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https://socratic.org/questions/5995250611ef6b429fd6779c | Chemistry
Topics
# What is the concentration of hydroxyl ions in a solution of pH 8 ?
Feb 5, 2018
${10}^{- 6}$ mol per litre
#### Explanation:
$p H$ is $- {\log}_{10}$ of the hydrogen ion concentration (actually hydronium ${H}_{3} {O}^{+}$ concentration).
$p O H$ is $- {\log}_{10}$ of the hydroxyl ($O {H}^{-}$) ion concentration.
To say $p H + p O H = 14$ is to say that the product of the ${H}_{3} {O}^{+}$ and $O {H}^{-}$ ion concentrations is ${10}^{-} 14$.
If a solution has $p H = 8$, then its $p O H$ is $14 - 8 = 6$ and the corresponding concentration of hydroxyl $O {H}^{-}$ ions is ${10}^{- 6}$ mol per litre.
##### Impact of this question
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https://www.imsi.institute/videos/interleaving-by-parts-for-persistence-in-a-poset/ | This was part of Topological Data Analysis
## Interleaving by parts for persistence in a poset
Woojin Kim, Duke University
Tuesday, April 27, 2021
Abstract: Metrics in computational topology are often either (i) themselves in the form of the interleaving distance d(F,G) between certain order-preserving maps F and G between posets or (ii) admit d(F,G) as a tractable lower bound. In this talk, assuming that the target poset of F and G admits a join-dense subset, we propose certain join representations of F and G which facilitate the computation of d(F,G). We leverage this result in order to (i) elucidate the structure and computational complexity of the interleaving distance for poset-indexed clusterings (i.e. poset-indexed subpartition-valued functors), (ii) to clarify the relationship between the erosion distance by Patel and the graded rank function by Betthauser, Bubenik, and Edwards, and (iii) to reformulate and generalize the tripod distance by the second author. This is joint work with Facundo Memoli and Anastasios Stefanou. https://arxiv.org/abs/1912.04366 | {"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": 0.849931538105011, "perplexity": 1991.5660976853123}, "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/1656104054564.59/warc/CC-MAIN-20220702101738-20220702131738-00048.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/20031-what-kind-problem-print.html | # What kind of a problem is this?
• Oct 5th 2007, 08:36 AM
alexthedon
What kind of a problem is this?
I have the following problem to solve but don't recognise what type of problem this actually is let alone how to solve it:
Determine those values of k for which
| k k | = 0
| 8 4k |
the | are supposed to be continuous lines.
Is this comething to do with eigenvectors/eigenvalues? or maybe complex numbers?
Cheers,
Alex.
• Oct 5th 2007, 08:43 AM
Plato
Given the matrix $\left( {\begin{array}{cc} a & b \\ c & d \\\end{array}} \right)$ then its determinant is $\left| {\begin{array}{cc} a & b \\ c & d \\
\end{array}} \right| = ad - bc$ | {"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": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9232490658760071, "perplexity": 1356.3427695375522}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738660350.13/warc/CC-MAIN-20160924173740-00069-ip-10-143-35-109.ec2.internal.warc.gz"} |
http://physics.stackexchange.com/questions/907/type-of-stationary-point-in-hamiltons-principle/955 | # Type of stationary point in Hamilton's principle
In this question it is discussed why by Hamilton's principle the action integral must be stationary. Most examples deal with the case that the action integral is minimal: this makes sense - we all follow the path with the least resistance.
However, a stationary point can be a maximal or minimal extremum or even a point of inflexion (rising or falling). Can anybody give examples from theory or practice where the action integral takes on a maximal extremum or a point of inflexion?
-
The (timelike) geodesics in space-time (GR) are precisely the curves that maximize the proper time. Also, it depends on your definition: If you change the sign of the action (just changing your convention), what is a minimum turns out now to be a maximum – Ronaldo Nov 16 '10 at 13:17
Does Fermat's principle count? en.wikipedia.org/wiki/Fermat%27s_principle#Modern_version – KennyTM Nov 16 '10 at 13:50
If you look at (anti-)self-duality, you get exactly inflection points.
Also, you can think in terms of the WKB-approximation of Feynman Path Integrals, and classify your extrema accordingly. Behind this, there's a whole story about Morse Theory and manifold classification/decomposition, etc... but, the bottom line is that you can think of these WKBs of the Path Integral as the tool to classify the extrema of your Action.
-
If you have in mind a mechanical system and if you are really referring to the Hamilton's principle (which tells how to find the actual trajectory the system follows between the fixed initial and final times at fixed initial conditions), then I doubt you can find such an example. The reason is that a mechanical system contains the kinetic term in its lagrangian, which is unbounded from above. That is, you can always construct a trajectory $\{x(t), {\dot x}(t)\}$ with as large action as you wish; so there is no maximum, not even a local one.
In his answer above, Mark refers to an optical example that sort of maximizes, not minimizes the action. The example can be reformulated in mechanical form using small balls rolling on a surface instead of light rays, but I don't think it really refers to the Hamilton's principle. In this example you send light rays (or small balls) from a fixed initial point into different directions and search for caustics in balls' trajectories. That is, you vary the initial condition (the direction of initial velocity) and at the same time you assume that each trajectory (for each choice of the velocity) is already known. That's not what the Hamilton's principle is about (there you fix all the initial conditions and solve for the trajectory).
- | {"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": 0.9643988609313965, "perplexity": 392.39287322236885}, "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-2014-10/segments/1393999673298/warc/CC-MAIN-20140305060753-00085-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/107735/valid-use-of-laplaces-method?sort=oldest | # Valid use of Laplace's method?
I am trying to say something about the asymptotics of $$\int_{\mathbb{R}} e^{cx - x^{4/3}}dx$$ as $c \to +\infty$, and need a sanity check. As I understand it, Laplace's method is to write $$q(x) = x-c^{-1}x^{4/3}$$ and note that $q(x)$ has a global maximum at $x_{0} = \frac{27c^{3}}{81}$. Then it follows $$\int_{\mathbb{R}}e^{cq(x)}dx \lesssim e^{c q(x_{0}})\int_{\mathbb{R}} e^{-c|q''(x_{0})|(x-x_{0})^{2}/2}dx$$ And because the right hand integral is a Gaussian, in fact $$\int_{\mathbb{R}}e^{cq(x)}dx \lesssim e^{c q(x_{0})}\sqrt{\frac{2\pi}{c|q''(x_{0})|}}$$ So explicitly in my case $$\int_{\mathbb{R}} e^{cx - x^{4/3}}dx \lesssim \frac{9\sqrt{2\pi}}{8}ce^{\frac{27}{256}c^{4}}$$ My concern is about $q''(x) = -\frac{4}{9c}x^{-2/3}$. The proofs of Laplace's method I have seen only APPEAR to require $q''(x)$ be continuous in a neighborhood of $x_{0}$, with $q''(x_{0}) < 0$. This will certainly hold for me when $c > 0$. But am I missing something?
-
I've used Laplace's Method on a single function $q$ rather than a family varying with $c$. If you let $q$ depend on $c$, don't you need some sort of uniform bound on the rest of the function that is automatic when $q$ doesn't depend on $c$? By the way, I think that should be $\exp(\frac{27}{256}c^4)$ in the last line, as $\exp(cq(x_0)) \ne\exp(cx_0).$ – Douglas Zare Sep 21 '12 at 10:54
You can transform this to a problem with a fixed function q(x) by first substituting $x=c^3y$ and then applying Laplace's method to the transformed problem. – Michael Renardy Sep 21 '12 at 14:19
@Douglas - corrected, thank you! @Michael - true. I just ordered Bleistein and Handelsman's Asymptotic Expansions of Integrals, which will hopefully lead to a fuller understanding of this technique. – Michael Tinker Sep 21 '12 at 17:38
@Douglas - the issue of uniformity bothers me a lot as well, since I also need to estimate things like $\int_{\mathbb{R}}e^{\eta x -\sigma x^{2} - \tau x^{4}}dx$ where $\eta,\sigma \in \mathbb{R}$ and $\tau \in \mathbb{R}^{+}$. – Michael Tinker Sep 21 '12 at 17:52
In addition to @Michael's comment, this seems to be a standard Watson's lemma integral. – Igor Rivin Oct 6 '12 at 1:16
The version of Laplace's Method I know uses that $q(x)$ does not depend on $c$. If you try to extend it, you need some extra uniformity condition to show that the rest of the function does not contribute. Here is a counterexample when you only assume that the maximum is always at $x_0$ and the second derivative is constant there.
Let the interval be $[-1,1]$ and let $q(x,c) = \max (-x^2,-1/c^2).$ This means $q(x,c) = -x^2$ in a neighborhood of $0$ whose width depends on $c$, $[-1/c, 1/c]$ and it is flat outside. Let $f(x,c) = c q(x,c).$
For each $c$, $q(x,c)$ has a global maximum at $x=0$ and the second derivative with respect to x there is -2. However,
$$\int_{-1}^1 e^{f(x,c)} dx ~\large{\nsim} ~ e^{c q(x_0,c)} \sqrt{\frac{2\pi}{c|q''(x_0,c)|}}= \sqrt{\frac{\pi}{c}}.$$
In fact,
$$\lim_{c \to \infty}\int_{-1}^1 e^{f(x,c)} dx = 2.$$
The part of $\exp(f(x,c))$ near $0$ doesn't dominate the integral.
-
@Douglas - appreciated. My professor, when recommending Bleistein's book, said he had vague memories of such a uniformity condition. (I.e. a condition for when $\int_{\mathbb{R}}e^{f(x,c)}dx$ admits an asymptotic estimate in the parameter $c$.) In some days or weeks I will report back with a verdict on the accuracy of his recollection. :) – Michael Tinker Sep 22 '12 at 6:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 0.9191478490829468, "perplexity": 190.39512713492596}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644064590.32/warc/CC-MAIN-20150827025424-00294-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://math.answers.com/Q/What_is_the_equivalent_fraction_for_1824 | 0
# What is the equivalent fraction for 1824?
Wiki User
2017-11-28 17:57:45
1824 is an integer and so there is no sensible way of writing it as a fraction or mixed number.
Wiki User
2017-12-24 11:27:24
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151 Reviews | {"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": 0.8327124714851379, "perplexity": 4006.7682466354786}, "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/1679296949355.52/warc/CC-MAIN-20230330163823-20230330193823-00362.warc.gz"} |
https://mathoverflow.net/questions/157423/blowing-up-rational-singularities | # Blowing up rational singularities
Let $X$ be a projective surface embedded into $\mathbb{P}^n_{\mathbb{C}}$ having at most rational singularities. Let $\tilde{X} \to X$ be the minimal resolution of $X$. Is it possible to embed $\tilde{X}$ into $\mathbb{P}^n$ for the same integer $n$? If not, is there any additional condition we can impose on $X$ so that this is possible?
Certainly not. The simplest example is the quadratic cone in $\mathbb{P}^3$; the minimal resolution is the surface $\mathbb{F}_2$, which is not isomorphic to a smooth surface in $\mathbb{P}^3$. The same can be done with any rational double point. On the other hand any smooth surface can be embedded in $\mathbb{P}^5$, so the answer is positive for $n\geq 5$. But I do not see any relation between embeddings of $X$ and those of $\tilde{X}$. | {"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": 0.9797536134719849, "perplexity": 46.099342065538316}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107909746.93/warc/CC-MAIN-20201030063319-20201030093319-00614.warc.gz"} |
http://tex.stackexchange.com/questions/103469/defining-nextslide-in-overprint-environment-in-beamer | # Defining \nextslide in overprint environment in beamer
We all use the \onslide command inside the overprint environment in beamer. However, I find that assigning hard coded slide values, like \onslide<5> in this environment is really cumbersome. This situation becomes difficult whenever we have to insert a new slide in the middle of twenty some overlay slides. We have to manually renumber every subsequent slide after a new slide is inserted. That is why, I tried to define a \nextslide command in beamer.
The idea is very simple, the \nextslide command will increment a predefined counter and will use this value with the \onslide command.
\documentclass{beamer}
\newcounter{slidecounter}
\setcounter{slidecounter}{0}
\begin{document}
\begin{frame}{Title}
\begin{overprint}
\nextslide
On one
\nextslide
On two
\end{overprint}
\end{frame}
\end{document}
I understand that I still need to implement resetting the counter at the start of overprint environment once I am successful with this part.
However, whenever I run the code with pdflatex, it keeps on generating page after page, which I have to interrupt with Ctrl-C.
[1{/var/lib/texmf/fonts/map/pdftex/updmap/pdftex.map}] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31] [32] [33] [34] [35] [36] [37] [38] [39] [40] [41] [42] [43] [44] [45] [46] [47] [48] [49] [50] [51] [52] [53] [54] [55] [56] [57] [58] [59] [60] [61] [62] [63] [64] [65] [66] [67] [68] [69] [70] [71] [72] [73] [74]^C ! Interruption. \pgfsysprotocol@literalbuffered ...tocol@temp {{#1 \space }}\expandafter \pgf... l.20 \end{frame} ? x
The beamer code is too myriad for me to find out where this infinite loop is happening.
Any suggestion will be appreciated.
-
That's what the + overlay specification is made for (cf. the beamer user guide, section 9.6.4):
The effect of the +-sign is the following: You can use it in any overlay specification at any point where you would usually use a number. If a +-sign is encountered, it is replaced by the current value of the LaTeX counter beamerpauses, which is 1 at the beginning of the frame. Then the counter is increased by 1, though it is only increased once for every overlay specification, even if the specification contains multiple +-signs (they are replaced by the same number).
So your MWE can be solved with
\documentclass{beamer}
\begin{document}
\begin{frame}{Title}
\begin{overprint}
\onslide<+>
On one
\onslide<+>
On two
\onslide<+>
On three
\end{overprint}
\end{frame}
\end{document}
-
Oh, goodness, I feel like a fool. Still, I would like to know what caused the infinite loop. – Masroor Mar 20 '13 at 17:22
@MMA The problem is that the slide is compiled each time an overlay is typeset, so slidecounter is increased by 2 and such produces new overlays: For example, on the second overlay, \onslide<3> and \onslide<4> are executed, such creating two new pages etc. If you use \setcounter{slidecounter}{0} at the beginning of the frame, this won't happen and your MWE will produce the expected result. – diabonas Mar 20 '13 at 17:36 | {"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": 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": 0.849765956401825, "perplexity": 4035.320013288296}, "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-2014-42/segments/1414119645329.44/warc/CC-MAIN-20141024030045-00001-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://academic.oup.com/acn/article-abstract/10/5/463/1516/Construct-validity-of-the-WCST-in-normal-elderly | Abstract
Separate principle component analyses of the WCST were conducted on 187 normal elderly and 181 persons with Parkinson's disease (PD). Adequate construct validity was found for “conceptualization/problem solving” and “failure to maintain set” factors in both groups. Perseverative and nonperseverative errors were related for the normal group, but not for the PD subjects. This may reflect the frontal systems deficit observed in PD. Additional principle component analyses were conducted with the WCST and measures of memory and attention. In neither the normal nor the PD group did the WCST significantly load with the memory and attention measures. This suggests that the WCST provides information about problem solving relatively independent of memory and attention functions for elderly persons. | {"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": 0.8901035189628601, "perplexity": 4655.797158224436}, "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-2017-04/segments/1484560281263.12/warc/CC-MAIN-20170116095121-00056-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://philpapers.org/s/Gabor%20Hofer-Szabo | ## Works by Gabor Hofer-Szabo
14 found
Order:
1. In the paper it will be shown that Reichenbach’s Weak Common Cause Principle is not valid in algebraic quantum field theory with locally finite degrees of freedom in general. Namely, for any pair of projections A, B supported in spacelike separated double cones ${\mathcal{O}}_{a}$ and ${\mathcal{O}}_{b}$ , respectively, a correlating state can be given for which there is no nontrivial common cause (system) located in the union of the backward light cones of ${\mathcal{O}}_{a}$ and ${\mathcal{O}}_{b}$ and commuting with the both (...)
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2. This paper is the philosopher-friendly version of our more technical work. It aims to give a clear-cut definition of Bell's notion of local causality. Having provided a framework, called local physical theory, which integrates probabilistic and spatiotemporal concepts, we formulate the notion of local causality and relate it to other locality and causality concepts. Then we compare Bell's local causality with Reichenbach's Common Cause Principle and relate both to the Bell inequalities. We find a nice parallelism: both local causality and (...)
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3. Gábor Hofer-Szabó & Péter Vecsernyés (2013). Bell Inequality and Common Causal Explanation in Algebraic Quantum Field Theory. Studies in History and Philosophy of Science Part B: Studies in History and Philosophy of Modern Physics 44 (4):404-416.
Bell inequalities, understood as constraints between classical conditional probabilities, can be derived from a set of assumptions representing a common causal explanation of classical correlations. A similar derivation, however, is not known for Bell inequalities in algebraic quantum field theories establishing constraints for the expectation of specific linear combinations of projections in a quantum state. In the paper we address the question as to whether a ‘common causal justification’ of these non-classical Bell inequalities is possible. We will show that although (...)
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4. The aim of this paper is to give a sharp definition of Bell's notion of local causality. To this end, first we unfold a framework, called local physical theory, integrating probabilistic and spatiotemporal concepts. Formulating local causality within this framework and classifying local physical theories by whether they obey local primitive causality---a property rendering the dynamics of the theory causal, we then investigate what is needed for a local physical theory, with or without local primitive causality, to be locally causal. (...)
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5. Gábor Hofer-Szabó (2015). Relating Bell’s Local Causality to the Causal Markov Condition. Foundations of Physics 45 (9):1110-1136.
The aim of the paper is to relate Bell’s notion of local causality to the Causal Markov Condition. To this end, first a framework, called local physical theory, will be introduced integrating spatiotemporal and probabilistic entities and the notions of local causality and Markovity will be defined. Then, illustrated in a simple stochastic model, it will be shown how a discrete local physical theory transforms into a Bayesian network and how the Causal Markov Condition arises as a special case of (...)
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6. Gábor Hofer-Szabó & Miklós Rédei (2006). Reichenbachian Common Cause Systems of Arbitrary Finite Size Exist. Foundations of Physics 36 (5):745-756.
A partition $\{C_i\}_{i\in I}$ of a Boolean algebra Ω in a probability measure space (Ω, p) is called a Reichenbachian common cause system for the correlation between a pair A,B of events in Ω if any two elements in the partition behave like a Reichenbachian common cause and its complement; the cardinality of the index set I is called the size of the common cause system. It is shown that given any non-strict correlation in (...)
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7. Gábor Hofer-Szabó, Miklós Rédei & László E. Szabó (2002). Common-Causes Are Not Common Common-Causes. Philosophy of Science 69 (4):623-636.
A condition is formulated in terms of the probabilities of two pairs of correlated events in a classical probability space which is necessary for the two correlations to have a single (Reichenbachian) common-cause and it is shown that there exists pairs of correlated events probabilities of which violate the necessary condition. It is concluded that different correlations do not in general have a common common-cause. It is also shown that this conclusion remains valid even if one weakens slightly Reichenbach's definition (...)
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8. A partition $\{C_i\}_{i\in I}$ of a Boolean algebra $\cS$ in a probability measure space $(\cS,p)$ is called a Reichenbachian common cause system for the correlated pair $A,B$ of events in $\cS$ if any two elements in the partition behave like a Reichenbachian common cause and its complement, the cardinality of the index set $I$ is called the size of the common cause system. It is shown that given any correlation in $(\cS,p)$, and given any finite size $n>2$, the probability space (...)
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9. The Borel-Kolmogorov Paradox is typically taken to highlight a tension between our intuition that certain conditional probabilities with respect to probability zero conditioning events are well defined and the mathematical definition of conditional probability by Bayes' formula, which looses its meaning when the conditioning event has probability zero. We argue in this paper that the theory of conditional expectations is the proper mathematical device to conditionalize, and this theory allows conditionalization with respect to probability zero events. The conditional probabilities on (...)
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10. Gabor Hofer-Szabo, Miklos Redei & Laszlo E. Szabo (2002). Common-Causes Are Not Common Common-Causes. Philosophy of Science 69 (4):623-636.
A condition is formulated in terms of the probabilities of two pairs of correlated events in a classical probability space which is necessary for the two correlations to have a single (Reichenbachian) common-cause and it is shown that there exists pairs of correlated events probabilities of which violate the necessary condition. It is concluded that different correlations do not in general have a common common-cause. It is also shown that this conclusion remains valid even if one weakens slightly Reichenbach's definition (...)
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11. Gábor Hofer-Szabó (2015). On the Relation Between the Probabilistic Characterization of the Common Cause and Bell׳s Notion of Local Causality. Studies in History and Philosophy of Science Part B: Studies in History and Philosophy of Modern Physics 49:32-41.
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12. Gábor Hofer-Szabó (2007). Separate- Versus Common -Common-Cause-Type Derivations of the Bell Inequalities. Synthese 163 (2):199 - 215.
Standard derivations of the Bell inequalities assume a common common cause system that is a common screener-off for all correlations and some additional assumptions concerning locality and no-conspiracy. In a recent paper (Grasshoff et al., 2005) Bell inequalities have been derived via separate common causes assuming perfect correlations between the events. In the paper it will be shown that the assumptions of this separate-common-cause-type derivation of the Bell inequalities in the case of perfect correlations can be reduced to the assumptions (...)
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13. This paper aims to implement Bell’s notion of local causality into a framework, called local physical theory, which is general enough to integrate both probabilistic and spatiotemporal concepts and also classical and quantum theories. Bell’s original idea of local causality will then arise as the classical case of our definition. First, we investigate what is needed for a local physical theory to be locally causal. Then we compare local causality with Reichenbach’s common cause principle and relate both to the Bell (...)
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# Basic Algebra - Simplifying radicals
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See the attached file.
Problem 66.
y^1/3 y^1/3
Now we have to simplify the problem. Looking at the problem and identifying what every part of the problem is we get that "y" is the principal root and it is being raised to an nth root, the 1 in the exponent is the power and the 3 in the exponent is the root. So for my first problem I will have to use the product rule to add the exponents and then I would simplify for a final solution.
= y^2/3 Final simplified solution
Problem 84.
9^-1* 9^1/2
The second problem is a little different that the first one. For this one we will use the quotient rule. The 9 is the principle root, the 1 is the power and the 2 is the root. The negative symbol represents the reciprocal to get the negative away and make it a positive number we will make
9^-1 = 1/9 To continue with the problem now we have changed it too
1/9 * 9^1/2 Since it is a ½ we will look for the sqrt of 9.
1/9 * √9 = 1/9 * 3/1 Use cross multiply and result is 3/9
3/9 = 1/3 Simplified is the final answer.
That concludes my problems for week 3.
References
Dugopolski, M. (2012). Elementary and intermediate algebra (4th ed.). New York, NY: McGraw-Hill Publishing.
Simplifying Expressions Using the Rules of Exponentials
The following are the rules of exponential (Clark & Anfinson, 2011).
Rule one: To multiply expressions with bases, which are identical, add their exponents. This is known as the product rule.
Rule two: To divide expressions with identical bases, subtract their exponents. This is referred to Quotient rule.
Rule three: multiply the exponents for expressions with one base, but with two or more exponents.
To simplify the two expressions given, the above rules of exponentials shall be applied as illustrated below;
Number 42; 〖27〗^(〖-2〗_3 )/〖27〗^(〖-1〗_3 )
Solution
To divide identical bases, the rule of exponent stipulates that we subtract the respective exponents. Therefore, the above fraction will be simplified to;
=〖27〗^((-2)⁄3+1⁄3) Solving the exponential part of the fraction we obtain;
= -2/3+1/3=-1/3 Substituting the solution ( - 1/3) in the above equation we obtain〖27〗^((-1)⁄3). The reciprocal of this can be written as1/〖27〗^(1⁄3) .This is equivalent to finding the cube root of 1/27 i.e.( 1/(∛27) ). This will give;
1/(∛27)= 1/3
Therefore, the answer will be1⁄3, which is the principal cube root of 1⁄27. In this case, 3 is also the 3rd root of 27.
No. 101; (a^(1_2 ) b)^(1_2 ) (〖ab〗^(1⁄2) )
Solution
Applying rule three, the above expression can be simplified to;
a^(1⁄4) b^(1⁄2) ab^(1⁄2)
Rearranging the above equation we obtain;
a^(1⁄4) 〖ab〗^(1⁄2) b^(1⁄2)
Applying rule one, we obtain;
a^(1_4 ) +1b^(1_2+1_2 )
Simplifying the exponentials we obtain;
1/4 +1= ( 1+4)/4 = 5/4 and
1/2+1/2=2/2 = 1
Substituting in the above equation we obtain;
a^(5⁄4) b
Therefore the answer will be a^(5⁄4) b.
#### Solution Preview
See attachment.
Problem 66.
y^1/3 y^1/3
Now we have to simplify the problem. Looking at the problem and identifying what every part of the problem is we get that "y" is the principal root and it is being raised to an nth root, the 1 in the exponent is the power and the 3 in the exponent is the root. So for my first problem I will have to use the product rule to add the exponents and then I would simplify for a final solution.
= y^2/3 Final simplified solution
Problem 84.
9^-1* 9^1/2
The second problem is a little different that the first one. For this one we will use the quotient rule. The 9 is the principle root, the 1 is the power and the 2 is the root. The negative symbol represents the reciprocal to get the negative away and make it a ...
#### Solution Summary
The solution assists with simplifying radicals in basic algebra.
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https://www.physicsforums.com/threads/topology-proof.354912/ | # Topology Proof
1. Nov 15, 2009
1. The problem statement, all variables and given/known data
Prove that if X is compact and Y is Hausdorff then a continuous bijection $$f: X \longrightarrow Y$$ is a homeomorphism. (You may assume that a closed subspace of a compact space is compact, and that an identification space of a compact space is compact).
2. Relevant equations
A space X is compact if every open cover {$$\left{ U_{\lambda} | \lambda \in \Lambda \right}$$} , $$\cup U_{\lambda} = X$$.
A space is Hausdorff if for any pair of distinct points x,y in Y, there exist open sets separating them.
3. The attempt at a solution
Going on the hint of what I can assume, I define the equivalence class x~x' if f(x) = f(x'), with projection p(x) = [x]. Then f = gp: X -> X/~ -> Y. It may be easier to prove p and g are homeomorphisms than f. The identification space is compact, and so is any closed subset. We are given that f is a continuous bijection, so only need to show that f inverse is continuous, ie f(U) is open for any U open, or perhaps more easily that f(C) is closed for C closed in this case. A compact subset of a Hausdorff space is closed, so if I can prove that p maps a compact space to a compact space then I think I've proved p is a homeomorphism.
Then I would need to prove g.
Last edited: Nov 15, 2009
2. Nov 15, 2009
### rasmhop
I must admit that I don't really see why you would use the identification space. I don't think that your identification will help much because f is injective so f(x)=f(x') imply x=x' which means that ~ is just equality and therefore X is naturally homeomorphic to X/~ by the correspondence $x \leftrightarrow \{x\}$. Thus proving g a homeomorphism is pretty much the same as proving f a homeomorphism.
Anyway the way I would do it without identification spaces is to realize that a continuous function maps compact subspaces to compact subspaces in general (either refer to your book or show it directly by considering a covering of f(C) where C compact, taking pre-images of the covering, using compactness to find a finite subcover of the pre-images and then finally using this to get a finite subcover of f(C)) and then simply assume $C \subseteq X$ is closed and show:
C closed => C compact => f(C) compact => f(C) closed.
3. Nov 15, 2009
Yeah you're right about my identification space, I didn't think about that. I think I was already going in the direction of showing compact spaces map to compact spaces, I'll try you're suggestion.
4. Nov 15, 2009
Ok I copied the structure of an earlier proof and it seems straight forward:
Take a compact subspace C in X. Let U be an open covering of f(C) with sets $$U_{\lambda}$$. Define an open covering V of C by taking $$V_{\lambda} = f^{-1} \left( U_{\lambda} \right)$$. Note that the openness of V relies on the continuity of f, and the fact that it covers C relies on the fact that f is bijective. Now taking a finite refinement of V gives a finite refinement of U, so f(C) is compact.
If C is closed, then C is compact in X. If f(C) is compact then f(C) is closed in Y, since Y is Hausdorff. So f(C) is closed for every C closed. Hence f inverse is continuous.
Does that look ok?
5. Nov 15, 2009
### rasmhop
Yes except for one minor detail:
You don't use the bijectiveness here. If $x \in C$, then $f(x) \in f(C)$ so there exists some $\lambda$ such that $f(x) \in U_\lambda$ because U is a covering of f(C), and then $x \in f^{-1}(U_\lambda) = V_\lambda$. Thus the pre-images $\{V_\lambda\}$ cover C whether f is bijective or not. Apart from that it's good.
6. Nov 15, 2009
My definition of a covering is that the union of the $$V_{\lambda}$$ is strictly equal to C, not a subset. It may be that all the x in C are covered, but if the map is not injective then $$f^{\left (-1) \right}$$ could give out more elements than are in the set C. Otherwise the bijective property would be used. Now I realised I only used injectivity, I wonder if surjectivity is necessary?
7. Nov 15, 2009
### rasmhop
In that case: Yes you used it, but you don't need it. As you have observed $f^{-1}(U_\lambda)$ is open in X, so $V_\lambda' = f^{-1}(U_\lambda) \cap C$ is open in C given the subspace topology, so C is covered by $\{V_\lambda'\}$ but of course for this problem it doesn't really matter since you know that f is bijective.
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https://www.wheelofwellbeing.org/52fjcn/315036-spherical-harmonics-mathematica | SphericalHarmonicY. Hobson, E. W. The This is the convention They are given by , where are associated Legendre polynomials and and are the orbital and magnetic quantum numbers, respectively. They are orthogonal over with the weighting function 1 They are a higher-dimensional analogy of Fourier series, which form a complete basis for the set of periodic functions of a single variable ((functions on the circle Spherical harmonics are mathematical functions that are common in many physical situations, notably atomic orbitals, particle scattering processes, and antenna radiation patterns. New York: Cambridge the form, for . 253-272, 1991. For my lecture notes in theoretical physics (namely quantum mechanics) I'm trying to visualize the spherical harmonics. The Wigner D-matrix is a unitary matrix in an irreducible representation of the groups SU(2) and SO(3).The complex conjugate of the D-matrix is an eigenfunction of the Hamiltonian of spherical and symmetric rigid rotors.The matrix was introduced in 1927 by Eugene Wigner. For , . Curated computable knowledge powering Wolfram|Alpha. "Visualizing Atomic Orbitals." Trajectories in an LCAO Approximation for the Hydrogen Molecule H_2. Involving elementary functions of the direct function and elementary functions Involving elementary functions of the direct function and elementary functions of Toronto Press, 1946. Hints help you try the next step on your own. convention being used. Knowledge-based programming for everyone. 403 Citations; 24k Downloads; Part of the Lecture Notes in Mathematics book series (LNM, volume 17) Log in to check access. Central infrastructure for Wolfram's cloud products & services. USD 29.99 Instant download; Readable on all devices; Own it forever; Local sales tax included if applicable ; Buy Physical Book Learn about institutional subscriptions. Quaternions, rotations, spherical coordinates. For spin weight , the spin-weighted spherical harmonics become identical to the spherical harmonics.The case of spin weight is important for describing gravitational waves. and "Completeness of Spherical Harmonics with Respect to Square Integrable Functions." In mathematics and physical science, spherical harmonics are special functions defined on the surface of a sphere. The spherical harmonics are eigenfunctions of the angular part of the Laplace operator, known to physicists as the angular momentum operator: (11.9) Spherical Harmonic. https://physics.uwa.edu.au/pub/Orbitals. Browse other questions tagged special-functions mathematical-physics legendre-polynomials spherical-harmonics parity or ask your own question. In this Demonstration you can choose different values of the spin weight to see the angular distribution in space for different and modes.. 1) Introductory level reference: Spherical harmonics can be drawn, plotted or represented with a Computer Algebra System such as Mathematica by using the Mathematica built-in functions SphericalPlot3D[] and SphericalHarmonicY[] . Writing The Mathematica equivalent of the real spherical harmonic basis implemented in enoki/sh.h is given by the following definition: ... Evaluates the real spherical harmonics basis functions up to and including order order. Wolfram Research (1988), SphericalHarmonicY, Wolfram Language function, https://reference.wolfram.com/language/ref/SphericalHarmonicY.html. Decompose a spherical harmonic into a sum of products of two spherical harmonics: Apply angular momentum operators to spherical harmonics: Properties & Relations (2) Abbott, P. "2. Authors; Claus Müller; Book. add a comment. Polynomials: SphericalHarmonicY[n,m,theta,phi] (223 formulas)Primary definition (5 formulas) Functions, rev. and 18.4 in A coordinate). Amsterdam, Netherlands: North-Holland, is. Language as SphericalHarmonicY[l, SphericalPlot3D [ { r 1 , r 2 , … } , { θ , θ min , θ max } , { ϕ , ϕ min , ϕ max } ] generates a 3D spherical plot with multiple surfaces. Learn how, Wolfram Natural Language Understanding System, Functions for Separable Coordinate Systems. spherical harmonic coefficients of the volume element with the same surficial shape as some surface element. ]}. S 1). equation in spherical coordinates. Spherical harmonics can be drawn, plotted or represented with a Computer Algebra System such as Mathematica by using the Mathematica built-in functions SphericalPlot3D[] and SphericalHarmonicY[] . The Overflow Blog Ciao Winter Bash 2020! are the angular portion of the solution to Laplace's [1.4]-[1.6] it … In this entry, is taken as Some of these formulas give the "Cartesian" version. Spherical harmonics and some of their properties H. Hagemann 2008 Spherical harmonics 2 Previous knowlegde required • Group theory • Quantum mechanics • Spectroscopy Spherical harmonics 3 References • Brian L. Silver, « Irreducible Tensor methods An Introduction for chemists » Academic Press 1976 • D.A. 17 0. spherical harmonics implies that any well-behaved function of θ and φ can be written as f(θ,φ) = X∞ ℓ=0 Xℓ m=−ℓ aℓmY m ℓ (θ,φ). and as the azimuthal (longitudinal) coordinate There is no "sphere" per se.. it's like if you say "there is a value for every point on the unit circle", it means you trace a circle around the origin and give each point a value. Please note that this is not the behaviour one would get from a casual application of the function's definition. Byerly, W. E. "Spherical Harmonics." Special cases include, The above illustrations show normally used in physics, as described by Arfken (1985) and the Wolfram Associated polynomials are sometimes called Ferrers' functions (Sansone 1991, p. 246). In special functions, a topic in mathematics, spin-weighted spherical harmonics are generalizations of the standard spherical harmonics and—like the usual spherical harmonics—are functions on the sphere.Unlike ordinary spherical harmonics, the spin-weighted harmonics are U(1) gauge fields rather than scalar fields: mathematically, they take values in a complex line bundle. They are often employed in solving partial differential equations in many scientific fields. Spherical harmonic functions arise for central force problems in quantum mechanics as the angular part of the Schrödinger equation in spherical polar coordinates. rev. An Elementary Treatise on Spherical Harmonics and Subjects Connected with Them. Visualising the spherical harmonics is a little tricky because they are complex and defined in terms of angular co-ordinates, $(\theta, \phi)$. Oxford, England: Pergamon Press, 1967. Course in Modern Analysis, 4th ed. Spherical harmonics are implemented in the Wolfram University Press, 1996. coefficients). One can clearly see that is symmetric for a rotation about the z axis. The Overflow Blog Ciao Winter Bash 2020! and imaginary parts, Integrals of the spherical harmonics are given by, where is a Wigner 1980. Hence the application of (10) to the potential coefficients of a spherical cap (6) yields the potential coefficients of a spherical cone extending between radii rl and r,. gives, Using separation of variables by equating the -dependent portion to a constant gives, Plugging in (3) into (2) gives the equation for the -dependent portion, whose solution Sansone, G. "Harmonic Polynomials and Spherical Harmonics," "Integral Properties of Spherical Harmonics and the Addition Theorem for Legendre Polynomials," Kalf, H. "On the Expansion of a Function in Terms of Spherical Harmonics in Arbitrary Dimensions." The Theory of Potential and Spherical Harmonics, 2nd ed. Dover, pp. Harmonic Differential Equation, https://functions.wolfram.com/Polynomials/SphericalHarmonicY/, https://functions.wolfram.com/HypergeometricFunctions/SphericalHarmonicYGeneral/, https://physics.uwa.edu.au/pub/Computational/CP2/2.Schroedinger.nb. In special functions, a topic in mathematics, spin-weighted spherical harmonics are generalizations of the standard spherical harmonics and—like the usual spherical harmonics—are functions on the sphere. Involving functions of the direct function and elementary functions with respect to theta. Boston, MA: Academic Press, p. 129, Theory of Spherical and Ellipsoidal Harmonics. New York: Chelsea, 1955. English ed. Spherical Harmonics: An Elementary Treatise on Harmonic Functions, with Applications, 3rd ed. Particularly I'd like to understand why we should expect that harmonic functions on $\mathbb{R}^{n+1}$ restrict to eigenfunctions of the Laplacian on the sphere. https://www.ericweisstein.com/encyclopedias/books/SphericalHarmonics.html. Spherical harmonics & Mathematica Mathematica; Thread starter shetland; Start date Nov 21, 2005; Nov 21, 2005 #1 shetland. Satisfies Assigned Boundary Conditions at the Surface of a Sphere." Spherical harmonics are most commonly encountered by physicists and engineers in order to solve specific problems in three-dimensional space, and introductory physics and engineering textbooks typically do not devote a whole lot of time to develop Hilbert space theory. MacRobert, T. M. and Sneddon, I. N. Spherical Harmonics: An Elementary Treatise on Harmonic Functions, with Applications, 3rd ed. The spherical harmonics are orthonormal with respect to integration over the surface of the unit sphere. Buy eBook. The result is called a spherical harmonic and denoted (11.8) it is known to Mathematica, of course, as SphericalHarmonicY[l,m,theta,phi]. @misc{reference.wolfram_2020_sphericalharmonicy, author="Wolfram Research", title="{SphericalHarmonicY}", year="1988", howpublished="\url{https://reference.wolfram.com/language/ref/SphericalHarmonicY.html}", note=[Accessed: 11-January-2021 In a similar fashion, McAdoo (1981) develops the spherical harmonic expression for the potential due to a great circle ring source, and examines its spectral properties. For certain special arguments, SphericalHarmonicY automatically evaluates to exact values. I'm calculating the zz Component for the quadruple tensor. the polar (colatitudinal) coordinate with , Please note that this is not the behaviour one would get from a casual application of the function's definition. Some care must be taken in identifying the notational convention being used. SpinWeightedSpheroidalHarmonics Install this package! Cambridge University Press, pp. https://reference.wolfram.com/language/ref/SphericalHarmonicY.html. Caution Care must be taken in correctly identifying the arguments to this function: θ is taken as the polar (colatitudinal) coordinate with θ in [0, π], and φas the azimuthal (longitudinal) coordinate with φin [0,2π). Numerical The Contributed by: Stephen Wolfram (March 2011) Open content licensed under CC … The spherical harmonics are often represented graphically since their linear combinations correspond to the angular functions of orbitals. The spherical harmonics are the angular portion of the solution to Laplace's equation in spherical coordinates where azimuthal symmetry is not present. the Condon-Shortley phase is prepended Spherical Sakurai, Modern Quantum Mechanics, 2nd Ed. I will give some examples. Here, denotes the complex conjugate and is the Open tools for black hole perturbation theory. Ferrers, N. M. An Elementary Treatise on Spherical Harmonics and Subjects Connected with Them. Practice online or make a printable study sheet. 246-248, 1992. Weisstein, Eric W. "Spherical Harmonic." https://functions.wolfram.com/HypergeometricFunctions/SphericalHarmonicYGeneral/. Press, W. H.; Flannery, B. P.; Teukolsky, S. A.; and Vetterling, W. T. "Spherical Harmonics." Unlimited random practice problems and answers with built-in Step-by-step solutions. I got Mathematica to plot them an this is what I did: SphericalPlot3D[(Sqrt[... Visualizing the spherical harmonics. "SphericalHarmonicY." MAIN PROGRAM USAGE: The main programs are 'shana' and 'shsyn' for spherical harmonic analysis and synthesis (go from spatial data to spherical harmonic coefficients and vice versa, respectively). spherical cap, and a spherical rectangle located at the surface of a planet, and discusses the spherical harmonic spectra associated with these mass elements. Written in terms of Cartesian coordinates, The zonal harmonics are defined to be those of the form, The tesseral harmonics are those of The associated Legendre polynomials and generalize the Legendre polynomials and are solutions to the associated Legendre differential equation, where is a positive integer and , ..., .They are implemented in the Wolfram Language as LegendreP[l, m, x].For positive , they can be given in terms of the unassociated polynomials by Now the sphere comes from the idea that, SH functions, use the Legendre polynomials (but Legendre polynomials are 1D functions), and the specification of spherical … In this study we review the literature on the density-normalized spherical harmonics, clarify the existing notations, use the Paturle–Coppens method in the Wolfram Mathematicasoftware to derive the Cartesian spherical harmonics for l ≤ 20 and determine the density normalization coefficients to 35 significant figures, and computer-generate a Fortran90 code. Browse other questions tagged harmonic-analysis harmonic-functions spherical-geometry spherical-varieties derivations or ask your own question. §12.6 and 12.9 in Mathematical Similarly, we will be able to express spin-weighted spherical harmonics directly in terms of quaternions, though with a simple translation to and from standard spherical coordinates. Toronto: University Spherical Harmonics. §18.31 Harmonics, with Applications to Problems in Mathematical Physics. 680-685 ]}, @online{reference.wolfram_2020_sphericalharmonicy, organization={Wolfram Research}, title={SphericalHarmonicY}, year={1988}, url={https://reference.wolfram.com/language/ref/SphericalHarmonicY.html}, note=[Accessed: 11-January-2021 ..., , and is an associated Legendre polynomial. D stands for Darstellung, which means "representation" in German. Spherical harmonics give the angular part of the solution to Laplace's equation in spherical coordinates. The associated Legendre functions are part of the spherical harmonics, which are the solution of Laplace's equation in spherical coordinates. 1997. 391-395, 1990. 1 answer Sort by » oldest newest most voted. generates a 3D spherical plot over the specified ranges of spherical coordinates. Wolfram Language. Figure 1.1a shows a plot of the spherical harmonics where the phase is color coded. equation in spherical coordinates where Software engine implementing the Wolfram Language. Details. https://mathworld.wolfram.com/SphericalHarmonic.html. In the chapter, the spherical harmonics is connected with potential theory and cylindrical harmonics with the wave equation and its simplest solution—the monochromatic wave. Spherical harmonics satisfy the spherical harmonic differential equation, which is given by the angular part of Laplace's Ch. §6.8 in Numerical Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. SphericalHarmonicY. Mc Quarrie, J.D. Technology-enabling science of the computational universe. The sphere is colored according to the real or imaginary part of the spherical harmonic . If the situation warrants it, the cosine function can be replaced by the sine function. The For spin weight , the spin-weighted spherical harmonics become identical to the spherical harmonics.The case of spin weight is important for describing gravitational waves. Decompose a spherical harmonic into a sum of products of two spherical harmonics: Apply angular momentum operators to spherical harmonics: Properties & Relations (2) Kronecker delta. The spherical harmonics Y n m (theta, phi) are the angular portion of the solution to Laplace's equation in spherical coordinates where azimuthal symmetry is not present. (In this system, a point in space is located by three coordinates, one representing the distance from the origin and two others representing the angles of elevation and azimuth, as in astronomy.) They arise in many practical situations, notably atomic orbitals, particle scattering processes and antenna radiation patterns. Documentation is provided in the form of a live script with examples, as well as an HTML … edit retag flag offensive close merge delete. In this Demonstration you can choose different values of the spin weight to see the angular distribution in space for different and modes.. 1) Introductory level reference: Wolfram Language & System Documentation Center. Capabilities include the computation of surface/solid, complex/real and normalized/unnormalized spherical harmonics. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … The precision of the output tracks the precision of the input: Evaluate SphericalHarmonicY symbolically for integer orders: Evaluate SphericalHarmonicY symbolically for noninteger orders: Evaluate SphericalHarmonicY symbolically for : Find the first positive maximum of SphericalHarmonicY[2,2,θ,Pi/2]: Plot the SphericalHarmonicY function for various orders: Plot the absolute value of the SphericalHarmonicY function in three dimensions: SphericalHarmonicY is an even function with respect to θ and ϕ for even-order m: SphericalHarmonicY is an odd function with respect to θ and ϕ for odd-order m: SphericalHarmonicY is a periodic function with respect to θ and ϕ: SphericalHarmonicY has the mirror property : SphericalHarmonicY threads elementwise over lists: Plot the absolute values of the higher derivatives of with respect to : Formula for the derivative with respect to : Compute the indefinite integral using Integrate: General term in the series expansion using SeriesCoefficient: SphericalHarmonicY can be applied to a power series: SphericalHarmonicY is an eigenfunction of the spherical part of the Laplace operator: Use FunctionExpand to expand SphericalHarmonicY[n,m,θ,ϕ] for half-integers and : LegendreP ClebschGordan SphericalBesselJ ZernikeR. Spherical harmonics are a set of functions used to represent functions on the surface of the sphere S^2 S 2. This contribution includes a single MATLAB function ('harmonicY') that computes spherical harmonics of any degree and order, evaluated at arbitrary inclination, azimuth and radius. Wolfram Language. in this equation The first few spherical harmonics are. Thanks alot. The good news is that we can actually do the integral algebraically! (top), (bottom left), and https://www.ericweisstein.com/encyclopedias/books/SphericalHarmonics.html. For my lecture notes in theoretical physics (namely quantum mechanics) I'm trying to visualize the spherical harmonics. This assumes x, y, z, and r are related to and through the usual spherical-to-Cartesian coordinate transformation: {= = = In his derivation the product of the first two spherical harmonics is expanded using the Clebsch-Gordan Series (which is also proved) to get the following equation. Recipes in FORTRAN: The Art of Scientific Computing, 2nd ed. Soc. New York: The spherical harmonics are sometimes separated into their real Φ(φ) = cos mφ , (2) where m is an integer. Mathematical This is a table of orthonormalized spherical harmonics that employ the Condon-Shortley phase up to degree = 10. Sometimes (e.g., Arfken 1985), Sternberg, W. and Smith, T. L. The Theory of Potential and Spherical Harmonics, 2nd ed. (12) for some choice of coefficients aℓm. The output array must have room for (order + 1)*(order + 1) entries. They are a higher-dimensional analogy of Fourier series, which form a complete basis for the set of periodic functions of a single variable (((functions on the circle S 1). In special functions, a topic in mathematics, spin-weighted spherical harmonics are generalizations of the standard spherical harmonics and—like the usual spherical harmonics—are functions on the sphere.Unlike ordinary spherical harmonics, the spin-weighted harmonics are U(1) gauge fields rather than scalar fields: mathematically, they take values in a complex line bundle. Revolutionary knowledge-based programming language. The purely angular part of the Helmholtz equation i.e. p.216. Not so well-known are the transforms in the radial direction. Press, pp. Retrieved from https://reference.wolfram.com/language/ref/SphericalHarmonicY.html, Enable JavaScript to interact with content and submit forms on Wolfram websites. London: Macmillan, 1877. Walk through homework problems step-by-step from beginning to end. Knowledge-based, broadly deployed natural language. Spherical harmonics. and 698-700, 1985. Some care must be taken in identifying the notational where , , ..., 0, ))eim" New York: Dover, pp. The spherical harmonics Groemer, H. Geometric Applications of Fourier Series and Spherical Harmonics. $$Q_{zz} = 3cos^2\theta-1$$(r=1 in this case), and the $$Y_{lm}(\theta,\phi)$$ would be l=2, m=0. SphericalHarmonicY can be evaluated to arbitrary numerical precision. Methods for Physicists, 3rd ed. Cambridge, England: Not to be confused with spinor spherical harmonics. The spherical harmonics Y n m (theta, ... 2π] this implementation follows the convention used by Mathematica: the function is periodic with period π in θ and 2π in φ. The preeminent environment for any technical workflows. Simon Stevin 2, Spherical As of Version 9.0, vector analysis functionality is built into the Wolfram Language » represents the spherical coordinate system with default variables Rr , Ttheta , and Pphi . The angular parts of the transforms in 2D and 3D are therefore very familiar. Recipes in FORTRAN: The Art of Scientific Computing, 2nd ed. 361-380, 1995. The spherical harmonics are eigenfunctions of this operator with eigenvalue : The generalization of the Coulomb potential — the electric potential of a point charge — to n dimensions is: Since the charge density is only nonzero at the origin, the Laplacian must be equal to zero everywhere else: Main collaborators: Alfonso García-Parrado, Alessandro Stecchina, Barry Wardell, Cyril Pitrou, David Brizuela, David Yllanes, Guillaume Faye, Leo Stein, Renato Portugal, Teake Nutma, Thomas Bäckdahl. xAct: Efficient tensor computer algebra for the Wolfram Language José M. Martín-García, GPL 2002-2021. 1.3.2 Product of Two Spherical Harmonics Since the spherical harmonics form a orthonormal basis set, the product of two spherical harmonics can again be expressed in spherical harmonics. I will give some examples. Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. 195-218, 1959. the longitudinal coordinate and the colatitudinal If, they reduce to the unassociated polynomials. I got Mathematica to plot them an this is what I did: SphericalPlot3D[(Sqrt[... Stack Exchange Network. This will allow us to derive simple rotation laws for the SWSHs and modes of a general decomposition in terms of SWSHs. Wang, J.; Abbott, P.; and Williams, J. with . (bottom right). S^1). The spherical harmonics Y n m (theta, phi) are the angular portion of the solution to Laplace's equation in spherical coordinates where azimuthal symmetry is not present. Spherical harmonics are a set of functions used to represent functions on the surface of the sphere S 2 S^2 S 2. azimuthal symmetry is not present. For convenience, we list the spherical harmonics for ℓ = 0,1,2 and non-negative values of m. ℓ = 0, Y0 0 (θ,φ) = 1 √ 4π ℓ = 1, Y1 In this entry, is taken as the polar (colatitudinal) coordinate with , and as the azimuthal (longitudinal) coordinate with . Handbook of Differential Equations, 3rd ed. Another paper [Green 2003] has code for evaluating the RSH in spherical coordi- nates, but it is 2–3 orders of magnitude slower than the techniques presented in this paper. Involving functions of the direct function and elementary functions with respect to theta. Spherical harmonic functions arise when the spherical coordinate system is used. spherical harmonics are then defined by combining and , where the normalization is chosen such that. §3.18-3.20 in Orthogonal Whittaker, E. T. and Watson, G. N. "Solution of Laplace's Equation Involving Legendre Functions" and "The Solution of Laplace's Equation which Arfken, G. "Spherical Harmonics" and "Integrals of the Products of Three Spherical Harmonics." So, what is the mathematical relationship between Sage's spherical_harmonic and Mathematica's SphericalHarmonicY in terms of l and m? Geometric Applications of Fourier Series and Spherical Harmonics. reference-request harmonic-analysis harmonic-functions laplacian spherical-harmonics Bohm to Classical Trajectories in a Hydrogen Atom, Spherical Schrödinger Equation." Orlando, FL: Academic Press, pp. We know what the spherical harmonics are, so we can certainly just open Mathematica and do the integral; but for the specific example of a $$3d \rightarrow 2p$$ transition we brought up, there are 45 different integrals to do. Normand, J. M. A Lie Group: Rotations in Quantum Mechanics. As for the reason this expansion is usually not stated in the Hilbert space context, I suspect it is largely cultural. Methods for Physicists, 3rd ed. The sectorial Unfortunately it's littered with \[symbol] tags as in Mathematica I used some symbols for variables and shortcuts (which you can enter either in that form or as esc-symbol-esc). Physics 2. https://physics.uwa.edu.au/pub/Computational/CP2/2.Schroedinger.nb. 1988. Belg. From MathWorld--A Wolfram Web Resource. transform is called Spherical Harmonic (SH) transform and has been widely used in representation and registration of 3D shapes [8–10]. Harmonic on Constant Latitude or Longitude, Bohm Language (in mathematical literature, usually denotes A Lie Group: Rotations in Quantum Mechanics. An Elementary Treatise on Fourier's Series, and Spherical, Cylindrical, and Ellipsoidal And as the polar ( colatitudinal ) coordinate with, and as the angular portion the... 129, 1997 Start date Nov 21, 2005 ; Nov 21, 2005 # 1 tool for creating and! Can clearly see that is symmetric for spherical harmonics mathematica rotation about the z axis 0,...,. An associated Legendre functions are part of the Schrödinger equation in spherical coordinates representation in. Are then defined by combining and, where are associated Legendre polynomial E. W. the Theory Potential... And as the polar ( colatitudinal ) coordinate with of Eqs, and! The form, RELATED Wolfram SITES: https: //functions.wolfram.com/Polynomials/SphericalHarmonicY/, https //functions.wolfram.com/Polynomials/SphericalHarmonicY/. Spherical-Varieties derivations or ask your own question ask your spherical harmonics mathematica question Wolfram websites arise when the spherical harmonics.The of.: Academic Press, P. 129, 1997 symmetry is not present are set! Related Wolfram SITES: https: //functions.wolfram.com/HypergeometricFunctions/SphericalHarmonicYGeneral/ where,, and more that! On Fourier 's Series, and as the angular functions of the spherical harmonics, which means representation. 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Of the spherical harmonics that employ the Condon-Shortley phase is color coded Visualizing the spherical.... ) entries surface of the unit sphere the sectorial harmonics are implemented in the Wolfram function... I suspect it is largely cultural visualize the spherical coordinate System is used harmonic functions, with Applications, ed! ) coordinate with, and spherical harmonics mathematica bottom right ) the same surficial shape some. Of functions used to represent functions on the surface of the Schrödinger equation spherical. Questions tagged special-functions mathematical-physics legendre-polynomials spherical-harmonics parity or ask your own question, W. T. spherical!, ( bottom right ) Condon-Shortley phase is color coded, notably atomic orbitals, particle scattering processes and radiation. The expansion of a function in Terms spherical harmonics mathematica spherical harmonics in Arbitrary Dimensions. some of formulas! Central force problems in Mathematical Physics A. ; and Williams, J Treatise on spherical are... The Kronecker delta harmonics: an Elementary Treatise on harmonic functions arise for central force problems in mechanics! For my lecture Notes for Computational Physics 2. https: //functions.wolfram.com/HypergeometricFunctions/SphericalHarmonicYGeneral/ Helmholtz i.e. Casual application of the transforms in the Wolfram Language as SphericalHarmonicY [ l,,! Taken as the azimuthal ( longitudinal ) coordinate with been widely used in representation and registration 3D! M. and Sneddon, I. N. spherical harmonics. 1 tool for creating and... Are special functions defined on the surface of the direct function and Elementary with! Namely quantum mechanics 12.9 in Mathematical Physics hints help you try the next step on own!
Auto Mall Vaughan, Youtubers From Virginia, Victorian Garden Pictures, Herrera Elementary School Supply List, Ramsey Park Hotel Phone Number, Bumrah Ipl Team 2020 Price, Stage 4 Restrictions Vic, Ecu Meaning Car, | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 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": 0.8373258113861084, "perplexity": 1378.9004582368375}, "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/1634323585911.17/warc/CC-MAIN-20211024050128-20211024080128-00364.warc.gz"} |
https://testbook.com/question-answer/raman-and-raghav-went-for-an-interview-which-was-f--608bc49aecd1bbada0fd71df | # Raman and Raghav went for an interview which was for two vacancies for the same post. The probability of Raman's selection is 1/6 and the probability of Raghav's selection is 1/5. what is the probability of if both of them get selected and one of them is selected respectively?A.$$1\over30$$B.$$31\over61$$C.$$3\over10$$D.$$1\over6$$
1. Only A
2. Both A and C
3. Both B and C
4. Only B
Option 2 : Both A and C
## Detailed Solution
Given:
Probability of Raman to get selected: $$1\over6$$
Probability of Raghav to get selected: $$1\over5$$
Formula used:
nCr = n!/((n-r)! × r!)
P = (Number of favorable outcomes)/(Total number if outcomes)
Calculation:
Probability of both of them get selected: $$1\over6$$ $$\times$$ $$1\over5$$ = $$1\over30$$
Probability of Raman to not get selected: 1 - $$1\over6$$ = $$5\over6$$
Probability of Raghav to not get selected: 1 - $$1\over5$$ = $$4\over5$$
Probability of one of them is selected: ($$1\over6$$$$\times$$$$4\over5$$) + ($$1\over5$$$$\times$$$$5\over6$$) = $$3\over10$$
∴ Both A and C is the correct answer. | {"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": 0.8364178538322449, "perplexity": 1653.8649618623165}, "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/1637964363336.93/warc/CC-MAIN-20211207045002-20211207075002-00546.warc.gz"} |
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## sim activation website
Hi. I'm trying to activate a sim, but for the past three days I get through the whole process only for the website to advise it is experiencing difficulties. This happens both on my laptop or if I use my mobile. Anyone having the same problem?
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## Re: sim activation website
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## Re: sim activation website
FYI. Just to let you know I resolved this myself. I still have a telstra login which I used but it seems the website let me sign in but wouldn't activate the app. So the answer is to sign in as a guest. Once activated, the website then let me use my current login. Hope this helps someone else. | {"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": 0.8647140264511108, "perplexity": 2745.532943546172}, "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/1614178360853.31/warc/CC-MAIN-20210228115201-20210228145201-00434.warc.gz"} |
http://mathhelpforum.com/calculus/206482-integation-limits.html | 1. ## integation limits
find two numbers a and b with a<=b such that has the largest value
i have done the normal integration with expression coming in form of a,b what should i do after that
2. ## Re: integation limits
Originally Posted by prasum
find two numbers a and b with a<=b such that has the largest value
i have done the normal integration with expression coming in form of a,b what should i do after that
The function $f(x)=-x^2-x+6=-(x+3)(x-2)$ has zeros $x=-3 \quad x=2$
The function will only be above the $x-$axis on the set $(-3,2)$
3. ## Re: integation limits
Basically integrand has the largest value iff, it is positive in the interval. If you consider all interval where it is positive and integrate it, you will get the max value. Now, before becoming negative from positive, the polynomial changes sign, i.e, becomes zero. So, you have to find the roots, and integrate between the roots, between which it stays positive.
This is sufficient for quadratic polynomial. For others, you need to consider multiple intervals, and the integrand value on intermediate negative intervals also, and try to do an optimization.
Salahuddin
Maths online | {"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": 0.8375034332275391, "perplexity": 776.3500302483727}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982296020.34/warc/CC-MAIN-20160823195816-00079-ip-10-153-172-175.ec2.internal.warc.gz"} |
http://tutorial.math.lamar.edu/Classes/DE/SystemsDE.aspx | Paul's Online Notes
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### Section 5-4 : Systems of Differential Equations
In the introduction to this section we briefly discussed how a system of differential equations can arise from a population problem in which we keep track of the population of both the prey and the predator. It makes sense that the number of prey present will affect the number of the predator present. Likewise, the number of predator present will affect the number of prey present. Therefore the differential equation that governs the population of either the prey or the predator should in some way depend on the population of the other. This will lead to two differential equations that must be solved simultaneously in order to determine the population of the prey and the predator.
