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http://math.stackexchange.com/questions/291634/differentiation-question	# Differentiation question? How would I solve the following problem? Where would the function $\|2x-1\|$ not be differentiable? I am thinking it would not be differentiable at $x=1/2$ because there it would be zero. - 1 I presume abs is the absolute value. Then yes, there is only one point where it is not differentiable and it is $1/2$. – julien Jan 31 at 20:27 @Fernando Martinez That's correct. It might help to think about the composition of the fucntion $abs$ with the function $2x-1$, i.e., $abs\circ f$, where $f(x)=2x-1$. – Git Gud Jan 31 at 20:28 yes it is absolute value, my question is that my question aks me to explain why it can not be differentiable at x=1/2 is it because there y=0, I am not sure. – Fernando Martinez Jan 31 at 20:29 1 If you sketch the graph, you will see it has a 'kink' at that point. It is not because it is zero there. – copper.hat Jan 31 at 20:31 ## 2 Answers Given $f(x) = \|2x - 1\|$, The function is not differentiable at $x = 1/2$; you can check the explanation below, and view the graph of the function, to see why. When $x > 1/2$, $f(x) = 2x - 1$. When $x\lt 1/2$, $f(x) = 1 - 2x$. If you graph these lines, you'll seen that they form a "upward V" where the graph abruptly changes direction at $x = 1/2$, at the point $(1/2, 0)$. By non-differentiable, I mean $$\lim_{x \downarrow \large\frac{1}{2}}\frac{f(x) - f(\frac{1}{2})}{x-\frac{1}{2}} = \lim_{x \downarrow \large\frac{1}{2}}\frac{2x-1}{x-\frac{1}{2}} = +2,$$ while $$\lim_{x \uparrow \large\frac{1}{2}}\frac{f(x) - f(\frac{1}{2})}{x-\frac{1}{2}} = \lim_{x \uparrow \large \frac{1}{2}}\frac{1-2x}{x-\frac{1}{2}} = -2$$ Hence, $lim_{x \to \frac{1}{2}} \dfrac{f(x) - f(\frac{1}{2})}{x-\frac{1}{2}}$ does not exist, and it follows by defintion that $f(x)$ is therefore not differentiable at $x = 1/2$ Graph of $\;f(x) = \left\|2x - 1\right\|$: - the function has no discontinuity in $x=1/2$ – Emanuele Paolini Jan 31 at 20:59 So the function is not differentiable when $x = 1/2$, and it just happens to be the case that that is where $f(x) = y = 0$. It is not differentiable by definition at $x = 1/2$, not because $f(x) = 0$. – amWhy Jan 31 at 20:59 2 being continuous is not enough. A differentiable function (at a point or on an interval) must be continuous there, but being non-differentiable at a point doesn't necessarily mean it's not continuous: The absolute value function is continuous, but fails to be differentiable at x = 1/2 since the tangent slopes do not approach the same value from the left as they do from the right of $x = 1/2$. – amWhy Jan 31 at 21:30 2 Because $x \gt 1/2 \implies 2x - 1 > 0 \implies \|2x - 1\| = 2x - 1$. Likewise $x \lt 1/2 \implies 2x - 1 < 0 \implies 1 - 2x > 0 \implies \|2x - 1\| = 1 - 2x$ – amWhy Jan 31 at 21:35 1 Yes, exactly! You've got it. – amWhy Jan 31 at 21:41 show 6 more comments Let $f(x) = \|2x-1\|$. Then, if $x<\frac{1}{2}$, $f(x) = 1-2x$, if $x\geq \frac{1}{2}$, then $f(x) = 2x-1$. Hence $\lim_{x \downarrow \frac{1}{2}}\frac{f(x) - f(\frac{1}{2})}{x-\frac{1}{2}} = \lim_{x \downarrow \frac{1}{2}}\frac{2x-1}{x-\frac{1}{2}} = +2$, but $\lim_{x \uparrow \frac{1}{2}}\frac{f(x) - f(\frac{1}{2})}{x-\frac{1}{2}} = \lim_{x \uparrow \frac{1}{2}}\frac{1-2x}{x-\frac{1}{2}} = -2$. So, the limit $x \to \frac{1}{2}$ does not exist. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 35, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9312646389007568, "perplexity_flag": "head"}
http://mathoverflow.net/questions/95021/homotopy-limits-of-quasi-categories/95054	## Homotopy limits of quasi-categories ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Quasi-categories (or $\infty$-categories, as they are often called) are a very convenient setting for doing abstract homotopy theory. One of their amazing features is the following: Given a diagram of quasi-categories, we can form its homotopy limit, yielding a quasi-category again. For example, the inverse (homotopy) limit of the diagram `$\cdots \xrightarrow{\Omega} \mathcal{S}_* \xrightarrow{\Omega} \mathcal{S}_* \xrightarrow{\Omega} \mathcal{S}_$` (for $\mathcal{S}_$ the quasi-category of pointed spaces and $\Omega$ denotes the loop space) gives the quasi-category of spectra. These homotopy limits can be (abstractly) defined to be the homotopy limits in the Joyal model structure on simplicial sets, where the quasi-categories are just the fibrant objects. There is also a more explicit description given in Lurie's Higher Topos Theory (we come back to an example later in this question). Many important examples of quasi-categories are constructed from a simplicial model category $\mathcal{M}$ in the following way: The sub simplicial category $\mathcal{M}^\circ$ of bifibrant objects forms a Bergner fibrant simplicial category and taking the coherent nerve produces a quasi-category $N(\mathcal{M}^\circ)$. Thus, the following question seems to be natural: Can we reconstruct the homotopy limit of the coherent nerves of a diagram of model categories as the coherent nerve of a "homotopy limit" of model categories? A candidate is given in Julie Bergner's paper Homotopy limits of model categories and more general homotopy theories, Definition 3.1. We won't recall here the general definition, but only indicate it in the case that we index over a diagram with one object and a group $G$ as automorphism (i.e., we have a group action on our model category): Then an object in $holim_G \mathcal{M}$ is an object $X \in \mathcal{M}$ together with morphisms $f_g: X \to g\cdot X$ (for $g\in G$) such that $f_e = id_X$ and $f_{hg} = (h\cdot f_g)\circ f_h$ (i.e, objects with a twisted $G$-action). At least in this case, the homotopy limit has a model structure (Bergner mentions the injective one, but at least sometimes, it has also the projective one, which is Quillen equivalent); actually it is a simplicial one if $\mathcal{M}$ was a simplicial model category. Indeed, it is the simplicial subcategory of $G$-equivariant morphisms in $Fun(EG, \mathcal{M})$ where $EG$ denotes the contractible groupoid associated to $G$. Thus, more precisely, our question is: Is $N((holim_G \mathcal{M})^\circ)$ categorically equivalent to $holim_G N(\mathcal{M}^\circ)$ as quasi-categories? There are several pieces of evidence for this: 1. If I am not mistaken, the description of homotopy limits in Higher Topos Theory implies that the homotopy fixed points of a quasi-category $\mathcal{C}$ are given as $Map(N(EG), \mathcal{C})^G$ (where Map denotes the internal Hom of simplicial sets and $()^G$ denotes strict fixed points). Thus the question is equivalent to whether $Map(N(EG), N(\mathcal{M}^\circ))^G$ is categorically equivalent to $N((Fun(EG, \mathcal{M})^G)^\circ)$. By strictification of homotopy coherent diagrams, a similar statement holds if we don't take $G$-fixed points, but I don't see how to prove the statement involving the $G$-fixed points. 2. Even more convincingly, Julie Bergner shows in her paper that homotopy limits of model categories are compatible with homotopy limits in the complete Segal space model structure on simplicial spaces. More precisely, she shows that the classification diagram functor commutes with homotopy limits up to weak equivalence (Theorem 4.1). Now, one could get the impression that we are finished since the complete Segal space model structure and the Joyal model structure are Quillen equivalent. But this is not sufficient: one has to prove that the classification diagram functor is send under this Quillen equivalence to something weakly equivalent to the coherent nerve. Although one gets some compatibility results from the papers Quasi-categories vs. Simplicial Categories (by Andre Joyal) and Complete Segal spaces arising from simplicial categories (by Julie Bergner), I didn't quite find the right statement to make the comparision work. As a last word of motiviation, I want to add that I stumbled upon these questions when I thought about Galois descent, where one often considers objects with twisted group actions. - I am guessing that $\Omega$ means loop space? – Spice the Bird Apr 24 2012 at 14:06 May be I'm wrong, but it seems to me that it should be: $Map(N(EG),N\mathbf({\mathcal{M}}^{\circ}))^{G}$ equivalent to $N((RHom(EG,\mathcal{M})^{G})^{\circ})$, where RHom is the right derived internal Hom functor defined by B.Toën for $\mathrm{Ho}(\mathbf{Cat}_{\Delta})$. – Fedotov Apr 24 2012 at 14:55 In general, if we have a (say, presentable, or simplicially enriched) model category, then a (co)limit (in the $\infty$-categorical sense) of a diagram in the underlying infinity-category corresponds to a homotopy (co)limit in the model category. Since the category of simplicially enriched categories with the Bergner model structure is Quillen equivalent to the category of simplicial sets with the Joyal model structure, the underlying $\infty$-categories are equivalent. So an $\infty$-categorical (co)limit of $\ifnty$-categories corresponds to a homotopy colimit of simplicial sets which – Dylan Wilson Apr 24 2012 at 17:19 (contd) corresponds to a homotopy (co)limit of simplicially enriched categories. – Dylan Wilson Apr 24 2012 at 17:20 Somewhere in there lies an answer to your question, since at the end of the day a homotopy limit of the simplicially enriched model categories will correspond to a homotopy limit of some ordinary simplicially enriched categories... But I may be spouting nonsense. – Dylan Wilson Apr 24 2012 at 17:21 show 1 more comment ## 1 Answer I will address your second question: "one has to prove that the classification diagram functor is sent under this Quillen equivalence to something weakly equivalent to the coherent nerve". The answer is yes, this is true. First note that the simplicial category $M^\circ$ is equivalent as a simplicial category to $L^HM$, the hammock localization of M (at the weak equivalences). See here for references. So the answer to your second question (or is it a remark?) follows as a special case of this previous MO answer (which cites work of Barwick-Kan and Toen). So the classification diagram functor and the coherent nerve of $M^\circ$ are sent to equivalent things under the Quillen equivalence between quasicategories and complete Segal spaces (I prefer the term Rezk categories or Rezk $\infty$-categories). I believe that combined with your other comments (notably using Julie Bergener's results) this is enough to answer your main question in the affirmative. - This seems to work (with a few details added). Many thanks! – Lennart Meier May 8 2012 at 22:19	{"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": 40, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9048364162445068, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/218043/functional-analysis-normed-linear-space?answertab=oldest	# functional analysis-normed linear space Let $V$ be a norm linear space and let $x\in V\setminus\{0\}$. Also let $W$ be a linear subspace of $V$. Show that if there is $r>0$ such that $\{y \in V \mid \lVert y\rVert< r\}$ is a subset of $W$, then $\frac{rx}{2\lVert x\rVert}\in W$. - I have edited the question. However I am not sure what exactly was your question at the end. Please use LaTeX in the future and try and ask questions with a bit more clarity in your words.It'll help you get better answers. – Vishesh Oct 21 '12 at 14:51 Right, Davide got it clean. Thanks. – Vishesh Oct 21 '12 at 14:53 ## 2 Answers Let $B_r=\{y\in V: \|\|y\|\|<r\}$. Because $\|\|x\|\|\ne 0$, we get $$\|\|\frac{r}{2\|\|x\|\|}x\|\|=\|\frac{r}{2\|\|x\|\|}\|\cdot \|\|x\|\|=\frac{r}{2\|\|x\|\|}\cdot \|\|x\|\|=\frac{r}{2}<r.$$ This implies that $\displaystyle{\frac{r}{2\|\|x\|\|}x}\in B_r$ and because $B_r\subseteq W$, the result follows. Thats what Davide means... - Hint: what is the norm of $\frac r{2\lVert x\rVert}x$? What can we say about elements which have norm $<r$? - Sorry Davide,I didn't get your hint!can you please explain me further.. – ccc Oct 21 '12 at 15:10 What don't you get exactly? – Davide Giraudo Oct 21 '12 at 15:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "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}, "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}, "english_confidence": 0.9378751516342163, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=4173603	Physics Forums Page 7 of 8 « First < 4 5 6 7 8 > ## Showing that Lorentz transformations are the only ones possible Quote by Fredrik Those formulas do imply that ##F=0\Leftrightarrow \dot v=0##. $$\gamma=\frac{1}{\sqrt{1-v^2}},\qquad m=\gamma m_0$$ $$\dot\gamma=-\frac{1}{2}(1-v^2)^{-\frac{3}{2}}(-2v\dot v)=\gamma^3v\dot v$$ $$\dot m=\dot\gamma m_0=\gamma^3v\dot v m_0$$ \begin{align} F &=\frac{d}{dt}(mv)=\dot m v+m\dot v=\gamma^3v^2\dot v m_0+\gamma m_0\dot v =\gamma m_0\dot v(\gamma^2v^2+1)\\ & =\gamma m_0\dot v\left(\frac{v^2}{1-v^2}+\frac{1-v^2}{1-v^2}\right) =\gamma^3 m_0\dot v \end{align} A complete specification of a theory of physics must include a specification of what measuring devices to use to test the theory's predictions. In particular, a theory about space, time and motion must describe how to measure lengths. It's not enough to just describe a meter stick, because the properties of a stick will to some degree depend on what's being done to it. So the theory must also specify the ideal conditions under which the measuring devices are expected to work the best. It's going to be very hard to specify a theory without ever requiring that an accelerometer displays 0. I don't even know if can be done. So non-accelerated motion is probably always going to be an essential part of all theories of physics. In all of our theories, motion is represented by curves in the underlying set of a structure called "spacetime". I will denote that set by M. A coordinate system is a function from a subset of M into ℝ4. If ##C:(a,b)\to M## is a curve in M, U is a subset of M, and ##x:U\to\mathbb R^4## is a coordinate system, then ##x\circ C## is a curve in C. So each coordinate systems takes curves in spacetime to curves in ℝ4. If such a curve is a straight line, then object has zero velocity in that coordinate system. If a coordinate system takes all the curves that represent non-accelerating motion are take to straight lines, then it assigns a constant velocity to every to non-accelerating objects. Those are the coordinate systems we call "inertial". There's nothing particularly old-fashioned about that. Ok, well-done and -explained (thanks). But all this concerns only special relativity. Where do you see that the question asked by the OP (and recalled by friend) is imposing linearity? For me it only imposes the Christoffel's work; see the other discussion "O-S model of star collapse" post 109, Foundations of the GTR by A. Einstein and translated by Bose, [793], (25). My impression (perhaps false) is that SR is based on a coherent but circular way of thinking including "linearity" for easy understandable historical reasons. The preservation of a length element (which is the initial question here) does not impose a flat geometry. Don't you think so? Quote by Fredrik The problem I'm interested in is this: If space and time are represented in a theory of physics as a mathematical structure ("spacetime") with underlying set ℝ4, then what is the structure? When ℝ4 is the underlying set, it's natural to assume that those functions are defined on all of ℝ4.The axioms will then include the statement that those functions are bijections from ℝ4 into ℝ4 I find this confusing, if you start by assuming a spacetime structure that admits bijections from ℝ4 into ℝ4 (that is E^4 or M^4) as the underlying structure because it seems natural to you, you are already imposing linearity for the transformations that respect the relativity principle. This leaves only the two posible transformations you comment below. The second postulate of SR is what allows us to pick which of the two is the right transformation. But if you follow this path it is completely superfluous to prove anything about mapping straight lines to straight lines to get the most general transformation that does that and once you have it restrict it to the linear ones with a plausible physical assumption, since you are already starting with linear transformations. Quote by Fredrik What we find when we use the axioms is that this subgroup is either the group of Galilean boosts and proper and orthochronous rotations, or it's isomorphic to the restricted (i.e. proper and orthochronous) Lorentz group. In other words, we find that "spacetime" is either the spacetime of Newtonian mechanics, or the spacetime of special relativity. Those are really the only options when we take "spacetime" to be a structure with underlying set ℝ4. Just a minor correction the Lorentz transformations are locally isomorphic to the restricted group. Mentor Quote by TrickyDicky I find this confusing, if you start by assuming a spacetime structure that admits bijections from ℝ4 into ℝ4 (that is E^4 or M^4) as the underlying structure because it seems natural to you, you are already imposing linearity for the transformations that respect the relativity principle. This leaves only the two posible transformations you comment below. How am I "already imposing linearity"? I'm starting with "takes straight lines to straight lines", because that is the obvious property of inertial coordinate transformations, and then I'm using the theorem to prove that (when spacetime is ℝ4) an inertial coordinate transformation is the composition of a linear map and a translation. I don't think linearity is obvious. It's just an algebraic condition with no obvious connection to the concept of inertial coordinate transformations. Quote by TrickyDicky The second postulate of SR is what allows us to pick which of the two is the right transformation. Right, if we add that to our assumptions, we can eliminate the Galilean group as a possibility. But I would prefer to just say this: These are the two theories that are consistent with a) the idea that ℝ4 is the underlying set of "spacetime", and b) our interpretation of the principle of relativity as a set of mathematically precise statements about transformations between global inertial coordinate systems. Now that we have two theories, we can use experiments to determine which one of them makes the better predictions. Quote by TrickyDicky Just a minor correction the Lorentz transformations are locally isomorphic to the restricted group. How is that a correction? It seems like an unrelated statement. Quote by Fredrik How am I "already imposing linearity"? I'm starting with "takes straight lines to straight lines", because that is the obvious property of inertial coordinate transformations, and then I'm using the theorem to prove that (when spacetime is ℝ4) an inertial coordinate transformation is the composition of a linear map and a translation. I don't think linearity is obvious. It's just an algebraic condition with no obvious connection to the concept of inertial coordinate transformations. Right, if we add that to our assumptions, we can eliminate the Galilean group as a possibility. But I would prefer to just say this: These are the two theories that are consistent with a) the idea that ℝ4 is the underlying set of "spacetime", and b) our interpretation of the principle of relativity as a set of mathematically precise statements about transformations between global inertial coordinate systems. Now that we have two theories, we can use experiments to determine which one of them makes the better predictions. The experiment for the actual discussion here is the Morley and Michelson experiment. How is that a correction? It seems like an unrelated statement. Intuitively (I am not a specialist) this means that that isomorphism holds true only locally (on short distances around the observer). There is not really a global inertial coordinate system (except on the paper, in theory). And (as far I understand the generalized version of the theory) this is a crucial point. Among others things, this was forcing us (Weyl's work) to introduce the concept of parallel transport and of connection. Quote by Fredrik How am I "already imposing linearity"? The assumption of a spacetime that is globally R^4(not just locally wich is the weaker asumption) means your underlying geometry is flat(Minkowskian, Euclidean), do you agree? Given that space, the transformations that leave inertial coordinates invariant in the sense of SR first postulate must automatically be linear transformations, do you agree? Maybe this is not as obvious to see as I think, but I I think it is correct. Quote by Fredrik How is that a correction? It seems like an unrelated statement. Well, It just seemed important to make more precise that the isomorphism you were talking about is local. Mentor Quote by Blackforest But all this concerns only special relativity. And pre-relativistic classical mechanics. It concerns all theories with ℝ4 as spacetime. I think it's pretty cool that there are only two such theories that are consistent with a straightforward interpretation of the principle of relativity. Quote by Blackforest Where do you see that the question asked by the OP (and recalled by friend) is imposing linearity? For me it only imposes the Christoffel's work; see the other discussion "O-S model of star collapse" post 109, Foundations of the GTR by A. Einstein and translated by Bose, [793], (25). Someone who tries to argue that a transformation that satisfies the OP's condition must be a Lorentz transformation has probably already assumed that spacetime is ℝ4, and that the theory will involve global (i.e. defined on all of spacetime) inertial coordinate systems. That a transformation between two global inertial coordinate systems is a bijection and takes straight lines to straight lines is just a consequence of the definition of "global inertial coordinate system". The 4-dimensional version of the theorem I stated and proved in #98 shows that a bijection that takes straight lines to straight lines is affine (i.e. a composition of a linear map and a translation). So when we begin to consider the OP's condition, it's already a matter of determining which affine maps satisfy it. And the condition implies that 0 is taken to 0, so there's no translation involved, i.e. the transformation is linear. Quote by Blackforest My impression (perhaps false) is that SR is based on a coherent but circular way of thinking including "linearity" for easy understandable historical reasons. I don't think there's anything circular about it. It's perhaps naive to think that we should be able to use ℝ4 as our spacetime, and talk about global inertial coordinate systems. But it makes sense to first find all such theories, and then ask what other theories are worth considering. I might take a look at that problem when I have worked out all the details of the ℝ4 case. Mentor Quote by TrickyDicky The assumption of a spacetime that is globally R^4(not just locally wich is the weaker asumption) means your underlying geometry is flat(Minkowskian, Euclidean), do you agree? I don't agree. We don't have a geometry at that stage, because until we have chosen an inner product (or something similar), ℝ4 is just a set. (And in the case of Galilean transformations, we will never define anything like an inner product on ℝ4). The lines that we call "straight" are straight in the Euclidean sense, but we're not considering them because they're straight in the Euclidean sense, but because they describe motion with a constant velocity. We don't need an inner product to see that they do. Quote by TrickyDicky Given that space, the transformations that leave inertial coordinates invariant in the sense of SR first postulate must automatically be linear transformations, do you agree? Maybe this is not as obvious to see as I think, but I I think it is correct. They must automatically be affine maps, but it takes a non-trivial theorem* to see that, and you specifically said that there's no need to prove that theorem. ) This theorem is essentially "the fundamental theorem of affine geometry", stated in terms of vector spaces instead of affine spaces. Quote by TrickyDicky Well, It just seemed important to make more precise that the isomorphism you were talking about is local. But it's not. This is the 1+1-dimensional version of what I said, with all the details made explicit: For each K>0, the group ##G_K=\{\Lambda(v)\|v\in (-c,c)\}##, where ##c=1/\sqrt{K}## and $$\Lambda(v)=\frac{1}{\sqrt{1-Kv^2}}\begin{pmatrix}1 & -Kv\\ -v & 1\end{pmatrix}$$ is isomorphic to the restricted Lorentz group. There's nothing local about this. In fact, when K=1, this group is the restricted Lorentz group, and the isomorphism is the identity map. Quote by Fredrik I don't agree. We don't have a geometry at that stage, because until we have chosen an inner product (or something similar), ℝ4 is just a set. . Sorry, aren't we asuming inner product spaces? how can we even talk about transformation matrices otherwise? Quote by Fredrik But it's not. Well, it's not with your assumption of flat inner product space, but if you consider general manifolds the restricted Lorentz group is locally isomorphic to the Lorentz group. Mentor Quote by TrickyDicky Sorry, aren't we asuming inner product spaces? how can we even talk about transformation matrices otherwise? I'm not even mentioning matrices until later in the argument, after I've determined that we're dealing with linear operators. To associate a matrix with a linear operator, we only need a basis. You wanted to prove that linear transformations are the only ones possible if one wants use rigorously the first postulate of SR, you bring a R^4 vector space because you consider natural the assumption that the space must be globally R^4, not just locally like in general manifolds, and in this space you need to perform matrix multiplications like:##T(x)=\Lambda x## that looks like a matrix product to me so we are starting with an R^4 vector space with an inner product structure, no? That is called a Euclidean structure IMO. Here's a web page that talks about how Einstein and others justified the linearity of the transformations, and the extra assumptions necessary to exclude linear fractional transformations: http://www.mathpages.com/home/kmath659/kmath659.htm Mentor Quote by TrickyDicky You wanted to prove that linear transformations are the only ones possible if one wants use rigorously the first postulate of SR, you bring a R^4 vector space because you consider natural the assumption that the space must be globally R^4, not just locally like in general manifolds, and in this space you need to perform matrix multiplications like:##T(x)=\Lambda x## that looks like a matrix product to me so we are starting with an R^4 vector space with an inner product structure, no? That is called a Euclidean structure IMO. I'm not using the principle of relativity to prove that they're linear. The notation ##T(x)=\Lambda x+a## doesn't mean that ##\Lambda## is a matrix at this point. It only means that I'm using the standard convention to not write out parentheses when the map is known to be linear. We don't need an inner product to associate matrices with linear operators. We only need a basis for that. If U and V are vector spaces with bases ##A=\{u_i\}## and ##B=\{v_i\}## respectively, then the ij component of ##T:U\to V## with respect to the pair of bases (A,B) is defined as ##(Tu_j)_i##. The matrix associated with T (and the pair (A,B)) has ##(Tu_j)_i## (=the ith component of ##Tu_j##) on row i, column j. Spacetime is a structure with underlying set M. * We intend to use curves in M to represent motion. * There's a special set of curves in M that we can use to represent the motion of non-accelerating objects. * M can be bijectively mapped onto ℝ4. * A coordinate system on a subset ##U\subset M## is an injective map from U into ℝ4. * A global coordinate system on M is a coordinate system with domain M. * A global inertial coordinate system is a global coordinate system that takes the curves that represent non-accelerating motion to straight lines. * If x and y are global coordinate systems, then ##x\circ y^{-1}## represents a change of coordinates. I call these functions coordinate transformations. When both x and y are global inertial coordinate systems, I call ##x\circ y^{-1}## an inertial coordinate transformation. (I'm getting tired of saying "global" all the time). * These definitions imply that an inertial coordinate transformation is a bijection that takes straight lines to straight lines. * The fundamental theorem of affine geometry tells us that this implies that inertial coordinate transformations are affine maps. * This implies that an inertial coordinate transformation that takes 0 to 0 is linear. * The principle of relativity tells us among other things that the set of inertial coordinate transformations is a group. * This group has a subgroup G that consists of the proper and orthochronous inertial coordinate transformations that take 0 to 0. * We interpret the principle of relativity as imposing a number of other conditions on G. * Since the members of G are linear (we know this because they are affine and take 0 to 0), we can write an arbitrary member of G as a matrix. (This requires only a basis, not an inner product, and all vector spaces have a basis). * The conditions inspired by the principle of relativity determine a bunch of relationships between the components of that matrix. * Those relationships tell us that the group is either the restricted Galilean group without translations, or isomorphic to the restricted Lorentz group. (Restricted = proper and orthochronous). * This implies that the group of all inertial coordinate transformations is either the Galilean group or the Poincaré group. * We therefore define spacetime as a structure that has ℝ4 as the underlying set, and somehow singles out exactly one of these two groups as "special". * A nice way to define a structure that singles out the Poincaré group is to define spacetime as the pair (ℝ4,g), where g is a Lorentzian metric whose isometry group is the Poincaré group. * There's no equally nice way to handle the Galilean case. I think we either have to define spacetime as (ℝ4,G,g), where G is the Galilean group and G the metric on "space", or define it as a fiber bundle. (An ℝ3 bundle over ℝ, where each copy of ℝ3 is equipped with the Euclidean inner product). The former option is ugly. The latter is difficult to understand, unless you already understand fiber bundles of course. Recognitions: Gold Member I'm given to understand that dτ2=dt2-dx2 = dt'2-dx'2 when (t',x') are the Lorentz transformation of (t,x). Perhaps it's instructive to consider in what circumstances dτ should want to be considered invariant wrt to coordinate changes. Maybe those requirements are the driving force behind the necessity of the Lorentz transformations. For example, the most obvious use of dτ is in the calculation of the line integral, $$\int_{{\tau _0}}^\tau {d\tau '} = \tau - {\tau _0}$$ which is the length of a line measured in terms of segments marked off along the length of the line. Then, of course, we can always place this line in an arbitrarily oriented coordinate system and express τ in term of those coordinates. So the question is, when do we want to use the coordinates (t,x), and when would we want τ-τ0 to be invariant wrt to those coordinates? Usually, we specify a curve in space by parameterizing the space coordinates with an arbitrary variable, call it "t". But since the x and t coordinates are arbitrarily assigned, the length of the curve can depend on the (t,x) coordinates. But if you specify that the length of the curve is invariant, then this requires the Lorentz transformations between coordinate systems. But what requires the length of the curve to be invariant? Perhaps if we have a more fundamental requirement like $$\int_{{\tau _0}}^\tau {f(\tau - {\tau _0})d\tau } = a$$ this will require the length of τ-τ0 to be invariant wrt to coordinate changes in (t,x). For example, maybe ${f(\tau - {\tau _0})}$ might be a probability distribution along a path so that its integral along the path must be 1 in any coordinate system. Did I get this all right? I would appreciate comments. Thank you. Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Quote by TrickyDicky You wanted to prove that linear transformations are the only ones possible if one wants use rigorously the first postulate of SR, you bring a R^4 vector space because you consider natural the assumption that the space must be globally R^4, not just locally like in general manifolds, and in this space you need to perform matrix multiplications like:##T(x)=\Lambda x## that looks like a matrix product to me so we are starting with an R^4 vector space with an inner product structure, no? That is called a Euclidean structure IMO. Why do you think we need inner products to define matrix products?? Quote by micromass Why do you think we need inner products to define matrix products?? No, it's not needed, I thought Fredrik was assuming Euclidean geometry but he wasn't. Mentor Quote by Fredrik * Spacetime is a structure with underlying set M. * We intend to use curves in M to represent motion. * There's a special set of curves in M that we can use to represent the motion of non-accelerating objects. * M can be bijectively mapped onto ℝ4. * A coordinate system on a subset ##U\subset M## is an injective map from U into ℝ4. * A global coordinate system on M is a coordinate system with domain M. * A global inertial coordinate system is a global coordinate system that takes the curves that represent non-accelerating motion to straight lines. * If x and y are global coordinate systems, then ##x\circ y^{-1}## represents a change of coordinates. I call these functions coordinate transformations. When both x and y are global inertial coordinate systems, I call ##x\circ y^{-1}## an inertial coordinate transformation. (I'm getting tired of saying "global" all the time). * These definitions imply that an inertial coordinate transformation is a bijection that takes straight lines to straight lines. I have some concerns about this part. Maybe there is some circularity in the argument after all. It doesn't seem obvious* that the "special" curves in spacetime that represent non-accelerated motion should include curves that correspond to infinite speed in some inertial coordinate system. If we leave them out, then what I call an inertial coordinate transformation will be a map that takes finite-speed straight lines to finite-speed straight lines. Of course, inertial coordinate transformations in SR (i.e. Poincaré transformations) can take infinite-speed lines to finite-speed lines and vice versa. If inertial coordinate transformations can't do this, there's no relativity of simultaneity. So if we leave out the infinite-speed lines from the start, we will come to the conclusion that there's only one possibility: The group is the Galilean group. (Hm, maybe there will actually be infinitely many possibilities, distinguished by what exactly they're doing to infinite-speed lines). Do we have a reason to include infinite-speed lines other than that we know what we want the final answer to be? ) Recall that the main reason why we need spacetime to include that special set of curves is that they (or at least some of them) are to represent the motions of "observers" that are minimally disturbed by what's being done to them. (An "observer" here is not necessarily conscious. It could be a measuring device). Recognitions: Science Advisor Quote by Fredrik Do we have reason to include infinite-speed lines other than that we know what we want the final answer to be? ) Recall that the main reason why we need spacetime to include that special set of curves is that they (or at least some of them) are to represent the motions of "observers" that are minimally disturbed by what's being done to them. (An "observer" here is not necessarily conscious. It could be a measuring device). This sort of thing is one reason why I prefer to start from inertial observers defined as those that feel no acceleration. If one finds the maximal dynamical group applicable to the zero-acceleration equations of motion, the problematic case you mentioned can be handled by taking a limit afterwards. Page 7 of 8 « First < 4 5 6 7 8 > Thread Tools \| \| \| \| \|--------------------------------------------------------------------------------------\|------------------------------\|---------\| \| Similar Threads for: Showing that Lorentz transformations are the only ones possible \| \| \| \| Thread \| Forum \| Replies \| \| \| General Math \| 35 \| \| \| Special & General Relativity \| 5 \| \| \| Special & General Relativity \| 1 \| \| \| Special & General Relativity \| 4 \| \| \| Special & General Relativity \| 8 \|	{"extraction_info": {"found_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": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9302687048912048, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/251209/when-is-a-quotient-ring-of-the-gaussian-integers-an-integral-domain	# When is a quotient ring of the Gaussian integers an integral domain? Let $\mathbb Z[i]$ denote the ring of Gaussian integers. For which of the following values of $n$ is the quotient ring $\mathbb Z[i]/n \mathbb Z[i]$ an integral domain? $2,13,19,7$ How can I solve the problem? - 1 Why do you post two questions within 5 minutes with two different user names? – Phira Dec 5 '12 at 2:43 1 when n is a gaussian prime, so 19 and 17 (because they equal 3 mod 4). – user51427 Dec 5 '12 at 2:44 ## 3 Answers $Z[i]/n Z[i]$ is an integral domain iff $n$ is a Gaussian prime. No composite integer $n$ can be a Gaussian prime. $2$ is not a Gaussian prime because $2=(1+i)(1-i)=(1+i)^2(-i)$. Primes $p$ that are the sum of two squares are not Gaussian primes, because $p=a^2+b^2=(a+bi)(a-bi)$. So $p=2$ and $p\equiv 1\bmod 4$ are not Gaussian primes. This leaves us with the primes that are not the sum of two squares, which are exactly those that are congruent to $3 \bmod 4$. These are indeed Gaussian primes. - A small start: $(3+2i)(3-2i)$ is divisible by $13$. - Hint $\rm\ \, R = \Bbb Z[{\it i}\,]\cong \Bbb Z[x]/(x^2\!+\!1)\:$ so $\rm\:R/2 \cong \Bbb Z_2[x]/(x^2\!+\!1) \cong \Bbb Z_2[x]/(x\!+\!1)^2\:$ is not a domain. Similarly, we easily compute that $\rm R/13 \cong \Bbb Z_{13}[x]/(x^2\!+\!1) \cong \Bbb Z_{13}[x]/(x\!-\!5)(x\!+\!5)\:$ is not a domain. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 19, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8982937335968018, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/22201/can-a-static-magnetic-field-turned-into-a-static-electric-field-or-vice-versa?answertab=oldest	# Can a static magnetic field turned into a static electric field? or vice versa? Consider some positive charge that is distributed uniformly over a very long line along the z-axis. If I am stationary with respect to the line then there is only static electric field which has cylindrical symmetry. Assume now that I am moving with some constant velocity which has only a (positive) component along the z axis. With respect to me there is now current running down the (negative) z-axis, hence I expect to find a static magnetic field in my moving inertial frame. My question is, do I still find (a static) electric field as well in this case? [Because on one hand I find the answer to be NO, since static electric fields should come only from static charges, and the charges here will not be static. On the other hand, although the line carries current, the line is still charged (which is different from currents running in conducting wires where the wires are neutral at all times). That means the answer is YES and one expects to see static electric field because the line is continuously charged.] So is the answer yes or no (and why)? - ## 2 Answers By taking the contraction of the electromagnetic field tensor with itself, you can obtain the following Lorentz invariant quantity: $$B^2 - \frac{E^2}{c^2}$$ Now, in the original frame we have only an electric field and no magnetic field. Thus, the quantity above is negative. In the other frame, we know that we have a magnetic field. If we had no electric field, the resulting quantity $B^2 - E^2/c$ would be positive. But that cannot be, because if a quantity is invariant, it must stay the same after a Lorentz transformation, and it can surely not simply change sign! So there must still be a static electric field! - Wonderful. That answers automatically the question in the title of my post as well. So an electric field CANNOT turn into a magnetic field and vice versa, right? By the way, that Lorentz invariant quantity is true regardless of the fields are time dependent or not, right? – Revo Mar 11 '12 at 0:07 Well, it's true that the quantity is Lorentz invariant, but you might have to be careful about at what time you evaluate the fields, e.g. $B(t)$ in one frame and $B'(t')$ in another frame. – Lagerbaer Mar 11 '12 at 0:32 1 Won't the linear charge density simply increase (for the moving versus at-rest observer) according to the $\gamma$ factor? I thought charge itself was taken to be Lorentz invariant, since if you had a spaceship with a given charge all observers would agree on that charge. Obviously it's a little different for a line charge. – AlanSE Mar 11 '12 at 1:04 1 You can define the Lorentz covariant four-density $(c\rho, \vec{j})$, and that already suggests that if you have $(c\rho, \vec{0})$ in one system, you can get a non-zero current density in a system in relation to that. – Lagerbaer Mar 11 '12 at 15:33 Lagerbaer's answer is great, and convincingly demonstrates that it is impossible to eliminate a static electric field through a Lorentz transformation. (In fact, you only make it stronger!) Another argument uses Gauss's law. No matter how you Lorentz-transform your configuration of static charges, you cannot eliminate the charge; all you do is scale its density to account for the change in volume caused by length contraction. And if there is any charge at all in some frame, Gauss's law tells you there must be electric field in that frame. So: no way to get rid of the electric field. In your specific case: If you are travelling parallel to the line of charge, the line of charge will be contracted by the factor $\gamma$, so the density of charge will increase by the factor $\gamma$. Then, if you draw a cylinder around some length $L$ of charge, Gauss's law tells you that there must be net electric flux through this cylinder; in particular, $\gamma$ times the electric flux through a cylinder of length $L$ in the rest frame. If you knew that the electric field in the moving frame were perfectly radial, you could infer that the electric field in the moving frame would have to be $\gamma$ times the electric field in the rest frame. This is a little dicey, though -- since the charges are moving in a certain direction, you've lost a reflective symmetry which you used to ensure that the field was radial in the rest frame. As it turns out, the electric field is radial, even in the moving frame, so you get $E'=\gamma E$. Off the top of my head, I can't think of an easy way to see this, though. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9455974698066711, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/71447/large-n-asymptotic-of-int-0-infty-left-1-x-n-rightn-1-exp-x/71454	# Large $n$ asymptotic of $\int_0^\infty \left( 1 + x/n\right)^{n-1} \exp(-x) \, \mathrm{d} x$ While thinking of 71432, I encountered the following integral: $$\mathcal{I}_n = \int_0^\infty \left( 1 + \frac{x}{n}\right)^{n-1} \mathrm{e}^{-x} \, \mathrm{d} x$$ Eric's answer to the linked question implies that $\mathcal{I}_n \sim \sqrt{\frac{\pi n}{2}} + O(1)$. How would one arrive at this asymptotic from the integral representation, without reducing the problem back to the sum ([added] i.e. expanding $(1+x/n)^{n-1}$ into series and integrating term-wise, reducing the problem back to the sum solve by Eric) ? Thanks for reading. - what if expand brackets and write series in Gamma-functions? – Ilya Oct 10 '11 at 15:10 @Gortaur In that case we would arrive back to the sum Eric dealt with. I am hoping for something like a saddle point approximation approach. – Sasha Oct 10 '11 at 15:13 I see, sorry - didn't understand which sum you tried to avoid – Ilya Oct 10 '11 at 15:14 1 – J. M. Oct 10 '11 at 15:46 3 For what it's worth, this integral and its relationship to the sum in Question 71432 is mentioned in Flajolet, Grabner, Kirschenhofer, and Prodinger, "On Ramanujan's Q-Function" (Journal of Computational and Applied Mathematics 58 (1995), 103-116). – Mike Spivey Oct 10 '11 at 16:10 show 4 more comments ## 4 Answers With the change of variables $x\to(n-1)t-1$, we get $$\begin{align} &\int_0^\infty\left(1+\frac{x}{n}\right)^{n-1}e^{-x}\;\mathrm{d}x\\ &=ne\left(1-\frac{1}{n}\right)^n\int_\frac{1}{n-1}^\infty e^{-(n-1)(t-\log(1+t))}\;\mathrm{d}t\tag{1}\\ \end{align}$$ Since $$1+n\log\left(1-\frac{1}{n}\right)=-\frac{1}{2n}-\frac{1}{3n^2}+O\left(\frac{1}{n^3}\right)$$ exponentiating and multiplying by $n$, we get $$\begin{align} ne\left(1-\frac{1}{n}\right)^n &=ne^{-\frac{1}{2n}-\frac{1}{3n^2}+O\left(\frac{1}{n^3}\right)}\\ &=n-\frac{1}{2}-\frac{5}{24n}+O\left(\frac{1}{n^2}\right)\tag{2} \end{align}$$ Note that $$\begin{align} \int_0^\frac{1}{n-1} e^{-(n-1)(t-\log(1+t))}\;\mathrm{d}t &=\int_0^\frac{1}{n-1}\left(1-\tfrac{n-1}{2}t^2+\tfrac{n-1}{3}t^3+\tfrac{(n-1)^2}{8}t^4\right)\;\mathrm{d}t+O\left(\tfrac{1}{n^4}\right)\\ &=\frac{1}{n-1}-\frac{1}{6(n-1)^2}+\frac{13}{120(n-1)^3}+O\left(\frac{1}{n^4}\right)\\ &=\frac{1}{n}+\frac{5}{6n^2}+\frac{31}{40n^3}+O\left(\frac{1}{n^4}\right)\tag{3} \end{align}$$ Finally, setting $\frac{u^2}{2}=t-\log(1+t)$, so that $t=u+\frac{u^2}{3}+\frac{u^3}{36}-\frac{u^4}{270}+\frac{u^5}{4320}+\frac{u^6}{17010}+O(u^7)$, we get $$\begin{align} &\int_0^\infty e^{-(n-1)(t-\log(1+t))}\;\mathrm{d}t\\ &=\int_0^\infty e^{-(n-1)u^2/2}\;\mathrm{d}t\\ &=\int_0^\infty e^{-(n-1)u^2/2}\;(1+\frac{2u}{3}+\frac{u^2}{12}-\frac{2u^3}{135}+\frac{u^4}{864}+\frac{u^5}{2835}+O(u^6))\;\mathrm{d}u\\ &=\sqrt{\tfrac{\pi}{2(n-1)}}+\tfrac{2}{3(n-1)}+\sqrt{\tfrac{\pi}{288(n-1)^3}}-\tfrac{4}{135(n-1)^2}+\sqrt{\tfrac{\pi}{165888(n-1)^5}}+\tfrac{8}{2835(n-1)^3}+O\left(\tfrac{1}{n^{7/2}}\right)\\ &=\sqrt{\tfrac{\pi}{2n}}\left(1+\tfrac{7}{12n}+\tfrac{145}{288n^2}\right)+\left(\tfrac{2}{3n}+\tfrac{86}{135n^2}+\tfrac{346}{567n^3}\right)+O\left(\tfrac{1}{n^{7/2}}\right)\tag{4} \end{align}$$ Combining $(3)$ and $(4)$, we get $$\int_\frac{1}{n-1}^\infty e^{-(n-1)(t-\log(1+t))}\;\mathrm{d}t=\sqrt{\tfrac{\pi}{2n}}\left(1+\tfrac{7}{12n}+\tfrac{145}{288n^2}\right)-\left(\tfrac{1}{3n}+\tfrac{53}{270n^2}+\tfrac{3737}{22680n^3}\right)+O\left(\tfrac{1}{n^{7/2}}\right)$$ Including $(2)$, yields $$\int_0^\infty\left(1+\frac{x}{n}\right)^{n-1}e^{-x}\;\mathrm{d}x=\sqrt{\tfrac{n\pi}{2}}\left(1+\tfrac{1}{12n}+\tfrac{1}{288n^2}\right)-\left(\tfrac{1}{3}+\tfrac{4}{135n}-\tfrac{8}{2835n^2}\right)+O\left(\tfrac{1}{n^{5/2}}\right)$$ - Added a bit more of the asymptotic expansion – robjohn♦ Oct 10 '11 at 21:07 Your edited answer is close, but doesn't quite agree with the published result in Flajolet, et al, mentioned in my comment on the original question. (The integral is equal to $Q(n)$ in their paper; see Theorem 2.) – Mike Spivey Oct 10 '11 at 21:45 @Mike: Thanks! I found the error; I had multiplied two terms instead of adding them. I have double checked the rest, too. – robjohn♦ Oct 10 '11 at 23:45 You're welcome, and nice work, by the way! – Mike Spivey Oct 11 '11 at 0:07 Doing a numerical check, I found an error in the sixth order $\left(\frac{1}{n^2}\right)$ term. Now things seem correct. The error term is approximately $\frac{1}{300n^{5/2}}$. – robjohn♦ Oct 11 '11 at 13:51 A related result was given in the problems column of the American Mathematical Monthly not too long ago. This is problem 11353 whose solution was published in the January 2010 issue. Let $$g(s)=\int_0^\infty \left(1+{x\over s}\right)^se^{-x}\, dx-\sqrt{s\pi\over 2}.$$ Show that $g(s)$ decreases from $1$ to $2/3$ as $s$ ranges from $0$ to $\infty$. Note that the exponent in the integral is $s$, not $s-1$. - 3 Note that by-parts gives $$\int_0^\infty\left(1+\frac{x}{s}\right)^{s-1}e^{-x}dx=1+\int_0^\infty\left(1+\‌frac{x}{s}\right)^se^{-x}dx.$$ – anon Oct 10 '11 at 18:31 @anon: Since my answer gives $-\frac{1}{3}$, I checked and I believe that your limits on the boundary terms are switched. I get $$\int_0^\infty\left(1+\frac{x}{s}\right)^{s-1}e^{-x}dx=-1+\int_0^\infty\left(1+‌\frac{x}{s}\right)^se^{-x}dx.$$ – robjohn♦ Oct 10 '11 at 21:20 @robjohn: whoops, you are correct. – anon Oct 10 '11 at 21:26 @Byron: is their solution simpler in nature than mine? They needn't carry out as many terms as I did, but other than that, is the idea of theirs simpler? – robjohn♦ Oct 10 '11 at 21:27 @robjohn No, I wouldn't call their solution simple. It uses similar ideas to yours, but needs extra care in showing that $g(s)$ is decreasing. – Byron Schmuland Oct 10 '11 at 21:39 Interesting. I've got a representation $$\mathcal{I}_n = n e^n \int_1^\infty t^{n-1} e^{- nt}\, dt$$ which can be obtained from yours by the change of variables $t=1+\frac xn$. After some fiddling one can get $$2\mathcal{I}_n= n e^n \int_0^\infty t^{n-1} e^{- nt}\, dt+o(\mathcal{I}_n)= n^{-n} e^n \Gamma(n+1)+\ldots=\sqrt{2\pi n}+\ldots.$$ - Thank you, could you elaborate a little on the fiddling needed ? – Sasha Oct 10 '11 at 15:49 1 Briefly it's like follows. Consider function $g_n(t)=n e^n t^{n-1}e^{-nt}$, so $\mathcal{I}_n=\int_1^\infty g_n(t)\, dt\$. The maximum of $g_n$ tends to $1$n as $n\to\infty$. For $\varepsilon\in(0,1)$ $\int_{[0,+\infty]\backslash [1-\varepsilon,1+\varepsilon]\ }g_n(t)\, dt=o(1)\$, $n\to\infty\$. So it is enough to consider $\int_{1-\varepsilon}^{1+\varepsilon}g_n(t)$. Also $g(1-t)=e^{2 n t}\left(\frac{1-t}{1+t}\right)^{n-1}g(1+t)\ \;$. The factor of $g$ in the rhs is small for small values of $t$ so the graphic of $g(1+t)$ is sort of symmetric around $t=0$. – Andrew Oct 10 '11 at 16:54 Hence $$\int_{0}^{\varepsilon}g_n(1-t)=\int_0^{\varepsilon}g_n(1+t)+ \int_0^{\varepsilon}\left(1-e^{2 n t}\left(\frac{1-t}{1+t}\right)^{n-1}\right)g(1+t)=$$ $$\int_0^{\varepsilon}g_n(1+t)+O(\mathcal{I}_n).$$ – Andrew Oct 10 '11 at 16:54 I shifted the function by a unit since it won't effect the asymptotics and I'd like the global maximum to occur at $x=0$. $$\mathcal{I}_n \sim \int^{\infty}_0 \left( 1 + \frac{x-1}{n} \right)^{n-1} e^{-(x-1) } dx$$ $$\left( 1 + \frac{ x-1}{n} \right)^{n-1} e^{-(x-1) } = e \left( 1 - \frac{1}{n} \right)^{n-1} \left( 1 - \frac{x^2}{2(n-1)} + \cdots \right)$$ $$\approx e \left(1 - \frac{1}{n} \right)^{n-1} \exp\left(\frac{-x^2}{2(n-1)} \right)$$ so $$\mathcal{I}_n \sim e\left( 1 -\frac{1}{n}\right)^{n-1} \int^{\infty}_0 \exp\left( \frac{-x^2}{2(n-1)} \right) dx$$ $$= e\left( 1 -\frac{1}{n}\right)^{n-1} \sqrt{\pi(n-1)/2} \sim \sqrt{\pi n/2}$$ - But the fist integral diverges for all $n$ (for $x \to \infty$) – leonbloy Oct 10 '11 at 15:48 @RagibZaman Do you mean to write $\left(1 + \frac{\vert x-1\vert}{n}\right)^{n-1} \exp( - \vert x-1 \vert )$ ? Otherwise the first line contains a divergent integral for finite $n$. – Sasha Oct 10 '11 at 15:48 Sorry, I hadn't noticed that. Of course you are correct @Sasha, I will correct it now. – Ragib Zaman Oct 10 '11 at 15:51 The easiest fix up ended up being just to integrate over the positive reals, so I just did that in the end, @Sasha. – Ragib Zaman Oct 10 '11 at 15:56 @Ragib, I wonder if this is meant as a proof, or as an indication of what causes the asymptotics of $I_n$. – Did Oct 10 '11 at 17:56 show 5 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 45, "mathjax_display_tex": 20, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9354149699211121, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?p=4193617	Physics Forums ## Chiral symmetry and quark condensate I'm studying chiral symmetry in QCD. I understand that in order for a spontaneous symmetry breaking to occur, there must be some state with a vacuum expectation value different from zero. My question is: can someone prove that is the chiral symmetry is an exact symmetry of the QCD then necessarily: $$\langle 0 \| \bar{\psi}\psi \|0\rangle = 0$$ In understand that this has to be derived from the invariance of the vacuum but I can't prove it explicitely. Thanks PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus Recognitions: Gold Member Homework Help Science Advisor Quote by Einj I'm studying chiral symmetry in QCD. I understand that in order for a spontaneous symmetry breaking to occur, there must be some state with a vacuum expectation value different from zero. My question is: can someone prove that is the chiral symmetry is an exact symmetry of the QCD then necessarily: $$\langle 0 \| \bar{\psi}\psi \|0\rangle = 0$$ In understand that this has to be derived from the invariance of the vacuum but I can't prove it explicitely. Thanks It's actually easy to see that $\langle 0 \| \bar{\psi}\psi \|0\rangle$ is not invariant under chiral symmetry. For QCD with one flavor, $$\psi = \begin{pmatrix} u \\ d \end{pmatrix},$$ where we can consider the up and down quarks $u,d$ as Dirac spinors. Then we can write the chiral symmetry as $$\psi \rightarrow \exp\left[i\gamma^5 \left( \vec{\theta}\cdot\vec{\tau}\right) \right] \psi,$$ where the $\vec{\tau}$ are the generators of $SU(2)$ flavor transformations. Some algebra will show that the kinetic term $\bar{\psi} \gamma^\mu\partial_\mu \psi$ is chiral invariant, but $\bar{\psi} \psi$ is not, because the $\gamma^5$ in the chiral transformation anticommutes with the factor of $\gamma^0$ in the Dirac conjugate. Ok, I knew that. Actually my question was a little bit different (or maybe it's the same thing but I can't see it ). I was asking: IF the theory is invariant under chiral symmetry (i.e. the vacuum state is invariant) how can I show that necessarily $\langle \bar{\psi}\psi\rangle=0$?? Recognitions: Gold Member Homework Help Science Advisor ## Chiral symmetry and quark condensate Quote by Einj Ok, I knew that. Actually my question was a little bit different (or maybe it's the same thing but I can't see it ). I was asking: IF the theory is invariant under chiral symmetry (i.e. the vacuum state is invariant) how can I show that necessarily $\langle \bar{\psi}\psi\rangle=0$?? Zero is the only value that is invariant under the chiral symmetry. You can be as explicit as you like by picking a one-parameter transformation and using the $u,d$ parameterization. Show that, if $\rho = \langle \bar{\psi}\psi\rangle_0$, then $\delta\rho \neq 0$ unless $\rho =0$. Thread Tools \| \| \| \| \|-----------------------------------------------------------\|----------------------------------------\|---------\| \| Similar Threads for: Chiral symmetry and quark condensate \| \| \| \| Thread \| Forum \| Replies \| \| \| Advanced Physics Homework \| 0 \| \| \| High Energy, Nuclear, Particle Physics \| 3 \| \| \| Quantum Physics \| 9 \| \| \| High Energy, Nuclear, Particle Physics \| 4 \| \| \| High Energy, Nuclear, Particle Physics \| 8 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9394786357879639, "perplexity_flag": "middle"}
http://stats.stackexchange.com/questions/38328/hyperplane-problem-in-linear-congruent-generator	# Hyperplane problem in linear congruent generator From Wikipedia if an LCG is used to choose points in an n-dimensional space, the points will lie on, at most, $m^{1/n}$ hyperplanes (Marsaglia's Theorem, developed by George Marsaglia). This is due to serial correlation between successive values of the sequence $X_n$. 1. Mathematically, what is the serial correlation between successive values of the sequence $X_n$? 2. I was wondering how the points within one period are distributed to different hyperplanes, as they are generated one after another in a sequence? I guess it is unlikely that the points firstly fill out one hyperplane, and then fill out the next one, and will never visit the hyperplanes visited before in the single period? Is the time interval between every two consecutive visits to each fixed hyperplane fixed, and same for all hyperplanes? Is it connected to the period of the LCG? Does taking a small front portion of a sequence of a full period somehow overcome this drawback of LCG? That is the reason why I ask the above questions. 3. Note I think there is a typo. "at most, $m^{1/n}$ hyperplanes" should be "at most, $(n!m)^{1/n}$ hyperplanes". Am I correct? Thanks! -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9254377484321594, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Timsort	# Timsort Class Sorting algorithm Array $O(n\log n)$[1] $O(n)$ $O(n\log n)$ $O(n)$ Timsort is a hybrid sorting algorithm, derived from merge sort and insertion sort, designed to perform well on many kinds of real-world data. It was invented by Tim Peters in 2002 for use in the Python programming language. The algorithm finds subsets of the data that are already ordered, and uses that knowledge to sort the remainder more efficiently. This is done by merging an identified subset, called a run, with existing runs until certain criteria are fulfilled. Timsort has been Python's standard sorting algorithm since version 2.3. It is used to sort arrays in Java SE 7,[2] on the Android platform,[3] and in GNU Octave.[4] ## Operation Timsort was designed to take advantage of partial orderings that already exist in most real-world data. Timsort operates by finding runs, subsets of at least two elements, in the data. Runs are either non-descending (each element is equal to or greater than its predecessor) or strictly descending (each element is lower than its predecessor). If it is descending, it must be strictly descending, since descending runs are later reversed by a simple swap of elements from both ends converging in the middle. This method is stable if the elements are present in strictly descending order. After obtaining such a run in the given array, Timsort processes it, and then searches for the next run. ### Minrun Timsort algorithm searches for such ordered sequences, minruns, to perform its sort A natural run is a sub-array that is already ordered. Natural runs in real-world data may be of varied lengths. Timsort chooses a sorting technique depending on the length of the run. For example, if the run length is smaller than a certain value, insertion sort is used. Thus Timsort is an adaptive sort.[5] The size of the run is checked against the minimum run size. The minimum run size (minrun) depends on the size of the array. For an array of fewer than 64 elements, minrun is the size of the array, reducing Timsort to an insertion sort. For larger arrays, minrun is chosen from the range 32 to 64 inclusive, such that the size of the array, divided by minrun, is equal to, or slightly smaller than, a power of two. The final algorithm takes the six most significant bits of the size of the array, adds one if any of the remaining bits are set, and uses that result as the minrun. This algorithm works for all arrays, including those smaller than 64.[5] ### Insertion sort When an array is random, natural runs most likely contain fewer than minrun elements. In this case, an appropriate number of succeeding elements are selected, and an insertion sort increases the size of the run to minrun size. Thus, most runs in a random array are, or become, minrun in length. This results in efficient, balanced merges. It also results in a reasonable number of function calls in the implementation of the sort.[6] ### Merge memory The minruns are inserted in a stack. If X < Y + Z then X and Y are merged and then inserted into a stack. In this way, merging is continued until all arrays satisfy a) X > Y + Z and b) Y > Z Once run lengths are optimized, the runs are merged. When a run is found, the algorithm pushes its base address and length on a stack. A function determines whether the run should be merged with previous runs. Timsort does not merge non-consecutive runs, because doing this would cause the element common to all three runs to become out of order with respect to the middle run. Thus, merging is always done on consecutive runs. For this, the three top-most runs in the stack which are unsorted are considered. If, say, X, Y, Z represent the lengths of the three uppermost runs in the stack, the algorithm merges the runs so that ultimately the following two rules are satisfied: 1. X > Y + Z 2. Y > Z[5] For example, if the first of the two rules is not satisfied by the current run status, that is, if X < Y + Z, then, Y is merged with the smaller of X and Z. The merging continues until both rules are satisfied. Then the algorithm determines the next run.[6] The rules above aim at maintaining run lengths as close to each other as possible to balance the merges. Only a small number of runs are remembered, as the stack is of a specific size. The algorithm exploits the fresh occurrence of the runs to be merged, in cache memory. Thus a compromise is attained between delaying merging, and exploiting fresh occurrence in cache. ### Merging procedure Algorithm creates a temporary memory equal to size of smaller array. Then, it shifts elements in (say if X is smaller) X to the temporary memory and then sorts and fills elements in final order into combined space of X and Y Merging adjacent runs is done with the help of temporary memory. The temporary memory is of the size of the lesser of the two runs. The algorithm copies the smaller of the two runs into this temporary memory and then uses the original memory (of the smaller run) and the memory of the other run to store sorted output. A simple merge algorithm runs left to right or right to left depending on which run is smaller, on the temporary memory and original memory of the larger run. The final sorted run is stored in the original memory of the two initial runs. Timsort searches for appropriate positions for the starting element of one array in the other using an adaptation of binary search. Say, for example, two runs A and B are to be merged, with A as the smaller run. In this case a binary search examines A to find the first position larger than the first element of B (a'). Note that A and B are already sorted individually. When a' is found, the algorithm can ignore elements before that position while inserting B. Similarly, the algorithm also looks for the smallest element in B (b') greater than the largest element in A (a). The elements after b' can also be ignored for the merging. This preliminary searching is not efficient for highly random data, but is efficient in other situations and is hence included. ### Galloping mode Elements (pointed to by blue arrow) are compared and the smaller element is moved to its final position (pointed to by red arrow). Most of the merge occurs in what is called ‘one pair at a time’ mode, where respective elements of both runs are compared. When the algorithm merges left-to-right, the smaller of the two is brought to a merge area. A count of the number of times the final element appears in a given run is recorded. When this value reaches a certain threshold, MIN_GALLOP, the merge switches to 'galloping mode’. In this mode we use the previously mentioned adaptation of binary search to identify where the first element of the smaller array must be placed in the larger array (and vice-versa). All elements in the larger array that occur before this location can be moved to the merge area as a group (and vice-versa). The functions merge-lo and merge-hi increment the value of min-gallop (initialized to MIN_GALLOP), if galloping is not efficient, and decrement it if it is. If too many consecutive elements come from different runs, galloping mode is exited.[5] In galloping mode, the algorithm searches for the first element of one array in the other. This is done by comparing that first element (initial element) with the zeroth element of the other array, then the first, the third and so on, that is (2k - 1)th element, so as to get a range of elements between which the initial element will lie. This shortens the range for binary searching, thus increasing efficiency. Galloping proves to be more efficient except in cases with especially long runs, but random data usually has shorter runs. Also, in cases where galloping is found to be less efficient as compared to binary search, galloping mode is exited. All red elements are smaller than blue (here, 21). Thus they can be moved in a chunk to the final array. Galloping is not always efficient. One reason is due to excessive function calls. Function calls are expensive and thus when frequent, they affect program efficiency. In some cases galloping mode requires more comparisons than a simple linear search (one at a time search). While for the first few cases both modes may require the same number of comparisons, over time galloping mode requires 33% more comparisons than linear search to arrive at the same results. Moreover all comparisons in galloping mode are done by function calls. Galloping is beneficial only when the initial element of one run is not one of the first seven elements of the other run. This implies a MIN_GALLOP of 7. To avoid the drawbacks of galloping mode, the merging functions adjust the value of min-gallop. If the element is from the array currently that has been returning elements, min-gallop is reduced by one. Otherwise, the value is incremented by one, thus discouraging a return to galloping mode. When this is done, in the case of random data, the value of min-gallop becomes so large that galloping mode never recurs. In the case where merge-hi is used (that is, merging is done right-to-left), galloping starts from the right end of the data, that is, the last element. Galloping from the beginning also gives the required results, but makes more comparisons. Thus, the galloping algorithm uses a variable that gives the index at which galloping should begin. Timsort can enter galloping mode at any index and continue checking at the next index which is offset by 1, 3, 7,...., (2k - 1).. and so on from the current index. In the case of merge-hi, the offsets to the index will be -1, -3, -7,....[5] ## Performance According to information theory, no comparison sort can perform better than $\Theta(n \log n)$ comparisons in the average case. On real-world data, Timsort often requires far fewer than $\Theta(n \log n)$ comparisons, because it takes advantage of the fact that sublists of the data may already be ordered.[7] In the case of data that is completely reversed with respect to the sort order, Timsort approaches the theoretical limit of $\log(n!)$, which is in $\Theta(n \log n)$.[5] The following table compares the time complexity of timsort with other comparison sorts. Timsort Merge sort Quicksort Insertion sort Selection sort Smoothsort Best case $\Theta(n)$ $\Theta(n \log n)$ $\Theta(n \log n)$ $\Theta(n)$ $\Theta(n^2)$ $\Theta(n)$ Average case $\Theta(n \log n)$ $\Theta(n \log n)$ $\Theta(n \log n)$ $\Theta(n^2)$ $\Theta(n^2)$ $\Theta(n \log n)$ Worst case $\Theta(n \log n)$ $\Theta(n \log n)$ $\Theta(n^2)$ $\Theta(n^2)$ $\Theta(n^2)$ $\Theta(n \log n)$ The following table provides a comparison of the space complexities of the various sorting techniques. Note that for merge sort, the worst case space complexity is usually $O(n)$. Timsort Merge sort Quicksort Insertion sort Selection sort Smoothsort Space complexity $O(n)$ $O(n)$ $O(\log n)$ $O(1)$ $O(1)$ $O(1)$ Note, however, that the space complexity of both Timsort and merge sort can be reduced to $\log n$ at the cost of speed (see in-place merge sort). ## References 1. Peters, Tim. "[Python-Dev] Sorting". Python Developers Mailinglist. Retrieved 24 Feb 2011. "[Timsort] also has good aspects: It's stable (items that compare equal retain their relative order, so, e.g., if you sort first on zip code, and a second time on name, people with the same name still appear in order of increasing zip code; this is important in apps that, e.g., refine the results of queries based on user input). ... It has no bad cases (O(N log N) is worst case; N-1 compares is best)." 2. jjb. "Commit 6804124: Replace "modified mergesort" in java.util.Arrays.sort with timsort". Java Development Kit 7 Hg repo. Retrieved 24 Feb 2011. 3. "Class: java.util.TimSort<T>". Android JDK 1.5 Documentation. Retrieved 24 Feb 2011. 4. "liboctave/util/oct-sort.cc". Mercurial repository of Octave source code. Lines 23-25 of the initial comment block. Retrieved 18 Feb 2013. "Code stolen in large part from Python's, listobject.c, which itself had no license header. However, thanks to Tim Peters for the parts of the code I ripped-off." 5. 6. ^ a b 7. Martelli, Alex (2006). Python in a Nutshell (In a Nutshell (O'Reilly)). O'Reilly Media, Inc. p. 57. ISBN 0-596-10046-9.	{"extraction_info": {"found_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": 34, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9129919409751892, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/14820/calculating-uncertainty-in-the-final-result-combining-uncertainties?answertab=active	Calculating uncertainty in the final result (combining uncertainties) I'm struggling to determine the uncertainty in $F$ so it would match the textbook answer. The problem statement is: A force F is obtained using the equation: $F = \frac{mv^2}{2\pi(x_2 - x_1)}$. The readings taken were: $m = 54.0 \pm 0.5\ \mathrm{kg}$, $v = 6.3 \pm 0.2\ \mathrm{ms}^{-1}$, $x_2 = 4.7 \pm 0.1\ \mathrm{m}$, $x_1 = 3.9 \pm 0.1\ \mathrm{m}$. Calculate the value of F and determine the uncertainty in your value. Calculating the force: $F = \frac{(54.0\ \mathrm{kg}) \times (6.3\ \mathrm{ms}^{-1})^2}{2 \pi \times (4.7\ \mathrm{m} - 3.9\ \mathrm{m})} \approx 426.388\ \mathrm{N} = 430\ \mathrm{N} \text{(to 2 s.f.)}$. This agrees with the textbook answer. Now, let $X = x_2 - x_1$. Then $X = 4.7\ \mathrm{m} - 3.9\ \mathrm{m} = 0.8\ \mathrm{m}$. The uncertainty in $X$ is $\delta X = \sqrt{(\delta x_2)^2 + (\delta x_1)^2} = \sqrt{(0.1\ \mathrm{m})^2 + (0.1\ \mathrm{m})^2} \approx 0.1414\ \mathrm{m}$. Thus, $X=0.8 \pm 0.1\ \mathrm{m}$ and the formula becomes $F = \frac{mv^2}{2\pi \times X} = \frac{1}{2 \pi} \frac{m v v}{X}$. This means I can now use another standard formula to calculate the uncertainty in $F$: $\delta F = \sqrt{(\frac{\delta m}{m})^2 + (\frac{\delta v}{v})^2 + (\frac{\delta v}{v})^2 + (\frac{\delta X}{X})^2}$. And with values $\delta F = \sqrt{(\frac{0.5\ \mathrm{kg}}{54.0\ \mathrm{kg}})^2 + 2 \times (\frac{0.2\ \mathrm{ms}^{-1}}{6.3 \ \mathrm{ms}^{-1}})^2 + (\frac{0.1414\ \mathrm{m}}{0.8\ \mathrm{m}})^2} \approx 0.133$ or about 13%. But the bloody textbook says it is 40% and quotes the answer as $430 \pm 180\ \mathrm{N}$. I've tried some calculations with various values within the uncertainty and my result was always within 13% (or about 60 N) of 430 N, just as I would expect. Where have I got wrong? - The book is probably using rule-of-thumb uncertainty estimates, while you are using more accurate ones. Throw the book away. – Ron Maimon Sep 18 '11 at 21:58 @RonMaimon the odd thing is that the rule of thumb I have in this book supports my calculations. They say that percentage uncertainties should be added together and if there is a square of a value (like velocity in this case), then such percentage uncertainty should be added in twice. If I add percentage uncertainties for $m$, $v$, $x_1$ and $x_2$ I get about 12% as a result which is quite close to my calculation. They clearly had something in mind when they wrote 40% and I'd love to know what "rules of thumb" they have used as it might be helpful in the exam. – Gruffalo Sep 18 '11 at 22:04 2 Answers Both you and the book made a mistake, but the book's mistake is large, and an error of principle, while your mistake is just simple arithmetic. First, you should get a feel for the errors involved: the mass error and the v error is negligible, because they are of order a percent or two, while the error in the difference in x, value .8m, is .14m, as you calculated, it is about 15%. This is something you should be aware of--- when you subtract approximately equal quantities, the errors amplify, because the fractional error is what is important, and the quantity becomes smaller. ${\delta F\over F} = \sqrt{(\frac{0.5\ \mathrm{kg}}{54.0\ \mathrm{kg}})^2 + 2 \times (\frac{0.2\ \mathrm{ms}^{-1}}{6.3 \ \mathrm{ms}^{-1}})^2 + (\frac{0.1414\ \mathrm{m}}{0.8\ \mathrm{m}})^2} \approx 0.133$ You didn't get the right answer. The answer is almost exactly equal to the square root of the last term, or $${\delta F\over F} = {.14 \over .8} = .18$$ The actual error is 18%, not 13%. The remaining terms make this a little bit bigger, but not much. You made an error of arithmetic, which could have been avoided by noting that the last term, the error in $\Delta X$, is the only important one. But the book did the following brain-damaged error estimate: they took the two values of X, and treated the plus/minus error as something you add or subtract to the quantity to find the biggest and smallest value it can have. Then they took the "boundary" values by adding/subtracting .1 from each, to get a largest/smallest value $\Delta X$: $$\Delta X_s = (4.7 - .1) - ( 3.9 + .1 ) = .6$$ $$\Delta X_l = (4.7 + .1) - ( 3.9 - .1) = 1.0$$ This gives a 40% error. This procedure is wrong on principle, because the errors in the two x values are independent, and it is extremely unlikely that they will align to be exactly opposite. The correct estimate is that the error is 18%, for both $\Delta X$ and the final answer. - 1 Unless, of course, x1 and x2 are somehow not independent. – Colin K Sep 19 '11 at 3:31 Thank you! This explains it. – Gruffalo Sep 27 '11 at 17:42 see WP Resistance_measurement Example solution using 'WP interval arithmetic' with sagemath package (free and online server at http://sagenb.org) ````dM=0.5 dv=0.2 dx=0.1 M=RIF((54-dM,54+dM)) v=RIF((6.3-dv,6.3+dv)) x2=RIF((4.7-dx,4.7+dx)) x1=RIF((3.9-dx,3.9-dx)) F=M*v^2/(x2-x1)/2/math.pi F0=F.center() df=F0-F.lower() df/F0 ```` gives the result 0.182284511784079 18% is the answer A PSE similar problem was solved with the (free) Euler Toolbox that also has implemented the interval arithmetic. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9323158264160156, "perplexity_flag": "middle"}
