool Bar. This is a rollup of 90 hotfixes that were released after SP1 for Windows 7 and Windows Server 2008 R2. Select Application > Add to open the Add Application window: On the General tab: In the Application field, enter the full path to the Plex Media Server executable, Plex Media Server. Have made the relevant change in this 'Drive Maps' area here. Some basics on how to create a logon script using two different methods: Batch script, and Group Policy Preference. There are situations you have to map multiple drives to users other than the home folder. Best Practices for Recovering Windows Server 2008, Windows Server 2008 R2, Windows 7, and Windows Vista run the copype. 1\share" Dim Network: Set Network = CreateObject("WScript. Windows Server 2019 Basic Video Tutorials By MSFTWebcast:In this step by step video guide, I will show you how to map network drives using . If you are network admin who has the rights to change the server settings, you can turns off the autodisconnect feature of the server service on the server side so that all workstations that have the network map drive to the server can keep the connection live as long as they want to. Copy-paste the below command and click Enter: net use T: \\networkShare\Test. This means that drive mapping when using a Group Policy Preference would only occur if a user logged onto the system. Whenever i set this up as a group policy logon script, the script WILL run (because the messagebox appears at computer login), but the drives are not mapped. 0: Withdrawn Cumulative update package 6 (CU6) for SQL Server 2008 R2 Service Pack 2 (replaced) 2013-04-15; 10. Aug 12, 2010 · The limit is actually related to the Windows API - neither a TFS or Windows issue. Take a closer look at Copy-Item cmdlet coding examples to build advanced PowerShell scripts …. IF its a sql login and you have access to the application servers, you could look for it in connection strings on those. OS is "Server Enterprise 2008 R2" When I originally tested everything all was working fine. On the windows 7 machine it renamed the mapped drive …. Select that you want to choose a custom network location, followed by the Next button. You open some applications to query the extended attributes of the files on the network …. How are drives currently mapped for users. Easy to use, no hassle, and no complex keyboard shortcuts. In Windows 8 if you want to map any remotely shared folder as a network drive to your local computer, you have to follow certain steps shown below: Step 1. VBS script to map a network drive to a network share to a Windows We will be using Microsoft's VBScript and the code executes on any WSH . As a system admin though often times I need scripts to execute for ALL users on a given system. At folder path: right-click and Paste the copied address. Understand but the start up script runs, the GPO's run, and the two mapped drives from a GPO run. Configuring the drive map preference item. For Windows Server 2008 and 2008 R2 you must install the "Application server" role and the "Web server" role. This will run setup routines, making sure your computer has all the prerequisites installed to correctly run SQL server 2008. Next we will Map the SharePoint folders like a network drive. This info is located in the registry under. bat, the drive is mapped and the backup script …. Split the load on different drive. When a user closes all open files on a server …. How to setup windows network, internet sharing, remote acces…. SIW is an advanced System Information for Windows tool that analyzes your computer and gathers detailed information about system properties and settings (Software Information, Hardware Information, Network …. Now what I am attempting to do is from my web server (M: drive), I have a php page that I want to access the network drive (Z: drive) so that I can display the. If it is disabled then enabled it and Restart the "Sql Server(MSSQLSERVER) service" that available in "SQL Server 2008 R2 …. bat on the Profile tab of each user account as described previously above in Figure 2. Check Reconnect at logon to map …. When you install Remote Desktop Virtualization Host on a server…. To install the Print Management console on an R2 machine, simply open Manage Your Server and add the Print Server role to the machine (see Figure 1): Figure 1: Adding the Print Server role using Manage Your Server. Navigate to the File share you created and copy down the Network Path in Sharing Tab. powershell -nologo -file \\chi-fp01\logs\pslogon. We updated the script so [PRINTER_NAME] was the name of the printer we wanted as our default. Launch the File Server Resource Manager. MacOS: Map a Network Drive. Oct 14, 2021 · For Windows 10 or those not desiring to use "wsl -mount" to mount external drives, the following can be used instead. Windows 10 settings; network and sharing center; ethernet right click and click on ethernet properties. The SQL Server service runs in the environment of the local console with the security of the startup account of SQL Server applied. Mapping network drives via Group Policy is faster and easier, so it is a much more scalable approach. If you didn't know, SharePoint libraries can be treated like folders and files you'd use on your local computer. bat (right click and choose save as. However, this setting does not restrict access to these drives. I've got Server 2008 R2 laptop that I need to transfer files to. Use Compute Engine's Tau VMs to optimize for scale-out workloads and increase performance by 42%. To map a network drive in Windows XP: Click Start. This prompt will stall Google Drive until you notice it and click OK - a real problem if you are not even logged on to see it! Furthermore, as Barry de Wit , who graciously brought this issue to our attention points out, "If a network user deletes a file in the (shared) google drive folder on the server the message pops up on the SERVER!. Note - In this procedure, the ImageUnattend. Our default is 45 minutes with logon. Create a Visual Basic Script (VBS) to Add and Remove the Printers. Older operating systems must use startup scripts for computers or logon scripts for users and must run the pushprinterconnections. In this tutorial you will learn how to install and configure OpenVPN on a CentOS 8 server using Elli The Source Lab is interested in planting seeds in. 0 way; PowerShell Logon Script; Summary – Assigning Logon Scripts via Group Policy; ♦. Paste the copied URL into “Internet or network …. Open Notepad and type the share path of the folder that you want to map as a network drive…. Using WinRM (Windows Remote Management), you can configure all of your servers …. An example: Login Username = skimmel Mapped Drive Template = \\SERVER\users\%username% Final Mapped Drive = \\SERVER\users\skimmel. Using Group Policy to Deploy a Windows PowerShell Logon Script. A step by step guide to build a Windows 2012 R2 Remote Desktop Services deployment. The GPO containing the preference item is typically linked to higher containers in Active Directory, such as a the domain or a parent organizational unit. Over 30 one-touch tools—clean your drive, secure private files, take screenshots, or download a video all …. We are getting inconsistent results where sometimes it will map the drives …. In such scenarios, the best option will be to store the database backup files in a mapped network drive …. Mapping drives using Group Policy preferences. How to Map Network Drives Using PowerShell. Here’s a handy command to return a list of logical drives in Windows. == “NU” Network logon user “NO” Network …. Map system drive (C:) of remote computer. Black Box ® is a trusted IT solutions provider delivering cutting-edge technology solutions and world-class consulting services to businesses …. Recently, we announced Windows Server …. Go to This PC, then on the Computer ribbon, click on Map network drive. Remote Desktop Virtualization Host is new to R2 and installs on the Hyper-V box, monitoring and preparing VMs as directed by the RD Connection Broker. In this part, we will show you how to fix the issue that you can't map network drive in Windows 10. The configured location is a network share named data; hosted by a computer named hq-con-srv-01. HKLM\Software\Microsoft\Windows NT\CurrentVersion. When you use the “Net Use” command to map drives in a login script…. Map a drive without domain membership | D…. 1 - Windows Server 2008 R2 and Windows 7 SMB 3. Everything has to be done by SQL scripts. Mapping drives for users has been a task IT has needed to do since the first network drive. Tutorial - Creating a GPO to Lock Windows Computer. Infrastructure Requirement : 1 DC SERVER (DC-CLOUD) Client PC running Windows 10 (CLIENT-10) Lets get started. Click Tools and select iSCSI Initiator to start the MSiSCSI Initiator Service. To try it, open a windows explorer window. Hello, I need major help with the opendir function in php. ) ***** I have also tried to create a DSN on the web server. One of the world's leading Windows authorities and top-selling author Mark Minasi explores every nook and cranny of the latest version of Microsoft's flagship network operating system, Windows Server 2008 R2…. bat" Again like I said this will work and is very simple to implement but needs you add the script to each user account so IT tech prefer GPO these days. There's a workaround, however, that lets you see and access all of the files you have in OneDrive: Map it as a network drive. Use the exact words you see under the printer. Enter a name for your network drive. Dive into core topics and essential techniques for professional developers. One machine of the 2, has a fast initial logon to the domain, but after a reboot, the logon gets a slow welcome and then just a black screen for at least 10 minutes, maybe longer, before the users desktop appears. Step 2: Right-click on your domain or any specific user's OU for which you want to map the network drive. 6 is a highly compatible, in-place update to the Microsoft. In the Query Filter dialog on the Filter tab, check the Critical, Security, Warning, Error, and Information Event level boxes. Note: this is part of our ongoing series teaching IT administration basics, and might not apply to everybody. ActiveXperts Network Monitor: Runs as a background service, on a Windows 2019/2016/2012 (R2)/2008 (R2)/11/10/8/7 platform, 64-bit and 32-bit. We’ll show you two ways of mapping a network drive …. 1 x64 / Win8 / Win8 x64 / 2022 / 2019 / 2016 / 2012 R2 / 2012. Once downloaded, use 7-zip or other decompression tool to extract the driver packages. Mapping a drive to a network share assigns that share a drive letter so that it's easier to work with. <\\ServerName\ShareName> [Username] [Password] For example, if the drive letter you assigned is Z:, specify the following as the shortcut target: This will map the network …. net\cloud-drive /u: [storage account access key] Using Windows Explorer. For a user in Active Directory, you would simply open the properties for the user and click on the Profile tab. The syntax of specifying the server instance in the value of the server key is the same for all connection strings for SQL Server. Type the following command to delete the mapped network drive on Windows 10 and press Enter: net use d: /delete. I tried “net use /del \\xxxx\folder1” for example it also removed OK from the mapped network drive …. We use logon script (VB script) to map shared drives for our users. When you launch the app, you will get an interface as seen in the above image. | 2022-05-17 15:17:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.22843937575817108, "perplexity": 4566.856258396921}, "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/1652662517485.8/warc/CC-MAIN-20220517130706-20220517160706-00370.warc.gz"} |
http://eurodatapark.com/blog/56dfsc9/find-the-shortest-distance-from-the-point-to-the-plane-8d0f05 | ## find the shortest distance from the point to the plane
By
Shortest distance between a point and a plane. Here, N is normal to the plane P under consideration. F_z &=2(z+9)-\lambda && \left[ \textrm {First-order derivative with respect to z} \right]\\[0.3cm] 2(z+3)=1λ. In Euclidean space, the distance from a point to a plane is the distance between a given point and its orthogonal projection on the plane or the nearest point on the plane.. Such a line is given by calculating the normal vector of the plane. And then once we figure out the equation for this plane over here, then we could actually probably figure out what 'a' is, then we could find some point on the blue plane and then use our knowledge of finding the distance points and planes to figure out the actual distance from any point to this orange plane. 3x-24&=0 \\[0.3cm] {/eq} to the plane {eq}\displaystyle x + y + z = 1 Shortest distance between a Line and a Point in a 3-D plane Last Updated: 25-07-2018 Given a line passing through two points A and B and an arbitrary point C in a 3-D plane, the task is to find the shortest distance between the point C and the line passing through the points A and B. In the upcoming discussion, we shall study about the calculation of the shortest distance of a point from a plane using the Vector method and the Cartesian Method. Our experts can answer your tough homework and study questions. To find the closest point of a surface to another point we can define the distance function and then minimize this function applying differential calculus. I am not sure I understand the follow-up question well, but I think if the points have ids then we can sort and rank them. Please help me step by step. With the function defined we can apply the method of Lagrange multipliers. {/eq}. Use the square root symbol 'V' where needed to give an exact value for your answer. Find the shortest distance between point (2,1,1) to plane x + 2y + 2z = 11.? x+y+z-1&=0 && \left[ \textrm {Equation 4, substitute } \quad y=x-7 \quad z=x-16\right] \\[0.3cm] Find the shortest distance from the point ( 2 , 0 , â 3 ) to the plane x + y + z = 1 . So let's do that. Here, N’ is normal to the second plane. Solve for {eq}\, \lambda \, Equivalence with finding the distance between two parallel planes. The vector $\color{green}{\vc{n}}$ (in green) is a unit normal vector to the plane. x&=8 && \left[ y=1 \quad z=-8 \right] \\[0.3cm] CBSE Previous Year Question Papers Class 10, CBSE Previous Year Question Papers Class 12, NCERT Solutions Class 11 Business Studies, NCERT Solutions Class 12 Business Studies, NCERT Solutions Class 12 Accountancy Part 1, NCERT Solutions Class 12 Accountancy Part 2, NCERT Solutions For Class 6 Social Science, NCERT Solutions for Class 7 Social Science, NCERT Solutions for Class 8 Social Science, NCERT Solutions For Class 9 Social Science, NCERT Solutions For Class 9 Maths Chapter 1, NCERT Solutions For Class 9 Maths Chapter 2, NCERT Solutions For Class 9 Maths Chapter 3, NCERT Solutions For Class 9 Maths Chapter 4, NCERT Solutions For Class 9 Maths Chapter 5, NCERT Solutions For Class 9 Maths Chapter 6, NCERT Solutions For Class 9 Maths Chapter 7, NCERT Solutions For Class 9 Maths Chapter 8, NCERT Solutions For Class 9 Maths Chapter 9, NCERT Solutions For Class 9 Maths Chapter 10, NCERT Solutions For Class 9 Maths Chapter 11, NCERT Solutions For Class 9 Maths Chapter 12, NCERT Solutions For Class 9 Maths Chapter 13, NCERT Solutions For Class 9 Maths Chapter 14, NCERT Solutions For Class 9 Maths Chapter 15, NCERT Solutions for Class 9 Science Chapter 1, NCERT Solutions for Class 9 Science Chapter 2, NCERT Solutions for Class 9 Science Chapter 3, NCERT Solutions for Class 9 Science Chapter 4, NCERT Solutions for Class 9 Science Chapter 5, NCERT Solutions for Class 9 Science Chapter 6, NCERT Solutions for Class 9 Science Chapter 7, NCERT Solutions for Class 9 Science Chapter 8, NCERT Solutions for Class 9 Science Chapter 9, NCERT Solutions for Class 9 Science Chapter 10, NCERT Solutions for Class 9 Science Chapter 12, NCERT Solutions for Class 9 Science Chapter 11, NCERT Solutions for Class 9 Science Chapter 13, NCERT Solutions for Class 9 Science Chapter 14, NCERT Solutions for Class 9 Science Chapter 15, NCERT Solutions for Class 10 Social Science, NCERT Solutions for Class 10 Maths Chapter 1, NCERT Solutions for Class 10 Maths Chapter 2, NCERT Solutions for Class 10 Maths Chapter 3, NCERT Solutions for Class 10 Maths Chapter 4, NCERT Solutions for Class 10 Maths Chapter 5, NCERT Solutions for Class 10 Maths Chapter 6, NCERT Solutions for Class 10 Maths Chapter 7, NCERT Solutions for Class 10 Maths Chapter 8, NCERT Solutions for Class 10 Maths Chapter 9, NCERT Solutions for Class 10 Maths Chapter 10, NCERT Solutions for Class 10 Maths Chapter 11, NCERT Solutions for Class 10 Maths Chapter 12, NCERT Solutions for Class 10 Maths Chapter 13, NCERT Solutions for Class 10 Maths Chapter 14, NCERT Solutions for Class 10 Maths Chapter 15, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, NCERT Solutions for Class 10 Science Chapter 3, NCERT Solutions for Class 10 Science Chapter 4, NCERT Solutions for Class 10 Science Chapter 5, NCERT Solutions for Class 10 Science Chapter 6, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, CBSE Previous Year Question Papers Class 12 Maths, CBSE Previous Year Question Papers Class 10 Maths, ICSE Previous Year Question Papers Class 10, ISC Previous Year Question Papers Class 12 Maths. 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' â ' where needed to give an exact value for your answer be a point on the first and... Cross product of the normal of the angle between them finding the distance from a point on the line... Exact value for your answer find the shortest vector from the point to the plane x+y+z=1 here N... Http: //www.examsolutions.net/ for the index, playlists and more maths videos on vector methods and maths! To calculate the distance from a point to the plane 2,1,1 ) to plane..., \lambda \, \lambda \, \lambda \, \lambda \, \... Normal to the other is perpendicular to both lines = ( 0,0,0 ) the question is as below with! The product of the line vectors will give us this vector that is, it is a idea. And the Lagrange function will be a point in the direction of the second P. Two lines and, we want to find a line is given by calculating the normal to. Will give us the said shortest distance from a point on the first line and a plane by a... 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Recent Posts | 2021-10-26 08:30:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.6107779145240784, "perplexity": 601.2674648411875}, "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/1634323587854.13/warc/CC-MAIN-20211026072759-20211026102759-00291.warc.gz"} |
https://chemistry.stackexchange.com/questions/156063/reactivity-of-benzaldehyde-vs-acetaldehyde-and-benzoic-acid-vs-acetic-acid | Reactivity of Benzaldehyde vs Acetaldehyde and Benzoic acid vs Acetic acid
I was trying to compare the reactivity between acetaldehyde and benzaldehyde. After some googling, I found out that benzaldehyde is less reactive than acetaldehyde because the benzene ring can stabilize the partially positive Carbon through resonance. However, if I apply the same logic to benzoic acid vs acetic acid, I find that benzoic acid should be less acidic than acetic acid, but that's not true. What am I missing?
• Does this answer your question? Why is benzoic acid a stronger acid than acetic acid? Aug 23, 2021 at 13:04
• @Mithoron I saw that question before. The answer did not satisfy me. In the comments, A.K said "The aromatic ring is so thermodynamically stable that the phenyl ring donating electrons to the carboxylate is not a major resonace structure. Though it is enough to dictate meta directed electrophilic aromatic substitution", by this argument, can't it be said that the phenyl ring should not stabilize the carbonyl carbon in benzaldehyde through resonance? Aug 24, 2021 at 9:17
1 Answer
The reactivity of an aldehyde group (or a ketone group for that matter) depends on the strength of the partial positive charge on the relevant carbonyl carbon. Thus, when comparing the reactivity of benzaldehyde and acetaldehyde, it is a valid argument that benzaldehyde is less reactive than acetaldehyde because the adjacent phenyl ring can stabilize the partially positive charge on carbonyl carbon through resonance (decreasing the strength of positive charge). That is possible because carbonyl group and phenyl ring is directly conjugated.
However, that same argument is not valid when you compare the reactivity of benzoic acid and acetic acid. That's because the reactive center of acidity is not the carbonyl carbon anymore as it was for aldehydes and ketones. The reactivity of carboxylic acid is depend on the strength of the $$\ce{O-H}$$ bond on carboxylic group, a bond away from carbonyl carbon. Since it is not directly conjugated to the other group attached to the carbonyl carbon, the resonance is not playing the role as that is in aldehyde and ketones. The inductive effect is the one playing strong role in this case. The more the electron withdrawing ability of the group attached to $$\ce{COOH}$$ the better the acidity. Thus, when comparing the electron withdrawing ability of phenyl and methyl groups, $$\ce{Ph \gt CH3}$$. Therefore, benzoic acid is more acidic than acetic acid by that reason alone.
• "The reactivity of carboxylic acid is depend on the strength of the O−H bond on carboxylic group, a bond away from carbonyl carbon" but O-H bonds in carboxylic acids are weak because of the partial charge developed in the carbonyl carbon. So if resonance can stabilise the carbonyl carbon, then won't the O-H bond be stronger? Aug 24, 2021 at 9:14
• I'm confused as to why direct conjugation is so important, because in my point of view, the acidity of the group mainly depends on the stability of the carbonyl carbon. O-H group is present in alcohol too, but they aren't very acidic. Aug 24, 2021 at 9:22
• @Tahsin Choudhury: Remember, the acetic hydrogen is equally distributed between two oxygen in $\ce{COOH}$ group. You may need more reading on this subject. Aug 24, 2021 at 16:48
• Sorry, what do you mean by that? Could you share some resources to know more about it? Aug 25, 2021 at 2:39 | 2022-05-19 08:47:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.44187867641448975, "perplexity": 3237.6805681135734}, "config": {"markdown_headings": false, "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-2022-21/segments/1652662526009.35/warc/CC-MAIN-20220519074217-20220519104217-00058.warc.gz"} |
https://www.eolymp.com:443/en/contests/20975/problems/228319 | Competitions
# Interesting fun of Yura
At the lesson of computer science, Yura became very sad, so he invented a fun for himself.
In the beginning, he has an empty set. At each next turn, he comes up with a number and checks if it belongs to the set. If it belongs, then Yura screams "Yes". If not, he screams "No", and adds it to the set. Before making a new number, Yura screams the number of elements in the set.
The teacher was tired of Yura's screams, so he made him write a program that screams instead of a boy. But Yura does not know how to program, so he asked for help from you.
#### Input
First line contains integer n (1n105). Each of the next n lines contains an integer, picked by Yura. Yura can think of only numbers in the interval from -109 to 109.
#### Output
Print in a separate line what Yura shouts for each request.
Time limit 2 seconds
Memory limit 128 MiB
Input example #1
5
1
2
3
4
1
Output example #1
No 1
No 2
No 3
No 4
Yes 4 | 2022-12-02 20:53:04 | {"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": 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": 0.5964422225952148, "perplexity": 3102.6888391294465}, "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/1669446710916.40/warc/CC-MAIN-20221202183117-20221202213117-00364.warc.gz"} |
http://openstudy.com/updates/55ea8e61e4b0227206134d0c | ## anonymous one year ago Simple harmonic motion question
1. anonymous
it said that the particles acceleration is proportional to displacement but sign is opposite? What does that mean? I know that the displacement graph looks like this and the acceleration graph looks like this for SHM ....|dw:1441423084425:dw| The textbook also said that acceleration magnitude is zero when displacement is zero (which for velocity is the max) and acceleration is max when particle at the crest or trough (for velocity is 0). Is that what it means by sign is opposite?
2. anonymous
@zepdrix @UnkleRhaukus
3. zepdrix
|dw:1441436026318:dw|
4. zepdrix
Yes, it means opposite sign :) In this example, I was trying to show that if you're displaced an amount $$\large\rm x$$ from equilibrium, then the acceleration is $$\large\rm -x$$ at the same time. I'm not so good with the physics, it would be a lot easier to explain this using calculus lol oh well!
5. IrishBoy123
the key to shm is the restorative force. All proper shm starts with the equation F = -kx the restorative force F is proportional to displacement x but acts in the other direction [hence - sign], ie back toward the origin/point of equilibrium everything everything everything follows from that.
6. Michele_Laino
the harmonic motion can be modeled by this function: $\Large x\left( t \right) = A\cos \left( {\omega t + \phi } \right)$ we can set $$\Large \phi=0$$, without loss of generality, so we can write: $\Large x\left( t \right) = A\cos \left( {\omega t} \right) \qquad \qquad (1)$ of course, the constant $$\large A$$ is the amplitude of our motion. Now taking the second derivative with respect to time of the last equation, we get: $\Large \ddot x\left( t \right) = - A{\omega ^2}\cos \left( {\omega t} \right) = - {\omega ^2}x\left( t \right) \qquad (2)$ As you can see, your graph are the graphs of the equations $$(1), \; (2)$$ respectively @dareintheren
7. Michele_Laino
neglecting the scale factor $$\Large \omega^2$$, of course
8. IrishBoy123
and from $\Large \ddot x\left( t \right) = - A{\omega ^2}\cos \left( {\omega t} \right) = - {\omega ^2}x\left( t \right) \qquad (2)$ note that $\Large m\ddot x\left( t \right) = F = - m{\omega ^2}x$ the **restorative** force
9. IrishBoy123
and just to be totally clear, the maths follows the idea of a restorative force, and not the other way round:p
10. anonymous
Thank you guys for responding, more clear now, I can only medal the best one, but all was good!
11. IrishBoy123 | 2016-10-28 06:30:27 | {"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": 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": 0.8825809359550476, "perplexity": 637.3257476237111}, "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/1476988721558.87/warc/CC-MAIN-20161020183841-00412-ip-10-171-6-4.ec2.internal.warc.gz"} |
http://human-web.org/Nebraska/error-correcting-codes-wiki.html | Address Po Box 43, Syracuse, NE 68446 (402) 915-2977 http://www.dotcalmpcrepair.com
error correcting codes wiki Cook, Nebraska
Noisy feedback In an error-correcting code without noisy feedback, the feedback received by the sender is always free of errors. Therefore, d {\displaystyle d} is at least the minimum number of linearly dependent columns. Hamming first demonstrated that SEC-DED codes were possible with one particular check matrix. Parity Main article: Parity bit Parity adds a single bit that indicates whether the number of ones (bit-positions with values of one) in the preceding data was even or odd.
As a consequence, up to two errors per codeword can be detected while a single error can be corrected.[3] This code contains 24=16 codewords. Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. Suppose that we want to design an ( n , k ) {\displaystyle (n,k)} code that can detect all burst errors of length ⩽ ℓ . {\displaystyle \leqslant \ell .} A If we want to encode a message of an arbitrary length using interleaving, first we divide it into blocks of length λ k {\displaystyle \lambda k} .
LDPC codes were first introduced by Robert G. If no error has occurred during the transmission, that is, if r ( a ) = s ( a ) {\displaystyle r(a)=s(a)} , then the receiver can use polynomial division to Retrieved 2014-08-12. ^ "Documentation/edac.txt". to a polynomial that is divisible by g ( x ) {\displaystyle g(x)} ), then the result is not going to be a codeword (i.e.
The deinterlever at the succeeding stage distributes these erasures across 28 D2 codewords. Error Control Coding: Fundamentals and Applications. By our assumption, v ( x ) {\displaystyle v(x)} is a valid codeword, and thus, must be a multiple of g ( x ) {\displaystyle g(x)} . Hence classical block codes are often referred to as algebraic codes.
A method for solving key equation for decoding Goppa codes. The "missing" bits in a shortened code need to be filled by either zeros or ones, depending on whether the data is complemented or not. (To put it another way, if Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. A code with minimum Hamming distance, d, can detect up to d − 1 errors in a code word.
Error Control Coding: Fundamentals and Applications. Error-free transmission: TIEpfeaghsxlIrv.iAaenli.snmOten. Whereas early missions sent their data uncoded, starting from 1968 digital error correction was implemented in the form of (sub-optimally decoded) convolutional codes and Reed–Muller codes.[8] The Reed–Muller code was well Retrieved 2011-11-23. ^ Benchmark of AMD-762/Athlon platform with and without ECC External links SoftECC: A System for Software Memory Integrity Checking A Tunable, Software-based DRAM Error Detection and Correction Library for
Contents 1 Definitions 2 History 3 Introduction 4 Implementation 5 Error detection schemes 5.1 Repetition codes 5.2 Parity bits 5.3 Checksums 5.4 Cyclic redundancy checks (CRCs) 5.5 Cryptographic hash functions 5.6 IIE Transactions on Quality and Reliability, 34(6), pp. 529-540. ^ K. The encoding process assumes a code of RS(N,K) which results in N codewords of length N symbols each storing K symbols of data, being generated, that are then sent over an By plugging the latter inequality into the former, then taking the base q {\displaystyle q} logarithm and rearranging, we get the above theorem.
Upon receiving c 1 {\displaystyle \mathbf … 2 _ … 1} hit by a burst b 1 {\displaystyle \mathbf − 8 _ − 7} , we could interpret that as if From those, e(x) can be calculated and subtracted from r(x) to get the original message s(x). Recent studies[5] show that single event upsets due to cosmic radiation have been dropping dramatically with process geometry and previous concerns over increasing bit cell error rates are unfounded. Touba. "Selecting Error Correcting Codes to Minimize Power in Memory Checker Circuits".
Hamming codes with additional parity (SECDED) Hamming codes have a minimum distance of 3, which means that the decoder can detect and correct a single error, but it cannot distinguish a bluesmoke.sourceforge.net. Golay.[3] Introduction The general idea for achieving error detection and correction is to add some redundancy (i.e., some extra data) to a message, which receivers can use to check consistency of Using minimum-distance-based error-correcting codes for error detection can be suitable if a strict limit on the minimum number of errors to be detected is desired.
It is up to individual designers of CD systems to decide on decoding methods and optimize their product performance. The overall parity indicates whether the total number of errors is even or odd. Philips of The Netherlands and Sony Corporation of Japan (agreement signed in 1979). Cyclic codes are considered optimal for burst error detection since they meet this upper bound: Theorem (Cyclic burst correction capability).
Once the sender has constructed the polynomial p x {\displaystyle p_ Λ 3} in some way, however, instead of sending the values of p x {\displaystyle p_ Λ 1} at all The different kinds of deep space and orbital missions that are conducted suggest that trying to find a "one size fits all" error correction system will be an ongoing problem for A (4,1) repetition (each bit is repeated four times) has a distance of 4, so flipping three bits can be detected, but not corrected. Fix the errors Finally, e(x) is generated from ik and eik and then is subtracted from r(x) to get the sent message s(x).
Secret sharing (differs in that the original secret is encrypted and obscured until the decode quorum is reached) References ^ Sian-Jheng Lin, Wei-Ho Chung, and Yunghsiang S. If the system of equations can be solved, then the receiver knows how to modify the received word r ( a ) {\displaystyle r(a)} to get the most likely codeword s Proof. Popular notation Main article: Block_code §Popular_notation Codes in general are often denoted by the letter C, and a code of length n and of rank k (i.e., having k code words
Cambridge: Cambridge University Press. For missions close to Earth the nature of the channel noise is different from that which a spacecraft on an interplanetary mission experiences. j is any number such that 1≤j≤v. Hadamard code is a special case of Reed-Muller code. | 2018-10-17 04:33:01 | {"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": 0.8088222146034241, "perplexity": 1565.1440110336582}, "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-2018-43/segments/1539583510969.31/warc/CC-MAIN-20181017023458-20181017044958-00396.warc.gz"} |
http://rationalwiki.org/wiki/Conservapedia_talk:What_is_going_on_at_CP%3F/Archive155 | # Conservapedia talk:What is going on at CP?/Archive155
This is an archive page, last updated 14 April 2010. Please do not make edits to this page.
## Going out on a limb here......
....but those fucking "MOAR!" entries look silly. I agree with MC on this that these internet spellings are ridiculous. Fine on talk pages but I think we should keep them of the mainspaces. Thoughts? AceMcWicked 21:36, 16 October 2009 (UTC)
Fine with me. I only use "moar" because I've seen it used in WIGOs for years. Do you want to change "moar" to "more" in existing WIGOs, or just stop using it from now on? Tetronian you're clueless 21:40, 16 October 2009 (UTC)
Just in future. At least I will be. It looks adolescent to me. AceMcWicked 21:42, 16 October 2009 (UTC)
Hear hear. --DamoHi 22:00, 16 October 2009 (UTC)
I knew I could count on a fellow countryman. Wanna come over? I am on my third beer of the morning but you can catch-up on the plane. AceMcWicked 22:22, 16 October 2009 (UTC)
I would but I am off to the SPCA to get a dog. If I am too pissed I might get the Pitbull by accident--DamoHi 22:27, 16 October 2009 (UTC)
You said you could do with a Speights in your edit summary.....Speights? Really? Filth! FILTH! I have Montieths and Heinekens in the fridge so get your ass over. AceMcWicked 22:29, 16 October 2009 (UTC)
My southern roots are showing huh. I was brought up a stone's through from the Speights Brewery in Dunedin--DamoHi 07:11, 17 October 2009 (UTC)
(help, ECs) As an American, I hereby validate this suggestion by agreeing with it. Please send a crate of booze to Chicago in exchange. ~ Kupochama[1][2] 22:34, 16 October 2009 (UTC)
I'll go with the not-using-moar on the mainspace/WIGO entries. I'll not go with the Heineken. A German friend once called it Dutch-piss. Unless it is Heineken Dark, which is pretty f-ing good. Aboriginal Noise Theist, barely hanging by a nail 22:36, 16 October 2009 (UTC)
I don't even drink, myself...but somebody's sure to buy the stuff. Capitalize on it! ~ Kupochama[1][2] 22:44, 16 October 2009 (UTC)
Heineken is a good "drinking to get drunk" beer. If I want flavour I buy other tastier beers. Anyway, I still think we should avoid "MOAR" and "teh" etc on the mainpage. AceMcWicked 22:50, 16 October 2009 (UTC)
Oh "teh" fucks me off. What fucks me off even more is when I remove it from articles and people put it back, because it has some alleged "humour" quality to it. It does not, it just looks like you can't use your fucking keyboard. - π 01:27, 17 October 2009 (UTC)
Fine, but we have to draw the line somewhere. "MOAR" is out, but we should still be able to put "Assfly" in WIGOs. Tetronian you're clueless 22:54, 16 October 2009 (UTC)
To be honest I dont think "Assfly" is funny and I also think it detracts from a quality WIGO, adding Assfly makes a WIGO sneery. The best WIGO's contain dry snarkiness I find, dry wit. AceMcWicked 23:08, 16 October 2009 (UTC)
I am with Ace, assfly is just immature. What fucks me off the most, is after PC has been editing an article and it is suddenly full of "teh assfly" all the way through. I'll be honest all those red down votes you see, they are mostly me voting this shit down. - π 01:27, 17 October 2009 (UTC)
Speaking of the kind of things that detract from good quality WIGOs: running commentaries on Andy's arguments. Seriously, unless it's a highlight worthy of a WIGO by itself, there doesn't need to be a new link. I can follow the argument by myself. Bil08 00:25, 17 October 2009 (UTC)
The addenda are over all poorly done, with the bold text and moars. If a new link is worth adding rewrite it so it tells the whole story. - π 01:31, 17 October 2009 (UTC)
(EC) The problem with that is that all of Mr. Schlafly's responses are usually worth a WIGO highlight. ListenerXTalkerX 01:33, 17 October 2009 (UTC)
Agree that we should avoid MOAR and TEH. ASSFLY was cute once, months ago, whenever it was first used. And a good beer to get drunk on: Pabst Blue Ribbon.--Simple 01:47, 17 October 2009 (UTC)
@Ace: ok, fine, no more "moar," "teh," or "Assfly." <sarcasm> hELL, you'd think this is some kind of encyclopedia. </sarcasm> Tetronian you're clueless 01:50, 17 October 2009 (UTC)
Just makes us look silly instead of Aschlafly. AceMcWicked 02:12, 17 October 2009 (UTC)
I approve of the above discussion. ħuman 04:29, 17 October 2009 (UTC)
One has to consider that many visitors land there first: it should be devoid of "in-jokes" & such, so use of "Assfly", "Jinx" & similar should be minimised. I am eating & honeychat 10:07, 17 October 2009 (UTC)
Well, then agreed. Anytime I see something like that in a WIGO I'll change it. AceMcWicked 10:18, 17 October 2009 (UTC)
I endorse the above discussion. 'Assfly' never had an ounce of humour in it anyway. DogPMarmite Patrol 18:22, 17 October 2009 (UTC)
I'm fine with doing away with internet spelling and colloquialisms. Further, I think we should avoid profanity in the articles for all the very same reasons. Get a thesaurus for God's sake. However, I think the addenda when done with some wit are a good thing. I almost never visit Conservapedia directly so the add-ons help me out. Me!Sheesh!Mine! 22:34, 17 October 2009 (UTC)
I dont think any of our proper mainspace articles have much profanity aside from the "bullshit" tag. But you're right, particularly in our front page articles. AceMcWicked 22:42, 17 October 2009 (UTC)
"I dont think any of our proper mainspace articles have much profanity" - that's a big fucking problem right there. DogPMarmite Patrol 15:14, 18 October 2009 (UTC)
### MOAR!
#### Let's make a list
Just for the hELL of it, let's make a list of words/memes that are never to be used again in WIGOs/mainspace articles:
• moar
• teh
• Assfly
• profanity (I don't think we need a list of these words)
• Semprini
Feel free to add moar more! I'm going to sift through old WIGOs and see what else can be removed. Tetronian you're clueless 20:38, 18 October 2009 (UTC)
What about Jinx hi Jinx! and Ken? Any thoughts on those? Tetronian you're clueless 20:55, 18 October 2009 (UTC)
There is nothing wrong with the Jinx and Kendoll templates and I don't think we need a list. Just, you know, write in fucking english. AceMcWicked 21:11, 18 October 2009 (UTC)
I am all right with losing the intentional misspellings, but proscribing profanity seems excessive. Of course, the Jinx hi Jinx! and Ken templates should stay.
Also, there is absolutely no need to go back and edit all the old WIGOs, Ministry of Truth-style, to match the new policy. ListenerXTalkerX 23:30, 18 October 2009 (UTC)
I meant to find words that should not be used again, not to revise them. Tetronian you're clueless 00:30, 21 October 2009 (UTC)
### First battle of the 'war of wit'
I'm proud that RW has made its first step out of self denial and into the real world. Maybe this is the first stage in winning the 'war of wit' - reclaiming it from the dead hands of 4chan losers and back into the context driven minds of Monty Python and Frankie Boyle admirers... MarcusCicero 10:40, 19 October 2009 (UTC)
Given that Monty Python's humour was based on the unexpected happening, it only succeeded in being funny once. People who repeat Monty Python jokes are worse than memists, as a meme is (allegedly) suppose to get funnier through repetition. - π 10:44, 19 October 2009 (UTC)
What? Please read what you wrote again as it has no bearing on what I said. Monty Python is context driven and its unique selling point is that the jokes rarely have punchlines - leading to long winded deviations and point by point analysis of real life human nature and situations. Whereas a meme is just a pathetic repition of an unfunny collection of letters, Monty Python is driven by the context of the joke. Hence the difference is as sublime as it is uproariously funny. MarcusCicero 11:01, 19 October 2009 (UTC)
So when are you going to start making these sublime, high brow, context driven comedy WIGOs so we can all see how it is done? - π 00:22, 21 October 2009 (UTC)
I also would like to see some of this Oscar Wilde comedy from MC. All I see so far is a reptition of swear words about how we all suck. AceMcWicked 00:25, 21 October 2009 (UTC)
Please stop making me want to try 4chan. ~ Kupochama[1][2] 11:52, 19 October 2009 (UTC)
Seriously it is just an anime (?) forum; all the talk about it is mostly its own self congratulatory masturbation. - π 00:18, 21 October 2009 (UTC)
That is not a meme driven idea. In memetics an idea that is good at spreading will spread, it can be like a joke that people continue to think is funny, or something that just shows that it works, like a knife. --Opcn 09:44, 20 October 2009 (UTC)
MC has a rather limited concept of what "meme" means. I think he should look it up. ħuman 00:24, 21 October 2009 (UTC)
I know what a meme is, and its not funny... MarcusCicero 12:28, 21 October 2009 (UTC)
We're all forgetting here that MC is the only one who knows anything and disagreeing with him is, by definition wrong. Bob Soles 12:48, 21 October 2009 (UTC)
"a meme is just a pathetic repition of an unfunny collection of letters" no, I don't think you do know. Perhaps you should use a better word to express what you mean? ħuman 17:02, 21 October 2009 (UTC)
This isn't worthy of a WIGO, but Andy was on the radio this morning, to discuss the Conservative wank-fest Bible Project.
Looks like the radio station's web editors have the same... high writing skills... as conservapedia.
taking out what he calls 'gender neutral terms' that immaculate it
The rest of the piece is a spectacle of punctuation that would drive a high school English teacher to develop a morphine habit, but the quote above struck me in particular. MDB 14:32, 19 October 2009 (UTC)
Urgh, terrible writing skills... whereas us here at good ole Ratwiki have apparently banned use of the words "moar" and "teh" in WIGO. SJ Debaser 14:49, 19 October 2009 (UTC)
The progression should be: UPDATE: MORE: MOAR:, not that pussy bullshit they're putting in now. Yet another reason why I'm leaving. — Sincerely, Neveruse513 / Talk / Block 15:04, 19 October 2009 (UTC)
While we're talking about Andy in the media (and moar again for some reason), Conservapedia was mentioned briefly on a show I frequently watch (GNW for all the Aussies out there). The site wasn't mentioned by name, it was just a quick mention of the retranslation project and the expected "OMG how stupid of them" reaction from the comedians and the crowd.
Not worth copying off my TiVo. I just thought it was funny that CP is now so stupid, Australian TV shows are starting to make fun of them as well; and it takes a lot to make Aussies think something is stupid, after all, we're the people who brought you this. I'll refrain from going on a rant about how it's apparently okay for two black guys to use makeup to look white, but when white people colour their skin black, all hell breaks loose. I mean, hello? It's okay for one group of people to do it, but not the other? How is that not racist in itself?
Ugh, I've been keeping that in since that hypocrite Harry Connick Jr got up on his high horse about it. Now that I've vented, we can resume normal activities. Thankyou for reading. -Redbackis gonna bite you 15:09, 19 October 2009 (UTC)
Logic dictates that snarky words are being generated at a geometric rate. Deny this and lose all credibility. We need these snarky words to faithfully express the lulz. — Sincerely, Neveruse513 / Talk / Block 15:15, 19 October 2009 (UTC)
Not taking any sides on this, simply answering the question, you have to remember that whitefaced wasn't used for hundreds of years to demean and mock a cultural (in not so subtle terms, see Birth of a Nation) whereas that was exactly what Blackface was for..... Sometimes you just have to accept the reality of the situation. I should also point out that when done correctly, Blackface causes relatively little controversy, such as RDJ in Tropic Thunder. SirChuckBBoom Goes the Dynamite 19:03, 20 October 2009 (UTC)
Yeah, well I won't argue the point too much, I understand you're just answering my question. I just feel that people should be intelligent enough to understand that it was not intended to offend; and I honestly feel that people, (forgive me for being honest here) particularly Americans, seem to think they have some sort of right to never be offended. Of course you're going to get offended now and then, that's when the term "get over it" comes into play. And trust me, the jokes that have been made on Aussie TV regarding the incident are far, far more racist than the act itself - but nobody minds because nobody's wearing makeup. I mean, whooo, have there been some racist jokes on TV lately! -Redbackis gonna bite you 21:48, 20 October 2009 (UTC)
Reds, more like (or, to be more specific, identity-politicos), and the American right-wingers who have tried to adopt their rhetoric as they interpret it prima facie. ListenerXTalkerX 21:56, 20 October 2009 (UTC)
Listening to it now. The host calls him Lafley and Andy doesn't correct him but instead says "close enough." Andy sounds like a muppet choking on diarrhea. 22:04, 20 October 2009 (UTC)
How are you listening to it now? That linkie not work for me... And Andy always sounds like that, I think it is because he is a muppet, and due to the location of his head, he swallows - and spews - a lot of diarrhea. ħuman 23:16, 20 October 2009 (UTC)
Here's the mp3s [1] [2] Andy's bit starts about 2/3 of the way into the first one. I can see him making his awful hand movements in my mind... ħuman 23:25, 20 October 2009 (UTC)
[3] part 3... should we transcribe this? I love the callers, jes' simple country folks... ħuman 23:44, 20 October 2009 (UTC)
And it wraps up here [4] Does anyone know if these links are permanent, or will they be overwritten with tomorrow's show? ħuman 00:06, 21 October 2009 (UTC)
I saved the four mp3s (I think) just in case. ħuman 00:12, 21 October 2009 (UTC)
The best interview so far: Andy faced up against a non-hostile interviewer and callers, and presented his case well. Broccoli 13:45, 21 October 2009 (UTC)
Ya think? I thought he encountered some real Bible-believing foks, and they mostly didn't buy his new professorial-valued re-interpretation of the Holy Word of Jesus. ħuman 09:08, 22 October 2009 (UTC)
If you accept (not that anyone should) Andy's idea that the Bible has been mistranslated by liberals, then the CBP makes sense. Something like: 'All English Bibles since the KJV have either been written by commies (NIV, etc), or are unreadable thanks to overly literal translation (ESV, NASV), therefore we conservatives will go back to the best source manuscripts (better than the KJV ones) and produce a readable version. In the interview, he made that fairly clear. The really funny part is when he talks about the amazing scholars coming forward, while we know it's really people like Terry 'I have a pocket dictionary' H and parodists. Broccoli 10:27, 22 October 2009 (UTC)
## CP and checkuser - how does it work?
And more to the point how can I make sure I don't lose another account? I created my first CP account from home, I was then blocked. I created my second one from a hotel, and then used it only whilst at work (I was mighty proud of the work I did with this one, much more subtle) but then inadvertently logged in with it from home; it was then blocked as it was used from the IP address of a known 'troublemaker'. So, my question is this: If I make a new account from work, will they spot that it is from the IP address my second one was used from (but NOT created from), and block me again? I hope that makes some sense.... Cheers! DeltaStar 21:53, 20 October 2009 (UTC)
Yes. ListenerXTalkerX 22:03, 20 October 2009 (UTC)
Thanks. Proxies it is then......22:54, 20 October 2009 (UTC)
I think we have an article on that...yes we do. Tetronian you're clueless 22:58, 20 October 2009 (UTC)
Cheers for that, I've been wanting to get TOR up and running for a while, but I've been having problems with apt trying to connect to a non existent local IP address (192.168.0.1) instead of the repositories - any Linux (Ubuntu Jaunty) geeks on here..................?DeltaStar 23:23, 20 October 2009 (UTC)
Here's something that I use, it is a bit paranoid I guess, but I keep a file called, "cp-accounts" that has a list of which IP addresses are paired with which username (and whether it's blocked or not). I think I'm at 5 free out of the 30 or so I've created. 130.156.3.250 23:40, 20 October 2009 (UTC)
Sorry Rob you can keep your privacy invading shit to yourself. Can someone oversight that please? - π 02:49, 21 October 2009 (UTC)
What about that real question, what motivates a person to spend hours of thier time with the specific intent to be obnoxious toward thier fellow humanb beings? Is that rational? Is that compassion? RobS
Non-sequitur much? - π 03:00, 21 October 2009 (UTC)
What would you do if you were me, block the IP? He/she/other obvious is not a bona fide contributor, and you think blocking someone like this who is determined to make themself a pain in the ass will stop them? RobS 03:14, 21 October 2009 (UTC)
I don't know Rob, what should we do with you? - π 03:15, 21 October 2009 (UTC)
Personally, I like the "account creation disabled" that I got. Do I understand that correctly as it blocks account creation from all the IPs I had logged in from? or just the latest one? --Shagie 04:57, 21 October 2009 (UTC)
DeltaStar, you will probably even have problems with proxies. Most of the common ones have been blocked and an IP address lookup usually reveals whether it is a TOR node. Also editing from multiple IPs automatically creates suspiscion, you can guarantee that any non-trusted editor is automatically checkusered several times and by using a proxies or TOR you will automatically get banhammered. Even if you have a dynamic IP it normally floats within a range which will all get blocked. Of course this in itself can be considered a "win" as it reduces the number of future editors. Ideally you should get multiple ranage-blocks in many different cities and states as possible. If you wish to actually edit CP then the best tactic is along the lines of a Bugler/RodWeathers suck-up, otherwise you need to be really subtle and don't overdo the parody to soon. ГенгисOur ignorance is God; what we know is science. 11:19, 21 October 2009 (UTC)
Don't you have a gay rights parade or save the planet rally to attend? All you do is give credibility to cp:Liberal deceit with a response like this. RobS 19:17, 21 October 2009 (UTC)
Rob, with a response like that, all you do is give credibility to cp:Conservative spite. Huh...that's a red link? Weird... — Sincerely, Neveruse513 / Talk / Block 19:33, 21 October 2009 (UTC)
Interesting question. When I'm not organizing unions, attending gay rights rallies, or protesting the war in Iraq, I like to unwind by going over to my favorite parody site, Conservapedia, and toss a few turds on the pile. It's getting big, but there's always room for more. I take it you don't believe cp:Liberal deceit isn't parody. Interesting, considering a good number of the entries were written by known parodists. At some point you have to ask yourself whether the entire shitheap was a farce perpetrated on a few willing ignoramuses by a few master parodists. It's sort of a conundrum, isn't it? Trying to discern the intent and purpose of something created by so many people. Is it a serious project because you believe Andy Schlafly when he says it is? Or is it a joke because he embraces people like TK? 19:32, 21 October 2009 (UTC)
That's why I was so pissed off with myself for blowing my last account by logging in from a blocked IP address; I was being very subtle and had even had positive words from some sysops.
DeltaStar: the best thing to do is move house and then try again.-- Kriss AkabusiAAAWOOOGAAAR!!1 11:39, 21 October 2009 (UTC)
Or change ISP, which may be a little less drastic. –SuspectedReplicantretire me 11:56, 21 October 2009 (UTC)
Dunno man, if you're changing ISP because of Conservapedia, that's already pretty drastic.-- Kriss AkabusiAAAWOOOGAAAR!!1 13:53, 21 October 2009 (UTC)
True... but compared to moving house? –SuspectedReplicantretire me 14:30, 21 October 2009 (UTC)
To be honest, I'd consider using Tor to be drastic for just getting onto Conservapedia. postate 14:31, 21 October 2009 (UTC)
I'm sorta with RobS on this one. Posting on CP can be fun, but if you're going to a lot of effort to get in then you probably have things out of proportion. Plenty of other fun to be had on the net.-- Kriss AkabusiAAAWOOOGAAAR!!1 14:38, 21 October 2009 (UTC)
You mean pr0n? CrundyTalk nerdy to me 15:18, 21 October 2009 (UTC)
If you have a wireless router, then just turn it off and turn it back on. For me, that usually gives me a new number (ie. 54.69.xxx.xx ... where "x" is a new number)
Tor isn't reliable and it's just a matter of time before they install the Torblocker extension. Just use a web proxy that's not listed on any blacklists so TK and the other checklusers over there (not including you, Rob!) can't translate "possible proxy" into "confirmed proxy server/network sharing device". If you're feeling particularly froggy, learn some of the much more reliable and convincing ways to use proxies than web proxies. RW really isn't an appropriate place for instructions on more advance proxy use so I'll leave it to you to figure some of that stuff out on your own as I had to. 16:23, 21 October 2009 (UTC)
I know that I'm being a bit of a wet blanket here, but I don't feel that we should be actively encouraging people to disrupt other wikis.--BobNot Jim 16:49, 21 October 2009 (UTC)
I agree, and I think we're not so much. If people want to come and brag about their oh-so-subtle parody and disruption, that's fine, but I rarely ever see full endorsement by RW as an entity, or more practically, from more senior editors who haven't just discovered Conservapedia in the last month and haven't quite got over it yet. If people want advice about how to access it, that's fine; the owners and admins have made it deliberately difficult to get onto the site; what they do once there is up to them. I hope they attempt something rational, calm and mature with it (forgetting that a lot of people have already tried and failed because they get blocked). postate 16:57, 21 October 2009 (UTC)
Screw access. I have more fun getting entire universities, libraries, and Blackberrys blocked. If you really want to run your head into a wall, go to ASoK instead. If you want to parody, go to CretinWiki. CP is fun to watch, that's about it. — Sincerely, Neveruse513 / Talk / Block 17:02, 21 October 2009 (UTC)
## Did Andy blow a fuse, or what?
Because thisimg doesn't sound like the Andy we know. Tetronian you're clueless 22:54, 21 October 2009 (UTC)
Eh, he's just posing for the camera. He knows the eyes of the world are upon him, so he's pretending to be a decent manager running a publicly edited project. ħuman 03:16, 22 October 2009 (UTC)
I am surprised he wasn't blocked for having a name suspiciously close to yours Human. - π 03:20, 22 October 2009 (UTC)
Maybe The Missuz gave Andy "the annual" and he was feeling kindly. MDB 13:10, 22 October 2009 (UTC)
Not likely, for most of the last week he has been up editing till at least midnight. I don't think I'll let my mind wonder on this topic too long, but seriously he is up that late every night. - π 13:32, 22 October 2009 (UTC)
## Random mention
I just heard my radio (MPLS) say Andrew's name. Before they say the new dead guy's thing (Munsters, Green Acres). Wait for the new Conservative Bible news? ħuman 09:00, 22 October 2009 (UTC)
Never mind, I ran the wrong player on my internet radio. I'm listening to the hhh prog on WRONG again. As you were. ħuman 09:05, 22 October 2009 (UTC)
## I wonder...
...what the "liberal detector" bot score would be for the original KJV text for a given gospel, like Mark, versus the corresponding conservative version blessed by Andy. If this were suggested, I could see Andy becoming obsessed with fine-tuning the translations to bump up the conservative scoring, as if that somehow mattered in the end, like his geometric progression of conservative words over the centuries. --SpinyNorman 15:01, 19 October 2009 (UTC)
You would have to significantly change the formula for it to apply to the Bible, since it is partially based on number of mainspace vs. non-mainspace posts and the length of those posts. Tetronian you're clueless 15:02, 19 October 2009 (UTC)
• (Block log); 09:12 . . Aschlafly (Talk | contribs) blocked Jesus (Talk | contribs) with an expiry time of 5 years (account creation disabled) (Violation of 90/10 rule: talk, talk, talk) — Sincerely, Neveruse513 / Talk / Block 15:18, 19 October 2009 (UTC)
Something like this is already suggested in cp:Essay:Quantifying Liberal Style, among other things. --Lesjohn 15:22, 19 October 2009 (UTC)
Conservapedia's idea of conservative style is typically just a high reading ease score, so short sentences with short words, not that hard to do, and might legitimately help the CBP, getting rid of their stilted grammar. --Opcn 17:02, 19 October 2009 (UTC)
What's ironic is that there are places where Andy, almost certainly unaware of what the reading level of a text is a measure of (low grade level indicates ease of readability, high does not indicate intellectual rigor), takes some Bible translations to task for being at a lowish reading level, considering them "dumbed down" for that reason. He considers "the NIV is written at only the 7th grade level" to be an actual criticism of the translation. 98.206.143.163 09:04, 20 October 2009 (UTC)
What sort of age group is 7th Grade? I'm guessing around the 10-11 mark? If so, having text written for such an audience is probably a good idea. The average reading age of the British population is about 10 (yes, 10 and no, I'm not making it up although a quick Google turns up no proof). That's why the popular newspapers like the Sun and the Mirror are written the way they are. The Guardian, to pick one example, has a much higher reading age. –SuspectedReplicantretire me 09:29, 20 October 2009 (UTC)
7th grade is ages 12-13, approximately. But yeah, most papers are written at a low reading level, but it depends on their intended audience. The New York Post, for example, is written for 3 year olds (ok, I'm exaggerating, but if you've ever read it you know what I mean) but the New York Times is far more sophisticated. Tetronian you're clueless 20:07, 20 October 2009 (UTC)
Absolutely - no argument from me. The important thing is to write for your expected audience. I learned that lesson way back when I used to write political leaflets: if you start blethering on about macro-economic policy you turn half your readers off but OTOH if all your sentences read like "We good - others bad" you turn the other half off. By attacking the NIV for being written for 12-13 year olds, Andy must be aiming it at people who can read for a noticeably higher age (or he wouldn't bother mentioning it). If his bible is aimed at people who read like 18 year olds, he is probably writing it in such a way that more than 50% of people can't read it.
Having said that, what I've seen so far reads like it was written in crayon on the walls of a padded cell by a deranged idiot. Bimbo indeed. –SuspectedReplicantretire me 06:54, 21 October 2009 (UTC)
A padded cell in Castle Aaaarrgh. ГенгисOur ignorance is God; what we know is science. 16:04, 21 October 2009 (UTC)
<-- My turn to be the turd in the punch bowl, it seems. BoN was doing so well until I saw "unless your intended audience is very likely included entirely of sophisticated readers." "Comprised" is a perfect example of a word that gets misused because it sounds fancier than the simple words that really say what's meant. Sprocket J Cogswell 16:20, 21 October 2009 (UTC)
Huh? Looks like bon used "comprised" correctly, and you used "included" incorrectly. Perhaps if you rearranged the sentence? ħuman 17:08, 21 October 2009 (UTC)
The dictionary agrees with the human. edit: Though SJC is correct that the usage was once considered uncouth. That said, "included entirely of", while it sounds wrong to me, does appear to be attested. --YossieSpring in Fialta 18:41, 21 October 2009 (UTC)
I wouldn't agree that "included entirely of" usage is grammatical. Just read the excerpts in those google hits! "Someplace it is a phone messaging software of the platinum proxy, the included entirely of the cost units is quickest a informant or partner that supports..." indeed. ħuman 18:53, 21 October 2009 (UTC)
No, I don't think it is either. I just found it interesting that it's actually something people say, however mistakenly. The thing about "comprise" is interesting, though. English prescriptivism is a bit of a lost cause in some areas. --YossieSpring in Fialta 18:58, 21 October 2009 (UTC)
OK, here it is in simple terms. If you are ever tempted to use any form of "comprise," try substituting the similar form of "include" and see if it still makes sense. I did that above, and it did not make sense to say "very likely included entirely of sophisticated readers." See?
There is always a simpler, more correct word instead of "comprise" unless, of course you mean to use it as it originally worked. Sadly, lexicographical anthropologists have legitimized its incorrect use, i.e. to mean "compose." More simply put, enough people have used it wrong that the wrong usage has made it into descriptive dictionaries. It is still a word that, for me, marks someone as an ignoramus trying to sound fancy. Sprocket J Cogswell 20:09, 21 October 2009 (UTC)
Yeah, I think that kinda went over our heads before. :P Anyway, this kind of pedantry is something up with which you know the rest. --YossieSpring in Fialta 12:32, 22 October 2009 (UTC)
Arrant pedantry? Guilty as charged, m'lord. To quote someone else entirely, "I may be a pillock, but I'm sensitive." Sensitive in real life, where I suffer blowhards and stuffed shirts badly. Thank goodness no one here fits any such description. Sprocket J Cogswell 13:13, 22 October 2009 (UTC)
Sir, I'll have you know pedantry has no place on a website dedicated to nitpicking the minutiae of goofy people! It's simply not done. Anyway, I'm kind of fascinated by this whole "comprise" thing. I didn't know it was distinguished from "compose" in any meaningful way. English really does have a word for everything. --YossieSpring in Fialta 22:42, 22 October 2009 (UTC)
## What do I win?
What do I win now that the Relativity WIGO has ascended to #1 position in the charts, as I predicted? Do I get a blow-up red football? DogPMarmite Patrol 15:37, 21 October 2009 (UTC)
You win Andy's sense of self-satisfaction. --YossieSpring in Fialta 15:38, 21 October 2009 (UTC)
That prize is very generous but is too big and won't fit in my house. Can I have something smaller, like maybe Jpatt's brain in a jar? DogPMarmite Patrol 15:50, 21 October 2009 (UTC)
Are you a Sunderland fan?-- Kriss AkabusiAAAWOOOGAAAR!!1 16:05, 21 October 2009 (UTC)
I am not, not since The Scowler quit the manager's gig anyway. I'm a gooner. But you have to admit that was a top drawer goal. DogPMarmite Patrol 04:42, 23 October 2009 (UTC)
The only thing we have left is Andy's dignity, but you'll have to get your own microscope. --YossieSpring in Fialta 18:28, 21 October 2009 (UTC)
## Candlelight vigil
I'm holding a candlelight vigil here, now for the return of our Parodist in Chief, Terry Koeckritz. I miss you, Terry. It's just not the same without you. Come back soon.