The whole point of this is to notice that systems of differential equations can arise quite easily from naturally occurring situations. Developing an effective predator-prey system of differential equations is not the subject of this chapter. However, systems can arise from $$n^{\text{th}}$$ order linear differential equations as well. Before we get into this however, let’s write down a system and get some terminology out of the way.
We are going to be looking at first order, linear systems of differential equations. These terms mean the same thing that they have meant up to this point. The largest derivative anywhere in the system will be a first derivative and all unknown functions and their derivatives will only occur to the first power and will not be multiplied by other unknown functions. Here is an example of a system of first order, linear differential equations.
\begin{align*}{{x'}_1} & = {x_1} + 2{x_2}\\ {{x'}_2} & = 3{x_1} + 2{x_2}\end{align*}
We call this kind of system a coupled system since knowledge of $$x_{2}$$ is required in order to find $$x_{1}$$ and likewise knowledge of $$x_{1}$$ is required to find $$x_{2}$$. We will worry about how to go about solving these later. At this point we are only interested in becoming familiar with some of the basics of systems.
Now, as mentioned earlier, we can write an $$n^{\text{th}}$$ order linear differential equation as a system. Let’s see how that can be done.
Example 1 Write the following 2nd order differential equation as a system of first order, linear differential equations. $2y'' - 5y' + y = 0\hspace{0.25in}y\left( 3 \right) = 6\,\,\,\,\,\,\,\,\,y'\left( 3 \right) = - 1$
Show Solution
We can write higher order differential equations as a system with a very simple change of variable. We’ll start by defining the following two new functions.
\begin{align*}{x_1}\left( t \right) & = y\left( t \right)\\ {x_2}\left( t \right) & = y'\left( t \right)\end{align*}
Now notice that if we differentiate both sides of these we get,
\begin{align*}{{x'}_1} & = y' = {x_2}\\ {{x'}_2} & = y'' = - \frac{1}{2}y + \frac{5}{2}y' = - \frac{1}{2}{x_1} + \frac{5}{2}{x_2}\end{align*}
Note the use of the differential equation in the second equation. We can also convert the initial conditions over to the new functions.
\begin{align*}{x_1}\left( 3 \right) & = y\left( 3 \right) = 6\\ {x_2}\left( 3 \right) & = y'\left( 3 \right) = - 1\end{align*}
Putting all of this together gives the following system of differential equations.
\begin{align*}{{x'}_1} & = {x_2} & \hspace{0.25in}{x_1}\left( 3 \right) & = 6\\ {{x'}_2} & = - \frac{1}{2}{x_1} + \frac{5}{2}{x_2} & \hspace{0.25in}{x_2}\left( 3 \right) & = - 1\end{align*}
We will call the system in the above example an Initial Value Problem just as we did for differential equations with initial conditions.
Let’s take a look at another example.
Example 2 Write the following 4th order differential equation as a system of first order, linear differential equations. ${y^{\left( 4 \right)}} + 3y'' - \sin \left( t \right)y' + 8y = {t^2}\hspace{0.25in}y\left( 0 \right) = 1\,\,\,\,y'\left( 0 \right) = 2\,\,\,\,y''\left( 0 \right) = 3\,\,\,\,y'''\left( 0 \right) = 4$
Show Solution
Just as we did in the last example we’ll need to define some new functions. This time we’ll need 4 new functions.
\begin{align*}{x_1} & = y & \Rightarrow \hspace{0.25in}{{x'}_1} & = y' = {x_2}\\ {x_2} & = y' & \Rightarrow \hspace{0.25in}{{x'}_2} & = y'' = {x_3}\\ {x_3} & = y'' & \Rightarrow \hspace{0.25in}{{x'}_3} & = y''' = {x_4}\\ {x_4} & = y''' & \Rightarrow \hspace{0.25in}{{x'}_4}& = {y^{\left( 4 \right)}} = - 8y + \sin \left( t \right)y' - 3y'' + {t^2} = - 8{x_1} + \sin \left( t \right){x_2} - 3{x_3} + {t^2}\end{align*}
The system along with the initial conditions is then,
\begin{align*}{{x'}_1} & = {x_2} & \hspace{0.25in}{x_1}\left( 0 \right) & = 1\\ {{x'}_2} & = {x_3} & \hspace{0.25in}{x_2}\left( 0 \right) & = 2\\ {{x'}_3} & = {x_4} & \hspace{0.25in}{x_3}\left( 0 \right) & = 3\\ {{x'}_4} & = - 8{x_1} + \sin \left( t \right){x_2} - 3{x_3} + {t^2} & \hspace{0.25in}{x_4}\left( 0 \right) & = 4\end{align*}
Now, when we finally get around to solving these we will see that we generally don’t solve systems in the form that we’ve given them in this section. Systems of differential equations can be converted to matrix form and this is the form that we usually use in solving systems.
Example 3 Convert the following system to matrix from. \begin{align*}{{x'}_1} & = 4{x_1} + 7{x_2}\\ {{x'}_2} & = - 2{x_1} - 5{x_2}\end{align*}
Show Solution
First write the system so that each side is a vector.
$\left( {\begin{array}{*{20}{c}}{{{x'}_1}}\\{{{x'}_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{4{x_1} + 7{x_2}}\\{ - 2{x_1} - 5{x_2}}\end{array}} \right)$
Now the right side can be written as a matrix multiplication,
$\left( {\begin{array}{*{20}{c}}{{{x'}_1}}\\{{{x'}_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}4&7\\{ - 2}&{ - 5}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right)$
Now, if we define,
$\vec x = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right)$
then,
$\vec x' = \left( {\begin{array}{*{20}{c}}{{{x'}_1}}\\{{{x'}_2}}\end{array}} \right)$
The system can then be written in the matrix form,
$\vec x' = \left( {\begin{array}{*{20}{c}}4&7\\{ - 2}&{ - 5}\end{array}} \right)\vec x$
Example 4 Convert the systems from Examples 1 and 2 into matrix form.
Show Solution
\begin{align*}{{x'}_1} & = {x_2}\hspace{0.25in}{x_1}\left( 3 \right) = 6\\ {{x'}_2} & = - \frac{1}{2}{x_1} + \frac{5}{2}{x_2}\hspace{0.25in}{x_2}\left( 3 \right) = - 1\end{align*}
First define,
$\vec x = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right)$
The system is then,
$\vec x' = \left( {\begin{array}{*{20}{c}}0&1\\{ - \frac{1}{2}}&{\frac{5}{2}}\end{array}} \right)\vec x\hspace{0.25in}\vec x\left( 3 \right) = \left( {\begin{array}{*{20}{c}}{{x_1}\left( 3 \right)}\\{{x_2}\left( 3 \right)}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\{ - 1}\end{array}} \right)$
Now, let’s do the system from Example 2.
\begin{align*}{{x'}_1} & = {x_2}\hspace{0.25in}{x_1}\left( 0 \right) = 1\\ {{x'}_2} & = {x_3}\hspace{0.25in}{x_2}\left( 0 \right) = 2\\ {{x'}_3} & = {x_4}\hspace{0.25in}{x_3}\left( 0 \right) = 3\\ {{x'}_4} & = - 8{x_1} + \sin \left( t \right){x_2} - 3{x_3} + {t^2}\hspace{0.25in}{x_4}\left( 0 \right) = 4\end{align*}
In this case we need to be careful with the t2 in the last equation. We’ll start by writing the system as a vector again and then break it up into two vectors, one vector that contains the unknown functions and the other that contains any known functions.
$\left( {\begin{array}{*{20}{c}}{{{x'}_1}}\\{{{x'}_2}}\\{{{x'}_3}}\\{{{x'}_4}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{{x_2}}\\{{x_3}}\\{{x_4}}\\{ - 8{x_1} + \sin \left( t \right){x_2} - 3{x_3} + {t^2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{{x_2}}\\{{x_3}}\\{{x_4}}\\{ - 8{x_1} + \sin \left( t \right){x_2} - 3{x_3}}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}0\\0\\0\\{{t^2}}\end{array}} \right)$
Now, the first vector can now be written as a matrix multiplication and we’ll leave the second vector alone.
$\vec x' = \left( {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\{ - 8}&{\sin \left( t \right)}&{ - 3}&0\end{array}} \right)\vec x + \left( {\begin{array}{*{20}{c}}0\\0\\0\\{{t^2}}\end{array}} \right)\hspace{0.25in}\vec x\left( 0 \right) = \left( {\begin{array}{*{20}{c}}1\\2\\3\\4\end{array}} \right)$
where,
$\vec x\left( t \right) = \left( {\begin{array}{*{20}{c}}{{x_1}\left( t \right)}\\{{x_2}\left( t \right)}\\{{x_3}\left( t \right)}\\{{x_4}\left( t \right)}\end{array}} \right)$
Note that occasionally for “large” systems such as this we will go one step farther and write the system as,
$\vec x' = A\vec x + \vec g\left( t \right)$
The last thing that we need to do in this section is get a bit of terminology out of the way. Starting with
$\vec x' = A\vec x + \vec g\left( t \right)$
we say that the system is homogeneous if $$\vec g\left( t \right) = \vec 0$$ and we say the system is nonhomogeneous if $$\vec g\left( t \right) \ne \vec 0$$. | {"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": 0.9650986194610596, "perplexity": 509.32233702166076}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987773711.75/warc/CC-MAIN-20191021120639-20191021144139-00019.warc.gz"} |
https://math.stackexchange.com/questions/3015521/two-tangent-circles-are-inscribed-in-a-semicircle-one-touching-the-diameters-m | # Two tangent circles are inscribed in a semicircle, one touching the diameter's midpoint; find the radius of the smaller circle
I am unable to upload the image of my trials.
I assumed the radius of small circle is $$x,$$ horizontal distance between the centers of two circles is $$y.$$
I have joined the centers of the two circles and the length is $$(5+x).$$
I have drawn a vertical line from the center of the bigger circle to the center of the semi circle.
I have also drawn a horizontal line from the center of the small circle to the above line.
Then, by applying Pythagoras theorem, I get
$$(x+5)^2=(5-x)^2 + y^2.$$ I need one more equation to solve for $$x.$$
Intuitively I wonder if the radius of the small circle could be half of the big circle. I carefully constructed it and got $$2.5 cm$$ as the radius but I am not sure.
• Hey Ashwini. You should be able to type out your working-out in your trials. People do appreciate when you show effort in trying to solve the question and it helps viewers solve the answers for you. – MBorg Nov 27 '18 at 9:39
• Okay. Let me try – Ashwini Nov 27 '18 at 10:00
• Radius of small circle = x Horizontal distance between two circles= y. I have joined centers of the two circles and the length is (5+x). I have drawn a vertical line from the center of big circle to center of semi circle. I have also drawn horizontal line from center of small circle to the above line. Then ,by applying Pythagoras theorem, I get – Ashwini Nov 27 '18 at 10:06
• (x+5)^2=(5-x)^2 + y^2. I need one more equation to solve for x.. Intuitively I wonder if the radius of the small circle could be half of the big circle. I carefully constructed it and got 2.5 cm as the radius but I am not sure. – Ashwini Nov 27 '18 at 10:10
• You should edit your working-out into your question, to improve its quality and for easier-viewing :) – MBorg Nov 27 '18 at 23:35
In addition to using Pythagorean theorem, one can use circle inversion to figure out the radius of the small circle (let's call it $$r$$).
Let $$AB$$ be the base of the semicircle. Let $$O$$ be its midpoint. Let $$C$$ be the contact point between the small circle (blue) with the circular arc $$AB$$. Draw the line $$OC$$ and let $$D$$ be its other intersection with the small circle. $$CD$$ will be a diameter of the small circle.
Perform a circle inversion with respect to the circle centered at $$O$$ with radius $$10$$. The line $$AB$$ get mapped to itself. The big circle (green) get mapped to a line (green, dashed) parallel to $$AB$$ and at a distance $$10$$ from it. The small circle get mapped to a circle (blue, dashed) sandwiched between these two lines. So its diameter will be $$10$$. Let $$D'$$ be the image of $$D$$ under circle inversion. $$CD'$$ will be a diameter of the image of the small circle. We have
$$|CD'| = 10 \implies |OD'| = |OC|+|CD'| = 10 + 10 = 2|OC|$$ Circle invert $$OD'$$ back to $$OD$$, we find \begin{align}|OD| = \frac12|OC| &\implies |CD| = |OC| - |OD| = \frac12|OC|\\ &\implies r = \frac12|CD| = \frac14|OC| = \frac52 \end{align} The radius we seek is $$\frac52$$. One half of that of the big circle and a quarter of that of the semicircle.
• This means the radius of the small circle will be half the radius of big circle for any given radius. – Ashwini Nov 28 '18 at 13:34
• @Ashwini yes, under the assumption the radius of big circle is one half of that of the semi-circle, – achille hui Nov 28 '18 at 13:48
Let the radius of smaller circle be $$\displaystyle r$$ and x-coordinate of its center$$\displaystyle ( C)$$ is $$\displaystyle a$$. As the circle is touching x-axis, so the ordinate of center of cicle is equal to radius of the circle,i.e., $$\displaystyle r$$. Let the point of touching of semicircle and smaller circle be $$\displaystyle P_{1}$$ $$\displaystyle ( x_{1}\displaystyle ,y_{1})$$. As the semicircle and smaller circle touch each other so $$\displaystyle P_{1}$$, $$\displaystyle C,Origin$$ are collinear. $$\begin{gather*} \therefore \dfrac{r}{a} =\dfrac{y_{1}}{x_{1}} \ \ \ \ \ \ \ \ \ \ \ ( 1)\\ \end{gather*}$$
And the distance between $$\displaystyle C$$ and center of bigger circle is $$\displaystyle r+5$$ $$\begin{gather*} ( r-5)^{2} +a^{2} =( r+5)^{2}\\ or\ \ 20r=a^{2} \ \ \ \ \ \ \ \ \ \ ( 2) \end{gather*}$$ Also the point $$\displaystyle P_{1}$$ satisfies both the semicircle and the smaller circle. $$\begin{gather*} \therefore x^{2}_{1} +y^{2}_{1} =100\ \ \ \ \ \ \ ( 3)\\ And\ ( x_{1} -a)^{2} +( y_{1} -r)^{2} =r^{2}\\ or\ x^{2}_{1} +y^{2}_{1} +a^{2} -2ax_{1} -2ry_{1} =0\\ or\ 100+a^{2} -2ax_{1} -2ry_{1} =0\ \ \ \ \ \ \ \ \ ( 4) \end{gather*}$$ Solving these four equations, we get $$\displaystyle r=2.5\ units$$
• Yes. That means we should use coordinate geometry. And the radius happens to be half of the big circle. This maybe a symmetrical property of circle. – Ashwini Nov 27 '18 at 10:19
• @Ashwini I don't think so. It is just a coincidence that its half of radius of bigger circle – Dikshit Gautam Nov 27 '18 at 12:11
• @DikshitGautam this cannot be a coincidence. – user376343 Nov 29 '18 at 20:34
Let $$R$$ be the (known) radius of the large inscribed circle, $$r$$ the radius of the small inscribed circle, and $$(x,r)$$ the center of this small circle. Then one has the two equations $$x^2+(R-r)^2=(R+r)^2,\qquad\sqrt{x^2+r^2}+ r=2R$$ in the two unknowns $$r$$ and $$x$$.
• Yes I understand. there are 2 unknowns here r and x. how can I proceed further to get r? Or is it that there is insufficient information to solve this problem? – Ashwini Nov 27 '18 at 10:14
With hindsight - never any use, of course! - one can see that such a configuration must exist, because there exists an isosceles triangle whose sides are in the ratio of $$3:2$$ (whose height and angles one needn't know): | {"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": 50, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9004045724868774, "perplexity": 264.4262972629545}, "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-2019-22/segments/1558232257497.4/warc/CC-MAIN-20190524024253-20190524050253-00411.warc.gz"} |
http://mathhelpforum.com/calculus/129290-derivative-check.html | # Math Help - derivative check
1. ## derivative check
f(x) = (sqrt(x))/(x^3+1)
f'(x) = (-2.5x^5/2)+(1/2sqrt(x)) / (x^3+1)^2
f(x) = (cosx/t^3)
f'(x) = (-x^3sinx)-(3x²cosx)/(t^3)²
f(x) = [2-(1/x)]/(x-3)
f'(x) = (-2x²+2x-3)/(x²-3x)²
unsuree if i have done these correctly on my H/W...-thx
2. Originally Posted by maybnxtseasn
f(x) = (sqrt(x))/(x^3+1)
f'(x) = (-2.5x^5/2)+(1/2sqrt(x)) / (x^3+1)^2
f(x) = (cosx/t^3)
f'(x) = (-x^3sinx)-(3x²cosx)/(t^3)²
f(x) = [2-(1/x)]/(x-3)
f'(x) = (-2x²+2x-3)/(x²-3x)²
unsuree if i have done these correctly on my H/W...-thx
Here's a kickstart on the first one:
$f(x) = \frac{\sqrt x}{x^3+1}$
$f'(x) = \frac {(x^3+1) \frac{1}{2}x^{-\frac{1}{2}} - \sqrt{x} \cdot 3x^2}{(x^3+1)^2} = \frac {\frac{1}{2}x^{\frac{5}{2}} + \frac{1}{2}x^{-\frac{1}{2}} - 3x^{\frac{5}{2}}}{(x^3+1)^2} = \frac {-\frac{5}{2}x^{\frac{5}{2}} + \frac{1}{2}x^{-\frac{1}{2}}}{(x^3+1)^2}$
I can't read your Math clearly, but I think you got it right. By
(1/2sqrt(x))
, you mean $\frac{1}{2\sqrt{x}}$ right? Then, I think what you have is correct.
3. Originally Posted by maybnxtseasn
f(x) = (sqrt(x))/(x^3+1)
f'(x) = (-2.5x^5/2)+(1/2sqrt(x)) / (x^3+1)^2
f(x) = (cosx/t^3)
f'(x) = (-x^3sinx)-(3x²cosx)/(t^3)²
f(x) = [2-(1/x)]/(x-3)
f'(x) = (-2x²+2x-3)/(x²-3x)²
unsuree if i have done these correctly on my H/W...-thx
And for the second one:
$f(x) = \frac{cosx}{t^3} = t^{-3}cosx$
You are forgetting implicit differentiation here. You are differentiating with respect to x, so when you take the derivative of t it will be t' or $\frac {dt}{dx}$, but not 1.
$f'(x) = -t^{-3}sinx - 3t^{-4}cosx \cdot t'$
You also seem to be changing the t into an x sometimes. Make sure you understand what I did there. If you want me to explain it in more detail, I will.
Hope I helped you
4. Hello, maybnxtseasn!
$f(x) \:= \:\frac{x^{\frac{1}{2}}} {x^3+1}$
$f'(x) \:=\: \frac{-2.5x^{\frac{5}{2}} +{\color{red}\dfrac{1}{2\sqrt{x}}}}{(x^3+1)^2}$ . . Is this what you meant?
$f'(x) \;=\;\frac{(x^3+1)\cdot \frac{1}{2}x^{-\frac{1}{2}} - x^{\frac{1}{2}}(3x^2)}{(x^3+1)^2} \;=\;\frac{\frac{1}{2}x^{-\frac{1}{2}}(x^3+1) - 3x^{\frac{5}{2}}}{(x^3+1)^2}$
Multiply by $\frac{2x^{\frac{1}{2}}}{2x^{\frac{1}{2}}}$
. $f'(x) \;=\;\frac{2x^{\frac{1}{2}} \bigg[ \frac{1}{2}x^{-\frac{1}{2}}(x^3+1) - 3x^{\frac{5}{2}} \bigg]} {2x^{\frac{1}{2}}\cdot (x^3+1)^2}$
. . . . $=\;\frac{x^3+1 - 6x^3}{2x^{\frac{1}{2}}(x^3+1)^2} \;=\;\frac{-5x^3 + 1}{2\sqrt{x}\,(x^3+1)^2}$
. . which is equivalent to your result.
5. ## yo
Originally Posted by mathemagister
And for the second one:
$f(x) = \frac{cosx}{t^3} = t^{-3}cosx$
You are forgetting implicit differentiation here. You are differentiating with respect to x, so when you take the derivative of t it will be t' or $\frac {dt}{dx}$, but not 1.
$f'(x) = -t^{-3}sinx - 3t^{-4}cosx \cdot t'$
You also seem to be changing the t into an x sometimes. Make sure you understand what I did there. If you want me to explain it in more detail, I will.
Hope I helped you
i still dont see how u got that awnser....anyone else help? how come i dont get this awnser when i do the simple quotient rule with this? | {"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": 15, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9291619062423706, "perplexity": 1329.572604925111}, "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-2015-14/segments/1427131297281.13/warc/CC-MAIN-20150323172137-00127-ip-10-168-14-71.ec2.internal.warc.gz"} |
http://math.stackexchange.com/users/42814/soham-chowdhury | # Soham Chowdhury
less info
reputation
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bio website location age member for 1 year, 10 months seen Jul 20 at 5:56 profile views 168
# 23 Questions
8 Why is $\sqrt{2\sqrt{2\sqrt{2\cdots}}} = 2$? 8 Is this a new expression for $e$? 6 Guides/tutorials to learn abstract algebra? 5 Can this type of limit even be evaluated? 4 Evaluate $\int^{441}_0\frac{\pi\sin \pi \sqrt x}{\sqrt x} dx$
# 598 Reputation
+10 How do people come up with difficult math Olympiad questions? +2 How do I evaluate the definite integral $\int_{1}^{-17} x^{-\frac{1}{3}}dx$ +5 Champernowne constant - summation and behavior of terms in continued fraction expansion +2 If $T$ is the derivative operator and $(v_0,…,v_m)$ is a standard basis, how can I prove that $Tv_k \in {\rm span}(v_0,…,v_k)$?
5 Can we get just $3$ from $\pi$? 5 Solve $(x^2 + 5)^2 - 15(x^2 + 5) + 54 = 0$ 3 Evaluate $\int^{441}_0\frac{\pi\sin \pi \sqrt x}{\sqrt x} dx$ 2 How to prove that $\lim\limits_{x\to0}\frac{\tan x}x=1$? 2 How do people come up with difficult math Olympiad questions?
# 41 Tags
6 algebra-precalculus × 4 2 contest-math × 3 5 polynomials × 2 2 trigonometry × 2 3 improper-integrals × 2 2 alternative-proof 2 limits × 5 2 derivatives 2 soft-question × 4 2 calculus
# 21 Accounts
Mathematics 598 rep 212 Programming Puzzles & Code Golf 446 rep 210 Stack Overflow 324 rep 215 English Language & Usage 169 rep 116 Musical Practice & Performance 131 rep 3 | {"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": 0.8685944080352783, "perplexity": 1669.6754893634543}, "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-2014-23/segments/1406510274581.53/warc/CC-MAIN-20140728011754-00186-ip-10-146-231-18.ec2.internal.warc.gz"} |
https://bmasterz.com/2017/11/neco-gce-physics-obj-and-theory-bmasterz/ | NECO GCE PHYSICS OBJ AND THEORY – BMASTERZ
0
12
NECO GCE PHYSICS OBJ AND THEORY
Physics OBJ:
1-10: DEEBCDBCDC
21-30: EECACCBABC
31-40: BDBDDCBAAA
41-50: EBEACCBDED
51-60: CBCCDECCEA
1a)
I)
Regular maintenance is essential to keep machines and the work environment safe and reliable.
II)
It helps to eliminate workplace hazards.
1aii)
Electrolysis can be used for electroplating of metals.
+++++++++++++++++++++++
3a)
couple is a system of forces with a resultant (a.k.a. net or sum) moment but no resultant force.
+++++++++++++++++++++++
4a)
The statement: The linear expansivity of copper is 0.000017k^-1 means that, A unit length of copper will expand in length by a fraction 0.000017k^-1 of the original per Kelvin rise in temperature.
4b)
In plane mirror, if the object is
upright, the image is also upright, of same
size and does not change in size by moving
the mirror towards or away from the face.
*While* in concave mirror, if the object is
upright, the image is also upright but
magnified and increases in size by moving the
mirror away from the face.
+++++++++++++++++++++++
15a)
The degree to which a specified material conducts electricity, calculated as the ratio of the current density in the material to the electric field which causes the flow of current.
6a)wavelength of a is the spatial period of the wave I.e the distance over which the wave’s shape repeats, and thus the inverse of the spatial frequency.
7a
Critical angle c is the incident angle in the sender medium when the angle of refraction in the less medium is 90%.
7b
Sound waves are longitudinal wave(they travel in the same direction as the vibration)
While
Light wave are transverse waves (they travel at right angles to the vibration)
SHARE | {"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": 0.8427413702011108, "perplexity": 1863.6247799928447}, "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/1531676590199.42/warc/CC-MAIN-20180718135047-20180718155047-00631.warc.gz"} |
http://swmath.org/software/11034 | # StatWeave
Reproducible statistical analysis with multiple languages. This paper describes the StatWeave system for making reproducible statistical analyses. StatWeave differs from other systems for reproducible analysis in several ways. The two main differences are: (1) Several statistics programs can be in used in the same document. (2) Documents can be prepared using OpenOffice or LaTeX . The main part of this paper is an example showing how to use R and SAS together in an OpenOffice text document. The paper also contains some practical considerations on the use of literate programming in statistics
## Keywords for this software
Anything in here will be replaced on browsers that support the canvas element
## References in zbMATH (referenced in 1 article )
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https://aviation.stackexchange.com/questions/44510/how-can-i-calculate-the-optimum-loiter-airspeed-for-an-aircraft/44515 | # How can I calculate the optimum loiter airspeed for an aircraft?
I am in the conceptual phase of designing a 10 seater amphibian aircraft. I have done the initial sizing using Raymer and Roskam's methods. I have got my design point. To do a constraint analysis as well as to find the fuel fraction for loiter phase I need to calculate the optimum loiter velocity for a loiter time of 60 minutes at 7000ft. Breguet's range equation doesn't help since I don't know the fuel fraction of loiter phase. How can I calculate the loiter airspeed?
The optimum cruise speed is the speed at which fuel consumption is minimized per unit of velocity. So, you need to compute the derivative (C/v)' where C is rate of fuel consumption and v is rate of travel (airspeed). (I assume you want this by "loiter". If you want to minimize fuel consumption, that is called "maximum endurance" and that is explained below on the chart.)
Since fuel consumption is proportional to power, and power divided by velocity is thrust, we can graphically determine this point by the thrust required curve. The diagram below shows the relevant relationships:
So, to compute the optimum cruise speed, first draw the thrust-required curve, as shown in the lower part of the diagram above, then you draw a line from the origin tangential to the thrust required curve. The point of tangency lies above the optimum cruise speed on the x-axis.
Maximum Endurance
If by "loiter" your goal is to minimize the fuel flow while maintaining a particular altitude, then you use the power required chart (the upper chart in the figure above). The fuel flow is proportional to power, so the maximum endurance speed is found by locating the lowest point on the power required curve. In the example above, that point is located at 80 CAS. Mathematically, the maximum endurance is found by solving the equation P'=0, where P is the power required function.
• To loiter is to stand or wait around without apparent purpose. The loiter speed or holding speed is the same as the maximum endurance speed. So, why not turn it around and make the main answer about the maximum endurance speed and the "bonus" about maximum range. I'd vote you up for it :-) – DeltaLima Oct 9 '17 at 20:02
Would be great if you add your existing approach. Roskam has already an example towards an amphibious aircraft if I am not mistaken.
If you don't have other concerns and if your design is not already constrained by some other points (take off, max speed, etc) here's the first approach:
Several certification regulations enforce a minimum loiter speed that is 1.3 times the stall speed.
Remember that there's a trade off between max speed and loiter speed.
• Thanks for your help but I don't have stall speed for my aircraft. To calculate stall speed I will need Clmax and wing area, which I don't know because I haven't reached the wing design phase. To do wing design I need to do constraint analysis so that I can get W/S and hence S i.e wing area. So I am stuck in a loop. What I want to know is if there is any rule of thumb that can give me loiter airspeed when you know empty weight and max take off weight. – Anirudh Prabhu Oct 10 '17 at 9:02 | {"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": 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": 0.8690062165260315, "perplexity": 889.0771555428996}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487608702.10/warc/CC-MAIN-20210613100830-20210613130830-00532.warc.gz"} |
https://www.studyadda.com/current-affairs/9th-class/30 | # Current Affairs 9th Class
#### Number Systems
Number Systems
• Rational numbers (Q): The numbers of the form,$\frac{p}{q}$ where 'p' and 'q' are integers and are called rational number A numfaer of the form r is a fraction. So all fractions are rational numbers.
Note: A number of the form $\frac{\mathbf{a}}{\mathbf{b}}$ is a fraction. So all are rational numbers. in the fraction, ‘a’ is called the numbers and ‘b’ is called the denominator. e.g. $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{- }\frac{\mathbf{2}}{\mathbf{3}}\mathbf{,}\frac{\mathbf{7}}{\mathbf{6}}\mathbf{,}\frac{\mat hbf{6}}{\mathbf{11}}\mathbf{,-}\frac{\mathbf{2}}{\mathbf{9}}\mathbf{,}....$ (i) Zero is a rational number. Note: 0 by 0 is undefined. (ii) Every integer is a rational number. (iii) A rational number, may or may not be an integer. (iv) To write W distinct rational numbers between any two rational numbers 'a' and $'b'$, we write$a=\frac{{{P}_{1}}}{q}$ and$b=\frac{{{P}_{2}}}{q}$ such that $\left( {{P}_{2}}\text{ }- \text{ }{{P}_{1}} \right)$is a positive integer greater than $'n'$, (Here a < b); ${{P}_{1}},{{P}_{2}}$and q are integers$(q\ne 0)$. p- +1 p. + 2 Pi + n A set of n rational numbers can be written as $\frac{{{P}_{1}}+1}{q},\frac{{{P}_{2}}+2}{q},......,\frac{{{P}_{1}}+n}{q}$ i.e.,$a=\frac{{{p}_{1}}}{q}<\frac{{{p}_{1}}+1}{q}<\frac{{{p}_{1}}+2}{q}.....<\frac{{{p}_{1}}+ n}{q}<\frac{{{p}_{2}}}{q}=b$ Between two given rational numbers a and b, there are infinitely many rational numbers.
• Properties of rational number
(i) If $\frac{p}{q}$, is a rational number and $'m'$ is a non-zero integer, then $\frac{p}{q}=\frac{p\times m}{q\times m}$
• (ii) If q is a rational number and $'m'$ is a common divisor of p and q, then $\frac{p}{q}=\frac{p\div m}{q\div m}$
(iii) Two rational numbers are equivalent only when the product of the numerator of the first rational number and the denominator of the second is equal to the product of the denominator of the first and the numerator of the second.
• Thus,$\frac{p}{q}=\frac{r}{s}\operatorname{only}\,if\,p\times s=q\times r$
Note $\frac{\mathbf{-p}}{\mathbf{q}}\mathbf{=}\frac{\mathbf{p}}{\mathbf{-q}}\mathbf{=- }\frac{\mathbf{p}}{\mathbf{q}}$
• Representation of rational numbers on a number line:
(i) Rational numbers of the form $\frac{m}{n}$ where m < n are represented on the number line as shown below. (ii) Rational numbers of the form $\frac{m}{n}$ where m > n are represented on the number line as shown below.
• Irrational numbers (Q'): Any number which cannot be expressed in the form of. (Which is neither terminating nor repeating decimal) where p and q are integers and $q\ne 0$is said to be an irrational
Note:$\pi$ more...
#### Polynomials
Polynomials
• An expression of the form $p(\operatorname{x})=+{{a}_{n}}{{\operatorname{x}}^{n}}+{{a}_{n-1}}......+{{a}_{2}}{{\operatorname{x}}^{2}}+{{a}_{1}}{{\operatorname{x}}^{2}}+{{a}_{0'}}\,\operatorname{where}{{a}_{0}},{{a}_{1}},a{{ & }_{2}},......,$are real numbers $'n'$is a non-negative integer and ${{a}_{n}}\ne 0$ is called a polynomial of degree.
• Each of ${{a}_{n}}{{\operatorname{x}}^{n}},{{a}_{n-1}},......{{a}_{2}},{{x}^{2}},{{a}_{1}}\operatorname{x}\,and\,{{a}_{n}}\ne 0$and a with is called a term of the polynomial p(x).
Note: The power of variable in a polynomial must be a whole number.
• An expression of the form$\frac{p\left( \operatorname{x} \right)}{q\left( \operatorname{x} \right)}$ where p(x) and q(x) are polynomials and $q(\operatorname{x})\ne 0$is called a rational expression.
Note: Every polynomial is a rational expression, but every rational expression need not be a polynomial.
• A polynomial d(x) is called a divisor of a polynomial p(x) if p(x) = d(x).q(x) for some polynomial q(x).
• Polynomials of one term, two terms and three terms are called monomial, binomial and trinomial respectively.
• A polynomial of degree one is called a linear polynomial.
• A polynomial of degree two is called a quadratic polynomial.
• A polynomial of degree three is called a cubic polynomial.
• A polynomial of degree four is called a biquadratic polynomial.
• A real number 'a' is a zero of a polynomial p(x) if p (a) = 0. 'a' is also called the root of the equation p(x) = 0.
• Every linear polynomial in one variable has a unique zero.
• A non-zero constant polynomial has no zero.
• Every real number is a zero of the zero polynomial.
• The degree of a non-zero constant polynomial is zero.
• The degree of a zero polynomial is not defined.
• If p(x) and g(x) are two polynomials such that degree of p(x) $\ge$ degree of g(x) and g(x)$\ne$0, then we can find polynomials q(x) and r(x) such that p(x) = g(x) q(x) + r(x).
• Factor theorem:
• Let f(x) be a polynomial of degree in > 1 and 'a' be any real number. Then
(x - a) is a factor of f(x) if (a) = 0. (a) = 0 if (x - a) is a factor of f(x). If x - 1 is a factor of a polynomial of degree 'n' then the sum of its coefficients is zero.
• Remainder theorem:
• If p(x) is any polynomial of degree greater than or equal to 1 and p(x) is divided by the line polynomial x - a (where 'a' is any real number) then the remainder is p (a).
• We can express p(x) as p(x) = (x-a) q(x) +r(x) where q(x) is the quotient and r(x) is U remainder.
• The process of writing an algebraic expression as the product of two or more algebra expressions is called factorization.
• Some important identities:
• ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
• ${{\left( a\text{ }-\text{ }b \right)}^{2}}=\text{ }{{a}^{2}}-2ab+{{b}^{2}}$)
• $~\left( a+b \right)\text{ }\left( a-b \right)\text{ }=\text{ }{{a}^{2}}-{{b}^{2}}$
• ${{\left( a\text{ }+\text{ }b\text{ }+\text{ }c more... • #### Co-ordinate Geometry Co-ordinate Geometry • Co-ordinate Geometry: The branch of mathematics in which geometric problems are solved through algebra by using the coordinate system is known as coordinate geometry. • In coordinate geometry, every point is represented by an ordered pair, called coordinates of that point. • A pair of numbers 'a' and V listed in a specific order with 'a' at the first place and 'b' at the second place is called an ordered pair (a, b). Note :(i) (a, b) \[\ne$ (b, a) (ii) If (a, b) = (c, d) then a = a and b=d.
• The position of a point in a plane is determined with reference to two fixed mutually perpendicular lines called the coordinate axes.
• The horizontal line is called X-axis and the vertical line is called Y-axis.
• The point of intersection of the coordinate axes is called origin 0(0, 0).
• In a point P (a, b), 'a' is called x-coordinate or first coordinate or abscissa and 'b' is called y- coordinate or second coordinate or ordinate.
• The axes divide the plane into four quadrants,
(i) ${{Q}_{1}}$is the I quadrant? Here both x and y are positive i.e., x > 0 and y > 0. The ordered pair (a, b) belongs to this quadrant. (ii) ${{Q}_{2}}$is the II quadrant. Here x is negative and y is positive i.e., x < 0 and y > 0. The ordered pair (-a, b) belongs to this quadrant. (iii) ${{Q}_{3}}$is the III quadrant. Here both x and y are negative, i.e., x < 0 and y < 0. The ordered pair (-a,- b) belongs to the quadrant, (iv) ${{Q}_{4}}$is the IV quadrant. Here x is positive and y is negative i.e., x > 0 and y < 0. The ordered pair (a, -b) belongs to this quadrant.
• The coordinates of any point on X - axis is of the form (a, 0) [y-coordinate zero],
• The coordinates of any point on Y - axis is of the form (0, b) [x-coordinate zero].
#### Linear Equations in Two Variables
Linear Equations in Two Variables
• Equation: A statement of equality of two algebraic expressions involving a variable is called an equation.
• Simple linear equation: An equation which contains only one variable of degree 1 is called a simple linear equation.
• Solution of an equation: The value of the variable, which when substituted in the given equation, makes the two sides L.H.S (Left Hand Side) and R.H.S (Right Hand Side) of the equation equal is called the solution of that equation.
• Transposition: Any term of an equation may be taken to the other side with a change in its sign. This process is called transposition.
• Cross multiplication: If $\frac{\operatorname{ax}+b}{\operatorname{cx}+d}=\frac{p}{q}$then q (ax + b) = p (cx + d). This process is called cross multiplication.
• Linear equation in one variable: An equation of the form ax + b = 0 or ax = c, is a linear equation in one variable x, where $a\ne 0$and a, b and c are real numbers.
• Solution of a linear equation in one variable: The value of the variable (x) for which both the sides of the equation become equal is said to be the solution of the equation.
Note: A solution is also called the ‘root’ of the equation.
• Linear equation in two variables:
(i) An equation of the form ax + by + c = 0, where a, b and c are real numbers, such that 'a' and 'b' are not both zero, is called a linear equation in two variables. (ii) A linear equation in two variables has infinitely many solutions. (iii) The graph of every linear equation in two variables is a straight line. (iv) x = 0 is the equation of Y-axis. (v) y = 0 is the equation of X-axis. (vi) The graph of x = a is a straight line parallel to the Y-axis. (vii) The graph of y = a is a straight line parallel to the X-axis. (viii) An equation of the type y = mx represents ‘a line passing through the origin. (ix) Every point on the graph of a linear equation in two variables is a solution of the linear equation. Conversely, every solution of the linear equation is a point on the graph of the linear equation.
#### Introduction to Euclid's Geometry
Introduction to Euclid's Geometry
• Axioms: Axioms or postulates are the assumptions which are obvious universal truths and are not to be proved.
• Some of the axioms given by Euclid: (i) Things which are equal to the same thing are equal to one another. i.e., if a = c and b = c, then a = b.
(ii) If equals are added to equals, the wholes are equal. i.e., if a = b and c = d, then a + c = b + d. Also a = b $\Rightarrow$a+c=b+c. Here, a, b, c and d are same kind of things. (iii) If equals are subtracted from equals, the remainders are equal. (iv) The things which coincide with one another are equal to one another. (v) The whole is greater than the part. i.e., if a > b, then there exists 'c' such that a = b + c. Here, 'b' is a part of 'a' and therefore, 'a' is greater than 'b'. (vi) Things which are double the same things are equal to one another.
• (vii) Things which are halves of the same things are equal to one another.
• Euclid's five postulates:
(i) Postulate 1: A straight line may be drawn from any one point to any other point. (ii) Postulate 2: A terminated line (i.e., a line segment) can be produced indefinitely on either side to form a line. (iii) Postulate 3: A circle can be drawn with any centre and any radius. (vi) Postulate 4: All right angles are equal to one another, (v) Postulate 5: If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced a indefinitely, meet on that side on which the sum of angles is less than two right angles.
• Theorems or propositions are the properties which are to be proved, using definitions, axioms/postulates, previously proved statements and deductive reasoning.
• Any two distinct straight lines are either intersecting or parallel. If the two lines are intersecting, then they can be oblique to each other or perpendicular to each other.
• Intersecting lines: Two distinct straight lines which meet each other at a point are called intersecting straight lines.
• Perpendicular lines: Two intersecting lines are perpendicular to each other if one meets the other at a point and the angles made by the first with either side of the second at the point more...
#### Lines and Angles
Lines and Angles
• Angle: An angle is the union of two rays with a common initial point. An angle is denoted by symbol$\angle$. It is measured in degrees,
The angle formed by the two rays $\overline{AB}\,\,and\,\,\overline{AC}\text{ }is\text{ }\angle BAC\text{ }or\text{ }\angle CAB.~$called $\overline{AB}\,\,and\,\,\overline{AC}$are called the arms and the common initial point ‘A’ is called the vertex of the angle.
• Bisector of an angle: A line which divides an angle into two equal a parts is alled the bisector of the angle.
e.g., In the adjacent figure, the line OP divides $\angle$AOB into two Equal parts. $\angle AOP=\angle POB={{\operatorname{x}}^{o}}$ So, the line OP is ‘called the bisector of $\angle$AOB.
• Pairs of angles:
(i) Complementary angles: Two angles are said to be Complementary if the sum of their measures is equal to$~{{90}^{o}}$ Here$\angle x+\angle y={{90}^{o}},$therefore $\angle x\,\operatorname{and}\,\angle y$ Complementary angles: (ii) Supplementary angles: Two angles are said to be supplementary if the sum of their measures isequal to${{180}^{o}}$. Here $\angle x+\angle y={{180}^{o}},$therefore $\angle x\,\operatorname{and}\,\angle y$ Supplementary angles.
• Adjacent angles: Angles having the same vertex and a common arm, and the non-common arms lie on the opposite sides of the common arm are called adjacent angles.
• $\angle AOB\text{ }and\text{ }\angle COB$with common vertex 0 and common arm OB are adjacent angles,
Note: $\angle AOC=\angle AOB+\angle BOC$
• Linear pair of angles: Two adjacent angles make a linear pair of angles, if the non-common arms of these angles are two opposite rays (with same end point),
• In the adjacent figure, $\angle BAC\text{ }and\text{ }\angle CAD$form a linear pair of angles because the non -common arms AB and AD of the two angles are two opposite rays.
Moreover,$\angle \text{ }BAC\text{ }+\text{ }\angle more... #### Triangles Triangles • A triangle is a closed figure bounded by three straight lines. It is denoted by the symbol\[\Delta$.
$\Delta$ABC has three sides denoted by AB, BC and CA; three angles denoted by $\angle ~A,\angle B\text{ }and\text{ }\angle C\,;$and three vertices denoted by A, B and C.
• Two geometrical figures are said to be congruent if they have exactly the same shape and size. Congruence is denoted by the symbol =.
• Two triangles are congruent if the sides and angles of one triangle are equal to the corresponding sides and angles of the other triangle.
• The congruence of two triangles ABC and PQR under the correspondence $A\leftrightarrow P,B\leftrightarrow Q$
and $C\leftrightarrow R$is symbolically expressed as $\Delta ABC=\Delta PQR.$
• Two congruent figures are equal in area, but two figures having the same area need not be congruent.
• Congruence relation is an equivalence relation:
(i) Congruence relation is reflexive. $\Delta ABC=\Delta ABC.$ (ii) Congruence relation is symmetric. If $\Delta ABC\text{ }\cong \text{ }\Delta DEF,\text{ }then\text{ }\Delta DEF\text{ }\cong \text{ }ABC\text{ }.$ (iii)Congruence relation is transitive.
• If $\Delta ABC\text{ }=\text{ }\Delta DEF\text{ }and\text{ }\Delta DEF\text{ }=\text{ }\Delta XYZ\text{ }then\text{ }\Delta ABC\text{ }=\text{ }\Delta XYZ.$
• Criteria for congruence of triangles:
• (i) A.S. congruence rule: Two triangles are congruent if two sides and the included angle of
one triangle are equal to the two sides and the included angle of the other triangle. $\Delta ABC=\Delta PQR$ $Since=PQ=7cm,\text{ }\angle C=PR=5cm\,\,and\,\angle A=\angle P=50{}^\circ .$corresponding sides and angles of the other triangle. (ii) A.S.A. congruence rule: Two e.g., triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of the other triangle. $\Delta ABC\cong \Delta DEF$ $\angle B=\angle E={{45}^{o}},\angle C=\angle F=30{}^\circ \text{ }and\text{ }BC\text{ }=\text{ }EF\text{ }=\text{ }5cm.$ (iii) S.S.S. congruence rule: If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent. $\Delta ABC=\Delta XYZ$ Since AB = XY = 5 cm, BC = YZ = 7 cm and CA = ZX = 6 cm.
• H.S.congruencerule: fin two right triangles more...
• A quadrilateral in which the measure of each angle is less than 180° is called a convex quadrilateral.
• A quadrilateral in which the measure of at least one of the angles is more than ${{180}^{o}}$s known as a concave quadrilateral.
• The sum of the angles of a quadrilateral is ${{360}^{o}}$ (or) 4 right angles.
• When the sides of a quadrilateral are produced, the sum of the four exterior angles so formed ${{360}^{o}}$
(i) Trapezium: (a) A quadrilateral having exactly one pair of parallel sides is called a trapezium. (b) A trapezium is said to be an isosceles trapezium if its non-parallel sides are equal. ABCD is a trapezium in which AB || DC. This trapezium is said to be an isosceles trapezium if AB || DC and AD = BC. (ii) Parallelogram: A quadrilateral in which both pairs of opposite sides are parallel is called a parallelogram. ABCD is a parallelogram in which AB || DC and AD II BC. Properties: (a) In a parallelogram, any two opposite sides are equal. (b) In a parallelogram, any two opposite angles are equal. (c)In a parallelogram, the diagonals bisect each other. (d) In a parallelogram, each diagonal divides it into two congruent triangles. (e) In a parallelogram, any two adjacent angles have their sum equal to ${{180}^{o}}$ i.e. the adjacent angles are supplementary. (iii) Rhombus: A quadrilateral having all sides equal is called a rhombus. ABCD is a rhombus in which AB II DC, AD || BC and AB = BC = CD = DA. Properties: (a) The diagonals of a rhombus bisect each other at right angles. (b) Each diagonal of a rhombus divides it into two congruent triangles. (c) Opposite angles of a rhombus are equal and the sum of any two adjacent angles is${{180}^{o}}$ (d) The opposite sides of a rhombus are parallel. (e) All the sides of a rhombus are equal. (iv) Rectangle: A parallelogram whose angles are all right angles is called a rectangle. ABCD is a rectangle in which, AD || BC and AB || CD and $\angle A=\angle B=\angle C=\angle D={{90}^{o}}$
• Properties:
(a) Opposite sides of a rectangle are equal and opposite angles of a rectangle are equal. (b) The diagonals of a rectangle bisect each other. (c) Each diagonal divides the rectangle into two congruent triangles. (d) The diagonals of a rectangle are equal. (v) Square: A parallelogram having all sides equal and each angle equal to a right angle is called a square. ABCD is a square in which AB || DC, AD more...
#### Area of Parallelograms and Triangles
Areas of Parallelograms and Triangles
• A polygonal region is the union of a polygon and it’s interior. For e.g., the union of a triangle and its interior is called the triangular region.
• Every polygonal region has area. Area of a figure is a number associated with the part of the plane enclosed by that figure.
• Two congruent figures have equal areas but figures with equal areas need not be congruent.
• If ABCD is a rectangle with AB = $l$ m and BC = b m, then the area of the rectangular region ABCD is$lb\text{ }sq.\text{ }m\text{ }or\text{ l}b\text{ }{{m}^{2}}$.
• If A and B are two regions having at the most a line segment common between them two, then the area of their combined region S is equal to the sum of their areas taken separately. i.e, ar (S) = ar (A) +ar (B).
• The area of a parallelogram is the product of any of its sides and the corresponding altitude.
• Two figures are said to be on the same base and between the same parallels, if they have a common base (side) and the vertices, (or the vertex) opposite to the common base of each figure lie on the same line parallel to the base.
• A diagonal of a parallelogram divides it into two triangles of equal areas.
• Parallelograms on the same (or equal) base and between the same parallel lines are equal in area.
• Parallelograms on the same base (or equal bases) and having equal areas lie between the same parallels.
• Parallelograms on equal bases and between the same parallel lines are equal in area.
• Area of a triangle is half the product of any of its sides and the corresponding altitude.
• Triangles on the same base and between the same parallel lines are equal in area.
• Two triangles having the same base (or equal bases) and equal areas lie between the same parallel lines.
• If a triangle and a parallelogram are on the same base and between the same parallel lines, then the area of the triangle is equal to half that of the parallelogram.
• The area of a trapezium is half the product of its height and the sum of the parallel sides.
• Triangles with equal areas and having one side of one triangle, equal to one side of the other, have their corresponding altitudes equal.
• A rectangle and a parallelogram on the same base and between the same parallels have more...
#### Circles
Circles
• A circle is a closed figure in a plane formed by
The collection of all the points in the plane which centre are at a constant distance from a fixed point in the plane. The fixed point is called the centre of length of radlus the circle and the constant distance is called the radius of the circle.
• The plane region inside the circle is called tne interior of the circle,
• If a circle is drawn in the plane X (infinite dimensions), then the part of the plane region outside the circular region is called the exterior of the circle.
• The circumference of a circle is the length of the complete circular curve constituting the circle, given by circumference, $C=\text{2}\pi r$where r is the radius of the circle.
• Any two points A and B of a circle, divide the circle into two parts. The smaller part is called a minor arc of the circle denoted by $\overset\frown{AB}$(read as arc AB).The larger part is called a major arc denoted by q$\overset\frown{APB}$or BA (read as arc$\overset\frown{BA}$).
• A line segment joining two points on the circumference of the circle, is called a chord of the circle. In the figure, AB is a chord.
• A chord passing through the centre O of the circle is called a diameter of the circle.
A diameter of a circle is the longest chord of the circle and its length is twice that of the radius of the circle. Diameter, d = 2r where r is the radius of the circle.
• A diameter of a circle divides it into two equal arcs. Each of these two arcs is called a semicircle. In the figure AB is a semicircle.
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http://mathhelpforum.com/calculus/62469-series-convergence-divergence.html | # Thread: series convergence and divergence
1. ## series convergence and divergence
hey first of all i d like to start this problem off with saying i have no idea what the ! means so i cant even start this problem.. even if the ! wasnt there i d assume this was a geometric series since its in that section of the book. Other then that i know if it converges i need to find the sum. any help appreciated.
2. n! ... is read "n" factorial.
e.g.
$\displaystyle 5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$
$\displaystyle n! = n(n-1)(n-2)(n-3)...(3)(2)(1)$
familiar with the ratio test for convergence/divergence ?
3. Originally Posted by Legendsn3verdie
hey first of all i d like to start this problem off with saying i have no idea what the ! means so i cant even start this problem.. even if the ! wasnt there i d assume this was a geometric series since its in that section of the book. Other then that i know if it converges i need to find the sum. any help appreciated.
$\displaystyle \frac{n^n}{n!}=\frac{\overbrace{n\cdot{n}\cdot{n}\ cdots}^{n\text{ number of times}}}{\underbrace{n\cdot(n-1)\cdots}_{n\text{ number of times}}}=\frac{n}{n}\cdot\frac{n}{n-1}\cdots\geqslant{1\cdot{1}\cdots=1}$
So we can see that
$\displaystyle \sum_{n=1}^{\infty}1\to\infty\leqslant\sum_{n=1}^{ \infty}\frac{n^n}{n!}$
4. $\displaystyle \sum_{n=1}^{\infty}\frac{n^{n}}{n!}$
The ratio test may be a decent choice here.
$\displaystyle {\rho}=\lim_{n\to +\infty}\frac{u_{n+1}}{u_{n}}=\lim_{n\to +\infty}\frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n ^{n}}$
$\displaystyle =\lim_{n\to +\infty}\frac{(n+1)^{n}}{n^{n}}$
$\displaystyle =\lim_{n\to +\infty}\left(1+\frac{1}{n}\right)^{n}=e$
Since e>1, then it diverges.
If you do not recognize it, that last limit is a very famous one. Therefore, one does not have to bother proving it every time. | {"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": 0.9619584679603577, "perplexity": 402.367854960645}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267868135.87/warc/CC-MAIN-20180625150537-20180625170537-00021.warc.gz"} |
https://enricosanino.wordpress.com/2015/12/08/ntc-measurements/comment-page-1/ | NTC Measurements (Pt. 1)
When dealing with low-cost, home made termometers, the choice often fall on the NTCs. These sensors are resistors which are varying their resistance in function of the temperature, with a negative trend: Negative Temperature Coefficient is their name, also known as Thermistor. That means higher the temperature, lower the resistance:
We observe that this cheap sensor have the price to be non-linear at all. There are various methods to derive a method to interpret the correct temperature. Here I will go through the ones that are used to achieve quite reasonable precision without having/paying a calibration laboratory.
Look-Up Table approach
Usually the datasheet provides a set of values which are sampled from a sensor and correspond to our being inside their stated tolerance:
In this picture you can see the stated resistance at a certain temperature, and its tolerance. An immediate drawback, if no other tables are available, is that this table can bring a precision of up to 5°C only. The parameter “B” is called normally $\beta$.
Moreover, a sensor with a tolerance of 5%, refers to 25°C only. This error of the resistance must be added with the “$\frac{\Delta R}{R}$ due to β toll.” factor in the table above, from now on called ΔR/β for simplicity, which is given from the datasheet. Associated with it, there is the temperature coefficient, TCR (also known as $\alpha$), which describes how steep is the curve. You may understand that with a very steep curve, so an high TCR, there is a little deviation of the temperature in the X axis, associated with a high variation of resistance on the Y axis therefore we have an high sensitivity. The situation is the opposite when reading high temperatures with a low TCR, the sensitivity drops rapidly.
The thermistor error of a punctual read is then devised as follow:
$\frac{\Delta R_{therm}}{R_{therm}} = (((1 + \frac{\Delta R_{25^{\circ}C}}{R_{25^{\circ}C}}) \cdot (1 + \frac{\Delta R}{\beta})) - 1) \cdot 100$ (eq. 1)
where $R_{25^{\circ}C}$ is the resistance tolerance at the reference temperature (specified in the datasheet, here is 25°C), and $\beta$ is a coefficient which characterize the NTC material, devised by measuring two different temperature (and is specified in the datasheet). If we need to keep the right tolerance after changing the sensor with another of the same model without calibration, we need to observe these tolerances. Combineing them with the TCR, we obtain the temperature’s punctual error:
$\Delta T = \frac{\Delta R_{therm}}{R_{therm}} / TCR$ (eq. 2)
Equation approach
There are a lot of methods which are used to linearize the behaviour , by linearizing the model. The most famous is the Steinhart-Hart equation, which uses a set of coefficients which are provided in two ways: from manufacturer, or can be devised by measuring 3 different temperatures and solving 3 equations for 3 unknowns. If these coeffients are not provided from manufacturer, one can use a more precise termometer, measure 3 temperatures and its resistance temperature, solving this equation for the S-H of 3rd order:
The sensor’s manufacturer adopted to experiment provides the coefficients up to 4th order, both for reversed anddirect measurement:
With a sensor stated to have 5% of tolerance, one can actually use the coefficient without the full decimal precision instead used (ideally) from the manufacturer, because the error provided by such formula is in the order of mK, while the final reading, due to various errors, is higher than 1.5K. I neglect this S-H error.
Error estimation
Assuming a negligible error from the S-H calculus that uses the parameters given by manufacturer, usually we use a microcontroller with an ADC and a voltage divider. The error from a common ADC is half LSB, so that
$\frac{\Delta ADC}{ADC} \% = \frac{1}{2^{(N+1)}} = 0.5 LSB$ (eq. 3)
The error of the voltage divider tends to be double of the two resistors used if are too much different, otherwise tends to be the mean of the two relative errors of the two resistors, like $\frac{\Delta R1 + \Delta R2}{2}$. See the graph, where 100% is the mean value between errors of the 2 resistors, and 200% represent the sum of the 2 errors:
These two resistors are used in this way, in which one is the Thermistor:
where at the reference temperature (provided by manufacturer and usually 25°C) the Rref and Rtherm have the same value (so we bought a matched thermistor with a certain resistance to reduce errors). Depending on your temperature range, you can se how varies the the ratio and see how greater can be considered the total error.
How greater is the error in my temperature range? The previous value of $\frac{\Delta R_{therm}}{R_{therm}}$ and its TCR will lead to a temperature error like this:
in which the ADC and condition circuitry errors are NOT considered. But an idea of performance can be made if no calibration is performed (see later). If the resistance at the extremes of my range is not so different from the reference resistor (a normal fixed resistor in the schematic above), then it is not a mandatory to sum it up both errors of the two resistors, but can be a little less.
If I need to measure between 0°C and 100°C, the datasheet provides the additional error of the resistance due to the tolerance of β parameter, called Δβ/β. We will find that at 100°C there is low TCR and high relative error. ONLY NOW we can apply the worst case total error $\Delta T = \frac{\Delta R}{R} / TCR$ with $\frac{\Delta R}{R} = \frac{\Delta R_{therm 100^{\circ}C}}{R_{therm 100^{\circ}C}} + \frac{\Delta R_{ref}}{R_{ref}}$, where $R_{therm 100^{\circ}C}$ is the (eq. 1) at temperature of 100°C using the table from manufacturer, while Rref is the the fixed resistor in the schematic above and $\frac{\Delta R_{ref}}{R_{ref}}$ its relative error.
TCR will be chosen to achieve the higher relative error, so will be the TCR at 100°C (as said before, the higher temperature of the range), along with the estimated resistance value at that temperature (of course..). You may see how the error can be greatly reduced if reading values with higher TCR at lower temperatures, and how small is it at 25°C. But the boundaries must contains the greater error tolerance, allowing the user to change the sensor in the field without recalibration.
With the calibration using the set of 3 equations above, all these errors are compensated, voltage divider included. The remaining one will be truncation error of the S-H coefficients due to the finite machine precision (whether is a PC or an MCU used to make the calibration), the errors of the reference termometer and the intrinsic errors of the S-H model, the quantization error of the ADC (half LSB) and for sure others that I have missed. It is not trivial to quantify everything. And quantification, when talking about measures, is almost everything.
We have found how greater is the error of the analog quantities. Now where is the least sensible part of the NTC curve? The one at the higher temperature, as said before (lower TCR). Until now I have estimated a certain error of the total voltage divider’s resistance.
Now is needed to find how an ADC error can mismatch the resistance. Let’s go at 100°C, using an S-H estimation or the Look-Up table, then calculate a sort of manual derivative, let’s say the value $R_{therm} = R_{100^{\circ}C} + \Delta R$, where ΔR is the immediate available step to achieve a temperature lower of a step equal to the required precision (if I want a precision of 1°C, then is the resistance at 99°C if I have the S-H equation, or it is the resistance at 95°C if I have a rough Look-Up table like the one in this article).
From the circuit of the voltage divider, we have $V_{adc1} = V_{refADC} \cdot \frac {R_{100^{\circ}C}}{R_{100^{\circ}C}+R_{REF}}$ and $V_{adc2} = V_{ref ADC} \cdot \frac {R_{100^{\circ}C} + \Delta R}{R_{100^{\circ}C} + \Delta R +R_{REF}}$. The $\frac{V_{adc1} - V_{adc2}}{V_{LSB}}$ is how many discrete steps can be sampled inside a ΔR variation. E.g., if 1LSB = 3mV (ADC provides 3mV/bit) and from 100°C to 99°C the variation read from ADC is 6mV, I can’t have an accuracy higher than 2LSB, meaning 0.5°C (2LSB to represent 1°C). If I am lower than 1LSB, I can’t discern my prefixed step lower than 1°C.
