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http://unapologetic.wordpress.com/2011/01/13/standard-polytabloids-are-independent/?like=1&source=post_flair&_wpnonce=c1c9c39881	# The Unapologetic Mathematician ## Standard Polytabloids are Independent Now we’re all set to show that the polytabloids that come from standard tableaux are linearly independent. This is half of showing that they form a basis of our Specht modules. We’ll actually use a lemma that applies to any vector space $V$ with an ordered basis $e_\alpha$. Here $\alpha$ indexes some set $B$ of basis vectors which has some partial order $\preceq$. So, let $v_1,\dots,v_m$ be vectors in $V$, and suppose that for each $v_i$ we can pick some basis vector $e_{\alpha_i}$ which shows up with a nonzero coefficient in $v_i$ subject to the following two conditions. First, for each $i$ the basis element $e_{\alpha_i}$ should be the maximum of all the basis vectors having nonzero coefficients in $v_i$. Second, the $e_{\alpha_i}$ are all distinct. We should note that the first of these conditions actually places some restrictions on what vectors the $v_i$ can be in the first place. For each one, the collection of basis vectors with nonzero coefficients must have a maximum. That is, there must be some basis vector in the collection which is actually bigger (according to the partial order $\preceq$) than all the others in the collection. It’s not sufficient for $e_{\alpha_i}$ to be maximal, which only means that there is no larger index in the collection. The difference is similar to that between local maxima and a global maximum for a real-valued function. This distinction should be kept in mind, since now we’re going to shuffle the order of the $v_i$ so that $e_{\alpha_1}$ is maximal among the basis elements $e_{\alpha_i}$. That is, none of the other $e_{\alpha_i}$ should be bigger than $e_{\alpha_1}$, although some may be incomparable with it. Now I say that $e_{\alpha_i}$ cannot have a nonzero coefficient in any other of the $v_i$. Indeed, if it had a nonzero coefficient in, say, $v_2$, then by assumption we would have $e_{\alpha_1}\prec e_{\alpha_2}$, which contradicts the maximality of $e_{\alpha_1}$. Thus in any linear combination $\displaystyle c_1v_1+\dots+c_mv_m=0$ we must have $c_1=0$, since there is no other way to cancel off all the occurrences of $e_{\alpha_1}$. Removing $v_1$ from the collection, we can repeat the reasoning with the remaining vectors until we get down to a single one, which is trivially independent. So in the case we care about the space is the Young tabloid module $M^\lambda$, with the basis of Young tabloids having the dominance ordering. In particular, we consider for our $v_i$ the collection of polytabloids $e_t$ where $t$ is a standard tableau. In this case, we know that $\{t\}$ is the maximum of all the tabloids showing up as summands in $e_t$. And these standard tabloids are all distinct, since they arise from distinct standard tableaux. Thus our lemma shows that not only are the standard polytabloids $e_t$ distinct, they are actually linearly independent vectors in $M^\lambda$. ## 3 Comments » 1. [...] Przeczytaj artykuł: Standard Polytabloids are Independent [...] Pingback by \| January 16, 2011 \| Reply 2. [...] of shape . But these polytabloids are not independent. We’ve seen that standard polytabloids are independent, and it turns out that they also span. That is, they provide an explicit basis for the Specht [...] Pingback by \| January 21, 2011 \| Reply 3. [...] now we can go back to the lemma we used when showing that the standard polytabloids were independent! The are a collection of vectors in . [...] Pingback by \| February 9, 2011 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 39, "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": 0.9295424818992615, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/tagged/real-analysis?sort=faq&pagesize=30	# Tagged Questions Theoretical foundations of calculus: limits, convergence of sequences, construction of the real numbers, least upper bound property, and related analysis topics such as continuity, differentiation, integration through the fundamental theorem of calculus... 6answers 4k views ### How can you prove that a function has no closed form integral? I've come across statements in the past along the lines of "function $f(x)$ has no closed form integral", which I assume means that there is no combination of the operations: addition/subtraction ... 11answers 3k views ### Zero to the zero power - Is $0^0=1$? Could someone provide me with good explanation of why $0^0 = 1$? My train of thought: $x > 0$ $0^x = 0^{x-0} = 0^x/0^0$, so $0^0 = 0^x/0^x = ?$ Possible answers: $0^0 * 0^x = 1 * 0^x$, so ... 5answers 3k views ### Completion of rational numbers via Cauchy sequences Can anyone recommend a good self-contained reference for completion of rationals to get reals using Cauchy sequences? 3answers 938 views ### Set of continuity points of a real function I have a question about subsets $$A \subseteq \mathbb R$$ for which there exists a function $$f : \mathbb R \to \mathbb R$$ such that the set of continuity points of $f$ is $A$. Can I characterize ... 2answers 712 views ### bijective map from $\mathbb{R}^3\rightarrow \mathbb{R}$ could any one give an example of a bijective map from $\mathbb{R}^3\rightarrow \mathbb{R}$? Thank you. 3answers 2k views ### Why is $1^{\infty}$ considered to be an indeterminate form From Wikipedia: In calculus and other branches of mathematical analysis, an indeterminate form is an algebraic expression obtained in the context of limits. Limits involving algebraic operations are ... 4answers 1k views ### Sine function dense in $[-1,1]$ We know that the sine function takes it values between $[-1,1]$. So is the set $$A = \{ \sin{n} \ : \ n \in \mathbb{N}\}$$ dense in $[-1,1]$. Generally, for showing the set is dense, one proceeds, by ... 8answers 2k views ### Self-Contained Proof that $\sum\limits_{n=1}^{\infty} \frac1{n^p}$ Converges for $p > 1$ To prove the convergence of $$\sum_{n=1}^{\infty} \frac1{n^p}$$ for $p > 1$, one typically appeals to either the Integral Test or the Cauchy Condensation Test. I am wondering if there is a ... 2answers 1k views ### Convergence a.e. and of norms implies that in Lebesgue space Let $1\leq p < \infty$. Suppose that $\{f_k\} \subset L^p$ (the domain here does not necessarily have to be finite), $f_k \to f$ almost everywhere, and $\\|f_k\\|_{L^p} \to \\|f\\|_{L^p}$. Why is it ... 3answers 2k views ### Construction of a Borel set with positive but not full measure in each interval I was wondering how one can construct a Borel set that doesn't have full measure on any interval of the real line but does have positive measure everywhere. To be precise, if $\mu$ denotes Lebesgue ... 7answers 993 views ### Is $\lim\limits_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} = 0$? Is $\lim\limits_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} = 0$ I.e - does the second half of the harmonic series go to zero? I know that for a finite number of terms the limit of the sum is ... 3answers 718 views ### Continuity of the metric function Let $(X,d)$ be a metric space. How to prove that for any closed $A$ a function $d(x,A)$ is continuous - I know that it is even Lipschitz continuous, but I have a problem with the proof: \|d(x,a) - ... 2answers 4k views ### How can I prove $\sup(A+B)=\sup A+\sup B$ if $A+B=\{a+b\mid a\in A, b\in B\}$ If $A,B$ non empty, upper bounded sets and $A+B=\{a+b\mid a\in A, b\in B\}$, how can I prove that $\sup(A+B)=\sup A+\sup B$? 4answers 2k views ### How do I define a bijection between $(0,1)$ and $(0,1]$? How do I define a bijection between $(0,1)$ and $(0,1]$? Or any other open and closed intervals? If the intervals are both open like $(-1,2)\text{ and }(-5,4)$ I do a cheap trick (don't know if ... 4answers 1k views ### Calculating the integral $\int_{0}^{\infty} \frac{\cos x}{1+x^2}\mathrm{d}x$ without using complex analysis Suppose that we do not know anything about the complex analysis (numbers). In this case, how to calculate the following integral in closed form? $$\int_0^\infty\frac{\cos\;x}{1+x^2}\mathrm{d}x$$ 3answers 566 views ### A vector space over $R$ is not a countable union of proper subspaces I was looking for alternate proofs of the theorem that "a vector space $V$ of dimension greater than $1$ over an infinite field is not a union of fewer than $\|\mathbf{F}\|$ proper subspaces" and ... 5answers 1k views ### Nonzero $f \in C([0, 1])$ for which $\int_0^1 f(x)x^n dx = 0$ for all $n$ As the title says, I'm wondering if there is a continuous function such that $f$ is nonzero on $[0, 1]$, and for which $\int_0^1 f(x)x^n dx = 0$ for all $n \geq 1$. I am trying to solve a problem ... 11answers 2k views ### $\lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite How do I prove that $\displaystyle\lim_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite? 1answer 1k views ### Proving that the sequence $F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}$ is boundedly convergent on $\mathbb{R}$ Here is an exercise, on analysis which i am stuck. How do I prove that if $F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}$, then the sequence $\{F_{n}(x)\}$ is boundedly convergent on ... 3answers 2k views ### Is every Lebesgue measurable function on $\mathbb{R}$ the pointwise limit of continuous functions? I know that if $f$ is a Lebesgue measurable function on $[a,b]$ then there exists a continuous function $g$ such that $\|f(x)-g(x)\|< \epsilon$ for all $x\in [a,b]\setminus P$ where the measure of ... 5answers 945 views ### Perfect set without rationals Give an example of a perfect set in $\mathbb R^n$ that does not contain any of the rationals. (Or prove that it does not exist). 4answers 1k views ### Limit of the nested radical $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$ [duplicate] Possible Duplicate: On the sequence $x_{n+1} = \sqrt{c+x_n}$ Where does this sequence converge? $\sqrt{7},\sqrt{7+\sqrt{7}},\sqrt{7+\sqrt{7+\sqrt{7}}}$,... 5answers 865 views ### $\lim_{n\rightarrow\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$ I'm supposed to calculate: $$\lim_{n\rightarrow\infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}$$ By using W\|A, i may guess that the limit is $\frac{1}{2}$ that is a pretty interesting and nice result. ... 2answers 283 views ### Methods to evaluate $\int _{a }^{b }\!{\frac {\ln \left( tx + u \right) }{m{x}^{2}+nx +p}}{dx}$ Today I saw a question with an answer that made me rethink of the following question, since it's not the first time I try to find an answer to it. If you look at the answer of Mhenni Benghorbal here ... 3answers 958 views ### Proving that $m+n\sqrt{2}$ is dense in R I am having trouble proving the statement: Let $S = \{m + n\sqrt 2 : m, n \in\mathbb Z\}$. Prove for every $\epsilon > 0$, The intersection of $S$ and $(0, \epsilon)$ is nonempty. 2answers 923 views ### Does uncountable summation, with a finite sum, ever occur in mathematics? Obviously, “most” of the terms must cancel out with opposite algebraic sign. You can contrive examples such as the sum of the members of R being 0, but does an uncountable sum, with a finite sum, ... 3answers 556 views ### Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$ Let $$A(p,q) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(p)}_k}{k^q},$$ where $H^{(p)}_n = \sum_{i=1}^n i^{-p}$, the $n$th $p$-harmonic number. The $A(p,q)$'s are known as alternating Euler sums. ... 7answers 1k views ### Is there any geometric way to characterize $e$? Let me explain it better: after this question, I've been looking for a way to put famous constants in the real line in a geometrical way -- just for fun. Putting $\sqrt2$ is really easy: constructing ... 5answers 3k views ### Does $\int_0^{\infty}\frac{\sin x}{x}dx$ have an improper Riemann integral or a Lebesgue integral? In this wikipedia article for improper integral, $$\int_0^{\infty}\frac{\sin x}{x}dx$$ is given as an example for the integrals that have an improper Riemann integral but do not have a (proper) ... 4answers 3k views ### Continuous bijection from $(0,1)$ to $[0,1]$ Does there exist a continuous bijection from $(0,1)$ to $[0,1]$? Of course the map should not be a proper map. 5answers 975 views ### Proof that $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$ Thomson et al. provide a proof that $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$ in this book. It has to do with using an inequality that relies on the binomial theorem. I tried to do an alternate proof ... 1answer 1k views ### Cardinality of Borel sigma algebra It seems it's well known that if a sigma algebra is generated by countably many sets, then the cardinality of it is either finite or $c$ (the cardinality of continuum). But it seems hard to prove it, ... 3answers 433 views ### When $L_p = L_q$? As we know that $L_p \subseteq L_q$ when $0 < p < q$ for probability measure, I was wondering when $L_p = L_q$ is true and why. Is it to impose some restriction on the domain space? Thanks! 2answers 785 views ### examples of measurable functions on $\mathbb{R}$ Could give some examples of nonnegative measurable function $f:\mathbb{R}\to[0,\infty)$, such that its integral over any bounded interval is infinite? 3answers 877 views ### No maximum(minimum) of rationals whose square is lesser(greater) than $2$. Suppose $A$ is the set of all rational numbers $p$ such that $p^2 <2$ and $B$ is the set of all rational numbers $p$ such that $p^2 > 2$. We want to show that $A$ contains no largest element and ... 2answers 643 views ### What can we say about $f$ if $\int_0^1 f(x)p(x)dx=0$ for all polynomials $p$? This question was motivated by another question in this site. As explained in that problem (and its answers), if $\displaystyle f$ is continuous on $\displaystyle [0,1]$ and \$\displaystyle \int_0^1 ... 3answers 233 views ### uniform convergence of few sequence of functions Pick out the sequences $\{f_n\}$ which are uniformly convergent. (a) $f_n(x) = nxe^{−nx}$ on $(0,∞)$. (b)$f_n(x) = x^n$ on $[0, 1]$. (c)$f_n(x) = \frac{\sin(nx)}{\sqrt{n}}$ on $\mathbb{R}$. (d) ... 1answer 302 views ### If $f(x)$ is continuous on $[a,b]$ and $M=\max \; \|f(x)\|$, is $M=\lim \limits_{n\to\infty} \left(\int_a^b\|f(x)\|^n\,\mathrm dx\right)^{1/n}$? Let $f(x)$ be a continuous real-valued function on $[a,b]$ and $M=\max\{\|f(x)\| \; :\; x \in [a,b]\}$. Is it true that: $$M= \lim_{n\to\infty}\left(\int_a^b\|f(x)\|^n\,\mathrm dx\right)^{1/n} ?$$ ... 1answer 379 views ### continuous functions on $\mathbb R$ such that $g(x+y)=g(x)g(y)$ Let $g$ be a function on $\mathbb R$ to $\mathbb R$ which is not identically zero and which satisfies the equation $g(x+y)=g(x)g(y)$ for $x$,$y$ in $\mathbb R$. $g(0)=1$. If $a=g(1)$,then $a>0$ ... 1answer 1k views ### How discontinuous can a derivative be? There is a well-known result in elementary analysis due to Darboux which says if $f$ is a differentiable function then $f'$ satisfies the intermediate value property. To my knowledge, not many ... 4answers 4k views ### When can you switch the order of limits? Suppose you have a double sequence $\displaystyle a_{nm}$. What are sufficient conditions for you to be able to say that \$\displaystyle \lim_{n\to \infty}\,\lim_{m\to \infty}{a_{nm}} = \lim_{m\to ... 3answers 1k views ### Can there be two distinct, continuous functions that are equal at all rationals? Akhil showed that the Cardinality of set of real continuous functions is the same as the continuum, using as a step the observation that continuous functions that agree at rational points must agree ... 4answers 1k views ### Finding Value of the Infinite Product $\prod \Bigl(1-\frac{1}{n^{2}}\Bigr)$ While trying some problems along with my friends we had difficulty in this question. True or False: The value of the infinite product $$\prod\limits_{n=2}^{\infty} \biggl(1-\frac{1}{n^{2}}\biggr)$$ ... 2answers 844 views ### Comparing the Lebesgue measure of an open set and its closure Let $E$ be an open set in $[0,1]^n$ and $m$ be the Lebesgue measure. Is it possible that $m(E)\neq m(\bar{E})$, where $\bar{E}$ stands for the closure of $E$? 3answers 327 views ### A log improper integral Evaluate : $$\int_0^{\frac{\pi}{2}}\ln ^2\left(\cos ^2x\right)\text{d}x$$ I found it can be simplified to $$\int_0^{\frac{\pi}{2}}4\ln ^2\left(\cos x\right)\text{d}x$$ I found the exact value in the ... 3answers 534 views ### $f$ uniformly continuous and $\int_a^\infty f(x)\,dx$ converges imply $\lim_{x \to \infty} f(x) = 0$ Trying to solve $f(x)$ is uniformly continuous in the range of $[0, +\infty)$ and $\int_a^\infty f(x)dx$ converges. I need to prove that: $$\lim \limits_{x \to \infty} f(x) = 0$$ Would ... 5answers 175 views ### Given $y_n=(1+\frac{1}{n})^{n+1},n \in \mathbb{N},n \geq 1$ Show that $\lbrace y_n \rbrace$ is a decrasing sequence Given $$y_n=\left(1+\frac{1}{n}\right)^{n+1}\hspace{-6mm},\qquad n \in \mathbb{N}, \quad n \geq 1.$$ Show that $\lbrace y_n \rbrace$ is a decrasing sequence. Anyone can help ? I consider the ratio ... 4answers 428 views ### Showing the inequality $\|\alpha + \beta\|^p \leq 2^{p-1}(\|\alpha\|^p + \|\beta\|^p)$ I have a small question that I think is very basic but I am unsure how to tackle since my background in computing inequalities is embarrasingly weak - I would like to show that, for a real number \$p ... 1answer 732 views ### how to show this is a dense set? I am not able to prove that this set is dense in $\mathbb{R}$. Will be pleased if you help in a easiest way, $\{a+b\alpha: a,b\in \mathbb{Z}\}$ where $\alpha\in\mathbb{Q}^c$ is a fixed irrational. 2answers 465 views ### limit of $\frac{a_{n+1}}{a_n}$ Today my lecturer put up on the board that: If $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}$ exists and $a_n>0$ then \$\displaystyle ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 161, "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": 0.9387878775596619, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/83966-find-unit-vector-angel.html	# Thread: 1. ## Find Unit Vector and Angel Suppose that you are climbing a hill whose shape is given by , and that you are at the point (90, 50, 300). In which direction (unit vector) should you proceed initially in order to reach the top of the hill fastest? , If you climb in that direction, at what angle above the horizontal will you be climbing initially (radian measure)? __________. help me understand and solve this one. 2. Originally Posted by DMDil Suppose that you are climbing a hill whose shape is given by , and that you are at the point (90, 50, 300). In which direction (unit vector) should you proceed initially in order to reach the top of the hill fastest? , If you climb in that direction, at what angle above the horizontal will you be climbing initially (radian measure)? __________. help me understand and solve this one. The surface can be though of $F(x,y,z)=0.04x^2+0.02y^2+z-674$ The gradient will give the the direction of fastest increase $\frac{\nabla F(90,50,300)}{\|\nabla F(90,50,300)\|}$	{"extraction_info": {"found_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": 0.951677680015564, "perplexity_flag": "middle"}
http://www.math.uah.edu/stat/poisson/Poisson.html	$$\newcommand{\P}{\mathbb{P}}$$ $$\newcommand{\E}{\mathbb{E}}$$ $$\newcommand{\R}{\mathbb{R}}$$ $$\newcommand{\N}{\mathbb{N}}$$ $$\newcommand{\bs}{\boldsymbol}$$ $$\newcommand{\var}{\text{var}}$$ $$\newcommand{\sd}{\text{sd}}$$ $$\newcommand{\skew}{\text{skew}}$$ $$\newcommand{\kurt}{\text{kurt}}$$ ## 4. The Poisson Distribution #### The Counting Variables Recall that in the Poisson model, $$\bs{X} = (X_1, X_2, \ldots)$$ denotes the sequence of interarrival times, and $$\bs{T} = (T_0, T_1, T_2, \ldots)$$ denotes the sequence of arrival times. Thus $$\bs{T}$$ is the partial sum process associated with $$\bs{X}$$: $T_n = \sum_{i=0}^n X_i, \quad n \in \N$ Based on the strong renewal assumption, that the process restarts at each fixed time and each arrival time, independently of the past, we now know that $$\bs{X}$$ is a sequence of independent random variables, each with the exponential distribution with parameter $$r \gt 0$$. We also know that $$T_n$$ has the gamma distribution with rate parameter $$r$$ and scale parameter $$n$$. The probability density function of $$T_n$$ is $f_n(t) = r^n \frac{t^{n-1}}{(n-1)!} e^{-r\,t}, \quad 0 \le t \lt \infty$ Recall that for $$t \ge 0$$, $$N_t$$ denotes the number of arrivals in the interval $$(0, t]$$, so that $$N_t = \max\{n \in \N: T_n \le t\}$$. We refer to $$\bs{N} = (N_t: t \ge 0)$$ as the counting process. We can find the distribution of $$N_t$$ because of the inverse relation between $$\bs{N}$$ and $$\bs{T}$$. In particular, recall that $N_t \ge n \iff T_n \le t, \quad t \in (0, \infty), \; n \in \N_+$ since both events mean that there are at least $$n$$ arrivals in $$(0, t]$$. For $$t \in [0, \infty)$$, the probability density function of $$N_t$$ is given by $\P(N_t = n) = e^{-r\,t} \frac{(r\,t)^n}{n!}, \quad n \in \N$ Proof: Using the inverse relationship noted above, and integration by parts, we have $\P(N_t \ge n) = \P(T_n \le t) = \int_0^t f_n(s) \, ds = 1 - \sum_{k=0}^{n-1} e^{-r\,t} \frac{(r\,t)^k}{k!}, \quad n \in \N$ For $$n \in \N$$ we have $$\P(N_t = n) = \P(N_t \ge n) - \P(N_t \ge n + 1)$$. Simplifying gives the result. Note that the distribution of $$N_t$$ depends on the paramters $$r$$ and $$t$$ only through the product $$r\,t$$. The distribution is called the Poisson distribution with parameter $$r\,t$$ and is named for Simeon Poisson. In the Poisson experiment, vary $$r$$ and $$t$$ with the scroll bars and note the shape of the probability density function. For various values of $$r$$ and $$t$$, run the experiment 1000 times and watch the apparent convergence of the relative frequency function to the probability density function. The Poisson distribution is one of the most important in probability. In general, a random variable $$N$$ taking values in $$\N$$ is said to have the Poisson distribution with parameter $$c \gt 0$$ if it has the probability density function $g(n) = e^{-c} \frac{c^n}{n!}, \quad n \in \N$ To check our work, note that $$g$$ is a valid probability density function. Proof: Recall that $$\sum_{n=0}^\infty c^n / n! = e^c$$. The Poisson distribution is unimodal. 1. $$g(n - 1) \lt g(n)$$ if and only if $$n \lt c$$. 2. If $$c \notin \N_+$$, there is a single mode at $$\lfloor c \rfloor$$. 3. If $$c \in \N_+$$, there are consecutive modes at $$c - 1$$ and $$c$$. Suppose that requests to a web server follow the Poisson model with rate $$r = 5$$ per minute. Find the probability that there will be at least 8 requests in a 2 minute period. Answer: 0.7798 Defects in a certain type of wire follow the Poisson model with rate 1.5 per meter. Find the probability that there will be no more than 4 defects in a 2 meter piece of the wire. Answer: 0.8153 #### Moments Suppose that $$N$$ has the Poisson distribution with parameter $$c \gt 0$$. Naturally we want to know the mean, variance, and probability generating function of $$N$$. The easiest moments to compute are the factorial moments. For this result, recall the falling power notation for the number of permutations of size $$k$$ chosen from a population of size $$n$$: $n^{(k)} = n \, (n - 1) \cdots [n - (k + 1)]$ The factorial moments of $$N$$ of order $$k \in \N$$ is $$\E[N^{(k)}] = c^k$$. Proof: Using the standard change of variables formula for expected value, $\E[N^{(k)}] = \sum_{n=0}^\infty n^{(k)} e^{-c} \frac{c^n}{n!} = e^{-c} c^k \sum_{n=k}^\infty \frac{c^{n-k}}{(n-k)!} = e^{-c} c^k e^c = c^k$ The mean and variance of $$N$$ are the parameter $$c$$. 1. $$\E(N) = c$$ 2. $$\var(N) = c$$ Proof: Part (a) follows directly from the first factorial moment: $$\E(N) = \E[N^{(1)}] = c$$. For part (b), note that $$\E(N^2) = \E[N^{(2)}] + \E(N) = c^2 + c$$. The probability generating function of $$N$$ is $P(s) = \E(s^N) = e^{c(s - 1)}, \quad s \in \R$ Proof: Using the change of variables formula again, $\E(s^N) = \sum_{n=0}^\infty s^n e^{-c} \frac{c^n}{n!} = e^{-c} \sum_{n=0}^\infty \frac{(c\,s)^n}{n!} = e^{-c} e^{c\,s}, \quad s \in \R$ Returning to the Poisson process $$\bs{N} = (N_t: t \ge 0)$$ with rate parameter $$r$$, it follows that $$\E(N_t) = r\,t$$ and $$\var(N_t) = r\,t$$ for $$t \ge 0$$. Once again, we see that $$r$$ can be interpreted as the average arrival rate. In an interval of length $$t$$, we expect about $$r\,t$$ arrivals. In the Poisson experiment, vary $$r$$ and $$t$$ with the scroll bars and note the location and size of the mean/standard deviation bar. For various values of $$r$$ and $$t$$, run the experiment 1000 times and watch the apparent convergence of the sample mean and standard deviation to the distribution mean and standard deviation, respectively. Suppose that customers arrive at a service station according to the Poisson model, at a rate of $$r = 4$$. Find the mean and standard deviation of the number of customers in an 8 hour period. Answer: 32, 5.657 #### Additional Properties Let's explore the basic renewal assumption of the Poisson model in terms of the counting process $$\bs{N} = (N_t: t \ge 0)$$. Recall that $$N_t$$ is the number of arrivals in the interval $$(0, t]$$. If $$s, \; t \in [0, \infty)$$ with $$s \lt t$$, then $$N_t - N_s$$ is the number of arrivals in the interval $$(s, t]$$. Of course, the arrival times have continuous distributions, so the probability that an arrival occurs at a specific point $$t$$ is 0. Thus, it does not really matter if we write the interval above as $$(s, t]$$, $$(s, t)$$, $$[s, t)$$ or $$[s, t]$$. The process $$\bs{N}$$ has stationary, independent increments. 1. If $$s, \; t \in [0, \infty)$$ with $$s \lt t$$ then $$N_t - N_s$$ has the same distribution as $$N_{t-s}$$, namely Poisson with parameter $$r\,(t - s)$$. 2. If $$t_1, \; t_2, \; t_3 \ldots \in [0, \infty)$$ with $$t_1 \lt t_2 \lt t_3 \lt \cdots$$ then $$(N_{t_1}, N_{t_2} - N_{t_1}, N_{t_3} - N_{t_2}, \ldots)$$ is an independent sequence. The results in the last theorem can be stated more elegantly in terms of our more general counting process. Recall that for $$A \subseteq [0, \infty)$$ (measurable of course), $$N(A)$$ denotes the number of random points in $$A$$: $N(A) = \#\{n \in \N_+: T_n \in A\}$ Recall also that $$\lambda$$ denotes the standard length (Lebesgue) measure on $$[0, \infty)$$. The stationary property states that the distribution of $$N(A)$$ depends only on $$\lambda(A)$$; in fact, the distribution is Poisson with parameter $$r \, \lambda(A)$$. The independence property states that if $$(A_1, A_2, \ldots)$$ is a sequence of disjoint subsets of $$[0, \infty)$$ then $$(N(A_1), N(A_2), \ldots)$$ is a sequence of independent variables. Suppose that $$N$$ and $$M$$ are independent random variables, and that $$N$$ has the Poisson distribution with parameter $$c$$ and $$M$$ has the Poisson distribution with parameter $$d$$. Then $$N + M$$ has the Poisson distribution with parameter $$c + d$$. Proof: There are several ways to prove this result, but the one that gives the most insight is a probabilistic proof based on the Poisson process. Thus, suppose that we start with a sequence $$\bs{X} = (X_1, X_2, \ldots)$$ of independent random variables, each with the exponential distribution with parameter 1. This sequence leads to a Poisson process with rate 1, and in particular the counting process $$\bs{N} = (N_t: t \ge 0)$$. We can associate $$N$$ with $$N_c$$ and $$M$$ with $$N_{c + d} - N_c$$, so that $$N + M$$ is $$N_{c + d}$$ Simple analytic proofs can be constructed using probability generating functions or probability density functions. Recall that the PGF of $$N + M$$ is the product of the PGFs of $$N$$ and $$M$$. The PDF of $$N + M$$ is the convolution of the PDFs of $$N$$ and $$M$$. From the last theorem, it follows that for each $$n \in \N_+$$, the Poisson distribution is the distribution of the sum of $$n$$ independent, identically distributed variables. A distribution with this property is said to be infinitely divisible. Again, the Poisson process gives an explicit decomposition. Suppose that $$(N_t: t \ge 0)$$ is a Poisson process with rate 1. For fixed $$t \gt 0$$, $$N_t$$ has the Poisson distribution with parameter $$t$$, and $N_t = \sum_{i=1}^n [N_{i\,t / n} - N_{(i-1)t / n}]$ The variables in the sum are independent, and each has the Poisson distribution with parameter $$t / n$$. Other infinitely divisible distributions that we have studied are the normal distribution and the Cauchy distribution. #### Normal Approximation Because of the representation as a sum of independent, identically distributed variables, it's not surprising that the Poisson distribution can be approximated by the normal. Suppose that $$N_t$$ has the Poisson distribution with parameter $$t \gt 0$$. Then the distribution of the variable below converges to the standard normal distribution as $$t \to \infty$$. $Z_t = \frac{N_t - t}{\sqrt{t}}$ Proof: As usual, we can assume that $$(N_t: t \ge 0)$$ is the Poisson counting process with rate 1. Note that $$Z_t$$ is simply the standard score associated with $$N_t$$. For $$n \in \N_+$$, $$N_n$$ is the sum of $$n$$ independent variables, each with the Poisson distribution with parameter 1. Thus, from the central limit theorem, the distribution of $$Z_n$$ converges to the standard normal distribution as $$n \to \infty$$. For general $$t \in [0, \infty)$$, it's possible to write $$Z_t = Z_n + W_t$$ where $$n = \lfloor t \rfloor$$ and $$W_t \to 0$$ as $$t \to \infty$$ (in probability and hence in distribution). Thus, if $$N$$ has the Poisson distribution with parameter $$c$$, and $$c$$ is large, then the distribution of $$N$$ is approximately normal with mean $$c$$ and standard deviation $$\sqrt{c}$$. When using the normal approximation, we should remember to use the continuity correction, since the Poisson is a discrete distribution. In the Poisson experiment, set $$r = t = 1$$. Increase $$t$$ and note how the graph of the probability density function becomes more bell-shaped. In the Poisson experiment, set $$r = 5$$ and $$t = 4$$. Run the experiment 1000 times and compute the following: 1. $$\P(15 \le N_4 \le 22)$$ 2. The relative frequency of the event $$\{15 \le N_4 \le 22\}$$. 3. The normal approximation to $$\P(15 \le N_4 \le 22)$$. Answer: 1. 0.6157 2. 0.6025 Suppose that requests to a web server follow the Poisson model with rate $$r = 5$$ per minute. Compute the normal approximation to the probability that there will be at least 280 requests in a 1 hour period. Answer: 0.8818 #### Conditional Distributions Consider again the basic Poisson model with rate $$r \gt 0$$. As usual, $$\bs{T} = (T_0, T_1, \ldots)$$ denotes the arrival time sequence and $$\bs{N} = (N_t: t \ge 0)$$ the counting process. For $$t \gt 0$$, the conditional distribution of $$T_1$$ given $$N_t = 1$$ is uniform on the interval $$(0, t]$$. Proof: Given $$N_t = 1$$ (one arrival in $$(0, t]$$) the arrival time $$T_1$$ takes values in $$(0, t]$$. From independent and stationary increments properties, $\P(T_1 \le s \mid N_t = 1) = \P(N_s = 1, N_t - N_s = 0 \mid N_t = 1) = \frac{\P(N_s = 1, N_t - N_s = 0)}{\P(N_t = 0)} = \frac{\P(N_s = 1) \P(N_t - N_s = 0)}{\P(N_t = 1)}$ Hence using the Poisson distribution, $\P(T_1 \le s \mid N_t = 1) = \frac{e^{-r\,s} s e^{-r(t - s)}}{e^{-r\,t} t} = \frac{s}{t}, \quad 0 \lt s \le t$ More generally, for $$t \gt 0$$ and $$n \in \N_+$$, the conditional distribution of $$(T_1, T_2, \ldots, T_n)$$ given $$N_t = n$$ is the same as the distribution of the order statistics of a random sample of size $$n$$ from the uniform distribution on the interval $$(0, t]$$. Note that the conditional distribution in the last exercise is independent of the rate $$r$$. This result means that, in a sense, the Poisson model gives the most random distribution of points in time. Suppose that requests to a web server follow the Poisson model, and that 1 request comes in a five minute period. Find the probability that the request came during the first 3 minutes of the period. Answer: 0.6 Suppose that $$0 \lt s \lt t$$ and $$n \in \N_+$$. Then the conditional distribution of $$N_s$$ given $$N_t = n$$ is binomial with trial parameter $$n$$ and success parameter $$p \ s / t$$. Proof: Note that given $$N_t = n$$, the number of arrivals $$N_s$$ in $$(0, s]$$ takes values in $$\{0, 1 \ldots, n\}$$. Again, the stationary and independent increments properties are critical for the proof. $\P(N_s = k \mid N_t = n) = \frac{\P(N_s = k, N_t = n)}{\P(N_t = n)} = \frac{\P(N_s = k, N_t - N_s = n - k)}{\P(N_t = n)} = \frac{\P(N_s = k) \P(N_t - N_s = n - k)}{\P(N_t = n)}$ Subsitituting into the Poisson PDFs gives $\P(N_s = k \mid N_t = n) = \frac{\left(e^{-r\,s}(r\,s)^k / k!\right)\left(e^{-r(t-s)}[(r(t - s)]^{n-k}/(n - k)!\right)}{e^{-r\,t}(r\,t)^n / n!} = \frac{n!}{k! (n - k)!} \left(\frac{s}{t}\right)^k \left(1 - \frac{s}{t}\right)^{n-k}$ Note again that the conditional distribution in the last Theorem does not depend on the rate $$r$$. Suppose that requests to a web server follow the Poisson model, and that 10 requests come during a 5 minute period. Find the probability that at least 4 requests came during the first 3 minutes of the period. Answer: 0.9452 #### Estimating the Rate In many practical situations, the rate $$r$$ of the process in unknown and must be estimated based on observing data. For fixed $$t \gt 0$$, a natural estimator of the rate $$r$$ is $$N_t / t$$. The mean and variance of $$N_t / t$$ are 1. $$\E(N_t / t) = r$$ 2. $$\var(N_t / t) = r / t$$ Proof: These result follow easily from $$\E(N_t) = \var(N_t) = r\,t$$ and basic properties of expected value and variance. Part (a) means that the estimator is unbiased. Since this is the case, the variance in part (b) gives the mean square error. Since $$\var(N_t)$$ decreases to 0 as $$t \to \infty$$, the estimator is consistent. In the Poisson experiment, set $$r = 3$$ and $$t = 5$$. Run the experiment 100 times. 1. For each run, compute the estimate of $$r$$ based on $$N_t$$. 2. Over the 100 runs, compute the average of the squares of the errors. 3. Compare the result in (b) with the variance in previous exercise. Suppose that requests to a web server follow the Poisson model with unknown rate $$r$$ per minute. In a one hour period, the server receives 342 requests. Estimate $$r$$. Answer: $$r = 5.7$$ per minute	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 220, "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": 0.8857414126396179, "perplexity_flag": "head"}
http://mathoverflow.net/questions/109868?sort=newest	## Finding linearly independent columns of a large sparse rectangular matrix ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I have a problem that necessitates solving a large non-negative least-squares problem. My matrix A is large, sparse, highly rectangular (num rows >> num cols) and nearly binary. However, A is not necessarily of full column-rank, causing my non-negative least-squares solver (http://www.jasoncantarella.com/webpage/index.php?title=Tsnnls) to fail. Is there an efficient algorithm that will allow me to select a maximum cardinality set of linearly-independent columns from A so that I can solve my least-squares problem? - ## 1 Answer By nearly binary do you mean most entries are $0$ or $1$ but a few are not? I will assume that all entries are $0,1.$ My main comments involve merely noting zero and non-zero. To be sure of having maximum rank you will need to look at each column. You can start with any independent set (such as the set consisting of a single column) and then examine the rest of the columns in some order. If the next column is independent of the current set adjoin it as a new member otherwise discard it. This will result in a maximum size set (a basis of the column space). Depending on how sparse it is, it could be pretty easy to decide when the next column can be added, at least for a while. Here are a few simple ideas to boost efficiency: • If the matrix is very sparse then use efficient sparse matrix algorithms and representations, these might involve storing it as a set of ordered tuples $(i,j)$ or $(i,j,a_{ij})$ showing the location (and value) of the non-negative entries. There is bound to be highly efficent and well validated existing code which will provide you with a basis for the column space of a (sparse) matrix. • Let the depth of a column be the first row with a non-zero entry. Sort the columns according to depth (don't bother to break ties by looking further.) So far this might not require you to look at all the entries, just those down to the first non-zero entry. If the matrix is quite sparse then one might expect many different heights. • Pick one column from each depth class, This will give you a starting independent set which might include a relatively large number of columns. Assume that this has been done. • Go through the current set of independent columns, look for any rows which are all zero (for the selected column vectors) Now look through the undiscarded columns and try to find one which is non-zero in that position. If one is found then add it to the set. This will enlarge the set of columns. - Thanks for the response Aaron! By binary, I do indeed mean that most entries (but not necessarily all) are 0 or 1. If the final method necessitates a binary matrix, then the non-zero entries can be set to 1 without too much cause for concern. To solve the general problem, my current approach is to consider the procedure you outlined below to obtain an initial set of independent columns, then to proceed via a Gram-Schmidt procedure. Orthogonalize this set of columns, and for each remaining column, try to add it to the basis, dropping it if it's orthogonalized norm is 0. I think this works. – Rob Oct 17 at 18:07 Using Gram-Schmidt means using floating point computations. Especially if the entries are all integers or even $0,1$ just do column reduction. But again, there are sure to be optimized algorithms for this exact problem in general and in the case of sparse matrices. – Aaron Meyerowitz Oct 18 at 7:42	{"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": 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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9091442227363586, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/176353-center-sl-n-f.html	Thread: 1. center of SL(n,F) Show that Z(SL(n,F)) $\cong$ {a $\in$ F* \| a^n = 1}. I know that if X $\in$ Z(SL(n,F)) then det(X) = det(aI) = a^n which has to equal 1 since it is in SL(n,F). Any Help greatly appreciated. 2. Hi! You write X=aI, so i suppose you know that a matrix commutes with any other matrix iff it is a multiple of the identity. The same holds for SL(n,F). Banach 3. if X Z(SL(n,F)) then det(X) = det(aI) = a^n which has to equal 1 since it is in SL(n,F). So does this count as a proof. 4. Goku: What you've written doesn't quite make sense. The center will $\{aI_n : a^n=1\}$. Clearly this set is contained in the center. Conversely, let $A=(a_{i j})$ be in the center. Let $E_{i j}$ be the matrix with 1 in the $(i,j)$th position and 0 elsewhere. Then $I_n+E_{i j}\in SL(n, F)$ if $i\ne j$. Now use the fact that this matrix must commute with $A$.	{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.9258264899253845, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-applied-math/20836-fourier-series.html	# Thread: 1. ## Fourier series i, I have a question about Fourier Series(FS) in my textbook which is persentted like this: The uniqueness of a FS means that if we can find the FS of a waveform, we are assured that there is no other waveform with that FS, except for waveforms differing from the waveform under consideration only over an inconsequential set of values of the independent variable. With this assitance, find the following trigonometric FS without doing any integration: x(t)= $cos^3(20\pit)$[ $1-sin^2(10\pit)$] The solution to this is supposed to be: x(t)=(5/8) $cos(20\pit)$+(5/16) $cos(60\pit)$+(1/16) $cos(100\pit)$ all the powers have been eliminated. I tried to use the Euler theorem and the trigonometric identity , but I could not find the solution.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9137589931488037, "perplexity_flag": "middle"}
http://snikolov.wordpress.com/2012/11/14/early-detection-of-twitter-trends/	Early detection of Twitter trends explained / Next » A couple of weeks ago on Halloween night, I was out with some friends when my advisor sent me a message to check web.mit.edu, right now. It took me a few seconds of staring to realize that an article about my masters thesis work on a nonparametric approach to detecting trends on Twitter was on the homepage of MIT. Over the next few days, it was picked up by Forbes, Wired, Mashable, Engadget, The Boston Herald and others, and my inbox and Twitter Connect tab were abuzz like never before. There was a lot of interest in how this thing works and in this post I want to give an informal overview of the method Prof. Shah and I developed. But first, let me start with a story… A scandal On June 27, 2012, Barclays Bank was fined \$450 million for manipulating the Libor interest rate, in what was possibly the biggest banking fraud scandal in history. People were in uproar about this, and many took their outrage to Twitter. In retrospect, “#Barclays” was bound to become a popular, or “trending” topic. But how soon could one have known this with reasonable certainty? Twitter’s algorithm detected “#Barclays” as a trend at 12:49pm GMT following a big jump in activity around noon (Figure 1). Figure 1 But is there something about the preceding activity that would have allowed us to detect it earlier? It turns out that there is. We detected it at 12:03, more than 45 minutes in advance. Overall, we were able to detect trends in advance of Twitter 79% of the time, with a mean early advantage of 1.43 hours and an accuracy of 95%. In this post I’ll tell you how we did it. But before diving into our approach, I want to motivate the thinking behind it by going over another approach to detecting trends. The problem with parametric models A popular approach to trend detection is to have a model of the type of activity that comes before a topic is declared trending, and to try to detect that type of activity. One possible model is that activity is roughly constant most of the time but has occasional jumps. A big jump would indicate that something is becoming popular. One way to detect trends would be to estimate a “jumpiness” parameter, say p, from a window of activity and declare something trending or not based on whether p exceeds some threshold. Figure 2 This kind of method is called parametric, because it estimates parameters from data. But such a “constant + jumps” model does not fully capture the types of patterns that can precede a topic becoming trending. There could be several small jumps leading up to a big jump. There could be a gradual rise and no clear jump. Or any number of other patterns (Figure 3). Figure 3 Of course, we could build parametric models to detect each of these kinds of patterns. Or even one master parametric model that detects all of them. But pretty soon, we get into a mess. Out of all the possible parametric models one could use, which one should we pick? A priori, it is not clear. We don’t need to do this — there’s another way. A data-driven approach Instead of deciding what the parametric model should be, we take a nonparametric approach. That is, we let the data itself define the model. If we gather enough data about patterns that precede trends and patterns that don’t we can sufficiently characterize all possible types of patterns that can happen. Then instead of building a model from the data, we can use the data directly to decide whether a new pattern is going to lead to a trend or not. You might ask: aren’t there an unlimited number of patterns that can happen? Don’t we need an unreasonable amount of data to characterize all these possibilities? It turns out that we don’t, at least in this case. People acting in social networks are reasonably predictable. If many of your friends talk about something, it’s likely that you will as well. If many of your friends are friends with person X, it is likely that you are friends with them too. Because the underlying system has, in this sense, low complexity, we should expect that the measurements from that system are also of low complexity. As a result, there should only be a few types of patterns that precede a topic becoming trending. One type of pattern could be “gradual rise”; another could be “small jump, then a big jump”; yet another could be “a jump, then a gradual rise”, and so on. But you’ll never get a sawtooth pattern, a pattern with downward jumps, or any other crazy pattern. To see what I mean, take a look at this sample of patterns (Figure 4) and how it can be clustered into a few different “ways” that something can become trending. Figure 4: The patterns of activity in black are a sample of patterns of activity leading up to a topic becoming trending. Each subsequent cluster of patterns represents a “way” that something can become trending. Having outlined this data-driven approach, let’s dive into the actual algorithm. Our algorithm Suppose we are tracking the activity of a new topic. To decide whether a topic is trending at some time we take some recent activity, which we call the observation $\mathbf{s}$, and compare it to example patterns of activity $\mathbf{r}$ from topics that became trending in the past and topics that did not. Each of these examples takes a vote $V(\mathbf{r}, \mathbf{s})$ on whether the topic is trending or not trending (Figure 5). Positive, or trending examples ($\mathcal{R}_+$ in Figure 5) vote “trending” and negative, or non-trending examples ($\mathcal{R}_-$ in Figure 5) vote “non-trending”. The weight of each vote depends on the similarity, or distance $d(\mathbf{r}, \mathbf{s})$ between the example and the observation according to a decaying exponential $V(\mathbf{r}, \mathbf{s}) = e^{-\gamma d(\mathbf{r}, \mathbf{s})}$ where $\gamma$ is a scaling parameter that determines the “sphere of influence” of each example. Essentially, each example says, with a certain confidence, “The observation looks like me, so it should have the same label as me.” We used a Euclidean distance between activity patterns. Figure 5 Finally, we sum up all of the “trending” and “non-trending” votes, and see if the ratio of these sums is greater than or less than 1. $\displaystyle \sum_{r \in \mathcal{R}_+} e^{-\gamma d(\mathbf{r}, \mathbf{s})} / \displaystyle \sum_{r \in \mathcal{R}_-} e^{-\gamma d(\mathbf{r}, \mathbf{s})} > 1?$ One could think of this as a kind of weighted majority vote k-nearest-neighbors classification. It also has a probabilistic interpretation that you can find in Chapter 2 of my thesis. In general, the examples $\mathbf{r}$ will be much longer than the observations $\mathbf{s}$. In that case, we look for the “best match” between $\mathbf{s}$ and $\mathbf{r}$ and define the distance $d(\mathbf{s}, \mathbf{r})$ to be the minimum distance over all $dim(\mathbf{s})$-sized chunks of $\mathbf{r}$. This approach has some nice properties. The core computations are pretty simple, as we only compute distances. It is scalable since computation of distances can be parallelized. Lastly, it is nonparametric, which means we don’t have to decide what model to use. Results To evaluate our approach, we collected 500 topics that trended in some time window (sampled from previous lists of trending topics) and 500 that did not (sampled from random phrases in tweets, with trending topics removed). We then tried to predict, on a holdout set of 50% of the topics, which one would trend and which one would not. For topics that both our algorithm and Twitter’s detected as trending, we measured how early or late our algorithm was relative to Twitter’s. Our most striking result is that we were able to detect Twitter trends in advance of Twitter’s trend detection algorithm a good percent of the time, while maintaining a low rate of error. In 79% percent of cases, we detected trending topics earlier than Twitter (1.43 hours earlier), and we managed to keep an error rate of around 95% (4% false positive rate, 95% true positive rate). Naturally, our algorithm has various parameters (most notably the scaling parameter $\gamma$ and the length of an observation signal) that affect the tradeoff between the types of error and how early we can detect trends. If we are very aggressive about detecting trends, we will have a high true positive rate and early detection, but also a high false positive rate. If we are very conservative, we will have a low false positive rate, but also a low true positive rate and late detection. And there are various tradeoffs in between these extremes. Figure 6 shows a scatterplot of errors in the FPR(false positive rate)-TPR(true positive rate) plane, where each point corresponds to a different combination of parameters. The FPR-TPR plane is split up into three regions corresponding to the aggressive (“top”), conservative (“bottom”), and in between (“center”) strategies. Figure 6 also shows histograms of detection times for each of these strategies. Figure 6 Conclusion We’ve designed a new approach to detecting Twitter trends in a nonparametric fashion. But more than that, we’ve presented a general time series analysis method that can be used not only for classification (as in the case of trends), but also for prediction and anomaly detection (cartooned in Figure 7). Figure 7 And it has the potential to work for a lot more than just predicting trends on Twitter. We can try this on traffic data to predict the duration of a bus ride, on movie ticket sales, on stock prices, or any other time-varying measurements. We are excited by the early results, but there’s a lot more work ahead. We are continuing to investigate, both theoretically and experimentally, how well this does with different kinds and amounts of data, and on tasks other than classification. Stay tuned! ________________________________________________________ Notes: Thanks to Ben Lerner, Caitlin Mehl, and Coleman Shelton for reading drafts of this. I gave a talk about this at the Interdisciplinary Workshop on Information and Decision in Social Networks at MIT on November 9th, and I’ve included the slides below. A huge thank you to the many people who listened to dry runs of it and gave me wonderful feedback. For a less technical look, Prof. Shah gave a great talk at the MIT Museum on Friday, November 9th: 29 Comments 1. Pingback: Early detection of Twitter trends explained \| My Daily Feeds 2. How did you get your tweet data? Without firehose access (maybe even with), is it possible there is an inherent bias in twitter’s infrastructure that your algorithm has learned? • You can get the tweet data through the twitter API, which gives you a random sample of the firehose (though I have no idea if it is truly uniform). 3. Great work! Want to become able to do such things BTW, how what tools did you use to draw such pretty figures as 2, 3, 5, 7? • Thanks! Believe it or not, Powerpoint (for mac)! 4. Uzair Isn’t this incredibly basic? It’s basically photometric similarity (except you haven’t mentioned the importance of making the data scale-free here). I thought this was fairly common for comparing time series? • Indeed, the method is surprisingly simple. As far as making the data scale-free, I bet this would improve performance in practice (better accuracy given the amount of data, or less data needed for a desired accuracy). In theory, though, if we have enough data, the data itself should sufficiently cover all time scales, without doing any extra normalization. 5. acoulton This is a really interesting approach. I could also see it being useful with software/systems metrics (given access to a wide enough range of data). For example, predicting when to scale up or down clusters based on what’s about to happen (rather than waiting until the servers are already somewhat overloaded). Or alerting an engineer when the system might be about to crash (rather than waiting for the metrics that indicate it has already) so they can take preventative access or at least be at their desk when it happens. Lots of possibilities where currently we’re limited to rough guesses based on point in time values and again where there’s a range of semi-predicatable patterns leading up to an event. • Yes! I’d love to apply this to other domains that are serious pain-points for people. Monitoring a complex software system or forecasting cluster usage would be two excellent things to try. 6. Pingback: pinboard November 18, 2012 — arghh.net 7. Thanks for the great explanation of a classic time-series technique, using moving average (“MA”) features with k-nearest neighbors. Have you experimented with other (nearly) non-parametric models, such as decision tree ensembles? With trees you may not need to keep as much of the training data available. In the past I have found that kNN models like this are very sensitive to the choice of distance metric. So sensitive that the distance metric itself becomes something like a “parameter.” Have you tested other distance metrics, like simple Euclidean or Chebyshev? Also, how did you fit your gamma (example weight) parameters? • Thanks for the suggestions! I have not had a chance to play with decision tree ensembles. What kinds of decisions do you think would be appropriate at each node in this case? I have not yet tested other distance metrics, but this would be interesting to try. It is possible that the distance metric or the “features” (whatever you are measuring) have to be adjusted for different tasks. To pick parameters, I did parameter exploration for various combinations of parameters, and estimated detection earliness and error rates using cross-validation with each parameter setting. This showed the tradeoffs between early/late and types of error. One can then pick a parameter setting manually based on one’s desired specifications. • bgimpert With decision tree ensembles (i.e. random forests), you typically use a measure of feature importance at each decision node to build each path “down” the tree. When doing non-parametric classification with a decision tree ensemble (this problem), I have had good luck ranking features by information gain. That is, the reduction in Shannon entropy you get from “knowing” the value of each feature. Be careful with cross-validation in a time-series context like this one, because it is easy to accidentally peek into the future when training your model and/or fitting your parameters via simulation. I have not looked at your code on Github yet, but staying strictly “out-of-sample” when training or optimizing parameters is crucial for not being overly confident in a model. 8. zhabanet What if the “hot” pattern is “compressed” or “stretched” in time? Did you observe such phenomena in the rejected by your approach positive patterns? • In principle, with enough data, all the stretched versions of a pattern should be represented. In practice, it might make sense to apply something like Dynamic Time Warping in the distance metric. • zhabanet Sure, with enough data, we may see those, but I would guess that DTW will bring in more noise by being flexible; I asked because I wonder – if there exist at all slow and rapid pre-trending topics patterns, and if they would relate with the abundance of “followers”. • We have not looked at the effect of number of followers, network structure, or any other micro scale things on the pre trend pattern, but it would be an interesting question for future study. 9. Li Hi, I find your approach highly interesting. Would you to explain the algorithm behind it in more detail? Or where one can find it? Could also shoot me an email. Thanks in advance. • Hi Li, You can find my thesis on this topic at http://web.mit.edu/snikolov/Public/trend.pdf. Chapter 2 discusses the theory, Chapter 3 has the algorithm and pseudocode, and Chapter 4 talks about the application to Twitter data. • Li Great! Thanks a lot. 10. zhabanet Can you make the data available? It would be nice to see it and to try couple of thoughts… I have some experience with data from this archive http://www.cs.ucr.edu/~eamonn/time_series_data, and would like to try my toolkit on the twitter data. • Unfortunately not, but you can build a similar dataset using the Twitter streaming (https://dev.twitter.com/docs/streaming-apis) and trends (https://dev.twitter.com/docs/api/1/get/trends/%3Awoeid) APIs. • zhabanet I can, but how would I compare the performance? It would be much easier to tinker having the both – a working code and a sample set(s). 11. Yaks Hello, congratulations this looks really interesting. Is there anyway we can play with the algorithm? Are you planning to realease the code or should I implement it from the data provided in your thesis to give it a try? Thanks a lot! • Thanks! I have the code up at https://github.com/snikolov/rumor/. At the moment, it needs a lot of cleaning up. The main detection routine is called ts_shift_detect in https://github.com/snikolov/rumor/blob/master/processing.py if you want to take a look. You might be better off starting from scratch though, since the core algorithm itself is pretty straightforward. • thanks a lot for everything! I will take a look at it and if I can clean it up or help in someway I will 12. Jules Regarding your example with Twitter; how would your algorithm fare if your process of selecting the topics were truly online so that we can use them as queries ? 13. Very cool. I’m curious how you would ‘burn-in’ this algorithm. That is, in the absence of existing labeled trends, do you have any thoughts as to how you would start? The current algorithm seems like it might be at risk for only detecting trends that are similar to pre-labeled trends, which I suppose is the idea. 14. Man Regarding the trends topic: - how do you config your alpha ? - why are you using euler number since we can approximate it with e^y ~ 1+y Thanks %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 17, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.940976619720459, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/241583/computing-a-derivative-using-logarithmic-differentiation?answertab=active	Computing a derivative using logarithmic differentiation Using the Logarithmic Differentiation find the derivative of $y=\sqrt{x(x-1)/(x-2)}$...so I tried,but the result is not correct..can you show me a hint? so $\ln y= 0.5\ln[x(x-1)/(x-2)]$ $$\ln y=0.5[\ln x+\ln(x-1)-\ln(x-2)]$$ $$\ln y=0.5\ln x+0.5\ln(x-1)+0.5\ln(x-2)$$ $$y'=0.5\ln x+0.5\ln(x-1)-0.5\ln(x-2)[\sqrt{x(x-1)/(x-2)}]$$ - I gave your question a more descriptive title, and added the `calculus` tag. – Nate Eldredge Nov 20 '12 at 21:04 Check the plusses and minuses in the third row. Also, it'd be helpful if said what you got when you went on, instead of just saying it's not the same as what you want... – Micah Nov 20 '12 at 21:21 Your last line should be $y'=\frac 12 \left(\frac 1x+\frac 1{x-1}-\frac 1{x-2}\right)\sqrt{x(x-1)/(x-2)}$ (you forgot to differentiate the $\ln$ terms and the parenthesis). – Raymond Manzoni Nov 20 '12 at 22:05 2 Answers Your error is that you didn't actually differentiate $\ln x$ and the other logarithms on the right side. $$\frac{d}{dx} (0.5\ln x) = 0.5\cdot\frac1x = \frac{0.5}x.$$ - Hint: Remember that the derivative of $\ln y$ is $\frac{y'}{y}$, by the chain rule. Also, you seem to have flipped a sign in your last step; the last term of the third line should be $-0.5 \ln(x-2)$. Edit: After seeing what you did, you forgot to differentiate the terms with $x$! Suppose we have $\ln y = g(x)$. Then, differentiating both sides, $\frac{y'}{y} = g'(x)$, or $y'(x) = g'(x) y(x)$. You forgot to differentiate $g$, which in this case is $g(x) = \frac12[\ln x + \ln(x-1) - \ln(x-2)]$. - wait,i will edit it.. – AAS Nov 20 '12 at 21:21	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.928002119064331, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/124123/number-of-divisions-using-trial-division-to-factorize-the-product-of-two-k-digit	# Number of divisions using trial division to factorize the product of two k-digit primes I'm more of a programmer than a Mathematician so please bear with me if my question is too trivial. I am looking up RSA specifically the key generation bit. Using Trial division, I know that it would take roughly `(2*(n^0.5))/ln n` divisions to find the factors of n. However I'd like to represent this in a generalized manner i.e. how many divisions would it take to factorize the product of two k-digit primes ? Any help would be much appreciated - ## 1 Answer I will assume that both primes are approximately $5\times 10^k$. This is in fact the worst-case scenario for trial division. How long trial division will take depends on how exactly you implement it. If you have a table of all primes with up to $5\times 10^k$, then the number of trial divisions will be equal to the number of primes up to $5\times 10^k$, which is $$\pi(5\times 10^k)\approx\frac{5\times 10^k}{\ln 5\times 10^k}=\frac{5\times 10^k}{\ln 5 + k\ln 10}$$ with the approximation following from the prime number theorem. However, if you do not have such a table (which is likely the case if $k$ is large enough for the primes to be used for RSA) then you will need to perform $5\times 10^k$ trial divisions. - Thank you that helped :) – Jonny Mar 25 '12 at 12:25	{"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": 5, "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": 0.9554336667060852, "perplexity_flag": "head"}