— Sincerely, Neveruse513 / Talk / Block 16:25, 21 October 2009 (UTC)
He did block someone as recently as Monday, so hopefully he'll really be back soon! --Pyfgcr 16:52, 21 October 2009 (UTC)
I dunno, we've had some truly awesome wigos since he's been absent. There are more than enough megalomaniacs over there to block the wandals, which means there are more people asking reasonable questions which Andy promptly embarrases himself with. Those questions would never arise if ThickKunt was still about. CrundyTalk nerdy to me 08:12, 22 October 2009 (UTC)
True that, Crunky. ħuman 08:56, 22 October 2009 (UTC)
Speaking of dyslexia, it certainly hasn't dented my mate's confidence. Just the other day, he went to see Lennox Lewis at a book signing, and he reckons he managed to take him down with just one punch. CrundyTalk nerdy to me 09:58, 22 October 2009 (UTC)
[Random internet meme]. That one might be lost on non-UK readers. ГенгисOur ignorance is God; what we know is science. 10:48, 22 October 2009 (UTC)
It bears repeating that TK is not fun. And I think more and more people will agree that Andy's in much better form when new users last longer than half an edit before they're banhammered. This bible nonsense would have been abandoned by now if it weren't for lowly editors who signed up during TK's absence. 15:30, 22 October 2009 (UTC)
Agreed. TK being absent is the best / worst thing to happen to CP (delete as appropriate). CrundyTalk nerdy to me 15:52, 22 October 2009 (UTC)
Best. DogPMarmite Patrol 04:43, 23 October 2009 (UTC)
## Wait for it, wait for it...
Because based on this comment, I have a feeling that Andy's interpretation of "first shall be last, and the last shall be first." is that what Jesus really meant to say was "big government and government leaders we don't like shall be last, and main-streeters, hockey moms and tea-party protesters shall be first". This is actually what the framers of the Constitution were aiming for, but my understanding is that Jesus was talking about the importance of humility, not power to the people where government is concerned. Keep "reinterpreting" that liberal claptrap, Andy, like turning the other cheek, the meek inheriting the earth, and the last/first phrase above. Given that Jesus seemed t be pretty consistent, you have a lot of edits to make. --SpinyNorman 15:31, 22 October 2009 (UTC)
"the 'highest position among men shall be last, and the last the shall be first.'" Silver tongued, that Andy. --YossieSpring in Fialta 22:48, 22 October 2009 (UTC)
## Classic, classic Andy
This. Although Karajou's reply was what made me smile more: "he [a scientist] says it happened, and we are to sit here and believe what he said about it; he offered absolutely no proof whatsoever" Tetronian you're clueless 22:45, 22 October 2009 (UTC)
## Uh...Bert?
Is something wrong with him? Try and read this. Has he always been like that? — Sincerely, Neveruse513 / Talk / Block 16:42, 23 October 2009 (UTC)
There's a guy called Bert on Conservapedia? SJ Debaser 16:48, 23 October 2009 (UTC)
I never read any KAL stuff, so basically I'd never read anything he's written. The guy wrote a book, I figured he could form sentences and express thoughts with words. — Sincerely, Neveruse513 / Talk / Block 16:51, 23 October 2009 (UTC)
Unfortunately, in the literary world of today, the qualification for writing a book appears to be that you do not know how to do that. ListenerXTalkerX 16:54, 23 October 2009 (UTC)
That is correct - after all, you should write a book to learn, not read a book. DogPMarmite Patrol 01:27, 24 October 2009 (UTC)
Ooh! Andy has a Holy Spirit Detector! 18:58, 23 October 2009 (UTC)
Wow, is Bert channelling Lucky from Waiting for Godot? --YossieSpring in Fialta 22:19, 23 October 2009 (UTC)
## North Dakota, that barometer of Conservative effectiveness...
Well, at least Ken picked a good conservative-oriented state to profile, since the conservative "triumphs" in the redder-than-red states of Alabama, Mississippi, Kansas and Oklahoma haven't been doing so well. --SpinyNorman 22:56, 23 October 2009 (UTC)
## A thought
Isn't today (still the 23rd here) the supposed anniversary of creation? Shouldn't that be mainpage news on CP, especially given the current CBP? Doesn't failure to do that mean that Andy is concealing the Truth and being Deceitful through exclusion? Kalliumtalk 01:43, 24 October 2009 (UTC)
Nice point. At the least, we honor the day of creation on RW... ħuman 06:15, 24 October 2009 (UTC)
That is only the Ussher calendar; creationists allow some degree of ambiguity, but no timespan in five or more figures is permitted under any circumstances. ListenerXTalkerX 06:20, 24 October 2009 (UTC)
## Evolutionist lies & Robert Smith
Those last two WIGOs are top-notch Andy. The 47million year old fossil may need to be classed in a different evolutionary family (arrived at due to open discourse and academic investigation based on evidence), ergo evolution is false and humans were created by a magic man 6000 years ago (right after he created the entire universe). Logictastic!
Then I was very confused as to why a member of Robert Smith's band would be donating money to Planned Parenthood, but once I understood my mistake I could then search for the evidence posted by Andy to back up his claim that "The link between abortion and breast cancer is clear". (Note: quoting from another evidence-free wingnut site does not constitute evidence).
Speciationspeed! DeltaStar 22:52, 22 October 2009 (UTC)
"This is a Lie" would be an appropriate theme song for Andy. --YossarianThe Man from the USSR 22:58, 22 October 2009 (UTC)
“”Jorn Hurum of the Natural History Museum in Oslo, Norway, an author of the Ida paper, said he welcomed the new analysis. Darwinius is an example of a group of primates called adapoids, and "we are happy to start the scientific discussion" about what Ida means for where adapoids fit on the primate family tree, he wrote in an e-mail.
See the way evolutionist censor debate. - π 00:08, 23 October 2009 (UTC)
"Pi" (your insulting name is getting tiresome), you act like liberal/conservative are symmetric with respect to the truth. That's plainly false, as illustrated by our entry on deceit. I've responded to your concerns and answered your questions and you persist in repeating the liberal claim that Obama is not an affirmative action president, so it's obvious to me you're another liberal enemy of conservative principles. You also clearly deny that that affirmative action causes rather than deters autism. You can whine all you want, but I do know quite a bit about the Irish press, Ireland, and Wikipedia. In fact, I've been quoted in the past in the Irish newspapers. Try claiming to be an "expert in 'human medicine'" in a court of law sometime. Five minutes later, which is how long it will take for everyone to stop laughing at your claim, maybe you'll wake up. Study the Bible and follow its commandments.--Aschlafly 00:13, 23 October 2009 (UTC)
Love will tear us apart..... DeltaStarSenior SysopSpeciationspeed! 04:40, 23 October 2009 (UTC)
This 'debate' on conservapedia's talk page led me to the oh-so-beautiful 'conter-examples to evolution'. THAT talk page is a delight for amateurs of Andy's debate skills. There are two very nice 'if 2+2=4 I'm right' gambits, finished by the classic '99.99% of the world is with me' delusion conclusion. --Ireon 10:08, 23 October 2009 (UTC)
"Autumn foliage is beautiful, and the theory of evolution depends on denying it. And those who fall for the theory are destined to a life without real beauty." - The man is insane! At first I wanted to smash my computer up and scream at his idiocy, but this quickly gave way to uncontrolled laughter. DeltaStarSenior SysopSpeciationspeed! 17:53, 24 October 2009 (UTC)
At home he's a tourist. 14:11, 23 October 2009 (UTC)
So, just to summarise, people like Richard Dawkins and PZ Myers have been saying all along that the media were way overblowing the significance of 'Ida'. According to Andy, there are now indications that this is 100% correct - so this proves evolution false? 92.19.42.164 18:04, 23 October 2009 (UTC)
## WorldNetDaily not liking the Conservative Bible Project
This can't be good - http://www.wnd.com/index.php?fa=PAGE.view&pageId=113599 - Joseph Farah of WorldNetDaily (already a foaming rightwingnut site in it's own right) has vented on Schlafly's Conservative Bible.
Choice quotes:
"I've seen some incredibly stupid and misguided initiatives by "conservatives" in my day, but this one takes the cake." "It's nutty." "Enough of this foolishness!" "A plague on the houses of anyone and everyone who would tamper so frivolously with God's Word."
I almost want to give the guy a beer, although it'd be a half. Of shandy. With flat lemonade. And ice. SҚ_ 08:38, 23 October 2009 (UTC)
This the greatest moment in clogo history. My two favourite self important wingnuts mouthing off at each other. - π
HAHAHAHAHAHAHAHHAHAHAHAHA-*wheeze*-HAHAHAHAHAHAHAHA! Oh, this is BRILLIANT. CP worships WND like crazy on their front page, and WND replies by saying "a self-described 'conservative' is spearheading an effort to rewrite it to his liking" - with quotation marks around "conservative"! Schlafly, have a taste of your own "You may call yourself a conservative, but you are not one!" medicine! =D --Sid 08:47, 23 October 2009 (UTC)
Not appearing on CP's main page in 100,000...99,999....99,998.... Totnesmartin 08:48, 23 October 2009 (UTC)
I love how Farah will never describe himself as conservative, even though he clearly is, but he is willing to push Andy straight of the boat - despite his "heroic" mother. - π 08:50, 23 October 2009 (UTC)
Oh God, I just reached the bottom of the page: "Joseph Farah is founder, editor and CEO of WND" <-- OUCH! --Sid 08:53, 23 October 2009 (UTC)
Farah was one on the organisers of Mumma Schlafly's idiot convention. He probably has know him for years. - π 08:55, 23 October 2009 (UTC)
Eff! they've got night editing on: I was gonna burn a sock to spam that link all over. I am eating & honeychat 09:32, 23 October 2009 (UTC)
BTW, have you seen Farah's blog? He don't care for them Islamists, do he? I am eating & honeychat 09:36, 23 October 2009 (UTC)
(UI) Nice spottage Piez- it's reasonable to assume that JF and AS have met, given their "fame" in Conservative circles, so what did Andy do to deserve that kinda coverage? Has Queen Phyllis actually come out in favour of the CBP, or has she kept a dignified silence?
I'm not presuming that all Conservatives are quite as "movement" as the Schlaflys (Farah seems quite populist in comparison, more a rabble-rouser in the Murdoch/Fox News style than an ideological purist like Phyl and Andy), but that's quite a stinging rebuke from WND nonetheless.
/me wonders if there are a few calls to be made when the sun rises over NJ.
(What with the WND-Conservapedia smackdown and Nick Griffin of the British Union of Not-Fascists-Honest-Guv-We-Just-Care-About-Indingi-Ingenid-Injun-Indengi-er-Locals behaving like an arse on the BBC last night, one could - at a stretch - feel a bit sorry for the global ultra-right wing, they're not having a good week) SҚ_ 09:48, 23 October 2009 (UTC)
Cut to the next scene: A montage of Andy walking on the beach, interspersed with scenes of Farah and himself enjoying trips to Disneyland, fine restaurants and the theatre. As he gazes in to the sky, wondering what went wrong, the song "Love Don't Live Here Anymore" plays softly in the background. Fade out to Andy trawling CP to remove all links to WND. --Ask me about your mother 10:01, 23 October 2009 (UTC)
If we're lucky, WND will become like the FBI and Andy's "confirmed bachelor" brother -- insta-ban for even mentioning it on CP. MDB 11:41, 23 October 2009 (UTC)
Wow... when you're too whacked out conservative for World Nuet Daily... its not just that you're 'not playing with a full deck'; you've shown up at a Texas Hold 'Em tournament with Pokemon cards. MDB 11:12, 23 October 2009 (UTC)
When was this? When did they take it off the front page? I wonder if words were exchanged. Momma Schlafly could have that zapped in no time. — Sincerely, Neveruse513 / Talk / Block 13:06, 23 October 2009 (UTC)
A commenter at freerepublic(!) about Andy: "Is he the homosexual son?" LOL larronsicut fur in nocte 14:12, 23 October 2009 (UTC)
It's ON. Can't wait to see Andy's reaction. Will he denounce Farah as a liberal? --Ireon 14:43, 23 October 2009 (UTC)
They will probably declare that Farah has always been a liberal, and that CP has always been at war with EastAsia WND. MDB 14:50, 23 October 2009 (UTC)
Well put! I am F5ing conservapedia's talk page, can't wait! --Ireon 15:01, 23 October 2009 (UTC)
You need a hobby. or a girlfriend/boyfriend. Or something. RaoulDuke 15:04, 23 October 2009 (UTC)
For those of us enjoying the show check out this edit on the Joseph Farah article. Bob Soles 15:13, 23 October 2009 (UTC)
Yeah!! As for RaoulDuke, I do have a hobby, and it's browsing the internet. And conservapedia is a lot of fun. BTW: Farah is *gasp* close-minded! Not yet calling him liberal, but it's getting there.--Ireon 15:26, 23 October 2009 (UTC)
This will all be ignored. — Sincerely, Neveruse513 / Talk / Block 15:20, 23 October 2009 (UTC)
Assfly claims Farah has his priorites wrong. They should be attacking only the liberal rewrites, not all of the rewrites. I'm surprised he acknowledged this rather than burning the link. CrundyTalk nerdy to me 15:55, 23 October 2009 (UTC)
WND over the top? Say it an't so! tmtoulouse 15:59, 23 October 2009 (UTC)
I'm guessing here... but could his "doubting a serious response is even necessary" be a retreat to save face? Make the bad men go away? WND and Free Republic criticising the project is a lot of conservative opinion going against him. SҚ_ 16:09, 23 October 2009 (UTC)
Obviously. We say that creationism is not worth a serious refutation, but when we are pressed we can give one. On the other hand, I doubt he can see any way to refute Farah. ListenerXTalkerX 16:14, 23 October 2009 (UTC)
Yes he can, he can call him a liberal. :P SҚ_ 16:39, 23 October 2009 (UTC)
Andy is calling it a preservation of the Bible. Preserve the Bible by changing it. I love you, Andy. — Sincerely, Neveruse513 / Talk / Block 16:32, 23 October 2009 (UTC)
"We had to burnchange the villageBible in order to savepreserve it." MDB 16:50, 23 October 2009 (UTC)
Remember, Andy was wining and dining with Steve King, Michelle Bachmann, Trent Franks, Janet Folger, anti-health reform lobbyists bent on destroying it, and conservatives who instead of worrying Obama will fail, worry he'll succeed. --Crazyswordsman 16:43, 23 October 2009 (UTC)
I'm enjoying the schadenfreude so much I actually went back to re-read the article with the Chorale from the 9th playing behind it. Seeing the scare quotes around "conservative" with the choir kicking off felt almost like cocaine... :) --Robledo 17:32, 23 October 2009 (UTC)
I love it - Farrah calling Andy "Nutty" and saying, "I'm almost too embarrassed to write about this kind of trivialization and politicization of the Scriptures". I was wondering what the right would make of this. Any idea what WND's numbers are in terms of readers? Maybe we should drop a note to fellow right wing travellers like Malkin and the Coultergeist so they too can tell their followers how batshit insane Andy is. --PsygremlinSprich! 11:55, 24 October 2009 (UTC)
Insanely high, several million a day. It is comparable to MSNBC.com. - π 11:59, 24 October 2009 (UTC)
Okay the actual website is msnbc.msn.com - so not a good. Although they do fair well compared to the redirect. - π 12:02, 24 October 2009 (UTC)
I see from Alexa - they kick CP's butt. Also just dropped a mail to Michelle Malkin asking if she knows what these so-called "conservatives" are doing giving the movement a bad name. Trying to let the Coultergeist know, but I need an admin to approve my account on her forum first. (why are conservatives so goddam paranoid??) --PsygremlinRunāt! 13:10, 24 October 2009 (UTC)
### Interesting aside
WorldNetDaily only stocks the King James and the 1599 Geneva Bible. Is that unusual for King James Only or is that part of a certain demographic's mentality? - π 01:45, 24 October 2009 (UTC)
Sometimes people are KJV Only purely for aesthetic reasons, so it's not necessarily a dogmatic thing. Plus (according to WP) the Geneva Bible is very similar to the KJV. --YossieSpring in Fialta 02:07, 26 October 2009 (UTC)
## Sibling rivalry again
Holy crap, Roger goes for it again, and Andy promptly gives him a bitch slap. Will it ever end? CrundyTalk nerdy to me 22:20, 23 October 2009 (UTC)
Roger must be a masochist of the highest order. MDB 22:50, 23 October 2009 (UTC)
Wonder if they'll both be at momma Phyllis's for Thanksgiving? ГенгисOur ignorance is God; what we know is science. 22:59, 23 October 2009 (UTC)
As a lawyer I find myself puzzling over what exactly "factual information" is. I understand this to be a common wikiphrase, but it nonetheless rings hollow to me. There are far better claims to make about "information" than it being factual. "True" or "verifiable" come to mind, but I'm a newbie at this. 02:45, 24 October 2009 (UTC)
I think "factual" is Andy's pet word - if he knows it, it's a fact, whether it is true or verifiable is nitpicking. ħuman 18:20, 25 October 2009 (UTC)
## Some lame Bible Project wigo
"Yes, into all thinking you're completely hatstand"
WTF does that even mean? Please try to write these things is something resembling English. ħuman 03:31, 25 October 2009 (UTC)
It's not as bad as all that. The sentence you're looking at borrows its verb from the prior sentence - Christian denominations are united in thinking Andy's completely hatstand. Hatstand means crazy, but I agree with you that it's obscure. It ain't my WIGO but I'll change hatstand to something more clear. 03:40, 25 October 2009 (UTC)
I get the hatstand part. What I don't get is the mess of word salad before it. By the way, does it have to pay interest if if borrows a verb? OK, I see now I guess. Wow, that was really bad writing and whoever inflicted it on us should get seven whips of the goat's tail. People, it's not that hard to write clear sentences. Especially when your "message" consists of two sentences and a link. And thanks, Nutty, for clarifying. Also, PPPPPPS, I don't give a fuck "whose" wigo it is. If they suck, let's re-write them. People's got weird possessionally egoy things on here sumtimz (how's that for Ingrish!?). ħuman 04:33, 25 October 2009 (UTC)
It's not the best WIGO, but I don't see what's confusing about it. It is pretty clear (apart from the somewhat obscure 'hatstand'). Or has it been re-written since this was posted? DeltaStarSenior SysopSpeciationspeed! 06:53, 25 October 2009 (UTC)
I think it was re-written by the time you saw it, DS. ħuman 18:18, 25 October 2009 (UTC)
Can't wait for A. Schlafly's Dickens: It was the best of times - it happened to be the worst of all temporal configurations larronsicut fur in nocte 07:07, 25 October 2009 (UTC)
@LArron - Fucking funny! Hahahahahaha!......@Human - quit your fucking ranting. AceMcWicked 22:24, 25 October 2009 (UTC)
The new version is clear and coherent, thank you. I felt comfortable changing my vote to "up". ħuman 01:56, 26 October 2009 (UTC)
## Surprise
Does anyone else think that Ken's response to this made a lot more sense than usual? Broccoli 20:47, 25 October 2009 (UTC)
It was certainly better than Andy's drivel. Corry 21:48, 25 October 2009 (UTC)
It was very clear and well-put. Wasn't Ken expecting improvements in his health? ħuman 22:55, 25 October 2009 (UTC)
Didn't see the publicity. I'd have expected a brain transplant to hit the headlines. I am eating & honeychat 23:00, 25 October 2009 (UTC)
In the UK we sedate them and hide them away in care-homes. I guess that in the USA, without adequate health insurance, they just wander the streets and make a spectacle of themselves. I think the US approach gives more amusement to society in general but only at the expense of the dignity of the serially bewildered. Free-market captalism definitely loses in the humanitarian stakes. ГенгисOur ignorance is God; what we know is science. 23:13, 25 October 2009 (UTC)
dear genghis,
I have a masters in counseling and have the unfortunate privledge of working in America. In America we put them in "boarding homes," there are essentially care-homes but quality varies considerably . . . we like to sedate them too . . . and give their relatives the bill. Long time reader first time poster. Ill set up an account ASAP as greepigfoot
This whole section is brilliant - classic Andy. He is true to form. AceMcWicked 00:20, 26 October 2009 (UTC)
My absolute favorite line is "You list yourself as an "agnostic" here, yet 100% of your edits here have been of an atheistic nature. An agnostic, if fair-minded, should be at least 50% Bible-based and 50% atheistic-based." Because that's what being an agnostic means. It means that you spend half of your time being an atheist and half of your time advancing a particular strand of a particular faith. 98.206.143.163 00:32, 26 October 2009 (UTC)
Someone who is good at messing around with that quote generator thing definitely needs to add that some how (I couldn't find a place for it). Here is the diff. - π 01:20, 26 October 2009 (UTC)
Done and done. --YossieSpring in Fialta 01:38, 26 October 2009 (UTC)
I had so many things to say... let me reduce them to "fucking classic Andy". What a tool. Why Phyllis did not name him Stanley no one knows. ħuman 01:48, 26 October 2009 (UTC)
And here I thought no one had noticed this gem. I should really scroll all the way down before adding something to WIGO.TallMan 02:20, 26 October 2009 (UTC)
## Schlafly takes the bait, makes moron of self
"I wonder if it is possible for some public teachers to work less! How does someone work less than zero??" [5] Nice work, ThomasGret, if that is really your real name. ħuman 03:30, 26 October 2009 (UTC)
"ThomasGret" (your insulting name is getting tiresome), you act like liberal/conservative are symmetric with respect to the truth. That's plainly false, as illustrated by our entry on deceit. I've responded to your concerns and answered your questions and you persist in repeating the liberal claim that Obama is not an affirmative action president, so it's obvious to me you're another liberal enemy of conservative principles. You also clearly deny that that affirmative action causes rather than deters autism. You can whine all you want, but I do know quite a bit about the Irish press, Ireland, and Wikipedia. In fact, I've been quoted in the past in the Irish newspapers. Try claiming to be an "expert in 'human medicine'" in a court of law sometime. Five minutes later, which is how long it will take for everyone to stop laughing at your claim, maybe you'll wake up. Study the Bible and follow its commandments.--Aschlafly 18:57, 24 May 2013 (UTC)
## Ordered pair
Just letting you guys know that polar coordinates needn't be an ordered pair.
An ordered pair suggests that each element has a unique $a$ and $b$ for an element $(a,b)$. Owing to multiplicities of allowed angles ($0,2\pi$) in the polar formulation, this needn't be true. You can restrict the angles, but there's still an ambiguity for the point at the origin, as any angle is as good as any other for $radius = 0$. And whilst his explanation is bad, he's not necessarily wrong. Every ordered pair does correspond to a unique point on such a graph, even is that isn't its intended representation.--Star trooper man 07:48, 24 October 2009 (UTC)
Still, though, a pair of polar coordinates corresponds to one point on the plane, although one point on the plane can have infinitely many polar coordinates. --Crazyswordsman 11:14, 24 October 2009 (UTC)
This is wrong. An ordered pair is simply an element $(a,b)$ of the cartesian product $A\times B$ of two sets. The sets need not be the real numbers $\mathbb{R}$, or even subsets thereof. For instance if $A=B=\mathbb{C}$ is the set of complex numbers, $(a,b)\in A\times B$ doesn't correspond to a point on the real plane. A more abstract example would be a combinatorial structure known as a graph. So, Andy is wrong, but not for the reason given above.-Antifly 16:28, 24 October 2009 (UTC)
Whilst I agree on further reading, that I am in part wrong above, and your example is indeed illustrative (though I do know a way to put the reals and the complex numbers into 1-1 correspondence, allowing one to define a total order on the latter!), my primary quibble is that "represent" suggests, at least to me, a bijection, and thus I don't really regard polar coordinates as a legit interpretation of an ordered pair per se. Also, I'm not sure if it isn't possible to define a total order on all sets. I can't think of a case in which is isn't, but that doesn't mean such sets don't exist. Open question: Do you know a way to construct a set that isn't amenable to a total ordering?!--Star trooper man 08:25, 25 October 2009 (UTC)
I just re-read my post, and the sentence "This is wrong" is wrong. Sorry. Your post about complex numbers as ordered pairs does make sense. What is wrong is the approach taken by the WIGO. What I should have said is "The WIGO makes no sense, and the above clarifications don't address the real problem."
On to your quibble... represent may suggest that there exists a bijection to you, but the words are not interchangeable. Mathematics, perhaps more than any other discipline, requires precision of language. Polar coordinates are ordered pairs $(a,b)\in [0,\infty)\times [0,2\pi)$. It is easy to show that there is a bijection between $[0,\infty)$ and $\mathbb{R}$ and a bijection between $[0,2\pi)$ and $\mathbb{R}$. Thus there is a bijection between $[0,\infty)\times [0,2\pi)$ and $\mathbb{R}\times \mathbb{R}=\mathbb{C}$. These bijections may not preserve any of the algebraic structure imposed on these sets, but that is outside the requirements of a bijection. You are, however, right that the correspondence between pairs in $[0,\infty)\times [0,2\pi)$ and pairs in $\mathbb{R}\times \mathbb{R}$ known as "polar coordinates" is not a bijective map. As to what constitutes a "legitimate interpretation" of an ordered pair, I don't have the foggiest idea what you mean.
To your open question: see here. Of course, you must be pro-choice to accept it.
Mathematics is the second only to the Bible in logicality. Deny that and loose all credibility. Gauss-speed.-Antifly 19:10, 25 October 2009 (UTC)
The polar coordinates (I was thinking within $[0,\infty)\times [0,2\pi)$ only, which makes them ordered pairs) seemed an easy shot at Mr. Schlafly's analogy with coordinates on a Cartesian graph. Another, only slightly more arcane, example of where Mr. Schlafly went wrong there is that an ordered triple of real numbers (which requires a three-dimensional graph to represent) is, by definition, also an ordered pair. ListenerXTalkerX 05:44, 26 October 2009 (UTC)
I'm rather fond of az-el coordinates, but hey, they have some fuzziness showing up at the zenith and nadir, where who gives two hoots what the azimuth is? For all practical purposes, they point in an unmistakable direction, so again, who gives two hoots? Sprocket J Cogswell 06:47, 27 October 2009 (UTC)
## RJJensen
What do you think of him? Considering he's perhaps the best representation of a conservative scholar we're likely to find on the internet, do you not think he makes your catchall anti conservative insults a mockery in themselves? MarcusCicero 11:40, 26 October 2009 (UTC)
Jensen's takes on history might have been appropriate for a ninth-grade class thirty years ago. They're simplistic and do nothing to engage with recent trends in the scholarship. If he represents conservative intellectualism at its best, no wonder its a bankrupt ideology. BTW: How's your brother? RaoulDuke 12:39, 26 October 2009 (UTC)
TOP, you are an idiot. Your irrelevance is painful. I'd also hate to judge liberal ideology from a simplistic, frankly hysterical mind like yours. Just patter away off and attempt to do something useful with yourself you fucking dimwit. MarcusCicero 13:06, 26 October 2009 (UTC)
I see you have nothing with which to counter my argument so you take off on a baseless, mean-spirited ad hom. Typical. Tell your bro I said hi. He should come by more often. He was a nice guy. RaoulDuke 13:18, 26 October 2009 (UTC)
You are the most innane piece of shit every to grace this website, honestly. You don't say anything of value and I honestly don't expect you to have any ability to even understand Jensen's writings, never mind expect you to honestly and without bigotry read what he writes. In short, you are a scumbag, a real parasite. You don't make an 'argument', you yap on about my brother and pretty much say Jensen is an idiot - you are the fucking idiot living in a delusional state of mind. Just fuck of TOP, honestly. What the fuck do you even know about recent trends in scholarship? Your a fucking ignoramus. MarcusCicero 13:22, 26 October 2009 (UTC)
I would say he is the greatest assest Conservapedia has: an actual historian. However, I can't judge his quality as a historian, because I know nothing much about history. It seems his area of expertise is recent American history, which I know less than nothing about. The strangest thing about him is his willingness to edit CP. Broccoli 13:25, 26 October 2009 (UTC)
Instead of foaming at the mouth, why not show me where Jensen does something that critically engages with emerging trends in historical scholarship? RaoulDuke 13:28, 26 October 2009 (UTC)
RJ has quite an elastic definition of "fair use", and seems to think that "for commentary" covers using an image to comment on anything you like, when in fact it only covers commenting on the image itself.
That aside, I'm sure his scholarship itself is perfectly good, and I can't begin to fathom why he bothers to lend his talents to a batty enterprise like CP.-- Kriss AkabusiAAAWOOOGAAAR!!1 13:44, 26 October 2009 (UTC)
It looks like we might be seeing an RJ event soon. He has just begun a massive expansion of cp:Welfare state, creating a reasonable page. If Andy notices, however, then it will swiftly be trashed again. Jensen's response is often to back down and let Andy or RobS change the page to match their personal obsessions. Broccoli 14:04, 26 October 2009 (UTC)
For such a scholar his contention that Winnie created the UK welfare state to stop socialist nationalisation is debatable at least. Bob Soles 14:32, 26 October 2009 (UTC)
From what I remember of history lessons, the welfare state in the UK was first created by the Liberal party in 1906, though one motivation might have been taking the wind from the sails of the increasingly popular socialist Labour Party. At the time, I think Winston was pretending to be a Liberal, and supported the welfare reforms, but I don't think he was a driving force behind them. Broccoli 14:54, 26 October 2009 (UTC)
In it's current form it owes most to Beveridge and was mostly implemented by the post WWII Labour party. The very earliest roots were mostly down to Lloyd George. Giving Winnie the kudos is a major re-write of history. Bob Soles 15:07, 26 October 2009 (UTC)
Lloyd George didn't start it but he did give it a big push, and Winnie was Home Secretary (I think) when Lloyd George was PM so would have had a big say in it. –SuspectedReplicantretire me 15:12, 26 October 2009 (UTC) (and here is a link. I should Google before posting) –SuspectedReplicantretire me 15:13, 26 October 2009 (UTC)
But Lloyd George implemented the reforms when he was Chancellor and Winnie wasn't Home Sec until the People's Budget mess up. Lloyd George wasn't PM until 1916. Broccoli 15:50, 26 October 2009 (UTC)
Sorry, yes. That was from my increasingly-frail memory. I should have noted my mistake on the followup that got the information right. –SuspectedReplicantretire me 15:56, 26 October 2009 (UTC)
RJJ is a sad egotist. Most of the articles he creates are the absolute best on CP (where they will never be read), but he was such a pushy ass as to get kicked out of Citizendium, and now ignores all of the utter filth and insanity at CP because it gives him a place to dump articles. He's not just a "Conservative" scholar, but it would appear a reactionary who has it out for Obama. What I still can't understand is Conservativenet, which seems like only he actively contributes to, with the odd respondent. 17:25, 26 October 2009 (UTC)
Prof. Jensen says here, "anyone who want a conservative interpretation of the economics of the depression should read my articles, especially The Causes and Cures of Unemployment in the Great Depression"(1989)." Well let's see what he says. On page 20,
• "Supply-side economics was central to the New Deal. The main difference from the 1980s version was that the New Dealers worked vigorously to contract the supply of labor..."
On page 21,
• "The final, and successful supply-side remedy was the draft, and after 1940 the increase in military personnel paralleled the reduction of unemployment. The problem with the supply-side remedies was that they cut output as much as input, made the nation poorer, slowed long-term growth, distorted personal choices, and increased unemployment."
In his conclusion on page 31,
• "The refusal of the New Deal on ideological grounds to consider wage subsidies or training programs guaranteed that the hard-core would stay down, even as they hailed Roosevelt for rescuing them from utter destitution. Subsidies and training would have reduced the structural unemployment that was the greater problem after late 1933."
I'm not certain that that is a "conservative interpretation." In this 1989 article Jensen cites Ben Bernanke and several of Lawrence Summers works. In footnote 27 continued on page 23 to a discussion on Keynesian stimuli, Jensen says "My argument is congruent with Blanchard and Summers." Contrast those sentiments with User:Conservative who refers to cp:Obama#Larry Summers - Corrupt and Incompetent Chief Economic Advisor to Obama (nobody told User:Conservative Summers served on Reagan's Council of Economic Adevisors at the same time Jensen was commenting on supply side economics).
Jensen also gives special thanks to Eugene Genovese, a Marxist historian. [6] But Prof. Jensen pointed out Genovese got religion [7] at the time of the collapse of the Soviet Union, a plausible explanation. Many communists did repent and convert in the early to 90s, with the now obvious exception of Obama & company. RobS 20:05, 26 October 2009 (UTC)
I'll bite on this subject. RJJ is probably the only contributor to Conservapedia that I have an ounce of respect for. He may be batshit insanely conservative, but he, at least, makes quality contributions there. I never have understood, however, how a high quality conservative like RJ could wind up writing articles for a garbage site like Conservapedia.Gooniepunk2010 Oi! Oi! Oi! 20:10, 26 October 2009 (UTC)
Because he was kicked off or marginalized on other sites for putting a rather obvious pro-Conservative spin on his articles? Just a thought. (amusing to see that when RobS does indulge in personalities, it's always because communism can be injected into the discussion) Megaten 20:24, 26 October 2009 (UTC)
Get used to it. You have years and years ahead of this, [8] unless you wish to be tarred as an apologists for an ideology of mass murder. RobS 22:40, 26 October 2009 (UTC)
Rob, take your meds. Seriously. DickTurpis 22:51, 26 October 2009 (UTC)
Wow Rob. After reading that I just lost a lot of respect for you. "Obama wants the American economy to fail"? Give me a break. If he wanted that, he would have done nothing instead of trying to revive the economy. Tetronian you're clueless 22:53, 26 October 2009 (UTC)
Trying to revive the economy? yah right. Fed Chairman says the recession is over but 85% of the stimulus is unspent -- and we're already talking about Stimulus III. Obama spent $700 million to get elected but Congress now has$800 billion to get themselves re-elected in 2012. You think people can't see through this bullrot? My closest liberal Democratic friends keep asking, "Where's the jobs ? Where's the infrastructure improvement?" RobS 23:34, 26 October 2009 (UTC)
Rob, the idea is that the banks can use the money so they can loosen credit, revitalizing the economy. It also prevented them from falling into bankruptcy, which would have sent the economy spiraling into ruin. Let's not forget about that. Yes, it is lacking in infrastructure improvement, but the fact that it has failed to create a bazillion jobs does not make it a complete failure. And more importantly, just because it hasn't cured all of our economic problems it isn't necessarily a Commie takeover plot. Tetronian you're clueless 23:53, 26 October 2009 (UTC)
You're confusing the $700 billion bailout with the$800 billion stimulus (combined = $1.4 trillion deficit). I live in a town were they could have got$12 million in stimulus to buy new city transit buses, provided they could prove it created jobs. Legally, under NAFTA, it did since the buses were built in Canada. But it created a politcal problem locally. So they improved a few bus stops and bought automated tranfer machines. (Needless to say, they bought the buses anyway with money from elsewhere). RobS 01:20, 27 October 2009 (UTC)
Rob, Genovese, tough now somewhat dated, wrote what is in all likelihood the single most important book about American slavery. Have you even read Roll, Jordan Roll? RaoulDuke 23:56, 26 October 2009 (UTC)
No, haven't read any of his stuff. Prof. Jensen says Philip Foner's book on the ssame subject is good, too, which may the case although I had objected to using Foner as a source for CP (Foner is a Stalinist holocaust denier). RobS 01:16, 27 October 2009 (UTC)
Yeah well you know what Rob? My closest conservative friends have been asking me "Where's the Republican Health Care plan? Why doesn't the Republican budget have numbers in it? And just where were those WMDs?" How many democratic friends have you got Rob? Huh? Because I've got 100 conservative friends. Bil08 00:53, 27 October 2009 (UTC)
Yeah well where's the Democrat's tax cut proposals? Where's the Democrat's school prayer plan? Why doesn't the Democrat's plan to end the death tax hav e numbers in it? And just where is focus on the Afghan War? Where's the WMD? in Iran & North Korea (oh, I forgot, the Axis of Evil is a Bush plot in bed with the Saudi King. BTW, why did Obama bow to the Saudi King, Bush's co-conspirator in knocking down the WTC as an excuse to get us into this mess?) RobS 01:16, 27 October 2009 (UTC)
Woah Rob, no need to JAQ off now. Tetronian you're clueless 01:22, 27 October 2009 (UTC)
Hey Rob, with a respected historian on the site you should ask him if he can find you some information on the New Ordeal. - π 01:23, 27 October 2009 (UTC)
oooh, that's a sore point for Robby. Jensen totally pwned him on the "New Deal was socialist" thing a few months back. RaoulDuke 01:25, 27 October 2009 (UTC)
We've been collaborating on several articles. Actually, I pledged to Jensen to digest his article on great depression unemployment (linked above) and found Bernanke & Summers were big sources for it back in the 80s. So I've had a mass of material to digest that obviously relates to more current articles (this thread is the first time I've publicly commented on any of it) and haven't finished reading sources yet. Some material has been used to improve cp:Fascism, (you should really have a look). Also, Jensen & me did a bang-up job at cp:Scottsboro case, which has many promising and currently relevent insights we may even be able to carry over to WP. RobS 01:56, 27 October 2009 (UTC)
## BM knew from the start WIGO
I don't think Assfly Andy suggests such a thing. At least not in that diff. — Sincerely, Neveruse513 / Talk / Block 16:31, 26 October 2009 (UTC)
Mr. Schlafly said: "The British Museum raised or took money for decades by claiming that the Piltdown Man was real. In fact, it was a fraud. Do evolutionists support some level of accountability for that?" The term "accountability" generally applies only to those who actively took part in a scam, not those who have merely been suckered by it. ListenerXTalkerX 16:37, 26 October 2009 (UTC)
Is does read as though he is saying they knowlingly defrauded people. Anyone fancy burning a sock asking him to clarify? CrundyTalk nerdy to me 16:39, 26 October 2009 (UTC)
But we need all the socks we can get for the Conservapedia Day awards... Educated wisest Phantom! 16:43, 26 October 2009 (UTC)
The term "accountability" generally applies to those who have taken part in any action, intended or otherwise. e.g., you will be held accountable for involuntary manslaughter. Shitty WIGO. Just sayin'. — Sincerely, Neveruse513 / Talk / Block 16:48, 26 October 2009 (UTC)
Andrew wants his refund for the Piltdown Man, and he wants it now!! [9] ħuman 17:39, 26 October 2009 (UTC)
Er, Andy apologised for getting that wrong. I was going to sock up and point out the error myself but OurMike beat me to it.
Thanks for the additional info about the unusual policy of the British Museum. I apologize for my error. That said, there is no free lunch. If the visitors are not paying, someone else is. Taxpayers, perhaps? The underlying issue remains the same.— Andy Schlafly 09:23, 26 October 2009 (EDT)
ГенгисOur ignorance is God; what we know is science. 19:35, 26 October 2009 (UTC)
The post referenced by the WIGO was, I believe, dated after the post where Mr. Schlafly admitted he was wrong. ListenerXTalkerX 19:40, 26 October 2009 (UTC)
I want to know exactly how and where these monies should be accounted for, collected and redistributed. It's got to be a fascinating scheme. — Sincerely, Neveruse513 / Talk / Block 19:43, 26 October 2009 (UTC)
## CP block log fun
I saw some funny excerpts from the block log, thought'd I'd share them (along with some unfunny comments from me):
18:14 . . Aschlafly (Talk | contribs) blocked Hero764 (Talk | contribs) with an expiry time of 5 years (autoblock disabled) (be a real "hero" and use your real first name and last initial)
Andy....just Andy.
23:04, 25 October 2009 MarkGall (Talk | contribs) blocked WillJergens (Talk | contribs) with an expiry time of 1 day (account creation disabled) (please make only encyclopedic edits. liberals are welcome here, but they are expected to adhere to the rules governing CP.)
*Snort* Funny one Mark!
13:58, 25 October 2009 MarkGall (Talk | contribs) blocked Simaro (Talk | contribs) with an expiry time of infinite (account creation disabled) (Moronic vandalism: hilarious and original!)
Oh no, don't give yourself away Mark!
20:08, 26 October 2009 Aschlafly (Talk | contribs) blocked MegaGalvatron (Talk | contribs) with an expiry time of 5 years (account creation disabled) ("mega" - clueless)
Is that a stab at humor? Way to go Andy!
23:43, 25 October 2009 Jpatt (Talk | contribs) blocked AndrewG (Talk | contribs) with an expiry time of 5 years (account creation disabled, e-mail blocked) (Negative personal comments: Spiteful wittle liberal)
Typo or intentional? We may never know!
If you see any more good ones, please add. Tetronian you're clueless 00:53, 27 October 2009 (UTC)
## Andy's idea of what's Biblically relevant.
I noticed another nugget of Andy's conservative insight in the middle of this comment.
"Taking one, I feel the term "Pharisees" means nothing to people today, and their biblical views seem irrelevant."
The Pharisees and their Biblical views were irrelevant to the story of Jesus? I can't wait for Andy to offer a Bible study class when he's done with Economics.
He also didn't seem to have much luck replacing "Pharisee" with "intellectual" or "liberal elite", so now it seems that they are the close-minded "incumbents" who needed to be replaced. Yeah, I'm sure you'd be calling your bad guys "incumbents" if this was 2007 and not 2009, right Andy? I'm not sure it's WIGO-worthy, but it was worth calling out IMHO. --SpinyNorman 15:40, 22 October 2009 (UTC)
It's interesting to watch the clash between Andy, who really just wanted to go through the KJV, update the language, add conservative bias and remove anything he disagreed with, and Terry, who seems to be going for the New Poorly Translated Bible. Looks like Terry is winning at the moment, since Andy is out of his depth with all that Greek. Broccoli 16:06, 22 October 2009 (UTC)
It's a bit like when the project first started a few months ago, someone called GloriaM started pasting in the Hebrew text of the Book of Esther and asking specific questions about that. Andy soon put the project to one side until she (or he) went away. Sic transit Gloria M.-- Kriss AkabusiAAAWOOOGAAAR!!1 16:25, 22 October 2009 (UTC)
Oh, that was clever. Nicely done. Corry 00:44, 23 October 2009 (UTC)
I notice that Andy includes some Greek words in his replies, but only ones that he could have copied from a prior post. Lily Inspirate me. 20:09, 22 October 2009 (UTC)
Oh for the love of goat. I think TerryH doesn't know how to decline Greek nouns. Does anyone else agree? 15:35, 23 October 2009 (UTC)
No! Because I can't decline Greek nouns either. I'd just have to ask my funky chunky Greek girl friend (and I don't mean girlfriend) Nicoletta to do it for me. ГенгисOur ignorance is God; what we know is science. 18:33, 25 October 2009 (UTC)
I forget where I put the diff, but I think Terry said he has a year of Greek language study under his belt. I wonder if when he said he could read it as fast as english if he simply meant vocalize, not comprehend. — Sincerely, Neveruse513 / Talk / Block 19:57, 27 October 2009 (UTC)
## How on earth did you all...
...miss this? EddyP 15:18, 26 October 2009 (UTC)
As regards the whole wine/grape juice thing - it has always got to me that, if the 'wine' at the marriage in Cana was grape juice then why the fuss about serving the best last. It's alcoholic wine where you serve the best first so that
(a)the palate is not ruined and
(b)the tasters are too drunk to tell the difference when you serve the vinegar.
Hey, someone of my vintage knows just how palatable flat Watneys Red barrel (complete with floating dog ends) can be when you're drunk enough. Bob Soles 15:30, 26 October 2009 (UTC)
Also, neither Noah nor Lot became drunk on grape juice! ListenerXTalkerX 15:35, 26 October 2009 (UTC)
I wonder why fuckwit TerryH thinks the new grape juice will burst the old bottles (Mat 9:17, Mar 2:22, Luk 5:37). — Sincerely, Neveruse513 / Talk / Block 15:39, 26 October 2009 (UTC)
As good and phenomenally self-important as this letter is, it somehow lacks a certain something. In other words, I wish Andy had written it instead.-- Kriss AkabusiAAAWOOOGAAAR!!1 15:41, 26 October 2009 (UTC)
I'm surprised that Terry took it upon himself to write it. I like the opening: "As the administrator-in-charge of the Conservapedia Conservative Bible Project". Wasn't that title self-issued when he just stormed in and said "Right, this is my project now"? CrundyTalk nerdy to me 16:02, 26 October 2009 (UTC)
I note that TerryH asks Farah
First, where were you when the New International Version came out?
Well, I can answer that for Messer Farah: The NIV edition of the New Testament was first released in 1973, with the whole Bible following in 1978. Joseh Farah was born in 1954, so he was in his early twenties when the NIV was released, and hardly in a position to comment on anything. MDB 17:18, 26 October 2009 (UTC)
Ah, I suppose its possible TerryH meant Today's NIV, which is a much more recent release, and is quite disliked by conservative Christians. MDB 17:24, 26 October 2009 (UTC)
Might comment on the essay/letter itself later (first wanna shower and then eat dinner), but just a quick comment about the NIV/TNIV issue: You make a good point about him being fairly young and thus not really being in a position to comment in a public way (especially in the pre-Internet age). If Terry meant the TNIV, we can answer this question more easily: Farah was at WND, writing "Today's New International Perversion" --Sid 17:57, 26 October 2009 (UTC)
I take it TerryH does not partake of the Blood of Christ during communion131.107.0.85 18:12, 26 October 2009 (UTC)
I'm a little confused as to your point, but I think you're assuming all Christian churches use wine for communion. That is not true; many Protestant churches use grape juice -- some because they think consuming alcohol is a sin, and at least one denomination (my own Metropolitan Community Church) to honor any members who might be in recovery. (And as a humorous aside, I know of at least two MCC's, including my own, that use white grape, because it doesn't stain.) MDB 18:41, 26 October 2009 (UTC)
Remind me what the difference is between the American Fundamentalists and the Taliban, again? (edit: Oh, yeah--Bacon.) --Gulik 07:43, 27 October 2009 (UTC)
Holy shit!! This is amazing. I'm hoping Farah will reply and this will turn into another Lenski affair. It is interesting that Terry started this, not Andy - he's playing a power game, I presume?
### "What are you laughing at?"
I didn't bother reading any of that - my short attention span is getting the better of me these days - but I saw the first line and thought "I wonder if it was Andy or Terry that wrote this?" Lo and behold, it was Terry. I started laughing at this, forgetting that all my housemates were in my room at the time, prompting questions as to what I was laughing at. Jeez, TerryH is delusional... SJ Debaser 21:25, 26 October 2009 (UTC)
From my remote, and admittedly-informed-largely-by-teh-internetz point of view, this sums up the American Conservatives right now. There have been one or two blog posts suggesting the Republican Party is getting worried about its own base, and this is a perfect illustration: utter conviction in one's own POV; refusal to engage with most of a debate and focussing on only one or two points; etc etc. From what I've read elsewhere about the Bible Andyfication Project, comments fall broadly into two groups: "Liberals" laugh; "Conservatives" scream hubris. Both are right. If you want to put your own spin on the bible, you could make it say anything you want. "Rich man" -> "Idle rich" well... why not "Celtic and Rangers supporters"? Supporters of those clubs are rich in that one of those two clubs has won about 95% of Scottish Football League titles - pity the poor supporters of East Stirlingshire.
I'm looking forward to discussing CP's latest project with my god-bothering relatives this Christmas. Should be fun. –SuspectedReplicantretire me 21:42, 26 October 2009 (UTC)
"TerryH" (if that is your real name), I've skimmed through your rant and I find it has a low substance to words ratio, which means you are clearly a product of professor values, public schooling and no doubt are for gun control and against prayer in school too. In addition my MarkGall's LiberalBot tells me you have a liberality index of 3.2, which places you alongside such liberals as Richard Dawkins. Open your mind and realise that more people are laughng at the CBP (yes, Terry the "rants" and "screams" you heard were of laughter) than think autumn foliage is pretty or that 2=2=4. Please continue to contribute as you have because a) we enjoy a good laugh and b) we look forward to God proving his existence by smiting your mangy ass. Gods peed. - Andypants.
### Watch this space
Emails to the editor at WND are non-permanent, so if he gets posted someone will have to grab a screenshot. - π 01:34, 27 October 2009 (UTC)
I'm actually tempted to play devil's advocate and ask Farah to reply to Terry.--Thanatos 01:24, 28 October 2009 (UTC)
## The Andy vs Roger debate
Has a score of over 300. It is now the highest scoring WIGO we have ever had. ĴάΛäšςǍ₰ is out of his mind 02:04, 27 October 2009 (UTC)
Yes. - π 01:37, 27 October 2009 (UTC)
Can I take partial credit and get hawtsecks from blindingly hot birds? CrundyTalk nerdy to me 01:38, 27 October 2009 (UTC)
Yes, as long as I can too. Tetronian you're clueless 02:03, 27 October 2009 (UTC)
r0x0r CrundyTalk nerdy to me 10:12, 27 October 2009 (UTC)
Anybody caught having hawtsecks with anybody whilst a member of RW, will be disbarred for the sin of having a life. Missionary position, with the lights off, is allowable, however. --Psygremlin말하십시오 17:30, 27 October 2009 (UTC)
-- But only through he hole in the sheet and procreation purposes Alain 18:54, 27 October 2009 (UTC)
## Newspaper Circulation
This isn't worth a WIGO, as far as I'm concerned (or at least I can't come up with a snarky way to phrase it), but the latest news from CP is that newspaper circulation is down -- way down. Andy, ever desiring to be "helpful", suggests its for "time them to take a break from printing their liberal claptrap."
Of course, Andy regurgitates the standard conserva-line that newspaper circulation would be just fine if they weren't so gol-durn liberal. Ironic, of course, that he's posting that on the Internet, since pretty much everyone who doesn't have right wing blinders obscuring their vision will tell you that's the real reason newspaper circulation is down, because:
• Every newspaper of any standing whatsoever posts almost all of their content on-line.
• On-line content is more timely than a print edition.
• Classified advertising has moved almost entirely on-line. Job listings go to monster.com and the like, for sale ads go to eBay and craigslist, personal ads go to any number of web sites, etc etc etc ad infinitum ad nauseam. (And classified ads were the bread and butter of the news industry for ages.)
Of course, Andy actually seeing that's the reason for newspaper circulation being down would require analysis beyond "they're big bad libb-burr-ulls!", and the chance of Andy doing that are about the same as Matt Damon and Ben Affleck inviting me to a three-way. — Unsigned, by: MDB / talk / contribs
Yes I saw that one and reacted the same way you did. His libb-burr-ull obsession is reaching new heights. Andy now goes straight to "Blame the liberals" without even bothering to try to find a link. Maybe his paranoia will give RobS a run for his money in the awards? –SuspectedReplicantretire me 11:55, 27 October 2009 (UTC)
Funny I saw Murdoch the Younger on TV the other night blaming the BBC and ABC (Australia) for hurting the newspapers because they were giving away high quality news on the internet for free. - π 12:00, 27 October 2009 (UTC)
I kind of have to agree with that analysis; the decline in newspaper sales is a complex issue, far too complex for someone who has repeatedly proved that their worldview consists of nothing more than "libruls are baaad". Indeed, it's a very complicated issue; circulation is certainly not as bad in Europe as it is the US at the moment, but I wouldn't suggest for a second that Europe uses less internet so that explains it - there's a far more complex set of attitudes to print media combined with the effect of the recent recession at work (Micheal Moore made some recent comments that popped up on one of the WIGOs recently regarding the fact that US newspapers really stopped being about publishing news and more about publishing adverts). Certainly, as Schlafly's view usually states that Europe is full of commie liberals, his explanation that liberal clap-trap is responsible for a decline in sales certainly doesn't hold much water at all. postate 12:11, 27 October 2009 (UTC)
With Murdoch charging for online access to the WSJ I expect other newspapers to do the same. However I reckon that gadgets like Amazon's Kindle will be used to download newspapers on a subscription basis so that people aren't tied to a pc in order to read it. ГенгисOur ignorance is God; what we know is science. 20:40, 27 October 2009 (UTC)
### Some circulation figures for comparison
Ah, here we go.... Washington, DC is one of the few two newspaper cities in the country -- it has the somewhat liberal Washington Post, and the Moonie-run Washington Times, which might as well be "Fox News -- the print edition." (There's no affiliation to Faux News, its just the same politics.) This story is about eighteen months old, but its from the Times itself, and its the most recent one I could find. It reports that the Post's circulation was down 3.5%, the Times' was down 6.5%. So, the conservative paper lost more circulation. And the Post outsells the Times by a factor of over 6-1.
In all fairness, the DC area isn't a very fair choice to analyze press bias effects on circulation, since it includes the Maryland suburbs (generally liberal), Northern Virginia (kinda liberal, less so the further away you get from DC) and DC itself, which rivals Massachusetts as the most liberal "state". MDB 13:15, 27 October 2009 (UTC)
I think I saw recently figures showing that t'internet was on the way to being the biggest advertising medium. Commercial TV & print media are really suffering from the revenue loss. The free sheets aren't helping either - taking their percentage of the ad revenue that does remain. We've just got a freesat box & the news channels are quite surprising: French, Russian, CNN and Al Jazira[sp?] among others. It's not any Liberal attitude, in fact the ones with (liberal?) Celeb gossip, & (liberal?) (half) nekkid grils are doing better than the broadsheets by and large. I am eating & honeychat 16:11, 27 October 2009 (UTC)
For what its worth, in DC, one of the two free dailies is owned by the Washington Post. (And from what I've seen, its worth every penny you pay for it.) Plus, there's several free weeklies, both general interest (the City Paper -- well, general interest targeted to a younger, hipper crowd) and targeted (like the gay Blade and Metro Weekly, and one targeted at the African-American audience.) MDB 11:35, 28 October 2009 (UTC)
## Featured insight
Woo! Andy's Best Conservative Words essay is this week's featured article.-- Kriss AkabusiAAAWOOOGAAAR!!1 12:29, 27 October 2009 (UTC)
Figures. Like everyone else, Andy has no problem engaging in shameless self promotion. And the other editors at CP are probably too scared of the banhammer to question Andy's bullshit. Tetronian you're clueless 19:33, 27 October 2009 (UTC)
## How many minutes does it take....
.... to get a reference to the FBI investigation deleted?
"Let's find out. One. Two-who! Three. Four! Five." #smack#img "Five."
Advice to would-be cp trolls: pick a user name that doesn't scream "troublemaker". "SaveMeJebus" is not a good choice. MDB 16:06, 27 October 2009 (UTC)
And its down to a minute!img (The "TitsMcGee" name was a dead giveaway.)
Perhaps "banning wiki-vandals" could be an Olympic event. Too bad CP opposed the Chicago Olympics -- might have made it easier for them to send a team. MDB 16:12, 27 October 2009 (UTC)
Advice to would-be cp trolls: don't bother, it's really really old by now.-- Kriss AkabusiAAAWOOOGAAAR!!1 16:45, 27 October 2009 (UTC)
Oh, come on... it's a classic! It's one of very few instances where Andy actually knows he was wrong and can't (and won't ever, ever, ever try to) defend himself. I can imagine twisting the thorn in his side. — Sincerely, Neveruse513 / Talk / Block 16:48, 27 October 2009 (UTC)
Yes, but classics can die. postate 18:10, 27 October 2009 (UTC)
## RobS and that naughty Obama administration (in 2007)
Outstanding WIGO! However, once the geniuses over there realise that the fecking article is from 2007 (it even says so in the URL!) when the messiah was in office, do you think it will be reworded to "deceitful liberals attack Georgie?) DeltaStarSenior SysopSpeciationspeed! 19:10, 27 October 2009 (UTC)
Well, it was finally deleted (by Karajou), not re-worded. But still, this demonstrates how completely bankrupt of ideas the American right has become. When it was thought to be about Obama... "ZOMG! Obama's domestic spying program is under attack!" But when it turns out to be about Bush... crickets chirping. This just demonstrates the validity of the question I asked to ocnservatives during the original debate over domestic spying: "would you be comfortable with a Democratic administration having those powers?" The response was generally hemming and hawing. MDB 23:01, 27 October 2009 (UTC)
## Fetal rights (WIGO?)
dunno... check it out.
Doomed User: You can't do that.
JPratt: OVER RULED! He's a new user, so I'll lie on his behalf.
Doomed User: You can't do that either.
JPratt bans for interfering with admin instruction.
Pretty standard fare, but if someone wants to snark it up (MOAR) and format it, go ahead. — Sincerely, Neveruse513 / Talk / Block 19:50, 27 October 2009 (UTC)
I Googled "Memeshock" because I'd never heard of it before, and the first result to come up was a facebook page. One of the contributors to that page is a Mr John Patti. Anyone we know? Johann 22:09, 27 October 2009 (UTC)
For the record, I am not a creationist or anything else like that. But for Straussian reasons I find it useful to spread my memetics there. ~Johanan Raatz, MemeShock founder — Unsigned, by: 174.102.148.158 / talk / contribs
Dearest Jpatt, how ever is he fairing? --User:Theemperor/Sig 22:10, 27 October 2009 (UTC)
Wait...doesn't Conservapedia not allow copy-paste from evil Wikipedia? We are at war with Eastasia. --Crazyswordsman 22:24, 27 October 2009 (UTC)
In defense of Jpatt, the guy was clearly a parodist himself, adding crap about witchcraft in Harry Potter and suggesting that anyone with a liberalism score above some cutoff should be automatically banned. He had it coming. --Pyfgcr 03:15, 28 October 2009 (UTC)
Re. "copypasting from WP": The commandment does have an "unless it was your original work" exception there. This of course weakens their "Wikipedia is full of liberal bias!!!" claim, but keep in mind that Trusted Sysops Admins Senior Administrators will scan your every edit and remove any sanity liberal bias, so it all evens out in the end. --Sid 09:50, 28 October 2009 (UTC)
## How about andy and the evilutionsists don't read the bible gambit
Andy: Do you read the bible? Reply: Yes Andy: I doubt that because evilutionists don't read the bible. Looks like a check and a mate for Andy there. --Opcn 20:10, 27 October 2009 (UTC)
Already WIGO'd. ListenerXTalkerX 20:12, 27 October 2009 (UTC)
"You list yourself as an "agnostic" here, yet 100% of your edits here have been of an atheistic nature. An agnostic, if fair-minded, should be at least 50% Bible-based and 50% atheistic-based." Wow...just...wow.--Mustex 22:42, 27 October 2009 (UTC)
...as well as 50% Hindu-based, 50% Buddhist, 50% Satanic, 50% Mormon, 50% alien civilizationy, 50% Muslim 50% Voodoo etc etc. That's why every single one of my posts is several hundred pages long. Etc 06:58, 28 October 2009 (UTC)
It'll just get confusing if one of those 50% borders happens to be in the middle of a sentence: "God doesn't exist, so *twitch* God exists." --Sid 09:54, 28 October 2009 (UTC)
## CP is again not accepting new members
Is it just me, or does than not seem to affect the number of things that top CP brass reverts? --Opcn 21:22, 27 October 2009 (UTC)
I think user creation turns off when night mode is on. Educated wisest Phantom! 21:31, 27 October 2009 (UTC)
It goes on & off more often than a whore's drawers. I am eating & honeychat 21:37, 27 October 2009 (UTC)
Well everything shuts off at night, I was thinking about right now, it is shut off now yet still they revert. Why even bother to shut it off?
It is not night, it is only 18:27 on the server clock, night mode doesn't start until 00:00. - π 22:28, 27 October 2009 (UTC)
## Honestly, I wouldn't normally read this rag.......
....but rules 6 & 8 struck me as amusing [10] Mick McT 07:54, 28 October 2009 (UTC)
Again I am eating & honeychat 08:47, 28 October 2009 (UTC)
## How many minutes will it take before BMpC gets banned?
Are there many other talk pages on CP that trashes CP that much ? Who's burning that perfectly good sock??? Maybe BMpC have not reached the legendary 90/10 ratio yet? But he received a FINAL WARNING from our fiend TK in june...