Saying the same more mathematically: consider the reference voltage applied to voltage divider to be 3V. And the datasheet provides a certain TCR at 100°C. Then the resistance at 99°C will be:
$\Delta R_{1^{\circ}C} = R_{100^{\circ}C} - R_{100^{\circ}C} \cdot \frac{TCR_{100^{\circ}C}}{100}\cdot 1^{\circ}C$
so that:
$1.5mV \leq 3V \cdot \frac{R_{100^{\circ}C}}{R_{100^{\circ}C} + R_{REF}} - 3V \cdot \frac{R_{100^{\circ}C} - \Delta R_{1^{\circ}C}}{R_{100^{\circ}C} - \Delta R_{1^{\circ}C} + R_{REF}} = \Delta V_{1^{\circ}C}$
Finally, the additional temperature error from the ADC is, in the worst case:
$\Delta T_{ADC} = \frac {1^{\circ}C}{\frac{\Delta V_{1^{\circ}C}}{0.5LSB[V]}}$ (eq. 4)
Conclusions
The final precision, from (eq.2) and (eq. 4), is:
$\Delta T = \frac{\frac{\Delta{R}}{R}}{TCR_{100^{\circ}C}} + \Delta T_{ADC}$
One can try and find out that with an ADC of N = 10 bit, and components at 5%, included the NTC, in range between 0°C-100°C, hardly can be achieved a precision lower than 3°C/4°C, despite the accuracy can be around 0.5°C/1°C, without calibration. But note that this low precision is due to the consideration of the range up to its most imprecise extreme: reducing the range to, let’s say, 60°C the precision can be improved a lot. Just keep that in mind when you read 25°C, or 150°C using an NTC.
Annunci | {"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": 22, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8757056593894958, "perplexity": 1006.9168313282403}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125948426.82/warc/CC-MAIN-20180426164149-20180426184149-00193.warc.gz"} |
https://www.physicsforums.com/threads/magnetic-pinches-and-mhd.366912/ | # Magnetic pinches and MHD
1. Jan 3, 2010
### Crusoe
Can someone give me an intuitive explanation why plasmas or conductive fluids tend to follow magnetic flux lines?
E.g. http://en.wikipedia.org/wiki/Magnetohydrodynamics" [Broken] entry on this says:
If I take a stab at a qualitative explanation, is it because magnetic flux lines are equipotential lines, therefore if the conductor deviates from the flux lines, induced currents are produced which oppose the motion in accordance with Lenz's law?
Also, I know the explanation for Lenz's law is conservation of energy, but it just seems odd that induced currents so happen to be set up to precisely keep energy conserved.
What causes magnetic pinching, e.g. in lightning bolts or even lightning rods (below) then? I would have thought an axial current along a conductive fluid would cause charged particles within it to follow a circular orbit. Where do they get the centripetal force to both follow such an orbit, and in fact exceed that for a stable orbit, and end up reducing their orbital radii?
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http://wicomdsp.blogspot.com/2015/03/frequency-error-estimation-range-and.html | ## Pages
### Frequency Error Estimate Range and Resolution Determination
Frequency Error Estimation Range and Resolution Determination Frequency error estimation in IEEE 802.11 systems is done over preamble Short Training Field (STF) and Long Training Field (LTF) signals.
Time domain signal of STF and LTF has period 0.8us and 3.2us respectively. With this knowledge, frequency error can be computed through shifted cross-correlation operation.
STF signal assist in computing coarse estimate of frequency error and a fine estimate can be obtained from LTF.
Frequency Error Estimate Range
Range of frequency errors that can be computed from STF will be [-1/(2*0.8u) to 1/(2*0.8u)] or [-625khz to +625khz]
Range of frequency errors that can be computed from LTF will be [-1/(2*3.2u) to 1/(2*3.2u)] or [-156.25khz to +156.25khz]
Above ranges can be obtained by equating the angle of $e^{j2\pi f T_p}$ to $+/- \pi$. Where $T_p$ is the period of STF or LTF.
Frequency Error Estimate Resolution
Resolution of frequency error estimate has significance in real time systems. Resolution of fixed point implemenation determine the resolution of the frequency error estimation algorithm.
A $'n'$ bit precision fixed point implementation will have an ampitude resolution of $2^{-n}$. To know the resolution of frequency error estimate, it is required to know the smallest possible phase represented in the implemented fixed point.
Let us say $\cos\left(x\right)$ is well represented by the fixed point. If $\Delta = (2 \pi f T_p)$ be the possible phase resolution by the fixed point, then $\cos\left(x +\Delta\right)~ - ~\cos\left(x\right) = -2^{-n}$ gives rise to $\Delta \sin\left(x\right) = 2^{-n}$.
Similarly, Let us say $\sin\left(x\right)$ is well represented by the fixed point, then $\sin\left(x +\Delta\right)~ - ~\sin\left(x\right) = 2^{-n}$ gives rise to $\Delta \cos\left(x\right) = 2^{-n}$.
From the above $\Delta = +/- \sqrt{2} . 2^{-n}$ or $f = +/- \sqrt{2} . 2^{-n} . \frac{1}{2 \pi T_p}$ | {"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": 0.977301299571991, "perplexity": 1251.6859037824074}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676591543.63/warc/CC-MAIN-20180720061052-20180720081052-00530.warc.gz"} |
http://math.stackexchange.com/questions/256886/a-p-adic-jacobi-sum-with-an-unramified-character | # A $p$-adic “Jacobi sum” with an unramified character
Working over a $p$-adic field with absolute value $|\cdot|$, let $\chi$ be a character on ${\mathfrak o}^\times$ with conductor $n\ge 1$, meaning that $n$ is the smallest integer such that $\chi$ is trivial on $U_n=1+{\mathfrak p}^n$. I'm trying to calculate the integral, with ${\rm Re}(s)\gg 1$, $$J(\chi,|\cdot|^s)=\int_{\mathfrak o^\times}\chi(x)|x-1|^s\ dx$$ Note that if we replace $|\cdot|^s$ with a character $\chi'$ with conductor $n$, this would reduce to the typical Jacobi sum, and we'd have the formula $$J(\chi,\chi')={G(\chi,\psi)G(\chi',\psi)\over G(\chi\chi',\psi)}$$ where $\psi$ is an additive character with conductor $n$ and $G(\chi,\psi)$ is the Gauss sum $$G(\chi,\psi)=\int_{\mathfrak o^\times}\chi(x)\psi(x)\ dx$$ Of course, this formula wouldn't work for $\chi'$ unramified.
Is anyone aware of how to calculate this sort of Jacobi sum?
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I think I've figured out how to do it. It's essentially the same argument as in the "classical" case, just interpreted slightly differently. Here's the sketch, which may lose some change-of-measure constants. $$J(\chi,|\cdot|^s)=\int_{\mathfrak o^\times}\chi(x)|x-1|^s\ dx$$ Extend $\chi$ by zero to all of $k$ and apply Fourier inversion to get $$=\int_{\mathfrak o^\times}\int_k |x-1|^s\psi(-ax)\int_{\mathfrak o^\times}\chi(y)\psi(ay)\ dy\ da\ dx$$ where $\psi$ has conductor $0$. The integral over $y$ vanishes unless ${\rm ord}(a)=-n$, so the integral becomes $$=\int_{\mathfrak o^\times}\int_{\mathfrak o^\times}|x-1|^s\psi(-\varpi^{-n}ux)\int_{\mathfrak o^\times}\chi(y)\psi(\varpi^{-n}uy)\ dy\ du\ dx$$ $$=\int_{\mathfrak o^\times}\int_{\mathfrak o^\times}|x-1|^s\psi(-\varpi^{-n}ux)\bar\chi(u)\int_{\mathfrak o^\times}\chi(y)\psi(\varpi^{-n}y)\ dy\ du\ dx$$ $$=G(\chi,\psi_{\varpi^{-n}})\int_{k}\int_{\mathfrak o^\times}{\bf 1}_{\mathfrak o^\times}(x)|x-1|^s\psi(-\varpi^{-n}ux)\bar\chi(u)\ du\ dx$$ $$=G(\chi,\psi_{\varpi^{-n}})\int_{k}\int_{\mathfrak o^\times}{\bf 1}_{\mathfrak o^\times}(1+x)|x|^s\psi(-\varpi^{-n}ux)\psi(-\varpi^{-n}u)\bar\chi(u)\ du\ dx$$ $$=q^{-ns-1}G(\chi,\psi_{\varpi^{-n}})\int_{k}\int_{\mathfrak o^\times}{\bf 1}_{\mathfrak o^\times}(1-\varpi^{n}u^{-1}x)|x|^s\psi(x)\bar\psi(\varpi^{-n}u)\bar\chi(u)\ du\ dx$$ Now, as a function of $x$, the characteristic function ${\bf 1}_{\mathfrak o^\times}(1-\varpi^{n}u^{-1}x)$ differs from ${\bf 1}_{\mathfrak o}(\varpi^{n}x)$ only on a set of measure zero, so we have $$J(\chi,|\cdot|^s)=q^{-ns-1}\big|G(\chi,\psi_{\varpi^{-n}})\big|^2\int_{\varpi^{-n}\mathfrak o}|x|^s\psi(x)\ dx$$ This final integral can be calculated using the well-known formula for $\int_{|x|=p^i}\psi(x)\ dx$, giving (if I did everything correctly), $$J(\chi,|\cdot|^s)=q^{-ns-1}\big|G(\chi,\psi_{\varpi^{-n}})\big|^2{2-q^s\over 1-q^{-s}}$$
And, of course, the norm-square of the Gauss sum is $q^{-n}$. – B R Dec 12 '12 at 22:52 | {"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": 0.9972295761108398, "perplexity": 86.3756651671992}, "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-2014-10/segments/1393999655040/warc/CC-MAIN-20140305060735-00073-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/58709/understanding-the-definition-of-dense-sets | # Understanding the definition of dense sets
I am recently confused about the definition of dense sets. What I learned is as the following
In topology and related areas of mathematics, a subset $A$ of a topological space $X$ is called dense (in $X$) if any point $x$ in $X$ belongs to $A$ or is a limit point of $A$.
The point is that when we say "a set $A$ is dense in a topological space $X$" we should first have the fact that $A$ is a subset of $X$. However, there is a notion that "a set $E$ is dense in a ball $X$" as I mentioned in this question. In this notion, $E$ is not necessarily a subset of $X$.
[Edit: How should I understand the definition of dense sets?] Can anybody tell me a reference for the most general definition for the dense sets?
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It isn’t so much a matter of needing another definition as of getting the hang of how mathematical language evolves and is actually used ‘in the wild’. In the example in question, ‘$E$ is dense in the ball $B$’ can be thought of as verbal shorthand for ‘$E \cap B$ is a dense subset of the ball $B$’ $-$ which, when you think about it, is just about the only thing that it could reasonably mean.
How does such shorthand arise? Let’s say that you have the idea of a set $D$ in a topological space $X$ being dense in $X$ according to the definition that you gave. Any subset $Y$ of $X$ can be viewed as a space in its own right, with the subspace topology inherited from $X$, so you can certainly talk about a subset of $Y$ being dense in $Y$. But somewhere along the line you prove that $D$ is dense in $X$ iff $\operatorname{cl}D = X$; this characterization is far too useful to be omitted. What happens when you try to apply this definition to the subspace $Y$? It’s certainly true that a set $D \subseteq Y$ is dense in $Y$ iff $\operatorname{cl}_Y D = Y$, but if you’re thinking of $Y$ as a subset of $X$ rather than as a space in its own right, it’s more convenient to look at $\operatorname{cl}_X D$, which isn’t necessarily $Y$: it may be a proper superset of $Y$. (Take $X$ to be $E^1$, $Y$ to be $(0,1)$, and $D$ to be $\mathbb{Q} \cap (0,1)$.)
Thus, if we don’t want to bother with different closures in different sets, it’s much more convenient to say simply that a subset $D$ of a set $Y$ is dense in $Y$ iff $\operatorname{cl}D \supseteq Y$, where the closure operator is $\operatorname{cl}_X$, the closure in the whole space $X$. And once you get to that point, it’s sometimes convenient to separate the notion of dense in $Y$ from the notion of subset of $Y$ altogther. That is, we might as well say that $D$ is dense in $Y$ iff $\operatorname{cl}D \supseteq Y$, irrespective of whether $D \subseteq Y$. Then, for instance, we can say simply that the rationals are dense in $(0,1)$; we don’t have to say explicitly ‘the rationals in $(0,1)$’.
Many textbooks, including some very good ones (Munkres, Dugundji, Willard) define dense subset of a space but (so far as I can quickly tell) silently leave it to the reader to make the generalization to dense subset of a set in a space and the further generalization to dense in a set in a space, or to pick them up from context. Of course, one can define dense more generally. For example, John Greever’s Theory and Examples of Point-Set Topology gives this definition:
(2.14) Definition. If $(X,\mathscr{T})$ is a topological space and $A,B \subset X$, then $A$ is dense in $B$ if and only if $\overline{A} \supset B$.
Notice that he doesn’t require $A$ to be a subset of $B$: right from the start he’s using the most general of the three forms that I considered above. Had you learned the concept from Greever, you’d not have had a problem with the dense in a ball language. But whether with this concept or with another, you’re eventually going to encounter standard usages that require a little interpretation of or extrapolation from the ones that you’ve been taught.
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Dear Brian, I'm a bit confused by the assertion that cl $D \supseteq Y$ iff cl $D\cap Y \supseteq Y$. Couldn't $D \cap Y = \emptyset$, e.g. if $X = \mathbb R$, $D = \mathbb Q$, and $Y = \mathbb R\setminus \mathbb Q$? Or have I gotten things muddled? Best wishes, – Matt E Mar 11 '14 at 4:07
Brian, this is actually the same question as above. If we take the reals with the standard topology and put $Y=\{1\}$ and $D=(0,1)$, then $Y\subseteq\mathrm{cl}\,D$, yet $\mathrm{cl}\,(D\cap Y)=\emptyset$. – Mad Hatter Aug 17 '15 at 9:01
Indeed, if $Y\not\subset \overline{\operatorname{int}_X Y}$, then $Y\subset \overline{D}$ does not imply $Y\subset \overline{Y\cap D}$. – Daniel Fischer Aug 17 '15 at 14:56
In the second paragraph, $cl_Y(D)=cl_X(D)\cap Y$. Hence $cl_Y(D)=Y$ iff $cl_X(D)\cap Y=Y$ iff $cl_X(D)\supset Y$. I think this might be a reason why we don't need to bother with different closures in different sets. – Jack Sep 30 '15 at 14:36
@MadHatter: You and Matt E are quite right; fixed now. (I’m not sure how I missed yours before.) – Brian M. Scott Sep 30 '15 at 14:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9512075185775757, "perplexity": 130.34866144919187}, "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/1466783399117.38/warc/CC-MAIN-20160624154959-00142-ip-10-164-35-72.ec2.internal.warc.gz"} |
https://tel.archives-ouvertes.fr/tel-01819086v4 | Long-Moody functors and stable homology of mapping class groups
Abstract : Among the linear representations of braid groups, Burau representations are recovered from a trivial representation using a construction introduced by Long in 1994, following a collaboration with Moody. This construction, called the Long-Moody construction, thus allows to construct more and more complex representations of braid groups. In this thesis, we have a functorial point of view on this construction, which allows find more easily some variants. Moreover, the degree of polynomiality of a functor measures its complexity. We thus show that the Long-Moody construction defines a functor LM, which increases the degree of polynomiality. Furthermore, we define analogous functors for other families of groups such as mapping class groups of surfaces and 3-manifolds, symmetric groups or automorphism groups of free groups. They satisfy similar properties on the polynomiality. Hence, Long-Moody functors provide twisted coefficients fitting into the framework of the homological stability results of Randal-Williams and Wahl for the afore mentioned families of groups. Finally, we give a comparison result for the stable homology with coefficient given by a functor F and the one with coefficient given by the functor LM(F), obtained applying a Long-Moody functor. This thesis has three chapters. The first one introduces Long-Moody functors for braid groups and deals with their effect on the polynomiality. The first one deals with the generalisation of Long-Moody functors for other families of groups. The last chapter touches on stable homology computations for mapping class group.
Keywords :
Document type :
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Cited literature [76 references]
https://tel.archives-ouvertes.fr/tel-01819086
Contributor : Abes Star <>
Submitted on : Monday, May 6, 2019 - 10:39:06 AM
Last modification on : Thursday, August 29, 2019 - 2:43:16 PM
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Arthur Soulié. Long-Moody functors and stable homology of mapping class groups. Algebraic Topology [math.AT]. Université de Strasbourg, 2018. English. ⟨NNT : 2018STRAD016⟩. ⟨tel-01819086v4⟩
Record views | {"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": 0.9079681634902954, "perplexity": 1286.0600650731378}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573465.18/warc/CC-MAIN-20190919081032-20190919103032-00338.warc.gz"} |
http://tex.stackexchange.com/questions/127601/setting-the-size-of-an-image-using-tkz-fct | # Setting the size of an image using tkz_fct
I'm working on our coding system which combines a number of software possibilities. Here is what I have thus far.
### TOPIC 1
\topic = Sullivan 10.2 easy: parabolas
|@|
### MODEL 1.1, QUESTION 1
\versions = 10
|@|
\code =
my $a = randint(-9,9,1,(0)); my$b = 4*$a; my$Xmin=-4*$a; my$Xmax=4*$a; my$Xstep=$a; my$Ymin=-4*$a; my$Ymax=4*$a; my$Ystep=$a; |@| \qutext=What is the focus of the parabola$x^2={#b}y? |@| \mode = point |@| \answer = (0,#a) |@| \soltext=Compare the equation $$x^2={#b}y$$ with the form $$x^2=4ay$$ and note that: \begin{align*} 4a&=#b\\ a&=#a \end{align*} The parabola having the formx^2=4ay$opens up or down, depending on the value of$a$. In this case,$a=#a$, so the focus is at the point$(0,#a)\$.
\begin{center}
\begin{tikzpicture}
\tkzInit[xmin=#Xmin,xmax=#Xmax,xstep=#Xstep,ymin=#Ymin,ymax=#Ymax,ystep=#Ystep]
\tkzGrid
\end{tikzpicture}
\end{center}
|@|
\endquestion|@|
As you can see near the bottom, I'm trying to use tkz_fct to draw a graph based on some of the perl settings in the \code section above. What I am having trouble with is the fact that the size of the final grid on screen will change with different values of a. I'd like to scale the picture so that it is 6cm by 6cm. Can you make any suggestions?
Thanks.
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Still working on this? Perhaps tex.stackexchange.com/questions/6388/… can be of help. – Torbjørn T. Jan 4 at 23:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 0.9320187568664551, "perplexity": 1587.2127631749224}, "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-2014-10/segments/1394011237821/warc/CC-MAIN-20140305092037-00068-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://phys.libretexts.org/Bookshelves/Relativity/Book%3A_General_Relativity_(Crowell)/10%3A_Appendices/10.02%3A_Appendix_A_(Part_2) | $$\require{cancel}$$
# 10.2: Appendix A (Part 2)
## § 4. Physical Meaning of the Equations Obtained in Respect to Moving Rigid Bodies and Moving Clocks
We envisage a rigid sphere26 of radius R, at rest relative to the moving system k, and with its centre at the origin of coordinates of k. The equation of the surface of this sphere moving relative to the system K with velocity v is
$\xi^{2} + \eta^{2} + \zeta^{2} = R^{2} \ldotp$
Note
That is, a body possessing spherical form when examined at rest.—AS
The equation of this surface expressed in x, y, z at the time t = 0 is
$\frac{x^{2}}{\left(\sqrt{1 - \dfrac{v^{2}}{c^{2}}} \right)^{2}} + y^{2} + z^{2} = R^{2} \ldotp$
A rigid body which, measured in a state of rest, has the form of a sphere, therefore has in a state of motion—viewed from the stationary system—the form of an ellipsoid of revolution with the axes
$R \sqrt{1 - \frac{v^{2}}{c^{2}}}, R, R \ldotp$
Thus, whereas the Y and Z dimensions of the sphere (and therefore of every rigid body of no matter what form) do not appear modified by the motion, the X dimension appears shortened in the ratio 1 : $$sqrt{1 − \frac{v^{2}}{c^{2}}}$$, i.e., the greater the value of v, the greater the shortening. For v = c all moving objects—viewed from the “stationary” system—shrivel up into plane figures.27 For velocities greater than that of light our deliberations become meaningless; we shall, however, find in what follows, that the velocity of light in our theory plays the part, physically, of an infinitely great velocity.
Note
That is, a body possessing spherical form when examined at rest.—AS
It is clear that the same results hold good of bodies at rest in the “stationary” system, viewed from a system in uniform motion.
Further, we imagine one of the clocks which are qualified to mark the time t when at rest relative to the stationary system, and the time $$\tau$$ when at rest relative to the moving system, to be located at the origin of the coordinates of k, and so adjusted that it marks the time $$\tau$$. What is the rate of this clock, when viewed from the stationary system?
Between the quantities x, t, and $$\tau$$, which refer to the position of the clock, we have, evidently, x = vt and
$\tau = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} \left( t - \dfrac{vx}{c^{2}} \right) \ldotp$
Therefore,
$\tau = t \sqrt{1 - \frac{v^{2}}{c^{2}}} = t - \left( 1 - \sqrt{1 - \dfrac{v^{2}}{c^{2}}} \right) t$
whence it follows that the time marked by the clock (viewed in the stationary system) is slow by 1 − $$\sqrt{1 − \frac{v^{2}}{c^{2}}}$$ seconds per second, or—neglecting magnitudes of fourth and higher order—by $$\frac{1}{2} \frac{v^{2}}{c^{2}}$$.
From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by $$\frac{1}{2} \frac{tv^{2}}{c^{2}}$$ (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B.
It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide.
If we assume that the result proved for a polygonal line is also valid for a continuously curved line, we arrive at this result: If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the travelled clock on its arrival at A will be $$\frac{1}{2} \frac{tv^{2}}{c^{2}}$$ second slow. Thence we conclude that a spring-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions.28
Note
Einstein specifies a spring-clock (“unruhuhr”) because the effective gravitational field is weaker at the equator than at the poles, so a pendulum clock at the equator would run more slowly by about two parts per thousand than one at the north pole, for nonrelativistic reasons. This would completely mask any relativistic effect, which he expected to be on the order of $$\frac{v^{2}}{c^{2}}$$, or about 10−13. In any case, it later turned out that Einstein was mistaken about this example. There is also a gravitational time dilation that cancels the kinematic effect. See example 10. The two clocks would actually agree.—BC
## § 5. The Composition of Velocities
In the system k moving along the axis of X of the system K with velocity v, let a point move in accordance with the equations
$\xi = w_{\xi} \tau, \quad \eta = w_{\eta} \tau, \quad \zeta = 0,$
where $$w_{\xi}$$ and $$w_{\eta}$$ denote constants.
Required: the motion of the point relative to the system K. If with the help of the equations of transformation developed in § 3 we introduce the quantities x, y, z, t into the equations of motion of the point, we obtain
$\begin{split} x &= \frac{w_{\xi} + v}{1 + \frac{vw_{\xi}}{c^{2}}} t, \\ y &= \frac{\sqrt{1 - \frac{v^{2}}{c^{2}}}}{1 + \frac{vw_{\xi}}{c^{2}}} w_{\eta} t, \\ z &= 0 \ldotp \end{split}$
Thus the law of the parallelogram of velocities is valid according to our theory only to a first approximation. We set29
$\begin{split} V^{2} &= \left(\dfrac{dx}{dt} \right)^{2} + \left(\dfrac{dy}{dt} \right)^{2}, \\ w^{2} &= w^{2}_{\xi} + w^{2}_{\eta}, \\ a &= \tan^{-1} \frac{w_{\eta}}{w_{\xi}}, \end{split}$
a is then to be looked upon as the angle between the velocities v and w. After a simple calculation we obtain
$V = \frac{\sqrt{(v^{2} + w^{2} + 2vw \cos a) - (vw \sin \frac{a}{c})^{2}}}{1 + vw \cos \frac{a}{c^{2}}} \ldotp$
Note
This equation was incorrectly given in Einstein’s original paper and the 1923 English translation as a = tan−1 $$\frac{w_{y}}{w_{x}}$$.—JW
It is worthy of remark that v and w enter into the expression for the resultant velocity in a symmetrical manner. If w also has the direction of the axis of X, we get
$V = \frac{v + w}{1 + \frac{vw}{c^{2}}} \ldotp$
It follows from this equation that from a composition of two velocities which are less than c, there always results a velocity less than c. For if we set v = c − $$\kappa$$, w = c − $$\lambda$$, $$\kappa$$ and $$\lambda$$ being positive and less than c, then
$V = c \frac{2c - \kappa - \lambda}{2c - \kappa - \lambda + \frac{\kappa \lambda}{c}} < c \ldotp$
It follows, further, that the velocity of light c cannot be altered by composition with a velocity less than that of light. For this case we obtain
$V = \frac{c + w}{1 + \frac{w}{c}} = c \ldotp$
We might also have obtained the formula for V, for the case when v and w have the same direction, by compounding two transformations in accordance with § 3. If in addition to the systems K and k figuring in § 3 we introduce still another system of coordinates k' moving parallel to k, its initial point moving on the axis of $$\Xi$$30 with the velocity w, we obtain equations between the quantities x, y, z, t and the corresponding quantities of k', which differ from the equations found in § 3 only in that the place of “v” is taken by the quantity
$\frac{v + w}{1 + \frac{vw}{c^{2}}};$
from which we see that such parallel transformations—necessarily—form a group.
Note
“X” in the 1923 English translation.—JW
We have now deduced the requisite laws of the theory of kinematics corresponding to our two principles, and we proceed to show their application to electrodynamics.31
Note
The remainder of the paper is not given here, but can be obtained from John Walker’s web site at www.fourmilab.ch.—BC | {"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": 0.866108238697052, "perplexity": 414.7400330440029}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141737946.86/warc/CC-MAIN-20201204131750-20201204161750-00293.warc.gz"} |
http://mathhelpforum.com/calculus/168481-attempt-solve-confusing-integral-x-3-e-x-2-dx-see-attached-scanned-paper.html | # Math Help - Attempt to solve confusing integral $x^3 *e^[x^2] * dx (See attached scanned paper) 1. ## Attempt to solve confusing integral$x^3 *e^[x^2] * dx (See attached scanned paper)
Here it is
2. Originally Posted by Riazy
Here it is
Put $\displaystyle{u = x^2\,; \;u'=2x\,,\,\,v'=xe^{x^2}\,;\;v=\frac{1}{2}e^{x^2} }$ , so integrating by parts:
$\displaystyle{\int x^3e^{x^2} dx=\int x^2(xe^{x^2})dx=\frac{1}{2}x^2e^{x^2}-\int xe^{x^2}dx=\frac{e^{x^2}}{2}\left(x^2-1)+C$ .
Tonio
3. $\displaystyle \int{x^3e^{x^2}\,dx} = \int{\frac{1}{2}x^2\cdot 2x\,e^{x^2}\,dx}$.
Now use integration by parts with $\displaystyle u = \frac{1}{2}x^2 \implies du = x\,dx$ and $\displaystyle dv = 2x\,e^{x^2}\,dx \implies v = e^{x^2}$ and the integral becomes
$\displaystyle \frac{1}{2}x^2e^{x^2} - \int{x\,e^{x^2}\,dx}$
$\displaystyle = \frac{1}{2}x^2e^{x^2} - \frac{1}{2}\int{2x\,e^{x^2}\,dx}$
$\displaystyle = \frac{1}{2}x^2e^{x^2} - \frac{1}{2}e^{x^2} + C$. | {"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": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9998133778572083, "perplexity": 4256.630559821661}, "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-2015-14/segments/1427131292683.3/warc/CC-MAIN-20150323172132-00242-ip-10-168-14-71.ec2.internal.warc.gz"} |
https://hal-insu.archives-ouvertes.fr/insu-01182158 | # Discovery of a Spin-down State Change in the LMC Pulsar B0540-69
Abstract : We report the discovery of a large, sudden, and persistent increase in the spin-down rate of B0540-69, a young pulsar in the Large Magellanic Cloud, using observations from the Swift and RXTE satellites. The relative increase in the spin-down rate $\dot{\nu }$ of 36% is unprecedented for B0540-69. No accompanying change in the spin rate is seen, and no change is seen in the pulsed X-ray emission from B0540-69 following the change in the spin-down rate. Such large relative changes in the spin-down rate are seen in the recently discovered class of "intermittent pulsars," and we compare the properties of B0540-69 to such pulsars. We consider possible changes in the magnetosphere of the pulsar that could cause such a large change in the spin-down rate.
Document type :
Journal articles
Cited literature [15 references]
https://hal-insu.archives-ouvertes.fr/insu-01182158
Contributor : Nathalie Pothier <>
Submitted on : Thursday, June 9, 2016 - 2:09:33 PM
Last modification on : Thursday, October 15, 2020 - 4:06:49 AM
### File
apjl_807_2_L27.pdf
Publisher files allowed on an open archive
### Citation
F. E. Marshall, Lucas Guillemot, A. K. Harding, P. Martin, D. A. Smith. Discovery of a Spin-down State Change in the LMC Pulsar B0540-69. The Astrophysical journal letters, Bristol : IOP Publishing, 2015, 807 (2), pp.L27. ⟨10.1088/2041-8205/807/2/L27⟩. ⟨insu-01182158⟩
Record views | {"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": 0.8622301816940308, "perplexity": 4954.486931073942}, "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/1603107890273.42/warc/CC-MAIN-20201026031408-20201026061408-00615.warc.gz"} |
http://math.stackexchange.com/questions/45074/finding-sets-that-do-not-intersect-approximation-algorithm | # Finding sets that do not intersect. Approximation algorithm
I am trying to solve the following problem. Let $S$ be a set and $F \subset {S \choose k}$ that is $F$ is a subset of the set of all $k$-sets of $S$.
A set $M \subset F$ such that no two elements of $M$ intersect is called a solution for the problem.
I am looking for an algorithm $A$ that finds a solution to the above problem such that if the given solution is a set $M$ and the largest set that is a solution for the same instance is $M^*$ then
$$\frac{|M|}{|M^*|} \geq \frac1{k}\qquad (1)$$
The only algorithm I can see for the above problem is to pick an element $P$ from $F$, remove all the elements in $F$ that intersect with $P$ and continue.
However I am not able to analyze the above procedure to obtain the given bound (if even attainable).
Any hints on this problem?
-
This problem is usually called the unweighted set-packing problem. As a side-note, this problem is NP-complete; in fact, as I came to know only today (thanks, OP!), it is one of Karp's gang of 21!
Here's a simple proof that the heuristic sketched in the question gives a set $M$ that is at most a factor $k$ from the optimum, call this $M^*$. Create a bipartite graph with the sets in $M$ as the left vertices, and the sets in $M^*$ as the right vertices. Add an edge between $A \in M$ and $B \in M^*$ iff $A$ intersects $B$. Here are some basic observations:
• Every right vertex has at least one left neighbor. Otherwise, we could add this set safely to $M$ to get a bigger family, contradicting the maximality of $M$.
• Every left vertex has at most $k$ right neighbors. Otherwise two of the right neighbors will be forced to "share" a common element, contradicting the assumption that $M^*$ is a disjoint family of sets.
Putting these two together, you should be able to deduce that $|M^*| \leq |M|/k$. (Note that we do not have strict inequality.) Btw, the set-packing problem for the case $k=2$ is the well-known maximum matching problem. If you have seen matchings before, I encourage you to work out the proof for this specific case; it's quite illuminating.
This bound is tight, as seen from the following example: let $S$ be the set of $k^2$ points arranged as a $k \times k$ grid. Define $M$ to be just the leftmost column. Define $M^*$ to be the family of $k$ rows, each of which contains $k$ points. Take $F$ to be $M \cup M^*$. Clearly $|M| =1$ and $|M^*|=k$.
Better heuristics. Apparently, better local-heuristics heuristics are known for this problem, achieving a factor $k/2+\epsilon$ approximation. See e.g.: "Approximating Discrete Collections via Local Improvements" by Halldorsson.
-
What is wrong with partitioning $S$ into $k$ element disjoint sets?
-
Ross, that doesn't work since the $k$-sets in $M$ are required to be picked from $F$. – Srivatsan Jul 23 '11 at 23:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.9259423613548279, "perplexity": 172.6134377010868}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207930866.20/warc/CC-MAIN-20150521113210-00250-ip-10-180-206-219.ec2.internal.warc.gz"} |
http://sci-gems.math.bas.bg/jspui/browse?type=author&sort_by=1&order=ASC&rpp=20&etal=-1&value=Liu%2C+F.&starts_with=N | ## Browsing by Author "Liu, F."
Jump to: 0-9 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z or enter first few letters:
Sort by: In order: Results/Page Authors/Record:
Showing results 1 to 2 of 2
Issue DateTitleAuthor(s)Publisher
2005 Numerical Approximation of a Fractional-In-Space Diffusion Equation, IIlic, M.; Liu, F.; Turner, I.; Anh, V.Institute of Mathematics and Informatics Bulgarian Academy of Sciences
2006 Numerical Approximation of a Fractional-In-Space Diffusion Equation (II) – with Nonhomogeneous Boundary ConditionsIlic, M.; Liu, F.; Turner, I.; Anh, V.Institute of Mathematics and Informatics Bulgarian Academy of Sciences
Showing results 1 to 2 of 2 | {"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": 0.9316165447235107, "perplexity": 3402.3930108411037}, "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-2017-30/segments/1500549428300.23/warc/CC-MAIN-20170727142514-20170727162514-00468.warc.gz"} |
http://mathoverflow.net/questions/847/is-any-representation-of-a-finite-group-defined-over-the-algebraic-integers/882 | # Is any representation of a finite group defined over the algebraic integers?
Apologies in advance if this is obvious.
-
I'm pretty sure it's not obvious...my guess is there isn't, but I can't think of any counterexamples right off the bat. – Ben Webster Oct 17 '09 at 14:51
So you are asking if there exist a basis for every GIVEN finite subgroup, not one basis that works for every finite subgroup, right? Coz then the answer is no. – shenghao Oct 17 '09 at 19:55
Yes, any particular subgroup. – Qiaochu Yuan Oct 17 '09 at 21:39
The subset of GL_n(\bar{Q}) consisting of matrices such that the matrix AND its inverse have algebraic integer entries is a subgroup, and obviously a group representation with algebraic integer entries would have to land there. – Ben Webster Oct 18 '09 at 1:54
Somewhat related: groupprops.subwiki.org/wiki/… Unfortunately, not every ring of algebraic integers (or even cyclotomic integers) is a PID. – Vipul Naik Apr 26 '11 at 20:12
This is not really an answer, but is too long for a comment. The proof given by Moonface above is given in more or less that form in the 1962 book of Curtis and Reiner. As far as I know, it is still open whether all irreducible representations of a finite group $G$ can be realized over $\mathbb{Z}[\omega]$, where $\omega$ is a complex primitive $|G|$-th roots of unity, though I think the paper of Cliff,Ritter and Weiss settles the questions for finite solvable groups. The paper of Serre ( the three letters to Feit) give counterexamples to a slightly different question: they show (among other things) that a representation of a finite group can be realised over some number fields, but might not be able to be realised over the ring of algebraic integers of that field. Brauer's characterization of characters/Brauer's induction theorem show that all representations of the finite group $G$ may be realised over $\mathbb{Q}[\omega]$ for $\omega$ as above ( $|G|$ can be replaced by the exponent of $G$ if desired). As I said, realizability over $\mathbb{Z}[\omega]$ is a different matter. | {"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": 0.9230245351791382, "perplexity": 232.46350443076835}, "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-2016-18/segments/1461860118790.25/warc/CC-MAIN-20160428161518-00215-ip-10-239-7-51.ec2.internal.warc.gz"} |
http://www.mathcaptain.com/statistics/mean.html | Even though mean refers to one of the measures of average in statistics, it is the most commonly used measure of central tendency, because it is based on all the data values and least affected by fluctuations in sampling. Mean is of three types based on the method of computation:
1. Arithmetic Mean
2. Geometric Mean
3. Harmonic Mean
## Definition of Mean in Statistics
Mean is a central measure which is used in describing a data distribution.
Indeed a data set is described by mean along with a measure of dispersion like range or variance.
Mean in statistics is also called Expectation or expected value when computed for random variables associated with probability experiments. In such cases the mean is essential to compute the probabilities of the variable assuming a specific discrete value or a range of values in an interval.
## Arithmetic Mean
The arithmetic mean of a data set is often referred to as mean of the data set. It is the mathematical average of the data values.
The arithmetic mean of n observations x1, x2, .......xn is given by
x = $\frac{x_{1}+x_{2}+......x_{n}}{n}$ or using sigma notation = $\frac{\sum_{i=1}^{n}x_{i}}{n}$
When the data is represented by a frequency distribution where f1, f2,......fn are the corresponding frequencies of data values x1, x2, xn, then the arithmetic mean is computed using the formula
x = $\frac{\sum_{i=1}^{n}x_{i}f_{i}}{N}$ where N = f1 + f2 + .........+ fn.
While x denotes the mean of sample data the letter μ is used to represent the population mean.
The value of arithmetic mean is also used in other computations of statistic measures like variance.
Even though the arithmetic mean is easy to compute and based on all data values, it's value is influenced by outliers or extreme values in the data set and hence lead to misinterpretations.
Example: The AM (x ) of the given five numbers $10,12,8,14,2$ is
$\frac{10+12+8+14+2}{5}$ = $9.2$
## Geometric Mean
Geometric Mean of n data values is the nth root of their product.
The formula for finding the geometric mean of n values x1 , x2, .......xn is given by
GM = $(x_{1}x_{2}......x_{n})$$^{\frac{1}{n}} which can be written using Pi notation as (\Pi_{i=1}^{n} x_{i})$$^{\frac{1}{n}}$
If f1, f2,....fn are the corresponding frequencies associated with the above data values the GM is given by,
GM = $(x_{1}^{f_{1}}x_{2}^{f_{2}}....x_{n}^{f_{n}})$$^{\frac{1}{N}} where N = f1 + f2 + .........+ fn. Example: The GM of the given five numbers 10, 12, 8, 14, 2 is (10\times 12\times 8\times 14\times 2)$$^{\frac{1}{5}}$ = $7.689$
## Harmonic Mean
Harmonic mean of n observations x1 , x2, .......xn is defined by
HM = $n(\Pi_{i=1}^{n}$$\frac{1}{x_{i}}$)$^{-1}$ = $\frac{n}{\frac{1}{x_{1}}+\frac{1}{x_{2}}+......+\frac{1}{x_{n}}}$
Harmonic mean formula for a frequency distribution is as follows:
HM = $\frac{N}{\frac{f_{1}}{x_{1}}+\frac{f_{2}}{x_{2}}+......\frac{f_{n}}{x_{n}}}$
where N = f1 + f2 + .........+ fn.
Example: The HM of the given five numbers $10, 12, 8, 14, 2$ is
$\frac{5}{\frac{1}{10}+\frac{1}{12}+\frac{1}{8}+\frac{1}{14}+\frac{1}{2}}$ = $\frac{5}{0.8797}$ = $5.68335$
### Median
Arithmetic Mean Geometric Mean Harmonic Mean
Arithmetic Means Geometrical Mean Harmonic Means Integration Meaning Intercept Meaning Mean Proportion Meaning of Quadrant parabola meaning A Negative Correlation Means that Binomial Distribution Mean Calculus Mean Value Theorem Central Tendency Mean
Calculator Mean Calculate Mean Deviation Calculate Sample Mean | {"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": 0.9787490963935852, "perplexity": 878.1662574647758}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422122238694.20/warc/CC-MAIN-20150124175718-00057-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/conceptual-question-on-angular-momentum.870998/ | # Conceptual Question on Angular Momentum.
• #1
65
15
## Homework Statement
A Person sitting firmly over a rotating stool has his arms stretched. If he fold his arms, his angular momentum about the axis of rotation :
A.) Increases
B.) Decreases
C.) Remains Unchanged
D.) doubles
## Homework Equations
[/B]Conservation of Angular Momentum
## The Attempt at a Solution
Since External Torque =0
Then final angular momentum = initial angular momentum
=> (C.)
But I wanted to further ask that in the Equation L=I*(omega)
since L has to be constant(if external torque = 0) then:
If I increases then omega decreases to keep L constant
or if omega increases then I decreases to keep constant
Will folding his arms increase I( Moment of Inertia ) or decrease it??
• #2
963
214
But I wanted to further ask that in the Equation L=I*(omega)
since L has to be constant(if external torque = 0) then:
If I increases then omega decreases to keep L constant
or if omega increases then I decreases to keep constant
by folding his arms the new moment of inertia of the system will increase /decrease?
how moment of inertia depends on mass distribution? if a mass is farther from axis of rotation its moment of inertia should be larger as I= m.r^2
L= I . angular velocity therefore a decrease/increase in I should result in the angular velocity to increse/decrease
Likes Sahil Kukreja
• #3
65
15
by folding his arms the new moment of inertia of the system will increase /decrease?
how moment of inertia depends on mass distribution? if a mass is farther from axis of rotation its moment of inertia should be larger as I= m.r^2
L= I . angular velocity therefore a decrease/increase in I should result in the angular velocity to increse/decrease
Ok. Thanks! so if the mass is extremely close and external torque=0, then angular velocity will reach very high.
• #4
963
214
Ok. Thanks! so if the mass is extremely close and external torque=0, then angular velocity will reach very high.
the example is ice-skating sport -there the rotation speed of skaters is increased or decreased by them using their arms folded or spreading out.
Likes Sahil Kukreja
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1K | {"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": 0.940527081489563, "perplexity": 3792.128353027124}, "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/1656104692018.96/warc/CC-MAIN-20220707124050-20220707154050-00521.warc.gz"} |
http://www.gradesaver.com/textbooks/science/physics/physics-for-scientists-and-engineers-a-strategic-approach-with-modern-physics-3rd-edition/chapter-10-energy-exercises-and-problems-page-271/1 | ## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)
We first find the kinetic energy of the bullet; $KE = \frac{1}{2}mv^2$ $KE = \frac{1}{2}(0.010~kg)(500~m/s)^2$ $KE = 1250~joules$ We then find the kinetic energy of the bowling ball; $KE = \frac{1}{2}mv^2$ $KE = \frac{1}{2}(10~kg)(10~m/s)^2$ $KE = 500~joules$ The bullet has the larger kinetic energy. | {"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": 0.9035917520523071, "perplexity": 402.16194612992734}, "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/1524125946077.4/warc/CC-MAIN-20180423144933-20180423164933-00132.warc.gz"} |
http://math.stackexchange.com/questions/147190/birational-map-between-product-of-projective-varieties | # Birational map between product of projective varieties
What is an example of a birational morphism between $\mathbb{P}^{n} \times \mathbb{P}^{m} \rightarrow \mathbb{P}^{n+m}$?
-
The subset $\mathbb A^n\times \mathbb A^m$ is open dense in $\mathbb P^n\times \mathbb P^m$ and the subset $\mathbb A^{n+m}$ is open dense in $\mathbb P^n\times \mathbb P^m$.
Hence the isomorphism $\mathbb A^n\times \mathbb A^m\stackrel {\cong}{\to} \mathbb A^{n+m}$ is the required birational isomorphism.
why $\mathbb{A}^{n} \times \mathbb{A}^{m}$ and $\mathbb{A}^{n+m}$ are open in $\mathbb{P}^{n} \times \mathbb{P}^{m}$? Can you please explain more this part? – user31509 May 19 '12 at 23:34
Dear user, $\mathbb A^N=\mathbb P^N \setminus H$ where $H$ is the hyperplane $z_0=0$ ( $\mathbb P^N$ has coordinates $[z_0:z_1:...:z_N]$). Since $z_0$ is a homogeneous polynomial (of degree one) , $H$ is closed and its complement $\mathbb A^N$ is open. For your first question use this result plus the fact that the product of two open sets is open : the Zariski topology for $\mathbb P^n\times \mathbb P^m$ is finer than the product topology. – Georges Elencwajg May 19 '12 at 23:49 | {"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": 0.962980329990387, "perplexity": 146.56865230996948}, "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-2015-40/segments/1443736678861.8/warc/CC-MAIN-20151001215758-00231-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://www.groundai.com/project/trisections-of-4-manifolds-via-lefschetz-fibrations/2 | Trisections of 4-manifolds via Lefschetz fibrations
# Trisections of 4-manifolds via Lefschetz fibrations
Nickolas A. Castro and Burak Ozbagci Department of Mathematics, UC Davis, Davis, CA 95616 Department of Mathematics, UCLA, Los Angeles, CA 90095 and Department of Mathematics, Koç University, 34450, Istanbul, Turkey
###### Abstract.
We develop a technique for gluing relative trisection diagrams of -manifolds with nonempty connected boundary to obtain trisection diagrams for closed -manifolds. As an application, we describe a trisection of any closed -manifold which admits a Lefschetz fibration over equipped with a section of square , by an explicit diagram determined by the vanishing cycles of the Lefschetz fibration. In particular, we obtain a trisection diagram for some simply connected minimal complex surface of general type. As a consequence, we obtain explicit trisection diagrams for a pair of closed -manifolds which are homeomorphic but not diffeomorphic. Moreover, we describe a trisection for any oriented -bundle over any closed surface and in particular we draw the corresponding diagrams for and using our gluing technique. Furthermore, we provide an alternate proof of a recent result of Gay and Kirby which says that every closed -manifold admits a trisection. The key feature of our proof is that Cerf theory takes a back seat to contact geometry.
###### 2000 Mathematics Subject Classification:
The second author was partially supported by a research grant of the Scientific and Technological Research Council of Turkey.
## 1. Introduction
Recently, Gay and Kirby [12] proved that every smooth, closed, oriented, connected -manifold admits a trisection, meaning that for every , there exist non-negative integers such that is diffeomorphic to the union of three copies of the -dimensional -handlebody , intersecting pairwise in -dimensional handlebodies, with triple intersection a closed, oriented, connected -dimensional surface of genus . Such a decomposition of is called a -trisection or simply a genus trisection, since is determined by using the fact that , where denotes the Euler characteristic of .
Moreover, they showed that the trisection data can be encoded as a -tuple , which is called a -trisection diagram, such that each triple , , and is a genus Heegaard diagram for . Furthermore, they proved that trisection of (and its diagram) is unique up to a natural stabilization operation.
On the other hand, various flavors of Lefschetz fibrations have been studied extensively in the last two decades to understand the topology of smooth -manifolds. Suppose that a closed -manifold admits a Lefschetz fibration over , whose regular fiber is a smooth, closed, oriented, connected surface of genus . The fibration induces a handle decomposition of , where the essential data can be encoded by a finite set of ordered simple closed curves (the vanishing cycles) on a surface diffeomorphic to . The only condition imposed on the set curves is that the product of right-handed Dehn twists along these curves is isotopic to the identity diffeomorphism of .
In addition, every -manifold with nonempty boundary has a relative trisection and under favorable circumstances also admits an achiral Lefschetz fibration over with bounded fibers. The common feature shared by these structures is that each induces a natural open book on . To exploit this feature in the present paper, we develop a technique to obtain trisection diagrams for closed -manifolds by gluing relative trisection diagrams of -manifolds with nonempty connected boundary. The precise result is stated in Proposition 2.12, which is too technical to include in the introduction. Nevertheless, our gluing technique has several applications — one of which is the following result.
Theorem 3.7. Suppose that is a smooth, closed, oriented, connected -manifold which admits a genus Lefschetz fibration over with singular fibers, equipped with a section of square . Then, an explicit -trisection of can be described by a corresponding trisection diagram, which is determined by the vanishing cycles of the Lefschetz fibration. Moreover, if denotes the -manifold obtained from by blowing down the section of square , then we also obtain a -trisection of along with a corresponding diagram.
In particular, Theorem 3.7 provides a description of a -trisection diagram of the Horikawa surface (see [14, page 269] for its definition and properties), a simply connected complex surface of general type which admits a genus Lefschetz fibration over with singular fibers, equipped with a section of square . This section is the unique sphere in with self-intersection so that by blowing it down, we obtain a trisection diagram for the simply connected minimal complex surface of general type.
To the best of our knowledge, none of the existing methods in the literature can be effectively utilized to obtain explicit trisection diagrams for complex surfaces of general type. For example, Gay and Kirby describe trisections of , , , closed -manifolds admitting locally trivial fibrations over or (including of course , and ) and arbitrary connected sums of these in [12].
Note that, by Freedman’s celebrated theorem, the Horikawa surface is homeomorphic to (and also to the elliptic surface ), since it is simply connected, nonspin and its Euler characteristic is , while its signature is . On the other hand, since is a simply connected complex surface (hence Kähler) with , it has non-vanishing Seiberg-Witten invariants, while has vanishing Seiberg-Witten invariants which follows from the fact that . Hence, we conclude that is certainly not diffeomorphic to .
As a consequence, we obtain explicit -trisection diagrams for a pair of closed -manifolds, the Horikawa surface and , which are homeomorphic but not diffeomorphic. Note that has a natural -trisection diagram (obtained by the connected sum of the standard -trisection diagrams of and ), which can be stabilized four times to yield a -trisection diagram.
More generally, Theorem 3.7 can be applied to a large class of -manifolds. A fundamental result of Donaldson [6] says that every closed symplectic -manifold admits a Lefschetz pencil over . By blowing up its base locus the Lefschetz pencil can be turned into a Lefschetz fibration over , so that each exceptional sphere becomes a (symplectic) section of square . Conversely, any -manifold satisfying the hypothesis of Theorem 3.7 must carry a symplectic structure where the section of square can be assumed to be symplectically embedded. Therefore, is necessarily a nonminimal symplectic -manifold.
In [11, Theorem 3], Gay describes a trisection for any closed -manifold admitting a Lefschetz pencil, although he does not formulate the trisection of in terms of the vanishing cycles of the pencil (see [11, Remark 9]). He also points out that his technique does not extend to cover the case of Lefschetz fibrations on closed -manifolds [11, Remark 8].
We would like to point out that Theorem 3.7 holds true for any achiral Lefschetz fibration equipped with a section of square . In this case, is not necessarily symplectic. We opted to state our result only for Lefschetz fibrations to emphasize their connection to symplectic geometry.
Next we turn our attention to another natural application of our gluing technique where we find trisections of doubles of -manifolds with nonempty connected boundary. It is well-known (see, for example, [14, Example 4.6.5]) that there are two oriented -bundles over a closed, oriented, connected surface of genus : the trivial bundle and the twisted bundle . The former is the double of any -bundle over with even Euler number, while the latter is the double of any -bundle over with odd Euler number. We obtain trisections of these -bundles by doubling the relative trisections of the appropriate -bundles. In particular, we draw the corresponding -trisection diagram for and the -trisection diagram for using our gluing technique.
For any , the twisted bundle is not covered by the examples in [12], while our trisection for has smaller genus compared to that of given in [12]. We discuss the case of oriented -bundles over nonorientable surfaces in Section 4.5.
Finally, we provide a simple alternate proof of the following result due to Gay and Kirby.
Theorem 5.1. Every smooth, closed, oriented, connected -manifold admits a trisection.
Our proof is genuinely different from the two original proofs due to Gay and Kirby [12], one with Morse -functions and one with ordinary Morse functions, since not only contact geometry plays a crucial role in our proof, but we also employ a technique for gluing relative trisections.
After the completion of our work, we learned that Baykur and Saeki [3] gave yet another proof of Theorem 5.1, setting up a correspondence between broken Lefschetz fibrations and trisections on -manifolds, using a method which is very different from ours. In particular, they prove the existence of a -trisection on a 4-manifold which admits a genus Lefschetz fibration over with Lefschetz singularities — generalizing the first assertion in our Theorem 3.7, but without providing the corresponding explicit diagram for the trisection. They also give examples of trisections (without diagrams) on a pair of closed -manifolds (different from ours) which are homeomorphic but not diffeomorphic. In addition, for any , they give small genus trisections (again without the diagrams) for .
Conventions: All -manifolds are assumed to be smooth, compact, oriented and connected throughout the paper. The corners which appear in gluing manifolds are smoothed in a canonical way.
## 2. Gluing relative trisections
We first review some basic results about trisections and their diagrams (cf. [12]). Let denote the standard genus Heegaard splitting of obtained by stabilizing the standard genus Heegaard splitting times.
###### Definition 2.1.
A -trisection of a closed -manifold is a decomposition such that for each ,
1. there is a diffeomorphism and
2. taking indices mod , and .
It follows that is a closed surface of genus . Also note that and determine each other, since the Euler characteristic is equal to , which can be easily derived by gluing and first and then gluing .
Suppose that each of and is a collection of disjoint simple closed curves on some compact surface . We say that two such triples and are diffeomorphism and handleslide equivalent if there exists a diffeomorphism such that is related to by a sequence of handleslides and is related to by a sequence of handleslides.
###### Definition 2.2.
A -trisection diagram is an ordered -tuple such that
1. is a closed genus surface,
2. each of , and is a non-separating collection of disjoint, simple closed curves on ,
3. each triple , and is diffeomorphism and handleslide equivalent to the standard genus Heegaard diagram of depicted in Figure 1.
According to Gay and Kirby [12], every closed -manifold admits a trisection, which in turn, can be encoded by a diagram. Conversely, every trisection diagram determines a trisected closed -manifold, uniquely up to diffeomorphism.
Next we recall the analogous definitions of relative trisections and their diagrams for -manifolds with nonempty connected boundary (cf. [12, 5]).
Suppose that is a -manifold with nonempty connected boundary . We would like to find a decomposition , such that each is diffeomorphic to for some fixed . Since , it would be natural to require that part of each contribute to . Hence, we need a particular decomposition of to specify a submanifold of to be embedded in . With this goal in mind, we proceed as follows to develop the language we will use throughout the paper.
Suppose that are non-negative integers satisfying and
g+p+b−1≥k≥2p+b−1.
Let and for , and , .
We denote by a fixed genus surface with boundary components. Let
D={reiθ∈C|r∈[0,1]and−π/3≤θ≤π/3}
be a third of the unit disk in the complex plane whose boundary is decomposed as , where
∂−D ={reiθ∈∂D|θ=−π/3} ∂0D ={eiθ∈∂D} ∂+D ={reiθ∈∂D|θ=π/3}
The somewhat unusual choice of the disk will be justified by the construction below. We set , which is indeed diffeomorphic to . Then, inherits a decomposition , where and .
Let be the standard genus one Heegaard splitting of . For any , let , where the boundary connected sum is taken in neighborhoods of the Heegaard surfaces, inducing the standard genus Heegaard splitting of . Stabilizing this Heegaard splitting times we obtain a genus Heegaard splitting of .
We set and . Note that we can identify , where the boundary connected sum again takes place along the neighborhoods of points in the Heegaard surfaces. We now have a decomposition of as follows:
∂Zk=Yk=Y+g,k;p,b∪Y0g,k;p,b∪Y−g,k;p,b,
where and
###### Definition 2.3.
A -relative trisection of a -manifold with non-empty connected boundary is a decomposition such that for each ,
1. there is a diffeomorphism and
2. taking indices mod , and
As a consequence, is diffeomorphic to , which is a genus surface with boundary components. Note that the Euler characteristic is equal to , which can be calculated directly from the definition of a relative trisection. We also give alternate method to calculate in Corollary 2.10.
According to [12], every -manifold with nonempty connected boundary admits a trisection. Moreover, there is a natural open book induced on , whose page is diffeomorphic to , which is an essential ingredient in our definition of .
Informally, the contribution of each to is one third of an open book. This is because the part of each that contributes to is diffeomorphic to
Y0g,k;p,b=∂0U=(P×∂0D)∪(∂P×D),
where is one third of the truncated pages, while is one third of the neighborhood of the binding. In other words, not only we trisect the -manifold , but we also trisect its boundary . Conversely, if an open book is fixed on , then admits a trisection whose induced open book coincides with the given one.
###### Definition 2.4.
A -relative trisection diagram is an ordered -tuple such that
1. is a genus surface with boundary components,
2. each of and is a collection of disjoint, essential, simple closed curves,
3. each triple , and is diffeomorphism and handleslide equivalent to the diagram depicted in Figure 2.
It was shown in [5] that every relative trisection diagram determines uniquely, up to diffeomorphism, (i) a relatively trisected -manifold with nonempty connected boundary and (ii) the open book on induced by the trisection. Moreover, the page and the monodromy of the open book on is determined completely by the relative trisection diagram by an explicit algorithm, which we spell out below.
Suppose that is a -relative trisection diagram, which represents a relative trisection of a -manifold with nonempty connected boundary. The page of the induced open book on is given by , which is the genus surface with boundary components obtained from by performing surgery along the curves. This means that to obtain , we cut open along each curve and glue in disks to cap off the resulting boundaries.
Now that we have a fixed identification of the page of as , we use Alexander’s trick to describe the monodromy of . Namely, we cut into a single disk via two distinct ordered collections of arcs, so that for each arc in one collection there is an arc in the other collection with the same endpoints. As a result, we get a self-diffeomorphism of that takes one collection of arcs to the other respecting the ordering of the arcs, and equals to the identity otherwise. This diffeomorphism uniquely extends to a self-diffeomorphism of the disk, up to isotopy. Therefore, we get a self-diffeomorphism of fixing pointwise, which is uniquely determined up to isotopy. Next, we provide some more details (see [5, Theorem 5]) about how to obtain the aforementioned collection of arcs.
Let be any ordered collection of disjoint, properly embedded arcs in disjoint from , such that the image of in cuts into a disk. We choose a collection of arcs , and a collection of simple closed curves disjoint from in such that is handleslide equivalent to , and is handleslide equivalent to . This means that arcs are obtained by sliding arcs over curves, and is obtained by sliding curves over curves. Next we choose a collection of arcs , and a collection of simple closed curves disjoint from in such that is handleslide equivalent to , and is handleslide equivalent to . This means that arcs are obtained by sliding arcs over curves, and is obtained by sliding curves over curves. Finally, we choose a collection of arcs , and a collection of simple closed curves disjoint from in such that is handleslide equivalent to , and is handleslide equivalent to . This means that arcs are obtained by sliding arcs over curves, and is obtained by sliding curves over curves. It follows that is handleslide equivalent to for some collection of arcs disjoint from in .
###### Definition 2.5.
We call the triple a cut system of arcs associated to the diagram .
Now we have two ordered collections of arcs and in , such that each of their images in cuts into a disk. Then, as we explained above, there is a unique diffeomorphism , up to isotopy, which fixes pointwise such that .
###### Remark 2.6.
It is shown in [5] that, up to isotopy, the monodromy of the resulting open book is independent of the choices in the above algorithm.
Next, we give a very simple version of the general gluing theorem [4] for relatively trisected -manifolds. Here we present a different proof — where we use the definition of a relative trisection as given in [5] instead of [12] — for the case of a single boundary component.
###### Lemma 2.7.
Suppose that and are -manifolds such that and are both nonempty and connected. Let and be - and -relative trisections with induced open books and on and , respectively. If is an orientation-reversing diffeomorphism which takes to (and hence and ), then the relative trisections on and can be glued together to yield a -trisection of the closed -manifold , where and .
###### Proof.
Since there is an orientation-reversing diffeomorphism which takes the open book to the open book , we have and . Let and be - and -relative trisections, respectively. Then, by Definition 2.3, there are diffeomorphisms and for .
Let and be the bindings of and , where and are the projection maps of these open books, respectively. Since the gluing diffeomorphism takes to by our assumption, we have , for all Moreover, we may assume that
fi=φ′i∘f∘φ−1i:Y0g,k;p,b→Y0g′,k′;p,b
is a diffeomorphism for each . Informally, we identify each third of on with the appropriate third of on via the gluing map . This allows us to define
Xi=Wi∪fW′i=Wi∪W′i/∼
where if and , where .