http://mathhelpforum.com/pre-calculus/126338-graphing-function-window-print.html	# Graphing a function in the window? Printable View Show 40 post(s) from this thread on one page • January 30th 2010, 05:14 PM krzyrice Graphing a function in the window? I need help with this problem. So do I start out with plugging in random values for x? Such as -2, -1, 0, 1, 2 and so on. And what do they mean by the "Graph the following function in the window -10 < x < 10, -10 < y < 10?" Thank you. http://i50.tinypic.com/8yuzb7.jpg • January 30th 2010, 05:19 PM Prove It Quote: Originally Posted by krzyrice I need help with this problem. So do I start out with plugging in random values for x? Such as -2, -1, 0, 1, 2 and so on. And what do they mean by the "Graph the following function in the window -10 < x < 10, -10 < y < 10?" Thank you. http://i50.tinypic.com/8yuzb7.jpg Expand it out, so that you have $y = -\frac{1}{900}x^2 + \frac{10}{9}x$. You should know how to graph quadratics by finding $x$ and $y$ intercepts and turning points. • January 30th 2010, 05:25 PM krzyrice So like I said do I just plug in random values of x now and find the y values of it? Then I graph the equation and find the average rate of change to see if it is a line or not? Thanks for helping me so far. • January 30th 2010, 05:50 PM Prove It Is a Quadratic function ever a line? Like I said, the easiest way to sketch a graph is to find the $x$ and $y$ intercepts and the turning point. This gives you an idea of the shape of the graph. And then yes, you can plot points to make your graph a bit more accurate. • January 30th 2010, 05:57 PM krzyrice Nope a quadratic it's never a line. But What I don't get is what is the -10 < x < 10 and -10 < y < 10 for? What am I suppose to do with those? • January 30th 2010, 06:09 PM mr fantastic Quote: Originally Posted by krzyrice Nope a quadratic it's never a line. But What I don't get is what is the -10 < x < 10 and -10 < y < 10 for? What am I suppose to do with those? You are being told what scale to use on your x and y-axes (potentially, you're meant to graph this on a Graphing calculator and you're being told what window to use). • January 30th 2010, 06:36 PM skeeter screenshots ... • January 30th 2010, 07:27 PM krzyrice I thought the shape was suppose to be a parabola. Dang this problem is giving me a hard time. Sorry I don't have a graphing calculator so I don't really know how to solve this. And my teacher makes us write out the answer and explain how we got it also. (Worried) • January 30th 2010, 07:46 PM skeeter Quote: Originally Posted by krzyrice I thought the shape was suppose to be a parabola. Dang this problem is giving me a hard time. Sorry I don't have a calculator so I don't really know how to solve this. And my teacher makes us write out the answer and explain how we got it also. (Worried) the graph is a parabola ... you're only looking at a very small section of the graph in the given window. here's how it looks in an expanded window ... • January 30th 2010, 07:48 PM skeeter free download and easy to use ... Graph • January 30th 2010, 08:59 PM krzyrice I just tried graphing it on a graphing calculator and it gave me this: http://i45.tinypic.com/mcsz0n.jpg • January 31st 2010, 02:34 AM HallsofIvy Quote: Originally Posted by krzyrice I just tried graphing it on a graphing calculator and it gave me this: http://i45.tinypic.com/mcsz0n.jpg Okay, what window did you use? What does the graph look like if you set the window to $-10\le x\le 10$, $-10\le y\le -10$ as in your problem? • January 31st 2010, 04:48 AM skeeter Quote: Originally Posted by krzyrice I just tried graphing it on a graphing calculator and it gave me this: http://i45.tinypic.com/mcsz0n.jpg that is not the function the exercise gave you ... $y = -x\left(\frac{x-1000}{900}\right)$ looks like you graphed $y = x^2$ • January 31st 2010, 10:41 AM krzyrice In the calculator I typed in -x(x-1000/900). Should I have typed it differently? I also tried your way -x(x-1000)/900 and got a straight line. And even if I select zoomfit it would still show a straight line. http://i45.tinypic.com/wvvdqu.jpg • January 31st 2010, 10:45 AM krzyrice Quote: Originally Posted by HallsofIvy Okay, what window did you use? What does the graph look like if you set the window to $-10\le x\le 10$, $-10\le y\le -10$ as in your problem? This is what it looks like when I use $-10\le x\le 10$, $-10\le y\le -10$ http://i48.tinypic.com/ipuqsm.jpg Also I input the equation in the calculator as -x((x-1000/900)) Show 40 post(s) from this thread on one page All times are GMT -8. The time now is 05:33 AM. Copyright © 2005-2013 Math Help Forum. All rights reserved. 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": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": 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": 0.9492079615592957, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/288319/what-is-a-tangent-bundle-aubin/288334	# What is a tangent bundle? (Aubin) Here's what I read in A Course in Differential Geometry by Thierry Aubin. 2.5. Definition. The tangent bundle $T(M)$ is $\bigcup_{P\in M} T_P(M).$ And then 2.6. Definition. Let $\Phi$ be a differentiable map of $M_n$ into $W_p$ (two differentiable manifolds). Let $P\in M_n,$ and set $Q=\Phi(P).$ The map $\Phi$ induces a linear map $(\Phi_)_P$ of the tangent bundle $T_P(M)$ into $T_Q(W)$ defined by $$[(\Phi_)_PX](f)=X(f\circ\Phi);$$ here $X\in T_P(M),\;(\Phi_)_PX\in T_Q(W)$ and $f$ is a differentiable function in a neighbourhood $\theta$ of $Q.$ We call $(\Phi_)_P$ the linear tangent mapping of $\Phi$ at $P.$ I don't understand why the author calls $T_P(M)$ a tangent bundle in the second definition. Is it a mistake? From the first definition, a tangent bundle is the union of all tangent spaces over all points of the manifold. And $T_P(M)$ is just one tangent space, at a particular point $P$. And an additional question: Should I be worried whether the union in the first definition is disjoint or not? After a moment's thought, I believe it might turn out not to be according to the previous definitions. - ## 3 Answers Definition 2.6 has a typo; $(\Phi_\ast)_P$ is a map of tangent spaces, not tangent bundles (although all of the $(\Phi_\ast)_P$ combine to form a bundle map between tangent bundles). The tangent bundle is the disjoint union of the tangent spaces: $$TM = \coprod_{P \in M} T_P M.$$ It has the topology of a smooth manifold in the following manner. Let $(U_\alpha, \phi_\alpha)$ be an atlas for $M$, and let $\pi: TM \longrightarrow M$ be the natural projection, i.e. if $(P, v) \in T_P M \subset TM$, then $\pi(P, v) = P$. Then we get an atlas $(\pi^{-1}(U_\alpha), \tilde{\phi}_\alpha)$ for $TM$, where $$\tilde{\phi}_\alpha(P, v) = (\phi_\alpha(P), v).$$ - I will begin by attempting to address the question in the subject and definition 2.5. The tangent bundle of a differentiable manifold $M$ is way to organize all of the (point-wise) tangent spaces of $M$ into a formal geometric object (in fact, the tangent bundle $T(M)$ turns out to be a differentiable manifold in its own right). When first learning the material, I would not be too worried about whether you should have a disjoint union or a union in the formal definition; I would be concerned about what the definition is trying to tell you. One way to think of the definition of the $T(M)$ is that it is family of vector spaces parametrized by the manifold $M$. At each point $P \in M$, you get an $n$-dimensional vector space associated to $P$. In this case, the vector space associated to the point $P$ is the tangent space $T_{P}M$. In formal bundle''terminology, the tangent space $T_{P}M$ associated to a point $P \in M$ is called the fiber over $P$. In definition 2.6, the author is trying to tell you that a differentiable map $\Phi : M \to W$ induces a mapping $\Phi_{} : T(M) \to T(W)$ that maps the fiber over $P$ (i.e. $T_{P}M$) to the fiber over the image of $P$ (i.e. $T_{Q}W$, where $\Phi(P) = Q$). It does appear that the use of the word tangent bundle'' in definition 2.6 is not quite correct (or at least, could be confusing). Maybe something along the following lines would work better: . . . At each $P \in M$, $\Phi$ induces a linear mapping $(\Phi_{})_{P}: T_{P}M \to T_{Q}W$ defined by, . . .'' It should be noted that when all of the fiber wise maps are pieced together, however, one does indeed obtain a linear map $\Phi_{*} : T(M) \to T(W)$. There is a more general definition of a vector bundle over a manifold $M$, of which the tangent bundle is the prototypical example. (I find Spivak's Comprehensive Introduction to Differential Geometry (I forget which volume), or John Lee's Introduction to Smooth Manifolds to be quite helpful on these matters.) - It looks to me like Aubin isn't assuming you know the general definition of a bundle. So strictly speaking he can't really talk about a "linear map of bundles", only a linear map of each individual tangent space, and he's trying to finesse that point. I'd say you're right that it's poorly phrased, though. (The reason you shouldn't worry too much about it is that a linear map of bundles is pretty much just a linear map on each tangent space separately, with some added continuity/differentiability conditions that are more or less automatic in this case.) For your additional question, you should definitely take the union in the first definition to be disjoint, or bad things will happen. If it makes you feel better, you can think of $TM$ as consisting of ordered pairs $(p,v)$ where $v \in T_p M$. - Thank you very much. How can there be linear maps between of bundles? It's true that I don't know the general definition of a bundle, but in this case I don't see a good way to make a tangent bundle a vector space generally. Is there a way to make a disjoint union of vector spaces a vector space? – Bartek Jan 27 at 21:03 @Bartek: There is not a way to make it into a vector space. A linear map between bundles is a different object than a linear map between vector spaces (albeit closely related); you have to define what it means separately. (On the other hand, if you have that definition, it's not hard to see that the disjoint union of all of Aubin's maps satisfies it, which I'm guessing is what was in his mind when he wrote that sentence.) – Micah Jan 27 at 21:16	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 58, "mathjax_display_tex": 3, "mathjax_asciimath": 3, "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": 0.9434433579444885, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/tagged/modular-arithmetic?page=1&sort=votes&pagesize=50	# Tagged Questions Modular arithmetic (clock arithmetic) is a system of arithmetic of integers. The basic ingredient is the congruence relation $a \equiv b \bmod n$ which means that $n$ divides $b-a$. In modular arithmetic one can add, subtract, multiply and exponentiate but not divide in general. The Euclidean ... 14answers 2k views ### 'Linux' math program with interactive terminal? Are there any open source math programs out there that have an interactive terminal and that work on linux? So for example you could enter two matrices and specify an operation such as multiply and ... 6answers 558 views ### Show that $\frac{(3^{77}-1)}{2}$ is odd and composite The question given to me is: Show that $\large\frac{(3^{77}-1)}{2}$ is odd and composite. We can show that $\forall n\in\mathbb{N}$: 3^{n}\equiv\left\{ \begin{array}{l l} 1 & \quad ... 2answers 296 views ### Find a composite number $n$ satisfies $(2+3I)^n≡2-3I\pmod{n}$ As we know if $p$ is an odd prime number then $$(a+bI)^p\equiv a+(-1)^\frac{p-1}2bI\pmod{p},$$ where $I=\sqrt{-1}$. However, is there any composite number $n$ that satisfies ... 2answers 239 views ### A puzzle with powers and tetration mod n A friend recently asked me if I could solve these three problems: (a) Prove that the sequence $1^1, 2^2, 3^3, \dots \pmod{3}$ in other words $\{n^n \pmod{3} \}$ is periodic, and find the length of ... 2answers 461 views ### nth powers modulo all primes Let $a \in \mathbb{Z}$, $n \in \mathbb{N}^$ be integers, and set $P=X^n - a$. Let us consider the three following statements : 1) $P$ has a root in $\mathbb{Z}$ (i.e. $a$ is an nth power) 2) $P$ ... 5answers 193 views ### Why are there$736$ $2\times 2$ matrices $(M)$ over $\mathbb{Z}_{26}$ for which it holds that $M=M^{-1}$? I'm currently trying to introduce myself to cryptography. I'm reading about the Hill Cipher currently in the book Applied Abstract Algebra. The Hill Cipher uses an invertible matrix $M$ for ... 1answer 7k views ### Finding a primitive root of a prime number How would you find a primitive root of a prime number such as 761? How do you pick the primitive roots to test? Randomly? Thanks 2answers 194 views ### Solve congruence: $45x \equiv 15 \pmod{78}$ (What am I doing wrong?) Question about solving congruence. I've worked out how to solve them for the most part except for the following problem I'm having: $$45x \equiv 15 \pmod{78}$$ By the euclidean algorithm, I work out ... 3answers 2k views ### Calculate which day of the week a date falls in using modular arithmetic In Summer Wars the main character (he is a mathematician) calculates the day of the week of someone's birthday (19/07/1992 is Sunday). I know (very) basic modular arithmetic but I can't figure out how ... 2answers 157 views ### Is my shorter expression for $s_m(n)= 1^m+2^m+3^m+\cdots+(n-1)^m \pmod n$ true? I'm considering the following sums for natural numbers n,m $$s_m(n)= \sum_{k=1}^{n-1} k^m =1^m+2^m+3^m+\cdots+(n-1)^m$$ modulo n . Looking at odd n first, I found by analysis of the pattern of ... 5answers 1k views ### If $n$ is composite, then $n$ divides $(n-1)!$. I have a proof and need some feedback. It seems really obvious that the statement is true but it is always the obvious ones that are a little trickier to prove. So I would appreciate any feedback. ... 2answers 226 views ### Formula for occurrence of leap years in the Jewish calendar Over at Judaism.SE, there was a discussion about a formula to determine leap years in the Jewish calendar. Basically, the calendar follows a 19-year cycle, and seven of those years -- 3, 6, 8, 11, 14, ... 7answers 654 views ### Pattern to last three digits of power of $3$? I'm wondering if there is a pattern to the last three digits of a a power of $3$? I need to find out the last three digits of $3^{27}$, without a calculator. I've tried to find a pattern but can not ... 1answer 158 views ### Elementary Number Theory; prove existence Prove that there exists a positive integer $n$ such that $$2^{2012}\;\|\;n^n+2011.$$ I was wondering if you could prove this somehow with induction (assume that $n$ exists for $2^k\|n^n+2011$ ... 2answers 441 views ### The modular curve X(N) I have a question about the modular curve X(N), which classifies elliptic curves with full level N structure. (A level N structure of an elliptic curve E is an isomorphism from $Z/NZ \times Z/NZ$ to ... 2answers 583 views ### Very simple question, but what is the proof that x.y mod m == ((x mod m).y) mod m? I apologise for this question, as it is no doubt very simple, but I've never been very confident with proofs. Our lecturer today (in a course related to maths but not mathematical itself) was playing ... 1answer 403 views ### Why are the only numbers $m$ for which $n^{m+1}\equiv n \pmod{m}$ is true also unique for $\displaystyle\sum_{n=1}^{m}{n^m}\equiv 1 \bmod m$? It can be seen here that the only numbers for which $n^{m+1}\equiv n \pmod{m}$ is true are 1, 2, 6, 42, and 1806. Through experimentation, it has been found that ... 3answers 164 views ### Why do the gaussian integers have only 2 congruence classes mod 1+i? If we consider $Z[i]$ modulo $1+i$, why are there only two congruence classes? 4answers 329 views ### A “fast” way to manually compute $3^{41}+7^{41} \pmod{13}$ The problem: Find the remainder of $3^{41}+7^{41}$ when divided by $13$. My approach is by utilizing the cyclicity of remainders for examples $3^1,3^2,3^3,3^4,3^5 \text{ and }3^6$ when divided ... 2answers 208 views ### $3x^2 ≡ 9 \pmod{13}$ What is $3x^2 ≡ 9 \pmod{13}$? By simplifying the expression as $x^2 ≡ 3 \pmod{13}$ and applying brute force I can show that the answers are 4 and 9, but how to approach this in a more efficient way? ... 3answers 392 views ### How to find the solutions for the n-th root of unity in modular arithmetic? $$\begin{align} x^n\equiv1&\pmod p\quad(1)\\ x^n\equiv-1&\pmod p\quad(2)\end{align}$$ Where $n\in\mathbb{N}$,$\quad p\in\text{Primes}$ and $x\in \{0,1,2\dots,p-1\}$. How we can find the ... 3answers 207 views ### Number of integers not divisible by $p$ and $q$ Here's a part of question from Siklos' "Advanced Problems in Core Mathematics": How many integers greater than or equal to zero and less than 1000 are not divisible by 2 or 5? What is the average ... 4answers 237 views ### A solution to $y^5 \equiv 2\pmod{251}$ I need to show that the following equation has a solution. (I am not asked for the answer, which I know by Mathematica to be $y=43$. ) $y^5 \equiv 2 \pmod{251}.$ I know that the order of 2 is 50, ... 1answer 265 views ### Is there formal name and proof for this formula/theorem (for now I'm calling it Orbital Collinearity Theorem)? Suppose you are given a question that goes like this: Consider three planets that revolve around a star on separate circular orbits sharing the same orbital plane. The first planet, A, takes ... 5answers 315 views ### Basic question about mod I'm having a tough time understanding why $(a^x \bmod p)^y \bmod p$ is equal to $a^{xy}\bmod p$. Does this have a mathematical proof? Please advise. 4answers 263 views ### How can I show $e^2 \equiv 1 \bmod 24$, given that $\gcd(e, 24) = 1$? The problem comes from a practice final for a final exam I have later today. It says "Show that if $\gcd(e, 24) = 1$ then $e^2 \equiv 1 \bmod 24$". I found that Euler's totient function \$\phi(24) = ... 4answers 2k views ### Modular exponentiation using Euler’s theorem How can I calculate $27^{41}\ \mathrm{mod}\ 77$ as simple as possible? I already know that $27^{60}\ \mathrm{mod}\ 77 = 1$ because of Euler’s theorem: $$a^{\phi(n)}\ \mathrm{mod}\ n = 1$$ and ... 4answers 485 views ### calculating n choose k mod one million I am working on a programming problem where I need to calculate 'n choose k'. I am using the relation formula $${n\choose k} = {n\choose k-1} \frac{n-k+1}{k}$$ so I don't have to calculate huge ... 6answers 124 views ### Number Theory and Congruency I have the following problem: $$2x+7=3 \pmod{17}$$ I know HOW to do this problem. It's as follows: $$2x=3-7\\ x=-2\equiv 15\pmod{17}$$ But I have no idea WHY I'm doing that. I don't really even ... 3answers 216 views ### Find $n\geq1$ such that 7 divides $n^n-3$ Find $n\geq1$ such that 7 divides $n^n-3$. Here is what I found: $n\equiv 0 \mod7, n^n\equiv 0 \mod7,n^n-3\equiv -3 \mod7$ no solution. \$ n\equiv 1 \mod7, n^n\equiv 1 \mod7,n^n-3\equiv -2 \mod7 ... 4answers 453 views ### How do I compute $a^b\,\bmod c$ by hand? How do I efficiently compute $a^b\,\bmod c$: When $b$ is huge, for instance $5^{844325}\,\bmod 21$? When $b$ is less than $c$ but it would still be a lot of work to multiply $a$ by itself $b$ times, ... 3answers 143 views ### Number of $k^p \bmod q$ classes when $q\%p > 1$ I want to show that when $p, q$ are primes, $k^p\bmod q$ takes on $q-1$ distinct values (as $k$ ranges over positive integers) if and only if $q \not\equiv 1 \pmod p$. (It is easy to verify this ... 2answers 282 views ### Find all linearly dependent subsets of this set of vectors I have vectors in such form (1 1 1 0 1 0) (0 0 1 0 0 0) (1 0 0 0 0 0) (0 0 0 1 0 0) (1 1 0 0 1 0) (0 0 1 1 0 0) (1 0 1 1 0 0) I need to find all linear ... 1answer 62 views ### Is there always a primitive m-th root of unity with imaginary part bigger than 1/2 Let $m$ be a positive integer. I need the existence of a primitive $m$-th root of unity $\zeta_m$ such that its imaginary part is strictly greater than $1/2$. We can write \$\zeta_m = \exp(2\pi i ... 2answers 495 views ### Tribonacci sequence modulo X The Tribonacci sequence satisfies $$T(n) = T(n-1) + T(n-2) + T(n-3)$$ with $T(0)=0$, $T(1)=1$, $T(2)=1$. I need to calculate $T(y) \mod 10000$ for $y > 2^{40}$. How can I make this faster? I ... 1answer 86 views ### Tell wether $(1234657)! +1 \equiv_{11111} (7654321)! -1$ is true or false I have to tell if the following is true or false: $$(1234657)! +1 \equiv_{11111} (7654321)! -1$$ so by definition we can rewrite the previous equivalence as: \$(1 \cdot 2 \cdot \ldots \cdot 11110 ... 1answer 437 views ### Modular arithmetic problem I need some hints for this problem from Dummit and Foote. (edit: added the full question verbatim for context) Let $p$ be an odd prime and let $n$ be a positive integer. Use the binomial theorem to ... 1answer 108 views ### A proof in number theory dealing with modular congruences. So we are asked to show that $$(p-1)(p-2)\cdots(p-r)\equiv (-1)^{r}r! \pmod{p}$$ for $r=1,2,...,p-1$. I worked on it and I want to know if my proof suffices to show what is being asked. I would also ... 5answers 209 views ### Proving that $2^{2^n} + 5$ is always composite by working modulo 3 By working modulo 3, prove that $2^{2^n} + 5$ is always composite for every positive integer n. No need for a formal proof by induction, just the basic idea will be great. 3answers 222 views ### How can I prove that an order (“$<$” say) on $\mathbb Z_n$ cannot be defined? I'm trying to show why it isn't possible to define an order of magnitude on $\mathbb Z_n$ (modular arithmetic) that satisfies the ordering properties of $\mathbb Z$. By letting addition to be ... 5answers 250 views ### How to solve $100x +19 =0 \pmod{23}$ How to solve $100x +19 =0 \pmod{23}$, which is $100x=-19 \pmod{23}$ ? In general I want to know how to solve $ax=b \pmod{c}$. 5answers 209 views ### Why is $x^2 \pmod{16}$ always $0, \ 1,\ 4,\ 9$? With a simple piece of code I could deduce that for any non-negative integer $x$ the value of $x^2 \pmod{16}$ is always a number from the set $\{0, 1, 4, 9\}$. However, the math behind it evades me. ... 6answers 134 views ### Show that $x^4 +1$ is reducible over $\mathbb{Z}_{11}[x]$ and splits over $\mathbb{Z}_{17}[x]$. Reduction into linear factors $\mathbb{Z}_{17}[x]$: This part is not too hard: $x^4 \equiv -1$ mod 17 has solutions: 2, 8, 9, 15 so \$(x-2)(x-8)(x-9)(x-15) = x^4 -34 x^3 +391 x^2-1734 x+2160 \equiv ... 3answers 216 views ### How to compute $7^{7^{7^{100}}} \bmod 100$? How to compute $7^{7^{7^{100}}} \bmod 100$? Is $$7^{7^{7^{100}}} \equiv7^{7^{\left(7^{100} \bmod 100\right)}} \bmod 100?$$ Thank you very much. 2answers 129 views ### How to prove that for all $m,n\in\mathbb N$, $\ 56786730 \mid mn(m^{60}-n^{60})$? How to prove $\forall m,n\in\mathbb N$: $$56786730 \mid mn(m^{60}-n^{60})?$$ Thanks in advance. 4answers 999 views ### How do you calculate the modulo of a high-raised number? I need some help with this problem: $$439^{233} \mod 713$$ I can't calculate $439^{223}$ since it's a very big number, there must be a way to do this. Thanks. 2answers 139 views ### $(1+p)^n$ is not $1 \pmod {p^r}$ when $n < p^{r-1}$ Let $p$ be an odd prime. I know that $(1+p)^{p^{r-1}}\equiv 1 \pmod {p^r}$ but how can I prove that if $n < p^{r-1}$ then $(1+p)^n$ is not $1 \pmod {p^r}$. I tried to prove using the properties of ... 2answers 157 views ### Efficient way to find $a$ in $c = 6a\mod n$ Given $c$ and $n$ in $c = 6a\mod n$, how can I find the lowest positive integer value for $a$? I could find it iteratively by rewriting as $\displaystyle a = \frac{c + xn}{6}$ and increasing $x$ ... 3answers 435 views ### How to calculate $3^{45357} \mod 5$? I wrote some code, here is what it gives: \begin{align} 3^0 \mod 5 = 1 \\ 3^1 \mod 5 = 3 \\ 3^2 \mod 5 = 4 \\ 3^3 \mod 5 = 2 \\\\ 3^4 \mod 5 = 1 \\ 3^5 \mod 5 = 3 \\ 3^6 \mod 5 = 4 \\ 3^7 \mod 5 = 2 ... 4answers 57 views ### $20^{15} + 16^{18}$ is divided by 17 What is the reminder, when $20^{15} + 16^{18}$ is divided by 17. I'm asking the similar question because I have little confusions in MOD. If you use mod then please elaborate that for beginner. ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 147, "mathjax_display_tex": 15, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.932000994682312, "perplexity_flag": "head"}
http://mathematica.stackexchange.com/questions/tagged/functions+manipulate	Tagged Questions 6answers 258 views Manipulate and Turning Expressions into Functions I've been trying to use Manipulate to do interactive plotting, but I've been running into a few problems with saved expressions. I have an expression saved as "func" and I want to work with it and ... 1answer 217 views Cournot Equilibrium I'm trying to determine reaction functions of a Cournot equilibrium for $n$ firms using the same optimal condition, using the fixed point method. This is the method I'm using: ... 2answers 113 views A command for outputting a list of parameter values in a Manipulate interface I am looking for a command that will output a list of values that correspond to the current parameter values manually set on a particular Manipulate interface. For ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.846612811088562, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/10349?sort=oldest	## Smoothness of Symmetric Powers ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Here's something that's been bothering me, and that's come up again for me recently while reading some stuff about Hilbert schemes of points (Nakajima's lectures, specifically): Let $C$ be an algebraic curve. Define $S^nC$ to be $C\times\ldots\times C/S_n$, the symmetric power. Now, over $\mathbb{C}$, I can show that $C$ a complex manifold implies that $S^nC$ is, and that if $X$ is a variety with $S^nX$ smooth, then $X$ is one dimensional, but the argument I have involves looking in analytic open sets and reducing to the case of $\mathbb{C}^n$, and additionally is fairly unhelpful for identifying the total space (ie, that $S^n\mathbb{P}^1\cong\mathbb{P}^n$) So here's my question: how can we, in a fairly quick and natural way, show that 1. If $C$ is a smooth, 1-dimensional variety over an algebraically closed field $k$,, then $S^nC$ is smooth. 2. If $X$ is a smooth variety over an algebraically closed field $k$, and $S^nX$ is smooth, then $X$ is one dimensional. Now, I don't want any projectivity hypotheses here, and I'm curious, with the more arithmetically inclined, is this still true over an arbitrary field? - ## 4 Answers Just convince yourself that if $(C,P_1,\dots,P_n)$ and $(D,Q_1,\dots,D_n)$ are analytically isomorphic -- $C$ at the point $P_i$, $D$ at the point $Q_i$ -- then $Sym^n C$ at the point $(P_1,\dots,P_n)$ is analytically isomorphic to $Sym^n D$ at the point $(Q_1,\dots,Q_n)$. For this, you need to see that completion commutes with taking $S_n$-invariants. Let me hedge and say that this is clear if $P_2,\dots,P_n$ are smooth points. (When I say "analytically isomorphic", I don't mean working over $\mathbb C$. Instead, as customary, I mean that the completion of the corresponding rings are isomorphic, for example $\widehat{\mathcal O}_{C,P_i}\simeq \widehat{\mathcal O}_{D,Q_i}$.) Thus for 1 you can reduce to $D=\mathbb A^1$, and you are done. 1 is true over any field, without restrictions on characteristic. Indeed, the ring corresponding to $(\mathbb A^1)^n/S_n$ is the ring of invariants $k[x_1,\dots, x_n]^{S_n}$, and it is a polynomial ring on $n$ variables, the elementary symmetric polynomials in $x_i$. That is true over any ring $k$ (commutative, with identity). Part 2 is reduced to the case of $\mathbb A^2$ by the same trick, and then it is an explicit computation. I will illustrate the $(\mathbb A^2)^n/S_n$ case. We need to find the ring of invariants $k[x_1,y_1,x_2,y_2,\dots,x_n,y_n]^{S_n}$. Now there are $2n$ elementary polynomials in $x_i$, resp. in $y_i$, of course. But there is more. For example $x_1y_1 + \dots + x_ny_n$ is an invariant. To prove that $X=(\mathbb A^2)^n/S_n$ is singular at the point $P=(0,\dots,0)$ is equivalent to showing that $\dim T_{P,X} = \dim m/m^2 > 2n = \dim X$, where $m$ is the maximal ideal in $\mathcal O_{P,X}$. It is an exercise that the above $2n+1$ polynomials are linearly equivalent in $m/m^2$, which would complete the proof. - So then after using completions (always forget about those...need to work on that) it reduces to affine space, and then symmetric polynomials prove the first claim for the affine line over any field. Now, for the second, you're taking the points to all be distinct, but what about on the diagonals? – Charles Siegel Jan 1 2010 at 17:08 I misunderstood question 2. I will fix the solution. The previous version proved that $X$ singular implies $Sym^n X$ singular (which is trivial of course). – VA Jan 1 2010 at 19:56 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Smoothness is an etale local property, and you can check that symmetric power sends etale maps to etale maps. So checking on $\mathbb{A}^n$ is fine, since all smooth varieties over an algebraically closed field is etale locally isomorphic to it. Of course, this is secretly the same as VA's answer, since stuff which makes sense etale locally is exactly the stuff remembered by the completion of the local ring, but I like the sound of it better. - I also wanted to mention a `high-technology' answer to (1). Namely, if $C$ is a smooth algebraic curve, its $n$-th symmetric power coincides with the variety of all degree $n$ effective divisors on $C$ (that is, we consider $n$-element subsets of $C$, and then allow points to merge). In a fancier language, we are looking at the Hilbert scheme of length $n$ subschemes of $C$. Note that this agrees with the explicit choice of coordinates if $C={\mathbb A}^1$ (and also $C={\mathbb P}^1$): namely, if you want to parametrize $n$-element subsets of ${\mathbb A}^1$ with multiplicities, you do so (essentially by unique factorization) by saying that such a subset is the zero locus of a degree $n$ monic polynomial $p$, so it is uniquely determined by coefficients of $p$ (which of course are the elementary symmetric polynomials in terms of original $n$ points). OK, so why do finite subschemes of $C$ form a smooth variety? Because deformation theory says that at a particular finite subscheme $D\subset C$, the obstructions to smoothness belong to $Ext^2_C(O_D,O_D)$. This vanishes because $C$ is a smooth curve, and so there are no $Ext$'s beyond first (OK, no local $Ext$'s, but $D$ is finite, so it's the same thing). I think this is a useful point of view. For instance, if $S$ is a smooth surface, $Sym^n S$ is singular, but one can check that the Hilbert scheme $Hilb_n S$ of degree $n$ finite subschemes is still smooth; it provides a resolution of singularities of $Sym^n S$. One can thus argue that the Hilbert scheme is a 'better behaved' object compared to the symmetric power... - Edit: It seems appropriate to recall here Chevalley-Shephard-Todd's Theorem. It says the quotient of $\mathbb A^n_k$ by a finite linear group $G$ with order prime to the characteristic of $k$ is smooth (i.e. the algebra of invariants is polynomial) if and only if $G$ is generated by pseudo-reflections (codimension one fixed point set). Once one localizes the problem at a point, as VA and Ben Webster did, this settles both 1 and 2 over arbitrary fields of characteristic zero. Of course VA argument is preferable as it is more direct/elementary. Original answer. Below is my original answer commented by David Speyer below: Over the category of smooth real manifolds the symmetric power of smooth curves is not smooth in general. The second symmetric power is a smooth surface with boundary and starting from the third symmetric power what we get are varieties with corners at the boundary. Symmetric powers of smooth surfaces are still smooth as they are locally diffeomorphic to complex curves. If nothing else these examples show that smoothness might mean different things for algebraic geometers and differential geometers in algebraic geometry over $\mathbb R$ and in differential geometry. - 1 I disagree. These examples show that "curve" and "surface" mean something different in algebraic and differential geometry, because algebraic geometers usually describe things by their complex dimension. What you are calling surfaces are being called curves in the other answers. But we agree about which objects are smooth. – David Speyer Jan 1 2010 at 14:59 I was thinking on curves over $\mathbb R$, or more precisely on its real points. According to VA's answer the symmetric powers of $\mathbb A^1$ are smooth while the symmetric powers of $\mathbb A^1(\mathbb R)=\mathbb R$, as real manifolds, are not. For $\mathbb A^2$ and $\mathbb A^2(\mathbb R) = \mathbb R^2$ we have the reverse phenomenon. – jvp Jan 2 2010 at 1:03	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 93, "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": 0.9344316720962524, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/8149/collision-time-of-brownian-particles/9552	# Collision time of Brownian particles Let's assume two spherical particles $p_1$ and $p_2$ of finite radius $r_1$ and $r_2$, which are at locations $(\pm\frac{d}{2},0,0)$ a distance $d$ apart at initial time $t$. These particles diffuse with coefficients $D_1$ and $D_2$, respectively. How can I obtain the probability distribution of the collision time (that is, if the time at which they collide is $t + \Delta t$, I would like to know the probability density function of the random variable $\Delta t$), or at least some of its moments, in these two cases: 1. when the domain is infinite, e.g., $\mathbb{R}^3$, and 2. when it is bounded (inside a sphere with surface area $A$ and the origin $(0,0,0)$ as its center)? I tried to look up for an answer in Van Kampen's book, but I couldn't really come up with an answer. - 3 Is this the same (at least for the infinite domain) as a single particle $p$ starting at a distance $d$ from the origin and you want to know when it first comes within distance $r_1+r_2$ of the origin? – Henry Apr 6 '11 at 18:02 1 @Georg: collision time is expectation of time of first meeting. Care must be taken when formulating this problem but the question certainly makes perfect sense, at least in some cases. E.g. for random walk of two particles on $\mathbb Z^2$ the difference in position is another random walk and because in two dimension the random walk is recurrent, so the collision time is finite too. By letting the lattice spacing go to zero, one expects similar fact to hold in continuous case, depending on precise formulation. – Marek Apr 6 '11 at 19:09 2 What is the collision time? Expected collision time? If so, it is infinite even for 1D random walk on Z – Holowitz Apr 7 '11 at 1:48 3 The relative position of two particles, each one taking a Brownian walk, is in itself a Brownian walk. So you're asking for the recurrence time for a Brownian walk to return close to the origin. This is a well-studied problem whose answer you should be able to find. (I'm too lazy to look the answer up, but if I recall correctly, in two dimensions, the particle will with probability 1 return close to the origin, but the expected time to do so is infinite.) – Peter Shor Apr 7 '11 at 13:55 1 @Peter: if you do stand by it, then post it as a genuine answer, not just a comment. And perhaps you could illuminate for the less knowledgeable of us how you obtained your result. – Marek Apr 7 '11 at 21:38 show 8 more comments ## 3 Answers The previous answers do not exhaust this question, so I will add a solution to the problem as posted. The question is: what is the probability distribution for the meeting time of two diffusing spheres of radii $r_1$ and $r_2$ starting at initial separation R. By transforming to relative coordinates (as Peter Shor said), you reduce the problem to one diffusing point particle which is absorbed on a sphere of radius $A=r_1+r_2$. The following is true in 3d • The particles will not certainly collide. The probability of eventual collision is A/R. The probability of reaching infinity is $1-A/R$. • If you ask the question--- do the particles collide first or reach a separation of B first, the probability of colliding first is ${ {1\over R} - {1\over B} \over {1\over A} - {1\over B} }$ • There is the usual 3d miracle of Laplacians, that there is a change of variables which normalizes the radial part to a free 1d Laplacian, which allows a closed form expression for the probability density of colliding at any finite time. This gives a complete answer to the question: the probability of not having collided at time t is given by $P(t) = (1-{A\over R}) + {A\over R} S(\Delta,t)$, where $\Delta= R-A$, and $S(\Delta,t)$ is the probability that a 1 dimensional Brownian motion starting at position $\Delta$ has not hit the origin by time t. $S(\Delta,t)$ is a simple function which is essentially just the error function: $$S(\Delta,t) = ( {2\over \sqrt{\pi}} \mathrm{erf}({\Delta \over \sqrt{2t}}))$$ It's only a function of ${\Delta \over\sqrt{t}}$ by scaling. The probability density of a collision at time t is found by differentiating the above expression, and it only involves elementary functions: $$\rho(t) = {A\over R} {1\over 4\sqrt{\pi}}{\Delta e^{-{\Delta^2\over 2t}} \over t^{3\over 2}}$$ The answer is strange, because up to the factor of A/R which accounts for the fact that not all walks collide, it is exactly the same as the probability density for a 1d Brownian motion to collide with the origin at time t. ### Image charge magnitude is collision probability Diffusion is by the Laplacian. If you introduce a steady stream of particles at position R, they will come to a steady equilibrium where some are running off to infinity, and some are absorbed by the sphere of radius A. If you introduce these particles at a random angular position on the sphere of radius R, you get a spherically symmetric solution of Laplace's equation at steady state $\phi(r)$, which represents the steady-state density of particles. The gradient of $\phi$ is the particle steady-state flow, it is the electric field in an electrostatic analog. The source on the sphere at R is a charge Q uniformly distributed on the sphere, which produces an outgoing electric field, an outgoing flow of particles. The particle sink on the sphere of radius A is a potential 0 surface, a metal grounded to infinity. The solution to this places an image sphere at $A^2/R$ which has charge as required to cancel the potential. The charge is QA/R. This means that the flux into the metal is a fraction A/R of the flux introduced at R, which means that of all the outgoing flux, a fraction A/R is captured by the sphere. This gives the collision probability. In one and two dimensions, the image charge magnitude in a sphere is always the same as the charge magnitude, and this says that 2d walks are recurrent. ### Time-independent Problems are Easy The solution to the problem of collision with an inner sphere vs. an outer sphere can be solved in the same way. The sphere at A is now is grounded on the outer sphere at a radius B, so that, matching the potentials: $${Q \over R} - {Q_i \over A} = {Q-Q_i \over B}$$ Where $Q$ is the charge and $Q_i$ is the image charge. The image charge magnitude is $$Q_i = Q { {1\over R} - {1\over B} \over {1\over A} - {1\over B} }$$ This is the probability of reaching A before reaching B. It asymptotes to the correct answer for $B=\infty$, of course. This is a time-independent problem, so it has a simple answer using grounded conductors. This gives you a qualitative indication of what the distribution for first colllsion is, By using B as a surrogate for $\sqrt{t}$. Expanding in B to leading order, and substituting $\sqrt{t}$ for B gives that the probability of surviving is $$P(t) = {A\over R} (1 - {R-A\over \sqrt{t}})$$ Which is to be compared to the exact answer below. ### Time dependent problem The time dependent problem, the original question, also has a closed form solution. The first thing to note is that the distance between the two particles is undergoing a diffusion too. This distance prefers to go out rather than in, according to the extra phase space volume of being at a larger radius. The form of the effective one dimensional diffusion can be worked out using a simple trick--- the uniform 3d distribution must be the Boltzmann distribution for diffusion in the effective potential on r. The resulting equation for the radial diffusion is: $${\partial\rho\over\partial t} = {\partial \over\partial r} ( {\partial\rho\over \partial r} - {(d-1)\over r} \rho )$$ Where $d=3$ is the case of interest, and the boundary conditions you enforce are $\rho(A)=0$ and $\rho(\infty)=0$. This has a simple solution in the usual 3d way: you write $\rho(r,t) = rf(r,t)$ and then $${\partial f\over \partial t} = {\partial^2 f\over \partial^2 r}$$ Which is the usual miracle of three dimensions, radial Laplacians become free 1d Laplacians once you extract a 1/r factor. In this case, since the distribution $\rho$ is already mutliplied by $r^2$ relative to the radially symmetric solution of Laplace's equation, you extract an r instead. I will digress to clear up a few confusions: • How is it possible for a 3d Laplacian, which conserves the 1d integral $r^2\rho(r)$, to be equivalent to a 1d Laplacian, which conserves the integral $f(r)$ without powers of r? The reason is that diffusion conserves two different moments: the total probability, and the center of mass. The center of mass for f is the total probability for $\rho$. This is important, because f is not to be interpreted as a probability directly, it's first moment is the total probability. • The location of images in 1d and in 3d are totally different! In 1d, if you have a Dirichlet condition, you always reflect at that point to find the image position. In 3d, you have to reflect in a sphere. How could the two problems be related? There is an important note regarding this: if you just map a free diffusion in 3d to 1d, the 1d problem has absorbing boundary at the origin. The reason is that the 1d diffusion must be 0 at 0, because it comes from a 3d diffusion which has a finite density at 0 (or from a 1d radial diffusion whose density vanishes as $r^2$ near 0--- same thing). The correct images to use when there is a Dirichlet boundary condition on a sphere are the 1d images. These give the correct 1d solution. Although this is a trivial point, considering the previous section, it is confusing when you take the limits to extract the behavior at long times. The solution for a delta function at r=R is $$f_D(r,t) = {1\over \sqrt{2\pi t}} (e^{-{(r-R)^2\over 2t}} - e^{-{(r-(2A-R))^2\over 2t}})$$ Which is the image-charge method for producing a solution which is zero at $r=A$. This solution will decay, because of the negative probability coming from the image, or in real life, because of the absorption of the probability into the boundary at r=A, i.e. because of collisions. The leftover probability density is the answer to the question. So the answer is the first moment of $f_D(r,t)$: $${1\over N} \int_A^\infty r f_D(r,t) = {1\over N} \int_A (r-A) f_D(r,t) + A f_D(r,t)$$ Where the normalization constant N is so that the answer will be 1 at $t=0$. $$N = \int_A^\infty r f_D(r,0) = \int_A^\infty r \delta(r-R) = R$$ Using the variable $u=r-A$, the first integral is symmetric between u and -u, and becomes half the integral from minus infinity to infinity of something simple to do. $${1\over 2R} \int_{-\infty}^\infty u f_d(u,t) = {A\over R}$$ The second integral $\int_0^\infty A f_D(r,t)$ is simply the probability of a 1d walk to avoid the origin up to time t, and it is multiplied by ${A\over R}$. - thanks a lot, i couldn't upvote enough for your answer :) – Greg Dec 6 '11 at 9:21 Marek suggested I post my comment (which doesn't completely answer the question) as an answer. Here it is: Suppose you have two Brownian motions with diffusion coefficients $D_1$ and $D_2$, which start at the same point at $t=0$. Let $x_i$ be the average displacement vector for particle $i$ after time $t$. Then, $\langle x_i^2 \rangle = 2D_it$, where $x_i^2$ means the square of the length of vector $x_i$. Now, consider the relative position of these two random walks after time $t$ (assuming they start out at the same place). The expected value of the relative displacement will be $$\langle (x_1 - x_2)^2 \rangle = \langle x_1^2 \rangle + \langle x_2^2 \rangle = 2 (D_1 + D_2)t,$$ where you can ignore the cross terms since the expected inner product of the two displacement vectors is 0. Hopefully, this argument convinces you that the relative position of two points undergoing Brownian motion is still Brownian. You can make it completely rigorous if you want. So your question boils down to: what is the probability distribution for the time it takes Brownian motion starting at distance $d$ from the origin to first reach a disc of radius $r=r_1+r_2$ around the origin. It's straightforward to find the probability that Brownian motion is in that disc after time $t$, but computing the probability distribution for the first time it's reached is harder. I am sure this has been done, but right now I don't have time to do a literature search for this. - – Raskolnikov Apr 9 '11 at 13:27 – Itamar Sep 20 '11 at 10:57 Here is an answer in the first case (unbounded domain). As already noted, the difference between the positions of the two particles performs a Brownian motion starting from point $(d,0,0)$ with diffusion coefficient $D=D_1+D_2$ and one is interested in the time $\tau$ of the first hitting of the ball $B(r)$ around zero with radius $r=r_1+r_2$ when $d>r$. It happens that in three or more dimensions, the Brownian motion is transient. This means that the particle goes to infinity almost surely, in the sense that for any radius $R$, there will exist an almost surely finite time $L_R$ such that after time $L_R$, the particle is not in $B(R)$. The same holds if one replaces $B(R)$ by any bounded domain: after a random time, finite with full probability, the particle wanders out of the domain. Another consequence is that, starting from a point not in $B(r)$, for example from the point $(d,0,0)$, the particle has a nonzero probability to never hit $B(r)$. Thus, it happens with positive probability that the two particles the OP is considering never collide in $\mathbb{R}^3$ (and of course this probability depends on $d$, $r_1$ and $r_2$, but for any $d>r_1+r_2$ it is positive). In particular: The collision time is infinite with nonzero probability. All its (positive) moments are infinite. Two more remarks. First, in one dimension the collision time $\tau$ would be almost surely finite, with the well known density $$P(\tau\in\mathrm{d}t)=\frac{a}{\sqrt{2\pi}}\mathrm{e}^{-a^2/2t}\frac{\mathrm{d}t}{t^{3/2}},\qquad a=d-r_1-r_2.$$ Note that, since the density above is equivalent to a multiple of $t^{-3/2}$ when $t\to+\infty$, the mean of $\tau$ is infinite. More precisely, for positive $\gamma$, $\tau^\gamma$ is integrable if $\gamma<1/2$ but not if $\gamma\ge1/2$. Beware that we followed the mathematicians' convention for Brownian motion, that is, the formula above applies to a one dimensional Brownian motion $(B_t)$ with transition probabilities $$P(B_t\in\mathrm{d}x\vert B_0=y)=\frac1{\sqrt{2\pi t}}\mathrm{e}^{-(x-y)^2/2t}\mathrm{d}x.$$ By scaling, it should be straightforward to deduce the distribution of $\tau$ for a one dimensional Brownian motion with diffusion constant $D$ since $\tau$ for a diffusion constant $D$ is distributed like $\tau^{(1)}/D$ where $\tau^{(1)}$ corresponds to the first hitting time for a diffusion constant $1$. Second, in case 2 (bounded domain), one can expect the collision time to be almost surely finite and integrable (and even exponentially integrable) but to make it possible to answer the question, one should explain what happens to the particles when one of them hits the boundary of the domain before their collision time. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 101, "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": 0.9355053901672363, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/28254-need-help-here-derivatives.html	# Thread: 1. ## Need help here with derivatives Hey guys. I'm having trouble with derivatives and I need someone's help with my homework. Could you guys please explain to me how to solve these problems? 1. Given $y = f(x) = \frac{1}{x^2} - \sqrt[3]{x}$ Find the second derivative $f"(x)=\frac{d^2y}{dx^2}$ 2. Given $f(x) = \frac{x^2}{1-3x}$ Find the derivative f'(x) 3. Given $\Delta y = 2x^2(\Delta x+3x(\Delta x)^2-2(\Delta x)$ Find the derivative = $Lim \frac{f(x+\Delta x)-f(x)}{\Delta x}$ And I'm also stuck at this problem: 4. Given $y=f(x)=2x^2-x$ Find $\Delta y=f(x+\Delta x)-f(x)$ Help is greatly appreciated. Thanks guys! --RS 2. 1.) can be written as f(x) = x^(-2) - x^(1/3), so the derivative will be: f '(x) = -2x^(-3) - (1/3)x^(-2/3) 2.) you can you the quotient rule, where dy/dx = [V(du/dx) - U(dv/dx)] / V^2 and u = x^2, v = (1-3x) 3. Originally Posted by Greengoblin 1.) can be written as f(x) = x^(-2) - x^(1/3), so the derivative will be: f '(x) = -2x^(-3) - (1/3)x^(-2/3) 2.) you can you the quotient rule, where dy/dx = [V(du/dx) - U(dv/dx)] / V^2 and u = x^2, v = (1-3x) If it is okay by you, could you explain to me how exactly did you do that? I'm sorry, derivatives are one of a math topic I'm stuck with and my professor is those types that knows his stuff but can't really teach it to us. 4. Sure. First of all, the derivative of a sum of functions is the sum of derivatives of those functions, so: y = f(x) + g(x) + h(x) dy/dx = f '(x) + g '(x) + h '(x) The power rule tells us that for a function like x^n: dy/dx = nx^(n-1) In other words we bring the power down in front of x, and then decrease the power by one. when you have the inverse of a function x^n you have 1/ x^n, and this can also be written as x^-n. For the first term I did it like this, and then used the power rule, to get -nx^(-n-1). For the second term you can wite square roots as fractions. For example sqrt(x) can be written x^(1/2), the cubed root can be written x^(1/3), and the nth root can be written x^(1/n). Once you have written it like this you can simply apply the power rule to it again. so than if f(x)= nth root of x: f(x) = x^(1/n) dy/dx = (1/n)x^(1/n - 1) When you have a problem like this, it's much easier to write inverses out as negative powers, and square roots out as fractional powers first. Knowing this, you could probably make a good attempt at the second derivative? If not I'll try it for you. Cheers! 5. Originally Posted by RedSpades [snip] 2. Given $f(x) = \frac{x^2}{1-3x}$ Find the derivative f'(x) [snip] The quotient rule says that if y = u(x) v(x), then $\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$. In your problem, $u = x^2 \Rightarrow \frac{du}{dx} = 2x \,$ and $v = 1 - 3x \Rightarrow \frac{dv}{dx} = -3$. You substitute these four things into the rule, simplifying where necessary. So it's nothing more than basic algebra 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": 11, "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": 0.9475585222244263, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/76687/list	## Return to Answer 2 added 120 characters in body I had a minor thought. Did you ever look the paper Multiplicity of the special fiber of blowups by Corso,Polini, Vasconcelos. In particular, they give an upper bound on the number (I realize that you are interested in lower bounds, but maybe some of the ideas are related / or could be useful) by in particular, bounding the multiplicity at the origin of the Rees algebra fiber ring (in other words, if you compute the blow-up by computing Proj R[It], $R[It]$, then mod out by an ideal from $R$, you get some graded ring corresponding to the fiber over the ideal you modded. Then you can study the blow-up by studying properties of the ring , and I think the multiplicity they study should give you an upper bound on the number of components). Of course, the number of minimal associated primes gives you some bound on the components of the pre-image of $Z$. 1 I had a minor thought. Did you ever look the paper Multiplicity of the special fiber of blowups by Corso,Polini, Vasconcelos. In particular, they give an upper bound on the number (I realize that you are interested in lower bounds, but maybe some of the ideas are related / or could be useful) by in particular, bounding the multiplicity at the origin of the Rees algebra fiber ring (in other words, if you compute the blow-up by computing Proj R[It], then you can study the blow-up by studying properties of the ring, I think the multiplicity they study should give you an upper bound on the number of components). Of course, the number of minimal associated primes gives you some bound on the components of the pre-image of $Z$.	{"extraction_info": {"found_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": 0.9393574595451355, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/52028/radiation-pressure-on-a-dyson-sphere	# Radiation pressure on a Dyson sphere To find the outward pressure from the sun's light on an enveloping spherical shell (Dyson sphere), one can simply divide the insolation by $c^2$. Using the entire system, we can specify the power of the sun in watts, $W$, leading to a pressure of $P=W/(4 \pi R^2 c^2)$. The problem is that a Dyson sphere structure emits its own thermal radiation. So then one would think that the pressure from the thermal radiation on the outside surface of the Dyson sphere could just cancel with the inside radiation pressure from the sun since the power rates are the same. There's one more problem, the outer surface's thermal radiation has a different directionality. There would be a net pressure pushing outward since the sun's light is preferentially directed outward. What would that be? - Great question. Are you considering a vanishingly thin sphere, where light is absorbed and the re-emitted isotropically, with possible paths back into the sphere? Or are you imagining that the sphere has finite, non-negligible thickness, with different temperatures on the inner and outer surface? – kleingordon Jan 24 at 6:55 @kleingordon Ah ha, you ask of the addition outward radiation pressure due to the exchange of photons from points on the inside the sphere to other points on the inside of the sphere. That should be included to be most realistic, because you can't make the inner surface cooler except with active cooling. I would approximate the inner and outer surfaces to be the same temperature myself. However, I think the central challenge of the problem will be solved with or without that detail. – AlanSE Jan 24 at 12:57 ## 2 Answers The basic assumption here is isotropic emission on the surface of the sphere, which is exactly $W$, the power production of the star. I thought that would be a shockingly simple conversion, and I was right. Using spherical angles of $\theta$ for azimuthal and $\phi$ for polar angle, the fraction the photon momentum is reduced by due to the angle factor is $\cos{(\phi)}$. Let it be known here that I'm using the Wolfram Mathworld spherical convention. As per that reference, a unit of solid angle will be $dA = \sin{(\phi)} d\phi d\theta$. Now, I care nothing about the total number of photons emitted as I'm calculating the correction factor for angle. So for this factor, I divide the outward momentum by the integral of $1$ over the solid angular interval. This solid angle interval is a $2 \pi$ half circle pointing outward on the surface of the Dyson sphere, and $\phi=0..\pi/2$, $\theta=0..2 \pi$. $$\frac{\text{momentum}}{\text{number}} = \frac{ \int_{2\pi} \cos{(\phi)} dA }{\int_{2\pi} dA }= \frac{\int_0^{2\pi} \int_0^{\pi/2} \cos{(\phi)} \sin{(\phi)} d\phi d\theta }{\int_0^{2\pi} \int_0^{\pi/2} \sin{(\phi)} d\phi d\theta} = \frac{1}{2}$$ What is this arguing? It is arguing that the inward pressure from the emission of $W$ on the outside surface of the Dyson sphere is $1/2$ the outward pressure from the absorption of sunlight on the inside surface. Let's assume that the inner surface is the same temperature as the outer surface. In that case, both sides of the thin Dyson Sphere will exert a pressure of $W/(2 c^2 A)$ from thermal emission. These would cancel each other out, but the inside surface has re-absorbs 100% of what it emits (as per comments). So here are the additions and subtractions to outward (or lifting) pressure. • outside surface radiation: $-W/(2 c^2 A)$ • inside surface radiation: $W/(2 c^2 A)$ • inside surface re-absorption:$W/(2 c^2 A)$ • sun's radiation absorption: $W/(c^2 A)$ Now, for the total, we have a correction factor of $-1/2+1/2+1/2+1=1.5$. This is the factor to correct the $P$ value I gave in the question in order to get the correct radiation lifting pressure on a Dyson Sphere. - 1 Well-reasonsed. By the way, I think both here and in the question itself, you should be dividing by $c$ instead of $c^2$ (check the units). I'm currently trying to come up with an answer that works at the photon-by-photon level to see if I get the same answer. If I don't, I want to try to figure out what the error in my reasoning is. More to come soon – kleingordon Jan 28 at 0:02 @kleingordon That's right. I should have been thinking $E=pc$ whereas I was stupidly thinking some form of E=mc2 that didn't apply. – AlanSE Jan 28 at 3:02 I suspect there's a fallacy lurking somewhere in the following argument, so I want people to point it out for me in the comments. I post this purely for self-pedagogical purposes. Let's think about this in a photon-by-photon manner. We assume the temperature of the sphere remains constant in time. The only heating and cooling mechanisms are radiative. Then, for a given interval in time, the sphere must emit exactly as many photons as it absorbs. I therefore assume that the time between an absorption event and the corresponding emission event to be effectively zero. Consider a photon emitted by the sun. It will deposit its full momentum $p$ on the sphere over a time interval $R/c$, resulting in a force of $pR/c$. The absorption of this photon will lead immediately to a re-emission, as discussed above. With a probability of $1/2$, the emitted photon will be sent out of the sphere, carrying average outward momentum $p/2$ as discussed in Alan's answer. The net momentum transferred to the sphere is then $p(1 - 1/2)$ = $p/2$, over a time interval of $R/c$, so the force is $pc/(2R)$. If we let $n$ denote the number of re-emissions into the sphere as the result of the absorption of an initial stellar photon, then in this case $n=0$, the force is $pc/(2R)$, and this outcome has probability $Pr_0 = 1/2$. Now consider the $n=1$ case: the absorption of a photon from the sun results in a re-emission into the sphere, this re-emitted photon is then absorbed back by the sphere, and that the subsequent emission is out of the sphere. I'm treating the chance that the photon runs into the star at the center to be negligible. Note that $Pr_1 = 1/4$ because there's a 1/2 chance that $n=0$, then a $1/2 * 1/2$ chance that $n>1$. To find the momentum per time deposited in the sphere when the inward-directed photon is emitted and re-absorbed, one must integrate over all cords of the sphere emanating from the emission point, taking into account the angle of the photon with respect to the normal direction of the sphere. I can go into more detail for this step if requested, but the result is an average momentum/time (force) of $2 p c /R$. Then, the total force exerted by the photon for the $n =1$ event is $pc/(2R) + 2 p c /R$. We can now sum up the forces over all $n$, weighted by $Pr_n$. I claim that $Pr_n = (1/2)^{n+1}$ ; you can think of it as the probability of flipping a coin $n+1$ times and getting tails the first $n$ times, then heads on the last flip. The formula is \begin{equation} {\rm Force \:per \:photon} = \frac{pc}{R}\sum_0^\infty(1/2)^{n+1}(1/2 + 2n) = \frac{5}{2}\frac{pc}{R} \end{equation}. Let $N_\ast$ denote the number of photons emitted by the star per time. We can then relate $N$ to the luminosity of the star, $W$, by $N_\ast = W/(p c)$. The rate of increase of force on the sphere with time is then $N_\ast \frac{5}{2} \frac{pc}{R} = \frac{5}{2} \frac{W}{R}$. If we imagine instantaneously constructing the sphere at time $t = 0$, or similarly "turning on" the star at $t=0$, then the total outward force on the sphere would be \begin{equation} F_{\rm total} = \frac{5}{2} \frac{W}{R} t \end{equation} In summary, this argument seems to suggest that the force on the sphere grows in an unbounded manner over time. The temperature of the sphere remains constant, but the temperature of the radiation field inside the sphere continues to grow with time, eventually cooking everything inside (unless they figure out a way to store/use the radiative energy quickly enough). What went wrong in this analysis? - For one to give the force per photon, they must answer "at what time", and because of this I think one would do better to quote the impulse per photon. The infinite series is a neat way to look at it, but this non-steady-state force summing may result in a huge math problem. The time of first wall hit is deterministic (R/c), but subsequent hits are distributed at times 0 to 2R/c. This results in infinite sums of piecewise functions familiar to particle transport. Even if you can find a way around that, force per photon should be a decreasing function of time and not a constant. – AlanSE Jan 28 at 3:24 Every time I use the word force, I'm talking about an average momentum deposition over a time interval. I can see how this might be a problem considering that for some of these averages, the time interval must extend to an arbitrarily large value of time (for photons that remain trapped in the sphere). This might indeed pose a problem for my analysis. However, what if one were to focus on energy for a moment, and not momentum? What is wrong with the photon-by-photon argument that the radiation energy density inside the sphere grows without bound over time? – kleingordon Jan 28 at 3:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 57, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9369746446609497, "perplexity_flag": "head"}
http://mathoverflow.net/questions/14683/can-the-quantum-torus-be-realized-as-a-hall-algebra/14744	## Background ### The Quantum Torus Let $q$ be an arbitrary complex number, and define (the algebra of) the quantum torus to be $$T_q:=\mathbb{C}\langle x^{\pm 1},y^{\pm 1}\rangle/xy-qyx$$ For $q=1$, this is the commutative ring of functions on the torus $\mathbb{C}^\times\times \mathbb{C}^\times$; hence, for general $q$, this is regarded as a quantization of the torus. ### Hall Algebras Consider a small abelian category $A$, with the property that $Hom_A(M,N)$ and $Ext^i_A(M,N)$ are always finite sets for any $M,N\in A$ and $i\in \mathbb{Z}$. Let $\overline{A}$ denote the set of isomorphism classes in $A$, and let $$H(A)=\oplus_{[M]\in \overline{A}}\mathbb{C}[M]$$ denote the complex vector space spanned by $\overline{A}$. Endow $H(A)$ with a multiplication by the formula $$[M]\cdot [N]=\sqrt{\langle [M],[N]\rangle)}\sum_{[R]\in \overline{A}}\frac{a_{MN}^R}{\|Aut(M)\|\|Aut(N)\|}[R]$$ where $a_{MN}^R$ is the number of short exact sequences $$0\rightarrow N\rightarrow R\rightarrow M\rightarrow 0$$ and $$\langle [M],[N]\rangle = \sum (-1)^i \|Ext^i_A(M,N)\|$$ is the Euler form. This multiplication makes $H(A)$ into an associate algebra called the Hall algebra of $A$; the proof can be found e.g. here. ### Finite Fields and Quantization The categories $A$ appearing in the construction of a Hall algebra are usually linear over some finite field $\mathbb{F}_q$. Often, it is possible to simultaneously define a category $A_q$ for each finite field $\mathbb{F}_q$; usually by considering modules on the $\mathbb{F}_q$-points of some scheme over $\mathbb{Z}$. The corresponding Hall algebras $H(A_q)$ will then usually be closely related, and can often be defined by relations that are functions in $q$. ## The Question I know that there are cases where an algebra is deformed by a parameter $q$, and then the resulting family of algebras `magically' coincides with a family of Hall algebras $H(A_q)$ in the special cases when $q$ is a prime power. I think this happens in the case of the Hecke algebra (discussed here), and the case of quantum universal enveloping algebras (discussed here). I somewhat understand that this is a symptom of a related convolution algebra on the scheme used to define $A_q$. Is there a family of categories $A_q$ such that the corresponding Hall algebras $H(A_q)$ are isomorphic to the Quantum Torus $T_q$ for all $q$ a prime power? If so, is there a convolution algebra realization of the Quantum Torus? - ## 2 Answers As far as I understand, the Hall algebra of a category (say, with finite length of objects) is graded by the Grothendieck monoid of this category, spanned by simple objects over $\Bbb Z_+$, and it must have the ground field in degree $0$. The quantum torus algebra does not seem to have such a grading (it has a $\Bbb Z^2$-grading, not a $\Bbb Z_+^m$-grading). Maybe one should ask this question for the q-Weyl algebra $xy=qyx$ (not allowing negative powers of x and y)? Note that this algebra appears as a subalgebra of a Hall algebra (the Hall algebra for the quiver $A_2$ is $U_q(n_+)$, where $n_+$ is the nilpotent subalgebra of $sl(3)$; the q-Weyl algebra is generated by $e_{12}$ and $e_{13}$ inside this algebra). - ### 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. [EDIT: ignore this post. I wrote it last night, and this morning I see that it's nonsense.] - How about element $1$? It is not a commutator. – Pavel Etingof Feb 9 2010 at 13:23 Fair enough. The product of things that are trivial in the Grothendieck group are still trivial in the Grothendieck group, so being a product of commutators is enough to show something goes to 0 in the Grothendieck group. – Ben Webster♦ Feb 9 2010 at 13:42 I must admit I also did not understand the argument with commutators. For example, I don't see any connections between extensions of $X$ by $Y$ and extensions of $Y$ by $X$. Also I don't understand what your map to the Grothendieck groups is. Do you rather mean that the Hall algebra is graded by the Grothendieck group (and even by the Grothendieck semigroup?) – Pavel Etingof Feb 9 2010 at 13:54 You're right, that didn't make any sense. It sounded right last night, but now that I look, it's complete nonsense. – Ben Webster♦ Feb 9 2010 at 14:47	{"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": 48, "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": 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": 0.936131477355957, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=4024320	Physics Forums ## How did "v" come out of this derivative? 1. The problem statement, all variables and given/known data 3. The attempt at a solution I basically understood how to attack the problem and my answer was very close to the key. The only thing bothering me is that v on the top. I cannot see how that could come up anywhere. I know that the "r" is r = r(t), so that I am only taking the derivative of $$\ln(1 + w/r)$$. I don't see where the v actually is 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 The flux is expressed as a function of r. You're taking the derivative with respect to t. So, recall the chain rule of calculus: first take the derivative with respect to r and then multiply by the derivative of r with respect to t. Nope, implicit differentiation. I had forgotten about that Recognitions: Gold Member Homework Help ## How did "v" come out of this derivative? Quote by Dens Nope, implicit differentiation. Here's the chain rule: If $y=f[u]$ and $u = g[x]$ are differentiable functions, then $\frac{dy}{dx}$=$\frac{dy}{du}$$\cdot$$\frac{du}{dx}$ You have the flux expressed as a function of r: $\Phi[r]$ and r is some function of time: $r[t]$. So, the chain rule says $\frac{d\Phi}{dt}$ = $\frac{d\Phi}{dr}$$\cdot$$\frac{dr}{dt}$ This get's you the answer and it shows why the speed v appears. How do you get the answer using implicit differentiation? Quote by TSny Here's the chain rule: If $y=f[u]$ and $u = g[x]$ are differentiable functions, then $\frac{dy}{dx}$=$\frac{dy}{du}$$\cdot$$\frac{du}{dx}$ You have the flux expressed as a function of r: $\Phi[r]$ and r is some function of time: $r[t]$. So, the chain rule says $\frac{d\Phi}{dt}$ = $\frac{d\Phi}{dr}$$\cdot$$\frac{dr}{dt}$ This get's you the answer and it shows why the speed v appears. How do you get the answer using implicit differentiation? No you are right, I forget the $\phi$ on the LHS. I was thinking something like Something = r as a function of t ==> differentiate both sides wrt t ==> implicitly differentiate RHS http://www.wolframalpha.com/input/?i=D[ln%281+%2B+w%2Fr%28t%29%29%2Ct] Thread Tools \| \| \| \| \|---------------------------------------------------------------\|----------------------------\|---------\| \| Similar Threads for: How did "v" come out of this derivative? \| \| \| \| Thread \| Forum \| Replies \| \| \| Computers \| 14 \| \| \| General Physics \| 3 \| \| \| Calculus & Beyond Homework \| 3 \| \| \| Calculus & Beyond Homework \| 1 \| \| \| Calculus & Beyond Homework \| 9 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "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": 0.9393141865730286, "perplexity_flag": "middle"}