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Is the magnetic field produced then because the ... 1answer 52 views ### Any suggestions for units conversion? [duplicate] Possible Duplicate: Photometer: measured Irradiance L converted to photon rate I am conducting a experiment where stimulus output of $470\ nm$ is measured by a radiometer at \$30\ \mu W\ ... 3answers 158 views ### Compton Scattering Compton Scattering essentially states that when a photon of a given wavelength hits an electron the energy level of the electron changes and the photon has its wavelength changed. This seems to be ... 1answer 82 views ### Graviton and photons interaction If one believes in the theory of gravitons then by viewing a black hole you see gravitons affect photons. This in turn leads to the conclusion that force carrier's mass equivalences allow them to be ... 1answer 102 views ### Specific electron energy gap values $E_{i+1}-E_i$ vs. photons with arbitrary energy $\hbar \omega$ The energy levels of electrons in an atom are quantized $E_i$. 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He derives this by ... 1answer 76 views ### How can the 5-photon absorption coefficient be estimated? Imagine a large bandgap material which is irradiated by an intense laser beam. If the photon energy is only high enough for 1/5 of the bandgap, is there a way to approximate the absorption by 5-photon ... 1answer 70 views ### Photoionization equation I really need help understanding this equation ,i am new to quantum mechanics and i cant understand the math, so i need every single symbol to be explained or given a value if it is a constant , ( ... 3answers 182 views ### Why do Photons want to be together? So I've heard that when a photon flies by a atom excited enough to release a photon there's a good chance it will. Because Photons want to be together and have the same direction etc? Is this true? ... 1answer 134 views ### Does photon possesses no time to cover any arbitrary distance? Photon travel 8 minutes (with speed $c$) from the sun to reach the earth. Any particle (or space-ship) with velocity $0.99 c$ covers the same distance (93 millions km) within less than 2 minutes ... 2answers 210 views ### Why does $\mathcal L = -\frac14 F^{\mu\nu} F_{\mu\nu}$ imply Photons are massless? The Lagrangian $\mathcal L = -\frac14 F^{\mu\nu} F_{\mu\nu}$ with $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ results in the four-potential's equation of motion \underbrace{\partial^\mu ... 0answers 102 views ### Polarization photon and Stokes parameters I have the following situation: About the polarization of the photon, I introduce the basis: Horizontal polarization $\|\leftrightarrow>=\binom{1}{0}$ Vertical polarization ... 2answers 610 views ### Do photons lose energy while travelling through space? Or why are planets closer to the sun warmer? My train of thought was the following: The Earth orbiting the Sun is at times 5 million kilometers closer to it than others, but this is almost irrelevant to the seasons. Instead, the temperature ... 2answers 169 views ### How does scattering work? Why is the sky blue? I was always taught in high school that light with wavelength $\lambda$ acts like a little particle that wiggles up and down through space (in proportion to its magnitude). I was ... 1answer 107 views ### Photons and Relativity Consider a Photon from Sun and travels with a velocity $c$. Now think we are that photon. For us, it looks like Sun is moving away from us with a velocity $c$. So, why don't we get attracted back ... 2answers 196 views ### What do massive photons have to do with superconductivity? I keep reading that the idea of massive photons leads to an explanation of the Meissner effect but I fail to see how photons are involved with the repulsion of fields inside a superconductor. 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So I was taught that: Kinetic Energy (of electron) = Energy (of photon) - Ionization Energy If E(photon) = IE, then KE=0 of the electron. What does this physically/theoretically mean? My current ... 3answers 205 views ### Mechanism for the gravitational field generated by photons This question follows from a schooling I received in this thread. I figured that photons do not interact with gravity, except when they've spontaneously converted into a particle-antiparticle pair. ... 1answer 211 views ### What really is Planck's constant and what are its origins? In the physics texts I have read and other info online, they says Planck's constant is the quantum of action or that it is a constant of the ratio of the energy of a particle to its frequency. Im ... 1answer 107 views ### How to relate photon's higher frequency to time dilation? The usual explanation for photon's higher frequency in lower altitudes (higher gravity), when the photon is going downward towards a massive body, is that gravitational potential energy is converted ... 0answers 45 views ### Modeling the probability of a photodiode measuring photons targets at a neighbor In current digital cameras, sensors are arrays of photodiodes which "transform" photons energy to electrons. I am aware that the probability of a photon to generate an electron is modeled by a Poisson ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 26, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.931054949760437, "perplexity_flag": "middle"}
http://mathhelpforum.com/differential-equations/118487-mass-spring-problem.html	# Thread: 1. ## Mass on a spring problem Can someone explain how to solve this? Cheers. 2. You will need to solve the homogeneous problem $\frac{d^2x}{dt^2} + w^2x = 0$ Then you will find a particular solution you can do this using the method of undetermined coefficients. Then you add the two solutions together. 3. Can you explain how to find the C.F.? Apparently its x=cos(wt)+sin(wt) ? I don't know how they got that answer though. 4. Okay, I will go through the particular solution first. I will let $x_p$ denote the particular solution. Since the nonhomogeneous part is a trigonometric function we will let the particular solution be of the form $x_p = A\sin \gamma t + B\cos \gamma t$. Now we must determine the coefficients $A,B$. To do this we will plug $x_p$ back into the original DEQ. $x''_p = -\gamma^2 A \sin \gamma t - \gamma^2 B \cos \gamma t$ and so $-\gamma^2 A \sin \gamma t - \gamma^2 B \cos \gamma t + w^2 \cdot \left( A\sin \gamma t + B\cos \gamma t \right) = \cos \gamma t$ $\left(w^2 -\gamma^2\right) A \sin \gamma t + \left(w^2 - \gamma^2\right) B \cos \gamma t = \cos \gamma t$. Thus we have $\left(w^2 - \gamma^2\right)A \sin \gamma t = 0$ and $\left(w^2 - \gamma^2\right)B \cos\gamma t = \cos \gamma t$. Therefore, we have $\left(w^2 - \gamma^2\right)A = 0$ and $\left(w^2 - \gamma^2\right)B = 1$, so $A = 0$ $B = \frac{1}{w^2 - \gamma^2}$. Finally, we see that our particular solution is $x_p = \frac{1}{w^2 - \gamma^2} \cos \gamma t$. Deleted mistake see following post. 5. Sorry I made a mistake. The homogeneous solution is wrong. The solution of the characteristic equation is not $\lambda = \pm w$. The solution is $\lambda = \pm w i$. Therefore, the homogeneous solution is of the form $x_h = C_1\cos w t + C_2 \sin w t$. Finally, the general solution should be $x(t) = C_1\cos w t + C_2 \sin w t + \frac{1}{w^2 - \gamma^2} \cos \gamma t$. To determine the constants $C_1,C_2$ we must have initial conditions.	{"extraction_info": {"found_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": 20, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.9019896388053894, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/36975/how-many-different-proofs-can-a-theorem-have	# How many different proofs can a theorem have? I notice some problems has many different proofs, do all theorems have multiple proofs, is there some theorems which has only 1 way to prove it? $n$ ways? infinite? - 4 How would you define "one way"? You can switch the order a bit, do you then have a new proof? – Jonas Teuwen May 4 '11 at 16:48 10 – Raeder May 4 '11 at 16:49 – Yuval Filmus May 4 '11 at 17:06 3 The (many!) proofs of quadratic reciprocity are quite varied, I'm told. – J. M. May 4 '11 at 17:16 1 The collection of all proofs of a theorem does not form a set. It at least forms a category with nontrivial isomorphisms, possibly an $\infty$-category... – Qiaochu Yuan May 4 '11 at 18:43 show 2 more comments ## 2 Answers If one looks at any of the formal definitions of proof in books on logic, there is a countable infinity of proofs for any theorem (if, as usual, the language has a finite or countably infinite set of symbols). This may not answer the question, which was not explicitly about formal proof. Informally, there are examples of famous results (like the Arithmetic Mean/Geometric Mean Inequality) that have quite a number of proofs based on essentially different ideas. - I guess one way to interpret the question is how many proofs of a given theorem can be published, under the (somewhat dubious) assumption that two proofs have to be significantly different in order for both of them to get published. So I direct your attention to Murray Gerstenhaber's paper, The 152nd proof of the law of quadratic reciprocity, Amer. Math. Monthly 70 (1963) 397–398, MR0150097 (27 #100) (but I warn you that the title was somewhat tounge-in-cheek). See also Elisha Loomis, The Pythagorean Proposition, which has 365 proofs of that well-known result. There have been many "proofs" of P = NP, and roughly an equal number of "proofs" of P $\ne$ NP. GJ Woeginger keeps track of them at http://www.win.tue.nl/~gwoegi/P-versus-NP.htm -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9534265398979187, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/247177/limit-value-problem	# Limit value problem Let $f_n(x)=n^2 x(1-x^2)^n$ ($0 \le x \le 1, n=1,2,3...$) For $0<x\le1$, we have $\lim_{n→\infty}f_n(x)=0$ by the theorem. Theorem: If $p>0$ and $\alpha$ is real, then $\lim_{n→\infty}\frac{n^\alpha}{(1+p)^n}=0$ I have no idea how to apply this theorem to this limit value. I think it's impossible to apply that theorem, becaus in here, if we put $n=-n, \alpha=2, p=-x^2$ then $-x^2<0$. - ## 2 Answers HINT: Use the theorem in the following form: If $0\leq r < 1$ then $n^2\cdot r^n \xrightarrow[n\to \infty]{}0$ (i.e. $a=2$ and $p=\frac{1}{r}-1$, for $r\neq 0$). - ## Did you find this question interesting? Try our newsletter Wait. Use the following: $$1-x^2 \leq \frac{1}{1+x^2}$$ Then your problem is: $$0\leq n^2x(1-x^2)^n \leq n^2x\left(\frac{1}{1+x^2}\right)^n$$ The hint follows because $1-x^4 \leq 1$ – Gautam Shenoy Nov 29 '12 at 11:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8687013387680054, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/109751/list	## Return to Answer 3 added 230 characters in body; added 1 characters in body; added 30 characters in body If every initial segment of the order has only countably many predecessors, then every countable subset of the order would be bounded (for otherwise the whole order would be a countable union of countable sets). In this case, we may therefore construct an increasing $\omega_1$ sequence in the order by recursion: start by choosing any point. After any countable number of stages, you've picked at most countably many points, which are bounded, and so there remains points above for you to pick. Thus, by recursion, you find an increasing $\omega_1$ sequence from the order. That In case anyone objects to the fact that the previous argument used uses the axim axiom of choice, and this let me point out that one cannot prove the result without any use is necessary, in view of the followingaxiom of choice. It To see this, recall that it is known to be relatively consistent with ZF that $\omega_1$ is singular, the limit of countably many countable ordinals. Suppose that $\omega_1=\sup_n\alpha_n$, where $\alpha_0\lt\alpha_1\lt\cdots$ and so on. Let $L$ be the linear order obtained by chopping $\omega_1$ into the $\omega$ many blocks corresponding intervals $[\alpha_n,\alpha_{n+1})$, but swapping turning each of these blocks individually intervals upside down, to have the reverse order. So we have an $\omega$ increasing sequence of blocks, each is anti-well-ordered with countable order type. This is an uncountable order, and every initial segment is countable, because it is a finite union of countable sets. But there is no increasing $\omega_1$ sequence , in $L$, because any well-ordered sequence can have at most finitely many members of each (anti-well-ordered block), and there are $\omega$ many blocks in order. So every increasing suborder has finite order type or order type $\omega$ at most. Thus, AC it is required for consistent with ZF that the desired result about orders is false. 2 AC required If every initial segment of the order has only countably many predecessors, then every countable subset of the order would be bounded (for otherwise the whole order would be a countable union of countable sets). In this case, we may therefore construct an increasing $\omega_1$ sequence in the order by recursion: start by choosing any point. After any countable number of stages, you've picked at most countably many points, which are bounded, and so there remains points above for you to pick. Thus, by recursion, you find an increasing $\omega_1$ sequence from the order. That argument used the axim of choice, and this use is necessary, in view of the following. It is consistent with ZF that $\omega_1$ is singular, the limit of countably many countable ordinals. Suppose that $\omega_1=\sup_n\alpha_n$, where $\alpha_0\lt\alpha_1\lt\cdots$ and so on. Let $L$ be the linear order obtained by chopping $\omega_1$ into $\omega$ many blocks $[\alpha_n,\alpha_{n+1})$, but swapping these blocks individually to have the reverse order. So we have an $\omega$ increasing sequence of blocks, each is anti-well-ordered with countable order type. This is an uncountable order, and every initial segment is countable. But there is no increasing $\omega_1$ sequence, because any well-ordered sequence can have at most finitely many members of each (anti-well-ordered block), and there are $\omega$ many blocks in order. So every increasing suborder has finite order type or order type $\omega$ at most. Thus, AC is required for the result. 1 If every initial segment of the order has only countably many predecessors, then every countable subset of the order would be bounded (for otherwise the whole order would be a countable union of countable sets). In this case, we may therefore construct an increasing $\omega_1$ sequence in the order by recursion: start by choosing any point. After any countable number of stages, you've picked at most countably many points, which are bounded, and so there remains points above for you to pick. Thus, by recursion, you find an increasing $\omega_1$ sequence from the order.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 29, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9602063298225403, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/tagged/rotational-dynamics+friction	# Tagged Questions 1answer 109 views ### Angular momentum conservation while internal frictional torque is present So this appears in a problem which looks simple enough in its context; It's something like this: Two discs, A and B, are mounted coaxially on a vertical axle. The discs have moments of inertia $I$ ... 0answers 94 views ### What determines the angle of the cushion on a pool table? If you look at the cushions (bumpers) on a pool table, you'll see that they're not vertical. They're tilted inwards. About 10 years ago, I came across a physics exam in which one of the problems ... 1answer 82 views ### Approximating Rolling/Sliding in 2D Shape I'm trying to find more information on how a 2D shape (could be defined by a function, such as ellipse, or by a polygon) will roll across a surface. The shape could be nearly circular or quite ... 1answer 261 views ### How does the resistance force on a rolling ball depend on the ball radius? A billiard ball set gently rolling on a billiard table slows and stops, because it is decelerated by resistance forces at the contact between the ball and table. I expect the magnitude of the ... 1answer 164 views ### What controls whether a ball will skid or roll? A billard ball is struck with a cue. The line of action of the applied impulse is horizontal and passes through the center of the ball. The initial velocity $v_0$ of the ball, its radius $R$, its mass ... 3answers 137 views ### Effect of surface treatment on fair dice If I have a perfectly balanced and thus fair cubic die, then polish 3 adjacent faces (so that their coefficient of friction is effectively zero) and roughen the remaining faces (so that their ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9370076656341553, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/5027/why-is-the-answer-to-this-diffusion-example-unintuitive	# Why is the answer to this diffusion example unintuitive? Imagine a linear decrease in concentration from left to right. Using Fick's first law, $J = -D \frac{d \psi}{d x}$ for all x, from left to right, we have the same flux amount because the decrease is linear. So $J(x) = m$ According to Fick's second law, $\frac{d \psi}{d t} = -D \frac{d^2 \psi}{d x^2} = \frac{d J}{d x}$ So $\frac{d J(x)}{d x} = 0$ so dJ/dx is just 0 since the 2nd derivative of a line is 0. Yet this seems unintuitive. I would expect as long as there is a concentration gradient, there should be a change in concentration at each point until the concentration is completely uniform. There must be an error in my math or reasoning, where is it? EDIT: To clarify the boundary conditions, imagine a closed box with no out flow or in flow at the edges. - 1 If there is a concentration gradient, the concentration will change - but only later. At the very same time, there's just nonzero "acceleration", and it takes nonzero time for it to acquire a velocity. Your complaint is completely analogous to the complaint that $d^2 h/dt^2 = -g$ for the height of an apple is wrong because the apple should always fall down, even if it has a zero velocity. Well, velocity is independent of the acceleration. It's only the acceleration that is equal to $-g$, and the initial velocity may be 0 or anything. The same with velocity of the concentration. – Luboš Motl Feb 12 '11 at 7:14 ## 2 Answers As Ted Bunn said, the linear concentration profile is only a steady state if there is a steady inflow at one end and a steady outflow at the other. This net flow is what preserves the concentration gradient. With the "closed box" boundary condition instead, there is indeed an error in your reasoning because the linear profile is no longer a steady state. So, to make things explicit, you should have instead: $$\left.J(x)\right\|_{t=0} = m$$ $$\left.\frac{\partial\psi}{\partial t}\right\|_{t=0} = 0$$ for all x in the interior of the box. However, these results do not imply that $\psi(x)$ is always constant. At time $t=0$, there is a constant flow from left to right, but because the box is closed this means that the concentration at the left edge of the box is decreasing and the concentration at the right edge is increasing (even though it hasn't yet begun to change anywhere in the interior - If you want, you can say that $\partial\psi/\partial t(t=0)$ has the form of two Dirac delta functions). The only way I know to get the full solution is expanding in a Fourier series. For concreteness, say the box extends from $x=-1/2$ to $x=1/2$. The correct basis of eigenfunctions to use for this boundary condition contains functions whose derivative is zero at the edges of the box, namely $\sin(n \pi x)$ for odd n and $\cos(n \pi x)$ for even n. Since the initial condition is an odd function, the cosines don't appear. Also, for convenience, set the initial slope equal to $\pi^2/4$. $$\psi(x,t=0) = \frac{\pi^2}{8} - \frac{\pi^2}{4} x = \frac{\pi^2}{8} - \sum_{n\text{ odd}} \frac{\sin(n\pi x)}{n^2}$$ (where the last equality is from the well-known Fourier series of a triangle wave) $$\psi(x,t) = \frac{\pi^2}{8} - \sum_{n\text{ odd}} \frac{\sin(n\pi x)}{n^2} e^{-nt/\tau}$$ where $\tau$ is a time scale that depends on the diffusion constant and dimensions (if you want, I can work out what it actually is, but it's irrelevant for the discussion). If you plot this function at different $t$ values increasing from zero, you can clearly see that the concentration is becoming smoothed out and tending toward a uniform concentration of the average value, $\pi^2/8$. Thus, even though at $t = 0$ it seems like the concentration isn't changing anywhere ($\partial\psi/\partial t$ = 0), it immediately begins changing, and diffusion does eventually lead to a uniform concentration. Here are some plots I made, using terms through $n=15$: - 1 +1 excellent answer! – David Zaslavsky♦ Feb 12 '11 at 6:44 Thanks for the great answer. I have to remember to keep in mind the interesting stuff tends to happen at the boundary conditions so I always need to consider them for these types of physics problems. – bmillare Feb 12 '11 at 16:09 One thing that might help your intuition is to think about what happens at the edges of the region under consideration. There must be a steady inflow from one end and a steady outflow from the other end. Fluid is continually flowing "downhill" (from high concentration to low), but the source at the "top" constantly replenishes things, so the concentration gradient remains. I don't know if that helps, but it's the best I can do. - 1 Perhaps it's tidier to focus explicitly on the edges, and rephrase the problem/solution as a boundary value problem? – genneth Feb 12 '11 at 1:31 2 I thought (although I guess it isn't explicitly stated) that the question was considering diffusion in a closed box with no inflow or outflow through the edges. – David Zaslavsky♦ Feb 12 '11 at 2:34 1 @David, thats a good point, I didn't state the boundary conditions, but what I had in mind is what you said, a closed box with no inflow or outflow through the edges. I'll update the question. – bmillare Feb 12 '11 at 5:28 I see. I was thinking that the $d\psi/dx$ retained its constant value even at the edges, which implies the inflow / outflow. Now that I understand the intended situation correctly, I agree with Keenan Pepper's excellent answer. – Ted Bunn Feb 12 '11 at 14:36	{"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": 21, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9580690264701843, "perplexity_flag": "head"}
http://mathoverflow.net/questions/23143?sort=oldest	## What theorem constructs an initial object for this category? (Formerly “Integrability by abstract nonsense”) ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Tom Leinster has a note here about how you can realize L^1[0,1] as the initial object of a certain category. You should really read his note because it is only 2.5 pages and is much more charming than what I am going to write below as background, but if you don't want to click on the link here is the idea: We work in the category of Banach spaces with contractive maps, where we are defining $X \oplus Y$ to have the norm $\|\| (x,y) \|\| = \frac{1}{2}(\|\|x\|\| + \|\|y\|\|)$. Consider triples $(X, \xi, u)$ where $X$ is a Banach space, $u \in X$ has norm at most 1, $\xi:X \oplus X \to X$ is a map of Banach spaces with $\xi(u,u) = u$. A morphism of such triples is a map of Banach spaces commuting with all structure in sight. The it turns out that the initial object in this category is $(L^1[0,1], \gamma, 1)$ where $\gamma(f,g)$ smushes $f$ and $g$ by a factor of two horizontally and then puts them side by side. Essentially this is because once you know where the constant function 1 goes, you can determine where any piecewise constant function whose discontinuities are at dyadic rationals goes, and then by density you get a unique map out of $L^1$. Leinster mentions that there are some abstract results which actually construct an initial object for a category like this. I have looked through Barr and Wells, but I do not see what exactly I should be using here. The only general initial object construction I know (The one at the beginning of the adjoint functor theorem chapter of Categories for the Working Mathematician), doesn't seem to apply (maybe it does but I am not seeing it). Does anyone know such a general construction which applies here? How does that construction look when you apply it to this situation? Does it look anything like the usual construction of $L^1[0,1]$? - 1 Unless I've misunderstood your question, it is not really about "integrability by abstract nonsense", but about finding a reference for & explanation of the general result which constructs an initial object. If that is the case then (a) perhaps you could consider amending the title of your question (b) the note you link to says that such a result should be somewhere in Barr and Wells' book, which is freely available online, see tac.mta.ca/tac/reprints/articles/12/tr12abs.html – Yemon Choi Apr 30 2010 at 22:25 2 You forgot $\xi(u,u) = u$. Also the exact norm for $X \oplus X$ is crucial. I would say: it is not actually $[0,1]$ we are "constructing" but it is $\{0,1\}^\mathbb{N}$ with i.i.d. Bernoulli product measure. (Which is essentially the same thing.) – Gerald Edgar May 1 2010 at 0:06 While I do think TL's note is interesting & thought-provoking (I still feel I should have guessed correctly when the question was posed at the start of that talk!) it feels like something better discussed in a blog post. OTOH, the closing question of this post seems a decent one for MO: I don't know offhand of the initial algebra theorem that's being used, but I'm sure passing category-theorists will. – Yemon Choi May 1 2010 at 0:29 2 Upvoted for turning me on to this awesome note. ^_^ – Vectornaut May 1 2010 at 3:35 1 @Yemon: I've been to Saskatoon in January. Clearly, just one beer is insufficient compensation for this type of weather experience. – François G. Dorais♦ May 1 2010 at 17:42 show 3 more comments ## 2 Answers I hope this answers the question a bit more explicitely as the other one. The general theorem which applies here is the following: Theorem: Let $C$ be a category which as an initial object and colimits of $\omega$-chains. Then for every functor $F : C \to C$ which preserves these colimits, there exists an initial $F$-algebra. Namely, you can take the colimit of $0 \stackrel{i}{\rightarrow} F(0) \stackrel{F(i)}{\rightarrow} F^2(0) \stackrel{F^2(i)}{\rightarrow} \cdots$. Remark the similarity to other fixed point theorems, such as by Banach (for Banach spaces) or Tarski (for CPOs). The proof is easy and straight forward. In our case, we have the category of Banach spaces $B$ and consider the comma category $C=\mathbb{K} / B$. It has products and also colimits of $\omega$-chains (take the completion of the colimit of the underlying normed spaces) and the functor $F : C \to C,~ X \mapsto X \times X$ preserves them. Thus we can apply the Theorem. The sequence $\mathbb{K} \to \mathbb{K}^{\{0,1\}^1} \to \mathbb{K}^{\{0,1\}^2} \to \dotsc$ identifies with the sequence of inclusions $X_0 \subseteq X_1 \subseteq X_2 \subseteq \dotsc$, where $X_n$ is the space of step functions $[0,1[ \to \mathbb{K}$ which are constant on every intervall of the $n$-th dyadic subdivision of the interval. The norm becomes the usual $\mathrm{L}^1$-norm. The union $\cup_n X_n$ consists of those step functions which are locally constant with respect to some dyadic subdivision and is endowed with the $\mathrm{L}^1$-norm. All step functions can be approximated by these dyadic step functions. The completion is thus, indeed, $\mathrm{L}^1[0,1]$. Also some inspection gives us that the algebra structure consists of the constant function $1$ and the map $(f,g) \mapsto f \star g$, which squeezes and juxtaposes. So basically this construction is very similar to the usual one of $\mathrm{L}^1[0,1]$, but we don't have to care about any ambiguity of choices since all the involved universal properties ensure this automatically. Besides, it avoids the general measure theory. - I missed this question when it first appeared. Yes, Martin, that's exactly the construction I had in mind. Thanks. The initial algebra theorem is the simplest of all initial algebra theorems. The only complication is that you have to apply it to an endofunctor of a coslice of Ban, not to an endofunctor of Ban itself. – Tom Leinster Oct 16 2011 at 13:50 2 +1: This is indeed an easier proof. A few remarks: (1) the general theorem has been attributed to Adamek. (2) $\mathbb{K}$ denotes the ground field (usually $\mathbb{R}$ or $\mathbb{C}$, but sometimes another local field), and $B$ is, as before, the category of Banach spaces appropriate to that ground field, with morphisms contractive maps. (3) For the notion of algebra and initial algebra of an endofunctor, one may consult the nLab: nlab.mathforge.org/nlab/show/initial+algebra – Todd Trimble Oct 16 2011 at 14:10 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. The category you described can be written as a lax limit of a diagram in the 2-category of categories. The diagram in question consists of accessible categories and accessible functors, so its limit is again accessible by a theorem of Makkai and Paré. It is obvious from the construction that the category has all small limits (they are computed as in the category of Banach spaces and nonexpanding maps), so it is in fact locally presentable. This means that it has all colimits and in particular an initial object. Here are some more details and references. Let $\mathbf{Ban}_1$ be the category of Banach spaces and nonexpanding maps. This category is locally $\aleph_1$-presentable (see e.g. Borceux, Handbook of Categorical algebra, Volume II, 5.2.2.e). Let $F \colon \mathbf{Ban}_1 \rightarrow \mathbf{Ban}_1$ be the functor which sends a Banach space $X$ to $\mathbb{R}+X\oplus X$, where + stands for the coproduct. A morphism f of Banach spaces gets sent to $\mathrm{id}_{\mathbb{R}}+f\oplus f$. Let $U \colon \mathbf{Ban}_1 \rightarrow \mathbf{Set}$ be the functor which sends a Banach space to its underlying set. Since $\aleph_1$-filtered colimits are computed as in the category of sets (see e.g. Borceux, Handbook of Categorical algebra, Volume II, 5.2.2.e) we know that the composite UF preserves $\aleph_1$-filtered colimits. By the open mapping theorem it follows that $F$ preserves $\aleph_1$-filtered colimits. By the theorem of Makkai and Paré (see e.g. Adámek, Rosický, Locally Presentable and accessible categories, Theorem 2.77), the inserter $\mathcal{C}$ of $F$ and the identity functor on $\mathbf{Ban}_1$ is again an accessible category. The objects of $\mathcal{C}$ are triples $(X,\xi,u)$ where $\xi\colon X\oplus X \rightarrow X$ and $u\colon \mathbb{R} \rightarrow X$ are nonexpanding (i.e., u corresponds to an element of $X$ of norm less than or equal one). The morphisms are morphisms of Banach spaces which are compatible with $u$ and $\xi$. Thus $\mathcal{C}$ is almost the category we are interested in; the only thing that's missing is the requirement that $\xi(u,u)=u$. Let $G \colon \mathcal{C} \rightarrow \mathbf{Ban_1}$ be the functor which sends every object to $\mathbb{R}$ and every morphism to $\mathrm{id}_{\mathbb{R}}$. This is clearly a functor which preserves $\aleph_1$-filtered colimits. Let $H \colon \mathcal{C} \rightarrow \mathbf{Ban_1}$ be the functor which sends $(X,\xi,u)$ to $X$ and a morphism to itself; this is again a functor which preserves $\aleph_1$-filtered colimits. There are two natural transformations $\alpha,\beta \colon G \Rightarrow H$, whose component at $(X,xi,u)$ is given by $u$ and $\xi(u,u)$ respectively. The equifier of $G$ and $H$ is precisely the category we are looking for, and from our construction we can conclude that it is accessible. The obvious forgetful functor to Banach spaces creates limits, so this category is complete. Thus it is locally presentable, and therefore also cocomplete. In particular, it follows that our category has an initial object. - Thank you for the incredibly informative answer. I do not really understand it currently, but the references you cite seem like good places to learn about this stuff. Thanks! – Steven Gubkin May 13 2010 at 0:35 A side question would be: is THIS simple enough to combine with the remarks in the question to make a construction of the integral that turns out to be simpler than the standard one? – Gerald Edgar May 13 2010 at 12:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 79, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9442950487136841, "perplexity_flag": "head"}
http://nrich.maths.org/7283	### Adding in Rows List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it? ### Magic Sums and Products How to build your own magic squares. ### Weekly Problem 2 - 2013 Weekly Problem 2 -2013 # Perimeter Expressions ##### Stage: 3 Challenge Level: Charlie took a sheet of paper and cut it in half. Then he cut one of those pieces in half, and repeated until he had five pieces altogether. He labelled the sides of the smallest rectangle, $a$ for the shorter side and $b$ for the longer side. Here is a shape that Charlie made by combining the largest and smallest rectangles: Check you agree that the perimeter is $10a + 4b$. Alison combined the largest and smallest rectangles in a different way. Her shape had perimeter $8a + 6b$. Can you find how she might have done it? Charlie and Alison made sure their rectangles always met along an edge, with corners touching. Can you combine the largest and smallest rectangles in this way to create other perimeters? Create some other shapes by combining two or more rectangles, making sure they meet edge to edge and corner to corner. What can you say about the areas and perimeters of the shapes you can make? If you have a friend to work with, you could each create a shape and work out the area and perimeter. Can you recreate each other's arrangement if you only know the area and perimeter? Here are some questions to consider: What's the largest perimeter you can make using ALL the pieces? Can you make two different shapes which have the same area and perimeter as each other? Can you make two different shapes which have the same perimeter but different areas? How do you combine any set of rectangles to create the largest possible perimeter? Charlie thinks he has found a shape with the perimeter $7a + 4b$. Can you find his shape? What can you say about the perimeters it is possible to make, if $a$ and $b$ are the dimensions of one of the other rectangles? This problem is based on an idea shared by Sue Southward. The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9365944862365723, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/205353-finding-phase-characteristic-transfer-function.html	# Thread: 1. ## Finding a phase characteristic from a transfer function? Given a transfer function: $H(s)=\frac{-1}{s+1}$ Show that the phase characteristic is $\varphi(\omega)=\pi-Arctan \omega , \omega>0$ My text book only briefly covers this subject so I really have no idea what to do. I know the formula is: $\varphi(\omega)=Arg H(i\omega) , \omega>0$ So I have that: $H(i\omega)=\frac{-1}{i\omega+1}$ So I assume I need to find the argument of this fraction? 2. ## Re: Finding a phase characteristic from a transfer function? My attempt: $Arg\left ( \frac{-1}{i\omega+1} \right )=Arg(-1)-Arg(i\omega+1)$ I looked at this Wikipedia article under the section "Computation" where it says that: $Arg(x+iy)=\pi/2-arctan(x/y)$ for $y>0$ I'd assume this is the definition I need to look at because in the formula, it says that $\omega>0$ and the omega is the imaginary part. So that leaves me with: $\varphi(\omega)=\pi/2-arctan(1/\omega)$ So I've figured out this much. I'm a bit confused as what to do next. What about $Arg(-1)$? 3. ## Re: Finding a phase characteristic from a transfer function? Nevermind, I think I solved it on my own. Arg(-1)=pi Arg(ib+1)=arctan(b) So subtracting them with each other gives the answer. 4. ## Re: Finding a phase characteristic from a transfer function? Originally Posted by MathIsOhSoHard My attempt: $Arg\left ( \frac{-1}{i\omega+1} \right )=Arg(-1)-Arg(i\omega+1)$ I looked at this Wikipedia article under the section "Computation" where it says that: $Arg(x+iy)=\pi/2-arctan(x/y)$ for $y>0$ I'd assume this is the definition I need to look at because in the formula, it says that $\omega>0$ and the omega is the imaginary part. So that leaves me with: $\varphi(\omega)=\pi/2-arctan(1/\omega)$ So I've figured out this much. I'm a bit confused as what to do next. What about $Arg(-1)$? You can take the $i$ out of the denominator by multiplying the numerator and denominator of $\frac{-1}{i\omega+1}$ by $1-i\omega$ to get, $Arg\left ( \frac{-1}{1-\omega^2} + i \frac{\omega}{1-\omega^2} \right )=\frac{\pi}{2}-tan^{-1}(\frac{-1}{\omega})$ Using the properties of arctan, $tan^{-1}(\frac{1}{x})=-\frac{\pi}{2}-tan^{-1}(x)$ and $tan^{-1}(-x)=-tan^{-1}(x)$ results in the answer. EDIT:You've already done it. Well done	{"extraction_info": {"found_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": 22, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.9329378008842468, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/18570/why-are-tensors-a-generalization-of-scalars-vectors-and-matrices/18586	## Why are tensors a generalization of scalars, vectors, and matrices? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Take two vector spaces $V$ and $W$ over a field $F$. One may form the tensor product $V\otimes W$ and it fulfills an universal property. Elements of $V\otimes W$ are called tensors and they are linear combinations of elementary tensors $v\otimes w$, the elementary tensors generate $V\otimes W$. People from physics think of a tensor as a generalization of scalars, vectors, and matrices, I think and I have seen them tensoring matrices with matrices as entries with matrices and so on. What does this mean and what has it to do with the definition from above? What "is" a tensor? - For the "tensoring of matrices", see en.wikipedia.org/wiki/Kronecker_product – Willie Wong Mar 18 2010 at 12:13 Community wiki, by the way? This question is also a little vague.. – Harry Gindi Mar 18 2010 at 13:48 21 The question is not vague at all. Being puzzled by tensors is a very natural thing when most people first see them. Asking what tensors "are" is pretty reasonable. – KConrad Mar 18 2010 at 16:08 would like to add that the grasping of the fundamental sense for these objects and properties, are implanted around the generalization of calculus: differential forms and its applications... – ivane May 31 2010 at 15:47 ## 12 Answers I have heard it said that tensor products are the hardest thing in mathematics. Of course that's not really true, but certainly a fluent understanding of how to work with tensor products is one of the dividing lines in your education from basic to advanced mathematics. Disclaimer: What I will discuss here are tensor products in the sense of linear algebra, so only tensor products of individual vector spaces rather than tensor fields (which is what the physicists mean by tensor product). For a long time I could not understand how the physicists could work with tensors by thinking about them as "quantities that transform in a certain way under a change in coordinates". The only way I could come to terms with them is by their characterization as something that satisfies a universal mapping property. Do not think about what tensors (elements of a tensor product space) are but rather what the whole construction of a tensor product space can do for you. It's sort of like quotient groups (only harder), where if you focus all your energy on trying to understand cosets you kind of miss the point of quotient groups. What makes tensor product spaces harder to come to terms with than quotient groups is that most elements of a tensor product space are not elementary tensors $v \otimes w$ but only sums of these things. The whole (mathematical) point of tensor products of vector spaces is to linearize bilinear maps. A bilinear map is a function $V \times W \rightarrow U$ among $F$-vector spaces $V, W$, and $U$ which is linear in each coordinate when the other one is kept fixed. There are tons of bilinear maps in mathematics, and if we can turn them into linear maps then we can use constructions related to linear maps on them. The tensor product $V \otimes_F W$ of two $F$-vector spaces provides the most extreme space, so to speak, which is a domain for the linearization of all bilinear maps out of $V \times W$ into all vector spaces (over $F$). It is a particular vector space together with a particular bilinear map $V \times W \rightarrow V \otimes_F W$ such that any bilinear map out of $V \times W$ into any vector space naturally (!) gets converted into a linear map out of this new space $V \otimes_F W$. Some notes I wrote on tensor products for an algebra course are at http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf, and in it I address questions like "what does $v \otimes w$ mean?" and "what does it mean to say $$v_1 \otimes w_1 + \cdots + v_k \otimes w_k = v_1' \otimes w_1' + \cdots + v_k' \otimes w_k'?"$$ Right from the start I allow tensor products of modules over a ring, not just vector spaces over a field. There are some aspects of tensor products which appear in the wider module context that don't show up for vector spaces (particularly since modules need not have bases). So you might want to skip over, say, tensor products involving ${\mathbf Z}/m{\mathbf Z}$ over $\mathbf Z$ on a first pass if you don't know about modules. As for the question of how tensor products generalize scalars, vector spaces, and matrices, this comes from the natural (!) isomorphisms $$F \otimes_F F \cong F, \ \ \ F \otimes_F V \cong V, \ \ V \otimes_F V^* \cong {\rm Hom}_F(V,V).$$ On the left side of each isomorphism is a tensor product of $F$-vector spaces, and on the right side are spaces of scalars, vectors, and matrices. In the link I wrote above, see Theorems 4.3, 4.5, and 5.9. You can also tensor two matrices as a particular example in a tensor product of two spaces of linear maps. Spaces of linear maps are vector spaces (with some extra structure to them), so they admit tensor products as well (with some extra features). Returning to the physicist's definition of tensors as quantities that transform by a rule, what they always forget to say is "transform by a multilinear rule". I discuss the transition between tensor products from the viewpoint of mathematicians and physicists in section 6 of a second set of notes at http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod2.pdf. - 4 "Do not think about what tensors (elements of a tensor product space) are but rather what the whole construction of a tensor product space can do for you." I really like that witty exhortation ! – Georges Elencwajg Mar 18 2010 at 20:02 Good response and even better notes,K-thanks for making them freely available. We need as many good,free sources of information on this somewhat impenetrable topic as possible for beginners. – Andrew L Apr 3 2010 at 19:06 1 You have written up some really nice stuff here math.uconn.edu/~kconrad Thank you so much for making it freely available! – Predrag Punosevac Aug 2 2011 at 16:40 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. There is a notion of multiplication of tensors, because the physicist tensors come in covariant, contravariant and mixed flavour. And r-fold covariant and s-fold contravariant tensor is mathematically speaking an element of $V^{\otimes r}\otimes (V^\ast)^{\otimes s}$ (or maybe the other way). A matrix is a mixed (1,1)-tensor if you interpret it as a linear map, i.e. an element of $V\otimes V^\ast$. A matrix is a pure (0,2)-tensor if you interpret it as a bilinearform. This yields an element of $V^\ast \otimes V^\ast$. There are two kinds of multiplication of tensors: The tensorproduct itself taking a (r,s) and a (r',s') tensor and returning a (r+r',s+s') tensor. But if you want to multiply a (r,s) and a (s,t) tensor, there is another way: applying the linearform from the $V^\ast$-parts of the first tensor to the vectors from the $V$-parts in the second tensor. For matrices (=(1,1)-tensors) this is $(v\otimes \alpha)\circ (w\otimes \beta) := v\otimes (\alpha(w)\cdot \beta)=\alpha(w)\cdot (v\otimes \beta)$. This second multiplication is a generalization of the matrix multiplication to tensors. - And you can also contract some covariant indices in the first 'factor' with the same number of contravariant indices in the second one, and get 'partial' multiplications. – Mariano Suárez-Alvarez Mar 18 2010 at 13:17 1 Small addition: the reason elements of $V\otimes V^$ correspond to linear maps $V \to V$ (and therefore to matrices): an element of $V\otimes V^$ has the form $v_1 \otimes v_1^* + \cdots + v_n \otimes v_n^$. If $w \in V$ is a vector, plugging it into each $v_i^$ gives a number by definition of dual vectors. So, plugging it in the whole expression will give $(number) \cdot v_1 + \ldots (number)\cdot v_n$ which is some vector in V. By the properties of tensor product, the map that takes $v \in V$ and spits out this vector is linear. Any linear map can be obtained this way (nice exercise). – Ilya Grigoriev Mar 18 2010 at 17:46 Also, for the same reason elements of $V\otimes W^$ correspond to linear maps $W \to V$. This is all very standard, but I thought it could be useful to state explicitly for some people - there was I time I was confused about it... – Ilya Grigoriev Mar 18 2010 at 17:48 And another one: The map $V\otimes W\to Hom(W,V)$, $v\otimes w^\ast \mapsto (w\mapsto w^\ast(w)\cdot v)$ is an isomorphism onto the set of linear maps with finite rank, i.e. the identification $V\otimes W^\ast = Hom(W,V)$ works only if at least one of the two vector spaces has finite dimension. – Johannes Hahn Mar 18 2010 at 20:53 Suppose you have a pair of elements in vector spaces, $v\in V$ and $w\in W$. Now suppose that at some future point you're going to compute $f(v,w)$ where $f$ is a bilinear function. For example, when $V=W=\mathbb{R}^3$, $f$ might be the familiar dot or cross product. But it might be something else entirely. In fact, suppose you don't know in advance what $f$ is going to be. The tensor product precisely answers the question "what information do I need about $v$ and $w$ in order to be able to compute $f(v,w)$ at some future point, whatever $f$ turns out to be?" You could say "knowing $v$ and $w$ is enough information". But that that's more information than you need. $v\otimes w$ contains less information than $(v,w)$ and actually contains the least information you can get away with, and still be able to compute $f(v,w)$ for any bilinear $f$. I don't know if this will be helpful for you, but when I realised this everything suddenly became crystal clear. It's really just a restatement of the universal property. - If you think in terms of bases, then the basis for the tensor product is $\{e_i \otimes e_j\}$. You may think of this as an $n\times m$ matrix with exactly one non-zero entry (1) at the (i,j)th position. The tensor product of $V$ and $W$, then is isomorphic to the space of $n\times m$ matrices. Meanwhile, the tensor product of linear maps can be thought of as an index array with four indices --- i.e. as a hypercubical array. There is hardly no advantage in trying to multiply hyper-cubical arrays along edges, faces, cubes etc. So a very efficient way of doing so is to begin thinking of things as "abstract tensors." These are boxes with strings emanating from the tops and bottoms. They are notational short hands for the resulting multi-indexed objects such as vectors, and matrices. An alternative answer can in as I was typing this ... - I'd like to borrow a little from several of the answers given so far to give a "practical" perspective. Let $V$ be an $\mathbb{R}$-vector space. A tensor of type $(m, n)$ is an element of $V^{\otimes m} \otimes (V^{})^{\otimes n}$, which translates for small $m, n$ into the following: • A scalar is a tensor of type $(0, 0)$, • A vector (in the strict sense of an element of $V$) is a tensor of type $(1, 0)$, • A covector is a tensor of type $(0, 1)$, • A bilinear form is a tensor of type $(0, 2)$, and • A matrix is a tensor of type $(1, 1)$. Now, why would we be interested in higher-order tensors? Well, let $f : V \to \mathbb{R}$ be a sufficiently nice function. Then it has a Taylor expansion $$f(v) = f(0) + v^T \nabla f(0) + v^T H(0) v + ...$$ where the first term is a tensor of type $(0, 0)$, the second is a tensor of type $(0, 1)$ described by the gradient, and the third is a (symmetric) tensor of type $(0, 2)$ described by the Hessian. All this is standard calculus material. But how does the Taylor expansion continue? As it turns out, the next terms are described by (symmetric) tensors of type $(0, 3)$, type $(0, 4)$, ... etc. So that's one sense in which tensors naturally generalize familiar vector- and matrix-like objects (the gradient and the Hessian). - To pick bones: a matrix is only a tensor of type (1,1) if it is the basis representation of a linear operator from V to itself. While in an ideal world this will be the case, practically any (introductory/undergraduate/applied) linear algebra textbook will include the basis representation of a bilinear form as an example of a matrix. I get annoyed teaching that part. – Willie Wong Mar 18 2010 at 17:02 In old times, a tensor was 'something' whose components transformed in a specific way when you changed coordinates. Nowadays, well, it is the same thing with different language. - 1 You could also say that a tensor "eats" however many vectors/covectors its type would suggest, resulting in a scalar. – Steve Huntsman Mar 18 2010 at 12:11 ... whose components transformed in a multilinear way, not merely a specific way. I wish the physicists would give their transformation rule a real name and not just call it a specific kind of transformation. – KConrad Mar 18 2010 at 16:10 Well, multilinear is not specific enough! :) By 'specific' I meant 'in a very well defined way which I do not want to write down in detail', not 'in a random way' (I do not know what the proper way of phrasing this would be, actually...) – Mariano Suárez-Alvarez Mar 18 2010 at 16:19 I had in mind a tensor product of one k-tuple of vector spaces, so for me "multilinear rule" would have been more enlightening than the usual "specific rule" the physicists say. I agree the physicists/geometers want to go further and include partial derivatives in their multilinear change-of-coordinate formulas since their tensors belong to a tensor field. – KConrad Mar 18 2010 at 18:47 I remember being thoroughly confused by this when I first tried to learn this from older math and physics books as a highschool student. From this point of view a tensor was something with many indices which transformed according to specific and rather complicated rules. It was never clear to me what a tensor actually was. Later on, when I learned the modern viewpoint, things (eventually) made more sense. In modern language, it goes like this. Start with an open set X of Euclidean space or more generally a manifold. A tensor (field) on X of type (m,n) assigns to each $x\in X$ an element of the tensor product $T_x\otimes \ldots T_x\otimes T_x^\otimes \ldots T_x^$ where one has m factors of the tangent space $T_x$ and n factors of the cotangent space $T_x^$. After choosing local coordinates, one can write everything out in bases and compare it to the classical notation. Specializing $(m,n)=(0,0)$ corresponds to a scalar field, and $(m,n)=(1,0)$ to a vector field. Also when $m+n=2$, we can view a tensor (locally) as a matrix valued function. I hope that helps. - If you take a (finite dimensional) vector space V then there is an isomorphism $V^ \otimes V \simeq \text{Hom}(V,V)$ So the first tensor product is precisely linear maps from $V$ to itself (viz. matrices, once you pick a basis for $V$). If you have some extra structure on V (say a non-degenerated inner product or the choice of a basis) then you can identify V with $V^$, thus having an identification $V \otimes V \simeq V^ \otimes V \simeq \text{Hom} (V,V)$. Everything else is just a (straightforward) generalisation of the above. - I think many of the answers assume you know what tensor product is... A particularly illuminating example might help here. Before I give it, I'd like to mention that mathematicians don't acknowledge the origin of the name tensor. The meaning was probably lost in over use, but is preserved in translations in other languages (I'm Chinese). Tensor looks like tension, and (I imagine) was used first to describe tension of a membrane or something. (Then Riemann took it up as something that has more than two indices, for his curvature for example. I could be wrong about the history.) If you want to tensor a plane with another plane, that's not gonna be particularly illuminating. It'll be a four-dimensional space, but one hardly sees what it has to do with the two planes. Let $V$ be the vector space of polynomials of degree $\leq n$, and let $W$ be the vector space of polynomials of degree $\leq m$. As you know, the tensor product $V\otimes W$ should consist of all things of the form $v\otimes w$, for each $v\in V$ and each $w\in W$, and all (finite) linear combinations of it. Luckily in this case, we have a good candidate for what $v\otimes w$ is (or is being identified with). For $v=p(x)$ a polynomial in $x$, and $w=q(y)$ a polynomial in (another variable) $y$, then $v\otimes w$ is simply the product $p(x)q(y)$ as a polynomial of two variables. Now you see what the tensor product is: $V\otimes W$ is the vector space of polynomials of two variables, whose degree in $x$ is at most $n$, and degree in $y$ is at most $m$. Obviously $\dim V\otimes W = \dim V\times\dim W$. It might be worthwhile to think about what happens to $V\otimes V$ when you make a change of basis on $V$. It would be a sin not to mention the extension in infinite-dimensional case. $k[x]\otimes k[y]=k[x,y]$, as you can check using the universal property. I hope it helps. - Tensor in the sense of tension probably refers to the stress tensor in physics. In a solid body, one can consider the force acting on any infinitesimal two-dimensional section, and this force may be any vector but depends linearly on the normal vector to the section. Considering the normal vector to be a dual vector, this gives a tensor of what the physicists call (I think) type (0,2): an element of $V \otimes V$, where $V = \mathbb{R}^3$ in this case. Wikipedia tells me that this idea is due to Cauchy and (of course) Euler, so has a chance of being the original source of the word. – Ryan Reich May 24 2011 at 16:25 The following discussion is for finite-dimensional vector spaces only: The tenor product of two vector spaces $V$ and $W$ arises, because you want to work with bilinear, rather than just linear, maps and functions of $V \times W$. The key observation is that the space of all possible bilinear scalar-valued functions or vector-valued maps is itself a vector space that is spanned by simple bilinear functions constructed by multiplying a linear function on $V$ by a linear function on $W$. Let $T$ denote the vector space of all scalar-valued bilinear functions on $V \times W$ and observe that there is a natural map $V^* \times W^* \rightarrow T$. Moreover, it is easy to see that $V^* \times W^$ generates all of $T$ in the sense that its image does not lie in any proper subspace of $T$. So this leads to the notation $T = V^ \otimes W^$ The next observation is that there is a natural bilinear map $V\times W \rightarrow T^$ that corresponds to evaluation of the given $(v,w)$ with an element of $T$. Again, the image "spans" all of $T^$ in the sense that it does not lie in any proper subspace. Moreover, the linear duality between $T$ and $T^$ corresponds exactly to the evaluation of a bilinear function on an element in $V\times W$. So it is reasonable to denote $T^$ by $V \otimes W$. So, morally speaking, $V\otimes W$ is the "smallest" vector space such that there is a bilinear injective map $V \times W \rightarrow V\otimes W$. Of course, this statement can be made precise by defining $V\otimes W$ as a universal object. Finally, it is not difficult to prove the rather cool (and for me not so obvious) observation that the space of linear maps $V \rightarrow W$ is naturally isomorphic to $W\otimes V^$. In this sense, tensors generalize the matrices viewed as linear transformations. - Tensor in physical means is kind of linear operator. It acts on tangent/cotangent space of (usually) differential manifold and gives us at the end some invariants in term of scalars which means that such values are independent of certain class of coordinate changes. As main change on the manifold one have to concern is change of the map, such objects in order to have physical meaning ( which is simple independence of coordinate change) should have certain properties, during this change. So basically example here are volume, energy, mass. The other cause of tonsorial calculus is that evolution operator for many dynamical systems may be prescribed infinitesimally as linear operator on tangent space, and in fact this nation is connected to Lie algebra of group acting on differential manifold of system configurations. So such linear "transports" on manifold configuration space acts on various quantities defined for a system, in infinitesimal as tensors - in fact they are coefficients of multidimensional Taylor expansion of such quantities. Such examples arises mainly in chemical kinetics of reactions for example where usually one concern linear approximations, the stress tensors of various continuous materials, or in General Relativity. It is worth of note that not all physical objects are tensors (although all coordinate independent are, however sometimes usual physical quantity may be for example a part of bigger tensorial object. In this cases such quantity is coordinate dependent, see: see "relativistic mass" or time dilatation discussions). So basically geometry is behind tensors and it is beautiful one. - More generally, you can form tensor products $V_1\otimes \cdots \otimes V_k$ of an arbitrary number of vector spaces, and a tensor refers to an element of one of these spaces, not just the case $k=2$. If $k=1$ then obviously a tensor is the same as a vector in $V_k$. Scott Carter explained how tensors for $k=2$ (i.e. rank 2 tensors) correspond to matrices. Here's another viewpoint on the same thing. If `$W^$` is the dual space of $W$ (i.e. the vector space of linear functionals from $W$ to the base field) then `$V\otimes W^$` can be identified with $L(W,V)$, the space of linear maps from $W$ to $V$, by setting $(v\otimes f)(w) = f(w) v$, for $v\in V$, $f\in W^$, and $w\in W$. (I'm assuming all vector spaces are finite dimensional for simplicity; also there's some work to be done to show this identification really is well-defined.) Of course, given bases of $W$ and $V$, $L(W,V)$ can also be identified with a space of matrices. In a similar way, `$V^ \otimes V^*$` is identified with bilinear maps on $V$. I was about to type something about multiplication of tensors, but since Johannes Hahn just wrote an answer about that so I'll refer you there. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 163, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.941904604434967, "perplexity_flag": "head"}