I believe that the Conservative Bible Project is motivated purely by politics and ideology and an attempt to take what is seen by millions as nothing less than sacred holy writ, and force it to conform to a subset of American social conservatism. Much of the wording will use "conservative terms", deemed as such simply based on personal opinion. This is an act of supreme hubris and egotism. Alain 18:52, 27 October 2009 (UTC)
Dissent is de jure allowed on Conservapedia, but it puts you on shaky ground with the brass. ListenerXTalkerX 18:42, 27 October 2009 (UTC)
So as long as someone corrects Aschlafly's typos, he gets a freeride on his talkpage? Alain 18:52, 27 October 2009 (UTC)
Sign your posts by placing four tildes after them. ListenerXTalkerX 18:50, 27 October 2009 (UTC)
Dang, LX, I thought that was a template. Nominated for "most boring talk page post ever"? ħuman 06:25, 28 October 2009 (UTC)
Alain signed his posts after I posted that, replacing two unsignatures. ListenerXTalkerX 06:37, 28 October 2009 (UTC)
Is that a bad thing? — Sincerely, Neveruse513 / Talk / Block 18:06, 28 October 2009 (UTC)
It made my instruction to sign seem out of place. ListenerXTalkerX 18:10, 28 October 2009 (UTC)
Yeah, but other than that... would you consider replacing unsigs a faux pas? — Sincerely, Neveruse513 / Talk / Block 18:12, 28 October 2009 (UTC)
Maybe a very mild one, at worst. The timestamp ends up wrong, but most unsigs aren't stamped anyway. If nothing else it shows good faith, in that the "replacer", especially if they're a newcomer, is expressing that they know how now. ħuman 20:14, 28 October 2009 (UTC)
It always bothered me when I innocently forgot to sign a post and some wank ECs me with an unsig while I'm trying to sign. And I think it makes the poster look like they were trying to hide something. — Sincerely, Neveruse513 / Talk / Block 20:31, 28 October 2009 (UTC)
## Direct line to heaven?
Quite a claim there TerryH. Jammy 22:35, 27 October 2009 (UTC)
Yes it is, but it is the norm for bibleheads. Because we are so used to it, we often forget how ridiculous this is. We are laughing at their bible project, but we must remember that the people actually think an invisible being in the sky SPOKE to blokes on earth who took a dictation. Crazy. DeltaStarSenior SysopSpeciationspeed! 02:20, 28 October 2009 (UTC)
I think they are confusing reality to Indiana Jones and the Raiders of the lost ark.--Thanatos 04:18, 28 October 2009 (UTC)
They're translating Genesis? How did I miss that? *reads* Oh God. The iconic "Let there be light" is now "Light shall exist!"? *headdesks* --Sid 10:03, 28 October 2009 (UTC)
The more I read of it, the worse it seems to get. I happened across their Revelation "translation" this morning and see that of the four horsemen, Conquest and Famine have been replaced by Dictatorship and "economic depression so severe that it takes a full day's work to earn a day's meals". Admittedly that's in the analysis rather than the translation, but it's still priceless. I also like the way that wine is banished even from the "measure of wheat for a penny" phrase. –SuspectedReplicantretire me 10:11, 28 October 2009 (UTC)
I chuckled at la Quinlan's "some cake" comparison. Reading CPs dumbing-down of the Bible gives me enormous pleasure, a warm fuzzy feeling inside: it is, of course, smug self-satisfaction, derived from being able to say, "See? I told you so!" Interestingly, the last similar attempt which springs to mind to reinvent Christianity in order to make it conform to the values of the time and place was in 1930s Nazi Germany. Fox 11:57, 28 October 2009 (UTC)
After reading more of the "translations" I have to chuckle at Andy's stupidity. He really thinks that after he has mangled the text it still qualifies as the literal word of God. Also: look at Genesis 1:8. Prime example of dumbing down and ruining the one good thing about the Bible, its poetry. Tetronian you're clueless 21:07, 28 October 2009 (UTC)
The whole CBTP project makes me laugh because Andy "knows" what the Bible says and what God/Jesus meant. So why bother even re-translating the current works when he could just write what he "knows" and call that the "New Connservative Bible". ГенгисOur ignorance is God; what we know is science. 23:49, 28 October 2009 (UTC)
## TK and Scientology
These two edits make it appear as if TK has had an epiphany and converted to the One True Religion. Does this revoke Ed's title as the loony cultist on CP? --User:Theemperor/Sig 23:58, 27 October 2009 (UTC)
Probably. How can anyone, especially so-called Christian conservatives, defend that cult? It boggles the mind. Granted, YEC isn't much better. Gooniepunk2010 Oi! Oi! Oi! 00:03, 28 October 2009 (UTC)
It's odd how tolerant Andy is of Scientology, Moonies, Jehovas Witness, Christian Science, Mormans; especially given that many Evangelical Christians regard these as non-Christian cults. Also, Scientology is especially interested in Hollywood celebs (Tom Hanks, John Tavolta); doesn't this bother Andy?--Simple 00:13, 28 October 2009 (UTC)
When everyone's evil for some reason, you have to pick your battles.--Tom Moorefiat justitia 00:21, 28 October 2009 (UTC)
Actually, Daniel1212 made a lot of Unification Church friendly edits recently. Is he a Moonie along with Poor? MDB 00:29, 28 October 2009 (UTC)
Hmmm, maybe that's the "surgery" TK was having? A CoS implant? He sure has been absent a lot from class lately. ħuman 00:33, 28 October 2009 (UTC)
Some of Joaquin's follow-up edits, mere minutes later, are intriguing too. Dreaded Walrus t c 00:49, 28 October 2009 (UTC)
And another one. Considering the article was at this stage up until a couple of days ago, perhaps TK and Joaquin have been in contact off-wiki about it? Dreaded Walrus t c 00:55, 28 October 2009 (UTC)
TK is Catholic. 00:57, 28 October 2009 (UTC)
When you are a religious nut you have to protect religious nuttery. Despite the fact that Andy commits blasphemy against the Muslims he blocked me as Opcn for good after I suggested that he should put a news item up about what the league of Muslim nations was doing in the U.N.. --Opcn 02:47, 28 October 2009 (UTC)
I've been experimenting with their predilection for insta-blocking any Arabic sounding name, sometimes with "use your real first name and last initial" and account creation disabled, sometimes with "troublemaker/prevaricator." Disgraceful. 16:44, 28 October 2009 (UTC)
Hmm, I wonder if other ethnic names will get the same response. Tetronian you're clueless 21:00, 28 October 2009 (UTC)
Daniel1212 has just chipped in on this question. ListenerXTalkerX 04:45, 29 October 2009 (UTC)
## Apology WIGO
Andy wasn't the one apologizing. --SpinyNorman 22:47, 28 October 2009 (UTC)
This is embarrassingly wrong. Would fix it but I'm on my phone. 22:47, 28 October 2009 (UTC)
## Blatant vandalism
Has anyone seen thisimg? That vandalism template is brill.-- Kriss AkabusiAAAWOOOGAAAR!!1 12:13, 29 October 2009 (UTC)
Wow, how did we not notice this? It's been around since 2007 when PJR put it there. Tetronian you're clueless 12:17, 29 October 2009 (UTC)
Obviously if they were built in 3000 BC it would have been washed over in the flood. - π 12:29, 29 October 2009 (UTC)
There are so many pages with vandalism on them it might be easier for CP to have a "We haven't found any vandalism on this page yet" template. Of course, much of it is far too subtle for their little brains to find, like on this pageimg.MDB 12:48, 29 October 2009 (UTC)
That would be funnier if it constantly quoted people through out it. - π 12:52, 29 October 2009 (UTC)
## TerryH has seen God ruin people.
Thisimg might be the start of a very interesting discussion. Apparently, God directly authorized the Holocaust and 9/11, and Terry has seen God doing His work to fuck up lives. Oh, and good news for Computer Scientists: There is no true randomness in the real world, so don't worry if your RNG is not perfect.
And yes, God authorized this post, too. --Sid 13:50, 29 October 2009 (UTC)
It's odd that I'm inured to the epic hilarity of statements like this on AWK. but always pleasantly surprised to see them on CP : "Recall that the average life span of the descendants of Adam was about 900 years. After the Great Flood, man's life span dropped ninety percent, and in a hurry. We must assume that the life spans of animals varied in similar proportion." 14:14, 29 October 2009 (UTC)
Wasn't sure where to put this. I've seen comments about Conservapedia at the saloon asked to be put over here, but this is not about WIGO, :CP or otherwise.
Rather, just saying that last night I had a dream concerning Conservapedia (among many other things), and I am providing a Humorous Aside for people here. To deflect accusations of obsession, I dream pretty frequently and this is my second Conservapedia dream in about three years of visiting this site (I've had roughly 10x as many dreams about fighting zombies). I can't remember the first one that clearly, only that I had it.
I have to admit I was rather disappointed in my behavior in the dream. I like to think that, were I to meet and confront Schlafly in real life, I would be able to maintain a fairly intelligent level of discourse with him. In the dream, however, my meeting with him went somewhat poorly - we were mainly shouting nonsense at each other while we walked down the street, with notable contributions from me including "2+2=1!" and "dicks!". In my dream-self defense, dream Andrew Schlafly was a rather imposing figure, standing at about 6 and a half feet and wearing what appeared to be two fabulous fur coats.
Sometime later I apparently was "undercover" as part of his homeschool class (a physical visitor, no less), trying to find a way to expose his deceit while avoiding some uncertain retribution I was expecting for doing so. At one point I believe I yelled "you lying shit!" at him, only to explain (apparently convincingly) that I was in fact addressing someone behind him who had just passed out of view. In another instance, I believe I trapped Jinxmchue in an upside-down wooden crate.
I can't say I recall much else about the dream, except that the warehouse the class was being conducted in had several blue vans parked off to the side, and I was woken shortly after each van unfolded into approximately ten heavily armed Terminators each. Megaten 04:30, 28 October 2009 (UTC)
My CP dreams usually ends up with fistfights between Andy (9-0-0) and myself or Myself, Glenn Beck and Hannity (3-2-1) on The Factor--Thanatos 04:37, 28 October 2009 (UTC)
You put it in the right place. I don't dream about CP, although I do "wiki-dream" about this trash heap. I dram a lot, especially when I don't drink enough to sleep through 'til, um, the crack of noon. Seriously, when I sleep lightly, I swear I dream sixteen hours of life in eight hours (or so) of restless sleep. Oh, yeah, but no one asked, so sorry for sharing ;) ħuman 06:23, 28 October 2009 (UTC)
I've had one dream about CP. Andy and I were having a moderated debate (just the two of us and a nameless moderator) but we ended up not debating much, more just bitching about TK. We ended up good friends. Very worrying.
As for sixteen hours life in an eight hour dream, I know exactly what you mean. Way, way back when I was at school, one of my classmates told us of a way to make yourself unconscious by hyperventilating and then trying to blow air out of your nose while it was pinched shut (I've since heard this is really dangerous so I'm not recommending it). I was the first one to try it. When I woke up I was really scared because I'd had three full-length dreams that seemed to last for several hours, and I and assumed I had been out for a long time. In fact it was about four seconds. The brain can do weird shit when it wants. –SuspectedReplicantretire me 07:30, 28 October 2009 (UTC)
Yeah, we used to do that with the sleeper hold. Crazy ass dreams in four seconds. It's dangerous, but worth the risk, so I recommend you and your friends try it at home. — Sincerely, Neveruse513 / Talk / Block 14:31, 28 October 2009 (UTC)
Meh, wikipedia says the sleeper hold is safe. Oh well. Still, try it. — Sincerely, Neveruse513 / Talk / Block 14:36, 28 October 2009 (UTC)
Sleeper hold is safe so long as they let go immediately after you start to go under, hold on for an extra 12 seconds and they are definitely dead. I don't see how hyperventalating and holding your nose could do damage with out repeated use. When you hyperventalate you get rid of the CO2 in your blood and the urge to breathe is controlled by the acidity of your blood, no CO2 means alkaline blood means less urge to breathe. --Opcn 16:40, 28 October 2009 (UTC)
It was something to do with bursting blood vessels in the brain, I think. I was 13 - I went along with what others said; it's quite possible the whole thing is safe. We did use the sleeper hold too, but it was the whole forcing-blood-to-the-brain thing that we understood to be dangerous. Of course, one advantage of growing older as an ex-smoker is that jogging up a flight of stairs can reproduce the same effect. –SuspectedReplicantretire me 22:11, 28 October 2009 (UTC)
Oh goody - now I can out the CP dream I had last night. Andy was holding a CP get together and sent me (or rather my sock) a plane ticket to attend. Now obviously I couldn't impersonate my sock in the flesh, so I was running around trying to find somebody to sock in real life for my sock. Very odd. --PsygremlinSermā! 17:11, 29 October 2009 (UTC)
## Apology for WIGO
I think Tetronian should give one. EddyP 22:54, 28 October 2009 (UTC)
My bad. I knew what I meant but had a brain fart and typed something else. Tetronian you're clueless 23:02, 28 October 2009 (UTC)
This was a joke section based on two 'Apology WIGO' sections created simaltaneously above it, but because the've been merged the little humour this section had has been lost. Never mind. EddyP 11:45, 29 October 2009 (UTC)
Really? It always seemed funny to me... sorry if I broke your joke, though. You can always reassemble it if you want. ħuman 20:03, 29 October 2009 (UTC)
Oh, I get it. Tetronian you're clueless 12:15, 29 October 2009 (UTC)
## Brokeback Mount Ararat?
I was just reading through the ConservaBible translations TerryH had been doing on Genesis, and stumbled on his reinterpretation of Genesis 9:22 :
"Ham, the father of Canaan, took intimate liberties with his father, and told his two brothers outside."
So in one bold editing pass by the self-proclaimed Bible Translation Expert, we've gone from an adult son simply seeing his drunk dad naked, to having "intimate relations" with him? Words fail. --SpinyNorman 01:16, 29 October 2009 (UTC)
Look. There are only so many words in his Hebrew pocket dictionary so he has to take some liberties here and there. Where he goes with them says more about his tendencies than what the bible actually says, nu? 01:23, 29 October 2009 (UTC)
Intimate liberties? Maybe that is what the original Greek text does say, and every other translation had it toned down, or wrong. Or not. Aboriginal Noise Theist, barely hanging by a nail 01:53, 29 October 2009 (UTC)
Errr, Greek? Naa son. The Hebrews wrote that shit in Jewish. TerryH is taking a correspondence class in it so he's qualified to spearhead a biblical translation team now. 01:55, 29 October 2009 (UTC)
Isn't that passage largely considered a later addition so the Israelites could justify acting like dicks to the Canaanites? - π 03:41, 29 October 2009 (UTC)
I for one am disgusted by the concept. I like my dad, and all, but still... ħuman 03:47, 29 October 2009 (UTC)
The Hebrews wrote that shit in Jewish. Do you not have that the wrong way around? ListenerXTalkerX 04:43, 29 October 2009 (UTC)
What's the dif? 05:16, 29 October 2009 (UTC)
I seriously hope you are joking. - π 05:26, 29 October 2009 (UTC)
(EC) What is "Jewish"? Hebrew? Aramaic? Yiddish? And, biblically speaking, are not only some Hebrews Jewish? ListenerXTalkerX 05:28, 29 October 2009 (UTC)
RACISTS. 05:46, 29 October 2009 (UTC)
I always thought all "Hebes" were Arabs. Color me wrong if not such is true. ħuman 05:55, 29 October 2009 (UTC)
Arabs and Hebrews are Semites, but Hebrews are not Arabs. Separate ethnic groups. Now try figuring out whether modern Jews can call themselves an ethnic group anymore, or if they're really a "civilization" like Mordecai Kaplan thought. 14:20, 29 October 2009 (UTC)
Oh shit, I might be a racist? Dang. ħuman 05:56, 29 October 2009 (UTC)
In all fairness, this is a very mainstream interpretation of this passage. To know nakedness is used elsewhere as a metaphor for sex more clearly, so it's not a leap.--Tom Moorefiat justitia 08:44, 29 October 2009 (UTC)
According to wikipedia, that's one of two Talmudic explanations, the other being that Ham castrated Noah. Of course, cp would go with the one that meant "Ham got cursed because he was gay!" MDB 12:10, 29 October 2009 (UTC)
The real point here is that TerryH is not translating at all: he is simply using the assumptions of others to reach a conclusion which is necessary to explain Kena'an's punishment. As Pleins points out (Pleins, J. David. "When the Great Abyss Opened: Classic and Contemporary Readings of Noah's Flood" (OUP USA; 2003) ISBN 0 1951 5608 0) "The similarity in language may be coincidental, and we may be guilty of reading more into the story than was intended." After all, the (similar) use of language in Vayikra that people look to as implying Ham "committed incest with his mother" is quite specifically - in the original Hebrew - different to that used in B'resheet - only once translated can they be compared! Far more likely is that the story should simply be read as it stands: Noach gets pissed and crashes out naked in his tent. Ham goes in and sees this. Rather than being a good and considerate son, and hastily covering him up and keeping shtum, he instead goes off laughing and tells his brothers, "Hey, the daft old bugger's pissed up in his tent with his wrinkly old dick on show to the world!" Shem and Yefet are appalled and rush off to cover him up before anybody else sees him. Occam's razor in action... This is why CBTP is full of epic fail; this is why Bible translations are carried out painstakingly by experts in the field, not by some guys on the internet. Fox 13:32, 29 October 2009 (UTC)
"Noach gets pissed and ... wrinkly old dick on show to the world!" Awesome. You should go improve the CBTP with that wording. ħuman 20:07, 29 October 2009 (UTC)
Maybe we should have our own version? Nobody has actually annotated our annotated bible, we could have the RationalWiki rewrite and rambling about crap bible project. - π 22:10, 29 October 2009 (UTC)
I found my stunning contribution to RationalWiki's annotated bible. Imagine more thought provoking ideas like that. - π 03:16, 30 October 2009 (UTC)
I seem to recall that "uncovering your father's nakedness" was a euphemism for sleeping with your mother in Leviticus, and that this was Ham's crime. Stile4aly 21:45, 29 October 2009 (UTC)
## Scientology
Is Joaquín Martínez a Scientologist? I've looked further and some of what he writes is critical. Proxima Centauri 09:16, 29 October 2009 (UTC)
Unlikely, given that he is Mexican I would hazard a guess he is Catholic. Just because you edit an article doesn't make you that. Are you a creationist if you edit a creation article? - π 11:15, 29 October 2009 (UTC)
"See Also: Charles Manson" WTF? What does Manson have to do with this? Hactar 16:23, 29 October 2009 (UTC)
## Huh?
I just added the Hoffman WIGO with a brand-new ID number, and the score started at minus 3 instead of zero. This kind of thing happen to anyone else, or did the wikiware thing my entry sucks that much ? --SpinyNorman 15:53, 29 October 2009 (UTC)
You put in the wrong number; 266 instead of 2656. Then I tried to correct it and put in 2666. ListenerXTalkerX 15:58, 29 October 2009 (UTC)
2666 is the name of a book that I am halfway through.-- Kriss AkabusiAAAWOOOGAAAR!!1 16:18, 29 October 2009 (UTC)
Would that be equivalent to having completed reading 1333? MDB 16:29, 29 October 2009 (UTC)
Thanks for the fix - I must have looked at the numbers three times and still not caught on. --SpinyNorman
For what it's worth, the second link in the WIGO is great reading, and the photo of Hoffman kind of resembles what an older Andy would look like. One of my favorite quotes:
"A flustered and ill-at-ease Mr. Hoffman objected to the heated questioning, saying he should have been provided a list of questions he might be asked. He was, if he had taken the time to read the Thursday morning Times editorial raising the very same questions."
No wonder Andy can relate to the guy. --SpinyNorman 16:36, 29 October 2009 (UTC)
This entire election is what I've predicted would happen for some time -- the hard core right wing faction of the Republican is spinning off into its own party. And its basically one that demands complete ideological purity on a handful of issues -- essentially the three G's of God, guns and gays -- and if you stray one iota from the Approved Conservative View (as did the Republican candidate in the election, who doesn't hate teh gheyz enough) -- they'll turn on you.
To which I say, "keep it up!" They're just helping the Democrats more and more. (Especially when the conservative candidate is a complete loser.) MDB 16:51, 29 October 2009 (UTC)
Perhaps this new spin-off party will unite with the Anti-Constitution Party. ListenerXTalkerX 16:56, 29 October 2009 (UTC)
Its not new -- the Conservative Party has been around in New York State for years. They just generally endorse the Republican candidate, and exist mainly to see that the Republican candidate doesn't get too liberal. MDB 17:04, 29 October 2009 (UTC)
## Thou shalt not...
... question TK. MDB 16:59, 29 October 2009 (UTC)
No no, he isn't a "block monger", but the truth does sound the same. CrundyTalk nerdy to me 11:50, 30 October 2009 (UTC)
## OMG: no soap, radio.
Karajou uses the discredited soap allegation favored by Holocaust denialists, in an anti-abortion propaganda posting. Mmm, skin cream, now with extra aborted fetuses (except that it uses a fetal cell line, not actual fetuses....) Godwin, anyone? - Poor Excuse 23:28, 29 October 2009 (UTC)
His source is WND, which is hardly surprising. To their credit though, they do mention that it's a cell line - not babies being minced up for each new batch of cream. There's an irony in Karajou typing that drivel on technology that almost certainly has origins in the work done by those immoral homosexuals. I encourage him to follow his brave stand to its logical conclusion, and ditch that immoral computer! - Ask me about your mother 23:43, 29 October 2009 (UTC)
## all-time favorite
Kinda bored with CP, so I revisited by all-time favorite pages: the Richard Dawkin's talk archives. With the possible exception of the Lenski Dialog, this has got to be the greatest ever. Anyone have one better? — Sincerely, Neveruse513 / Talk / Block 15:05, 28 October 2009 (UTC)
The whole Essay and Mystery sections are my favorite when-all-else-fails reading material. "Jesus caused comedy" was fun, of course, and there're various others. ~ Kupochama[1][2] 16:35, 28 October 2009 (UTC)
Wow, that "marry a conservative" article is a gem I hadn't seen before - especially its made-up anecdotes high-quality evidence. Thanks for the link. Cantabrigian 16:43, 28 October 2009 (UTC)
Oh, great googaly moogaly... the "no humor before Jesus" debate. When Andy was presented with example after example after example after example after example... of pre-Christian humor, he basically just denied that they were humor, turning it into "there was nothing I thought was appropriate humor before Christ." And then he goes on to site Trading Places as an example of humor, an odd choice for a conservative family friendly type like Andy, considering it stars the notoriously foul-mouthed Eddie Murphy, includes drug use, and has a class warfare theme. MDB 17:09, 28 October 2009 (UTC)
You forgot something (nsfw). — Sincerely, Neveruse513 / Talk / Block 17:27, 28 October 2009 (UTC)
The Lenski Dialog, to date, has been my personal favorite Andy bugger up. While it did not reach quite the level of point and laugh profile as the CBP has I always imagine Andy going into that one with a pre-emptive smug look, confident he is going to completely discredit or upset the evilutionist only to be utterly and completely smacked down by no less then the Professor himself. Lenski's second response was so well crafted, articulate, AND amusing it should serve as a template for any scientist getting scoffed at by creationist luddites. - Tygrehart
It is really hard to beat the Lenski affair. The severe pwning Lenski gave the assfly was nothing less than epic. EternalCritic 18:11, 28 October 2009 (UTC)
The Lenski affair was great. The old debates were much more fun because there was actually debate- banhammering took a little longer back then. Corry 23:47, 28 October 2009 (UTC)
this is one of many favourite pages of mine. Andy at his best. AceMcWicked 03:35, 29 October 2009 (UTC)
"We don't allow ideological censorship here. Try Wikipedia if you want to insist on censoring criticism of evolution, though I oppose censorship by Wikipedia also.--Andy Schlafly 21:51, 7 March 2009 (EST)" - classic, and not hard to find. A random section from Ace's link. ħuman 03:49, 29 October 2009 (UTC)
"Autumn foliage is beautiful, and the theory of evolution depends on denying it." Oh that is magnificent, it really is. The whole of evolutionary science (I have come to the conclusion there is no theory of evolution, just as their is no theory of physics) come crashing down because Andy finds autumn leaves pretty. It is simply beyond comprehension how he came up with that. - π 03:59, 29 October 2009 (UTC)
I am sure, user "pI" - what a silly name - that you are as equally unconvinced that 2+2=4 as that the LORD made autumn foliage pritty for us to enjoy. Deny that and lose all credibility. PS, I'm sure that not only do you not read the Bible (the most logical book ever written), but also seek to censor prayer on crashing planes. Sully would be embarrassed to have you on his airbarge, I'm sure. Begone, and be hence, and be silent. Goatspeed, - An DyShfly. ħuman 04:05, 29 October 2009 (UTC)
What was that parody page related to counter-examples of evolution? That one was about the only piece of parody I liked, it just took the idea of "EVOLUTION CAN'T EXPLAIN BEAUTIFUL FOLIAGE" and ran with it. Also, this classic gives a whole new level to "crazy moon talk". --GTac 09:43, 29 October 2009 (UTC)
Hardly my favorite, but related to that talk page (and another interesting example of what people were able to say without being instantly banhammered and reverted): Talk:Origin of the Moon --Sid 11:12, 29 October 2009 (UTC)
(undent) I think Andy easily takes the top ranks for Favorite CP Moments, but there are some non-Andy moments I found noteworthy:
• JM's definition and example image of a paintingimg was quite interesting indeed, especially when you check the article history (and the revert wars over the Christo installation) and read through the talk pageimg there.
• TerryH briefly exposes the full WTFitude of his beliefs and suggests that Tiktaalik was possibly "an antediluvian laboratory chimera and hence an out-of-place artifact of a society already known to have been inordinately cruel". This suggestion stands for roughly one hour before he realizesimg that this went even beyond the craziness levels CP had been known for back then. And then he tries to spin this slip-up into some sort of "See? People mocked me for this! This just proves that evolution is indeed more vulnerable than ever!"img
• "Two meters" by the Grand Master of Stubs.
Probably more, but I'd have to dig through my memory and ancient talk pages. --Sid 11:30, 29 October 2009 (UTC)
I have added captures just in case. If we lost the Tiktaalik one I don't know what I would do. That was golden. - π 11:37, 29 October 2009 (UTC)
I can't believe people always over look the liberal denial talkpage, it is hilarious. Every time someone questions Andy they are proving him more right, the trap he set was brilliant. - π 11:43, 29 October 2009 (UTC)
Someone mentioned the "Counterexamples to evolution" page... here is that classic piece of vandalism made to it. 194.6.79.200 12:46, 29 October 2009 (UTC)
Thanks BoN! --GTac 16:31, 29 October 2009 (UTC)
Thisimg addition to "Homosexual Agenda" was still a classic. Especially the 'Break for lunch' comment. Kapchured for posterior tea. Also, wasn't there some strange discussion involving aliens and UFOs (and Jinx?). I can't remember what it was exactly, it was just very odd. --PsygremlinHable! 17:39, 29 October 2009 (UTC)
That pretty much made my morning. May the FSM bless you with his noodly appendage. MDB 11:14, 30 October 2009 (UTC)
The Dawkins talk page archives are amazing! "I'm open to information about the belief system of the discoverers of DNA ... at the time they made the discovery. Personally, I wouldn't rank that discovery as among the greatest scientific discoveries in history, but I'm open to discussing it." - Assfly. CrundyTalk nerdy to me 11:48, 30 October 2009 (UTC)
Oh wow! This is brilliant! Go to this page (same discussion), and scroll down a bit. This is classic creationist debating. So Roger says that most prominent 20th century scientists who made great discoveries were atheists. Assfly calls him out on it asking for examples. PJR gives the example of Crick and Watson, which assfly dismisses because the discovery of DNA isn't important. 'Graham' comes along and gives 11 (count 'em) examples of great scientists who were athesists. At this point assfly ducks out and PJR goes onto perform the biggest Gish gallop I have ever seen, pages and pages of irrelevant nonsense. Extraordinary. CrundyTalk nerdy to me 12:01, 30 October 2009 (UTC)
Actually, come to think about it... there were some interesting Philip-Andy debates. For example the talk pages of "Gun Control" (plus archive), "Obama" featuring "ANDY, I AM NOT A LIBERAL. STOP CALLING ME ONE AND APOLOGISE FOR DOING SO." (plus archives, I guess) and of course two not-quite-mainstream ones: "Catholic views on creation" (a giant trainwreck centered around the "evolution is more than a/one hypothesis" Pope translation question) and "Archaeopteryx" (a.k.a. Andy's hard-on for Fred Hoyle). --Sid 13:25, 31 October 2009 (UTC)
## Gay Charity Work
After reading this I wondered what Andy would do if a bunch of homosexuals turned up on his doorstep asking for a donation to a particular cause or offered to paint his house. AceMcWicked 20:33, 30 October 2009 (UTC)
I think you've put the wrong link in. Jammy 20:43, 30 October 2009 (UTC)
About to say similar. I am eating & honeychat 20:44, 30 October 2009 (UTC)
nvm the diff, I'm just curious as to what Andy would do if a bunch of homosexuals turned up on his doorstep heh Fox 20:46, 30 October 2009 (UTC)
Link is fixed. AceMcWicked 21:09, 30 October 2009 (UTC)
That has made my day: "Undermining the resolve of latent homosexuals so that their will becomes too weak to resist the temptations of homosexuality." Er, is that semi-confessional, Andy? Fox 21:17, 30 October 2009 (UTC)
That's good, but I'm more amused by the "fact" that the "gay lobby" tells "seven- or eight-year-old boys" that if they "only like girls, maybe [they] are lesbian". And fail to recognize the existence of "cooties". - Poor Excuse 22:26, 30 October 2009 (UTC)
## Does this look like parody?
this got GeoffreyW blocked from CP, but it doesn't seem that provocative to me, things like changing bought to read and mentioning that he has a new book out don't seem very parodyesque to me. --Opcn 23:55, 30 October 2009 (UTC)
Meh, I was at a loose end so I blocked him using my Jinxmchue account, just for something to do. Fox 00:06, 31 October 2009 (UTC)
## Think of the Children
Ok, I'm just going to spit this out. None of the news articles I have read have mentioned that Andy is a homeschool teacher. None. I think that if word were to get out that Andy is a homeschool teacher who uses the Bible in class, more people would turn away from the Conservative Bible Project. After all, what better place to indoctrinate people into his ideology than his own class? Their fuckin' parents pay for it and I cannot see Andy using the King James Bible when he has his version ready to be forced upon the students. I would personally like to see Andy confronted by the parents over this kind of issue.
I think that WE could deal a serious blow to Andy if we were publish an article or editorial about this. We know Andy best. No one else is as qualified as we are. I dunno, we have been reporting on the CBP since before it was cool and I feel that we are overlooked when ever someone writes about it. Please someone either join me or talk me down.--Thanatos 02:48, 30 October 2009 (UTC)
Huh. I hadn't even thought about him using his new bible in class. He often asks for biblical reference in the answers to his questions, so eventually the Conservative Bible could become required reading for his class. That would be interesting. I wonder how he'd react to parents being angry at the idea (if they even got angry, hell, they could be his strongest supporters, I don't know). X Stickman 11:25, 30 October 2009 (UTC)
He won't use his own Bible, he'll continue to use the KJV. His own Bible will never see print, making it a good thing to reference on occasion (maybe assigning readings from it), but also providing a convenient excuse to avoid the conflict that would come from jettisoning the highly-regarded KJV.--Tom Moorefiat justitia 11:34, 30 October 2009 (UTC)
Assuming that he ever finishes it. --Sid 11:56, 30 October 2009 (UTC)
His attention span's pretty short. But he's got a few people helping, and the New Testament's not far off. He could just publish that.--Star trooper man 12:06, 30 October 2009 (UTC)
I fear he'll be bloody minded enough to see it through to spite the evil lie-burr-uls who insulted his new baby. –SuspectedReplicantretire me 12:55, 30 October 2009 (UTC)
I haven't predicted that far ahead, but saying he'll use it in his class...? Thing is, I can't think of a lot more RationalWiki could do to get noticed, unless we just go and troll the newspaper sites and tell them about us, which at any rate would be quite sad. SJ Debaser 13:11, 30 October 2009 (UTC)
I think the "method" would be to write and submit letters to the editor or even op-ed pieces, not so much to spam RW links as to tell the "whole story" behind the man behind the CBTP. A bit like that crummy email I sent Mike Malloy when he did his 5 minutes on the CBTP - I mentioned Phyllis, CP, BHO, and RW, filling in a whole bunch of missing links, basically. Sound clip is on my talk page. ħuman 18:23, 30 October 2009 (UTC)
I work on the assumption that anyone who uses Andy for homeschooling their kids has fully bought in to AndyLand and would, at the very least, be in the know about what he's doing. They might well approve. Bob Soles 13:15, 30 October 2009 (UTC)
There was some disquiet over the whole giving-girls-easier-exams thing (diffs gone; use the images here - it's at 141) and don't forget just how popular the KJV is. I would imagine that if the parents read the drivel about grape juice and bimbos they'll be pretty upset. You may be right though: if you've done so little research into Andy that you're prepared to have him ruin your kids' brains you're probably happy with whatever he says. –SuspectedReplicantretire me 14:07, 30 October 2009 (UTC)
I would assume (please!) that Andy's only giving part of the education to these poor kids. History & Economics: no sign of the promised Critical Maths lessons yet anyhow. The parents are probably awed by the Son Of Phyllis aura. I am eating & honeychat 14:17, 30 October 2009 (UTC)
I'm not keen of the idea of spamming RW. Sorry if it came out that way. But I think that we should raise concerns about this. If we could get a few media outlets to comment on the homeschooling bit, it could get the CBP harsher commentary. The fear of indoctrination is strong in the US (I need not remind of the Obama song thing from FoxNews) and maybe Andy could be classified as a cult leader. Other Christians are against him, as well as other conservatives. Who else would he preach his false scripture too?--Thanatos 18:45, 30 October 2009 (UTC)
(undent)
Well, my irony meter just popped after reading this thread - good thing I have the easy-reset model. Given how often Andy references the Bible in his online classes, it wouldn't be surprising if he started directing his students to use his ConservaBible as the approved classroom reference version. Thinking about that after all the CP references to communist leaders indoctrinating their students using their little red books popped the meter, and now my office smells like ozone. --SpinyNorman 19:36, 30 October 2009 (UTC)
I can't imagine him finishing his bible and then using the KJV still in class. He firmly believes he is making the bible better and more accurate, why the hell wouldn't he use it in class? CrundyTalk nerdy to me 20:14, 31 October 2009 (UTC)
I can't imagine him still going to church with this going on. The way those evil liberal heretics look at him while the preacher spews liberal propaganda. No place for a man who claims to be divinely inspired--Thanatos 23:57, 31 October 2009 (UTC)
The way that that sort of Puritan-descended church works, he just makes a new one - 'The First Church of Andy Magnificent', or something like that. It's inherent to fundamentalism - if you claim to all be very very close to the text, and the true meaning, then natural variations and disagreements will split you up into different congregations. --מְתֻרְגְּמָן וִיקִי שְׁלֹום!
Before or after he gets excommunicated?--Thanatos 00:14, 1 November 2009 (UTC) | 2013-05-24 18:57:08 | {"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": 20, "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": 0.47797104716300964, "perplexity": 4839.193031017992}, "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/1368704986352/warc/CC-MAIN-20130516114946-00046-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://publ.beesbuzz.biz/blog/381-In-better-news-v0-1-18 | # In better news, v0.1.18
Posted (2 years ago)
So aside from the Dreamhost issues, I would like to share what’s new in the latest version as part of my big “get my website online” push:
• PERFORMANCE: Improved the threading mechanisms around image renditions for better stability and performance
• BUGFIX: Made markup tagging work consistently between image types
• FEATURE: view.link() now allows overriding category
• FEATURE: You can now mark an entry with an Entry-Status of DELETED or GONE, which results in a 410 error instead of a 404 (be the envy of your web-developer nerd friends!)
• FEATURE: Error templates will automatically use the x00 error code as a potential fallback (e.g. error code 503 will also fall back to a template for 500)
• FEATURE: Entry titles can now have Markdown in them, and it usually works most of the time!
• BUGFIX: Now when running in debug mode you don’t end up with two watchdog threads
• FEATURE/QUALITY: Refactored the way error pages are handled, and now if you get a 404-type error on a page while the index is being asynchronously scanned, it’ll turn into a 503 with a Retry-After disposition in case it’s just something that hasn’t been indexed yet
• UX: View pagination URLs now all use id as the query parameter rather than a miasma of contextually-dependent start, last, or first which made no sense anyway | 2020-02-24 03:01:59 | {"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": 0.4362030327320099, "perplexity": 6250.120646025122}, "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/1581875145869.83/warc/CC-MAIN-20200224010150-20200224040150-00072.warc.gz"} |
https://followtutorials.com/2012/03/how-to-make-vertex-table-edge-table-and-surface-table-to-store-geometric-information-of-an-object.html | # How to make Vertex Table, Edge Table and Surface Table to store geometric information of an object??
Listing the geometric data in three tables provides a convenient reference to the individual components of each object. Also, the object can be displayed efficiently be using data from the edge table to draw the component lines. Here I discuss about how to create those tables with the help of C++ code.
Vertex table consists of the array of vertices with the coordinate value of each vertex. In C++, a vertex table can be created by creating a class that has . For example
class vertex
{
private:
Point3D *points;
public:
vertex()
{
points = new Point3D();
}
};
Edge table consists of pointers back into vertex to identify the edges for each polygon. For example,
class edge
{
public:
vertex *ver1;
vertex *ver2;
bool done; //1 if edge not to be drawn:
//0 if edge to be drawn, avoids multiple drawing of the edges
edge()
{
ver1 = new vertex();
ver2 = new vertex();
done = false;
}
};
Surface table consists of pointers of edges associated with the surface. Similarly it consists of Coefficients of Plane equations A, B, C, D and color intensity value of each surface. Surface table has great importance in Surface Rendering and Visible surface detection. The surface table can be created in C++ as
class surface
{
public:
edge *edges;
float A, B, C, D;
int icolor;
surface()
{
edges = new edge();
A = 0.0f;
B = 0.0f;
C = 0.0f;
D = 0.0f;
icolor = 0;
}
}
Rules for creating Geometric table:
1. Every vertex is listed as an end point for at least two edges.
2. Every edge is part of at least one polygon
3. Each polygon has at least one shared edge.
4. Every surface is close.
SHARE How to make Vertex Table, Edge Table and Surface Table to store geometric information of an object?? | 2021-04-19 06:44:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.24844209849834442, "perplexity": 2127.546643876123}, "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/1618038878326.67/warc/CC-MAIN-20210419045820-20210419075820-00591.warc.gz"} |
http://easyadbuilder.com/witcher-regis-thh/article.php?90b27b=the-area-of-quadrant-is-154-find-its-perimeter | Click hereðto get an answer to your question ï¸ Find the area of a quadrant of a circle whose circumference is 22 cm. If the diameter of a semicircular protractor is 14 cm, then find its perimeter. Find the length of the corresponding arc of the sector. 3. A quadrant is a quarter of a circle. Apart from the stuff given above, if you need any other stuff, please use our google custom search here. Question 9. Answer: a) 25 cm. (2) From a square cardboard, a circle of biggest area was cut out. Find the area sy Formulas, explanations, and graphs for each calculation. Question 9. Find the area of quadrant with radius 7 cm. Area of the circle = Ïr² = Ï × 3 × 3 = 9Ï cm² We hope the given MCQ Questions for Class 10 Maths Areas Related to Circles with Answers will help you. Question 7. a) 8 units. Sign in; ui-button; ui-button. We have provided Perimeter and Area Class 7 Maths MCQs Questions with Answers to help students understand the concept very well. It has a path of width 2 metre all around it inside. Here we are going to see the how to find area and perimeter of quadrant. Now let's move on to the program. Solution: Area of the circle = 154 cm 2. Perimeter = 23 cm,Area = 33 cm2 24. So to work out the area of a quadrant, first work out the area of the whole circle (use the formula A = Ï ×r²) and then divide the answer by 4. follow me please . b) n units c) 4 units d) 2 units. Find the area of the roads. Find an answer to your question the area of quadrant is 154sq cm find its perimeter 1. All the vertices of a rhombus lie on a circle.Find the area of rhombus,if the area of circle is 1256 cm² Solution: Question 14. hardeepsinghguler123 hardeepsinghguler123 ... 154×7 âr . If the height of a triangle is 19cm and its base length is 12cm. Area of shaded region = Area of semicircle BEC â (Area of quadrant ABDC â Area of Î ABC) Area quadrant ABDC Radius = 14 cm Area o (2pi.r) = 2×14 +(1/4)(2×22×14/7) =28+22 =50 cm. Solution: The perimeter of the circle = 176 cm. 50 - 2Ï = 0. In the figure, O is the centre of a circular arc and AOB is a straight line. Regular polygon area formula: P = n * a; Perimeter of a square formula. 31. Question 8. The largest triangle is inscribed in a semi-circle of radius 4 cm. f you need any other stuff, please use our google custom search here. â¦, express the percentage as ratio( 1).108:100....(2) 225%.explain step by step.â, 3]x' + 5y = - (3 - x); Write equation in standard form and write valueof a, b, c.â, Find the area of the shaded region.5 cma)b)4 cm4 cm3 cm2 cmI cm2 cmcm4 cm2 cmâ. len indicates length. Infront of a house, flower plants are grown in a circular quadrant shaped pot whose radius is 2 feet. Find the area of the path. To calculate its perimeter, all you need to do is to multiply side length by 4: Square Perimeter = a + a + a + a = 4a. Find the area of quadrant with radius 64 cm. Find its area (346.5 sqcm) 13) A circular disc of 6cm radius is divided into 3 sectors with central angles 120Ë, 150Ëand 90Ë.Find the ratio of the areas of 3 Sectors Ask your question. Find its area. Find the area of the minor segment of a circle of radius 14 cm, when its central angle is 60°. Find the area of the shaded region. Perimeter of a quadrant=2r+1/4. 12) The perimeter of a sheet of paper in the shape of a quadrant of a circle is 75 cm. A square has four sides of equal length. 11) Find the area of a quadrant of a circle whose circumference is 44cm (38.5cm2) 12) The perimeter of a sheet of paper in the shape of a quadrant of a circle is 75 cm. Question 8. Find its area (346.5 sqcm) 13) If the perimeter of the protractor is 72cm, calculate its area ( 308cm2) 14) A circular disc of 6cm radius is divided into 3 sectors with central angles 120Ë, 150Ëand 90Ë.Find the ratio of the areas of 3 Sectors Solving linear equations using elimination method, Solving linear equations using substitution method, Solving linear equations using cross multiplication method, Solving quadratic equations by quadratic formula, Solving quadratic equations by completing square, Nature of the roots of a quadratic equations, Sum and product of the roots of a quadratic equations, Complementary and supplementary worksheet, Complementary and supplementary word problems worksheet, Sum of the angles in a triangle is 180 degree worksheet, Special line segments in triangles worksheet, Proving trigonometric identities worksheet, Quadratic equations word problems worksheet, Distributive property of multiplication worksheet - I, Distributive property of multiplication worksheet - II, Writing and evaluating expressions worksheet, Nature of the roots of a quadratic equation worksheets, Determine if the relationship is proportional worksheet, Trigonometric ratios of some specific angles, Trigonometric ratios of some negative angles, Trigonometric ratios of 90 degree minus theta, Trigonometric ratios of 90 degree plus theta, Trigonometric ratios of 180 degree plus theta, Trigonometric ratios of 180 degree minus theta, Trigonometric ratios of 270 degree minus theta, Trigonometric ratios of 270 degree plus theta, Trigonometric ratios of angles greater than or equal to 360 degree, Trigonometric ratios of complementary angles, Trigonometric ratios of supplementary angles, Domain and range of trigonometric functions, Domain and range of inverse trigonometric functions, Sum of the angle in a triangle is 180 degree, Different forms equations of straight lines, Word problems on direct variation and inverse variation, Complementary and supplementary angles word problems, Word problems on sum of the angles of a triangle is 180 degree, Domain and range of rational functions with holes, Converting repeating decimals in to fractions, Decimal representation of rational numbers, L.C.M method to solve time and work problems, Translating the word problems in to algebraic expressions, Remainder when 2 power 256 is divided by 17, Remainder when 17 power 23 is divided by 16, Sum of all three digit numbers divisible by 6, Sum of all three digit numbers divisible by 7, Sum of all three digit numbers divisible by 8, Sum of all three digit numbers formed using 1, 3, 4, Sum of all three four digit numbers formed with non zero digits, Sum of all three four digit numbers formed using 0, 1, 2, 3, Sum of all three four digit numbers formed using 1, 2, 5, 6, Practical Problems on Relations and Functions in Set Theory, Relations and Functions Class 11 Questions. Find the area of a quadrant of a circle, where the circumference of circle is 44 cm. (Ï = 3.14) Solution: Central angle of the quadrant = 90° Radius of the circle = 2 feet Area of the quadrant = 3.14 sq. Find the circumference of the quadrant with radius 14 cm. Area of quarter circle = (1/4) Î r² Perimeter of a quadrant = ((Î /2) + 2)r It has 90 degree at the center. Solution: Radius = 3.5 cm Circumference = 2Ïr. (Use Ï = $$\frac{22}{7}$$] (2011OD) Solution: Circumference of a circle = 44 cm â 2Ïr = 44 cm. Perimeter of a rectangular field = 151 m Breadth = 32 m Length = â Breadth = â 32 = = 43.5 m and area = l × b = 43.5 × 32 m 2 = 1392 m 2. [Ï = $$\frac{22}{7}$$] Answer/ Explanation. find its perimeter. (False) Correct: Area = Base × Corresponding altitude. asked Apr 20 in Areas Related To Circles by Vevek01 (47.2k points) areas related to circles; ... Perimeter and Area of Plane Figures (326) Areas Related ⦠Question 41. 2. â¦and to obtain the maximum of a function, you must find when the derivative is zero. Answer/ Explanation. Find its area 1. An easy to use, free perimeter calculator you can use to calculate the perimeter of shapes like square, rectangle, triangle, circle, parallelogram, trapezoid, ellipse, octagon, and sector of a circle. To calculate area of a square in C++ programming, you have to ask from user to enter the side length of square. Join now. The perimeter of a square is 480 cm. there is no such thing as a perimeter of a circle, since the term used will be CIRCUMFERENCE. Perimeter of a triangle calculation using all different rules: SSS, ASA, SAS, SSA, etc. You can see that you obtain the maximum when: Ï = â, i.e. Find Its Area. Area of quadrant = 1/4× area of circle = 1/4×("Ï"r2 ) Now, we need to find r. 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Circle whose circumference is 44 cm for each calculation here we are going to see the how find. A sheet of tin in the figure, O is the centre of a function, will! = 176 cm, is = 28 cm, then find its perimeter a function, you will come many... A path of width 2 metre all around it inside square cardboard, a circle is 154 cm.... Perimeter 1 11.5 cm and 8.8 cm respectively no such thing as a perimeter a... Answers to help students understand the concept very well perimeter = 23,... Is 19cm and its Base length is 12cm a square cardboard, circle! The quadrant with radius 4.2 cm ) ( 22/7 ).r^2=154 r^2=154×7×4/22 r=14.! Given four options ( 3 to 14 ): Question 13 4 cm linear equations represent parallel lines -. Answer to your Question the area of a circle, where the circumference of quadrant..., you must find when the derivative is zero ) from a circular arc AOB. Programming, you will come across many shapes such as circle, where circumference... Area with Answers were prepared based on the latest exam pattern ; perimeter a... ) correct: area = Base × corresponding altitude cut out, SSA, etc, where the of. Units c ) 4 units d ) 2 units understand the concept very well is 154sq find. ) n units c ) 14 cm is 154 cm sq one fourth of any.... Radius 4.2 cm find when the derivative is zero height of a circular quadrant shaped pot radius... Rules: SSS, ASA, SAS, SSA, etc - â! And AOB is a straight line a semi-circle of radius 14cm is 154 cm 2 in programming. Pdf free download the plants grow â this is the centre of a quadrant a. With radius 64 cm pot in which the plants grow Class 7 Maths MCQs Questions Answers... Corresponding arc of the minor segment of a triangle is inscribed in a semi-circle of 3cm. Then find its circumference decimal place is 44 cm to one decimal place respectively.: SSS, ASA, SAS, SSA, etc ) ( 22/7 ).r^2=154 r=14! Shape of a quadrant of a clock is 14 cm is 154 cm.. And 60 metre broad also find the area of quadrant with radius 14 cm cm respectively circular arc and is. 2Pi.R ) = 2×14 + ( 1/4 ) ( 22/7 ).r^2=154 r^2=154×7×4/22 r=14 cm ) n units )..., a circle of radius 4 cm and the area of a quadrant of a quadrant a! Provided perimeter and area have to ask from user to enter the length... 3Cm is removed, AB = 28 cm, then find its perimeter 1 = n * a ; of! × corresponding altitude = 0 as circle, since the term used will be circumference is... You solve the equation: a ' = 0 ' = 0 the latest exam pattern, ASA,,. Sas, SSA, etc the plants grow are 11.5 cm and 8.8 cm respectively going! 176 cm, pentagon, octagon, etc â, i.e pot whose radius 2! Paper in the figure, O is the derivative is zero square has 625! Understand the concept the area of quadrant is 154 find its perimeter well NCERT MCQ Questions for Class 7 Maths Chapter 11 perimeter area! Whose circumference is 22 cm 3 to 14 ): Question 1 decimal places perimeter 1 places... Correct answer from the stuff given above, if you need any other stuff, please use our custom.: a ' = 0 shape of a rectangle are 11.5 cm 8.8... The latest exam pattern arc and AOB is a radius of the corresponding arc of the circle 44! The pot in which the plants grow ( False ) correct: area of a semi-circle 77. Circumference of quadrant 625 sq.m., then the diameter of the circle is 154 2..., explanations, and graphs for each calculation ofradius7cmandAPBisasemicircle.Find the area of the shaded region correct two... Is removed the park excluding cross roads = 28 cm, find perimeter... See the how to find area and perimeter of a circle is 154 cm.! With radius 3.5 cm circumference = 2Ïr =50 cm in the shape of a circle are equal,. 4 cm pot in which the plants grow is 3.5 cm, pentagon, octagon,.!: in right ÎABC, AB = 28 cm, when its central angle is 60° circular quadrant shaped whose! Cardboard, a circle is 176 cm, area = 33 cm2 24 of. ) ( 2×22×14/7 ) =28+22 =50 cm quad-rant with radius 4.2 cm 2×22×14/7 ) =50... A sheet of radius 3cm is removed the following pair of linear equations parallel! False ) correct: area = Base × corresponding altitude cm find perimeter... A quadrant of a quadrant=1/4.pi.r^2 ( 1/4 ) ( 22/7 ).r^2=154 r^2=154×7×4/22 r=14.! Find area and perimeter of a quadrant of a triangle is inscribed in a semi-circle be 77 sq.m octagon etc... Width of a semi-circle of radius 7cm the largest triangle is inscribed in a semi-circle 77... 60 metre broad: in right ÎABC, AB = 28 cm, find its length and area 7! To enter the side length of the quadrant of a circle of radius 3cm removed! Decimal places shape a quadrant of a circle whose circumference is 44 cm, square, pentagon, octagon etc. Area formula: P = n * a ; perimeter of the corresponding major segment solution in. Maths MCQs Questions with Answers Pdf free download ) 49 cm c 14... Answers were prepared based on the latest exam pattern find an answer to your the. Is 3.5 cm free download 154 cm 2 units d ) 66 cm to enter the side of... Is 12.5 cm of linear equations represent parallel lines 12.5 cm = 21 cm please use our google search! -2 a quadrant of a rectangle are 11.5 cm and 8.8 cm respectively ) + 2 ] r. the... Tin in the figure, O is the centre of a circle, since the term will. Plants grow = product of lengths of its two adjacent sides =50 cm shape of clock... Square has area 625 sq.m., then the diameter of the pot in which the plants grow an answer your! Π/2 ) + 2 ] r. find the perimeter and area Class 7 Maths Chapter 11 perimeter and the of! Its radius, AB = 28 cm, find its length and area Class 7 Maths with to... From user to enter the side length of the quadrant of a circle of biggest area was cut out cm. | 2021-07-24 14:12:02 | {"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": 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": 0.6335639357566833, "perplexity": 1061.6424127494545}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046150266.65/warc/CC-MAIN-20210724125655-20210724155655-00599.warc.gz"} |
http://math.stackexchange.com/questions/176185/prove-that-this-is-a-banach-space?answertab=active | # Prove that this is a Banach space
Let $I=[0,1]$ and let $\displaystyle X:=\left\{f: I\times \mathbb R\to \mathbb R\colon \sup_{(t,x)}\frac{|f(t,x)|}{1+|x|}<\infty\right\}$. Prove that $X$, equipped with the norm $\displaystyle \|f\|:=\sup_{(t,x)}\frac{|f(t,x)|}{1+|x|}$ is a Banach space.
My first attempt was to use the characterization that $X$ is a Banach space if and only if every absolutely convergent series converges, but no success.
Then I've noticed that I can prove convergence on every Ball of arbitrarily large radius, however still i cannot conclude on the whole $I\times \mathbb R$. Maybe Ascoli Arzela, but, honestly, i don't know.
Hope you can help me. Thank you.
-
But Arzelà-Ascoli is a theorem about $C(K)$ endowed with the sup norm. Here you have a different norm. So I'm not sure how to apply Arzelà-Ascoli in this case. – Rudy the Reindeer Jul 28 '12 at 14:32
• The fact that $X$ is a normed space follows from the properties of supremum.
• Take $\{f_k\}\subset X$ a Cauchy sequence. In particular, for each $(t,x)\in [0,1]\times \Bbb R$, the sequence $\{f_k(t,x)\}$ is Cauchy, hence converges to some $f(t,x)$. Fix $\varepsilon>0$, and take $N=n(\varepsilon)$ such that the norm of $f_k-f_j$ is $\leq \varepsilon$ whenever $j,k\geq N$. We have for all $(t,x)\in I\times \Bbb R$ and $k,j\geq N$ that $$|f_j(t,x)-f_k(t,x)|\leq (1+|x|)\varepsilon.$$ We can take the limit in the last equation to show that $f\in X$. Further more, we can show that this gives the convergence of $f_j$ to $f$.
An alternative way is the following. First show that if $X$ is a non-emptyset, the set of bounded real-valued functions defined on $X$, $B(X)$, endowed with the supremum norm is a Banach space. In this particular case, show that $X$ is a closed subspace of $B(X)$. | 2015-07-28 21:57:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9686377644538879, "perplexity": 63.55114558773652}, "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-2015-32/segments/1438042982745.46/warc/CC-MAIN-20150728002302-00233-ip-10-236-191-2.ec2.internal.warc.gz"} |
http://www.windows-commandline.com/xcopy-command-syntax-examples/ | # Xcopy command : syntax and examples
by on February 19, 2011
What’s Xcopy?
Xcopy is a built in command on Windows OS. Copy is the other command we have on Windows, it can copy files from one directory to another. But it does not work for copying directories. Using xcopy we can copy directories also.
Copy all the files in one directory to another directory.
Xcopy /I Source_Directory Destination_directory
This would copy only the immediate files in the source directory to the destination. It would not copy files from sub directories. Adding /I switch will avoid the question to the user ‘Does the destination directory specify a file name or directory name on the target‘.
Copy a directory including all its files and sub directories to another location on the same drive or a different drive.
Xcopy /S /I /E Source_Directory Destination_directory
For example to copy all the files from D:\data\documents to the folder E:\Newfolder\documents we need to run the below command.
Xcopy /S /I /E D:\data\documents E:\Newfolder\documents
This command creates the folder E:\Newfolder\documents if it already does not exist. It also creates the same directory/file structure in the destination folder. If the destination folder already exists you do not need to add /I switch. If the folder does not exist and you do not specify /I you will be prompted to confirm if the destination is a filer or folder. This would cause issues if you want to run unattended or automated copying of files.
Copy files including hidden and system files.
The above command excludes hidden and system files from copying. If you want to these files also then you need to add /H option to your command. Then the command would be
Xcopy /S /I /E /H D:\data\documents E:\Newfolder\documents
/E option causes to copy empty sub directories to the destination folder.
If you add /H option the the command would also copy hidden and system files to the destination folder.
If you are trying to overwrite an existing file then add the option /Y so that you will not be prompted for confirmation for overwriting the file.
Copy files based on archive attribute
If you want to copy only the files that have archive attribute set, you can use /A and /M switches. Archive attribute indicates whether the file has been modified since the time the attribute was reset.
The command for this is:
Xcopy /A /I /S source_directory destination_directory.
The above command keeps the archive attribute set; It does not reset the attribute.
If you want to reset the archive attribute, you can use /M switch instead of /A. The command for this case would be:
Xcopy /M /I /S source_directory destination_directory
Exclude files in the copying
If you want to copy a directory to another location but want to exclude some files then you can use /EXCLUDE switch with Xcopy command. You need to put the list of the files to be excluded in a file and then specify this file with the /EXCLUDE switch.
Xcopy /I Sourcedir Destdir /EXCLUDE:filename
Example:
Copy the directory D:\docs to E:\newdocs\ excluding all pdf and mp3 files:
C:\>type 1.txt
.pdf
.mp3
C:\>Xcopy D:\docs E:\newdocs /EXCLUDE:1.txt
You can also specify the full names of the files. Each file name/pattern should be in a separate line.
The commands shown above can be used from batch files also. This command is available in Windows 7, XP, Vista, Server 2003 and Server 2008 editions.
Copy files based on modified date
Xcopy /D:dd-mm-yy /I sourcedir destinationdir
Example: To copy all the files in the directory ‘E:data’ that are modified on or after 1 st February 2011 to the folder ‘E:\backup
Xcopy /D:01-02-11 /I E:\data E:\backup
Related Posts:
Backup files using xcopy command
Srini May 3, 2011 at 11:02 am
More examples would have been good for excluding files from copying. For example, If I need to exclude certain file types in xcopy, say .doc, How can I do it.
admin September 3, 2011 at 11:14 am
Thanks for the comment. Added examples for those cases also.
Riddle June 14, 2011 at 8:23 am
Does this work on all Windows releases? I am running Vista and can’t seem to have this command,
admin September 3, 2011 at 11:13 am
Yes, it works on XP, Vista and Windows 7.
Aust September 3, 2011 at 4:20 am
How do I copy files that have not been modified in the last 7 days. I mean without specifying any date. I need to use xcopy command in a batch file and whenever I run the batch file, it should copy only the files that have been modified in the last 7 days. Any suggestion for this?
Jonathan September 3, 2011 at 11:31 am
Anyway to specify the excluded file types as xcopy arguments rather than mentioning in a file? I am getting ‘Can’t find file’ error if I do ‘xcopy /exclude:.mp3’.
Bruce November 6, 2011 at 5:37 pm
I am getting the below error with xcopy.
Invalid number of parameters.
0 File(s) copied
Can you tell me which cases one would see this error.
Douglas July 17, 2014 at 9:31 pm
source_directory and destination_directory need to be surrounded by quotation marks
ie: xcopy “C:\directory\*.*” “D:\directory\”
tom graham December 3, 2013 at 2:27 am
I have a whole lot of .JPGfiles that are in different folders with other files.
I want to extract (or just copy) all the .JPG files into one folder without the other stuff.
How can I do this?
I have tried various xcopy command strings but I still get the containing folders and the other files in that folder.
I have windows XP 64 bit.
Tom
admin December 7, 2013 at 2:51 am
You can use move and forfiles commands. Try below command.
forfiles /S /M *.JPG /C "cmd /c move @file c:\folder"
JohnW December 29, 2013 at 1:58 pm
Good tip nearly always missed for xcopy:
xcopy does not like split directory/folder or file names, eg : Documents and Settings , and will give an invalid number of parameters’ error message –
Just put the name inside quotes: “Documents and Settings” – problem solved
uche April 4, 2014 at 11:07 am
does it mean that it will not work in windows 8
admin April 7, 2014 at 5:14 am
This command should work on Win8 also. The article was published pre-Win8.
Fred April 21, 2014 at 2:30 am
Is it possible to copy a file or group of files with a variable path to them? e.g., certain flash settings files are always stored in the same location, save one folder in the path which name is changed after cache is cleared and flash has been reloaded.
Fred April 21, 2014 at 7:49 pm
Never mind. PowerShell is *exactly* what I’m looking for. Cheers!
sandip December 9, 2014 at 4:07 am
Thanks a ton for these examples.
How to exclude files which starts with specific name?
admin December 9, 2014 at 6:02 am
You can do that with /EXCLUDE switch. Put the name in a file and pass that file name with /EXCLUDE switch. For example, if you want to exclude all files starting with ‘abc’ you would create a text file(say patterns.txt) and put the text “abc”(without quotes).
Next, run the command below.
Xcopy srcdir destdir /EXCLUDE:patterns.txt
Ken January 11, 2015 at 6:15 pm
I am trying to copy three directories and have three xcopy statements but it only copies the first one. How can I have multiple statements execute after it deletes all the old files? Thanks!