http://stats.stackexchange.com/questions/tagged/regularization?sort=active&pagesize=15	# Tagged Questions The regularization tag has no wiki summary. 1answer 215 views ### matlab gmdistribution.fit 'Regularize' - what regularization method? I am wondering what is behind matlab 'Regularize' option for method gmdistribution.fit. If it is simply adding a 'little' value to diagonal elements of covariance matrix, so as to make covariance ... 0answers 8 views ### Do I use regularizer when I measure validation error? I have a cost function with a regularizer term, and I'd like to find to optimal hyperparameters for the regularizer term. So I train with different parameter, but when I measure the validation error, ... 0answers 38 views ### Questions regarding predict.glmpath() I'm trying to do LASSO in R with the package glmpath. However, I'm not sure if I am using the accompanying prediction function predict.glmpath() correctly. 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http://mathoverflow.net/revisions/47777/list	## Return to Answer 3 added 66 characters in body The "correct" abstract definition of a derived functor involves the theory of "homotopical categories". An abelian category $\cal A$ does not have enough structure alone to consider homotopical properties. To this end, we usually look at $Ch^+(\cal A)$ or $s\cal A$ of chain complexes in $\cal A$ or simplicial objects in $\cal A$ respectively. We equip these larger categories with the structure of a homotopical category by fixing the lluf subcategories consisting of quasi-isomorphisms or weak homotopy equivalences or quasi-isomorphisms respectively (by the Dold-Kan correspondence, these are essentially identical). In the terminology of Dwyer-Hirschhorn-Kan-Smith, left-exact functors are deformable in the sense that they preserve weak equivalences (that is to say, they are homotopical) on a deformation retract of our original category. So in truth, a derived functor is simply the evaluation of a functor on a suitable approximation of our original object by a homotopically equivalent one on which the functor is well-behaved. When $\cal A$ has enough structure, we can upgrade the homotopical structure on our homotopical version of $\cal A$ (that is, chain complexes or simplicial objects) to a model structure, which makes it substantially easier to compute derived functors, since the existence of approximations is guaranteed, and further, such approximations can usually be taken to be functorial. For Quillen functors between model categories (these are functors that become homotopical on the deformation retracts consisting of cofibrant or fibrant objects), we can, in the presence of functorial factorization, define the derived functors to be the composition of the functorial approximation and the original functor. These approximations have the property that they descend to the usual Kan extensions at the level of the derived categories (for a proof, see Dwyer-Hirschhorn-Kan-Smith Homotopy Limit Functors on Model Categories and Homotopical Categories). That is, in your case, we approximate the chain complex $X$ (concentrated in degree zero, so it is constant) by an injective approximation (fibrant for the injective model structure) and evaluate the functor on this new complex. In fact, if injective approximations exist for all chain complexes, we see that when we compose with an appropriate functorial injective approximation, our functor becomes homotopical (it preserves quasi-isomorphisms) and therefore descends uniquely by the universal property of localization to a total derived functor between derived categories of chain complexes. 2 added 602 characters in body; added 9 characters in body; deleted 3 characters in body The "correct" abstract definition of a derived functor involves the theory of "homotopical categories". An abelian category $\cal A$ does not have enough structure alone to consider homotopical properties. To this end, we usually look at $Ch^+(\cal A)$ or $s\cal A$ of chain complexes in $\cal A$ or simplicial objects in $\cal A$ respectively. We equip these larger categories with the structure of a homotopical category by fixing the lluf subcategories consisting of weak homotopy equivalences or quasi-isomorphisms respectively. In the terminology of Dwyer-Hirschhorn-Kan-Smith, left-exact functors are deformable in the sense that they preserve weak equivalences (that is to say, they are homotopical) on a deformation retract of our original category. So in truth, a derived functor is simply the evaluation of a functor on a suitable approximation of our original object by a homotopically equivalent one on which the functor is well-behaved. When $\cal A$ has enough structure, we can upgrade the homotopical structure on our homotopical version of $\cal A$ (that is, chain complexes or simplicial objects) to a model structure, which makes it substantially easier to compute derived functors, since the existence of approximations is guaranteed, and further, such approximations can usually be taken to be functorial. For Quillen functors between model categories (these are functors that become homotopical on the deformation retracts consisting of cofibrant or fibrant objects), we can, in the presence of functorial factorization, define the derived functors to be the composition of the functorial approximation and the original functor. These approximations have the property that they descend to the usual Kan extensions at the level of the derived categories (for a proof, see Dwyer-Hirschhorn-Kan-Smith Homotopy Limit Functors on Model Categories and Homotopical Categories). That is, in your case, we approximate the chain complex $X$ (concentrated in degree zero, so it is constant) by an injective approximation (fibrant for the injective model structure) and evaluate the functor on this new complex. In fact, if injective approximations exist for all chain complexes, we see that when we compose with an appropriate functorial injective approximation, our functor becomes homotopical (it preserves quasi-isomorphisms) and therefore descends uniquely by the universal property of localization to a total derived functor between derived categories of chain complexes. 1 The "correct" definition of a derived functor involves the theory of "homotopical categories". An abelian category $\cal A$ does not have enough structure alone to consider homotopical properties. To this end, we usually look at $Ch^+(\cal A)$ or $s\cal A$ of chain complexes in $\cal A$ or simplicial objects in $\cal A$ respectively. We equip these larger categories with the structure of a homotopical category by fixing the lluf subcategories consisting of weak homotopy equivalences or quasi-isomorphisms respectively. In the terminology of Dwyer-Hirschhorn-Kan-Smith, left-exact functors are deformable in the sense that they preserve weak equivalences (that is to say, they are homotopical) on a deformation retract of our original category. So in truth, a derived functor is simply the evaluation of a functor on a suitable approximation of our original object by a homotopically equivalent one on which the functor is well-behaved. When $\cal A$ has enough structure, we can upgrade the homotopical structure on our homotopical version of $\cal A$ (that is, chain complexes or simplicial objects) to a model structure, which makes it substantially easier to compute derived functors, since the existence of approximations is guaranteed, and further, such approximations can usually be taken to be functorial. For Quillen functors between model categories (these are functors that become homotopical on the deformation retracts consisting of cofibrant or fibrant objects), we can, in the presence of functorial factorization, define the derived functors to be the composition of the functorial approximation and the original functor. These approximations have the property that they descend to the usual Kan extensions at the level of the derived categories (for a proof, see Dwyer-Hirschhorn-Kan-Smith Homotopy Limit Functors on Model Categories and Homotopical Categories).	{"extraction_info": {"found_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": 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": 0.912083625793457, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/21997/odds-ratio-for-continuous-variable	# Odds ratio for continuous variable I have a question How would you interpret the odds ratios for a continuous variable? - 1 – Macro Feb 1 '12 at 2:27 ## 1 Answer It's not completely clear from your question, but I'm assuming you're talking about the situation where you have a single binary response $Y$ and a continuous predictor $X$ and fit a logistic regression model: $$\log \left( \frac{ P(Y_{i}=1\|X_{i}) }{P(Y_{i}=0\|X_{i})} \right) = \beta_{0} + \beta_{1} X_{i}$$ Then the odds ratio, $e^{\beta_{1}}$, is the odds ratio associated with a one unit increase in $X$. Essentially you can think of it as the odds ratio between $Y$ and the dummy variable $B$ defined such that $B=1$ if $X=x+1$ and $B=0$ if $X=x$. Note: The assumption underlying the logistic model is that this odds ratio does not depend on $x$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9482734203338623, "perplexity_flag": "head"}
http://mathoverflow.net/questions/23829/solutions-to-the-continuum-hypothesis/50654	## Solutions to the Continuum Hypothesis ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) (A related MO question: http://mathoverflow.net/questions/14338/what-is-the-general-opinion-on-the-generalized-continuum-hypothesis ) ## Background The Continuum Hypothesis (CH) posed by Cantor in 1890 asserts that $\aleph_1=2^{\aleph_0}$. In other words, it asserts that every subset of the set of real numbers that contains the natural numbers has either the cardinality of the natural numbers or the cardinality of the real numbers. It was the first problem on the 1900 Hilbert's list of problems. The generalized continuum hypothesis asserts that there are no intermediate cardinals between every infinite set X and its power set. Cohen proved that the CH is independent from the axioms of set theory. (Earlier Goedel showed that a positive answer is consistent with the axioms). Several mathematicians proposed definite answers or approaches towards such answers regarding what the answer for the CH (and GCH) should be. ## More background I am aware of the existence of 2-3 approaches. One is by Woodin described in two 2001 Notices of the AMS papers (part 1, part 2). Another by Shelah (perhaps in this paper entitled "The Generalized Continuum Hypothesis revisited "). See also the paper entitled "You can enter Cantor paradize" (Offered in Haim's answer.); There is a very nice presentation by Matt Foreman discussing Woodin's approach and some other avenues. Another description of Woodin's asnwer is by Lucca Belloti (also suggested by Haim). The proposed asnwer $2^{\aleph_0}=\aleph_2$ goes back according to François to Goedel. It is (perhaps) mentioned in Foreman's presentation. (I heard also from Menachem Magidor that this answer might have some advantages.) François G. Dorais mentioned an important paper by Todorcevic's entitled "Comparing the Continuuum with the First Two Uncountable Cardinals". There is also a very rich theory (pcf theory) of cardinal arithmetic which deals with what can be proved in ZFC. ### Remark: I included some information and links from the comments and answer in the body of question. What I would hope most from an answer is some friendly elementary descriptions of the proposed solutions. There are by now a couple of long detailed excellent answers (that I still have to digest) by Joel David Hamkins and by Anders Caicedo and several other useful answers. (Unfortunately, I can accept only one answer.) Update (Fenruary 2011): A new detailed answer was contributed by Justin Moore. - Another noteworthy reference is Gödel's unpublished note 1970a where he collects evidence in favor of $2^{\aleph_0} = \aleph_2$ - books.google.com/… – François G. Dorais♦ May 7 2010 at 8:00 1 Can you please edit your second sentence starting by "In other words" ? "containing of real numbers" does not seem very clear. Also small typos like "now intermediate" -> "no intermediate" "Eralier"=>"Earlier" Another pedantic remark is that one does not "solve" an hypothesis. One adopts it or one rejects it, eventually replacing it by another. Perhaps should you slightly rephrase your question. – ogerard May 8 2010 at 12:40 3 Gil, you're absolutely right that using "solution" in this and related contexts is common usage, but IMHO it's bad English. One solves a problem, answers a question, and proves or disproves a hypothesis or conjecture. Whenever I hear that a "conjecture" has been "solved" I don't know whether the speaker means the conjecture is now known to be true, false, or undecidable (and I've seen examples in which each of the three was meant). But I'm grumpier about such things than most. – Mark Meckes May 10 2010 at 13:36 1 Gil, I agree with your characterization of what the question is. But to my ears, a "hypothesis" is a statement, not a question. You're asking about answers to the question (or solutions to the problem) of whether the Continuum Hypothesis should be accepted; what I object to is using "Continuum Hypothesis" as a shorthand name for that question. – Mark Meckes May 13 2010 at 23:52 1 The presentation was very delightful. – To be cont'd May 18 2010 at 11:30 show 5 more comments ## 9 Answers Since you have already linked to some of the contemporary primary sources, where of course the full accounts of those views can be found, let me interpret your question as a request for summary accounts of the various views on CH. I'll just describe in a few sentences each of what I find to be the main issues surrounding CH, beginning with some historical views. Please forgive the necessary simplifications. Cantor. Cantor introduced the Continuum Hypothesis when he discovered the transfinite numbers and proved that the reals are uncountable. It was quite natural to inquire whether the continuum was the same as the first uncountable cardinal. He became obsessed with this question, working on it from various angles and sometimes switching opinion as to the likely outcome. Giving birth to the field of descriptive set theory, he settled the CH question for closed sets of reals, by proving (the Cantor-Bendixon theorem) that every closed set is the union of a countable set and a perfect set. Sets with this perfect set property cannot be counterexamples to CH, and Cantor hoped to extend this method to additional larger classes of sets. Hilbert. Hilbert thought the CH question so important that he listed it as the first on his famous list of problems at the opening of the 20th century. Goedel. Goedel proved that CH holds in the constructible universe $L$, and so is relatively consistent with ZFC. Goedel viewed $L$ as a device for establishing consistency, rather than as a description of our (Platonic) mathematical world, and so he did not take this result to settle CH. He hoped that the emerging large cardinal concepts, such as measurable cardinals, would settle the CH question, and as you mentioned, favored a solution of the form $2^\omega=\aleph_2$. Cohen. Cohen introduced the method of forcing and used it to prove that $\neg$CH is relatively consistent with ZFC. Every model of ZFC has a forcing extension with $\neg$CH. Thus, the CH question is independent of ZFC, neither provable nor refutable. Solovay observed that CH also is forceable over any model of ZFC. Large cardinals. Goedel's expectation that large cardinals might settle CH was decisively refuted by the Levy-Solovay theorem, which showed that one can force either CH or $\neg$CH while preserving all known large cardinals. Thus, there can be no direct implication from large cardinals to either CH or $\neg$CH. At the same time, Solovay extended Cantor's original strategy by proving that if there are large cardinals, then increasing levels of the projective hierarchy have the perfect set property, and therefore do not admit counterexamples to CH. All of the strongest large cardinal axioms considered today imply that there are no projective counterexamples to CH. This can be seen as a complete affirmation of Cantor's original strategy. Basic Platonic position. This is the realist view that there is Platonic universe of sets that our axioms are attempting to describe, in which every set-theoretic question such as CH has a truth value. In my experience, this is the most common or orthodox view in the set-theoretic community. Several of the later more subtle views rest solidly upon the idea that there is a fact of the matter to be determined. Old-school dream solution of CH. The hope was that we might settle CH by finding a new set-theoretic principle that we all agreed was obviously true for the intended interpretation of sets (in the way that many find AC to be obviously true, for example) and which also settled the CH question. Then, we would extend ZFC to include this new principle and thereby have an answer to CH. Unfortunately, no such conclusive principles were found, although there have been some proposals in this vein, such as Freilings axiom of symmetry. Formalist view. Rarely held by mathematicians, although occasionally held by philosophers, this is the anti-realist view that there is no truth of the matter of CH, and that mathematics consists of (perhaps meaningless) manipulations of strings of symbols in a formal system. The formalist view can be taken to hold that the independence result itself settles CH, since CH is neither provable nor refutable in ZFC. One can have either CH or $\neg$CH as axioms and form the new formal systems ZFC+CH or ZFC+$\neg$CH. This view is often mocked in straw-man form, suggesting that the formalist can have no preference for CH or $\neg$CH, but philosophers defend more subtle versions, where there can be reason to prefer one formal system to another. Pragmatic view. This is the view one finds in practice, where mathematicians do not take a position on CH, but feel free to use CH or $\neg$CH if it helps their argument, keeping careful track of where it is used. Usually, when either CH or $\neg$CH is used, then one naturally inquires about the situation under the alternative hypothesis, and this leads to numerous consistency or independence results. Cardinal invariants. Exemplifying the pragmatic view, this is a very rich subject studying various cardinal characteristics of the continuum, such as the size of the smallest unbounded family of functions $f:\omega\to\omega$, the additivity of the ideal of measure-zero sets, or the smallest size family of functions $f:\omega\to\omega$ that dominate all other such functions. Since these characteristics are all uncountable and at most the continuum, the entire theory trivializes under CH, but under $\neg$CH is a rich, fascinating subject. Canonical Inner models. The paradigmatic canonical inner model is Goedel's constructible universe $L$, which satisfies CH and indeed, the Generalized Continuum Hypothesis, as well as many other regularity properties. Larger but still canonical inner models have been built by Silver, Jensen, Mitchell, Steel and others that share the GCH and these regularity properties, while also satisfying larger large cardinal axioms than are possible in $L$. Most set-theorists do not view these inner models as likely to be the "real" universe, for similar reasons that they reject $V=L$, but as the models accommodate larger and larger large cardinals, it becomes increasingly difficult to make this case. Even $V=L$ is compatible with the existence of transitive set models of the very largest large cardinals (since the assertion that such sets exist is $\Sigma^1_2$ and hence absolute to $L$). In this sense, the canonical inner models are fundamentally compatible with whatever kind of set theory we are imagining. Woodin. In contrast to the Old-School Dream Solution, Woodin has advanced a more technical argument in favor of $\neg$CH. The main concepts include $\Omega$-logic and the $\Omega$-conjecture, concerning the limits of forcing-invariant assertions, particularly those expressible in the structure $H_{\omega_2}$, where CH is expressible. Woodin's is a decidedly Platonist position, but from what I have seen, he has remained guarded in his presentations, describing the argument as a proposal or possible solution, despite the fact that others sometimes characterize his position as more definitive. Foreman. Foreman, who also comes from a strong Platonist position, argues against Woodin's view. He writes supremely well, and I recommend following the links to his articles. Multiverse view. This is the view, offered in opposition to the Basic Platonist Position above, that we do not have just one concept of set leading to a unique set-theoretic universe, but rather a complex variety of set concepts leading to many different set-theoretic worlds. Indeed, the view is that much of set-theoretic research in the past half-century has been about constructing these various alternative worlds. Many of the alternative set concepts, such as those arising by forcing or by large cardinal embeddings are closely enough related to each other that they can be compared from the perspective of each other. The multiverse view of CH is that the CH question is largely settled by the fact that we know precisely how to build CH or $\neg$CH worlds close to any given set-theoretic universe---the CH and $\neg$CH worlds are in a sense dense among the set-theoretic universes. The multiverse view is realist as opposed to formalist, since it affirms the real nature of the set-theoretic worlds to which the various set concepts give rise. On the Multiverse view, the Old-School Dream Solution is impossible, since our experience in the CH and $\neg$CH worlds will prevent us from accepting any principle $\Phi$ that settles CH as "obviously true". Rather, on the multiverse view we are to study all the possible set-theoretic worlds and especially how they relate to each other. I should stop now, and I apologize for the length of this answer. - 2 it seems that the multiverse view is the beginning of plurality in set theory. This is analogous to how there used to be only one geometry -- Euclidean -- but the investigation of PP and not PP led to a multiverse of geometries. – Colin Tan May 19 2010 at 8:32 1 I agree, Colin; the analogy with geometry is very strong and extends to many facets of how we think about the various geometries. – Joel David Hamkins May 19 2010 at 16:18 1 Is there any name for something like "formalism with a preferred model", where one acknowledges that there is no reason apart from personal preference to study a given system of axioms, but where one also acknowledges that one has a preferred system? It seems somewhere in between the formalism and pragmatism and seems to be a view partially justified by advances in topos theory. – Harry Gindi May 20 2010 at 7:21 1 Because it admits a preferred model, that view seems much closer to Platonism or realism than formalism. It is similar to what occurs in (first order) arithmetic, where we have the standard model of the natural numbers. In set theory, however, we seem to lack such a standard model. Topos theory does not provide a standard model, since all the diverse models of set theory that have been constructed by forcing and so on each have their own incompatible version of Set. The nature and truths of any absolute set-theoretic background is what is at stake and it remains disapointingly murky. – Joel David Hamkins May 20 2010 at 12:23 2 "Formalist view. Rarely held by mathematicians, although occasionally held by philosophers ..." I recently read the following, written by a distinguished philosopher of mathematics. "The formalist's central thought is that arithmetic is not ultimately concerned with an extralinguistic domain of things. Rather, insofar as arithmetic has a proper subject matter, it is the language of arithmetic itself and certain formal relations among its sentences." This was accompanied by a footnote: "The view has few contemporary adherents among philosophers, though mathematicians often find it congenial." – gowers Feb 6 2011 at 17:28 show 7 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. (1) Patrick Dehornoy gave a nice talk at the Séminaire Bourbaki explaining Hugh Woodin's approach. It omits many technical details, so you may want to look at it before looking again at the Notices papers. I think looking at those slides and then at the Notices articles gives a reasonable picture of what the approach is and what kind of problems remain there. You can find the slides here, under "Recent results about the Continuum Hypothesis, after Woodin". (In true Bourbaki fashion, I heard that the talk was not well received.) Roughly, Woodin's approach shows that in a sense, the theory of $H(\omega_2)$ decided by the usual set of axioms, ZFC and large cardinals, can be "finitely completed" in a way that would make it reasonable to expect to settle all its properties. However, any such completion implies the negation of CH. It is a conditional result, depending on a highly non-trivial problem, the $\Omega$-conjecture. If true, this conjecture gives us that Cohen's technique of forcing is in a sense the only method (in the presence of large cardinals) that is required to established consistency. (The precise statement is more technical.) $H(\omega_2)$, that Dehornoy calls $H_2$, is the structure obtained by considering only those sets $X$ such that $X\cup\bigcup X\cup\bigcup\bigcup X\cup\dots$ has size strictly less than $\aleph_2$, the second uncountable cardinal. Replacing $\aleph_2$ with $\aleph_1$, we have $H(\omega_1)$, whose theory is completely settled in a sense, in the presence of large cardinals. If nothing else, one can think of Woodin's approach as trying to build an analogy with this situation, but "one level up." Whether or not one considers that settling the $\Omega$-conjecture in a positive fashion actually refutes CH in some sense, is a delicate matter. In any case (and I was happy to see that Dehornoy emphasizes this), Woodin's approach gives strength to the position that the question of CH is meaningful (as opposed to simply saying that, since it is independent, there is nothing to decide). (2) There is another approach to the problem, also pioneered by Hugh Woodin. It is the matter of "conditional absoluteness." CH is a $\Sigma^2_1$ statement. Roughly, this means that it has the form: "There is a set of reals such that $\phi$", where $\phi$ can be described by quantifying only over reals and natural numbers. In the presence of large cardinals, Woodin proved the following remarkable property: If $A$ is a $\Sigma^2_1$ statement, and we can force $A$, then $A$ holds in any model of CH obtained by forcing. Recall that forcing is essentially the only tool we have to establish consistency of statements. Also, there is a "trivial" forcing that does not do anything, so the result is essentially saying that any statement of the same complexity as CH, if it is consistent (with large cardinals), then it is actually a consequence of CH. This would seem a highly desirable ''maximality'' property that would make CH a good candidate to be adopted. However, recent results (by Aspero, Larson, and Moore) suggest that $\Sigma^2_1$ is close to being the highest complexity for which a result of this kind holds, which perhaps weakens the argument for CH that one could do based on Hugh's result. A good presentation of this theorem is available in Larson's book "The stationary tower. Notes on a Course by W. Hugh Woodin." Unfortunately, the book is technical. (3) Foreman's approach is perhaps the strongest opponent to the approach suggested by Woodin in (1). Again, it is based in the technique of forcing, now looking at small cardinal analogues of large cardinal properties. Many large cardinal properties are expressed in terms of the existence of elementary embeddings of the universe of sets. This embeddings tend to be "based" at cardinals much much larger than the size of the reals. With forcing, one can produce such embeddings "based" at the size of the reals, or nearby. Analyzing a large class of such forcing notions, Foreman shows that they must imply CH. If one where to adopt the consequences of performing these forcing constructions as additional axioms one would then be required to also adopt CH. I had to cut my answer short last time. I would like now to say a few details about a particular approach. (4) Forcing axioms imply that $2^{\aleph_0}=\aleph_2$, and (it may be argued) strongly suggest that this should be the right answer. Now, before I add anything, note that Woodin's approach (1) uses forcing axioms to prove that there are "finite completions" of the theory of $H(\omega_2)$ (and the reals have $\aleph_2$). However, this does not mean that all such completions would be compatible in any particular sense, or that all would decide the size of the reals. What Woodin proves is that all completions negate CH, and forcing axioms show that there is at least one such completion. I believe there has been some explanation of forcing axioms in the answer to the related question on GCH. Briefly, the intuition is this: ZFC seems to capture the basic properties of the universe of sets, but fails to account for its width and its height. (What one means by this is: how big should power sets be, and how many ordinals there are.) Our current understanding suggests that the universe should indeed be very tall, meaning there should be many many large cardinals. As Joel indicated, there was originally some hope that large cardinals would determine the size of the reals, but just about immediately after forcing was introduced, it was proved that this was not the case. (Technically, small forcing preserves large cardinals.) However, large cardinals settle many questions about the structure of the reals (all first order, or projective statements, in fact). CH, however, is "just" beyond what large cardinals can settle. One could say that, as far as large cardinals are concerned, CH is true. What I mean is that, in the presence of large cardinals, any definable set of reals (for any reasonable notion of definability) is either countable or contains a perfect subset. However, this may simply mean that there is certain intrinsic non-canonicity in the sets of reals that would disprove CH, if this is the case. (A word of caution is in order here, and there are candidates for large cardinal axioms [presented by Hugh Woodin in his work on suitable extender sequences] for which preservation under small forcing is not clear. Perhaps the solution to CH will actually come, unexpectedly, from studying these cardinals. But this is too speculative at the moment.) I have avoided above saying much about forcing. It is a massive machinery, and any short description is bound to be very inaccurate, so I'll be more than brief. An ordinary algebraic structure (a group, he universe of sets) can be seen as a bi-valued model. Just as well, one can define, for any complete Boolean algebra ${\mathbb B}$, the notion of a structure being ${\mathbb B}$-vaued. If you wish, "fuzzy set theory" is an approximation to this, as are many of the ways we model the world by using a probability density to decide the likelihood of events. For any complete Boolean algebra ${\mathbb B}$, we can define a ${\mathbb B}$-valued model $V^{\mathbb B}$ of set theory. In it, rather than having for sets $x$ and $y$ that either $x\in y$ or it doesn't, we assign to the statement $x\in y$ a value $[x\in y]\in{\mathbb B}$. The way the construction is performed, $[\phi]=1$ for each axiom $\phi$ of ZFC. Also, for each element $x$ of the actual universe of sets, there is a copy $\check x$ in the ${\mathbb B}$-valued model, so that the universe $V$ is a submodel of $V^{\mathbb B}$. If it happens that for some statement $\psi$ we have $[\psi]>0$, we have established that $\psi$ is consistent with ZFC. By carefully choosing ${\mathbb B}$, we can do this for many $\psi$. This is the technique of forcing, and one can add many wrinkles to the approach just outlined. One refers to ${\mathbb B}$ as a forcing notion. Now, the intuition that the universe should be very fat is harder to capture than the idea of largeness of the ordinals. One way of expressing it is that the universe is somehow "saturated": If the existence of some object is consistent in some sense, then in fact such object should exist. Formalizing this, one is led to forcing axioms. A typical forcing axiom says that relatively simple (under some measure of complexity) statements that can be shown consistent using the technique of forcing via a Boolean algebra ${\mathbb B}$ that is not too pathological, should in fact hold. The seminal Martin's Maximum paper of Foreman-Magidor-Shelah identified the most generous notion of "not too pathological", it corresponds to the class of "stationary set preserving" forcing notions. The corresponding forcing axiom is Martin's Maximum, MM. In that paper, it was shown that MM implies that the size of the reals is $\aleph_2$. The hypothesis of MM has been significantly weakened, through a series of results by different people, culminating in the Mapping Reflection Principle paper of Justin Moore. Besides this line of work, many natural consequences of forcing axioms (commonly termed reflection principles) have been identified, and shown to be independent of one another. Remarkably, just about all these principles either imply that the size of the reals is $\aleph_2$, or give $\aleph_2$ as an upper bound. Even if one finds that forcing axioms are too blunt a way of capturing the intuition of "the universe is wide", many of its consequences are considered very natural. (For example, the singular cardinal hypothesis, but this is another story.) Just as most of the set theoretic community now understands that large cardinals are part of what we accept about the universe of sets (and therefore, so is determinacy of reasonably definable sets of reals, and its consequences such us the perfect set property), it is perhaps not completely off the mark to expect that as our understanding of reflection principles grow, we will adopt them (or a reasonable variant) as the right way of formulating "wideness". Once/if that happens, the size of the reals will be taken as $\aleph_2$ and therefore CH will be settled as false. The point here is that this would be a solution to the problem of CH that does not attack CH directly. Rather, it turns out that the negation of CH is a common consequence of many principles that it may be reasonable to adapt in light of the naturalness of some of their best known consequences, and of their intrinsic motivation coming from the "wide universe" picture. (Apologies for the long post.) Edit, Nov. 22/10: I have recently learned about Woodin's "Ultimate L" which, essentially, advances a view that theories are "equally good" if they are bi-interpretable, and identifies a theory ("ultimate L") that, modulo large cardinals, would work as a universal theory from which to interpret all extensions. This theory postulates an $L$-like structure for the universe and in particular implies CH, see this answer. But, again, the theory is not advocated on grounds that it ought to be true, whatever this means, but rather, that it is "richest" possible in that it allows us to interpret all possible "natural" extensions of ZFC. In particular, under this approach, only large cardinals are relevant if we want to strengthen the theory, while "width" considerations, such as those supporting forcing axioms, are no longer relevant. Since the approach I numbered (1) above implies the negation of CH, I feel I should add that one of the main reasons for it being advanced originally depended on the fact that the set of $\Omega$-validities can be defined "locally", at the level of $H({\mathfrak c}^+)$, at least if the $\Omega$-conjecture holds. However, recent results of Sargsyan uncovered a mistake in the argument giving this local definability. From what I understand, Woodin feels that this weakens the case he was making for not-CH significantly. - 1 Thanks very much for this answer, Andres. I had heard that Woodin proved an extremal property of CH, but I didn't know what it was. It is presumably your item (2). – John Stillwell May 18 2010 at 23:47 Could you give some more details about the connection with the recent work of Sargsyan? – Simon Thomas Nov 23 2010 at 13:08 Hi Simon. I am not too certain of all the details; I believe that the issue is this: In the context of $AD^+$, say that $\alpha$ is a "local $\Theta$" if it is the $\Theta$ of a hod-model. Woodin had an argument showing that no such $\alpha$ could be overlapped by a strong cardinal. This put serious limitations on the strength of large cardinals that hod-models could contain. In particular, this was the reason why it was expected that "CH + there is an $\omega_1$-dense ideal on $\omega_1$" and "$AD_{\mathbb R}+\Theta$ regular" were expected to have really high consistency strength (continued) – Andres Caicedo Nov 23 2010 at 15:33 (2) Woodin's local definability argument depended on this limitation of hod models. (I am not sure of the details here.) Grigor's analysis in the context of the core model induction (there are slides of a talk at Boise on his website, and I can email you his thesis, let me know) showed that these "local overlaps" are possible, and deduced as a corollary that "$AD_{\mathbb R}+\Theta$ regular has much lower consistency strength than expected. Grigor in fact has a very detailed analysis of hod models. Without the overlap limitation, the set of $\Omega$-validities ends up being harder to define. – Andres Caicedo Nov 23 2010 at 15:42 (3) It can be shown that it is $H(\delta_0^+)$-definable, where $\delta_0$ is the smallest Woodin of $V$. But the argument pro-not-CH went by a level by level analysis of the $H(\kappa)$-levels, and this jump (from ${\mathfrak c}$ to a Woodin) is too high to overlook. As far as I understand, this is the nature of the issue. – Andres Caicedo Nov 23 2010 at 15:45 show 3 more comments Regarding Shelah's approach, I believe that the following paper should be quite accessible to non-professionals: http://arxiv.org/pdf/math.LO/0102056 Now, I have no idea how to explain Woodin's approach to CH without relying on some cryptic terminology, but I believe that the following page might be useful: http://www2.units.it/episteme/L&PS_Vol3No1/bellotti/bellotti-html/bellotti_L&PS_Vol3No1.html - Thanks Haim, I think I vaguely had this paper of Shelah also in mind but did not find it when I wrote the question. – Gil Kalai May 8 2010 at 8:54 First I'll say a few words about forcing axioms and then I'll answer your question. Forcing axioms were developed to provide a unified framework to establish the consistency of a number of combinatorial statements, particularly about the first uncountable cardinal. They began with Solovay and Tennenbaum's proof of the consistency of Souslin's Hypothesis (that every linear order in which there are no uncountable families of pairwise disjoint intervals is necessarily separable). Many of the consequences of forcing axioms, particularly the stronger Proper Forcing Axiom and Martin's Maximum, had the form of classification results: Baumgartner's classification of the isomorphism types of $\aleph_1$-dense sets of reals, Abraham and Shelah's classification of the Aronszajn trees up to club isomorphism, Todorcevic's classification of linear gaps in $\mathcal{P}(\mathbb{N})/\mathrm{fin}$, and Todorcevic's classification of transitive relations on $\omega_1$. A survey of these results (plus many references) can be found in both Stevo Todorcevic's ICM article and my own (the later can be found here). These are accessible to a general audience. What does all this have to do with the Continuum Problem? It was noticed early on that forcing axioms imply that the continuum is $\aleph_2$. The first proof of this is, I believe, in Foreman, Magidor, and Shelah's seminal paper in Martin's Maximum. Other quite different proofs were given by Caicedo, Todorcevic, Velickovic, and myself. An elementary proof which is purely Ramsey-theoretic in nature is in my article " Open colorings, the continuum, and the second uncountable cardinal" (PAMS, 2002). Since it is often the case that the combinatorial essence of the proofs of these classification results and that the continuum is $\aleph_2$ are similar, one is left to speculate that perhaps there may some day be a classification result concerning structures of cardinality $\aleph_1$ which already implies that the continuum is $\aleph_2$. There is a candidate for such a classification: the assertion that there are five uncountable linear orders such that every other contains an isomorphic copy of one of these five. Another related candidate for such a classification is the assertion that the Aronszajn lines are well quasi-ordered by embeddability (if $L_i$ $(i < \infty)$ is a sequence of Aronszajn lines, then there is an $i < j$ such that $L_i$ embeds into $L_j$). These are due to myself and Carlos Martinez, respectively. See a discussion of this (with references) in my ICM paper. - The question of whether or not $2^{\aleph_{0}} = \aleph_{1}$ is not even considered in Shelah's approach. In fact, this question is regarded as part of the "white noise" which has distracted the attention of set theorists from some striking $ZFC$-results about cardinal exponentiation $\kappa^{\lambda}$ when you consider relatively small exponents $\lambda$ and relatively large bases $\kappa$. - This is a nice survey: Koellner, Peter (2011). "The Continuum Hypothesis". Exploring the Frontiers of Independence (Harvard lecture series). (Pasted from Wikipedia article on CH.) - Shelah's approach from his paper in his "The Generalized Continuum Hypothesis revisited", concerns mainly the generalized continuum hypothesis. His main theorem addresses an appealing variation of the GCH which is based on a revised notion of "power." Let me explain what is this notion of $\lambda^{[\kappa]}$ (in words: $\lambda$ to the revised power of $\kappa$,) which is central to his approach and is also of independent interest. $\lambda^{[\kappa]}$ is the minimum size of a family of subsets of size $\kappa$ of a set $X$ of cardinality $\lambda$ such that every subset of cardinality $\kappa$ of $X$ is covered by less than $\kappa$ members of the family. (Of course we need that $\kappa < \lambda$ and also we need that $\kappa$ is a regular cardinal.) The introduction to the paper is readable and motivated. - There is a recent (and unfinished) attempt to view the Shelah's approach to Continuum Hypothesis in terms of homotopy theory. - I believe Gödel and Cohen's arguments are not acceptable. My reasoning is a common sense one: A mathematical proof is a text whose spelling correctness can be checked by a machine. The text in itself has no meaning. Only an interpretation of it can have any meaning. So, at best, Gödel and Cohen proved some mathematical statements which can be interpreted as implying that the continuum hypothesis is undecidable (in a given axiom system). It's surprising that Bourbaki, in his Set Theory book, takes Gödel and Cohen's arguments for granted. I think that Bourbaki's approach, coherently followed, leads to a rejection of these arguments. http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Continuum_Hypothesis/ I think many people share this opinion. (For some reason they don't express it.) In another answer I gave three quotations from Cohen's book "Set theory and the Continuum Hypothesis". For the reader's convenience I'll paste them below. I think we'll all find them interesting. Quotation 1 (pages 26-27) It should be emphasized that these functions are "real" mathematical objects and not objects of any formal system ... Question. What is a "real mathematical object"? (And what is an "unreal" or "irreal" mathematical object?) Quotation 2 (page 39) The theorems of the previous section are not results about what can be proved in particular axiom systems; they are absolute statements about functions. Questions. What is an "absolute statement"? Is there an "axiom free" mathematics? Quotation 3 (page 41) We have now arrived at a rather peculiar situation. On the one hand $\sim A$ is not provable in $Z_1$ and yet we have just given an informal proof that $\sim A$ is true. (There is no contradiction here since we have merely shown that the proofs in $Z_1$ do not exhaust the set of all acceptable arguments.) Question. What is an "acceptable argument"? It seems that "true" is defined as "following from an acceptable argument". Here is another quotation (page 13): If a set of statements S has a model then it is consistent. If the underlying set theory is inconsistent, then any set of statements has a model. I suggest the following three books to the interested reader. (1) P. Cohen, Set theory and the Continuum Hypothesis. (2) S.C. Kleene, Introduction to Metamathematics. (3) N. Bourbaki, Theory of Sets. French original: Théorie des ensembles. Let me ask explicitly the question which is implicit in this post: Is the Continuum Hypothesis true, false or undecidable in Bourbaki's set theory? There is no mathematical questions whose answer I more ardently wish I knew. Thank you to all the people in charge of MathOverflow for everything you're doing! - If one is a pragmatic formalist (believing that formal systems are no more nor less than working models of other systems), one may still ask about formal systems modeling the reachability of propositions in other formal system, such as ZFC. My impression is that these are precisely the sorts of results that Gödel and Cohen produced with respect to CH; that is, they indeed prove that a mathematician who works correctly cannot prove CH in ZFC. This is no more objectionable than proving that an infinite graph is disconnected, even though one can never prove this by exhausting all possible paths. – Niel de Beaudrap May 20 2010 at 8:00 Dear Niel - Thanks for your comment. This is just to see if I understand you correctly. Your opinion is (as I understand it) that Gödel and Cohen proved some mathematical statements which can be interpreted as implying that the continuum hypothesis is undecidable (in ZFC to quote your comment). Please correct me if I'm wrong. – Pierre-Yves Gaillard May 20 2010 at 9:50 6 It is true that Goedel and Cohen proved that if ZF is consistent, then so are ZFC+CH and ZFC+$\neg$CH. But their proofs provided more than just this relative consistency result, as a formal syntactic result. Rather, what they provided were construction methods for building new models of set theory from old models---from any universe of sets in which ZFC is true, we can build closely related models where ZFC+CH and ZFC+$\neg$CH hold. So the methods are more semantic than syntactic, and perhaps the situation is like interpreting the various Euclidean and hyperbolic geometries within one another. – Joel David Hamkins May 20 2010 at 12:31	{"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": 136, "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": 0.958989679813385, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/44833?sort=newest	## If $k[S]$ is noetherian, is S finitely generated? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $S$ be a semigroup. If $S$ is abelian, then it follows that the semigroup algebra $k[S]$ is finitely generated if and only if $S$ is. What if we relax the condition on $k[S]$, so that $k[S]$ is only noetherian. Does it in this case follow that $S$ is finitely generated? - ## 2 Answers It is an open problem (or was, last time I checked!) whether the noetherianity of $k[S]$ implies finite generation of $S$, when $S$ is not abelian. This is discussed in chapter 5 of Noetherian semigroup algebras by Eric Jespers and Jan Okniński, along with various cases where we know that $S$ is finitely generated. They prove, for example, that this is so if $k[S]$ satisfies a polynomial identity, and this gives the case in which $S$ is abelian as a corollary. - That's interesting, I was actually expecting some strange counterexample. Thanks for the reference! – J.C. Ottem Nov 4 2010 at 23:33 ### 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. No, at least in the case that $k$ is infinite, and not prime - for an indeterminate $t$ take $S$ generated by ${t, t/a, t/a^2, ...}$ with $0 \ne a \in k$ of infinite (multiplicative) order not in the prime field. Then $k[S] = k[t]$ but $S$ is not finitely generated. - Your $k[S]$ is not the group algebra as in it the elements of $S$ are by definition linearly independent. Also $S$ is not a semigroup. – Torsten Ekedahl Nov 4 2010 at 19:19 Sorry, I guess I mean the semigroup generated by the $t/a^n$. I've edited my post to reflect this change. And I'm missing something, but I don't quite understand what you mean by the elements of $S$ are by definition linearly independent? – Justin Shih Nov 4 2010 at 19:25 Oh, I see now. I guess I'm taking $k[t]$ and then finding a sub-semigroup inside there, instead of taking a semigroup $S$ and then making $k[S]$. – Justin Shih Nov 4 2010 at 19:27 @Justin: By definition, $k[S]$ is the $k$-vector space with basis given by elements of $S$. (Multiplication is induced by the semigroup structure on $S$.) So I'm afraid the edit still doesn't fix this... – Dave Anderson Nov 4 2010 at 19:27 Right. So then can we make an increasing chain of ideals by picking $s_1 \in S$, setting $I_1 = (s_1)$, and then by induction picking $s_{k+1} \in$S\backslash\I_k$and setting$I_{k+1} = I_k + (s_{k+1})$? This chain eventually terminates, which I think means that$S\$ is finitely generated? – Justin Shih Nov 4 2010 at 19:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 38, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9369643330574036, "perplexity_flag": "head"}
http://www.haskell.org/haskellwiki/index.php?title=User:Michiexile/MATH198/Lecture_3&diff=30605&oldid=30322	# User:Michiexile/MATH198/Lecture 3 ### From HaskellWiki (Difference between revisions) \| \| \| \| \| \|-----------\|----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\|-----------\|----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\| \| \| \| \| \| \| Line 136: \| \| Line 136: \| \| \| \| * Prove the propositions in the section on natural transformations. \| \| * Prove the propositions in the section on natural transformations. \| \| \| * Prove that <hask>listToMaybe :: [a] -> Maybe a</hask> is a natural transformation from the list functor to the maybe functor. Is <hask>catMaybes</hask> a natural transformation? Between which functors? Find three more natural transformations defined in the standard Haskell library. \| \| * Prove that <hask>listToMaybe :: [a] -> Maybe a</hask> is a natural transformation from the list functor to the maybe functor. Is <hask>catMaybes</hask> a natural transformation? Between which functors? Find three more natural transformations defined in the standard Haskell library. \| \| \| \| + \| * Write a functor instance for <hask>data F a = F (Int -> a)</hask> \| \| \| \| + \| * Write a functor instance for <hask>data F a = F ((Int -> a) -> Int)</hask> \| \| \| \| + \| * * We could write a pretty printer, or XML library, using the following data type as the core data type: \| \| \| \| + \| <haskell> \| \| \| \| + \| data ML a = Tag a (ML a) \| \| \| \| \| + \| Str String \| \| \| \| \| + \| Seq [ML a] \| \| \| \| \| + \| Nil \| \| \| \| + \| </haskell> \| \| \| \| + \| :With this, we can use a specific string-generating function to generate the tagged marked up text, such as, for instance: \| \| \| \| + \| <haskell> \| \| \| \| + \| prettyprint (Tag a ml) = "<" ++ show a ++ ">" ++ prettyprint ml ++ "</" ++ show a ++ ">" \| \| \| \| + \| prettyprint (Str s) = s \| \| \| \| + \| prettyprint (Seq []) = "" \| \| \| \| + \| prettyprint (Seq (m:ms)) = prettyprint m ++ "\n" ++ prettyprint (Seq ms) \| \| \| \| + \| prettyprint Nil = "" \| \| \| \| + \| </haskell> \| \| \| \| + \| :Write an instance of Functor that allows us to apply changes to the tagging type. Then, using the following tagging types: \| \| \| \| + \| <haskell> \| \| \| \| + \| data HTMLTag = HTML \| BODY \| P \| I \| H1 deriving (show) \| \| \| \| + \| data XMLTag = DOCUMENT \| HEADING \| ABSTRACT \| TEXT deriving (show) \| \| \| \| + \| </haskell> \| \| \| \| + \| :write a function <hask>htmlize :: ML XMLTag -> ML HTMLTag</hask> and use it to generate a html document out of: \| \| \| \| + \| <haskell> \| \| \| \| + \| Tag DOCUMENT \| \| \| \| + \| Seq [ \| \| \| \| + \| Tag HEADING \| \| \| \| + \| String "Nobel prize for chromosome find", \| \| \| \| + \| Tag ABSTRACT \| \| \| \| + \| String "STOCKHOLM (Reuters) - Three Americans won the Nobel prize for medicine on Monday for revealing the existence and nature of telomerase, an enzyme which helps prevent the fraying of chromosomes that underlies aging and cancer.", \| \| \| \| + \| Tag TEXT \| \| \| \| + \| String "Australian-born Elizabeth Blackburn, British-born Jack Szostak and Carol Greider won the prize of 10 million Swedish crowns (\$1.42 million), Sweden's Karolinska Institute said.", \| \| \| \| + \| Tag TEXT \| \| \| \| + \| String "'The discoveries ... have added a new dimension to our understanding of the cell, shed light on disease mechanisms, and stimulated the development of potential new therapies,' it said.", \| \| \| \| + \| Tag TEXT \| \| \| \| + \| String "The trio's work laid the foundation for studies that have linked telomerase and telomeres -- the small caps on the end of chromosomes -- to cancer and age-related conditions.", \| \| \| \| + \| Tag TEXT \| \| \| \| + \| String "Work on the enzyme has become a hot area of drug research, particularly in cancer, as it is thought to play a key role in allowing tumor cells to reproduce out of control.", \| \| \| \| + \| Tag TEXT \| \| \| \| + \| String "One example, a so-called therapeutic vaccine that targets telomerase, in trials since last year by drug and biotech firms Merck and Geron, could yield a treatment for patients with tumors including lung and prostate cancer.", \| \| \| \| + \| Tag TEXT \| \| \| \| + \| String "The Chief Executive of Britain's Medical Research Council said the discovery of telomerase had spawned research of 'huge importance' to the world of science and medicine.", \| \| \| \| + \| Tag TEXT \| \| \| \| + \| String "'Their research on chromosomes helped lay the foundations of future work on cancer, stem cells and even human aging, areas that continue to be of huge importance,' Sir Leszek Borysiewicz said in a statement.", \| \| \| \| + \| </haskell> \| ## Revision as of 17:23, 5 October 2009 IMPORTANT NOTE: THESE NOTES ARE STILL UNDER DEVELOPMENT. PLEASE WAIT UNTIL AFTER THE LECTURE WITH HANDING ANYTHING IN, OR TREATING THE NOTES AS READY TO READ. ## Contents ### 1 Functors We've spent quite a bit of time talking about categories, and special entities in them - morphisms and objects, and special kinds of them, and properties we can find. And one of the main messages visible so far is that as soon as we have an algebraic structure, and homomorphisms, this forms a category. More importantly, many algebraic structures, and algebraic theories, can be captured by studying the structure of the category they form. So obviously, in order to understand Category Theory, one key will be to understand homomorphisms between categories. #### 1.1 Homomorphisms of categories A category is a graph, so a homomorphism of a category should be a homomorphism of a graph that respect the extra structure. Thus, we are led to the definition: Definition A functor $F:C\to D$ from a category C to a category D is a graph homomorphism F0,F1 between the underlying graphs such that for every object $X\in C_0$: • $F_1(1_X) = 1_{F_0(X)}$ • F1(gf) = F1(g)F1(f) Note: We shall consistently use F in place of F0 and F1. The context should be able to tell you whether you are mapping an object or a morphism at any given moment. ##### 1.1.1 Examples and non-examples • Monoid homomorphisms • Monotone functions between posets • Pick a basis for every vectorspace, send $V\mapsto\dim V$ and $f:V\to W$ to the matrix representing that morphism in the chosen bases. #### 1.2 Interpreting functors in Haskell One example of particular interest to us is the category Hask. A functor in Hask is something that takes a type, and returns a new type. Not only that, we also require that it takes arrows and return new arrows. So let's pick all this apart for a minute or two. Taking a type and returning a type means that you are really building a polymorphic type class: you have a family of types parametrized by some type variable. For each type a , the functor dataFa = ... will produce a new type, F a . This, really, is all we need to reflect the action of F0. The action of F1 in turn is recovered by requiring the parametrized type F a to implement the Functor typeclass. This typeclass requires you to implement a function fmap::(a -> b) -> F a -> F b . This function, as the signature indicates, takes a function f :: a -> b and returns a new function fmap f :: F a -> F b . The rules we expect a Functor to obey seem obvious: translating from the categorical intuition we arrive at the rules • fmap id = id and • fmap (g . f) = fmap g . fmap f Now, the real power of a Functor still isn't obvious with this viewpoint. The real power comes in approaching it less categorically. A Haskell functor is a polymorphic type. In a way, it is an prototypical polymorphic type. We have some type, and we change it, in a meaningful way. And the existence of the Functor typeclass demands of us that we find a way to translate function applications into the Functor image. We can certainly define a boring Functor, such as ```data Boring a = Boring instance Functor Boring where fmap f = const Boring``` but this is not particularly useful. Almost all Functor instances will take your type and include it into something different, something useful. And it does this in a way that allows you to lift functions acting on the type it contains, so that they transform them in their container. And the choice of words here is deliberate. Functors can be thought of as data containers, their parameters declaring what they contain, and the fmap implementation allowing access to the contents. Lists, trees with node values, trees with leaf values, Maybe , Either all are Functor s in obvious manners. ```data List a = Nil \| Cons a (List a) instance Functor List where fmap f Nil = Nil fmap f (Cons x lst) = Cons (f x) (fmap f lst) data Maybe a = Nothing \| Just a instance Functor Maybe where fmap f Nothing = Nothing fmap f (Just x) = Just (f x) data Either b a = Left b \| Right a instance Functor (Either b) where fmap f (Left x) = Left x fmap f (Right y) = Right (f y) data LeafTree a = Leaf a \| Node [LeafTree a] instance Functor LeafTree where fmap f (Node subtrees) = Node (map (fmap f) subtrees) fmap f (Leaf x) = Leaf (f x) data NodeTree a = Leaf \| Node a [NodeTree a] instance Functor NodeTree where fmap f Leaf = Leaf fmap f (Node x subtrees) = Node (f x) (map (fmap f) subtrees)``` ### 2 The category of categories We define a category Cat by setting objects to be all small categories, and arrows to be all functors between them. Being graph homomorphisms, functors compose, their composition fulfills all requirements on forming a category. It is sometimes useful to argue about a category CAT of all small and most large categories. The issue here is that allowing $CAT\in CAT_0$ opens up for set-theoretical paradoxes. #### 2.1 Isomorphisms in Cat and equivalences of categories The definition of an isomorphism holds as is in Cat. However, isomorphisms of categories are too restrictive a concept. To see this, recall the category Monoid, where each object is a monoid, and each arrow is a monoid homomorphism. We can form a one-object category out of each monoid, and the method to do this is functorial - i.e. does the right thing to arrows to make the whole process a functor. Specifically, if $h:M\to N$ is a monoid homomorphism, we create a new functor $C(h):C(M)\to C(N)$ by setting C(h)[0]( * ) = * and C(h)[1](m) = h(m). This creates a functor from Monoid to Cat. The domain can be further restricted to a full subcategory OOC of Cat, consisting of all the 1-object categories. We can also define a functor $U:OOC\to Monoid$ by U(C) = C[1] with the monoidal structure on U(C) given by the composition in C. For an arrow $F:A\to B$ we define U(F) = F[1]. These functors take a monoid, builds a one-object category, and hits all of them; and takes a one-object category and builds a monoid. Both functors respect the monoidal structures - yet these are not an isomorphism pair. The clou here is that our construction of C(M) from M requires us to choose something for the one object of the category. And choosing different objects gives us different categories. Thus, the composition CU is not the identity; there is no guarantee that we will pick the object we started with in the construction in C. Nevertheless, we would be inclined to regard the categories Monoid and OOC as essentially the same. The solution is to introduce a different kind of sameness: Definition A functor $F:C\to D$ is an equivalence of categories if there is a functor $G:D\to C$ and: • A family $u_C:C\to G(F(C))$ of isomorphisms in C indexed by the objects of C, such that for every arrow $f:C\to C': G(F(f)) = u_{C'}\circ f\circ u_C^{-1}$. • A family $u_D:D\to F(G(D))$ of isomorphisms in D indexed by the objects of D, such that for every arrow $f:D\to D': F(G(f)) = u_{D'}\circ f\circ u_D^{-1}$. The functor G in the definition is called a pseudo-inverse of F. #### 2.2 Natural transformations The families of morphisms required in the definition of an equivalence show up in more places. Suppose we have two functors $F:A\to B$ and $G:A\to B$. Definition A natural transformation $\alpha:F\to G$ is a family of arrows $\alpha a:F(a)\to G(a)$ indexed by the objects of A such that for any arrow $s:a\to b$ in $G(s)\circ\alpha a = \alpha b\circ F(s)$ (draw diagram) The commutativity of the corresponding diagram is called the naturality condition on α, and the arrow αa is called the component of the natural transformation α at the object a. Given two natural transformations $\alpha: F\to G$ and $\beta: G\to H$, we can define a composition $\beta\circ\alpha$ componentwise as $(\beta\circ\alpha)(a) = \beta a \circ \alpha a$. Proposition The composite of two natural transformations is also a natural transformation. Proposition Given two categories C,D the collection of all functors $C\to D$ form a category Func(C,D) with objects functors and morphisms natural transformations between these functors. ### 3 Properties of functors The process of forming homsets within a category C gives, for any object A, two different functors Hom(A, − ): $X\mapsto Hom(A,X)$ and $Hom(-,A): X\mapsto Hom(X,A)$. Functoriality for Hom(A, − ) is easy: Hom(A,f) is the map that takes some $g:A\to X$ and transforms it into $fg:A\to Y$. Functoriality for Hom( − ,A) is more involved. We can view this as a functor either from Cop, or as a different kind of functor. If we just work with Cop, then no additional definitions are needed - but we need an intuition for the dual categories. Alternatively, we introduce a new concept of a contravariant functor. A contravariant functor $F:C\to D$ is some map of categories, just like a functor is, but such that F(1[X]) = 1[F(X)], as usual, but such that for a $f:A\to B$, the functor image is some $F(f):F(B)\to F(A)$, and the composition is F(gf) = F(f)F(g). The usual kind of functors are named covariant. • Full • Faithful ### 4 Applications for functors • CS applications • State machines and monoid actions ### 5 Homework • * Recall that a category is called discrete if it has no arrows other than the identities. Show that a small category A is discrete if and only if every set function $A_0\to B_0$, for every small category B, is the object part of a unique functor $A\to B$. Analogously, we define a small category B to be indiscrete if for every small category A, every set function $A_0\to B_0$ is the object part of a unique functor $A\to B$. Characterise indiscrete categories by the objects and arrows they have. • Show that the category of vectorspaces is equivalent to the category with objects integers and arrows matrices. • Prove the propositions in the section on natural transformations. • Prove that listToMaybe :: [a] -> Maybe a is a natural transformation from the list functor to the maybe functor. Is catMaybes a natural transformation? Between which functors? Find three more natural transformations defined in the standard Haskell library. • Write a functor instance for data F a = F (Int -> a) • Write a functor instance for data F a = F ((Int -> a) -> Int) • * We could write a pretty printer, or XML library, using the following data type as the core data type: ```data ML a = Tag a (ML a) \| Str String \| Seq [ML a] \| Nil``` With this, we can use a specific string-generating function to generate the tagged marked up text, such as, for instance: ```prettyprint (Tag a ml) = "<" ++ show a ++ ">" ++ prettyprint ml ++ "</" ++ show a ++ ">" prettyprint (Str s) = s prettyprint (Seq []) = "" prettyprint (Seq (m:ms)) = prettyprint m ++ "\n" ++ prettyprint (Seq ms) prettyprint Nil = ""``` Write an instance of Functor that allows us to apply changes to the tagging type. Then, using the following tagging types: ```data HTMLTag = HTML \| BODY \| P \| I \| H1 deriving (show) data XMLTag = DOCUMENT \| HEADING \| ABSTRACT \| TEXT deriving (show)``` write a function htmlize :: ML XMLTag -> ML HTMLTag and use it to generate a html document out of: ```Tag DOCUMENT Seq [ Tag HEADING String "Nobel prize for chromosome find", Tag ABSTRACT String "STOCKHOLM (Reuters) - Three Americans won the Nobel prize for medicine on Monday for revealing the existence and nature of telomerase, an enzyme which helps prevent the fraying of chromosomes that underlies aging and cancer.", Tag TEXT String "Australian-born Elizabeth Blackburn, British-born Jack Szostak and Carol Greider won the prize of 10 million Swedish crowns ($1.42 million), Sweden's Karolinska Institute said.", Tag TEXT String "'The discoveries ... have added a new dimension to our understanding of the cell, shed light on disease mechanisms, and stimulated the development of potential new therapies,' it said.", Tag TEXT String "The trio's work laid the foundation for studies that have linked telomerase and telomeres -- the small caps on the end of chromosomes -- to cancer and age-related conditions.", Tag TEXT String "Work on the enzyme has become a hot area of drug research, particularly in cancer, as it is thought to play a key role in allowing tumor cells to reproduce out of control.", Tag TEXT String "One example, a so-called therapeutic vaccine that targets telomerase, in trials since last year by drug and biotech firms Merck and Geron, could yield a treatment for patients with tumors including lung and prostate cancer.", Tag TEXT String "The Chief Executive of Britain's Medical Research Council said the discovery of telomerase had spawned research of 'huge importance' to the world of science and medicine.", Tag TEXT String "'Their research on chromosomes helped lay the foundations of future work on cancer, stem cells and even human aging, areas that continue to be of huge importance,' Sir Leszek Borysiewicz said in a statement.",```	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 38, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.8850173354148865, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/162781/possible-subgroups-of-mathbbz-36-mathbbz-oplus-mathbbz-35-mathbbz-o?answertab=oldest	# Possible subgroups of $\mathbb{Z}/3^6\mathbb{Z} \oplus\mathbb{Z}/3^5\mathbb{Z}\oplus\mathbb{Z}/3^2\mathbb{Z}$ $G \cong \mathbb{Z}/3^6\mathbb{Z} \oplus\mathbb{Z}/3^5\mathbb{Z}\oplus\mathbb{Z}/3^2\mathbb{Z}$ $H\leq G$ so that $G/H \cong \mathbb{Z}/3^2\mathbb{Z}\oplus\mathbb{Z}/3\mathbb{Z}$ Find all possible $H$ up to isomorphism. This is what I did: (1) $H$ must be of the form $\mathbb{Z}/3^{k_{1}}\mathbb{Z} \oplus\mathbb{Z}/3^{k_{2}}\mathbb{Z}\oplus\mathbb{Z}/3^{k_{3}}\mathbb{Z}$ (2) WLOG, $0 \leq k_{1} \leq 6$, $0 \leq k_{2} \leq 5$, $0 \leq k_{3} \leq 2$, (The order obiously doesn't matter because a change of order will be isomorphic) Also, $\|G/H\|=3^3=3^{13}/(3^{{k_{1}+k_{2}+k_{3}}}) \Rightarrow k_{1}+k_{2}+k_{3}=10$ Thus, by choosing $k_{i}$ we can find all possibilities. However, I have a few questions: 1. Is (1) justified? Why $\mathbb{Z}/3^{k_{1}}\mathbb{Z} \oplus\mathbb{Z}/3^{k_{2}}\mathbb{Z}\oplus\mathbb{Z}/3^{k_{3}}\mathbb{Z}\oplus\mathbb{Z}/3^{k_{4}}\mathbb{Z}$ isn't possible? 2. Is (2) justified? Why $k_{1}=7$ is not possible? Thank you! - (1) is justified by the general result that any subgroup of an abelian group with $n$ generators can be generated by at most $n$ elements. (2) is justified by the fact that all elements of $G$, and hence also of $H$, have order dividing $3^6$. – Derek Holt Jun 25 '12 at 11:54 WOLG? Not WLOG? – Gerry Myerson Jun 25 '12 at 12:43 @ Gerry - you're right (: @ Derek - perhaps it was a too easy example. Why $k_{3}=3$ is not possible? – Roy Jun 25 '12 at 12:56 $k_3=3$ is not possible because ${\mathbb Z}/3^3{\mathbb Z} \oplus {\mathbb Z}/3^3{\mathbb Z} \oplus {\mathbb Z}/3^3{\mathbb Z}$ has more elements of order dividing 27 than $G$ does. – Derek Holt Jun 25 '12 at 16:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": 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": 0.9537529945373535, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/21892?sort=oldest	## Formal deformations of algebras over not necessarily commutative rings ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In Iain Gordon's survery article "Symplectic reflection algebras" the concept of formal deformations of algebras over semisimple artinian (not necessarily commutative) rings is summarized (chapter 2). Unfortunately, deformations are needed in this generality and there are a few general things I don't understand: 1. An algebra over a semisimple artinian $\mathbb{C}$-algebra $k$ is "defined" as a $k$-bimodule $A$ with a $k$-bimodule morphism $A \otimes_k A \rightarrow A$. Is this a standard definition and is it correct that there is no associativity or unity assumption? I could not find a single book defining an algebra over a not necessarily commutative ring. 2. In the definition of a formal deformation given in the survey there also seems to be no associativity or unity assumption. Is this a standard definition and does Hochschild cohomology also work in this setting with the same interpretation? My problem is that when I don't restrict my deformations to be associative or have a unit, I might get a lot more deformations. Is there some literature discussing this in more detail? - ## 1 Answer I recall your question on a related topic... I am sure that associativity is assumed here (and just omitted because the audience is unlikely to think of any other algebras); as for unitality, won't it be preserved automatically exactly because of that yoga "when we deform A we don't want to deform k"? - I remember my question, but there it was about algebras over commutative rings. That's fine! – S1 Apr 19 2010 at 21:35 2 For associative rings there is some noncommutative deformation theory of Laudal but also a nice unpublished preprint of Michael Artin from 1994. Without associativity you would tend to have to big things to have useful artinianess; besides even in the commutative case the existence of the unit element in a ring is responsible for the fact that every artinian unital ring is also noetherian. – Zoran Škoda Apr 20 2010 at 4:03	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9309549331665039, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/80759/how-to-prove-that-x4x3x23x3-is-irreducible-over-ring-mathbbz-of-i/80762	How to prove that $x^4+x^3+x^2+3x+3$ is irreducible over ring $\mathbb{Z}$ of integers? Which criterion (test) one can use in order to prove that $x^4+x^3+x^2+3x+3$ is irreducible over ring $\mathbb{Z}$ of integers ? Neither of Eisenstein's criterion and Cohn's criterion cannot be applied on this polynomial. I know that one can use factor command in Wolfram Alpha and show that polinomial is irreducible but that isn't point of this question. - Is semi brute force OK? If so, there is an easy argument. It actually doesn't care very much about the coefficient of $x$, as long as integer root is avoided. – André Nicolas Nov 10 '11 at 5:35 @AndreNicolas,Actually I would like to know if there is some good computational-algebra algorithm – pedja Nov 10 '11 at 5:52 @There are good canned procedures for computing Galois groups. Overkill, probably. – André Nicolas Nov 10 '11 at 6:53 2 Answers Let your polynomial be $f$. Clearly it has no linear factor, since it has no root in $\mathbb{Z}$. Hence if it factors, it factors as the product of two irreducible quadratics $f_1, f_2$. Now looking mod $2$ we get a factorization $f=x^4+x^3+x^2+x+1=f_1f_2$ in $\mathbb{F}_2[x]$. Now since $f$ has no root mod $2$, $f_1$ and $f_2$ are also irreducible quadratics in $\mathbb{F}_2[x]$. But the only irreducible quadratic in $\mathbb{F}_2[x]$ is $x^2+x+1$. This would imply $x^4+x^3+x^2+x+1 = (x^2+x+1)^2$ in $\mathbb{F}_2[x]$, which is false. - 2 +1 Another way of seeing that $x^4+x^3+x^2+x+1$ is irreducible in $F_2[x]$ is to observe that it is a factor of $(x^5-1)=(x-1)(x^4+x^3+x^2+x+1)$. Thus any zero would a root of unity of order five. But $q=16$ is the lowest power of two with the property that $5\mid q-1$, so those roots of unity live in $GF(16)$. Thus their primitive polynomial has to be quartic. Your solution may be simpler. – Jyrki Lahtonen Nov 10 '11 at 6:54 Hint: If a quartic is reducible it has either a linear factor or a quadratic factor. It is easy to check that your polynomial above has no linear solutions, and you can work out a contradiction if you assume that it can be factored into the product of two (monic) quadratics. -	{"extraction_info": {"found_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": 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": 0.9450201392173767, "perplexity_flag": "head"}
