Datasets:

keirp
/

tiny_math

Dataset card Data Studio Files Files and versions Community

Split (1)

train · 17.1k rows

url stringlengths 17 172	text stringlengths 44 1.14M	metadata stringlengths 820 832
http://mathoverflow.net/questions/119965/the-terminology-for-a-nodes-number-of-in-links-in-weighted-directed-graph/119970	## The terminology for a node’s number of in-links in weighted directed graph ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Given a weighted graph $G = (V,E)$, the in-degree of a node is defined as $k_{in}(i) = \sum_{j:j \rightarrow i} A_{ji}$ where $A_{ji}$ is the weight of edge from node $j$ to $i$. My question is if there are $m$ nodes pointing to $i$, what's the terminology for $m$? Is it the number of in-links? - ## 2 Answers The quantity you are asking about is usually called the "fan-in" of node $i$. The dual quantity which counts the number of nodes to which $i$ points is called the "fan-out". The terms are standard at least in the circuit design and computer science literature, but I can't seem to find them in graph theory texts at the moment. Searching google for something like directed graph "fan in" will reveal many papers that use the term. - Thank you Vel. I googled it and this is the exact word I am looking for. "fan−in" describes the situation vividly. – drk Jan 27 at 19:09 ### 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. I've seen papers where $m$ is called the in-degree and $k_{in}$ is called something else (such as weighted in-degree). Using in-degree and fan-in as Vel Nias suggests would be fine too. But you need to define it as there isn't universal agreement on how to interpret these names in the weighted graph context. - Thank you Brendan. I am also confused by the different definitions in differnt papers. I will take your advice to explicitly define $k_{in}$ and "fan-in" to avoid ambiguity. – drk Jan 27 at 19:14	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "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": 0.9547709822654724, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-statistics/175340-correlation-coefficient.html	# Thread: 1. ## Correlation coefficient Suppose i have 50 pairs that are independent X,Y. The observations are paired. I have the following estimator for the correlation coefficient. $r = \frac{\sum ^n _{i=1}(X_i - \bar{X})(Y_i - \bar{Y})}{\sqrt{\sum ^n _{i=1}(X_i - \bar{X})^2} \sqrt{\sum ^n _{i=1}(Y_i - \bar{Y})^2}}.$ How can I estimate the distribution of this estimator without knowing anything about the bivariate distriubution? 2. $r\sqrt{\frac{n-2}{1-r^2}}$ follows a $t$ distribution with $n-2$ df. 3. Thanks for answer. How can I show that? Is it possible to derive it? #### Search Tags View Tag Cloud Copyright © 2005-2013 Math Help Forum. All rights reserved.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.9001693725585938, "perplexity_flag": "head"}
http://mathoverflow.net/questions/103101/frobenius-eigenvalues-of-abelian-variety	## Frobenius eigenvalues of abelian variety ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $A$ be an abelian variety over a finite field `$\mathbb{F}_q$` and $x_i$ the Frobenius eigenvalues on $H^1$. Does $x_i \mapsto q/x_i$ permute the $x_i$, and why? It should follow from Poincare duality. - This sounds like homework. If so, you should ask it on math.stackexchange.com. That said, try the Riemann hypothesis instead of Poincare duality. – Will Sawin Jul 25 at 14:17 maybe PD with polarization. – shenghao Jul 25 at 18:03	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8164697885513306, "perplexity_flag": "middle"}
http://quant.stackexchange.com/questions/2000/when-should-you-build-your-own-equity-risk-model/2006	# When should you build your own equity risk model? Commercial risk models (e.g., Barra, Axioma, Barclays, Northfield) have evolved to a very high level of sophistication. However, all of these models attempt to solve a very broad set of problems. The optimal risk model for, say, risk attribution in a fundamental portfolio may differ substantially from the optimal risk model for downside risk estimation of an optimized quantitative strategy or for hedging unwanted exposures in a pure relative value play. Suppose that one already subscribes to a decent risk model provider, so that cost is not an issue. For what applications is it most appropriate to build your own equity risk model? What are the main benefits of a customized risk model? When is it worth the time and effort to replicate the increasingly sophisticated data cleaning/analysis and statistical methods to reap these benefits? - ## 2 Answers Great question. We would expect 3rd party risk providers to have specialized expertise (robust regression techniques, factor research, data cleansing etc.). We might grant them these advantages but still find weakness in the product design. Let's start off with the different uses of risk models and the procedure or metric which is maximized to solve for that use case. What we will see is that solving for a particular objective diminishes our ability to achieve other objectives. 1. Portfolio construction = If you want to construct a minimum variance portfolio, for example, then the key here is developing a covariance matrix (of factor returns) that is invertible and stable. So we could use procedures that develop well-conditioned cleansed covariance matrices. This conflicts with #3 2. Estimate beta for the purposes of hedging = here you care about the yet-to-be-realized return on a security you wish to hedge, and yet-to-be-realized returns on a basket which you will use to hedge. So if you wanted to create a market-neutral constraint, then you would want to use betas from a time-series regression so that the estimation error for each particular beta can be diversified away. You would also want to maximize accuracy (at the expense of interpretation) perhaps using statistical factor methods such as asymptotic PCA. 3. Performance reporting (risk and return decomposition) = here you have some contemporaneous regression specification (i.e. the time-index is the same on the left and right-hand side of the regression). Your concern is interpretability of factor exposures at the expense of accuracy. 4. Estimate marginal factor returns = use a cross-sectional regression to explain the returns accruing to a factor after controlling for all other factors. The technique is quite popular and used to explain the cross-section of returns or measuring the risk premiums for various factors. However, there is a substantial errors-in-variables problem. The errors in the estimated betas for such a security cannot be diversified away, unlike a time-series regression so it is risky to apply this model to other use cases. This conflicts with #2. 5. Risk forecasting = here you have a forecasting specification (left-hand side time index is $t+1$, right-hand side time index is $t$). This conflicts with #3. 6. Some people use risk models to make systematic factor bets. It can be difficult to develop a variant perception if you are using the same risk framework as everyone else. 7. Predict volatility. At shorter-horizons a stochastic volatility model would be appropriate, whereas at longer horizons a factor-model makes more sense. Any risk model that excels at one of these objectives will have severe weaknesses in some other areas. You could have multiple risk models (indeed Axioma has one for fundamentals that is easily interpretable, and another based on statistical methods for accuracy) but this can be confusing to clients. - Great answer! I see that essentially your answer is to be sure to use the right risk model for the task at hand. Thus my question may actually be six separate questions of which is better, build or buy, for the six use cases you outlined. I wonder, though, if there are any general principles for whether to build or buy, assuming that you would buy the appropriate risk model. – Tal Fishman Sep 23 '11 at 18:16	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9302241206169128, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/93867/coskeleta-of-simplicial-principal-bundles/93883	## Coskeleta of simplicial principal bundles ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $X_{\bullet}$ be a simplicial topological space. There is a truncation functor $tr^n \colon Fun(\Delta^{op}, Top) \to Fun(\Delta_n^{op},Top)$ (where $\Delta_n$ is the full subcategory of $\Delta$ that has the objects $[0], \dots, [n]$) which is just given by pullback with the inclusion. The functor $tr^n$ has a left adjoint called the skeleton $sk^n$ and a right adjoint called the coskeleton $cosk^n$. If I am not mistaken, we can use the following inductive definition of $cosk^n$: $cosk^n(X)_m = X_m$ if $m \leq n$ and `$cosk^n(X)_m = \{ (x_0, \dots, x_m) \ \| \ x_i \in cosk^n(X)_{m-1} \ {\rm with }\ d_i(x_j) = d_{j-1}(x_i)\ \text{if}\ i < j \}$` for $m > n$. Now suppose I have a simplicial principal bundle $P_{\bullet} \to X_{\bullet}$ for a simplicial topological group $G_{\bullet}$. By this I mean that $P_{\bullet}$ carries a simplicial action of the simplicial group $G_{\bullet}$, such that $P_m \to X_m$ is a principal $G_m$-bundle at each level $m$. If I apply $cosk^n$ to all spaces, I get a simplicial space $cosk^n(P)_{\bullet}$ with a projection map `$cosk^n(P)_{\bullet} \to cosk^n(X)_{\bullet}$` and a simplicial action by the group `$cosk^n(G)_{\bullet}$`, since $cosk^n$ is well-behaved with respect to products. Are there sensible conditions on $X$, $P$ and $G$ that allow to conclude that `$cosk^n(P)_{\bullet} \to cosk^n(X)_{\bullet}$` is a $cosk^n(G)_{\bullet}$-principal bundle? In particular, when is this map locally trivial? - ## 1 Answer If you interpret locally trivial as follows, then the answer is always, though it will actually be a $Cosk^{n-1}G_{\bullet}$ bundle. The interpretation of locally trivial which I want to suggest is that $P_{\bullet}\rightarrow X_{\bullet}$ is locally trivial if there exists a hypercover $c:U_{\bullet}\rightarrow X_{\bullet}$, such that $c^P_{\bullet}\rightarrow U_{\bullet}$ fits into a pullback diagram (which I'll describe since I can't figure out how to draw it in MathJax). The diagram expresses $f^P_{\bullet}$ as the pullback of the universal $G_{\bullet}$ bundle $WG_{\bullet}\rightarrow \bar{W}G_{\bullet}$ along a cocycle $U_{\bullet}\rightarrow\bar{W}G_{\bullet}$. With this definition of locally trivial, we can see that $Cosk^n$ preserves local triviality as follows: namely, $Cosk^n$ preserves limits (it's a right adjoint), so applying it to the pullback square, we get another pullback square. Now, a calculation in the definitions of $W$, $\bar{W}$ and the combinatorics of simplices shows that $Cosk^n\bar{W}=\bar{W}Cosk^{n-1}$. Working backward through the definitions, we see $Cosk^nP_{\bullet}$ is a $Cosk^{n-1}G_{\bullet}$ bundle over $Cosk^nX_{\bullet}$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 42, "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": 0.9196189641952515, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/27326/uniqueness-of-eigenvector-representation-in-a-complete-set-of-compatible-observa/27328	# Uniqueness of eigenvector representation in a complete set of compatible observables [duplicate] Possible Duplicate: Uniqueness of eigenvector representation in a complete set of compatible observables Sakurai states that if we have a complete, maximal set of compatible observables, say A,B,C... Then, an eigenvector represented by \|a,b,c....> , where a,b,c... are respective eigenvalues, is unique. Why is it so? Why can't there be two eigenvectors with same eigenvalues for each observable? Does maximality of the set has some role to play in it? I asked this question on Physics SE and was not satisfied with answers. Hope that I get help here. - 1 I am not sure this question is a right fit for this site, I think physics.se is a much better fit. Besides, seems to me that genneth already gave an excellent answer over there. – user566 Dec 6 '11 at 23:12 1 @Moshe: I didn't bother looking at the Physics.SE link before answering, but now you've pointed it out I agree that genetth's answer was perfect. – Joe Fitzsimons Dec 7 '11 at 3:00 ## marked as duplicate by Qmechanic♦Jan 21 at 13:15 This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question. ## 2 Answers Yes, since it is the maximal set of compatible observables, it includes all observables for which $\|a\rangle$, $\|b\rangle$, $\|c\rangle$, etc. are the eigenvectors (I'll use the notation $\|\psi_1\rangle$, $\|\psi_2\rangle$, $\|\psi_3\rangle$ etc instead). Hence this includes the observable $D = \sum_k k \|\psi_k\rangle \langle \psi_k\|$ . However $D$ has a unique set of eigenvectors, and hence so does the any compatible set of observables which contains $D$. - Two questions. Are your psi_1, psi_2 etc. all different numbers? If yes, then what if the spectrum is degenerate? If no, then I think your proof won't work. Second, why is D an observable? Isn't the definition of observable that it is a physical quantity that can be measured? All I can see is that D is a Hermitian Operator. Why should it be an observable? – Lakshya Bhardwaj Dec 6 '11 at 14:48 $\psi_i$ above are a basis for the Hilbert space in which all measurements are diagonal. If the set of measurements is maximal then it necessarily contains $D$ for some specific choice of basis. Since you specify the set of observables by their eigenvectors, you can explicitly construct $D$. – Joe Fitzsimons Dec 6 '11 at 14:54 4 Secondly, in quantum mechanics observable and Hermitian operator are synonymous. You can construct a physical measurement (in principle at least) for any Hermitian operator, and any physical observable is Hermitian. – Joe Fitzsimons Dec 6 '11 at 14:56 You can and you do sometimes have degenerate eigenvalues. This might just be a question of definitions: "a complete set of commuting observables (CSCO) is a set of commuting operators whose eigenvalues completely specify the state of a system (Gasiorowicz 1974, p. 119)." [1] That said, if you give me some cooked up Hamiltonian with at least one degenerate eigenvalue, perhaps one might be able to prove no observable commutes with it. - 2 That is incorrect. The Hamiltonian itself is an observable. Further, if you assign an arbitrary set of unique eigenvalues to the same eigenvectors (picking a basis for each degenerate subspace), thus lifting the degeneracy, this produces an observable which is simultaneously diagonalizable with the Hamiltonian, and hence commutes with it, but which has no degenerate eigenspaces. – Joe Fitzsimons Dec 9 '11 at 9:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "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": 0.9216806292533875, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/38874/derivability-conditions-for-robinson-arithmetic/38924	## Derivability conditions for Robinson arithmetic ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Two pieces of hearsay I have encountered about Robinson's Q: 1. Q fails to satisfy the Löb derivability conditions; 2. Pudlák criticised the Löb derivability conditions and suggested rival, weaker conditions. Which leads to three questions, if the above are right: 1. Which derivability condition(s) does Q not satisfy; 2. What were Pudlák's rival conditions and what was his complaint with the Löb conditions; and 3. Does Q satisfy the rival conditions? These questions arose from Carl Mummert's answer to a math.sx question of mine, Can Robinson's Q prove Presburger arithmetic consistent?. - I can take blame for passing along part (1) of the hearsay. When I went back to look for where I had seen it, I found published claims that Q is "too weak" to prove the derivability conditions, but not an explicit example nor a reference. Perhaps those authors were only referring to the fact that the usual proofs of the derivability conditions don't go through in Q, and I read too much into their statements. So I'm also interested to learn whether there is a published counterexample to the provability of the derivability conditions in Q. – Carl Mummert Sep 15 2010 at 21:42 ## 1 Answer I am not sure if this answers your question, but it was at least too long for a comment. First, note than one can interpret Sam Buss's bounded arithmetic theories like $S^1_2$ in $Q$, so it is not as weak as it seems at first sight in expressing and proving theorems. One can use a reasonable formula to exclude those non-standard numbers which are too pathological and prove consistency of $L$ (if $S^1_2$ proves consistency of $L$). I am not sure but you might find Pudlak's criticism in the last part of Hajek and Pudlak, and his alternative condition that you are looking for might be "sequential theory". Also take a look at this article which cites the Bezboruah et al. 1976 paper. (By the way, Bezboruah et al. (1976) seems to be a decade before Nelson's Predicative Arithmetic (1986) which shows that one can interpret $I\Delta_0$ in Q.) -	{"extraction_info": {"found_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": 0.9557866454124451, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/22084/list	## Return to Answer 2 fixed bugged latex Claim: Suppose that $G$ is a connected bounded open set in $\mathbb R^n$ such that for every $x\in\partial G$, there exist $\exists r>0$ and a half-space $S$ such that $x\in\partial S$ and $G\cap B(x,r)\subset S$. Then $G$ is convex. Proof: Step 1. Suppose that $f:G\to \mathbb R$ is a continuous function such that for every $x\in G$, there exists $r>0$ and a linear function $L_x$ satisfying $L_x(x)=f(x)$ and $f(y)<L_x(y)$ for all $y\ne x$ with $\|y-x\|<r$. Then $f$ is concave in the sense that if $a,b\in G$ and the whole interval $[a,b]$ is contained in $G$, then $f(ta+(1-t)b)\ge tf(a)+(1-t)f(b)$ for $t\in[0,1]$. Proof: Suppose not. Then $\min_t[f(ta+(1-t)b)-tf(a)+(1-t)f(b)]<0$. Take $s\in(0,1)$ to be the point where it is attained and let $x=sa+(1-s)b$. Then the linear function $L_x(ta+(1-t)b)-tf(a)+(1-t)f(b)$ has a strict local minimum at $t=s$, which is impossible. Step 2. We can replace the strict inequality in the conditions of Step 1 by a nonstrict one keeping the conclusion. Proof: Just subtract $\delta\|x\|^2$ with small $\delta>0$. Step 3: The distance to the boundary function satisfies the conditions of Step 1. Proof: Let $x\in G$. Let $y$ be the boundary point closest to $x$. Let $r$ and $S$ be the radius and the half-space for $y$. Then $L_x(z)=\text{dist}(z,\partial S)$ and $r$ work for $x$. Step 4: $G$ is convex. Proof: Take any 2 points $a,b$ in $G$. Suppose that the interval $[a,b]$ is not contained in $G$. Start moving $b$ towards $a$ along some path connecting them in $G$. Somewhere on the way, you'll get the situation when $a$ and $b$ are deep inside $G$ (that is true all the time) but $[a,b]$ is just barely inside $G$. Then the distance to the boundary dips on $[a,b]$, which is impossible due to the concavity just proved. The whole thing is certainly well-known and in good old times all of this would be written in most standard calculus textbooks (possibly, as an exercise). Unfortunately, nowadays we have to teach students to add fractions instead. Nevertheless, the textbooks in convex geometry and analysis written before 1980 would be your best bet if you want a reference. I would try something like Rockafellar's "Convex Analysis" to start with. 1 Claim: Suppose that $G$ is a connected bounded open set in $\mathbb R^n$ such that for every $x\in\partial G$, there exist $r>0$ and a half-space $S$ such that $x\in\partial S$ and $G\cap B(x,r)\subset S$. Then $G$ is convex. Proof: Step 1. Suppose that $f:G\to \mathbb R$ is a continuous function such that for every $x\in G$, there exists $r>0$ and a linear function $L_x$ satisfying $L_x(x)=f(x)$ and $f(y)<L_x(y)$ for all $y\ne x$ with $\|y-x\|<r$. Then $f$ is concave in the sense that if $a,b\in G$ and the whole interval $[a,b]$ is contained in $G$, then $f(ta+(1-t)b)\ge tf(a)+(1-t)f(b)$ for $t\in[0,1]$. Proof: Suppose not. Then $\min_t[f(ta+(1-t)b)-tf(a)+(1-t)f(b)]<0$. Take $s\in(0,1)$ to be the point where it is attained and let $x=sa+(1-s)b$. Then the linear function $L_x(ta+(1-t)b)-tf(a)+(1-t)f(b)$ has a strict local minimum at $t=s$, which is impossible. Step 2. We can replace the strict inequality in the conditions of Step 1 by a nonstrict one keeping the conclusion. Proof: Just subtract $\delta\|x\|^2$ with small $\delta>0$. Step 3: The distance to the boundary function satisfies the conditions of Step 1. Proof: Let $x\in G$. Let $y$ be the boundary point closest to $x$. Let $r$ and $S$ be the radius and the half-space for $y$. Then $L_x(z)=\text{dist}(z,\partial S)$ and $r$ work for $x$. Step 4: $G$ is convex. Proof: Take any 2 points $a,b$ in $G$. Suppose that the interval $[a,b]$ is not contained in $G$. Start moving $b$ towards $a$ along some path connecting them in $G$. Somewhere on the way, you'll get the situation when $a$ and $b$ are deep inside $G$ (that is true all the time) but $[a,b]$ is just barely inside $G$. Then the distance to the boundary dips on $[a,b]$, which is impossible due to the concavity just proved. The whole thing is certainly well-known and in good old times all of this would be written in most standard calculus textbooks (possibly, as an exercise). Unfortunately, nowadays we have to teach students to add fractions instead. Nevertheless, the textbooks in convex geometry and analysis written before 1980 would be your best bet if you want a reference. I would try something like Rockafellar's "Convex Analysis" to start with.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 104, "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": 0.9476820826530457, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/6363/tachyonic-complex-structure-directions-in-flux-vacua	# Tachyonic complex structure directions in flux vacua In flux compactifications to 4D, e.g. Type IIB on a CY orientifold $X$, one uses fluxes to stabilize the axio-dilaton $\tau$ and the complex structure moduli $z_a$ - the periods of the holomorphic three-form $\Omega$ over the basis three-cycles: $z_a=\int_{\alpha_a}\Omega$, by solving a system of equations for a supersymmetric extremum: $\partial_{\tau}W+W\partial_{\tau}K=0$, $\partial_{z_a}W+W\partial_{z_a}K=0$, where $W$ is the flux superpotential $W=\int_X G_3\wedge\Omega$ and $K$ is the Kahler potential. Since in Type IIB the supergravity scalar potential has a no-scale structure, i.e. $V=e^KD^iW{\bar D_i \bar W}$ where $i$ runs over the complex structure and the axio-dilaton, the potential is positive-definite and therefore the supersymmetric extremum appears as an actual minimum, although the Kahler moduli are still unfixed. However, once we include the non-perturbative corrections to $W$ to stabilize the Kahler moduli and break supersymmetry by e.g. an anti-D3 brane, the scalar potential is no longer positive-definite. So, my question is: is there a proof that the eigenvalues of the Hessian of the scalar potential with respect to all the moduli are positive-definite once all the moduli are fixed by the combination of fluxes and non-perturbative terms? In particular, there seems to be no good reason to believe that there are no tachyons in any of the complex structure directions. - ## 3 Answers Dear stringpheno, in the KKLT model, the uplifting by the anti-D3-brane means an addition of a positive term $D/\sigma^3$ to the potential. Here, $\sigma$ is what appears in $\exp(-K/3)$. It creates a new minimum for very small value of $\sigma$ because the coefficient $D$ is also expected (or required) to be small, and because the original second derivative near the zero was positively definite (a supersymmetric AdS vacuum can't have tachyons by supersymmetry) and $D/\sigma^3$ also has a positive second derivative, the second derivative at the new minimum - which is near the original $\sigma=0$ point - remains positively definite, too. Of course, if one allows $D$ to become large, the sketched derivation above won't necessarily hold. There can be local maxima somewhere in the configuration space - like the electroweak-symmetry-preserving point of the Higgs potential - which have tachyonic directions but they're not considered "uplifting by anti-D3-branes". - Dear @Luboš Motl, thank you very much for your answer. However, my question was more about the possibility of tachyons in the complex structure directions. – stringpheno Mar 4 '11 at 18:38 I think this is already implicit in Lubos's discussion, but to try to be more explicit: the assumption that's being made here is essentially just the usual logic of effective field theory and decoupling. The complex structure moduli will get large masses (which are maybe roughly of order the string scale times some small couplings), and those masses are not tachyonic to the extent that breaking of no-scale structure is small, by the argument you gave in your question. So you should imagine that you can supersymmetrically integrate them out in a consistent way, and then ignore them. The KKLT discussion proceeds after this point: it begins with a superpotential for the remaining Kähler moduli, assuming that there are small numbers involved ($W_0 \ll 1$ by tuning, and $A e^{-aT} \ll 1$ at the minimum dynamically, with similar smallness assumptions on SUSY breaking as Lubos mentioned). It is only to the extent that these remaining moduli fields are light compared to the complex structure moduli that you can consistently assume the CS moduli have been integrated out. (You can always play with a toy model where you keep a heavy modulus around and study how its VEV and mass change after turning on the no-scale breaking $Ae^{-aT}$ term; you will find that, provided you have a large hierarchy of scales in your model, the changes are small.) - – stringpheno Mar 11 '11 at 14:21 However, if we move one modulus, for concreteness the dilaton, away from its stabilised value, the resulting positive definite contribution $\frac{e^{K_{cs}}}{V^2}G^{\tau\bar\tau}D_{\tau}W\bar D_{\bar\tau}\bar W$ is only of the same order as the minimum. Moving the dilaton alters the value of $e^{K_{cs}}$ and thus may increase the numerator of the negative term. As the positive and negative contributions are of the same order, we see that depending on the magnitude of $G^{\tau\bar\tau}D_{\tau}W\bar D_{\bar\tau}\bar W$, this may in general decrease the overall value of the potential. – stringpheno Mar 11 '11 at 14:22 Therefore it is necessary to check explicitly for each choice of fluxes that the resulting potential has a minimum. – stringpheno Mar 11 '11 at 14:22 If you have a concrete model for $W$ and $K_{cs}$ that you're concerned about, we could calculate. But I maintain that if the complex structure directions are not minima at the point where the Kähler modulus is stabilized, it means they're being strongly influenced by the no-scale breaking $e^{-aT}$ terms and thus you never had a hierarchy of scales. – Matt Reece Mar 11 '11 at 15:37 The question is also dependent upon supersymmetry. For the case of $E_6$, relevant for the current “mini-twistor revolution,” positive condition appears straight forwards. The first Kahler form $K_1~=~ln(W(z)$ is computed by a three-form (or three $1,1$) forms from $h_{1,1}$ in the Hodge diamond, one element for a ${\bf\bar 27}$ of the $E_6$ $$W(z)~=~\int\omega\wedge\omega\wedge\omega$$ The holomorphic metric on the $E_6$ is given by this potential $$g_{a{\bar b}}~=~-\partial_{z^a}\partial_{{\bar z}^b}W(z)$$ The $h_{1,2}$ and $h_{2,1}$, given by $\bf 27$ of $E_6$ determine a second Kahler form $$e^{-iK_2}~=~\int\Omega_3\wedge\Omega^*_3$$ and the $\bf 27$ metric in generality is $g’_{a{\bar b}}~=~exp((k_2-K_1)/3) g_{a{\bar b}}$. The $3,0$ forms determine a superpotential as well, which is related to $K_2$ as the third Betti number is given by $h_{2,1}$, which is the cycle $int\Omega_{3,0}~=~w$ $K_2$ may be computed as well as $log(Im(z^a\partial_aw(z)))$, since $b_3~=~2h_{2,1}~+~2$. The superpotential is then the holomorphic function $$W(z)~=~\phi^a\phi^b\phi^c\frac{\partial^3w(z)}{\partial z^a\partial z^b\partial z^c}$$ Supersymmetry in its unbroken phase will have a zero energy, which will hold for any topology of the CY manifold. Broken supersymmetry will have positive energy, which is then mapped to the CY manifold. The holomorphic functions are then given by the topology of the CY manifold. However, in the Hodge diamond elements are chiral fields of $E_6$ which are positive. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 51, "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": 0.9457536339759827, "perplexity_flag": "head"}
http://mathoverflow.net/questions/85547/extensions-of-the-koebeandreevthurston-theorem-to-sphere-packing/85551	## Extensions of the Koebe–Andreev–Thurston theorem to sphere packing? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The Koebe–Andreev–Thurston theorem states that any planar graph can be represented "in such a way that its vertices correspond to disjoint disks, which touch if and only if the corresponding vertices are adjacent" (to quote Günter Ziegler, Lectures on Polytopes, Springer, 1995 p.117. (See also the Wikipedia article, "Circle packing theorem.") (Image due to David Eppstein, here.) What is the corresponding statement for spheres in $\mathbb{R}^3$? Every graph $G$ satisfying property $X$(?) can be represented by touching spheres. This is surely known—Thanks for pointers! - I do not see how to put the complete graph on 6 vertices, $K_6,$ in $\mathbb R^3.$ A regular tetrahedron with an extra point in the center gives $K_5.$ – Will Jagy Jan 13 2012 at 4:05 there also seems to be a related square tiling – John Mangual Jan 13 2012 at 4:10 It is a little uncertain, but the articles about recognizing ball-touching graphs appear to restrict to unit radius... citeseerx.ist.psu.edu/viewdoc/… – Will Jagy Jan 13 2012 at 5:53 ## 5 Answers Yes, certain restrictions are well known. One reference is Kuperberg & Schramm here ("Average kissing numbers for non-congruent sphere packings", 1994) which says that such graphs would have to have average degree <15. A more recent reference is Benjamini & Schramm here ("Lack of Sphere Packing of Graphs via Non-Linear Potential Theory", 2009) which shows that certain low degree infinite graphs are not realizable this way. - Also, $K_6$ cannot be a subgraph, from Ian's answer. – Will Jagy Jan 13 2012 at 6:12 Yeah, this is Exc. 20.12 (and a solution at the end) in my book math.ucla.edu/~pak/geompol8.pdf – Igor Pak Jan 13 2012 at 6:28 I see, part (c) , page 191, answer page 395, exactly what Ian came up with. Do you think we can always place a graph with $n$ vertices in $\mathbb R^{n-2}?$ – Will Jagy Jan 13 2012 at 7:00 Well, by Ian's argument, you can send two spheres to tangent hyperplanes, at which point you are asking if you can place a $K_{n-2}$ in $\mathbb{R}^{n-3},$ which you can, by the magic of regular simplices. – Igor Rivin Jan 13 2012 at 9:21 Igor, this seems to be the induction step in your book's answer. I am still uncertain about graphs on $n$ vertices that are not complete, in $\mathbb R^{n-1}$ or possibly $R^{n-2}.$ Steve Carnahan feels that there is enough room in $\mathbb R^{n-1}$ to slightly perturb the positions of the balls (so as to erase edges from a complete graph and realize the graph we are actually given), perhaps keeping all radii the same. Note that we can do all graphs with 5 vertices in $\mathbb R^3,$ if not complete then planar, if complete then a regular tetrahedron with a smaller ball at center. – Will Jagy Jan 13 2012 at 20:11 ### 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. I don't know of any results on this. There are some reformulations in terms of Gram matrices, but I don't know if they help. For $K_6$, there is no realization by touching spheres. Suppose you had such a realization. The property of having tangent spheres realizing a graph is invariant under Mobius transformations. In particular, one may perform a Mobius transformation sending the tangency between two spheres to infinity in $\mathbb{R}^3$. The two spheres are sent to parallel planes, and the other 4 spheres are tangent to both of these planes and to each other. In particular, the midplane between these two planes intersects the other 4 spheres in 4 tangent circles of equal radius. But 4 equal radius circles in the plane cannot be simultaneously tangent. So $K_6$ cannot be realized by tangent spheres. - Beautiful argument! Thanks! – Joseph O'Rourke Jan 13 2012 at 12:43 Looks like according to Pak this was discovered by Benjamini in 2010. – Agol Jan 13 2012 at 15:58 According to corollary 4.6 of "Representing Graphs by Disks and Balls (a survey of recognition-complexity results)" these graphs are NP hard to recognize. - 1 This is clearly THE answer, interesting though @Igor Pak's is. – Igor Rivin Jan 13 2012 at 12:42 I can make a pretty good case for sphere packing of a finite graph in some $\mathbb R^n.$ From http://en.wikipedia.org/wiki/Circle_packing_theorem#Generalizations_of_the_circle_packing_theorem we learn that a nonplanar graph still induces a circle packing on a compact orientable surface of larger genus, the surface having constant curvature. For example, for planar graphs we could realize them either as circle packings in the plane, then simply make those into spheres (thereby done), or we could make a circle packing on $\mathbb S^2$ and then ask whether we can blow up those circles into spheres with the same tangency relationships. The answer is yes, for each circle, take the sphere that intersects $\mathbb S^2$ orthogonally in precisely that circle. For torus graphs, I am betting on 3-spheres in $\mathbb R^4,$ where we can take the flat Clifford torus. For larger genus, from Nash embedding we can take our compact surface with constant curvature $-1$ in some $\mathbb R^n.$ Well, somebody just posted an answer, it says so at the top of the page, maybe they actually know something. The circle packing theorem generalizes to graphs that are not planar. If G is a graph that can be embedded on a surface S, then there is a constant curvature Riemannian metric d on S and a circle packing on (S, d) whose contacts graph is isomorphic to G. If S is closed (compact and without boundary) and G is a triangulation of S, then (S, d) and the packing are unique up to conformal equivalence. If S is the sphere, then this equivalence is up to Möbius transformations; if it is a torus, then the equivalence is up to scaling by a constant and isometries, while if S has genus at least 2, then the equivalence is up to isometries. EDIT: I suspect it is worth trying to disprove $\mathbb R^3$ for, say, $K_7,$ which is a torus graph. See Topological Graph Theory by Jonathan L. Gross and Thomas W. Tucker. If $K_7$ works in $\mathbb R^3$ try $K_8$ and $K_9,$ the dog graph. EDIT TOO: some anecdotal evidence, we can always place $K_n$ as the regular simplex in $\mathbb R^{n-1}.$ So now it is a question of how to selectively erase edges in order to get the graph we are actually given...Note that the articles on ball-touching in $\mathbb R^3$ all seem to be about balls of fixed unit radius. - In $\mathb{R}^{n-1}$, you have enough room to suitably perturb the simplex by some small $\epsilon$ without changing the sizes of the discs. If I'm not mistaken, you may also surround such a formation of discs with a $n+1$st disc turned inside-out (i.e., you can get a valid formation by a Möbius transformation). – S. Carnahan♦ Jan 13 2012 at 7:34 @S. Carnahan Thanks, I asked Igor the same thing, or tried to, I don't see that his argument applies to any possible incomplete graph on $n$ vertices. I think he was just showing me the induction step in the answer in his book, which is about the complete graphs. – Will Jagy Jan 13 2012 at 19:30 These are ball-touching graphs; according to some google sources they seem to be NP hard to recognize. -	{"extraction_info": {"found_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": 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": 0.9246017336845398, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/102322?sort=votes	## Analogues of fibrations ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Recall the following analogy Serre fibrations : Kan fibrations in spaces and simplicial sets respectively, related by the singular simplices functor and geometric realisation. There are other sorts of fibrations on each side. Can anyone fill in the following analogies ... quasifibrations : ?? ?? : inner fibrations ... if they do indeed exist. I'm not pedantic about using the specific adjunction $S\dashv \|-\|$. - The analogies I'm after may not be symmetric... – David Roberts Jul 16 at 6:19 Peter's and Charles's answers are excellent. It seems to me that to have a topological analogue of inner fibrations you'd need to use some sort of directed spaces. I thought about something like that for a while once, but didn't really get anywhere. – Mike Shulman Jul 16 at 19:48 ## 2 Answers I have a preprint http://arxiv.org/abs/math/9811038 where I develop a bit about the simplicial analogue of quasifibrations; they are called "sharp maps" there. (There is actually a kind of local-to-global principle for these too.) - ### 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. Quasifibrations can be defined simplicially since one can develop the analogue of the mapping class fibration and ask for an equivalence from the fiber to the homotopy fiber. (Edit: on second thought this isn't quite that obvious.) It is pointless to do so since the value of quasifibrations is their local to global topological characterization, which makes it easy to recognize them when you see them. Of course inner fibrations make no sense topologically, since all topological horns are created equal. It is the valuable idiosyncrasies of different Quillen equivalent categories that make it worthwhile having them. -	{"extraction_info": {"found_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": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9315648674964905, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/64216/list	## Return to Answer 2 fixed a statement For $R^2$, connected open subspaces are precisely the non-compact orientable surfaces with genus zero (see this question for more info.) For $R^3$, there is no homotopy criterion. The Whitehead manifold embeds in $R^3$ with complement the Whitehead continuum. The Whitehead manifold may be constructed as a union of solid tori, each one homotopically trivial in the next. One may modify the usual construction slightly by embedding each solid torus in the next with a little knot tied in it. Then one can show that this union does not embed in $R^3$. However, such manifolds are contractible, and therefore all homotopy invariants vanish. To see that it doesn't embed in $R^3$ (or $S^3$), suppose you had an embedding in $S^3$. Then the complement of each solid torus is a knot complement, and each knot is a satellite of the next one. If each such knot were non-trivial, then you would have a knot whose complement contained infinitely many incompressible tori, contradicting Haken finiteness. So infinitely many of these solid tori must have trivial complement (this is exactly what you see in the embedding of the Whitehead manifold). But one may see that the little knot we tie in each embedding of a solid torus in the next must make the resulting knot complement have the little knot as a connect summand, giving a contradiction to embedding in $S^3$. 1 For $R^2$, connected open subspaces are precisely the orientable surfaces with genus zero (see this question for more info.) For $R^3$, there is no homotopy criterion. The Whitehead manifold embeds in $R^3$ with complement the Whitehead continuum. The Whitehead manifold may be constructed as a union of solid tori, each one homotopically trivial in the next. One may modify the usual construction slightly by embedding each solid torus in the next with a little knot tied in it. Then one can show that this union does not embed in $R^3$. However, such manifolds are contractible, and therefore all homotopy invariants vanish. To see that it doesn't embed in $R^3$ (or $S^3$), suppose you had an embedding in $S^3$. Then the complement of each solid torus is a knot complement, and each knot is a satellite of the next one. If each such knot were non-trivial, then you would have a knot whose complement contained infinitely many incompressible tori, contradicting Haken finiteness. So infinitely many of these solid tori must have trivial complement (this is exactly what you see in the embedding of the Whitehead manifold). But one may see that the little knot we tie in each embedding of a solid torus in the next must make the resulting knot complement have the little knot as a connect summand, giving a contradiction to embedding in $S^3$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "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": 0.9432258605957031, "perplexity_flag": "head"}
http://mathoverflow.net/questions/80407/continuation-of-a-smooth-function	## Continuation of a smooth function ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Setting Suppose I have two bounded open domains $\Omega' \subset \Omega \subset \mathbb{R}^n$ (I'm particularly interested in case n = 2 or n = 3). We assume that all boundaries of domains are $C^\infty$-smooth and that inner domain is lying properly (with it closure) inside the outer: $\bar{\Omega'} \subset \Omega$. Suppose I'm given a smooth function $f \in C^\infty\left(\overline{\Omega \setminus \Omega'}; \mathbb{R}\right)$. We could assume that $L^\infty$ norm of all derivatives of $f$ is bounded. Question Is it possible to continue it smoothly inside $\Omega'$? What are the constructive ways to do it (or may be just with finite smoothness)? Example If we seek for just a continuous continuation of $f$, then we could put a "rubber film" over inner domain, or, formally, continue $f$ with the solution of the following Poisson equation: $\Delta u = 0$ in $\Omega'$ with Dirichlet boundary conditions: $u_{\partial \Omega'} = f_{\partial \Omega'}$. Update By constructive I mean "numerically friendly", i.e. easy to code. References and comments are appretiated! - 2 Look in Stein, "Singular integrals and differentiability properties of functions" – Deane Yang Nov 8 2011 at 17:14 3 For finite order smoothness, the Babitch extension does the job. It is used to extend Sobolev functions $u\in H^s(\Omega)$ to $H^s({\mathbb R}^n)$. – Denis Serre Nov 8 2011 at 17:20 @Dean Yang Thanks, the parity of unity answers the question of existence. Although from computational point of view it doesn't look very helpful . – Kirill Shmakov Nov 8 2011 at 17:52 If you know more about the domains (like explicit formulas or global co-ordinates), you don't need to use a partition of unity. You can probably do what Stein does locally but do it globally. – Deane Yang Nov 8 2011 at 18:10 ## 2 Answers This answer may not be practically useful to you, but I think it's nice from a conceptual point of view. The extension could be done in two steps. I'm presuming that you are defining a smooth function on the closed set $K=\overline{\Omega\setminus\Omega'}$ as one that is smooth on a small neighborhood $K'$ (which I'll also take as closed) $K$. First, you could extend $f$ continuously from the closed set $K'$ to all of $\Omega$. Second, you could smooth the extension, preserving it on the interior of $K$. The continuous extension is a direct application of the Tietze extension theorem, while the smoothing is an application of the Steenrod approximation theorem. - @Igor Not exactly: function is given to me as a result of numerical evaluation of Navier-Stokes. I don't know anything specific about it, so the continuation on a neighborhood is as hard as a total continuation. – Kirill Shmakov Nov 9 2011 at 23:58 1 @Kirill, then yor hypotheses should say that $f$ is only smooth onthe interior of $\Omega\setminus\Omega'$. Also if you are not dealing with questions of principle, but only discretized approximations, you could just break up your domain info simplices and use splines. If you are then interested in the combinatorics of precisely how to do that, that's a different question. – Igor Khavkine Nov 10 2011 at 8:36 @Igor Ok, I mean let it has this smooth continuation. The question then is how to compute it inside (numerically)? As I understand splines give polynomial continuation, but may be there is still possibilities to do it either more elegant or computationally simple. – Kirill Shmakov Nov 10 2011 at 17:53 1 @Kirill, you are now asking a very different question than originally posted. If you put any set of values on a finite grid, you can fit them with splines of sufficiently high polynomial order. That's what they are for: to construct finitely differentiable fits to arbitrary data. The construction of the splines will of course depend on the underlying grid. I realize that you probably won't find this answer satisfactory, but it's impossible to provide a better one without more constraints on the extension. Otherwise, as you can see, there are too many solutions to choose from. – Igor Khavkine Nov 10 2011 at 20:44 @Igor, I agree that the question in thew form I posted it useless and I need to change it. Anyway, thank you for comments and answering. – Kirill Shmakov Nov 13 2011 at 20:54 ### 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 has basically already been answered, but because all your boundaries are smooth (or, more importantly, $C^1$), and your domains bounded, then if you want an explicit smooth continuation, you could do the following: for a sufficiently small $\epsilon > 0$, you can construct a subset $V \subset \Omega'$ such that for any $x \in \partial V$, dist($x,\partial \Omega'$) = $\epsilon$. Then, "radially" along these lines of length $\epsilon$ connecting the boundary of $V$ to the boundary of $\Omega'$, you can have your continuation of $f$ smoothly vanish to 0 using a scaling of the form $e^{-1/x}$. Then set $f = 0$ on $V$. Therefore this continuation of $f$ will "mostly" vanish in $\Omega'$. You're basically just constructing a smooth mollification of $f \chi_{\Omega \setminus \Omega'}$. The problem with this construction is that, from an application point of view, it's not very helpful. By making $f$ vanish on the interior set, you've basically lost all the information that was encoded in $f$ in the outer set. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 41, "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": 0.9471787214279175, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/17574/how-can-i-find-the-potential-created-by-spherical-capacitor-with-dielectric-mate?answertab=active	# how can I find the potential created by spherical capacitor with dielectric material If we have a spherical capacitor with inner radius or r1 and outer radius of r2, with charges (+/-)q on them and there is a dielectric material (with constant e) in between them with. What kind of a potential would this create outside the entire capacitor? in the region with the dielectric? and inside the entire thing? - ## 2 Answers You can write down these fields directly without calculating anything because the field of a charged sphere is the same as that of a point charge outside the sphere and zero inside the sphere. http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potsph.html - This is quite simple as this is a standard equation for the capacitance. You will have $$C=\epsilon C_0$$ being $C_0$ the capacitance in vacuum. For a full set of formulas in different geometrical configurations see 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": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.933580219745636, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/47154/null-sequences-ii	# Null Sequences (II) I came across the following problems on null sequences during the course of my self-study of real analysis. Let $x_n = \sqrt{n+1}- \sqrt{n}$. Is $(x_n)$ a null sequence? Consider $y_n = \sqrt{n+1}+ \sqrt{n}$. Then $x_{n}y_{n} = 1$ for all $n$. So either $(x_n)$ or $(y_n)$ is not a null sequence. It seems $(y_n)$ is not a null sequence. I think $(x_n)$ is a null sequence because $\sqrt{n+1} \approx \sqrt{n}$ for large $n$ which implies that $x_n \approx 0$ for large $n$. If $(x_n)$ is a null sequence and $y_n = (x_1+ x_2+ \dots + x_n)/n$ then $(y_n)$ is a null sequence. Suppose $\|x_n\| \leq \epsilon$ for all $n >N$. If $n>N$, then $y_n = y_{N}(N/n)+ (x_{N+1}+ \dots+ x_n)/n$. From here what should I do? If $p: \mathbb{R} \to \mathbb{R}$ is a polynomial function without constant term and $(x_n)$ is a null sequence, then $p((x_n))$ is null. We know that $\|x_n\| \leq \epsilon$ for all $n>N$. We want to show that $\|p(x_n)\| \leq \epsilon$ for all $n>N_1$. We know that $p(x) = a_{d}x^{d} + \cdots+ a_{1}x$. So $$\|p(x_n)\| \leq a_{d} \epsilon^{d} + \cdots+ a_{1} \epsilon$$ for all $n>N_1$. - For the first question, the fact that $\sqrt{n+1}\overset{+\infty}{\sim}\sqrt n$ doesn't imply the result since we have for example $n+1\overset{+\infty}{\sim}n$. But you noticed that $x_n =\frac 1{y_n}$. Did you compute $\lim_{n\to +\infty}y_n$? – Davide Giraudo Jun 23 '11 at 14:38 Since $x_ny_n=1$, what is $x_n$....For the second question, what can you say about each bracket? Second one is less than $\epsilon$, while in the first one one $n$ changes... – N. S. Jun 23 '11 at 14:38 For the third question, maybe you should denote the degree of $p$ by $d$ instead of $n$. You can only check the definition for $\varepsilon <1$. Show that for $n\geq N$ we have $\displaystyle \|p(x_n)\|\leq \sum_{k=1}^d\|a_k\|\varepsilon$. – Davide Giraudo Jun 23 '11 at 14:42 I'm not sure about the meaning of $\approx$ but I think your argumentation i wrong. From $x_ny_n=1$ follows $x_n=\frac{1}{y_n}$. Perhaps you can show that $\frac{1}{y_n}$ is a null sequence? – miracle173 Jun 23 '11 at 14:44 ## 2 Answers You have made pretty good progress on all three problems. For problem 1: note that in fact $\lim_{n \rightarrow \infty} y_n = \lim_{n \rightarrow \infty} \sqrt{n+1} + \sqrt{n} = \infty + \infty = \infty$, so $\lim_{n \rightarrow \infty} x_n = \lim_{n \rightarrow \infty} \frac{1}{y_n} = \frac{1}{\infty} = 0$. For problem 2: you have $y_n = y_N (N/n) + (x_{N+1} + \ldots + x_n)/n$. It's enough to show that both terms of the right hand side can be made arbitrarily small as $n$ gets arbitrarily large. The first term is a constant divided by $n$: this goes to zero with $n$. The second term is a sum of at most $n$ things each one of which is in absolute value at most $\frac{\epsilon}{n}$, so the sum is in absolute value at most $\epsilon$. So you're basically done. For problem 3: If you choose $\epsilon \leq 1$ then $\epsilon^n \leq \epsilon$ for all $n \geq 1$, so $\|a_d \epsilon^d + \ldots + a_1 \epsilon\| \leq \|a_d + \ldots + a_1\| \epsilon$, a quantity which goes to zero with $\epsilon$. - So we want to show that $\|p(x_n)\| \leq \epsilon$ for all $n > N_1$. We know that $\|x_n\| \leq \epsilon$ for all $n >N$. Can we choose $N_1 = N$? – Damien Jun 23 '11 at 15:06 @Damien: no, we cannot. Think about the easy case in which $p(x) = a x$: we will be "off by a constant". In general, we want to choose $N$ such that for all $n \geq N$, we have $\|a_1 + \ldots + a_d\| \|x_n\| < \epsilon$. I am starting to get the impression that you are good at finding the right algebraic manipulations to do these proofs but are a little shaky on the logic behind them. If so, perhaps you could ask further questions with more of a focus in this direction. – Pete L. Clark Jun 23 '11 at 15:46 Try this. $$x_n = {(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1} + \sqrt{n})\over \sqrt{n+1} + \sqrt{n}} = {1\over \sqrt{n+1} + \sqrt{n}}.$$ This shows $x_n\to 0$ as $n\to\infty$. - $y_n$ should be $x_n$, of course. (+1 anyway.) – Shai Covo Jun 23 '11 at 14:49 1 Thanks, Shai. I fixed the x-rated problem. y-not? – ncmathsadist Jun 23 '11 at 15:08	{"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": 74, "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": 0.9460114240646362, "perplexity_flag": "head"}
