qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
102,721 | <p>This is probably a very simple question, but I couldn't find a duplicate.</p>
<p>As everybody knows, <code>{x, y} + v</code> gives <code>{x + v, y + v}</code>. But if I intend <code>v</code> to represent a vector, for example if I am going to substitute <code>v -> {vx, vy}</code> in the future, then the result <code>{x + v, y + v}</code> is meaningless.</p>
<p>How I can indicate to Mathematica that <code>v</code> is not a scalar and functions like <code>Plus</code> should not treat it as one? I tried setting <code>$Assumptions = v ∈ Vectors[2]</code> but that didn't help.</p>
| Community | -1 | <p>If there's a specific addition operation one wants to block, one can just apply a custom function instead, for example</p>
<pre><code>{x, y} ~plus~ v
(* plus[{x, y}, v] *)
</code></pre>
<p>and replace it with <code>Plus</code> when one is ready:</p>
<pre><code>{x, y} ~plus~ v /. v -> {vx, vy} /. plus -> Plus
(* {vx + x, vy + y} *)
</code></pre>
|
2,072,059 | <blockquote>
<p>Let $x$ be an element of the group $G$. Prove that $x^2 = 1$ <em>iff</em> $|x|$ is either $1$ or $2$</p>
</blockquote>
<p>Now I know that $G$ is implicitly a group under multiplication, therefore $G = (A, \ \cdot \ )$, for some underlying set $A$.</p>
<p>The problem I'm facing is that $A$ could be any set. So now if we take $A = \mathbb{R}$, then $x^2 = 1 \implies |x| = 1$, but it certainly does not imply that $|x| = 2$. So how is this problem even well defined?</p>
<p>This problem is taken from <em>Chapter 1: Group Theory - Exercise 16 in 'Abstarct Algebra' by Dummit and Foote</em></p>
<p>Is there a mistake in this problem? If not how would one proceed to prove that $x^2 = 1$ <em>iff</em> $|x|$ is either $1$ or $2$, despite the counterexample I gave above?</p>
| Eric Wofsey | 86,856 | <p>In the context of this problem, $|x|$ presumably means the order of $x$ as an element of $G$, not the absolute value of $x$. That is, $|x|$ is the least positive integer $n$ such that $x^n=1$ (where $x^n$ means $x$ multiplied with itself $n$ times using the group operation of $G$, and $1$ means the identity element of $G$).</p>
|
951 | <p>I think that complex analysis is hard because graphs of even basic functions are 4 dimensional. Does anyone have any good visual representations of basic complex functions or know of any tools for generating them?</p>
| Kevin H. Lin | 263 | <p>For Moebius transformations, check out <a href="http://www.youtube.com/watch?v=JX3VmDgiFnY">this nice YouTube video</a>.</p>
|
951 | <p>I think that complex analysis is hard because graphs of even basic functions are 4 dimensional. Does anyone have any good visual representations of basic complex functions or know of any tools for generating them?</p>
| Isaac | 72 | <p>The graphs in the middle of <a href="http://mathworld.wolfram.com/ConformalMapping.html" rel="nofollow">the MathWorld page on Conformal Mapping</a> show examples of the first method in Michael Lugo's answer as well as something somewhat similar to the second method in that answer.</p>
|
3,443,082 | <p>Find all the positive integral solutions of, <span class="math-container">$\tan^{-1}x+\cos^{-1}\dfrac{y}{\sqrt{y^2+1}}=\sin^{-1}\dfrac{3}{\sqrt{10}}$</span></p>
<p>Assuming <span class="math-container">$x\ge1,y\ge1$</span> as we have to find positive integral solutions of <span class="math-container">$(x,y)$</span></p>
<p><span class="math-container">$$\tan^{-1}x=\tan^{-1}3-\tan^{-1}\dfrac{1}{y}$$</span></p>
<p>As <span class="math-container">$3>0$</span> and <span class="math-container">$\dfrac{1}{y}>0$</span>
<span class="math-container">$$\tan^{-1}x=\tan^{-1}\left(\dfrac{3-\dfrac{1}{y}}{1+\dfrac{3}{y}}\right)$$</span>
<span class="math-container">$$\tan^{-1}x=\tan^{-1}\dfrac{3y-1}{y+3}$$</span></p>
<p><span class="math-container">$$x=\dfrac{3y-1}{y+3}$$</span></p>
<p><span class="math-container">$y+3\in[4,\infty)$</span> as <span class="math-container">$y\ge1$</span>, <span class="math-container">$3y-1\in [2,\infty)$</span> as <span class="math-container">$y\ge1$</span></p>
<p>For <span class="math-container">$x$</span> to be positive integer, <span class="math-container">$3y-1$</span> should be multiple of <span class="math-container">$y+3$</span></p>