http://nrich.maths.org/6416	### Lunar Leaper Gravity on the Moon is about 1/6th that on the Earth. A pole-vaulter 2 metres tall can clear a 5 metres pole on the Earth. How high a pole could he clear on the Moon? ### Which Twin Is Older? A simplified account of special relativity and the twins paradox. ### Whoosh A ball whooshes down a slide and hits another ball which flies off the slide horizontally as a projectile. How far does it go? # Go Spaceship Go ##### Stage: 5 Challenge Level: A spaceship driven by an infinite source of power accelerates with constant power through deep space. We watch it fly away from Earth and record its speed and progress. Show that after a certain time of travelling it will take longer than one day for the velocity of the spaceship to increase by $1 \mathrm{\ m\ s^{-1}}$. Find this time of travelling (in years) if the spaceship starts travelling from rest and has a power to weight ratio of $500 \mathrm{\ W\ kg^{-1}}$. How fast would the spaceship be travelling at this time? NOTES AND BACKGROUND Of course, this problem uses entirely classical, Newtonian physics. A proper analysis would need to take into account special relativity, and it is worth considering how important this might be given the answer. It is also worth noting that the spaceship is actually quite powerful, given that a typical sportscar has a power to weight ratio of around $200 \mathrm{\ W\ kg^{-1}}$ . 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": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9368430972099304, "perplexity_flag": "middle"}
http://cstheory.stackexchange.com/questions/376/using-lambda-calculus-to-derive-time-complexity	# Using lambda calculus to derive time complexity? Are there any benefits to calculating the time complexity of an algorithm using lambda calculus? Or is there another system designed for this purpose? Any references would be appreciated. - ## 6 Answers Ohad is quite right about the problems that the lambda calculus faces as a basis for talking about complexity classes. There has been a fair bit of work done on characterising complexity of reducibility in the lambda calculus, particularly around the work on labelled and optimal reductions from Lèvy's PhD thesis. Generally speaking, good cost models for the lambda calculus should not assign a constant weight to all beta reductions: intuitively, substituting a large subterm into many, differently scoped places should cost more that contracting a small K redex, and if one wants a certain amount of invariance of cost under different rewrite strategies, this becomes essential. 1. Lawall & Mairson, 1996, Optimality and inefficiency: what isn't a cost model of the lambda calculus? (.ps.gz) – Seminal survey of issues bearing on choice of cost model, and why many plausible ideas don't work. 2. Dal Lago & Martini, 2008, The weak lambda calculus as a reasonable machine – Offers a cost model for the call-by-value lambda calculus, together with good discussion of the literature. - Interesting references, I didn't know about these works. – Iddo Tzameret Aug 22 '10 at 21:21 There are quantitative results concerning $\lambda$-calculus, in the form of measuring the length of reductions in (typed-)lambda calculus. But this is of course far from saying anything about the complexity of algorithms (especially that the bounds obtained are fast-growing). See for example: Arnold Beckmann, Exact bounds for lengths of reductions in typed $\lambda$-calculus, Journal of Symbolic Logic 2001, 66(3): 1277-1285. For something closer to your question, there is a current project that develops and studies a type-system (a functional programing language) which by static-analysis can determine (polynomial) run-time bounds of programs (as well as other resources used by programs). So in some sense, this might hint that there might be some advantage in using functional-programing for analyzing run-time complexity. The project homepage is here. A possibly representative paper of this project is: Jan Hoffmann, Martin Hofmann. Amortized Resource Analysis with Polynomial Potential - A Static Inference of Polynomial Bounds for Functional Programs. In Proceedings of the 19th European Symposium on Programming (ESOP'10).link - There is a very interesting line of work based on linear logic, called implicit complexity theory, which characterizes various complexity classes by imposing various type disciplines on the lambda calculus. IIRC, this work began when Bellantoni and Cook, and Leivant figured out how to use the type system to bound primitive recursion to capture various complexity classes. In general, the attraction to working with lambda calculi is that it is sometimes possible to find more extensional (ie, more mathematically tractable) characterizations of various intensional features that give models like Turing machines their power. For example, one difference between Turing machines and pure lambda calculus is that since Turing receive codes of programs, a client can manually implement timeouts, to implement dovetailing -- and hence can compute parallel-or. However, timeouts can also be modelled metrically, and Escardo has conjectured (I don't know its status) that metric space models of the lambda calculus are fully abstract for PCF + timeouts. Metric spaces are very well-studied mathematical objects, and it is very nice to be able to make use of that body of theory. However, the difficulty of using lambda calculus is that it forces you to confront higher-order phenomena right from the starting gate. This can be very subtle, since the Church-Turing thesis fails at higher type -- natural models of computation differ at higher type, since they differ in what you are permitted to do with the representations of computations. (Parallel-or is a simple example of this phenomenon, since it exhibits a difference between LC and TMs.) Moreover, there isn't even a strict inclusion between the different models, since the contravariance of the function space means that more expressive power at one order implies less expressive power one order higher. - As far as I know, lambda calculus is ill suited for this purpose, as the notion of time/space complexity is hard to formulate in lambda calculus. What is 1 unit of time complexity? A beta reduction? What about the units of space complexity? The length of the string? Lambda calculus is more suitable for abstract manipulation of algorithms, as it is much more readily composable than Turing machines. - You could also look up calculi of explicit substitutions which break up the meta-level substitution of the lambda-calculus into a series of explicit reduction steps. This touches on Charles' point that all substitutions should not be considered the same when considering time complexity. - See Nils Anders Danielsson, Lightweight Semiformal Time Complexity Analysis for Purely Functional Data Structures which is implemented as a library in Agda. The citations given in the paper also look very promising. One key takeaway for me is that it is appropriate/useful/reasonable/semi-automatable to derive the time complexity of algorithms in the simply typed lambda calculus especially if those algorithms are easily expressible in it (i.e. purely functional) and very especially if those algorithms make essential use of, e.g., call-by-name semantics. Along with this is the probably obvious point that one does not calculate complexity just "in the lambda calculus" but in the lambda calculus under a given evaluation strategy. -	{"extraction_info": {"found_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": 0.927821695804596, "perplexity_flag": "middle"}
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Is the degree of the Hessian \$H_f = V(\det ... 1answer 412 views ### Find point on sphere with directional tangent vector Say a sphere equation like this: $x^2+y^2+z^2=5$. I want to find a point on the sphere whose tangent vector is perpendicular to the vector $\begin{bmatrix} 2\\ 3\\ 4 \end{bmatrix}$. I go ... 1answer 54 views ### Can any smooth planar curve which is closed, be a base for a 3 dimensional cone? A cone in 3 dimensions has a vertex and a base. The contour of the base is a circle which is a smooth closed planar curve. Can there be a more general cone which can have any smooth closed planar ... 1answer 184 views ### References for the basic theory of surfaces of revolution, cylinders and cones I'm looking for references to books were the following types of problems about finding the equation defining a surface of revolution, a cylinder or a cone are treated. These are problems that are ... 1answer 369 views ### Locus of osculation of concentric ellipses (elliptic pond ripples) If you dropped two rocks in a pond, the concentric circles emanating from the two spots would osculate $\infty$ times. The locus of osculating points would form a line. Now imagine that instead of ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 34, "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": 0.9128078818321228, "perplexity_flag": "head"}
http://mathoverflow.net/questions/56603?sort=newest	Quasi-isometries vs Cayley Graphs Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The following questions might be trivial, however, I couldn't solve them: Let $G$ be generated by a finite symmetric set $S.$ Suppose that $\Gamma(G,S)$ is the corresponding right Cayley graph of $G.$ $X$ is a metric space(or, maybe a topological space with some nice structure). (1) Is there a way to check the following property of a space $X:$ $X$ is not quasi-isometric to a space $Z$ which is quasi-isometric to a(hence, every) Carley graph $\Gamma(G,S)$ of some f.g group $G.$ I.e, If we partition the space of spaces upto quasi-isometric equivalence then does every equivalence class contain a space which is quasi-isometric to a Cayley graph of some f.g group $G?$ (2) By Stalling's theorem, # of ends is a geometric property of the group. Does this mean that # of ends is a quasi-isometric invariant of the spaces which are quasi-isometric to Cayley graphs? If the answer of question #2 is affirmative and equivalence class question above fails; i.e, there is an equivalence class whose elements are not quasi-iso. to any Cayley graph ; then what is the example of spaces $W_1, W_2$ which are not quasi-iso. to any Cayley graph but $W_1$ is quasi-isometric to $W_2$ ,however, # of ends of $W_1$ is different than the # of ends of $W_2.$ Thank you. - 1 There are a lot of conditions missing in order for these questions to make a little sense. Make the spaces geodesic, "almost homogeneous" and of bounded geometry for a start. – Theo Buehler Feb 25 2011 at 7:43 Could you precise what definition of the number of ends of a metric space you use? – Benoît Kloeckner Feb 25 2011 at 7:56 @Buehler: Yes, I couldn't figure out the minimal conditions on $X.$ I am interested in the invariance of number of ends. And the end is the number of components of $X-B(n)$ as n->\infity. I know that it doesn't make sense for general metric spaces. But, according to answer given below, this number should make sense for a large collection of spaces also. – Niyazi Feb 25 2011 at 9:32 1 @Niyazi: the problem is that the number of ends defined in your way is infinite for an unbounded discrete space (in particular it sis far from being a quasi-isometric invariant, even for spaces quasi-isometric to a Cayley graph), so you should really make explicit the conditions on $X$, as asked by Theo Buehler. – Benoît Kloeckner Feb 25 2011 at 15:43 2 Niyazi, Stallings's Theorem does not say that the number of Ends is a quasi-isometric invariant of the group. It gives a condition under which a fg group has infinitely many ends. – HW Feb 25 2011 at 16:07 2 Answers I guess that a star (a tree with $n$ infinite branches issued from a single vertex) should answer at least your first question. It should have $n$ ends, whatever meaningful definition you use, an we know that a group has $1$, $2$ or an infinity of ends. Since quasi-isometry is an equivalence relation, you do not need to invoke a space $Z$ in your first question and the answer of your second question is obviously positive. - 1 My comment was intended to exclude such examples as you give in the first paragraph (among other things). With any of the definitions I know, $n$ parallel lines in $\mathbb{R}^2$ with the Euclidean metric will have $2n$ ends. – Theo Buehler Feb 25 2011 at 8:24 @Kloeckner: So, if $Z$ is quasi-isometric to 5-star then can we conclude that Z has 5 ends? I know that the number of ends is a quasi-iso. invariant only for cayley graphs, is it again an invariant of non-Cayley but quasi-isometric spaces? Also, according to your answer examples of $W_1$ and $W_2$ exist. Is it correct? Thank you. – Niyazi Feb 25 2011 at 9:24 Benoit answers both questions. E.g. the union of the $x$ and $y$ axes in the plane is a metric space with 4 ends. It is therefore not q-i to any group (which must have 0, 1, 2, or infinity ends). It is also not q-i to the union of the $x$, $y$, and $z$ axes in $R^3$, which has 6 ends. In other words, yes, the number of ends is a q-i invariant of metric spaces, not just of groups. (It is easy to prove: a q-i induces a bijection on the ends.) – aaron Feb 25 2011 at 12:45 1 @aaron: Either you're working with a coarse notion of ends which I don't know or you're a bit simplistic here. I agree with your argument for proper geodesic spaces. Could you please elucidate your argument with a precise definition of your notion of an end? With the usual topological definition of an end, a quasi-geodesic doesn't need to lie in a single end, see my first comment to this answer. – Theo Buehler Feb 25 2011 at 14:10 1 Oh right, certainly one needs a coarse notion of ends. The integers should have 2 ends regardless of whether or not you draw in the edges connecting $n$ to $n+1$. – aaron Feb 25 2011 at 21:39 show 4 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. There was a conjecture by Woess that every infinite vertex-transitive graph is quasi-isometric to a Cayley graph. A slightly more sophisticated counter-example for your first question is the counter-example to this conjecture that was proposed by R.Diestel and I. Leader in "A conjecture concerning a limit of non-Cayley graphs". It was later proved by A. Eskin, D. Fisher, and K. Whyte in "Quasi-isometries and rigidity of solvable groups". - @Zaimi: Does it make sense to define number of edges for Diestel-Leader graphs,or, for their quasi-isometric copies? Is this number again invariant? – Niyazi Feb 25 2011 at 12:40 Do you mean "ends" for D-L graphs? If so then they have infinitely many ends, which is clear from their construction as horocycle products of homogeneous trees. – Gjergji Zaimi Feb 25 2011 at 13:24	{"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": 38, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.934066116809845, "perplexity_flag": "head"}
http://mathoverflow.net/questions/114020/how-to-understand-the-harish-chandra-isomorphism/114029	## How to understand the Harish-Chandra isomorphism? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The Harish-Chandra isomorphism describe the center $Z(\mathfrak{g})$ of $U(\mathfrak{g})$ as invariants of $\text{Sym}^\mathfrak{h}$ under the action of the Weyl group. (One need to twist the action or make a change of coordinate on the affine space `$\mathfrak{h}^$` for this isomorphism work.) My question is, how does one understand this isomorphism, and what's the geometric context of it? Why should one expect something like this might be true? - @36min just notifier - answered your quest on mathoverflow.net/questions/60108/… – Alexander Chervov Nov 21 at 11:10 ## 2 Answers Harish-Chandra isomorphism is defined for semisimple algebras. Let me try to give informal explanation for classical simple Lie algebras, take gl_n as main example. We are interested in the center of U(gl_n) - the same as invariants of gl in that space. First let us look on S(gl_n) - symmetric algebra of gl_n - as a vector space and moreover as module of g, it is isomorphic to U(gl_n). So you need to describe the invariants of S(gl). Same as invariants of group GL on the space of all matrices where GL acts by conjugation Actually everybody knows the answer - invariants are coefficients of the characteristic polynomial ! Coefficients of the char.pol are symmetric functions of eigenvalues - so you get the S(gl)^GL = Sym(h^)^W. (As for me this simple fact is main "motivation" for HC-isomorphism). Now, what is NON-obvious ? We need to pass from S(gl) to U(gl) and this passing is actually related to twisting the action Weyl group by shift on rho/2. I do not think there is simple explanation of this, but there is some general context which explains at least why ZU(g) is isomorphic as commutative algebra to S(g)^g. Moreover for any Lie algebra "g" not just semisimple one. This is what is called Duflo isomorphism. Moreover it was generalized by M. Kontsevich to quantization of arbitrary Poisson manifold. So roughly speaking the "general context" is the following - "quantization is GOOD" i.e. it many algebraic structures which we can see on the classical level of Poisson algebras can be transfered to quantum level isomorphically. This is formalized by "formality theorems" like Kontsevich one, and later ones. Now what about why the Weyl group becomes shifted ? I do not know the answer - it is clear how one can guess this - just take Casimir for sl(2) and you will see it. But conceptual meaning is not clear for me. If it is clear for someone - then please would you be so kind to comment what happens in the case of affine gl ? See my question: http://mathoverflow.net/questions/93626/harish-chandra-isomorphism-for-loop-algebras-is-the-image-invariant-with-respect Similar reasons works for other classical semisimple Lie algebras, and may be not only for classical. PS There is generalization of Harish-Chandra isomorphism http://arxiv.org/abs/0912.1100 A generalized Harish-Chandra isomorphism Sergey Khoroshkin, Maxim Nazarov, Ernest Vinberg (to confess S. Khoroshkin was my supervisor). - ### 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. Here is one algebraic/representation theory perspective on why such a morphism might exist, although I don't think historically this is the way it went at all. Let's admit for a moment that one might be interested in Verma modules. It is easy to see that the center acts via scalars on these. Verma modules are parametrized by $\mathfrak{h}^$ and in this way one gets an algebra morphism $Z(\mathfrak{g}) \to Sym(\mathfrak{h})$. Now it is also relatively straightforward to see (if I remember correctly) that for each simple reflection $s$, the Verma module $M(s\cdot \lambda)$ occurs as a submodule of the Verma module $M(\lambda)$, for $\lambda$ integral. It follows that the algebra morphism constructed earlier lands in $W$-invariants. - Verma modules are parametrized by `$\mathfrak{h}^∗$` and in this way one gets an algebra morphism `$Z(\mathfrak{g})\to \text{Sym}(\mathfrak{h})$`... How do you see one gets a regular function on `$\mathfrak{h}^*$`? – 36min Nov 21 at 5:41 Hmmm, one needs a tiny bit more knowledge about Verma modules plus the triangular decomposition: the triangular decomposition for the enveloping algebra shows that on a highest weight vector an element in the center acts as the weight that the Verma corresponds to evaluated on the Cartan piece appearing in the triangular decomposition of the element. Thus, it's a regular function. – Reladenine Vakalwe Nov 21 at 5:54 In the comment above, by Cartan piece' I really mean purely Cartan piece'. I hope what I mean by this is clear? – Reladenine Vakalwe Nov 21 at 5:56	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9105588793754578, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/231848/prove-that-a-function-is-a-total-function-as-opposed-to-a-partial-function/236121	# Prove that a function is a total function (as opposed to a partial function) I've got as part of an assignment to determine whether a given function is total, and if so, to say whether it's injective, surjective, or bijective. I can tell the answer by looking at it, but I feel that I need to make some kind of concrete proof that the function is total. $$f:\mathbb{N}\to \mathbb{N},f(x)=x^2+4$$ I can tell by looking at it that it's a one-to-one total function. But is there a satisfactory way I can prove that this is a total function? I'm extremely new to this whole idea of functions outside of algebraic functions (such as total and partial functions)... - What is your definition of "partial function"? – wj32 Nov 7 '12 at 2:32 – agent154 Nov 7 '12 at 2:34 1 If your definition of partial function is the usual one ($f\subseteq X\times Y$ is a partial function from $X$ to $Y$ if $f$ is a function from some subset of $X$ to some subset of $Y$), then this $f$ is clearly total: it’s defined on all of $\Bbb N$. – Brian M. Scott Nov 7 '12 at 2:35 @agent154: I don't think I made my point very well :). What I was suggesting was for you to go back to the definition and see why it is so easy to "tell the answer by looking at it". – wj32 Nov 7 '12 at 2:39 Or in other words: is it true that $f(n)$ is defined for every $n\in\Bbb N$? – Brian M. Scott Nov 7 '12 at 2:41 show 4 more comments ## 2 Answers To formally prove that it's a total function, you need to look up the definitions/basic theorems about $\mathbb{N}$ and multiplication (or squaring) and addition, and check that 1. $x^2+4$ has a fixed value for each $x\in\mathbb{N}$. 2. That value is actually in $\mathbb{N}$. (This is something to check if your definitions only ensure it's in $\mathbb{Z}$, or something like that.) Since I don't have your definitions, and you may not even have definitions for some of this stuff, I can't help you further except to say "if you don't have definitions for something, just argue intuitively for that part". - That does seem a little beyond me at this point... But then again, this whole question was beyond what I should have been doing at this point in the class. This professor likes to ask questions on assignments for which he hasn't even covered the material. I ended up passing in the assignment last week and forgot to mark this question as answered. Thanks anyhow though. – agent154 Nov 13 '12 at 1:56 Well, you're probably not expected to worry about formally proving it completely rigorously at this level. But you seemed worried about rigor, so I described what more rigor would be like. If you feel this question is answered you can either accept my answer or add an answer of your own and accept that (since you had an answer first), or I think you can vote to delete your question if you really want. – Mark S. Nov 13 '12 at 1:59 I went with Brian's suggestion from above comments: You have the idea. How about: Let $n\in \mathbb{N}$. $\mathbb{N}$ is closed under multiplication and addition, so $f(n)=n^2+4\in \mathbb{N}$. Therefore f is total. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 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": 0.9712137579917908, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/150025-why-set-subspace.html	# Thread: 1. ## Why is this set a subspace ?? Why is the following set a subspace of $M_{n \times n}^R, n \geq 2$ $B = \{A \in M_{n \times n}^C \| A_\downarrow_1 = (1-2i)A_\downarrow_2 \}$ I immediately said it is not a subspace since complex numbers aren't in real numbers. It's the other way around. But I am wrong apparently... 2. Originally Posted by jayshizwiz Why is the following set a subspace of $M_{n \times n}^R, n \geq 2$ $B = \{A \in M_{n \times n}^C \| A_\downarrow_1 = (1-2i)A_\downarrow_2 \}$ I immediately said it is not a subspace since complex numbers aren't in real numbers. It's the other way around. But I am wrong apparently... This is NOT a vector space over the real numbers but it IS a vector space over the complex numbers, Just as the original space, $M_{nxn}^C$ is a vector space over the complex numbers, not the real numbers. 3. This is NOT a vector space over the real numbers but it IS a vector space over the complex numbers, Just as the original space, is a vector space over the complex numbers, not the real numbers. So is it a mistake that the question says it is a subspace over $M_{n \times n}^R$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 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": 0.9360498189926147, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/106579/list	## Return to Answer 3 added 1 characters in body Your example of $f(x,y)=\left(y-T_5(x)\right)\left(x-T_5(y)\right)$ is ```$$ 256\,{x}^{5}{y}^{5}-320\,{x}^{5}{y}^{3}-320\,{x}^{3}{y}^{5}-16\,{x}^{6 }+80\,{x}^{5}y+400\,{x}^{3}{y}^{3}+80\, x{y}^{5}\\-16\,{y}^{6}+20\,{x}^{4 }-100\,{x}^{3}y-100\,x{y}^{3}+20\,{y}^{4}-5\,{x}^{2}+26\,xy-5\,{y}^{2} $$``` If I count correctly, it has $28$ regionswould . Would that qualify as maximum degree $10?$ If so, then as Pietro points out, $g(x,y)=\prod_{i=1}^{10}\left(a_ix+b_iy+c_i\right)$ will have $56$ regions (if $a_i,b_i,c_i$ are such that the $10$ lines are in general position: no two parallel and no three meeting at a common point). Similar things (as he says) can be done with hyperplane arrangements in higher dimension. You can color the regions $g /\gt 0$ white and $g \lt 0$ black so that each region is bounded by regions of the opposite color. If you want the curve itself to have many disjoint connected components then $g(x,y)+\epsilon$ and $g(x,y)-\epsilon$ are nice to look at. Then all the regions of one color fuse together but each of the others becomes a nicely bordered region. I think that (in the two variable case with a projective viewpoint, at any rate) these achieve that bound given by Harnack's theorem. 2 added 376 characters in body Your example of $f(x,y)=\left(y-T_5(x)\right)\left(x-T_5(y)\right)$ is ```$$ 256\,{x}^{5}{y}^{5}-320\,{x}^{5}{y}^{3}-320\,{x}^{3}{y}^{5}-16\,{x}^{6 }+80\,{x}^{5}y+400\,{x}^{3}{y}^{3}+80\, x{y}^{5}\\-16\,{y}^{6}+20\,{x}^{4 }-100\,{x}^{3}y-100\,x{y}^{3}+20\,{y}^{4}-5\,{x}^{2}+26\,xy-5\,{y}^{2} $$``` If I count correctly, it has $28$ regions would that qualify as maximum degree $10?$ If so, then as Pietro points out, $g(x,y)=\prod_{i=1}^{10}\left(a_ix+b_iy+c_i\right)$ will have $56$ regions (if $a_i,b_i,c_i$ are such that the $10$ lines are in general position: no two parallel and no three meeting at a common point). Similar things (as he says)can says) can be done with hyperplane arrangements in higher dimension. You can color the regions $g /gt 0$ white and $g \lt 0$ black so that each region is bounded by regions of the opposite color. If you want the curve itself to have many disjoint connected components then $g(x,y)+\epsilon$ and $g(x,y)-\epsilon$ are nice to look at. Then all the regions of one color fuse together but each of the others becomes a nicely bordered region. I think that (in the two variable case with a projective viewpoint, at any rate) these achieve that bound given by Harnack's theorem. 1 Your example of $f(x,y)=\left(y-T_5(x)\right)\left(x-T_5(y)\right)$ is ```$$ 256\,{x}^{5}{y}^{5}-320\,{x}^{5}{y}^{3}-320\,{x}^{3}{y}^{5}-16\,{x}^{6 }+80\,{x}^{5}y+400\,{x}^{3}{y}^{3}+80\, x{y}^{5}\\-16\,{y}^{6}+20\,{x}^{4 }-100\,{x}^{3}y-100\,x{y}^{3}+20\,{y}^{4}-5\,{x}^{2}+26\,xy-5\,{y}^{2} $$``` If I count correctly, it has $28$ regions would that qualify as maximum degree $10?$ If so, then as Pietro points out, $g(x,y)=\prod_{i=1}^{10}\left(a_ix+b_iy+c_i\right)$ will have $56$ regions (if $a_i,b_i,c_i$ are such that the $10$ lines are in general position: no two parallel and no three meeting at a common point). Similar things (as he says)can be done with hyperplane arrangements in higher dimension. If you want the curve itself to have many disjoint components then $g(x,y)+\epsilon$ and $g(x,y)-\epsilon$ are nice to look at.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 31, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9675045013427734, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/263101/prove-every-odd-integer-is-the-difference-of-two-squares/263108	# Prove every odd integer is the difference of two squares I know that I should use the definition of an odd integer ($2k+1$), but that's about it. Thanks in advance! - Hi, papercuts. You might want to start thinking about upvoting answers that are helpful (click on the upward arrow (grey) to the left of the answer. You can also accept one answer, which you can do by clicking on the "greyed-out" check-mark to the left of the answer you want to accept. upvoting and accepting answers encourages people to take the time needed to answer your questions, and is a way to show appreciation. – amWhy Jan 16 at 2:59 Ohhhhhh I see, gotcha thanks for the tips yo! – papercuts Jan 16 at 3:01 ## 4 Answers Step 1: pick an odd number (like $n=13$ here) Step 2: bend it in "half" (any odd number $n$ can be written as $2k+1$, and $13=2\cdot 6 + 1$) Step 3: fill in the blank space Step 4: Count squares. (Here, the blue square has area $36=6^2$, while the whole square has area $49=7^2$) - 20 +1, very nice illustration. – Eric♦ Dec 21 '12 at 8:27 +1 Very very nice! – Nameless Dec 21 '12 at 8:33 + Very nice colour – Grijesh Chauhan Dec 21 '12 at 12:09 6 +1 nice straight lines and corners, no wobbliness or smudges. – Graham Borland Dec 21 '12 at 13:53 1 It's great to see a visual presentation. I never thought about it this way. – krikara Dec 21 '12 at 22:32 show 1 more comment Hint: Consider the difference of two consecutive squares. What is $(k+1)^2-k^2$? - 1 Oh, haha well that was easy, but how'd you know to use consecutive squares? – papercuts Dec 21 '12 at 7:57 9 @papercuts Probably mostly experience. There's not much substitute for it. – Alex Becker Dec 21 '12 at 8:00 2 It's also one of the simplest sorts of differences of two squares to analyze. – Hurkyl Dec 21 '12 at 9:35 3 @papercuts Try it with the difference of any two squares: $(k+n)^2-k^2$. Then let n=1. – Griffin Dec 21 '12 at 18:10 HINT: $$\begin{align} &2k + 1 \\= & 1\cdot(2k + 1) \\ =& \left(k + 1 - k \right)\left(k + 1 + k\right) \\ = & \cdots\end{align}$$ - Eric and orlandpm already showed how this works for consecutive squares, so this is just to show how you can arrive at that conclusion just using the equations. So let the difference of two squares be $A^2-B^2$ and odd numbers be, as you mentioned, $2k+1$. This gives you $A^2-B^2=2k+1$. Now you can add $B^2$ to both sides to get $A^2=B^2+2k+1$. Since $B$ and $k$ are both just constants, they could be equal, so assume $B=k$ to get $A^2=k^2+2k+1$. The second half of this equation is just $(k+1)^2$, so $A^2=(k+1)^2$, giving $A = ±(k+1)$, so for any odd number $2k+1$, $(k+1)^2-k^2=2k+1$. - 1 Re "Since B and k are both just constants, they can be treated as equal": no: you already fixed both. – msh210 Dec 21 '12 at 15:30 @msh210 Neither of them are "fixed" yet; if they were, they would have set values and wouldn't be very usable for proofs (and, barring certain specific values, would probably be referred to with that value instead). I edited my answer to try and clarify (since it's more of an assumption than a known). – MyNameIsNotMcThomasJohannson Dec 21 '12 at 15:37	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "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": 0.9423003196716309, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/10933/why-do-we-say-that-the-earth-moves-around-the-sun	# Why do we say that the earth moves around the sun? In history we are taught that the Catholic Church was wrong, because the Sun does not move around the Earth, instead the Earth moves around the Sun. But then in physics we learn that movement is relative, and it depends on the reference point that we choose. Woulnd't the Sun (and the whole universe) move around the Earth if I place my reference point on Earth? Was movement considered absolute in physics back then? PS: I'm not religious - Historian are not physicists. Why would you search physic in history books? on the other hand physics wants to be absolute in some sense (read answer of physics.stackexchange.com/questions/10078/…) Theory of Relativity, is a physics theory and (besides it works pretty well) it has in its assumptions a constant light speed and the principle of relativity "physics is the same for any frame of reference" , as long as measures fit, we can suspect that perhaps those premise are not so bad. – HDE Jun 9 '11 at 12:15 In classical physics there is the notion of an inertial frame of which the sun is a good approximation generally, and the earth only sometimes. It all depends on what you want to measure in the first place. – ja72 Jun 9 '11 at 13:40 Actually the entire universe rotates about me - but nobody else seems to see that! – Martin Beckett Aug 26 '12 at 20:54 ## 11 Answers Imagine two donut-shaped spaceships meeting in deep space. Further, suppose that when a passenger in ship A looks out the window, they see ship B rotating clockwise. That means that when a passenger in B looks out the window, they see ship A rotating clockwise as well (hold up your two hands and try it!). From pure kinematics, we can't say "ship A is really rotating, and ship B is really stationary", nor the opposite. The two descriptions, one with A rotating and the other with B, are equivalent. (We could also say they are both rotating a partial amount.) All we know, from a pure kinematics point of view, is that the ships have some relative rotation. However, physics does not agree that the rotation of the ships is purely relative. Passengers on the ships will feel artificial gravity. Perhaps ship A feels lots of artificial gravity and ship B feels none. Then we can say with definity that ship A is the one that's really rotating. So motion in physics is not all relative. There is a set of reference frames, called inertial frames, that the universe somehow picks out as being special. Ships that have no angular velocity in these inertial frames feel no artificial gravity. These frames are all related to each other via the Poincare group. In general relativity, the picture is a bit more complicated (and I will let other answerers discuss GR, since I don't know much), but the basic idea is that we have a symmetry in physical laws that lets us boost to reference frames moving at constant speed, but not to reference frames that are accelerating. This principle underlies the existence of inertia, because if accelerated frames had the same physics as normal frames, no force would be needed to accelerate things. For the Earth going around the sun and vice versa, yes, it is possibly to describe the kinematics of the situation by saying that the Earth is stationary. However, when you do this, you're no longer working in an inertial frame. Newton's laws do not hold in a frame with the Earth stationary. This was dramatically demonstrated for Earth's rotation about its own axis by Foucalt's pendulum, which showed inexplicable acceleration of the pendulum unless we take into account the fictitious forces induced by Earth's rotation. Similarly, if we believed the Earth was stationary and the sun orbited it, we'd be at a loss to explain the Sun's motion, because it is extremely massive, but has no force on it large enough to make it orbit the Earth. At the same time, the Sun ought to be exerting a huge force on Earth, but Earth, being stationary, doesn't move - another violation of Newton's laws. So, the reason we say that the Earth goes around the sun is that when we do that, we can calculate its orbit using only Newton's laws. In fact, in an inertial frame, the sun moves slightly due to Earth's pull on it (and much more due to Jupiter's), so we really don't say the sun is stationary. We say that it moves much less than Earth. (This answer largely rehashes Lubos' above, but I was most of the way done when he posted, and our answers are different enough to complement each other, I think.) - 2 In Your 1st paragraph You should define what rotation is meant.(Around which axis) The donut shape of the craft lures me to think of rotation about the donuts symmety axis. – Georg Jun 9 '11 at 9:43 3 @Georg That is what I meant. – Mark Eichenlaub Jun 9 '11 at 14:24 A much more scientifically correct answer that Lubos', imho. Thanks for posting it. – KPM Jan 11 at 22:49 yes, you may describe the motion from any reference frame, including the geocentric one, assuming that you add the appropriate "fictitious" forces (centrifugal, Coriolis, and so on). But the special property of the reference frame associated with the Sun - more precisely, with the barycenter (center of mass) of the Solar System, which is just a solar radius away from the Sun's center - is that this system is inertial. It means that there are no centrifugal or other inertial forces. The equations of physics have a particularly simple form in the frame associated with the Sun. $$M_1 d^2 / dt^2 \vec x = G M_1 M_2 (\vec r_1-\vec r_2) / r^3 + \dots$$ There are just simple inverse-squared-distance gravitational forces entering the equations for the acceleration. For other frames, e.g. the geocentric one, there are many other inertial/centrifugal "artificial" terms on the right hand side that can be eliminated by going to the more natural solar frame. In this sense, the heliocentric frame is more true. - 5 @Lubosh: "In this sense, the heliocentric frame is more true." I don't agree with the phrasing of this statement - it is more convenient perhaps, but it is equally as 'true' as any other choice of reference frame. If the geocentric equations of motion correctly predict the motion of all celestial bodies then surely they too are 'true,' albeit a little more complicated. – qftme Jun 9 '11 at 9:24 5 @qftm: little more complicated? By choosing arbitrary coordinates you can get arbitrarily huge complexity in description! On the other hand, there is a certain minimum bound on complexity you can achieve in certain nice systems and this minimum is attained precisely in inertial frames. So, yes, these frames are natural and canonical. I don't find anything strange in Luboš's formulation. – Marek Jun 9 '11 at 13:14 2 @Marek: Complexity of the equations isn't really the issue here. I was merely contending the use of the phrase "more true". In that I thought 'more convenient', 'more sensible' or 'a more natural choice' would be a more physically correct statement. – qftme Jun 9 '11 at 13:34 5 Fine, @qftme, but you may make any statement in science equally relative. For example, one may describe the origin of species by God creating the world 6,000 years ago including all the fossils whose distribution happened to be dictated by the same patterns as if the fossils were leftovers from some insanely long pre-genesis history of literally billions of years. Both models are by construction equivalent. It follows that creationism is on par with evolution, doesn't it? ;-) Well, it's not. In science, if one may undo a simple transformation to get a more uniform description, one does it. – Luboš Motl Jun 9 '11 at 15:12 1 – AlanSE Jun 13 '11 at 4:14 This was going to be a comment on Luboš Motl's answer, but it would be more appropriate as a full answer now. His answer says: Laws of physics can be written more simply for the solar system's center of mass (barycenter) than for a point on Earth (geocentric). Just one thing! One mustn't neglect the non-idealities of the barycenter itself, which has a location in the Milky Way that biases it gravitationally at least. On the surface this is splitting hairs, but the greater point is that the idealness of any reference frame is also relative, and no "ultimate" frame exists. Likewise, choosing a point on the skin of an elephant over a geocentric point is sacrificing universality just as much as choosing a geocentric point over the barycenter is. To a flee however, consideration of physics formulated at a point beyond the surface of the elephant may be just "academic". Sound familiar? - Yes, the proposition: "the sun moves around the earth" had the earth immobile. This suited the theology of the times which was completely anthropocentric and that is why it prevailed over other theories coming from antiquity, like Aristarchos', who had a heliocentric proposal. The relativity of motion was explored, as Lubos describes, when equations could be written down, and one chooses the heliocentric for its beauty and simplicity. The epicycles exist if one plots the solutions in a geocentric system, but they are so cumbersome and "ugly" as a shorthand of physics. - Well, epicycles are just a form of describing motion as a superposition of circular orbits, so something like a simple representation in Fourier space -- indeed Copernicus used them in his original heliocentric theory to compensate the eccentricity of orbits. – mbq♦ Jun 9 '11 at 7:06 3 – Mark Eichenlaub Jun 9 '11 at 7:40 @Mark: Great video-link. :: chuckles :: – qftme Jun 9 '11 at 9:38 @Mark and @mbq the video was fun, but do keep in mind that the epicycles appear as solutions of the graviational equations anyway, when they are transformed to the geocentric system, one to one correspondence. It is not an approximation. I first became clear of this when discussing planetarium models, and somebody who had a program for the solar system, showed the epicycles by changing the coordinate system. – anna v Jun 9 '11 at 10:28 @annav An an experimentalist, surely you can appreciate the good reason everyone had for not accepting heliocentrism. It had nothing to do with theology, as the Greeks themselves rejected Aristarchos well over 2000 years ago, noting that his theory made a prediction - stellar parallax - which was simply not observed. Theoretically, heliocentrism only made sense after Newton, who himself came after Galileo and Copernicus, and experimentally it was only directly confirmed in the early 19th century. – Chris White Feb 26 at 8:17 show 1 more comment There may be a confusion : it is wrong to say that the Earth is the centre of the Universe, that is, the (unique) point from which the Universe is to be (fundamentally) described (the fact that the Sun turns around the Earth is only a consequence of this) ; what actually matters is that there is no centre of the Universe : there is no such point ; the description of the Universe from any point is equivalent to the description of the Universe from any other (then you are allowed to describe motions either from the Earth or from the Sun). Mathematically, in classical mechanics, the Universe is said to be an affine space. - Both Sun and Earth move in circles around their barycenter i.e. centre of mass. The trick is that since Sun is too massive, the center of mass is too close to the sun, actually beneath the surface of the Sun, which makes the motion of Sun negligible. And, we say that Earth moves around the Sun. - The sun, moon, earth (and so on) all move around each other. The reason we say the earth moves around the sun is because the effects are more visible on a macro scale, and easier to predict with reasonable precision. Yes, it's most correct to say that all motion is relative, but it gets a lot more complicated to explain it if you're speaking to a layman. - Actually, ‘move around each other’ is a misleading phrase. They don't move around each other (in the sense, the Sun moves around the Earth and the Earth moves around the Sun), they move around their barycentre. Caution about language-induced pitfalls! – KPM Jan 11 at 22:53 In ancient times, the mechanics of orbital motion due to gravitational attraction wasn't known. What was known, though, was that if Earth orbits Sun, then stars would display a cyclical motion called "parallax." The Greeks actually predicted this, but didn't have the technology to observe it. This was a main reason the geocentric solar system model stood for so long. The parallax is real though, and observable, and provides direct observational evidence that Earth orbits Sun. - There are experimental evidences of absolute motion of the Earth around the Sun. There is a dipole anisotropy in fine measures of the Background Radiation temperature that is known from the analysis of the COBE satellite measures, in the early 90s. See for instance this paper. In order to make the adequate corrections, so that the Cosmic Background Radiation "seems" isotropic, the absolute velocity of the Local Group against the Cosmic Background Radiation must be accounted for, but that correction depends on the month of the year, because a small part of the correction comes from the orbital speed of the Earth around the barycentre of the Solar System (among other terms). That small part of the corrections needed is exactly what you would expect if you assumed that is the Earth who is going around the Sun, and not vice versa. (the cosmic background dipole anisotropy, image from map.gsfc.nasa.gov) Here is an extract from the abstract of the quoted paper: We present a determination of the cosmic microwave background dipole amplitude and direction from the COBE Differential Microwave Radiometers (DMR) first year of data (...) The implied velocity of the Local Group with respect to the CMB rest frame is $v_{LG}=627 \pm 22 km s^{-1}$ toward (...). DMR has also mapped the dipole anisotropy resulting from the Earth's orbital motion about the Solar System barycenter, yielding a measurement of the monopole CMB temperature (...) $T_0=2.75 \pm 0.05 K$ This doesn't mean however, that there is an absolute reference frame in the Universe. Other comoving observers will detect another dipole anisotropy. The Last Scattering Surface, as well as the cosmological horizons are different for different comoving observers. But nevertheless it proves that it is the Earth that moves around the Sun, and not vice versa. Since the 90s this is no more a philosophical issue: WE are going certainly, absolutely, surely and gloriously, around the Sun. - – Eduardo Guerras Valera Feb 26 at 3:55 The argument given by Luboš is essentially the same as Copernicus, namely that the calculations are much simpler for a Sun-centred reference frame. I am talking however about a very different concept, this is an observational evidence, for the first time in history, but it has not yet transcended to the general public or even to scientist of other branches, because it was eclipsed by the confirmation of the Big Bang, that happened too because of the COBE measures of the CMB spectrum. – Eduardo Guerras Valera Feb 26 at 4:50 Nobody has paid attention to the philosophical implications of the orbital terms in the CMB dipole correction, but in essence, it is the first empirical confirmation in centuries, that Copernicus was right, independent of Occam's Razor or any other reasoning. It makes no practical difference, but philosophically it is very deep. – Eduardo Guerras Valera Feb 26 at 4:56 The statement it is the Earth, not the Sun, that moves with respect to the background stars was already shown observationally in 1838 (I think that's the year) with stellar parallax. Thus the CMB evidence is most profound only for those who believed the stars were conspiring to imitate motion in an annual cycle, but that the surface of last scattering can be trusted to not have yearly motion of its own. – Chris White Feb 26 at 8:25 Well, 1838 is the date given in Carrol & Ostlie for Bessel's first parallax measure, but it is only a bunch of very nearly stars until photographic plates in the first years of the XX century were used. In any case, yes, stellar parallax is another evidence, but the question is that whoever believed the outer planets going around the Sun and the Sun going around the Earth, would probably have no trouble adding nearby stars too, and the CMB is sort of a global evidence, it is the whole Universe, and that makes a philosophical difference: to consider the whole universe going around the Earth – Eduardo Guerras Valera Feb 26 at 8:34 show 3 more comments Every experiment ever designed to detect the motion of the Earth has failed to detect earth's motion and/or distinguish it from relative counter motion of the Universe. The Michelson-Morley experiment for one: http://2.bp.blogspot.com/-_17XBXum_q4/TrNpb9tJ2sI/AAAAAAAAC8M/CgXR1PXr6do/s1600/392959_xlarge.jpg Additionally There was a group of early "scientists" that decided that their heliocentric pronouncements would be proved and accepted. They conducted various experiments to prove that the earth moves around the sun. Unsurprisingly, these experiments FAILED--and they knew it--but they covered this information up. False science could not and can not prove the motion of the earth. Those early false scientists are the authors of the textbooks that today's "scientists" and authors consult. – The Michelson-Gale experiment (Reference - Astrophysical Journal 1925 v 61 pp 140-5 - I forgot to put this reference in) This detected the aether passing the surface of the earth with an accuracy of 2% of the speed of the daily rotation of the earth! Thus, the Michelson-Morely experiment detected no movement of the earth around the sun, yet the Michelson-Gale experiment measured the earth's rotation (or the aether's rotation around the earth!) to within 2%! This surely speaks volumes for geocentricity. "Airey's failure" (Reference - Proc. Roy. Soc. London v 20 p 35). Telescopes have to be very slightly tilted to get the starlight going down the axis of the tube because of the earth's "speed around the sun". Airey filled a telescope with water that greatly slowed down the speed of the light inside the telescope and found that he did not have to change the angle of the telescope. This showed that the starlight was already coming in at the correct angle so that no change was needed. This demonstrated that it was the stars moving relative to a stationary earth and not the fast orbiting earth moving relative to the comparatively stationary stars. If it was the telescope moving he would have had to change the angle. The Sagnac experiment (Reference - Comptes Rendus 1913 v157 p 708-710 and 1410-3) Sagnac rotated a table complete with light and mirrors with the light being passed in opposite directions around the table between the mirrors. He detected the movement of the table by the movement of the interference fringes on the target where they were recombined. This proved that there IS an aether that the light has to pass through and this completely destroys Einstein's theory of Relativity that says there is no aether. It is for this reason that this experiment is completely ignored by scientists. More recently Kantor has found the same result with similar apparatus. All these experiments are never taught at universities, so consequently, scientists, and students are ignorant of this evidence for geocentricity. In order to save the dying heliocentric theory from the conclusive geocentric experiments performed by Michelson, Morley, Gale, Sagnac, Kantor and others, establishment master-mind Albert Einstein created his Special Theory of Relativity which in one philosophical swoop banished the absolute aether/firmament from scientific study and replaced it with a form of relativism which allowed for heliocentricism and geocentricism to hold equal merit. If there is no universal aetheric medium within which all things exist then philosophically – one can postulate complete relativism with regard to the movement of two objects (such as the Earth and sun) Nowadays, just like the theory of heliocentricism, Einstein's theory of relativity is accepted worldwide as gospel truth, even though he himself admitted geocentricism is equally justifiable! Finally, I quote Albert Einstein - "The struggle, so violent in the early days of science, between the views of Ptolemy and Copernicus would then be quite meaningless. Either coordinate system could be used with equal justification. The two sentences, 'the sun is at rest and the earth moves,' or 'the sun moves and the earth is at rest,' would simply mean two different conventions concerning two different coordinate systems." -Albert Einstein. If one accepts the unintuitive, but very imaginative heliocentric model, then one accepts (even though it goes against observation, experimental evidence and common sense) that the Earth is actually spinning around its axis at 1,000 miles per hour, revolving around the sun at 67,000 miles per hour, while the entire solar system rotates around the Milky Way galaxy at 500,000 miles per hour, and the Milky Way speeds through the known Universe at over 670,000,000 miles per hour! I'd like to know if we're really being subject to all those forces/motions, then why hasn't anyone in all of history ever felt it? How is it that all the centrifugal, gravitational, inertial and kinematic forces somehow cancel each other out perfectly so that no one has ever felt the slightest bit of motion or resistance? Why aren't there world-wide perturbations of our smooth rotation after earthquakes or meteor strikes? Why can I still feel the slightest breeze on my face, but not the air displacement from all this motion? If the Earth is spinning beneath us, why can't I just hover in a helicopter, wait until my destination reaches me, and then land when it comes? Most people answer (though they can't explain how) that this is because the Earth's atmosphere supposedly rotates precisely along with the Earth. But if that's the case then heliocentric dogmatists run into a whole other host of problems. For instance, if both the Earth and its atmosphere are spinning 1,000 miles per hour West to East, then why don't pilots need to make 1,000 mph compensation acceleration when flying East to West? If thousand mile per hour atmosphere is constantly flowing Eastward, why don't North/South bound pilots have to set diagonal courses to compensate? If thousand mile per hour atmosphere is constantly flowing Eastward, how do you explain the casual yet unpredictable movement of clouds, wind patterns and weather formations every which way? If the atmosphere is constantly being pulled along with the Earth's rotation, then why can I feel the slightest Westward breeze but not the Earth's 1,000 mile per hour Eastward spin? "In short, the sun, moon, and stars are actually doing precisely what everyone throughout all history has seen them do. We do not believe what our eyes tell us because we have been taught a counterfeit system which demands that we believe what has never been confirmed by observation or experiment. That counterfeit system demands that the Earth rotate on an 'axis' every 24 hours at a speed of over 1000 MPH at the equator. No one has ever, ever, ever seen or felt such movement (nor seen or felt the 67,000MPH speed of the Earth's alleged orbit around the sun ... or its 500,000 MPH alleged speed around a galaxy ... or its retreat from an alleged 'Big Bang' at over 670,000,000 MPH!). Remember, no experiment has ever shown the earth to be moving. Add to that the fact that the alleged rotational speed we've all been taught as scientific fact MUST decrease every inch or mile one goes north or south of the equator, and it becomes readily apparent that such things as accurate aerial bombing in WWII (down a chimney from 25,000 feet with a plane going any direction at high speed) would have been impossible if calculated on an earth moving below at several hundred MPH and changing constantly with the latitude." I HOPE THIS IS INSIGHTFUL AND HELPFUL FOR YOU :-) - 1 Is that you, Ron? – Robert Harvey Feb 25 at 22:56 Since we cannot accept the absence of a 1000 mph wind, or our 500,000 mph speed around the centre of the galaxy, we instead need to accept a much simpler consequence: the rest of the universe spins around us at far greater speeds. Any star further away than 1/3 light day is moving at more than lightspeed to make one orbit per day around the earth. And the nearest star is only about 5000 times further away than that, with correspondingly greater speed... – hdhondt Feb 26 at 0:23 @RobertHarvey, this is absolutely not RM. It's simply somebody that thinks very much but studies very little in books, therefore trying to re-derive by himself some three centuries of science. No wonder (s)he writes this. Somebody could give him/her simply the information (s)he is missing instead of downvoting: all that speed components (s)he mentions (in case they are true, I have not paid attention to the numers) are very nearly rectilinear due to the enormous curvature radii, and uniform in speed, so we feel no accelerations for them. That's all. – Eduardo Guerras Valera Feb 26 at 1:48 Not so due to the fact that no one can prove how far away a star is nor how far away the Sun is for that matter. There is not a measurement tool that can accurately perform this task. All estimations are speculations and are subjective at best. – The Steevee B Feb 26 at 1:50 @TheSteeveeB, dear friend, there are indeed very precise tools (like radar waves) to accurately measure distances in the inner Solar System and thus scale the models. Triangulation based on the AU leads to measures to nearby stars. Spectral measures compared to that nearby stars determinations allow to extend the range to the Galaxy. Intrinsic relation between Cepheids intrinsic bright and pulsation periods leads to determination of nearby galaxy distances. Linear Redshift-distance relation extends the measures up to some Mpc. Cosmological models allow further estimations. It is the so-called – Eduardo Guerras Valera Feb 26 at 1:59 show 13 more comments I'm putting down a much shorter answer: The sun doesn't move, the earth (together with the other revolving planets in our galaxy) does. The earth basically rotates around the sun in a ring (and its axis, but that's besides the point). Besides, the churches have always made many wrong claims (especially in the middle ages), such as that the earth was flat. They though so because we 'stand on top of the earth' whilst there really is no up or down. - 2 Welcome to SE.com but I'm afraid your answer contains a few errors. Grammar: "churches" implies the buildings whereas 'the Church' implies the organisation. Spelling: "though-t" ends with a 't'. Physics: the Sun rises in the East and sets in the West, thus it does move around the Earth (Galilaen relativity[1] is sufficient for this and has been understood since the 17th century,) and the Earth and Sun orbit around the barycenter[2] of the solar system. [1]: en.wikipedia.org/wiki/Galilean_invariance [2]: en.wikipedia.org/wiki/… – qftme Jun 9 '11 at 13:23 1 @qftme Galilean relativity doesn't apply to a rotating frame of reference – Random832 Jun 9 '11 at 15:14 @Random: You're right of course (+1). I mentioned it purely due to its explanation of how either of two reference frames moving relative to eachother may be considered to be 'at rest'. – qftme Jun 9 '11 at 15:25 ## protected by Qmechanic♦Feb 26 at 1:09 This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.	