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http://mathoverflow.net/questions/103175/what-are-the-fallacies-that-this-rh-inequality-may-fail-at-most-finitely-often/103181	## What are the fallacies that this RH inequality may fail at most finitely often? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) According to "EQUIVALENCES TO THE RIEMANN HYPOTHESIS p.4 Let $g(n)$ be the maximal order of a permutation of n objects RH Equivalence 3.3. The Riemann Hypothesis is equivalent to $\log{g(n)} < Li^{-1} (n)$ for n large enough. $Li$ is strictly increasing for $n>1$ and $\log{g(n)} \sim \sqrt{n \log{n}}$. $\log{g(n)} \ge \sqrt{n \log{n}}$ for $n \ge 906$ and $\log{g(n)} \gg 1$. Some equivalences using the fact that $Li$ and squaring preserve inequalities (because they are strictly increasing in the relevant intervals, easy to show) are: $$\log{g(n)} < \sqrt{Li^{-1}(n)} \iff (\log{g(n))^2 < Li^{-1}(n) \iff Li( (\log{g(n)})^2}) < n \qquad (1)$$ for $n$ large enough. In Effective Bounds for the Maximal Order of an Element in the Symmetric Group Theorem 2, p. 2 the following bound for $g(n)$ is given unconditionally for $n \ge 3$: $$\log{g(n)} \le F(n) = \sqrt{n \log{n}} \left( 1 + \frac{\log{\log{n}} - 0.975}{2 \log{n}}\right)$$ $$\log{g(n)} \le F(n) \iff (\log{g(n)})^2 \le F(n)^2 \iff Li((\log{g(n)})^2) \le Li(F(n)^2) \qquad (1a)$$ From (1) and (1a) showing $Li( (\log{g(n)) ^2}) \le Li(F(n)^2) < n$ (if true) will prove (1). Let $$G(n) = Li(F(n)^2) - n =$$ $$-n + {Li}\left(\frac{1}{4} \, {\left(\frac{\log\left(\log\left(n\right)\right) - 0.975}{\log\left(n\right)} + 2\right)}^{2} n \log\left(n\right)\right) =$$ $$-n + {\rm Ei}\left(\log\left(\frac{1}{4} \, {\left(\frac{\log\left(\log\left(n\right)\right) - 0.975}{\log\left(n\right)} + 2\right)}^{2} n \log\left(n\right)\right)\right) - {\rm Ei}\left(\log\left(2\right)\right)$$ We must show $$G(n)<0 \qquad (2)$$ (if true) for $n$ large enough. According to both sage and maple (using the $\rm Ei$ expression) $\lim_{n \to \infty} G(n) = -\infty$ so (2) may fail at most finitely often ($n \in \mathbb{N}$). The derivative of $G(n)$ is of elementary functions only. What are the mistakes and logical errors in this? - ## 2 Answers I suspect the mistake is in relying on Sage or Maple in the last step. Instead use the asymptotic expansion of li(x) ( http://en.wikipedia.org/wiki/Logarithmic_integral_function ) $$\operatorname{li}(x)=\frac x {\log x}+\frac{x}{\log^2 x}+O \left(\frac x {\log^3 x} \right)$$ to obtain an asymptotic expression of $G(n)$. Letting $$x=\frac 1 4 \left(\frac {\log \log n-0.975}{\log n}+2 \right)^2 n \log n$$ we see that $$x = n \log n \left(1+\frac{\log \log n}{\log n}- \frac{0.975}{\log n} +O \left( \frac{\log \log n}{\log^2 n} \right) \right)$$ and that $$\log x=\log n+\log \log n+\frac{\log \log n}{\log n}-\frac{0.975}{\log n}+O \left(\frac {\log \log n}{\log^2 n} \right).$$ With some more calculation we get $$\frac x {\log x}=n-\frac{0.975 n}{\log n}+O \left(\frac {n \log \log n}{\log^2 n} \right)$$ By the first two terms in the asymtotic expansion of li(x) we get that $$\operatorname{li} (x)=n+\frac{0.025n}{\log n}+O \left(\frac {n \log \log n}{\log^2 n} \right)$$ and thus $$G(n)=\frac{0.025 n}{\log n}+O \left(\frac {n \log \log n}{\log^2 n} \right)$$ and $G(n) \to \infty$ as $n\to \infty$. - Thank you ${}{}{}$ – joro Jul 27 at 8:52 ### 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. Don't blame Maple (blame the user). - Thank you! I confirm this on version 13. Can you try without "MultiSeries" and writing Li in terms of Ei (that is what worked for me). – joro Jul 27 at 8:51 $$\tt{> G := -n+Ei(1/4((log(log(n))-0.975)/log(n)+2)^2n*log(n))-Ei(log(2)):}$$ $$\tt{> limit(G,n=infinity);}$$ $$\tt{Float(infinity)}$$ – Gerald Edgar Jul 27 at 14:13 Thank you. The last result is certainly different in version 13. – joro Jul 27 at 14:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 22, "mathjax_display_tex": 17, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9036542177200317, "perplexity_flag": "middle"}
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http://unapologetic.wordpress.com/2007/10/06/spans-and-cospans/?like=1&source=post_flair&_wpnonce=7b62a7b6bf	# The Unapologetic Mathematician ## Spans and Cospans I was busy all yesterday with my talk at George Washington, so today I’ll make up for it by explaining one of the main tools that went into the talk. Coincidentally, it’s one of my favorite examples of a weak 2-category. We start by considering a category $\mathcal{C}$ with pullbacks. A span in $\mathcal{C}$ is a diagram of the form $A_1\leftarrow B\rightarrow A_2$. We think of it as going from $A_1$ to $A_2$. It turns out that we can consider these to be the morphisms in a weak 2-category $\mathbf{Span}(\mathcal{C})$, whose objects are just the objects of $\mathcal{C}$. Now since this is supposed to be a 2-category we need 2-morphisms to go between spans. Given spans $A_1\leftarrow B\rightarrow A_2$ and $A_1\leftarrow B'\rightarrow A_2$ a 2-morphism from the first to the second will be an arrow from $B$ to $B'$ making the triangles on the sides of the following diagram commute: We compose 2-morphisms in the obvious way, stacking these diamond diagrams on top of each other and composing the arrows down the middle. The composition for 1-morphisms is where it gets interesting. If we have spans $A_1\leftarrow B_1\rightarrow A_2$ and $A_2\leftarrow B_2\rightarrow A_3$ we compose them by overlapping them at $A_2$ and pulling back the square in the middle, as in the following diagram: where the middle square is a pullback. Then the outer arrows form a span $A_1\leftarrow C\rightarrow A_3$. Notice that this composition is not in general associative. If we have three spans $A_1\leftarrow B_1\rightarrow A_2$, $A_2\leftarrow B_2\rightarrow A_3$, and $A_3\leftarrow B_3\rightarrow A_4$, we could pull back on the left first, then on the right, or the other way around: But the object $D$ at the top of each diagram is a limit for the $A_1\leftarrow B_1\rightarrow A_2\leftarrow B_2\rightarrow A_3\leftarrow B_3\rightarrow A_4$ along the bottom of the diagram. And thus the two of them must be isomorphic, and the isomorphism must commute with all the arrows from $D$ to an object along the bottom. In particular, it commutes with the arrows from $D$ to $A_1$ and to $A_4$, and thus gives a canonical “associator” 2-morphism. I’ll leave the straightforward-but-tedious verification of the pentagon identity as an exercise. What is the identity span? It’s just $A\leftarrow A\rightarrow A$, where the arrows are the identity arrows on $A$. When we pull back anything along the identity arrow on $A$, we get the same arrow again. Thus the 2-morphisms we need for the left and right unit are just the identities, and so the triangle identity is trivially true. Thus we have constructed a 2-category $\mathbf{Span}(\mathcal{C})$ from any category with pullbacks. The concept of a cospan is similar, but the arrows (predictably) run the other way. A cospan in a category $\mathcal{C}$ with pushouts (instead of pullbacks) is a diagram of the form $A_1\rightarrow B\leftarrow A_2$. Everything else goes the same way — 2-morphisms are arrows in the middle making the side triangles commute, and composition of cospans goes by pushing out the obvious square. Unusual as it is for this weblog, I’d like to make one historical note. Spans and cospans were introduced by Bénabou in the early 1960s. This was first pointed out to me by Eugenia Cheng (of The Catsters) and John Baez (of The n-Category Café) at a conference at Union College two years ago. Until that point I’d been calling cospans “$\Lambda$-diagrams” because I’d invented them out of whole cloth to solve a problem that I’ll eventually get around to explaining. The name seemed appropriate because I tend to draw them (as I have in the above diagrams) as wedges with the arrows on the sides pointing up into the center, sort of like a capital lambda. They’ve shown up in a number of contexts, but as far as I can tell I’m the first (and so far only) to use them in knot theory like I do. However, Jeff Morton has noted that cobordisms are a sort of cospan, and my use of tangles is analogous to that. ### Like this: Posted by John Armstrong \| Category theory ## 6 Comments » 1. [...] there is a (bi)category , where spans are composed using pullbacks. John Armstrong recently posted about spans, describing , which has the same objects as , and morphisms which are spans in [...] Pingback by \| October 9, 2007 \| Reply 2. [...] and Cospans II There’s something we need to note about spans that will come in extremely handy as we start trying to add structure to our categories of [...] Pingback by \| October 9, 2007 \| Reply 3. [...] Structures on Span 2-Categories Now we want to take our 2-categories of spans and add some 2-categorical analogue of a monoidal structure on [...] Pingback by \| October 10, 2007 \| Reply 4. [...] Useless After All! Today on the arXiv, we find a posting of an old paper, which uses spans of “reflexive graphs” to give an algebraic framework for describing partita doppia [...] Pingback by \| March 18, 2008 \| Reply 5. Hey John, you’ve got a dangling latex above the second image. Comment by Jon McKenzie \| March 18, 2008 \| Reply 6. Thanks. fixing it. Comment by \| March 18, 2008 \| 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": 31, "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": 0.9421244859695435, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/82196/elliptic-orbital-integral/82201	## Elliptic orbital integral ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $F$ be a local field and $G = GL(n,F)$. Let $f$ be an element $C_c^\infty(G)$. Let $\gamma$ be an elliptic element of $G$ with irreducible characteristic polynomial. What are strategies to compute $$\int\limits_{G_\gamma \backslash G} \phi(g^{-1}\gamma g) \mathrm{d} g?$$ Due to a comment of Paul Broussous: Assume that $\phi$ is bi $GL(n,o)$ invariant (respective $O(n)$ o $U(n)$ invariant at real/complex places). Please give a reference. I know that Drinfeld has computed this for certain functions of specific type for $F$ non archimedean. Can the general computations be deduced from this? How does the space $G_\gamma \backslash G / GL(2, o)$ look? - 1 I guess that $x=\gamma$. I think your first question is not precise enough. What type of function $\phi$ are you interested in ? If $\phi$ is e.g. the characteristic function of ${\rm GL}({\mathfrak o}_F )$, then you have computations of the integral made by Langlands, Kottwitz, Rogawsky in very particular cases. I don't think that one can expect a nice general formula. But I may be wrong. – Paul Broussous Nov 29 2011 at 17:18 I edited the question. In fact, I need the computation only slightly more general then for the characteristic function of $K$. Can you give some papers, which contain the computations> I guess with Kottwitz you meant his paper "on Tamagawa numbers", right? – Marc Palm Nov 30 2011 at 7:27 ## 2 Answers You have introductions to the building of ${\rm GL}(n)$ in Brown's book "Buildings", or in Paul Garrett's book. The extended building is just the product of the non extended building $X$ with the real line $\mathbb R$, with the action $$g.(x,r)=(\ g.x \ , \ r+val_F (det(g)) )$$ It is the geometric realization of a simplicial complex whose vertex set is in equivariant bijection with the set of lattice in $F^n$. In the non extended building the stabilizer of a vertex is conjugate to $F^{\times} {\rm GL}(n,{\mathfrak o})$. In the extended building the stabilizer of a vertex is conjugate to ${\rm GL}(n,{\mathfrak o})$. When $\phi$ is the characteristic function of ${\rm GL}(n,{\mathfrak o})$, the value of the orbital integral is related to the number of lattices in $F^n$ fixed by $\gamma$, so to a set of fixed vertices in the extended building. Examples of calculations of orbital integrals using the building may be found: -- in the PhD thesis of J. Rogawski that you can find online : http://www.math.ucla.edu/~jonr/eprints.html -- in Langlands's book "base change for ${\rm GL}(2)$". -- in Kottwitz's PhD thesis : "Orbital integrals on ${\rm GL}_{3}$" Amer. J. Math. 102 (1980), no. 2, 327–384. - ### 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 coset space $G/{\rm GL}(n,{\mathfrak o}_F)$ may be seen as the set of vertices of the extended Bruhat-Tits building of ${\rm GL}(n,F)$. There is a second point of view that amounts to consider this set as the set of $k_F$-rational points of a variety defined over $k_F$ (the residue field of $F$) : an affine flag variety. This is the point of view used by Ngo Bao Chau to prove the Fundamental Lemma. - Is this a typo, you mean $G_\gamma \backslash G / GL(n,o)$? I am not familiar with Bruhat-Tits buildings, extended or not. Is there a intro specifically dealing with $GL(n)$? What you mention with the flag structure over $k_F$ seems really interesting to me, and also fits within my concept. Would you care to include a reference for this, with page if possible? Thanks a lot for your help. – Marc Palm Nov 30 2011 at 7:25	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 37, "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": 0.917614758014679, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/298122/solving-forward-equation?answertab=active	# Solving Forward Equation I've currently started reading 'Lectures on Partial Differential Equations' by Faris. Page 44 he states the following forward equation: $$J=a(y)p-\frac{1}{2}\frac{\partial \sigma(y)^2p}{\partial y} = 0$$ I understand how to solve this and obtain the following: $$I(y) = \int_{y_0}^{y}\frac{2}{\sigma(z)^2}a(z)dz$$ i.e. integrate once and separate the variables. He then states solving for $p$ produces: $$p(y)=C\frac{2}{\sigma(y)^2}exp(I(y))$$ I don't understand where the $\frac{2}{\sigma(y)^2}$ comes from. Can someone guide me in the right direction please? - ## 1 Answer It's been four days and you've managed to solve your own question. I'll now share it with you. Note $J=0$ and $\sigma(y)$ is a function on of variable (i.e. there is no time variable). 1. Rearrange the Forward Equation and use standard differential operators: $$a(y)p = \frac{1}{2}\frac{d(\sigma(y)^2p)}{dy}$$ 2. Multiply through by 2 and use the product rule: $$2a(y)p=\frac{d(\sigma(y)^2)}{dy}p+\frac{d(p)}{dy}\sigma(y)^2$$ 3. Now divide through by p and divide through by $\sigma(y)^2$: $$\frac{2a(y)}{\sigma(y)^2}=\frac{1}{\sigma(y)^2}\frac{d(\sigma(y)^2)}{dy}+\frac{1}{p}\frac{d(p)}{dy}$$ 4. Use an integration 'trick': $$\int_{}{}\frac{2a(y)}{\sigma(y)^2}dy=\int_{}{}\frac{1}{\sigma(y)^2}\frac{d(\sigma(y)^2)}{dy}dy+\int_{}{}\frac{1}{p}\frac{d(p)}{dy}dy$$ 5. Simplify! $$\int_{}{}\frac{2a(y)}{\sigma(y)^2}dy=\int_{}{}\frac{d(\sigma(y)^2)}{\sigma(y)^2}+\int_{}{}\frac{d(p)}{p}$$ 6. Perform rhs integration: $$\int_{}{}\frac{2a(y)}{\sigma(y)^2}dy=\ln \sigma(y)^2 +\ln P +\ln C$$ 7. Exponential and rearrange: $$p(y) = C\frac{1}{\sigma(y)^2}exp(\int_{}{}\frac{2a(y)}{\sigma(y)^2}dy)$$ Note: I am not sure where Mr Faris obtained the '2' in the final solution. Perhaps someone more knowledgable will be able to explain. I believe it is a mistake and the constant of integration should be correct. -	{"extraction_info": {"found_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": 10, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.9376962780952454, "perplexity_flag": "middle"}
http://nrich.maths.org/7405/clue	nrich enriching mathematicsSkip over navigation Calendar Capers Choose any three by three square of dates on a calendar page. Circle any number on the top row, put a line through the other numbers that are in the same row and column as your circled number. Repeat this for a number of your choice from the second row. You should now have just one number left on the bottom row, circle it. Find the total for the three numbers circled. Compare this total with the number in the centre of the square. What do you find? Can you explain why this happens? Adding All Nine Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself! Rotating Triangle What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle? What Numbers Can We Make? Stage: 3 Challenge Level: Begin by exploring what happens when you add two, three, four... numbers chosen from a set of bags containing $2$s, $4$s, $6$s and $8$s. Can you explain your findings? 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": 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": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9312363862991333, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/117923/list	## Return to Question 4 added 3 characters in body The following questions occurred to me while browsing this site and looking at Exercise 20 here. Question 1. Let $n>1$. Does there exist a countable dense subset $A\subset\mathbb{R}^n$ for which the complement $\mathbb{R}^n\setminus A$ is connecteddisconnected? EDIT. I deleted an erroneous paragraph. Let me add two more questions, the first one being Gerald Edgar's comment here, the second one correcting the erroneous paragraph. Question 2. Let $n>1$. Is it true that for any subset $A\subset\mathbb{R}^n$ of Hausdorff dimension less than $n-1$ the complement $\mathbb{R}^n\setminus A$ is connected? It seems that Joel's argument answers this in the affirmative as well. Question 3. Let $n>1$. Are there two countable dense subsets $A,B\subset\mathbb{R}^n$ whose complements are not homeomorphic? 3 further errors removed; added 23 characters in body The following questions occurred to me while browsing this site and looking at Exercise 20 here. Let $n>1$: Question 1. Let $n>1$. Does there exist a countable dense subset $A\subset\mathbb{R}^n$ for which the complement $\mathbb{R}^n\setminus A$ is connected? EDIT. I deleted an erroneous paragraph. Let me add two more questions, the first one being Gerald Edgar's comment here, the second one correcting the erroneous paragraph. Question 2. Let $n>1$. Is it true that for any subset $A\subset\mathbb{R}^n$ of Hausdorff dimension less than $n-1$ the complement $\mathbb{R}^n\setminus A$ is connected? It seems that Joel's argument answers this in the affirmative as well. Question 3. Let $n>1$. Are there two countable dense subsets $A,B\subset\mathbb{R}^n$ whose complements are not homeomorphic? According to Exercise 20 here such subsets exist, but I cannot see an example right away. 2 added 403 characters in body; added 1 characters in body; edited title # Connectedness of the complement of acountabledensesubsetsmallsubsets(extendedquestion) The following question questions occurred to me while browsing this site:. Let $n>1$: Question 1. Does there exist a countable dense subset $A\subset\mathbb{R}^n$ for which the complement $\mathbb{R}^n\setminus A$ is connected? For sets of EDIT. I deleted an erroneous paragraph. Let me add two more questions, the form first one being Gerald Edgar's comment here, the second one correcting the erroneous paragraph. Question 2. Is it true that for any subset $A=B^n$ (i.e. A\subset\mathbb{R}^n$of Hausdorff dimension less than$B\subset\mathbb{R}$is countable and dense) n-1$ the complement $\mathbb{R}^n\setminus A$ is homeomorphic to connected? It seems that Joel's argument answers this in the Baire space, hence it is totally disconnectedaffirmative as well.On the other hand, not all complements Question 3. Are there two countable dense subsets $\mathbb{R}^n\setminus A$ above A,B\subset\mathbb{R}^n\$ whose complements are not homeomorphicto the Baire space, at least according ? According to Exercise 20 here whose solution such subsets exist, but I would also appreciatecannot see an example right away. 1 # Connectedness of the complement of a countable dense subset The following question occurred to me while browsing this site: Question. Does there exist a countable dense subset $A\subset\mathbb{R}^n$ for which $\mathbb{R}^n\setminus A$ is connected? For sets of the form $A=B^n$ (i.e. $B\subset\mathbb{R}$ is countable and dense) the complement $\mathbb{R}^n\setminus A$ is homeomorphic to the Baire space, hence it is totally disconnected. On the other hand, not all complements $\mathbb{R}^n\setminus A$ above are homeomorphic to the Baire space, at least according to Exercise 20 here whose solution I would also appreciate.	{"extraction_info": {"found_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": 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": 0.9264333248138428, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/45635/list	## Return to Answer 2 fixing typos, and hyperlinking the title. I posted a paper on the arXiv: , Group Width which answers this question for manifolds of dim >3 $>3$ with sufficiently complicated fundamental group (there will be no finite set of blocks). As Greg Kupeberg Kuperberg said, there are many interesting variations which remain open and are a nice chalange challenge to technique, e.g. the case of simply connected manifolds. 1 I posted a paper on the arXiv: Group Width which answers this question for manifolds of dim >3 with sufficiently complicated fundamental group (there will be no finite set of blocks). As Greg Kupeberg said, there are many interesting variations which remain open and are a nice chalange to technique, e.g. the case of simply connected manifolds.	{"extraction_info": {"found_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": 0.9250079393386841, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/118667/open-problems-in-the-theory-of-compact-quantum-groups/118670	## Open problems in the theory of compact quantum groups ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) What are the important open problems in the theory of compact quantum groups? Or conjectures? Here is an example from An De Rijdt's Ph.D. thesis: Is every compact quantum group with the fusion rules of $SO(3)$ of the the form $A_{aut}(B,\varphi)$ for a finite dimensional $C^*$-algebra $B$ and a $\delta$-form $\varphi$ on $B$? See page 83 at the end of Chapter 3 in An De Rijdt's Ph.D. thesis, available online from https://perswww.kuleuven.be/~u0018768/students/derijdt-phd-thesis.pdf - ## 7 Answers I'm late to the party! Here are some I momentarily mused on. Sorry I couldn't find explicit literature stating these as open problems, but I think these are natural developments to the famous results. • Determine the von Neumann algebraic type (and the factoriality) of $A_o(F)$. Is anything known beyond the results of Vaes-Vergnioux (Duke Math. J., 2007)? • Prove the strong Baum-Connes conjecture a la Meyer-Nest for the (edit: discrete dual of) q-deformation of compact simply connected simple group $G$, beyond $SU_q(2)$ due to Voigt (Adv. Math. 2011). Even at the 'classical' limit, it seems to require the understanding of the $G_{\mathbb C}$-equivariant KK-theory of $G_{\mathbb C}/B$, which is hard as far as I understand. - Thanks! The party still goes on, as far as I am concerned. – Uwe Franz Feb 19 at 15:59 ### 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. An De Rijdt describes another open problem in her thesis, see the summary of Chapter 3 on page 2: The study of ergodic action of compact quantum groups. Here we have the results of Wassermann for compact groups saying, e.g., that $SU(2)$ admits ergodic action only on von Neumann algebras of finite type I. It is open if this is also true for, say, $SU(3)$. There is work by Boca, Landstad, Tomatsu, An De Rijdt for compact quantum groups. - 3 Kenny De Commer and Makoto Yamashita have classified all operator algebras which allow some $SU_q(2)$ as an ergodic symmetry group, in their paper on "Tannaka-Krein duality for compact quantum homogeneous spaces. II. Classification of quantum homogeneous spaces for quantum SU(2)", see lanl.arxiv.org/abs/1212.3413. – Uwe Franz Jan 16 at 13:23 Probably not important in any sense, but something I thought about very briefly recently, and asked originally by Woronowicz in the Pseudo Group paper: Let $(A,\Delta)$ be a CQG, and say that $a\in A$ is central if $\Delta(a)=\sigma\Delta(a)$ where $\sigma$ is the tensor swap map. You can show the all characters (traces of corepresentations) are central. Is the linear span of the characters norm dense in the central elements of A? For a compact group, I'd use an averaging argument to prove this (approximate a by arbitrary matrix elements of coreps, and then average into characters). But I think the naive version of this works iff your Kac. Does anyone know any progress on this? - 1 @Matthew: Yes, my student Francois Lemeux wrote this argument up in a preprint on the Haagerup property for the complex reflection quantum groups. But it only works in the Kac case. I haven't seen any progress in the general case. – Uwe Franz Jan 17 at 12:19 Finally on ArXiv --- here is a link to the preprint arxiv.org/abs/1303.2151. – Uwe Franz Mar 18 at 16:22 Questions concerning coamenability Here are few more open questions : a) Reiji Tomatsu stated in Reiji Tomatsu, Amenable discrete quantum groups, Journal of the Mathematical Society of Japan Vol. 58 (2006) No. 4 P 949-964, see also http://arxiv.org/abs/math/0302222 that nuclear compact Kac algebras (=compact quantum groups of Kac type, i.e. with a tracial Haar state) are coamenable, and gave as on open problem, if this true more generally for all compact quantum groups? Since Doplicher, Longo, Roberts, and Zsido (Reviews in Mathematical Physics, Vol. 14, Nos. 7 & 8 (2002) 787–796) have shown that coamenability implies nuclearity, the question is whether nuclearity and coamenability are equivalent for compact quantum groups? Does anybody know if there has been progress on this question since 2006? b) Under what conditions (like compact, discrete, Kac) is it true that a locally compact quantum group $G$ is amenable if and only if it's dual locally compact quantum group $\hat{G}$ is coamenable? Can anybody give references that contain the proofs for the new results? Many thanks in advance! In which cases is the question still open? - 2 As for (b), it's trivial for compact, shown by Reiji for discrete, and I think still wide open for Kac...? – Matthew Daws Jan 17 at 9:04 Links to few questions that have already been published on MO: a) http://mathoverflow.net/questions/86575/why-is-the-quantum-lorentz-group-not-connected Or: What does it mean for a a (compact) quantum group to be connected? (Ok, the (quantum) Lorentz group is not compact, but I find the question interesting). b) http://mathoverflow.net/questions/43331/quantum-group-calculations-in-mathematica Or more generally: What software is available for "quantum group calculations" (in Mathematica, Maple, Sage, GAP, etc.)? c) http://mathoverflow.net/questions/29810/weyl-character-formula-for-quantum-groups d) http://mathoverflow.net/questions/17861/bicovariant-calculi-on-the-quantum-unitary-groups e) http://mathoverflow.net/questions/118124/matrix-model-or-cocycle-twist-construction-for-q-deformations-of-compact-simple-l Or: Find exemples of compact quantum groups for which are "close" to classical compact groups, e.g. can be constructed via cocycle twists or can be described as matrix-valued functions on classical groups. - Well, but we still have to find more good reasons why these are the right numbers. Can they be linked, e.g., to the free Jacobi process? – Uwe Franz Jan 16 at 18:32 I found another open CQG problem on MO, there is even a reward of 3 bottles of champagne offered for solving it: J.-B. Zuber offered respectively 1, 2 and 3 bottles of Champagne for the classification of finite quantum subgroups of SUq(3), SUq(4), and SUq(5), respectively. Ocneanu solved the first two cases, but as far as I know the problem for SUq(5) is still open. See http://mathoverflow.net/questions/66084/open-problems-with-monetary-rewards/66133#66133 Is the case of $SU_q(5)$ really still open? Does anybody know references for Ocneanu's solutions? - I think it appeared in one "Contemporary Mathematics" under the name "The classification of subgroups of quantum SU(N)", but I've never looked much at it. – Amin Feb 19 at 7:33 Maybe the study (and explicite computation) of the 6j-Symbols/Wigner-Racah coefficients can also be considered as a problem in the theory of compact quantum groups... see http://mathoverflow.net/questions/15800/calculating-6j-symbols-aka-racah-wigner-coefficients-for-quantum-groups -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 22, "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": 0.8714815974235535, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/109680/does-zf-bound-countable-unions-of-countable-sets	## Does ZF bound countable unions of countable sets? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) ZF proves that whenever a countable union of countable sets can be well ordered then its cardinality is at most $\aleph_1$. But what if it cannot be well ordered? The Feferman-Levy model shows the continuum can be a countable union of countable sets. Is there a ZF proof that a countable union of countable sets must be smaller than or equal to that in size? - I think that should be $\aleph_0$ in the first sentence. – Michael Greinecker Oct 15 at 7:21 2 Michael, no. It should be $\aleph_1$. If $\aleph_1$ is singular then it is the countable union of countable ordinals; but $\aleph_2$ (even if singular) can never be the countable union of countable ordinals. – Asaf Karagila Oct 15 at 7:47 ## 1 Answer In Jech The Axiom of Choice, problem 14 on chapter 5 states: Let $M$ be a transitive model of ZFC, there exists $M\subseteq N$ with the same cardinals as $M$ [read: initial ordinals] and the following statement is true in $N$: For every $\alpha$ there exists a set $X$ such that $X$ is a countable union of countable sets, and $\mathcal P(X)$ can be partitioned into $\aleph_\alpha$ nonempty sets. Note that $\mathcal P(X)$ can only be mapped onto set many ordinals, so this ensures us that there is indeed a proper class of cardinalities which can be written as countable unions of countable sets. Jech adds and points out that this $N$ is not an inner model of any transitive model of ZFC, as that would imply $2^{\aleph_0}$ can be partitioned into any number of sets. The reference given is to Morris from 1970: D. B. Morris, A model of ZF which cannot be extended to a model of ZFC without adding ordinals, Notices Am. Math. Soc. 17, 577. I haven't sat down to verify all the details, but it should probably work: Using an Easton symmetry-like class forcing, we may obtain model add for every regular $\kappa$ a countable collection of countable subsets of $2^\kappa$ whose union is not countable, let us denote this union $A_\kappa$. Genericity arguments should give us that for $\kappa\neq\kappa'$, $\|A_\kappa\|$ is incomparable with $\|A_{\kappa'}\|$ - namely there are no injections between those two sets. This shows that there is a proper class of different sets which can be represented as countable unions of countable sets. So no upper-bound is possible. - I am working on something similar in nature, so I will have the answer whether or not this argument is correct some time in the future. I will post an update when I'm done. – Asaf Karagila Oct 15 at 8:01 1 mathoverflow.net/questions/33028/… $\:$ – Ricky Demer Oct 15 at 8:59 Ricky, and? Smaller cardinality need not be countable. – Asaf Karagila Oct 15 at 9:10 1 Oh yeah, I misread the question. $\:$ – Ricky Demer Oct 15 at 9:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9205268025398254, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/4827/is-the-total-energy-conserved-in-a-moving-reference-frame/4828	# Is the total energy conserved in a moving reference frame? [closed] I have a ball attached to a spring and the spring is attached to a wall. There is no gravity for simplicity. In the rest RF the oscillating ball energy is conserved: T + U = const. In a moving RF it is not conserved. I would like to see the shortest answer to the question "Why?". \| <-- --> \|/\/\/\/\/\/\/O \| =====> $V_{RF}-->$ Edit: For those who has doubt - I choose the moving RF velocity equal to the maximum ball velocity, for example. Then the ball with its maximum velocity in the rest RF looks as still and does not have any potential energy in the moving RF (the maximum velocity is attained at the equilibrium position where no force acts). So T = 0 and U = 0. At the farthest right position the ball looks as moving with $-V_{RF}$ and having a potential energy $kx^2/2$, both energy terms are positive ($v^2 > 0, x^2 > 0$). So now the total energy of the ball is positive and thus is not conserved. It oscillates from zero to some maximum value. Why? - 4 Downvoters, give your short objections, please. – Vladimir Kalitvianski Feb 8 '11 at 20:18 6 Minus one point for abusing the space that is reserved for questions for spreading primitive misconceptions about physics. Of course that energy is conserved in any - moving - reference frame. You may need to include the terms from the wall, too. But energy is conserved in any system with a time-translational invariance. – Luboš Motl Feb 8 '11 at 20:19 1 Thanks, Lubosh, you are so kind as usual. But I do not want to include the wall. I ask another question. So it is -1 to you. – Vladimir Kalitvianski Feb 8 '11 at 20:21 1 Maybe it is a good idea to rephrase the question and e.g. tag it as a [riddle]. Because it makes perfectly sense to consider such a RF and then address the apparent contradictions. One can learn a lot from studying such special situations. – Gerard Feb 8 '11 at 22:41 1 I deleted the tail of the comment thread, since it was getting out of control. If there continue to be argumentative comments posted, I'll lock the question. – David Zaslavsky♦ Feb 9 '11 at 0:15 show 8 more comments ## closed as not constructive by pho, Moshe, Matt Reece, Kostya, gigacyanFeb 8 '11 at 22:33 As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or specific expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, see the FAQ for guidance. ## 3 Answers Energy is not a Lorentz scalar (or Galilean scalar). In different reference frames, the values of energies wil also be different, but this does not mean that there is no energy conservation. The energy is still conserved in each reference frame. - 1 -1 for not answering my question. The energy is not only different but is not conserved. – Vladimir Kalitvianski Feb 8 '11 at 21:06 2 @Vlad You're right that the answerer did not directly answer your question, but the name-calling is unnecessary. – Mark Eichenlaub Feb 8 '11 at 21:58 We must consider the kinetic energy of the wall to see energy conservation. Taking the limit where the wall's mass goes to infinity does not alleviate this responsibility because although the wall's oscillations become infinitely small in that case, the oscillations in its energy do not. To see this roughly, imagine that the wall is very massive and moving sinusoidally. In the center of mass frame, the maximum speed of the wall is $s$. Then the fluctuations in its kinetic energy are proportional to $M s^2$. Thus, in the rest frame, we can make the wall's energy fluctuations arbitrarily small by increasing $M$, because $s^2$ will decrease more rapidly. Now go into a moving frame where the average speed of the wall is $v$. The fluctuations in its energy are now proportional to $M(v+s)^2 - Mv^2 \approx 2Mvs$ in the limit $v>>s$. For fixed $v$, we can make $s$ smaller and smaller by increasing $M$, but as we do so the fluctuations in the energy of the wall don't change size because $s \propto 1/M$.' Next note that the fluctuations in the energy of the ball in this frame are proportional to $mvs'$, with $s'$ the speed of the ball in this frame. By conservation of momentum, $Ms = ms'$, so the fluctuations in energy of the ball and wall are about the same size in the moving reference frame. Thus, we must account for both of them. I'll give a fuller answer in one dimension for simplicity. I'll first find the motion in the center of mass frame, then transform it and show that energy is conserved in the tranformed frame. Let the ball have mass $m$ and the wall have mass $M$. The spring constant is $k$. Let the ball's position from equilibrium be $x$ and the wall's position be $X$. The Lagrangian is $$\mathcal{L} = \frac{1}{2}\left(M\dot{X}^2 + m\dot{x}^2 - k(X-x)^2\right)$$ The equations of motion are $$-k(X-x) = M\ddot{X}$$ $$k(X-x) = m\ddot{x}$$ Adding these gives the conservation of momentum $$\frac{\textrm{d}}{\textrm{d}t}(M\dot{X} + m\dot{x}) = 0$$ In the center of mass frame this momentum is zero, so $$\dot{x} = \frac{M\dot{X}}{m}$$ With the initial condition $x(0) = X(0) = 0$ we can integrate this to $$X = -\frac{m}{M}x$$ The equation of motion for the small mass becomes $$k(\frac{1}{M} + \frac{1}{m}) x = \ddot{x}$$ with the solution $$x = A \sin(\omega t + \phi)$$ $$X = -\frac{m}{M}A\sin(\omega t + \phi)$$ with $\omega^2 = k(\frac{1}{M} + \frac{1}{m})$. Now boost to a frame moving at speed $v$. $$x = A \sin(\omega t + \phi) - vt$$ $$X = -\frac{m}{M}A\sin(\omega t + \phi) - vt$$ The potential energy is in the spring $$U = \frac{1}{2}k(X-x)^2 = \frac{1}{2}k \left(A (1+\frac{m}{M})(\sin(\omega t + \phi) \right)^2$$ The kinetic energy is in the motion of the ball and wall $$T = \frac{1}{2} \left(A \omega \cos(\omega t + \phi)\right)^2 \left(m + M(\frac{m}{M})^2\right)$$ Add these and applying the identity $\sin^2x+\cos^2x = 1$ gives $$E = \frac{1}{2}k\left(A(1+\frac{m}{M})\right)^2$$ which is constant in time. - 1 @Vlad did you down-vote? The fact that it wasn't the language you wanted doesn't make it incorrect or deserve a down vote. – Mark Eichenlaub Feb 8 '11 at 22:13 Yes, it was me. I said many times I do not want to consider the wall. – Vladimir Kalitvianski Feb 8 '11 at 22:16 4 @Vlad It's there. It's reasonable to consider it. That was the point of my answer. Your behavior is annoying. – Mark Eichenlaub Feb 8 '11 at 22:20 1 Sorry to hear that. – Vladimir Kalitvianski Feb 8 '11 at 22:25 3 @Vlad That's just a general feature of mechanics. It's completely unenlightening in this scenario and as vacuous as explaining why something is accelerating by saying "because there is a force on it". – Mark Eichenlaub Feb 8 '11 at 22:59 show 3 more comments Suppose that the wall has some large but finite mass. Then the actual motion of the system involves oscillations of the wall as well as oscillations of the mass. If you work out the total energy of the entire system, you'll find that it's conserved. - I know, Ted, I know. Where is your short answer to the post question? – Vladimir Kalitvianski Feb 8 '11 at 21:16 If you're looking for an answer to the question "why is the energy of the ball not conserved?" then the answer is that there's no reason to expect the energy of just one part of a system to be conserved. Conservation of energy applies to the total energy. – Ted Bunn Feb 8 '11 at 21:22 And I apply it to the total energy E = T + U of the ball. In the rest RF it is conserved, in a moving RF it is not. – Vladimir Kalitvianski Feb 8 '11 at 21:31 The simple problem here is in sharp contrast to the "big science" (at first sight) in Vladimirs pages. I recommend to have a visit there. – Georg Feb 8 '11 at 21:57 1 I don't understand the question and frankly doubt your seriousness in asking it. I'm done. – Ted Bunn Feb 8 '11 at 22:48 show 7 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": 27, "mathjax_display_tex": 14, "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": 0.9408189058303833, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/74963/can-you-formulate-a-phi-theta-restriction-in-spherical-coordinates-for-a/75015	# Can you formulate a $\phi, \theta$ restriction in spherical coordinates for a great circle? Further to this question Quaternion rotation has a nice property that you can trace any great circle you like. You specify the axis of rotation, and you will automatically follow the great circle when rotating. However spherical coordinates only trace a great circle when $\theta (elevation) = 0$, and $\phi (azimuth)$ is allowed to travel $0.. 2\pi$. So my question is, is there a way to formulate a $\theta, \phi$ restriction that will allow me to trace an arbitrary great circle in spherical coordinates? - There are at least two ways to start with constructing the parametric equations of a great circle: one involves finding the plane through the origin where your great circle lies in, and the other involves finding two points where your great circle passes through. – J. M. Oct 23 '11 at 2:01 Do you have a little more detail? I've got 4 points in each plane, so can do both! – bobobobo Oct 23 '11 at 2:05 ## 1 Answer Since you've got four points in the plane of the great circle, you can find its normal $n$. Then the equation for the spherical coordinates is simply $$\pmatrix{\sin\theta\cos\phi\\\sin\theta\sin\phi\\\cos\theta}\cdot n=0\;.$$ If you like, you can solve this for $\theta$ in terms of $\phi$: $$\theta=-\arctan\left(\frac{n_z}{n_x\cos\phi+n_y\sin\phi}\right)\;,$$ where you have to add $\pi$ to negative values if you want $\theta$ in the usual range $[0,\pi]$. -	{"extraction_info": {"found_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": 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": 0.8978074789047241, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/5338/why-are-these-specific-values-used-to-initialise-the-hash-buffer-in-sha-512?answertab=oldest	# Why are these specific values used to initialise the hash buffer in SHA-512? I'm reading the book Network Security Essentials written by William Stallings. To create a message digest with SHA-512, we have to go through some steps: 1. append padding bits. 2. append length 3. initialize hash buffer 4. ... so on. In the book is written for step 3: A 512 bit buffer is used to hold intermediate & final results of the hash function. The buffer can be represented as eight 64-bit registers (a,b,c,d,e,f,g,h). These registers are initialized to the following 64-bit integers (here in hexadecimal values): ```` a = 6A09E667F3BCC908 b = BB67AE8584CAA73B ... ```` up to register `h`. Why are these specific hexadecimal values used, is there any particular reason behind this? I know why they are necessary, but why these specific ones? - ## 1 Answer The initial hash values for SHA-512 are the 64-bit binary expansion of the fractional part of the square root of the 9th through 16th primes (23, 29, 31, ..., 53). That is: $$I_0 = \left \lfloor \mathrm{frac} \left (\sqrt{23} \right ) · 2^{64} \right \rfloor$$ $$I_1 = \left \lfloor \mathrm{frac} \left (\sqrt{29} \right ) · 2^{64} \right \rfloor$$ $$\cdots$$ And obviously the reason they use primes is that their square root is always irrational, which makes for a nice, random-looking number. Taking the square root of four would be less interesting, as it would give 0. It also means the constants are relatively independent from each other, as none of them can be easily expressed in terms of any other, which might be a weakness - but that probably doesn't matter as much. They are chosen this way because while they generally don't need to be anything in particular, as long as the chosen numbers don't have some special properties (i.e. setting them to zero might be a bad choice), the designers of SHA-512 wanted people to know that the numbers weren't chosen so as to stealthily introduce a backdoor in the algorithm (for instance, if a certain combination of initial values were chosen, it may have been possible for the NSA to break the hash function more easily) which would be very difficult to confirm or refute as it would require months, perhaps years of cryptanalysis. Using publicly available numbers that anyone can recreate, here the fractional parts of some prime square roots, solves this problem - there is no way anyone can change these numbers (nobody can change the square root of 23, for instance), and it is considerably harder to design an algorithm to have a backdoor under set constants, than to design the algorithm first and work out those constants later on. The first few digits of $\pi$ or $e$ are popular choices too, but only if you use the first digits. If you were using, say, the digits starting at position 94858991, I would get very suspicious. See Nothing up my sleeve number, also this similar question. - 1 In addition: the design of SHA-2 hashes (and SHA-1, MD5..) pads the end of the message using two redundant methods (padding with a 1 bit then some number of 0 bits; and padding with the message length). Using only the first of these two methods would leave these hashes open to an attack by the party that decided the constant (possibility to insert a secret prefix without changing the hash). This shows that it is sound to use "nothing up my sleeve numbers" here. – fgrieu Nov 12 '12 at 11:55	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "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": 0.9447052478790283, "perplexity_flag": "middle"}
http://openwetware.org/index.php?title=User:Pranav_Rathi/Notebook/OT/2010/08/18/CrystaLaser_specifications&direction=prev&oldid=654989	# User:Pranav Rathi/Notebook/OT/2010/08/18/CrystaLaser specifications ### From OpenWetWare Revision as of 19:05, 14 November 2012 by Pranav Rathi (Talk \| contribs) ## Specifications We are expecting our laser any time. To know the laser more we are looking forward to investigate number of things. These specifications are already given by the maker, but we will verify them. ### Polarization Laser is TM (transverse magnetic) or P or Horizontal linearly polarized (in the specimen plane laser is still TM polarized; when looking into the sample plane from the front of the microscope). We investigated these two ways: 1) by putting a glass interface at Brewster’s angle and measured the reflected and transmitted power. At this angle all the light is transmitted because the laser is P-polarized, 2) by putting a polarizing beam splitter which uses birefringence to separate the two polarizations; P is reflected and S is transmitted, by measuring and comparing the powers, the desired polarizability is determined. We performed the experiment at 1.8 W where P is 1.77 W and S is less than .03 W* ### Beam waist at the output window We used knife edge method (this method is used to determine the beam waist (not the beam diameter) directly); measure the input power of 1.86W at 86.5 and 13.5 % at the laser head (15mm). It gave us the beam waist (Wo) of .82mm (beam diameter =1.64mm). ### Possible power fluctuations if any The power supply temperature is really critical. Laser starts at roughly 1.8 W but if the temperature of the power supply is controlled very well it reaches to 2 W in few minutes and stay there. It’s really stupid of manufacturer that they do not have any fans inside so we put two chopper fans on the top of it to cool it and keep it cool. If no fans are used then within an hour the power supply reaches above 50 degrees of Celsius and then, not only the laser output falls but also the power supply turns itself off after every few minutes. ### Mode Profile Higher order modes had been a serious problem in our old laser, which compelled us to buy this one. The success of our experiments depends on the requirement of TEM00 profile, efficiency of trap and stiffness is a function of profile.So mode profiling is critical; we want our laser to be in TEM00. I am not going to discuss the technique of mode profiling; it can be learned from this link: [1] [2]. As a result it’s confirmed that this laser is TEM00 mode. Check out the pics: A LabView program is written to show a 3D Gaussian profile, it also contains a MatLab code[3]. ## Specs by the Manufacturer All the laser specs and the manual are in the document: [Specs[4]] ## Beam Profile The original beam waist of the laser is .2mm, but since we requested the 4x beam expansion option, the resultant beam waist is .84 at the output aperture of the laser. As the nature of Gaussian beam it still converges in the far field. We do not know where? So there is a beam waist somewhere in the far field. There are two ways to solve the problem; by using Gaussian formal but, for that we need the beam parameters before expansion optics and information about the expansion optics, which we do not have. So the only way we have, is experimentally measure the beam waist along the z-axis at many points and verify its location for the minimum. Once this is found we put the AOM there. So the experimental data gives us the beam waist and its distance from the laser in the z-direction. We use scanning knife edge method to measure the beam waist. ### Method • In this method we used a knife blade on a translation stage with 10 micron accuracy. The blade is moved transverse to the beam and the power of the uneclipsed portion is recorded with a power meter. The cross section of a Gaussian beam is given by: $I(r)=I_0 exp(\frac {-2r^2}{w_L^2})$ Where I(r) is the Intensity as function of radius (distance in transverse direction), I0 is the input intensity at r = 0, and wL is the beam radius. Here the beam radius is defined as the radius where the intensity is reduced to 1/e2 of the value at r = 0. This can be seen by letting r = wL. setup Power Profile The experiment data is obtained by gradually moving the blade across from point A to B, and recording the power. Without going into the math the intensity at the points can be obtained. For starting point A $\mathbf{I_A(r=0)}=I_0 exp(-2)=I_0.865$ For stopping point B $\mathbf{I_B}=I_0 (1-.865)$ By measuring this distance the beam waist can be measured and beam diameter is just twice of it: $\mathbf{\omega_0}=r_{.135}-r_{.865}$ this is the method we used below. • Beam waist can also be measured the same way in terms of the power. The power transmitted by a partially occluding knife edge: $\mathbf {p(r)}=\frac{P_0}{\omega_0} \sqrt{\frac{2}{\pi}} \int\limits_r^\infty exp(-\frac{2r^2}{\omega^2}) dr$ After integrating for transmitted power: $\mathbf {p(r)}=\frac{P_0}{2}{erfc}(2^{1/2}\frac{r}{\omega_0})$ Now the power of 10% and 90% is measured at two points and the value of the points substituted here: $\mathbf{\omega_0}=.783(r_{.1} - r_{.9})$ The difference between the methods is; the first method measures the value little higher than the second method (power), but the difference is still under 13%. So either method is GOOD but the second is more accurate. Here is a link of a LabView code to calculate the beam waist with knife edge method[5]. #### Data We measured the beam waist at every 12.5, 15 and 25mm, over a range of 2000mm from the output aperture of the laser head. The measurement is minimum at 612.5 mm from the laser, thus the beam waist is at 612.5±12.5mm from the laser. And it is to be 1.26±.1 mm. #### Analysis Here the plot beam diameter Vs Z is presented. Experimental data is presented as blue and model is red. As it can be seen that model does not fit the data. Experimental beam expands much faster than the model; this proves that the beam waist before the expansion optics must be relatively smaller. And also we are missing an important characterization parameter. The real word lasers work differently in a way that their beam do not follow the regular Gaussian formula for large propagation lengths (more than Raleigh range). So that's why we will have introduce a beam propagation factor called M2. ##### M2 The beam propagation factor M2 was specifically introduced to enable accurate calculation of the properties of laser beams which depart from the theoretically perfect TEM00 beam. This is important because it is quite literally impossible to construct a real world laser that achieves this theoretically ideal performance level. M2 is defined as the ratio of a beam’s actual divergence to the divergence of an ideal, diffraction limited, Gaussian, TEM00 beam having the same waist size and location. Specifically, beam divergence for an ideal, diffraction limited beam is given by: $\theta_{th}=\frac{\lambda}{\pi w_0}$ this is theoretical half divergence angle in radian. $\theta_{ac}=M^2\frac{\lambda}{\pi w_0}$ this is actual half divergence angle in radian. Where: λ is the laser wavelength w0 is the beam waist radius (at the 1/e2 point) M2 is the beam propagation factor This definition of M2 allows us to make simple change to optical formulas by taking M2 factor as multiplication, to account for the actual beam divergence. This is the reason why M2 is also sometimes referred to as the “times diffraction limit number”. The more information about M2 is available in these links:[6][7]	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9346017837524414, "perplexity_flag": "middle"}
http://mathhelpforum.com/pre-calculus/170602-finding-equation-lines.html	# Thread: 1. ## finding equation of lines The points (0,0), (a,0) and (B,C) are the vertices of an arbitrar triangle which is placed in a coninesvenient position relative to the coordinate system. a) find the equation of the line through each vertex perpendicular to the opposite side, and show algebraically that these 3 lines intersect at a single point. b) find the equation of the perpendicular bisector of each side, and show algebraically that these 3 lines intersect at a single point. Why is this fact geometrically obvious? c) find the equation of the line through each vertex and the midpoint of the opposite side, and show algebraically that these 3 lines intersect at a single point. Also verif that this point is 2/3rds of the way from each vertex to the midpoint of the opposite side Im not really good at answering questions like this. Im better at like, just solving an equation with numbers in my face. How do i show this algebraicly? like get the slopes of each line and find a way to substitute them into each other for the point somehow? 2. I suppose you know how to find the equation of the line when two points of the line are given or when the slope of the line and a point on the line are given. a)find the equation of the lines of the sides of the triangle, from that derive the slopes of the required lines ( $m.m_1=-1$). Now you have slope and a point of the line which enables you to find the equations of the equired lines. b) same as the first one find the slope and a point which the line goes through c) You can find two points which the required line crosses (vertex and the midpoint of the opposite side).	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9226263165473938, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/114053/are-topological-vector-spaces-completely-regular	# Are topological vector spaces completely regular? From the first two, we can say every topological vector space is completely regular, can't we? Does a TVS need to be hausdorff in order to be completely regular? Thanks! - I think this depends on your definitions. Some people require TVS to have the condition $T_0$, in which case they are automatically Hausdorff, so I assume you don't require this. Some people use the term completely regular to mean $T_{3\frac{1}{2}}$ in which case Hausdorff is needed. It seems if completely regular is the definition where you don't require Hausdorff, then it should be true. – Matt Feb 27 '12 at 17:32 @Matt: Thanks! (1) Could you point out what "completely regular" means in the first one and the third one? (2) Do "topological vector space" in the last two include $T_0$ in their definitions (Wiki also mentions this, but its definition doesn't assume $T_0$ in TVS definition). – Tim Feb 27 '12 at 17:59 In the first and second ones, completely regular does not include $T_0$; in the third it does (or might as well, since Hausdorff implies $T_0$). – Brian M. Scott Feb 28 '12 at 0:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9669145345687866, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/90441/confusing-in-seal-software-optimized-encryption-algorithm-mathematic-notation	# confusing in SEAL (Software-Optimized Encryption Algorithm) mathematic notation I'm developing an encryption software based on SEAL algorihm for my research. I found the paper in here My Question is what is the meaning of $$H_{i \operatorname{mod} 5}^i$$ in page 5? Thanks in advance. - 4 Please do not cross-post a question through several Stack Exchange sites (Stack Overflow, Security). – Paŭlo Ebermann Dec 11 '11 at 13:10 1 I have closed the one on Security. Migrating it to math does not make sense as there is already a copy here. – Hendrik Brummermann Dec 11 '11 at 13:39 thanks for correcting me Paulo and Hendrik..:) – David Dec 11 '11 at 16:22 ## 3 Answers I had a look at the paper. To really understand it, we need to read the surrounding two paragraphs: We specify the tables using a function $G$. For $a$ a $160$-bit string and $i$ an integer, $0 \leq i < 2^{32}$, $G_a (i)$ is a $160$-bit value. The function G is just the compression function of the Secure Hash Algorithm SHA-1. For completeness, its definition is given in Appendix A. Let us re-index $G$ to construct a function $\Gamma$ whose images are $32$-bit words instead of $160$-bit ones. The function $\Gamma$ is defined by $\Gamma_a(i) = H^i_{i\,\bmod\,5}$, where $$H^{5j}_0 \|\| H^{5j+1}_1 \|\| H^{5j+2}_2 \|\| H^{5j+3}_3 \|\| H^{5j+4}_4 = G_a(j),$$ for $j = \lfloor i/5\rfloor$. Thus a table of $\Gamma$-values is exactly a table for $G$-values read left-to-right, top-to-bottom. So to find $\Gamma_a(i) = H^i_{i\,\bmod\,5}$, we will calculate $j = \lfloor i/5 \rfloor$ (the integer part of the quotient of $i$ and $5$) and then $i \bmod 5$ (the remainder of this same division operation), calculate $G_a(j)$, split it in five $32$-bit blocks, and use $i \bmod 5$ to select the right one of them. I think the following way of defining $\Gamma$ would be clearer: $\Gamma_a(i) = H^{\lfloor i/5\rfloor}_{i\,\bmod\,5},$ where $H^j_0 \|\| H^j_1 \|\| H^j_2 \|\| H^j_3 \|\| H^j_4 = G_a(j).$ - The entire $H^i_{i\;\bmod\;5}$ notation seems confusing and non-standard, and doesn't seem to be used anywhere else in the main body of the paper. (Appendix A uses a somewhat similar but less confusing notation, omitting the upper indices.) If I were to rewrite the article, I'd just leave it out entirely, and simply define $\Gamma_a$ as a function mapping integers to 32-bit strings, such that $$G_a(j) = \Gamma_a(5j)\;\\|\;\Gamma_a(5j+1)\;\\|\;\Gamma_a(5j+2)\;\\|\;\Gamma_a(5j+3)\;\\|\;\Gamma_a(5j+4)$$ for all integers $j$. Alternatively, if we were to introduce some kind of bit-slicing notation, such that e.g. $S_{i .. j}$ stood for the bitstring formed by the $i$-th to $j$-th bits of $S$, then we could define $\Gamma_a$ directly as $$\Gamma_a(i) = G_a(\lfloor i/5 \rfloor)_{32(i\;\bmod\;5)\,..\,32(i\;\bmod\;5)+31}$$ In any case, I'd suggest ignoring the $H$ notation completely. The underlying concept it's meant to define is simple enough: we take each 160-bit block $G_a(j)$ and split it into five 32-bit blocks called $\Gamma_a(5j)$ to $\Gamma_a(5j+4)$. That's all there is to it. - I think I've got the answer.. $$H_{i \operatorname{mod} 5}^i$$ That function means that $$H_i$$ is assigned by $$H_{i{mod} 5}$$ -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 39, "mathjax_display_tex": 7, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9327284097671509, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-applied-math/57309-forces-walls.html	# Thread: 1. ## Forces in walls? Hi! I have a question about forces inside of walls. When you concider the walls of a house, they have a certain density $\delta$ and they create a force downwards, which becomes bigger closer to the ground. Assume we don't need to care about the weight of the roof or that of the atmosphere. The force in each wall, will be $\overbrace{\underbrace{w\cdot h\cdot t}_\texttt{total volume}\cdot \delta}^\texttt{total weight}\cdot g$ where w is the width of the wall, h is the heigth up to the top of the wall, and t is the thickness of the wall. So, independent of the width and the thickness of the wall, the pressure create by the wall above will be $h\cdot\delta\cdot g$ Now to my question: Is the pressure uniform? That is, will the pressure be the same in all directions, vertically as horizontally? The pressure vertically will be $h\cdot\delta\cdot g$, since when the material gets squeezed from the top and the bottom, it gets compressed vertically, so it creates a pressure vertically since it wants to expand in that direction. Besides, it needs to support its own weigth. But what about horizontally, does it want to expand it that direction as well? How big will the pressure be in that direction? Near to the pressure vertically, or almost zero? Does it depend on the material of the wall? (wood/concrete/metal?) Thanks in advance! #### 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": 0.9503591060638428, "perplexity_flag": "head"}