<p><span class="math-container">$$3y-1=m(y+3) \text { where } m\in Z^{+}$$</span>
<span class="math-container">$$3y-my=3m-1$$</span>
<span class="math-container">$$(3-m)y=3m-1$$</span></p>
<p>Here R.H.S is positive, so L.H.S should also be positive.</p>
<p>So <span class="math-container">$3-m>0$</span>, hence <span class="math-container">$m<3$</span></p>
<p>So possible values of <span class="math-container">$m$</span> are {<span class="math-container">$1$</span>,<span class="math-container">$2$</span>}.</p>
<p>For <span class="math-container">$m=1$</span>, <span class="math-container">$$3y-1=y+3$$</span>
<span class="math-container">$$2y=4$$</span>
<span class="math-container">$$y=2$$</span></p>
<p><span class="math-container">$$x=\dfrac{3\cdot2-1}{2+3}$$</span>
<span class="math-container">$$x=1$$</span></p>
<p>For <span class="math-container">$m=2$</span>, <span class="math-container">$$3y-1=2(y+3)$$</span>
<span class="math-container">$$3y-1=2y+6$$</span>
<span class="math-container">$$y=7$$</span></p>
<p><span class="math-container">$$x=\dfrac{3\cdot7-1}{7+3}$$</span>
<span class="math-container">$$x=\dfrac{20}{10}$$</span>
<span class="math-container">$$x=2$$</span></p>
<p>Is there some other nicer way to solve this problem.</p>
| Certainly not a dog | 691,550 | <p>Your method seems decent. Here I propose another way to ensure the integer-ness of <span class="math-container">$x$</span> and <span class="math-container">$y$</span>.</p>
<p>Note from a rough sketch and limits that the range of <span class="math-container">$f:\mathbb{R^+} \mapsto X,~f(y)= {3y-1\over y+3}$</span> is <span class="math-container">$\left[-\frac 13, 3\right)$</span>. The only positive integers that lie in this range are <span class="math-container">${1,2}$</span>. You may use the inverse process <span class="math-container">$y={3x+1\over 3-x}$</span> and,use <span class="math-container">$x\in \{1,2\}$</span> to see for which of these <span class="math-container">$y$</span>, <span class="math-container">$x$</span> is also a positive integer. </p>
<p>For <span class="math-container">$x=1$</span>, <span class="math-container">$y=\frac 42 = 2$</span>, which is a positive integer. So one solution is <span class="math-container">$(x,y)\equiv (1,2)$</span>. For <span class="math-container">$x=2$</span>, <span class="math-container">$y=\frac 71 = 7$</span>, also a positive integer. So another solution is <span class="math-container">$(2,7)$</span>. We have exhausted all possibilities for <span class="math-container">$y$</span>. Our solution set is <span class="math-container">$\left\{(1,2),(2,7)\right\}$</span>. </p>
|
209,869 | <p>I am very interested in the maximum number of triangles could a connected graph with $n$ vertices and $m$ edges have. For example, if $m\leq n−1$, this number is $0$, if $m=n$, this number is $1$, if $m=n+1$, this number is $2$, and if $m=n+2$, this number is $4$. </p>
| Igor Rivin | 11,142 | <p>This question (together with massive generalizations) is answered in <a href="http://arxiv.org/abs/math/0111106" rel="nofollow">I. Rivin's 2001 paper.</a></p>
|
3,700,938 | <p>I know that the equations are equivalent by doing the math with the same value for x, but I don't understand the rules for changing orders or operations.<br>
When it is not the first addition or subtraction happening in the equation, parentheses make the addition subtraction and vice versa? Are there any other rules?</p>
<p><span class="math-container">$x^2+x-x-1 = (x^2+x)-(x+1)$</span>? </p>
<p>What if you put the parentheses around the two <span class="math-container">$x$</span>s in the middle?<br>
<span class="math-container">$x^2+(x-x)+1$</span>? Should that be (x+x) in the middle?</p>
| J. W. Tanner | 615,567 | <p>Think of subtracting <span class="math-container">$(x+1)$</span> as adding <span class="math-container">$-1\cdot(x+1)$</span>. </p>
<p>Then apply the distributive property, so it's adding <span class="math-container">$-1\cdot x + -1\cdot 1=-x-1$</span>.</p>