{"extraction_info": {"found_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": 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": 0.9447761178016663, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/186214/computing-an-integral	# Computing an integral. How to compute $\displaystyle\iint\limits_{\Sigma} z\sqrt{1+z^2}dS$, where $\Sigma=\Big\{(x,y,z)\Big\| \cfrac{x^2}{2}+\cfrac{y^2}{2}+z^2=1,z\ge 0\Big\}$? - ## 3 Answers Since in $\Sigma$, $z>0$ so our domain is $D:\sqrt{1-(1/2)x^2-(1/2)y^2}>0$ or $D: x^2+y^2<1$. Clearly $S: f=x^2+y^2+2z^2=2$ which is an ellipsoid which we regard just the upper part of it (top of the $x-y$ plane in $\mathbb R^3$). Now $$\displaystyle\iint\limits_{\Sigma} z\sqrt{1+z^2}dS=\displaystyle\iint\limits_{D}z\sqrt{1+z^2}d\sigma$$ where $d\sigma=\frac{\|\|\nabla f\|\|}{\|\partial f/\partial z\|}dxdy=\sqrt{1+z_x^2+z_y^2}dxdy$. I think you can evaluate the second double integral over $D$. - I don't think this is a good method. too much computing work to do. – Leitingok Aug 24 '12 at 14:30 @Leitingok: I wrote here what I had learnt about the surface integral. Your function in integrand is notable and it takes time to be done properly. :) – Babak S. Aug 24 '12 at 14:50 you are right. I worked out – Leitingok Aug 26 '12 at 11:00 @Leitingok: Just out of curiosity, I wonder whether you worked out a numerical value for your integral starting from the indications in Babak's post, and, if you did, what is this numerical value. – Did Aug 26 '12 at 11:38 1 @did: Thanks. As I have seen through questions, your answers have been of the neat and perfect ones. I love this way of doing Maths. I am just at the beginning of this way here. :) – Babak S. Aug 28 '12 at 13:55 show 5 more comments The surface $\Sigma$ has equation $x^2+y^2=r(z)^2$ for $0\leqslant z\leqslant1$, with $r(z)=\sqrt{2}\sqrt{1-z^2}$. For each $z$, this is the equation of a circle with radius $r(z)$, whose length is $2\pi r(z)$, hence the integral is $$\int_0^1z\sqrt{1+z^2}\cdot2\pi r(z)\cdot\mathrm dz.$$ The change of variable $z=\cos^2t$ yields the numerical value $\frac14\pi^2\sqrt2$ (to be checked). - 1 +1 Numerical value of $\frac{\pi^2}{2\sqrt{2}}$ confirmed. – Sasha Aug 25 '12 at 0:15 @Sasha Thanks for the appreciation and for the confirmation. – Did Aug 25 '12 at 8:15 @did, by Babak Sorouh's way, I get diffenrent answer form yours. – Leitingok Aug 26 '12 at 4:54 @Leitingok: Well, too bad. What do you suggest I should do about it? – Did Aug 26 '12 at 7:52 In Ellipsoidal coordinates, surfaces with constant $\lambda$ have: $$\frac{x^{2}}{a^{2} + \lambda} + \frac{y^{2}}{b^{2} + \lambda} + \frac{z^{2}}{c^{2} + \lambda} = 1$$ So that you can take for example: $a^2 = b^2 = 1, c^2 = 0$ and integrate over the other two coordinates (after applying the Jacobian of course). -	{"extraction_info": {"found_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": 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": 0.9141138792037964, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/43969/relativistic-momentum	# Relativistic momentum I have been trying to derive why relativistic momentum is defined as $p=\gamma mv$. I set up a collision between 2 same balls ($m_1 = m_2 = m$). Before the collision these two balls travel one towards another in $x$ direction with velocities ${v_1}_x = (-{v_2}_x) = v$. After the collision these two balls travel away from each other with velocity ${v_1}_y = (-{v_2}_y) = v$. Coordinate system travells from left to right with velocity $u=v$ at all times (after and before collision). Please see the pictures below where picture (a) shows situation before collision and picture (b) after collision. Below is a proof that Newtonian momentum $mv$ is not preserved in coordinate system $x'y'$. I used $[\, \| \,]$ to split $x$ and $y$ components. $p_z'$ is momentum before collision where $p_k'$ is momentum after collision. $$\scriptsize \begin{split} p_z' &= \left[ m_1 {v_1}_x' + m_2 {v_2}_x'\, \biggl\| \, 0 \right] = \left[ m_1 0 + m_2 \left( \frac{{v_2}_x - u}{1-{v_2}_x\frac{u}{c^2}} \right)\, \biggl\| \, 0 \right]= \left[ m \left( \frac{-v - v}{1+ v \frac{v}{c^2}} \right) \, \biggl\| \, 0 \right] \\ p_z' &= \left[ - 2mv \left( \frac{1}{1+ \frac{v^2}{c^2}}\right) \, \biggl\| \, 0 \right] \end{split}$$ $$\scriptsize \begin{split} p_k' &= \left[-2mv \, \biggl\| \,m_1 {v_1}_y' + m_2 {v_2}_y'\right]=\left[ -2mv \, \biggl\| \, m_1 \left( \frac{{v_1}_y}{\gamma \left(1 - {v_1}_y \frac{u}{c^2}\right)} \right) + m_2 \left( \frac{{v_2}_y}{\gamma \left(1 - {v_2}_y \frac{u}{c^2}\right)} \right) \right]\\ p_k' &= \left[ -2mv \, \biggl\| \, m \left( \frac{v}{\gamma \left(1 - v \frac{v}{c^2}\right)} \right) - m \left( \frac{v}{\gamma \left(1 - v \frac{v}{c^2}\right)} \right)\right]\\ p_k' &= \left[ -2mv \, \biggl\| \, 0 \right] \end{split}$$ It is clear that $x$ components differ by factor $1/\left(1+\frac{v^2}{c^2}\right)$. QUESTION: I want to know why do we multiply Newtonian momentum $p=mv$ by factor $\gamma = 1/ \sqrt{1 - \frac{v^2}{c^2}}$ and where is the connection between $\gamma$ and factor $1/\left(1+\frac{v^2}{c^2}\right)$ which i got? - Is there only an $x$ component to the velocities? Because then in the $x-y$ frame, the initial momentum $p_{i} = m_{1}v_{1x} + m_{2}v_{2x} = mv - mv = 0$, and there cannot be a $y$ component to the velocities after collision. – Kitchi Nov 14 '12 at 13:45 ## 4 Answers Assume that the relativistic momentum is the same as the nonrelativistic momentum you used, but multiplied by some unknown function of velocity $\alpha(v)$. $$\mathbf{p} = \alpha(v)\,\, m \mathbf{v}$$ Then in the primed frame, the total momentum before the collision is just what you had, but multiplied by $\alpha(v_i)$, with $v_i$ the speed before collision. The momentum after the collision is again what you had, but multiplied by $\alpha(v_f)$, with $v_f$ the speed after the collision. In order to conserve momentum we must have $$\alpha(v_i) \frac{-2mv}{1+v^2} = -2mv \,\alpha(v_f)$$ For simplicity, I'm suppressing factors of $c$. After the collision, you have a mistake in your velocity transformations. The vertical speed is just $v/\gamma$. That makes the speed of each ball $v_f = (v^2 + (v/\gamma)^2)^{1/2} = v \left(2-v^2\right)^{1/2}$ Plugging in $v_i$ and $v_f$ into the previous equation and canceling some like terms we have $$\alpha\left(\frac{2v}{1+v^2}\right) \frac{1}{1 + v^2} = \alpha\left(v[2-v^2]^{1/2}\right)$$ If you let $\alpha(v) = \gamma(v)$ and crunch some algebra you'll see that the identity above is satisfied. As for your original point, a desire to understand why momentum has a factor $\gamma$ in it, analyzing situations like this one is helpful, but ultimately it is probably best to understand momentum as the spatial component of the energy-momentum four-vector. Since it is a four-vector, it must transform like any other four-vector, $\gamma$'s and all. - Why should the Newtonian momentum be multiplied by some function of the velocity? – Physiks lover Nov 11 '12 at 23:25 in the context of this question, because if we do that, we get a conserved quantity – Mark Eichenlaub Nov 11 '12 at 23:28 I don't think my transformation in $y$ direction is wrong. Here is the derivation of transformation: $$\begin{split}v_y' &= \frac{dy'}{dt'}=\frac{d y}{\gamma \left(d t - d x \frac{u}{c^2} \right)} = \\ &= \frac{\frac{dy}{dt}}{\gamma \left(\frac{dt}{dt} - \frac{dx}{dt} \frac{u}{c^2} \right)}\\ &\boxed{v_y' = \frac{v_y}{\gamma \left(1 - v_x \frac{u}{c^2} \right)}}\end{split}$$ – 71GA Nov 12 '12 at 9:01 dude, v_x is zero after the collision... – Mark Eichenlaub Nov 12 '12 at 14:05 – Mark Eichenlaub Nov 12 '12 at 14:09 show 11 more comments Relativistic momentum is not defined as $p= \gamma m v$. For instance, this expression does not apply to photons, which are massless particles []. In the Lagrangian formalism of mechanics the momentum is defined as $$p\equiv \frac{\partial L}{\partial v}$$ Using the relativistic Lagrangian for a free massive particle $$L = - m c^2 \sqrt {1 - \frac{v^2}{c^2}}$$ we obtain $p= \gamma m v$ from the definition. The conservation laws can be obtained by using the ordinary Lagrangian formalism. For Lagrangians invariant to coordinate transformations, the momentum (as defined above) is automatically conserved. The above Lagrangian for one particle does not depend on coordinates (only on velocity) and, thus, the particle momentum $p$ is conserved. The Lagrangian for two particles does not depend on coordinates (only on velocities) and, thus, the total momentum $p_1 + p_2$ is conserved. If you use non-relativistic Lagrangians you obtain that it is the total non-relativistic momentum which is conserved. [] For a photon the relativistic momentum is given by $\|p\|= E/c^2$. - In relativistic mechanics, it isn't the 3 dimensional $\vec p$ that is invariant, but the four vector $\bf p$. Therefore, the four momentum $${\bf p} = m {\bf v}$$ where $\bf v$ is the four velocity, and $m$ is the rest mass of the object. Here, the four velocity is defined as $${\bf v} = \frac {d\vec x}{d\tau}$$ where $\vec x$ is a four vector and $$\tau = \gamma t$$ where $\tau$ is the proper time, which is a Lorentz invariant (the frame-dependent time $t$ and it's time intervals $dt$ are not invariant). Now, the four momentum $\bf p$ has components $$(E/c, p_{x}, p_{y}, p_{z}) = (\gamma mc, \gamma mv_{x}, \gamma mv_{y}, \gamma mv_{z})$$ so we can get back new (relativistically correct) expressions for three momentum and the energy : $$E = \gamma mc^{2}$$ $$\vec p = \gamma m \vec v$$ where $\vec v$ is the three velocity. I think the error in your calculations arises from the fact that you have to consider the four momentum as invariant, since the three momentum is not invariant under Lorentz transformations. - Yes, 3-momentum is conserved. 3-momentum and 4-momentum are both conserved. – Mark Eichenlaub Nov 14 '12 at 12:23 Yeah, sorry. I meant to write out not invariant. I've edited to correct that. – Kitchi Nov 14 '12 at 12:53 Okay. I'm not sure how this answers the question at all, since the OP does not consider 3-momentum invariant. In fact, if he had not made some simple errors, he could have found the correct expression for 3-momentum, as I showed in my answer. – Mark Eichenlaub Nov 14 '12 at 13:06 It does explain why you multiply newtonian momentum by a factor of $\gamma$. I haven't derived it, like your answer, but it's a valid formulation nevertheless. – Kitchi Nov 14 '12 at 13:31 I find it easier to start with notions of four-vectors (momentum and velocity). Then you can massage these quantities into the usual Newtonian momentum and velocity. Building up four-momentum the other way around (from $p=mv$) is not obvious at all. The four-velocity $u = ds/d\tau$ is constrained to have magnitude $c$ under the Minkowski metric. Take the metric to be $(-,+)$ for $(e_t, e_x)$, basis vectors for a 1+1 space. Full 3+1 isn't necessary to illustrate what's going on. We can imagine that a valid four-velocity will have the form $u = ae_t + b e_x$. Forcing that $u \cdot u = -c^2$ implies that $-a^2 + b^2 = -c^2$. This is satisfied, without any loss of generality, by $a = c \cosh \phi$ and $b = c \sinh \phi$. That means we have $$u = \frac{ds}{d\tau} = e_t \frac{d(ct)}{d\tau} + e_x \frac{dx}{d\tau} = c \left( e_t \cosh \phi + e_x \sinh \phi \right)$$ From this we conclude that $dx/d\tau = c \sinh \phi$ and $dt/d\tau = \cosh \phi$. By the chain rule, it follows that $$\frac{dx}{dt} = \frac{dx}{d\tau} \frac{d\tau}{dt}$$ By the inverse function theorem, it follows that $d\tau/dt = (dt/d\tau)^{-1}$, so that we have $$\frac{dx}{dt} = c \frac{\sinh \phi}{\cosh \phi} = c \tanh \phi$$ We then identify that $c \tanh \phi = v$, or $\tanh \phi = \beta$, per the standard terminology in special relativity, for we have found an expression for the conventional three-velocity in terms of $\phi$. From here, all that remains is to do algebra and use hyperbolic trig identities. See that $$\cosh^2 \phi - \sinh^2 \phi = 1 \implies 1 - \tanh^2 \phi = 1/\cosh^2 \phi$$ Put in $\beta = \tanh \phi$ to get $$\cosh^2 \phi = \frac{1}{1-\beta^2}$$ We choose to define $\gamma = \cosh \phi$, which results in $\sinh \phi = \gamma \beta$. This makes the four-velocity read $$u = c (\gamma e_t + \gamma \beta e_x)$$ It's not hard to get to four-momentum from here. All of this follows naturally from the hyperbolic geometry of spacetime and the requirement that four-velocity have magnitude $c$ for massive objects. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 73, "mathjax_display_tex": 20, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9198241829872131, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/295918/simple-topology-question-about-bases-of-topologies	# simple topology question about bases of topologies For a topological space $(X,T)$ with a basis $B$, is every basis element of $B$ an open set of $X$ (i.e. in $T$)? (Forgive me for the dumb question, just trying to learn the basics) - 2 yes, $B$ is a subset of $T$. – Damien L Feb 6 at 0:52 ## 2 Answers By definition $B$ is a base for $T$ if and only if $T$ is the set of all unions of subsets of $B$. For any $b\in B$, $\{b\}$ is a subset of $B$, so its union is in $T$. But its union is just the set $b$, so $b\in T$. Thus, $B\subseteq T$. - One last question...I'm trying to prove that the lower limit topology of the reals (T') is strictly finer than the standard topology on R (T), and first, I want to show that I can pick any element of T and that is in T'. So does picking an open interval (a,b) basis count as picking any element of T? Or are there elements in T that aren't in the basis for R? – Allison Cameron Feb 6 at 1:02 @Allison: There are many $T$-open sets in $\Bbb R$ besides the open intervals: $(0,1)\cup(2,3)$ is a very simple example. However, you can show that every $(a,b)$ with $a<b$ is in $T'$ by showing that it’s a union of sets of the form $[x,b)$, and it will then follow that every $U\in T$ is in $T'$: it’s a union of open intervals, they’re in $T'$, and topologies are by definition closed under taking unions. – Brian M. Scott Feb 6 at 1:08 Thank you for the help!!! – Allison Cameron Feb 6 at 1:12 @Allison: You’re welcome! – Brian M. Scott Feb 6 at 1:15 (just so the question does not remain unanswered) By definition, any arbitrary union has to be in $T$ hence also a `union' with one element. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 29, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": 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": 0.9499810338020325, "perplexity_flag": "head"}
http://nrich.maths.org/2365/solution	### Golden Thoughts Rectangle PQRS has X and Y on the edges. Triangles PQY, YRX and XSP have equal areas. Prove X and Y divide the sides of PQRS in the golden ratio. ### Ladder and Cube A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground? ### Days and Dates Investigate how you can work out what day of the week your birthday will be on next year, and the year after... # The Spider and the Fly ##### Stage: 4 Challenge Level: You sent in a large number of solutions to this problem but many of them only considered the spider moving horizontally or vertically but not diagonally across the sides of the room. I have included below a net of the room with the path of the spider as it takes the shortest route over the ceiling. I hope it helps you to see what was going on. However the spider could walk around the walls to get to the fly or along the floor - risking a human foot! Andrei of Tudor Vianu National College calculated the distance the spider must travel in the original problem and has worked out when it is best to go via the floor instead of the ceiling. First, I observed that the shortest distance can be found on the net of the box. In the figures below, the distances on the box are written, if the spider travels by the ceiling (figure 1) or by the lateral wall (figure 2). In figure 1 - I marked the position of the fly (F), and the spider (S). The next step is to calculate the distance travelled by the spider for each case. Applying Pythagoras Theorem in Figure 1, I obtain: $FS^2 = 7.25^2 + 1.5^2$ $FS = 7.40355 m$ In Figure 2, I have: $FS^2 = 7.5^2 + 0.25^2$ $FS = 7.50417 m$ The shortest distance is $7.40355$ m, i.e. the first situation. This is because the dimension "wide" is larger than the dimension "high"and the spider starts from the middle of the wall. Now I analyze the situation when the fly goes down the wall. In the first case, with the fly fixed, it was situated in the upper middle of the face, so it was better for the spider to go on the top of the box. When the fly arrives at the middle (height) of the box, it is the same for the spider to go over the top or over the bottom of the box. When the fly goes still further to the bottom, the spider should go on the bottom of the box. However in some cases it is quicker to go via the side wall instead of either the floor or the ceiling. Gillian sent us in the following diagrams to help explain how she worked out when this is the case: My first diagram shows the measurements when the spider is crawling along the ceiling. She has to go along $(8.75 - h)$ metres and down $1.5$ metres, so the shortest distance she can travel is the square root of $(8.75-h)^2 + 1.5^2$. As the height, $h$, of the fly decreases, the distance increases. If she crawls along the floor, the shortest distance she can travel is the square root of $(6.25+h)^2 + 1.5^2$. As h descreases, the distance decreases. These two results could be seen just by common sense. My second diagram shows the measurements when the spider is crawling along the wall closest to the fly. This way, she goes along $7.5$ metres and up $(h - 1.25)$ metres. The shortest distance she can crawl is the square root of $(7.5^2 + (h-1.25)^2$. At $h = 1.25$ metres, it is quicker for her to crawl along the wall than either the floor or the ceiling. (Distance $= 7.5$m, as opposed to $7.65$m if she were going along the floor or ceiling.) To see when it's quicker to go along the wall than the ceiling, I solved the inequality $(8.75 - h)^2 + 1.5^2 \geq 7.5^2 + (h - 1.25)^2$ $(8.75 - h)^2 - (h - 1.25)^2 \geq 7.5^2 - 1.5^2$ $7.5\times (10 - 2h) \geq 54$ $h \leq 1.4$ When the fly is less than $1.4$m from the ground, it is quicker to go by the wall than by the ceiling. Similarly, when it is higher than $1.1$m, it is quicker to go by the wall than the floor. 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": 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": 0.18, "absolute_threshold": 10, "end_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": 0.9512465596199036, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/tagged/ads-cft	# Tagged Questions The ads-cft tag has no wiki summary. 0answers 37 views ### Construction of the supergravity side explicitly in gauge/gravity dualities Although the motivation of this question comes from the AdS/CFT correspondence, it actually is related to a more general principle of gauge/gravity duality. We know from Maldacena's conjecture that a ... 2answers 107 views ### What is the exact relationship between on-shell amplitudes and off-shell correlators in AdS/CFT? In this answer to a question, it is mentioned that in the AdS/CFT correspondence, on-shell amplitudes on the AdS side are related to off-shell correlators on the CFT side. Can somebody explain this ... 1answer 59 views ### AdS/RCFT examples? RCFT's (rational conformal field theories) in two dimension are very well-studied, but it seems that there is no any gravity dual of such theories has been found. 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In their ... 0answers 68 views ### Good introductory books on AdS/CFT correspondence [duplicate] Possible Duplicate: Introduction to AdS/CFT Since my question in a similar topic was deleted, I'll ask away and hope ppl won't come here telling me: this was already asked! :\ I have a ... 2answers 175 views ### Why is a stack of N D-branes equivalent to an extremal black brane? A stack of N D-branes can have open strings ending on them. There is a U(N) brane gauge field, and r adjoint Higgs fields, with r equal to the number of transverse spatial dimensions. The eigenvalues ... 0answers 81 views ### Examples of manifolds and fluxes coming from generalized complex geometry The paramount object in generalized gomplex geometry is the Courant algebroid $TM\oplus T^\star M$, where the manifold $M$ is called background geometry I think (I am not sure). 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Also some people were ...	{"extraction_info": {"found_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": 0.9134585857391357, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/4458/how-do-i-map-a-spherical-triangle-to-a-plane-triangle	# How do I map a spherical triangle to a plane triangle? My goal here is to make my own custom "polyhedral map" of Earth. If you print out something from the "Map Fold-outs" page, you will have something almost exactly like what I'm trying to make. I have 2 triangles. One is a spherical triangle drawn on a 3D globe. By definition, each edge of a spherical triangle is part of a great circle. When you look at that 3D globe, there are a bunch of cities, coastlines, etc. that are (hopefully) accurately plotted on that 3D globe, inside that spherical triangle. The other triangle is a flat, plane (2D), straight-edged, Euclidean triangle. On paper. At the moment the interior of that triangle is blank white paper, but eventually I want to draw a copy of all those cities, coastlines, etc. into that area. Every map projection will map that 3D spherical triangle to a 2D image. Then it's easy to (in 2D) rotate and slide and shrink, and perhaps skew, that image until the 3 corners exactly line up with the 3 corners of my plane triangle. If I stack the results of a bunch of different map projections on top of each other, even though I've forced the 3 corners to exactly line up, each projection will put the cities in a slightly different location. Unfortunately, many projections take cities that are slightly inside the spherical triangle and draw them slightly outside the flat 2D triangle. (One symptom of this problem is that the sides of the spherical triangle are mapped to plane (2D) curves whose endpoints match the corners properly, but they bulge outward slightly from perfectly straight lines drawn between the corners). That leads to the city being completely missing from my polyhedral map. I'd rather not run the risk of some Polish mathematician getting upset that Warsaw is inexplicably missing from the flat, plane (2d), paper map that I've made :-). So I'm looking for map projections that "keep it inside the lines". I need 3 specific great circles (the edges of the spherical triangle) to be mapped to straight lines on paper. I don't care about other great circles -- straight, non-straight, whatever. What map projections meet that criteria? And can you give me a link to the (x,y) = f(lat, long) equations for that projection? I hear that the map projection used for the Dymaxion map meets that criteria; and someone told me that he thought it used the Chamberlin trimetric projection. Alas, when I use my (extremely rough and probably buggy) implementation of Chamberlin trimetric projection to map the spherical triangle formed by its 3 base points to the plane, I seem to be getting a shape that is almost a triangle, but the three "lines" curve and bulge out. Is that a bug in my code, or is it supposed to do that? So apparently either (a) I'm using the wrong equations -- so where can I find the right equations? Or (b) Dymaxion actually uses some other projection -- so where can I find the equations for that projection? I'm looking for answers of the form "The gnomonic projection meets your criteria. The gnomonic projection equations." I'm trying to get a list of several map projections that meet that critera. I know that the gnomonic projection is not the only one, because the Collignon projection and the Peirce quincuncial projection can map an octant (a spherical triangle with 90 degree corners) to a straight-edged plane triangle. (Forgive me for reposting this question from StackOverflow. I hope that a few people here will know a few answers to this that no one at StackOverflow seems to know). EDIT: Several responses have mentioned that the only way to make every straight line on the plane map correspond to a great circle on the globe is to use the gnomonic projection (or some linear affine transformation of it). That's technically correct, but irrelevant to my question. I'm not asking for a projection that maps every straight line on the plane map correspond to a great circle on the globe. I'm asking for projections that map 3 particular straight lines -- the boundaries of one facet of my map -- to 3 particular great circles on the globe -- the great circles connecting a few particular carefully-chosen locations on Earth. While certainly the gnomonic projection is one way to map the cities, coastlines, etc. in a spherical triangle to a plane triangle, it is not the only way. As an existence proof, as I mentioned before, the Collignon projection and the Peirce quincuncial projection also map the octant -- a spherical triangle -- to a straight-edged plane triangle. (What are the equations for the Dymaxion map projection? I hear that it is yet another way to map a spherical triangle to a plane triangle). Alas, some people do not like the area distortion, length distortion, and the angle distortion of the gnomonic projection. I'm hoping that by allowing a little distortion in the paths of some great circles (mapping them to slightly curved paths on the plane), I will end up with less area distortion, length distortion, and angle distortion. (I see that the Peirce quincuncial projection has zero angle distortion at most points). "there are infinitely many smooth mappings from the sphere to the plane that satisfy that criterion. ... you can always take any such mapping and "distort" it a little in the interior." Yes, but most of them make area distortion, length distortion, and angle distortion worse, and have horrifically complicated equations describing them. I'm looking for projections that are better than the gnomonic projection, preferably with relatively simple equations describing the projection. I hesitate to say that I'm looking for the projection that minimizes distortion, because I don't want several pages of horrifically complicated equations that happen to minimize some specific arbitrarily-weighted average of area distortion, length distortion, and angle distortion. - A number of (perhaps not-so-authoritative) sources I found online, including Principles of Meteorological Analysis by Saucier, claim that the gnomonic projection is the only one that maps all great circles to straight lines, but I haven't been able to locate a proof. On the other hand, all you really need is to map the three great circles that form your spherical triangle to straight lines, but there is an infinite number of smooth functions satisfying that criterion. – Rahul Narain Sep 12 '10 at 5:15 – J. M. Sep 12 '10 at 5:19 – J. M. Sep 12 '10 at 5:46 Sorry, I hadn't noticed you had already said in your question that "I don't care about other great circles -- straight, non-straight, whatever." But as I said, there are infinitely many smooth mappings from the sphere to the plane that satisfy that criterion. You can use a gnomonic projection. You can use barycentric coordinates. You can use a perspective warp of either and affinely realign it. More generally, you can always take any such mapping and "distort" it a little in the interior. So an exhaustive characterization of such projections is simply not possible. – Rahul Narain Sep 13 '10 at 13:56 – stevenvh Sep 13 '10 at 16:20 ## 2 Answers You might want to look at the relatively recent book: Portrait of the Earth: A Mathematician Looks at Maps by Timothy Freeman American Mathematical Society, Providence, 2002 - The uniqueness of gnomonic projections is a consequence of the uniqueness of line-preserving transformations of the plane. If you have map projections $P_1$ and $P_2$ that take all arcs of great circles to segments of lines, then $P_1 \circ P_2^{-1}$ is a transformation of the plane (or of one part of the plane onto another) that sends line segments to line segments. [In case the $\circ$ notation is not clear, I mean the transformation that results from taking the target of $P_2$, un-projecting it back to the sphere, and projecting that pre-target from the sphere back to the plane using $P_1$.] It is known that, within the class of topology-preserving (continuous and one-to-one) transformations between connected regions in the plane, any collinearity-preserving transformation is the restriction of a single linear transformation. If you allow points at infinity (for the gnomonic projection these are the projection of points on the equator of the sphere) then it is also true that any collinearity-preserving transformation is the restriction of a single projective transformation. The consequence for gnomonic projections is that if you consider the projection plane to be a rigid object, with an x-y coordinate system drawn on it, but that coordinatized plane can be moved around in space, then any map projection from a sphere (or parts of the sphere) to the plane that takes arcs of great circles to segments or lines, is a gnomonic projection onto some position of the plane. - The Collignon projection and the Peirce quincuncial projection each map a few arcs of great circles to line segments. However, they are not gnomonic projections. – David Cary Sep 13 '10 at 16:20 The proof above is for projections that map all great circle arcs (within some patch of sphere) to line segments. The same argument shows how to construct a multitude of projections that map only the boundary arcs of a spherical triangle onto the sides of a plane triangle, by reducing it to mapping of a planar triangle onto itself. For example, you could map barycentric coordinates (a,b,c) of points in the triangle to (a^t, b^t, c^t) and then normalize to get a set of barycentric coordinates (divide by a^t + b^t + c^t). There is also a unique projection that is a conformal mapping. – T.. 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http://physics.stackexchange.com/questions/3534/dont-heavier-objects-actually-fall-faster-because-they-exert-their-own-gravity/3556	# Don't heavier objects actually fall faster because they exert their own gravity? The common understanding is that, setting air resistance aside, all objects dropped to Earth fall at the same rate. This is often demonstrated through the thought experiment of cutting a large object in half, the halves of which clearly can't then fall more slowly just by being sliced in two. However, I believe the answer is that when two objects fall together, attached or not, they do "fall" faster than an object of less mass alone does. This is because not only does the Earth accelerate the objects toward itself but the objects also accelerate the Earth toward themselves. Considering the formula: $F_g = G m_1 m_2/d^2$ We can see that the force of gravity is dependent on BOTH the masses, not just that of the more massive object. Of course in everyday situations, we can for all practical purposes treat objects as falling at the same speed. But I'm hoping not for a discussion of practicality or what's measurable or observable, but what we think is actually happening. Am I right or wrong? What really clinched this for me was considering dropping a small Moon-massed object close to the Earth and a small Earth-massed object close to the Earth. This made me realize that falling isn't one object moving toward some fixed frame of reference, but that the Earth is just another object and falling is multiple objects mutually attracting in space. - 2 Now that's one piece of trivia to blast unsuspecting nerds :D – SF. Dec 8 '12 at 21:17 1 The classic experiment, as performed by Galileo and David Scott, involves dropping two objects simultaneously. In that case, the pull of the larger mass also affects the falling time of the smaller mass, as the Earth falls up and meets both objects slightly sooner. With <handwave>appropriate assumptions</handwave>, do they still hit simultaneously, or is there some tiny third-order effect I'm not allowing for? – Keith Thompson Dec 8 '12 at 21:45 @Keith I haven't been considering the case of dropping both dissimilar-mass objects simultaneously--the question is considering the two objects separately. The accepted answer says that the difference in the times between two objects could be on the order of 1 part in a trillion trillion. So there won't be much to measure, and according to human perception & instrumentation we would always call it "simultaneous." Dropping the items together would take even less time. But one way to eliminate the complexity of the number of items is just to consider each atom individually. – ErikE Dec 8 '12 at 21:59 I removed the CW status on your question, but keep in mind that this is something we try to do quite sparingly. Just try to be careful with your edits on future questions :-) – David Zaslavsky♦ Dec 9 '12 at 2:02 – ErikE Dec 9 '12 at 3:09 show 2 more comments ## 9 Answers Using your definition of "falling," heavier objects do fall faster, and here's one way to justify it: consider the situation in the frame of reference of the center of mass of the two-body system (CM of the Earth and whatever you're dropping on it, for example). Each object exerts a force on the other of $$F = \frac{G m_1 m_2}{r^2}$$ where $r = x_2 - x_1$ (assuming $x_2 > x_1$) is the separation distance. So for object 1, you have $$\frac{G m_1 m_2}{r^2} = m_1\ddot{x}_1$$ and for object 2, $$\frac{G m_1 m_2}{r^2} = -m_2\ddot{x}_2$$ Since object 2 is to the right, it gets pulled to the left, in the negative direction. Canceling common factors and adding these up, you get $$\frac{G(m_1 + m_2)}{r^2} = -\ddot{r}$$ So it's clear that when the total mass is larger, the magnitude of the acceleration is larger, meaning that it will take less time for the objects to come together. If you want to see this mathematically, multiply both sides of the equation by $\dot{r}\mathrm{d}t$ to get $$\frac{G(m_1 + m_2)}{r^2}\mathrm{d}r = -\dot{r}\mathrm{d}\dot{r}$$ and integrate, $$G(m_1 + m_2)\left(\frac{1}{r} - \frac{1}{r_i}\right) = \frac{\dot{r}^2 - \dot{r}_i^2}{2}$$ Assuming $\dot{r}_i = 0$ (the objects start from relative rest), you can rearrange this to $$\sqrt{2G(m_1 + m_2)}\ \mathrm{d}t = -\sqrt{\frac{r_i r}{r_i - r}}\mathrm{d}r$$ where I've chosen the negative square root because $\dot{r} < 0$, and integrate it again to find $$t = \frac{1}{\sqrt{2G(m_1 + m_2)}}\biggl(\sqrt{r_i r_f(r_i - r_f)} + r_i^{3/2}\cos^{-1}\sqrt{\frac{r_f}{r_i}}\biggr)$$ where $r_f$ is the final center-to-center separation distance. Notice that $t$ is inversely proportional to the total mass, so larger mass translates into a lower collision time. In the case of something like the Earth and a bowling ball, one of the masses is much larger, $m_1 \gg m_2$. So you can approximate the mass dependence of $t$ using a Taylor series, $$\frac{1}{\sqrt{2G(m_1 + m_2)}} = \frac{1}{\sqrt{2Gm_1}}\biggl(1 - \frac{1}{2}\frac{m_2}{m_1} + \cdots\biggr)$$ The leading term is completely independent of $m_2$ (mass of the bowling ball or whatever), and this is why we can say, to a leading order approximation, that all objects fall at the same rate on the Earth's surface. For typical objects that might be dropped, the first correction term has a magnitude of a few kilograms divided by the mass of the Earth, which works out to $10^{-24}$. So the inaccuracy introduced by ignoring the motion of the Earth is roughly one part in a trillion trillion, far beyond the sensitivity of any measuring device that exists (or can even be imagined) today. - The paradox appears because the "rest frame" of the Earth is not an inertial reference frame, it is accelerating. Keep yourself in the CM reference frame and, at least for two bodies, there is no paradox. Given an Earth of mass M, a body of mass $m_i$ will fall towards the center of mass $x_{CM}=(M x_M + m_i x_i)/(M+m_i)$ with an acceleration $GM/(x_i-x_M)^2$. Note that $\ddot x_{CM}=0$ Really we have only hidden the paradox, because of course $x_{CM}$ is different for each $m_i$. But this is a first step to formulate the problem in a decent inertial frame. The paradox resurfaces again if you want to get rid of $(x_i-x_M)$. In most applications, now that you are in a non accelerating reference system, you want to consider distances related to it, ie $x_i-X_{CM}$. The solution is to redefine the mass. As $x_i-x_{CM}= M (x_i - x_M) /(M+m_i)$, we can say that the object $i$ falls into the Mass Center with an acceleration $G{M^3 \over (M+m_ i)^2}{1 \over (x_i-x_{CM})^2}$ You could say that the actual mass of the "earth at center of mass" is this correction. Once you are into the trick of changing the value of the mass, you can still stick to the reference frame of the earth. In this reference frame the quotient between force and acceleration is $Mm_i/M+m_i$ You can claim that this is the actual mass of the body during the calculation. This is called the reduced mass $m_r$ of the system, and you can see that for small $m_i$, it is almost equal to $m_i$ itself. You can ever write some of the previous formulae using the reduced mass $m_r$ in combination with the original masses, for instance the above ${M^3 \over (M+m_ i)^2}= M {m_r^2\over m^2}$, but I am not sure of how useful it is. In any case, you see that you were right about the "heavier implies faster" but that it is perfectly managed. For three objects, m_1 and m_2 falling into M, the question is how to compare the case to m_1+m_2 falling into M. You separate the forces between internal, between 1 and 2, and external, against M. Look at the point $x_0= {m_1 x_1 + m_2 x_2 \over m_1+m_2}$ . This point it is not accelerated by the internal forces. And the external forces move them as $$\ddot x_0={1 \over m_1+m_2} (m_1 {G M \over (x_1-x_M)^2} + m_2 {G M \over (x_2-x_M)^2})={F_1+F_2 \over m_1 + m_2}$$ This is becoming long... ¡I can not put all the Principia in a single answer!. So you can forget all the previous stuff, consider it is just to a mean to fix notation and get some practice, an read the answer: If the two bodies are at the same distance $x$ of the "external" earth, they suffer the same external acceleration $g=GM/(x-x_M)^2$, and the same happens with $x_0$. If both bodies are in an approximation where $g$ can be considered constant, which was the case originally considered by Galileo (and the modern $g=9.8m/s^2$), then they have the same acceleration -and also the combined position $x_0$-. If they are not at the same distance nor in an approximation of constant -equal everywhere- field, then you can still save the movement of $x_0$ to work as if it were a gravitational force for some single mass $m_T$, but then the manipulation of the equations will produce in the relative positions of $x_1$ and $x_2$ some accelerations of the order of $1/(x_0-x_M)^3$. Such forces are the "tidal forces". - What is "CM reference frame" please? – ErikE Jan 21 '11 at 21:25 Center of Mass reference frame – Greg P Jan 21 '11 at 21:28 In any case, I feel that my answer is not honest enough... but I am in hurry, sorry. Come back later tomorrow. – arivero Jan 21 '11 at 21:45 @Emtucitor I am back, but a bit drunken. Anyway. My answer could go along the lines now of explaining the concept of reduced mass (thus resusciting the paradox) and then about the decomposition of every collective movement into CM plus local. But at the end we should go to the difference between inertial and accelerating systems, and we would do a lenghtly mathematical thing, while it seems that really you are not into maths. So my short answer is, forget gravity, the point is about if all the bodies fall freely in a similar way, and that is not Newton. – arivero Jan 22 '11 at 3:22 @Emtucitor so Try "two new sciences", which has a very light math level (no differential calculus!), and is freely available in the internet. and the arguments therein, without invoking any accelerating frame of reference. Good night! – arivero Jan 22 '11 at 3:23 show 1 more comment The new version of the question is much clearer than before, and I removed my down vote accordingly. The answer is yes: in principle there is such an effect. When the mass of the dropped object is small compared to the mass of the planet, the effect is very small, of course, but in principle it's there. - Yes, a heavy object dropped from the same height will fall faster then a lighter one. This is true in the rest frame of either object. You can see this from $F=GmM/r^2=ma=md^2r/dt^2$. The fastest "falling" (since we are reredefining falling) object however is a photon, which has no mass. - Hmmm... where did M go in your second formula? – ErikE Jan 21 '11 at 21:02 A simple explanation is that it takes more FORCE to ACCELERATE an object of greater MASS. A=F/M....or with a constant FORCE, the ACCELERATION is inversely proportional to the MASS. The is a result of inertia. A larger mass dropped (on a massive objects) requires a GREATER force to accelerate; additionally a GREATER mass exerts a GREATER gravitaional force. Setting newtons second law to the gravitional force law cancles mass out and therefore renders MASS not related to the ACCELERATION of two gravitationally related objects. - Halston, you performed a presto-change-o in the middle of your answer. You started out with accelerating one mass, but suddenly we're accelerating two. My point is that even with just dropping "one mass" there are really two masses involved, two separate accelerations, each imparted on one object by the other. You can't just cancel one out. The Earth imparts 9.80665 ms^2 of acceleration. But each object imparts its own acceleration on the Earth. Falling is not just acceleration of a single mass but the two toward each other. I think you have missed the point. – ErikE Feb 19 '11 at 1:19 Adding a third mass to the equation (the "second" object) changes things enough that the closing acceleration between the Earth and the objects will not be the same as either mass alone. – ErikE Feb 19 '11 at 1:21 The same will apply to your "two mass". There is a force exerted on object one, there is a force exerted on object two. In fact, it is an action-reaction force. The book exerts a gravitational force on the earth, the earth exerts a gravitational force on the book. The book seems to be accelerating only because its mass (and inertia) is insignificant in comparison to the earth. Now, if the book had a mass of equal magnitude to the earth, the equal masses would demonstrate equal (in magnitude) and opposite acceleration. – HAL Feb 19 '11 at 4:41 What do you mean, "seems to accelerate?" it does, in fact. I doubt I can explain it at this point. Do you really think that if you had a moon-massed object the size of a basketball that it would only close with the Earth at 9.8 m/s/s? – ErikE Feb 19 '11 at 4:53 the acceleration of the book differs from the acceleration of the earth towards the book. seemingly just simply notes the availability of observation. I dont know if you are saying that the force law of gravity changes with a given mass. If you are invalidating Newton's law then perhaps. – HAL Feb 19 '11 at 5:00 show 1 more comment Drop a 5 lb iron barbell and a 25 lb iron barbell simultaneously. The'll hit the ground simultaneously. The only possible caveat is the recoil effect Ted Bunn brings up above. - It seems you didn't read carefully. Please see section "Practically Speaking". I'd appreciate an answer that is more on topic and responds to the points I made in my question. – ErikE Jan 21 '11 at 20:52 2 @Emtucifor: At this point, why not fac tor in the gravitational radiation of the bodies? Or the one or two electron ionization of the falling masses, or a million other effects? If the effect you're talking about is completely unmeasurable, is it an effect at all? And I did, in fact, mention the recoil effect, which is going to be the leading order correction to the equivalence principle, anyway. – Jerry Schirmer Jan 21 '11 at 22:12 Now that I shortened my answer, I should clarify that in my post I made it clear that I know normal observation would seem to indicate all objects fall at the same rate, but that wasn't what I was interested in. Now to Jerry's last comment: Ted made no mention of any recoil effect. Could you tell me more about that? Also, I'm quite interested in any other factors that could affect falling. Would you care to elaborate on gravitational radiation, electron ionization, and so on? – ErikE Jan 21 '11 at 22:34 @Emtucifor I get the feeling that you're seeking replies more for attention than any other reason. You mention you have a 20 year old high school education. That is fine. But one needs to recognize what their limitations are or they just end up sounding silly. – user346 Jan 22 '11 at 3:45 @space Your feeling is misplaced. And I'll thank you kindly to not discuss my supposed mental limitations. It's all very well for you to claim I am misunderstanding, but please give me the benefit of the doubt and post an answer to help disabuse me of my mistaken notions. Otherwise you're just here to insult and try to feel good about yourself. I dare you: set me (and us all) straight with superior understanding, in a proper answer. Please. – ErikE Jan 22 '11 at 10:23 show 1 more comment The free fall time of two point masses is $t = \frac{\pi}{2} \sqrt{ \frac{r^3}{2 G(m1+m2)}}$. The free fall time is dependent on the sum of the two masses. For a given total mass, the free fall time is independent of the ratio of the two masses. The free fall time is the same whether m1 = m2, or m1 >> m2. When a body is picked up to a certain height and then dropped, the time to fall to the Earth does not depend on the mass of the object. If you lift a ping-pong ball and then drop it, it will take the same time to fall to the Earth as a bowling ball. Splitting the Earth into two masses does not change the sum of those masses, or the free fall time. However, when an external body is brought to a certain height above the Earth and then dropped, the free fall time does depend on the mass of the external body. Because the sum of the Earth and the external body obviously does depend on the mass of the external body. "Most bodies fall at the same rate on earth, relative to the earth, because the earth's mass M is extremely large compared with the mass m of most falling bodies. The body and the earth each fall toward their common center of mass, which for most cases is approximately the same as relative to the earth. In principle, the results of a free fall experiment depend on whether falling masses originate on earth, are extraterrestrial, are sequential or concurrent, or are simultaneous for coincident or separated bodies, etc. When falling bodies originate from the earth, all bodies fall at the same rate relative to the earth because the sum m + M remains constant. -- ArXiv:Deterrents to a Theory of Quantum Gravity Assumptions: The Earth is isolated (there is no moon, sun, etc). The Earth is non-rotating. The Earth has no atmosphere. (A hot air balloon would fall upwards because it is less dense than the atmosphere it displaces.) - This makes no sense at all. There is no sensible distinction between local and external masses. Please read the accepted answer which provides the proper proof with formulas. – ErikE Dec 8 '12 at 18:12 @ErikE There are two scenarios. In the first, the total mass of the system is constant (you split the Earth into two pieces). In the second, the total mass of the system increases (you introduce new mass). – Nick Dec 8 '12 at 20:45 It makes no difference. It's a human reference frame to imagine the Earth to be still. But that is illogical. The Earth is not fixed in space. It moves due to acceleration imparted on it by other objects! Stop thinking about when the second object is introduced. Calculate everything after that. The scenario is: two objects in an otherwise empty universe, one very massive, one of unknown mass, are in airless free fall, held apart by a force. When the force separating them is removed, they both accelerate towards each other. The mass of #2 affects the time until impact. – ErikE Dec 8 '12 at 21:09 I always thought the reason a penny falls faster than a feather is air resistance. There is real physics involved, but it's aerodynamics, not tides or gravity. - It seems you didn't read carefully. Please see section "Guidelines", point 5. I'd appreciate an answer that is more on topic and responds to the points I actually raised in my question. – ErikE Jan 21 '11 at 20:54 Now that I shortened my answer, I should clarify that in my original post I asked answerers to completely ignore wind resistance. – ErikE Jan 21 '11 at 22:35 F=ma F = GMm/r^2 a = F/m = GM/r^2 Hence to first approximation a one kilogram weight falls down at exactly the same speed as a two kilogram weight. However, the earth "falls up" twice as fast for the two kilogram weight, and would "fall up” one hundred times as fast for a one hundred kilogram weight. Of course if the one kilogram effect is of order one part in 10^24 as suggested byanother poster, the one hundred kilogram effect would still be only one part in 10^22 – Jim Graber Jan 27 '11 at 11:31 With gravity and wind resistance as factors, you may already know that an object dropped from a point of rest goes 9 meters per second per second. This differs in a vacuum environment such as the moon. - The $9\,\mathrm{m}\,\mathrm{s}^2$ acceleration is due to gravity alone, not air resistance. Moreover, the point of the question is that the fact that this acceleration is constant is only approximation because the Earth will also move a bit due to gravity from the falling object. – Ondřej Černotík Feb 4 at 19:19 I said "setting air resistance aside"... – ErikE Feb 4 at 23:14 ## protected by Qmechanic♦Feb 4 at 18:55 This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 44, "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": 0.18, "absolute_threshold": 10, "end_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": 0.9445734024047852, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/92589-limit-rational-expression-above-below.html	Thread: 1. Limit of Rational Expression Above or Below How can you tell if the limit is from above or below? 2. Originally Posted by Yale How can you tell if the limit is from above or below? Do you mean from the left or the right? Do you mean f(x) tendind to -infinity or +infinity? I'm having trouble understanding your question... 3. Originally Posted by VonNemo19 Do you mean from the left or the right? Do you mean f(x) tendind to -infinity or +infinity? I'm having trouble understanding your question... Perhaps they are more commonly referred to as left or right but the text I am looking at calls them above or below. What you posted there does seem to match up however. How would you go about determining that? 4. Originally Posted by Yale Perhaps they are more commonly referred to as left or right but the text I am looking at calls them above or below. What you posted there does seem to match up however. How would you go about determining that? Generally try direct substitution. If that does not work, try to rewrite the expression by simplifying, rationalizing, etc.. until direct substitution works. If the function is piece wise, pay carefull attention to which piece of the function to use with which variable. This is really hard to discuss without an example. 5. Originally Posted by VonNemo19 Generally try direct substitution. If that does not work, try to rewrite the expression by simplifying, rationalizing, etc.. until direct substitution works. If the function is piece wise, pay carefull attention to which piece of the function to use with which variable. This is really hard to discuss without an example. I shall do my best to post an example then: lim (5x^2 -6x - 1) / (2x^2 + 6) x -> positive infinity I know how to solve to get 5 / 2 but I am not sure if it is above or below (left or right). 6. Originally Posted by Yale Perhaps they are more commonly referred to as left or right but the text I am looking at calls them above or below. What you posted there does seem to match up however. How would you go about determining that? Post the question that has prompted your question. 7. Originally Posted by Yale I shall do my best to post an example then: lim (5x^2 -6x - 1) / (2x^2 + 6) x -> positive infinity I know how to solve to get 5 / 2 but I am not sure if it is above or below (left or right). This is easy as long as you understand that $\lim_{x\to+\infty}\frac{1}{x}=0$ Intuitively this makes sense because as x becomes larger and larger $\frac{1}{x}$ becomes smaller and smaller. So, whenever we can rewrite a function to look like something similar to $\frac{1}{x}$ then we can take the limit. Like so... $\frac{5x^2-6x-1}{2x^2+6}=\frac{x^2(5-\frac{6}{x}-\frac{1}{x^2})}{x^2(2+\frac{6}{x^2})}$ $=\frac{5-\frac{6}{x}-\frac{1}{x^2}}{2+\frac{6}{x^2}}$ So, by making use of our limit properties, we can "distribute" the limit throughout the terms. What happens to $\frac{6}{x}$ as x tends to intinity? It goes to zero. The same is true for the terms: $\frac{6}{x^2}=6\cdot\frac{1}{x}\cdot\frac{1}{x}$ as x tends to infinity we have $6\cdot0\cdot0=0$ So evaluating the limit we have $\lim_{x\to+\infty}\frac{5x^2-6x-1}{2x^2+6}=\lim_{x\to+\infty}\frac{5-\frac{6}{x}-\frac{1}{x^2}}{2+\frac{6}{x^2}}=$ $=\frac{5-0-0}{2+0}=\frac{5}{2}$ How can you tend to positive infinty from the right? 8. Originally Posted by Yale I shall do my best to post an example then: lim (5x^2 -6x - 1) / (2x^2 + 6) x -> positive infinity I know how to solve to get 5 / 2 but I am not sure if it is above or below (left or right). How can you be above infinity? The only way you can approach infinity is from below! IF you had something like " $\lim_{x\rightarrow 2} \frac{5x^2- 6x- 1}{2x^2+ 6}$", in order for that to exist you must get the same thing from both sides. IF you have something like " $\lim_{x\rightarrow 2^-}\frac{5x^2- 6x- 1}{2x^2+ 6}$" that is "from below" or "from the left" because of that little "-" on the 2. IF you have something like " $\lim_{x\rightarrow 2^+}\frac{5x^2- 6x- 1}{2x^2+ 6}$" that is "from above" or "from the right" because of that little "+" on the 2. In this case because the limit itself exists, both are the same.	{"extraction_info": {"found_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": 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": 0.9504122138023376, "perplexity_flag": "middle"}
http://programmingpraxis.com/2013/02/15/facebook-hacker-cup-2013-round-1-problem-1/2/	# Programming Praxis A collection of etudes, updated weekly, for the education and enjoyment of the savvy programmer ## Facebook Hacker Cup 2013, Round 1, Problem 1 ### February 15, 2013 The first step is to understand the problem. Consider the example with N = 4, K = 3, and a = [3, 6, 2, 8]. There are four different triples that can be formed from the input array: [3, 6, 2], [3, 6, 8], [3, 2, 8] and [6, 2, 8]. The maximums of those triples are 6, 8, 8 and 8, and the sum of the maximums is 30, which is the desired output. In general, if we sort the array, the smallest K – 1 items in the array will never be maximums, so the desired output will be the sum of the product of each of the highest array items times the number of different ways to choose the lower items. In the example, there is 1 maximum of 6 and 3 maximums of 8, producing a sum of 30. The number of ways to choose the lower items is given by the binomial theorem, the function nCr where n is the number of items available and r is the number of items to choose. In the example, nCr(2,2)=1 is the number of ways that 3 items can be chosen from a list of 3, and nCr(3,2)=3 is the number of ways 3 items can be chosen from a list of 4. Note that, for our problem, c is K-1 and n ranges from K-1 to N-1. The nCr function can be computed as $\frac{n!}{r! (n-r)!}$ and implemented like this (nCr is normally pronounced “choose”, because it is the number of ways to choose r items out of n): ```(define (choose n r) (if (zero? r) 1 (* n (/ r) (choose (- n 1) (- k 1)))))``` However, that doesn’t work for this problem because of the requirement that solutions are given modulo 1000000007; there is no division in modular arithmetic. What we can do instead is multiply by the modular inverse, and since the limit on K is 10000, we can precompute the modular inverses. Here are some convenient constants for the problem limits, a function to compute the modular inverse, a function that precomputes the 10000 needed inverses, and a function that computes the choose function: `(define m 1000000007)` `(define n 10000)` ```(define (inverse x m) (let loop ((x x) (a 0) (b m) (u 1)) (if (positive? x) (let ((q (quotient b x)) (r (remainder b x))) (loop (modulo b x) u x (- a (* q u)))) (if (= b 1) (modulo a m) #f))))``` ```(define inv (let ((inverses (make-vector (+ n 1) 0))) (do ((x 1 (+ x 1))) ((< n x)) (vector-set! inverses x (inverse x m))) (lambda (x) (vector-ref inverses x))))``` ```(define (choose n k) (if (zero? k) 1 (modulo (* n (inv k) (choose (- n 1) (- k 1))) m)))``` With that, the function that computes the output for a given set of inputs is simple; it uses `drop` from the Standard Prelude: ```(define (f n k as) (let loop ((i (- k 1)) (as (drop (- k 1) (sort < as))) (s 0)) (if (null? as) s (loop (+ i 1) (cdr as) (+ s (* (choose i (- k 1)) (car as)))))))``` This just loops with i counting the number of items available to be chosen, as listing the successive maximums, and s is the accumulating sum. Here are some examples: ```> (f 4 3 '(3 6 2 8)) 30 > (f 5 2 '(10 20 30 40 50)) 400 > (f 6 4 '(0 1 2 3 5 8)) 103 > (f 2 2 '(1069 1122)) 1122 > (f 10 5 '(10386 10257 10432 10087 10381 10035 10167 10206 10347 10088)) 2621483``` The Facebook problem requires some specific input and output formatting, but we’ll ignore that. You can run the program at http://programmingpraxis.codepad.org/S0qgV4pB. ### Like this: Pages: 1 2 Posted by programmingpraxis Filed in Exercises 6 Comments » ### 6 Responses to “Facebook Hacker Cup 2013, Round 1, Problem 1” 1. February 15, 2013 at 9:06 AM [...] Pages: 1 2 [...] 2. February 15, 2013 at 11:01 AM [...] today’s Programming Praxis exercise, our goal is to solve the first problem of the 2013 Facebook hacker [...] 3. Remco Niemeijer said February 15, 2013 at 11:01 AM ```import Data.List facts :: [Integer] facts = scanl () 1 [1..] facebook :: Int -> [Int] -> Integer facebook size = foldr (\(i,x) a -> mod (a + x choose (size-1) i) 1000000007) 0 . drop (size - 1) . zip [0..] . map fromIntegral . sort where choose k n = div (facts!!n) (facts!!k * facts!!(n-k)) ``` 4. Paul said February 15, 2013 at 2:39 PM A Python version. The gmpy library is used for speed. ```import gmpy def sumup(arr, k): arr.sort(reverse=True) n = len(arr) return sum(ai * gmpy.bincoef(n - 1 - i, k - 1) for i, ai in enumerate(arr[:n - k + 1])) % 1000000007 ``` 5. razvan said February 16, 2013 at 1:53 PM Here’s my Java solution: ```package cardgame; import java.io.File; import java.io.FileNotFoundException; import java.util.Arrays; import java.util.Scanner; /** * This class solves the card game problem from Facebook Hacker Cup 2013 round 1. * http://programmingpraxis.com/2013/02/15/facebook-hacker-cup-2013-round-1-problem-1/ * / public class CardGame { private static final int MOD_FACTOR = 1000000007; private int[][] decks; private int[] handSizes; /* * Main method. Expects one command-line argument: the name of the input file. * @param args / public static void main(String[] args) { if(args.length > 0) { try { CardGame cg = new CardGame(); cg.parseFile(args[0]); int[] strengths = cg.totalHandStrengths(); for(int i=0; i<strengths.length; i++) { System.out.println("Case #" + (i + 1) +": " + strengths[i]); } } catch (FileNotFoundException e) { System.out.println("File not found"); } } else { System.out.println("Supply file name as argument"); } } /* * This method parses the input file. * The input's format is described in the problem statement. * @param fileName name of input file * @throws FileNotFoundException / public void parseFile(String fileName) throws FileNotFoundException { Scanner sc; sc = new Scanner(new File(fileName)); int t = sc.nextInt(); decks = new int[t][]; handSizes = new int[t]; for(int i=0; i<t; i++) { int n = sc.nextInt(); handSizes[i] = sc.nextInt(); decks[i] = new int[n]; for(int j=0; j<n; j++) { decks[i][j] = sc.nextInt(); } } } /* * This method computes the total strength of all hands * for each (hand, hand size) pair in the input file. / public int[] totalHandStrengths() { int [] strengths = new int[decks.length]; for(int i=0; i<decks.length; i++) { Arrays.sort(decks[i]); strengths[i] = handStrengths(decks[i], handSizes[i], decks[i].length); } return strengths; } /* * This method computes the total hand strength for the first n cards in deck, where hands * have size k. * @param deck The deck of cards. This is assumed to be sorted. * @param k The size of a hand of cards. * @param n The number of cards in the deck. * @return Integer representing sum of scores of hands in deck. / private int handStrengths(int[] deck, int k, int n) { if (k == n) { return deck[n-1]; } // Make as many k-tuples you can that include the maximum card // Then, remove maximum and recurse. return (deck[n-1] choose(n-1, k-1) + handStrengths(deck, k, n-1)) % MOD_FACTOR; } /** * This method computes n choose k. * (How many subsets of size k there are in a set of size n) / private int choose(int n, int k) { int answer = 1; for(int i=n-k+1; i<=n; i++) { answer = answer i; } for(int i=1; i<=k; i++) { answer /= i; } return answer; } } ``` 6. Luke said March 3, 2013 at 9:37 PM Easier to fill out 10001 rows of Pascal’s triangle, nCr is the rth column of the nth row. %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 1, "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": 0.8724403381347656, "perplexity_flag": "middle"}
http://nrich.maths.org/434/index?nomenu=1	Let $N$ be a six digit number with distinct digits. Find the number $N$ given that the numbers $N$, $2N$, $3N$, $4N$, $5N$, $6N$, when written underneath each other, form a latin square (that is each row and each column contains all six digits).	{"extraction_info": {"found_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": false, "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": 0.8218600749969482, "perplexity_flag": "head"}