Bob UK February 11, 2015 at 6:06 pm
I’m trying to use these lines in a batch file running on a Win8.1 64 bit but it does not work.
echo.
if not exist \\BACKUP\Backup\Bob\""." goto nodrive
xcopy e:\Bob \\BACKUP\Backup\Bob /d /e /h /i /y
echo
I can see the remote Backup machine in the network but it is not found with the “if not exist” or xcopy command nor the dir c ommand on the command prompt. It works fine if I use it to copy updated files to a local disk/ Are these commands supported in 64 bit Win 8.1?
Bob UK February 11, 2015 at 6:15 pm
sorry should be
if not exist \\BACKUP\Backup\Bob\*.* goto nodrive
but this does not work in 64 bit Win8.1 to the remote \\BACKUP machine but works if I use a local drive.
James Tew April 17, 2015 at 8:33 am
try pointing the “if not exist” at a specific file within the target location instead of “*.*”
James Tew April 17, 2015 at 8:30 am
I would like to use a single xcopy command to copy files to two different locations. Is this possible?
admin April 29, 2015 at 4:47 am
Not sure why you can’t run xcopy twice, once for each of the locations. But apparently xcopy does not allow to specify two destination folders for the copy.
srikanta May 24, 2015 at 12:28 pm
I have a requirement where i need to exclude some specific file types in child directory rather than in root directory. How can I achieve it using xcopy command with the help of exclude switch ? Please suggest.
My arrangement is like below,
D:\Source
D:\Source\abc.rpt
D:\Source\Child\abc1.rpt
D:\Source\Child\xyz.rpt
D:\Source\Child\def.txt
D:\Destination
I have used xcopy command as below
xcopy D:\Source D:\Destination /S /Y /Exclude:D:\Source\ExclusionList.txt
ExclusionList.txt has below exclusion list
————————————————–
child\.rpt
After executing xcopy command everything is copied from source to destination directory.
I want the output in destination directory as below
D:\Destination\abc.rpt
D:\Destination\def.txt
admin May 26, 2015 at 2:46 am
There does not seem to be any straight solution using xcopy. But one workaround is to mark all those files which should be excluded as hidden files. Xcopy by default would ignore hidden files from copying. So you can run the below commands.
attrib +H child\*.rpt
and then
xcopy D:\Source D:\Destination /S /Y`
xcopy June 29, 2015 at 2:07 pm
Hello,
I have i bit problem to use xcopy with variable, when i put source and destinaton in variable the don’t work , without variable all work as exepected
Could y get an eye and see the mistake please.
echo off
SET SRC = C:\Projects\\source
SET DEST = C:\MBE\Projects\TESTLIVREAISON\destination
SET EXCLUDEDFILES = C:\MBE\Projects\TESTLIVREAISON\excludedfiles.txt
XCOPY %SRC% %DEST% /E /F /Y /EXCLUDE:%EXCLUDEDFILES%
pause
jann July 2, 2015 at 1:53 pm
Hello,
I would like to copy a set of files listed in a text file in the same folder structure to another location. Could you please help me on this.
keepthesefiles.txt has the list of files that need to be copied (A.txt)
src : folder1\folder2\A.txt
destination folder should look like this : folder1\folder2\A.txt
Thanks,
Jann
Yatin August 1, 2015 at 1:18 pm
Hi ,
How can i copy files and folder from one drive to another drive scheduled basis and i also want that when job will run next scheduled time it will only copy the NEW data (folder and file) from the source to destination?
satendra August 20, 2015 at 10:32 am
Dear all,
I have a question.
If i have some folders as well as some computers with different host names.
example:
computer names like: ABC-PC, DEF-PC, GHI-PC, JKL-PC, QRS-PC, XYZ-PC.
Folder names like: ABC_abc, DEF_def, GHI_ghi, JKL_jkl, QRS_qrs, XYZ_xyz.
is it possible to make a batch file which copy only that folder, which is according to the computer name.
if yes then what will be the command.
zaboykie October 25, 2015 at 8:23 am
Hi.
For regular backups, can I run a command that will copy all updated and new files in all directories and sub-directories to my backup drive, as opposed to copying everything as I’m doing now (takes forever)?
Looking forward to responses. Thank, guys!
Sameer November 6, 2015 at 11:11 pm
Hello,
I want to use xcopy tool to copy all drivers that are located in different folders to a one single folder in the destination. E.g. I have drivers extracted in a few folders with different file extensions and now I want to copy only files from those folders to the destination in ONE FOLDER, is this possible? | 2015-11-29 14:03:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.5092613697052002, "perplexity": 4179.47365227381}, "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-2015-48/segments/1448398458511.65/warc/CC-MAIN-20151124205418-00188-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://www.zamanianlab.org/ZamanianLabDocs/protocols/Parasitic_Nematodes/Mosquito_Blood_Feeding/Mosquito_Blood_Feeding/ | # Infecting Mosquitoes with Microfilaremic Blood
## Materials
• Glass slides
• Glass cover slips
• Defibrinated Sheep's Blood (Hemostat Laboratories DSB100)
• Bleach
• Glass or electric blood feeders
• Parafilm
• Ring stand (glass feeders only)
## Preparing blood for blood feeding
• All pipette tips, slides, serological pipettes, tubes, and other materials that come into contact with blood throughout this protocol must be treated with bleach prior to disposal. A bleach solution can be prepared in the bleach-specific pitcher in the insectary (labeled with yellow tape) by adding ~½ cup of bleach and 1 L of tap water.
Important: Remove the sucrose pad(s) at least 2 hr before feeding for LVP and AaSD mosquito species.
### Starting with microfilaremic blood
1. Invert the tube of infected blood several times to mix the microfilariae (mf).
Note: If the microfilariae are shipped in multiple glass vials from the same subject, combine all of the vials into a single 50 mL conical tube and mix by inversion.
2. Add 20 µL of infected blood to 50 µL of dH2O on a glass slide and stir with the pipette tip to mix.
3. Add a cover slip to the slide. Adding the cover slip beginning at an angle and slowly lowering it will reduce bubbles in the middle of your slide.
4. Return the mf back to the incubator while counting.
5. Place the slide on a compound microscope and count the mf. Start at a corner of the cover slip and count the mf in the fluid outside of the coverslip. Once the perimeter has been counted, count the mf underneath the cover slip area using an s-pattern. (See figure below)
6. Calculate the mf concentration in the microfilaremic blood stock.
7. In a 15 mL or 50 mL conical tube, dilute the microfilaremic blood with Defibrinated Sheep's Blood to the desired blood feeding concentration. Use C1V1=C2V2 to calculate how much microfilaremic blood and how much control blood to add to the conical tube.
Note: The blood feeders used for feeding hold 10 mL of blood. When calculating final volume use 11 mL for 1 mL overage. Multiply this by number of feeders you will need.
8. Appropriate concentrations for blood feeding:
• B. pahangi: 120 – 160 mf / 20 µL
• D. immitis: 80 – 160 mf / 20 µL
• B. malayi:
9. Re-enumerate the mf in blood in duplicate to verify that it is within target concentration for the designated species (see target concentrations above). If concentration is off, re-calculate dilutions based on the new numbers and add defibrinated sheep's blood as needed. Additionally, verify mf have recovered (are moving) before feeding; otherwise they may be damaged or dead.
### Starting with mf in IP fluid
1. Invert the tube in order to evenly distribute the mf throughout the IP fluid.
2. Add 20 µL of microfilaremic IP fluid to a glass slide and place a coverslip on top. Count the number of mf in the sample on a compound microscope using the pattern described above. This should be done in duplicate.
Note: It may be necessary to dilute the sample if the mf are too numerous to accurately count. A 1:10 dilution can be carried out by adding 20 µL of microfilaremic IP fluid to 180 µL RPMI-1640 medium, mixing, and dispensing 20 µL on a glass slide for quantification.
1. Determine how much volume of IP fluid you will need to get the desired mf concentration for blood feeding by using C1V1=C2V2. If a 1:10 dilution was used to count the microfilariae, multiply the concentration of mf/20 µL observed in step 2 by 10 to account for the dilution.
Note: The blood feeders used for feeding hold 10 mL of blood. When calculating final volume use 11 mL for 1 mL overage. Multiply this by number of feeders you will need.
2. Add the calculated amount of IP fluid to a 15 mL conical tube. Spin at 900 rpm for 10 min. You should see a pellet of mf at the bottom of the tube. Remove all supernatant without disturbing pellet. Add the final volume of blood you desire (11 mL x the number of feeders).
3. Re-enumerate the mf in blood in duplicate to verify that it is within target concentration for the designated species (see target concentrations above). If concentration is off, re-calculate dilutions based on the new numbers and add defibrinated sheep's blood as needed. Additionally, verify mf have recovered (are moving) before feeding; otherwise they may be damaged or dead.
## Assembling glass blood feeders and feeding mosquitoes
1. Use the large glass blood feeders. Stretch a 2-square piece of parafilm pulled once in both directions and place it on the bottom of the glass feeder. Avoid wrinkles in the parafilm and secure the parafilm by smoothing it along the glass rim.
2. Cut a second piece of parafilm (2 squares wide, half square tall) and stretch it around the bottom edge of the glass feeder.
3. Secure the parafilmed glass feeder upside down on a metal ring stand and place on a cart in the insectary.
4. Setup the water circulator on a cart in the insectary by filling it with dH2O to the black fill line.
5. Attach the water circulator tubes to each inlet and outlet arm on the side of the glass feeders and wrap a piece of parafilm (2 squares wide, half square tall) around the connecting points. This will help control any water leaks.
6. Turn the water circulator on and set the temperature to 37°C.
7. In these feeders carefully place ~10 mL of diluted, infected blood into the top of the blood feeders using a disposable pipet. You need to make sure that the feeders are level so that the blood doesn’t pool in one side and unevenly distribute the mf.
8. Place a small strip of parafilm over the opening to the inner chamber where the infected blood was loaded into the glass feeder.
9. Let the water, glass feeders, and blood equilibrate to 37°C prior to feeding.
10. Place a carton of mosquitoes under a membrane feeder and slowly lower the feeder so that it rests atop the carton and is flush with the mesh of the carton.
11. Allow the carton of females to feed for 30 – 45 min. or until a majority of females have fed.
12. While the mosquitoes are feeding, print out a carton label according to the example below:
13. When feeding is complete, remove an aliquot of blood to confirm that the microfilariae are still alive and active. Do not separate blood fed from non-blood fed mosquitoes at conclusion of feeding.
14. Affix the carton label to the corresponding carton and place blood fed mosquitoes in the insect incubator.
Note: Mosquito cartons blood fed with B. pahangi or B. malayi must be double contained by placing the fed cartons inside the mesh cage located inside the incubator.
1. Allow the infection to incubate for 14-15 days before you extract L3 from the mosquitoes.
2. Other important notes:
• Maintain the sucrose pads on the cartons daily. It will be important to communicate with the insectary manager or undergraduates to make sure sucrose pads are moistened every day including on the weekends.
• If you notice an accumulation of dead mosquitoes in the bottom of the carton, suck out the dead mosquitoes, empty the suction device into a plastic container and freeze the container overnight. The mosquitoes can then be placed in the trash the next day after freezing.
## Assembling electric blood feeders and feeding mosquitoes
1. Use the metal electric membrane feeder with the large screw hole. Stretch a 2-square piece of parafilm pulled once in both directions and place it on the indented side of the metal feeder. Avoid wrinkles in the parafilm and secure the parafilm by smoothing it along the rim and around the sides of the metal feeder.
2. Cut a second piece of parafilm (2 squares wide, half square tall) and stretch it around the edge of the metal electric feeder.
3. Cut off excess parafilm so the feeder will lay flat against the electric heating unit.
4. Fill the unit with microfilaremic blood until the bottom of the membrane is evenly covered. It will take 10-20 mLs of blood.
5. Once the feeder is filled with blood, place the plugs in the port holes and ensure they are flush with the metal feeder. If they are not flush, the feeder will not fit correctly on the electric unit and the blood will not be kept warm while feeding the mosquitoes.
6. Set up the electrical unit in the insectary and screw on the blood-filled membrane feeder. Turn the unit on and set the feeder on its side while the blood warms up for about 5 min.
7. Gently set the blood-filled feeder on top of the mosquito carton and allow the mosquitoes to feed for about an hour or until a majority of females have fed.
8. While the mosquitoes are feeding, print out a carton label according to the example below:
9. When feeding is complete, remove an aliquot of blood to confirm that the microfilariae are still alive and active. Do not separate blood fed from non-blood fed mosquitoes at conclusion of feeding.
Note: Mosquito cartons blood fed with B. pahangi or B. malayi must be double contained by placing the fed cartons inside the mesh cage located inside the incubator.
10. Allow the infection to incubate for 14-15 days before you extract L3 from the mosquitoes.
11. Other important notes:
• Maintain the sucrose pads on the cartons daily. It will be important to communicate with the insectary manager or undergraduates to make sure sucrose pads are moistened every day including on the weekends.
• If you notice an accumulation of dead mosquitoes in the bottom of the carton, suck out the dead mosquitoes, empty the suction device into a plastic container and freeze the container overnight. The mosquitoes can then be placed in the trash the next day after freezing.
## Note about feeding SD Mosquitoes
These mosquitoes are fragile and the following precautions should be taken to achieve the best yield:
- SD should be panned out on a Friday and separated/counted on a Thursday
- Mosquitoes should not be blood fed on a Wednesday; if mf are arriving on a Wednesday because of a holiday, request they be sent a different week
- Only starve mosquitoes for four hours prior to blood feeding
- L3s should be extracted on day 13 after blood-feeding | 2022-11-27 08:20:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.4248266816139221, "perplexity": 4946.299823837346}, "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/1669446710218.49/warc/CC-MAIN-20221127073607-20221127103607-00206.warc.gz"} |
http://www.mailund.dk/index.php/2017/04/11/ | ## Evaluating queue data structures
Working on my book on functional data structures in R, I have implemented three versions of functional queues, two that are not really functional—they need to update a data structure as side effects—and one that is. As far as I can tell, they should have roughly the same performance, but they don’t.
Asymptotically, they all work in $$O(1)$$ time for all operations, this nice figure explains why:
The implementations have two lists and need to move elements from one to the other when you access the front of the queue, so you can imagine all enqueue operations costing twice as many computrones as the other operations and then that can pay for the movement of elements later.
However, the three implementations I have in R have different performances when I measure them. Two of them modify an environment and are not persistent; the third remembers one element more than the others and gives me a persistent data structure. For some reason, the last version has a performance that is about twice as bad as the other two.
I am stumped as to why.
You can reproduce the experiments with the code below—there might be a beer in it for you if you can explain to me why the functional version is so much slower.
devtoolsinstall_github("mailund/ralgo")
library(ralgo)
library(microbenchmark)
library(tibble)
library(ggplot2)
setup < - function(n) n
evaluate <- function(empty) function(n, x)
elements <- 1:n
queue <- empty
for (elm in elements)
queue <- enqueue(queue, elm)
}
for (i in seq_along(elements))
queue <- dequeue(queue)
}
}
ns <- seq(5000, 10000, by = 1000)
performance \<- rbind(get_performance("explicity environment", ns, setup, evaluate(empty_env_queue())),
get_performance("closure environment", ns, setup, evaluate(empty_closure_queue())),
get_performance("functional queue", ns, setup, evaluate(empty_extended_queue())))
ggplot(performance, aes(x = as.factor(n), y = time / n, fill = algo)) +
geom_boxplot() +
scale_fill_grey("Data structure") +
xlab(quote(n)) + ylab(expression(Time / n)) + theme_minimal()
## Just sold Meta-programming in R
I just sold Meta-programming in R to Apress. This will then be the third book I publish with them. That is nice—I got the contract yesterday, at the same day as I got confirmation that my copies of Functional Programming in R are in the mail. I will pull down Meta-Programming from Amazon tonight after work, so if you want a copy, you should probably hurry up and get it. | 2018-06-25 19:24:20 | {"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": 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": 0.356648713350296, "perplexity": 1931.7272075248825}, "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-2018-26/segments/1529267868876.81/warc/CC-MAIN-20180625185510-20180625205510-00150.warc.gz"} |
https://access.openupresources.org/curricula/our6-8math-v1/8/students/1/6.html | # Lesson 6: Describing Transformations
Let’s transform some polygons in the coordinate plane.
## 6.1: Finding a Center of Rotation
Andre performs a 90-degree counterclockwise rotation of Polygon P and gets Polygon P’, but he does not say what the center of the rotation is. Can you find the center?
## 6.2: Info Gap: Transformation Information
Your teacher will give you either a problem card or a data card. Do not show or read your card to your partner.
If your teacher gives you the problem card:
1. Silently read your card and think about what information you need to answer the question.
2. Ask your partner for the specific information that you need.
3. Explain to your partner how you are using the information to solve the problem.
4. Solve the problem and explain your reasoning to your partner.
If your teacher gives you the data card:
1. Silently read the information on your card.
2. Ask your partner “What specific information do you need?” and wait for your partner to ask for information. Only give information that is on your card. (Do not figure out anything for your partner!)
3. Before telling your partner the information, ask “Why do you need that information?”
4. After your partner solves the problem, ask them to explain their reasoning and listen to their explanation.
Pause here so your teacher can review your work. Ask your teacher for a new set of cards and repeat the activity, trading roles with your partner.
## Summary
The center of a rotation for a figure doesn’t have to be one of the points on the figure. To find a center of rotation, look for a point that is the same distance from two corresponding points. You will probably have to do this for a couple of different pairs of corresponding points to nail it down.
When we perform a sequence of transformations, the order of the transformations can be important. Here is triangle $ABC$ translated up two units and then reflected over the $x$-axis.
Here is triangle $ABC$ reflected over the $x$-axis and then translated up two units.
Triangle $ABC$ ends up in different places when the transformations are applied in the opposite order! | 2021-05-12 02:06:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.36685261130332947, "perplexity": 808.855820103491}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991693.14/warc/CC-MAIN-20210512004850-20210512034850-00441.warc.gz"} |
https://www.micronanoflows.ac.uk/hydrodynamic-wall/ | Close
# Hydrodynamic Wall
The "hydrodynamic position" of the wall Z_H is defined as the position at which the boundary condition on the macroscopic velocity field has to be applied.
(frac{partial v_{x,y}(mathbf{r},t)}{partial z})_{z=Z_H}=frac{1}{L_s}v_{x,y}(mathbf{r},t)|_{z=Z_H}
where v_{x,y} is the slip velocity, L_s is the slip length. The hydrodynamic position Z_H fluctuates as it is defined by the molecular trajectories as a result of collision with the wall.
Fluid in the dense state undergo frequent collisions with the wall as compared to a rarefied state. As a result the fluctuating "hydrodynamic wall" observed in a small time interval is density dependent for a given fluid-wall interaction.
Click on the images to view how hydrodynamic wall looks for rarefied and dense cases.
Rarefied case
• Dense case | 2019-04-23 02:28:55 | {"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": 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": 0.6190723776817322, "perplexity": 2264.808153397802}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578584186.40/warc/CC-MAIN-20190423015050-20190423041050-00126.warc.gz"} |
https://socratic.org/questions/how-would-you-create-a-x-y-table-for-the-equation-y-2x-1 | # How would you create a (x,y) table for the equation y=2x-1?
Nov 7, 2014
To crate an $\left(x , y\right)$ table for $y = 2 x - 1$, you could plug in a bunch of $x$ values into that equation and find the corresponding $y$ values and then tabulate the results. This will give you an idea of what the graph will look like. Note that the graph of $y = 2 x - 1$ is a line that goes on forever and ever in the $y$ and $x$ directions, but by plugging in a few values of choice, you can see a certain part of the graph and then put arrows on the endpoints you get to indicate that the line continues.
So to make a table, you have to pick a few values to plug in. I think that 5 values would be good to start with. Let's say those values are
$x = - 4 , - 2 , 0 , 2 , 4$
Note that you can pick any values you want. I just picked these because they're easy to work with.
So,
$x = - 4 , y = 2 \left(- 4\right) - 1 = - 9$
$x = - 2 , y = - 5$
$x = 0 , y = 2 \left(0\right) - 1 = - 1$
$x = 2 , y = 2 \left(2\right) - 1 = 3$
$x = 4 , y = 2 \left(4\right) - 1 = 7$
You can then tabulate these values with x and y as headings, and whatever inputs you choose under x, and whatever outputs you get under y. It can look something like this:
x -4 -2 0 2 4
y-9 -5 -1 3 7
You can then plot these points to see what the graph will look like. The corresponding points are (-4, -9), (-2,-5), (0,-1), (2,3), and (4,7)
The graph of $y = 2 x - 1$ looks like this: | 2022-08-10 17:04:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 14, "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": 0.6135140657424927, "perplexity": 192.66467112275035}, "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/1659882571198.57/warc/CC-MAIN-20220810161541-20220810191541-00240.warc.gz"} |
https://quantumcomputing.stackexchange.com/questions/11699/why-should-we-measure-in-x-z-basis-for-z-x-errors-in-steane-syndrome-extraction/21404 | # Why should we measure in X/Z basis for Z/X errors in Steane syndrome extraction?
From the Steane syndrome extraction of quantum error correcting code, we use ancilla qubit prepared in logical X/Z basis to detect logical Z/X errors in the logical data state (The CNOT is transversal).
Every material tell me that it works because X errors will propogate to the left part of ancilla and Z errors will propogate to the right part. However, I'm confused how the measurement works. For example in the left part, even there is no error (Or applying a stabilizer), measuring logical + at Z basis will cause uncertain results, and X error just make thing more complex. Can somebody explain how the measure work and why it can get the syndrome?
• Where did you get this circuit from? Apr 27 '20 at 7:41
• arxiv.org/abs/1605.05647v4, there is a lot of other similar cricuit if you search steane syndrome extraction Apr 27 '20 at 10:49
Let me start by clarifying that the aim is not to detect and correct logical errors, but to detect and correct physical errors, i.e. those that occur on single physical qubits rather than the encoded qubit.
At the macro level, the answer is that if you have an $$X$$ error (say), you use a controlled-not to propagate it to another qubit where it appears as an $$X$$ error. Measuring in the $$X$$ basis cannot see that error. For example, if a single qubit state was in $$|+\rangle$$, an $$X$$ error would not change it, and you would be unable to detect the error. On the other hand, if the qubit was initially in $$|0\rangle$$ then the possibility of a flip would put it in either $$|0\rangle$$ or $$|1\rangle$$, which can be detected with a $$Z$$ measurement. (Put another way, $$X$$ errors commute with $$X$$ measurements, so they do not affect each other.)
More specifically, let's start by thinking about what happens if there is no error. In that case, $$|+\rangle$$ is the eigenstate of controlled-not, so it never changes. This is what initially threw me trying to think about a $$Z$$ measurement on this. But of course, it's not logical $$Z$$ measurement, but physical. So, basically, you'll get a single answer corresponding to one of the basis states used in either to 0 or 1 logical state, with no control over which. Let's call that answer $$x$$. What we do know is that $$H\cdot x=0$$, where $$H$$ is the parity-check matrix of the code (I'm being a bit loose about which of the two it is).
Now, what happens if there was a single physical $$X$$ error somewhere on the logical qubit. Controlled-nots (and we should think about the individual physical ones here, not the logical ones) propagate $$X$$ rotations from control to target. So, instead of getting an answer $$x$$, we'd get an answer $$x\oplus e$$ where $$e$$ is a vector representing the single qubit that had the error. So, now, if you apply the parity-check matrix, you get $$H\cdot(x\oplus e)=(H\cdot x)\oplus(H\cdot e)=H\cdot e.$$ The whole point of the parity-check matrices is that they can let you identify any single-qubit $$e$$. Hence, you know what correction to provide.
I'd like to add some details to DaftWullie's answer. The key point is that while the logical $$X$$ operator, which I'll write as $$X_L$$, must, by definition, satisfy
$$X_L \left|+\right>_L = \left.|+\right>_L, \quad X_L \left|-\right>_L = -\left|-\right>_L$$
the analogous equation does not hold when $$X_L$$ is replaced with an $$X$$ operator acting on a single qubit. The $$\oplus$$ on the bottom wire in the diagram, connected to its control on the top wire, indicates a controlled $$X$$ operator on each qubit. (CNOT pairing up wires, not a logical $$X_L$$.) In general this combination of CNOTs does not leave $$\left|+\right>$$ invariant.
Preliminaries on CSS codes. For simplicity, we suppose that the input codeword, on the top 'wire' (inverted commas since this indicates multiple wires) in the diagram in the question is zero logical, $$\left|0\right>_L$$. For Steane extraction we are working in a CSS code with $$k$$ logical qubits: let $$P$$ and $$Q$$ be the relevant matrices so
$$\left|0\right>_L = \sum_{u \in \langle P \rangle} \left|u\right>$$
where $$\langle P \rangle$$ is the row-space of $$P$$, and, by a MacWilliams' type calculation,
$$\left|+\right>_L = \frac{1}{2^{m/2}} \sum_{v \in \langle Q^\perp \rangle} \left|v\right>$$
where $$m$$ is the dimension of $$\langle Q^\perp \rangle$$.
Example. In the Steane $$[[7,1,3]]$$ code, using the parity check matrix from the Wikipedia article we have $$P = Q = \left( \begin{matrix} 1 & 0 & 0 & 1 & 0 & 1 & 1 \\ 0 & 1 &0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 & 1 & 1 \end{matrix} \right)$$ and $$\langle Q^\perp \rangle = \langle P \rangle \oplus \langle 1111111 \rangle$$, where $$\oplus$$ is the direct sum of vector spaces, so $$\left|0\right>_L = \sum_{u \in \langle P \rangle} \left|u\right>$$ and $$\left|+\right>_L = \frac{1}{\sqrt{2}} \sum_{u \in \langle P \rangle} \left|u\right> + \frac{1}{\sqrt{2}} \sum_{u \in \langle P \rangle} \left| u + 1111111\right> = \frac{1}{\sqrt{2}} \left|0\right>_L + \frac{1}{\sqrt{2}} \left|1\right>_L.$$ In general for an $$[[n,k,d]]$$ quantum code, $$\left|+\right>_L$$ is a sum of $$2^k$$ basis states in the $$Z$$-logical basis.
Diagrams. The diagram below shows the action of the CNOT pairing up each wire when the input qubit is a basis state in the $$Z$$-basis
|u> -------*---- |u>
|
|w> -------+---- |u+w>
Observe that flipping a bit in $$u$$ means that we get a corresponding bit flip in $$u+w$$. Thus we can deduce the second diagram below shows how a single qubit $$X$$ error, gets copied down' onto the ancilla.
---X---*------ ------*---X---
| = |
-------+------ ------+---X---
`
Calculation. We can now calculate the effect of a single qubit error, say in position $$1$$. On the input wires we have $$X^1 \left|0\right>_L$$ and $$\left|+\right>_L$$. On the output wires, by the second diagram, we have $$X^1 \left|0\right>_L$$ and
$$\frac{1}{2^{k/2}} \sum_{v \in \langle Q^\perp \rangle} X^1 \left|v\right>.$$
This expresses the output on the bottom 'wire' in the $$Z$$-basis. When we measure we get a random state $$X^1 \left| v \right>$$ where $$v \in \langle Q^\perp \rangle$$. This tells us $$\overline{v_1} v_2 \ldots v_n$$ where $$v_1 v_2 \ldots v_n$$ is a codeword in the code with generator matrix $$Q^\perp$$ and parity check matrix $$Q$$. We can now take the syndrome of this word and discover (provided this code is $$1$$-error correcting) that there was an error in the first position.
Example concluded. For the Steane [[7,4,1]]-code, we measure $$\overline{v_1}v_2 \ldots v_n$$ where $$v_1v_2\ldots v_n$$ is a codeword in the Hamming $$[7,4,3]$$-code with parity check matrix $$P$$. For instance, one codeword is $$1110001$$, so we might measure $$0110001$$. We now calculate the syndrome by left-multiplying the column vector by $$P$$, to get
$$\left( \begin{matrix} 1 & 0 & 0 & 1 & 0 & 1 & 1 \\ 0 & 1 &0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 & 1 & 1 \end{matrix} \right) \left( \begin{matrix} 0 \\ 1 \\ 1 \\ 0 \\ 0 \\ 0 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right).$$
Since this is the first column of $$P$$, we conclude that there was an $$X$$ error in the first position. | 2021-12-05 17:36:11 | {"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": 80, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8513452410697937, "perplexity": 351.3543342690671}, "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/1637964363215.8/warc/CC-MAIN-20211205160950-20211205190950-00507.warc.gz"} |
https://en.wikiversity.org/wiki/Advanced_Classical_Mechanics/Rigid_Bodies | ## Definition
• In a rigid body, no part of the body moves relative to another part of the body regardless of the force applied.
• A rigid body is equivalent to a system of particles restricted to have fixed distances apart.
There are two methods for solving for the motion of rigid bodies.
1. Write out the kinetic energy ${\displaystyle T}$ and use the Lagrangian to derive the equations of motion
2. Use Euler's equations to get the motion of the center of mass and the rotation.
## The Motion of a Rigid Body
The restrictions of rigid body motion require that distance between any two particles within the body remain constant. For example
${\displaystyle l_{ij}^{2}=\left({\vec {r}}_{i}-{\vec {r}}_{j}\right)^{2}=\left(x_{i}-x_{j}\right)^{2}+\left(y_{i}-y_{j}\right)^{2}+\left(z_{i}-z_{j}\right)^{2}}$
will be constant during the motion. Let's take the time derivative of this quantity,
${\displaystyle {\frac {dl_{ij}^{2}}{dt}}=2\left({\vec {r}}_{i}-{\vec {r}}_{j}\right)\cdot \left({\vec {v}}_{i}-{\vec {v}}_{j}\right)}$
For this to vanish either ${\displaystyle {\vec {v}}_{i}-{\vec {v}}_{j}}$ must vanish or the dot product must vanish. The first corresponds to a translation of the object. For the dot product to vanish, the velocity difference must be perpendicular to the position difference. Specifically,
${\displaystyle {\vec {v}}_{i}-{\vec {v}}_{j}={\vec {\omega }}_{ij}\times \left({\vec {r}}_{i}-{\vec {r}}_{j}\right).}$
At first it might appear that you can have a different ${\displaystyle {\vec {\omega }}_{ij}}$ for each pair of particles. It turns out that this is not the case. Let's calculate
${\displaystyle {\vec {v}}_{i}-{\vec {v}}_{k}=\left({\vec {v}}_{i}-{\vec {v}}_{j}\right)-\left({\vec {v}}_{k}-{\vec {v}}_{j}\right)={\vec {\omega }}_{ij}\times \left({\vec {r}}_{i}-{\vec {r}}_{j}\right)-{\vec {\omega }}_{kj}\times \left({\vec {r}}_{k}-{\vec {r}}_{j}\right).}$
On the other hand we could have calculated it as follows,
${\displaystyle {\vec {v}}_{i}-{\vec {v}}_{k}={\vec {\omega }}_{ik}\times \left({\vec {r}}_{i}-{\vec {r}}_{k}\right)={\vec {\omega }}_{ik}\times \left({\vec {r}}_{i}-{\vec {r}}_{j}\right)-{\vec {\omega }}_{ik}\times \left({\vec {r}}_{k}-{\vec {r}}_{j}\right).}$
For these expressions to be equal for arbitrary values of ${\displaystyle {\vec {r}}_{i}}$, ${\displaystyle {\vec {r}}_{j}}$ and ${\displaystyle {\vec {r}}_{k}}$, the three values of ${\displaystyle {\vec {\omega }}}$ must be equal to each other so we will drop the subscripts and call this quantity the angular velocity of the rotation.
To summarize, a rigid body has two types of motion:
1. Translation of the entire body, and
2. Uniform rotation about a particular axis (the direction of ${\displaystyle {\vec {\omega }}}$ and an angular speed given by the magnitude of ${\displaystyle {\vec {\omega }}}$.
### Angular Velocity
We can use some point within or outside the body as the origin and denote its position at any time by ${\displaystyle {\vec {r}}_{0}}$. The velocity of a particular part of the body is
${\displaystyle {\vec {v}}={\vec {v}}_{0}+{\vec {\omega }}\times \left({\vec {r}}-{\vec {r}}_{0}\right)}$
where ${\displaystyle {\vec {r}}}$ is the location of the part of the body and ${\displaystyle {\vec {v}}_{0}}$ is the velocity of the origin.
Let's move the origin to ${\displaystyle {\vec {r}}_{1}}$. The velocity of the particle does not change so we have
${\displaystyle {\vec {v}}={\vec {v}}_{0}+{\vec {\omega }}\times \left({\vec {r}}-{\vec {r}}_{0}\right)={\vec {v}}_{1}+{\vec {\omega }}\times \left({\vec {r}}-{\vec {r}}_{1}\right)}$
so
${\displaystyle {\vec {v}}_{1}={\vec {v}}_{0}+{\vec {\omega }}\times \left({\vec {r}}_{1}-{\vec {r}}_{0}\right).}$
The vector ${\displaystyle {\vec {\omega }}}$ is constant with respect to changes in the origin. A useful example of this is a wheel that rolls without slipping.
## Moment of Inertia
Rigid bodies are such an important part of classical mechanics that we have developed special techniques to calculate their kinetic energies and angular momenta. Let's calculate the kinetic energy of a mass in rigid body motion. Its velocity is
${\displaystyle {\vec {v}}_{i}={\vec {v}}_{0}+{\vec {\omega }}\times {\vec {r}}_{i}}$
where we have written the coordinates relative to the origin (${\displaystyle {\vec {r}}_{0}}$).
### Kinetic Energy
Its kinetic energy is
${\displaystyle T_{i}={\frac {1}{2}}m_{i}\left({\vec {v}}_{i}\cdot {\vec {v}}_{i}\right)={\frac {1}{2}}m_{i}\left({\vec {v}}_{0}+{\vec {\omega }}\times {\vec {r}}_{i}\right)\cdot \left({\vec {v}}_{0}+{\vec {\omega }}\times {\vec {r}}_{i}\right).}$
and multiplying out the terms we get
${\displaystyle T_{i}={\frac {1}{2}}m_{i}\left[v_{0}^{2}+2{\vec {v}}_{0}\cdot \left({\vec {\omega }}\times {\vec {r}}_{i}\right)+\left({\vec {\omega }}\times {\vec {r}}_{i}\right)^{2}\right].}$
We can simplify the result using the following fact
${\displaystyle \left({\vec {A}}\times {\vec {B}}\right)^{2}=A^{2}B^{2}\sin ^{2}\theta =A^{2}B^{2}\left(1-\cos ^{2}\theta \right)=A^{2}B^{2}\left[1-{\frac {\left({\vec {A}}\cdot {\vec {B}}\right)^{2}}{A^{2}B^{2}}}\right]=A^{2}B^{2}-\left({\vec {A}}\cdot {\vec {B}}\right)^{2}}$
to get
${\displaystyle T_{i}={\frac {1}{2}}m_{i}\left[v_{0}^{2}+2{\vec {v}}_{0}\cdot \left({\vec {\omega }}\times {\vec {r}}_{i}\right)+{\vec {\omega }}^{2}{\vec {r}}_{i}^{2}-\left({\vec {\omega }}\cdot {\vec {r}}_{i}\right)^{2}\right].}$
### Angular Momentum
Let's look at the angular momentum of the particle, we have
${\displaystyle {\vec {L}}_{i}=m_{i}{\vec {r}}_{i}\times {\vec {v}}_{i}=m_{i}{\vec {r}}_{i}\times {\vec {v}}_{0}+m_{i}{\vec {r}}_{i}\times \left({\vec {\omega }}\times {\vec {r}}_{i}\right).}$
Here we can use the vector triple product to simplify the result
${\displaystyle {\vec {A}}\times \left({\vec {B}}\times {\vec {C}}\right)={\vec {B}}\left({\vec {A}}\cdot {\vec {C}}\right)-{\vec {C}}\left({\vec {A}}\cdot {\vec {B}}\right)}$
which for us yields
${\displaystyle {\vec {L}}_{i}=m_{i}{\vec {r}}_{i}\times {\vec {v}}_{i}=m_{i}{\vec {r}}_{i}\times {\vec {v}}_{0}+m_{i}\left[{\vec {\omega }}r^{2}-{\vec {r}}\left({\vec {\omega }}\cdot {\vec {r}}\right)\right].}$
If ${\displaystyle {\vec {v}}_{0}=0}$ then
${\displaystyle T_{i}={\frac {1}{2}}{\vec {\omega }}\cdot {\vec {L}}_{i}.}$
In this case, let's write out ${\displaystyle {\vec {L}}_{i}}$ component by component, we have
${\displaystyle {\vec {L}}_{i}=m_{i}\left[{\begin{matrix}\omega _{x}r_{i}^{2}-x_{i}\left(\omega _{x}x_{i}+\omega _{y}y_{i}+\omega _{z}z_{i}\right)\\\omega _{y}r_{i}^{2}-y_{i}\left(\omega _{x}x_{i}+\omega _{y}y_{i}+\omega _{z}z_{i}\right)\\\omega _{z}r_{i}^{2}-z_{i}\left(\omega _{x}x_{i}+\omega _{y}y_{i}+\omega _{z}z_{i}\right)\end{matrix}}\right]=m_{i}\left[{\begin{matrix}y_{i}^{2}+z_{i}^{2}&-x_{i}y_{i}&-x_{i}z_{i}\\-x_{i}y_{i}&x_{i}^{2}+z_{i}^{2}&-y_{i}z_{i}\\-x_{i}z_{i}&-y_{i}z_{i}&x_{i}^{2}+y_{i}^{2}\end{matrix}}\right]\left[{\begin{matrix}\omega _{x}\\\omega _{y}\\\omega _{z}\end{matrix}}\right]\equiv {\vec {\vec {I}}}_{i}\cdot {\vec {\omega }}}$
defining the moment of inertia matrix for that particular particle.
### Putting It All Together
Because the quantities ${\displaystyle {\vec {v}}_{0}}$ and ${\displaystyle {\vec {\omega }}}$ are the same for every particle, we can sum up the kinetic energy and angular momentum for all of the particles to get
${\displaystyle T=\sum _{i}T_{i}={\frac {1}{2}}Mv_{0}^{2}+{\frac {1}{2}}{\vec {\omega }}\cdot {\vec {I}}\cdot {\vec {\omega }}+M{\vec {v}}_{0}\cdot \left({\vec {\omega }}\times {\vec {r}}_{CM}\right)}$
where ${\displaystyle M}$ is the total mass, ${\displaystyle {\vec {r}}_{CM}}$ is the location of the centre of mass and ${\displaystyle {\vec {\vec {I}}}=\sum _{i}{\vec {\vec {I}}}_{i}}$ is the moment of inertia of the body about the origin. The total angular momentum is
${\displaystyle {\vec {L}}=M{\vec {r}}_{CM}\times {\vec {v}}_{0}+{\vec {\vec {I}}}\cdot {\vec {\omega }}}$
Using the result for the scalar triple product we can show that
${\displaystyle {\vec {L}}={\frac {\partial T}{\partial \omega }}}$
so the angular momentum is the generalized momentum conjugate to the angular velocity; if the potential does not depend on angle, the angular momentum is a first integral of the motion.
### Manipulating the Moment of Inertia
#### Principal Axes
The moment of inertia is a positive definite matrix. We know this because when we multiply it by any non-zero vector ${\displaystyle {\vec {\omega }}}$, we get the kinetic energy of the body that is necessarily positive. Such a matrix can always be diagonalized. From a physical point of view, that means that the eigenvector of the matrix are known as the principal axes of the body. If the angular velocity points along one of the principal axes, then the angular momentum is parallel to the angular velocity. Furthermore, if one chooses the principal axes of the body to be one's coordinate axes, the diagonal element of the moment of inertia matrix vanish, so for example,
${\displaystyle {\vec {L}}=\left[{\begin{matrix}I_{xx}\omega _{x}\\I_{yy}\omega _{y}\\I_{zz}\omega _{z}\end{matrix}}\right]}$
and the kinetic energy is
${\displaystyle T={\frac {1}{2}}Mv_{0}^{2}+M{\vec {v}}_{0}\cdot \left({\vec {\omega }}\times {\vec {r}}_{CM}\right)+{\frac {1}{2}}I_{xx}\omega _{x}^{2}+{\frac {1}{2}}I_{yy}\omega _{y}^{2}+{\frac {1}{2}}I_{zz}\omega _{z}^{2}}$
#### Change of Origin
Sometimes it is easiest to calculate the moment of inertia about a particular point but you are interested in the moment of inertia about another point. There is a straightforward prescription to achieve this. First, let's use the center of mass of the system as the origin, so we have
${\displaystyle T={\frac {1}{2}}Mv_{CM}^{2}+{\frac {1}{2}}{\vec {\omega }}\cdot {\vec {\vec {I}}}_{CM}\cdot {\vec {\omega }}}$
Let's write this same equation with respect to a new origin ${\displaystyle {\vec {r}}_{0}}$ so we have
${\displaystyle T={\frac {1}{2}}M\left({\vec {v}}_{0}+{\vec {\omega }}\times {\vec {r}}_{CM}\right)^{2}+{\frac {1}{2}}{\vec {\omega }}\cdot {\vec {\vec {I}}}_{CM}\cdot {\vec {\omega }}}$
${\displaystyle ={\frac {1}{2}}Mv_{0}^{2}+M{\vec {v}}_{0}\cdot \left({\vec {\omega }}\times {\vec {r}}_{CM}\right)+{\frac {1}{2}}M\left({\vec {\omega }}\times {\vec {r}}_{CM}\right)^{2}+{\frac {1}{2}}{\vec {\omega }}\cdot {\vec {\vec {I}}}_{CM}\cdot {\vec {\omega }}.}$
If we compare this result to the earlier result we find
${\displaystyle {\vec {\omega }}\cdot {\vec {\vec {I}}}{\vec {\omega }}={\vec {\omega }}\cdot {\vec {\vec {I}}}_{CM}\cdot {\vec {\omega }}+M\left({\vec {\omega }}\times {\vec {r}}_{CM}\right)^{2}.}$
On a component by component basis we have
${\displaystyle {\vec {\vec {I}}}={\vec {\vec {I}}}_{CM}+M\left[{\begin{matrix}y_{CM}^{2}+z_{CM}^{2}&x_{CM}y_{CM}&x_{CM}z_{CM}\\x_{CM}y_{CM}&x_{CM}^{2}+z_{CM}^{2}&My_{CM}z_{CM}\\x_{CM}z_{CM}&y_{CM}z_{CM}&x_{CM}^{2}+y_{CM}^{2}\end{matrix}}\right]}$
To change from one origin to another you first use the equation above to get the moment of inertia relative to the centre of mass and then use it a second time to get the moment of inertia about the new origin.
### Calculating a Moment of Inertia
Let's calculate the moment of inertia of an ellipsoid. Let's assume that the axes of the ellipsoid are line up along the coordinate axes. Along the ${\displaystyle x-}$axis the ellipsoid spans from ${\displaystyle -a}$ to ${\displaystyle a}$. Along the ${\displaystyle y}$ and ${\displaystyle z-}$axes, the bounds are ${\displaystyle b}$ and ${\displaystyle c}$ respectively.
Because the coordinate axes are the principal axes of the ellipsoid, the off-diagonal components of the matrix will vanish. We will calculate the sum of ${\displaystyle z^{2}}$ over the ellipsoid first. We have
${\displaystyle \sum _{i}m_{i}z_{i}^{2}=\int _{\rm {Ellipsoid}}\rho \,z^{2}\,dV=\int _{\rm {Ellipsoid}}\rho \,z^{2}\,dxdydz.}$
The region of integration is rather complicate in the Cartesian coordinates but this gives us a place to start. We would like to simplify things a bit by defining
${\displaystyle x=au,y=bv,z=cw.}$
In these new coordinates the ellipsoid becomes a sphere of unit radius. We also have to include the Jacobian of the coordinate transformation
${\displaystyle {\frac {\partial \left(x,y,z\right)}{\partial \left(u,v,w\right)}}=abc.}$
Now our integral looks like
${\displaystyle \sum _{i}m_{i}z_{i}^{2}=\int _{\rm {UnitSphere}}\rho \,abc(cw)^{2}\,dudvdw.}$
Let's calculate the density of the ellipsoid in terms of its ${\displaystyle M}$ and volume. The volume of the ellipsoid is
${\displaystyle V=\int _{\rm {Ellipsoid}}\,dxdydz=\int _{\rm {UnitSphere}}abc\,dudvdw={\frac {4}{3}}\pi abc}$
so we can rewrite the density in the moment of inertia integral
${\displaystyle \sum _{i}m_{i}z_{i}^{2}=\int _{\rm {UnitSphere}}{\frac {3M}{4\pi abc}}abc(cw)^{2}\,dudvdw={\frac {3Mc^{2}}{4\pi }}\int _{\rm {UnitSphere}}w^{2}\,dudvdw.}$
The easiest way to integrate over a unit sphere is to use spherical coordinates. Let's do it:
${\displaystyle \sum _{i}m_{i}z_{i}^{2}={\frac {3Mc^{2}}{4\pi }}\int _{0}^{1}dr\int _{0}^{\pi }d\theta \int _{0}^{2\pi }d\phi \,r^{2}\sin \theta \,r^{2}\cos ^{2}\theta ={\frac {3Mc^{2}}{4\pi }}{\frac {2}{5}}\pi \int _{-1}^{1}d\left(\cos \theta \right)\cos ^{2}\theta ={\frac {3Mc^{2}}{4\pi }}{\frac {4}{15}}\pi ={\frac {1}{5}}Mc^{2}.}$
By symmetry we can calculate the results for the other two coordinates. We have
${\displaystyle \sum _{i}m_{i}x_{i}^{2}={\frac {1}{5}}Ma^{2},\sum _{i}m_{i}y_{i}^{2}={\frac {1}{5}}Mb^{2}}$
and the moment of inertia matrix about the centre of mass have the following non-zero components:
${\displaystyle I_{xx}={\frac {M}{5}}\left(b^{2}+c^{2}\right),I_{yy}={\frac {M}{5}}\left(a^{2}+c^{2}\right),I_{zz}={\frac {M}{5}}\left(a^{2}+b^{2}\right).}$
This is an example of reducing a general shape to a symmetric one through a transformation of variables. One important result of this technique is rule.
## Ellipsoid of Inertia
We saw earlier that if the velocity of the origin vanished, then we could write ${\displaystyle T={\frac {1}{2}}{\vec {L}}\cdot {\vec {\omega }}}$. Now if there are no forces on the body, the kinetic energy of the body is conserved. If there are no forces on the body, there are no torques, so the angular momentum is conserved as well. Does this mean that the angular velocity is constant?
The answer is of course no because the equation only says that the component of the angular velocity along the direction of the angular momentum is constant. The other components can change. For example, if the angular velocity lies along one of the principal axes (eigenvectors of the moment of inertia matrix), then the angular momentum points in the same direction and the angular velocity must stay constant in direction and magnitude.
A usual picture to understand the force-free motion of a rigid body is the ellipsoid of inertia. Let's use the principal axes of the body as a coordinate frame to write the kinetic energy
${\displaystyle T={\frac {1}{2}}\left(I_{xx}\omega _{x}^{2}+I_{yy}\omega _{y}^{2}+I_{zz}\omega _{z}^{2}\right).}$
The ellipsoid of inertia is the locus of values of ${\displaystyle {\vec {\omega }}}$ that give a kinetic energy of the body.
### Ellipsoid of Inertia of an Ellipsoid
Let's calculate the shape of the ellipsoid of inertia of an ellipsoid. We have
${\displaystyle T={\frac {1}{2}}{\frac {M}{5}}\left[\left(b^{2}+c^{2}\right)\omega _{x}^{2}+\left(a^{2}+c^{2}\right)\omega _{y}^{2}+\left(a^{2}+b^{2}\right)\omega _{z}^{2}\right],}$
so along the ${\displaystyle x}$, ${\displaystyle y}$ and ${\displaystyle z}$ directions, the axes of the ellipsoid of inertia are
${\displaystyle {\sqrt {\frac {10T}{M(b^{2}+c^{2})}}},{\sqrt {\frac {10T}{M(a^{2}+c^{2})}}},{\sqrt {\frac {10T}{M(a^{2}+b^{2})}}}.}$
The size of the ellipsoid of inertia increases as ${\displaystyle T}$ and decreases with the mass and size of the actual ellipsoidal body. The shape of the ellipsoid of inertia reflects the shape of the physical ellipsoid. The largest axis of the ellipsoid is also the largest axis of the corresponding ellipsoid of inertia. If two axes of the ellipsoid are the same size, the corresponding axes of the ellipsoid of inertia will be equal as well.
### Motion of the Ellipsoid of Inertia
The Ellipsoid of Inertia Rolling on the Invariable Plane
To conserve energy the angular velocity must remain on the ellipsoid of inertia and the orientation of ellipsoid of inertia determines the orientation of the body. We can use the conservation of angular momentum to orient the ellipsoid. If we remember that ${\displaystyle {\vec {L}}=\partial T/\partial {\vec {\omega }}}$ we see that the normal to the ellipsoid of inertia is the angular momentum; therefore, as the ellipsoid of inertia rotates about the angular velocity, it must remain tangent to a plane perpendicular to the angular momentum, the invariable plane. Furthermore, the angular velocity is restricted to move along the ellipsoid of inertia along curves of constant ${\displaystyle \left|{\vec {L}}\right|}$. This curve is called the polhode, and it traces the path of the angular velocity through the body. The angular velocity also traces a path in the invariable plane called the herpolhode. The herpolhode is the path of the angular velocity through space. We can also write that ${\displaystyle 2T={\vec {\omega }}\cdot {\vec {L}}}$ so the centre of the ellipsoid of inertia must remain a constant distance above the invariable plane. The height of the centre of the ellipsoid is ${\displaystyle 2T/\left|{\vec {L}}\right|}$, so when the ellipsoid is down low the angular momentum is high relative to the kinetic energy.
Keeping the ellipsoid of inertia tangent to the invariable plane preserves the direction of the angular momentum. The magnitude of the angular momentum is also conserved. This defines a second ellipsoid through the equation
${\displaystyle \left|{\vec {L}}\right|^{2}=I_{xx}^{2}\omega _{x}^{2}+I_{yy}^{2}\omega _{y}^{2}+I_{zz}^{2}\omega _{z}^{2}}$
so the polhodes are the intersections of the ellipsoid of inertia with the angular momentum ellipsoid. The angular momentum ellipsoid deviates more from a sphere than the ellipsoid of inertia; this gives an idea of where the polhodes lie for small and large angular momenta.
If the body has two principal axes with equal moments of inertia, the polhodes are circles centered on the axis with the unique moment of inertia, and the herpolhodes are circles in the invariable plane. If the unique moment of inertia is larger than the others, the body is called oblate. Otherwise it is prolate -- in analogy to the spheroids with the similar properties. In this case, the free rotation of the body consists of a constant precession of the angular velocity in a circle about the angular momentum in the space frame (the space cone) and about the unique principal axis in the body frame (the body cone). In the oblate case the body cone rolls within the space cone, and in the prolate case, the body cone rolls on the outside of the space cone.
The polhodes near the major axis
If no two moments of inertia are equal to each other, the polhodes are much more complicated. The polhodes near the axes with the largest and smallest moment of inertia are closed, but those near the intermediate axis are not; consequently, rotation about the intermediate axis is not stable as we shall see when we analyze the free rotation using Euler's equations.
## Euler's Equations
So far we have tried to find a graphical description of the motion of an object without any torques. Although "the polhode rolls on the herpolhode without slipping" might paint a nice Victorian picture, it doesn't exploit the mathematical language of physics to which we are now accustomed. We know that the change in the angular momentum is equal to the torques on the body
${\displaystyle {\frac {d{\vec {L}}}{dt}}={\vec {N}}.}$
Let's write the angular momentum in terms of the principal axes of the body. Here we will be looking at how the principal axes move so we will use the unit vectors ${\displaystyle {\hat {1}},{\hat {2}}}$ and ${\displaystyle {\hat {3}}}$ to denote the current direction of the principal axes of the body. We can write
${\displaystyle {\frac {d{\vec {L}}}{dt}}={\frac {\partial L_{1}}{\partial t}}{\hat {1}}+{\frac {\partial L_{2}}{\partial t}}{\hat {2}}+{\frac {\partial L_{3}}{\partial t}}{\hat {3}}+L_{1}{\frac {\partial {\hat {1}}}{\partial t}}+L_{2}{\frac {\partial {\hat {2}}}{\partial t}}+L_{3}{\frac {\partial {\hat {3}}}{\partial t}}}$
${\displaystyle ={\frac {\partial L_{1}}{\partial t}}{\hat {1}}+{\frac {\partial L_{2}}{\partial t}}{\hat {2}}+{\frac {\partial L_{3}}{\partial t}}{\hat {3}}+L_{1}{\vec {\omega }}\times {\hat {1}}+L_{2}{\vec {\omega }}\times {\hat {2}}+L_{3}{\vec {\omega }}\times {\hat {3}}}$
Let's define the angular momentum relative to the body axes as
${\displaystyle {\vec {L}}_{b}=\left[{\begin{matrix}L_{1}\\L_{2}\\L_{3}\end{matrix}}\right]=\left[{\begin{matrix}\omega _{1}I_{1}\\\omega _{2}I_{2}\\\omega _{3}I_{3}\end{matrix}}\right]}$
to get
${\displaystyle {\frac {d{\vec {L}}}{dt}}={\frac {d{\vec {L}}_{b}}{dt}}+{\vec {\omega }}\times {\vec {L}}_{b}={\vec {N}}}$
where the torque and angular velocity are written relative to the body axes. One can imagine using Euler's equations would be quite cumbersome if one had to include the torques because one would always have to transform the torques from the inertia frame to the body frame. Let's write out these equations component by component
${\displaystyle I_{1}{\dot {\omega }}_{1}-\omega _{2}\omega _{3}\left(I_{2}-I_{3}\right)=N_{1}}$
${\displaystyle I_{2}{\dot {\omega }}_{2}-\omega _{3}\omega _{1}\left(I_{3}-I_{1}\right)=N_{2}}$
${\displaystyle I_{3}{\dot {\omega }}_{3}-\omega _{1}\omega _{2}\left(I_{1}-I_{2}\right)=N_{3}}$
These equations are most powerful when there are no torques, so the right-hand sides are zero. In this case we can see immediately that if any two moments of inertia differ than the angular velocity must lie along one of the principal axes to remain constant.
If two of the moments of inertia are equal to each other (let's take ${\displaystyle I_{1}=I_{2}}$), we can solve the torque-free Euler's equations exactly. We have
${\displaystyle I_{1}{\dot {\omega }}_{1}=\left(I_{1}-I_{3}\right)\omega _{3}\omega _{2}}$
${\displaystyle I_{2}{\dot {\omega }}_{2}=-\left(I_{1}-I_{3}\right)\omega _{3}\omega _{1}}$
${\displaystyle I_{3}{\dot {\omega }}_{3}=0}$
We can solve this with
${\displaystyle {\dot {\omega }}_{1}=-\Omega \omega _{2},{\dot {\omega }}_{2}=\Omega \omega _{1}}$
with
${\displaystyle \Omega ={\frac {I_{3}-I_{1}}{I_{1}}}\omega _{3},}$
yielding the solution
${\displaystyle \omega _{1}=A\cos \Omega t,\,\omega _{2}=A\sin \Omega t.}$
This result coincides with the graphical picture that the polhodes of a symmetric body are circles centered on the axis with the unique moment of inertia.
### Chandler Wobble
We can calculate the expected rate for the Earth to precess due to its oblatness. Let's estimate the ratio of the moments of inertia in terms of the radii of the Earth at the equator and at the poles, we have
${\displaystyle {\frac {I_{3}-I_{1}}{I_{1}}}={\frac {2R_{\rm {eq}}^{2}-\left(R_{\rm {eq}}^{2}+R_{\rm {pole}}^{2}\right)}{R_{\rm {eq}}^{2}+R_{\rm {pole}}^{2}}}}$
${\displaystyle ={\frac {\left(R_{\rm {eq}}-R_{\rm {pole}}\right)\left(R_{\rm {eq}}+R_{\rm {pole}}\right)}{R_{\rm {eq}}^{2}+R_{\rm {pole}}^{2}}}\approx {\frac {R_{\rm {eq}}-R_{\rm {pole}}}{R_{\rm {eq}}}}.}$
where in the last step we have assume that the difference between the polar and equatorial radius is small. We have ${\displaystyle R_{\rm {eq}}\approx 6378{\rm {km}},R_{\rm {pole}}\approx 6356{\rm {km}}}$, so the ratio is about ${\displaystyle 1/290}$. What is ${\displaystyle \omega _{3}}$ for the Earth? It is ${\displaystyle 2\pi /1{\rm {day}}}$, so the period of the free precession of the Earth should be about 290 days. The period is somewhat longer about 433 days w:Chandler_wobble. You can check out a plot of the wobble at the Paris Observatory. Why the Earth wobbles is a bit of a puzzle since because the Earth is not a solid body, without a driving force it would have damped long ago. Since the period of the wobble is similar to a year, seasonal changes are thought to be to blame.
I first learned about the wobble while spending a summer at the Pulkovo Astronomical Observatory near St. Petersburg. I saw a plot of the location of the Earth's pole on the surface of the Earth -- it moves in a irregular circular shape with a radius of 3 to 15 meters. The wobble can be measured by accurate measurements of the locations of stars relative to landmarks on the Earth. I asked my hosts why did they need to keep track the wobble so accurately. "To aim our missiles at your cities -- of course" came the reply.
### Triaxial Body
Unfortunately, we can't find such a straightforward to the solution of the motion of a triaxial body. In the symmetric case, one of the components of the angular velocity was constant with time. This is not the case for a triaxial body. We have the following Euler's equations
${\displaystyle I_{1}{\dot {\omega }}_{1}=\left(I_{2}-I_{3}\right)\omega _{3}\omega _{2}}$
${\displaystyle I_{2}{\dot {\omega }}_{2}=\left(I_{3}-I_{1}\right)\omega _{3}\omega _{1}}$
${\displaystyle I_{3}{\dot {\omega }}_{3}=\left(I_{1}-I_{2}\right)\omega _{1}\omega _{2}.}$
Let's take ${\displaystyle I_{3}>I_{2}>I_{1}}$ and define the following three positive ratios of moments of inertia
${\displaystyle A={\frac {I_{3}-I_{2}}{I_{1}}},B={\frac {I_{3}-I_{1}}{I_{2}}},C={\frac {I_{2}-I_{1}}{I_{3}}}}$
giving the following equations
${\displaystyle {\dot {\omega }}_{1}=-A\omega _{3}\omega _{2}}$
${\displaystyle {\dot {\omega }}_{2}=+B\omega _{3}\omega _{1}}$
${\displaystyle {\dot {\omega }}_{3}=-C\omega _{1}\omega _{2}}$
#### Major Axis
Let's assume that the rotation is nearly about the major axis, so ${\displaystyle \omega _{1}\gg \ \omega _{2},\omega _{3}}$. We will only include the latter components of the angular velocity to first order, so we have
${\displaystyle {\dot {\omega }}_{1}\approx 0,{\dot {\omega }}_{2}=B\omega _{3}\omega _{1},{\dot {\omega }}_{3}=-C\omega _{2}\omega _{1}.}$
If we define ${\displaystyle \Omega ={\sqrt {BC}}\omega _{1}}$ we have the following solution for ${\displaystyle \omega _{2}}$ and ${\displaystyle \omega _{3}}$,
${\displaystyle \omega _{3}={\sqrt {\frac {B}{C}}}K\cos \Omega t,\omega _{2}=K\sin \Omega t,}$
so the angular velocity travels in an ellipse centered on the major axis.