<p>When you add <span class="math-container">$(x-x)$</span>, it should be <span class="math-container">$(x-x)$</span>, not <span class="math-container">$x+x$</span>, because there's no <span class="math-container">$-1$</span> to distribute.</p>
|
243,336 | <p>This question was asked to me in an interview, I still cannot think of its solution. Can anyone help? Following is the question:</p>
<blockquote>
<p>Given an infinite number of ropes of length $R$, you have to calculate the
minimum number of cuts required to make $N$ ropes of length $C$. You can
append ropes one after the other to make ropes of length $2R, 3R, 4R, \ldots$</p>
</blockquote>
| Hagen von Eitzen | 39,174 | <p>Wlog. $R=1$.
If $C$ is an integer, no cuts are required.
Otherwise let $c=C-\lfloor C\rfloor$, a real number between $0$ and $1$.
Each rope must finally contain at least one cut end, thus the number of cuts is at least $\frac N2$ and it is easily solvable with $N$ cuts.</p>
<p>Indeed, $\lceil \frac N2 \rceil$ is enough if $c=\frac12$.</p>
<p>If $c=\frac23$, one can produce $\lceil \frac23 N\rceil$ pieces of $\frac23$ and combine the $\frac13$ rests for the remaining ropes, hence $\lceil \frac23 N\rceil$ cuts suffice.</p>
<p>If $c=\frac13$, then $\lceil \frac23 N\rceil$ cuts suffice again.</p>
<p>For $c=\frac34$ or $c=\frac14$, I can do it with $\lceil \frac34 N\rceil$ cuts, for $\frac k 5$ with $1\le k\le 4$, I can do it with $\lceil \frac45N\rceil$ cuts.</p>
<p>A pattern sems to emerge here, but I'm not sure if it is really optimal: $c=\frac pq$ requires $\lceil (1-\frac1q)N\rceil$ cuts. And if $c$is irrational, $N$ cuts are needed, I suppose.</p>
<p><em>Remark:</em> If the final result is correct, there is no need to go for $c$, one can directly say that $\lceil (1-\frac1q)N\rceil$ cuts are needed if $\frac CR=\frac pq$.</p>
<hr>
<p><strong>Edit:</strong> <em>(Revisited this answer after a few years in order to add proof)</em></p>
<p>Allowed solutions consist of an algorithm that describes a sequence of cuts and (at no cost) concatenation operations. I tis clear that any such solution can be brought into this standard form:</p>
<ol>
<li>Cut some of the unit length input ropes into pieces.</li>
<li>Combine the pieces into the $N$ output ropes (plus possibly some waste)</li>
</ol>
<p>It doesn't harm to allow the following more general form:</p>
<ol>
<li>Combine unit length input ropes into some integer length intermediate ropes.</li>
<li>Cut the intermediate ropes into pieces</li>
<li>Combine the pieces into the $N$ output ropes and combine the waste pieces (if there are any) into a "waste rope"</li>
</ol>
<p>Our task is to find an algorithm that minimizes the cuts performed in step 2. To do so we consider <em>all</em> algorithms that achieve the minimal number of cuts, and among these consider the one that minimizes the number of intermediate ropes produced in step 1.</p>
<p>If two of the pieces an intermediate rope is cut into end up in the same output rope, we can rearrange the locations of the pieces on the uncut rope in such a manner that they are adjacent, which allows us to save a cut. By the minimality of our algorithm, this does not happen. That is: Each intermediate rope contains at most one contiguous piece from each output rope; note that this applies also to the waste rope.</p>
<p>Assume two distinct intermediate ropes each contain a piece of the same output rope, say one intermediate rope is cut into pieces of lengths $a_0, a_1,\ldots, a_r$ by $r$ cuts and the other is cut into pieces of lengths $b_0,b_1,\ldots,b_s$ by $s$ cuts and that the pieces of length $a_0,b_0$ both end up in the same output rope. Then we can instead combine these two intermediate ropes into one and cut it into $a_0+b_10, a_1,\ldots, a_r, b_1,\ldots,b_s$ with the same number $r+s$ of cuts, but with less intermediate ropes. By our choice of algorithm, this is impossible. Therefore: Each output rope is contained in exactly one intermediate rope; again, this also applies to the waste rope. In particular, there is at most one intermediate rope containing waste. We may additionally assume (by rearranging parts) that the waste part occurs at the end in its intermediate rope.</p>
<p>A wasteless intermediate rope as a positive length $\in \Bbb Z\cap C\Bbb Z$.