http://xianblog.wordpress.com/2012/10/03/estimating-a-constant/	# Xi'an's Og an attempt at bloggin, from scratch… ## estimating a constant Paulo (a.k.a., Zen) posted a comment in StackExchange on Larry Wasserman‘s paradox about Bayesians and likelihoodists (or likelihood-wallahs, to quote Basu!) being unable to solve the problem of estimating the normalising constant c of the sample density, f, known up to a constant $f(x) = c g(x)$ (Example 11.10, page 188, of All of Statistics) My own comment is that, with all due respect to Larry!, I do not see much appeal in this example, esp. as a potential criticism of Bayesians and likelihood-wallahs…. The constant c is known, being equal to $1/\int_\mathcal{X} g(x)\text{d}x$ If c is the only “unknown” in the picture, given a sample x1,…,xn, then there is no statistical issue whatsoever about the “problem” and I do not agree with the postulate that there exist estimators of c. Nor priors on c (other than the Dirac mass on the above value). This is not in the least a statistical problem but rather a numerical issue.That the sample x1,…,xn can be (re)used through a (frequentist) density estimate to provide a numerical approximation of c $\hat c = \hat f(x_0) \big/ g(x_0)$ is a mere curiosity. Not a criticism of alternative statistical approaches: e.g., I could also use a Bayesian density estimate… Furthermore, the estimate provided by the sample x1,…,xn is not of particular interest since its precision is imposed by the sample size n (and converging at non-parametric rates, which is not a particularly relevant issue!), while I could use importance sampling (or even numerical integration) if I was truly interested in c. I however find the discussion interesting for many reasons 1. it somehow relates to the infamous harmonic mean estimator issue, often discussed on the’Og!; 2. it brings more light on the paradoxical differences between statistics and Monte Carlo methods, in that statistics is usually constrained by the sample while Monte Carlo methods have more freedom in generating samples (up to some budget limits). It does not make sense to speak of estimators in Monte Carlo methods because there is no parameter in the picture, only “unknown” constants. Both fields rely on samples and probability theory, and share many features, but there is nothing like a “best unbiased estimator” in Monte Carlo integration, see the case of the “optimal importance function” leading to a zero variance; 3. in connection with the previous point, the fascinating Bernoulli factory problem is not a statistical problem because it requires an infinite sequence of Bernoullis to operate; 4. the discussion induced Chris Sims to contribute to StackExchange! ### Share: This entry was posted on October 3, 2012 at 12:12 am and is filed under Books, Statistics with tags All of Statistics, Bayesian Analysis, Bernoulli factory, Chris Sims, cross validated, harmonic mean estimator, Larry Wasserman, numerical analysis, StackExchange, statistical inference. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site. ### 15 Responses to “estimating a constant” 1. [...] Robert discussed the problem on his blog. If I understand what Christian has written, he claims that this cannot be considered a statistical [...] 2. Christian I don’t know why you keep calling it “known.” Known means: you know it. But you don’t know it! To estimate the normalizing constant of a posterior from simulated values, what method do you use? Frequentist or Bayes? If bayes, how do you do it? By the way, credit where credit is due: the example is due to Ed! Best wishes Larry • When I am using simulation, or Monte Carlo methods, I rely on the law of large numbers and the stabilisation of frequencies. So Monte Carlo is a ‘frequentist” method, granted, but it is not connected with a statistical issue, which is why I find the debate “frequentist vs. Bayesian” rather vacuous here… 3. JohnDoe Says: October 3, 2012 at 4:23 am Agree it’s a terrible example. Larry W’s claim there is no Bayesian estimate of the normalizing constant seems equivalent to claiming there is no Bayesian estimate of the 5th decimal place of pi. • Indeed, I think that, in both cases, we should not use the wording estimate, but instead the wording approximation. 4. Christian: A parameter theta is a fixed constat. But until you know it, as a Bayesian, you treat it as a random variable. Why not the same for c? –Larry • Hi, Larry! The constant c is known in this case, which makes all the difference with an unknown parameter driving the distribution of an observed sample. For instance, I cannot make observations about c, I cannot build a likelihood on c, &tc… At a more subtle level, we should get back to what Persi suggested as Bayesian numerical analysis (in the 1992 Purdue Symposium?). 5. Dan Simpson Says: October 3, 2012 at 1:01 am “But if we still try to play this artificial game introducing a random normalization constant C,” may actually be the strangest thing I’ve ever seen on the internet. How is this even possibly nearly an argument? It blows my mind. I know that we often say that in Bayesian statistics “everything is random”, but we don’t actually mean it! Further down the thread “Suppose you can’t do the numerical integral”………. Is Wasserman actually saying “suppose f is not in L2″? • Dan Simpson Says: October 3, 2012 at 1:08 am Actually on further thought, this is only completely trivial if the density is absolutely continuous. If it isn’t then the summation in the normalising constant can be tricky. That being said, there are at least 3 methods that I can think of (as a non-expert) for doing non-paradoxical inference with uncomputable normalising constants. And those constants still aren’t random! • You never know what the next strangest thing on the internet can be..! As a Bayesian I do not say that everything is random, rather that using a probability distribution (or a measure) is the optimal way to represent my uncertainty about things, like non-random parameters driving a random phenomena. In the current case, my prior distribution is a Dirac mass. • Keith O'Rourke Says: October 10, 2012 at 4:42 pm Might help to distinguish (uncertainty of) actualities from potentialities (CS Peirce vocabulary) as for instance actualities are (must be) discrete and finite (e.g. non-random parameters driving a random phenomena in _this_ universe) while potentialities must be any logically allowed possibly continuous quantity (exactly). I would agree that statistical problems and estimates should be about (uncertainty of) actualities, but the real importance of vocabulary is that the community can agree on it. Hopefully people might agree that there should be such a distinction (actualities from potentialities) if only to avoid a lot of confusion. 6. Anonymous Says: October 3, 2012 at 12:30 am As a subjective Bayesian, can you elicit the prior that the 100-th digit of pi is 7? For me this is the same problem, but I’ve heard some Bayesians saying it is fine to do it. • If you only consider the 100th digit of π, I would indeed say it is a similar issue: using a series representation of π eventually leads to this 100th digit. And there is no clear “observation” I can gather about this 100th digit… If now you are interested in the distribution of the digits of $\pi$, this is another issue: I have observations and I can put a prior of the distribution. (Not that it will answer the on-going mathematical question about the randomness of the digits of π, obviously!) • Anonymous Says: October 3, 2012 at 3:34 pm Well, we can “estimate” pi by generating data in a square and counting the proportion of points that fall inside the circle inside the square http://yichuanshen.de/blog/2012/01/06/monte-carlo-pi/ We could do this in a Bayesian way as well by putting a prior on pi. The Binomial likelihood here is defined for every value of pi (which is not true in Larry’s example), even though there is only one true value. • Yes, this is an example I used in my public lecture in Australia (and that my daughter also learned in secondary school). However, this is not a statistical problem, in my opinion, rather a stochastic approximation to this unique and known number. Cancel	{"extraction_info": {"found_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": 4, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.916990339756012, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/39192/proof-of-inequality	# Proof of inequality I have problems with proving inequality : $${a^{2}}+b^2+c^2+\frac{2}{5}abc<50$$ where $a,b,c$ are the lengths of triangle's sides, and the circumference of the triangle is $10$. Thanks. - what do u mean by a circumference of a triangle. – M.Subramani May 15 '11 at 11:33 Presumably by "circumference" what's meant is what I'd call "perimeter". – Gerry Myerson May 15 '11 at 12:11 ## 1 Answer Consider the polynomial $(x-a)(x-b)(x-c)$. Multiplied out, this is $$\begin{eqnarray} (x-a)(x-b)(x-c) &=& x^3-(a+b+c)x^2+(ab+ac+bc)x-abc \\ &=& x^3-10x^2+(ab+ac+bc)x-abc\;. \end{eqnarray}$$ We also have $$10^2=(a+b+c)^2=(a^2+b^2+c^2) + 2(ab+ac+bc)\;,$$ $$ab+ac+bc=\frac{100-(a^2+b^2+c^2)}{2}\;,$$ and thus $$(x-a)(x-b)(x-c) = x^3-10x^2+\frac{100-(a^2+b^2+c^2)}{2}x-abc\;.$$ Now since $a$, $b$ and $c$ form a triangle with perimeter $10$, they must all be less than $5$. Thus the value of the polynomial for $x=5$ is positive, that is, $$5^3-10\cdot5^2+\frac{100-(a^2+b^2+c^2)}{2}\cdot5-abc>0\;,$$ which upon rearrangement becomes your inequality. - 2 Bravo!${}{}{}{}$ – Gerry Myerson May 15 '11 at 12:19	{"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": 10, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9175384044647217, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/109717?sort=oldest	## Functional/variational derivative and the Leibniz rule ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I am currently trying to understand the BV-formalism, which makes heavy use of the functional derivative. Let us consider the functional derivative, as defined in for example its Wikipedia article. Let $F$ be a functional, i.e. a map from, say, $C^\infty(\mathbb{R})$ to $\mathbb{R}$, and suppose it may be written as $F[\phi] = \int f\big(x,\phi(x),\phi'(x),\dots,\phi^{(n)}(x)\big)\,dx$ for some function $f$ which depends on the derivatives of $\phi$ up to order $n$. Then the functional derivative of $F$ is $\displaystyle \frac{\delta F}{\delta \phi} = \sum_{i=1}^n(-1)^i\frac{d^i}{dx^i}\frac{\partial f}{\partial \phi^{(i)}}$. Now, my background is that of differential equations and differential geometry, i.e. jet spaces and variational calculus and the like. In that area, the latter operator, $\sum_{i}(-1)^i\frac{d^i}{dx^i}\frac{\partial}{\partial \phi^{(i)}}$, is well known; it is called the variational derivative. Summarizing, then, we seem to have that the functional derivative of a functional is the variational derivative of (one of its) densities. Since the variational derivative involves lots of derivatives, it certainly does not satisfy the Leibniz rule, i.e. it is not a derivation. In various places, however, I've come across the statement that the functional derivative does satisfy the Leibniz rule. (That already seems unexpected to me: how can an operator which is so intimately connected to a decidedly non-derivation be a derivation?) There are various ways to prove it, but I would like to understand this fact in terms of the variational derivative, if possible. So: how can the Leibniz rule of the functional derivative related to variational derivative; can the former be expressed somehow in terms of the latter? - 1 1) Maybe you should provide a precise statement of the Leibniz rule you're referring to. 2) I don't see any difference between what you call the "functional derivative" and what is often called the "variational derivative". Either way, it tells you what the directional derivative $\left. F[\phi + t\dot\phi]\right\|_{t=0}$ is, where you've integrated by parts to shift all the derivatives off $\dot\phi$. – Deane Yang Oct 15 at 13:57 The difference is that the variational derivative (as I understand it anyway) acts on ordinary functions, such as $f$, by the operator described above; while the functional derivative acts on functionals, such as $F$. – Sietse Ringers Oct 15 at 14:05 And what is the variational derivative (acting on an ordinary function) used for? – Deane Yang Oct 15 at 14:15 The Leibniz rule probably holds on the level of the functionals, which leads to the question: is the product $F[\varphi]G[\varphi]$ of two such functionals still a functional which can be written as the integral over a function $F[\varphi]G[\varphi]=\int h(\ldots)$? – Michael Oct 15 at 14:45 Deane: the variational derivative is defined on the space of functions on jet space more or less as a jet space-analog of the functional derivative, but without the actual functional aspect (as described above and below). In short, it is important in for example the geometry of jet spaces (in particular in the horizontal cohomology, where it comes from the de Rham differential along the fibers); and in mathematical physics, since one can express Euler-Lagrange equations in terms of it. – Sietse Ringers Oct 16 at 7:44 ## 1 Answer Connection of functional derivative with variational derivative: $\frac{\delta}{\delta\phi(x)} F[\phi] = \frac{\delta F[\phi]}{\delta\phi}(x)$. Note that the variational derivative carries an extra coordinate variable dependence. It helps to make it explicit when there is similar confusion. Functional derivative Leibniz rule: $\frac{\delta}{\delta\phi(x)} F[\phi] G[\phi] = \frac{\delta F[\phi]}{\delta\phi}(x) G[\phi] + F[\phi] \frac{\delta G[\phi]}{\delta\phi}(x)$. Special case: $F_x[\phi] = \phi(x)$, $G_{i,y}[\phi] = (\partial_i\phi)(y)$, and $$\frac{\delta}{\delta\phi(z)} F_x[\phi] G_{i,y}[\phi] = \delta(x-z) (\partial_i\phi)(y) - \phi(x) \frac{d}{dz_i}\delta(y-z)$$. Notice the distributional coefficients in the derivatives. There is no way to get away from them if you wish to consider $\phi(x)$ and such as functionals in their own right. If you are interested in the BV formalism in the physics formalism, where the distinction between the functional and variational derivatives is barely remarked, I recommend the reviews by Henneaux and by Gomis, París and Samuel: doi:10.1016/0920-5632(90)90647-D, doi:10.1016/0370-1573(94)00112-G. If you are interested in the BV formalism purely from the point of view of jets, without bringing functionals into the picture, other than peripherally, I recommend the early paper of McCloud and this sequence of papers by Barnich, Brandt and Henneaux: arXiv:hep-th/9307022, arXiv:hep-th/9405109, arXiv:hep-th/9405194, arXiv:hep-th/0002245. If you are more interested in the BV formalism more from the functional point of view, with the appropriate level of functional analysis included, and with jets appearing only peripherally, I recommend the papers by Fredenhagen and Rejzner, as well as Rejzner's thesis: arXiv:1101.5112, arXiv:1110.5232, arXiv:1111.5130. - I should have asked for an explanation for what "BV formalism" is, so thank you for answering this anyway. – Deane Yang Oct 15 at 15:36 Thank you, Igor, for this answer. It was not precisely what I was looking for, but that's to be expected because I don't think I managed to write down exactly what I was looking for. – Sietse Ringers Oct 16 at 7:36	{"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": 21, "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}, "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}, "english_confidence": 0.9418619871139526, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/51180/relation-between-field-strength-and-potential/51208	# Relation between field strength and potential? In terms of gravity and electric fields, I'm not sure what the difference is between field strength and potential is and how they are related? Both using maths and not. - ## 6 Answers The field strength is the negative first derivative of the potential. For example, in Cartesian coordinates, with electric potential $V = V(x,y,z)$, the electric field is: $$\boldsymbol E =-\nabla V= -\frac{\partial V}{\partial x}\boldsymbol i - \frac{\partial V}{\partial y}\boldsymbol j - \frac{\partial V}{\partial z}\boldsymbol k$$ Note that you can add or subtract an arbitrary constant potential to $V$ and it will not affect the field strength, since its derivative will be 0. For the field/potential, you can calculate the force/energy by just multiplying by a charge. In terms of gravitational field, use $\boldsymbol g$ instead of $\boldsymbol E$, and you multiply by mass, rather than charge, to get $\boldsymbol F$. A "physical" way to think of the derivatives is that they show how rapidly the potential changes in space; big changes in potential over a short distance make for big fields in the direction of maximum change. Note that the relation between electric field and potential above is true only for static fields. The more general formula is: $$\boldsymbol E=-\nabla V-\frac{\partial\boldsymbol A}{\partial t}$$ where $\boldsymbol A$ is the magnetic vector potential. - 1 Hi Will. Welcome to Physics.SE. Thanks for your answer. As a part of your contribution, please use TeX markups (MathJax) in equations of your answers. If you could spend some time, have a look over here or our FAQ. Feel free to respond to guys who comment (for instance, me) on your posts using `@` before the commentator's name :-) – Ϛѓăʑɏ βµԂԃϔ Jan 14 at 14:50 Thanks, will do! – thatnerd Jan 14 at 14:56 @CrazyBuddy D'oh! How about this? At any rate, I'm researching TeX and the FAQ before I go too crazy on comments & answers. – thatnerd Jan 14 at 15:22 The units of electric field should be a big hint here: Volts/meter. It is literaly the electric potential per unit length along the field vector. - I don't understand what it means by a meter the field is an area (or really a volume), but a meter is one dimensional. And that would seem to suggest the field is constant? – Jonathan. Jan 14 at 12:53 The meter here is measured along the direction of the field. So the field strength is a measure of the difference in potential along one meter of field line. – dmckee♦ Jan 14 at 12:57 So the field strength defines the potential at each point? (or the difference in potential between two point?) – Jonathan. Jan 14 at 12:58 And what is the units for gravitational field? I can't find such a clear answer as there is for electric fields? – Jonathan. Jan 14 at 12:59 You can add or substract an arbitray constant from the potential without affecting the physics (a simple gauge transform), so the field tells you about potential differences rather than absolute potential. – dmckee♦ Jan 14 at 12:59 show 2 more comments In terms of electromagnetism, the potential is given by the gauge field $A_\mu$, whereas the field strength is given by $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu.$ The main difference is that the field itself is can be changed to a different value by a so-called gauge transformation (which is given by an element of the group $U(1)$), while the field strength itself is invariant (not changed) by it. For gravity, one can find a similar structure: here the field strength corresponds to the Riemann tensor ${R^\mu}_{\nu\rho\sigma}$, from which one can construct a quantity that is invariant under a change of coordinates (which corresponds to the electromagnetic gauge transformation), i.e. the Ricci scalar $R$. The coordinates are represented by the metric tensor $g_{\mu\nu}$, and the relations between the aforementioned quantities are given by \begin{equation}{R^\mu}_{\nu\rho\sigma}=\partial_\rho{\Gamma^\mu}_{\nu\sigma}-\partial_\sigma{\Gamma^\mu}_{\nu\rho}+{\Gamma^\mu}_{\lambda\rho}{\Gamma^\lambda}_{\nu\sigma}-{\Gamma^\mu}_{\lambda\sigma}{\Gamma^\lambda}_{\nu\rho},\end{equation} \begin{equation}{\Gamma^\mu}_{\nu\rho}=\frac12g^{\mu\sigma}(\partial_\rho g_{\nu\sigma}+\partial_\nu g_{\sigma\rho}-\partial_\sigma g_{\nu\rho}),\end{equation} \begin{equation}R_{\mu\nu}={R^\lambda}_{\mu\lambda\nu}\end{equation} and \begin{equation}R=R_{\mu\nu}g^{\mu\nu}.\end{equation} $R_{\mu\nu}$ is the Ricci tensor and ${\Gamma^\mu}_{\nu\rho}$ is the Christoffel symbol. The latter can be interpreted as a gauge field for gravity. The analogy is even clearer if one considers the field strength of a non-abelian gauge symmetry. For example, the field strength of $SU(3)$ is given by \begin{equation}F_{\mu\nu}^{ab}=\partial_\mu A_\nu^{ab}-\partial_\nu A_\mu^{ab}+iA_\mu^{ac}A_\nu^{cb}-iA_\nu^{ac}A_\mu^{cb},\end{equation} where $a$ and $b$ are indices that transform under SU(3). - 1 For someone who is clearly starting, and doesn't "see" the difference, this may not be the best explanation. – MyUserIsThis Jan 14 at 17:39 While I agree that my answer might be a little over the top, I'm convinced that it is on topic and useful to some. – Frederic Brünner Jan 14 at 17:41 Field strength determines the force in the Newton equation. It is responsible for the velocity changes. The potential has no such a meaning - its absolute value is not important if kept constant throughout consideration. So its absolute value is a matter of convenience. It is involved in such a thing as the total energy. And the velocity absolute value will change only if the potential changes. A particle never feels the potential itself, but a force (a local potential difference in the direction of maximum potential change). The local potential difference $\phi(x+\delta x)-\phi(x)\;$ divided by $\delta x\;$ with small $\delta x\;$ determines the field strength at $x$. - You have to view the field as a vector field, that is, for every point, you will have a vector, that is $E$, that represents the force that a unit charge will feel at that point, and the direction of the force. If you don't feel comfortable with Volts/meter units, then use N/C (Newtons/Coulomb) That if: force per unit charge, that happens to be the same as volts/meter. The potential talks about energies. The potential is now a scalar field, that is: for every point in the space we have a number, that number represents the energy a unit charge would have if it was located there (potential electric energy). By definition the potentials are defined so their gradient is the force: $E=-\nabla V$, as someone said before, in units, we have that $N/C=[V]/m$, being $N$=newtons, $C$=coulombs, $m$=meter, and $[V]$ the units of our new potential unit, and we decide to call that volts, as you see, the measure of the electric field is the same as volts/meter: $V/m$ See it as the slope of the potential, that is how it's defined: $-\nabla V$, the field goes down/up $x$ volts per meter in point $P$. - In a region with electric potential $V$ that varies with position, the electric field $E$ points along the potential's direction of greatest decrease. A positive charge will naturally go towards points of lower electric potential, and so at any given position, it will feel a force proportional to the electric field at that point, for the $E$-field points downhill, so to speak, exactly where the particle is compelled to go. You can imagine the $V$ field as determining a landscape with hills and valleys--the value of $V$ determines the heights of these hills and valleys. The $E$-field points along the unique direction that most quickly goes downhill. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 38, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9202390313148499, "perplexity_flag": "head"}
http://windowsontheory.org/2012/11/16/tennis-for-the-people-ii/?like=1&_wpnonce=aaa135d5d2	## A Research Blog by I continue the discussion from last post. We are trying to add unpredictability to tennis by looking for a monotone, transitive and balanced function $f$ such that $E_p(f)$ has a wide threshold window, that is, we want the range of $p$ where $E_p(f)$ is, say, between 0.01 and 0.99, to be as large as possible. A formal treatment of this problem could be found in the seminal paper of Kalai and Freidgut. The major simplifying step we take is to look at the derivative of $E_p(f)$ at the point $p=1/2$ as a proxy for the width of the window. Think of it as the first order linear approximation for the (inverse of the) width of the threshold window: the smaller the derivative the wider the window. The main tool to help us reason about the derivative is Russo’s formula, but first a definition: given ${x_1,.. ,x_n}$ we say that bit $x_i$ is pivotal if flipping $x_i$ changes the value of $f$. The total influence of the function is the expected number of pivotal bits when $x_1,..,x_n$ are drawn uniformly at random. Russo’s formula states that for monotone functions the derivative of $E_p(f)$ at $p=1/2$ is exactly the total influence of $f$! One way to see why Russo’s formula makes sense is to think about the definition of a derivative and observe that increasing $p$ by a small amount could be thought of as picking a bit at random and setting it to 1. Now, the probability that $f$ changes its value is proportional to the number of pivotal bits. We now face a well-defined challenge of coming up with a function satisfying our constraints and with small total influence. As a sanity check it is easy to observe that the total influence of the majority function is indeed roughly $\sqrt n$. The reason is that with probability about $1/\sqrt n$ the majority is by exactly one point, in which case half the bits are pivotal. What about majority of majorities, or better yet , recursive-majority of three values? Well, the total influence of recursive majority can be computed to be roughly $n^{0.37}$, which is an improvement over simple majority (not surprising since simple majority could be shown to have the maximal total influence of all monotone Boolean functions). But is it the best? The answer is given in the celebrated theorem by Kahn, Kalai and Linial which implies in our case that the total influence is $\Omega(\log n)$. A tight example was provided by Ben-Or and Linial in the TRIBES function. In the TRIBES function the bits are partitioned to sets (‘tribes’) of roughly $\log n$ – $\log{ \log n}$ bits each. Player ‘1’ wins if there is a tribe with all its bits set to 1. Player ‘0’ wins otherwise. The function is clearly transitive and monotone. Exact parameters can be set so that it is approximately balanced. To see that the total influence is small observe that for a bit to be pivotal it has to belong to a tribe where all the other bits are 1, which happens with probability about $\log n/ n$ so the total influence is $O(\log n)$. So is TRIBES a good candidate for an athletic game? I don’t think so, for several reasons. First, it is not exactly balanced, I can’t imagine an athlete happy with a game where he is unlikely to win even when competing on equal grounds point by point. Second, there is a difference between the role of player 1, which plays offense trying to obtain a tribe, and player 0 which plays defense trying to block player 1. Both problems are rectified by adding another requirement. We want the function to be symmetric, that is $f(1-x)=1-f(x)$. This means that when flipping all bits the outcome also flips and since $f$ is also monotonic it also means it is balanced. Cycle-Run Here is a suggestion for a transitive, monotone, symmetric function, with low total influence. We call it Cycle-Run and it could be viewed as a symmetric version of TRIBES. Call a consecutive sequence of ‘1’’s a 1-run. Similarly a consecutive sequence of zeros is a ‘0-run’. We allow runs to wrap around, so if a run reaches $x_n$ it may continue with $x_1$. The value of $f$ is determined as follows: 1. Check which player has the longest run. 2. In case of tie check which player has a larger number of maximal runs. 3. In case of tie check the total length of segments between maximal runs, where a segment starting from a 1-run clockwise is counted for the 1-player and a segment starting at a 0-run clockwise is counted for the 0-player. The player that has a larger total count is declared the winner. In the example below both players have a single maximal run of length 3, but player 0 wins the tie breaker 9-2. Monotonicity and symmetry are readily verifiable. Transitivity is obtained via rotations of the cycle. Why is the total influence logarithmic? One way to see it is to observe that the expected number of runs of length $k$ is $n2^{-k}$, so the expected number of runs longer than $\log n$ is smaller than 1. Further work is required, but basically this is the reason. If you want to do the calculations yourself you may find this survey useful. So is this useful? In practice most tennis matches consist of roughly 200-400 points. Below we plot Majority, Recursive Majority and Cycle-Run for n=243. We see that indeed Majority has the narrowest threshold window and Cycle-Run the widest. The difference between Cycle-Run and Recursive-Majority is not that big. When $p=0.4$ the probability of an upset rises from roughly 3% to 5.3%. Asymptotics however don’t lie and when n is increased to 2187 the differences are quite striking. Now, when $p=0.45$, for simple majority the probability of an upset is negligible. For recursive majority it is less than 2% and for cycle run it is more than 11%. What about real life? I have the data of three tennis matches. In these three examples all rules have the same result. An anecdotal observation is that at least in these cases the winning run tends to be very close to the end of the match, suggesting that the mental aspect of the game plays an important role. So should we petition the ATP for a rule change? Credits: This post is the outcome of discussions with Parikshit Gopalan, Yuval Peres, Omer Reingold and Kunal Talwar. The data was given to me through the generosity of Daniele Paserman and the facilitation of Ran Abramitzky. Professional tennis is a good setting for researching human decision-making when payoffs are well defined. If you want to read some cool papers checkout a paper by Daniele and a paper by Ran. ### Like this: from → Uncategorized 4 Comments leave one → 1. Yevgeniy November 17, 2012 12:04 am Excellent posts, Udi. One thing slightly complicating the tennis application is that the player serving has an advantage, so it is a bit too simplistic to simply say that the game consists of a sequence of “same” bits x_i, as the decision must be made who is serving for each i. I suppose one simple rule is to make the players alternate their serves, although this is slightly not nice as people like to serve for a while; gives them a sense of continutiy. To fix this, we can play n/2 points as follows: have player 1 start serving, and then the winner of the point serves the next point. Then, after n/2 points the same is repeated with player 2 starting. And then compare the maximum runs in both cases with something ugly in case of a tie. (Or, more generally, have have 2t “sets” of n/2t point each with something to aggrgate the 2t maximal runs.) This is also nice as it gives explicit reward for “keep holding my serve”. Another advantage of this is that the rules become very clean. Serve until you lose serve, and try to maximize the number of points you win consecutively. (Here I ignore a slight asymetry at the beginning, but I suspect it’s effect if negligible if n is large.) With these tules, we get something VERY similar to badminton and voleyball (where new set has a different player starting, or maybe a loser, I forgot). The main differences are: (a) maximum total number of sets is odd, not even (so no need for “ugly” rule 3); (b) more importantly, you are trying to maximize the length of the run, not the number of sets you win (which is important for incluence); (c) not important, but in both badminton and voleyball the server is actually at DISADVANTAGE, so the runs are likely very short, even when one polayer is much better. (does it mean we should change it, and the winner receives in these games? Not sure…) Overall, the main question is if your simplified analysis through influence stills holds when there are two probabilities involved (Pr(servring 1 wins)=p and Pr(receiving 1 wins)=q). What do you think? Yevgeniy • Udi Wieder November 17, 2012 5:22 am I agree that the model above simplifies the game considerably, for many reasons. By the way, one technical difficulty with your suggestion is that the number of points is even whereas we must have an odd number of points. Regarding your question, I guess it is possible to push through an analysis with the product space you describe, though I haven’t done it. 2. Mohit Singh November 17, 2012 12:06 am One nice things about majority (or majority of majorities as in Tennis) is the fact that lopsided games end early. Thus for small p, we do not need to draw out all n tosses. I guess here we might have to do all tosses except O(log n/p) of them. Is there a simple modification to deal with this issue? • Udi Wieder November 17, 2012 1:40 am One way is to end the game if a run of say log n + 3 is obtained. We lose the transitivity property, but in a reasonable way. A more subtle issue I think is the distribution of ‘important points’. I wonder what is a sensible definition for that and what are requirements for a crowd pleasing game. %d bloggers 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": 0, "wp_latex": 34, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9444994330406189, "perplexity_flag": "middle"}
http://jellymatter.com/2011/04/17/	# Jellymatter The blog that is not afraid of equations… or bees \| April 17, 2011 ## A scientist modelling a scientist modelling science by Lucas Wilkins The is a follow up from Nathaniel’s post. One of the ways that the probabilities of probabilities can be used is in asking what experiments would be best for a scientist to do. We can do this because scientists would like to have a logically consistent system that describes the world but make measurements which are not completely certain – the interpretation of probability as uncertain logic is justified. Lets make a probabilist model of scientific inquiry. To do this, the first component we need is a model of “what science knows”, or equally, “what the literature says”. For the purposes here, I will only consider what science knows about one statement: “The literature says X is true”. I’ll write this as $p(X\|L)$ and its negation as $p(\bar{X}\|L) = 1 - p(X\|L)$. This is a really minimal example. Posted in Actual Science \| 2 Comments » \|	{"extraction_info": {"found_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": 2, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9450459480285645, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/83183/example-of-2-dimensional-hypersurface-singularity-whose-exceptional-locus-of-mini/83196	## Example of 2-dimensional hypersurface singularity whose exceptional locus of minimal resolution is not normal crossing ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $U= (f=0) \subset \mathbb{C}^3$ be an isolated hypersurface singularity of dimension $2$. Let $\mu: \tilde{U} \rightarrow U$ be its minimal resolution. Question Is there an example of $U$ such that the exceptional locus $E$ of $\mu$ is not normal crossing? - ## 1 Answer Yes. Take $f=z^2+(x^3+y^3)(y^3+x^4)$. To compute the resolution, consider the projection $U\to {\mathbb C}^2$ onto the $x,y$ coordinates: it is a double cover branched on the curve $B:={(x^3+y^3)(y^3+x^4)=0}$. Blow up the origin in ${\mathbb C}^2$, and let $U'$ be the surface obtained by base change and normalization: $U'$ is a double cover branched on the strict transform $B'$ of $B$ and it is smooth since $B'$ is smooth. The exceptional locus of the resolution $U'\to U$ is the inverse image $Z\subset U'$ of the exceptional curve of the blow up. It is a standard computation to show that $Z$ is irreducible with $Z^2=-2$, $p_a(Z)=2$ and $Z$ has a simple cusp as singularity. So the resolution is minimal and $Z$ is not normal crossings. - Thank you for the example. – tarosano Dec 27 2011 at 5:09	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "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}, "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}, "english_confidence": 0.9168760180473328, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/129319/algebra-abstract-symmetry-counting	Algebra - abstract symmetry - counting I am reading Michael Artin's Algebra, encountered paragraphs( 7.3) that I don't quite understand. so there's the citation The elementary formula which uses the partition of $S$ into orbits to count its elements. We label the different orbits which make up $S$ in someway, say as $O_1, ..... , O_k$ then $\|S\| = \|O_1\|+.....\|O_k\|$, the formula hs great number of applications. Example, consider the group G of orientation-preserving symmetries of a regular dodecahedron $D$. these symmetries are $11$ rotations. it is tricky to count them without error. Consider the action of $G$ on the set $S$ of the faces $D$. the stablilizer of a face is the group of ortations by multiples of $2\pi/5$ about a perpendicular through the center of $S$. So the order of $G_s$ is $5$. There are $12$ faces, and $G$ acts transitively on them, thus $\|G\| = 5\times 12 = 60$ or $G$ operates trasnsitively on the vertices $v$ of $D$. There are three rotations including 1 fix a vertex, so $\|G_v\| = 3$. There are $20$ vertices hence $\|G\| =3\times 20 = 60$, which checks. There is a similar computation for edges. If $e$ is an edge then $\|G_e\| = 2$, so since $60 = 2\times 30$, the dodecahedron has $30$ edges. I don't quite understand when counting using vertex what are the three rotations, and when counting using edge what is exactly $G_e$ meant for ? let $S$ be the set of $12$ faces of the dodecahedron, and let $H$ be the stablilizer of a particular face $S$, then $H$ also fixes the face opposite to $S$, and so there are two $H$-orbits of order 1, The remaining faces make up two orbits of order $5$. In this case it reads as follows $12 = 1+1+5+5$. I really don't understand how the order of H-orbits are derived? and infact don't even understand, what are the gemometric meaning of these H-orbits? - 1 Answer We used Michael Artin's text last semester, I remember reading this bit. Your first question: "I don't quite understand when counting using vertex what are the three rotations, and when counting using edge what is exactly $G_e$ meant for ?" Answering this is going to be difficult because you need a scaled model of the dodecahedron to really see what I'm talking about. At a vertex, there are three faces that meet to give a vertex (each face is a pentagon). Now suppose you rotate the dodecahedron about a vertex. You want to rotate it by an angle such that after the rotation, you get the exact same configuration of the dodecahedron. The clue to what angle you must rotate it by is in the number of faces that meet at a vertex. This number is three so that the angle is $360^{\circ}/3$ which is $120^{\circ}$. So the rotations that fix a vertex are rotation by $120^{\circ}$, $240^{\circ}$ and doing nothing (or $360^{\circ}$), which is three in total. This also means to say that the order of the stabiliser of a vertex is three. When counting the stabiliser of an edge, $G_e$ is just purely notational to mean "stabiliser of an edge", the subscript "e" for "edge". Does this answer your first question? For the second part of your question, can your provide more context (i.e. copy more of what the text says before and after that paragraph)? This is because I believe we need to look more carefully at what now is acting on the set of 12 faces of the dodecahedron. It can't be the whole group because that acts transitively on the dodecahedron. Perhaps it is being acted on by some subgroup of the whole group? I don't have the book with me now. $\textbf{Edit:}$ I get your second question now: Fix a face $H$ of your dodecahedron. Now the stabiliser of $H$, $G_H$ has order 5. Now when we let $G_H$ act on the dodecahedron, trivially $G_H$ acting on $H$ will not move it to any other face. So we get one orbit here of order 1. Similarly, $G_H$ acting on the opposite face $H'$ will not move it to any other face so that we get another orbit here of the same order. Now consider the 5 faces of the dodecahedron that are directly adjacent to $H$. When $G_H$ acts on them, they simply rotate about the line perpendicular to $H$ so that $G_H$ acts transitively on these 5 faces. In other words, these 5 faces give us another orbit. Similarly the set of 5 faces adjacent to $H'$ gives us another orbit. So since $5 + 5 + 1 + 1 = 12,$ this illustrates why the sum of the size or orbits must be equal to the size of the whole set (I believe that's what Artin is trying to illustrate). Does this help? - to clarify, if we have a group action of G on a set X, then any subgroup H of G also acts on that same set X using the very same action. Artin is using H-orbits, to distinguish them from the G-orbits (the orbits of the whole group). given the context, it might have been better to use a different letter besides "H" for the face in question. – David Wheeler Apr 8 '12 at 15:57 @DavidWheeler Thanks for the clarification. I vaguely remember something like that in Artin's text, but as I said I don't have it now. – BenjaLim Apr 9 '12 at 1:28 @BenjaminLim thanks very much for the explanation. they help a lot, I think I need that mini model of dodecahedron to help understand this. for "what Ge means, I was also asking for the rotations Artion mentioned when rotating around edge". I think the main issue is I don't get the rotation axis when thinking about the rotation. – zinking Apr 9 '12 at 3:12 @zinking Take an edge, draw a line through that edge until it meets the opposite edge at right angles. Then if you rotate your dodecahedron about that edge, you should see the only rotations that preserve the dodecahedron are $180$ degrees of the full 360. That's why the order of the stabiliser is 2. Hint: I suggest making paper models of these things. If you are here in australia I would gladly pass you mine! Does this answer all your queries now? Please accept my answer above if it does. – BenjaLim Apr 9 '12 at 3:21 yes, thanks a lot. accepted, not familiar with MathExchange. – zinking Apr 9 '12 at 10:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 59, "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}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9500939846038818, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/19165/relating-milliampere-hours-to-watt-hours-for-batteries	# Relating milliampere-hours to watt-hours for batteries I've seen many batteries that are measured in milliampere hours (mAh), while others are measured in watt hours (wh). How can I convert them between each other so that I can actually compare them? It's often the case that computers or smartphones use one or the other and would be great to have a metric. - ## 2 Answers To the best of my knowledge, wattage is just the product of voltage x amperage. So if you know the output voltage of the battery, you should be able to convert mA to w, and vice versa simply by multiplying or dividing, and taking the "milli" prefix into account. You can drop the "hour" variable since it is in both units. In equation form: $$\text{mA}\times\text{V}\times 1000 = \text{W}$$ and $$\frac{\text{W}}{\text{V}}\times \frac{1}{1000} = \text{mA}$$ - To get an approximate result for energy, you need to multiply the charge in mAh by the voltage in volts. If you have a battery with specs 10V and 100mAh, you have the energy estimate of 1000mWh, or 1Wh, or 0.001kWh. However, this estimate is approximate, as voltage decreases in the course of discharge. -	{"extraction_info": {"found_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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.946672260761261, "perplexity_flag": "middle"}
http://mathhelpforum.com/differential-equations/83238-riccati-equation-existence-uniqueness-relation.html	# Thread: 1. ## Riccati Equation: existence/uniqueness relation What does the existence/uniqueness theorem say about solutions to initial value problems for Riccati equations of the form This is what I came up with, but I'm not sure. A Riccati Equation of a particular form has a unique solution satisfies the following conditions: It is to exist for large values of the independent variable Its graph must stay above a certain line for large values of the variable 2. Originally Posted by tn2k7 What does the existence/uniqueness theorem say about solutions to initial value problems for Riccati equations of the form This is what I came up with, but I'm not sure. A Riccati Equation of a particular form has a unique solution satisfies the following conditions: It is to exist for large values of the independent variable Its graph must stay above a certain line for large values of the variable First we need to isolate the derivative $\frac{dy}{dx}=f(x,y)$ so we get $f(x,y)=y'=-(y^2+1)$ Since both $f(x,y)$ and $\frac{\partial f}{\partial y}$ are continous on a rectnagle containing (0,1) the solution exists and is unique. 3. Hi, thanks for your help. I was wondering if you (or anyone else) could help me with this similar problem. What does the existence/uniqueness theorem say about solutions to initial value problems for Riccati equations of the form $y'(x) + y(x)^{2} + Ay(x) + B = 0.$ Thanks!	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9100672602653503, "perplexity_flag": "head"}
http://mathoverflow.net/questions/116367/open-question-non-commutative-site-following-grothendieck-quillen-connes-and-c	## Open question: non-commutative site following Grothendieck, Quillen, Connes and Crane for quantum gravity. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) This is an open question and it's to find out who is interested in this kind of thing, who can benefit from thinking about this. It is very brief but hopefully will only be unclear to people who are not into this anyway. Briefly, we can think of a Morita category (with object C-algebras and morphisms bimodules) as a "weak" groupoid. It seems to me one can put a topology on the object space by thinking of it as a Banach bundle or continuous field of C-algebras, which comes with a topological structure. This should satisfy axioms of a Grothendieck site (a topology on a category). So I think one can use this to define a weak topological groupoid. What about a weak differentiable groupoid or Lie groupoid? We can consider comparing this structure (with a certain sheaf over the site of objects) to a spectral triple with the geometrical data in the Dirac operator. I will stop to keep this short. This should lead to something closely related to Quillen's point of view on cyclic cohomology where the spectral triple algebra is also related to a Grothendieck site. The motivation is coming from Crane's "quantum geometries" for quantum gravity and I think there are unexplored conceptual intersections between category theory and non-commutative geometry. Thanks for reading! Sorry if this is not a math overflow type question but there is an indication that we may ask open questions. - 11 The ideas in your question sound interesting, but I have a hard time understanding what question you are asking. So I will point you to the following "how to ask" advice and hope that a revised version of your question receives more attention: mathoverflow.net/howtoask – Manny Reyes Dec 14 at 16:11 2 I'm 'interested in this kind of thing', but, i don't get what you mean... what do you mean by "weak" groupoid. what is for you a 'Morita category', the whole 2-category of C^* algebra - bimodules - map between bimodule ? – Simon Henry Dec 14 at 17:58 1 This is not really an answer to your question, so it is a comment. I see the problem as to how to use the power of strict higher groupoids in the context you mention - see my question mathoverflow.net/questions/86617/… . In homotopy theory, the use of strict multiple groupoids has given rise to new nonabelian calculations of $n$-ad homotopy groups and so of absolute homotopy groups of some complexes, and of k-invariants. One would seem to need strict structures to make explicit calculations, which is surely what you need for physics (??). – Ronnie Brown Dec 14 at 21:46 Thank you for reading my question and taking the time to leave comments :) In the coarse-grain view you have a category where arrows are Morita equivalence bimodules (objects are C-algebras). Bertozzini et al called this a Morita category if the bimodules are considered up to isomorphism. This is really like a "weak" groupoid i.e. it is a groupoid but with equalities replaced by isomorphisms. Yes this is really a bicategory :) ... For a generalised notion of path or geodesic in a non-com space, one can consider elements of these bimodules, (now we're strict) in a richer context. – Rachel Dec 17 at 16:06 1 You should realise the difference between a Grothendieck topology and an ordinary topology - they really are nothing alike. If you start out with a continuous field of $C^$-algebras, then you might just be defining a topological groupoid outright, but starting from an arbitrary category whose objects are $C^$-algebras, you don't really have anything to grab onto. In any case, a 'Morita category' is really the shadow of some 2-category of stacks or similar, via the correspondence between (certain) locally compact groupoids and (certain) $C^$-algebras. – David Roberts Jan 23 at 23:29 show 1 more comment	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "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}, "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}, "english_confidence": 0.9290244579315186, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/44435/could-some-red-and-blue-shifts-be-the-result-of-light-passing-through-dark-matt/44436	# Could some Red and Blue shifts be the result of light passing through “dark matter”? As i see it, light behaves in certain ways, as the Double Slit experiement shows, So when light comes into contact with dark matter, it becomes both a wave and a particle, the wave is bent around the dark matter, which we see as gravitational lensing, but the particle will go through the dark matter, but will also leave our frame of reference, follow the particle through dark matter and its speed would not change staying at $c$, but only relative to the the space it travels through, it would eventually pass through dark matter, we would see this first as a red or blue shift as it slowly emerges (from our perspective), and continues on it way, Never knowing it had passed through any medium, where as the wave had passed around the dark matter thousands, millions, or billions of years before, depending upon the mass of the dark matter. - 6 This isn't the way light behaves. There isn't a wave and a particle that follow separate paths. – Peter Shor Nov 17 '12 at 23:28 – Qmechanic♦ May 4 at 18:57 ## 3 Answers Light interacts with dark matter only through gravitation. That is why we try to map dark matter distribution in the universe by statistically studying the small distortions in the images of background galaxies (weak lensing regime). Whole redshift-sliced 3D maps of dark matter have been traced for some small regions of the sky, by means of weak lensing distortion. In all cases, the observed properties of the background objects are in a statistical sense the same (apart from shape distortion) for objects observed through, and not through, dark matter clumps. That is one of the proofs for dark matter not interacting with photons. Another proof is in the strong lensing regime of multiple lensed quasars. The Hubble parameter can be derived by measuring time delays between flux changes of a background quasar, as detected in two lensed images. That time delays happen because the photons travel through different paths, both through the dark matter halo of the lens galaxy and the Universe. But the measured line waveleght redshifts agree perfectly with the optical paths derived from the geometry of the lens, and no additional frequency shift has ever been observed. As expected from GR, photons leaving a dark matter clump behind recover the colour they had when they entered it. That is, dark matter affects light only by means of the metric tensor, but doesn't interact with the photons. Otherwise it would had been already observed. - 3 After reading your question again, I definitively don't know what your question is... – Eduardo Guerras Valera Nov 17 '12 at 22:22 2 Come on @MarkEmery, Eduardo has given you quite a nice answer (+1), even though reading your question it is hard to guess what you are up to. – Dilaton Nov 18 '12 at 11:48 1 @Mark Emery, dark matter is as "folded space" as normal matter, and that is why we observe different images of the same background quasar when a galactic dark matter halo happens to be in the line of sight. The phenomenon is studied with real data since 1979 with the discovery of the first lensed quasar Q0957+561. Dark matter DOES slow down the photons as you well say, but that is indeed the way all lenses work (glass lenses too). You are right, but there is no othe way we could be observing lensed images... – Eduardo Guerras Valera Nov 18 '12 at 13:07 1 And for the images to reach us, the photons must have already crossed the halos. Otherwise we wouldn`t see the background images at all! And there is no statistical differences in surface flux or redshift between images that reach us after travelling across dark matter and the others. That is, there is lot of empirical evidence of photons having crossed dark matter and not have suffered any scattering or interaction further than the gravitational deflection. It is not dogma, but just overwhelming empirical evidence. – Eduardo Guerras Valera Nov 18 '12 at 13:14 1 The magnification of lensed images is a consequence of their bigger apparent size, but mean surface flux is unchanged, as well as the frequency of the light. Dark matter has no cross-section for photons, or even for baryonic matter, it is collisionless and only produces curvature. That is why many still are searching for alternative gravity theories. Because there is observational evidences far beyond any doubt, that photons do cross dark matter and do reach us unscattered. – Eduardo Guerras Valera Nov 18 '12 at 13:25 show 8 more comments This is another version of the "tired light" hypothesis. This doesn't work for a variety of reasons, among which are the facts that there are no absorption or emission peaks evident in light, there is no blurring of the light from distant objects (stars look like points, not globs), and there appears to be no dependence on the angular view from the sky when looking for redshifts (while dark matter DOES appear to be clumped). - light passing through dark matter would be similar to light passing through fiber optics, there is no reason for "blurring",dark matter is folded space, which would from our perspective focus light into a very narrow beam. – Mark Emery Nov 21 '12 at 20:14 @MarkEmery: scattering is scattering. And fiber optics certainly do blur the light--the index of refraction depends on the wavelength of the light passing through it. If the interaction with dark matter is independent of wavelength, then it would be quite unique. – Jerry Schirmer Nov 21 '12 at 20:31 As you say scattering is scattering, and there is no blurring of light from distant objects, but there should be,it is dark matter that refocuses these objects. – Mark Emery Nov 21 '12 at 20:51 @MarkEmery: what? Are you talking about gravitational lensing? That is one of the few effects that is NOT frequency-dependent (so long as the frequency of the light is much less than the curvature of the relevant region of space). If you have light bouncing off of/through dark matter, you have to explain why there is no dispersion of the light--i.e., answer why is the DM/matter interaction frequency-independent. – Jerry Schirmer Nov 21 '12 at 21:00 Our universe is thought to be made of 85% dark matter, plus galaxy clusters, alot of gravitational lensing going on, but still we see "no blurring" from distant objects,what we are really seeing is the photon particle that as travelled through the D/M, which as not scattered, the wave that went around the dark matter has G/L as been lost in the maze of D/M, we only see the odd effect that as not yet been completely distorted, Once the photon leaves D/M it continues as a wave and a particle.but each time on coming in contact with D/M, the photon wave goes around, and the particle goes through. – Mark Emery Nov 22 '12 at 1:49 show 1 more comment Dark matter--as the term is understood--does not interact electromagnetically. Light only interacts electromagnetically. So, no it could not. - Sorry we do not yet fully understand Dark matter, you are only stateing speculation yourself. "So,no it could not", is what i expect from one with no ability to see other than what they are told. Not long ago it was thought that light could not be a wave and a particle, but we move on. – Mark Emery Nov 17 '12 at 18:45 This question is for those who can think outside the box, – Mark Emery Nov 17 '12 at 18:56 I'm not speculating. Dark matter is known to not interact electromagnetically because if it did we would be able to see it. Essentially nothing else is known about the stuff, but that is an observational fact. – dmckee♦ Nov 17 '12 at 19:09 We can not see it, because once the particle enters dark matter it leaves "our space time", dark matter is higher dimensional, even though the "particle" would still be travelling at c within this medium, – Mark Emery Nov 17 '12 at 22:12 1 Light interacts gravitationally too, as is explained in the other answer. – Dilaton Nov 18 '12 at 0:00 show 1 more comment	{"extraction_info": {"found_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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9485268592834473, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/50990/list	## Return to Question 2 deleted 2 characters in body; edited tags 1 # Non-standard enlargements, $\zeta(s)$ and analytic continuation Consider an extension of the Riemann zeta function $\zeta(s)$ where $s$ now runs over a non-standard enlargement of the complex plane. Observe that if $s=\sigma + it$ with $\sigma>1$ real and finite (or at least infinitesimally close thereto), but $t$ infinite, then the summands $n^s= n^\sigma(\cos \ln(n)t + i\sin \ln(n)t)$ still have values infinitesimally close to finite complex numbers. Indeed, by fixing an infinite real $T$, we can obtain from $\zeta(s+iT)$, by passing to standard parts, a convergent standard Dirichlet series. At least with a sufficiently saturated non-standard enlargement, I believe these same Dirichlet series arise from starting with the standard Euler product and shifting, arbitrarily and independently, all the various factors vertically by various amounts. My questions: What can one say about the possibility of finding analytic continuations for any or all of these Dirichlet series to larger domains? Does the functional equation speak to this matter? If any of these functions have natural boundary at $s=1$ on the standard view, but analytic continuation one the non-standard sense, I would welcome any insight into how such could happen. (Corrections welcome if my question betrays any basic misunderstanding!)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8884303569793701, "perplexity_flag": "middle"}