http://mathoverflow.net/questions/96976/kakeya-problem-and-arithmetic-progressions	## Kakeya problem and arithmetic progressions ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Here's something I also posted on Stackexchange recently. It's very related to the Kakeya problem, yet I fail to see why this is true. It goes like this: Let $r > 2$ be an integer parameter. Let $K \subset [N]=1,\ldots,N$ be a set containing arithmetic progressions of length $r$ in at least $N' \leq N/r$ (up to some constant) different dirrections ($b$ is the integer of the arithmetic progression $a,a+b,a+2b,\ldots$). Then, for every $r$, show that there exists a set $K$ as above of size at most $N^{1-\epsilon_{r}}$ with $\epsilon_{r} > 0$ a constant depending only on $r$. I am kind of confused b/c I read some time ago about some result of J. Bourgain which basically said that a similar conclusion would give the validity of the general Kakeya, so I feel this is very hard, yet I got it as an exercise, so I'm probably missing something. Also, it would be very nice if someone can give some references about Bourgain's result which I'm talking about - I can't find it anywhere. Thanks PS. can't login into the old account. - Kakeya (for Minkowski dimension) would follow from the assertion that it is impossible to make $\epsilon_r$ a constant independent of $r$. – Boris Bukh May 15 2012 at 10:42 1 Someone has flagged this question, claiming that it is a homework problem. I don't know enough to tell if that is a reasonable conclusion. (What was your old account?) – S. Carnahan♦ May 15 2012 at 11:13 1 Can you edit in the link to the stackexchange post? Wouldn't want to have any duplication of effort. – Gerry Myerson May 15 2012 at 12:18	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "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": 0.9582653641700745, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/107387-how-factorise-expression-without-sollution.html	# Thread: 1. ## How to factorise expression without sollution? I've got an expression: $x^4+5x^2+6$. How can I factorise it? All methods I know are based on finding one solution and simplifying the expression to (x-a)(...). But this one is always positive, no matter x, so it's not possible here. It's a polynomial of degree 4, so it must be possible to at least make it to the form $a(x^2+bx+c)(x^2+dx+e)$ How can I do that? (only real numbers, please) 2. Are you sure that there are two variables in your expression ( $x$ and $s$)? I think they want $\left(2+x^2\right) \left(3+x^2\right)$... 3. Right, that's a typos. $x^4+5x^2+6$ 4. Try look under Application to higher-degree equations. Quadratic equation - Wikipedia, the free encyclopedia 5. hello put $x^{2}=u$,and factorize. 6. The problem is that in this way I get $x^2<0$ and I'd rather not mess up with complex numbers. edit: To be specific: $x^2=-2 \lor x^2=-3$ if I've got it right. 7. ## Re: How to factorise expression without sollution? take x^2 = a a^2 + 5a + 6 = 0 solve this simple quadratic eqn and then put the value of a = x^2 and solve further 8. well, $x^4+5x^2+6$ has only complex roots. 9. Originally Posted by Raoh well, $x^4+5x^2+6$ has only complex roots. Yes, but as far as I know any polynomial can be written in form of multiplied $(x-a)$ and $(ax^2+bx+c)$ (in this case it would be only the latter). Or at least that's what I was taught. 10. well in $\mathbb{R}$,the only factorization you can get is $(x^2+2) (x^2+3)$. 11. in your $4^{th}$ post, $(a,b,c) \in \mathbb{R}^{3}$. 12. Originally Posted by Raoh well in $\mathbb{R}$,the only factorization you can get is $(x^2+2) (x^2+3)$. Ahh, what a fool I am! That's what I was looking for	{"extraction_info": {"found_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": 19, "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": 0.9526845812797546, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/calculator?sort=active&pagesize=15	# Tagged Questions Questions regarding calculators such as the TI-83 (Plus), HP 50g, and other calculator models. learn more… \| top users \| synonyms 0answers 31 views ### TI83+ emulator for Windows 7? (Old XP one doesn't work) [closed] I just upgraded from XP and my old TI83+ emulator/ROM does not seem to work anymore. I understand this is probably not the right place to ask, but I know someone here will know the answer, which is ... 2answers 215 views ### How do I find, algorithmically, which parts of a given function are interesting to graph? I'm building a program that does 2D graphing, and was wondering: How can I determine the default zoom level and x/y extents to display on screen, in such a way as to maximise the 'interesting' parts ... 1answer 51 views ### A calculator's solution to irrational exponent An irrational number cannot be represented by $\frac{p}{q}$ where $p$ and $q$ are integers. And when we encounter exponents with decimal points, it is a possible way and a rather simple one to turn ... 1answer 45 views ### Checking for symmetry in a graph on the TI-84 Plus I was using the TI-84 Plus calculator to check for Y-Axis symmetry in the graph of "cos(2X+1)" and as I had to do it fast (I was in the middle of a test), I quickly noted that it was symmetrical. I ... 2answers 64 views ### Expecting a Discontinuity in Piecewise Function on TI-89 but finding Unexpected Results When I look at the following function $f(x)$, it would seem to me that because there are no $\le$ or $\ge$ inequalities included where the function's formula changes (for the parts relating to ... 2answers 80 views ### How can I generate a random number on my ti-84 without the rand function Is there an equation that is possible where I enter a seed, and it will give me the number from the seed. I want it to give me the same results as the rand function would give with the seed as 0. I ... 1answer 61 views ### Getting the Total value from the Nett I just cant figure out this formula, I thought I will take expert advice to write formula for my php script. I have a value of $\$80$.$\$80$ has come from a total sale of $\$100$and$ 20\%\$ ... 0answers 140 views ### ti 89 complex eigenvectors [closed] I have a TI-89 calculator and when creating a certain matrix A and then use the command eigenvl(A) it gives me a Error-vectors contain complex numbers. How can I find my eigenvectors of a matrix ... 1answer 1k views ### Calculating the percentage of the difference between two values If I have the following two different values for x: x = 0.022 x = 0.020 And, if I want to calculate the percentage of the difference (change) in x, will it be? ---> 0.022 - 0.020 = 0.002 ---> ... 0answers 57 views ### equivalent calculator keys to: inv ln What are the equivalent keys on a calculator to inv + ln? My calculator doesn't have a inv key: 2answers 170 views ### What happens as you take repeated square roots, starting with 8? What does the answer approach as you take more and more square… a) What happens as you take repeated square roots, starting with 8? b) What does the answer approach as you take more and more square roots? c) Would the answer be the same if you started with any ... 0answers 93 views ### How does my $10\times10$ abacus work? How does this $10\times10$ abacus work? More specific: Counting: is it common to count from 1 to 100, or from 1 to 9.999.999? How does addition work? How does multiplication work? Are there any ... 2answers 61 views ### Calculator question involving $\log_2$? I have a question, I have a calculator that does $\log$ but I think it does it it in a base ten format for example $\log_{10}(100)=2$ I am wondering how I can solve $\log$ using a base of 2 for ... 0answers 76 views ### Weibull Distribution Calculation Sequence I am in the process of calculating a Weibull Distribution Calculation as follows: $1-e^{-(6.5/6.0)^2}$ Note: Last figure is a squared value. I am having trouble however remembering the correct input ... 2answers 47 views ### Calculate on which side of straign line is dot located I am a programmer without really good knowledge in math. :/ So I have to write an algorithm that changes the color of pixel(dot) P to opposite if it's on left side of the straigt line in coordinate ... 2answers 82 views ### In a Calculator, Which Button Should be Bigger: $\sinh()$ or $\rm asin()$? I am making a calculator app and would like to know: Which of the following functions would be used more often: $\sin()$ $\sinh()$ $\text{arsin}()$ ($\sin^{-1}$) $\text{arsinh}()$ ($\sinh^{-1}$) ... 0answers 70 views ### TI-89 calculating incorrect power regression. I have a TI-89 and use the "Stats/List Editor" to calculate power regression. My numbers are as follows (X , Y): 10 169 15 120 20 94 25 78 30 67 35 58 When ... 0answers 121 views ### Is there any derivative that the Ti-Nspire can't do? A fellow student was showing off by quickly have his ti-nspire solve the homework problems. Just wondering if there is any derivative that the ti-nspire would mess up on. Just where is the limit ... 2answers 102 views ### Is it possible to get any number from two numbers? Is it possible to get any give number with a minimum value and a maximum value? to output any number you desire? for example ... 0answers 132 views ### Calculating confidence intervals for variance and standard deviation on TI 89/Stat List Editor We're learning about problems like the ones below in my stat class right now, but my professor has no idea how to use a TI 89, so I'm usually stuck in class. Would anyone know how to solve the ... 1answer 44 views ### Help using TI-83 to find max and min… This may be somewhat of a basic question, but I need to know... I'm trying to find the minimum and maximum for the following: $$y = \frac{\cos 3x}{5}+\sin 3x$$ How is this done on a TI-83? 2answers 852 views ### Casio fx-85GT PLUS calculator. How do I access & then use the quadratic equation formula? Have looked in the user guide - cannot find information on use of the quadratic formula. 3answers 221 views ### How to determine end behavior with my TI-83 So, I need to determine the end behavior of $P$ for this equation: $$P(x) = −x^5 + 3x^2 + x.$$ If possible, how is this done? 1answer 106 views ### Why does calculator say that $\sqrt{4} = 2$? [duplicate] Possible Duplicate: Square roots — positive and negative I know that the square roots of four are 2 because $2^2=4$ and $-2$ because $(-2)^2=4$ Are there any conventions in ... 2answers 376 views ### Gr. 11 Functions - Limiting domain and range on a graphing calculator I'm trying to do a homework assignment on my graphing calculator app for my ipad (Quick graph+). (Edit by Willie: The images are originally posted here http://imgur.com/a/XRj4g I've included them ... 1answer 441 views ### How to get complex mode on a Casio fx-83GT plus scientific calculator [closed] questions all in the title. My first question on this site! Can anyone help me? 1answer 65 views ### Is $\log_{5}{-3} = \frac{\log(3)+\pi i}{\log(5)}$? Why does my calculator return false when I input $\log_{5}{-3} = \frac{\log(3)+\pi i}{\log(5)}$ but W\|A returns true? I'm thinking my calculator is wrong because I know that \$\displaystyle ... 1answer 151 views ### Subtraction of numbers with arbitrary bases [duplicate] Possible Duplicate: How to do +, -, *, / with number in a base b? I am reading research papers in the category of recreational mathematics on the topic of numbers similar to Kaprekar ... 1answer 164 views ### Integration from 0 to 0 - why does my calculator say “undefined” in one case, and “0” in another? I have the following question: Using my calculator, $\displaystyle\int_0^{0} \frac{1}{x}dx$ is "undefined". But when I type $\displaystyle\int_0^{0} - \frac{\ln(1-t)}{t} dt$, the result is 0. What ... 1answer 162 views ### Drawing on calculator, Parametric Equation. What does the “pitch” do? Casio So under options "view window" on my calculator I can't understand what the value under pitch does to the drawing. I'm just drawing simple 2D vectors. This is high school stuff, so it isn't hard. But ... 3answers 254 views ### What is the definition of $2.5!$? (2.5 factorial) I was messing around with my TI-84 Plus Silver Edition calculator and discovered that it will actually give me values when taking the factorial of any number $n/2$ where $n$ is any integer greater ... 2answers 121 views ### Calculators: Algebraic Equations and Exams Kind of a stupid question, but I don't have much experience with calculators and thought some of you might. I'm sitting an exam and would like to use my calculator, in particular this one: ... 2answers 274 views ### How do calculators handle $\pi$? When the calculator displays the digits of $\pi$, how does it arrive at that answer? Also, at what digit does the approximation of $\pi$ stop at? 0answers 114 views ### How to get the LinReg(a+bx) function to display the correlation At the beginning of the semester, when I wanted to find the correlation between two variables I would use the LinReg(a+bx) function on my TI-83 Plus calculator (you get to this function by Stat -> ... 2answers 621 views ### How do you plot $x = y^2 - 6$ on a graphing calculator? I have a graphing calculator app (Graphing Calculator+) that only allows me to enter $x$ as a variable, but I need to graph $x = y^2 - 6$. I haven't used a graphing calculator in awhile. Is this ... 2answers 386 views ### Base-2 exponentiation bug on Citizen SR-270X for $2^{255}$? According to my calculator, $2^{255} = 0.5$. Also, for the range $2^x, x\in[254.5, 255.5)$ the result spans $[0.3535533906, 0.7071067812]$. Outside of that, the answers seem correct. Curiosly, ... 1answer 615 views ### Casio fx-82MS: n-th logarithm I have got a Casio fx-82MS calculator, and I want to calcuate various logarithms with it, especially $\log₂(x)$-style logarithms. Unfortunately, its user interface contains buttons for just $\ln$ and ... 8answers 558 views ### Computing a very large number I've been making a program that can delay for any number of nanoseconds 0-65534 to any power 0-65534. That is to say, the shortest possible delay is 0^0 nanoseconds (1) and the longest possible delay ... 1answer 70 views ### Finding the range with a statistics calculator Is there a button on statistics calculators that easily finds the range of the numbers? I am using window's pre-installed calculator. 3answers 664 views ### How can I evaluate an expression like $\sin(3\pi/2)$ on a calculator and get an answer in terms of $\pi$? I have an expression like this that I need to evaluate: $$16\sin(2\pi/3)$$ According to my book the answer is $8\sqrt{3}$. However, when I'm using my calculator to get this I get an answer like ... 1answer 220 views ### Negative × Negative = Positive… right? The wife and I were doing homework together, and we noticed something really strange when charting quadratics with a TI-series graphing calculator: ... 2answers 313 views ### Do calculators have floating point error? As a programmer, we have been told about floating points errors on computer. Do Calculators have floating point error too? Example. ... 3answers 3k views ### Calculator algorithms Does there exist a good reference on the algorithms used by calculators, especially on the trigonometric and transcendental functions? I would still like to know how Casio generates its random ... 7answers 2k views ### How to calculate $e^x$ with a standard calculator Is there a simple method for calculating the $e^x$ ($x\in\mathbb{R}$) with a basic add/subtract/multiply/divide calculator that converges in reasonable time, preferably without having to memorize ... 1answer 42 views ### Calculating new score weight See I have a test combined from 3 segments. The first two are 15% each, and the third is 70%. I can't do the second part, so they told me the scores would be 20% and 80% instead. Now, the scores I ... 1answer 2k views ### Financial Calculator HP 10bII - Standard Deviation with Probability I have a financial calculator and am trying to use it to calculate the standard deviation when probability is involved. For a standard deviation question say on the numbers 1, 5, 6, 10 ... 2answers 247 views ### Numbers too large for my calculator, probability. Old calculator or should it be done another way? It's basic binomial probability, the problem is that the numbers are to large for my calculator. Is this the point of the task, to force me to use another method? It doesn't seem lightly, because n ... 3answers 5k views ### TI-83 Plus: Is it possible to find the eigenvalues on this calculator? I searched the Internet and can't seem to find a source on how to find the eigenvalues on a TI-83 Plus calculator. Can anyone give me a hand? 1answer 229 views ### What does the hyp checkbox do on calculators? What does the hyp checkbox do on calculators (for example on the Microsoft Windows calculator)? 0answers 205 views ### how do you define the decimals indicator: E-3? Sorry for the terrible question, but how do you define E-3, which is used in the calculators to indicate that the first 3 decimals in the number are not displayed ? ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 56, "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": 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": 0.9257921576499939, "perplexity_flag": "middle"}
http://en.m.wikibooks.org/wiki/Cryptography/Meet_In_The_Middle_Attack	# Cryptography/Meet In The Middle Attack An extremely specialized attack, meet in the middle is a known plaintext attack that only affects a specific class of encryption methods - those which achieve increased security by using one or more "rounds" of an otherwise normal symmetrical encryption algorithm. An example of such a compound system is 3DES. However, to explain this attack let us begin with a simpler system defined as follows: Two cryptographic systems denoted $encrypt_\alpha$ and $encrypt_\beta$ (with inverse functions $decrypt_\alpha$ and $decrypt_\beta$ respectively) are combined simply (by applying one then the other) to give a composite cryptosystem. each accepts a 64 bit key (for values from 0 to 18446744073709551615) which we can call $key_\alpha$ or $key_\beta$ as appropriate. So for a given plaintext, we can calculate a cryptotext as $cryptotext=encrypt_\beta( key_\beta , encrypt_\alpha(key_\alpha,plaintext))$ and correspondingly $plaintext=decrypt_\alpha( key_\alpha , decrypt_\beta(key_\beta,cryptotext))$ Now, given that each has a 64 bit key, the amount of key needed to encrypt or decrypt is 128 bits, so a simple analysis would assume this is the same as a 128 bit cypher. However, given sufficient storage, you can reduce the effective key strength of this to a few bits larger than the largest of the two keys employed, as follows. 1. Given a plaintext/cyphertext pair, apply $encrypt_\alpha$ to the plaintext with each possible key in turn, generating $2^{64}$intermediate cryptotexts $cryptotext_1$$\rightarrow$$cryptotext_n$ where $n=2^{64}$ 2. Store each of the $n$ cryptotexts in a hash table so that each can be referenced by its cryptotext, and give the key used to generate that cryptotext 3. Apply $decrypt_\beta$ to the ciphertext for each possible key in turn, comparing the intermediate plaintext to the hash table calculated earlier. this gives a pair of keys (one for each of the two algorithms employed, $\alpha$ and $\beta$) 4. Taking the two keys from stage 3, test each against a second plaintext/cryptotext pair. if this also matches, odds are extremely high you have a valid keypair for the message - not in $2^{128}$ operations, but a "mere" $2x2^{64}$ operations (which nonetheless are significantly longer due to the hash table operations, but not so much as to add more than a couple of extra bits worth of time to the complexity of the task) The downside to this approach is storage. Assuming you have a 64 bit key, then you will need at least $2^{64}$units of storage - where each unit is the amount of space used by a single hash record. Even given a minimal implementation (say, 64 bits for the key plus four bits hash collision overhead), if you implemented such a system using 160GB hard drives, you would need close to one billion of them to store the hash table alone.	{"extraction_info": {"found_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": 21, "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": 0.9232715964317322, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/45347/a-question-about-an-example-on-flat-families-from-hartshorne-in-particular-i	# A question about an example on flat families from Hartshorne. In particular, is this local ring reduced? Is the local ring $R_p$ reduced, where $p=(a,x,y,z)$ and $R=k[a,x,y,z]/I$ and $I=(a^2(x+1)-z^2,ax(x+1)-yz,xz-ay,y^2-x^2(x+1))$ ? This comes from example III.9.8.4 in Hartshorne's algebraic geometry. There he explains that $R/(a)$ has a non-reduced local ring at $p=(x,y,z)$. My question is motivated by the following (all reference numbers are from Hartshorne): • I think that $X=Spec R$ is not reduced at $(0,0,0,0)$ (corresponding to $p$). • There is a map $X_{red}\to X$ that gets rid of the non-reduced structure. • The composition $X_{red}\to X\to Spec\quad k[a]$ should be the family he mentions in example III.9.10.1. • III.9.7 says that a reduced and irreducible family X over a smooth Y is flat (X dominating Y). • The last two bullets contradict each other since example III.9.10.1 precisely says that this family is not flat. Any help in sorting out where the problem in my reasoning is welcome! -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "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": 0.9310157299041748, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2007/08/14/enriched-categories-ii/?like=1&_wpnonce=34114a6dfa	# The Unapologetic Mathematician ## Enriched Categories II So we have the basic data of a category $\mathcal{C}$ enriched over a monoidal category $\mathcal{V}$. Of course, what I left out were the relations that have to hold. And they’re just the same as those from categories, but now written in terms of $\mathcal{V}$ instead of $\mathbf{Set}$: associativity and identity relations, as encoded in the following commutative diagrams: Notice how these are very similar to the axioms for a monoidal category or a monoid object. And this shouldn’t be unexpected by now, since we know that a monoid is just a (small) category with only one object. In fact, if we only have one object in a $\mathcal{V}$-enriched category we get back exactly a monoid object in $\mathcal{V}$! Now, often we’re thinking of our hom-objects as “hom-sets with additional structure”. There should be a nice way to forget that extra structure and recover just a regular category again. To an extent this is true, but for some monoidal categories $\mathcal{V}$ the “underlying set” functor isn’t really an underlying set at all. For now, though, let’s look at a familiar category of “sets with extra structure” and see how we get the underlying set out of the category itself. Again, the good example to always refer back to for enriched categories is $\mathbf{Ab}$, the category of abelian groups with tensor product as the monoidal structure. We recall that the functor giving the free abelian group on a set is left adjoint to the forgetful functor from abelian groups to sets. That is, $\hom_\mathbf{Ab}(F(S),A)\cong\hom_\mathbf{Set}(S,U(A))$. We also know that we can consider an element of the underlying set $U(A)$ of an abelian group as a function from a one-point set into $U(A)$. That is, $\hom_\mathbf{Set}(\{*\},U(A))\cong U(A)$. Putting these together, we see that $U(A)\cong\hom_\mathbf{Ab}(\mathbb{Z},A)$, since $\mathbb{Z}$ is the free abelian group on one generator. But $\mathbb{Z}$ is also the identity object for the tensor product! The same sort of argument goes through for all our usual sets-with-structure, telling us that in all these cases the “underlying set” functor is represented by the monoidal identity $\mathbf{1}$, which is the free object on one generator. We take this as our general rule, giving the representable functor $V(\underline{\hphantom{X}})=\hom_{\mathcal{V}_0}(\mathbf{1},\underline{\hphantom{X}}):\mathcal{V}_0\rightarrow\mathbf{Set}$. In many cases (but not all!) this is the usual “underlying set” functor, but now we’ve written it entirely in terms of the monoidal category $\mathcal{V}$! As time goes by, we’ll use this construction to recover the “underlying category” of an enriched category. The basic idea should be apparent, but before we can really write it down properly we need to enrich the notions of functors and natural transformations. ### 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": 18, "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": 0.9235060214996338, "perplexity_flag": "head"}
http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.dmj/1082138585	### A cup product in the Galois cohomology of number fields William G. McCallum and Romyar T. Shafiri Source: Duke Math. J. Volume 120, Number 2 (2003), 269-310. #### Abstract Let K be a number field containing the group μn of nth roots of unity, and let S be a set of primes of K including all those dividing n and all real archimedean places. We consider the cup product on the first Galois cohomology group of the maximal S-ramified extension of K with coefficients in μn, which yields a pairing on a subgroup of $K\sp \mathsf{x}$ containing the S-units. In this general situation, we determine a formula for the cup product of two elements that pair trivially at all local places. Our primary focus is the case in which $K=\mathbb {Q}(\mu\sb p)$ for n=p, an odd prime, and S consists of the unique prime above p in K. We describe a formula for this cup product in the case that one element is a pth root of unity. We explain a conjectural calculation of the restriction of the cup product to p-units for all p≤10,000\$ and conjecture its surjectivity for all p satisfying Vandiver's conjecture. We prove this for the smallest irregular prime p=37 via a computation related to the Galois module structure of p-units in the unramified extension of K of degree p. We describe a number of applications: to a product map in K-theory, to the structure of S-class groups in Kummer extensions of S, to relations in the Galois group of the maximal pro-p extension of $\mathbb {Q}(mu\sb p)$ unramified outside p, to relations in the graded ℤp-Lie algebra associated to the representation of the absolute Galois group of ȑA in the outer automorphism group of the pro-p fundamental group of $\mathbf {P}\sp 1(\overline \mathbb {Q})-\{0,1,\infty\}$, and to Greenberg's pseudonullity conjecture. First Page: Primary Subjects: 11R34 Secondary Subjects: 11R23, 11R29, 11R70, 20F34 We're sorry, but we are unable to provide you with the full text of this article because we are not able to identify you as a subscriber. If you have a personal subscription to this journal, then please login. If you are already logged in, then you may need to update your profile to register your subscription. Read more about accessing full-text Permanent link to this document: http://projecteuclid.org/euclid.dmj/1082138585 Digital Object Identifier: doi:10.1215/S0012-7094-03-12023-2 Mathematical Reviews number (MathSciNet): MR2019977 Zentralblatt MATH identifier: 1047.11106	{"extraction_info": {"found_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": 0.18, "absolute_threshold": 10, "end_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": 0.8494000434875488, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/61367/two-specific-sets-of-sums-closed	## Two Specific sets of Sums [closed] ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I've been looking into two very specific forms that are related to $\sum _{n=1}^{\infty } (-1)^n \left(n^{\frac{1}{n}}-1\right)$ cf http://mathworld.wolfram.com/MRBConstant.html. They are $\sum _{n=1}^{\infty } (-1)^n \left(n^{\frac{u}{n}}-1\right)$ and $\sum _{n=1}^{\infty } (-1)^n \left(n^{\frac{u}{n}}-u\right)$. As the amateur that I am, I would like to ask academic (or other) professionals if they have ever come across these forms, or are they clearly localized? To see where I get these two forms you can cf http://www.mapleprimes.com/posts/104016-A-Very-False-Graph. - 1 So, the question is to come up with other questions? And how will it be judged whether the questions we come up with are the right questions? Please have a look at the faq and see whether you think your question is suitable for this site. – Gerry Myerson Apr 12 2011 at 5:43 This is my first time to know about the MRB constant，does it play a very important part in number theory? – yaoxiao Apr 12 2011 at 6:26 1 Is MRB your acronym? Anyway, maybe start by asking yourself why you are interested in that series, and what you are looking for. – Pietro Majer Apr 12 2011 at 7:38 I think that "in which problems does this series come up?" is a legitimate question. Voting to reopen. – algori May 31 2011 at 3:53	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9551506042480469, "perplexity_flag": "middle"}
http://www.aimath.org/textbooks/beezer/Darchetype.html	Archetype D Archetype D Summary: System with three equations, four variables. Consistent. Null space of coefficient matrix has dimension 2. Coefficient matrix identical to that of Archetype E, vector of constants is different. ◊ A system of linear equations (Definition SLE): \begin{align} 2x_1 + x_2 + 7x_3 - 7x_4 &= 8 \\ -3x_1 + 4x_2 -5x_3 - 6x_4 &= -12 \\ x_1 +x_2 + 4x_3 - 5x_4 &= 4 \end{align} ◊ Some solutions to the system of linear equations (not necessarily exhaustive): $x_1 = 0, x_2 = 1, x_3 = 2, x_4 = 1$ $x_1 = 4, x_2 = 0, x_3 = 0, x_4 = 0$ $x_1 = 7, x_2 = 8, x_3 = 1, x_4 = 3$ ◊ Augmented matrix of the linear system of equations (Definition AM): \begin{bmatrix} 2 & 1 & 7 & -7 & 8\\ -3 & 4 & -5 & -6 & -12\\ 1 & 1 & 4 & -5 & 4 \end{bmatrix} ◊ Matrix in reduced row-echelon form, row-equivalent to augmented matrix: \begin{bmatrix} \leading{1} & 0 & 3 & -2 & 4 \\ 0 & \leading{1} & 1 & -3 & 0\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} ◊ Analysis of the augmented matrix (Notation RREFA): \begin{align} r=2&&D=\set{1,\,2}&&F=\set{3,\,4,\,5} \end{align} ◊ Vector form of the solution set to the system of equations (Theorem VFSLS). Notice the relationship between the free variables and the set $F$ above. Also, notice the pattern of 0's and 1's in the entries of the vectors corresponding to elements of the set $F$ for the larger examples. $\colvector{x_1\\x_2\\x_3\\x_4}= \colvector{4\\0\\0\\0} + x_3\colvector{-3\\-1\\1\\0}+ x_4\colvector{2\\3\\0\\1}$ ◊ Given a system of equations we can always build a new, related, homogeneous system (Definition HS) by converting the constant terms to zeros and retaining the coefficients of the variables. Properties of this new system will have precise relationships with various properties of the original system. \begin{align} 2x_1 + x_2 + 7x_3 - 7x_4 &= 0 \\ -3x_1 + 4x_2 -5x_3 - 6x_4 &= 0 \\ x_1 +x_2 + 4x_3 - 5x_4 &= 0 \end{align} ◊ Some solutions to the associated homogenous system of linear equations (not necessarily exhaustive): $x_1 = 0, x_2 = 0, x_3 = 0, x_4=0$ $x_1 = -3, x_2 = -1, x_3 = 1, x_4=0$ $x_1 = 2, x_2 = 3, x_3 = 0, x_4=1$ $x_1 = -1, x_2 = 2, x_3 = 1, x_4=1$ ◊ Form the augmented matrix of the homogenous linear system, and use row operations to convert to reduced row-echelon form. Notice how the entries of the final column remain zeros: \begin{bmatrix} \leading{1} & 0 & 3 & -2 & 0 \\ 0 & \leading{1} & 1 & -3 & 0\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} ◊ Analysis of the augmented matrix for the homogenous system (Notation RREFA). Notice the slight variation for the same analysis of the original system only when the original system was consistent: \begin{align} r=2&&D=\set{1,\,2}&&F=\set{3,\,4,\,5} \end{align} ◊ Coefficient matrix of original system of equations, and of associated homogenous system. This matrix will be the subject of further analysis, rather than the systems of equations. \begin{bmatrix} 2 & 1 & 7 & -7\\ -3 & 4 & -5 & -6\\ 1 & 1 & 4 & -5 \end{bmatrix} ◊ Matrix brought to reduced row-echelon form: \begin{bmatrix} \leading{1} & 0 & 3 & -2\\ 0 & \leading{1} & 1 & -3\\ 0 & 0 & 0 & 0 \end{bmatrix} ◊ Analysis of the row-reduced matrix (Notation RREFA): \begin{align} r=2&&D=\set{1,\,2}&&F=\set{3,\,4} \end{align} ◊ Matrix (coefficient matrix) is nonsingular or singular? (Theorem NMRRI) at the same time, examine the size of the set $F$ above. Note that this property does not apply to matrices that are not square. Matrix is not square, so the question does not apply. ◊ This is the null space of the matrix. The set of vectors used in the span construction is a linearly independent set of column vectors that spans the null space of the matrix (Theorem SSNS, Theorem BNS). Solve the homogenous system with this matrix as the coefficient matrix and write the solutions in vector form (Theorem VFSLS) to see these vectors arise. $\spn{ \set{\colvector{-3\\-1\\1\\0},\,\colvector{2\\3\\0\\1}} }$ ◊ Column space of the matrix, expressed as the span of a set of linearly independent vectors that are also columns of the matrix. These columns have indices that form the set $D$ above. (Theorem BCS) $\spn{ \set{\colvector{2\\-3\\1},\,\colvector{1\\4\\1}}}$ ◊ The column space of the matrix, as it arises from the extended echelon form of the matrix. The matrix $L$ is computed as described in Definition EEF. This is followed by the column space described by a set of linearly independent vectors that span the null space of $L$, computed as according to Theorem FS and Theorem BNS. When $r=m$, the matrix $L$ has no rows and the column space is all of $\complex{m}$. $L=\begin{bmatrix}1&\frac{1}{7}&-\frac{11}{7}\end{bmatrix}$ $\spn{ \set{\colvector{\frac{11}{7}\\0\\1},\,\colvector{-\frac{1}{7}\\1\\0}}}$ ◊ Column space of the matrix, expressed as the span of a set of linearly independent vectors. These vectors are computed by row-reducing the transpose of the matrix into reduced row-echelon form, tossing out the zero rows, and writing the remaining nonzero rows as column vectors. By Theorem CSRST and Theorem BRS, and in the style of Example CSROI, this yields a linearly independent set of vectors that span the column space. $\spn{ \set{\colvector{1\\0\\\frac{7}{11}},\,\colvector{0\\1\\\frac{1}{11}}}}$ ◊ Row space of the matrix, expressed as a span of a set of linearly independent vectors, obtained from the nonzero rows of the equivalent matrix in reduced row-echelon form. (Theorem BRS) $\spn{ \set{\colvector{1\\0\\3\\-2},\,\colvector{0\\1\\1\\-3}}}$ ◊ Inverse matrix, if it exists. The inverse is not defined for matrices that are not square, and if the matrix is square, then the matrix must be nonsingular. (Definition MI, Theorem NI) $\null$ ◊ Subspace dimensions associated with the matrix. (Definition NOM, Definition ROM) Verify Theorem RPNC. \begin{align} \text{Matrix columns: }4&& \text{Rank: }2&& \text{Nullity: }2 \end{align} ◊ Determinant of the matrix, which is only defined for square matrices. The matrix is nonsingular if and only if the determinant is nonzero (Theorem SMZD). The determinant is the product of all eigenvalues. $\text{Determinant} = \null$	{"extraction_info": {"found_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": 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": 0.7741267085075378, "perplexity_flag": "head"}
http://mathhelpforum.com/discrete-math/76008-solved-boolean-algebra.html	# Thread: 1. ## [SOLVED] Boolean Algebra Assume that B is a Boolean algebra with operations + and ·. For each true statement below, give a detailed proof. For each false statement below, write out its negation, then give a proof of the negation. (∀a, b ∈ B) (a + b = 1 ↔ b · ā = ā) (∀a, b ∈ B) (a · b = 0 ↔ b + ā = ā) (∀a, b ∈ B) (a + b = a + c → b = c) for the last one, i had the same proof as http://www.mathhelpforum.com/math-he...b-b-c-b-c.html but it just doesn't seem right to me... completely stumped on the first two . thanks!!! 2. ## Boolean Algebra Proofs Hello starman_dx Originally Posted by starman_dx Assume that B is a Boolean algebra with operations + and ·. For each true statement below, give a detailed proof. For each false statement below, write out its negation, then give a proof of the negation. (∀a, b ∈ B) (a + b = 1 ↔ b · ā = ā) (∀a, b ∈ B) (a · b = 0 ↔ b + ā = ā) (∀a, b ∈ B) (a + b = a + c → b = c) for the last one, i had the same proof as http://www.mathhelpforum.com/math-he...b-b-c-b-c.html but it just doesn't seem right to me... completely stumped on the first two . thanks!!! Here are the proofs you need. I'll leave you to supply the names of the Laws I'm using at each stage. (1) $a+b= 1$ $\rightarrow (a+b)\cdot \bar{a} = 1\cdot\bar{a}$ $\rightarrow a\cdot\bar{a} +b\cdot\bar{a} = \bar{a}$ $\rightarrow 0+b\cdot\bar{a} = \bar{a}$ $\rightarrow b\cdot\bar{a} = \bar{a}$ And $b\cdot\bar{a} = \bar{a}$ $\rightarrow b\cdot\bar{a} + a = \bar{a}+a$ $\rightarrow (b+a)\cdot(\bar{a}+a) = 1$ $\rightarrow (a+b).1 = 1$ $\rightarrow a+b = 1$ (2) $a\cdot b=0$ $\rightarrow a\cdot b + \bar{a} = 0 + \bar{a}$ $\rightarrow (a+\bar{a})\cdot(b+\bar{a}) = \bar{a}$ $\rightarrow 1\cdot(b+\bar{a}) = \bar{a}$ $\rightarrow b+\bar{a} = \bar{a}$ And $b+\bar{a} = \bar{a}$ $\rightarrow a\cdot(b+\bar{a}) = a\cdot\bar{a}$ $\rightarrow a\cdot b + a\cdot \bar{a} = 0$ $\rightarrow a\cdot b + 0 = 0$ $\rightarrow a\cdot b = 0$ (3) is not true. The negation is $\exists a, b, c \in B, (a+b = a+c) \wedge (b \ne c)$ As a counter-example, consider sets $A = \{1\}, B = \{2\}$ and $C = \{1, 2\}$. Then $A \cup B = A \cup C$, but $B \ne C$. Grandad 3. you are a life saver thank you!!!!	{"extraction_info": {"found_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": 0.18, "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": 0.8786089420318604, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/tagged/brst+ghosts	# Tagged Questions 1answer 194 views ### Clarification on “central charge equals number of degrees of freedom” It's often stated that the central charge c of a CFT counts the degrees of freedom: it adds up when stacking different fields, decreases as you integrate out UV dof from one fixed point to another, ... 3answers 122 views ### Is BRST ghost number conserved in quantum gravity? Quantum gravity needs Faddeev-Popov ghosts. Feynman showed that. Take a black hole. Hawking pair production of ghost-antighost pair. One ghost falls into the hole and hits the singularity. The ghost ... 2answers 404 views ### What is the ontological status of Faddeev Popov ghosts? We all know Faddeev-Popov ghosts are needed in manifestly Lorentz covariant nonabelian quantum gauge theories. We also all know they decouple from the rest of matter asymptotically, although they ... 1answer 112 views ### Who added $\frac{3}{2} \partial^2 c$ to the virasoro BRST current (and why)? I've been looking at the literature on quantizing the bosonic string, and I noticed that there was a change made in the definition of the BRST current around 1992. However, I haven't found any ... 2answers 130 views ### Is ghost-number a physical reality/observable? One perspective is to say that one introduced the ghost fields into the Lagrangian to be able to write the gauge transformation determinant as a path-integral. Hence I was tempted to think of them as ...	{"extraction_info": {"found_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": 0.9292420744895935, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/18693/list	## Return to Answer 2 Multiplication by p should be purely inseparable, not just inseparable... In characteristic $p$, every map $E_1 \to E_2$ factors as a power of the Frobenius $\varphi_r \colon E_1 \to E_1^{(p^r)}$ followed by a separable morphism $E_1^{(p^r)} \to E_2$, and we find $r$ by looking at the inseparable degree of our map (if the map is separable, then $r=0$, as Pete pointed out). Now, in the case of interest, if $E$ is supersingular, $\widehat{\varphi}$ is inseparable (as this is equivalent to multiplication by $p$ being purely inseparable). But then $\widehat{\varphi} \colon E^{(p)} \to E$ factors as $E^{(p)} \to E^{(p^2)} \to E$ by comparing degrees, where the first map is the Frobenius and the second is an isomorphism. It then follows that $j(E) = j(E^{(p^2)}) = j(E)^{p^2}$ so $j(E) \in \mathbb{F}_{p^2}$. 1 In characteristic $p$, every map $E_1 \to E_2$ factors as a power of the Frobenius $\varphi_r \colon E_1 \to E_1^{(p^r)}$ followed by a separable morphism $E_1^{(p^r)} \to E_2$. Now, in the case of interest, if $E$ is supersingular, $\widehat{\varphi}$ is inseparable (as this is equivalent to multiplication by $p$ being inseparable). But then $\widehat{\varphi} \colon E^{(p)} \to E$ factors as $E^{(p)} \to E^{(p^2)} \to E$ by comparing degrees, where the first map is the Frobenius and the second is an isomorphism. It then follows that $j(E) = j(E^{(p^2)}) = j(E)^{p^2}$ so $j(E) \in \mathbb{F}_{p^2}$.	{"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": 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": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9394903779029846, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2011/04/25/spheres-as-submanifolds/?like=1&source=post_flair&_wpnonce=7b62a7b6bf	# The Unapologetic Mathematician ## Spheres as Submanifolds With our extension of the implicit function theorem in hand, we have another way of getting at the sphere, this time as a submanifold. Start with the Euclidean space $\mathbb{R}^{n+1}$ and take the smooth function $f:\mathbb{R}^{n+1}\to\mathbb{R}$ defined by $f(x)=\langle x,x\rangle$. In components, this is $\sum_i\left(x^i\right)^2$, where the $x^i$ are the canonical coordinates on $\mathbb{R}^{n+1}$. We can easily calculate the derivative in these coordinates: $f_{x}(v)=2\langle x,v\rangle$. This is the zero function if and only if $x=0$, and so $f_{x}$ has rank $1$ at any nonzero point $x$. The point $x=0$ is a critical point, and every other point is regular. On the image side, we see that $f(0)=0$, so the only critical value is $0$. Every other value is regular, though $f^{-1}(y)$ is empty for $y<0$. For $f^{-1}(a^2)$ we have a nonempty preimage, which by our result is a manifold of dimension $(n+1)-1=n$. This is the $n$-dimensional sphere of radius $a$, though we aren’t going to care so much about the radius for now. Anyway, is this really the same sphere as before? Remember, when we first saw the two-dimensional sphere as an example, we picked coordinate patches by hand. Now we have the same set of points — those with a fixed squared-distance from the origin — but we might have a different differentiable manifold structure. But if we can show that the inclusion mapping that takes each of our handcrafted coordinate patches into $\mathbb{R}^3$ is an immersion, then they must be compatible with the submanifold structure. We only really need to check this for a single patch, since all six are very similar. We take the local coordinates from our patch and the canonical coordinates on $\mathbb{R}^3$ to write out the inclusion map: $\displaystyle g(x,y)=\left(x,y,\sqrt{1-x^2-y^2}\right)$ Then we use these coordinates to calculate the derivative $\displaystyle\begin{aligned}g_{*(x,y)}(u,v)&=\left(1,0,\frac{-x}{\sqrt{1-x^2-y^2}}\right)u+\left(0,1,\frac{-y}{\sqrt{1-x^2-y^2}}\right)v\\&=\left(u,v,\frac{-xu-yv}{\sqrt{1-x^2-y^2}}\right)\end{aligned}$ This clearly always has rank $2$ for $x^2+y^2<1$, and so the inclusion of our original sphere into $\mathbb{R}^3$ is an immersion, which must then be equivalent to the inclusion of the submanifold $f^{-1}(1)$, since they give the same subspace of $\mathbb{R}^3$. ### Like this: Posted by John Armstrong \| Differential Topology, Topology 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": 29, "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": 0.918688178062439, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-geometry/93644-convergence-convolutions-approximate-convolution-identity.html	Thread: 1. Convergence of Convolutions (Approximate Convolution Identity) I'm trying to show that if a sequence of functions, (f_n) is uniformly convergent to a function f, and \phi_n (do people use LaTeX notation here?) is an approximation to the dirac delta (an approximate convolution identity), then f_n\phi_n, i.e. the sequence of convolutions, is also convergent uniformly to f (for my purposes we can restrain ourselves to showing this on a compact set luckily!). This intuitively makes sense, but I'm having a doggone hard time showing it rigorously. Anyone have any ideas? Is this the right place for this post? 2. Let me make this simpler. Can we at least show that f_n\phi_n is pointwise convergent to f? 3. Originally Posted by jon.s.beardsley I'm trying to show that if a sequence of functions, (f_n) is uniformly convergent to a function f, and \phi_n (do people use LaTeX notation here?) is an approximation to the dirac delta (an approximate convolution identity), then f_n*\phi_n, i.e. the sequence of convolutions, is also convergent uniformly to f (for my purposes we can restrain ourselves to showing this on a compact set luckily!). This intuitively makes sense, but I'm having a doggone hard time showing it rigorously. Anyone have any ideas? Is this the right place for this post? Have you tried to write $f_n\ast\phi_n-f=(f_n-f)\ast \phi_n + (f\ast\phi_n-f)$? Then $\\|(f_n-f)\ast \phi_n\\|_\infty \leq \\|f_n-f\\|_\infty \int\phi_n(t)dt=\\|f_n-f\\|_\infty$ tends to 0 (by uniform convergence of $(f_n)_n$ to $f$), and $\\|f\ast \phi_n - f\\|_\infty$ tends to 0 as well (usual result about convolution with an approximation of delta). nb: click on the formulas to see what I typed to get them. 4. So I'm assuming we can use $\\|_\infty$ because on a compact set the function is necessarily bounded? For $f$ the only condition I have is that it is continuous (I could use locally integrable instead). Also, is there any way you might clarify the algebra in creating that first inequality you have? I understand rewriting the convolution (very clever!) but I have trouble seeing how we then immediately go to the inequality with the norms in it. Thanks so much. 5. Originally Posted by jon.s.beardsley So I'm assuming we can use $\\|_\infty$ because on a compact set the function is necessarily bounded? For $f$ the only condition I have is that it is continuous (I could use locally integrable instead). Also, is there any way you might clarify the algebra in creating that first inequality you have? I understand rewriting the convolution (very clever!) but I have trouble seeing how we then immediately go to the inequality with the norms in it. Thanks so much. The first inequality is: for every $x$, $\|((f_n-f)\ast \phi_n(x)\|\leq \int \|(f_n-f)(x-t)\| \|\phi_n(t)\| dt$ $\leq \int \\|f_n-f\\|_\infty \|\phi_n(t)\| dt = \\|f_n-f\\|_\infty \int\phi_n(t)dt$ (I bound $\|(f_n-f)(x-t)\|$ by its maximum value, and $\|\phi_n(t)\|=\phi_n(t)$ because $\phi_n$ is positive (?)), hence we have an upper bound independent of $x$. This gives what I wrote. Since there is uniform convergence, by definition $\\|f_n-f\\|_\infty$ is finite and converges to 0. The compacity is involved in the justification of the second term: why $f\ast \phi_n$ converges uniformly to $f$. 6. Dearly sorry about the trouble but The first inequality is: for every $x$, [LaTeX Error: Image is too big (644x37, limit 600x220)] 7. nevermind, i have the TeX in my e-mail. Thanks! 8. How is it that $\\|f\ast\phi_n\\|_\infty$ converges uniformly? As far as I understood, pointwise convergence on a compact set does not necessarily imply uniform convergence. 9. Originally Posted by jon.s.beardsley How is it that $\\|f\ast\phi_n\\|_\infty$ converges uniformly? As far as I understood, pointwise convergence on a compact set does not necessarily imply uniform convergence. I guess you mean $f\ast\phi_n$. Since this is the result in the case when $f_n=f$ for all $n$, I supposed you already knew that $f\ast\phi_n$ converges uniformly to $f$. (So that you were asked for a generalization of this fact) This is a usual property of approximations of delta, a quick one but not a trivial one. On a compact set, if $f$ is continuous, $\\|f\ast\phi_n-f\\|_\infty\to 0$. The idea for the proof is to write $(f\ast\phi_n)(x)-f(x)=\int (f(t)-f(x))\phi_n(x-t) dt$, and to use the fact that $\phi_n$ is 0 away from 0 (precisions depend on your own definition) so that the integral reduces to a segment around $x$ of small width. Using the uniform continuity of $f$ (that's were compacity is involved), we conclude. (Use an $\varepsilon>0$ for the proof) I guess you already met this proof in your lecture, didn't you? 10. Thankyou for your help. I have not had any sort of lecture or class on this material. I am attempting to prove some properties of generalized functions for an undergraduate research thesis, and I'm having to figure much out my own (i.e. measure theory, distributions, a lot of functional analysis). I'm trying to show that a class of generalized functions form a sheaf (you know, those things Grothendieck liked) over locally compact spaces, and am starting with $\mathbb{R}^n$. Right now I'm working on a glueing property. Thanks again for all the help. 11. I cannot see how $<br /> (f\ast\phi_n)(x)-f(x)=\int (f(t)-f(x))\phi_n(x-t) dt<br />$ is true. 12. Originally Posted by jon.s.beardsley I cannot see how $<br /> (f\ast\phi_n)(x)-f(x)=\int (f(t)-f(x))\phi_n(x-t) dt<br />$ is true. Because $f\ast\phi_n(x)=\int f(t)\phi_n(x-t)dt$( and $=\int f(x-t)\phi_n(t)dt$ as well) and, since $\int\phi_n(t)dt=1$ (approximation of delta...), $f(x)=f(x)\int \phi_n(t) dt=f(x)\int \phi_n(x-t) dt=\int f(x)\phi_n(x-t)dt$ (change of variable) 13. All is clear now! Thankyou!	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 41, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.9418429136276245, "perplexity_flag": "head"}
http://mathoverflow.net/questions/69436?sort=oldest	## How to construct log-canonical (or Calabi-Yau), non-Cohen-Macaulay singularities of low codimensions? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) (EDIT 07/06/11: although the question has not been settled definitely, Sándor's excellent answer and the comments by Angelo and ulrich have highlighted many potential obstructions to the constructions I wanted. Thank you all! I am still very interested in any leads, so please keep them coming if you have some more.) I would like to know examples of log-canonical singularities of low codimension which is normal but non-Cohen-Macaulay. A silly way to do it may be just adjoining variables, so here is the precise question: Fix a number $c$, for what $n$ can one construct an affine variety $X \subseteq \mathbb A^n_{\mathbb C}$ such that: $X$ is indecomposable and normal of codimension $c$, $X$ has at worst log-canonical singularity but $X$ is not Cohen-Macaulay? Given $c$, can we construct such $X$ for all $n$ big enough? The case $c=1$ is easy, there are no example since hypersurfaces are Cohen-Macaulay, so let's begin with $c=2$. Motivation/Comments: I am actually looking for $F$-pure rings (i.e., the Frobenius is a pure morphism), but conjecturally my question above is virtually the same. Karl Schwede told me one can also try to look for (projective) Calabi-Yau varieties with some non-vanishing middle cohomolgy and low codimension embedding, then take their cones. But not being a geometer, I do not know how to construct such things. - 2 There is a conjecture of Hartshorne that says that a smooth projective variety in $\mathbb{P}^n$ of codimension $<n/3$ should be a complete intersection. Very little is known about it but it does suggest that it is unlikely that taking cones over smooth projective varieties will give examples of fixed codimension $c$ for large $n$. – ulrich Jul 4 2011 at 6:30 @Dear ulrich: that is a very good point. Actually, it reminded me that even for normal projective varieties of small codimensions, there are some restrictions: the Picard group is $\mathbb Z$. That probably rules out quite a few candidates. – Hailong Dao Jul 4 2011 at 13:47 Oops, thank you Sándor! – Hailong Dao Jul 6 2011 at 14:54 No problem. (I removed the incriminating evidence, too....) – Sándor Kovács Jul 6 2011 at 14:55 ## 1 Examples, smallest $n$ Let $A$ be an abelian variety and $X$ a cone over $A$. Then $X$ is log canonical, but as soon as $\dim A\geq 2$, then $X$ is not Cohen-Macaulay. (This you can see by computing the local cohomology at the vertex). For the $c=1$ case: those are obviously CM. As ulrich points out, there exist abelian surfaces in $\mathbb P^4$, so those give you $c=2$ with $3$-dimensional singularities, that is, $n=5=2c+1$. For $c=2$ this is the smallest $n$ you can get. Here is why: For $n\leq 4$ anything of codimension $c=2$ would be of dimension at most $2$ and hence if it is normal, it is $S_2$ and in particular CM. So, let's assume that $c>2$. As any quasi-projective variety $A$ of dimension $d$ maybe embedded in $\mathbb P^{2d+1}$ (Embed in some $\mathbb P^N$ and notice that the closure of the secant variety of $A$ is of dimension $2d+1$, so as long as $N>2d+1$, one may find a point and a projection that gives an embedding of $X$ into $\mathbb P^{N-1}$. Repeat.), you can do that with abelian varieties as well. This gives you $c=d+1$, so turning it around, for any $c>2$ you can find an $X$ that you're looking for in $\mathbb A^n$ with $n=2c-1$. ## 2 Non-indecomposables (Note: this is here, because I originally did not read the requirements carefully. Then I didn't feel like erasing it. ) If you did not required an indecomposable singularity, then you could take the product of this $X\subset \mathbb A^{2c-1}$ and an arbitrary $\mathbb A^r$. Of course, this is why you made that requirement. Anyway, you'd get $$X_{c,r}:=X\times \mathbb A^r\subset \mathbb A^{2c-1+r}$$ of codimension $c$ with $n=2c-1+r$. In other words, yes, you can construct such an $X$ for all $n$ big enough. I am not sure whether the bound $2c-1$ is optimal, but I have a feeling that you can't get much better than that. Note that this construction works for $c=2$ as well, so you can get examples in all dimensions $n\geq 5=2\cdot 2+1$. If you wanted indecomposable singularities, I would expect that the codimension is actually increasing with the dimension. In other words, I would expect low codimensional examples in low dimension and not in (arbitrarily) high dimension. ## 3 Indecomposable vs. isolated The previous point shows why Hailong assumed indecomposable. I claim that one may as well also assume isolated. At least to start. First, let $X$ be an example as required, of dimension $d=\dim X$ and of codimension $c$. Let $s=\mathrm{Sing} X$ the dimension of the singular set of $X$ and assume that $\mathrm{Sing} X$ is irreducible. Take a general complete intersection of codimension $s$. This will have dimension $d-s$, codimension $c$ and it is an isolated non-CM log canonical singularity. Next, let $X$ be a codimension $c$ and it is an isolated non-CM log canonical singularity and consider a $\mathbb Q$-Gorenstein deformation of $X$ over a base of dimension $s$. (A $\mathbb Q$-Gorenstein deformation means that the relative dualizing sheaf of the family is a $\mathbb Q$-line bundle and its line bundle powers restrict to the appropriate power of the dualizing sheaf of the members of the family). The total space of the deformation will be an example of the kind you want. (For an indecomposable example you need a deformation without a trivial component). The fact that this gives you an example that you want is non-trivial. It is log canonical by inversion of adjunction see the main result of Kawakita's paper and it is non-CM at every point of the singular set by Corollary 1.3 of this paper. From this you can conclude that any example you get will be a deformation of an isolated example. A priori you get that at the generic points of the singular set. At non-generic points those are still degenerations (i.e., non-small deformations) of the general ones. The problem you run into is that you need non-trivial deformations of these singularities and they have finite dimensional versal deformation spaces, so you can't get too far with this idea. (Angelo will correct me if this is wrong, since he is one of the ultimate experts on this. See also Artin's extended work on this topic.) This suggests that given your restriction of being indecomposable, regarding your "every big enough $n$" question, it seems that in order for that to happen you really need low codimensional isolated examples, which will be hard to construct since you can't get too far with cones (cf. ulrich's and Angelo's comments). On the other hand we do get new examples out of this: A flat family of polarized abelian/CY/etc varieties will give you a flat family of the cones over them. As long as the family has maximal variation (i.e., the moduli map of the base is generically finite) the resulting singularity will be what you want. So, this way you can get higher dimensional examples, but it seems that all constructions are limited in dimension. All of this suggests that the answer to your big enough question will be "no". ## 4 Better examples (manageability over low codimension) To get an example that you want, you do not need to have an abelian variety for the cone construction. If $A$ is a smooth projective variety of dimension $d$ such that $\omega_A\simeq \mathscr O_A$ and there exist two integers $i,m\in\mathbb Z$ such that `$0<i<d$` and `$$H^i(A,\mathscr O_A(m))\neq 0,$$` then the cone over $A$ has non-CM log canonical singularities. In other words, to get more examples you just need to find such subvarieties of low codimension. Of course, complete intersections do not satisfy this, but for example the product of any two CYs do. So, you could take, say, a CY hypersurface $H$ and an elliptic curve $E$ and then $A=H\times E$ satisfies to condition (and $A$ is generally neither CY nor abelian). The obvious embedding via Segre gives larger codimension than what you get from the above procedure, so this does not give you smaller codimensional examples, but they may be more manageable since you have the product of a hypersurface and a plane curve and even if the codimension is high, you know the embedding pretty well. So, from a practical point of view these are perhaps better examples after all. - Dear S\'andor, how can we make sure that $X$ has low codimension? – Hailong Dao Jul 4 2011 at 3:10 And how can I type your name correctly in the comments? Thanks. – Hailong Dao Jul 4 2011 at 3:11 4 An abelian variety of dimension $n$ cannot be embedded in $\mathbb{P}^m$ for $m < 2n$. See, for example, Fulton's Intersection Theory, Example 3.2.15. Also, there do exist abelian surfaces in $\mathbb{P}^4$ (constructed by Horrocks and Mumford) but I'm not sure about the higher dimensional case. (Of course one can always embed an $n$ dimensional abelian variety in $\mathbb{P}^{2n+1}$.) – ulrich Jul 4 2011 at 5:31 3 Van De Ven proved that an abelian variety of dimension $n$ can not be embedded in $\mathbb{P}^{2n}$, as soon as $n \geq 3$ (On the embedding of abelian varieties in projective spaces, Annali di Matematica Pura ed Applicata 103, 127-129). – Angelo Jul 4 2011 at 8:29 1 The Barth-Lefschetz theorem implies that for any smooth projective variety $X$ of dimension $m$ in $\mathbb{P}^n$, $H^1(X, \mathbb{Z}) = 0$ if $2m > n$. So one can get an improvement of at most $1$ by using a product of an elliptic curve and a CY hypersurface rather than an abelian variety. – ulrich Jul 4 2011 at 9:40 show 10 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": 102, "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": 0.9330567717552185, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-equations/175409-formulating-bv-problem-print.html	# Formulating the BV problem Printable View • March 22nd 2011, 12:25 PM dwsmith Formulating the BV problem In a rod length L with insulated lateral surface and thermal constants $c, \ \rho, \ \kappa, \ K$ heat is generated at a rate uniformly proportional to the temperature, that is, at a rate $ru(x,t)$ per unit volume per unit time, where r is a constant and $u(x,t)$ is the temperature function of the rod. The ends of the rod are maintained at temperature 0 and initially the rod has uniform temperature 1. Formulate an initial-BV problem for determining $u(x,t)$ $\text{D.E.}: \ u_t=ru_{xx}$ $\displaystyle\text{B.C.}=\begin{cases}u(0,t)=0\\u( L,t)=0\end{cases}$ $\text{I.C.}: \ u(x,0)=1$ I am not sure what to do with the r so I just put it in front of u_{xx}. Is that correct? Also, I need a K, c, and rho somewhere but not sure where and why. • March 22nd 2011, 01:49 PM Aryth r does belong there. It is known as Thermal Diffusivity. It relates thermal conductivity and volumetric heat capacity together. It specifically shows how conductivity changes in comparison to the thermal bulk. As to why it belongs there, I don't know. Maybe seeing the equation will help you figure it out. $r = \frac{K}{\rho c_p}$ • March 22nd 2011, 01:51 PM dwsmith Quote: Originally Posted by Aryth r does belong there. It is known as Thermal Diffusivity. It relates thermal conductivity and volumetric heat capacity together. It specifically shows how conductivity changes in comparison to the thermal bulk. As to why it belongs there, I don't know. Maybe seeing the equation will help you figure it out. $r = \frac{K}{\rho c_p}$ But by making that substitution, I will lose r. In the final solution, there is an r, rho, K, and c. • March 24th 2011, 12:41 PM dwsmith The DE is $\displaystyle\text{D.E.}: \ u_t=ku_{xx}+\frac{r}{c\rho}u$ All times are GMT -8. The time now is 03:53 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": 10, "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": 0.9228976368904114, "perplexity_flag": "middle"}
http://stats.stackexchange.com/questions/24937/how-to-understand-standardized-residual-in-regression-analysis	# How to understand standardized residual in regression analysis? I have a stupid question. According to textbook, the residual is the difference between response and predicted value, then it is said that every residual has different variance, so we need to consider standardized residuals. But the variance is for a group of values, how could a single value have variance? - 2 It would help to quote the textbook directly or (if it is available online) to provide a link to it. Much can get lost if even a single word is taken out of order or out of context. (For instance, residuals are usually defined as the difference between prediction and response, not the other way around.) – whuber♦ Mar 20 '12 at 15:56 Single random variables have variances. Residuals are random variables - they are functions of the data. So, single residuals (standardized or not) have variances. – guest Mar 20 '12 at 23:09 #whuber The textbook is "Regression.Analysis.by.Example", page,89. It discussed kinds of residuals. ordinary residual is response-prediction. @guest " Single random variables have variances", this is what I dont understand, variables is a property for a sample, isnt it? why single value in a sample (such as a residual) have variance? – hiberbear Mar 21 '12 at 8:08 Does the book have an author...? That usually makes it easier to find. I think that you are getting sample variance and population variance confused. The residual is unknown before the experiment is carried out. The response is random and so is the residual, since it is a function of the response. When we speak of the variance of the residual, we talk about the variance of the underlying random variable. – MånsT Mar 21 '12 at 10:07 sorry for the inconvenience, the authors are SAMPRIT CHATTEFUEE and ALI S. HADI, Regression Analysis by Example, fourth edition. – hiberbear Mar 21 '12 at 15:55 ## 2 Answers In contrast to @Glen_b, I would say that an individual number (such as a residual), which resulted from a random draw from a probability distribution, is a realized value, not a random variable. Likewise, I would say that the set of $N$ residuals, calculated from your data and your model fit using $\bf{e}=\bf{y}-\bf{\hat{y}}$, is a set of realized values. This set of numbers may be loosely conceptualized as independent draws from an underlying distribution $\epsilon$ ~ $\mathcal{N}(\mu,\sigma^2)$. (Unfortunately however, there are several additional complexities here. For example, you do not actually have $N$ independent pieces of information, because the residuals, $\bf{e}$, must satisfy two conditions: $\sum e_i=0$, and $\sum x_ie_i=0$.) Now, given some set of numbers, be they residuals or whatever, it is certainly true that they have a variance, $\sum(e_i-\bar{e})^2/N$, but this is uninteresting. What we care about is being able to say something about the data generating process (for instance, to estimate the variance of the population distribution). Using the preceding formula, we could give an approximation by replacing the $N$ with the residual degrees of freedom, but this may not be a good approximation. This is a topic that can get very complicated very fast, but a couple of possible reasons could be heteroscedasticity (i.e., that the variance of the population differs at different levels of $x$), and the presence of outliers (i.e., that a given residual is drawn from a different population entirely). Almost certainly, in practice, you will not be able to estimate the variance of the population from which an outlier was drawn, but nonetheless, in theory, it does have a variance. I suspect something along these lines is what the authors had in mind, however, I should note that I have not read that book. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "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": 0.9501990675926208, "perplexity_flag": "head"}