If $C$ is <strong>irrational</strong>, this is not possible. Hence in this case there is only one intermediate rope and it requires
$$N $$
cuts at positions $C, 2C, \ldots, NC$.</p>
<p>If $C=\frac pq$ with $\gcd(p,q)=1$ is <strong>rational</strong>, the length $\ell$ of a wasteless intermediate rope must be $\in \Bbb Z\cap C\Bbb Z=p\Bbb Z$ and it is cut at positions $C, 2C, 3C,\ldots$. If $\ell>p$, we could save a cut by replacing it with intermediate ropes of lengths $p$ and $\ell-p$. By minimality, this is not possible, hence each wasteless rope has length $p$
and needs $q-1$ cuts to produce $q$ output ropes of length $\frac pq$.
By the same argument, the number of output ropes in the wasteful intermediate rope must be $<q$ and requires just as many cuts.
We thus have described an algorithm that for $N=qM+R$, $ 0\le R<q$, requires
$$(q-1)M+R =N-\left\lfloor\frac Nq\right\rfloor $$
cuts and is optimal.</p>
|
1,750,592 | <p>Can anyone help with this? I got a wrong answer. </p>
<p>Problem: Joe's French poodle, FooFoo, is tied to the corner of the barn which measures 40 x 30. FooFoo's rope is 50 long. In terms of π, over how many square feet can FooFoo wander?</p>
<p><a href="https://i.stack.imgur.com/Nmrc4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Nmrc4.jpg" alt="enter image description here"></a></p>
| Ross Millikan | 1,827 | <p>Hint: If there weren't a building in the way, it would be a circle of radius $50$. If FooFoo is left of the right wall of the building and above the bottom wall, the rope will be flat against the right wall, reducing its effective length. You should get an area that is parts of three circles. If you show your reasoning, we can find the error, but without that, we can't</p>
|
729,054 | <p>Let $f$ be continuous and $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$.</p>
<p>Suppose $|f(x)-x| \leq 2$ holds for all $x$. Is $f$ surjective?</p>
| Marc van Leeuwen | 18,880 | <p>You can do this by Brouwer's fixed-point theorem. Let $a\in\Bbb R^2$ be arbitrary, and define $g:\Bbb R^2\to\Bbb R^2$ by $g(x)=x-f(x)+a$. By hypothesis, $g$ is continuous and has image contained in the disk of radius$~2$ and center$~a$. Restricting to that disk, $g$ has a fixed point $p$ in it by mentioned theorem. But $g(p)=p$ means $f(p)=a$, and since $a$ was arbitrary, $f$ is surjective.</p>
|
207,266 | <p>I am looking for the integrability condition of the following system of pde:</p>
<p>$$\partial_{[\nu}\Gamma^\kappa_{\mu]\lambda}+\Gamma^\kappa_{[\nu|\rho|}\Gamma^\rho_{\mu]\lambda}=\frac{1}{2}R_{\mu\nu\lambda}{}^{\kappa},\,\,\,\,\,\,\,\,\,(1)$$</p>
<p>given sufficiently smooth functions $R_{\mu\nu\lambda}{}^{\kappa}$ on some smooth manifold $M^n$, with $R_{(\mu\nu)\lambda}{}^{\kappa}=0$. This equation is the definition of the Riemann-Christoffel curvature of a <em>given</em> connection $\Gamma^\lambda_{\mu\nu}$. But I am interested in the reverse questions: </p>
<ol>
<li><p><em>Given functions $R_{\mu\nu\lambda}{}^{\kappa}$, with $R_{(\mu\nu)\lambda}{}^{\kappa}=0$, what are the conditions such that they form the compoments of the Riemann-Christoffel curvature of some connection?</em></p></li>
<li><p><em>If the given functions $R_{\mu\nu\lambda}{}^{\kappa}$ do satisfy those integrability conditions, then what will be general form of the connection coefficients $\Gamma^\lambda_{\mu\nu}$?</em> (When $R_{\mu\nu\lambda}{}^{\kappa}=0$, it is well-known that there exists an invertible matrix field $[A_{\mu\nu}]$ (with inverse $[A^{\mu\nu}]$) such that $\Gamma^\gamma_{\mu\nu}=A^{\gamma\lambda}\partial_{\nu} A_{\lambda\mu}$.)</p></li>
</ol>