http://cms.math.ca/Competitions/MOCP/2002/prob_apr.mml	Canadian Mathematical Society www.cms.math.ca \| \| Français Site map \| CMS store location: # PROBLEMS FOR APRIL The first five problems appeared on the annual University of Toronto Undergraduate Competition. Please send your solutions to Professor E.J. Barbeau Department of Mathematics University of Toronto Toronto, ON M5S 3G3 no later than May 15, 2002. 139. Let $A$, $B$, $C$ be three pairwise orthogonal faces of a tetrahedran meeting at one of its vertices and having respective areas $a$, $b$, $c$. Let the face $D$ opposite this vertex have area $d$. Prove that ${d}^{2}={a}^{2}+{b}^{2}+{c}^{2} .$ 140. Angus likes to go to the movies. On Monday, standing in line, he noted that the fraction $x$ of the line was in front of him, while $1/n$ of the line was behind him. On Tuesday, the same fraction $x$ of the line was in front of him, while $1/\left(n+1\right)$ of the line was behind him. On Wednesday, the same fraction $x$ of the line was in front of him, while $1/\left(n+2\right)$ of the line was behind him. Determine a value of $n$ for which this is possible. 141. In how many ways can the rational $2002/2001$ be written as the product of two rationals of the form $\left(n+1\right)/n$, where $n$ is a positive integer? 142. Let $x,y>0$ be such that ${x}^{3}+{y}^{3}\le x-y$. Prove that ${x}^{2}+{y}^{2}\le 1$. 143. A sequence whose entries are $0$ and $1$ has the property that, if each $0$ is replaced by $01$ and each $1$ by $001$, then the sequence remains unchanged. Thus, it starts out as $010010101001\dots $. What is the $2002$th term of the sequence? 144. Let $a$, $b$, $c$, $d$ be rational numbers for which $\mathrm{bc}\ne \mathrm{ad}$. Prove that there are infinitely many rational values of $x$ for which $\sqrt{\left(a+\mathrm{bx}\right)\left(c+\mathrm{dx}\right)}$ is rational. Explain the situation when $\mathrm{bc}=\mathrm{ad}$. © Canadian Mathematical Society, 2013 © Canadian Mathematical Society, 2013 : http://www.cms.math.ca/	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 38, "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}, "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}, "english_confidence": 0.9727843403816223, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2008/08/18/	# The Unapologetic Mathematician ## Power Series Prodded by some comments, I think I’ll go even further afield from linear algebra. It’s a slightly different order than I’d originally thought of, but it will lead to some more explicit examples when we’re back in the realm of linear algebra, so it’s got its own benefits. I’ll note here in passing that mathematics actually doesn’t proceed in a straight line, despite the impression most people get. The lower-level classes are pretty standard, yes — natural-number arithmetic, fractions, algebra, geometry, calculus, and so on. But at about this point where most people peter out, the subject behaves more like an alluvial fan — many parallel rivulets carry off in different directions, but they’re all ultimately part of the same river. So in that metaphor, I’m pulling a bit of an avulsion. Anyhow, power series are sort of like polynomials, except that the coefficients don’t have to die out at infinity. That is, when we consider the algebra of polynomials $\mathbb{F}[X]$ as a vector space over $\mathbb{F}$ it’s isomorphic to the infinite direct sum $\displaystyle\mathbb{F}[X]\cong\bigoplus\limits_{k=0}^\infty\mathbb{F}X^k$ but the algebra of power series — written $\mathbb{F}[[X]]$ — is isomorphic to the infinite direct product $\displaystyle\mathbb{F}[[X]]\cong\prod\limits_{k=0}^\infty\mathbb{F}X^k$ It’s important to note here that the $X^i$ do not form a basis here, since we can’t write an arbitrary power series as a finite linear combination of them. But really they should behave like a basis, because they capture the behavior of every power series. In particular, if we specify that $\mu(X^m,X^n)=X^{m+n}$ then we have a well-defined multiplication extending that of power series. I don’t want to do all the fine details right now, but I can at least sketch how this all works out, and how we can adjust our semantics to talk about power series as if the $X^i$ were an honest basis. The core idea is that we’re going to introduce a topology on the space of polynomials. So what polynomials should be considered “close” to each other? It turns out to make sense to consider those which agree in their lower-degree terms to be close. That is, we should have the space of tails $\displaystyle\bigoplus\limits_{k=n+1}^\infty\mathbb{F}X^k$ as an open set. More concretely, for every polynomial $p$ with degree $n$ there is an open set $U_p$ consisting of those polynomials $q$ so that $X^{n+1}$ divides the difference $q-p$. Notice here that any power series defines, by cutting it off after successively higher degree terms, a descending sequence of these open sets. More to the point, it defines a sequence of polynomials. If the power series’ coefficients are zero after some point — if it’s a polynomial itself — then this sequence stops and stays at that polynomial. But if not it never quite settles down to any one point in the space. Doesn’t this look familiar? Exactly. Earlier we had sequences of rational numbers which didn’t converge to a rational number. Then we completed the topology to give us the real numbers. Well here we’re just doing the same thing! It turns out that the topology above gives a uniform structure to the space of polynomials, and we can complete that uniform structure to give the vector space underlying the algebra of power series. So here’s the punch line: once we do this, it becomes natural to consider not just linear maps, but continuous linear maps. Now the images of the $X^k$ can’t be used to uniquely specify a linear map, but they will specify at most one value for a continuous linear map! That is, any power series comes with a sequence converging to it — its polynomial truncations — and if we know the values $f(X^k)$ then we have uniquely defined images of each of these polynomial truncations since each one is a finite linear combination. Then continuity tells us that the image of the power series must be the limit of this sequence of images, if the limit exists. ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now. • ## Feedback Got something to say? Anonymous questions, comments, and suggestions at Formspring.me!	{"extraction_info": {"found_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": 17, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.91425621509552, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Yaw_(rotation)	# Yaw (rotation) For other uses, see Yaw. Yaw motion in an aircraft A yaw rotation is a movement around the yaw axis of a vehicle that changes the direction the vehicle is facing, to the left or right of its direction of motion. The yaw rate or yaw velocity of a car or other rigid body is the angular velocity of this rotation, or rate of change of the heading angle when the aircraft is horizontal. It is commonly measured in degrees per second or radians per second. Another important concept is the yaw moment, or yawing moment, which is the projection of a given torque over the yaw axis. It is specially important in road vehicles because pitch and roll moments are limited by the floor reaction. ## Measurement Yaw velocity can be measured by measuring the ground velocity at two geometrically separated points on the body, or by a gyroscope, or it can be synthesized from accelerometers and the like. It is the primary measure of how drivers sense a car's turning visually. Axes of a ship and rotations around them It is important in electronic stabilized vehicles. The yaw rate is directly related to the lateral acceleration of the vehicle turning at constant speed around a constant radius, by the relationship tangential speed*yaw velocity = lateral acceleration = tangential speed^2/radius of turn, in appropriate units The sign convention can be established by rigorous attention to coordinate systems. In a more general manoeuvre where the radius is varying, and/or the speed is varying, the above relationship no longer holds. ## Yaw rate control The yaw rate can be measured with accelerometers in the vertical axis. Any device intended to measure the yaw rate is called a yaw rate sensor. Yaw rate control is a system that maintains the yaw rate of a vehicle, for example a car, according to the values requested by the driver. It differs from simple Yaw control because is not the heading angle the controlled variable, but its variation rate instead. It is one of the main targets of electronic stability control systems. Arrows, darts, rockets and airships have tail surfaces to achieve stability. A road vehicle does not have elements specifically designed to maintain stability, but relies primarily on the distribution of mass. ## Road vehicles Studying the stability of a road vehicle requires a reasonable approximation to the equations of motion. Dynamics of a road vehicle The diagram illustrates a four wheel vehicle, in which the front axle is located a metres ahead of the centre of gravity and the rear axle is b metres aft of the cg. The body of the car is pointing in a direction $\theta$ (theta) whilst it is travelling in a direction $\psi$ (psi). In general, these are not the same. The tyre treads at the region of contact point in the direction of travel, but the hubs are aligned with the vehicle body, with the steering held central. The tyres distort as they rotate to accommodate this mis-alignment, and generate side forces as a consequence. From Directional stability study, denoting the angular velocity $\omega$, the equations of motion are: $\frac{d\omega}{dt}=2k\frac{(a-b)}{I}\beta-2k\frac{(a^2+b^2)}{VI}\omega$ $\frac{d\beta}{dt}=-\frac{4k}{MV}\beta+(1-2k)\frac{(b-a)}{MV^2}\omega$ The coefficient of $\frac{d\beta}{dt}$ will be called the 'damping' by analogy with a mass-spring-damper which has a similar equation of motion. By the same analogy, the coefficient of $\beta$ will be called the 'stiffness', as its function is to return the system to zero deflection, in the same manner as a spring. The form of the solution depends only on the signs of the damping and stiffness terms. The four possible solution types are presented in the figure. The only satisfactory solution requires both stiffness and damping to be positive. If the centre of gravity is ahead of the centre of the wheelbase ($(b>a)$, this will always be positive, and the vehicle will be stable at all speeds. However, if it lies further aft, the term has the potential of becoming negative above a speed given by: $V^2=\frac{2k(a+b)^2}{M(a-b)}$ Above this speed, the vehicle will be directionally (yaw) unstable. Corrections for relative effect of front and rear tyres and steering forces are available in the main article. ## Relationship with other rotation systems These rotations are intrinsic rotations and the calculus behind them is similar to the Frenet-Serret formulas. Performing a rotation in an intrinsic reference frame is equivalent to right-multiply its characteristic matrix (the matrix that has the vector of the reference frame as columns) by the matrix of the rotation. ## History The first aircraft to demonstrate active control about all three axes was the Wright brothers' 1902 glider.[1] ## References 1. "Aircraft rotations". Retrieved 2008-08-04.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.9363293647766113, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/37677/divergence-of-cross-product-of-transverse-component	# Divergence of cross product of transverse component If I define the vector as $V_i=V^T_i+V^L_i$ and the transverse part is defined by $$V^T_i=\Big(\delta_{ij}-\frac{\partial_i\partial_j}{\partial^2}\Big)V_j$$ then is is obvious that $\nabla.V^T=0$ as well as $\nabla \times V^L=0$. What happened if I took the the divergence of the cross product of two different vectors with only transverse component? Are they zero too? For example is $\nabla.(A^T\times V^T)=0?$ - ## 1 Answer The identity you're looking for is $$\nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B})$$ so indeed, if the curl of both factors in the cross product vanish, the divergence of the cross product vanishes, too. - +1, but one should add that these identities are easier to identify in k-space, since then they are algebraic k identities rather than differential identities (although the two are obviously the same, psychologically, I find k-identifies slightly easier to internalize than cross-product identities). – Ron Maimon Sep 18 '12 at 13:23 Actually using this identity give me the term above! I would like to vanish it somehow! Using this identity again will just take me to the previous step. – aries0152 Sep 18 '12 at 13:53 I don't understand what you want, aries, or more likely, you don't understand something. You assumed $\nabla\times A=0$ and $\nabla\times B=0$, see the first line below your question's displayed equation where $A\to A^T$ and $B\to V^T$, so if those vanish, the combination in the identity vanishes, too, doesn't it? – Luboš Motl Sep 18 '12 at 14:46 1 If I understand well from the question, aries assumes that $\nabla\cdot A=0$ and $\nabla\cdot B=0$, and not that $\nabla\times A=0$ and $\nabla\times B=0$. – Cristi Stoica Sep 18 '12 at 15:09 Oh, I see, then it doesn't hold, obviously. – Luboš Motl Sep 18 '12 at 15:10	{"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": 12, "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}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9550936222076416, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/42452-area-rectangle-under-cosine-curve-print.html	# Area of a rectangle under a cosine curve? Printable View • June 25th 2008, 04:39 PM Myung Area of a rectangle under a cosine curve? http://www.cubeupload.com/files/d01fa3capture.jpg What is the equation of the rectangle under the curve? The curve is 3cos(x). • June 25th 2008, 04:47 PM Mathstud28 Quote: Originally Posted by Myung http://www.cubeupload.com/files/d01fa3capture.jpg What is the equation of the rectangle under the curve? The curve is 3cos(x). Equation of the rectangle? If you are asking what $3\int\cos(x)dx=3\sin(x)+C$ • June 25th 2008, 04:51 PM Myung Is that all? My guess is 6xcos(x) because x+x times 3cos(x) with something like -pi/2 < x < pi/2 as the restrictions, is that incorrect? • June 25th 2008, 04:56 PM Mathstud28 Quote: Originally Posted by Myung Is that all? My guess is 6xcos(x) because x+x times 3cos(x) with something like -pi/2 < x < pi/2 as the restrictions, is that incorrect? I have no idea what you area asking, are you asking fo the area between $\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$? Given by $3\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\cos(x)$? • June 25th 2008, 05:00 PM Myung I'm looking for an equation for the area of the rectangle inscribed under the curve 3cos(x) in the picture, the rectangle represents the dotted lines from -x to x. I don't really know where to start. • June 25th 2008, 05:10 PM Chris L T521 Quote: Originally Posted by Myung http://www.cubeupload.com/files/d01fa3capture.jpg What is the equation of the rectangle under the curve? The curve is 3cos(x). Quote: Originally Posted by Myung I'm looking for an equation for the area of the rectangle inscribed under the curve 3cos(x) in the picture, the rectangle represents the dotted lines from -x to x. I don't really know where to start. The Area of the rectangle is $A(x)=BaseHeight=6x\cos(x)$ Are we supposed to find the value of x that maximizes the area? Thus, $\frac{dA}{dx}=-6x\sin(x)+6\cos(x)$ Setting this equal to zero, we get that $\cot(x)=x$. Using your calculator, we see that this is the case for $x=\pm.86 \ x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ We take the positive value. Thus, the area of the rectangle is $A(.86)=6(.86)\cos(.86)\approx \color{red}\boxed{3.37}$. I believe this is the answer. Hope that this makes sense to you! :D --Chris • June 25th 2008, 05:12 PM Mathstud28 Quote: Originally Posted by Myung I'm looking for an equation for the area of the rectangle inscribed under the curve 3cos(x) in the picture, the rectangle represents the dotted lines from -x to x. I don't really know where to start. This is a maximum problem, you know the width is $2x$ and the heigth is $3\cos(x)$ since cosine is symetric So you need to maximize $A=6x\cos(x)$ • June 25th 2008, 05:18 PM Myung Quote: Originally Posted by Chris L T521 The Area of the rectangle is $A(x)=BaseHeight=6x\cos(x)$ Are we supposed to find the value of x that maximizes the area? Thus, $\frac{dA}{dx}=-6x\sin(x)+6\cos(x)$ Setting this equal to zero, we get that $\cot(x)=x$. Using your calculator, we see that this is the case for $x=\pm.86 \ x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ We take the positive value. Thus, the area of the rectangle is $A(.86)=6(.86)\cos(.86)\approx \color{red}\boxed{3.37}$. I believe this is the answer. Hope that this makes sense to you! :D --Chris I think this might be it, not completely sure but i'll check in class. Thanks All times are GMT -8. The time now is 01:00 PM.	{"extraction_info": {"found_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": 16, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9403055310249329, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/105477/what-are-the-limits-of-the-erds-rankin-method-for-covering-intervals-by-arithmet	## What are the limits of the Erdős-Rankin method for covering intervals by arithmetic progressions? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) To construct gaps between primes which are marginally larger than average, Erdős and Rankin covered an interval $[1,y]$ with arithmetic progressions with prime differences. A nice short exposition is here, but I'll summarize. The classic construction $z!+2, z!+3, ...z!+z$ constructs an interval of composite numbers which is embarrassingly short. It is shorter than the average distance between primes of size $z!$, $\log z! \sim z \log z,$ but we only constructed an interval of length $z.$ Slightly better is to replace $z!$ with the product of primes up to $z$, but this only produces a gap of about the average distance between primes. The method of Erdős and Rankin constructs a slightly larger gap, not quite $g \log g$ where $g$ is the size of an average gap. The point of covering an interval with arithmetic progressions is as follows. If you have a covering of $[1,y]$ by arithmetic progressions $a_p \mod p$ with $p\lt z \lt y$ then by the Chinese Remainder Theorem choose $n$ between $z$ and $z+\prod_{p\lt z} p$ so that $n \equiv -a_p \mod p.$ Then $n+k\in \lbrace n+1, n+2, ... n+y \rbrace$ is divisible by the difference of whichever arithmetic progression covers $k$. If you can cover an interval of length $y$ which is much longer than $z$, then you can construct a gap which is much longer than average. One ingredient in the construction is to choose $a_p = 0$ for primes between well-chosen $z_1 \lt z_2 \lt z$ with $z_1z_2 \gt y$. The point is that we get no collisions, so the arithmetic progression corresponding to each prime $p \in [z_1,z_2]$ covers a different $\lfloor \frac y p \rfloor$ integers in $[1,y].$ Second, use a greedy algorithm for small primes, choosing $a_p$ for $p \lt z_1$ so that each arithmetic progression covers as many uncovered integers in $[1,y]$ as possible. By the pigeonhole principle, you can reduce the number of uncovered integers by a factor of $(1-\frac 1 p).$ Use Mertens' Theorem, that $$\prod_{p\lt z_1}(1-\frac1p) \sim \frac{e^{-\gamma}}{\log z_1}.$$ Third, use the larger primes $p \gt z_2$ to eliminate the remaining uncovered integers, using each prime to cover at least one integer until everything is covered. Optimizing $z_1$ and $z_2$ is a bit messy, but using arithmetic progressions whose differences are primes up to $z$, they covered an interval of length at least $c \frac{z \log z \log\log\log z}{(\log\log z)^2} = o(z \log z).$ My question is what upper bounds are known for the effectiveness of this type of construction. I suspect that Erdős and Rankin couldn't have done much better by this technique. ## If you take arithmetic progressions whose differences are the primes up to $z$, must there be an integer smaller than $O(z^2)$ which is not covered by any arithmetic progression? $O(z^{3/2})?$ $O(z \log z)$? If there must be an uncovered integer smaller than $z^2$ then a different technique, perhaps not a constructive one, would be needed to establish that the existence of gaps of the conjectured size $z^2$ between primes of size about $\exp(z)$. - I am still working through the literature myself, so I don't know the answer. I take it you know of the further advances on prime gap lower bounds (Pomerance, Maier, Pintz, I think?), and that they bear no resemblance to Rankin's method? Also, have you checked Hagedorn's 2009 paper on computing Jacobsthal's function to make sure there is nothing you want there? Gerhard "Just Checking On The Obvious" Paseman, 2012.08.25 – Gerhard Paseman Aug 25 at 21:46 Also, Westzynthius uses a similar argument to get bounds close to what Rankin and Erdos have. I will review the paper and post something summarizing the differences between W's method and the one you outline above (which may very well be no difference). Gerhard "Ask Me About System Design" Paseman, 2012.08.25 – Gerhard Paseman Aug 25 at 22:18 If I remember correctly, Erdös himself was positive that this construction can't be improved easily (he called it hopeless even), which is the reason he offered a large prize for it. – Woett Aug 25 at 23:58 I wouldn't be shocked if it could be improved by a really clever trick, but I'd still like to know if there is some clear obstruction to improving it all of the way to $z^2$, say. If so, then to prove there are large prime gaps one has to use other techniques than covering intervals by arithmetic progressions, although that still looks like a natural problem on its own. @Gerhard, I'm not very familiar with the recent progress on this problem, and I'll look into the work you mention. Thanks. – Douglas Zare Aug 26 at 0:22 1 Maier-Pomerance indicate that they expect $z(\log z)^2$ as the limit (see 1.5), if one knew prime $k$-tuples. They basically use an on-average version of that (in AP), in the paper improving the constant. Thus for large primes, they can't show that any of them individually sieves out more than 1 number, but on average they can show at least 1.31, and Pintz 2. When knowing prime $k$-tuples, at least with uniformity enough, the large primes would then be shown to be more optimal, in sieving out. – Junkie Oct 25 at 2:37 show 1 more comment	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 46, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.948560893535614, "perplexity_flag": "head"}
http://mathoverflow.net/questions/38991?sort=oldest	## Criterion for open morphisms without constructible sets? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The following theorem is proved in EGA IV 2.4.6: Every morphism of schemes, which is flat and locally of finite presentation, is open. I've already seen some applications of this theorem, so I want to understand the proof. But it is based on the whole theory of (ind)constructible sets, whose development seems to be quite long and (sorry!) uninteresting in EGA. So I want to know: Is it possible to give a direct proof? We may reduce to the affine case, so perhaps it's an observation from commutative algebra? - See Proposition 28.8 in the Commutative Algebra chapter of the stacks project, math.columbia.edu/algebraic_geometry/stacks-git/…. – mdeland Sep 16 2010 at 16:54 This also uses constructible sets. But I think it is more concise as EGA. – Martin Brandenburg Sep 16 2010 at 19:27 ## 1 Answer You can of course assume that the base scheme is affine. Then it works in two steps : morally the result is really a result for morphisms locally of finite type (or finite presentation, as you wish) between locally noetherian schemes ; but a morphism locally of finite presentation comes by base change from a noetherian base scheme (remember we have an affine base). More precisely : 1) There is a nice proof without constructibility in Milne's Etale Cohomoloy, theorem 2.12, for morphisms locally of finite type. That does the job in case the base scheme is locally noetherian. 2) In the general case, there is a technical result in EGA (see in particular EGA IV, corollaire 11.2.6.1 and proposition 11.3.9) that says if $f:X\to Spec(A)$ is locally of finite presentation and flat then there exists a finitely generated sub-$\mathbb{Z}$-algebra $A_0\subset A$ (hence a noetherian ring) and an $A_0$-scheme $X_0$ which is locally of finite presentation (or finite type, as you wish) and flat such that $X\simeq X_0\otimes_{A_0} A$. Then you are in case 1). I understand that you want to avoid too much technical stuff, but I think that this very point can not be avoided, unless you are OK with locally noetherian schemes which after all is quite reasonable. I am preparing notes for a course on schemes with this theorem included. If you wish, I can send you the relevant pages (it's in french). -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9421283602714539, "perplexity_flag": "head"}
http://johncarlosbaez.wordpress.com/2013/01/17/game-theory-part-3/	# Azimuth ## Game Theory (Part 3) Last time we started looking at 2-player normal form games. The idea is that player A has $m$ different choices of which move to make, while player B has $n$ choices, and we have two $m \times n$ matrices of payoffs, $A$ and $B.$ If player A makes choice $i$ and player B makes choice $j,$ then the payoff to player A is $A_{i j},$ while the payoff to player B is $B_{i j}.$ What should you do if you’re playing a game like this? As we saw when playing rock-paper-scissors, sometimes the best thing is to adopt a strategy with some built-in randomness. In this game, if your opponent knows what you’re going to do, he can exploit that and beat you! A strategy involving randomness is called a mixed strategy. They’re very important, and we’ll talk about them a lot. But today we’ll only consider pure strategies, where we definitely choose one of the choices available, with no randomness. Notice: there are as many pure strategies as there are choices! For each choice, there’s a pure strategy where you always make that choice when playing the game. So, a lot of people say ‘pure strategy’ for what I call a ‘choice’. Other people use the word ‘action’. Anyway: let’s try to figure out the best pure strategy for each player, if we can. Sometimes this is impossible. For example, rock-paper-scissors has no best pure strategy. Could playing rock be best? No, because paper beats rock. Could playing paper be best? No, because scissors beats paper. Could playing scissors be best? No, because rock beats scissors. But sometimes we can find a best pure strategy, and the ideas we meet will help us later when we study mixed strategies. So, let’s dive in! ### Nash equilibria Suppose player A makes choice $i$ and player B makes choice $j$. Then we say $(i,j)$ is a Nash equilibrium if neither player can improve their payoff by changing their choice. For example, look at this game: \| \| \| \| \|----\|---------\|---------\| \| \| 1 \| 2 \| \| 1 \| (10,10) \| (9,-1) \| \| 2 \| (-1,1) \| (-1,-1) \| Here we are writing the matrices $A$ and $B$ together in one table, like last time, with $A$ in black and $B$ in red. But we could also separate them out: $A = \left( \begin{array}{rr} 10 & 9 \\ -1 & -1 \end{array} \right) , \qquad B = \left( \begin{array}{rr} 10 & -1 \\ 1 & -1 \end{array} \right)$ It’s just a different way of conveying the same information. Either way, it’s pretty clear what the players should do in this game! Both players should make choice 1. That way, they both win 10 points. Furthermore, this pair of choices $(i,j) = (1,1)$ is certainly a Nash equilibrium. In other words: neither player can improve their payoff by unilaterally changing their choice! If player A switches to choice 2, their payoff drops to -1 points. If player B switches to choice 2, their payoff drops to 9 points. Let’s give an official definition of Nash equilibrium and then look at some trickier examples. Remember, player A’s choices are $i = 1, 2, \dots , m,$ while player B’s choices are $j = 1, 2, \dots, n.$ Definition. Given a 2-player normal form game, a pair of choices $(i,j)$ is a Nash equilibrium if: 1) For all $1 \le i' \le m$, $A_{i'j} \le A_{ij}.$ 2) For all $1 \le j' \le n$, $B_{ij'} \le B_{ij}.$ Condition 1) says that player A can’t improve their payoff by switching their choice from $i$ to any other choice $i'.$ Condition 2) says that player B can’t improve their payoff by switching their choice from $j$ to any other choice $j'.$ ### Examples Let’s look at more examples of Nash equilibria, to see some funny things that can happen. First let’s modify the last game a little: \| \| \| \| \|----\|---------\|---------\| \| \| 1 \| 2 \| \| 1 \| (10,10) \| (-1,10) \| \| 2 \| (10,-1) \| (-1,-1) \| Is there a Nash equilibrium? Yes—and just as before, it happens when both players pick pure strategy 1. Now player A’s payoff doesn’t get any worse when they switch to pure strategy 2. But it doesn’t improve, either! It stays equal to 10. Similarly, player B’s payoff doesn’t get any worse when they switch to pure strategy 2… but it doesn’t improve. So, neither player is motivated to change their strategy. I said a Nash equilibrium is a situation where neither player can improve their payoff by changing their choice. But it might be clearer to say: neither player can improve their payoff by unilaterally changing their choice. What do I mean by ‘unilaterally’ changing their choice? I mean that one player changes their choice while the other player does not change their choice. But if both players change their choices simultaneously, sometimes they can improve both their payoffs! Lets see an example of that: \| \| \| \| \|----\|---------\|---------\| \| \| 1 \| 2 \| \| 1 \| (10,10) \| (9,-1) \| \| 2 \| (-1,8) \| (20,20) \| Now it looks like both players will be happiest if they pick pure strategy 2. And it’s true! Moreover, this is a Nash equilibrium. Check and see. But what if both players pick choice 1? This is also a Nash equilibrium! Shocking but true. If they both use pure strategy 1, neither player can improve their payoff by unilaterally changing their choice. If player A changes her choice, her payoff drops to -1. And if player B changes his choice, his payoff drops to -1. Of course, they can improve their payoff if they both simultaneously change their choice to 2. But that’s not what the concept of Nash equilibrium is about. This raises the question of whether Nash equilibrium is a good definition of ‘the best’ choice for each player. The big problem is figuring out what ‘best’ means—we’ll have to talk about that more. But we’re also seeing some problems with the word ‘the’. There may be more than one Nash equilibrium, so we can’t always talk about ‘the’ Nash equilibrium. In other words, our example shows there isn’t always a unique Nash equilibrium. Furthermore, a Nash equilibrium doesn’t always exist! Check out the game of rock-paper-scissors: \| \| \| \| \| \|----------\|--------\|--------\|----------\| \| \| rock \| paper \| scissors \| \| rock \| (0,0) \| (-1,1) \| (1,-1) \| \| paper \| (1,-1) \| (0,0) \| (-1,1) \| \| scissors \| (-1,1) \| (1,-1) \| (0,0) \| It’s easy to see that there’s no Nash equilibrium. No matter what both players choose, at least one of them can always improve their payoff by switching to a different choice. If one of them wins the game, the loser can improve their payoff by switching to a different choice. If it’s a tie, either player can improve their payoff by switching to a different choice. So: Nash equilibria are interesting, but they don’t always exist— at least not when we consider only pure strategies, as we’ve been doing. And even when they do, they’re not always unique! This is a bit frustrating. We’d like game theory to tell us how to play games well. We can improve the situation by allowing mixed strategies, where a player picks different choices with different probabilities. If we do this, there is always at least one Nash equilibrium! This result was proved by—you guessed it!—John Nash. He did it in 1950, in this astoundingly short paper: • John F. Nash, Jr., Equilibrium points in n-person games, Proceedings of the National Academy of Sciences 36 (1950), 48–9. He eventually won the Nobel prize in economics for this work. In fact, Nash was not the first to think about Nash equilibria; this idea goes back to the French economist Antoine Cournot, who wrote about it way back in 1838! However, Nash was the first to consider Nash equilibria for mixed strategies for general simultaneous multi-player games, and prove they always exist. (As we’ll see later, von Neumann and Morgenstern had already done this for the zero-sum 2-player games.) ### John Nash If you haven’t seen the movie about Nash called A Beautiful Mind, you should! Here’s the scene where Nash figures out something about multiplayer games: It’s not very clear, mathematically speaking… but oh well. It’s a good movie and it gives you some sense of how Nash battled with mental illness for much of his life. He has now largely recovered and spends his time at Princeton University. Here’s the young Nash—click for more details: ### A ‘practical’ question A student in my class asked what’s the best book or website on how to play blackjack well. I asked this on Google+ and got over 50 replies, which you can read here. If you know more, please tell me! This entry was posted on Thursday, January 17th, 2013 at 4:59 pm and is filed under game theory, mathematics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site. ### 9 Responses to Game Theory (Part 3) 1. Last time we talked about Nash equilibria for a 2-player normal form game. We saw that sometimes a Nash equilibrium doesn’t exist! Sometimes there’s more than one! But suppose there is at least one Nash equilibrium. How do we find one? [...] 2. Joan says: Possible typo: in the definition of Nash equilibrium, for the second condition, might it be 1<=j'<=n (instead of m) ? I recall that the second player can choose between n strategies, not m. Thanks for these nice and useful entries. • John Baez says: Thanks! I fixed that mistake. 3. Ali says: Nice post—thanks. In the second example, is there more than one Nash equilibrium? E.g., cell (1, 2) with $A_{12} = -1$ and $B_{12} = 10$: if A switches from 1 to 2, then $A_{22} = -1 \le A_{12}$ and if B switches from 2 to 1, then $B_{11} = 10 \le B_{12}$. So neither can improve with a unilateral move. Is that right? (I suspect I’m confused about something….) • John Baez says: Nice question! I’ll let someone else answer, so they can get a bit of practice. Either a student in the class, or someone reading this blog… • Jesse says: It looks like you are right. Good observation! 4. We’ll have to look at examples to understand this stuff better, but let me charge ahead and define ‘Nash equilibria’ for mixed strategies. The idea is similar to the idea we’ve already seen. A pair of mixed strategies, one for A and one for B, is a Nash equilibrium if neither player can improve the expected value of their payoff by unilaterally changing their mixed strategy. [...] 5. Jesse says: There might be a couple of mix-ups: Shortly before the definition of Nash equilibrium, regarding the first example… “If player A switches to choice 2, their payoff drops to 9 points. If player B switches to choice 2, their payoff drops to 1 point.” It looks like player A’s payoff would drop to -1 by switching to choice 2; on the other hand, player B’s payoff would drop to 9 by switching to choice 2. Similarly in the third example… “If player A changes her choice, her payoff drops to 9. And if player B changes his choice, his payoff drops to 8.” It looks like if player A changes to choice 2, her payoff drops to -1; if player B changes to choice 2, his payoff drops to -1. • John Baez says: Thanks! Fixed, I hope! I really appreciate all these corrections.	{"extraction_info": {"found_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": 33, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9431806206703186, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2007/08/31/an-awful-lot-of-natural-maps/?like=1&_wpnonce=401286028f	# The Unapologetic Mathematician ## An awful lot of natural maps There are a bunch of natural (in both senses) maps we can consider now. Some look all but tautological over $\mathbf{Set}$, and we may have used them in the past without comment. In the enriched context, though, we should go over them. 1. The family of arrows $\eta_C:\mathbf{1}\rightarrow\hom_\mathcal{D}(S(C),T(C))$ is extraordinarily $\mathcal{V}$-natural exactly when $\eta_C:S(C)\rightarrow T(C)$ is $\mathcal{V}$-natural. 2. For a $\mathcal{V}$-functor $T:\mathcal{C}\rightarrow\mathcal{D}$, the map $T_{A,B}:\hom_\mathcal{C}(A,B)\rightarrow\hom_\mathcal{D}(T(A),T(B))$ is natural in both $A$ and $B$. 3. Similarly, if $T:\mathcal{A}\otimes\mathcal{B}\rightarrow\mathcal{C}$ is a $\mathcal{V}$-functor, then $T(\underline{\hphantom{X}},1_B):\hom_\mathcal{A}(A,A')\rightarrow\hom_\mathcal{C}(T(A,B),T(A',B))$ is $\mathcal{V}$-natural. And it’s also natural in $B)$. 4. In particular, $\hom_\mathcal{C}(C,\underline{\hphantom{X}})_{A,B}:\hom_\mathcal{C}(A,B)\rightarrow\hom_\mathcal{C}(C,B)^{\hom_\mathcal{C}(C,A)}$ is natural in all three variables. 5. Putting together naturalities 1 and 4 tells us that for a morphism $f:\mathbf{1}\rightarrow\hom_\mathcal{C}(A,B)$, the transformation $\hom_\mathcal{C}(1_C,f):\hom_\mathcal{C}(C,A)\rightarrow\hom_\mathcal{C}(C,B)$ is natural. 6. The “evaluation” $e:Z^Y\otimes Y\rightarrow Z$ is natural in both variables. 7. Since we built compositions from evaluations and the closure adjunction, the arrow $\hom_\mathcal{C}(B,C)\otimes\hom_\mathcal{C}(A,B)\rightarrow\hom_\mathcal{C}(A,C)$ is natural in all variables. 8. The identity functor on $\mathcal{C}$ has an identity natural transformation, so by naturality 1 we see that $i_C:\mathbf{1}\rightarrow\hom_\mathcal{C}(C,C)$ is natural. 9. All the monoidal structural isomorphisms — $\alpha_{A,B,C}:(A\otimes B)\otimes C)\rightarrow A\otimes(B\otimes C)$, $\lambda_A:\mathbf{1}\otimes A\rightarrow A$, $\rho:A\otimes\mathbf{1}\rightarrow A$, and $\gamma_{A,B}:A\otimes B\rightarrow B\otimes A$ — are natural in all variables. 10. We can start with $(Z^Y)^X$ and hit it with $\underline{\hphantom{X}}\otimes Y$ to get $(Z^Y\otimes Y)^{X\otimes Y}$. Then we can evaluate to get $Z^{X\otimes Y}$. This is an isomorphism we call $p_{X,Y,Z}^{-1}$, corresponding to the adjunction $\hom_\mathcal{C}(A\hom_\mathcal{C}(B,C))\cong\hom_\mathcal{C}(A\otimes B,C)$, and it’s natural in all variables. 11. We can compose the following arrows: • $\lambda_X^{-1}:X\rightarrow\mathbf{1}\otimes X$ • $j_{X\otimes Y}\otimes1_X:\mathbf{1}\otimes X\rightarrow(X\otimes Y)^{X\otimes Y}\otimes X$ • $p_{X,Y,X\otimes Y}\otimes1_X:(X\otimes Y)^{X\otimes Y}\otimes X\rightarrow((X\otimes Y)^Y)^X\otimes X$ • $e_{X,(X\otimes Y)^Y}:((X\otimes Y)^Y)^X\otimes X\rightarrow(X\otimes Y)^Y$ and we get the “coevaluation” $d_{X,Y}$ — the counit of the closure adjunction. And thus the coevaluation is $\mathcal{V}$-natural in all variables. 12. We can compose the coevaluation $d_{X,\mathbf{1}}:X\rightarrow(X\otimes\mathbf{1})^\mathbf{1}$, and then use the right unit isomorphism to get a $\mathcal{V}$-natural isomorphism $X\rightarrow X^\mathbf{1}$. 13. In general, a family $f:T(D,D,A,B)\otimes S(E,E,A,C)\rightarrow R(F,F,B,C)$ is $\mathcal{V}$-natural in any of its variables if the corresponding variables are natural in $\overline{f}:T(D,D,A,B)\rightarrow R(F,F,B,C)^{S(E,E,A,C)}$ Whew. That’s a mouthful. It can be instructive to sit down and try to interpret some of these in the context of categories enriched over $\mathbf{Set}$, so when you recover I’d advise taking a look at that. ### Like this: Posted by John Armstrong \| Category theory No comments yet. « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_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": 45, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8733177185058594, "perplexity_flag": "middle"}
http://crypto.stackexchange.com/questions/2127/what-are-the-methods-to-construct-a-primitive-binary-nonlinear-feedback-shift-re	# What are the methods to construct a primitive binary nonlinear feedback shift register (NLFSR)? Given a binary shift register of $n$ bits, a primitive binary nonlinear feedback shift register will generate a sequence with a period of $2^n - 1$. While I am unable to find a paper which directly describes the methods, the paper A List of Maximum Period NLFSRs has just came up on eprint, which lists all NLFRSs with a period of $2^n - 1$, for $n < 25$. Does anyone have links to papers describing the construction methods for such primitive binary nonlinear feedback shift registers? - What would be the definition of a primitive NLFSR? Aren't you looking for something more like a maximal-length NLFSR? – fgrieu Mar 19 '12 at 11:16 Yes, that would be another way of describing it. Thanks for clarifying. – Bluemilk Mar 19 '12 at 15:45 1 Try googling maximum-length NLFSR – fgrieu Mar 19 '12 at 17:42 1 Please update and expand the question with a common terminology (maximal-length), or a definition of primitive in the context; references to the constructs that you considered; and explanation on why they are not fit, or more generally your goal. – fgrieu Mar 20 '12 at 6:41 – Bluemilk Mar 30 '12 at 15:20 show 2 more comments ## 2 Answers As far as I know, there is no general method to construct Non-Linear Feeback Shift Registers with $n$ bits and period $2^n-1$, beside basically trying one by simulating the NLFSR for that number of steps, with cost $O(2^n)$ when done naively. Large speedups are possible depending on the construct; in particular, if it is possible to explicitly compute the states that could reach the original state within $s$ steps, and performs $s$ steps as fast as one step, we can save a factor of $s$. But still the cost of establishing that a NLFSR with $n$ bits is near-maximal-period would be, it seems, at least $O(2^{n/2})$ steps with $O(2^{n/2})$ memory. However it is clearly easy to go much further that $n=25$ bits. Caveat: this is out of my head, I have no reference to quote. In practice, it seems open cryptography seldom use maximal-length NLFSR, probably for that very lack of known method to study their period; they could however make some sense in a cascaded construct, similar to the Alternating Step Generator or Shrinking Generator. - The Motorola website has application notes for older pseudo-digital chipsets (CVDSM, etc). Look for shift registers and drill down from there. Opencores dot org is also an excellent resource. AFAIK, they are not secure because of LSB change bias which is relatively predictable (the lesser significant bits have more change bias in the long run). Hope that helps, Iceberg -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9291658997535706, "perplexity_flag": "middle"}
http://www.territorioscuola.com/wikipedia/en.wikipedia.php?title=Ultraviolet_catastrophe	More results on: Download PDF files on: Download Word files on: Images on: Video/Audio on: Download PowerPoint on: More results from.edu web: Map (if applicable) of: Ultraviolet catastrophe - Wikipedia, the free encyclopedia Ultraviolet catastrophe The Ultraviolet catastrophe is the error at short wavelengths in the Rayleigh–Jeans law (depicted as "classical theory" in the graph) for the energy emitted by an ideal black-body. The error, much more pronounced for short wavelengths, is the difference between the black curve (the wrong curve predicted by the Rayleigh–Jeans law) and the blue curve (the correct curve predicted by Planck's law). The ultraviolet catastrophe, also called the Rayleigh–Jeans catastrophe, was a prediction of late 19th century/early 20th century classical physics that an ideal black body at thermal equilibrium will emit radiation with infinite power. The term "ultraviolet catastrophe" was first used in 1911 by Paul Ehrenfest, although the concept goes back to 1900 with the first derivation of the $\lambda^{-4}$ dependence of the Rayleigh–Jeans law; the word "ultraviolet" refers to the fact that the problem appears in the short wavelength region of the electromagnetic spectrum. Since the first appearance of the term, it has also been used for other predictions of a similar nature, as in quantum electrodynamics and such cases as ultraviolet divergence. Problem The ultraviolet catastrophe results from the equipartition theorem of classical statistical mechanics which states that all harmonic oscillator modes (degrees of freedom) of a system at equilibrium have an average energy of $kT$. An example, from Mason's A History of the Sciences,1 illustrates multi-mode vibration via a piece of string. As a natural vibrator, the string will oscillate with specific modes (the standing waves of a string in harmonic resonance), dependent on the length of the string. In classical physics, a radiator of energy will act as a natural vibrator. And, since each mode will have the same energy, most of the energy in a natural vibrator will be in the smaller wavelengths and higher frequencies, where most of the modes are. According to classical electromagnetism, the number of electromagnetic modes in a 3-dimensional cavity, per unit frequency, is proportional to the square of the frequency. This therefore implies that the radiated power per unit frequency should follow the Rayleigh–Jeans law, and be proportional to frequency squared. Thus, both the power at a given frequency and the total radiated power is unlimited as higher and higher frequencies are considered: this is clearly unphysical as the total radiated power of a cavity is not observed to be infinite, a point that was made independently by Einstein and by Lord Rayleigh and Sir James Jeans in the year 1905. Solution Max Planck solved the problem by postulating that electromagnetic energy did not follow the classical description, but could only be emitted in discrete packets of energy proportional to the frequency, as given by Planck's law. This has the effect of reducing the number of possible excited modes with a given frequency in the cavity described above, and thus the average energy at those frequencies. The radiated power eventually goes to zero at infinite frequencies, and the total predicted power is finite.2 The formula for the radiated power for the idealized system (black body) was in line with known experiments, and came to be called Planck's law of black body radiation. Based on past experiments, Planck was also able to determine the value of its parameter, now called Planck's constant. The packets of energy later came to be called photons, and played a key role in the quantum description of electromagnetism.2 Historical inaccuracies Many popular histories of physics, as well as a number of physics textbooks, present an incorrect version of the history of the ultraviolet catastrophe. In this version, the "catastrophe" was first noticed by Planck, who developed his formula in response. In fact Planck never concerned himself with this aspect of the problem, because he did not believe that the equipartition theorem was fundamental – his motivation for introducing "quanta" was entirely different. That Planck's proposal happened to provide a solution for it was realized much later, as stated above.3 Though this has been known by historians for many decades, the historically incorrect version persists, in part because Planck's actual motivations for the proposal of the quantum are complicated and difficult to summarize to a lay audience.4 In popular culture Ultraviolet catastrophe has appeared in popular culture as the title to a children's book by New Zealand author Margaret Mahy. The reference in the title isn't directly to the physics concept but rather to the catch-phrase of the main character's great-uncle who would have been roughly contemporary with the discovery of Planck's constant. References 1. Mason, MA, PhD, Stephen (1962). A History of the Sciences. New York, NY: Collier Books. p. 550. 2. ^ a b 3. Kragh, Helge (December 2000). "Max Planck: The reluctant revolutionary". Physics World. 4. For some of the historiographical debates over what actually motivated Planck, see Kuhn, Thomas (1978). . Clarendon Press, Oxford. ISBN 0-226-45800-8. Galison, Peter (1981). "Kuhn and the Quantum Controversy". British Journal for the Philosophy of Science 32 (1): 71–85. doi:10.1093/bjps/32.1.71. • Kroemer, Herbert; Kittel, Charles (1980). . W. H. Freeman Company. Chapter 4. ISBN 0-7167-1088-9. • Cohen-Tannoudji, Claude; Diu, Bernard; Laloë, Franck (1977). . Hermann, Paris. pp. 624–626. ISBN 0-471-16433-X. HPTS - Area Progetti - Edu-Soft - JavaEdu - N.Saperi - Ass.Scuola.. - TS BCTV - TSODP - TRTWE TSE-Wiki - Blog Lavoro - InterAzioni- NormaScuola - Editoriali - Job Search - DownFree ! TerritorioScuola. Some rights reserved. Informazioni d'uso ☞	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.920762836933136, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/158542/isomorphism-from-homv-w-to-m-m-times-nf/158625	# Isomorphism from $Hom(V,W)$ to $M_{m\times n}^F$ Our book states that there is an isomorphism from $Hom(V,W)$, the vector space of all linear transformations from $V$ to $W$, to the matrix space $M_{m\times n}^F$ which is defined by $T:\rightarrow [T]^B_C$ where $B$ and $C$ are bases of $V$ and $W$ respectively. There are actually two questions here which I think are related. 1. It's not entirely clear to me what the nature of the object in $Hom(V,W)$ are. Are they the explicit formulas for $T$ according to the given basis? For example T(x,y)=(2x+y,x+y) according to the given basis? 2. This is related to another question I asked. Can the explicit formula for a linear transformation be according to anything other than the standard basis? - The matrices in $M_{m\times n}$ are explicit formulas in terms of bases. The objects in $Hom$ are abstract maps without reference to bases :) – rschwieb Jun 15 '12 at 11:58 ## 3 Answers $\,(1)\,\,\,$The objects in $\,\hom(V,W)\,$ are abstract linear transformations, without any need of specify basis, though one can use them to give a little more concrete form to the maps. $\,(2)\,\,\,$If by "explicit formula" you mean something like what you wrote with coordinates, the answer is yes: you can choose any basis in the domain $\,V\,$ and the range $\,W\,$ and give a explicit formula for any map wrt these basis' coordinates in both spaces Added Take the $\,3-$dimensional vector space $$\mathcal{P}_2:=\{f(x)\in\mathbb{R}[x]\;\|\; \deg f\leq 2\}$$ First problem: what is "the standard basis" here? Let's take the basis $\,\{1,x,x^2\}\,$ and the derivation linear map $$D:\,\mathcal P_2\to\mathcal P_2\,\,,\,D(f(x)):=f'(x)$$The matrix of $\,D\,$ wrt the above basis is $$\begin{align}D(1)&=0=0\cdot 1 +0\cdot x+0\cdot x^2\\D(x)&=1=1\cdot 1+0\cdot x+0\cdot x^2\\D(x^2)&=2x=0\cdot 1+2\cdot x+0\cdot x^2\end{align}\Longrightarrow [D]=\begin{pmatrix}0&1&0\\0&0&2\\0&0&0\end{pmatrix}$$ If you choose any other basis in the domain/range you'll get something else, of course. - Could you give me an example of how (2) works? – Robert S. Barnes Jun 17 '12 at 6:22 @Robert, I added to my question – DonAntonio Jun 17 '12 at 10:14 Elements of $\mathrm{Hom}(V,W)$ are just all $F$-linear maps $\varphi: V \to W$. The set of those again form a vectorspace over $F$ where scalarmultiplication and addition are given poitwise, i.e., for $f$ and $g$ such maps define $(f+g)(v) := f(v) + g(v)$ and for $\lambda \in F$ define $(\lambda\cdot f)(v):= \lambda\cdot(f(v))$. All of this is independent of any basis you could choose. To your second question I would like to say two things. It is one of the first things one shows about linear maps of vectorspaces that whenever you choose any basis of $V$ you can define a linear map by sending your basis elements to anything you want in $W$. (This is also referred to as "Vectorspaces are free"). In particular it does not have to be the standard basis if $V = F^n$. What is very important is to note that if $V$ is any $F$-vectorspace then there is no such thing as a standard basis of $V$ in general. In particular the second part of your second question does not really make sense for vectorspaces which are not equal to $F^n$ (of course they are isomorphic, but for this you have to choose a basis!). I hope this helps. - The other solutions are already doing a good job, but I think there is a little confusion here about thinking of bases of $V$ and $F^n$. There is no "standard basis" of $V$, they are all on equal footing. I think standard basis usually refers to the basis of $F^n$ consisting of vectors with exactly one $1$ in a given position. Every time you pick a basis of $V$, that gives rise to an isomorphism with $F^n$ in which a basis element for $V$ goes to one of those "standard basis" elements of $F^n$. (The isomorphism I'm thinking of sends each element of $V$ to the coefficients that represent it in that basis.) Try to sort these two bases out! -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 49, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9428361058235168, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/17711/why-do-we-need-to-prove-euv-euev	# Why do we need to prove $e^{u+v} = e^ue^v$? In this book I'm using the author seems to feel a need to prove $e^{u+v} = e^ue^v$ By $\ln(e^{u+v}) = u + v = \ln(e^u) + \ln(e^v) = \ln(e^u e^v)$ Hence $e^{u+v} = e^u e^v$ But we know from basic algebra that $x^{a+b} = x^ax^b$. Earlier in the chapter the author says that you should not assume $e^x$ "is an ordinary power of a base e with exponent x." This is both a math and pedagogy question then, why does he do that? So 2 questions really 1. Do we need to prove this for such a basic property? 2. If we don't need to, then why does he do it? Fun? To make it memorable? Establish more neural connections? A case of wildly uncontrolled OCD? Also I've always taken for granted the property that $x^{a+b} = x^a x^b$. I take it as an axiom, but I actually don't know where that axiom is listed. - 4 How is the function $e^x$ defined in the text? – Jon Jan 16 '11 at 16:00 4 @bobobobo: How is $\text{ln}(x)$ defined? – Isaac Jan 16 '11 at 16:12 12 @bobobobo: in other words, I think your confusion here is linguistic. You know there is an operation called exponentiation which satisfies x^{a+b} = x^a x^b, and the author is defining an operation which he calls exponentiation, so you assume it also satisfies x^{a+b} = x^a x^b. But you can't assume this. If I call something red, and you call something red, we aren't necessarily talking about the same color. We have to prove it, e.g. by agreeing on a unit of length and measuring the wavelengths of the appropriate type of light. Only then can we agree on the meaning of a statement like... – Qiaochu Yuan Jan 16 '11 at 17:19 16 -1 for asking a question of the form "Why did the author of my book do this?" without identifying the book in question. This makes things unnecessarily difficult and can waste a lot of people's time as they are forced to guess what the text might say. – Pete L. Clark Jan 17 '11 at 3:43 6 @bobobobo: there's absolutely nothing wrong with being a "math noob", and of course everyone who knows some advanced mathematics was at one point a beginner. I removed my downvote and added an upvote since you identified the book. @Jonas: "There's nothing wrong with inexpensive math books." No kidding. I'll complement your observation with: "There's something wrong with expensive math books." (Namely, they're expensive!) – Pete L. Clark Jan 19 '11 at 0:24 show 9 more comments ## 6 Answers In fact, such a proof is often necessary, which is why many authors write the function $e^x$ as $\exp(x)$ until they establish that it's just a "normal" exponent. For instance, if the original definition is given as $$\exp(x) = \lim_{n \to \infty} \left(1+\frac{x}{n}\right)^n,$$ then proving that $\exp(x + y) = \exp(x) \exp(y)$ is non-obvious, and certainly necessary. - This is very true, though defining the exponential function in this way and proceeding from there is slightly masochistic! – Noldorin Jan 18 '11 at 16:29 5 @Noldorin: why? It is an excellent intuitive definition from the point of view of differential equations: it's exactly what you get if you use Euler's method to approximate the solution to y' = y with step size 1/n and take n to infinity. Certainly it's better motivated from first principles than the power series definition, even if the latter is easier to prove things about. – Qiaochu Yuan Jan 18 '11 at 16:39 @Quiaochu: Oh, it's a lovely definition; only, it makes the following proof less trivial than other definitions. :) – Noldorin Jan 18 '11 at 16:39 I remember in Analysis I my instructor defined $e^x$ by its Taylor series. It is easy to check with this definition that this function is increasing, $f(1)=e$ and $f(x+y)=f(x)f(y)$. It follows imediatelly from these three simple properties that this is what we would call $e^x$... I liked that approach. – N. S. Oct 12 '11 at 23:09 Mathematicians have a habit of using the same notation to denote many different concepts. To justify their overloading, they tend to point to similarities between properties of those concepts. You may think of the exponential as one concept, but it is actually a large family of related concepts: • The exponentials $x^n$ where $x$ is an element of an arbitrary monoid and $n$ a non-negative integer, • The exponentials $x^n$ where $x$ is an element of an arbitrary group and $n$ an integer, • The exponentials $x^n$ where $x$ is an element of an arbitrary group and $n$ a rational (which are not guaranteed to exist in general, and are also not guaranteed to be unique in general), • The exponential $x^n$ where $x$ is an element of a topological group where $x^n$ for $n$ rational is unique, and $n$ a real number (defined by limits as in Rudin), • The exponential $e^x = 1 + x + \frac{x^2}{2} + ...$ where $x$ is an element of a topological ring containing $\mathbb{Q}$ and the series converges, • etc. It is easy to be fooled into thinking that these are all the same concept because they define essentially the same operation on $\mathbb{R}$, but 1) this is not obvious and requires proof, and 2) they generalize in different directions, so should be regarded as different in full generality. Note that the condition for $x^n$ to be defined where $x$ is an element of a topological group and $n$ a real number is quite strong and rarely satisfied. Note also that your proof using logarithms runs into unnecessary subtleties when $u, v$ are taken to be complex numbers, since in this setting the logarithm is not uniquely defined. Nevertheless, the exponential law still holds in this setting; it is one of the basic properties that mathematicians point to to justify calling something an exponential in the first place. In higher mathematics, the exponential further generalizes to: • The exponential map in Lie theory, • The exponential map in Riemannian geometry, • The exponential map in category theory, • etc. So it is good to keep in mind that "exponential" does not just refer to one thing. - 6 You've probably lost the reader when talking about group theory and topology - you certainly lost me. Best to stick to the level of the question, in my view. – Noldorin Jan 16 '11 at 16:17 21 @Noldorin: answers are not just for the questioner. The reason questions stick around after the OP accepts an answer is that questions and answers are supposed to be searchable and preserved for future users, who may have a very different background from the OP but may be wondering the same thing. This is an enormous difference between the internet and your grandmother's living room. Like I said, since I figured several other people would be giving answers at the appropriate level, I thought I would try something different. And if your grandmother asked about the properties of silicon... – Qiaochu Yuan Jan 16 '11 at 16:38 7 Saying "A,B,C are several concepts in higher mathematics that differ from your usual view of D, so you shouldn't take the properties of D for granted" is not itself an argument in 'higher mathematics' even though it refers to it. If Qiaochu used technical details of the latest set-theoretic constructions of group exponential functions by ultrafilters, I might agree with you; but since he just refers to the idea as an example I'd say it's at the appropriate level. – Michael Burge Jan 16 '11 at 16:48 15 @Noldorin: I can't help but feel like you're missing the point. The existence of people who would appreciate low-level answers does not say anything about the existence of people who would appreciate high-level answers. I guessed that several other people would provide simple answers, and they did - three of them. This question didn't need a fourth one, at least not until the OP clarified what his definitions were. Meanwhile, the comments in this answer apply regardless of what definition of exponentiation the OP is using. I don't know why you think I wrote this answer because I have a – Qiaochu Yuan Jan 16 '11 at 17:12 6 This discussion reminds me of Mordell's review of Lang's book on Diophantine geometry (projecteuclid.org/DPubS/Repository/1.0/…) and Lang's later review of Mordell's book on Diophantine equations (projecteuclid.org/DPubS/Repository/1.0/…). – KCd Jan 17 '11 at 4:49 show 16 more comments This may seem like a pretty pointless proof, at least on the surface, but I suspect there's some subtlety in the way the author's defined things here. (It may even appear circular at first, considering that the logarithm is often introduced as the inverse of the exponential function. Saying that, it can be derived the other way round, and this can sometimes be enlightening.) A rigorous proof for integer exponents is very straightforward indeed, and follows simply from the definition of the exponential function (of arbitrary base). For arbitrary exponents, things get slightly more complicated. I present a more complete proof below. So, let us suppose that the author began by defining the (natural) logarithm function, $$\ln a = \int_1^a \frac{dx}{x} .$$ We can then prove the addition property of logarithms, $\ln (ab) = \ln a + \ln b$, by considering $$\ln (ab) = \int_1^{ab} \frac{1}{x} \; dx = \int_1^a \frac{1}{x} \; dx \; + \int_a^{ab} \frac{1}{x} \; dx =\int_1^{a} \frac{1}{x} \; dx \; + \int_1^{b} \frac{1}{t} \; dt = \ln (a) + \ln (b)$$ The exponential function can of course be defined as the inverse of the logarithm, i.e. $$exp(ln(a)) = a$$ Now, to prove the property of exponentials, $e^{u+v} = e^u e^v$, we start as follows. $$\text{Let}\ u = \ln a, v = \ln b .$$ Then, using this property of logarithms and the definition of the inverse, consider $$e^{u+v} = e^{\ln a + \ln b} = e^{\ln (ab)} = ab = e^u e^v .\ \square$$ That should hopefuly be straightforward enough to follow. There are of course other equivalent definitions of $exp$ and $ln$. (You can for example define the Taylor series of $exp$, use the Cauchy product, and then simplify, but that's slightly trickier.) - 1 It's also quite common to define $ln(x)$ as the integral $\int_1^a \frac{1}{x} \, dx$, which might be the case here. – Jon Jan 16 '11 at 16:01 1 @Jon Yes, he did define $ln(x)$ like that previously – bobobobo Jan 16 '11 at 16:07 @Jon: Didn't notice that comment, but I was just writing that in my update as you made it. ;) – Noldorin Jan 16 '11 at 16:12 2 @bobobobo: Then this is your answer. It would have been great if you had given that definition of $\ln(x)$ in your question after Isaac in his comment asked for it. – Hendrik Vogt Jan 18 '11 at 16:16 The necessity of a proof depends very much on how each of those things is defined and the domain over which the variables may vary. For example, if $e^x$ is defined by the power series $\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{n!}$ for all $x \in \mathbb{C}$, then there is a very real need to prove that $e^{x + y} = e^x e^y$, and the proof is non-trivial. On the other hand, if $e$ is some constant defined elsewhere and $x$ and $y$ are positive integers, then there's essentially nothing to prove. I would also like to add that the law doesn't hold everywhere. If $A$ and $B$ are square matrices (or endomorphisms of a vector space) that do not commute, then in general $\exp(A + B) \ne \exp(A) \exp(B)$. This fact allows us to study certain non-commutative groups using the tools of analysis and differential geometry — this is the field of study known as Lie theory. - @ Zhen:"On the other hand, if e is some constant defined elsewhere and x and y are positive integers, then there's essentially nothing to prove." Why is it there is nothing to prove? How do we know that $x^a x^b = x^{a+b}$ in general $\forall x,a,b$? My line of thought is if I can prove that $e^{a_1 + b_1} = e^{a_1}e^{b_1}$ and $(e^{c})^d = e^{cd}$ then I can choose $a_1 = a \log_e(x)$ and $b_1 = b \log_e(x)$ and prove that $x^ax^b = x^{a+b}$. Or you could prove it for natural numbers, extend it to integers, to rationals to reals and to complex. In either case,there is some proof needed right? – user17762 Jan 17 '11 at 16:31 1 @Sivaram: If $x^a x^b = x^{a+b}$ for any number $x$ and any positive integers $a$ and $b$ essentially by definition of the notation. The only assumption is that multiplication is associative, so in fact, as Qiaochu alludes to, this is true for $x$ in any arbitrary monoid. – Zhen Lin Jan 17 '11 at 17:22 One advanced reason for giving a proof is that in some related contexts such a formula breaks down. On the 2-adic numbers, the power series $A(x) = \exp(2x^2 - 2x)$ when expanded out and written in standard form turns out to converge on ${\mathbf Z}_2$, so in particular $A(1)$ is defined. One can prove $A(1)^2 = 1$, but although naively one may expect $A(1) = 1$ by just plugging 1 directly into the original definition I gave for $A(x)$, in fact $A(1) = -1$. (The point here is that the naive calculation $A(1) = \exp(2 - 2) = \exp(2)\exp(-2) = 1$ is wrong since $\exp(2)$ doesn't make sense 2-adically.) - One definition of $e^x$ is $\displaystyle \lim_{n \rightarrow \infty}(1 + \frac{x}{n})^n$. From this definition, it doesn't automatically follow that $e^x e^y = e^{x+y}$. In fact, it doesn't even follow immediately that $e^x = \displaystyle \lim_{n \rightarrow \infty}(1 + \frac{x}{n})^n = (\displaystyle \lim_{n \rightarrow \infty}(1 + \frac{1}{n})^n)^x = (e)^x$. What this means is $e^x$ is just a short hand notation for the limit which after some analysis we realize it as $(e)^x$. By limit arguments, we can now show that $\displaystyle \lim_{n \rightarrow \infty}(1 + \frac{x}{n})^n = 1 + \sum_{k=1}^{\infty} \frac{x^k}{k!}$, $\forall x \in \mathbb{R}$. Now $e^x \times e^y = (1 + \sum_{k=1}^{\infty} \frac{x^k}{k!}) \times (1 + \sum_{k=1}^{\infty} \frac{y^k}{k!})$. Now we need to realize that we can rearrange the terms in the series and multiply terms of the two series since both of them converge absolutely. Hence $e^x \times e^y = (1 + \sum_{k=1}^{\infty} \frac{x^k}{k!}) \times (1 + \sum_{k=1}^{\infty} \frac{y^k}{k!}) = 1 + (x+y) + (\frac{x^2 + 2xy + y^2}{2!}) + (\frac{x^3 + 3x^2y + 3xy^2 + y^3}{3!}) + \cdots$ Now make use of the binomial theorem to get $$e^{x} e^{y} = e^{x+y}$$ PS: Though I have taken care to make sure the line of thought is right, you need to be careful when writing down the argument as to when you can interchange terms in an infinite series, multiply out two infinite series etc etc. - 3 One definition of e^x is what you have written. – Qiaochu Yuan Jan 17 '11 at 17:57 @Qiaochu: I prefer to define $e^x$ as $\displaystyle \lim_{n \rightarrow \infty}(1 + \frac{x}{n})^n$ (at least for complex numbers) and then derive all the other formulas like the infinite series etc from this definition. – user17762 Jan 17 '11 at 18:53 1 yes, but your use of the word "the" is misleading. The crux of this question is that it is not trivial to show that multiple definitions are equivalent. Your second argument doesn't work, either; like I mentioned in the comments, you still have to show that this function exists. – Qiaochu Yuan Jan 17 '11 at 19:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 90, "mathjax_display_tex": 7, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9534127116203308, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/275414/distribution-of-largest-sample-from-normal-distribution	# Distribution of largest sample from normal distribution. Given $n$ independent random variables $X_i$ with normal distribution, mean $\mu$, variance $\sigma^2$, what is the distribution of $\max\limits_{i=1}^n(X_i)$ ? In particular I am interested in whether it would be Normal or close to Normal as well, and what the mean and variance would be. - ## 1 Answer The random variable $Z = \max_{i=1}^n(X_i)$ is known as order statistics, and is sometimes denoted as $X_{n:n}$. The cumulative density function of $Z$ is easy to find: $$F_Z(z) = \mathbb{P}\left(Z \leqslant z\right) = \mathbb{P}\left(\max_{i=1}^n(X_i) \leqslant z\right) = \mathbb{P}\left( X_1 \leqslant z, X_2 \leqslant z, \ldots, X_n \leqslant z\right)$$ using independence: $$F_Z(z) = \left(F_X(z)\right)^n$$ Thus the density function is $$f_Z(z) = n f_X(z) F_X^{n-1}(z)$$ In particular, it follows that $Z$ is not normal. Expected values of $Z$ are known in closed form for $n=1,2,3,4,5$ (asking Mathematica): ````In[31]:= Table[ Mean[OrderDistribution[{NormalDistribution[m, s], n}, n]], {n, 1, 5}] Out[31]= {m, m + s/Sqrt[Pi], m + (3 s)/(2 Sqrt[Pi]), m + (6 s ArcTan[Sqrt[2]])/Pi^(3/2), m - (5 s)/(2 Sqrt[Pi]) + (15 s ArcTan[Sqrt[2]])/Pi^(3/2)} ```` Large $n$ asymptotics is discussed here. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8834216594696045, "perplexity_flag": "middle"}