We claim that is a -trisection, where
G=g+g′+b−1andK=k+k′−(2p+b−1).
In order to prove our claim, we first need to describe, for each , a diffeomorphism . The diffeomorphism is essentially obtained by gluing the diffeomorphisms and using the diffeomorphism , as we describe below.
To construct the desired diffeomorphism , it suffices to describe how to glue with to obtain by identifying with using the gluing map .
By definition, , and similarly , where and . Note that is diffeomorphic to via . To glue to we identify with via the diffeomorphism . Next, we observe that by gluing and along the aforementioned parts of their boundaries using , we get , where .
To see this, we view as , and similarly as , where . We glue with and then take its product with . To glue with we identify with using . The result of this identification is diffeomorphic to . However, to complete the identification dictated by , we have to take the quotient of by the relation for all . Note that we suppressed here since we have already identified with via . The result is still diffeomorphic to the handlebody . Therefore, the gluing of and is diffeomorphic to , since it is a thickening of by taking its product with .
As a consequence, the result of gluing to is diffeomorphic to
(♮lS1×B3)♮Vn♮Vn′≅♮l+n+n′S1×B3≅♮KS1×B3≅ZK,
since .
To finish the proof of the lemma, we need to show that taking indices mod ,
Φi(Xi∩Xi+1)=Y+G,KandΦi(Xi∩Xi−1)=Y−G,K,
where is the standard genus Heegaard splitting of .
We observe that
Φi(Xi∩Xi+1)=φi(Wi∩Wi+1)∪fiφ′i(W′i∩W′i+1)=((Y+g,k;p,b∪Y+g′,k′;p,b)/∼)⊂YK
where if and .
Note that by definition, and similarly . Since the boundary connected sums are taken along the interior of Heegaard surfaces, the identification does not interact with and and hence
But we see that is diffeomorphic to , by exactly the same argument used above when we discussed the gluing of with . Therefore, we have
(Y+g,k;p,b∪Y+g′,k′;p,b)/∼ ≅((∂+U∪∂+U′)/∼)♮(∂+sVn♮∂+s′Vn′) ≅(♮lS1×B2)♮(∂+sVn♮∂+s′Vn′) ≅(♮lS1×B2)♮s+n+s′+n′S1×B2 ≅♮GS1×B2
since . Similarly, we have
Φi(Xi∩Xi−1)=φi(Wi∩Wi−1)∪fiφ′i(W′i∩W′i−1)=(Y−g,k;p,b∪Y−g′,k′;p,b)/∼)⊂YK
where if and . Thus we obtain . Moreover,
Therefore, and gives the standard genus Heegaard splitting of , as desired.
To summarize, we showed that there is a diffeomorphism , for each , and moreover, taking indices mod , and . Therefore, we conclude that is a -trisection. ∎
Here is an immediate corollary of Lemma 2.7.
###### Corollary 2.8.
Suppose that is a -relative trisection of a -manifold with nonempty connected boundary. Let denote the double of obtained by gluing and (meaning with the opposite orientation) by the identity map of the boundary . Then admits a -trisection.
###### Proof.
If is a -relative trisection of with the induced open book on , then is a -relative trisection of with the induced open book on , where is obtained from by reversing the orientation of the pages. Since the identity map from to is an orientation-reversing diffeomorphism which takes to , we obtain the desired result about by Lemma 2.7. ∎
The point of Corollary 2.8 is that one does not need to know the monodromy of the open book on to describe a trisection on .
###### Example 2.9.
Let denote the -bundle over with Euler number . In [5], there is a description of a -relative trisection of for , and a -relative trisection of . Since the double of is or depending on modulo , we get a -trisection of (resp. ) for any even (resp. odd) nonzero integer , by Corollary 2.8.
In particular, by doubling the -relative trisection of , we obtain a -trisection of which is smaller compared to the -trisection presented in [12], provided that . Similarly, by setting , we obtain a -trisection for , which is not covered by the examples in [12], except for . Note that there is also a -relative trisection of given in [5] for each . Since the double of is or depending on modulo , we get infinitely many -trisections of and .
###### Corollary 2.10.
If is a -relative trisection of a -manifold with nonempty connected boundary, then the Euler characteristic is equal to
###### Proof.
Using Corollary 2.8, we compute
χ(W)=12χ(DW)=12(2+2g+b−1−3(2k−2p−b+1)))=g−3k+3p+2b−1.
One can of course derive the same formula directly from the definition of a relative trisection. ∎
Since every relatively trisected -manifold with connected boundary is determined by some relative trisection diagram, it would be desirable to have a version of Lemma 2.7, where one “glues” the relative trisection diagrams corresponding to and to get a diagram corresponding to the trisection . This is the content of Proposition 2.12, but first we develop some language to be used in its statement.
Let and be - and -relative trisection diagrams corresponding to the relative trisections and , with induced open books and on and | {"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": 0.9532490968704224, "perplexity": 488.2388832937321}, "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-2020-34/segments/1596439738777.54/warc/CC-MAIN-20200811115957-20200811145957-00337.warc.gz"} |
https://homework.zookal.com/questions-and-answers/1-which-of-the-followings-are-true-a-h2o-contains-642666057 | 1. Science
2. Chemistry
3. 1 which of the followings are true a h2o contains...
# Question: 1 which of the followings are true a h2o contains...
###### Question details
1. Which of the followings are true:
A. H2O contains polar covalent bonds B. H2O can disrupt other hydrogen bonds in aqueous solutions C. Non-polar molecules can not have polar covalent bonds D. H2O has a higher boiling point than methane primarily due to the hydrogen bond formation E. Usually hydrogen bonds are weaker than covalent bonds
* The answer A, D, E is incorrect | {"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": 0.8518059253692627, "perplexity": 2780.5269895582596}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178359624.36/warc/CC-MAIN-20210227234501-20210228024501-00629.warc.gz"} |
http://mathhelpforum.com/calculus/136855-newtons-method-finding-solutions.html | 1. ## newtons method(finding solutions)
we first used newtons method to find the approximate value of 200^(1/7) and got it equal to the awnser in the caclulator which is = 2.131663117
i need some help in the right direction on how to solve the question after solving the above equation using newtons method!
Question:
the fundamental theorem of algebra guarantees that the equation x^7-200 = 0 must have 7 solutions. You have just found one of these. Are there any more real solutions? Use your tools of calculus to sustain your awnser.
2. Do you know Descartes rule of sign?
3. ## k
ok so in other words using decartes rule states that
x^(7)-200=0
has only 1 possible positive real zero? the other 6 are imaginary/complex values?
4. Yes because we obtain 1 (+), 0 (-), so the last solutions are 3 pairs of complex congujates
5. ## y
thx
6. I don't think you need DesCarte's rule of signs. Just observe that the derivative of $x^7- 200$, $7x^6$ is always non-negative so it is a "one to one" function on the real numbers.
7. Originally Posted by dwsmith
Yes because we obtain 1 (+), 0 (-), so the last solutions are 3 pairs of complex congujates
Descartes rule of signs tells you there is exactly one positive solution. Substitute $u=-x$ and you get $-u^7-1$ and then Descartes rule of signs tells you there are no positive solutions for $u$ and so no negative solutions for $x$ to the original equation. Hence there is exactly 1 real solution to $x^7-1=0$.
But the question requires that you use HallsofIvy's method not the rule of signs.
CB | {"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": 0.9392415881156921, "perplexity": 488.4234649341294}, "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/1484560281332.92/warc/CC-MAIN-20170116095121-00167-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://meangreenmath.com/tag/microsoft-excel/ | # Exponential growth and decay (Part 9): Amortization tables
This post is inspired by one of the questions that I pose to our future high school math teachers during our Friday question-and-answer sessions. In these sessions, I play the role of a middle- or high-school student who’s asking a tricky question to his math teacher. Here’s the question:
A student asks, “My father bought a house for $200,000 at 12% interest. He told me that by the time he fi nishes paying for the house, it will have cost him more than$500,000. How is that possible? 12% of $200,000 is only$24,000.”
Without fail, these future teachers don’t have a good response to this question. Indeed, my experience is that most young adults (including college students) have never used an amortization table, which is the subject of today’s post.
In the past few posts, we have considered the solution of the following recurrence relation, which is often used to model the payment of a mortgage or of credit-card debt:
$A_{n+1} = r A_n - k$
With this difference equation, the rate at which the principal is reduced can be simply computed using Microsoft Excel. This tool is called an amortization schedule or an amortization table; see E-How for the instructions of how to build one. Here’s a sample Excel spreadsheet that I’ll be illustrating below: Amortization schedule. My personal experience is that many math majors have never seen such a spreadsheet, even though they are familiar with compound interest problems and certainly have the mathematical tools to understand this spreadsheet.
Here’s a screen capture from the spreadsheet:
The terms of the loan are typed into Cells B1 (length of loan, in years), B2 (annual percentage rate), and B3 (initial principal). Cell B4 is computed from this information using the Microsoft Excel command $\hbox{PMT}$:
$=\hbox{PMT}(\hbox{B2}/12,\hbox{B1}*12,-\hbox{B3})$
This is the amount that must be paid every month in order to pay off the loan in the prescribed number of years. Of course, there is a formula for this:
$M = \displaystyle \frac{Pr}{12 \displaystyle \left[1 - \left( 1 + \frac{r}{12} \right)^{-12t} \right]}$
I won’t go into the derivation of this formula here, as it’s a bit complicated. Notice that this formula does not include escrow, points, closing costs, etc. This is strictly the amount of money that’s needed to pay down the principal.
The table, beginning in Row 8 of the above picture, shows how quickly the principal will be paid off. In row 8, the interest that’s paid for that month is computed by
$=\hbox{B8} * \ \hbox{B}\\hbox{2}/12$
Therefore, the amount of the monthly payment that actually goes toward paying down the principal will be
$= \\hbox{B}\\hbox{4} - \hbox{C8}$
Column E provides an opportunity to pay something extra each month; more on this later. So, after taking into account the payments in columns D and E, the amount remaining on the loan is recorded in Cell F8:
$= \hbox{B8} - \hbox{D8} - \hbox{E8}$
This amount is then copied into Cell B9, and then the pattern can be filled down.
The yellow graph shows how quickly the balance of the loan is paid off over the length of the loan. A picture is worth a thousand words: in the initial years of the loan, most of the payments are gobbled up by the interest, and so the principal is paid off slowly. Only in the latter years of the loan is the principal paid off quickly.
So, it stands to reason that any extra payments in the initial months and years of the loan can do wonders for paying off the loan quickly. For example, here’s a screenshot of what happens if an extra $200/month is paid only in the first 12 months of the loan: A definite bend in the graph is evident in the initial 12 months until the normal payment is resumed in month 13. As a result of those extra payments, the curve now intersects the horizontal axis around 340. In other words, 20 fewer months are required to pay off the loan. Stated another way, the extra payments in the first year cost an extra $\200 \times 12 = \2400$. However, in the long run, those payments saved about $\536.82 \times 20 \approx \10,700$! # Exponential growth and decay (Part 8): Paying off credit-card debt via recurrence relations The following problem in differential equations has a very practical application for anyone who has either (1) taken out a loan to buy a house or a car or (2) is trying to pay off credit card debt. To my surprise, most math majors haven’t thought through the obvious applications of exponential functions as a means of engaging their future students, even though it is directly pertinent to their lives (both the students’ and the teachers’). You have a balance of$2,000 on your credit card. Interest is compounded continuously with a rate of growth of 25% per year. If you pay the minimum amount of $50 per month (or$600 per year), how long will it take for the balance to be paid?
In previous posts, I approached this problem using differential equations. There’s another way to approach this problem that avoids using calculus that, hypothetically, is within the grasp of talented Precalculus students. Instead of treating this problem as a differential equation, we instead treat it as a first-order difference equation (also called a recurrence relation):
$A_{n+1} = r A_n - k$
The idea is that the amount owed is multiplied by a factor $r$ (which is greater than 1), and from this product the amount paid is deducted. With this approach — and unlike the approach using calculus — the payment period would be each month and not per year. Therefore, we can write
$A_{n+1} = \displaystyle \left( 1 + \frac{0.25}{12} \right) A_n - 50$
Notice that the meaning of the 25% has changed somewhat… it’s no longer the relative rate of growth, as the 25% has been equally divided for the 12 months. The solution of this difference equation is
$A_n = r^n P - k \left( \displaystyle \frac{1 - r^n}{1-r} \right)$
A great advantage of using a difference equation to solve this problem is that the solution can be easily checked with a simple spreadsheet. (Indeed, pedagogically, I would recommend showing a spreadsheet like this before doing any of the calculations of the previous few posts, so that students can begin to wrap their heads around the notion of a difference equation before the solution is presented.)
To start the spreadsheet, I wrote “Step” in Cell A1 and “Amount” in Cell B1. Then I entered the initial conditions: $0$ in Cell A2 and $2000$ in Cell B2. (In the screenshot below, I changed the format of column B to show dollars and cents.) Next, I entered $=\hbox{A2}+1$ in Cell A3 and
$=\hbox{B2}*(1+0.25/12)-50$
in Cell B3. Finally, I copied the pattern in Cells A3 and B3 downward. Here’s the result:
After the formula the algebraic solution of the difference equation has been found, this can be added to the spreadsheet in a different column. For example, I added the header “Predicted Amount” in Cell D1. In Cell D2, I typed the formula
$=2000*\hbox{POWER}(1+0.25/12,\hbox{A2})-50*(1-\hbox{POWER}(1+0.25/12,\hbox{A2}))/(1-(1+0.25/12))$
Finally, I copied this pattern down the Column D. Here’s the result:
Invariably, when I perform a demonstration like this in class, I elicit a reaction of “Whoa…. it actually works!” Even for a class of math majors. Naturally, I tease them about this… they didn’t believe me when I used algebra, but now it has to be true because the computer says so.
Here’s the spreadsheet that I used to make the above pictures: CreditCardDebt.
# Day One of my Calculus I class: Part 5
In this series of posts, I’d like to describe what I tell my students on the very first day of Calculus I. On this first day, I try to set the table for the topics that will be discussed throughout the semester. I should emphasize that I don’t hold students immediately responsible for the content of this lecture. Instead, this introduction, which usually takes 30-45 minutes, depending on the questions I get, is meant to help my students see the forest for all of the trees. For example, when we start discussing somewhat dry topics like the definition of a continuous function and the Mean Value Theorem, I can always refer back to this initial lecture for why these concepts are ultimately important.
I’ve told students that the topics in Calculus I build upon each other (unlike the topics of Precalculus), but that there are going to be two themes that run throughout the course:
1. Approximating curved things by straight things, and
2. Passing to limits
We are now trying to answer the following problem.
Problem #2. Find the area under the parabola $f(x) = x^2$ between $x=0$ and $x=1$.
Using five rectangles with right endpoints, we find the approximate answer of $0.44$. With ten rectangles, the approximation is $0.385$. With one hundred rectangles (and Microsoft Excel), the approximation is $0.33835$. This last expression was found by evaluating
$0.01[ (0.01)^2 + (0.02)^2 + \dots + (0.99)^2 + 1^2]$
At this juncture, what I’ll do depends on my students’ background. For many years, I had the same group of students for both Precalculus and Calculus I, and so I knew full well that they had seen the formula for $\displaystyle \sum_{k=1}^n k^2$. And so I’d feel comfortable showing my students the contents of this post. However, if I didn’t know for sure that my students had at least seen this formula, I probably would just ask them to guess the limiting answer without doing any of the algebra to follow.
Assuming students have the necessary prerequisite knowledge, I’ll ask, “What happens if we have $n$ rectangles?” Without much difficulty, they’ll see that the rectangles have a common width of $1/n$. The heights of the rectangles take a little more work to determine. I’ll usually work left to right. The left-most rectangle has right-most $x-$coordinate of $1/n$, and so the height of the leftmost rectangle is $(1/n)^2$. The next rectangle has a height of $(2/n)^2$, and so we must evaluate
$\displaystyle \frac{1}{n} \left[ \frac{1^2}{n^2} + \frac{2^2}{n^2} + \dots + \frac{n^2}{n^2} \right]$, or
$\displaystyle \frac{1}{n^3} \left[ 1^2 + 2^2 + \dots + n^2 \right]$
I then ask my class, what’s the formula for this sum? Invariably, they’ve forgotten the formula in the five or six weeks between the end of Precalculus and the start of Calculus I, and I’ll tease them about this a bit. Eventually, I’ll give them the answer (or someone volunteers an answer that’s either correct or partially correct):
$\displaystyle \frac{1}{n^3} \times \frac{n(n+1)(2n+1)}{6}$, or $\frac{(n+1)(2n+1)}{6n^2}$.
I’ll then directly verify that our previous numerical work matches this expression by plugging in $n=5$, $n= 10$, and $n = 100$.
I then ask, “What limit do we need to take this time?” Occasionally, I’ll get the incorrect answer of sending $n$ to zero, as students sometimes get mixed up thinking about the width of the rectangles instead of the number of rectangles. Eventually, the class will agree that we should send $n$ to plus infinity. Fortunately, the answer $\displaystyle \frac{(n+1)(2n+1)}{6n^2}$ is an example of a rational function, and so the horizontal asymptote can be immediately determined by dividing the leading coefficients of the numerator and denominator (since both have degree 2). We conclude that the limit is $2/6 = 1/3$, and so that’s the area under the parabola.
# Reminding students about Taylor series (Part 4)
I’m in the middle of a series of posts describing how I remind students about Taylor series. In the previous posts, I described how I lead students to the definition of the Maclaurin series
$f(x) = \displaystyle \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!} x^k$,
which converges to $f(x)$ within some radius of convergence for all functions that commonly appear in the secondary mathematics curriculum.
Step 4. Let’s now get some practice with Maclaurin series. Let’s start with $f(x) = e^x$.
What’s $f(0)$? That’s easy: $f(0) = e^0 = 1$.
Next, to find $f'(0)$, we first find $f'(x)$. What is it? Well, that’s also easy: $f'(x) = \frac{d}{dx} (e^x) = e^x$. So $f'(0)$ is also equal to $1$.
How about $f''(0)$? Yep, it’s also $1$. In fact, it’s clear that $f^{(n)}(0) = 1$ for all $n$, though we’ll skip the formal proof by induction.
Plugging into the above formula, we find that
$e^x = \displaystyle \sum_{k=0}^{\infty} \frac{1}{k!} x^k = \sum_{k=0}^{\infty} \frac{x^k}{k!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$
It turns out that the radius of convergence for this power series is $\infty$. In other words, the series on the right converges for all values of $x$. So we’ll skip this for review purposes, this can be formally checked by using the Ratio Test.
At this point, students generally feel confident about the mechanics of finding a Taylor series expansion, and that’s a good thing. However, in my experience, their command of Taylor series is still somewhat artificial. They can go through the motions of taking derivatives and finding the Taylor series, but this complicated symbol in $\displaystyle \sum$ notation still doesn’t have much meaning.
So I shift gears somewhat to discuss the rate of convergence. My hope is to deepen students’ knowledge by getting them to believe that $f(x)$ really can be approximated to high precision with only a few terms. Perhaps not surprisingly, it converges quicker for small values of $x$ than for big values of $x$.
Pedagogically, I like to use a spreadsheet like Microsoft Excel to demonstrate the rate of convergence. A calculator could be used, but students can see quickly with Excel how quickly (or slowly) the terms get smaller. I usually construct the spreadsheet in class on the fly (the fill down feature is really helpful for doing this quickly), with the end product looking something like this:
In this way, students can immediately see that the Taylor series is accurate to four significant digits by going up to the $x^4$ term and that about ten or eleven terms are needed to get a figure that is as accurate as the precision of the computer will allow. In other words, for all practical purposes, an infinite number of terms are not necessary.
In short, this is how a calculator computes $e^x$: adding up the first few terms of a Taylor series. Back in high school, when students hit the $e^x$ button on their calculators, they’ve trusted the result but the mechanics of how the calculator gets the result was shrouded in mystery. No longer.
Then I shift gears by trying a larger value of $x$:
I ask my students the obvious question: What went wrong? They’re usually able to volunteer a few ideas:
• The convergence is slower for larger values of $x$.
• The series will converge, but more terms are needed (and I’ll later use the fill down feature to get enough terms so that it does converge as accurate as double precision will allow).
• The individual terms get bigger until $k=11$ and then start getting smaller. I’ll ask my students why this happens, and I’ll eventually get an explanation like
$\displaystyle \frac{(11.5)^6}{6!} < \frac{(11.5)^6}{6!} \times \frac{11.5}{7} = \frac{(11.5)^7}{7!}$
but
$\displaystyle \frac{(11.5)^{11}}{11!} < \frac{(11.5)^{11}}{11!} \times \frac{11.5}{12} = \frac{(11.5)^{12}}{12!}$
At this point, I’ll mention that calculators use some tricks to speed up convergence. For example, the calculator can simply store a few values of $e^x$ in memory, like $e^{16}$, $e^{8}$, $e^{4}$, $e^{2}$, and $e^{1} = e$. I then ask my class how these could be used to find $e^{11.5}$. After some thought, they will volunteer that
$e^{11.5} = e^8 \cdot e^2 \cdot e \cdot e^{0.5}$.
The first three values don’t need to be computed — they’ve already been stored in memory — while the last value can be computed via Taylor series. Also, since $0.5 < 1$, the series for $e^{0.5}$ will converge pretty quickly. (Some students may volunteer that the above product is logically equivalent to turning $11$ into binary.)
At this point — after doing these explicit numerical examples — I’ll show graphs of $e^x$ and graphs of the Taylor polynomials of $e^x$, observing that the polynomials get closer and closer to the graph of $e^x$ as more terms are added. (For example, see the graphs on the Wikipedia page for Taylor series, though I prefer to use Mathematica for in-class purposes.) In my opinion, the convergence of the graphs only becomes meaningful to students only after doing some numerical examples, as done above.
At this point, I hope my students are familiar with the definition of Taylor (Maclaurin) series, can apply the definition to $e^x$, and have some intuition meaning that the nasty Taylor series expression practically means add a bunch of terms together until you’re satisfied with the convergence.
In the next post, we’ll consider another Taylor series which ought to be (but usually isn’t) really familiar to students: an infinite geometric series.
P.S. Here’s the Excel spreadsheet that I used to make the above figures: Taylor. | {"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": 87, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.860878586769104, "perplexity": 384.8480899227849}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195526064.11/warc/CC-MAIN-20190719053856-20190719075856-00290.warc.gz"} |
http://www.physicsforums.com/showthread.php?t=385204 | # Negative acceleration and instantaneous velocity on an object.
by CaptFormal
Tags: acceleration, instantaneous, negative, object, velocity
P: 33 1. The problem statement, all variables and given/known data The position of an object along a straight tunnel as a function of time is plotted below: 1. Are there any time intervals in which the object has a negative acceleration? If so, list them and explain how you know. 2. Calculate the instantaneous velocity at t=20s and t-45s. 2. Relevant equations N/A 3. The attempt at a solution So, for the first question I stated that at 30s to 37s the object's velocity is decreasing, thus the acceleration is negative. However, I was told that this is only part of the interval. What am I missing? For the second question I did the following calculations: at 20s = (6 m)/(20s) = 0.3 m/s at 45s = (16m)/(45s) = 0.36 m/s I did get the instantaneous velocity at 20s correct but not the instantaneous velocity at 45 s. Where am I going wrong? Thanks for your assistance.
P: 1,877 I would say that the velocity is decreasing from about 27 seconds (at the inflection point) to about 46s (at the other inflection point). Draw tangent lines to the graph, and note the parts where those slopes are getting smaller and smaller (the slope of the slopes). I think you must be close for the one at 45 seconds. It should be around what you got, I'd say more around .43m/s though it's an estimate when you are using a graph rather than the function.
HW Helper
P: 2,324
Quote by CaptFormal 1. The problem statement, all variables and given/known data So, for the first question I stated that at 30s to 37s the object's velocity is decreasing, thus the acceleration is negative. However, I was told that this is only part of the interval. What am I missing?
The velocity is decreasing up until 45 s. Remember that velocity can be negative, so if velocity goes from -1 m/s to -2 m/s, it's decreasing.
For the second question I did the following calculations: at 20s = (6 m)/(20s) = 0.3 m/s at 45s = (16m)/(45s) = 0.36 m/s I did get the instantaneous velocity at 20s correct but not the instantaneous velocity at 45 s. Where am I going wrong?
The velocity should be negative, not positive. Also, you're calculating average velocity. The question is looking for instantaneous velocity, which is the slope of the tangent of the graph.
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https://www.physicsforums.com/threads/a-physics-question.268489/ | # Homework Help: A Physics Question
1. Oct 31, 2008
### unstoppable
1. The problem statement, all variables and given/known data
2) A researcher claims that she can convert a circularly polarized plane wave in air into a linearly polarized one by simply reflecting it from a suitably chosen lossless nonmagnetic dieletric. Assume the dielectric she will use has refractive index n. Is this feat possible for normail incidence, as she claims? Calculate the reflected wave to justify your answer. If she is right, is there a mininum value of n for this to work? If she is wrong; briefly explain why, based on your calculation.
2. Relevant equations
3. The attempt at a solution
2. Nov 1, 2008
### malawi_glenn
attempt to solution? | {"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": 0.8728813529014587, "perplexity": 1472.4197792060202}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267860684.14/warc/CC-MAIN-20180618164208-20180618184208-00264.warc.gz"} |
https://openclimb.io/practice/p3/q11/ | Let $I_n=\int_{-\pi/2}^{\pi/2} \cos^n\!x \,dx$ for any non-negative integer $n.$ $\;(i)$ Find a recursive expression for $I_n.$ $(ii)$ Find the simplest non-recursive expression for $I_{2n}$ that contains $\binom{2n}{n},$ where $\binom{n}{k}=\frac{n!}{k!(n-k)!}.$
1. Evaluate $\int_{-\pi}^{\pi}x^2\,dx.$
2. Find $\int\ln x \,dx.$
3. A geometric sequence starts with $\frac{1}{5}, \frac{2}{15}, \frac{4}{45} \ldots.$ Find a recursive formula for the sequence and solve it to get the closed form.
Try to split $\cos^n x$ into some terms, one of which you can integrate easily.
Have you tried integrating by parts?
Integrate and simplify $I_n$ as much as you can.
Try to rewrite $I_n$ as a recurrence relation.
… you may find a trigonometric identity helpful.
Which initial term can you find to help solve the recurrence for $I_{2n}?$
Try writing out and unwinding $I_{2n}.$
… remembering that your expression should contain $\frac{(2n)!}{(n!)^2}.$
Writing $\cos^n x = \cos x \cdot \cos^{n-1} x$ allows us to use integration by parts to get $I_n = [\sin x \cdot \cos^{n-1} x]_{-\pi/2}^{\pi/2} + \int_{-\pi/2}^{\pi/2} (n-1) \sin^2\!x \cos^{n-2}\!x \,dx.$ Simplifying using a trigonometric identity yields $I_n = (n-1) \int_{-\pi/2}^{\pi/2} (\cos^{n-2} x-\cos^{n} x) \,dx,$ so $I_n = (n-1)(I_{n-2}-I_{n}).$ Rearranging gives $I_{n} = \frac{n-1}{n} I_{n-2}.$
To solve this recurrence relation for $I_{2n},$ we first find $I_0=\int_{-\pi/2}^{\pi/2}dx=\pi.$ Unwinding the recursion we get $I_{2n} = \frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\cdot\frac{2n-5}{2n-4}\cdots\frac{1}{2}\cdot\pi$ which has the product of all odd terms in the numerator and the product of all even terms in the denominator. To get $(2n)!$ in the numerator, we need to fill in the even terms, so we multiply both the numerator and denominator with the denominator, which yields $I_{2n} = \pi \cdot \frac{(2n)!}{2^2 4^2 6^2 \cdots (2n)^2}.$ Factorising $2^2$ from each term in the denominator gives $I_{2n}=\pi \cdot \frac{(2n)!}{2^{2n} (n!)^2}=\binom{2n}{n}\frac{\pi}{4^n}.$ | {"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": 0.9905195236206055, "perplexity": 157.5870265413389}, "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-2021-04/segments/1610703497681.4/warc/CC-MAIN-20210115224908-20210116014908-00044.warc.gz"} |
https://dirkmittler.homeip.net/blog/archives/tag/fluids | ## What Is A Plasma?
The fact that blood plasma exists in Medicine, should not be confused with the fact that Plasmas exist, that are defined in Physics, and which all matter can be converted to. In short, a Plasma is what becomes of a gas, when its temperature is too hot, for it to be a gas.
The long form of the answer is a bit more complex. In Elementary School, Students are taught that there exist three familiar phases of a given substance: Solid, Liquid and Gas. But according to slightly more advanced knowledge in Physics, there is no real guarantee, that there will always be these three phases. A gas first results when the thermal agitation between molecules becomes stronger – i.e. their temperature hotter – than the force that holds individual molecules together. At that point, the molecules separate and a gas results, the physical behavior of which is approximately what one would obtain, if a swarm of particles was to exist through collisions but through few other interactions.
Similarly, Liquids will form, when the molecules are forced from occupying fixed positions, but when they still don’t expand.
Well, as the degree of thermal agitation (of a Gas) is increased further, first, molecules become separated into atoms, and then, the electrons get separated from their nuclei, as a result of ordinary collisions with other atoms. This results in the negative particles – electrons – following different trajectories than the positive particles – the nuclei. And the result of that is that the collective behavior of the fluid changes, from that of a gas.
When a charged particle crosses the lines of force, of a magnetic field, a force is generated which is perpendicular to both the velocity vector and the magnetic field vector. As a result, the particles can travel without restriction along the lines of magnetic force, but their motion at right angles to it is deflected, and becomes helical. Not only that, but the direction in which the paths of the particles becomes curved, is opposite for the negative and positive particles.
For this reason, Plasmas can be confined by magnetic fields, except along the lines of the magnetic field. Increasing the strength of an applied field will also cause a Plasma to become compressed, as these helices become narrower.
A good natural example of this type of Plasma, is what becomes of the substance of the Sun. Its temperatures are easily hot enough to cause the transition from Gas to Plasma, especially since the temperature inside the Sun is much higher, than the temperatures which are observed at its surface. At 5000K, gasses are still possible. But at hundreds of thousands Kelvin, or at a Million degrees Kelvin, the bulk of the Sun’s substance becomes a Plasma.
Now, if the reader is a skeptic, who has trouble believing that ‘other phases’ can exist, than Solid, Liquid and Gas, there is an example that takes place at lower temperatures, and that involves Oxygen, namely, O2. We’re aware of gaseous O2 as well as liquid O2 that gets used in rocketry. But as the O2 is cooled further, to 54.36K at 1 atmosphere, it solidifies. Thus, it has already demonstrated the 3 phases which we’re taught about in Elementary School. But, if we cool already-solid O2 below an even lower temperature, 43.8K at 1 atmosphere, its phase changes again, into yet another phase, which is also a solid one. It’s currently understood that solid O2 has 6 phases. (:1)
At the same time, many fluids are known to exhibit Supercritical Behavior, which is most commonly, a behavior of a fluid which is normally differentiated between Liquid and Gaseous, losing this differentiation, due to its critical pressure being exceeded, but at temperatures at which fluids are commonly boiled. This has nothing to do with Plasmas, but without any distinction between Liquid and Gaseous, a substance which is ordinarily though to have three phases – such as water – ends up demonstrating only two: Fluid and Non-Fluid.
So there is no ultimate reason for which matter needs to be in one out of three phases.
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http://physics.stackexchange.com/questions/35658/how-can-we-define-bf-theory-on-a-general-4-manifold/38356 | How can we define BF theory on a general 4-manifold?
(I have rewritten the question some, with new understanding)
4d BF theory is classically presented as the TFT arising from the Lagrangian
$B\wedge F$,
where $B$ is an abelian 2-connection (locally a real 2-form) and $F$ is the curvature of a connection $A$. People often throw about this theory on general 4-manifolds, but it seems the naive definition from the Lagrangian above is anomalous for most situations. If one restricts to the simpler case of $A$ and $B$ being globally defined forms (not considering them as connections) the phase space is $H^1(\Sigma_4,\mathbb{R}) + H^2(\Sigma_4,\mathbb{R})$ with $A$ as the first coordinate and $B$ as the second. $A$ and $B$ are conjugate from the point of view of the above Lagrangian, but this has no chance of being symplectic if the first and second Betti numbers are not equal. If $A$ and $B$ are connections, however, the only solution to the equations of motion (which not only set $dA=dB=0$ but also their holonomies) is the trivial one, and it is not a problem
However, what I'd really like to consider is $nB\wedge F$. With $A$ and $B$ full-fledged connections. Let us first integrate out $B$. $B$ is a (2-)connection, so $dB$ can be any integral 3-form. We can thus write it as $d\beta + \sum_k m^k \lambda_k$, where $\beta$ is an ordinary real 2-form, and $\lambda_k$ form a basis of integral harmonic 3-forms. Then the action becomes $dA \wedge \beta + \sum_k n m^k A\wedge\lambda_k$. The first term sets $dA=0$ after integrating over $\beta$ and the second term sets the holonomies of $A$ to be nth roots of unity. If our 4-manifold doesn't have torsion, there are precisely $n^{b_1}$ such $A$s.
If we do the same thing, integrating out $A$ instead, we get Dirac deltas setting $dB=0$ and the holonomies of $B$ to be nth roots of unity. Again, if we have no torsion, there are $n^{b_2}$ such $B$s.
It looks like the determinants for these Dirac deltas are all 1, so from the naive calculation above, $Z_n = n^{b_1}$ but also $Z_n = n^{b_2}$. This is a problem on a general 4-manifold, but like the simpler situation above, there is no issue in quantizing on something of the form $\Sigma_3 \times \mathbb{R}$. What I think must be happening is that when I am integrating out $A$ or $B$, I am really only integrating out the non-flat parts. There should still be some more factors that correct the discrepancy above. I think I should be more careful in defining the above integral. Perhaps it is possible to define it sensibly on 4-manifolds which bound a 5-manifold a la Chern-Simons theory.
How does one define this theory on general 4-manifolds?
Thanks.
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BF theory secretly has another name - $Z_n$ gauge theory in the deconfined limit. The parameter $n$ is what appears in front of the $BF$ action. $Z_n$ gauge theory can be defined on any manifold you want - just introduce a lattice approximation of the manifold and compute the lattice gauge theory partition function. By taking the extreme deconfined limit of the lattice gauge theory one can verify that this construction is independent of the way you approximated the manifold.
Basic examples
In the deconfined limit the $Z_n$ flux through every plaquette of the lattice gauge theory is zero. (This is like the constraint imposed by integrating out $B$.) We have a residual freedom to specify the holonomy around all non-contractible loops. Hence $Z_n(M) = Z_n(S^4) n^{b_1(M)}$ where $Z_n(S^4)$ is a normalization constant. Requiring that $Z_n(S^3 \times S^1) =1$ gives $Z_n(S^4) = 1/n$. This condition, $Z_n(S^3 \times S^1) = 1$, is the statement that the theory has one unique ground state on $S^3$. In general, $Z_n(\Sigma^3 \times S^1)$ is $\text{tr}(e^{-\beta H})$, but since $H=0$ we are simply counting ground states.
As a quick check, $Z_n(S^2 \times S^1 \times S^1) = n$, which is the number of ground states on $S^2 \times S^1$, and $Z_n(T^3\times S^1) = n^3$, which is the number of ground states on $T^3 = (S^1)^3$.
Further relations
Another check on the value of $Z_n(S^4)$: this is the renormalized topological entanglement entropy of a ball in the $3+1$d topological phase described by deconfined $Z_n$ gauge theory. More precisely, the topological entanglement entropy of a ball is $-\ln{Z_n(S^4)}$ which gives $-\log{n}$ in agreement with explicit wavefunction calculations.
We can also consider defects. The $BF$ action is $\frac{n}{2\pi} \int B \wedge d A$. Pointlike particles (spacetime worldlines) that minimally couple to $A$ carry $Z_n$ charge of $1$. Similarly, string excitations (spacetime worldsheets) that minimally couple to $B$ act like flux tubes carry $Z_n$ flux of $2\pi/n$. Now when a minimal particle encircles a minimal flux, we get a phase of $2 \pi/n$ (AB phase), but this also follows from the $BF$ action. Without getting into two many details, the term in the action like $B_{12} \partial_t A_3$ fixes the commutator of $A_3$ and $B_{12}$ to be $[A_3(x),B_{12}(y)] = \frac{2\pi i}{n} \delta^3(x-y)$ (flat space). The Wilson-line like operators given by $W_A = e^{i \int dx^3 A_3}$ and $W_B = e^{i \int dx^1 dx^2 B_{12}}$ thus satisfy $W_A W_B = W_B W_A e^{2 \pi i /n}$ which is an expression of the braiding flux above that arises since the particle worldline pierced the flux string worldsheet.
Conservative thoughts
If I understand you correctly, what you want to do is sort of argue directly from the continuum path integral and show how the asymmetry you mentioned arises. I haven't done this calculation, so I can't be of direct help on this point right now. I'll ponder it though.
That being said, it's not at all clear to me that treating the action as $\int A dB$ leads one to just counting 2-cycles. Of course, I agree that 2-cycles are how you get non-trivial configurations of $B$, but in a conservative spirit, it's not at all clear to me, after gauge fixing and adding ghosts and whatever else you need to do, that the path integral simply counts these.
The way that I know the path integral counts 1-cycles is that I have an alternative formulation, the lattice gauge theory, that is easy to define and unambiguously does the counting. However, I don't know how it works for the other formulation. I suppose a naive guess would be to look at lattice 2-gauge theory for $B$, but this doesn't seem to work.
Ground states
One thing I can address is the issue of ground states. In fact, I favor this approach because one doesn't have to worry about so many path integral subtleties. All you really needs is the commutator.
Take the example you raise, $S^2 \times S^1$. There are two non-trivial operators in this case, a $W_B$ for the $S^2$ and a $W_A$ for the $S^1$. Furthermore, these two operators don't commute, which is why there are only $n$ ground states. If we define $|0\rangle$ by $W_A |0 \rangle = |0\rangle$, then the $n$ states $\{|0\rangle, W_B |0\rangle, ..., W_B^{n-1} | 0 \rangle\}$ span the ground state space and are distinguished by their $W_A$ eigenvalue. Importantly, you can also label states by $W_B$ eigenvalue, but you still only get $n$ states. Poincare duality assures you that this counting of states will always work out no matter how you do it.
Furthermore, the operator $W_B$ has a beautiful interpretation in terms of tunneling flux into the non-contractible loop measured by $W_A$. It's easier to visualize the analogous process in 3d, but it still works here.
You can also see the difference between the theory in 4d and 4d. Since both $B$ and $A$ are 1-forms you have different possibilities. The analogue of $S^2 \times S^1$ might be $S^1 \times S^1$ and this space does have $n^2$ states. However, that is because both $A$ and $B$ can wrap both cycles.
The remarkable thing is that the $Z_n$ gauge theory formulation always gets it right. You simply count the number of one-cycles and that's it.
Getting at other spaces
The state counting approach gets you everything of the form $Z_n(\Sigma^3 \times S^1)$, but even spaces like $S^2 \times S^2$ can be accessed. You can either do the direct euclidean gauge theory computation, or you can regard $Z_n(S^2 \times S^2) = |Z(S^2 \times D^2)|^2$ i.e. the inner product of a state on $S^2 \times S^1 = \partial (S^2 \times D^2)$ generated by imaginary time evolution.
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Thanks, Physics Monkey. This is a good answer so far, but it does not address the inconsistency that I'm talking about. If one takes the other polarization ($B$ as coordinates, integrating out $A$), the theory is [ $U(1)$-n->$U(1)$ ] 2-gauge theory (roughly speaking, instead of holonomies around 1-cycles, we have holonomies around 2-cycles, also in $\mathbb{Z}_n$). This interpretation gives the calculation $Z_n = n^{b_2}$. – Ryan Thorngren Sep 5 '12 at 19:52
Let me be more particular with my complaint. The number of ground states on $S^2 \times S^1$ in my mind should be $n^2$. Once you pick two elements of $\mathbb{Z}_n$ this uniquely defines for you a flat 2-connection $B$ (that is, a global 2-form with $dB=0$) and a flat 1-connection $A$ each with the prescribed holonomies around the sphere and circle factor, respectively. That is to say, $dB=0$ does not imply that $B$ has null periods. I guess this really comes down to what we choose our gauge transformations to be. – Ryan Thorngren Sep 5 '12 at 19:56
Here's what I mean by my last comment. The theory is symmetric under $B\rightarrow B+\beta$ and $A\rightarrow A+\alpha$ for a flat 2-connection $\beta$ and a flat 1-connection $\alpha$. There is a unique ground state up to these gauge transformations. – Ryan Thorngren Sep 5 '12 at 19:59
Ok, I think I understand more what you're getting at. I'll edit my answer above. – Physics Monkey Sep 7 '12 at 1:09
Thanks for the updated answer. I certainly see what you mean for $S^2 \times S^1$ and agree with you now (also, my third comment is wrong, those are not actually symmetries). The $A$ and $B$ coordinates are conjugate is another way to look at why $W_A$ and $W_B$ must not commute (indeed, the linking number prescription is like a topological Heisenberg algebra). I am continuing to think about more general 4-manifolds. I will get back to you hopefully soon with some resolution or maybe more questions. – Ryan Thorngren Sep 7 '12 at 3:38
I figured this out a little while ago, so as not to leave any of you hanging, here is a (almost complete) resolution.
First, there is a lattice formulation of the 2-gauge theory that agrees with the 1-gauge theory up to the topological factor $n^{\chi(\Sigma_4)}$. One has a $\mathbb{Z}_n$ variable for each 2-cell, gauge transformations on 1-cells, and importantly, 2-gauge transformations (gauge transformations between gauge transformations) on 0-cells. The number of configurations is $n^{b_2}$. Then we divide by the volume of the gauge group, which is the number of gauge transformations divided by the number of 2-gauge transformations. So the partition function of this lattice theory is $n^{b_2-b_1+b_0}$. This gives the same answer as the $A$-theory on any 4-manifold fibered over $S^1$, in particular the 4-torus @Physics Monkey and I were discussing.
To get this answer from the path integral, I gauge fix $B$ using the Lorentz gauge $d*B=0$, introducing a Lagrange multiplier term $\langle \pi_1, \delta B\rangle$ as well as some ghost terms not involving $B$ or $\pi_1$. $\pi_1$ has gauge symmetries, being a 1-form, which I also fix, introducing a term $\langle E_0, \delta \pi_1\rangle$ and some more ghost terms which do not involve $B$, $\pi_1$, or $E_0$. Now I scale $B\rightarrow B/n$, $\pi_1 \rightarrow n\pi_1$, and $E_0\rightarrow E_0/n$ to reduce the integrand to the case with $n=1$ (that integral can be shown to be 1).
I just need to find how the path integral measure scales under these transformations. $DB$ will scale by $n$ to the power of the dimension of the $B$s being integrated over. It's important to notice that the zero modes of the integrand are precisely the harmonic 2-forms, a space of dimension $b_2$. Then if $B_2$ is the (infinite) dimension of all 2-forms, $DB$ scales by $n^{b_2-B_2}$. Terms like $n^{B_k}$ can all be removed by some suitable regularization a la Witten's paper on abelian S-duality. What we are left with is the answer above, $n^{b_2-b_1+b_0}$.
Maybe there is some way to understand the topological factor that appears?
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Excuse me, is the k-cycle used in your context is the same as this definition? i.e. The length of a cycle, is the number of elements of its orbit of non-fixed elements. A cycle of length k is also called a k-cycle. – mysteriousness Jan 10 at 20:43
I mean cycle in the homology sense. Let me know if you have more questions. I understand this topic quite a bit more now that it's been over a year. :) – Ryan Thorngren Jan 12 at 5:48
How do you understand the $B$ as a n-form field in terms of $\mathbb{Z}_N$ gauge group? In the sense of group cohomology, a 1-form field $A$ can be realize as a group element $a \in \mathbb{Z}_N$. The derivative of 1-form $A$ as $dA$ can be realize as $(a+b)-[a+b]$mod$\mathbb{Z}_N$. Is that $B$ as n-form field still realize as $b \in \mathbb{Z}_N$? or something more? – mysteriousness Jan 21 at 1:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9606262445449829, "perplexity": 256.3422048031206}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997894289.49/warc/CC-MAIN-20140722025814-00189-ip-10-33-131-23.ec2.internal.warc.gz"} |
https://www.futurelearn.com/info/courses/unlocking-design-potential/0/steps/266446 | # SimScale: Importing geometry
In this video, Dr. Ico Broekhuizen explains how to import your model into SimScale and generate a virtual wind tunnel for simulation.
2.2
Hi, in this video, I will start to show you how to perform wind simulation of urban areas using the online SimScale platform. The first thing we need to do is import a model that we’ve created previously, for example, in Revit or SketchUp. So here on the left hand side, you can click the plus icon on the geometries and upload almost any file, any model file that you may have, and then SimScale imports this and shows you a quick representation of what the model looks like. And you have some control over this view and what it will look like using the buttons here at the top.
37.9
The first thing we then need to do is to translate this model of the buildings to a model of the wind tunnel that we will actually be simulating. So we will edit this model in the CAD mode. So once you go there, you see the same model as before. The first thing we will do is to define an external flow volume. And we can then specify the dimensions that this wind tunnel should have. So the bottom will usually be at zero meters, so it matches that, so it matches the bottom of the buildings that we have.
74.3
And for the height of the model, there is a rule of thumb that says that if the highest building is, say, 20 meters high, you use six times that height as the height of the wind tunnel. And in the same way, you can also specify the other directions so that the - If the wind will be coming from the right hand side of the view here, we want five times that height of the tallest building on the right hand side, on the bottom, and at the top. And we want around 15 times the height and the downwind side of the model.
116.1
So with those basic dimensions set up, we can then use this seed face option and we can simply select a roof of a building. This isn’t always necessary, but sometimes we need to do this so SimScale can figure out how to create this volume. And then we simply click apply and it will create this box representing the wind tunnel that we will be simulating.
138.5
So then we just need to get rid of all the buildings that we added because they won’t be part of the simulation, so you can either use this view on the right here to select all of the buildings, or you can also simply select the volume of the wind tunnel and then right click to invert that selection. So if you look underneath, you can see that all the buildings are selected. And with all the buildings selected, we can simply delete them from the model.
171.8
So as you’ll see the buildings have now disappeared, but the imprints of the buildings are still present in the wind volume that we will be simulating later on. And you can simply click Export to return to the CFD environment. And there you will once again see the wind tunnel volume that we have just created. In the next video, I’ll go over all the settings you need to put up before we can start the simulation.
SimScale allows you to import models that you have previously created in 3D modelling programs such as Revit or Sketchup. The CAD mode in SimScale then allows you to convert a model of buildings to a model of the wind volume around the buildings, which needs to be sufficiently large to get accurate simulation results. | {"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": 0.8777242302894592, "perplexity": 350.05521294865747}, "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/1669446711042.33/warc/CC-MAIN-20221205164659-20221205194659-00630.warc.gz"} |
http://www.emathhelp.net/notes/differential-equations/nth-order-linear-ode-with-constant-coefficients/variation-of-parameters/ | # Variation of Parameters
Variation of parameters, like method of undetermined coefficients, is another method for finding a particular solution of the nth-order linear differential equation L(y)=phi(x) once the solution of the associated homogeneous equation L(y) = 0 is known.
Recall from linear independence note, that if y_1(x),\ y_2(x),\ ...,\ y_n(x) are n linearly independent solutions of L(y)=0, then the general solution of L(y)=0 is y_h=c_1 y_1(x)+c_2 y_2(x)+...+c_n y_n(x) .
A variation of parameter method assumes that particular solution has form of homogeneous solution except that constants c_i (i=1, 2, ..., n) are functions of x (that's why it is called variation of parameter): y_p=c_1(x) y_1(x)+c_2(x)y_2(x)+...+c_n(x)y_n(x) .
We know y_i(x) (i=1, 2, ..., n) but we need to determine c_i(x) (i=1, 2, ..., n).
For this set up following linear system of equations:
{(c_1'y_1+c_2'y_2+...+c_n'y_n=0),(c_1'y_1'+c_2'y_2'+...+c_n'y_n'=0),(...),(c_1'y_1^((n-1))+c_2'y_2^((n-1))+...+c_n'y_n^((n-1))=0),(c_1'y_1^((n))+c_2'y_2^((n))+...+c_n'y_n^((n))=0):}
Then integrate each c_i' to obtain c_i , disregarding all constants of integration. This is permissible because we are seeking only one particular solution.
Since y_1(x),\ y_2(x),\ ...,\ y_n(x) are n linearly independent solutions of the same equation L(y) = 0, their Wronskian is not zero. This means that the system has a nonzero determinant and can be solved uniquely for c_1'(x),\ c_2'(x),\ ...,\ c_n'(x) .
The method of variation of parameters can be applied to all linear differential equations. It is therefore more powerful than the method of undetermined coefficients, which is restricted to linear differential equations with constant coefficients and particular forms of phi(x) . Nonetheless, in those cases where both methods are applicable, the method of undetermined coefficients is usually the more efficient and, hence, preferable.
As a practical matter, the integration of v_i'(x) may be impossible to perform. In such an event, other methods (in particular, numerical techniques) must be employed.
Example 1. Solve y''-2y'+y=(e^x)/x .
Homogeneous solution is y_h=c_1e^(x)+c_2xe^x .
To find particular solution, set up system:
{(c_1'e^x+c_2'xe^x=0),(c_1'(e^x)'+c_2'(xe^x)'=(e^x)/x):}
or
{(c_1'e^x+c_2'xe^x=0),(c_1'e^x+c_2'(xe^x+e^x)=(e^x)/x):}
Subtract first equation from second: c_2'e^x=(e^x)/x or c_2'=1/x .
From first equation c_1'=-c_2'x=-1/x x=-1 .
So, c_2=int 1/x dx=ln(|x|) and c_1=int -1dx=-x .
So, y_p=c_1e^x+c^2xe^x=-xe^x+ln(|x|)xe^x .
Finally, general solution is y=y_h+y_p=c_1e^x+c_2xe^x-xe^x+ln(|x|)xe^x=c_1e^x+c_3xe^x+ln(|x|)xe^x where c_3=c_2-1 .
Example 2. Solve 2y''-3y'+y=e^x .
Corresponding homogeneous equation is 2y''-3y'+y=0 . Characteristic equation is 2r^2-3r+1=0 which has roots r_1=1, r_2=1/2 .
So, y_h=c_1e^x+c_2e^(1/2 x) .
Now, be careful with applying variation of parameters. To apply it correctly coefficient near the highest derivative must be 1, so first divide both sides of differential equation by 2: y''-1.5y'+0.5y=1/2 e^x .
Now, set up system:
{(c_1'e^x+c_2'e^(1/2 x)=0),(c_1'(e^x)'+c_2'(e^(1/2 x))'=1/2 e^x):}
Or
{(c_1'e^x+c_2'e^(1/2 x)=0),(c_1'e^x+1/2 c_2'e^(1/2 x)=1/2 e^x):}
Subtract first equation from second: -1/2c_2'e^(1/2 x)=1/2e^x or c_2=-e^(1/2 x) .
From first equation c_1'=-c_2'e^(-1/2x)=-(-e^(1/2 x))e^(-1/2 x)=1 .
So, c_1=int 1dx=x and c_2=int -e^(1/2 x)dx=-2e^(1/2 x) .
Therefore, y_p=c_1e^x+c_2e^(1/2 x)=xe^x-2e^x .
Finally, y=y_h+y_p=c_1e^x+c_2e^(1/2 x)+xe^x-2e^x=c_3e^x+c_2e^(1/2 x)+xe^x where c_3=c_1-2 .
Example 3. Find general solution of y'''+y'=1/(cos(x)) .
Solution of the corresponding homogeneous equation is y_h=c_1+c_2cos(x)+c_3sin(x) .
To find particular solution set up system:
{(c_1'*1+c_2'cos(x)+c_3'sin(x)=0),(c_1'*(1)'+c_2'(cos(x))'+c_3'(sin(x))'=0),(c_1'*(1)''+c_2'(cos(x))''+c_3'(sin(x))''=1/(cos(x))):}
Or
{(c_1'+c_2'cos(x)+c_3'sin(x)=0),(-c_2'sin(x)+c_3'cos(x)=0),(-c_2'cos(x)-c_3'sin(x)=1/(cos(x))):}
Solving this system gives c_1'=1/(cos(x)),\ c_2'=-1,\ c_3'=-tan(x) .
So,
c_1=int 1/(cos(x))dx=ln(|(1+sin(x))/(cos(x))|)
c_2=int -1dx=-x
c_3=int -tan(x)dx=ln(|cos(x)|) .
Thus, y_p=c_1+c_2cos(x)+c_3sin(x)=ln(|(1+sin(x))/(cos(x))|)-xcos(x)+ln(|cos(x)|)sin(x) .
Finally, general solution is
y=y_h+y_p=c_1+c_2cos(x)+c_3sin(x)+ln(|(1+sin(x))/(cos(x))|)-xcos(x)+ln(|cos(x)|)sin(x) | {"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": 0.9698055386543274, "perplexity": 4382.074233307808}, "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/1454701164289.84/warc/CC-MAIN-20160205193924-00341-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://mathhelpforum.com/algebra/156023-not-sure-if-algebra.html | # Thread: not sure if it is algebra...
1. ## not sure if it is algebra...
but im not sure how i would solve this (question in attachment!)
could someone please point me in the right direction to solving this question!?!
Thanks!!
2. Originally Posted by andyboy179
but im not sure how i would solve this (question in attachment!)
could someone please point me in the right direction to solving this question!?!
Thanks!!
$\displaystyle \frac {2x+1} {3x-7}= 2$
multiply all with 3x-7
$2x+1= 2\cdot (3x-7)$
$2x+1 = 6x-14$
$4x = 15 \Rightarrow x= 15/4$
3. if i multiply all by 3x-7 it would be, 6x+7/3x-7=2? then what would i do?
4. Originally Posted by andyboy179
but im not sure how i would solve this (question in attachment!)
could someone please point me in the right direction to solving this question!?!
Thanks!!
Dear andyboy179,
First multiply both sides of the equation by 3x-7. Afterwards you can substract 2x from both sides. Hope you can continue.
5. so it would be, 6x+7/ 9x-49=2 then 4x+7/ 7x-49=2???
6. Originally Posted by andyboy179
if i multiply all by 3x-7 it would be, 6x+7/3x-7=2? then what would i do?
$\displaystyle \frac {2x+1} {3x-7}= 2 \Rightarrow
\frac {(2x+1)\cdot (3x-7)} {(3x-7) }= 2\cdot (3x-7 ) \Rightarrow 2x+1= 2\cdot (3x-7)$
so you will lose fraction after dividing $\displaystyle \frac {(2x+1)\cdot (3x-7)}{(3x-7)} = (2x+1)$
do you understand ?
P.S. when said multiply all .... means that you multiply and left and right side of equation !!!
7. i really don't understand this!! could i do box method with 2x+1 and 3x-7? to get 6x^2-x-6?
8. Originally Posted by andyboy179
i really don't understand this!! could i do box method with 2x+1 and 3x-7? to get 6x^2-x-6?
let's look at this this way ...
$2x = 2 \Rightarrow x = 2/2 = 1$
multiply both sides of equation with 2 and you will have
$4x= 4 \Rightarrow x = 4/4 = 1$
do you see that nothing is changed ?
now with fractions ....
$\frac {2x}{2} = 1$
because of that under fraction is 2 (can be anything) to lose that fraction you multiply both sides with that whatever is under fraction ....
in this case multiply with 2
$\frac {2x}{2}\cdot 2 = 1 \cdot 2 \Rightarrow \frac {2x \cdot 2 }{2} = 1\cdot 2 \Rightarrow 2x = 2 \Rightarrow x =2/2 = 1$
9. i kinda understand now. you say multiply both sides with 2 so how would i do 2x+1x2? and 3x-7x2?
10. Originally Posted by andyboy179
i kinda understand now. you say multiply both sides with 2 so how would i do 2x+1x2? and 3x-7x2?
you don't multiply by two in your problem there... i just do that with my example to show you that nothing changes if you multiply both sides (or add or divide or ... ) Whatever is "down" in fraction (can't figure out how to write it on English ) you multiply both sides with THAT in your case it's 3x-7
so you will be free of the fraction
P.S. when multiplying fraction with number you don't multiply down expression
$\displaystyle \frac {x+y}{a+b} \cdot 10 = \frac {10x +10y}{a+b}$
but only if you are multiplying fraction with fraction ...
$\displaystyle \frac {x+y}{a+b} \cdot \frac {10}{3} = \frac {10x+10y}{3a+3b}$
11. OHHH so would it be 6x^2 - 10x -6?
12. Originally Posted by andyboy179
OHHH so would it be 6x^2 - 10x -6?
what would that be ... ????
lol... don't know what to do now .... probably someone who write better in English can help you maybe that's the problem
sorry...
13. would the answer be 2.3?? don't worry i can understand the english but im just finding it hard to pick up how to do these sort of questions!
14. Originally Posted by andyboy179
would the answer be 2.3?? don't worry i can understand the english but im just finding it hard to pick up how to do these sort of questions!
you have solution in my first post (#2)
it's not problem for you with English ... it's mine
P.S. you need to get familiar with basic operations with fractions, and on fractions (as multiplying , divide, add , subtract .... ) so you will not have problem piking up stuff like this | {"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": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9394155740737915, "perplexity": 951.5945907711488}, "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/1476988719564.4/warc/CC-MAIN-20161020183839-00042-ip-10-171-6-4.ec2.internal.warc.gz"} |
http://mathhelpforum.com/geometry/78932-sphere-cone-geometry.html | # Math Help - Sphere in Cone Geometry
1. ## Sphere in Cone Geometry
That's a question on my textbook.... i can't solve it and my teacher is away. Anyone help?
Thanks.
2. Originally Posted by anonymous_maths
...
That's a question on my textbook.... i can't solve it and my teacher is away. Anyone help?
Thanks.
1. The cross-section of the cone is an equilateral triangle with AM = R.
In the top right triangle you get:
$R \cdot \tan(30^\circ) = r~\implies~R=\dfrac r{\tan(30^\circ)} = \dfrac r{\frac13 \sqrt{3}}~\implies~\boxed{R=r \cdot \sqrt{3}}$
2. Let denote x the distance in question. Use Pythagorean theorem:
$R^2+r^2=(r+x)^2~\implies~3r^2+r^2 = (r+x)^2~\implies~2r=r+x~\implies~x=r$
Unfortunately I have to leave now ...
Attached Thumbnails
3. Originally Posted by anonymous_maths
...
That's a question on my textbook.... i can't solve it and my teacher is away. Anyone help?
Thanks.