<p>In Schouten (Chapter III, equation 5.19), it is stated that the integrability condition of (1) are given by the Bianchi identities </p>
<p>$$\nabla_{[\omega}R_{\mu\nu]\lambda}{}^\kappa=2S_{[\omega\nu}{}^\sigma\, R_{\mu]\sigma\lambda}{}^\kappa,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$$
where $S_{\mu\nu}{}^\kappa$ is the torsion of the connection $\Gamma^\kappa_{\mu\nu}$.</p>
<p>It is clear that (2) involves covarient derivative wirh respect to the connection $\Gamma$ and also the torsion $S$ of $\Gamma$, but to start with no such $\Gamma$ is given. We are <em>looking for</em> such a $\Gamma$. I fail to understand in what sense, (2) is an integrability condition of (1). I'm also not sure, whether to determine a $\Gamma$ satisfying (1), some (skew-symmetric) functions $S_{\mu\nu}{}^\lambda$ should also be given <em>a priori</em> so that the desired connection has also these specified fuctions as its torsion.</p>
<p><strong>Notes:</strong> 1. No metric structure is given on the manifold $M^n$.</p>
<ol start="2">
<li>It would be great if it can be answered at least for $n=3$.</li>
</ol>
<p>Thanks in advance.</p>
| Zurab Silagadze | 32,389 | <p>It seems a partial answer (the solution is found with some additional restrictions)
is given in the paper <a href="http://link.springer.com/article/10.1007%2FBF00759087" rel="nofollow">http://link.springer.com/article/10.1007%2FBF00759087</a> (Determination of the metric from the curvature, by Hernando Quevedo)</p>
<p>P.S. Of course Deane Yang's answer/comment above is more relevant for your question. If you not already found it, see also <a href="http://www.sciencedirect.com/science/article/pii/S0021782409001317" rel="nofollow">http://www.sciencedirect.com/science/article/pii/S0021782409001317</a> (The prescribed curvature problem in dimension four, by J.M. Masqué, L.M. Pozo Coronado and I. Sánchez Rodríguez).</p>
|
207,266 | <p>I am looking for the integrability condition of the following system of pde:</p>
<p>$$\partial_{[\nu}\Gamma^\kappa_{\mu]\lambda}+\Gamma^\kappa_{[\nu|\rho|}\Gamma^\rho_{\mu]\lambda}=\frac{1}{2}R_{\mu\nu\lambda}{}^{\kappa},\,\,\,\,\,\,\,\,\,(1)$$</p>
<p>given sufficiently smooth functions $R_{\mu\nu\lambda}{}^{\kappa}$ on some smooth manifold $M^n$, with $R_{(\mu\nu)\lambda}{}^{\kappa}=0$. This equation is the definition of the Riemann-Christoffel curvature of a <em>given</em> connection $\Gamma^\lambda_{\mu\nu}$. But I am interested in the reverse questions: </p>
<ol>
<li><p><em>Given functions $R_{\mu\nu\lambda}{}^{\kappa}$, with $R_{(\mu\nu)\lambda}{}^{\kappa}=0$, what are the conditions such that they form the compoments of the Riemann-Christoffel curvature of some connection?</em></p></li>
<li><p><em>If the given functions $R_{\mu\nu\lambda}{}^{\kappa}$ do satisfy those integrability conditions, then what will be general form of the connection coefficients $\Gamma^\lambda_{\mu\nu}$?</em> (When $R_{\mu\nu\lambda}{}^{\kappa}=0$, it is well-known that there exists an invertible matrix field $[A_{\mu\nu}]$ (with inverse $[A^{\mu\nu}]$) such that $\Gamma^\gamma_{\mu\nu}=A^{\gamma\lambda}\partial_{\nu} A_{\lambda\mu}$.)</p></li>
</ol>
<p>In Schouten (Chapter III, equation 5.19), it is stated that the integrability condition of (1) are given by the Bianchi identities </p>
<p>$$\nabla_{[\omega}R_{\mu\nu]\lambda}{}^\kappa=2S_{[\omega\nu}{}^\sigma\, R_{\mu]\sigma\lambda}{}^\kappa,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$$
where $S_{\mu\nu}{}^\kappa$ is the torsion of the connection $\Gamma^\kappa_{\mu\nu}$.</p>