http://mathhelpforum.com/new-users/199422-help-calculus-undergraduate-mathematics.html	# Thread: 1. ## Help with Calculus (undergraduate mathematics) Hi, I'm a mathematics student and I need help with a simple problem: Given a C1 function f:[a,b]-->R, with a right-hand derivative f'(a) on a. Proof if the following statement is true or false. If f'(a)=0, then f has a local maximum or a local minimum on a. Thanks very much!! 2. ## Re: Help with Calculus (undergraduate mathematics) Originally Posted by yannt Hi, I'm a mathematics student and I need help with a simple problem: Given a C1 function f:[a,b]-->R, with a right-hand derivative f'(a) on a. Proof if the following statement is true or false. If f'(a)=0, then f has a local maximum or a local minimum on a. Think about $f(x)=x^3$ on $[0,1]$. 3. ## Re: Help with Calculus (undergraduate mathematics) That is exactly what I did in the first place, but the teacher wrote the following: You have to place the function on an interval of the type [0,b], in your example, there is a local minimum on 0 in the interval [0,b] 4. ## Re: Help with Calculus (undergraduate mathematics) Originally Posted by yannt That is exactly what I did in the first place, but the teacher wrote the following: You have to place the function on an interval of the type [0,b], in your example, there is a local minimum on 0 in the interval [0,b] The function $f(x) = \left\{ {\begin{array}{rl} {{x^2}\sin \left( {{x^{ - 2}}} \right),}&{0 < x \leqslant b} \\ {0,}&{x = 0}\end{array}} \right.$ has that property. $f(0)=0~\&~f'(0)=0$ but $f(0)$ is neither a min nor a max in any neighborhood of $0$. 5. ## Re: Help with Calculus (undergraduate mathematics) Ok, I can see it now. Thanks!! 6. ## Re: Help with Calculus (undergraduate mathematics) need some help in this please!! is given the function {y=sin2x/x for x different from 0} or {y=2a for x=o}. find "a" if this function is continuous everywhere.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8927378058433533, "perplexity_flag": "middle"}
http://motls.blogspot.com/2012/04/schrodinger-has-never-met-newton.html?showComment=1334938206154&m=0	# The Reference Frame ## Friday, April 20, 2012 ... ///// ### Schrödinger has never met Newton Sabine Hossenfelder believes that Schrödinger meets Newton. But is the story about the two physicists' encounter true? Yes, these were just jokes. I don't think that Sabine Hossenfelder misunderstands the history in this way. Instead, what she completely misunderstands is the physics, especially quantum physics. She is in a good company. Aside from the authors of some nonsensical papers she mentions, e.g. van Meter, Giulini and Großardt, Harrison and Moroz with Tod, Diósi, and Carlip with Salzman, similar basic misconceptions about elementary quantum mechanics have been promoted by Penrose and Hameroff. Hameroff is a physician who, along with Penrose, prescribed supernatural abilities to the gravitational field. It's responsible for the gravitationally induced "collapse of the wave function" which also gives us consciousness and may be even blamed for Penrose's (not to mention Hameroff's) complete inability to understand rudimentary quantum mechanics, among many other wonderful things; I am sure that many of you have read the Penrose-Hameroff crackpottery and a large percentage of those readers even fail to see why it is a crackpottery, a problem I will try to fix (and judging by the 85-year-long experience, I will fail). It's really Penrose who should be blamed for the concept known as the Schrödinger-Newton equations So what are the equations? Sabine Hossenfelder reproduces them completely mindlessly and uncritically. They're supposed to be the symbiosis of quantum mechanics combined with the Newtonian limit of general relativity. They say:\[ \eq{ i\hbar \pfrac {}{t} \Psi(t,\vec x) &= \zav{-\frac{\hbar^2}{2m} \Delta + m\Phi(t,\vec x)} \Psi(t,\vec x) \\ \Delta \Phi(t,\vec x) &= 4\pi G m \abs{\Psi(t,\vec x)}^2 } \] Don't get misled by the beautiful form they take in $\rm\LaTeX$ implemented by MathJax; superficial beauty of the letters doesn't guarantee the validity. Sabine Hossenfelder and others immediately talk about mechanically inserting numbers into these equations, and so on, but they never ask a basic question: Are these equations actually right? Can we prove that they are wrong? And if they are right, can they be responsible for anything important that shapes our observations? Of course that the second one is completely wrong; it fundamentally misunderstands the basic concepts in physics. And even if you forgot the reasons why the second equation is completely wrong, they couldn't be responsible for anything important we observe – e.g. for well-defined perceptions after we measure something – because of the immense weakness of gravity (and because of other reasons). Analyzing the equations one by one So let us look at the equations, what they say, and whether they are the right equations describing the particular physical problems. We begin with the first one,\[ i\hbar \pfrac {}{t} \Psi(t,\vec x) = \zav{-\frac{\hbar^2}{2m} \Delta + m\Phi(t,\vec x)} \Psi(t,\vec x) \] Is it right? Yes, it is a conventional time-dependent Schrödinger equation for a single particle that includes the gravitational potential. When the gravitational potential matters, it's important to include it in the Hamiltonian as well. The gravitational potential energy is of course as good a part of the energy (the Hamiltonian) as the kinetic energy, given by the spatial Laplacian term, and it should be included in the equations. In reality, we may of course neglect the gravitational potential in practice. When we study the motion of a few elementary particles, their mutual gravitational attraction is negligible. For two electrons, the gravitational force is more than $10^{40}$ times weaker than the electrostatic force. Clearly, we can't measure the transitions in a Hydrogen atom with the relative precision of $10^{-40}$. The "gravitational Bohr radius" of an atom that is only held gravitationally would be comparably large to the visible Universe because the particles are very weakly bound, indeed. Of course, it makes no practical sense to talk about energy eigenstates that occupy similarly huge regions because well before the first revolution (a time scale), something will hit the particles so that they will never be in the hypothetical "weakly bound state" for a whole period. But even if you consider the gravity between a microscopic particle (which must be there for our equation to be relevant) such as a proton and the whole Earth, it's pretty much negligible. For example, the protons are running around the LHC collider and the Earth's gravitational pull is dragging them down, with the usual acceleration of $g=9.8\,\,{\rm m}/{\rm s}^2$. However, there are so many forces that accelerate the protons much more strongly in various directions that the gravitational pull exerted by the Earth can't be measured. But yes, it's true that the LHC magnets and electric fields are also preventing the protons from "falling down". The protons circulate for minutes if not hours and as skydivers know, one may fall pretty far down during such a time. An exceptional experiment in which the Earth's gravity has a detectable impact on the quantum behavior of particles are the neutron interference experiments, those that may be used to prove that gravity cannot be an entropic force. To describe similar experiments, one really has to study the neutron's Schrödinger equation together with the kinetic term and the gravitational potential created by the Earth. Needless to say, much of the behavior is obvious. If you shoot neutrons through a pair of slits, of course that they will accelerate towards the Earth much like everything else so the interference pattern may be found again; it's just shifted down by the expected distance. People have also studied neutrons that are jumping on a trampoline. There is an infinite potential energy beneath the trampoline which shoots the neutrons up. And there's also the Earth's gravity that attracts them down. Moreover, neutrons are described by quantum mechanics which makes their energy eigenstates quantized. It's an interesting experiment that makes one sure that quantum mechanics does apply in all situations, even if the Earth's gravity plays a role as well, and that's where the Schrödinger equation with the gravitational potential may be verified. I want to say that while the one-particle Schrödinger equation written above is the right description for situations similar to the neutron interference experiments, it already betrays some misconceptions by the "Schrödinger meets Newton" folks. The fact that they write a one-particle equation is suspicious. The corresponding right description of many particles wouldn't contain wave functions that depend on the spacetime, $\Psi(t,\vec x)$. Instead, the multi-particle wave function has to depend on positions of all the particles, e.g. $\Psi(t,\vec x_1,\vec x_2)$. However, the Schrödinger equation above already suggests that the "Schrödinger meets Newton" folks want to treat the wave function as an object analogous to the gravitational potential, a classical field. This totally invalid interpretation of the objects becomes lethal in the second equation. Confusing observables with their expectation values, mixing up probability waves with classical fields The actual problem with the Schrödinger-Newton system of equations is the second equation, Poisson's equation for the gravitational potential,\[ \Delta \Phi(t,\vec x) = 4\pi G m \abs{\Psi(t,\vec x)}^2 \] Is this equation right under some circumstances? No, it is never right. It is a completely nonsensical equation which is nonlinear in the wave function $\Psi$ – a fatal inconsistency – and which mixes apples with oranges. I will spend some time with explaining these points. First, let me start with the full quantum gravity. Quantum gravity contains some complicated enough quantum observables that may only be described by the full-fledged string/M-theory but in the low-energy approximation of an "effective field theory", it contains quantum fields including the metric tensor $\hat g_{\mu\nu}$. I added a hat to emphasize that each component of the tensor field at each point is a linear operator (well, operator distribution) acting on the Hilbert space. I have already discussed the one-particle Schrödinger equation that dictates how the gravitational field influences the particles, at least in the non-relativistic, low-energy approximation. But we also want to know how the particles influence the gravitational field. That's given by Einstein's equations,\[ \hat{R}_{\mu\nu} - \frac{1}{2} \hat{R} \hat{g}_{\mu\nu} = 8\pi G \,\hat{T}_{\mu\nu} \] In the quantum version, Einstein's equations become a form of the Heisenberg equations in the Heisenberg picture (Schrödinger's picture looks very complicated for gravity or other field theories) and these equations simply add hats above the metric tensor, Ricci tensor, Ricci scalar, as well as the stress-energy tensor. All these objects have to be operators. For example, the stress-energy tensor is constructed out of other operators, including the operators for the intensity of electromagnetic and other fields and/or positions of particles, so it must be an operator. If an equation relates it to something else, this something else has to be an operator as well. Think about Schrödinger's cat – or any other macroscopic physical system, for that matter. To make the thought experiment more spectacular, attach the whole Earth to the cat so if the cat dies, the whole Earth explodes and its gravitational field changes. It's clear that the values of microscopic quantities such as the decay stage of a radioactive nucleus may imprint themselves to the gravitational field around the Earth – something that may influence the Moon etc. (We may subjectively feel that we have already perceived one particular answer but a more perfect physicist has to evolve us into linear superpositions as well, in order to allow our wave function to interfere with itself and to negate the result of our perceptions. This more perfect and larger physicist will rightfully deny that in a precise calculation, it's possible to treat the wave function as a "collapsed one" at the moment right after we "feel an outcome".) Because the radioactive nucleus may be found in a linear superposition of dictinct states and because this state is imprinted onto the cat and the Earth, it's obvious that even the gravitational field around the (former?) Earth is generally found in a probabilistic linear superposition of different states. Consequently, the values of the metric tensors at various points have to be operators whose values may only be predicted probabilistically, much like the values of any observable in any quantum theory. Let's now take the non-relativistic, weak-gravitational-field, low-energy limit of Einstein's equations written above. In this non-relativistic limit, $\hat g_{00}$ is the only important component of the metric tensor (the gravitational redshift) and it gets translated to the gravitational potential $\hat \Phi$ which is clearly an operator-valued field, too. We get\[ \Delta \hat\Phi(t,\vec x) = 4\pi G \hat\rho(t,\vec x). \] It looks like the Hossenfelder version of Poisson's equation except that the gravitational potential on the left hand side has a hat; and the source $\hat\rho$, i.e. the mass density, has replaced her $m \abs{\Psi(t,\vec x)}^2$. Fine. There are some differences. But can I make special choices that will produce her equation out of the correct equation above? What is the mass density operator $\hat\rho$ equal to in the case of the electron? Well, it's easy to answer this question. The mass density coming from an electron blows up at the point where the electron is located; it's zero everywhere else. Clearly, the mass density is a three-dimensional delta-function:\[ \hat\rho(t,\vec x) = m \delta^{(3)}(\hat{\vec X} - \vec x) \] Just to be sure, the arguments of the field operators such as $\hat\rho$ – the arguments that the fields depend on – are ordinary coordinates $\vec x$ which have no hats because they're not operators. In quantum field theories, whether they're relativistic or not, they're as independent variables as the time $t$; after all, $(t,x,y,z)$ are mixed with each other by the relativistic Lorentz transformations which are manifest symmetries in relativistic quantum field theories. However, the equation above says that the mass density at the point $\vec x$ blows up iff the eigenvalue of the electron's position $\hat X$, an eigenvalue of an observable, is equal to this $\vec x$. The equation above is an operator equation. And yes, it's possible to compute functions (including the delta-function) out of operator-valued arguments. Semiclassical gravity isn't necessarily too self-consistent an approximation. It may resemble the equally named song by Savagery above. Clearly, the operator $\delta^{(3)}(\hat X - \vec x)$ is something different than Hossenfelder's $\abs{\Psi(t,\vec x)}^2$ – which isn't an operator at all – so her equation isn't right. Can we obtain the squared wave function in some way? Well, you could try to take the expectation value of the last displayed equation:\[ \bra\Psi \Delta \hat\Phi(t,\vec x)\ket\Psi = 4\pi G m \abs{\Psi(t,\vec x)}^2 \] Indeed, if you compute the expectation value of the operator $\delta^{(3)}(\hat X - \vec x)$ in the state $\ket\Psi$, you will obtain $\abs{\Psi(t,\vec x)}^2$. However, note that the equation above still differs from the Hossenfelder-Poisson equation: our right equation properly sandwiches the gravitational potential, which is an operator-valued field, in between the two copies of the wave functions. Can't you just introduce a new symbol $\Delta\Phi$, one without any hats, for the expectation value entering the left hand side of the last equation? You may but it's just an expectation value, a number that depends on the state. The proper Schrödinger equation with the gravitational potential that we started with contains the operator $\hat\Phi(t,\vec x)$ that is manifestly independent of the wave function (either because it is an external classical field – if we want to treat it as a deterministically evolving background field – or because it is a particular operator acting on the Hilbert space). So they're different things. At any rate, the original pair of equations is wrong. Nonlinearity in the wave function is lethal Those deluded people are obsessed with expectation values because they don't want to accept quantum mechanics. The expectation value of an operator "looks like" a classical quantity and classical quantities are the only physical quantities they have really accepted – and 19th century classical physics is the newest framework for physics that they have swallowed – so they try to deform and distort everything so that it resembles classical physics. An arbitrarily silly caricature of the reality is always preferred by them over the right equations as long as it looks more classical. But Nature obeys quantum mechanics. The observables we can see – all of them – are indeed linear operators acting on the Hilbert space. If something may be measured and seen to be equal to something or something else (this includes Yes/No questions we may answer by an experiment), then "something" is always associated with a linear operator on the Hilbert space (Yes/No questions are associated with Hermitian projection operators). If you are using a set of concepts that violate this universal postulate, then you contradict basic rules of quantum mechanics and what you say is just demonstrably wrong. This basic rule doesn't depend on any dynamical details of your would-be quantum theory and it admits no loopholes. Two pieces of the wave function don't attract each other at all You could say that one may talk about the expectation values in some contexts because they may give a fair approximation to quantum mechanics. The behavior of some systems may be close to the classical one, anyway, so why wouldn't we talk about the expectation values only? However, this approximation is only meaningful if the variations of the physical observables (encoded in the spread of the wave function) are much smaller than their characteristic values such as the (mean) distances between the particles which we want to treat as classical numbers, e.g.\[ \abs{\Delta \vec x} \ll O(\abs{\vec x_1-\vec x_2}) \] However, the very motivation that makes those confused people study the Schrödinger-Newton system of equations is that this condition isn't satisfied at all. What they typically want to achieve is to "collapse" the wave function packets. They're composed of several distant enough pieces, otherwise they wouldn't feel the need to collapse them. In their system of equations, two distant portions of the wave function attract each other in the same way as two celestial bodies do – because $m \abs{\Psi}^2$ enters as the classical mass density to Poisson's equation for the gravitational potential. They write many papers studying whether this self-attraction of "parts of the electron" or another object may be enough to "keep the wave function compact enough". Of course, it is not enough. The gravitational force is extremely weak and cannot play such an essential role in the experiments with elementary particles. In Ghirardi-Rimini-Weber: collapsed pseudoscience, I have described somewhat more sophisticated "collapse theories" that are trying to achieve a similar outcome: to misinterpret the wave function as a "classical object" and to prevent it from spreading. Of course, these theories cannot work, either. To keep these wave functions compact enough, they have to introduce kicks that are so large that we are sure that they don't exist. You simply cannot find any classical model that agrees with observations in which the wave function is a classical object – simply because the wave function isn't a classical object and this fact is really an experimentally proven one as you know if you think a little bit. But what the people studying the Schrödinger-Newton system of equations do is even much more stupid than what the GRW folks attempted. It is internally inconsistent already at the mathematical level. You don't have to think about some sophisticated experiments to verify whether these equations are viable. They can be safely ruled out by pure thought because they predict things that are manifestly wrong. I have already said that the Hossenfelder-Poisson equation for the gravitational potential treats the squared wave function as if it were a mass density. If your wave function is composed of two major pieces in two regions, they will behave as two clouds of interplanetary gas and these two clouds will attract because each of them influences the gravitational potential that influences the motion of the other cloud, too. However, this attraction between two "pieces" of a wave function definitely doesn't exist, in a sharp contrast with the immensely dumb opinion held by pretty much every "alternative" kibitzer about quantum mechanics i.e. everyone who has ever offered any musings that something is fundamentally wrong with the proper Copenhagen quantum mechanics. There would only be an attraction if the matter (electron) existed at both places because the attraction is proportional to $M_1 M_2$. However, one may easily show that the counterpart of $M_1M_2$ is zero: the matter is never at both places at the same time. Imagine that the wave function has the form\[ \ket\psi = 0.6\ket \phi+ 0.8 i \ket \chi \] where the states $\ket\phi$ and $\ket\chi$ are supported by very distant regions. As you know, this state vector implies that the particle has 36% odds to be in the "phi" region and 64% odds to be in the "chi" region. I chose probabilities that are nicely rational, exploiting the famous 3-4-5 Pythagorean triangle, but there's another reason why I didn't pick the odds to be 50% and 50%: there is absolutely nothing special about wave functions that predict exactly the same odds for two different outcomes. The number 50 is just a random number in between $0$ and $100$ and it only becomes special if there is an exact symmetry between $p$ and $(1-p)$ which is usually not the case. Much of the self-delusion by the "many worlds" proponents is based on the misconception that predictions with equal odds for various outcomes are special or "canonical". They're not. Fine. So if we have the wave function $\ket\psi$ above, do the two parts of the wave function attract each other? The answer is a resounding No. The basic fact about quantum mechanics that all these Schrödinger-Newton and many-worlds and other pseudoscientists misunderstand is the following point. The wave function above doesn't mean that there is 36% of an object here AND 64% of an object there. (WRONG.) Note that there is "AND" in the sentence above, indicating the existence of two objects. Instead, the right interpretation is that the particle is here (36% odds) OR there (64% odds). (RIGHT.) The correct word is "OR", not "AND"! However, unlike in classical physics, you're not allowed to assume that one of the possibilities is "objectively true" in the classical sense even if the position isn't measured. On the other hand, even in quantum mechanics, it's still possible to strictly prove that the particle isn't found at both places simultaneously; the state vector is an eigenstate of the "both places" projection operator (product of two projection operators) with the eigenvalue zero. (The same comments apply to two slits in a double-slit experiment.) The mutually orthogonal terms contributing to the wave function or density matrix aren't multiple objects that simultaneously exist, as the word "AND" would indicate. You would need (tensor) products of Hilbert spaces and/or wave functions, not sums, to describe multiple objects! Instead, they are mutually excluding alternatives for what may exist, alternative properties that one physical system (e.g. one electron) may have. And mutually excluding alternatives simply cannot interact with each other, gravitationally or otherwise. Imagine you throw dice. The result may be "1" or "2" or "3" or "4" or "5" or "6". But you know that only one answer is right. There can't be any interaction that would say that because both "1" and "6" may occur, they attract each other which is why you probably get "3" or "4" in the middle. It's nonsense because "1" and "6" are never objects that simultaneously exist. If they don't simultaneously exist, they can't attract each other, whatever the rules are. They can't interact with one another at all! While the expectation value of the electron's position may be "somewhere in between" the regions "phi" and "chi", we may use the wave function to prove with absolute certainty that the electron isn't in between. The proponents of the "many-worlds interpretation" often commit the same trivial mistake. They are imagining that two copies of you co-exist at the same moment – in some larger "multiverse". That's why they often talk about one copy's thinking how the other copy is feeling in another part of a multiverse. But the other copy can't be feeling anything at all because it doesn't exist if you do! You and your copy are mutually excluding. If you wanted to describe two people, you would need a larger Hilbert space (a tensor product of two copies of the space for one person) and if you produced two people out of one, the evolution of the wave function would be quadratic i.e. nonlinear which would conflict with quantum mechanics (and its no-xerox theorem), too. These many-worlds apologists, including Brian Greene, often like to say (see e.g. The Hidden Reality) that the proper Copenhagen interpretation doesn't allow us to treat macroscopic objects by the very same rules of quantum mechanics with which the microscopic objects are treated and that's why they promote the many worlds. This proposition is what I call chutzpah. In reality, the claim that right after the measurement by one person, there suddenly exist several people is in a striking contradiction with facts that may be easily extracted from quantum mechanics applied to a system of people. The quantum mechanical laws – laws meticulously followed by the Copenhagen school, regardless of the size and context – still imply that the total mass is conserved, at least at a 1-kilogram precision, so it is simply impossible for one person to evolve into two. It's impossible because of the very same laws of quantum mechanics that, among many other things, protect Nature against the violation of charge conservation in nuclear processes. It's them, the many-worlds apologists, who are totally denying the validity of the laws of quantum mechanics for the macroscopic objects. In reality, quantum mechanics holds for all systems and for macroscopic objects, one may prove that classical physics is often a valid approximation, as the founding fathers of quantum mechanics knew and explicitly said. The validity of this approximation, as they also knew, is also a necessary condition for us to be able to make any "strict valid statements" of the classical type. The condition is hugely violated by interfering quantum microscopic (but, in principle, also large) objects before they are measured so one can't talk about the state of the system before the measurement in any classical language. In Nature, all observables (as well as the S-matrix and other evolution operators) are expressed by linear operators acting on the Hilbert space and Schrödinger's equation describing the evolution of any physical system has to be linear, too. Even if you use the density matrix, it evolves according to the "mixed Schrödinger equation" which is also linear:\[ i\hbar \ddfrac{}{t}\hat\rho = [\hat H(t),\hat \rho(t)]. \] It's extremely important that the density matrix $\hat \rho$ enters linearly because $\hat \rho$ is the quantum mechanical representation of the probability distribution, even the initial one. And the probabilities of final states are always linear combinations of the probabilities of the initial states. This claim follows from pure logic and will hold in any physical system, regardless of its laws. Why? Classically, the probabilities of final states $P({\rm final}_j)$ are always given by\[ P({\rm final}_j) = \sum_{i=1}^N P({\rm initial}_i) P({\rm evolution}_{i\to j}) \] whose right hand side is linear in the probabilities of the initial states and the left hand side is linear in the probabilities of the final states. Regardless of the system, these dependences are simply linear. Quantum mechanics generalizes the probability distributions to the density matrices which admit states arising from superpositions (by having off-diagonal elements) and which are compatible with the non-zero commutators between generic observables. However, whenever your knowledge about a system may be described classically, the equation above strictly holds. It is pure maths; it is as questionable or unquestionable (make your guess) as $2+2=4$. There isn't any "alternative probability calculus" in which the final probabilities would depend on the initial probabilities nonlinearly. If you carefully study the possible consistent algorithms to calculate the probabilities of various final outcomes or observations, you will find out that it is indeed the case that the quantum mechanical evolution still has to be linear in the density matrix. The Hossenfelder-Poisson equation fails to obey this condition so it violates totally basic rules of the probability calculus. Just to connect the density matrix discussion with a more widespread formalism, let us mention that quantum mechanics allows you to decompose any density matrix into a sum of terms arising from pure states,\[ \hat\rho = \sum_{k=1}^M p_k \ket{\psi_k}\bra{\psi_k} \] and it may study the individual terms, pure states, independently of others. When we do so, and we often do, we find out that the evolution of $\ket\psi$, the pure states, has to be linear as well. The linear maps $\ket\psi\to U\ket\psi$ produce $\hat\rho\to U\hat\rho \hat U^\dagger$ for $\hat\rho=\ket\psi\bra\psi$ which is still linear in the density matrix, as required. If you had a more general, nonlinear evolution – or if you represented observables by non-linear operators etc. – then these nonlinear rules for the wave function would get translated to nonlinear rules for the density matrix as well. And nonlinear rules for the density matrix would contradict some completely basic "linear" rules for probabilities that are completely independent of any properties of the laws of physics, such as\[ P(A\text{ or }B) = P(A)+P(B) - P(A\text{ and }B). \] So the linearity of the evolution equations in the density matrix (and, consequently, also the linearity in the state vector which is a polotovar for the density matrix) is totally necessary for the internal consistency of a theory that predicts probabilities, whatever the internal rules that yield these probabilistic predictions are! That's why two pieces of the wave function (or the density matrix) can never attract each other or otherwise interact with each other. As long as they're orthogonal, they're mutually exclusive possibilities of what may happen. They can never be interpreted as objects that simultaneously exist at the same moment. The product of their probabilities (and anything that depends on its being nontrivial) is zero because at least one of them equals zero. And the wave functions and density matrix cannot be interpreted as classical objects because it's been proven, by the most rudimentary experiments, that these objects are probabilistic distributions or their polotovars rather than observables. These statements depend on no open questions at the cutting edge of the modern physics research; they're parts of the elementary undergraduate material that has been understood by active physicists since the mid 1920s. It now trivially follows that all the people who study Schrödinger-Newton equations are profoundly deluded, moronic crackpots. And that's the memo. Single mom: totally off-topic Totally off-topic. I had to click somewhere, not sure where (correction: e-mail tip from Tudor C.), and I was led to this "news article"; click to zoom in. Single mom Amy Livingston of Plzeň, 87, is making \$14,000 a month. That's not bad. First of all, not every girl manages to become a mom at the age of 87. Second of all, it is impressive for a mom with such a name – who probably doesn't speak Czech at all – to survive in my hometown at all. Her having 12 times the average salary makes her achievements even more impressive. ;-) Posted by Luboš Motl \| Other texts on similar topics: philosophy of science #### snail feedback (3) : reader Ervin Goldfain said... Lubos, Your points are well taken: the Schrodinger-Newton equation is fundamentally flawed. Expanding on these issues, I'd like to know your views on the validity of: 1)the WKB approximation, 2)semiclassical gravity, 3)quantum chaos and quantization of classically chaotic dynamical systems? Cheers, Ervin reader Luboš Motl said... Dear Ervin, thanks for your listening. All the entries in your systems are obviously legitimate and interesting approximations (1,2) or topics that may be studied (3). That doesn't mean that all people say correct things about them and use them properly, of course. ;-) The WKB approximation is just the "leading correction coming from quantum mechanics" to classical physics. Various simplified Ansaetze may be written down in various contexts. Semiclassical gravity either refers to general relativity with the first (one-loop) quantum corrections; or it represents the co-existence of quantized matter fields with non-quantized gravitational fields. This is only legitimate if the gravitational fields aren't affected by the matter fields - if the spacetime geometry solve the classical Einstein equations with sources that don't depend on the microscopic details of the matter fields and particles which are studied in the quantum framework. The matter fields propagate on a fixed classical background in this approximation but they don't affect the background by their detailed microstates. Indeed, if the dependence of the gravitational fields on the properties of the matter fields is substantial or important, there's no way to use the semiclassical approximation. Some people would evolve the gravitational fields according to the expectation values of the stress-energy tensor but that's the same mistake as discussed in this article in the context of the Poisson-Hossenfelder equation. Classical systems may be chaotic - unpredictable behavior very sensitive on initial conditions. Quantum chaos is about the research of the complicated wave functions etc. in systems that are analogous to (hatted) classically chaotic systems. reader Ervin Goldfain said... Thanks Lubos. I also take classical approximations with a grain of salt. For instance, mixing classical gravity with quantum behavior is almost always questionable a way or another. Here is a follow up question. What would you say if experiments on carefully prepared quantum systems could be carried out in highly accelerated frames of references? Could this be a reliable way of falsifying predictions of semiclassical gravity, for example? Ervin \| Subscribe to: all TRF disqus traffic via RSS [old] To subscribe to this single disqus thread only, click at ▼ next to "★ 0 stars" at the top of the disqus thread above and choose "Subscribe via RSS". Subscribe to: new TRF blog entries (RSS)	{"extraction_info": {"found_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": 0.18, "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": 0.9417586922645569, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/25345/why-do-rhombus-diagonals-intersect-at-right-angles	# Why do rhombus diagonals intersect at right angles? I've looked all over and I can't find a good proof of why the diagonals of a rhombus should intersect at right angles. I can intuitively see its true, just by drawing rhombuses, but I'm trying to prove that the slopes of the diagonals are negative reciprocals and its not working out. I'm defining my rhombus as follows: $[(0,0), (a, 0), (b, c), (a+b, c)]$ I've managed to figure out that $c = \sqrt{a^2-b^2}$ and that the slopes of the diagonals are $\frac{\sqrt{a^2-b^2}}{a+b}$ and $\frac{-\sqrt{a^2-b^2}}{a-b}$ What I can't figure out is how they can be negative reciprocals of one another. EDIT: I mean to say that I could not find the algebraic proof. I've seen and understand the geometric proof, but I needed help translating it into coordinate form. - 5 You've looked all over?? I typed "rhombus" into google, and the first hit was the wikipedia article on rhombus, which suggests a proof using high school geometry: the diagonals divide the rhombus into four triangles. The top two triangles are congruent by the "side-side-side" criterion and their congruence shows that the bottom angle in the left triangle has the same measure as the bottom angle in the right triangle. But these are supplementary angles, so they must both be right angles. – Pete L. Clark Mar 6 '11 at 19:41 1 By the way, here's a suggestion to show that two numbers $X$ and $Y$ are negative reciprocals: show that $XY = -1$. In your case, the identity $a^2-b^2 = (a+b)(a-b)$ will be useful. – Pete L. Clark Mar 6 '11 at 19:42 – Pete L. Clark Mar 6 '11 at 19:46 @Pete L. Clark I meant to say that I couldn't find the algebraic proof. I had found the geometric proof, but I was having trouble translating it into an algebraic form. – quanticle Mar 6 '11 at 20:21 Thanks for editing. I have removed my downvote. – Pete L. Clark Mar 6 '11 at 23:08 ## 4 Answers Another way to say that the slopes are opposite reciprocals is to say that their product is $-1$. $$\begin{align} \frac{\sqrt{a^2-b^2}}{a+b}\cdot\frac{-\sqrt{a^2-b^2}}{a-b} &=\frac{-(\sqrt{a^2-b^2})^2}{(a+b)(a-b)} \\ &=\frac{-(a^2-b^2)}{a^2-b^2} \\ &=-1 \end{align}$$ - Thanks! That's exactly what I was looking for. My problem was that I was looking for something that looked like a reciprocal without going ahead and multiplying out my two slopes. – quanticle Mar 6 '11 at 20:11 @quanticle: When you're looking for things that look like reciprocals and you've got radicals, doing some algebra may make the connection clearer. For example: $$\frac{\sqrt{a^2-b^2}}{a+b}\cdot\frac{\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}=\frac{a^2‌-b^2}{(a+b)\sqrt{a^2-b^2}}=\frac{(a+b)(a-b)}{(a+b)\sqrt{a^2-b^2}}=\frac{a-b}{\sqr‌t{a^2-b^2}}$$ – Isaac Mar 6 '11 at 20:17 n.b. in both my answer and my immediately-previous comment, $a>b$, so $$(\sqrt{a^2-b^2})^2=\left\|a^2-b^2\right\|=a^2-b^2.$$ – Isaac Mar 6 '11 at 20:18 You don't have to work through square roots if you use the properties of the vector dot product and the parallelogram law to construct the rhombus. I.e one of the rhombus's diagonals can be identified with $\mathbf{a + b}$ where $\mathbf{b}$ is a vector added head-to-tail to vector $\mathbf{a}$, according to the parallelogram law. Similarly the other diagonal is given by $\mathbf{b - a}$. The constraint for a rhombus is $\lVert \mathbf{a} \rVert^2 = \lVert \mathbf{b} \rVert^2$. Two vectors $\mathbf{u, v}$ are perpendicular iff $\mathbf{u} \cdot \mathbf{v} = 0$, and as $\mathbf{(a + b)} \, \mathbf{\cdot} \, \mathbf{(b - a)}$ = $\lVert \mathbf{b} \rVert^2 - \lVert \mathbf{a} \rVert^2 = 0$, the two diagonals are perpendicular. - Hint: Multiply the slopes together and simplify. - One more way of looking at it is by factoring $a^2-b^2$ while it's inside the square root. Then $$\begin{align} \frac{\sqrt{a^2-b^2}}{a+b} &= \frac{\sqrt{(a-b)(a+b)}}{a+b} = \frac{\sqrt{a-b}}{\sqrt{a+b}} \\ - \frac{\sqrt{a^2-b^2}}{a-b} &= \frac{\sqrt{(a-b)(a+b)}}{a-b} = -\frac{\sqrt{a+b}}{\sqrt{a-b}} \end{align}$$ From here, it's easier to see that they have the correct relationship. -	{"extraction_info": {"found_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": 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": 0.9605067372322083, "perplexity_flag": "head"}
http://cstheory.stackexchange.com/questions/14726/does-conways-primegame-generate-all-prime-powers-of-2	# Does Conway's PRIMEGAME generate all prime powers of 2? Most sites I have visited reading on this interesting topic state something along the lines "the only powers of two (other than 2 itself) that occur in this sequence are those with prime exponent" (MathWorld) or "After 2, this sequence contains the following powers of 2: [...] which are the prime powers of 2." (Wikipedia) These careful formulations would imply that the set of powers of 2 generated in the sequence is a subset of prime powers of 2. However, the OEIS seems absolutely certain that the two sets are equal: http://oeis.org/A034785 This result is also cited on other sites I do not consider very reliable for exact wording, like http://esolangs.org/wiki/Fractran. Honestly, I have not understood the internal mechanics of PRIMEGAME enough to answer my own question yet. However, I think it makes a significant difference in the interestingness of PRIMEGAME. Why would sites like MathWorld not state the full fact? - – Chris Pressey Dec 14 '12 at 12:41 2 Yes, PRIMEGAME outputs 2^k if and only if k is prime. Here's one explanation by Conway himself: mathematik.uni-bielefeld.de/~sillke/NEWS/fractran See also oeis.org/wiki/Conway's_PRIMEGAME The original paper is well worth reading if you can track it down. – JɛﬀE Dec 14 '12 at 14:59 3 @JɛﬀE comment->answer ? – Suresh Venkat♦ Dec 14 '12 at 16:21 note [complexity-theory angle] its very inefficient. in analysis article decomposing it into subroutine primitives, "Using these, the author shows that the Conway program is equivalent to a well-known (although highly inefficient) procedure for inspecting the next number for primality. His running analysis shows that inspecting the thousandth prime (8831) would require 468 056 052 atomic steps." Richard K. Guy, Math. Mag. 56 (1983), no. 1, 26--33. – vzn Dec 14 '12 at 20:09 ## 1 Answer Yes, PRIMEGAME outputs $2^k$ if and only if $k$ is prime. Conway's original paper is well worth reading if you can track it down. You can also find a very clear exposition in Richard Guy's paper Conway's prime producing machine (Mathematics Magazine 56(1):26–33, 1983), including the wonderful cartoon below. (Yes, that's Conway with the Alexander horns.) Conway himself posted a concise proof on the math-fun mailing list. There's also a brief explanation at the OEIS blog. -	{"extraction_info": {"found_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": 0.913295328617096, "perplexity_flag": "head"}
http://mathoverflow.net/questions/37495/random-noncrossing-chords-of-a-circle	## Random noncrossing chords of a circle ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Suppose you generate random chords of a circle, with endpoints selected uniformly over the circumference, rejecting any chord that crosses a previously generated chord. The disk is then partitioned into regions bounded by chords alternating with circular arcs. For example, here are $n{=}100$ random noncrossing chords, with a region bounded by 5 chords highlighted (in green). I am interested in the statistics of the structure of the dual trees for these regions. Assign each region a node, and connect two nodes by an edge if they share a chord. In the example above, the highlighted region's node has degree 5. Example questions: What is the expected maximum degree of a node for $n$ chords? Making a max-degree node the root, what is the expected height of the tree? (In the example above, the height is 21.) Etc. Has anyone encountered this model before? Or a model sufficiently analogous to help establish these statistics? Thanks for any pointers! Edit. Many thanks for the wealth of information provided by the community! I have not yet absorbed all the information in the cited papers, but so far I have not found this specific question answered (although it is likely implied, perhaps in the papers they cite): What is the expected maximum degree of a node as $n \rightarrow \infty$? What brought me to this topic in the first place is that I wondered if it might be near 3. - 2 are you aware of the work by David Aldous on random triangulations of the circle? there's a nice American Mathematical Monthly article of his from 1991 reviewing that construction. He considers triangulations of regular n-gons as n goes to infinity, and chooses triangulations uniformly from that set. in this case the dual trees are binary trees and there is a series of bijections to positive walks from 0 to 2(n-1) which in the large n limit tend to Brownian excursions after rescaling. – jc Sep 2 2010 at 13:52 @jc: No, I was not familiar. Must be this paper: "Triangulating the Circle at Random." Amer. Math. Monthly 101 (1994) 223-233. I will investigate. Thanks! – Joseph O'Rourke Sep 2 2010 at 13:57 oops, I misstated the result: the positive walks are those starting at 0 and first returning to 0 after 2(n-1) steps of +1 or -1 – jc Sep 2 2010 at 14:05 ## 2 Answers The article "Random recursive triangulations of the disk via fragmentation theory" discusses many properties of the model you describe. The search word is random geodesic lamination. - Ah, thanks! I would never have hit upon "random geodesic laminations"! – Joseph O'Rourke Sep 2 2010 at 13:55 2 The height of that tree is conjectured (by Nicolas Curien in a talk I saw) to be of the order n^{(sqrt(17)-3)/2} -- if you pick two uniformly random nodes, this will be the rough distance between them. But an argument is missing to show that there is no small exceptional set of nodes at greater distance from one another. – Louigi Addario-Berry Sep 2 2010 at 14:21 @louigi: Cool! That evaluates to approx. 13 for $n{=}100$. – Joseph O'Rourke Sep 2 2010 at 14:34 2 But there are constants. Together Theorem 1.1 (ii) and Proposition 4.1 imply that, viewing the circle as the unit circle in the complex plane, the expected distance from 1 to $e^{2 \pi i u}$ is about $2.0(n u (1-u))^{\beta}$, where $\beta = (\sqrt{17}-3)/2$. (The 2.0 here is a pretty good approximation to what is in fact a ratio of Gamma functions.) This gives expected distance a little over 12 between $1$ and $-1$, for example. – Louigi Addario-Berry Sep 2 2010 at 15:16 ### 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. Louigi is absolutely right. We control "typical" height not absolute one. Note that a similar discrete model has been investigated by physicists see http://www.phys.ens.fr/~wiese/pdf/hiraRNA.pdf I think the maximal degree after n steps is logarithmic, but I don't have any exact expression for the expected value... - Thanks, your paper that Gjergji cited is exactly what I was looking for! I will enjoy learning the source of the magic number $\frac{1}{2} (\sqrt{17} -3)$! – Joseph O'Rourke Sep 2 2010 at 15:09	{"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": 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": 0.92633056640625, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/116892/list	## Return to Answer 2 added 18 characters in body By way of penance for my earlier "answer": Take $A=\pmatrix{1&0\cr x&0\cr}$ and $B=\pmatrix{1&y\cr 0&0\cr}$. Then the eigenvectors of $M=AA'+BB'$ and $N=A'A+B'B$ are in general different. As $x$ goes to 0, the eigenvectors of $M$ go off to zero and infinity while the eigenvectors of $N$ can be anything; as $y$ goes to 0, the eigenvectors of $N$ go off to zero and infinity while the eigenvectors of $M$ can be anything. 1 By way of penance for my earlier "answer": Take $A=\pmatrix{1&0\cr x&0\cr}$ and $B=\pmatrix{1&y\cr 0&0\cr}$. Then the eigenvectors of $M=AA'+BB'$ and $N=A'A+B'B$ are in general different. As $x$ goes to 0, the eigenvectors of $M$ go off to infinity while the eigenvectors of $N$ can be anything; as $y$ goes to 0, the eigenvectors of $N$ go off to infinity while the eigenvectors of $M$ can be anything.	{"extraction_info": {"found_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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9350727200508118, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/152769-ratio-root-tests.html	# Thread: 1. ## Ratio and Root Tests Good morning All, I am working on ratio and root tests for infinite series. I would like to post what I have done, and see if anyone agrees with it. Please see below: $<br /> \displaystyle\sum_{n=2}^{\infty}(-1)^n\frac{123^n}{n!}<br />$ By the Ratio Test... $<br /> \displaystyle\lim_{n\rightarrow\infty}\mid\frac{(-1)^{(n+1)}123^{(n+1)}}{(n+1)!}\frac{n!}{(-1)^{n}123^{n}}\mid<br />$ And... $<br /> \displaystyle\lim_{n\rightarrow\infty}\frac{123n!} {(n+1)!} = \lim_{n\rightarrow\infty}\frac{123n!}{(n+1)n!} = \lim_{n\rightarrow\infty}\frac{123}{n+1} = 0<br />$ Therefore... The orignal series $\displaystyle\sum_{n=2}^{\infty}(-)^n\frac{123^n}{n!}$ is Absolutely Convergent. Are we good here? My second problem is: $\displaystyle\sum_{n=1}^{\infty}(-1)^n(\sqrt{n+1}-\sqrt{n})$ I'm not really sure where to start here. Would I apply the root test to this problem? 2. The first problem looks good to me, and you have the right idea for the second. What do you get when you try to take the $n$-th root? 3. Thanks! When I try to take the nth root i get the following: $\displaystyle\sum_{n=1}^{\infty}(-1)^n(\sqrt{n+1}-\sqrt{n})$ $\displaystyle\lim_{n\rightarrow\infty}\|[(-1)^n(\sqrt{n+1}-\sqrt{n})]^{\frac{1}{n}}\|$ $=\displaystyle\lim_{n\rightarrow\infty}\|(-1)^{\frac{n}{n}}((n+1)^{\frac{1}{2n}}-(n)^{\frac{1}{2n}})\|$ $=\displaystyle\|(-1)^1((n+1)^0-(n)^0)\|$ $=\displaystyle\|(-1)((1)-(1))\|$ $=\displaystyle\|-1+1\|$ $=\displaystyle\|0\|$ $=\displaystyle0$ That can't be right??? I have three problems; one is supposed to be divergent, one conditionally convergent, and one absolutely convergent. Since my first problem was absolutely convergent (and I was able to follow the entire process for the first problem) I would have thought that the second problem would be either conditonally convergent or divergent. ...so, where did I go wrong? 4. Have you studied the alternating series test? If so that is what is to be used. 5. I have reviewed the alternating series test. I might be making this harder on myself than it needs to be, but since the section deals with ratio and root tests, I would like to use one of those methods on this problem. Is this reasonable? 6. Originally Posted by MechEng I have reviewed the alternating series test. I might be making this harder on myself than it needs to be, but since the section deals with ratio and root tests, I would like to use one of those methods on this problem. Is this reasonable? Did you notice that $\displaystyle x_n=(-1)^n\left(\sqrt{n+1}-\sqrt{n}\right)=\frac{(-1)^n}{\sqrt{n+1}+\sqrt{n}\right }?$ 7. Originally Posted by Plato Did you notice that $\displaystyle x_n=(-1)^n\left(\sqrt{n+1}-\sqrt{n}\right)=\frac{(-1)^n}{\sqrt{n+1}+\sqrt{n}\right }?$ Shoot... I did not notice that. Let me try this again... 8. When I try to take the nth root i get the following: $\displaystyle\sum_{n=1}^{\infty}(-1)^n(\sqrt{n+1}-\sqrt{n}) = \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{(\sqrt{n+1}+\sqrt{n})}$ $=\displaystyle\lim_{n\rightarrow\infty}\|[\frac{(-1)^n}{(\sqrt{n+1}+\sqrt{n})}]^{\frac{1}{n}}\|$ $=\displaystyle\lim_{n\rightarrow\infty}\|\frac{(-1)^\frac{n}{n}}{(n+1)^\frac{1}{2n}+(n)^\frac{1}{2n }}}\|$ $=\displaystyle\|\frac{(-1)^1}{(n+1)^0+(n)^0}\|$ $=\displaystyle\|\frac{-1}{1+1}\| = \frac{1}{2}$ Am I losing it again somewhere? Or, are you implying that I use the Ratio Test? Thanks for the help. 9. Use the alternating series test. 10. Ok, I am waiting to hear back with respect to whether or not using the alternating series test is acceptable. In the interim, I have worked out the last of three problems and concluded that: $\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{(2n)!}{3^n(n!)}$ By Ratio Test... $\displaystyle\lim_{n\rightarrow\infty}\|\frac{(-1)^{(n+1)}(2(n+1))!}{3^{(n+1)}((n+1)!)}\frac{3^n(n !)}{(-1)^n(2n)!}\|$ $=\displaystyle\lim_{n\rightarrow\infty}\frac{4n^2+ 6n+2}{3n+3} = \infty$ Therefore... By the Ratio test, the series Diverges So, this leaves me with my remaining series that is conditionally convergent. And, this means that if I am able to use the Ratio or Root Test, I should wind up with a limit equal to one... right? 11. Plato, I am noticing a trend with the lesson plans that I am following; the questions seem to be quite difficult using the methods covered in that section, but following section will introduce a method that makes the problem quite simple to do. I assume this one is the same. I am not a huge fan of the way the text book is organized, but i am nearly done with it so it's not worth complaining now. I appreciate your patience. I used to tutor physics in college, so I know it can be trying when someone keeps mising the fundamental ideas. Thank you for continuing to work with me. 12. All tests are now available to use on this problem, so I will heed you advice, Plato, and try the alternating series test... 13. You should find that it in conditional but not absolute. To do the later, use the limit comparison test using $\dfrac{1}{\sqrt{n}}$ 14. Easy question I hope, but... Since one of the requirements for the limit comparison test is that the series are both positive, does that mean that my original series becomes... $\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n+1 }+\sqrt{n}}$ 15. Please correct me if I am out of whack here... Let $\displaystyle\sum a_n = \frac{1}{\sqrt{n+1}+\sqrt{n}}$ So... $\displaystyle\lim_{n\rightarrow\infty}\frac{1}{\sq rt{n+1}+\sqrt{n}}=0$ By Alternating Series Test... Convergent Let $\displaystyle\sum b_n = \frac{1}{\sqrt{n}}$ A Divergent P-series So... $\displaystyle\lim_{n\rightarrow\infty}\frac{1}{\sq rt{n+1}+\sqrt{n}}\frac{\sqrt{n}}{1}$ $=\displaystyle\lim_{n\rightarrow\infty}\frac{1}{\s qrt{n+1}}=0$ By Limit Comparison Test... Divergent This doesn't make sense, does it? Would it make ore sense if the order of the tests was reversed?	{"extraction_info": {"found_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": 31, "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": 0.9530503153800964, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/46883?sort=newest	## Examples of using physical intuition to solve math problems ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) For the purposes of this question let a "physical intuition" be an intuition that is derived from your everyday experience of physical reality. Your intuitions about how the spin of a ball affects it's subsequent bounce would be considered physical intuitions. Using physical intuitions to solve a math problem means that you are able to translate the math problem into a physical situation where you have physical intuitions, and are able to use these intuitions to solve the problem. One possible example of this is using your intuitions about fluid flow to solve problems concerning what happens in certain types of vector fields. Besides being interesting in its own right, I hope that this list will give people an idea of how and when people can solve math problems in this way. (In its essence, the question is about leveraging personal experience for solving math problems. Using physical intuitions to solve math problems is a special case.) These two MO questions are relevant. The first is aimed at identifying when using physical intuitions goes wrong, while the second seems to be an epistemological question about how using physical intuition is unsatisfactory. - 1 You have two links to only one question. – J. M. Nov 22 2010 at 0:45 J.M.: Thanks. Fixed the link. – Jack Lemon Nov 22 2010 at 1:12 1 Another [relevant question](mathoverflow.net/questions/38909/…) is about translating a physical argument into a mathematical one. – Ramsay Nov 22 2010 at 2:17 ## 10 Answers A paradigmatic example is Riemann's original "proof" of his mapping theorem in complex analysis. He gave an heuristic argument using Dirichlet's principle which was motivated by electrostatics in the plane. - ### 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. Read the following paper for some striking examples. MR2587923 Atiyah, Michael; Dijkgraaf, Robbert; Hitchin, Nigel Geometry and physics. Philos. Trans. R. Soc. Lond. Ser. A Math. Phys. Eng. Sci. 368 (2010), no. 1914, 913–926. - Here is a proof of Pick's area theorem $\mu(P)=i +{b\over2}-1$ "using physical intuition": Assume that at time 0 a unit of heat is concentrated at each lattice point. This heat will be distributed over the whole plane by heat conduction, and at time $\infty$ it is equally distributed on the plane with density 1. In particular, the amount of heat contained in $P$ will be $\mu(P)$. Where does this amount of heat come from? Consider a segment $e$ between two consecutive boundary lattice points. The midpoint $m$ of $e$ is a symmetry center of the lattice, so at each instant the heat flow is centrally symmetric with respect to $m$. This implies that the total heat flux across $e$ is 0. As a consequence, the final amount of heat within $P$ comes from the $i$ interior lattice points and from the $b$ boundary lattice points. To account for the latter, orient $\partial P$ so that the interior is to the left of $\partial P$. The amount of heat going from a boundary lattice point into the interior of $P$ is a half, minus the turning angle of $\partial P$ at that point, measured in units of $2\pi$. Since the sum of all turning angles for a simple polygon is known to be one full turn, we arrive at the stated formula. - Beautiful! How does one make this rigorous? Gauss-Bonnet? – Qiaochu Yuan Nov 22 2010 at 14:11 Archimedes gave exact proofs as well as mechanically motivated explanations for results like the quadrature of the parabola or the volume of spheres. - Franz: Do you know any sources that detail this? – Jack Lemon Nov 22 2010 at 20:25 I seem to remember that this is described in "Archimedes. What did he do besides cry eureka?" by Sh. Stein. – Franz Lemmermeyer Nov 23 2010 at 6:53 Let $X$ be a random variable taking on $n$ distinct values with probabilities $p_1,\dots,p_n$. The entropy of $X$ is defined by $H(X)=\sum p_i \log_2(1/p_i)$. An early theorem is that $H(X) \leq \log_2(n)$, and here's a physical proof. Place a point with mass $p_i$ at $(x_i,y_i)=(1/p_i,\log(1/p_i))$. The center of mass $$(\bar x,\bar y) = \frac{\sum (m_ix_i, m_iy_i)}{\sum m_i} = (n,H(X))$$ of the $n$ points must lie in the convex hull of the points (this is the physical intuition part). But since $y=\log(x)$ is concave, the convex hull is completely below (or on) the curve $y=\log(x)$. That is, $H(X) \leq \log_2(n)$. - Polya's Induction and Analogy in Mathematics has a chapter on this, along with some great examples. It's not just physical intuition influencing mathematics; it's more a powerful synergy between physical and mathematical intuition. I'll summarize some of it: 1. Suppose we have two points A and B on the same side of some line L in the plane. What's the shortest path from A to L and then to B? The solution is obvious once we reflect one of the points (and its segment of the path) across L. That solution seems tricky in the abstract, but it's very intuitive if we imagine a reflecting ray of light and think about looking at things in a mirror. 2. Now suppose A and B are on different sides of L, and a particle moves from A to B, and its speed is different on the two sides of L. What's the shortest path (in time)? (This problem is to a refracting ray of light as the previous one is to a reflecting ray.) It turns out this can be solved by reducing it to a physical problem involving a system of weights and pulleys at equilibrium. I won't try to describe it here, but it might be fun to try to reinvent it. 3. Now let's take a serious math problem: what plane path minimizes the time an object takes to move from point A (at rest) to point B, assuming constant gravity? (This is the famous "brachistochrone" problem.) By conservation of energy, the speed of the object at a point on the curve depends only on its height (defined relative to its starting point and with respect to the direction of gravity). Thus, we're led to consider light moving in a very particular heterogeneous refracting medium, where the index of refraction depends in a specific way on the height. To find the path taken by light, we simply apply the law of refraction to this medium to obtain a differential equation for the path, which we can then solve. The interplay between mathematical and physical intuition is very interesting here. The first problem is mathematical, but in trying to solve it, it's natural to draw an analogy to optics. The second problem is suggested by optics, but we solve it by analogy with mechanics. The third problem is basically mechanical, but we solve it by analogy with optics, and we actually use the solution to the second problem! - I assign question one as a calculus problem with coordinates given for the two points. Of course, everyone tries to solve this using calculus, but the algebra of finding the zeros of the derivative of the length function is a little tricky. Then when someone asks me to solve the problem, I show them the conceptual solution. – Steven Gubkin Nov 22 2010 at 12:59 It's nearly trivial if you find the minimum of the square of the length rather than the length (they are related by a monotonic function, hence they have the same minima). – Chris Taylor Apr 1 2011 at 11:19 For electrical network intuition/applications to random walks see the beautiful little book of Doyle and Snell http://arxiv.org/abs/math/0001057 - I gave an answer based on surface tension (which I did not invent) to the napkin ring problem - The first and second laws of thermodynamics allow you to recover the inequality between the arithmetic and the geometric means: Bring together n identical heat reservoirs with heat capacity C and temperatures T_1,...T_n and allow them to reach a final temperature T. The first law of thermodynamics tells you that T is the arithmetic mean of the T_i. The second law of thermodynamics demands the non-negativity of the change in entropy, which is Cn Log(T/G) where G is the geometric mean. It follows that T > G. I believe this argument was first made by P.T. Landsberg (no relation!). - 13 Warmest congratulations for this cool example. – Georges Elencwajg Nov 22 2010 at 13:10 2 Does this argument basically say conversely that the first law of thermodynamics conversely boils down to the same kind of convexity as in the arithmetic-geometric mean? – Phil Isett Sep 8 2011 at 2:38 Mark Levi's book The Mathematical Mechanic is full of elementary and beautiful examples of this kind. Some examples are also given in this blog post by Yan Zhang. A classic example is a "proof" that there exist non-constant meromorphic functions on a compact Riemann surface, which I think is due to Klein: see this MO question. - +1: You beat me to it -- I heard Mark Levi give a number of talks along those lines while I was at Penn State, and they always ranked among the more interesting talks I've attended. – Vaughn Climenhaga Nov 22 2010 at 1:33 This is officially now one of my all time favorite books and not only am I inspired to write a similar and more advanced text someday,but to use this book and others of its ilk in my teaching! – Andrew L Nov 22 2010 at 6:28 +1: You beat me too! (yet why giving +1 to whom beats you?) ;-) – Pietro Majer Nov 22 2010 at 13:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 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": 0.9372287392616272, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/14026/estimation-of-accuracy-of-measurements	# Estimation of accuracy of measurements? When stating an estimation of the accuracy of my measurements (of electric current, in this case) would I state that it is equal to the residual error, or the limit of reading? (in this case, ±0.2 or ±0.01). Why? This is for pretty simple high school physics - so nothing too complex. Thanks! - Just a few things about your question; the title is not descriptive enough, the $\pm$ numbers you give don't have units or the measurement value and thus aren't helpful, and lastly, we don't know what your setup is, it's not clear what your specific difficulty is. If you correct those things I'll upvote your question. – AlanSE Aug 28 '11 at 0:49 ## 1 Answer It depends on the situation and what the expectation of the teacher is. If the teacher specifically asked you to record error, then s/he likely provided methodology somewhere along the line. If you are making note of the error because you think it is the right thing to do, I would recommend using the residual error and making note of how you calculated it in your lab report. There are at least two reasons I would take this route: 1. If your residual error is an order of magnitude larger than the readability of the device, chances are very good that the experiment is complex enough that there are many sources of error you are overlooking. Given the accuracy of most lab equipment (even in a high school setting), there are aspects of your experiment that add to your uncertainty more than the equipment readability. 2. The residual error is likely based on a well recognized phenomenon with an accepted mathematical model. If the standard model says that you should get a certain value, that's what you should get. The chances that you have discovered some strange exception to the recognized physics are remote. If you are interested in high end metrology and would like to learn more, the internationally recognized best practice for calculating and citing uncertainties can be found here. Good luck with your class! - 1 I would add: take a look at the data sheet of the measuring device to see its quoted accuracy. – nibot Aug 27 '11 at 21:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9577774405479431, "perplexity_flag": "middle"}