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http://mathoverflow.net/questions/81209/galois-theory-for-polynomials-in-several-variables/81230	Galois theory for polynomials in several variables Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I feel a bit ashamed to ask the following question here. What is (actually, is there) Galois theory for polynomials in $n$-variables for $n\geq2$? I am preparing a large audience talk on Lie theory, and decided to start talking about symmetries and take Galois theory as a "baby" example. I know that Lie groups are somehow to differential equations what discrete groups are to algebraic equations. But I nevertheless would expect Lie (or algebraic) groups to appear naturally as higher dimensional analogs of Galois groups. Namely, the Galois group $G_P$ of a polynomial $P(x)$ in one variable can be defined as the symmetry group of the equation $P(x)=0$ (very shortly, the subgroup of permutations of the solutions/roots that preserves any algebraic equation satisfied by them). Then one of the great results of Galois theory is that $P(x)=0$ is solvable by radicals if and only if the group $G_P$ is solvable (meaning that its derived series reaches `$\{1\}$`). I was wondering what is the analog of the story in higher dimension (i.e. for equations of the form $P(x_1,\dots,x_n)=0$. I would naively expect algebraic group to show up... I googled the main key words and found this presentation: on the last slide it is written that the task at hand is to develop a Galois theory of polynomials in two variables This convinced me to anyway ask the question EDIT: the first "idea" I had I first thought about the following strategy. Consider $P(x,y)=0$ as an polynomial equation in one variable $x$ with coefficients in the field $k(y)$ of rational functions in $y$, and consider its Galois group. But then we could do the opposite...what would happen? - Take the étale fundamental group of the corresponding scheme? – Qiaochu Yuan Nov 17 2011 at 23:49 That doesn't seem like the correct generalization to me. The Galois group of $f(T)$ is the quotient of the etale fundamental group of Spec $k$ by the etale fundamental group of Spec $k[T]/(f(T))$. In particular, we would like to define it as the automorphism group of something. The Galois group of a polynomial, though, is not the automorphism group of $k[T]/f(T)$ but of $k[T_1,T_2,...,T_n]/f(T_1),etc.,etc.$. One could take the limit of the automorphism groups of $X$, some subset of $X \times X$, some subset of $X\times X\times X$, et cetera... – Will Sawin Nov 18 2011 at 2:07 1 I don't understand why I see on this question a vote to close. I like this question, and vote NOT to close, and whoever has cast a vote is being very rude by not saying why. – Theo Johnson-Freyd Nov 18 2011 at 4:08 1 @Theo I voted to close as not a real question. I have nothing further to add. – Felipe Voloch Nov 18 2011 at 10:20 1 I should also mention that the étale fundamental group can indeed be seen as a generalization of the (absolute) Galois group for objects of dimension >0 (although this may not be the generalization you're looking for). – François Brunault Nov 18 2011 at 14:14 show 5 more comments 2 Answers (This should really be a comment I think, but I'm not highly rated enough to leave one, so please bear with me) A Galois Theoretic condition for a polynomial in two variables to be solvable by radicals is found in the following paper: http://arxiv.org/abs/math/0305226. It seems to indicate that something similar can be done for higher variables. Perhaps I'll ask Jochen next time I see him about this. - Thank you for the link. I'll have a look as soon as possible. – DamienC Dec 9 2011 at 8:58 You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. This will not answer the question but is more than a comment in addition it may be very naive! (This is a hard question not a soft question!!!) I wonder if given the Galois group <-> étale fundamental group link works for dimension 1, should there not be a link '2-Galois thingie'<->étale 2-type, and hence a link with Grothendieck's Pursuing Stacks and his letters to Breen in 1975. The sought after model might be a profinite (?) crossed module. These are able to be seen as automorphism 2-groups of groupoids, so although they are automorphism things, there is a gap to bridge before the link would work well. I have also met a similar idea when working with orbifolds, and related ideas but have not any definite reply to the particular question, rather more an addition to the question! (I hope this helps... or inspires someone to think 'outside the box'.) There would be then a similar idea for polynomials in n-variable and models for n-types??? (This may be all rubbish but it is nice to dream sometimes!) - 3 I don't think that increasing the number of variables means that we have to use higher category theory. – Martin Brandenburg Nov 18 2011 at 8:40 @Martin May be not, but the possibility is there. Again it is sometimes not a question of have to' but maybe 'might'. NB. In fact <i>I</i> did not mention higher category theory as the models for 2-types are quite standard simplicial things and those are classical' (due to Whitehead 1950 or Reidemeister 1930s, and Peiffer 1940s). :-) The automorphism 2-group of group is very simply the inner automorphism morphism from G to Aut(G), so is not per se higher category theory – Tim Porter Nov 18 2011 at 10:02 David Corfield kindly reminded me of this n-cat café posting:golem.ph.utexas.edu/category/2009/12/… Kim has a lot more to say on this area, but it seems that it is a very deep and hard area. In his talk at the INI that David pointed out to me, he mentioned Pursuing Stacks and the anabelian theory mentioned therein, so perhaps (more by chace than by good knowledge) I was nearer the mark than I thought! – Tim Porter Nov 20 2011 at 16:35	{"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": 20, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9501016139984131, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/51778/show-that-partial-nu-t-mu-nu-j-nu-f-mu-nu/51801	# Show that $\partial_\nu T^{\mu\nu} = - j_\nu F^{\mu\nu}$ In a theoretical physics homework problem, I have to show the following: $$\partial_\nu T^{\mu\nu} = - j_\nu F^{\mu\nu}$$ Where $T$ is the Energie-Momentum-Tensor, $j$ the generalized current and $F$ the Field-Tensor. We use the $g$ for the metric tensor, I think in English the $\eta$ is more common. I know the following relationships: • Current and magnetic potential with Lorenz gauge condition: $$\mathop\Box A^\mu = \mu_0 j^\mu$$ • Energy-Momentum-Tensor: $$T^{\mu\nu} = \frac1{\mu_0} g^{\mu\alpha} F_{\alpha\beta} F^{\beta\nu} + \frac1{4\mu_0} g^{\mu\nu} F_{\kappa\lambda} F^{\kappa\lambda}$$ • Field-Tensor: $$F^{\mu\nu} = 2 \partial^{[\mu} A^{\nu]} = \partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu}$$ • d'Alambert operator: $$\mathop\Box = \partial_\mu \partial^\mu$$ • Bianchi identity: $$\partial^{[\mu} F^{\nu\alpha]} = 0$$ So far I have set all the definitions into the formula I have to show, but I only end up a lot of terms from antisymmetrisation and product rule. I also drew all what I have in Penrose graphical notation, but I still cannot see how to tackle this problem. Could somebody please give me a hint into the right direction? - Look at $F_{\alpha \beta}$ in $T_{\mu\nu}$ I think that $\beta$ is not right because the free indices are $\mu \nu$ and you have an extra free index $\beta$ – Nivalth Jan 21 at 13:59 1 The first term in the expression for $T^{\mu\nu}$ should be something like $F^{\mu\alpha}F^{\nu}_{\alpha}$ – twistor59 Jan 21 at 14:20 Indeed, I fixed it. I just typed it wrong here, that was not source of my confusion so far. – queueoverflow Jan 21 at 15:21 2 I think you're missing the most important equation of all: that $\partial_\mu F^{\mu \nu} = \mu_0 j^\nu$. – Muphrid Jan 21 at 15:43 2 @queueoverflow By the way, in English $g$ is used for any general metric, while $\eta$ is reserved for the Minkowski metric. – Chris White Jan 21 at 19:53 show 3 more comments ## 1 Answer Let's look at different terms from differentiating $T^{\mu\nu}$. The first from differentiating $g^{\mu\alpha} F_{\alpha\beta} F^{\beta\nu}$ is $$\partial_\nu g^{\mu\alpha} F_{\alpha\beta} F^{\beta\nu}= g^{\mu\alpha} F_{\alpha\beta} (\partial_\nu F^{\beta\nu}) +(\partial^\nu F^{\mu\beta}) F_{\beta\nu}= - \mu_0 F_{\alpha\beta} j^\beta +(\partial^\nu F^{\mu\beta}) F_{\beta\nu}$$ The first term is exactly what you want, the second cancels against the stuff you get from differentiating $g^{\mu\nu} F_{\kappa\lambda} F^{\kappa\lambda}$: $$\partial^\mu F_{\kappa\lambda} F^{\kappa\lambda}=2 F_{\kappa\lambda} (\partial^\mu F^{\kappa\lambda})=-2 F_{\kappa\lambda} (\partial^\kappa F^{\lambda\mu}+\partial^\lambda F^{\mu\kappa}) =-4 (\partial^\nu F^{\mu\beta}) F_{\beta\nu}$$ where in the second equality sign we have used Bianchi identity and in the last equality we have used $$F_{\kappa\lambda} \partial^\kappa F^{\lambda\mu} \underset{\text{relabel indecies}}= F_{\nu\beta}\partial^\nu F^{\beta \mu} \underset{\text{antisym. of $F$}}= F_{\beta\nu}\partial^\nu F^{\mu\beta}$$ This exactly cancels the second term in the first equation. - With Murphrid's comment in mind, I am able to follow your answer, except for the very last equality sign. I renamed the indices in my notes to match yours, but I do not see why $\partial^\beta F^{\nu\mu} = \partial^\nu F^{\mu\beta}$ holds. – queueoverflow Jan 21 at 19:33 1 It doesn't. You should split the two terms (including the $F_{\kappa \lambda}$) and relabel the dummy indices on the second term. – Vibert Jan 21 at 20:20 Now I see it. Thanks! – queueoverflow Jan 22 at 15:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.915252685546875, "perplexity_flag": "head"}
http://mathoverflow.net/questions/12861/representations-in-characteristic-p/63448	## Representations in characteristic p ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let G be a finite group and let F be an algebraically closed field. If the characteristic of F is 0, then the number of irreducible F-representations of G is given by the number of conjugacy classes of elements of G. A paper I'm reading says that if the characteristic of F is p>0, then the number of F-irreps of G is the same as the number of conjugacy classes of elements whose order is not divisible by p. If G is abelian, it seems to me that this should say that the p-sylow subgroup of G acts trivially on every characteristic p irrep. This is because I can split G into G'x P (non-p and p-sylow subgroups), and then any irrep of G' extends to one of G by letting P act trivially. Since the formula mentioned above would say that they have the same number of irreps these must be all of them. My question is: Is this true? If not, then where is my reasoning going off track? - 1 Could you give the reference to this paper that you are reading, containing the claim that the number of F-irreps is the number of conjugacy classes of elements of order prime to p? – Leonid Positselski Jan 24 2010 at 22:06 1 'Iwasawa theory and projective modules' by Ralph Greenberg. – Joel Dodge Jan 24 2010 at 22:13 ## 3 Answers Yes, you are correct. The point is that a $p$-group acting in char. $p$ always has a fixed point (and so acts trivially on an irrep.). So every irrep. of $G$ in char. $p$ factors through $G'$, as you anticipated. The proof of the claim about $p$-groups is not hard. One approach (in general, even when $P$ is not nec. abelian) is to prove it by induction on the order of $P$, and so (using the fact that the descending central series is non-trivial) reduce to the case when $P$ is cyclic of order $p$. In this case, one can for example look at the group ring $k[P] = k[t]/(t^p - 1) = k[t-1]/(t-1)^p,$ and observe that it has a unique maximal ideal, and hence that $P$ has a unique irrep., namely the trivial one. - 2 Great thanks! If F is a finite field, then the claim about p-groups also follows from standard facts about groups acting on sets: If G is a p-group acting on a set X, then \|X\| = \|X^G\| (mod p). – Joel Dodge Jan 24 2010 at 19:56 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. The fact that the number of isomorphism types of irreducible $FG$ modules is the number of conjugacy classes of elements of $G$ of order prime to $p$ (or number of $p$-regular conjugacy classes) when $F$ is an algebraically closed field of characteristic $p$ is a theorem of Richard Brauer. Proofs can be found in almost any text which treats modular representations of finite groups (eg Curtis and Reiner,Representation Theory of Finite Groups and Associative Algebras, Wiley, 1962). It is a reasonably elementary exercise in finite group theory that, letting $O_{p}(G)$ denote the (unique) largest normal $p$-subgroup of $G$, there is a one-to-one correspondence between $p$-regular conjugacy classes of $G$ and $p$-regular conjugacy classes of $G/O_{p}(G)$. This is one way to see that $O_{p}(G)$ must act trivially on all irreducible $FG$-modules, a special case of which you had observed. A more representation theoretic way to see this is to note that if $V$ is an irreducible $FG$-module, then the subspace $V^{O_{p}(G)}$ of $O_{p}(G)$-fixed points on $V$ is a $G$-invariant supspace; as observed in comments above, this space is non-zero, so by irreducibility, it must be all of $V$. (This can be seen as a particular application of Clifford's Theorem). - Brauer's proof that the number of similarity classes of irreducible representations of $G$ over an algebraically closed field of characteristic $p$ is equal to the number of $p$-regular conjugacy classes of $G$ is ring-theoretic in flavor, and rather tricky. There is also an easy character theoretic proof based on the following ideas. First, the set IBr$(G)$ of irreducible Brauer characters is in bijective correspondence with the irreducible representations, and this set of functions lives in the space $V$ of all complex-valued class functions defined on the set of p-regular elements. Since $\dim(V)$ equals the number of $p$-regular classes, it suffices to show that IBr($G$) is a basis for $V$. The linear independence of IBr$(G)$ is a standard result. To see that IBr$(G)$ spans, use the facts that Irr$(G)$ spans the space of all class functions and that on each $p$-regular class, the value of an ordinary character is a linear combination (and in fact, a sum) of values of Brauer characters. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 48, "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}, "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}, "english_confidence": 0.9291551113128662, "perplexity_flag": "head"}
http://mathoverflow.net/questions/53134/what-sums-of-equal-powers-of-consecutive-natural-numbers-are-powers-of-the-same-o/53142	## What sums of equal powers of consecutive natural numbers are powers of the same order? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Trivially $n^1=n^1$, and everyone knows that $3^2+4^2=5^2$. Denis Serre quoted $3^3+4^3+5^3=6^3$ in a recent MathOverflow question (which prompted this one). Are any other examples known? - 6 I've always liked $1^2 + 2^2 + 3^2 + ... + 24^2 = 70^2$. – J.C. Ottem Jan 24 2011 at 21:47 1 Tito Piezas III has been collecting examples for more general sums at sites.google.com/site/tpiezas/Home . If he doesn't know of an example to your question, it's probably because no computer has found it yet. Gerhard "Ask Me About System Design" Paseman, 2011.01.24 – Gerhard Paseman Jan 24 2011 at 22:12 this one is already there but...: $(-2)^3+(-1)^3+0^3+1^3+ \cdots + 5^3 = 6^3.$ – Luis H Gallardo Jan 24 2011 at 22:22 @Luis: Negative integers aren't natural numbers. – John Bentin Jan 24 2011 at 22:27 2 This page has many examples, and speculates that there are none for exponents greater than 3. mathpages.com/home/kmath147.htm – JSE Jan 24 2011 at 22:41 ## 6 Answers There is a good discussion at http://www.mathpages.com/home/kmath147.htm along with some nice examples, e.g., $6^3 + 7^3 + \dots + 69^3 = 180^3$, $1134^3 + \dots + 2133^3 = 16830^3$, which apparently are part of an infinite family (starting with $3^3+4^3+5^3=6^3$). There is a table of sums of consecutive cubes equal to a cube, not coming from this infinite family. The author of this page (Kevin Brown, if I'm not mistaken) writes, "If we go on to consider sums of higher powers, it appears that there are no sums of two or more consecutive 4th powers equal to a 4th power, or in general sums of two or more consecutive $n$th powers equal to an $n$th power for any $n\gt3$. Can anyone supply a proof, reference, or counter-example?" I suspect that any proof will be too big too fit in the margin. - While the same holds to be true for the Fermat's case. – awllower Feb 11 2011 at 10:25 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Let S be the set of integers k such that there exists a sequence of k consecutive squares whose sum is a square. According to the paper "Squares Expressible as Sum of Consecutive Squares" by L. Beekmans, S is known to be infinite and density 0; the citation is to problem 6552 in the American Math Monthly. If F(x) is the sum of the first x squares, then you are really asking about integral solutions to the Diophantine equation () F(x) - F(y) = z^2 which is a double cover of the plane branched at a cubic curve (Even a reducible cubic curve, since x-y \| F(x) - F(y).) Heuristically, you would expect about N^{1/2} solutions as x and y range over a box of size N. It would be interesting to ask: a) whether the geometry of this surface is so easy to describe that you can say something about its integral points; and b) whether () has a solution in polynomials in one variable (or, what's the same -- does the surface contain an affine line?) - Well, if we consider n consecutive 4th powers with initial a, F(a,n) = a^4 + (a+1)^4 + (a+2)^4 + ... + (a+n-1)^4 or, equivalently, F(a,n) = (n/30)(-1+30a^2-60a^3+30a^4+30a(1-3a+2a^2)n+10(1-6a+6a^2)n^2+(-15+30a)n^3+6n^4) it is easy to check that F(a,n) = y^4 (or even just y^2) has NO solution in the positive integers with BOTH {a,n} < 1000, with the exception of the trivial n = 1. (I had checked this with Mathematica some time back.) If we relax your question and allow n 4th powers in arithmetic progression d equal to some kth power, then the smallest I found was 64 4th powers with common difference d = 2 starting with, 29^4 + 31^4 + 33^4 + ... + 155^4 = 96104^2 P.S. The closed-form formula for general d is available, but I find it too tedious to include in this post. - You mean it's easy to check these 1000 Diophantine equations by computer? Even though you can find a closed form of $F(a,n)-y^4$, it is not so easy to see that this has no integer solutions. – J.C. Ottem Feb 11 2011 at 10:14 Yes, actually its 1000x1000 = 10^6 equations, since you go through two variables {a,n} < 1000. (Disregarding the trivial case n = 1). The closed form is, F(a,n) = (n/30)(-1+30a^2-60a^3+30a^4+30a(1-3a+2a^2)n+10(1-6a+6a^2)n^2+(-15+30a)n^3+6n^4 A simple Mathematica code can show that, with the exception of the trivial case n = 1, F(a,n) = y^2 has no positive integer solution with BOTH {a,n} < 1000. Of course, beyond that is another matter. – Tito Piezas III Feb 11 2011 at 11:44 @Tito: (+1) Thanks for the answer. – John Bentin Feb 11 2011 at 15:46 There are 126 pairs $i\lt x\le 1000$ such that $i^2+(i+1)^2+...+x^2$ is a square. If you fix $i$ then the sum $i^2+...+x^2$ is a cubic polynomial $f_i(x)$ in $x$. So you are looking for integer points on the elliptic curve $y^2=f_i(x)$. For example for $i=3$, the first of these are $(4,5), (580, 8075), (963,17267)$. I hope number theorists here can give more information. See also the comment by JSE below. - Not quite -- the rational points form a finitely generated group, but only finitely many of these are integral (Siegel's theorem) – JSE Jan 24 2011 at 22:31 JSE: yes, I forgot. Thanks! I have edited the answer. – Mark Sapir Jan 24 2011 at 22:35 J. C. Ottem's example $1^2 + ... + 24^2 = 70^2$ in the comments is of particular mathematical interest; it is one way to construct the Leech lattice, and is therefore somehow mysteriously related to other appearances of the number $24$ in mathematics (see, e.g. John Baez's thoughts). - While browsing the site http://sites.google.com/site/tpiezas/Home mentioned in the comments above, I found this on the page for cubes: "There are many particular cubic equations with this property, one of which is \$9^3+13^3+19^3+23^3 = 28^3, (9+23 = 13+19) as well as those in a nice arithmetic progression like, 11^3+12^3+13^3+14^3 = 20^3 31^3+33^3+35^3+37^3+39^3+41^3 = 66^3" . You might ask Mr. Piezas directly for more examples. Gerhard "Ask Me About System Design" Paseman, 2011.01.24 -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "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}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9165434837341309, "perplexity_flag": "middle"}
http://planetmath.org/SturmsTheorem	# Sturm’s theorem This root-counting theorem was produced by the French mathematician Jacques Sturm in 1829. ###### Definition 1. Let $P(x)$ be a real polynomial in $x$, and define the Sturm sequence of polynomials $\big(P_{0}(x),P_{1}(x),\ldots\big)$ by \| \| \| \| \|----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\| \| $\displaystyle P_{0}(x)$ \| $\displaystyle=$ \| $\displaystyle P(x)$ \| \| $\displaystyle P_{1}(x)$ \| $\displaystyle=$ \| $\displaystyle P^{{\prime}}(x)$ \| \| $\displaystyle P_{n}(x)$ \| $\displaystyle=$ \| $\displaystyle-\mathrm{rem}(P_{{n-2}},P_{{n-1}}),n\geq 2$ \| Here $\mathrm{rem}(P_{{n-2}},P_{{n-1}})$ denotes the remainder of the polynomial $P_{{n-2}}$ upon division by the polynomial $P_{{n-1}}$. The sequence terminates once one of the $P_{i}$ is zero. ###### Definition 2. For any number $t$, let $var_{P}(t)$ denote the number of sign changes in the sequence $P_{0}(t),P_{1}(t),\ldots$. ###### Theorem 1. For real numbers $a$ and $b$ that are both not roots of $P(x)$, $\#\{\textrm{distinct real roots of }P\textrm{ in }(a,b)\}=var_{P}(a)-var_{P}(b)$ In particular, we can count the total number of distinct real roots by looking at the limits as $a\rightarrow-\infty$ and $b\rightarrow+\infty$. The total number of distinct real roots will depend only on the leading terms of the Sturm sequence polynomials. Note that deg $P_{n}<$ deg $P_{{n-1}}$, and so the longest possible Sturm sequence has deg $P+1$ terms. Also, note that this sequence is very closely related to the sequence of remainders generated by the Euclidean Algorithm; in fact, the term $P_{i}$ is the exact same except with a sign changed when $i\equiv 2$ or $3\;\;(\mathop{{\rm mod}}4)$. Thus, the Half-GCD Algorithm may be used to compute this sequence. Be aware that some computer algebra systems may normalize remainders from the Euclidean Algorithm which messes up the sign. For a proof, see Wolpert, N., “ Proof of Sturm’s Theorem” Type of Math Object: Theorem Major Section: Reference ## Mathematics Subject Classification 11A05 Multiplicative structure; Euclidean algorithm; greatest common divisors 26A06 One-variable calculus ## Recent Activity May 17 new image: sinx_approx.png by jeremyboden new image: approximation_to_sinx by jeremyboden new image: approximation_to_sinx by jeremyboden new question: Solving the word problem for isomorphic groups by mairiwalker new image: LineDiagrams.jpg by m759 new image: ProjPoints.jpg by m759 new image: AbstrExample3.jpg by m759 new image: four-diamond_figure.jpg by m759 May 16 new problem: Curve fitting using the Exchange Algorithm. by jeremyboden new question: Undirected graphs and their Chromatic Number by Serchinnho ## Info Owner: rspuzio Added: 2004-07-22 - 20:04 Author(s): rspuzio ## Corrections theorem? by CWoo ✓ ## Versions (v14) by rspuzio 2013-03-22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 31, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8531388640403748, "perplexity_flag": "middle"}
http://skullsinthestars.com/2012/05/16/the-secret-molecular-life-of-soap-bubbles-1913/	The intersection of physics, optics, history and pulp fiction ## The secret molecular life of soap bubbles (1913) Posted on May 16, 2012 by Nature can be extremely devious in the way it hides its secrets. Sometimes the most remarkable and profound insights are staring us right in the face every day in the most mundane phenomena. For instance, we have all seen the spectacular colors that can appear in soap bubbles: Image from Microscopy-uk.org.uk, by Michael Reese Much. Borrowing his lovely images until I can produce my own! These colors are produced by optical interference, as we will discuss below; the “thin film optics” that creates bright colors in soap films also results in the bright colors of oil slicks. A rainbow of color produced by white light reflecting off of a thin layer of diesel fuel on water, via Wikipedia. Most of us would look at a soap film image and marvel at the beautiful rainbow colors; others would investigate the optics underlying them. But it took an exceptional physicist, Jean Baptiste Perrin (1870-1942), to realize that these colors concealed something more: direct evidence that matter consists of discrete atoms and molecules! Today we take for granted that all material objects in the universe are comprised of discrete “bits” of matter, which we call atoms; however, even up until the early 20th century there were still proponents of the continuum hypothesis, in which all matter is assumed to be infinitely divisible. This is quite surprising, because the first atomic hypothesis goes back to ancient Greece and the philosophical speculations of Democritus and his teacher Leucippus; the modern atomic theory dates to 1804, when John Dalton observed his law of multiple proportions. This law is stated as follows: If two elements form more than one compound between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will be ratios of small whole numbers. That is, to use the example from Wikipedia, it is possible to react carbon with oxygen in two ways, one of which forms carbon monoxide and one of which forms carbon dioxide. We can react 100 grams of carbon with 133 grams of oxygen to form CO, and that same mass of carbon can react with 266 grams of oxygen to form CO2. The ratio of masses of oxygen in the two compounds is CO2 : CO = 2 : 1, indicating that oxygen can only join with carbon in discrete amounts. This strongly suggested to Dalton that matter such as oxygen exists in discrete quantities: the “atoms” of his theory. However, other researchers of the time argued that this law could be explained by the continuum hypothesis as well, with a little theoretical gymnastics. Arguments between the atomists and the continuum proponents continued through the 19th century. The fundamental difficulty for the atomic theory? Nobody could actually see atoms or molecules, or find any direct experimental evidence for their existence; without even an estimate of their size, the idea of atoms was purely academic. The continuum hypothesis was finally put to rest when, in 1905, Albert Einstein proposed an explanation for the phenomenon of Brownian motion, which had mystified scientists for nearly one hundred years. In 1827, the biologist Robert Brown noticed that minute particles ejected by pollen grains suspended in water were moving around in an irregular, jittery motion. An example of this motion is shown in the video below. Einstein explained this motion as due to the rapid and near continuous bombardment of the visible particles by the invisible molecules of the water surrounding them. Einstein’s theory explained Brownian motion both physically and mathematically, and furthermore the theory gave for the first time a method to calculate the density and size of the molecules. Jean Baptiste Perrin Einstein’s theory still needed to be confirmed experimentally, however, and Jean Perrin was perfectly suited in both skills and interests for the task! In the early 1890s he had been deeply involved in the investigations of mysterious cathode rays (now known to be streams of electrons) and the even more mysterious Röntgen rays (now called X-rays). In 1897 he earned his doctorate in science and became a lecturer of physical chemistry at the Sorbonne in Paris. With his new position, he became fascinated with the “molecular hypothesis” (more or less another term for the atomic theory) and began extensive investigations into the subject. When Einstein’s theory of Brownian motion was put forth, Perrin devoted his efforts to confirming the theory and determining an estimate of molecular size; his results were published in 1910 and contributed to Perrin’s Nobel Prize in Physics for 1926. Even though the molecular hypothesis was now on solid footing thanks to Einstein’s theory and Perrin’s experiments, Perrin was not completely satisfied. The Brownian theory provided an indirect measurement of the size of molecules, in that the size could be deduced from Einstein’s formulas. Perrin wanted to know if there was a way to directly measure the molecular size; as he recollected in his Nobel lecture,* The objective reality of molecules and atoms which was doubted twenty years ago, can today be accepted as a principle the consequences of which can always be proved. Nevertheless, however sure this new principle may be, it would still be a great step forward in our knowledge of matter, and for all that a certitude of a different order, if we could perceive directly these molecules the existence of which has been demonstrated. To make this direct measurement, Perrin turned to a phenomenon that had fascinated him since childhood: the colors of soap bubbles! The spectacular colors of soap films come from a wave interference effect, as illustrated below. An example of destructive interference in thin films. The two reflected waves (shown horizontally offset for clarity) partially cancel out. Let’s first imagine that a light wave comes in on a thin film which is two wavelengths λ thick — this means that two complete “ups and downs” of the wave exactly fit inside the film. Part of the wave will be reflected from the exterior surface of the film (shown in the middle) and part of the wave will be reflected from the interior of the film (shown on the right). Glossing over some subtlety in the nature of reflection from the inside and the outside, we can see that the two reflected waves are out of phase — where one is waving left, the other is waving right, and vice-versa. The two waves destructively interfere, and the net result is that very little light reflects at that particular wavelength. In a slightly different situation, let us imagine that a light wave comes in on a thin film which is two and one quarter wavelengths thick! An example of constructive interference in thin films. The two reflected waves are in phase and constructively interfere. Now the two reflected waves are in phase — they both wave left and right at the same time. These waves constructively interfere, and the reflection of the wave is increased. It turns out that the reflection of light from such a thin film will be enhanced whenever the thickness d of the film is an odd number of quarter wavelengths: $d = \lambda/4,3\lambda/4,5\lambda/4,7\lambda/4,\ldots$ This enhancement is the source of the bright colors of a thin film! Because a thin film will only reflect light of certain wavelengths, and the reflected color depends on the film thickness, the reflected color is a direct measure of the film thickness. Looking again at the soap bubble picture from the beginning of the post, we can now interpret what is going on. This film is oriented vertically, so gravity is pulling down on it, and it is therefore on average thickest at the bottom and thinnest at the top. Perrin himself describes the colors to be seen as follows: You know the properties of thin laminae: each ray reflected from such a lamina is formed by the superposition of a ray reflected from the front side of the lamina on a ray reflected from the rear side. For each elementary colour these rays add together or subtract from one another according to a classical formula, depending on whether they are in phase or out of phase; in particular, there is extinction when the thickness of the lamina is an even multiple of one quarter of the wavelength, and there is maximum reflection when it is an odd multiple. If, therefore, white light strikes a lamina which has a thickness increasing continuously from zero, the reflected light is at first non-existent (black lamina), then weak (grey lamina), then lively and still almost white, becoming successively straw yellow, orange yellow, red, violet, blue (tints of the first order), then again (but with different tints) yellow, red, violet, blue, green (second order); and so on, the reflected colour becoming continuously more complex and more off-white up to the “white of a higher order” (the spectrum is furrowed with black grooves the number of which increases with the thickness of the lamina). All these tints will be present at the same time on a lamina which has not a uniform thickness and which will be black or grey in its thinnest region, straw yellow in a thicker region, red in an even thicker region, and so forth. At the top of the film where it appears black, it is in fact almost perfectly transparent: the film is so thin that it reflects almost no light at all, and we see the black screen behind it. It is to be noted, in the figure above, that near the top of the film, there are regions of bright and nearly constant color, bounded by lines where the colors change dramatically. Perrin realized that something interesting was happening at these transition lines. Having examined a large number of stratified laminae, it occurred to me, before I made any measurement, that the difference in thickness between two adjacent bands cannot fall below a certain value and that this elementary minimum difference, a kind of “step of a staircase”, is included a whole number of times in each band. Similarly, if we throw playing-cards on the table, the thickness at each point is that of a whole number of cards, without all possible thicknesses being necessarily present, since two or three cards may remain stuck together. The stratified liquid strips would, therefore, be formed by the piling up of identical sheets, more or less overlapping each other, their liquid state imposing on the free contours the form of arcs of a circle (which are fixed at their extremities on globules or on the non-stratified periphery, according to conditions so far unknown). In short: where the soap bubble is at its thinnest, it can only be a finite number of molecules thick. At any point where it gets thicker or thinner, the thickness must change by a discrete number of molecular thicknesses. Perrin likens this to a random pile of playing cards, whose thickness on the table can only change by the thickness of one card; I crudely illustrate the idea for layers of soap molecules below. By measuring the reflectivity of the thin film for the different regions of uniform colors, Perrin was able to deduce how much the thickness of the film changed between regions. He collected a large number of these differences in thickness; assuming each must be a multiple of the molecular thickness, he was able to determine the size of the soap molecules! The measurements confirmed this impression. From 1913 onwards I found a value ranging between 4.2 and 5.5 mµ. And since then, precise photometric determinations made under my direction in 1921 by P.V. Wells, who otherwise had to overcome serious experimental difficulties, have fully established what we can call a law of multiple thicknesses. We first of all applied simply the classical relationship between the thickness of the lamina and the intensity of the reflected light, using monochromatic lighting. On the first-order band 120 measurements were made, giving thicknesses grouped according to the law of chances around 4.4 mµ. It is certainly the best measurement made so far of the thickness of the “black spot ” for which Johonnott gave 6 mµ. The extreme thinness of this band, the faintness of the reflected light, and the difficulties due to parasitic lights make this determination particularly interesting. The set of the measurements for the first fifteen bands give similarly thicknesses which are, within several hundredths, of the successive multiples of 4.5 mµ. Perrin therefore found a value of 4.5 nanometers (4.5 billionths of a meter) for the size of soap molecules. This number may seem much too large to those who know a little atomic physics: atomic sizes are usually measured in angstroms (0.1 nanometers), and the soap molecule is over a factor of ten larger. Perrin’s result is not surprising, however, because soap molecules are fashioned as large chains of other molecules, as shown below. Two chemical representations of sodium stearate, a typical soap molecule. Via Wikipedia. Soap derives its effective cleaning properties due to the fact that one end of this chain is repelled by water (hydrophobic) and the other end is attracted to water (hydrophilic). The right side of the molecule shown above, the polar end, is attracted to water; a film of soap therefore consists of a collection of soap molecules standing at attention, side by side. Monolayer of soap molecules encasing a thin film of water. The red spheres represent the polar ends of the soap molecules. It is interesting to note that Perrin’s experiment is similar in philosophy to the Millikan oil drop experiment, performed a few years earlier in 1909. In the Millikan experiment, microscopic droplets of oil were imbued with a small amount of electric charge, with presumably only a small and countable number of electrons in each droplet. Though it was not possible to reliably imbue a droplet with a single electron charge, Millikan measured the charge for a large number of drops and demonstrated that the difference between droplets was always an integer multiple of a fundamental value: the charge of the electron. Similarly, Perrin demonstrated that the difference in soap film thicknesses was always an integer multiple of the size of the soap molecule. Perrin was aware of Millikan’s work, and it seems not unreasonable to assume that he was inspired by it, whether he realized it or not. Perrin’s result was the first direct measurement of molecular size. It seems to have been largely forgotten today for a number of reasons: • It was not the first strong evidence of molecular existence; Einstein’s Brownian motion explanation was the deciding factor. • It was a technique of limited usefulness: the method works pretty much only for thin films of soap. • With modern technology such as atomic force microscopy, we can produce images of objects, including molecules, with sub-nanometer resolution! Though Perrin’s work with soap films did not fundamentally alter our understanding of the natural world, it is an imaginative and beautiful result that demonstrates that profound insights can come from even the most mundane observations. Who knows what other scientific discoveries are hidden in plain sight, right before our eyes? ************************* * Quotes on the soap bubble experiments will be taken from Perrin’s Nobel lecture, as his original 1913 paper is brief and less detailed. The original paper is J. Perrin, “Observations sur les lames minces,” Archives des Sciences Physiques et Naturelles 35 (1913), 384-385. ** Perrin himself notes, however, that a similar technique was used to determine the molecular size of mica, a solid which can be divided into thin planar sheets. ### Like this: This entry was posted in History of science, Optics. Bookmark the permalink. ### 8 Responses to The secret molecular life of soap bubbles (1913) 1. Chemjobber says: Beautiful pic, nice stearate structure. (Magnesium stearate is used in pills/tablets as a lubricant for the powder formulation (so it doesn’t hang up in the chutes, etc. during processing.) 2. Pingback: The Blobologist 3. Bill says: I think that you’ll find that we are seeing Destructive interference, not Constructive, as the colors are the complementary (Cyan, Magenta, Yellow) rather than the Primary (ROYGBV) • skullsinthestars says: Well, we can’t be seeing destructive interference — the light has destructively interfered, i.e. cancelled out in reflection! There is a complementary color effect, however: those colors that don’t reflect are transmitted, and vice versa. If white light is used to illuminate a thin film, the transmitted and reflected light are perfect complements of one another. 4. Pingback: Digital Photos Guide » The Scienceblogging Weekly (May 18, 2012) 5. Kaleberg says: Perrin wrote the extremely influential book Les Atomes which put together maybe a dozen methods of determining Avogadro’s Number. That was the book that more or less clinched atomic theory. I’m not sure if he used the soap bubble method in the book. I think he used a variety of manufactured colloids and measured their pressure half heights to estimate atomic sizes. I’m glad he’s getting some new press. • skullsinthestars says: Yes, he was quite the scientist! His Nobel lecture describes the variety of ways in which he (and others) measured Avogadro’s number. He doesn’t seem to be as well known as many other Nobel laureates, though his contributions were clearly of fundamental importance. I suspect that this is, in part, due to the fact that he did so many things that it is hard to describe in a simple manner his contributions. 6. Charan Rathod says: can any one please let me know which color corresponds to what thickness ? • ### Search Skulls in the Stars: • The author of Skulls in the Stars is an associate professor of physics, specializing in optical science, at UNC Charlotte. 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http://motls.blogspot.cz/2013/03/bubbles-support-10-gev-or-50-gev-dark.html?m=1	# The Reference Frame Our stringy Universe from a conservative viewpoint ## Friday, March 01, 2013 ### Bubbles support $$10\GeV$$ or $$50\GeV$$ dark matter March 2013 is expected to be a great dark matter month, especially due to the eagerly expected results from AMS-02 that may emerge as early as the next week (ANTARES has seen nothing a few days ago). Joseph S. has brought my attention to an excellent astro-ph paper by Tracy Slatyer (IAS) and Dan Hooper (FNAL) Two Emission Mechanisms in the Fermi Bubbles: A Possible Signal of Annihilating Dark Matter that eliminates all doubts that the authors belong among the very top of the world's astroparticle physics. They looked at the spectrum of the Fermi bubbles – that Tracy co-discovered – and decided to write down the most natural model(s) that explain(s) the observer spectrum. What the models depend upon – and what the observations should therefore clarify – is what is the spectrum of electrons, the radiation, and masses and dominant decay channels of hypothetical dark matter particles that team up to produce the spectrum. I think that they show their ability to split the data into regions that seem to be dominated by different effects, explain the partial datasets, and design economic theories that are able to explain several features of some partial datasets simultaneously. What does it mean in practice? In practice, it means that they divided the picture of the galaxy to two regions, according to the latitude (angular distance from the galactic center). For the "tropical" and "polar" regions, they found two different spectra and two different models that explain them. What are they? Far from the galactic plane, at least 30 degrees from the equator, they see that the spectrum ceases to depend on the latitude. The gamma rays over there may apparently be described by inverse Compton scattering. Note that the inverse Compton scattering is the process \[ e^-+\gamma\to e^-+\gamma \] in which the photon's energy increases (the photon steals some energy from the initially fast electron): that's why it's inverse. This description works if the electrons obey a power law spectrum and if they collide with an interstellar radiation field. Cutely enough, this model may also explain the microwave "haze" assuming that the electrons are moving in a magnetic field whose intensity is a microgauss or so. For the "tropical" region, they need a different description and a different mechanism, however. The reason is that the spectrum seems to have a peak near $$1-4\GeV$$ if $$E^2\,\dd N/\dd E$$ is graphed against $$E$$. The inverse Compton scattering is no good to produce such peaks. Instead, they decide that this part of the dataset is similar to results previously reported from the Galactic Center. The luminosity per volume seems to decrease as the $$r^{-2.4}$$ power law with the distance from the Galactic Center. The chief NASA administrator and a rapper gives a not terribly comprehensible but sufficiently stringy introduction to dark matter. An important part of the answer is that this radiation seems to be consistent with the annihilation of dark matter particles! It's either due to $$10\GeV$$ dark matter particles pair-annihilating into lepton pairs or due to $$50\GeV$$ dark matter particles annihilating into quark-antiquark pairs. They seem to propose two comparably likely scenarios for the possible mass and dominant interactions of the dark matter particles. Note that none of these two scenarios should be new to the TRF readers. A possible $$10\GeV$$ dark matter particle has been discussed many times in the context of the "Is Dark Matter Seen" war between the direct search experiments. The allies in the "Dark Matter Is Seen" coalition do generally claim that they see collisions with numerous $$10\GeV$$ or sub-$$10\GeV$$ dark matter particles. The "Dark Matter Is Not Seen" axis is vehemently rejecting all these assertions. However, even dark matter particles around $$50\GeV$$ have previously been spotted by careful TRF readers. In January 2012, I mentioned that Virgo favored a $$20-60\GeV$$ dark matter particle. This was based on a fresh preprint by Han et al. who looked into Virgo, Fornax, and Coma clusters through the Fermi satellite and concluded something remarkably similar to Slatyer and Hooper: there is either a $$20-60\GeV$$ dark matter particle annihilating into $$b\bar b$$ quark pairs, or a $$2-10\GeV$$ dark matter particles annihilating into $$\mu^+\mu^-$$. Because this Han et al. paper doesn't seem to be cited by Tracy and Dan, you could view the agreement as a strong piece of circumstantial evidence that the possible masses and dominant annihilation channels of the dark matter are pretty much what Slatyer and Hooper say now. This sociological argument has only one delicate problem, aside from its being sociological and therefore worthless: in July 2012, Han et al. II wrote a new paper in which they added several new point sources. It seems they thought that they had to add them. Their original signal got contaminated and all the peaks of "extended emission" went away. Now, another half a year later, Slatyer and Hooper are reviving a statement very similar to the Han et al. statement from January 2012. Cosmology entered a high-precision era 15 years ago but even when it depends on solid fellow disciplines such as astroparticle physics, you may see that it sometimes takes a lot of time to pick the right answers in similar uncertain situations and to choose the winners in assorted dark matter wars. In these wars, March 2013 could turn out to be an analogous month to June 1944. I hope that some readers know what the D-Day means ;-) – the Battle of Stalingrad wasn't good enough for the previous sentence because it was too long and its date was therefore too fuzzy. At any rate, stay tuned. Things could get very exciting and very convincing very soon. #### 11 comments: 1. Gene Day This is getting really exciting! It’s about time we observed the major matter constituent of the universe. I do think you meant to say “microgauss” rather than “milligauss” fields in order for cyclotron radiation to account for the microwave haze. 2. Wow, Gene! First, thanks, and second, your catching of the milli/microgauss error is amazing. Was it due to your pure on-top-of-your-head knowledge of this physics or were you actually studying the abstract/paper? ;-) I suspect it's the former and one should stay breathless. Needless to say, I have no intuition about the difference between microgausses and milligausses in similar phenomena. But yes, if you ask me and force me to think, it is quite certainly true that the energy density from an astronomical scale milligauss field would be too much. 3. Gene Day A milligauss field seemed too large intuitively so I did a very rough calculation, which confirmed it. I have done a lot of work with magnetic fields, including my PhD thesis, so they are very familiar to me. I think the microwave “haze”comes from synchrotron radiation 4. kneemo Not to mention the abstract states, "The same electrons in the presence of microgauss-scale magnetic fields can also generate the the observed microwave 'haze'." :) 5. Dilaton This looks good, I would be very happy if things could be settled this March such that the DM wars come to a happy end :-) These nice explanations of the paper and its nicely written abstract makes me want to read the whole thing ... :-D To my shame, I had to google what the D-day is ... :-P Cheers 6. David Nataf Time for NASA to authorize construction of Super-Fermi. Would be neat fibe-tuning if dark matter turned out to be due to a few equally important particles. 7. jitter If dark matter particles are in the Gev range then why haven't they been created and their decay detected in particle accellerators? 8. For the most motivated models with this light new particle, your complaint is solid and it is a genuine contradiction. However, one may think of models in which the dark matter particle, while light, is so weakly interacting with the quarks and gluons and electrons - that made up the existing and current colliders - that only an undetectably small number of these particles are produced during the career of the accelerators. 9. jitter Ok another thought, isn't dark matter supposedtobe in a halo around our galaxy? How would these two lobes affect gravity and the orbit of stars? 10. Dear jitter, the two lobes don't represent the ordinary density of matter (mass density). They represent a more complicated thing, one could say "just an optical illusion". Certain radiation appears in the lobes which doesn't mean that this is where the mass is exclusively concentrated. If the dark matter halo had this infinity-symbol-like shape, there could still be things orbiting in between. However, their orbits would be unstable, trying to get caught by one of the lobes. In fact, the whole situation would be unstable - the lobes would probably wish to merge in some way, unless stabilized by some extra force someone would have to propose first. 11. Even so, the Anomalous Single Photon experiment at PEP should have been exquisitely sensitive to a 10 GeV dark matter particle that annihilates to a lepton pair. At 29 GeV, PEP would have been sitting just above the onset of pair production. Radiative processes (particularly off the t-channel propagator) would show this resonance as a sharp cutoff in the high-transverse-momentum single photon spectrum around 9 GeV, just the sort of thing ASP was designed to see with negligible background. In other words, they were looking right at the sweet spot for the exactly the right signature, and saw nothing. Granted, the coupling must be small, but not so small (well away from any resonance) as to make the Fermi bubbles disappear. ## Who is Lumo? Luboš Motl Pilsen, Czech Republic View my complete profile ← by date	{"extraction_info": {"found_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": 17, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9558207988739014, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/33273/combinations-of-multisets-with-finite-multiplicities	## Combinations of multisets with finite multiplicities ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The question may be of little interest to most people here on MathOverflow, but after browsing a pile of books in combinatorics, I had to ask it somewhere: What are the most efficient formulae for calculating the number of $k$-combinations (and $k$-permutations) of multisets with finite multiplicities (i.e. combinations and permutations with repetition, but with restrictions on the number of repetition)? I know that generating functions are often used for solving this kind of problems, but there has been a number of formulae used for such counting, such as Percy MacMahon's one ($m_i$ denotes multiplicities of $n$ different elements in the multiset): $$C(k;m_{1},m_{2},\ldots,m_{n})=\sum_{p=0}^{n}(-1)^{p}\sum_{1\le i_{1}\le i_{2}\le\cdots\le i_{p}\le n}{n+k-m_{i_{1}}-m_{i_{2}}-\ldots-m_{i_{p}}-p-1 \choose n-1}$$ Are you aware of other formulae for it, or useful references in literature? EDIT: Clearing up the statement: a $k$-combination means simply picking $k$ elements from the multiset (order not important). $k$-permutation is basically the same, but order is important. In the example above, the multiset is ${ m_1\cdot a_1,m_2\cdot a_2,\ldots m_n\cdot a_n}$, $a_i$ being the elements, $m_i$ being the multiplicities. - I am not sure I understand what a k-combination or a k-permutation is. – Qiaochu Yuan Jul 25 2010 at 16:39 Oh, sorry for not making it clear: a $k$-combination means simply picking k terms from the multiset (order not important). $k$-permutation is basically the same, but order is important. In the example above, the multiset is $\{m_1\cdot a_1, m_2\cdot a_2,\ldots,m_n\cdot a_n\}$, $a_i$ being the elements, $m_i$ being the multiplicities. – Harun Šiljak Jul 25 2010 at 16:59 Have you looked at Knuth's Volume 4 fascicle 3, in particular exercise 39 in section 7.2.1.4? – András Salamon Jul 25 2010 at 20:22 @András: looking it right now - but I don't see how a formula for combinations I'm looking for could directly follow from a formula for partitions given there? – Harun Šiljak Jul 25 2010 at 21:26	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8890790343284607, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/12534/list	## Return to Answer 1 [made Community Wiki] @steve: I'm not sure I understand your objection, but how about this modification. Let A be defined by $A(n) = \sum_{k=0}^{n-1} A(k)$ for every nonnegative integer n. Then by strong induction we can show that A(n) = 0. That's a trivial example, but if we wanted to, we could make it less trivial: $B(n) = 1 - n + \sum_{k=0}^{n-1} B(k)$, which has solution $B(n) = 1$.	{"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": 3, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9401866793632507, "perplexity_flag": "head"}
http://mathoverflow.net/questions/119913/what-is-the-difference-between-a-function-and-a-morphism/119926	## What is the difference between a function and a morphism? [closed] ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Say I have a function $f$. Then, if I understand correctly, $f$ can be regarded as a morphism in a suitably chosen category which has the domain of $f$ and the range of $f$ as objects. So, the other direction. Say I have a morphism $f$ in some category from one object $A$ to another object $B$. Can I not regard $f$ as a function whose domain is $\{A\}$ and range $\{B\}$? I'm interested in looking for contexts in which the latter move is not appropriate for some reason. - 4 Think of a one object category which had more than one morphism... – Yemon Choi Jan 26 at 4:49 Don't you really mean "whose domain is $A$ and range $B$" ? (then, of course, you have to think of the objects of your category as sets). But of course, the answer is negative, as different morphisms can correspond to the same function. – Feldmann Denis Jan 26 at 5:59 @Feldmann - No, I wanted the domain of $f$ to be the set containing $A$, simply because $A$ may not be a set, and the domain of a function is a set by definition. – Benjamin Braun Jan 26 at 7:36 ## 4 Answers You're losing a lot of information. Let's say we have two morphisms $f,g\in \mathrm{Hom}(A,B)$. Using your perspective, we see that both $f$ and $g$ agree at all points (as the only point is $B$). Thus by eta-reduction we should have $f=g$, but this is not necessarily the case. So thinking of morphisms as functions from $\{A\}$ to $\{B\}$ is the wrong perspective. - Got it. I've resolved my issue by thinking of morphisms as just triples of the form $(x,y,z)$ where $x$ and $z$ are objects of the category and $y$ is some mathematical object. So in your example, $f$ would be $(A,u,B)$ where $g$ would be $(A,v,B)$ where $u$ and $v$ are what is making $f$ different from $g$. – Benjamin Braun Jan 26 at 5:08 3 @Benjamin I'm not really sure how that differs from saying $f,g\in \mathrm{Hom}(A,B)$. You're just using slightly different notation with $u$ instead of $f$ and $v$ instead of $g$. – Alex Becker Jan 26 at 5:18 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. There are various natural categories in topology where morphisms are not functions. For instance, you may define for every $n$ the category whose objects are all compact closed $n$-manifolds and whose morphisms are the $(n+1)$-dimensional cobordisms, i.e. compact $(n+1)$-manifolds $W$ with boundary, where each boundary component of $W$ is coloured with a "+" or "-" sign. Of course you cannot interprete a cobordism $W$ from two $n$-manifolds $M$ and $N$ as a function $M\to N$. However, you can naturally define a "composition" of morphisms by gluing the manifolds. The "identity" is just the product manifold $W = M \times [0,1]$. You can also construct embedded versions of that. For instance: objects are compact 1-submanifolds of $\mathbb R^2$ (i.e. finitely many disjoint circles) and morphisms are proper subsurfaces of $\mathbb R^2 \times [0,1]$. You might decide to see morphisms only up to isotopy fixing their boundary. - 2 There is also the category whose objects are all topological spaces and whose morphisms are homotopy classes of functions. Maybe this is even the first "non set theory" functor students will encounter, at least in topology. – Lee Mosher Jan 26 at 15:39 or also, the obvious construction of "opposite category". – Pietro Majer Jan 26 at 17:12 Another interesting reason why categories cannot be identified always with categories having functions for morphisms is given in this paper, by Peter Freyd in which is proven that there are some categories which aren't concrete: i.e. which don't have any faithful functor from the category in $\mathbf{Set}$ (the category of sets and functions). Having a such functor is necessary condition to have functions as morphisms. - A main point is that considering objects and morphism (more generally than sets with some structure and functions that preserve it), is both a clarifying and fruitful abstraction. There are situations in which a category may be represented by a sub-category of the Set category (though it can be more complicated than what you suggested. If you are interested, you may start from the Yoneda lemma). But, in order to understand universal properties and to make categorical constructions it could even be counter-productive, like e.g. insisting in metrization with topological spaces, or coordinates with vector spaces. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 54, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.953153133392334, "perplexity_flag": "head"}
http://mathhelpforum.com/discrete-math/75991-induction-sequences.html	# Thread: 1. ## Induction for sequences OK, I am completely stupid when it comes to induction. Any help? I shouldn't have taken this class. 2. Use Mathematicla induction for part 1. let $S_n$be a statement that $a_n$ is odd Assume $S_{k-2}$ and $S_{k-1}$ are true and then prove $S_k$is true. $a_k = 2a_{k-2} +3a_{k-1}$ Now $2a_{k-2}$ is even and $3a_{k-1}$ is odd since odd multiplied by odd is odd. Sum of an even number and odd number is always odd. so $a_k$is odd. Hence $S_k$ is proved true. base case $a_1$ and $a_2$are already give odd.	{"extraction_info": {"found_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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9302178025245667, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/91253/list	## Return to Answer 3 added 5 characters in body If your homology theory is of the form $H_n(X) = H_n(S(X))$ where $S$ is some functor from your original category to non-negative chain complexes, then the Dold-Kan correspondence gives you a corresponding simplicial abelian group $\Gamma(S(X))$ and hence (by realization) a topological space in which you are "counting holes". 2 removed backticks on discovering they worsen display problem If your homology theory is of the form $H_n(X) = H_n(S(X))$ H_n(S(X)) $where$S$is some functor from your original category to non-negative chain complexes, then the Dold-Kan correspondence gives you a corresponding simplicial abelian group$\Gamma(S(X))$\Gamma(S(X))$ and hence (by realization) a topological space in which you are "counting holes". 1 If your homology theory is of the form $H_n(X) = H_n(S(X))$ where $S$ is some functor from your original category to non-negative chain complexes, then the Dold-Kan correspondence gives you a corresponding simplicial abelian group $\Gamma(S(X))$ and hence (by realization) a topological space in which you are "counting holes".	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9076836705207825, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/recurrence-relations?page=11&sort=newest&pagesize=50	# Tagged Questions Questions regarding functions defined recursively, such as the Fibonacci sequence. 1answer 185 views ### Finding ($2012$th term of the sequence) $\pmod {2012}$ Let $a_n$ be a sequence given by formula: $a_1=1\\a_2=2012\\a_{n+2}=a_{n+1}+a_{n}$ find the value: $a_{2012}\pmod{2012}$ So, in fact, we have to find the value of ... 1answer 242 views ### Why do the Fibonacci numbers recycle these formulas? The Fibonacci numbers $F_n = 0, 1, 1, 2, 3, 5, 8, 13, \dots$ obey the following recurrence relations, \$ \begin{aligned} &F_{n}-\;F_{n-1}-F_{n-2} = 0\\[1.5mm] &F_{n-1}^3-F_{n}^3-F_{n+1}^3 = ... 0answers 103 views ### A recurrence relation Motivated by a specific example, I have a rather general question to ask: suppose $a_n$ is a sequence defined by the relation $a_{n+1}=f_na_n+g_na_{n-1}$, $a_0=a>0$, $a_1=b>0$, where both $f_n$, ... 2answers 1k views ### Why is solving non-linear recurrence relations “hopeless”? I came across a non-linear recurrence relation I want to solve, and most of the places I look for help will say things like "it's hopeless to solve non-linear recurrence relations in general." Is ... 3answers 916 views ### What is the upper bound on $T(n) = 3T(n/2) + n$? 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I ... 1answer 196 views ### Towers of Hanoi - are there configurations of $n$ disks that are more than $2^n - 1$ moves apart? This is an exercise from Chapter 1 of "Concrete Mathematics". It concerns the Towers of Hanoi. Are there any starting and ending configurations of $n$ disks on three pegs that are more than \$2^n - ... 0answers 97 views ### How to solve the recursive relation in Kalman filter? I was wondering how to solve the Kalman filter's recursive equation (also see the appendix at the end of this post) for the estimated state $\hat{\textbf{x}}_{n\|n}$ at time $n$, over discrete times ... 1answer 354 views ### Interesting Recurrence Relation $T(n) = T(\sqrt{n}) + T(n-\sqrt{n}) + n$ I found an interesting recurrence that I do not know how to solve. I think this has to do with quicksort with pivots at rank $\sqrt{n}$. I do not know how to approach this problem nor found any ... 3answers 108 views ### How do you go about solving difference equations? 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http://mathoverflow.net/questions/7317/computation-of-joins-of-simplicial-sets/7323	## Computation of Joins of Simplicial Sets ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It turns out that joins of simplicial sets are fairly easy to define, but hard to manage. In lots of cases, we'd like to compute what a join is, does it look like a horn?, a boundary?, etc? and identify it as such, so we can figure out when our morphisms from the join have certain nice properties like being anodyne, having lifting properties, and all of that wonderful stuff. For example, consider the join, $\Lambda^n_j \star \Delta^m$. The problem that I currently face is, I can't tell what this thing looks like from the definition. Consider an even simpler case, $\Delta^n \star \partial \Delta^m$. From the definition, we get a very nasty definition of this join, and I'm having trouble applying it and computing the join in terms of nicer simplicial sets. I ask this, because on p.62 of Higher Topos Theory by Lurie, for example, he states that for some $0 < j \leq n$ $$\Lambda^n_j \star \Delta^m \coprod_{\Lambda^n_j \star \partial \Delta^m} \Delta^n \star \partial \Delta^m$$ and says that we can identify this with the horn $\Lambda^{n+m+1}_j$. Unraveling the definitions seems to make it harder to understand, and I just don't see how this result was achieved. However, my aim here is to understand how the computation was actually carried out, since it is completely omitted. For convenience, here is the definition of the join of $S$ and $S'$ for each object $J \in \Delta$ $$(S\star S')(J)=\coprod_{J=I\cup I'}S(I) \times S'(I')$$ Where $\forall (i \in I \land i' \in I') i < i'$, which implies that $I$ and $I'$ are disjoint. EDIT AFTER ANSWER: Both Reid and Greg provided good answers to the question, and I only accepted the one that I did because Greg commented more recently. So for anyone reading this at some point in the future, read both answers, as they are both good. - By $\delta \Lambda^m$ do you mean $\partial \Delta^m$? – Reid Barton Dec 1 2009 at 0:04 That I did, ser. Fixed now, hopefully. – Harry Gindi Dec 1 2009 at 0:11 ## 3 Answers Since the join of simplicial sets is associative and $\Delta^m = \Delta^0 \star \cdots \star \Delta^0$ ($m+1$ times), we should start by trying to understand things like $\Lambda^n_j \star \Delta^0$, a.k.a. the "final" cone on $\Lambda^n_j$. It's not too hard to see that this is the subcomplex of $\Delta^{n+1}$ consisting of those faces which do not contain the (codimension 2) face $\{0, \ldots, r-1, r+1, \ldots, n\}$. In other words, we are missing the face opposite $r$ and $n+1$, because we were originally missing the face opposite $r$ of $\Delta^n$, as well as the three other faces (including the interior of $\Delta^{n+1}$) it contains. Similarly $\Delta^0 \star \partial \Delta^n$ is the horn $\Lambda^{n+1}_0$ (we are missing the interior of $\Delta^{1,\ldots,n}$ and the cone on it). In general all the simplicial sets that come up have the form of the subcomplex of $\Delta^N$ consisting of those faces which do not contain a fixed face $\Delta^S$, $S \subset \{0, \ldots, N\}$. Forming the cone (on either side) on such a space results in another such space with $N$ replaced by $N+1$ and $S$ unchanged (as a subset of the vertices of the original $\Delta^N$, which if we formed a cone on the left, means we increment each index in $S$). After doing these sorts of computations, I expect that $\Lambda^n_j \star \Delta^m$ and $\Delta^n \star \partial \Delta^m$ will be two subcomplexes of $\Delta^{n+m+1}$ each characterized by avoiding faces containing a certain face, and that $\Lambda^n_j \star \partial \Delta^m$ is their intersection and $\Lambda^{n+m+1}_j$ is their union, from which the claim would follow. - I kinda see what you mean, but what happens when we take the next join in the first thing, since you split it up and all. – Harry Gindi Dec 1 2009 at 0:38 I elaborated a little on what happens in general. – Reid Barton Dec 1 2009 at 1:01 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. It may be helpful to consider the geometric join operation for compact topological spaces or simplicial complexes. If $A$ and $B$ are discrete sets, then their join $A \star B$ is the complete bipartite graph connecting $A$ to $B$. The join in general is the natural generalization of that to topological spaces, or simplicial complexes, or simplicial sets, or (if you like) CW complexes. In fact it always is the complete bipartite graph, but with a topology or a simplicial geometry on the set of line segments. If $A$ and $B$ are simplices, their join is another simplex, whose vertices are the disjoint union of the vertices of $A$ and $B$. A simplicial set $S$ has a small geometric realization consisting of the non-degenerating simplices glued together. In many early examples, such as for instance horns, the small geometric realization is just a finite simplicial complex with locally ordered vertices. (That is, the vertices of each simplex are compatibly ordered.) For instance, the horn `$\Lambda_j^2$` is a V, and the horn `$\Lambda_j^3$` is a triangular hat (or horn, hence the name and the clever symbol). (As best I can tell, the subscript referring to the apex of the horn `$\Lambda^n_j$` is not intrinsic to it as a simplicial set, but rather comes from its inclusion into the simplex $\Delta^n$.) (Edit: Even though a horn is hollow, the local orderings of its faces induce a total ordering of its vertices when $n \ge 2$, or a partial ordering when $n=1$. The subscript indicates the position of the apex.) The join of $A$ with point $\Delta^0$ is also a cone with base $A$. So as Reid said, the cone $\Delta^0 \star \partial \Delta^n$ over a hollow simplex is a horn. It's also easy to see, by drawing a picture, that the cone over a horn is the next horn. [Edit: It's a good idea to draw the picture of the cone over of a horn, to see that it isn't a horn. :-)] (I apologize if this geometric discussion is too close to what Reid already said somewhat more algebraically.) - 2 Just a minor point--the various horns of the n-simplex are nonisomorphic as simplicial sets, since the vertices of a simplex are distinguished (effectively, ordered). This doesn't matter much when you use simplicial sets to model spaces, but it does matter when you use them as quasicategories since the direction of a 1-simplex corresponds to the direction of a morphism in a category. – Reid Barton Dec 1 2009 at 5:38 No, I just goofed, as the new parenthetical explains. – Greg Kuperberg Dec 1 2009 at 5:44 Ah, I see. I guess what you wrote is true for n = 1. :) – Reid Barton Dec 1 2009 at 5:46 No, even then it is only partly true. – Greg Kuperberg Dec 1 2009 at 5:50 I am not entirely sure what you're asking. Do you mean, how does the geometric description of the join match your coproduct of products formula for $S$ and $S'$? The answer is that each non-degenerate simplex of $S$ and each non-degenerate simplex of $S'$ join together into a simplex of $S \star S'$. This exactly matches your formula. – Greg Kuperberg Dec 1 2009 at 7:05 show 7 more comments If one thinks of joins as most naturally applicable to augmented simplicial sets, then many of the questions asked take a (I hope) clearer aspect. The join of augmented simplicial sets is part of a monoidal category structure that explicitly uses ordinal sums and hence has a directional aspect. The ordinal sum of two finite ordinals works well but $[m]\oplus [n]$ and $[n]\oplus [m]$ although isomorphic are linked in a subtler way. Phil Ehler attacked this in his thesis at Bangor using ideas of Duskin and van Osdol (ultimately derivable from Lawvere). I wrote up Phil's thesis material on this an there are two published papers available on ordinal sum and its relationship with join and also on links with anodyne extensions and quasicategories. ( I can give pointers if it would help.) - I mean, I've mostly resolved the problem, but I still don't really have an intuitive feeling for the join of arbitrary simplicial sets. – Harry Gindi Jan 24 2010 at 10:39 My intuition comes from extending the join in simplicial complexes which is described in a classical source such as Spanier. The examples that Phil Ehlers looked at in his thesis again made a lot of sense and showed how the augmentation aspect comes in. (I can send you a link to Phil's work but first have a look at Joins for (augmented) simplicial sets, Jour. Pure Applied Algebra, 145 (2000) 37-44 to see if that helps. – Tim Porter Jan 24 2010 at 12:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 63, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9548649787902832, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/123254/how-to-show-this-ideal-is-not-principal	# How to show this ideal is not principal I have been brushing up on cubic number fields. Specifically, let $s$ be a root of the polynomial $x^3 + x^2 + 3x + 17$, and consider $K = \mathbb{Q}(s)$; we have $\mathcal{O}_K = \mathbb{Z}[s]$, and $$2\mathcal{O}_K = (2,s+1)^3,$$ this by the theorem of Dedekind. My favourite computer algebra system tells me that $(2,s+1)$ is not principal, but I would like to justify this myself. How may I show that this ideal is not principal? I can see how my question is equivalent to asking why the ring $\mathbb{Z}[s]$ has no element of norm $\pm 2$, or why 2 is irreducible in this ring. In principle I could answer this by writing down the norm of a general element $a + bs + cs^2$ for $a,b,c \in \mathbb{Z}$; but this is something that I really do not want to do. Rather I'm looking for a way to answer the question whilst keeping my hands clean. - ## 1 Answer In general, principal ideal tests require knowledge of the unit group, which means that you cannot expect to keep your hands clean. Quite likely even writing down the norm form is not going to help you (in the sense that you should expect a contradiction modulo some integer). There are exceptions when genus theory is strong enough to tell you something; this requires completely ramified prime ideals. Assume e.g. that $(p) = P^3$ and write $(2,s+1) = (\alpha)$. Then $\alpha \equiv a \bmod P$ for some integer $a$, hence $a^3 \equiv \pm 2 \bmod p$ in the integers. If you're lucky, $2$ is not a cube modulo $p$, and then the ideal is not principal. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "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}, "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}, "english_confidence": 0.9545174241065979, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/93423-constructing-graph.html	# Thread: 1. ## constructing a graph hi here's my question : Construct an example of a polynomial so that its graph goes through the points $\left ( -1,5 \right )$ and $\left ( 3,-2 \right )$ thanks 2. Originally Posted by Raoh here's my question : Construct an example of a polynomial so that its graph goes through the points $\left ( -1,5 \right )$ and $\left ( 3,-2 \right )$ what degree of polynomial? 3. thanks for your reply,but i don't know what degree is the polynomial,that was the question . thanks again. 4. Originally Posted by Raoh thanks for your reply,but i don't know what degree is the polynomial,that was the question. o.k. , let's keep it simple. here is a polynomial of degree 1 that works ... 7x + 4y = 13 5. If you want a quadratic, this is how to do it: Pick one of the points as the vertex -- though it does matter which one you pick, because you get different answers otherwise, the problem is only asking for an example. Let's pick (3, -2). The vertex form of the parabola is $y = a(x - h)^2 + k$ Plug in the vertex for (h, k): $y = a(x - 3)^2 - 2$ Plug in the other point for (x, y) to solve for a: $\begin{aligned}<br /> 5 &= a(-1 - 3)^2 - 2 \\<br /> 5 &= (-4)^2 a - 2 \\<br /> 5 &= 16a - 2 \\<br /> 7 &= 16a \\<br /> a &= \frac{7}{16}<br /> \end{aligned}$ So the equation of the parabola would be $y = \frac{7}{16}(x - 3)^2 - 2$ 01 6. thanks a lot guys if i wanted to give an example of polynomial of degree 1,it will have always the form : $7x+4y = k$ and $k$could be anything,am i right ? thanks again 7. Originally Posted by Raoh thanks a lot guys if i wanted to give an example of polynomial of degree 1,it will have always the form : $7x+4y = k$ and $k$could be anything,am i right ? thanks again no ... not if they pass through the two given points (-1,5) , (3,-2) 8. Originally Posted by Raoh thanks a lot guys if i wanted to give an example of polynomial of degree 1,it will have always the form : $7x+4y = k$ and $k$could be anything,am i right ? thanks again No, k can't be just anything. To find a linear equation, you need to find the slope: $\begin{aligned}<br /> m &= \frac{y_2 - y^1}{y_2 - y^1} \\<br /> &= \frac{-2 - 5}{3 - (-1)} \\<br /> &= -\frac{7}{4}<br /> \end{aligned}$ Use point-slope form: $\begin{aligned}<br /> y - y_1 &= m(x - x_1) \\<br /> y - 5 &= -\frac{7}{4}(x + 1) \\<br /> 4y - 20 &= -7(x + 1) \\<br /> 4y - 20 &= -7x - 7 \\<br /> 7x + 4y - 20 &= -7 \\<br /> 7x + 4y &= 13 \\<br /> \end{aligned}$ So you see, it does matter what the constant term is. 01 9. Originally Posted by skeeter no ... not if they pass through the two given points (-1,5) , (3,-2) of course they have to pass through the two points $(1.5),(3.-2)$ sorry i forgot to mention that.does this make my previous post right ? thanks. 10. Originally Posted by Raoh of course they have to pass through the two points $(1.5),(3.-2)$ sorry i forgot to mention that.does this make my previous post right ? thanks. The answer is still no. A linear equation in standard form is $Ax + By = C$ where A, B, and C are constants. Depending on the two given points, you'll have different sets of values for A, B, and C. 01 11. i don't get it (sorry),if a line passes by those two points,it will have always the same slope $(m = \frac{-7}{4} )$,right ? thanks for helping me. 12. Originally Posted by Raoh i don't get it (sorry),if a line passes by those two points,it will have always the same slope $(m = \frac{-7}{4} )$,right ? thanks for helping me. Yes. 01 13. oooops ! sorry.. my mistake ,in your previous post you wrote : $Ax + By = C$,the $C$constant depends in $A$and $B$,right ? 14. ok,thanks guys i got 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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 25, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.949946939945221, "perplexity_flag": "middle"}