<p>It is clear that (2) involves covarient derivative wirh respect to the connection $\Gamma$ and also the torsion $S$ of $\Gamma$, but to start with no such $\Gamma$ is given. We are <em>looking for</em> such a $\Gamma$. I fail to understand in what sense, (2) is an integrability condition of (1). I'm also not sure, whether to determine a $\Gamma$ satisfying (1), some (skew-symmetric) functions $S_{\mu\nu}{}^\lambda$ should also be given <em>a priori</em> so that the desired connection has also these specified fuctions as its torsion.</p>
<p><strong>Notes:</strong> 1. No metric structure is given on the manifold $M^n$.</p>
<ol start="2">
<li>It would be great if it can be answered at least for $n=3$.</li>
</ol>
<p>Thanks in advance.</p>
| Robert Bryant | 13,972 | <p>Part of the difficulty in providing an answer to your question is the fact that the expression "the integrability condition" is a somewhat vague notion, and it's used in slightly different senses in different contexts. </p>
<p>The usual, somewhat imprecise, sense is that, for a given system of PDE, its 'integrability condition' is the set of necessary and sufficient conditions that it have local solutions. Slightly more precise usage would be, given a PDE system that depends on some data (in your case, the equations (1), where the data is the curvature $R$), to find the necessary and sufficient conditions <em>on the data</em> that ensure that the PDE system have local solutions.</p>
<p>In this sense, the Bianchi identities (your equation (2)) are usually not <em>the</em> integrability condition for the system (1) but are only <em>part</em> of the integrability condition, and they have to be interpreted appropriately. The point is that, when $R$ is specified, equation (2) is an <em>algebraic</em> equation $\Gamma$ must satisfy if it is to satisfy the first order PDE system (1). [It may not be immediately apparent that (2) is an algebraic equation on $\Gamma$, but if you look at the definition of $\nabla$, you'll see that the left hand side of (2) only involves the unknown coefficients of $\Gamma$ linearly with no differentiation of them, and the right hand side involves the unknown $S$, which is computed linearly from the unknown $\Gamma$.] Thus, a necessary condition <em>on the data $R$</em> that (1) have a solution is that there be at least one solution $\overline\Gamma$ to the inhomogeneous linear algebraic system (2) (whose coefficients come from $R$ and its first derivatives). Generally, this necessary condition is <em>not</em> sufficient, though, as examples in dimensions $4$ and above show.</p>
<p>However, what Deane Yang's reference eudml.org/doc/74779 (DeTurck & Talvacchia, 1987) shows is that, when (i) $n=3$, (ii) $R$ is sufficiently generic (in an appropriate sense) while satisfying the integrability condition (2), <em>and</em> (iii) $R$ is real-analytic, then the system (1) is locally solvable. The system of PDE for $\Gamma$ that has to be solved, though, is highly nonlinear and, in a sense, overdetermined, so that the Cartan-Kähler Theorem or one of its modern versions has to be applied. There is no uniqueness and no way explicitly to solve the equations for $\Gamma$ for a given generic $R$ that satisfies the integrability condition. In that sense, your second 'reverse question' has no good answer.</p>
<p>Finally, let me remark that one <em>could</em> conceivably want to answer the more restrictive problem of specifying <em>both</em> the curvature $R$ and torsion $S$ of a connection $\Gamma$ on a manifold $M$. This problem is overdetermined even when $n=3$ and, generally, has no solution. This problem also has an additional set of integrability conditions of the form $\nabla S = F(R,S)$ (for an explicit expression $F(R,S)$ that one can write down easily) and which are also inhomogeneous linear algebraic in $\Gamma$, in addition to the Bianchi conditions $\nabla R = G(R,S)$ that you already have written down as (2). In general, these combined integrability conditions are <em>not</em> sufficient for the generic pair $(S,R)$ that satisfy them in order for a solution $\Gamma$ to exist.</p>