http://polymathprojects.org/2009/11/20/proposals-tim-gowers-polynomial-dhj-and-littlewoods-problem/?like=1&source=post_flair&_wpnonce=d42062ed33	The polymath blog November 20, 2009 Proposals (Tim Gowers): Polynomial DHJ, and Littlewood’s problem Filed under: polymath proposals — Gil Kalai @ 10:06 am Tim Gowers described two additional proposed polymath projects. One about the first unknown cases of the polynomial Density Hales Jewett problem. Another about the Littelwood’s conjecture. I will state one problem from each of these posts: 1) (Related to polynomial DHJ) Suppose you have a family $\cal F$ of graphs on n labelled vertices, so that we do not two graphs in the family $G,H$ such that $H$ is a subgraph of $G$ and the edges of $G$ which are not in $H$ form a clique. (A complete graph on 2 or more vertices.) can we conclude that $\|{\cal F}\|$ =$2^{o({{n} \choose {2}}}$? (In other words, can we conclude that $\cal F$ contains only a diminishing fraction of all graphs?) Define the “distance” between two points in the unit cube as the product of the absolute value of the differences in the three coordinates. (See Tim’s remark below.) 2) (Related to Littlewood) Is it possible to find n points in the unit cube $[0,1]^3$ so that the “distance” between any two of them is at least $1/100000000000000000000n$? A negative answer to Littlewood’s problem will imply a positive answer to problem 2 (with some constant). So the pessimistic saddle thought would be that the answer to Problem 2 is yes without any bearing on Littlewood’s problem. 7 Comments » 1. It is important to clarify that the problem numbered (2) is not itself Littlewood’s problem but rather a related problem. Secondly, the “distance” is not what one might think. In fact, it is not even a distance. We define $d(x,y)$ to be $\|x_1-y_1\|\|x_2-y_2\|\|x_3-y_3\|.$ (Obviously if we were using the normal Euclidean distance then we could get $Cn^3$ points and that would be best possible.) Also, I find the problem (2) interesting in itself, so am not put off by the possibility that there might exist such a set of points even if the Littlewood conjecture is true. It would even, in a modest way, shed some light on the Littlewood conjecture, since it would demonstrate that the problem was “genuinely number-theoretic”. I say “in a modest way” because I get the impression that the experts more or less take for granted that it is a problem in number theory, so a conclusion of that kind would not change the way people think. But it would also provide some kind of explanation for why the problem is hard. Comment by — November 20, 2009 @ 10:13 am • Dear Tim, yes, I overlooked the distance definition for a short time but then realized it cannot be the ususal distance and looked back in the post. I agree that problem (2) is very interesting! (In any direction it might go!)And if the answer is no it may be, as you explained, harder than proving LP but still could be useful. Comment by — November 20, 2009 @ 12:10 pm 2. I think I can solve the problem about points in the cube. One starts with this basis (2,1,-2,) (2,-2,1)(1/4,1/2,1/,2) normalize it then scale it down by a factor of 1/n^1/3 then construct it by a factor of two then we should be able to fit about 8n of these points in the cube and the closest we can get two points is where one point difference from another by one coordinate and then the product of the difference of the coordinates will be much larger than 10^-14 n^3 if I am counting the zeros correctly Comment by — November 20, 2009 @ 9:10 pm 3. The above doesn’t work as I can add vectors to get one coordinate zero making the product zero. I don’t think there is a lattice that is going to work. Comment by — November 20, 2009 @ 9:42 pm 4. According to a comment by Tim Gowers on his blog the answer to the question about points on the cube is yes. I think this is problem 5 on his blog. Here is a link to the comment with the solution to the problem: http://gowers.wordpress.com/2009/11/17/problems-related-to-littlewoods-conjecture-2/#comment-4377 Comment by — November 25, 2009 @ 5:47 am • The answer to Problem 5 is yes, but that is with the dyadic distance in each coordinate (that is, $2^{-k}$, where $k$ is the first place where the two binary expansions differ). For the ordinary Euclidean distance I do not know the answer — that’s Problem 3. Comment by — December 2, 2009 @ 9:25 am 5. Wow! This can be one particular of the most beneficial blogs We’ve ever arrive across on this subject. Basically Wonderful. I’m also a specialist in this topic so I can understand your effort. Comment by — August 12, 2012 @ 7:27 am RSS feed for comments on this post. TrackBack URI Theme: Customized Rubric. 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http://mathhelpforum.com/advanced-algebra/123272-bilinear-forms-print.html	# Bilinear Forms Printable View • January 11th 2010, 08:47 AM adam63 Bilinear Forms f:VxV -> F we call f a withered form if there's a vector $v\ne 0$ for which f(v,u)=0 for all u in V. Could f be not-withered, while there still is a $v\ne 0$ for which f(u,v)=0 for all u in V? (BTW : $f(u,v)=[u]^t_{B}[f]_B[v]_B$ for any base B of V) • January 11th 2010, 10:54 AM tonio Quote: Originally Posted by adam63 f:VxV -> F we call f a withered form if there's a vector $v\ne 0$ for which f(v,u)=0 for all u in V. Could f be not-withered, while there still is a $v\ne 0$ for which f(u,v)=0 for all u in V? (BTW : $f(u,v)=[u]^t_{B}[f]_B[v]_B$ for any base B of V) Perhaps I'm missing something, but I'm afraid your question is nonsensical. You're given a definition (If A then B) and then you ask: could it be that not B but still A? Of course not! The definition tells you that if A (=exists a non zero vector s.t. f(u,v) = 0 ...etc.) then B(=the bil. form is called withered)...!! Tonio • January 11th 2010, 11:28 AM adam63 Quote: Originally Posted by tonio Perhaps I'm missing something, but I'm afraid your question is nonsensical. You're given a definition (If A then B) and then you ask: could it be that not B but still A? Of course not! The definition tells you that if A (=exists a non zero vector s.t. f(u,v) = 0 ...etc.) then B(=the bil. form is called withered)...!! Tonio Please notice the difference, it's f(v,u) once, then f(u,v). • January 11th 2010, 11:15 PM NonCommAlg Quote: Originally Posted by adam63 f:VxV -> F we call f a withered form if there's a vector $v\ne 0$ for which f(v,u)=0 for all u in V. Could f be not-withered, while there still is a $v\ne 0$ for which f(u,v)=0 for all u in V? (BTW : $f(u,v)=[u]^t_{B}[f]_B[v]_B$ for any base B of V) the answer is no. to see this, suppose $\dim_F V = n$ and identify $V$ with $F^n.$ then $f(X,Y)=X^TAY,$ for some $A \in \mathbb{M}_n(F)$ and all $X,Y \in F^n.$ let $E_i$ be the $n \times 1$ vector with $1$ in the $i$th row and $0$ everywhere else. see that $E_i^T A$ is the $i$th row of $A$ and, for any $Y \in F^n,$ the $i$th row of $AY$ is $E_i^TAY.$ now suppose there exists $0 \neq Y_0 \in F^n$ such that $X^TAY_0 = 0$ for all $X \in F^n.$ so $E_i^T A Y_0=0,$ for all $i.$ thus $AY_0 = 0$ and so $\det A = 0.$ but then $\det A^T =0$ and hence $A^T X_0 = 0,$ for some $0 \neq X_0 \in F.$ thus $X_0^T A=0$ and so $X_0^TAY=0,$ for all $Y \in F^n,$ i.e. $f$ is withered. All times are GMT -8. The time now is 10:35 AM.	{"extraction_info": {"found_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": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9259039759635925, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/16076?sort=newest	## Is the cohomology of a topological operad a cooperad? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) For cohomology with coefficients in a field $F$ the map $H^\cdot(X;F) \otimes H^\cdot(Y;F) \to H^\cdot(X \times Y;F)$ of the Kunneth theorem is an isomorphism of algebras over $F$. I am correct in thinking that this together with the fact that cohomology with a contravariant functor, implies that the cohomology of a topological operad is a cooperad, the dual of a notion of a cooperad? If this is not true, where does my reasoning fail? Does it at least hold on the level of vectorspaces over $F$? If this is true, why aren't there many references about this construction? It seems that the additional algebra structure would give you some additional information. Furthermore, using the Thom isomorphism for the cohomology, would this also imply that there is a notion of string cohomology dual to string topology? - ## 2 Answers There is one little caveat, though: you need either to assume your spaces in the operad are of finite type, or work in the category of topological vector spaces and completed tensor products. The reason is that the Künneth theorem in cohomology only holds for spaces of finite type if you use the regular tensor product. Consider, for a counterexample, an infinite wedge of spheres. - Do you a reference this way of considering cohomology groups as topological vector spaces? What is the natural topology on them? Is it the weak-* topology with respect to the homology-cohomology pairing? – skupers Feb 23 2010 at 14:57 It's the topology of a pro-finite dimensional vector space. You have to take field coefficients. Let's assume we work in the category of CW-complexes. Every CW-complex X is the directed colimit of its finite subcomplexes, and this gives an inverse system after applying cohomology. The limit of this system is the cohomology of X. (There are no phantom maps.) – Tilman Feb 24 2010 at 6:38 ### 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. Yes the cohomology of an operad over a field is a cooperad . Indeed over a field F, the cohomology functor H* is a monoidal functor (in a contravariant sense.) More precisely the category of topological spaces Top equipped with the cartesian product × is a symmetric monoidal category (the unit being the one point space *). Similarly the category of graded vector spaces is symmetric monoidal when equipped with the tensor product ⊗ (the unit being the ground field concentrated in degree 0 that we denote by F). Then Kunneth theorem tells you that you have a map H∗(X)⊗ H∗(X) --> H∗(XxY), and moreover it is an isomorphism. Also H∗(∗) is isomorphic to the field F. This is exactly the meaning of your functor being monoidal (which here as to be interpreted in a contravariant sense.) In other words your functor commutes, up to canonical isomorphisms, with the two monoidal structures × and ⊗. A consequence of this suppose given a topological operad, that is a sequence (X(n))n≥0 of spaces together with structure maps μ:X(k)xX(n1)x...xX(nk) --> X(n1+...+nk) satisfying the associativity condition for an operad (and also a unit belonging to X(1) and some compatible action of the symmetric groups). Then, applying the monoidal functor H∗ you get a cooperad (H∗X(n))n≥0 in the category of graded vector spaces. Indeed composing H∗(μ) with the Kunneth isomorphism you get a map H(X(n1+...+nk)) --> H(X(k)xX(n1)x...xX(nk)) ≅ H(X(k))⊗H(X(n1))⊗...⊗H(X(nk)) satisfying the coassociativity condition of a cooperad (with also a counit and Σ-equivariance.) You can also try to do the same at the level of (co)chains. However the cochain functor is not monoidal (it is weakly "comonoidal"). So it is not true that the cochains of an operad is a genuine cooperad (although it is amost a cooperad). But the functor of singular chains is monoidal and so it is true that the chains of a topolgical operad is an operad of chain complexes. Maybe it is because this lack of genuine structure at the cochain level that one do not consider often the cooperad structure on the cohomology. Anyway it would not give more structure that the homology and it is often easier to work with this. The reason for which this is not in the litterature is that it is folklore that a monoidal covariant/contravariant functor turns operad into operads/cooperads. - You consider the symmetric monoidal category of graded vector spaces, thereby forgetting about the cup product. Does the statement fail if we use the category of algebras over $F$ with tensor product, trying to remember the cup product? – skupers Feb 23 2010 at 14:52 You can also do tis in the category of graded algebra, so you get indeed a cooperad of algebras using the cup products. Notice however that if you do this in homology you get an operad of coalgebras (the coalgebra structure in homology is dual to the algebra structure of cohomology). Therefore you dot not really gain more structure passing from homology to cohomology. As Tilman pointed out homology is actually better when you deal with space of not finite type (whch is actually rare in the usual applications) – Pascal Lambrechts Feb 24 2010 at 8:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9110220670700073, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/283444/problem-related-to-a-square-matrix	# Problem related to a square matrix Let $A$ be an $n\times n$ matrix with real entries such that $A^{2}+I=\mathbf{0}$. Then: (A) $n$ is an odd integer. (B) $n$ is an even integer. (C) $n$ has to be $2$ (D) $n$ could be any positive integer. I was thinking about the problem.I noticed for a $2\times 2$ matrix $A$ of the form $$\begin{pmatrix} 1 &-2 \\ 1& -1 \end{pmatrix},$$ the given condition holds good.So option (C) is a possibility.But I am not sure about other options. Is there any convenient way to tackle it?With regards.. - @amWhy: Don't you think when we are supposed to ruled out any options in a multiple choices like above, we need just an example for doing this? Do you think, we have to prove this theoretically? Thanks. – Babak S. Jan 21 at 14:08 Hint: Think about how one multiplies block matrices. – peoplepower Jan 21 at 14:11 ## 3 Answers Note: $n=1\,$ is ruled out, since e.g., $A$ consists of the single scalar entry $1$: $A = [1],\; I = I_1,\;, A^2 + I = 2.\;$ Indeed there is no real scalar $\,k\,\neq 0\,$ (in the case $n = 1, A = k\,$) such that $\,k^2 = -1.\,$ So option (A) is ruled out, since $\,n = 1\,$ is odd, and option (D) is ruled out, since $\,n = 1 >0\, n \in \mathbb{Z}^+$. What remains is to decide between (B) and (C). You know for $n = 2\,$, the equality is satisfied (hence "(C)" is in the "running") but $n = 2\,$ is also even: so "(B)" has a chance. You can rule out (C) if there exists any $\,n= 2k,\; k\in \mathbb{Z}^+$, $n > 2,\,$ such that $\,A_{n\times n}^2 + I_n\, = 0$. • Hint: try $n = 4$: construct a $4\times 4$ matrix made of $4$-square block $2 \times 2$ matrices, using your matrix for each of the two block entries on the diagonal, and zero blocks off the diagonal.) Let $A = \begin{pmatrix} 1 &-2 \\ 1& -1 \end{pmatrix}. \quad$ So using $A^2 + I = 0$, construct $A_{4 \times 4} = \begin{pmatrix} A & 0 \\ 0 & A \end{pmatrix}$ $$A_{4\times 4}^2 + I_4 = \begin{pmatrix} A &0 \\ 0& A \end{pmatrix} \cdot \begin{pmatrix} A & 0 \\ 0 & A \end{pmatrix} + I_4 = \begin{pmatrix} A^2 & 0 \\ 0& A^2 \end{pmatrix} + \begin{pmatrix} I_2 & 0\\ 0 & I_2 \end{pmatrix}$$ $$= \begin{pmatrix} A^2 + I_2 & 0\\ 0 & A^2 + I_2 \end{pmatrix}= 0$$ - You did rule out those poor A and D. :D & +1 – Babak S. Jan 21 at 14:23 @amWhy so from your calculation,i see that $(B)$ is the right choice..Thanks a lot sir for the detailed clarification. – user33640 Jan 21 at 15:45 user33640: Yes indeed! B is correct: Your welcome! – amWhy Jan 21 at 15:50 As $A$ is real, so is $\det A$. Yet $A^2+I=0$ implies that $(\det A)^2=\det(-I)=(-1)^n$. Therefore $n$ has to be an even integer and (A), (D) are incorrect. Now, if there is a $2\times 2$ matrix $A$ such that $A^2+I=0$, then $\tilde{A}=\begin{pmatrix}A\\&A\end{pmatrix}$ would be a $4\times4$ matrix such that $\tilde{A}^2+I=0$. Therefore (C) is incorrect too. To verify that (B) is indeed the correct answer, consider the aforementioned $\tilde{A}$ with $A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$. - Hint: Let's $n>1$ an interger positive and even. Set the $n\times n$ matrix $$J_{n\times n}=\begin{pmatrix} 0 & \ldots & 0 & \ldots &-1 \\ \vdots & & \vdots & & \vdots \\ 0 & \ldots & -1 & \ldots & 0 \\ \vdots & & \vdots & & \vdots \\ -1 & \ldots & 0 & \ldots &0 \\ \end{pmatrix}.$$ Note that $I_{n\times n}=J_{n\times n}^2$ for if $n$ is even and $$A^2+I_{n\times n}= ( A+J_{n\times n})(A-J_{n\times n})=0.$$ Then we have $A=\pm J$. If for n=1 we are a conter exemple. It's exclude (D). Then the answer is (C). - Not n = 1!.............. – amWhy Jan 21 at 14:46 1 "Then n could be any positive integer." Not true: you have not established this is true for $n = 1$. Indeed, $A^2 + I = 0 \implies n \neq 1$. Hence $n$ cannot be any positive integer, nor can it be any odd integer, as $n = 1 > 0; \, n \text{ is odd.}$ – amWhy Jan 21 at 14:52 @amWhy, Tank's for you coments – Elias Jan 21 at 14:53	{"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": 54, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9302038550376892, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/1019/common-false-beliefs-in-physics/30836	Common false beliefs in Physics [closed] Well, in Mathematics there are somethings, which appear true but they aren't true. Naive students often get fooled by these results. Let me consider a very simple example. As a child one learns this formula $$(a+b)^{2} =a^{2}+ 2 \cdot a \cdot b + b^{2}$$ But as one mature's he applies this same formula for Matrices. That is given any two $n \times n$ square matrices, one believes that this result is true: $$(A+B)^{2} = A^{2} + 2 \cdot A \cdot B +B^{2}$$ But eventually this is false as Matrices aren't necessarily commutative. I would like to know whether there any such things happening with physics students as well. My motivation came from the following MO thread, which many of you might take a look into: - 4 Community wiki? – Marek Nov 17 '10 at 23:10 1 @MArek: i didnt find the option. if anyone can do it they are welcome – Chandrasekhar Nov 17 '10 at 23:15 2 @Chandru: AFAIK StackExchange recently changed its rules about this matter so that only moderators can make a question community wiki (the rationale being that the CW option is being misused at StackOverflow). – Marek Nov 17 '10 at 23:26 22 While i perceive the question interesting, I think the example with matrices is rather 'a common silly mistake' than 'a common false belief'. – Piotr Migdal Nov 17 '10 at 23:48 6 It seems a bit odd to be accepting a single answer to a soft question... – David Zaslavsky♦ Dec 1 '10 at 0:21 show 3 more comments locked by David Zaslavsky♦Sep 30 '12 at 20:06 This question exists because it has historical significance, but it is not considered a good, on-topic question for this site, so please do not use it as evidence that you can ask similar questions here. This question and its answers are frozen and cannot be changed. More info: FAQ. closed as not constructive by David Zaslavsky♦Sep 30 '12 at 20:05 As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or specific expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, see the FAQ for guidance. 49 Answers That the PI Team is the best group to design and implement their own data systems, rather than bring in people experienced with IT security, data modeling for reuse by other groups, and other data informatics issues. Related misconception: that their data is so different from other data that a data system must be designed from the ground up for each new experiment. - Notion of simultaneity. Because of speed of light is so big, it looks true in our day to day affairs. But it really is a non existent thing [due to special relativity]. 2 people in 2 different places can't say "at the same instant". - I find people think that "a bullet and a ball shot and dropped repectively from the same height will not hit the ground at the same time". - 3 ...because that's a true statement. A shot bullet is going to start out with quite a bit higher vertical velocity than a dropped ball. – BlueRaja - Danny Pflughoeft Jan 12 '11 at 22:45 A common misconception is that motion of massive objects with superluminal velocity is prohibited in General Relativity. This is not true as only local superluminal motion is prohibited. - How about these three misconceptions: The big bang theory tells us the universe started from a point Since Einstein we know all is relative The speed of light is a fundamental constant - This is also a misconception of science in general, but I've heard many people say, "Physics (science) has proven. . ." or "Can we use Physics (science) to prove this?" The misconception is that the scientific method can prove something with 100% certainty. This is certainly not the case; experiments only validate a law in settings similar to that in which the experiments were conducted. Granted, we can often reasonably generalize our results far beyond particular settings, but we can only hold laws with certainty where they have been tested and verified. - The constancy of the speed of light postulated by Albert Einstein in 1905 was motivated by the null result of the Michelson-Morley experiment. This is wrong. See Einstein was mainly motivated by the results of Fizeau's experiment measuring the speed of light in moving water - show 2 more comments The greenhouse effect (not the atmospherical one) : One common misbelieve is, that the reason for this effect is, that sunlight comes in and is transformed to infra red radiation that can't go out. But the main reason is a lack of air exchange (see the section "Real greenhouses" in http://en.wikipedia.org/wiki/Greenhouse_effect). - show 1 more comment A common misconception is that in a double slit experiment, the electrons or the photons go through both slits at the same time and interfere with themselves. Nothing of the sort ever happens and that idea doesn’t come from quantum mechanics. It is just meaningless to ask the question (from which slit did the particle pass?) in the context of the particular experiment and even more meaningless to say that it went through both. The particle always passes from the one or the other slit if you setup an experiment that asks the question and in the case that you shoot the particles one at a time you always get one hit on the screen and not an interference of portions of the particle. I think that the misconception has its roots in the wave analogy of the wave function description of quantum mechanics, where you must have a wave passing from both slits in order to have interference on the other side. Of course that picture doesn’t translate to having a particle passing from both slits, but it only states that there exists interference between the wave functions of the two independent events that translates to a distribution of probability for finding a particle on a particular point on the screen. That is a statement of Born’s interpretation that has been recently tested with a triple slit experiment. One could ask at this point, “Ok, I get it for the electrons, but light behaves like a wave in everyday experience. What happens in that case?”, and the answer is that light behaves like a wave only when you have a large flux of photons and you can go to the continuous field approximation. It is then that the wave-like properties become apparent both for photons and electrons. Update: A very interesting video on Quantum Mechanics Update2: QM in your face Update3: The Feynman Lectures on Physics vol3: Scattering from a crystal (neutrons). ... Let’s review the physics of this experiment. If we could, in principle, distinguish the alternative final states (even though you do not bother to do so), the total, final probability is obtained by calculating the probability for each state (not the amplitude) and then adding them together. If you cannot distinguish the final states even in principle, then the probability amplitudes must be summed before taking the absolute square to find the actual probability. The thing you should notice particularly is that if you were to try to represent the neutron by a wave alone, you would get the same kind of distribution for the scattering of a down spinning neutron as for an up spinning neutron. You would have to say that the “wave” would come from all the different atoms and interfere just as for the up spinning one with the same amplitude. But we know that this is not the way it works. So as we stated earlier, we must be careful not to attribute too much reality to the waves in space. They are useful for certain problems but not for all. - 1 Actually, what you said is not true. First, electron go through both slits at once (if you want to say something contrary, it's much more about taste or interpretation, that 'fighting with a false belief'). Second, while properties of a single photon are interesting, I would not say it does not behave like a wave (it certainly have many wave-like properties). – Piotr Migdal Nov 20 '10 at 9:56 3 Very strange answer. Actually, it only depends on interpretation whether you think that electron passes through both slits. So this means that it is not a misconception at all. Actually, the most attractive interpretation to me (Feynman's path integral) tells you that the electron travels through all paths and all of them are equally important in determining the final amplitude. In particular, the paths through left and right slits interfere. – Marek Dec 1 '10 at 0:04 2 It seems to me like you are trying to propose ancient preconceptions (like the ones Einstein had) that quantum effects appear just as statistics after letting many electrons through the slits. Well, this is obviously not the case. Every single electron behaves quantum mechanically and travels through all the paths (in path integral view) or is a wave that passes through both slits (in Schroedingerian view). – Marek Dec 1 '10 at 0:08 3 Sorry to step in abruptly, but QM does not tell us ANYTHING about the path of the electron. The ONLY objective, measurable and sustainable thing that QM tells us is the OUTCOME of the MEASUREMENTS. I.e. that electrons will hit the screen in some pattern. All the rest is (non-physical) interpretation and subjective. – Sklivvz♦ Dec 3 '10 at 21:16 1 @Sklivvz: but you are implicitly using one such interpretation yourself. Namely, one that tells you that it's unphysical to ask anything except measurements. You have arbitrarily chosen to make measurement the basis of what is physical (and seems you are not even aware of it). The point is, I can take another interpretation, one that tells me where the electrons went and it also perfectly reproduces all the measurements, so there is no problem with it. It's as valid (and physical) an interpretation as yours is ;-) – Marek Dec 4 '10 at 16:12 show 28 more comments Here's a false belief. Everything your physics teacher says is true. Because physics teachers would never mislead you over how many states of matter there are or how lift is generated on a wing. - How about this - that parity is one of the most fundamental symmetries (which is not). I know the experiment(s) showing parity violation. However, I guess what I still don't understand is the following, rotational symmetry is fundamental. And parity is just rotating 180 degrees. So if we say that all rotational symmetry is fundametal, why would a subset (rotating with a very specific angle - 180 degrees) not being fundamental? Isn't that self contradicting? - 1 That's only the case in two dimensions. In 3D space, a parity transformation is not equivalent to any rotation. – David Zaslavsky♦ Jan 13 '11 at 22:59 show 2 more comments I highly recommend reading the article "Quantum mechanics: Myths and facts" by H. Nikolic http://arxiv.org/abs/quant-ph/0609163 Some topics include wave-particle duality, time-energy uncertainty relation and fundamental randomness. I've discussed this article in other communities and it seems reliable. - Misconception - The uncertainty principle is a statement about "our" ability to make measurements. Correction - The uncertainty principle is a result of the nature of the particles themselves and refers to the ability of anything to "make the relevant measurements". It's not just us that can't determine the simultaneous values of incompatable observables, God can't either. - 1 yes, you are right. – Ron Maimon Sep 23 '11 at 22:31 show 2 more comments Earth revolves around the Sun. It is wrong to say that the Sun revolves around the Earth. FACT: Motion is relative. There is nothing wrong in saying the Sun revolves around the Earth. The former is more frequently mentioned because the Sun (or more accurately, barycenter) is a better inertial frame, and other planets revolve in a near circle around the Sun as well, but in a bizarre way around the Earth. By traveling faster than light in the vacuum one can go back to the past. FACT: No explanation needed. - show 1 more comment The misconception: Naive people consider an object will float if it is placed in a vacuum container regardless of the existence of gravity. The correct fact: The absence of gravity makes the object float. - It's a common false belief among cynical physicists that there is no physical meaning in asking "why is there something rather than nothing". The question of whether there is an inevitable, self-consistent, self-referencial mathematical law that mandates the universe to exist is a real and legitimate one, a unified theory would be a step in the right direction. Research in number theory, prime numbers, infinity, etc, also plays into this. - I know it's an old question, but it's too much fun to pass this one up. There are quite a few examples of wrong ideas that have taken hold, but for the centrality to physics and the degree to which it is wrong and misleading, it is hard to find a better example than the Bohr atom. When it came out, it electrified the physics world on account of its revolutionary viewpoint and stunning success. But eleven years later it became completely obsolete. Nevertheless, in that short time it gripped the imagination of both the physics world and the general public to the point where it remains today the iconic picture of the atom in the minds of just about everyone. The misleading influences of this model are extremely far reaching. One can begin with the idea of the "quantum leap", which begat wave function collapse, which begat multiple universes et cetera. But perhaps the most persistent holdover of the Bohr atom is the notion that the ordinary laws of electromagnetics must be suspended at the atomic level, otherwise the atom would be unstable. Of course, nothing of the sort is true in modern quantum mechanics, least of all for the hydrogen atom. The greatest triumph of the Schroedinger equation was to show that the motion of electric charge could be tracked through time, and that the stable atomic configurations were precisely those with no accelerating charge distributions. This was an immediate and obvious consequence of Schroedinger's solution for the hydrogen atom. I would claim that it is not merely true that the stable configurations in QM are those with no accelerating charges. I would go further and suggest that in those cases where the charges do accelerate, the rate at which the kinetic energy of motion is converted into electromagnetic energy is exactly in agreement with that which would be calculated in the ordinary way by applying Maxwell's equations. So, for example, in the case of a black body radiator, if one simply takes the vibrating atoms, accounting for the accelerating motion of the charges, and applies classical antenna theory, you will get the correct black-body spectrum. - show 5 more comments This is tongue-in-cheek! That an emptying tank with a nozzle pointing downwards would actually experience a force! (see http://arxiv.org/pdf/physics/0312087v3) :-) - Something cannot come from nothing Yes, it can: In quantum field theories, the vacuum is not empty. - 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": 1, "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": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9442663192749023, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/26723/does-smooth-target-space-and-smooth-fibers-imply-smooth-total-space/26745	## Does smooth target space and smooth fibers imply smooth total space? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Suppose that $f: X \rightarrow Y$ is a morphism between algebraic varieties. If $Y$ is smooth, and the fibers of $f$ over closed points of $Y$ are proper and nonsingular, does it follow that $X$ is smooth? Update: The answer to the question as posed, is NO. See a comment by Karl Schwede below for a counterexample. Modified question: Let $f$ be a surjective morphism of algebraic varieties (reduced, irreducible, separated schemes, finite type over an algebraically closed field). Let $x \in X$ be a closed point and let $y = f(x)$. Just because the fiber $f^{-1}(y)$ is smooth does not mean $X$ is smooth at $x$. What if $X \times_Y Spec \mathcal{O}_y/m^n$ is smooth over $Spec \mathcal{O}_y/m^n$ for every positive integer n - is $X$ smooth at $x$? Here $m$ is the maximal ideal of the local ring at $y$. Is there any condition on $f$ or the fibers which will guarantee smoothness of the total space? Flatness plus smooth fibers is one, is there anything weaker? - In the modified question, the scheme-theoretic fibre of f over $\mathcal{O}_{y}/\mathfrak{m}^n$ will be nonreduced for n>1, and hence not a regular scheme --- am I misinterpreting the question? – Mike Roth Jun 1 2010 at 19:53 No, it wasn't carefully thought out. Thanks - now it is more open ended. – unknown Jun 1 2010 at 20:31 ## 4 Answers No. The blow up of a point on the plane provides a counterexample. You need to add flatness. Added: It seems I answered something different from what was asked. Perhaps someone can answer the actual question, which isn't so clear to me. 10 seconds later: It looks like Karl Schwede has a counterexample below. - The blowup of a smooth variety along a smooth subvariety is still smooth though? – unknown Jun 1 2010 at 14:41 I'm asking about the domain X, not the morphism f. – unknown Jun 1 2010 at 15:00 OK, sorry. I misread your question. – Donu Arapura Jun 1 2010 at 15:06 3 Modifying Donu's example very slightly, blow up the ideal $(x^2, y^2)$. You have two charts, one is $k[x, y, (y/x)^2] = k[a,b,c]/(a^2*c - b)$ and the other is the other obvious (and symmetric) one. This chart isn't smooth, the fiber over the closed point $(x, y)$ is $k[c]$. – Karl Schwede Jun 1 2010 at 15:11 Ah, I think you are right (though you mean b^2 in your formula). Thanks, is there any other condition on the situation (besides flatness) which would guarantee that the domain is smooth or is it just hopeless? – unknown Jun 1 2010 at 15:33 show 3 more 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. An even more basic example: take $X$ to be any singular affine variety, and $f$ to be the inclusion of $X$ into the affine space $\mathbb{A}^N$. - That is a good point, I should probably include surjetive. – unknown Jun 1 2010 at 19:14 I think the answer is no. Consider the case where $X$ is the two coordinate axes in $\mathbb{A}^2$ (corresponding to the ring $\mathbb{C}[x,y]/(xy)$) and $f$ is the projection onto the first axis (corresponding to $\mathbb{C}[x] \to \mathbb{C}[x,y]/(xy)$). Then the fibers of this map are a point, except over zero where the fiber is an $\mathbb{A}^1.$ I realize that this map is not proper, but I'm sure you could modify this example so that the map is proper. - Nice example, but for me, an algebraic variety includes irreducible... – unknown Jun 1 2010 at 15:37 Here is an example where all the spaces involved are irreducible. Let Y = variety of nilpotent 2 by 2 matrices. X = variety of pairs (N, F) where N is in Y and F is a line preserved by N. Let f : X -> Y be the natural projection. Now X is certainly smooth (as the projection to P^1 is a smooth morphism) and the fibres of f are points or P^1's. But Y is not regular. - That's backwards. He wants an example where Y is smooth but X is not. – David Speyer Jun 1 2010 at 18:08	{"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": 33, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9388189911842346, "perplexity_flag": "head"}
http://mathoverflow.net/questions/16533/slicing-the-fibres-of-a-meromorphic-function-with-the-zero-set-of-a-section-of-an	## Slicing the fibres of a meromorphic function with the zero set of a section of an ample line bundle ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm going through a proof of a vanishing theorem by Sommese ($H^{p,q}(X,L) = 0$ for $p+q > n+k$ if $L$ is $k$-ample) and have hit the following brick wall: I've got a complex projective manifold $X$, a meromorphic map $f : X \to \mathbb P^n$ and an ample line bundle $L$ over $X$. I want to show that for any big multiple $mL$ there is a global section $\sigma$ of $mL$ which is not identically zero on any $f^{-1}(y)$ for $y \in f(X \setminus (\text{singularities of f}))$. Basically we're proving this vanishing theorem by induction on $k$ and such a section gives the induction step. Now, for any fixed $y$ we let $Z$ be the closure of an irreducible component of $f^{-1}(y)$ and find an $N_0$ such that $$0 \to H^0(X, \mathcal I_Z \otimes mL) \to H^0(X, mL) \to H^0(Z, mL\|_Z) \to 0$$ is exact for all $m \geq N_0$, which gives us what we want. The problem is that this $N_0$ depends on $y$. Sommese says in his book that we get around this by invoking the Hilbert scheme on $\mathbb P^n$ and a result by Grothendieck. He also gives a reference to an article by Hironaka where this result is proved. Both methods are very general and make heavy use of schemes, which I don't know well enough to be able to follow. I've mostly convinced myself that I can reduce this to where $f$ is holomorphic, and I think some sort of a deformation argument might be possible if $f$ can be made behave nicely enough, but I don't see how to proceed. I'd really like some comments or suggestions on if it's possible to do this with analytic methods; a bit violently if you will. My thesis advisor suggested this theorem as a topic to present to other students with a rather diffeo-geometric background, and diving into schemes just to prove this one lemma seems like too big a detour for that audience. -	{"extraction_info": {"found_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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9568663835525513, "perplexity_flag": "head"}
http://mathoverflow.net/questions/19477/chow-ring-of-moduli-space-of-abelian-varieties	## Chow Ring of Moduli Space of Abelian Varieties ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Is there a good reference for the structure of the Chow ring of $\mathcal{A}_g$, the moduli space of complex principally polarized abelian varieties? More generally, references for the intersection theory, enumerative geometry, and vector bundles on $\mathcal{A}_g$ would be nice. - i would think faltings and chai would be a good place to start... i guess you've already looked at that and not found what you want? – Max Flander Mar 27 2010 at 6:09 Isn't the point of F-C getting everything (and in particular the theory of compactifications of A_g) to work over Z? I think the questioner is more interested in questions over C; F-C might be a bit daunting with their 14-tuples of degeneration data... – Kevin Buzzard Mar 27 2010 at 7:27 1 Yes, I'm interested in things over $\mathbb{C}$, so Faltings-Chai is a bit terrifying to me. – Charles Siegel Mar 27 2010 at 13:33 ## 1 Answer Van der Geer has written a paper computing what he calls the tautological subring of the chow ring of $\mathcal{A}_g$. He also computes the tautological ring for a smooth toroidal compactification. G. van der Geer, Cycles on the Moduli Space of Abelian Varieties, in "Moduli of Curves and Abelian Varieties (The Dutch Intercity Seminar on Moduli)", p. 65-89 (Carel Faber and Eduard Looijenga, editors), Aspects of Mathematics, Vieweg, Wiesbaden 1999. It is available on the van der Geer's website here Regarding intersection theory, Erdenberger, Grushevsky, and Hulek have been working on this for the toroidal compactifications, mostly for small values of $g$. For example, see the following references. C. Erdenberger, S. Grushevsky, K. Hulek, Intersection theory of toroidal compactifications of `$\mathcal{A}_4$`. Bull. London Math. Soc. 38 (2006), no. 3, 396--400. C. Erdenberger, S. Grushevsky, K. Hulek, Some intersection numbers of divisors on toroidal compactifications of `$\mathcal{A}_g$`. J. Algebraic Geom. 19 (2010), no. 1, 99--132. S. Grushevsky, Geometry of `$\mathcal{A}_g$` and its compactifications. Algebraic geometry---Seattle 2005. Part 1, 193--234, Proc. Sympos. Pure Math., 80, Part 1, Amer. Math. Soc., Providence, RI, 2009. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8416401743888855, "perplexity_flag": "middle"}
http://mathhelpforum.com/pre-calculus/100947-domain-range.html	# Thread: 1. ## Domain & Range 1. y=2e^-x -3 2. y=2sin(3x+pi) - 1 3. y= {sqrt(-x) -4 < x < 0 {sqrt(x) 0 < x < 4 domain is asking for allowable x-values and range asks for resulting y-values, but i'm not sure how to do it for these. 2. Originally Posted by fezz349 1. y=2e^-x -3 2. y=2sin(3x+pi) - 1 3. y= {sqrt(-x) -4 < x < 0 {sqrt(x) 0 < x < 4 domain is asking for allowable x-values and range asks for resulting y-values, but i'm not sure how to do it for these. if you are unfamiliar with the behavior of each parent function and its subsequent transformation, then I recommend that you graph each equation using technology to better "see" the domain and range. 3. I know I can easily graph them in my calculator and come up with the answers but I'm interested in the algebraic breakdown. 4. Originally Posted by fezz349 I know I can easily graph them in my calculator and come up with the answers but I'm interested in the algebraic breakdown. ok, here is the algebra thought process involved for your first function $y = 2e^{-x} - 3$ ... what is the domain and range of the parent function $y = e^{x}$ ? what transformation occurs to $y = e^x$ if the sign of x is changed? how does that transformation affect the domain and range? how does the transformation of multiplying $e^{-x}$ by $2$ affect the domain and range? finally, how does the final transformation of subtracting $3$ from $2e^{-x}$ affect the domain and range?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": 0.9036584496498108, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/13/learning-about-lie-groups/	## Learning about Lie groups ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Can someone suggest a good book for teaching myself about Lie groups? I study algebraic geometry and commutative algebra, and I like lots of examples. Thanks. - 1 I haven't seen Taylor's "Several Complex Variables" mentioned, but coming from an algebraic geometry background myself, I found that book very nice. It does use a lot of analysis though (a lot for me, anyway). – Steve D Jul 21 2010 at 16:33 ## 16 Answers There's also Fulton & Harris "Representation Theory" (a Springer GTM), which largely focusses on the representation theory of Lie algebras. Everything is developed via examples, so it works carefully through $sl_2$, $sl_3$ and $sl_4$ before tackling $sl_n$. By the time you get to the end, you've covered a lot, but might want to look elsewhere to see the "uniform statements". An excellent book. - @Scott I agree and it's one of the few introductory books that develops why Lie groups are important as algebraic objects in thier own right rather then just tools in differential geometry. – Andrew L Jun 24 2010 at 19:28 ### 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 realize this answer is rather late, but I just wanted to mention a fairly recent book on Lie theory that offers a gentle introduction to the basics: John Stillwell's Naive Lie Theory. It does not cover representation theory, but might be a pleasant step up to a book that does. The level is advanced undergraduate. - For someone with algebraic geometry background, I would heartily recommend Procesi's Lie groups: An approach through invariants and representations. It is masterfully written, with a lot of explicit results, and covers a lot more ground than Fulton and Harris. If you like "theory through exercises" approach then Vinberg and Onishchik, Lie groups and algebraic groups is very good (the Russian title included the word "seminar" that disappeared in translation). However, if you want to learn about the "real" side of Lie groups, both in linear and abstract manifold setting, my favorite is Godement's "Introduction à la théorie des groupes de Lie". Several of the books mentioned in other answers are devoted mostly or entirely to Lie algebras and their representations, rather than Lie groups. Here are more comments on the Lie group books that I am familiar with. If you aren't put off by a bit archaic notation and language, vol 1 of Chevalley's Lie groups is still good. I've taught a course using the 1st edition of Rossmann's book, and while I like his explicit approach, it was a real nightmare to use due to an unconscionable number of errors. In stark contrast with Complex semisimple Lie algebras by Serre, his Lie groups, just like Bourbaki's, is ultra dry. Knapp's Lie groups: beyond the introduction contains a wealth of material about semisimple groups, but it's definitely not a first course ("The main prerequisite is some degree of familiarity with elementary Lie theory", xvii), and unlike Procesi or Chevalley, the writing style is not crisp. An earlier and more focused book with similar goals is Goto and Grosshans, Semisimple Lie algebras (don't be fooled by the title, there are groups in there!). - Brian Hall's "Lie Groups, Lie Algebras and Representations: An Elementary Introduction" specializes to matrix Lie groups, so it makes for an accessible introduction. Like Fulton & Harris, it's got plenty of worked examples. It also has some stuff about Verma modules that's not in Fulton & Harris. I think it'd be a great book for a first course. Knapp's "Lie Groups: Beyond an Introduction" might be good for a second course (it has more of the "uniform statements" Scott mentioned) and is handy to have around as a reference. It has an appendix with historical notes and a ton of suggestions for further reading. It also has a lot more on Lie groups themselves than most books do. - Knapp's book is a lot harder then either F&H or Hall. Hall's really more for physics majors,but it's a nice book nevertheless. – Andrew L Jun 24 2010 at 19:29 I like Humphreys' book, Introduction to Lie Algebras and Representation Theory, which is short and sweet, but doesn't really talk about Lie groups (just Lie algebras). I also sometimes find myself looking through Knapp's Lie Groups: Beyond an Introduction. If the material was covered in the Spring 2006 Lie groups course at Berkeley, then I prefer the presentation in this guy's notes. - Before the 2006 course, there was Allen Knutson's 2001 course, from which there are several sets of notes, e.g. math.berkeley.edu/~allenk/courses/spr02/261b/… – Scott Morrison♦ Sep 28 2009 at 23:36 1 Also, Theo Johnson-Freyd has some notes from Mark Haiman's Fall 2008 course here: math.berkeley.edu/~theojf/LieGroupsBook.pdf – Anton Geraschenko♦ Sep 29 2009 at 0:00 1 Finding a reasonably elementary book on Lie groups with lots of examples is challenging. What makes the subject attractive is that it's the crossroads for many subjects. My book definitely wasn't about Lie groups (and has too few examples) but does get somewhat into "modern" representation theory. Knapp is reliable but somewhat advanced. Fulton-Harris is also not a Lie group book and doesn't introduce infinite dimensional representations, but covers a lot of concrete classical examples plus symmetric groups. Free online notes can be a safe starting point, but shop around. – Jim Humphreys Jun 24 2010 at 14:04 @Anton You and Theo should be both be very proud of your TeX-ed notes,particularly on Lie theory. – Andrew L Jun 24 2010 at 19:37 Dan, knowing your tastes, I think you will like Fulton-Harris very much. However, if I recall correctly, Fulton-Harris doesn't go into much depth about some important (and really cool) theorems in Lie groups, such as Peter-Weyl and Borel-Weil-Bott. But of course, you can learn these theorems elsewhere. I think the book "Compact Lie Groups" by Sepanski is nice, and it does cover P-W and B-W-B. I also found this note on B-W-B to be useful in the past: http://www-math.mit.edu/~lurie/papers/bwb.pdf - Nice link on BWB. I had not seen that before. – B. Bischof Dec 18 2009 at 15:34 Just to add one more to the already mentioned. I find the book of Bump on Lie groups very good, as well as the other ones. - The book "Introduction to Lie groups and Lie algebras" by A. Kirillov, Jr., is quite nice, and seems to be free online. It might be a good starting point, and it has an excellent annotated bibliography. (Edit: On further inspection, the .pdf I linked to just seems to be a draft. The actual book has the good bibliography.) - I haven't carefully read that one,but a lot of my friends who work in Lie theory swear by it. – Andrew L Jun 24 2010 at 19:30 Although perhaps not from the point of view of someone interested in algebraic geometry and commutative algebra, others of different persuasions might enjoy the following books: • Lectures on Lie groups, by J. Frank Adams • Representations of compact Lie groups, by Theodor Bröcker and Tammo tom Dieck • Lie groups: an introduction through linear groups, by Wulf Rossmann Adam's book is a classic and has a very nice proof of the conjugacy theorem of maximal tori using algebraic topology (via a fixed point theorem). Bröcker-tom Dieck is a good companion to Adams, as it often reads like an expanded version of it. At any rate, it goes into more detail. Rossmann's book is reviewed by Knapp in http://www.math.sunysb.edu/~aknapp/pdf-files/BakerRossmann.pdf - Adams's book is my favorite - it is a real gem. – Jeffrey Giansiracusa Jun 24 2010 at 10:14 I really like Goodman & Wallach. This is a new revised version of their old book which was called, "Representations and Invariants of the Classical Groups". It is really clearly written and covers a lot of material. It might suit your interests, since it's a bit bent towards the algebraic groups part of Lie theory, but it does also cover the analytic side. - Roger Godement-Introduction a la theorie des groupes de Lie-Springer(only in french as far as I know).An introduction to Lie groups via linear groups(with John von Neumann in backstage..) and a touch of Hilbert 5th problem...Very fun, as always with Godement - My favourite reference is Serre, Lie algebras and Lie groups. It's a tour of Bourbaki's Lie groups and Lie algebras that is concise and, being Serre, of course, very clear. - As an elementary introduction with lots of examples you may take a look at A.Baker,"Matrix Groups. An Introduction to Lie Group Theory" which appeared in Springer's Undergraduate Texts in Mathematics. After this a very good book with lot of results and almost self-contained, but rather demanding is M.M.Postnikov "Lie Groups and Lie Algebras" (it was published by "Mir" in English). - There are many courses, including something about Lie groups at J.Milne's page: jmilne.org - Nobody mentioned "Gilmore: Lie Groups, Physics, and Geometry" yet. A very down to earth introduction with many examples and clear explanations. Especially targeted at physicists, engineers and chemists. If you follow the above link you can read some sample chapters. The cover summarizes the set up of the book quite neatly: "Describing many of the most important aspects of Lie group theory, this book presents the subject in a ‘hands on’ way. Rather than concentrating on theorems and proofs, the book shows the relation of Lie groups with many branches of mathematics and physics, and illustrates these with concrete computations. Many examples of Lie groups and Lie algebras are given throughout the text, with applications of the material to physical sciences and applied mathematics. The relation between Lie group theory and algorithms for solving ordinary differential equations is presented and shown to be analogous to the relation between Galois groups and algorithms for solving polynomial equations. Other chapters are devoted to differential geometry, relativity, electrodynamics, and the hydrogen atom. Problems are given at the end of each chapter so readers can monitor their understanding of the materials. This is a fascinating introduction to Lie groups for graduate and undergraduate students in physics, mathematics and electrical engineering, as well as researchers in these fields. Robert Gilmore is a Professor in the Department of Physics at Drexel University, Philadelphia. He is a Fellow of the American Physical Society, and a Member of the Standing Committee for the International Colloquium on Group Theoretical Methods in Physics. His research areas include group theory, catastrophe theory, atomic and nuclear physics, singularity theory, and chaos." - In my opinion, the best quick introduction to Lie group and algebra theory is in chapter 12 of E. B. Vinberg's A Course In Algebra. It is short, geometric and deep with all the essential facts and theorems presented. There's a similar presentation in Artin's Algebra, but that one is done entirely in terms of matrix groups. The Vinberg chapter is on general Lie theory. By the way, it's mostly drawn from the Vinberg/Onischick book mentioned by Victor above -- but it's a little gentler and more detailed, being pitched at beginners. The Vinberg book is one of those texts you read over and over because every time you look at it, you realize a little more just how damn good it 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": 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": 0.9484040141105652, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/91217/a-point-in-the-weak-closure-but-not-in-the-weak-sequential-closure/91321	## A point in the weak closure but not in the weak sequential closure ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm trying to find a proof of this counterexample by von Neumann: Let $x_{mn}\in \ell^2$ be defined by $$x_{mn}(m)=n \quad,\quad x_{mn}(n)=m \quad,\quad x_{mn}(k)=0 \hbox{ otherwise,}$$ and let `$S=\{ x_{mn} : m, n\geq 1\}$`. Von Neumann shows that $0$ is in the weak closure of this set but no sequence in $S$ convergess weakly to $0$. - 1 I have taken the liberty of rewriting and retitling the question, hopefully preserving the original sense. That said, I am not sure if the question really belongs on MO – Yemon Choi Mar 14 2012 at 21:30 ## 2 Answers As Aaron pointed out, "von Neumann's" example is really a non example. To salvage the problem, restate it as: construct a sequence in $\ell_2$ which has $0$ in its weak closure, but no subsequence converges weakly to $0$. First note that such a sequence must be unbounded (by Eberlein-Smulian). Secondly, observe that it is enough to have for each $\epsilon > 0$ a (necessarily bounded) subsequence that converges weakly to a point whose norm is at most $\epsilon$ (and, of course, no subsequence that converges weakly to $0$). With these "hints", it is easy to construct an example: Let $x_{nm}(k)$ be $1/n$ if $k=1$, $n$ if $k=m>1$, and $0$ otherwise. With the "obvious" definition, $0$ is in the $2$-weak sequential closure of $x_{nm}$ but not in the $1$-weak sequential closure. From this beginning it is natural to define for each countable ordinal $\alpha$ the $\alpha$-weak sequential closure and to state an obvious problem. Another (not very difficult once you understand the example above) problem is to build a sequence in $\ell_2$ whose norms tend to infinity and yet $0$ is in the weak closure of the sequence. Another striking example of the phenomena sought by the OP is the following. Take a dense sequence in the unit sphere of $\ell_1$. Then $0$ is in the weak closure of the sequence but no subsequence converges weakly to $0$ because $\ell_1$ has the Shur property. - Did you mean "unit sphere of $\ell_1$"? – Philip Brooker Mar 15 2012 at 20:03 Thanks, Philip. I corrected the typo. – Bill Johnson Mar 15 2012 at 20:18 Another small typo: I think in the definition of $x_{nm}(k)$, you want $n=m > 1$ to read $k=m > 1$. – Aaron Tikuisis Mar 15 2012 at 21:21 Thanks, Aaron; I corrected it. – Bill Johnson Mar 15 2012 at 21:26 ### 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 believe that $0$ is a weak cluster point of this set. For example, consider $y \in \ell^2$ defined by $$y(k) = 1/k.$$ Then we have, for any $m,n$ that $$\langle x_{m,n}, y \rangle = m/n + n/m \geq 2.