http://medlibrary.org/medwiki/Partial_derivatives	# Partial derivatives Welcome to MedLibrary.org. For best results, we recommend beginning with the navigation links at the top of the page, which can guide you through our collection of over 14,000 medication labels and package inserts. For additional information on other topics which are not covered by our database of medications, just enter your topic in the search box below: Calculus Definitions Concepts Rules and identities Integral calculus Definitions Integration by Formalisms Definitions • Partial derivative • Multiple integral • Line integral • Surface integral • Volume integral • Jacobian Specialized calculi In mathematics, a partial derivative of a function of several variables is its derivative with respect to one of those variables, with the others held constant (as opposed to the total derivative, in which all variables are allowed to vary). Partial derivatives are used in vector calculus and differential geometry. The partial derivative of a function f with respect to the variable x is variously denoted by $f^\prime_x,\ f_{,x},\ \partial_x f, \text{ or } \frac{\partial f}{\partial x}$ The partial-derivative symbol is ∂. One of the first known uses of the symbol in mathematics is by Marquis de Condorcet from 1770, who used it for partial differences. The modern partial derivative notation is by Adrien-Marie Legendre (1786), though he later abandoned it; Carl Gustav Jacob Jacobi re-introduced the symbol in 1841.[1] ## Introduction Suppose that ƒ is a function of more than one variable. For instance, $z = f(x, y) = \,\! x^2 + xy + y^2.\,$ A graph of z = x2 + xy + y2. For the partial derivative at (1, 1, 3) that leaves y constant, the corresponding tangent line is parallel to the xz-plane. A slice of the graph above at y= 1 The graph of this function defines a surface in Euclidean space. To every point on this surface, there is an infinite number of tangent lines. Partial differentiation is the act of choosing one of these lines and finding its slope. Usually, the lines of most interest are those that are parallel to the xz-plane, and those that are parallel to the yz-plane (which result from holding either y or x constant, respectively.) To find the slope of the line tangent to the function at P(1, 1, 3) that is parallel to the xz-plane, the y variable is treated as constant. The graph and this plane are shown on the right. On the graph below it, we see the way the function looks on the plane y = 1. By finding the derivative of the equation while assuming that y is a constant, the slope of ƒ at the point (x, y, z) is found to be: $\frac{\partial z}{\partial x} = 2x+y$ So at (1, 1, 3), by substitution, the slope is 3. Therefore $\frac{\partial z}{\partial x} = 3$ at the point. (1, 1, 3). That is, the partial derivative of z with respect to x at (1, 1, 3) is 3. ## Definition ### Basic definition The function f can be reinterpreted as a family of functions of one variable indexed by the other variables: $f(x,y) = f_x(y) = \,\! x^2 + xy + y^2.\,$ In other words, every value of x defines a function, denoted fx, which is a function of one variable.[2] That is, $f_x(y) = x^2 + xy + y^2.\,$ Once a value of x is chosen, say a, then f(x,y) determines a function fa which sends y to a2 + ay + y2: $f_a(y) = a^2 + ay + y^2. \,$ In this expression, a is a constant, not a variable, so fa is a function of only one real variable, that being y. Consequently, the definition of the derivative for a function of one variable applies: $f_a'(y) = a + 2y. \,$ The above procedure can be performed for any choice of a. Assembling the derivatives together into a function gives a function which describes the variation of f in the y direction: $\frac{\partial f}{\partial y}(x,y) = x + 2y.\,$ This is the partial derivative of f with respect to y. Here ∂ is a rounded d called the partial derivative symbol. To distinguish it from the letter d, ∂ is sometimes pronounced "del" or "partial" instead of "dee". In general, the partial derivative of a function f(x1,...,xn) in the direction xi at the point (a1,...,an) is defined to be: $\frac{\partial f}{\partial x_i}(a_1,\ldots,a_n) = \lim_{h \to 0}\frac{f(a_1,\ldots,a_i+h,\ldots,a_n) - f(a_1,\ldots, a_i, \dots,a_n)}{h}.$ In the above difference quotient, all the variables except xi are held fixed. That choice of fixed values determines a function of one variable $f_{a_1,\ldots,a_{i-1},a_{i+1},\ldots,a_n}(x_i) = f(a_1,\ldots,a_{i-1},x_i,a_{i+1},\ldots,a_n)$, and by definition, $\frac{df_{a_1,\ldots,a_{i-1},a_{i+1},\ldots,a_n}}{dx_i}(x_i) = \frac{\partial f}{\partial x_i}(a_1,\ldots,a_n).$ In other words, the different choices of a index a family of one-variable functions just as in the example above. This expression also shows that the computation of partial derivatives reduces to the computation of one-variable derivatives. An important example of a function of several variables is the case of a scalar-valued function f(x1,...xn) on a domain in Euclidean space Rn (e.g., on R2 or R3). In this case f has a partial derivative ∂f/∂xj with respect to each variable xj. At the point a, these partial derivatives define the vector $\nabla f(a) = \left(\frac{\partial f}{\partial x_1}(a), \ldots, \frac{\partial f}{\partial x_n}(a)\right).$ This vector is called the gradient of f at a. If f is differentiable at every point in some domain, then the gradient is a vector-valued function ∇f which takes the point a to the vector ∇f(a). Consequently, the gradient produces a vector field. A common abuse of notation is to define the del operator (∇) as follows in three-dimensional Euclidean space R3 with unit vectors $\mathbf{\hat{i}}, \mathbf{\hat{j}}, \mathbf{\hat{k}}$: $\nabla = \bigg[{\frac{\partial}{\partial x}} \bigg] \mathbf{\hat{i}} + \bigg[{\frac{\partial}{\partial y}}\bigg] \mathbf{\hat{j}} + \bigg[{\frac{\partial}{\partial z}}\bigg] \mathbf{\hat{k}}$ Or, more generally, for n-dimensional Euclidean space Rn with coordinates (x1, x2, x3,...,xn) and unit vectors ($\mathbf{\hat{e}_1}, \mathbf{\hat{e}_2}, \mathbf{\hat{e}_3}, \dots , \mathbf{\hat{e}_n}$): $\nabla = \sum_{j=1}^n \bigg[{\frac{\partial}{\partial x_j}}\bigg] \mathbf{\hat{e}_j} = \bigg[{\frac{\partial}{\partial x_1}}\bigg] \mathbf{\hat{e}_1} + \bigg[{\frac{\partial}{\partial x_2}}\bigg] \mathbf{\hat{e}_2} + \bigg[{\frac{\partial}{\partial x_3}}\bigg] \mathbf{\hat{e}_3} + \dots + \bigg[{\frac{\partial}{\partial x_n}}\bigg] \mathbf{\hat{e}_n}$ ### Formal definition Like ordinary derivatives, the partial derivative is defined as a limit. Let U be an open subset of Rn and f : U → R a function. The partial derivative of f at the point a = (a1, ..., an) ∈ U with respect to the i-th variable ai is defined as $\frac{ \partial }{\partial a_i }f(\mathbf{a}) = \lim_{h \rightarrow 0}{ f(a_1, \dots , a_{i-1}, a_i+h, a_{i+1}, \dots ,a_n) - f(a_1, \dots, a_i, \dots ,a_n) \over h }$ Even if all partial derivatives ∂f/∂ai(a) exist at a given point a, the function need not be continuous there. However, if all partial derivatives exist in a neighborhood of a and are continuous there, then f is totally differentiable in that neighborhood and the total derivative is continuous. In this case, it is said that f is a C1 function. This can be used to generalize for vector valued functions (f : U → R'm) by carefully using a componentwise argument. The partial derivative $\frac{\partial f}{\partial x}$ can be seen as another function defined on U and can again be partially differentiated. If all mixed second order partial derivatives are continuous at a point (or on a set), f is termed a C2 function at that point (or on that set); in this case, the partial derivatives can be exchanged by Clairaut's theorem: $\frac{\partial^2f}{\partial x_i\, \partial x_j} = \frac{\partial^2f} {\partial x_j\, \partial x_i}.$ ## Examples The volume of a cone depends on height and radius The volume V of a cone depends on the cone's height h and its radius r according to the formula $V(r, h) = \frac{\pi r^2 h}{3}.$ The partial derivative of V with respect to r is $\frac{ \partial V}{\partial r} = \frac{ 2 \pi r h}{3},$ which represents the rate with which a cone's volume changes if its radius is varied and its height is kept constant. The partial derivative with respect to h is $\frac{ \partial V}{\partial h} = \frac{\pi r^2}{3},$ which represents the rate with which the volume changes if its height is varied and its radius is kept constant. By contrast, the total derivative of V with respect to r and h are respectively $\frac{\operatorname dV}{\operatorname dr} = \overbrace{\frac{2 \pi r h}{3}}^\frac{ \partial V}{\partial r} + \overbrace{\frac{\pi r^2}{3}}^\frac{ \partial V}{\partial h}\frac{\operatorname d h}{\operatorname d r}$ and $\frac{\operatorname dV}{\operatorname dh} = \overbrace{\frac{\pi r^2}{3}}^\frac{ \partial V}{\partial h} + \overbrace{\frac{2 \pi r h}{3}}^\frac{ \partial V}{\partial r}\frac{\operatorname d r}{\operatorname d h}$ The difference between the total and partial derivative is the elimination of indirect dependencies between variables in partial derivatives. If (for some arbitrary reason) the cone's proportions have to stay the same, and the height and radius are in a fixed ratio k, $k = \frac{h}{r} = \frac{\operatorname d h}{\operatorname d r}.$ This gives the total derivative with respect to r: $\frac{\operatorname dV}{\operatorname dr} = \frac{2 \pi r h}{3} + \frac{\pi r^2}{3}k$ Which simplifies to: $\frac{\operatorname dV}{\operatorname dr} = k\pi r^2$ Similarly, the total derivative with respect to h is: $\frac{\operatorname dV}{\operatorname dh} = \pi r^2$ Equations involving an unknown function's partial derivatives are called partial differential equations and are common in physics, engineering, and other sciences and applied disciplines. ## Notation For the following examples, let f be a function in x, y and z. First-order partial derivatives: $\frac{ \partial f}{ \partial x} = f_x = \partial_x f.$ Second-order partial derivatives: $\frac{ \partial^2 f}{ \partial x^2} = f_{xx} = \partial_{xx} f.$ Second-order mixed derivatives: $\frac{\partial^2 f}{\partial y \, \partial x} = \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) = f_{xy} = \partial_{yx} f.$ Higher-order partial and mixed derivatives: $\frac{ \partial^{i+j+k} f}{ \partial x^i\, \partial y^j\, \partial z^k } = f^{(i, j, k)}.$ When dealing with functions of multiple variables, some of these variables may be related to each other, and it may be necessary to specify explicitly which variables are being held constant. In fields such as statistical mechanics, the partial derivative of f with respect to x, holding y and z constant, is often expressed as $\left( \frac{\partial f}{\partial x} \right)_{y,z}.$ ## Antiderivative analogue There is a concept for partial derivatives that is analogous to antiderivatives for regular derivatives. Given a partial derivative, it allows for the partial recovery of the original function. Consider the example of $\frac{\partial z}{\partial x} = 2x+y$. The "partial" integral can be taken with respect to x (treating y as constant, in a similar manner to partial derivation): $z = \int \frac{\partial z}{\partial x} \,dx = x^2 + xy + g(y)$ Here, the "constant" of integration is no longer a constant, but instead a function of all the variables of the original function except x. The reason for this is that all the other variables are treated as constant when taking the partial derivative, so any function which does not involve $x$ will disappear when taking the partial derivative, and we have to account for this when we take the antiderivative. The most general way to represent this is to have the "constant" represent an unknown function of all the other variables. Thus the set of functions $x^2 + xy + g(y)$, where g is any one-argument function, represents the entire set of functions in variables x,y that could have produced the x-partial derivative 2x+y. If all the partial derivatives of a function are known (for example, with the gradient), then the antiderivatives can be matched via the above process to reconstruct the original function up to a constant. ## Notes 1. Jeff Miller (2009-06-14). "Earliest Uses of Symbols of Calculus". Earliest Uses of Various Mathematical Symbols. Retrieved 2009-02-20. 2. This can also be expressed as the adjointness between the product space and function space constructions. Content in this section is authored by an open community of volunteers and is not produced by, reviewed by, or in any way affiliated with MedLibrary.org. Licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported License, using material from the Wikipedia article on "Partial derivatives", available in its original form here: http://en.wikipedia.org/w/index.php?title=Partial_derivatives • ## Finding More You are currently browsing the the MedLibrary.org general encyclopedia supplement. To return to our medication library, please select from the menu above or use our search box at the top of the page. In addition to our search facility, alphabetical listings and a date list can help you find every medication in our library. • ## Questions or Comments? 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You may also get in touch directly with the original material provider. • ## About This site is provided for educational and informational purposes only and is not intended as a substitute for the advice of a medical doctor, nurse, nurse practitioner or other qualified health professional.	{"extraction_info": {"found_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": 38, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9037783145904541, "perplexity_flag": "head"}
http://en.wikibooks.org/wiki/GCSE_Science/Kinetic_Energy	# GCSE Science/Kinetic Energy The cars of a roller coaster reach their maximum kinetic energy when at the bottom of their path. When they start rising, the kinetic energy begins to be converted to gravitational potential energy, but the total amount of energy in the system remains constant; assuming negligible friction and other energy conversion factors. The kinetic energy of an object is the extra energy which it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its current velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. Negative work of the same magnitude would be required to return the body to a state of rest from that velocity. ## Etymology The adjective "kinetic" to the noun energy has its roots in the Greek word for "motion" (kinesis). The terms kinetic energy and work and their present scientific meanings date back to the mid 19th century. Early understandings of these ideas can be attributed to Gaspard-Gustave Coriolis who in 1829 published the paper titled Du Calcul de l'Effet des Machines outlining the mathematics of kinetic energy. William Thomson, later Lord Kelvin, is given the credit for coining the term kinetic energy c. 1849. ## Introduction There are various forms of energy : chemical energy, heat, electromagnetic radiation, potential energy (gravitational, electric, elastic, etc.), nuclear energy, rest energy. These can be categorized in two main classes: potential energy and kinetic energy. Kinetic energy can be best understood by examples that demonstrate how it is transformed from other forms of energy and to the other forms. For example, a cyclist will use chemical energy that was provided by food to accelerate a bicycle to a chosen speed. This speed can be maintained without further work, except to overcome air-resistance and friction. The energy has been converted into the energy of motion, known as kinetic energy but the process is not completely efficient and heat is also produced within the cyclist. The kinetic energy in the moving bicycle and the cyclist can be converted to other forms. For example, the cyclist could encounter a hill just high enough to coast up, so that the bicycle comes to a complete halt at the top. The kinetic energy has now largely been converted to gravitational potential energy that can be released by freewheeling down the other side of the hill. (Since the bicycle lost some of its energy to friction, it will never regain all of its speed without further pedaling. Note that the energy is not destroyed; it has only been converted to another form by friction.) Alternatively the cyclist could connect a dynamo to one of the wheels and also generate some electrical energy on the descent. The bicycle would be traveling more slowly at the bottom of the hill because some of the energy has been diverted into making electrical power. Another possibility would be for the cyclist to apply the brakes, in which case the kinetic energy would be dissipated through friction as heat energy. Like any physical quantity which is a function of velocity, the kinetic energy of an object does not depend only on the inner nature of that object. It also depends on the relationship between that object and the observer (in physics an observer is formally defined by a particular class of coordinate system called an inertial reference frame). Physical quantities like this are said to be not invariant. The kinetic energy is co-located with the object and contributes to its gravitational field. ### Calculations There are several different equations that may be used to calculate the kinetic energy of an object. In many cases they give almost the same answer to well within measurable accuracy. Where they differ, the choice of which to use is determined by the velocity of the body or its size. Thus, if the object is moving at a velocity much smaller than the speed of light, the Newtonian (classical) mechanics will be sufficiently accurate; but if the speed is comparable to the speed of light, relativity starts to make significant differences to the result and should be used. If the size of the object is sub-atomic, the quantum mechanical equation is most appropriate. ## Newtonian kinetic energy ### Kinetic energy of rigid bodies In classical mechanics, the kinetic energy of a "point object" (a body so small that its size can be ignored), or a non rotating rigid body, is given by the equation $E_k = \begin{matrix} \frac{1}{2} \end{matrix} mv^2$ where m is the mass and v is the speed of the body. For example - one would calculate the kinetic energy of an 80 kg mass traveling at 18 meters per second (40 mph) as $\begin{matrix} \frac{1}{2} \end{matrix} \cdot 80 \cdot 18^2 = 12,960 \ \mathrm{joules}$. Note that the kinetic energy increases with the square of the speed. This means, for example, that an object traveling twice as fast will have four times as much kinetic energy. As a result of this, a car traveling twice as fast requires four times as much distance to stop (assuming a constant braking force. See mechanical work). Thus, the kinetic energy can be calculated using the formula: $E_k = \frac{1}{2}mv^2$ where: Ek is the kinetic energy in joules m is the mass in kilograms, and v is the speed in meters per second. For the translational kinetic energy of a body with constant mass m, whose center of mass is moving in a straight line with speed v, as seen above is equal to $E_t = \begin{matrix} \frac{1}{2} \end{matrix} mv^2$ where: m is mass of the body v is speed of the center of mass of the body. Thus kinetic energy is a relative measure and no object can be said to have a unique kinetic energy. A rocket engine could be seen to transfer its energy to the rocket ship or to the exhaust stream depending upon the chosen frame of reference. But the total energy of the system, i.e. kinetic energy, fuel chemical energy, heat energy etc., will be conserved regardless of the choice of measurement frame. The kinetic energy of an object is related to its momentum by the equation: $E_k = \frac{p^2}{2m}$ ### Derivation and definition The work done accelerating a particle during the infinitesimal time interval dt is given by the dot product of force and displacement: $\mathbf{F} \cdot d \mathbf{x} = \mathbf{F} \cdot \mathbf{v} d t = \frac{d \mathbf{p}}{d t} \cdot \mathbf{v} d t = \mathbf{v} \cdot d \mathbf{p} = \mathbf{v} \cdot d (m \mathbf{v})$ Applying the product rule we see that: $d(\mathbf{v} \cdot \mathbf{v}) = (d \mathbf{v}) \cdot \mathbf{v} + \mathbf{v} \cdot (d \mathbf{v}) = 2(\mathbf{v} \cdot d\mathbf{v})$ Therefore (assuming constant mass), the following can be seen: $\mathbf{v} \cdot d (m \mathbf{v}) = \frac{m}{2} d (\mathbf{v} \cdot \mathbf{v}) = \frac{m}{2} d v^2 = d \left(\frac{m v^2}{2}\right)$ Since this is a total differential (that is, it only depends on the final state, not how the particle got there), we can integrate it and call the result kinetic energy: $E_k = \int \mathbf{F} \cdot d \mathbf{x} = \int \mathbf{v} \cdot d \mathbf{p}= \frac{m v^2}{2}$ This equation states that the kinetic energy (Ek) is equal to the integral of the dot product of the velocity (v) of a body and the infinitesimal change of the body's momentum (p). It is assumed that the body starts with no kinetic energy when it is at rest (motionless). ### Kinetic energy of systems For a single point, or a rigid body that is not rotating, the kinetic energy goes to zero when the body stops. However, for systems containing multiple independently moving bodies, which may exert forces between themselves, and may (or may not) be rotating; this is no longer true. This energy is called 'internal energy' The kinetic energy of the system at any instant in time is simply the sum of the kinetic energies of the masses- including the kinetic energy due to the rotations. An example would be the solar system. In the center of mass frame of the solar system, the Sun is (almost) stationary, but the planets and planetoids are in motion about it. Thus even in a stationary center of mass frame, there is still kinetic energy present. However, recalculating the energy from different frames would be tedious, but there is a trick. The kinetic energy of the system from a different inertial frame can be calculated simply from the sum of the kinetic energy in the center of mass frame and adding on the energy that the total mass of bodies in the center of mass frame would have if it were moving at the relative speed between the two frames. This may be simply shown: let V be the relative speed of the frame k from the center of mass frame i : $E_k = \int \frac{v_k^2 dm}{2} = \int \frac{(v_i + V)^2 dm}{2} = \int \frac{(v_i^2 + 2 v_i V + V^2) dm}{2} = \int \frac{v_i^2 dm}{2} + V \int v_i dm + \frac{V^2}{2} \int dm$ However, let $\int \frac{v_i^2 dm}{2} = E_i$ the kinetic energy in the center of mass frame, $\int v_i dm$ would be simply the total momentum which is by definition zero in the center of mass frame, and let the total mass: $\int dm = M$. Substituting, we get:[1] $E_k = E_i + \frac{M V^2}{2}$ The kinetic energy of a system thus depends on the inertial frame of reference and it is lowest with respect to the center of mass reference frame, i.e., in a frame of reference in which the center of mass is stationary. In any other frame of reference there is an additional kinetic energy corresponding to the total mass moving at the speed of the center of mass. ### Rotating bodies If a rigid body is rotating about any line through the center of mass then it has rotational kinetic energy ($E_r$) which is simply the sum of the kinetic energies of its moving parts, and thus it is equal to: $E_r = \int \frac{v^2 dm}{2} = \int \frac{(r \omega)^2 dm}{2} = \frac{\omega^2}{2} \int{r^2}dm = \frac{\omega^2}{2} I = \begin{matrix} \frac{1}{2} \end{matrix} I \omega^2$ where: ω is the body's angular velocity. r is the distance of any mass dm from that line I is the body's moment of inertia$= \int{r^2}dm$ (In this equation the moment of inertia must be taken about an axis through the center of mass and the rotation measured by ω must be around that axis; more general equations exist for systems where the object is subject to wobble due to its eccentric shape). ### Rotation in systems It sometimes is convenient to split the total kinetic energy of a body into the sum of the body's center-of-mass translational kinetic energy and the energy of rotation around the center of mass rotational energy: $E_k = E_t + E_r \,$ where: Ek is the total kinetic energy Et is the translational kinetic energy Er is the rotational energy or angular kinetic energy in the rest frame Thus the kinetic energy of a tennis ball in flight is the kinetic energy due to its rotation, plus the kinetic energy due to its translation. ## Relativistic kinetic energy of rigid bodies In special relativity, we must change the expression for linear momentum. Integrating by parts, we get: $E_k = \int \mathbf{v} \cdot d \mathbf{p}= \int \mathbf{v} \cdot d (m \gamma \mathbf{v}) = m \gamma \mathbf{v} \cdot \mathbf{v} - \int m \gamma \mathbf{v} \cdot d \mathbf{v} = m \gamma v^2 - \frac{m}{2} \int \gamma d (v^2)$ Remembering that $\gamma = (1 - v^2/c^2)^{-1/2}\!$, we get: $E_k = m \gamma v^2 - \frac{- m c^2}{2} \int \gamma d (1 - v^2/c^2) = m \gamma v^2 + m c^2 (1 - v^2/c^2)^{1/2} + C$ And thus: $E_k = m \gamma (v^2 + c^2 (1 - v^2/c^2)) + C = m \gamma (v^2 + c^2 - v^2) + C = m \gamma c^2 + C\!$ The constant of integration is found by observing that $\gamma = 1\!$ when $\mathbf{v }= 0$, so we get the usual formula: $E_k = m \gamma c^2 - m c^2 = \frac{m c^2}{\sqrt{1 - v^2/c^2}} - m c^2$ If a body's speed is a significant fraction of the speed of light, it is necessary to use relativistic mechanics (the theory of relativity as expounded by Albert Einstein) to calculate its kinetic energy. For a relativistic object the momentum p is equal to: $p = \frac{m v}{\sqrt{1 - (v/c)^2}}$, where m is the rest mass, v is the object's speed, and c is the speed of light in vacuum. Thus the work expended accelerating an object from rest to a relativistic speed is: $E_k = \frac{m c^2}{\sqrt{1 - (v/c)^2}} - m c^2$. The equation shows that the energy of an object approaches infinity as the velocity v approaches the speed of light c, thus it is impossible to accelerate an object across this boundary. The mathematical by-product of this calculation is the mass-energy equivalence formula—the body at rest must have energy content equal to: $E_\mbox{rest} = m c^2 \!$ At a low speed (v<<c), the relativistic kinetic energy may be approximated well by the classical kinetic energy. This is done by binomial approximation. Indeed, taking Taylor expansion for square root and keeping first two terms we get: $E_k \approx m c^2 \left(1 + \frac{1}{2} v^2/c^2\right) - m c^2 = \frac{1}{2} m v^2$, So, the total energy E can be partitioned into the energy of the rest mass plus the traditional Newtonian kinetic energy at low speeds. When objects move at a speed much slower than light (e.g. in everyday phenomena on Earth), the first two terms of the series predominate. The next term in the approximation is small for low speeds, and can be found by extending the expansion into a Taylor series by one more term: $E \approx m c^2 \left(1 + \frac{1}{2} v^2/c^2 + \frac{3}{8} v^4/c^4\right) = m c^2 + \frac{1}{2} m v^2 + \frac{3}{8} m v^4/c^2$. For example, for a speed of 10 km/s the correction to the Newtonian kinetic energy is 0.07 J/kg (on a Newtonian kinetic energy of 50 MJ/kg) and for a speed of 100 km/s it is 710 J/kg (on a Newtonian kinetic energy of 5 GJ/kg), etc. For higher speeds, the formula for the relativistic kinetic energy [2] is derived by simply subtracting the rest mass energy from the total energy: $E_k = m \gamma c^2 - m c^2 = m c^2\left(\frac{1}{\sqrt{1 - (v/c)^2}} - 1\right)$. The relation between kinetic energy and momentum is more complicated in this case, and is given by the equation: $E_k = \sqrt{p^2 c^2 + m^2 c^4} - m c^2$. This can also be expanded as a Taylor series, the first term of which is the simple expression from Newtonian mechanics. What this suggests is that the formulas for energy and momentum are not special and axiomatic, but rather concepts which emerge from the equation of mass with energy and the principles of relativity. ## Quantum mechanical kinetic energy of rigid bodies In the realm of quantum mechanics, the expectation value of the electron kinetic energy, $\langle\hat{T}\rangle$, for a system of electrons described by the wave function $\vert\psi\rangle$ is a sum of 1-electron operator expectation values: $\langle\hat{T}\rangle = -\frac{\hbar^2}{2 m_e}\bigg\langle\psi \bigg\vert \sum_{i=1}^N \nabla^2_i \bigg\vert \psi \bigg\rangle$ where $m_e$ is the mass of the electron and $\nabla^2_i$ is the Laplacian operator acting upon the coordinates of the ith electron and the summation runs over all electrons. Notice that this is the quantized version of the non-relativistic expression for kinetic energy in terms of momentum: $E_k = \frac{p^2}{2m}$ The [density functional formalism of quantum mechanics requires knowledge of the electron density only, i.e., it formally does not require knowledge of the wavefunction. Given an electron density $\rho(\mathbf{r})$, the exact N-electron kinetic energy functional is unknown; however, for the specific case of a 1-electron system, the kinetic energy can be written as $T[\rho] = \frac{1}{8} \int \frac{ \nabla \rho(\mathbf{r}) \cdot \nabla \rho(\mathbf{r}) }{ \rho(\mathbf{r}) } d^3r$ where $T[\rho]$ is known as the Von Weizsacker kinetic energy functional. ## Some examples Spacecraft use chemical energy to take off and gain considerable kinetic energy to reach orbital velocity. This kinetic energy gained during launch will remain constant while in orbit because there is almost no friction. However it becomes apparent at re-entry when the kinetic energy is converted to heat. Kinetic energy can be passed from one object to another. In the game of billiards, the player gives kinetic energy to the cue ball by striking it with the cue stick. If the cue ball collides with another ball, it will slow down dramatically and the ball it collided with will accelerate to a speed as the kinetic energy is passed on to it. Collisions in billiards are effectively elastic collisions, where kinetic energy is preserved. Flywheels are being developed as a method of energy storage (see article flywheel energy storage). This illustrates that kinetic energy can also be rotational. Note the formula in the articles on flywheels for calculating rotational kinetic energy is different, though analogous. ## Notes 1. In Einstein's original Über die spezielle und die allgemeine Relativitätstheorie (Zu Seite 41) and in most translations (e.g. Relativity - The Special and General Theory) kinetic energy is defined as $m c^2 / \sqrt{1 - v^2/c^2}$. ## References • Serway, Raymond A.; Jewett, John W. (2004). Physics for Scientists and Engineers (6th ed. ed.). Brooks/Cole. ISBN 0-534-40842-7. • Tipler, Paul (2004). Physics for Scientists and Engineers: Mechanics, Oscillations and Waves, Thermodynamics (5th ed. ed.). W. H. Freeman. ISBN 0-7167-0809-4. • Tipler, Paul; Llewellyn, Ralph (2002). Modern Physics (4th ed. ed.). W. H. Freeman. ISBN 0-7167-4345-0. • School of Mathematics and Statistics, University of St Andrews (2000). "Biography of Gaspard-Gustave de Coriolis (1792-1843)". Retrieved 2006-03-03. • Oxford Dictionary, Oxford Dictionary 1998	{"extraction_info": {"found_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": 42, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.919903576374054, "perplexity_flag": "head"}
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http://electronics.stackexchange.com/questions/34482/optocoupler-relay-type-for-isolating-two-dc-circuits	# Optocoupler/relay type for isolating two DC circuits I have recently started experimenting with an Arduino and would like to control a seperate circuit (for example, a small and cheap digital video camera powered by battery) from my arduino. Basically my plan is to solder two wires to either side of a tact switch on the target device's PCB. Since I am new to the world of electronics and do not wish to risk any possible damage to my Arduino, I would like to isolate the two circuits from one another. I have done some research to figure out how to switch one circuit from the other and the options I have come across are relays and optocouplers. I think I prefer the idea of using an optocoupler because they are small and completely separate the two circuits. I can also order some free samples to try them out. My understanding of optocouplers is that I would not need to connect the ground of the arduino to the ground of my target device (operating on it's own battery source). Is this correct? Also, considering I plan on using this to control only low voltage DC circuits, what would be a good choice of optocoupler? I plan on ordering several so that I can use them for various breadboard experiments where needed. Thanks - I have a similar question regarding the use of an Arduino board to measure voltage of a DC-powersupply without sharing grounds. So there should be a galvanic separation between the circuitry of the Arduino and the 48VDC-powersupply. A friend of mine told me to use the LM321 (Voltage to frequency converter) but it can't handle a voltage in excess of 40 VDC. The next problem is, that if the LM321 circuitry uses the same powersupply as the supply it needs to probe, the frequency-value doesn't change when the voltage drops. So how am I going to monitor the 48VDC powersupply with an Arduino, withou – Miech Mar 15 at 20:52 @Miech welcome to the StackExchange community! This site is not a forum, and if you have a question you should use the button on the homepage. Comments (like this) can be used to ask or provide more information about questions and answers. – clabacchio♦ Mar 15 at 22:00 ## 4 Answers If the circuit on the target device does not require too much current to trigger, using an optocoupler will probably work fine. Here's the pinout of a fairly typical optocoupler - 4N35: Arduino gets connected to the pins on the left side, the target device is connected to the pins on the right side. You are correct - the two devices should not share a common ground. On the Arduino side, wiring and controlling the optocoupler is as simple as wiring and flashing a LED. You just need to place a current-limiting resistor in series with either the anode or the cathode. For example: PIN -> R_limit -> Anode ... Cathode -> GND. Resistor values between 330 ohm and 1k ohm should be OK. At the target device, the collector pin is connected to the node with higher voltage and the emitter pin is connected to the node with lower voltage. When the optocoupler is activated, the transistor at its output starts conducting, effectively closing the circuit. - 1 You don't mention a pull-up resistor for the transistor. And why should resistor values between 330 ohm and 1k ohm be OK? Did you calculate it, which is something you have to do for optocouplers? The word is CTR. – stevenvh Jun 25 '12 at 15:13 Just some practical values, which are likely to work, without delving into too much calculations. R=330 ohm gives a forward current of ~10mA and an output current of similar value. The optocoupler is to bypass an existing push-button. If the target circuit needed a pull-up, it would already have one in place, I'd presume. – DimKo Jun 25 '12 at 16:50 1 If you use the 1k you have 4 mA LED current, and then CTR is only 40 %, so it will only sink 1.6 mA. There probably is a pull-up, but that may be a high value, and the transistor's leakage current may cause a too high voltage drop on low supply voltages. – stevenvh Jun 25 '12 at 16:59 1 – stevenvh Jun 25 '12 at 17:28 1 What's improper in my answer? Judging from an extensive experience as a novice in electronics, I'd say that most of the time when dealing with a new component, it's more important to first get it working in a simple way, without being overwhelmed by tons and tons of parameters, graphs and curves. The OP has already done some research, he can further dig in the details, little by little, if/when he needs to. – DimKo Jun 25 '12 at 17:43 show 2 more comments Optocouplers are not designed to both conduct significant current and (while doing so) drop a very low voltage. I suggest you opt for a relay. There are several topics here on driving a relay from a microcontroller pin, or you can use an existing relay shield. - Optoisolators are fine for switching low current, low voltage DC circuits. However, if you're thinking of supplying the power for the camera or other device via the optocoupler, I think you'd be better with a relay - it will provide isolation between the circuits and the output is just a mechanical switch with an easily understandable current/voltage rating. In any case, I think you should read up on what "isolation" means, if you're interconnecting any kind of mains powered equipment, even if they have a low voltage output. Both relays and optoisolators have limits on how much isolation they can safely provide. - I only plan on switching a button on the camera, which is DC. Basically I want to simulate the action of pressing a button on the camera by closing the circuit. – user1092697 Jun 25 '12 at 3:07 For switching a button on a camera I would opt for the relay: just like a button it can switch anything and has a low on-resistance. You don't have to worry if $V_{CE(SAT)}$ is low enough, collector dark current too high, and CTR high enough, things which need attention when using an optocoupler. This reed relay can be controlled directly from the Arduino's output (needs only 5 mA). - lang-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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9311240911483765, "perplexity_flag": "middle"}
http://quant.stackexchange.com/questions/1118/correct-way-to-find-the-mean-of-annual-geometric-returns-of-monthly-returns?answertab=votes	# Correct way to find the mean of annual geometric returns of monthly returns? Say I'm given I set of monthly returns over 10 years on a monthly basis. What is the correct way to find the geometric returns of this data? I ask because a classmate and I are on different sides of the coin. I found the cumulative returns of each year, then found the geometric mean of the 10 years. He found the cumulative returns of the entire time period, then took the (months in a year / total months) root of that data. The numbers turn out to be very close, but if I remember correctly mine are slightly lower over all the funds we measured. Or is there a different way and we're both wrong? - 1 Did you mean he took the (months in a year / total months) power? Taking the root with that would lead to a huge number. – chrisaycock♦ May 4 '11 at 22:49 ## 3 Answers If I understand you correctly, your question is whether this is true: \begin{equation} \sqrt[10]{\prod_{i=1}^{10}{Y_i}} < \sqrt[10]{A} \end{equation} where $Y$ is the yearly cumulative returns (your method), and $A$ is the absolute cumulative return (your classmate's method). The question then becomes whether you find this relationship: \begin{equation} \prod_{i=1}^{10}{Y_i} < A \end{equation} But that can't be! The absolute cumulative return must be equal to the product of the yearly cumulative returns. So if your yearly returns don't multiply to be his absolute return, then one of you has made a mistake. If you believe that your and his math are both correct, then the culprit is most likely a rounding error. - @chrisaycock already gave you a correct answer, but I thought I would add a more verbose version (and practice some MathJax by the way). In fact when I began answering I thought it was going to be a straightforward answer, but having spent some more time with this question I see there are some potential traps you can fall in. Especially since some of the steps you name are not 100% clear, I assumed the worst-case scenario (AKA everything wrong). I suppose some of them are just shorthand notions. Sorry if you already do it the right way and it's obvious it's wrong making my explanations ridiculous, but at least one of the steps is to blame as you are getting different results. • `Say I'm given a set of monthly returns over 10 years on a monthly basis.` Let's call them $$r_{1_{jan}}, \ ...,\ r_{1_{dec}}, \ ...,\ r_{10_{jan}}, \ ...,\ r_{10_{dec}} \ [eq. 1]$$ What you do is: • `I found the cumulative returns of each year` Your cumulative return for a year is a product of monthly returns: $$R_{i} = (1+r_{i_{jan}}) * \ ... \ * (1+r_{i_{dec}}) - 1 \ [eq. 2]$$ OK, straightforward. Not that many options here. • `then found the geometric mean of the 10 years` if you mean that literally (I warned you I would take the worst approach possible, sorry), as in found the geometric mean of those 10 returns: $$R_{G} = \sqrt[10]{R_{1} * R_{2} * \ ... \ * R_{10}} \ [eq. 3]$$ we have our first problem. While technically you can calculate anything (as long as it's not negative), it doesn't make sense. We are looking for a geometric average rate of return instead: $$R_{G} = \sqrt[10]{(1 + R_{1}) * (1 + R_{2}) * \ ... \ * (1 + R_{10})} - 1 \ [eq. 4]$$ OK, done, should be the correct answer. • `He found the cumulative returns of the entire time period,` He calculated it either this way: $$AR = (1+r_{1_{jan}}) * \ ... \ * (1+r_{1_{dec}}) * \ ... \ * (1+r_{10_{jan}}) * \ ... \ * (1+r_{10_{dec}}) - 1 \ [eq. 5]$$ or just used $\frac{P_{last}}{P_{first}} - 1$ which is the same. No problem here. • `then took the (months in a year / total months) root of that data.` First assumption - I suppose you meant power here (or `total months / months in a year` root), because otherwise it wouldn't make much sense. Now, if we literally take the root out of our accumulated returns ($AR$): $$\sqrt[\frac{120}{12}]{AR} = \sqrt[10]{(1+r_{1_{jan}}) * \ ... \ * (1+r_{1_{dec}}) * \ ... \ * (1+r_{10_{jan}}) * \ ... \ * (1+r_{10_{dec}}) - 1} \ [eq. 6]$$ using $[eq. 2]$ we get: $$= \sqrt[10]{(1+R_{1})(1+R_{2}) \ ... \ * (1+R_{10}) - 1}$$ Oops, seems similar to $[eq. 4]$, but it's not the same. We did something wrong. In fact we wanted it this way (remembering that we're looking for annual returns): $$R_{G} = \sqrt[10]{1 + AR} - 1 \ [eq. 7]$$ Now plugging $[eq. 5]$ and $[eq. 2]$: $$= \sqrt[10]{1 + (1+R_{1})(1+R_{2}) \ ... \ * (1+R_{10}) - 1} -1$$ $$= \sqrt[10]{(1+R_{1})(1+R_{2}) \ ... \ * (1+R_{10})} - 1$$ and this is the same as $[eq. 4]$ This way you see that both methods should give equivalent results. If not, then either it's a calculation mistake/rounding issue or you're using different methods and someone is not calculating an actual geometric average rate of return. I hope now you can find where the issue was. - Hmm, I see there's no easy way to deal with long MathJax lines. And there are some issues with multi-line equations too, which I will report on meta. But it was a nice primer on using LaTeX on SE. I didn't realize I forgot that much syntax. ;-) – Karol Piczak May 5 '11 at 12:31 I assume you have net simple montly returns. 12 months and 10 years gives you 120 monthly returns $r_1, r_2,...,r_{120}$. You want to know the annual geometric return. Then solve for $r_g:$ $$(1+r_1)\times(1+r_2)\times \dots \times(1+r_{120})=(1+r_g)^{10}$$ The order of the multiplication on the LHS is important, that is, you should start multiplying with the oldest return ($r_1$). - 2 I don't think in this case your LHS ordering is important. It's simple multiplication only. – Karol Piczak May 5 '11 at 12:33 You are right of course. – Dmitrii I. May 5 '11 at 13:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 11, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9411337971687317, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/5707/characteristics-for-2nd-order-differential-equations?answertab=active	# Characteristics for 2nd order differential equations If I have an equation $p(x)\frac{\partial^2u}{\partial x^2} + r(x)\frac{\partial^2u}{\partial x\partial y} + q(x)\frac{\partial^2 u}{\partial y^2}=f(x,y,u)$ Where $f$ maybe contains first partial derivatives for $u$. Can anyone give me a worked example of how to solve this using the method of characteristics. Thanks in advance. - ## 1 Answer See these lecture notes on the method of characteristics for second-order PDEs, where the method is applied to steady isentropic flow (gasdynamics) -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9213210940361023, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/226376/inverse-fourier-transform-of-fx-frac-textux-sqrt1-x2/226399	Inverse fourier transform of $f(x)=\frac{\text{u}(x)}{\sqrt{1-x^2}}$ How to calculate (or derive) the inverse Fourier transform of $$f(x)=\frac{\text{u}(x)}{\sqrt{1-x^2}}$$ where $u(x)$ is the rectangular function? I know that $f(x)$ is the Fourier transform of the Bessel function, so its inverse must be the Bessel function, but how to show? - Well, have you tried computing $g(k) = \tfrac{1}{{\sqrt {2\pi } }}\int_{ - \infty }^\infty {f(x){e^{ixk}}dx}$? – glebovg Oct 31 '12 at 23:21 1 Answer I assume "the rectangular function" is the indicator function of the interval $[-1,1]$. Thus you want to calculate $$F(s) = \int_{-1}^1 \frac{e^{isx}}{\sqrt{1-x^2}} \ dx$$ Expand it in a power series: $e^{isx} = \sum_{n=0}^\infty (isx)^n/n!$ (converging uniformly on this interval for any fixed $s$). Since $1/\sqrt{1-x^2}$ is integrable, we can interchange integral and sum. Now let $$L_n = \int_{-1}^{1} \frac{x^n \ dx}{\sqrt{1-x^2}}$$ By symmetry, $L_n = 0$ for odd $n$. So now assume $n = 2m$ is even. The substitution $x = \cos(t)$ gives us $$L_{2m} = \int_0^{\pi} \cos(t)^{2m}\ dt = \frac{1}{2} \int_0^{2\pi} \left(\dfrac{e^{it} + e^{-it}}{2}\right)^{2m}\ dt = 2^{-2m-1} \sum_{j=0}^{2m} {2m \choose j} \int_0^{2\pi} e^{2(j-m)it} \ dt$$ All the integrals on the right side are $0$ except for the case $j=m$. Thus we get $$L_{2m} = 2^{-2m} \pi {2m \choose m} = 2^{-2m} \pi \dfrac{(2m)!}{(m!)^2}$$ We conclude that $$F(s) = \sum_{m=0}^\infty 2^{-2m} \pi \frac{(-s^2)^{m}}{(m!)^2} = \pi J_0(s)$$ -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8614902496337891, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/190905-dimension-basis-solution-set-homogeneous-system.html	Thread: 1. Dimension of and Basis for Solution Set of Homogeneous System This is probably pretty simple I just want to make sure I'm doing it right: Find the dimension of and a basis for the solution set of the following homogeneous systems of equations: 1. x1+2x2-3x3+x4=0 2.x1+2x2=0 x1-x2=0 Any help would be appreciated! Thanks! 2. Re: Dimension of and Basis for Solution Set of Homogeneous System Originally Posted by divinelogos Find the dimension of and a basis for the solution set of the following homogeneous systems of equations: 1. x1+2x2-3x3+x4=0 Solving the system: $\begin{bmatrix}{x_1}\\{x_2}\\{x_3}\\{x_4} \end{bmatrix}=\begin{bmatrix}{-2\lambda+2\mu-\gamma}\\{\lambda}\\{\mu}\\{\gamma}\end{bmatrix}= \lambda \begin{bmatrix}{-2}\\{1}\\{0}\\{0}\end{bmatrix}+\mu \begin{bmatrix}{2}\\{0}\\{1}\\{0}\end{bmatrix}+ \gamma \begin{bmatrix}{-1}\\{0}\\{0}\\{1}\end{bmatrix}$ Those three vectors span the subspace and are linearly independent. Try 2.	{"extraction_info": {"found_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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8879530429840088, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/116282/limits-at-infinity-division-by-highest-degree	# Limits at infinity & Division by highest degree I believe to solve expressions like: $$\lim_{x\to \infty} (2x^2 + 1)$$ is: $\lim_{x \to \infty} (2 + \frac{1}{x^2}) = 2+0=2$ Now for: $$\lim_{n \to \infty} \frac{-1}{\sqrt{n} + 2}$$ The answer given is $\frac{-1}{\infty}=0$. What I did was $$\lim_{x \to \infty} \frac{-1}{1 + \frac{2}{\sqrt{n}}}=\frac{-1}{1}=-1$$ Whats wrong? - 2 $2x^2 + 1 \neq (2+1/x^2)$, but $2x^2 + 1 = x^2 (2 + 1/x^2)$. Same thing for the second limit: you're forgetting the term you're factoring for. – Andy Mar 4 '12 at 10:54 1 We solve equations; we evaluate expressions. – Gerry Myerson Mar 4 '12 at 11:19 Yeah jiew be careful with the algebraic manipulation. That thinking leaded you to $-1=0$ or $1=0$ which implies every number $x=0$ . – checkmath Mar 4 '12 at 13:40 @Jiew Meng: It is useful to look before calculating. For example, it is clear that when $n$ is big, the number $\frac{-1}{\sqrt{n}+2}$ is negative but close to $0$. – André Nicolas Mar 4 '12 at 14:55 ## 2 Answers The rule you are referring to in your title "divide by the highest power" means that you divide every term in the given "rational expression" by $x^k$ where $k$ is the highest power of $x$ in the entire limit expression. This includes terms in both the numerator and the denominator. If no denominator is explicitly shown, then the denominator is 1. For example, you could apply this rule to $$\lim_{x\rightarrow\infty} (x^2+1)=\lim_{x\rightarrow\infty} {x^2+1\over 1}$$ to obtain $$\tag{0}\lim_{x\rightarrow\infty} {x^2+1\over1} =\lim_{x\rightarrow\infty} {1+{1\over x^2}\over{ 1\over x^2}} = \lim_{x\rightarrow\infty}x^2 (1+\textstyle{1\over x^2} ).$$ The important observation is: Note that you are just algebraically manipulating the limit expression; but, you cannot change the limit expression, as you did when you replaced $x^2+1$ by non-equivalent expression $1+{1\over x^2}$. You can however replace $x^2+1$ by the equivalent expression (as far as taking limits is concerned) $x^2(1+{1\over x^2})$, as was done above. Dividing everything in the limit expression in the left hand side of $(0)$ by $x^2$ gives an equivalent expression because this amounts to multiplying both top and bottom by the same quantity (namely $1/x^2$). I will also caution here that this rule is applicable in the specific scenario where $\ \$1) you are taking a limit at infinity or minus ininifty and $\ \$2) the limit expression is of the form $P(x)/Q(x)$ where both $p$ and $Q$ are linear combinations of powers of $x$ and the largest power of $x$ appearing in the entire expression is nonnegative. In this case, the rule says to: find the highest power of $x$ in the entire expression and then divide both $P(x)$ and $Q(x)$ by $x^k$. Note that this will lead to an equivalent expression since it amounts to multiplying bot top and bottom by the same quantity. Moreover, after doing this and simplifying, the limit will be easily determined. Let's pause a second. There really is no reason to use this "divide by the highest power rule" when evaluating $$\lim_{x\rightarrow\infty} (x^2+1),$$ as this limit is clearly infinite. However, if you had to find $$\tag{1} \lim_{x\rightarrow\infty} (x^2-x^3)$$ then things aren't so clear. Here, you could use the "divide by the highest power rule" (and note then, there is a 1 downstairs); but I think it's more clear to instead (and equivalently) factor the highest power of $x$ in the entire expression from both top and bottom (if there is one) and cancel (if you can). So for $(1)$, we would write: $$\lim_{x\rightarrow\infty} (x^2-x^3)= \lim_{x\rightarrow\infty} x^3 (\textstyle{1\over x}-1)$$ And now we can see that the limit is $-\infty$. Or, consider the following example: $$\tag{2} \lim_{x\rightarrow\infty} {2x^3-x+2\over 3x^3-2x+1}= \lim_{x\rightarrow\infty} {x^3(2-{1\over x^2}+{2\over x^3})\over x^3( 3-{2\over x^2}+{1\over x^3})}= \lim_{x\rightarrow\infty} { 2-{1\over x^2}+{2\over x^3} \over 3-{2\over x^2}+{1\over x^3} } ={2\over3}.$$ Note that the last limit in $(2)$ could have been obtained by "dividing everything by $x^3$"; but it's of utmost importance that you divide all terms in both the numerator and denominator by $x^3$. Again, you can't change the limit expression. For your second problem, dividing by $\sqrt n$ means you have to divide the $1$ in the numerator by $\sqrt n$ too $$\lim_{n\rightarrow\infty}{-1\over \sqrt n +2} =\lim_{n\rightarrow\infty}{-1/\sqrt n\over 1 +{2\over\sqrt n}} ={0\over 1}=0.$$ As with the first example, though, this is a bit much for this limit. The limit is clearly 0... As mentiond by Andy, the method is better suited for something like: $$\tag{3} \lim_{n\rightarrow\infty}{-n^2\over \sqrt n +n^3}.$$ Using the "divide trick", you would divide everything by $n^3$: $$\lim_{n\rightarrow\infty}{-n^2\over \sqrt n +n^3} =\lim_{n\rightarrow\infty}{-{n^2\over n^3}\over {\sqrt n\over n^3} +1} =\lim_{n\rightarrow\infty}{-{1\over n }\over {1\over n^{5/2}} +1} ={0\over 0+1}=0.$$ Equivalently, use the "factor method" described above: $$\lim_{n\rightarrow\infty}{-n^2\over \sqrt n +n^3} =\lim_{n\rightarrow\infty}{ n^3\cdot -{1\over n}\over n^3({1\over n^{5/2}}+1))} =\lim_{n\rightarrow\infty}{-{1\over n }\over {1\over n^{5/2}} +1} ={0\over 0+1}=0.$$ - It's usually hard for people -who are not seeing math on an every-day basis- to get that factoring is just dividing. Good job at conveying that! – Andy Mar 4 '12 at 15:54 What you are saying in your calculations is $2x^2 +1 = (2 + 1/x^2)$, which is wrong. What you want it $2x^2 +1 = x^2(2 + 1/x^2)$, thus $$\lim_{x \to \infty} 2x^2 +1 = \lim_{x \to \infty} x^2(2+ 1/x^2) = \lim_{x\to \infty} 2x^2 = \infty$$ for your second limit you are making the same mistake: $\sqrt{n} + 2 \neq 1+ 2/\sqrt{n}$, but $\sqrt{n} + 2 = \sqrt{n}(1+2/\sqrt{n})$. Thus, $$\lim_{x \to \infty} \frac{-1}{\sqrt{n}+2} = \lim_{x \to \infty} \frac{-1}{\sqrt{n}(1+2/\sqrt{n})} = \lim_{x \to \infty} \frac{-1}{\sqrt{n}} = 0$$ This method is usually better suited for indeterminate cases, like $\lim_{x \to \infty} \frac{2x^2 +1}{3x^2 + 4}$ - – Jiew Meng Mar 4 '12 at 15:08 @Jiew Meng: The numerator is $2x^2 + 4 = x^2(2+4/x^2)$ and in the denominator you factor $x^4$ inside both square roots, take it out to get $x^2$ (in both cases), then you factor again and simplify with the numerator. All this is done implicitly in your image. – Andy Mar 4 '12 at 15:17 To simplify, they divided top and bottom by $x^2$. – Mike Mar 4 '12 at 16:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 50, "mathjax_display_tex": 15, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9527772665023804, "perplexity_flag": "head"}
http://mathhelpforum.com/number-theory/25248-bernoulli-numbers.html	# Thread: 1. ## Bernoulli numbers Say we try to define the Bernoulli numbers using Faulhaber's formula... $\sum_{k=0}^{m-1}k^n=\frac{1}{1+n}\sum_{k=0}^n\binom{n+1}{k}B_km^ {n+1-k}$ how do we show that the values $\{B_n\}$ are uniquely defined? That is, in the coefficients of $m^{n+1}, m^n, m^{n-1}\cdots$, $\frac{B_0}{1+n}, B_1, B_2 \frac{n}{2}, \cdots$ the values $B_0, B_1, B_2, \cdots$ will always be the same regardless of the power $n$.	{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.6997509598731995, "perplexity_flag": "head"}