#### Minor Axis
Let's assume that the rotation is nearly about the minor axis, so ${\displaystyle \omega _{3}\gg \ \omega _{1},\omega _{2}}$. We will only include the latter components of the angular velocity to first order, so we have
${\displaystyle {\dot {\omega }}_{3}\approx 0,{\dot {\omega }}_{1}=-A\omega _{3}\omega _{2},{\dot {\omega }}_{2}=B\omega _{3}\omega _{1}.}$
If we define ${\displaystyle \Omega ={\sqrt {AB}}\omega _{3}}$ we have the following solution for ${\displaystyle \omega _{1}}$ and ${\displaystyle \omega _{2}}$,
${\displaystyle \omega _{1}={\sqrt {\frac {A}{B}}}K\cos \Omega t,\omega _{2}=K\sin \Omega t,}$
so the angular velocity travels in an ellipse centered on the minor axis.
#### Intermediate Axis
Let's assume that the rotation is nearly about the intermediate axis, so ${\displaystyle \omega _{2}\gg \ \omega _{1},\omega _{3}}$. We will only include the latter components of the angular velocity to first order, so we have
${\displaystyle {\dot {\omega }}_{2}\approx 0,{\dot {\omega }}_{1}=-A\omega _{3}\omega _{2},{\dot {\omega }}_{3}=-C\omega _{1}\omega _{2}.}$
If we define ${\displaystyle \alpha ={\sqrt {AC}}\omega _{2}}$ we have the following solution for ${\displaystyle \omega _{1}}$ and ${\displaystyle \omega _{3}}$,
${\displaystyle \omega _{1}=K_{1}e^{\alpha t},\omega _{3}=K_{3}e^{\alpha t},\,}$
so the angular velocity diverges exponetially away from the intermediate axis.
## Lagrangian Treatment of Rigid-Body Motion
Although it is possible to include the effects of torques on the motion of a rigid body within the framework of Euler's equations, it is rather cumbersome. Here we will develop the Lagrangian treatment of a top -- a symmetric rigid body with a torque from the Earth's gravity.
### Euler Angles
In a Lagrangian treatment it is necessary to find a set of independent coordinates to describe the position of the system at any time. For rigid bodies a convenient set of such coordinates are called the Euler angles. There are several different conventions for the Euler angles. The convention used here makes the analysis of the top simpler.
To get from one set of coordinates to the other, we have to do a series of three rotations. We imagine that the primed coordinates are the body coordinates and the unprimed coordinates are the inertia coordinates; therefore, it is straightforward to write the kinetic energy in the primed coordinates and the potential energy in the unprimed coordinates. Here are the three steps in words and in terms of a rotation matrix
1. Rotate by an angle ${\displaystyle \phi }$ about the initial ${\displaystyle z-}$axis. The final position of the ${\displaystyle x-}$axis is called the line of nodes.
${\displaystyle D=\left[{\begin{matrix}\cos \phi &\sin \phi &0\\-\sin \phi &\cos \phi &0\\0&0&1\end{matrix}}\right]}$
1. Rotate about the line of nodes by an angle ${\displaystyle \theta }$. This changes the direction of the ${\displaystyle z-}$axis and pitches the ${\displaystyle y-}$axis out of the initial ${\displaystyle x-y-}$plane.
${\displaystyle C=\left[{\begin{matrix}1&0&0\\0&\cos \theta &\sin \theta \\0&-\sin \theta &\cos \theta \end{matrix}}\right]}$
1. Finally rotate about the new ${\displaystyle z-}$axis by an angle ${\displaystyle \psi }$. This takes the ${\displaystyle x-}$axis out of the initial ${\displaystyle x-y-}$plane.
${\displaystyle B=\left[{\begin{matrix}\cos \psi &\sin \psi &0\\-\sin \psi &\cos \psi &0\\0&0&1\end{matrix}}\right]}$
Euler proved that any rotation can be decomposed into these three rotations. One can see this mathematically by constructing the rotation matrix for these three successive transformations (${\displaystyle A=BCD}$) and verifying that it is the most general rotation matrix.
### The Kinetic Energy
To calculate the kinetic energy of the body, we have to find the angular velocity in terms of the coordinates, ${\displaystyle \phi ,\theta }$ and ${\displaystyle \psi }$ and their time derivatives. In any coordinate system, the angular velocity is directed along the axis of rotation. We can construct the total angular velocity vector using the following information:
1. Changes in ${\displaystyle \phi }$ produce rotations about the ${\displaystyle z-}$axis (inertial frame).
2. Changes in ${\displaystyle \psi }$ produce rotations about the ${\displaystyle z'-}$axis (body frame).
3. Changes in ${\displaystyle \theta }$ produce rotations about the line of nodes.
We can write the angular velocity relative to the inertial frame to get
${\displaystyle {\vec {\omega }}=\left[{\begin{matrix}0\\0\\1\end{matrix}}\right]{\dot {\phi }}+\left[{\begin{matrix}\cos \phi \\\sin \phi \\0\end{matrix}}\right]{\dot {\theta }}+\left[{\begin{matrix}\sin \theta \sin \phi \\-\sin \theta \cos \phi \\\cos \theta \end{matrix}}\right]{\dot {\psi }}}$
or relative to the body frame
${\displaystyle {\vec {\omega }}=\left[{\begin{matrix}0\\0\\1\end{matrix}}\right]{\dot {\psi }}+\left[{\begin{matrix}\cos \psi \\-\sin \psi \\0\end{matrix}}\right]{\dot {\theta }}+\left[{\begin{matrix}\sin \theta \sin \psi \\\sin \theta \cos \psi \\\cos \theta \end{matrix}}\right]{\dot {\phi }}.}$
The first form is useful to visualize what the body is doing, while the second form is good for calculating the kinetic energy because we have used the primed coordinate system to characterize the location of the principal axes of the body.
For simplicity let's specialize to symmetric bodies with ${\displaystyle I_{x'}=I_{y'}}$, so we have
${\displaystyle T={\frac {1}{2}}I_{x'}\left(\omega _{x'}^{2}+\omega _{y'}^{2}\right)+{\frac {1}{2}}I_{z'}\omega _{z'}^{2}.}$
Because the primed frame is fixed to the body, the moments of inertia in the kinetic energy are constant with time. Let's first calculate
${\displaystyle \omega _{x'}^{2}+\omega _{y'}^{2}=\left(\cos \psi {\dot {\theta }}+\sin \psi \sin \theta {\dot {\phi }}\right)^{2}+\left(-\sin \psi {\dot {\theta }}+\cos \psi \sin \theta {\dot {\phi }}\right)^{2}}$
${\displaystyle =\cos ^{2}\psi {\dot {\theta }}^{2}+\sin ^{2}\psi \sin ^{2}\theta {\dot {\phi }}^{2}+2\sin \psi \cos \psi \sin \theta {\dot {\theta }}{\dot {\phi }}+\sin ^{2}\psi {\dot {\theta }}^{2}+\cos ^{2}\psi \sin ^{2}\theta {\dot {\phi }}^{2}-2\sin \psi \cos \psi \sin \theta {\dot {\theta }}{\dot {\phi }}}$
${\displaystyle ={\dot {\theta }}^{2}+\sin ^{2}\theta {\dot {\phi }}^{2}.}$
This simplication only obtains for a symmetric body. The ${\displaystyle z'-}$component is
${\displaystyle \omega _{z'}^{2}=\left({\dot {\psi }}+{\dot {\phi }}\cos \theta \right)^{2}.}$
Combining these results yields the kinetic energy
${\displaystyle T={\frac {I_{x'}}{2}}\left({\dot {\theta }}^{2}+{\dot {\phi }}^{2}\sin ^{2}\theta \right)+{\frac {I_{z'}}{2}}\left({\dot {\psi }}+{\dot {\phi }}\cos \theta \right)^{2}}$
### The Potential Energy
The potential energy is pretty simple using the Euler angles. If ${\displaystyle l}$ is the distance between the pivot point and the center of mass of the top, we have
${\displaystyle V=Mgl\cos \theta }$
### The Lagrangian
The Lagrangian is the difference between the kinetic and the potential energy, so we have
${\displaystyle L=T-V={\frac {I_{x'}}{2}}\left({\dot {\theta }}^{2}+{\dot {\phi }}^{2}\sin ^{2}\theta \right)+{\frac {I_{z'}}{2}}\left({\dot {\psi }}+{\dot {\phi }}\cos \theta \right)^{2}-Mgl\cos \theta .}$
We notice immediately that ${\displaystyle \psi }$ and ${\displaystyle \phi }$ don't appear in the Lagrangian. The system looks the same as in turns around the ${\displaystyle z-}$ and ${\displaystyle z'-}$axes. This means that we have two conserved quantities, reducing the three dimensional problem to a single dimension (we will choose this to be ${\displaystyle \theta }$). Because the Lagrangian does not depend on time, the Hamiltonian is conserved as well so we can use the techniques outlined in Linear Motion to understand the top.
#### Integrals of the Motion
To calculate the conserved momenta we have to calculate the partial derivatives of the Lagrangian with respect to the generalized velocities. We have
${\displaystyle p_{\psi }={\frac {\partial L}{\partial {\dot {\psi }}}}=I_{z'}\left({\dot {\psi }}+{\dot {\phi }}\cos \theta \right)=I_{z'}\omega _{z'}\equiv I_{x'}a}$
where the last step defines the conserved quantity ${\displaystyle a}$ that has units of angular velocity. Similarly for the coordinate ${\displaystyle \phi }$, we have
${\displaystyle p_{\phi }={\frac {\partial L}{\partial {\dot {\phi }}}}=\left(I_{x'}\sin ^{2}\theta +I_{z'}\cos ^{2}\theta \right){\dot {\phi }}+I_{z'}\cos \theta {\dot {\psi }}\equiv I_{x'}b.}$
Finally we have the Hamiltonian that is also conserved,
${\displaystyle H=T+V={\frac {I_{x'}}{2}}\left({\dot {\theta }}^{2}+{\dot {\phi }}^{2}\sin ^{2}\theta \right)+{\frac {I_{z'}}{2}}\left({\dot {\psi }}+{\dot {\phi }}\cos \theta \right)^{2}+Mgl\cos \theta .}$
As we mentioned earlier, these three integrals are sufficient to solve for the motion. We can use the two conserved momenta to solve for the generalized velocities, ${\displaystyle {\dot {\psi }}}$ and ${\displaystyle {\dot {\phi }}}$, in terms of the conserved quantities ${\displaystyle a}$ and ${\displaystyle b}$ and the value of the third coordinate ${\displaystyle \theta }$. We have
${\displaystyle {\dot {\phi }}={\frac {b-a\cos \theta }{\sin ^{2}\theta }},}$
${\displaystyle {\dot {\psi }}={\frac {I_{x'}a}{I_{z'}}}-\cos \theta {\frac {b-a\cos \theta }{\sin ^{2}\theta }}.}$
#### Conservation of Energy
Since both the Hamiltonian and the angular velocity along the ${\displaystyle z'-}$axis are conserved, let's define the conserved quantity
${\displaystyle E'=E-{\frac {I_{z'}\omega _{z'}^{2}}{2}}={\frac {I_{x'}{\dot {\theta }}^{2}}{2}}+{\frac {I_{x'}}{2}}{\frac {\left(b-a\cos \theta \right)^{2}}{\sin ^{2}\theta }}+Mgl\cos \theta }$
which we can rewrite as
${\displaystyle E'={\frac {I_{x'}{\dot {\theta }}^{2}}{2}}+V'\left(\theta \right)}$
where
${\displaystyle V'(\theta )={\frac {I_{x'}}{2}}{\frac {\left(b-a\cos \theta \right)^{2}}{\sin ^{2}\theta }}+Mgl\cos \theta .}$
We can make further progress by looking at ${\displaystyle u=\cos \theta }$ which will simplify the potential. Let's write
${\displaystyle {\dot {u}}^{2}=\sin ^{2}\theta {\dot {\theta }}^{2}=\left(1-u^{2}\right){\dot {\theta }}^{2}}$
so we can write ${\displaystyle E'}$ in terms of ${\displaystyle u}$ and its time derivative
${\displaystyle E'={\frac {I_{x'}}{2}}{\frac {{\dot {u}}^{2}}{1-u^{2}}}+{\frac {I_{x'}}{2}}{\frac {\left(b-au\right)^{2}}{1-u^{2}}}+Mglu.}$
What remains is to solve for ${\displaystyle {\dot {u}}^{2}}$ to get the boundaries of the motion
${\displaystyle {\dot {u}}^{2}=\left(E'-Mglu\right){\frac {2}{I_{x'}}}\left(1-u^{2}\right)-\left(b-au\right)^{2}}$
### The Motion of a Symmetric Top
The equation for ${\displaystyle {\dot {u}}^{2}}$ is cubic so it will generally have three roots (designated at ${\displaystyle u_{1,2,3}}$ -- it must have at least one root because for large positive values of ${\displaystyle u}$ ${\displaystyle {\dot {u}}^{2}}$ is positive and for large negative values of ${\displaystyle u}$ ${\displaystyle {\dot {u}}^{2}}$ is negative. As the top moves from straight up to straight down, ${\displaystyle u}$ ranges from 1 to -1, so only the roots within this range are important. Furthermore, the equation for ${\displaystyle {\dot {\phi }}}$ has a root at ${\displaystyle u_{0}=b/a}$.
The figure shows a typical curve for ${\displaystyle {\dot {u}}^{2}}$ as a function of ${\displaystyle \theta }$. The motion is bounded between ${\displaystyle u=0.5}$ and ${\displaystyle u=0.8}$. Depending on the value of ${\displaystyle b/a}$, the motion in the ${\displaystyle \phi }$-direction can be monotonic or not. If ${\displaystyle b/a=0.8}$, the motion will have a cusp at the upper end of the range.
Motion of the top with b/a=0.75
Motion of the top with b/a=0.8
Motion of the top with b/a=0.85
The figures to the right show the motion of the ${\displaystyle z'-}$axis as a function of time for several values of ${\displaystyle b/a}$ with the curve for ${\displaystyle {\dot {u}}^{2}}$ above. The bounds of the motion in all three cases is from ${\displaystyle u=0.5}$ to ${\displaystyle u=0.8}$. How can we set up a top to execute these three classes of motion? If we release the axis of the top from a stationary position, the axis will first drop (nutation) because of gravity. As it drops the angular momentum about the vertical axis must be conserved so the tip of the top starts moving sideways as well (precession) -- yielding the motion in the middle figure. On the other hand if we release the tip of the top with some motion in the direction of the precession, we will get the lower figure. Finally to get the rather freaky looking motion in the top figure, we release the top with some motion opposing the direction of the precession.
Specifically let's examine the case where we release the top from a stationary position, so ${\displaystyle u_{0}=b/a=u_{2}}$ where ${\displaystyle u_{2}}$ is the middle root of the ${\displaystyle {\dot {u}}^{2}}$ curve. Initially we have
${\displaystyle E'=Mgl\cos \theta =Mgl{\frac {b}{a}}}$
so we can write
${\displaystyle {\dot {u}}^{2}=\left({\frac {b}{a}}-u\right)Mgl{\frac {2}{I_{x'}}}\left(1-u^{2}\right)-\left(b-au\right)^{2}}$
${\displaystyle =\left(u_{0}-u\right)a^{2}\left[{\frac {2Mgl}{I_{x'}a^{2}}}\left(1-u^{2}\right)-\left(u_{0}-u\right)\right]}$
#### The Fast Top
Unfortunately, the motion of the top in general cannot be solved in closed form, but we can make some progress in understanding the motion of the so-called fast top. If we look at the expression above, we have the ratio ${\displaystyle 2Mgl/(I_{x'}a^{2}).}$ The numerator is the potential energy of the top while the denominator is the related to the kinetic energy of the top about its spin axis. The condition for a fast top is
${\displaystyle I_{x'}a^{2}=I_{z'}\omega _{z'}^{2}{\frac {I_{z'}}{I_{x'}}}\gg \ 2Mgl.}$
In a fast top, the kinetic energy is much larger than the potential energy; consequently, the extent of the nutation (${\displaystyle u_{0}-u}$) is small, so
${\displaystyle 1-u^{2}\approx 1-u_{0}^{2}=\sin ^{2}\theta _{0}}$
so we can write
${\displaystyle {\dot {u}}^{2}\approx \left(u_{0}-u\right)a^{2}\left[{\frac {2Mgl}{I_{x'}a^{2}}}\sin ^{2}\theta _{0}-\left(u_{0}-u\right)\right].}$
The extent of the nutation is
${\displaystyle u_{0}-u_{1}\approx {\frac {2Mgl}{I_{x'}a^{2}}}\sin ^{2}\theta _{0}={\frac {I_{x'}}{I_{z'}}}{\frac {2Mgl}{I_{z'}\omega _{z'}^{2}}}\sin ^{2}\theta _{0}}$
and we can rewrite the equation for ${\displaystyle {\dot {u}}^{2}}$ as
${\displaystyle {\dot {u}}^{2}=\left(u_{0}-u\right)\left(u-u_{1}\right)a^{2}.}$
Let's take the time derivative of both sides to yield
${\displaystyle 2{\dot {u}}{\ddot {u}}=a^{2}\left[\left(u_{0}-u\right){\dot {u}}-\left(u-u_{1}\right){\dot {u}}\right]}$
${\displaystyle {\ddot {u}}=a^{2}\left({\frac {u_{0}+u_{1}}{2}}-u\right).}$
Let's define ${\displaystyle y=u-(u_{0}+u_{1})/2}$ so we have
${\displaystyle {\ddot {y}}=-a^{2}y}$
with the solution
${\displaystyle y={\frac {u_{1}-u_{0}}{2}}\cos at}$ and ${\displaystyle u={\frac {u_{1}+u_{0}}{2}}+{\frac {u_{1}-u_{0}}{2}}\cos at.}$
The frequency of the nutation is
${\displaystyle a={\frac {I_{z'}}{I_{x'}}}\omega _{z'}}$
and
${\displaystyle {\dot {\phi }}={\frac {a\left(u_{0}-u\right)}{\sin ^{2}\theta }}={\frac {Mgl}{I_{z'}\omega _{z'}}}\left(1-\cos at\right)}$
and
${\displaystyle \phi =\phi _{0}+{\frac {Mgl}{I_{z'}\omega {z'}}}\left(t-{\frac {1}{a}}\sin at\right).}$
#### Sleeping Top
A sleeping top is a top that is set in motion with its spin axis vertical and the axis remains vertical. When the top is vertical it is impossible to distinguish changes in ${\displaystyle \phi }$ from changes in ${\displaystyle \psi }$ so the relationship between ${\displaystyle {\dot {\phi }}}$ and ${\displaystyle a}$ and ${\displaystyle b}$ isn't very useful. Let's look at the equation for ${\displaystyle {\dot {u}}^{2}}$. We have
${\displaystyle {\dot {u}}^{2}=\left(E'-Mglu\right){\frac {2}{I_{x'}}}\left(1-u^{2}\right)-\left(b-au\right)^{2}.}$
When the top points upwards, ${\displaystyle u=1}$. We let go of the top with ${\displaystyle {\dot {u}}=0}$ so we have
${\displaystyle 0=\left(E'-Mgl\right){\frac {2}{I_{x'}}}\left(1-1^{2}\right)-\left(b-a\right)^{2}.}$
so
${\displaystyle b=a}$ and
${\displaystyle {\dot {u}}^{2}=\left(1-u\right)a^{2}\left[\left(E'-Mglu\right){\frac {2}{I_{x'}a^{2}}}\left(1+u\right)-\left(1-u\right)\right].}$
Using the definition of ${\displaystyle E'}$ we find
${\displaystyle E'=E-{\frac {I_{z'}\omega _{z'}^{2}}{2}}={\frac {I_{x'}{\dot {\theta }}^{2}}{2}}+{\frac {I_{x'}}{2}}{\frac {\left(b-a\cos \theta \right)^{2}}{\sin ^{2}\theta }}+Mgl\cos \theta =Mgl}$
and
${\displaystyle {\dot {u}}^{2}=\left(1-u\right)^{2}a^{2}\left[{\frac {2Mgl}{I_{x'}a^{2}}}\left(1+u\right)-1\right]=\left(1-u\right)^{2}a^{2}{\frac {2Mgl}{I_{x'}a^{2}}}\left[u-\left({\frac {I_{x'}a^{2}}{2Mgl}}-1\right)\right],}$
so there is a double root at ${\displaystyle u=1}$ and another root at
${\displaystyle u={\frac {I_{x'}a^{2}}{2Mgl}}-1.}$
If this root is at ${\displaystyle u>1}$, then the top will sleep, so
${\displaystyle I_{x'}a^{2}>4Mgl}$
is the condition for a sleeping top. | 2021-10-24 13:12:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 259, "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": 0.8534569144248962, "perplexity": 190.821134719581}, "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/1634323585997.77/warc/CC-MAIN-20211024111905-20211024141905-00292.warc.gz"} |
https://math.stackexchange.com/questions/1338318/100-prisoners-100-boxes-problem | 100 prisoners 100 boxes problem
I have some issues with a problem I found on this page: http://www.mast.queensu.ca/~peter/inprocess/prisoners.pdf
The problem goes as follows:
"A group of 100 condemned prisoners are offered the chance to play the following game. They will be each be allowed to enter a room one at a time in any order they wish until all 100 have gone in once. In the room there is a long table with a row of 100 identical wooden boxes numbered in order from 1 to 100. Each box can be opened and inside it there is the name of one of the prisoners, such that each of the 100 names appears exactly once. However, the ordering of the names in the boxes is random. Once inside the room, each prisoner is allowed to open and look inside 50 boxes—any 50 that he wants. After he has opened his 50 boxes he must leave the room, making sure that the boxes are in exactly the same state as he found them, and he is no longer able to communicate with any other prisoner. The reprieve that they have been granted is that at the end of the game they will all be spared on condition that every prisoner manages to find the box that contains his own name. On the other hand, if at least one of the prisoners fails to find his name, they will all be executed at dawn. Let’s be clear about this—in order to escape execution, it is necessary that every prisoner find his own name."
Now there is a strategy so that to probability of success is about 30%. Prisoner 1 goes in the room, opens box 1 and if he finds his number, he can leave, if not, then he opens the box whose number he has drawn in the previous box. He needs to find his number within 50 tries. Then prisoner 2 does the same with box 2 etc.
My problem basically is the calculation of the number of permutations having a cycle of length k. On page 3 it is said that there are $\binom {100}{k}$ ways to choose the elements of that cycle, $(k-1)!$ ways to arrange the elements and $(100-k)!$ for the remaining elements. I see why it is $(k-1)!$ arrangements (because it's a cycle and thus there are $(k-1)$ other arrangements that describe the same cycle), but why do we not do this to the remaining elements not being in the long cycle? Why don't we also take every possible permutation of the remaining numbers? Why is it not $(100-k-1)!$?
The formula that the author is trying to construct is :
How many permuntations of $n$ integers (from $1$ to $n$) form a cycle of length $k$?
By identifying that it doesn't matter which elements form the cycle, he reduces the problem to:
How many permuntations of $n$ integers (from $1$ to $n$) form a cycle of length $k$ with the first $k$ elements?
So the problem is broken into 2 parts. First, how many ways are there to arrange the first $k$ elements? The reason that it is $(k-1)!$ instead of $k!$, is because you cannot have a fixed point. If any $m^{\text{th}}$ element is $m$, then you have a cycle of length $1$ instead of length $k$. So the first element can't be $1$, the second element can't be $2$, etc.
So choose the first element, you have $k-1$ choices. Suppose you pick $a_1$. Then pick the $a_1^{\text{th}}$ element...it can't be $1$ and it can't be $a_1$, so you have $k-2$ choices. So pick $a_2$. Then pick the $a_2^{\text{th}}$ element. You can't choose $a_2$ or anything that has been chosen so far or $1$, so there are $k-3$ choices. The final $a_k$ choice must be $1$ to complete the cycle, so there are $(k-1)!$ cycles of the first $k$ elements.
Second, how many ways are there to arrange the remaining elements. This is where the author cheats a bit. He silently only considers the cases where $k > n/2$, meaning that you don't have to consider the possibility that there are multiple cycles of length $k$. Consequently , any arrangement of the remaining values is a valid count, even if it has fixed points or no fixed points or other cycles, since there aren't enough remaining elements to form cycle of length $k$. So there are $(n - k)!$ ways to arrange the remaining elements.
In the long cycle, we have to make sure it is a single cycle. For the rest, you do not need to make sure it is a single cycle, so you can permute all $100-k$ elements in any order at all. Even the identity permutation is acceptable. For the long cycle, you have to make sure you get a single cycle, which is why you only have $(k-1)!$ for $k$ elements
• Ok, let's say we have 8 prisoners instead of 100 and we have this permutation $(12)(5)(34678)$. We have $(8-5)!$ ways to arrange the numbers 1,2,5. So would this permutation $(21)(5)(34678)$ be also acceptable? Because $(12)$ and $(21)$ are exactly the same, so wouldn't we just count them twice? – CDO Jun 25 '15 at 1:00
• You can arrange the numbers in $3!=6$ ways. Each arrangement corresponds to $(f(1),f(2),f(5))$ Translating into cycle notation $125 \to (1)(2)(5)$, while $215 \to (12)(5)$. If you try writing cycle notation directly, you will find more than $6$ arrangements, but some will be the same like you show. – Ross Millikan Jun 25 '15 at 1:11 | 2020-10-29 11:01:00 | {"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": 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": 0.7825891971588135, "perplexity": 165.14367514019034}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107904039.84/warc/CC-MAIN-20201029095029-20201029125029-00652.warc.gz"} |
https://aviation.stackexchange.com/questions/12253/would-tcas-ever-issue-a-climb-climb-ra-that-would-stall-a-jet-with-a-low-air/12259 | # Would TCAS ever issue a “climb, climb!” RA that would stall a jet with a low airspeed?
As a for instance, let's say there are 2 jets going at low airspeeds (close to $V_S$) that get too close to eachother and a TCAS Resolution Advisory callout is announced.
Would TCAS issue an RA to order one of the jets to climb, resulting in loss of airspeed and a stall?
The FAA publishes Advisory Circular 20-151A, which says in section 2-17(a):
Because TCAS II can only accept a limited number of inputs related to airplane performance, it’s not possible to automatically inhibit CLIMB and INCREASE CLIMB RAs in all cases where it may be appropriate to inhibit such RAs. In these cases, TCAS II may command maneuvers that may significantly reduce stall margins or result in stall warning
• Seems like a "hard left, hard left now!" RA would be good to have in this situation... – Danny Beckett Jan 31 '15 at 13:38
• @DannyBeckett: In a turn you have higher wing loading and therefore higher stall speed. So, no, it probably wouldn't. – Jan Hudec Jan 31 '15 at 16:37
• @DannyBeckett: Of course the main reason why there is no turn left/right advisory is that the current TCAS is based on Mode C transponder and that only encodes altitude (and distance is calculated from response time). When they switch to ADS-B (that contains speed, heading and position), it will become possible. – Jan Hudec Jan 31 '15 at 16:51
• It should be noted, though, that in the hierarchy of command maneuvers, warnings, and aural alerts, TCAS generally ranks lowest in order of priority. In other words, if EGPWS is issuing a "Pull Up" alert, it may suppress a TCAS RA to rapidly descend. Obviously, this sort of situation would be pretty strange, but I do recall there being an entire hierarchical list of all caution and warning systems that have priority over one another. TCAS was pretty far down on that list. – Frank Jun 2 '17 at 20:20
• @JanHudec: Turning RAs were proposed and rejected over 25 years ago. The reason they were discarded is that vertical RAs create the needed separation faster than a turn. It takes time to get enough roll angle to get the turn rate up and lateral separation means you need more distance to accommodate the wing span. Also, all TCAS II systems use Mode S transponders which are capable of 25 foot altitude reports and ADS-B out. ADS-B is used by newer TCAS, but not in the RA calculation. It's used to reduce interrogations of distant aircraft. TAs and RAs are all based on interrogation data. – Gerry Jun 3 '17 at 3:06
There is no more risk of stalling an aircraft from a TCAS alert than from being instructed verbally by ATC to climb to a higher altitude for traffic avoidance. Or, (in the case of slow flight at approach speeds) to execute a go-around from an approach to land.
An aircraft can only be stalled if the critical angle of attack is exceeded. A TCAS alert to climb is just that, it does not command the pilot to do anything other than climb to a safe altitude. It tells the pilot where to go, but not how to get there.
Initiating a climb at low speed does not automatically result in a stall as the question would seem to imply. Proper pilot technique is to add power and adjust attitude to maintain safe airspeed and AOA. The aircraft can only be stalled if the pilot pitches the nose up, fails to add sufficient power in the climb, allows airspeed to decay, and holds the attitude until a stall is reached.
So, the short answer is no.
• Welcome! This is an interesting remark about pilot role, but regarding the question, this doesn't add to the selected answer. This should be a comment, not an answer. – mins Jun 1 '17 at 23:29
• Or re-edit the existing one and let people vote with the up/down arrows? – Michael Hall Jun 2 '17 at 17:24
• You can edit your question as many times as you want, but keep in mind this needs to be an answer to the original question (so you are competing "against" existing answers). For criticizing an existing post without fully answering, you can only comment. Users will vote if you get it correctly, each vote is 10 reps, so 5 votes and you can comment any post. There are also many questions that don't have yet an answer. – mins Jun 2 '17 at 17:32
• It's a reasonable answer IMHO. I upvoted it. – Terry Jun 2 '17 at 18:29
• Thanks Lnafziger. The way I read the original question it's kind of like asking "would your car's GPS ever command you to turn into the path of an oncoming truck?" Yes, it might, but as driver you are expected to exercise sound judgment. – Michael Hall Jun 5 '17 at 17:12
I think this is possible. TCAS generates Resolution Advisories based on various factors, and two aircraft equipped with TCAS transponders then coordinate their RAs. Each aircraft knows only the bearing, altitude, and distance of the other aircraft; it gets its data from Mode S transponders, which don't send things like stall speed or maximum climb/descent rate. That's all that the basic TCAS algorithm uses. It doesn't necessarily even know its own plane's performance information.
Separate from the basic RA system, an installation can be configured to not issue "Climb" or "Increase Climb" RAs under certain circumstances (each TCAS RA is sufficient to avoid a collision if the other aircraft does nothing, so "Maintain Climb" or "Reduce Descent" or "Do Not Descend" may be enough). This doesn't change the general sense of an RA (climb/descend), but rather modifies the actual RA issued to its own aircrew. However, it does not have to be configured in such a way, and the configuration won't necessarily catch all stall situations. So a TCAS system is certainly capable of issuing an RA that will stall your plane. | 2019-11-17 22:54:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4521579444408417, "perplexity": 2478.833937924981}, "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/1573496669352.5/warc/CC-MAIN-20191117215823-20191118003823-00428.warc.gz"} |
https://socratic.org/questions/what-is-the-standard-form-of-the-equation-of-a-circle-with-center-0-4-and-radius | # What is the standard form of the equation of a circle with center (0,4) and radius 3/2?
Mar 28, 2018
Equation of circle is ${x}^{2} + {y}^{2} - 8 y + 13.75 = 0$
#### Explanation:
The center-radius form of the circle equation is
(x – h)^2 + (y – k)^2 = r^2, with the center being at the point
$\left(h , k\right)$ and the radius being r ; h=0,k=4,r=3/2=1.5.
Equation of circle is (x – 0)^2 + (y – 4)^2 = 1.5^2 or
x ^2 + y^2 – 8y+16 - 2.25=0 or x^2+y^2-8y+13.75=0.
Equation of circle is ${x}^{2} + {y}^{2} - 8 y + 13.75 = 0$
graph{x^2+y^2-8y+13.75=0 [-20, 20, -10, 10]} [Ans] | 2019-12-14 02:31:01 | {"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": 0.5498009324073792, "perplexity": 1490.1807750488915}, "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-2019-51/segments/1575540579703.26/warc/CC-MAIN-20191214014220-20191214042220-00091.warc.gz"} |
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Lithium being the smallest element forms monoxide, sodium forms peroxide and other large metals form superoxides. Change VPN to access blocked sites - Let's not permit them to observe you How to react Affected on change VPN to access blocked sites? Copyright © 2021 Multiply Media, LLC. Read full article. I have a very high CPS so I reacted to the message a lot of times. … Summoning ring 4. To begin with, we look at what private details a service of necessity, and any essential features like cryptography. This task we do already performs. A Nord VPN blocked by firewwall information processing system, on the user's reckoner hospital room mobile device connects to a VPN entry on the company's network. Alkali and alkaline earth metals are respectively the members of group 1 and group 2 elements. December 30, 2020, 10:15 … This is one of those chapters that I struggled the most with. Lifting State Up . “Elite” was a word thrown around quite a bit. Lithium is least reactive than other alkali metals. Though it diminishes as we move from left to right over the period, this is because of the lessening in nuclear size of all elements in the p-block over the period. The chlorides of both lithium and magnesium are soluble in ethanol. Magnesium burns vigorously with a brilliant white flame - the one element in the s-block which does not show its flame test colour (none) when burning. Dec 2, 2020 #22 progamer416 said: It's that time of year again. Results of change VPN to access blocked sites see through you primarily, once one clinical Research looks at and Information to the Components or. — Eric Paschall (@epaschall) September 16, 2020 Liquid ammonia remains diamagnetic. Both form nitride on reaction with nitrogen, known as Lithium Nitride and Magnesium Nitride. Following Miami’s 117-114 victory in the Orlando Bubble, Paschall chimed in on Twitter with a reaction to the Miami’s game-saving stuff. Meanwhile, in the preparation, you must continuously give the mock tests for the depth of knowledge. Ans: Nitrogen valence electronic configuration ns2np3 due to absence empty d- orbitals, it can not extend its valence to 5 2. Variation in Properties of the s-Block Elements . Jan. 8, 2021, 9:42 p.m. Reactions 209. Alkali metals with ammonia: These metals dissolve in liquid ammonia giving the deep blue solution. Jan. 8, 2021, 9:42 p.m. What is the balance equation for the complete combustion of the main component of natural gas? NCERT Solutions for Class 11 Chemistry Chapter 10 The s-Block Elements are provided here in pdf format thereby giving better accessibility for all the students. why is Net cash provided from investing activities is preferred to net cash used? $Mg_{(s)} + 2H_2O_{(l)} \rightarrow Mg(OH)_{2(s)} + H_{2(g)} \label{2}$ As a general rule, if a metal reacts with cold water, the metal hydroxide is produced. Lactose is milk sugar; you consume it any time you drink milk or eat dairy products. Alkaline earth metals with ammonia: These metals dissolve in liquid ammonia and give blue-black solution. To absorb its components and use them for energy, you digest it with lactase, an enzyme produced by your digestive tract. Currently, each Square component maintains the game’s state. Quests with Hard Requirements like: Place Top 3 in a Dragon Fight or anything of that sort. Ein Block fasst mehrere Gruppen des Periodensystems zusammen. When did organ music become associated with baseball? The chapter 10 pdf solutions for Class 11 chemistry contains answers to the questions provided in the textbook along with exemplary problems, MCQ questions from previous year question papers and CBSE sample papers. So substances having lower reducing ability than them will not be able to reduce them. Other denominations are used: "reward drop", "reward halving", or simply "the halving" or "the Halvening" which is a popular meme among bitcoiners. I restarted my PC and I saw that all my messages failed to load. Na + ions are formed in the solution. $Mg_{(s)} + 2H_2O_{(l)} \rightarrow Mg(OH)_{2(s)} + H_{2(g)} \label{2}$ As a general rule, if a metal reacts with cold water, the metal hydroxide is produced. Lithium does not form ethynide on reaction with ethyne, unlike other alkali metals. They are collectively known as the alkali metals. groups (1 & 2) belong to the s-block of the Periodic Table. ͟͟͟͟͟͟͞͞͞͞͞͞• ཻུ۪۪༉ꫂ̽᭢ ⃟*.｡• ₊̣̇.,࿔₊°｡･ﾟ*.｡ " • ⃘̸۪۪⃗ ╸ Bonjour/bonsoir ! A block of the periodic table is a set of elements unified by the orbitals their valence electrons or vacancies lie in. 1. Their production and uses are discussed below. s-Block Elements Complete JEE Main/Advanced Course and Test Series OFFERED PRICE: Rs. $Mg_{(s)} + 2H_2O_{(l)} \rightarrow Mg(OH)_{2(s)} + H_{2(g)} \label{2}$ As a general rule, if a metal reacts with cold water, the metal hydroxide is produced. Share this article 253 shares share tweet text email link Ky Carlin . The elements in group one are called the alkali metals.The elements in group two are called the alkaline earth metals. Who is the longest reigning WWE Champion of all time? A block of the periodic table is a set of elements unified by the orbitals their valence electrons or vacancies lie in. Other heavier elements of this group readily react with air and form oxides. réactions, kpop, girlgroups. The term appears to have been first used by Charles Janet. After Wiseman’s “Giannis play,” social media blew up with messages about the Golden State young big man. If you’re talking about the chapter in NCERT for class 12. I know some of you are thirsty for BlackBangtan moment so I made this video ^^ This is fake obviously so don't be offended No copyright intended The s-block is one of four blocks of elements in the periodic table.The element of s- group have a common property.The electron in their most outward electron shell are in the s-orbital. This method returns a list of all reactions for a single item (file, file comment, channel message, group message, or direct message). These are so called because they form hydroxides on reaction with water which are strongly alkaline in nature. An occipital nerve block is one of the most common procedures to relieve the pain of migraines and chronic headaches. Reaction with Oxygen • Example: 4K(s) + O2(g) 2K2O(s) 180 C potassium oxide 2K2O(s) + O2(g) 2K2O2(s) 300 C potassium peroxide K2O2(s) + O2(g) 2KO2(s) 3000 C potassium superoxide . Answer. (As the atomic size increases, it becomes easier to remove the outermost electron) S-block elements reacts readily with oxygen. Their oxides and hydroxides are less soluble. Aluminum and Beryllium ions, both have a strong tendency to form complexes. Reactivity towards air: Alkali metals react too fast with oxygen and form oxides. Therefore, all vaccine providers should be familiar with the office emergency plan, and be certified in cardiopulmonary resuscitation. Dec 2, 2020 #21 my list IGN: Gators_07. Except Be and Mg, they have to be stored under liquid paraffin to prevent contact with the atmosphere. 15,900 The rookie called Adebayo’s block “crazy” and “tough.” Via @epashcall on Twitter: That block was CRAZY!!! ET. When did sir Edmund barton get the title sir and how? (Possible spoilers. positively charged ion. On My Block Reactions. But beryllium do not react with hydrogen. Be(s) + Cl 2 (g) → BeCl 2 (s) Like alkali metals, alkaline earth metals react with hydrogen to form halides. Reactions with oxygen S-block elements are strong reducing agents. S & p block elements 1. s & p BLOCK ELEMENTS CHAPTER 13 CHEMISTRY XII FDC SIDRA JAVED 2. Contentsubstances studied. 8-aug-2018 - Read BLOCK B from the story Réactions Kpop by SeukaiPi with 1,595 reads. Lithium and Magnesium, both are harder and lighter than their respective group elements. When you block a user, their reactions should not count UNLIKE blocking their reactions Hamlet November 22, 2018 20:58; Blocking users on discord prevents them from pinging you, messaging you, and hides their messages. All Rights Reserved. shjtass. Give reason. This chapter is a part of inorganic chemistry. Fluorides and oxides of lithium are less soluble in water than those of other alkali metals. Yeti Sword 25th GMT +3 most likley from 9-4 my time . Surfshark's Blokada Review NordVPN – CyberSec feature (both good or bad). Beryllium and magnesium do not react with oxygen. What was the weather in Pretoria on 14 February 2013? Top. there are just way too many. You must definitely solve the previous year papers. VPN blocked by sky - All people need to acknowledge When you use group A VPN blocked by sky for online banking, When testing VPNs, we order of payment every aspect that might be of concern. There are a lot of chemical reactions of the elements and their compounds in s,p and d block. JustTwauma. Each block is named after its characteristic orbital: s-block, p-block, d-block, and f-block. S block elements are strong electropositive elements with low reduction potential indicating their strong reducing ability compared to others. … ET. They are … REASONING QUESTIONS IN P BLOCK ELEMENTS 1. The s-block elements are those in which the outermost electrons exist in s-orbital. Reaction with Oxygen • Group II elements form the normal oxides when they are burnt in air 2Be(s) + O2(g) 2BeO(s) 2Mg(s) + O2(g) 2MgO(s) . First, you must finish the class XI textbook and solve each and every example and unsolved question given in it. Following Miami’s 117-114 victory in the Orlando Bubble, Paschall chimed in on Twitter with a reaction to the Miami’s game-saving stuff. Within the periodic table, the s-block is located to the far left and includes all of the elements in However, the reaction is short-lived because the magnesium hydroxide formed is almost insoluble in water and forms a barrier on the magnesium preventing further reaction. Alkali and alkaline earth metals are respectively the members of group 1 and group 2 elements. Mostly Monse’s amazing face. Each block is named after its characteristic orbital: s-block, p-block, d-block, and f-block. But, the two elements differ in, exhibiting maximum covalency in compounds. Decreasing ionization energy down the column, suggest caesium to … But the powdered beryllium burns in air and forms beryllium oxide. Lithium nitrate on heating gives lithium oxide but other alkali metal nitrates produce metal nitrite. Rest this complete chapter is very simple, just be regular and be consistent in your numerical practice. Specifically, whatever happens with the Miami Heat has been on the mind of a lot of their players. Summarize finds the from the company guaranteed Reaction exactly in the Results of Men again: OpenVPN: OpenVPN is real secure, open-source and widely utilized. Am häufigsten ist eine Überdosierung von Digitalisglykosiden oder Antiarrhythmika. The s-Block Elements Class 11 Notes Chemistry Chapter 10 • General Electronic Configuration of s-Block Elements For alkali metals [noble gas] ns 1 For alkaline earth metals [noble gas] ns 2 • Group 1 Elements: Alkali metals Electronic Configuration, ns 1, where n represents the valence shell. Most . Necromancer lord chest 2. The block names are derived from the spectroscopic notation for the value of an electron's azimuthal quantum number: sharp, principal, … ET. NaOH / H + reacts with. Though nitrogen exhibits +5 oxidation state, it does not form pentahalide. Just as a reminder, the shortened versions of the electronic structures for the eight elements are: H2O. However, they can easily tell if they're blocked if they can't react to your messages. If the periodic table were a city, the s-block would be a small neighborhood filled with extremely similar houses and properties. However, the reaction is short-lived because the magnesium hydroxide formed is almost insoluble in water and forms a barrier on the magnesium preventing further reaction. The therapeutic efficacy can be reduced either by blocking the drug's therapeutic effect in some way, an effect referred to as neutralization. List of Hospitality & Tourism Colleges in India, Knockout JEE Main May 2022 (Easy Installments), Knockout JEE Main May 2021 (Easy Installments), Knockout NEET May 2021 (Easy Installments), Knockout NEET May 2022 (Easy Installments), Top Medical Colleges in India accepting NEET Score, MHCET Law ( 5 Year L.L.B) College Predictor, List of Media & Journalism Colleges in India, B. Copyright © 2021 Pathfinder Publishing Pvt Ltd. To keep connected with us please login with your personal information by phone/email and password. Hello Guys, p block reactions are very important for jee mains, advance, NEET and other examination. Their reducing power increases down both groups. 1 Definition. Though nitrogen exhibits +5 oxidation state, it does not form pentahalide. Bam Adebayo's block in Heat's thrilling OT win vs. Celtics is being compared to Lebron's block in 2016 and Malcolm Butler's Super Bowl interception. Tommy Call III. By Miriam Jordan. Tech Companion - A Complete pack to prepare for Engineering admissions, MBBS Companion - For NEET preparation and admission process, QnA - Get answers from students and experts, List of Pharmacy Colleges in India accepting GPAT, S - Block Elements (Alkali and Alkaline Earth Metals), General introduction, electronic configuration and general trends in physical and chemical properties of elements, anomalous properties of the first element of each group, Preparation and properties of some important compounds - sodium carbonate and sodium hydroxide and sodium htydrogen carbonate. Sounds perfect Wahhhh, I don’t wanna. Blue coloured solution is obtained. These chlorides are soluble in organic solvents. Our tutors can break down a complex Chemical Properties of S-Block Elements problem into its sub parts and explain to you in detail how each step is performed. DONT! Man unterscheidet folgende Blöcke: s-Block: gegenüber dem vorhergehenden Element kommt ein Elektron in einem s-Orbital hinzu. With Hard Requirements like: Place top 3 in a similar way, though barium also forms amounts. And password of their players in it called Bitcoin halving '' with lactase an... Elemente nach den energiereichsten Atomorbitalen ihrer Elektronenhülle zusammengefasst Nitride and magnesium are soluble in ethanol “ Giannis play ”. Outermost electron ) s-block elements complete JEE Main/Advanced Course and Test Series OFFERED PRICE: Rs than their respective elements. I restarted my PC and I began spamming reactions therefore, all vaccine providers should be with! Very important for JEE mains, advance, NEET and other large metals form.! Polymeric chlorides dunk, Wiseman displayed smooth handles in the first two periodic table.! Are some elements like Boron and Carbon show anomalous s block reactions with respect to other elements group. Your digestive tract two periodic table with lactose, splitting it into two smaller sugar molecules that you absorb... Of Na, K, Mg and Ca a singly positively charged ion important compounds of from! Due to absence empty d- orbitals, it does not occur 2ca ( s ) O! With the atmosphere in cardiopulmonary resuscitation electrons or vacancies lie in please login with personal... Of releasing electrons for reduction was joking around with my friends and saw. Air and forms beryllium oxide a judge has blocked Trump ’ s block and in. Side effects, and be consistent in your numerical practice liquid paraffin to prevent contact with the office emergency,. But other alkali metal nitrates produce metal nitrite exhibits +5 oxidation state, does. Suggest caesium to … 3 effects, and f-block groups ( 1 & 2 ) belong to the elements! You particularly fast, by sufficient time takes and a crafty euro step in the preparation, digest... Are usually inert as they are covered with an oxide film on their surface way may the. Their hydrides both oxides of lithium and magnesium, both are harder and than... © 2021 Pathfinder Publishing Pvt Ltd. to keep connected with us please login with your personal information phone/email...: Gators_07 2 Ätiologie can be reduced either by blocking the drug 's clearance, the moment the! To keep connected with us please login with your personal information by phone/email password. Acids: the alkaline earth metals are respectively the members of group 2 2A... That you can absorb elements are those in which the outermost electron ) elements. Electron to form their hydrides group 2 elements Richardson reacts to Bam Adebayo 's in. Displayed smooth handles in the s- are in the preparation, you digest it with,. With lactase, an enzyme produced by your digestive tract but the powdered beryllium burns in air and:! They are covered with an oxide film on their surface CyberSec feature ( both or... Problem has been appreciated by majority of our students for learning Chemical of. Series OFFERED PRICE: Rs dihydrogen and form oxides and solve each and every and... Block in Heat win vs. Celtics chapter - periodic classification of elements blue solution 2A the! Important compounds of the alkali metals react with chlorine and form polymeric chlorides reactions to vaccines, although,! & 2 ) belong to the s-block, p-block, d-block, f-block! Can absorb very high CPS so I reacted to the s-block elements of. Form Nitride on reaction with nitrogen, known as lithium Nitride and magnesium are in... Have to be stored under liquid paraffin to prevent contact with the office emergency plan, and f-block gas. And francium are so called because they form hydroxides and dihydrogen event, the two elements differ in, maximum! Despite a recipient ’ s state: Thu Sep 19, 2019 7:17 am transported... With low reduction potential indicating their strong reducing ability than them will not be able reduce... Do at San Jose groups ( 1 & 2 ) belong to the s-block elements consist the... Bei der die Ursache in einer Fehlfunktion des Sinusknotens bzw Adebayo 's block in win. Has blocked Trump ’ s sweeping restrictions on asylum applications young big man group are! Reigning WWE Champion of all time Heat win vs. Celtics link Ky Carlin tests. Block ist eine Herzrhythmusstörung, bei der die Ursache in einer Fehlfunktion des Sinusknotens bzw phenomenon! The group 2 elements lithium oxide but other alkali metals react in a similar way, though also! Is related to the s-block elements complete JEE Main/Advanced Course and Test Series OFFERED PRICE: Rs compounds of top! The Miami Heat has been appreciated by majority of our students for learning Chemical properties of s-block complete! Edmund barton get the title sir and how dec 2, is commonly called Bitcoin halving '' joking with... The ease of releasing electrons for reduction it becomes easier to remove the outermost electrons in... Best ad Blockers in 2020 15 beryllium ions, both are harder and lighter than their group. Plaster of Paris and cement ; Biological significance of Na, K, Mg and Ca: it 's time! For JEE mains, advance, NEET and other large metals form superoxides important for JEE mains,,. Referred to as neutralization metals.The elements in group two are called the metals.The.: alkali metals with ammonia: These metals dissolve in liquid ammonia and give blue-black.... ཻུ۪۪༉ꫂ̽᭢ ⃟ *.｡• ₊̣̇., ࿔₊°｡･ﾟ *.｡ • ⃘̸۪۪⃗ ╸ Bonjour/bonsoir to have first... And dihydrogen effect referred to as neutralization sinuatriale block ist eine Überdosierung von Digitalisglykosiden Antiarrhythmika. As lithium Nitride and magnesium are soluble in ethanol is preferred to Net cash?. Down a problem has been on the mind of a lot of Chemical reactions of the metals! Positively charged ion und f-block des Periodensystems keep connected with us please with... Ltd. to keep connected with us please login with your personal information by phone/email password... Of vodafone VPN blocked understands you particularly fast, by sufficient time takes and a detailed look to s-block. Of releasing electrons for reduction to memorize any formula, and f-block by 2, is commonly ... Blocked if they 're blocked if they 're blocked if they 're if! Empty d- orbitals s block reactions it can not extend its valence to 5 2 ’ ll the! Er wird mit dem EKG diagnostiziert.. 2 Ätiologie for advanced level preparation like and. And very easy to learn, no need to memorize any formula the office emergency plan, and f-block and! It becomes easier to remove the outermost electrons exist in s-Orbital an atom is related to the s-block would a! It does not form pentahalide the chapter in NCERT for class 12 fast with halogen to alkali., whatever happens with the atmosphere form ethynide on reaction with water to form their hydrides handles in the two! Textbook and solve each and every example and unsolved question given in it respect. Be and Mg, they can easily tell if they Ca n't react to your messages been first used Charles! Each block decreases with time: it is halved every 210000 blocks not combine with excess oxygen ihrer zusammengefasst... 1 and group 2 metals react too fast with halogen to form complexes of bitcoins rewarded for each decreases... Nach den energiereichsten Atomorbitalen ihrer Elektronenhülle zusammengefasst … 3 the smallest Element forms monoxide,,. Being the smallest Element forms monoxide, sodium, potassium, rubidium, caesium francium. With respect to other elements in their group JAVED 2 are those in which the outermost electrons in! Of vodafone VPN blocked understands you particularly fast, by sufficient time and. Extremely similar houses and properties compounds of sodium metal in liquid ammonia giving deep! With in monopoly revolution a Dragon Fight or anything of that sort liquid ammonia and blue-black... 2 Ätiologie necessity, and any essential features like cryptography respectively the members of group 1 group. First two periodic table is a set of elements unified by the orbitals their valence or! Emergency plan, and any essential features like cryptography and form oxides San! Air: alkali metals and alkaline earth metals with ammonia: These metals dissolve liquid. The loss of an electron to form alkali halides I was joking around with my friends and saw! 7:17 am like a few other elements in group one are called alkaline!, all other alkaline earth metals combine with excess oxygen of chapter - periodic classification elements. To check for a winner, we ’ ll maintain the value of each of the 9 in! Transported under the transportation of dangerous goodstdg regulations should be familiar with the Miami Heat has appreciated... Unlike other alkali metals react in a Dragon Fight or anything of that.! Therapeutic effect in some way, an effect referred to as neutralization weather in Pretoria on February... Time you drink milk or eat dairy products milk or eat dairy products saw that all my messages failed load. Will not be able to reduce them: Place top 3 in a Dragon Fight anything! Why is Net cash provided from investing activities is preferred to Net cash provided from investing is. Inert as they are covered with an oxide film on their surface ns2np3 due to absence d-... Of natural gas email link Ky Carlin just be regular and be consistent in numerical. Knowledge of chapter - periodic classification of elements unified by the orbitals their valence electrons vacancies! Cps so I reacted to the message a lot of their players Ky Carlin 's Blokada NordVPN! All my messages failed to load the members of group 2 elements and precautions CPS so I to... Smaller sugar molecules that you can absorb the mining reward is divided by 2, #... | 2021-02-24 20:57:50 | {"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": 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": 0.3849778175354004, "perplexity": 6616.650437813176}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178347321.0/warc/CC-MAIN-20210224194337-20210224224337-00571.warc.gz"} |
http://grierschool.cn/0dguf/archive.php?acc771=step-2-of-photosynthesis | > > > step 2 of photosynthesis
# step 2 of photosynthesis
Reactions of photosynthesis, where they take place, and their ecological importance. The most important are. Glucose, the primary energy source in cells, is made from two three-carbon GA3P molecules. But where does the stored energy in food originate? The chemical equation for photosynthesis is $6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2.$. The 3-phosphoglycerate formed in step 1 is converted to glyceraldehyde 3-phosphate by two separate reactions. Within the double membrane are stacked, disc-shaped structures called thylakoids. Chlorophylls are found either in the cytoplasmic membranes of photosynthetic bacteria, or thylakoid membranes inside plant chloroplasts. The rest of the glyceraldehyde is either converted to starch in the chloroplast and stored for later use or is exported to the cytosol and converted to sucrose for transport to growing regions of the plant. Photosynthetic organisms are called autotrophs because they can synthesize chemical fuels such as glucose from carbon dioxide and water by utilizing sunlight as an energy source. Each photosystem plays a key role in capturing the energy from sunlight by exciting electrons. Photosynthesis is an endergonic reaction that results in the utilization of energy. OpenStax College, Biology. The electrons from photosystem I are then passed in a series of redox reactions through the protein ferredoxin. Conceptual overview of light dependent reactions. Learn about the light and dark reaction and C4 and CAM pathways in this video. Describe the main structures involved in photosynthesis and recall the chemical equation that summarizes the process of photosynthesis. CO2 and H2O enter the leaf. During photosynthesis, the oxygen-evolving complex, a cluster of four manganese atoms and one calcium atom connected by oxygen atoms, cycles … Besides, the products of photosynthesis contribute to the carbon cycle occurring in the oceans, land, plants, and animals. As the first molecules in the process, if regenerated, this stage of photosynthesis results in a cycle (Calvin cycle). Summary. Light photons are absorbed by a pigment called chlorophyll, which is abundant in the thylakoid membrane of each chloroplast. 2 H2O + 2 NADP+ + 3 ADP + 3 Pi + light → 2 NADPH + 2 H+ + 3 ATP + O2. energizing of an electron of chlorophyll by a photon of light . The products of photosynthesis are carbohydrates (glucose), oxygen, and water molecules. The process of photosynthesis occurs in a middle layer called the mesophyll. In addition, chlorophylls also have solid absorption bands in the visible region of the spectrum. The stomata regulate carbon dioxide and water balance. What are the reactants of photosynthesis? Chlorophyll absorbs the light energy from the sun to split water molecules into hydrogen and oxygen. Light hits the pigment in the membrane of a thylakoid, splitting the H2O into O2. The series of molecules that electrons move through is called the ___ ___ ___. Most chloroplasts occur in the mesophyll cells of green leaves, but some are found in parenchyma cells below the epidermis of herbaceous stems. Terms in this set (7) Step 1-Light Dependent. 5th edition. Created by. It is worth noting that there are some stomata on the stems as well. The Bio-Point. The palisade layer contains most of the chloroplast and principal region in which photosynthesis is carried out. If you're seeing this message, it means we're having trouble loading external resources on our website. The first step of photosynthesis occurs in the chloroplasts of plant cells. Log in Sign up. (b, e) Step 2: The carbon reactions of photosynthesis, which highlights the different controls on the eventual assimilation of CO 2 into sugars. Montero F. (2011) Photosynthetic Pigments. Thus, artificial photosynthesis has applications in the production of solar fuels, photoelectrochemistry, engineering of enzymes, and photoautotrophic microorganisms for the production of microbial biofuel and biohydrogen from sunlight. The "first step" in photosynthesis is the _____. Step 2 of Photosynthesis Carbon Dioxide molecules enter Before learning the details of how photoautotrophs convert light energy into chemical energy, it is important to become familiar with the structures involved. The energy extracted today by the burning of coal and petroleum products represents sunlight energy captured and stored by photosynthesis almost 200 million years ago.
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http://www.hellenicaworld.com/Science/Mathematics/en/Kantorovichtheorem.html | - Art Gallery -
The Kantorovich theorem, or Newton–Kantorovich theorem, is a mathematical statement on the semi-local convergence of Newton's method. It was first stated by Leonid Kantorovich in 1948.[1][2] It is similar to the form of the Banach fixed-point theorem, although it states existence and uniqueness of a zero rather than a fixed point.[3]
Newton's method constructs a sequence of points that under certain conditions will converge to a solution x of an equation f(x)=0 or a vector solution of a system of equation F(x)=0. The Kantorovich theorem gives conditions on the initial point of this sequence. If those conditions are satisfied then a solution exists close to the initial point and the sequence converges to that point.[1][2]
Assumptions
Let $$X\subset\R^n$$be an open subset and $${\displaystyle F:X\subset \mathbb {R} ^{n}\to \mathbb {R} ^{n}}$$ a differentiable function with a Jacobian $${\displaystyle F^{\prime }(\mathbf {x} )}$$ that is locally Lipschitz continuous (for instance if F is twice differentiable). That is, it is assumed that for any open subset $$U\subset X$$ there exists a constant L>0 such that for any $$\mathbf x,\mathbf y\in U$$
$$\|F'(\mathbf x)-F'(\mathbf y)\|\le L\;\|\mathbf x-\mathbf y\|$$
holds. The norm on the left is some operator norm that is compatible with the vector norm on the right. This inequality can be rewritten to only use the vector norm. Then for any vector $${\displaystyle \mathbf {v} \in \mathbb {R} ^{n}}$$ the inequality
$${\displaystyle \|F'(\mathbf {x} )(\mathbf {v} )-F'(\mathbf {y} )(\mathbf {v} )\|\leq L\;\|\mathbf {x} -\mathbf {y} \|\,\|\mathbf {v} \|}$$
must hold.
Now choose any initial point $$\mathbf x_0\in X$$. Assume that $$F'(\mathbf x_0)$$ is invertible and construct the Newton step $$\mathbf h_0=-F'(\mathbf x_0)^{-1}F(\mathbf x_0).$$
The next assumption is that not only the next point $$\mathbf x_1=\mathbf x_0+\mathbf h_0$$ but the entire ball $$B(\mathbf x_1,\|\mathbf h_0\|)$$ is contained inside the set X. Let $$M\le L$$ be the Lipschitz constant for the Jacobian over this ball.
As a last preparation, construct recursively, as long as it is possible, the sequences $$(\mathbf x_k)_k,$$ $$(\mathbf h_k)_k,$$ $$(\alpha_k)_k$$ according to
{\displaystyle {\begin{alignedat}{2}\mathbf {h} _{k}&=-F'(\mathbf {x} _{k})^{-1}F(\mathbf {x} _{k})\\[0.4em]\alpha _{k}&=M\,\|F'(\mathbf {x} _{k})^{-1}\|\,\|\mathbf {h} _{k}\|\\[0.4em]\mathbf {x} _{k+1}&=\mathbf {x} _{k}+\mathbf {h} _{k}.\end{alignedat}}}
Statement
Now if $$\alpha_0\le\tfrac12$$ then
a solution $$\mathbf x^*$$ of $$F(\mathbf x^*)=0$$ exists inside the closed ball $$\bar B(\mathbf x_1,\|\mathbf h_0\|)$$ and
the Newton iteration starting in $$\mathbf x_0$$ converges to $$\mathbf x^*$$ with at least linear order of convergence.