$$ Therefore, the weak neighbourhood `$$ \{x \in \ell^2: \|\langle x, y\rangle\| < 1\} $$` of $0$ does not intersect $S$. - It is possible that I made an error in transcription when reformatting the original question, but unless the OP turns up thete's no way of knowing – Yemon Choi Mar 15 2012 at 18:25	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 37, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9405761361122131, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/32194/what-are-ell-p-norms-and-how-are-they-relevant-to-regularization	# What are $\ell_p$ norms and how are they relevant to regularization? I have been seeing a lot of papers on sparse representations lately, and most of them use the $\ell_p$ norm and do some minimization. My question is, what is the $\ell_p$ norm, and the $\ell_{p, q}$ mixed norm? And how are they relevant to regularization? Thanks - ## 1 Answer $\ell_p$ norms are functions that take vectors and return nonnegative numbers. They're defined as $$\\|\vec x\\|_p = \left(\sum_{i=1}^d \|x_i\|^p\right)^{1/p}$$ In the case where $p=2$, this is called the Euclidean norm. You can define the Euclidean distance as $\\|\vec x - \vec y\\|_2$. When $p = \infty$, this just means $\\|\vec x\\|_\infty = \sup_i x_i$ (or $\max_i x_i$). Strictly speaking, $p$ must be at least one for $\\|\vec x\\|_p$ to be a norm. If $0 < p < 1$, then $\\|\vec x\\|_p$ isn't really a norm, because norms must satisfy the triangle inequality. (There are also $L_p$ norms, which are defined analagously, except for functions instead of vectors or sequences -- really this is the same thing, since vectors are functions with finite domains.) I'm not aware of any use for a norm in a machine learning application where $p > 2$, except where $p = \infty$. Usually you see $p = 2$ or $p = 1$, or sometimes $1 < p < 2$ where you want to relax the $p = 1$ case; $\\|\vec x\\|_1$ isn't strictly convex in $\vec x$, but $\\|\vec x\\|_p$ is, for $1 < p < \infty$. This can make finding the solution "easier" in certain cases. In the context of regularization, if you add $\\|\vec x\\|_1$ to your objective function, what you're saying is that you expect $\vec x$ to be sparse, that is, mostly made up of zeros. It's a bit technical, but basically, if there is a dense solution, there's likely a sparser solution with the same norm. If you expect your solution to be dense, you can add $\\|\vec x\\|_2^2$ to your objective, because then it's much easier to work with its derivative. Both serve the purpose of keeping the solution from having too much weight. The mixed norm comes in when you're trying to integrate several sources. Basically you want the solution vector to be made up of several pieces $\vec{x}^j$, where $j$ is the index of some source. The $\ell_{p,q}$ norm is just the $q$-norm of all the $p$-norms collected in a vector. I.e., $$\\|\vec x\\|_{p,q} = \left( \sum_{j = 1}^m \left( \sum_{i=1}^d \|x_i^j\|^p\right)^{q/p}\right)^{1/q}$$ The purpose of this is not to "oversparsify" a set of solutions, say by using $\\|\vec x\\|_{1,2}$. The individual pieces are sparse, but you don't risk nuking a whole solution vector by taking the $1$-norm of all of the solutions. So you use the $2$-norm on the outside instead. Hope that helps. See this paper for more details. - 1 +1 for the explanation of mixed norms. I never understood them myself. – Suresh Venkatasubramanian Jul 13 '12 at 4:55 (+1) Nice answer. Welcome to CrossValidated, John! – MånsT Jul 13 '12 at 8:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 36, "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": 0.9563779234886169, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/740/useful-examples-of-pathological-functions/116623	Useful examples of pathological functions What are some particularly well-known functions that exhibit pathological behavior at or near at least one value and are particularly useful as examples? For instance, if $f'(a) = b$, then $f(a)$ exists, $f$ is continuous at $a$, $f$ is differentiable at $a$, but $f'$ need not be continuous at $a$. A function for which this is true is $f(x) = x^2 \sin(1/x)$ at $x=0$. - 15 – Larry Wang Jul 26 '10 at 21:46 Nice! I can't believe I didn't know such a book existed. – Isaac Jul 26 '10 at 21:58 15 Answers Have also a look here: http://mathoverflow.net/questions/22189/what-is-your-favorite-strange-function - – Mechanical snail Nov 27 '12 at 19:09 The Weierstrass function is continuous everywhere and differentiable nowhere. The Dirichlet function (the indicator function for the rationals) is continuous nowhere. A modification of the Dirichlet function is continuous at all irrational values and discontinuous at rational values. The Devil's Staircase is uniformly continuous but not absolutely. It increases from 0 to 1, but the derivative is 0 almost everywhere. - I think you mean "discontinuous at all rational functions" – BlueRaja - Danny Pflughoeft Jul 27 '10 at 4:51 Should the modification of the Dirichlet function refer to rational/irrational numbers (not functions)? – Isaac Jul 27 '10 at 6:54 @BlueRaja: @Isaac: You're both right. I copy/pasted that description from somewhere and didn't double-check it. – Larry Wang Jul 27 '10 at 10:45 The devil's staircase has zero derivative almost everywhere, not at all but countably many points. Any continuous function with zero derivative at all but countably many points is constant. – George Lowther Aug 25 '10 at 20:50 Here's an example of a strictly increasing function on ℝ which is continuous exactly at the irrationals. Pick your favorite absolutely convergent series ∑an in which all the terms are positive (mine is ∑1/2n) and your favorite enumeration of the rationals: ℚ={q1,q2,...}. For a real number x, define f(x) to be the sum of all the an for which qn ≤ x. - Also Conway base 13 function. This function has the following properties: 1. On every closed interval $[a, b]$ take every real value. 2. Is continuous nowhere. - The Dirac delta "function." It's not a "function," strictly speaking, but rather a very simple example of a distribution that isn't a function. - The bump function $\psi\colon \mathbb{R} \to \mathbb{R}$ given by $\psi(x) = e^{-1/(1-x^2)}$ for $\|x\|\leq 1$ and $\psi(x) = 0$ for $\|x\| > 1$ is not exactly pathological per se, but it's very useful for (at least) two things: 1. It's an example of a function that is smooth (all derivatives exist) but not analytic (not equal to its Taylor series in a neighbourhood of every point) -- look at the points $x=\pm 1$. 2. You can use it to build partitions of unity on manifolds, which is important when you want to build a global object (like a Riemannian metric) out of local data (defined in each chart). - +1, and to add a more practical use for this function: this, as well as its integral, is sometimes used in conjunction with the trapezoidal rule as a variable substitution for handling badly-behaved improper integrals. Its "singularity-flattening" property at the points 1 and -1 works well in tandem with the Euler-Maclaurin error of the trapezoidal rule. – J. M. Aug 8 '10 at 10:41 – kahen Oct 25 '10 at 18:09 I like $f(x,y) = \frac{xy}{x^{2}+y{2}}$ for $(x,y) \ne (0,0)$ and $f(x,y)=0$ for $(x,y)=(0,0)$. Here $f$ has partial derivatives at (0,0) but it is not differentiable at $(0,0)$ - 3 A less trivial exercise is to find a function with partial derivatives in every direction at a point but which is still not differentiable there. – Qiaochu Yuan Aug 7 '10 at 18:01 The constant function $0$ is an extremely pathological function: it has all kinds of properties that almost none of the functions $\mathbb R\to\mathbb R$ have: it is everywhere continuous, differentiable, analytic, polynomial, constant (not all of those are independent of course...), you name it. By contrast many of the answers given here involve properties that almost all functions have, or that almost all functions with the mentioned pathologies (e.g., being everywhere continuous for the Weierstrass function) have. In fact being a definable function is a pathology that all answers share, but this is admittedly hard to avoid in an answer. - +1 for the last comment. – Mechanical snail Nov 27 '12 at 19:09 $\displaystyle\frac{\sin(x)}{x}$ is useful; it has a singularity at $x=0$, but if you take the union of $\displaystyle y=\frac{\sin(x)}{x}$ with the point $(x=0,y=1)$ then you get $\text{sinc}(x/\pi)$. The $\text{sinc}$ function has a lot of applications in signal processing and diffraction. - The Cauchy distribution is another example. This distribution pops up frequently in statistical reasoning. For example, the ratio of two independent standard normal random variables is a standard Cauchy variable. We calculate ratios all the time, but often forget to consider that their distribution does not follow a simple normal distribution. This is interesting for several reasons: 1. The mean and variance for the Cauchy distribution are undefined (--> infinity). If one is trying to estimate these parameters for the ratio of two normal variables, the results may blow up to rather large values that are hard to interpret. 2. In such a situation, one would therefore prefer to use more robust estimators of the central tendency (such as the median) and scale (median absolute deviation). When designing and testing robust estimators (en.wikipedia.org/wiki/Robust_statistics), one way to test them is to try them out on a Cauchy distribution, and make sure that they don't blow up like the usual formulas for mean and variance. - Also used to justify the finite variance requirement of CLT, and to motivate a more general form. – Larry Wang Jul 27 '10 at 23:30 When I was first learning calculus, the fact that $sin(1/x)$ is continuous on the set $(0,\infty)$ gave me a headache. - 2 – Qiaochu Yuan Aug 7 '10 at 18:03 The function f(x) = x over the rationals and 2x over the irrationals is locally increasing in 0 but it is neither increasing nor decreasing. - Increasing at x=0, but neither increasing nor decreasing for any interval? – Isaac Jul 29 '10 at 11:15 The Cauchy functionals, which satisfy the very simple equation f(a+b) = f(a)+f(b) for all real a,b. These are either a line through the origin (the "nice" ones) or really "ugly" functions that are discontinuous and unbounded in every interval. The latter are possible because the Axiom of Choice implies (actually is equivalent to) that infinite dimensional vector spaces have bases; i.e. the reals over the rationals have a Hamel basis. A great explanation of all this (including the nice/ugly terminology) is in Horst Herrlich's monograph The Axiom of Choice. - (1/x)^(1/x). Try graphing it if you dare (including over the negatives) - How to graph over the negatives? It's not real there for most x's. – KennyTM Jul 27 '10 at 13:31 3 @KennyTM: Exactly – Casebash Jul 27 '10 at 20:53 1 I think, depending on how you define exponentiation of negative bases with rational exponents, it might be possible to interpret it to have real values on a dense subset of the negative reals. – Isaac Jul 27 '10 at 21:26 1 – J. M. Aug 9 '10 at 10:08 Three examples suffice to show why some modes of convergence don't imply other modes of convergence: pointwise convergence, Lp norm convergence, convergence in measure, etc. See the counterexamples section 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": 33, "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": 0.9191279411315918, "perplexity_flag": "middle"}
http://nrich.maths.org/5937/note	nrich enriching mathematicsSkip over navigation ### Game of PIG - Sixes Can you beat Piggy in this simple dice game? Can you figure out Piggy's strategy, and is there a better one? ### Distribution Maker This tool allows you to create custom-specified random numbers, such as the total on three dice. ### The Random World Think that a coin toss is 50-50 heads or tails? Read on to appreciate the ever-changing and random nature of the world in which we live. # Scale Invariance ### Why do this problem? This problem offers a fascinating exploration into probability density functions for real world data. Whilst the individual steps are quite simple, the problem draws together many strands from distribution theory. The results can be tested on any set of data from any geography book, giving an interesting relevance to the mathematics. ### Possible approach The first obstacle to overcome is that of notation: can the students understand what is being asked? The question involves little computation but requires clear thinking of the ideas. This might be facilitated in a group discussion, but might also require individual work. ### Key questions • If a function is to be a probability density function, what is the major property it must possess? • What ranges of values will start with a digit $1$? ### Possible extension Consider carefully why this problem involves 'scale invariance'. Consider the restriction of scale invariance on real world data. Which sets of real world data do you think will be modelled by this distribution? Why? ### Possible support Skip the first part and provide students with the scale invariant functions. Also, first use the range 1< x < 10 in the last part of the question. 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": 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": 0.3, "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": 0.8954493999481201, "perplexity_flag": "middle"}
http://crypto.stackexchange.com/questions/3505/what-is-the-length-of-an-rsa-signature/3508	# What is the length of an RSA signature? Is it the same as the bits of the key (So a 2048 bit system will yield a 2048 bit signature)? At most as the key? Or something else entirely? - As a sidenote: While the signature size will correspond to the key size, that doesn't mean that the size of the signed message is the size of the message plus the signature size. It's possible to embed part of the message into the signature itself, making the combined size a bit smaller. This is called message recovery. – CodesInChaos Aug 10 '12 at 19:13 ## 3 Answers If $d,N$ is the private key, signing a message $m$ is computed as $m^d\bmod N$. The $\bmod N$ makes it so that the signed message is between $0$ and $N$. So, it is no larger than $N$. In most applications, however, there is usually some encoding or protocol fields that will make it larger. - Thanks. And when we say 2048 bit encryption - Do we mean that N itself is 2048 bit? – ispiro Aug 10 '12 at 15:06 1 – ispiro Aug 10 '12 at 15:26 PKCS#1, "the" RSA standard, describes how a signature should be encoded, and it is a sequence of bytes with big-endian unsigned encoding, always of the size of the modulus. This means that for a 2048-bit modulus, all signatures have length exactly 256 bytes, never more, never less. PKCS#1 is the most widely used standard, but there are other standards in some areas which may decide otherwise. Mathematically, a RSA signature is an integer between $1$ and $N-1$, where $N$ is the modulus. In some protocols, there can be some wrapping around the signature, e.g. to indicate which algorithm was used, or to embed certificates. For instance, in CMS, a "signature" contains the RSA value itself, but also quite a lot of additional data, for a virtually unbounded total size (I have seen signatures of a size of several megabytes, due to inclusion of huge CRL). - The RSA signature size is dependent on the key size, the RSA signature size is equal to the length of the modulus in bytes. This means that for a "n bit key", the resulting signature will be exactly n bits long. Although the computed signature value is not necessarily n bits, the result will be padded to match exactly n bits. Now here is how this works: The RSA algorithm is based on modular exponentiation. For such a calculation the final result is the remainder of the "normal" result divided by the modulus. Modular arithmetic plays a large role in Number Theory. There the definition for congruence is (I'll use 'congruent' since I don't know how to get those three-line equal signs) $m \equiv n \mod k$ if $k$ divides $m - n$ Simple example: Let $n = 2$ and $k = 7$, then $2 \equiv 2 \mod 7$ ($7$ divides $2 - 2$) $9 \equiv 2 \mod 7$ ($7$ divides $9 - 2$) $16 \equiv 2 \mod 7$ ($7$ divides $16 - 2$) ... $7$ actually does divide $0$, the definition for division is An integer $a$ divides an integer $b$ if there is an integer $n$ with the property that $b = n·a$. For $a = 7$ and $b = 0$ choose $n = 0$. This implies that every integer divides $0$, but it also implies that congruence can be expanded to negative numbers (won't go into details here, it's not important for RSA). So the gist is that the congruence principle expands our naive understanding of remainders, the modulus is the "number after mod", in our example it would be $7$. As there are an infinite amount of numbers that are congruent given a modulus, we speak of this as the congruence classes and usually pick one representative (the smallest congruent integer $\geq 0$) for our calculations, just as we intuitively do when talking about the "remainder" of a calculation. In RSA, signing a message $m$ means exponentiation with the "private exponent" $d$, the result $r$ is the smallest integer with $0 \leq r < n$ so that $$m^d \equiv r \bmod n.$$ This implies two things: 1. The length of $r$ (in bits) is bounded by the length of $n$ (in bits). 2. The length of $m$ (in bits) must $\leq$ length($n$) (in bits, too). To make the signature exactly $n$ bits long, some form of padding is applied. Cf. PKCS#1 for valid options. The second fact implies that messages larger than n would either have to be signed by breaking $m$ in several chunks $< n$, but this is not done in practice since it would be way too slow (modular exponentiation is computationally expensive), so we need another way to "compress" our messages to be smaller than $n$. For this purpose we use cryptographically secure hash functions such as SHA-1 that you mentioned. Applying SHA-1 to an arbitrary-length message m will produce a "hash" that is 20 bytes long, smaller than the typical size of a RSA modulus, common sizes are 1024 bits or 2048 bits, i.e. 128 or 256 bytes, so the signature calculation can be applied for any arbitrary message. The cryptographic properties of such a hash function ensures (in theory - signature forgery is a huge topic in the research community) that it is not possible to forge a signature other than by brute force. - But since the signature is $m^d \bmod n$ - It seems that the length can even be $0$. Or are you referring to a system that will pad it as an extra feature? – ispiro Aug 10 '12 at 15:05 yes, I'm referring it as an extra feature – K Kiran Aug 10 '12 at 15:09 Actually, modular exponentiation works just fine for numbers greater than $n$. It's just that this would trivially give us signature collisions (as $m + n \equiv m \mod n$, we also have $(m + n)^d \equiv m^d \mod n$), which is one of the reasons to use a hash first. – Paŭlo Ebermann♦ Aug 11 '12 at 10:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 54, "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": 0.9267765879631042, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/67800-dimension.html	# Thread: 1. ## dimension Let V be a finite dimensional vector space. Show that if W1,....,Wn are subspaces of V such that none of these subspaces are qeual to V, then Union of all these subspaces does not equal V. i am thinking induction on dim V might help but do not really have idea how to go bout it. so please help. 2. Originally Posted by Kat-M Let V be a finite dimensional vector space. Show that if W1,....,Wn are subspaces of V such that none of these subspaces are qeual to V, then Union of all these subspaces does not equal V. The union of two subpaces is a subspace if and only one is contained in another. Thus if $W_1\cup W_2$ is a subspace thus $W_1\subseteq W_2$ and $W_2\subseteq W_1$. And so $\dim (W_1\cap W_2) = \max \{ \dim (W_1),\dim (W_2) \} < \dim (V)$. 3. Originally Posted by ThePerfectHacker The union of two subpaces is a subspace if and only one is contained in another. The reason that is so: Suppose u is in subspace $U_1$ but NOT subspace $U_2$ and v is in subspace $U_2$ but not in $U_1$. The u+ v cannot be in $U_1\cup U_2$. If it were, then it would have to be in either $U_1$ or $U_2$ (or both). If u+ v were in $U_1$, then, because $U_1$ is closed under addition and scalar multiplication u+ v+ (-u)= v would be in $U_1$, a cotradiction. If u+ v were in $U_2$, simlarly u+ v+ (-v)= u would be in $U_2$, again 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": 16, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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": 0.9452162981033325, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/69326-i-need-help-asvab.html	Thread: 1. i need help for asvab i am 34 and getting ready for the asvab and i do not know how to answer some of the questions. they all look like there should be a formula for them but i cant figure this out it may be basic but i need help...not the answers but how to do them here are some of the questions ............ If Steven can mix 20 drinks in 5 minutes, Sue can mix 20 drinks in 10 minutes, and Jack can mix 20 drinks in 15 minutes, how much time will it take all 3 of them working together to mix the 20 drinks?..................Jim can fill a pool carrying buckets of water in 30 minutes. Sue can do the same job in 45 minutes. Tony can do the same job in 1 ½ hours. How quickly can all three fill the pool together? ..................... If Sam can do a job in 4 days that Lisa can do in 6 days and Tom can do in 2 days, how long would the job take if Sam, Lisa, and Tom worked together to complete it?..............thx hope someone can help me 2. Whatever you do, do not freak out when you see these problems on the asvab. I have taken this test before and it seems difficult at first but all these problems are, are just word problems. Usually you do not need to know any special formulas. If Steven can mix 20 drinks in 5 minutes, Sue can mix 20 drinks in 10 minutes, and Jack can mix 20 drinks in 15 minutes, how much time will it take all 3 of them working together to mix the 20 drinks? In this problem, you need to think of proportions. I believe that you add up all of the minutes together and divide by the number of drinks. 3. Originally Posted by patpatrickc i am 34 and getting ready for the asvab and i do not know how to answer some of the questions. they all look like there should be a formula for them but i cant figure this out it may be basic but i need help...not the answers but how to do them here are some of the questions ............ If Steven can mix 20 drinks in 5 minutes, Sue can mix 20 drinks in 10 minutes, and Jack can mix 20 drinks in 15 minutes, how much time will it take all 3 of them working together to mix the 20 drinks?..................Jim can fill a pool carrying buckets of water in 30 minutes. Sue can do the same job in 45 minutes. Tony can do the same job in 1 ½ hours. How quickly can all three fill the pool together? ..................... If Sam can do a job in 4 days that Lisa can do in 6 days and Tom can do in 2 days, how long would the job take if Sam, Lisa, and Tom worked together to complete it?..............thx hope someone can help me drinks/minutes Jim 20/5= 4 Sue 20/10= 2 Tony 20/15= 4/3 4+ 2+ 4/3 = 20/x 22/3=20/x 22x= 60 x=2.72 minutes correct me if i am wrong If Sam can do the entire job in 4 days, Tom in 2 days and Lisa in 6 days then: job completed per day will be: Sam= 1/4 Tom= 1/2 Lisa= 1/6 together= x to find out how much they can do the job together per day, add what they can do per day that is: 1/4 + 1/2 + 1/6 = (3+6+2)/12 = 11/12 let x be how long they will take to do the job together. they can do 1/x per day, 11/12= 1/x 11x=12 x=1.09 day 4. ok these work but thanks for the help i understand all of that now but i am not so bright how do i convert the 2.72 into seconds i understand 2 minutes but how many secs is .72 5. Originally Posted by patpatrickc thanks for the help i understand all of that now but i am not so bright how do i convert the 2.72 into seconds i understand 2 minutes but how many secs is .72 $(.72) minutes * \frac{60seconds}{1minute}$ $= 43.2 seconds$ you can cancel minutes that leave seconds as the unit.	{"extraction_info": {"found_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": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9069865345954895, "perplexity_flag": "middle"}
http://mathhelpforum.com/number-theory/19276-complex-prime-numbers.html	# Thread: 1. ## Complex prime numbers? Hello! I just want to know if there is anything like complex prime numbers, and what work has been done in the subject. Ordinary prime numbers is just there if you want to factorise real integer numbers, not even zero or negative number is factorable, case 0 and -1 is missing prime numbers in that case. You start with the multiplicative identity, 1, so the number 1 doesn't need to have any prime numbers to be achieved. But if one want to get 0, multiplication by 0 is necessary, but zero is not a prime number. And if one want to get a negative number, multiplication by -1 is necessary, wich is not a prime number either. But if we want every complex number with integer coordinates to be factorable, we'll have to introduce other prime numbers. 0 is one of them, i is another. Cause, to reach the negative numbers, you could multiply by -1, but to reach the imaginary numbers, one will have to multiply by i, and i*i = -1. But i can't be achieved by multiplying other prime numbers with each other, possible if you introduced -i instead, but having two different prime numbers which can be factorized by each other isn't such a good idea, and i is more natural than -i. Now every number in the 2:nd, 3:d or 4:th quadrant, can be written by a number in the first quadrant times a power to i. Every negative number or imaginary number can be written as a positive number times a power to i. Clearly the first number in the first quadrant we hit is (1, 1), so that can be made a prime number. Now, (2, 1) and (1, 2) can be made prime numbers as well. In fact, every number (a, 1) can be made a prime number. But what about the numbers (1, a), are those prime numbers? What about the rest of the numbers in the first quadrant? Is there some method to get those? This is my conclusions: The ordinary prime numbers still exist at the possitive real axis. No imaginary prime number more than i and 0 exists, since the rest of the numbers at the imaginary axis can be written as the same real number times i if the number is possitive, and times i^3 if the number is negative. The same rule still exists that no prime number shall be able to be written as a product of other prime numbers. On the other hand, prime number i can always be written as i^5. The other way around, a big or a negative exponent of i can always be reduced to an exponent that lays in some specific interval, like [0, 4[, or maybe ]-2, 2]. i however along with zero is the only prime number which can be factorized with another set of prime factors than the set only containing the original prime number. 2. Originally Posted by TriKri The ordinary prime numbers still exist at the positive real axis. That is clearly true! Every real number is a complex number. So prime in the reals then prime in the complex. No problem there, correct? What do you understand a prime complex number to be? You may want to look-up Gaussian-integers. See if you can use that concept to forward your question. 3. There is actually a very abstract notion of what a prime number is. But I do not know if you have the background to understand it. Have you ever studied basic abstract algebra and basic field theory? Definition 1: Let $D$ be an integral domain we say $a\|b$ (reads $a$ divides $b$) iff there exists $c$ so that $b = ac$. Definition 2: A "unit" in $D$ is an element $u$ so that $u\|1$. Definition 3: A non-unit non-zero number is called "irreducible" iff it cannot be factored as a product of two non-unit numbers. Definition 4: A non-unit non-zero element $p$ is called "prime" iff $p\|(ab)\implies p\|a \mbox{ or }p\|b$. Theorem 1: If $p$ is prime then it is irreducible. We ask is the converse true? In fact it is not! Definition 5: An integral domain is "principal" (called "principal ideal domain") iff every ideal is principal. Theorem: For every PID any irreducible elements are prime elements. So as you can see there is a very abstract approach to the meaning of what "prime" is. As an example, you can work in $\mathbb{Z}[i] = \{ a+ bi \| a,b\in \mathbb{Z} \}$ called the "Gaussian integers". And look for primes. The two great mathematicians that should be credited for this abstract approach to factorization are Kummer and his student Kronecker.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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": 0.9405336380004883, "perplexity_flag": "head"}
http://mathhelpforum.com/number-theory/197883-convergent-series-p-adic-completion-q.html	# Thread: 1. ## Convergent series in the p-adic completion of Q Hi there, I've got a question on p-adic numbers which make pretty much no sense to me so I have no idea what to do or why, any help appreciated. Prove that if the sequence $\{t_k\}_{k \in N} \subset \mathbb{Q}$ has the property that $\|t_k\|_p \to 0$ as $p \to \infty$, then the series $\sum_{k=1}^{\infty}t_k$ converges in the p-adic completion $\mathbb{Q}_p$ of $\mathbb{Q}$.	{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9551370739936829, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=3721925	Physics Forums Page 2 of 2 < 1 2 ## Circuit to retain only the positive frequency components in a signal? Method using sine and cosine as carrier. Frequency spectra. http://ocw.mit.edu/resources/res-6-0...7S11_lec13.pdf Recognitions: Gold Member Science Advisor That would work fine at low power but I think the (200kW??) transmitter I have seen used just one transmitting valve and achieved the result by AM and PM in a single unit. Hang on a minute. How would you combine your two signals losslessly from two high power amplifiers? I think that could be a problem. I only know the theory behind it, I never designed one :D Quote by Bassalisk Well there is no such thing as negative frequency. In real life. That doesn't make sense. I only think math behind the fourier transform, and generally math tools are made in that way so that we do operate with negative frequencies. Cosine is represented with 2 deltas right? (amplitude spectra) Eliminate one of them, and you no longer have a cosine... Specifically lets consider this. $F (cos(\omega _0 t)) = \pi\left[\delta (\omega -\omega _0)+\delta (\omega +\omega _0)\right]$ Eliminate the negative frequency and you get: $\pi\left[ \delta (\omega -\omega _0)\right]$ Inverse Fourier transform of this is no longer a cosine. Its complex exponential: $F^{-1}(\pi\left[ \delta (\omega -\omega _0)\right])=\frac{1}{2}\cdot e^{j\omega _0 t}$ So its not even a real function anymore. You can go with using a complex exponential as your carrier. Design a system, which transfers in parallel a Sine and a Cosine. Bassalisk, How do i go about generating the complex exponential e^(jwt)? You said a system which transfers sine and cosine waves in parallel. Do you mean sin(wt)+cos(wt)? shouldnt it be cos(wt)+j sin (wt)? Quote by sanjaysan Bassalisk, How do i go about generating the complex exponential e^(jwt)? You said a system which transfers sine and cosine waves in parallel. Do you mean sin(wt)+cos(wt)? shouldnt it be cos(wt)+j sin (wt)? There wouldn't be imaginary current or potential in an actual circuit. The circuit depicted carries two separate products with sine and cosine waves respectively, to represent multiplying by a complex exponential. You should know that real signals have Hermitian symmetry in the frequency domain. The component at a negative frequency is the complex conjugate of the component at the corresponding positive frequency. This has to do with the symmetry of cosine and the "odd" symmetry of sine. Consider what happens if we make the frequency negative. cos(ωt) = cos(-ωt) but sin(ωt) = -sin(-ωt) Anyway, removing the negative frequencies seemingly would imply two outputs instead of one, based on Hermitian symmetry requirement for the frequency domain of real signals. Quote by sanjaysan Bassalisk, How do i go about generating the complex exponential e^(jwt)? You said a system which transfers sine and cosine waves in parallel. Do you mean sin(wt)+cos(wt)? shouldnt it be cos(wt)+j sin (wt)? Well, imaginary numbers are a pair of real numbers. So that system that I posted in the post above, explains it nicely. When you go deeper into Signals and Systems, these things are pretty easy to understand :D Recognitions: Gold Member Science Advisor I think we are confusing real, time varying signals with a convenient mathematical representation of them. Neither the exponential notation or the 'cis' notation are any more than models, and 'half' of that model is not relevant to the real world. afaik, one should really prefix the final result of any 'complex' jiggery pokery with the words "The Real Part Of . . . " if you want to get a proper answer. I think you can do the analysis (albeit in a more lumpy way) without using i at all. Quote by sophiecentaur I think we are confusing real, time varying signals with a convenient mathematical representation of them. Neither the exponential notation or the 'cis' notation are any more than models, and 'half' of that model is not relevant to the real world. afaik, one should really prefix the final result of any 'complex' jiggery pokery with the words "The Real Part Of . . . " if you want to get a proper answer. Well put. Recognitions: Gold Member Science Advisor That's where that paper on "negative Frequencies in Modulation" seems to be skating on thin ice. Quote by sophiecentaur That's where that paper on "negative Frequencies in Modulation" seems to be skating on thin ice. Yes its somewhat like talking about negative time. But nevertheless, this thread was really interesting! Quote by Bassalisk Well, imaginary numbers are a pair of real numbers. So that system that I posted in the post above, explains it nicely. When you go deeper into Signals and Systems, these things are pretty easy to understand :D So, is x(t)cos(wt)+y(t)sin(wt) the required complex exponential modulated signal? Quote by sanjaysan So, is x(t)cos(wt)+y(t)sin(wt) the required complex exponential modulated signal? x(t)cos(wt)+jx(t)sin(wt) *** y(t) would have a real part x(t)cos(wt) y(t) would have an imaginary part x(t)sin(wt) They come in pairs. http://ocw.mit.edu/resources/res-6-0...me-modulation/ This video is very much addressing that subject. Recognitions: Gold Member Science Advisor If I have a room of 12 square metres area and I want a carpet for it, I would choose one with dimensions 3m by 4m. This would be a REAL carpet that had Real dimensions, made of Real Wool. The Maths would tell me that a carpet with dimension -3m by -4m would also do the job. Only a loony would go out to look for one of those in a shop. Why look for anything more significant when some Maths suggests that a Negative Frequency component could exist for a signal? At the beginning of the thread there was a question about the existence of Power in this 'Mirror' signal. Clearly not. A square wave oscillator takes power from its (DC) power supply (real, measurable Joules from a battery). This Power is exactly the same as the power in the square wave - as you can see by heating up an element or by integration. There is no other power in the system. The 'negative frequency' component is just an artifact of the Maths - just like the negatively dimensioned carpet. The same applies to a sinewave. So what about the two sidebands which are generated in AM? They are at (absolute) positive frequencies and carry energy - along with the carrier and, if you add up the three Powers, you get the value of Power which the power supply delivers (less a measurable / calculable factor due to the efficiency of the modulator. The power in each sideband can be used to carry other information and the power of the carrier can be reduced to zero - giving you two SSB transmissions, each of which has the same SNR as the original dsbam signal, half the signal spectrum occupancy and may save the power of a carrier, depending on how you actually produce the ssb signal. When you draw diagrams of signal spectra and show them shifting around and being filtered, there is no explicit mention of the Power involved. So the diagram proves nothing about the existence or otherwise of 'components'. I, personally, was a bit disappointed with that MIT Movie. It would be very easy to get some wrong messages from it, I think. OK, as far as it went but strictly an undergraduate treatment of the topic, I should say, aimed at getting predictable results from a comms system. Quote by chroot sanjaysan, the negative frequency components are redundant, in a sense. Consider your time domain signal, $cos(2\pi \omega t)$. The angular frequency, $\omega$, could be either positive or negative, and the resulting wave would look the same in the time domain. That ambiguity leads to the two-sided, symmetric spectrum. You can move to a one-sided spectrum if you wish, with no loss of generality, but that's just a mathematical trick. You don't need to design any real, physical device to discard the negative frequencies; they're all in your head from the beginning! - Warren Then how do you explain the recovery of baseband signal from single sided passband signal. Suppose we have only upper sideband of a signal then in recovery of message signal the mirrored band of the signal contributes to form the spectrum of the message signal. What do you think happens physically here........ Recognitions: Gold Member Science Advisor Look at the block diagram of an ssb receiver. A local oscillator at the original carrier frequency will mix with the sideband and produce a baseband signal - and other mixing products, of course but they will be at non-baseband frequencies and their power level is not relevant. If you're concerned about the SNR of the demodulated signal then it would be 3dB lower than when both sidebands are demodulated because the noise bandwidth would be half but the demodulated signal would be half the level - giving 3dB net loss. BUT there was 3dB less power transmitted for a start so there would is overall disadvantage (as long as the transmitter can be made efficient. It is important to get the Maths and the Physical World reconciled properly. Like I said earlier. All the answers to the calculations should start off with "The real part of". Page 2 of 2 < 1 2 Tags electronics, hilbert transform, modulation Thread Tools \| \| \| \| \|--------------------------------------------------------------------------------------------\|----------------------------------------------\|---------\| \| Similar Threads for: Circuit to retain only the positive frequency components in a signal? \| \| \| \| Thread \| Forum \| Replies \| \| \| Electrical Engineering \| 3 \| \| \| Calculus & Beyond Homework \| 0 \| \| \| Engineering Systems & Design \| 2 \| \| \| Engineering, Comp Sci, & Technology Homework \| 5 \| \| \| Introductory Physics Homework \| 7 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9365985989570618, "perplexity_flag": "middle"}
http://mathhelpforum.com/algebra/176493-help-sample-proble-college-entrance-exam.html	# Thread: 1. ## Help with this sample proble for college entrance exam!! Hi need help understanding how to simplify this expression with steps i do not know how these online math solvers got their answers. sqrt= square route ^means to the 1/3 power just in case you were confused. expression: 4ln(sqrt(x))+6ln(x^(1/3)) math solver got ln(x^4) any help will be appriciated thanks 2. You need to make use of the formulas $\displaystyle n\log{(x)} = \log{\left(x^n\right)}$ and $\displaystyle \log{(a)} + \log{(b)} = \log{(ab)}$. 3. ok so how does that apply here i already got that the 6ln(x^(1/3))=2ln(x) right?? and i look for examples for a problem with the sqrt but cannot find any its been like 8 or 9 years sincer i was last using math like this. how would i simplify the 4ln(sqrt(x))??? 4. ohhhhhhhhhhhhhhhhhhh i think i got it so heres what i got for showing my work let me know 4ln(sqrt(x))+6ln(x^(1/3))------------------>square the 4, and divide the 6 by 3 to isolate the x's 2ln(x) + 2ln(x) 4ln(x)----------------> using that rule n log(x) = log(x^n) thus getting the ln(x^4) 5. You actually halve the 4, but yes, you have the general idea. Well done.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 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": 0.9144083261489868, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/81429?sort=oldest	## Is the set of undecidable problems decidable? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I would like to know if the set of undecidable problems (within ZFC or other standard system of axioms) is decidable (in the same sense of decidable). Thanks in advance, and I apologize if the question is too basic. - 1 I think that this was addressed somewhere on math.stackexchange.com however, I cannot find it. – Asaf Karagila Nov 20 2011 at 16:27 3 Perhaps it is worth clarifying that the term "undecidable" refers to "independece in ZFC", while the term "decidable" refers to "computable (also called recursive)" – boumol Nov 20 2011 at 16:59 12 This set in non-computable. Why? Let us suppose that the set of independent statements is computable. Then, so is its complement, i.e., the set of formulas $\varphi$ such that either $ZFC \vdash \varphi$ or $ZFC \vdash \neg \varphi$. Using this it easily follows that the set of consequences of ZFC is decidable, which is well-known to be false. – boumol Nov 20 2011 at 17:04 ## 3 Answers No, in fact the situation is even worse than that. The set $T$ of all (Gödel codes for) sentences that are provable from ZFC is computably enumerable; the set $F$ of all sentences that refutable from ZFC is also computably enumerable. These two sets $T$ and $F$ form an inseparable pair: $T \cap F = \varnothing$ but there is no computable set $C$ such that $T \subseteq C$ and $F \cap C = \varnothing$. In other words, there is no finite algorithm that can reliably tell whether a sentence is "not provable" or "not refutable." This is much weaker than asking for an algorithm to tell us whether a sentence is independent or not (in which case we could determine whether it is provable or refutable by waiting until it enters $T$ or $F$). This remains true if ZFC is replaced by PA and some still weaker theories. - ### 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 believe that one can expand on boumol's answer, as follows. The spirit of the OP's question attempts to regain Eden after the Turing-Godel expulsion. One might attempt to repair the OP's attempt in a more general way by adjusting the mathematician's aim from settling questions (with a proof of either yes or no) to resolving questions (say, with proof of yes, or no, or independence (relative to consisistency), independence of independence, etc). One can even allow, in many different ways, stronger and stronger reasonable axioms (iterated consistency or reflection principles) as one pursues all these statements of iterated independence. Trouble always ensues because a programmer can always dovetail a countable number of uniformly specified computer searches, making the program terminate if any one of those searches turns out successful. Then having every formula in the theory fall into one class or another again makes the set of consequences of ZFC decidable, contradiction. - For every consistent recursively axiomatizable theory $T$ (one of which $ZFC$ is widely believed to be) there exists an integer number $K$ such that the following Diophantine equation (where all letters except $K$ are variables) has no solutions over non-negative integers, but this fact cannot be proved in $T$: \begin{align}&(elg^2 + \alpha - bq^2)^2 + (q - b^{5^{60}})^2 + (\lambda + q^4 - 1 - \lambda b^5)^2 + \\ &(\theta + 2z - b^5)^2 + (u + t \theta - l)^2 + (y + m \theta - e)^2 + (n - q^{16})^2 + \\ &((g + eq^3 + lq^5 + (2(e - z \lambda)(1 + g)^4 + \lambda b^5 + \lambda b^5 q^4)q^4)(n^2 - n) + \\ &(q^3 - bl + l + \theta \lambda q^3 + (b^5-2)q^5)(n^2 - 1) - r)^2 + \\ &(p - 2w s^2 r^2 n^2)^2 + (p^2 k^2 - k^2 + 1 - \tau^2)^2 + \\ &(4(c - ksn^2)^2 + \eta - k^2)^2 + (r + 1 + hp - h - k)^2 + \\ &(a - (wn^2 + 1)rsn^2)^2 + (2r + 1 + \phi - c)^2 + \\ &(bw + ca - 2c + 4\alpha \gamma - 5\gamma - d)^2 + \\ &((a^2 - 1)c^2 + 1 - d^2)^2 + ((a^2 - 1)i^2c^4 + 1 - f^2)^2 + \\ &(((a + f^2(d^2 - a))^2 - 1) (2r + 1 + jc)^2 + 1 - (d + of)^2)^2 + \\ &(((z+u+y)^2+u)^2 + y-K)^2 = 0. \end{align} The set of numbers with this property is not recursively enumerable (let alone decidable). But curiously enough, one can effectively construct an infinite sequence of such numbers from the axioms of $T$. The equation is derived from Universal Diophantine Equation by James P. Jones. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 19, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8997398614883423, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/58186/positive-element-in-c-tensor-product/58200	## positive element in C* tensor product ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let A, B be two C-Algebras and $A\otimes B$ denote their minimal tensor product(I don't know whether C-norm matters or not, but for simplicity we can assume that one of them is nuclear so all C-norm coincide). Let x be a non-zero positive element in $A\otimes B$, can we always find a single tensor $0\neq x_1\otimes x_2$, where $x_1$ and $x_2$ are positive elements in A and B respectively, such that $x_1\otimes x_2\leq x$? It's fairly easy to see that if both C-algebras are communicative or one of them is a finite dimensional C-Algebra(Sorry this is false), then the above assertion is true. So it's tempting to think that more general case should hold. I asked a similar question before, where the stronger assertion that any positive element in tensor algebra is a sum of tensors of positive elements, is false. See the following link: - In view of Jesse Peterson's example, your claim that such elements exist when one of your C-algebras is finite-dimensional seems to be false... – Yemon Choi Mar 12 2011 at 4:19 Right, I made a mistake – Qingyun Mar 12 2011 at 6:55 ## 2 Answers The same answer as before, the matrix ```$$ a=\begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{bmatrix} $$``` in $M_2(\mathbb{C})\otimes M_2(\mathbb{C})$, also works here since it is twice a rank one projection and so any smaller positive matrix must be a scalar multiple of $a$. - Got it, thanks! – Qingyun Mar 12 2011 at 7:45 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. There is a result of Kirchberg that comes close to giving a positive answer to this question. Given $x\geq 0$ as in the question, there exists $z\neq 0$ such that $z^z=x_1\otimes x_2$ and $zz^ \leq x$. See Lemma 4.1.9 of Rordam's book "Classification of nuclear C*-algebras". -	{"extraction_info": {"found_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": 0.8938550353050232, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/218856/is-it-possible-to-make-a-graph-eulerian-by-adding-exactly-one-node	# Is it possible to make a graph eulerian by adding exactly one node? Let $G=(V,E)$ denote a connected graph with $\|V\|\geq 2$. Is it possible to add a new node $v$ with corresponding edges $e_k=\{v,w\}$ with $w\in V$(1) such that $(V\cup\{v\},E^\prime)$ contains an eulerian cycle where $E^\prime=E\cup\left(\bigcup\{e_k\}\right)$? Prove if possible; otherwise, show a counterexample. (1) It is not mandatory to connect the new node $v$ to every other node. One is allowed to omit some of them. Thoughts. Basically I think it is possible, however I don't know exactly how to argument. I was thinking about comparing the degrees of all nodes and when adding new edges connecting the new node to the old graph to argument when degrees get even/odd to show, that in the end it is possible that all nodes will have even degree, no matter which graph we had at the beginning. This is the result I got when adding one edge and inspecting that there can't be a circle, and obviously the degree of the new node is 1. That way I would have to add an even number of edges to achieve what has to be done. Question. Can anyone give me some hints on how to start and what exactly to prove? - 2 – Martin Sleziak Oct 22 '12 at 17:38 This lemma and Michaels answer helped me out! – Christian Ivicevic Oct 22 '12 at 17:51 ## 1 Answer Use the property: A connected graph has an Eulerian path if and only if it has at most two vertices with odd degree. Then look at the number of odd degree vertices in $G$, and figure out the correct edges to use to make $(V \cup \{v\},E^\prime)$ have at most two vertices with odd degree. Edit: If you want an Euler cycle, then you must make $(V \cup \{v\},E^\prime)$ have no vertices of odd degree. We'd then want to connect the new vertex to every odd degree vertex in $G$ and then argue that the new vertex has even degree. - I know that a connected graph has an Eulerian path iff all vertices have even degree. I even proved that yesterday, isn't that contradicting with your mentioned property? – Christian Ivicevic Oct 22 '12 at 17:41 It has an Eulerian cycle iff every vertex has even degree. Paths are less restrictive. – Michael Biro Oct 22 '12 at 17:43 I am sorry, then I have translated it wrong. I have to work with a cycle - editing it now! – Christian Ivicevic Oct 22 '12 at 17:45 1 This statement should also work for Eulerian cycles if you use the parity argument for vertices of odd degree. See Martin Sleziak's comment. – Michael Biro Oct 22 '12 at 17:46 The handshaking lemma and your edit gave me a good clue to solve this. Thanks! – Christian Ivicevic Oct 22 '12 at 17:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9413614869117737, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/20927/what-are-low-lying-energy-levels	# What are Low-lying energy levels? I am reading about some canonical transformations of the Hamiltonian (of a system consisting of an electron interacting with an ionic lattice) due to Tomanaga and Lee, Low and Pines. One of the important considerations should be translation invariance (momentum conservation). In a paper by Lee, Low and Pines ("Motion of Slow Electrons in a Polar Crystal", 1953), there is a frequent mention of low-lying energy levels. Intuitively, I would like to think they are referring to energy states close to the ground-state, but I don't think that is quite correct. Do low-lying energy levels have to do with, perhaps, energy of the electron with small momentum? The title of the paper by Lee, Low, and Pines is after all about "Slow Electrons". I'm confused. Some clarification could be helpful. - ## 1 Answer Lee, Low, and Pines mention "low lying energy states of the system ($P^2/2m\ll\omega$, where $P$ is the total momentum of the system, $m$ is the mass of the electron, and $\omega$ is the frequency of lattice oscillations)." So yes, on the one hand, these states are "close to the ground state", on the other hand, they have electrons "with small momentum" in the conduction band. They discuss dielectrics, as far as I can understand, so the conduction band is empty in the ground state, so electrons can have low momentum in the low-lying energy states (the momentum of these electrons is low compared to the Debye momentum and, therefore, compared to the Fermi momentum as well.) - Not sure what Debye and Fermi momentum are but I think your answer helps with some intution. thanks. -R – r.g. Feb 17 '12 at 0:47 You may wish to google "Debye frequency" and "Fermi energy". The relevant momenta are defined by the standard formulas. For example, Fermi momentum is, roughly speaking, the largest momentum of electron for the ground state of the crystal. – akhmeteli Feb 17 '12 at 3:00	{"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": 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": 0.9364278316497803, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Hilbert's_third_problem	# Hilbert's third problem The third on Hilbert's list of mathematical problems, presented in 1900, is the easiest one. The problem is related to the following question: given any two polyhedra of equal volume, is it always possible to cut the first into finitely many polyhedral pieces which can be reassembled to yield the second? Based on earlier writings by Gauss,[which?] Hilbert conjectured that this is not always possible. This was confirmed within the year by his student Max Dehn, who proved that the answer in general is "no" by producing a counterexample. The answer for the analogous question about polygons in 2 dimensions is "yes" and had been known for a long time; this is the Bolyai–Gerwien theorem. ## History and motivation The formula for the volume of a pyramid, $\frac{\text{base area} \times \text{height}}{3},$ had been known to Euclid, but all proofs of it involve some form of limiting process or calculus, notably the method of exhaustion or, in more modern form, Cavalieri's principle. Similar formulas in plane geometry can be proven with more elementary means. Gauss regretted this defect in two of his letters. This was the motivation for Hilbert: is it possible to prove the equality of volume using elementary "cut-and-glue" methods? Because if not, then an elementary proof of Euclid's result is also impossible. ## Dehn's answer Dehn's proof is an instance in which abstract algebra is used to prove an impossibility result in geometry. Other examples are doubling the cube and trisecting the angle. We call two polyhedra scissors-congruent if the first can be cut into finitely many polyhedral pieces which can be reassembled to yield the second. Obviously, any two scissors-congruent polyhedra have the same volume. Hilbert asks about the converse. For every polyhedron P, Dehn defines a value, now known as the Dehn invariant D(P), with the following property: • If P is cut into two polyhedral pieces P1 and P2 with one plane cut, then D(P) = D(P1) + D(P2). From this it follows • If P is cut into n polyhedral pieces P1,...,Pn, then D(P) = D(P1) + ... + D(Pn) and in particular • If two polyhedra are scissors-congruent, then they have the same Dehn invariant. He then shows that every cube has Dehn invariant zero while every regular tetrahedron has non-zero Dehn invariant. This settles the matter. A polyhedron's invariant is defined based on the lengths of its edges and the angles between its faces. Note that if a polyhedron is cut into two, some edges are cut into two, and the corresponding contributions to the Dehn invariants should therefore be additive in the edge lengths. Similarly, if a polyhedron is cut along an edge, the corresponding angle is cut into two. However, normally cutting a polyhedron introduces new edges and angles; we need to make sure that the contributions of these cancel out. The two angles introduced will always add up to π; we therefore define our Dehn invariant so that multiples of angles of π give a net contribution of zero. All of the above requirements can be met if we define D(P) as an element of the tensor product of the real numbers R and the quotient space R/(Qπ) in which all rational multiples of π are zero. For the present purposes, it suffices to consider this as a tensor product of Z-modules (or equivalently of abelian groups).[further explanation needed] However, the more difficult proof of the converse (see below) makes use of the vector space structure: Since both of the factors are vector spaces over Q, the tensor product can be taken over Q. Let ℓ(e) be the length of the edge e and θ(e) be the dihedral angle between the two faces meeting at e, measured in radians. The Dehn invariant is then defined as $\operatorname{D}(P) = \sum_{e} \ell(e)\otimes (\theta(e)+\mathbb{Q}\pi)$ where the sum is taken over all edges e of the polyhedron P. ## Further information In light of Dehn's theorem above, one might ask "which polyhedra are scissors-congruent"? Sydler (1965) showed that two polyhedra are scissors-congruent if and only if they have the same volume and the same Dehn invariant. Børge Jessen later extended Sydler's results to four dimensions. In 1990, Dupont and Sah provided a simpler proof of Sydler's result by reinterpreting it as a theorem about the homology of certain classical groups. Debrunner showed in 1980 that the Dehn invariant of any polyhedron with which all of three-dimensional space can be tiled periodically is zero. ## Original question Hilbert's original question was more complicated: given any two tetrahedra T1 and T2 with equal base area and equal height (and therefore equal volume), is it always possible to find a finite number of tetrahedra, so that when these tetrahedra are glued in some way to T1 and also glued to T2, the resulting polyhedra are scissors-congruent? Dehn's invariant can be used to yield a negative answer also to this stronger question. ## References • Max Dehn: "Über den Rauminhalt", Mathematische Annalen 55 (1901), no. 3, pages 465–478. • Sydler, J.-P. "Conditions nécessaires et suffisantes pour l'équivalence des polyèdres de l'espace euclidien à trois dimensions", Comment. Math. Helv. 40 (1965), pages 43–80 • Johan Dupont and Chih-Han Sah: "Homology of Euclidean groups of motions made discrete and Euclidean scissors congruences", Acta Math. 164 (1990), no. 1–2, pages 1–27 • Hans E. Debrunner: "Über Zerlegungsgleichheit von Pflasterpolyedern mit Würfeln", Arch. Math. (Basel) 35 (1980), no. 6, pages 583–587 • Rich Schwartz: "The Dehn-Sydler Theorem Explained"	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9063040018081665, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/24487/homotopy-colimits-of-cyclic-spaces/24507	## Homotopy colimits of cyclic spaces ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $\Lambda$ denote Connes's cyclic category. It is an extension of the simplex category $\Delta$ (of nonempty finite linearly ordered sets) obtained by adding an automorphism of order $n+1$ to the object $\textbf{n}$. Question: Suppose `$X: \Lambda^{op} \to Top$` is a cyclic space. What is a description of the homotopy colimit of this functor? Just to put this in a bit of context, if $Y: \Delta^{op} \to Top$ is a simplicial space then it has a geometric realisation $\|Y\|$. One can also take the homotopy colimit of $Y$, and under some reasonable hypotheses there will be an equivalence `$\mathrm{hocolim}_{\Delta^{op}} Y \simeq \|Y\|$`. There is an inclusion $\Delta \to \Lambda$, so a cyclic space $X$ can be considered as a simplicial space and one can thus make a geometric realisation $\|X\|$. This space is supposed to have a circle action. I suppose my question should be: How is the hocolim of $X$ over $\Delta^{op}$ related to the hocolim of $X$ over $\Lambda^{op}$. - ## 1 Answer The homotopy theory of cyclic spaces is equivalent to that of spaces over $BS^1$ (Dwyer-Hopkins-Kan). The colimit over the simplicial category is as you say a space $X$ with $S^1$ action, and the colimit over the cyclic category is the quotient (Borel construction) $X/S^1$ as a space over $BS^1$. - Thanks - I suspected this was the case. – Jeffrey Giansiracusa May 13 2010 at 15:29 1 More precisely, a localization of cyclic spaces is equivalent to spaces over $BS^1$. Without the localization, it's more like spaces with an $S^1$ action, with actual fixed point spaces for subgroups (but not for the whole group). also see MO's suggestions. Spalinski: ams.org/mathscinet-getitem?mr=1325168 – Ben Wieland May 13 2010 at 16:45 Hi Ben! I'm not sure I understand your comment -- with the model structure that DHK give (as your link mentions) you do get the $\infty$-category of spaces over $BS^1$ on the nose. So sure, you can give them a more refined model structure (which Spalinski does) keeping track of extra fixed point data, and this is what is discussed in Reid's question, but in what sense does this is a "more precise" answer to the question as opposed to a different homotopy theory? – David Ben-Zvi May 13 2010 at 22:17	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9183160662651062, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/31042-dimension-basis.html	# Thread: 1. ## Dimension and basis My text is not helping at all and the wikipedia explanations of the concepts: basis and dimension are way over my head. Let V be the space of polynomials in x and y of degree <= 10. Specify a basis of V and compute dimV. thanks jblorien 2. To be a basis it must span and be linearly independent. The polynomials $1,x,x^{2},x^{3},.....,x^{10}$ span the vector space of $P_{n}$ since each p in $P_{10}$ can be written as $p=a_{0}+a_{1}x+....+a_{10}x^{10}$ which is a linear combination of $S=\left[1,x,x^{2},.....,x^{10}\right]$. $dim(P_{n})=n+1$. The basis has n+1 vectors. In other words, it has dimension 11. #### 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": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": 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": 0.8884430527687073, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/34692/list	## Return to Question 2 added 19 characters in body In the late 70s, Cuntz and Behncke had a paper H. Behncke and J. Cuntz, Local Completeness of Operator Algebras, Proceedings of the American Mathematical Society, Vol. 62, No. 1 (Jan., 1977), pp. 95- 100 the about the following question. Let $A$ be a $C^\star$-algebra and let $B \subset A$ be some $$-subalgebra. Let $B$ be dense and such that every maximal abelian subalgebra of $B$ is norm-closed. Is it true that $B=A$? They proved this in various cases. I want to ask a related question in the von-Neumann-setting. More precisely: Let $A$ be a von Neumann algebra and $B \subset A$ be some ultra-weakly dense $\star$-subalgebra such every MASA of $B$ is ultra-weakly closed in $A$. Is it true that $B=A$? A result of Gert Pedersen implies that once $B$ is a $C^\star$-algebra, then $B=A$. Hence, the two questions are closely related. Taking the work of Behncke-Cuntz and Pedersen together, it is known that $B=A$ if $A$ has no $II_1$-part. Question: Is it true for every von Neumann algebra? or a little more modest Question: Is it true for the hyperfinite $II_1$-factor? Another strenghtening of the the assumption (which could help) would be the following: Question: Let $A$ be a $II_1$-factor and $B \subset A$ be a ultra-weakly dense $\star$-subalgebra such that for every hyperfinite subalgebra $R \subset A$ one has that $R \cap B$ is ultra-weakly closed in $A$. Is it true that $B=A$? 1 # Subalgebras of von Neumann algebras In the late 70s, Cuntz and Behncke had a paper H. Behncke and J. Cuntz, Local Completeness of Operator Algebras, Proceedings of the American Mathematical Society, Vol. 62, No. 1 (Jan., 1977), pp. 95- 100 the about the following question. Let $A$ be a $C^\star$-algebra and let $B \subset A$ be some $$-subalgebra. Let $B$ be dense and such that every maximal abelian subalgebra of $B$ is norm-closed. Is it true that $B=A$? They proved this in various cases. I want to ask a related question in the von-Neumann-setting. More precisely: Let $A$ be a von Neumann algebra and $B \subset A$ be some $\star$-subalgebra such every MASA of $B$ is ultra-weakly closed in $A$. Is it true that $B=A$? A result of Gert Pedersen implies that once $B$ is a $C^\star$-algebra, then $B=A$. Hence, the two questions are closely related. Taking the work of Behncke-Cuntz and Pedersen together, it is known that $B=A$ if $A$ has no $II_1$-part. Question: Is it true for every von Neumann algebra? or a little more modest Question: Is it true for the hyperfinite $II_1$-factor? Another strenghtening of the the assumption (which could help) would be the following: Question: Let $A$ be a $II_1$-factor and $B \subset A$ be a ultra-weakly dense $\star$-subalgebra such that for every hyperfinite subalgebra $R \subset A$ one has that $R \cap B$ is ultra-weakly closed in $A$. Is it true that $B=A$?	{"extraction_info": {"found_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": 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": 0.9264647364616394, "perplexity_flag": "head"}