|
3,415,266 | <p>I cannot find how or why this,</p>
<p><span class="math-container">$$5\sin(3x)−1 = 3$$</span></p>
<p>Has one of two solutions being this,</p>
<p><span class="math-container">$$42.29 + n \times 120^\circ$$</span></p>
<p>I am lost on how to get this solution. I have found the other one, so I will not mention it.</p>
<p>Can someone help explain this and show how to get this solution? Thank you!</p>
| Quanto | 686,284 | <p>Note that <span class="math-container">$\sin 3x =\sin (180^\circ - 3x)= \frac45=\sin53.1^\circ$</span>, which gives two sets of solutions, </p>
<p><span class="math-container">$$3x=53.1^\circ + n\>360^\circ$$</span></p>
<p><span class="math-container">$$3x=(180-53.1)^\circ + n\>360^\circ$$</span></p>
<p>with <span class="math-container">$n$</span> being integers. Thus, the complete sultions are</p>
<p><span class="math-container">$$x= 17.7^\circ + n\>120^\circ,\>\>\>\>x= 42.3^\circ + n\>120^\circ$$</span></p>
|
2,017,993 | <p>Is $0$ an eigenvalue for a compact normal operator?</p>
<p>Many texts mention that compact normal operators have a complete orthonormal basis of eigenvectors. If they do, then what about the kernel of the operator? The elements in the kernel, may not be eigenvectors.</p>
<p>Where is the mistake in my understanding?</p>
| Fred | 380,717 | <p>If $X$ is a normed space and $\dim X < \infty$, then each linear operator $K:X \to X$ is compact. Thus $0$ may or may not be an eigenvalue.</p>
|
2,017,993 | <p>Is $0$ an eigenvalue for a compact normal operator?</p>
<p>Many texts mention that compact normal operators have a complete orthonormal basis of eigenvectors. If they do, then what about the kernel of the operator? The elements in the kernel, may not be eigenvectors.</p>
<p>Where is the mistake in my understanding?</p>
| Gyu Eun Lee | 52,450 | <p>A compact normal operator need not have zero as an eigenvalue. But nothing rules that possibility out.</p>
<p>For easy examples, we have the operator <span class="math-container">$A$</span> represented by the following matrix in the standard basis on <span class="math-container">$\mathbb{R}^2$</span>:
<span class="math-container">$$
[A] = \begin{pmatrix}
1 & 0\\
0 & 0
\end{pmatrix}
$$</span>
Then <span class="math-container">$A:\mathbb{R}^2\to\mathbb{R}^2$</span> is a Hermitian operator, hence normal. It is also continuous and compact, because all linear operators on finite-dimensional normed vector spaces are continuous and compact. Finally, it has <span class="math-container">$1$</span> and <span class="math-container">$0$</span> as eigenvalues, and an obvious orthonormal basis of eigenvectors.</p>
<p>On the other hand, the identity <span class="math-container">$I:\mathbb{R}^2\to\mathbb{R}^2$</span> is also normal, continuous, and compact, but does not have <span class="math-container">$0$</span> as an eigenvalue.</p>
|
643,851 | <p>Let the ODE
$$\frac{dy}{dx}=\frac{y+x-2}{y+x-4}$$</p>
<p>I got the general (implicit) solution:
$$y=\ln|x+y-3|+x+A$$ A is arbitrary constant.</p>
<p>My question is:
is $3=y+x$ a solution of this ODE? I know it's not contained in the general solution.</p>
| LecSka | 122,425 | <p>Let $n\geq 100$ an even number. Consider the quantities $n-91$, $n-93$ and $n-95$, one of these is a multiple of 3, and not exactly 3 cause $100-95>3$, then is a composite odd number.