http://mathhelpforum.com/discrete-math/73013-generating-function.html	Thread: 1. generating function So there are three shareholders and 2n shares to split among them let $a_n$ be the number of ways the shares can be split such that no single person controls half the total or more The question asks me to find the generating function A(x) for the sequence $\{a_n\}$ I know $a_n = {{n-1}\choose{2}}$ but I need to find the generating function not the actual closed formula I've reverse engineered it to be $\frac{x^3}{(1-x)^3}$ But I need to find it coming from the other direction haha Could anyone explain how to do this? thanks JB	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.8841738700866699, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2011/09/07/integrals-over-manifolds-part-2/?like=1&source=post_flair&_wpnonce=18b5244a97	# The Unapologetic Mathematician ## Integrals over Manifolds (part 2) Okay, so we can now integrate forms as long as they’re supported within the image of an orientation-preserving singular cube. But what if the form $\omega$ is bigger than that? Well, paradoxically, we start by getting smaller. Specifically, I say that we can always find an orientable open cover of $M$ such that each set in the cover is contained within the image of a singular cube. We start with any orientable atlas, which gives us a coordinate patch $(U,x)$ around any point $p$ we choose. Without loss of generality we can pick the coordinates such that $x(p)=0$. There must be some open ball around $0$ whose closure is completely contained within $x(U)$; this closure is itself the image of a singular cube, and the ball obviously contained in its closure. Hitting everything with $x^{-1}$ we get an open set — the inverse image of the ball — contained in the image of a singular cube, all of which contains $p$. Since we can find such a set around any point $p\in M$ we can throw them together to get an open cover of $M$. So, what does this buy us? If $\omega$ is any compactly-supported $n$ form on an $n$-dimensional manifold $M$, we can cover its support with some open subsets of $M$, each of which is contained in the image of a singular $n$-cube. In fact, since the support is compact, we only need a finite number of the open sets to do the job, and throw in however many others we need to cover the rest of $M$. We can then find a partition of unity $\Phi=\{\phi\}$ subordinate to this cover of $M$. We can decompose $\omega$ into a (finite) sum: $\displaystyle\omega=\sum\limits_{\phi\in\Phi}\phi\omega$ which is great because now we can define $\displaystyle\int\limits_M\omega=\sum\limits_{\phi\in\Phi}\int\limits_M\phi\omega$ But now we must be careful! What if this definition depends on our choice of a suitable partition of unity? Well, say that $\Psi=\{\psi\}$ is another such partition. Then we can write $\displaystyle\sum\limits_{\phi\in\Phi}\int\limits_M\phi\omega=\sum\limits_{\phi\in\Phi}\int\limits_M\sum\limits_{\psi\in\Psi}\psi\phi\omega=\sum\limits_{\psi\in\Psi}\int\limits_M\sum\limits_{\phi\in\Phi}\phi\psi\omega=\sum\limits_{\psi\in\Psi}\int\limits_M\psi\omega$ so we get the same answer no matter which partition we use. ### Like this: Posted by John Armstrong \| Differential Topology, Topology ## 4 Comments » 1. [...] Without loss of generality, we may assume that is supported within the image of a singular cube . If not, we break it apart with a partition of unity as usual. [...] Pingback by \| September 8, 2011 \| Reply 2. [...] we can assume that the support of fits within some singular cube , for if it doesn’t we can chop it up into pieces that do fit into cubes , and similarly chop up into pieces that fit within [...] Pingback by \| September 12, 2011 \| Reply 3. [...] is such a manifold of dimension , and if is a compactly-supported -form, then as usual we can use a partition of unity to break up the form into pieces, each of which is supported within [...] Pingback by \| September 16, 2011 \| Reply 4. [...] many such singular cubes, and the integral on each is well-defined. Using a partition of unity as usual this shows us that the integral over all of exists and, further, must be strictly positive. In [...] Pingback by \| November 24, 2011 \| Reply « 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": 25, "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": 0.9463765025138855, "perplexity_flag": "head"}
http://programmingpraxis.com/2010/01/26/primality-checking-revisited/?like=1&source=post_flair&_wpnonce=e5382928c8	# Programming Praxis A collection of etudes, updated weekly, for the education and enjoyment of the savvy programmer ## Primality Checking, Revisited ### January 26, 2010 [ There is a bug in the solution of this exercise. See the revised version of the exercise for a proper solution. ] We examined the Miller-Rabin probabilistic primality checker in a previous exercise. Today, we examine a primality checker that combines the Miller-Rabin test with a test on Lucas pseudoprimes, devised by Robert Baillie and described by Baillie and Wagstaff in their article “Lucas Pseudoprimes” in Mathematics of Computation, Volume 35, Number 152, pages 1391-1417, October 1980; see also Thomas Nicely’s web page devoted to Baillie’s test. This is the same algorithm used in the `PrimeQ` function in Mathematica. Lucas numbers are defined by a recurrence formula similar to Fibonacci numbers, where $L_n = L_{n-1} + L_{n-2}$ with $L_1 = 1$ and $L_2 = 3$; the Lucas numbers are 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, … (Sloane’s A000204). Lucas numbers have the rather startling property that, if n is prime, $L_n \equiv 1 \pmod{n}$. But the converse is not true, and composite numbers n such that $L_n \equiv 1 \pmod{n}$ are known as Lucas pseudoprimes; the first few Lucas pseudoprimes are 705, 2465, 2737, 3745, 4171, … (Sloane’s A005845). Lucas numbers are a special case of Lucas sequences. If P and Q are integers such that the discriminant $D = P^2 - 4Q$, then the roots of $x^2 - P x + Q = 0$ are $a = \frac{P + \sqrt{D}}{2}$ and $b = \frac{P - \sqrt{D}}{2}$. There are two Lucas sequences $U_n(P, Q) = \frac{a^n - b^n}{a - b}$ and $V_n(P, Q) = a^n + b^n$ for $n \geq 0$, which can be computed by the recurrence equations $U_m(P,Q) = P U_{m-1}(P,Q) - Q U_{m-2}(P,Q)$ and $V_m(P,Q) = P V_{m-1}(P,Q) - Q V_{m-2}(P,Q)$. The Lucas numbers are given by the sequence $V(1,-1)$ and the Fibonacci numbers are given by the sequence $U(1,-1)$. We will want to compute the nth element of a Lucas sequence, mod n, for large n. Rather than computing the entire recurrence in time proportional to n, it is possible to use doubling and halving, in the same way as the exercise on Three Binary Algorithms, to compute the nth element in log n time. Such a computation is known as a Lucas chain. Thus, to test whether an odd number n is a Lucas pseudoprime, we choose sequence parameters P and Q so that the the discriminant D is non-square and the Legendre number $\left( \begin{array}{c} D \\ n\end{array} \right) = -1$ (otherwise the modular arithmetic would fail). Then we construct the Lucas chain; if the nth element is zero modulo n, then n is either prime or is a Lucas pseudoprime. Recall that the Miller-Rabin test used strong pseudoprime tests on fifty bases to check primality. Using Lucas pseudoprimes, we can reduce the number of tests substantially. It turns out that combining two strong pseudoprime tests, with bases 2 and 3, with a Lucas pseudoprime test, there are no known pseudoprimes; the test has been performed exhaustively on all numbers less than 1016, and no counter-examples have been found in over twenty-five years of use (in addition to fifteen minutes of fame, there is a monetary reward from the original authors for the finder of a counter-example). Your task is to write a function that checks primality using the algorithm described above. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below. ### Like this: Pages: 1 2 Posted by programmingpraxis Filed in Exercises 1 Comment » ### One Response to “Primality Checking, Revisited” 1. Mike said April 12, 2010 at 11:55 PM In python: ```from math import sqrt def bits_of( n ): '''generator that returns the bits in n, from MSB to LSB.''' bit = 1 while bit < n: bit <<= 8 while bit > n: bit >>= 1 while bit: yield 1 if ( bit & n ) else 0 bit >>= 1 def euclid(a, b): '''Extended Euclid's algorithm.''' lastx, x = 1, 0 lasty, y = 0, 1 while b: q = a/b a, b = b, a%b lastx, x = x, lastx - qx lasty, y = y, lasty - qy return lastx, lasty, a def gcd(m, n): '''Return greatest common divisor or m and n.''' if m < n: m,n = n,m while n: m, n = n, m % n return m def inverse(x, m): a, b, g = euclid(x, m) if g != 1: raise ValueError("x and m must be coprime.") return a % m def is_square( xx ): if xx&7==1 or xx&31==4 or xx&127==16 or xx&191==0: x = sqrt( xx ) return x == int( x ) return False def jacobi(a, n): if a == 0: return 0 elif a == 1: return 1 elif a == 2: return 1 if (n%8) in (1,7) else -1 elif a&1 == 0: return jacobi(2,n) * jacobi(a/2, n) elif n < a: return jacobi(a%n, n) elif a%4 == 3 and n%4 == 3: return -jacobi(n, a) else: return jacobi(n, a) legendre = jacobi # well, for odd primes anyway def lucas_test( n ): a, b = 11, 7 while True: d = a * a - 4 * b if is_square(d): a += 2, b += 1 elif gcd( n, 2 * a * b * d ) != 1: a += 2 b += 2 else: break x1 = ( a * a * inverse(b, n) - 2 ) % n m = ( n - legendre(d, n) ) / 2 u = 2 v = x1 for bit in bits_of( m ): if bit: u = ( u * v - x1 ) %n v = ( v * v - 2 ) % n else: v = ( u * v - x1 ) %n u = ( u * u - 2 ) % n return ( x1 * u - 2 * v ) % n == 0 def miller_rabin_test(n, a): '''checks if n is prime base a. returns True is a is likely a composite. ''' r, s = 0, n-1 while s&1 == 0: r += 1 s >>= 1 if pow(a, s, n) == 1: return True # possibly prime for j in range(r): if pow(a, s, n) == n-1: return True # possibly prime s *= 2 return False # definitely composite small_primes = set([ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 ]) def is_prime(n): if n < 100: return n in small_primes return all(n%p for p in small_primes) and \ miller_rabin_test( n, 2 ) and \ miller_rabin_test( n, 3 ) and \ lucas_test( n ) ``` %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": 17, "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": 0.8914188742637634, "perplexity_flag": "middle"}
http://mathhelpforum.com/pre-calculus/110531-deduce-q.html	# Thread: 1. ## Deduce Q Given $\cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta})$ and $\sin\theta=\frac{1}{2i}(e^{i\theta}-e^{-i\theta})$ Deduce $1+2\cos\theta+2\cos2\theta+...+2\cos n\theta=\frac{\sin(n+\frac{1}{2})\theta}{\sin\frac {1}{2}\theta}$ Use this result to show $1+(2\cos\theta)^2+(2\cos2\theta)^2+...+(2\cos n\theta)^2=\frac{\sin(2n+1)\theta}{\sin\theta}$ I was thinking geometric series but cant seem to figure it out. It can easily be done using sums to products but i was wondering if there was a way which makes use of the given result. thanks 2. Originally Posted by vuze88 Given $\cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta})$ and $\sin\theta=\frac{1}{2i}(e^{i\theta}-e^{-i\theta})$ Deduce $1+2\cos\theta+2\cos2\theta+...+2\cos n\theta=\frac{\sin(n+\frac{1}{2})\theta}{\sin\frac {1}{2}\theta}$ Use this result to show $1+(2\cos\theta)^2+(2\cos2\theta)^2+...+(2\cos n\theta)^2=\frac{\sin(2n+1)\theta}{\sin\theta}\ {\color{red}{}+2n}$ I was thinking geometric series but cant seem to figure it out. It can easily be done using sums to products but i was wondering if there was a way which makes use of the given result. $1+2\cos\theta+2\cos2\theta+...+2\cos n\theta= e^0 + (e^{i\theta}+ e^{-i\theta}) + \ldots + (e^{in\theta} + e^{-in\theta}) = \sum_{k=-n}^ne^{ik\theta}.$ This is a geometric series with common ratio $e^{i\theta}$ and sum $\frac{e^{-in\theta}(e^{i(2n+1)\theta}-1)}{e^{i\theta}-1}$. Divide top and bottom of that by $2ie^{i\theta/2}$ and you will have something that you can express in terms of sines. For the second part (which should have an extra term +2n as shown), take the formula for the first part, replace $\theta$ by $2\theta$ and use the formula $\cos2\phi = 2\cos^2\phi-1$.	{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 15, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.9778805375099182, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/3871/is-causality-a-formalised-concept-in-physics	# Is causality a formalised concept in physics? I have never seen a “causality operator” in physics. When people invoke the informal concept of causality aren’t they really talking about consistency (perhaps in a temporal context)? For example, if you allow material object velocities > c in SR you will be able to prove that at a definite space-time location the physical state of an object is undefined (for example, a light might be shown to be both on and off). This merely shows that SR is formally inconsistent if the v <= c boundary condition is violated, doesn’t it; despite there being a narrative saying FTL travel violates causality? Note: this is a spinoff from the question: The transactional interpretation of quantum mechanics. - I voted to close the question as "subjective and argumentative" because its author tries to force the visitors to give him a philosophically pleasing answer and totally ignores all the deep physics answers that are actually being written. – Luboš Motl Jan 25 '11 at 22:16 It seems to me that what you are asking boils down to a philosophical question about the nature of causality, which is a central one in philosophy of science starting from Hume. I'd suggest to pick a text on philosophy of science (e.g the one by Alex Rosenberg is good) and familiarize yourself with the debate, it's good stuff. – user566 Jan 25 '11 at 22:22 2 @Lubos you can't retroactively decide a question is "subjective and argumentative" based on your negative view of the OP's response to the answers, including your own. @Moshe you seem to assume the OP doesn't already have a familiarity with the philosophy of science or has not already given the question of causality some consideration. Perhaps you could give @Nigel the benefit of doubt since he does state that he has a Phd in AI. – user346 Jan 26 '11 at 6:11 4 Dear @Lubos it is hard to have a meaningful discussion with you when you dismiss viewpoints that you disagree with as "nonsensical philosophical gibberish". – user346 Jan 26 '11 at 7:30 1 I'll try to summarise the overall discussion in my own answer tomorrow. This question originated from the concept of "retrocausality" in the previous cited question, where it was flagged as a rebuttal of the TIQM. So it's hard to justify taking an agnostic position and booting the question over to "the philosophers" as if physicists don't need to worry. (I see a lot of worrying in some of the answers!). Anyway, more later. – Nigel Seel Jan 26 '11 at 21:58 show 6 more comments ## 9 Answers Dear Nigel, causality is not an observable (quantity) with a value and a unit; so it is not identified with any operators. Causality is a principle. In a broader scientific and colloquial context, causality is any property of the relationship between the cause and its effect. However, in physics, we mean something more particular by causality. In classical physics, we mean the following proposition: If a cause takes place at time $t_1$ and its effect at time $t_2$, then $t_1<t_2$ must hold. In other words, the cause precedes its effect. It's obvious that if the principle above would be violated, the world would become logically inconsistent. Events at time $t_1$ could cause some events at time $t_2$ which would generally cause different events at time $t_1$, producing contradictory answers to the question what happened at time $t_1$. Looking at causality from a relativistic viewpoint In the special theory of relativity, the statement above must still hold for the history of spacetime to be free of logical contradictions. However, special relativity is based on the principle of relativity that says Laws of physics take the same form in all inertial frames - those that are in uniform motion relatively to one chosen inertial system. This must be true for all laws, including the principle of causality itself. If this principle of relativity is combined with the principle of locality above, we may actually derive a stronger statement. In relativity, the delay between two events depends on the inertial system: simultaneity of events is relative, we say. So two events may be chronologically ordered in the opposite way if you switch into a different inertial system. However, spacelike separated events remain spacelike separated events; and an event in the future (or past) light cone of another event stays in the same cone from the viewpoint of all inertial frames. Applying the principle of relativity to the principle of causality, we may derive a stronger, relativistic principle of causality: If a cause takes place at point $P_1$ in spacetime and if its effect takes place at point $P_2$ in spacetime, then $P_2$ must belong to the future light cone of $P_1$. This is a stronger statement than the original one (about the ordering of $t_1$ and $t_2$): the relativistic causality implies the ordinary causality, but something more (it implies the non-relativistic condition from the viewpoint of all relativistic inertial frames). A cause is not only unable to affect its past, like in the non-relativistic causality; it is unable to affect the spacelike-separated points in the spacetime, too. Any violation of the relativistic causality - which means that causes may only influence their future light cones - would lead to the same logical contradictions that I explained in the non-relativistic context. In particular, you wrote: This merely shows that SR is formally inconsistent if the $v \leq c$ boundary condition is violated, doesn’t it; despite there being a narrative saying FTL travel violates causality? Well, indeed. However, special relativity is demonstrably a valid theory of our spacetime (at least locally). So the "mere" inconsistency of special relativity that you mentioned, in a somewhat incomprehensibly dismissive tone, would automatically mean an inconsistency of the whole Universe which is a pretty serious problem. There's no doubt that there can't be any signals that move faster than light. Logical consistency is an omnipresent and unquestionable assumption in all of physics (and maths), so one is always allowed - and encouraged - to assume it. When we assume it, we may easily show that faster-than-light motion violates causality. In fact, relativistic causality is exactly what bans faster-than-light motion. I am convinced that this text explains - and fully unmasks - all deeper and more foundational facts and arguments behind the notion of causality in physics. - The problem that I have with this answer is that you are using natural-language terms (cause, effect) as a metalanguage to talk about the physics. Where in the mathematics is there some entity which you can point to and say 'that's intrinsically the concept of a 'cause' or 'that's the concept of an effect'? Formally speaking, how could this work when we can (for the purposes of this discussion) just time-reverse everything in our fundamental theories? I could be led to believe that our everyday concept of causality has more to do with the arrow of time and increasing entropy :-). – Nigel Seel Jan 25 '11 at 20:54 It is you who began to talk about causality, so I just replied. As I said at the very beginning, "causality" is not any quantity that can be expressed by a variable or an equation. Causality is a principle addressing natural-language terms. It's exactly what I wrote. Everyone who writes that it is something completely different is lying to you. ... And you can't reverse things in physics: it's the very point of causality that you can't. The fundamental laws are CPT-symmetric but the system of predictions in physics requires an arrow of time to define a globally consistent law of causality. – Luboš Motl Jan 25 '11 at 22:03 "I am convinced that this text explains - and fully unmasks - all deeper and more foundational facts and arguments behind the notion of causality in physics." .... @Lubos by "this text" do you mean your answer? – user346 Jan 26 '11 at 6:02 @Lubos The word locality appeared once in your answer, is it supposed to be causality instead of locality? – Revo Aug 23 '11 at 2:05 Lubos, In the line "If this principle of relativity is combined with the principle of locality above", I think you mean "causality above" ? – Revo Aug 11 '12 at 12:26 Causality becomes much more subtle whenever theories are statistical or probabilistic. When we see a correlation, it may be that one event caused the other, but it may be that there is a common cause or that there is just a chance correlation that would disappear if we do more of the same data gathering. For the notion of "common cause", the standard old-time reference is Reichenbach. Try the "Reichenbach's common cause principle" entry in the Stanford Encyclopedia of Philosophy, and other entries on causality therein. Often quite a good reference for Philosophy of Science. The mathematics of causality in modern statistical physics is very simple, but it has subtle consequences. Quantum field theory distinguishes between the brute random fluctuations of the vacuum state (which are caused, if they are caused by anything, by the random fluctuations that were there in the past, which were caused, if they were caused ...) and causal relationships between measurements. If two measurements are associated with regions of space-time that are at space-like separation from each other, quantum field theories predict that correlations will be observed in the recorded data, but we say that ideal measurement devices do not cause such correlations. If we don't have measurement devices that are close enough to this ideal, we may have to make allowances for the non-ideal details of the real devices to make the theory match the data. Somewhat non-standardly, I would say that ideal quantum measurements that are at time-like separation from each other do cause some component of the correlations we observe in the recorded data from such measurements. That's different from ideal classical measurement devices, which record data while changing neither the physical state nor the data recorded by other ideal classical measurement devices. But this is a research project. Apropos of the Transactional Interpretation starting point, the SEP entry "Action at a Distance in Quantum Mechanics" might be interesting. - I think this is a very interesting comment and one which genuinely engages with what it even means to say that event A "causes" event B within the framework of a particular physical theory (QFT in your example). It's good that we're now focussing on operational concepts such as measurements and correlations: +1. – Nigel Seel Jan 26 '11 at 22:05 In the axiomatic approach to quantum field theory, sometimes also called local or algebraic quantum field theory, pioneered by Araki, Haag, Kastler, Bogoljiobov et. alt., causality is formalized as an axiom, most often called the "locality" axiom. The idea is this: To every bounded open subset of Minkowski spacetime we associate an operator algebra, all selfadjoint elements of this algebra represent all observables of this region, that is everything that is measurable in this region. Then algebras associated to two spacelike separated regions are assumed to commute, this is the locality or causality axiom. When two observables aka selfadjoint operators commute, this means of course that measuring aka observing the first will have no effect on measuring aka observing the second and vice versa, therefore there cannot be any causal relationship of the events of measuring them. BTW: The Reeh-Schlieder theorem seems - intuitively - to violate causality/locality, so it is interesting to note that it is possible to prove this theorem without invoking the locality axiom. The reason for this is that the Reeh-Schlieder theorem is about entanglement effects which don't violate locality in the sense of SR. - Interesting. However what you're describing seems to be the formalisation of the concept of a spacelike interval within a variant of QFT and then labelling this axiom with the word "causality" (or "non-causality" in fact). My question was intended to explore whether causality has a deeper, more foundational role in physics than as a synonym for non-spacelike separation. – Nigel Seel Jan 25 '11 at 20:31 @Nigel: No, as far as I know there is no deeper formalization of causality in physics, as Lubos already said. Maybe one could add that the time evolution of physical systems is described by hyperbolic evolution equations and that the causal relationship of events is preserved via the appropriate representations of the Poincare group, and that's it. – Tim van Beek Jan 26 '11 at 19:46 I don't think spacelike commutativity is the same thing as causality. Probably a much closer characterization is the primitive causality axiom stating that the algebra associated with the causal completion of a region is the same as the algebra associated with the region. – QGR Jan 27 '11 at 16:42 Perhaps the question is too broad. The fundamental principle underlying causality is locality. We can observe causality indirectly by observing locality. By this I mean that the very existence of a coordinate system is the very manifestation of locality. Think of it this way; the motion of an observer 'causes' lengths to appear smaller. In other words, the 'effect' is the shortening of measured length. You can imagine that the relative motion is a cause and that length contraction is an effect. In order to measure this effect we postulate a coordinate system. When we do this, it turns out, that we predict a finite maximal speed. Which happens to be the speed of light. In Quantum Mechanics the issue of locality is completely defined by the Heisenberg Uncertainty Principle (HUP) which is equivalent to the treatment of locality in General Relativity. Namely, the HUP allows you to measure precisely the position of a particle/field. In other words, the HUP allows you to define a coordinate system. This is sufficient to define locality and hence sufficent to embody causality. Think of it this way; the existence of an observer causes a coordinate system to appear. In QM, the observer is the 'cause' and the coordinate system is an 'effect'. - As the originator of this question I have reviewed and learned something from all the answers posted so far. I would like to summarise my own views here. 1. Where did this query come from? From a question about the Transactional Interpretation of Quantum mechanics (TIQM), where said theory’s reliance upon “retrocausality” (‘causality’ backwards in time) was held to be a fatal defect. This kind of causality-argument is common in physics: we say that faster-than-light travel in SR is ruled out because it would violate causality. 2. Such causality arguments are conducted in what you might call the “metalanguage of physics”: technical English which supports and explains formal results. However, the arbiter in the end is the maths, so how do we interpret the notion of causality within the formalism? 3. Physical theories are defined by mathematical relationships between entities (observed and unobserved) usually expressed by equations (think Schrodinger, Dirac, the Lorentz transformation). If we say that event E1 “causes” event E2, several answers here suggest that the interpretation of causality in the formal theory is that: (i) if E1 is postulated to occur then the theory logically implies that E2 must occur as well; (ii) E2 is within or on the future light-cone of E1 (we say “cause precedes effect”). 4. However, it’s possible that condition (ii) is too stringent. While logical entailment is obviously an essential part of any formalised theory, our smuggling in of the word “future” is already an extra assumption. Our fundamental theories do not impose a specific past-future direction on the time dimension. This means that if you reverse the film, the events you see are still consistent with our fundamental theories. 5. Sometimes people use causality-like words in the physics metalanguage without conventional time-ordering condition (ii). For example, a possible Feynman diagram for electron-positron scattering has a narrative that an electron travelling backwards in time from the future encounters a (normal) electron, they exchange a virtual photon and continue on their way scattered. The 'cause' of the scattering event was the arrival of the future electron. Many textbooks mention this way of thinking but we don’t mind because the underlying theory gives consistent results which accord with observation. Perhaps TIQM is like this despite its narrative of retrocausation. 6. So my conclusion is that we have to be careful about arguments concerning a theory’s validity relying upon causation arguments couched in physics’ metalanguage. It’s not a slam-dunk. Sometimes if a theory violates conventional “cause precedes effect” causation it indicates a breakdown in the underlying mathematics, normally inconsistency. At other times a 'causation' argument is just a way of talking about the entailment of the theory in an innovative or whimsical way, and the theory is actually OK. Go look at the maths. NOTE: there is a whole separate discussion about why, in natural language, we think so naturally in terms of cause and effect. It links to discussions about the arrow of time and why we do seem to be unhappy about running the film backwards as a valid picture of reality. That is a whole separate issue but still, I suspect, part of physics judging by the number of recent books on the subject. - 1 Good clarifying points: I think though that a Stack lesson is that if the question was really about "retrocausality in TQM" that could have been made explicit in the question description. I have examined Cramer's 1986 paper in the light of your answer and found he covers these points too in defence of TQM. Specifically he has introduced "strong" and "weak" causality. Put simply he admits that TQM has retro strong causality but has no retro weak causality. He also mentions the arrow of time and the K meson problem associated with it. I'll amend my answer slightly too. – Roy Simpson Jan 27 '11 at 12:34 The Causality structure of a Spacetime might be best modelled by a partial order between events. An event is a primitive concept in spacetime theory, though one is welcome to question that primitiveness in a yet more fundamental theory. Penrose's Twistor Theory is another theory (this time geometric) in which spacetime points (events) are derived, not fundamental. However given that we assume events, we can introduce a partial order between them. Introduce some conditions - or derive a null cone structure from the partial order. So one can ask for conditions on the PO to derive a null cone from it. Then from a Global perspective one can introduce further conditions to ensure that this PO does not have x < y and y < x. Of course this is the Closed Timelike Curve condition in General Relativity. Now in GR/SR a "cause" of x is just anything in the past lightcone of x. In Quantum Theory this gets more interesting, particularly in the context of the TQM: here something needs to be "emitted" and "received" - much like a telephone message. TQM might not be correct, but this telecom analogy shows that one could go deeper into what a "cause" really is. I suspect that the place for such a theory is in the "Fundamental Framework for Quantum Gravity" - should such a revolutionary Framework come about. Roy. EDIT: It has become clear that this question was motivated by a criticism of retrocausality in a related post on Transactional Quantum Mechanics (TQM). Rather than continue with a general discussion of Causality in Physics I shall note that Cramer addresses this question in his papers (Transactional Interpretation of Quantum Mechanics, 1986 and cites therein). He introduces two notions of Causality: Strong Causality Cause always precedes effect in any reference frame. Weak Causality Ditto, but only for macroscopic observations and observer-to-observer communication. Cramer's cancellation argument implies that there is no Weak Causality Violation in his theory/interpretation. If we accept this, then the question is whether it is acceptable to have strong causality violation in a theory of physics. I can envisage further debate on this. - From a statistical perspective, whenever you can establish that two type of events (described as states (e1,e2) of a pair systems S1xS2 ) have high correlation (correlations basically stands for the probability of system S1 be in state e1 and S2 be in state e2 simultaneously), you can say that both events are 'causally related' Being said that, there is no formal way to establish (at least from a purely statistical point of view) which event causes which one, other than your willingness to accept a particular convention of time order. In QFT, There is a very formal notion of causality established by the conmutator of field operators as functions of space-time, for instance: $$\left[ \phi(x_\mu),\phi(x'_\mu)\right]=\theta (x_\mu x'_\mu)$$ when this conmutator is zero it means there is no possible correlation between physical events. the quantity in the right side is different from zero when the two space-time points are time-like to each other (inside the light cone of influence) - 1 lurscher, try $[ \phi(x_\mu) , \phi(x'_\mu) ] = \Sigma ( x_\mu x'_\mu )$ – Roy Simpson Jan 25 '11 at 21:59 One interesting question here is whether the commutator is ever different from zero when the events are spacelike (also null) separated? Another complicating factor is whether one can use points (events) at all in an axiomatic formulation, but must use "test functions". As test functions are spread out the definition of timelike might be more subtle. – Roy Simpson Feb 4 '11 at 13:49 i agree, the current formalism is not entirely satisfactory. But it is a formal concept of causality nonetheless. Thanks for the edits! – lurscher Feb 4 '11 at 15:28 Great question. I'll try to answer. In quantum-mechanical world the timeline is completely reversible, the system undergoes unitary evolution and there is no preferred time direction. There is no concept of causalty either. That's why it is possible to move faster than light in the quantum world. For example, electron (and photon) can travel faster than light with some probability. But it can be interpreted as a virtual electron-positron pair appearing ahead of the propagating electron and then the positron from the pair annihilates with the propagating electron (the trick is that information cannot be transferred such way because you cannot be sure whether you spotted just a virtual particle from a vacuum fluctuation or an actual signal). The causality problem arises from asymmetry of time in classical world. And the asymmetry of time arises from the irreversibility of collapse of the wavefunction. That is the FTL travel can lead to a controversy only when measurements are taken. It is impossible to measure the quantum value and leave it as it is without collapse. Now imagine one can travel in time, measure the value in the future, then return to present. It would appear as if he made a measurement without collapsing the WF. In fact the WF collapse is the only existing fundamental law which is not symmetric against the time direction. It is the cause of all other observed irreversible processes, including the second law of thermodynamics. - "And the asymmetry of time arises from the irreversibility of collapse of the wavefunction." uh, no. Nothing arises from the collapse of the wavefunction because it is completely unphysical. Asymmetry arises from the simple observation that we know past and not future. It cannot be explained, it's just the only possibility to make sense of physics (or indeed, anything in life). – Marek Feb 21 '11 at 8:21 "Nothing arises from the collapse of the wavefunction because it is completely unphysical." - It is physical. At least in Copenhagen interpretation. "Asymmetry arises from the simple observation that we know past and not future." - we know? Asymmetry is already in the thermodynamics. Thermodynamical system cannot know anything. – Anixx Feb 21 '11 at 13:24 What is physical can't depend on the interpretation. If you are familiar with decoherence you should understand this basic point. Add asymmetry: yes, we know past, at least in principle (it's called history, by the way). But we don't know future. We can only predict it using the knowledge of the past. This fundamental assymetry (knowledge of past and prediction of future) underlies all the physics. Nothing to do with QM or TD in particular. And by the way, please use @-sign, so that I know you left a comment. – Marek Mar 28 '11 at 11:41 "What is physical can't depend on the interpretation." No. What is OBSERVED can't depend on interpretation. What is physical varies between interpretations heavily. "Add asymmetry: yes, we know past, at least in principle (it's called history, by the way). But we don't know future." Actually "knowledge of the past" is a Classical process, and cannot happen without measurement and wavefunction collapse. That is why this assymetry. Quantum system without collapse has no knowledge of the past and time-symmetric. – Anixx Mar 28 '11 at 13:11 Anixx is not wrong, just taking a nonstandard philosophical path through a certain well-trod thicket. – Ron Maimon Aug 15 '11 at 5:10 Causality as a consequence of finite energy: 0 - in the universe there is more than one object with finite energy content each. "Reductio ad absurdum': 1 - If there is no limit on the speed of transmission of energy (light, electric force, gravitational ,...) 2 - then at the same time a moving object (with finite energy content) is here and there. You need twice the energy. 3 - then at the same time an object is here and in any other place. 4 - then that object must have an infinite amount of energy distributed throughout the universe. 5 - then there is no room or energy for more than that object in the universe. 5 contradicts 0, then 1 is false. -	{"extraction_info": {"found_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": 0.9440777897834778, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2008/04/18/improper-integrals-i/?like=1&source=post_flair&_wpnonce=8cb43e0c56	# The Unapologetic Mathematician ## Improper Integrals I We’ve dealt with Riemann integrals and their extensions to Riemann-Stieltjes integrals. But these are both defined to integrate a function over a finite interval. What if we want to integrate over an infinite ray, like all positive numbers? As a specific example, let’s consider the function $f(x)=\frac{1}{x^2}$, and let it be defined on the ray $\left[1,\infty\right)$. For any real number $b>1$ we can pick some $a'>a$. In the interval $\left[1,b'\right]$ the function $f$ is continuous and of bounded variation (in fact it’s decreasing), and so it’s integrable with respect to $x$. Then it’s integrable over the subinterval $\left[1,b\right]$. Why not just start by saying it’s integrable over $\left[1,b\right]$? Because now we have a function on $\left[1,b'\right]$ defined by $\displaystyle F(x)=\int\limits_1^x\frac{1}{t^2}dt$ Since $t$ is differentiable and $\frac{1}{t^2}$ is continuous at $a$, we see that $F$ is differentiable here, and its derivative is $F'(b)\frac{1}{b^2}$. This result is independent of the $b'$ we picked. Since we can do this for any $b>1$ we get a function $F(b)$ defined for $b\in\left(1,\infty\right)$. Its derivative must be $\frac{1}{b^2}$, and we can check that $\frac{-1}{b}$ also has this derivative, so these two functions can only differ by a constant. Clearly we want $F(1)=0$, since at that point we’re “integrating” over a degenerate interval consisting of a single point. This fixes our function as $F(b)=1-\frac{1}{b}$. Now our question is, what happens as we take $a$ to get larger and larger? Our intervals $\left[1,b\right]$ get bigger and bigger, trying to fill out the whole ray $\left[1,\infty\right)$. And for each one we have a value for the integral: $1-\frac{1}{b}$. So we take the limit as $b$ approaches infinity: $\lim\limits_{b\rightarrow\infty}1-\frac{1}{b}=1$. This will be the value of the integral over the entire ray. We turn this rubric into a definition: given a function $f$ that is integrable with respect to $\alpha$ over the interval $\left[a,b\right]$ for all $b>a$, we can define a function $F$ on $\left[a,\infty\right)$ by $\displaystyle F(b)=\int\limits_a^bfd\alpha$ We define the improper integral to be the limit $\displaystyle\int\limits_a^\infty fd\alpha=\lim\limits_{b\rightarrow\infty}\int_a^bfd\alpha$ if this limit exists. Otherwise we say that the integral diverges. We can similarly define improper integrals for leftward rays as $\displaystyle\int\limits_{-\infty}^bfd\alpha=\lim\limits_{a\rightarrow-\infty}\int_a^bfd\alpha$ And over the entire real line by choosing an arbitrary point $c$ and defining $\displaystyle\int\limits_{-\infty}^\infty fd\alpha=\int\limits_{-\infty}^cfd\alpha+\int\limits_c^\infty fd\alpha$ That is, we take the two bounds of integration to go to their respective infinities separately. It must be noted that the limit where they go to infinity together: $\displaystyle\int\limits_{-\infty}^\infty fd\alpha=\lim\limits_{b\rightarrow\infty}\int\limits_{-b}^bfd\alpha$ may exist even if the improper integral diverges. In this case we call it the “Cauchy principal value” of the integral, but it is not the only justifiable value we could assign to the integral. For example, it’s easy to check that $\displaystyle\lim\limits_{b\rightarrow\infty}\int\limits_{-b}^bxdx=\lim\limits_{b\rightarrow\infty}0=0$ so the Cauchy principal value is ${0}$. However, we might also consider $\displaystyle\lim\limits_{b\rightarrow\infty}\int\limits_{1-b}^{1+b}xdx=\lim\limits_{b\rightarrow\infty}2b$ which diverges. ### Like this: Posted by John Armstrong \| Analysis, Calculus ## 8 Comments » 1. [...] Tests for Improper Integrals We have a few tests that will come in handy for determining if an improper integral converges. In all of these we’ll have an integrator on the ray , and a function which is [...] Pingback by \| April 21, 2008 \| Reply 2. [...] on the ray , and be any function integrable with respect to through the whole ray. Then if the improper integral converges, then so does [...] Pingback by \| April 22, 2008 \| Reply 3. Is there some significance attached with the Cauchy Principal Value? I mean since it has been given a name, it most have encountered at more places. Comment by Nilay \| April 23, 2008 \| Reply 4. I’m not sure what I’ll do with it yet, but I may as well give its name now, and if it comes up later I can refer back to it. Comment by \| April 23, 2008 \| Reply 5. [...] the particular case of an improper integral, we have . Then . Our condition then [...] Pingback by \| April 23, 2008 \| Reply 6. [...] Then as we let go to infinity, goes to infinity with it. Thus the sum of the series is the same as the improper integral. [...] Pingback by \| April 24, 2008 \| Reply 7. [...] can forge a direct connection between the sum of an infinite series and the improper integral of a function using the famed integral test for [...] Pingback by \| April 29, 2008 \| Reply 8. [...] I get the title of the post wrong? No. It turns out that I covered half of this topic two years ago as I prepared to push into infinite series. Back then, I dealt with what happened when we wanted to [...] Pingback by \| January 15, 2010 \| Reply « 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": 0.921056866645813, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/305297/a-question-about-prime-ideals	A question about prime ideals suppose $p$ a prime ideal. If $p=a \cap b$, $a,b$ ideals. Is it true that we have to have $p=a$ or $p=b$? - Do you know about co-maximal ideals? – Mathematician Feb 16 at 5:30 3 Answers True. Suppose that $\mathfrak{p}$ is neither $\mathfrak{a}$ nor $\mathfrak{b}$. Pick $x \in \mathfrak{a} \backslash \mathfrak{p}$, $y \in \mathfrak{b} \backslash \mathfrak{p}$. Consider $xy \in \mathfrak{a} \cap \mathfrak{b} = \mathfrak{p}$. This implies that $x$ or $y$ must lie in $\mathfrak{p}$, contradiction. - Simple and clear, thanks. – user62508 Feb 16 at 5:37 An ideal $I$ satisfying this condition -- namely that if for ideals $\mathfrak{a}$ and $\mathfrak{b}$ of $R$, $I = \mathfrak{a} \cap \mathfrak{b} \implies I = \mathfrak{a}$ or $I = \mathfrak{b}$ -- is called irreducible. Irreducible ideals occur in the theory of primary decomposition. For the bare rudiments of this theory, see e.g. Chapter 10 of my commutative algebra notes. A few samples: In a PID, an ideal is irreducible iff it is a prime power. Prime ideals are irreducible (the OP's question). In a Noetherian ring, irreducible ideals are primary. A proper ideal in a Noetherian ring is a finite intersection of irreducible ideals. Combining the last two results one gets Noether's theorem, the basic result on primary decomposition. - +1, nice answer. Is there any relation between irreducible ideals and irreducible elements of the ring? – Eric♦ Feb 16 at 5:53 @Eric: I don't see one. An ideal in a GCD domain generated by an irreducible element is prime hence irreducible, but even in a PID there are nonprime irreducible ideals. (Needless to say, the terminology "irreducible ideal" was not my doing!) – Pete L. Clark Feb 16 at 8:00 Browsing through §10 of Pete's notes I am once again amazed by the quality of his text. – Georges Elencwajg Feb 16 at 8:20 Since $ab\subset a\cap b$, we see that $ab\subset p$. Now use the definition of prime to see that $a\subset p$ or $b\subset p$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 27, "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": 0.9186428189277649, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/33796/existence-of-a-pseudo-polynomial-time-algorithm-for-a-counting-problem	## Existence of a pseudo-polynomial time algorithm for a counting problem. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let T={1,...,n} be a set of tasks. Each task i has associated a non negative processing time p_i and a deadline d_i. A feasible schedule of the tasks consists of a permutation of n elements pi, such that \sum_i=1^k p_(pi(i)) <= d_(pi(i)) for all k=1,...n. Does there exists a pseudo-polynomial time algorithm for computing the total number of feasible schedules? A pseudo-polynomial time algorithm is an algorithm whose running time is bounded by a polynomial on the size of the input, given that the input is written in unary notation (2=II, 3 =III). (e.g., the size of a number n in unary notation is O(n), and not O(log(n)). This is an open question from an article published in 2009 at Operations Research Letters. - 1 Usually asking open problems is not really considered appropriate for MO; see the FAQ. That said, you might ask something like "what is the current state of knowledge about this problem?" – Daniel Litt Jul 29 2010 at 14:57 Should the RHS read $d_{\pi(k)}$ instead of $d_{\pi(i)}$? I also agree with Daniel, it seems inappropriate to ask an open question (especially such a recent one) on MO. – Artem Kaznatcheev Aug 27 2010 at 13:00 1 At the time I write this comment, the question has been edited and now makes no sense whatsoever – Yemon Choi Sep 13 2010 at 15:40 1 I've rolled the question back to it's previous version, which at least made sense. I don't know anything about this question, but it seems perfectly reasonable to ask whether any progress has been made on it. – David Speyer Sep 13 2010 at 15:51 1 Actually, David, could you further roll it back to the version by Gerry, which had the same content but had grammar corrected and used LaTeX? – JBL Sep 13 2010 at 17:51 show 1 more comment ## 1 Answer If you need to compute all solution for a specific instance, you could generate a an IP formulation of the problem and use a lattice point enumeration code such as LattE. This might be a good problem for the operations research QA -	{"extraction_info": {"found_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": 0.934169352054596, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/23644/is-there-a-simple-way-to-derive-a-t-s-diagram-from-a-p-v-diagram-for-arbitrary-p	# Is there a simple way to derive a T-S diagram from a p-V diagram for arbitrary processes? Often, for thermodynamic processes only a p-V diagram is shown. Even without hard figures, the shape of the curve can be helpful to evaluate the process. However, it is hard to figure out for real processes. Are there good rules of the thumb to arrive at a T-S diagram from a p-V diagram? - – David Zaslavsky♦ Apr 23 '12 at 20:05 yes. Possible to merge? – mart Apr 24 '12 at 6:59 sure, I'd be happy to do that. Just change the email address on the other account to match the one on this account to confirm the identity, and as soon as you do that I'll merge them together. – David Zaslavsky♦ Apr 24 '12 at 15:00 can't change the address on the other to match this cause this one is already registered. – mart Apr 26 '12 at 11:51 Oh, OK. Well if you can just edit something into the profile field of the other account to indicate that it's you ("please merge me" or something would work), that would be fine as well. – David Zaslavsky♦ Apr 26 '12 at 15:56 ## 2 Answers For a general system (unknown equation of state/energy) this will not be possible. For example a $(p,V)$-curve with points $(p,\frac{c}{p})$ could be an isotherm because maybe $V(p)=\frac{nRT}{p}$. An isotherm corresponds to a horizontal line in a $(T,S)$-diagram. But if you don't know you're dealing with an ideal gas situation then it might look different. As a general orientation, the stability criteria for reasonable thermodynamic systems and related relations might be helpful to guess the signs of the rates of change. For example in a typical situation, you'd expect the compressibility $$\beta=-\frac{1}{V}\frac{\partial V}{\partial p}$$ to be positive. The adiabatic and isothermal compressibilities are defined for curves of constant $S$ and $T$ respectively, so the corresponding adiabatic or isothermal line in the $(T,S)$-diagram will be associated with a curve $(p,V(p))$ in the $(p,V)$-diagram with $$\frac{\partial V(p)}{\partial p}<0.$$ You see this in the ideal gas example (see picture), where the isotherm is $V\propto \frac{1}{p}$ and the adiabate is also some negative power of pressure $V(p)\propto \frac{1}{p^{\frac{1}{\gamma}}}$, with $\frac{1}{\gamma}>0$. Positivity of the heat capacities $C_V$ or $C_p$ at constant volume or pressure is a dual example, which relates to properties of $S(T)$ in similar way. Furthermore, the Maxwell relations cross correlate the slopes. Lastly, there are some principles, following from the laws of thermodynamics, which forbid some curve configurations. E.g. two different adiabates can never cross twice, because that would mean a possibility for work without heat flow. But as I said, besides that rough reasoning regarding the shapes of the cyclic processes, if you don't know $T(p,V)$ etc. and you can't compute heat or work, how would you know which line in the $(p,V)$ diagram corresponds to the isotherm or the adiabate? In the other case, i.e. if know know the equation of state or some other relation, then the dependencies can of course be explicitly computed. - The equations of state of a Gibbsian substance are usually given in the form $$T=f(p,V), \quad S=g(p,V)$$ for suitable functions $f$ and $g$. I understand the query as asking when we can use them to express $p$ and $V$ in terms of $S$ and $T$. Mathematically speaking, this means that you are asking for a representation of the inverse of the corresponding mapping $$(p,V)\mapsto (S,T)$$ of the plane or a part thereof. For simple models, this can be done by solving the resulting equation for $p$ and $V$. Thus for the van der Waals gas in the form $$T=\left (p + \frac 1 {V^2}\right )(V-b), \quad S = \frac 1 {\gamma-1} (\ln p + \gamma \ln V)$$ we can easily compute that $$V=b+e^S T^{\frac 1 {1-\gamma}}\quad p=e^{-S}T^{\frac \gamma {\gamma-1}}-\frac a{(b+e^ST^{\frac 1 {\gamma-1}})^2}.$$ (The ideal gas is the special case $a=b=0$). For more complicated models, it may not be possible to compute this inverse explicitly (for example, for a model due to Feynman which allows for the fact that the adiabatic exponent depends on temperature) but there are numerical methods available (for example, in Mathematica one has the option NSolve). On the more theoretical side, the inverse function theorem and the fact that the Jacobian determinant of the above transformation is $1$ (Maxwell relation), ensures that this inverse exists (at least locally). -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 27, "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": 0.9261308908462524, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/tagged/special-relativity+time	# Tagged Questions 1answer 129 views +50 ### Is period of rotation relative? My question is inspired by the following answer by voix to another problem: "There is a real object with relativistic speed of surface - millisecond pulsar. The swiftest spinning pulsar currently ... 3answers 112 views ### Do velocity and acceleration time dilation factors add? For a spinning space station such as in 2001, A Space Odyssey, what would be the time slowing in the perimeter of the spinning space station with respect to the center axis of the station? The ... 4answers 171 views ### What is the exact mechanism by which time dilates? What is the exact mechanism by which time dilates for a fast moving object? Can the time dilation be explained by any theory other than relativity? 1answer 110 views ### Cancelling special & general relativistic effects We know that for a GPS we need to make a correction for both general and special relativity: general relativity predicts that clocks go slower in a higher gravitational field (the clock aboard a GPS ... 1answer 52 views ### Time Dilation in relation to Acceleration What I am looking for is a layman's explanation on the equations required to work out Time Dilation at high speeds including acceleration and deceleration of velocity. Or I would greatly appreciate it ... 2answers 182 views ### Will observers moving on a sphere experience time dilation? A single source of light exists at a fixed point in space relative to two observers. The two observers move on the surface of a shell with a fixed radius with the light source at its centre. They move ... 1answer 112 views ### Life of a photon [duplicate] I am a student of class 12th and as far as i know when anything reaches about 99.99% of the speed of light it starts traveling in time or time for it slows down so that it don't breaks the speed ... 3answers 305 views ### Why do clocks measure arc-length? Apologies in advance for the long question. My understanding is that in GR, massive observers move along timelike curves $x^\mu(\lambda)$, and if an observer moves from point $x^\mu(\lambda_a)$ to ... 1answer 93 views ### The status / acceptance of block time? What is the current status or acceptance of block time as it relates to Einstein's theory of relativity? Has quantum mechanics ruled it out or is it still the favored view of the world? Perhaps there ... 1answer 147 views ### The real meaning of time dilation Is this true or false: If A and B have clocks and are traveling at relative velocity to each other, then to B it APPEARS that A's clock moving slower, but A sees his own clock moving at normal speed. ... 1answer 99 views ### Why does only one twin travel in the twin paradox? The wikipedia page repeatedly says that the twin travelling in space is the only one which travels, and also is the only one which faces acceleration and deceleration. So it does not age, while the ... 3answers 265 views ### Is time dilation an illusion? It is said that we can verify time dilation by flying a very accurate clock on a fast jet or spaceship and prove that it registers less time than the clocks on earth. However, the clocks on earth ... 1answer 103 views ### What is the maximum time dilation between two objects, if one is standing still and the other is moving at $c$? What is the maximum ratio in the rate of change in time in reference to object $A$ which is standing still and object $B$ which is moving at the speed of light? 1answer 96 views ### Confusion about time shift in special relativity I have never really found a way to comfortably comprehend the idea of time shift even though I know its not the hard part of relativity theory. In that light, can someone point out what is wrong or ... 1answer 223 views ### What is the speed of time When we measure the speed of a moving element we do it with the help of a reference frame. Now if we need to measure the speed of time, is it possible? Is time really has speed? Thanks in advance. 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 112 views ### Can time dilation be explained by limitations on computing power? Are there any ideas of explaining the time dilatation as limits in "computing power"? What I mean is basically that the greater is a concentrated mass, the harder is to "compute" what happens in such ... 2answers 376 views ### What's the difference between space and time? I'm having a hard time understanding how changing space means changing time. In books I've read people are saying "space and time" or "space-time" but never explain what the difference is between the ... 1answer 114 views ### Time Contraction This is my first time posting on this site. I am a computer programmer that stumbled across a physics text book and have a question on special relativity. So firstly, I understand that there is no ... 1answer 86 views ### How does light travel create time travel violating causality? Saw a question about faster than light travel... I still have the same question though none of the answers offered any resolution for me. It is so summarily assumed by all physicists and commentaries ... 2answers 140 views ### Is there a “present state” of distant stars if simultaneity is relative? Special relativity theory says simultaneity is relative, meaning that different observers will not agree on what happened first and what second. Does it then make sense to say that looking at distant ... 0answers 76 views ### How would an observer perceive movement on a train that's travelling near the speed of light? Person A is on Earth and a train (or whatever you want to imagine) travels past him at near the speed of light. How would person A perceive movement on the ship? If time is slowed on the ship from the ... 2answers 159 views ### When moving fast Time slows down Vs speeds up I was watching an old cartoon movie where a scientist makes a gadget, which when bound on the wrist, freezes the movement of the whole world. So, that one may do 100s of things in a single second. ... 9answers 444 views ### How to explain (pedagogically) why there is 4 spacetime dimensions while we see only the 3 spatial dimesions? I have been asked this question by a student, but I was able and in the same time incapable to give a good answer for this without equations, so do you have ideas how one can explain this in a simple ... 2answers 119 views ### Speed of Entropy change If time in systems moving with different speed goes differently, does speed of entropy change differ in these systems? (is "speed of entropy change" a valid term? can we compare them?) 3answers 739 views ### What is the length of 1 second in meters If time is treated as a fourth dimension of spacetime, what is relation between length and time units? Or in other words, how can I convert time units to length units, for instance seconds to meters? ... 1answer 75 views ### A clock devoid of motion My understanding is that every clock mechanism we have depends on motion of something w.r.t the observer. From atomic transitions to clockwork gears. So, does this property/constraint makes every ... 