Let h denote the height of the cone. Then:
$h^2+R^2=(2R^2)~\implies~h=R\cdot \sqrt{3}$
The volume of a cone is calculated by:
$V=\dfrac13 \cdot \pi \cdot R^2 \cdot h$
That means:
$V=\dfrac13 \cdot \pi \cdot R^2 \cdot R \cdot \sqrt{3} = \dfrac{\pi \sqrt{3}}3 R^3$
------------------------------------------------------------------------------------
From a) you know that
$R=r \cdot \sqrt{3}~\implies~r=\dfrac{\sqrt{3}}3 R$
The volume of the sphere is calculated by:
$V=\dfrac43 \pi r^3$
That means:
$V=\dfrac43 \pi \left(\dfrac{\sqrt{3}}3 R\right)^3 = \dfrac{4 \pi \sqrt{3}}{27} R^3$
------------------------------------------------------------------------------------
The percentage is calculated by $percentage = \dfrac{partial\ amount}{total\ amount} \cdot 100$
$p=\dfrac{\dfrac{4\pi \sqrt{3}}{27} R^3}{\dfrac{\pi \sqrt{3}}3 R^3}\cdot 100 = \dfrac49 \cdot 100 \approx 44.4\%$ | {"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": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9902644753456116, "perplexity": 1918.6839113104306}, "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-2014-15/segments/1397609537186.46/warc/CC-MAIN-20140416005217-00367-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://mathhelpforum.com/algebra/4348-need-help-variation.html | # Math Help - Need HELP: Variation
1. ## Need HELP: Variation
The Problem
(1)
The time T needed to mill a channel varies directly as the length, width, and depth of the channel, assuming the cutting tool and material being machined stay constant.
a) Write this as an equation using a constant of proportionality;
b) A channel 30 cm long, 5.0cm wide and 4.5 cm deep took 36 minutes to mill. Use this information to determine the constant proportionality, with units.
c) How long will it take to mill a channel in the same material, 45 cm long, 6.4 cm wide, and 3.75 cm deep?
d) How long would the machining job of b) above take, if the length, width, and depth were all doubled
(2) It takes three men three hours to assemble three shelving units. How long will it take nine men to assemble twelve shelving units? Prove your answer
(3) The time taken to machine out the interior of a hollow cylinder in a shaft on a lathe varies directly as the length of the cylinder and directly as the square of the internal diameter.
a) Write this as an equation using a constant of proportionality;
b) It took 22 minutes to machine out a cylinder 14.5 cm long and 3.45 cm in diameter. Use this information to determine the constant of proportionality, with units.
c) How long would it take to machine out a cylinder 9.6 cm long, and 5.90 cm in diameter?
(4) The cutting speed of a milling cutter varies directly as the diameter of the cutter and directly as the RPM of the cutter.
a) Write this as an equation using a constant of proportionality;
b) Turing a milling cutter 8.5 cm in diamter at 6250 RPM results in a cutting speed of 2.78 m/s. Use this information to determine the constant of proportionality, with units.
c) Calculate the needed RPM to produce a recommended cutting speed of 4.25 m/s (in another material), using a 12.5 diameter cutter.
2. Originally Posted by bigstarz
The Problem
(1)
The time T needed to mill a channel varies directly as the length, width, and depth of the channel, assuming the cutting tool and material being machined stay constant.
a) Write this as an equation using a constant of proportionality;
$T=\kappa \times l \times w \times d$
b) A channel 30 cm long, 5.0cm wide and 4.5 cm deep took 36 minutes to mill. Use this information to determine the constant proportionality, with units.
Given the above data we have:
$36= \kappa \times 30 \times 5 \times 4.5$
so:
$\kappa=\frac{36}{30 \times 5 \times 4.5}\approx 0.0533 \mbox{ min/cm^3}$
c) How long will it take to mill a channel in the same material, 45 cm long, 6.4 cm wide, and 3.75 cm deep?
d) How long would the machining job of b) above take, if the length, width, and depth were all doubled
I'm sure you can now do these two parts yourself.
RonL
3. Originally Posted by bigstarz[B
(2) [/B] It takes three men three hours to assemble three shelving units. How long will it take nine men to assemble twelve shelving units? Prove your answer
If it takes three men three hours to assemble three shelving units, it will
take one man three hours to assemble one shelving unit. therefore each
man can assemble 1/3 of a shelving unit per hour.
Twelve shelving units is 36 thirds of a shelving unit, so 36 man hours of
work is required to assemble them, which is 4 hours of work by nine men.
RonL
4. Originally Posted by bigstarz
(4) The cutting speed of a milling cutter varies directly as the diameter of the cutter and directly as the RPM of the cutter.
a) Write this as an equation using a constant of proportionality;
b) Turing a milling cutter 8.5 cm in diamter at 6250 RPM results in a cutting speed of 2.78 m/s. Use this information to determine the constant of proportionality, with units.
$\kappa=\frac{\mbox{speed}}{\mbox{diameter}}$
Thus,
$\kappa=\frac{\frac{2\pi r}{t}}{2\pi r}=\frac{1}{t}$
Notes:
Where "t" is time for revolution.
To find the speed you divide distance by time.
Distance in this case the circumfurence of circle.
5. Originally Posted by ThePerfectHacker
[QUOTE=bigstarz
(4) The cutting speed of a milling cutter varies directly as the diameter of the cutter and directly as the RPM of the cutter.
a) Write this as an equation using a constant of proportionality;
b) Turing a milling cutter 8.5 cm in diamter at 6250 RPM results in a cutting speed of 2.78 m/s. Use this information to determine the constant of proportionality, with units.
$\kappa=\frac{\mbox{speed}}{\mbox{diameter}}$[/QUOTE]
The cutting speed of a milling cutter varies directly as the diameter of the cutter and directly as the RPM of the cutter.
So you should have:
$a)\ speed=\kappa \times diameter \times RPM$,
then:
$
\kappa=\frac{speed}{diameter \times RPM}
$
,
and so:
$
b)\ \kappa=\frac{2.78}{8.5 \times 6250}=0.0000523\ \mbox{(m/s)/(cm/s)}
$
RonL | {"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": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8116240501403809, "perplexity": 1497.177155959702}, "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/1424936462206.12/warc/CC-MAIN-20150226074102-00155-ip-10-28-5-156.ec2.internal.warc.gz"} |
https://mca2017.org/fr/prog/session/fid | # Systèmes hamiltoniens en dimension finie et infinie
Tous les résumés
Taille : 94 kb
Organisateurs :
• Carlos García Azpeitia (Universidad Nacional Autónoma de México)
• Renato Calleja (Universidad Nacional Autónoma de México)
• Andrés Contreras Marcillo (New Mexico State University)
• Walter Craig (McMaster University)
• Geordie Richards
University of Rochester
On invariant Gibbs measures for generalized KdV
Résumé en PDF
Taille : 39 kb
We will discuss some recent results on invariant Gibbs measures for the periodic generalized KdV equations (gKdV). Proving invariance of the Gibbs measure for gKdV is nontrivial due to the low regularity of functions in the support of this measure. Bourgain proved this invariance for KdV and mKdV, which have quadratic and cubic nonlinearities, respectively. Previously, we proved invariance of the Gibbs measure for the quartic gKdV by exploiting a nonlinear smoothing induced by initial data randomization. More recently, in joint work with Tadahiro Oh (Edinburgh) and Laurent Thomann (Lorraine), we have established this invariance for gKdV with any odd power (defocusing) nonlinearity. The proof relies on a probabilistic construction of solutions using the Skorokhod representation theorem.
• Robert L. Jerrard
University of Toronto
dynamics of nearly-parallel vortex filaments in the Gross-Pitaevskii equations
Résumé en PDF
Taille : 39 kb
We study the motion of thin, nearly parallel vortex filaments in 3d solutions of the Gross-Pitaevskii equations. In particular, we show that in a certain scaling limit, these filaments are governed by a system of nonlinear Schrödinger equations formally derived by Klein, Majda, and Damodaran in the mid '90s in the context of the Euler equations. This is the first rigorous justification of the Klein-Majda-Damodaran model in any setting. This is joint work with Didier Smets.
• Luis Vega
Sans titre
Résumé en PDF
Taille : 25 kb
• Andres Contreras
New Mexico State University
Eigenvalue preservation for the Beris-Edwards system
Résumé en PDF
Taille : 38 kb
Eigenvalue preservation in the Beris-Edwards system is the property that ensures that Q-tensors remain physical along the flow. The Beris-Edwards system is a simplified model for the evolution of nematic liquid crystals. The preservation of eigenvalues property is known to hold for the evolution in the whole space thanks to the work of Xiang-Zarnescu. In this talk I present a simpler and shorter proof that also applies to the bounded domain case. This is joint work with Xiang Xu and Wujun Zhang.
• Cheng Yu
University of Texas at Austin
Sans titre
Résumé en PDF
Taille : 25 kb
• Stefan Le Coz
Institut de Mathématiques de Toulouse
Stability of multi-solitons for the derivative nonlinear Schrödinger equation
Résumé en PDF
Taille : 38 kb
The nonlinear Schrödinger equation with derivative cubic nonlinearity (dNLS) is a model quasilinear dispersive equation. It admits a family of solitons, which are orbitally stable in the energy space. After a review of the many interesting properties of dNLS, we will present a result of orbital stability of multi-solitons configurations in the energy space, and some ingredients of the proof.
• Walter Craig
McMaster University
A Hamiltonian and its Birkhoff normal form for water waves
Résumé en PDF
Taille : 39 kb
A 1968 paper by VE Zakharov gives a formulation of the equations for water waves as a Hamiltonian dynamical system, and shows that the equilibrium solution is an elliptic stationary point. This talk will discuss two aspects of the water wave equations in this context. Firstly, we generalize the formulation of Zakharov to include overturning wave profiles, answering a question posed to the speaker by T. Nishida. Secondly, we will discuss the question of Birkhoff normal forms for the water waves equations in the setting of spatially periodic solutions. We transform the water wave problem with nonzero surface tension to third order Birkhoff normal form, and in the case of zero surface tension in deep water, to fourth order Birkhoff normal form. The result includes a discussion of the dynamics of the normal form, and a quantification of the function space mapping properties of these transformations.
• Renato Calleja
Symmetries and choreographies in families that bifurcate from the polygonal relative equilibrium of the n-body problem
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Taille : 40 kb
In my talk I will describe numerical continuation and bifurcation techniques in a boundary value setting used to follow Lyapunov families of periodic orbits. These arise from the polygonal system of n bodies in a rotating frame of reference. When the frequency of a Lyapunov orbit and the frequency of the rotating frame have a rational relationship, the orbit is also periodic in the inertial frame. We prove that a dense set of Lyapunov orbits, with frequencies satisfying a Diophantine equation, correspond to choreographies. We present a sample of the many choreographies that we have determined numerically along the Lyapunov families and bifurcating families, namely for the cases n=4,6,7,8 and, 9. We also present numerical results for the case where there is a central body that affects the choreography, but that does not participate in it. This is joint work with Eusebius Doedel and Carlos García Azpeitia.
• Michela Procesi
Universitá di Roma Tre
Finite dimensional invariant tori in PDEs
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I shall discuss the existence and stability of quasi-periodic invariant tori for classes of evolution PDEs, both in Hamiltonian and reversible setting,trying to give an idea of the general strategy and the main difficulties in problems of this kind. I will also discuss the related problem of unstable solutions.
• Yannick Sire
John Hopkins University
A posteriori KAM for PDEs
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Taille : 37 kb
I will describe recent results in collaboration with R. de la Llave on an a posteriori KAM method for PDEs. In particular, our methods uses very little of symplectic geometry and does not use transformation theory. It applies to ill-posed equations in the Hadamard sense and we will give applications to the so-called Boussinesq equation by constructing periodic solutions for it.
• Slim Ibrahim
University of Victoria
Ground State Solutions of the Gross Pitaevskii Equation Associated to Exciton-Polariton Bose-Einstein Condensates
Résumé en PDF
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We investigate the existence of ground state solutions of a Gross-Pitaevskii equation modeling the dynamics of pumped Bose Einstein condensates (BEC). The main interest in such BEC comes from its important nature as macroscopic quantum system, constituting an excellent alternative to the classical condensates which are hard to realize because of the very low temperature required. Nevertheless, the Gross Pitaevskii equation governing the new condensates presents some mathematical challenges due to the presence of the pumping and damping terms. Following a self-contained approach, we prove the existence of ground state solutions of this equation under suitable assumptions: This is equivalent to say that condensation occurs in these situations. We also solve the Cauchy problem of the Nonlinear Schr\"{o}dinger equation and prove some corresponding laws
• Rosa María Vargas Magaña
Whitham-Boussinesq model for variable depth topography. Results on normal and trapped modes for non trivial geometries.
Résumé en PDF
Taille : 42 kb
The water-wave problem describes the evolution of an incompressible ideal, irrotational fluid with a free surface under the influence of gravity. A significant development in water-wave theory was the discovery by Zakharov in 1968 that the problem has a Hamiltonian structure and later W. Craig and C. Sulem introduced the Dirichlet-Neumann operator explicitly on the Hamiltonian. In this talk I will present a joint work with Prof. Panayotis Panayotaros and Prof. Antonmaria Minzoni from Universidad Nacional Autónoma de México, we propose a simplified long wave model combining a variable depth generalization of the exact nonlocal dispersion with the standard Boussinesq nonlinearity. The model relies on an approximate Dirichlet-Neumann operator that preserves some key structural properties of the exact operator and is simpler than alternative perturbative or implicit expressions. We examine the accuracy of this approximation by studying linear (2-D) normal modes and (3-D) longitudinal and Ursell modes for some geometries for which there are exact results.
• Livia Corsi
Georgia Institute of Technology
A non-separable locally integrable Hamiltonian system
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Taille : 72 kb
A mechanical Hamiltonian on ${\mathbf{T}}^2\times{\mathbf{R}}^2$ is said to be "separable" if in some coordinate system $(q,p)$ it has the form $$H(q,p) =\frac{1}{2} |p|^2+V_1(q_1)+V_2(q_2).$$ Clearly such an Hamiltonian is globally Liouville-integrable. I will show that there exists an analytic, non-separable, mechanical Hamiltonian which is only locally integrable. Precisely I will show that $H$ is integrable on an open subset $\mathcal U$ of the energy surface ${\mathcal S}:=\{H = 1/2\}$, whereas on $\mathcal S \setminus \mathcal U$ it exhibits cahotic behavior. This is a joint work with V. Kaloshin
• Nicholas Faulkner
University of Ontario Institute of Technology
Equivariant KAM
Résumé en PDF
Taille : 47 kb
Kolmogorov-Arnold-Moser (KAM) theory has a rich and well developed history. In this talk we present a KAM theory for $\Gamma$-equivariant Hamiltonian systems. Hamiltonian systems with discrete symmetry groups $\Gamma$ arise naturally in many settings including for instances the $N$-body problem. If $\Gamma$ is Abelian, then KAM theorem applies, but for $\Gamma$ non-Abelian, $1\colon1$ resonance effects lead to small divisor problems. These problems can be overcome by combining the isotypic decomposition of phase space with a detailed study of $\Gamma$ and Torus invariants, all within the classical iterative proof structure. This is joint work with Dr. Luciano Buono. | {"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": 0.8583533763885498, "perplexity": 1035.4978357974671}, "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/1581875146127.10/warc/CC-MAIN-20200225172036-20200225202036-00220.warc.gz"} |
https://astronomy.stackexchange.com/questions/13544/does-every-object-in-the-universe-have-gravity-space-has-no-gravity-why | # Does every object in the Universe have gravity? Space has no gravity, why?
My science teacher told me that every object in Universe has gravity but space is a part of Universe too, so, why does the space have no gravity?
• Are you saying that "space" is an "object"? I don't think that's what your teacher meant. Mar 10, 2018 at 19:28
• When you say "space has no gravity" do you mean that empty space doesn't pull things towards it (which is more or less true) or are you referring to the way astronauts (and other things) float around in space and are not pulled towards the floor of the ISS or whatever. That's because the floor and the astronaut are being pulled equally be Earth's gravity and constantly falling around the Earth. There is a gravitational force between the astronaut and the ISS (or anything else in it) but it is too weak to notice without sensitive instruments. Can you clarify? Mar 10, 2018 at 19:32
A quick search up on google you would find this:
Gravity causes every object to pull every other object toward it. Some people think that there is no gravity in space. In fact, a small amount of gravity can be found everywhere in space. Gravity is what holds the moon in orbit around Earth.
Why is this? That's because:
Outer space is the closest known approximation to a perfect vacuum. It has effectively no friction, allowing stars, planets and moons to move freely along their ideal orbits. However, even the deep vacuum of intergalactic space is not devoid of matter, as it contains a few hydrogen atoms per cubic meter.
If there's matter in space, thus this makes this statement valid.
Yes, everything that has mass will cause a gravitational pull on other objects.
• This isn't really an answer. The OP states that the universe has "objects" and "space" and wants to know if the "space" has gravity. Your answer boils down to "there are objects throughout space, and objects have gravity". Feb 5, 2016 at 15:59
In the intergalactic medium — the most dilute regions of the Universe between the galaxies — as CipherBot writes you'll find roughly one hydrogen per cubic meter, i.e. the density is $\sim10^{-6}\,\mathrm{atoms}\,\mathrm{cm}^{-3}$, or $\sim10^{-30}\,\mathrm{g}\,\mathrm{cm}^{-3}$, or, in terms of energy (since mass and energy are equivalent through $E=mc^2$), $\sim10^{-9}\,\mathrm{erg}^\dagger$. In addition to this, you'll find 5–6 times as much dark matter.
Space itself can't really be considered "an object". Nevertheless, even ignoring the normal and dark matter, space does have energy: the so-called dark energy. We don't really know much about it, but we can measure its presence through its effect on the expansion of the Universe. But whereas normal and dark matter decelerates this expansion, dark energy has the opposite effect of accelerating the expansion. And since the energy density of dark energy is more than twice that of the other two components together, it actually dominates the dynamics of the Universe.
So, in this regard you can say that space does have gravity, although it's a "negative gravity". This phrasing is a bit misinterpreted, though. The dark energy is thought to be a negative pressure. Pressure has a energy density associated with is, so negative pressure has a negative energy density.
$^\dagger$$1\,\mathrm{erg}\equiv10^{-7}\,\mathrm{Joule}$. | {"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": 0.8651577234268188, "perplexity": 453.65413046382326}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662520817.27/warc/CC-MAIN-20220517194243-20220517224243-00302.warc.gz"} |
https://www.physicsforums.com/threads/efficiency-of-an-infinite-series-of-carnot-cycles.76225/ | # Efficiency of an infinite series of Carnot cycles
1. May 19, 2005
### Iraides Belandria
Dear people of Physics Forums, I would like to have your opinion about the following problem: Suppose we are interested to evaluate the average efficiency ( e ) of an infinite series of carnot cycles. The temperature of the hot reservoir of each carnot Cycle of the series, Ta, is at 990K, but the cold reservoir of each cycle is at different temperatures ( Tb ) ranging from 295 K to 752K , in such a way that, the cold reservoir of the first carnot cycle is at 295 K and the cold reservoir of the last is at 752 K. ¿ What is the average efficiency of these series of Carnot cycles?. Now, we know that the efficiency of each Carnot cycle can be determined with the equation e= 1-Tb/Ta. Since we know the temperatures we can calculate the efficiency of each Carnot cycle of the series ranging from 295 to 752 K(e1, e2, e3, e4, e5, e6,e7............. ). ?How can we get the average efficiency from these set of individual values?. I notice that if I plot the efficiency of each Carnot Cycle versus Tb ( the cold resrvoir temperature of each cycle) I get an straight line. Then I applied the mean value theorem to get the average efficiency. I found that the average efficiency is 0.47. ? Is this procedure correct?.
2. May 20, 2005
### Andrew Mason
The average efficiency would be the total work (output) divided by the total heat input:
Qh is the same for each cycle. Qc is proportional to Tc (Qc/Tc = Qh/Th). Since, W = Qh - Qc,
$$\eta = \frac{W}{Q_h} = \frac{\sum_{i=1}^n Q_h - Q_c(i)}{nQ_h}$$
$$\eta = \frac{\sum_{i=1}^n 1 - Q_c(i)/Q_h}{n}$$
$$\eta = \frac{\sum_{i=1}^n 1 - T_c(i)/T_h}{n}$$
$$\eta = \frac{1}{n}\sum_{i=1}^n 1 - T_c(i)/T_h$$
$$\eta = \frac{1}{n}\sum_{i=1}^n \eta(i)$$
So the total efficiency over n cycles is 1/n times the sum of the efficiencies of each cycle.
AM
3. May 20, 2005
### Iraides Belandria
Thanks Andrew for your help, but I still have a question because I have to sum infinite cycles and divide by n. ¿How can I get a numerical answer?. Because of this I use the mean value theorem of calculus. ¿Is that correct?
4. May 20, 2005
### Iraides Belandria
Again, Andrew, ¿Why do you assume that each cycle receive the same amount of heat Qa? ¿ Will the average efficiency change if each cycle receive a different amount of heat, Qa1, Qa2, Qa3,......?. Thanks
5. May 20, 2005
### Andrew Mason
You would have to work out the series and determine the limit as $n \rightarrow \infty$
$$\eta = \frac{1}{n}\sum_{i=1}^n 1 - T_c(i)/T_h = \frac{1}{n}(n - \frac{1}{T_h}\sum_{i=1}^n T_c(i))$$
Now if Tc is linear and the increments of Tc are 1 degree at a time then n = $T_{cend} - T_{cbeg} + 1$ so
$$\sum_{i=1}^n T_c(i) = \frac{(T_{cend} + T_{cbeg})(T_{cend} - T_{cbeg} + 1)}{2}$$
So:
$$\eta = \frac{1}{n}(n - \frac{1}{T_h}\frac{(T_{cend} + T_{cbeg})(T_{cend} - T_{cbeg} + 1)}{2}) = (1 - \frac{(T_{cend} + T_{cbeg})(T_{cend} - T_{cbeg} + 1)}{2nT_h})$$
Substitute for $n = T_{cend} - T_{cbeg} + 1$:
$$\eta = (1 - \frac{(T_{cend} + T_{cbeg})(T_{cend} - T_{cbeg} + 1)}{2T_h(T_{cend} - T_{cbeg} + 1)}) = (1 - \frac{(T_{cend} + T_{cbeg})}{2T_h})$$
As you increase n and decrease the increments, the series should converge to something very very close to this (if it is linear).
AM
Last edited: May 20, 2005
6. May 20, 2005
### Andrew Mason
Why would Qh depend upon Qc?
AM
7. May 20, 2005
### Iraides Belandria
Excuse me, I don understand your answer to the second question.¿Would the average efficiency depends upon the heat received by each carnot cycle of the series or it is independen of it?
8. May 21, 2005
### Andrew Mason
The average efficiency would depend on the total heat received, which is the sum of the heats received in each carnot cycle. I am assuming they are the same but they may not be. I was asking you why they would depend on Qc.
AM
9. May 21, 2005
### Iraides Belandria
I appreciate very much your interest in trying to solve my doubts about this problem. Indeed I am not sure if Qh depend upon Qc. My principal doubt at this moment is if the average efficiency of the series is independent of the amount of heat received by each Carnot Cycle. In your solution you assume that each cycle receives the same amount of heat. But what happen, if the case is more general, and we assume different values for each carnot Cycle?. Would we obtain the same value?. Thanks
10. May 23, 2005
### Iraides Belandria
Other mathematical question is related to find an average for a set of infinite values of efficiencies. As I have stated at the beginning of the thread, we can estimate the numerical eficciency of each Carnot cycle of the series without problem, because I know the temperatures of each Cycle. ?How can we evaluate the average of these values of efficiencies of the infinite carnot cycles?.We also can see that these set of values of efficiency varies linearly with Tc. ¿If this is the case, would be correct to say that the average coincides with the value giving by the limit of above series?.
11. May 23, 2005
### Andrew Mason
As I said before, the average efficiency is the total Work done divided by the total heat input. You can't assume that the average efficiency is the average of all the individual efficiencies just as you cannot determine average speed by averaging speeds (ie. average speed it is total distance divided by total time not an average of speeds over the journey).
AM
12. May 25, 2005
### Iraides Belandria
Dear Mason, If we assume that each carnot cycle receive differents amounts of heat from the hot reservoir, Qh1, Qh2, Qh3,....Qhi, we get the following expression for the average efficiency (Using above procedure)
n = (Qh1/Qh) n1+ (Qh2/Qh) n2 + (Qh3/Qh) n3 + .....(Qhi/Qh) ni
Where Qh is the total heat received fom the hot reservoir equal to Qh=Qh1+Qh2+ Qh3+ ......+Qhi+.... , and ni are efficiencies of each cycle
This equation shows that the average eficciency, n , depends on the amount of heat received by the cycles. In a certain way each Qhi/Qh is a weigth to average the efficiency of the series of cycles. In your deduction you use a uniform weigth to get the average efficiency , Qhi /Qh = N , Where N is the number of carnot cycles. ¿Is this the weigth that we must use in order to compare the performance of any cycle against a series of carnot cycles?. ¿Statiscally , what will be the best weigth to compare an specific Cycle? ? What weight is fair, just ?
13. Jun 2, 2005
### Iraides Belandria
Dear Andrew Mason, I have tried to understand how you obtain the following expresions in the limit of the serie proposed by you
n= Tcend -Tcbeg + 1
and
summatory Tc(i)= [(Tcend + Tc beg) ( Tcend - Tc beg + 1)] / 2
Could you explain these equations with more details?
Thanks
14. Jun 2, 2005
### Andrew Mason
It is just the sum of an arithmetic series with increment of 1.
$$S = (n / 2) * [2i + (n-1)]$$ where n = Tcend - Tcbeg + 1 and i = Tcbeg
eg:
$$5 + 6 + 7 + 8 + 9 + 10$$
n = 6 (10-5+1); i = 5; S = (6/2) * (10 + 5) = 3 * 15 = 45
AM
15. Jun 3, 2005
### Iraides Belandria
Andrew, Thank you, very much. I appreciate your cooperation, and I would like to say that my empirical results coincide with your proposition, as you may verify in the begining of this thread. In any case, the most important consequence of this discussion is that I have found a thermodynamic cycle, operating between the temperatures levels proposed in this post, more efficient than an infinite series of carnot cycles. Because of this reason, I have insisted so much in the solution of this problem, from differents points of view. All of the alternatives, I have explored coincide with your results. thanks to god. Now, I think I am confident on my results. I am going to try to publish an article related to this cycle, and I am going to use this thread as a bibliographycal reference. ¿ Can I do this? . In the future, I will inform you about it. | {"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": 0.951261043548584, "perplexity": 793.8716119216236}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280888.62/warc/CC-MAIN-20170116095120-00491-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://kb.osu.edu/dspace/handle/1811/17649 | A FORBIDDEN BAND IN THE FAR INFRARED SPECTRUM OF $H^{1 3}CNO$
Please use this identifier to cite or link to this item: http://hdl.handle.net/1811/17649
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1988-TB-11.jpg 112.1Kb JPEG image
Title: A FORBIDDEN BAND IN THE FAR INFRARED SPECTRUM OF $H^{1 3}CNO$ Creators: Winnewisser, M.; Wagner, G.; Winnewisser, B. P. Issue Date: 1988 Publisher: Ohio State University Abstract: The far infrared spectrum of HCNO exhibits a widely spread set of perpendicular transitions in the manifold of the quasilinear bending mode, $\nu_{o}$. and the corbiration manifold $\nu_{4} - n\nu$ where $\nu_{4}$ is the skeletal bending mode. We have analyzed the spectrum of the $^{13}C$ substituted species, and found a Q branch without any P or R branches and a P. R. subband without any Q branch. Neither of these features are observed in the spectrum of the parent species. They have been assigned to the vibrational transitions $(v_{i}v_{a})^{\ell} = (02)^{\prime\prime} (11)^{0*}$ and $(02)^{0*}= (11)^{0*}$ respectively. These transitions are only allowed due to a Conclis interaction between the levels $(02)^{0e}$ and $(10)^{1e}$ which was observed as a relatively weak interaction in the parent species. Description: Author Institution: Physikalich-Chemisches Institut, Justut Liebig Universitat URI: http://hdl.handle.net/1811/17649 Other Identifiers: 1988-TB-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": 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": 0.8747866153717041, "perplexity": 2579.2047897992384}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042988650.6/warc/CC-MAIN-20150728002308-00137-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://www.ourxanadu.net/info/1115/18725.htm | Exact overlaps on the projections of four corner Cantor set
The four corner Cantor set C is a planar self-similar set generated by the IFS {(x/4, y/4), ((x+3)/4,y/4), (x/4,(y+3)/4), ((x+3)/4,(y+3)/4)}. In this talk I will give a complete characterization on the set of ts in which the projection C_t:={x+ty: (x,y)\in C} is a self-similar set having an exact overlap; the distribution of these ts in the real line is also considered. This is joint work with Beibei Sun. | {"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": 0.8559809923171997, "perplexity": 1389.7897102680668}, "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-2023-06/segments/1674764500294.64/warc/CC-MAIN-20230205224620-20230206014620-00814.warc.gz"} |
http://papers.neurips.cc/paper/5226-near-optimal-density-estimation-in-near-linear-time-using-variable-width-histograms | # NIPS Proceedingsβ
## Near-Optimal Density Estimation in Near-Linear Time Using Variable-Width Histograms
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### Abstract
Let $p$ be an unknown and arbitrary probability distribution over $[0 ,1)$. We consider the problem of \emph{density estimation}, in which a learning algorithm is given i.i.d. draws from $p$ and must (with high probability) output a hypothesis distribution that is close to $p$. The main contribution of this paper is a highly efficient density estimation algorithm for learning using a variable-width histogram, i.e., a hypothesis distribution with a piecewise constant probability density function. In more detail, for any $k$ and $\eps$, we give an algorithm that makes $\tilde{O}(k/\eps^2)$ draws from $p$, runs in $\tilde{O}(k/\eps^2)$ time, and outputs a hypothesis distribution $h$ that is piecewise constant with $O(k \log^2(1/\eps))$ pieces. With high probability the hypothesis $h$ satisfies $\dtv(p,h) \leq C \cdot \opt_k(p) + \eps$, where $\dtv$ denotes the total variation distance (statistical distance), $C$ is a universal constant, and $\opt_k(p)$ is the smallest total variation distance between $p$ and any $k$-piecewise constant distribution. The sample size and running time of our algorithm are both optimal up to logarithmic factors. The approximation factor'' $C$ that is present in our result is inherent in the problem, as we prove that no algorithm with sample size bounded in terms of $k$ and $\eps$ can achieve $C < 2$ regardless of what kind of hypothesis distribution it uses. | {"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": 0.9851012825965881, "perplexity": 214.40910679741413}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027321160.93/warc/CC-MAIN-20190824152236-20190824174236-00390.warc.gz"} |
https://www.physicsforums.com/threads/period-of-oscillation-of-a-rocking-semi-cylinder.289850/ | # Period of oscillation of a rocking semi cylinder?
1. Feb 4, 2009
### jaron_denson
1. The problem statement, all variables and given/known data
a thin rectangular plate is bent into a semicircular cylinder, dertermine the period of oscillation if it is allowed to rock on a horizantal surface.
3. The attempt at a solution
So I can figure out the rest of the problem if I know what the moment of inertia of a half cylinder is. My side note is a guess, but not sure if it is correct. I checked around online couldn't find anything that helpful
Can you offer guidance or do you also need help?
Draft saved Draft deleted
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https://arxiv.org/abs/hep-ph/0408191 | hep-ph
(what is this?)
# Title: Seesaw induced electroweak scale, the hierarchy problem and sub-eV neutrino masses
Abstract: We describe a model for the scalar sector where all interactions occur either at an ultra-high scale L_U ~ 10^{16}-10^{19} GeV or at an intermediate scale L_I ~ 10^{9}-10^{11} GeV. The interaction of physics on these two scales results in an SU(2) Higgs condensate at the electroweak (EW) scale, L_{EW}, through a seesaw-like Higgs mechanism, L_{EW} ~ L_I^2/L_U, while the breaking of the SM SU(2)XU(1) gauge symmetry occurs at the intermediate scale L_I. The EW scale is, therefore, not fundamental but is naturally generated in terms of ultra-high energy phenomena and so the hierarchy problem is alleviated. We show that the class of such seesaw Higgs'' models predict the existence of sub-eV neutrino masses which are generated through a two-step'' seesaw mechanism in terms of the same two ultra-high scales: m_nu ~ L_I^4/L_U^3 ~ L_{EW}^2/L_U. The neutrinos can be either Dirac or Majorana, depending on the structure of the scalar potential. We also show that our seesaw Higgs model can be naturally embedded in theories with tiny extra dimensions of size R ~ 1/L_U ~ 10^{-16} fm, where the seesaw induced EW scale arises from a violation of a symmetry at a distant brane; in particular, in the scenario presented there are 7 tiny extra dimensions.
Comments: latex, 25 pages, no figures. Version 2 is a shorter version (as accepted in EPJC), not including the discussion on the heavy seesaw Higgs model Subjects: High Energy Physics - Phenomenology (hep-ph) Journal reference: Eur.Phys.J.C45:219-225,2006 DOI: 10.1140/epjc/s2005-02403-x Report number: BNL-HET-04/10 Cite as: arXiv:hep-ph/0408191 (or arXiv:hep-ph/0408191v2 for this version)
## Submission history
From: Shaouly Bar-Shalom [view email]
[v1] Tue, 17 Aug 2004 18:56:24 GMT (25kb)
[v2] Mon, 19 Sep 2005 19:26:06 GMT (15kb) | {"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": 0.9215267896652222, "perplexity": 4048.0819219281702}, "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-2017-43/segments/1508187820487.5/warc/CC-MAIN-20171016233304-20171017013304-00306.warc.gz"} |
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# Mayer, L (2009). Formation of supermassive black hole binaries and massive seed SMBHs in gas-rich mergers. In: Hunting for the Dark: the Hidden Side of Galaxy Formation, Qawra, Malta, 19 October 2009 - 23 October 2009, 181-194.
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## Abstract
We review the results of multi-scale, hydrodynamical simulations of major mergers between galaxies with or without central supermassive black holes (SMBHs) to investigate the orbital decay of SMBH pairs in galactic nuclei and the formation of massive SMBH seeds via direct gas collapse. Both SPH simulations and AMR simulations are carried out. The complex balance between heating and cooling is modeled via an effective EOS with varying adiabatic index γ apparopriate for the conditions of an intense nuclear starburst such as that expected during and after the merger. Prominent gas inflows due to tidal torques produce nuclear disks at the centers of merger remnants whose properties depend sensitively on the details of gas thermodynamics. In parsec-scale resolution simulations starting with two SMBHs originally at the centers of the two galaxies, a SMBH binary forms very rapidly, less than a million years after the merger of the two galaxies, owing to the drag exerted by the surrounding gaseous nuclear disk. Binary formation is significantly suppressed if heating, by e.g. radiative feedback from the accreting SMBHs, renders cooling negligible.
The nuclear disk rearranges its mass distribution in response to a second, internal gas inflow occurring while the binary sinks. The inflow is driven by spiral instabilities imprinted by the final collision between the two galactic cores. In simulations with 0.1 pc resolution, the gas inflow continues all the way down to the center and peaks at >104 Msolar/yr, producing a Jeans-unstable supermassive central cloud (with mass a few times 108 Msolar) only 105 yr after the merger. If the collapse continues, the cloud could form a massive black hole seed (Mseed>105 Msolar) after prior formation of a supermassive star or quasi-star. The massive SMBH seed can grow up to a billion solar masses in less than a billion years by accreting the surrounding nuclear gas. If the gas-rich merger occurs at z>8, this is then a new, attractive way to explain the rapid emergence of the bright QSOs discovered by the Sloan Digital Sky survey at z>6, which does not require the assumption of primordial gas composition in order to suppress cooling below 104 K and star formationas in models starting from unstable, isolated protogalactic disks. If there is a pre-existing pair of SMBHs their orbital decay stalls at parsec scales because, as a result of the formation of the supermassive cloud, the nuclear disk density decreases outside the cloud, yielding much weaker dynamical friction. We envision a new scenario in which direct formation of massive black hole seeds and SMBH binary formation are mutually exclusive; if a SMBH is already present in the nuclear disk it can stabilize it and weaken the secondary inflow via its energetic feedback, maintaining a high enough density where the SMBHs are located and assisting their sinking. | {"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": 0.8468769192695618, "perplexity": 2644.094505712746}, "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/1438042982013.25/warc/CC-MAIN-20150728002302-00021-ip-10-236-191-2.ec2.internal.warc.gz"} |
http://psjd.icm.edu.pl/psjd/element/bwmeta1.element.bwnjournal-article-appv92z410kz | PL EN
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## Acta Physica Polonica A
1997 | 92 | 4 | 695-698
Article title
### Two-Electron Quantum Dots in Magnetic Field
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EN
A theoretical description is given for electronic properties of semiconductor quantum dots in a magnetic field. A two-electron model is applied for electrons in a cylindrical quantum dot with a parabolic confinement potential. The eigenvalue problem is solved by the variational method with the trial wave function proposed in the form of linear combination of S-type and P-type Gaussians. The energy levels of singlet and triplet states with arbitrary radial and magnetic quantum numbers have been calculated as a function of the applied magnetic field. The calculated cyclotron transition energies agree well with those measured for InGaAs/GaAs quantum dots. It is shown that the electron-electron interaction has a small influence on the transition energy.
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695-698
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1997-10
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• Faculty of Physics and Nuclear Techniques, Technical University (AGH), Al. Mickiewicza 30, 30-059 Kraków, Poland
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• Faculty of Physics and Nuclear Techniques, Technical University (AGH), Al. Mickiewicza 30, 30-059 Kraków, Poland
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Identifiers | {"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": 0.9317781329154968, "perplexity": 1602.3071965978834}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038461619.53/warc/CC-MAIN-20210417162353-20210417192353-00078.warc.gz"} |
http://math.stackexchange.com/questions/292859/matrix-algebra-true-or-false | # Matrix Algebra - True or False?
I have 5 T-F statements right here about symmetric matrices, and to the right are my attempts. I have a feeling some of them are wrong, though.
(a) Symmetric matrices must be square. (T)
There can be a rectangular matrix which is symmetrical.
(b) A scalar is symmetric. (T)
I don't know why
(c) If $A$ is symmetric, then $\alpha A$ is symmetric. (T)
What is $\alpha A$?
(d) The sum of symmetric matrices is symmetric.
I think this is false as not all sums will be symmetric.
(e) If $(A')' = A$ , then $A$ is symmetric.
I think this is true.
Could I encourage that, if you know one part of the answer, please go ahead to answer them? I know that it will be very, very hard to find one answer with all 5.
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It should not be that hard...no worries. – 1015 Feb 2 '13 at 16:01
I guess everybody here will be able to answer, but I somewhat feel that answering your question will deprive you of the chance to learn for youself. Why don't you tell us which definition you are using and what you have to check exactly for every single question. I bet you will find some answers yourself. Also: make sure you understand how you can consider a scalar as a matrix. What is $A'$ ? do you mean the transpose? – Simon Markett Feb 2 '13 at 16:06
I'll take $A=A^t$ (where $A^t$ denotes the transposed matrix) for the definition of $A$ to be symmetric.
(a) True. If $A$ is an $m\times n$ matrix and is equal to its transpose which is an $n\times m$ matrix, then $m=n$.
(b) True. A scalar is a $1\times 1$ matrix, so equal to its transpose. More generally, what is the transpose of a scalar matrix? Itself.
(c) True. Note that $\alpha A$ is the matrix obtained from $A$ by multiplying each coefficient by $\alpha$.
(d) True (symmetric matrices of same size). Essentially because $(A+B)^t=A^t+B^t$.
(e) False. It $A'$ denotes $A^t$ the transpose of $A$.... Note that $(A^t)^t=A$ for every matrix $A$. And not every matrix is symmetric.
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I Finally reviewed my notes and understand what you're talking about...thank you so much. – bryansis2010 Feb 13 '13 at 15:29
@bryansis2010 Great! You're welcome. – 1015 Feb 13 '13 at 15:31
Hints:
(a) This is true apparently directly from the very definition of symmetric matrix, but also because one the main characteristics of a symmetric matrix $\,A\,$ is that $\,A^t=A\,$ , and this equality forces $\,A\,$ .
(b) You seem to have meant "a scalar matrix is symmetric", and a scalar matrix is of the form
$$\alpha I=\begin{pmatrix}\alpha&0&0&...&0\\0&\alpha&0&...&0\\...&...&...&...&...\\0&0&0&...&\alpha\end{pmatrix}$$
So what say you? Is that symmetric or not?
(c) $\,A=(a_{ij})\,$ is symmetric iff $\,a_{ij}=a_{ji}\,\,\forall\,i\neq j\,$ , and $\,\alpha A=(\alpha a_{ij})\,$ ...
(d) This now follows at once from the above, and about (e) I'm not sure what you mean by $\,A'\,$...
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+1. Don. I have learned much from your answers here. – S. Snape Feb 4 '13 at 9:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.8871986269950867, "perplexity": 384.35053601193937}, "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/1454701157262.85/warc/CC-MAIN-20160205193917-00150-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/arrow-of-time-entropy-probability.119086/ | Arrow of time/ entropy probability
1. Apr 28, 2006
Tsunami
In Popper's autobiography, published in his contribution of Library of Philosophers, I read the following, after Popper's presentation of Boltzmann's ideas about entropy:
A bit further this is formulated as such:
Is this all accepted? Any additional comments? (The arrow of time always sounded valid to me.)
2. Apr 28, 2006
Tsunami
edit: via search functions I did find the two pages of that paper.
It seems to me though as if Prigogine seems to discard Popper's point. What Popper tried to show, was that states of more entropy are more likely, regardless of the direction of time. If you imagine a curve of entropy states in function of time, it is easy to imagine a state of low entropy being surrounded by states of higher entropy... thus, you get a sort of gaussian curve with the most optimal result in the center. (It doesn't have to be gaussian, just an high medium which is lower in both directions starting from this point.)
Popper's idea is, I suppose, that if we could track what happens to this situation of entropy states when it moves from low entropy to high entropy, then this situation could be reversed by backtracking everything that happens.
One should note that Popper also refused the idea of having information as a counter-force against entropy. (If people want to use the information argument to tackle Popper's point, I'll look up his ideas regarding this so this can be tackled as well if necessary)
Last edited: Apr 28, 2006
3. Apr 28, 2006
Andrew Mason
I have no idea what the scientific basis for this could be. I am not aware of a principle of physics that:
Not only that, I think it is an incorrect statement. A gas that expands from a smaller to larger space will not compress by itself back to the smaller space. Nor will the forces of nature suddenly combine (eventually) to compress it.
Now, this does not necessarily mean that the conclusion (that the arrow of time is not the result of ever-increasing entropy) is incorrect. Popper was certainly at the top of his field in philosophy, but he was not a scientist. Philosophy should not be confused with science.
It seems to me that the arrow of time is partly about probability (of the universe self-organizing itself to a precise state that existed). Entropy is partly about probability. To say there is no connection seems to be more than a little presumptuous.
I would say that the arrow of time is simply the result of logic. Time can only go in one direction by definition. If the universe is in a certain state, changes and then returns to the precise previous state, the initial state is still in the past.
AM
4. Apr 28, 2006
Physics Monkey
The theorem being referred to is a famous one by Poincare called the Poincare Recurrence Theorem which roughly states that an isolated and bounded mechanical system will eventually return arbitrarily close to any initial state. This theorem does quite explicitly state that if you had a perfectly isolated classical gas in which the gas started in one side of the container, you would find at a later time that the gas was again all in one side of the container. There is no conflict with the laws of thermodynamics since such a recurrence would take far too long to occur even if one had happened to have on hand a perfectly isolated classical gas.
5. Apr 28, 2006
Physics Monkey
And to address Tsunami's original question
Yes, this stuff, by which I mean the reconciliation of the reversible laws of mechanics with the apparent irreversability of thermodynamics, is all pretty much accepted. One particular interesting and transparent test of these ideas is to look at "gases" of cellular automata that have reversible dynamics. One can watch these purely reversible systems start in one side of the container, expand, and eventually fill the whole container. With a sufficiently large number of the little guys running around, you never see them all return to one side of the container. In short, the laws of thermodynamics appear to hold good. However, if you go and perfectly reverse all the velocities, sure enough, the gas goes dutifully right back into one side of the container.
6. Apr 29, 2006
SizarieldoR
Hay, physics isn't the only science (although it is the basic natural science), if memory serves, philosophy can be called a science as well.
Does entropy decrease with the action of forces? I mean, does the entropy in a given amount of lava decrease, when the lava crystallizes?
Also, thermal energy is the vibration of atoms/molecules hitting each other (correct me if i'm wrong). So, if we can calculate how two billiard balls hit each other (and the change in their momentums), we should be able to calculate how two atoms/molecules hit each other as well in the limits of the uncertainty principle. Of course, we will need some freakin-powerful supercomputers and stuff, not to mention taking the London dispersion force into account, but doesn't that mean that entropy isn't an increase in disorder?
7. Apr 29, 2006
vanesch
Staff Emeritus
There is some controversy over this, but the most "down to earth" view is this: the arrow of time (= the increase in entropy) comes about from the special initial condition (low-entropy state of the early universe).
In fact, the initial state of the universe (just after the big bang) is a *highly peculiar state* of very low entropy, and we're still evolving towards equilibrium in this view.
The asymmetry is not in the time symmetric evolution laws (which, to our knowledge, are time reversible), but in the initial condition.
This displaces the question of course to why there was this special initial condition :-)
Now, to come back to Poincare recurrence, this comes in fact down to saying that in a closed classical system, motion is "essentially" periodic ; however, this period is extremely long! If you take any initial state, you will have the motion essentially divided in 3 parts:
a part where there is "increase in entropy" ; a part where there is "decrease in entropy" and a VERY LONG PART where there is equilibrium ; this then cycles on and on, and depending on where you are on this trajectory, you observe increase, decrease, or equilibrium.
In the first case, the "special state" is closer in the past than in the future. In the last case, the "special state" is closer in the future than in the past (just before a "big crunch" ?). And in between there is an extremely long part of equilibrium.
Now, it seems that we are now in a phase of "increase of entropy", because the big bang (special state) is closer to us in the past than in the future.
Now, you could even say that even if it were different, and the special state were closer to us in the "future" than the past, we would probably EXPERIENCE time in the other sense, and call this future the past, and vice versa. But that's less of a standard view. But fun to think about.
However, one thing is sure: entropy changes can only occur in "the neighbourhood" of a special (initial or final) state. If not, you are in equilibrium, which is by far the state that prevails during the longest time. At least in a purely classical system.
8. Apr 29, 2006
Tsunami
Ok, that makes sense.
I still want to solve one question though (just trying to get my fundamentals straight ):
If, in principle, we can move to a state of lower entropy by reversing all velocities, why don't we?
I can see how this can be tackled by saying that we lack sufficient information to know the exact reversal of motion, but that doesn't satisfy me. Surely, the second law must be stronger than that?
9. Apr 29, 2006
Andrew Mason
Entropy of any system can increase or decrease in a process. But the entropy of the universe will always increase in the process.
Entropy is not really about disorder.
The second law of thermodynamics simply says that the energy of the universe flows in a way that makes it more disperse.
Entropy is really about energy dispersion and probability. Consider the break in game of pool (assume a frictionless surface, perfectly elastic collisions and rebounds from the cushions). The relatively high energy of the cue ball is dispersed into all 16 balls, with each ball having a portion of that original energy. The probability that the balls will all collide with the cue ball in such a way that they all stop and the cue ball regains its original energy is very low. It would not violate the laws of physics. It would just be extremely improbable.
AM
10. Apr 29, 2006
Andrew Mason
"In science there is only physics; all the rest is stamp collecting."
Ernest Rutherford (before he won the Nobel Prize for Chemistry)
AM
11. Apr 30, 2006
Andrew Mason
Poincaré Recurrence Theorem - Zeno's Paradox in reverse
The Poincaré Recurrence theorem states that any possible event, no matter how improbable, will occur in a time interval $\Delta t$ if we make $\Delta t$ arbitrarily large.
Zeno's paradox says that two objects separated by a distance s and approaching at a fixed speed v will never hit each other in an infinite number of time intervals $\Delta t_i$ if we make the time intervals arbitrarily small (so that $\sum \Delta t_i$ < s/v).
It seems to me that both are correct mathematically. But both have nothing to do with the real world.
Similarly Quantum Mechanics has nothing to do with the lunar orbit. There is a finite probability that the moon will stop obeying Newton's laws of motion and pass through the earth and out the otherside. This would violate Newton's laws of motion but would not violate QM. It is just that the probability of this happening is very small - so small that it will not likely happen in 10^10^10^10^10^10^10^10^10 lifetimes of the universe. This does not invalidate Newton's laws of motion.
So to the suggestion that the Poincaré Recurrence theorem shows that entropy of the universe can and will decrease spontaneously, I say: 'not in this universe'.
AM
Last edited: Apr 30, 2006
12. May 3, 2006
Tsunami
However, you can make an adjusted Zeno's paradox that suits the real world: you can place a finite lower boundary on the amount of divisions you make, something like "the turtle cannot walk half a distance if this half is smaller than one of his feet"
Anyway, doing the same for Poincare probably means saying : the finite upper boundary we can think of is the duration of conscious life in the universe, so it doesn't matter.
So considering practically everything can be explained using an infinite duration of time and any improbable probability : I get your point.
13. May 15, 2006
vanesch
Staff Emeritus
As we don't know the laws of this universe (but only some aspects of it), we can't tell but you're probably right. The point was not this, the point was that even in a classical MODEL where Poincare recurrence holds, and where the microdynamics is time-reversible, a second law can hold during a time lapse which is "shortly" after a special initial condition.
And now when you look at where, in this world, we DO get our second law from, then this clearly points out to a similar reason: we get our entropy increase potential essentially from the influx of low-entropy radiation from the sun, and the outflux of high-entropy radiation into the blackness of space. Both are related to the state of the early universe and its expansion. If the universe were in thermodynamic equilibrium, then we wouldn't observe a second law (we wouldn't be there either).
Now whether there is some cyclic equivalent of the Poincare recurrence theorem or not for the "real" laws of this universe I don't know of course, but that was not the point. The point was that the apparent paradox between reversible microlaws and the second law can be solved by considering that you are evolving away from a special initial state. It might turn out that microlaws are, in the end, not time-reversible, who knows. But this is not necessary to explain the second law.
14. May 16, 2006
Andrew Mason
I don't follow you there. What do you mean by "where the microdynamics is time-reversible"?
But the universe would only move into thermodynamic equilibrium because of the second law: the state of equilibrium being the most probable state.
As a great politician once observed, there are many ways to change but only one way to remain the same. That, in a nutshell, sums up the second law.
AM
Last edited: May 16, 2006
15. May 16, 2006
vanesch
Staff Emeritus
That the dynamical laws are time-symmetric (that there is a symmetry when replacing t -> -t in the dynamical laws of classical mechanics).
Well, the point was, that this "evolution towards a more probable state" (= second law) comes about only, because we come from a state which is NOT very probable (probable in the sense of Boltzmann entropy, where the available phase space is chunked up in boxes of "equivalent macrostates", where we don't care for high-order correlations), and that there is no specific conservation law which keeps us in the small, =unprobable, box. So the time evolution of the classical state will wander from small boxes into big boxes, most of the time, because there is no specific relationship between the delimitation of the boxes, and the time evolution of the microstate. Given this a priori independence, a stochastical argument can do. It can happen that it wanders from a big box into a small box, but that is a highly peculiar happening (which happens nevertheless, due to Poincare recurrence, at the end of a "cycle"), because the size of the boxes, and the time evolution, are not really related. All this in a classical MODEL universe.
The point is simply that the timescales on which this could happen (Poincare recurrence time) is vastly longer than the time since a fixed initial condition, so the "slope of increase of box size" is still very steep, hence a clearly defined second law.
That's a nice quote :-).
However, "staying the same" is microscopically as involved as "changing". It is only that the low-order correlation functions (like matter densities and so on) do not change anymore in what we call 'equilibrium'. The high-order correlation functions change as much during "equilibrium" as during "irreversible evolution" in a classical universe.
16. May 17, 2006
Farsight
Doublepost deleted.
Last edited: May 17, 2006
17. May 17, 2006
Farsight
Time is to do with motion over a distance. The direction doesn't matter, so changing the direction won't reverse the arrow of time. What you need for that, is a negative distance. There's one in here, thankyou google. But apologies if it's bunk, I haven't read it through.
http://www.comcity.com/distance-time/Photon%20Kinematics.html [Broken]
Last edited by a moderator: May 2, 2017
18. May 17, 2006
lalbatros
Tsunami,
Already in 70's some people made computer simulations showing that entropy can decrease (close to its original value) after a long time. This is just unavoidable when the microscopic laws are reversible. Such simulations of course need to consider a small number of particles. The "recurrence time" that can simulated in this way increases very fast with the number of particles. This recurrence time also increases -of course- with the precision of the recurrence, a perfect recurrence requires generically an infinite time.
Some models can make things easier to visualise, like particles in a box with periodicity conditions: then there may even be no need to calculate the trajectories since they are analytically simple.
More obvious simulations were dealing with velocity-reversal. In such simulations, the entropy evolves exactly in the opposite directions than in the forward simulation (it decreases) until the initial condition is recovered, and thereafter the entropy increases again as usual.
Many people consider that these simulations don't represent physical reality, and that therefore something is missing in the physics to account for irreversibility. The interpretation is simple however. Preparing a system that could exhibit anti-thermodynamics behaviour is theoretically possible from the microscopic law of physics. But this preparation is extremely difficult and such systems are "fragile": they loose their exceptional properties by any small perturbation.
That's an interresting and fundamental topic. But I have not seen that it delivered any real progress in physics, till now.
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https://www.cut-the-knot.org/Curriculum/Geometry/GeoGebra/FlankTriangles4.shtml | # Two Properties of Flank Triangles - A Proof with Complex Numbers
Bottema's configuration of two squares that share a vertex is naturally embedded into Vecten's configuration of three squares erected on the sides of a triangle. The latter generalizes the first of Euclid's proofs of the Pythagorean theorem, so it rightfully refers to the Bride's Chair.
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### References
1. R. Honsberger, Mathematical Diamonds, MAA, 2003, 63-64
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# In the xy-plane, if line k has negative slope and passes
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In the xy-plane, if line k has negative slope and passes [#permalink]
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29 Feb 2012, 13:18
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Whenever I see a similar question, my mind freezes and I go into fetal position. I am unable to picture them and even if sketch them, I am still confused as to how approach them.
I know how to get a slope.
I know the basic equation of a line.
I know how a prep bisector works.
I know the distance between two points.
I know how to get the mid between two points.
I am just unable to lump all these together and solve these questions quick enough.
Here are some examples of these questions:
In the xy-plane, if line k has negative slope and passes through the point (−5,r ), is the x-intercept of line k positive?
(1) The slope of line k is –5.
(2) r > 0
Official answer for this one is :
[Reveal] Spoiler:
E
In the rectangular coordinate system, are the points
(r,s) and (u,v ) equidistant from the origin?
(1) r + s = 1
(2) u = 1 – r and v = 1 – s
OA:
[Reveal] Spoiler:
C
If line k in the xy-plane has equation y = mx + b, where
m and b are constants, what is the slope of k ?
(1) k is parallel to the line with equation
y = (1 – m)x + b + 1.
(2) k intersects the line with equation y = 2x + 3 at
the point (2,7).
OA :
[Reveal] Spoiler:
A
In the XY plane, region R consists of all the points (x,y) such that 2x+3y<=6. Is the point (r,s) in region R?
1. 3r+2s=6
2. r<=3 & s<=2
OA :
[Reveal] Spoiler:
E
These questions seem to pop up on every GMAT prep I took.
Any helps or tricks are appreciated. If I can't thank you in a post, I will make sure I kudos you.
Last edited by Bunuel on 29 Feb 2012, 13:59, edited 1 time in total.
Topic is locked. The links to the open discussions of these questions are given in the posts below.
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Re: These questions really scare me - coordinate Geom. [#permalink]
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29 Feb 2012, 13:34
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1. In the xy-plane, if line k has negative slope and passes through the point (-5,r), is the x-intercept of line k positive?
This question can be done with graphic approach (just by drawing the lines) or with algebraic approach.
Algebraic approach:
Equation of a line in point intercept form is $$y=mx+b$$, where: $$m$$ is the slope of the line, $$b$$ is the y-intercept of the line (the value of $$y$$ for $$x=0$$), and $$x$$ is the independent variable of the function $$y$$.
We are told that slope of line $$k$$ is negative ($$m<0$$) and it passes through the point (-5,r): $$y=mx+b$$ --> $$r=-5m+b$$.
Question: is x-intercept of line $$k$$ positive? x-intercep is the value of $$x$$ for $$y=0$$ --> $$0=mx+b$$ --> is $$x=-\frac{b}{m}>0$$? As we know that $$m<0$$, then the question basically becomes: is $$b>0$$?.
(1) The slope of line $$k$$ is -5 --> $$m=-5<0$$. We've already known that slope was negative and there is no info about $$b$$, hence this statement is insufficient.
(2) $$r>0$$ --> $$r=-5m+b>0$$ --> $$b>5m=some \ negative \ number$$, as $$m<0$$ we have that $$b$$ is more than some negative number ($$5m$$), hence insufficient, to say whether $$b>0$$.
(1)+(2) From (1) $$m=-5$$ and from (2) $$r=-5m+b>0$$ --> $$r=-5m+b=25+b>0$$ --> $$b>-25$$. Not sufficient to say whether $$b>0$$.
Graphic approach:
If the slope of a line is negative, the line WILL intersect quadrants II and IV. X and Y intersects of the line with negative slope have the same sign. Therefore if X and Y intersects are positive, the line intersects quadrant I; if negative, quadrant III.
When we take both statement together all we know is that slope is negative and that it crosses some point in II quadrant (-5, r>0) (this info is redundant as we know that if the slope of the line is negative, the line WILL intersect quadrants II). Basically we just know that the slope is negative - that's all. We can not say whether x-intercept is positive or negative from this info.
Below are two graphs with positive and negative x-intercepts. Statements that the slope=-5 and that the line crosses (-5, r>0) are satisfied.
$$y=-5x+5$$:
Attachment:
1.png [ 9.73 KiB | Viewed 9878 times ]
$$y=-5x-20$$:
Attachment:
2.png [ 10.17 KiB | Viewed 9870 times ]
More on this please check Coordinate Geometry chapter of Math Book: math-coordinate-geometry-87652.html
In case of any question please post it here: in-the-xy-plane-if-line-k-has-negative-slope-and-passes-110044.html
Hope it helps.
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Re: These questions really scare me - coordinate Geom. [#permalink]
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29 Feb 2012, 13:35
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2. In the rectangular coordinate system, are the points (r,s) and (u,v) equidistant from the origin?
(1) r + s = 1
(2) u = 1 - r and v = 1 - s
Distance between the point A (x,y) and the origin can be found by the formula: $$D=\sqrt{x^2+y^2}$$.
Basically the question asks is $$\sqrt{r^2+s^2}=\sqrt{u^2+v^2}$$ OR is $$r^2+s^2=u^2+v^2$$?
(1) $$r+s=1$$, no info about $$u$$ and $$v$$;
(2) $$u=1-r$$ and $$v=1-s$$ --> substitute $$u$$ and $$v$$ and express RHS using $$r$$ and $$s$$ to see what we get: $$RHS=u^2+v^2=(1-r)^2+(1-s)^2=2-2(r+s)+ r^2+s^2$$. So we have that $$RHS=u^2+v^2=2-2(r+s)+ r^2+s^2$$ and thus the question becomes: is $$r^2+s^2=2-2(r+s)+ r^2+s^2$$? --> is $$r+s=1$$? We don't know that, so this statement is not sufficient.