http://nrich.maths.org/481	### I'm Eight Find a great variety of ways of asking questions which make 8. ### 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? ### GOT IT Now For this challenge, you'll need to play Got It! Can you explain the strategy for winning this game with any target? # Pair Sums ##### Stage: 3 Challenge Level: Five numbers are added together in pairs to produce the following answers: $0, 2, 4, 4, 6, 8, 9, 11, 13, 15$ What are the five numbers? 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": 0.18, "absolute_threshold": 10, "end_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": 0.9224658012390137, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/80955?sort=newest	## mean value theorem for operators ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) This might be a trivial question but I am not very familiar with the subject matter. I was wondering if some sort of mean value theorem works for operators on function spaces. Say $F: \mathcal{S_1} \to \mathcal{S_2}$ is an operator on the function spaces $\mathcal{S_{1,2}}$ then for every $f,g \in \mathcal{S_1}$ there exist $h$ such that \begin{align} F(f) - F (g) = [DF(h)] (f - g), \end{align} or something like that! What is a good reference to look at if I want to learn more. - This question needs some work before it makes sense. Do you mean linear operators? You also presumably mean $DF(h)$, and in the linear case the derivative should be $F$ interpreted correctly, I believe. – Otis Chodosh Nov 15 2011 at 5:40 F is not linear and by D[F] i meant the differential operator – Nima Nov 15 2011 at 6:43 actually what i am interested in is a bit different. mean value theorem of this form might help me to prove that. I want to find a function $c(x,y)$ (or prove it exists) such that $[F(f)] (x) -[F(g)] (x) = \int c(x,y) (f(y) - g(y))$ for all x. – Nima Nov 15 2011 at 6:51 ## 2 Answers Here is a nice list (by John H. Mathews) of articles of various authors on the theme of extending the validity of the Mean Value Theorem to vector values function. However, as Dieudonné remarks (Foundations of Modern Analysis) the main point of the classical MVT, even in the case of a one variable real valued function, is not the identity $$f(b)-f(a)=f'(\xi)(b-a)\ ,$$ also because we usally can say nothing about the point $\xi$, apart the fact that it is strictly betweeen $a$ and $b$. Rather, it is the inequality it implies: $$\|f(b)-f(a)\|\le \sup_{a < \xi < b} \|f'(\xi)\| \|b-a\| \ ,$$ and this is also the statement that generalizes naturally in the Banach setting, and has the most important consequences, as it is the key tool of most fundamental theorems of differential calculus (to quote some: the symmetry of higher order differentials, the Lagrange's remainder form in Taylor's formula, the theorem of the total differential, the theorem of limit under the sign of derivative, the inverse and the implicit function theorem,...&c.) - Thanks for the list – Nima Nov 15 2011 at 10:54 +1 for the remark about the inequality. – Nate Eldredge Nov 15 2011 at 19:23 One application for which you need the full MVT (and not just the inequality you mention) is in the proof of l'Hopital's rule (for which you actually need to strengthen the MVT a bit). However, I don't think there's any generalization of l'Hopital's rule to higher dimensions, so I guess that's nothing to be worried about. – Mark Schwarzmann Nov 15 2011 at 21:10 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Even for functions from ${\mathbb R}$ to ${\mathbb R}^2$ the Mean Value Theorem fails. Thus it is possible to go from $(0,0)$ to $(1,0)$ in time 1 and have the velocity vector never equal to $(1,0)$. - this is not correct. for vector valued functions the integral form of mean value theorem does hold – Nima Nov 15 2011 at 6:45 5 I think Robert Israel refers to the form of the MVT you quoted in the question, and that certainly does not hold for vector valued functions (e.g. $\exp(it)$ takes the same values at $0$ and $2\pi$) – Pietro Majer Nov 15 2011 at 7:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 19, "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": 0.935232937335968, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/293580/do-there-exist-bijections-between-the-following-sets	# Do there exist bijections between the following sets? Let $A$ be an infinite set. Do there exist bijections between the following sets? 1. $A$ and $A\setminus B$ where $B$ is a finite subset 2. $A$ and $A\times \{1, 2, \dots, n\}$ 3. $A$ and $A\times A$ - Yes to the three of them. What have you tried? Where are you stuck? – DonAntonio Feb 3 at 12:35 For $\mathbb{N}$ I know they hold because I can order $\mathbb{N}$. But for a general infinite set I tried to order it in some way and got stuck. – Montez Feb 3 at 12:38 2 @Montez Assume the Axiom of Choice, you can well-order to every set. – tetori Feb 3 at 12:40 I assume Cantor-Bernstein and/or cardinal arithmetic cannot be used? – DonAntonio Feb 3 at 12:53 I do know Cantor-Bernstein but do not know cardinal arithmetic. – Montez Feb 3 at 13:00 ## 2 Answers All the answers depend on your definitions of "infinite" and whether or not you assume the axiom of choice. If your definition of infinite means "not finite" then all answers are positive assuming the axiom of choice, and all answers might be "no" if the axiom of choice fails. However if we assume the axiom of choice the following holds: 1. Every infinite set has a countably infinite subset. Namely $A=A'\cup\{a_n\mid n\in\mathbb N\}$ where $A'$ is disjoint of the aforementioned sequence. If we removed $x_1,\ldots,x_n$ then by rearranging we may assume $x_i=a_i$ for all $i$, and then $$f(x)=\begin{cases} x& x\in A'\\a_{k+n} & x=a_k\end{cases}$$ is a bijection between $A$ and $A\setminus\{a_1,\ldots,a_n\}$. 2. This argument is slightly more difficult in the general case, so instead of proving this directly I will use the next part and deduce the following: $$\|A\|\leq\|A\times\{0,\ldots,n\}\|\leq\|A\times A\|=\|A\|$$ and therefore equality holds. 3. This is a non-trivial theorem whose proof is usually given by well-ordering the set $A$ and proving that every infinite well-ordered set has a bijection with its square. You can find the details here: About a paper of Zermelo Of course even without any assumptions all the three claims are true for $\mathbb N$ and $\mathbb R$, and other well-behaved objects. In some introductory courses they omit the axiom of choice, and others do mention it. The axiom itself is not difficult, and it allows us to "tame" (to a certain degree) the behavior of infinite sets. If you are unfamiliar with it, don't worry. It will certainly come into action if you continue with mathematics. - 1. Yes. The set $A\setminus B$ is infinite, hence there is an injective map $f\colon \mathbb N\to A\setminus B$. Also there is some bijection $g\colon \{1,\ldots,n\}\to B$. Define $$\begin{align}h\colon A\setminus B &\to A\\x&\mapsto\begin{cases} g(k)&\text{if }x=f(k), k\le n\\f(k-n)&\text{if }x=f(k), k>n\\x&\text{otherwise.}\end{cases}\end{align}$$ -	{"extraction_info": {"found_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": 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": 0.9254545569419861, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Pushdown_automaton	# Pushdown automaton In computer science, a pushdown automaton (PDA) is a type of automaton that employs a stack. The PDA is used in theories about what can be computed by machines. It is more capable than a finite-state machine but less capable than a Turing machine. Because its input can be described with a formal grammar, it can be used in parser design. The deterministic pushdown automaton can handle all deterministic context-free languages while the nondeterministic version can handle all context-free languages. The term "pushdown" refers to the fact that the stack can be regarded as being "pushed down" like a tray dispenser at a cafeteria, since the operations never work on elements other than the top element. A stack automaton, by contrast, does allow access to and operations on deeper elements. Stack automata can recognize a strictly larger set of languages than deterministic pushdown automata.[citation needed] A nested stack automaton allows full access, and also allows stacked values to be entire sub-stacks rather than just single finite symbols. ## Operation a diagram of the pushdown automaton Pushdown automata differ from finite state machines in two ways: 1. They can use the top of the stack to decide which transition to take. 2. They can manipulate the stack as part of performing a transition. Pushdown automata choose a transition by indexing a table by input signal, current state, and the symbol at the top of the stack. This means that those three parameters completely determine the transition path that is chosen. Finite state machines just look at the input signal and the current state: they have no stack to work with. Pushdown automata add the stack as a parameter for choice. Pushdown automata can also manipulate the stack, as part of performing a transition. Finite state machines choose a new state, the result of following the transition. The manipulation can be to push a particular symbol to the top of the stack, or to pop off the top of the stack. The automaton can alternatively ignore the stack, and leave it as it is. The choice of manipulation (or no manipulation) is determined by the transition table. Put together: Given an input signal, current state, and stack symbol, the automaton can follow a transition to another state, and optionally manipulate (push or pop) the stack. In general, pushdown automata may have several computations on a given input string, some of which may be halting in accepting configurations. If only one computation exists for all accepted strings, the result is a deterministic pushdown automaton (DPDA) and the language of these strings is a deterministic context-free language. Not all context-free languages are deterministic. As a consequence of the above the DPDA is a strictly weaker variant of the PDA and there exists no algorithm for converting a PDA to an equivalent DPDA, if such a DPDA exists. If we allow a finite automaton access to two stacks instead of just one, we obtain a more powerful device, equivalent in power to a Turing machine. A linear bounded automaton is a device which is more powerful than a pushdown automaton but less so than a Turing machine. ## Relation to backtracking Nondeterministic PDAs are able to handle situations where more than one choices of action are available. In principle it is enough to create in every such case new automaton instances that will handle the extra choices. The problem with this approach is that in practice most of these instances quickly fail. This can severely affect the automaton's performance as the execution of multiple instances is a costly operation. Situations such as these can be identified in the design phase of the automaton by examining the grammar the automaton uses. This makes possible the use of backtracking in every such case in order to improve performance. ## Formal Definition We use standard formal language notation: $\Gamma^{}$ denotes the set of strings over alphabet $\Gamma$ and $\varepsilon$ denotes the empty string. A PDA is formally defined as a 7-tuple: $M=(Q,\ \Sigma,\ \Gamma,\ \delta, \ q_{0},\ Z, \ F)$ where • $\, Q$ is a finite set of states • $\,\Sigma$ is a finite set which is called the input alphabet • $\,\Gamma$ is a finite set which is called the stack alphabet • $\,\delta$ is a finite subset of $Q \times (\Sigma \cup\{\varepsilon\}) \times \Gamma \times Q \times \Gamma^$, the transition relation. • $\,q_{0}\in\, Q$ is the start state • $\ Z\in\,\Gamma$ is the initial stack symbol • $F\subseteq Q$ is the set of accepting states An element $(p,a,A,q,\alpha) \in \delta$ is a transition of $M$. It has the intended meaning that $M$, in state $p \in Q$, with $a \in \Sigma \cup\{\varepsilon\}$ on the input and with $A \in \Gamma$ as topmost stack symbol, may read $a$, change the state to $q$, pop $A$, replacing it by pushing $\alpha \in \Gamma^$. The $(\Sigma \cup\{\varepsilon\})$ component of the transition relation is used to formalize that the PDA can either read a letter from the input, or proceed leaving the input untouched. In many texts the transition relation is replaced by an (equivalent) formalization, where • $\,\delta$ is the transition function, mapping $Q \times (\Sigma \cup\{\varepsilon\}) \times \Gamma$ into finite subsets of $Q \times \Gamma^$. Here $\delta(p,a,A)$ contains all possible actions in state $p$ with $A$ on the stack, while reading $a$ on the input. One writes $(q,\alpha) \in \delta(p,a,A)$ for the function precisely when $(p,a,A,q,\alpha) \in\delta$ for the relation. Note that finite in this definition is essential. Computations a step of the pushdown automaton In order to formalize the semantics of the pushdown automaton a description of the current situation is introduced. Any 3-tuple $(p,w,\beta) \in Q \times \Sigma^* \times \Gamma^$ is called an instantaneous description (ID) of $M$, which includes the current state, the part of the input tape that has not been read, and the contents of the stack (topmost symbol written first). The transition relation $\delta$ defines the step-relation $\vdash_{M}$ of $M$ on instantaneous descriptions. For instruction $(p,a,A,q,\alpha) \in \delta$ there exists a step $(p,ax,A\gamma) \vdash_{M} (q,x,\alpha\gamma)$, for every $x\in\Sigma^$ and every $\gamma\in \Gamma^$. In general pushdown automata are nondeterministic meaning that in a given instantaneous description $(p,w,\beta)$ there may be several possible steps. Any of these steps can be chosen in a computation. With the above definition in each step always a single symbol (top of the stack) is popped, replacing it with as many symbols as necessary. As a consequence no step is defined when the stack is empty. Computations of the pushdown automaton are sequences of steps. The computation starts in the initial state $q_{0}$ with the initial stack symbol $Z$ on the stack, and a string $w$ on the input tape, thus with initial description $(q_{0},w,Z)$. There are two modes of accepting. The pushdown automaton either accepts by final state, which means after reading its input the automaton reaches an accepting state (in $F$), or it accepts by empty stack ($\varepsilon$), which means after reading its input the automaton empties its stack. The first acceptance mode uses the internal memory (state), the second the external memory (stack). Formally one defines 1. $L(M) = \{ w\in\Sigma^ \| (q_{0},w,Z) \vdash_M^* (f,\varepsilon,\gamma)$ with $f \in F$ and $\gamma \in \Gamma^* \}$ (final state) 2. $N(M) = \{ w\in\Sigma^* \| (q_{0},w,Z) \vdash_M^* (q,\varepsilon,\varepsilon)$ with $q \in Q \}$ (empty stack) Here $\vdash_M^$ represents the reflexive and transitive closure of the step relation $\vdash_M$ meaning any number of consecutive steps (zero, one or more). For each single pushdown automaton these two languages need to have no relation: they may be equal but usually this is not the case. A specification of the automaton should also include the intended mode of acceptance. Taken over all pushdown automata both acceptance conditions define the same family of languages. Theorem. For each pushdown automaton $M$ one may construct a pushdown automaton $M'$ such that $L(M)=N(M')$, and vice versa, for each pushdown automaton $M$ one may construct a pushdown automaton $M'$ such that $N(M)=L(M')$ ## Example The following is the formal description of the PDA which recognizes the language $\{0^n1^n \mid n \ge 0 \}$ by final state: PDA for $\{0^n1^n \mid n \ge 0\}$ (by final state) $M=(Q,\ \Sigma,\ \Gamma,\ \delta, \ p,\ Z, \ F)$, where $Q = \{ p,q,r \}$ $\Sigma = \{0, 1\}$ $\Gamma = \{A, Z\}$ $q_{0} = p$ $F = \{r\}$ $\delta$ consists of the following six instructions: $(p,0,Z,p,AZ)$, $(p,0,A,p,AA)$, $(p,\epsilon,Z,q,Z)$, $(p,\epsilon,A,q,A)$, $(q,1,A,q,\epsilon)$, and $(q,\epsilon,Z,r,Z)$. In words, in state $p$ for each symbol $0$ read, one $A$ is pushed onto the stack. Pushing symbol $A$ on top of another $A$ is formalized as replacing top $A$ by $AA$. In state $q$ for each symbol $1$ read one $A$ is popped. At any moment the automaton may move from state $p$ to state $q$, while it may move from state $q$ to accepting state $r$ only when the stack consists of a single $Z$. There seems to be no generally used representation for PDA. Here we have depicted the instruction $(p,a,A,q,\alpha)$ by an edge from state $p$ to state $q$ labelled by $a; A/\alpha$ (read $a$; replace $A$ by $\alpha$). ## Understanding the computation process accepting computation for $0011$ The following illustrates how the above PDA computes on different input strings. The subscript $M$ from the step symbol $\vdash$ is here omitted. (a) Input string = 0011. There are various computations, depending on the moment the move from state $p$ to state $q$ is made. Only one of these is accepting. (i) $(p,0011,Z) \vdash (q,0011,Z) \vdash (r,0011,Z)$. The final state is accepting, but the input is not accepted this way as it has not been read. (ii) $(p,0011,Z) \vdash (p,011,AZ) \vdash (q,011,AZ)$. No further steps possible. (iii) $(p,0011,Z) \vdash (p,011,AZ) \vdash (p,11,AAZ) \vdash (q,11,AAZ)$ $\vdash (q,1,AZ) \vdash (q,\epsilon,Z)$ $\vdash (r,\epsilon,Z)$. Accepting computation: ends in accepting state, while complete input has been read. (b) Input string = 00111. Again there are various computations. None of these is accepting. (i) $(p,00111,Z) \vdash (q,00111,Z) \vdash (r,00111,Z)$. The final state is accepting, but the input is not accepted this way as it has not been read. (ii) $(p,00111,Z) \vdash (p,0111,AZ) \vdash (q,0111,AZ)$. No further steps possible. (iii) $(p,00111,Z) \vdash (p,0111,AZ) \vdash (p,111,AAZ) \vdash (q,111,AAZ)$ $\vdash (q,11,AZ) \vdash (q,1,Z)$ $\vdash (r,1,Z)$. The final state is accepting, but the input is not accepted this way as it has not been (completely) read. ## PDA and Context-free Languages Every context-free grammar can be transformed into an equivalent pushdown automaton. The derivation process of the grammar is simulated in a leftmost way. Where the grammar rewrites a nonterminal, the PDA takes the topmost nonterminal from its stack and replaces it by the right-hand part of a grammatical rule (expand). Where the grammar generates a terminal symbol, the PDA reads a symbol from input when it is the topmost symbol on the stack (match). In a sense the stack of the PDA contains the unprocessed data of the grammar, corresponding to a pre-order traversal of a derivation tree. Technically, given a context-free grammar, the PDA is constructed as follows. 1. $(1,\varepsilon,A,1,\alpha)$ for each rule $A\to\alpha$ (expand) 2. $(1,a,A,1,\varepsilon)$ for each terminal symbol $a$ (match) As a result we obtain a single state pushdown automaton, the state here is $1$, accepting the context-free language by empty stack. Its initial stack symbol equals the axiom of the context-free grammar. The converse, finding a grammar for a given PDA, is not that easy. The trick is to code two states of the PDA into the nonterminals of the grammar. Theorem. For each pushdown automaton $M$ one may construct a context-free grammar $G$ such that $N(M)=L(G)$. ## Generalized Pushdown Automaton (GPDA) A GPDA is a PDA which writes an entire string of some known length to the stack or removes an entire string from the stack in one step. A GPDA is formally defined as a 6-tuple: $M=(Q,\ \Sigma,\ \Gamma,\ \delta, \ q_{0}, \ F)$ where Q, $\Sigma\,$, $\Gamma\,$, q0 and F are defined the same way as a PDA. $\,\delta$: $Q \times \Sigma_{\epsilon} \times \Gamma^{} \longrightarrow P( Q \times \Gamma^{} )$ is the transition function. Computation rules for a GPDA are the same as a PDA except that the ai+1's and bi+1's are now strings instead of symbols. GPDA's and PDA's are equivalent in that if a language is recognized by a PDA, it is also recognized by a GPDA and vice versa. One can formulate an analytic proof for the equivalence of GPDA's and PDA's using the following simulation: Let $\delta$(q1, w, x1x2...xm) $\longrightarrow$ (q2, y1y2...yn) be a transition of the GPDA where $q_1, q_2 \in Q$, $w \in\Sigma_{\epsilon}$, $x_1, x_2,\ldots,x_m\in\Gamma^{}$, $m\geq0$, $y_1, y_2,\ldots,y_n\in\Gamma^{*}$, $n\geq 0$. Construct the following transitions for the PDA: $\delta^{'}$(q1, w, x1) $\longrightarrow$ (p1, $\epsilon$) $\delta^{'}$(p1, $\epsilon$, x2) $\longrightarrow$ (p2, $\epsilon$) $\vdots$ $\delta^{'}$(pm-1, $\epsilon$, xm) $\longrightarrow$ (pm, $\epsilon$) $\delta^{'}$(pm, $\epsilon$, $\epsilon$ ) $\longrightarrow$ (pm+1, yn) $\delta^{'}$(pm+1, $\epsilon$, $\epsilon$ ) $\longrightarrow$ (pm+2, yn-1) $\vdots$ $\delta^{'}$(pm+n-1, $\epsilon$, $\epsilon$ ) $\longrightarrow$ (q2, y1) ## References • Michael Sipser (1997). Introduction to the Theory of Computation. PWS Publishing. ISBN 0-534-94728-X. Section 2.2: Pushdown Automata, pp.101–114. • Jean-Michel Autebert, Jean Berstel, Luc Boasson, Context-Free Languages and Push-Down Automata, in: G. Rozenberg, A. Salomaa (eds.), Handbook of Formal Languages, Vol. 1, Springer-Verlag, 1997, 111-174.	{"extraction_info": {"found_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": 159, "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": 0.8952814340591431, "perplexity_flag": "middle"}
http://mathhelpforum.com/differential-geometry/123731-uniform-continuity.html	# Thread: 1. ## uniform continuity Let $f:\mathbb{R}^{2} \to \mathbb{R}$ be defined as $f(x,y)=\text{max}\{\|x\|,\|y\|\}$. Prove that f is uniformly continuous 2. Originally Posted by Chandru1 Let $f:\mathbb{R}^{2} \to \mathbb{R}$ be defined as $f(x,y)=\text{max}\{\|x\|,\|y\|\}$. Prove that f is uniformly continuous Okay, what is the definition of "uniform continuity"? Also, I would suggest dividing this into 4 cases: $x\ge 0$, $\|y\|\le \|x\|$; x< 0, \|y\|< \|x\|; $x\ge 0$, $\|y\|\ge \|x\|$, and x< 0, \|y\|< \|x\|. 3. You can also show that the function is Holder continuous on $\mathbb{R}^2$; that is, $\|f(\mathbf{y})-f(\mathbf{x})\| \le C\|\mathbf{y}-\mathbf{x}\|^{\lambda}$ $\forall \mathbf{x}, \mathbf{y} \in \mathbb{R}^2,$ where $C$ and $\lambda$ are nonnegative real constants. If a function is Holder continuous, then it is uniformly continuous. #### 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": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": 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": 0.9499346017837524, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/27383/a-question-on-finding-the-intersecting-line-between-two-planes	A question on finding the intersecting line between two planes According to my math book, in order to find the intersecting line between two planes we need to: 1. Find the vector product of the direction normals of the two planes 2. Write the equations of the planes in Cartesian form. 3. Assume that $z=0$ since the line has to intersect this plane. 4. Solve simultaneously for a point on the line and write down a vector equation of the line What is meant by point 3? I'm having problems visualizing it. Please be explicit. - 1 Answer To find the line of intersection, you need two pieces of information. The direction of the line as a vector and a point on the line. Step 3 and the first part of step 4 is about locating a single point on the line of intersection. Since you are only looking for a point, you might as well assume $z=0$, to reduce your two plane equations to two equations in two unknowns which can be solved simultaneously. Now this might not always work. Sometimes the line of intersection happens to be parallel to the $z=0$ plane. In that case you could try $y=0$ or $x=0$. (One of these is sure to work.) - Thank you for your answer. I realize why we assume $z=0$ equation-wise, but I didn't really get the vector-space explanation. If by plane they mean the $z$-axis, the intersecting line does NOT always have to intersect it (in case of one of the planes being parallel to the $y$-axis for instance). Then you have to try with either $x=0$ or $y=0$ like you said. If there really isn't anything more to it, than I guess the problem here is the vague(incorrect) explanation in my math book - am I correct? – Milosz Wielondek Mar 16 '11 at 11:24 1 I agree with you except that $z=0$ represents the $xy$-plane, not the $z$-axis. Still, if your book says that $z=0$ always works, then it is not quite right. – Grumpy Parsnip Mar 16 '11 at 11:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.9632168412208557, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/tagged/energy+spring	# Tagged Questions 1answer 74 views ### Simple harmonic oscillator system and changes in its total energy Suppose I have a body of mass $M$ connected to a spring (which is connected to a vertical wall) with a stiffness coefficient of $k$ on some frictionless surface. The body oscillates from point $C$ to ... 2answers 307 views ### Dynamics of a Vertical Mass-Spring Simple Harmonic Oscillator with Gravity I am having some trouble obtaining the elastic potential energy and gravitational potential energy of a simple mass spring system. In this experiment, masses attached to a spring were dropped from a ... 0answers 132 views ### Potential energy of a spring I have a little problem with the potential energy of a spring... I hope you can help me! I have two coupled pendula, given by two masses $m$ fixed to two rigid bars (that haven't any mass) and with ... 2answers 406 views ### What is the formula for max kinetic and max potential energy of a spring? What is the formula for max kinetic and max potential energy of a spring? 1answer 420 views ### Maximum Kinetic energy of a spring The block in the figure below lies on a horizontal frictionless surface and is attached to the free end of the spring, with a spring constant of 35 N/m. Initially, the spring is at its relaxed ... 1answer 281 views ### Two Blocks of mass M1 and M2 are connected by a spring of force constant k [closed] If Block 1 is elongated towards right to a distance $x_1$ and Block 2 is elongated towards left to a distance of $x_2$ simultaneously, what is the work done by the spring on each of these blocks ... 1answer 88 views ### Why are there two ways to solve for energy of a spring? I can find the energy of a spring using $F = -kx$, or by using the formula $e = 1/2mv^2 + 1/2I\omega^2 + mgh + 1/2kx^2$. The first way, I get $mg/k = x$, but the second way, I get $2mg/k = x$. Which ... 1answer 303 views ### Work and Area under a Curve relating to Hooke's Law If it takes work W to stretch a Hooke’s-law spring (F = kx) a distance d from its unstressed length, determine the extra work required to stretch it an additional distance d (Hint: draw a graph and ... 4answers 612 views ### 17 Joules of Energy From a Mouse Trap Do you think it would be possible to get 17 joules out of a standard size mouse trap. By my math, it is a torsion coefficient of 3.45 or so out of the spring. 2answers 39 views ### Energy transferring question If I have a spring being compressed by two bodies, A and B, with different masses, how much energy would be transferred to each one when they are released and the spring expanded? 1answer 116 views ### In continuum mechanics, what is work potential in the context of total potential energy? I'm reading a book on the finite element method. Specifically I'm looking at the background material where they are discussing potential energy, equilibrium, and the Rayleigh-Ritz method. The book ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9370060563087463, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/158168/how-to-show-that-f-is-an-odd-function?answertab=votes	# How to show that $f$ is an odd function? An entire function $f$ takes real $z$ to real and purely imaginary to purely imaginary. We need to show that $f$ is an odd function. well, $f=\sum_{n=0}^{\infty}a_nz^n$ what I can say is $f(\mathbb{R})\subseteq\mathbb{R}$ and $f(\mathbb{iR})\subseteq\mathbb{iR}$ How to proceed, please give me hint. - ahhh! I just got it since $f(\mathbb{R})\subseteq\mathbb{R}$ so each $a_n$ is real again since $f(\mathbb{iR})\subseteq\mathbb{iR}$ each $a_n=0$ for even $n$ am I right? – Taxi Driver Jun 14 '12 at 9:08 You may be expected to show that the two conditions respectively force (i) the $a_n$ to be real and (ii) the even coefficients to be $0$. – André Nicolas Jun 14 '12 at 9:28 @Mex that is correct. – TenaliRaman Jun 14 '12 at 9:30 – TenaliRaman Jun 14 '12 at 9:32 1 @Mex: I encourage you to answer your own question and accept it, if you think you have the correct answer. – mixedmath♦ Jun 14 '12 at 10:38 show 3 more comments ## 1 Answer I am writing, enlarging and enhancing (hopefully...) Mex's answer to his own question (kudos!) and I'll be happy to erase this answer of mine if he decides to write down his. We have that $$f(z)=\sum_{n=0}^\infty a_nz^n$$ because $\,f\,$ is entire, and by the given conditions we have$$(1)\,\,f(r)=\sum_{n=0}^\infty a_nr^n\in\mathbb{R}\,,\,\,\,r\in\mathbb{R}$$$$(2)\,\,f(ir)=\sum_{n=0}^\infty a_n(ir)^n\in i\mathbb{R}\,\,,\,\,r\in\mathbb{R}$$but we have that $$\sum_{n=0}^\infty a_n(ir)^n=\sum_{n=0}^\infty i^n (a_nr^n)=\sum_{n=0}^\infty (-1)^na_{2n}r^{2n}+i\sum_{n=0}^\infty (-1)^na_{2n+1}r^{2n+1}$$and as the above is purely imaginary we get that $\,a_{2n}=0\,,\,\forall n\in\mathbb{N}\,$ , so the power series of the function has zero coefficients for the even powers of $\,z\,$ and is thus a sum of odd powers and trivially then an odd function. - Don :) my great pleasure, thank you very much. – Taxi Driver Jun 14 '12 at 14:42 (1)You've very much welcome, @Mex, and (2) thanks for giving the solution to your own question. – DonAntonio Jun 14 '12 at 14:50	{"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": 17, "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": 0.9333441853523254, "perplexity_flag": "middle"}
http://gilkalai.wordpress.com/2008/11/24/sarkarias-proof-of-tverbergs-theorem-1/?like=1&source=post_flair&_wpnonce=1c9f85538d	Gil Kalai’s blog ## Sarkaria’s Proof of Tverberg’s Theorem 1 Posted on November 24, 2008 by Helge Tverberg Ladies and gentlemen, this is an excellent time to tell you about the beautiful theorem of Tverberg and the startling proof of Sarkaria to Tverberg’s theorem (two parts). A good place to start is Radon’s theorem. ## 1. The theorems of Radon, Helly, and Caratheodory. ### Radon’s theorem Radon’s theorem: Let $x_1,x_2,\dots, x_m$ be points in $R^d$, $m \ge d+2$. Then there is a partition $S,T$ of $\{1,2,\dots,m\}$ such that $conv(x_i: i \in S) \cap conv(x_i: i \in T)$ $\ne \emptyset$. (Such a partition is called a Radon partition.) Proof: Since $m>d+1$ the points $x_1,x_2, \dots, x_m$ are affinely dependent. Namely, there are coefficients, not all zero, $\lambda_1,\lambda_2,\dots,\lambda_m$ such that $\sum _{i=1}^m \lambda_i x_i = 0$ and $\sum _{i=1}^m \lambda_i=0$. Now we can write $S=\{i:\lambda_i >0\}$ and $T = \{i: \lambda_i \le 0\}$, and note that () $\sum_{i \in S} \lambda_i x_i = \sum _{i \in T} (-\lambda_i) x_i$, and also $\sum _{i \in S} \lambda_i = \sum_{i \in T} (-\lambda_i)$. This last sum is positive because not all the $\lambda$s are equal to zero. We call it $t$. To exhibit a convex combination of $x_i: i\in S$ which is equal to a convex combination in $x_i: i \in T$ just divide relation () by $t$. Walla. This trick of basing a partition on the signs of the coefficients repeats in other proofs. Take note! Radon used his theorem to prove Helly’s theorem. ### Helly’s theorem Helly’s theorem: For a family $K_1,K_2,\dots, K_n$, $n \ge d+1$, of convex sets in $R^d$, if every $d+1$ of the sets have a point in common then all of the sets have a point in common. Proof: It is enough to show that when $n> d+1$ if every $n-1$ of the sets have a point in common then there is a point in common to them all. Let $x_k \in \cap\{K_j: 1\le j\le n, j \ne k\}$. In words, $x_k$ is a point that belongs to all the $K_j$s except perhaps to $K_k$. We assumed that such a point $x_k$ exists for every $k$. Now we can apply Radon’s theorem: Consider the Radon partition of the points, namely a partition $S,T$ of $\{1,2,\dots,m\}$ such that $conv(x_i: i \in S) \cap conv(x_i: i \in T)$ $\ne \emptyset$. Let $z \in$ $conv(x_i: i \in S) \cap conv (x_i: i \in T)$. Since $z$ belongs to the convex hull of $conv (x_i: i \in S)$ and every $x_i$ for $i \in S$ belongs to every $K_j$ for every $j$ not in $S$ we obtain that $z$ belongs to every $K_j$ for $j$ not in $S$. By the same argument $z$ belongs to every $K_j$ for $j$ not in $T$. Since $S$ and $T$ are disjoint, the point $z$ belongs to all $K_j$s. Ahla. The proof of Helly’s theorem from Radon’s theorem as described on the cover of a book by Steven R. Lay ### Caratheodory’s theorem Caratheodory’s theorem: For $S \subset R^d$, if $x \in conv (S)$ then $x \in conv (R)$ for some $R \subset S$, $\|R\| \le d+1$. Like for Radon’s theorem, there is a simple linear algebra proof. We can assume that $S$ is finite; we start with a presentation of $x$ as a convex combination of vectors in $S$, and if $\|S\|>d+1$ we can use an affine dependency among the vectors in $S$ to change the convex combination, forcing one coefficient to vanish. Without further ado here is Tverberg’s theorem. ## 2. Tverberg’s Theorem Tverberg Theorem (1965):Let $x_1,x_2,\dots, x_m$ be points in $R^d$, $m \ge (r-1)(d+1)+1$. Then there is a partition $S_1,S_2,\dots, S_r$ of $\{1,2,\dots,m\}$ such that $\cap _{j=1}^rconv (x_i: i \in S_j) \ne \emptyset$. So Tverberg’s theorem is very similar to Radon’s theorem except that we want more than two parts! The first thing to note is that Tverberg’s theorem is sharp. If you have only $(r-1)(d+1)$ points in $R^d$ in a “generic” position then for every partition into $r$ parts even the affine spans of the points in the parts will not have a point in common. (It is not entirely easy to prove it; and “generic” should be more than just “in general position.”) The first proof of this theorem appeared in 1965. It was rather complicated and was based on the idea to first prove the theorem for points in some special position and then show that when you continuously changethe location of the points the theorem remains true. A common dream was to find an extension of the proof of Radon’s theorem, a proof which is based on the two types of numbers - positive and negative. Somehow we need three, four, or $r$ types of numbers. In 1981 Helge Tverberg found yet another proof of his theorem. This proof was inspired by Barany’s proof of the colored Caratheodory theorem (mentioned below) and it was still rather complicated. It once took me 6-7 hours in class to present it. What could be the probability of hearing two new simple proofs of Tverberg’stheorem on the same day? While visiting the Mittag-Leffler Institute in 1992, I met Helge one day around lunch and he told me about a new proof that he found with Sinisa Vrecica. This is a proof that can be presented in class in 2 hours! It appeared (look here) along with a far-reaching conjecture (still unproved). Later in the afternoon I met Karanbir Sarkaria and he told me about a proof he found to Tverberg’s theorem which was absolutely startling. This is a proof you can present in a one hour lecture; it also somehow goes along with the dream of having $r$ “types” of numbers replacing the role of positive and negative real numbers. Another very simple proof of Tverberg’s theorem was found by Jean-Pierre Roudneff in 1999. (Here is Sarkaria’s homepage.) ## 3. Colorful Caratheodory The Colorful Caratheodory Theorem (Barany): Let $S_1, S_2, \dots S_{d+1}$ be $d+1$ sets in $R^d$. Suppose that $x \in conv(S_1) \cap conv (S_2) \cap conv (S_3) \cap \dots \cap conv (S_{d+1})$. Then there are $x_1 \in S_1$, $x_2 \in S_2$, $\dots x_{d+1} \in S_{d+1}$ such that $x \in conv (x_1,x_2,\dots,x_{d+1})$. Note that when all the sets $S_i$ are equal we return to Caratheodory’s theorem. (Why is it called colorful? Because we can think of the different $S_i$s as representing different colors and the conclusion will be that $x$ is in the convex hull of a set of points containing one point of each color. Indeed, colorful.) Proof: We will consider the minimum distance $D$ between $x$ and $conv (x_1,x_2,\dots,x_{d+1})$ where $x_1 \in S_1$, $x_2 \in S_2$, $\dots x_{d+1} \in S_{d+1}$. Suppose that $z \in conv (x_1,x_2, \dots, x_{d+1})$ realizes this distance, and suppose by contradiction that $z \ne x$, i.e., $D>0$. Write $S = conv (x_1,x_2,\dots,x_{d+1})$. Now consider the hyperplane $H$ through the point $z$ which is perpendicular to the line through latex $z$ and $x$. The point $x$ belongs to the interior of one closed halfspace $H^-$ and we first note that $S$ belongs to the other closed halfspace$H^+$. Indeed, if $u$ was a point in $S$ in the interior of $H^-$ then the anglebetween [$u,z$] and [$x,z$] would be less than 180 degrees and a point on the interval [$z,u$] very close to $z$ would be at a shorter distance from $x$ than $z$. This is impossible. It follows that $z \in conv \{x_i: 1 \le i \le d+1, x_i \in H \}$. By Caratheodory’s theorem (in dimension $d-1$), $z$ is in the convex hull of $d$ of the $x_i$s. In other words: $z \in conv (x_1,x_2,\dots,x_{j-1}$, $x_{j+1},\dots,x_{d+1})$, for some $j$. Let’s look now at $S_j$. Is it possible that $S_j \subset H^+$? No! This is impossible since $x \in conv(S_j)$. Therefore, there is $x'_j \in S_j$ such that $x'_j$ belongs to the interior of $H^-$. Consider now $S' = conv (\{x_1,x_2,\dots ,x_{j-1},x'_j,x_{j+1},x_{d+1}\})$. Both $z$ and $x'_j$ belong to $S'$. Since $x'_j$ is in the interior of $H^-$ so is the half open interval $(z,x'_j]$, and a point on this interval very close to $z$ will be closer to $x$ than $z$. This is a contradiction. Sababa! Imre Barany ### Like this: This entry was posted in Convexity and tagged Helly type theorems, Tverberg's theorem. Bookmark the permalink. ### 13 Responses to Sarkaria’s Proof of Tverberg’s Theorem 1 1. Pingback: Tverberg’s theorem « Konrad Swanepoel’s blog 2. NH says: Beautiful post, thank you. Small typo: In the last paragraph, when you say “z \in conv(S_j)”, this should be “x \in conv(S_j)” 3. Gil Kalai says: Thanks, NH! Corrected. 4. Pingback: Seven Problems Around Tverberg’s Theorem « Combinatorics and more 5. a says: Thanks Gil !!!! the Ahla, Sababa and Walla are very cute too 6. Pingback: Colloquium at Berlin « Combinatorics and more 7. Pingback: Colorful Caratheodory Revisited « Combinatorics and more 8. Pingback: Radon Related Problem 3 « Euclidean Ramsey Theory 9. Austin says: In the proof of the Colorful Carathéodory Theorem, is there necessarily some “z” achieving the minimum distance as stated? I think not, unless we assume that the sets S_i are finite. However, we should be able to do this with no loss of generality: since x is a convex combination of points from S_i (for each i), we may assume S_i to consist of only those points which contribute to that combination. Yes? 10. Gil says: Right! 11. Phyllis Hazlegrove says: I found this site from searching on Google and just wanted to say thank you for this interesting webpage on weight loss. Thanks again! 12. Natalya Vivona says: Thank you very much for the post! Many guys always have no idea why they couldn’t burn fat or have a better figure. The point is they finding for a magic bullet which brings them what they desire immediately meanwhile all they should do is reading useful posts like this as well as implement the training. 13. Pingback: Colorful Caratheodory Theorem \| Math by Matt • ### Blogroll %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 141, "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": 0.9167280197143555, "perplexity_flag": "head"}
http://nrich.maths.org/6541	### F'arc'tion At the corner of the cube circular arcs are drawn and the area enclosed shaded. What fraction of the surface area of the cube is shaded? Try working out the answer without recourse to pencil and paper. ### Do Unto Caesar At the beginning of the night three poker players; Alan, Bernie and Craig had money in the ratios 7 : 6 : 5. At the end of the night the ratio was 6 : 5 : 4. One of them won \$1 200. What were the assets of the players at the beginning of the evening? ### Plutarch's Boxes According to Plutarch, the Greeks found all the rectangles with integer sides, whose areas are equal to their perimeters. Can you find them? What rectangular boxes, with integer sides, have their surface areas equal to their volumes? # The Greedy Algorithm ##### Stage: 3 Challenge Level: This problem follows on from Keep it Simple and Egyptian Fractions So far you may have looked at how the Egyptians expressed fractions as the sum of different unit fractions. You may have started by considering fractions with small numerators, such as $\frac{2}{5}$, $\frac{3}{7}$, $\frac{4}{11}$, etc. But how would the Egyptians have coped with fractions with large numerators such as $\frac{115}{137}$? They might have written $\frac{115}{137} = \frac{1}{137} + \frac{1}{137} + \frac{1}{137}$.... and then used Jamie's method to make them all different, but this would have made an extremely lengthy calculation! #### Fibonacci found an alternative strategy, called the Greedy Algorithm: At every stage, write down the largest possible unit fraction that is smaller than the fraction you're working on. For example, let's start with $\frac{11}{12}$: The largest possible unit fraction that is smaller than $\frac{11}{12}$ is $\frac{1}{2}$ $\frac{11}{12} - \frac{1}{2} = \frac{5}{12}$ So $\frac{11}{12} = \frac{1}{2} + \frac{5}{12}$ The largest possible unit fraction that is smaller than $\frac{5}{12}$ is $\frac{1}{3}$ $\frac{5}{12} - \frac{1}{3} = \frac{1}{12}$ So $\frac{11}{12} = \frac{1}{2} + \frac{1}{3} + \frac{1}{12}$ #### Choose a fraction of your own and apply the Greedy Algorithm to see if you can finish up with a string of unit fractions. Does the greedy algorithm always work? Can all fractions be expressed as a sum of different unit fractions by applying the Greedy Algorithm? Can you convince yourself of this? Why do you think it is called the Greedy Algorithm? What do these words mean in a mathematical context? 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": 14, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.9415193796157837, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/141133/how-to-prove-lim-x-y-to0-0-fracxyxy-0	# How to prove $\lim_{(x,y)\to(0,0)} \frac{xy}{x+y} = 0$ I want to find $\displaystyle\lim_{(x,y)\to(0,0)} \frac{xy}{x+y} = 0$. After trying different ways to approach $(0,0)$, I am fairly convinced the limit is $0$, but I need to prove it by definition, and I seem to be stuck. I want to prove that $\forall \epsilon >0, \exists \delta > 0$ such that $0<\parallel(x,y)\parallel<\delta \implies \|\frac{xy}{x+y}\| < \epsilon$. I'm having trouble with the denominator. I know that to get $\|\frac{xy}{x+y}\|$ to be less than something, I have to show that $\|x+y\|$ can be made greater than something, but I don't know what. Any suggestions? - What if you (can you?) approach along the line $y = -x$? – Dylan Moreland May 4 '12 at 21:03 @Dylan: Well, the function is not even defined along that line. Does that matter? – Javier Badia May 4 '12 at 21:05 @Javier: Well, it matters in-so-far as that means that your definition cannot be satisfied, since you can find points $(x,y)$ with $0\lt\lVert (x,y)\rVert \lt\delta$ but for which $\left\|\frac{xy}{x+y}\right\|\lt\epsilon$ does not hold... – Arturo Magidin May 4 '12 at 21:08 consider approaching on this curve $y=-x+x^3$, when $x$ is relatively small. – Yimin May 4 '12 at 21:10 @ArturoMagidin: Oh, I thought you could just sort of ignore it. This means that the limit doesn't exist, then? – Javier Badia May 4 '12 at 21:13 show 1 more comment ## 1 Answer If you approach $(0,0)$ along the line $x=0$ the function has constant value $0$ and the limit is $0$. But now suppose that you approach along a curve like $y=x^2-x$. Then $$\frac{xy}{x+y} = \frac{x^3-x^2}{x^2} =x - 1$$ and the limit as $x\to 0$ is... - I guess this is a simpler way to know. How did you think of approaching along $y=x^2-x$? – Javier Badia May 4 '12 at 21:17 1 @Javier: Experience, I'm afraid, not much more. – Arturo Magidin May 4 '12 at 21:23 4 @JavierBadia Clearly approach along a linear function won't do the job. So lets try the quadratic $y=ax^2 + bx$. Then we get that $\frac{ax^3 + bx^2}{ax^2 + (b+1)x} = \frac{ax^2 + bx}{ax +(b+1)}$. Note that if $b \neq -1$, then the limit will be zero. If $b=-1$, then we get $\frac{ax^2 - x}{ax} = \frac{ax-1}{a}$ and the limit is $-\frac1{a}$. So moving along the curve $y = ax^2 -1$ gives the limit $-\frac1a$ – user17762 May 4 '12 at 21:25	{"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": 29, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9578173160552979, "perplexity_flag": "head"}
http://poker.stackexchange.com/questions/1063/the-importance-of-mathematics-in-poker/1074	# The Importance of Mathematics in Poker Poker is an awesome and very difficult game. But as a probability/strategy game is a subject of mathematics. However, I found a lot a rounders, even pros, that are completely blind to the power of mathematics in this game. The most evident example is in heads up play, that is completely resolved from the point of view of Nash equilibria. Looking closer at heads up strategy of the majority of pro players, it is possible to find a lot of trivial strategic errors. And yet, even on this useful website, it is still not possible to use latex to explain probabilistic formula! Why are people so reluctant to use mathematics in poker? - ## 5 Answers Your premise is off here. It isn't that high level players ignore math, it's that a lot of other concepts start to become more important. The incomplete nature of the information that we are given in poker means that it will never be a purely mathematical challenge. Figuring out what flaws exist in your opponent's game and then exploiting those flaws will result in better results than an approach based purely on the math of the game can provide. As you face more advanced opponents, the importance of these types of concepts only increases. Further, as IHars mentioned, even in the heads-up arena, math doesn't solve the game, because neither player ultimately plays according to the mathematical optimum. And for the one who is best at adjusting, they will end up achieving better results than a Nash-based approach would yield. - To add to this very good answer, it can also be a mathematically sound decision to make a sub-optimal play (one not supported by the mathematics of the hand) if it will allow you to achieve higher value in future hands because of it. (ie, if the other players perceive you as a "bluffer" because of how you played a few hands early, when you stop bluffing and start value betting in a way that looks like a bluff, it can be very profitable.) – lnafziger Feb 3 at 22:32 I'm not sure the idea that because we have incomplete information that means that it will never be a purely mathematical challenge. I'm not saying that poker is a purely mathematical challenge but there is a lot (most of applied maths) maths dedicated to treating situations where we have incomplete information. – hmmmm Mar 9 at 12:56 Heads Up is very special type of game... Adjustment your style to the opponent, bluffing is more important than following to some mathematically optimal strategy. Look at Viktor Blom (Isildur1). His hyper-aggressive and "non-optimal" style baffles most opponents on high stakes. - 1 Bluffing correctly would actually be part of a mathematically optimal strategy, so although it might be more frequent, without extra information about profitability, it's problematic to assume that it's more or less important than any single part of a total strategy. – Toby Booth♦ Jan 30 at 20:02 The point of winning in poker is to maximize your EV. Math is great tool for analyzing multiple situations, and so part of your decision process. Many successful players, not knowing math, are really good at estimating their optimal play without using it. That is all about it, deciding your optimal moves better then your opponent. If you can do it w/o math does not matter. - Well my answer for you question comes in a few parts. First off I'm not too sure if you understand what you mean when you say that "heads up play is completely solved from the point of view of nash equilibria". Yes you can google for a couple of charts and get a good idea of some basic heads up strategy but saying "completely solved" is a bit misleading. Being in a Nash equilibria for a two player game means that player A cannot change his strategy unilaterally, in the knowledge of players B's strategy to make more profit and vice versa. One problem here is that player A needs to know player B's strategy and vice versa, in poker this is simply not the case. If both players are playing heads up play completely according to Nash equilibria then great but that isn't really the case and if it was then one player would surely adjust to make their play more explotive. The thing to notice about the Nash equilibria is that it give exploitative play but not optimal play. So saying that heads up play is completely solved by Nash equilibria is maybe a bit misleading (there is a Nash solution but it's not optimal). So I would say that my first point is that some people do seem to overestimate the importance of mathematics in poker. You can't just look up a few charts and have the game solved and maybe more importantly people sometimes fail to notice when mathematics is giving an explotive solution not an optimal one. Bringing me onto my second point which is maybe a bit more in response to some of the answers here is that people also seem to underestimate the importance of mathematics in poker. In the above answers I can see maybe see some misconceptions already (maybe not, I could just be reading them wrong). In reference to Isuldur1's hyper agressive play being "mathematically poor" and non-optimal, I would say is maybe a bit naive. Looking at that sort of strategy with some pretty basic mathematical tools we can see some advantages of it. (from phil gordon's little green book of poker) If you consider a heads up match with a hyper aggresive player (HAP) and player A playing regular poker, consider the following stacks are $5000, preflop pot$500: Player A (A, k) HAP (7,7) (Ts, 8s) ( 8s,5h) (5h,9s) Flop (Ah,5s,6s) HAP moves all in. What does player A do? Well, if you are in the knowledge that you are playing a hyper aggresive player with top pair on a flop such as this you may be thinking great, easy call but lets look at a rough table: Hand A chances A equitiy 1 (Ts, 8s) 52 5,481 2 (8s,5h) 63 6,615 3 (7,7) 1.6 168 4.(5h,9s) 76.3 8012 Player A total equitiy vs HAP: -\$1448 Now that is interesting. Playing very aggressively with these hand can actually give you a positive equity against a strong made hand. Now I am not attempting to argue that this in any way clears the matter up or that this is an in depth analysis- it clearly isn't but what it does show is a particular example of where some (very basic) math can give us a good strategy and some idea of where Isuldur1 and other players hyper aggresive strategy comes from. Of course if you want some more in depth analysis then you need to use some more advanced mathematics (it's easy to come up with a strategy that counters the above type of play) but as you start to look in more generality the math becomes more vague. This does not make it less useful and (I would argue contrary to some of the above answers) it does not make it less applicable in higher level games it just means you need some higher level maths. For anyone that is interested in maths and poker I would reccomend Bill Chen's Mathematics of Poker as the best book by far. I may need to come back and edit this post as it is a bit long. Also I would also like to say that we really should get latex (mathjax or something) for this forum as it would be extremely useful! (I might bring it up on meta as a suggestion) - Nice post, but I think the specific example you give is incorrect. maybe a typo? The matchup you state give's the Hero ~82% equity at this point. Otherwise, the post brings up some interesting points. – Toby Booth♦ Jan 30 at 19:56 Up to a point, mathematics is important. A good grasp of it will allow to beat average players. Beyond that point, psychology is key. "Good" players have a grasp of the basics. Then it resolves into, "he knows that I know that he knows," and then things resolve themselves into how deep the chain of thinking goes (how many "ply") in mathematics. That is something that say, a computer cannot be taught. -	{"extraction_info": {"found_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": 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": 0.9669521450996399, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/23577-proving-function-one-one.html	# Thread: 1. ## Proving a function is one-to-one Problem: Prove that the function f: Q ---> R, f(a/b) = (2^a)(3^b) is one to one, assuming that gcd(a,b) = 1, that is the fraction a/b is reduced. I know f is one to one if it never maps 2 different elements to the same place, i.e. f(a) does not equal f(b) whenever a does not equal b. I have tried looking at how other problems are proved to be one to one and am attempting to do it in a similar way: Suppose f(a/b) = f(c/d) then (2^a)(3^b) = (2^c)(3^d) I'm not sure what to do though. I don't think I'm doing this right. Please help me with this problem. 2. Originally Posted by hannahs Problem: Prove that the function f: Q ---> R, f(a/b) = (2^a)(3^b) is one to one, assuming that gcd(a,b) = 1, that is the fraction a/b is reduced. . Say $f(a_1/b_1)=f(a_2/b_2)\implies 2^{a_1}3^{b_1}=2^{a_2}3^{b_2}$ by unique factorization of the integers $a_1=a_2,b_1=b_2$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 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": 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": 0.9407211542129517, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/205273/dense-subset-of-the-cantor-set	# Dense subset of the Cantor set Prove that the set of endpoints of removed intervals in the Cantor middle thirds set is a dense subset of the Cantor set. Attempt at proof: Since each subinterval is of length $(1/3)^n$, any point contained in $K_n$ is at a distance of less than or equal to $(1/3)^n$ from an endpoint in $K_n$ for all $n$. Thus there exists a sequence of points in $K_n$ that converge to an endpoint in $K_n$. - You haven't told us what $K_n$ is, nor what it has to do with the Cantor middle-thirds set. – Gerry Myerson Oct 1 '12 at 5:46 @GerryMyerson I'm pretty sure he means $K_n$ to be the $n$-th approximation of the Cantor set you get, after removing the "middle thirds" $n$ times. – nullUser Oct 2 '12 at 23:09 @null, you may be right. It would be nice of OP to come back to clarify, but perhaps Heisenberg is uncertain. – Gerry Myerson Oct 3 '12 at 1:53 ## 1 Answer I assume that $K_0=[0,1]$, and that for $n\ge 1$, $K_n$ is what’s left after the open middle $(1/3)^n$-intervals have been removed from $K_{n-1}$. If you have already proved that $K_n$ is the union of $2^n$ closed intervals of length $(1/3)^n$, then the first sentence of your argument is fine. The second, however, goes badly astray: you want to show that each point of the Cantor set is the limit of a sequence of endpoints, not that each endpoint is the limit of a sequence of something. To fix this, let $p$ be any point of the Cantor set. Then for each $n\ge 1$ there is a unique closed interval of length $(1/3)^n$ in $K_n$ that contains $p$. At least one of the endpoints of that interval is an endpoint of a removed interval; choose one that is, and call it $e_n$. (You can’t guarantee that both are: one might be $0$ or $1$.) Now use your first observation to argue that $\langle e_n:n\in\Bbb Z^+\rangle$ is a sequence of endpoints of removed intervals that converges to $p$, so the set of endpoints of removed intervals must be dense in the Cantor 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": 29, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9445194602012634, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/23934/understanding-the-oscillating-part-of-the-gutzwiller-trace?answertab=votes	# understanding the oscillating part of the Gutzwiller trace given the density of states according to Gutzwiller's trace formula $g(E)= g_{smooth}(E)+ g_{osc}(E)$ i know that the 'smooth' part comes from $g_{smooth}(E)= \iint dxdp \delta(E-p^{2}-V(x))$ for one dimensional system however how it is the oscillating part of the trace obtained ?? :D i mean the sum over lenghts of the orbit (in the phase space) also how does the condition for WKB energies appear ?? $\oint _{C} p.dq= 2\pi \hbar (n+ \alpha)$ from the Gutzwiller trace ?? - ## 1 Answer The oscillatory part is nothing but Thomas-Fermi approximation or more riguresly, this is a version (someone should correct me if I am wrong) Weyl's formula Regrading on how to obtain the WKB from the trace formula: You can read the 2 papers by Berry and Tabor on how they derived a trace formula (like that of Gutzwiller) but to the case of integrable systems. From the derivation there you can see how the EBK pop up... -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 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": 0.8980500102043152, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/216984/finding-the-number-of-subset-pairs-of-a-set/216985	# Finding the number of subset pairs of a set If a set $S$ has $n$ elements, how many such pairs $(A,B)$ can be formed where $A$ and $B$ are subsets of $S$ and $A \cap B = \emptyset$? - ## 2 Answers Hint: If you don't insist that $A \cup B = S$, each element has three places it can go: into $A$, into $B$ or neither. - Knowing your answers, Ross, and seeing the upvotes, I'm convinced this is a good hint. But for the life of me I cannot see how it works :) I think I'm just not a combinatorics person... – rschwieb Oct 19 '12 at 17:06 1 So would it just be $3^n$? – user26069 Oct 19 '12 at 17:15 @rschwieb: think about picking up each element of $S$ in turn and putting into one of three boxes. How many ways are there to do that? Then the ones you put into first box are $A$, the second $B$ – Ross Millikan Oct 19 '12 at 17:18 @user46221: that's right. It is like the argument that the number of subsets of $S$ is $2^n$-each element can be in or out. – Ross Millikan Oct 19 '12 at 17:19 @rschwieb: What do you mean "no matter what sets $A,B$ you chose"? It is the very task to count the ways to chose $A,B$, not count some number for possibly different choices of $A,B$. – Hagen von Eitzen Oct 19 '12 at 17:33 show 3 more comments Hint: If you consider fixed $A$ for a moment, then $B$ must be a subset of the complement of $A$, and so there are $2^{n-i}$ choices for $B$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 24, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9333704113960266, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/30929/intersection-of-2-spheres-and-a-cube/30942	Intersection of 2 spheres and a cube In the Cartesian coordinate system, given 3 geometrical solid objects (interior plus boundary): spheres S1(x1,y1,z1, R1), S2(x2,y2,z2, R2) and a cube (which is orthogonal with coordinate system) at the center C (x3,y3,z3) with size L x L x L. The question is “Whether the 3 objects have any common volumetric intersection (Yes or No) ” ? I need a math solution or computing algorithm for this problem. - It depends on the values of your parameters. – Ronaldo Apr 4 '11 at 17:33 3 "Sphere" and "cube" could refer to either the solid object (interior plus boundary) or the surface (boundary only). Which specific type of intersection are you looking for? – mjqxxxx Apr 4 '11 at 17:45 2 Answers At first we need to check whether 2 spheres intersect or not. if (x1-x2)^2+ (y1-y2)^2+ (z1-y2)^2 > (R1+R2)^2 ---> flag of intersection=FALSE. else ( (i) Define x4=(x1R2+x2R1)/(R1+R2) y4=(y1R2+y2R1)/(R1+R2) z4=(z1R2+z2R1)/(R1+R2) (ii) if ( (abs(x4-x3)<=L/2) & ( abs(y4-y3)<=L/2) & ( abs(z4-z3)<=L/2) & ---->flag of intersection=TRUE ) else ( // they have intersection if and only if the volumetric intersection of the 2 spheres intersects with at least one of 6 faces of the cube ( e.g. bottom face z= x3-L/2, x3-L/2<= x <= x3+L/2 & y3-L/2<= y <= y3+L/2) // this criteria is quite straightforward to check since it is only in the plane ( e.g. bottom face z= x3-L/2) ) ) - Probably the easiest way to solve this problem is to reduce it to linear programming. Note that this formulation would work for checking if the intersection of any number of spheres/convex polytopes is non-empty. To begin with, we would first rewrite the cube as an intersection of halfspaces. Next we need to replace the spheres with linear halfspaces as well. This is done by lifting them along the map from $\mathbb{R}^3$ to $\mathbb{R}^{4,1}$ where, $(x, y, z) \mapsto (1, x, y, z, \frac{1}{2} ( x^2 + y^2 + z^2 ) )$ Where we fix the metric that, $\langle (o_1, x_1, y_1, z_1, n_1), (o_2, x_2, y_2, z_2, n_2) \rangle = x_1 x_2 + y_1 y_2 + z_1 z_2 - o_1 n_2 - n_1 o_2$ So, a linear halfspace of the form, $a x + b y + c z \leq d$ Can be written as: $\langle (0, a, b, c, d), (1, x, y, z, \frac{1}{2}(x^2 + y^2 + z^2) ) \rangle \leq 0$ And a spherical halfspace, $(x - a)^2 + (y - b)^2 + (z - c)^2 \leq r^2$ Becomes: $\langle (1, a, b, c, \frac{1}{2} (a^2 + b^2 + c^2 - r^2) ), (1, x, y, z, \frac{1}{2}(x^2 + y^2 + z^2) ) \rangle \leq 0$ As a result, our problem is now to determine if a system of linear inequalities has a solution. This is of course linear programming, and you can resolve it either by a brute-force determinant test (aka apply Helly's theorem), or use whatever efficient algorithm you would prefer. In this instance, since it is so small, brute force may be preferrable. For larger numbers of primitives, Seidel's algorithm should perform better. - Hi Mikola, thank you your answer, however I need only an algorithm in this particular case ( 2 spheres and 1 cube) and easy to be implemented for computer code. – user9107 Apr 5 '11 at 10:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 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": 0.8551796674728394, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/55288/how-to-show-modularity-of-an-elliptic-curve/55292	## How to show modularity of an elliptic curve? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In the days before [W, TW, BCDT], how did people show that specific elliptic curves over $\mathbb{Q}$ were modular? For instance, I was reading through a paper of Buhler, Gross and Zagier from 1985 on the curve 5077a, and they say that modularity can be checked by a finite computation in the 422-dimensional space of cuspforms of weight 2 and level 5077 (and remark at the end that Serre and Mestre have checked it). A google search brought up the name "Faltings-Serre method": was this the technique of choice? Also, are there any good references for it? - 1 Could you please explain who do you mean by W, TW and BCDT? I just recoqnize Brian Conrad and Wiles ... – Martin Brandenburg Feb 13 2011 at 12:33 3 I'm referring to the papers by Wiles, Taylor-Wiles, and Breuil-Conrad-Diamond-Taylor. – Abhinav Kumar Feb 13 2011 at 14:05 ## 3 Answers They explicitly computed quotients of $X_0(N)$ and identified them with elliptic curves. Suppose you can compute the space $S_2(\Gamma_0(N),\mathbf{C})$ of modular forms. An (isogeny class of) elliptic curves of conductor $N$ corresponds (by modularity) to a normalized new Hecke eigenform with coefficients in $\mathbf{Z}$. Given such an $f$, one can compute the periods of $f$, which allows one to write down a Weierstrass equation for $f$. If one can do this in such a way to guarantee that the coefficients of the Weierstrass equation are integers, this allows one to computationally determine exactly the modular elliptic curves of conductor $N$. All this is very well explained in Cremona's book, which is available free online: http://www.warwick.ac.uk/~masgaj/book/amec.html The particular method referred to in the paper of [BGZ] was presumably the "method of graphs", see for example here: http://modular.math.washington.edu/msri06/refs/mestre-method-of-graphs/mestre-en.pdf The Faltings-Serre method is slightly different; it allows you to determine, given two Galois representations $\rho_1$ and $\rho_2$ from $G_{K}$ to (say) $\mathrm{GL}_n(\mathbf{Z}_p)$ such that: 1. $\overline{\rho}_1 \simeq \overline{\rho}_2$, 2. $\rho_1$ and $\rho_2$ are both unramified outside some finite (given) set of places $S$ of $K$. whether $\rho_1 \simeq \rho_2$. You could use it to prove that an elliptic curve $E/\mathbf{Q}$ is modular by comparing the Galois representation attached to the $p$-adic Tate module of $E$ and the Galois representation attached to the conjecturally corresponding eigenform of level $N$ (EDIT: this works because the Galois representation determines the isogeny class of $E$ by Faltings' proof of the Tate conjecture for abelian varieties). However, this wouldn't be so efficient. However, there are other situations (for example, elliptic curves over other number fields) where the corresponding automophic form is not enough to compute the periods. In this case, the Faltings-Serre method can be used to prove modularity. Taylor uses this to prove that a certain elliptic curve over $\mathbf{Q}(\sqrt{-3})$ is modular. - 1 For those who prefer to read Mestre in French, William Stein's site has the original too: boxen.math.washington.edu/msri06/refs/… – Chandan Singh Dalawat Feb 15 2011 at 13:58 There is also a paper of Dieulefait, Guerberoff and Pacetti (see arxiv.org/abs/0804.2302) where they use the Faltings-Serre method to prove modularity of some elliptic curves over imaginary quadratic fields. – Nicolas B. Nov 23 2011 at 8:50 ### 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 general "standard approach" was to use the suite of algorithms described in Cremona's book (which is now freely available). The basic idea is to use modular symbols to write down a basis for the homology of the modular curve $X_0(N)$, then use the action of Hecke operators to compute all of the rational cuspidal newforms of level $N$. To each newform there is a corresponding optimal elliptic curve quotient of $X_0(N)$, whose equation can be analytically computed exactly (this uses Edixhoven's remark that the Manin constant is an integer); once that optimal curve has been computed there are standard formulas to list all curves isogenous to it. Thus to prove that a given elliptic curve $E$ is modular, first compute its conductor $N$ (using Tate's algorithm), then enumerate all the modular elliptic curves of conductor $N$ (via Cremona's algorithm), and finally note that $E$ is in this list. If you're interested in efficiency there are shortcuts, e.g., instead of finding all rational newforms, just find the one that looks like it comes from your curve $E$. - Thanks for your answers, William and Wanax! William: it seems I can only accept one answer, so accepting Wanax's because of its multiple references. Hope you don't mind! – Abhinav Kumar Feb 13 2011 at 6:32 1 Regarding complexity, I just mentioned this post to Cremona and he remarks: "...and doing level 5077 takes under 1 second now" – William Stein Feb 13 2011 at 6:52 2 @Abhinav: Amusingly, Wanax's one extra reference is to a paper hosted on my website that I got a student to translate into English. :-) (I don't care about the points stuff.) – William Stein Feb 13 2011 at 22:48 The answer is outlined in Don Zagier's 1985 paper Modular points, modular curves, modular surfaces and modular forms. -	{"extraction_info": {"found_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": 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": 0.9184978008270264, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/106150?sort=votes	## Commutator of algebraic subgroups is connected ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $G$ be an algebraic group over an algebraically closed field. If $H$ and $K$ are closed subgroups and one of them is connected, then their commutator $[H,K]$ is also connected. Is there an easy way to see this fact? The proof that I see in Springer's book on Linear Algebraic Groups is very long-winded. (Of course, he obtains many results on the way). My question is, is the given statement true for any topological group? If yes, does that proof apply to the algebraic groups' case? (Remark: The definition of a topological group includes the Hausdorff property, whereas algebraic groups are not so). - ## 4 Answers The case for $G$ a topological group doesn't look hard; it just comes down to two facts: • A union of connected sets which have a point in common is also connected. For example, suppose $K$ is a connected subgroup and $H$ is any set. For each fixed $h \in H$, the set of commutators $[h, K]$ is the image of $K$ under the continuous map $G \to G$ that takes $g$ to $[h, g]$; this image is connected and contains the identity (since $g = e$ belongs to $K$). But then the set of all commutators $[h, k]$ is a union of connected sets $[h, K]$ each containing the identity, so it too is connected. • If $S \subset G$ is a connected set containing the identity, then so is the group it generates. This is on the same principles as before, that the image of a connected set under a continuous map is connected, and the union of connected sets with an element in common is connected. For example, the set $T = S \cup S^{-1}$ is connected, and then each $T \cdot T \cdot \ldots \cdot T$ is connected, being the image of the connected set $T \times T \times \ldots \times T$ under the multiplication map $G^n \to G$, and finally the union of all these sets gives the group generated by $S$. Now apply the second fact to the connected set of commutators coming from the first fact, to show that the subgroup $[H, K]$ is connected. - That answers my question, thanks! – Abhishek Parab Sep 2 at 3:45 1 It is somewhat subtle, but an algebraic group is not a topological group for the Zariski topology. This is because for a topological group one requires that the multiplication map $\mu:G \times G \to G$ be continuous for the product topology, while for algebraic groups one puts on the Zariski topology on $G \times G$, which does not coincide with the product topology in general. So it is not clear to me why the argument for topological groups should imply the statement for algebraic groups. – Guntram Sep 2 at 5:24 1 Right, I was only addressing one of the questions of the OP. There is no reason I can think of why the statement for topological groups should imply the statement for algebraic groups. – Todd Trimble Sep 2 at 6:04 I missed the subtle point that Guntram mentioned. Thanks for pointing it out. – Abhishek Parab Sep 2 at 15:31 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. There are actually three questions here. The first is: Is there an easy way to see this fact?, to which the answer is (almost certainly) no. Keep in mind that both Chevalley and Borel came to algebraic groups from a background in Lie groups and were therefore well aware of what worked for a topological group. The third question has the same answer: whether the topological group result applies to the algebraic group case. "Long-winded" answers may be a necessary evil here. Keep in mind too that the applications of the foundational results behind your questions deal with whether certain subgroups of algebraic groups are closed and/or connected. There is some essential interaction between those properties. Todd has worked out the answer to the second question, concerning topological groups. Here the basic results are fairly old, so probably a similar proof is written down somewhere in the literature. But I'll elaborate on some of the other answers and comments about the essential difference in the algebraic group setting. 1) Algebraic groups (here assumed affine) are given the Zariski topology; in particular, irreducible sets are the more natural refinement of the topological notion of connected. In fact, Chevalley's original arguments about commutator groups and such just used the term "irreducible". It's true that for an algebraic group, which has only finitely many irreducible components (all disjoint), the notions "irreducible" and "connected" coincide. But this can cause confusion at times. In any case, the Zariski product topology isn't the usual one, so all topological arguments involving products and continuous maps have to be rethought in algebraic geometry. 2) In his 1951 second volume in French on Theorie des groupes de Lie, Chevalley tried out a framework for algebraic groups which proved later to be inadequate in prime characteristic especially (so he changed gears). But he did rethink all the foundational material related to connectedness, which led classically to connectedness of familiar linear groups. His II.7 contains the prototype, hard to read now, of the basic argument. 3) The notes by Bass of Borel's 1968 Columbia lectures Linear Algebraic Groups adopted more modern algebraic geometry but avoided most scheme language due to time constraints. Chevalley's "long-winded" argument is recast here in I.2.2 and I.2.3 (where part (a) is used to get part (b)). These ideas are crucial for several applications. Coming this early in the theory and used for example to treat solvability, the proofs necessarily rely on first principles. (Borel's expanded second edition, Springer GTM 126, leaves this material unchanged.) 4) No substantive changes are made in my 1975 book GTM 21 and in Springer's 1981 Birkhauser text. In my book, see 7.5 and 17.2, while in Springer's book see 2.2.6-2.2.8. - Thank you very much for your enlightening views. Point (1) was especially instructive. But in (4) - references, your book has no 7.5 and 17.2 is about universal enveloping algebras. Are you sure you wanted to point to that? – Abhishek Parab Sep 3 at 18:25 @Abhishek: Sorry for the wrong GTM number, which I've edited. (I meant my second book in 1975, with the same title Linear Algebraic Groups as the books by Borel and Springer.) – Jim Humphreys Sep 3 at 20:52 Using classical varieties (and classical points only), since $G^n$ does not have the product topology (in the algebraic group setting) it isn't clear what can be useful for $G$ concerning "topological" statements (as in Todd's answer) concerning the product topology on the subset $T \times T$ inside $G \times G$ when $T$ is just some random subset of $G$ (not yet known to be constructible). So although topological groups provide valuable intuition that can sometimes be transported to the case of algebraic groups (which are of course not themselves topological groups in general), in this case the central issue is not addressed by thinking about topological groups. The purpose of the longer delicate arguments one finds in the basic textbooks on algebraic groups is that the commutator subgroup is reached in "finitely many steps" (even without connected hypotheses on $H$ or $K$, which is very important for applications) and so is constructible. It is for constructible $T$ that $T \times T$ with the "right" topology (inherited from $G \times G$) is connected when $T$ is connected, etc. The hard part therefore involves a problem which doesn't arise in the topological group setting (unless one poses finer topological question, such as closedness of $(H,K)$ under some reasonable hypotheses, which is a deeper problem than mere connectedness). For an arbitrary (not necessarily constructible) connected subset $T$ of $G$ is the subset $T \times T$ inside $G \times G$ (the latter given the Zariski topology) connected? - What do you mean by `constructible'? – Abhishek Parab Sep 2 at 15:30 Chevalley's theorem on preservation of constructibility under images pervades the beginning of the theory of algebraic groups. The constructible subsets of a noetherian topological space are the finite unions of locally closed sets. Check Wikipedia under "constructible set" (though the example there is weak, since it is even locally closed; a better example is the subset of the plane given by the union of the origin and the complement of the $x$-axis). By the way, the end of your posted question made me think that you were well aware that algebraic groups aren't topological groups. – grp Sep 2 at 16:07 It seems to me that in a topological group $G$ if $H$ is a connected subset containing the identity and $K$ is any subset then the subgroup generated by all commutators $hkh^{-1}k^{-1}$ is connected. $G$ does not to be Hausdorff, and $H$ and $K$ do not have to be closed or to be subgroups. The ingredients in the proof are: (1) The image of a connected set under a continuous map is connected. (2) The union of a set of connected sets is connected if they all have some point in common. (3) The subgroup generated by a connected set is connected. - Sorry, Tom, I didn't see your answer while I was typing. – Todd Trimble Sep 2 at 3:33 What are the connected sets whose union you are considering in (2)? – Abhishek Parab Sep 2 at 3:37 Actually, is what you wrote true if the connected subset does not contain the identity? If for example $H$ and $K$ were both 1-element sets, then it seems $[H, K]$ could be a discrete set, unless I'm missing something. – Todd Trimble Sep 2 at 3:39 Todd, yes, I was unconsciously assuming $H$ contained the identity. Fixed now. – Tom Goodwillie Sep 2 at 12:12 The proof that I had in mind was exactly the one that Todd gives in detail in his answer. – Tom Goodwillie Sep 2 at 15:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 55, "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": 0.9501049518585205, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2010/06/08/	# The Unapologetic Mathematician ## Mean Convergence is Complete I evidently have to avoid saying $L^1$ in post titles because WordPress’ pingbacks can’t handle the unicode character ¹… Anyhow, today we can show that the $L^1$ norm gives us a complete metric structure on the space of all integrable functions on $X$. But first: If $\{f_n\}$ is a mean Cauchy sequence of integrable simple functions converging in measure to $f$, then it converges in the mean to $f$ as well. Indeed, for any fixed $m$ the sequence $\{\lvert f_n-f_m\rvert\}$ is a mean Cauchy sequence of integrable simple functions converging in measure to $\lvert f-f_m\rvert$. Thus we find $\displaystyle\int\lvert f-f_m\rvert\,d\mu=\lim\limits_{n\to\infty}\int\vert f_n-f_m\rvert\,d\mu$ But the fact that $\{f_n\}$ is mean Cauchy means that the integral on the right gets arbitrarily small as we take large enough $m$ and $n$. And so on the left, the integral gets arbitrarily small as we take a large enough $m$. That is, $\displaystyle\lim\limits_{m\to\infty}\int\vert f-f_m\rvert\,d\mu=0$ and $\{f_n\}$ converges in the mean to $0$. As a quick corollary, given any integrable function $f$ and positive number $\epsilon$, there is some integrable simple function $g$ with $\lVert f-g\rVert_1<\epsilon$. Indeed, since $f$ is integrable there must be some sequence $\{f_n\}$ of integrable simple functions converging in measure to it, and we can pick $g=f_n$ for some sufficiently large $n$. Now, if $\{f_n\}$ is a mean Cauchy sequence of integrable functions — any integrable functions — then there is some integrable function $f$ to which the sequence converges in the mean. The previous corollary tells us that for every $f_n$ there’s some integrable simple $g_n$ so that $\lVert f_n-g_n\rVert_1<\frac{1}{n}$. This gives us a new sequence $\{g_n\}$, which is itself mean Cauchy. Indeed, for any $\epsilon>0$ choose an $N>\frac{3}{\epsilon}$ large enough so that $\lVert f_n-f_m\rVert_1<\frac{\epsilon}{3}$. Then for $n,m\geq N$ we have $\displaystyle\lVert g_n-g_m\rVert_1\leq\lVert g_n-f_n\rVert_1+\lVert f_n-f_m\rVert_1+\lVert f_m-g_m\rVert_1\leq\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$ Since $\{g_n\}$ is Cauchy in measure it converges in measure to some function $f$. Then our first result shows that $\{g_n\}$ also converges in the mean to $f$ — that $\lVert g_n-f\rVert_1$ goes to zero as $n$ goes to $\infty$. But we can also write $\displaystyle\lVert f_n-f\rVert_1\leq\lVert f_n-g_n\rVert_1+\lVert g_n-f\rVert_1\leq\frac{1}{n}+\lVert g_n-f\rVert_1$ so if we choose a large enough $n$, this will get arbitrarily small as well. That is, $\{f_n\}$ also converges in the mean to $f$. This shows that any mean Cauchy sequence of integrable functions is also mean convergent to some function, and thus the space of integrable functions equipped with the $L^1$ norm is a Banach space. Posted by John Armstrong \| Analysis, Measure Theory \| 5 Comments ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 48, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.91106116771698, "perplexity_flag": "head"}
http://mathoverflow.net/questions/12628/applications-of-euler-cauchy-odes	## Applications of Euler-Cauchy ODEs ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The Euler-Cauchy ODE (2nd order, homogeneous version) is: $$x^2 y'' + a x y' + b y = 0$$ Looking in various books on ODEs and a random walk on a web search (i.e. I didn't click on every link, but tried a random sample) came up with no actual applications but just lots of vague "This is really important."s. The closest actual application was on Wikipedia's page, which says: The second order Euler–Cauchy equation appears in a number of physics and engineering applications, such as when solving Laplace's equation in polar coordinates. Is there a more direct application of this ODE? Ideally, I'd like something along the lines of deriving the corresponding ODE with constant coefficients from considering springs or pendula. My motivation is pure and simple that I'd like to be able to say something in class a little more motivating than: "We study this ODE simply because we can actually write down a solution, and it's quite amazing that we can do so." - I've only ever come across it when solving Laplace's equation in cylindrical coordinates by separation of variables. – jc Jan 22 2010 at 13:28 2 Frankly, solving Laplace's equation is already rather significant application. I agree with you that it'd be nice to have more, but don't sweep that one under the rug! – Tom LaGatta Jan 22 2010 at 15:06 I don't want to sweep Laplace's equation under the rug - far from it - but it's too long from that to this ODE for this particular course. If you think about the derivation of the ODE with constant coefficients from considering the mechanics of a spring and compare that with deriving the Euler-Cauchy from Laplace's equation (a PDE!) which you first need to motivate from physics ... it's too long! Hence I'm looking for something quicker. – Andrew Stacey Jan 22 2010 at 15:26 ## 3 Answers Well, it is the simplest case of a second order equation with regular singular points, so it is a pretty good example in trying to figure out what to expect from the general case. In fact, in a way, it contains essentially all the complexities of the general case (as long as you care about what happens locally at the singular point), which is something quite nice for a 'simplest example'! - ### 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. Well, if your purpose is simply to have a direct mechanical example motivating the introduction of this equation in the class, you can use the equation describing time-harmonic vibrations of a thin elastic rod: $$\frac{\partial}{\partial x}\left(E(x)\frac{\partial u}{\partial x}\right)+\rho\omega^2u=0$$ in which $E(x)$ is Young's modulus, $\rho$ is the material density and $\omega$ the angular frequency of vibrations (see e.g. Graff's "Wave motion in elastic solids"). Now, if you assume that the rod is made of inhomogeneous material such that $E(x)=E_0x^2$ (a bit unusual, but very possible) then you'll get precisely your equation with $a=2$ and $b=\rho\omega^2/E_0$. - While probably not the application you were looking for, they are of pedagogical/theoretical significance for being (explicitly solvable) differential equations for which the supposition of a non-zero real analytic solution (i.e., solving by Taylor series) fails quite dramatically. This seems to fulfill the "say something interesting about them" clause of the question (though I should note that I don't give them much emphasis in my DEs class). And, as Mariano points out, they're a gateway set of D.E.s to understanding what's going on in much more complicated scenarios -- the vanishing set of the leading term of a linear PDE (e.g., our Euler equation's regular singular point) defined on a manifold cuts out the characteristic variety in the manifold's cotangent bundle. As I understand it (and I'm talking a little over my head here), generalizing the solvability of Cauchy-Euler equations to this much more general setting via the characteristic variety is/was a prominent goal of Algebraic Analysis and the theory of D-modules. -	{"extraction_info": {"found_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": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_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": 0.9502680897712708, "perplexity_flag": "middle"}
http://stats.stackexchange.com/questions/24263/model-stability-in-cross-validation-of-regression-models	# Model stability in cross-validation of regression models Given multiple cross-validation folds of a logistic regression, and the resulting multiple estimates of each regression coefficient, how should one measure whether or not a predictor (or set of predictors) is/are stable and meaningful based on the regression coefficient(s)? Is this different for linear regression? - @BGreene Very clever. Why not post that as an answer? You're also making me wonder whether the ensemble learning literature holds something relevant. – Jack Tanner Feb 28 at 17:14 thanks, will do. – BGreene Feb 28 at 18:48 When you say "multiple cross-validation", do you mean that you run $m$ times a $k$-fold cross-validation? – andrea Mar 11 at 9:22 @andrea, I say "multiple cross-validation folds", i.e., $k$ folds. – Jack Tanner Mar 11 at 19:48 ## 2 Answers You could treat the regression coefficients resulting from each test fold in the CV as independent observations and then calculate their reliability/stability using intra-class correlation coefficient (ICC) as reported by Shrout & Fleiss. - I assume you in your cross-validation you divide the data in two parts, a training set and a test set. In one fold you fit a model from the training set and use it to predict the response of the test set, right? This will give you an error rate for the whole model, not for a single predictor. I do not know if it is possible to find p-values for predictors using something like the F-tests used in ordinary linear regression. You can try remove predictors from the model using for example backward or forward selection if that is your aim. You could instead of CV use bootstrap to find a confidence interval for each predictor and then see how stable it is. How many folds do you use in your CV, is it leave-one-out cross-validation? Perhaps more details of what your aim is would help to answer this question. - Assume this is leave-one-out. Each predictor in each fold already has a confidence interval, e.g., from a Bayesian posterior CI or std err from `glm(..., family="binomial")` in R. What do I do with the intervals for each predictor across the leave-one-out runs? – Jack Tanner Apr 27 '12 at 3:22	{"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": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8910489082336426, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/16261/quotient-of-a-reductive-group-by-a-non-smooth-central-finite-subgroup/16273	## Quotient of a reductive group by a non-smooth central finite subgroup ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I need a construction in linear algebraic groups which uses taking quotient by a central finite group subscheme. My question is, whether it goes through in bad'' characteristics, when this group subscheme is not smooth. First I write this construction in a special case, and then in the general case. Let $G$ be a connected semisimple $k$-group over a field $k$ of characteristic $p>0$. We may assume that $k$ is algebraically closed. Assume that the corresponding adjoint group $G^{ad}$ is $PGL_n$. In general, my group $G$ is not special (recall that a $k$-group $G$ is called special if $H^1(K,G)=1$ for any field extension $K/k$). I want to construct a special $k$-group $H$ related to $G$. For this end I consider the universal covering $G^{sc}$ of $G$, then $G^{sc}=SL_n$. Let $Z$ denote the center of $G^{sc}$, then $Z=\mu_n$. We have a canonical epimorphism $\varphi\colon SL_n \to G$. We denote by $C$ the kernel of $\varphi$. Then $C$ is a group subscheme of $Z$, defined over $k$. Since $Z=\mu_n$, there is a canonical embedding $Z\hookrightarrow \mathbb{G}_m$ into the multiplicative group $\mathbb{G}_m$. Thus we obtain an embedding $C\hookrightarrow \mathbb{G}_m$. Consider the diagonal embedding $$C\hookrightarrow SL_n\times \mathbb{G}_m.$$ I would like to define $H:=(SL_n\times \mathbb{G}_m)/C$. Is such a quotient defined, when char($k$) divides $n$ and $C$ is not smooth? Note that $SL_n$ embeds into $H$, and we have a short exact sequence $$1\to SL_n \to H \to \mathbb{G}_m \to 1$$ In this exact sequence both $SL_n$ and $\mathbb{G}_m$ are special, and from the Galois cohomology exact sequence we see that $H$ is special as well. In the general case I assume that $G^{ad}$ is a product of groups $PGL_{n_i}$, $i=1,\dots s$. Then $G^{sc}$ is the product of $SL_{n_i}$. Let $C$ denote the kernel of the canonical epimorphism $\varphi\colon G^{sc}\to G$, then $C$ is contained in the center $Z$ of $G^{sc}$. We have $Z=\prod_{i=1}^s \mu_{n_i}$. Again we embed diagonally $$C\hookrightarrow (\prod_{i=1}^s SL_{n_i}) \times (\mathbb{G}_m)^s$$ and denote by $H$ the quotient. Again $H$ is special (if it is defined), and again my question is, whether this construction makes sense when char($k$) divides $n_i$ for some $i$. Any help is welcome! - ## 3 Answers This is an instance of what I believe is called the $z$-construction, and it is a very useful trick in the arithmetic theory of algebraic groups. (Small correction: your diagonal embedding should really be "anti-diagonal". You are really computing a "central pushout".) However, you may need to restrict your ground field to be local or global (with some caveats in case of real places) to get the cohomological vanishing. In general, let $G$ be any split connected semisimple group over a field $k$, and $G' \rightarrow G$ the $k$-split simply connected central cover. Let $Z$ be the center of $G'$.This is a $k$-group of multiplicative type, possibly not etale. It is $k$-split since $G'$ is $k$-split, so we can choose a $k$-subgroup inclusion of $Z$ into a $k$-split $k$-torus $T$ (seen using character group). Now just form the central pushout of $G'$ along this inclusion. More precisely, the antidiagonal map $$Z \rightarrow G' \times T$$ is a central $k$-subgroup scheme, and in SGA3, VI$_{\rm{A}}$ quotients are constructed and studied for arbitrary finite type group schemes over a field modulo closed subgroup schemes of finite type (and even more generally). We can therefore form the quotient by this central subgroup scheme, and it has all of the reasonable properties one would want for a quotient. This pushout $H$ is a $k$-group containing $G'$ as a normal $k$-subgroup, and the quotient $H/G'$ is $T/Z$. Being a quotient of $k$-split torus, it is again a $k$-split torus. Provided the ground field is a local or global function field (with some further restrictions in case of a real place), all simply connected semisimple groups have vanishing degree-1 Galois cohomology. So that does the job for such fields. One can also do a variant for split connected reductive groups, and even a variant without a split hypothesis (but then the conclusion is a little different). - I should also mention that the appendix by T. Ono in Weil's book "Adeles and algebraic groups" also discusses the z-construction, where the point of the construction is to promote the derived group to be simply connected. That is the reason it is so useful over general ground fields. In my answer above, I should have noted that G' is indeed the derived group of H. – BCnrd Feb 24 2010 at 16:12 The "quotient by a $p$-Lie algebra" construction in section 17 of Borel's book can be applied several times to also cover the quotient construction which is being used above. It is Borel's way of handling infinitesimal kernels without directly using infinitesimal group schemes. This is due to Serre, I have been told. – BCnrd Feb 24 2010 at 16:37 Brian, I think that the notion of a $z$-construction (there called a $z$-extension) may have been introduced by Kottwitz, in his 1982 Duke J. paper "Rational conjugacy classes in reductive groups". – L Spice Feb 24 2010 at 22:46 Loren, do you mean the terminology rather than the concept? The trick of using central extension pushout to promote the derived group to be simply connected is there in the appendix to Weil's book. I was told that this is what is called the z-construction (I'm not sure why; due to "central" subgroups, or a "Z" in the diagram of arrows?). The appendix suggests it goes back at least to Weil. – BCnrd Feb 25 2010 at 1:45 Brian, I'm sorry; I misread your quote about Weil's book as discussing a later addition, rather than part of the book itself. The Kottwitz paper was my first exposure to it, but, as you point out, it is certainly not the first occurrence of the concept. (For what it's worth, Kottwitz cites Langlands's "stable conjugacy" paper, which I don't have to hand, but does not mention Weil.) I'd always taken the $z$ to refer to the centre, but wondered (as I guess you do, too) why then '$z$' instead of '$Z$'. – L Spice Feb 25 2010 at 17:23 ### 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 standard isomorphism theorems in abstract group theory all hold for group schemes of finite type over a field. This is implicit in SGA (a key point is the statement mentioned by Ekedahl) and is explicit in the notes on algebraic groups,... appearing on my website (Section 7 of Chapter I). This makes a lot of things obvious (including your questions). [The isomorphism theorems fail when you don't allow nilpotents, which is why the standard expositions on algebraic groups are so complicated.] - The quotient by any finite flat subgroup scheme always exists (see for example SGA 3:Exp V, Thm 7.1). In your case the subgroup scheme is of multiplicative type, the dual of a finite abelian group $A$, so an action of it on an affine $k$-scheme is just an $A$-grading of the coordinate ring on the scheme. The quotient is just the spectrum of the degree $0$-part. If $R$ is an $A$-graded commutative Hopf algebra (over $k$), it is clear that the degree $0$-part of $R$ is a commutative Hopf algebra. This gives a reasonably concrete description of the quotient in that case. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 98, "mathjax_display_tex": 4, "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": 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": 0.9425082206726074, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/158281/pattern-of-orders-of-elements-in-a-cyclic-group/158286	# Pattern of orders of elements in a cyclic group Doodling as I was contemplating another recent question, I picked out the orders of elements of the cyclic group $Z_{15}$ namely: 1 element of order 1 2 elements of order 3 4 elements of order 5 8 elements of order 15 In a group of order 3 you have 1 element of order 1 and two elements of order 3. Are there other examples of this same pattern (powers of 2), and can anyone show and prove a general rule. - Great question. Thanks Mark. Amazing! – Babak S. Jun 14 '12 at 15:42 ## 3 Answers The number of elements of order $d$, where $d$ is a divisor of $n$, is $\varphi(d)$. These values of $\varphi(d)$ are powers of $2$ for all divisors $d$ of $n$ precisely when $n$ is a power of $2$ times a product of distinct Fermat primes, that is, primes of the form $2^{2^k}+1$. The power of $2$ may be $2^0$, and the product may be the empty one. The proof is immediate from the usual formula for $\varphi$ in terms of the prime power factorization. So the answer is essentially the same as the classical one of which regular polygons are Euclidean-constructible. Sadly, there are not many Fermat primes known. At this time, we only have $3$, $5$, $17$, $257$, and $65537$ to play with. - So 255 and 65535 work. $255 = (2-1)(2+1)(2^2+1)(2^{2^2}+1)$ and 65535 has a similarly nice decomposition, and this works so neatly because $2^1-1=1$. Then we have 4,294,967,295 with 65537 as an additional factor. It looks to me as though these are the only cases where you get a complete sequence of powers of 2 starting at 1 and with no repetitions. – Mark Bennet Jun 14 '12 at 21:54 Let $G$ be a cyclic group of order $n$, and let $a\in G$ be a generator. Let $d$ be a divisor of $n$. Certainly, $a^{n/d}$ is an element of $G$ of order $d$ (in other words, $\langle a^{n/d}\rangle$ is a subgroup of $G$ of order $d$). If $a^t\in G$ is an element of order $d$, then $a^{td}=e$, hence $n\mid td$, and thus $\frac{n}{d}\mid t$. This shows that $a^t\in \langle a^{n/d}\rangle$, and thus $\langle a^t\rangle=\langle a^{n/d}\rangle$ (since they are both subgroups of order $d$). Thus, there is exactly one subgroup, let's call it $H_d$, of $G$ of order $d$, for each divisor $d$ of $n$, and all of these subgroups are themselves cyclic. Any cyclic group of order $d$ has $\phi(d)$ generators, i.e. there are $\phi(d)$ elements of order $d$ in $H_d$, and hence there are $\phi(d)$ elements of order $d$ in $G$. Here, $\phi$ is Euler's phi function. This can be checked to make sense via the identity $$\sum_{d\mid n}\phi(d)=n.$$ - Here's te easiest pattern I know For any cyclic group $\mathbb{Z}_n \cong \left<a\right>$ for some $a$ with order $n$ Then $\left<a\right>=\{a^k \mid k=1 \ldots n \}$ and the order $\left\|a^k\right\|=\frac{n}{\text{gcd}(k,n)}$ for $k=1\ldots n$ -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 61, "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": 0.9329342246055603, "perplexity_flag": "head"}
http://nrich.maths.org/5616/clue	### Flexi Quads A quadrilateral changes shape with the edge lengths constant. Show the scalar product of the diagonals is constant. If the diagonals are perpendicular in one position are they always perpendicular? ### Flexi Quad Tan As a quadrilateral Q is deformed (keeping the edge lengths constnt) the diagonals and the angle X between them change. Prove that the area of Q is proportional to tanX. ### Air Routes Find the distance of the shortest air route at an altitude of 6000 metres between London and Cape Town given the latitudes and longitudes. A simple application of scalar products of vectors. # Pythagoras on a Sphere ##### Stage: 5 Challenge Level: All angles are in radians. (1) Without loss of generality take coordinate axes so that $A$ is the point$(0,0,1)$, the xz-plane contains the point $C$ and the yz-plane contains the point $B$. (2) Thinking of $A$ as the North Pole then $C$ has latitude $u$ and longitude 0 and $B$ has latitude $v$ and longitude $\pi/2$. (3) Find the 3D coordinates of $B$ and $C$. Where the origin O is the centre of the sphere ${\bf OA, OB}$ and ${\bf OC}$ are vectors of unit length. (4) Use scalar products and vectors ${\bf OA, OB}$ and ${\bf OC}$ to find the lengths of the arcs $AB, BC$ and $CA$ in terms of $u$ and $v$. The required result follows. 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": 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": 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": 0.8796863555908203, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/43573/negative-sign-of-acceleration	# Negative sign of acceleration [closed] This is the problem from our Physics textbook : A player throws a ball upwards with an initial speed of 29.4 m s–1. (a) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion. Answer is : x > 0 (upward and downward motion); v < 0 (upward), v > 0 (downward), a > 0 throughout; a > 0 is what is bothering me and I have done hours of searching and trying to understand. Kindly explain all the parts of the answer. Thanks in advance. - We don't do particular instance of homework questions. There is a meta thread that explains how to ask basic questions in a way that we will answer. – dmckee♦ Nov 7 '12 at 3:30 ## closed as too localized by dmckee♦Nov 7 '12 at 3:30 This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, see the FAQ. ## 2 Answers The acceleration is a vector $\mathbf{g}$ throughout the motion, and $\mathbf{g}$ is always pointing downward. Since you choose positive $x$ to be vertically downward, so $\mathbf{g}$ is along positive $x$ if we draw out the Cartesian coordinate, then $\mathbf{g}$ must have positive value, $\mathbf{g}=g\,\hat{\mathbf{x}}$. If you choose vertically upward to be $x>0$, then acceleration $\mathbf{g}$ has negative sign, $\mathbf{g}=-g\,\hat{\mathbf{x}}$. It's just the matter how you choose the $x>0,y>0$ directions. Draw a diagram of $v$, $g$, force on the ball with $(x,y)$ coordinates. - Since top is x=0, so position remains positive through out, and why velocity sign is - when going up and + when going down, as object never enters negative territory which is beyond x=0, so entire action takes place in the plus zone. – pokrate Nov 7 '12 at 5:37 a > 0 is what is bothering me The only acceleration in the problem is the acceleration of gravity, correct? If you agree with that, then I think you'll agree that the acceleration is always pointing downward towards the Earth and thus, the acceleration does not change sign for this problem. Now, from the problem statement: vertically downward direction to be the positive direction So, what is the sign of the acceleration? -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9419971704483032, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2012/02/16/ohms-law/?like=1&source=post_flair&_wpnonce=671450efa2	# The Unapologetic Mathematician ## Ohm’s Law When calculating the potential energy of the magnetic field, we calculated the power needed to run a certain current around a certain circuit. Let’s look into that a little more deeply. We start with Ohm’s law, which basically says that — as a first approximation — the electromotive force around a circuit is proportional to the current around it; push harder and you’ll move charge faster. As a formula: $\displaystyle V=IR$ The electromotive force — or “voltage” — on the left is equal to the current around the circuit times the “resistance”. What’s the resistance? Well, here it’s basically just a constant of proportionality, which we read as “how hard is it to push charge around this circuit?” But let’s dig in a bit more. A current doesn’t really flow around an infinitely-thin wire; it flows around a wire with some thickness. The thicker the wire is — the bigger its cross-sectional area — the easier it should be to push charge around, while the longer the circuit is, the harder. We’ll write down our resistance in the form $\displaystyle R=\eta\frac{l}{A}$ where $l$ is the length of the wire, $A$ is its cross-sectional area, and $\eta$ is a new proportionality constant we call “resistivity”. Putting this together with the first form of Ohm’s law we find $\displaystyle V=\eta\frac{l}{A}I$ But look at this: the current is made up of a current density flowing along the wire, integrated across a cross-section. If the wire is running in the $z$ direction and the current density in that direction is constantly $J_z$, then $I=JA$. Further — at least to a first approximation — the electromotive force is the $z$-component of the electric field $E_z$ times the length $l$ traveled in that direction. Thus we conclude that $E_z=\eta J_z$. But since there’s nothing really special about the $z$ direction, we actually find that $\displaystyle E=\eta J$ which is Ohm’s law again, but now in terms of fields and current distributions. But what about the power? We’ve got a battery pushing a current around a circuit and using power to do it; where does the energy go? Well, if we think about pushing little bits of charge around the wire, they’re going to hit parts of the wire and lose some energy in the process. The parts they hit get shaken up, and this appears as heat energy; the process is called “Ohmic” or “Joule” heating, the latter from Joule’s own experiments using a resistive wire to heat up a tub of water. If we have a current $I$ made up of $N$ bits of charge $q$ per unit time, then each bit takes an energy of $qV$ to go around the circuit once. This happens $N$ times per unit time, so the total power expenditure is $\displaystyle P=NqV=IV$ just as we said last time. But now we can do the same trick as above and write $\displaystyle P=IV=(J\cdot E)Al$ or $\displaystyle\frac{P}{Al}=E\cdot J$ which measures the power per unit volume dissipated through Joule heating in the circuit. About these ads Like Loading... ## 1 Comment » 1. [...] second term on the right is the energy density lost to Joule heating per unit time. The only thing left is this vector [...] Pingback by \| February 17, 2012 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now. • ## RSS Feeds RSS - Posts RSS - Comments • ## Feedback Got something to say? Anonymous questions, comments, and suggestions at Formspring.me! %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 23, "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": 0.9317700266838074, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/tagged/wave-particle-duality?sort=unanswered&pagesize=50	# Tagged Questions The wave-particle-duality tag has no wiki summary. I am trying to solve a problem that includes a function of the light hitting a certain area. My question is, how would I change a function $G(x)$ of photons hitting a certain area to include just ...	{"extraction_info": {"found_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": 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": 0.9273574352264404, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/143055/propositional-logic-substitution	propositional logic - substitution Prove: $\varphi_1 =\!\mathrel\|\mathrel\|\!= \varphi_2 \implies \varphi_1[\psi/p] =\!\mathrel\|\mathrel\|\!= \varphi_2[\psi/p]$. We've proven that $\varphi_1 =\!\mathrel\|\mathrel\|\!= \varphi_2 \implies \psi[\varphi_1/p] =\!\mathrel\|\mathrel\|\!= \psi[\varphi_2/p]$. Maybe you could give me a hint what way of proving I should use? I don't see over wich formula or rank I could use induction in a useful way.. ($\phi =\!\mathrel\|\mathrel\|\!= \psi$ means $\phi \models \psi$ and $\psi \models \phi$) Let $\varphi_1 =\!\mathrel\|\mathrel\|\!= \varphi_2$. Let $v$ be so that $\|p\|_v$ = 1. If now $\|\psi\|_v = 1$ then clearly $\varphi_1[\psi/p] =\!\mathrel\|\mathrel\|\!= \varphi_2[\psi/p]$ (is this clear?). Accordingly: Let $v$ be so that $\|p\|_v$ = 0. If now $\|\psi\|_v = 0$ then clearly $\varphi_1[\psi/p] =\!\mathrel\|\mathrel\|\!= \varphi_2[\psi/p]$ (is this clear?). I don't know how to approach the cases when $\|p\|_v$ = 0 while $\|\psi\|_v = 1$ and $\|p\|_v$ = 1 while $\|\psi\|_v = 0$. Other attempt: Show the equivalent statement $\quad \|\!\!\!= \varphi_1 \leftrightarrow \varphi_2 \implies \|\!\!\!= \varphi_1[\psi/p] \leftrightarrow \varphi_2[\psi/p]$ Notice, that $\quad \varphi_1[\psi/p] \leftrightarrow \varphi_2[\psi/p] \quad \equiv \quad (\varphi_1 \leftrightarrow \varphi_2)[\psi/p]$ Let now $\|\!\!\!= \varphi_1 \leftrightarrow \varphi_2$. Then $\varphi_1 \leftrightarrow \varphi_2$ is a tautology and hence is true for any valuation $v$. In particular $\varphi_1 \leftrightarrow \varphi_2$ is true independent of whether $\|p\|=0$ or $\|p\|=1$. It follows (is this right?) that $(\varphi_1 \leftrightarrow \varphi_2)[\psi/p]$ is also true for any valuation $v$ as well. The claim follows. What on earth does $\models \models$ mean? – Chris Eagle May 9 '12 at 13:10 What on earth does $=\| \models$ mean? – Chris Eagle May 9 '12 at 13:14 `$=\!\mathrel\|\mathrel\|\!=$` yields $=\!\mathrel\|\mathrel\|\!=$ – joriki May 9 '12 at 13:19 I guess $a =\!\mathrel\|\mathrel\|\!= b$ means that both $a\vDash b$ and $b\vDash a$ ? – MJD May 9 '12 at 14:43	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 34, "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": 0.9075546860694885, "perplexity_flag": "head"}
http://simple.wikipedia.org/wiki/Golden_ratio	# Golden ratio If a person has one number a and another smaller number b, he can make the ratio of the two numbers by dividing them. Their ratio is a/b. He can make another ratio by adding the two numbers together a+b and dividing this by the larger number a. The new ratio is (a+b)/a. If these two ratios are equal to the same number, then that number is called the golden ratio. The Greek letter $\varphi$ (phi) is usually used as the name for the golden ratio. For example, if b = 1 and a/b = $\varphi$, then a = $\varphi$. The second ratio (a+b)/a is then $(\varphi+1)/\varphi$. Because these two ratios are equal, this is true: $\varphi = \frac{\varphi+1}{\varphi}$ One way to write this number is $\varphi = \frac{1 + \sqrt{5}}{2}$ $\sqrt{5}$ is like any number which, when multiplied by itself, makes 5(or which number is multiplied): $\sqrt5\times\sqrt5=5$. The golden ratio is an irrational number. If a person tries to write it, it will never stop and never be the same again and again, but it will start this way: 1.6180339887... An important thing about this number is that a person can subtract 1 from it or divide 1 by it. Either way, he will find the same number: $\begin{array}{ccccc} \varphi-1 &=& 1.6180339887...-1 &=& 0.6180339887...\\ 1/\varphi &=& \frac{1}{1.6180339887...} &=& 0.6180339887... \end{array}$ ## Golden rectangle The large rectangle BA is a golden rectangle; that is, the proportion b:a is 1:$\varphi$. For any such rectangle, and only for rectangles of that specific proportion, if we remove square B, what is left, A, is another golden rectangle; that is, with the same proportions as the original rectangle. If the length of a rectangle divided by its width is equal to the golden ratio, then the rectangle is a "golden rectangle". If a square is cut off from one end of a golden rectangle, then the other end is a new golden rectangle. In the picture, the big rectangle (blue and pink together) is a golden rectangle because $a/b=\varphi$. The blue part (B) is a square. The pink part by itself (A) is another golden rectangle because $b/(a-b)=\varphi$. The big rectangle and the pink rectangle have the same form, but the pink rectangle is smaller and is turned. ## Fibonacci numbers The Fibonacci numbers are a list of numbers. A person can find the next number in the list by adding the last two numbers together. If a person divides a number in the list by the number that came before it, this ratio comes closer and closer to the golden ratio. Fibonacci number divided by the one before ratio 1 1 1/1 = 1.0000 2 2/1 = 2.0000 3 3/2 = 1.5000 5 5/3 = 1.6667 8 8/5 = 1.6000 13 13/8 = 1.6250 21 21/13 = 1.6154... 34 34/21 = 1.6190... 55 55/34 = 1.6177... 89 89/55 = 1.6182... ... ... ... $\varphi$ = 1.6180... ## Golden ratio in nature Using the golden angle will optimally use the light of the sun. This is a view from the top. A leaf of common ivy, showing the golden ratio In nature, the golden ratio is often used for the arrangement of leaves or flowers. These use the golden angle of approximately 137.5 degrees. Leaves or flowers arranged in that angle best use sunlight.	{"extraction_info": {"found_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": 13, "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": 0.9065207839012146, "perplexity_flag": "middle"}
http://cdsmith.wordpress.com/2009/07/20/calculating-multiplicative-inverses-in-modular-arithmetic/?like=1&source=post_flair&_wpnonce=57db9114c5	software, programming languages, and other ideas July 20, 2009 / cdsmith I’m sure plenty of people already know this, but I ran into it today, and it’s interesting… so here it is. A common and very easy result in abstract algebra is that the ring $\mathbb{Z}_q = \mathbb{Z}/q\mathbb{Z}$ (where $q$ is any positive integer) contains a multiplicative inverse for $p$ (with $0 < p < q$), if and only if $\gcd(p,q) = 1$. But what if you want to know what the multiplicative inverse is? Well, for small values of $q$, you could just try numbers until one of them works. It turns out that $p^{-1}$ also has a multiplicative inverse (namely, $p$), so it must also be relatively prime to $q$… that helps when we’re doing the math mentally, but less so with a computer, since every possibility still requires some amount of work. It turns out there’s a slightly faster way to find multiplicative inverses in $\mathbb{Z}_q$, and it ban be done by doing the calculation of $\gcd(p,q)$ by the Euclidean algorithm, and tracing our own calculation through the steps. For those who may not remember, the Euclidean algorithm for the greatest common divisor depends on the fact that the greatest common divisor of $q$ and $p$ is the same as the greatest common divisor of $p$ and the remainder when $q$ is divided by $p$. So, for example: $\gcd(12,8) = \gcd(8, 4) = \gcd(4, 0) = 4$. In our case, we start with the assumption that $\gcd(q,p) = 1$, so ultimately, the process will end with $\ldots = \gcd(?, 1) = \gcd(1,0) = 1$. That’s all the setup we need. Here are the two cases for our calculation: Case 1: $p = 1$. Clearly, the muliplicative inverse of one is one, so we are done. Case 2: $p > 1$. Here, we reduce the problem to the next step in the chain of the Euclidean algorithm. Restating the problem, we wish to find a $p'$ such that $pp' = nq + 1$, for any integer $n$. Using simple division, we know that we can write $q = mp + r$, where $m,r$ are integers and $0 \leq r < p$. But, and this is important, we also know that $\gcd(q,p) = \gcd(p,r)$ (the next step in the Euclidean algorithm), so we can actually refine our inequality to $0 < r < p$. Now, by solving a smaller instance of this same problem, we can find $r'$ such that $rr' \equiv 1 \mod p$. Another way of putting this is that $p$ divides $rr' - 1$. Now we are done. We simply choose $n = p - r'$. Then $nq + 1 = (p - r')(mp + r) + 1 = (mp + r - mr')p - (rr' - 1)$. By examining each term separately, it’s clear that this is evenly divisible by $p$, so we have $p' = (nq + 1) / p$, and we’re done. Here’s some code to compute this in Haskell: ``` inverse :: Integral a => a -> a -> a inverse q 1 = 1 inverse q p = (n * q + 1) `div` p where n = p - inverse p (q `mod` p)``` And there you have it, multiplicative inverses in $\mathbb{Z}_q$. ### Like this: Filed under Uncategorized #### 10 Comments 1. programmingpraxis / Jul 21 2009 6:59 am Here is my normal computation of the modular inverse in Scheme, using a method very similar to yours: (define (inverse x m) (let loop ((x (modulo x m)) (a 1)) (cond ((zero? x) #f) ((= x 1) a) (else (let ((q (- (quotient m x)))) (loop (+ m (* q x)) (modulo (* q a) m))))))) I discussed modular arithmetic, including the inverse, at http://programmingpraxis.com/2009/07/07/modular-arithmetic/. 2. Aaron / Jul 21 2009 2:56 pm Can you do it for finite fields? • lpsmith / Jul 21 2009 3:24 pm Yes. 3. lpsmith / Jul 21 2009 3:22 pm Indeed, this is often referred to as the Extended Euclidean Algorithm, or rather, it’s a simple application of it. 4. Joe / Jul 21 2009 6:36 pm My wife informs me, that if your intentions was to say ‘The opinions of cdsmith’ then you need to put ‘cdsmith’ into the genitive case which is ‘cdsmithi’, so it should say ‘Sententia cdsmithi’. • BMeph / Jul 21 2009 6:52 pm Unless it’s third declension (often used for foreign, i.e. non-Latin) words, in which case it’d be “Sententia cdsmithis” • cdsmith / Jul 23 2009 9:18 am Actually, there’s about a 15 year long story around the blog name… far too involved and silly to tell here. So I understand it’s bad grammar, but that’s just how it is. 5. BMeph / Jul 21 2009 6:41 pm Also, the extended Euclidean (which I call “Egcd”) is usually done to give both the inverse of p in Z(q) and the inverse of q in Z(p) – at the same time! Here’s my work-up, taken from a utility module of mine: eGCD a b = eGCDWorker a b (1,0) (0,1) where eGCDWorker _ 0 (m1,n1) _ = (m1,n1) eGCDWorker 0 _ _ (m2,n2) = (m2,n2) eGCDWorker m n (m1,n1) (m2,n2) = eGCDWorker n r (m2,n2) (m3,n3) where (q , r) = quotRem m n (m3,n3) = (m1 – qm2, n1 – qn2) 6. Dave / Jul 22 2009 8:26 am When q is prime, the multiplicative group is cyclic, meaning that one can find a generator x such that all nonzero elements are of the form x^i for some i. For extensive computations in a fixed field, it pays to store two tables i -> x^i and i -> x^i + 1, and carry out all computations in terms of the representation i. Versions of the computer algebra system GAP used this approach, for example. When q is not prime, this idea is less useful. x^i +1 is less frequently a unit, for example. It makes a good exercise in first year algebra to decide nevertheless the structure of the group of multiplicative units. 7. Renaya / Mar 1 2012 8:57 am what is the multiplicative inverse of negative 3 and 2 %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 38, "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": 0.8717691898345947, "perplexity_flag": "middle"}
http://mathhelpforum.com/pre-calculus/148538-limits-point.html	# Thread: 1. ## Limits at a point Hey guys, I'm trying to understand how to determine the left and right side of this limit: limit as x -> 2 \|x-2\|/(x-2) Do I have to graph it, or is there an algebraic way to do this? Cheers. 2. Originally Posted by Glitch Hey guys, I'm trying to understand how to determine the left and right side of this limit: limit as x -> 2 \|x-2\|/(x-2) Do I have to graph it, or is there an algebraic way to do this? Cheers. Recall that $\|x - 2\| = \begin{cases}\phantom{-(}x-2\phantom{)}\textrm{ if }x-2\geq 0\\-(x - 2)\textrm{ if }x-2<0\end{cases}$ $\|x - 2\| = \begin{cases}x-2\textrm{ if }x\geq 2\\2-x\textrm{ if }x<2\end{cases}$ When you are are approaching $x = 2$ from the left, $x < 2$. So $\lim_{x \to 2^{-}}\frac{\|x-2\|}{x-2} = \lim_{x\to 2^{-}}\frac{2-x}{x-2}$ $= \lim_{x\to 2^{-}}\frac{-(x-2)}{x-2}$ $= \lim_{x\to 2^{-}}(-1)$ $= -1$. When you are approaching $x=2$ from the right, $x>2$. So $\lim_{x \to 2^{+}}\frac{\|x-2\|}{x-2} = \lim_{x\to 2^{+}}\frac{x-2}{x-2}$ $=\lim_{x \to 2^{+}}(1)$ $= 1$. Clearly, since the left hand limit is not the same as the right hand limit, the limit does not exist. 3. Ahh, I see. I was thinking that that'd be 0/0 (when you sub in 2) which is undefined. Thanks.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 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": 0.9555320739746094, "perplexity_flag": "middle"}
http://jlebl.wordpress.com/2012/11/09/numerical-range/	# The Spectre of Math ## November 9, 2012 ### Numerical range Filed under: Hacking,Mathematics — jlebl @ 12:06 am I was fiddling with numerical range of two by two matrices so I modified my root testing python program to do this. The numerical range of $A$ is the set of all values $\frac{v^* A v}{v^* v}$ for all nonzero vectors $v$. This set is a compact set (it can be seen as the image via the mapping $v \mapsto v^* A v$ of vectors on the unit sphere $v^* v = 1$ which is a compact set). It’s convex which is harder to show. For two by two it is an elliptic disc (could be degenerate). See the result here, it plugs in random vectors and shows the result. Here’s an example plot for the matrix $A=\begin{bmatrix} 1 & i \\ 1 & -1 \end{bmatrix}$. The code is really inefficient and eats up all your cpu. There’s no effort to optimize this.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 6, "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": 0.9218195080757141, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/84954/list	2 edited title 1 # Can you make the cotangent bundle to a complex The cotangent bundle of a manifold has a canonical symplectic form and if we choose a riemannian metric on $M$, we can give it an almost complex structure. Is this structure integrable, and if it isn't in general, what are the conditions on the manifold for it to be integrable? Can we give $T^*M$ a complex structure in some other way? Furthermore, what are the conditions on $M$ to ensure that the symplectic form is a Kaehler form? I read this thread: http://mathoverflow.net/questions/26776/kahler-structure-on-cotangent-bundle, but I honestly didn't understand, how far it answers these questions.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8844164609909058, "perplexity_flag": "head"}
http://lucatrevisan.wordpress.com/2010/04/22/cs254-lecture-1-introduction/	in theory "Marge, I agree with you - in theory. In theory, communism works. In theory." -- Homer Simpson # CS254 Lecture 1 – Introduction April 22, 2010 in CS254 \| Tags: CS254 2010 Last revised 4/29/2010 This course assumes CS154, or an equivalent course on automata, computability and computational complexity, as a prerequisite. We will assume that the reader is familiar with the notions of algorithm, running time, and of polynomial time reduction, as well as with basic notions of discrete math and probability. We will occasionally refer to Turing machines. In this lecture we give an (incomplete) overview of the field of Computational Complexity and of the topics covered by this course. Computational complexity is the part of theoretical computer science that studies 1. Impossibility results (lower bounds) for computations. Eventually, one would like the theory to prove its major conjectured lower bounds, and in particular, to prove ${{\bf P} \neq {\bf NP}}$, implying that thousands of natural combinatorial problems admit no polynomial time algorithm 2. Relations between the power of different computational resources (time, memory randomness, communications) and the difficulties of different modes of computations (exact vs approximate, worst case versus average-case, etc.). A representative open problem of this type is ${{\bf P} = {\bf BPP}}$, meaning that every polynomial time randomized algorithm admits a polynomial time deterministic simulation. Ultimately, one would like complexity theory to not only answer asymptotic worst-case complexity questions such as the P versus NP problem, but also address the complexity of specific finite-size instances, on average as well as in worst-case. Ideally, the theory would be eventually able to prove results such as • The smallest boolean circuit that solves 3SAT on formulas with 300 variables has size more than ${2^{70}}$, or • The smallest boolean circuit that can factor more than half of the 2000-digit integers, has size more than ${2^{80}}$. At that stage, the theory will help develop unconditionally secure cryptosystems, it will give us an understanding what makes certain instances harder than other, thus helping develop more efficient algorithms, and it will provide the mathematical language to talk not just about computatations performed by computers, but about the behavior of any discrete system that evolves according to well-defined law. It will be the ideal mathematical tool to reason about the working of the cell, of the brain, natural evolution, economic systems, and so on. It will probably involve some of the most interesting mathematics of the 23rd century and beyond. For the time being, complexity theorists have had some success in proving lower bounds for restricted models of computations, including models that capture the behavior of general algorithmic approaches. Some of the most interesting, and surprising, results in complexity theory regard connections between seemingly unrelated questions, yielding considerable “unification” to the field. 1. Some Examples of Lower Bound Results in Complexity Theory It remains open to rule out the possibility that every problem in ${{\bf NP}}$ is solvable by ${O(n)}$ size circuits on inputs of length ${n}$, even though the existence of ${{\bf NP}}$ problems requiring circuits of size ${2^{\Omega(n)}}$ is considered plausible. There has been some success, however, in dealing with restricted models of computations. Some examples follow. 1.1. Communication Complexity In this set-up, several parties each hold a part of the input to a computational problem that they wish to solve together. We ignore the complexity of the computations that the various parties perform, and we focus on how much communication they need in order to solve the problem. In the most basic set-up of this type, our computational problem is to compute a function ${f(\cdot,\cdot)}$. There are only two parties, holding respectively a string ${x}$ and a string ${y}$ (say, of equal length), and they wish to collaboratively compute ${f(x,y)}$. For many interesting functions ${f(\cdot,\cdot)}$ tight bounds on the communication complexity can be established. For example, if the parties wish to determine whether ${x=y}$, then the communication complexity is linear if the parties run deterministic algorithms, but sub-linear if they are allowed to use randomness. If ${x}$ and ${y}$ are bit-vector representation of two sets, and the parties wish to determine if the two sets have non-empty intersection, then linear communication is required even if the two parties use randomized algorithms. A very useful feature of this model is that, in several set-ups in algorithm design and data structure design, a good solutions implies an efficient protocol for a related communication complexity problems. If one is able to establish a lower bound for the communication complexity problem, then one derives a lower bound for the algorithmic or data structure problem. For example, algorithms in the streaming model, which is a useful model for applications in data bases and networks, are allowed to only make one pass over their input, and the goal is to design algorithms that use a limited amount of memory. Now note that, if we have a streaming algorithm using space complexity ${s(n)}$ to solve a problem on inputs ${x=(x_1,\ldots,x_n)}$ of length ${n}$, then we also have a communication protocol of communication complexity ${s(n)}$ for two parties who know, respectively, the first ${n/2}$ bits of the input and the last ${n/2}$ bits of the input. The first party simply runs the streaming algorithm on the first ${n/2}$ parts of the input, then sends the state of the algorithm to the second party, who has now enough information to complete the computation. A very useful communication complexity set-up is the “number on the forehead” model. Here ${k}$ parties want to jointly compute ${f(x_1,\ldots,x_k)}$, where each ${x_i}$ is ${n/k}$ bits long; the ${i}$-th party in the protocol knows the value of ${x_j}$ for all ${j\neq i}$. (The name of the model comes from the image of party ${i}$ having the value of ${x_i}$ written on his forehead, so that it is the only value that he cannot see.) Every problem can clearly be solved with communication complexity ${n/k}$, but only ${n/2^k}$ lower bounds are known for explicit functions. Proving a ${n/k^{O(1)}}$ lower bound for a polynomial time computable functions is a major open problem. Its solution would have several applications, including circuit lower bounds. 1.2. Proof complexity Proof complexity lower bounds are inspired by the conjecture that ${{\bf NP} \neq co{\bf NP}}$. If so, then if we consider the ${co{\bf NP}}$-complete problem of testing if a boolean formula is unsatisfiable, and if we fix any formalism to write down mathematical proofs (provided that the validity of a given proof can be checked in polynomial time), there must be families of unsatisfiable formulas such that their shortest proof of unsatisfiability in the formalism grows more than polynomially with the size of the formula — otherwise we would have an ${{\bf NP}}$ algorithm for unsatisfiability which simply guesses a polynoial-length proof of unsatisfiability for the given formula and then verifies the validity of the proof. For general formal languages for mathematical proofs (or “proof systems”), it remains open to construct unsatisfiable boolean formulas whose shortest proofs of unsatisfiability have superpolynomial length, but such “proof complexity” lower bounds are known for several specialized proof system there are known super-polynomial, and even exponential, lower bounds. Such lower bounds have implications for the performance of SAT-solvers. If an algorithm for SAT is complete (meaning that it always finds a satisfying assignment when given a satisfiable formula), then, when the algorithm outputs no satisfying assignment, the sequence of steps of its computation is a proof of unsatisfiability of the given instance. If one can model such a proof within a proof system for which there are known lower bounds, then such lower bounds apply to the running time of the SAT solver as well. Backtracking-based solvers such as DPLL are complete and, when they fail to output a satisfying assignment after ${t}$ steps, one can construct a tree-like resolution proof of unsatisfiability of size about ${t}$ for the given formula. Families of unsatisfiability formulas are known whose shortest resolution proofs of unsatisfiability are of exponential length, and so DPLL type algorithms must take exponential time on such instances. 1.3. Integrality gaps A common approach to find approximations to ${{\bf NP}}$-hard combinatorial optimization problems is to relax the problem to a convex optimization problem in which the set of feasible solutions, instead of being the set of binary strings satisfying a certain condition, is a larger convex set. Under general conditions, optimizing over a convex set of feasible solutions can be done in polynomial time, and one can often derive an approximate solution to the original combinatorial problem by using the optimal solution of the convex relaxation. The integrality gap of a convex relaxation is the worst-case ratio (over all instances) between the optimum of the combinatorial problem and the optimum of the convex relaxation. For relaxations whose integrality gap is very far from one (very small for maximization problems, or very large for minimization problems), we can conclude that they are not useful to derive approximation algorithms. Recent results of this type rule out very general classes of relaxations, and they apply to infinite families of relaxations which add sub-exponentially many auxiliary variables and constraints. 