1answer 172 views ### Does the Earth's revolution around the Sun affect radioactive decay? Premises: The radioactivity is either hastened or slowed inside a fast moving aircraft. Speed of fastest aircraft: 3,529.6 km/h. The earth's revolution is: 107278.87 km/h. The earth's ... 3answers 148 views ### Time slowing down problem When someone moves, time slows down for him. Let, a man standing still and another moving very very very fast, this happens for an hour (as measured by the standing man). Time has moved slower for the ... 4answers 332 views ### Why are objects at rest in motion through spacetime at the speed of light? I read that an object at rest has such a stupendous amount of energy, $E=mc^2$ because it's effectively in motion through space-time at the speed of light and it's traveling through the time dimension ... 5answers 487 views ### Why isn't the symmetric twin paradox a paradox? Two twin sisters synchronize their watches and simultaneously (from the earth frame) depart earth in different directions. Following a predetermined flight plan, each sister accelerates identically to ... 1answer 205 views ### Two identical rockets, time dilation, and possible weirdness [duplicate] Possible Duplicate: Why isn't the symmetric twin paradox a paradox? Suppose there are two identical rockets, each carrying one of two identical clocks and one of two identical ... 3answers 173 views ### Twin paradox - observers counter orbiting Earth Imagine three observers - one (A) stationary on the surface of Earth (latitude 0 deg) and two others orbiting the planet in the same circular equatorial orbit just in the opposite direction. When the ... 2answers 586 views ### How can time be relative? I don't understand how time can be relative to different observers, and I think my confusion is around how I understand what time is. I have always been told (and thought) that time is basically a ... 1answer 133 views ### Approximate Time Dilation at Rocket Speeds How do you calculate the time dilation effect experienced by a traveler traveling at a relatively low speed? Specifically, how much time dilation would a traveller moving at $v=0.0007 c$ (speed of ... 2answers 337 views ### Time dilation as an observer in special relativity I've been having a discussion regarding time dilation relating to special relativity and how it should be observed from the FoR (Frame of Reference) of "the person moving" : I assert- If we have a ... 1answer 283 views ### Einstein Relative Motion and Time Order of Events According to Einstein, do observers in relative motion agree on the time order of all events? I don't think they would agree on the timing of events, but I am having trouble figuring out why they ... 3answers 136 views ### How can we know, today, that there's something from 100 light-years from here? In my understanding, to take a picture of something that is 100 light-years from here, our "camera" would have to travel 100 years at light speed, take the picture, send to us, and 100 years later we ... 4answers 817 views ### That 10km/day error predicted if GPS satellite clocks not corrected for relativity Some authorities have stated publicly and without explanation that if the theories of Special and General Relativity were not taken into account in the design of the GPS (by building the satellite ... 13answers 1k views ### What are the mechanics by which Time Dilation and Length Contraction occur? What are the mechanics of time dilation and length contraction? Going beyond the mathematical equations involving light and the "speed limit of the universe", what is observed is merely a phenomenon ... 2answers 359 views ### Why is time special? In Special Relativity, the spacetime interval between two events is $s^2 = -(c{\Delta}t)^2+({\Delta}x)^2+({\Delta}y)^2+({\Delta}z)^2$ giving the Minkowski metric \$\eta_{\mu\nu}=\text{diag}(-1, 1, 1, ... 4answers 722 views ### Relativity in Real Life I have no formal training at all in relativity. But I was just wondering if our perception of time is altered by different events. Here are some examples I had in mind: If I touch a stove for 10 ... 3answers 549 views ### Time in special relativity and quantum mechanics The time is treated differently in special relativity and quantum mechanics. What is the exact difference and why relativistic quantum mechanics (Dirac equation etc.) works? 2answers 202 views ### Special Relativity and time [duplicate] Possible Duplicate: Question about Time Dilation.. I have a question about special relativity which was bothering me for a while now. I know that as one approaches the speed of light, time ... 4answers 2k views ### How is the classical twin paradox resolved? I read a lot about the classical twin paradox recently. What confuses me is that some authors claim that it can be resolved within SRT, others say that you need GRT. Now, what is true (and why)? 2answers 554 views ### Symmetrical twin paradox Take the following gedankenexperiment in which two astronauts meet each other again and again in a perfectly symmetrical setting - a hyperspherical (3-manifold) universe in which the 3 dimensions are ...	{"extraction_info": {"found_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": 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": 0.944191575050354, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/184884/what-is-the-name-of-this-curve-figure-inside	# What is the name of this curve (figure inside)? A can be written as... $y = a$ B can be written as... $y = bx + a$ C can be written as... $y = cx^2 + bx + a$ 1) How can I write D? I was looking at implementing bezier curves into some code but is there something similar which could be computed much quicker? 2) C is a quadratic... what is the name for D? Edit: Perhaps if I explain what I'm doing.... I am trying to build a decent servo controller. I have a servo which will move from servo position 30 to 0. I want it to move in a more elegant way than B (above)... ideally like in the picture. I will come up with a way of defining the movement between 2 positions when I send a command to the controller. Perhaps this is a quadratic equation which could be calculated easily for each x (time) between y (servo angles)... this would mean I would only need to pass a, b and c to the controller along with the time (length) and both positions. I figured it would be better to have it able to be more irregular... but still super quick to compute. I considered bezier curves as a definition for the line, but they seem overcomplicated and slow to compute... if there anything else I could use? - It's hard to tell. There are a lot of curves that locally look like that. If you want it to look like a rotated parabola, for example, that's pretty easy to do. – Potato Aug 21 '12 at 4:14 If you fix the two endpoints of curve $D$, and you know the slope of the curve at those endpoints, you can build the Hermite interpolating polynomial that more or less looks like curve $D$. – J. M. Aug 21 '12 at 4:14 I googled hermite interpolation. looks promising. seems fairly heavy on computation, though. – Beakie Aug 21 '12 at 4:41 @J.M. i think u answered my question.... please add it as answer and i will mark it as such – Beakie Aug 21 '12 at 4:44 For a rotated parabola, $y$ would not be a function of $x$ (globally). – Robert Israel Aug 21 '12 at 4:45 show 1 more comment ## 2 Answers One can find a cubic, $y=ax^3+bx^2+cx+d$, looking like $D$. EDIT: If you want a curve that looks like the blue curve in the edited version of the question, you can pick a few points you want the curve to go through and then join them up with cubic splines. There is a lot of info about these on the web and in the more applied Linear Algebra textbooks and elsewhere, and while some of the formulas may look complicated, there's really nothing there that a computer can't handle quickly and efficiently. - please see above edit – Beakie Aug 21 '12 at 4:30 Well, part of it will look like $D$. Then somewhere outside the picture, the graph will turn down. – Robert Israel Aug 21 '12 at 4:47 To me, $D$ looks more like a branch of a hyperbola, which could have an equation $y = c x + \sqrt{a^2 x^2 + b}$ with $b > 0$, $\|c\| < a$. This is asymptotic to $y = (a+c) x$ as $x \to +\infty$ and $y = (-a+c) x$ as $x \to -\infty$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "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": 0.9651498794555664, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/35148/equivalences-for-mci-doses-and-rads/35157	# Equivalences for mCi doses and Rads? What would be the magnitude of radiation received for a 50mCi dose of radioiodine I-31 as compared to say RADS? I read that it might be on the order of about 35 RADS. Does this make sense? - ## 2 Answers There is no direct conversion for what you are asking. Curies are a measure of activity equal to $3.7 \cdot 10^7$ decays per second. Rad are a measure of absorbed dose equal to 1 Joule per kilogram. In order to determine how much radiation is received, you need to know things like the amount of time of exposure, distance between source and target, sheilding if applicable (and intervening material regardless), solid angle of target, and mass of the target. I assume that you meant to write $^{131}I$ and I expect that you are implying that the source was ingested and so all of the radiation is absorbed. $^{131}I$ always decays by $\beta^-$ emission and about 90% of these decays have an average energy of 191 keV, 7% have an average energy of 96 keV, and 2% have an average energy of 69 keV. Thus, on average, the $\beta^-$ have an energy of about 180 keV which is about $2.9 \cdot 10^{-14} J$. A 50 mCi source gives off $1.85\cdot 10^9 \beta^-$ particles per second so, in this case, we are looking at about 0.003 Joules per minute of exposure. If you divide this by the weight of the patient, it will tell you the absorbed dose in rad. Now, $^{131}I$ has a biological half-life of about 100 days. Biological half-life works about the same as radioactive half-life, there is an exponential decay of the effects. In other words, about 50 days after ingestion, you will be recieving only half of the exposure you would receive if the sample were, say, sitting on a table. Keep in mind, however, that the radioactive half-life of $^{131}I$ is only 8 days. You can combine radioactive half-life and biological half-life to determine the expected absorbed dose over time but I will not do it here. One other thing to consider, however, is that the exposure to radiation in this case is not uniform because Iodine migrates to specific places in the body so that the residence time depends on which organ you are considering. Additionally, for health purposes we typically use dose equivalent rather than absorbed dose which is measured in Sieverts rather than rads. Dose equivalent is determined by multiplying the absorbed dose by a quality factor Q as recommended by the International Commission on Radiological Protection (ICRP). This accounts for the difference resulting from the energy of the incident particles. To make matters even more confusing, there is often also an "equivalent dose" (not the same as dose equivalent) that includes an additional weighting factor to account for the different nature of radiation from varous kinds of particles. On top of that, there is an "effective dose" that is different still and includes an additional weighting factor to account for the fact that a dose to the gonads is worse than one to bone. In conclussion, yes, the suggested result you heard of 35 rad as an absorbed dose received from $^{131}I$ sounds reasonable. However, simplifying radiation exposure is frought with difficulty, every exposure situation is unique and needs individual and competant medical attention. Many of us in the industry wish the situation were simpler because the current situation is a public relations nightmare when something like Fukushima happens. Unfortunately, when you are dealing with the effect a stochastic process on biological systems, it just isn't that simple. - Thank you so much. I am asking as a patient (thyroid) and am trying to determine the magnitude of the ingested dose I received. One of the personnel told me that I probably absorbed the same amount of radiation as you would normally receive in about a year. This seems like it must be incorrect, and that it must be much higher. – S. Danielson Aug 29 '12 at 18:07 I'm not sure I would say it is "much" higher. 35 rad would contribute on the order of magnitude of 300 mSv which is certainly quite high but that takes a lot of assumptions into account. If you are very concerned, do a bit of research; I work in the nuke power industry and am not a health physicist at that so my answer is more hueristic than quantitative result. – AdamRedwine Aug 29 '12 at 18:20 1 Are you sure, for example, that the activity was milicuries and not microcuries? – AdamRedwine Aug 29 '12 at 18:22 That's interesting. I have tried to look up the information but as there is no direct conversion as you said, its difficult for a layman to get an idea of magnitude compared to things we have a sense of already, like X rays. I'm guessing you answered as you did because the dose to the thyroid in RAI therapy is much higher than to the rest of the body, even though that would be still higher than what would be normally absorbed in a year? I wondered about this too because there are so many precautions advised for the people around you when you have the dose, yet they insist that the treatment is – S. Danielson Aug 29 '12 at 19:00 1 @S.Danielson Hi, and welcome to Physics Stack Exchange! As dmckee told you earlier, please do not respond to other people's answers by posting another answer of your own. Use the "add comment" link which appears just below this message at the left. – David Zaslavsky♦ Aug 29 '12 at 19:54 show 4 more comments Hi: I apologize but I am not seeing the "add comment" option to the left. Could you give me some guidance? I wanted to tell Adam Redwine that a radiation officer from the UK just told me in an email that the dose to the bladder wall from 50 mCi is probably about 400 times what would be received from a year's background radiation dose. Thanks and I apologize again. - Hello Daniel, You'd be able to earn the privilege once you attain 50 reputation score..! – Ϛѓăʑɏ βµԂԃϔ Sep 22 '12 at 14:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 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": 0.9670834541320801, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/112294/arbitrary-union-and-intersection-of-closed-and-open-sets	# Arbitrary Union and Intersection of Closed and Open Sets I have four quick questions and have listed them below. I am seeking for corroboration of the first three and a bit of insight on the fourth, as I have hit a solid brick wall. Definition. A set $A\subseteq\mathbb{R}$ is called an $F_\sigma$ set if it can be written as the countable union of closed sets. A set $B\subseteq\mathbb{R}$ is called a $G_\delta$ set if it can be written as the countable intersection of open sets. 1. Argue that a set $A$ is a $G_\delta$ set if and only if its complement is an $F_\sigma$ set. Proof. Let $A$ be a $G_\delta$ set. Then$$A=\bigcap_{n=1}K_n,$$where each $K_n$ is an open set. By De Morgan's law, it follows that$$A^c=\left(\bigcap_{n=1}K_n\right)^c=\bigcup_{n=1}K_n^c,$$which is an $F_\sigma$ set, since it is a countable union of closed sets. The converse statement can be proven in a similar fashion. $\square$ 2. Show that a closed interval $[a,b]$ is a $G_\delta$ set. Proof. Take$$\bigcap_{n=1}^\infty\left(a-\frac{1}{n},b+\frac{1}{n}\right)=[a,b].$$Therefore, $[a,b]$ is a $G_\delta$ set. $\square$ 3. Show that the half-open interval $(a,b]$ is both a $G_\delta$ and an $F_\sigma$ set. Proof. Take$$\bigcap_{n=1}^\infty\left(a,b+\frac{1}{n}\right)=(a,b],$$and$$\bigcup_{n=4}^\infty\left[a-\frac{a-b}{n},b\right]=(a,b].$$Therefore, $(a,b]$ is both a $G_\delta$ and an $F_\sigma$ set. $\square$ 4. Show that $\mathbb{Q}$ is an $F_\sigma$ set, and the set of irrationals $\mathbb{I}$ forms a $G_\delta$ set. I am oblivious as to how to tackle this problem. I know that I need to show that $\mathbb{Q}$ can be written as the countable union of closed sets, and that upon doing this, I can apply De Morgan's law to it to prove the second statement. However, I do not know how to begin. Do you guys have any ideas? Thanks in advance! Edit 1: I followed Robert's advice, and this is what I managed to weave: Proof. Since $\mathbb{Q}$ is countable, we can write it as $\mathbb{Q}=\{r_1,r_2,r_3,\cdots\}$, where each $r_n$ is a unique rational number. It follows that we can express $\mathbb{Q}$ as the union of singletons $[r_n]$, which are closed sets, as follows:$$\mathbb{Q}=\bigcup_{n=1}^\infty[r_n].$$Therefore, $\mathbb{Q}$ is an $F_\sigma$ set, since it is a countable union of closed sets. Furthermore, it naturally follows that$$\mathbb{I}=\mathbb{Q}^c=\left(\bigcup_{n=1}^\infty[r_n]\right)^c=\bigcap_{n=1}^\infty[r_n]^c=\bigcap_{n=1}^\infty(-\infty,r_n)\cup(r_n,\infty)$$is a $G_\delta$ set, since it is a countable intersection of open sets. $\square$ Is this sound? - 6 Hint: a one-point set is closed. – Robert Israel Feb 23 '12 at 4:00 I think in your definition of $G_{\delta}$ set you meant "countable union of open sets", since that's what you're using later on. – Patrick Da Silva Feb 23 '12 at 4:18 another hint: $\mathbb Q$ is countable. – azarel Feb 23 '12 at 4:19 A further hint: how many rational numbers are there? (And since no one’s actually said so: your first three are fine.) – Brian M. Scott Feb 23 '12 at 4:19 I just thought we all noticed the same thing and the answer would've gotten out at some point so I answered it. – Patrick Da Silva Feb 23 '12 at 4:21 show 7 more comments ## 1 Answer You just need to notice that $\mathbb Q$ is countable and that singletons are closed sets, hence $$\mathbb Q = \bigcup_{x \in \mathbb Q} \{ x \}.$$ Hope that helps, -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 41, "mathjax_display_tex": 8, "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": 0.9703747630119324, "perplexity_flag": "head"}
http://mathoverflow.net/questions/5436?sort=newest	## Are mapping spaces paracompact? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let X be a (finite dimensional) manifold. Consider smooth mapping space $$PX = C^\infty(I, X)$$ where I = [0,1] is the closed interval. Is this space paracompact? What if we fix a point x in X and consider the pointed path space, is this space paracompact? - ## 1 Answer What topology do you want on $C^\infty(I,X)$? One natural topology, the one of uniform convergence of all derivatives, is metrizable, so paracompact. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.8337647914886475, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2010/12/24/permutations-and-polytabloids/?like=1&source=post_flair&_wpnonce=d0199d491e	# The Unapologetic Mathematician ## Permutations and Polytabloids We’ve defined a bunch of objects related to polytabloids. Let’s see how they relate to permutations. First of all, I say that $\displaystyle R_{\pi t}=\pi R_t\pi^{-1}$ Indeed, what does it mean to say that $\sigma\in R_{\pi t}$? It means that $\sigma$ preserves the rows of the tableau $\pi t$. And therefore it acts trivially on the tabloid $\{\pi t\}$. That is: $\sigma\{\pi t\}=\{\pi t\}$. But of course we know that $\{\pi t\}=\pi\{t\}$, and thus we rewrite $\sigma\pi\{t\}=\pi\{t\}$, or equivalently $\pi^{-1}\sigma\pi\{t\}=\{t\}$. This means that $\pi^{-1}\sigma\pi\in R_t$, and thus $\sigma\in\pi R_t\pi^{-1}$, as asserted. Similarly, we can show that $C_{\pi t}=\pi C_t\pi^{-1}$. This is slightly more complicated, since the action of the column-stabilizer on a Young tabloid isn’t as straightforward as the action of the row-stabilizer. But for the moment we can imagine a column-oriented analogue of Young tabloids that lets the same proof go through. From here it should be clear that $\kappa_{\pi t}=\pi\kappa_t\pi^{-1}$. Finally, I say that the polytabloid $e_{\pi t}$ is the same as the polytabloid $\pi e_t$. Indeed, we compute $\displaystyle e_{\pi t}=\kappa_{\pi t}\{\pi t\}=\pi\kappa_t\pi^{-1}\pi\{t\}=\pi\kappa_t\{t\}=\pi e_t$ ## 2 Comments » 1. [...] polytabloids is a submodule, we must see that it’s invariant under the action of . We can use our relations to check this. Indeed, if is a polytabloid, then is another polytabloid, so the subspace spanned [...] Pingback by \| December 27, 2010 \| Reply 2. [...] use our relations to [...] Pingback by \| December 28, 2010 \| Reply « 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": 16, "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": 0.9008460640907288, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/18553/proper-distance-and-embedding-diagrams	Proper distance and embedding diagrams? I'm trying to understand proper distance equation in Schwarzschild spacetime. $d\sigma=\frac{dr}{\left(1-\frac{R_{S}}{r}\right)^{1/2}}$. I'm sure I'm missing something really obvious here, but how do I use this to find the coordinate distance $r$ for a particular proper distance $\sigma$ . For example, if I found the proper circumference of circle going round the Sun that roughly coincides with the Earth's orbit. Then I move radially inwards one proper mile, how would I then find the circumference of the circle I now find myself on. This example is also in the context of trying to understand the spacing of concentric circles in embedding diagrams. Thank you - 1 Answer You have the Schwarzschild metric $ds^2=(1-R_s/r)c^2dt^2-(1-R_s/r)^{-1}dr^2-r^2(d\theta^2+sin^2\theta d\phi^2)$ For an equatorial orbit, put $\theta=\pi/2; d\theta=0$. The proper distance between two events is defined as the integral of $ds$ along a spacelike path between them. I'm guessing the two events we're interested in are the (identical) start and end point of an elliptical path, where the path is traversed in zero coordinate time (so $dt=0$). The only things that vary on the path are $r$ and $\phi$. This path will be a function like $r=a(1-e^2)/(1-e.cos\phi)$ where a and e are a couple of fixed parameters. Since spacelike separations are negative in this signature, we apply an extra minus sign and get $d\sigma = \sqrt{(1-R_s/r)^{-1}dr^2-r^2d\phi^2}$ Using the formula for the ellipse, you can get $dr$ in terms of $d\phi$ and take the square root, leaving just a $d\phi$ on the RHS. Integrating from 0 to 2$\pi$ should then give you your proper distance. When you want to compare this with the value "one mile in", you need to decide what your measure of the radius of the elliptical path is (maybe average r), and adjust the parameters accordingly. Edit: I just noticed you're using circular orbits. That simplifies it a bit ! In fact, doesn't it make it trivial, since $dr=0$ on your orbit ? - – Peter4075 Dec 21 '11 at 9:31 – Peter4075 Dec 21 '11 at 9:37 thought I should point out that at my level nothing is trivial. – Peter4075 Dec 21 '11 at 9:45 Ah, OK, looking at that reference, I think the point he was trying to make was that if, instead of the coordinate radius r, you use the proper radius defined as $\sigma$ in the formula in your question, then the relation between circumference and radius is no longer $C=2\pi. radius$. Proper radius is the distance between two events for which $t, \theta and \phi$ are the same. – twistor59 Dec 21 '11 at 10:31 Thanks. I quite like that picture of the two concentric circles because I can then easily see how it builds into the kind of upside down witch's hat shape of an embedding diagram. But how does he calculate the circumference of the inner circle to be 0.99999999x2pi miles less than the outer circle? Does it involve some nasty integral of the proper distance equation I gave in my question? My definition of "nasty integral" is pretty all encompassing. – Peter4075 Dec 21 '11 at 11:00 show 2 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": 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": 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": 0.931063711643219, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2009/01/30/the-trace-of-a-linear-transformation/?like=1&source=post_flair&_wpnonce=fe7f791e1e	The Unapologetic Mathematician The Trace of a Linear Transformation Again, let’s take a linear endomorphism $T$ on a vector space $V$ of finite dimension $d$. We know that its characteristic polynomial can be defined without reference to a basis of $V$, and so each of the coefficients of $V$ is independent of any choice of basis. The leading coefficient is always ${1}$, so that’s not very interesting. The constant term is the determinant, which we’d known from other considerations before. There’s one more coefficient we’re interested in, partly for the interesting properties we’ll explore, and partly for its ease of computation. This is the coefficient of $\lambda^{d-1}$. So, let’s go back to our formula for the characteristic polynomial: $\displaystyle\sum\limits_{\pi\in S_d}\mathrm{sgn}(\pi)\prod\limits_{k=1}^d(\lambda\delta_k^{\pi(k)}-t_k^{\pi(k)})$ Which terms can involve $\lambda^{d-1}$. Well, we can get one factor of $\lambda$ every time we have $k=\pi(k)$, and so we need this to happen at least $d-1$ times to get $d-1$ factors of $\lambda$. But if the permutation $\pi$ sends all but one index back to itself, then the last index must also be fixed, since there’s nowhere else for it to go! So we only have to look at the term corresponding to the identity permutation. This will simplify our lives immensely. Now we’re considering the product $\displaystyle\prod\limits_{k=1}^d(\lambda-t_k^k)$ When we multiply this out, we make $d$ choices. At each step we can either take the $\lambda$, or we can take the $t_k^k$. We’re interested in the terms where we take the $\lambda$ $d-1$ times. There are $d$ ways of making this choice, corresponding to which one of the indices we don’t take the $\lambda$. Incidentally, we could also think of this in terms of combinations, as $d=\binom{d}{d-1}$. Anyhow, for each choice of one index to use the matrix entry instead of the variable, we’ll have a term $-t_k^k\lambda^{d-1}$. We add all of these up, summing over $k$ — as our notation suggests we should! And now we have the second coefficient of the characteristic polynomial. We drop the negative sign and call this the “trace” of $T$: $\displaystyle\mathrm{Tr}(T)=\sum\limits_{k=1}^dt_k^k=t_k^k$ where in the last formula we’re using the summation convention again. Incidentally, “trace” should be read as referring to a telltale sign that $T$ has left behind, like a hunted animal’s.. um.. “leavings”. Anyhow, we can now write out a few of the terms in the characteristic polynomial: $\det(\lambda1_V-T)=\lambda^d-\mathrm{Tr}(T)\lambda^{d-1}+...+(-1)^d\det(T)$ Like this: Posted by John Armstrong \| Algebra, Linear Algebra 5 Comments » 1. [...] we can define to be the trace of this matrix, and to be its determinant. If , we’ve got an eigenpair. Possibly related [...] Pingback by \| March 13, 2009 \| Reply 2. [...] is the trace of an endomorphism. Given a matrix, it’s the sum of the diagonal entries. Since the [...] Pingback by \| October 13, 2010 \| Reply 3. [...] is, the character is “the trace of the representation”. But why this is interesting is almost completely opaque at this [...] Pingback by \| October 14, 2010 \| Reply 4. [...] with dimension and start with . Inside this, we consider the subalgebra of endomorphisms whose trace is zero, which we write and call the “special linear Lie algebra”. This is a subspace, [...] Pingback by \| August 8, 2012 \| Reply 5. [...] for all in the cases of and the special linear Lie algebra — the latter because the trace is invariant under a change of [...] Pingback by \| August 18, 2012 \| Reply « 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": 30, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.917596161365509, "perplexity_flag": "head"}
http://gowers.wordpress.com/2009/07/04/a-mathematician-watches-tennis/?like=1&_wpnonce=9e687b4e3a	# Gowers's Weblog Mathematics related discussions ## A mathematician watches tennis Because the French Open and Wimbledon have been available on the BBC website I’ve been watching a lot of tennis recently. And as I do so I can’t help thinking about whether mathematics has anything to say about the tactics that the players should adopt in various situations. And the more I think (or rather, idly muse) about this question, the more it becomes clear that the modelling problem it presents is a pretty hard one. Most of this post will be a discussion of questions rather than a serious attempt to supply answers. Just to make the discussion more concrete, here are a couple of more specific questions, which I’ll come back to later. The first one is fairly simple. 1. It is generally held to be a slight advantage to serve first in a set. The reasoning goes like this. Let’s suppose (for simplicity) that the game goes with serve till 4-4. If you are serving first, then you will be in a very dangerous position if your serve is broken, since you will then have to break back immediately or lose the set. However, at least you won’t have lost. By contrast, if you are serving second and the score is 4-5, then you can’t afford to be broken — if you are broken then you lose the set and do not get even a small chance to redeem yourself. And if you have just broken your opponent so that it’s 5-4, then you still have the task of serving for the set. However, a simple model would suggest that this reasoning is flawed. If you have a probability $p$ of winning a game on your serve and a probability $q$ of winning it on your opponent’s serve, then over the next two games you have a probability $pq$ of winning both, $p(1-q)+q(1-p)$ of winning one, and $(1-p)(1-q)$ of losing both, and the order the games are played in makes no difference. I’ll come back to the (not particularly interesting) solution of this “paradox” in a moment, but before that let me turn to a second and rather general piece of conventional tennis wisdom. 2. It seems pretty obvious that different circumstances call for different styles of play. For example, if you are a break up and 40-0 on your serve, then it feels as though you can afford to take a few risks, whereas if you are serving a second serve at match point down, then you should play it a little bit safe, since a double fault would cost you the match (and similarly, if you are going for a winner, you shouldn’t aim too close to the line, and so on). But a very simple argument suggests that a different piece of tennis wisdom, the one that says, “Forget where you are in the game and just take things point by point” must be correct. After all, the best strategy cannot be anything other than to maximize the probability that you win the point, so if it ever makes sense to serve a reasonably powerful second serve and risk serving a double fault, then it makes sense even if you are match point down. I have more mathematics-of-tennis questions that I want to mention, but first let’s dispose of these two, by thinking about what the model is that underlies the second argument in each case and what important factors it fails to take into account. It is a model much loved by setters of questions in elementary probability: on each point, if player A is serving then there is a probability $p$ that A will win and a probability $1-p$ that B will win, whereas if player B is serving then these probabilities are $q$ and $1-q$. Moreover, the outcomes of all the points are independent. If that were a realistic model, then there would indeed be no advantage in serving first, and one should play the same tactics on every point. However, there are two (at least) very important things that it does not take into account. One is that the probabilities change according to how nervous a player is (and also, it has to be said, appear to ebb and flow during a match for no apparent reason). So for example it might be a good strategy to serve a rather lame second serve when you are match point down, since if you do a normal one then nerves are more likely to cause you to mess it up, and your opponent’s nerves may cause them to mess up their return even if under normal circumstances they could hit a winner off it. Similarly, the pressure of serving at 4-5 down is greater than the pressure of serving at 5-4 up, so perhaps there really is some justification for the belief that serving first is an advantage. Unfortunately, there doesn’t seem to be an easy way of realistically incorporating players’ psychological states into a model, so let’s set that question aside (unless someone feels like having a go at it). But there is a second consideration that does lend itself to a mathematical analysis, and that is the fact that if you play precisely the same tactic every point, then you will be handing an advantage to your opponent. For instance, you may win the point with probability $p$ if you serve out wide, and with probability $q<p$ if you serve near the T, but if you serve out wide every time, then your opponent can take advantage of this by standing wider, at which point the probability of winning the point will no longer be $p$ but something lower that may well be less than $q$. I think (without having checked) that one can deal with the second consideration by defining a strategy to be something that tells you not what shot to play at each stage, but what probability distribution to choose your shot from: e.g., it might tell you to serve out wide 50% of the time, for the T 30% of the time, and straight at the body 20% of the time. And the optimal strategy would be the probability distribution that maximized your overall probability of winning the point. That sounds to me like fairly standard game theory. A third factor I won't discuss here is conservation of energy — not the basic law of physics, but the more homely physical principle that says that if you run around too much then you will get tired later on. That might tell you that if you can just get a shot back by running all the way across the court, but can't do anything with it when you do except gently pat it over the net straight at your opponent, then it might be better not to bother, even if it increases your chances of winning the point from 0 to 1/100. There's one more question I want to mention, and it's the main one I wondered about. It is how one should model the tactical decisions that are made during a rally. As a warm-up, here is a special case. It's your first serve. Is it better to go all out for an ace, or should you try for a serve that has a slightly higher chance of going in and still leaves you in a very strong position? Under suitable assumptions, that question is pretty easy once one puts some numbers in. Let's suppose that if you go for an ace, then you'll serve one with probability $p$ and you'll serve a fault with probability $1-p$. (That ignores the possibility that your opponent might occasionally be able to get your serve back, but let us indeed ignore that.) And let's suppose that if you go for a serve that is merely very hard to return, then you'll succeed with probability $q$ and serve a fault with probability $1-q$, and furthermore that if you succeed then your opponent will give you the opportunity to smash, and that your smash will win the point with probability $r$ (which is very close to $1$) and will go out with probability $1-r$. Then if you go for an ace, you win with probability $p$, and if you go for a win on your second shot then you win with probability $qr$. After that trivial calculation, the question that remains is whether $p$ is likely to be higher or lower than $qr$, and rather boringly I don't have any idea. By roughly how much do professional tennis players increase their first-serve percentage if they make their serves very slightly slower, or very slightly less close to the lines? And how much can they afford to do that before their serves become easy to return? Obviously it depends a lot on who is playing, so a different question might be whether it is in fact the case that there are professional tennis players who are trying for aces when it is clearly not the best strategy. (Of course, the mixing-it-up principle could complicate this question and lead to the conclusion that one should sometimes try for an ace and sometimes go for a good first serve that leaves one in a strong position in the ensuing rally.) Another simple question: is it conceivable that it might sometimes be good tactics to attempt first-serve-type serves all the time? For instance, if you are Ivo Karlovich on a good day and your first-serve percentage is 75% (as happened at times during Wimbledon), and if you are winning 95% of the points when your first serve goes in, then your probability of losing the point if you go for first serves every time is $0.95\times 0.75+0.25(0.95\times 0.75)$, which is $0.95\times(0.75\times 1.25)$, which is 19/20 times 15/16, which is certainly at least 7/8. That gives your opponent almost no chance of winning a game. Maybe I should change that question to a related one: why is it that nobody ever plays that tactic, except perhaps on the occasional point? This is a fairly simple question. Let's suppose that with my best strategy, if I have just one serve left then I have a probability $r$ of winning the point. And now suppose that it is in fact my first serve. Let's suppose that I can choose the probability $p$ that my first serve goes in, and that there is some function $q(p)$, which tells me the probability that I will win the point if I do a probability-$p$ serve. So for example, if I go for an ace, then $p$ may be fairly low, but $q(p)=1$, whereas if I go for a safer first serve then $p$ is higher and $q(p)$ correspondingly lower. If I have just one serve left, then I have a very simple optimization problem: I just need to maximize $pq(p)$, and we are assuming that its maximal value is $r$. But if I have two serves left, then the quantity I am trying to maximize is $pq(p)+(1-p)r$. What difference does that make? Well, we are assuming that $q(p)$ is a decreasing function of $p$. If $p$ is chosen to maximize $pq(p)$, then, at least if $q$ is differentiable, decreasing $p$ by a small $\delta$ will decrease $pq(p)$ by $o(\delta)$ (since the derivative of $q$ at $p$ is zero) but will also increase $(1-p)r$ by $\delta r$. Therefore, the maximum of $pq(p)+(1-p)r$ will occur at a lower value of $p$, which corresponds to a riskier serve. (I've made various assumptions there that I can't be bothered to make explicit.) Finally, here is the general case of the question that intrigued me most. It is to try to find a model that would justify the following piece of reasonable-sounding tennis advice: that if you are in the middle of trading groundstrokes with your opponent, then you should not go directly for a winner, but should instead try to play yourself into a stronger and stronger position until you can hit a winner with less risk. That advice may not be universally applicable, but it does appear to be the case that some players sometimes go for winners when they would have been better advised to be patient. (Equally, if you are playing Roger Federer, you may realize that you have no hope of winning a long rally so you are more or less forced to go for a risky shot that will be a winner if it comes off.) How does one even think about such questions mathematically? I'm more concerned about that than about the actual answers that a model might yield. One might try something like this. In any given situation, you have a range of shots you can attempt. They have various probabilities of success, and if they do succeed then your opponent has a range of possibilities that depends on the shot you do. It is natural to simplify the model as follows. In any given situation, you have a function $q(p)$ like the one discussed earlier. That is, you can choose a shot that will go in with probability $p$, and if it goes in you will win the rally with probability $q(p)$. But hang on, how do we have any idea what $q(p)$ is? Well, by hitting your probability-$p$ shot you give your opponent a choice similar to yours: a function $r$ that says "If I hit a probability-$s$ shot then I will win the rally with probability $r(s)$." But things are more complicated, because $r$ depends not just on $s$ but also on how risky your shot was — that is, on $p$. In other words, your aim is clearly to maximize $pq(p)$, but this can be broken down further as a wish to maximize $p(1-\max sr(p,s))$. We can take this further. Let $p_1,p_2,p_3\dots$ measure the riskiness of the shots and let $r(p_1,\dots,p_k)$ denote my probability of winning if I chose to play a $p_1$-shot, you chose to play a $p_2$-shot, and so on. (This is a slight change because it focuses on my probability of winning and saves me having to write "$1-$".) Then it looks as though my best strategy is to maximize $p_1r(p_1)$, which is the same as maximizing $p_1(1-\max_{p_2} p_2(1-r(p_1,p_2))=p_1\min_{p_2}(q_2+p_2r(p_1,p_2))$, where $q_2=1-p_2$, which is the same as maximizing $p_1\min_{p_2}(q_2+p_2\max_{p_3}r(p_1,p_2,p_3))$, and so on. Let me just write out the $k$th version of this expression in full gory detail. My probability of winning appears to be (I haven't checked this carefully) $\max_{p_1}p_1\min_{p_2}(q_2+p_2\max_{p_3}p_3\dots\min_k(q_k+p_kr(p_1,\dots,p_k)))$ if $k$ is even and a similar expression ending with $\max_{p_k}p_kr(p_1,\dots,p_k)$ if $k$ is odd. But $k$ isn't bounded, so it seems as though we need to take some kind of limit as $k$ tends to infinity. Does anyone have any idea what sort of thing that limit is? It makes my head spin like the ball on a heavily sliced second serve. And that is just the question of modelling the situation in the first place: actually solving the optimization problem given the function $r$ looks pretty unpleasant. So are there simplifying assumptions that would at least allow one to justify some of the advice given to players about when to attack, when to go for winners, and so on? One final remark: I do not for one moment think that anything in this post, even when fully developed, would be of the slightest use to a tennis player. I just thought I'd better make that clear. ### Like this: This entry was posted on July 4, 2009 at 6:24 pm and is filed under Uncategorized. 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. ### 26 Responses to “A mathematician watches tennis” 1. roland Says: July 4, 2009 at 9:05 pm \| Reply Andy Roddick sometimes adopts your proposed tactic of going for first-serve-type serves only. That fits with your calculation as he is one of the best servers (inferior to Karlovic though). But in real life the tactic is dangerous. Players go through ups and downs and if one misses his first service that alone indicates that one might miss the next first-serve-type too. Another reason migth be that the strategy does not work in tie-breaks, where slower but more reliable second serve-type serve are sensible. So for every player it makes sense to get into the rhythm of winning points with slower second serves during the set. 2. roland Says: July 4, 2009 at 9:26 pm \| Reply Andy Roddick sometimes adopts your proposed tactic of going for first-serve-type serves only. That fits with your calculation as he is one of the best servers (inferior to Karlovic though). But in real life the tactic is dangerous. Players go through ups and downs and if one misses his first service that alone indicates that one might miss the next first-serve-type too. Another reason migth be that the strategy does not work in tie-breaks, where slower but more reliable second serve-type serve are sensible. So for every player it makes sense to get into the rhythm of winning points with slower second serves during the set. Forgot to add good post! Looking forward to seeing the next one! 3. Theo Says: July 4, 2009 at 10:26 pm \| Reply I have only epsilon to add to your very nice post. Well, epsilon + epsilon^2. The epsilon^2 part is to point out a sign error (I believe) in the paragraph that starts “If I have just one serve left,”. In particular: increasing p decreases (1-p)r, not increases it, so you want a riskier serve on your first serve, which is how the players actually play, unless I’ve made a sign error. The epsilon part is the following, with the caveat that I am no expert in any part of probability, and also haven’t spent more time than it took to read your post to think about any of your questions. But I believe that there is some work by Josh Tenenbaum (a mathematical cognitive psychologist at MIT) on other parts of probability theory, and in one way the flavor is different. In particular, if memory serves, then Tenenbaum has a model of probality requiring a similar limit, but shows that the contributions of large-k probabilities vanish fairly quickly: you can truncate the sequence and replace the tail with an arbitrary probability as long as your arbitrary number was not 1 or 0. I should condition the above paragraph more strongly. My knowledge of Tenebaum’s work comes from one lunch-time casual conversation with his partner, rather than from reading his papers. Moreover, he was looking at a very different question, and it’s likely that his models are sufficiently different that my connection of the two is purely superficial. • gowers Says: July 5, 2009 at 12:55 am Thanks for pointing out the sign mistake, which I’ve now corrected. I was thinking about my question about large $k$ after posting this post and thought that something along the lines of what you say should be correct: that is, that one should be able to get a good model by taking a large but finite $k$. I think all that would be needed is some assumption like that on any given shot there is at least a 0.0001 probability that it will go out (or if that seems unrealistic, try going for three consecutive shots — if you do a ludicrously safe shot, then your opponent will surely do a shot that leaves you with at least a 1% chance of missing the one after that) so that the probability that a rally lasts for at least $k$ shots tends to zero as $k$ tends to infinity. 4. Mark Bennet Says: July 4, 2009 at 10:28 pm \| Reply Is it possible that first serve = greatest chance of winning point with a defined minimum energy (so there is a constraint on effort, and that if the point is won, it is cheaply won) Second serve = greatest chance of winning point at a higher energy cost. Except that the statistics seem to show that second serves lose more often than first ones. I have had similar thoughts – why not first serves all the time? 5. Rajiv Says: July 4, 2009 at 11:11 pm \| Reply Moreover, if if points are independently decided, there’s a winning streak factor – a player who is able to win a tough point is often seen to steamroll the next point or two or even the entire set. On the other hand, winning a freak point can instill a false sense of bravado and the player could make the silliest of unforced error in the next point. In short, it may be useful to learn models that don’t treat points independently. 6. Harrison Says: July 5, 2009 at 1:26 am \| Reply Have there been statistical analyses done on “streaks” in tennis? I know, for instance, that there don’t appear to be any such things outside of what’s predicted assuming independence. Tennis might be different, of course, but we humans are notoriously bad at seeing patterns where there aren’t any. • Harrison Says: July 5, 2009 at 1:27 am Sorry, my first sentence there should end with “… in basketball or baseball.” • Timothy Chow Says: July 5, 2009 at 4:59 am The fascinating book “Anthology of Statistics in Sports” (Albert, Bennett, Cochran) has an article by Jackson and Mosurski, “Heavy Defeats in Tennis: Psychological Momentum or Random Effect?” that discusses precisely this question. The paper shows that there is some evidence of non-independence in tennis, unlike basketball for instance. 7. roland Says: July 5, 2009 at 2:32 am \| Reply I think your trick to get rid of the 1-’s is flawed. For example take the formula after “which is the same as maximizing ..” (first occurence). We minimize the probability that my opponent get his shot in times the probability that that shot makes me win the point. The outcome should be the must unprobable shot that leaves me next to no chance to get the ball back. • gowers Says: July 5, 2009 at 9:06 am You’re quite right of course. I’ll change it when I get a spare moment. [Now changed -- I still haven't checked it quite as carefully as I should but I think I'm closer this time.] 8. speedy Says: July 5, 2009 at 3:57 am \| Reply I have read this discussion before in John Haigh’s taking chances http://www.amazon.co.uk/Taking-Chances-Probability-John-Haigh/dp/0198526636 9. gowers Says: July 5, 2009 at 9:09 am \| Reply I should have a look at that. Another book that probably discusses it is How to take a penalty: the hidden mathematics of sport, by Rob Eastaway and, again, John Haigh. 10. oz Says: July 5, 2009 at 3:25 pm \| Reply 1. Interestingly, in tie-breakers the serving order is different. While in the ‘regular’ match the serving alternates: ABABAB .. in tie-breaker it is: ABBAABBAA .. Why were the rules chosen to be so? the only difference is that tie-breaker each ‘mini-game’ has p closer to half than in a game (since in a game you have to win 4-points by a 2 point lead, which sort of ‘pushes’ the single-point p away from half). Suppose both players are of the same quality (so p=q) – then in both a ‘standard set’ setting and in tie-breaker both players have prob. 1/2 to win. Still, psychologically, I can’t help but feeling that the tie-breaker system is ‘more fair’ – maybe since in this system each player has advantage in terms of the number of games served so far. 2. Another interesting issue which could be modeled mathematically is when to challenge the calls. Each player has 3 incorrect challenge in each set. Lets say that you estimate a call to be incorrect (against you) to be alpha. Then a naive strategy is to challenge whenever alpha>=r for some threshold r (lets say setting r=90%, so you challenge only if you’re 90% sure that the call is wrong). But surely in the last games you should adjust your calling according to how many challenges you’ve got left, the position in the game etc. (e.g. trailing 0-40 on your opponent’s serve it might be not so productive to challenge even if you’re almost sure you are correct, since you’re probably gonna lose the game anyway) 11. Elsinor Says: July 6, 2009 at 7:16 am \| Reply A similar fun piece of mathematics is applicable to the game of squash. In squash, the winner of a rally is the server for the next point, and only a server can win a point. A game or set is won by the first player to reach 9 points, typically by 2 or more points. However, if the score reaches 8-8 then the receiver can choose to play wither first to 9 points or first to 10 points wins the game. The receiver should choose “10 ponts” if their (fixed) probability of winning any rally is larger than 0.38. 12. Richard Gill Says: July 6, 2009 at 7:39 am \| Reply Gerard Sierksma (operations research, Univ. Groningen) is a sportsman and mathematician who also has a company whose software is used by several big football clubs, the Dutch Olympic committee etc, for tactics and strategy in a number of games. I don’t know if he tackled tennis yet. 13. Richard Gill Says: July 6, 2009 at 7:39 am \| Reply http://www.rug.nl/staff/g.sierksma/index 14. Antonis Says: July 6, 2009 at 1:49 pm \| Reply Jan Vecer of Columbia University has been working in this direction: http://www.columbia.edu/cu/news/04/01/janVecer.html 15. Kevin O'Bryant Says: July 8, 2009 at 9:10 pm \| Reply I once heard Agassi answer the question “how is it that Federer is able to raise his game so much on the big points?” Agassi’s answer was that good players will notice a weakness in the opponent (maybe foot position correlating with type of serve, or standing too far out when receiving a serve), but won’t exploit it immediately. They’ll save the weakness for a crucial moment in the match (when behind a few points in the second set tiebreak against Roddick, for example). Interestingly, this possibility arises only because of the peculiar scoring system of tennis. There wouldn’t be any advantage in soccer or basketball, that I see, to not exploiting any advantage immediately. This isn’t really different from the Allied powers not using information (and allowing ships to sink and men to die) in order to conceal the breaking of the enigma code. That is, use the info often enough to gain advantage, but seldom enough that the opponent can dismiss it as a fluke. • gowers Says: July 8, 2009 at 10:43 pm That’s very interesting, but I’m not sure I agree that the same idea couldn’t be applied to soccer. Suppose, for example, that you are playing a soccer match where your team needs to win and the opposing team needs at least a draw. And just to make the example very extreme, suppose you noticed such a glaring weakness in your opponents’ tactics that you had a guaranteed goal whenever you wanted it, but as soon as you cashed in they would realize their mistake and correct it. Finally, suppose your team was not very good in defence. Then it would make sense not to exploit the loophole but just play conventionally and hope that you were not behind near the end of the game, and then score the guaranteed extra goal right near the end when the opposing team wouldn’t have time to react to it by switching to a more attacking style of play. But like the tennis example, this is exploiting a particular feature of football — that the game ends after 90 minutes. If the aim were to be the first team to get to three goals, or the first to be two goals ahead, with no time limit, then I don’t see any reason not to exploit the advantage immediately. Also, a variant of what Agassi talked about could apply in soccer, which is that if you noticed a weakness in one of the opposing team’s defenders, then you might decide not to exploit it unless there was a good chance of going on to score a goal. That is, the equivalent for soccer of playing a big point in tennis would be something like being near to the other team’s goal with not many of their defenders in the way. • Mark Bennet Says: July 9, 2009 at 8:25 am I once heard Gary Lineker not quite denying that during the first part of a game he would deliberately allow himself to be caught offside in positions which might otherwise be advantageous in order to induce a significant weakness which could later be exploited. I guess the same could be done in tennis too. 16. Can we take games seriously? « Subramanian Ramamoorthy’s Weblog Says: July 12, 2009 at 5:19 pm \| Reply [...] practical problems under discussion. In this context, I was pleasantly surprised and pleased to see this excellent post by the eminent mathematician Tim Gowers about our difficulty in modelling such games [...] 17. Mariano Beguerisse Says: July 20, 2009 at 2:28 pm \| Reply Very nice post! It touches the very bottom of my Maths/Tennis-loving heart!! Recently, as part of my research on network science, I told a collaborator that I wanted to do some work on tennis and he pointed out a paper for me to read. I think that it could be worth sharing here (I’ve just began to read it so I don’t know how good it is): Monte Carlo Tennis Paul K. Newton and Kamran Aslam SIAM Review Volume 48 , Issue 4 (November 2006) Pages: 722 – 742 http://dx.doi.org/10.1137/050640278 Greetings to all, Mariano. 18. Steve Lawford Says: July 24, 2009 at 12:44 pm \| Reply Jan Magnus (Tilburg) has done a lot of interesting work with his co-authors on the statistics of tennis, including the probability of a player winning the match, conditional on being at a particular stage of the match. See Section 14 of http://center.uvt.nl/staff/magnus/subjects.pdf for links to papers. 19. Alyssa Grossen Says: September 10, 2009 at 5:14 am \| Reply This is post is really interesting! For one thing, those who read it will hopefully be encouraged to view sports with a more mathematical eye, such as what probabilities and statisitics are involved with certain strategies. Not only can the numerical ideas in this post be applied to tennis, but they can be applied to other sports as well. Stephen R. Clarke has also done research regarding to mathematics in tennis that I also found interesting. Clarke uses tree diagrams in order to connect the probabilites of a player winning a match to the current score of the match.	{"extraction_info": {"found_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": 86, "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": 0.9649844169616699, "perplexity_flag": "middle"}