http://www.reference.com/browse/moving-coil+galvanometer	Related Searches Definitions # Galvanometer [gal-vuh-nom-i-ter] /ˌgælvəˈnɒmɪtər/ A galvanometer is a type of ammeter; an instrument for detecting and measuring electric current. It is an analog electromechanical transducer that produces a rotary deflection, through a limited arc, in response to electric current flowing through its coil. The term has expanded to include uses of the same mechanism in recording, positioning, and servomechanism equipment. ## History Deflection of a magnetic compass needle by current in a wire was first described by Hans Oersted in 1820. The phenomenon was studied both for its own sake and as a means of measuring electrical current. The earliest galvanometer was reported by Johann (Johan) Schweigger of Nuremberg at the University of Halle on 16 September 1820. André-Marie Ampère also contributed to its development. Early designs increased the effect of the magnetic field due to the current by using multiple turns of wire; the instruments were at first called "multipliers" due to this common design feature. The term "galvanometer", in common use by 1836, derives from the surname of Italian electricity researcher Luigi Galvani, who discovered that electric current could make a frog's leg jerk. Originally the instruments relied on the Earth's magnetic field to provide the restoring force for the compass needle; these were called "tangent" galvanometers and had to be oriented before use. Later instruments of the "astatic" type used opposing magnets to become independent of the Earth's field and would operate in any orientation. The most sensitive form, the Thompson or mirror galvanometer, was invented by William Thomson (Lord Kelvin). Instead of a compass needle, it used tiny magnets attached to a small lightweight mirror, suspended by a thread; the deflection of a beam of light greatly magnified the deflection due to small currents. Alternatively the deflection of the suspended magnets could be observed directly through a microscope. The ability to quantitatively measure voltage and current allowed Georg Ohm to formulate Ohm's Law, which states that the voltage across an element is directly proportional to the current through it. The early moving-magnet form of galvanometer had the disadvantage that it was affected by any magnets or iron masses near it, and its deflection was not linearly proportional to the current. In 1882 Jacques-Arsène d'Arsonval developed a form with a stationary permanent magnet and a moving coil of wire, suspended by coiled hair springs. The concentrated magnetic field and delicate suspension made these instruments sensitive and they could be mounted in any position. By 1888 Edward Weston had brought out a commercial form of this instrument, which became a standard component in electrical equipment. This design is almost universally used in moving-vane meters today. ## Operation The most familiar use is as an analog measuring instrument, often called a meter. It is used to measure the direct current (flow of electric charges) through an electric circuit. The D'Arsonval/Weston form used today is constructed with a small pivoting coil of wire in the field of a permanent magnet. The coil is attached to a thin pointer that traverses a calibrated scale. A tiny torsion spring pulls the coil and pointer to the zero position. When a direct current (DC) flows through the coil, the coil generates a magnetic field. This field acts against the permanent magnet. The coil twists, pushing against the spring, and moves the pointer. The hand points at a scale indicating the electric current. Careful design of the pole pieces ensures that the magnetic field is uniform, so that the angular deflection of the pointer is proportional to the current. A useful meter generally contains provision for damping the mechanical resonance of the moving coil and pointer, so that the pointer settles quickly to its position without oscillation. The basic sensitivity of a meter might be, for instance, 100 microamperes full scale (with a voltage drop of, say, 50 millivolts at full current). Such meters are often calibrated to read some other quantity that can be converted to a current of that magnitude. The use of current dividers, often called shunts, allows a meter to be calibrated to measure larger currents. A meter can be calibrated as a DC voltmeter if the resistance of the coil is known by calculating the voltage required to generate a full scale current. A meter can be configured to read other voltages by putting it in a voltage divider circuit. This is generally done by placing a resistor in series with the meter coil. A meter can be used to read resistance by placing it in series with a known voltage (a battery) and an adjustable resistor. In a preparatory step, the circuit is completed and the resistor adjusted to produce full scale deflection. When an unknown resistor is placed in series in the circuit the current will be less than full scale and an appropriately calibrated scale can display the value of the previously-unknown resistor. Because the pointer of the meter is usually a small distance above the scale of the meter, parallax error can occur when the operator attempts to read the scale line that "lines up" with the pointer. To counter this, some meters include a mirror along the markings of the principal scale. The accuracy of the reading from a mirrored scale is improved by positioning one's head while reading the scale so that the pointer and the reflection of the pointer are aligned; at this point, the operator's eye must be directly above the pointer and any parallax error has been minimized. ## Types Extremely sensitive measuring equipment once used mirror galvanometers that substituted a mirror for the pointer. A beam of light reflected from the mirror acted as a long, massless pointer. Such instruments were used as receivers for early trans-Atlantic telegraph systems, for instance. The moving beam of light could also be used to make a record on a moving photographic film, producing a graph of current versus time, in a device called an oscillograph. Galvanometer mechanisms are used to position the pens of analog chart recorders such as used for making an electrocardiogram. Strip chart recorders with galvanometer driven pens might have a full scale frequency response of 100 Hz and several centimeters deflection. In some cases (the classical polygraph of movies or the electroencephalograph), the galvanometer is strong enough to move the pen while it remains in contact with the paper; the writing mechanism may be a heated tip on the needle writing on heat-sensitive paper or a fluid-fed pen. In other cases (the Rustrak recorders), the needle is only intermittently pressed against the writing medium; at that moment, an impression is made and then the pressure is removed, allowing the needle to move to a new position and the cycle repeats. In this case, the galvanometer need not be especially strong. ### Tangent galvanometer A tangent galvanometer is an early measuring instrument used for the measurement of electric current. It works by using a compass needle to compare a magnetic field generated by the unknown current to the magnetic field of the Earth. It gets its name from its operating principle, the tangent law of magnetism, which states that the tangent of the angle a compass needle makes is proportional to the ratio of the strengths of the two perpendicular magnetic fields. It was first described by Claude Servais Mathias Pouillet in 1837. A tangent galvanometer consists of a coil of insulated copper wire wound on a circular non-magnetic frame. The frame is mounted vertically on a horizontal base provided with levelling screws. The coil can be rotated on a vertical axis passing through its centre. A compass box is mounted horizontally at the centre of a circular scale. It consists of a tiny, powerful magnetic needle pivoted at the centre of the coil. The magnetic needle is free to rotate in the horizontal plane. The circular scale is divided into four quadrants. Each quadrant is graduated from 0° to 90°. A long thin aluminium pointer is attached to the needle at its centre and at right angle to it. To avoid errors due to parallax a plane mirror is mounted below the compass needle. In operation, the instrument is first rotated until the magnetic field of the Earth, indicated by the compass needle, is parallel with the plane of the coil. Then the unknown current is applied to the coil. This creates a second magnetic field on the axis of the coil, perpendicular to the Earth's magnetic field. The compass needle responds to the vector sum of the two fields, and deflects to an angle equal to the tangent of the ratio of the two fields. From the angle read from the compass's scale, the current could be found from a table. The current supply wires have to be wound in a small helix, like a pig's tail, otherwise the field due to the wire will affect the compass needle and an incorrect reading will be obtained. #### Theory When current is passed through the tangent galvanometer a magnetic field is created at its corners given by $B=\left\{mu_0 nIover 2r\right\}$ where I is the current in ampere, n is the number of turns of the coil and r is the radius of the coil. If the galvanometer is set such that the plane of the coil is along the magnetic meridian i.e., B is perpendicular to $B_H$ ($B_H$ is the horizontal component of the Earth's magnetic field), the needle rests along the resultant. From tangent law, $B = B_H tantheta$, i.e. $\left\{mu_0 nIover 2r\right\} = B_H tantheta$ or $I=left\left(frac\left\{2rB_H\right\}\left\{mu_0 n\right\}right\right)tantheta$ or $I=K tantheta$, where K is called the Reduction Factor of the tangent galvanometer. The value of $theta$ is taken at 45 degrees for maximum accuracy. #### Geomagnetic field measurement A tangent galvanometer can also be used to measure the magnitude of the horizontal component of the geomagnetic field. When used in this way, a low-voltage power source, such as a battery, is connected in series with a rheostat, the galvanometer, and an ammeter. The galvanometer is first aligned so that the coil is parallel to the geomagnetic field, whose direction is indicated by the compass when there is no current through the coils. The battery is then connected and the rheostat is adjusted until the compass needle deflects 45 degrees from the geomagnetic field, indicating that the magnitude of the magnetic field at the center of the coil is the same as that of the horizontal component of the geomagnetic field. This field strength can be calculated from the current as measured by the ammeter, the number of turns of the coil, and the radius of the coils. ## Uses A major early use for galvanometers was for finding faults in telecommunications cables. They were superseded in this application late in the 20th century by time-domain reflectometers. Since the 1980s, galvanometer-type analog meter movements may be displaced by analog to digital converters (ADCs) for some uses. A digital panel meter (DPM) contains an analog to digital converter and numeric display. The advantages of a digital instrument are higher precision and accuracy, but factors such as power consumption or cost may still favor application of analog meter movements. Most modern uses for the galvanometer mechanism are in positioning and control systems. These are used in laser marking and projection, and in imaging application such as Optical Coherence Tomography (OCT) retinal scanning. Mirror galvanometer systems are used as beam positioning elements in laser optical systems. These are typically high power galvanometer mechanisms used with closed loop servo control systems. The newest generation of galvanometers designed for beam steering applications can have frequency responses over 10 kHz with appropriate servo technology. Examples of manufacturers of such systems are Cambridge Technology Inc. (www.camtech.com) and General Scanning (www.gsig.com). A galvanometer appeared in an episode of the television medical drama House to function as an electrocardiogram for a patient whose severe and extensive burns prevented use of the normal electrodes. ## External links • Galvanometer - Interactive Java Tutorial National High Magnetic Field Laboratory • Selection of historic galvanometer in the Virtual Laboratory of the Max Planck Institute for the History of Science	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 8, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9397522807121277, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/54144/self-avoiding-walk-pair-correlation	Self Avoiding Walk Pair Correlation Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $\gamma(i)$ be a self avoiding walk (SAW) on a 2D lattice $L$ (a square lattice for example) starting at a predefined origin ( $\gamma(0)=(0,0)$ ) and having length $n:=\ell(\gamma)$. Furthermore, suppose $\gamma$ ends at a point $u$, so that $\gamma(n)=u$. Let $N(u)$ be the set of neighboring vertices of $u$ (that are connected to $u$). If I were to further condition the SAW by clamping the second-to-last-step, i.e. $\gamma(n-1)=v$, what can be said about the probability that the walk visited any of the other neighbors of $u$. Naturally, the length of the SAW and the distance of $u$ from the origin will be interlocked in a tug of war. I am primarily interested in seeing when the probability of visiting adjacent neighbors tends to zero. - 1 Answer If the walks are sampled uniformly at random, then the probability of another neighbour of $u$ being visited is related to the "atmosphere" of a walk, and the connective constant $\mu$. The mean number of additional neighbours approaches $3-\mu \approx 0.361841469\cdots$ for SAWs on $Z^2$, while the probability of there being at least one additional neighbour approaches $1-0.71114 = 0.28886$ (from eq. 12, Owczarek and Prellberg, ref. below). A recent article gave probabilities for having 0, 1, 2, and 3 unoccupied neighbours, A L Owczarek and T Prellberg, "Scaling of the atmosphere of self-avoiding walks", J. Phys. A: Math. Theor. 41 (2008) 375004 (6pp). The most accurate estimate of $\mu$ for $Z^2$ comes from self-avoiding polygon enumerations by Iwan Jensen, "A parallel algorithm for the enumeration of self-avoiding polygons on the square lattice", J. Phys. A: Math. Gen. 36 (2003) 5731–5745, also on the arXiv: http://arXiv.org/abs/cond-mat/0301468v1 If you're wondering about the scaling behaviour as $\|u\|$ is varied, then I think that given $x = \|u\|/\langle R_e^2 \rangle^{1/2}$, the scaling function $P(x)$ for the probability of more than one neighbour being occupied must go to zero as $x \rightarrow \infty$, but I don't have any intuition yet as to how it will approach zero (I think this is a nice question). - Thanks! I didn't know atmosphere was the correct terminology. – Alex R. Feb 3 2011 at 1:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 24, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9250178337097168, "perplexity_flag": "head"}
http://meetings.aps.org/Meeting/HAW05/SessionIndex3/?SessionEventID=38116	# Bulletin of the American Physical Society ## 2005 2nd Joint Meeting of the Nuclear Physics Divisions of the APS and The Physical Society of Japan Sunday–Thursday, September 18–22, 2005; Maui, Hawaii ### Session KK: Nuclear Theory III Sponsoring Units: DNP JPS Chair: Thomas Luu, Los Alamos National Laboratory Room: Ritz-Carlton Hotel Maui Thursday, September 22, 2005 2:00PM - 2:15PM KK.00001: Critical Point Symmetry in A Fermion Monopole and Quadrupole Pairing Model Joseph N. Ginocchio Recent interest in symmetries at a critical point of phase transitions in nuclei prompts a revisit to the fermion monopole and quadrupole pairing model [1]. This model has an exactly solvable symmetry limit that is transitional between spherical nuclei and gamma unstable deformed nuclei. The eigenenergies, eigenfunctions, pairing strength and quadrupole transition rates in this limit are derived. Comparison with empirical quadrupole transition rates suggests that the Xenon isotopes may have this symmetry [2]. 1. Joseph N. Ginocchio, Ann. Phys. 126, 234 (1980). 2. Joseph N. Ginocchio, Phys. Rev. C (2005). Thursday, September 22, 2005 2:15PM - 2:30PM KK.00002: All-order core polarization for shell-model effective interactions Jason D. Holt , Jeremy Holt , T.T.S. Kuo , G.E. Brown , Scott Bogner Although core polarization, calculated to second-order in perturbation theory, has been successful in describing a range of nuclear observables, the effect of high-order diagrams has been a long-standing issue. In this talk we present an all-order summation of a large class of core polarization diagrams using the low-momentum NN interaction $V_{low-k}$. Our calculation, based on the elegant formalism of Kirson and Babu-Brown, involves solving a set of coupled non-linear equations in which the vertex functions are generated self-consistently. By using $V_{low-k}$, which is energy independent, and true Green functions in the particle-particle and particle-hole channels, we can simplify the solution and include a class of diagrams whose calculation has been previously intractable. We apply this procedure to the $sd$-shell effective interactions and find that the all-order calculation serves to mildly suppress the second order results, typically by less than 10\%. Thursday, September 22, 2005 2:30PM - 2:45PM KK.00003: Very-low Momentum Nucleon-Nucleon Interaction Based upon Chiral Perturbation Theory Ruprecht Machleidt , Luigi Coraggio , David Entem Recently, several groups have constructed low-momentuum nucleon-nucleon (NN) interactions that have become known as $V_{\rm low-k}$. One starts from a conventional high-momentum NN potential and applies renormalization group techniques that preserve the (half)-on-shell T-matrix to obtain a new potential that is charcterized by a low-momentum cutoff, typically around 2 fm$^{-1}$. The general justification for this proceedure comes from low-energy effective field theory (EFT). This fact suggests that there may be a more efficient way to construct a $V_{\rm low-k}$. Namely, instead of taking the detour through a high-momentum NN potential, one may as well construct a low-momentum potential from scratch---and this is what our contribution is about. We use chiral perturbation theory at next-to-next-to-next-to-leading order (N3LO) and apply a sharp cutoff at 2.1 fm$^{-1}$. This potential reproduces the NN phase shifts up to about 300 MeV lab energy and the deuteron properties. While the $V_{\rm low-k}$ constructed in the past allow only for a rather cumbersome numerical representation, our low-momentum potential is given in analytic form. Moreover, the low-energy constants are explicitly known such that the chiral three-nucleon forces consistent with our NN potential can be properly defined. Thursday, September 22, 2005 2:45PM - 3:00PM KK.00004: Fermi liquid theory and Kuo-Brown effective interactions Jeremy W. Holt , G.E. Brown , J.D. Holt , T.T.S. Kuo , S.K. Bogner We study the properties of nuclear matter using the low-momentum nucleon-nucleon interaction $V_{\rm low-k}$ and Landau's theory of normal Fermi liquids. The Landau $f$-function, which describes the quasiparticle-quasihole interaction at the Fermi surface, can be expanded in Legendre polynomials whose coefficients are directly related to the effective mass, symmetry energy, and compression modulus of nuclear matter. It is found that in the single-bubble approximation to the induced interaction of Babu and Brown, the compression modulus is much too repulsive compared with experiment. This is remedied by solving the Babu-Brown equation self-consistently using $V_{\rm low-k}$ as the driving term. The result is a reasonable agreement with experiment, both for the compression modulus and the remaining Fermi liquid parameters. In addition, we discuss the effect of high-order direct and exchange terms in the quasiparticle scattering amplitude. Thursday, September 22, 2005 3:00PM - 3:15PM KK.00005: Shell Model Analysis of the $^{56}$Ni Spectrum in the Full $pf$ Model Space Mihai Horoi , B. Alex Brown , T. Otsuka , M. Honma , T. Mizusaki We present a full $pf$-shell spectroscopy of the low-lying states of $^{56}$Ni using the GXPF1A interaction.[M. Honma et al., Proceedings of ENAM, 2004] Both, the ground state band and the first deformed band, as well as the transition probabilities compares favorably with the experimental data.[D. Rudolpf et al., Phys. Rev. Lett. {\bf 82}, 5763 (1999)] We analyze the significance of the $np-nh$ contributions to the full model calculations, similar to the analysis done for $^{28}$Si in the $sd$-shell some twenty years ago.[Brown and Wildenthal, Ann. Rev. Nucl. Part. Sci. {\bf 38}, 29 (1988)] Thursday, September 22, 2005 3:15PM - 3:30PM KK.00006: Exactly Solvable Nuclear Models and Richardson-Gaudin Algebraic Structures V.G. Gueorguiev , J. Dukelsky , P. Van Isacker , S.S. Dimitrova We discuss the exact solution of the isovector pairing (T=1) in nuclei within the SO(5) Richardson-Gaudin model. $^{64}$Ge is used to illustrate the parameter space and shed light on the solutions of the Richardson-Gaudin models. Basic properties of the integrable Richardson-Gaudin models are summarized and their possible applications to variety of nuclear physics models are emphasized. This work was partially performed under the auspices of the U. S. Department of Energy by the University of California, Lawrence Livermore National Laboratory under contract No. W-7405-Eng-48. Thursday, September 22, 2005 3:30PM - 3:45PM KK.00007: The Redmond Formula with Seniority Larry Zamick , Alberto Escuderos As we get to heavier nuclei, we find more states with different seniorities and several states of a given seniority. There is a recursion formula by Redmond that relates an $n \to (n+1)$ coefficient of fractional parentage (cfp) to that of $(n-1) \to n$. However, this involves an {\it overcomplete} set of principal parent (pp) cfp's. For example, for a 3-particle system, we can form basis states $[ [12]^{J_0} 3]^J$, where $J_0$ is the pp; we then antisymmetrize and normalize $\Psi [J_0]=N[J_0] (1-P_{12}-P_{13}) \left[ [12]^{J_0} 3\right]^J$, and form a ppcfp expansion $\Psi[J_0]=\sum_{J_1} [j^2 (J_1) j \|\} j^ 3 [J_0] J] \left[ [12]^{J_1} 3\right]^J$. But for, say, $J=j=9/2$, there are five $\Psi[J_0]$'s, but only two independent wave functions, one with seniority 1 and one with seniority 3. We note that $[j^2 (J_0) j \|\} j^3 [J_0] J]=1/(3 N[J_0])$. We are able then to obtain the following relation between overcomplete ppcfp's and complete orthonormal cfp's: $A=B=C$, where $$A=(n+1)[j^n(J_0 v_0) j \|\} j^{n+1} [J_0 v_0] J] \; \; [j^n (J_1 v_1) j \|\} j^{n+1} [J_0 v_0] J],$$ $$B=(n+1) \sum_v [j^n (J_0 v_0) j \|\} j^{n+1} J v] \; \; [j^n (J_1 v_1) j \|\} j^{n+1} J v],$$ $$C=\delta_{J_0 J_1} \delta_{v_0 v_1} + n (-1)^{J_0+J_1} \sqrt{(2J_0+1)(2J_1+1)} \sum_{v_2 J_2} \begin{Bmatrix} J_2 & j & J_1 \\ J & j & J_0 \end{Bmatrix} \times$$ $$\times [j^{n-1} (J_2 v_2) j \|\} j^n J_0 v_0] \; \; [ j^{n-1} (J_2 v_2) j \|\} j^n J_1 v_1].$$ Thursday, September 22, 2005 3:45PM - 4:00PM KK.00008: Applications of RPA in the nuclear shell model Calvin Johnson , Ionel Stetcu In recent work we have described a computational implementation of the random phase approximation (RPA) in the interacting shell model. Such an implementation is computationally much cheaper than full scale diagonalization, and provides a reasonable approximation, to binding energies and transitions, including charge-changing transitions. Here we discuss the latest applications of our computer code, SHERPA, with an emphasis on astrophysics.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 29, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8831079006195068, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/131013/matrix-exponential-convergence/131016	# Matrix exponential convergence Help me please to prove that matrix exponential which defined as: $e^{A}=\sum\limits_{k=0}^{\infty }\frac{A^{k}}{k!}$ converges for all matrix $A$ Thanks beforehand. - ## 3 Answers Since the space $E:=\mathcal M_n(\mathbb C)$ of all $n \times n$ complex matrices is a finite-dimensional vector space, all norms define the same topology. So we can take a sub-multiplicative norm, that is, a norm $\lVert\cdot\rVert$ such that $\lVert AB\rVert\leq \lVert A\rVert \cdot\lVert B\rVert$. (For example, we can take $\lVert\cdot\rVert$ to be the operator norm on $E$.) As a finite dimensional vector space, $E$ is complete, so it's enough to show normal convergence. We have that, for each integer $n \geq 0$, $$0\leq \lVert\frac{A^n}{n!}\rVert\leq \frac{\lVert A\rVert^n}{n!},$$ and we know that, for each real number $x$, the series $\sum_{n=0}^{+\infty}\frac{x^n}{n!}$ converges (it defines the exponential function). Therefore, for any $A \in E$, the series $\sum_{n=0}^{+\infty}\frac{A^n}{n!}$ converges. (We also got the additional result that $\lVert e^A\rVert\leq e^{\lVert A\rVert}$ for any $A \in E$.) - Thank you a lot! – Mushka May 13 '12 at 13:46 Hint: Show that $\sum \\|A_n\\|<\infty$ implies that $\sum A_n$ is convergent. - Thank you!!!.... – Mushka May 13 '12 at 13:45 HINT: Use one of the sub-multiplicative matrix norms to show that the series of real numbers in each position of $\sum_{k\ge 0}\frac{A^k}{k!}$ converges. - Thank you!!!.... – Mushka May 13 '12 at 13:45	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 19, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8564390540122986, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/26895/random-walk-randomly-reflected/26897	Random Walk Randomly Reflected Hi I am not specialist in probability so I will not be surprised if the answer for this question is just a simple consequence of well known results from the random walk theory. In this case, I will be happy if you can tell me the "magic words" to find the material related to this problem. Let $\sum_{i=1}^n X_i$ a random walk in $\mathbb{Z}^d$ defined in the following way: In each step, the walker look at your current position and as long as its distance to the origin does not belong to the set $\{d_1,d_2,\ldots\}$ ( we assume that $d_n\in\mathbb{N}$ and $d_i<d_{i+1}$ ) it jumps for one of its nearest neighbor with probability $1/2d$. For the other hand if $\\|\sum_{i=1}^n X_i\\|\in \{d_1,d_2,\ldots\}$ then the walker toss a biased coin if the result is head, which occurs with probability $p$, it goes back to the previous position, that is, $\sum_{i=1}^{n-1} X_i$. If the result is tails it chose randomly one of its neighbors with probability $1/2d$. I would like to know if there exist a non-trivial $p_c$ such that this random walk is transient for $p<p_c$ in dimension $d=3$. Of course, for this question to be well posed we have to define the sequence $(d_n)$. I would like to chose it randomly but I have no idea for what kind of distribution this problem is tractable. So at this time we can assume that $d_n$ is deterministic sequence and, for example, has a polynomial growth as $d_n=n^2$. Any references or comments are very welcome. PS: Could an user with more rep. place random walk tag for this question ? - How are you measuring distance from the oracle? Is it the Euclidean norm? – Joe Fitzsimons Oct 6 '11 at 6:16 Hi Joe you right, I should fixed a norm in $\mathbb{Z}^d$. So I will pick the $\ell^1$ norm, that is, $\\|(x_1,\ldots,x_d)\\|=\|x_1\|+\ldots+\|x_d\|$. – Leandro Oct 6 '11 at 6:54 Thanks for the tag Piotr. – Leandro Oct 6 '11 at 6:55 You want your walk to be reflected both when it hits the barrier coming from inside or from outside, or just when coming from inside? – Yvan Velenik Oct 6 '11 at 7:21 Good point Velenik. I guess that just inner reflection it was how I initially thought about this problem. But maybe both reflections create more symmetries and helps one give an argument. – Leandro Oct 6 '11 at 12:16 2 Answers If $p=1$ and $d$ is non-empty, then clearly the random walk is not transient, because since you use the $\ell^1$ norm, there is a closed surface which always reflects the walker back into the enclosed finite region. If, however, $p = 0$ then the walk is transient, since simple random walks in 3 dimensions are transient. Now, if there is a $p_c<1$ its existence is dependent on the specifics of the set $d$. Certainly, there exist some choices for $d$ which are transient: if the set $\{d_n\}$ is finite, then clearly there is a finite probability of ending up outside the final shell (unless $p=1$). Since the random walk on the region external to this shell is transient, then the probability of returning to the interior of the shell infinitely many times is zero, since the shell has a finite boundary, and the probability of returning to a specific site on the boundary site an infinite number of times is zero. Thus $p_c = 1$. Further in the case where $d_n = n$, then every site is reflective and hence it acts as a random walk on a 3d lattice with a finite probability not to hop for two steps (a hop plus a hop back are the same to simply waiting two time steps), and is hence still transient, unless $p_c = 1$. Indeed, this seems to be a very general principle (that there is no $p_c < 1$). One way to see why this should be true is to consider what happens when the walker enters the region between $d_i$ and $d_{i+1}$. The walker will be constrained to the region for longer, but this both keeps it from returning to the origin, and from jumping to the next level. When $p$ is extremely high, we expect the walker to have encountered each site within the region roughly an equal number of times before it eventually manages to exit the region. In this case the transition probability is given by the ratio of the surface areas of the two relevant reflecting surfaces, and hence it's probability of jumping to the region between $d_{i+1}$ and $d_{i+2}$ is linearly greater than the probability of jumping back into the region closer to the origin. This is exactly the same probability as we get when we consider the walker to start from a random position within that region when $p=0$. Thus, the reflection term is only increasing the amount of time spent in a region, not the probability of returning to it. At the increase in time spent in a region is constant for fixed $p \neq 1$, the walk remains transient. Thus I believe there is no choice of $d$ such that $p_c < 1$. Conversely if $p=1$ then the walk is not transient unless $d$ is empty. Lastly, if $d$ is empty, then there is no $p_c$. - Hi Joe thanks for have thought about the problem. I start reading your answer and it seems nice. Anyway I will give you another feedback tomorrow, now it is around 04:00am in Brazil and I need to rest a little bit. – Leandro Oct 11 '11 at 7:18 @Leandro: No problem. – Joe Fitzsimons Oct 11 '11 at 7:32 The fourth paragraph gave me the intuition of what one need for do not have a trivial model. It seems that we have to launch the walker back more than one site. Anyway this requires another question. Thanks Joe. – Leandro Oct 12 '11 at 5:09 @Leandro: No problem. As far as I can see this question actually amounts to a random walk on a weighted graph with a certain structure. The most obvious generalization would be a walk on a weighted directed graph, which would give you the direction dependent reflectors. – Joe Fitzsimons Oct 12 '11 at 6:49 @Leandro: The argument in this answer is using the symmetric reflection condition. You do not need to modify the question. I gave a separate answer. – Ron Maimon Oct 12 '11 at 8:42 show 1 more comment Your walk is always transient when you use the symmetric condition of reflection in both directions, and the argument is essentially given in the previous answer. But when the condition of reflection is asymmetric, so that you have reflection only when you are going out, an infinite number of reflectors will give recurrent behavior for arbitrarily small reflection probability in Euclidean space (LATER EDIT: so long as the $R_k$ do not grow exponentially fast--- in this case, you can have a transition--- you also have a good transition on the Cayley graph). Electrostatic criterion for recurrence The recurrence/nonrecurrence of a random walk is a time-independent problem, and can be solved by finding a steady-state solution. This has a simple electrostatic analog, see this answer: http://physics.stackexchange.com/questions/8149/collision-time-of-brownian-particles/14837#14837 The basic principle is as follows: consider a small sphere and a large sphere which absorb Brownian particles. Inject particles at a random position in a sphere of unit radius, and ask what is the probability that they are absorbed by the large sphere as opposed to the small one. Recurrence is when the small sphere always absorbs first, in the limit of an infinite large sphere. In steady state, the particle density obeys Laplace's equation with the boundary condition that the potential function is zero on the small inner sphere, and on the outer sphere, with a kink on the unit sphere which represents the incoming flux of injected particles. The flux of particles which are first absorbed at infinity is given by the integrated gradient of the potential (the electric flux) at large distances, and the flux which is first absorbed by the small sphere is given by the gradient of the potential at small distances (the electric flux at small distances). The flux is normalized by the values of the electrostatic potential at the small and large radius (because the potential has to vanish on these two spheres, because they are absorbing). The upshot of this is that the walk is recurrent if and only if the potential from a small sphere is divergent at infinity (Later edit: this is true for homogenous lattices--- see below). This is true in 1d, where it is linearly divergent, and in 2d, where it is log divergent, but fails in 3d and above, where you approach a constant asymptotic limit at long distances. The constant limit makes the flux at infinity nonzero, because the normalization from the zero potential metal boundary condition is not infinite. One Way Reflectors When you add two-way reflectors, the steady state distribution is unchanged, because if you have an infinitesimally thin reflecting plane, in order for there to be zero flux through it, the density of walkers on the left and on the right of the plane have to be equal. This means that the plane does not change the flux, and there is no difference from the no-reflector case. The walk is still transient with two-way reflectors (this is the content of the previous answer). When you have a one way reflector, however, the condition is different. Now the density on the interior has to be bigger right on the reflector by a ratio of (1/1-p) compared to the exterior to allow the flux through the plane to balance. Call the small ratio $A_p = 1-p$. In order to have zero flux to infinity in steady state, the distribution of particles with a unit source must have zero r-derivative away from the jumps at the reflectors. This means that all the density falloff must come at the jumps, and the condition that the sphere at infinity is at zero potential gives the restriction: $$\prod_k A_{p_k} = \prod (1 - p_k) = 0$$ When this infinite product is zero the walk is recurrent. When it is not zero, the walk is transient. EDIT: Or so I thought! The last equation is obviously completely wrong, as pointed out by Peter Shor, and seconded by anonymous downvoters. Thank you for catching the mistake I was blind to. There is nothing wrong with the method above, it is only that asymptoting to zero potential at large distances does not guarantee that an infinitesimal flux will go to zero at a finite time (this was counterintuitive for me). If you asymptote to zero slowly enough, you can still have a nonzero flux at large radii without ever hitting zero. The correct condition on the big conductor is that it must have zero flux through it when it is very large. For normal growth rates, to get zero flux, all you need is the potential to go to zero. But for exponentially fast growing $R_k$, to zeroing out the potential on a large sphere, you can still require a nonzero flux even if the sphere starts out at an infinitesimal potential, just because it is so enormously huge (thanks again to Peter Shor for discovering this stupid error--- it does not require a modification to the method, only to the faulty analysis at the very end). So one asks, given that there is an outward flux $q$, what is the radius of the sphere at which the potential is first zero? If there is no such radius for small enough $q$ (it is allowed to asymptote to zero, so long as it never reaches it--- this was my mistake of before) then the random walk is recurrent. The outward flux is the electric charge, so the solution going out is of the form $$\phi(r) = {q\over r} + A$$ At $R_1$, it is attenuated by $a_1 = 1-p_1$, so that $$\phi(R_1) = a_1 ( {q\over r} + A)$$ It continues along for larger r with the same q, but starting at the attenuated height, so that the solution for $r>R_1$ is: $$\phi(r) = ({q_1\over r} + A')$$ Where the constant A' is given by $$A' = a_1 A + (1-a_1) (- {q\over R_1})$$ i.e., it is the weighted average of the previous value of A with the negative value in parentheses, with weight given by $a_1$. The q can be factored out if you redefine A multiplicatively to absorb it. The value of A after each of the transitions is given by a similar weighted average $$A_{n+1} = a_n A_n + (1-a_n) ( - {1\over R_n} )$$ This linear difference equation can be solved by standard methods, in particular defining $$A_n = B_n \prod_{k=0}^n a_k$$ Then $$B_{n+1} - B_n = { (1-a_n)\over \prod_{k=0}^n a_k } ( - {1\over R_n} )$$ And the condition that A becomes negative after a finite number of steps for arbitrarily small q tells you that the above series is divergent: $$\sum_{n=1}^\infty {(1-a_n) \over \prod_{k=0}^n a_k } {1\over R_n} = \infty$$ The divergence of this series is the condition for recurrence. In the special case of constant $a_k = 1-p$, then the series has terms $${p\over (1-p)^n} {1\over R_n}$$ Which is a diverging exponential unless R_n is growing faster than $1\over (1-p)^n$. So for exponentially growing $R_n$, you do get a nontrivial phase transition, contrary to what I wrote initially. MORE OF LATER EDIT: Cayley Graph On the Cayley graph (infinite binary tree), instead of 3 space, the problem does have a phase transition in p, because by having every radius be one-way reflecting, and tuning p, you can make the radial walk unbiased after a path-dependent time reparametrization. There is a steady stream outward with probability 2/3, but if p is 1/2, then half of the outgoing 2/3 is brought back on the next step, so that you get a 1/3 probability for going inward, a 1/3 probability for going outward, and a 1/3 probability of coming back after two steps, which is an standard unbiased diffusion after you merge the two steps on the returning path into one. So for p<1/2 you have transience, and for p>1/2 you have recurrence, and the boundary is exactly at p=1/2. The Cayley graph example shows the approximate strength of constant reflectors--- they counteract exponential growth. Further, the steady state distribution on the Cayley graph can be analyzed by either merging all the vertices at distance r from the origin into one big vertex, in which case the steady state distribution at zero flux has exponential growth, or by not merging, in which case the steady state distribution at zero flux is constant on all the vertices. The two different descriptions shows clearly that the condition of zero flux at infinity is not just that the steady state distribution decays to zero. - 1 It would help if you defined $p_k$ in your answer. – Peter Shor Oct 12 '11 at 10:45 @Peter: p_k is the probability of reflection of the k-th concentric sphere (if all the p's are the same, the problem is trivial--- you always get recurrence). I am using Brownian motion, and spherical reflectors, plus a nonzero ratio between the density of the steady-state distribution in the interior and exterior of the sphere. – Ron Maimon Oct 12 '11 at 16:55 2 No, it does. Suppose you choose your $p$ so each sphere is half as dense as the one before it, and you choose $d_n = 2^n$. Then, in the steady state distribution in 3d, the probability that you're in the $k$th sphere is proportional to $4^k$ (8 times the volume, and 1/2 the density of the $k-1$st sphere), which diverges. So there is no steady state distribution. If $d_n$ grows polynomially, you're O.K. – Peter Shor Oct 12 '11 at 17:51 1 My argument was indeed total nonsense, but I still think I'm right. The transience of a random walk in 3 and higher dimensions is a fairly robust phenomenon, and should not be spoiled if you put a one-way partially reflecting barrier at positions $2^{2^k}$ for all $k$. Consider the probability that you return to the $k-1$st sphere once you reach the $k+1$st sphere. – Peter Shor Oct 15 '11 at 12:44 1 @PeterShor: Thank you for finding the error, and insisting on your correct intuition. I had the same intuition, but dismissed it, because I could not understand how a decaying solution asymptoting to zero could have a finite flux without crossing zero. The Cayley graph made it obvious that it was possible, and then I see that you were right all along (although your earlier argument was no good, but the intuition was correct). – Ron Maimon Oct 17 '11 at 20:35 show 5 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 68, "mathjax_display_tex": 10, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.943481981754303, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/80525-integral-test-series-problem.html	# Thread: 1. ## Integral test for series problem I am not sure how I need to solve this. Can someone get me started? How many terms of the following series would you need to add to find its sum to within 0.01? $\sum^{\infty}_{n=2} \frac{4}{n(ln(n))^9}$ a) $n > e^{(25)^{\frac{1}{4}}}$ b) $n > e^{(50)^{\frac{1}{8}}}$ c) $n > e^{(100)^{\frac{1}{8}}}$ d) $n > e^{(25)^{\frac{1}{8}}}$ e) $n > e^{(100)^{\frac{1}{4}}}$ 2. integral test, after you let $u=\ln x$ you will get a p-integral. You can prove that $\sum {1\over n(\ln n)^p}$ converges whenever p exceeds one. Now we need to be careul where we start this series, that's why I left out the n=2.... 3. Originally Posted by matheagle integral test, after you let $u=\ln x$ I know, but will that give me the answer? I just wasn't sure how to test it for a specific value. 4. I only proved it converged. To estimate the error, well I guess you should split the sum into two parts. One from 2 to N and the other from N+1 to infinity. You can then bound that second sum (which is the remainder via the integral) $\sum_{n=2}^{\infty} {4\over n(\ln n)^9}=\sum_{n=2}^N {4\over n(\ln n)^9}+<br /> \sum_{n=N+1}^{\infty} {4\over n(\ln n)^9}$. Call this $\sum_{n=2}^{\infty} {4\over n(\ln n)^9}=S_N+R_N$ You want a bound on $R_N=\sum_{n=N+1}^{\infty} {4\over n(\ln n)^9}$. And you can play the same game with the boxes and use the integral $\int_N^{\infty} {4dx\over x(\ln x)^9}={4\over 8(\ln N)^8}<.01$, solve for N, giving me (b). The you can sum up to that 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": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9168300032615662, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/3932/in-laymans-terms-how-does-shors-algorithm-work/4004	# In layman's terms, how does Shor's algorithm work? I've just been reading up on Shor's algorithm, and I find it both fascinating and baffling. I don't really understand much about it, other than that it can factor semiprimes in polynomial time. Could someone provide a layman's terms explanation of how it works, and why it is reliant on quantum computing? Keep in mind that whilst I kinda understand the basics of quantum computing (i.e. it uses photons instead of electrons, and bits are replaced with quibits that can be 0, 1 or a superposition of both) I don't know anything in-depth about it. I just know it's supposedly super-fast, compared to classical computing mechanisms. - 1 – Huck Bennett Oct 3 '12 at 15:56 1 – ericball Oct 4 '12 at 12:28 I think Shor's algorithm would work even if its input wasn't a semiprime. $\:$ – Ricky Demer Oct 6 '12 at 10:12 @RickyDemer Maybe, but isn't the whole scary part that it can factor semiprimes in polynomial time, thus rendering a good portion of our current asymmetric ciphers broken? – Polynomial Oct 6 '12 at 10:25 No, because it also solves discrete logarithm efficiently. $\:$ – Ricky Demer Oct 7 '12 at 20:38 show 2 more comments ## 1 Answer The question to answer is "Is N the product of PQ?" I believe that the easiest way to understand Shor is to imagine two sine waves, one length P and one length Q. Assuming that P and Q are co-prime, then the question above can also be answered "At what point does the harmony of P overlapped with Q repeat itself?" And the answer can be determined quickly, based upon the simple observation that we can figure out the phase of each wave at any given point on the number line. (The phase, remember, is "how far along the pattern is this point?" and is often measured in degrees; 90 degrees = 1/4 rotation) If we look at the point N then the phase of P and Q must be 0 if they are factors of N (or else there would be a remainder of the division N/Q or N/P). Shor just tests whether the phase of P == the phase of Q == 0 at point N. It turns out that phase estimation is pretty easy for quantum computers, so it's a good tool to build into a factoring machine. See http://www.wikipedia.org/wiki/Quantum_phase_estimation_algorithm for more information. - That's actually a really great way of explaining it. – Polynomial Oct 10 '12 at 7:31 The question to answer is "Of which P and Q at least 2 is N the product of?" rather than "Is N the product of PQ?". For the later we have very efficient algorithms, polynomial in Log(N) on classical computers. – fgrieu Oct 10 '12 at 10:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9468562602996826, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/29745/semisimple-ish-rings/29748	## Semisimple-ish rings! ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let S be the class of all rings R which have 1 and satisfy this condition: for every "non-zero" right ideal I of R there exists a "proper" right ideal J of R such that I + J = R. (The + here is not necessarily direct.) All semisimple rings are in S and (commutative) local rings which are not fields are not in S. The ring of integers Z is also in S and so S properly contains the class of semisimple rings. My questions: Will this condition by itself force an element of S to have any (known, interesting) structure? A more important question: What about simple rings which are in S? For example, do they have to be semisimple? (Unlikely!) - ## 1 Answer By Zorn's lemma, each right ideal is contained in a maximal right ideal, therefore if $I+J = R$ then $I+M = R$ where $M$ is a maximal right ideal. If $I+M\ne R$ for all maximal right ideals $M$ then $I\subseteq M$ for all maximal ideals $M$. Thus $I\subseteq J(R)$, the Jacobson radical of $R$ which is the intersection of all maximal right ideals of $R$. Hence condition $S$ is equivalent to $J(R)=0$. A ring with vanishing Jacobson ideal is called semiprimitive. As $J(R)$ is also the intersection of the maximal left ideals of $R$ then the property of semiprimitivity is left-right symmetric. There are plenty of examples of semiprimitive rings which are not semisimple. For instance every simple ring is semiprimitive and every subdirect product of semiprimitive rings is semiprimitive ($\mathbb{Z}$ is a subdirect product of finite fields). As a reference see Section 10.4 of P. M. Cohn Algebra (2nd ed. vol 3) Wiley 1991. - 1 That's great! Thank you very much. – carlos Jun 28 2010 at 7:06 The type of right ideals which do not have such a complement are exactly the superfluous (or small) right ideals. As the excellent answer above shows, rings with $J(R)=0$ are the rings without small right or left ideals. Going one step further, semisimple rings (right Artinian +$J(R)=0$) are the rings without essential (or large) right ideals. It's interesting that "no essential right ideals" implies it's dual relative "no superfluous right ideals". It's someone akin to right Artinian implying right Noetherian in rings. – rschwieb Dec 17 2011 at 12:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9023562073707581, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=612088	Physics Forums Page 1 of 2 1 2 > ## Spiral motion & angular acceleration (Question) Hello, I'm in desperate need of some help with a problem regarding spiral motion. If a particle moves along a spiral path by degrees θ in time t with intial angular velocity ωo, can the rotational kinematic equation (θ = ωot + (1/2)$\alpha$t2) still be applied? I'm trying to solve for angular acceleration $\alpha$. (See link for spiral picture) http://mathworld.wolfram.com/ArchimedesSpiral.html So in other words, is spiral motion any different than standard rotational motion with regards to calculating angular velocity/acceleration? It seems I would need to account for arc length (s) and curvature (k) using the equations in the posted link. Any help would be greatly appreciated. PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus When it completes a revolution it will have also gained 'height' as it will have moved 'up' the spiral. So you're angular velocity will have to account for that direction as well. However with your angular acceleration, as long as it is moving 'up' the spiral at constant velocity the equations shouldn't change (if being referenced in the x-y plane that is). just my 2c though In the link I posted above there are three equations: the polar equation, one for curvature (k) and one for arc length (s). In all equations there is a constant "a." What is this constant exactly and how is it measured(in a real application)? r = aθ k(θ) = [2+θ2]/[a(1+θ2)3/2 s(θ) = (1/2)a[θ√(1+θ2) + ln(θ+√(1+θ2)) Recognitions: Homework Help Science Advisor ## Spiral motion & angular acceleration (Question) Quote by clevermetal When it completes a revolution it will have also gained 'height' as it will have moved 'up' the spiral. So you're angular velocity will have to account for that direction as well. However with your angular acceleration, as long as it is moving 'up' the spiral at constant velocity the equations shouldn't change (if being referenced in the x-y plane that is). Sounds like you're thinking of a helix, not a spiral. The question relates only to 2D. Recognitions: Homework Help Science Advisor Quote by Fjolvar If a particle moves along a spiral path by degrees θ in time t with intial angular velocity ωo, can the rotational kinematic equation (θ = ωot + (1/2)$\alpha$t2) still be applied? I'm trying to solve for angular acceleration $\alpha$. That formula is only valid for constant angular acceleration. Do you know that to be the case here? If not, what determines the motion? Woops sorry, I was thinking of something along the lines of: z(t) = (acos(t), bsin(t), ct) which I now realise is completely wrong. Quote by Fjolvar In the link I posted [Archimedes' Spiral] above there are three equations: the polar equation, one for curvature (k) and one for arc length (s). In all equations there is a constant "a." What is this constant exactly and how is it measured(in a real application)? r = aθ k(θ) = [2+θ2]/[a(1+θ2)3/2 s(θ) = (1/2)a[θ√(1+θ2) + ln(θ+√(1+θ2)) If a=1, then after one rotation (Δθ=2pi) r increases by 2pi. If a=2, then after one rotation (Δθ=2pi) r increases by 4pi. etc... To measure a, you find how many rotations there have been total, and you measure r. Once you find a, you can solve for k and s. Quote by Fjolvar Hello, I'm in desperate need of some help with a problem regarding spiral motion. If a particle moves along a spiral path by degrees θ in time t with intial angular velocity ωo, can the rotational kinematic equation (θ = ωot + (1/2)$\alpha$t2) still be applied? I'm trying to solve for angular acceleration $\alpha$. Only if the angular acceleration is constant. Quote by Fjolvar So in other words, is spiral motion any different than standard rotational motion with regards to calculating angular velocity/acceleration? It seems I would need to account for arc length (s) and curvature (k) using the equations in the posted link. Any help would be greatly appreciated. By spiral motion, I suppose that you mean the distance from the particle to the origin is increasing with time. The physical effect of this would be the moment of inertia is increasing with time. If the external torque is constant and the moment of inertial is increasing in time, the angular acceleration must be decreasing in time. Thus, your formula is wrong if the torque is constant. For planar motion, Torque=(Moment of inertia)(Angular acceleration). If the Torque is constant while the moment of inertia is increasing with time, then the angular momentum has to be decreasing with time. Thus, you can't use a formula that is valid for constant angular acceleration. Quote by Darwin123 Only if the angular acceleration is constant. By spiral motion, I suppose that you mean the distance from the particle to the origin is increasing with time. The physical effect of this would be the moment of inertia is increasing with time. If the external torque is constant and the moment of inertial is increasing in time, the angular acceleration must be decreasing in time. Thus, your formula is wrong if the torque is constant. For planar motion, Torque=(Moment of inertia)(Angular acceleration). If the Torque is constant while the moment of inertia is increasing with time, then the angular momentum has to be decreasing with time. Thus, you can't use a formula that is valid for constant angular acceleration. The angular acceleration is decreasing as the particle approaches the center, but it stops before it reaches the center with a final angular velocity of zero (the particle hits a wall). So if I know the initial velocity, the distance or degrees travelled, and final velocity of zero.. what equation could I use to calculate the deceleration? I realize now that the moment of inertia is changing, but I'm not sure how to translate this into a calculation. Also I need to calculate the deceleration of the particle as it approaches a point on one of the outer spirals (also hitting a wall and stopping) in the reverse direction, but that should be the same concept. Recognitions: Homework Help Science Advisor Fjolvar, the conditions are still not clear to me. What is controlling the motion of the particle? As far as I can tell from your posts, this is irrelevant: the particle is being made to move in some way, and all you want is an expression for the angular acceleration about the origin. That's easy: it's the second derivative of the angle wrt to time. That's true whether it's on a spiral, a circle or a straight line. Or maybe you mean acceleration in the tangential direction? That's $2\dot{r}\dot{\theta}+r\ddot{\theta}$. Given that $r = a\theta$, you can eliminate $r$ or $\theta$ from that. Meanwhile, some of the posts bring up momentum and torque, which will be relevant if the particle is moving subject to known forces. So which is it? Quote by haruspex Fjolvar, the conditions are still not clear to me. What is controlling the motion of the particle? As far as I can tell from your posts, this is irrelevant: the particle is being made to move in some way, and all you want is an expression for the angular acceleration about the origin. That's easy: it's the second derivative of the angle wrt to time. That's true whether it's on a spiral, a circle or a straight line. Or maybe you mean acceleration in the tangential direction? That's $2\dot{r}\dot{\theta}+r\ddot{\theta}$. Given that $r = a\theta$, you can eliminate $r$ or $\theta$ from that. Meanwhile, some of the posts bring up momentum and torque, which will be relevant if the particle is moving subject to known forces. So which is it? So to clarify some, my ultimate goal is to calculate the torque/force of the particle's collision with the wall, which is in place near the end of the spiral. The particle starts from the outer ring of the spiral, has constant velocity ωo, and moves inwards until about half way when it begins to decelerate and then eventually collides into a wall near the origin bringing it to a complete stop. The driving force moving the particle is irrelevant in this case. I'm trying to satisfy the equation: Torque = Moment of Inertia * Angular Acceleration (Deceleration in this case), but I'm not sure if this equation can be used now since the moment of inertia varies with time. Since the moment of inertia is changing, I'm stuck trying to derive an equation to make this calculation. Hopefully I've explained this situation clearly enough. Feel free to ask any more questions, as I still have yet to derive the proper equation =/ Recognitions: Homework Help Science Advisor Quote by Fjolvar my ultimate goal is to calculate the torque/force of the particle's collision with the wall, which is in place near the end of the spiral. You cannot calculate those. How a force varies over the duration of an impact depends on subtle considerations of the elasticity, plasticity and density of the materials. What you usually can can calculate is momentum and/or angular momentum of the impact. The particle starts from the outer ring of the spiral, has constant velocity ωo, and moves inwards until about half way when it begins to decelerate and then eventually collides into a wall near the origin bringing it to a complete stop. The driving force moving the particle is irrelevant in this case. I'm trying to satisfy the equation: Torque = Moment of Inertia * Angular Acceleration (Deceleration in this case), but I'm not sure if this equation can be used now since the moment of inertia varies with time. Is the deceleration constant in magnitude? Constant in terms of angular acceleration? If not, what determines the speed of the particle at impact? Why would there necessarily be any remaining speed? [/QUOTE] Read below for my engineering project, which I tried to describe above in terms of a particle hitting a wall, since the same principles can be applied.. I thought it would've been easier to explain that way. There is a spinning wheel about an axle with a spiral groove on the side (where the rim of the wheel would be). The wheel is spinning at a constant velocity, controlled by an electric current. Then the current is cut off and the wheel begins to decelerate spinning freely losing velocity. A locking bolt is then engaged perpendicular into the spiral until it hits a hard stop/wall located within the spiral groove, near the center of the spiral. I am trying to calculate the torque/force exerted on this hard stop. I only know the initial velocity before the current is cut off, the distance travelled to the hard stop after, and the time it takes to reach the hard stop. Maybe it would have been easier to have explained it this way to begin with. Also the answer doesn't have to be 100% exact, just a close estimate, so we can ignore friction for now. So I'm looking for an equation that takes the following into consideration: 1. Intial angular velocity 2. Angular deceleration 3. Changing moment of inertia with respect to time 4. Torque This is very important to me, so any help is greatly appreciated, even a hint at how to derive such an equation. Recognitions: Homework Help Science Advisor OK, if I've understood that correctly: - the wheel decelerates steadily for reasons unconnected with the bolt; - the bolt's progress is not considered to affect the wheel until it hits the stop; that means we're ignoring both friction and the inertial mass of the bolt; - the bolt slides both in the groove on the wheel and in a stationary groove set along a radius of the wheel Since the spiral is arithmetic (r = Aθ), and the wheel's rotation rate ω = ω0 - Bt, the bolt's radius r = r0 - Aω0t + ABt2/2. You can solve using the final value of r to find the travel time, and hence the final value of ω. The impulse, moment of inertia of wheel × ω, comes from the wheel being brought abruptly to a halt. As I said, you cannot easily calculate a torque or force. If there is very little give in the system you will get a very high torque and would be in danger of shearing the bolt or damaging the groove. A better solution might be to attach a spring to the bolt so that it helps to decelerate the wheel while as it slides in. Quote by haruspex OK, if I've understood that correctly: - the wheel decelerates steadily for reasons unconnected with the bolt; - the bolt's progress is not considered to affect the wheel until it hits the stop; that means we're ignoring both friction and the inertial mass of the bolt; - the bolt slides both in the groove on the wheel and in a stationary groove set along a radius of the wheel Since the spiral is arithmetic (r = Aθ), and the wheel's rotation rate ω = ω0 - Bt, the bolt's radius r = r0 - Aω0t + ABt2/2. You can solve using the final value of r to find the travel time, and hence the final value of ω. The impulse, moment of inertia of wheel × ω, comes from the wheel being brought abruptly to a halt. As I said, you cannot easily calculate a torque or force. If there is very little give in the system you will get a very high torque and would be in danger of shearing the bolt or damaging the groove. A better solution might be to attach a spring to the bolt so that it helps to decelerate the wheel while as it slides in. All is correct in your understanding haruspex. So with these givens, you are saying the torque cannot be calculated then, correct? Recognitions: Homework Help Science Advisor It's a more elaborate version of a question that comes up every week or three on this forum: if a mass m travelling at speed v hits the ground, what is the force of impact? Answer: there is not a single value of the force. If you watched events microsecond by microsecond you'd see the force increase, more-or-less linearly at first, up to some maximum. It might stay at the max a little while before dying away. The integral of the force over the duration is mv, the change in momentum, but the peak value etc. depend very much on how long the process lasts and the force versus time profile. That depends on the elasticities, plasticities, and densities of the impacting bodies. Consider dropping an egg a metre onto concrete versus onto hay. Page 1 of 2 1 2 > Tags angular acceleration, angular velocity, archimedes, inertia, spiral motion Thread Tools \| \| \| \| \|----------------------------------------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: Spiral motion & angular acceleration (Question) \| \| \| \| Thread \| Forum \| Replies \| \| \| Introductory Physics Homework \| 1 \| \| \| Introductory Physics Homework \| 2 \| \| \| Introductory Physics Homework \| 2 \| \| \| Introductory Physics Homework \| 9 \| \| \| Introductory Physics Homework \| 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": 14, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9372373223304749, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/209320-when-integreating-parts-how-choose-u-v.html	3Thanks • 1 Post By MarkFL • 1 Post By HallsofIvy • 1 Post By Soroban # Thread: 1. ## When integreating by parts, how to choose u and v'? Hello, I never really understood how to choose the u and the v' when integrating by parts. I know if you choose the wrong one sometimes you end up in a loop. For the question Integrate ln x dx It becomes : Integrate 1 * ln x dx Now do I set u' as 1 or as ln x? And how do you decide which one is which? Thank you 2. ## Re: When integreating by parts, how to choose u and v'? A general rule for which function to choose as u is given by the mnemonic LIATE: Integration by parts - Wikipedia, the free encyclopedia 3. ## Re: When integreating by parts, how to choose u and v'? oh thank you , I have been guessing all this time. 4. ## Re: When integreating by parts, how to choose u and v'? Sorry one more thing. How come in this question they set u' as the exponential rather than the algebraic 5. ## Re: When integreating by parts, how to choose u and v'? We are given: $\int xe^x\,dx$ I was taught: $\int u\,dv=uv-\int v\,du$ So I would choose: $u=x\,\therefore\,du=dx$ $dv=e^x\,dx\,\therefore\,v=e^x$ giving: $\int xe^x\,dx=xe^x-\int e^x\,dx=e^x(x-1)+C$ It appears in the image you attached, that they are using the reverse of the variables I was taught. 6. ## Re: When integreating by parts, how to choose u and v'? "Guessing" is not a bad method, as long as you check! In general, because $\int u dv= uv- \int vdu$ you want to choose u that can be differentiated, dv that can be integrated, and then check to see if you can integrate $\int vdu$. 7. ## Re: When integreating by parts, how to choose u and v'? Hello, anees7112! I never really understood how to choose the $u$ and the $dv$ when integrating by parts. I know if you choose the wrong one sometimes you end up in a loop. Here's the rule that I was taught: Let $dv$ be the "hardest" part you can integrate. Example: . $I \:= \int\!\! x\,e^{3x}dx$ We can integrate $x$ and we can integrate $e^{3x}.$ . . The harder one is $e^{3x}.$ We have: . $\begin{Bmatrix} u &=& x && dv &=& e^{3x}dx \\ du &=& dv && v &=& \frac{1}{3}e^{3x}\end{Bmatrix}$ Then: . $I \;=\;\tfrac{1}{3}xe^{3x} - \tfrac{1}{3}\!\int\! e^{3x}dx$ . . . . . . . $=\;\tfrac{1}{3}xe^{3x} - \tfrac{1}{9}e^{3x} + C \;=\;\tfrac{1}{9}e^{3x}(3x-1) + C$ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Same example: . $I \;=\;\int x\,e^{3x}\,dx$ Watch what happens if you do it "the other way" . . . You have: . $\begin{Bmatrix}u &=& e^{3x} && dv &=& x\,dx \\ du &=& 3e^{3x}dx && v &=& \frac{1}{2}x^2\end{Bmatrix}$ $\text{Then: }\:I \;=\;\tfrac{1}{2}x^2e^{3x} - \tfrac{3}{2}\!\underbrace{\int\!x^2e^{3x}\,dx}_{?? }$ The new integral is worse than the original integral! 8. ## Re: When integreating by parts, how to choose u and v'? Thank you all for the help I probably should have asked this question a few years ago when we learnt integration by parts but it feels good to finally get it . 9. ## Re: When integreating by parts, how to choose u and v'? I have always just "guessed", or more accurately, set u and dv to what seemed easy to differentiate and integrate. It wasn't until I actually taught Calculus that I learned about the LIATE acronym. One of my students told me about it. - Hollywood	{"extraction_info": {"found_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": 20, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9349915981292725, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/23264/are-c-and-barq-p-isomorphic/23272	## Are $C$ and $\bar{Q_p}$ isomorphic? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) There is a famous passage on the third page of Deligne's second paper on the Weil conjectures where he expresses his dislike of the axiom of choice, as manifested in the isomorphism between $C$ and $\bar{Q_p}$. The proof of said isomorphism runs as follows. Both $C$ and $\overline{Q_p}$ have transcendence bases, $S$ and $T$. Then $C\simeq \overline{Q(S)}$ and $\overline{Q_p}\simeq \overline{Q(T)}$. But $C$ and $\overline{Q_p}$ have the same cardinality, and hence, so do $S$ and $T$. Therefore, $Q(S)\simeq Q(T)$ and, from there, $C\simeq \overline{Q_p}$. For myself, this proof is quite convincing. Recently, Torsten Ekedahl expressed his opinion to the contrary, and this led to the following exchange: So I wondered about other expert opinions on this matter. Do you find the isomorphism unbelievable and, if so, why? - 1 The theorem is true given the axiom of choice. In fact any two fields of a given characteristic that are algebraically closed and share the same cardinality are isomorphic (this is a property of the "theory" - I forget what this property is called, but it has something to do with completeness, which this theory is also. I'll let the model theorists be more specific). Is the problem that you didn't find a proof that was... rigorous enough? It's surely "believable". You can find much less believable iso., like: C is iso. to the ultraproduct of all the F_p bars. – H. Hasson May 2 2010 at 17:09 2 Considering injective resolutions, I never really understood AC objections if one uses derived functors in a substantial way. I agree with Minhyong about the way one uses trasncendence bases, so I have no problem with that field isomorphism. However, as a matter of style it shouldn't be invoked if not actually needed. In the case of Weil II, the issue can be avoided at the end of the proof because only countably many complex numbers actually arise (from stalks at closed points for countably many Weil sheaves on finitely many schemes of f.t. over Fbar_p): so only countable AC arises. – BCnrd May 2 2010 at 17:15 1 @H.Hasson: the property of ACF_p that you mention is called being "uncountably categorical". (this corrects my previous comment which just said categorical, which is not true.) For the connection to completeness, se the last paragraph of en.wikipedia.org/wiki/Morley's_categoricity_theorem – Dan Petersen May 2 2010 at 17:34 2 @Georges: Yes, the correct invariant is the transcendence degree. The theory ACF_p is only uncountably categorical. – François G. Dorais♦ May 2 2010 at 20:44 2 H. Hasson, that is not quite said correctly. In the countable case the transcendence degree DOES have to be countable, but the issue is that there are infinitely many different countable cardinalities: all the finite cardinalities, plus the countably infinite cardinalitiy. Each of these occurs as a transcendence degree, leading to non-categoricity. – Joel David Hamkins May 4 2010 at 10:49 show 6 more comments ## 4 Answers First, let me observe that it is consistent with ZF + DC that there is no such isomorphism. (This follows from this answer of mine.) However, as I commented on Torsten's post, the existence of such an isomorphism is a relatively harmless since one can force the existence of such an isomorphism without adding new points to C or Qp. Consequently, any purely field-theoretic fact that can be proved using this generic isomorphism can also be proved without (usually with more work). Since forcing is not widely understood, I will explain this in terms of sheaves instead. (If you're more familiar with forcing and you don't care about sheaves, simply observe that the poset P below is countably closed and ignore the rest of this post.) Let P be the poset of field isomorphisms p:A→B where A is a countable subfield of C and B is a countable subfield of the algebraic closure of Qp, and p ≤ q iff p ⊇ q (i.e. q is a restriction of p). This ordering is slightly counterintuitive, but it is more convenient than the opposite. The poset P can be viewed as a category where there is one and only one arrow between any two objects p and q iff p ≤ q. The poset P then becomes a Cartesian category where the terminal element is the isomorphism between the two copies of Q in each field, and the product of p and q are is the intersection of the (graphs of) p and q. There are many Grothendieck topologies that one could define on P. The relevant one for our context is the smallest Grothendieck topology S on P such that, for all x in C and all y in the algebraic closure of Qp, the sieves {q ≤ p : x ∈ dom(q)} and {q ≤ p : y ∈ rng(q)} are both covering sieves at p. (Any larger Grothendieck topology will do; for forcing one uses the double negation topology which includes this one.) Note that the points of (the locale associated to) the site (P, S) are in one-to-one correspondence with isomorphisms between C and the algebraic closure of Qp. Now, the isomorphisms between C and the algebraic closure of Qp correspond precisely with geometric morphisms Set → Sh(P, S). Whatever is preserved by this geometric morphism can be done equally well on either side. In other words, many things that can be done in Set using such an isomorphism can also be done in Sh(P, S) without this assumption. Of course, this heavily depends on what needs to be done, but there are known ways to carry out this kind of analysis. Since the site (P, S) is relatively nice, this analysis is far from impossible. It's interesting to see how this formalizes Emerton's view. Objects of Sh(P, S) are functors F:Pop→Set, subject to the usual continuity requirements. One can think of F as a set which evolves along P. This makes sense since we should think of partial isomorphisms p ∈ P as approximations to the desired isomorphism from C onto the algebraic closure of Qp. As more and more information is packed into p, we gain more and more information about the stalk of F at the given point. Although he only considers the first few approximations in his answer, Emerton's view corresponds precisely to working in Set while keeping in mind that the work being done could be done equally well in Sh(P, S) instead. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Dear Minhyong, I am quite happy with this isomorphism, but maybe not so much because of the proof using the axiom of choice (although I don't particularly object to AC) but rather because my sense is that, whenever this is used, what is really being used is a choice of isomorphism between the algebraic closure of $\mathbb Q$ in $\mathbb C$ and the algebraic closure of $\mathbb Q$ in $\overline{\mathbb Q}_{\ell}$ (and I have absolutely no objection to identifying these two algebraic closures). Anytime one uses such an isomorphism in arithmetic, and it isn't ultimately being used to identify algebraic numbers in the two fields, I think it is fairly meaningless. (E.g., for modular forms of wt. $k \geq 1$, I am happy to identify the space over such over $\mathbb C$ with the analogous space over $\mathbb Q_{\ell}$, since the normalized cupsidal eigenforms have algebraic integer coefficients, and so these spaces have a natural underlying $\overline{\mathbb Q}$-structure. But to take non-algebraic Maass eigenforms, and to think of their Fourier coefficients as numbers in $\overline{\mathbb Q}_{\ell}$, while technically possible, is conceptually meaningless.) In my own papers I often fix such an isomorphism (or even one for each $\ell$), but I don't think of it as having any significance beyond the identification of the two copies of $\overline{\mathbb Q}$. Added: The comments below have forced me to think a little harder about my position. Here is an attempt to refine it: Any countably generated extension of $\mathbb Q$ can be embedded into either $\mathbb C$ or $\overline{\mathbb Q}_{\ell}$, and when I invoke, or seen invoked, an isomorphism between the latter two fields, I think of it as a short-hand for something like the following: in the given proof, a countably generated subfield of $\mathbb C$ will appear (e.g. the field generated by the Hecke eigenvalues of a Maass form). Having fixed the isomorphism between $\mathbb C$ and $\overline{\mathbb Q}_{\ell}$, we have in particular fixed an embedding of this field into $\overline{\mathbb Q}_{\ell}$, and hence have chosen an extension of the $\ell$-adic absolute value to this field. (Of course, one could switch the roles of $\mathbb C$ and $\overline{\mathbb Q}_{\ell}$ here.) By virtue of fixing the isomorphism between $\mathbb C$ and $\overline{\mathbb Q}_{\ell}$, one is ensuring that any such extensions are compatible, if along the way we encounter different subfields of $\mathbb C$, and that is one big advantage, when writing an argument, of fixing such an isomorphism once and for all. But in practice I don't know that one encounters anything more serious than one single countably generated subfield that contains all the complex numbers appearing in the proof. And hence one doesn't use anything like the full strength of the isomorphism. I guess this does put me in Deligne's camp: the isomorphism is convenient, but one could get by with something much weaker, just involving countably generated subfields of $\mathbb C$. - Hmm. I'm not sure what you mean by meaningless. It sounds like you and a number of other people agree with Deligne, that is, to be happy with the isomorphism in situations where you can do without it. Is this correct? (On the other hand, you say you're fine with AC, in which case I suppose you should be happy with the isomorphism regardless of the situation.) – Minhyong Kim May 2 2010 at 20:21 1 Matt, the situation is as you say after Deligne's proof is done, but during the proof we don't know the algebraicity, so the framework of Weil sheaves and their twistings using possibly non-algebraic numbers is important in the method. That's why I pointed out in my comment that only countably many such numbers actually come up in the proof (due to countability of set of closed points), so even for the proof itself only a countably generated subfield of C intervenes. Last year when Akshay and I did a student seminar on Weil II he hoped it could be avoided, but by the end he agreed with me. – BCnrd May 2 2010 at 20:50 I don't think one should try to determine whether to accept or reject the axiom of choice or any other independent axiom by appealing to "believability" of some consequences of it. With infinite sets, our intuition is just too often misleading. We get used to certain "paradoxa" like Hilbert's hotel because we see them very early in our mathematical life, but nobody should ever claim that he has a complete intuition for set theory. As for the example, $\bar{\mathbf{Q}}_p$ and $\mathbf{C}$ are isomorphic if the axiom of choice is true, and that's that. Both are constructed using a completion, which makes them topological fields, and they are not homeomorphic or normed isomorphic, that's probably why it feels a bit wrong to some of us. - I'm afraid this was a bit too long for a comment. It seems that people don't object too much when we assert that $\mathbb{C}$ and $\overline{\mathbb{Q}_p}$ are isomorphic as sets or as abelian groups. Somehow the use of choice when establishing a ring-theoretic isomorphism bothers mathematicians much more, and I suspect it is because the implications are more at odds with intuition we build up from considering finite extensions of fields, or geometric structure that is normally attached to the fields. When considering purely ring-theoretic maps, we are still forgetting a lot of structural baggage, e.g., an affirmative answer to the similar question about existence of an embedding of fields $\mathbb{C}(t) \hookrightarrow \mathbb{C}$ throws away anything we know about genus zero curves. I would personally answer your question with "yes" although I would not argue the point with much conviction. I would be interested to know if there were a logical way of chopping up choice so that the isomorphisms of sets and groups were okay but the isomorphisms of rings were not. - 3 It seems to me that proving $\mathbb{C}$ and $\overline{\mathbb{Q}}_p$ to be bijective as sets is rather straightforward and does not require any axiom of choice at all. This is like proving the unit interval and the Cantor set to be bijective. A little use of the Cantor-Bernstein theorem and you are done. On the other hand, I would be surprised if a proof of group isomorphism were possible without AC. – Leonid Positselski May 2 2010 at 17:55 1 You are right. Even group isomorphism of $\mathbb{R}$ and $\mathbb{R}^2$ requires AC. – Gerald Edgar May 3 2010 at 0:15	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 50, "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}, "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}, "english_confidence": 0.9582645297050476, "perplexity_flag": "head"}
http://mathoverflow.net/questions/40453/are-there-unique-geodesics-in-the-nil-and-sol-geometry/40503	## Are there unique geodesics in the NIL and SOL geometry? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Is there a unique geodesics between any two points in the NIL (resp. SOL) geometry? If so, is there a nice way of parametrizing them? For example geodesics in $S^3$ can be parametrized using the embedding in $\mathbb{R}^4$ and $\sin , \cos$ functions. Geodesics in hyperbolic space can be parametrized using the hyperboloid model and the functions sinh,cosh. - 1 Look at [Kaplan, On the geometry of groups of Heisenberg type. Bull. London Math. Soc. 15 (1983), no. 1, 35--42], as well as the papers referring to it in mathscinet. – Igor Belegradek Sep 29 2010 at 13:31 ## 3 Answers The geodesics between points are not unique in both cases. Moreover the following is true: if $M$ is a universal cover of a compact Riemannian manifold whose fundamental group is virtually solvable but not virtually abelian, then there are conjugate points on some geodesics in $M$ and hence geodesics between some points are not unique. See Croke and Schroeder "The fundamental group of compact manifolds without conjugate points", Comment. Math. Helv. 61 (1986), no. 1, 161--175, MR847526, for the case when the metric is analytic, and Lebedeva "On the fundamental group of a compact space without conjugate points", here, for the general case. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. There are not unique geodesics between points in Sol geometry. The Sol metric on $\mathbb{R}^3$ may be given as $e^{-z}dx^2+e^zdy^2+dz^2$. The claim is that there is not a unique geodesic between $(0,0,0)$ and $(t,t,0)$ for $t$ large enough. There is a rotational isometry $(x,y,z)\overset{\varphi}{\mapsto}(y,x,-z)$. This leaves invariant the line $l=\{(t,t,0)\|t\in \mathbb{R}\}$, and therefore $l$ is a geodesic. For $t$ small enough, $(t,t,0)$ will lie in a normal coordinate patch about $(0,0,0)$, so $l$ will be the unique shortest geodesic between these points. However, for $t$ large, there are much shorter paths connecting $(0,0,0)$ and $(t,t,0)$. The length of the geodesic $l$ is linear in $t$. But one may take a piecewise geodesic path, starting as a geodesic in the hyperbolic plane $x=0$ connecting $(0,0,0)$ and $(0,t,0)$, and then a geodesic in the hyperbolic plane $y=t$ connecting $(0,t,0)$ and $(t,t,0)$. Since the path $(0,u,0), 0\leq u\leq t$ is a horocycle in the hyperbolic plane $x=0$, and similarly $(u,t,0),0\leq u\leq t$ is a horocycle in the hyperbolic plane $y=t$, the length of this piecewise geodesic path is on the order of $C\log(t)$. Since the metric space is complete, there is a minimal length geodesic connecting $(0,0,0)$ and $(t,t,0)$ which is not invariant under $\varphi$, and thus there are at least two minimal length geodesic paths (and at least three geodesics) connecting the two points. - Despite their non-uniqueness, a lot is known about geodesics of left-invariant metrics on Heisenberg groups. For example, Jang-Park [Conjugate points on 2-step nilpotent groups, Geom. Dedicata 79 (2000), no. 1, 65--80] describe conjugate points for all geodesic passing through the idenity element of a simply-connected 2-step nilpotent Lie group with 1-dimensional center. Earlier Kaplan and Eberlein obtained explicit equations for geodesics (see references in the above paper). -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 36, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8998638391494751, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/2384/security-equivalence-proofs-for-breaking-rsa?answertab=oldest	# Security equivalence proofs for breaking RSA It is my understanding that while a practical solution to the factoring problem will definitely break RSA, it has never been proven that the security of RSA is equivalent to factoring. In otherwords, theoretically, someone could mathematically break RSA without breaking the factoring problem. Are there any problems for which breaking RSA would be equivalent to (besides the obvious RSA Problem)? - 2 – Henrick Hellström Apr 16 '12 at 23:00 1 @HenrickHellström: Do you (or someone else) want to post a summary of the results in this paper (where they relate to the question) as an answer? – Paŭlo Ebermann♦ Apr 17 '12 at 20:08 ## 3 Answers The summary in the Rivest-Kaliski paper does indicate that two specific problems are equivalent to the RSA problem itself. 1. Due to the algebraic properties of the field $Z_n$, if there exists any small fraction of weak cipher texts (i.e. a particular subset of cipher texts $C$ for which finding the solution $M$ to $C = M^e \pmod N$ is easy), then the RSA Problem is at most as hard as the running time of the adversarial procedure for decrypting the weak cipher texts times a polynomial factor. IOW: The RSA Problem guarantees that all cipher texts are equally hard to invert. 2. A similar argument applies to each individual bit of a RSA cipher text. If an adversary has a non trivial advantage to determine any bit of a RSA plain text $M$ given a RSA cipher text $C$, then all bits of $M$ are equally easy to determine. - Actually, this is an equivalence (see http://www.jscoron.fr/publications.html#joc2007) when ed <= N^2. - 2 Actually, no, that paper doesn't claim that 'if you can break RSA (that is, find the $e$'th roots modulo a composite of unknown factorization), you can factor the composite.' Instead, the paper shows that, given $N$, $e$ and $d$, you can factor $N$ (which we knew already with the probabilistic algorithm). The paper does not show that 'if you could break RSA, you can recover $d$'. – poncho Apr 17 '12 at 14:38 3 Tukikun: Welcome to Crypto Stack Exchange! The confusion here is that one can see two ways of Breaking RSA: The key recovery one (i.e. from $N$ and $e$ and maybe some messages/signatures, recover $d$) is (which shows the paper you linked) equivalent to factoring of $N$. But there is a more general way of defining breaking RSA encryption, which simply is plaintext recovery, and this is called the RSA problem ... and this could be easier than key recovery/factoring. (cc @poncho) – Paŭlo Ebermann♦ Apr 17 '12 at 20:01 It seems that the RSA Problem stands alone (at least for now, right?). The paper pointed to by Henrick outlines the following. 1. The RSA Problem is no harder than factoring, but (especially for small encryption exponents) it is likely "easier" than factoring. 2. Computing the private key from the public key has been shown to be equivalent to factoring. 3. There are no easy-to-break ciphertexts. 4. If an adversary can break a single bit given the ciphertext, the whole ciphertext can be revealed. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9012213349342346, "perplexity_flag": "middle"}
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http://crypto.stackexchange.com/questions/3705/proof-that-lottery-does-not-know-outcome-of-draw/3717	# Proof that lottery does not know outcome of draw Could a variable participant lottery system cryptographically prove that they have zero knowledge of the outcome of a draw? Participants do not choose numbers in this lottery and winning numbers are not drawn. Instead, they simply enrol with a wager. Winners are selected by taking all wagers and building an array of 'tickets' where the number of ticket entries per participant is proportional to their wager. A random index is chosen from this array, this is the winner. I found a similar question but the answers assume that the number of participants is fixed. I'd also like to keep the participants from having to perform bit commitment as some have suggested - for trusting participants this is a PITA. - 2 A zero-knowledge proof is not a proof that the prover has no knowledge, but a proof which does not give the verifier any additional knowledge beyond what is to be proved. But this is still an interesting question. I suppose the only possible way would be to include some (big enough) entropy in the calculation which will only be available after the closing of the draw, and can't be influenced by anyone interested. – Paŭlo Ebermann♦ Sep 2 '12 at 21:18 ## 2 Answers Protocols for selecting uninfluenced random numbers typically fall into two camps: random beacons and coin-tossing protocols. A random beacon is a source of randomness that is agreed by everyone to be unpredictable. For example, you can funnel a large amount of financial data into a small random number. While you cannot prove you didn't know what the closing prices would be before they actually closed, can prove that someone who possesses such knowledge can make a lot of money. A coin-tossing protocol is a protocol where everyone chooses random numbers that are combined to form the master random number. This is what is mentioned in the other answer you found. As you point out, both of these produce a number of predetermined length. But that is ok, once the lottery closes, 1. Calculate the number of bits you need to select $m$ out of $n$ tickets using a standard selection algorithm (for example, you can run the Fisher–Yates shuffle until you've pulled out $m$ items). Say it takes $N$ bits. 2. Use either a random beacon or coin-tossing protocol to generate a cryptographic seed for a pseudo-random generator (PRG). 3. Use the PRG to generate $N$ bits and run the selection algorithm. Note selection algorithms may be non-deterministic in the number of bits they need, so you can skip step 1 and just run the PRG until the selection has completed. You must publish beforehand the exact protocol (selection algorithm, beacon/coin-tossing, PRG) that will be used. The only variable will be the secret seed. - The third camp being trusted third parties. – Zsbán Ambrus Sep 5 '12 at 17:02 As I understand it, you're asking for a protocol where every player contributes some "entropy" to the overall random mechanism, so they can be assured that nobody else gamed the system. But in such a system you would assume that every player has the opportunity to select a value giving them the best chance of winning. Assuming the players know the algorithm in use, the advantage always goes to the last participant, who would have the ability to select a number that would give them the winning selection. For starters, let's simplify the discussion and ignore the complexities introduced by the "wagers", "participants", "tickets", and "indexes". That's all noise that covers up the root of this problem, which is that you're looking for a fair 1-out-of-N choice lottery. Whatever protocol you come up with, you can figure out how tickets are sold or who gets paid later. To make it easy for discussion, I'll assume that "player" means one entry or ticket, and that "contribution" means that player's chosen value for entropy. What is needed is a cryptographically secure random number to be generated after the lottery is closed and all participants have made their choices. From there, you can do anything to mix in all the player's entropy contributions to yield a single number. You could simply concatenate each contribution together, including the random number at the end, and take a SHA hash of the result. Divide the hash digest by the number of players, and the remainder gives your winning index. You could be much more complex using something like the Tor protocol, where each player's contribution is used as a key to encrypt or hash the previous player's contribution, and that hash is used to determine the next "random" player, and so on. But all these are just different ways of stirring the same pot - none of them prove the pot is fair. The problem is these schemes don't "prove" that the secure random number was generated fairly. If I were running the lottery, I could rig the random number generator to emit a non-random value, then give my collaborator-player all the other player's values, and allow him to make the last choice. When the pre-determined non-random number appears, my collaborator wins and splits the prize with me. The algorithms chosen don't matter, as long as I let my collaborator go last. The entire lottery and every player in it could be a sham designed to take money from one unsuspecting victim. As the crooked operator, my collaborators could include every player except one: the mark. And if that's the case, then there is no value to allowing players to contribute entropy. It mostly provides an avenue for a collaborator to make a choice that would sway the lottery in their direction. Ultimately, every lottery boils down to "can the lottery operator be trusted?" And that's a society problem, not a math problem. Instead of asking the players for entropy, your best approach is to use an agreed-upon future public information source as the source of entropy for your secure random number generator. XKCD did this with "geohashing", which uses the value of the previous day's NYSE closing value as the input to a hash algorithm to come up with a random latitude / longitude. It's non-deterministic (nobody can control the value to the last penny) and everyone can do the computation for themselves to see that it was fairly computed. You'd obviously need more than one source of data, and they'd have to be pretty public to avoid tampering. Perhaps you could use something like the combined hash of each of the Dow Jones 100 values at the previous day's closing, using the DJIA as a randomizer to pick the order in which they are hashed. - Sounds like you might be unfamiliar with multi-party secure computation? You might want to look at distributed coin-flipping protocols. See, e.g., How to fairly select a random number for a game without trusting a third party? and how to generate a community random value and Is there some way to generate a non-predictable random number in a decentralised network?. – D.W. Sep 7 '12 at 8:30 I derived two assumed requirements from his request. His remark about bit-commitment being a PITA made it seem like the parties did not want to participate in a multi-phase commitment process, so complex interaction protocols like coin-tossing were out. He also said the participants were to contribute randomness, I assumed that was because they did not trust the lottery administrator. That pretty much leaves a pseudo random algorithm with an agreed-upon future seed as the remaining approach. The only work a player might do is to verify the results, and that's optional if they trust it. – John Deters Sep 8 '12 at 5:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9532334804534912, "perplexity_flag": "middle"}
http://openwetware.org/index.php?title=User:Timothee_Flutre/Notebook/Postdoc/2011/11/10&oldid=658104	# User:Timothee Flutre/Notebook/Postdoc/2011/11/10 ### From OpenWetWare Revision as of 23:19, 21 November 2012 by Timothee Flutre (Talk \| contribs) Project name Main project page Previous entry Next entry ## Bayesian model of univariate linear regression for QTL detection See Servin & Stephens (PLoS Genetics, 2007). • Data: let's assume that we obtained data from N individuals. We note $y_1,\ldots,y_N$ the (quantitative) phenotypes (e.g. expression levels at a given gene), and $g_1,\ldots,g_N$ the genotypes at a given SNP (encoded as allele dose: 0, 1 or 2). • Goal: we want to assess the evidence in the data for an effect of the genotype on the phenotype. • Assumptions: the relationship between genotype and phenotype is linear; the individuals are not genetically related; there is no hidden confounding factors in the phenotypes. • Likelihood: we start by writing the usual linear regression for one individual $\forall i \in \{1,\ldots,N\}, \; y_i = \mu + \beta_1 g_i + \beta_2 \mathbf{1}_{g_i=1} + \epsilon_i \text{ with } \epsilon_i \overset{i.i.d}{\sim} \mathcal{N}(0,\tau^{-1})$ where β1 is in fact the additive effect of the SNP, noted a from now on, and β2 is the dominance effect of the SNP, d = ak. Let's now write the model in matrix notation: $Y = X B + E \text{ where } B = [ \mu \; a \; d ]^T$ This gives the following multivariate Normal distribution for the phenotypes: $Y \| X, \tau, B \sim \mathcal{N}(XB, \tau^{-1} I_N)$ The likelihood of the parameters given the data is therefore: $\mathcal{L}(\tau, B) = \mathsf{P}(Y \| X, \tau, B)$ $\mathcal{L}(\tau, B) = \left(\frac{\tau}{2 \pi}\right)^{\frac{N}{2}} exp \left( -\frac{\tau}{2} (Y - XB)^T (Y - XB) \right)$ • Priors: we use the usual conjugate prior $\mathsf{P}(\tau, B) = \mathsf{P}(\tau) \mathsf{P}(B \| \tau)$ A Gamma distribution for τ: $\tau \sim \Gamma(\kappa/2, \, \lambda/2)$ which means: $\mathsf{P}(\tau) = \frac{\frac{\lambda}{2}^{\kappa/2}}{\Gamma(\frac{\kappa}{2})} \tau^{\frac{\kappa}{2}-1} e^{-\frac{\lambda}{2} \tau}$ And a multivariate Normal distribution for B: $B \| \tau \sim \mathcal{N}(\vec{0}, \, \tau^{-1} \Sigma_B) \text{ with } \Sigma_B = diag(\sigma_{\mu}^2, \sigma_a^2, \sigma_d^2)$ which means: $\mathsf{P}(B \| \tau) = \left(\frac{\tau}{2 \pi}\right)^{\frac{3}{2}} \|\Sigma_B\|^{-\frac{1}{2}} exp \left(-\frac{\tau}{2} B^T \Sigma_B^{-1} B \right)$ • Joint posterior (1): $\mathsf{P}(\tau, B \| Y, X) = \mathsf{P}(\tau \| Y, X) \mathsf{P}(B \| Y, X, \tau)$ • Conditional posterior of B: $\mathsf{P}(B \| Y, X, \tau) = \mathsf{P}(B, Y \| X, \tau)$ $\mathsf{P}(B \| Y, X, \tau) = \frac{\mathsf{P}(B, Y \| X, \tau)}{\mathsf{P}(Y \| X, \tau)}$ $\mathsf{P}(B \| Y, X, \tau) = \frac{\mathsf{P}(B \| \tau) \mathsf{P}(Y \| X, B, \tau)}{\int \mathsf{P}(B \| \tau) \mathsf{P}(Y \| X, \tau, B) \mathsf{d}B}$ Here and in the following, we neglect all constants (e.g. normalization constant, YTY, etc): $\mathsf{P}(B \| Y, X, \tau) \propto \mathsf{P}(B \| \tau) \mathsf{P}(Y \| X, \tau, B)$ We use the prior and likelihood and keep only the terms in B: $\mathsf{P}(B \| Y, X, \tau) \propto exp(B^T \Sigma_B^{-1} B) exp((Y-XB)^T(Y-XB))$ We expand: $\mathsf{P}(B \| Y, X, \tau) \propto exp(B^T \Sigma_B^{-1} B - Y^TXB -B^TX^TY + B^TX^TXB)$ We factorize some terms: $\mathsf{P}(B \| Y, X, \tau) \propto exp(B^T (\Sigma_B^{-1} + X^TX) B - Y^TXB -B^TX^TY)$ Let's define $\Omega = (\Sigma_B^{-1} + X^TX)^{-1}$. We can see that ΩT = Ω, which means that Ω is a symmetric matrix. This is particularly useful here because we can use the following equality: Ω − 1ΩT = I. $\mathsf{P}(B \| Y, X, \tau) \propto exp(B^T \Omega^{-1} B - (X^TY)^T\Omega^{-1}\Omega^TB -B^T\Omega^{-1}\Omega^TX^TY)$ This now becomes easy to factorizes totally: $\mathsf{P}(B \| Y, X, \tau) \propto exp((B^T - \Omega X^TY)^T\Omega^{-1}(B - \Omega X^TY))$ We recognize the kernel of a Normal distribution, allowing us to write the conditional posterior as: $B \| Y, X, \tau \sim \mathcal{N}(\Omega X^TY, \tau^{-1} \Omega)$ • Posterior of τ: Similarly to the equations above: $\mathsf{P}(\tau \| Y, X) \propto \mathsf{P}(\tau) \mathsf{P}(Y \| X, \tau)$ But now, to handle the second term, we need to integrate over B, thus effectively taking into account the uncertainty in B: $\mathsf{P}(\tau \| Y, X) \propto \mathsf{P}(\tau) \int \mathsf{P}(B \| \tau) \mathsf{P}(Y \| X, \tau, B) \mathsf{d}B$ Again, we use the priors and likelihoods specified above (but everything inside the integral is kept inside it, even if it doesn't depend on B!): $\mathsf{P}(\tau \| Y, X) \propto \tau^{\frac{\kappa}{2} - 1} e^{-\frac{\lambda}{2} \tau} \int \tau^{3/2} \tau^{N/2} exp(-\frac{\tau}{2} B^T \Sigma_B^{-1} B) exp(-\frac{\tau}{2} (Y - XB)^T (Y - XB)) \mathsf{d}B$ As we used a conjugate prior for τ, we know that we expect a Gamma distribution for the posterior. Therefore, we can take τN / 2 out of the integral and start guessing what looks like a Gamma distribution. We also factorize inside the exponential: $\mathsf{P}(\tau \| Y, X) \propto \tau^{\frac{N+\kappa}{2} - 1} e^{-\frac{\lambda}{2} \tau} \int \tau^{3/2} exp \left[-\frac{\tau}{2} \left( (B - \Omega X^T Y)^T \Omega^{-1} (B - \Omega X^T Y) - Y^T X \Omega X^T Y + Y^T Y \right) \right] \mathsf{d}B$ We recognize the conditional posterior of B. This allows us to use the fact that the pdf of the Normal distribution integrates to one: $\mathsf{P}(\tau \| Y, X) \propto \tau^{\frac{N+\kappa}{2} - 1} e^{-\frac{\lambda}{2} \tau} exp\left[-\frac{\tau}{2} (Y^T X \Omega X^T Y + Y^T Y) \right]$ We finally recognize a Gamma distribution, allowing us to write the posterior as: $\tau \| Y, X \sim \Gamma \left( \frac{N+\kappa}{2}, \; \frac{1}{2} (Y^T X \Omega X^T Y + Y^T Y + \lambda) \right)$ • Joint posterior (2): sometimes it is said that the joint posterior follows a Normal Inverse Gamma distribution: $B, \tau \| Y, X \sim \mathcal{N}IG(\Omega X^TY, \tau^{-1}\Omega, \frac{N+\kappa}{2}, \frac{\lambda^\ast}{2})$ where $\lambda^\ast = (Y^T X \Omega X^T Y + Y^T Y + \lambda)$ • Marginal posterior of B: we can now integrate out τ: $\mathsf{P}(B \| Y, X) = \int \mathsf{P}(B, \tau \| Y, X) \mathsf{d}\tau$ $\mathsf{P}(B \| Y, X) = \frac{\frac{\lambda^\ast}{2}^{\frac{N+\kappa}{2}}}{(2\pi)^\frac{3}{2} \|\Omega\|^{\frac{1}{2}} \Gamma(\frac{N+\kappa}{2})} \int \tau^{\frac{N+\kappa+3}{2}-1} exp \left[-\tau \left( \frac{\lambda^\ast}{2} + (B - \Omega X^TY)^T \Omega^{-1} (B - \Omega X^TY) \right) \right] \mathsf{d}\tau$ Here we recognize the formula to integrate the Gamma function: $\mathsf{P}(B \| Y, X) = \frac{\frac{\lambda^\ast}{2}^{\frac{N+\kappa}{2}} \Gamma(\frac{N+\kappa+3}{2})}{(2\pi)^\frac{3}{2} \|\Omega\|^{\frac{1}{2}} \Gamma(\frac{N+\kappa}{2})} \left( \frac{\lambda^\ast}{2} + (B - \Omega X^TY)^T \Omega^{-1} (B - \Omega X^TY) \right)^{-\frac{N+\kappa+3}{2}}$	{"extraction_info": {"found_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": 35, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8426439166069031, "perplexity_flag": "middle"}
http://www.wall.org/~aron/blog/the-ten-symmetries-of-spacetime/	comments on physics and theology # The Ten Symmetries of Spacetime by Posted on November 5, 2012 Previously, I described the main formula of Special Relativity: This formula tells us the amount of distance squared between two points (if $s^2 > 0$) or the amount of duration squared (if $s^2 < 0$). (By using some trigonometry we can also use this formula to figure out the size of angles, so this encodes everything about the geometry). All the crazy time dilation and distance contraction effects you've probably heard about are encoded in this formula. Today I want to talk about the symmetries of spacetime. What I mean by a symmetry is this: a way to change the coordinates $(t,\,x,\,y,\,z)$ of spacetime in a way that leaves the laws of physics the same. Now I haven't told you what the laws of physics are, but the important thing is that they depend on the geometry of spacetime. So that means that we need to check in what ways we can change the coordinates of spacetime without changing the formula for $s^2$. The first kind of symmetry is called a translation. This consists of simply shifting the coordinate system e.g. one meter to the right, or one second to the future. This doesn't affect the formula for $s^2$ since it only depends on the coordinate differences $\Delta t$, $\Delta x$ etc. We can write a time translation like this: i.e. the new time parameter $t^\prime$ equals the old one plus some number $a$. Similarly, the three possible kinds of spatial translations are: By choosing the numbers a, b, c, d, arbitrarily, one obtains a four dimensional space of possible translation symmetries. The second kind of symmetry is more complicated, but you've certainly heard of it before—it's called a rotation. If we have two spatial coordinates, then we can rotate them by some angle $\theta$ (measured in radians), which leaves all the distances the same. The algebraic formula for a rotation looks like this: That involves some trigonometry, but things look a bit simpler if we take the angle $\theta$ to be a really tiny parameter $\epsilon$, and just consider the resulting infinitesimal coordinate changes $\delta x \approx (x^\prime - x)$: Translated into English, that says that if you rotate the y-axis of your coordinate chart a little bit towards the x-axis, you have to rotate the x-axis a little bit away from the y-axis (or vice versa if $\epsilon$ is negative). I'm too lazy to draw this, but if for some reason you can't visualize it, a little bit of figeting with any rigid flat object should convince you. Now actually we have three different spatial coordinates: x, y, and z. That means that you can actually rotate in 3 different ways: along the x-y plane, the y-z plane, and the z-x plane. Of course there are other angles you can rotate at as well, but they are all just combinations of those three; in other words the space of possible rotations is 3-dimensional. But now, what about the time direction? It would feel terribly lonely if it were left out, and in fact it is also possible to rotate spacetime about the t-x plane, the t-y plane, and the t-z plane. However, remember how time is not quite the same as space? Instead, it's just like space except for a funny minus sign. So not surprisingly, the formula for a rotation also has a funny minus sign—or rather, a funny absence of a minus sign: So if you rotate the t-axis towards the x-axis (which corresponds to changing your coordinate system so that you are travelling at a constant speed), then the x-axis has to rotate towards the t-axis (which means that your notion of simultaneity has to change as well). If you know how to integrate this with calculus, you can get the effects of a finite "rotation" in space (called a Lorentz boost) through an "angle" $\chi$: In the above, cosh and sinh are functions similar to cosine and sine but defined using hyperbolas instead of circles. So this rotation has some wierd properties: It describes a crazy world (ours!) in which things rotate in hyperbolas instead of circles. That's because of the minus sign in the formula for $s^2$ above, which makes it so the points of equal distance (or duration) correspond to hyperbolas instead of circles. This has some additional consequences: 1) Because hyperbolas are infinitely long, the "hyperbolic angle" $\chi$ ranges from $-\infty$ to $+\infty$, unlike circular angles which come back to where you started after you rotate through $2\pi$ radians. 2) Because the two axes both move towards (or both move away) from each other, when you do a really big rotation it scrunches everything up towards $t = x$ or $t = -x$. What this means is that when you accelerate objects more and more, they don't go arbitrarily fast. Instead they just get closer and closer to the speed of light. In conclusion, spacetime has 10 kinds of symmetry: 4 kinds of translations and 6 kinds of rotations. The space of possible symmetries is 10 dimensional. It is called the Poincaré group. P.S. In this whole discussion I have ignored the possibility of reflection symmetries such as $t \to -t$ or $x \to -x$. These are also symmetries of the formula for $s^2$, but they are discrete rather than continuous—there's no such thing as a "small" reflection the way you can have a small rotation. Adding these in doesn't change the fact that the Poincare group is 10 dimensional. However, these transformations are actually NOT symmetries of Nature. They are violated by our theory of the weak force. The only discrete symmetry like this which is preserved by the weak force is CPT: the combination of time reflection, space reflection, and switching matter and antimatter. ## About Aron Wall I am a postdoctoral researcher studying quantum gravity and black hole thermodynamics at UC Santa Barbara. Before that, I studied the Great Books program at St. John's college Santa Fe, and got my Ph.D. in physics from U Maryland. This entry was posted in Physics. Bookmark the permalink.	{"extraction_info": {"found_math": true, "script_math_tex": 25, "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": 32, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9201024770736694, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/8550/escape-velocity-of-asteriod-243-ida/8553	# Escape Velocity of Asteriod 243 Ida I was reading about this asteroid (apparently, it has a moon, isn't that awesome?) and I started thinking about if I was on this asteroid, and I jumped, would I fall off? It's been a while since I did something like this, so I went to Wikipedia to get some formulas. There I found $$v_e = \sqrt{\frac{2GM}{r}}$$ Where • $G$ is the gravitational constant ($6.673 × 10^{-11}\ \mathrm{m}^3\mathrm{kg}^{-1}\mathrm{s}^{-2}$) • $M$ is mass of the body (found to be $4.2 × 10^{16}\ \mathrm{kg}$) • $r$ distance from center of mass (found to be $53.6\ \mathrm{km}$) $$v_e = \sqrt{\frac{2(6.673\times 10^{-11})(4.2\times 10^{16})}{53600}} = \sqrt{104} = 10.22\,\frac{\mathrm{m}}{\mathrm{s}}$$ I'm a fairly fit guy, about 73kg and .6m vertical leap here on earth. I found a formula for the speed of a free falling object to be $$v = \sqrt{2gd}$$ If I assume that my speed when I hit the ground is the same as when I jump up, that means my jump speed is about 3.4 m/s, which is significantly less than the required 10.4 m/s. So it would seem I would not be able to jump off of this asteroid. Are my formulas, assumptions, and calculations correct here? Also, how would I use this information to determine how high I would jump (would it be enough to freak me out? Well I'd be jumping off an asteroid, so it would freak me out anyway... although I imagine I would be going slow enough for (my ship?) to pick me back up.) - ## 2 Answers Looks as if your arithmetic is correct, and you could not escape 243 Ida. Given your leap velocity capability of 3.4 m/s, you would be able to leap 593 m above 243 Ida's surface, since its surface gravitational acceleration is only about 9.75E-3 m/s^2. - Wow, I'd jump half a km up before falling back down? That would make for a fun ride! :-) – corsiKa Apr 13 '11 at 14:44 1 If I'm not mistaken, the total time you'd be in flight would be almost 700 seconds. Lot's of time to enjoy it. Scarey. – Michael Luciuk Apr 13 '11 at 17:24 "So it would seem I would not be able to jump off of this asteroid." That APPEARS to be correct. There are two adjustments that need to be made to the formula you used, plus other corrections; but I don't think they would get you permanently off 243 Ida. "Are my formulas, assumptions, and calculations correct here?" Calculations appear to be good, but there are two 'mis-assumptions' which invalidate the formulas you used, plus one error for constants (radius), plus plausibly one error in initial velocity: 1.) The geometry of 243 Ida is not even close to spherical. 2.) The formula you chose for speed of a falling object assumes a constant gravitational field, which is far from the truth. I think the easiest to use formula would be: (v^2)/2 = Integral of (g(h))dh, where 'h' is height above asteroid surface, and 'g(h)' is a function of h. 3.) the dimesions of 243 Ida are roughly 54km x 24km x 15km; so the largest value for 'r' would be 54km/2 = 27km. 'r' could be as small as 7.5km. 4.) Your jump velocity of 3.4m/s is in Earth's gravity. That means you are fighting your own weight. The excess force OVER and above your weight is what is used to generate your jump velocity through the mechanism of F=ma. What do you think your initial jump velocity would be if you were hooked up to a bungie cord which lifted you with a force of (your body weight minus 1 ounce)? "Also, how would I use this information to determine how high I would jump (would it be enough to freak me out? Well I'd be jumping off an asteroid, so it would freak me out anyway... although I imagine I would be going slow enough for (my ship?) to pick me back up.)" The near field gravity on the surface of 243 Ida varies greatly, depending on where you are. To understand this better, look at a pic of the rock. It is roughly barbell-shaped. Try to imagine if we took this to the extreme, and 243 Ida was shape like two identical spheres, with a very small connecting waist. What would be the gravity at that waist? Very close to zero, since the forces of gravity of the two spheres would cancel each other out. I'm not going to try to work out the integral, here; but will make a few assumptions which will show how much the answer given by the standard formula could be off: The standard answer for surface gravity on Ida 243 is 1.1 cm/s^2 down to 0.3 cm/s^2. Assuming it is 0.3 cm/s^2 at the waist and remains constant (which is not true, but wait...) AND you jumped with an inital velocity of 5m/s (a conservative estimate), you would attain a height of 4167 meters. Now, assuming r = 7500m at the waist, you might be tempted to think that gravitational pull would fall off by a factor of (4167/7500)^2 = 0.31, and so your jump would actually be higher; BUT, because of the geometry of the problem (two point sources of gravity instead of one), I think gravity would actually increase with height, as the two lobes of the asteroid started appearing more as a point source. - The potential from two point sources is actually pretty easy to do. Also the asteroid is likely spinning, and you would want to jump in the direction of the spin. One possibility for the reason there are a lot of binary asteroids, is that they can get spun up by assymetries in solar heating and radiative cooling forces until some of the stuff falls off and becomes a satellite. – Omega Centauri Apr 13 '11 at 18:11 Wow, Jeff, this is a very thorough post! I was assuming the gravity would be lowest "at the end of the barbell" so to speak, which is why I used it as the radius, and was assuming I'd jump straight up (as in, away from the center of mass) but your point about the extreme scenario is an interesting one. Also your point about initial velocity means I very well may be able to accelerate myself to 10 m/s or more! I would have to hope to bounce into another asteroid to get back. – corsiKa Apr 13 '11 at 18:52	{"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": 6, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9692764282226562, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/235432/math-card-probability-questions/235448	# Math card probability questions 2)Three cards are drawn at random from a standard deck without replacement. What is the probability that all three cards are hearts? 3)Three cards are drawn at random from a standard deck with replacement. What is the probability that exactly two of the three cards are red? 4)Three cards are drawn at random from a standard deck without replacement. What is the probability that exactly two of the three cards are red? 5)Three cards are drawn at random from a standard deck with replacement. What is the probability that at least two of the three cards are red? I am having trouble finding the correct answer when I am having to do the math to figure out the with replacement and without replacement. And I also don't understand the difference in the math I need to do to figure out the exactly two of the three cars and the at least two cards. - Can you show us what you have written? – The Substitute Nov 12 '12 at 4:51 For number 2 i figured you would do 13/26 x 12/25 x 11/24 – susie q Nov 12 '12 at 5:01 and then for number 3 I thought youd just do 13/26x13/26x13/26 since you are replacing the card each time – susie q Nov 12 '12 at 5:02 where are you getting the 26, 25, 24 for number 2? There are 52 cards in the deck, and there are 13 hearts. Therefore, the probability of drawing a heart on the first draw is $\frac{13}{52}$. – The Substitute Nov 12 '12 at 21:49 ## 2 Answers 2) $P(\text{3 hearts})=P(\text{heart on first draw})\cdot P(\text{heart on second draw given that we drew a heart on first draw})\cdot P(\text{heart on third draw given that we drew two hearts})=\frac{13}{52}\cdot \frac{12}{51}\cdot \frac{11}{50}=\frac{11}{850}$ - Here's a related example to show how you can approach the replacement question. • Two cards are drawn at random from a standard deck without replacement. What is the probability that both are even? The total number of ways to draw two cards is $C(52, 2)$ (also known as "52 choose 2"). The total number of ways to draw two cards with both even is $C(26, 2)$, since you're drawing from the even cards only. So the probability is $C(26, 2) / C(52, 2) = (26 \times 25) / (52 \times 51)$, which is about $.245$. • Two cards are drawn at random from a standard deck with replacement. What is the probability that both are even? Each time you draw a card, the probability is $1 / 2$. Each draw is independent. So the probability of doing it twice is $1/2 * 1/2 = 1 / 4$. See the difference? In the first case you have to think about drawing a pair of cards, but in the second case the two draws are independent. In the first case, notice that if one card is even, that makes it slightly harder to get an even card the second time (since there's one fewer even cards to choose from), so the probability is slightly lower for the first case than for the second. I hope that helps you apply the idea to red cards or hearts now. - Thank you! It just seems so much easier when you only have two cards. Hopefully I'll get this figured out – susie q Nov 12 '12 at 4:59	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.967140257358551, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-geometry/188036-complex-analysis-convergence.html	Thread: 1. Complex Analysis -- convergence Let M be a complete metric space. Suppose that $f_n$is a sequence of continuous functions in $(\mathcal{C}(U, M), \rho)$ which converges to f and $z_n$ is a sequence in U which converges to a point z in U. Show that $\lim f_n(z_n) = f(z).$ I'm not sure where to start with this one. Clearly, since the functions are continuous, $\lim f(z_n) = f(z)$. I need a way to make sense of the $\rho$ metric so that I can pass the correct limit, but I can't seem to be able to do it. Can anyone help? Thank you in advance. 2. Re: Complex Analysis -- convergence What is the definition of $\rho$? 3. Re: Complex Analysis -- convergence $\rho(f,g) = \sum_{n=0}^\infty\frac{1}{2^n}\cdot{\frac{\rho_n(f ,g)}{1+\rho_n(f,g)}$. Here, $\rho_n(f,g) = \sup_{z\in K_n} d(f(z),g(z))$ where $K_n = \{z : \|z\| \leq n\} \cap \{ z : d(z,\mathbb{C}\backslash U) \geq \frac{1}{n}\}$ f,g are defined on an open set U in the complex plane and take values in a complete metric space M with metric d. 4. Re: Complex Analysis -- convergence We can find a $k_0$ such that $z_n\in K_{k_0}$ for all $n$, since the map $z\mapsto d(z,\mathbb C\setminus U)$ is continuous. We get $\frac 1{2^{k_0}}\frac{d(f_n(z_n),f(z_n))}{1+d(f_n(z_n),f (z_n))}\leq \frac 1{2^{k_0}}\frac{\rho_{k_0}(f_n,f)}{1+\rho_{k_0}(f_ n,f)}\leq \rho(f,f_n)$ and given $\varepsilon>0$, we can find $N(\varepsilon)$ such that for $n\geq N(\varepsilon)$ we have $\frac{d(f_n(z_n),f(z_n))}{1+d(f_n(z_n),f(z_n))}\le \frac{\varepsilon}{1+\varepsilon}$, hence $d(f_n(z_n),f(z_n))\le \varepsilon$. 5. Re: Complex Analysis -- convergence Thanks very much for this! The first inequality is the key one I was missing.	{"extraction_info": {"found_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": 20, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9252246022224426, "perplexity_flag": "head"}
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I have two questions: What topological restrictions do Einstein's equations ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9080707430839539, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/94119/what-is-the-formal-definition-of-a-dual-category?answertab=oldest	# What is the formal definition of a dual category? I'm reading this book as an introduction to category and I have a problem with the definition of the dual category given on page 25. The right way to get the dual category can be described by turning around arrows and permuting the order of their composition. However I don't know how this works out with the statement $$\text{hom}_{\bf{A^{OP}}}(A,B)=\text{hom}_{\bf{A}}(B,A).$$ The set of morphisms $\text{hom}_{\bf{A}}(B,A)$ is already defined and the category $\bf{A}^{OP}$ is sopposed to contain morphisms, which are turned around, i.e. morphisms with different domain and codomain. How can these new ones equal the one from the first category? I remember reading this before in a book, and although the english wikipedia doesn't use this expression, the russian (?) one seems to use this definition as well. So say I have a category $\bf{C}$ with only two objects $a,b$, as well as a single morphism $f$ from $a$ to $b$. From the description I think the dual category $\bf{C}^{OP}$ is the one with $a,b$ and an arrow from $b$ to $a$. How does the formula work if this new morphism isn't contained in $\bf{C}$, so that $\text{hom}_{\bf{C}}(b,a)$ is essentially empty? - The equal sign means that there is a natural identification between the two hom-sets (given by reversing the arrow back). That's all. – Qiaochu Yuan Dec 25 '11 at 21:58 1 Just like two sets can contain the same object, two categories can have the same morphism; and it does not have to be between the same objects. – sdcvvc Dec 25 '11 at 22:04 @sdcvvc: I disagree. Thinking like that is not thinking categorically. The hom-sets in two different categories are two different types of mathematical objects, and it shouldn't make sense to directly compare them for equality. – Qiaochu Yuan Dec 25 '11 at 22:05 @Qiaochu: It depends on the way of thinking. Formally you can define them to be equal (just like you can define $(x,y)=\{\{x,y\},\{y\}\}$ on formal level). You don't have to do this, but then, you have still to tell what exactly $\text{hom}_{\bf{A^{OP}}}(A,B)$ is, or what is that "natural identification" etc. – sdcvvc Dec 25 '11 at 22:20 I hope you didn't disapprove if I interfere in your speech: I think that the objection of sdcvvc is legitimate because while is true that this is thinking categorically it is set theoretic thinking, and because the definitions used since here are think in a set theory all the objections hold. – ineff Dec 25 '11 at 23:36 show 1 more comment ## 2 Answers The point is that source and target of morphisms aren't an intrinsic property of the morphisms themselves, but are more like a property of the category. In general you can see the same element of a set as a morphism for two different categories and this morphism can have different source and target in these categories. When you say that $\hom_{\mathbf A^\text{op}}(A,B)=\hom_\mathbf{A}(B,A)$, you're building up a new category in which the morphisms are the same elements/morphisms of $\mathbf C$ but in this new category you see these morphisms with reversed direction. If you prefer you can think of the opposite category of category $\mathbf C$ as a new category such that for each morphism $f \colon A \to B$ in $\mathbf C$ there exists a unique $f^\text{op} \colon B \to A$ in $\mathbf C$ and such that the composition is such that $g^\text{op} \circ f^\text{op}=(f \circ g)^\text{op}$, where the composition on the left is the one in the opposite category, while the one on the right is that of the category $\mathbf C$. In this way you have defined the opposite category $\mathbf C^\text{op}$ up to isomorphism, thus you can think of the definition that you gave as a model, i.e. a concrete representation, of the opposite category. Hope this can help. - I think I get the picture, but the first sentence is strange. If I have a function like $f(x,y):=e^{x-y}$ then isn't the source and target (some spaces) pretty much encoded in the function itself? And by "same element of a set" you want to point out that it's just one right? – Nick Kidman Dec 25 '11 at 23:03 Nikolaj: Even though $f$ might be a function $\mathbb R \times \mathbb R \to \mathbb R$, in any arbitrary category $A$ it does not have to be a morphism between those two objects. You can create a category out of any two objects $a,b$ and make $f$ a morphism between them. In fact, the morphisms need not be functions at all. In category theory the exact nature of morphisms is not important (just like in group theory, you do not care about exact nature of objects in a group, it's only up to isomorphism). – sdcvvc Dec 25 '11 at 23:33 @sdcvvc: Yeah, they don't have to be functions, but that doesn't really answer my question. In my example, if the objects are $\mathbb{R}$ etc. as you say and if I define my function $f$ as above, it seems to me that $f$ already says everything about source and target. How is this not intrinsic to it? – Nick Kidman Dec 25 '11 at 23:41 1 As I and sdcvvc have already pointed out source and target of a morphism aren't inherently connected to the said morphism, you have to think in a more abstract way. Remeber that morphisms doesn't have to be functions (or structure preserving functions), they are simply some elements of a given collection or type on which we pose some structure (we give to each of them a source and a target, a left and right identity and composites): for instance think to monoids that are just categories with one object. – ineff Dec 25 '11 at 23:47 @NikolajKunst. Even your particular function $f(x,y) = e^{x-y}$, thought as a morphism in the category of topological spaces, has no intrinsic domain and codomain. You can say that the domain is $\mathbb{R}^2$ (since it is defined over all couples of real numbers) and the codomain is $\mathbb{R}$ (as sdcvvc says). But, in fact, the range of $f$ is just the positive real numbers, isn't it? And I could thought, if I want, $f$ as a function with domain the unit disk and codomain, say, $\mathbb{R}^2$. Why not? The domain and codomain of a morphism are part of the morphism. – Agustí Roig Dec 26 '11 at 8:51 show 3 more comments I think MacLane's "Categories for the working mathematician" says it better than your book: "To each category $\cal{C}$ we also associate the opposite category $\cal{C}^{op}$. The objects of $\cal{C}^{op}$ are the objects of $\cal{C}$, the arrows of $\cal{C}^{op}$ are arrows $f^{op}$ in one-one correspondence $f \mapsto f^{op}$ with the arrows $f$ of $\cal{C}$. For each arrow $f: a \longrightarrow b$ of $\cal{C}$, the domain and codomain of the corresponding arrow $f^{op}$ are as in $f^{op}: b \longrightarrow a$ (the direction reversed). The composite $f^{op}g^{op} = (gf)^{op}$ is defined in $\cal{C}^{op}$ exactly when the composite $gf$ is defined in $\cal{C}$. -	{"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": 57, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.958526611328125, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?t=414061	Physics Forums ## How is the entropy of the universe increasing when entropy is simply transferred? For example, an increase in the entropy of the system will be exactly equal to the entropy decrease of the surroundings. So the net change in the entropy of the system and its surroundings is zero. Putting this in perspective to all the systems and environments in our universe, how is the universe always increasing in entropy? Thanks in advance! ninjarawr PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug I'm not really sure the if the entropy of a system increases, the entropy of the surroundings will decrease. There's no conservation of entropy-law. If you have a closed system, it's entropy has nothing to do with the surroundings. Quote by ninjarawr For example, an increase in the entropy of the system will be exactly equal to the entropy decrease of the surroundings. Oh, really? Prove it ## How is the entropy of the universe increasing when entropy is simply transferred? I think what may have confused you is that if you look at a 'non-closed' system you can observe an entropy decrease, but this means that entropy somewhere else was increased. So entropy is still increasing overall. Perhaps you read something like that and generalized it to a conservation law.. However, it doesn't work the other way around. So an entropy increase does not have to accompany an entropy decrease. Quote by Gear.0 I think what may have confused you is that if you look at a 'non-closed' system you can observe an entropy decrease, but this means that entropy somewhere else was increased. So entropy is still increasing overall. Perhaps you read something like that and generalized it to a conservation law.. However, it doesn't work the other way around. So an entropy increase does not have to accompany an entropy decrease. The statement "an increase in the entropy of the system will be exactly equal to the entropy decrease of the surroundings. So the net change in the entropy of the system and its surroundings is zero" is straight of a kaplan physics review book for the MCAT. It didn't make sense to me because it does imply a conservation law, and then it confused me even more on the concept of entropy....if the equation for change in entropy is delta S = Q/T, if the environment loses heat to the system, wouldn't the environment have a negative delta S (and the system positive)? Does that mean entropy is decreased in the environment, and increased by the same amount by the system? Recognitions: Homework Help Fom Wikipwdia: "In systems held at constant temperature, the change in entropy, ΔS, is given by the equation $$\Delta S = \frac{Q}{T}$$, where Q is the amount of heat absorbed by the system in an isothermal and reversible process in which the system goes from one state to another, and T is the absolute temperature at which the process is occurring." ehild Thread Tools \| \| \| \| \|--------------------------------------------------------------------------------------------------------\|-------------------\|---------\| \| Similar Threads for: How is the entropy of the universe increasing when entropy is simply transferred? \| \| \| \| Thread \| Forum \| Replies \| \| \| General Astronomy \| 0 \| \| \| Classical Physics \| 7 \| \| \| Classical Physics \| 4 \| \| \| Classical Physics \| 9 \| \| \| Quantum Physics \| 10 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9447767734527588, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/94308/pushforwards-of-stacks-of-algebras	## Pushforwards of stacks of algebras? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) This is a refined/sheafified version of this previos question of mine. Let $(X,\mathcal{O}_X)$ be a ringed space or more in general a ringed stack, where the structure sheaf $\mathcal{O}_X$ is a sheaf of $\mathbb{K}$-algebras for some field $\mathbb{K}$. Then a global section of $\mathcal{O}_X$ can be thought of as a "scalar field" on $X$. In the "categorical progression" of higher $\mathbb{K}$-vector spaces, the field $\mathbb{K}$ is the 0th level. 1st level are classical $\mathbb{K}$-vector spaces, and so the "1st level version" of a section of $\mathcal{O}_X$ is (at least roughtly) a field of $\mathbb{K}$-vector spaces on $X$. A natural formalization of this rough idea is that of going from the sheaf $\mathcal{O}_X$ to the stack $\mathcal{O}_X$-Mod of sheaves of $\mathcal{O}_X$-modules on $X$ (maybe with suitable assumptions, e.g., one could be considering only quasi-coherent sheaves of $\mathcal{O}_X$-modules). A remarkable property of $\mathcal{O}_X$-modules is that they can be pushed forward along morphisms of ringed spaces. Now, the structure sheaf $\mathcal{O}_X$ we started with is in particular an $\mathcal{O}_X$-module, and pushing it forward along the terminal morphism $\pi:X\to {pt}$ we precisely get sections of $\mathcal{O}_X$ mentioned above. Now we can make a further step, and go from the stack of $\mathcal{O}_X$-modules to the 2-stack of $\mathcal{O}_X$-algebras (with $\mathcal{O}_X$-$\mathcal{O}_X$-bimodules as 1-morphisms and morphisms of bimodules as 2-morphisms). My question is: can $\mathcal{O}_X$-algebras be pushed forward along morphisms of ringed spaces? (under which hypothesis?). In particular, considering $\mathcal{O}_X$ as an $\mathcal{O}_X$-algebra, what is the $\mathbb{K}$-algebra `$\pi_\mathcal{O}_X$`? (the prototypical conjectural example of this in my mind is the following: if $G$ is a finite group, then `$\pi_\mathcal{O}_{pt//G}$` is Morita equivalent to $\mathbb{K}[G]$, the group algebra of $G$). In the above, an $\mathcal{O}_X$-algebra is to be thought as of a placeholder for its category of modules, so in a non-affine situation it is actually deceitful to reason in terms of algebras and a better description would be thinking of the "second step" as $\mathcal{O}_X$-linear categories, with additive functors and natural transformations of these. In this more general setting, the pushforward to a point of $\mathcal{O}_X$ would be the "pushforward of the 2-category of $\mathcal{O}_X$-linear categories" and this should be a $\mathbb{K}$-linear category, but not necessarily the category of representations of an algebra. In particular, by abstract nonsense I would expect this pushforward to be the category of $\mathcal{O}_X$-modules. Note that going one step backwards instead of one step forward, the existence of a pushforward is nontrivial: the pushforward of a section of $\mathcal{O}_X$ along $X\to {pt}$ is an integration of fuctions on $X$, so it requires additional data to be defined (a "measure"). - Given a morphism of ringed spaces, the pushforward of module sheaves is lax monoidal and therefore restricts to the category of algebras. See for example EGA I 4.2. But I assume that you know that and I've misunderstood the question ... – Martin Brandenburg Apr 17 2012 at 17:50 Hi Martin, thanks. Here I'm thinking of $\mathcal{O}_X$-algebras as the 2-category having $\mathcal{O}_X$-$\mathcal{O}_X$-bimodules as 1-morphisms and bimodule morphisms as 2-morphisms, not of 𝕆X-algebras as a subcategory of $\mathcal{O}_X$-modules. But I may be misunderstanding your comment: let me know about this. – domenico fiorenza Apr 17 2012 at 18:02 Ok, but the objects are just algebras? But doesn't the usual pushforward extend to your $2$-category? – Martin Brandenburg Apr 17 2012 at 18:10 Yes, but in this context an algebra is to be thought as a placeholder for its category of modules. Now that I write it I see that in the non-affine case one should not expect an algebra just like a quasi-coherent $\mathcal{O}_X$-module on $X$ is not the sheaf of modules associated with an $\mathcal{O}_X(X)$-module. I'll now edit accordingly my question. Thanks again. – domenico fiorenza Apr 17 2012 at 19:21 ## 1 Answer Yes, there are several different formalisms that achieve this - for example it's treated in Lurie's DAG XI or Toen-Vezzosi Caractères de Chern, traces équivariantes et géométrie algébrique dérivée in the derived context, and in many places (eg Gaitsgory The notion of category over an algebraic stack) in the underived context. Basically as I understand your question you're looking at an appropriate class of quasicoherent sheaves of categories on a variety or stack $X$ - which, given enough compact objects (see eg Toen's derived Azumaya paper) are monadic over $QC(X)$, i.e. are the same as module categories for $O_X$-algebras. But in fact such objects can be said much more concretely: all geometric stacks (eg quasicompact with affine diagonal) are "affine" from the point of view of these sheaves, ie there's no difference between such a quasicoherent sheaf of categories and its global sections (this is in DAG XI). So one can just take the following as definition: we're just looking at module categories for the symmetric monoidal (dg or $\infty$- if you like) category $QC(X)$ of quasicoherent sheaves on $X$. In this language pushforward and pullback of the kind you ask are super easy, just as they are for modules over ordinary rings: given a map $X\to Y$ you have a "homomorphism" (symmetric monoidal functor) $QC(Y)\to QC(X)$ (pullback), and you can use this to forget $QC(X)$ modules down to $QC(Y)$, i.e. push forward, or (the left adjoint of this) tensor a $QC(Y)$-module by $QC(X)$ (pullback). Moreover under some hypotheses ($X\to Y$ faithfully flat or $X\to Y$ proper and $X,Y$ smooth) descent holds - i.e. quasicoherent sheaves of categories on $Y$ are the same as those on $X$ with descent data. Said another way there's a Morita equivalence of monoidal categories between $QC(Y)$ and $QC(X\times_Y X)$ (theorem of Francis, Nadler and myself which will be posted, at least on my webpage, sometime in the next couple of weeks). In the case of $pt \to pt/G$ this recovers a Morita equivalence between categories with$G$ action and categories over $BG$. - Thanks a lot, David! – domenico fiorenza Apr 18 2012 at 5:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 72, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.903963565826416, "perplexity_flag": "head"}

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