(1)+(2) From (2) question became: is $$r+s=1$$? And (1) says that this is true. Thus taken together statements are sufficient to answer the question.
In case of any question please post it here: in-the-rectangular-coordinate-system-are-the-points-r-s-92823.html
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Re: These questions really scare me - coordinate Geom. [#permalink]
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29 Feb 2012, 13:37
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If Line k in the xy-plane has equation y = mx + b, where m and b are constants, what is the slope of k?
$$y=mx+b$$ is called point-intercept form of equation of a line. Where: $$m$$ is the slope of the line; $$b$$ is the y-intercept of the line; $$x$$ is the independent variable of the function $$y$$.
So we are asked to find the value of $$m$$.
(1) k is parallel to the line with equation y = (1-m)x + b +1 --> parallel lines have the same slope --> slope of this line is $$1-m$$, so $$1-m=m$$ --> $$m=\frac{1}{2}$$. Sufficient.
(2) k intersects the line with equation y = 2x + 3 at the point (2, 7) --> so line k contains the point (2,7) --> $$7=2m+b$$ --> can not solve for $$m$$. Not sufficient.
In case of any question please post it here: if-line-k-in-the-xy-plane-has-equation-y-mx-b-where-m-100295.html
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Re: These questions really scare me - coordinate Geom. [#permalink]
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29 Feb 2012, 13:41
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In the xy-plane, region R consists of all the points (x, y) such that $$2x + 3y =< 6$$ . Is the point (r,s) in region R?
I'd say the best way for this question would be to try boundary values.
Q: is $$2r+3s\leq{6}$$?
(1) $$3r + 2s = 6$$ --> very easy to see that this statement is not sufficient:
If $$r=2$$ and $$s=0$$ then $$2r+3s=4<{6}$$, so the answer is YES;
If $$r=0$$ and $$s=3$$ then $$2r+3s=9>6$$, so the answer is NO.
Not sufficient.
(2) $$r\leq{3}$$ and $$s\leq{2}$$ --> also very easy to see that this statement is not sufficient:
If $$r=0$$ and $$s=0$$ then $$2r+3s=0<{6}$$, so the answer is YES;
If $$r=3$$ and $$s=2$$ then $$2r+3s=12>6$$, so the answer is NO.
Not sufficient.
(1)+(2) We already have an example for YES answer in (1) which valid for combined statements:
If $$r=2<3$$ and $$s=0<2$$ then $$2r+3s=4<{6}$$, so the answer is YES;
To get NO answer try max possible value of $$s$$, which is $$s=2$$, then from (1) $$r=\frac{2}{3}<3$$ --> $$2r+3s=\frac{4}{3}+6>6$$, so the answer is NO.
Not sufficient.
Number picking strategy for this question is explained here: in-the-xy-plane-region-r-consists-of-all-the-points-x-y-102233.html#p795613
In case of any question pleas post it here: in-the-xy-plane-region-r-consists-of-all-the-points-x-y-102233.html
_________________
Re: These questions really scare me - coordinate Geom. [#permalink] 29 Feb 2012, 13:41
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Display posts from previous: Sort by | {"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": 0.8670945167541504, "perplexity": 1243.334495441037}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257836399.81/warc/CC-MAIN-20160723071036-00284-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://math.ncku.edu.tw/research/talk_detail.php?id=1497 | NCTS(South)/ NCKU Math Colloquium DATE 2014-10-30¡@16:10-17:00 PLACE R204, 2F, NCTS, NCKU SPEAKER ¤ý®¶¨k ±Ð±Â¡]¥x¤j¼Æ¾Ç¨t¡^ TITLE Quantitative Uniqueness Estimates, Landis' Conjecture, and Related Questions ABSTRACT In the late 60's, E.M. Landis conjectured that if $\Delta u+Vu=0$ in $\R^n$ with $\|V\|_{L^{\infty}(\R^n)}\le 1$ and $\|u\|_{L^{\infty}(\R^n)}\le C_0$ satisfying $|u(x)|\le C\exp(-C|x|^{1+})$, then $u\equiv 0$. Landis' conjecture was disproved by Meshkov who constructed such $V$ and nontrivial $u$ satisfying $|u(x)|\le C\exp(-C|x|^{\frac 43})$. He also showed that if $|u(x)|\le C\exp(-C|x|^{\frac 43+})$, then $u\equiv 0$. A quantitative form of Meshkov's result was derived by Bourgain and Kenig in their resolution of Anderson localization for the Bernoulli model in higher dimensions. It should be noted that both $V$ and $u$ constructed by Meshkov are \emph{complex-valued} functions. It remains an open question whether Landis' conjecture is true for real-valued $V$ and $u$. In view of Bourgain and Kenig's scaling argument, Landis' conjecture is closely related to the estimate of the maximal vanishing order of $u$ in a bounded domain. In this talk, I would like to discuss my recent joint work with Kenig and Silvestre on Landis' conjecture in two dimensions. | {"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": 0.9205628633499146, "perplexity": 616.804044089492}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141189141.23/warc/CC-MAIN-20201127044624-20201127074624-00408.warc.gz"} |
https://zbmath.org/?q=an%3A1073.34072 | ×
# zbMATH — the first resource for mathematics
Higher order abstract Cauchy problems: their existence and uniqueness families. (English) Zbl 1073.34072
The authors deal with the abstract Cauchy problem for higher-order linear differential equations $u^{(n)}(t)+\sum^{n-1}_{k=0}A_ku^{(k)}(t)=0,\;t\geq 0,\quad u^{(k)}(0)=u_k, \;0\leq k\leq n-1,\tag{1}$ and its inhomogeneous version, where $$A_0,\dots,A_{n-1}$$ are linear operators in a Banach space $$X$$. The authors introduce a new operator family of bounded linear operators from a Banach space $$Y$$ into $$X$$, called an existence family for (1), so that the existence and continuous dependence on initial data can be studied and some basic results in a quite general setting can be obtained. Necessary and sufficient conditions, ensuring (1) to possess an exponentially bounded existence family, are presented in terms of Laplace transforms. As applications, two concrete initial value problems for partial differential equations are studied.
##### MSC:
34G10 Linear differential equations in abstract spaces 47D06 One-parameter semigroups and linear evolution equations 35K90 Abstract parabolic equations
Full Text: | {"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": 0.9151487946510315, "perplexity": 433.26966114240554}, "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/1618038468066.58/warc/CC-MAIN-20210418043500-20210418073500-00267.warc.gz"} |
https://www.physicsforums.com/threads/higgs-and-fermion-masses.184355/ | # Higgs and fermion masses
1. Sep 13, 2007
### arivero
Let me see if I get it right or I dreamed it: in order to give mass to a quark or a lepton the higgs field must be in the same isospin representation that the fermion, must it? IE, can a particle in a isospin triplet get mass from the minimal higgs? Or in the reverse, should a triplet higgs contribute to the mass of a standard model quark?
2. Sep 13, 2007
### BenTheMan
arivero---
The only requirement is that you make a singlet under the gauge group out of the higgs and two fermions---this is the requirement that the Lagrangian be gauge invariant. I may be wrong, but isospin isn't what you should be thinking of. You should be thinking of standard model quantum numbers, i.e. SU(3)xSU(2)xU(1).
I wrote a rather long post on mass terms and the higgs that may answer some of your questions on another forum. I'll link to it here, but I don't know if linking to another forum is exactly kosher :)
3. Sep 13, 2007
### arivero
Ah yes, I call isospin to the quantum numbers of the electroweak SU(2). Some old books name it "weak isospin" and I got hang of the name. Point is, given that the left fermion is a SU(2) doublet, are we forced to put the higgs field also in a SU(2) doublet or are there other solutions?
Similar Discussions: Higgs and fermion masses | {"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": 0.9450588822364807, "perplexity": 710.855446878211}, "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-2017-30/segments/1500549428257.45/warc/CC-MAIN-20170727122407-20170727142407-00253.warc.gz"} |
https://math.stackexchange.com/questions/1642872/how-to-find-the-index-of-following-subgroup | # how to find the index of following subgroup?
if I denotes the principal congurence group of level 2 i.e. $I=\{ M \in SL(2,Z) ; \:M \:\:\text{congruent to I} \mod(2)\}$.
or I= \begin{bmatrix}2\mathbb{Z}+1&2\mathbb{Z}\\2\mathbb{Z}&2\mathbb{Z}+1\end{bmatrix}
H be its subgroup \begin{bmatrix}2\mathbb{Z}+1&4\mathbb{Z}\\2\mathbb{Z}&2\mathbb{Z}+1\end{bmatrix}
how to prove above subgroup is of index 2 in I.
it seems that $4\mathbb{Z}$ and $4\mathbb{Z}+2$ will be two members of the cosets where entries written by me are a12 position of the matrices?
am i right?
$X\sim Y$ iff $XH=YH$ iff $XY^{-1}\in H$. If $X=\begin{pmatrix}p&q\\r&s\end{pmatrix},Y=\begin{pmatrix}u&v\\w&x\end{pmatrix}\in I$, then $X\sim Y$ iff $-pv+qu=0 \mod 4$.
Let $H_1$ be the equivalence class of $X=I_2$: the condition on $Y$ is $v=0\mod 4$. Let $H_2$ be the equivalence class of $X=\begin{pmatrix}1&2\\0&1\end{pmatrix}$: the condition on $Y$ is $-v+2u=0\mod 4$. Since $2u=2\mod 4$, the condition reduces to $v=2\mod 4$. Since $v$ is even, $H_1,H_2$ is a partition of $I$ and we are done.
take an element of I not in H.
a=\begin{bmatrix}1&2\\0&1\end{bmatrix}
writing I=H ∪ Ha
H=\begin{bmatrix}2\mathbb{Z}+1&4\mathbb{Z}\\2\mathbb{Z}&2\mathbb{Z}+1\end{bmatrix}
Ha=\begin{bmatrix}2\mathbb{Z}+1&2+2\mathbb{Z}\\2\mathbb{Z}&2\mathbb{Z}+1\end{bmatrix} as for any x in I
x=\begin{bmatrix}2a+1&2b\\2c&2d+1\end{bmatrix}
we have only 2 possibilities for b
if b is even then x is element of H and if it is odd then x is element of Ha proving I as union of two cosets
proving further they are disjoint is easy as
so clearly intersection is empty. | {"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": 0.9789243936538696, "perplexity": 158.60201033929278}, "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-2020-05/segments/1579251783000.84/warc/CC-MAIN-20200128184745-20200128214745-00163.warc.gz"} |
http://math.stackexchange.com/questions/599875/how-can-i-compare-three-different-variable | # How can I compare three different variable?
I have to verify the presence or the the absence of a phenomenon in 3 different cases (A,B,C) and then representing the results in a single bar graph in order to compare the the results of the three cases together. The phenomenon to detect is constituted by four indicators ($I_1$, $I_2$, $I_3$, $I_4$) and in order to verify if and in which levels this phenomenon is present in each case, I have to calculate if and how much this indicators are present in every single case. In order to fulfill my aims (the verification of the presence of the phenomenon and the comparison) I thought to proceed as following:
1. calculating (separately for each indicator) in which percentage the $I_1, I_2, I_3, I_4$ are present in the cases A, B, C. For example, in the case A: $I_1=70%$, $I_2=20%$, $I_3=10%$, $I_4=20%$.
2. considering that the indicators have not the same weight in the determination of the phenomenon, I have used the operation of "pondering" by giving to each indicator the correct weight and then multiply the "weight" for all indicators' values of each case. For example, $I_1= 3$; $I_2=1$; $I_3=1$; $I_4=5$. So, in the case A: $I_1=70*3$; $I_2=20*1.....$
3. at this point I though that it seemed right to do the operation of "normalization" in order to make possible the comparison of all the resulting of the three cases in a single bar graph. The operation that I have do is the following: *$I_1(A)-X_m$(possible minimum value of $I_1A$)$/XM$(possible maximum value of $I_1$A)$-X_m$ . In this case $X_1=$ each value that each indicator takes in the single cases.
4. after having gathered all the resulting data normalized, I have have summed them for each case ($I_1A+I_2A+I_3A+I_4A$) and then convert them in a bar graph where the y-axis represent the levels in which the phenomenon is present in the case, and the x-axis represent every case.
I would to know if, by doing the normalization, I can make comparable all the three cases in a graph which act like an "index" of the presence of the phenomenon or if I made something wrong.
Blockquote
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http://tug.org/pipermail/texhax/2007-May/008425.html | # [texhax] Problem with column width
Samuel Lelievre samuel.lelievre.tex at free.fr
Mon May 21 22:13:20 CEST 2007
Kenneth Cabrera wrote:
> Hi TeX users:
>
> Why do I obtain a table with the last column wider than the rest?
> Thank you for your help.
>
> I am attaching the file code.
>
> Thank you for your help.
>
> --
> Kenneth Roy Cabrera Torres
> Cel 315 504 9339
Hi Kenneth,
but since there is a multicolumn on the last 4 columns, the last
of them is expanded to match the length needed for the entry in
multicolumn.
If you make the multicolumn entry shorter (eg by splitting it in
two lines), this problem will disappear. (cf. code below).
I don't know a way to have the extra space spread evenly among
the last four columns (but I'd be interested in knowing!).
Apart from that, you might be interested in the package booktabs
for professional-looking rules (horizontal lines) in tables.
The package longtable is also useful when your tables need to be
allowed to spread over a page break (I use it by default for any
table that stands on its own, i.e. not included in a paragraph).
Best,
Samuel
----- one way to avoid extra spacing in last column -----
\documentclass{article}
\usepackage[spanish]{babel}
\usepackage[latin1]{inputenc}
\usepackage{graphicx}
\usepackage{amsmath,amssymb,latexsym,amsthm}
\usepackage{fancyvrb}
\usepackage[margin=2cm]{geometry}
\setlength{\parindent}{0pt}
\setlength{\parskip}{1.5ex plus 0.5ex minus 0.2ex}
\begin{document}
\begin{table}[ht]
\begin{center}
\caption{Tabla X}\label{tb}
\vspace{2ex}
\begin{tabular}{ccccccc}
\hline
\multicolumn{3}{c}{} &
\multicolumn{4}{c}{N\'umero de viajes} \\
Tipo & Capacidad & N\'umero &
\multicolumn{4}{c}{diarios en la ruta} \\
\cline{4-7}
de avi\'on & (pasajeros) & de aviones & 1 & 2 & 3 & 4 \\
\hline
1 & 50 & 5 & 3 & 2 & 2 & 1 \\
2 & 30 & 8 & 4 & 3 & 3 & 2 \\
3 & 20 & 10 & 5 & 5 & 4 & 2 \\
\hline
\multicolumn{3}{l}{N\'umero de clientes diarios} &
1000 & 2000 & 900 & 1200 \\
\hline
\end{tabular}
\end{center}
\end{table}
\end{document}
----- end of modified example file -----
Kenneth Cabrera wrote:
>Hi TeX users:
>
>Why do I obtain a table with the last column wider than the rest? | {"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": 0.9102591276168823, "perplexity": 1633.681066226393}, "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/1508187824733.32/warc/CC-MAIN-20171021095939-20171021115939-00131.warc.gz"} |
https://www.physicsforums.com/threads/is-this-all-the-evidence-for-quarks.303164/ | # Is this all the evidence for quarks?
1. Mar 28, 2009
### dangerbird
1-electron scattering
2-collider data
3- ? is there anything else which supports the quark model or is it just those 2. from what ive read so far its just those 2 but could be incorrect so correct me if my primative understandings off
THANKS
2. Mar 28, 2009
### malawi_glenn
Deep inelastisc scattering have used more probes than just electrons. One has used neutrinos and muons too for instance. (just deep inelastic scattering is enough to say that quarks exists)
Just saying "Collider data" is quite non precise - there are a lot of different data, which measures different things, and also many different collider experiment. It is like saying that the only evidence we have for Z bosons are collider data.
I would say the existence of the top quark as the strongest proofs of them all to call for the existence of quarks. The top quark does not hadronize, and the signal for top quarks is really interesting.
So you should be beware of that these proves are really good and are quite astonishing, and also some of these explorations have been awarded Nobel Prize.
Also when someone says "from what ive read", WHAT have you read? maybe there was a misunderstanding? etc.
3. Mar 28, 2009
### dangerbird
so basically anything that can enter the nucleus + can be measured when it deflects has been used then.... and the results of the paterns of deflections suggest theres 3 quarks per hadrun? that seems rather clever
4. Mar 28, 2009
### malawi_glenn
oh, no, there are more than 3 quarks in a proton, there are sea quarks as well. I just gave that answer to you in the thread you created. https://www.physicsforums.com/showthread.php?t=302385
I think you need to calm down, and get a good book on particle physics, and try to study it. You have again some misconceptions. First of all, there are hadrons which are made up on 2 valence quarks (one quark and one anti-quark).
5. Mar 28, 2009
### clem
The quark model was popular for five years before the first DIS experiments. This is because
3- A large number of the static properties (mass, spin, charge, muliplicity, magnetic moment)
of hadrons were correlated and predicted by assuming that baryons are composed of three quarks, and mesons of a quark-antiquark pair.
6. Mar 28, 2009
### humanino
Quite frankly to me the original question is not far from "what evidence do we have for quarks apart from physics ?". If you are clever enough to devise another experiment, please go ahead. But be aware that those two points you mention actually cover many different experiments. If only in DIS, historical measurements were done inclusively, by detecting only the scattered lepton. Nowadays we perform semi-inclusive measurements, where another particle at least is detected, and exclusive measurements, where all particles in the final state are detected. We find that all those phenomena are described by the same universal "wave functions" in terms of quark-gluon degrees of freedom.
As for collider data, again there are many different observations. You may collide two leptons or two hadrons for instance. If you collide two hadrons, you have theorems to deal with phenomena at high transverse momenta, or you have lepton pair productions (Drell-Yann). In fact it is very difficult to make an exhaustive list.
There is no definitive evidence that anything goes wrong with the partonic picture. On the contrary, the more we try to apply it to new situation, the more confident we become that our understanding is correct.
7. Mar 28, 2009
### Dmitry67
Hadronize???
Could you give more details or where to look?
8. Mar 28, 2009
### malawi_glenn
what? you don't know what hadronization is or what the top-quark signal looks like?
9. Mar 28, 2009
### Dmitry67
Yes, I am very stupid
So you claim that t quark do not form any bound systems with other quarks or what?
10. Mar 28, 2009
### malawi_glenn
no you are not stupid, I asked what you was asking for.
No, the top quark does not form any bound states, it is too short-lived, it decays before reaching out of the "perturbative scale".
11. Mar 28, 2009
### Dmitry67
I see... And what is so special about the signal?
BTW I heard that there is a Higgs-less model where t-anti t pairs play the role of Higgs, does it make sense?
12. Mar 28, 2009
### malawi_glenn
the signal is that you will have a b-quark etc which will give a invariant mass peak of about 170GeV for a fermion.
I have not heard of it, so I let someone else answer about that model.
13. Mar 28, 2009
### humanino
It depends which model exactly you are talking about. Gribov tried to do this, but he passed away before he could convince the community, and now his specific views seem out of fashion. But in any case, people still think around those kind of ideas and we should have more to test with LHC.
Electroweak symmetry breaking: to Higgs or not to Higgs
14. Mar 28, 2009
### Staff: Mentor
When talking about this time period, it's important to distinguish between quarks, the entities postulated by Gell-Mann and others, to account for the patterns of properties among hadrons, and partons, the hard point-like entities inside nucleons that were postulated by Feynman and others to account for the results of deep inelastic scattering experiments.
In the 1970s, it was by no means certain that quarks and partons were the same thing. One of the main goals of deep inelastic scattering experiments of the time (including the neutrino experiments that I worked on as a graduate student), was to test what was then called the "quark-parton model," which is now a basic part of the "standard model." One of the professors in my research group warned me not to get too attached to the quark-parton model, because it might turn out to be wrong.
15. Mar 29, 2009
### granpa
16. Mar 29, 2009
### dangerbird
alright, but now what im mainly wondering is how the DIS experiments support the quark model. i dont know how by shooting particle at the nucleus that it can differenciated that theres 3 quarks in a hadrun vs there being 9999. just by the paths of the deflections?
Last edited: Mar 29, 2009
17. Mar 29, 2009
### humanino
This very same question you asked just a few hours earlier on this forum. It might be more constructive if you do not ignore the answers you were already given.
It so happens that neutrinos respond differently to matter and anti-matter. So counting the difference between matter and antimatter inside a proton can be done by comparing the scattering of neutrinos and antineutrinos. It is by no means simple. But the observations agree with the theory.
18. Mar 29, 2009
### Vanadium 50
Staff Emeritus
A few comments:
As pointed out, there are a great many DIS experiments (including some at colliders - ZEUS and H1) and they clearly indicate the presence of three valance quarks - one type with charge +2/3 and the other with -1/3. It is, however, difficult to explain the details of how this can be extracted before someone understands (and by "understands", I mean "can calculate") the basics of DIS. Oh, and it's deeply inelastic scattering.
While baryon magnetic moments are often touted as a success of the quark model, it's not the best example. Any theory that has SU(3) flavor symmetry will make the same predictions as the quark model. So while it's evidence in favor of the quark model (and evidence against alternatives), it's not as compelling as it's usually advertised.
It's believed true that the top quark decays before it hadronizes, so one actually does observe a bare quark. However, there's no experimental evidence of this at the moment. One would need to study the angular correlations between polarized top quark pairs, and there just aren't enough of them out there to make a convincing measurement. We just have to wait.
I have no idea what granpa is talking about with the Delta. It's a hadron, to be sure, and it's therefore made of quarks, but it was not a particularly important stepping stone on the road to the quark model. The better example was the Omega-minus baryon, which was a state predicted by the quark model and (at the time) was undiscovered. Nick Samios and collaborators looked for it, and discovered it with exactly the predicted properties.
A powerful case for quarks is, in my mind, the energy levels of quarkonium - bound states of a heavy quark and an antiquark. These have energy levels similar to that of a hydrogen atom, and as such illustrate the dynamics of quark behavior. These measurements show that there are actual physical objects with the quark quantum numbers moving around inside the hadron.
19. Mar 29, 2009
### dangerbird
no i didnt
thats impossible neutrinos go through protons
20. Mar 29, 2009
### Vanadium 50
Staff Emeritus
So, JTBell, who worked on exactly these experiments (see above) is wrong? Or lying?
What makes you think that you know better than someone who actually did the experiment?
21. Mar 29, 2009
### dangerbird
Im just gona go out on a limb here, maybe neutrinos just naturally move arround like that and it has nothing to do with it bouncing off of some nucleus or proton? theres many possabilities
plus my IQ is 129
22. Mar 30, 2009
### Staff: Mentor
Going out on a limb is OK for professionals who are familiar with the field and who know what others have done before them. For others it's arrogance.
All theories and models are subject to being superseded by something better, but it happens only after solid experimental evidence or testable theoretical considerations, not some random musings about "many possibilities."
23. Mar 30, 2009
### Vanadium 50
Staff Emeritus
I agree with the sentiment, but might drop the word "professionals". The key is understanding what's gone before. If an amateur has taken the time and expended the effort to understand that, it's possible that their criticism is valid. Evidence is the key here, and I think the key is whether or not one can recognize it.
Of course, "you're wrong because I have some random musings about other possibilities - and a high IQ" is a non-starter.
24. Mar 30, 2009
### Staff: Mentor
Yes, I agree. I didn't intend to exclude serioius amateurs who have knowledge equivalent to a "real physicist."
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | {"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": 0.843756377696991, "perplexity": 1284.3457025754442}, "config": {"markdown_headings": true, "markdown_code": false, "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-2018-30/segments/1531676592861.86/warc/CC-MAIN-20180721223206-20180722003206-00475.warc.gz"} |
https://reperiendi.wordpress.com/2007/09/19/the-problem-of-evil-and-cartesian-categories/ | # reperiendi
## Cartesian categories and the problem of evil
Posted in Category theory, Math by Mike Stay on 2007 September 19
How many one-element sets are there? Well, given any set $S,$ we can construct the one-element set $\{S\},$ so the collection of one-element sets has to be a proper class, a mindbogglingly enormous collection far larger than any mere set could be. However, they’re all the same from the point of view of functions coming out of them: the one element, no matter what it is, maps to a point in the range. The internal nature of a given one-element set is completely irrelevant to the way it behaves as a member of the category Set of all sets and functions.
For a category theorist, making a distinction between one-element sets is evil. Instead of looking inside an object to see how it’s made, we should only care about how it interacts with the world around it. There are certain kinds of objects that are naturally special because of the way they interact with everything else; we say they satisfy universal properties.
Just as it is evil to dwell on the differences between isomorphic one-element sets, it is evil to care about the inner workings of ordered pairs. Category theory elevates an ordered pair to a primitive concept by ignoring all details about the implementation of an ordered pair except how it interacts with the rest of the world. Ordered pairs are called “products” in category theory.
A product of the objects $G$ and $H$ in the category $C$ is
• an object of $C$, which we’ll label $G\times H,$ together with
• two maps, called projections
• $\pi_G:G\times H\to G$
• $\pi_H:G\times H\to H$
that satisfy the following universal property: for any triple $(X, f_G:X\to G, f_H:X \to H),$ there is a unique map from $X$ to $G\times H$ making the following diagram commute:
In particular, given two different representations of ordered pairs, there’s a unique way to map between them, so they must be isomorphic.
A category will either have products or it won’t:
——————
1. The category Set has the obvious cartesian product.
2. The trivial category has one object and one morphism, the identity, so there’s only one choice for a triple like the one in the definition:
$(X, 1_X, 1_X),$
and it’s clearly isomorphic to itself, so the trivial category has products.
3. A preorder is a set $S$ equipped with a $\le$ relation on pairs of elements of $S$ that is transitive and reflexive. Given any preorder, we can construct a category whose
• objects are the elements of $S$ and
• there is an arrow from $x$ to $y$ if $x \le y.$
So a product in a preorder is
• an element $z = x \times y$ of $S$ together with maps
• $\pi_x:z\to x$ (that is, $z \le x$)
• $\pi_y:z \to y$ (that is. $z \le y$)
such that for any other element $w \in S, \, w\le x, \, w \le y,$ we have $w \le z.$
In other words, a product $x \times y$ in a preorder is the greatest lower bound of $x, y$. For example, in the preorder $(\mathbb{R}, \le)$, the product of two numbers $x, y$ is min($x, y$). In the preorder $(\mbox{Set}, \subseteq)$, the product is $x \cap y$. In the preorder $(\mathbb{N}, |)$, where “|” is “divides”, the product is gcd($x, y$).
Exercise: in what preorder over $\mathbb{R}$ is the cartesian product the same as multiplication?
——————
A cartesian category is a category with products. (Note the lowercase ‘c:’ you know someone’s famous if things named after them aren’t capitalized. C.f. ‘abelian.’) You can think of the cartesian coordinate system for the plane with its ordered pair of coordinates to remind you that you can make ordered pairs of objects in a cartesian category.
### 6 Responses
1. Roshan said, on 2011 March 17 at 9:24 am
Nice writeup.
Btw you did mean z <= x in \pi_x instead of z < x (and similarly in \pi_y) didn't you?
2. a b said, on 2014 December 10 at 12:00 pm
Can I get a hint on the exercise?
• Mike Stay said, on 2014 December 10 at 12:37 pm
I don’t know if there *is* a preorder where the cartesian product is multiplication. In the category of finite sets and all functions between them, the cartesian product of two sets with cardinality m and n has cartinality mn. Groupoid cardinality (http://ncatlab.org/nlab/show/groupoid+cardinality) generalizes cardinality to a function from groupoids to nonnegative real numbers. We can generalize to other categories as described in http://arxiv.org/abs/math.CT/0212377 and get complex cardinalities; in all of these, the cardinality of the product is the product of the cardinalities. I think there may be a way to consider a real number to be a cardinality-equivalence class of such objects and get an arrow between two numbers if there’s a morphism between any two objects in the corresponding classes, but I haven’t proven it.
3. Joost Winter said, on 2015 January 28 at 1:53 pm
Let’s see… Assume there is such a preorder on R.
By reflexivity, we must have 4<=4 in the preorder. Because 2*2 equals 4, 4<2 must hold in the preorder. By a similar argument, 8<4 should hold in the preorder as 4*2 equals 8, however as we already know that 4<=2 and 4<=4 must hold, it follows that 4 is a lower bound of 2 and 4, and as 8<4 holds by one of our assumption, 8 cannot be the glb of 2 and 4. It thus follows that there cannot be a preorder on R (or on N, Q, Z for that matter) s.t. cartesian product/glb is multiplication.
4. Joost Winter said, on 2015 January 28 at 2:18 pm
A simpler counterexample is of course given by the fact that in any preorder with glbs, the glb of an element x with itself is always x… so the glb of 2 and 2 is always 2 (or some object isomorphic to 2) in any preorder with domain N/Z/Q/R… | {"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": 38, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8649474382400513, "perplexity": 385.8775541672958}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027317339.12/warc/CC-MAIN-20190822172901-20190822194901-00484.warc.gz"} |
https://arxiver.wordpress.com/2016/09/08/a-new-correlation-with-lower-kilohertz-quasi-periodic-oscillation-frequency-in-the-ensemble-of-low-mass-x-ray-binaries-heap/ | # A new correlation with lower kilohertz quasi-periodic oscillation frequency in the ensemble of low-mass X-ray binaries [HEAP]
We study the dependence of kHz quasi-periodic oscillation (QPO) frequency on accretion-related parameters in the ensemble of neutron star low-mass X-ray binaries. Based on the mass accretion rate, $\dot{M}$, and the magnetic field strength, $B$, on the surface of the neutron star, we find a correlation between the lower kHz QPO frequency and $\dot{M}/B^{2}$. The correlation holds in the current ensemble of Z and atoll sources and therefore can explain the lack of correlation between the kHz QPO frequency and X-ray luminosity in the same ensemble. The average run of lower kHz QPO frequencies throughout the correlation can be described by a power-law fit to source data. The simple power-law, however, cannot describe the frequency distribution in an individual source. The model function fit to frequency data, on the other hand, can account for the observed distribution of lower kHz QPO frequencies in the case of individual sources as well as the ensemble of sources. The model function depends on the basic length scales such as the magnetospheric radius and the radial width of the boundary region, both of which are expected to vary with $\dot{M}$ to determine the QPO frequencies. In addition to modifying the length scales and hence the QPO frequencies, the variation in $\dot{M}$, being sufficiently large, may also lead to distinct accretion regimes, which would be characterized by Z and atoll phases.
M. Erkut, S. Duran, O. Catmabacak, et. al.
Thu, 8 Sep 16
58/60
Comments: 11 pages, 5 figures, 3 tables, accepted for publication in The Astrophysical Journal | {"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": 0.9758796095848083, "perplexity": 1086.6643294413207}, "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-2018-51/segments/1544376825916.52/warc/CC-MAIN-20181214140721-20181214162221-00045.warc.gz"} |
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http://www.ck12.org/physical-science/Calculating-Acceleration-from-Force-and-Mass-in-Physical-Science/asmtpractice/Calculating-Acceleration-from-Force-and-Mass-in-Physical-Science-Practice/r1/ | <meta http-equiv="refresh" content="1; url=/nojavascript/">
# Calculating Acceleration from Force and Mass
The acceleration of an object equals the net force acting on it divided by its mass.
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Calculating Acceleration from Force and Mass Practice
SCI.PSC.214.31 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9049716591835022, "perplexity": 3171.6225148001017}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1433195036776.1/warc/CC-MAIN-20150601214356-00090-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://link.springer.com/article/10.1007%2Fs10035-016-0624-2 | Granular Matter
, 18:58
# Memory of jamming–multiscale models for soft and granular matter
• Nishant Kumar
• Stefan Luding
Open Access
Original Paper
Part of the following topical collections:
1. Micro origins for macro behavior of granular matter
## Abstract
Soft, disordered, micro-structured materials are ubiquitous in nature and industry, and are different from ordinary fluids or solids, with unusual, interesting static and flow properties. The transition from fluid to solid—at the so-called jamming density—features a multitude of complex mechanisms, but there is no unified theoretical framework that explains them all. In this study, a simple yet quantitative and predictive model is presented, which allows for a changing jamming density, encompassing the memory of the deformation history and explaining a multitude of phenomena at and around jamming. The jamming density, now introduced as a new state-variable, changes due to the deformation history and relates the system’s macroscopic response to its micro-structure. The packing efficiency can increase logarithmically slow under gentle repeated (isotropic) compression, leading to an increase of the jamming density. In contrast, shear deformations cause anisotropy, changing the packing efficiency exponentially fast with either dilatancy or compactancy as result. The memory of the system near jamming can be explained by a micro-statistical model that involves a multiscale, fractal energy landscape and links the microscopic particle picture to the macroscopic continuum description, providing a unified explanation for the qualitatively different flow-behavior for different deformation modes. To complement our work, a recipe to extract the history-dependent jamming density from experimentally accessible data is proposed, and alternative state-variables are compared. The proposed simple macroscopic constitutive model is calibrated from particles simulation data, with the variable jamming density—resembling the memory of microstructure—as essential novel ingredient. This approach can help understanding predicting and mitigating failure of structures or geophysical hazards, and will bring forward industrial process design and optimization, and help solving scientific challenges in fundamental research.
## Keywords
Jamming Structure Anisotropy Dilatancy Creep/relaxation Memory Critical state
## 1 Introduction
Granular materials are a special case of soft-matter with micro-structure, as also foams, colloidal systems, glasses, or emulsions [1, 2, 3]. Particles can flow through a hopper or an hour-glass when shaken, but jam (solidify) when the shaking stops [4]. These materials jam above a “certain” volume fraction, or jamming density, referred to as the “jamming point” or “jamming density” [3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23], and become mechanically stable with finite bulk- and shear-moduli [8, 9, 12, 15, 24, 25, 26, 27]. Notably, in the jammed state, these systems can “flow” by reorganizations of their micro-structure [28, 29]. Around the jamming transition, these systems display considerable inhomogeneity, such as reflected by over-population of weak/soft/slow mechanical oscillation modes [11], force-networks [10, 30, 31], diverging correlation lengths and relaxation time-scales [9, 13, 22, 32, 33, 34, 35], and some universal scaling behaviors [36, 37]. Related to jamming, but at all densities, other phenomena occur, like shear-strain localization [12, 16, 38, 39, 40], anisotropic evolution of structure and stress [7, 9, 11, 13, 30, 31, 38, 39, 40, 41, 42, 43, 44, 45, 46], and force chain inhomogeneity [7, 19, 28]. To gain a better understanding of the jamming transition concept, one needs to consider both the structure (positions and contacts) and contact forces. Both of them illustrate and reflect the transition, e.g., with a strong force chain network percolating the full system and thus making unstable packings permanent, stable and rigid [7, 19, 47, 48, 49].
For many years, scientists and researchers have considered the jamming transition in granular materials to occur at a particular volume fraction, $$\phi _J$$ [50]. In contrast, over the last decade, numerous experiments and computer simulations have suggested the existence of a broad range of $$\phi _J$$, even for a given material. It was shown that the critical density for the jamming transition depends on the preparation protocol [12, 18, 22, 23, 36, 51, 52, 53, 54, 55, 56, 57, 58], and that this state-variable can be used to describe and scale macroscopic properties of the system [26]. For example, rheological studies have shown that $$\phi _J$$ decreases with increasing compression rate [8, 57, 59, 60] (or with increasing growth rate of the particles), with the critical scaling by the distance from the jamming point ($$\phi - \phi _J$$) being universal and independent of $$\phi _J$$ [20, 36, 51, 61, 62] Recently, the notion of an a-thermal isotropic jamming “point” was challenged due to its protocol dependence, suggesting the extension of the jamming point, to become a J-segment [42, 60, 63, 64]. Furthermore, it was shown experimentally, that for a tapped, unjammed frictional 2D systems, shear can jam the system (known as “shear jamming”), with force chain networks percolating throughout the system, making the assemblies jammed, rigid and stable [7, 29, 47, 48, 65, 66], all highlighting a memory that makes the structure dependent on history H. But to the best of our knowledge, quantitative characterization of the varying/moving/changing transition points, based on H, remains a major open challenge.
### 1.1 Application examples
In the fields of material science, civil engineering and geophysics, the materials behave highly hysteretic, non-linear and involve irreversibility (plasticity), possibly already at very small deformations, due to particle rearrangements, more visible near the jamming transition [67, 68, 69, 70]. Many industrial and geotechnical applications that are crucial for our society involve structures that are designed to be far from failure (e.g. shallow foundations or underlying infrastructure), since the understanding when failure and flow happens is not sufficient, but is essential for the realistic prediction of ground movements [71]. Finite-element analyses of, for example tunnels, depend on the model adopted for the pre-failure soil behavior; when surface settlement is considered, the models predicting non-linear elasticity and history dependence become of utmost importance [72]. Design and licensing of infrastructure such as nuclear plants and long span bridges are dependent on a robust knowledge of elastic properties in order to predict their response to seismic ground motion such as the risk of liquefaction and the effect of the presence of anisotropic strata. (Sediments are one example of anisotropic granular materials of particles of organic or inorganic origin that accumulate in a loose, unconsolidated form before they are compacted and solidified. Knowing their mechanical behavior is important in industrial, geotechnical and geophysical applications. For instance, the elastic properties of high-porosity ocean-bottom sediments have a massive impact on unconventional resource exploration and exploitation by ocean drilling programs.)
When looking at natural flows, a complete description of the granular rheology should include an elastic regime [73], and the onset of failure (flow or unjamming) deserves particular attention in this context. The material parameters have a profound influence on the computed deformations prior to failure [74, 75], as the information on the material state is usually embedded in the parameters. Likewise, also for the onset of flow, the state of the material is characterized by the value of the macroscopic friction angle, as obtained, e.g., from shear box experiments or tri-axial tests. Since any predictive model must describe the pre-failure deformation [76] as well as the onset of flow (unjamming) of the material, many studies have been devoted to the characteristics of geomaterials (e.g., tangent moduli, secant moduli, peak strength) and to the post-failure regime [77] or the steady (critical) state flow rheology, see Refs. [40, 78] and references therein.
### 1.2 Approach of this study
Here, we consider frictionless sphere assemblies in a periodic system, which can help to elegantly probe the behavior of disordered bulk granular matter, allowing to focus on the structure [3], without being disturbed by other non-linearities [7, 29, 79] (as e.g. friction, cohesion, walls, environmental fluids or non-linear interaction laws). For frictionless assemblies, it is often assumed that the influence of memory is of little importance, maybe even negligible. If one really looks close enough, however, its relevance becomes evident. We quantitatively explore its structural origin in systems where the re-arrangements of the micro-structure (contact network) are the only possible mechanisms leading to the range of jamming densities (points), i.e. a variable state-variable jamming density.
In this study, we probe the jamming transition concept by two pure deformation modes: isotropic compression or “tapping” and deviatoric pure shear (volume conserving), which allow us to combine the J-segment concept with a history dependent jamming density.1 Assuming that all other deformations can be superimposed by these two pure modes, we coalesce the two concepts of isotropic and shear induced jamming, and provide the unified model picture, involving a multiscale, fractal-type energy landscape [18, 80, 81, 82]; in general, deformation (or the preparation procedure) modify the landscape and its population; considering only changes of the population already allows to establish new configurations and to predict their evolution. The observations of different $$\phi _J$$ of a single material require an alternative interpretation of the classical “jamming diagram” [5].
Our results will provide a unified picture, including some answers to the open questions from literature: (i) What lies in between the jammed and flowing (unjammed) regime? As posed by Ciamarra et al. [63]. (ii) Is there an absolute minimum jamming density? As posed by Ciamarra et al. [63]. (iii) What protocols can generate jammed states? As posed by Torquato et al. [56]. (iv) What happens to the jamming and shear jamming regime in 3D and is friction important to observe it? As posed by Bi et al. [7]. Eventually, accepting the fact that the jamming density is changing with deformation history, significant improvement of continuum models is expected, not only for classical elasto-plastic or rheology models, but also, e.g., for anisotropic constitutive models [41, 69, 83, 84], GSH rate type models [85, 86], Cosserat micro-polar or hypoplastic models [87, 88, 89] or continuum models with a length scale and non-locality [90, 91]. For this purpose we provide a simple (usable) analytical macro/continuum model as generalization of continuum models by adding one isotropic state-variable. Only allowing $$\phi _J(H)$$ to be dependent on history H [64, 92], as key modification, explains a multitude of reported observations and can be significant step forward to solve real-world problems in e.g. electronic industry related novel materials, geophysics or mechanical engineering.
Recent works showed already that, along with the classical macroscopic properties (stress and volume fraction), the structural anisotropy is an important [41, 45, 46, 93, 94, 95, 96] state-variable for granular materials, as quantified by the fabric tensor [43, 69] that characterizes, on average, the geometric arrangement of the particles, the contacts and their network, i.e. the microstructure of the particle packing. Note that the anisotropy alone is not enough to characterize the structure, but also an isotropic state-variable is needed, as is the main message of this study.
### 1.3 Overview
The paper continues with the simulation method in Sect. 2, before the micromechanical particle- and contact-scale observations are presented in Sect. 3, providing analytical (quantitative) constitutive expressions for the change of the jamming density with different modes of deformation. Section 4 is dedicated to a (qualitative) meso-scale stochastic model that explains the different (slow versus fast) change of $$\phi _J(H)$$ for different deformation modes (isotropic versus deviatoric/shear). A quantitative predictive macroscale model is presented in Sect. 5 and verified by comparison with the microscale simulations, before an experimental validation procedure is discussed in Sect. 6 and the paper is summarized and conclusions are given in Sect. 7.
## 2 Simulation method
Discrete Element Method (DEM) simulations are used to model the deformation behavior of systems with $$N = 9261$$ soft frictionless spherical particles with average radius $$\langle r \rangle = 1$$ (mm), density $$\rho = 2000$$ (kg/m3), and a uniform polydispersity width $$w = r_{\mathrm {max}}/r_{\mathrm {min}}= 3$$, using the linear visco-elastic contact model in a 3D box with periodic boundaries [44, 69]. The particle stiffness is $$k = 10^8$$ (kg/s2), contact viscosity is $$\gamma = 1$$ (kg/s). A background dissipation force proportional to the moving velocity is added with $$\gamma _b= 0.1$$ (kg/s). The particle density is $$\rho = 2000$$ (kg/m3). The smallest time of contact is $$t_c = 0.2279$$ ($$\upmu$$s) for a collision between two smallest sized particles [41].
### 2.1 Preparation procedure and main experiments
For the preparation, the particles are generated with random velocities at volume (solid) fraction $$\phi =0.3$$ and are isotropically compressed to $$\phi _t=0.64$$, and later relaxed. From such a relaxed, unjammed, stress free initial state with volume fraction, $$\phi _t= 0.64 < \phi _J$$, we compress isotropically further to a maximum volume fraction, $$\phi ^{\mathrm {max}}_i$$, and decompress back to $$\phi _t$$, during the latter unloading $$\phi _J$$ is identified. This process is repeated over M (100) cycles, which provides different isotropic jamming densities (points) $$\phi _J=:{}^M\phi _{J,i}$$, related with $$\phi ^{\mathrm {max}}_i$$ and M (see Sect. 3.1).
Several isotropic configurations $$\phi$$, such that $$\phi _t< \phi < {}^1\phi _{J,i}$$ from the decompression branch are chosen as the initial configurations for shear experiments. We relax them and apply pure (volume conserving) shear (plane-strain) with the diagonal strain-rate tensor $${{\dot{\varvec{\mathrm E}}}}= \pm {\dot{\epsilon }}_\mathrm {d} \left( -1,1,0\right)$$, for four cycles.2 The x and y walls move, while the z wall remain stationary. The strain rate of the (quasi-static) deformation is small, $${\dot{\epsilon }}_\mathrm {d} t_c < 3 \times 10^{-6}$$, to minimize transient behavior and dynamic effects.3
### 2.2 Macroscopic (tensorial) quantities
Here, we focus on defining averaged tensorial macroscopic quantities—including strain-, stress- and fabric (structure) tensors—that provide information about the state of the packing and reveal interesting bulk features.
From DEM simulations, one can measure the ‘static’ stress in the system [97] as
\begin{aligned} \varvec{\sigma }=\left( {1}/{V}\right) \sum _{c\in V}\mathbf {l}^{c}\otimes \mathbf {f}^{c}, \end{aligned}
(1)
average over all the contacts in the volume V of the dyadic products between the contact force $$\mathbf {f}^{c}$$ and the branch vector $$\mathbf {l}^{c}$$, where the contribution of the kinetic fluctuation energy has been neglected [41, 93]. The dynamic component of the stress tensor is four orders of magnitude smaller than the former and hence its contribution is neglected. The isotropic component of the stress is the pressure $$P= \mathrm {tr}(\varvec{\sigma })/3$$.
In order to characterize the geometry/structure of the static aggregate at microscopic level, we will measure the fabric tensor, defined as
\begin{aligned} \mathbf {F}=\frac{1}{V}\sum _{{\mathcal {P}}\in V}V^{{\mathcal {P}}}\sum _{c\in {\mathcal {P}}}\mathbf {n}^{c}\otimes \mathbf {n}^{c}, \end{aligned}
(2)
where $$V^{\mathcal {P}}$$ is the volume relative to particle $${\mathcal {P}}$$, which lies inside the averaging volume V, and $$\mathbf {n}^{c}$$ is the normal unit branch-vector pointing from center of particle $${\mathcal {P}}$$ to contact c [93, 98, 99]. Isotropic part of fabric is $$F_\mathrm {v}= \mathrm {tr}(\mathbf {F})$$. The corrected coordination number [7, 41] is $$C^{*}= {M_4}/{N_4},$$ where, $$M_4$$ is total contacts of the $$N_4$$ particles having at least 4 contacts, and the non-rattler fraction is $$f_\mathrm {NR}= N_4/N$$. C is the ratio of total non-rattler contacts $$M_4$$ and total number of particles N, i.e., $$C=M_4/N = \left( M_4/N_4\right) \left( N_4/N\right) = C^{*}f_\mathrm {NR}$$, with corrected coordination number $$C^{*}$$ and fraction of non-rattlers $$f_\mathrm {NR}$$. The isotropic fabric $$F_\mathrm {v}$$ is given by the relation $$F_\mathrm {v}= g_3 \phi C$$, as taken from Imole et al. [41], with $$g_3\cong 1.22$$ for the polydispersity used in the present work. For any tensor $${\mathrm {\varvec{\mathrm Q}}}$$, its deviatoric part can be defined as $$Q_\mathrm {d} = \mathrm {sgn}\left( q_{yy} - q_{xx}\right) \sqrt{ 3 q_{ij}q_{ij}/2}$$, where $$q_{ij}$$ are the components of the deviator of $${\mathrm {\varvec{\mathrm Q}}}$$, and the sign function accounts for the shear direction, in the system considered here, where a more general formulation is given in Ref. [69]. Both pressure $$P$$ and shear stress $$\varGamma$$ are non-dimensionalized by $${2\langle r \rangle }/{k}$$ to give dimensionless pressure p and shear stress $$\uptau$$.
Table 1
Parameters used in Eq. (3) and Eqs. (911), where ‘*’ represents slightly different values than from Imole et al. [41], modified slightly to have more simple numbers, without big deviation, and without loss of generality
Quantity
Isotropic
Shear
p
$$p_0=0.042$$; $$\gamma _p=0\pm 0.1$$*
$$p_0=0.042$$; $$\gamma _p=0\pm 0.1$$*
$$C^{*}$$
$$C_1=8.5 \pm 0.3$$*; $$\theta =0.58$$
$$C_1=8.5 \pm 0.3$$*; $$\theta =0.58$$
$$f_\mathrm {NR}$$
$$\varphi _c=0.13$$; $$\varphi _v=15$$
$$\varphi _c=0.16$$; $$\varphi _v=15$$
## 3 Micromechanical results
### 3.1 Isotropic deformation
In this section, we present a procedure to identify the jamming densities and their range. We also show the effect of cyclic over-compression to different target volume fractions and present a model that captures this phenomena.
#### 3.1.1 Identification of the jamming density
When a sample is over-compressed isotropically, the loading and unloading paths are different in pressure $$p$$. This difference is most pronounced near the jamming density $$\phi _J$$, and for the first cycle. It brings up the first question of how to identify a jamming density, $$\phi _J$$. The unloading branch of a cyclic isotropic over-compression along volume fraction $$\phi$$ is well described by a linear relation in volumetric strain, with a tiny quadratic correction [44, 100, 101]:
\begin{aligned} p=\frac{\phi C}{\phi _J}p_0 (-\varepsilon _\mathrm {v}) \left[ 1-\gamma _p (-\varepsilon _\mathrm {v}) \right] , \end{aligned}
(3)
where $$p_0$$, $$\gamma _p$$, as presented in Table 1, and the jamming density $$\phi _J$$ are the fit parameters, and $$-\varepsilon _\mathrm {v}=\log (\phi /\phi _J)$$ is the true or logarithmic volumetric strain of the system, defined relative to the reference where $$p\rightarrow 0$$, i.e. the jamming volume fraction.
Equation (3), quantifies the scaled stress and is proportional to the dimensionless deformation (overlap per particle size), as derived analytically [100] from the definition of stress and converges to $$p\rightarrow 0$$ when $$\phi \rightarrow \phi _J$$.
We apply the same procedure for different over-compressions, $$\phi ^{\mathrm {max}}_i$$, and many subsequent cycles M to obtain $${}^M\phi _{J,i}$$, for which the results are discussed below. The material parameter $$p_0$$ is finite, almost constant, whereas $$\gamma _p$$ is small, sensitive to history and contributes mainly for large $$-\varepsilon _\mathrm {v}$$, with values ranging around $$0\pm 0.1$$; in particular, it is dependent on the over-compression $$\phi ^{\mathrm {max}}_i$$ (data not shown). Unless strictly mentioned, we shall be using the values of $$p_0$$ and $$\gamma _p$$ given in Table 1.
Figure 1a shows the behavior of $$p$$ with $$\phi$$ during one full over-compression cycle to display the dependence of the jamming density on the maximum over-compression volume fraction and the number of cycles. With increasing over-compression amplitude, e.g. comparing $$\phi ^{\mathrm {max}}_i = 0.68$$ and $$\phi ^{\mathrm {max}}_i = 0.82$$, the jamming density, as realized after unloading, is increasing. Also, with each cycle, from $$M=1$$ to $$M=100$$, the jamming density moves to larger values. Note that the difference between the loading and the unloading curves becomes smaller for subsequent over-compressions. Fig. 1b shows the scaled pressure, i.e., $$p$$ normalized by $$\phi C/\phi _J$$, which removes its non-linear behavior. $$p$$ represents the average deformation (overlap) of the particles at a given volume fraction, proportional to the distance from the jamming density $$\phi _J$$.4 In the small strain region, for all over-compression amplitude and cycles, the datasets collapse on a line with slope $$p_0\sim 0.042$$. Only for very strong over-compression, $$-\varepsilon _\mathrm {v}>0.1$$, a small deviation (from linear) of the simulation data is observed due to the tiny quadratic correction in Eq. (3).
#### 3.1.2 Isotropic cyclic over-compression
Many different isotropic jamming densities can be found in real systems and—as shown here—also for the simplest model material in 3D [64]. Figure 2a shows the evolution of these extracted isotropic jamming densities $${}^M\phi _{J,i}$$, which increase with increasing M and with over-compression $$\phi ^{\mathrm {max}}_i$$, for subsequent cycles M of over-compressions, the jamming density $${}^M\phi _{J,i}$$ grows slower and slower and is best captured by a Kohlrausch-Williams-Watts (KWW) stretched exponential relation:
\begin{aligned} \begin{aligned} {}^M\phi _{J,i}:&= \phi _J(\phi ^{\mathrm {max}}_i, M) \\&= {}^{\infty }\phi _{J,i}- \left( {}^{\infty }\phi _{J,i}- \phi _{c}\right) \exp \left[ {-\left( {M}/{\mu _i}\right) ^{\beta _i}}\right] , \end{aligned} \end{aligned}
(4)
with the three universal “material”-constants $$\phi _{c}= 0.6567$$ (Sect. 3.2.2), $$\mu _i =1$$, and $$\beta _i=0.3$$, the lower limit of possible $$\phi _J$$’s, the relaxation (cycle) scale and the stretched exponent parameters, respectively. Only $${}^{\infty }\phi _{J,i}$$, the equilibrium (steady-state or shakedown [102]) jamming density limit (extrapolated for $$M \rightarrow \infty$$), depends on the over-compressions $$\phi ^{\mathrm {max}}_i$$. $$\phi _{c}$$ is the critical density in the zero pressure limit without previous history, or after very long shear without temperature (which all are impossible to realize with experiments or simulation–only maybe with energy minimization).
Very little over-compression, $$\phi ^{\mathrm {max}}_i \gtrsim \phi _{c}$$, does not lead to a significant increase in $$\phi _{J,i}$$, giving us information about the lower limits of the isotropic jamming densities achievable by shear, which is the critical jamming density $$\phi _{c}= 0.6567$$. With each over-compression cycle, $${}^M\phi _{J,i}$$ increases, but for larger M, it increases less and less. This is analogous to compaction by tapping, where the tapped density increases logarithmically slow with the number of taps. The limit value $${}^{\infty }\phi _{J,i}$$ with $$\phi ^{\mathrm {max}}_i$$ can be fitted with a simple power law relation:
\begin{aligned} {}^{\infty }\phi _{J,i}= \phi _{c}+ \alpha _\mathrm{max} \left( \phi ^{\mathrm {max}}_i / \phi _{c}- 1 \right) ^{\beta } , \end{aligned}
(5)
where the fit works perfect for $$\phi _{c}< \phi ^{\mathrm {max}}_i \le 0.9$$, with parameters $$\phi _{c}=0.6567$$, $$\alpha _\mathrm{max}=0.02\pm 2\,\%$$, and $$\beta =0.3$$, while the few points for $$\phi ^{\mathrm {max}}_i \sim \phi _{c}$$ are not well captured. The relation between the limit-value $${}^{\infty }\phi _{J,i}$$ and $${}^1\phi _{J,i}$$ is derived using Eq. (4):
\begin{aligned} {}^{\infty }\phi _{J,i}-\phi _{c}= \frac{{}^1\phi _{J,i}-\phi _{c}}{1-e^{-1}} \cong 1.58 \left( {{}^1\phi _{J,i}-\phi _{c}} \right) , \end{aligned}
(6)
only by setting $$M=1$$, as shown in Fig. 2b, with perfect match. With other words, using a single over-compression, Eq. (6) allows to predict the limit value after first over-compression $${}^1\phi _{J,i}$$ (or subsequent over-compression cycles, using appropriate M).
Thus, the isotropic jamming density $$\phi _J$$ is not a unique point, not even for frictionless particle systems, and is dependent on the previous deformation history of the system [63, 82, 103], e.g. over-compression or tapping/driving (data not shown). Both (isotropic) modes of deformation lead to more compact, better packed configurations [7, 47, 104]. Considering different system sizes, and different preparation procedures, we confirm that the jamming regime is the same (within fluctuations) for all the cases considered (not shown). All our data so far, for the material used, are consistent with a unique limit density $$\phi _{c}$$ that is reached after large strain, very slow shear, in the limit of vanishing confining pressure. Unfortunately this limit is vaguely defined, since it is not directly accessible, but rather corresponds to a virtual stress-free state. The limit density is hard to determine experimentally and numerically as well. Reason is that any slow deformation (e.g. compression from below jamming) also leads to perturbations (like tapping leads to granular temperature): the stronger the system is perturbed, the better it will pack, so that usually $$\phi _J > \phi _{c}$$ is established. Repeated perturbations lead to a slow stretched exponential approach to an upper-limit jamming density $$\phi _J\rightarrow \phi _J^{\mathrm {max}}$$ that itself increases slowly with perturbation amplitude, see Fig. 2b. The observation of different $$\phi _J$$ of a single material, was referred to as J-segment [63, 103], and requires an alternative interpretation of the classical “jamming diagram” [5, 7, 66], giving up the misconception of a single, constant jamming “density”. Note that the J-segment is not just due to fluctuations, but it is due to the deformation history, and with fluctuations superposed. The state-variable $$\phi _J$$ varies due to deformation, but possibly has a unique limit value that we denote for now as $$\phi _{c}$$. Jammed states below $$\phi _{c}$$ might be possible too, but require different protocols [105], or different materials, and are thus not addressed here. Next, we discuss the concept of shear jammed states [7] below $$\phi _J$$.
### 3.2 Shear deformation
To study shear jamming, we choose several unjammed states with volume fractions $$\phi$$ below their jamming densities $${}^1\phi _{J,i}$$, which were established after the first compression-decompression cycle, for different history, i.e., various previously applied over-compression to $$\phi ^{\mathrm {max}}_i$$. Each configuration is first relaxed and then subjected to four isochoric (volume conserving) pure shear cycles (see Sect. 2.1).
#### 3.2.1 Shear jamming below $$\phi _J(H)$$
We confirm shear jamming, e.g., by a transition in the coordination number $$C^{*}$$, from below to above its isostatic limit, $$C^*_0= 6$$, for frictionless grains [13, 31, 38, 41]. This was consistently (independently) reconfirmed by using percolation analysis [7, 30], allowing us to distinguish the three different regimes namely, unjammed, fragile and shear jammed states during (and after) shear [66], as shown in Fig. 3a. We study how the $$k-$$cluster, defined as the largest force network, connecting strong forces, $$f \ge k f_\mathrm {avg}$$ [109, 110], with $$k=2.2$$, different from $$k=1$$ for 2D frictional systems [7], percolates when the initially unjammed isotropic system is sheared. More quantitatively, for an exemplary volume fraction $$\phi \left( \phi ^{\mathrm {max}}_i = 0.82, M=1\right) = 0.6584$$, very close to $$\phi _{c}$$, Fig. 3b shows that $$f_\mathrm {NR}$$ increases from initially zero to large values well below unity due to the always existing rattlers. The compressive direction percolating network $$\xi _y/L_y$$ grows faster than the extension direction network $$\xi _x/L_x$$, while the network in the non-mobile direction, $$\xi _z/L_z$$, lies in between them. For $$f_\mathrm {NR}>0.82\pm 0.01$$, we observe that the growing force network is percolated in all three directions (Fig. 3a), which is astonishingly similar to the value reported for the 2D systems [7]. The jamming by shear of the material corresponds (independently) to the crossing of $$C^{*}$$ from the isostatic limit of $$C^*_0=6$$, as presented in Fig. 3b.
From this perspective, when an unjammed material is sheared at constant volume, and it jams after application of sufficient shear strain, clearly showing that the jamming density has moved to a lower value. Shearing the system also perturbs it, just like over-compression; however, in addition, finite shear strains enforce shape- and structure-changes and thus allow the system to explore new configurations; typically, the elevated jamming density $$\phi _J$$ of a previously compacted system will rapidly decrease and exponentially approach its lower-limit, the critical jamming density $$\phi _{c}$$, below which no shear jamming exists. Note that we do not exclude the possibility that jammed states below $$\phi _{c}$$ could be achieved by other, special, careful preparation procedures [111].