http://mathhelpforum.com/number-theory/6552-difference-squares-two-integers.html	# Thread: 1. ## Difference of the squares of two integers Which one of the following numbers cannot be expressed as the difference of the squares of two integers? A:314159265 B:314159266 C:314159267 D:314159268 E:314159269 For this one, I'm just completley clueless. 2. Originally Posted by ceasar_19134 Which one of the following numbers cannot be expressed as the difference of the squares of two integers? A:314159265 B:314159266 C:314159267 D:314159268 E:314159269 For this one, I'm just completley clueless. Numbers of the form 4k,4k+1,4k+3 can always be expressed as a difference of squares. Remove the odd numbers for they can always be expressed as difference of squares are you have, 314159266 314159268 Divide them by 4 and find the one which leaves a remainder of 2. Since both are even they leave either no remiander or 2. Thus, find the number not divisible by 4. The trick is to look at the last two digits, 66----> Not Divisible 68----> Yes Divisible Thus, the answer is, 314159266 3. Originally Posted by ceasar_19134 Which one of the following numbers cannot be expressed as the difference of the squares of two integers? A:314159265 B:314159266 C:314159267 D:314159268 E:314159269 For this one, I'm just completley clueless. The prime factorisation of 314159266 is: 314159266=2x157079633 Now if N=314159266 is the difference of two square integers there exists integers m > n such that: N = m^2-n^2 = (m+n)(m-n). Now if (m-n) is even then both n and m are even or both are odd. But as we can see from the factorisation (m-n)=2, and (m+n)= 157079633, that is one of (m+n) and (m-n) is odd and the other even which is not possible so N is not expressible as the difference of two squares. RonL 4. Originally Posted by ThePerfectHacker Numbers of the form 4k,4k+1,4k+3 can always be expressed as a difference of squares. Remove the odd numbers for they can always be expressed as difference of squares are you have, 314159266 314159268 Divide them by 4 and find the one which leaves a remainder of 2. Since both are even they leave either no remiander or 2. Thus, find the number not divisible by 4. The trick is to look at the last two digits, 66----> Not Divisible 68----> Yes Divisible Thus, the answer is, 314159266 If I may ask, what does k stand for? Edit: I didnt know where the quote button would put it, sorry. 5. Originally Posted by ceasar_19134 If I may ask, what does k stand for? Edit: I didnt know where the quote button would put it, sorry. any integer, as k varies you get all numbers which leave remainders of 0, 1 or 3 from 4k,4k+1,4k+3. RonL 6. So we subsitute the numbers in as k and then what? And why does the remainder have any significance? My teacher does not bother teaching us how to solve any of these problems. 7. Hello, ceasar_19134! Which one of the following numbers cannot be expressed as the difference of the squares of two integers? A: 314159265 B: 314159266 C: 314159267 D: 314159268 E: 314159269 It can be shown that any odd number can be expressed as the difference of two squares. The only suspects are B and D. To have an even difference of squares, . . either (1) both squares are even or (2) both squares are odd. (1) If both $a^2$ and $b^2$ are even, then both $a$ and $b$ are even. . . Then $a = 2m$ and $b = 2n$ for some integers $m$ and $n.$ Hence: . $a^2-b^2\:=\:(2m)^2 - (2n)^2 \:=\:4m^2 - 4n^2\:=\:4(m^2-n^2)$ . . Therefore: $a^2-b^2$ is a multiple of 4. (2) If both $a^2$ and $b^2$ are odd, then both $a$ and $b$ are odd. . . Then $a = 2m+1$ and $b = 2n+1$ for some integers $m$ and $n.$ Hence: . $a^2-b^2\:=\:(2m+1)^2-(2n+1)^2\:=\:4m^2 + 4m + 1 - 4n^2 - 4n - 1 \:=\:4(m^2+m-n^2-n)$ . . Therefore: $a^2-b^2$ is a multiple of 4. We have just established a theorem. . . The difference of two squares is either odd or a multiple of 4. We can now "eyeball" the list and determine the answer. The even choice not divisible by 4 is: $B:\;314159266$ . . [Its last two digits is "66", which is not divisible by 4.] ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Two consecutive squares always differ by an odd number . . because: . $(n+1)^2 - n^2\:=\:2n+1$ ... an odd number Given an odd difference, we can find the two squares. Example: $d = 37$ Subtract 1 and divide by 2: . $\frac{37-1}{2} = 18$ The two squares are: $18^2$ and $19^2$ Given answer-choice $A:\;314,159,265$ . . we can confidently state that: . $314,159,265\;=\;157,079,633^2 - 157,079,632^2$	{"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": 28, "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": 0.9158017039299011, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/233246/periodic-orbits/265599	# Periodic orbits Let $x\in\mathbb R$ be a periodic point with lenght 2 of the recursion $x_{n+1}=f(x_n)$ My book about Dynamic systems says that this recursion has a fixed point now, because a periodic point with length 2 exists. May you could help me to prove this statement? - What does length 2 mean? – Amr Nov 9 '12 at 0:49 Does it mean that there exists n such that f(f(x n))=x n – Amr Nov 9 '12 at 0:50 Yes, exactly. $f(f(x_2))=x_2$ – Montaigne Nov 9 '12 at 0:51 ## 2 Answers In general if $f$ is continuous, and there exists an integer k>1 and a real number $a$ such that $f^k(a)=a$ then there exists a real number $r$ such that $f(r)=r$. Proof: Let $g(x)=f(x)-x$. Since: $g(a)+g(f(a))+g(f^2(a))+...+g(f^{k-1}(a))=f^k(a)-a=0$, therefore either one of the numbers $g(a),g(f(a)),g(f^2(a)),...,g(f^{k-1}(a))$ is zero (in this case we are done) or one of these numbers is positive and another number is negative. Assume WLOG that the second case holds. Let $g(f^i(a))g(f^j(a))<0$ (it means that they have different signs). Since, $g$ has a sign change in the interval $[\min(f^i(a),f^j(a)),\max(f^i(a),f^j(a))]$, thus $g$ has a root. - This works because g(a)=-g(f(a)). Thus if a is different from f(a) from a, we know that g(x) has a sign change on the interval [min{a,f(a)},max{a,f(a)}]. Thus g has a root r in that interval. Therefore g(r)=0, hence f(r)=r – Amr Nov 9 '12 at 0:57 Thank you for your answer. Why do we know that g(a)=-g(f(a)) ? – Montaigne Nov 9 '12 at 1:01 Because a-f(a)=-(f(a)-a) – Amr Nov 9 '12 at 1:02 Thanks, now everything is clear. – Montaigne Nov 9 '12 at 1:02 Also, if you like hitting flies with sledgehammers, this directly follows from Sharkovskii's theorem. - This is a nice theorem. – Amr Dec 29 '12 at 15:47	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "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": 0.8898386359214783, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/166962/need-to-resize-file-system-to-98-of-itself/166983	# need to resize file system to 98% of itself. A computer problem...but really math in disguise. I have a file system that has used 105076 of space. I need to resize the file system so that 105076 is 98% of x. How do I find x? - $\frac{98}{100}\ast x=105076\Rightarrow x=\frac{105076\ast 100}{98}\Rightarrow x=\frac{105076\ast 50}{49}$ – user1729 Jul 5 '12 at 10:36 ## 2 Answers The easiest way to think about problems like these is to first work out what one per cent is: you have $98\%$ so you just need to divide by $98$. Then you want to find out what $100\%$ is, so you multiply by $100$. There are slightly more direct ways to do it when you understand percentages better. But the above way is intuitive and flexible. - Since $105076\;$ is $98\%$ of $x$, then $0.98*x = 105076\;\;$. This means $x = 105076/0.98 = 107220.4\;\;\;$. -	{"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": 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": 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": 0.9469595551490784, "perplexity_flag": "head"}
http://rigtriv.wordpress.com/2012/12/09/the-gauss-map/	# Rigorous Trivialities December 9, 2012 ## The Gauss Map Posted by Charles Siegel under AG From the Beginning, Algebraic Geometry, Complex Analysis, Differential Geometry 1 Comment Posting is slowing down a bit, I’ve got a paper I’m trying to get out, and a couple of projects that are hitting some preliminary results, plus, I’m getting ready for holiday travel, and then two months at Humboldt. Trying out an experiment with more rigid personal scheduling, and hopefully I’ll post more often. Also, I’m reviewing Atiyah-Macdonald, Eisenbud, and Schenck so that perhaps in March I can begin a “Commutative Algebra from the Beginning” series, or perhaps just a series on geometric interpretation of commutative algebra theorems. However, for today, we’re going to take something most of us first saw in differential geometry (I first met this map in do Carmo‘s book) and translate it into algebraic geometry. We will start in the absolute least general way possible, following do Carmo. Let $S\subset \mathbb{R}^3$ be a surface. Then there’s a map $N:S\to \mathbb{S}^2$ to the unit sphere taking each point to its unit normal vector. This is the Gauss map, and it’s a REALLY useful tool, as anyone who has gone through this book can attest. For instance, if you want to define the curvature of a surface in $\mathbb{R}^3$, the Gauss map is essential. For instance, the Gaussian curvature is the determinant of the derivative of $N$, and in fact it would be redundant to go through everything about the Gauss map for surfaces in $\mathbb{R}^3$ because there’s a whole chapter in do Carmo titled “The Geometry of the Gauss Map!” We’re going to generalize and then algebraize. First, let’s just drop the orientation on our surface. To forget that, we can replace the normal vector with the normal line. So then instead of getting a point in $\mathbb{S}^2$ we get a pair of antipodal points, or just a point of $\mathbb{RP}^2$ from our surface. Then we can see that the map is really given by taking $\iota:S\to \mathbb{R}^3$ the inclusion, then we have $d\iota_p:T_p(S)\to T_p(\mathbb{R}^3)$, and then taking the line perpendicular to the image plane. Taking the union of these maps, we just have the map $d\iota:TS\to T\mathbb{R}^3$. Then, using the Riemannian metric on $\mathbb{R}^3$, we can make this a map $N:N_{S/\mathbb{R}^3}\to T\mathbb{R}^3$, and follow it up with the fact that $T\mathbb{R}^3\cong \mathbb{R}^3\times\mathbb{R}^3$, and project down, to get the map $N_{S/\mathbb{R}^3}\to \mathbb{R}^3$, and then we can rewrite it by taking each point to the line in $\mathbb{R}^3$, which gives us a map $S\to \mathbb{RP}^2$, giving the usual Gauss map. So how can we simplify and generalize this? Step 1 is to replace the normal vector with the tangent space, which gives a point in the dual projective space. Then we want to generalize dimension. It’s easy to handle hypersurfaces in $\mathbb{R}^n$, we just get a map to $\mathbb{RP}^{n-1}$ (or rather, to its dual). In general, if we allow non-hypersurfaces, we get maps to Grassmannians, so $M^k\to \mathbb{R}^n$ gives us a Gauss map $N:M\to Gr(k,n)$. Now, we’re going to let the target space vary. We just need a space $Y$ such that $TY\cong Y\times V$ where $V$ is the tangent space at some specific point. If we have trivial tangent bundle, we can identify all the fibers and then the derivative of our map actually gives us a map from the domain to a Grassmannian. What are some spaces that have this property? Lie groups! It’s important that we have Lie groups, not just homogeneous spaces, because of the unique way that we can identify fibers. Now, if we try to algebraize, the first thing we get is a Gauss map for affine varieties $X\to \mathbb{A}^n$. We can even get rid of $\mathbb{A}^n$ and replace it with an algebraic group $G$, but most of the algebraic groups that immediately come to mind are affine, things like $GL(n), SO(n), Sp(n)$ etc, the classical groups. Plus $\mathbb{G}_m$ and $\mathbb{G}_a$ and products of these. But that still only gives us affine varieties, nothing projective or complete. Fortunately, there’s one remaining option that are commonly studied: abelian varieties. Though we can’t do much with rational varieties (as there are no maps from them into abelian varieties other than constant maps), we can get a lot of mileage out of the Gauss map on abelian varieties, as we’ll see in the next post.	{"extraction_info": {"found_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": 29, "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": 0.8864160180091858, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/6216/whats-the-worst-thing-about-the-tomographic-approach-to-qm/6240	# What's the worst thing about the tomographic approach to QM? I saw a paper on arXiv that referenced this approach to an ontology of QM: Phys.Lett. A213 (1996) 1, S. Mancini, V. I. Man'ko, P.Tombesi Symplectic tomography as classical approach to quantum systems http://arxiv.org/abs/quant-ph/9603002 An introduction to the tomographic picture of quantum mechanics http://arxiv.org/abs/0904.4439 - Do you mean "worst thing"? – Mark Eichenlaub Mar 2 '11 at 6:26 Yep. I haven't looked at it, but since I've never heard of it, it's got to have something seriously wrong with it, eh? – Carl Brannen Mar 2 '11 at 6:28 ## 3 Answers The worst thing about it is that it is one of a large number of misguided papers that try to deny that the world is not and cannot be classical. The Wigner function may be calculated from a wave function or a density matrix and is approximately related to the probability distribution but not exactly. The Wigner function is constrained and the precise probabilities that an observable will have a particular value have to be calculated as the expectation value of the corresponding projection operator (a linear operator on a Hilbert space) - and can never be fully encoded in a distribution function that is a function of several commuting variables simply because the real variables, as quantum mechanics and the experiments verifying it guarantees, do not commute. There are several basic facts about quantum mechanics - the probabilistic character and the nonzero commutators of the observables (i.e. the uncertainty principle) are two important examples - and any paper denying those facts is simply wrong and doesn't deserve a further discussion. To emphasize that they won't be convinced by any evidence, they motivate their paper - in the second sentence of this paper - by a "permanent wish to understand quantum mechanics in terms of classical probabilities." Sorry, this is not science. This wish has been ruled out and it can never be "unruled out" again. In science, hypotheses are never "permanent". Hypotheses are only acceptable as long as they haven't been falsified. Falsification ends the debate and it's the case of the hypothesis that the microscopic physics may be described by classical physics. - So far, this seems to be the only answer that actually answers the question. In the absence of alternatives, I'll eventually mark it as "accepted". – Carl Brannen Mar 4 '11 at 0:12 The answer by Lubos and this unfortunate sentence in page one of the older paper (as quoted by Lubos) "permanent wish to understand quantum mechanics in terms of classical probabilities" gives the impression that this Tomographic approach is trying to do something known to be impossible. In fact the field has become the basis of a wide ranging series of practical techniques as discussed in the Wikipedia article on Quantum Tomography. Amongst its applications are quantum optics and practical quantum computation. So I will just mention a few points of clarification here. Clearly any simplistic attempt to "reconstruct" $\Psi$ from knowledge of any (say) position (X) probability distribution cannot work. Even knowing the momentum (P) probability distribution as well is not always enough. So a price has to be paid to reformulate $\Psi$ in this kind of way. In summary one needs to know these probability distributions: $T(\mu, \nu) = \mu X + \nu P$ As you can see this is a parameterised family with 2 dimensions worth of parameters. Using Radon transforms, Wigner functions and the like the theory is able to reconstruct (well not $\Psi$) but the density matrix $\rho$. Now there are still some issues here. 1. Q. How to actually obtain this data? A. By doing multiple measurements on similarly prepared quantum states. This is how Tomography has found applications in Optics, where there will be plenty of similar source photons to use. 2. Q. Are $\mu$ and $\nu$ not continuous, leading to error? A. Yes. So some form of statistical sampling logic needs to be added in practice. 3. Q. What about other observables apart from X and P? A. Yes the theory can be generalised to other classes of observables. This is one area where the theory may have current limits though. 4. Q. From a Hilbert space point of view what is going on? A. (I think) that what is happening is that alternative representations (to X, P, etc) are being constructed. One example they derive which is familiar to me is the coherent state representation. In this representation $\Psi$ becomes a function of Z=(X,P) as in Z = X + iP. Every coherent state is defined by a single (complex - hence 2D) point in this representation and the vacuum is 0 = origin. It is well known that the coherent states form an overcomplete basis for $\Psi$ and this is represented in their framework. 5.Q. Is Tomography really an Interpretation of QM? A. I dont think so, just a set of mathematical tools to be used in practical and maybe theoretical analysis of QM systems. So what is the "worst thing"? - lack of a good coherent presentation of the entire area and its research philosophy (at least if those cited papers are anything to go by). - The second paper is available as Phys. Scr. 79 (2009) 065013 (29pp). An editor and a referee or two think there are not too many things wrong with it, or that it has enough good things about it. One assumes Luboš was not a referee. The whole community doesn't vote on whether a paper is accepted into a journal, of course, so a pinch of salt, but as of now there are 26 citations to the 2009 paper in published papers, according to Web of Science (which is not shabby at all). The editor required them to cut down the paper significantly, 64pp→29pp, so one of the worst;-) things about the arXiv version is mitigated in the published paper. I take this kind of approach to QM to be on the edge of engineering and interpretation. There is an engineering problem that is addressed by tomography — how do we determine what quantum observables we are measuring (and the relationships between them, expressed in terms of the mathematical ansatz we use to model them) and what quantum state we have prepared, from the basic classical raw data points (that are written in lab note books or stored in computer memory) and whatever statistics of that raw data we may construct? Note that the engineering problem begins with classical information and ends with quantum observables and a quantum state, and, as such, is intimately concerned with the interpretation of QM, at least in the old-fashioned vaguely formulated terms of the Copenhagen and Bohr's interpretations. For more modern interpretations, which often layer some kind of metaphysics of the quantum state on top of the raw data, the relevance is not as clear-cut. The worst thing about the second paper may not be something for which it can be hung out to dry, just because no one else has achieved definitude: it does not construct a compelling account. Specifically, it does not compel Luboš to take it seriously. This is an extraordinarily difficult balance to achieve, however it looks to me that if current progress is maintained someone will find an adequate combination of simplicity and mathematical and philosophical sophistication in the next 10 years. Having written part of this before Roy Simpson's Answer, I have up-voted his Answer and commend it to other people as a counterpoint to the approach I have taken. I commend the papers cited in the Question as useful for engineering purposes and as a valiant attempt at interpretation. - 1 I think that one of the lessons here is how a "research programme" is actually presented. Your view is "not compelling", mine is "poorly presented"; and Lubos is "misguided". Yet the authors see new things to do... – Roy Simpson Mar 2 '11 at 17:43 I see them as doing something interesting both in engineering terms and in interpretation terms, although I think their engineering is more successful. I'm sure that if I diverted my attention towards their work I would find new things to do, perhaps even things that I might be able to do, as others have (those citations!), but we make our choices. – Peter Morgan Mar 2 '11 at 18:11 – Carl Brannen Mar 2 '11 at 23:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "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": 0.9535933136940002, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/109691-help-legrange.html	# Thread: 1. ## Help with Legrange Problem: Find max and/or min values of function f given the contraints: f(x,y,z) = x^2 +y^2 +z^2 x + y + z = 1 x + 2y + 3z = 6 I know how to use legrange generally. When I solve using both constraints, I get values for x,y,z that when plugged into the function f gives 25/3. The answer in the book gives: "No Maximum, minimum: 25/3" My question is how do I know the value i got (25/3) is the minimum and how do I know there is no maximum for the function? 2. It sounds like you are doing the math work in the probelm right, which is good. One way to check your critical points is using the second partials of the function, which can be a pain, but in this case isnt too bad. Just remember to substitute a value for $z$ in terms of $x$ and $y$ in the equation before you try to take a derivative of it. Equation: $D=FxxFyy-(Fxy)^2$ -If Fxx and D(x,y,z) are positve then its a min. -If Fxx is negative and D(x,y,z) is positive then the point is a max. -If D(x,y,z) is less than 0 then the point is a saddle point. *note: where $Fxx$ means take the partial derivative of F with respect to x twice. 3. Yes but these equations have 3 variables so wouldnt I have to do something with the partial derivative with respect to z? 4. Correct, you would need to solve for Z in terms of x and y. Then, you would put that in for F(x), then start the second partials. After that your equatoin will be in terms of only x and y.	{"extraction_info": {"found_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": 10, "end_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": 0.9293075799942017, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/74113/problem-solving-a-set-of-quadratic-equations	Problem solving a set of quadratic equations Sorry for being a newbie barging in with a question, but I'm facing a rather trivial problem which I seem unable to solve... Not being a matemathician (but an engineer with a bit of knack for math), I managed to formulate it in a way that seemed solveable to me, but when I try to solve it seemingly I chase my own tail. I'm not familiar with LaTeX, so I'll hopefully be excused for enclosing the picture with starting equations and what I'm having trouble solving: (I'm not allowed to post pictures, so I'll try with a link to picture instead: link to picture with equations) Edit: Here are the equations as pictured: $$\begin{align} &r_{T_i}\sqrt{1-(\overrightarrow{n_G}\cdot\overrightarrow{n_{W_i}})^2}=\overrightarrow{n_G}\cdot\overrightarrow{W_{C_i}}+D, \qquad i=1,2,3\tag{1}\\ &\|\overrightarrow{n_G}\|=1\tag{2}\\ (1)\qquad\dots\qquad&aa_i\cdot x^2+bb_i\cdot y^2+cc_i\cdot z^2+ab_i\cdot x\cdot y+bc_i\cdot y\cdot z+ca_i\cdot z\cdot x+\\ &\qquad+a_i\cdot x\cdot D+b_i\cdot y\cdot D+c_i\cdot z\cdot D+D^2-(r_i)^2 = 0\\ (2)\qquad\dots\qquad&x^2+y^2+z^2=1 \end{align}$$ If the origin of the problem is of any consequence (in case anyone knows a simpler way to solve it)- I'm trying to get the plane tangent to three circles (ground plane defined by being tangential to three wheels). In those equations $n_G$ is the vector normal to the ground plane and $D$ it's distance from origin; $W_C$ are center points of wheels and $n_W$ are their axes (perpendicular to the wheel plane). In 2nd set of equations $x$, $y$ and $z$ are components of vector $n_G$... Any help would be appreciated. Thanks in advance. - Unless I'm missing something, I see only two equations in three unknowns. We expect to find a solution set that is "generically" one-dimensional. In other words, we won't be able to find a finite set of solutions. – Shaun Ault Oct 19 '11 at 20:56 Please let me know if I copied any of the equations incorrectly. – Brian M. Scott Oct 19 '11 at 21:09 Sorry for not explaining my notation- it's set of four equations with four unknowns: the equations (1) have index i (i=1 to 3) for each of three circles. In the first (vector) representation $n_G$ and $D$ are unknowns, but $n_G$ consists of three components ($x$, $y$ and $z$ in second representation)... edit: Thanks Brian, they're picture perfect and accurate. – Tomislav Petričević Oct 19 '11 at 21:12	{"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": 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": 0.9599956274032593, "perplexity_flag": "middle"}
http://ams.org/bookstore?fn=20&arg1=gsmseries&ikey=GSM-37	New Titles \| FAQ \| Keep Informed \| Review Cart \| Contact Us Quick Search (Advanced Search ) Browse by Subject General Interest Logic & Foundations Number Theory Algebra & Algebraic Geometry Discrete Math & Combinatorics Analysis Differential Equations Geometry & Topology Probability & Statistics Applications Mathematical Physics Math Education Return to List Theta Constants, Riemann Surfaces and the Modular Group: An Introduction with Applications to Uniformization Theorems, Partition Identities and Combinatorial Number Theory Hershel M. Farkas, The Hebrew University, Jerusalem, Israel, and Irwin Kra, State University of New York, Stony Brook, NY SEARCH THIS BOOK: Graduate Studies in Mathematics 2001; 531 pp; hardcover Volume: 37 ISBN-10: 0-8218-1392-7 ISBN-13: 978-0-8218-1392-8 List Price: US\$80 Member Price: US\$64 Order Code: GSM/37 There are incredibly rich connections between classical analysis and number theory. For instance, analytic number theory contains many examples of asymptotic expressions derived from estimates for analytic functions, such as in the proof of the Prime Number Theorem. In combinatorial number theory, exact formulas for number-theoretic quantities are derived from relations between analytic functions. Elliptic functions, especially theta functions, are an important class of such functions in this context, which had been made clear already in Jacobi's Fundamenta nova. Theta functions are also classically connected with Riemann surfaces and with the modular group $$\Gamma = \mathrm{PSL}(2,\mathbb{Z})$$, which provide another path for insights into number theory. Farkas and Kra, well-known masters of the theory of Riemann surfaces and the analysis of theta functions, uncover here interesting combinatorial identities by means of the function theory on Riemann surfaces related to the principal congruence subgroups $$\Gamma(k)$$. For instance, the authors use this approach to derive congruences discovered by Ramanujan for the partition function, with the main ingredient being the construction of the same function in more than one way. The authors also obtain a variant on Jacobi's famous result on the number of ways that an integer can be represented as a sum of four squares, replacing the squares by triangular numbers and, in the process, obtaining a cleaner result. The recent trend of applying the ideas and methods of algebraic geometry to the study of theta functions and number theory has resulted in great advances in the area. However, the authors choose to stay with the classical point of view. As a result, their statements and proofs are very concrete. In this book the mathematician familiar with the algebraic geometry approach to theta functions and number theory will find many interesting ideas as well as detailed explanations and derivations of new and old results. Highlights of the book include systematic studies of theta constant identities, uniformizations of surfaces represented by subgroups of the modular group, partition identities, and Fourier coefficients of automorphic functions. Prerequisites are a solid understanding of complex analysis, some familiarity with Riemann surfaces, Fuchsian groups, and elliptic functions, and an interest in number theory. The book contains summaries of some of the required material, particularly for theta functions and theta constants. Readers will find here a careful exposition of a classical point of view of analysis and number theory. Presented are numerous examples plus suggestions for research-level problems. The text is suitable for a graduate course or for independent reading. Readership Graduate students, research mathematicians interested in complex analysis and number theory. Reviews "Can be useful to experts and novices alike, ... details are abundant and developments mainly self-contained, ... the book can be read with profit by anyone with a sufficient background in complex analysis, ... Farkas and Kra have exposed a great deal of beautiful mathematics, all of it solidly grounded in the classics of our tradition, and yet much of it new. ... this elevates their work to a model of exposition ... that could be emulated to the benefit of the entire mathematical community." -- Mathematical Reviews • The modular group and elliptic function theory • Theta functions with characteristics • Function theory for the modular group $$\Gamma$$ and its subgroups • Theta constant identities • Partition theory: Ramanujan congruences and generalizations • Identities related to partition functions • Combinatorial and number theoretic applications • Bibliography • Bibliographical notes • Index AMS Home \| Comments: [email protected] © Copyright 2012, American Mathematical Society Privacy Statement	{"extraction_info": {"found_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": 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": 0.8894893527030945, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/71952?sort=oldest	## Do the elementary properties of mixed volume characterize it uniquely? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Background Take 2 convex sets in $\mathbb{R}^2$, or 3 convex sets in $\mathbb{R}^3$, or generally, $n$ convex sets in $\mathbb{R}^n$. "Mixed volume" assigns to such a family `$A_1, \ldots, A_n$` a real number `$V(A_1, \ldots, A_n)$`, measured in $\mathrm{metres}^n$. As I understand it, mixed volume is a kind of cousin of the determinant. I'll give the definition in a moment, but first here are some examples. 1. $V(A, \ldots, A) = \mathrm{Vol}(A)$, for any convex set $A$. 2. More generally, suppose that `$A_1, \ldots, A_n$` are all scalings of a single convex set (so that $A = r_i B$ for some convex $B$ and $r_i \geq 0$). Then `$V(A_1, \ldots, A_n)$` is the geometric mean of `$\mathrm{Vol}(A_1), \ldots, \mathrm{Vol}(A_n)$`. 3. The previous examples don't show how mixed volume typically depends on the interplay between the sets. So, taking $n = 2$, let $A_1$ be an $a \times b$ rectangle and $A_2$ a $c \times d$ rectangle in $\mathbb{R}^2$, with their edges parallel to the coordinate axes. Then ```$$ V(A_1, A_2) = \frac{1}{2}(ad + bc). $$``` (Compare and contrast the determinant formula $ad - bc$.) 4. More generally, take axis-parallel parallelepipeds `$A_1, \ldots, A_n$` in $\mathbb{R}^n$. Write `$m_{i1}, \ldots, m_{in}$` for the edge-lengths of $A_i$. Then ```$$ V(A_1, \ldots, A_n) = \frac{1}{n!} \sum_{\sigma \in S_n} m_{1, \sigma(1)} \cdots m_{n, \sigma(n)}. $$``` (Again, compare and contrast the determinant formula.) The definition of mixed volume depends on a theorem of Minkowski: for any compact convex sets `$A_1, \ldots, A_m$` in $\mathbb{R}^n$, the function ```$$ (\lambda_1, \ldots, \lambda_m) \mapsto \mathrm{Vol}(\lambda_1 A_1 + \cdots + \lambda_m A_m) $$``` (where $\lambda_i \geq 0$ and $+$ means Minkowski sum) is a polynomial, homogeneous of degree $n$. For $m = n$, the mixed volume `$V(A_1, \ldots, A_n)$` is defined as the coefficient of `$\lambda_1 \lambda_2 \cdots \lambda_n$` in this polynomial, divided by $n!$. Why pick out this particular coefficient? Because it turns out to tell you everything, in the following sense: for any convex sets `$A_1, \ldots, A_m$` in $\mathbb{R}^n$, ```$$ \mathrm{Vol}(\lambda_1 A_1 + \cdots + \lambda_m A_m) = \sum_{i_1, \ldots, i_n = 1}^m V(A_{i_1}, \ldots, A_{i_n}) \lambda_{i_1} \cdots \lambda_{i_n}. $$``` Properties of mixed volume Formally, let $\mathscr{K}_n$ be the set of nonempty compact convex subsets of $\mathbb{R}^n$. Then mixed volume is a function $$V: (\mathscr{K}_n)^n \to [0, \infty),$$ and has the following properties: 1. Volume: $V(A, \ldots, A) = \mathrm{Vol}(A)$. (Here and below, the letters $A$, $A_i$ etc. will be understood to range over $\mathscr{K}_n$, and $\lambda$, $\lambda_i$ etc. will be nonnegative reals.) 2. Symmetry: $V$ is symmetric in its arguments. 3. Multilinearity: ```$$ V(\lambda A_1 + \lambda' A'_1, A_2, \ldots, A_n) = \lambda V(A_1, A_2, \ldots, A_n) + \lambda' V(A'_1, A_2, \ldots, A_n). $$``` (These first three properties closely resemble a standard characterization of determinants.) 4. Continuity: $V$ is continuous with respect to the Hausdorff metric on $\mathscr{K}_n$. 5. Invariance: `$V(gA_1, \ldots, gA_n) = V(A_1, \ldots, A_n)$` for any isometry $g$ of Euclidean space $\mathbb{R}^n$ onto itself. 6. Multivaluation: ```$$ V(A_1 \cup A'_1, A_2, \ldots, A_n) = V(A_1, A_2, \ldots) + V(A'_1, A_2, \ldots) - V(A_1 \cap A'_1, A_2, \ldots) $$``` whenever `$A_1, A'_1, A_1 \cup A'_1 \in \mathscr{K}_n$`. 7. Monotonicity: `$V(A_1, A_2, \ldots, A_n) \leq V(A'_1, A_2, \ldots, A_n)$` whenever `$A_1 \subseteq A'_1$`. There are other basic properties, but I'll stop there. Questions Is $V$ the unique function $(\mathscr{K}_n)^n \to [0, \infty)$ satisfying properties 1--7? If so, does some subset of these properties suffice? In particular, do properties 1--3 suffice? If not, is there a similar characterization involving different properties? (Partway through writing this question, I found a recent paper of Vitali Milman and Rolf Schneider: Characterizing the mixed volume. I don't think it answers my question, though it does give me the impression that the answer might be unknown.) - Looking at your papers, it appears that you are in contact with some of the top experts in this area. The only other person I suggest asking is Monika Ludwig. If she and the other who you know already don't know the answer to this, I doubt anybody else does. But I do like the question. – Deane Yang Aug 3 2011 at 2:30 Thanks, Deane. It's a pleasant surprise to be told I'm so well-connected, as I'm genuinely a total amateur at this stuff. – Tom Leinster Aug 3 2011 at 2:33 The other obvious person to ask is Dan Klain. – Deane Yang Aug 3 2011 at 3:42 ## 2 Answers I think the first three properties do indeed characterize mixed volume. For example, in two dimensions they imply that $V(A_1, A_2) = \frac{1}{2}(V(A_1 + A_2, A_1 + A_2) - V(A_1, A_1) - V(A_2,A_2))$ $= \frac{1}{2}(Vol(A_1 + A_2) - Vol(A_1) - Vol(A_2)),$ which gives the formula of mixed volume in terms of volume. You can perform the same trick to get in 3 dimensions: $V(A_1,A_2, A_3) = \frac{1}{6}(Vol(A_1+A_2+A_3) - Vol(A_1+A_2) - Vol(A_2+A_3)$ $- Vol(A_3+A_1) + Vol(A_1) + Vol(A_2) + Vol(A_3))$ In general I believe you get something like: $V(A_1, \ldots,A_n) = \frac{1}{n!}(Vol(A_1 + \cdots + A_n) - \sum_{i=1}^n Vol(A_1 + \cdots \hat A_i + \cdots + A_n)$ $+ \cdots +(-1)^{n-1}\sum_{i=1}^n Vol(A_i))$ I learned of this from Bernstein's paper that contains his famous result that the number of solutions in $(\mathbb{C}^*)^n$ of $n$ generic Laurent polynomials is precisely the mixed volume of their Newton polytopes. - Great! Thanks. I see that formula in Lemma 5.1.3 of Schneider's book Convex Bodies, but he doesn't say there that it holds for all convex bodies (just ones in a certain class), and he also doesn't say there that it follows from properties 1-3. – Tom Leinster Aug 3 2011 at 15:00 ### 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. Sorry to answer my own question, but asking this in public seems to have spurred me into thought. As auniket suspected, the answer is "yes" in the strongest sense I'd hoped: properties 1-3 do characterize mixed volume. In fact, something slightly stronger is true: $V$ is the unique function $(\mathscr{K}_n)^n \to \mathbb{R}$ satisfying 1. $V(A, \ldots, A) = Vol(A)$ 2. $V$ is symmetric 3. `$V(A_1 + A'_1, A_2, \ldots, A_n) = V(A_1, A_2, \ldots, A_n) + V(A'_1, A_2, \ldots, A_n)$`. In other words, we don't need multilinearity, just multiadditivity. The proof is along the lines suggested by auniket. Fix $n$ and `$A_1, \ldots, A_n \in \mathscr{K}_n$`. Write `$\mathbf{n} = \{1, \ldots, n\}$`, and for sets $R$ and $S$, write $\mathrm{Surj}(R, S)$ for the set of surjections $R \to S$. I claim that for all subsets $S$ of $\mathbf{n}$, ```$$ \sum_{f \in \mathrm{Surj}(\mathbf{n}, S)} V(A_{f(1)}, \ldots, A_{f(n)}) $$``` is uniquely determined by the properties above. The proof will be by induction on the cardinality of $S$. When $S = \mathbf{n}$, this sum is ```$$ n! V(A_1, \ldots, A_n), $$``` so this claim will imply the characterization theorem. To prove the claim, take $S \subseteq \mathbf{n}$. Then ```$$ Vol(\sum_{i \in S} A_i) = \sum_{f: \mathbf{n} \to S} V(A_{f(1)}, \ldots, A_{f(n)}) $$``` by the three properties. This in turn is equal to ```$$ \sum_{R \subseteq S} \sum_{f \in \mathrm{Surj}(\mathbf{n}, R)} V(A_{f(1)}, \ldots, A_{f(n)}). $$``` By the inductive assumption, all but one of the summands in the first summation - namely, $R = S$ - is uniquely determined. Hence the $S$-summand is uniquely determined too. This completes the induction, and so completes the proof. The proof makes it clear that `$V(A_1, \ldots, A_n)$` is some rational linear combination of ordinary volumes of Minkowski sums of some of the $A_i$s. It must be possible to unwind this proof and get an explicit expression; and that expression must be the one auniket gave (which also appears in Lemma 5.1.3 of Schneider's book Convex Bodies: The Brunn-Minkowski Theory). This all seems rather easy, and must be well-known, though I'm a bit surprised that this characterization isn't mentioned in some of the things I've read. Incidentally, I now understand why it doesn't appear in the paper of Milman and Schneider mentioned in my question: they explicitly state that they want to avoid assuming property 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": 66, "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": 0.9197235107421875, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/31345-simplifying-complex-numbers-i.html	# Thread: 1. ## Simplifying Complex Numbers (i) Okay, I think I did this right, but I'm going to post it here just to make sure. Simplify the complex number i^31 as much as possible. Here's what I did: i^31 =i^30+1 =i^65+1 =i^1 =i So would my final answer just be i? (assuming I did that right) 2. Hello, The method is correct, but not the result : i=i i²=-1 i^3 = -i i^4 = 1 3. I don't understand how the result is incorrect. Shouldn't it be simply $i$? What isn't right? 4. =i^65+1 =i^1 This step is strange. This means you suppose i^(6*5) = 1, which is false. You may write that $i^{31}=i^{28}i^3 = (i^4)^7 i^2 i$ $i^4 = 1$ as i've shown you. So what's the result ? 5. So would it be (1^7)i ? 6. Learn these four numbers $i,\,i^2 = - 1,\,i^3 = - i,\,i^4 = 1$. Now divide the exponent by 4 and take the remainder. Thus $i^{31} = i^3 = - i$ because 31 divided by 4 leaves a remainder of 3. Here why that works. $i^{31} = i^{28 + 3} = \left( {i^4 } \right)^7 \left( {i^3 } \right) = \left( 1 \right)^7 \left( {i^3 } \right) = - i$ 7. Originally Posted by eraser851 So would it be (1^7)i ? The very first property of i is that i²=-1 8. eraser, Plato is right. When dealing with Imaginary numbers, learn those 4 powers. If the power you are to raise i to exceeds 4, you want to use $i^4$ and see what the remainder of powers is after you factor out a 4. Either 4 divides in evenly or it doesn't. If it does not, you are left with either $i^1$, $i^2$, or $i^3$. See here: Imaginary Numbers	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.9527903199195862, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/90656/are-banach-manifolds-intrinsically-interesting	## Are Banach Manifolds intrinsically interesting? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In the introduction to 'A convenient setting for Global Analysis', Michor & Kriegl make this claim: "The study of Banach manifolds per se is not very interesting, since they turnout to be open subsets of the modeling space for many modeling spaces." But finite-dimensional manifolds are found to be interesting even though they can be embedded in some euclidean space (of larger dimension). (Actually this seems to me, to make the above claim intuitively plausible, so that claim should be no more than we should expect). But they do go on to say that "Banach manifolds are not suitable for many questions of Global Analysis, as...a Banach Lie group acting eﬀectively on a ﬁnite dimensional smooth manifold it must be ﬁnite dimensional itself.", which does seem a rather strong limitation. - 8 There is a big difference between the finite-dimentional case and the other: in the latter, manifolds are open subsets of the modeling space, while in the former manifolds rarely are open sets of the modeling space (they embed in a euclidean space of a generally much larger dimension, and even more rarely as open sets thereof) – Mariano Suárez-Alvarez Mar 9 2012 at 4:02 6 Usually the infinite dimensional objects one encounters in geometry are Fréchet manifolds (e.g. diffeomorphism groups or spaces of maps with the smooth topology) and these are really the objects of interest. Banach manifolds are a useful tool for studying infinite dimensions and actually proving anything because there you have the implicit function theorem. In particular if you have an elliptic problem, e.g. holomorphic curves, it's much easier to think of the solution space inside a Banach manifold of maps and then use elliptic regularity it prove it's really inside the smooth locus. – Jonny Evans Mar 9 2012 at 7:09 @suarez-alvarez: got you, thanks. – Mozibur Ullah Dec 4 at 21:05 @Evans: you mean infinite-dimensional vector spaces? – Mozibur Ullah Dec 4 at 21:06 ## 2 Answers In his remarkable thesis Douady proved that, given a compact complex analytic space $X$, the set $H(X)$ of analytic subspaces of $X$ has itself a natural structure of analytic space . If $X=\mathbb P^n(\mathbb C)$ for example, then $H(X)$ is the Hilbert scheme $Hilb(\mathbb P^n(\mathbb C))$. However the problem is much more difficult for non algebraic $X$. Douady solved it by massive use of Banach analytic manifolds, the most important of them being the grassmannian of complemented closed subspaces of a Banach space. The thesis starts with the candid statement of its aim: "Le but de ce travail est de munir son auteur du grade de docteur-ès-sciences mathématiques et l'ensemble H(X) des sous espaces analytiques compacts de X d'une structure d'espace analytique", that is to endow its author with the title of doctor in mathematics and the the set H(X) of compact analytic subspaces of X with the structure of analytic space. - ### 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. I am of two minds on this topic. It is much easier to work on Banach manifolds because the implicit function theorem on such spaces has a simple formulation. On the other hand, as the examples of gauge theory or the theory of pseudo-holomorphic curves show, in these contexts one works not with one Banach manifold, but with several, determined by stronger and stronger Sobolev norms. One important part of the game is to conclude that objects with a priori weaker Sobolev regularity are in fact smooth. This feels very much like we are implicitly working on a Frechet manifold. One draw back of Banach spaces is that they do not have many smooth functions on them, and the notion of real analycity on such spaces is problematic. Let's take the example of Seiberg-Witten equations. These are quadratic equations in its variables, so intuitively they ought to be real analytic, though I do not know how to formulate this rigorously in a Sobolev context. Why do I care about real analycity? In the real analytic context one can formulate an intersection theory involving not necessarily smooth objects. For example, the point $0\in\mathbb{R}$ is a solution of the quadratic equation $x^2=0$. It is a degenerate zero, and from the point of view of intersection theory it has multiplicity $0$. My hope is that this real analytic point of view would allow one to deal with mildly degenerate solutions of the the Seiberg-Witten equations, and assign multiplicities to such solutions. - Multiplicity 0? Not 2? – Allen Knutson Mar 11 2012 at 14:23 In the real case the multiplicity is zero. To see this look at the equation $x^2-\varepsilon=0$. For $\varepsilon>0$ it has two solutions, one counter with multiplicity $1$ and the other with multiplicity $-1$ and thus the intersection number of the graph of $x^2-\varepsilon$ with the $x$-axis is zero. If $\varepsilon <0$, then the equation has no solution, and the above intersection number is also zero. – Liviu Nicolaescu Mar 11 2012 at 16:53 Got it. $\ \ \$ – Allen Knutson Mar 12 2012 at 1:29	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "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": 0.9060971736907959, "perplexity_flag": "head"}
http://mathoverflow.net/questions/19740?sort=oldest	## How do you recover the structure of the upper half plane from its description as a coset space? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) This is maybe a dumb question. $SL_2(\mathbb{R})$ has a natural action on the upper half plane $\mathbb{H}$ which is transitive with stabilizer isomorphic to $SO_2(\mathbb{R})$. For this reason, people sometimes write $\mathbb{H}$ as the coset space $SL_2(\mathbb{R})/SO_2(\mathbb{R})$. Now, it's clear how this description recovers the topology of $\mathbb{H}$: it's just the quotient topology. But can you recover either the Riemann surface structure or the hyperbolic metric on $\mathbb{H}$ from this description? How much of the structure of $SL_2(\mathbb{R})$ and $SO_2(\mathbb{R})$ do you need to do this, if it's possible? - 2 @Qiaochu: that's a highly non-dumb question! It comes up in the theory of moduli spaces. It's not hard to check that the points of SL_2(Z) \ H parametrise elliptic curves over the complexes, but why should such a set be a complex manifold? I think there's a very fancy answer to do with variation of Hodge structures but I don't think I ever understood the details of that point of view well enough to be able to explain them :-( . I somehow feel that Deligne's axioms for a Shimura variety should somehow help, but on some level I've never understood these either :-( – Kevin Buzzard Mar 29 2010 at 17:49 "t's not hard to check that the points of SL_2(Z) \ H parametrise elliptic curves over the complexes, but why should such a set be a complex manifold?" I am not sure if I understood you correctly. What's wrong with the usual proof (say, in Silverman's Advanced Topics in Elliptic Curves") that shows that after giving the appropriate charts and adding an extra point at infinity (one needs to be careful about the elliptic points) this is biholomorphic to the Riemann sphere? – Idoneal Mar 29 2010 at 18:21 1 Sorry Idoneal, I didn't explain myself well. I claim that there is a canonical bijection between the set of isomorphism classes of elliptic curves over the complexes, and SL_2(Z) \ H. If you're prepared to believe that H is a complex manifold, then this set becomes a complex manifold. However if you didn't know H existed, and just had a set of isomorphism classes of elliptic curves, however would you put a complex structure on it? It can be done! You need to consider holomorphic families of ell curves (i.e. maps M-->S of cx mfds whose fibres are ell curves)... – Kevin Buzzard Mar 29 2010 at 18:27 1 ...and the point is that you want M-->S to be a holomorphic family of elliptic curves iff the induced map from the complex manifold S to the set of iso classes of ell curves (sending s to the fibre above s) is holomorphic. This gives you the structure of a complex manifold on SL_2(Z) \ H. But now one has to prove that it's the same one as the one coming from the upper half plane, and this genuinely needs proof. – Kevin Buzzard Mar 29 2010 at 18:28 1 @Mariano: it's not obvious that SL_2(R)/SO_2(R) has a complex structure, and it's not obvious that SL_2(Z)\SL_2(R)/SO_2(R) has a complex structure, but I'm sure you can see that one has a complex structure iff the other one does. He asked about the first but I mentioned the second because I am pretty sure he knows SL_2(Z)\H is a moduli space and I'm less sure if he knows that H is (it's also another moduli space but it's of elliptic curves plus more structure). – Kevin Buzzard Mar 29 2010 at 20:27 show 2 more comments ## 3 Answers Edit: I should have put a short version of the answer in the beginning, so here is how the various structures are recovered. To get a smooth manifold structure on the quotient, you use the fact that `$SL_2(\mathbb{R})$` is a real Lie group and `$SO_2(\mathbb{R})$` is a closed subgroup. To get a hyperbolic structure, you use the fact that `$SL_2(\mathbb{R})$` is isomorphic to an orthogonal group of signature (n,1) for some n (giving a transitive action on hyperbolic n-space). To get a complex structure, you use the fact that `$SL_2(\mathbb{R})$` is isomorphic to an orthogonal group of signature (2,m) for some m (giving an action on a hermitian symmetric space). As others have noted, you can get a bijection on points using the Iwasawa decomposition, and you can get a hyperbolic structure using the exceptional isomorphism `$PSL_2(\mathbb{R}) \cong SO_{2,1}^+(\mathbb{R})$`. First, I'd like to clean up the Iwasawa treatment a bit. Any element of $SL_2(\mathbb{R})$ can be uniquely decomposed as BK, where K is a rotation and B is upper triangular with positive diagonal. Any rotation K fixes i, so we should consider what elements B do. A bit of fiddling shows that `$\begin{pmatrix} \sqrt{y} & x/\sqrt{y} \\ 0 & 1/\sqrt{y} \end{pmatrix} \cdot i = x+iy$`. We can view the exceptional isomorphism in another way that makes the complex structure more apparent, by viewing the hyperbolic plane as the Grassmannian `$O_{2,1}(\mathbb{R})/(O_2(\mathbb{R}) \times O_1(\mathbb{R}))$`. From the standpoint of special relativity, this is the space of timelike lines through the origin in $\mathbb{R}^{2,1}$. Taking a quotient of the total space of these lines (minus origin) by positive rescaling, we find that this space is isomorphic to the space of pairs of antipodal points of norm -1. In particular, we have an isomorphism of the Grassmannian with the quotient of the hyperboloid with two sheets (i.e., solutions of the equation $x^2 + y^2 - z^2 = -1$) by the antipodal automorphism. One way to explain the origin of the complex structure is by the fact that all Grassmannians of the form $O(2,n)/(O(2) \times O(n))$ are hermitian symmetric spaces, and the hyperbolic plane is just the case $n=1$. The 2 in $O(2)$ is essential, because the orthogonal group action is what yields the ninety degree rotation in the tangent space of any point, and this is what endows the quotient with an almost complex structure. If you want to see more about hermitian symmetric spaces than the Wikipedia blurb, I recommend looking in chapter 1 of Milne's introduction to Shimura varieties. Finally, I'd like to point out Deligne's description of the upper half plane as a moduli space of structured elliptic curves. Points on H parametrize elliptic curves with an oriented basis of first homology (as mentioned a few times in our class). If you want to say it is a fine moduli space, you need a functor that it represents, and it is unfortunately a bit complicated. The functor takes as input the category of complex analytic spaces, and for any such space S, it gives the set of isomorphism classes of elliptic curves over S (i.e., diagrams $E \underset{\pi}{\leftrightarrows} S$ of complex analytic spaces, where $\pi$ is smooth and proper with one-dimensional genus one fibers and the leftward map is a section) equipped with an isomorphism `$R^1\pi_\underline{\mathbb{Z}} \cong \underline{\mathbb{Z} \times \mathbb{Z}}$` that induces the canonical identity `$R^2\pi_\underline{\mathbb{Z}} \cong \underline{\mathbb{Z}}$` on exterior squares. Here, the underscore indicates a constant sheaf. The functor also takes morphisms to "the evident diagrams". To be honest, I have never seen a complete proof that this functor is represented by the complex upper half plane, although it seems to be more a question of doing lots of writing than an honest theoretical problem. You can probably do it using the fact that H is a classifying space of polarized Hodge structures, as Kevin Buzzard mentioned in the comments. - ### 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. Its easy to check that every matrix in $SL_2(\mathbb{R})$ can be wrtten uniquely as $\begin{pmatrix} \lambda & \alpha \ 0 & \lambda^{-1}\end{pmatrix}\begin{pmatrix} \cos{\theta} & -\sin{\theta} \ \sin{\theta} & \cos{\theta} \end{pmatrix}$ where $\lambda>0$. This is exactly written then as a coset representative of the quotient you wrote down above. You can arrive at the hyperbolic metric by following the definition of the pushforward metric from a left invariant metric on $SL_2(\mathbb{R})$. Notice $(\alpha,\lambda)$ is a point in the upperhalf plane. The latex misbehaved, those should be $2\times 2$ matrices. - Thanks for your answer! What's an example of a left-invariant metric on SL_2(R)? Does it matter which one I choose if there are more than one? – Qiaochu Yuan Mar 29 2010 at 17:14 1 Quaochu, pick any metric on the Lie algebra---that is, the tangent space at the identity element---and translate it. – Mariano Suárez-Alvarez Mar 29 2010 at 17:26 Cool. So there remains another question which is maybe better asked as a separate question: what is the connection between the metric structure on H and the Riemann surface structure? – Qiaochu Yuan Mar 29 2010 at 18:15 2 A Riemann surface structure is the same thing as a conformal class of metric structures. – Mariano Suárez-Alvarez Mar 29 2010 at 18:49 In general left-invariant metrics on a Lie group, even a simply connected one, can be non-isometric -- Tam Nguyen Phan showed me a nice example of different left-invariant metrics on S^3. But the pushforward metric on SL(2,R)/SO(2) is homogeneous (acted on transitively by isometries), and thus of constant curvature, and so is either S^2, R^2, or H^2. You can then check (in any number of ways) that it's not flat, so must be H^2. – Tom Church Mar 29 2010 at 20:21 There is a very physical picture to this, if you are willing to work with the disk model of hyperbolic space, instead of the upper half plane, to which it is related by an isometry. The Lie group $\mathrm{SL}(2,\mathbb{R})$ is a double cover of the identity component $\mathrm{SO}_0(2,1)$ of $\mathrm{O}(2,1)$, which is the Lorentz group in 3 dimensions. In other words, $\mathrm{O}(2,1)$ is the subgroup of $\mathrm{GL}(3,\mathbb{R})$ which preserves a symmetric inner product $\eta$ of signature $(2,1)$: $$\eta = \begin{pmatrix} 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & -1 \end{pmatrix}.$$ Now consider the two-sheeted hyperboloid in $\mathbb{R}^3$ defined by $x^2 + y^2 - z^2 = -1$. The upper sheet -- let's call it $\mathbb{D}$ -- with $z>0$ is topologically a disk. It inherits a riemannian metric from the Minkowski metric on $\mathbb{R}^3$ defined by $\eta$. The identity component of $\mathrm{O}(2,1)$ acts on $\mathbb{D}$ as isometries. The isotropy at the point $(0,0,1)$ consists of rotations in the $x,y$-plane, whence it is isomorphic to $\mathrm{SO}(2)$. Hence $\mathbb{D} = \mathrm{SO}_0(2,1)/\mathrm{SO}(2)$. Notice that it is is $\mathrm{SO}_0(2,1)$ (a.k.a. $\mathrm{PSL}(2,\mathbb{R})$) which acts effectively on $\mathbb{D}$ and not $\mathrm{SL}(2,\mathbb{R})$. Added I forgot to relate the disk to the upper half plane. If you think of $\mathbb{D}$ as the unit disk in the complex plane, then the map $\mathbb{D} \to \mathbb{H}$ is given by the following Möbius transformation: $$z \mapsto \frac{z-i}{z+i}$$ - I like this explanation very much - I remember Scott Carnahan explaining this to us once - but one has to produce the action of O(2, 1) on a "unit ball" in order to do this. Given only the abstract structure of O(2, 1) (in whatever category is necessary), can one recover its action on Minkowski space? – Qiaochu Yuan Mar 29 2010 at 18:33 What is O(2,1) except the linear isometries of Minkowski space? – Tom Church Mar 29 2010 at 20:00 I guess I'm not explaining myself well. Suppose I'm given a nice topological group G and a nice subgroup H. Then G/H is a topological space. What are general situations in which G/H comes equipped with extra geometric structure, and how much about G and H do I have to know to find that structure? (In other words, is it enough to have a black box which tells you group-theoretic things about G and H such as their other subgroups?) – Qiaochu Yuan Mar 29 2010 at 20:49 2 Yes, if you include the structure of Lie algebras as part of your group theoretic things. When I was a student I really enjoyed Helgason's Differential Geometry, Lie Groups and Symmetric Spaces, and Joe Wolfe's Spaces of Constant Curvature. – Charlie Frohman Mar 29 2010 at 21:05 1 The adjoint representation of O(2,1) on its Lie algebra preserves the Killing form, which has signature (2,1). This is as canonical as it gets. This is very particular to this signature, though. – José Figueroa-O'Farrill Mar 29 2010 at 21:58	{"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": 48, "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": 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": 0.940636396408081, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/3974-need-help-calculating-limit.html	# Thread: 1. ## need help on calculating a limit. lim [sqrt(1+x)-sqrt(x-1)] x->infinity i tried to move it to the denominator like this: lim 1/[sqrt(x+1)+sqrt(x-1)] but here im stuck, i think it should be zero, but hoe do i prove it? thanks in advance. 2. Originally Posted by DangerMan lim [sqrt(1+x)-sqrt(x-1)] x->infinity i tried to move it to the denominator like this: lim 1/[sqrt(x+1)+sqrt(x-1)] but here im stuck, i think it should be zero, but hoe do i prove it? thanks in advance. A simple (nonrigorous) way to do it is to look at $\sqrt{x \pm 1}$. When x is large, the 1 is so small as to be ignorable. Thus $\lim_{x \to \infty} \left ( \sqrt{1+x} - \sqrt{x-1} \right ) \approx \lim_{x \to \infty} \left ( \sqrt{x} - \sqrt{x} \right ) \to 0$ -Dan 3. but i asked for a proof, do you know of a rigorous way to prove it? thanks. 4. Depends on "how much rigour" you want... You could always go back to the definition, but I wouldn't do that unless necessary. Multiply numerator and denominator with the complement of the numerator to get: $<br /> \frac{{\left( {\sqrt {x + 1} - \sqrt {x - 1} } \right)\left( {\sqrt {x + 1} + \sqrt {x - 1} } \right)}}{{\left( {\sqrt {x + 1} + \sqrt {x - 1} } \right)}} = \frac{2}{{\left( {\sqrt {x + 1} + \sqrt {x - 1} } \right)}}<br />$ Now the numerator is constant and positive and the denominator clearly goes to +infinity, yielding 0 as limit. 5. You can do even this, note that for $x\geq 1$ we have, $0\leq \sqrt{x+1}-\sqrt{x-1} \leq \frac{1}{\sqrt{.5x}}$ But, $\lim_{x\to\infty} 0=0$ and, $\lim_{x\to\infty} \frac{1}{\sqrt{.5 x}}=0$ Thus, you have a function, $f(x)=\sqrt{x+1}-\sqrt{x-1}$ squeezed between two functions which have the same limit. Conclude from the squeeze theorem that, $\lim_{x\to\infty}\sqrt{x+1}-\sqrt{x-1}=0$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "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": 0.9032368063926697, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/114251/stable-infinity-categories-vs-dg-categories	## Stable infinity categories vs dg-categories ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) What is the relation between dg-categories and stable $\infty$-categories? Given a dg-category one can form its dg-nerve and get a $\infty$-category (which will be stable if the dg-category is?). Can one turn a stable $\infty$-category into a dg-category or $A_\infty$-category somehow? I have heard the statement that at least over a field of characteristic zero the theories of stable $\infty$-categories and dg-categories are "equivalent". What would be a precise formulation of this statement and what would be a reference? - 4 What's a stable $\infty$-category over a field (of characteristic zero)? – Fernando Muro Nov 23 at 15:40 ## 1 Answer Here are a few observations... 1. I think there exist stable infinity categories that are not the dg-nerve (resp. $A_\infty$-nerve) of a dg-category (resp. $A_\infty$ category). In particular, the category of spectra should not arise in this way. I think Keller has a paper on differential graded categories that answers this question; he notes at some point that the homotopy category of spectra is not "algebraic" but that homotopy categories of differential graded categories are (and in fact sort of encompass all such algebraic categories.) Basically it comes down to something like the existence of Hopf maps. Now- could one define somehow the "closest dg-category approximation" to a given stable infty category? Probably. I don't know how. Or maybe I could come up with how, but I'm not sure how useful this would be if the functor wasn't an equivalence? 2. To answer Fernando's question, see DAG X.5 or DAG VII.6.2. That is, a stable $\infty$-category over a field $k$ is a presentable, stable $\infty$-category "equipped with an action of the monoidal $\infty$-category of $k$-module spectra". Unless I'm mistaken I think this basically implies that it is enriched and tensored over k-module spectra. 3. Here would be a precise formulation of the statement about categories over a field of characteristic zero: The dg-nerve functor induces an equivalence of $\infty$-categories between the $\infty$-category underlying the model category of dg-categories over k and the $\infty$-category of stable, k-linear $\infty$-categories. (I don't mean to overwhelm with "infinities", I would state this in the perhaps friendlier world of model categories, but I'm not sure what precisely the model category is that corresponds to stable, k-linear $\infty$-categories.) I don't know of a reference for a proof, though Lurie alludes to this a lot. It would be great if someone wrote this down! - 1 I should mention that the "Hopf maps" proof that the homotopy category of spectra does not arise from algebraic procedures is due, I believe, to Fernando Muro :) – Dylan Wilson Nov 24 at 6:20 Another point is that I think the proof of (3) would basically reduce to showing that the $\infty$-category of k-module spectra is the same as the $\infty$-category of chain complexes over $k$. This is probably in the literature in the language of model categories, maybe even in Shipley's paper "HZ-algebra spectra are differential graded algebras". Maybe in general it's true that R-linear spectra are the same as DG R-modules when R is a discrete ring? I haven't read the paper so I don't know... – Dylan Wilson Nov 24 at 6:24 ... so maybe the characteristic zero part only enters the picture when we want to relate this to the $A_\infty$ story. But I really don't know about this stuff, so can someone else fill in here? I'm interested! – Dylan Wilson Nov 24 at 6:24 2 HR-linear spectra are the same as dg R-modules for any R. What's special about characteristic $0$ is that if the homotopy groups of a spectrum are $\mathbb{Q}$-modules, that automatically implies that the spectrum is an $H{\mathbb Q}$-module (that is to say, the localization of the sphere spectrum $S_{\mathbb Q}$ is the same as $H{\mathbb Q}$). Thus if you have a stable $\infty$-category in which the groups of maps between objects are always ${\mathbb Q}$-modules, it automatically is "dg" or "algebraic". – Eric Wofsey Nov 24 at 8:14 1 Also, in response to Jan's point: a dg-category is basically a category enriched in chain complexes ($H \mathbb{Z}$-module spectra). However, that doesn't mean that the associated $\infty$-category is actually stable. You probably want to say that pull-back squares are push-outs, or something equivalent to that. – Akhil Mathew Nov 24 at 16:12 show 3 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": 31, "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": 0.9294451475143433, "perplexity_flag": "middle"}