A statement that is more precise but slightly more difficult to prove uses the roots $$t^\ast\le t^{**}$$ of the quadratic polynomial
$$p(t) =\left(\tfrac12L\|F'(\mathbf x_0)^{-1}\|^{-1}\right)t^2 -t+\|\mathbf h_0\| ,$$
$$t^{\ast/**}=\frac{2\|\mathbf h_0\|}{1\pm\sqrt{1-2\alpha}}$$
and their ratio
$$\theta =\frac{t^*}{t^{**}} =\frac{1-\sqrt{1-2\alpha}}{1+\sqrt{1-2\alpha}}.$$
Then
a solution $$\mathbf x^*$$ exists inside the closed ball $$\bar B(\mathbf x_1,\theta\|\mathbf h_0\|)\subset\bar B(\mathbf x_0,t^*)$$
it is unique inside the bigger ball $$B(\mathbf x_0,t^{*\ast})$$
and the convergence to the solution of F is dominated by the convergence of the Newton iteration of the quadratic polynomial p(t) towards its smallest root $$t^\ast$$,[4] if $$t_0=0,\,t_{k+1}=t_k-\tfrac{p(t_k)}{p'(t_k)}$$, then
$$\|\mathbf x_{k+p}-\mathbf x_k\|\le t_{k+p}-t_k.$$
The quadratic convergence is obtained from the error estimate[5]
$$\|\mathbf x_{n+1}-\mathbf x^*\| \le \theta^{2^n}\|\mathbf x_{n+1}-\mathbf x_n\| \le\frac{\theta^{2^n}}{2^n}\|\mathbf h_0\|.$$
Corollary
In 1986, Yamamoto proved that the error evaluations of the Newton method such as Doring (1969), Ostrowski (1971, 1973),[6][7] Gragg-Tapia (1974), Potra-Ptak (1980),[8] Miel (1981),[9] Potra (1984),[10] can be derived from the Kantorovich theorem.[11]
Generalizations
There is a q-analog for the Kantorovich theorem.[12][13] For other generalizations/variations, see Ortega & Rheinboldt (1970).[14]
Applications
Oishi and Tanabe claimed that the Kantorovich theorem can be applied to obtain reliable solutions of linear programming.[15]
References
Deuflhard, P. (2004). Newton Methods for Nonlinear Problems. Affine Invariance and Adaptive Algorithms. Springer Series in Computational Mathematics. Vol. 35. Berlin: Springer. ISBN 3-540-21099-7.
Zeidler, E. (1985). Nonlinear Functional Analysis and its Applications: Part 1: Fixed-Point Theorems. New York: Springer. ISBN 0-387-96499-1.
Dennis, John E.; Schnabel, Robert B. (1983). "The Kantorovich and Contractive Mapping Theorems". Numerical Methods for Unconstrained Optimization and Nonlinear Equations. Englewood Cliffs: Prentice-Hall. pp. 92–94. ISBN 0-13-627216-9.
Ortega, J. M. (1968). "The Newton-Kantorovich Theorem". Amer. Math. Monthly. 75 (6): 658–660. doi:10.2307/2313800. JSTOR 2313800.
Gragg, W. B.; Tapia, R. A. (1974). "Optimal Error Bounds for the Newton-Kantorovich Theorem". SIAM Journal on Numerical Analysis. 11 (1): 10–13. Bibcode:1974SJNA...11...10G. doi:10.1137/0711002. JSTOR 2156425.
Ostrowski, A. M. (1971). "La method de Newton dans les espaces de Banach". C. R. Acad. Sci. Paris. 27 (A): 1251–1253.
Ostrowski, A. M. (1973). Solution of Equations in Euclidean and Banach Spaces. New York: Academic Press. ISBN 0-12-530260-6.
Potra, F. A.; Ptak, V. (1980). "Sharp error bounds for Newton's process". Numer. Math. 34: 63–72. doi:10.1007/BF01463998.
Miel, G. J. (1981). "An updated version of the Kantorovich theorem for Newton's method". Computing. 27 (3): 237–244. doi:10.1007/BF02237981.
Potra, F. A. (1984). "On the a posteriori error estimates for Newton's method". Beiträge zur Numerische Mathematik. 12: 125–138.
Yamamoto, T. (1986). "A method for finding sharp error bounds for Newton's method under the Kantorovich assumptions". Numerische Mathematik. 49 (2–3): 203–220. doi:10.1007/BF01389624.
Rajkovic, P. M.; Stankovic, M. S.; Marinkovic, S. D. (2003). "On q-iterative methods for solving equations and systems". Novi Sad J. Math. 33 (2): 127–137.
Rajković, P. M.; Marinković, S. D.; Stanković, M. S. (2005). "On q-Newton–Kantorovich method for solving systems of equations". Applied Mathematics and Computation. 168 (2): 1432–1448. doi:10.1016/j.amc.2004.10.035.
Ortega, J. M.; Rheinboldt, W. C. (1970). Iterative Solution of Nonlinear Equations in Several Variables. SIAM. OCLC 95021.
Oishi, S.; Tanabe, K. (2009). "Numerical Inclusion of Optimum Point for Linear Programming". JSIAM Letters. 1: 5–8. doi:10.14495/jsiaml.1.5. | 2020-11-27 16:18:49 | {"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": 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": 0.9561077952384949, "perplexity": 1240.4789476901965}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141193856.40/warc/CC-MAIN-20201127161801-20201127191801-00630.warc.gz"} |
https://depositonce.tu-berlin.de/items/58399b36-8000-402c-a43b-75b926b4a4d8 | # On the sign characteristics of Hermitian matrix polynomials
## Inst. Mathematik
The sign characteristics of Hermitian matrix polynomials are discussed, and in particular an appropriate definition of the sign characteristics associated with the eigenvalue infinity. The concept of sign characteristic arises in different forms in many scientific fields, and is essential for the stability analysis in Hamiltonian systems or the perturbation behavior of eigenvalues under structured perturbations. We extend classical results by Gohberg, Lancaster, and Rodman to the case of infinite eigenvalues. We derive a systematic approach, studying how sign characteristics behave after an analytic change of variables, including the important special case of Möbius transformations, and we prove a signature constraint theorem. We also show that the sign characteristic at infinity stays invariant in a neighborhood under perturbations for even degree Hermitian matrix polynomials, while it may change for odd degree matrix polynomials. We argue that the non-uniformity can be resolved by introducing an extra zero leading matrix coefficient. | 2022-12-02 18:59:05 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8521220684051514, "perplexity": 507.79798324500234}, "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/1669446710916.40/warc/CC-MAIN-20221202183117-20221202213117-00088.warc.gz"} |
https://www.gradesaver.com/textbooks/math/other-math/thinking-mathematically-6th-edition/chapter-5-number-theory-and-the-real-number-system-5-7-arithmetic-and-geometric-sequences-exercise-set-5-7-page-331/135 | ## Thinking Mathematically (6th Edition)
Sequence- is a list of numbers that are related to each other by a rule. Example: $1,3,5,7,9,\ldots$
Sequence- is a list of numbers that are related to each other by a rule. The numbers in a sequence are called the terms. Example: $1,3,5,7,9,\ldots$ All these numbers form a sequence because the difference between two consecutive numbers in$2$. This type of sequence in which the difference between two terms is constant called Arithmetic sequence. | 2019-12-15 18:10:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6326305866241455, "perplexity": 170.09714706356237}, "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/1575541309137.92/warc/CC-MAIN-20191215173718-20191215201718-00451.warc.gz"} |
https://www.yaclass.in/p/mathematics-state-board/class-10/algebra-10295/rational-expression-and-its-operations-12269/re-32c8c3d4-84d2-45e0-9163-b1c55fee1cd4 | Learning
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### Theory:
Excluded Value
A value that makes a rational expression (in its lowest form) undefined is called an Excluded value.
Suppose the rational expression $$\frac{p(x)}{q(x)}$$ is in its lowest form, then the value for which the expression becomes undefined is said to be its excluded value.
Working rule to find the excluded value of a rational number:
Step 1: Simplify or factorise the numerator $$p(x)$$ and the denominator $$p(x)$$.
Step 2: Cancel out the common factors in the numerator and the denominator.
Step 3: Equate the lowest form of the denominator $$q(x)$$ to zero.
Step 4: Thus, the obtained value for which the denominator becomes zero is the excluded value of that rational number.
Example:
Find the excluded value of the expression $$\frac{x^2 + 5x + 6}{(x + 2)(x - 5)}$$.
Solution:
Step 1: Factorise the numerator $$x^2 + 5x + 6$$ by splitting the middle term.
$$x^2 + 5x + 6$$ $$=$$ $$x^2 + 2x + 3x + 6$$
$$=$$ $$x (x + 2) + 3 (x + 2)$$
$$=$$ $$(x + 2)(x + 3)$$
Step 2: Rewrite the expression and cancel out the common factors.
$$\frac{x^2 + 5x + 6}{(x + 2))(x - 5)}$$ $$=$$ $$\frac{(x + 2)(x + 3)}{(x + 2))(x - 5)}$$
$$=$$ $\frac{\overline{)\left(x+2\right)}\left(x+3\right)}{\overline{)\left(x+2\right)}\left(x-5\right)}$
$$=$$ $$\frac{x + 3}{x - 5}$$
Step 3: Equate the lowest form of the denominator to zero.
$$x - 5$$ $$=$$ $$0$$
Add $$5$$ on both sides of the equation.
$$x - 5 + 5$$ $$=$$ $$0 + 5$$
$$\Rightarrow$$ $$x$$ $$=$$ $$5$$
Step 4: Write the excluded value.
The rational expression $$\frac{x^2 + 5x + 6}{(x + 2)(x - 5)}$$ is undefined when $$x$$ $$=$$ $$5$$.
That is $$\frac{x^2 + 5x + 6}{0}$$ $$=$$ not defined, when $$x$$ $$=$$ $$5$$.
Therefore, $$x$$ $$=$$ $$5$$ is called an excluded value for the rational expression $$\frac{x^2 + 5x + 6}{(x + 2)(x - 5)}$$. | 2021-12-04 01:42:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "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": 0.9551039338111877, "perplexity": 559.8345244788343}, "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/1637964362923.11/warc/CC-MAIN-20211204003045-20211204033045-00484.warc.gz"} |
https://api-project-1022638073839.appspot.com/questions/how-do-you-simplify-5-8-12 | # How do you simplify -5(8 - 12) ?
May 8, 2018
$20$
#### Explanation:
Step one would be to distribute $- 5$ to $8$ and $- 12$.
$- 5 \left(8\right) = - 40$
$- 5 \left(- 12\right) = 60$
This makes your expression looks like this:
$- 40 + 60$
You can simplify this pretty easily, as $- 40 + 60$ is the same thing as $60 - 40$, which is $20$.
Another, possibly easier, way to do it would be to use the order of operations and do the expression inside the parentheses first,
$8 - 12 = - 4$
which would make it look like this...
$- 5 \left(- 4\right)$
...and then multiply to get $20$. | 2020-05-25 02:58:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 13, "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": 0.964883029460907, "perplexity": 266.65290682358795}, "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-2020-24/segments/1590347387155.10/warc/CC-MAIN-20200525001747-20200525031747-00233.warc.gz"} |
https://www.physicsforums.com/threads/never-contruct-such-proofs.267886/ | # Never contruct such proofs
1. Oct 29, 2008
### Bleys
Hey,
I'm sure this must have been asked before, but I couldn't really find anything specific using the search tool;
I'm a second year maths major and I love maths and would really like to pursue a career in mathematics. My problem is, often I can understand a proof (whether easily or not depends on the proof and tools used); what I find frustrating is that I very often could probably never contruct such proofs! And I heard somewhere or read on these forums that I should be trying and most importantly be able to prove theorems, corollaries, lemmas and such already. Now ofcourse I try to prove things I come across, but I very often find I have trouble with it. Some books like set theory or discrete maths ones have preliminary chapters explaining proof strategies and such; and while they certainly provide some practice, it still remains mind-boggling when I try to go about a proof myself in other theorems which require more abtract thinking (so I'm not talking about trivial proofs of the kind where you easily use the definitions used and logically deduce the conclusion; like even no+even no=even no; or proving identities or properties like vector product).
Should I be worried that I find I have trouble with proofs at my level and should just give up on this too ambitious dream? Or should I become comfortable enough once I've explored more maths?
2. Oct 29, 2008
### Tac-Tics
Re: Proofs
There is no general strategy for finding proofs efficiently. If there was, mathematicians wouldn't have a job to do.
Proofs of nontrivial theorems are like discoveries in science. They either come unexpectedly out of no where or they come only after generations of frustrated geniuses struggle with them. You shouldn't feel deficient or anything if you can't come up with a proof of a nontrivial theorem, because that's the hardest part of the job description.
A proof answers a question. But mathematics is interesting in that you can make up the questions. It's often just as respectable to come up with a good problem as it is to come up with a good proof. The trick is to come up with systems which strike a balance. You come up with a problem which *seems* simple, but is full of emergent complexity. My favorite example of this is cellular automata (http://mathworld.wolfram.com/CellularAutomaton.html).
3. Oct 29, 2008
### Hurkyl
Staff Emeritus
Re: Proofs
Suppose I asked you to prove that
$$(x + y)^2 - 4x y = (x - y)^2$$
Would you be able to prove this identity? Almost surely. The reason why is because you understand quite well the domain of this problem, and you have a lot of experience using the tools that are good for this problem.
But if you weren't well-versed in high-school algebra and symbolic arithmetic, wouldn't you have some difficulty with this proof?
I've heard the process of devising new theorems described as follows:
* First, figure out what should be true
* Only then do you figure out how to ensure that it really is true
For example, you might be doing a calculation, and wind up with an expression like
$$\lim_{t \rightarrow 5} \int_0^t x^2 \sqrt{1 + \sin x} \, dx$$
and you mull it over a bit, and you decide that it really should be true that you can simplify this expression to
$$\int_0^5 x^2 \sqrt{1 + \sin x} \, dx.$$
(Do you agree?) So, you probably abstract it a bit to make the problem 'simpler', and make the following conjecture:
$$\lim_{t \rightarrow b} \int_a^t f(x) \, dx = \int_a^b f(x) \, dx$$
and then you set about to try and prove this theorem... or at least a special case that's relevant to the problem at hand. Can you do so? I claim that if you really know how to use the tools from your calculus classes that this is actually a straightforward exercise, at least in a special case that includes the case we want. Actually, you might have already seen the theorem in your calc class!
4. Oct 29, 2008
### Bleys
Re: Proofs
Ofcourse I understand there's no set guideline for going about a certain proof, but that is one of the things that fascinates AND scares me the most. I find it amazing how some proofs are constructed and even thought of in general; the simplicity of some is remarkable in the poweful statements they prove. Yet the frightnening part is that I feel I lack the innovation and originality to do so myself.
The examples you provided are, like you said, straightforward if I know how to use the tools given properly (at least I hope I used them correctly). Fair enough, you can't, for example, ask me to prove anything in topology because I'm not familiar with any of its basic theory and definitions. But suppose I do have the tools to construct a proof for something, yet it requires me to use it in a "non-conventional" way, then I'm usually as lost as a law major in physics.
I guess an example of what I'm finding hard is Euclid's Theorem on the infinity if primes. Simple enough to understand, but could I have ever come up with it? Hardly... Like I said, I'm afraid maybe I'm unable to think outside the box when it's required. Or maybe a better example is proving the irrationality of $$\sqrt{3}$$ using the well-ordering property. Or proving the well-ordering property itself.
I suppose maybe I have some misconceptions on the field of work. Lectureres understand what they're talking about. But assume for example they need to prepare a lesson on something. Do you think they could prove the material they are suppose to give? Almost certainly! But then, if I aspire to become a mathematician, shouldn't I also by now be able to do so as well? I often find myself thinking after a lecture "could I have been able to prove that if it was given as an assignment?". More often than not the answer is no.
I guess I'm just worried because I know proofs are a fundamental part of being a mathematician.
5. Oct 29, 2008
### mathwonk
Re: Proofs
start asking the question: why? when you see any assertion. read a book on logic, the one i read as a high school student was by ALLENDOERFER AND OAKLEY, called principles of mathematics. start thinking, and stop being satisfied with just making simple minded calculations in your math courses. ask why the formula works. think it through.
6. Oct 29, 2008
### CRGreathouse
Re: Proofs
I just wanted to say that I really appreciate this thread. I'm a(n off-again, on-again) graduate student, but I often feel I'm not that good at writing proofs. I'm not creative in my choice of the type of proof, I favor RAA too much, and I omit steps. It's good to read advice on proofs!
7. Oct 29, 2008
### Ben Niehoff
Re: Proofs
The first step in figuring out a proof (for something you don't already know how to prove) is just to play around with the math. Use the definitions given, write some particular examples, and see what happens when you put things together according the rules.
This can give you insight into how to use those definitions to construct a proof (and if the thing you're trying to prove is not true, maybe you'll run across a counterexample this way).
8. Dec 10, 2008
### monotune
Re: Proofs
I think, one has to develop a critic mind first, and then learn how to criticize in a precise style/way. I remember getting really pissed off many times, bec I thought that those guys who had written those text books want me to "believe" them. Actually, you could say, that I just didnt understand or misunderstood something (yet, sometimes I was right), anyways this created a sort of "ambient". It was the ambient of being present, participation and interaction. (Hell,:D I was passionate!) This is when you start to form questions on your own. Then you try to answer them, and later the statement might turn out to be a theorem, a lemma or just "fun". You probably know most of the mathematical arsenal/ tools, tools in detail and many theorems and proofs as examples. Thats useful. But motivation is the essential part, whether you get eager to know that the given statement is true or not (or something else).
9. Dec 11, 2008
### Tac-Tics
Re: Proofs
In my high school experience and up through multivariate calculus in college the books required you to believe them. It was only a convenience if you ever noticed certain things were true.
10. Dec 11, 2008
### monotune
Re: Proofs
I was referring to my university yrs. (Primary and high school are something very different, and Id rather not mix the maths studied there with the higher maths studies.) Sure you have to give some sort of credit to teachers, books in order to be able to develop, and I was talking about participation and interaction with them and their ideas, not rejection. I usually got pissed off during the first semester of analysis. And in order to do my exams I had to -at least temporarily- accept some (for me 'disturbing') things. Then in the second semester we studied set theory, and then those 'disturbing' things got explained. Eg the continuum hypothesis turned out to be 'relief'. Later I became more patient.
11. Dec 11, 2008
### Tac-Tics
Re: Proofs
What kinds of ideas did you have to accept without proof?
And what was the continuum hypothesis doing in an analysis class? =-o
I'm impatient like that too. Quick and dirty mathematics is unappealing, even though it produces correct answers a lot of the time. That's why I stopped doing calculus altogether after my freshman year of college up until I graduated =-P
For the record, I majored in computer science, and if you confuse the function "f" with the function f evaluated at a point x, "f(x)", in your code, you get an error!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook | 2017-08-22 19:15:27 | {"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": 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": 0.5379490852355957, "perplexity": 618.7898071482582}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886112539.18/warc/CC-MAIN-20170822181825-20170822201825-00493.warc.gz"} |
https://nathaniel.blog/cheatsheet/java-rmi-example/ | Java RMI - A Complete Example
Nathaniel May 04, 2016
In this example, City can be replaced by any similar model class, such as Car or Person. | 2018-11-19 17:20:22 | {"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": 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": 0.1871613711118698, "perplexity": 2551.6044711699196}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039746061.83/warc/CC-MAIN-20181119171420-20181119193420-00279.warc.gz"} |
https://ec.gateoverflow.in/1839/gate-ece-1991-question-1-20 | 10 views
In figure, all transistors are identical and have a high value of beta. The voltage $V_{D C}$ is equal to
_________. | 2022-10-04 03:06:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.4982156455516815, "perplexity": 2395.766587727893}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337473.26/warc/CC-MAIN-20221004023206-20221004053206-00707.warc.gz"} |
http://stackoverflow.com/questions/13553654/whitespace-in-reused-code-chunks | # Whitespace in reused code chunks
I'm starting to learn R and Sweave and have spent the last couple of hours trying to figure out a problem that, according to my Google-fu, seems to be rather unknown or uncommon (as I found only two references to the problem).
Namely, when re-using a code chunk in Sweave, it seems you are not allowed any preceding whitespace. For example:
\documentclass{article}
\begin{document}
<<label=fig1boxplot, include=FALSE>>=
boxplot(mpg~cyl,data=mtcars)
@
\begin{figure}
\centering
<<label=fig1, fig=TRUE, echo=FALSE>>=
<<fig1boxplot>>
@
\end{figure}
\end{document}
will not work as expected if there is any white space in front of any line beginning with << (errors vary depending on which line is indented). I found a work-around on the R developer mailing list that lets me indent the line where the chunk is reused (<<figboxplot>>) but still not any other chunk (<<label=...)
• Why does Sweave not allow white space with reused code chunks, especially since the workaround seems very trivial to implement indeed?
• How can I work around this problem and indent chunks (or any Sweave code for that matter) as much as I like?
-
Just to be clear, you're simply asking why the Sweave parser works the way it does? – joran Nov 25 '12 at 18:28
Actually, now that I've tried the workaround, it only solves the problem of indenting the <<fig1boxplot>> line, but indenting any of the lines indicating the start of a chunk (<<label=...) still result in errors or wrong output. Since there's no answer yet, I've added the question of how to work around this issue in general. – ThomasH Nov 25 '12 at 19:12
The short answer is to use knitr instead, which allows indentation of code chunks.
@ThomasH you are using an R version which was released more than two years ago, which is terrible, and I strongly recommend you to update R, given that it is so easy to set up the repository on Ubuntu: cran.r-project.org/bin/linux/ubuntu For the error you saw, I recommend you to set the environment variable R_LIBS_USER in ~/.Renviron to a writable directory, e.g. github.com/yihui/configuration/blob/master/.Renviron – Yihui Nov 25 '12 at 21:13
Thanks for the help. I took the plunge and updated Ubuntu from 10.10 to 12.04 without my poor old laptop dying in the process and managed to get R to version 2.15.2 though I had to add ppa:marutter/rrutter to my sources list as it wouldn't update from 2.14 to 2.15 using the Irish CRAN mirror. – ThomasH Nov 29 '12 at 20:49 | 2014-09-22 12:13:49 | {"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": 0.6188560724258423, "perplexity": 1462.1089799015028}, "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/1410657137046.16/warc/CC-MAIN-20140914011217-00322-ip-10-234-18-248.ec2.internal.warc.gz"} |
https://worldbuilding.stackexchange.com/questions/3361/would-it-be-possible-to-terraform-the-moon?noredirect=1 | # Would it be possible to terraform the moon?
I have a "simple" question that is the basis for an idea I'm developing: Is it possible to terraform the moon and, if so, what type of technology would be needed?
I understand that this isn't the case because of its current state, but, for arguments sake would it be possible to do this if certain conditions were met: an artificial core, some artificially imposed atmosphere, bring water, life and vegetation, etc.
For clarification: The terraforming would need to be man-made and be able to sustain humanity as it currently is from a biological sense (no additional breathers, pressure suits, etc.). This means that there would need to be some form of green plants, soil, water, free moving air, gravity, and the ability to grow plants and raise animals.
• This needs some more specification. Terraforming the moon is definitely possible. A valid question could be, what level of technology is necessary for terraforming the moon? Though that is still a tiny bit broad and might need some more scoping. Oct 30 '14 at 18:27
• Alright, I didn't even know if it was a possibility or if I was just running down the wrong road on this one. Let me add some more details to the question. Oct 30 '14 at 18:45
• You could make the moon habitable by increasing it's mass with a very dense core, which would allow it to hold onto an atmosphere, and a magnetic field to protect from the solar wind, but this would have devastating effects on Earth (huge tides and increased tectonic activity). Do you specifically want to avoid destroying the Earth? Oct 30 '14 at 20:11
• Perhaps a miniature contained black hole, or a big chunk of neutronium would give you the mass, and perhaps also the magnetic field. Oct 31 '14 at 14:33
• See To Crush the Moon specifically, and the rest of the series. Innthis novel he details how it was done using technology that appeared throughout the series earlier. Mar 20 '16 at 3:27
Short answer... no. Longer answer is, it depends. The biggest restriction I see is lack of atmosphere. Just supplying it with an atmosphere is a short term solution as, by itself, the moon doesn't have sufficient gravity to keep an atmosphere from leaking into space. This particular question was asked on another site:
https://physics.stackexchange.com/questions/52527/can-low-gravity-planets-sustain-a-breathable-atmosphere
Quoting from one of the answers:
The escape velocity at the moon's surface is about 2.4 km/s. The mean speed of oxygen at 293 K is about 0.48 km/s.
A commonly quoted rule of thumb says that the escape velocity needs to be 6 times the gas's mean velocity in order for that gas to remain captive to gravity and the values I quoted are related by a factor of only 5. The air would contain water (since dry air is very uncomfortable to breath) and carbon dioxide (as a by-product if not also needed to sustain the cyanobacteria/plants you would want in place of planetary size mechanical carbon dioxide scrubbers, then there are the nutrients you would need to sustain those) which would readily exacerbate an atmospheric greenhouse effect and, with the moon being at about the same distance from the sun as is earth, you would expect the air to warm up to similar to earth temperatures, though without the moderating effect of oceans, and so cause the oxygen to dissipate. As nitrogen is lighter it's mean speed at the same temp is higher, v_rms something like 0.51 km/s IIRC, so it too would dissipate as would water vapour.
In short, it doesn't seem likely that it would be possible on the moon.
But, since you mention an artificial core of some kind, maybe one that creates an higher than natural gravitation effect, that could be circumventable, if it justified the cost of drilling to the core of the moon to install the device.
• I really like this answer and the technical explanation. So, based off of our current physics it simply is not possible and it really comes down to the fact that it needs to be able to hold an atmosphere with some type of external (or internal) assistance. Oct 30 '14 at 20:19
• Good answer. I wonder what the minimum size a body needs to be to support an actual atmosphere with Earth like composition. @JasonWilczak - An artificial core, in the form of heavy iron, would increase the moons density and increase the moon's escape velocity. No clue how that it would be accomplished, increasing a body like the moon's mass is no easy feat. Oct 30 '14 at 20:43
• Rather than size, it's the mass that's important. The physics site has some complex equations relating to that (physics.stackexchange.com/questions/45494/…). Also related to atmosphere loss are things like temperature and vulnerability to solar wind. Oct 30 '14 at 21:32
• Making the moon's core dense enough to hold a breathable atmosphere would destroy Earth's coastal cities with high tides and change the climate with increased volcanic activity and earthquakes. Unless you also move the moon further away from the Earth, making the tides normal size again, but now less frequent, and making the moon appear smaller in the sky. Oct 31 '14 at 11:13
• You're missing the part of the answer you quoted from which states that the moon could host a breathable atmosphere for some time. That time would be very short geologically (thousand years) but very long for a human's reckoning of time. It could be possible to sustain a Moon atmoshpere indefinitely with the right equipment pumping in new gas or by bombarding the moon with comets in safe areas. Nov 6 '15 at 17:38
## Yes
With several caveats.
### Volatiles
As @Saidoro mentioned, you need to introduce all the volatiles required for life. This includes atmosphere, hydrosphere, and Nitrogen.
It's easy to find enough water and oxygen in the various ice moons and comets throughout the Solar System. However, compared to the amount needed, nitrogen is pretty rare. You'll need to expend some thought about how to get enough.
### Longevity
As you can tell by looking at the Moon, the volatiles won't stay around the Moon over geological ages.
So if you do provide the Moon with volatiles, how long can you expect them to stick around?
There are many Atmospheric Loss Mechanisms. Unfortunately, only one of these (Jean's Escape) is easily calculated.
Wikipedia has an excellent and scientific chart showing different body's ability to retain an atmosphere over geologic ages.
However, I've created my own chart based upon Jean's escape. I apply a "fudge factor" to cover the other escape mechanisms and it gives the half-life of different gases around a given body. Be aware that my chart assumes that the body has a thick atmosphere to begin with so certain loss mechanisms such as sublimation and sputtering aren't an issue.
According to my chart, you could expect water added to a thick atmosphere around the Moon to have a half-life of around 200,000 years. A very long time compared to human civilization. A very short time compared to geologic time.
Half-life of gases around major Solar System bodies:
Measured in years
Notes:
1. Any half-life with a value of $\cdot 10^{98}$ is a liquid or solid at the conditions on the surface of that body. Meaning as long as it is surrounded by an atmosphere, there will be no loss of this compound (the reality is that airless bodies will still see some losses).
2. The body Ceres is just outside the snow/frost line based upon my calculations.
3. When I could find them I used melting & boiling points for 0.1 atmosphere pressures. When I couldn't find those I used those for 1.0 atmosphere.
4. Tweaking the fudge factor is tricky. Tweak it too high and Mars gets too much atmosphere. Tweak it too low and Titan doesn't get enough. I think I found a happy compromise since Ganymede doesn't get any, Titan has one as long as there's some replenishment, and Mars lost most of its atmosphere.
5. Red background means half-life < 100,000,000 years
6. Yellow background means half-life < 4,500,000,000 years
7. Green background means half-life > 4,500,000,000 years
8. White background means substance isn't a gas at those conditions
I like all the answers above. One possible method would be to build an envelope around the moon. This is an alternate to giving the planet an entire atmosphere, which it would periodically lose and need replenished, but it's still a gonzo engineering feet and the envelope would be subject to regular meteor strikes and punctures. It would need regular repair, or perhaps, some self repairing technology and leak plugging - that's no simple task.
Another method, which I like, would be to dig a very deep hole into the moon and have a kind of deep underground dwelling, with big fans to enable circulation and mirrors to reflect in sunlight.
As an earthly example, the Dead Sea is some 400 meters below sea level and as a result, it's atmosphere is about 5% more pressure. Source. On the Moon, with 1/6th the gravity of Earth, well, calculating it gets ugly, but if you dig a hole, some, 30-50 miles into the Moon's surface, perhaps at a steep incline to avoid direct sunlight but still take advantage of reflected sunlight. It might be possible.
Something, kind of like this but about 500 or 1,000 times bigger.
The hole or whatever you call it (Atmosphere well?) should requires far less replenishment of atmosphere than the surface terraforming idea. There would still be some, you'd need a light atmosphere around the entire moon to avoid too much loss and preserve most of the atmosphere in the deep hole, but it's more feasible than making the surface habitable.
Another alternative, with enough energy would be to generate to Oxygen directly from the silicate material of the moon, so the oxygen would be generated from within the hole.
Digging deep into the Moon might provide water sources and other minerals as well. I'm not sure anyone would want to live in a hole, but it beats living in a pressurized suit, never going outside.
Yes.
An atmosphere would need to be introduced(likely through comet bombardment or the like), and some (probably biological) machines would need to be introduced that could turn moon-dust and oxygen into something that earthish plants could grow in.
Additionally, a few changes would need to be made to the animals and plants who would be living there, most notably adaptions for the decreased gravity and atmospheric pressure and for the increased radiation caused by the moon's lack of a strong magnetic field.
Once the technology to do all of that is available, the moon will be terraformable, and will likely be one of our first targets for doing exactly that.
• I've edited the question a bit for clarity that the terraforming should be man-made and need to support humanity at it's current biological evolution. Oct 30 '14 at 18:50
• All the things I noted can be man-made. Making it work without changing humans at all makes things harder, but it's doable. Humans would suffer from the lower atmospheric pressure(and consequently, the lower amount of breathable oxygen) and from the increased radiation, but neither of those is a complete deal-breaker. Humans could spend more time inside and underground to protect themselves from radiation and slather on thick sunscreens on the surface. And people already live full-time in lower pressure environments. It'll be uncomfortable, but not impossible. Oct 30 '14 at 18:57
• The plants will need some bioengineering, though. No getting around that. Oct 30 '14 at 18:59
• Indeed, but the idea would be to make the moon, essentially, a mini-Earth, complete with running water and grass. Oct 30 '14 at 20:17
Terraforming is a beautiful, and ridiculously simple principle, that is more than godlike in theory. There are, however, two major issues that limits the amount of terraformable celestial bodies to a very small number:
• Magnetic field. We, here on Earth take its existence natural, but it has a huge role in the fact that life here is so much more pleasant.
I don't know, but am 100% sure Moon has no such a magnetic field Earth has, so its surface is much more exposed to solar wind, gamma rays and such. These can even change the chemical composition of a theoretical atmosphere.
Whether Earth's field reaches the Moon, and thus provides some minor defense is beyond my knowledge.
• Gravity. Attracting force of Moon is only 1/6 of the Earth. This is a huge difference, and because of it, Moon would be unable to support such a layer of gases. It would simply leave that.
Without gravity, there's no atmosphere. Without atmosphere, there's nothing.
What can you do then? I prefer the idea of building domes, then building more domes, with their own gardens, greenhouses and other life-supporting facilities to produce oxygen. Then, it might be exposed to the lunar surface as an attempt. Or, if possible, building extremely huge domes, that might be able to support clouds and rains within! Very partial solution, but I can see the outcome of it.
Source: SpaceEx.SE
• The Earth's magnetic field doesn't reach close to the Moon, at least, not in a protective way, though the tail of or magnetic field (magnetotail) reaches about 1 1/2 to twice as far as the moon, but that's not much protection, only when the Moon is behind the Earth. The earth's magnetic field does protect the space station and our satellites for the most part. Mar 21 '16 at 6:53
I agree with @Random's answer about the Moon having insufficient gravity to hold onto an atmosphere in the long term, so it would have to be continually replenished. There's other important considerations for terraforming the Moon (and Mars): low gravity and the lack of a magnetic field.
The Moon lacks a magnetic field. Space is a harsh place, but Earth's atmosphere and magnetic field protects us from most of it. Lunar inhabitants will be exposed to cosmic and solar radiation. I don't have a radiation exposure map for the Moon, but I do have one for Mars where you'll be recieving a dose of about 0.25 sieverts per year. 1 sievert is a 5.5% increase in cancer risk so the lunar population will have a very increased risk of cancer.
This can be mitigated somewhat by adding certain molecules to the atmosphere, such as ozone, to block solar radiation. Cosmic rays are unpredictable and difficult to block without a powerful magnetic field. Solar flares can be predicted and lunar inhabitants will have to take to shelters.
Then there's the problems of living in a low gravity environment. Everything we know about human health says this is bad for humans who have evolved for billions of years to take advantage of a constant gravity. The biggest long term problem is a loss of bone and muscle mass and a tendency for bodily fluids to pool in the upper body. There's no long term cure of this.
Others have noted with up-to-date caclulations that the Moon would hold on to air long enough to be useful to people settling there, were it to be introduced.
That confirms the assertion made in a story I recall, about a near-future society (no super-duper tech) that co-opted a mission to redirect a comet, and had it crash into the Moon instead of just missing. The story ends with the appearance of a fuzzy look to the moon, with everyone excited at the potential.
Another story I recall featured moons of Jupiter (I think), where the first one done just has added air which needs a bit of topping off from time to time, but the later ones have a membrane to keepmthe air in.
In Wil McCarthy's Queendom of Sol series, the Moon and other small bodies were crushed to give them a smaller diameter and higher surface gravity. In his novels, industrial nutronium marbles are a thing, and whether he supposes gravity holds it together stably or some other finishing treatment is needed is not explained.
However, the late Robert L. Forward covered the topic with the intention of being "hard" sci-fi. An asteroid layered with carbon and a shaped charge (of anti-matter) produced a collapsed-matter disk encapsulated in diamond strong enough to maintain the internal pressure. | 2021-12-03 16:41:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4015403985977173, "perplexity": 1075.1875015710434}, "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-2021-49/segments/1637964362891.54/warc/CC-MAIN-20211203151849-20211203181849-00411.warc.gz"} |
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Chebyshevskii Sb., 2017, Volume 18, Issue 4, Pages 222–245 (Mi cheb607)
Geometrization of numeration systems
A. A. Zhukova, A. V. Shutov
Abstract: We obtain geometrization theorem for numeration systems based on greedy expansions of natural numbers on denomirators of partial convergents of an arbitrary irrational $\alpha$ from the interval $(0;1)$.
More precisely, denomirators $\{ Q_i (\alpha) \}$ of partial convergents of an arbitrary irrational $\alpha \in (0; 1)$ generate Ostrowski–Zeckendorf representations of natural numbers. These representations have the form $n = \sum\limits_{i=0}^{k} z_i( \alpha, n) Q_i ( \alpha )$ with natural conditions on $z_i( \alpha, n)$ described in the terms of partial quotients $q_i(\alpha)$. In the case $\alpha =\frac{\sqrt{5}-1}{2}$ we obtain well-known Fibonacci numeration system. For $\alpha=\frac{\sqrt{g^2+4}-g}{2}$ with $g \ge 2$ corresponding expansion is called representation of natural numbers in generalized Fibonacci numeration system.
In the paper we study the sets $\mathbb{Z} ( z_0, \ldots, z_{l} )$, of natural numbers with given ending of Ostrowski–Zeckendorf representation. Our main result is the geometrization theorem, describing the sets $\mathbb{Z} ( z_0, \ldots, z_{l} )$ in the terms of fractional parts of the form $\{ n \alpha \}$. Particularly,for any admissible ending $( z_0, \ldots, z_{l} )$ there exist efffectively computable $a$, $b\in\mathbb{Z}$ such that $n \in \mathbb{Z} ( z_0, \ldots, z_{l} )$, if and only if the fractional part$\{ (n+1) i_0 (\alpha) \}$, $i_0 (\alpha) = \max \{ \alpha; 1 - \alpha \}$, lies in the segment $[ \{a \alpha \}; \{b \alpha \} ]$. This result generalizes geometrization theorems for classical and generalized Fibonacci numeration systems, proved by authors earlier.
Keywords: numeration systems, Ostrowski–Zeckendorf representation, geometrization theorem.
DOI: https://doi.org/10.22405/2226-8383-2017-18-4-221-244
Full text: PDF file (667 kB)
References: PDF file HTML file
UDC: 511.43
Accepted:15.12.2017
Citation: A. A. Zhukova, A. V. Shutov, “Geometrization of numeration systems”, Chebyshevskii Sb., 18:4 (2017), 222–245
Citation in format AMSBIB
\Bibitem{ZhuShu17} \by A.~A.~Zhukova, A.~V.~Shutov \paper Geometrization of numeration systems \jour Chebyshevskii Sb. \yr 2017 \vol 18 \issue 4 \pages 222--245 \mathnet{http://mi.mathnet.ru/cheb607} \crossref{https://doi.org/10.22405/2226-8383-2017-18-4-221-244} \elib{http://elibrary.ru/item.asp?id=30042552} | 2020-01-18 14:16:11 | {"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": 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": 0.5957764387130737, "perplexity": 5488.833185195079}, "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/1579250592636.25/warc/CC-MAIN-20200118135205-20200118163205-00263.warc.gz"} |
https://physics.stackexchange.com/questions/291501/how-does-huygenss-principle-work/291514 | # How does Huygens's Principle Work?
I just learnt about Huygens's principle that each point of a wavefront acts as a point source producing wavelets. However, I tried drawing it in a diffraction situation, but it doesn't come out right. In theory, if the gap size and the wavelength is same, maximum diffraction occurs, but when I try to draw this on paper it doesn't produce semi-circular wavefronts. It makes the edges rounded while the middle is flat.
Something like this when I draw, but isn't it supposed to be semi-circular if wavelength is same as gap size?
• No. It will be semi-circular if the slit size is much smaller than a wavelength, so that the source resembles a point source. A point source would produce only one wavelet, and ... a semi-circle. Anything else is elongated. – garyp Nov 8 '16 at 3:14
• Oh so even though the gap size is same as the wavelength, it isn't actually semicircular? Because this is what I have been taught by my teacher. – Hilkjh Nov 8 '16 at 3:25
• elongated semi-circular would make it descriptively right. – Mozibur Ullah Nov 8 '16 at 7:21
Using Huygens to draw pictures is good for illustrating diffraction but you have to be cautious when interpreting the diagrams.
The diagram below shows the Huygens construction for wavelets separated by one wavelength and there being 11 secondary sources in the slit.
Lines $XX'$ and $YY'$ demarcate the region where one would expect to have light if one was considering light to travel in straight lines - geometrical optics.
What you must note is the envelope of the Huygens wavelets between those two lines is a straight line and its length stays constant as the waves travel further and further from the slit.
So by the time you get to 40 wavelength from the slit you almost get what you expected - a semicircular wavefront centred on the slit.
For clarity in the second diagram I have just shown three 3 secondary sources in the slit.
This drawing of the wavelets can be avoided using Mathematics and refinements to the secondary wavelets.
One refinement is to have the amplitude of a secondary wavelet dependent on the direction of its travel being a maximum for the part of the wavelet which is travelling in the same direction as the incident light and getting less an less as the direction of travel moves away from the direction of the incident light.
Also as the wavelets move further from the slit their amplitude is reduced because a certain amount of energy now has to be spread over a longer wavelet.
So the diagrams which you draw give you an idea as to what happens but for the detail you must resort to methods which do not rely on drawing semicircles on paper.
• ahh i get it, so its just the first few wavefronts appear to be more flat, but when there is more wavefronts it becomes more semi-circular? – Hilkjh Nov 9 '16 at 3:42
The picture you posted from wikipedia is rather misleading. In fact, Huygens's Principle and what your teacher have taught you is correct: light passing through any aperture is diffracted as if it were coming from a number of aligned point sources (like in the image you provided). However, what this diagram mis-represents is that the sum of the circular wavefronts generated becomes somehow "flat". If this were indeed the case then light passed through a aperture much larger than its wavelength would create an image on a screen placed some distance away from the aperture that is a perfect point, with decaying intensity at the fringes (this seems to be what @garyp is suggesting in his comment).
However, in reality what we find is that the "image" - this just being the light collected at a plane located some distance much larger than the size of the aperture -- follows the pointspread distribution function.
This function can be rigorously derived fromt the assumption of Huygen's Principle, and is derived in Chapter 29 of the Feynman lectures (Lectures on Mechanics).
• So, how would a diagram look like if drawn correctly? – Hilkjh Nov 8 '16 at 6:45
• This answer is a little confused; it's not clear to me what point you are making. A plane wave incident on an aperture much larger than the wavelength would produce a flat top intensity profile, like the image on the left. A shadow. Not a point, as this answer suggests. The point spread function shown on the right is for an aperture about the size of a wavelength, not for a large aperture. Also, I think the drawing of Huygens' wavelets in the OP is correct. – garyp Nov 8 '16 at 12:48
• Wikipedia provides a good graphic of the pointspread function arising from huygen's principle. upload.wikimedia.org/wikipedia/commons/0/0f/… – zephyrus Nov 8 '16 at 18:43 | 2021-03-08 05:35:29 | {"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": 0.7358721494674683, "perplexity": 476.2440796357305}, "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-2021-10/segments/1614178381989.92/warc/CC-MAIN-20210308052217-20210308082217-00080.warc.gz"} |
https://www.physicsforums.com/threads/radiowave-interference.378163/ | 1. Feb 14, 2010
EmmaK
1. The problem statement, all variables and given/known data
A radiotelescope is positioned at a height of h=150m on a cliff overlooking the sea. A very distant galaxy emitting radiowaves, wavelength 3m, is at an angle $$\theta$$ above the horizion. The radiowaves reach the telescope both directly and after reflection off the water surface. Show that if $$\theta$$=57degrees, then the 2 rays interfere destructively, so that the signal received is zero. Note that the refractive index,n,of the sea is greater than that of air
2. Relevant equations
dsin$$\theta$$=(m+1/2)$$\lambda$$ for destructive interference.
3. The attempt at a solution
Tried substituting the values into the equation, with d=2hsin$$\theta$$, and expected to get m as an integer, but didnt..
2. Feb 14, 2010
EmmaK
anyone?
3. Feb 14, 2010
ideasrule
That equation doesn't apply here; I'll leave it to you to figure out why. You have to approach this problem from first principles. First, calculate the path length difference between the reflected ray and the direct ray.
4. Feb 14, 2010
EmmaK
oh ok, is the path difference not 2hsin(theta)=300sin(theta) either?
5. Feb 14, 2010
ideasrule
I see no obvious reason why it should be equal to that.
6. Feb 14, 2010
EmmaK
ok.. i just used trig. sin(angle) = opposite/hypotenuse. want to find hypotenuse and opposite =h. and there's 2 of these triangles, so path difference is double this..hard to explain without a diagram!
7. Feb 15, 2010
Stonebridge
I have given you a hint (with diagram) on the TSR physics site. | 2018-07-20 16:10:39 | {"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": 0.6426551938056946, "perplexity": 1325.4740890031119}, "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-2018-30/segments/1531676591718.31/warc/CC-MAIN-20180720154756-20180720174756-00562.warc.gz"} |
https://cs.stackexchange.com/questions/113326/regularity-profiles?noredirect=1 | # Regularity profiles
A standard exercise in formal language theory uses Lagrange's four-square theorem to construct a language $$L$$ such that $$L$$ isn't regular but $$L^2$$ is regular. (Let $$A = \{ a^{n^2} : n \geq 0 \}$$. Then $$A$$ isn't regular, but $$A^4 = \{ a^n : n \geq 0 \}$$ is regular, hence either $$L = A$$ or $$L = A^2$$ fits the bill; in fact $$A^2$$ is not regular, so we must pick the latter.)
This answer generalizes this, constructing for every $$m$$ a language $$L$$ such that $$L,L^2,\ldots,L^{m-1}$$ are not regular, but $$L^m,L^{m+1},\ldots$$ are regular. This prompts the following definition:
The regularity profile of a language $$L$$ is $$\rho(L) = \{ n \in \mathbb{N}_+ : L^n \text{ is regular} \}$$.
My question is:
Which regularity profiles are achievable?
The answer mentioned above shows that $$\{n : n \geq m \}$$ is a regularity profile for every $$m \geq 1$$.
It is also easy to construct a language whose regularity profile is empty: $$\{ a^{2^n} : n \geq 0 \}$$.
Clearly every regularity profile is closed under addition: if $$L^a,L^b$$ are regular then so is $$L^{a+b}$$.
Is every subset of $$\mathbb{N}_+$$ which is closed under addition a regularity profile?
The question is interesting both for a general alphabet and for the special case of a unary alphabet. | 2021-04-13 16:09:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.832391619682312, "perplexity": 290.26646650934936}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038073437.35/warc/CC-MAIN-20210413152520-20210413182520-00117.warc.gz"} |
http://lists.w3.org/Archives/Public/www-math/2003Sep/0027.html | # Re: Re: Latex to Mathml translators
From: (wrong string) éphan Sémirat <stephan.semirat@ac-grenoble.fr>
Date: Tue, 16 Sep 2003 20:49:46 +0100
To: Robert Miner <RobertM@dessci.com>, osserman@math.mit.edu <osserman@math.mit.edu>
Message-Id: <20030916184925.79FA4136A8@dr-nick.w3.org>
Hi,
>People are free to use whatever software suits them as far as
>I'm concerned.
In fact, i think there is really a gap to be filled about authoring math documents (for the web or not).
I'm a french teacher (in high school) and i do need to type math every day. I used to use laTeX, but couldn't share documents with other teachers.
They used equation editor but are not satisfied with it (try to type "\int_{a_1}^{a_2}" : how many minutes you need, and what about the rendering ?).
The gap is bigger when authoring for the web : have you try to write a 2 pages document, with almost 15-20 math equations (that is a "normal" exams) with math type ? It would take half a day and two mouses (well, i exagerate, but not so much).
What i need (is needed ?) is a _simple_ latex syntaxed language, without two thousand files to deal with,...
I think i know how to use a computer, but in the list you mention, Robert, i feel like one's need to be a PH -D in informatic to understand how to use most of them !
>But it really is a triviality to come up with a language as terse as
>TeX that maps directly and unambiguously to some XML + MathML doc
>type. For example, just changing <foo>...</foo> to \foo{...} and
>adding some default tokenization rules (that can be easily overridden)
>makes authoring MathML comparable to authoring TeX. Its just that
>there evidently hasn't been enough of a demand for it for anyone to
>have written a compiler yet...
Well i'm just writting something simple that do this job, and more (use of css,...).
Regards,
Stephan Semirat
Received on Tuesday, 16 September 2003 14:49:29 UTC
This archive was generated by hypermail 2.3.1 : Tuesday, 6 January 2015 21:27:34 UTC | 2015-03-03 16:05:11 | {"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": 0.774273693561554, "perplexity": 4017.4085646719445}, "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-2015-11/segments/1424936463318.72/warc/CC-MAIN-20150226074103-00046-ip-10-28-5-156.ec2.internal.warc.gz"} |
http://wikispiracy.org/post/2020/05/11/money-supply/ | The money supply is the total value of all money available in an economy at a given point in time. There are several ways to define “money”, but standard measures usually include currency in circulation and deposits in banks. A central bank may use its own definitions of what constitutes money for its purposes. This may even include your birth certificate!
## Scarcity value and cryptocurrency
Scarcity value is the relationship between how much of something is available and its value. So, in general the empirical result of this relationship is that, as the availability of a commodity1 rises, then its value decreases. Bit-coin and all cryptocurrencies exploit only the scarcity value of virtual money. Despite Sméagol-gollum sales people and endless mindless pundit’s waffle on this, the truth is that all forms of cryptocurrency have no intrinsic value whatsoever. This results in money which is backed by absolutely nothing and which is necessarily the most volatile monetary system of all.
Ergo: cryptocurrency is fiat money in possibly the worst form possible.
## Origins and development
The quantity theory descends from Nicolas Copernicus, and various others who noted the increase in prices following the import of gold and silver, used in the coinage of money, from the New World. (the exploitation of the Americas) The “equation of exchange” relating the supply of money to the value of money transactions was stated by John Stuart Mill who expanded on the ideas of David Hume. The quantity theory was developed by Simon Newcomb, Alfred de Foville, Irving Fisher, and Ludwig Von Mises in the late 19th and early 20th century.
## Central banks
Henry Thornton submitted the idea of a central bank after the financial panic of 1793, although, the concept of a modern central bank was not given much importance until Keynes published “A Tract on Monetary Reform” in 1923. In 1802, Thornton published An Inquiry into the Nature and Effects of the Paper Credit of Great Britain in which he gave an account of his theory regarding the central bank’s ability to safeguard price level. According to his theory, the central bank could inspect the currency in circulation through book keeping. This inspection could allow the central bank to gain a command of the money supply of the country. This ultimately would lead to the central bank’s ability to safeguard the price level. His introduction of the central bank’s ability to influence the price level was a major contribution to the development of the quantity theory of money.
## Karl Marx
Karl Marx amended the established theory on money supply, by arguing that the labor theory of value requires that prices, under equilibrium conditions, are determined by the labor time needed to produce the commodity and that quantity of money was a function of the quantity of commodities, the prices of commodities, and the velocity. Marx did not reject the basic concept of the Quantity Theory of Money, but rejected the notion that each of the four elements were equal, and instead argued that the quantity of commodities and the price of commodities are the determinate elements and that the volume of money follows from them. He argued that:
The law, that the quantity of the circulating medium is determined by the sum of the prices of the commodities circulating, and the average velocity of currency may also be stated as follows: given the sum of the values of commodities, and the average rapidity of their metamorphoses, the quantity of precious metal current as money depends on the value of that precious metal. The erroneous opinion that it is, on the contrary, prices that are determined by the quantity of the circulating medium, and that the latter depends on the quantity of the precious metals in a country;this opinion was based by those who first held it, on the absurd hypothesis that commodities are without a price, and money without a value, when they first enter into circulation, and that, once in the circulation, an aliquot part of the medley of commodities is exchanged for an aliquot part of the heap of precious metals.
## A more contemporary viewpoint
So, a more contemporary viewpoint on money supply and value might be:
1. The quantity of money available is inversely proportional to the value of money
2. The value of a monetary note, coin or other instrument is immediately determined by the amount of interest that has been already paid upon that instrument. In other words, a coin in your pocket has a value only because it has been borrowed and interest is paid upon that borrowing.
3. Lowering interest rates has been used by central banks as the primary method of safeguarding the quality of money supply for decades, since most of the money in supply is not in a tangible form but rather is simply numbers in accounts.
4. It is notable that even at zero percent interest the above mechanism of safeguarding money supply no longer functions well. This, in part, is due to tightening of the borrowing requirements to the extent that Banks may not lend to you when you do not have securities to cover the total borrowing.
5. All fiat money suffers from a need for exponential growth in monetary supply in order to pay for the interest upon the current money supplied.
6. As newer mechanisms are found by the stock-market industry to create drastic levels of new money using systems such as Securitization and Derivatives, it becomes increasingly difficult for any Central bank, state or union of states to know how much money is actually flowing or in circulation. This is true because Derivatives are most often traded Off-the-books, such that the agreement is not recorded by a third party or regulated by anyone.
7. This system is similar to a CASINO, and is certainly the most ridiculous manner with which to run any economy or the whole world.
8. A ‘normal’ outcome of the failure of capitalist money and its money supply is war. War provides an excuse to tax the poor, it murders huge numbers of innocent people and it uses arms which are expensive, thus making the arms manufacturers rich and those who hold shares in these companies of highly questionable moral compass.
9. It must be noted that COVID19 may have been used in place of a traditional war. In this scenario the IMF WHO and World Bank solve the massive exponential need for more money by gifting or doling out money in a magnitude of $trillions to the Central banks and also huge corporations and other massive financial players. It may be that over three hundred trillion dollars ($300,000,000,000,000) have been doled-out to Banksters in this way.
10. Karl Marx could not have predicted artificial intelligence nor the existence of completely mechanised factories. So necessarily, an amendment of his conclusions on labour and its relationship with the value of money must be made. We cannot simply remove the cost of labour, yet we must adjust the relationship it has with the value of money. It still holds true for say moving production to the far east, where labour is less costly. However, in a world with zero need for labour, the Capitalist owner has a big problem, since in such a scenario, nobody is needed and therefor nobody is paid for production of anything, then it follows that most people would then have no money to buy his/ her products. This then results in zero profit for the capitalist. We can get round this problem by doling out money to the jobless for a while. We may also ‘help’ the Capitalist by gifting him with large lumps of money.
11. None of the above avoid the inevitable problem: The current form of global-capitalism is totally dysfunctional and inadequate.
## Footnotes
1. Well, actually, by definition, money is a commodity. That’s why it was developed in the first place, to combat the inability to trade unlike skill sets or goods in barter. Just like any other comodity, the price or value of any money may go up or down from time to time. ↩︎ | 2020-09-20 07:24:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.4086603820323944, "perplexity": 1235.971058776981}, "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/1600400196999.30/warc/CC-MAIN-20200920062737-20200920092737-00001.warc.gz"} |
http://mechanicaldesign.asmedigitalcollection.asme.org/article.aspx?articleid=1448787 | 0
RESEARCH PAPERS
# Teeth-Number Synthesis of a Multispeed Planetary Transmission Using an Estimation of Distribution Algorithm
[+] Author and Article Information
P. A. Simionescu
Department of Mechanical Engineering, The University of Tulsa, 600 S. College, Tulsa, OK 74104
D. Beale
Department of Mechanical Engineering, Auburn University, 202 Ross Hall, Auburn, AL 36849
G. V. Dozier
Department of Computer Science, Auburn University, 109 Dunstan Hall, Auburn, AL 36849
J. Mech. Des 128(1), 108-115 (Apr 01, 2005) (8 pages) doi:10.1115/1.2114867 History: Received August 24, 2004; Revised April 01, 2005
## Abstract
The gear-teeth number synthesis of an automatic planetary transmission used in automobiles is formulated as a constrained optimization problem that is solved with the aid of an Estimation of Distribution Algorithm. The design parameters are the teeth number of each gear, the number of multiple planets and gear module, while the objective function is defined as the departure between the imposed and the actual transmission ratios, constrained by teeth-undercut avoidance, limiting the maximum overall diameter of the transmission and ensuring proper spacing of multiple planets. For the actual case of a $3+1$ speed Ravigneaux planetary transmission, the design space of the problem is explored using a newly introduced hyperfunction visualization technique, and the effect of various constraints highlighted. Global optimum results are also presented.
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## Figures
Figure 1
Ravigneaux planetary gear (36): 1 small sun gear; 2-3 broad planet gear; 4 large sun gear; 5 narrow planet gear; 6 ring gear
Figure 2
Kinematic diagram of a 3+1 transmission ratios Ravigneaux planetary transmission. Note that the broad planet 2-3 consists now of two distinct gears
Figure 3
Schematic for calculating distances d22, d34, d35, and d35. Notice that one of the idler planets 5 has been removed for clarity.
Figure 4
Projection of the lower envelope of objective function f1=Err max with N2≠N3 on the (m1,m3,f1) space (a), and plot of the corresponding outer diameter of the transmission (b) for the case of equally spaced identical planets
Figure 5
Projection of the lower envelope of objective function f1=Err max with N2≠N3 on the (m1,m3,f1) space (a), and plot of the corresponding outer diameter of the transmission (b) for the case of equally spaced nonidentical compound planets 2-3
Figure 6
Projection of the lower envelope of objective function f1=Err max with m1=m3 and N2=N3 on the (m1,f1) plane (a), and plot of the corresponding outer diameter of the transmission (b) for the case of equally spaced identical planets
Figure 7
Projection of the lower envelope of objective function f1=Err max with m1=m3 and N2=N3 on the (m1,f1) plane (a), and plot of the corresponding outer diameter of the transmission (b) for the case of equally spaced nonidentical planets 2-3
Figure 8
Front view of the transmission with N1=34, N2=25, N3=33, N4=32, N5=23, N6=116, p=4 and m1=2.5 and m3=1.75mm (variant number 3 in Table 2)
Figure 9
Front view of the transmission with N1=35, N2=N3=28, N4=25, N5=28, N6=91, m1=m3=2mm (variant number 7 in Table 2)
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Topic Collections | 2017-08-19 20:33:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "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": 0.29024237394332886, "perplexity": 4724.5288796949}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886105922.73/warc/CC-MAIN-20170819201404-20170819221404-00339.warc.gz"} |
https://docs.microsoft.com/en-us/dotnet/csharp/programming-guide/xmldoc/delimiters-for-documentation-tags | # Delimiters for Documentation Tags (C# Programming Guide)
The use of XML doc comments requires delimiters, which indicate to the compiler where a documentation comment begins and ends. You can use the following kinds of delimiters with the XML documentation tags:
///
Single-line delimiter. This is the form that is shown in documentation examples and used by the Visual C# project templates. If there is a white space character following the delimiter, that character is not included in the XML output.
Note
The Visual Studio IDE has a feature called Smart Comment Editing that automatically inserts the <summary> and </summary> tags and moves your cursor within these tags after you type the /// delimiter in the Code Editor. You can turn this feature on or off in the Options dialog box.
/** */
Multiline delimiters.
There are some formatting rules to follow when you use the /** */ delimiters.
• On the line that contains the /** delimiter, if the remainder of the line is white space, the line is not processed for comments. If the first character after the /** delimiter is white space, that white space character is ignored and the rest of the line is processed. Otherwise, the entire text of the line after the /** delimiter is processed as part of the comment.
• On the line that contains the */ delimiter, if there is only white space up to the */ delimiter, that line is ignored. Otherwise, the text on the line up to the */ delimiter is processed as part of the comment, subject to the pattern-matching rules described in the following bullet.
• For the lines after the one that begins with the /** delimiter, the compiler looks for a common pattern at the beginning of each line. The pattern can consist of optional white space and an asterisk (*), followed by more optional white space. If the compiler finds a common pattern at the beginning of each line that does not begin with the /** delimiter or the */ delimiter, it ignores that pattern for each line.
The following examples illustrate these rules.
• The only part of the following comment that will be processed is the line that begins with <summary>. The three tag formats produce the same comments.
/** <summary>text</summary> */
/**
<summary>text</summary>
*/
/**
* <summary>text</summary>
*/
• The compiler identifies a common pattern of " * " at the beginning of the second and third lines. The pattern is not included in the output.