Next, we present the evolution of the strong force networks in each direction during cyclic shear, as shown in Fig. 4, for the same initial system. After the first loading, at reversal $$f_\mathrm {NR}$$ drops below the 0.82 threshold, which indicates the breakage/disappearance of strong clusters, i.e. the system unjams. The new extension direction $$\xi _y/L_y$$ drops first with the network in the non-mobile directions, $$\xi _z/L_z$$, lying again in between the two mobile direction. With further applied strains, $$f_\mathrm {NR}$$ increases and again, the cluster associated with the compression direction grows faster than in the extension direction. For $$f_\mathrm {NR}$$ above the threshold, the cluster percolates the full system, leading to shear jammed states again. At each reversal, the strong force network breaks/fails in all directions, and the system gets “soft” or even unjams temporarily. However, the network is rapidly re-established in the perpendicular direction, i.e., the system jams and the strong, anisotropic force network again sustains the load. Note that some systems with volume fraction higher and away from $$\phi _{c}$$ can resist shear strain reversal as described and modeled in Sect. 5.1.3.
#### 3.2.2 Relaxation effects on shear jammed states
Here, we will discuss the system stability by looking at the macroscopic quantities in the saturation state (after large shear strain), by relaxing them sufficiently long to have non-fluctuating values in the microscopic and macroscopic quantities. Every shear cycle after defining e.g. the $$y-$$direction as the initial active loading direction, has two saturation states, one during loading and, after reversal, the other during unloading. In Fig. 5, we show values attained by the isotropic quantities pressure $$p$$, isotropic fabric $$F_\mathrm {v}$$ and the deviatoric quantities shear stress $$\uptau$$, shear stress ratio $$\uptau /p$$, and deviatoric fabric $$F_\mathrm {d}$$ for various $$\phi$$ given the same initial jamming density $$\phi _J\left( \phi ^{\mathrm {max}}_i = 0.82, M=1\right) =: {}^1\phi _{J,i}= 0.6652$$. Data are shown during cyclic shear as well as at the two relaxed saturation states (averaged over four cycles), leading to following observations:
1. (i)
With increasing volume fraction, $$p$$, $$F_\mathrm {v}$$ and $$\uptau$$ increase, while a weak decreasing trend in stress ratio $$\uptau /p$$ and deviatoric fabric $$F_\mathrm {d}$$ is observed.
2. (ii)
There is almost no difference in the relaxed states in isotropic quantities, $$p$$ and $$F_\mathrm {v}$$ for the two directions, whereas it is symmetric about zero for deviatoric quantities, $$\uptau$$, $$\uptau /p$$, and $$F_\mathrm {d}$$. The decrease in pressure during relaxation is associated with dissipation of kinetic energy and partial opening of the contacts to “dissipate” the related part of the contact potential energy. However, $$F_\mathrm {v}$$ remains at its peak value during relaxation. It is shown in Sect. 2.2 that $$F_\mathrm {v}= g_3 \phi C$$, as taken from Imole et al. [41], with $$g_3\cong 1.22$$ for the polydispersity used in the present work. Thus we conclude that the contact structure is almost unchanged and the network remains stable during relaxation, since during relaxation $$\phi$$ does not change.
3. (iii)
For small volume fractions, close to $$\phi _{c}$$, the system becomes strongly anisotropic in stress ratio $$\uptau /p$$, and fabric $$F_\mathrm {d}$$ rather quickly, during (slow) shear (envelope for low volume fractions in Fig. 5d, e), before it reaches the steady state [49].
4. (iv)
It is easy to obtain the critical (shear) jamming density $$\phi _{c}$$ from the relaxed critical (steady) state pressure $$p$$, and shear stress $$\uptau$$, by extrapolation to zero, as the envelope of relaxed data in Fig. 5a, c.
We use the same methodology presented in Eq. (3) to extract the critical jamming density $$\phi _{c}$$. When the relaxed $$p$$ is normalized with the contact density $$\phi C$$, we obtain $$\phi _{c}= 0.6567 \pm 0.0005$$ by linear extrapolation. A similar value of $$\phi _{c}$$ is obtained from the extrapolation of the relaxed $$\uptau$$ data set, and is consistent with other methods using the coordination number $$C^{*}$$, or the energy [112]. The quantification of history dependent jamming densities $$\phi _J(H)$$, due to shear complementing the slow changes by cyclic isotropic (over)compression in Eq. (4), is discussed next.
### 3.3 Jamming phase diagram with history H
We propose a jamming phase diagram with shear strain, and present a new, quantitative history dependent model that explains jamming and shear jamming, but also predicts that shear jamming vanishes under some conditions, namely when the system is not tapped, tempered or over-compressed before shear is applied. Using $${\varepsilon }_{d}$$ and $$\phi$$ as parameters, Fig. 6a shows that for one initial the history dependent jamming state at $${}^1\phi _{J,i}$$, there exist sheared states within the range $$\phi _{c}\le \phi \le \phi _J(H)$$, which are isotropically unjammed. After small shear strain they become fragile, and for larger shear strain jam and remain jammed, i.e., eventually showing the critical state flow regime [45, 46], where pressure, shear stress ratio and structural anisotropy have reached their saturation levels and forgotten their initial state (data not shown). The transition to fragile states is accompanied by partial percolation of the strong force network, while percolation in all directions indicates the shear jamming transition. Above jamming, the large fraction of non-rattlers provides a persistent mechanical stability to the structure, even after shear is stopped.
For $$\phi$$ approaching $$\phi _{c}$$, the required shear strain to jam $${\varepsilon }_{d}^{SJ}$$ increases, i.e., there exists a divergence “point” $$\phi _{c}$$, where ‘infinite’ shear strain might jam the system, but below which no shear jamming was observed. The closer the (constant) volume fraction $$\phi$$ is to the initial $${}^1\phi _{J,i}$$, the smaller is $${\varepsilon }_{d}^{SJ}$$. States with $$\phi \ge {}^1\phi _{J,i}$$ are isotropically jammed already before shear is applied.
Based on the study of many systems, prepared via isotropic over-compression to a wide range of volume fractions $$\phi ^{\mathrm {max}}_i \ge \phi _{c}$$, and subsequent shear deformation, Fig. 6b shows the strains required to jam these states by applying pure shear. A striking observation is that independent of the isotropic jamming density $${}^1\phi _{J,i}$$, all curves approach a unique critical jamming density at $$\phi _{c}\sim 0.6567$$ (see Sect. 3.2.2). When all the curves are scaled with their original isotropic jamming density $${}^M\phi _{J,i}$$ as $$\phi _{sc}= \left( \phi -\phi _{c}\right) /\left( {}^M\phi _{J,i}-\phi _{c}\right)$$ they collapse on a unique master curve
\begin{aligned} \left( {\varepsilon }_{d}^{SJ}/{\varepsilon }_{d}^{0}\right) ^{\alpha } ={-} \log {\phi _{sc}} = {-} \log {\left( \frac{\phi -\phi _{c}}{{}^M\phi _{J,i}-\phi _{c}}\right) }, \end{aligned}
(7)
shown in the inset of Fig. 6b, with power $$\alpha =1.37 \pm 0.01$$ and shear strain scale $${\varepsilon }_{d}^{0}=0.102 \pm 0.001$$ as the fit parameters. Hence, if the initial jamming density $${}^M\phi _{J,i}$$ or $$\phi _J(H)$$ is known based on the past history of the sample, the shear jamming strain $${\varepsilon }_{d}^{SJ}$$ can be predicted.
From the measured shear jamming strain, Eq. (7), knowing the initial and the limit value of $$\phi _J$$, we now postulate its evolution under isochoric pure shear strain:
\begin{aligned} \phi _J({\varepsilon }_{d}) = \phi _{c}+ \left( \phi - \phi _{c}\right) \exp \left[ \left( \frac{ \left( {\varepsilon }_{d}^{SJ}\right) ^\alpha - \left( {\varepsilon }_{d}\right) ^\alpha }{ \left( {\varepsilon }_{d}^{0}\right) ^\alpha }\right) \right] . \end{aligned}
(8)
Inserting, $${\varepsilon }_{d}=0$$, $${\varepsilon }_{d}={\varepsilon }_{d}^{SJ}$$ and $${\varepsilon }_{d}=\infty$$ leads to $$\phi _J= {}^M\phi _{J,i}$$, $$\phi _J= \phi$$ and $$\phi _J= \phi _{c}$$, respectively. The jamming density evolution due to shear strain $${\varepsilon }_{d}$$ is faster than exponential (since $$\alpha > 1$$) decreasing to its lower limit $$\phi _{c}$$. This is qualitatively different from the stretched exponential (slow) relaxation dynamics that leads to the increase of $$\phi _J$$ due to over-compression or tapping, see Fig. 7a for both cases.
## 4 Meso-scale stochastic slow dynamics model
The last challenge is to unify the observations in a qualitative model that accounts for the changes in the jamming densities for both isotropic and shear deformation modes. Over-compressing a soft granular assembly is analogous to small-amplitude tapping [21, 47, 104] of more rigid particles, in so far that both methods lead to more compact (efficient) packing structures, i.e., both representing more isotropic perturbations, rather than shear, which is deviatoric (anisotropic) in nature. These changes are shown in Fig. 2a, where the originally reported logarithmically slow dynamics for tapping [107, 108, 113] is very similar to our results that are also very slow, with a stretched exponential behavior; such slow relaxation dynamics can be explained by a simple Sinai-Diffusion model of random walkers in a random, hierarchical, fractal, free energy landscape [106, 114] in the (a-thermal) limit, where the landscape does not change—for the sake of simplicity.
The granular packing is represented in this picture by an ensemble of random walkers in (arbitrary) configuration space with (potential) energy according to the height of their position on the landscape. (Their average energy corresponds to the jamming density and a decrease in energy corresponds to an increase in $$\phi _J(H)$$, thus representing the “memory” and history dependence with protocol H.) Each change of the ensemble represents a rearrangement of packing and units in ensemble represent sub-systems. Perturbations, such as tapping with some small-amplitude (corresponding to “temperature”) allow the ensemble to find denser configurations, i.e., deeper valleys in the landscape, representing larger (jamming) densities [22, 82]. Similarly, over-compression is squeezing the ensemble “down-hill”, also leading to an increase of $$\phi _J$$, as presented in Fig. 7b. Larger amplitudes will allow the ensemble to overcome larger barriers and thus find even deeper valleys. Repetitions have a smaller chance to do so—since the easy reorganizations have been realized previously—which explains the slow dynamics in the hierarchical multiscale structure of the energy landscape.
In contrast to the isotropic perturbations, where the random walkers follow the “down-hill” trend, shear is anisotropic and thus pushing parts of the ensemble in “up-hill’ direction’. For example, under planar simple shear, one (eigen) direction is extensive (up-hill) whereas an other is compressive (down-hill). If the ensemble is random, shear will only re-shuffle the population. But if the material was previously forced or relaxed towards the (local) land-scape minima, shear can only lead to a net up-hill drift of the ensemble, i.e., to decreasing $$\phi _J$$, referred to as dilatancy under constant stress boundary conditions.
For ongoing over-compression, both coordination number and pressure slowly increase, as sketched in Fig. 8, while the jamming density drifts to larger values due to re-organization events that make the packing more effective, which moves the state-line to the right (also shown in Fig. 7a). For decompression, we assume that there are much less re-organization events happening, so that the pressure moves down on the state-line, until the system unjams. For ongoing perturbations, at constant volume, as tapping or a finite temperature, $$T_g$$, both coordination number and pressure slowly decrease (data not shown), whereas for fixed confining pressure the volume would decrease (compactancy, also not shown).
For ongoing shear, the coordination number, the pressure and the shear stress increase, since the jamming density decreases, as sketched in Fig. 9 until a steady state is reached. This process is driven by shear strain amplitude and is much faster than the relaxation dynamics. For large enough strain the system will be sufficiently re-shuffled, randomized, or “re-juvenated” such that it approaches its quenched, random state close to $$\phi _{c}$$ (see Fig. 7a).
If both mechanisms, relaxation by temperature, and continuous shear are occurring at the same time, one can reach another (non)-“equilibrium” steady state, where the jamming density remains constant, balancing the respective increasing and decreasing trends, as sketched in Fig. 9e.
## 5 Macroscopic constitutive model
In this section, we present the simplest model equations, as used for the predictions, involving a history dependent $$\phi _J(H)$$, as given by Eq. (4) for isotropic deformations and Eq. (8) for shear deformations. The only difference to Imole et al. [41], where these relations are taken from, based on purely isotropic unloading, is the variable $$\phi _J= \phi _J(H)$$.
### 5.1 Presentation and model calibration
#### 5.1.1 During cyclic isotropic deformation
During (cyclic) isotropic deformation, the evolution equation for the corrected coordination number $$C^{*}$$ is:
\begin{aligned} C^{*}= C_0 + C_1\left( \frac{\phi }{\phi _J(H)}-1 \right) ^\theta , \end{aligned}
(9)
with $$C_0=6$$ for the frictionless case and parameters $$C_1$$ and $$\theta$$ are presented in Table 1. The fraction of non-rattlers $$f_\mathrm {NR}$$ is given as:
\begin{aligned} f_\mathrm {NR}= 1 - \varphi _c\mathrm {exp}\left[ -\varphi _v \left( \frac{\phi }{\phi _J(H)}-1 \right) \right] , \end{aligned}
(10)
with parameters $$\varphi _c$$ and $$\varphi _v$$ presented in Table 1. We modify Eq. (3) for the evolution of $$p$$ together with the history dependent $$\phi _J=\phi _J(H)$$ so that,
\begin{aligned} p=\frac{\phi C}{\phi _J(H)}p_0 (-\varepsilon _\mathrm {v}) \left[ 1-\gamma _p (-\varepsilon _\mathrm {v}) \right] , \end{aligned}
(11)
with parameters $$p_0$$ and $$\gamma _p$$ presented in Table 1, and the true or logarithmic volume change of the system is $$-\varepsilon _\mathrm {v} = \log (\phi /\phi _J(H))$$, relative to the momentary jamming density. The non-corrected coordination number is $$C = C^{*}f_\mathrm {NR}$$, as can be computed using Eqs. (9) and (10). Also the parameters $$C_1$$, $$\theta$$ for $$C^{*}$$, $$\varphi _c$$, $$\varphi _v$$ for $$f_\mathrm {NR}$$, and $$p_0$$, $$\gamma _p$$ for pressure p are similar to Imole et al. [41], with the second order correction parameter $$\gamma _p$$ most sensitive to the details of previous deformations; however, not being very relevant since it is always a small correction due to the product $$\gamma _p (-\varepsilon _\mathrm {v})$$.
The above relations are used to predict the behavior of the isotropic quantities: dimensionless pressure $$p$$ and coordination number $$C^{*}$$, during cyclic isotropic compression, as well as for the fraction of non-rattlers for cyclic shear, with corresponding parameters presented in Table 1. Note that during isotropic deformation, $$\phi _J(H)$$ was changed only during the compression branch, using Eq. (4) for fixed $$M=1$$ using $$\phi ^{\mathrm {max}}_i$$ as variable, but is kept constant during unloading/expansion.
The above relations are used to predict the behavior of the isotropic quantities: dimensionless pressure $$p$$ and coordination number $$C^{*}$$, by only adding the history dependent jamming density $$\phi _J(H)$$ to the constitutive model, as tested below in Sect. 5.2.
#### 5.1.2 Cyclic (pure) shear deformation
During cyclic (pure) shear deformation, a simplified equation for the shear stress ratio $$\uptau /p$$ is taken from Imole et al. [41], where the full model was introduced as rate-type evolution equations, and further calibrated and tested by Kumar et al. [69]:
\begin{aligned} \uptau /p= {\left( \uptau /p\right) }^{\mathrm {max}} - \left[ {\left( \uptau /p\right) }^{\mathrm {max}} - \left( \uptau /p\right) ^{\mathrm {0}} \right] \exp \left[ -\beta _s {\varepsilon }_{d}\right] ,\nonumber \\ \end{aligned}
(12)
with $$\left( \uptau /p\right) ^{\mathrm {0}}$$ and $$\left( \uptau /p\right) ^{\mathrm {max}}$$ the initial and maximum (saturation) shear stress ratio, respectively, and $$\beta _s$$ its growth rate.5 Similarly, a simplified equation for the deviatoric fabric $$F_\mathrm {d}$$ can be taken from Refs. [41, 69] as:
\begin{aligned} F_\mathrm {d}= {F_\mathrm {d}}^{\mathrm {max}} - \left[ {F_\mathrm {d}}^{\mathrm {max}} - {F_\mathrm {d}}^{\mathrm {0}} \right] \exp \left[ -\beta _F {\varepsilon }_{d}\right] , \end{aligned}
(13)
with $${F_\mathrm {d}}^{\mathrm {0}}$$ and $${F_\mathrm {d}}^{\mathrm {max}}$$ the initial and maximum (saturation) values of the deviatoric fabric, respectively, and $$\beta _F$$ its growth rate. The four parameters $$\left( \uptau /p\right) ^{\mathrm {max}}$$, $$\beta _s$$ for $$\uptau /p$$ and $${F_\mathrm {d}}^{\mathrm {max}}$$, $$\beta _F$$ for $$F_\mathrm {d}$$ are dependent on the volume fraction $$\phi$$ and are well described by the general relation from Imole et al. [41] as:
\begin{aligned} Q = Q_a + Q_c \exp \left[ -\varPsi \left( \frac{\phi }{\phi _J(H)}-1 \right) \right] , \end{aligned}
(14)
where $$Q_a$$, $$Q_c$$ and $$\varPsi$$ are the fitting constants with values presented in Table 2.
For predictions during cyclic shear deformation, $$\phi _J(H)$$ was changed with applied shear strain $${\varepsilon }_{d}$$ using Eq. (8). Furthermore, the jamming density is set to a larger value just after strain-reversal, as discussed next.
Table 2
Parameters for Eqs. (12) and (13) using Eq. (14), with slightly different values than from Imole et al. [41], that are extracted using the similar procedure as in Imole et al. [41], for states with volume fraction close to the jamming volume fraction
Evolution parameters
$$Q_{a}$$
$$Q_{c}$$
$$\varPsi$$
$$\left( \uptau /p\right) ^{\mathrm {max}}$$
0.12
0.091
7.9
$$\beta _s$$
30
40
16
$${F_\mathrm {d}}^{\mathrm {max}}$$
0
0.17
5.3
$$\beta _F$$
0
40
5.3
#### 5.1.3 Behavior of the jamming density at strain reversal
As mentioned in Sect. 3.2, there are some states below $$\phi _J$$, where application of shear strain jams the systems. The densest of those can resist shear reversal, but below a certain $$\phi _\mathrm {cr}\approx 0.662<\phi _J$$, shear reversal unjams the system again [116]. With this information, we postulate the following:
1. (i)
After the first phase, for large strain pure shear, the system should forget where it was isotropically compressed to before i.e., $${}^M\phi _{J,i}$$ is forgotten and $$\phi _J=\phi _{c}$$ is realized.
2. (ii)
There exists a volume fraction $$\phi _\mathrm {cr}$$, above which the systems can just resist shear reversal and remain always jammed in both forward and reverse shear.
3. (iii)
Below this $$\phi _\mathrm {cr}$$, reversal unjams the system. Therefore, more strain is needed to jam the system (when compared to the initial loading), first to forget its state before reversal, and then to re-jam it in opposite (perpendicular) shear direction. Hence, the strain necessary to jam in reversal direction should be higher than for the first shear cycle.
4. (iv)
As we approach $$\phi _{c}$$, the reverse strain needed to jam the system increases.
We use these ideas and measure the reversal shear strain $${\varepsilon }_{d}^{SJ,R}$$, needed to re-jam the states below $$\phi _\mathrm {cr}$$, as shown in Fig. 10. When they are scaled with $$\phi _\mathrm {cr}$$ as $$\phi _{sc}= \left( \phi -\phi _{c}\right) /\left( \phi _\mathrm {cr}-\phi _{c}\right)$$, they collapse on a unique master curve, very similar to Eq. (7):
\begin{aligned} \left( {\varepsilon }_{d}^{SJ,R}/{\varepsilon }_{d}^{0,R}\right) ^{\alpha } ={-} \log {\phi _{sc}} = {-} \log {\left( \frac{\phi -\phi _{c}}{\phi _\mathrm {cr}-\phi _{c}}\right) }, \end{aligned}
(15)
shown in the inset of Fig. 6b, with the same power $$\alpha =1.37 \pm 0.01$$ as Eq. (7). Fit parameter strain scale $${\varepsilon }_{d}^{0,R}=0.17 \pm 0.002 > {\varepsilon }_{d}^{0} = 0.102$$, is consistent with the above postulates (iii) and (iv).
The above relations are used to predict the isotropic and the deviatoric quantities, during cyclic shear deformation, as described next, with the additional rule that all the quantities attain value zero for $$\phi \le \phi _J(H)$$. Moreover, for any state with $$\phi \le \phi _\mathrm {cr}$$, shear strain reversal moves the jamming density to $$\phi _\mathrm {cr}$$, and the evolution of the jamming density follows Eq. (15).
Any other deformation mode, can be written as a unique superposition of pure isotropic and pure and axial shear deformation modes [117]. Hence the combination of the above can be easily used to describe any general deformation, e.g. uniaxial cyclic compression (data not presented) where the axial strain can be decomposed in two plane strain modes.
### 5.2 Prediction: minimal model
Finally, we test the proposed history dependent jamming density $$\phi _J(H)$$ model, by predicting $$p$$ and $$C^{*}$$, when a granular assembly is subjected to cyclic isotropic compression to $$\phi ^{\mathrm {max}}_i = 0.73$$ for $$M=1$$ and for $$M=300$$ cycles, with $${}^{\infty }\phi _{J,i}=0.667$$, as shown in Fig. 11a, b. It is observed that using the history dependence of $$\phi _J(H)$$, the hysteretic behavior of the isotropic quantities, $$p$$ and $$C^{*}$$, is very well predicted, qualitatively similar to isotropic compression and decompression of real 2D frictional granular assemblies, as shown in by Bandi et al. [58] and Reichhardt and Reichhardt [22].
In Fig. 11c, we show the evolution of the deviatoric quantities shear stress ratio $$\uptau /p$$ and deviatoric fabric $$F_\mathrm {d}$$, when a system with $$\phi =0.6584$$, close to $$\phi _{c}$$, and initial jamming density $$\phi _J(0)= 0.6652$$, is subjected to three shear cycles (lowest panel). The shear stress ratio $$\uptau /p$$ is initially undefined, but soon establishes a maximum (not shown) and decays to its saturation level at large strain. After strain reversal, $$\uptau /p$$ drops suddenly and attains the same saturation value, for each half-cycle, only with alternating sign. The behavior of the anisotropic fabric $$F_\mathrm {d}$$ is similar to that of $$\uptau /p$$. During the first loading cycle, the system is unjammed for some strain, and hence $$F_\mathrm {d}$$ is zero in the model (observations in simulations can be non-zero, when the data correspond to only few contacts, mostly coming from rattlers). However, the growth/decay rate and the saturation values attained are different from those of $$\uptau /p$$, implying a different, independent stress- and structure-evolution with strain—which is at the basis of recently proposed anisotropic constitutive models for quasi-static granular flow under various deformation modes [41]. The simple model with $$\phi _J(H)$$, is able to predict quantitatively the behavior the $$\uptau /p$$ and $$F_\mathrm {d}$$ after the first loading path, and is qualitatively close to the cyclic shear behavior of real 2D frictional granular assemblies, as shown in Supplementary Fig. 7 by Bi et al. [7].
At the same time, also the isotropic quantities are very well predicted by the model, using the simple equations from Sect. 5.1, where only the jamming density is varying with shear strain, while all material parameters are kept constant. Some arbitrariness involves the sudden changes of $$\phi _J$$ at reversal, as discussed in Sect. 5.1. Therefore, using a history dependent $$\phi _J(H)$$ gives hope to understand the hysteretic observations from realistic granular assemblies, and also provides a simple explanation of shear jamming. Modifications of continuum models like anisotropic models [41, 69], or GSH type models [85, 86], by including a variable $$\phi _J$$, can this way quantitatively explain various mechanisms around jamming.
## 6 Towards experimental validation
The purpose of this section is two-fold: First, we propose ways to (indirectly) measure the jamming density, since it is a virtual quantity that is hard to measure directly, just as the “virtual, stress-free reference state” in continuum mechanics which it resembles. Second, this way, we will introduce alternative state-variables, since by no means is the jamming density the only possibility.
Measuring $$\phi _J$$ from experiments Here we show the procedure to extract the history dependent jamming density $$\phi _J(H)$$ from measurable quantities, indirectly obtained via Eqs. (9), (10), (11), and directly from Eq. (8). There are two reasons to do so: (i) the jamming density $$\phi _J(H)$$ is only accessible in the unloading limit $$p\rightarrow 0$$, which requires an experiment or a simulation to “measure” it (however, during this measurement, it might change again); (ii) deducing the jamming density from other quantities that are defined for an instantaneous snapshot/configuration for $$p > 0$$ allows to indirectly obtain it—if, as shown next, these indirect “measurements” are compatible/consistent: Showing the equivalence of all the different $$\phi _J(H)$$, proofs the consistency and completeness of the model and, even more important, provides a way to obtain $$\phi _J(H)$$ indirectly from experimentally accessible quantities.
For isotropic compression Figure 12 shows the evolution of $$\phi _J(H)$$, measured from the two experimentally accessible quantities: coordination number $$C^{*}$$ and pressure p, using Eqs. (9) and (11) respectively for isotropic over-compression to $$\phi ^{\mathrm {max}}_i=0.82$$ over two cycles. Following observations can be made: (i) $$\phi _J$$ for isotropic loading and unloading can be extracted from $$C^{*}$$ and p, (ii) it rapidly increases and then saturates during loading, (iii) it mimics the fractal energy landscape model in Fig. 4 from Luding et al. [114] very well, (iv) while is was assumed not to change for unloading, it even increases, which we attribute to the perturbations and fluctuations (granular temperature) induced during the quasi-static deformations, (v) the indirect $$\phi _J$$ are reproducible and follow the same master-curve for first over-compression as seen in Figs. 12, independent of the maximum—all following deformation is dependent on the previous maximum density.
For shear deformation Figure 13 shows the evolution of $$\phi _J(H)$$, measured from the two experimentally accessible quantities: coordination number $$C^{*}$$ and pressure p, using Eqs. (9) and (11) respectively during volume conserving shear with $$\phi =0.66$$, and the initial jamming density $$\phi _J\left( \phi ^{\mathrm {max}}_i = 0.82, M=1\right) =: {}^1\phi _{J,i}= 0.6652 > \phi$$ and shows good agreement with the theoretical predictions using Eq. (8) after shear jamming. Thus the indirect measurements of $$\phi _J(H)$$ can be applied if $$\phi _J(H)<\phi$$; the result deduced from pressure fits the best, i.e., it interpolates the two others and is smoother.
## 7 Summary, discussion and outlook
In summary, this study presents a quantitative, predictive macroscopic constitutive model that unifies a variety of phenomena around and above jamming, for quasi-static deformation modes. The most important ingredient is a scalar state-variable that characterizes the packing “efficiency” and responds very slowly to (isotropic, perturbative) deformation. In contrast, it responds fast, saturating exponentially with finite shear deformation. This different response to the two fundamentally different modes of deformation (isotropic or deviatoric, shear) is (qualitatively) explained by a stochastic (meso-scale) model with fractal (multiscale) character. All simulation results considered here are quantitatively matched by the macroscopic model after including both the isotropic and the anisotropic microstructure as state-variables. Discussing the equivalence of alternative state-variables and ways to experimentally measure the model parameters concludes the study and paves the way to apply the model to other, more realistic materials. The following subsections wrap up some major aspects of this study and also add some partly speculative arguments about the wider consequences of our results for rheology as well as an outlook.
The questions posed in the introduction can now be answered: (i) The transition between the jammed and flowing (unjammed) regimes is controlled by a single new, isotropic, history dependent state-variable, the jamming density $$\phi _J(H)$$ (with history H as shorthand place-holder for any deformation path), which (ii) has a unique lower critical jamming density $$\phi _{c}$$ when $$p \rightarrow 0$$, reached after long shear without temperature $$T_g$$, so that (iii) the history (protocol dependence) of jamming is completely encompassed by this new state-variable, and (iv) jamming, unjamming and shear jamming can all occur in 3D without any friction, only by reorganizations of the micro-structure.
### 7.2 Lower limit of jamming
The multiscale model framework implies now a minimum $$\phi _{c}$$ that represents the (critical) steady state for a given sample in the limit of vanishing confining stress, i.e., the lower limit of all jamming densities. This is nothing but the mean lowest stable random density a sheared system “locally” can reach due to continuously ongoing shear, in the limit of vanishing confining stress.
This lower limit is difficult to access in experiments and simulations, since every shear also perturbs the system leading at the same time to (slow) relaxation and thus a competing increase in $$\phi _J(H)$$. However, it can be obtained from the (relaxed) steady state values of pressure, extrapolated to zero, i.e., from the envelope of pressure in Fig. 5. Note that either fluctuations, special deformation modes or careful preparation procedures e.g. energy minimization techniques or manual construction [9, 23] may lead to jammed states at even lower density than $$\phi _{c}$$, from which starting to shear would lead to an increase of the jamming density (a mechanism which we could not clearly identify from our frictionless simulations due to very long relaxation times near jamming for soft particles). This suggests future studies in the presence of friction so that one has a wider range of jamming densities and lower density states will be much more stable as compared to the frictionless systems. In this work, we focused on fixed particle size polydispersity with uniform size distribution. We expect the effects of polydispersity [44] will have similar order of explorable jamming range as in this work, whereas friction etc. will cause larger explorable jamming ranges [92] and bigger changes in the calibrated parameters.
### 7.3 Shear jamming as consequence of a varying $$\phi _J(H)$$
Given an extremely simple model picture, starting from an isotropically unjammed system that was previously compressed or tapped (tempered), shear jamming is not anymore a new effect, but is just due to the shift of the state-variable (jamming density) to lower values during shear. In other words, shear jamming occurs when the state-variable $$\phi _J(H)$$ drops below the density $$\phi$$ of the system.
Even though dilatancy is that what is typically expected under shear (of a consolidated packing), also compactancy is observed in some cases [41] and can be readily explained by our model. Given a certain preparation protocol, typically a jamming density $$\phi _J> \phi _{c}$$ will be established for a sample, since the critical limit $$\phi _{c}$$ is very difficult to reach. When next a shear deformation is applied, it depends e.g. on the strain rate whether dilatancy or compaction will be observed: if the shear mode is “slower” than the preparation, or if $$\phi _J> \phi _{c}$$, dilatancy is expected as a consequence of the rapidly decreasing $$\phi _J$$ of the sample. In contrast, for a relatively “fast”, violent shear test (relative to the previous preparation and possibly relaxation procedure), compactancy also can be the result, due to an increase of $$\phi _J$$ during shear.
### 7.4 Rheology
The multiscale models presented in this study, based on data from frictionless particle simulations, imply that a superposition of the two fundamental deformation modes (isotropic and deviatoric, i.e. plane strain pure shear) is possible or, with other words, that the respective system responses are mostly decoupled as shown for the non-Newtonian rheology of simple fluids in Ref. [117]. Even though this decoupling is mostly consistent with our present data (the responses to isotropic and deviatoric deformations are mostly independent and can be measured independently), this separation and superposition cannot be taken for granted for more realistic granular and powder systems.
Nevertheless, the meso-scale model presented here, as based on a multi-scale energy landscape, explains compactancy and dilatancy, at constant confining stress, as caused by an increasing jamming density, or a decreasing jamming density, respectively (not shown). Similarly, at constant volume, the pressure either decreases or increases (pressure-dilatancy) due to an increasing or decreasing jamming density, respectively.
The model also allows to explain other rheological phenomena as shear-thinning (e.g., due to an increasing jamming density, at constant volume) or shear-thickening (e.g., due to a decreasing jamming density, at constant volume). As generalization of the present work, also the (granular) temperature (fluctuations of kinetic energy) can be considered, setting an additional (relaxation) time-scale, which effects the interplay between (shear) strain-rate and the evolution of the jamming density, so that even in a presumed “quasi-static” regime interesting new phenomena can be observed and explained.
### 7.5 Towards experimental validation
The history dependent jamming density $$\phi _J(H)$$ is difficult to access directly, but can consistently be extracted from other, experimentally measurable quantities, e.g. pressure p, coordination number $$C^{*}$$ or fraction of non-rattlers $$f_\mathrm {NR}$$. We explain the methodology to extract $$\phi _J(H)$$ experimentally, and confirm by indirect measurement, as detailed in Sect. 6, that the jamming density is indeed increasing during isotropic deformation and decreasing during shear, consistently also when deduced from these other quantities.
With other words, we do not claim that the jamming density is the only choice for the new state-variable that is needed. It can be replaced by any other isotropic quantity as, e.g. the isotropic fabric, the fraction of non-rattlers, the coordination number, or an empirical stress-free state that is extrapolated from pressure (which can be measured most easily), as long as this variable characterizes the packing “efficiency”.
Since an increased packing efficiency could be due to ordering (crystallization), we tried to, but could not trace any considerable crystallization and definitely no phase-separation. We attribute this to the polydispersity of the sizes of the particles used being in the range to avoid ordering effects, as studied in detail in Ref. [118]. Quantities like the coordination number, which can tremendously increase due to crystallization, did not display significant deviation from the random packing values and, actually, it even decreases in the unloading phases, relative to the initial loading phase, see Fig. 11b. This is not a proof that there is no crystallization going on, it is just not strong enough to be clearly seen. The reasons and micro-structural origin of the increased packing efficiency, as quantified by the new state-variable, are subject of ongoing research.
### 7.6 Outlook
Experiments should be performed to calibrate our model for suspended soft spheres (e.g. gels, almost frictionless) and real, frictional materials [119, 120, 121]. Over-compression is possible for soft materials, but not expected to lead to considerable relaxation due to the small possible compressive strain for harder materials. However, tapping or small-amplitude shear can take the role of over-compression, also leading to perturbations and increasing $$\phi _J$$, in contrast, large-amplitude shear leads to decreasing $$\phi _J$$ and can be calibrated indirectly from different isotropic quantities. Note that the accessible range of $$\phi _J- \phi _{c}$$ is expected to much increase for more realistic systems, e.g., with friction, for non-spherical particle shapes, or for cohesive powders.
From the theoretical side, a measurement of the multiscale energy landscape, e.g. the valley width, depth/shapes and their probabilities [81] should be done to verify our model-picture, as it remains qualitative so far. Finally, applying our model to glassy dynamics, ageing and re-juvenation, and frequency dependent responses, encompassing also stretched exponential relaxation, see e.g. Lieou and Langer [122], is another open challenge for future research. All this involves the temperature as a source of perturbations that affect the jamming density, and will thus also allow to understand more dynamic granular systems where the granular temperature is finite and not negligible as implied in most of this study for the sake of simplicity. A more complete theory for soft and granular matter, which involves also the (granular) temperature, is in preparation.
Last, but not least, while the macro/continuum model predicts a smooth evolution of the state variables, finite-size systems display (system-size dependent) fluctuations that only can be explained by a meso-scale stochastic model as proposed above, with particular statistics as predicted already by rather simple models in Refs. [28, 123, 124].
## Footnotes
1. 1.
Tapping or compression may not be technically equivalent to the protocol isotropic compression. In soil mechanics, the process of tapping may involve anisotropic compression or shear. The process of compression may be either isotropic or anisotropic or even involving shear. For example, a typical soil tests may include biaxial compression, conventional triaxial compression and true triaxial compression. In this work, in the context of compression, we always mean true isotropic in strain. In the context of tapping, we assume that the granular temperature, which is often assumed isotropic, does the work, even though the tapping process is normally not isotropic. So this is an oversimplification, and subject to future study since it was not detailed here.
2. 2.
This deformation mode represents the only fundamental deviatoric deformation motion (complementary to isotropic deformation), since axial strain can be superposed by two plane-strain modes, and because the plane-strain mode allows to study the non-Newtonian out-of-shear-plane response of the system (pressure dilatancy), whereas the axial mode does not. If superposition is allowed, as it seems to be the case for frictionless particles, studying only these two modes is minimal effort, however, we cannot directly extrapolate to more realistic materials.
3. 3.
For the isotropic deformation tests, we move the (virtual) walls and for the shear tests, we move all the grains according to an affine motion compatible with the (virtual) wall motion. For the case where only the (virtual) walls move some arching near the corners can be seen when there is a huge particle size dispersity or if there is a considerable particle friction (data not shown). For the small polydispersity and the frictionless spheres considered in this work, the system is and remains homogeneous and the macroscopic quantities are indistinguishable between the two methods, however, this must not be taken for granted in the presence of friction or cohesion, where wall motions other than by imposed homogeneous strain, can lead to undesired inhomogeneities in the periodic representative volume element.
4. 4.
The grains are soft and overlap $$\delta$$ increases with increasing compression ($$\phi$$). For a linear contact model, it has been shown in Refs. [100, 101] that $$\langle \delta \rangle /\langle r \rangle \propto \mathrm {ln}\left( \phi /\phi _J\right) = -\varepsilon _\mathrm {v}$$ (volumetric strain).
5. 5.
Note that the model in the form used here is ignoring the presence of kinetic energy fluctuations, referred to as granular temperature $$T_g$$, or fields like the so-called fluidity [90, 91, 115], that introduce an additional relaxation time-scale, as is subject of ongoing studies.
## Notes
### Acknowledgments
We thank Robert Behringer, Karin Dahmen, Itai Einav, Ken Kamrin, Mario Liu, Vitaliy Ogarko, Corey O’Hern, and Matthias Sperl, for valuable scientific discussions; critical comments and reviews from Vanessa Magnanimo and Olukayode Imole are gratefully acknowledged. This work was financially supported by the European Union funded Marie Curie Initial Training Network, FP7 (ITN-238577), PARDEM (www.pardem.eu) and the NWO STW-VICI project 10828.
### Conflict of interest
The authors declare no conflict of interest.
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Dahmen, K.A., Ben-Zion, Y., Uhl, J.T.: A simple analytic theory for the statistics of avalanches in sheared granular materials. Nat. Phys. 7, 554–557 (2011) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9174265265464783, "perplexity": 3341.842452456835}, "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/1570986661296.12/warc/CC-MAIN-20191016014439-20191016041939-00007.warc.gz"} |
http://www.physicsforums.com/showthread.php?t=441187 | This isn't really a homework problem. In lab we had an unknown buffer solution and we had to to titrate it with NaOH and HCl to try to identify what the buffer is. In the end I got my pka value equal to 4.52 I can't identify the buffer. My Ka is 3E-5 and I can't match any Ka values to this. Did I just totally screw up my lab??
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Admin Is there a list of buffers that you have to select from? Or can it be anything?
Nope, we weren't given a list to choose from so I'm guessing it could be anything?
Probably. Although some compounds are much more likely than others.
How did you get your pKa?
I required 19.92 mL of NaOH to neutralize a 10 mL portion of my buffer. I did the same thing using another 10 mL portion of the buffer but this time with HCl and it took me 17 mL. I then used the henderson-hasselbalch equation to find my pKa. I measured my pH of my buffer with a pH meter and got it to be 4.40. I had already standardized my acid and base previously and got 0.1472 M NaOH and 0.1325 M HCl. Also I made sure to change the ml to L for the following calculations. For the concentration of my base and acid I did this: [A] = (19.92 ml NaOH x 0.1472 M NaOH)/10 mL = 0.293 [B] = (17ml HCl x 0.1325 M HCl)/10 ml =0.225 4.40= pka + log [0.225]/[0.293] pka= 4.52
Posted something before thinking it was right... still wrong.
Quote by kooombaya I required 19.92 mL of NaOH to neutralize a 10 mL portion of my buffer. I did the same thing using another 10 mL portion of the buffer but this time with HCl and it took me 17 mL.
What do you mean by "neutralize the buffer"?
Quote by Borek What do you mean by "neutralize the buffer"?
This was how the question was stated in the lab book.
I found out what I did wrong by the way. Thanks for your time.
Quote by kooombaya This was how the question was stated in the lab book.
Can you explain what they meant? I have never seen something like that, even if it is wrong, I have nothing against knowing.
Quote by Borek Can you explain what they meant? I have never seen something like that, even if it is wrong, I have nothing against knowing.
From what I understood it goes something like this:
Say we added 10 mL of 0.1 M NaOH. Then this is the amount of acid in the buffer solution that reacted with the NaOH to reach a new equivalence point.
It's the same for HCl, except the HCl acts with the base in the buffer solution to reach a new equivalence point.
Admin So it was just shifting pH of the buffer by addition of strong acid or base. There was an acid base reaction involved (which can be technically called neutralization), but it didn't end with neutral solution. Not the best wording if you ask me. -- buffer calculator, concentration calculator pH calculator, stoichiometry calculator
Quote by Borek So it was just shifting pH of the buffer by addition of strong acid or base. There was an acid base reaction involved (which can be technically called neutralization), but it didn't end with neutral solution. Not the best wording if you ask me.
Yup exactly. Thanks again. | {"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": 0.808948814868927, "perplexity": 1703.5455868245642}, "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/1368702718570/warc/CC-MAIN-20130516111158-00067-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://askmetips.com/standard-error/standard-error-for-sample-mean.php | Home > Standard Error > Standard Error For Sample Mean
# Standard Error For Sample Mean
## Contents
And it turns out, there is. So, in the trial we just did, my wacky distribution had a standard deviation of 9.3. If we magically knew the distribution, there's some true variance here. With statistics, I'm always struggling whether I should be formal in giving you rigorous proofs, but I've come to the conclusion that it's more important to get the working knowledge first my review here
You're just very unlikely to be far away if you took 100 trials as opposed to taking five. When the sampling fraction is large (approximately at 5% or more) in an enumerative study, the estimate of the standard error must be corrected by multiplying by a "finite population correction"[9] The standard deviation is computed solely from sample attributes. Required fields are marked *Comment Name * Email * Website Find an article Search Feel like "cheating" at Statistics?
## Standard Error Of Mean Calculator
The proportion or the mean is calculated using the sample. Popular Articles 1. To understand this, first we need to understand why a sampling distribution is required.
So if this up here has a variance of-- let's say this up here has a variance of 20. Now, if I do that 10,000 times, what do I get? Now, to show that this is the variance of our sampling distribution of our sample mean, we'll write it right here. Standard Error Regression Step 6: Take the square root of the number you found in Step 5.
Because of random variation in sampling, the proportion or mean calculated using the sample will usually differ from the true proportion or mean in the entire population. Standard Error Of Sample Mean Formula So that's my new distribution. So I think you know that, in some way, it should be inversely proportional to n. http://vassarstats.net/dist2.html But let's say we eventually-- all of our samples, we get a lot of averages that are there.
Standard Error of Sample Estimates Sadly, the values of population parameters are often unknown, making it impossible to compute the standard deviation of a statistic. Standard Error Of The Mean Excel With n = 2 the underestimate is about 25%, but for n = 6 the underestimate is only 5%. Then you do it again, and you do another trial. The standard error can be computed from a knowledge of sample attributes - sample size and sample statistics.
## Standard Error Of Sample Mean Formula
The distribution of these 20,000 sample means indicate how far the mean of a sample may be from the true population mean. https://www.khanacademy.org/math/statistics-probability/sampling-distributions-library/sample-means/v/standard-error-of-the-mean The standard error is computed from known sample statistics. Standard Error Of Mean Calculator And if it confuses you, let me know. Standard Error Vs Standard Deviation Because the 5,534 women are the entire population, 23.44 years is the population mean, μ {\displaystyle \mu } , and 3.56 years is the population standard deviation, σ {\displaystyle \sigma }
And then you now also understand how to get to the standard error of the mean.Sampling distribution of the sample mean 2Sampling distribution example problemUp NextSampling distribution example problem menuMinitab® 17 SupportWhat is http://askmetips.com/standard-error/standard-error-of-the-sample.php Then the mean here is also going to be 5. This is your standard deviation. √(68.175) = 8.257 Step 6: Divide the number you calculated in Step 6 by the square root of the sample size (in this sample problem, the Sampling from a distribution with a large standard deviation The first data set consists of the ages of 9,732 women who completed the 2012 Cherry Blossom run, a 10-mile race held Standard Error Of The Mean Definition
I don't necessarily believe you. Step 1:Add up all of the numbers: 12 + 13 + 14 + 16 + 17 + 40 + 43 + 55 + 56 + 67 + 78 + 78 + And we've seen from the last video that, one, if-- let's say we were to do it again. http://askmetips.com/standard-error/standard-deviation-standard-error-sample-size.php Expected Value 9.
The mean age was 23.44 years. Standard Error Mean parameters) and with standard errors you use data from your sample. Working...
## Standard error of mean versus standard deviation In scientific and technical literature, experimental data are often summarized either using the mean and standard deviation or the mean with the standard error.
The standard deviation of the age was 3.56 years. Sample question: If a random sample of size 19 is drawn from a population distribution with standard deviation α = 20 then what will be the variance of the sampling distribution In fact, data organizations often set reliability standards that their data must reach before publication. Standard Error Of Proportion The standard error is important because it is used to compute other measures, like confidence intervals and margins of error.
jbstatistics 82,575 views 7:25 standard error.wmv - Duration: 3:27. Step 2: Divide the variance by the number of items in the sample. It's one of those magical things about mathematics. useful reference And you do it over and over again.
For the runners, the population mean age is 33.87, and the population standard deviation is 9.27. You're becoming more normal, and your standard deviation is getting smaller. Because the 9,732 runners are the entire population, 33.88 years is the population mean, μ {\displaystyle \mu } , and 9.27 years is the population standard deviation, σ. Back to Top How to Find the Sample Mean Watch the video or read the steps below: How to Find the Sample Mean: Overview Dividing the sum by the number of
Use the standard error of the mean to determine how precisely the mean of the sample estimates the population mean. Test Your Understanding Problem 1 Which of the following statements is true. | {"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": 0.9593684673309326, "perplexity": 476.3247809015298}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676590051.20/warc/CC-MAIN-20180718041450-20180718061450-00108.warc.gz"} |
https://shapeout.readthedocs.io/en/0.9.6/sec_qg_mixed_effects.html | # Comparing datasets with LMM¶
Consider the following datasets. A treatment is applied three times at different time points. For each treatment, a control measurement is performed. For each measurement day, a reservoir measurement is performed additionally for treatment and control.
• Day1:
• one sample, called “Treatment I”, measured at flow rates of 0.04, 0.08 and 0.12 µl/s and one measurement in the reservoir
• one control, called “Control I”, measured at flow rates 0.04, 0.08 and 0.12 µl/s and one measurement in the reservoir
• Day2:
• two samples, called “Treatment II” and “Treatment III”, measured at flow rates 0.04, 0.08 and 0.12 µl/s and one measurement in the reservoir
• two controls, called “Control II” and “Control III”, measured at flow rates 0.04, 0.08 and 0.12 µl/s and one measurement in the reservoir
Linear mixed models (LMM) allow to assign a significance to a treatment (fixed effect) while considering the systematic bias in-between the measurement repetitions (random effect).
We will assume that the datasets are loaded into Shape-Out and that invalid events have been filtered (see e.g. Excluding invalid events). The Analyze configuration tab enables the comparison of an experiment (control and treatment) and repetitions of the experiment using LMM [HKP+17], [HMMO18].
• Basic analysis:
Assign which measurement is a control and which is a treatment by choosing the option in the dropdown lists under Interpretation. Group the pairs of control and treatment done in one experiment, by choosing an index number, called Repetition. Here, Treatment I and Control I are one experiment – called Repetition 1, Treatment II and Control II are a repetition of the experiment – called Repetition 2, Treatment III and Control III are another repetition of the experiment – called Repetition 3.
Press Apply to start the calculations. A text file will open to show the results.
The most important numbers are:
• Fixed effects:
(Intercept)-Estimate
The mean of the parameter chosen for all controls.
treatment-Estimate
The effect size of the parameter chosen between the mean of all controls and the mean of all treatments.
• Full coefficient table: Shows the effect size of the parameter chosen between control and treatment for every single experiment.
• Model-Pr(>Chisq): Shows the p-value and the significance of the test.
• Differential feature analysis:
The LMM analysis is only applicable if the respective measurements show little difference in the reservoir for the feature chosen. For instance, if a treatment results in non-spherical cells in the reservoir, then the deformation recorded for the treatment might be biased towards higher values. In this case, the information of the reservoir measurement has to be included by means of the differential deformation [HMMO18]. This can be achieved by selecting the respective reservoir measurements in the dropdown menu. | {"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": 0.9380010962486267, "perplexity": 2354.8401471480947}, "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/1575540514893.41/warc/CC-MAIN-20191208202454-20191208230454-00308.warc.gz"} |
https://math.libretexts.org/TextMaps/Precalculus/Map%3A_Precalculus_(Stitz-Zeager)/10%3A_Foundations_of_Trigonometry/10.4%3A_Trigonometric_Identities | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 10.4: Trigonometric Identities
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
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In Section \ref{CircularFunctions}, we saw the utility of the Pythagorean Identities in Theorem \ref{pythids} along with the Quotient and Reciprocal Identities in Theorem \ref{recipquotid}. Not only did these identities help us compute the values of the circular functions for angles, they were also useful in simplifying expressions involving the circular functions. In this section, we introduce several collections of identities which have uses in this course and beyond. Our first set of identities is the Even / Odd' identities.\footnote{As mentioned at the end of Section \ref{TheUnitCircle}, properties of the circular functions when thought of as functions of angles in radian measure hold equally well if we view these functions as functions of real numbers. Not surprisingly, the Even / Odd properties of the circular functions are so named because they identify cosine and secant as even functions, while the remaining four circular functions are odd. (See Section \ref{GraphsofFunctions}.)}
Note: Even / Odd Identities
For all applicable angles $$\theta$$:
• $$\cos(-\theta) = \cos(\theta)$$
• $$\sec(-\theta) = \sec(\theta)$$
• $$\sin(-\theta) = -\sin(\theta)$$
• $$\csc(-\theta) = -\csc(\theta)$$
• $$\tan(-\theta) = -\tan(\theta)$$
• $$\cot(-\theta) = -\cot(\theta)$$
In light of the Quotient and Reciprocal Identities, Theorem \ref{recipquotid}, it suffices to show $$\cos(-\theta) = \cos(\theta)$$ and $$\sin(-\theta) = -\sin(\theta)$$. The remaining four circular functions can be expressed in terms of $$\cos(\theta)$$ and $$\sin(\theta)$$ so the proofs of their Even / Odd Identities are left as exercises. Consider an angle $$\theta$$ plotted in standard position. Let $$\theta_ { o}$$ be the angle coterminal with $$\theta$$ with $$0 \leq \theta_ { o} < 2\pi$$. (We can construct the angle $$\theta_ { o}$$ by rotating counter-clockwise from the positive $$x$$-axis to the terminal side of $$\theta$$ as pictured below.) Since $$\theta$$ and $$\theta_ { o}$$ are coterminal, $$\cos(\theta) = \cos(\theta_ { o})$$ and $$\sin(\theta) = \sin(\theta_ { o})$$.
We now consider the angles $$-\theta$$ and $$-\theta_ { o}$$. Since $$\theta$$ is coterminal with $$\theta_ { o}$$, there is some integer $$k$$ so that $$\theta = \theta_ { o} + 2\pi \cdot k$$. Therefore, $$-\theta = -\theta_ { o} - 2\pi \cdot k = -\theta_ { o} + 2\pi \cdot(-k)$$. Since $$k$$ is an integer, so is $$(-k)$$, which means $$-\theta$$ is coterminal with $$-\theta_ { o}$$. Hence, $$\cos(-\theta) = \cos(-\theta_ { o})$$ and $$\sin(-\theta) = \sin(-\theta_ { o})$$. Let $$P$$ and $$Q$$ denote the points on the terminal sides of $$\theta_ { o}$$ and $$-\theta_ { o}$$, respectively, which lie on the Unit Circle. By definition, the coordinates of $$P$$ are $$(\cos(\theta_ { o}),\sin(\theta_ { o}))$$ and the coordinates of $$Q$$ are $$(\cos(-\theta_ { o}),\sin(-\theta_ { o}))$$. Since $$\theta_ { o}$$ and $$-\theta_ { o}$$ sweep out congruent central sectors of the Unit Circle, it follows that the points $$P$$ and $$Q$$ are symmetric about the $$x$$-axis. Thus, $$\cos(-\theta_ { o}) = \cos(\theta_ { o})$$ and $$\sin(-\theta_ { o}) = -\sin(\theta_ { o})$$. Since the cosines and sines of $$\theta_ { o}$$ and $$-\theta_ { o}$$ are the same as those for $$\theta$$ and $$-\theta$$, respectively, we get $$\cos(-\theta) = \cos(\theta)$$ and $$\sin(-\theta) = -\sin(\theta)$$, as required. The Even / Odd Identities are readily demonstrated using any of the common angles' noted in Section \ref{TheUnitCircle}. Their true utility, however, lies not in computation, but in simplifying expressions involving the circular functions. In fact, our next batch of identities makes heavy use of the Even / Odd Identities.
Note: Sum and Difference Identities for Cosine
For all angles $$\alpha$$ and $$\beta$$:
• $$\cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)$$
• $$\cos(\alpha - \beta) = \cos(\alpha) \cos(\beta) + \sin(\alpha) \sin(\beta)$$
We first prove the result for differences. As in the proof of the Even / Odd Identities, we can reduce the proof for general angles $$\alpha$$ and $$\beta$$ to angles $$\alpha_ { o}$$ and $$\beta_ { o}$$, coterminal with $$\alpha$$ and $$\beta$$, respectively, each of which measure between $$0$$ and $$2\pi$$ radians. Since $$\alpha$$ and $$\alpha_ { o}$$ are coterminal, as are $$\beta$$ and $$\beta_ { o}$$, it follows that $$\alpha - \beta$$ is coterminal with $$\alpha_ { o} - \beta_ { o}$$. Consider the case below where $$\alpha_ { o} \geq \beta_ { o}$$.
Since the angles $$POQ$$ and $$AOB$$ are congruent, the distance between $$P$$ and $$Q$$ is equal to the distance between $$A$$ and $$B$$.\footnote{In the picture we've drawn, the \underline{tri}angles $$POQ$$ and $$AOB$$ are congruent, which is even better. However, $$\alpha_ { o} - \beta_ { o}$$ could be $$0$$ or it could be $$\pi$$, neither of which makes a triangle. It could also be larger than $$\pi$$, which makes a triangle, just not the one we've drawn. You should think about those three cases.} The distance formula, Equation \ref{distanceformula}, yields
$\begin{array}{rcl} \sqrt{(\cos(\alpha_ { o}) - \cos(\beta_ { o}))^2 + (\sin(\alpha_ { o}) - \sin(\beta_ { o}))^2 } & = & \sqrt{(\cos(\alpha_ { o} - \beta_ { o}) - 1)^2 + (\sin(\alpha_ { o} - \beta_ { o}) - 0)^2} \\ \end{array}$
Squaring both sides, we expand the left hand side of this equation as
$\begin{array}{rcl} (\cos(\alpha_ { o}) - \cos(\beta_ { o}))^2 + (\sin(\alpha_ { o}) - \sin(\beta_ { o}))^2 & = & \cos^2(\alpha_ { o}) - 2\cos(\alpha_ { o})\cos(\beta_ { o}) + \cos^2(\beta_ { o}) \\ & & + \sin^2(\alpha_ { o}) - 2\sin(\alpha_ { o})\sin(\beta_ { o}) + \sin^2(\beta_ { o}) \\ & = & \cos^2(\alpha_ { o}) + \sin^2(\alpha_ { o}) + \cos^2(\beta_ { o}) + \sin^2(\beta_ { o}) \\ & & - 2\cos(\alpha_ { o})\cos(\beta_ { o}) - 2\sin(\alpha_ { o})\sin(\beta_ { o}) \end{array}$
From the Pythagorean Identities, $$\cos^2(\alpha_ { o}) + \sin^2(\alpha_ { o}) = 1$$ and $$\cos^2(\beta_ { o}) + \sin^2(\beta_ { o}) = 1$$, so
$\begin{array}{rcl} (\cos(\alpha_ { o}) - \cos(\beta_ { o}))^2 + (\sin(\alpha_ { o}) - \sin(\beta_ { o}))^2 & = & 2 - 2\cos(\alpha_ { o})\cos(\beta_ { o}) - 2\sin(\alpha_ { o})\sin(\beta_ { o}) \end{array}$
Turning our attention to the right hand side of our equation, we find
$\begin{array}{rcl} (\cos(\alpha_ { o} - \beta_ { o}) - 1)^2 + (\sin(\alpha_ { o} - \beta_ { o}) - 0)^2 & = & \cos^2(\alpha_ { o} - \beta_ { o}) - 2\cos(\alpha_ { o} - \beta_ { o}) + 1 + \sin^2(\alpha_ { o} - \beta_ { o}) \\ & = & 1 + \cos^2(\alpha_ { o} - \beta_ { o}) + \sin^2(\alpha_ { o} - \beta_ { o}) - 2\cos(\alpha_ { o} - \beta_ { o}) \\ \end{array}$
Once again, we simplify $$\cos^2(\alpha_ { o} - \beta_ { o}) + \sin^2(\alpha_ { o} - \beta_ { o})= 1$$, so that
$\begin{array}{rcl} (\cos(\alpha_ { o} - \beta_ { o}) - 1)^2 + (\sin(\alpha_ { o} - \beta_ { o}) - 0)^2 & = & 2 - 2\cos(\alpha_ { o} - \beta_ { o}) \\ \end{array}$
Putting it all together, we get $$2 - 2\cos(\alpha_ { o})\cos(\beta_ { o}) - 2\sin(\alpha_ { o})\sin(\beta_ { o}) = 2 - 2\cos(\alpha_ { o} - \beta_ { o})$$, which simplifies to: $$\cos(\alpha_ { o} - \beta_ { o}) = \cos(\alpha_ { o})\cos(\beta_ { o}) + \sin(\alpha_ { o})\sin(\beta_ { o})$$. Since $$\alpha$$ and $$\alpha_ { o}$$, $$\beta$$ and $$\beta_ { o}$$ and $$\alpha - \beta$$ and $$\alpha_ { o}- \beta_ { o}$$ are all coterminal pairs of angles, we have $$\cos(\alpha - \beta) = \cos(\alpha) \cos(\beta) + \sin(\alpha) \sin(\beta)$$. For the case where $$\alpha_ { o} \leq \beta_ { o}$$, we can apply the above argument to the angle $$\beta_ { o} - \alpha_ { o}$$ to obtain the identity $$\cos(\beta_ { o} - \alpha_ { o}) = \cos(\beta_ { o})\cos(\alpha_ { o}) + \sin(\beta_ { o})\sin(\alpha_ { o})$$. Applying the Even Identity of cosine, we get $$\cos(\beta_ { o} - \alpha_ { o}) = \cos( - (\alpha_ { o} - \beta_ { o})) = \cos(\alpha_ { o} - \beta_ { o})$$, and we get the identity in this case, too.
To get the sum identity for cosine, we use the difference formula along with the Even/Odd Identities
$\cos(\alpha + \beta) = \cos(\alpha - (-\beta)) = \cos(\alpha) \cos(-\beta) + \sin(\alpha) \sin(-\beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)$
We put these newfound identities to good use in the following example.
Example $$\PageIndex{1}$$: Cosine Sum and Difference
1. Find the exact value of $$\cos\left(15^{\circ}\right)$$.
2. Verify the identity: $$\cos\left(\frac{\pi}{2} - \theta\right) = \sin(\theta)$$.
Solution
1. In order to use Theorem \ref{cosinesumdifference} to find $$\cos\left(15^{\circ}\right)$$, we need to write $$15^{\circ}$$ as a sum or difference of angles whose cosines and sines we know. One way to do so is to write $$15^{\circ} = 45^{\circ} - 30^{\circ}$$.