http://nrich.maths.org/7431/solution	Number Round Up Arrange the numbers 1 to 6 in each set of circles below. The sum of each side of the triangle should equal the number in its centre. Numbers as Shapes Use cubes to continue making the numbers from 7 to 20. Are they sticks, rectangles or squares? How Odd This problem challenges you to find out how many odd numbers there are between pairs of numbers. Can you find a pair of numbers that has four odds between them? Largest Even Stage: 1 Challenge Level: Well done for giving clear solutions to this problem. Many of you were able to explain a general strategy. Tom and Jeevan from Devonshire Primary School said: If you are trying to make the largest even number using an odd number you put the odd number in the tens column and in the units column you put $8$. But if you're doing it with an even number then you put $9$ in the tens column and then put the even number in the units column. Peter, Henry, Charlie, Lulu and Samuel at Weald C of E Primary School assumed that you could also get a zero. They went on to think about how you would make the smallest even number: Largest Even 1.If you get an odd number, put an eight in the units column. 2.If you get an even number, put a nine in the tens column. 3.If you get a zero, put a nine in the tens column. Smallest even 1.If you get an odd number, add a zero in the units column. 2.If you get an even number, add a one in the tens column 3.If you get a zero, put a one in the tens column. Homeroom 7 at Kororoit Creek Primary School, Australia looked at a strategy for making the largest odd number too: The largest two-digit odd number we could make is $97$. We could not make $99$ because we only have one $9$. If you pulled out an odd number first this should always go in the 'ones' column. Unless you pull out a number $9$. The $9$ will always go first because $97$ is the biggest number we can make. e.g. $97$, $79$, $59$, $39$, $19$. If you pull out an even number first it should always go in the 'tens' columns, otherwise your two-digit number will not be odd. You should still always choose the number $9$ as your second card to make the largest number possible. e.g. $89$, $69$, $49$, $29$. Well done too to Robert from West Hoathly and Sophie from Greenacre who also sent very full responses. 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": 18, "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": 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": 0.9249377846717834, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/170373/how-do-we-know-taylors-series-works-with-complex-numbers	# How do we know Taylor's Series works with complex numbers? Euler famously used the Taylor's Series of $\exp$: $$\exp (x)=\sum_{n=0}^\infty \frac{x^n}{n!}$$ and made the substitution $x=i\theta$ to find $$\exp(i\theta) = \cos (\theta)+i\sin (\theta)$$ How do we know that Taylor's series even hold for complex numbers? How can we justify the substitution of a complex number into the series? - 3 – J. M. Jul 13 '12 at 15:23 I think it would help if you asked yourself: what does it mean that "Taylor's series hold ..."? A series is just a series; it doesn't "hold", but it may or may not converge. – Marc van Leeuwen Jul 13 '12 at 17:11 ## 2 Answers The series $f(z)=\sum_{n=0}^{\infty} \frac{z^n}{n!}$ is convergent on $C$, and thus it defines an Analytic function. Now there are few different ways to convince yourself that this has to be $e^z$. For once, it is the only Analytic continuation of $e^x$ to the complex plane... Or, alternately, you can prove that $f(z_1+z_2)=f(z_1)f(z_2)$ and $f(1)=e$. Also, you can show that it is the only differentiable function satisfying these two relations. If you prefer differential equations, $f'(z)=f(z)$ and $f(1)=e$ uniquely determine a solution, and bot $e^z$ and $f(z)$ are solutions.... - You define a complex function by the formula $f(z)=\sum_{n=0}^{\infty} \frac{z^n}{n!}$. You prove that it converges everywhere and defines a holomorphic function of $z$. Then you prove that for $z=x \in \mathbb{R}$ it agrees with the usuall exponential. -	{"extraction_info": {"found_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": 0.9286657571792603, "perplexity_flag": "head"}
http://www.purplemath.com/learning/viewtopic.php?f=8&t=668&p=2081	# The Purplemath Forums Helping students gain understanding and self-confidence in algebra. ## Need help solving weirdly phrased quadratic equations Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc. 5 posts • Page 1 of 1 ### Need help solving weirdly phrased quadratic equations by algebraid on Thu Jul 02, 2009 6:57 pm Since this site has already helped me a lot I thought I would seek your help with a quadratic equation that I can't wrap my head around. The problem is phrased as follow (but I am translating from dutch): In the quadratic equation 2x^2 - 3x + p = 0 one of the answers is equal to 2. Calculate p and the other answer. Is it just me or is this pretty weirdly phrased? So I tried solving it with the quadratic formula(x), but there is no answer that equals 2.(or is there?) And I have no idea how to calculate p. The final answer using the quadratic formula is: $\LARGE x=\frac{3\pm\sqrt{-8p+9}}{4}$ If anyone could shed some light on this I would be very grateful. algebraid Posts: 3 Joined: Thu Jul 02, 2009 6:32 pm Sponsor Sponsor by stapel_eliz on Thu Jul 02, 2009 8:01 pm algebraid wrote:In the quadratic equation 2x^2 - 3x + p = 0 one of the answers is equal to 2. Calculate p and the other answer. The Quadratic Formula is probably the best way to go. It gives you: . . . . .$x\, =\, \frac{3\, \pm\, \sqrt{9\, -\, 8p}}{4}$ For this to come out "even", which a nice whole-number value for the solution, the value of 9 - 8p needs to be a perfect square. The squares are 0, 1, 4, 9, 16, 25, 36, etc. Let's see if we can narrow the list down a bit by applying what we know about solving radical equations.... . . . . .$\frac{3\, \pm\, \sqrt{9\, -\, 8p}}{4}\, =\, 2$ . . . . .$3\, \pm\, \sqrt{9\, -\, 8p}\, =\, 8$ . . . . .$\pm\sqrt{9\, -\, 8p}\, =\, 5$ . . . . .$9\, -\, 8p\, =\, 25$ . . . . .$-16\, =\, 8p$ What then is the value of "p"? Plugging this into the Quadratic Formula, what is the other solution? stapel_eliz Posts: 1701 Joined: Mon Dec 08, 2008 4:22 pm ### Re: Need help solving weirdly phrased quadratic equations by algebraid on Fri Jul 03, 2009 10:00 am First, thank you very much for your help. Second, I think the answer to the question is that the solution that is equal to 2 is p = -2. The other answer would be p=0. So p=0,-2. In the original equation p = 2x^2 +3x, so it makes sense to do x first and go from there to calculate p. Can you tell me if this is correct? algebraid Posts: 3 Joined: Thu Jul 02, 2009 6:32 pm by stapel_eliz on Fri Jul 03, 2009 12:53 pm If p = -2, then the quadratic equation becomes 2x2 - 3x - 2 = 0. If x = 0, then the equation is -2 = 0, which is not true. So the other solution cannot be "x = 0". You are given that one solution is "x = 2". This was used to find the value of p. Now that you have found the (one and only) value of p, you need to factor the quadratic, or else use the Quadratic Formula, to confirm the one (given) solution and find the other (requested) solution. stapel_eliz Posts: 1701 Joined: Mon Dec 08, 2008 4:22 pm ### Re: Need help solving weirdly phrased quadratic equations by algebraid on Fri Jul 03, 2009 8:20 pm So if I have this correct: p = -2 x = -0.5, 2 Again thank you so much for your help. I'm really glad I've found you to help me with this, because I'm doing a self-study for an upcoming entrance exam (important!) and I don't have anyone else I can ask to help me with this. Maybe I will speak/write to you again. algebraid Posts: 3 Joined: Thu Jul 02, 2009 6:32 pm 5 posts • Page 1 of 1 Return to Intermediate Algebra • Board index • The team • Delete all board cookies • All times are UTC	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 7, "/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": 0.9382575154304504, "perplexity_flag": "middle"}
http://crypto.stackexchange.com/questions/3367/how-do-i-calculate-crc32-mathematically	# How do I calculate CRC32 mathematically? I want to calcuate the CRC32 (algorithm http://www.libpng.org/pub/png/spec/1.2/PNG-CRCAppendix.html ) using polynomials directly but I don't know how. I found the generating polynomial listed here https://en.wikipedia.org/wiki/Cyclic_redundancy_check (this corresponds to 0xedb88320 in the example code). So please can someone define a mathematical specification that calculates the same result as this algorithm? - 1 – Thomas Jul 29 '12 at 11:04 @Thomas, yes but this does not help, thank you though. – user2558 Jul 29 '12 at 11:09 2 – David Cary Jul 30 '12 at 16:28 ## 1 Answer I guess that CRC is borrowed from the $32$-bit Frame Check Sequence in the 1988 edition of CCITT V.42, section 8.1.1.6.2, available here, which gives a mathematical definition (note: remove the obviously spurious $1$ after $x^{30}$ in the English edition). I prefer this alternate definition with some of the math on polynomial replaced by equivalent operations on bits: 1. Consider the message as a sequence of $n$ bits in chronological order (if the message is structured in words or bytes: with low-order bit first unless otherwise specified). 2. Append $32$ one bits, forming a sequence of $n+32$ bits. 3. Complement the first $32$ bits of that sequence. 4. Form the binary polynomial of degree (at most) $n+31$, with term $x^{n+32-j}$ present (resp. absent) when the $j$th bit in the result of the previous step is one (resp. zero), for $j$ with $0<j\le n+32$. 5. Compute the remainder of the polynomial division of that polynomial by the binary polynomial $x^{32}+x^{26}+x^{23}+x^{22}+x^{16}+x^{12}+x^{11}+x^{10}+x^8+x^7+x^5+x^4+x^2+x+1$, forming a binary polynomial of degree (at most) $31$. 6. Form the $32$-bit sequence with the $j$th bit one (resp. zero) when the term $x^{32-j}$ is present (resp. absent) in that polynomial, for $j$ with $0<j\le 32$. 7. Append that $32$-bit sequence to the ORIGINAL message (if it is to be converted to bytes: the first bit of that sequence shall be the low-order bit of the first byte unless otherwise specified). Note: Inserting $32$ bits at step 2 allows a receiver to process bits of the message and $32$-bit sequence uniformly as they are being received, without knowing the frontier between the message and the final 32-bit sequence until after the end of that sequence. Step 3 makes it likely that suppression of bits in the message is detected including for zero bits at the beginning of the message. Note: In term of binary polynomials (according to the conventions in steps 4 and 6), the combination of steps 2 and 3 changes $M(x)$ to $M(x)\cdot x^{32}+\Sigma_{i=n}^{n+31}{x^i}+\Sigma_{i=0}^{31}{x^i}$. Note: Steps 4, 5 and 6 can be replaced by: • repeat until the sequence has exactly 32 bits: • if the first bit is one: • complement the 7th, 10th, 11th, 17th, 21th, 22th, 23th, 25th, 26th, 28th, 29th, 31th, 32th and 33th bit. • remove the first bit. -	{"extraction_info": {"found_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": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.882468581199646, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2009/10/15/clairauts-theorem/?like=1&source=post_flair&_wpnonce=a082172354	# The Unapologetic Mathematician ## Clairaut’s Theorem Now for the most common sufficient condition ensuring that mixed partial derivatives commute. If $f$ is a function of $n\geq2$ variables, we can for the moment hold the values of all but two of them constant. We’ll only consider two variables at a time, which will simplify our notation. For the moment, then, we write $f(x,y)$. We will also assume that $f$ is real-valued, and deal with vector values one component at a time. I assert that if the partial derivatives $D_xf$ and $D_yf$ are continuous in a neighborhood of the point $(a,b)$, and if the mixed second partial derivative $D_{y,x}f$ exists and is continuous there, then the other mixed partial derivative $D_{x,y}f$ exists at $(a,b)$, and we have the equality $\displaystyle\left[D_{x,y}f\right](a,b)=\left[D_{y,x}f\right](a,b)$ By definition, within the neighborhood in the statement of the theorem the partial derivative $\frac{\partial f}{\partial y}$ is given by the limit $\displaystyle\left[D_yf\right](x,y)=\lim\limits_{k\to0}\frac{f(x,y+k)-f(x,y)}{k}$ So the numerator of the difference quotient defining the desired mixed partial derivative is $\displaystyle\begin{aligned}\left[D_yf\right](a+h,b)-\left[D_yf\right](a,b)=&\lim\limits_{k\to0}\frac{f(a+h,b+k)-f(a+h,b)}{k}\\-&\lim\limits_{k\to0}\frac{f(a,b+k)-f(a,b)}{k}\end{aligned}$ For a fixed $k$, we define the function $\displaystyle g_k(t)=f(a+t,b+k)-g(a+t,b)$ We compute the derivative of $g_k$ as $\displaystyle g_k'(t)=\left[D_xf\right](a+t,b+k)-\left[D_xf\right](a+t,b)$ so we can apply the mean value theorem to write $\displaystyle\left[D_yf\right](a+h,b)-\left[D_yf\right](a,b)=\lim\limits_{k\to0}\frac{g_k(h)-g_k(0)}{k}=\lim\limits_{k\to0}\frac{hg_k'(\bar{h})}{k}$ for some $\bar{h}$ between ${0}$ and $h$. We use the above expression for $g_k'$ to write the difference quotient $\displaystyle\frac{\left[D_yf\right](a+h,b)-\left[D_yf\right](a,b)}{h}=\lim\limits_{k\to0}\frac{\left[D_xf\right](a+\bar{h},b+k)-\left[D_xf\right](a+\bar{h},b)}{k}$ In a similar trick to the one above, we can see that $\left[D_xf\right](a+\bar{h},b+s)$ is differentiable as a function of $s$ with derivative $\left[D_{y,x}f\right](a+\bar{h},b+s)$. And so the mean value theorem tells us that we can write our difference quotient as $\displaystyle\frac{\left[D_yf\right](a+h,b)-\left[D_yf\right](a,b)}{h}=\lim\limits_{k\to0}\left[D_{y,x}f\right](a+\bar{h},\bar{y})$ for some $\bar{y}$ between $b$ and $b+k$. And so we come to try taking the limit $\displaystyle\lim\limits_{h\to0}\lim\limits_{k\to0}\left[D_{y,x}f\right](a+\bar{h},\bar{y})=\left[D_{y,x}f\right](a,b)$ If $\bar{h}$ didn’t depend in its definition on $k$, this would be easy. First we could let $k$ go to zero, which would make $\bar{y}$ go to $b$, and then letting $h$ go to zero would make $\bar{h}$ go to zero as well. But it’s not going to be quite so easy, and limits in two variables like this usually call for some delicacy. Given an $\epsilon>0$, there (by the assumption of continuity) is some $\delta>0$ so that $\displaystyle\lvert\left[D_{y,x}f\right](x,y)-\left[D_{y,x}f\right](a,b)\rvert<\frac{\epsilon}{2}$ for $(x,y)$ within a radius $\delta$ of $(a,b)$. As long as we keep $\lvert h\rvert$ and $\lvert k\rvert$ below $\frac{\delta}{2}$, the point $(a+\bar{h},\bar{y})$ will be within this radius. So we can keep $h$ fixed at some small enough value, and find that $\lvert k\rvert<\frac{\delta}{2}$ implies the inequality $\displaystyle\lvert\left[D_{y,x}f\right](a+\bar{h},\bar{y})-\left[D_{y,x}f\right](a,b)\rvert<\frac{\epsilon}{2}$ Now we can take the limit as $k$ goes to zero. As we do so, the inequality here may become an equality, but since we kept it below $\frac{\epsilon}{2}$, we still have some wiggle room. So, if $\lvert h\rvert<\frac{\delta}{2}$, we have the inequality $\displaystyle\left\lvert\lim\limits_{k\to0}\left[D_{y,x}f\right](a+\bar{h},\bar{y})-\left[D_{y,x}f\right](a,b)\right\rvert\leq\frac{\epsilon}{2}<\epsilon$ which gives us the limit we need. Of course we could instead assume that the second mixed partial derivative exists and is continuous near $(a,b)$, and conclude that the first one exists and is equal to the second. ### Like this: Posted by John Armstrong \| Analysis, Calculus ## 11 Comments » 1. [...] student. In reading this great expository blog on mathematics, I finally found a reference to a named theorem about it. And the name is not Young’s Theorem, as I have been taught. I learn something every day. [...] Pingback by \| October 16, 2009 \| Reply 2. Is this all you do everyday? Write dumb posts that you can get in any book? What is the usefulness of this in real life? Its not helping cure cancer… Comment by John \| October 16, 2009 \| Reply 3. John, I think your comment deserves a thorough answer. I’m going to creat a new post to address it. Comment by \| October 16, 2009 \| Reply • Ok thanks. Bear in mind this is just my opinion. Comment by John \| October 16, 2009 \| Reply 4. [...] Curing Cancer After today’s post on Clairaut’s theorem, a commenter named “John” took offense. Although I’m not sure that’s his [...] Pingback by \| October 16, 2009 \| Reply 5. [...] though, that we have not contradicted Clairaut’s theorem here. Indeed, as long as and all the have continuous second partial derivatives, then so will . [...] Pingback by \| October 19, 2009 \| Reply 6. [...] With Clairaut’s theorem comes the first common example of a smoothness assumption. It’s a good time to say just what [...] Pingback by \| October 21, 2009 \| Reply 7. [...] I said above, this is a bilinear form. Further, Clairaut’s theorem tells us that it’s a symmetric form. Then the spectral theorem tells us that we can find an [...] Pingback by \| November 24, 2009 \| Reply 8. Sorry that this is perhaps too late. In one of the last steps, why do you claim that the inequality may become an equality? Comment by Kiera \| June 24, 2012 \| Reply 9. That’s a good question. It’s a pretty common occurrence with limits. As an example, consider the limit of the sequence $\frac{1}{n}$; each term is strictly greater than zero, but the limit is zero. As a higher level view (if you’re up for it), it’s really about topology. Strict inequalities ($<$) define open sets, while non-strict inequalities ($\leq$) define closed sets. The closure of a set consists of the set itself along with all limit points, so a sequence contained in an open set (satisfying a strict inequality) may have a limit in its closure (satisfying a non-strict inequality). Comment by \| June 24, 2012 \| Reply • Thank you very much for such a quick response! Comment by Kiera \| June 24, 2012 \| Reply « 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": 60, "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": 0.9503232836723328, "perplexity_flag": "head"}
http://mathhelpforum.com/geometry/2319-word-problem.html	# Thread: 1. ## Word problem A box (rectangular/no top) is to have a volume of 64 cubic meters. The width of the base must be 4 meter to insert into a vault. When trimmed to size, material cost \$12.00 per square meter for the base and \$6.00 per square meter for the sides. What is the cost for the box? Thanks 2. Originally Posted by roseh A box (rectangular/no top) is to have a volume of 64 cubic meters. The width of the base must be 4 meter to insert into a vault. When trimmed to size, material cost \$12.00 per square meter for the base and \$6.00 per square meter for the sides. What is the cost for the box? Thanks The width of the box is $4$m, let the height be $h$m and the depth be $d$m. Then to make the volume $64 m^3$ we have: $<br /> 4\times h \times d=64<br />$ The area of the base is $4\times d$ so the cost of the base is: $<br /> 6 \times 4 \times d<br />$ The area of the sides id $(2 \times 4+2\times d)\times h$, so the cost of the sides is: $<br /> 12 \times (2 \times 4+2\times d)\times h =96+24\times d \times h<br />$ but $4\times h \times d=64$, so the cost of the sides is: $<br /> 96+24\times d \times h=96+6 \times (4 \times d \times h)=\$ 480<br />$. Hence the total cost is: $<br /> 480+24 \times d<br />$ dollars and you don't know enough to simplify further. RonL	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 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": 0.8058311939239502, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/293993/lie-bracket-on-vector-fields	# Lie bracket on vector fields I started reading about Lie derivative on vector fields and its properties, found an exercise, but i have doubts about my solution. Given are two vector fields $X_{1}=\frac{\partial }{\partial x_{1}} + x_{2}\frac{\partial }{\partial x_{3}} + x_{2}^{2}\frac{\partial }{\partial x_{4}}$ and $X_{2}=\frac{\partial }{\partial x_{2}}$. Calculate $\left [ X_{1},\left [ X_{1}, X_{2} \right ] \right ]$. OK, i start with $\left [ X_{1},\left [ X_{1}, X_{2} \right ] \right ] = \left [ X_{1}, \left [ X_{1}X_{2}-X_{2} X_{1}\right ] \right ] = X_{1}X_{1}X_{2} - X_{1}X_{2}X_{1} - X_{1}X_{2}X_{1} + X_{2}X_{1}X_{1} = \left [ X_{1}, X_{1}X_{2} \right ]- \left [ X_{1}, X_{2}X_{1} \right ]$ The book with the answers says that the result of $\left [ X_{1},\left [ X_{1}, X_{2} \right ] \right ]$ should be $0$. Is my idea correct? Can we replace $X_{1}X_{2}$ with $X_{2}X_{1}$? I am not sure about this. The other thing that bothers me is that i didn't use the definitions of $X_{1}$ and $X_{2}$ the way they were given in the exercise. Can somebody help me? Thank you very much - ## 2 Answers Your calculation is correct so far, but to proceed further you need to substitute in the specific $X_1$ and $X_2$ that were given. So far your calculation works for any $X_1$ and $X_2$. Remember that $\partial/\partial x_i$ and $\partial/\partial x_j$ commute. - Thank you very much, Ted! – Lullaby Feb 3 at 22:16 Is it true that for these specific $X_1$ and $X_2$, (writing $\partial_z$ for $\frac\partial{\partial z}$), we have $$\begin{align} X_2X_1 &= 1\cdot \partial_{x_3}+2x_2\cdot\partial_{x_4} \\ X_1X_2 &= 0 & ? \end{align}$$ – Berci Feb 3 at 22:18 @Berci That can't be right... $X_2 X_1$ is a second-order operator. It has terms like $\partial_{x_2} \partial_{x_1}$ in it, not $\partial_{x_3}$. – Ted Feb 3 at 23:13 You must understand the definiton: For smooth vector fields $X$ and $Y$, $XY:C^\infty(M)\to C^\infty(M)$ is the linear transformation (not necessarily a vector field) $$f\to X(Yf)$$ (remember that $Yf\in C^\infty(M)$). Hence, as @Ted said, $\frac{\partial}{\partial x_i}\frac{\partial}{\partial x_j}=\frac{\partial}{\partial x_i}\frac{\partial}{\partial x_j}$. The formula $$[X,Y] = \sum_{j=1}^n\sum_{i=1}^n\left(X_i\frac{\partial Y_j}{\partial x_i} - Y_i\frac{\partial X_j}{\partial x_i}\right)\frac{\partial}{\partial x_j},$$ for smooth vector fields $$X=\sum_{i=1}^nX_i\frac{\partial}{\partial x_i}\quad\text{ and }\quad\sum_{i=1}^nY_i\frac{\partial}{\partial x_i}$$ and smooth functions $X_i$ and $Y_i\in C^\infty(M)$, follows from this definition. In fact, for $f\in C^\infty(M)$, we have that $$\begin{array}{rcl} [X,Y]f & = & X(Yf) - Y(Xf) \\ & = & X\left(\sum_{j=1}^nY_j\frac{\partial f}{\partial x_j}\right) - Y\left(\sum_{j=1}^nX_j\frac{\partial f}{\partial x_j}\right) \\ & = & \sum_{i=1}^nX_i\frac{\partial}{\partial x_i}\left(\sum_{j=1}^nY_j\frac{\partial f}{\partial x_j}\right) - \sum_{i=1}^nY_i\frac{\partial}{\partial x_i}\left(\sum_{j=1}^nX_j\frac{\partial f}{\partial x_j}\right) \\ & = & \sum_{i=1}^nX_i\sum_{j=1}^n\left(\frac{\partial Y_j}{\partial x_i}\frac{\partial f}{\partial x_j} + Y_j\frac{\partial^2 f}{\partial x_i\partial x_j}\right) - \sum_{i=1}^nY_i\sum_{j=1}^n\left(\frac{\partial X_j}{\partial x_i}\frac{\partial f}{\partial x_j} + X_j\frac{\partial^2 f}{\partial x_i\partial x_j}\right) \\ & = & \sum_{i=1}^n\sum_{j=1}^nX_i\frac{\partial Y_j}{\partial x_i}\frac{\partial f}{\partial x_j} - \sum_{i=1}^n\sum_{j=1}^nY_i\frac{\partial X_j}{\partial x_i}\frac{\partial f}{\partial x_j} \\ & = & \sum_{j=1}^n\sum_{i=1}^n\left(X_i\frac{\partial Y_j}{\partial x_i} - Y_i\frac{\partial X_j}{\partial x_i}\right)\frac{\partial f}{\partial x_j} \\ & = & \left(\sum_{j=1}^n\sum_{i=1}^n\left(X_i\frac{\partial Y_j}{\partial x_i} - Y_i\frac{\partial X_j}{\partial x_i}\right)\frac{\partial}{\partial x_j}\right)f. \end{array}$$ Using this formula in your case, we get $[X_1,X_2] = 0$ and, hence, $[X_1,[X_1,X_2]]=0$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 33, "mathjax_display_tex": 5, "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": 0.9520232677459717, "perplexity_flag": "head"}
http://mathoverflow.net/questions/32413/when-is-a-fixed-point-of-fn-a-fixed-point-of-f	## When is a fixed point of f^n a fixed point of f? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $E$ be a Banach space and $f:E\to E$ be a continuous map. By $f^n$ we denote the $n$-th iterate of $f$, i.e. $f^n:=\underbrace{f\circ f\circ\cdots \circ f}_{\text{n times}}$. Let $x_0$ denote a fixed point of $f^n$. 1) Is it true that the question as to when $x_0$ is a fixed point of $f$ has been resolved in the case of $E$ being reflexive? (I would be also thankful for some reference.) 2) What are considered to be the main obstacles in the case of $E$ being not reflexive when dealing with the above question (besides the obvious)? (As far as I understand, the problem is still open for general Banach spaces. Again, I would be thankful for a reference.) A side remark: I realize that my questions may appear a bit too general. I am also aware of the fact that the field (fixed-point theory in banach spaces) is overflowed with publications of questionable value/quality/contribution like no other, which makes things even harder particularly for someone who is coming from a different field (e.g. number theory). Thus MO seems to be the only reasonable place to get a clear and compact answer from someone who has the overlook over the field. Thanks! - I think retagging this to "non-linear-functional-analysis" and/or "fixed-point-theory" would be more suitable, but unfortunately I cannot create new tags. – geeky Jul 18 2010 at 22:15 ## 1 Answer Actually I can't see the role of reflexivity for an answer to the question as it is, unless further properties on $f$ are assumed. In general, to start with the obvious case: if $f$ is a contraction, it has a fixed point, which is also the unique fixed point of the contraction $f^n$, therefore it is $x_0$, so $x_0$ is a fixed point of $f$. Of course, if $f$ is not a contraction, like e.g. the map $x\mapsto -x$, a fixed point of $f^n$ need not be a fixed point of $f$. Also, in a certain sense, the case of contractions essentially covers all cases where the answer is affirmative: by a result of Bessaga of around '60, if a map $f$ on a set $S$ is such that for all $n\in \mathbb{N}$ the iterated $f^n$ has a unique fixed point, then $f$ is a contraction with respect to a suitable complete metric on $S$. - It is a trivial remark (not due to me) that if a mapping $f$ on a set $X$ (no metric required) has a unique fixed point, then this is also a fixed point for any mapping which commutes with $f$. This implies immediately that a mapping on a complete metric space has a fixed point whenever some iterate is a contraction. No continuity assumption is required. – jbc Sep 18 at 19:48 (or, also: if $f^n$ is a contraction on a complete metric space, it has a unique fixed point $x$; but $f(x)$ is a fixed point of $f^n$ too, so by the uniqueness $x=f(x)$.) – Pietro Majer Jan 10 at 14:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9396603107452393, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/1957/what-is-the-difference-between-a-white-object-and-a-mirror/16944	# What is the difference between a white object and a mirror? I was taught that something which reflects all the colors of light is white. The function of a mirror is the same, it also reflects all light. What's the difference? Update: But what if the white object is purely plain and does not scatter light? Will it be a mirror? - 1 No, it's an excellent question! Galileo analyzed this question in his Dialogue and uses the scattering answer given below to arrive at the conclusion that the Moon's surface is not like a shiny mirror (as people believed at that time). – recipriversexclusion Apr 6 '12 at 15:08 ## 6 Answers The difference is the direction the light is emitted in. Mirrors 'bounce' light in a predictable direction, white objects scatter light. - +1 You beat me by a few seconds with a clear explanation in one sentence. – Frédéric Grosshans Dec 15 '10 at 17:03 1 @dan, @Frédéric: you both beat me by few seconds but I realized that there is something missing in both your answers, so I posted mine anyway. I hope no one will mind :-) – Marek Dec 15 '10 at 17:10 1 @Marek, If you find you're not fastest, most full seems like a reasonable stance. – dan_waterworth Dec 15 '10 at 17:12 2 @LifeH2O, I think the problem you have understanding is because of your concept of colour. A colour is a set of frequencies of light. When you see a green object, it's because all of the other frequencies have been absorbed by the object and only green light is re-emitted. With a white object, all frequencies of light are re-emitted. By this definition, pure mirrors are white because they re-emit all frequencies of light. With this in mind I think you may understand some of the answers a little better. – dan_waterworth Dec 16 '10 at 10:19 1 Thanks, You should add this detail in your answer. – LifeH2O Dec 16 '10 at 20:36 show 1 more comment A white object reflects the light in all the directions, independently of the original direction. It is called a diffuse reflection. If you shine a beam of light onto a white surface, it is scattered in all the directions. On the other hand, a mirror reflect the light symmetrically to the input direction, with no scattering. This is what allows us to see "through" the mirror. Edit to add: "Normal" coloured object all have a diffuse reflection, their colour being given by the mix of wavelength they reflect. To better understand the difference between the two kind of reflections, you can also read the related question : "Why can you have shiny black objects?" (and the answers !) - 1 The physical surface may not necessarily have to be rough, most materials are actually translucent, i.e. the light penetrates some depth before being scattered. Think of snow. Sunlight usually penetrates several centimeters before being scattered out. So it is not the surface roughness, but the photon by photon variability of the scattering depth (plus many will be scattered multiple times before exiting). Or for a larger example, light being scattered from a fog bank. A highly conductive material, like metal prevents this from happening and reflects a plane wave with a plane wave. – Omega Centauri Dec 15 '10 at 18:31 I think there is some other difference, the thing on the backside of mirror which cause reflection will continue to reflect even if it is not plain. Right? – LifeH2O Dec 16 '10 at 10:02 @Omega: very good point. But you'd still need the surface to be plain. So I am saying this is most important feature. A metallic surface which is not polished will not act as a mirror. This statement can be verified by just looking around. – Marek Dec 16 '10 at 10:16 1 @Omega, as for the snow: it has fractal structure visible by a naked eye (if you'll observe the snowflake really closely). So it is definitely bumpy from the point of view of the visible light and therefore the light will be scattered already on the surface, notwithstanding any penetration. Therefore it's not a good counter-example against "roughness criterion" at all. – Marek Dec 16 '10 at 10:19 1 @ Omega A real white has to have a rough surface. If You have eg a glossy white laquer, there is a combination of specular reflectance at the surface and diffuse from "inside" the laquer. Take for example practical white standards: some powder (often MdO) is compressed against a glas plate, the latter is then removed. This MgO-"tablet" then is used as a reference when measuring "whiteness" of paper or pulp. – Georg Nov 14 '11 at 20:18 Other answers are correct except that they forget to mention why white objects behave differently than mirrors. Well, generic white objects have very rough bumpy surfaces. This causes scattering. On the other hand, mirror has a very even and polished surface so that the simple law for specular reflection of incoming rays holds. Of course, you can also obtain any kind of roughness in between. Actually, you can produce a mirror by starting with a generic white object and polishing it. - 3 People say I'm a generic white object. Can I become a mirror? – Mark Eichenlaub Dec 15 '10 at 17:23 @Mark: I hate these trick questions :-) – Marek Dec 15 '10 at 18:27 You mean if the white object is purely plain like a mirror, it will act like a mirror? – LifeH2O Dec 16 '10 at 10:01 1 Well, the question doesn't make sense because such an object wouldn't be white anymore. But if you mean white as "reflective in visible range" then my answer is yes. And to clarify: plain from the point of view of the visible light (that is on the scale of microns). It will still be rough on the atomic level, of course. – Marek Dec 16 '10 at 10:11 To complicate matters, a reasonably smooth flat surface, white or otherwise, becomes a mirror at grazing incidence (the angle of incidence, and the angle of reflection, are close to 90$^\circ$ from the normal direction). This effect often finds application in focusing x-rays, but it works at visible and other wavelengths too. You can see this by putting your eyelevel just above a desk or table, and looking for the reflection of an object placed a cm above the surface at a distance of a few tens of cm away. - I think that maybe its not really on how smooth the surface is because white paint can be as smooth and yet not act like a mirror. Then again snow acts like shattered glass where broken individual snowflakes reflect sunlight making snow look white. It's mainly in the absorption spectrum or the molecular/bonding structure that determines the color of an object, and maybe how the object scatters light, which the electron levels either absorb, reflect, transparent, and/or refract a little bit, some, most, and/or all light. I do think that maybe white objects reflect and scatter white light in all and random directions while mirrors reflect all light in an evenly straight direction. But I do notice this. Let's say white light coming from the original light source, such as the sun or a light bulb, comes into a plant. The chlorophyll in the leaves will absorb most of the colors in the light and will reflect and scatter mainly green light. When the green light reaches our eyes we see the plant that it is green. If the plant is next to a mirror then the mirror will only reflect the green light from the plant in an evenly straight direction and once it reaches our eyes then we see another image of the green plant in the mirror. If the plant is next to a white object such as a white wall then the white object may absorb the green light reflecting out of the plant, unlike a mirror, but the white object will also reflect and scatter white light from the original light source. So we see the wall as white but we don't see another image of the green plant in the wall. Maybe there are other ways that a mirror is different from a white object but I hope I answered the question. - White color is associated with reflected (not absorbed) light. White paint usually includes a titanium oxide component, whose absorption is in UV. The difference between a shiny surface and a Lambertian surface is its roughness, if light is reflected in a collective manner it look shiny. Any smooth surface is shiny given a grazing angle. A mirror is usually coated with a metal surface, silver for example. Silver's metallic behavior makes most of the light reflected back (about %95 percent reflection, which is why it has a greyish tint), however it becomes transparent for UV light. A mirror can be engineered to have more reflectance than silver by using multi-layer reflection and interference, using dielectric materials instead of metal. They are called dielectric mirrors. To sum up, white color is due to non absorbed white light, reflection is due to non transmitted light. - 1 Welcome to Physics Stack Exchange! – Manishearth♦ Apr 6 '12 at 8:26 Thank you very much! – Ertuğrul Karademir Apr 6 '12 at 13:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9319361448287964, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/coloring+combinatorics	# Tagged Questions 0answers 106 views ### (3n,n)-Turán graph [closed] I'm working on a problem regarding (kn,n)-Turán graphs. The (2n,n)-Turán graph, also known as the cocktail party graph, has a closed formula for its number of spanning trees. I want to know if there ... 4answers 140 views ### Colouring a chessboard How can I demonstrate that I can colour a $2n\times\binom{2n}{2}$ chessboard, with $n$ different colours, such that there aren't $4$ separate unit squares of the same colour, the centers of which are ... 1answer 71 views ### Coloring numbers from $1$ to $1000$ I mostly just need someone to explain to me this problem: Prove that it is possible to $2$-color the integers from $1$ to $1000$ so that no monochromatic arithmetic progression of length $17$ is ... 1answer 48 views ### example of a finite coloring without infinite monochromatic set closed under addition I am studying some theorems on combinatorial set theory, especially Ramsey theorem and Hindman's theorem. I think I am going to ask a silly question, but I am too much involved in the subject to think ... 1answer 94 views ### Triangle free graphs with large chromatic number I am trying to understand the proof of Theorem 2 given here. (Page 5) The theorem states that $\forall k\exists$ a triangle free graph $G$ with $\chi(G)>k$. The proof constructs such a $G$ as ... 1answer 173 views ### Planar graph with a chromatic number of 4 where all vertices have a degree of 4. Is it possible to have a planar graph with a chromatic number of $4$ such that all vertices have degree $4$? Every time I try to make the degree condition to work on a graph, it loses its planarity. 1answer 99 views ### Can a planar graph without two triangles that share an edge have a chromatic number larger than 3? Let G be a square with one diagonal. Are there any planar graphs without G as a subgraph that are not 3-colourable? 1answer 84 views ### Coloring of positive integers Suppose $f:\mathbb{Z}^+\longrightarrow X$ is a function, with $X$ a finite set. Is it true that there are $a,b\in\mathbb{Z}^+$ such that $f(a)=f(b)=f(a+b)$. 1answer 63 views ### Coloring points in a cycle I have a question that relates to the Widom-Rowlinson model of statistical physics. Take a cycle on $n$ vertices. How many ways are there to color the $n$ vertices with the colors \$\{\text{Red, ... 0answers 65 views ### Monochromatic degenerate triangles in a two-coloring of the plane In a similar vein to a question I asked a few days ago: Do all two-colorings of $\mathbb{R}^2$ contain three points of the same color which form the vertices of a degenerate triangle of side-lengths ... 0answers 96 views ### Monochromatic triangles in a two-coloring of the plane A problem posed to me by a friend: Show that any two-coloring of $\mathbb{R}^2$ that contains a monochromatic equilateral triangle of side-lengths 1 also contains monochromatic triangles of all side ... 1answer 130 views ### Edge colorings of complete graphs without tricolored triangles Please prove the following theorem from Gallai : Theorem .In every coloring of a complete graph with three colors that avoiding rainbow triangle , at least one of the color classes must be ... 2answers 392 views ### Checkerboard-Coloring $\mathbb{Z}^2$ If every square of the unit square lattice in the plane is colored black or white according to a set of rules, is there a way to find the maximum asymptotic ratio $r_n$ of the number of black squares ... 1answer 67 views ### References for analogues of chromatic polynomials where colorings which differ only by permutation of colors are counted as the same It's well-known that chromatic polynomials count colorings which differ by permutations of colors. What is known about their analogues which don't count such colorings as distinct? 2answers 141 views ### Coloring points on an n-gon Given an $n$-sided polygon, how many ways can you color the vertices using $k$ colors so that no two adjacent vertices have the same color? (Inspired by 2011 AMC 12 A #16 – I'm able to do this for ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 27, "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": 0.9371161460876465, "perplexity_flag": "head"}
http://en.wikiversity.org/wiki/Introduction_to_Elasticity/Torsion_of_triangular_cylinder	# Introduction to Elasticity/Torsion of triangular cylinder From Wikiversity ## Example: Equilateral Triangle Torsion of a cylinder with a triangular cross section The equations of the three sides are $\begin{align} \text{side}~\partial S^{(1)} ~:~~ & f_1(x_1,x_2) = x_1 - \sqrt{3} x_2 + 2a = 0 \\ \text{side}~\partial S^{(2)} ~:~~ & f_2(x_1,x_2) = x_1 + \sqrt{3} x_2 + 2a = 0\\ \text{side}~\partial S^{(3)} ~:~~ & f_3(x_1,x_2) = x_1 - a = 0 \end{align}$ Let the Prandtl stress function be $\phi = C f_1 f_2 f_3 \,$ Clearly, $\phi = 0\,$ at the boundary of the cross-section (which is what we need for solid cross sections). Since, the traction-free boundary conditions are satisfied by $\phi\,$, all we have to do is satisfy the compatibility condition to get the value of $C\,$. If we can get a closed for solution for $C\,$, then the stresses derived from $\phi\,$ will satisfy equilibrium. Expanding $\phi\,$ out, $\phi = C (x_1 - \sqrt{3} x_2 + 2a)(x_1 + \sqrt{3} x_2 + 2a)(x_1 - a)$ Plugging into the compatibility condition $\nabla^2{\phi} = 12 C a = -2\mu\alpha$ Therefore, $C = -\frac{\mu\alpha}{6a}$ and the Prandtl stress function can be written as $\phi = -\frac{\mu\alpha}{6a} (x_1^3+3ax_1^2+3ax_2^2-3x_1x_2^2-4a^3)$ The torque is given by $T = 2\int_S \phi dA = 2\int_{-2a}^{a} \int_{-(x_1+2a)/\sqrt{3}}^{(x_1+2a)/\sqrt{3}} \phi dx_2 dx_1 = \frac{27}{5\sqrt{3}} \mu\alpha a^4$ Therefore, the torsion constant is $\tilde{J} = \frac{27 a^4}{5\sqrt{3}}$ The non-zero components of stress are $\begin{align} \sigma_{13} = \phi_{,2} & = \frac{\mu\alpha}{a}(x_1-a)x_2 \\ \sigma_{23} = -\phi_{,1} & = \frac{\mu\alpha}{2a}(x_1^2+2ax_1-x_2^2) \end{align}$ The projected shear stress $\tau = \sqrt{\sigma_{13}^2+ \sigma_{23}^2}$ is plotted below Stresses in a cylinder with a triangular cross section under torsion The maximum value occurs at the middle of the sides. For example, at $(a,0)$, $\tau_{\text{max}} = \frac{3\mu\alpha a}{2}$ The out-of-plane displacements can be obtained by solving for the warping function $\psi$. For the equilateral triangle, after some algebra, we get $u_3 = \frac{\alpha}{x_2}{6a} (3x_1^2 - x_2^2)$ The displacement field is plotted below Displacements $u_3\,$ in a cylinder with a triangular cross section.	{"extraction_info": {"found_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": 21, "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": 0.8422389030456543, "perplexity_flag": "middle"}
http://mathhelpforum.com/algebra/4970-graphing-inequalities.html	# Thread: 1. ## Graphing inequalities x>4 - graph the solution for each set of inequalities 3x[U]>2x-4 - solve and graph the solution set of each inequalities I just need for someone to show and explain how to do these and I can try to do the rest on my own. Any help will be greatly appreciated THANKS 2. Do you mean on a graph or number line? 3. Originally Posted by Quick Do you mean on a graph or number line? Yes --- In the first problem just, draw a circle on top of 4 (to show you exclude 4) and draw a lined arrow to the right. In the second one, solve for all x, $3x>2x-4$ $3x-2x>2x-2x-4$ $x>-4$ That means a circle above -4 and draw an arrowed line to the right. 4. Hello, foofergutierrez! You should be familiar with Solving Inequalities. Solve for the variable as you would with an equation except: . . when multiplying or dividing by a negative, reverse the inequality. Graph the solution for: . $x > 4$ This is the set of all numbers greater than 4. . . . $- - - - \circ\!\!=\!=\!=\!=\!=\!=$ . . . . . . . . . $4$ The circle $(\circ)$ indicates that the $4$ is not included. Graph the solution for: . $3x \,\geq \,2x - 4$ Solve for $x:\;\;3x\:\geq \:2x - 4\quad\Rightarrow\quad x \:\geq \:-4$ This is the set of all numbers greater than or equal to $-4.$ . . . $- - \bullet\!\!=\!=\!=\!=\!=\!=$ . . . . . $-4$ The solid dot $(\bullet)$ indicates that the $-4$ is included. 5. ## THANKS but I need more help Ok I kind of understand what to do but here are some other practice equations I would like to see done before my attempt to work my problems so that I know that I am doing them right. 7(x-3>5x-14 2(x+28)<6(x) I also have this weird word problem that I cannot make any sense of. The cost for a long distance call is \$.36 for the first minute and \$.21 for each additional minute. Write an equality representing the number of minutes a person could talk without exceeding \$3.00. Again, I thank all who can help me and if I could I would pay ou back somehow! You guys are awesome!!! 6. Originally Posted by foofergutierrez 7(x-3)>5x-14 Open parantheses, $7x-21>5x-14$ Remove the x's $2x-21>-14$ Add 21 to both sides, $2x>7$ Divide by 2, $x>3.5$ --- 2(x+28)<6(x)] Open parantheses, $2x+56<6x$ Remove the x's (subtract 2x), $56<4x$ Divide by 4, $14<x$ 7. ## Thank you Perfect Hacker Thanks a million! It makes perfect sense to me now, you are a lifesaver . 8. Originally Posted by foofergutierrez you are a lifesaver I know. From where I come from everyone call me "God". 9. Hello again, foofergutierrez! The cost for a long distance call is \$0.36 for the first minute and \$0.21 for each additional minute. Write an equality representing the number of minutes a person could talk without exceeding \$3.00. Let $x$ be the number of minutes we can talk for \$3.00 or less. We are charged $36\!\!\not\!c$ for the first minute. We are charged $21\!\!\not\!c$ each for the other $x - 1$ minutes. . . This costs us: . $21(x - 1)$ cents. Hence, the total charge is: . $36 + 21(x - 1)$ cents Since we do not want to exceed \$3 $(300\!\!\not\!c)$ . . we have: . $36 + 21(x - 1)\:\leq \:300$ Solve for $x:\;\;36 + 21x - 21 \:\leq\:300$ . . . . . . . . . . . . . $21x + 15\:\leq\:300$ . . . . . . . . . . . . . . . . $21x \:\leq \:285$ . . . . . . . . . . . . . . . . . . $x\:\leq \:13.5714...$ Therefore, we can talk up to 13 minutes without going over the \$3 limit.	{"extraction_info": {"found_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": 34, "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": 0.8987412452697754, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/eulers-constant+probability	Tagged Questions 1answer 63 views exponential population growth models using $e$? Im trying to understand this write up [1] of cell population growth models and am confused about the use of natural logarithms. If cells double at a constant rate starting from 1 cell, then their cell ...	{"extraction_info": {"found_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": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8918845057487488, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/207096/relation-average-squared-distance-to-all-points-and-squared-distance-to-ce	# Relation: average (squared) distance to all points, and (squared) distance to centroid Suppose a set of $n$ high-dimensional points is given. It is known that the sum of all pair-wise squared Euclidean distances is proportional to sum of squared distances of all points to the centroid. However, given a specific point $a$, in what relation is the sum of squared Euclidean distances from $a$ to all other points, and the square Euclidean distance of $a$ from the centroid of all points (including $a$). - ## 1 Answer Since $(x-a)^2=(x-m+m-a)^2=(x-m)^2+(m-a)^2+2(x-m)(m-a)$ and the sum of the last term over all points $x$ vanishes by the definition of the centroid $m$, the sum over the squared distances from $a$ is the sum over the squared distances from the centroid plus $n$ times the square of the distance from $a$ to the centroid. This holds irrespective of whether $a$ is itself one of the points. See also the parallel axis theorem. -	{"extraction_info": {"found_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": 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": 0.9371837973594666, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2007/06/12/equalizers-and-coequalizers/?like=1&_wpnonce=9501e873bf	# The Unapologetic Mathematician ## Equalizers and coequalizers Let’s consider another construction from set theory. If we have sets $S$ and $T$, and functions $f:S\rightarrow T$ and $g:S\rightarrow T$, then we can talk about the subset $E=\{s\in S\|f(s)=g(s)\}$. This is what we want to generalize. First off, we know that subsets are subobjects, which are monomorphisms. More to the point, we can look at this subset and take its inclusion function $e:E\rightarrow S$. Then we see that $f\circ e=g\circ e$. Furthermore, if any other function $h:X\rightarrow S$ satisfies $f\circ h=g\circ h$ then its image must land in $E$. That is, the function $h$ must factor through $e$. So, given arrows $f$ and $g$, each from the object $S$ to the object $T$ in the category $\mathcal{C}$, we construct a new category. The objects are arrows $h:X\rightarrow S$ satisfying $f\circ h=g\circ h$, and a morphism from $(X_1,h_1)$ to $(X_2,h_2)$ is an arrow $m:X_1\rightarrow X_2$ so that $h_1=h_2\circ m$. Now we define the “equalizer” of $f$ and $g$ to be a couniversal object in this category, if one exists. To be a bit more explicit, look at this diagram: The equalizer is the pair $(E,e)$ so that for any other $(X,h)$ there is a unique arrow $X\rightarrow E$ making the triangle commute. We write it as $\mathrm{Equ}(f,g)$. Since it’s defined by a universal property, the equalizer is unique up to isomorphism when it exists. If it exists for all pairs of morphisms between the same two objects in the category $\mathcal{C}$, we say that $\mathcal{C}$ “has equalizers”. Now, as indicated above an equalizer is a monomorphism into $S$. Indeed, let’s say we’ve got two arrows $h_1:X\rightarrow E$ and $h_2:X\rightarrow E$ so that $e\circ h_1=e\circ h_2$. Then clearly $f\circ(e\circ h_1)=g\circ(e\circ h_1)$, since $e$ is the equalizer of $f$ and $g$. So there is a unique morphism $m:X\rightarrow E$ so that $e\circ m=e\circ h_1=e\circ h_2$, and both $h_1$ and $h_2$ must be this unique morphism. Since we can cancel $e$ on the left of $e\circ h_1=e\circ h_2$, $e$ is a monomorphism. The dual notion of an equalizer is a coequalizer. This uses the following diagram: Go through the above discussion of equalizers and dualize it. Describe a category whose universal object will be the coequalizer. Give an interpretation to this diagram. Prove that the coequalizer of two morphisms is an epimorphism. Try to give a description of coequalizers in $\mathbf{Set}$, or show that $\mathbf{Set}$ does not have coequalizers. ### Like this: Posted by John Armstrong \| Category theory ## 2 Comments » 1. [...] on the source category, limits give whole families of useful constructions, including (multiple) equalizers, (multiple) products, (multiple) pullbacks, and more we haven’t talked about yet. The dual [...] Pingback by \| June 19, 2007 \| Reply 2. [...] we say that a presheaf is a sheaf if and only if this diagram is an equalizer for every open cover . For it to be an equalizer, first the arrow on the left must be a [...] Pingback by \| March 17, 2011 \| Reply « 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": 48, "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": 0.9134513735771179, "perplexity_flag": "head"}