1.4. Restricted circuits Ideally, we would like to show that 3SAT cannot be solved by circuits of size ${2^{o(n)}}$ on inputs with ${n}$ variables, but this is completely out of reach for now; even “just” proving that 3SAT cannot be solved by circuits of polynomial size would imply ${{\bf P}\neq{\bf NP}}$. Indeed, currently, it is an open question to even prove that 3SAT, or any other problem in ${{\bf NP}}$ cannot be solved by circuits of size ${6n}$ on inputs of length ${n}$. There has been some success in proving lower bounds against special types of circuits. Unlike the models mentioned in the previous sections, such circuits do not model realistic algorithms, but such lower bounds have interesting applications as well. Monotone circuits. A boolean function ${f(\cdot)}$ is monotone if changing a zero to a one in the input cannot change the output from a one to a zero. It is easy to see that a monotone boolean function can always be computed by a circuit consisting only of AND gates and of OR gates (without NOT gates), and that every function computed by a circuit of this type is monotone. Hence, circuits made only of AND gates and OR gates are called monotone circuits. Razborov proved in the 1980s that ${CLIQUE_{\sqrt n}}$, the problem of deciding if a given ${n}$-vertex graph has a clique of size at least ${\sqrt n}$, cannot be solved by monotone circuits of polynomial size. Note that, if we represent graphs as adjacency matrices, then, as a boolean function of the input matrix, ${CLIQUE_{\sqrt n}}$ is a monotone function. It was conjectured that every polynomial time computable monotone function can also be computed by a monotone circuit of polynomial size and, if so, it would follow that ${CLIQUE_{\sqrt n}}$ is not solvable in polynomial time and hence ${{\bf P} \neq {\bf NP}}$. Unfortunately, it was soon proved that checking if a given graph has a perfect matching (a monotone function computable in polynomial time) cannot be done with polynomial size monotone circuits, and so the conjecture is false. AC0. ${AC0}$ is the class of decision problems solvable by polynomial size circuits that have NOT gates, AND and OR gates of unlimited fan-in, and have only constant depth (independent of the input length). This class captures “constant time on a parallel computer with a polynomial number of processors” and contains a few non-trivial boolean functions. It is known that PARITY, the problem of checking if the number of ones in the given input is odd, cannot be computed in AC0. Modular gates. To see how robust is the AC0 lower bound, it is natural to “hard-wire” into AC0 circuits the ability to compute parity, and see if it is still possible to prove a lower bound. Indeed, one of the proofs that PARITY is not in AC0 can be adapted to show that MOD3 (checking if the number of ones in the given input is a multiple of 3) cannot be computed by a constant depth, polynomial size, family of circuits with NOT gates, and AND, OR, and PARITY gates of unlimited fan-in. The latter class is called ${ACC0[2]}$, with the two in square brackets standing for the MOD2 gates which are allowed in the model. Similarly, one can define ${ACC0[m]}$ as the class of problems solvable by polynomial size, constant depth, families of circuits with NOT gates and unlimited-fanin AND, OR and MODm gates. It is possible to show that if ${p}$ and ${q}$ are distinct primes then MODq is not computable in ${ACC0[p]}$. It is open, however, whether there is a problem in ${{\bf NP}}$ not solvable by linear size ${ACC0[6]}$ circuits. (Or by linear size ${ACC0[m]}$ circuits, where ${m}$ is any composite with two distinct factors.) 2. Some Examples of “Connections” and “Unification” in Complexity Theory So far, unconditional lower bounds have been proved only against restricted classes of algorithms (or for problems of very high complexity. Most of the work of contemporary complexity theory is on connections about questions. For example: • NP-completeness is a prototypical example. We know thousands of ${{\bf NP}}$-complete problems, and although we do not know their complexity yet, we know that understanding the complexity of any one of them will imply understanding the complexity of all of them. If we consider, for every ${{\bf NP}}$-complete problem ${L}$, the question “does ${L}$ have a polynomial time algorithm?” then we don’t know the answer to all those thousands of questions, but we know that all the answers are the same. • One-way functions. A function ${f()}$ is one way if computing ${f(x)}$ given ${x}$ is easy, but finding an ${x'}$ such that ${f(x')=y}$ is hard given ${y=f(x)}$. (The difficulty has to hold on average, for a random ${x}$.) If one-way functions do not exist, then no cryptographic problem is solvable, except those that have simple information-theoretic solutions. For example one can encrypt one message no longer than a shared secret key using one-time pad, but it is not possible to encrypt messages longer than the shared key. If one-way functions exist, however, then all problems in private-key cryptography, as well as the problem of creating digital signatures, have solutions with extremely high security guarantees. This means that even though we don’t know whether there provably exist secure signature schemes, secure authentication schemes, secure encryption schemes and so on, we know that the questions of whether such protocols exist are all the same, and are all equivalent to the question of whether one-way functions exist. • Probabilistically Checkable Proofs (PCPs) provide a characterization of ${{\bf NP}}$ which is a convenient starting point to prove hardness of approximation of combinatorial optimization problems. Via the PCP theorem and various reductions, we know that for several optimization problems it is as hard to improve the performance of known polynomial time approximation algorithms as it is to solve the problem optimally in polynomial time. • Derandomization is the task of reducing the amount of randomness used by randomized algorithms. Ideally, one would like to show that every randomized algorithm can be simulated with no use of randomness whatsoever, in a purely deterministic way. It is known that if there is a problem in solvable in time ${2^{O(n)}}$ and having circuit complexity ${2^{\Omega(n)}}$ then such polynomial time deterministic simulations of randomized algorithms are possible. Note that this a connection of a different nature than the ones described above. Previously, we discussed results showing that if a certain computational problem (solving a problem in ${{\bf NP}}$, inverting a one-way function, optimally solving an optimization problem) is hard, then other computational problems are also hard (respectively, solving any ${{\bf NP}}$-complete problem, breaking certain encryption, authentication and signature schemes, approximately solving certain optimization problems). Derandomization results, however, turn a hardness assumption into an algorithm. • Worst-case versus average-case. The result on derandomization mentioned above is the combination of two main results: (i) one can simulate deterministically with polynomial slowdown all randomized algorithms, provided that there is a problem solvable in time ${2^{O(n)}}$ that is hard on average for circuits of sub-exponential size; (ii) if there is a problem solvable in time ${2^{O(n)}}$ that is hard in the worst case for circuits of sub-exponential size, then there is also a problem solvable in time ${2^{O(n)}}$ that is hard on average for circuits of sub-exponential size. The second theorem is part of a more general theory showing that for certain problems and complexity classes one can turn worst-case hardness assumptions into seemingly stronger (but in fact equivalent) average-case hardness assumptions. 3. The Plan for this Course 3.1. The Basics We will start by looking at the basic models in complexity theory, and consider deterministic, non-deterministic, randomized, non-uniform and memory-bounded algorithms, and the known relations between their power. 3.2. Reingold’s algorithm We will then give an overview of Reingold’s algorithm, which proves that undirected connectivity is solvable with ${O(\log n)}$ bits of memory deterministically. 3.3. PCP and Inapproximability We will continue with Probabilistically Checkable Proofs, the model that allows to prove NP-hardness results for approximability, and sketch Dinur’s proof of the PCP Theorem, the main result in this theory. 3.4. Parity not in AC0 We will then see the proof that computing the parity of the number of ones in an ${n}$-bit input cannot be computed by polynomial size, constant-depth, circuits made of AND, OR and NOT gates. The proof that we will see has a relatively easy extension to show that there is no polynomial size constant depth circuit made of AND, OR, NOT, and MOD${p}$ gates that checks whether the number of ones in the given input is a multiple of ${q}$, where ${p,q}$ are distinct primes. It remains open to prove lower bounds for the variation of this model in which MOD${m}$ gates are allowed, where ${m}$ is a composite with distinct prime factors. (E.g. ${m=6}$.) 3.5. Derandomization, Pseudorandomness, and Average-case Complexity We define pseudorandom generators, see how one can obtain derandomization results from the existence of strong pseudorandom generators, state the results of Impagliazzo, Nisan and Wigderson on derandomization based on worst-case and on average-case complexity assumptions, and look at the GGM approach to increase the stretch of pseudorandom generators. 3.6. Natural Proofs Razborov and Rudich have identified a bottleneck that applies to all known techniques to prove circuit lower bounds in restricted model. All known techniques allow us to efficiently distinguish the truth table of a function of low circuit complexity in the model from a random truth table. If strong pseudorandom generators exist, however, it is possible to generate truth-tables of functions that are computable by small circuits but that cannot be efficiently distinguished from random truth-tables. 3.7. Quantum Complexity Theory We will describe the computational model of quantum computers, and describe one of the two famous quantum algorithms, an algorithm that is able to search over a space of size ${2^n}$ in time ${2^{n/2}}$. Time permitting, we will also show that, with no further assumption on the search space, time ${2^{n/2}}$ is best possible. ## 11 comments How future-proof is this field? Say we believe the hype that quantum computers are coming (maybe a hundred years from now, maybe ten years, I don’t care). Are these results still interesting, or will they be a historical curiosity, people studying the wrong model? It seems obvious that NP-completeness will still be interesting, but I don’t know enough about the rest of the area. By the way, thank you for posting this. It is very clear. Sorry, can’t resist– half of the 2000 digit integers are easy to factor — they are even! I guess a proper wording would have to mention trivial factors. What is a good definition for a trivial factor? sorry, I didnt read that carefully, 2^80 is mentioned By “factor” I mean “find all prime factors” not “find a prime factor.” 2^80 is the ciruit size. So minimum non-trivial factor still would need definition if it was a real hypothesis. OK, I get it, all factors With some good probability a random number will be a prime or a prime multiplied by a small factor. 2^{-40} is too small. As Omid says, about a 2^{-12} fraction of 2000-digit numbers are prime. So it looks like we can knock off one of those problems. all right, and it took only 4 hours, instead of 300 years! That is a very nice introduction. You mentioned that one can construct a tree-like resolution proof of size ~t from a run of a DPLL-based SAT-solver on an unsatisfiable CNF that needed t steps. In general, this is not quite true. The form of the resolution proof depends on the features that are used in the SAT-solver: - basic DPLL gives tree-like resolution proofs - DPLL with clause learning and restarts (~ competitive solvers) gives dag-like resolution proofs - DPLL with clause learning and without restarts gives “regular resolution trees with lemmas” There remain interesting questions on what a SAT-solver needs to do in order to simulate full dag-like resolution (restart and learning strategies) and if there are examples where restarts yield significantly better results (separations). Two recent papers on the topic are (more references are in there): Pipatsrisawat and Darwiche – On the Power of Clause-Learning SAT Solvers with Restarts (CP 2009) Buss et al. – Resolution trees with lemmas: Resolution refinements that characterize DLL algorithms with clause learning (LMCS 2008) %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 95, "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": 0.9305010437965393, "perplexity_flag": "head"}
http://mathoverflow.net/questions/120545?sort=newest	Compute roots of sum_i c_i/(a_i + b_i x)^p Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) How to compute the (real) roots of $$\sum_{i=1}^n \frac{c_i}{(a_i + b_i \cdot x)^p}$$ for given reals $a_i, b_i, c_i$, and positive integers $n, p$? The cases $p=1, ..., 5$ and $n=6, ..., 20$ would already be very useful for me. I actually just need any root in a given interval. Multiplying by the denominators, this task can be reduced to finding roots of a polynomial, but this only works for very small $n$ whereas even for $n=8$ the coefficients in the polynomial are numerically unstable. The only other method I could think of is using binary search (aka the bisection method). But this is too slow. Is there a faster method that is numerically stable? - 2 If $p$ is even, I don't think you'll find many real roots! – Barry Cipra Feb 1 at 20:40 Sorry, I had set $c_i = 1$ to simplify the question. I have now put it back in so there may be roots even for even $p$. – Emanuele Viola Feb 1 at 20:48 With a $c_i$ in the numerator, you can simplify the denominator to just $(x-d_i)^p$. – Barry Cipra Feb 1 at 20:52 1 did you try Newton-Raphson? – S. Sra Feb 1 at 20:53 Thanks. I have not tried it because I was not sure what starting point to choose/whether the method would work in general. Do you see it? – Emanuele Viola Feb 1 at 21:49 show 1 more comment 1 Answer This recent master thesis by Leonardo Robol treats the case $p=1$ in a numerically sound way. I think they are going to release some code soon, so you might want to contact the author. - Thanks for the useful pointer. I wonder if something like that can be done for $p > 1$? – Emanuele Viola Feb 19 at 14:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.9370940327644348, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/206376/how-to-plot-n-points-on-the-surface-of-a-d-dimensional-sphere-roughly-equidistan?answertab=votes	# How to plot N points on the surface of a D-dimensional sphere roughly equidistant apart? Let's say I have a D-dimensional sphere with a radius R. I want to plot N number of points evenly distributed (equidistant apart from each other) on the surface of the sphere. It doesn't matter where those points are exactly, just that they are ROUGHLY equidistant from each other. How would I do this? - – Sasha Oct 3 '12 at 5:49 For the last method on that page, how would I generate Gaussian random variables so that the radius of the sphere is always the same for each generated point? – user644337 Oct 3 '12 at 6:10 2 Generate vector $\{Z_1,Z_2,\ldots, Z_{d+1}\}$ then divide each of its components by the Euclidean length of that vector and multiply by $R$. This rescales the vector to have length $R$. – Sasha Oct 3 '12 at 11:50 I guess N coordinates must be generated with any Gaussian random and after that resulting vector have to be normalized and multiplied on the R. – KvanTTT Oct 3 '12 at 11:52 1 To have N points drawn uniformly on a region is quite different to have N points approximately equidistant on that region – leonbloy Oct 3 '12 at 13:59 show 3 more comments ## 1 Answer If $N \le D + 1$ all points must be located on $N-1$ dimension equilateral triangle. UPDATE So, I thought today at this problem and invented method, contains following steps: 1. Generating $N$ random points on sphere $R$ (that is generating points with coordinates with Gaussian distribution). 2. Building Convex Hull (or triangulation on hypersphere even better) with generated on first step points. This step can be solved with MIConvexHull library if you familar with C#, similar library on your favorite language or your own code. 3. Using Genetic algorithm, Simulated annealing or another method of global optimization. This method then have to be applied to variance value of all edge lengths from convex hull from step 2. - What is an example of a 4D sphere with one equilateral triangle? How can it be generalised to NDimensional Sphere and triangle? – Arjang Oct 3 '12 at 5:22 – user644337 Oct 3 '12 at 5:46 No, I mean $N \le D + 1$, that is all simplexes with such dimensions. For example, for 3D sphere segment there are segment (2 vertices), triangle (3 vertices) and tetrahedron (4 vertices). But you are right: this method not suitable for $N > D + 1$ case. – KvanTTT Oct 3 '12 at 7:24	{"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": 9, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9053246974945068, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=42938	Physics Forums ## Gravity forces converted to energy Can anyone help me on this? If I have a very large ball (ball A) and a tiny ball (ball B) as such that ball B is attracted by the center of ball A, then theorically, if ball A is a perfect sphere I could expect ball B to start spinning on ball A and never stop as ball B won't have anywhere to rest on ball A. Am I right? In fact, this would be the opposite of anti-gravity and by multiplying that force we could extract an energy that could eventually move an object. PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus Mentor Blog Entries: 9 Actually we routinely extract energy, and nearly endless from a combination of the suns heat and gravity. The sun evaporates sea water, it falls in the mountains as rain. We dam the streams and generate electricity. As for your balls, I have no clue what you are talking about but I can be sure that it will not work. Simply because you are talking about a free lunch. TANSTAAFL. What you think is wrong because ball B WILL rest on ball A, because the gravitational forces will not help ball B to roll along the side of ball A. The force on ball B is drawn towards the centre of ball A, which makes the two stick. I don't see why ball B should start spinning around. Recognitions: Homework Help Science Advisor ## Gravity forces converted to energy The Earth is like a big ball. Place a small ball such as a bowling ball someplace on the Earth's surface and let us know whether it spins out of control! :-) ya, the ball would not spin around the other ball since the force between the two balls is directly perpendicular to the intersection tangent. Theregore, there would be no force to push the ball in one direction or another. since the force between the two balls is directly perpendicular to the intersection tangent ...a very long way to say towards the centers of masses $$v(average) = \frac{s}{t}$$ $$a=\frac{v-u}{t}$$ $$s=ut + \frac{1}{2} a t^2$$ $$v^2=u^2 + 2as$$ but what if I push two balls against each other? They can't possibly stick together. This is what I mean. As long as you still apply force to both balls directly towards each other they will touch at a tangent point and stay that way unless the force changes direction. Thread Tools \| \| \| \| \|---------------------------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: Gravity forces converted to energy \| \| \| \| Thread \| Forum \| Replies \| \| \| General Physics \| 15 \| \| \| Special & General Relativity \| 10 \| \| \| Introductory Physics Homework \| 1 \| \| \| General Physics \| 4 \| \| \| General Discussion \| 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": 0, "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": 0.9325041770935059, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/57719/algebraic-closure-of-commuting-pairs-of-matrices	## algebraic closure of commuting pairs of matrices ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $F$ be an arbitrary field of characteristic $0$, $K$ its algebraic closure. Define $M=\{ (x,y)\in M_n(F)×M_n(F) \mid [x,y]=0\}$ and let $N$ be the Zariski closure of $M$ in $K^{2n^2}$. How can one show that $N$ contains the set $\{(axa^{-1},aya^{-1}) \mid (x,y)\in N, a\in \mathrm{GL}(n,K)\}$? Thank you. - 2 It seems to me that $M$ itself is Zariski closed. – Johannes Ebert Mar 7 2011 at 19:46 1 From the way this problem is phrased, it seems to be designed to test understanding of basic algebraic geometry. In the absence of other information, I think it very likely that it is coursework. Voting to close. – Victor Protsak Mar 7 2011 at 20:01 1 @Victor: Sorry, I was reading this in an article where it was left as "easy to check". – spelas Mar 7 2011 at 20:18 1 My comment was wrong. If $F \neq K$, then $M$ is not Zariski closed. – Johannes Ebert Mar 7 2011 at 20:33 ## 1 Answer Note that $M$ is invariant under the $GL_n(F)$-action given by $a \cdot (x,y):=(axa^{-1},aya^{-1})$. It follows that its closure $N$ is also invariant under $GL_n(F)$. Since $F$ is infinite and $GL_n$ is reductive, the rational points $GL_n(F)$ are Zariski-dense in $GL_n(K)$ by Borel, Linear Algebraic Groups, Corr. 18.3. (This is probably the piece of information you were missing). Then it follows that $N$ is $GL_n(K)$-invariant, as claimed. - This helps a lot, thank you. – spelas Mar 7 2011 at 22:48	{"extraction_info": {"found_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": 0.951245129108429, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-geometry/127612-fort-space-topology.html	# Thread: 1. ## Fort Space Topology Does anybody know which sets are both open and closed, and which sets are neither open nor closed for the Fort Space Topology? 2. Hello, Just write down what open and closed sets are in this topology. Let X be the infinite set. Fix $p\in X$. Open sets A : 1.a) $p\not\in A$ 1.b) or $A'=X\setminus A$ is finite Closed sets B : 2.a) $p\in B$ (complement of a set in which p is not) 2.b) or $B$ is finite. So now try to combine... It's obvious that a set can't satisfy both 1.a) and 2.a), and can't satisfy both 1.b) and 2.b) Now is it possible for a set to satisfy 1.a) and 2.b) ? Of course ! A finite subset of X which doesn't contain p will satisfy the two conditions. Thus it's both closed and open. Similarly, it's possible to find a set that satisfies 1.b) and 2.a)... A set that is neither open nor closed ? No it's not possible, because either p is in the set, either it's not (that's the law of excluded middle ). If it is, then the set is closed, and if it's not, then the set is open ! 3. Thanks for this Moo. Am I right in thinking that the neigbourhoods of the point p, are just the open sets U of p whose complement is finite? 4. Originally Posted by Cairo Thanks for this Moo. Am I right in thinking that the neigbourhoods of the point p, are just the open sets U of p whose complement is finite? Yes, the neighbourhood U has to contain an open set, say A, which contains p. So the complement of A will be finite (since it doesn't satisfy 1.a), it has to satisfy 1.b)). And since $A \subset U$, $X\setminus U \subset X\setminus A$. Thus $X\setminus U$ is finite... 5. After a sleepless night and much scribbling, I am still unable to prove this result for the Fort space topology. Let (a_n) be a sequence in X, such that the set of the sequence, {a_n : n in N} is infinite. Using the definition of convergence in topological spaces, prove that (a_n) has a subsequence which converges to p. If it were a metric space, I think i could do it, but as it stands I can't. Could anybody offer a proof please?	{"extraction_info": {"found_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": 8, "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": 0.9575027227401733, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/132569/characterizing-open-closed-compact-sets-in-the-metric-space-mathbbzn-d	# Characterizing Open/Closed/compact sets in the metric space $(\mathbb{Z}^n,d)$ What is an open set in the metric space $(\mathbb{Z}^n,d)$, where $d$ is the Euclidean distance in $\mathbb{R}$? As far as I know, in a metric space an open set $O$ is defined as follows: For each point $x\in O$ it exists an $\varepsilon$ such that that $B_\varepsilon(x)\subset O$ is true. Since in $\mathbb{Z}^n$ I just have some sort of layer I thought of defining the $\varepsilon$ as follows: $$\varepsilon=\sqrt{\sum_{i=1}^{d}x_{i}^{2}}$$ This way I can characterize all the $\varepsilon$-Balls $B_\varepsilon(x)$ for each $x\in\mathbb{Z}^n$. So if $d(x,y)<\varepsilon$ for each $y\in B_\varepsilon(x)$ the set is open. And the same for $d(x,y)\le\varepsilon$ implies a closed set. For compactness I could say that, like in $\mathbb{R}^n$, each closed set which is bounded by a constant (using the euclidean metric here) is compact. What about the single points in $\mathbb{Z}^n$? Are they open sets, too? - 1 Single points are open. Everything is open, everything is closed. – André Nicolas Apr 16 '12 at 16:23 ## 1 Answer With the euclidean distance, $\mathbb{Z}^n$ is a discrete space. Every subset is open. Every subset is closed. Only the finite subsets are compact. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 19, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": 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": 0.9464584589004517, "perplexity_flag": "head"}
http://mathoverflow.net/questions/12483/uniqueness-of-a-polygon	## Uniqueness of a polygon ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Suppose I have two $n$-sided polygons A and B. Is there a non-trivial upper bound on the number of parameters (eg. area, perimeter, etc) of the two polygons, that need to be the same, for A and B to be identical? (For example, if we consider a simpler case to the problem when say A is constrained to lie inside B, then if area(A) = area(B), then A and B are identical. Hence area is the only non-trivial parameter in this case) - ## 3 Answers I guess you are looking for a nice answer, but here is a stupid one. On the other hand I'm sure that there is no "nice answer". A polygon can is uniquely determined by length of sides $\ell_i$ and angles $\alpha_i$. Thus we have to find a complete set of invariants for sequence $(\alpha_1,\ell_1,\alpha_2,\ell_2,\dots\alpha_n,\ell_n)$ which survive after even cyclic shifts and reversing order. Then you prepare symmetric polynomials for your group. Say take all monomials of degree at most one in each $\alpha_i$ and $\ell_i$ and take its mean value it along the group. You obtain a big collection of polynomial expressions in $\alpha_i$ and $\ell_i$ which gives complete invariant (perimeter will be one of them). - I liked the first version of your answer better --- somehow symmetric polynomials of complex coordinates of vertices seem more natural to me. – t3suji Jan 21 2010 at 5:21 @t3suji. Well, it was not quite correct... – Anton Petrunin Jan 21 2010 at 15:42 ### 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 moduli space of $n$-gons up to orientation preserving similarity can be identified with $\mathbb{C}P^{n-2}$, so I would say $2n-3$. See On the moduli space of polygons in the Euclidean plane by Kapovich and Millson for more about this space. - In what sense do two parameters suffice to pick out a triangle? Naively one needs three (say, the lengths of the sides). – Michael Lugo Jan 21 2010 at 3:37 Oh, sorry, it's the moduli space up to orientation preserving similarity. So two angles suffice for a triangle. – Richard Kent Jan 21 2010 at 4:03 Edited to fix this. – Richard Kent Jan 21 2010 at 4:08 There are many structures that mathematicians study because of their intrinsic interest, simplicity and being quick starting. Examples that come to mind are plane polygons and graphs. Plane polygons have a variety of properties that one can look at: area, perimeter, minimum number of vertex guards, number of reflex angles, number of right angles, etc. An individual graph can have a variety of "invariants" that can be studied: coloring number, clique number, being eulerian, or hamiltonian, etc. For two graphs there is no list of invariants that guarantees that the two graphs are isomorphic. The complexity of checking when two graphs are isomorphic is still a dynamic area to investigate. I do not know any list of properties that guarantee that two polygons are congruent going beyond specifying the lengths of the sides and the measure of the angles. Finding new combinatorial/geometric properties of polygons seems to continue to be very worthwhile. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9058282971382141, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/tagged/distributions+quantum-mechanics	# Tagged Questions 3answers 135 views ### Hilbert space of harmonic oscillator: Countable vs uncountable? Hm, this just occurred to me while answering another question: If I write the Hamiltonian for a harmonic oscillator as $$H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2$$ then wouldn't one set of ... 2answers 72 views ### Translator Operator In Modern Quantum Mechanics by Sakurai, at page 46 while deriving commutator of translator operator with position operator, he uses $$\left\| x+dx\right\rangle \simeq \left\| x \right\rangle.$$ But for ... 4answers 266 views ### Is the momentum operator diagonal in position representation? The matrix elements of the momentum operator in position representation are: $$\langle x \| \hat{p} \| x' \rangle = -i \hbar \frac{\partial \delta(x-x')}{\partial x}$$ Does this imply that \$\langle x ... 1answer 115 views ### Schrodinger equation in term of Fokker-Planck equation From Wikipedia on the Fokker-Planck equation: \tag{1}\frac{\partial }{\partial t}f\left( x^{\prime },t\right) ~=~\int_{-\infty}^\infty dx\left( \left[ D_{1}\left( x,t\right) \frac{\partial ... 3answers 450 views ### Matrix elements of momentum operator in position representation I have two related questions on the representation of the momentum operator in the position basis. The action of the momentum operator on a wave function is to derive it: \hat{p} ... 3answers 683 views ### Don't understand the integral over the square of the Dirac delta function In Griffiths' Introduction to Quantum Mechanics he gives the eigenfunctions of the Hermitian operator $\hat{x}=x$ as being $$g_{\lambda}\left(x\right)~=~B_{\lambda}\delta\left(x-\lambda\right).$$ ... 1answer 164 views ### State normalization in Dirac's formulation of quantum mechanics Let us divide the time $T$ into $N$ segments each lasting $δt = T/N$. Then we write $\langle q_F \| e^{−iHT} \|q_I \rangle = \langle q_F \| e^{−iHδt} e^{−iHδt} . . . e^{−iHδt} \|q_I \rangle$ Our ... 2answers 219 views ### Is the braket notation of the Dirac delta function symmetric? I have a book saying, $\int \delta(x-x')\psi(x)dx = \psi(x')$ where $\psi(x) = \langle x\lvert\psi\rangle$, so our definition of delta function would be $\langle x'\lvert x\rangle = \delta(x-x')$. ... 1answer 78 views ### When does the “norm of quasi-eigenvectors” matter in calculations? For which physical results are these even used? Which physical system in nonrelativistic quantum mechanics is actually described by a model, where the norm of the "position eigenstate" (i.e. the delta distribution as limit of vectors in 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": 10, "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": 0.3, "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": 0.8536309599876404, "perplexity_flag": "middle"}
http://nrich.maths.org/1131/note	### 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? ### Reverse to Order Take any two digit number, for example 58. What do you have to do to reverse the order of the digits? Can you find a rule for reversing the order of digits for any two digit number? ### Card Trick 2 Can you explain how this card trick works? # Back to the Planet of Vuvv ### Why do this problem? This problem one which requires some knowledge of both place value and different bases. Working in another base can help with real understanding of our base-$10$ number system. You could start this by either explaining or re-visiting counting in a base such as $6$. Base $7$ could be introduced using the days of a week as an example. ### Key questions If Zios count in $3$s, what will their first 2-digit number be in human numbers? If Zepts count in $7$s, what will their first 2-digit number be in human numbers? What is $122, 22, 101, 41$ in Zio counting? What is $122, 22, 101, 41$ in Zept counting? Would drawing a sketch help with sorting out the four compass points? ### Possible extension Learners could make a similar puzzle for themselves, or go on to this similar problem: Basically. ### Possible support Suggest trying Alien Counting instead which is a simpler problem of the same type. The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9245089292526245, "perplexity_flag": "middle"}
http://psychology.wikia.com/wiki/Natural_number?interlang=all	# Natural number Talk0 31,725pages on this wiki Assessment \| Biopsychology \| Comparative \| Cognitive \| Developmental \| Language \| Individual differences \| Personality \| Philosophy \| Social \| Methods \| Statistics \| Clinical \| Educational \| Industrial \| Professional items \| World psychology \| Statistics: Scientific method · Research methods · Experimental design · Undergraduate statistics courses · Statistical tests · Game theory · Decision theory In mathematics, a natural number is either a positive integer (1, 2, 3, 4, ...) or a non-negative integer (0, 1, 2, 3, 4, ...). The former definition is generally used in number theory, while the latter is preferred in set theory and computer science. Natural numbers have two main purposes: they can be used for counting ("there are 3 apples on the table"), and they can be used for ordering ("this is the 3rd largest city in the country"). Properties of the natural numbers related to divisibility, such as the distribution of prime numbers, are studied in number theory. Problems concerning counting, such as Ramsey theory, are studied in combinatorics. ## History of natural numbers and the status of zero The natural numbers presumably had their origins in the words used to count things, beginning with the number one. The first major advance in abstraction was the use of numerals to represent numbers. This allowed systems to be developed for recording large numbers. For example, the Babylonians developed a powerful place-value system based essentially on the numerals for 1 and 10. The ancient Egyptians had a system of numerals with distinct hieroglyphs for 1, 10, and all the powers of 10 up to one million. A stone carving from Karnak, dating from around 1500 BC and now at the Louvre in Paris, depicts 276 as 2 hundreds, 7 tens, and 6 ones; and similarly for the number 4,622. A much later advance in abstraction was the development of the idea of zero as a number with its own numeral. A zero digit had been used in place-value notation as early as 700 BC by the Babylonians, but it was never used as a final element.1 The Olmec and Maya civilization used zero as a separate number as early as 1st century BC, apparently developed independently, but this usage did not spread beyond Mesoamerica. The concept as used in modern times originated with the Indian mathematician Brahmagupta in 628 AD. Nevertheless, zero was used as a number by all medieval computists (calculators of Easter) beginning with Dionysius Exiguus in 525, but in general no Roman numeral was used to write it. Instead, the Latin word for "nothing," nullae, was employed. The first systematic study of numbers as abstractions (that is, as abstract entities) is usually credited to the Greek philosophers Pythagoras and Archimedes. However, independent studies also occurred at around the same time in India, China, and Mesoamerica. In the nineteenth century, a set-theoretical definition of natural numbers was developed. With this definition, it was more convenient to include zero (corresponding to the empty set) as a natural number. This convention is followed by set theorists, logicians, and computer scientists. Other mathematicians, primarily number theorists, often prefer to follow the older tradition and exclude zero from the natural numbers. The term whole number is used informally by some authors for an element of the set of integers, the set of non-negative integers, or the set of positive integers. ## Notation Mathematicians use N or $\mathbb{N}$ (an N in blackboard bold) to refer to the set of all natural numbers. This set is infinite but countable by definition. To be unambiguous about whether zero is included or not, sometimes an index "0" is added in the former case, and a superscript "" is added in the latter case: N0 = { 0, 1, 2, ... } ; N = { 1, 2, ... }. (Sometimes, an index or superscript "+" is added to signify "positive". However, this is often used for "nonnegative" in other cases, as R+ = [0,∞) and Z+ = { 0, 1, 2,... }, at least in European literature. The notation "*", however, is quite standard for nonzero or rather invertible elements.) Less frequently, W or $\mathbb{W}$ is used for the set of "whole numbers", which are sometimes identified with the natural numbers as defined here, sometimes with the integers. Set theorists often denote the set of all natural numbers by ω. When this notation is used, zero is explicitly included as a natural number. ## Formal definitions Historically, the precise mathematical definition of the natural numbers developed with some difficulty. The Peano postulates state conditions that any successful definition must satisfy. Certain constructions show that, given set theory, models of the Peano postulates must exist. ### Peano axioms • There is a natural number 0. • Every natural number a has a natural number successor, denoted by S(a). • There is no natural number whose successor is 0. • Distinct natural numbers have distinct successors: if a ≠ b, then S(a) ≠ S(b). • If a property is possessed by 0 and also by the successor of every natural number which possesses it, then it is possessed by all natural numbers. (This postulate ensures that the proof technique of mathematical induction is valid.) It should be noted that the "0" in the above definition need not correspond to what we normally consider to be the number zero. "0" simply means some object that when combined with an appropriate successor function, satisfies the Peano axioms. There are many systems that satisfy these axioms, including the natural numbers (either starting from zero or one). ### Constructions based on set theory #### The standard construction A standard construction in set theory is to define the natural numbers as follows: We set 0 := { } and define S(a) = a U {a} for all a. The set of natural numbers is then defined to be the intersection of all sets containing 0 which are closed under the successor function. Assuming the axiom of infinity, this definition can be shown to satisfy the Peano axioms. Each natural number is then equal to the set of natural numbers less than it, so that • 0 = { } • 1 = {0} = {{ }} • 2 = {0,1} = {0, {0}} = {{ }, {{ }}} • 3 = {0,1,2} = {0, {0}, {0, {0}}} = {{ }, {{ }}, {{ }, {{ }}}} and so on. When you see a natural number used as a set, this is typically what is meant. Under this definition, there are exactly n elements (in the naïve sense) in the set n and n ≤ m (in the naïve sense) iff n is a subset of m. Also, with this definition, different possible interpretations of notations like Rn (n-tuples vs. mappings of n into R) coincide. #### Other constructions Although the standard construction is useful, it is not the only possible construction. For example: one could define 0 = { } and S(a) = {a}, producing 0 = { } 1 = {0} = {{ }} 2 = {1} = {{{ }}}, etc. Or we could even define 0 = {{ }} and S(a) = a U {a} producing 0 = {{ }} 1 = {{ }, 0} = {{ }, {{ }}} 2 = {{ }, 0, 1}, etc. Arguably the oldest set-theoretic definition of the natural numbers is the definition commonly ascribed to Frege and Russell under which each concrete natural number n is defined as the set of all sets with n elements. This may appear circular, but can be made rigorous with care. Define 0 as $\{\{\}\}$ (clearly the set of all sets with 0 elements) and define $\sigma(A)$ (for any set A) as $\{x \cup \{y\} \mid x \in A \wedge y \not\in x\}$. Then 0 will be the set of all sets with 0 elements, $1=\sigma(0)$ will be the set of all sets with 1 element, $2=\sigma(1)$ will be the set of all sets with 2 elements, and so forth. The set of all natural numbers can be defined as the intersection of all sets containing 0 as an element and closed under $\sigma$ (that is, if the set contains an element n, it also contains $\sigma(n)$). This definition does not work in the usual systems of axiomatic set theory because the collections involved are too large (it will not work in any set theory with the axiom of separation); but it does work in New Foundations (and in related systems known to be consistent) and in some systems of type theory. ## Properties One can recursively define an addition on the natural numbers by setting a + 0 = a and a + S(b) = S(a) + b for all a, b. This turns the natural numbers (N, +) into a commutative monoid with identity element 0, the so-called free monoid with one generator. This monoid satisfies the cancellation property and can therefore be embedded in a group. The smallest group containing the natural numbers is the integers. If we define S(0) := 1, then S(b) = S(b + 0) = b + S(0) = b + 1; i.e. the successor of b is simply b + 1. Analogously, given that addition has been defined, a multiplication × can be defined via a × 0 = 0 and a × S(b) = (a × b) + a. This turns (N, ×) into a commutative monoid with identity element 1; a generator set for this monoid is the set of prime numbers. Addition and multiplication are compatible, which is expressed in the distribution law: a × (b + c) = (a × b) + (a × c). These properties of addition and multiplication make the natural numbers an instance of a commutative semiring. Semirings are an algebraic generalization of the natural numbers where multiplication is not necessarily commutative. If we interpret the natural numbers as "excluding 0", and "starting at 1", the definitions of + and × are as above, except that a + 1 = S(a) and a × 1 = a. For the remainder of the article, we write ab to indicate the product a × b, and we also assume the standard order of operations. Furthermore, one defines a total order on the natural numbers by writing a ≤ b if and only if there exists another natural number c with a + c = b. This order is compatible with the arithmetical operations in the following sense: if a, b and c are natural numbers and a ≤ b, then a + c ≤ b + c and ac ≤ bc. An important property of the natural numbers is that they are well-ordered: every non-empty set of natural numbers has a least element. While it is in general not possible to divide one natural number by another and get a natural number as result, the procedure of division with remainder is available as a substitute: for any two natural numbers a and b with b ≠ 0 we can find natural numbers q and r such that a = bq + r and r < b The number q is called the quotient and r is called the remainder of division of a by b. The numbers q and r are uniquely determined by a and b. This, the Division algorithm, is key to several other properties (divisibility), algorithms (such as the Euclidean algorithm), and ideas in number theory. ## Generalizations Two generalizations of natural numbers arise from the two uses: ordinal numbers are used to describe the position of an element in an ordered sequence and cardinal numbers are used to specify the size of a given set. For finite sequences or finite sets, both of these properties are embodied in the natural numbers. Other generalizations are discussed in the article on numbers. ## Footnote ¹ "... a tablet found at Kish ... thought to date from around 700 BC, uses three hooks to denote an empty place in the positional notation. Other tablets dated from around the same time use a single hook for an empty place." [1]af:Natuurlike getal ar:عدد طبيعي bg:Естествено число ca:Nombre natural cs:Přirozené číslo da:Naturligt tal de:Natürliche Zahl et:Naturaalarv es:Número natural eo:Natura nombro eu:Zenbaki arrunt fa:اعداد طبیعی fo:Teljital fr:Entier naturel ko:자연수 hr:Prirodan broj id:Bilangan asli ia:Numero natural is:Náttúrlegar tölurhe:מספר טבעי la:Numerus naturalis lt:Natūralieji skaičiai jbo:rarna'u hu:Természetes számok nl:Natuurlijk getalno:Naturlig tall nn:Naturleg talpt:Número natural ro:Număr natural ru:Натуральное число sq:Numrat natyral scn:Nùmmuru naturali simple:Natural number sk:Množina prirodzených čísel sl:Naravno število sr:Природан број sh:Prirodan broj fi:Luonnollinen luku sv:Naturliga tal ta:இயல்பெண் th:จำนวนธรรมชาติuk:Натуральне число zh:自然数 ## Read more • Assessment \| Biopsychology \| Comparative \| Cognitive \| Developmental \| Language \| Individual... Real number • Ordinal number • Function (mathematics) # Photos Add a Photo 6,465photos on this wiki • by Dr9855 2013-05-14T02:10:22Z • by PARANOiA 12 2013-05-11T19:25:04Z Posted in more... • by Addyrocker 2013-04-04T18:59:14Z • by Psymba 2013-03-24T20:27:47Z Posted in Mike Abrams • by Omaspiter 2013-03-14T09:55:55Z • by Omaspiter 2013-03-14T09:28:22Z • by Bigkellyna 2013-03-14T04:00:48Z Posted in User talk:Bigkellyna • by Preggo 2013-02-15T05:10:37Z • by Preggo 2013-02-15T05:10:17Z • by Preggo 2013-02-15T05:09:48Z • by Preggo 2013-02-15T05:09:35Z • See all photos See all photos >	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8916589617729187, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/41689?sort=votes	## Why symmetric spaces? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In Bar-Natan's "Knots at Lunch" seminar at the University of Toronto, we are currently discussing a talk by Alekseev at Montpellier about Rouvière's expansion of the Duflo isomorphism to the setting of symmetric spaces. We understand the definition of a symmetric space, and we know that people have written books about them; but we I don't understand in what sense symmetric spaces are useful and interesting mathematical objects. In particular: Are there significant (analytic? geometric? algebraic?) techniques which work for symmetric spaces, but not for more general classes of homogenous spaces? What I'm trying to understand (at least vaguely) is the role of symmetric spaces in the Kashiwara-Vergne picture, and the conceptual reason one might expect Duflo's isomorphism to generalize to this specific class of mathematical objects. There must be a conceptual explanation why symmetric spaces are the natural class of objects to consider in such contexts. A closely related question is THIS. - While hardly an answer, I'd like to advertise the following connection. A simple kind of symmetric space is G/H where H is the fixed-point set of an inner involution, and even simpler, the centralizer Z(t) of an involution. Brauer (I believe) proved that there are finitely many finite simple G with fixed isomorphism classes of Z(t), and proposed that finite simple groups be classified using them. Eventually, they were! – Allen Knutson Oct 21 2010 at 2:56 ## 3 Answers The algebra of invariant differential operators on a symmetric space is commutative, and this is certainly not true for an arbitrary homogeneous space. While it is not true that the commutativity of the algebra $D^{G}$ of $G$ invariant differential operators on a homogeneous space $X = G/H$ implies that $X$ is symmetric, if $G$ is reductive it is true that $D^{G}$ is commutative if and only if $G/H$ is weakly symmetric in the sense defined by Selberg (see E.B. Vinberg's survey in Russ. Math. Surveys 56(1)). - ### 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. If one accepts curvature as a measure of complexity of a Riemannian manifold (which one might or might not agree with), then the "simplest" Riemannian manifolds are those of constant curvature. Unfortunately, there are not so many of these; besides Euclidean space the only simply-connected examples are the spheres (constant positive curvature) and the hyperbolic spaces (constant negative curvature). (Of course there is a rich theory of non-simply-connected spaces of constant negative curvature, but never mind.) So, how can one weaken the notion of constant curvature to obtain a larger class of interesting, but not "too complicated" spaces? Well, it seems natural to ask that the covariant derivative of the curvature should be 0. In this case, one has many more 1-connected examples, and these are precisely symmetric spaces. They are nice in various senses: • They can be classified, so there are not too many of them (but still sufficiently many to be "interesting"). • They admit a description in terms of Lie groups. This allows for very explicit computations, e.g. of curvature and characteristic classes. • They admit a natural duality (compact/non-neg. curved vs. non-compact/non-pos. curved) if one ignores flat factors. This allows for the transfer of ideas between two different worlds (see e.g. Hirzebruch proportionality). What more can one ask for? On the other hand, one should stress that they are really rare and special objects, just slightly less rare than manifolds of constant curvature. - Would symmetric spaces be the widest class of homogeneous spaces for which explicit computations of curvature and characteristic classes could be made, and for which Hirzebruch proportionality holds? (the answer to the second question is surely no: kurims.kyoto-u.ac.jp/~toshi/texpdf/hirz-the.pdf) – Daniel Moskovich Oct 10 2010 at 17:18 Good news for anyone interested in the subject is the fact that Wolff's classic text,SPACES OF CONSTANT CURVATURE,which is the most in-depth source on this important area of geometry,will be reissued by the AMS early next year,after over a decade in limbo where copies of the 5th edition were going for over 300 dollars a piece online. That's good news indeed. – Andrew L Oct 10 2010 at 22:25 I'm interested in the reference to duality between non-pos and non-neg curved spaces. Could you point me towards some references for this ? – Suresh Venkat Oct 11 2010 at 2:40 Symmetric spaces are certainly not the widest classes of homogeneous spces for which explicit computations of curvature and characteristic classes can be made, but they are the ones for which one gets arguably the nicest and easiest formulas. As an example, consider the rational cohomology: For general homogeneous spaces U/K this depends on the projection p: U -> U/K and the classifying map f: U/K ->BK in a rather complicated way. In the symmetric case, everything splits nicely and one gets H^(U/K; Q) = p^H(U/K; Q) \otimes f^H(BK; Q), i.e. the contribution from characteristic classes splits – Tobias Hartnick Oct 11 2010 at 10:32 Concerning Suresh's question, one should emphasize that to the best of my knowledge there is no such thing as a general duality between non-pos and non-neg curved spaces. This is something very special to symmetric spaces (and some of there generalizations). In this context, instances of duality can be found in every book on symmetric spaces (e.g. Helgason). As far as duality of char. classes is concerned, the above cite article of Kobayashi-Ono is state of the art, as far as I know. – Tobias Hartnick Oct 11 2010 at 10:34 Here is one example. When $G$ is a compact Lie group and $H$ is a Lie subgroup, the real cohomology of the homogeneous space $G/H$ is the same as the relative cohomology $H^(g,h,\mathbf{R})$ where $g$ and $h$ are the Lie algebras of $G$, respectively $H$. This can be proven by averaging, just as in the case when $H$ is trivial. In general the differential in the relative cochain complex is not zero, but when $G/H$ is symmetric, it is, for a simple reason: the symmetric involution acts as $(-1)^d$ on the degree $d$ part of the complex; since this action should commute with the differential, the differential must be 0; for more details see Felix, Halperin, Thomas, Rational homotopy theory, p. 162. So symmetric spaces are formal. However, in general compact homogeneous spaces need not be formal; e.g. $SU(n)/Sp(n)$ is not formal for $n\geq 5$, see Greub, Halperin, Vanstone, Curvature, connections and cohomology. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 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": 0.9188638925552368, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/3618/solving-an-equation-with-irrational-exponents	# Solving an equation with irrational exponents Is there any theory (analogous to Galois theory) for solving equations with irrational exponents like: $x^{\sqrt{2}}+x^{\sqrt{3}}=1$ ? - Well, for one thing, this is now a transcendental equation instead of an algebraic one, and the theory of seeing whether a transcendental equation has (unique) solutions is a lot less developed than in the algebraic case (i.e., one usually proceeds on a case-to-case basis for determining if a transcendental equation has solutions). – J. M. Aug 30 '10 at 0:54 In my physics classes, we would just graph these things and read off the answer. I don't know if there is a better way, but I think this is an interesting question. There must be some transcendental equations with an analytic solution, but I suspect the situation could be like Diophantine equations, where there is just no hope for a general method. – Matt Calhoun Aug 30 '10 at 13:25 – Matt Calhoun Aug 30 '10 at 13:48 The key point is "some"; the difficulty is in finding general analytic solutions for transcendental functions. – J. M. Aug 30 '10 at 13:48 Matt: the thing to understand with Galois theory is that one has to keep adding "tools" to your repertoire to (analytically) solve polynomial equations of increasing degree. For linears, you only need division; for quadratics to quartics, you need radicals; for quintics you need to use hypergeometric or theta functions, and even higher degree polynomial equation need even more complicated functions. Newton is an approximation method; it does not give analytic solutions. – J. M. Aug 30 '10 at 13:52 show 1 more comment ## 1 Answer The study of such equations is not "abstract algebra" as it is usually understood. The reason is that to even define the function $x^{\sqrt{2}}$, for example, requires analysis; one has to prove certain properties of $\mathbb{R}$ to ensure that such a function exists. This is in marked contrast to the case of integer or rational powers, where one has a purely algebraic definition and the background theory is equational. To define the function $x^{\sqrt{2}}$ one has to either define $e^x$ and the logarithm or consider a limit of functions $x^{p_n}$ where $p_n$ form a sequence of rational approximations to $\sqrt{2}$, and this is irreducibly non-algebraic stuff. In particular, while polynomials can be studied in an absurdly general setting, transcendental equations like those you describe are more or less restricted to $\mathbb{R}$ (or $\mathbb{C}$ if you really want to pick a branch of the logarithm). The LHS is an increasing function of $x$, so there is at most one root, which probably one can really only compute numerically if it exists. (Its nonexistence can be ruled out by computing local minima in $(0, 1)$.) This is another question which touches on a theme which has come up several times on math.SE, which is that exponentiation should really not be thought of as one operation. Instead, it is a collection of related operations with various degrees of generality and applicability which happen to share the same algebraic properties, and one should not infer too much about how similar these operations are. - 6 One should be careful about making strong broad statements such as so-and-so "cannot be called abstract algebra". A large part of elementary calculus can be algebraicized. That's how computer algebra systems work. One has differential algebra and Galois theory for computing integrals and solving ODEs, Hardy-Rosenlicht fields and transseries for computing limits and asymptotics, etc. – Gone Aug 30 '10 at 1:30 Fair point; I'll modify the wording. – Qiaochu Yuan Aug 30 '10 at 1:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.939218282699585, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/229652/existence-of-a-sequence-which-is-good-for-mean-convergence-but-not-good-for-poin?answertab=oldest	# Existence of a sequence which is good for mean convergence but not good for pointwise convergence The ergodic theorems talk about the limit behaviour of the ergodic averages. For example, if $(X,\chi,\mu,T)$ is a measure preserving system, where $\mu$ is a probability measure on $X$, then we have the mean ergodic theorem: $$\text{If f}\in L^1(X),\text{ then we have that} \frac{1}{n}\sum_{i=1}^nf(T^ix) \text{ converges in } L^1.$$ and the pointwise ergodic theorem: $$\text{If f}\in L^1(X),\text{ then we have that } \frac{1}{n}\sum_{i=1}^nf(T^ix) \text{ converges almost everywhere.}$$ Now we can consider a more general ergodic average of the form: $$\frac{1}{n}\sum_{i=1}^nf(T^{a(i)}x)$$ where $\{a(i)\}$ is a sequence of natural numbers with $a(1)<a(2)<\ldots$. If the averages converges in $L^1$, then we say that $\{a(i)\}$ is good for mean convergence, and similarly for pointwise convergence. In general not every sequence is good for the two kinds of convergence. Now my question is that: does there exist a sequence which is good for mean convergence but not good for pointwise convergence? I think such one exists but can not give an example. - I do not understand your question. Are you asking for a counterexample to a theorem which is known to be true? – Ray Yang Feb 7 at 12:21	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9397369623184204, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/114989-help-trig-limit.html	# Thread: 1. ## Help with trig limit Can anyone show me how the limit of $\frac{sin 5x}{x}$ as x goes to 0 equals 5? I'm only supposed to use the fact that $\frac{sin x}{x}$ goes to 1 as x goes to 0. Any tips? 2. Originally Posted by paupsers Can anyone show me how the limit of $\frac{sin 5x}{x}$ as x goes to 0 equals 5? I'm only supposed to use the fact that $\frac{sin x}{x}$ goes to 1 as x goes to 0. Any tips? If we let $z=5x$ then our limit becomes $\lim_{z\to0}\frac{\sin(z)}{\tfrac{z}{5}}$........so	{"extraction_info": {"found_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": 0.9503770470619202, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2011/09/20/pseudo-riemannian-metrics/?like=1&source=post_flair&_wpnonce=a832ba55c5	# The Unapologetic Mathematician ## (Pseudo-)Riemannian Metrics Ironically, in order to tie what we’ve been doing back to more familiar material, we actually have to introduce more structure. It’s sort of astonishing in retrospect how much structure comes along with the most basic, intuitive cases, or how much we can do before even using that structure. In particular, we need to introduce something called a “Riemannian metric”, which will move us into the realm of differential geometry instead of just topology. Everything up until this point has been concerned with manifolds as “shapes”, but we haven’t really had any sense of “size” or “angle” or anything else we could measure. Having these notions — and asking that they be preserved — is the difference between geometry and topology. Anyway, a Riemannian metric on a manifold $M$ is nothing more than a certain kind of tensor field $g$ of type $(0,2)$ on $M$. At each point $p\in M$, the field $g$ gives us a tensor: $\displaystyle g_p\in\mathcal{T}_p^M\otimes\mathcal{T}_p^M\cong\left(\mathcal{T}_pM\otimes\mathcal{T}_pM\right)^*$ We can interpret this as a bilinear function which takes in two vectors $v_p,w_p\in\mathcal{T}_pM$ and spits out a number $g_p(v_p,w_p)$. That is, $g_p$ is a bilinear form on the space $\mathcal{T}_pM$ of tangent vectors at $p$. So, what makes $g$ into a Riemannian metric? We now add the assumption that $g_p$ is not just a bilinear form, but that it’s an inner product. That is, $g_p$ is symmetric, nondegenerate, and positive-definite. We can let the last condition slip a bit, in which case we call $g$ a “pseudo-Riemannian metric”. When equipped with a metric, we call $M$ a “(pseudo-)Riemannian manifold”. It’s common to also say “Riemannian” in the case of negative-definite metrics, since there’s little difference between the cases of signature $(n,0)$ and $(0,n)$. Another common special case is that of a “Lorentzian” metric, which is signature $(n-1,1)$ or $(1,n-1)$. As we might expect, $g$ is called a metric because it lets us measure things. Specifically, since $g_p$ is an inner product it gives us notions of the length and angle for tangent vectors at $p$. We must be careful here; we do not yet have a way of measuring distances between points on the manifold $M$ itself. The metric only tells us about the lengths of tangent vectors; it is not a metric in the sense of metric spaces. However, if two curves cross at a point $p$ we can use their tangent vectors to define the angle between the curves, so that’s something. ## 19 Comments » 1. as we move from patch to patch for the manifold, does g (tensor field) need any restrictions? does the form change value depending on which piece of the atlas we’re using? thanks for this blog series! Comment by scot \| September 21, 2011 \| Reply 2. That’s the neatest thing: the metric $g$ is defined as a geometric object — a tensor field — so it doesn’t depend on the local coordinate patches at all! All that the patches matter is when you want to represent the inner products with matrices with respect to some basis of the (co)tangent vector space. Comment by \| September 21, 2011 \| Reply 3. [...] now that we’ve introduced the idea of a metric on a manifold, it’s natural to talk about mappings that preserve them. We call such maps [...] Pingback by \| September 27, 2011 \| Reply 4. [...] that we can define the inner product of two vectors using a metric , we want to generalize this to apply to vector [...] Pingback by \| September 30, 2011 \| Reply 5. [...] next step after using a metric to define an inner product on the module of vector spaces over the ring of smooth functions is to [...] Pingback by \| October 1, 2011 \| Reply 6. [...] say that is an orientable Riemannian manifold. We know that this lets us define a (non-degenerate) inner product on differential forms, [...] Pingback by \| October 6, 2011 \| Reply 7. [...] Armstrong: (Pseudo)-Riemannian Metrics, Isometries, Inner Products on 1-Forms, The Hodge Star in Coordinates, The Hodge Star on [...] Pingback by \| October 8, 2011 \| Reply 8. [...] want to start getting into a nice, simple, concrete example of the Hodge star. We need an oriented, Riemannian manifold to work with, and for this example we take , which we cover with the usual coordinate [...] Pingback by \| October 11, 2011 \| Reply 9. [...] continue our example considering the special case of as an oriented, Riemannian manifold, with the coordinate -forms forming an oriented, orthonormal basis at each [...] Pingback by \| October 12, 2011 \| Reply 10. [...] some examples will quickly shed some light on this. We can even extend to the pseudo-Riemannian case and pick a coordinate system so that , where . That is, any two are orthogonal, and each either [...] Pingback by \| October 18, 2011 \| Reply 11. [...] why do we care about this particularly? In the presence of a metric, we have an equivalence between -forms and vector fields . And specifically we know that the [...] Pingback by \| October 21, 2011 \| Reply 12. how can it help to our daily life ? Comment by \| October 24, 2011 \| Reply 13. [...] does this look like when we have a metric and we can rewrite the -form as a vector field ? In this case, is exact if and only if is [...] Pingback by \| October 24, 2011 \| Reply 14. [...] we want another way of viewing this orientation. Given a metric on we can use the inverse of the Hodge star from on the orientation -form of , which gives us a [...] Pingback by \| October 27, 2011 \| Reply 15. [...] a moment, let’s return to the case of Riemannian manifolds; the vector field analogue of an exact -form is called a “conservative” [...] Pingback by \| December 15, 2011 \| Reply 16. [...] -form, not a vector field, but remember that we’re working in our standard with the standard metric, which lets us use the Hodge star to flip a -form into a -form, and a -form into a vector field! [...] Pingback by \| January 11, 2012 \| Reply 17. И здесь Comment by charlesse \| January 16, 2012 \| Reply 18. [...] which, though familiar to many, are really heavy-duty equipment. In particular, they rely on the Riemannian structure on . We want to strip this away to find something that works without this assumption, and [...] Pingback by \| February 22, 2012 \| Reply 19. [...] in hand, we need to properly define the Hodge star in our four-dimensional space, and we need a pseudo-Riemannian metric to do this. Before we were just using the standard , but now that we’re lumping in time we [...] Pingback by \| March 7, 2012 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 27, "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": 0.937698245048523, "perplexity_flag": "head"}

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