http://mathoverflow.net/questions/94769?sort=votes	## Motivation for the Sprague-Grundy theorem ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The Sprague-Grundy theorem states that every impartial combinatorial game under the normal play convention is equivalent to a (unique) nimber. What does the equivalence relation thus defined tell us about a certain game? (e.g. does a high SG-number imply that the position is hard to compute?) What are some specific uses of the theorem (besides making the calculation of N and P positions considerably easier)? - ## 3 Answers There is no direct link with difficulty of computation -- for example a single Nim heap of $n$ counters has nim-value $n$. Of course, if a game is equivalent to $n$ it must have at least $n$ options, so in that sense it is more complex. As for applications, the hope generally is that it may be easier to analyse a game in general by working out the exact nim-values rather than just seeking the partition into next player wins ($\mathcal{N}$) and previous player wins ($\mathcal{P}$). Unfortunately, this hope is not often realised! For taking and breaking games there has been considerable work done in trying to determine both the actual sequences of nim-values, and the types of behaviours these sequences can exhibit (various forms of periodicity most prominently). See for instance the wikipedia article on octal games. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. It is quite typical that a game may be decomposed into a sum of rather easy games. The Sprague-Grundy theorem essentially tells us what the Sprague-Grundy function of this sum is just the nim-sum of the Sprage-Grundy functions of the summands, each being very easy to compute. Then one can determine the $\mathcal{N}$ and $\mathcal{P}$-positions of the sum by finding the zeros. This method is illustrated by some examples in section I.4 in Ferguson's Game Theory. - Thanks for your answer! I was aware of this, since I'm reading Ferguson's book; I just wanted to know if the equivalence relation tells us something about each class (e.g. if all games in a class have something in common). – Fernando Martin Apr 25 2012 at 3:31 Sorry. Actually I've only read Part I of his book and apart from that I have no background in game theory. I don't know a good answer to your question. But isn't it great that we can classify impartial combinatorial games by natural numbers? In Conway's theory these games embed into a larger class of "numbers", including surreal numbers ... – Martin Brandenburg Apr 25 2012 at 7:34 Martin: Indeed, it's a very nice result. Thanks for your help anyway! – Fernando Martin Apr 25 2012 at 18:06 The Sprague-Grundy theorem provides a surjective homomorphism $\mathcal{G}$ from the commutative monoid of symmetric games onto the group $On_2$ (the ordinals with the nim sum). The point of this quotient is that $\ker \mathcal{G}$ is exactly the class of $\mathcal{P}$ games. So you can see how the Sprague-Grundy theorem defines a partition on all games, which are many. In fact, you can think of a symmetric game as a well founded set (formally a biset with equal side-sets, as presented in On Numbers and Games by J. H. Conway); in this light we are partitioning the whole Von Neumann hierarchy $\mathbb{V}$, where each class is indeed very well populated (it is a proper class). This bridge theorem, as already outlined, allows to compute the outcome of a game just by computing the nimber of each of its components; this is the main use and its original purpose. But it can also be useful the other way round: If $n_1 + \cdots + n_k = t \neq 0$, where $+$ is the nim sum, then $$\exists i \in \{ 1, \dots ,k \} \mbox{ such that } n_i + t < n_i.$$ A proof is as follows: since $t \neq 0$, $n_1 + \cdots + n_k$ is a $\mathcal{N}$-position in Nim so there must be a winning move from, say, $n_1$ to $\bar{n}_1$ such that $\bar{n}_1 + n_2 + \cdots + n_k =0$. Since in Nim the only legal moves are to decrease numbers it follows that $n_1 > \bar{n}_1$. But, since $On_2$ satisfies $\forall x\ x+x=0$, adding $n_2 + \cdots + n_k + t$ to each side of the previous equation yields (after cancellations) $\bar{n}_1 + t = n_2 + \cdots + n_k + t = n_1$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 26, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9461444020271301, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/36278/how-is-this-equation-read	# How is this equation read? I want to understand this paper on brain tumour segmentation. How is this equation read? I'm guessing $q_i(t_i)$ represents the likelihood of tumour on voxel i.Is q usually used to represent likelihoods? What does the equal sign with the triangle mean? I understand what each term on the conditional probability represents, but I don't get why it's called proportional to the product of the summation. I'm looking for a reading reference here. - The equal sign with triangle usually means "is defined as". – mark999 Sep 14 '12 at 7:00 1 The equation might be erroneously including a summation sign. If the observation is $y$, then the likelihood of $p(y\|t_i, k_i)$, the joint probability is the likelihood times the prior probability $p(t_i,k_i)$. Thus, $p(y) = \sum_j p(y\|t_j, k_j)p(t_j,k_j)$ is the probability of the observation and is the sum above. The posterior probability is $$p(t_i\|y) = \frac{p(y\|t_i, k_i)p(t_i,k_i)}{p(y)}$$ and is thus proportional (with proportionality factor $[p(y)^{-1}$) to the joint probability $p(y\|t_i,k_i)p(t_i,k_i)$. It is not proportional to, but rather inversely proportional to the sum $p(y)$. – Dilip Sarwate Sep 14 '12 at 13:21 ## 1 Answer I think the use of q for the function is the authors choice and not anything standard. I agree with mark999 about the triangle over the equal sign. From reviewing the article I see that the function p on the left side of the proportionality sign is a component of the posterior distribution for the t$_i$s and the proportionality piece is just the well-known relation dues to Bayes theorem, that the posterior distribution is proportion to the prior times the likelihood and in this case both sides are decomposible into products for the i$_t$$_h$ t. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9312129020690918, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/95205/list	## Return to Question 2 added 4 characters in body; edited tags Let $K/{\mathbb Q}$ be an extension of degree $d$. Let $S$ be the set of primes $p$ which split completely in $K$. What can one say about the analytic properties of $$\zeta_{K, S}(s) : = \prod_{p \in S} \frac{1}{1-p^{-s}}.$$ More generally, one can define a similar partial Euler product for any splitting type, and ask the same question about the analytic properties of the resulting function. Any advice would be greatly appreciated. 1 # A question about partial Euler products Let $K/{\mathbb Q}$ be an extension of degree $d$. Let $S$ be the set of primes $p$ which split completely in $K$. What can one say about the analytic properties of $$\zeta_{K, S}(s) : = \prod_{p \in S} \frac{1}{1-p^{-s}}.$$ More generally, one can define a similar partial Euler product for any splitting type, and the same question about the analytic properties of the resulting function. Any advice would be greatly appreciated.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9120215177536011, "perplexity_flag": "head"}
http://mathoverflow.net/questions/115447?sort=oldest	The Riemann zeros and the heat equation Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The Riemann xi function $\Xi(x)$ is defined, with $s=1/2+ix$, as $$\Xi(x)=\frac12 s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s)=2\int_0^\infty \Phi(u)\cos(ux) \, du,$$ where $\Phi(u)$ is defined as $$2\sum_{n=1}^\infty\left(2\pi^2n^4\exp(9u/2)-3\pi n^2\exp(5u/2)\right)\exp(-n^2\pi\exp(2u)).$$ This arises from integration by parts after writing $\Xi$ as the Mellin transform of the theta function, and then a change of variables from multiplicative to additive notation. In 1950 de Bruijn (building on work of Polya) introduced a deformation parameter $t$: $$\Xi_t(x)=\int_0^\infty \exp(t u^2)\Phi(u)\cos(ux)\, du,$$ so that for $t=0$, $\Xi_0(x)$ is just $\Xi(x)/2$. de Bruijn proved the following theorem about the zeros in $x$: (i) For $t\ge 1/2$, $\Xi_t(x)$ has only real zeros. (ii) If for some real $t$, $\Xi_t(x)$ has only real zeros, then $\Xi_{t^\prime}(x)$ also has only real zeros for any $t^\prime>t.$ In 1976 Newman showed that there exists a real constant $\Lambda$, $-\infty<\Lambda\le 1/2$, such that (i) $\Xi_t(x)$ has only real zeros if and only if $t\ge\Lambda$. (ii) $\Xi_t(x)$ has some complex zeros if $t<\Lambda$. The constant $\Lambda$ is known as the de Bruijn-Newman constant. The Riemann hypothesis is the conjecture that $\Lambda\le 0$. Newman made the complementary conjecture that $\Lambda\ge 0$, with the often quoted remark "This new conjecture is a quantitative version of the dictum that the Riemann hypothesis, if true, is only barely so." Given the significance of the de Bruijn-Newman constant $\Lambda$, much work has gone into estimating lower bounds, and the current record (Saouter et. al.) is $-1.14\times 10^{-11}<\Lambda.$ A breakthrough occurred in the work of Csordas, Smith and Varga, "Lehmer pairs of zeros, the de Bruijn-Newman constant, and the Riemann Hypothesis", Constructive Approximation, 10 (1994), pp. 107-129. They realized that unusually close pairs of zeros of the Riemann zeta function, the so-called Lehmer pairs, could be used to give lower bounds on $\Lambda$. The idea of the proof is that the function $\Xi_t(x)$ satisfies the backward heat equation $$\frac{\partial \Xi}{\partial t}+\frac{\partial^2 \Xi}{\partial x^2}=0,$$ from which they are able to draw conclusions about the differential equation satisfied by the $k$-th gap between the zeros as the deformation parameter $t$ varies. They mention this PDE in a rather offhand way, as a remark on an alternate proof to one of the lemmas. In fact, it does not seem to be well known that the de Bruijn-Newman constant can be interpreted as a time variable in a heat flow. Is this well known? Or put more concretely, does anyone have a citation prior to 1994 which mentions this fact? - 1 Answer ````Q. Does anyone have a citation prior to 1994 which mentions this fact? ```` 1988: Numer. Math. 52, 483-497 (the differential equation is given in a slightly different form on page 493). - 1 In that reference, do you mean the equation: $$H_\lambda(x)=F_\lambda(D)H_0(x),\qquad D=d/dx,$$ where $$F_\lambda(z)=\sum_{m=0}^\infty (-1)^m\lambda^m z^{2m}/m!$$ This is not the heat equation. – Stopple Dec 5 at 0:04 to convert it to the 1994 heat equation, just take the derivative with respect to $\lambda$ of both sides of the 1988 equation, and then note that $\partial F_\lambda/\partial\lambda=-z^2 F_\lambda$. Since $z=d/dx$, you arrive at $\partial H_\lambda/\partial\lambda=-\partial^2 H_\lambda/\partial x^2$, which is the (backward) heat equation (with $\lambda$ representing time). – Carlo Beenakker Dec 5 at 11:39 We could debate whether the derivation above means that the 1988 formula is the heat equation. But regardless I think this answer misses the spirit of the original question, of whether the connection to the heat equation is well known. The word 'heat' does not appear. – Stopple Dec 5 at 16:10	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 35, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "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": 0.929696261882782, "perplexity_flag": "head"}
http://mathoverflow.net/questions/5892/what-is-convolution-intuitively/18923	## What is convolution intuitively? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) If random variable $X$ has a probability distribution of $f(x)$ and random variable $Y$ has a probability distribution $g(x)$ then $(fg)(x)$, the convolution of $f$ and $g$, is the probability distribution of $X+Y$. This is the only intuition I have for what convolution means. Are there any other intuitive models for the process of convolution? - See my previous question: mathoverflow.net/questions/1977/… – Kim Greene Nov 18 2009 at 1:35 3 There are some nice interpretations listed in the Wikipedia article: en.wikipedia.org/wiki/Convolution#Applications – Qiaochu Yuan Nov 18 2009 at 2:54 ## 20 Answers I remember as a graduate student that Ingrid Daubechies frequently referred to convolution by a bump function as "blurring" - its effect on images is similar to what a short-sighted person experiences when taking off his or her glasses (and, indeed, if one works through the geometric optics, convolution is not a bad first approximation for this effect). I found this to be very helpful, not just for understanding convolution per se, but as a lesson that one should try to use physical intuition to model mathematical concepts whenever one can. More generally, if one thinks of functions as fuzzy versions of points, then convolution is the fuzzy version of addition (or sometimes multiplication, depending on the context). The probabilistic interpretation is one example of this (where the fuzz is a a probability distribution), but one can also have signed, complex-valued, or vector-valued fuzz, of course. - 6 Also, Gaussian Blur is a convolution filter on some image manipulation programs that is often used to test computer speed. – S. Carnahan♦ Nov 18 2009 at 4:41 ### 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. What is the operator $C_f\colon g\mapsto fg$? Consider the translation operator $T_y$ defined by $T_y(g)(x)=g(x-y)$, and look at $fg(x)=\int_{\mathbb{R}}f(y)g(x-y)dy$. Rewriting this as an operator by taking out $g$, you end up with the operator equation $$C_f=\int_{\mathbb{R}}f(y)T_ydy.$$ This is only formally correct of course, but it roughly says that convolution with $f$ is a linear combination of translation operators, the integral being a sort of generalized sum. Tying this in with Terry Tao's answer, which came in while I was writing the above, if $f$ is a bump function, say nonnegative, with integral equal to 1 and concentrated near the origin, then $fg$ is a (generalized) linear combination of translates of $g$, each one translated just a short distance, hence the blurryness of the result. - I think one's standards of intuitiveness depend strongly on one's background. Even if a picture seems unintuitive at first, it can be helpful later. 1. If you're an algebraist, I'd suggest the multiplication operation in group rings or monoid rings (easiest example: polynomial rings). 2. If you like differential or integral operators, I'd suggest convolving with derivatives of delta and Heaviside functions to realize derivatives and integrals. 3. If you like multiplying large numbers (or power series) together, convolution in the places (possibly with a carry) is the process by which this is done. - 5 Convolution of probability distributions which are supported on the integers is a special case of multiplying power series together; it corresponds to multiplication of the probability generating functions. – Michael Lugo Nov 18 2009 at 3:31 The two things that first come to mind when I think 'convolution' are: 1. It's the thing that corresponds to multiplication on the other side of the Fourier transform. (This was already mentioned by John D. Cook) It works both ways, of course, $\mathcal F (fg)=\mathcal F f\cdot \mathcal F g$ and $\mathcal F (f\cdot g)=\mathcal F f \mathcal F g$. This fact is useful when used in combination with other simple facts about the Fourier transform (such as the fact that a rectangular function corresponds to sinc and, in the limit, a Dirac impulse corresponds to a constant function). 2. Imagine a black box that receives one number $x_n$ every second and must output a number $y_n$ every second. (DSP people call it a 'filter' and it's used, for example, to process audio signals in a mobile phone in real-time.) The simplest thing the box could do is to output some function of the current input. The natural next step is to remember the last k inputs and output some function of those k values. One of the simplest functions is a linear combination $$y_n=\sum_i c_i x_{n-i}$$ where $c_i$ is non-zero only for `$0\le i<k$`. That's a convolution! To generalize, you make the filter remember all previous values and even be clairvoyant. That is, you extend the support of $[c_n]$. Then, if you want, you replace digital circuits with analog ones. That is, you go from summing to integration. As an example of combining these two points, if the filter always outputs the average of the last k inputs then that's a convolution with the rectangular function in the time domain so it must be a multiplication with a sinc in the frequency domain. Therefore, averaging the last k values attenuates high frequencies. (Hardly surprising, but at least you see immediately that the frequency response is not monotonic and there are only a few frequencies that are completely filtered out.) - Among things that it's good to know about convolution is that the identity element for convolution is the Dirac delta function $\delta$. Another is that if you convolve a function $f$ with $\delta'$, the derivative of the delta function, you get $f'$. Since convolution is associative, that implies that $f'g = fg'$. Another is that often the convolution of two functions is as well-behaved as the better-behaved one of the two. If you convolve something with a smooth function, you get a smooth function; if you convolve something with a polynomial, you get a polynomial. In other words, many classes of "well-behaved" functions are ideals in a ring whose multiplication is convolution. So if you convolve $f$ with a smooth approximation to Dirac's delta function, you get a smooth approximation to $f$. Thinking about why that works can probably shed a lot of intuitive light on the nature of convolution. - I want to expand on a special case of Terry's answer which I think is particularly intuitive. Suppose there is a function $f$ that you want to understand, but perhaps it is not smooth. Convolution gives you a way to construct new, possibly nicer functions which approximate $f$. If you let $g$ be a bump function centered at the origin, then the convolution $fg$ is a new function whose value at $x$ is given by averaging the values of $f$ around $x$. What do we mean exactly by "averaging"? Well, you use $g$ as your measure; translate it over so that it is centered at $x$, and then the integral $$f g(x) = \int_{\mathbb{R}^n} f(y)g(y - x) dy$$ in the convolution corresponds to the $g$-weighted average of the values of $f$ around $x$ (i.e. in the small ball where $g$ doesn't vanish). The convolution $fg$ in this case has the advantage that it is much smoother than $f$. Intuitively, this should be not surprising since the value of $fg(x)$ was gotten by averaging nearby $f$-values of $x$. Furthermore, you can approximate $f$ by smooth(er) things by considering a sequence of convolutions $f(g_n)$ where $g_n$ is a sequence of bump functions which are more and more concentrated at the origin. If you think of the second function $g$ in the convolution $fg$ as a measure, then you can think of convolutions as $g$-weighted averages of $f$. - Have a look here - its rather helpful! http://answers.yahoo.com/question/index?qid=20070125163821AA5hyRX - Maybe it would help your intuition to think about the discrete case first where the convolution is a sum rather than an integral. (fg)(x) is the sum of f(i) g(j) over all (i, j) that sum to x. Or maybe you could think of convolution as a kind of multiplication. Convolution makes certain function spaces into algebras. Or you could think in terms of Fourier transforms: the Fourier transform of fg is the product of the Fourier transforms of f and g. - The way you have described is the best way to think about convolution. More generally, if you have a group and a class of square-integrable functions (really I should say "half-densities") on it, then the convolution product precisely extends the group product. - This is really following on from John's answer, but is a bit long for a comment, so I thought I'd write it out here for extra clarity. Say you have a semigroup S, and you take the free (complex) vector space this generates, call it $C^S$. By construction/definition this has a distinguished basis, indexed by elements of the semigroup; and since we have a semigroup structure, that means we can multiply basis elements to get other basis elements. But that now gives us a way to multiply two vectors $a$ and $b$ together: write $a$ and $b$ as linear combinations of basis elements, and then define their product as the obvious (bi)linear extension of the multiplication on basis elements. If you do all this starting with $S={\bf Z}$, the group of integers, then what we've done is defined multiplication of trigonometric polynomials, or convolution of finitely supported sequences. The thing I like about this point of view is that it immediately generalises to $l^1(S)$, and makes $l^1(S)$ into a Banach algebra. If $S$ is a topological group with a Haar measure on it -- such as the real line with Lebesgue measure -- then the same idea gives us the usual Banach algebra structure on $L^1(S)$, which in the case $S={\bf R}$ is precisely the convolution of integrable functions in the usual sense. (At this point someone -- often me -- usually wants to mention forgetful functors from algebras to vector spaces and from semigroups to sets, but that's probably getting a bit OTT for the question at hand.) - If you convolve an image with a discrete matrix of values - so like a function that is zero outside a few pixles then you can create almost an unlimited number of filtering effects around each point. Fof example you can do some kind of averaging or weighted integration - which looks like blurring as Professor Tao mentions if you use a matrix whose values drop off smoothly, radially from the centre - a bump. You can also compute directional derivatives, look for edges, circles, blobs, steps - basically anything you like. The interpretation in terms of multiplication of Fourier coeficients is interesting and makes applications of the above in reality fast, especially if the filter is fixed, because you can use the Fast Fourier Transform on both images but you only need to update one of them. However I a not sure how intuitive it is! I'm not sure I have added much additional information but I hope this helps anyway, Ivan - I like the answer you gave when you asked the question. More generally, the convolution of two measures $\mu$ and $\nu$ is the pushforward of $\mu \times \nu$ by multiplication. In probability, that means you independently draw from $\mu$ and $\nu$ and add the resulting random vector. It's something that you can visualize to a certain extent if you do think of measures as fuzzy versions of points (like Terry Tao said). One point of view of measures is that they are linear combinations of points (or limits of things you can get from linear combinations of points). If you take this point of view, then convolution is simply the extension of the addition law by linearity to the case of measures. Since you can translate functions as well as measures, you can convolve, say, a probability measure with a function by randomly translating the function, giving the averaged out function $\int f(x-y) d\mu(y)$ which generally looks like a smoothed out version of your function $f$ -- $\mu$ tells you which translations you use and how to average. Again, you can view this as the extension of the operation of translating functions by linearity/continuity to the case of measures. The Lebesgue measure allows you to identify functions with measures, $g \mapsto g(x) dx$, so you can also convolve functions with other functions, but you might think of this operation is a bit less basic. Actually, the process of convolution extends by continuity to more than just measures but also to distributions. For example, you can approximate a tangent vector at $0$ (giving the distribution $u(x) = \sum_i c^i \partial_i \delta_0$) by differences of point masses, so convolution extends to distributions as well, but you can even get differential operators this way (in this example, $u \ast f$ is the derivative of $f$ in the $u$ direction). The technical difference here is that the approximation is only valid when integrated against $C^k$ functions (rather than $C^0$ functions in the case of measures). But the principle is the same -- it's the extension of the addition law by linearity and limits. - The fundamental theorem of calculus says that $\frac{d}{dx} \displaystyle \int_a^x f(t)dt = f(x)$. In other words, $f(x) \approx \displaystyle \int_{x-h}^{x+h} \frac{1}{2h} f(t)dt$. Letting $g_h(u) = \frac{1}{2h}$ on the interval $(-h,h)$ and 0 elsewhere, we see by pure algebraic manipulation that $f(x) \approx \displaystyle \int_{-\infty}^\infty g_h(x-t)f(t)dt$. So the fundamental theorem of calculus can very naturally be rephrased in terms of convolution with a bump function. Differentiation under the integral sign immediately gives the differentiation formula for convolutions, and thus that convolutions of two functions are at least as smooth as both factors. Thus finding good smooth approximations to the rectangular bump functions $g_h$ automatically gives us smooth approximations to any integrable function we like, just by convolving against these "smooth molifiers". Pretty cool stuff. As mentioned in other answers they really start to shine when you start thinking about fourier analysis, but that is a another story. If you google "low pass filter" you will find some pretty snazzy applications of the fact that the fourier transform turns convolution into multiplication. - It looks I am a bit late to the party; but hopefully not too late. I can contribute one instance where the application of convolution is very enlightening. Look at this in the light of Harald Hanche-Olsen's answer. Let us consider the operation of a signal processing system, or an electrical network, or a control system. Let $i(t)$ be the response of the system to the unit impulse function, ie the Dirac delta function, $\delta(t)$. Now we give an arbitrary signal $f(t)$ as input to the system. Then, the response of the system to $f(t)$ is $(f * i) (t)$, i.e. the convolution of $i$ and $f$. Next, in the spirit of rgrig's answer, I add that convolution becomes multiplication in the frequency domain. In that domain, it is like you multiply individually at each frequency component and add up again(ie integrate). - 1 You may want to add that you talk about a linear system. – rgrig Mar 21 2010 at 13:45 Thanks. The systems considered in electrical engineering and signal processing are usually linear. So it escaped my mind to mention it explicitly. – Anweshi Mar 21 2010 at 13:50 2 More specifically, linear and time-invariant. – Noah Stein Nov 19 2010 at 18:10 There is a nice relationship between convolution of probability measures and random walks which is very clear on finite groups. For a particularly concrete example, suppose you are shuffling a deck of cards. You can model this as picking elements of $S_{52}$ according to some probability measure $P$ on $S_{52}$. This generates a Markov chain with transition matrix whose $(s,t)$-th entry is given by $P(ts^{-1})$ --- if I am permitted to abuse notation somewhat, the element $ts^{-1}$ is the shuffle that takes the deck from ordering $s$ to ordering $t$. If one wants to know the transition matrix for two shuffles, this corresponds to the square of the original matrix. One can then check that this new matrix corresponds to constructing the transition matrix generated by $PP$, that is the matrix whose $(s,t)$-th entry is given by $(PP)(ts^{-1})$, and in fact $k$ shuffles corresponds to the the $k$-fold convolution of $P$ with itself. Convolving two different probability measures then corresponds to shuffling your deck according to one technique and then a different technique. - Here is my take on this. $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\bR}{\mathbb{R}}$ Discretize the real axis and thing of it as the collection of point $\Lambda_\hbar:=\hbar \bZ$, where $\hbar>0$ is a small number. Then a function $f:\Lambda_\hbar\to \bR$ is determined by its generating function, i.e., the formal power series $\newcommand{\ii}{\boldsymbol{i}}$ $$G^\hbar_f(t)=\sum_{n\in\bZ}f(n\hbar)t^n.$$ Then $$G^\hbar_{f_0\ast f_1}(t)= G^\hbar_{f_0}(t)\cdot G^\hbar_{f_1}(t).\tag{1}$$ Observe that if we set $t=e^{-\ii\xi \hbar}$, then $$G^\hbar_f(t)=\sum_{x\in\Lambda_\hbar} f(x) e^{-\ii \xi x}.$$ Moreover $$\hbar G^\hbar_f(e^{-\ii\xi \hbar})=\sum _{n\in \bZ} \hbar f(n\hbar) e^{-\ii\xi(n\hbar)}, \tag{2}$$ and the expression in the right hand sum is a "Riemann sum" approximating $$\int_{\bR} f(x)^{-\ii\xi x} dx.$$ Above we recognize the Fourier transform of $f$. If we let $\hbar\to 0$ in (\ref{2}) and we use (\ref{1}) we obtain the wellknown fact that the Fourier transform maps the convolution to the usual pointwise product of functions. (The fact that this rather careless passing to the limit can be rigorous is what the Poisson formula is all about.) The above argument shows that we can regard $\hbar G_f^\hbar(1)$ as an approximation for $\int_{\bR} f(x) dx$. Denote by $\delta(x)$ the Delta function 9concentrated at $0$. The Delta function concentrated at $x_0$ is then $\delta(x-x_0)$. What could be the generating function of $\delta(x)$, $G\delta^\hbar$? First, we know that $\delta(x)=0$, $\forall x\neq 0$ so that $$G_\delta^\hbar(t) =ct^0=c.$$ The constant $c$ can be determined from the equality $$1= \int_{\bR} \delta(x) dx=\hbar G_\delta^\hbar(1)=\hbar c$$ Hence $\hbar G_\delta^\hbar(1)=1$. Similarly $$G^\hbar_{\delta(\cdot-n\hbar)} =\frac{1}{\hbar} t^n.$$ Putting together all of the above we obtain an equivalemt description for the generating functon af a function $f:\Lambda_\hbar\to\bR$. More precisely $$G_f(t)=\hbar\sum_{\lambda\in\Lambda_\hbar}f(\lambda) G_{\delta(\cdot-\lambda)}(t).$$ The last equality suggests an interpretation for the generating function as an algebraic encoding of the fact that $f:\Lambda_\hbar\to\bR$ is a superposition of $\delta$ functions concentrated along the points of the lattice $\Lambda_\hbar$. - Arthur Mattuck gives an interesting lecture on the convolution, its construction, and its applications: http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-21-convolution-formula/ - As far as I understand, in simple words, considering the simple moving average algorithm, when you convolve F with G, then G defines how you are going to do the weightings to get the average. G can be seen as a component for defining the weighting policy. - To complement/supplement other answers, it may be worthwhile to note that the question itself blurs two substantially different mechanisms. Namely, there is, first, for any group representation of a topological group $G$ on a topological vector space $V$, an action of compactly-supported continuous functions on $G$ on $V$, by (e.g., Gelfand-Pettis/weak) integrals $f\cdot v=\int_G f(g)\cdot gv\,dg$. It is of some moment to note that this does not depend on $v$ being in any sort of natural function-space. The second point is that $f\cdot (g\cdot v)=(fg)\cdot v$, where $$ denotes the convolution. That is, the notion of convolution is externally determined by being what it has to be for (for example) compactly-supported continuous functions to act (associatively) on any representation space. Depending on one's outlook, this may reduce some element of seeming whimsy in "defining" convolution, since, in a larger context, _there_is_no_choice_. - Algebraic meaning of convolution in analysis can be seen from the following formula: $$\int_{-\infty}^{+\infty} f(x)g(x)\ dx=\left(\int_{-\infty}^{+\infty} f(x)\right)\left(\int_{-\infty}^{+\infty} g(x)\right)$$ So, in short, convolution is the product of integrals. This is also true in discrete calculus where discrete convolution is used when multiplying sums, and in particular n-times repeated convolution is used in binomial theorem to express the power of a binomial. - 1 Ignoring the extraneous first $x$, this formula is only correct if the functions are integrable. Convolution makes sense if both functions are square-integrable but not integrable, in which case this equation is incorrect (since Fubini does not apply). But in any case, what intuition do you glean from knowing, for a fixed $f$, that $T_f(g)=fg$ has the property you stated? The operator $S_f(g)=g(x)\int_{-\infty}^{\infty}f(t)dt$ has the same property but is just a constant times the identity... – Peter Luthy Jun 16 2011 at 20:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 146, "mathjax_display_tex": 13, "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": 0.9317183494567871, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?t=162967	Physics Forums ## nonlinearity and dispersion in Kdv equation? Hi all. I am referring to the Kdv equtaion as follows: u_t = u u_x + u_(xxx) A fundamental point concerning the KdV equation is that it exhibits two opposing tendencies: 1. "nonlinear convection", uu_x, which tends to -steepen- wavecrests, 2. "dispersion", u_(xxx), which tends to -flatten- wave crests. However, I don't quite understand why the nonlinear term and the disperion term would tend to steepen the wave profile? Could I see this through the geometrical meaning of uu_x and u_(xxx) or what? how could one make such a statment? PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Recognitions: Gold Member Homework Help Science Advisor By the method of characteristics you may show that the non-linear term typically will steepen the wave crest until the point of wave-breaking. how about the dispersive term? I know dispersive wave flatten the wave profile but why the third derivative of the dimensionaless amplitude implies dispersivity? ## nonlinearity and dispersion in Kdv equation? Quote by hanson Hi all. I am referring to the Kdv equtaion as follows: u_t = u u_x + u_(xxx) A fundamental point concerning the KdV equation is that it exhibits two opposing tendencies: 1. "nonlinear convection", uu_x, which tends to -steepen- wavecrests, 2. "dispersion", u_(xxx), which tends to -flatten- wave crests. However, I don't quite understand why the nonlinear term and the disperion term would tend to steepen the wave profile? Could I see this through the geometrical meaning of uu_x and u_(xxx) or what? how could one make such a statment? __________________________ I would say the nonlinear + dispersion terms tend to maintain the waveform as opposed to steepening as u propose... maybe you check up references on solitons/solitary waves and stuff - start with basic texts like Agrawal's Nonlinear Effects in Optical Fibers... good luck and please do come back to let us know what u find Recognitions: Gold Member Homework Help Science Advisor Quote by hanson how about the dispersive term? I know dispersive wave flatten the wave profile but why the third derivative of the dimensionaless amplitude implies dispersivity? First of all, call it the LINEAR dispersive term, not dispersive! Let us look at the linearized equation: $$u_{t}=u_{xxx}$$ Let us consider a monochrome wave train as our trial solution, $$u(x,t)=Ae^{i(kx-wt)}$$ The dispersion relation is of the form w(k), and we have non-dispersive situation if the phase velocity c=w/k is independent of k. Let us insert our trial solution in our diff.eq; in order to have NON-TRIVIAL solutions, we must have a specific relation between w and k, i.e the disp. relation. We get $$-iwAe^{i(kx-wt)}=-ik^{3}Ae^{i(kx-wt)}$$ If this is to be satisfied for all t and x, A non-zero, we must have: $$w=w(k)=k^{3}$$ The phase velocity c is therefore a function of the wave-number: $$c=k^{2}$$ This shows that high wave-number components of a complex wave will rush off from the low wave-number components, leaving long flattened-out waves moving slowly behind them. (Since most of the energy will typically be contained in the low/mid-range wavenumbers, it folllows that the amplitude of the fast-moving ripples is so small that most of them won't be seen, since the amplitude is typically proportional to the root of the energy contained in the wave-component.) You can find a nice treatment of the approximation (viscocity=0, the term multiplied to the third derivative) in Whitham's book on nonlinear waves. There are proofs of all statements made above. thanks so much for the explantation Arildno... May you please show me how to solve the following NLSE pde: $$iU_{z} + dU_{tt} = 0$$ where 1. $$d [\tex] = constant, and 2. [tex]U(z=0,t)=e^{(-t^{2})}$$ It's the NLSE, the nonlinear terms and the loss terms are here considered negligible. Why don't you propose a traveling wave? Let $U(z,t)=f(z-ct)$, find an ode for $f$ and the value of $c$. P.S. The TeX command \qquad works like the indent you are using. Recognitions: Science Advisor I think the OP is probably reading Drazin and Johnson, Solitons: an Introduction, which IMO explains all these issues very clearly. I'd add that of course not all interesting exact solutions of the KdV are obtained by assuming that the solution we want is a traveling wave! Jacobi elliptic functions are your friend, and all that. In addition to the book I just mentioned, I'd also highly recommend Olver, Applications Of Lie Groups To Differential Equations. Dear all, I am wondering if you could inkindly read and give some response to my thread below: http://www.physicsforums.com/showthread.php?t=163709 Recognitions: Science Advisor Quote by hanson Dear all, I am wondering if you could inkindly read and give some response to my thread below: http://www.physicsforums.com/showthread.php?t=163709 It seems to me that Mute answered your basic question (and your later question about how $u u_x$ drops out to give the "linearized KdV"). You imply that you are disappointed in the response to your questions. Part of the problem might be that you seem to have started several distinct threads on the KdV equation. Another might be that you didn't explain what book(s) you are reading (not just Drazin and Johnson?) or why the author was discussing perturbation expansions; rather, you in effect asked us to guess. We can probably make some guesses based on knowledge of the literature on solitons, but it would be helpful if you provided more context, I think. Thanks for mentioning your background, since this is also helpful. But to answer an easy question: yes, in $u_t + 6 u u_x + u_{xxx} =0$ the nonlinearity is due to the term $u u_x$, the term which drops out when we form the linearized KdV $u_t + u_{xxx} = 0$. Chris Hillman, I am grateful for Mute's nice reply but I have got some follow-up questions for Mute. However, it has been a long time that my follow-up questions are not replied by Mute. So I am wondering if some other people would kindly help answer the questions. Where does all that sudden interest in solitons come from? Earthquakes,tsunami,what? 2-3 years ago you could rarely even hear of a word "soliton". Now,solitons this solitons that! Recognitions: Homework Help Quote by hanson Chris Hillman, I am grateful for Mute's nice reply but I have got some follow-up questions for Mute. However, it has been a long time that my follow-up questions are not replied by Mute. So I am wondering if some other people would kindly help answer the questions. Sorry, I was going to reply, but I wanted to make sure I didn't say anything that wasn't quite correct, so I delayed the response, but I've hit exam season and so studying for that has taken up the bulk of my time and I hadn't had time to carefully think about what to say. In essence, though, you can think of "weakly nonlinear" as meaning the nonlinear term in a given PDE as being more like a perturbation to the linear equation, so it's the linear terms that dominate the behaviour of the solution. Typically this means that the order of the nonlinear term is lower than the order of the nonlinear terms. (This point was the one I wanted to think about more, since the scale on which the time and spatial derivatives change can cause u_t and u_xxx to be small - so, depending on the x scaling, the dispersive term could be quite small, but this isn't necessarily the case in the weakly nonlinear regime, I think). But I think that is essentially the point - in the weakly nonlinear regime, the order of the nonlinear terms in a PDE is smaller than the linear terms, and so the nonlinearity acts more like a perturbation to the linear equation. To answer the last question from that other thread, you assume a perturbation expansion like that because you expect the first few terms of the expansion to dominate the dynamics, with further terms just contributing higher order corrections. As I should in the other thread, the leading order term of the expansion obeyed the linearized KdV (assuming time and spatial derivates did not change the order of the derivative terms), which is obviously easier to solve than the nonlinear PDE. This was another thing I wanted to think about before replying - the way we've done things in the course I took this semester was to assume a Fourier series expansion for u(x,t) in which two or three of the modes dominated the dynamics, and the rest were negligible. I think the same sorts of things go for the kind of expansion you were using, but I wasn't quite certain so I had neglected to say anything about it before now. Recognitions: Science Advisor Quote by zoki85 Where does all that sudden interest in solitons come from? Earthquakes,tsunami,what? 2-3 years ago you could rarely even hear of a word "soliton". Now,solitons this solitons that! Actually, people have always been very interested in solitons. However, if you are reading the arxiv, eprints in the math section are often motivated by issues arising from better understanding the beauteous inverse scattering transform, or generalized symmetries and infinite sequences of conserved quantities, or connections (no pun intended) between Lax pairs and differential geometry. Some interesting equations currently touted in mathematical physics turn out to have the same exciting properties as the KdV and friends. There are some excellent review papers you can look for which discuss all these points. You might also see the lovely picture book by by E. Atlee Jackson, Perspectives of nonlinear dynamics, Cambridge University Press, 1989. Thread Tools \| \| \| \| \|-------------------------------------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: nonlinearity and dispersion in Kdv equation? \| \| \| \| Thread \| Forum \| Replies \| \| \| Differential Equations \| 3 \| \| \| Special & General Relativity \| 1 \| \| \| Introductory Physics Homework \| 0 \| \| \| Mechanical Engineering \| 1 \| \| \| Classical Physics \| 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": 7, "mathjax_display_tex": 7, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9527587294578552, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/112447/can-we-efficiently-compute-a-third-nash-equilibrium-given-two	## Can we efficiently compute a third Nash Equilibrium, given two? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) A finite, two-player, nondegenerate, symmetric game is defined by a nondegenerate $n \times n$ payoff matrix $A$. If player 1 plays strategy $i$ and player 2 plays strategy $j$, then player 1's payoff is $A_{ij}$ and player 2's payoff is $A_{ji}$. It is well known that the problem of computing a symmetric Nash Equillibrium in such a game is PPAD-complete (PPAD lies between P and NP but is probably intractable). Wilson's Oddness Theorem states that there are an odd number of symmetric Nash Equilibria in such games. This gives rise to my question. Suppose we have found two equilibria of $A$. Given these, what is the computational complexity of computing one more? Or, more generally - given $2k$ equilibria, what is the complexity of computing another? - 1 Relevant reading: public.iastate.edu/~riczw/MEGliter/Book/… (see especially around Corollary 3.7 on pdf p.83/775) – Benjamin Dickman Nov 15 at 4:54 You might get more answers on cstheory.stackexchange – Josh Nov 19 at 15:06	{"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": 8, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9097061157226562, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-statistics/22404-conditional-probability.html	# Thread: 1. ## conditional probability I'm having trouble solving a problem. I've been staring at it for almost an hour now, and have had hardly any progress... I'm trying to do problems 1 and 4 from here. Thanks in advance. 2. Hello, valosn! Here's #1 . . . 1. A factory produces items in boxes of two. Over the long run: . . (A): 92% of the boxes contain 0 defective items, . . (B): 5% of the boxes contain 1 defective item, and . . (C): 3% of the boxes contain 2 defective items. A box is picked at random from production, then an item is picked at random from that box. Given that the item is defective, what is the probability that the second item in the box is defective? We are concerned with two events: . . $1D$: the first item is defective, and . . $2D$: the second item is defective. We want: . ${\color{blue}P(2D\,\|\,1D) \;=\;\frac{P(1D \wedge 2D)}{P(1D)}}$ .[1] The numerator is: $P(1D \wedge 2D)$, the probability that both are defective ... case (C). . . We are told that this happens: $3\%$ of the time. Hence: . ${\color{blue}P(1D \wedge 2D) \:=\:\frac{3}{100}}$ The denominator is: $P(1D)$, the probability that the first is defective. This can happen in two ways: (i) Case (B) which happens $\frac{5}{100}$ of the time . . and we pick the defective item, probability $\frac{1}{2}$. Thus, this has probability: $\frac{5}{100}\cdot\frac{1}{2} \:=\:\frac{1}{40}$ (ii) Case (C) which happens $\frac{3}{100}$ of the time . . and we pick a defective item, probability $1.$ Thus, this has probability: $\frac{3}{100}(1) \:=\:\frac{3}{100}$ Hence: . $P(1D) \:=\:\frac{1}{40} + \frac{3}{100} \quad\Rightarrow\quad{\color{blue}P(1D) \:=\:\frac{11}{200}}$ Substitute into [1]: . $P(2D\,\|\,1D) \;=\;\frac{\dfrac{3}{100}}{\dfrac{11}{200}} \;=\;\boxed{\frac{6}{11}}$ 3. Here's 4 recall that we say two events are independent if knowledge that one occurs does not change the probability of the other occurring. (a) here, there are different numbers of each type of ball in both boxes. however, the numbers are proportional. no matter which box we select, the probability of selecting a black ball from that box is the same, and the probability of selecting a red ball from that box is the same. thus, yes, the color of the ball is independent of which box is chosen. verify this using the formulas for conditional probability and show that you get the same result for selecting, say, a black ball given that you have selected box 1, and then, the probability of selecting a black ball given that you have selected box 2. do this for the red ball also (b) the events are now dependent. knowledge of which box we've selected changes the probability of a black ball being chosen. verify this using the formulas. 4. Thanks a lot! It was the denominator that was giving me trouble, but you explained it very clearly. Thank you! As for the 4th problem, I was having trouble with part b), but I figured it out. Again, thanks a lot for the help! 5. I have that book in front of me. I reviewed an early edition for Springer. For #4, first find P(R). $\begin{array}{rcl}<br /> P(R) & = & P(R \cap 1) + P(R \cap 2) \\ <br /> & = & P(R\|1)P(1) + P(R\|2)P(2) \\ <br /> & = & \frac{2}{5}\frac{1}{2} + \frac{8}{{20}}\frac{1}{2} \\ <br /> & = & \frac{2}{5} \\ \end{array}$ Now we know that $P(R\|1) = \frac{2}{5}\frac{1}{2} = \frac{1}{5}\,\& \,P(R\|2) = \frac{8}{{20}}\frac{1}{2} = \frac{1}{5}$ Do you think they are independent?	{"extraction_info": {"found_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": 17, "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": 0.9503013491630554, "perplexity_flag": "middle"}
http://physics.aps.org/articles/large_image/f1/10.1103/Physics.1.39	Illustration: Alan Stonebraker Figure 1: By coupling a nonlinear system, such as an atom, to the electromagnetic field, it is possible to create Fock states (eigenstates of the harmonic oscillator). (Top) Brune et al. send atoms (left) into a cavity (center). The atoms are prepared with pulse $P1$ to be in a superposition of states $\|e〉$ and $\|g〉$ before they enter the cavity. The relative phase between these states, which is converted to probability amplitudes for $\|e〉$ and $\|g〉$ with pulse $P2$ when the atoms exit the cavity, depends on the number of photons in the cavity. (Bottom) In place of a cavity, Wang et al. create an electromagnetic field in a microwave resonator (blue). A superconducting qubit, acting as an artificial atom, couples to the center conductor.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9005295038223267, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/163486-help-determinants-eigenvalues-print.html	# Help with determinants and eigenvalues. Printable View • November 16th 2010, 04:31 PM pantsaregood 1 Attachment(s) Help with determinants and eigenvalues. Attachment 19739 For #1, I currently have that the two values are equal if A is either a matrix whose determinant is zero or a 1x1 matrix. I don't really have any direction for the others. I see that the determinant of a square matrix whose columns add to zero is 0, but I'm not sure of why. Can it somehow be reduced to the zero matrix? Edit: Okay, messed around with row reduction. Quickly found that all columns adding to zero forces at least one row to row-reduce to a zero row, forcing a determinant of zero. • November 16th 2010, 04:37 PM dwsmith det(cA)=c^ndet(a) where n is the number of rows so as long as you have an odd number of rows and the scalar is -1, then det(-A)=-det(A) • November 16th 2010, 04:45 PM pantsaregood I'm not following. -det(A) = det(-A) and det(-A) = -1^ndet(A) don't immediately seem to be equivalent statements. In that case, a 2x2 matrix would have the property det(A) = det(-A). Oh, okay. Nevermind. Odd number of rows. • November 16th 2010, 05:34 PM Drexel28 Quote: Originally Posted by pantsaregood I'm not following. -det(A) = det(-A) and det(-A) = -1^n*det(A) don't immediately seem to be equivalent statements. In that case, a 2x2 matrix would have the property det(A) = det(-A). Oh, okay. Nevermind. Odd number of rows. 2) If $c_1,\cdots,c_n$ are the columns it's asking for what $\det\left[\begin{array}{c\|c\|c}& & &\\ c_1 & \cdots & c_n\\ & & &\end{array}\right]$ given that $\displaystyle \sum_{j=1}^{n}c_j=\begin{bmatrix}0\\ \vdots\\ 0\end{bmatrix}$. Let me ask you this: what is the determinant of a matrix whose columns are linearly dependent equal? Moreover, how does that apply to this question? 3) This is pretty straightforward. I don't know how to give a decent hint that doesn't give it away. Merely note that $(AB)\bold{x}=A\left(B\bold{x}\right)$ and go from there. 4) If $\lambda$ is an eigenvalue for $A$ note that $\bold{0}=A^k\bold{x}=\lambda A^{k-1}\bold{x}=\cdots=\lambda^k\bold{x}$...so... All times are GMT -8. The time now is 10:53 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": 7, "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": 0.9178259372711182, "perplexity_flag": "middle"}
http://mathhelpforum.com/differential-geometry/170680-conclude-closure-union-union-closures-print.html	conclude that the closure of a union is the union of the closures. Printable View • February 9th 2011, 11:10 AM alice8675309 conclude that the closure of a union is the union of the closures. For this question, we are allowed to assume the following is true and has been proven: Let A,B $\subseteq$R and p $\in$R. Prove that p is an accumulation point of A $\cup$B if and only if p is an accumulation point of A or p is an accumulation point of B: (A $\cup$B)'=A' $\cup$ B'. Now since we assumed the above is true, Conclude that the closure of a union is the union of the closures: (A $\cup$ B)^cl=A^cl $\cup$ B^cl. How would I go about proving this? • February 9th 2011, 11:26 AM Plato The closure of a set is the ‘smallest’ closed that contains the set. We know that $A\subseteq\overline{A}~\&~ B\subseteq\overline{B}$. The union of two closed sets is a closed set. That forces $\overline{A\cup B}\subseteq\overline{A}\cup \overline{B}$. Why and how? Can you do the converse? • February 13th 2011, 06:50 PM alice8675309 After working on this for a little, is this proof correct? Assume p $\notin$A^cl and that p $\notin$ B^cl. That is, there exists $\epsilon$ $_1$>0 such that A has no point in (p- $\epsilon$ $_1$,p+ $\epsilon$_1) and there exists $\epsilon$ $_2$>0 such that B has no point in (p- $\epsilon$ $_2$,p+ $\epsilon$_2). Let $\epsilon$ $_3$=min{ $\epsilon$ $_1$, $\epsilon$ $_2$}, and clearly $\epsilon$ $_3$>0 and satisfies the requirement. Then Assume p $\in$A^cl $\cup$ B^cl. case 1: p $\in$A^cl which means that every neighborhood of p contains a point of A. Hence every neighborhood contains a point of A $\cup$B . case 2: p $\in$B^cl which means that every neighborhood of p contains a point of B. Hence every neighborhood contains a point of A $\cup$B . Thus, (A $\cup$B)^cl=A^cl $\cup$B^cl All times are GMT -8. The time now is 01:00 AM.	{"extraction_info": {"found_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": 37, "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": 0.938601553440094, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/56041/list	## Return to Answer 2 added 2 characters in body; added 2 characters in body; edited body The answer is YES, the volume of the moduli space is finite with respect to the Teichmuller metric. The reason is the theorem of Royden, that the Kobayashi metric on Teich(S) coincides with the Teichmuller metric, and the fact that the moduli space $M(S)$ associated to S has a nice compactification $\overline{M(S)}$, the Deligne-Mumford compactification. The argument goes as follows: For a stable curve Z in $\overline{M(S)}$ with k nodes, you can find a neighborhood U of Z such that U is locally $\Delta^n /G$, where $\Delta$ is the unit disc in $\mathbb{C}$ and $G$ is a finite group. Then $U\cap M(S)$ is locally isomorphic to $((\Delta^{})^{k} ((\Delta^{})^{k} \times \Delta^{n-k} )/ G$. The volume of $(\Delta^{})^{k} (\Delta^{})^{k} \times \Delta^{n-k}$ near the origin is finite in the Kobayashi metric. Since inclusion contracts inclusions contract the Kobayashi metric it follows that there is a small neighborhood V of $Z \in \overline{M(S)}$ such that volume of $V\cap M(S)$ is finite. The result now follows by compactness of $\overline{M(S)}$. You can look at Curt McMullen's paper : http://www.math.harvard.edu/~ctm/papers/home/text/papers/kahler/kahler.pdf for more details and references. (Proof of Theorem 8.1) 1 The answer is YES, the volume of the moduli space is finite with respect to the Teichmuller metric. The reason is the theorem of Royden, that the Kobayashi metric on Teich(S) coincides with the Teichmuller metric, and the fact that the moduli space $M(S)$ associated to S has a nice compactification $\overline{M(S)}$, the Deligne-Mumford compactification. The argument goes as follows: For a stable curve Z in $\overline{M(S)}$ with k nodes, you can find a neighborhood U of Z such that U is locally $\Delta^n /G$, where $\Delta$ is the unit disc in $\mathbb{C}$ and $G$ is a finite group. Then $U\cap M(S)$ is locally isomorphic to $((\Delta^{})^{k} \times \Delta^{n-k} )/ G$. The volume of $(\Delta^{})^{k} \times \Delta^{n-k}$ near the origin is finite in the Kobayashi metric. Since inclusion contracts the Kobayashi metric it follows that there is a small neighborhood V of $Z \in \overline{M(S)}$ such that volume of $V\cap M(S)$ is finite. The result now follows by compactness of $\overline{M(S)}$. You can look at Curt McMullen's paper : http://www.math.harvard.edu/~ctm/papers/home/text/papers/kahler/kahler.pdf for more details and references. (Proof of Theorem 8.1)	{"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": 26, "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": 0.9144170880317688, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/35939/why-is-the-anode-in-a-device-that-consumes-power-in-one-that-provides?answertab=active	# Why is the anode (+) in a device that consumes power & (-) in one that provides power? I was trying to figure out the flow of electrons in a battery connected to a circuit. Conventionally, current is from the (+) terminal to the (-) terminal of the battery. Realistically it flows the other way round; from the (-) terminal to the (+) terminal. My question is, assuming electron flow is from the (-) terminal, would the battery's cathode be located at the (+) terminal and it's anode at the (-) terminal or would it be vice versa? One other question? Why would the anode positive in a device that consumes power and negative in a device that provides power? - Because Benjamin Franklin screwed up. :) – user404153 Sep 8 '12 at 19:10 ## 1 Answer Electric current is the rate of flow of electric charges across any cross-sectional area of a conductor. The direction of electric current is taken as the direction of flow of positive ions or opposite to the direction of flow of free electrons. Your assumption is not necessary here... Electrons always flow from negative terminal to positive terminal. $$i=\frac{dq}{dt}$$ When current flows through an electrolytic solution or during the process of electrolysis, The plate towards which positive ions (cations) flow is called the cathode and the plate towards which negative ions (anions) flow is called the anode. Wikipedia says clearly, In an electrochemical cell, The electrode at which electrons leave the cell and oxidation occurs is called anode and the electrode at which electrons enter the cell and reduction occurs is called cathode. Each electrode may become either the anode or the cathode depending on the direction of current through the cell. A bipolar electrode is an electrode that functions as the anode of one cell and the cathode of another 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": 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": 0.9469240307807922, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-statistics/68572-any-one-good-standard-deviations-please-help.html	# Thread: 1. ## Any one good with standard deviations please help A sample of 15 observations has a standard deviation of 4. The sum of the squared deviations from the sample mean is? 1.) 19 2.)56 3.)60 4.) 224 5.)240 I'm stuck on this problem is you can please help me I would much appreciate it, thanx! 2. $s_x = \sqrt{\frac{1}{n-1} \sum_i (x_i - x_{\text{mean}})^2}$ $4 = \sqrt{\frac{1}{15-1} \sum_i (x_i - x_{\text{mean}})^2}$ $16 = \frac{1}{14} \sum_i (x_i - x_{\text{mean}})^2$ $16\cdot 14 = 224 = \sum_i (x_i - x_{\text{mean}})^2$ 3. Originally Posted by vincisonfire $s_x = \sqrt{\frac{1}{n-1} \sum_i (x_i - x_{\text{mean}})^2}$ $4 = \sqrt{\frac{1}{15-1} \sum_i (x_i - x_{\text{mean}})^2}$ $16 = \frac{1}{14} \sum_i (x_i - x_{\text{mean}})^2$ $16\cdot 14 = 224 = \sum_i (x_i - x_{\text{mean}})^2$ The sample SD uses a divisor of N not N-1, the latter is the usual estimator for the population SD. So this changes your post to: $s_x = \sqrt{\frac{1}{n} \sum_i (x_i - x_{\text{mean}})^2}$ $4 = \sqrt{\frac{1}{15} \sum_i (x_i - x_{\text{mean}})^2}$ $16 = \frac{1}{15} \sum_i (x_i - x_{\text{mean}})^2$ $16\times 15 = 240 = \sum_i (x_i - x_{\text{mean}})^2$ 4. For a sample you must use a divisor of N-1 to have an unbiased estimation of the real standard deviation. For a population you must use a divisor of N to get the real standard deviation. http://en.wikipedia.org/wiki/Standar...dard_deviation 5. Originally Posted by vincisonfire For a sample you must use a divisor of N-1 to have an unbiased estimation of the real standard deviation. For a population you must use a divisor of N to get the real standard deviation. Standard deviation - Wikipedia, the free encyclopedia Does the question ask for an estimate of a population variance based on the sample variance .....? That would be a completely different question and using N - 1 as the divisor would be appropriate. When calculating a sample variance in the context of plain and simple univariate statistics, a divisor of N is used. The wikipedia link says this. 6. Isn't it always the case that we are trying to figure out what the standard deviation of the population is? I may be wrong but I've never seen a sample variance written with a divisor of N. 7. Originally Posted by vincisonfire Isn't it always the case that we are trying to figure out what the standard deviation of the population is? [snip] No. Eg. You might have data from a test done by a class of 20 students and you're interested in the mean and the standard deviation ...... A key quote from the Wiki link: "Note that if the above data set represented only a sample from a greater population, a modified standard deviation would be calculated (explained below) to estimate the population standard deviation, which would give 6.93 for this example." 8. I understand what you mean. But as I learned it, in your example, the 20 students would be a population. 9. Originally Posted by vincisonfire I understand what you mean. But as I learned it, in your example, the 20 students would be a population. So then you're not estimating the variance of a population from a sample, you're actually getting the 'exact' varience of the population.	{"extraction_info": {"found_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": 0.3, "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": 0.9230148196220398, "perplexity_flag": "head"}