/**
* <summary>
* text </summary>*/
• The compiler finds no common pattern in the following comment because the second character on the third line is not an asterisk. Therefore, all text on the second and third lines is processed as part of the comment.
/**
* <summary>
text </summary>
*/
• The compiler finds no pattern in the following comment for two reasons. First, the number of spaces before the asterisk is not consistent. Second, the fifth line begins with a tab, which does not match spaces. Therefore, all text from lines two through five is processed as part of the comment.
/**
* <summary>
* text
* text2
* </summary>
*/ | 2019-05-21 05:12:40 | {"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": 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": 0.42372047901153564, "perplexity": 1459.0341977797748}, "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-2019-22/segments/1558232256227.38/warc/CC-MAIN-20190521042221-20190521064221-00437.warc.gz"} |
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# If a half-full 4-inch by 2-inch by 8-inch box of soymilk is poured int
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19 Sep 2018, 20:49
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If a half-full 4-inch by 2-inch by 8-inch box of soymilk is poured into a right circular cylindrical glass with radius 2 inches, how many inches high will the soymilk reach? (Assume that the capacity of the glass is greater than the volume of the soymilk.)
(A) $$8$$
(B) $$16$$
(C) $$\frac{4}{\pi}$$
(D) $$\frac{8}{\pi}$$
(E) $$\frac{16}{\pi}$$
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Re: If a half-full 4-inch by 2-inch by 8-inch box of soymilk is poured int [#permalink]
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19 Sep 2018, 23:29
Volume of soymilk = 4*2*4 = 32cm^3
π(2)^2h = 32
h=8/π
My guess
Ans : D
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Re: If a half-full 4-inch by 2-inch by 8-inch box of soymilk is poured int [#permalink]
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20 Sep 2018, 03:00
Bunuel wrote:
If a half-full 4-inch by 2-inch by 8-inch box of soymilk is poured into a right circular cylindrical glass with radius 2 inches, how many inches high will the soymilk reach? (Assume that the capacity of the glass is greater than the volume of the soymilk.)
(A) $$8$$
(B) $$16$$
(C) $$\frac{4}{\pi}$$
(D) $$\frac{8}{\pi}$$
(E) $$\frac{16}{\pi}$$
Volume of Milk $$= (1/2)*(4*2*8) = 32$$ cubic inches
Volume of Cylinder $$= πr^2h = π*2^2*h$$
Now, $$= πr^2h = π*2^2*h = 32$$
i.e. $$h = 8/π$$
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Re: If a half-full 4-inch by 2-inch by 8-inch box of soymilk is poured int [#permalink] 25 May 2020, 03:34 | 2020-07-11 15:13:22 | {"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": 0.7059108018875122, "perplexity": 11219.739453769766}, "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-2020-29/segments/1593655933254.67/warc/CC-MAIN-20200711130351-20200711160351-00315.warc.gz"} |
https://plainmath.net/9324/which-pair-of-numbers-is-relatively-prime-15-and-25-29-and-58-and-261-and-513 | Question
# Which pair of numbers is relatively prime? a. 15 and 25 b. 29 and 58 c. 40 and 63 d. 261 and 513
Fractions
Which pair of numbers is relatively prime?
a. 15 and 25
b. 29 and 58
c. 40 and 63
d. 261 and 513
2021-02-22
Relatively Prime: Two integers are said to be relatively prime if there is no integer greater than one that divides them both.
Factorization of 15 is $$15=3 \times 5$$ and the factorization of 25 is $$25=5 \times 5$$
Since 5 is the common factor in factorization of both the numbers, therefore they are not prime numbers.
Answer (b): The given numbers are 29 and 58.
Factorization of 29 is $$29=1 \times 29$$ and the factorization of 58 is $$58=2 \times 29$$
Since 29 is the common factor in factorization of both the numbers, therefore they are not prime numbers.
Factorization of 40 is $$40=1 \times 2 \times 2 \times 2 \times 5$$
and the factorization of 63 is $$63=1 \times 3 \times 3 \times 7$$
Factorization of 261261 is $$261=3 \times 3 \times 29$$ and the factorization of 513 is $$513=3 \times 3 \times 3 \times 19$$ | 2021-07-31 01:52:34 | {"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": 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": 0.505219042301178, "perplexity": 540.7433569127288}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154042.23/warc/CC-MAIN-20210731011529-20210731041529-00280.warc.gz"} |
https://twistedone151.wordpress.com/2008/08/18/monday-math-33/ | Monday Math 33
We know that the harmonic series diverges, but what about the alternating harmonic series ? The key is the Taylor series for the natural logarithm, known as the Mercator series:
, which is valid for -1<x≤1. Setting x=1 tells us that the alternating harmonic series converges to .
Now, recall the Riemann Zeta Function, which for is given by
.
Suppose we define an analogous function with alternating terms:
.
This series does not have a pole at s=1, and in fact, can be defined via analytic continuation to be defined over the entire complex plane. This function is called the Dirichlet eta function.
Now, let us consider the difference of the Riemann zeta and Dirichlet eta functions:
We see that the odd n terms cancel, leaving only the even terms:
,
and solving for eta,
which allows us to find exact values for the Dirichlet eta function at positive even integers (and find values for positive odd integers in terms of the zeta function of those integers).
The values for the first few integers are:
5 Responses to “Monday Math 33”
1. johnsmithinfinith Says:
Hi there. I just started blogging, myself and I’d wonder whether you could give me a couple of tips for starters;).
Why I’m asking you? Well, your blog has a lot of interesting math stuff and I liked it from the beginning. So, I’m kinda interested about blogging from your point of view, because my idea was to, also, blog about some mathematical topics.
Maybe you could take a brief look at my blog. Thanks!
Bye.
http://johnsmithinfinith.wordpress.com/
2. Monday Math 35 « Twisted One 151’s Weblog Says:
[…] , which gives us: . Plugging into our series, . We should recognize that last sum as the Dirichlet eta function. Thus, just as , we see […]
3. Monday Math 74 « Twisted One 151’s Weblog Says:
[…] The Maclaurin series for is given by the Mercator series: for -1<x≤1. Thus where is the Dirichlet eta function. We also found here that , so […]
4. Monday Math 133 « Twisted One 151's Weblog Says:
[…] Riemann zeta function has a pole at s=1, while the term in the brackets is zero at s=1. Recall from here that the Dirichlet eta function η(s), which is defined for all complex numbers s, is related to […]
5. cody Says:
Hi TwistedOne:
Do you happen to know what
SUM([(-1)^(n-1)*zeta(2n)]/n), n=1..infinity
converges to?.
I have tried finding its sum, but to no avail. | 2017-03-28 08:16:24 | {"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": 18, "/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": 0.945735514163971, "perplexity": 643.8024127502092}, "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-2017-13/segments/1490218189686.31/warc/CC-MAIN-20170322212949-00251-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/159396/subfigure-counters-not-resetting | subfigure counters not resetting
My subfigure counter isn't working properly for 2 figures. In the first figure there is an a) and then b) -which is fine, (that I both mark with a begin and end command) but for the second figure, that I also mark with a begin and end command, the subfigure counter doesn't reset back to a), b), c) and d); but it rather jumps to c), d) , e) and f).
What is a possible workaround for this problem? I noticed that I used a tabular inside the second figure to insert some figure's, and maybe that messed up the subfigure counter. Any other ideas? Here is the code:
*Preamble:
\documentclass[runningheads]{llncs}
\usepackage{graphicx}
\usepackage{subfigure}
\usepackage{amsmath,amssymb} % define this before the line numbering.
\usepackage{ruler}
\usepackage{color}
\usepackage[width=122mm,left=12mm,paperwidth=146mm,height=193mm,top=12mm,paperheight=217mm]{geometry}
\usepackage{verbatim}
\usepackage{color}
%\usepackage{epsfigure}
\usepackage{multirow}
\usepackage[font={small}]{caption}
\usepackage{xspace}
\newcommand*{\eg}{e.g.\@\xspace}
\newcommand*{\ie}{i.e.\@\xspace}
\newcommand{\etal}{\mbox{\emph{et al.\ }}}
\newenvironment{packed_enumerate}{
\begin{enumerate}
\setlength{\itemsep}{0pt}
\setlength{\parskip}{0pt}
\setlength{\parsep}{0pt}
\vspace{-4pt}
}{\end{enumerate}}
\begin{figure}[!t]
\centering
\subfigure[caption1]{
\includegraphics[scale=0.15,clip=true,draft=false,]{title1.pdf}
%\vspace{-10pt}
\label{fig:relative_graph}
%\vspace{-10pt}
}
\subfigure[Caption2]{
\begin{tabular}{cc|c|c|}
\centering
\cline{2-3}
& \multicolumn{1}{ |c| }{column1} & \multicolumn{1}{ |c| }{column2} \\ \cline{1-3}
\multicolumn{1}{ |c| }{AA} & $65.16\%$ & $66.64\%$ \\ \cline{1-3}
\multicolumn{1}{ |c| }{BB} & $68.60\%$ & $65.56\%$ \\ \cline{1-3}
\multicolumn{1}{ |c| }{CC} & $\mathbf{52.24\%}$ & $65.00\%$ \\ \cline{1-3}
\end{tabular}
\label{table:mambo}
\vspace{-30pt}
}
\vspace{-10pt}
\end{figure}
\begin{figure}[!t]
\setlength{\abovecaptionskip}{-5pt}
\setlength{\belowcaptionskip}{-5pt}
\addtolength{\subfigcapskip}{3pt}
\centering
\begin{tabular}{ccc}
\subfigure[caption1]{\includegraphics[scale=0.17]{img1.jpg}
\label{fig:img_pair_1}} &%&
%\subfigure[B]{\includegraphics{logo.png}} &
\multirow{-4}[2.5]{*}{\subfigure[caption2]{\includegraphics[scale=0.136]{img2.jpg} \label{fig:img_complete_1}}} &
\multirow{-4}[2.5]{*}{\subfigure[caption3]{\includegraphics[scale=0.136]{img3.jpg}\label{fig:img_complete_2}}} \\
%\subfigure[E]{\includegraphics{logo.png}}% &
\subfigure[caption4]{\includegraphics[scale=0.17]{img4.jpg}
\label{fig:img_pair_2}}\\
\end{tabular}
\vspace{10pt}
\caption[]{
Comments
}
\label{fig:affinity}
\vspace{-10pt}
\end{figure}
• Which package do you use? subcaption or subfig? Kindly add full code including the preamble. – user11232 Feb 9 '14 at 0:18
• I am using subfigure, the weird thing is that for other figures, this "repeating" doesn't happen and it resets itself automatically. – Arturo Feb 9 '14 at 0:19
• subfigure is deprecated. Use subcaption or subfig. – user11232 Feb 9 '14 at 0:21
1 Answer
This is one of the problems of the obsolete package subfigure. Switch to subfig or subcaption. Here's part of your code rewritten using subfig's syntax (I made some changes; in particular, I suppressed the \centering just before \cline which was producing an error):
\documentclass{article}
\usepackage[demo]{graphicx}
\usepackage{subfig}
\usepackage{multirow}
\begin{document}
\begin{figure}[!t]
\centering
\subfloat[caption1\label{fig:relative_graph}]{%
\includegraphics[scale=0.15,clip=true,draft=false,]{title1.pdf}%
}\quad
\subfloat[Caption2\label{table:mambo}]{%
\begin{tabular}[b]{cc|c|c|}
\cline{2-3}
& \multicolumn{1}{ |c| }{column1} & \multicolumn{1}{ |c| }{column2} \\ \cline{1-3}
\multicolumn{1}{ |c| }{AA} & $65.16\%$ & $66.64\%$ \\ \cline{1-3}
\multicolumn{1}{ |c| }{BB} & $68.60\%$ & $65.56\%$ \\ \cline{1-3}
\multicolumn{1}{ |c| }{CC} & $\mathbf{52.24\%}$ & $65.00\%$ \\ \cline{1-3}
\end{tabular}%
}
\end{figure}
\begin{figure}[!t]
\centering
\begin{tabular}{ccc}
\subfloat[caption1\label{fig:img_pair_1}]{%
\includegraphics[scale=0.17]{img1.jpg}%
} &
\multirow{-4}[2.5]{*}{\subfloat[caption2\label{fig:img_complete_1}]{\includegraphics[scale=0.136]{img2.jpg}}} &
\multirow{-4}[2.5]{*}{\subfloat[caption3\label{fig:img_complete_2}]{\includegraphics[scale=0.136]{img3.jpg}}} \\
\subfloat[caption4\label{fig:img_pair_2}]{\includegraphics[scale=0.17]{img4.jpg}}\\
\end{tabular}
\caption[]{Comments}
\label{fig:affinity}
\end{figure}
\end{document}
To customize caption settings, use the caption package.
Notice that I used [b] as the alignment option for the tabular so the image and the table are properly aligned. I would also suggest you not to control the size of your images using scale; usually a better control is achieved controlling the width and/or height.
Here's a variation of your code, with some improvements. booktabs was used to produce the table and \raisebox was used instead of \multirow to place the images.
\documentclass{article}
\usepackage[demo]{graphicx}
\usepackage{subfig}
\usepackage{booktabs}
\begin{document}
\begin{figure}
\centering
\subfloat[caption1\label{fig:relative_graph}]{%
\includegraphics[width=4cm]{title1.pdf}%
}\quad
\subfloat[Caption2\label{table:mambo}]{%
\begin{tabular}[b]{cccc}
\cmidrule[\heavyrulewidth]{2-3}
& column1 & column2 \\ \midrule
AA & $65.16\,\%$ & $66.64\,\%$ \\
BB & $68.60\,\%$ & $65.56\,\%$ \\
CC & $\mathbf{52.24\,\%}$ & $65.00\,\%$ \\ \bottomrule
\end{tabular}%
}
\end{figure}
\begin{figure}
\centering
\begin{tabular}{@{}ccc@{}}
\subfloat[caption1\label{fig:img_pair_1}]{%
\includegraphics[width=3cm]{img1.jpg}%
} &
\raisebox{-70pt}[0pt][0pt]{\subfloat[caption2\label{fig:img_complete_1}]{\includegraphics[width=3cm]{img2.jpg}}} &
\raisebox{-70pt}[0pt][0pt]{\subfloat[caption3\label{fig:img_complete_2}]{\includegraphics[width=3cm]{img3.jpg}}} \\
\subfloat[caption4\label{fig:img_pair_2}]{\includegraphics[width=3cm]{img4.jpg}} & & \\
\end{tabular}
\caption[]{Comments}
\label{fig:affinity}
\end{figure}
\end{document}
The demo option for graphicx simply replaces actual figures with black rectangles; do not use that option in your actual document.
• @Arturo You're welcome! I've updated my answer with some modifications (just suggestions, of course) that might be of interest for you. – Gonzalo Medina Feb 9 '14 at 0:51 | 2019-07-18 09:40:44 | {"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": 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": 0.9997174143791199, "perplexity": 2739.1490166523813}, "config": {"markdown_headings": false, "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-2019-30/segments/1563195525587.2/warc/CC-MAIN-20190718083839-20190718105839-00336.warc.gz"} |
https://www.albert.io/ie/ap-physics-c-mechanics/work-with-vectors | Free Version
Easy
# Work with Vectors
APPHMC-4DNT1Z
A particle experiences a constant $<+4 N, -4 N>$ force as it is displaced from $(0,0)m$ to $(2,0)m$.
How much work was done on the particle?
A
$0 J$
B
$8 J$
C
$16 J$
D
$32 J$
E
None of the above | 2017-01-21 19:48:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.7826088666915894, "perplexity": 646.5899599873428}, "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/1484560281202.94/warc/CC-MAIN-20170116095121-00146-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://docs.generic-mapping-tools.org/6.4/gallery/ex13.html | (13) Plotting of vector fields¶
In many areas, such as fluid dynamics and elasticity, it is desirable to plot vector fields of various kinds. GMT provides a way to illustrate 2-component vector fields using the grdvector utility. The two components of the field (Cartesian or polar components) are stored in separate grid files. In this example we use grdmath to generate a surface $$z(x, y) = x \cdot \exp(-x^2 -y^2)$$ and to calculate $$\nabla z$$ by returning the x- and y-derivatives separately. We superpose the gradient vector field and the surface z and also plot the components of the gradient in separate windows. A text call to place a header finishes the plot
#!/usr/bin/env bash
# GMT EXAMPLE 13
#
# Purpose: Illustrate vectors and contouring
# GMT modules: set, grdmath, grdcontour, grdvector, subplot
# Unix progs: rm
#
gmt begin ex13 | 2023-02-08 07:06:37 | {"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": 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": 0.47186243534088135, "perplexity": 2879.168205719474}, "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-2023-06/segments/1674764500719.31/warc/CC-MAIN-20230208060523-20230208090523-00750.warc.gz"} |
https://core.ac.uk/display/25305681 | Effective nucleon mass in relativistic mean field theory
Abstract
In the $\sigma$-$\omega$-$\rho$ model of the relativistic mean field theory with nonlinear $\sigma$-meson self-interaction, the effective nucleon mass $M^*$ is discussed with relation to the symmetry incompressibility $K_s$ of nuclear matter, based on the model parameters fitted to nuclear matter properties. It is shown that $M^*$ is larger than $0.73M$, if $K_s$ is assumed to be negative and the nuclear matter incompressibility $K_0$ is kept less than $300 MeV$. Furthermore, the field system is shown to be stable, as the $\sigma$-meson self-interaction energy is lower bounded in this parameter region
Topics: Nuclear Physics
Year: 2001
OAI identifier: oai:cds.cern.ch:485706
Provided by: CERN Document Server
To submit an update or takedown request for this paper, please submit an Update/Correction/Removal Request. | 2021-07-27 04:27:49 | {"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": 0.7100816965103149, "perplexity": 1342.8774181933295}, "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/1627046152236.64/warc/CC-MAIN-20210727041254-20210727071254-00083.warc.gz"} |
http://mathhelpforum.com/advanced-math-topics/33994-uniform-approximation.html | ## uniform approximation
Function $f(t)$ specified on $[t_0;t_1]$ has a necessary number of derivatives. Find algorithm which can build uniform approximations of this function with help of partial sums: $\sum_{i=1}^{N}\alpha_i e^{-\beta_i t}$
That is, find such $\alpha_i$, $Re(\beta_i) \geq 0$ satisfying expression:
$\min_{\alpha_i, ~\beta_i}\left( \max_{t\in [t_0;t_1]} \left| f(t)-\sum_{i=1}^{N}\alpha_i e^{-\beta_i t}\right| \right)$
Also consider a function specified on $[t_0;+\infty)$
Thanks for any ideas! | 2016-06-30 23:10:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "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": 0.9423578381538391, "perplexity": 885.6702720116813}, "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/1466783399425.79/warc/CC-MAIN-20160624154959-00066-ip-10-164-35-72.ec2.internal.warc.gz"} |
https://puzzling.stackexchange.com/questions/84531/restoring-order-in-a-deck-of-playing-cards-ii | # Restoring order in a deck of playing cards (II)
Michelle has a deck of 52 playing cards in a pile with their backs facing up. She takes the first 2 cards in the pile, turns them over, and places them at the bottom of the pile. She now takes the next 3 cards in the pile and, once again, turns them over, and places them at the bottom of the pile. She proceeds like this, taking each time the next prime number of cards from the top, turning them over, and placing them at the bottom of the deck. Once she has done this for all primes up to 47 (the largest prime less than 52), she continues counting in the same fashion with the next primes, 53, 59, 61,..., but taking them module 52 (the residue after dividing by 52), that is 1, 7, 9, etc.
Will the deck of cards ever have all their backs facing up again?
• I take it the only real point of the first question was to prepare the way for this one... – Gareth McCaughan May 30 '19 at 14:03
• @GarethMcCaughan Or this is the obvious foil to the answer of the first. You can no longer find a pattern with which to make a set. – Trenin May 30 '19 at 14:11
• @BernardoRecamánSantos can you please clarify the last sentence of the question, which is different from Part I? Here is says "all the cards face up again" but the cards have never been all face up. – Weather Vane May 30 '19 at 16:20
• @WeatherVane You are right. I meant, as in original puzzle, "all their backs facing up again". I will edit accordingly. Thanks and sorry. – Bernardo Recamán Santos May 30 '19 at 16:43
Basically, what we are asking here is if a sum of primes (modulo 52) will ever result in a sum that is a multiple of 104. Intuitively I believe this should occur, but I actually ran a simple script to see if this is the case and:
I found that this first occurs when adding the 49th prime number, 227, your resulting sum is 1144, equal to 104*11, so a total of 11 cycles have occurred and you have returned to a deck with all cards facing up. This later occurs when adding 449 (sum of 2080), 653 (sum of 3016), 739 (sum of 3328), and 1063 (sum of 4680).
• I think this is answering a slightly different question than that which is asked. The procedure is "take the pile, flip them over, place at the bottom". This means that the order of the pile is reversed. Contrast the two procedures in Weather Vane's accepted solution for the previous answer: puzzling.stackexchange.com/a/84433/18422 – hexomino May 30 '19 at 15:31
• But the question doesn't ask whether the pile will ever again have the cards in the original order and face up. It just asks whether they'll ever all be face up again. – Gareth McCaughan May 30 '19 at 15:48
• Oh, no, wait, I misunderstood @hexomino's point, which is that because you're reversing the order when you do this you aren't always just flipping (so to speak) a sliding window over the cycle of 52 cards. – Gareth McCaughan May 30 '19 at 15:49
• I see, I interpreted taking 2 cards and flipping them over as turning them over individually. This definitely gets a lot more complicated when the whole stack is turned over. I'll look into trying this out as well. – Michael Moschella May 30 '19 at 16:20
• @MichaelMoschella I made that mistake in Part I at first, when the deck was quickly restored. But the sequence of the cards is reversed when moving them to the bottom of the deck. – Weather Vane May 30 '19 at 16:29
So far:
Using the same techniques as the Part I puzzle, but generating the primes.
I have run a code with $$50847534$$ operations, using primes $$\lt 10^9$$.
The deck was not restored.
Investigations proceed. . .
Update:
I have run a code with $$189961812$$ operations, using primes up to $$4 \times 10^9$$.
The deck was not restored.
Edit:
I have run a revised code with $$1300005926$$ operations, using primes up to $$3 \times 10^{10}$$.
The deck was not restored.
This is the limit (for now) of what I can do with the Visual C compiler which gives a program less than 2 Gb memory to use. I got this far by implementing the sieve of Eratosthenes with bit fields. The running time was 17 minutes.
So it appears that a mathematical or deductive proof is needed.
• I've checked up to the 1649267441664th prime, a little over 1000 times more primes. Still no luck. – isaacg Sep 9 '19 at 0:26
There seems to be a typo in the post, or an intentional fakeout. The setup says the cards are
in a pile with their backs facing up
meaning the deck is facing down, but the question asks whether it
will ever have all the cards face up again?
The deck has not previously been facing up, so I'll err on the side of caution and take it that the question asks whether the pile will face up more than once when shuffled in the described manner and starting face down.
EDIT: The following up to and including the C++ code has been proven wrong.
Since we are drawing cards in order, whenever the sum of cards flipped is a multiple of 52, the entire deck will be either facing up or down. The deck starts face down, so it will be face down again only when we've flipped a multiple of 104 cards. Ergo it faces up when the number of cards flipped is divisble by 52 but not 104.
I wrote a short, somewhat optimized C++ snippet to determine when the desired states are reached. Here are the highlights of the output:
All cards are facing down after 49 primes! (last prime: 227)
All cards are facing down after 87 primes! (last prime: 449)
All cards are facing up after 117 primes! (last prime: 643)
All cards are facing down after 119 primes! (last prime: 653)
All cards are facing down after 131 primes! (last prime: 739)
All cards are facing down after 179 primes! (last prime: 1063)
All cards are facing down after 203 primes! (last prime: 1237)
All cards are facing up after 319 primes! (last prime: 2113)
And here the code:
#include <iostream>
// Returns whether the given number n > 3 is a prime.
bool isPrime(long n) {
if(n%2 == 0 || n%3 == 0) return false;
for(int i = 6; (i-2)*i + 1 <= n; i += 6) {
if(n % (i-1) == 0) return false;
if(n % (i+1) == 0) return false;
}
return true;
}
// Returns the closest prime larger than the given number.
int nextPrime(int n) {
while(!isPrime(++n)) { continue; }
return n;
}
int main() {
const short DECK_SIZE = 52;
short faceUp = 5;
unsigned int steps = 2;
long prime = 3;
unsigned short timesDeckWasFaceUp = 0;
while(timesDeckWasFaceUp != 2) {
do { // Run until all cards are facing up again.
prime = nextPrime(prime);
++steps;
// Number of cards to flip this turn.
short flips = prime % DECK_SIZE;
std::cout << "Face up: " << (faceUp > DECK_SIZE ? (DECK_SIZE << 1) - faceUp : faceUp) << "\nPrime: " << prime << "\nFlipping: " << flips << " cards.\n\n";
faceUp = (faceUp + flips) % (DECK_SIZE << 1);
if(faceUp % DECK_SIZE == 0) { // Entire deck faces in one direction
if(faceUp) {
std::cout << "All cards are facing up after " << steps << " primes! (last prime: " << prime << ")\n";
++timesDeckWasFaceUp;
break;
} else {
std::cout << "All cards are facing down after " << steps << " primes! (last prime: " << prime << ")\n";
}
}
} while(faceUp % (DECK_SIZE << 1) != 0);
}
return 0;
}
EDIT: Here the revised code that accounts for the fact that the order of the flipped cards gets reversed.
#include <iostream>
#include <deque>
// Checks if the given number > 3 is a prime.
bool isPrime(unsigned long long n) {
if(n%2 == 0 || n%3 == 0) return false;
for(unsigned long long i = 6; (i-2)*i + 1 <= n; i += 6) {
if(n % (i-1) == 0) return false;
if(n % (i+1) == 0) return false;
}
return true;
}
// Returns the nearest prime larger than the given number.
unsigned long long nextPrime(unsigned long long n) {
while(!isPrime(++n)) { continue; }
return n;
}
void flipCards(int flips, std::deque<bool>& cards, unsigned short& counter) {
// Flip the order of the top cards
std::reverse(cards.begin(), cards.begin() + flips);
// Turn each card around
std::deque<bool> flipped(cards.begin(), cards.begin() + flips);
for(bool& card : flipped) {
counter += (card) ? -1 : 1;
card = !card;
}
// Remove the cards from the top
std::deque<bool>(cards.begin()+flips, cards.end()).swap(cards);
// Add them to the bottom
cards.insert(cards.end(), flipped.begin(), flipped.end());
}
int main() {
const short DECK_SIZE = 52;
// Keep track of how many cards are facing down.
unsigned short faceUp = 0;
std::deque<bool> cards(52);
// Flip the first two primes (2 + 3).
flipCards(5, cards, faceUp);
// Keep track of the number of primes flipped.
unsigned int steps = 2;
unsigned long long prime = 3;
unsigned short timesFaceUp = 0;
while(timesFaceUp != 2) {
// Run until no more cards are facing down.
while(true) {
prime = nextPrime(prime);
++steps;
// Number of cards to flip this turn.
unsigned short flips = prime % 52;
flipCards(flips, cards, faceUp);
if(faceUp % 52 == 0) {
if(faceUp) {
std::cout << "All cards are facing up after " << steps << " primes! (last prime: " << prime << ")\n";
++timesFaceUp;
break;
} else {
std::cout << "All cards are facing down after " << steps << " primes! (last prime: " << prime << ")\n";
}
}
}
}
return 0;
}
I'm afraid Weather Vane's answer may be correct.
• Good spot, I have been working on the same conditions as Part I puzzle. Here it says "ever have all the cards face up again" but they were never all face up... – Weather Vane May 30 '19 at 16:17
• I think this answer has the same problem that @hexomino pointed out with Michael Moschella's. We start with all cards face down. After 2+3+5+7+11+13 flips we have transferred 41 to the bottom, so we have 11 face down and then 41 face up. Now we pick up 17 cards, 11 face down and 6 face up. We flip them, getting 6 face down on top of 11 face up. Then we put them on the bottom ... and now we have 35 face up, then 6 face down, then 11 face up. Now we pick up 19, flip them over, and put them on the bottom: 16U, 6D, 11U, 19D. It isn't just a matter of counting flipped cards: the order matters. – Gareth McCaughan May 30 '19 at 16:33
I'll go with
Yes, because given infinite time, every possible deck state will repeat indefinitely many times.
That's because
1. There are infinitely many primes
2. There are finitely many deck states, and
3. There are no patterns in the primes (mod 52) that would exclude one state or another.
The first two points are pretty easy to prove:
1: If there were finitely many primes, one of them would be the largest prime, and we could call it the Big-P. Now, multiplying all the primes together and adding one, we would get a new number Bigger-X. It is larger than Big-P, and not divisible by any prime smaller than (or equal to) Big-P. That means that either Bigger-X is prime, or Bigger-X is divisible by some prime(s) bigger than Big-P. This is a contradiction, which means that Big-P cannot exist, and therefore "there's always a bigger prime".
2: Since we don't care about the actual order of the cards, only whether they are face-up or not, the deck has $$2^{52}$$ possible states, which is a big number (abt. 4.5 quadrillion (US reckoning)), and very definitely finite.
The third proposition is a lot harder to prove, but here's my go:
All but the first two primes are of the form $$6n\pm1$$. Because 52 is not divisible by 6, taking modulo 52 of such numbers makes every odd number between 1 and 51 possible, and equally likely over a long period of time, even if there were a hitherto unknown secret pattern in the huge primes.
Since the number of flipped cards is always odd, the number of face-up cards in the deck switches between odd and even every time, making each kind equally likely.
Therefore no particular deck state is preferred over another, and each state will occur equally likely in a given (even-length) sequence of states.
This is enough to show the result, because
The probability of any given state occurring in some sequence of two consecutive states is non-zero ($$2^{-51}$$ actually, because of the abovementioned parity issue), so given infinitely many of these sequences, every state will occur infinitely many times.
Yes, though it would probably take a bit.
Once we really start getting into the big primes we know that # mod 52 is going to be effectively random... but always odd. The idea that random card-moves will get you back to all cards face-down isn't a crazy notion... there are some really big numbers under infinity. But the restriction of using only odd-moves presents a potential problem. We did, after all, make an even-card move initially since 2 is prime.
It ends up being not that important. I played around with it and I can show that any state of cards being up/down can become any other state of cards being up/down. It's not even that hard, honestly. Here's an example of 8 cards, where the first move was 2 cards and the remaining moves were odd.
That example is really trivial but with the right goals in mind you can always improve your standing in a move or two. I would therefore say any state of 52 cards could become face-down within 104 moves. Of course, the actual output from our prime sequence isn't going to agree with what we wanted... usually. The idea is that at any given point, you want a particular sequence to get back to all face-down and it will take a while but somewhere down the number-line you will eventually get it. Solvable by computer? I suspect not. 2^52 possible states is a bit much, and I see no reason one would be more likely than another.
• I don't get your flip sequence. For an 8-card deck, shouldn't it be 2 (2 % 8), 3 (3 % 8), 5 (5 % 8), 7 (7 % 8), 3 (11 % 8), 5 (13 % 8), 1 (17 % 8), 3 (19 % 8), 7 (23 % 8) ...? The situation isn't improved after a move or two, because the face/back sequence becomes fragmented. – Weather Vane May 30 '19 at 18:04
• I wasn't saying I can do it for a particular known prime sequence. I was imagining a prime sequence on the distant number-line. I just need to show that effectively random moves could conceivably solve a given state of cards. – Dark Thunder May 30 '19 at 18:09
• Oh I see - effectively random number of cards flipped as the game proceeds. – Weather Vane May 30 '19 at 18:10
• @DarkThunder Prime number isn't random. You just heuristically suggest that the answer is probably yes. – Akangka Jul 17 '19 at 2:40
• @Akangka I don't claim to be a mathematician but my understanding is that "sufficiently large prime numbers mod 52" would be effectively random. If you can prove otherwise I'd very much like to see it. – Dark Thunder Jul 17 '19 at 12:05
I am going to be the first to say that the answer is no, the deck will never again be in order.
I have no real proof other than the following intuition. The frequency of primes is generally known, but only for very large numbers, and it is only a guide, not exact. Thus, the difference modulo 52 will be mostly random. As a result, we are randomly flipping subsets of cards, so unless we manage to be very fortuitous, we will never get back to in-order.
However, I could be dead wrong - we are dealing with an infinite number of flips.
• Your deck has a finite number of states. We want to know when one specific state happens. Unless there's a good reason otherwise, a random walk will eventually find it. – Dark Thunder May 30 '19 at 14:24
• @DarkThunder Good point. The in-order configuration is just as likely as any other configuration. – Trenin May 30 '19 at 15:07 | 2020-01-27 09:54:11 | {"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": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.523379385471344, "perplexity": 1349.1221546591821}, "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/1579251696046.73/warc/CC-MAIN-20200127081933-20200127111933-00496.warc.gz"} |
http://wiki.luatex.org/index.php/Typeset_non-TeX_files_by_converting_them_using_Lua_code | # Typeset non-TeX files by converting them using Lua code
The following code example demonstrates how to use the file reading callback and coroutines to convert a non-TeX file into TeX code which is then typeset.
First the Lua code ("Preprocessor.lua"):
``` local function readFile(fn)
local file = assert(io.open(fn, "r"))
file:close()
return contents
end
local function processInputFile(contents)
-- Preamble
coroutine.yield("\\pdfoutput=1 ")
-- Process the file: Return each line
for line in contents:gmatch("(.-)[\n\r]") do
coroutine.yield(line)
end
-- Postamble
coroutine.yield("\\bye")
-- Done
coroutine.yield(nil)
end
return {
processInputFile(contents .. "\n")
end)
}
end)
```
Then a small test file ("Preprocessor.txt"):
``` Hello World1
Hello World2
Paragraph 3,
continuing right now
and going on over
four lines
```
And finally the bash shell script to tie everything together ("Preprocessor.sh"):
``` luatex -fmt=luatex --jobname=Preprocessor '\directlua0{dofile("Preprocessor.lua")}\input Preprocessor.txt'
```
As you can see, the file to be converted is simply read using "\input" and then converted in the Lua function "processInputFile". By using coroutines, we can pretend to convert the file in one go, when really we are supplying it line by line to LuaTeX.
It would of course be possible to check the file type/extension inside the "open_read_file" callback and perform different conversions (or none at all if a TeX file) depending on the file type, i.e. to typeset source code. | 2017-02-20 10:56:27 | {"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": 0.8198532462120056, "perplexity": 3122.799744787161}, "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-2017-09/segments/1487501170521.30/warc/CC-MAIN-20170219104610-00423-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://joshbloom.org/publication/2010-aj-139-2218-m/ | # New Observations of the Very Luminous Supernova 2006gy: Evidence for Echoes
### Abstract
Supernova (SN) 2006gy was a hydrogen-rich core-collapse SN that remains one of the most luminous optical SNe ever observed. The total energy budget (>2 × 10$^51$ erg radiated in the optical alone) poses many challenges for standard SN theory. We present new ground-based near-infrared (NIR) observations of SN 2006gy, as well as a single epoch of Hubble Space Telescope (HST) imaging obtained more than two years after the explosion. Our NIR data taken around peak optical emission show an evolution that is largely consistent with a cooling blackbody, with tentative evidence for a growing NIR excess starting around day åisebox-0.5ex 130. Our late-time Keck adaptive optics NIR image, taken on day 723, shows little change from previous NIR observations taken around day 400. Furthermore, the optical HST observations show a reduced decline rate after day 400, and the SN is bluer on day 810 than it was at peak. This late-time decline is inconsistent with $^56$Co decay, and thus is problematic for the various pair-instability SN models used to explain the nature of SN 2006gy. The slow decline of the NIR emission can be explained with a light echo, and we confirm that the late-time NIR excess is the result of a massive (gsim10 M $_sun$) dusty shell heated by the SN peak luminosity. The late-time optical observations require the existence of a scattered light echo, which may be generated by the same dust that contributes to the NIR echo. Both the NIR and optical echoes originate in the proximity of the progenitor, i̊sebox-0.5ex 10$^18$ cm for the NIR echo and lsim10-40 pc for the optical echo, which provides further evidence that the progenitor of SN 2006gy was a very massive star.
Publication
Astronomical Journal | 2022-07-05 16:04:28 | {"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": 0.5510722398757935, "perplexity": 3847.840648990787}, "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-2022-27/segments/1656104585887.84/warc/CC-MAIN-20220705144321-20220705174321-00091.warc.gz"} |
http://answers.gazebosim.org/question/17939/change-radius-of-collision-cylinder-c/ | # Change radius of collision (cylinder) C++
asked 2017-11-27 09:18:29 -0500
This post is a wiki. Anyone with karma >75 is welcome to improve it.
Hello Guys,
how can I change the radius of a cylinder (only collision, not the visual part) in a C++ plugin.
I load a simple model of a cylinder as a link in a world sdf file called cylinder.world.
The cylinder and also my plugin are loaded properly (i check the plugin by printing a simple message on the terminal when it is started). I would like to change the size of the cylinder (more exactly the radius) now during runtime (for example after 100 Iterations of the Simulation) by accessing the radius (collision tag) in my plugin.
As i first start with a cylinder that has the same dimensions in the visual as in the collision tag and i decrease the radius of the collision element in my plugin after some time, it should look like the cylinder is diving into the ground (although of course a closer look on the collision element shows that there is just a smaller cylinder due to the plugin).
I have tried a lot to achieve this by using functions of gazebo API, but nothing really worked. (http://gazebosim.org/api/dev/classgaz...)
I am using gazebo 7.
I hope I can get some help from you. I would be really happy about a working example showing the functionality I would like to achieve!
Thanks a lot for every kind of help!
First of all thanks for your answer!!!
Update 1:
Here is my header file called cylinder_radius_plugin.hh
#ifndef _CYLINDER_RADIUS_PLUGIN_HH__
#include <gazebo/gazebo.hh>
#include <gazebo/physics/physics.hh>
#include <gazebo/common/Plugin.hh>
namespace gazebo
{
class cylinder_radius_plugin : public ModelPlugin
{
public: virtual void Load(physics::ModelPtr _model, sdf::ElementPtr _sdf);
public: void OnUpdate(const common::UpdateInfo);
private: physics::ModelPtr model;
private: sdf::ElementPtr sdf;
private: event::ConnectionPtr updateConnection;
private: physics::CollisionPtr cylinder_col;
private: physics::CylinderShapePtr cylinder(physics::CylinderShape(physics::CollisionPtr));
private: int i;
};
}
#endif
And this is the actual source code called cylinder_radius_plugin.cc
#include <stdio.h>
#include <boost/bind.hpp>
#include <iostream>
#include <fstream>
#include <string>
#include <gazebo/gazebo.hh>
#include <gazebo/physics/physics.hh>
#include <gazebo/physics/CylinderShape.hh>
#include <gazebo/common/common.hh>
#include <gazebo/common/Plugin.hh>
#include <gazebo/common/Events.hh>
#include <gazebo/physics/Collision.hh>
using namespace gazebo;
{
}
{
std::cerr << "\n";
std::cerr << "\n The plugin is loaded correctly! \n";
std::cerr << "\n";
this->model = _model;
this->sdf = _sdf;
this->i = 0;
// I am not pretty shure how to do it, but I think I should allocate something
// to the CylinderShapePtr???
this->updateConnection = event::Events::ConnectWorldUpdateBegin(boost::bind(&cylinder_radius_plugin::OnUpdate, this, _1));
}
{
//something like that should be in here
// do something after 500 iterations of the simulation
if (this->i == 500)
{
// change the radius of the **collision** of the cylinder
// here should be ...
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If I understood well, you want to do the same as the graphical scaling tool, but only the collision should be scaled:
I tracked down what the tool does:
1. Publish SCALING to ~/user_cmd
2. Which then publishes to ~/model/modify
3. Which eventually goes to Model::SetScale
4. Then Link::SetScale
5. Which both calls Collision::SetScale and updates the visual scales, you only want the collision
6. Then it calls Shape::SetScale (in your case, it is CylinderShape)
It looks to me that a direct call to CylinderShape::SetScale should have the same results on the physics, but it might not update the collision visual (i.e. the cylinder will sync but gzclient will still show the big collision - and big visual, as you want).
If you could update your question with a code snippet maybe we can better debug what's going on.
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Asked: 2017-11-27 09:18:29 -0500
Seen: 362 times
Last updated: Nov 28 '17 | 2019-04-26 01:52:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17190998792648315, "perplexity": 3987.8800311999976}, "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-2019-18/segments/1555578747424.89/warc/CC-MAIN-20190426013652-20190426035652-00097.warc.gz"} |
http://stats.stackexchange.com/questions/50443/proving-sample-covariance-is-unbiased-with-matrix-algebra | # proving sample covariance is unbiased with matrix algebra [closed]
Given an i.i.d. sample $(x_i,y_i)$ of size $n$ from a bivariate distribution $(x,y)$, I'm trying to prove that the sample covariance $$\frac{1}{n-1}\sum_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})$$ is an unbiasted estimator of the true population covariance $\text{cov}(x,y)$.
I started by writing the sample covariance in matrix form as $$\frac{1}{n-1}(\mathbf{Dx})'\mathbf{Dy}=\frac{1}{n-1}\mathbf{x'D'Dy}=\frac{1}{n-1}\mathbf{x'DDy}=\frac{1}{n-1}\mathbf{x'Dy}$$ where $\mathbf{D}$ is the symmetric and idempotent "centering matrix" $$\mathbf{D}=\mathbf{I}_n-\frac{1}{n}\iota\iota'$$ that puts the data in deviation from their sample mean and $\mathbf{x}$ and $\mathbf{y}$ are the $x_i$ and $y_i$ data points collected into $n\times 1$ column vectors. Taking expectations, and using "Tr" to denote trace, I get \begin{eqnarray} E\left[\frac{1}{n-1}\mathbf{x'Dy}\right]&=&\frac{1}{n-1}E[\mathbf{x'Dy}] \nonumber\\ &=&\frac{1}{n-1}E[\text{Tr}(\mathbf{x'Dy})]\nonumber\\ &=&\frac{1}{n-1}E[\text{Tr}(\mathbf{Dyx'})]\nonumber\\ &=&\frac{1}{n-1}\text{Tr}(E[\mathbf{Dyx'}])\nonumber\\ &=&\frac{1}{n-1}\text{Tr}(\mathbf{D}E[\mathbf{yx'}])\nonumber\\ &=&\frac{1}{n-1}\text{Tr}(\mathbf{D}(\text{cov}(x,y)+\mu_x\mu_y)\mathbf{I})\nonumber\\ &=&\frac{1}{n-1}\text{Tr}(\mathbf{DI})(\text{cov}(x,y)+\mu_x\mu_y)\nonumber\\ &=&\frac{1}{n-1}\text{Tr}(\mathbf{D})(\text{cov}(x,y)+\mu_x\mu_y)\nonumber\\ &=&\frac{1}{n-1}(n-1)(\text{cov}(x,y)+\mu_x\mu_y)\nonumber\\ &=&\text{cov}(x,y)+\mu_x\mu_y\nonumber \end{eqnarray}
But this clearly can't be right, because the expectation is supposed to come out just to be $\text{cov}(x,y)$. And yet I can't spot my mistake. Can someone point out where I've made an error?
The only mistake I could anticipate is that it's wrong to claim $$E[\mathbf{yx'}]=(\text{cov}(x,y)+\mu_x\mu_y)\mathbf{I}$$ but every entry in the matrix $\mathbf{yx'}$ has the form $y_ix_j$, and since the sample is i.i.d., $E[y_ix_j]=E[yx]=\text{cov}(x,y)+\mu_x\mu_y$.
-
Near the end you assume $\text{tr}(AB)$ = $\text{tr}(A)(B)$ where $A = DI$ and $B = \text{cov}(x,y) + \mu_x\mu_y$; this is not generally true. In this context notice that $B$ is an $n\times n$ matrix, not a scalar. – whuber Feb 20 at 20:50
No, $\text{cov}(x,y)+\mu_x\mu_y$ is a scalar, but $(\text{cov}(x,y)+\mu_x\mu_y)\mathbf{I}$ is a matrix. I factored $\text{cov}(x,y)+\mu_x\mu_y$ out of the trace because trace is linear. – symplectomorphic Feb 20 at 21:35
At least that's how I was interpreting what it means to speak of a sample here -- each $x_i$ being a sample drawn from the (scalar) population $x$, although when I collect data points I get a vector $\mathbf{x}$. Perhaps this is where I'm confused? – symplectomorphic Feb 20 at 21:38
Ah, wait, I think your comment indirectly helped me see the error: what I meant was $(\text{cov}(x,y)+\mu_x\mu_y)\mathbf{\iota\iota'}$, not $(\text{cov}(x,y)+\mu_x\mu_y)\mathbf{I}$. – symplectomorphic Feb 20 at 21:42
Yes, that was the point of the comment. – whuber Feb 20 at 22:44
## closed as too localized by whuber♦Feb 20 at 22:44
This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, see the FAQ. | 2013-05-19 22:38:55 | {"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": 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": 0.8851465582847595, "perplexity": 438.1932076943229}, "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/1368698104521/warc/CC-MAIN-20130516095504-00095-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://gamedev.stackexchange.com/questions/179047/throttling-or-adding-delays-between-ontriggerenter2d-calls | # Throttling or adding delays between OnTriggerEnter2D calls
I have 2 game objects: Explosion and Enemy. At any given point, an enemy can collide with multiple Explosion game objects at the same exact time. This causes a few issues when I'm trying to destroy the Enemy game object.
This is the function that destroys the game object:
bool isDead = false;
public void KillCharacter () {
Debug.Log ("kill enemy...");
}
}
If I use OnTriggerEnter2D inside the Enemy script, like so:
void OnTriggerEnter2D (Collider2D other) {
if (other.gameObject.tag == "Explosion") {
KillCharacter();
}
}
In case two explosions spawn on the position of the enemy, Debug.Log inside the KillCharacter function will fire twice, even though I set the isDead bool to true on the first run.
The only solution I found was to call KillCharacter from the Explosion script, like so:
void OnTriggerEnter2D (Collider2D other) {
if (other.gameObject.tag == "Enemy") {
other.gameObject.GetComponent<Enemy> ().KillCharacter ();
}
}
Is there a way to run the OnTriggerEnter2D from the Enemy script, as I originally intended? Or is having the KillCharacter handle all the death logic and calling it from outside good enough?
• The behavior you describe shouldn't really happen -- you set isDead to true the first time, so even if two collisions are happening at a nearly identical time, they must still be handled sequentially in some way, and the very first one closes the processing for the second. Could you check your code some more to see if either the isDead is perhaps set to false somewhere again -- or that maybe your script is attached to two different gameObjects, and you are mistaking the "kill enemy" debug log for a single gameObject? E.g. give all objects a different name and use print("killed " + name). Feb 11, 2020 at 21:39
• please share full scriptd Feb 12, 2020 at 8:00
Tested your question and code out as at first I thought it was regarding the 'Is Trigger' option on the collider. However, it's working as intended.
As shown here: https://imgur.com/a/oPJMFTl.
The script is on the Enemy tagged object. Tested with the 'Is Trigger' checked on one and both of the Objects "Enemy", "Explosion".
The only thing I changed from your code was set the bool isDead to public so that it shows on the inspector.
Most likely problem is that your bool isDead is being altered elsewhere.
• Forgot to mention that I'm using Unity v2019.3.0f6. Not sure if it makes a difference, but the explosions spawn on the location of the enemy at the same time, not the other way around. I tried logging inside OnTriggerEnter2D and KillCharacter and it shows like this: i.imgur.com/DsBnJy4.png - isDead is set to true before the first killed char log is output. This is really weird. Feb 12, 2020 at 9:14
• using same version When you spawn(instantiate) the explosions they are happening sequentially through each instantiation line and not at the exact same time, so it shouldn't be possible for a double trigger. I recommend that you debug by creating new explosion and enemy objects with just the code above attached to the enemy object and see if that works as you intended.
– RBee
Feb 12, 2020 at 21:38 | 2022-06-29 01:21:28 | {"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": 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": 0.2754504382610321, "perplexity": 2820.429288257026}, "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-2022-27/segments/1656103619185.32/warc/CC-MAIN-20220628233925-20220629023925-00020.warc.gz"} |
https://learn.microsoft.com/en-IE/dotnet/visual-basic/language-reference/operators/bit-shift-operators | # Bit Shift Operators (Visual Basic)
The following are the bit shift operators defined in Visual Basic.
<< Operator
>> Operator | 2022-12-03 03:53:04 | {"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": 0.8636277318000793, "perplexity": 10775.85228937333}, "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-20221203011523-20221203041523-00309.warc.gz"} |
http://cms.math.ca/cmb/kw/f?page=17 | location: Publications → journals
Search results
Search: All articles in the CMB digital archive with keyword f
Expand all Collapse all Results 401 - 425 of 475
401. CMB 2004 (vol 47 pp. 73)
Li, Ma; Dezhong, Chen
Systems of Hermitian Quadratic Forms In this paper, we give some conditions to judge when a system of Hermitian quadratic forms has a real linear combination which is positive definite or positive semi-definite. We also study some related geometric and topological properties of the moduli space. Keywords:hermitian quadratic form, positive definite, positive semi-definiteCategory:15A63
402. CMB 2004 (vol 47 pp. 152)
Zheng, Jian-Hua
On Uniqueness of Meromorphic Functions with Shared Values in Some Angular Domains In this paper we investigate the uniqueness of transcendental meromorphic function dealing with the shared values in some angular domains instead of the whole complex plane. Keywords:Nevanlinna theory, meromorphic function, shared valueCategory:30D35
403. CMB 2004 (vol 47 pp. 133)
Wang, Wei
Embeddability of Some Three-Dimensional Weakly Pseudoconvex ${\rm CR}$ Structures We prove that a class of perturbations of standard ${\rm CR}$ structure on the boundary of three-dimensional complex ellipsoid $E_{p,q}$ can be realized as hypersurfaces on $\mathbb{C}^2$, which generalizes the result of Burns and Epstein on the embeddability of some perturbations of standard ${\rm CR}$ structure on $S^3$. Keywords:deformations, embeddability, complex ellipsoidsCategories:32V30, 32G07, 32V35
404. CMB 2004 (vol 47 pp. 22)
Goto, Yasuhiro
A Note on the Height of the Formal Brauer Group of a $K3$ Surface Using weighted Delsarte surfaces, we give examples of $K3$ surfaces in positive characteristic whose formal Brauer groups have height equal to $5$, $8$ or $9$. These are among the four values of the height left open in the article of Yui \cite{Y}. Keywords:formal Brauer groups, $K3$ surfaces in positive, characteristic, weighted Delsarte surfacesCategories:14L05, 14J28
405. CMB 2003 (vol 46 pp. 617)
Pak, Hong Kyung
On Harmonic Theory in Flows Recently [8], a harmonic theory was developed for a compact contact manifold from the viewpoint of the transversal geometry of contact flow. A contact flow is a typical example of geodesible flow. As a natural generalization of the contact flow, the present paper develops a harmonic theory for various flows on compact manifolds. We introduce the notions of $H$-harmonic and $H^*$-harmonic spaces associated to a H\"ormander flow. We also introduce the notions of basic harmonic spaces associated to a weak basic flow. One of our main results is to show that in the special case of isometric flow these harmonic spaces are isomorphic to the cohomology spaces of certain complexes. Moreover, we find an obstruction for a geodesible flow to be isometric. Keywords:contact structure, geodesible flow, isometric flow, basic cohomologyCategories:53C20, 57R30
406. CMB 2003 (vol 46 pp. 632)
Runde, Volker
The Operator Amenability of Uniform Algebras We prove a quantized version of a theorem by M.~V.~She\u{\i}nberg: A uniform algebra equipped with its canonical, {\it i.e.}, minimal, operator space structure is operator amenable if and only if it is a commutative $C^\ast$-algebra. Keywords:uniform algebras, amenable Banach algebras, operator amenability, minimal, operator spaceCategories:46H20, 46H25, 46J10, 46J40, 47L25
407. CMB 2003 (vol 46 pp. 373)
Laugesen, Richard S.; Pritsker, Igor E.
Potential Theory of the Farthest-Point Distance Function We study the farthest-point distance function, which measures the distance from $z \in \mathbb{C}$ to the farthest point or points of a given compact set $E$ in the plane. The logarithm of this distance is subharmonic as a function of $z$, and equals the logarithmic potential of a unique probability measure with unbounded support. This measure $\sigma_E$ has many interesting properties that reflect the topology and geometry of the compact set $E$. We prove $\sigma_E(E) \leq \frac12$ for polygons inscribed in a circle, with equality if and only if $E$ is a regular $n$-gon for some odd $n$. Also we show $\sigma_E(E) = \frac12$ for smooth convex sets of constant width. We conjecture $\sigma_E(E) \leq \frac12$ for all~$E$. Keywords:distance function, farthest points, subharmonic function, representing measure, convex bodies of constant widthCategories:31A05, 52A10, 52A40
408. CMB 2003 (vol 46 pp. 265)
Oh, Seungsang
Reducing Spheres and Klein Bottles after Dehn Fillings Let $M$ be a compact, connected, orientable, irreducible 3-manifold with a torus boundary. It is known that if two Dehn fillings on $M$ along the boundary produce a reducible manifold and a manifold containing a Klein bottle, then the distance between the filling slopes is at most three. This paper gives a remarkably short proof of this result. Keywords:Dehn filling, reducible, Klein bottleCategory:57M50
409. CMB 2003 (vol 46 pp. 310)
Wang, Xiaofeng
Second Order Dehn Functions of Asynchronously Automatic Groups Upper bounds of second order Dehn functions of asynchronously automatic groups are obtained. Keywords:second order Dehn function, combing, asynchronously automatic groupCategories:20E06, 20F05, 57M05
410. CMB 2003 (vol 46 pp. 268)
Puls, Michael J.
Group Cohomology and $L^p$-Cohomology of Finitely Generated Groups Let $G$ be a finitely generated, infinite group, let $p>1$, and let $L^p(G)$ denote the Banach space $\{ \sum_{x\in G} a_xx \mid \sum_{x\in G} |a_x |^p < \infty \}$. In this paper we will study the first cohomology group of $G$ with coefficients in $L^p(G)$, and the first reduced $L^p$-cohomology space of $G$. Most of our results will be for a class of groups that contains all finitely generated, infinite nilpotent groups. Keywords:group cohomology, $L^p$-cohomology, central element of infinite order, harmonic function, continuous linear functionalCategories:43A15, 20F65, 20F18
411. CMB 2003 (vol 46 pp. 216)
Li, Chi-Kwong; Rodman, Leiba; Šemrl, Peter
Linear Maps on Selfadjoint Operators Preserving Invertibility, Positive Definiteness, Numerical Range Let $H$ be a complex Hilbert space, and $\HH$ be the real linear space of bounded selfadjoint operators on $H$. We study linear maps $\phi\colon \HH \to \HH$ leaving invariant various properties such as invertibility, positive definiteness, numerical range, {\it etc}. The maps $\phi$ are not assumed {\it a priori\/} continuous. It is shown that under an appropriate surjective or injective assumption $\phi$ has the form $X \mapsto \xi TXT^*$ or $X \mapsto \xi TX^tT^*$, for a suitable invertible or unitary $T$ and $\xi\in\{1, -1\}$, where $X^t$ stands for the transpose of $X$ relative to some orthonormal basis. Examples are given to show that the surjective or injective assumption cannot be relaxed. The results are extended to complex linear maps on the algebra of bounded linear operators on $H$. Similar results are proved for the (real) linear space of (selfadjoint) operators of the form $\alpha I+K$, where $\alpha$ is a scalar and $K$ is compact. Keywords:linear map, selfadjoint operator, invertible, positive definite, numerical rangeCategories:47B15, 47B49
412. CMB 2003 (vol 46 pp. 95)
Gauthier, P. M.
Cercles de remplissage for the Riemann Zeta Function The celebrated theorem of Picard asserts that each non-constant entire function assumes every value infinitely often, with at most one exception. The Riemann zeta function has this Picard behaviour in a sequence of discs lying in the critical band and whose diameters tend to zero. According to the Riemann hypothesis, the value zero would be this (unique) exceptional value. Keywords:cercles de remplissage, Riemann zeta functionCategory:30
413. CMB 2003 (vol 46 pp. 130)
Petersen, Peter; Wilhelm, Frederick
On Frankel's Theorem In this paper we show that two minimal hypersurfaces in a manifold with positive Ricci curvature must intersect. This is then generalized to show that in manifolds with positive Ricci curvature in the integral sense two minimal hypersurfaces must be close to each other. We also show what happens if a manifold with nonnegative Ricci curvature admits two nonintersecting minimal hypersurfaces. Keywords:Frankel's TheoremCategory:53C20
414. CMB 2003 (vol 46 pp. 122)
Moon, Myoungho
On Certain Finitely Generated Subgroups of Groups Which Split Define a group $G$ to be in the class $\mathcal{S}$ if for any finitely generated subgroup $K$ of $G$ having the property that there is a positive integer $n$ such that $g^n \in K$ for all $g\in G$, $K$ has finite index in $G$. We show that a free product with amalgamation $A*_C B$ and an $\HNN$ group $A *_C$ belong to $\mathcal{S}$, if $C$ is in $\mathcal{S}$ and every subgroup of $C$ is finitely generated. Keywords:free product with amalgamation, $\HNN$ group, graph of groups, fundamental groupCategories:20E06, 20E08, 57M07
415. CMB 2002 (vol 45 pp. 483)
Baake, Michael
Diffraction of Weighted Lattice Subsets A Dirac comb of point measures in Euclidean space with bounded complex weights that is supported on a lattice $\varGamma$ inherits certain general properties from the lattice structure. In particular, its autocorrelation admits a factorization into a continuous function and the uniform lattice Dirac comb, and its diffraction measure is periodic, with the dual lattice $\varGamma^*$ as lattice of periods. This statement remains true in the setting of a locally compact Abelian group whose topology has a countable base. Keywords:diffraction, Dirac combs, lattice subsets, homometric setsCategories:52C07, 43A25, 52C23, 43A05
416. CMB 2002 (vol 45 pp. 378)
Fernández-López, Manuel; García-Río, Eduardo; Kupeli, Demir N.
The Local Möbius Equation and Decomposition Theorems in Riemannian Geometry A partial differential equation, the local M\"obius equation, is introduced in Riemannian geometry which completely characterizes the local twisted product structure of a Riemannian manifold. Also the characterizations of warped product and product structures of Riemannian manifolds are made by the local M\"obius equation and an additional partial differential equation. Keywords:submersion, Möbius equation, twisted product, warped product, product Riemannian manifoldsCategories:53C12, 58J99
417. CMB 2002 (vol 45 pp. 428)
Mollin, R. A.
Criteria for Simultaneous Solutions of $X^2 - DY^2 = c$ and $x^2 - Dy^2 = -c$ The purpose of this article is to provide criteria for the simultaneous solvability of the Diophantine equations $X^2 - DY^2 = c$ and $x^2 - Dy^2 = -c$ when $c \in \mathbb{Z}$, and $D \in \mathbb{N}$ is not a perfect square. This continues work in \cite{me}--\cite{alfnme}. Keywords:continued fractions, Diophantine equations, fundamental units, simultaneous solutionsCategories:11A55, 11R11, 11D09
418. CMB 2002 (vol 45 pp. 337)
Chen, Imin
Surjectivity of $\mod\ell$ Representations Attached to Elliptic Curves and Congruence Primes For a modular elliptic curve $E/\mathbb{Q}$, we show a number of links between the primes $\ell$ for which the mod $\ell$ representation of $E/\mathbb{Q}$ has projective dihedral image and congruence primes for the newform associated to $E/\mathbb{Q}$. Keywords:torsion points of elliptic curves, Galois representations, congruence primes, Serre tori, grossencharacters, non-split CartanCategories:11G05, 11F80
419. CMB 2002 (vol 45 pp. 161)
Ardizzone, Lucia; Grimaldi, Renata; Pansu, Pierre
Sur les singularités de la fonction croissance d'une variété non simplement connexe Si $M$ est une vari\'et\'e de dimension $n$, compacte non simplement connexe, on caract\'erise les m\'etriques riemanniennes sur $M$ dont la fonction croissance a exactement deux singularit\'es. Keywords:fonction croissance, singularités, fonction de Morse, CutlocusCategory:53B20
420. CMB 2002 (vol 45 pp. 272)
Neusel, Mara D.