$\begin{array}{rcl} \cos\left(15^{\circ}\right) & = & \cos\left(45^{\circ} - 30^{\circ} \right) \\ & = & \cos\left(45^{\circ}\right)\cos\left(30^{\circ} \right) + \sin\left(45^{\circ}\right)\sin\left(30^{\circ} \right) \\ & = & \left( \dfrac{\sqrt{2}}{2} \right)\left( \dfrac{\sqrt{3}}{2} \right) + \left( \dfrac{\sqrt{2}}{2} \right)\left( \dfrac{1}{2} \right)\\ & = & \dfrac{\sqrt{6}+ \sqrt{2}}{4} \\ \end{array}$
1. In a straightforward application of Theorem \ref{cosinesumdifference}, we find
$\begin{array}{rcl} \cos\left(\dfrac{\pi}{2} - \theta\right) & = & \cos\left(\dfrac{\pi}{2}\right)\cos\left(\theta\right) + \sin\left(\dfrac{\pi}{2}\right)\sin\left(\theta \right) \\ & = & \left( 0 \right)\left( \cos(\theta) \right) + \left( 1 \right)\left( \sin(\theta) \right) \\ & = & \sin(\theta) \\ \end{array}$
The identity verified in Example $$\PageIndex{1}$$, namely, $$\cos\left(\frac{\pi}{2} - \theta\right) = \sin(\theta)$$, is the first of the celebrated cofunction' identities. These identities were first hinted at in Exercise \ref{cofunctionforeshadowing} in Section \ref{TheUnitCircle}. From $$\sin(\theta) = \cos\left(\frac{\pi}{2} - \theta\right)$$, we get:
$\sin\left(\dfrac{\pi}{2} - \theta\right) = \cos\left(\dfrac{\pi}{2} -\left[\dfrac{\pi}{2} - \theta\right]\right) = \cos(\theta),$
which says, in words, that the co'sine of an angle is the sine of its co'mplement. Now that these identities have been established for cosine and sine, the remaining circular functions follow suit. The remaining proofs are left as exercises.
Note: Cofunction Identities
For all applicable angles $$\theta$$:
• $$\cos\left(\dfrac{\pi}{2} - \theta \right) = \sin(\theta)$$
• $$\sin\left(\dfrac{\pi}{2} - \theta \right) = \cos(\theta)$$
• $$\sec\left(\dfrac{\pi}{2} - \theta \right) = \csc(\theta)$$
• $$\csc\left(\dfrac{\pi}{2} - \theta \right) = \sec(\theta)$$
• $$\tan\left(\dfrac{\pi}{2} - \theta \right) = \cot(\theta)$$
• $$\cot\left(\dfrac{\pi}{2} - \theta \right) = \tan(\theta)$$
With the Cofunction Identities in place, we are now in the position to derive the sum and difference formulas for sine. To derive the sum formula for sine, we convert to cosines using a cofunction identity, then expand using the difference formula for cosine
$\begin{array}{rcl} \sin(\alpha + \beta) & = & \cos\left( \dfrac{\pi}{2} - (\alpha + \beta) \right) \\ & = & \cos\left( \left[\dfrac{\pi}{2} - \alpha \right] - \beta \right) \\ & = & \cos\left(\dfrac{\pi}{2} - \alpha \right) \cos(\beta) + \sin\left(\dfrac{\pi}{2} - \alpha \right)\sin(\beta) \\ & = & \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta) \\ \end{array}$
We can derive the difference formula for sine by rewriting $$\sin(\alpha - \beta)$$ as $$\sin(\alpha + (-\beta))$$ and using the sum formula and the Even / Odd Identities. Again, we leave the details to the reader.
Sum and Difference Identities for Sine
For all angles $$\alpha$$ and $$\beta$$, \index{Difference Identity ! for sine} \index{Sum Identity ! for sine}
• $$\sin(\alpha + \beta) = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta)$$
• $$\sin(\alpha - \beta) = \sin(\alpha) \cos(\beta) - \cos(\alpha) \sin(\beta)$$
Example $$\PageIndex{1}$$:
1. Find the exact value of $$\sin\left(\frac{19 \pi}{12}\right)$$
2. If $$\alpha$$ is a Quadrant II angle with $$\sin(\alpha) = \frac{5}{13}$$, and $$\beta$$ is a Quadrant III angle with $$\tan(\beta) = 2$$, find $$\sin(\alpha - \beta)$$.
3. Derive a formula for $$\tan(\alpha + \beta)$$ in terms of $$\tan(\alpha)$$ and $$\tan(\beta)$$.
Solution
1. As in Example \ref{cosinesumdiffex}, we need to write the angle $$\frac{19 \pi}{12}$$ as a sum or difference of common angles. The denominator of $$12$$ suggests a combination of angles with denominators $$3$$ and $$4$$. One such combination is $$\; \frac{19 \pi}{12} = \frac{4 \pi}{3} + \frac{\pi}{4}$$. Applying Theorem \ref{sinesumdifference}, we get
$\begin{array}{rcl} \sin\left(\dfrac{19 \pi}{12}\right) & = & \sin\left(\dfrac{4 \pi}{3} + \dfrac{\pi}{4} \right) \\ & = & \sin\left(\dfrac{4 \pi}{3} \right)\cos\left(\dfrac{\pi}{4} \right) + \cos\left(\dfrac{4 \pi}{3} \right)\sin\left(\dfrac{\pi}{4} \right) \\ & = & \left( -\dfrac{\sqrt{3}}{2} \right)\left( \dfrac{\sqrt{2}}{2} \right) + \left( -\dfrac{1}{2} \right)\left( \dfrac{\sqrt{2}}{2} \right) \\ & = & \dfrac{-\sqrt{6}- \sqrt{2}}{4} \\ \end{array}$
1. In order to find $$\sin(\alpha - \beta)$$ using Theorem \ref{sinesumdifference}, we need to find $$\cos(\alpha)$$ and both $$\cos(\beta)$$ and $$\sin(\beta)$$. To find $$\cos(\alpha)$$, we use the Pythagorean Identity $$\cos^2(\alpha) + \sin^2(\alpha) = 1$$. Since $$\sin(\alpha) = \frac{5}{13}$$, we have $$\cos^{2}(\alpha) + \left(\frac{5}{13}\right)^2 = 1$$, or $$\cos(\alpha) = \pm \frac{12}{13}$$. Since $$\alpha$$ is a Quadrant II angle, $$\cos(\alpha) = -\frac{12}{13}$$. We now set about finding $$\cos(\beta)$$ and $$\sin(\beta)$$. We have several ways to proceed, but the Pythagorean Identity $$1 + \tan^{2}(\beta) = \sec^{2}(\beta)$$ is a quick way to get $$\sec(\beta)$$, and hence, $$\cos(\beta)$$. With $$\tan(\beta) = 2$$, we get $$1 + 2^2 = \sec^{2}(\beta)$$ so that $$\sec(\beta) = \pm \sqrt{5}$$. Since $$\beta$$ is a Quadrant III angle, we choose $$\sec(\beta) = -\sqrt{5}$$ so $$\cos(\beta) = \frac{1}{\sec(\beta)} = \frac{1}{-\sqrt{5}} = -\frac{\sqrt{5}}{5}$$. We now need to determine $$\sin(\beta)$$. We could use The Pythagorean Identity $$\cos^{2}(\beta) + \sin^{2}(\beta) = 1$$, but we opt instead to use a quotient identity. From $$\tan(\beta) = \frac{\sin(\beta)}{\cos(\beta)}$$, we have $$\sin(\beta) = \tan(\beta) \cos(\beta)$$ so we get $$\sin(\beta) = (2) \left( -\frac{\sqrt{5}}{5}\right) = - \frac{2 \sqrt{5}}{5}$$. We now have all the pieces needed to find $$\sin(\alpha - \beta)$$:
$\begin{array}{rcl} \sin(\alpha - \beta) & = & \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta) \\ & = & \left( \dfrac{5}{13} \right)\left( -\dfrac{\sqrt{5}}{5} \right) - \left( -\dfrac{12}{13} \right)\left( - \dfrac{2 \sqrt{5}}{5} \right) \\ & = & -\dfrac{29\sqrt{5}}{65} \\ \end{array}$
We can start expanding $$\tan(\alpha + \beta)$$ using a quotient identity and our sum formulas
$\begin{array}{rcl} \tan(\alpha + \beta) & = & \dfrac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} \\ & = & \dfrac{\sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta)}{\cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)} \\ \end{array}$
Since $$\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}$$ and $$\tan(\beta) = \frac{\sin(\beta)}{\cos(\beta)}$$, it looks as though if we divide both numerator and denominator by $$\cos(\alpha) \cos(\beta)$$ we will have what we want
$\begin{array}{rcl} \tan(\alpha + \beta) & = & \dfrac{\sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta)}{\cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)} \cdot\dfrac{\dfrac{1}{\cos(\alpha) \cos(\beta)}}{\dfrac{1}{\cos(\alpha) \cos(\beta)}}\\ & & \\ & = & \dfrac{\dfrac{\sin(\alpha) \cos(\beta)}{\cos(\alpha) \cos(\beta)} + \dfrac{\cos(\alpha) \sin(\beta)}{\cos(\alpha) \cos(\beta)}}{\dfrac{\cos(\alpha) \cos(\beta)}{\cos(\alpha) \cos(\beta)} - \dfrac{\sin(\alpha) \sin(\beta)}{\cos(\alpha) \cos(\beta)}}\\ & & \\ & = & \dfrac{\dfrac{\sin(\alpha) \cancel{\cos(\beta)}}{\cos(\alpha) \cancel{\cos(\beta)}} + \dfrac{\cancel{\cos(\alpha)} \sin(\beta)}{\cancel{\cos(\alpha)} \cos(\beta)}}{\dfrac{\cancel{\cos(\alpha)} \cancel{\cos(\beta)}}{\cancel{\cos(\alpha)} \cancel{\cos(\beta)}} - \dfrac{\sin(\alpha) \sin(\beta)}{\cos(\alpha) \cos(\beta)}}\\ & & \\ & = & \dfrac{\tan(\alpha) + \tan(\beta)}{1 -\tan(\alpha) \tan(\beta)}\\ \end{array}$
Naturally, this formula is limited to those cases where all of the tangents are defined.\qed
The formula developed in Exercise \ref{sinesumanddiffex} for $$\tan(\alpha + \beta)$$ can be used to find a formula for $$\tan(\alpha - \beta)$$ by rewriting the difference as a sum, $$\tan(\alpha + (-\beta))$$, and the reader is encouraged to fill in the details. Below we summarize all of the sum and difference formulas for cosine, sine and tangent.
Note
Sum and Difference Identities:} For all applicable angles $$\alpha$$ and $$\beta$$, \index{Difference Identity ! for tangent} \index{Sum Identity ! for tangent} \index{Difference Identity ! for cosine} \index{Sum Identity ! for cosine} \index{Difference Identity ! for sine} \index{Sum Identity ! for sine}
• $$\cos(\alpha \pm \beta) = \cos(\alpha) \cos(\beta) \mp \sin(\alpha) \sin(\beta)$$
• $$\sin(\alpha \pm \beta) = \sin(\alpha) \cos(\beta) \pm \cos(\alpha) \sin(\beta)$$
• $$\tan(\alpha \pm \beta) = \dfrac{\tan(\alpha) \pm \tan(\beta)}{1 \mp \tan(\alpha) \tan(\beta)}$$
In the statement of Theorem \ref{circularsumdifference}, we have combined the cases for the sum $+$' and difference \)-$' of angles into one formula. The convention here is that if you want the formula for the sum $+$' of two angles, you use the top sign in the formula; for the difference, \)-$', use the bottom sign. For example, $\tan(\alpha - \beta) = \dfrac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha) \tan(\beta)}$
If we specialize the sum formulas in Theorem \ref{circularsumdifference} to the case when $$\alpha = \beta$$, we obtain the following Double Angle' Identities.
Note Double Angle Identities
For all applicable angles $$\theta$$, \index{Double Angle Identities}
• $$\cos(2\theta) = \left\{ \begin{array}{l} \cos^{2}(\theta) - \sin^{2}(\theta)\\ [5pt] 2\cos^{2}(\theta) - 1 \\ [5pt] 1-2\sin^{2}(\theta) \end{array} \right.$$
• $$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$$
• $$\tan(2\theta) = \dfrac{2\tan(\theta)}{1 - \tan^{2}(\theta)}$$
The three different forms for $$\cos(2\theta)$$ can be explained by our ability to exchange' squares of cosine and sine via the Pythagorean Identity $$\cos^{2}(\theta) + \sin^{2}(\theta) = 1$$ and we leave the details to the reader. It is interesting to note that to determine the value of $$\cos(2\theta)$$, only \textit{one} piece of information is required: either $$\cos(\theta)$$ or $$\sin(\theta)$$. To determine $$\sin(2\theta)$$, however, it appears that we must know both $$\sin(\theta)$$ and $$\cos(\theta)$$. In the next example, we show how we can find $$\sin(2\theta)$$ knowing just one piece of information, namely $$\tan(\theta)$$.
Example $$\PageIndex{1}$$:
1. Suppose $$P(-3,4)$$ lies on the terminal side of $$\theta$$ when $$\theta$$ is plotted in standard position. Find $$\cos(2\theta)$$ and $$\sin(2\theta)$$ and determine the quadrant in which the terminal side of the angle $$2\theta$$ lies when it is plotted in standard position.
2. If $$\sin(\theta) = x$$ for $$-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$$, find an expression for $$\sin(2\theta)$$ in terms of $$x$$.
3. \label{doubleanglesinewtan} Verify the identity: $$\sin(2\theta) = \dfrac{2\tan(\theta)}{1 + \tan^{2}(\theta)}$$.
4. Express $$\cos(3\theta)$$ as a polynomial in terms of $$\cos(\theta)$$.
Solution
1. Using Theorem \ref{cosinesinecircle} from Section \ref{TheUnitCircle} with $$x = -3$$ and $$y=4$$, we find $$r = \sqrt{x^2+y^2} = 5$$. Hence, $$\cos(\theta) = -\frac{3}{5}$$ and $$\sin(\theta) = \frac{4}{5}$$. Applying Theorem \ref{doubleangle}, we get $$\cos(2\theta) = \cos^{2}(\theta) - \sin^{2}(\theta) = \left(-\frac{3}{5}\right)^2 - \left(\frac{4}{5}\right)^2 = -\frac{7}{25}$$, and $$\sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2 \left(\frac{4}{5}\right)\left(-\frac{3}{5}\right) = -\frac{24}{25}$$. Since both cosine and sine of $$2\theta$$ are negative, the terminal side of $$2\theta$$, when plotted in standard position, lies in Quadrant III.
2. If your first reaction to $\sin(\theta) = x$' is No it's not, $$\cos(\theta) = x!' then you have indeed learned something, and we take comfort in that. However, context is everything. Here, x' is just a variable - it does not necessarily represent the \(x$$-coordinate of the point on The Unit Circle which lies on the terminal side of $$\theta$$, assuming $$\theta$$ is drawn in standard position. Here, $$x$$ represents the quantity $$\sin(\theta)$$, and what we wish to know is how to express $$\sin(2\theta)$$ in terms of $$x$$. We will see more of this kind of thing in Section \ref{ArcTrig}, and, as usual, this is something we need for Calculus. Since $$\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$$, we need to write $$\cos(\theta)$$ in terms of $$x$$ to finish the problem. We substitute $$x = \sin(\theta)$$ into the Pythagorean Identity, $$\cos^{2}(\theta) + \sin^{2}(\theta) = 1$$, to get $$\cos^{2}(\theta) + x^2 = 1$$, or $$\cos(\theta) = \pm \sqrt{1-x^2}$$. Since $$-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$$, $$\cos(\theta) \geq 0$$, and thus $$\cos(\theta) = \sqrt{1-x^2}$$. Our final answer is $$\sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2x\sqrt{1-x^2}$$.
We start with the right hand side of the identity and note that $$1 + \tan^{2}(\theta) = \sec^{2}(\theta)$$. From this point, we use the Reciprocal and Quotient Identities to rewrite $$\tan(\theta)$$ and $$\sec(\theta)$$ in terms of $$\cos(\theta)$$ and $$\sin(\theta)$$:
$\begin{array}{rcl} \dfrac{2\tan(\theta)}{1 + \tan^{2}(\theta)} & = & \dfrac{2\tan(\theta)}{\sec^{2}(\theta)}= \dfrac{2 \left( \dfrac{\sin(\theta)}{\cos(\theta)}\right)}{\dfrac{1}{\cos^{2}(\theta)}}= 2\left( \dfrac{\sin(\theta)}{\cos(\theta)}\right) \cos^{2}(\theta) \\ & = & 2\left( \dfrac{\sin(\theta)}{\cancel{\cos(\theta)}}\right) \cancel{\cos(\theta)} \cos(\theta) = 2\sin(\theta) \cos(\theta) = \sin(2\theta) \\ \end{array}$
1. In Theorem \ref{doubleangle}, one of the formulas for $$\cos(2\theta)$$, namely $$\cos(2\theta) = 2\cos^{2}(\theta) - 1$$, expresses $$\cos(2\theta)$$ as a polynomial in terms of $$\cos(\theta)$$. We are now asked to find such an identity for $$\cos(3\theta)$$. Using the sum formula for cosine, we begin with
$\begin{array}{rcl} \cos(3\theta) & = & \cos(2\theta + \theta) \\ & = & \cos(2\theta)\cos(\theta) - \sin(2\theta)\sin(\theta) \\ \end{array}$
Our ultimate goal is to express the right hand side in terms of $$\cos(\theta)$$ only. We substitute $$\cos(2\theta) = 2\cos^{2}(\theta) -1$$ and $$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$$ which yields
$\begin{array}{rcl} \cos(3\theta) & = & \cos(2\theta)\cos(\theta) - \sin(2\theta)\sin(\theta) \\ & = & \left(2\cos^{2}(\theta) - 1\right) \cos(\theta) - \left(2 \sin(\theta) \cos(\theta) \right)\sin(\theta) \\ & = & 2\cos^{3}(\theta)- \cos(\theta) - 2 \sin^2(\theta) \cos(\theta) \\ \end{array}$
Finally, we exchange $$\sin^{2}(\theta)$$ for $$1 - \cos^{2}(\theta)$$ courtesy of the Pythagorean Identity, and get
$\begin{array}{rcl} \cos(3\theta) & = & 2\cos^{3}(\theta)- \cos(\theta) - 2 \sin^2(\theta) \cos(\theta) \\ & = & 2\cos^{3}(\theta)- \cos(\theta) - 2 \left(1 - \cos^{2}(\theta)\right) \cos(\theta) \\ & = & 2\cos^{3}(\theta)- \cos(\theta) - 2\cos(\theta) + 2\cos^{3}(\theta) \\ & = & 4\cos^{3}(\theta)- 3\cos(\theta) \\ \end{array}$
and we are done.
In the last problem in Example \ref{doubleangleex}, we saw how we could rewrite $$\cos(3\theta)$$ as sums of powers of $$\cos(\theta)$$. In Calculus, we have occasion to do the reverse; that is, reduce the power of cosine and sine. Solving the identity $$\cos(2\theta) = 2\cos^{2}(\theta) -1$$ for $$\cos^{2}(\theta)$$ and the identity $$\cos(2\theta) = 1 - 2\sin^{2}(\theta)$$ for $$\sin^{2}(\theta)$$ results in the aptly-named `Power Reduction' formulas below.
Power Reduction Formulas
For all angles $$\theta$$, \index{Power Reduction Formulas}
• $$\cos^{2}(\theta) = \dfrac{1 + \cos(2\theta)}{2}$$
• $$\sin^{2}(\theta) = \dfrac{1 - \cos(2\theta)}{2}$$
Example $$\PageIndex{1}$$:
Rewrite $$\sin^{2}(\theta) \cos^{2}(\theta)$$ as a sum and difference of cosines to the first power.
Solution
We begin with a straightforward application of Theorem \ref{powerreduction}
$\begin{array}{rcl} \sin^{2}(\theta) \cos^{2}(\theta) & = & \left( \dfrac{1 - \cos(2\theta)}{2} \right) \left( \dfrac{1 + \cos(2\theta)}{2} \right) \\ & = & \dfrac{1}{4}\left(1 - \cos^{2}(2\theta)\right) \\ & = & \dfrac{1}{4} - \dfrac{1}{4}\cos^{2}(2\theta) \\ \end{array}$
Next, we apply the power reduction formula to $$\cos^{2}(2\theta)$$ to finish the reduction
$\begin{array}{rcl} \sin^{2}(\theta) \cos^{2}(\theta) & = & \dfrac{1}{4} - \dfrac{1}{4}\cos^{2}(2\theta) \\ & = & \dfrac{1}{4} - \dfrac{1}{4} \left(\dfrac{1 + \cos(2(2\theta))}{2}\right) \\ & = & \dfrac{1}{4} - \dfrac{1}{8} - \dfrac{1}{8}\cos(4\theta) \\ & = & \dfrac{1}{8} - \dfrac{1}{8}\cos(4\theta) \\ \end{array}$
Another application of the Power Reduction Formulas is the Half Angle Formulas. To start, we apply the Power Reduction Formula to $$\cos^{2}\left(\frac{\theta}{2}\right) $\cos^{2}\left(\dfrac{\theta}{2}\right) = \dfrac{1 + \cos\left(2 \left(\frac{\theta}{2}\right)\right)}{2} = \dfrac{1 + \cos(\theta)}{2}.$ We can obtain a formula for \(\cos\left(\frac{\theta}{2}\right)$$ by extracting square roots. In a similar fashion, we may obtain a half angle formula for sine, and by using a quotient formula, obtain a half angle formula for tangent. We summarize these formulas below.
Half Angle Formulas
For all applicable angles $$\theta$$:
• $$\cos\left(\dfrac{\theta}{2}\right) = \pm \sqrt{\dfrac{1 + \cos(\theta)}{2}}$$
• $$\sin\left(\dfrac{\theta}{2}\right) = \pm \sqrt{\dfrac{1 - \cos(\theta)}{2}}$$
• $$\tan\left(\dfrac{\theta}{2}\right) = \pm \sqrt{\dfrac{1 - \cos(\theta)}{1+\cos(\theta)}}$$
where the choice of $$\pm$$ depends on the quadrant in which the terminal side of $$\dfrac{\theta}{2}$$ lies.
Example $$\PageIndex{1}$$:
1. Use a half angle formula to find the exact value of $$\cos\left(15^{\circ}\right)$$.
2. Suppose $$-\pi \leq \theta \leq 0$$ with $$\cos(\theta) = -\frac{3}{5}$$. Find $$\sin\left(\frac{\theta}{2}\right)$$.
3. Use the identity given in number \ref{doubleanglesinewtan} of Example \ref{doubleangleex} to derive the identity $\tan\left(\dfrac{\theta}{2}\right) = \dfrac{\sin(\theta)}{1+\cos(\theta)}$
Solution
1. To use the half angle formula, we note that $$15^{\circ} = \frac{30^{\circ}}{2}$$ and since $$15^{\circ}$$ is a Quadrant I angle, its cosine is positive. Thus we have
$\begin{array}{rcl} \cos\left(15^{\circ}\right) & = & + \sqrt{\dfrac{1+\cos\left(30^{\circ}\right)}{2}} = \sqrt{\dfrac{1+\frac{\sqrt{3}}{2}}{2}}\\ & = & \sqrt{\dfrac{1+\frac{\sqrt{3}}{2}}{2}\cdot \dfrac{2}{2}} = \sqrt{\dfrac{2+\sqrt{3}}{4}} = \dfrac{\sqrt{2+\sqrt{3}}}{2}\\ \end{array}$
Back in Example \ref{cosinesumdiffex}, we found $$\cos\left(15^{\circ}\right)$$ by using the difference formula for cosine. In that case, we determined $$\cos\left(15^{\circ}\right) = \frac{\sqrt{6}+ \sqrt{2}}{4}$$. The reader is encouraged to prove that these two expressions are equal.
1. If $$-\pi \leq \theta \leq 0$$, then $$-\frac{\pi}{2} \leq \frac{\theta}{2} \leq 0$$, which means $$\sin\left(\frac{\theta}{2}\right) < 0$$. Theorem \ref{halfangle} gives
$\begin{array}{rcl} \sin\left(\dfrac{\theta}{2} \right) & = & -\sqrt{\dfrac{1-\cos\left(\theta \right)}{2}} = -\sqrt{\dfrac{1- \left(-\frac{3}{5}\right)}{2}}\\ & = & -\sqrt{\dfrac{1 + \frac{3}{5}}{2} \cdot \dfrac{5}{5}} = -\sqrt{\dfrac{8}{10}} = -\dfrac{2\sqrt{5}}{5}\\ \end{array}$
1. Instead of our usual approach to verifying identities, namely starting with one side of the equation and trying to transform it into the other, we will start with the identity we proved in number \ref{doubleanglesinewtan} of Example \ref{doubleangleex} and manipulate it into the identity we are asked to prove. The identity we are asked to start with is $$\; \sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^{2}(\theta)}$$. If we are to use this to derive an identity for $$\tan\left(\frac{\theta}{2}\right)$$, it seems reasonable to proceed by replacing each occurrence of $$\theta$$ with $$\frac{\theta}{2} $\begin{array}{rcl} \sin\left(2 \left(\frac{\theta}{2}\right)\right) & = & \dfrac{2\tan\left(\frac{\theta}{2}\right)}{1 + \tan^{2}\left(\frac{\theta}{2}\right)} \\ \sin(\theta) & = & \dfrac{2\tan\left(\frac{\theta}{2}\right)}{1 + \tan^{2}\left(\frac{\theta}{2}\right)} \\ \end{array}$ We now have the \(\sin(\theta)$$ we need, but we somehow need to get a factor of $$1+\cos(\theta)$$ involved. To get cosines involved, recall that $$1 + \tan^{2}\left(\frac{\theta}{2}\right) = \sec^{2}\left(\frac{\theta}{2}\right)$$. We continue to manipulate our given identity by converting secants to cosines and using a power reduction formula
$\begin{array}{rcl} \sin(\theta) & = & \dfrac{2\tan\left(\frac{\theta}{2}\right)}{1 + \tan^{2}\left(\frac{\theta}{2}\right)} \\ \sin(\theta) & = & \dfrac{2\tan\left(\frac{\theta}{2}\right)}{\sec^{2}\left(\frac{\theta}{2}\right)} \\ \sin(\theta) & = & 2 \tan\left(\frac{\theta}{2}\right) \cos^{2}\left(\frac{\theta}{2}\right) \\ \sin(\theta) & = & 2 \tan\left(\frac{\theta}{2}\right) \left(\dfrac{1 + \cos\left(2 \left(\frac{\theta}{2}\right)\right)}{2}\right) \\ \sin(\theta) & = & \tan\left(\frac{\theta}{2}\right) \left(1+\cos(\theta) \right) \\ \tan\left(\dfrac{\theta}{2}\right) & = & \dfrac{\sin(\theta)}{1+\cos(\theta)} \\ \end{array}$
Our next batch of identities, the Product to Sum Formulas,\footnote{These are also known as the Prosthaphaeresis Formulas and have a rich history. The authors recommend that you conduct some research on them as your schedule allows.} are easily verified by expanding each of the right hand sides in accordance with Theorem \ref{circularsumdifference} and as you should expect by now we leave the details as exercises. They are of particular use in Calculus, and we list them here for reference.
Note: Product to Sum Formulas
For all angles $$\alpha$$ and $$\beta$$, \index{Product to Sum Formulas}
• $$\cos(\alpha)\cos(\beta) = \frac{1}{2} \left[ \cos(\alpha - \beta) + \cos(\alpha + \beta)\right]$$
• $$\sin(\alpha)\sin(\beta) = \frac{1}{2} \left[ \cos(\alpha - \beta) - \cos(\alpha + \beta)\right]$$
• $$\sin(\alpha)\cos(\beta) = \frac{1}{2} \left[ \sin(\alpha - \beta) + \sin(\alpha + \beta)\right]$$
Related to the Product to Sum Formulas are the Sum to Product Formulas, which we will have need of in Section \ref{TrigEquIneq}. These are easily verified using the Product to Sum Formulas, and as such, their proofs are left as exercises.
Note: Sum to Product Formulas:
For all angles $$\alpha$$ and $$\beta$$:
1. $$\cos(\alpha) + \cos(\beta) = 2 \cos\left( \dfrac{\alpha + \beta}{2}\right)\cos\left( \dfrac{\alpha - \beta}{2}\right)$$
2. $$\cos(\alpha) - \cos(\beta) = - 2 \sin\left( \dfrac{\alpha + \beta}{2}\right)\sin\left( \dfrac{\alpha - \beta}{2}\right)$$
3. $$\sin(\alpha) \pm \sin(\beta) = 2 \sin\left( \dfrac{\alpha \pm \beta}{2}\right)\cos\left( \dfrac{\alpha \mp \beta}{2}\right)$$
Example $$\PageIndex{1}$$:
1. Write $$\; \cos(2\theta)\cos(6\theta) \;$$ as a sum.
2. \Write $$\; \sin(\theta) - \sin(3\theta) \;$$ as a product.
Solution
1. Identifying $$\alpha = 2\theta$$ and $$\beta = 6\theta$$, we find
$\begin{array}{rcl} \cos(2\theta)\cos(6\theta) & = & \frac{1}{2} \left[ \cos(2\theta - 6\theta) + \cos(2\theta + 6\theta)\right]\\ & = & \frac{1}{2} \cos(-4\theta) + \frac{1}{2}\cos(8\theta) \\ & = & \frac{1}{2} \cos(4\theta) + \frac{1}{2} \cos(8\theta), \end{array}$
where the last equality is courtesy of the even identity for cosine, $$\cos(-4\theta) = \cos(4\theta)$$.
1. Identifying $$\alpha = \theta$$ and $$\beta = 3\theta$$ yields
$\begin{array}{rcl} \sin(\theta) - \sin(3\theta) & = & 2 \sin\left( \dfrac{\theta - 3\theta}{2}\right)\cos\left( \dfrac{\theta + 3\theta}{2}\right) \\ & = & 2 \sin\left( -\theta \right)\cos\left( 2\theta \right) \\ & = & -2 \sin\left( \theta \right)\cos\left( 2\theta \right), \\ \end{array}$
where the last equality is courtesy of the odd identity for sine, $$\sin(-\theta) = -\sin(\theta)$$.
The reader is reminded that all of the identities presented in this section which regard the circular functions as functions of angles (in radian measure) apply equally well to the circular (trigonometric) functions regarded as functions of real numbers. In Exercises \ref{idengraphfirst} - \ref{idengraphlast} in Section \ref{TrigGraphs}, we see how some of these identities manifest themselves geometrically as we study the graphs of the these functions. In the upcoming Exercises, however, you need to do all of your work analytically without graphs.
### Contributors
• Carl Stitz, Ph.D. (Lakeland Community College) and Jeff Zeager, Ph.D. (Lorain County Community College) | {"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": 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": 0.9800750613212585, "perplexity": 244.89768183798046}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676591216.51/warc/CC-MAIN-20180719183926-20180719203926-00115.warc.gz"} |
https://tex.stackexchange.com/questions/208743/how-to-redefine-that-is-backslash-space | # How to redefine “\ ” (that is “<backslash> <space>”)?
In (La)TeX, is there a way to redefine the command \ (that is "<backslash> <space>")?
I have lots of texts where \ is used as non-breakable space after "." characters like in Dr.\ or Mr.\ and I want to make that space in the output just a bit shorter than the normal space.
Please note, I do not want to change the texts, but rather redefine the \ macro in the preamble. Is this possible?
• The \ macro inserts ordinary (breakable) space, not non-breakable space. For non-breakable space, you should use (and suitably modify, as necessary) the ~ symbol. – Mico Oct 24 '14 at 5:59
I would strongly recommend against this, but it can be done. The command \ is a primitive meaning 'a normal space' so shows up in various places, in particular the definition of \nonbreakspace. Thus a 'safe' redefinition of \ must at least deal with that:
\documentclass{article}
\let\hardspace\ %
\DeclareRobustCommand*\nobreakspace{\leavevmode\nobreak\hardspace}
%\let\ ~
\begin{document}
Some text to show that this is now a non-breaking space in a demo:
Mr.\ Black.
\let\ ~
Some text to show that this is now a non-breaking space in a demo:
Mr.\ Black.
\end{document}
I've commented out \let\ ~ in the preamble in the above so that the demo shows the effect of the change, but in a real case you'd apply it to everything. As pointed out by others, you really should use the correct mark-up to differentiate between a 'forced' normal space and a non-breaking space.
• As the poster I have to explain what the reason for my question was: I write my papers in Markdown and convert them with pandoc via LaTeX to pdf or to html. Markdown has basckslash-space as non-breakable space and I though that the same is in TeX, but of course not: So pandoc converts the Markdown-backslash-space into TeX-tilde, so if anything I redefine tilde in LaTeX as a little narrower space. Done that and it works great. But your answer is of course perfectly valid for the backslash-space redefinition. – halloleo Nov 1 '14 at 1:24
For example, starting from the definition of ~:
\makeatletter
\def\hallospace{\penalty\@M \kern0.3em} % say, 0.3em
\let\oldspace=\ %
\let\ =\hallospace
\makeatother
(And remember to use ~ as an unbreakable space in the future). | {"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": 0.9701371788978577, "perplexity": 2435.2919328762005}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703565376.63/warc/CC-MAIN-20210125061144-20210125091144-00255.warc.gz"} |
https://www.physicsforums.com/threads/proof-that-a-vector-space-w-is-the-direct-sum-of-ker-l-and-im-l.582462/ | Proof that a vector space W is the direct sum of Ker L and Im L
1. Feb 29, 2012
nilwill
Hi there. I'm a long time reader, first time poster. I'm an undergraduate in Math and Economics and I am having trouble in Linear Algebra. This is the first class I have had that focuses solely on proofs, so I am in new territory.
1. The problem statement, all variables and given/known data
note Although the question doesn't state it, I think P is supposed to be a projection.
Let W be a vector space. Let P:W→W be a linear map s.t. P2=P.
Show that W= KerP + ImP and KerP$\cap$ImP
namely, W is the direct sum of KerP and ImP
Hint: To show W is the sum, write an element of W in the form w=w-P(w)+P(w)
2. Relevant equations
I am unsure of any relevant equations.
3. The attempt at a solution
KerP={z|P(z)=0}
ImP={P(v)|v$\in$W}
I am kind of fuzzy on the meaning of P2=P, and this is where I am stuck.
Would an example be something like:
P=$\left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right)$
P2=$\left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right) \times \left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right) = \left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right)$ =P?
From this example: for KerP=P(z)=0 , we would need to satisfy z+0=0, so z=0
For some v$\in$W, the ImP is P(v)=v.
KerP$\cap$ImP=0→ P(v)=P(z) iff v=0 so z=v where the two intersect.
I am unsure where to go from here, or if I'm even doing this correctly. If someone could nudge me along towards and answer without giving the me the full proof, it would be much appreciated. I've been stuck on this problem for 4 hours over two days and I can't seem to figure it out.
2. Feb 29, 2012
Deveno
P2 = P is sometimes used as the definition of a projection.
we can always write w = P(w) + (w - P(w))
no matter what w is.
P(w) is clearly in Im(P), so all you have to do is show that
w - P(w) is in ker(P).
what is P(w - P(w))?
3. Feb 29, 2012
nilwill
P:V→V and P2=P
Let me see if I can define this better: so if V is the direct sum of U + W
let P(v)=u
let Im P be a complete subset of U; if u is an element of U P(u)=u, so Im P =U
let Ker P be a complete subset of W; if v is an element of V and v is an element of W, the v=0+w for some element w in W, so Ker P =W
let v=P(v)+(v-P(v))
Since P(v) is in the Im P, then (v-P(v)) is in the Ker P
So P(v-P(v))=P(v)-P2(v)=P(v)-P(u)=u-u=0
therefore, v-P(v) is in the Ker P
Is this correct so far? Sorry for not using latex. I have only used it in creating this post and was stretched for time.
I think I can do the rest if I have set this up correctly.
4. Feb 29, 2012
Deveno
we're not worried about "direct sum" just yet, just the sum (or "join").
none of this makes much sense. and it's unnecessary.
you don't need to "let" v = P(v) + (v-P(v)) it ALWAYS is, by the rules of vector addition.
you don't KNOW this yet, you're trying to prove it. proofs start with what you know. in this case, what you know is:
v = P(v) + (v-P(v)) (simple algebra, always true, since P(v) + -(P(v)) is the 0-vector) and:
P(v) is in Im(P) (by the definition of what Im(P) is Im(P)= {u in W: u = P(v) for some v in W}.) and:
P2 = P (so P(P(v)) = P(v), you'll use this later).
why bring "u" into this?
in my last post, i asked you a question:
what is P(w - P(w))? you don't need any "extra letters" to answer this, you just need one certain fact about P.
Similar Discussions: Proof that a vector space W is the direct sum of Ker L and Im L | {"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": 0.9206842184066772, "perplexity": 1096.1495384887871}, "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-2017-47/segments/1510934804666.54/warc/CC-MAIN-20171118055757-20171118075757-00560.warc.gz"} |
https://core.ac.uk/display/2181680 | Location of Repository
## Deformation of $\ell$-adic sheaves with Undeformed Local Monodromy
### Abstract
Let $X$ be a smooth connected algebraic curve over an algebraically closed field $k$. We study the deformation of $\ell$-adic Galois representations of the function field of $X$ while keeping the local Galois representations at all places undeformed.Comment: 21 pages. To appear in Journal of Number Theor
Topics: Mathematics - Algebraic Geometry, Mathematics - Number Theory, 14D15
Year: 2012
OAI identifier: oai:arXiv.org:1103.1093 | {"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": 0.9007775783538818, "perplexity": 705.0475818277414}, "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-2018-51/segments/1544376828507.57/warc/CC-MAIN-20181217113255-20181217135255-00125.warc.gz"} |
http://mathhelpforum.com/differential-geometry/189190-complex-differentiation.html | 1. Complex differentiation
Hey, I have this problem from a tute i was given,
Its all in the picture,
For part a i just showed the derivatives exist and that they were continuous, then used laplace to show that they are harmonic du/dx - du/dy = 0,
so that work, then to find v i just found the harmonic conjugate and i got -3y^2x+x^3 + c = v
so u+iv = y^3 - 3x^2y +i(3y^2x+x^3 + c )
Is that the complete solution to the question? I feel like I am missing somthing, more of a proof or somthing.
Part b i showed in the picture, i think the first part seems ok, then showing for elsewhere i just let z0 be any point, and tried to find if there were any points besides zero where the limit exists but lim z0/delz is infinity, so is that sufficient to say that that its only differentiable at zero?
We did somthing different aswell, let it be z= u+iv, then f(z) = u^2 + ivu cause Re(z) = u, then let these u^2 and vu be two new variables and applied C-R using implicite differentiation and found somthing different, that it is differentiable when Re(z) = constant or Im(z) = 0
Am i on the right track at all?
2. Re: Complex differentiation
Originally Posted by Daniiel
Hey, I have this problem from a tute i was given,
Its all in the picture,
For part a i just showed the derivatives exist and that they were continuous, then used laplace to show that they are harmonic du/dx - du/dy = 0,
so that work, then to find v i just found the harmonic conjugate and i got -3y^2x+x^3 + c = v
so u+iv = y^3 - 3x^2y +i(3y^2x+x^3 + c ) Note the missing minus sign.
Is that the complete solution to the question? I feel like I am missing something, more of a proof or something.
If z = x+iy and f(z) = u+iv, then I think they want you to express f(z) as a function of z (rather than x and y). With the change of sign that I indicated, you have $f(z) = ix^3 - 3x^2y - 3ixy^2 + y^3.$ Can you see how to write that as a function of z? [Think: Binomial theorem.]
Originally Posted by Daniiel
Part b i showed in the picture, i think the first part seems ok, then showing for elsewhere i just let z0 be any point, and tried to find if there were any points besides zero where the limit exists but lim z0/delz is infinity, so is that sufficient to say that that its only differentiable at zero?
We did somthing different as well, let it be z= u+iv, then f(z) = u^2 + ivu cause Re(z) = u, then let these u^2 and vu be two new variables and applied C-R using implicite differentiation and found somthing different, that it is differentiable when Re(z) = constant or Im(z) = 0
Am i on the right track at all?
The first part of b) is ok. For the second part, you are right to use the C–R equations, but you are getting the notation a bit mixed. If z = x+iy then $f(z) = x^2 + ixy.$ So you should take $u = x^2$ and $v = xy.$ Then apply the C–R equations.
3. Re: Complex differentiation
oh sweet thanks very much, so in terms of z for a, f(z)= iz^3 +ic right?
So for part b, because both C-R equations don't hold it is not differentiable elsewhere? only if x=y=0 | {"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": 0.9566742777824402, "perplexity": 829.5459795374439}, "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-2016-50/segments/1480698542009.32/warc/CC-MAIN-20161202170902-00328-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/density-of-an-iceberg.268414/ | # Density of an iceberg
1. Oct 31, 2008
### Vuldoraq
1. The problem statement, all variables and given/known data
a) If the part of an iceberg above sea level is one ninth of the whole, what is the density of ice?
b)How much of the iceberg would show if it moved into a fresh water region?
2. Relevant equations
Density of sea water=1025kgm^-3
Density of fresh water 1000kgm^-3
Weight displaced=upthrust
Force due to gravity = mg
3. The attempt at a solution
For part a) I equated the upthrust and the force due to gravity, by applying Newtons first law. However I am confused as to whether I should also take into account the atmospheric pressure pressing the part of the iceberg above sea level down. In all the stuff I've read no one seems to take it into account when finding the density of an iceberg. Is there a reason for this?
In part b) I think you again apply Newtons first law and use the ice density calculated in part a),
$$\rho_{ice}*g*v_{ice}=\rho_{water}*g*v_{water}$$
$$\frac{\rho_{ice}}{\rho_{water}}=\frac{v_{water}}{v_{ice}}$$
Which gives the proportion of ice under the water.
2. Oct 31, 2008
Staff Emeritus
If you pick up a piece of paper, do you feel the 1400 pounds of force that air pressure exerts on it?
3. Oct 31, 2008
### Office_Shredder
Staff Emeritus
I do :(
4. Oct 31, 2008
### Vuldoraq
I geuss not, but isn't that because as soon as you pick it up air rushes underneath the paper, very quickly, and the air underneath exerts an equal and oppisite force to the air above, thus we don't feel the pressure force. In the sea this situation is clearly impossible.
Last edited: Oct 31, 2008 | {"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": 0.92151939868927, "perplexity": 780.2298467951233}, "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/1476988719041.14/warc/CC-MAIN-20161020183839-00303-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://direct.mit.edu/neco/article-abstract/30/1/125/8332/Capturing-Spike-Variability-in-Noisy-Izhikevich?redirectedFrom=fulltext | To understand neural activity, two broad categories of models exist: statistical and dynamical. While statistical models possess rigorous methods for parameter estimation and goodness-of-fit assessment, dynamical models provide mechanistic insight. In general, these two categories of models are separately applied; understanding the relationships between these modeling approaches remains an area of active research. In this letter, we examine this relationship using simulation. To do so, we first generate spike train data from a well-known dynamical model, the Izhikevich neuron, with a noisy input current. We then fit these spike train data with a statistical model (a generalized linear model, GLM, with multiplicative influences of past spiking). For different levels of noise, we show how the GLM captures both the deterministic features of the Izhikevich neuron and the variability driven by the noise. We conclude that the GLM captures essential features of the simulated spike trains, but for near-deterministic spike trains, goodness-of-fit analyses reveal that the model does not fit very well in a statistical sense; the essential random part of the GLM is not captured. | {"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": 0.8173683285713196, "perplexity": 951.6760025898499}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662510097.3/warc/CC-MAIN-20220516073101-20220516103101-00562.warc.gz"} |
http://math.stackexchange.com/questions/143575/construct-locally-lipschitz-map-from-a-bounded-one | # Construct locally lipschitz map from a bounded one
Let $X$ be a Banach space and $BC(X)$ the space of all bounded closed subsets in $X$. It can be shown that $(BC(X),d_H)$ is a complete metric space (see this page for a definition of $d_H$). If $f:X\to X$ bounded. Does this imply that $F:BC(X)\to BC(X)$ given by $F(A)=\overline{\{f(a):a\in A\}}$ is locally Lipschitz continuous?
And if the map $f: X\to X$ is locally Lipschitz countinuous, is it automaticaly bounded?
-
Does this hold for $f:\mathbb R\to\mathbb R$ given by $f(x)=\sqrt[3]{x}$, which has arbitrarily large derivative near $0$? – Alex Becker May 10 '12 at 17:28
Take $X=\mathbb R$ and $f$ any continuous function on $\mathbb R$. Consider $A$, $B$ given by single points $A= \{a\}$, $B=\{b\}$. Then $F(A) = \{f(a)\}$, $F(B) = \{f(b)\}$, and $d_H(F(A),F(B)) = |f(a) - f(b)|$ while $d_H(A,B) = |a-b|$. So if $F$ is locally Lipschitz, $f$ must also be locally Lipschitz; if $F$ is bounded, $f$ must also be bounded. Any $f$ that is bounded but not locally Lipschitz is a counterexample to your first question. Any $f$ that is locally Lipschitz but not bounded is a counterexample to your second question (assuming the "it" there refers to $F$). | {"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": 0.9971151947975159, "perplexity": 81.9943403395691}, "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-2014-52/segments/1418802772757.23/warc/CC-MAIN-20141217075252-00090-ip-10-231-17-201.ec2.internal.warc.gz"} |
http://www.physicsforums.com/showthread.php?s=2e4c767208829914170bafd658d099cc&p=4583590 | # The role of phonons in momentum conservation
by hokhani
Tags: conservation, momentum, phonons, role
P: 265 In an indirect transition from the valence band maximum to conduction band minimum, the momentum of electron and hole would not change but the crystal momentum would change and this change is supplied by phonons.I have two questions here: 1) phonons don't carry momentum so how they can transfer thier momentum to the crystal? 2) phonons are part of the crystal. why do we separate their momentum from crystal momentum?
Sci Advisor P: 3,560 Phonons don't carry momentum but they carry crystal momentum which are two completely different things.
P: 265
Quote by DrDu Phonons don't carry momentum but they carry crystal momentum which are two completely different things.
Ok, but crystal momentum's change is sum of electron momentum's change and momentum change of the crystal;Ok? If yes, in the explained situation electron momentum is not changed so the momentum of crystal must be changed by phonons!
Sci Advisor P: 3,560 The role of phonons in momentum conservation In an indirect transition, the electrons crystal momentum changes this is compensated by a change of crystal momentum of the phonon. I don't see any problem here. The true momentum of the electron doesn't interest anyone in that context, as it isn't in a momentum eigenstate anyhow.
P: 265
Quote by DrDu The true momentum of the electron doesn't interest anyone in that context, as it isn't in a momentum eigenstate anyhow.
Yes, But what I meant was the mean value of true momentum of electron which is "mass of electron times its group velocity".
Sci Advisor P: 3,560 As I already explained in another thread, the lattice itself (as opposed to the phonons) can take up arbitrary amounts of momentum, so momentum conservation is always trivial.
Related Discussions Atomic, Solid State, Comp. Physics 18 Quantum Physics 9 General Physics 1 Classical Physics 2 General Engineering 0 | {"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": 0.8050955533981323, "perplexity": 1328.6109491052678}, "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-2014-23/segments/1406510264270.11/warc/CC-MAIN-20140728011744-00194-ip-10-146-231-18.ec2.internal.warc.gz"} |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=18&t=22049&p=63695 | ## Post-Module Assessment Q. 34
$\lambda=\frac{h}{p}$
William Lan 2l
Posts: 73
Joined: Fri Sep 29, 2017 7:07 am
### Post-Module Assessment Q. 34
If an electron (mass 9.11 x 10^-31 kg) has an associated wavelength of 7.28 x 10^-9 m, what is its speed? Is your answer reasonable, why?
A. 1.00 x 10-5 m.s-1. Yes. 1.00 x 10-5 m.s-1 is reasonable for e- as it is less than the speed of light, c = 3.0 x 108m.s-1.
B.1.00 x 10-5 m.s-1. No. 1.00 x 10-5 m.s-1 is not reasonable for e- as it is significantly slower than the speed of light, c = 3.0 x 108 m.s-1.
C. 1.00 x 105 m.s-1 . Yes. 1.00 x 105 m.s-1 is reasonable for e- as it is less than the speed of light, c = 3.0 x 108m.s-1.
D. 1.00 x 105 m.s-1. No. 1.00 x 105 m.s-1 is not reasonable for e- as it is significantly slower than the speed of light, c = 3.0 x 108 m.s-1.
So how exactly do you do this problem? I used lambda = h/mv equation and I switched it around so that v= h/(m x lambda). When I solved for V, I got 99939 m/s, which is none of the answers.
Chem_Mod
Posts: 17949
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 406 times
### Re: Post-Module Assessment Q. 34
When you do the calculation you did, you should get 1*10^5 m/s which is one of the answers given. Your formula is correct, but there is a mistake with the calculation.
Yashaswi Dis 1K
Posts: 56
Joined: Fri Sep 29, 2017 7:04 am
### Re: Post-Module Assessment Q. 34
I think your answer is like correct, meaning approx. close to the answers given b/c when you round the answer you got and change it to scientific notation, you should get 1.00 * 10^5 m/s and since it's way less than speed of light, which if the fastest speed known so far, answer should be reasonable. Again, check the video module maybe because it has more specifics on that. Hope this helps!
Lily Guo 1D
Posts: 64
Joined: Fri Sep 29, 2017 7:03 am
### Re: Post-Module Assessment Q. 34
Yashaswi Dis H wrote:I think your answer is like correct, meaning approx. close to the answers given b/c when you round the answer you got and change it to scientific notation, you should get 1.00 * 10^5 m/s and since it's way less than speed of light, which if the fastest speed known so far, answer should be reasonable. Again, check the video module maybe because it has more specifics on that. Hope this helps!
How do you know that the speed of the electron is reasonably fast or not? Is there a specific range of values that the speed should fall in? I got ~1.00 x 10^5 m/s, I'm just not sure if this is a reasonable speed or not since it's so much less than the speed of light.
Yashaswi Dis 1K
Posts: 56
Joined: Fri Sep 29, 2017 7:04 am
### Re: Post-Module Assessment Q. 34
I am not exactly sure but as far as I know, the speed of light is the fastest speed on earth, unless research shows otherwise. Thus, if you get a speed less than 3.00 * 10^8 m/s, I am pretty sure it's reasonable b/c it's less than speed of light, which is the fastest speed known so far. That's my thought process when I answered the module questions and it helped me out...so hope this helps! | {"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": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8728423118591309, "perplexity": 1155.0839212297847}, "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/1575540528490.48/warc/CC-MAIN-20191210180555-20191210204555-00255.warc.gz"} |
http://mathhelpforum.com/discrete-math/7398-countable-set.html | # Math Help - Countable set
1. ## Countable set
hello.
I cant seem to prove the following:
$A,B$ are two different sets such that $A \subset B$
prove that if there exists an injective function $f:A\to B$.
then there exists a countable infinite set $C \subset A$
(C and A can overlap)
can someone give me some hints?
thanks.
2. Originally Posted by parallel
hello.
I cant seem to prove the following:
$A,B$ are two different sets such that $A \subset B$
prove that if there exists an injective function $f:A\to B$.
then there exists a countable infinite set $C \subset A$
(C and A can overlap)
can someone give me some hints?
thanks.
There's something wrong here. First of all there is nothing in your statement to force A to be at least countably infinite, which is a requirement for C to be countably infinite.
For a counter-example let A = {1, 2, 3} and B = {1, 2, 3, 4, 5, 6, 7}. Define an injection $f: A \to B: a \mapsto a$. Then $A \subset B$ and f is an injection, but there exists no set $C \subset A$ that is countably infinite.
-Dan
3. I'm sorry,I typed it incorrectly
should be:
prove that if there exists an injective function $f:B\to A$.
then there exists a countable infinite set $C \subset A$
(C and A can overlap).
thanks
4. Originally Posted by parallel
I'm sorry,I typed it incorrectly
should be:
prove that if there exists an injective function $f:B\to A$.
then there exists a countable infinite set $C \subset A$
(C and A can overlap).
thanks
Which is again not true if both these sets are finite.
5. o.k so I think we can assume they are infinite(I think it's obvious),although I dont even have a clew,how to even start proving this.
6. If we were given that B is infinite then the statement is true.
We can prove that any infinite set has a countablely infinite subset call it $
D \subseteq B$
.
The subset of A you want is $\overrightarrow f (D)$, that is the image of D under f.
7. If $B$ is infinite this implies that $A$ is also infinite and at least as large as $B$ because of the injective map. Now, the property of the integers say there are contained up to cardinality in any infinite set, thus, there exist an injection
$i:\mathbb{Z}\to C$
Then the image of the function,
$i[\mathbb{Z}]\subseteq C$
(I hope that is what you mean by overlap).
8. I dont understand it.
I know that inorder to prove a set is countable,I need to show that there exists an injective function from this set to N(positive integers)
thanks
9. Originally Posted by parallel
I dont understand it.
I know that inorder to prove a set is countable,I need to show that there exists an injective function from this set to N(positive integers)
thanks
I assume that you are told $B$ is an infinite set.
1) $A$ is an infinite set also because the injection from $f:B\to A$ implies that $|B|\leq |A|$ in cardinality. So it must be infinite becuase it is larger than another infinite set (it cannot be finite because any infinite set is larger than any finite set).
2)Therefore, there exists an injection map $i: \mathbb{Z}\to A$. Because $|\mathbb{Z}|$ is the smallest possible infinite set which implies its size is contained in any infinite set. Thus, we state that in terms of an injection (since injection shows that one set is less than another in cardinality).
3)The image of the injection $i [ \mathbb{Z}]$ (image of a function, you should know what that is) is an infinite set that is the size of the integers and is contained in $A$. Thus, $i[\mathbb{Z}]\subset A$ which proves what you were asking.
10. Originally Posted by ThePerfectHacker
1) $A$ is an infinite set also because the injection from.
While that is perfectually true, I think that is the point of this problem.
In other words, I think that is what the student is asked to prove.
11. Thank you very much Plato.
12. Originally Posted by Plato
While that is perfectually true, I think that is the point of this problem.
In other words, I think that is what the student is asked to prove.
I can prove that! If that is what he wants.
I shall show there cannot exist a surjection,
$s: F\to I$
where $F$ is finite and $I$ is infinite.
Assume there is one.
Then since $I$ is infinite $\exists I' \subset I, |I'|=|I|$ (definition of infinity).
Then, the inverse image,
$s^{-1}[I']$ is a proper subset of $F$ (because that set was proper in the infinite set, the inverse image preserve this). But since $|I|=|I'|$ we have $|s^{-1}[I']|=|F|$. Thus, there exists a proper subset of a finite set having the same cardinality which is impossible by the definition of what finiteness is.
13. This is another case where the uniformity of mathematical definitions fails.
In the above, the definition of ‘infinite set’ was used.
Well which definition. This makes hard to help with knowing the text being followed.
There are at least three or maybe four popular definitions:
A set is infinite if it is not finite. A finite set is equipotent to a natural number.
A set is infinite if it is equipotent to a proper subset of itself (called Dedekind infinite).
A set is infinite if it contains a copy of the natural numbers.
I suspect that the purpose of the question was to show that “If an infinite set is mapped injectively of another set then the final set must also be infinite.” From the wording it is difficult to know what definition is being used. In any case my first response works.
14. Originally Posted by Plato
This is another case where the uniformity of mathematical definitions fails.
In the above, the definition of ‘infinite set’ was used.
Well which definition. This makes hard to help with knowing the text being followed.
There are at least three or maybe four popular definitions:
A set is infinite if it is not finite. A finite set is equipotent to a natural number.
A set is infinite if it is equipotent to a proper subset of itself (called Dedekind infinite).
A set is infinite if it contains a copy of the natural numbers.
I suspect that the purpose of the question was to show that “If an infinite set is mapped injectively of another set then the final set must also be infinite.” From the wording it is difficult to know what definition is being used. In any case my first response works.
But all of these definitions are equivalent!
Makes no difference what to use.
(I am only familar with Dedekind infinite).
15. Originally Posted by ThePerfectHacker
But all of these definitions are equivalent!
Makes no difference what to use.
But the definition in use changes the approach to the proof.
That was my point. | {"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": 41, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9790800213813782, "perplexity": 280.7844783896968}, "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/1435375096944.75/warc/CC-MAIN-20150627031816-00015-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://www.ipam.ucla.edu/abstract/?tid=8800&pcode=CMAWS2 | ## A Geometric Interpretation of the Characteristic Polynomial of a Hyperplane Arrangement
#### Caroline KlivansUniversity of ChicagoMathematics and Computer Science
We consider projections of points in $R^n$ onto chambers of real linear hyperplane arrangements. We show that the coefficients of the characteristic polynomial are proportional to the average spherical volumes of the sets of points that are projected onto faces of a given dimension. As a corollary we obtain that for real finite reflection arrangements the coefficients of the characteristic polynomial precisely give the spherical volumes of points projected onto faces of a fixed dimension of the fundamental chamber. The connection between projection volumes and the characteristic polynomial is established by considering angle sums of the associated zonotope.
This talk reflects joint work with Mathias Drton and Ed Swartz.
Back to Workshop II: Combinatorial Geometry | {"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": 0.9568727612495422, "perplexity": 534.1239231051809}, "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/1590347396300.22/warc/CC-MAIN-20200527235451-20200528025451-00348.warc.gz"} |
https://www.gradesaver.com/textbooks/math/calculus/university-calculus-early-transcendentals-3rd-edition/chapter-2-section-2-6-limits-involving-infinity-asymptotes-of-graphs-exercises-page-108/63 | ## University Calculus: Early Transcendentals (3rd Edition)
- Horizontal asymptote: $y=0$ - Vertical asymptote: $x=1$ As $x\to\pm\infty$, the dominant term is $0$. As $x\to1$, the dominant term is $1/(x-1)$.
$$y=\frac{1}{x-1}$$ We are interested in the behavior of function $y$ as $x\to\pm\infty$ as well as the behavior of $y$ as $x\to1$, which is where the denominator is $0$. We can rewrite the function into a polynomial with a remainder as follows: $$y=\frac{1}{x-1}=0+\frac{1}{x-1}$$ - As $x\to\pm\infty$, $(x-1)$ would get infinitely large, meaning that the curve will approach the line $y=0$, which is also the horizontal asymptote. Also, as $x$ becomes infinitely large, $1/(x-1)$ will become closer to $0$. So we can say the dominant term is $0$ as $x\to\pm\infty$. - As $x\to1$, $(x-1)$ would approach $0$, meaning that the curve will approach $\infty$. So $x=1$ is the vertical asymptote. And as $x\to1$, $1/(x-1)$ becomes infinitely large. So the dominant term as $x\to1$ is $1/(x-1)$. The graph is shown below. The red curve is the graph of $y$, while the blue line is the horizontal asymptote $y=0$ and the green one is the vertical asymptote $x=1$. | {"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": 0.9924293160438538, "perplexity": 86.7490633108112}, "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/1573496665985.40/warc/CC-MAIN-20191113035916-20191113063916-00397.warc.gz"} |
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