http://physics.aps.org/articles/v6/29	# Viewpoint: Dissipative Stopwatches , Institute of Photonic Sciences, Av. Carl Friedrich Gauss, 3, 08860 Castelldefels, Barcelona, Spain Published March 11, 2013 \| Physics 6, 29 (2013) \| DOI: 10.1103/Physics.6.29 Contact with the environment usually destroys the delicate operations of a quantum computer, but engineered dissipative processes may allow for a more robust preparation and processing of quantum states. #### Precisely Timing Dissipative Quantum Information Processing M. J. Kastoryano, M. M. Wolf, and J. Eisert Published March 11, 2013 \| PDF (free) When riffle-shuffling a deck of $52$ cards, seven shuffles are necessary to arrive at a distribution of playing cards that is, to a large degree, independent of the initial ordering. The fact that initial correlations survive if the deck is only shuffled a few times and disappear suddenly after seven shuffles is well known by magicians, who use this phenomenon in card tricks to amaze their audience. In a paper in Physical Review Letters [1], Michael Kastoryano of the Free University of Berlin, Germany, and colleagues show how this effect can be leveraged for quantum information processing. Quantum information science uses phenomena such as superpositions and entanglement to devise quantum devices capable of performing tasks that cannot be achieved classically. These applications are typically based on unitary dynamics (that is, time evolutions that are governed by the Schrödinger equation). One big practical problem hindering the operation of such devices in the quantum regime is dissipation caused by the interaction of the system with its environment. In the last several years, a new approach to quantum information processing has led to a rethinking of the traditional concepts that rely on unitary dynamics alone and avoid dissipation unconditionally: instead, these new protocols harness dissipative processes for quantum information science. Actively using dissipation in a controlled way opens up interesting new possibilities and has important advantages: dissipative protocols are robust and, as explained below, allow one to prepare a desired quantum state, irrespective of the initial state of the system. However, the underlying processes are intrinsically probabilistic and time independent. In general, it is therefore not clear how to incorporate them in the existing framework of unitary quantum information processing. One route around this difficulty is to embrace the probabilistic and time-independent nature of these processes and use specifically designed dissipative architectures (see, for example, Ref. [2]), but so far there are very few, and these schemes are conceptually very different from unitary protocols. Protocols based on unitary dynamics typically require precise timing and operations, which are conditioned on previous ones. The work by Kastoryano et al. shows how dissipative processes can be timed and used in a conventional way without losing the specific advantages of dissipative schemes [1]. The unitary time evolution (Fig. 1) of a pure quantum state $\|ψ〉$ under a Hamiltonian $Ĥ$ is governed by the Schrödinger equation $iħ\|Ψ·(t)〉=Ĥ\|Ψ(t)〉$. Accordingly, a unitary time evolution $U(t)=e-iĤtħ$ transforms a pure state $\|Ψin〉$ always into another pure state $\|Ψout〉=U(t)\|Ψin〉$. Consider, for example, a system with spin states $\|↑〉$ and $\|↓〉$. The Hamiltonian $Ĥ=ħκ(\|g〉〈e\|+\|g〉〈e\|)$ causes the spin to flip. If acting for a time $t=π/2κ$, it transforms $\|↑〉$ into $\|↓〉$ and $\|↓〉$ into $\|↑〉$. Dissipative processes, in contrast, can turn a pure state into a mixed one that is described by a density matrix $ρ$, representing a statistical mixture. A dissipative time evolution is governed by a master equation (i.e., an equation of motion for the reduced density operator of a subsystem that interacts with an environment; the dynamics of the system is obtained by tracing out the degrees of freedom of the environment). Here we consider Markov processes (that is, “memoryless” ones) that are described by a time-independent Liouvillian master equation $ρ·=L(ρ)$ with $L(ρ)=Γ(2âρâ+-â+âρ-ρâ+â$), with jump operator $â$ and rate $Γ$. It is instructive to consider, for example, the jump operator $â=\|↓〉〈↑\|$. The corresponding master equation causes a system in state $\|↑〉$ to relax to $\|↓〉$ at a rate $Γ$. A dissipative process of this type can be understood as applying the jump operator $â$ with certain probability $ρ$, which is determined by $Γ$. If we prepare our system in state $\|↑〉$ and wait a short while, we don’t know whether there has already been a quantum jump $\|↑〉→\|↓$⟩ or not, hence the resulting quantum state is a mixed one $ρmixed=p\|↓〉〈↓\|+(p-1)\|↑〉〈↑\|$. (The steady state can be still a pure one, $\|↓〉$ in this example.) Because of this intrinsically probabilistic feature, dissipative processes are difficult to time. Kastoryano and colleagues develop new tools that allow one to use dissipative processes in such a way that the desired transitions occur at very well-defined points in time (Fig. 2). The key to dissipative protocols is to tailor the interaction between the system and a bath such that a specific desired jump operator $â$ is realized. This jump operator is chosen such that the target state $ρfin$ is the unique steady state of the dissipative evolution $L(ρfin)=0$. For dissipative quantum computing [3], the result of the calculation is encoded in $ρfin$, and if the goal is quantum state engineering, $ρfin$ can be, for example, an entangled [4] or topological state [5]. This state is reached regardless of the initial state of the system. Should the system be disturbed, the dissipative dynamics will bring it back to the steady state $ρfin$. This is impossible for unitary dynamics. In the unitary case, imperfect initialization inevitably leads to deviations from the desired final state. Therefore “reliable state preparation,” the second of the five criteria that DiVincenzo established for a scalable quantum computer [6], was for a long time considered a fundamental requirement for quantum information processing. Kastoryano et al. extend the toolbox of dissipative quantum information processing by introducing devices for timing this type of process exactly. More specifically, they introduce schemes for (i) preparing a quantum state during a specified time window and (ii) for triggering dissipative operations at specific points in time. The authors use these tools for demonstrating a dissipative version of a one-way quantum computation scheme [7]. The central ingredient that is used here is a very interesting mechanism called the “cutoff phenomenon.” Stochastic processes that have this cutoff quality exhibit a sharp transition in convergence to stationarity. They do not converge smoothly to the stationary distribution during a certain period (if the initial state is far away from the stationary state) but instead converge abruptly (exponentially fast in the system size) at a specific point in time. This behavior was first recognized in classical systems [8]. An intriguing classical example is the shuffling of playing cards outlined above. This type of mechanism has now been investigated in the quantum setting. In earlier work [9] by some of the authors of the new paper, the cutoff phenomenon is introduced for quantum Markov processes using the notions of quantum information theory. This provides a quantitative tool to study the convergence of quantum dissipative processes. In Ref. [1], Kastoryano et al. employ this quantum-version of the cutoff phenomenon to start and end dissipative processes at specific points in time, thus allowing for the integration in the framework of regular quantum information architectures. In future, these kinds of tools might also become important for new dissipative quantum error correcting schemes and may point towards new insights regarding passive error protection. Quantum reservoir engineering is a young and rapidly growing area of research. Several protocols have been developed including dissipative schemes for quantum computing [3], quantum state engineering [10], quantum repeaters [2], error correction [11], and quantum memories [12]. Dissipative schemes for quantum simulation [13] and entanglement generation [4] have already been experimentally realized. Actively using dissipative processes is a conceptually interesting direction with important practical advantages. The results presented in Ref. [1] add new tools for exploiting and engineering dissipative processes and provide the linking element for incorporating dissipative methods into the regular framework of quantum information processing. ### References 1. M. J. Kastoryano, M. M. Wolf, and J. Eisert, “Precisely Timing Dissipative Quantum Information Processing,” Phys. Rev. Lett. 110, 110501 (2013). 2. K. G. H. Vollbrecht, C. A. Muschik, and J. I. Cirac, “Entanglement Distillation by Dissipation and Continuous Quantum Repeaters,” Phys. Rev. Lett. 107, 120502 (2011). 3. F. Vertraete, M. Wolf, and J. I. Cirac, “Quantum Computation and Quantum-State Engineering Driven by Dissipation,” Nature Phys. 5, 633 (2009). 4. H. Krauter, C. A. Muschik, K. Jensen, W. Wasilewski, J. M. Petersen, J. Ignacio Cirac, and E. S. Polzik, “Entanglement Generated by Dissipation and Steady State Entanglement of Two Macroscopic Objects,” Phys. Rev. Lett. 107, 080503 (2011). 5. S. Diehl, E. Rico, M. A. Baranov, and P. Zoller, “Topology by Dissipation in Atomic Quantum Wires,” Nature Phys. 7, 971 (2011). 6. D. P. DiVincenzo, “The Physical Implementation of Quantum Computation,” Fortschr. Phys. 48, 771 (2000). 7. R. Raussendorf and H. J. Briegel, “A One-Way Quantum Computer,” Phys. Rev. Lett. 86, 5188 (2003). 8. P. Diaconis, “The Cutoff Phenomenon in Finite Markov Chains,” Proc. Natl. Acad. Sci. U.S.A. 93, 1659 (1996). 9. M. J. Kastoryano, D. Reeb, and M. M. Wolf, “A Cutoff Phenomenon for Quantum Markov Chains,” J. Phys. A 45, 075308 (2012). 10. S. Diehl, A. Micheli, A. Kantian, B. Kraus, H. P. Büchler, and P. Zoller, “Quantum States and Phases in Driven Open Quantum Systems with Cold Atoms,” Nature Phys. 4, 878 (2008). 11. J. Kerckhoff, H. I. Nurdin, D. S. Pavlichin, and H. Mabuchi, “Designing Quantum Memories with Embedded Control: Photonic Circuits for Autonomous Quantum Error Correction,” Phys. Rev. Lett. 105, 040502 (2010); J. P. Paz and W. H. Zurek, “Continuous Error Correction,” Proc. R. Soc. London A 454, 355 (1998). 12. F. Pastawski, L. Clemente, and J. Ignacio Cirac, “Quantum Memories Based on Engineered Dissipation,” Phys. Rev. A 83, 012304 (2011). 13. J. T. Barreiro, M. Müller, P. Schindler, D. Nigg, T. Monz, M. Chwalla, M. Hennrich, C. F. Roos, P. Zoller, and R. Blatt, “An Open-System Quantum Simulator with Trapped Ions,” Nature 470, 486 (2011). ### About the Author: Christine Muschik Christine Muschik is an Alexander von Humboldt postdoctoral fellow at the Institute of Photonic Sciences (ICFO) in Barcelona, where she joined the quantum optics theory group led by Maciej Lewenstein. She received her Ph.D. at the Max Planck Institute for Quantum Optics, where she worked with Ignacio Cirac in close collaboration with Eugene Polzik at the Niels Bohr Institute. Her thesis on quantum information processing with atoms and photons was completed within the international Ph.D. program of excellence: QCCC (Quantum Computing, Control and Communication) supported by the ENB (Elite Network of Bavaria). Her current research interests are in theoretical quantum optics and quantum nanophotonics. ## Related Articles ### More Quantum Information Shedding Light on a Quantum Black Box Viewpoint \| Mar 25, 2013 Big Shifts on an Atomic Scale Synopsis \| Mar 14, 2013 ## New in Physics Wireless Power for Tiny Medical Devices Focus \| May 17, 2013 Pool of Candidate Spin Liquids Grows Synopsis \| May 16, 2013 Condensate in a Can Synopsis \| May 16, 2013 Nanostructures Put a Spin on Light Synopsis \| May 16, 2013 Fire in a Quantum Mechanical Forest Viewpoint \| May 13, 2013 Insulating Magnets Control Neighbor’s Conduction Viewpoint \| May 13, 2013 Invisibility Cloak for Heat Focus \| May 10, 2013 Desirable Defects Synopsis \| May 10, 2013	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 50, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.8805779814720154, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/34082/meaning-of-the-first-and-second-laws-of-the-photoelectric-effect?answertab=active	# Meaning of the first and second laws of the photoelectric effect I was reading the introduction to quantum mechanics in my physics book and it begins with a discussion of the photoelectric effect and energy quantas. The first law, the one that says that the intensity of the photoelectric saturation current is directly proportional to the flux of incident EM radiation, means that the more radiation will hit the plate the higher the intensity of the current produced (via displacement of electrons), right? And the second one, the one that says the kinetic energy of the emitted photoelectrons varies linearly with the frequency of incident EM radiation, and does not depend on the flux, means that the further up the EM spectrum the radiation is, the higher the voltage (electric potential) produced? - ## 1 Answer In the photoelectric effect one photon displaces one electron, so the way to understand it is to consider the properties of the photons. If you take light of a fixed frequency, $\nu$, then the energy of the photons is fixed at $h\nu$. That means the intensity of light is proportional to the number of photons, and because one photon = one displaced electron, the intensity of light is proportional to the number of photoelectrons. When a photon ejects a photoelectron the energy of the electron is equal to the energy of the photon, $h\nu$, minus the work function, $\phi$. This explains why the energy of the electrons, $E$, increases with frequency of the light, $\nu$. $$E = h\nu - \phi$$ In the last part of your question you say: means that the further up the EM spectrum the radiation is, the higher the voltage (electric potential) produced? The photoelectric effect ejects electrons, so whatever you are illuminating will get a positive charge and therefore a potential difference relative to ground. However the potential difference is just related to the total charge, i.e. the number of electrons ejected, so it isn't affected by the energy of the photoelectrons. - Thanks, I understand now. – andreas.vitikan Aug 14 '12 at 8:28 @Rennie What do you actually mean by "relative to ground"? Isn't voltage maintained by a battery (to supply the electrodes)? I've seen many experiments that don't use grounds... Are there any using grounds? I can't understand the 2nd line of your last paragraph... – Ϛѓăʑɏ βµԂԃϔ Aug 15 '12 at 3:05 The word "ground" generally just means the environment. If we start with no potential difference between the metal and it's environment, then eject some photoelectrons, the metal will become positively charged relative to it's environment and therefore there will be a potential difference. – John Rennie Aug 15 '12 at 6:15	{"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": 6, "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": 0.9227750301361084, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/87215/complicated-functional-integral	# Complicated “functional integral” I came across the following "functional" at work: $$\Pi [b]=\iint_0^{\lambda b(v,\lambda)} vf(v,\lambda) \; dv \; d\lambda$$ it's part of an optimization problem that tries to find $b$, subject to some constraints on $b$. I'm not familiar with that type of integral, where the solution function is actually in one of the bounds of the integral. Is there a specific name for that type of integral? Would the calculus of variations address that type of optimization problem? Or is there a field of functional analysis (calculus?) that would address it? Thanks! - Are you integrating with respect to $v$ or $\lambda$? Surely it can't be with respect to both? – Bruno Dec 1 '11 at 3:51 Yeah, I'm afraid it's with respect to both... But you are right, I need to fix the formula :-) Thanks! - the second integral is from 0 to +infinity. – Frank Dec 1 '11 at 4:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9526829123497009, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/72877/brownian-motion-and-fourier-series?answertab=oldest	# Brownian motion and Fourier series Let $(B_t)_{t \in [0, \infty)}$ be a Brownian motion. Can you prove me why it can be written as $$B_t= Z_0 \cdot t + \sum_{k=1}^{\infty} Z_k \frac{\sqrt{2} \cdot \sin(k \pi t)}{k \pi}$$ for some independent standard normal random variables $Z_0, Z_1,...$? - ## 1 Answer That's Karhunen-Loève decomposition of Gaussian process. Check wiki -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 0.8195779919624329, "perplexity_flag": "middle"}
http://mathematica.stackexchange.com/questions/tagged/inverse+replacement	# Tagged Questions 3answers 305 views ### Changing variables algebraically Suppose one has two functions, $y(x)$ and $z(x)$, and one seeks to obtain $y(z)$ by substituting $x(z)$ into $y(x)$. Can this be done in a single step? Or must $z(x)$ first be inverted independently? ...	{"extraction_info": {"found_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": 0.9344944953918457, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/102403/generating-function-identity-from-number-of-irreducible-monic-polynomials-in-m	Generating function identity from number of irreducible monic polynomials in $\mathbf{GF}(q)$. I denote by $m_n(q)$ the number of irreducible monic polynomials with degree $n$ over the finite field of order $q$. So the number of monic polynomials with degree is just $q^n$. From this, how does it follow that $$\prod_{n\geq 1}\frac{1}{(1-x^n)^{m_n(q)}}=\frac{1}{1-qx}.$$ I know $$\frac{1}{1-qx}=1+qx+q^2x^2+\cdots$$ but I don't see an obvious relation between the coefficients of one to the other. Thanks, 2nd try: Based on the help, I try to write this out in form $$(1+x+x^2+\cdots)^{m_1(q)}(1+x^2+x^4+c\dots)^{m_2(q)}\cdots=\sum_{f\text{ monic}}x^{\deg f}.$$ Due to unique factorization, distinct products of monic irreducibles of determine unique monic polynomials. So the coefficient of $x^{\deg f}$ is the number of monic polynomials of degree $\deg f$, But I'm having a hard time picturing how the expanded left hand side would count them for the identity to hold. Would someone be willing to explain this subtlety? - 1 Hint: Imagine expanding each term on the left, just like you did for $1/(1-qx)$. Then imagine multiplying out the expanded versions. For every $N$, you get a term in $x^N$ precisely for every product of monic irreducibles that have sum of degrees equal to $N$. It is a close analogue of the corresponding formula for the integers, and as for integers it uses unique factorization. – André Nicolas Jan 25 '12 at 19:25 Thanks @AndréNicolas. I tried to get a little further based on your hint. If you find the time, do you mind explaining how your method works? So far I'm not able to imagine how the expanded version multiply out correctly to get the identity. – Hailie Jan 29 '12 at 8:52 1 Answer $q^n$ is the number of monic polynomials of degree $n$ over $\mathbb{F}_q$, so we can write $\frac{1}{1 - qx}$ as $$\sum_{f \text{ a monic polynomial}} x^{\deg f}.$$ Now, monic polynomials can be uniquely factored into a product of irreducible polynomials, and $\deg$ satisfies $\deg fg = \deg f + \deg g$. What does that tell you about how the above generating function can be factored? (This is closely analogous to how we derive the Euler product for the Riemann zeta function.) - Thank you for answering Qiaochu Yuan. I tried to write down my thoughts based on your hint. If you find the time, could you expand on what you're getting at? If not, that's fine, thanks. – Hailie Jan 29 '12 at 7:00 @Hailie: are you familiar with the Euler product for the Riemann zeta function? It is very similar. – Qiaochu Yuan Jan 29 '12 at 7:14 – Hailie Jan 29 '12 at 7:26 @Hailie: you can't exactly identify them. The point is that $n \mapsto \frac{1}{n^s}$ maps is a multiplicative function in the same way that $f \mapsto x^{\deg f}$ is. Does that give you any ideas? – Qiaochu Yuan Jan 29 '12 at 7:34 Well, I see that $gf\mapsto x^{\deg fg}=x^{\deg f}x^{\deg g}$, but I don't see how this argument works. To me, $\sum_{f\text{ monic}}x^{\deg f}$ seems like it would just have form $1+x+x^2+x^3+\cdots$. I don't see any coefficients for the $x^{\deg f}$ monomials in the sum. – Hailie Jan 29 '12 at 7:49 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": 23, "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": 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": 0.9253811836242676, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/85625/how-can-i-prove-that-fx-frac11x-cdot-leftx-right-is-not-bounded-abo/85627	# How can I prove that $f(x):=\frac{1}{1+x\cdot\left\|x\right\|}$ is not bounded above? Let $f:\left]-1,\infty\right[ \to \mathbb{R}$ be defined by $$f(x) := \frac{1}{1+x \cdot \|x\|} .$$ I want to prove that $f$ is not bounded above. Here is my attempt: I assume that $f$ is upper bound by $k$: $$\exists k \gt 0 : f(x) \le k \qquad x \in \left]-1, \infty\right[$$ $$\Rightarrow \qquad \frac{1}{1+x\cdot\left\|x\right\|} \le k .$$ But at this point I don't know how to proceed. I can't find the contradiction. - I believe what you are trying to show is that the function $f(x) = \frac{1}{1 + x\|x\|}$ has no upper bound on the interval $(-1,\infty)$ (and yes I hate the notation $]a,b[$, despite how overloaded the parentheses notation is!). Anyways, is that what you are trying to show? – JavaMan Nov 25 '11 at 20:46 Could you be a bit more precise about what it is you're trying to prove, please? You want your $f$ to be an upper bound for what? – Henning Makholm Nov 25 '11 at 20:46 1 Graph the function, or if that is difficult have software do it for you. You will find that $f(x)$ is awfully big just to the right of $x=-1$. You should be able to translate that intuitive knowledge into a proof of the level of formality appropriate for your course. – André Nicolas Nov 25 '11 at 20:52 @JavaMan : Yes. – mcb Nov 25 '11 at 20:57 @André Nicolas: I think I have to prove it by finding a contradiction. – mcb Nov 25 '11 at 20:59 show 1 more comment ## 2 Answers If $x<0$, then $$\frac{1}{1+x\|x\|}=\frac{1}{1-x^2}.$$ You can make $\frac{1}{1-x^2}$ as large as you wish by selecting $x$ to be sufficiently close to $-1$. - I think I have to prove this by pointing out a contradiction. – mcb Nov 25 '11 at 21:12 @mcb: Is this a specific assignment for a class demanding that you use contradiction? With my formula above you can explicitly find an $x$ such that $f(x)$ is larger than $M$ for any $M>0$. – Joe Johnson 126 Nov 26 '11 at 2:01 We show that $f$ has no upper bound on $(-1,0)$, because we work with negative values you can simplify your equation to $$f(x)=\frac{1}{1-x^2}$$ assume that $f$ is bounded by some $k > 2$ then we have that $$-\frac{\sqrt{k}}{\sqrt{k+1}} \in (-1,0)$$ but $f(-\frac{\sqrt{k}}{\sqrt{k+1}})=k+1$ a contradiction. - How did you know it would be between -1 and 0? – mcb Nov 25 '11 at 21:43 2 Certainly its negative and certainly the denominator is bigger than the nominator. – Listing Nov 25 '11 at 21:47	{"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": 24, "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": 0.9530030488967896, "perplexity_flag": "head"}
http://mathhelpforum.com/discrete-math/198540-problem-graph-theory-algorithms.html	# Thread: 1. ## A problem on Graph theory and algorithms The problem is as follows: --------------------------------------------------------------------------------------------------------------------------------- Let G be a connected graph. For a vertex x of G we denote by G−x the graph formed by removing x and all edges incident on x from G. G is said to be good if there are at least two distinct vertices x,y in G such that both G − x and G − y are connected. (i) Show that for any subgraph H of G, H is good if and only if G is good. (ii) Given a good graph, devise a linear time algorithm to find two such vertices. (iii) Show that there exists a graph G such that we cannot find three distinct vertices u1,u2,u3such that each of G − u1,G − u2, and G − u3 is connected. --------------------------------------------------------------------------------------------------------------------------------- I have approached to some extent. Here is what I've done: If G is a connected graph then we can easily find its spanning tree and then removing any two leaves x and y from the tree one at a time gives us 2 another graphs G − x and G − y such that they are both connected. "Hence any connected graph G is a good graph." 1. I'm stuck on how to solve part (i). 2. I have designed the algorithm for the part(ii) which extracts the spanning tree from G and then finds two such leaves x and y. Clearly it is linear in terms of #vertices(=n) & #edges(=m) i.e. O(n+m) 3. I have also solved the third part with a graph as follows: O------------------O-----------------O In the above graph G we denote three vertices as u1, u2, u3. The edges are (u1,u2) and (u2,u3). Clearly G − u2 is not connected. Thanks in advance!! 2. ## Re: A problem on Graph theory and algorithms Show that for any subgraph H of G, H is good if and only if G is good. As written this claim is false. (Unless you define subgraph differently than what I am assuming) Counter: Let G be $P_5$ (a path on 5 vertices). Then clearly G is good. Let H be a subgraph created by removing a single vertex of degree 2 in G. Then H is disconnected and cannot be good. Hence the claim fails. This is true if you say: for any connected subgraph H of G, H is good if and only if G is good. But you already proved a stronger statement than this, that all connected graphs are good. 3. ## Re: A problem on Graph theory and algorithms Yes you are right...anyways thanks for pointing the counter eg ... 4. ## Re: A problem on Graph theory and algorithms How to prove that Hamming graph H(d,3) has a diameter d?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.9429603219032288, "perplexity_flag": "middle"}
http://mathforum.org/mathimages/index.php?title=Prime_spiral_(Ulam_spiral)&diff=32702&oldid=16118	Prime spiral (Ulam spiral) From Math Images (Difference between revisions) \| \| \| \| \| \|----------------------------------------\|---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\|------------------------------------------------------\|---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\| \| \| \| Current revision (15:53, 25 June 2012) (edit) (undo) \| \| \| (One intermediate revision not shown.) \| \| \| \| \| Line 1: \| \| Line 1: \| \| \| \| {{Image Description Ready \| \| {{Image Description Ready \| \| - \| \|ImageName=Iris \| + \| \|ImageName=Ulam Spiral \| \| \| \|Image=Ulam_spiral.png \| \| \|Image=Ulam_spiral.png \| \| \| \|ImageIntro=The Ulam spiral, or prime spiral, is a plot in which <balloon title="load:defprime">prime numbers</balloon><span id="defprime" style="display:none">A prime number is a natural number greater that is divisible only by 1 and itself. The first few prime numbers are :<math>2, 3, 5, 7, 11, 13,17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, \dots</math></span> are marked among positive integers that are arranged in a counterclockwise spiral. The prime numbers show a pattern of diagonal lines. \| \| \|ImageIntro=The Ulam spiral, or prime spiral, is a plot in which <balloon title="load:defprime">prime numbers</balloon><span id="defprime" style="display:none">A prime number is a natural number greater that is divisible only by 1 and itself. The first few prime numbers are :<math>2, 3, 5, 7, 11, 13,17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, \dots</math></span> are marked among positive integers that are arranged in a counterclockwise spiral. The prime numbers show a pattern of diagonal lines. \| \| Line 205: \| \| Line 205: \| \| \| \| Not much is discovered about the Ulam spiral. For instance, the reason for the diagonal alignment of prime numbers or the vertical and horizontal arrangement of non-prime numbers is not clear yet. Indeed, Ulam spiral is not heavily studied by mathematicians. However, Ulam spiral's importance lies on the fact that it shows a clear pattern among prime numbers. \| \| Not much is discovered about the Ulam spiral. For instance, the reason for the diagonal alignment of prime numbers or the vertical and horizontal arrangement of non-prime numbers is not clear yet. Indeed, Ulam spiral is not heavily studied by mathematicians. However, Ulam spiral's importance lies on the fact that it shows a clear pattern among prime numbers. \| \| \| \| \| \| \| - \| Some might suspect that we are seeing diagonal lines in the Ulam spiral because the human eye seeks patterns and groups even among random cluster of dots. However, we can compare [[#16\|Image 16]] and [[#17\|Image 17]] and see that the prime numbers actually have a distinct pattern of diagonal lines that random numbers do not have. [[#16\|Image 17]] is a Ulam spiral where the black dots denote for the prime numbers, and [[#17\|Image 17]] is a Ulam spiral of random numbers. \| + \| Some might suspect that we are seeing diagonal lines in the Ulam spiral because the human eye seeks patterns and groups even among random cluster of dots. However, we can compare [[#16\|Image 16]] and [[#17\|Image 17]] and see that the prime numbers actually have a distinct pattern of diagonal lines that random numbers do not have. [[#16\|Image 16]] is a Ulam spiral where the black dots denote for the prime numbers, and [[#17\|Image 17]] is a Ulam spiral of random numbers. \| \| \| \| \| \| \| \| {{Anchor\|Reference=17\|Link=[[Image:randomintegers.jpg\|Image 17\|thumb\|300px]]}} \| \| {{Anchor\|Reference=17\|Link=[[Image:randomintegers.jpg\|Image 17\|thumb\|300px]]}} \| Current revision Ulam Spiral Field: Number Theory Image Created By: en.wikipedia Ulam Spiral The Ulam spiral, or prime spiral, is a plot in which prime numbersA prime number is a natural number greater that is divisible only by 1 and itself. The first few prime numbers are :$2, 3, 5, 7, 11, 13,17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, \dots$ are marked among positive integers that are arranged in a counterclockwise spiral. The prime numbers show a pattern of diagonal lines. Basic Description Image 1 Image 2 The prime spiral was discovered by Stanislaw Ulam (1909-1984) in 1963 while he was doodling on a piece of paper during a science meeting. Starting with 1 in the middle, he wrote positive numbers in a grid as he spiraled out from the center, as shown in Image 1. He then circled the prime numbers, and the prime numbers showed patterns of diagonal lines as shown by the grid in Image 2. The grid in Image 2 is a close-up view of the center of the main image such that the green line segments and red boxes in the center of the main image line up with those in the grid. A larger Ulam spiral with 160,000 integers and 14,683 primes is shown in the main image. Black dots indicate prime numbers. In addition to diagonal line segments formed by the black dots, we can see white vertical and horizontal line segments that cross the center of the spiral and do not contain any black dots, or prime numbers. There are also white diagonal line segments that do not contain any prime numbers. Ulam spiral implies that there is some order in the distribution of prime numbers. A More Mathematical Explanation [Click to view A More Mathematical Explanation] From time immemorial, humans have tried to discover patterns among prime numbers. Currently, there is [...] [Click to hide A More Mathematical Explanation] From time immemorial, humans have tried to discover patterns among prime numbers. Currently, there is no known simple formula that yields all the primes. The diagonal patterns in the Ulam spiral gives some hint for formulas of primes numbers. Definition of Half-lines Many half-lines in the Ulam spiral can be described using quadratic polynomials A quadratic polynomial is a polynomial in which the exponent of the variables do not exceed 2. A quadratic polynomial typically has the form $ax^2+bx+c$ where $x$ is a variable and $a, b, c$ are coefficients.. A half-line is a line which starts at a point and continues infinitely in one direction. In this page, we only consider half-lines that are horizontal or vertical, or have a slope of -1 or +1. The half-lines that can be described using quadratic polynomials are the ones in which each entry of the diagonal is positioned on a different ring of the spiral. The ring of a spiral can be considered as the outermost layer of a concentric square or a rectangle centered around the center of the spiral, 1, as defined by the blue line in the grid. Although there is no exact place where we can determine the beginning and end of a ring, we will assume that entries that are positioned on squares or rectangles of the same sizes to be on the same ring. For instance, 24, 25, 26, 27, 28, which are shown in red boxes in Image 3, are on the same ring, whereas 48, 49, 50, 51, 52, as shown in blue boxes, are on a different ring because they are positioned on a bigger square. Image 3:Diagonal half-lines The green lines in the Image 3 qualify as diagonal half-lines whereas the red lines do not because the red lines cross the corner of the grid in a way that two entries of the red diagonal line are positioned on the same ring. Thus, diagonal line segments that are composed of prime numbers can also be expressed as outputs of quadratic polynomials. To learn more about the relation between the diagonal lines and quadratic polynomials, click below. Half-lines and quadratic polynomials First, we need to know the relation between quadratic polynomials and the difference table of sequences. Le [...] Half-lines and quadratic polynomials First, we need to know the relation between quadratic polynomials and the difference tableThe difference table lists the terms of a sequence in one row, and the differences between consecutive terms in the next row. It can include other higher order differences such as the second-differences and the third-differences. of sequences. Let's choose the diagonal $5, 19, 41, 71, 109, 155$, the diagonal indicated with green dotted line in Image 3, as our original sequence and create a difference table. As we can see from the table, the 2nd differences are constant, and this implies that the original sequence can be described by a second degree polynomial. For more information about difference tables and about finding specific polynomials for sequences, go to difference tables. Image 4:Constant 2nd differences Now, we will see that the second differences are always constant for any segment of a diagonal half-line. Let's choose the same diagonal with entries $5, 19, 41, 71, 109, 155$. (For a better understanding of the following paragraphs, the reader should wait until the animation shows three different blue rings and then watch the rest of the animation.) In Image 4, the number of the innermost light blue boxes, which is 14, indicate the distance from $5$ to $19$, or the difference between these two numbers. The number of darker blue boxes in the middle indicate the difference between the numbers $19$ and $41$. The darkest blue boxes on the outside indicate the difference between the entries $41$ and $71$. The number of all these blue boxes correspond to the first differences in the difference table. In Image 4, the light blue boxes are divided into pieces, and each piece is moved to on top of the darker blue boxes. We can see that there are $8$ more darker blue boxes than the innermost light blue boxes. Similarly, the blue boxes in the middle are divided into pieces and are moved to on top of the darkest blue boxes. Indeed, as Image 4 illustrates, regardless of the exact number of the blue boxes, there are $8$ more boxes in any given ring than there are in the ring that is one layer inwards from it. This means that the the second differences, or the differences between the first differences is always going to be constant at $8$. Thus, the sequence for any diagonal half-line can be described using a quadratic polynomial. Examples of quadratic polynomials for half-lines For instance, prime numbers $5, 19, 41, 71, 109$, which are aligned in the same diagonal, can be described through the output of the polynomial $4x^2+10x+5$ for $x=0, 1, 2, 3, 4$. Similarly, numbers $1, 3, 14, 31, 57, 91 \dots$, which are also aligned in a green diagonal in Image 5 starting from the center and continuing to the upper right corner, can be expressed by: $4x^2-2x+1$ for $x=0, 1, 2, 3, \dots$. (We will refer back to this polynomial in a later section) In fact, the part of the green diagonal that starts from the center and continues to the bottom left corner can also be described through Eq. (1) for $x=0, -1, -2, -3, \dots$. Image 5 In fact, even horizontal and vertical line segment in the grid can be described by quadratic polynomials, as long as the lines satisfy the condition that no two entries are positioned on the same ring. For instance, the blue horizontal line segment in Image 5, $10, 27, 52, 85$ can be described by a quadratic polynomial. However, the sequence $9, 10, 27, 85$ on the same horizontal line cannot be described by a polynomial because the entries 9 and 10 are on the same ring. Moreover, it is not hard to show that the green diagonal line from Image 5 that goes through the center and has a slope of +1 is the only line on which the Ulam numbers in both directions can be described by the same polynomial. Even the red diagonal line that goes through the center and has a slope of -1 cannot be described by one polynomial. Half of the red diagonal, the segment going up from the center, $5, 17, 37, 67, \dots$ can be described by $x^2+1$ for inputs that are even numbers, while the diagonal going down from the center, $1, 9, 25, 49, \dots$ is a sequence of perfect squares of odd numbers. Thus we cannot find one polynomial that generates both all entries of the red diagonal. Euler's Prime Generator Many have come up with polynomials in one variable that generate prime numbers, although none of these polynomials [...] Many have come up with polynomials in one variable that generate prime numbers, although none of these polynomials can generate all the prime numbers. One of the most famous polynomials is the one discovered by Leonhard Euler(1707-1783), which is : $x^2-x+41$ Euler's polynomial generates distinct prime numbers for each integer $x$ from $x=1$ to $x=40$. Image 6 As we can see in Image 6, we can start an Ulam's spiral with 41 at the center of the grid and get a long, continuous diagonal with 40 prime numbers. One interesting fact is that the 40 numbers in the diagonal line segment are the first 40 prime numbers that are generated through Euler's polynomial. Moreover, the prime numbers are not aligned in order of increasing values. In fact, with 41 in the center, other prime numbers alternate in position between the upper right and lower left part of the diagonal. We will show why an Ulam's spiral that starts with 41 generates entries aligned in a diagonal that can also be descried as outputs of Euler's polynomial. First, we found out in the previous section that Eq. (1) is the polynomial for the diagonal that goes through the center 1 and has a slope of +1 in the [...] First, we found out in the previous section that Eq. (1) is the polynomial for the diagonal that goes through the center 1 and has a slope of +1 in the Ulam's spiral. That is, $4x^2-2x+1$ describes the upper half of the diagonal as we plug in $x=0, 1, 2, 3, \dots$ and the lower half of the diagonal as we plug in $x=0, -1, -2, -3, \dots$. Thus, $x$'s are positioned in the diagonal as if the diagonal was a number line that starts with 0 at the center, with positive integers on the upper right and negative integers on the lower left direction, as shown in Image 7. Image 7 Then, if we start the Ulam's spiral at 41 and thus make 41 the center, each entry on the grid, including the entries on the diagonal segment, will increase by 40. Then, the diagonal can be described by the polynomial : $4x^2-2x+41$. In fact, we will see that this polynomial and Euler's polynomial are describing the same entries along the diagonal once we make some changes in numbering the position of $x$'s. Euler's polynomial can be found by reassigning the position of the entries of the diagonals so that $x'=1$ is at the center, $x'=2, 4, 6, 8, \dots$ are the positions up the diagonal line, $x'=1, 3, 5, 7, 9, \dots$ are positions of entries down the diagonal line from the center, as shown in Image 8. Thus, starting with $x'=1$ at the center, we alternate between the right and left of the center as we number the positions $x'=1, 2, 3, 4, 5, \dots$. Image 8 Then, for the upper half of the diagonal in Image 8, we can use $x'=2, 4, 6, 8, \dots$ instead of $x=1, 2, 3, 4, \dots$ from Image 7. Then, $x=\frac{1}{2} x'$ If we plug this value in to $4x^2-2x+41$ of Eq. (2), we get : $4(\frac{1}{2}x')^2-2(\frac{1}{2}x')+41$ $={x'}^2-x'+41$, which has the same form as Euler's polynomial. The same method works for the bottom half as well. We use $x'=1, 3, 5, 7, \dots$ in Image 8 while we used $x=0, -1, -2, -3, \dots$ in Image 7. Then, $x=\frac{-x'+1}{2}$. By substituting the $x$'s in the polynomial $4x^2-2x+41$ the same way we did above, we get: ${x'}^2-x'+41$ Thus, we can see that Eq. (2), in fact, have the same output as Euler's polynomial after we reassign the position of $x$'s. We have shown that Eq. (1), which describes the diagonal that goes through the center and has a slope of +1, can be transformed to Euler's polynomial that describes the same diagonal that has 41 as its center. Sacks Spiral Sacks spiral is a variation of the Ulam spiral that was devised by Robert Sacks in 1994. Sacks spiral places 0 in th [...] Image 9 Image 10 Sacks spiral is a variation of the Ulam spiral that was devised by Robert Sacks in 1994. Sacks spiral places $0$ in the center and places nonnegative numbers on an Archimedean spiralAn Archimedean spiral is a spiral traced by a point that is moving away from the center of the spiral with a constant speed along a line which is rotating with constant speed, or angular velocity. An Archimedean spiral typically has the form $r= a+b\theta$, where $a, b$ are real numbers. The image below is an example of an Archimedean spiral. , whereas the Ulam spiral places $1$ in the center and places other numbers on a square grid. Moreover, Sacks spiral makes one full counterclockwise rotation for each square number $4, 9, 16, 25, 36, 46, \dots$, as shown in Image 10. The darker dots indicate the prime numbers. We can also see in Image 10 that numbers that have blue check marks and are aligned in the left side of the spiral are pronic numbersA pronic number is the product of two consecutive integers. Pronic numbers have the form $x(x+1)$. For example, $2=1(1+1), 6=2(2+1), 12=3(3+1),\dots$. Image 11 Moreover, these numbers are aligned in positions that are a little less than half of one full rotation from one perfect square to the next perfect square, for instance, from 4 to 9, or 9 to 16. Let the $n$th perfect square be $n^2$. Then, going from the $n$th perfect square to the $(n+1)$st perfect square, the difference between the two numbers will be $2n+1$. We can show this by calculating the difference between two consecutive perfect squares, $(n+1)^2-n^2=n^2+2n+1-n^2=2n+1$. Because the pronic numbers are positioned a little less than half of one full rotation from the perfect squares, their position is a little less than $n+1/2$ from the $n$th perfect square as we go around the spiral. Indeed, any pronic number $n(n+1)=n^2+n$, and the pronic number with the form $n(n+1)$ is aligned at the $(n^2+n)$th position from the origin. From this, we can see that pronic number that has the form $n(n+1)$ appears as the $n$th number along the spiral from the $n$th perfect square. An interesting pattern can be discovered when we start the spiral at 41. As Image 11 shows, the red dots are the first 40 prime numbers generated by Euler's polynomial, and they are aligned in the center and positions where pronic numbers used to be in the original Sacks spiral. Other Numbers and Patterns Triangular Number A number n is a triangular number if n number of dots can be arranged into an equilateral triangle ev [...] A number $n$ is a triangular number if $n$ number of dots can be arranged into an equilateral triangle evenly filled with the dots. As shown in theimage below, the sequence of triangular numbers continue as $1, 3, 6, 10, 15, 21, \dots$. Image 12 The $n^{\rm th}$ triangular number, $T_n$, is given by the formula : $T_n=\frac{n(n+1)}{2}$ When we mark the triangular numbers in a Ulam spiral, a set of spirals are formed as shown in the image below. Image 13 Prime numbers in lines The Ulam spiral inspired the author of this page to create another table and find a pattern among prime numbers. Firs [...] Image 14 The Ulam spiral inspired the author of this page to create another table and find a pattern among prime numbers. First, we create a table that has 30 columns and write all the natural numbers starting from 1 as we go from left to right. Thus, each row will start with a multiple of 30 added by 1, such as 1, 31, 61, 91, 121, ... . When we mark the prime numbers in this table, we get Image 14. We can see that prime numbers appear only on certain columns that had 1, 7, 11, 13, 17, 19, 23, 29 on their first row. This image shows that all prime numbers have the form: $30n+(1, 7, 11, 13, 17, 19, 23, 29)$ Note that not all numbers generated by this form are prime numbers. Another point to notice in the picture is that the nonprime numbers that appear on prime-concentrated columns are all multiples of prime numbers larger than or equal to 7. For instance, $49=7\times 7, \quad 77=7\times 11, \quad 91=7\times 13, \quad 119=7\times 17, \quad 121=11\times 11, \quad 133=7\times 19 \dots$. In fact any combination of two prime numbers larger than or equal to 7 all appear on these prime-concentrated columns. To learn more about the reason for such alignment of prime numbers, click below. Image 15 shows the process of eliminating multiples of prime numbers from the table. First, we eliminate multiples of 2, and then we eliminate Image 15 shows the process of eliminating multiples of prime numbers from the table. First, we eliminate multiples of 2, and then we eliminate multiples of 3. Since 4 is a multiple of 2, any multiple of 4 was already eliminated. Thus, we next eliminate multiples of 5. We continue this process until we eliminate multiples of 19. (We do not eliminate numbers bigger than 19 because we only have numbers up to 390 in this table.) Note that when we eliminate multiples of 2, 3 or 5, the entire column that comes below any multiple of 2, 3, 5 on the first row get eliminated as well. Image 15 We can see from Image 15 that eliminating multiples of 2, 3, 5 leaves us with columns where prime numbers appear. All of these multiples of 2, 3, and 5, or all the numbers that are not in the prime-concentrated column, have the form $30n + 2^{a} 3^{b} 5^{c}$ or $(2\times 3 \times 5)n+2^{a} 3^{b} 5^{c}$, where $a \geq 0, b \geq 0, c \geq 0$. Thus, for a given number with the form of Eq. (3) we are able to factor out 2 or 3 or 5, or any combination of the three, which means that this number is divisible by 2, 3, or 5. However, we know that prime numbers are not multiples of 2 or 3 or 5. Then, these prime numbers will not have the same form as Eq. (3), or else the prime numbers will be divisible by 2 or 3 or 5. Indeed, we can see that the prime numbers are positioned at : $(2 \times 3 \times 5)+ (1, 7, 11, 13, 17, 19, 23, 29)$ Thus, prime numbers have to be positioned on the blue columns that do not have the same form as Eq. (3). Next, we look at nonprime numbers that are products of multiplication of two prime numbers or a perfect square of a prime number. These products will be divisible by the prime numbers they are composed of, but they will not be divisible by 2 or 3 or 5. For instance, 77, which is a nonprime number that appears on the column of prime numbers 117, 47, 107, 137, ..., is a product of two prime numbers, 7 and 11. Because each of these prime numbers do not have any divisors other than 1 and themselves, 77 is not divisible by 2 or 3 or 5. Thus, 77 and all other products of prime numbers also cannot have the same form as Eq. (3), and they have to appear on the blue columns that are concentrated with prime numbers. Why It's Interesting Image 16 Not much is discovered about the Ulam spiral. For instance, the reason for the diagonal alignment of prime numbers or the vertical and horizontal arrangement of non-prime numbers is not clear yet. Indeed, Ulam spiral is not heavily studied by mathematicians. However, Ulam spiral's importance lies on the fact that it shows a clear pattern among prime numbers. Some might suspect that we are seeing diagonal lines in the Ulam spiral because the human eye seeks patterns and groups even among random cluster of dots. However, we can compare Image 16 and Image 17 and see that the prime numbers actually have a distinct pattern of diagonal lines that random numbers do not have. Image 16 is a Ulam spiral where the black dots denote for the prime numbers, and Image 17 is a Ulam spiral of random numbers. Image 17 People are interested in the pattern among prime numbers because the pattern might give enough information for us to discover a new polynomial that will generate more prime numbers than previously-discovered polynomials. The discovery of formula for prime numbers can lead us to have better understanding of other mysterious conjectures and theories involving prime numbers, such as twin prime conjectureTwin prime conjecture states that there are infinitely many primes $p$ such that $p+2$ is also a prime. or Goldbach's conjecture. For more information about the twin prime conjecture or Goldbach's conjecture, go to Wolfram Math World :Twin Prime Conjecture or Wolfram Math World :Goldbach Conjecture. Teaching Materials There are currently no teaching materials for this page. Add teaching materials. References Pickover, Clifford A. (2009). The Math Book : From Pythagoras to the 57th Dimension, 250 Milestones in the History of Mathematics. London : Sterling Publishing Wikipedia (Ulam Spiral). (n.d.). Ulam Spiral. Retrieved from http://en.wikipedia.org/wiki/Ulam_spiral. Wikipedia (Sacks Spiral). (n.d.). Sacks Spiral. Retrieved from http://en.wikipedia.org/wiki/Sacks_spiral. Weisstein, Eric W. "Prime Spiral." In MathWorld--A Wolfram Web Resource. Retrieved from http://mathworld.wolfram.com/PrimeSpiral.html. Sacks, Robert. (2007) NumberSpiral.com. Retrieved from http://www.numberspiral.com/index.html. Weisstein, Eric W. "Prime-Generating Polynomial." In MathWorld--A Wolfram Web Resource. Retrieved from http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html Future Directions for this Page • An explanation for the patterns appearing among triangular numbers in Triangular number section • An helper page for Archimedean Spiral in Sacks Spiral section • An explanation of why the vertical and horizontal lines through the center do not contain any prime numbers • A dynamic development of the spiral with e.g. red = all products of 2; green = all products of 3; nextcolour = all products of nextprime (except for those that are already coloured). If we start with all numbers = black, then at the end, black=prime If you are able, please consider adding to or editing this page! Have questions about the image or the explanations on this page? Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page. 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http://mathoverflow.net/questions/38084/monoids-with-infinite-products/38111	## Monoids with infinite products ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Say a monoid $M$ has infinite products if, for any (possibly infinite) sequence $(m_1,m_2,\ldots)$ of elements of $M$, there exists an element $m_1m_2\cdots\in M$, satisfying some good properties. First, if the sequence is finite, it should coincide with the usual product on $M$. Second, concatenation of sequences results in multiplication of their products and is associative. Third, identities can be "thrown out," as can consecutive inverses ($m_{i+1}=m_i^{-1}$). (Other good properties to include?) Another way to phrase this is: "$M$ is closed under small ordinal colimits." That is, if $M$ is considered as a one-object category, then for any small ordinal $[\kappa]$ and functor $m\colon[\kappa]\to M$, the colimit of $m$ exists in $M$. Example: let ${\mathbb N}^+$ denote the monoid with underlying set ${\mathbb N}\cup${$\infty$} and whose operation on a sequence $m=(m_1,m_2,\ldots)$ is given by addition if $m$ has only finitely many non-zero elements, and by $\infty$ otherwise. Now suppose that $M$ is any monoid and I want to replace it by a monoid that has infinite products. I'm hoping there are two ways to do this. One would be to add colimits freely, and the other would be to add a single "$\infty$" element that served as a catch-all (as in the example above). Q: Do these both exist (functorially in $M$)? If so, can you describe them in elementary terms? For example, I'm worried about sequences like $1-1+1-1+\cdots$. So in a good answer I'd hope to see what happens with such infinite sums. - So far the question does not make much sense (for me). What do you mean by "inverses"? Consider the monoid given by one defining relation $ab=1$ (the polycyclic monoid; note that $ba\not = 1$ in this monoid). What do you want to do with it? If your original monoid is commutative, do you expect the result to be commutative too? – Mark Sapir Sep 8 2010 at 19:06 How about requiring that the product of (m_1, m_2, ...) is equal to the product of (1, m_1, m_2, ...)? – Qiaochu Yuan Sep 8 2010 at 19:50 I don't see how your "another way to phrase this" is the same thing... how do the axioms in the first paragraph force the infinite products to be colimits? – Peter LeFanu Lumsdaine Sep 8 2010 at 20:14 Sorry, I've been absent -- you all have good points. I'm not exactly sure what I want here; perhaps that is clear. I'm looking for the right notion. As for Mark's question -- you're right to ask about inverses: I was sloppy. I think the question of whether the addition of infinite products to a commutative monoid leaves it commutative is a matter of choice, at least in the "add products freely" functor. Under the "add a catch-all" it will remain commutative. – David Spivak Sep 8 2010 at 21:01 As for Qiaochu Yuan's question -- yes, this is what I mean by "identities can be thrown out." It also follows from the "concatenation of sequences" axiom. As for Peter's question -- good point. Just to make sure the idea is clear at least, the colimit of a finite sequence ($\kappa=n$ for some finite n) will include a canonical map from $0\in [n]$ and this map will be the product. So in the finite case, the ideas coincide. Colimits also take care of the other properties (think "final subcategory.") But that's not to say that I can show that colimits are really what I want here. – David Spivak Sep 8 2010 at 21:06 show 1 more comment ## 5 Answers If I understand the question, the short answer is "yes, you can freely and functorially adjoin infinite products to monoids". The basic idea is that algebraic theories can accommodate arbitrary arities (bounded above by some cardinal), and one can discuss relative free-forgetful adjunctions between categories of algebras in great generality. At first pass, let me just focus on the purely monoid-like aspects for now, because those are easier to visualize. Once the story for that is clear, one can work in inverses. As a warm-up, let's recall that one way of defining an ordinary monoid $M$ is as an algebra of the terminal nonpermutative operad. In plainer English, this means we a single operation $$\mu_n: M^n \to M$$ for each finite ordinal $n$, so that $$\mu_{n_1 + \ldots + n_k} = \mu_k(\mu_{n_1} \times \ldots \times \mu_{n_k})$$ (generalized associativity equation). Now let's generalize operads so as to allow operations of countably infinite arity. By "arities", I will really mean countable ordinals. Given an ordinal $k \lt \aleph_1$ and ordinals $n_j \lt \aleph_1$ for $j \lt k$, you can concatenate the $n_j$ to get a new ordinal $\sum_j n_j \lt \aleph_1$. Concatenation is associative in an evident sense. Now define an $\omega$-monoid to be a set $M$ equipped with operations $$\mu_k: \hom(k, M) \to M,$$ one for each $k \lt \aleph_1$, such that $\mu_{\sum_j n_j} = \mu_k (\prod_j \mu_{n_j})$. (I am not completely certain we have to go all the way up to $\aleph_1$, but if not it will be some suitable initial segment. Let's just say $\aleph_1$ for now.) This condition can be interpreted in any category with countable products, such as $Set$. The free $\omega$-monoid on a set $X$ will be $\sum_{k \lt \aleph_1} \hom(k, X)$. We get in this way a monad $T$ for a free-forgetful adjunction between $\omega$-monoids and sets. There is a general bit of nonsense that for any morphism of monads $\phi: S \to T$ on $Set$, there is a forgetful functor $Alg_T \to Alg_S$, and this forgetful functor has a left adjoint. This follows from an adjoint functor theorem, although if I'm not mistaken, in this particular scenario a more direct construction is available: if $S$ is the monad for ordinary monoids and $T$ is as above, the evident inclusion $S \to T$ induces a forgetful functor $$\omega-Mon \to Mon$$ which has a left adjoint $L$ described by the type of coequalizer familiar from tensor products: $$L(M) = coeq((\mu_T \circ \phi) M, T\theta: TSM \stackrel{\to}{\to} T M)$$ where $\mu_T: TT \to T$ is the monad multiplication and $\theta: S M \to M$ is the structure of $M$ as $S$-algebra. This left adjoint $L$ would correspond to what I think you were asking for with "freely adjoined colimits", and the left adjoint means we indeed have a functorial construction. If you want to work inverses in, you can do that too. Long story short: for any set of formal operation symbols of arbitrary arity, subject to any set of well-formed equations you jolly well please, you can form a monad whose algebras are precisely the models of for the corresponding algebraic theory. So: together with operations of countable arity as above, subject to generalized associativity equations, you can certainly toss in an unary inversion operation as well. I leave it to you to decide what, in addition to associativity, are the sensible equations to impose on inversion $i$, but it seems to me you might want to impose only $$\mu_2(id \times i) = id = \mu_2(i \times id)$$ and stop there. (Operations involving infinitely many instances of inversion are still permissible, but the equations would enforce only finitely many cancellations at a time.) You get in this way a monad for "$\omega$-groups", and again the forgetful functor from $\omega$-groups to groups admits a left adjoint, constructed in a way analogous to the above. - Thanks Todd. Is the forgetful functor $\omega-Mon\to Mon$ not also a left adjoint? If you work out your construction for the case of the free monoid on one generator ${\mathbb N}$, do you get this ${\mathbb N}^+$ thing I mention above, or something else? I was more hoping to have a "unique point at infinity," to sew it all up there, than to have so many different styles of infinite product under this completion process. Any chance of that? – David Spivak Sep 9 2010 at 12:52 David, I've nothing intelligent yet to say about the $\mathbb{N}^+$ thing, but I don't think that forgetful functor has a right adjoint. It would have to preserve coproducts. The coproduct of two copies of the free $\omega$-monoid on the 1-element set, $F(1)$, is the free $\omega$-monoid on the 2-element set, $F(2)$. You'd need the canonical map $F(1) + F(1) \to F(2)$ on the underlying monoids, where $+$ is monoid coproduct, to be invertible. So, each countable word in two letters would need to be a finite concatenation of countable words, each in one or the other letter. Which is false. :-( – Todd Trimble Sep 9 2010 at 21:33 While I'm here: on second thought I don't think $\omega$-groups are all that sensible, because of the swindle mentioned by both you and Theo. [The "logic" of an algebraic theory involving operations $p$ of infinite arity, like that of $\omega$-groups, means we must have $p(a_1, \ldots) = p(b_1, \ldots)$ whenever $a_i = b_i$. So the swindle is derivable in the theory.] In other words, while you can consider or construct the theory or monad of $\omega$-groups, it would unfortunately collapse to something trivial. Silly of me to have suggested otherwise. :-( – Todd Trimble Sep 9 2010 at 22:02 ### 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. Semigroups with an $\omega$-product play an important role in automata accepting infinite words and the second order monadic theory of the natural numbers with the successor operator. See the book Infinite Words: Automata, Semigroups, Logic and Games by Perrin and Pin. - Monoids with infinite products were considered by Kharlampovich, Myasnikov and Serbin. They use it in relation with the Tarski problems about free groups, and groups acting on $\Lambda$-trees. See, for example, http://uk.arxiv.org/PS_cache/arxiv/pdf/0911/0911.0209v1.pdf . Probably more details can be found in Serbin's thesis (McGill). - Depending on exactly what you want, if you take some monoid and force it to allow infinite products, you might end up with something rather trivial. In particular, the Mazur Swindle puts rather severe limits on the structure of monoids with infinite products and strongly-behaving inverses. Note that the Swindle requires, among other things, an infinite notion of associativity, which your proposal may or may not have. - Mazur's swindle depends strongly on the fact that you can embed the infinite sums in space. That allows you to do infinite cancellation, which is definitely not assumed here. – S. Carnahan♦ Sep 9 2010 at 1:05 My point was: if you do allow infinite cancellations, then you're hosed. The only options are either disallow infinite cancellations, or disallow inverses all together. – Theo Johnson-Freyd Sep 9 2010 at 14:21 I've also thought about monoids with infinite products in the context of this question. I just want to remark that every group, which has infinite products, which are invariant under permutations (this is a natural generalization of commutativity) is trivial: $a^\mathbb{N} = a a^{\mathbb{N} - \{0\}} = a a^{\mathbb{N}} \Rightarrow a=1$. Of course, this does not happen when you just impose invariance under permutations with finite support, but I don't think this is natural. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 79, "mathjax_display_tex": 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": 0.9414945840835571, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/74845/list	## Return to Answer Post Made Community Wiki by S. Carnahan♦ 1 There is a very elegant proof that there exists no continuous injection from the plane into the real line. The proof can basically be given by drawing a picture on the blackboard. Suppose there is such an injection $f$. Let $x$ and $y$ be distinct points in the plane and let $g_1$ and $g_2$ be paths from $x$ to $y$ such that $g_1(r_1)\neq g_2(r_2)$ for $r_1,r_2\in (0,1)$. Now this implies that $f\circ g_1((0,1))\cap f\circ g_2((0,1))=\emptyset$, contradicting the intermediate value 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": 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": 0.934077262878418, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/40744/curvature-and-edge-state?answertab=oldest	# Curvature and edge state If the boundary of quantum hall fluid has non-constant curvature, how will it affect the edge state which is usually described in chiral Luttinger fluid? - ## 1 Answer A naive guess would be that there isn't any real difference. The theoretical logic behind my guess is that at low-energies FQH states are described by 2+1D Chern-Simons theories, which are topological gauge theories. Although the bulk does not have any local degrees of freedom, the boundary does. This is because in the presence of a boundary $\partial M$ one has to impose boundary conditions and reduce the set of gauge transformations to those that respect this BC, therefore there will be an infinite number of states which are not gauge equivalent anymore and therefore correspond to physical degrees for freedom. More formally, the boundary dynamics are described by a Wess-Zumino-Witten theory which I think is nothing but a chiral Luttinger liquid in the simplest case. Now this is a conformal field theory and only depend on the conformal class of the boundary metric, not the metric itself. 2D manifolds, like the boundary $\partial M$, are however all conformally flat and therefore the boundary dynamics are insensitive to the curvature. This robustness against boundary curvature, impurity scattering and so on is a general feature of quantum Hall states. If you have access to Nature, see (here and here) the simulations done for photonic crystal analogs of IQHE boundary states. It is here seen that the light wave goes around any boundary defect, curvature or impurity without any reflections at all. It is quite non-intuitive that light can go around a mirror without any reflection! - 1 It's a good comment, but I do think this is indeed too naive. We already know that edge reconstruction comes into play when the boundary potential is not that steep (see e.g. arxiv.org/abs/cond-mat/0302344 ). What this effectively does is introduce higher order, non-linear terms in the otherwise linear CFT of the boundary. So coupling to the boundary potential really does matter. The experiment you mention really shows that the chirality of the system is not affected by these defects. You can't really make the same strong claim about the Hamiltonian though. – Olaf Oct 15 '12 at 10:09 @Olaf: Thanks for the very nice comment. Would that then mean that (assuming smooth but curved boundary) my argument is only correct if the length scales on which the boundary potential changes is very long? Since the arguments relies on Chern-Simons theory, which is only valid at very long wave-lengths? (I am not sure what "edge reconstruction" means here). And very good point, robustness of the chirality of the edge modes does not imply the detailed dynamics are robust! This is of course true. – Heidar Oct 15 '12 at 14:40	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 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": 0.9426307678222656, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-equations/142824-operator-method-help.html	# Thread: 1. ## Operator method help im trying to solve a system of differential equations of (x''(t ) + 5(x(t )) - 2y(t ) = 0 and, -2x(t ) + y''(t ) +2y(t ) = 0 but I keep getting the wrong answer.. 2. Originally Posted by Mist im trying to solve a system of differential equations of (x''(t ) + 5(x(t )) - 2y(t ) = 0 and, -2x(t ) + y''(t ) +2y(t ) = 0 but I keep getting the wrong answer.. Here's an outline of the solution. Write the equations as $x'' = -5x+2y$ $y'' = 2x-2y,$ or in vector form $\begin{bmatrix}x''\\y''\end{bmatrix} = \begin{bmatrix}-5&2\\2&-2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$. Now diagonalise that matrix $A = \begin{bmatrix}-5&2\\2&-2\end{bmatrix}$. You should find that the eigenvalues are –1 and –6, with corresponding eigenvectors $\begin{bmatrix}1\\2\end{bmatrix}$ and $\begin{bmatrix}-2\\1\end{bmatrix}$. Let P be the matrix having those eigenvectors as columns, and let D be the diagonal matrix with the eigenvalues –1 and –6 as its diagonal elements. Then $A = PDP^{-1}$ and so $P^{-1}\begin{bmatrix}x''\\y''\end{bmatrix} = DP^{-1}\begin{bmatrix}x\\y\end{bmatrix}$. Let $\begin{bmatrix}u\\v\end{bmatrix} = P^{-1}\begin{bmatrix}x\\y\end{bmatrix}$. Then $u'' = -u$ and $v'' = -6v$. Those are simple harmonic motion equations, with solutions $u = A\cos t + B\sin t$, $v = C\cos(\sqrt6t)+D\sin(\sqrt6t)$. Finally, $\begin{bmatrix}x\\y\end{bmatrix} = P\begin{bmatrix}u\\v\end{bmatrix}$, giving the solution $x(t) = A\cos t + B\sin t -2(C\cos(\sqrt6t)+D\sin(\sqrt6t)),$ $y(t) = 2(A\cos t + B\sin t) + C\cos(\sqrt6t)+D\sin(\sqrt6t).$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 16, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9056553840637207, "perplexity_flag": "middle"}