http://mathoverflow.net/questions/5321/flips-in-the-minimal-model-program/5459	## Flips in the Minimal Model Program ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In order get a minimal model for a given a variety , we can carry out a sequence of contractions in such a way that that every map contracts some curves on which the canonical divisor is negative. Here we have, at least, the following technical problem: in contracting curves, the resulting variety might have become singular. In order to fix this fact the people consider a flip. Here are my questions. What is the intuition to understand such a flip? Are there examples of such things in other contexts of math or is it an ad hoc construction? - Out of curiosity, does anyone know if this kind of flip has any relation to the "flips" considered in Mulmuley's geometric complexity theory program? – Harrison Brown Oct 4 2010 at 4:29 1 To answer Harrison's question in the above comment: I don't think there is any relation other than the word itself. If I understand correctly, in Mulmuley's work "flip" refers to a formal translation of a statement in logic so it has nothing to do with anything actually geometric. I suppose the fact that he uses algebraic geometry in complexity theory may be confusing, but I think his flips are not geometric. – Sándor Kovács Oct 10 2010 at 23:59 ## 5 Answers I am also just learning this stuff, and I'm partly writing this out for my own benefit. Experts, please correct and up/down vote as appropriate! The goal of the minimal model program is to give a standard, nonsingular, representative for each birational class of algebraic variety. As stated, this goal is too ambitious, but it will help us to understand the minimal model program if we think of it as a partially successful attempt at this goal. Let $X$ be a compact, smooth algebraic variety of dimension $n$. Let $\omega$ be the top wedge power of the holomorphic cotangent bundle. Then the vector space, $V:=H^0(X, \omega)$, of holomorphic $n$-forms on $X$ is a birational invariant of $X$. This means that we should be able to see $V$ from just the field of meromorphic functions on $X$; here is a sketch of how to do that. So we get a rational map $X \to \mathbb{P}(V^{})$ by the standard recipe. More generally, we can replace $\mathbb{P}(V)$ with Proj of the ring $\bigoplus H^0(X, \omega^{\otimes n})$. This is called the canonical ring; you may have heard of the recent breakthrough in proving that the canonical ring is finitely generated. We can map $X$ rationally to this Proj; the image is called the log model. This is a partial success: it is a canonical, birational construction, but it may not be birational to $X$ and may not be smooth. There are certain well understood rules of thumb for how various subobjects of $X$ behave in the log model. For example, if $X$ is a surface and $C$ a curve with negative self intersection, then $C$ will be blown down in the log model. Here is a more complicated example, which is relevant to your question. Let $Y$ be some variety that locally looks like the cone on the Segre embedding of $\mathbb{P}^1 \times \mathbb{P}^1$. So $Y$ is a $3$-fold with an isolated singularity. If you are familiar with the toric1 picture, it looks like the tip of a square pyramid. Inside $Y$, let $Z$ be the cone on one of the $\mathbb{P}^1$'s. This is a surface, but not a Cartier divisor. Let $X$ be $Y$ blown up along $Z$; so that the isolated singularity becomes a line. In the toric picture, the point of the pyramid has lengthened into a line segment, and two of the faces which used to touch at the point now border along an entire edge. In the log model, the line will blow back down to become a point. So the log model can turn a smooth variety, like $X$, into a singular one like $Y$. Now, birational geometers did not rest on their laurels when they had constructed the log model. They made other constructions, which are smoother but less canonical. Many of these constructions can be thought of as taking the log model and modifying it in some way. If the log model looks like the example of the previous paragraph, they want to take the singular point of $Y$ and replace it by a line, to look like $X$. But they have two ways they can do this; they can blow up one $\mathbb{P}^1$ or the other; giving either $X$ or $X'$. Often, replacing $X$ by $X'$ is crucial in order to improve the model somewhere else. The relationship between $X$ and $X'$ is called a flip, because we take the line inside $X$ and flip it around to point in a different direction. 1 Cautionary note: although the toric picture is excellent for visualizing what is going on locally, you shouldn't take $X$ itself to be a toric variety. There are no global sections of $\omega$ on a toric variety, so the log model is empty. You want $X$ to locally look like a toric variety, but have global geometry which is nontoric in a way that creates lots of sections of $\omega$. - 1 Hi. I'm trying to learn about these things too. Could you tell us what are the readings that you are following? – Franklin Oct 11 2010 at 11:05 ### 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 goal of the mmp is to find a representative in every birational class that for some reason may be considered nice. For curves the answer is clear, there is a unique smooth projective representative and by any consideration that is the one that represents the class best. For surfaces this gets complicated as there are non-trivial birational maps between smooth projective surfaces. However, since they is always a combination of blow-ups and blow-downs it is relatively easy to keep order. Observe that a $(-1)$-curve is usually defined as a curve isomorphic to $\mathbb P^1$ having self-intersection number $(-1)$. A perhaps better definition that points to higher dimensional equivalents is that a $(-1)$-curve is a curve isomorphic to $\mathbb P^1$ having an intersection number $(-1)$ against $K_X$ where $X$ is the surface on which the curve lives. These two definitions are equivalent by the adjunction formula, but the latter one has the advantage that it does not depend on $X$ being a surface. Let's take a look at a minimal model of a surface. Why do we pick that as our representative? In some sense there might be other ways to pick a representative, but one might argue that a minimal model is the "simplest" model that is still smooth (make a note of this, we will realize later that here smoothness is actually something else in disguise). Castelnuovo's theorem about blowing down $(-1)$-curves says that we can "get rid of them", so why not do that. Let's contract everything we can. It can be proven relatively easily that contracting a curve that is not a $(-1)$-curve will lead to singular points. OK, so the strategy is to contract as much stuff as we can and hope that this way we get a reasonable theory. The second definition of a $(-1)$-curve suggests that to find what we can contract is through $K_X$, that is, things that can be contracted and not cause too much trouble are $K_X$-negative. In fact there is a more precise way to say this, but let me not get into technical details now. So, either this way or already for surfaces one realizes that what makes a minimal model tick is that $K_X$ is nef, that is, intersecting with any proper curve gives a non-negative number. So, now you say that $\mathbb P^2$ is a minimal surface but $K_X$ is negative ample so this is pretty far from being nef. Yes, in the modern terminology of the mmp, $\mathbb P^2$ is actually not minimal. The claim is that every variety is birational to one that is a series of Fano fiber spaces over a minimal variety. Perhaps I should mention an interesting example here, I think it is due to Iitaka, or someone from his school: Take a $3$-dimensional abelian variety $A$ and mod out by the involution $(-1)\cdot$. Resolve the resulting $64$ double points and call the result $X$. Then it is relatively easy to prove that $X$ is not birational to a smooth projective variety with a nef canonical bundle. At the time this was thought of as proof that minimal models did not exist in higher dimensions, but then Reid and Mori realized that it only means that minimal models need not be smooth. (N.B.: The above accepted answer of David starts by saying that a minimal model should be non-singular. He says it is too ambitious, but it may not be absolutely clear to everyone that this means impossible--as stated. And I promised a comment about why $2$-dimensional minimal models are smooth. The thing is, minimal models have no worse than terminal singularities. It turns out that terminal singularities are smooth in codimension $2$, so in particular a $2$-dimensional terminal singularity is actually smooth. So, one could argue that even minimal models of surfaces have terminal singularities, that is, that's the natural class of singularities for a minimal model. It just so happens that in dimension $2$, these singularities are indistinguishable from smooth points.) Anyway, so we want $K_X$ to be nef and to obtain this we want to contract curves that are $K_X$-negative. It so happens that this can be done, but this is the result of some very deep results by Mori, Kollár, Kawamata, Reid, Shokurov and others. Now, already in dimension $2$ we get more than just blowing down $(-1)$-curves: the ruling map of a ruled surface and $\mathbb P^2$ mapping to a point are both contractions of $K_X$-negative curves. In general this is how we might end up with a Fano fibre space. It is possible that the contraction of a $K_X$-negative curve is not birational, but that's OK. This really means that the cycle class of that curve covers the entire $X$ and in particular it is uniruled and will never have a minimal model in the sense of $K_X$ being nef. If the contraction is birational, then there are still two possibilities: it is a divisorial contraction or a small contraction. The former means that the exceptional set is a divisor, the latter that it is smaller than that. Now, already the former can bring in singularities, but they are not so bad and the program can continue. When the contraction is small, there are several problems. Simply put the singularities become too bad. The badness mainly manifests itself in the singularity being non $\mathbb Q$-Gorenstein, that is, $K$ will no longer be $\mathbb Q$-Cartier which is otherwise needed. And it's not that this may be so, but it will be so for certain: if the target had a $\mathbb Q$-Cartier $K$, it could be pulled back, at least numerically (or some power could be pulled back). The pull-back would have to agree with $K$ upstairs since the map is an isomorphism in codimension $1$. However, a pull-back is necessarily trivial on the fiber of the map, but the fiber was chosen to be $K$-negative. This is a contradiction, so the target cannot have a $\mathbb Q$-Cartier canonical sheaf. Flips were invented to remedy this situation: the original reason for wanting to contract was to "get rid" of this $K$-negative curve, so let's get rid of it a different way. Being $K$-negative is really a curvature condition and it says something about the normal bundle of the curve inside the variety. (OK, you have to adjust this slightly for singularities, but I am not writing a precise paper here). So, the idea of the flip is this: let's change the normal bundle of the curve. So, let's "cut it out" and put it back with the opposite normal bundle, so in a "flipped" way. (Remark: this is the $3$-dimensional picture, in higher dimensions it's not just curves that get flipped, but this may be better delegated to another place). I guess I wrote a whole bunch of things just to say that and some people have said similar things already, but perhaps this little essay gives some new insight. To answer your question about whether a similar construction exists elsewhere, the answer is "yes". A "flip" is like a "surgery" in topology. But I am no expert on that. Actually, just to include a disclaimer: I am not claiming to be an expert on flips either. - Thank you --- this is indeed very helpful. – Emerton Oct 4 2010 at 4:57 This is a comment to Charles's answer, but I need more room than what comments allow. Anyway, what "glue back differently" means is that the curve is "glued back" with its normal bundle "reversed". There is also an algebraic way to think about flips: If $f:X\to Y$ is a contraction, then $X$ can be considered as ${\rm Proj}_Y\sum_{m=0}^\infty f_\mathcal O_X(-mK_X)$. Now if $f$ is small, then the flip of $f$ is given by the morphism $f^+: X^+={\rm Proj}_Y\sum_{m=0}^\infty f_*\mathcal O_X(mK_X)\to Y$. So, to prove the existence of a flip you "only" need to prove that the above algebra is finitely generated over $\mathcal O_Y$. This might not seem an intuitive way right away, but remember that Proj comes with a relatively ample divisor, so what's happening is that we make an $f$-anti-ample divisor into an $f^+$-ample one without changing it on the locus where $f$ was an isomorphism. If $X$ and $Y$ are $3$-dimensional and $f$ is a small Mori-contraction then it contracts a single rational curve and being ample is equivalent to the degree of the divisor on the curve being positive. Now the (anti-)ampleness of the canonical class is then governed by the normal bundle of the curve and hence "flipping" the positivity of $K_X$ on this curve is essentially the same as "flipping" the normal bundle. - This is very nice, thanks. – Hailong Dao Oct 15 2010 at 19:24 In order to run MMP, the variety must be Q-factorial or at least K(the canonical divisor) must be Q-Cartier so that we can check the nefness of K. (To run LMMP, you need to replace K with K+B.) There are two kinds of contractions performed in the process of MMP, divisorial contractions and small contractions. While the divisorial contractions preserve the condition, Q-factoriality, the small contractions don't. That means we can no longer check the nefness of K and we cannot resume MMP. Flips fix this problem. For a curve C in the extremal ray inducing a small contraction, a flip 'flips' the negative interesection K.C to positive. It is a condimension one surgery on X, which preserves the Q-factoriality. So we can run MMP again without worrying about the 'bad' curve C. Corti's 'What is....a flip?' may be helpful, too. http://www.ams.org/notices/200411/what-is.pdf - Thanks for the reference! that was fantastic! it answered several questions I had in mind! thanks a lot! – Csar Lozano Huerta Nov 14 2009 at 5:00 I've been told that the intuition behind flips are that when you get an extremely singular image (as jvp mentioned is REALLY the problem, not singularities as a rule) then it means you have a curve "in the wrong place" so you cut out a curve and glue it back in differently, roughly speaking. That is, you "flip" the curve around, so that when you do a contraction, things workout more nicely. Caveat: I'm learning this material right now, so this may be bad intuition, but it's what I've been told. - Thanks! the you answer sounds helpful, but what do you mean by "glue it back differently"? – Csar Lozano Huerta Nov 13 2009 at 16:19 Well, it's a birational map $X\to Y$ such that if you look at $X$ minus the curve and $Y$ minus the new curve, they're isomorphic, but such that $X$ and $Y$ are not. So you've sort of cut out the curve and repositioned it. Again, I'm just learning this, so I can't give that much more of a picture and certainly not much more rigor than this. – Charles Siegel Nov 14 2009 at 1:16	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 113, "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": 0.9596226811408997, "perplexity_flag": "head"}
http://mathoverflow.net/questions/20709/higher-order-axiomatisations-of-euclidean-geometry	## Higher-order axiomatisations of Euclidean Geometry? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I am currently thinking about the possibility to axiomatise Euclidean Geometry using higher-order axioms. The idea is that all objects are points, and that we only have two primitive notions: A three place predicate for betweenness, and a one-place third-order predicate indicating whether a function from a set of points to another set of points is a congruence mapping (so the existence of such a function between two given sets means that the sets are congruent in the intuitive way; the function maps points from the first set to corresponding points of the set congruent to it). $ab \equiv cd$ can then be defined to mean that there is a congruence mapping from {a,b} to {c,d}. The line through a and b can be defined to be {p\|p is between a and b or a is between p and b or b is between p and a} $\cup$ {a,b}. Using these two primitive notions of betwenness and congruence mappings, it seems to me that quite a short and simple list of axioms can specify Euclidean Geometry: Three simple betweenness properties, one completeness axiom (similar to that for $\mathbb{R}$, but using the betweenness relation rather than the less-than relation), one axiom ensuring the existence of enough congurence mappings, one axiom ensuring that relations between points are preserved by congurence mappings, and two axioms specifying the dimension of the space (one for the lower bound on the dimension, another for the upper bound). My interest in such an axiomatisation is mainly philosophical: The two primitive notions correspond to concepts from our natural intuition and expirience of space (congruence mappings correspond to moves of an object in space without the shape of the object changing). The axiomatisation seems simpler than first-order axiomatisations like those of Hilbert and Tarski. And the higher-order nature of the system makes it possible to define lines, circles, triangle, quadrilaterals and any other shapes, which is not possible in first-order systems. I have an idea for a proof that my axiomatisation specifies a space isomorphic to $\mathbb{R}^n$ (for the $n$ used in the dimension axioms); but before I actually work out the proof, I want to know whether a comparable higher-order axiomatisation of Euclidean Geometry already exists. Unfortunately, I couldn't find any higher-order axiomatisations of Euclidean Geometry. Does anyone know about such axiomatisations or attempts at them? - How are functions specified? Will your logic include the lambda-calculus (e.g., is a Henken-like calculus)? Do you want to have, say, all linear transformations? – Charles Stewart Apr 8 2010 at 12:08 The logical system that I would use for formalising my ideas is Church's Type Theory. But other systems for higher-order logic might also do the job. At any rate, I am interested in existing axiomatisations in any system of higher-order logic, not just in the Church's Type Theory. – Marcos Cramer Apr 8 2010 at 13:29 1 Why do you need the third order congruence predicate? It would seem that if you merely had a first order 4-ary predicate for congruence of line segments, then you would in effect have access to the metric, and you could express any kind of congruence in terms of it. – Joel David Hamkins Apr 8 2010 at 15:29 One reason for the third-order congruence predicate is that it makes it easy to express the following very powerful axiom: Given a set of points M, distinct points a,b in M, and distinct points c,d, there is a unique set N of points such that there is a congruence mapping f from M to N such that f(a)=c and a is between d and b. This intuitive axiom corresponds a lot to the cutting down of the number of axioms needed. Additionally, congruence between arbitrary set seems to me a very basic concept of our intuitions about space. Congruence between pairs of points is just a special case of this. – Marcos Cramer Apr 8 2010 at 19:50 2 Marcos, I'm not sure you stated the axiom you intended. But suitably corrected, I like it. My point, however, was that it seems to be second-order expressible, rather than third order, as long as you have a (first-order) way to say that line segments are congruent. That is, you could introduce your congruence predicate as a defined property of a set of pairs, state your axiom using it, and end up with a second-order axiomatization overall instead of third order. – Joel David Hamkins Apr 8 2010 at 22:27 ## 2 Answers Whether it fulfills your needs or not, you should definitely read A Formal System For Euclid's Elements by Jeremy Avigad, Edward Dean, and John Mumma, in Review of Symbolic Logic, Vol. 2, No. 4, 2009 [there was a discussion of this paper over at Lambda the Ultimate recently]. I would also check out the libraries of Coq, Isabelle, HOL-light and maybe PVS -- they are all based on higher-order logic too, and someone may well have done such a formalization already. - My work on Avigad's formal system is actually what originally got me thinking about the topic. Avigad's axiomatization is, however, a first-order axiomatization with a large number of axioms, so very much unlike what I am currently thinking about. Thanks for the idea to check out the libraries of Coq etc. I have just done this: the only things I found were a formalisation of Tarski's first-order theory in Coq, and a type-theoretic formalisation of $\mathbb{R}^n$ (for arbitrary n) in HOL-light (whereas I am thinking about a theory which doesn't depend on having defined the real numbers). – Marcos Cramer Apr 8 2010 at 19:37 ### 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. I thought Hilbert's postulates were a second-order axiomitization of Euclidean geometry, similar to what you are describing. - Hilbert's axioms are indeed second-order, because else they couldn't express Archimedes' axiom and non-extendibility. But the core of the axioms is in first-order language, and most proofs of geometric facts only require those first-order axiom. The axiom system I envisaged, on the other hand, would be inherently higher-order, because of the higher order primitive of a congruence mapping that is formalised in it. – Marcos Cramer Feb 12 2011 at 15:07	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9420913457870483, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/2040/why-are-functional-equations-important	## Why are functional equations important? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) People who talk about things like modular forms and zeta functions put a lot of emphasis on the existence and form of functional equations, but I've never seen them used as anything other than a technical tool. Is there a conceptual reason we want these functional equations around? Have I just not seen enough of the theory? - ## 10 Answers There are many reasons that functional equations are important. Some background: To most varieties/schemes occurring in arithmetic geometry, you can associate a zeta function/L-function. There are two main ways of constructing these, either from l-adic cohomology, or from counting solutions in various finite fields. Usually the word L-function is used for functions associated to varieties over number fields, constructed from l-adic cohomology, and the word zeta function is used for functions associated to schemes over Z or some other ring of integers, or over a finite field. However, there is some confusion about the terminology, and there is also some overlap between the two, since there is a close relation between the L-function of a variety and the zeta function of an integral model for the variety. Over finite fields, the functional equation is part of the famous Weil conjectures, proved by Deligne. One reason that the functional equation is cool is that it reflects Poincare duality in the etale cohomology of the variety, so it is in some sense a deep geometric statement. For background on the Weil conjectures, see for example Freitag and Kiehl: Etale cohomology and the Weil conjectures, and the survey of Mazur: Eigenvalues of Frobenius acting on algebraic varieties over finite fields, in some conference proceedings. For the definition of L-functions and zeta functions and lots of other background, see Manin and Panchishkin: Introduction to modern number theory, chapter 6. There are at least two deep reasons for being interested in the functional equations for these functions. Firstly, the existence of a functional equation seems to always be directly related to the L-function coming from an automorphic representation, and the idea that "every L-function from algebraic geometry (aka motivic L-function) also comes from an automorphic representation" is in some sense the number-theoretic incarnation of the global Langlands program. See for example Bump et al: An introduction to the Langlands program. The most famous cases where this has been proved is (1) Tate's thesis, which treats Hecke L-functions, and where the corresponding automorphic representation is one-dimensional, i.e. a character on the ideles, and (2) the work by Wiles and others related to Fermat's last theorem, where they show that the L-function associated to an elliptic curve over Q also comes from a modular form, and hence satisfies the expected functional equation. See the book of Diamond and Shurman on modular forms. The other deep reason for thinking about the functional equation is that some optimistic people dream of an "arithmetic cohomology theory", which would allow us to mimick the proof of the Weil conjectures, but for zeta functions over Z or L-functions over Q. Then the functional equation should be related to Poincare duality for this cohomology. All this is related to the Riemann hypothesis, noncommutative geometry, and the field with one element. See for example Deninger: Motivic L-functions and regularized determinants, the more recent survey available here, some slides of Paugam on the functional equation, and also the later chapters of Manin-Panchishkin. Some of the key names if you want to find more references: Deninger, Connes, Consani, Marcolli; most of them have lots of stuff on their webpages and on the arXiv. - ### 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 simplest reason functional equations have importance, for someone learning this stuff for the first time (not knowing about modular forms, automorphic representations, etc.) is that they can be used to verify the Riemann hypothesis numerically up to some height! To explain this, let's start off with a limitation of methods of complex analysis in detecting zeros of functions. There are theorems in complex analysis which tell you how to count zeros of an analytic function $f(s)$ inside a region by integrating $f'(s)/f(s)$ around the boundary (this is the argument principle). So we could integrate around the boundary of a box surrounding the critical strip up to some height and see there are, say, 10 zeros of the Riemann zeta-function up to that height (yeah, there's a pole on the boundary at $s = 1$ which messes up the argument principle integration, but don't worry about that right now). How can we prove the 10 zeros in the critical strip up to that height are on the critical line? Complex analysis provides no theorems that assure you an analytic function has zeros on a line! The functional equation comes to the rescue here. I'll illustrate for the Riemann zeta-function $\zeta(s)$. Its functional equation is most cleanly expressed in terms of $$Z(s) = \pi^{-s/2}\Gamma(s/2)\zeta(s)$$ and is the following: $$Z(1-s) = Z(s).$$ We also need another "symmetry": `$Z(s^)^ = Z(s)$`, where `${}^$` means complex conjugation. Where does this come from? For an entire function $f(s)$, the function `$f(s^)^$` is also entire: in fact its local power series expansion at any point a is the one whose coefficients are complex conjugate to the coefficients of $f(s)$ at $a^$. Or you could directly prove `$f(s^)^$` is complex-differentiable when $f(s)$ is. The significance of this is that if $f(s)$ is real-valued for some interval of real numbers then `$f(s^)^ = f(s)$` on that interval, so by the rigidity of analytic functions we must have `$f(s^)^ = f(s)$` everywhere when it is true on a real interval (not one point intervals, obviously). Lesson: an entire function $f(s)$ that is real-valued on some interval of the real line satisfies the formula `$f(s^)^ = f(s)$` for all complex numbers $s$. By the way, this also applies to meromorphic functions on $\mathbf C$ too (rigidity of meromorphic functions). Let's now return to the zeta-function. Because $\zeta(s), \pi^{-s/2}$, and $\Gamma(s/2)$ are real-valued for real $s > 1$, their product $Z(s)$ is real for $s > 1$, so `$Z(s^)^ = Z(s)$` for all complex $s$. In particular, for a number $s = \frac12 + it$ on the critical line (here $t$ is real), we have the key calculation `$$Z(s)^* = Z(1/2 + it)^* = Z(1 - (1/2 + it))^* = Z(1/2 - it)^* = Z((1/2 + it)^* )^* = Z(1/2 + it) = Z(s),$$` where we used $Z(s) = Z(1-s)$ in the second equation and `$Z(s^)^ = Z(s)$` in the second to last equation. This tells us the function $Z(s)$ is real-valued on the critical line. The Riemann zeta-function is not real-valued on the critical line, but this modified (completed) zeta-function $Z(s)$ is. Moreover, because $Z(s)$ differs from $\zeta(s)$ by factors that are finite and nonzero inside the critical strip ($\pi^{-s/2}$ is a nowhere-vanishing entire function and $\Gamma(s/2)$ is meromorphic with no zeros and only has poles at $s = 0, -2, -4, \dots$), the zeros of $\zeta(s)$ and $Z(s)$ inside the critical strip are the same thing. (In fact, nontrivial zeros of $\zeta(s)$ are exactly the same thing as all zeros of $Z(s)$, which is one reason $Z(s)$ is a nicer object that $\zeta(s)$: the Riemann hypothesis is a statement about all zeros of $Z(s)$!) So what? Well, we just showed in the key calculation above that the function $Z(1/2 + it)$ is real when $t$ is real, so by computing we can provably detect zeros of $Z(s)$ on the critical line Re($s$) = $\frac12$ by looking for sign changes of $Z(1/2 + it)$ as t runs through the real numbers. So here is a two-step procedure for proving the RH numerically up to height $T$ (i.e., in the box in the critical strip from the real axis up to height $T$): 1. Use complex analysis (the argument principle) to count how many zeros $Z(s)$ has in the critical strip up to height $T$ by integrating $Z'(s)/Z(s)$ around a box surrounding that region. (If the poles of $Z(s)$ at $s = 0$ and $s = 1$ bother you, recall the argument principle can account for poles or you might prefer to use $s(1-s)Z(s)$ in place of $Z(s)$ to be working with an entire function which satisfies the same functional equation as $Z(s)$ and is also real-valued on the critical line.) 2. Count sign changes for $Z(1/2 + it)$ when $0 \le t \le T$. There is a zero between any two sign changes, so $Z(s)$ has at least as many zeros on the critical line as the number of sign changes that were found. (Finding a sign change is a computable thing: if a function value at a point is approximately positive or negative then it is provably so by checking the error in your computation well enough, whereas proving a function value at a point is exactly zero with a computer is basically impossible.) If the counts in steps 1 and 2 match, then voila: all zeros of $Z(s)$ up to height $T$ in the critical strip are on the critical line, which confirms the Riemann hypothesis up to height $T$. This method will not work if there are any multiple zeros on the critical line: the argument principle counts each zero with its multiplicity, so if for instance there is a double zero then the argument principle may tell us $Z(s)$ has 10 zeros (with multiplicity!) up to some height while we find only 9 sign changes because one zero is a double zero so it doesn't give us a sign change. (Or if there were a triple zero we get 10 zeros with multiplicity from the argument principle but we find only 8 sign changes.) A graph may suggest that the mismatch in the numbers in the two steps is coming from a multiple zero, but it doesn't rigorously prove anything. Fortunately, this has never happened in practice with the Riemann zeta-function: the two counts always match. In fact the conjecture is that all nontrivial zeros of $\zeta(s)$ are simple zeros. What about more general L-functions $L(s)$? By multiplying $L(s)$ by suitable exponential and Gamma functions, you get a function $\Lambda(s)$ whose functional equation is `$$\Lambda(1-s^)^ = w\Lambda(s),$$` where $w$ is a constant with absolute value 1. (For the Riemann zeta-function, $\Lambda(s) = Z(s)$ and $w = 1$. For Dirichlet L-functions, $w$ is usually not equal to 1.) Let $u$ be one of the square roots of $w$, so $w = u^2$. Then using the above functional equation, the function $$\frac1u \Lambda(s)$$ is real-valued on the critical line ($s = 1/2 + it$ for real $t$), so we can detect its zeros there by looking for sign changes. The same method described above for detecting zeros of $\zeta(s)$ in the critical strip by using $Z(s)$ and its functional equation can be applied also to $L(s)$. This is basically the way all variants on the Riemann hypothesis are checked numerically (modulo important details of practical calculation that I don't get into), and the functional equation is an essential ingredient in justifying the method. What is crucial here is not just the idea of sign changes, but also the expectation that the zeros are all simple (so we can find all the zeros by sign changes and the argument principle). As with the Riemann zeta-function, it is expected that the nontrivial zeros of Dirichlet $L$-functions are all simple. But there are examples of $L$-functions with a multiple zero on its critical line, thanks to ideas from the Birch and Swinnerton-Dyer conjecture. This does not wreck this approach to verifying the Riemann hypothesis for such L-functions, because such multiple zeros are supposed to happen only at one (known) point which we think we understand. - Oy, I see now that I looked only at the question in its short title form and didn't notice the question in its full form was asking for a conceptual reason for the importance of functional equations. Boy, did I waste my time writing that non-conceptual answer... – KConrad Jan 15 2010 at 15:51 1 This answer is still very helpful, though; thank you! – Qiaochu Yuan Jan 15 2010 at 16:09 8 Thanks for the exposition anyway. – Douglas Zare Jan 15 2010 at 16:15 1 This is a wonderful answer! I edited math a bit (there's more about typing math at mathoverflow.net/faq#latex), which could bring some mistakes, watch out :) – Ilya Nikokoshev Jan 17 2010 at 9:08 As was mentioned in other answers, the functional equations are related to duality. In general, however, the "dual" object need not be the original object. It just so happens that the objects that give the Riemann zeta function and L-functions of modular forms are self-dual (up to a twist). For a general motive, M, for example, the functional equation (conjecturally) relates values of the L-function of M to values of the L-function of M*, the dual of M (see for example an article of Panchishkin's link text). For automorphic representations, its the contragredient (see section 14 of Borel's article on L-functions in Corvallis link text). So, conceptually, the functional equation is just an expression of this duality. As Andreas Holmstrom alluded to above, the functional equation is (conjecturally) related to which L-functions are automorphic. The type of result is called a "Converse theorem". A famous "classical" one is Weil's converse theorem which basically says that if you have two Dirichlet series such that for infinitely many Dirichlet characters, the twisted Dirichlet series have analytic continuation, are bounded in vertical strips AND are related by a functional equation, then they are Dirichlet series coming from a modular form (see Bump's book, theorem 1.5.1). For GL(n) see Cogdell's "L-functions and Converse theorems for GL(n)" link text). To add to the above paragraph, the existence of a functional equation is thought to be one of the characterizing properties of L-functions as expressed by the notion of the "Selberg class" (link text) - I am surprised no-one seems to have mentioned one key use for functional equations: they are a key input in converse theorems. If you have a power series in q that you're trying to show is a modular form of level 1, then one strategy is to show that the associated L-function has the right kind of functional equation (and a couple of other nice properties). The functional equation can translate into a relation between f(z) and f(-1/z), and the right kind of functional equation translates into the right kind of translation property for f, which is what you needed to show it was modular (the fact that it was a power series in q gives you the boundedness at infinity and f(z)=f(z+1)). This idea was generalsed by Weil, who proved that sufficiently many functional equations for your L-function and its twists will imply that it's the L-function of a modular form. Hence elliptic curves with complex multiplication are modular! (because their L-functions are L-functions of grossencharacters). That's an algebraic statement whose proof I outlined above makes essential use of the functional equation of the L-function. - Actually the relation to converse theorems was mentioned in my answer. Though the specific case of CM elliptic curves is a nice illustration. – Rob Harron Oct 31 2009 at 21:49 It seems worth mentioning another application of the functional equations. To fix ideas, I will suppose that $E$ is an elliptic curve over ${\mathbb Q}$, and let $L(E,s)$ be the $L$-function of $E$. There is then a functional equation relating $L(E,s)$ and $L(E,2-s)$; one feature of this functional equation is that it has a sign, i.e. has the form $L(E,s-2) = \pm (\text{ a positive constant })L(E,s),$ where $\pm$ is some fixed sign, depending on $E$, and easily computed (say using the modular form corresponding to $E$). Thus one find that the order of vanishing of $L(E,s)$ at $s = 1$ is odd (even) precisely if the sign is $-1$ ($+1$). Since the order of vanishing is conjectured to coincide with the rank of $E({\mathbb Q})$ (the BSD conjecture), this is pretty important arithmetic information which is obtained pretty easily from the functional equation. In particular, if the sign is $-1$, we always expect there to be a rational point of infinite order, and arranging things so that the sign is $-1$ (by twisting by a well-chosen character, say) is a common way of forcing the existence of rational points in various situations. (There is a large body of research focused around this expected relationship between signs of functional equations and the existence of rational points on elliptic curves; for example, it lies at the heart of the study of Heegner points by Gross and Zagier and Kolyvagin. Some important recent contributions are by Nekovar, Mazur and Rubin, and T. and V. Dokchitser.) - Apart from their conceptual significance, functional equations are useful when doing concrete estimates in analytic number theory. The usual textbook proof by the method of contour integrals for the bound \vartheta \leq 1/3 in the Dirichlet divisor problem relies on the functional equation of the Riemann zeta function. - Sometimes you get very concrete algebro-geometric facts from functional equations: Example 1: The functional equation relating Weierstrass P (for a lattice L in C) and it's derivative is the equation representing the elliptic curve C/L in Weierstrass form. Example 2: if F is a modular form on the Siegel upper half plane (the universal cover of the moduli of Abelian varieties Ag) with weight W, then the class of (F)_0 in Pic(Ag) is W times the Hodge class. - Hi Qiaochu, I would say that functional equations of zeta-like function are popular because it is one of the simplest non trivial example of functions having a symmetry only when analytically continued. But in itself the impact of such a functional equation is quite limited, even though some people are convinced that this equation is a key for solving the Riemann Hypothesis, which is misleading (this feeling is supported also by the Hamburger theorem that states that any Dirichlet function having an analytic continuation like zeta and obeying the same functional equation is proportional to zeta, but when considering L-functions it becomes false and the Davenport-Heilbronn L-function is an example of Dirichlet function with a L-function-like functional equation but also with infinitely many zeroes in any vertical strip included in the critical strip, therefore failing to obey a Reimann hypothesis). However, as for me these functional equations hide a much deeper phenomenon. The functional equation is an artefact of the action of the Fourier transform on zeta functions. What's more, the same symmetry which can be proved in a simple manner for local zeta functions extends to global ones, which is a highly non-trivial result (proved for general Hecke zeta functions by John Tate in his thesis). In addition, you can see the Fourier transform as the image of a particular linear operator by the Weil representation. And looking at the action of the Weil representation in a more general context leads to some generalized Poisson Formulas, and most probably there is still some unexplored stuff to discover in this field (but that's a personal point of view...). Best regards, Eric - It was my understanding that, in some cases, the functional equation is the only known way to prove that the analytic continuation of one of these number theoretic functions is entire. Even in the case of zeta, I don't know another argument that zeta continues past Re(s)=0. (I can get into the critical strip by writing zeta(s) = 1/(s-1) + \sum [ n^{-s} - (n^{-s+1} - (n+1)^{-s+1})/(s-1) ], the sum now converges uniformly for Re(s)>epsilon>0.) So maybe the question should be "do we care about the values of zeta (and other L-functions) to the left of the critical strip?" - 1 You can easily get the analytic continuation to the whole complex plane, strip by strip, by repeatedly integrating by parts in the representation obtained from the Euler-Maclaurin formula. – engelbrekt Oct 23 2009 at 15:20 This technique is originally due to Euler in his paper on the functional equation of the zeta function (note no notion of complex analysis yet!). – Noah Snyder Jan 15 2010 at 16:07 1 Tate's thesis gives a nice explanation of the prime factors and the Gamma function in the functional equation for Dedekind zeta functions. We wouldn't have that nice explanation without a functional equation. – Anweshi Jan 15 2010 at 16:57 1 "do we care about the values of zeta (and other L-functions) to the left of the critical strip?" Absolutely! The values of the zeta function at negative integers are rationals, and hence make sense as p-adic numbers. Furthermore, for n negative, the map sending n to zeta(n) is p-adically continuous (for n in a fixed congruence class mod p-1) and so we can define a p-adic zeta function! p-adic L-functions have been used to e.g. prove Leopoldt's conjecture for abelian extensions of the rationals. However pi isn't a p-adic number and so this trick wouldn't work if we only restricted to +ve vals. – Kevin Buzzard Jan 19 2010 at 16:59 Functional equations are useful to understand the "symmetries" of a given function, pretty much as periodicity. For example, they may be used in the continuation of functions to the complex plane. The functional equation for the Gamma function, Gamma(z+1) = zGamma(z), for instance, let us understand the continuation of this function to the left of the line Re(z)=0. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 95, "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": 0.9287300109863281, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Axiom_of_extensionality	# Axiom of extensionality In axiomatic set theory and the branches of logic, mathematics, and computer science that use it, the axiom of extensionality, or axiom of extension, is one of the axioms of Zermelo–Fraenkel set theory. ## Formal statement In the formal language of the Zermelo–Fraenkel axioms, the axiom reads: $\forall A \, \forall B \, ( \forall X \, (X \in A \iff X \in B) \Rightarrow A = B)$ or in words: Given any set A and any set B, if for every set X, X is a member of A if and only if X is a member of B, then A is equal to B. (It is not really essential that X here be a set — but in ZF, everything is. See Ur-elements below for when this is violated.) The converse, $\forall A \, \forall B \, (A = B \Rightarrow \forall X \, (X \in A \iff X \in B) )$, of this axiom follows from the substitution property of equality. ## Interpretation To understand this axiom, note that the clause in parentheses in the symbolic statement above simply states that A and B have precisely the same members. Thus, what the axiom is really saying is that two sets are equal if and only if they have precisely the same members. The essence of this is: A set is determined uniquely by its members. The axiom of extensionality can be used with any statement of the form $\exist A \, \forall X \, (X \in A \iff P(X) \, )$, where P is any unary predicate that does not mention A, to define a unique set $A$ whose members are precisely the sets satisfying the predicate $P$. We can then introduce a new symbol for $A$; it's in this way that definitions in ordinary mathematics ultimately work when their statements are reduced to purely set-theoretic terms. The axiom of extensionality is generally uncontroversial in set-theoretical foundations of mathematics, and it or an equivalent appears in just about any alternative axiomatisation of set theory. However, it may require modifications for some purposes, as below. ## In predicate logic without equality The axiom given above assumes that equality is a primitive symbol in predicate logic. Some treatments of axiomatic set theory prefer to do without this, and instead treat the above statement not as an axiom but as a definition of equality. Then it is necessary to include the usual axioms of equality from predicate logic as axioms about this defined symbol. Most of the axioms of equality still follow from the definition; the remaining one is $\forall A \, \forall B \, ( \forall X \, (X \in A \iff X \in B) \Rightarrow \forall Y \, (A \in Y \iff B \in Y) \, )$ and it becomes this axiom that is referred to as the axiom of extensionality in this context. ## In set theory with ur-elements An ur-element is a member of a set that is not itself a set. In the Zermelo–Fraenkel axioms, there are no ur-elements, but they are included in some alternative axiomatisations of set theory. Ur-elements can be treated as a different logical type from sets; in this case, $B \in A$ makes no sense if $A$ is an ur-element, so the axiom of extensionality simply applies only to sets. Alternatively, in untyped logic, we can require $B \in A$ to be false whenever $A$ is an ur-element. In this case, the usual axiom of extensionality would then imply that every ur-element is equal to the empty set. To avoid this consequence, we can modify the axiom of extensionality to apply only to nonempty sets, so that it reads: $\forall A \, \forall B \, ( \exist X \, (X \in A) \Rightarrow [ \forall Y \, (Y \in A \iff Y \in B) \Rightarrow A = B ] \, ).$ That is: Given any set A and any set B, if A is a nonempty set (that is, if there exists a member X of A), then if A and B have precisely the same members, then they are equal. Yet another alternative in untyped logic is to define $A$ itself to be the only element of $A$ whenever $A$ is an ur-element. While this approach can serve to preserve the axiom of extensionality, the axiom of regularity will need an adjustment instead. ## See also • Extensionality for a general overview. ## References • Paul Halmos, Naive set theory. Princeton, NJ: D. Van Nostrand Company, 1960. Reprinted by Springer-Verlag, New York, 1974. ISBN 0-387-90092-6 (Springer-Verlag edition). • Jech, Thomas, 2003. Set Theory: The Third Millennium Edition, Revised and Expanded. Springer. ISBN 3-540-44085-2. • Kunen, Kenneth, 1980. Set Theory: An Introduction to Independence Proofs. Elsevier. ISBN 0-444-86839-9.	{"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": 15, "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": 0.8996608257293701, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/147530-uniform-convergence-function-sequence.html	# Thread: 1. ## Uniform convergence of a function sequence Hey everyone Given function f(x) which is differentiable on R , and f'(x) is continuous on R, and a sequence (f_n(x)) such as : $f_n(x)=n(f(x+\frac{1}{n}))-f(x))$ I need to prove that f_n(x) is uniformly converges on each interval [a,b]. I have managed to show that (f_n(x)) approaches f'(x), but I couldn't get any further. I know nothing about the values of f(x), so I can't know if f_n(x) is monotonic or not , and therefore I can't use Dini's theorem. Plus , I can't say much about \|f_n(x)-f'(x)\| , so I can't figure out how do I show that for every epsilon > 0 , there is N such as for every n>N , \|f_n(x)-f'(x)\| < epsilon Any ideas how to get any further? Thanks people! 2. Originally Posted by Gok2 Hey everyone Given function f(x) which is differentiable on R , and f'(x) is continuous on R, and a sequence (f_n(x)) such as : $f_n(x)=n(f(x+\frac{1}{n}))-f(x))$ I need to prove that f_n(x) is uniformly converges on each interval [a,b]. I have managed to show that (f_n(x)) approaches f'(x), but I couldn't get any further. I know nothing about the values of f(x), so I can't know if f_n(x) is monotonic or not , and therefore I can't use Dini's theorem. Plus , I can't say much about \|f_n(x)-f'(x)\| , so I can't figure out how do I show that for every epsilon > 0 , there is N such as for every n>N , \|f_n(x)-f'(x)\| < epsilon Any ideas how to get any further? Thanks people! http://www.math.uchicago.edu/~ershov/16300/uniform2.pdf 3. Thank you , but I knew that. That's what I explained, I don't know nothing about \|f_n(x)-f'(x)\| , therefore I can't find a sequence c_n that \|f_n(x)-f'(x)\| <= c_n , and from the seam reason I can't find sup \|f_n(x)-f(x)\|, therefore I am kinda stuck... Any ideas people? 4. Anyone?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9568045735359192, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/36861-directional-derivatives.html	# Thread: 1. ## Directional derivatives Find a vector u for which Duf(1,1) =0 Thank you. Attached Thumbnails 2. Originally Posted by al1850 Find a vector u for which Duf(1,1) =0 Thank you. You ned to find a unit vector such that ${\rm{u}}.[(\nabla f (1,1)]=0$ RonL 3. Originally Posted by CaptainBlack You ned to find a unit vector such that ${\rm{u}}.[(\nabla f (1,1)]=0$ RonL But how? 4. well first for the definition of directional derivative: directional derivative of a scalar function, $f=f(x,y)$ in the direction of a unit vector, $\overrightarrow{u}=(u_1 , u_2)$ at a point $(x_0,y_0)$is given by: $D \overrightarrow {u} f(x_0,y_0) = \bigtriangledown f(x_0 , y_0) \cdot \overrightarrow {u} = u_1 \frac {\partial f}{\partial x} + u_2\frac {\partial f}{\partial y} <br />$ $<br /> f(x,y) = \frac{1}{x^2+y^2+1}<br />$ $<br /> \bigtriangledown f(x , y) = \frac {\partial f}{\partial x} +\frac {\partial f}{\partial y}<br />$ $<br /> = -\frac{2x}{(x^2+y^2+1)^2}\overrightarrow{i}-\frac{2y}{(x^2+y^2+1)^2}\overrightarrow{j}<br />$ at f(1,1): $<br /> = -\frac{2}{(1^2+1^2+1)^2}\overrightarrow{i}-\frac{2}{(1^2+1^2+1)^2}\overrightarrow{j}<br />$ $<br /> = -\frac{2}{9}\overrightarrow{i}--\frac{2}{9}\overrightarrow{j}<br />$ therefore from; $<br /> D \overrightarrow {u} f(x,y) <br />$ $<br /> D \overrightarrow {u} f(1,1) <br />$ $<br /> = (-\frac{2}{9}\overrightarrow{i}-\frac{2}{9}\overrightarrow{j})\cdot (\overrightarrow{u}) = 0<br />$ $<br /> = (-\frac{2}{9}\overrightarrow{i}-\frac{2}{9}\overrightarrow{j})\cdot (u_1\overrightarrow{i}+u_2\overrightarrow{j}) = 0<br />$ $<br /> -\frac{2}{9}u_1-\frac{2}{9} u_2 = 0<br />$ now to obtain a zero $u_1$ and $u_2$ must be equal in magnitude but opposite in sign: $<br /> \overrightarrow{u} = (n,-n)<br />$ n can take any value. Unit vector of $\overrightarrow{u}$ = $\frac{\overrightarrow{u}}{\|\overrightarrow{u}\|}$ = $\frac{u_1 \overrightarrow{i}+u_2\overrightarrow{j}}{\|\sqrt{{ u_1}^2+{u_2}^2}\|}$ = $\frac{u_1 \overrightarrow{i}+u_2\overrightarrow{j}}{\|\sqrt{n ^2+(-n)^2}\|}$ = $\frac{n \overrightarrow{i}-n\overrightarrow{j}}{\|\sqrt{2}n\|}$ = $\frac{n }{\|\sqrt{2}n\|}\overrightarrow{i}-\frac{n }{\|\sqrt{2}n\|}\overrightarrow{j}$	{"extraction_info": {"found_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": 0.18, "absolute_threshold": 10, "end_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": 0.8897212743759155, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/45238?sort=oldest	## scalar diffusions are reversible ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It is well known that under mild assumptions a scalar diffusion $dX_t = a(X_t) dt + \sigma(X_t) dW_t$ with invariant probability distribution $\pi$ is reversible. This is indeed not true for multidimensional diffusions. The usual proofs consists in writing down generators, speed functions etc... I am trying to intuitively understand this result, and the only (not very satisfying) argument that I have found is the following. It is straightforward to check that any Markov chain on $\mathbb{Z}$ that has an invariant probability $\pi$ and that can only make jumps of size $+1$ or $-1$ is reversible: notice for example that $$F(k) = \pi(k)p(k,k+1)-\pi(k+1)p(k+1,k)$$ is independent of $k$ and is thus equal to $0$. If $a(\cdot)$ and $\sigma(\cdot)$ are regular enough, a diffusion can be seen as a limit of such Markov chains on $\epsilon \mathbb{Z}$ so that this makes the result plausible. question: what are arguments/proofs/examples that could shed light on why a one dimensional ergodic diffusion is automatically reversible. - ## 1 Answer The identity you are referring to can be interpreted as saying that if you run your Markov chain on $\mathbb Z$ from the initial distribution $\pi$, then the probability of crossing the edge $[k,k+1]$ in the positive direction is the same as the probability of crossing it in the negative direction (as otherwise the measure $\pi$ won't be stationary). The same idea works in the general case as well. For any nice test function $f$ $$\lim \frac1t \int [f(\xi_t) - f(\xi_0)] d{\mathbf P}(\xi) = \int Df(x) d\pi(x) ,$$ where ${\mathbf P}$ is the measure in the space of sample paths $\xi$ of the diffusion process with the initial stationary distribution $\pi$, and $D$ is the diffusion generator. The left-hand side of the above equation is 0 by stationarity of the measure $\pi$, so that $\int Df(x)d\pi(x)=0$. In the same way it is true for the reverse generator $D^{\ast}$ as well. Now, the operators $D$ and $D^{\ast}$ may differ by their first-order parts only (denote these vector fields by $v$ and $v^{\ast}$, respectively). Taking a step-like function $f$ whose derivative is non-zero only on a small set where $v\neq v^*$ (it is at this place that dim $=1$ is used) then gives a contradiction. - thanks: this is exactly the kind of simple approaches that I am looking for. – Alekk Nov 8 2010 at 17:51	{"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": 30, "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": 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": 0.9331011772155762, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/34057/whats-the-role-of-field-equation-in-qft?answertab=active	# What's the role of field equation in QFT? For free field theory, it seems the solutions of a field equation are used to give a representation of Poincare group, and the field equation is still satisfied after quantization. However for an interacting theory, such as QED, the field equation: $$\partial_\nu F^{\nu \mu} ~=~ e \bar{\psi} \gamma^\mu \psi.$$ I don't see any possibility of satisfying this equation using operators since LHS will only act nontrivially on bosonic sector of the Hilbert space and RHS only on fermionic sector, and I don't think perturbation technique can fix this since it's qualitatively not satisfied. Yet nobody seems to be bothered by this, so I'm wondering what's the role of field equation in QFT, especially for interacting ones. - In QFT equations of motion are only satisfied in sense of expectation values i.e. expectation value of equation of motion in any given state should vanish. – user10001 Aug 13 '12 at 17:38 ## 1 Answer '' LHS will only act nontrivially on bosonic sector of the Hilbert space and RHS only on fermionic sector'': This would be the case in a free theory, but the field equation you wrote corresponds to an interacting theory. Its Hilbert space is not the Fock space one starts with, but is turned by renormalization into a different (and little understood) Hilbert space. (Actually, all separable Hilbert spaces in infinite dimensions are the same, but ''different'' conventially refers not to a structureless Hilbert space but to a Hilbert space with a good representation of the kinematic Lie algebra.) Actually, renormalization also changes the interpretation of the field equation, as the products of distributional fields at the same point on the right hand side must be interpreted with care. Edit: On a heuristic level, one can understand the ''mixing'' of bosonic and fermionic operators'' by considering naive (unrenormalized) perturbation theory of the operator equations. One finds that the solution of the field equations is a formal power series in the coupling constant, with the free field as the zero order part. The first and higher order part involves the interaction. Thus the free bosonic field acquires corrections involving the product of an even number of Fermionic fields, and therefore acts nontrivially on both the bosonic and the fermionic part of the state. Unforunately, this formal expansion is mathematically ill-defined, so this argument has only heuristic value. - Interesting, is there any article explaining the details? – Jia Yiyang Aug 14 '12 at 2:46 – Arnold Neumaier Aug 14 '12 at 8:48 I don't have the access to the journal article right now, but are you saying the field equation can actually be satisfied(perhaps in a more subtle way) after renormalization? – Jia Yiyang Aug 14 '12 at 11:01 In 1+1D and 1+2D, where one can make mathematically rigorous statements (though at present not for QED, unfortunately), they are satisfied weakly, i.e., on a dense subspace of the Hilbert space, and in an appropriate limit; i.e., the residual times a sufficiently nice state converges to zero as the nonlocal approximation gets local. – Arnold Neumaier Aug 14 '12 at 12:22 In the lower-dimension model you mentioned, are there fermion-boson interactions or just pure bosons? I am not able to imagine how to equate bosonic operators and fermionic operators, since I feel the distinction between bosons and fermions should be "sacred". – Jia Yiyang Aug 15 '12 at 12:01 show 4 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": 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": 0.9172074198722839, "perplexity_flag": "middle"}
http://mathhelpforum.com/differential-geometry/127740-solved-topology.html	# Thread: 1. ## [SOLVED] Topology Let $(A_\lambda)_{\lambda\in\mathbb{N}}$ be a collection of open sets of real numbers. Suppose $F\subset\mathbb{R}$ is such that $\overline{F\cap A_\lambda}=F$ $\forall \lambda\in\mathbb{N}$. Show that $\overline{F\cap\bigcap_{\lambda\in\mathbb{N}}A_\la mbda}=F$ I've managed to prove that $\overline{F\cap\bigcap_{\lambda\in\mathbb{N}}A_\la mbda}\subset F$. How can i prove the other inclusion? Thanks in advance. 2. Originally Posted by JoachimAgrell Let $(A_\lambda)_{\lambda\in\mathbb{N}}$ be a collection of open sets of real numbers. Suppose $F\subset\mathbb{R}$ is such that $\overline{F\cap A_\lambda}=F$ $\forall \lambda\in\mathbb{N}$. Show that $\overline{F\cap\bigcap_{\lambda\in\mathbb{N}}A_\la mbda}=F$ I've managed to prove that $\overline{F\cap\bigcap_{\lambda\in\mathbb{N}}A_\la mbda}\subset F$. How can i prove the other inclusion? This looks suspiciously like the Baire category theorem (see the section headed "Proof" in that link). 3. I actually used Baire category theorem. Here goes the solution: Let $A:=\bigcap_{\lambda\in\mathbb{N}}A_\lambda$. Since $F=\overline{A_\lambda\cap F}\text{ }\forall\lambda\in\mathbb{N}$, it follows that $A_\lambda\cap F$ is a dense subset of F for all lambda. Hence each $A_\lambda$ is a dense subset of F. Suppose $x\in F$ but $x\notin \overline{F\cap A}$. That means x is not an adherent point of $F\cap A$. Therefore $\exists \delta$ such that $I_{\delta}(x)\cap (F\cap A)=\emptyset$. But $x\in F$, so it must be true that $I_\delta(x)\cap A=\emptyset$. On the other hand, since each $A_\lambda$ is a dense subset of F, the category theorem guarantees that the countable intersection of these sets is a dense set in F. Thus $I_\delta(x)\cap A\neq\emptyset$, a contradiction.	{"extraction_info": {"found_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": 0.18, "absolute_threshold": 10, "end_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": 0.9254724979400635, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/tagged/brownian-motion+martingales	# Tagged Questions 1answer 97 views ### Show that $M$ is a martingale Let $B$ be typical Brownian motion with $\mu >0$ and $x \in \mathbb{R}$. $X(t):=x+B(t)+\mu t$, for each $t\geqslant 0$, Brownian motion with velocity $\mu$ that starts at $x$. For \$r \in ... 1answer 58 views ### Using a Brownian martingale to compute the second moment of a hitting time Prove $W_t=B_t^4 -6B_t^2t+3t^2$ is a martingale, and compute $E(T^2)$ where $T=\inf(t\ge0,B_t=-a, B_t=b)$ if $a=b$. Ok, if $0\lt t\lt s$, $W_t$ is a martingale if $E(W_s\|[B_r]_{r\le t})=W_t$ So ... 1answer 47 views ### Backward martingale property of quadratic variation Let $\pi_n$ denotes a refining sequence of partitions of a finite closed interval (refining means $\pi_n\subset\pi_{n+1}).$ And we denote $\pi_n B = \sum_{t_i\in \pi_n}(B_{t_{i+1}}-B_{t_i})^2$, where ... 1answer 110 views ### Checking for Martingales on Stochastic processes I am confused about how to check whether a process is Martingale. I know, I have to check for clear drift but a bit confused about to approach this problems. I need to apply Ito's first i think. For ... 0answers 136 views ### Are these processes martingales? Determine and prove if the following processes $Y(t)$ are martingales. Assume that $X(t)$ is the standard Brownian Motion $$Y(t) = e^{\sigma X(t)-0.5\sigma^2t}$$ Y(t) = e^{0.5t}\Bigg(1 - ... 1answer 120 views ### d-dimensional Brownian motion and martingales I was solving questions from the Martingales chapter in "Stochastic Processes" by Richard Bass. There was a question regarding d- dimensional Brownian motions(BM): Let $(W_t^1,...,W_t^d)$ be a d ... 1answer 77 views ### How to show that $X_t = \frac{1}{\left\| B_t -x\right\|}\mathbb{1}_{\left\{ B_t \neq x\right\}}$ (“inverse brownian”) is a martingale? Consider $$X_t = \frac{1}{\left\| B_t -x\right\|}\mathbb{1}_{\left\{ B_t \neq x\right\}}$$ where $\left(B_{t }\right)_{t \geq 0}$ is a $\mathcal F_t$- brownian motion in $\mathbb R ^3$, null at ... 1answer 161 views ### How to prove the martingale? How to prove that the integral $\int_{0}^{+\infty}\upsilon e^{-ru}S_{u}dW_{u}^{Q}$ is a martingale under Q where $S_{t}$ is a martingale under Q and \$\mathbb{E}^{Q}[\int_{0}^{+\infty}\|\upsilon ... 1answer 166 views ### How do you show this is a martingale? How do you show the following process is a martingale? My notes say it is a martingale by I can't work it out. $$E[e^{\sigma B(t) - \frac{\sigma ^2 t}{2}} \| \mathscr{F}(s)]$$ I tried to multiply ... 2answers 258 views ### Show that this process is a martingale Let $B_t$ be a Brownian motion and $M_t=\max_{0\leq s\leq t}B_s$. Show that: $$(M_t-B_t)^4-6t(M_t-B_t)^2+3t^2$$ is a martingale for $t\geq0$. 1answer 123 views ### Martingale representation theorem Trying to figure out how to solve problems on the 'form': Find a real number $z$ and a square integrable, adapted process $\psi(s,w)$ such that $$G(w) = z + \int \psi(s,w)\,dB_s(w)$$ for som ... 1answer 119 views ### Optional sampling exercise I came across the following exercise in Stochastic Calculus: Let $B=(B_t)_{t\geq0}$ be a standard Brownian motion. Let also $M$ be the following process: $M_t=B^4_t-6t(B^2_t-\dfrac{t}{3})$ for ... 1answer 105 views ### Wiener martingale Is $\frac{W^{2}(t)}{t}$ a martingale w.r.t. the usual filtration? $t>0$ and $W(t)$ is the Wiener process. What I have so far: Define $Z(t)=\frac{W^2(t)}{t}$. By Ito we get ... 2answers 61 views ### Certain transformation of Brownian motion is a submartingale I have a question about a proof in Protter. Let $B$ Brownian motion and $u$ a harmonic (subharmonic) function. Then $u(B)$ is a local martingale (submartingale). I was able to show the case of local ... 1answer 403 views ### Questions and Solutions in Brownian Motion and Stochastic Calculus? I am currently studying Brownian Motion and Stochastic Calculus. I believe the best way to understand any subject well is to do as many questions as possible. Unfortunately, I haven't been able to ... 1answer 190 views ### $\mathcal{F_t}$-martingales with Itô's formula? I need a little help with a problem. I am given some stochastic processes and supposed to show that they are $\mathcal{F_t}-$martingales. The first one is this, and they all look similar: ... 1answer 228 views ### $d$-Dimensional Brownian Motion Martingales Let $d > 1$ and let $W_t$ denote a standard $d$-dimensional Brownian motion starting at $x\neq 0$. Let $M_t = \log\|W_t\|$ for $d = 2$, and $M_t= \|W_t\|^{2-d}$ for $d > 2$. Show that $M_t$ is a ... 1answer 334 views ### Convergence of the exponential martingale How can we show that this martingale $$e^{aW_{t} - \frac{1}{2}a^2t}$$ converges to $0$ as $t \rightarrow \infty$ using law of iterated logarithm, for $a \neq 0$. 3answers 1k views ### The Laplace transform of the first hitting time of Brownian motion Let $B_t$ be the standard Brownian motion process, $a > 0$, and let $H_a = \inf \{ t : B_t > a \}$ be a stopping time. I want to show that the Laplace transform of $H_a$ is ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 62, "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": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8958935141563416, "perplexity_flag": "head"}