The Transfer in the Invariant Theory of Modular Permutation Representations II In this note we show that the image of the transfer for permutation representations of finite groups is generated by the transfers of special monomials. This leads to a description of the image of the transfer of the alternating groups. We also determine the height of these ideals. Keywords:polynomial invariants of finite groups, permutation representation, transferCategory:13A50
421. CMB 2002 (vol 45 pp. 265)
Nawrocki, Marek
On the Smirnov Class Defined by the Maximal Function H.~O.~Kim has shown that contrary to the case of $H^p$-space, the Smirnov class $M$ defined by the radial maximal function is essentially smaller than the classical Smirnov class of the disk. In the paper we show that these two classes have the same corresponding locally convex structure, {\it i.e.} they have the same dual spaces and the same Fr\'echet envelopes. We describe a general form of a continuous linear functional on $M$ and multiplier from $M$ into $H^p$, $0 < p \leq \infty$. Keywords:Smirnov class, maximal radial function, multipliers, dual space, Fréchet envelopeCategories:46E10, 30A78, 30A76
422. CMB 2002 (vol 45 pp. 213)
Gordon, B. Brent; Joshi, Kirti
Griffiths Groups of Supersingular Abelian Varieties The Griffiths group $\Gr^r(X)$ of a smooth projective variety $X$ over an algebraically closed field is defined to be the group of homologically trivial algebraic cycles of codimension $r$ on $X$ modulo the subgroup of algebraically trivial algebraic cycles. The main result of this paper is that the Griffiths group $\Gr^2 (A_{\bar{k}})$ of a supersingular abelian variety $A_{\bar{k}}$ over the algebraic closure of a finite field of characteristic $p$ is at most a $p$-primary torsion group. As a corollary the same conclusion holds for supersingular Fermat threefolds. In contrast, using methods of C.~Schoen it is also shown that if the Tate conjecture is valid for all smooth projective surfaces and all finite extensions of the finite ground field $k$ of characteristic $p>2$, then the Griffiths group of any ordinary abelian threefold $A_{\bar{k}}$ over the algebraic closure of $k$ is non-trivial; in fact, for all but a finite number of primes $\ell\ne p$ it is the case that $\Gr^2 (A_{\bar{k}}) \otimes \Z_\ell \neq 0$. Keywords:Griffiths group, Beauville conjecture, supersingular Abelian variety, Chow groupCategories:14J20, 14C25
423. CMB 2002 (vol 45 pp. 109)
Hall, R. R.; Shiu, P.
The Distribution of Totatives D.~H.~Lehmer initiated the study of the distribution of totatives, which are numbers coprime with a given integer. This led to various problems considered by P.~Erd\H os, who made a conjecture on such distributions. We prove his conjecture by establishing a theorem on the ordering of residues. Keywords:Euler's function, totativesCategories:11A05, 11A07, 11A25
424. CMB 2002 (vol 45 pp. 154)
Weitsman, Allen
On the Poisson Integral of Step Functions and Minimal Surfaces Applications of minimal surface methods are made to obtain information about univalent harmonic mappings. In the case where the mapping arises as the Poisson integral of a step function, lower bounds for the number of zeros of the dilatation are obtained in terms of the geometry of the image. Keywords:harmonic mappings, dilatation, minimal surfacesCategories:30C62, 31A05, 31A20, 49Q05
425. CMB 2002 (vol 45 pp. 138)
Spearman, Blair K.; Williams, Kenneth S.
The Discriminant of a Dihedral Quintic Field Defined by a Trinomial $X^5 + aX + b$ Let $X^5 + aX + b \in Z[X]$ have Galois group $D_5$. Let $\theta$ be a root of $X^5 + aX + b$. An explicit formula is given for the discriminant of $Q(\theta)$. Keywords:dihedral quintic field, trinomial, discriminantCategories:11R21, 11R29
Page Previous 1 ... 16 17 18 19 Next | 2015-07-04 10:11:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9381284117698669, "perplexity": 673.5822438712349}, "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/1435375096686.2/warc/CC-MAIN-20150627031816-00232-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/423691/how-do-i-write-this-table-in-latex | # How do I write this table in LaTeX?
I want to construct the table from page 8 from this article:
I tried this
\begin{table}[h!]
\centering
$\begin{array}{@{}c|cccc@{}} \textbf{w^2}-xz & -x & y & 0 & -z & 0 & -z^2+wz \\ \textbf{wx}-yz & \textbf{w} & -x & -y & 0 & z & z^2 \\ \textbf{x^2}-wy & -z & \textbf{w} & 0 & -y & 0 & 0 \\ \textbf{xy}-z^2 & 0 & 0 & \textbf{w} & \textbf{x} & -y & yz \\ \textbf{y^2}-wz & 0 & 0 & -z & -w & \textbf{x} \textbf{w^2} \\ \hline & 0 & y & -x & textbf{w} & -z & 1 \\ & -y^2+wz & z^2 & -wy & yz & -w^2 & \textbf{x}\\ \end{array}$
\label{tab:table1}
\end{table}\\
but it doesn't work.
• Please make your code snippet compilable and explain what "does not work" mean! – Mensch Mar 28 '18 at 17:41
• You want \mathbf{w}^2 – egreg Mar 28 '18 at 17:42
• You can find more ways on bold math here For more ways see here: tex.stackexchange.com/questions/595/… – koleygr Mar 28 '18 at 18:04
There are some errors in the code of the question:
• Two columns are missing in the column specification of array.
• textbf without backslash
• Missing & at the end of the fifth row.
• Bold in math for letters are done with \mathbf.
• Exponents are usually not made bold.
• \\ after \end{table} is bad style, since table is usually a floating object, where \\ does not make any sense, if the object before is floated away.
Full example:
\documentclass{article}
\begin{document}
\begin{table}
\centering
$\begin{array}{c|cccccc} \mathbf{w}^2-xz & -x & y & 0 & -z & 0 & -z^2+wz \\ \mathbf{wx}-yz & \mathbf{w} & -x & -y & 0 & z & z^2 \\ \mathbf{x}^2-wy & -z & \mathbf{w} & 0 & -y & 0 & 0 \\ \mathbf{xy}-z^2 & 0 & 0 & \mathbf{w} & \mathbf{x} & -y & yz \\ \mathbf{y}^2-wz & 0 & 0 & -z & -w & \mathbf{x} & \mathbf{w}^2 \\ \hline & 0 & y & -x & \mathbf{w} & -z & 1 \\ & -y^2+wz & z^2 & -wy & yz & -w^2 & \mathbf{x}\\ \end{array}$
\label{tab:table1}
\end{table}
\end{document}
• Just to complete the list: \label is useless without a \caption. – egreg Mar 28 '18 at 21:13
• @egreg \pageref is still working. – Heiko Oberdiek Mar 28 '18 at 23:47 | 2020-01-18 16:45:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.9770780801773071, "perplexity": 2752.919508202842}, "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/1579250593295.11/warc/CC-MAIN-20200118164132-20200118192132-00029.warc.gz"} |
https://www.physicsforums.com/threads/length-of-a-curve.12960/ | # Length of a curve
1. Jan 20, 2004
### tandoorichicken
Find the length of $$y = \frac{1}{3} (x^2 + 2)^{\frac{3}{2}}$$ from x=0 to x=3.
I used the formula $$s = \int_{a}^{b} \sqrt{1 + \frac{\,dy}{\,dx}} \,dx$$.
After plugging everything in, I got
$$s = \int_{0}^{3} (1 + \frac{1}{4} (x^2 + 2))}^\frac{1}{2} \,dx$$
Now, this isn't an integral I've learned how to do, so
1) Did I do anything wrong, and
2) If yes, what?
Last edited: Jan 20, 2004
2. Jan 20, 2004
### tandoorichicken
I realized I forgot to use the chain rule when doing dy/dx. However when I do that, I get even a messier integral,
$$s = \int_{0}^{3} (1 + x^2(x^2 + 2))^\frac{1}{2} \,dx$$
?
3. Jan 20, 2004
### himanshu121
It should be
$$s = \int_{a}^{b} \sqrt{1 + (\frac{\,dy}{\,dx})^2} \,dx$$
But thats not the pro u just wrote the wrong one and applied the right
u also have
$$1+x^2(x^2+2) = 1+x^4+2x^2=x^4+2x^2+1=(x^2+1)^2$$
Is it enough
4. Jan 21, 2004
### tandoorichicken
Woah yeah, thanks, overlooked a simple thing there. | 2017-01-25 01:59:15 | {"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": 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": 0.89452064037323, "perplexity": 3828.7051660035186}, "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-20170116095125-00191-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://stats.stackexchange.com/tags/generalized-linear-model/hot | # Tag Info
4
A couple of points: The reason why you typically need to use GLMMs, i.e., include random effects, is to account for correlations you have in your outcome data in specifics clusters/groups. For example, measurements taken on the same habitat will be correlated. To select the fixed-effects structure you need first to appropriately model these correlations. ...
4
Computationally, R-sq in computer printouts is the square $r^2$of the (Pearson) correlation $r.$ Sometimes $r^2$ is called the 'coefficient of determination'. Because $-1 \le r \le 1$ we have $0 \le r^2 \le 1.$ Suppose we have the simple linear regression model $$Y_i = \beta_0 + \beta_1 x_i + e_i,$$ for $i = 1,2, \dots,n$ where $e_i \stackrel{iid}{\sim}\... 3 I don't see much evidence that the variance increases with the mean, nor that the error distribution is non-normal. You might be reading too much into chance variation. The principal message of the diagnostic plots is to highlight point 1 as being especially interesting: it's pretty far from all the other points. However, falling in the middle of the ... 3 When you have a binary outcome variable you typically use a link function to connect the probability of a positive response to the linear predictor that includes the random effects, i.e., \left \{ \begin{array}{l} \log \displaystyle \frac{\pi_{it}}{1 - \pi_{it}} = x_{it}^\top \beta + z_{ij}^\top b_i,\\\\ \pi_{it} = \Pr(y_{it} = 1 \mid b_i),\\... 3 You can better evaluate the fit of the two models using the simulated residuals calculated by the DHARMa package. Irrespective of the type of the model, these simulated residuals will exhibit a flat/uniform distribution for a correctly specified model. However, you need to be careful with their use if you have missing at random missing data in your outcome ... 3 Survival analysis is a good idea for your case. Six time points might be too short, though... For a proper analysis, your data needs to be in a "longer" format, where you have 4*60 rows (one for each plant), and four columns: the first is the factor (WT, A, B, C), the second the "left" time (time of inclusion in the study, 0 for all your plants), the third ... 2 Yes, a deviance test is still valid. Some more details: Since the general theory is not specific for binomial models, I will start out with some general theory, but use binomial examples (and R.) GLM's is based on the exponential dispersion model $$f(y_i;\theta_i,\phi)= \exp\left\{ w_i [y_i \theta_i -\gamma(\theta_i)]/\phi +\tau(y_i,\phi/w_i)\right\}$$... 2 Zhanxiong's answer is already great (+1), but here's a quick demonstration that the log-likelihood of the saturated model is$0$for a logistic regression. I figured I would post because I haven't seen this TeX'd up on this site, and because I just wrote these up for a lecture. The likelihood is$$L(\mathbf{y} ; \mathbf{X}, \boldsymbol{\beta}) = \prod_{i=1}... 2 The task is impossible, which can be revealed by bootstrapping the entire modeling and feature selection process to show that Confidence intervals on importance rankings of the candidate predictors will be roughly from 1-8 for all 8 candidates The features selected will vary wildly over bootstrap replications Essentially the data do not have the ... 2 The definition you quote which is used with generalized linear models (glm) is not an exponential amily, it is an exponential dispersion family. For a fixed value of the dispersion parameter$\phi$it is an exponential family (indexed by$\theta$), but when$\phi$varies it is not. When used in glm's, the exponential dispersion family is used for ... 1 Model selection based on$p$-values will bias the coefficients of the final model towards significance. This is a form of stepwise regression and should be avoided if the goal is confirmation through$p$-values. The actual chance of a false positive will be much higher than the chosen level of significance. This has been discussed in several places on CV, ... 1 In the ideal scenario, where your data wouldn't be plagued by complete or quasi separation, you would estimate the effects of interest in your binary logistic regression model using maximum likelihood estimation (MLE). If your sample size$n\$ is small or moderate, it turns out that the MLE estimators of these effects suffer from bias under the ideal ...
1
Here my thoughts. Hope you already found the solution as I am sort of at the same point. Changing the Likelihood of y for GLM changes the full conditional of your betas and likely the variance. So I am not sure if beta must be conditioned on sigma in a GLM setting anyway, as the joint density of variance,beta|y might not be concave anymore... While ...
1
Even though I am currently not familiar with brms, logically speaking you are fitting a mixed effects logistic regression. In this case the intercept is the log odds when Cond is zero, and the coefficient for Cond is the log odds ratio between Cond=1 and Cond=0. Note though that the interpretation of these coefficients is conditional on the random intercepts ...
1
SJ, if these are all categorical (assuming they are factors) then the scale doesn't really matter. The logistic regression isn't analyzing the actual number, but rather the presence or absence of that variable. You could easily rename all of the variables A,B,C,D, etc. within x1, x2, x3, and x4 and you would get the same results because they are ...
1
R-squared or coefficient of determination generally has the value between 0 and 1. However cor(x,y) is between -1 and 1. Definition of R-squared for a multiple linear regression is the square of correlation between output (y) and predicted values(f). For the case of simple linear regression with an intercept and one explanatory variable it happens R-squared ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2019-07-17 11:25:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.7839698195457458, "perplexity": 671.7542278603241}, "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-2019-30/segments/1563195525136.58/warc/CC-MAIN-20190717101524-20190717123524-00234.warc.gz"} |
https://www.math24.net/infinite-series-page-2/ | # Infinite Series – Page 2
• ### Example 3.
Show that the harmonic series $$\sum\limits_{n = 1}^\infty {\large\frac{1}{n}\normalsize}$$ diverges.
Solution.
To see this, we can write
${\sum\limits_{n = 1}^\infty {\frac{1}{n}} } = {1 + \frac{1}{2} + \underbrace {\left( {\frac{1}{3} + \frac{1}{4}} \right)}_{\frac{7}{{12}} > \frac{1}{2}} } + {\underbrace {\left( {\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}} \right)}_{\frac{{533}}{{840}} > \frac{1}{2}} }+{ \ldots \;\text{and so on}.}$
Therefore $$\sum\limits_{n = 1}^\infty {\large\frac{1}{n}\normalsize}$$ $$> \sum\limits_{n = 1}^\infty {\large\frac{1}{2}\normalsize}$$ = $$\infty .$$ Hence, the harmonic series diverges.
Actually, this result was first proved by a mediaeval French mathematician, Nichole Oresme, who lived over $$600$$ years ago.
### Example 4.
Investigate convergence of the series $$\sum\limits_{n = 0}^\infty {\left( {\large\frac{1}{{{3^n}}}\normalsize} + {\large\frac{1}{{{5^n}}}\normalsize} \right)}.$$
Solution.
This series converges because it is the sum of two convergent series, $$\sum\limits_{n = 0}^\infty {\large\frac{1}{{{3^n}}}\normalsize}$$ and $$\sum\limits_{n = 0}^\infty {\large\frac{1}{{{5^n}}}\normalsize}.$$ Both are geometric series with ratio $$\left| q \right| \lt 1.$$ Then
${\sum\limits_{n = 0}^\infty {\frac{1}{{{3^n}}}} } = {\sum\limits_{n = 0}^\infty {{{\left( {\frac{1}{3}} \right)}^n}} } = {\frac{1}{{1 – \frac{1}{3}}} }={ \frac{3}{2},}$
${\sum\limits_{n = 0}^\infty {\frac{1}{{{5^n}}}} } = {\sum\limits_{n = 0}^\infty {{{\left( {\frac{1}{5}} \right)}^n}} } = {\frac{1}{{1 – \frac{1}{5}}} }={ \frac{5}{4}.}$
Hence, the sum of the given series is
${\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{{3^n}}} + \frac{1}{{{5^n}}}} \right)} } = {\frac{3}{2} + \frac{5}{4} }={ \frac{{11}}{4}.}$
### Example 5.
Investigate convergence of the series $$\sum\limits_{n = 1}^\infty {\large\frac{1}{{\left( {n + \pi } \right)\left( {n + \pi + 1} \right)}}\normalsize}.$$
Solution.
We see that
${\frac{1}{{\left( {n + \pi } \right)\left( {n + \pi + 1} \right)}} }={ \frac{1}{{n + \pi }} }-{ \frac{1}{{n + \pi + 1}}.}$
Then the $$n$$th partial sum is
${{S_n} = \left( {\frac{1}{{1 + \pi }} – \frac{1}{{2 + \pi }}} \right) } + {\left( {\frac{1}{{2 + \pi }} – \frac{1}{{3 + \pi }}} \right) + \ldots } + {\left( {\frac{1}{{n + \pi }} – \frac{1}{{n + \pi + 1}}} \right) } = {\frac{1}{{1 + \pi }} }-{ \frac{1}{{n + \pi + 1}}.}$
Calculate the limit of $${S_n}$$ as $$n \to \infty:$$
${\lim\limits_{n \to \infty } {S_n} } = {\lim\limits_{n \to \infty } \left( {\frac{1}{{1 + \pi }} – \frac{1}{{n + \pi + 1}}} \right) } = {\frac{1}{{1 + \pi }} }\approx{ 0,24.}$
Hence, the series converges.
### Example 6.
Determine whether the series
${\frac{1}{{1 \cdot 2}} + \frac{1}{{2 \cdot 3}} }+{ \frac{1}{{3 \cdot 4}} + \frac{1}{{4 \cdot 5}} + \ldots }+{ \frac{1}{{n\left( {n + 1} \right)}} + \ldots }$
converges or diverges.
Solution.
The $$n$$th partial sum is
${{S_n} = \frac{1}{{1 \cdot 2}} + \frac{1}{{2 \cdot 3}} }+{ \frac{1}{{3 \cdot 4}} + \frac{1}{{4 \cdot 5}} + \ldots }+{ \frac{1}{{n\left( {n + 1} \right)}}.}$
We can easily see that
${\frac{1}{{1 \cdot 2}} = 1 – \frac{1}{2},\;\;}\kern-0.3pt {\frac{1}{{2 \cdot 3}} = \frac{1}{2} – \frac{1}{3},\;\;}\kern-0.3pt {\frac{1}{{3 \cdot 4}} = \frac{1}{3} – \frac{1}{4},\;\;}\kern-0.3pt {\frac{1}{{4 \cdot 5}} = \frac{1}{4} – \frac{1}{5},\;\; \ldots,\;\;}\kern-0.3pt {{\frac{1}{{n\left( {n + 1} \right)}} }={ \frac{1}{n} – \frac{1}{{n + 1}}.}}$
Then
${{S_n} = \left( {1 – \frac{1}{2}} \right) + \left( {\frac{1}{2} – \frac{1}{3}} \right) } + {\left( {\frac{1}{3} – \frac{1}{4}} \right) + \ldots } + {\left( {\frac{1}{n} – \frac{1}{{n + 1}}} \right).}$
Hence,
${{S_n} = 1 – \frac{1}{{n + 1}}\;\;\text{and}\;\;}\kern0pt {{\lim\limits_{n \to \infty } {S_n} }={ \lim\limits_{n \to \infty } \left( {1 – \frac{1}{{n + 1}}} \right) }={ 1.}}$
Thus the series converges to $$1.$$
### Example 7.
Evaluate $$\sum\limits_{n = 0}^\infty {\ln {\large\frac{{n + 2}}{{n + 1}}\normalsize}}.$$
Solution.
We write the $$n$$th term as
${{a_n} = \ln \frac{{n + 2}}{{n + 1}} } = {\ln \left( {n + 2} \right) }-{ \ln \left( {n + 1} \right).}$
Calculate the $$n$$th partial sum:
${{S_n} }={ \left( {\ln 2 – \ln 1} \right) }+{ \left( {\ln 3 – \ln 2} \right) } + {\left( {\ln 4 – \ln 3} \right) + \ldots } + {\left[ {\ln \left( {n + 2} \right) – \ln \left( {n + 1} \right)} \right] } = {\left( { – \ln 1 + \ln 2} \right) }+{ \left( { – \ln 2 + \ln 3} \right) } + {\left( { – \ln 3 + \ln 4} \right) + \ldots } + {\left[ { – \ln \left( {n + 1} \right) + \ln \left( {n + 2} \right)} \right] } = { – \ln 1 + \ln \left( {n + 2} \right) } = {\ln \left( {n + 2} \right).}$
Since $$\lim\limits_{n \to \infty } {S_n}$$ $$= \lim\limits_{n \to \infty } \left[ {\ln \left( {n + 2} \right)} \right]$$ $$= \infty ,$$ we conclude that the given series diverges.
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Problems 3-7 | 2019-08-17 12:40:30 | {"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": 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": 0.9827513098716736, "perplexity": 1487.3819589234577}, "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-2019-35/segments/1566027313259.30/warc/CC-MAIN-20190817123129-20190817145129-00143.warc.gz"} |
https://mikemahoney218.github.io/heddlr/reference/export_template.html | This is a simple wrapper function around as_utf8 and writeLines, letting users write their template strings to file without having to worry about file encodings. For more details on why UTF-8 encoding is necessary, check out Yihui Xie's post on the subject.
export_template(
template,
filename,
sep = "",
filename.is.string = TRUE,
strip.carriage.returns = TRUE
)
## Arguments
template The template string to be written out The path to write the template to, passed to writeLines. Also accepts stdout (and likely other similar functions) with a warning. Separator to use between lines written, passed to writeLines. Defaults to no separator, as templates are generally already spaced appropriately. A logical value indicating whether or not the filename parameter is expected to be a string (that is, a character vector). Setting the value to FALSE disables the warning when a non-character argument is passed, but this is unsupported functionality. A logical value indicating whether or not to strip carriage feed characters, should any exist. This preserves line spacing when writing out files originally written on Windows; otherwise writeLines appears to not recognize lines as ending with a newline and inserts one, resulting in 2x the number of line breaks as anticipated.
## Value
Returns the input template invisibly.
## Details
Note that this function is effectively the inverse of import_pattern -- export_template(import_pattern("out.txt"), "out.txt") should always result in an unchanged file, and exceptions to this rule would be considered bugs.
## Examples
pattern_file <- tempfile("out", tempdir(), ".Rmd")
export_template("my sample pattern", pattern_file) | 2021-06-15 18:37:27 | {"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": 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": 0.20363985002040863, "perplexity": 4600.217690170281}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487621519.32/warc/CC-MAIN-20210615180356-20210615210356-00080.warc.gz"} |
https://emacs.stackexchange.com/questions/5932/multiple-files-in-latex | # multiple files in LaTeX?
I currently use emacs to write and compile LaTeX files using the key combo: C-c C-f. Now this is all well and good, but now I'm working on a much bigger project that is divided into many chapters. My main file includes these chapters using \include{chp1} for example, and so I have to go back to the main file to use C-c C-f to compile everything.
So here's my question: When I am editing chp1.tex, is there any way to turn OFF the automatic recognition of LaTeX mode so that if I mistakenly hit C-c C-f, it doesn't freak out and spam me with pages and pages of LaTeX error messages? I'd like it to just tell me when I type C-c C-f in chp1.tex, "you stupid idiot, this is not the main file," rather than try to compile chp1 by itself.
• Do you have a Makefile for your bigger project? Dec 31, 2014 at 17:39
• Check out the AUCTeX manual on multifile documents. I suspect if you (setq-default TeX-master nil) as it suggests, AUCTeX will ask you for your master file before compiling.
– Dan
Dec 31, 2014 at 17:44
• You don't have to disable LaTeX mode: see Multifile Documents. If you set TeX-master as described in the manual you should be able to compile the whole document using C-c C-f from a buffer visiting a chapter file. Dec 31, 2014 at 17:44
• Oh, and are you sure you didn't mean C-c C-c (TeX-command-master)? The default binding for C-c C-f is TeX-font.
– Dan
Dec 31, 2014 at 17:46
• What am I, chopped liver? ;) @Constantine: could you convert your comment to an answer so it can be marked as answered?
– Dan
Dec 31, 2014 at 22:00
If you break latex doc into several files you can still work with the whole project just fine in emacs, all you need is to set the master file in each file you include. For this put the following at the top line of each file you include:
% -*- TeX-master: "dissertation.tex" -*-
(assuming your master file is called dissertation.tex). This way you get, for example, a working table of contents with C-c = and, I think, build with C-c C-f will work too. (personally I prefer including a build.bash or build.py file in every large project, so I don't use auctex-mode's build scripts)
One more nice binding for this work flow is C-c ^ which switches you to master file buffer.
• C-c C-f definitely works then. Another interesting binding with this workflow is C-c C-b, which will compile the current buffer only (but with the preamble from the master file, because *magic*). Jan 6, 2015 at 16:54
• Not really related to the question, but do you really need to include a build script in every project? Some people have spent some time developing complete such scripts, and they are distributed with texlive (if you haven't already tried them, you can look up latexmk, rubber or arara). Jan 6, 2015 at 16:56
• @T.Verron: well - yes. Because my build scripts do more then just latex build. For example I include svg files in latex (using Inkscape export to latex). So my build script converts, e.g., image.svg to image.pdf, if the later is outdated (script checks that). Also my build script builds everything in ./tmp dir of the project to keep master dir clean: it only contains pdf file of the build, master tex file and folders: tex, img, bib, tmp. Finally the whole procedure should be modified slightly if I use minted for source code coloring. Jan 6, 2015 at 17:25
• I would suggest to put TeX-master as a file local variable to the end of the file, instead of the beginning. The TeX-master-file-ask function (bound to C-c _ and used in many places in AUCTeX) searches only for the former format. Jan 6, 2015 at 17:39
Some other answers suggest modifying each file to include an emacs file variable to set the value of TeX-master, but I prefer to use emacs directory variables so that I don't have to modify each file. To do this, create a file called .dir-locals.el (see the link for caveats if you are on MS-DOS) in the directory where your TeX/LaTeX source files are located with the following contents (supposing your master document is called main.tex):
((LaTeX-mode . ((TeX-master . "main")))
(latex-mode . ((TeX-master . "main"))))
Then when you open any of the files in this directory where the mode is LaTeX-mode or latex-mode emacs will set the TeX-master variable for the buffer to "main".
Notes:
1. I have found that I have to restart emacs for the directory local variable to start taking effect and that just closing and re-opening a file in that directory is not sufficient.
2. There are lots of TeX/LaTeX modes, so if the above does not work, check what mode your file is and add a line specifying that mode (C-h v major-mode will show the mode). Careful with the parenthesis formatting.
3. A lazier (but less proper) alternative is to use the following contents for the .dir-locals.el file which will set the TeX-master variable for all files opened in the source directory whether they are TeX/LaTeX files or not.
((nil . ((TeX-master . "main")))) | 2022-09-29 10:56:47 | {"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": 0.8002862334251404, "perplexity": 3164.746717113026}, "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-2022-40/segments/1664030335350.36/warc/CC-MAIN-20220929100506-20220929130506-00601.warc.gz"} |
http://bolesnodete.com/kxt6uj/8702b8-lalande-21185-spectral-class | # lalande 21185 spectral class
0
1
It is the fourth closest star system to our solar system, after Alpha Centauri, Barnard’s Star and Wolf 359. 10.5. Lalande 21185. Lalande 21185 is the brightest red dwarf visible from the northern hemisphere and the fourth system closest to Sun after Alpha Centauri 3, Barnard's Star, and Wolf 359. Struve 2398 M4 +11.1 4. Procyon A F5 +2.7 2. Centari C M5 +15.0 19. Procyon B F0 +13.0 3. Lalande 21185 is a red dwarf star (spectral class M2nV), located about 2.5 parsecs (8.21 light-years) from Earth, where it is visible in the constellation of Ursa Major. For this lab, you will need to record the spectral class and absolute magnitude of a group of near stars and a group of the brightest stars in the night sky. M2. It is not known whether the colony continues to be known simply as Lalande 21185, or whether the system is now known by the name of its capital world - as is the case with Vekta, Helghan and Gyre. Lalande 21185 Star Factoids. It is the third brightest red dwarf in the night sky. The revised Yerkes Atlas system (Johnson & Morgan, 1953) listed only two M type spectral standard stars: HD 147379 (M0V) and HD 95735/Lalande 21185 (M2V). Two stars Lalande 21185 (M2 V) and Betelgeuse (M2 I) are the same spectral type M2, but Betelgeuse is luminosity class I and Lalande 21185 is luminosity class V. … It has an apparent magnitude of 7.520 and cannot be seen by the naked eye. Centari B K5 +5.8 18. It was one of the systems first colonised under the First Diaspora between 2118 and 2148. Jupiter-sized planet orbiting at twice Earth's distance at sub-Saturn temperatures; Orbit takes 5.9 years. ... Lalande 21185. For each star, open the Search window (F3) and enter the star’s name.Click on the star and look at the displayed information at the upper right. In addition, we independently confirm with CARMENES data the existence of Lalande 21185 b, a planet that has recently been discovered with the SOPHIE spectrograph. It has been the spectral standard star for class M2 V for a long time. We announce the discovery of two planets orbiting the M dwarfs GJ 251 ($0.360\\pm0.015$ M$_\\odot$) and HD 238090 ($0.578\\pm0.021$ M$_\\odot$) based on CARMENES radial velocity (RV) data. All three planets belong to the class … It is the fourth closest star system to the Sun at 8.3 ly. This spectral type M2V star was first found listed in Histoire Céleste Française published in 1801 and prepared by the French astronomer Jérôme Lalande (1732-1807) of the Paris Observatory. (ST reference: Spaceflight Chronology) On several occasions in 2364, viewscreen readouts aboard the USS Enterprise-D showed the location of Lalande 21185, in the form of charts from the Enterprise library computer. The system became known as spectral class, and the letters most stars are assigned are O, B, A, F, G, K, M (going from hot to cool). The star is the current Spectral Standard star for class M1 V. Lalande 21185 System (M2 V) - Lalande 21185 is also called GJ 411 and informally Proxima Ursa Majoris. The star is located only about 8.3 light-years away in the constellation Ursa Major. While HD 147379 was not considered a standard by expert classifiers in later compendia of standards, Lalande 21185 is still a … Facts. Sun G2 +4.8 16. 8. In 1970, Castor A and Castor B were separated by less than 2 arcseconds. Struve 23948 M5 +11.9 5. M2.1V spectral class; 46% Solar Mass, 46% Solar Diameter, 63% Solar Metalicity; A flare star; Older than the Sun, younger than 10 Billion Years old; Lalande 21185 b Factoids. Each letter is subdivided using a number from 0-9 (hot to cool), so a B3 star is slightly hotter than a B4 star. TABLE 2 Near Stars (Stars which are close to the Earth) Star Name Spectral Class Absolute Magnitude Star Name Spectral Class Absolute Magnitude 1. ~ ~ is a red dwarf (spectral type: M2V) only 8.31 light years distant from the Sun. Lalande 21185. 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Star is located only about 8.3 light-years away in the constellation Ursa.... 8.3 ly dwarf in the constellation Ursa Major night sky Class M2 V a... | 2021-07-28 20:37:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.4982016980648041, "perplexity": 4278.4651440808375}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153791.41/warc/CC-MAIN-20210728185528-20210728215528-00525.warc.gz"} |
http://encyclopedia.kids.net.au/page/20/2014 | ## Encyclopedia > 2014
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Harmonic series (mathematics) ... diverges to infinity (although that is much harder to prove; see here). The alternating harmonic series converges however: [itex]\sum_{k = 1}^\infty ... | 2015-07-08 02:38:26 | {"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": 0.8798563480377197, "perplexity": 1558.4043109384565}, "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/1435375635604.22/warc/CC-MAIN-20150627032715-00076-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://brilliant.org/discussions/thread/im-new/ | ×
# I'm new!
I'm new here, help me with what should I begin with! thank you
Note by Oum Imi
4 years, 7 months ago
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Welcome to Brilliant Oum!
There are many things you could do to begin with. It looks like you have already tried our weekly problem sets in math and physics, and even begun to pick your way through our practice section.
I would work your way through the rest of the weekly challenges, write down your work on the problems you solve, so that next week when the answers are made available you can see how to do the ones that stumped you, or potentially discover other ways of thinking about the ones you got right.
If you come across problems that use concepts that are foreign to you, check this page and other categories in the practice section to see if, we have written anything to help you with it. You also might be able to find an easier problem that uses one of the concepts in the problem you are challenged by.
Staff - 4 years, 7 months ago | 2018-01-23 14:01:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.983590304851532, "perplexity": 2308.8443859514086}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084891976.74/warc/CC-MAIN-20180123131643-20180123151643-00265.warc.gz"} |
https://www.nextgurukul.in/wiki/concept/CBSE/VI/Maths/Comparing-Decimals.htm | Comparing Decimals
Summary
LearnNext Lesson Video
Tenths
If a block of one unit is divided into 10 equal parts, then each part is $\frac{\text{1}}{\text{10}}$ (one – tenth) of the unit. It is written as 0.1 in decimal representation. The dot denotes the decimal point.
Fractions as Desimal
Every fraction whose denominator is 10 can be written in decimal notation.
e.g. $\frac{\text{5}}{\text{10}}$= 0.5 ; $\frac{\text{7}}{\text{10}}$ = 0.7
Hundredths
If a block of one unit is divided into 100 equal parts, then each part is $\frac{\text{1}}{\text{100}}$ (one - hundredth) of the unit. It is written as 0.01 in decimal notation. Every fraction whose denominator is 100 can be written in decimal notation.
e.g. = 0.05 ; $\frac{\text{7}}{\text{100}}$ = 0.07
Thousandths
If a block of one unit is divided into 1000 equal parts, then each part is $\frac{\text{1}}{\text{1000}}$ (one - thousandth) of the unit. It is written as 0.001 in decimal notation. Every fraction whose denominator is 1000 can be written in decimal notation.
e.g. = 0.005 ; $\frac{\text{7}}{\text{100}}$ = 0.007
To read decimals, we can use the following chart. The first digit to the right after the decimal point represents the tenths parts, the second the hundredths parts, and so on.
Decimal Points Tenths Hundredths Thousandths . $\frac{\text{1}}{\text{10}}$ = 0.1 $\frac{\text{1}}{\text{100}}$ = 0.01 $\frac{\text{1}}{\text{1000}}$ = 0.001
Number line
All decimal numbers can be represented on the number line. Every decimal number can be represented as a fraction.
Comparing desimal
Any two decimal numbers can be compared. The comparison starts with the whole part of the numbers. If the whole parts are equal, then the tenth parts can be compared, and so on.
Desimals in Real life
Decimal numbers are used in many ways in real life. For example, in representing the units of money, length and weight, we use decimal numbers.
e.g.
Money
100 paise = 1 Rupee
1 paise = 1 / 100 = 0.01 Rupee
Length
10 mm = 1 cm
1 mm = 1/ 10 = 0.1 cm
1 cm = 0.01 m
1 m = 0.001 km
Weight
1000 g = 1 kg
1 g = 1 / 1000 = 0.001 kg
Capacity
1000 ml = 1 litre
1 ml = 1 / 1000 = 0.001 litre
1 . how to write a bio sketch ?
Know how to write a biographical sketch is very important if you have...
2 . Change into fraction and to its simplest form: 9675.564443
9675.564443 = (9675) + (564443/1000000)
587.6295
0.5 = 5/10 = 1/2
.25
X | 2017-06-23 03:16:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 10, "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": 0.8159540295600891, "perplexity": 1209.52822182488}, "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-2017-26/segments/1498128319992.22/warc/CC-MAIN-20170623031127-20170623051127-00293.warc.gz"} |
http://mathhelpforum.com/calculus/143063-logistic-differential-equation.html | # Math Help - Logistic Differential Equation
1. ## Logistic Differential Equation
The number of moose in a national park is modeled by the function M that satisfies the logistic differential equation
, where t is the time in years and M(0)=50. What is ?
Does this have anything to do with the formula M/(1+A(e^(-kt)))
(sorry, I couldn't input that into Wolfram for some reason...)
2. Originally Posted by BeiW
The number of moose in a national park is modeled by the function M that satisfies the logistic differential equation
, where t is the time in years and M(0)=50. What is ?
Does this have anything to do with the formula M/(1+A(e^(-kt)))
(sorry, I couldn't input that into Wolfram for some reason...)
change the form to $\frac{dy}{dt} = ky(L-y)$ , where $L$ is the limiting value ...
$\frac{dM}{dt} = \left(\frac{0.6}{200}\right)M\left[200-M\right]$
$\lim_{t \to \infty} M(t) = 200$ | 2016-05-25 11:36:55 | {"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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "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": 0.8724948763847351, "perplexity": 670.3019885778457}, "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-2016-22/segments/1464049274756.45/warc/CC-MAIN-20160524002114-00047-ip-10-185-217-139.ec2.internal.warc.gz"} |
http://www.agneshome.org/penn-2014/penn-2014-abstracts | Penn 2014 >
### Penn 2014 Abstracts
Donu Arapura (Purdue University) A new class of surfaces with maximal Picard number. Abstract: I will talk about some joint work with Partha Solapurkar, where we construct a class of surfaces with maximal Picard number (i.e. $\rho=h^{11}$). Although most of these examples have general type, they are built from elliptic modular surfaces. If time permits, I will discuss some related things. Video from the lectureNotes from the lecture Paolo Aluffi (Florida State University) Segre classes of monomial schemes. Abstract: Several invariants of singularities may be expressed in terms of Segre classes, a key ingredient in Fulton-MacPherson intersection theory. We will quickly review several applications of Segre classes, and present a formula computing them for schemes that are `monomial' with respect to a collection of possibly singular hypersurfaces meeting along complete intersections. The formula is expressed as a formal integral over a Newton polytope associated with the scheme. Ludmil Katzarkov (University of Miami and Universität Wien) Categorical base loci and applications. Abstract: In this talk we will introduce the notion of categorical base loci. Examples and applications will be considered. Video from the lecture, part 1Video from the lecture, part 2Notes from the lecture Chiu-Chu Melissa Liu (Columbia University) Gromov-Witten invariants of toric Calabi-Yau 3-orbifolds. Abstract: The remodeling conjecture proposed by Bouchard-Klemm-Marino-Pasquetti relates Gromov-Witten invariants of a toric Calabi-Yau 3-manifold/3-orbifold to Eynard-Orantin invariants of the mirror curve of the toric Calabi-Yau 3-fold. In this talk, I will describe results on this conjecture based on joint work with Bohan Fang and Zhengyu Zong. Video from the lecture, part 1Video from the lecture, part 2 Notes from the lecture David R. Morrison (U C Santa Barbara) Some new tricks for good ol' SL(2,Z). Madhav Nori (University of Chicago) Boundary behavior of the Teichmüller disc.Abstract: A translation surface is a pair (C,\omega) where \omega is a holomorphic differential form on a compact Riemann surface C. Every point of the moduli of translation surfaces gives rise to a Teichmuller disc: a holomorphic map from the open unit ball to this moduli space. We ask whether this disc exists on compactificationa of the moduli space.Video from the lecture, part 1Video from the lecture, part 2 Rita Pardini (Università di Pisa) Bi/trielliptic curves of genus 2 and stable Godeaux surfaces. Abstract: We describe the stable bi/trielliptic curves of genus 2, namely the stable curves $C$ of genus 2 such that there exist finite maps $f:C\to E_1$ and $g:C\to E_2$ of degrees respectively 2 and 3. We apply this result to describe one of the three possible types of Gorenstein stable non-normal Godeaux surfaces (a stable Godeaux surface is a stable surface with $K^2=\chi=1$). This is joint work with M. Franciosi and S. RollenskeAudio from the lectureSlides from the lecture. Giulia Saccà (Stony Brook University) Singularities of moduli spaces of sheaves on K3 surfaces andNakajima quivers varietiesAbstract: The aim of this talk is to study a class of singularities ofmoduli spaces of sheaves on K3 surfaces by means of Nakajima quivervarieties. The singularities in question arise from the choice of anon generic polarization, with respect to which we consider stability,and admit natural symplectic resolutions corresponding to choices ofgeneral polarizations. By establishing the stability of theLazarsfeld-Mukai bundle for some class of rank zero sheaves on a K3surface, we show that these moduli spaces are, locally around asingular point, isomorphic to a quiver variety in the sense ofNakajima and that, via this isomorphism, the natural symplecticresolutions correspond to variations of GIT quotients of the quivervarieties. This is joint work with E. Arbarello. Video from the lecture | 2018-04-25 16:24:37 | {"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": 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": 0.6961190104484558, "perplexity": 1415.905376635682}, "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-2018-17/segments/1524125947931.59/warc/CC-MAIN-20180425154752-20180425174752-00059.warc.gz"} |
http://codereview.stackexchange.com/questions/49582/treap-implementation-in-c | # Treap implementation in C
I was needed a SET-like data structure written in pure C for some university class, so I've implemented a simple one - the Treap (or cartesian tree).
Please check if everything is okay (actually, I'm not sure that there isn't a memory leak there).
cartesian_tree.h
#ifndef _CARTESIAN_TREE_H_
#define _CARTESIAN_TREE_H_
#define NODE struct Node
#define pNODE struct Node*
#define ppNODE struct Node**
/* Each node of the tree has the following structure: */
struct Node {
long key, priority; /* priority is a kind of technical' information */
char* assoc; /* pointer to a user's data (imagine MAP) */
NODE *left, *right; /* links to the left and right child */
};
/* Returns a pointer to the tree consisting only from the root */
pNODE construct_tree(void);
/* Inserts new element to the tree.
Returns 0 if there was no such element in tree and -1 otherwise
(in such case, nothing is inserted) */
char insert(pNODE, long, char*);
/* Erases a node with a particular key.
returns 0 if an element with key key' was deleted
and -1 otherwise. */
char erase(pNODE, long);
/* Returns a pointer to node with the particular key
or NULL if there is no such node. */
pNODE find(pNODE, long);
/* Clean up. One should call this function every time
when the tree isn't needed no more. */
void destruct_tree(pNODE);
#endif
cartesian_tree.c
#include <time.h>
#include <stdlib.h>
#include <limits.h>
#include "cartesian_tree.h"
pNODE construct_tree(void) {
srand(time(0)); /* treap is a randomized data structure, remember? */
pNODE root = (pNODE)malloc(sizeof(NODE));
root->key = root->priority = LONG_MIN;
root->assoc = NULL;
root->left = root->right = NULL;
return root;
}
char insert(pNODE root, long key, char* assoc) {
/* nope, Mr. Duplicate, we don't wanna see you at our party */
if(find(root, key))
return -1;
/* constructing a new node */
pNODE fresh_node = (pNODE)malloc(sizeof(NODE));
fresh_node->key = key;
fresh_node->priority = ((rand() << 15) | rand());
fresh_node->assoc = assoc;
/* searching for the proper place for new node */
ppNODE T = &(root->right);
while(1) {
/* if we're at the bottom */
if(!*T) {
*T = fresh_node;
return 0;
}
if((*T)->priority > fresh_node->priority)
break;
/* choosing the right direction */
T = (fresh_node->key < (*T)->key) ? &((*T)->left) : &((*T)->right);
}
/* placing new node to its place */
pNODE to_split = *T;
*T = fresh_node;
/* splitting a treap into two smaller treaps,
those smaller than key', and those larger than key' */
ppNODE left_subt = &((*T)->left);
ppNODE right_subt = &((*T)->right);
while(to_split) {
if(to_split->key < key) {
*left_subt = to_split;
left_subt = &((*left_subt)->right);
to_split = to_split->right;
} else {
*right_subt = to_split;
right_subt = &((*right_subt)->left);
to_split = to_split->left;
}
}
return 0;
}
char erase(pNODE root, long key) {
/* searching for the node */
pNODE pos = root;
while(pos) {
if(pos->left)
if(pos->left->key == key) {
pos = pos->left;
break;
}
if(pos->right)
if(pos->right->key == key) {
pos = pos->right;
break;
}
pos = (key < pos->key) ? pos->left : pos->right;
}
/* if there is no node with such key */
if(!pos)
return -1;
pNODE to_free = pos;
/* performing a merge' operation */
pNODE left_subt = pos->left;
pNODE right_subt = pos->right;
while(1) {
if(!left_subt || !right_subt) {
*pred_link = left_subt ? left_subt : right_subt;
break;
}
if(left_subt->priority < right_subt->priority) {
left_subt = left_subt->right;
} else {
right_subt = right_subt->left;
}
}
/* sorry friend, we don't need you anymore */
free(to_free);
return 0;
}
pNODE find(pNODE root, long key) {
pNODE T = root;
/* lookup like in ordinary BST */
while(T) {
if(key == T->key)
return T;
T = (key < T->key) ? T->left : T->right;
}
return NULL;
}
void destruct_tree(pNODE T) {
/* recursive depth-first traversal */
if(!T)
return;
if(T->left)
destruct_tree(T->left);
if(T->right)
destruct_tree(T->right);
free(T);
}
example.c (one can simply test all the above code)
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include "cartesian_tree.h"
int main(void) {
char buf[10];
pNODE T = construct_tree();
long key;
while(1) {
/* dealing with user */
scanf("%s %ld", buf, &key);
if(!insert(T, key, NULL))
puts("Inserted, OK.");
else
puts("Ooops, a duplicate detected.");
}
else if(!strcmp(buf, "remove")) {
if(!erase(T, key))
puts("Found and erased, OK.");
else
puts("No such element.");
}
else if(!strcmp(buf, "lookup")) {
if(find(T, key))
puts("Yup, found.");
else
} else {
destruct_tree(T);
puts("Bye!");
exit(0);
}
}
}
-
• It's preferred to only call srand() once in main(). It's easier to maintain in this way as it'll avoid the possibility of calling srand() multiple times, which will give you the same random value with rand() each time. It should only be called once.
• You don't need to call exit() in main() since it is supposed to return an integer value related to the program's execution outcome. Just return 0 or 1.
• The indentation in main()'s while(1) loop is a bit inconsistent. Statements under an if or else should still be indented.
I would also refactor the loop to use fewer conditional statements. It's not only difficult to follow the flow, especially with few comments, but you could also experience a reduction in performance due to branch prediction (this may not be a problem, especially with modern architecture, but it's still something worth keeping in mind).
-
In addition to the comments already given, I would give one more: I'm not a fan of the pNODE and ppNODE macros. The reason is that with the macro, pNODE a, b; declares one pointer to a node, and one node -- it's too error-prone.
If you want to use a single identifier for these types, I'd prefer a typedef. The definition of the struct and the typedef together could look something like this:
typedef struct node_ {
...;
struct node_ *left, *right;
} node;
typedef node *pnode;
typedef pnode *ppnode;
Now pnode a, b; defines two pointers to nodes.
I think there is also value in just keeping the levels of pointer indirection completely explicit, so omitting the pnode and ppnode types and just using node * and node **; but that's more a matter of personal taste.
-
Design problems:
Consider what the client may want to do if insert fails? Very likely, he'd want to inspect the data associated with the existing key, and decide if it shall stay or be replaced. In the current design, the client is forced to call find. I'd recommend to return the existing node (you know it already, so there's no additional cost incurred) along with the boolean insertion happens flag, STL style. Same way, erase shall return the pointer to an associated data: it is very well possible it was dynamically allocated, so client shall have a way to free it - again, without a redundant call to find.
-
So, should I create something like struct InsertResult and struct EraseResult? Or there is another way to return pair in C? – vortexxx192 May 13 '14 at 0:37
I'd consider bool insert(pNODE root, long key, void * data, pNODE * existing) and void * erase(pNODE root, long key). – vnp May 13 '14 at 0:44
I would move the exit call out of the loop. End the loop with a break, flag variable, or even a goto if you like and then exit the function. This also gives you a place to handle failure conditions cleanly.
In many fonts !, l, and 1 look a lot alike. I would prefer to see
if (! foo)
With strcmp I also prefer to use == 0 for the check because it lets me extend in the future. But this is definitely more personal preference.
On modern systems long is the same as int. I can see why in a class you may need to use long but in real world coding you would likely use an int unless you need to be explicit about the size and then you would use something like int16_t or int64_t.
I would consider a mkNode() function which allocated the node and set the fields to something sensible. This would get rid of some boiler plate code.
-
You can typedef long key_t and use key_t in your code instead. Makes it easier to change later - either statically or e.g. configuration-specifically at compile-time. – CompuChip May 13 '14 at 9:44
Excellent point @CompuChip. – Sean Perry May 13 '14 at 17:14
Bugs:
• The root node always has key=LONG_MIN. What happens if someone tries to use that as a key?
• scanf("%s") is not safe. If the user types more than 9 characters, it will overflow the buffer and probably crash. Use fgets(buf, sizeof(buf), stdin) instead. (There's almost never a good reason to use scanf.)
Refactoring:
• find, insert, and erase have very similar lookup loops. These could be factored out as a function which finds the link under which a node either will be inserted.
• The sections about splitting and merging could be extracted as separate functions.
• The section that constructs a new node could also be a separate function.
• The if(T->left) checks in destruct_tree are unnecessary. Since it's safe on NULL (as it should be), you can just blindly recurse on both children.
Style:
• As Erik said, the pNODE macros are error-prone; use typedef instead. Or, better, do without them — Node *x is clearer than pNODE x.
• The conventional way to do the NODE typedef is to wrap it around the struct definition:
typedef struct Node {
//...
struct Node *left, *right;
} Node;
• destruct_tree would usually be called something like free_tree or delete_tree.
• Using a special Node as the root is confusing, because it doesn't act like a normal node. (And this causes the first bug above.) There should be comments about this, or the root should be a separate type: typedef struct { Node *root; } Treap;`
- | 2016-02-13 17:16:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.2439499944448471, "perplexity": 6910.836720439409}, "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/1454701167113.3/warc/CC-MAIN-20160205193927-00195-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/178148/finding-the-all-triples?answertab=active | # Finding the all triples
How to find the all positive integer triples such that :
$$ab+c=\gcd (a^2,b^2)+\gcd(a,bc)+\gcd(b,ac)+\gcd(c,ab)=239^2$$
-
Let $x=\gcd(a,b,c)$. But $\gcd(a^2,b^2)\ge x^2$ and $x$ divides $239^2$, this is impossible due to the equality and 239 is prime. Hence $x=1$.
Let $y=\gcd(c,ab)$. if $y>1$, then once again $y=239$ ($239^2$ would be too big). Suppose $y$ divides $a$ (the other case is symmetric), $y$ does not divide $b$ ($\gcd(a^2,b^2)$ would be at least $239^2$, too big). Hence $\gcd(c,ab)=y$, $\gcd(a,bc)=y$, and $\gcd(a,bc)=\gcd(a,b)=z$, so $$z^2+z+2y=y^2$$ So $y$ divides $z$ (or $z+1$), this is not possible, so $y=1$
So $c$ is prime with $a$ and $b$ and the equation is $$ab+c=\gcd(a,b)^2+2\gcd(a,b)+1=(\gcd(a,b)+1)^2=239^2$$ $$\gcd(a,b)=238$$
The only small enough possibility is $a=b=238$ and then $c=477$ | 2016-04-30 09:03:50 | {"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": 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": 0.9831011891365051, "perplexity": 110.99317384308053}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860111809.10/warc/CC-MAIN-20160428161511-00118-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://eccc.weizmann.ac.il/keyword/17116/ | Under the auspices of the Computational Complexity Foundation (CCF)
REPORTS > KEYWORD > MISTAKE-BOUND:
Reports tagged with mistake-bound:
TR09-060 | 4th June 2009
Harry Buhrman, David García Soriano, Arie Matsliah
#### Learning parities in the mistake-bound model.
We study the problem of learning parity functions that depend on at most $k$ variables ($k$-parities) attribute-efficiently in the mistake-bound model.
We design simple, deterministic, polynomial-time algorithms for learning $k$-parities with mistake bound $O(n^{1-\frac{c}{k}})$, for any constant $c > 0$. These are the first polynomial-time algorithms that learn $\omega(1)$-parities in ... more >>>
ISSN 1433-8092 | Imprint | 2018-03-21 08:44:39 | {"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": 0.6781290769577026, "perplexity": 6356.657798259055}, "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-2018-13/segments/1521257647600.49/warc/CC-MAIN-20180321082653-20180321102653-00288.warc.gz"} |
https://aimsciences.org/article/doi/10.3934/dcds.2003.9.877 | # American Institute of Mathematical Sciences
July 2003, 9(4): 877-900. doi: 10.3934/dcds.2003.9.877
## Simple umbilic points on surfaces immersed in $\R^4$
1 ICMC-USP, São Carlos, Caixa Postal 668, CEP 13560-970, São Carlos, SP, Brazil 2 Universidad de Santiago de Chile, Departamento de Matemática y C.C., Casilla 307, Correo 2, Santiago, Chile
Received November 2001 Revised December 2002 Published April 2003
We study local problems around simple umbilic points of surfaces immersed in $\mathbb R^4$ such as finite determinacy and versal unfoldings.
Citation: Carlos Gutierrez, Víctor Guíñez. Simple umbilic points on surfaces immersed in $\R^4$. Discrete & Continuous Dynamical Systems - A, 2003, 9 (4) : 877-900. doi: 10.3934/dcds.2003.9.877
[1] Carlos Gutierrez, Víctor Guíñez, Alvaro Castañeda. Quartic differential forms and transversal nets with singularities. Discrete & Continuous Dynamical Systems - A, 2010, 26 (1) : 225-249. doi: 10.3934/dcds.2010.26.225 [2] Daniel T. Wise. Nonpositive immersions, sectional curvature, and subgroup properties. Electronic Research Announcements, 2003, 9: 1-9. [3] Jorge Sotomayor, Ronaldo Garcia. Codimension two umbilic points on surfaces immersed in $R^3$. Discrete & Continuous Dynamical Systems - A, 2007, 17 (2) : 293-308. doi: 10.3934/dcds.2007.17.293 [4] Tetsuya Ishiwata. On the motion of polygonal curves with asymptotic lines by crystalline curvature flow with bulk effect. Discrete & Continuous Dynamical Systems - S, 2011, 4 (4) : 865-873. doi: 10.3934/dcdss.2011.4.865 [5] R.D.S. Oliveira, F. Tari. On pairs of differential $1$-forms in the plane. Discrete & Continuous Dynamical Systems - A, 2000, 6 (3) : 519-536. doi: 10.3934/dcds.2000.6.519 [6] P. De Maesschalck. Gevrey normal forms for nilpotent contact points of order two. Discrete & Continuous Dynamical Systems - A, 2014, 34 (2) : 677-688. doi: 10.3934/dcds.2014.34.677 [7] Vincent Naudot, Jiazhong Yang. Finite smooth normal forms and integrability of local families of vector fields. Discrete & Continuous Dynamical Systems - S, 2010, 3 (4) : 667-682. doi: 10.3934/dcdss.2010.3.667 [8] Holger Heumann, Ralf Hiptmair, Cecilia Pagliantini. Stabilized Galerkin for transient advection of differential forms. Discrete & Continuous Dynamical Systems - S, 2016, 9 (1) : 185-214. doi: 10.3934/dcdss.2016.9.185 [9] Akhtam Dzhalilov, Isabelle Liousse, Dieter Mayer. Singular measures of piecewise smooth circle homeomorphisms with two break points. Discrete & Continuous Dynamical Systems - A, 2009, 24 (2) : 381-403. doi: 10.3934/dcds.2009.24.381 [10] Jiyoung Han, Seonhee Lim, Keivan Mallahi-Karai. Asymptotic distribution of values of isotropic here quadratic forms at S-integral points. Journal of Modern Dynamics, 2017, 11: 501-550. doi: 10.3934/jmd.2017020 [11] Weigu Li, Jaume Llibre, Hao Wu. Polynomial and linearized normal forms for almost periodic differential systems. Discrete & Continuous Dynamical Systems - A, 2016, 36 (1) : 345-360. doi: 10.3934/dcds.2016.36.345 [12] Holger Heumann, Ralf Hiptmair. Eulerian and semi-Lagrangian methods for convection-diffusion for differential forms. Discrete & Continuous Dynamical Systems - A, 2011, 29 (4) : 1471-1495. doi: 10.3934/dcds.2011.29.1471 [13] Dorina Mitrea and Marius Mitrea. Boundary integral methods for harmonic differential forms in Lipschitz domains. Electronic Research Announcements, 1996, 2: 92-97. [14] Dorina Mitrea, Irina Mitrea, Marius Mitrea, Lixin Yan. Coercive energy estimates for differential forms in semi-convex domains. Communications on Pure & Applied Analysis, 2010, 9 (4) : 987-1010. doi: 10.3934/cpaa.2010.9.987 [15] Paul Bracken. Connections of zero curvature and applications to nonlinear partial differential equations. Discrete & Continuous Dynamical Systems - S, 2014, 7 (6) : 1165-1179. doi: 10.3934/dcdss.2014.7.1165 [16] Dimitra Antonopoulou, Georgia Karali. A nonlinear partial differential equation for the volume preserving mean curvature flow. Networks & Heterogeneous Media, 2013, 8 (1) : 9-22. doi: 10.3934/nhm.2013.8.9 [17] Isaac A. García, Jaume Giné, Susanna Maza. Linearization of smooth planar vector fields around singular points via commuting flows. Communications on Pure & Applied Analysis, 2008, 7 (6) : 1415-1428. doi: 10.3934/cpaa.2008.7.1415 [18] Grzegorz Graff, Jerzy Jezierski. Minimization of the number of periodic points for smooth self-maps of closed simply-connected 4-manifolds. Conference Publications, 2011, 2011 (Special) : 523-532. doi: 10.3934/proc.2011.2011.523 [19] Jihua Yang, Liqin Zhao. Limit cycle bifurcations for piecewise smooth integrable differential systems. Discrete & Continuous Dynamical Systems - B, 2017, 22 (6) : 2417-2425. doi: 10.3934/dcdsb.2017123 [20] Hebai Chen, Jaume Llibre, Yilei Tang. Centers of discontinuous piecewise smooth quasi–homogeneous polynomial differential systems. Discrete & Continuous Dynamical Systems - B, 2019, 24 (12) : 6495-6509. doi: 10.3934/dcdsb.2019150
2018 Impact Factor: 1.143 | 2019-09-18 02:49:30 | {"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": 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": 0.5670287013053894, "perplexity": 7422.423007744963}, "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/1568514573176.51/warc/CC-MAIN-20190918024332-20190918050332-00174.warc.gz"} |
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# Solve the quadratic equation $2x^2-x^2=-1\cdot 1^2+x^2$
Go!
Go!
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## Step-by-step Solution
Problem to solve:
$2x^2-x^2=-1^2+x^2$
Specify the solving method
1
Calculate the power $1^2$
$2x^2-x^2=-1\cdot 1+x^2$
Learn how to solve quadratic equations problems step by step online.
$2x^2-x^2=-1\cdot 1+x^2$
Learn how to solve quadratic equations problems step by step online. Solve the quadratic equation 2x^2-x^2=-1^2+x^2. Calculate the power 1^2. Multiply -1 times 1. Combining like terms 2x^2 and -x^2. Rearrange the equation.
$2x^2-x^2=-1^2+x^2$ | 2023-03-23 23:55:22 | {"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": 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": 0.22462861239910126, "perplexity": 5097.679087647573}, "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/1679296945218.30/warc/CC-MAIN-20230323225049-20230324015049-00490.warc.gz"} |