http://quant.stackexchange.com/questions/4396/what-is-hestons-equation/4398	# What is Heston's equation? This paper mentions the elliptic Heston operator: $Av:= -\frac y2(v_{xx}+2\rho\sigma v_{xy} + \sigma^2v_{yy}) - (c_0 - q - \frac y2)v_x + \kappa(\theta -y)v_y + c_0v$. Then boundary value problem are discussed: $Au=f \text{ on } \Omega \\ u = g \text{ on } \partial\Omega$ I would like to know how people use such Dirichlet conditions in mathematical finance. - ## 2 Answers From this abstract: The Heston stochastic volatility process is a degenerate diffusion process where the degeneracy in the diffusion coefficient is proportional to the square root of the distance to the boundary of the half-plane. The generator of this process with killing, called the elliptic Heston operator, is a second-order, degenerate-elliptic partial differential operator, where the degeneracy in the operator symbol is proportional to the distance to the boundary of the half-plane. In mathematical finance, solutions to obstacle problem for the elliptic Heston operator correspond to value functions for perpetual American-style options on the underlying asset. A simple Google search shows that there are only a handful of academics who even use this term. Your best bet may be to contact one of them directly for support. (They are unlikely to entertain a broad "what do I use this for" question, however.) - Thank you! I read this article. Actually it interested me because the author used weighted Sobolev spaces which I study. – nikita2 Oct 23 '12 at 14:48 1 @nikita2 I do worry that the elliptic Heston operator is not a widely accepted term given that it doesn't appear in any peer-reviewed journal, and the few papers online seem to share the same particular co-author. If enough academics discover this work and agree with it, then the term may become more widely adopted; until then, you'll just get a lot of blank stares for using it in a sentence. – chrisaycock♦ Oct 23 '12 at 14:58 Expanding a bit on chrisaycock's answer, and noting in particular from the abstract In mathematical finance, solutions to obstacle problem for the elliptic Heston operator correspond to value functions for perpetual American-style options on the underlying asset. we can see that this would be used to price those few rare cases of perpetual options. The only traded examples I know of are perpetual convertible preferred securities, for example from Wells Fargo's offerings. Such securities are lightly traded by the market players and therefore not always analyzed using the full machinery of a stochastic vol model, even if they should be in principle. In practice, these "perps" are so bond-like that it is often more useful to think of them as fixed-income instruments. The main concern with them is that the issuer will stop paying the dividends or change capital structure, so it is a bit ridiculous to spend one's time on a fancy stochastic vol model when all the interesting stochastic events have to do with unrelated variables such as alterations in capital structure. - If I understood you correctly such models have little use and mathematicians just play with them. To do it clear, is there any sense to do research of PDE and SPDE in connection with mathematical finance? – nikita2 Oct 28 '12 at 10:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.3, "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": 0.9233353734016418, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/111578/another-exterior-differential-system-on-so3-mathbb-r-times-mathbb-r/111817	## ANOTHER Exterior differential system on $SO(3;\mathbb R) \times \mathbb R$ ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I have another exterior differential system for one forms $U^i$, where the $\theta^i$ are a cotangent basis on $SO(3)$, i.e. they satisfy $d \theta^i = \epsilon_{ijk} \theta^j \wedge \theta^k$ for the antisymmetric tensor $\epsilon$. This is related to a previous question here. \begin{array}{c} \text{dU}^1+\sqrt{3} \theta^1 \wedge U^3+\sqrt{3} \theta^2\wedge U^4 =0\\ \text{dU}^2+\theta^1 \wedge U^3-\theta^2 \wedge U^4+2 \cosh(\rho) \theta^3 \wedge U^5 =0\\ \text{dU}^3+\sinh(\rho)\theta^2\wedge U^5+\theta^3 \wedge U^4 =0\\ \text{dU}^4-\sinh(\rho)\theta^1\wedge U^5-\theta^3\wedge U^3 =0\\ -\sinh(\rho) \text{dU}^5 -\cosh(\rho) \text{d$\rho$}\wedge U^5-\theta^1\wedge U^4+\theta^2\wedge U^3 =0\\ \cosh(\rho) \text{dU}^5+\sinh(\rho) \text{d$\rho$}\wedge U^5+\theta^1\wedge U^4+\theta^2 \wedge U^3-2 \theta^3 \wedge U^2 =0\\ -\text{dU}^4+\cosh(\rho) \theta^1\wedge U^5+\sqrt{3} \theta^2\wedge U^1-\theta^2 \wedge U^2+\theta^3\wedge U^3 =0\\ -\text{dU}^3+\sqrt{3} \theta^1\wedge U^1+\theta^1\wedge U^2+\cosh(\rho) \theta^2\wedge U^5-\theta^3\wedge U^4=0 \end{array} FYI: this problem is related to my work on asymptotic symmetries of certain noncompact homogeneous spaces, i.e. I want to find diffeomorphisms which preserve a certain metric tensor asymptotically. It seems like it should be possible to solve systems like this (semi-) automatically with a computer algebra package. After all, the system reduces to an overdetermined system of first order partial differential equations, for which such tools already exist... but the EDS form is so much more convenient that I would hate to rewrite everything as 1st order PDEs! EDIT: As R.B. suggested in the comments, I forgot to mention that as in the other question, there is also a coordinate $\rho$ (of course), i.e. the one forms can be expanded as $\alpha = \alpha_a \theta^a + \alpha_\rho d\rho$ etc. Also, the summation convention is not implied in the expression for $d \theta^i$. EDIT2: Of course a good start would be to add the third and last equations to get rid of $dU^3$ and similarly for $dU^4$ and $dU^5$. The number of unknowns will then be reduced from 20 to 7, but the system still seems quite difficult to solve... I've been hacking away at it with Mathematica and some exterior differential and wedge product functions, but it's still a mess! - 2 Two questions: First, are you assuming that the $U^i$ are linear combinations of the $\theta^j$ and $d\rho$, as before? Second, in your structure equation for the $\theta$s, are you using the summation convention or not? I.e., with the summation convention, you'd have $d\theta^1 = 2\ \theta^2\wedge\theta^3$ while, without the summation convention, you'd have $d\theta^1 = \theta^2\wedge\theta^3$. – Robert Bryant Nov 5 at 20:51 Oops, should have been a bit more careful... edited the post accordingly! – H. Arponen Nov 5 at 21:44 I found an appropriate package for REDUCE: reduce-algebra.com/docs/eds.pdf Unfortunately I have virtually no experience with REDUCE... – H. Arponen Nov 5 at 21:48 5 I did some not-very-pretty Maple computations, and if I'm doing it right, it looks like there's a family of solutions parametrized by 2 arbitrary functions of 2 variables. I don't see any good way to write them down explicitly, though - is it enough for you to know that solutions exist? – Jeanne Clelland Nov 7 at 23:01 1 Jeanne: that is GREAT news, exactly what I was hoping!! Actually I also had a third equation which is a bit simpler than this one but more complicated than the other one, for which I managed to get solutions parametrized by one function of three variables. The result(s) mean that there's an infinite number of asymptotic symmetries of a certain noncompact homogeneous space. The problem is based on this one: arxiv.org/abs/1209.5597 – H. Arponen Nov 8 at 13:41 ## 1 Answer Jeanne's calculations give the right answer, i.e., that the solutions depend on two arbitrary functions of 2 variables. It turns out, though, that, with the right choice of variables, one can reduce the problem to an underdetermined system of $2$ linear equations for a pair of tensors of rank $2$ on the $2$-sphere. (This choice of variables is suggested by examining the characteristic variety and tableau of the EDS.) First of all, if one solves for the coefficients of the $U^i$ as the OP suggests, one does get $7$ parameters. One way to do this is as follows: $$\begin{align} U^1 &= -(\sqrt{3}/3)\bigl((v_4{-}(\cosh(\rho)-\sinh(\rho))v_1)\ \theta^1 -(v_2{+}(\cosh(\rho)+\sinh(\rho))v_3)\ \theta^2\bigr)\\ U^2 &= (\cosh(\rho)-\sinh(\rho))v_1\ \theta^1-(\cosh(\rho)+\sinh(\rho))v_3\ \theta^2\\ U^3 &= (\cosh(\rho)-\sinh(\rho))v_5\ \theta^1 + v_6\ \theta^2 + 2\sinh(\rho)v_3\ \theta^3 +v_4\ d\rho\\ U^4 &= v_7\ \theta^1-(\cosh(\rho)+\sinh(\rho))v_5\ \theta^2 + 2\sinh(\rho)v_1\ \theta^3 +v_2\ d\rho\\ U^5 &= (\cosh(\rho)-\sinh(\rho))v_2\ \theta^1-(\cosh(\rho)+\sinh(\rho))v_4\ \theta^2.\\ \end{align}$$ The $v_i$ are now $7$ unknowns, and substituting this into the above $2$-forms gives an involutive differential system for the $v_i$ that is generated by five $2$-forms and has characters $s_1=5$ and $s_2=2$, with $s_i=0$ for $i>2$. However, examining the characteristic variety and tableau of this system suggests that one should reparametrize the first four $v_i$ as $$\begin{align} v_1 &= -(2\sinh(\rho){+}\cosh(\rho))\ p_2-\cosh(\rho)\ q_2\\ v_2 &= (2\cosh(\rho)\sinh(\rho)+2\cosh^2(\rho)-1)(p_1{+}q_1)\\ v_3 &= +(2\sinh(\rho)-\cosh(\rho))\ p_1 + \cosh(\rho)\ q_1\\ v_4 &= (2\cosh(\rho)\sinh(\rho)-2\cosh^2(\rho)+1)(p_2{-}q_2).\\ \end{align}$$ for some functions $p_1,p_2,q_1,q_2$. When one does this, one finds that the differential equations imply $$\begin{align} dp_1 &= (p_0{+}p_3)\ \theta^1 + (p_4{+}p_5)\ \theta^2-3p_2\ \theta^3\\ dp_2 &= (p_4{-}p_5)\ \theta^1 + (p_0{-}p_3)\ \theta^2+3p_1\ \theta^3\\ dq_1 &= (q_0{+}q_3)\ \theta^1 + (q_4{+}q_5)\ \theta^2-1q_2\ \theta^3\\ dq_2 &= (q_4{-}q_5)\ \theta^1 + (q_0{-}q_3)\ \theta^2+1p_1\ \theta^3\\ \end{align}$$ for some functions $p_0,p_3,p_4,p_5,q_0,q_3,q_4,q_5$. This implies that the (complex-valued) linear form $Q = (q_1{-}iq_2)(\theta^1{+}i\theta^2)$ and the (complex-valued) cubic form $P = (p_1{-}ip_2)(\theta^1{+}i\theta^2)^3$ are well-defined on the $2$-sphere $S^2 = \mathrm{SO}(3)/\mathrm{SO}(2)$, where the $\mathrm{SO}(2)$-subgroup of $\mathrm{SO}(3)$ is the one that is an integral of the forms $\theta^1$ and $\theta^2$. Thus, $P$ and $Q$ are sections of natural complex line bundles over the $2$-sphere. One then finds that the integral manifolds of the differential system must satisfy $$\begin{align} v_5 &= -\sinh(\rho)\ (p_4{+}q_5)-\cosh(\rho)\ (p_5{+}q_4)\\ v_6 &= (\cosh^2(\rho){-}\cosh(\rho)\sinh(\rho){-}2)\ p_0 +(\cosh(\rho)\sinh(\rho){-}\cosh^2(\rho){+}1)\ p_3\\ &\quad +(\cosh(\rho)\sinh(\rho){-}\cosh^2(\rho){+}{\tfrac{1}{3}})\ q_0 +(\cosh^2(\rho){-}\cosh(\rho)\sinh(\rho))\ q_3\\ v_7 &= (\cosh^2(\rho){-}\cosh(\rho)\sinh(\rho){-}2)\ p_0 +(\cosh(\rho)\sinh(\rho){+}\cosh^2(\rho){-}1)\ p_3\\ &\quad +(\cosh(\rho)\sinh(\rho){+}\cosh^2(\rho){-}{\tfrac{1}{3}})\ q_0 +(\cosh^2(\rho){+}\cosh(\rho)\sinh(\rho))\ q_3\ .\\ \end{align}$$ Finally, once these are substituted into the $2$-forms, one finds that the differential system reduces to a pair of second order linear differential equations on $S^2$. Using notation that can be found in a paper of mine (TAMS 290 (1985), 259–271), this second order equation for the tensors $P$ and $Q$ can be written in the form $$Y(Y(P)) = \tfrac12\ Q + \tfrac23\ X(Y(Q)) - \tfrac13\ X(X({\overline{Q}})),$$ where $X$ and $Y$ are certain first-order differential operators that are invariant under $\mathrm{SO}(3)$ and that generalize $\partial$ and $\bar\partial$ to symmetric $(1,0)$-forms of arbitrary degree. This is an elliptic linear equation of second order that is underdetermined (it is $2$ equations for the $4$ components of $P$ and $Q$). It is invariant under the rotations of $S^2$ and equivalent to the original (overdetermined) EDS on $\mathrm{SO}(3)\times\mathbb{R}$. It is conceivable that this equation admits an explicit solution in terms of a potential (which would be a section of a rank $2$ vector bundle over $S^2$), but I have not tried to check whether this is true or not. Added Comment: I have now checked about the possibility of a potential and, miraculously, it turns out that there is a potential: One can show that the solutions of the above equation for $P$ and $Q$ are expressible in the form $$\begin{align} P &= X(X(L))\\ Q &= L + (XY{+}YX)L + X(X({\overline{L}})\bigr)\\ \end{align}$$ where $L=(L_1{+}iL_2)(\theta^1{+}i\theta^2)$ is an arbitrary complex-valued $(1,0)$-form on $S^2$. Thus, components of $L$ are the two arbitrary functions of $2$ variables predicted by the theory for the general solution. $L$ is not quite unique. It turns out that one can also add an expression of the form $X(a_1+ib_2)$ to $L$ where $a_1$ is the restriction to the $2$-sphere of a linear function in $\mathbb{R}^3$ and $b_2$ is the restriction to the $2$-sphere of a harmonic homogeneous quadratic function in $\mathbb{R}^3$. This describes the (local and global) ambiguity in the potential $L$ completely. Thus, with this construction and the above formulae, we have the complete description of the (local and global) integral manifolds of the original EDS in terms of the second and third derivatives of $L$ (which is arbitrary). - That is absolutely beautiful, thank you Robert!! That's something I need a while to digest, but I'll read your paper (seems it's open access). – H. Arponen Nov 9 at 20:14 2 You are welcome; it was interesting and a little bit surprising that it's really a 2-dimensional problem in disguise. The only part of my TAMS paper that is relevant to this problem is the first two pages of Section 1 (pp. 260–261), where $X$ and $Y$ are defined and Proposition 1.1 is proved, so ignore Proposition 1.2 and everything after. I could have written the equations out without using this notation, but, while the formulae aren't complicated, one might not immediately see that the two equations are elliptic. – Robert Bryant Nov 9 at 21:00 1 Dang, Robert, that's pretty awesome. :) – Jeanne Clelland Nov 9 at 23:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 86, "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": 10, "end_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": 0.93272864818573, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/696/brute-forcing-crc-32?answertab=votes	# Brute forcing CRC-32 I'm working on a cryptosystem which uses IDEA. The designer made the mistake of including a CRC-32B hash of the password unencrypted in the header, so that the system can quickly reject bad passwords. Of course, that opens it up to the obvious attack of brute forcing CRC32 over potential passwords until the right one is found. Which is what I'm trying to do. Questions: 1. Is the best way to do this simply to do an exhaustive search of the password space? It would seem to me that since CRC32(x) is related to x in manifest ways (eg XOR is preserved), it might be possible to do a construction where we actively direct the search in the right direction. I don't know enough about CRC32, but it would seem to me something like this is possible. Of course, it's not enough to find a string with the same CRC32. It has to be the real password used, otherwise the IDEA decrypt will be gibberish. So this might throw some water on this method. But, regardless, I'd like to know more about it. What do you say? 1. Failing that, I'll need fast code to brute force the CRC32B. What is the fastest code? My plan is to brute force candidates via CRC32, then try to decrypt via IDEA and measure the entropy to filter out all of the false positives (1 in 4 billion) that will still match CRC. - ## 2 Answers Assuming the n-bit CRC of an unknown bit string b is known, one can constructively rebuild any consecutive n bits of b from the rest of the bit string (and the definition of the CRC). Indeed, in the case described, that speeds up password search considerably. One can compute the last 32 bits of the password (likely, 4 characters) from the beginning of the password. If the password is constrained to characters in [20h..7Fh], that will further remove $49/50$ of the candidates. Many CRCs found in practice (including 32-bit CRC CCITT) are not "XOR-preserving" in the sense $CRC(x \oplus y) = CRC(x) \oplus CRC(y)$. However it allways holds that $CRC(x \oplus y \oplus z) = CRC(x) \oplus CRC(y) \oplus CRC(z)$ for any $x$, $y$, $z$ of identical length (including with $z$ all-zero); that (and a little linear algebra) is enough to organize the computation. At the end of the day, with $n=32$, one can compute the last 4 bytes by XOR-ing the values found in precomputed tables (one table for each preceding byte, with each table $256·32$ bits). - Fascinating, fgrieu. Can you give me more information? What do you mean "one can constructively rebuild any consecutive n bits of b from the rest of the bit string" - how much of b do you need to know? If you know the first 4 bytes, say, how do you build the rest of b? Remember, I'm looking for all candidates, not just one that happens to work. – S. Robert James Sep 16 '11 at 2:39 @S. Robert James, You need to know the rest of the string: i.e., all but 32 bits of it. You can do it by solving a system of 32 linear equations (linear over GF(2), i.e., with XOR instead of addition), but in the end, it can be optimized down to table lookups as fgrieu describes. – D.W. Sep 16 '11 at 6:35 Got it. CRC32 specifies 32 bits, so if I know everything but 32 bits, I can determine the missing 32. The tables show the CRC of one byte and the rest zeroes. I would assume that I need a separate set of tables if the length of the input changes (eg tables for last 4 of 8 bytes, tables for last 4 of 12 bytes). Where can I read more about this? – S. Robert James Sep 16 '11 at 9:17 1 @S. Robert James: Sorry I do not have a reference, but the linear algebra is easy. Assuming the relation I give holds (it does for any CRC variant regardless of polynomial, initialization, finalization details), you can use the CRC function as a "black box" without knowing its internals, and only need to invoke it for the n +1 distinct n bit strings with at most 1 bit set. The rest is Gaussian elimination restricted to bit coefficients, and caching into tables to avoid solving a linear system repeatedly. I won't help with (pseudo-)code for fear of being associated with something nefarious. – fgrieu Sep 16 '11 at 16:15 IDEA has a key size of 128 bits. Even if the 32-bit CRC fully leaks 32 key bits, that would still leave an effective key size of 96 bits, or 79 billion, billion, billion keys -- that's still too large to brute force. - 2 The OP seems to plan a brute force attack of the password from which the IDEA key is generated, helped by CRC32; then for each candidate password, build IDEA key, decrypt ciphertext with IDEA and check redundancy of the plaintext obtained. That works irrespective of the IDEA key size. – fgrieu Sep 16 '11 at 2:06 2 Exactly. This is especially so because in this implementation, the key is given as a string ("password"), and then internally hashed to 128 bits. But the space of expected strings is of course much smaller than $2^{128}$. – S. Robert James Sep 16 '11 at 2:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 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": 0.9225790500640869, "perplexity_flag": "middle"}
http://mathhelpforum.com/algebra/166970-help-arithmetic-series-print.html	# Help with arithmetic series. Printable View • December 27th 2010, 02:17 PM nelly1 Help with arithmetic series. Hi, I'm taking AS Maths Core 1 in January. I am stuck on a question in the text book. It goes like this: Calculate: 5 Sigma r(r+1) r = 0 The answer in the text book is 112 but I can't get that. I get 70 as an answer. I would really appreciate any help with how to solve this particular problem.] Thanks in advance, j • December 27th 2010, 02:27 PM Plato The answer to the problem you posted is 70. So check the problem again. Is it correct? If so, the given answer is wrong. It goes with $\sum\limits_{r = 0}^6 {r\left( {r + 1} \right)}$. • December 27th 2010, 02:29 PM snowtea The easiest way to do this summation is by using combinations: $\sum_{r=0}^5 r(r+1) = \sum_{r=0}^5 2\binom{r+1}{2} = 2\binom{7}{3} = 2\frac{7\times 6 \times 5}{3!} = 70$ You can always check by a direct sum 01 + 12 + 23 + 34 + 45 + 56 = 0 + 2 + 6 + 12 + 20 + 30 = 70 so you are absolutely correct :) However, $\sum_{r=0}^6 r(r+1) = 2\binom{8}{3} = 2\frac{8\times 7 \times 6}{3!} = 112$ • December 27th 2010, 02:35 PM nelly1 Thanks so much. It had my head chewed up all day. It is a misprint in the book. I checked so many things except for the upper limit. All times are GMT -8. The time now is 10:10 AM.	{"extraction_info": {"found_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": 3, "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": 0.9551634192466736, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/68999/why-is-the-determinant-of-the-following-matrix-zero?answertab=active	# Why is the determinant of the following matrix zero Let $a_1,\ldots,a_n$ be complex numbers and let $b_1,\ldots,b_n$ be complex numbers. Let $A$ be the matrix whose $(i,j)$-th entry is $A_{ij} = a_i b_j$. Then I think $\det A = 0$ when $n>1$. This is easy to compute when $n=2$. Question. Why is $\det A =0$? - 4 The matrix formed by the outer product of two vectors has rank 1... – J. M. Oct 1 '11 at 12:30 12 Isn't it obvious that the rows and columns are linearly dependent? They are multiples of $(a_1,\ldots,a_n)^T$ and $(b_1,\ldots,b_n)$ respectively. – t.b. Oct 1 '11 at 12:31 Use the multi-linearity of the determinant: $\det A=b_1\ldots b_n\det A'$ where $A'_{ij}=a_i$, and for $n>1$ $A'$ is a matrix whose columns are equal hence... – Davide Giraudo Oct 1 '11 at 12:33 – Dylan Moreland Oct 1 '11 at 13:02 1 Gooz/shaye: may I suggest registering your account? :) Then all your questions are in one place, and you don't have the trouble of multiple accounts... – J. M. Oct 1 '11 at 14:57 show 3 more comments ## 1 Answer Algebraically (mentioned by t.b.): The columns of the matrix are $b_1\mathbf{a}$, $b_2\mathbf{a}$, $\dots, b_n\mathbf{a}$. If all of $\mathbf{b}$'s components are zero then the result is trivial, while if at least one of the components is nonzero, say $b_1$ without loss of generality, then (for $n>1$) $$\color{Blue}{-(b_2+b_3+\cdots+b_n)}(b_1\mathbf{a})+\color{Blue}{b_1}(b_2\mathbf{a})+\color{Blue}{b_1}(b_3\mathbf{a})+\cdots+\color{Blue}{b_1}(b_n\mathbf{a})$$ $$=\big((\color{Blue}{-b_2}b_1+\color{Blue}{b_1}b_2)+(\color{Blue}{-b_3}b_1+\color{Blue}{b_1}b_3)\cdots+(\color{Blue}{-b_n}b_1+\color{Blue}{b_1}b_n)\big)\mathbf{a}$$ $$=(0+0+\cdots+0)\mathbf{a}=\mathbf{0}.$$ is a nontrivial linear combination of the matrix columns that evaluates to zero, hence the columns are linearly dependent and thus the matrix's determinant is zero. Shortcut (via David): Use the multlinearity of the determinant to reduce $$\det\begin{pmatrix}b_1\mathbf{a}&b_2\mathbf{a}&\cdots&b_n\mathbf{a}\end{pmatrix}=b_1b_2\cdots b_n \det\begin{pmatrix}\mathbf{a}&\mathbf{a}&\cdots&\mathbf{a}\end{pmatrix}.$$ For $n>1$, it is impossible for $n$ copies of a vector $\mathbf{a}$ to be linearly independent, since e.g. $$1\mathbf{a}+(-1)\mathbf{a}+0\mathbf{a}+\cdots+0\mathbf{a}=\mathbf{0}.$$ Hence the original determinant must also be zero. Geometrically: The parallelepiped formed by the matrix's columns are all contained in the one-dimensional subspace generated by $\mathbf{a}$: for $n>1$ this has zero $n$-dimensional content (hyper-volume), hence the determinant is zero. Geometrically/Algebraically: As I posted in a different answer (paraphrased here), We're assuming $\mathbf{a},\mathbf{b}\ne\mathbf{0}$. Then $\mathbf{b}^\perp$, the orthogonal complement of the linear subspace generated by $\mathbf{b}$ (i.e. the set of all vectors orthogonal to $\mathbf{b}$) is therefore $(n-1)$-dimensional. Let $\mathbf{c}_1,\dots,\mathbf{c}_{n-1}$ be a basis for this space. Then they are linearly independent and $$(\mathbf{a}\mathbf{b}^T)\mathbf{c}_i =\mathbf{a}(\mathbf{b}^T\mathbf{c}_i)= (\mathbf{b}\cdot\mathbf{c}_i)\mathbf{a}=\mathbf{0}.$$ Thus the eigenvalue $0$ has geometric multiplicity $n-1\qquad$ [...] The determinant is the product of the matrix's eigenvalues, so if one of those is $0$ the product is necessarily zero as well. Analytically/Combinatorially: Via Leibniz formula we have $$\det(\mathbf{a}\mathbf{b}^T)=\sum_{\sigma\in S_n}(-1)^{\sigma}\prod_{k=1}^na_{\sigma(k)}b_k$$ $$=\left(\sum_{\sigma\in S_n}(-1)^\sigma\right)\prod_{k=1}^na_kb_k=0.$$ Above we observe that $\sum (-1)^{\sigma}$ is zero because the permutations of even and odd parity are in bijection with each other (e.g. take an arbitrary transposition $\tau$ and define the map $\sigma\to\tau\sigma$). - 3 Wow. Way to put that to bed. There is one more that might be mentioned using rank. $A=a^Tb$ and the rank of $a$ is $1$, so the rank of $A$ has to be $1$, so $\det(A)=0$ unless it is a $1\times1$ matrix. – robjohn♦ Oct 1 '11 at 13:31	{"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": 46, "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": 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": 0.9033951163291931, "perplexity_flag": "head"}
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If I have the ...	{"extraction_info": {"found_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": 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": 0.9147292971611023, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/14574/your-favorite-surprising-connections-in-mathematics/15515	## Your favorite surprising connections in Mathematics ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) There are certain things in mathematics that have caused me a pleasant surprise -- when some part of mathematics is brought to bear in a fundamental way on another, where the connection between the two is unexpected. The first example that comes to my mind is the proof by Furstenberg and Katznelson of Szemeredi's theorem on the existence of arbitrarily long arithmetic progressions in a set of integers which has positive upper Banach density, but using ergodic theory. Of course in the years since then, this idea has now become enshrined and may no longer be viewed as surprising, but it certainly was when it was first devised. Another unexpected connection was when Kolmogorov used Shannon's notion of probabilistic entropy as an important invariant in dynamical systems. So, what other surprising connections are there out there? - 3 I think it could be interpreted as restricting the breadth of the list. I would suggest removing the tag to avoid this. – François G. Dorais♦ Feb 8 2010 at 0:30 show 2 more comments ## 56 Answers Monstrous Moonshine. I mean why should the Fourier series of the $j$-invariant have coefficients related to the dimensions of the representations of the largest sporadic simple group? And why should the proof of this fact drag in mathematics from String Theory? - ### 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. As well known as the connection is, I am constantly amazed by the power of analytical geometry (developed by Descartes and Fermat) to make connections between geometrical ideas and algebraic ideas. It seems remarkable to me that so much geometrical information (as for example in the case of the conic sections) can be represented so succinctly (via quadratic equations in two variables). The geometry suggests things to think about in algebra and the algebra suggests things to think about in geometry. It is just amazing!! - 24 This is the observation that should have occurred to everyone first! (It didn't to me either.) It is so familiar we forget how amazing it is. – SixWingedSeraph Feb 8 2010 at 3:02 From an essay of Arnol'd: Jacobi noted, as mathematics' most fascinating property, that in it one and the same function controls both the presentations of a whole number as a sum of four squares and the real movement of a pendulum. - 3 See also: "My Lunch with Arnol'd" which helps put this surprise in perspective by showing it through the eyes of an amateur mathematician... www.gomboc.eu/99.pdf – David White May 19 2011 at 20:08 Quillen's result that the ring of cobordism classes of (stably) complex manifolds is isomorphic to Lazard's ring (i.e. the universal ring classifying formal group laws). This seems so mysterious to me. Why should cobordism classes of complex manifolds have anything to do with the algebraic geometry of formal group laws? Nevertheless this has been one of the most important observations for modern homotopy theory. It is the driving force behind Chromatic Stable Homotopy which tries to build a dictionary between the algebraic geometry of FGLs and structures present in the stable homotopy category. It is shocking how successful this has been. - My favorite surprise, which is perhaps the record-holder for the longest time it took for the two ideas to be brought together, is the connection between regular n-gons and Fermat primes. The Greeks knew how to construct regular n-gons by ruler and compass for n=3,4,5,6. Fermat introduced numbers of the form $2^{2^m}+1$ around 1640 in the mistaken belief they were prime for all m. Then in 1796 Gauss discovered how to construct the regular 17-gon, and a few years later showed that the n in a constructible n-gon is the product of some power of 2 by distinct Fermat primes. - I'll recycle one I mentioned in a thread last week, connecting an elementary problem about polynomials to the classification of finite simple groups: Definition: A polynomial $f(x) \in \mathbb{C}[x]$ is indecomposable if whenever $f(x) = g(h(x))$ for polynomials $g$, $h$, one of $g$ or $h$ is linear. Theorem. Let $f, g$, be nonconstant indecomposable polynomials over $\mathbb C$. Suppose that $f(x)-g(y)$ factors in $\mathbb{C}[x,y]$. Then either $g(x) = f(ax+b)$ for some $a,b \in \mathbb{C}$, or $$\operatorname{deg} f = \operatorname{deg} g = 7, 11, 13, 15, 21, \text{ or } 31,$$ and each of these possibilities does occur. The proof uses the classification of the finite simple groups [!!!] and is due to Fried [1980, in the proceedings of the 1979 Santa Cruz conference on finite groups], following a the reduction of the problem to a group/Galois-theoretic statement by Cassels [1970]. [W. Feit, "Some consequences of the classification of finite simple groups," 1980.] - Here is one of my favorites. If you consider a singular node of an algebraic curve locally it looks like the curves $xy=0$ in $\mathbb{C}^2$, or $x^2+y^2=0$. This consists of two smooth arcs intersecting to each other transversally (reducible in particular). Now, one step further, if we consider a cusp which is analytically equivalent to the origin in the curve $y^2+x^3=0$ in $\mathbb{C}^2$, it is locally irreducible. However, here comes the interesting point, if we intersect the singularity with a small ball $$[(x,y)\in \mathbb C^2:\ \|x\|^2+\|y\|^2=\epsilon]\cong S^3$$ what we've got is that such an intersection is $$(ae^{2i\theta},a^{3/2}e^{3i\theta})\subset S^1\times S^1\subset S^3$$ which is contained in a torus winding two times in one direction in the torus and three times in the other direction, in other words, we have an trefoil knot. Now in the case of surfaces, all these facts give rise to an amazing relation between topology and algebraic geometry. The underlaying space topological space in $\mathbb C^4$ of $$x^2+y^2+z^2+w^3=0$$ is a manifold!! (note it is singular at the origin in the context of AG!). As far as I know, if one intersects a small ball with the singularity, as I did above, one gets a topological sphere whose differential structure is NOT the standard one. Even more, considering in $\mathbb C^5$ the following hypersurface $$x^2+y^2+z^2+w^3+t^{6k-1}=0$$ and carrying out the intersection with a small sphere around the origin, for $k=1,2,\ldots 28$ one may get all the 28 possible exotic differential structures on the 7-sphere that Milnor found. - Complex multiplication of elliptic curves and the explicit construction of the maximal abelian extension of a quadratic imaginary number field. - 2 If you evaluate (appropriately normalized) elliptic functions at points lying in quadratic imaginary fields, the values you obtain are algebraic numbers, lying in abelian extensions of said quadratic imaginary fields; and all such extensions can be obtained in this way. (Compare with: $e^{2\pi i z}$ evaluated at rational numbers gives algebraic numbers, which generate abelian extension of ${\mathbb Q}$, and all abelian extension of ${\mathbb Q}$ are obtained in this way. – Emerton Feb 8 2010 at 5:03 1 Another way of saying it is that (with some slight fiddling) coordinates of points of finite order on an elliptic curve with complex multiplication give abelian extensions of the appropriate quadratic imaginary field. This was Kronecker's Jugendtraum (dream of his youth). Only in few cases is this explicit description of abelian extensions possible. Why do I think it is surprising? Compare what Emerton said above, $\exp(2 \pi iz)$ generating abelian extensions of $\mathbb{Q}$. Tell this to someone and ask them to guess how you'd generalise! It really is surprising that it is possible at all. – Sam Derbyshire Feb 8 2010 at 8:37 11 Elliptic curves (over $\mathbb{C}$) have their origins in studying elliptic integrals. As analytic objects they are $\mathbb{C}$ modulo a lattice. It's not immediately obvious to me that this is an algebraic object, and that the Weierstrass $\mathcal{P}$ function, which is an infinite sum, should compute anything number theoretic. So perhaps the connection is between analysis and algebra/arithmetic from this point of view. – Zavosh Feb 8 2010 at 21:16 show 4 more comments It is possible to compute the Betti numbers of a smooth complex variety $X(\mathbb{C})$ by computing the cardinality of $X(\mathbb{F}_{p^n})$ for a prime $p$ with good reduction and a finite number of positive integers $n$; in other words, by brute force. The above claim is wrong, so I'll phrase it the other way around. The Betti numbers of a smooth complex variety control the behavior of the number of points on $X(\mathbb{F}_{p^n})$; for example, for a smooth projective curve of genus $g$ we have $\|\text{Card}(X(\mathbb{F}_q))\| - q - 1\| \le 2g \sqrt{q}$. Generally I find the relationship between the arithmetic and topological properties of varieties surprising, although maybe this is a temporary kind of surprise that arithmetic geometers are used to. Another example: if $X$ is a curve, then whether the curvature of $X(\mathbb{C})$ is positive, zero, or negative determines whether $X(\mathbb{Q})$ is rationally parameterizable, a finitely generated group, or finite (unless it's empty). - I think the disparity between the world-views in low-dimensional topology versus high-dimensional topology are surprising. Even after you learn the reasons why, IMO they should still be surprising. Examples: 1) Teichmuller space exists, yet hyperbolic manifolds in dimension $3$ and larger are rigid. There are many interesting connections here such as the link between conformal geometry, complex analysis and hyperbolic geometry in dimension 2. 2) Exotic smooth structures on $\mathbb R^4$ but not on $\mathbb R^n$ for $n\neq 4$. 3) Why the Poincare conjecture/hypothesis is "hard" in dimensions $3$ and $4$ yet relatively "easy" in other dimensions. 4) Geometry being particularly relevant to $2$ and $3$-dimensional manifolds yet less so in higher dimensions. I could go on. Some of these are connections, some I suppose are disconnections. But a connection is only a surprise if you have reason to think otherwise. :) - 1 @Ryan: is the fact that geometry is less useful in high dimensions an empirical observation, or is there more mathematical content to this? – Jim Conant Mar 13 2011 at 17:44 1 @Jim: From Hillman's "Four-manifolds, Geometries and Knots", up to homeomorphism there are only 11 geometric 4-dimensional manifolds with finite fundamental group. In dimension 4 a finite-volume hyperbolic manifold's volume is a function of its Euler characteristic. I see those as having a fair bit of content. Sorry for being slow to reply. – Ryan Budney Sep 1 2011 at 6:40 Ehud Hrushovski's proof, using model theory, of the geometric Mordell-Lang conjecture in algebraic geometry. - How about something simple: $e^{i\pi}=-1$. Like when you first hear that, what the hell does the ratio of circumference to diameter of circles have to do with the square root of negative one and the base of the natural exponent? - 6 In my opinion, this becomes a lot less mysterious as soon as you think of the exponential and trigonometric functions as eigenfunctions of the differentiation operator (respectively, its square), which is really the reason they're both so important. The basic properties and interrelationships of these functions - including the above identity - are natural consequences of this formulation. – Robin Saunders Jul 29 2011 at 3:06 show 1 more comment The ubiquity of Littlewood-Richardson coefficients. Given three partitions $\lambda, \mu, \nu$ each with at most $n$ parts, there is a combinatorial definition for a number $c^\nu_{\lambda, \mu}$ which is nonzero if and only if any of the following statements are true: • There exist Hermitian matrices $A, B, C$ whose eigenvalues are $\lambda, \mu, \nu$, respectively and $A + B = C$ (one can also replace Hermitian by real symmetric) • The irreducible representation of ${\bf GL}_n({\bf C})$ with highest weight $\nu$ is a subrepresentation of the tensor product of those irreducible representations with highest weights $\lambda$ and $\mu$. • Indexing the Schubert cells of the Grassmannian ${\bf Gr}(d,{\bf C}^m)$ (where $d \ge n$ and $m-d$ is at least as big as any part of $\lambda, \mu, \nu$) by $\sigma_\lambda$ appropriately, the cycle $\sigma_\nu$ appears in the intersection product $\sigma_\lambda \sigma_\mu$. • There exists finite Abelian $p$-groups $A,B,C$ and a short exact sequence $0 \to A \to B \to C \to 0$ such that $B \cong \bigoplus_i {\bf Z}/p^{\nu_i}$, $A\cong \bigoplus_i {\bf Z}/p^{\lambda_i}$, and $C\cong \bigoplus_i {\bf Z}/p^{\mu_i}$. And probably many more things. - The beautiful analogy between number fields and function fields (and in general, algebra and geometry) that one learns about in arithmetic algebraic geometry. Some specific examples: The idea that a Galois group and a fundamental group (one algebro-number theoretic, the other geometric/topological) are two instances of the same thing. The use of the term ramification in both number theory and geometry. Describing $\mathbb{Z}$ as simply connected because $\mathbb{Q}$ has no unramified extensions. The appearance of integral closure in both algebraic geometry and algebraic number theory. The integral closure, in the former case, actually corresponds to a distinct geometric idea: non-singularity. The idea of considering a prime number to be a point; then viewing localization at that prime, -adic completion at that prime, and the residue field of that prime as if they were the corresponding geometric objects. In particular, using the term "local" in number theory, as if we were talking about geometry! This idea is built into scheme theory. There are many more examples. This book looks deeply into the relationships between Galois groups and fundamental groups and eventually develops a theory which covers both. This book explores the beautiful relation between algebraic curves and algebraic number theory. - Connection between the typical number of solutions ($N$) of a system of equations $$f_1=f_2=\cdots=f_n=0,$$ where each $f_k$ is a polynomial in $n$ complex variables, and the mixed volume ($V$) of the Newton polytopes of $f_k$: $$N=\tfrac1{n!}V.$$ - The work of Nabutovsky and Weinberger applying computability theory (a.k.a. recursion theory) to differential geometry. For example one of their results is that if you consider the space of Riemannian metrics on a smooth compact manifold $M$ of dimension at least 5 and sectional curvature $K\le 1$, then there are infinitely many extremal metrics. This is a purely geometric statement, but the only known proof uses concepts from computability theory. Moreover the results from computability theory that are used in their work are very deep; prior to their work, some skeptics regarded this area of computability theory as being overly specialized and having no hope of being connected to other areas of mathematics. See the exposition of Robert Soare (available on his website) for more information. - Here is a copypaste of something I've already mentioned in this question: http://mathoverflow.net/questions/12804/large-cardinal-axioms-and-grothendieck-universes The fastest known solution of the word problem in braid groups originated from research on large cardinal axioms; the proof is independent of the existence of large cardinals, although the first version of the proof did use them. See Dehornoy, From large cardinals to braids via distributive algebra, Journal of knot theory and ramifications, 4, 1, 33-79. To me this is an absolute mystery! Large cardinals are usually considered an esoteric subject situated on the border of the observable universe. So why should they have any relevance to braids, a very down to earth part of mathematics? Let alone give an algorithm for distinguishing braids, and what's more, the fastest algorithm known. - 1 Slides of a recent talk by Dehornoy on the history of this braid group problem may be found at math.unicaen.fr/~dehornoy/Talks/DyfShort.pdf – John Stillwell Feb 19 2010 at 1:45 The Jones polynomial of knot theory and Feynman path integrals. - 2 I'm not sure why this is surprising. It was originally defined via subfactors, but the path integral formalism followed very closely behind. Also, I'm not sure that Feynman path integrals count as mathematics... – Daniel Moskovich Feb 8 2010 at 5:07 3 Well, you are right. I think "surprising" is a subjective property, perhaps an experience of facing one's own ignorance. I don't know much about subfactors and first saw this polynomial in the context of knot invariants, divorced from its origins. For this reason, the physics connection seemed like a big surprise. It appears that you are an expert and so it is not surprising to me that you are not surprised. – Zavosh Feb 8 2010 at 15:50 2 I always found it ironic that knot theory began with Lord Kelvins model of atoms as knots in ether (loosely speaking). After a 360 degree rotation (or make that 720 degree :-) we're at string theory now. – Hauke Reddmann Jul 25 2011 at 12:00 show 1 more comment The pair correlation function between Riemann zeta function zeros is the same as the pair correlation function between eigenvalues of random Hermitian matrices. - 3 A caveat, this is not a theorem. Montgomery showed this for test functions whose Fourier transform had restricted support (in fact, support [-2,2] iirc.) Montgomery conjectured the same holds for more general test functions. Odlyzko's computations provided spectacular numerical evidence. And Katz-Sarnak proved an analogous statement for function fields. – Stopple Mar 13 2011 at 20:02 McKay's observation that the special fiber in the desingularization of du Val singularities is a bunch of $\mathbb P^1$s linked according to the Dynkin diagram corresponding to the group of the singularity. - This is probably not the most serious of applications, but I found the equivalence (in game theory) of the determinacy of Nash's board game Hex with the Brouwer Fixed Point theorem to be a surprising, if somewhat lighthearted, connection. You can read David Gale's paper. - The shortest path between two truths in the real domain passes through the complex domain. Le plus court chemin entre deux vérités dans le domaine réel passe par le domaine complexe.Jacques Hadamard - 3 Why down vote? I thought Jacques Hadamard expressed in his quote that in his time the connection of prime numbers to the zeroes of the Riemann zeta function was surprising and much of a shortcut to proving the Prime Number Theorem. – To be cont'd Jun 2 2010 at 1:05 Special values of the Riemann zeta function and class numbers of cyclotomic fields. - The surprising applications of algebraic geometry to number theory, for instance evidenced in the work of Deligne in proving the Ramanujan conjectures. - 6 Or, Riemann's use of complex variables to prove the Prime number theorem. – Feb7 Feb 8 2010 at 0:16 12 Riemann did not prove the prime number theorem. – S. Carnahan♦ Feb 8 2010 at 4:50 4 Although he came pretty close! – Emerton Feb 8 2010 at 5:05 7 Deligne's work was about counting solutions to equations over finite fields. Ramanujan's conjecture was about bounding the absolute values of the Fourier coefficients of a certain complex analytically defined function. How is the connection possibly tautological? – Emerton Feb 8 2010 at 5:14 6 Indeed, the connection between the Weil conjectures (and in particular the Riemann hypothesis, the proof of which is the work of Deligne being referred to) and Ramanujan's conjecture was only made some time after both conjectures were formulated (by Serre, I believe). – Emerton Feb 8 2010 at 5:17 show 5 more comments The fact that the circumference of a unit circle is used to normalize the bell curve. Elementary compared to the other examples, yes, but how shocking was it when you first learned it? - 2 To me, this isn't really shocking. It's a natural consequence of the cute (and, yes, maybe even surprising) fact that the square of $\int e^{x^2}\;dx$ is equal to $\int e^{x^2 + y^2}\;dx\;dy$, the integral of a function whose level sets are circles. – Vectornaut Feb 1 2012 at 21:45 My personal favorite is Multiple Zeta Values $$\zeta(s_1,\ldots,s_d) = \sum_{n_1>\ldots>n_d} \frac{1}{n_1^{s_1}\ldots n_d^{s_d}}$$ They appears in relation with • Quantum groups (they are coefficient of Drinfeld's KZ associator) • Deformation quantization (Kontsevich's formula for the affine space) • Feynmann diagrams (a large class of diagrams evaluate to MZV's) • Kashiwara-Vergne conjecture (representation theory of Lie groups) • Modular forms (Zagier noticed that the space of relations in depth 2 is canonically isomorphic to the space of cusp forms on $SL_2$ through their period polynomials) • Moduli spaces of curves of genus 0 $\mathcal{M}_{0,n}$ the list goes on and on... the reason for all this lies in the theory of mixed Tate motives over $\mathbb{Z}$. - Ulam's problem on determining the length of the longest increasing subsequence of a random permutation. The solution and the full description of the answer brought together ideas from integrable systems, combinatorics, representation theory, probability (appearing in the form of polynuclear growth model for instance), and random matrix theory. - There exist two binary trees with rotation distance $2n-6$. The proof is unexpected and based on hyperbolic geometry (Sleator, Tarjan, Thurston (1988), "Rotation distance, triangulations, and hyperbolic geometry"). - I was recently amazed at a quick connection between two facts I've known since high school. The Euler characteristic of a sphere, thought of as #vertices + #faces - #edges on a polyhedron, buckyball, etc., is 2; I re-deduced this from the fact that the derivative of $f(x)=1/x$ is $f'(x)=-1/x^2$. The steps of the proof are as follows: construct the Riemann sphere using two complex charts, both C, with the holomorphic transition map $f(z)=1/z$ on each neighborhood minus its origin. Now we want to look at the Chern class of the cotangent bundle, which in standard orientation is the negative of the Euler class of the tangent bundle, i.e. the sphere. Well, assuming complex analysis, look at $df=\frac{-1}{z^2}dz$ to see the effect of the transition map on the cotangent bundles: as a holomorphic'' 1-form, that has a double pole at one point and no zeros. Thus we know that a section of the cotangent bundle of the sphere has divisor degree $-2$. So $\chi(S^2)=2$ and I now cannot separate this fact from $f'(x)=-1/x^2$ in my mind. It seem somehow more mysterious, ridiculous, and delightful that this connection is so short. (Everyone I've mentioned this to prefers their own proof and perhaps it's better to do this slightly more directly to get a self-intersection 2 for a section of the tangent bundle, i.e. vector fields vanish twice, which gives the Euler class in $H^2(S^2)$ as a multiple of the orientation class.) - Goppa’s construction of error-correcting codes from curves, leading to the Tsfasman-Vladut-Zink bound (the first improvement over the Gilbert-Varshamov bound). An error-correcting code may be regarded as a combinatorial structure, and I think that this is a surprising connection between algebraic geometry and combinatorics. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 96, "mathjax_display_tex": 8, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9375460743904114, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/5688/probability-that-an-attacker-wins-the-discrete-logarithm-game-when-exponents-are/5690	# Probability that an attacker wins the discrete logarithm game when exponents are drawn from a subset Suppose $g$ is a generator of an order $p$ cyclic group in which discrete logarithm is hard and $p$ is a prime (i.e., given $g^x$ for a random $x \in \{0,1,\ldots, p-1\}$, it is hard to recover $x$ except with $negl(\lambda)$, where $negl(\lambda)$ is a negligible function of the security parameter $\lambda$). My question is as follows: what happens if I am given $g^y$ such that $y$ is chosen randomly from $\{0, 1, \ldots, q-1\}$ for some $q < p$, is there a way to quantify the hardness of discrete logarithm in terms of $q$? perhaps, something like: probability that one recovers $y$ is at most $\frac{1}{q} + negl(\lambda)$ and the probability is taken over the choice of $y$, or is there no way to quantify this? I am mainly interested in knowing what happens in practice when one does this. For instance, in most libraries, the exponents are chosen to be 160 bits even though $p$ is 1024 bits. (I posted this on math.stackexchange, but no responses. So, I am moving the question here by deleting it from there.) - 1 The exponents are 160 bits in integer DL systems with 1024 bit $p$, because there we are actually working in a subgroup of size $q$, with $q$ a 160 bit prime. – Paŭlo Ebermann♦ Dec 13 '12 at 21:10 so, does it mean if I take q to be 160 bits integer in my example above, I can assume the attacker's success probability to be negligible (i.e., can be ignored in practice)? – user52914 Dec 14 '12 at 5:59 1 Actually, the integers mod p are (multiplicatively) not a cyclic group (not even a group at all, but excluding the zero we get a (normally non-cyclic) group). Then we chose a generator which generates a cyclic subgroup of size $q$ (where $q$ is a divisor of $p-1$). And yes, we normally assume that the discrete logarithm in this subgroup is hard. – Paŭlo Ebermann♦ Dec 14 '12 at 12:48 – D.W. Dec 18 '12 at 22:42 ## 2 Answers The answer depends upon the factorization of $p-1$. Let me explain the situation for two example cases: • Suppose $p-1=2p'$, where $p'$ is also prime. This is the best case (for the defender): it minimizes the information leakage. Then the best known attack that takes advantage of the reduced range of the exponents will solve the discrete logarithm in about $\sqrt{q/2}$ time. For instance, if $q$ is 160 bits, then the best attack known will take about $2^{80}$ time (assuming this is less than the time to solve the discrete logarithm modulo $p$ for an arbitrary exponent). • Suppose $p-1=2p'p''$, where $p',p''$ are prime and $p'$ is pretty small and $p''$ is pretty big. This situation is bad for the defender and good for the attacker. An attacker can first recover the value of the exponent $y$ modulo $2p'$ in $\sqrt{p'}$ time (using Pohlig-Hellman). Then, the attacker can learn the full value of $y$ in $\sqrt{q/(2p')}$ time. The total running time will be $\sqrt{p'}+\sqrt{q/(2p')}$. If $p' \approx \sqrt{q}$, then the running time of the attack is $O(q^{1/4})$, which is significantly faster than the above -- yet the general discrete logarithm problem (for unrestricted exponents) might still be very hard. What's going on here? Two things. First, if we know that the exponent comes from a consecutive range of $n$ possible values, then it's possible to recover the exponent in $\sqrt{n}$ time. Second, for any divisor $d$ of $p-1$, one can recover the value of the exponent $y$ modulo $d$ in $\sqrt{d}$ time (raise the group element to the $(p-1)/d$ power, then use Pohlig-Hellman to solve the resulting discrete log problem). You can combine both of these ideas. However, nobody knows how to do any better than that. So, if you pick the parameters right, using an exponent from a restricted range can be secure. However, you have to pick them carefully. A good choice is to select $p$ so that $(p-1)/2$ is prime, and select $q$ to be at least 160 bits in length. Even better yet, follow Paŭlo Ebermann's advice: select a prime $q$ that is 160 bits long or longer and a prime $p$ so that $q$ divides $p-1$, and then choose the generator $g$ to be a generator of the cyclic subgroup of order $q$ (e.g., let $g_0$ be a generator for the group of integers modulo $p$, then let $g=g_0^{(p-1)/q}$). - @user52914, as far as I can see, nothing terribly bad should happen. Releasing $y_1+y_3$ and $y_2+y_3$ doesn't reveal much more information than releasing $y_1-y_2$. And releasing $y_1-y_2$ does not help find $y_1$ or $y_2$ (though if you somehow manage to learn one of $y_1$ or $y_2$, obviously knowledge of $y_1-y_2$ reveals the other). – D.W. Dec 17 '12 at 22:36 @user52914, hire a cryptographer to prove it for you? At a meta level: this has now deviated pretty far from the original topic of this question, so I don't think it's on-topic on this question. FYI, Crypto.SE is not a discussion forum: it's a question-and-answer site. One question per question. If you have a question about the design of a particular system, I think you'll need to post a separate question. If you do that, make sure you post detailed enough information about the design of the system that analysis is possible. – D.W. Dec 18 '12 at 7:04 @user52914, again, that is off-topic. If you want to discuss it, please post a separate question. The more you discuss it here, the more sorry I am that I responded to your second question at all. As far as explaining the site -- well, it sounds like you are not clear on how to use it properly and so I do need to explain the site policies. And no, it is not this way to maximize reputation of users -- it is this way to maximize the utility of the site as an archive of quality answers to relevant questions. Off-topic comments reduce the ability of others to find this information. – D.W. Dec 18 '12 at 22:38 Assume there exists an adversary for your q-subset DL problem that will be successful with probability $\mathsf{succ(\lambda)}$. If you just use this adversary to attack the original DL problem, he will get an input that comes from the set it expects with probability $q/p$ and will therefore be succesfull with probability at least $q/p \cdot \mathsf{succ(\lambda)}$. However we know that the DL problem is hard. Thus, we can bound the success probability of the adversary: $$\mathsf{negl}(\lambda) \ge q/p \cdot \mathsf{succ(\lambda)}$$ $$p/q\cdot\mathsf{negl}(\lambda) \ge \mathsf{succ(\lambda)}$$ Thus we get a bound for the success probability against your q-subset DL problem. Whether this bound is helpful, depends on $q$. As long as $p/q$ is polynomial, the upper bound is still a negligible function. So at least in theory this is still a hard problem. In practice and for a fixed group size that is of course a completely different story. Of course, if $p/q$ is superpolynomial, then the all bets are off and this upper bound does not help you in any way. - – Alexandre Yamajako Dec 13 '12 at 17:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 78, "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": 0.9295800924301147, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/1558/counter-mode-secure-hash-algorithm?answertab=votes	# Counter mode secure hash algorithm Ever since the SHA-3 competition, I've been wondering if it is possible to create a hash algorithm that is easier to parallelize. The current algorithms all seem to require building a tree of hashes. This has however rather serious drawbacks: 1. as they requires a tree with certain parameters (such as the size of nodes and branches), it needs to be explicitly defined what the tree looks like 2. communicating parties should be programmed explicitly to allow hash trees 3. the tree is likely to be optimized for certain configurations, and not all parties will have the same configuration (e.g. an 8 or 16 bit processor checking the hash created by a 64 bit server machine with GPU acceleration) 4. if the leaves are too big, you need a whole lot of information before you could start with the next leaf 5. with the exception of Skein, none of the remaining candidates seem to define how the hash tree should even be constructed Now I am wondering if you could construct a hash algorithm using a PRF that has a counter as input. In this construct, you would have a relatively strong PRF that is fed both the counter and a block plain text, resulting in an output of a known size (step1). This output would then be compressed together with the other transformed blocks in (step 2). Finally, as usual for a hash, you would have a finalization step which avoids length extension attacks (step3). To allow for easy parallelism, I think that step 2 should be associative. In other words, it should not matter in which order the output of step 1 is given to the function in step 2. If it is associative, it would even be possible to distribute step 2 between the processes as well. In that case, step 2 may even be relatively heavy (sponge like?). The trick, obviously, is to create collision resistance for the combination of step 1 and step 2. This is where my perfect scheme begins to crumble, and I am wondering if it is even possible to create a solution for the problem. The advantages of such a scheme are obvious: it would have none of the drawbacks of the tree hash scheme. Anybody could generate a hash using as many threads or computing blocks as they want. Only the block size, the output size of step 1 and the state of step2 would be restraining implementations. The algorithm would of course not be able to use the Merkle–Damgård construction, as it assumes that each block is processed in a sequential fashion - which introduces the problem in the first place. It seems that Merkle–Damgård and variations on Merkle–Damgård still rule supreme in the SHA-3 competition (at least with most of the final candidates). [EDIT: By now Keccak has been chosen as winner of the SHA-3 competition, and one of the reasons is that it was the candidate within the 5 remaining candidates that was using a sponge instead of a Merkle–Damgård construction, so I guess that this paragraph is out of date by now] Unfortunately, I'm not introduced to mathematics enough to see the correct solution, I cannot find if it has been tried, and I'm certainly not able to prove that my scheme would be impossible. Hopefully, this is where you can give me some hints. - Btw. constructing a MAC with such a property is much easier. We know several such MACs. – CodesInChaos Mar 1 at 6:48 ## 3 Answers First of all, I think I want to correct you at one point; in step 2, you aren't actually that interested in whether the operation is commutative, what you're actually interested in is that the operation is associative, that is, if $(a \oplus b) \oplus c = a \oplus (b \oplus c)$. In essence, your operator $\oplus$ in step 2 turns out to be a group operation. Hence, your proposed hash function can be expressed as: $H(P_1, ..., P_n) = G( \sum_i F( P_i, i ) )$ where: • F is the your so-called 'cipher mimicking counter-mode encryption' (bad terminology, btw) in step 1, which outputs group members • $\sum$ is the group operation sum over all the group members • G is finalization operation in step 3, which takes as input a group member, and outputs the actual hash (and may also take the file length as input; that detail is irrelevant) Now, assuming that F and G are well-chosen (for example, that it's infeasible to find collisions or preimages in F), the obvious way to attack this construct is to find a collision at the summation step; • for second-preimage on a hash of $P_i$, find a distinct set of blocks $Q_i$ such that $\sum_i F(P_i, i) = \sum_i F(Q_i, i)$ • for a collision attack, find distinct sets of blocks $P_i$, $Q_i$ such that $\sum_i F(P_i, i) = \sum_i F(Q_i, i)$ Now, I have good news, and I have bad news. The good news is that there exists groups where the above problems are NP-hard (!), and even better, one such group is integer addition modulo a large number. There are also groups where the above problems are easy (such as the group of bitstrings where the group operation is bit-wise exclusive or's), so we'll need to be careful when choosing a group. Now, the bad news; just because the problem is NP-hard doesn't mean that there aren't shortcuts. One such short-cut to the second-preimage problem is a square-root attack which works by dividing the blocks $Q_1, ..., Q_n$ into two sets $Q_1, ..., Q_{n/2}$ and $Q_{n/2+1}, ..., Q_n$, collecting sums from both subsets, and scanning the sums for a pair that sum to the target value. This attack allows us to find a set that sums to the target N-bit value in $\sqrt N$ time; hence the group members must be twice as big in order to keep the second-preimage problem as difficult as expected. The update of this observation is that for an $N$ bit hash, the internal sum must be over $2N$ bit values; that means that your $F$ function must generate strong $2N$ bit values, and so ends up being considerably harder to compute than the leaf compression function in a hash-tree construction. - "hence the group members must be twice as big in order to keep the second-preimage problem as difficult as expected" - Actually, they have to be a lot bigger than that (for typical groups, such as addition modulo a large number). The birthday attack you describe is one attack, but it is not optimal. Generalized birthday attacks are even more efficient. Resisting them requires even larger group elements. Therefore, following the advice in your answer ("use twice-as-wide group elements") is likely to lead to an insecure scheme. See my answer for more. – D.W. Nov 29 '12 at 21:47 @D.W.: Did I say "must be twice as big"? My bad -- I ought to have said "at least twice as big"; as you point out, there's no indication that the attack I outlined is optimal – poncho Nov 29 '12 at 22:40 I think the problem lies generally with your commutative compression function. The simplest such function (XOR) gives an easy way to craft messages with arbitrary hash codes by collecting blocks and using linear algebra to chose a subset - I could imagine that this problem would show (with more complicated calculations) for other such functions, too. This is not to say that this has to be impossible, but commutativity in cryptographic functions gives much more structure to attack, and finding a good commutative compression function seems hard. - You should look at AdHash, by Bellare et al. It computes something like $\sum_i H(i,M_i) \bmod n$ where $M_i$ is the $i$th block of the message. I think this is almost exactly the sort of thing you were suggesting. The problem with AdHash is that, for it to be secure, $n$ needs to be quite large, e.g., 1600 bits or more. This hurts efficiency. (If $n$ is small, then it can be broken with a generalized birthday attack.) Therefore, AdHash might not be terribly attractive in practice. Summary: I know of good parallelized MAC algorithms, but I don't know (off the top of my head) of any good parallelized hash algorithms. References: • A New Paradigm for collision-free hashing: Incrementality at reduced cost. Mihir Bellare and Daniele Micciancio. EUROCRYPT 1997. [Introduces AdHash, and other similar hashing methods.] • A Generalized Birthday Problem. David Wagner. CRYPTO 2002. [Describes the generalized birthday attack on AdHash.] • Is digest=HASH(HASH(a)+HASH(b)) equivalent to publishing two digests? - Thanks for the additional info, I did hear about the hash functions that use large numbers to accomplish "my" requirements, but they do indeed seem to be too impractical to achieve performance benefits over a generic hash. That said, there are likely special benefits e.g. when some parts of the data is only available after a certain amount of time. – owlstead Nov 29 '12 at 21:36 @owlstead, thanks for your comments. Well, all I can say is: You asked "I cannot find if it has been tried". I'm telling you it has been tried ... and the results have been found to be unsatisfactory. As you say, the resulting schemes are too slow to be very practical. The takeaway is that the kind of approach you sketch does not appear to be promising -- or, at least, no one knows how to make it work in a way that is secure and efficient enough to be terribly useful. – D.W. Nov 29 '12 at 21:50 That said, I don't think the problem has been studied well enough that it doesn't merit more attention. But I also have to admit that my current knowledge dictates that it won't be me that is making a proposal anytime soon :). – owlstead Nov 29 '12 at 21:55 @owlstead: Look, let me put it another way. If you want to innovate, the starting point is to start by understanding what has already been done and what has already been tried. That is what I am trying to outline in my answer. What I'm saying is that there is existing research on this topic, and reading about the past work would be a good place to start for anyone who is interested in the topic. (Maybe you've heard the saying: "a month in the lab can save you a day in the library".) – D.W. Nov 29 '12 at 22:08 Again thanks, I'm trying to get access but the first promising link in Google seem to be down. Maybe I should push some cash some way to get to the paper. So be it. – owlstead Nov 29 '12 at 22:20 show 1 more comment	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9583666324615479, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/20977/equation-for-y-from-fracy-1y21-2-c	# Equation for $y'$ from $\frac{y'}{ [1+(y')^2]^{1/2}} = c$ In a book there is a derivation for y' that comes from $$\frac{y'}{[1+(y')^2]^{1/2}} = c,$$ where $c$ is a constant. The result they had was $$y' = \sqrt{\frac{c^2}{1-c^2}}.$$ How did they get this? I tried expanding the square, and other tricks, I cant seem to get their result. - This formula holds if $c\geq 0$. More precise result is $y'=\frac{c}{\sqrt{1-c^2}}$ – Norbert Jan 25 '12 at 21:43 ## 1 Answer Square both sides, solve for ${y^{\prime}}^2$ and then take the square root. - Id done that, I just forgot to reverse the division of y'^2's, so they would cancel out and give 1+1/y'^2 = 1/c^2. Now it works. – user6786 Feb 8 '11 at 11:31	{"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": 6, "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": 0.9539570808410645, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/71330/behaviour-of-morita-equivalence-in-families-of-sheaves	## Behaviour of Morita equivalence in families of sheaves ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Given a projective scheme $X$, say over $\mathbb{C}$, another $\mathbb{C}$-scheme $S$ and a coherent and torsion free (as an $O_X$-module) $M_n(O_X)$-module $F$. Now we can use Morita equivalence to get a sheaf of $O_X$-modules $G$, which is easier to handle. Given a deformation $\mathcal{G}$ of $G$ over $S$, i.e. a sheaf of coherent $O_{X\times S}$-modules, flat as an $O_S$-module. Can i get a deformation $\mathcal{F}$ of $F$ over $S$ from this? That is if $\pi: X\times S \rightarrow X$ is the projection, $\mathcal{F}$ should be a $\pi^{}M_n(O_X)=M_n(O_{X\times S})$-module, coherent as on $O_{X\times S}$-module and flat as an $O_S$-module. Can i then just take $\mathcal{F}:=\mathcal{G}\otimes_{O_{X\times S}} \pi^{}(O_X^n)$. This is a $\pi^{*}M_n(O_X)$-module, and flat over $S$, because $\mathcal{G}$ is. If $s_0$ is the point for which the fiber of $\mathcal{G}$ is $G$, i.e. $\mathcal{G}_{s_0}=G$, the fiber of $\mathcal{F}$ over $s_0$ should be $F$, because $G$ and $F$ are Morita equivalent. So this should give the desired deformation. But this seems to good to be true. Are there any pitfalls in this construction? Or does Morita equivalence really behave very good in flat families? I could't find anything in the literature about this subject. Does anybody know a reference for this? - ## 1 Answer Your answer is correct, no pitfalls on the way. - Thanks. That is indeed very good. – TonyS Jul 28 2011 at 15:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 35, "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": 0.8886600136756897, "perplexity_flag": "head"}
http://gauravtiwari.org/tag/change/	# MY DIGITAL NOTEBOOK A Personal Blog On Mathematical Sciences and Technology Home » Posts tagged 'change' # Tag Archives: change ## Study Notes Announcements and more… Thursday, September 22nd, 2011 19:12 / 2 Comments # Announcement Hi all! I know some friends, who don’t know what mathematics in real is, always blame me for the language of the blog. It is very complicated and detailed. I understand that it is. But MY DIGITAL NOTEBOOK is mainly prepared for my study and research on mathematical sciences. So, I don’t care about what people say (SAID) about the A Torus content and how many hits did my posts get. I feel happy in such a way that MY DIGITAL NOTEBOOK has satisfied me at its peak-est level. I would like to thank WordPress.com for their brilliant blogging tools and to my those friends, teachers and classmates who always encourage me about my passion. For me the most important thing is my study. More I learn, more I will go ahead. So, today (I mean tonight) I have decided to write some lecture-notes (say them study-notes, since I am not a lecturer) on MY DIGITAL NOTEBOOK. I have planned to write on Group Theory at first and then on Real Analysis. And this post is just to introduce you with some fundamental notations which will be used in those study-notes. # Notations Conditionals and Operators $r /; c$ : Relation $r$ holds under the condition $c$. $a=b$ : The expression $a$ is mathematically identical to $b$. $a \ne b$ : The expression a is mathematically different from $b$. $x > y$ : The quantity $x$ is greater than quantity $y$. $x \ge y$ : The quantity $x$ is greater than or equal to the quantity $y$. $x < y$ : The quantity $x$ is less than quantity $y$. $x \le y$ : The quantity $x$ is less than or equal to quantity $y$. $P := Q$ : Statement $P$ defines statement $Q$. $a \wedge b$ : a and b. $a \vee b$ : a or b. $\forall a$ : for all $a$. $\exists$ : [there] exists. $\iff$: If and only if. Sets & Domains $\{ a_1, a_2, \ldots, a_n \}$ : A finite set with some elements $a_1, a_2, \ldots, a_n$. $\{ a_1, a_2, \ldots, a_n \ldots \}$ : An infinite set with elements $a_1, a_2, \ldots$ $\mathrm{\{ listElement /; domainSpecification\}}$ : A sequence of elements `listElement` with some `domainSpecifications` in the set. For example, $\{ x : x=\frac{p}{q} /; p \in \mathbb{Z}, q \in \mathbb{N^+}\}$ $a \in A$ : $a$ is an element of the set A. $a \notin A$: a is not an element of the set A. $x \in (a,b)$: The number x lies within the specified interval $(a,b)$. $x \notin (a,b)$: The number x does not belong to the specified interval $(a,b)$. Standard Set Notations $\mathbb{N}$ : the set of natural numbers $\{0, 1, 2, \ldots \}$ $\mathbb{N}^+$: The set of positive natural numbers: $\{1, 2, 3, \ldots \}$ $\mathbb{Z}$ : The set of integers $\{ 0, \pm 1, \pm 2, \ldots\}$ $\mathbb{Q}$ : The set of rational numbers $\mathbb{R}$: The set of real numbers $\mathbb{C}$: The set of complex numbers $\mathbb{P}$: The set of prime numbers. $\{ \}$ : The empty set. $\{ A \otimes B \}$ : The ordered set of sets $A$ and $B$. $n!$ : Factorial of n: $n!=1\cdot 2 \cdot 3 \ldots (n-1) n /; n \in \mathbb{N}$ Other mathematical notations, constants and terms will be introduced as their need. For Non-Mathematicians: Don’t worry I have planned to post more fun. Let’s see how the time proceeds! ###### Some Other GoodReads: • Using math symbols in wordpress pages and posts – LaTex (1) (etidhor.wordpress.com) • Do complex numbers really exist? (math.stackexchange.com) • LaTeX Typesetting – Basic Mathematics (r-bloggers.com) 26.740278 83.888889	{"extraction_info": {"found_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": 58, "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": 0.850537896156311, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/37990?sort=oldest	## Definitions/Background Suppose $S$ is a scheme and $D\subseteq S$ is an irreducible effective Cartier divisor. Then $D$ induces a morphism from $S$ to the stack $[\mathbb A^1/\mathbb G_m]$ (a morphism to this stack is the data of a line bundle and a global section of the line bundle, modulo scaling). For a positive integer $k$, the root stack $\sqrt[k]{D/S}$ is defined as the fiber product $\begin{matrix} \sqrt[k]{D/S} & \longrightarrow & [\mathbb A^1/\mathbb G_m] \\ p\downarrow & & \downarrow \wedge k \\ S & \longrightarrow & [\mathbb A^1/\mathbb G_m] \end{matrix}$ where the map $\wedge k: [\mathbb A^1/\mathbb G_m]\to [\mathbb A^1/\mathbb G_m]$ is induced by the maps $x\mapsto x^k$ (on $\mathbb A^1$) and $t\mapsto t^k$ (on $\mathbb G_m$). The morphism $p:\sqrt[k]{D/S}\to S$ is a coarse moduli space and is an isomorphism over $S\smallsetminus D$. Moreover, there is a divisor $D'$ on $\sqrt[k]{D/S}$ such that $p^D$ is $kD'$. The data of a morphism from $T$ to $\sqrt[k]{D/S}$ is equivalent to the data a morphism $f:T\to S$ and a divisor $E$ on $T$ such that $f^D = kE$. ## The question Suppose $\mathcal X$ is a DM stack, that $f:\mathcal X\to S$ is a coarse moduli space, that $f$ is an isomorphism over $S\smallsetminus D$, and that $f^D = kE$ for an irreducible Cartier divisor $E$ on $\mathcal X$. Is the induced morphism $\mathcal X\to \sqrt[k]{D/S}$ an isomorphism? I get the strong impression that the answer should be "yes", at least if additional conditions are placed on $\mathcal X$. ## A counterexample Here's a counterexample to show that some additional condition needs to be put on $\mathcal X$. Take $G$ to be $\mathbb A^1$ with a doubled origin, viewed as a group scheme over $\mathbb A^1$. Then $\mathcal X=[\mathbb A^1/G]\to \mathbb A^1$ is a coarse moduli space ("there's a $B(\mathbb Z/2)$ at the origin"). If we take $D\subseteq \mathbb A^1$ to be the origin, then the pullback to $\mathcal X$ is the closed $B(\mathbb Z/2)$ with multiplicity 1. Yet the induced morphism from $\mathcal X$ to $\sqrt[1]{D/\mathbb A^1}\cong \mathbb A^1$ is not an isomorphism. In this case, $\mathcal X$ is a smooth DM stack, but has non-separated diagonal. - I should have mentioned that this question arose while trying to understand the proof of Theorem 5.2 of Fantechi-Mann-Nironi's paper "Smooth Toric DM Stacks": arxiv.org/abs/0708.1254. – Anton Geraschenko♦ Jul 17 2011 at 20:56 ## 1 Answer What do you mean by an irreducible Cartier divisor? Assume that $S$ and $D$ are regular, that $\mathcal X$ is normal and has finite inertia, and that $f^D = kE$ for a reduced divisor $E$ on $\mathcal X$. Also assume that $\mathcal X$ is tame in codimension 1. Then the induced morphism $\mathcal X\to \sqrt[k]{D/S}$ is proper, because both stacks are proper over $S$. It is also birational. I claim that is representable in codimension 1; this follows from the fact that $\mathcal X$ is ramified of degree $k$ at the generic point of each irreducible component of $D$ (this can be done, for example, by taking the strict henselization of $S$ at the generic point of such a component, thus reducing to the case that $S$ is an henselian trait, which is easy, using the tameness hypothesis). Thus $\mathcal X\to \sqrt[k]{D/S}$ is a proper morphism with finite fibers, $\mathcal X$ is normal, $\sqrt[k]{D/S}$ is regular, and is an isomorphism in codimension 1. By purity of branch locus, it must be étale; and then it must be an isomorphism. I think that all of the hypotheses are necessary. For example, already when $D$ is a nodal curve on a smooth surface $S$ there are counterexamples: there is a smooth stacks having $S$ as its moduli space, which is ramified of order $k$ along $D$ (this is different from $\sqrt[k]{D/S}$, because the latter is singular). For example, when $D$ is the union of two smooth curves intersecting transversally, you take the fiber product of the root stacks of the two curves. There are also counterexamples when $\mathcal X$ is not normal, or when it is not tame. - Thanks! By irreducible Cartier divisor, I meant one which is not the sum of two other effective divisors; I'm happy to take "reduced" if there is a problem with this notion of irreducible. I don't see how you get representability. More details would be appreciated, even though that's not usually the nature of an exercise :-). Once you have proper and representable, you combine that with birational (which was essentially given) and quasi-finite (since both are quasi-finite over $S$) and apply Zariski's Main Theorem. Is that what you had in mind, or is there a simpler way? – Anton Geraschenko♦ Sep 7 2010 at 20:04 My previous post was very rash. I edited it, I hope that now it is clearer, and correct (the first version was just plain wrong). – Angelo Sep 8 2010 at 5:22 I'm having some trouble digesting this answer. What does it mean for a morphism to be representable in codimension 1? How does it imply the map is an isomorphism in codimension 1? (This means it's an isomorphism when you pull back to any codimension 1 point, right?) After you apply purity to get that the map is etale, don't you still need representability to conclude that it's an isomorphism? This last part doesn't bother me so much; I know an argument that any etale map of orbifolds is representable. The main thing I don't understand is how to get etaleness. – Anton Geraschenko♦ Sep 13 2010 at 21:44 1 Martin Olsson helped clear things up for me at tea today. Here's what I got from our conversation. It's enough to show the map is an isomorphism at the generic point of $D$, so we base change by the strict hensilization of the DVR $\mathcal O_{S,D}$. Then $\mathcal X$ must be of the form $[A/G]$, where $G$ is a finite group and $A$ is a strictly henselian ring (so a DVR). (I'm not completely sure what hypotheses have been used here.) The action of $G$ on the tangent space of the closed point of $A$ must be faithful, so $G\hookrightarrow \mathbb G_m$, so $G$ is $\mu_n$ for some $n$. – Anton Geraschenko♦ Sep 14 2010 at 1:04 1 Counting ramification, we get $n=k$. By dancing around a bit, we can show that $A=L[[t]]$ with the action you'd expect, where $L$ is the function field of $D$. This is precisely the $k$-th root stack of $\mathcal O_{X,D}^{sh}$ along the closed point. So the map to the root stack is an isomorphism over the generic point of $D$, so we can apply purity to get etaleness. – Anton Geraschenko♦ Sep 14 2010 at 1:04 show 1 more comment	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 91, "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": 0.9469703435897827, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/287371/monotone-hermite-interpolation	# monotone Hermite interpolation Is there a reason why in particular it is so popular to use monotone cubic Hermite interpolation compare to say, quadratic? I understand the the order of the accuracy gets better with a higher degree, but quadratic splines seem already good enough with second order globally. I wonder if there are other reasons... - ## 1 Answer While it is possible to interpolate data exactly with $C^1$ quadratic splines, this does not allow us to interpolate monotone data with monotone $C^1$ quadratic splines. Consider one of the monotone example data sets given to your previous Question, $(0,0), (1,0), (2,1), (3,1)$. As with the monotone $C^1$ cubic splines, the only possible polynomial portions in the first and last subintervals are constants, so that to be $C^1$ on the middle interval imposes four conditions, the function and its derivative at both endpoints $x=1,2$. While a (monotone) cubic polynomial can be constructed to meet those four conditions, it is impossible for any quadratic polynomial to have a derivative with two roots. - your answers are very helpful. Thanks a lot! – Medan Feb 14 at 5:14	{"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": 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": 0.9101046323776245, "perplexity_flag": "head"}
http://mathoverflow.net/questions/61348/simple-groups-with-the-same-cardinality-as-psl-2z-p/62308	## Simple groups with the same cardinality as PSL_2(Z/p) ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In an undergrad honors algebra course it's sometimes shown that when $p$ is prime and $>3$ then $PSL_2(Z/p)$ is simple of of order $p(p-1)(p+1)/2$. But that this is the "only" simple group having that order is seldom or never (except when $p=5$) proved. I've worked out a fairly simple proof, using the Burnside transfer theorem, of this last result, but it's perhaps a little too intricate to present in class. QUESTION: Are there proofs of this result, on-line or in texts, that are appropriate for an undergrad honors algebra class? (If not, I might post an argument on arXiv). EDIT: Since no simple available proof has yet been found, I'll sketch the argument that I culled from classification arguments for Zassenhaus groups. Suppose G is simple of order p(p-1)(p+1)/2. First one shows that G has p+1 p-Sylows. Let S be the union of Z/p and {infinity}. An easy study of the conjugation action of G on the p-Sylows allows one to identify G with a doubly transitive group of (even) permutations of S, containing the p-cycle z-->z+1. Then the subgroup of G fixing both 0 and infinity is cyclic generated by z-->cz for some c. Once this is done, the key is in showing: A.--- The subgroup of elements that either fix 0 and infinity or interchange them is dihedral. Once A is shown it's not hard to show that z-->-1/z is in G, thereby identifying G with a fractional linear group. The proof of A is a counting argument when p=1 mod 4. But when p=3 mod 4 the situation is more delicate, and one uses Burnside tranfer. - 1 The Warwick wiki looks broken in rendering LaTeX, but it does $p=7$, as part of the standard "all simple groups less than order 500" exercise. wiki.dcs.warwick.ac.uk/ma3e2tidywiki/index.php/… The case of $p=9$ (not prime, but) is done a few places like Isaacs and an exercise in Rotman, usually in the guise of $A_6$. Cole does $p=11$ in his 1893 census. jstor.org/stable/2369516?seq=12 – Junkie Apr 12 2011 at 0:31 Burnside does $p=13$ in his 1895 paper (on the last page). plms.oxfordjournals.org/content/s1-26/1/… The proof is only a paragraph long, relying on known transitive groups of degree 14 it seems. – Junkie Apr 12 2011 at 0:43 1 I think it is well-known the exercise in Rotman is incorrect. – Steve D Apr 12 2011 at 0:45 1 That said, I would love to see a general argument using Burnside's Transfer Theorem!! – Steve D Apr 12 2011 at 0:50 1 @Junkie: Yes, but no simple group of order 360 has 6 Sylow 5-groups! – Steve D Apr 12 2011 at 2:00 show 4 more comments ## 4 Answers I seem to nave reinvented the wheel! I quote Gallian from his 1976 survey of the current state of the classification problem (Mathematics Magazine 1976): In 1902 Frobenius determined all transitive permutation groups of degree p+1 and order p(p^2-1)/2 where p is prime.... it follows ... that PSL(2,p) is the only simple group of order p(p^2-1)/2. This appears to be the first arithmetical characterization of an infinite family of simple groups. I haven't been able to access the Frobenius paper, but I expect that I've more or less duplicated his argument. Thanks Junkie for your comments which led me to the Frobenius paper and the Gallian article citing it. EDIT: My version of the Frobenius paper is now up on arXiv: (GR) 1107.4130, "Frobenius' result on simple groups of order (p^3-p)/2." I abandoned my Burnside transfer approach for p=3 mod 4, replacing it with a simplification of Frobenius' more elementary proof. I thank those here who led me to his elegant article. Those who would like to use my note in the classroom can of course do so, but I disclaim responsibility. - Are you sure your argument parallels his? It seems a lot of people would like to see a nice proof of this theorem (esp, in English); and that's my request for you to post a quick note on the arXive. – Dr Shello May 27 2011 at 15:36 Dr Shello--Apart from a few simplifications by me the argument is the same. I've written it up and can send you a copy by post if you give me an address. It will be another 6 weeks or so before I have a chance to get it up on the arXiv. – paul Monsky May 27 2011 at 16:55 ### 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 not sure that I know a really elementary proof of this fact, though I can imagine ways to start reconstructing one (Sylow theory and, as you say, transfer seem good places to start). But, in the end, you need a way to recognise this simple group. Seeing later comments, five pages sounds reasonable. There is a Theorem of M. Herzog (circa 1970) which (roughly speaking) characterizes finite groups of order less than $p^{3a}$ with a cyclic Sylow $p$-subgroup of order $p^{a}$, but this uses fairly sophisticated character theory, and (including background) needs a lot more than five pages. - I think the comments here converge toward the answer "no" to your basic question. While some very special small primes are treated in a few textbooks (using what seem to be ad hoc methods), it's much more natural to treat the entire family of groups at once in the wider context of the classification of finite simple groups. Here there should be some general methods in play, e.g., BN-pair structure, structure of "local" subgroups such as the normalizers of Sylow subgroups, centralizers of involutions, etc. Not to mention many techniques from character theory. I'm not aware of any treatment of this special family of rank 1 groups in a form suitable even for an honors course. The two books Danny Gorenstein wrote on finite groups, especially the second one published in 1982 when he and others felt the classification was complete, illustrate the difficulty at that time in organizing the subject in textbook form. In that later book the special result you want is embedded in a substantial treatment involving doubly transitive permutation groups, Zassenhaus groups, etc. I don't think it's feasible to extract from that textbook version an efficient proof for your purpose without bypassing the important underlying general ideas, though it's always interesting to see what can be written down in a more elementary and self-contained way. Anyway, the problem here is to work out in a very limited case the "recognition" theorem for finite simple groups: how do you argue that an unknown simple group (here specified just by an order formula) is isomorphic to some known group? - Jim--My argument just uses some of the elementary preliminaries to the classification of the Zassenhaus groups. It's 5 leisurely handwritten pages. I'm not sure if I'm bypassing the underlying ideas; if you'd like to see it I'll put a copy in the mail to you. – paul Monsky Apr 12 2011 at 18:18 Yes, it's interesting to see what can be done, especially because this isn't a standard textbook topic. My main question based on the comments made here is whether there is a good unified method applicable to the entire family of rank 1 groups. That seems to me the most natural approach to such a question concerning a Lie family. – Jim Humphreys Apr 12 2011 at 19:35 This being community wiki, I'll add another "answer" in response to Paul's recent post. It's clear that the desired uniqueness statement for simple groups of this special order is embedded in the CFSG, so the question is whether there is a way to isolate the result for an honors class in a self-contained treatment (modulo knowledge of Sylow theorems and the like). Probably what Paul has arrived at in his own way is about as efficient as possible. As he points out, one wants to show that a Sylow `$p$`-subgroup of the unknown `$G$` is of index `$(p-1)/2$` in its normalizer (which plays the role of Borel subgroup `$B$` in a BN-pair of rank 1). Then you are looking at a doubly transitive permutation group on the `$p+1$` elements of the coset space `$G/B$` (thought of as rational points of a projective line). I've just looked at the relevant Frobenius paper, being probably the only person in the U.S. with all three volumes of his collected papers on a shelf several feet from my home computer. (Not that I've read them all. It was an extra set owned by Wilhelm Magnus.) The 19 page 1902 paper Uber Gruppen des Grades `$p$` oder `$p+1$` was published in the Sitzungsberichte ...; it appears as number 66 in the third volume of collected papers containing later work on finite groups and character theory. The paper is somewhat electic, but the early Satz II affirms for any prime (except 7) the existence of a unique transitive permutation group of degree `$p+1$` and order `$p(p^2-1)/2$`. This requires four pages or so of heavy proof, with reference back to Sylow and others. It's hard to see at a glance how this translates into current terminology, but for example he sees very soon that the group permutes `$p+1$` objects which he denotes `$\infty, 0, 1, \dots, p-1$`. Once all the work is done, it follows as Satz III that for `$p>3$` there is a unique simple group of the given order. Is it true that no later textbook proof has actually been given? It's not a result to be taken up in a standard elementary course, or even a graduate course where there is too much else to cover, but it illustrates nicely the starting point for CFSG beyond alternating groups. -	{"extraction_info": {"found_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": 0.9444729685783386, "perplexity_flag": "middle"}
http://gauravtiwari.org/tag/set-theory/	# MY DIGITAL NOTEBOOK A Personal Blog On Mathematical Sciences and Technology Home » Posts tagged 'set theory' # Tag Archives: set theory ## A Trip To Mathematics: Part II Set(Basics) Friday, November 4th, 2011 05:55 / 3 Comments # Introduction In English dictionary, the word Set has various meanings. It is often said to be the word with maximum meanings (synonyms). But out of all, we should consider only one meaning: ”collection of objects” — a phrase that provides you enough clarity about what Set is all about. But It is not the exact mathematical definition of Set . The theory of Set as a mathematical discipline rose up with George Cantor, German mathematician. It is said that Cantor was working on some problems in Trigonometric series and series of real numbers, which accidently led him to recognise the importance of some distinct collections and intervals. And he started developing Set Theory. Well, we are not here to discuss the history of sets; but Mathematical importance. Cantor defined the set as a ‘plurality concieved as a unity’ (many in one; in other words, mentally putting together a number of things and assigning them into one box). Mathematically, a Set $S$ is ‘any collection’ of definite, distinguishable objects of our universe, concieved as a whole. The objects (or things) are called the elements or members of the set $S$. Some sets which are often termed in real life are, words like ”bunch”, ”herd”, ”flock” etc. The set is a different entity from any of its members. For example, a flock of birds (set) is not just only a single bird (member of the set). ‘Flock’ is just a concept with no material existence but ‘Bird’ or ‘birds’ are real. # Representing sets Sets are represented in two main ways: 1. Standard Method: In this method we use to write all elements of a set in a curly bracket ( { } ). For example: Flock of Birds := {Bird-1, Bird-2, …, Bird-100,…} or, $F:= \{B_1, B_2, \ldots, B_{100}, \ldots \}$ is a set. Here I have used first capital letter of each term to notate the example mathematically. We read this set as, A set F is defined by a collection of objects $B_1, B_2, \ldots$ etc. 2. Characteristic Method: In this method, we write a representative element and define that by a characteristic property. A characteristic property of a set is a property which is satisfied by each member of that set and by nothing else. For example, above set of Flock of birds can also be written as: $\mathrm{ F := \{ B : B \ is \ a \ bird \} }$ which has the same meaning at a wider extent. We read it as: ”A set F is defined by element B such that B is a bird.” # Standard Sets Some standard sets in Mathematics are: Set of Natural Numbers: It includes of the numbers, which we can count, viz. $\mathrm{ \{0,1,2,3,4,5,6,7, \ldots \}}$. The set of natural numbers is denoted by $\mathbb{N}$. Set of Integers: Integers includes of negatives of natural numbers and natural numbers itself. It is denoted by $\mathbb{Z}$. $-5, -4, 1, 2, 0$ …all are integers. The rigorous definition of integers be discussed in fourth part of the series. Set of Rational Numbers: Rational numbers are numbers which might be represented as $\frac{p}{q}$, where p and q both are integers and relatively prime to each other and q not being zero. The set of rational numbers is represented by $\mathbb{Q}$ and may include elements like $\frac{2}{3}, \frac{-5}{7}, 8$. The characteristic notation of the set of rational numbers is $\mathbb{Q} := \{ \dfrac{p}{q};/ p,q \in \mathbb{Z}, \ (p,q) \equiv 1 \ q \ne 0 \}$. The rigorous dicussion about rational numbers will be provided in fourth part of the series. Empty Set: It is possible to conceive a set with no elements at all. Such a set is variously known as an empty set or a void set or a vacuous set or a null set. An example of emptyset is the set $\{\mathrm{x:\ x \ is \ an \ integer \ and \ x^2=2} \}$, since there exists no integer which square is 2 —the set is empty. The unique empty set is denoted by $\emptyset$. Unit Set: A set with only one element is called the unit set. {x} is a unit set. Universal Set: A set which contains every possible element in the universe, is a universal set. It is denoted by $U$. # Two Sets Let $A$ and $B$ be two sets. We say that $A$ is a subset of $B$ (or $B$ is superset of $A$ or $A$ is contained in $B$ or $B$ contains $A$) if every element of $A$ is also an element of set $B$. In this case we write, $A \subseteq B$ or $B \supseteq A$ respectively, having the same meaning . Two sets are equal to each other if and only if each is a subset of the other. Subset word might be understood using ‘sub-collection’ or ‘subfamily’ as its synonyms. Operations on Sets: As Addition, Subtraction, Multiplication and Division are the most common mathematical operations between numbers; Union, Intersection, Complement, Symmetric difference, Cartesian Products are the same between sets. UNION OF SETS If A and B are two sets, then their union (or join) is the set, defined by another set $S \cup T$ such that it consists of elements from either A or B or both. If we write the sets A and B using Characteristic Method as, $\mathrm{A:= \{x : x \ is \ an \ element \ of \ set \ A\}}$. and,$\mathrm{B:=\{x : x \ is \ an \ element \ of \ set \ B\}}$ then the union set of A and B is defined by set J such that $\mathrm{J := \{x: x \ is \ an \ element \ of \ set \ A \ or \ set \ B \ or \ both \}}$. For practical example, let we have two sets: $A:= \{1,2,3,r,t,y\}$ and $B:=\{3,6,9,r,y,g,k\}$ be any two sets; then their union is $A \cup B :=\{ 1,2,3,6,9,r,t,y,g,k \}$. Note that it behaves like writting all the elements of each set, just caring that you are not allowed to write one element twice. Here is a short video explaining Unions of Sets: INTERSECTION OF SETS Intersection or meet of two sets A and B is similarly defined by ‘and’ connective. The set {x: x is an element of A and x is an element of B} or briefly $\mathrm { \{ x: x \in A \wedge x \in B \}}$. It is denoted by $A \cap B$ or by $A \cdot B$ or by $AB$. For example, and by definition, if A and B be two sets defined as, $A:=\{1,2,3,r,t,y\}$ $B:=\{3,6,9,r,g,k\}$ then their intersection set, defined by $A \cap B:= \{3,r\}$. In simple words, the set formed with all common elements of two or more sets is called the intersection set of those sets. Here is a video explaining the intersection of sets: If, again, A and B are two sets, we say that A is disjoint from B or B is disjoint from A or both A and B are mutually disjoint, if they have no common elements. Mathematically, two sets A and B are said to be disjoint iff $A \cap B := \emptyset$ . If two sets are not disjoint, they are said to intersect each other. PARTITION OF A SET A partition set of a set X is a disjoint collection of non-empty and distinct subsets of X such that each member of X is a member of exactly one member (subset) of the collection. For example, if $\{q,w,e,r,t,y,u\}$ is a set of keyboard letters, then $\{ \{q,w,e\}, \{r\}, \{t,y\},\{u\}\}$ is a partition of the set and each element of the set belongs to exactly one member (subset) of partition set. Note that there are many partition sets possible for a set. For example, $\{\{q,w\}, \{e,r\},\{t,y,u\}\}$ is also a partition set of set $\{q,w,e,r,t,y,u\}$. A Video on Partition of set: COMPLEMENT SET OF A SET The complement set $A^c$ of a set $A$ is a collection of objects which do not belong to $A$. Mathematically, $A^c := \{x: x \notin A \}$. The relative complement of set $A$ with respect to another set $X$ is $X \cap A^c$ ; i.e., intersection of set $X$ and the complement set of $A$. This is usually shortened by $X-A$, read X minus A. Thus, $X-A := {x : x \in X \wedge x \notin A}$, that is, the set of members of $X$ which are not members of $A$. The complement set is considered as a relatative complement set with respect to (w.r.t) the universal set, and is called the Absolute Complement Set. A Video on Complement of A Set: SYMMETRIC DIFFERENCE The symmetric difference is another difference of sets $A$ and $B$, symbolized $A \Delta B$, is defined by the union of mutual complements of sets $A$ and $B$, i.e., $A \Delta B := (A -B) \cup (B-A) = B \Delta A$. # Theorems on Sets 1. $A \cup (B \cup C) = (A \cup B) \cup C$ 2. $A \cap (B \cap C) = (A \cap B) \cap C$ 3. $A \cup B= B \cup A$ 4. $A \cap B= B \cap A$ 5. $A \cup (B \cap C)= (A \cup B) \cap (A \cup C)$ 6. $A \cap (B \cup C)= (A \cap B) \cup (A \cap C)$ 7. $A \cup \emptyset= A$ 8. $A \cap \emptyset= \emptyset$ 9. $A \cup U=U$ 10. $A \cap U=A$ 11. $A \cup A^c=U$ 12. $A \cap A^c=\emptyset$ 13. If $\forall A \ , A \cup B=A$ $\Rightarrow B=\emptyset$ 14. If $\forall A \ , A \cap B=A \Rightarrow B=U$ 15. Self-dual Property: If $A \cup B =U$ and $A \cap B=\emptyset \ \Rightarrow B=A^c$ 16. Self Dual: ${(A^c)}^c=A$ 17. ${\emptyset}^c=U$ 18. $U^c= \emptyset$ 19. Idempotent Law: $A \cup A=A$ 20. Idempotent Law: $A \cap A =A$ 21. Absorption Law: $A \cup (A \cap B) =A$ 22. Absorption Law: $A \cap (A \cup B) =A$ 23. de Morgen Law: ${(A \cup B)}^c =A^c \cap B^c$ 24. de Morgen Law: ${(A \cap B)}^c =A^c \cup B^c$ # Another Theorem The following statements about set A and set B are equivalent to one another 1. $A \subseteq B$ 2. $A \cap B=A$ 3. $A \cup B =B$ I trust that we are familiar with the basic properties of complements, unions and intersections. We should now turn to another very important concept, that of a function. So how to define a function? Have we any hint that can lead us to define one of the most important terms in mathematics? We have notion of Sets. We will use it in an ordered manner, saying that an ordered pair. First of all we need to explain the the notion of an ordered pair. If $x$ and $y$ are some objects, how should we define the ordered pair $(x,y)$ of those objects? By another set? Yes!! The ordered pair is also termed as an ordered set. We define ordered pair $(x,y)$ to be the set $\{\{x,y\},\{x\}\}$. We can denote the ordered pair $(x,y)$ by too, if there is a desperate need to use the small bracket ‘( )’ elsewhere. So, note that Ordered Pairs $(x,y) := \{\{x,y\},\{x\}\}$ and $(y,x) :=\{\{y,x\},\{y\}\} =\{\{x,y\},\{y\}\}$ are not identical. Both are different sets. You might think that if ordered pair can be defined with two objects, then why not with three or more objects. As we defined ordered pair (ordered double, as a term) $(x,y)$, we can also define $(x,y,z)$, an ordered triple. And similarly an ordered $n$ -tuple $(x_1, x_2, \ldots x_n)$ in general such that: $(x,y):=\{\{x,y\},\{x\}\}$ $(x,y,z):=\{\{x,y,z\},\{x,y\},\{x\}\}$ $(x_1, x_2, x_3, \ldots x_n) := \{\{x_1, x_2, x_3, \ldots x_n\}, \{x_1, x_2, x_3, \ldots x_{n-1}\}, \ldots, \{x_1, x_2\}, \{x_1\}\}$. Another important topic, which is very important in process to define function (actually in process to define ordered pair) is Cartesian Product (say it, Product, simply) of two sets. Let $A$ and $B$ be two sets. Then their Product (I said, we’ll not use Cartesian anymore) is defined to be the (another) set of an ordered pair, $(a,b)$, where $a$ and $b$ are the elements of set $A$ and set $B$ respectively. Mathematically; the product of two sets $A$ and $B$ $A \times B := \{(a,b) : a \in A, \ b \in B\}$. Note that $A \times B \ne B \times A$. The name as well as the notation is suggestive in that if $A$ has $m$ elements, $B$ has $n$ elements then $A \times B$ indeed has $mn$ elements. We see that if we product two sets, we get an ordered pair of two objects (now we’ll say them, variables). Similarly if we product more than two sets we get ordered pair of same number of variables. For example: $X \times Y := (x,y); x \in X, y \in Y$. $X \times Y \times Z :=(x,y,z); x \in X, y \in Y, z \in Z$. etc. The sets, which are being product are called the factor sets of the ordered pair obtained. When we form products, it is not necessary that the factor sets be distinct. The product of the same set $A$ taken $n$ times is called the $n$ -th power of $A$ and is denoted by $A^n$. Thus, $A^2$ is $A \times A$. $A^3$ is $A \times A \times A$. And so on. Now we are ready to define functions. The next part of this series will focus on functions. Keep Reading and Commenting. 26.740278 83.888889 1. # Cantor’s Concept of a set A set $S$ is any collection of definite, distinguishable objects of our intuition or of our intellect to be conceived as a whole. The objects are called the elements or members of set $S$ 2. # The intuitive principle of extension for sets Two sets are equal if and only if (iff) they have the same members. i.e., $X=Y \, \Leftrightarrow \,\forall x \in X \text{and} \ x \in Y$. 3. # The intuitive principle of abstraction A formula (syn: property) $P(x)$ defines a set $A$ by the convention that the members of $A$ are exactly those objects $a$ such that $P(a)$ is a true statement. $\Rightarrow a \in \{ x\|P(x) \}$. 4. # Operations with/for sets • Union (Sum or Join)$A \cup B= \{ x \| x \in A \, \text{or} \, x \in B \}$ • Intersection (Product or Meet) $A \cap B= \{ x\| x\in A \, \text{and} \, x\in B \}$ • Disjoint Sets $A$ and $B$ are disjoint sets iff $A \cap B=\emptyset : \text{an empty set}$ and they intersect iff $A \cap B \ne \emptyset$ • Partition of Sets A partition of a set $X$ is a disjoint collection $\mathfrak{X}$ of non-empty and distinct subsets of $X$ such that each member of $X$ is a member of some (and hence exactly one) member of $\mathfrak{X}$. For example: $\{ \{a,b\} \, \{c \} \, \{d, e\} \}$ is a partition of $\{a,b,c,d,e\}$. • Absolute Complement of a set $A$ is usually represented by $\Bar{A} = U-A = \{ x \| x \notin A \}$ where $U$ is universal set. • Relative Complement of a set $A \, \text{relative to another set} \, X$ is given by $X-A=X\cap \Bar{A}=\{ x \in X \| x \notin A\}$. 5. # Theorems on Sets 1. $A \cup (B \cup C) = (A \cup B) \cup C$ 2. $A \cap (B \cap C) = (A \cap B) \cap C$ 3. $A \cup B= B \cup A$ 4. $A \cap B= B \cap A$ 5. $A \cup (B \cap C)= (A \cup B) \cap (A \cup C)$ 6. $A \cap (B \cup C)= (A \cap B) \cup (A \cap C)$ 7. $A \cup \emptyset= A$ 8. $A \cap \emptyset= \emptyset$ 9. $A \cup U=U$ 10. $A \cap U=A$ 11. $A \cup \Bar{A}=U$ 12. $A \cap \Bar{A}=\emptyset$ 13. If $\forall A \ , A \cup B=A$ $\Rightarrow B=\emptyset$ 14. If $\forall A \ , A \cap B=A \Rightarrow B=U$ 15. Self-dual Property: If $A \cup B =U$ and $A \cap B=\emptyset \ \Rightarrow B=\Bar{A}$ 16. Self Dual: $\Bar{\Bar{A}}=A$ 17. $\overline{\emptyset}=U$ 18. $\Bar{U}= \emptyset$ 19. Idempotent Law: $A \cup A=A$ 20. Idempotent Law: $A \cap A =A$ 21. Absorption Law: $A \cup (A \cap B) =A$ 22. Absorption Law: $A \cap (A \cup B) =A$ 23. de Morgen Law: $\overline{A \cup B} =\Bar{A} \cap \Bar{B}$ 24. de Morgen Law: $\overline{A \cap B} =\Bar{A} \cup \Bar{B}$ 6. # Another Theorem The following statements about set A and set B are equivalent to one another 1. $A \subseteq B$ 2. $A \cap B=A$ 3. $A \cup B =B$ 7. # Functions Function is a relation such that no two distinct members have the same first co-ordinate in its graph. $f$ is a function iff 1. The members of $f$ are ordered pairs. 2. If ordered pairs $(x, y)$ and $(x, z)$ are members of $f$, then $y=z$ 8. Other words used as synonyms for the word ‘function’ are ‘transformation’, ‘map’, ‘mapping’, ‘correspondence’ and ‘operator’. 9. # Notations for functions A function is usually defined as ordered-pairs, see above, and $\text{ordered pair } (x,y) \in \text{function } f$ so that $xfy$ is (was) a way to represent where $x$ is an argument of $f$ and $y$ is image (value) of $f$. Other popular notations for $(x,y)\in f$ are: $y : xf$, $y=f(x)$, $y=fx$, $y=x^f$. 10. # Intuitive law of extension for Functions Two sets $f$ and $g$ are equal iff they have the same members (here, Domain and Range) $\Rightarrow f=g \Leftrightarrow D_f=D_g \ \text{and } \ f(x)=g(x)$ 11. # Into Function A function $f$ is into $Y$ iff the range of $f$ is a subset of $Y$. i.e., $R_f \subset Y$ 12. # Onto Function A function $f$ is onto $Y$ iff the range of $f$ is $Y$. i.e., $R_f=Y$ 13. Generally a mapping is represented by $f : X \rightarrow Y$. 14. # One-to-One function A function is called one-to-one if it maps distinct elements onto distinct elements. A function $f$ is one-to-one iff $x_1 \ne x_2 \Leftrightarrow f(x_1) \ne f(x_2)$ and $x_1 = x_2 \Leftrightarrow f(x_1)=f(x_2)$ 15. # Restriction of Function If $f : X \rightarrow Y$ and if $A \subseteq X$, then $f \cap (A \times Y)$ is a function on $A \ \text{into } \ Y$, called the restriction of $f$ to $A$ and $f \cap (A \times Y)$ is usually abbrevated by $f\|A$. 16. # Extension of function The function $f$ is an extension of a function $g$ iff $g \subseteq f$. 26.740278 83.888889	{"extraction_info": {"found_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": 241, "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": 0.9314488172531128, "perplexity_flag": "head"}
http://mathoverflow.net/questions/83753?sort=oldest	## Characterizing specific “concrete” mathematical objects by abstract general properties ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In this note by Tom Leinster the Banach space $\mathrm{L}^1[0,1]$ is recovered by "abstract nonsense" as the initial object of a certain category of (decorated) Banach spaces. So a function space, that would habitually be defined through the machinery of Lebesgue measure and integration, is uniquely described (up to isomorphism) in terms of abstract functional analysis and a bit of category theory. I would be curious to see more results, ideally in diverse areas of mathematics, in the spirit of the above one, in which a familiar and important "concrete" mathematical object is recovered by a universal property (in the technical categorical sense) or -more generally- by a characterizing property that is abstract and general or doesn't delve into the "concrete" habitual definition of that object. Community wiki, so put one item per answer please. - 1 Doesn't this amount to a collection of adjoint functor pairs? Yemon's answer has a different flavour, though. – Johannes Ebert Dec 18 2011 at 14:15 4 Not categorical, but the golden ratio is the hardest real to approximate by rationals. – Steve Huntsman Dec 18 2011 at 15:17 @J.E.: the answers don't need to be "categorical", the above example by S.H. fits perfectly in the kind of answers I'm expecting to get. – Qfwfq Dec 18 2011 at 18:30 @Qfwfq: so would other constants defined via extremal properties also fit? It seems like that might be too wide a scope if interpreted literally – Yemon Choi Dec 18 2011 at 19:46 1 Qfwfq: wouldn't the more interesting question be "provide examples of mathematical objects for which there's no known universal property that characterizes them"? It seems like there's almost no restriction on the objects that can appear in your list. – Ryan Budney Dec 20 2011 at 9:58 show 2 more comments ## 11 Answers The Stone-Cech compactification. Neither, Stone nor Cech was thinking about category theory at the time (since it didn't exist), but of course the Stone-Cech compactification is a left adjoint to the forgetful functor from compact Hausdorff spaces to completely regular Hausdorff spaces. If the general construction is not specific enough, then restrict my answer to $\beta \mathbb N$ which is a key object in Ramsey theory. - ### 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. Free groups. If I am not mistaken, they were first introduced by Dyck via the reduced words description. The modern universal property definition only came about later. - The real line as "the' complete Archimedean ordered field, as opposed to a bunch of Dedekind cuts. - 1 This description is redundant. The real line is the only complete ordered field; the Archimedean property follows. It is also the largest Archimedean ordered field. – Michael Hardy Dec 18 2011 at 1:28 4 Perhaps Yemon meant "complete" in the sense of convergence of Cauchy sequences. In this sense there are non archimedean complete ordered fields. – godelian Dec 18 2011 at 1:56 @Godelian: yes, that was what I had in mind (perhaps I am mis-using terminology?) So something like the hyper-reals would -- I think -- be complete and ordered in the sense I was thinking of – Yemon Choi Dec 18 2011 at 2:07 @Godelian: Cauchy completeness makes the sequence 1/n converge to 0 and this is logically equivalent to the Archimedean property. So how can there be a non Archimedean complete ordered fields? – Wouter Stekelenburg Sep 13 at 12:25 My impression is that most, if not all, ''natural objects'' in linear algebra, analysis or differential geometry, ..., are usefully characterized by some \emph{symmetry} property, for eaxmple ''The exterior derivative is, up to a constant multiple, the only linear operator from $k$-forms to $k+1$-forms such that for each open embedding $f:U \to M$ and each form $\omega \in \Omega^k (M)$, the idenity $f^{\ast} d \omega = d (f^{\ast}\omega)$ holds.'' - Slightly facetious one here: the 3-sphere is, up to diffeomorphism, the unique simply connected, closed, 3-manifold. - Probably, Johannes Ebert is right: (almost) all natural mathematical objects may be characterized by a universal property. The question is now what we understand exactly by the the fact that universal property is delving in the concrete habitual definition. More concrete, let consider the usual definition of a factor structure, let say a factor group (of $G$ modulo a normal subgroup $H$). There is also a universal one: A factor group is (up to an isomorphism) an epimorphism (i.e. a surjective group homomorphism) $G\to G'$. Does the second definition delve the first? I really don't know! Another example: Having two $R$-modules, $M$ of the right and $N$ of the left, one may define the tensor product as a factor of the free abelian group with the basis the cartezian product $M\times N$ modulo the relations which emphasize the bilinearity. Secondly, we may define the tensor $M\otimes_R-$ as the right adjoint of the functor $Hom_R(M,-)$, definition which may be extended for $M$ in a cocomplete abelian category. This time the possibility to change the settings leading to a more general definition stands as an argument that the universal definition is not delving in the usual one. - The natural numbers, maybe the oldest known mathematical obeject, have many universal properties in various categories. They are for example the free monoid on one generator, the initial rig, the free inductive set on one generator,... - The space of Radon measures on the closed unit interval is the free topological vector space over the interval. It has universal property that evey continuous function on the interval has a unique extension to a continuous linear mapping. This has zillions of generalisations---Radon measures on compacta, bounded Radon measures on a completely regular space, uniform measures on a uniform space and, and ... - The integers are the unique commutative ordered ring whose positive elements are well-ordered (thanks to Harry Altman). - The category $Set$ of sets is, up to equivalence, the only locally small category $C$ whose Yoneda embedding $y: C \to Set^{C^{op}}$ admits a string of adjoint functors $$u \dashv v \dashv w \dashv x \dashv y.$$ A precise treatment is given here. - But this characterization uses presheaves, and thus uses the category of sets. If we're allowed to use Set in the statement, a simpler characterization is "Set is, up to equivalence, the only category equivalent to Set". – Omar Antolín-Camarena Feb 14 at 12:03 Why don't you write the authors to tell them that? – Todd Trimble Feb 14 at 12:24 Well, for small categories, you can replace "$Set^{C^{op}}$" with the "free cocompletion of $C$", which doesn't explicitly use Set anymore. – Omar Antolín-Camarena Feb 14 at 15:20 1 Omar, unfortunately that's not going to work, since for large but locally small $C$, the functor category $Set^{Set^{op}}$ is not the free small cocompletion. (Also, words like "small" or "locally small" also refer to $Set$.) Anyway, I'm not very fussed about your objection. Compare for example the Gelfand-Mazur theorem, read as characterizing $\mathbb{C}$ as the unique complex Banach division algebra. We're allowed to say "complex" in the characterization. (This is the last comment I'm going to make about this. Feel free to downvote my answer, or contact the authors, or whatever you like.) – Todd Trimble Feb 14 at 17:50 I don't want to downvote this answer, I like this theorem (it's a lot fun to try to work out the adjunctions!), I didn't mean to make you feel defensive about it. – Omar Antolín-Camarena Feb 14 at 19:36 I happen to have just read Manes' theorem in the n-category café: Theorem The algebras for the ultrafilter monad are the compact Hausdorff spaces. The "ultrafilter monad" $X\mapsto \mathrm{U}(X)$ maps a set $X$ to the set of ultrafilters on it. The abstractness of the characterization of compact Hausdorff spaces lies in the fact that $\mathrm{U}$ is defined in purely set-theoretical (or, rather, category-theoretical) terms: it appears to be the "codensity monad" (don't ask me the meaning of this because I don't know!) of the inclusion $\mathrm{FinSet}\to\mathrm{Set}$. -	{"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": 30, "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": 0.9234814643859863, "perplexity_flag": "head"}

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