http://mathoverflow.net/questions/110955?sort=votes	## Amplitude and bigness issues ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) There are a couple of statements that I have read which are made as though they were trivial, but I am doubtful about them. 1. One is related to an example showing that the s-invariant of an ample line bundle on a projective variety X is an algebraic integer of degree $\leq dim X$. Recall that, given an ideal sheaf $\mathcal{I}\subset \mathcal{O}_X$ on a projective irreducible variety X, the s-invariant of $\mathcal{I}$ with respect to an ample line bundle L is $s_L(\mathcal{I})$ is the minimum $s\in \mathbb{R}$ such that $\mu^{\ast}(sL)-E$ is nef on $X'$, where $$\mu:X'=Bl_{\mathcal{I}}X\rightarrow X$$ is the blow-up along the ideal $\mathcal{I}$, with exceptional divisor E. The author claims that the class $s_L(\mathcal{I})L-E$ is nef (by definition) but not ample and then uses the Campana-Peternell theorem to conclude the result. How does the non-ampleness follow? 2. Now let L be a big and nef divisor on a smooth projective variety X of dimension n and assume that the Seshadri constant of L at some point $x\in X$ is $\epsilon(L;x)>2n$ (we hence have the same inequality at a very general point). Take two general points $x,y\in X$ and consider the blow-up $\mu:X'=Bl_{{x,y}}X\rightarrow X$, with corresponding exceptional divisors $E_x$ and $E_y$. The divisor $\mu^{\ast}(\frac{1}{2}L)-nE_x$ is then nef (by definition) and big. How does bigness follow? Thanks in advance for any insight. - 4 Regarding 1: the ample cone is open, and its closure is the nef cone. Since the class in question is on the boundary, it cannot be ample. – Piotr Achinger Oct 29 at 2:07 ## 1 Answer (1) was handled in the comments. Regarding (2): since $\epsilon(L;x) > 2n$ with strict inequality, in fact $\nu^\ast \left( \frac{1}{2} L \right) - (n+c) E_x$ is nef for some $c> 0$, where $\nu$ is the blow-up at just $x$. Pulling this back to $X^\prime$, we have $\mu^\ast \left( \frac{1}{2} L \right) - (n+c) E_x$ nef. But $\mu^\ast \left( \frac{1}{2} L \right) - n E_x$ lies on the segment between $\mu^\ast \left( \frac{1}{2} L \right) - (n+c) E_x$ and $\mu^\ast \left( \frac{1}{2} L \right)$ (strictly between the endpoints!), with the former nef and the latter big. So $\mu^\ast \left( \frac{1}{2} L \right) - n E_x$ is nef+big = big. Perhaps writing this up just obfuscates things -- just draw a picture of the cones and think about how $\mu^\ast \left( \frac{1}{2} L \right) - t E_x$ moves relative to the nef and pseudoeffective cones, and what $\epsilon$ means in the picture. -	{"extraction_info": {"found_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": 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": 0.9546529650688171, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/129850-matrix.html	# Thread: 1. ## Matrix Determine the value of for which the system has no solutions. How do i do this? im confused because how can you make a system have no solutions when it really does. I know it does because i did row operations until i got it down to the 1's in the diagonal and the rest of the entries were 0's any help will be much appreciated. thank you 2. Have you considered a Determinant? 2k + 18 - 8 - 12 + 24 - k = 0 I am really concerned with your opinion that this systems "really does" have a solution. You cannot know that until you determine the value of k. When you did those row operations, there was only one way to get rid of the k. That one way is to make an error. It's the only 'k' in the problem. Parity would suggest that you can't get rid of it. Whatever you do, there will always be a 'k'. Do you REALLY believe a system would be entirely independent of one of it's coefficients? No kidding, think about the implications for a minute. That '6y' is bothering me. Since the solution to the system doesn't care about that 6, let's just make it a 4. Awesome! That's much better. It is a mass of confusion. 3. You are right. you cant find a solution until you find k. that was my mistake. I didnt even think about finding a determinant. But what i am wondering is how do you find the determinate of this matrix because it has the column of solutions and is not just a 3x3 matrix. i havent learned about determinates yet in class but i know how to find it im just not sure how you got the 2k + 18 - 8 -12 + 24 - k = 0 because the matrix has the column of the 3 solutions 4. If you use proper row operations, you will find the same result. The determinant is not important. That's the thing with unique solutions, they don't care how you find them. As far as EXISTENCE, you do not need to be concerned with the constant side of things. Just the coefficients on the variables is enough. Determinant are not just parlor games. They are useful for things. This is one of those useful things. It's a rule. If the determinant is zero, there is no solution. Write it down. You do not have to see a 3x3 determinant out of the air. Feel free to use Expansion by Minors or however else you may have learned to do it. 5. I am stuck on my row operations I am supposed to be trying to get the matrix to reduced row echelon form correct? I am stuck at this 1 0 10 \| 6 0 1 -7 \| -4 0 0 k+22 \|1 6. No, you are not stuck. You are DONE! k+22 = 0 -- Solve for k. Compare this to my determinant equation. Good work. 7. Originally Posted by mybrohshi5 I am stuck on my row operations I am supposed to be trying to get the matrix to reduced row echelon form correct? I am stuck at this 1 0 10 \| 6 0 1 -7 \| -4 0 0 k+22 \|1 A little more- if k+ 22 is NOT 0, then you can divide the last row by that and continue your row reduction to get a unique solution to the equation. If k+22= 0, you cannot do that and, since $0\ne 1$ there is no solution. If k+ 22= 0 and the last number in that row were also 0, then there would be an infinite number of solutions. 8. Originally Posted by TKHunny No, you are not stuck. You are DONE! k+22 = 0 -- Solve for k. Compare this to my determinant equation. Good work. My k+22 doesnt = 0 though i have it equal to 1 so k would actually be -21 and thats wrong cause the answer is -22 as your determinate stated. I have checked my operations multiple times so i know my matrix where i am stuck at is correct i just dont know what to do now 9. Originally Posted by HallsofIvy A little more- if k+2 is NOT 0, then you can divide the last row by that and continue your row reduction to get a unique solution to the equation. If k+22= 0, you cannot do that and, since $0\ne 1$ there is no solution. If k+ 22= 0 and the last number in that row were also 0, then there would be an infinite number of solutions. I am a little confused on what you are trying to say. sorry What do yo mean by "if k+2 is NOT 0, then you can divide the last row by that " is that supposed to be k+22? and then what can i divide by? Thank you 10. 0 = 0 is fine. This normally would mean the system is under-defined and there are infinitely many solutions. 11. Originally Posted by mybrohshi5 My k+22 doesnt = 0 though i have it equal to 1 so k would actually be -21 and thats wrong cause the answer is -22 as your determinate stated. I have checked my operations multiple times so i know my matrix where i am stuck at is correct i just dont know what to do now No, no, no! You do NOT have k+22= 1! The last row of your reduced matrix corresponds to the equation (k+22)z= 1. That is not solvable if and only if k+ 22= 0. If k+22 is NOT 0, you can divide the equation by it: z= 1/(k+22) and complete the solution. In terms of the augmented matrix, to complete the row reduction, you would divide the last row by k+ 22- that is not possible if k+22= 0. (Yes, I meant k+ 22, not "k+2". I've gone back and edited so no one will ever know I made that mistake!)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9619712233543396, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/157686/probability-of-passing-an-exam-last-thoughts	# probability of passing an exam (last thoughts) So we have seen before (here: probability of passing an exam (continued)) that the average number of hours I would have worked for an exam, assuming I work one hour for an exam that I can pass with a probability p, and assuming I can try to pass it at most N times, is $\sum_{k=0}^{N-1}q^k$. (q=1-p) I know have one last question. What if this exam is actually composed of r tests, which follow the rules described above (so all tests can be taken at most N times), and the test i can only be passed if i-1 is taken (so if I reach N once, the whole exam is stopped, and thus my number of hours spent into this exam stops growing). What would then be the average number of hours spent into the exam ? If the question is unclear please state so and I'll try to reformulate. edit: maybe a little more explanation is needed as to why the exam requires on average $\sum_{k=0}^{N-1}q^k$ hours of work: $E(X)=\sum_{i=1}^{N-1}iP(X=i)+P(X=N)$, and $P(X=N)=pq^N$. The sum can be calculated by noticing that it's a sum of derivative, and it comes $E(X)=\frac{p}{(1-q)^2} \left( (N-1)q^{N}-Nq^{N-1}+1 \right)$+$pq^N$=(1-q^N)/(1-q) My question has still not been answered. - ## 1 Answer Well if $X$ is the random variable modelling the number of hours spent into the exam then $$E[X] = \sum_{i=1}^N P(X=i)i$$ $$E[X] = \sum_{i=1}^N q^{i-1}p i = p \frac{\partial}{\partial q}\left( \sum_{i=0}^N q^i \right) = \frac{p}{(1-q)^2} \left( Nq^{N+1}-(N+1)q^N+1 \right)$$ This is for your first problem. The second problem follows. - That's very good, thank you. I had actually already found that before, sorry I didn't add the link. I will do so now. Also, there's a slight mistake in your calculus, because the last try has a special treatment. In the end, you end up with a solution very much similar to yours, but which simplify to $(1-q^{N-1})/(1-q)$. So I'm actually more interested in the second part of the question. Please consider for the second part that $E[X]=\sum_{k=0}^{N-1} q^k$. – TimZEI Jun 13 '12 at 8:41 So how to actually solve the question I asked ? The answer above is only the answer to the previous question, which had already been answered... – TimZEI Jun 13 '12 at 13:09	{"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": 9, "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": 0.9759551882743835, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/48674/what-can-e-mc2-do?answertab=oldest	# What can $E=mc^2$ do? In the famous equation $E=mc^2$, the variables stand for: $E$ is energy, $m$ is mass, and $c$ is the speed of light (in vacuum). And I understand the equation fairly but limited in knowing in its effects/outputs in reality, honestly. • Some example? Could anybody please give me some sort of examples in real world which are done by this equation? - 6 Nuclear fission, and fusion. – ramanujan_dirac Jan 9 at 2:46 ## 6 Answers Equations don't do, they describe. This one describes the equivalence between a mass at rest and energy in a system. Some measurable mass gets converted to energy when some unstable atomic nuclei break apart into others. In other instances putting together of nuclei yields combinations which release excess energy by together having less rest mass. Those are fission reactions and fusion reactions, ie atom bombs and reactors and thermonuclear bombs and the sun respectively Even the much weaker breaking and forming of atomic bonds by electrons stores in mass and releases as energy from and to the surrounding but the changes in mass are not easily directly measured due the smallness of the mass changes. - So is there nothing it can do rather then bombs, mostly? – 夏期劇場 Jan 9 at 3:03 This equation is incredibly generic and describes many phenomena outside of nuclear phenomena. For example: place the the following setup in a box (a spring and some bars) and weigh them: . Now, loosen the spring and repeat. You should measure a smaller mass because you've removed some of the energy. In reality you couldn't possibly measure the difference in this experiment. The energy content of a typical spring is in the millijoules, so the mass difference is $E/c^2 = 10^{-20}\;\textrm{kg}$. That's the mass of a virus. However, in atomic system, such a difference is (barely) measurable. Mass of vacuum Now, for a much weirder example: the vacuum state. It's well known from quantum mechanics that the energy of a standing wave, such as a spring or photon in a box, has a minimum zero-point energy. An empty box is filled with many possible photon modes, even if it has no photons in it. Each of these modes has some energy. If that doesn't make any sense, that's okay, here's my point: quantum mechanics predicts that an empty box has energy. That means an empty box has mass. Is it a lot? Well, let's consider only photons. I want to calculate the total energy of all the modes in a box, so I integrate the number of standing waves per energy times the energy. $$E_\textrm{zero} = \int \rho(E) E \;dE = \frac{8\pi V}{(hc)^3} \int E^3\;dE$$ Now, this is the kind of integral you never want to see because it's badly divergent. Physicists usually try to bluff their way out of these problems by saying: "I know how physics works at low energy, but I have no idea what happens at very high energy. So, I'll just integrate until I get to the part where something else happens." So, we integrate up to $E_\textrm{cutoff}$ and find $E_\textrm{zero} = (2\pi V E_\textrm{cutoff}^4)/(hc)^3$. What do we use as a cutoff? An optimist says that we understand everything up to the Planck energy of $10^{28}\;\textrm{eV}$. A pessimist says we've got everything up to the LHC, which is $10^{14}\;\textrm{eV}$. $$E_\textrm{zero} = \left\{ \begin{array}{ll} 10^{111}\;\textrm{J/m}^3 & \textrm{optimist} \\ 10^{55}\;\textrm{J/m}^3 & \textrm{pessimist} \end{array}\right.$$ These are some very large numbers. Note that the optimist and pessimist disagree by 56 orders of magnitude! Do they have meaning? Well, yes and no. 1. Yes, when we look at differences. If there is an energy density, as shown here, there is also a force. You would expect that a box would feel a force inward. This has been observe and is called the Casimir Force. It turns out that the cutoff energy we used does not matter. 2. No, when we consider the actual value. If that's the energy density of the box, then the mass density is $10^{38} - 10^{94} \;\textrm{kg/m}^3$. (We're finally back to your question!) We know from astronomical observations of the universe's expansion that the actual mass density is something around $10^{-26} \;\textrm{kg/m}^3$. So, we're off by something around 64 to 120 orders of magnitude. This is an unbelievably large error (and error bar!). So, instead of telling you why $E=mc^2$ is useful, I've shown you an example where it's 120 orders of magnitude off, which happens when you compare a prediction from quantum mechanics with observations from cosmology. I hope that's more interesting. - The more general equation is: $$E=\sqrt {(mc^2)^2+(pc)^2}$$ where p is momentum $E=mc^2$ is a special case when momentum of system is zero. This popular equation says following things: • Energy has all properties of mass. It means, energy also posses momentum. And, it can be influenced by gravity. It can create gravity, too. • Mass has all properties of energy. Consequently, it can be converted to another form of energy. In fact, mass is most concentrated form of energy. Its applications: • New particles can be created with enormous amount of energy in other forms. Currently, its useful for research purpose only. But, who knows, one day we'd utilize it to synthesize objects using matter patterns stored in computer memory. • Matter can be converted to other forms of energy. Nuclear bombs, Sun etc do the same thing, but tapping out energy isn't easy thing. Nuclear bombs, Sun etc destroy less than 1% of their fuel to make other forms of energy. In acceretion disk of black holes, the number is 43% thanks to enormous gravity. But, there's one process in which the number is 100%: Matter-Antimatter annihilation. Matter-Antimatter annihilation can be used in energy sources of future mankind or in ultimate weapons of future. The problem is: We don't have access to antimatter repo. To create antimatter, we need more energy than annihilation output. Currently, antimatter is created for research purposes only. - I am answering to the comment in the answer of @Bobrebock So is there nothing it can do rather then bombs, mostly? – 4lvin 1. nuclear radiation is used for medical purposes, from x-rays to cancer treatment. It depends on this conversion 2. It is used to sterilize objects in large quantities. 3. It is a diagnostic method using the table of isotopes for innumerable needs, from paleontology ( + paleoclimate) to astrophysics, to medical tests, to checking underground sources etc And of course when fusion becomes commercial once ITER works, or some other competing method, it will supply all the energy needs of the world at low cost. So no, it is not only good for bombs. - That's not really a "real life" example, but something to give you an idea of how energy is related to speed. It pretty much describes the ultimate energy output from a perfect energy source. Let's assume, that unlike in fusion and fission where only a minor part of the energy is used up, we produce energy through annihilation of matter and antimatter where it all gets converted to energy. The old good "kinetic energy" is $$E={m_{payload}v^2 \over 2} \\ E=m_{fuel}c^2$$ Now to propel a mass of $m=1kg$ to "newtonian equivalent of speed of light" (subjectively speed of light from the point of view of the crew/payload), $v=c$ $${m_{payload} c^2 \over 2} = m_{fuel}c^2 \\ m_{payload} = 2 m_{fuel}$$ So to propel a ship of 10 tons to speed that would seem like speed of light from viewpoint of the crew, you'd need to "burn" 2.5 tons of matter and 2.5 tons of antimatter. (and that's not counting energy required to propel that fuel...) - "(subjectively speed of light from the point of view of the crew/payload)" What? From the point of view of the crew/payload they are at rest. Additionally, the speed of light as measured by the crew/payload is same the speed of light regardless of whether they are moving or not. You're also using the non-relativistic kinetic energy in what seems like a manifestly relativistic situation (I think - it's hard to tell from your description). I'm afraid this will just confuse the OP unnecessarily. – Michael Brown Jan 9 at 9:53 @MichaelBrown: No, if there's 4 light years from here to Proxima Centauri, and we put enough energy into accelerating the ship that it would travel at speed of light according to Newtonian physics (its kinetic energy is m*c^2/2), then exactly 4 years will pass on the crew's clocks before they reach the destination. And no, from their point of view they are definitely not at rest, from their point of view they are traveling the universe exactly at speed of light (or if you really love to nitpick, the y are at rest but the universe moves past them at that speed.) – SF. Jan 9 at 10:22 You're ignoring relativity which changes everything (and is required for the OP's question to even make sense), but I'll grant you that for the sake of argument. Your first sentence is okay. Your second is not. Even Newtonian mechanics is invariant under Galilean boosts. Their rest frame is their rest frame. They see the universe moving past them at the speed of light. In any case I don't see how this at all relevant to the OP's question. If you simply want to illustrate how much energy $mc^2$ is there are better ways to do that. – Michael Brown Jan 9 at 10:42 @MichaelBrown: You are nitpicking. If you're walking down the street, and a friend calls you asking you what are you doing, will you insist you just move your legs in place and the street is moving under you? It's the ages old conflict of strict definitions versus common language. Common language is not wrong, it's merely imprecise, but it's far easier to understand for layman, and explaining the problem in layman's terms is in place here. – SF. Jan 9 at 11:21 3 We're not talking about walking here. We're talking about interstellar space travel at near the speed of light. If you're not looking out the window you have no motivation whatsoever to say anything about your motion! Anyway, even if you forget about all of that you are still using non-relativistic mechanics to talk about a relativistic situation in the context of a relativistic question for reasons I can't understand. Relativistic math is not that much more complicated and has the advantage of being correct. I have nothing more to add to this conversation at this point. – Michael Brown Jan 9 at 11:30 I find particularly simple to understand the example of binding energy and in particular the so-called mass deficit of atoms. Mass change (decrease) in bound systems, particularly atomic nuclei, has also been termed mass defect, mass deficit, or mass packing fraction. The difference between the unbound system calculated mass and experimentally measured mass of nucleus (mass change) is denoted by Δm. It can be calculated as follows: $$\Delta m = \sum m_{unbound} - m_{measured}$$ i.e., (sum of masses of protons and neutrons) - (measured mass of nucleus) The difference in mass is actually due to a difference in energy. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "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": 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": 0.9516909122467041, "perplexity_flag": "middle"}
http://mathhelpforum.com/algebra/188244-simple-equation-help.html	# Thread: 1. ## Simple equation help? Hi there, This popped up in my exam the other day, absolutely no idea how to solve it using secondary school knowledge: 2^x = -x 2. ## Re: Simple equation help? You will need to solve it numerically, guess and check or technology. Graph $y = 2^x$ & $y= -x$ , to get a rough idea on where they intersect. 3. ## Re: Simple equation help? Originally Posted by eskimogenius Hi there, This popped up in my exam the other day, absolutely no idea how to solve it using secondary school knowledge: 2^x = -x The equation You have written is trascendental and its solution has to be found numerically. A non well popular but elegant way to solve it is to writre the equation as $f(x)= -2^{x}-x=0$ [th sign '-' has a precise scope...] and then compute for n large enough the n-th term of the sequence defined by the 'recursive relation' ... $x_{n+1}=-2^{x_{n}}\ ;\ x_{0}=0$ (1) In few step You arrive at the solution $x \sim -.641186$. The reason for which $x_{n}$ tends to the [real] solution of Your equation will be explained in a tutorial post written in the section 'Discrete mathematics'... Kind regards $\chi$ $\sigma$ 4. ## Re: Simple equation help? Originally Posted by chisigma The equation You have written is trascendental and its solution has to be found numerically. A non well popular but elegant way to solve it is to writre the equation as $f(x)= -2^{x}-x=0$ [th sign '-' has a precise scope...] and then compute for n large enough the n-th term of the sequence defined by the 'recursive relation' ... $x_{n+1}=-2^{x_{n}}\ ;\ x_{0}=0$ (1) In few step You arrive at the solution $x \sim -.641186$. The reason for which $x_{n}$ tends to the [real] solution of Your equation will be explained in a tutorial post written in the section 'Discrete mathematics'... Kind regards $\chi$ $\sigma$ Sorry, I have no idea what this means, or how to do this. I have graphed the lines, and realised that they intersect between 0 and -1, however, I don't know how to arrive at the solution. Can it be done using simple log laws? 5. ## Re: Simple equation help? the solution is probably transcendental. Do you know newton's method? 6. ## Re: Simple equation help? Originally Posted by eskimogenius Sorry, I have no idea what this means, or how to do this. I have graphed the lines, and realised that they intersect between 0 and -1, however, I don't know how to arrive at the solution. Can it be done using simple log laws? The equation... $f(x)=- 2^{x}-x=0$ (1) ... cannot be solved using simple log laws!... once You have found in graphical way that the solution is between -1 and 0, if You need better accuracy the only choice You have are numerical methods. Among them one of the most simple [and elegant...] even if not widely used is to consider the solution as the limit [if it exists...] of the sequence defined by the recursive relation... $\Delta_{n}= x_{n+1}-x_{n}= f(x_{n})$ (1) In Your case the (1) is... $x_{n+1}= -2^{x_{n}}$ (2) What You have to do now is to compute iteratively the $x_{n}$ till to an n 'large enough'. Starting with $x_{0}=0$ You easily find... $x_{1}= -2^{0}= -1$ $x_{2}= -2^{-1}= -.5$ $x_{3}= -2^{-.5}= -.707106...$ $x_{4}= -2^{-.707106...}= - .612547...$ $x_{5}= -2^{-.612547...}= -.65404...$ $x_{6}= -2^{-.65404...}= -.6354978...$ $x_{7}= -2^{-.6354978...}= -.6437186...$ $x_{8}= -2^{-.6437186...}= -64006102...$ ... and so one. The convergence to the solution $x \sim -.641186$ is evident... Kind regards $\chi$ $\sigma$ 7. ## Re: Simple equation help? Originally Posted by eskimogenius Hi there, This popped up in my exam the other day, absolutely no idea how to solve it using secondary school knowledge: 2^x = -x Clearly you are not expected to solve it by hand, for the simple fact that it cannot be solved exactly by hand or otherwise (unless you use a special function called the Lambert W-function). Did you have a graphics or CAS calculator in the exam? You are expected to know how to use your calculator to solve it. #### Search Tags View Tag Cloud Copyright © 2005-2013 Math Help Forum. All rights reserved.	{"extraction_info": {"found_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": 30, "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": 0.9430882930755615, "perplexity_flag": "middle"}
http://www.tutorhero.net/tag/number-theory	## Solution of FLT for $n=4$. April 24, 2011 I discussed FLT in an earlier post and stated a problem that if solved FLT for will be deduced from there. Below is a solution to that problem. Assume is a “non-trivial” triple of integers that satisfies (1) . By “non-trivial” we mean “". Obviously we can assume that and are positive. If we show for any such triple, there is another triple satisfying the same equation with a smaller , then using the method of “infinite descent” we get a contradiction, and therefore we can deduce that there is no “non-trivial” solutions for . Now one can show that and are relatively prime. Otherwise you can find a “smaller” solution for the equation (1).(more...) ## Fermat's Last Theorem April 21, 2011 One of the most important achievements of the 20th century in Math is a proof of the Fermat's Last Theorem. It states: Fermat's Last Theorem. Let and be 4 integers such that and . Then . In our previous post we discussed all solutions of this equation for as Pythagorean Triples. Proof of FLT involves advanced Mathematics. Here I would like to bring up a especial case of this theorem for . More generally one can prove the following. Problem. If for 3 integers and , then . One can prove the above statement using infinite descent and combining it with Pythagorean Triples. Think about it if you get a chance. A proof will be posted later. ## Pythagorean Triples April 20, 2011 Any three integers and that can be lengths of sides of a right triangle is a called Pythagorean triple. But we can also think of negative numbers as Pythagorean triples when they satisfy the equation . Here I would like to show you a simple method to find all Pythagorean triples. Asssume is the greatest common divisor of and , i.e. . If we divide and with obviously we get another Pythagorean triples. Therefore to find all Pythagorean triples it is enough to find all triples with g.c.d.=1. Therefore we may assume and have no common factors. One can check that one of or has the same parity as . (why?) Assume and have the(more...) ## Infinite Descent April 4, 2011 Infinite descent is a common mathematical method that we use in solving problems and proofs. It is based on a very simple and intuitive fact: every non-empty set of natural numbers has a smallest number. This means if you show a set of natural numbers has no smallest number that set should be empty. This method is commonly used to attack some problems in number theory. Especially for finding solutions to some specific Diophantine equations. I understand that wikipedia is not the most reliable source of information out there, but a lot of their articles are very useful, so I suggest you take a look at this wiki article to find out more about Infinite(more...) Do you have any questions that I can help you with? © 2010 Tutor Hero \| All Rights Reserved	{"extraction_info": {"found_math": true, "script_math_tex": 1, "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": 1, "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": 0.9274515509605408, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/54349/does-the-standard-model-have-a-landau-pole	# Does the Standard Model have a Landau pole? I have seen the statement that the Standard Model has a Landau pole, or at least it its believed that it does at $\sim 10^{34}$ GeV. Has this actually been proven (at least in perturbation theory, as in QED) or what kind of evidence is there to support this? - – Vibert Feb 19 at 15:29 @Vibert: The $\lambda$ result in that paper is interesting, but the $U(1)$ coupling shows no signs of slowing down. I think right now I'd bet that the Standard Model has at least one Landau pole. – user1504 Feb 20 at 13:16	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9668713212013245, "perplexity_flag": "head"}
http://mathoverflow.net/questions/68088?sort=votes	## Compatibility of adjoint action with comultiplication in a Hopf algebra ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm not especially well-versed in Hopf algebra theory, so apologies in advance if the following question has a very easy answer. Given a Hopf algebra $H$, let $\Delta$ denote the comultiplication, $\sigma$ the coinverse, and $$ the adjoint action of $H$ on itself. Then, explicitly, the adjoint action is given by $XY = \sum X' Y \sigma(X'')$, where $\Delta X = \sum X' \otimes X''$. We can extend the adjoint action to an action of $H$ on $H \otimes H$ in the standard way that one usually extends a left Hopf algebra action to a tensor product; I'll also call this action $$. In the case that $H$ is the hyperalgebra (or enveloping algebra) of the unipotent radical of a Borel of a semisimple algebraic group, it appears to be the case that $\Delta(XY) = X\Delta(Y)$ for all $X$ and $Y$. My question is: Are there simple conditions on a Hopf algebra, eg cocommutativity, that imply $\Delta(XY) = X\Delta(Y)$ for all $X$ and $Y$ in $H$? Or is this always true? I have a somewhat ugly ad-hoc proof for the case mentioned above, but this seems like the kind of thing that should follow easily from simple facts in Hopf algebra theory. - By "coinverse" you mean "antipode"? – darij grinberg Jun 17 2011 at 20:10 If yes, cocommutativity is indeed enough. I'll post a proof in a few minutes. – darij grinberg Jun 17 2011 at 20:12 Yes, "coinverse" does mean "antipode." – Chuck Hague Jun 18 2011 at 18:47 ## 1 Answer Yes, this is true for any cocommutative Hopf algebra. Let me rewrite your statement using my (or actually my professor's) brand of Sweedler notation: I write $x_{(1)}\otimes x_{(2)}$ for $\Delta\left(x\right)$, without sum sign. And I denote the antipode of the Hopf algebra $H$ by $S$. Also I will denote your action by $\rightharpoonup$ rather than by $$ (so I will write $X\rightharpoonup Y$ for your $XY$) because I prefer to use $$ for convolution. So we have $X\rightharpoonup Y = X_{(1)} Y S\left(X_{(2)}\right)$ for all $X,Y\in H$. Now, any $X,Y\in H$ satisfy $\Delta\left(X\rightharpoonup Y\right) = \Delta\left(X_{(1)} Y S\left(X_{(2)}\right)\right) = \Delta\left(X_{(1)}\right) \Delta\left(Y\right) \Delta\left(S\left(X_{(2)}\right)\right)$ (since $\Delta$ is an algebra homomorphism), whereas $X\rightharpoonup \Delta\left(Y\right) = X\rightharpoonup \left(Y_{(1)} \otimes Y_{(2)}\right) = \left(X_{(1)} \rightharpoonup Y_{(1)}\right) \otimes \left(X_{(2)} \rightharpoonup Y_{(2)}\right)$ `$= \left(X_{(1)}\right)_{(1)} Y_{(1)} S\left(\left(X_{(1)}\right)_{(2)}\right) \otimes \left(X_{(2)}\right)_{(1)} Y_{(2)} S\left(\left(X_{(2)}\right)_{(2)}\right) $` `$= X_{(1)} Y_{(1)} S\left(X_{(2)}\right) \otimes X_{(3)} Y_{(2)} S\left(X_{(4)}\right) $` `$= X_{(1)} Y_{(1)} S\left(X_{(4)}\right) \otimes X_{(2)} Y_{(2)} S\left(X_{(3)}\right) $` (since $X_{(1)} \otimes X_{(2)} \otimes X_{(3)} \otimes X_{(4)} = X_{(1)} \otimes X_{(4)} \otimes X_{(2)} \otimes X_{(3)}$, which is because $H$ is cocommutative) `$= \underbrace{\displaystyle \left(X_{(1)} \otimes X_{(2)}\right)}_{\displaystyle =\Delta\left(X_{(1)}\right)} \underbrace{\displaystyle \left(Y_{(1)} \otimes Y_{(2)}\right)}_{\displaystyle = \Delta\left(Y\right)} \underbrace{\displaystyle \left(S\left(X_{(4)}\right) \otimes S\left(X_{(3)}\right)\right)}_{\displaystyle = \Delta\left(S_{(3)}\right)\text{ (since }S\text{ is a anti-coalgebra homomorphism)}}$` $= \Delta\left(X_{(1)}\right) \Delta\left(Y\right) \Delta\left(S\left(X_{(2)}\right)\right) = \Delta\left(X\rightharpoonup Y\right)$, qed. This can be cast in diagram-chasing form, but that will make it totally unreadable. What I used are the facts that the antipode of a Hopf algebra is an anti-coalgebra homomorphism, and that if a coalgebra is cocommutative, then tensors of the form $X_{(1)} \otimes X_{(2)} \otimes ... \otimes X_{(n)}$ are symmetric. Should I prove any of these? - 1 For a cocommutative algebra, you can take it further away and simply write $x\otimes x$ :) – Mariano Suárez-Alvarez Jun 17 2011 at 20:50 Great, thanks! The main fact I was missing is that tensors of that form are symmetric in a cocommutative Hopf algebra. Is that easy to see? – Chuck Hague Jun 18 2011 at 18:45 1 Yes. It is enough to show that they are invariant under any transposition, i. e. that $X_{(1)}\otimes X_{(2)}\otimes ...\otimes X_{(n)} = X_{(1)}\otimes X_{(2)}\otimes ...\otimes X_{(i+1)} \otimes X_{(i)} \otimes ... \otimes X_{(n)}$ (on the right hand side, we have transposed $X_{(i)}$ and $X_{(i+1)}$ while leaving the rest of the tensor invariant). But the left hand side of this is equal to $X_{(1)}\otimes X_{(2)}\otimes ...\otimes X_{(i-1)} \otimes \Delta\left(X_{(i)}\right) \otimes X_{(i+1)} \otimes X_{(i+2)} \otimes ...\otimes X_{(n-1)}$, while the right hand side ... – darij grinberg Jun 18 2011 at 18:49 1 ... is equal to $X_{(1)}\otimes X_{(2)}\otimes ...\otimes X_{(i-1)} \otimes \left(\sigma\circ \Delta\right)\left(X_{(i)}\right) \otimes X_{(i+1)} \otimes X_{(i+2)} \otimes ...\otimes X_{(n-1)}$. (In fact, here we are using that $X_{(1)}\otimes X_{(2)}\otimes ...\otimes X_{(n)} = X_{(1)}\otimes X_{(2)}\otimes ...\otimes X_{(i-1)} \otimes \Delta\left(X_{(i)}\right) \otimes X_{(i+1)} \otimes X_{(i+2)} \otimes ...\otimes X_{(n-1)}$. This is a consequence of coassociativity. (Its proof depends on how exactly we define $X_{(1)}\otimes X_{(2)}\otimes ...\otimes X_{(n)}$...) – darij grinberg Jun 18 2011 at 19:15	{"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": 45, "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": 0.9009648561477661, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/119781/list	## Return to Answer 1 [made Community Wiki] Instead of writing $$\|x-y\|\le \varepsilon,$$ I used to write $$x\lessgtr y\pm \varepsilon.$$ You may read it as $x$ is more-or-less $y$ plus-minus $\varepsilon$. One may also write something like $$x\lessgtr e^{\pm\varepsilon}\cdot y$$ which is much better than $$\|\ln(y/x)\|\le\varepsilon$$ It is easier to read, especially if instead of $x$ and $y$ you have long expressions.	{"extraction_info": {"found_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": 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": 0.9551190733909607, "perplexity_flag": "head"}
http://medlibrary.org/medwiki/Fick's_law	# Fick's law 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: Molecular diffusion from a microscopic and macroscopic point of view. Initially, there are solute molecules on the left side of a barrier (purple line) and none on the right. The barrier is removed, and the solute diffuses to fill the whole container. Top: A single molecule moves around randomly. Middle: With more molecules, there is a clear trend where the solute fills the container more and more uniformly. Bottom: With an enormous number of solute molecules, randomness becomes undetectable: The solute appears to move smoothly and systematically from high-concentration areas to low-concentration areas. This smooth flow is described by Fick's laws. Fick's laws of diffusion describes diffusion and can be used to solve for the diffusion coefficient, D. They were derived by Adolf Fick in the year 1855. ## Fick's first law Fick's first law relates the diffusive flux to the concentration under the assumption of steady state. It postulates that the flux goes from regions of high concentration to regions of low concentration, with a magnitude that is proportional to the concentration gradient (spatial derivative). In one (spatial) dimension, the law is $\bigg. J = - D \frac{\partial Ci}{\partial x} \bigg.$ where • $J$ is the "diffusion flux" [(amount of substance) per unit area per unit time], example $(\tfrac{\mathrm{mol}}{ \mathrm m^2\cdot \mathrm s})$. $J$ measures the amount of substance that will flow through a small area during a small time interval. • $\, D$ is the diffusion coefficient or diffusivity in dimensions of [length2 time−1], example $(\tfrac{\mathrm m^2}{\mathrm s})$ • Ci (for ideal mixtures) is the concentration in dimensions of compound i[amount of substance per unit volume], example $(\tfrac{\mathrm {mol}}{\mathrm m^3})$ • $\, x$ is the position [length], example $\,\mathrm m$ $\, D$ is proportional to the squared velocity of the diffusing particles, which depends on the temperature, viscosity of the fluid and the size of the particles according to the Stokes-Einstein relation. In dilute aqueous solutions the diffusion coefficients of most ions are similar and have values that at room temperature are in the range of 0.6x10−9 to 2x10−9 m2/s. For biological molecules the diffusion coefficients normally range from 10−11 to 10−10 m2/s. In two or more dimensions we must use $\nabla$, the del or gradient operator, which generalises the first derivative, obtaining $\mathbf{J}=- D\nabla \phi$. The driving force for the one-dimensional diffusion is the quantity $- \frac{\partial \phi}{\partial x}$ which for ideal mixtures is the concentration gradient. In chemical systems other than ideal solutions or mixtures, the driving force for diffusion of each species is the gradient of chemical potential of this species. Then Fick's first law (one-dimensional case) can be written as: $J_i = - \frac{D c_i}{RT} \frac{\partial \mu_i}{\partial x}$ where the index i denotes the ith species, c is the concentration (mol/m3), R is the universal gas constant (J/(K mol)), T is the absolute temperature (K), and μ is the chemical potential (J/mol). If the primary variable is mass fraction ($y_i$, given, for example, in $\tfrac{\mathrm kg}{\mathrm kg}$), then the equation changes to: $J_i=- \rho D\nabla y_i$ where $\rho$ is the fluid density (for example, in $\tfrac{\mathrm kg}{\mathrm m^3}$). Note that the density is outside the gradient operator. ## Fick's second law Fick's second law predicts how diffusion causes the concentration to change with time: $\frac{\partial \phi}{\partial t} = D\,\frac{\partial^2 \phi}{\partial x^2}\,\!$ where • $\,\phi$ is the concentration in dimensions of [(amount of substance) length−3], example $(\tfrac{\mathrm{mol}}{m^3})$ • $\, t$ is time [s] • $\, D$ is the diffusion coefficient in dimensions of [length2 time−1], example $(\tfrac{m^2}{s})$ • $\, x$ is the position [length], example $\,m$ It can be derived from Fick's First law and the mass conservation in absence of any chemical reactions: $\frac{\partial \phi}{\partial t} +\,\frac{\partial}{\partial x}\,J = 0\Rightarrow\frac{\partial \phi}{\partial t} -\frac{\partial}{\partial x}\bigg(\,D\,\frac{\partial}{\partial x}\phi\,\bigg)\,=0\!$ Assuming the diffusion coefficient D to be a constant we can exchange the orders of the differentiation and multiply by the constant: $\frac{\partial}{\partial x}\bigg(\,D\,\frac{\partial}{\partial x} \phi\,\bigg) = D\,\frac{\partial}{\partial x} \frac{\partial}{\partial x} \,\phi = D\,\frac{\partial^2\phi}{\partial x^2}$ and, thus, receive the form of the Fick's equations as was stated above. For the case of diffusion in two or more dimensions Fick's Second Law becomes $\frac{\partial \phi}{\partial t} = D\,\nabla^2\,\phi\,\!$, which is analogous to the heat equation. If the diffusion coefficient is not a constant, but depends upon the coordinate and/or concentration, Fick's Second Law yields $\frac{\partial \phi}{\partial t} = \nabla \cdot (\,D\,\nabla\,\phi\,)\,\!$ An important example is the case where $\,\phi$ is at a steady state, i.e. the concentration does not change by time, so that the left part of the above equation is identically zero. In one dimension with constant $\, D$, the solution for the concentration will be a linear change of concentrations along $\, x$. In two or more dimensions we obtain $\nabla^2\,\phi =0\!$ which is Laplace's equation, the solutions to which are called harmonic functions by mathematicians. ### Example solution in one dimension: diffusion length A simple case of diffusion with time t in one dimension (taken as the x-axis) from a boundary located at position $x=0$, where the concentration is maintained at a value $n(0)$ is $n \left(x,t \right)=n(0) \mathrm{erfc} \left( \frac{x}{2\sqrt{Dt}}\right)$. where erfc is the complementary error function. The length $2\sqrt{Dt}$ is called the diffusion length and provides a measure of how far the concentration has propagated in the x-direction by diffusion in time t (Bird, 1976). As a quick approximation of the error function, the first 2 terms of the Taylor series can be used: $n \left(x,t \right)=n(0) \left[ 1 - 2 \left(\frac{x}{2\sqrt{Dt\pi}}\right) \right]$ If $D$ is time-dependent, the diffusion length becomes $2\sqrt{\int_0^{t'}D(t')dt'}$. This idea is useful for estimating a diffusion length over a heating and cooling cycle, where D varies with temperature. ### Generalizations 1. In the inhomogeneous media, the diffusion coefficient varies in space, $D=D(x)$. This dependence does not affect Fick's first law but the second law changes: $\frac{\partial \phi(x,t)}{\partial t}=\nabla\cdot (D(x) \nabla \phi(x,t))=D(x) \Delta \phi(x,t)+\sum_{i=1}^3 \frac{\partial D(x)}{\partial x_i} \frac{\partial \phi(x,t)}{\partial x_i}\$ 2. In the anisotropic media, the diffusion coefficient depends on the direction. It is a symmetric tensor $D=D_{ij}$. Fick's first law changes to $J=-D \nabla \phi \ , \mbox{ it is the product of a tensor and a vector: } \;\; J_i=-\sum_{j=1}^3D_{ij} \frac{\partial \phi}{\partial x_j} \ .$ For the diffusion equation this formula gives $\frac{\partial \phi(x,t)}{\partial t}=\nabla\cdot (D \nabla \phi(x,t))=\sum_{j=1}^3D_{ij} \frac{\partial^2 \phi(x,t)}{\partial x_i \partial x_j}\ .$ The symmetric matrix of diffusion coefficients $D_{ij}$ should be positive definite. It is needed to make the right hand side operator elliptic. 3. For the inhomogeneous anisotropic media these two forms of the diffusion equation should be combined in $\frac{\partial \phi(x,t)}{\partial t}=\nabla\cdot (D(x) \nabla \phi(x,t))=\sum_{i,j=1}^3\left(D_{ij}(x) \frac{\partial^2 \phi(x,t)}{\partial x_i \partial x_j}+ \frac{\partial D_{ij}(x)}{\partial x_i } \frac{\partial \phi(x,t)}{\partial x_j}\right)\ .$ 4. The approach based on the Einstein's mobility and Teorell formula gives the following generalization of Fick's equation for the multicomponent diffusion of the perfect components: $\frac{\partial \phi_i}{\partial t} =\sum_j {\rm div}\left(D_{ij} \frac{\phi_i}{\phi_j} {\rm grad} \, \phi_j\right) \, .$ where $\phi_i$ are concentrations of the components and $D_{ij}$ is the matrix of coefficients. Here, indexes i,j are related to the various components and not to the space coordinates. The Chapman-Enskog formulas for diffusion in gases include exactly the same terms. It should be stressed that these physical models of diffusion are different from the toy-models $\partial_t \phi_i = \sum_j D_{ij} \Delta \phi_j$ which are valid for very small deviations from the uniform equilibrium. Earlier, such terms were introduced in the Maxwell–Stefan diffusion equation. For anisotropic multicomponent diffusion coefficients one needs 4-index quantities, for example, $D_{ij\, \alpha \beta}$, where i, j are related to the components and α, β=1,2,3 correspond to the space coordinates. ## History In 1855, physiologist Adolf Fick first reported[1][2] his now-well-known laws governing the transport of mass through diffusive means. Fick's work was inspired by the earlier experiments of Thomas Graham, which fell short of proposing the fundamental laws for which Fick would become famous. The Fick's law is analogous to the relationships discovered at the same epoch by other eminent scientists: Darcy's law (hydraulic flow), Ohm's law (charge transport), and Fourier's Law (heat transport). Fick's experiments (modeled on Graham's) dealt with measuring the concentrations and fluxes of salt, diffusing between two reservoirs through tubes of water. It is notable that Fick's work primarily concerned diffusion in fluids, because at the time, diffusion in solids was not considered generally possible.[3] Today, Fick's Laws form the core of our understanding of diffusion in solids, liquids, and gases (in the absence of bulk fluid motion in the latter two cases). When a diffusion process does not follow Fick's laws (which does happen),[4][5] we refer to such processes as non-Fickian, in that they are exceptions that "prove" the importance of the general rules that Fick outlined in 1855. ## Applications Equations based on Fick's law have been commonly used to model transport processes in foods, neurons, biopolymers, pharmaceuticals, porous soils, population dynamics, nuclear materials, semiconductor doping process, etc. Theory of all voltammetric methods is based on solutions of Fick's equation. A large amount of experimental research in polymer science and food science has shown that a more general approach is required to describe transport of components in materials undergoing glass transition. In the vicinity of glass transition the flow behavior becomes "non-Fickian". It can be shown that the Fick's law can be obtained from the Maxwell-Stefan equations[6] of multi-component mass transfer. The Fick's law is limiting case of the Maxwell-Stefan equations, when the mixture is extremely dilute and every chemical species is interacting only with the bulk mixture and not with other species. To account for the presence of multiple species in a non-dilute mixture, several variations of the Maxwell-Stefan equations are used. See also non-diagonal coupled transport processes (Onsager relationship). ### Biological perspective The first law gives rise to the following formula:[7] $\text{Flux} = {-P (c_2 - c_1)}\,\!$ in which, • $\, P$ is the permeability, an experimentally determined membrane "conductance" for a given gas at a given temperature. • $\, c_2 - c_1$ is the difference in concentration of the gas across the membrane for the direction of flow (from $c_1$ to $c_2$). Fick's first law is also important in radiation transfer equations. However, in this context it becomes inaccurate when the diffusion constant is low and the radiation becomes limited by the speed of light rather than by the resistance of the material the radiation is flowing through. In this situation, one can use a flux limiter. The exchange rate of a gas across a fluid membrane can be determined by using this law together with Graham's law. ## Fick's flow in liquids When two miscible liquids are brought into contact, and diffusion takes place, the macroscopic (or average) concentration evolves following Fick's law. On a mesoscopic scale, that is, between the macroscopic scale described by Fick's law and molecular scale, where molecular random walks take place, fluctuations cannot be neglected. Such situations can be successfully modeled with Landau-Lifshitz fluctuating hydrodynamics. In this theoretical framework, diffusion is due to fluctuations whose dimensions range from the molecular scale to the macroscopic scale. [8] In particular, fluctuating hydrodynamic equations include a Fick's flow term, with a given diffusion coefficient, along with hydrodynamics equations and stochastic terms describing fluctuations. When calculating the fluctuations with a perturbative approach, the zero order approximation is Fick's law. The first order gives the fluctuations, and it comes out that fluctuations contribute to diffusion. This represents somehow a tautology, since the phenomena described by a lower order approximation is the result of a higher approximation: this problem is solved only by renormalizing fluctuating hydrodynamics equations. ### Semiconductor fabrication applications IC Fabrication technologies, model processes like CVD, Thermal Oxidation, and Wet Oxidation, doping, etc. use diffusion equations obtained from Fick's law. In certain cases, the solutions are obtained for boundary conditions such as constant source concentration diffusion, limited source concentration, or moving boundary diffusion (where junction depth keeps moving into the substrate). ## Derivation of Fick's 1st law in 1 dimension The following derivation is based on a similar argument made in Berg 1977 (see references). Consider a collection of particles performing a random walk in one dimension with length scale $\Delta x$ and time scale $\Delta t$. Let $N(x, t)$ be the number of particles at position $x$ at time $t$. At a given time step, half of the particles would move left and half would move right. Since half of the particles at point $x$ move right and half of the particles at point $x + \Delta x$ move left, the net movement to the right is: $-\frac{1}{2}\left[N(x + \Delta x, t) - N(x, t)\right]$ The flux, J, is this net movement of particles across some area element of area a, normal to the random walk during a time interval $\Delta t$. Hence we may write: $J = - \frac{1}{2} \left[\frac{ N(x + \Delta x, t)}{a \Delta t} - \frac{ N(x, t)}{a \Delta t}\right]$ Multiplying the top and bottom of the righthand side by $(\Delta x)^2$ and rewriting, we obtain: $J = -\frac{\left(\Delta x\right)^2}{2 \Delta t}\left[\frac{N(x + \Delta x, t)}{a (\Delta x)^2} - \frac{N(x, t)}{a (\Delta x)^2}\right]$ We note that concentration is defined as particles per unit volume, and hence $\phi (x, t) = \frac{N(x, t)}{a \Delta x}$. In addition, $\tfrac{\left(\Delta x\right)^2}{2 \Delta t}$ is the definition of the diffusion constant in one dimension, $D$. Thus our expression simplifies to: $J = -D \left[\frac{\phi (x + \Delta x, t)}{\Delta x} - \frac{\phi (x , t)}{\Delta x}\right]$ In the limit where $\Delta x$ is infinitesimal, the righthand side becomes a space derivative: $\bigg. J = - D \frac{\partial \phi}{\partial x} \bigg.$ ## Notes 1. A. Fick, Ann. der. Physik (1855), 94, 59, doi:10.1002/andp.18551700105 (in German). 2. A. Fick, Phil. Mag. (1855), 10, 30. (in English) 3. J. L. Vázquez (2006), The Porous Medium Equation. Mathematical Theory, Oxford Univ. Press. 4. A.N. Gorban, H.P. Sargsyan and H.A. Wahab (2011), Quasichemical Models of Multicomponent Nonlinear Diffusion, Mathematical Modelling of Natural Phenomena, Volume 6 / Issue 05, 184−262. 5. Taylor, Ross; R Krishna (1993). Multicomponent mass transfer. Wiley. 6. D. Brogioli and A. Vailati, Diffusive mass transfer by nonequilibrium fluctuations: Fick's law revisited, Phys. Rev. E 63, 012105/1-4 (2001) [1] ## References • W.F. Smith, Foundations of Materials Science and Engineering 3rd ed., McGraw-Hill (2004) • H.C. Berg, Random Walks in Biology, Princeton (1977) • R.B. Bird, W.E. Stewart, E.N. Lightfoot, Transport Phenomena, John Wiley & sons, (1976) • J. Crank, The Mathematics of Diffusion, Oxford University Press (1980) 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 "Fick's law", available in its original form here: http://en.wikipedia.org/w/index.php?title=Fick's_law • ## 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? If you have a question or comment about material specifically within the site’s encyclopedia supplement, we encourage you to read and follow the original source URL given near the bottom of each article. 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": 77, "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": 0.894914448261261, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/33282/list	## Return to Question 4 removed contentious phrase "freely available" Does there exist a set $A$ such that $A=\{A\}$ ? Edit(Peter LL): Such sets are called Quine atoms. Naive set theory By Paul Richard Halmos(freely available)Halmos On page three, the same question is asked. Using the usual set notation, I tried to construct such a set: First with finite number of brackets and it turns out that after deleting those finite number of pairs of brackets, we circle back to the original question. For instance, assuming $A=\{B\}$, we proceed as follows: $\{B\}=\{A\}\Leftrightarrow B=A \Leftrightarrow B=\{B\}$ which is equivalent to the original equation. So the only remaining possibility is to have infinitely many pairs of brackets, but I can't make sense of such set. (Literally, such a set is both a subset and an element of itself. Further more, It can be shown that it is singleton.) For some time, I thought this set is unique and corresponded to $\infty$ in some set-theoretic construction of naturals. To recap, my question is whether this set exists and if so what "concrete" examples there are. (Maybe this set is axiomatically prevented from existing.) 3 spelling Does there exist a set $A$ such that $A=\{A\}$ ? Edit(Peter LL): Such sets are called Quine atoms. Naive set theory By Paul Richard Halmos(feely avilableHalmos(freely available) On page three, the same question is asked. Using the usual set notation, I tried to construct such a set: First with finite number of brackets and it turns out that after deleting those finite number of pairs of brackets, we circle back to the original question. For instance, assuming $A=\{B\}$, we proceed as follows: $\{B\}=\{A\}\Leftrightarrow B=A \Leftrightarrow B=\{B\}$ which is equivalent to the original eqautionequation. So the only remaining possibility is to have infinitely many pairs of brackets, but I can't make sense of such set. (Literally, such a set is both a subset and an element of itself. Further more, It can be shown that it is singleton.) For some time, I thought this set is unique and corresponded to $\infty$ in some set-theoretic construction of naturals. To recap, my question is whether this set exists and if so what "concrete" examples there are. (Maybe this set is axiomatically prevented from existing.) 2 standard name Does there exist a set $A$ such that $A=\{A\}$ ? Edit(Peter LL): Such sets are called Quine atoms. Naive set theory By Paul Richard Halmos(feely avilable) On page three, the same question is asked. Using the usual set notation, I tried to construct such a set: First with finite number of brackets and it turns out that after deleting those finite number of pairs of brackets, we circle back to the original question. For instance, assuming $A=\{B\}$, we proceed as follows: $\{B\}=\{A\}\Leftrightarrow B=A \Leftrightarrow B=\{B\}$ which is equivalent to the original eqaution. So the only remaining possibility is to have infinitely many pairs of brackets, but I can't make sense of such set. (Literally, such a set is both a subset and an element of itself. Further more, It can be shown that it is singleton.) For some time, I thought this set is unique and corresponded to $\infty$ in some set-theoretic construction of naturals. To recap, my question is whether this set exists and if so what "concrete" examples there are. (Maybe this set is axiomatically prevented from existing.) 1 # Can we have A={A} ? Does there exist a set $A$ such that $A=\{A\}$ ? Naive set theory By Paul Richard Halmos(feely avilable) On page three, the same question is asked. Using the usual set notation, I tried to construct such a set: First with finite number of brackets and it turns out that after deleting those finite number of pairs of brackets, we circle back to the original question. For instance, assuming $A=\{B\}$, we proceed as follows: $\{B\}=\{A\}\Leftrightarrow B=A \Leftrightarrow B=\{B\}$ which is equivalent to the original eqaution. So the only remaining possibility is to have infinitely many pairs of brackets, but I can't make sense of such set. (Literally, such a set is both a subset and an element of itself. Further more, It can be shown that it is singleton.) For some time, I thought this set is unique and corresponded to $\infty$ in some set-theoretic construction of naturals. To recap, my question is whether this set exists and if so what "concrete" examples there are. (Maybe this set is axiomatically prevented from existing.)	{"extraction_info": {"found_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": 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": 0.9308905601501465, "perplexity_flag": "head"}

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.