Observing thet 91,93 and 95 are composite, you conclude that every $n\geq 100$ works. Now check directly the remaining numbers, and you have the solution. </p>
|
3,999,652 | <p>Let triangle <span class="math-container">$ABC$</span> is an equilateral triangle. Triangle <span class="math-container">$DEF$</span> is also an equilateral triangle and it is inscribed in triangle <span class="math-container">$ABC \left(D\in BC,E\in AC,F\in AB\right)$</span>. Find <span class="math-container">$\cos\measuredangle DEC$</span> if <span class="math-container">$AB:DF=8:5$</span>.</p>
<p><a href="https://i.stack.imgur.com/FZkBd.png" rel="noreferrer"><img src="https://i.stack.imgur.com/FZkBd.png" alt="enter image description here" /></a>
Firstly, I would be very grateful if someone can explain to me how I am supposed to draw the diagram. Obviously I have made it by sight.</p>
<p>Let <span class="math-container">$\measuredangle DEC=\alpha$</span>. We can note that <span class="math-container">$\triangle AEF \cong \triangle BFD \cong CDE$</span>. This is something we can always use in such configuration. So <span class="math-container">$$AE=BF=CD, $$</span> <span class="math-container">$$AF=BD=CE.$$</span> Let <span class="math-container">$AB=BC=AC=8x$</span> and <span class="math-container">$DF=DE=EF=5x$</span>. If we denote <span class="math-container">$CD=y,$</span> then <span class="math-container">$CE=AC-AE=AC-CD=8x-y$</span>. Cosine rule on <span class="math-container">$CED$</span> gives <span class="math-container">$$25x^2=(8x-y)^2+y^2-2y\cos60^\circ(8x-y)$$</span> which is a homogenous equation. I got that <span class="math-container">$\dfrac{y}{x}=4\pm\sqrt{3}.$</span> Now using the sine rule on <span class="math-container">$CED$</span> <span class="math-container">$$\dfrac{CD}{DE}=\dfrac{\sin\alpha}{\sin60^\circ}\Rightarrow \sin\alpha=\dfrac{\sqrt{3}}{10}\cdot\dfrac{y}{x}=\dfrac{4\sqrt3\pm3}{10}.$$</span> Now we can use the trig identity <span class="math-container">$\sin^2x+\cos^2x=1$</span> but it doesn't seem very rational. Can you give me a hint? I was able to find <span class="math-container">$\sin\measuredangle DEC$</span> in acceptable way, but I can't find <span class="math-container">$\cos\measuredangle DEC$</span>...</p>
| heropup | 118,193 | <p>Note that there are two distinct solutions given the conditions in the problem. This is because there are two admissible orientations of <span class="math-container">$\triangle DEF$</span>: one in which <span class="math-container">$AF > BF$</span>, and one in which <span class="math-container">$AF < BF$</span>. We can understand this by noting that since <span class="math-container">$\triangle DEF$</span> and <span class="math-container">$\triangle ABC$</span> share the same center, the circumcircle of <span class="math-container">$\triangle DEF$</span> will intersect <span class="math-container">$\triangle ABC$</span> at six points, unless the side length of <span class="math-container">$\triangle DEF$</span> is less than or equal to half the side length of <span class="math-container">$\triangle ABC$</span> (in which case there will be either exactly three intersection points in the tangent case, or zero), or greater than or equal to the side length of <span class="math-container">$\triangle ABC$</span>.</p>
<p>A simple way to solve for the required angle is to place the triangle on a coordinate system, say with <span class="math-container">$A = (0,0)$</span>, <span class="math-container">$B = (8,0)$</span>, <span class="math-container">$C = (4, 4 \sqrt{3})$</span>, and let <span class="math-container">$F = (x,0)$</span> for some <span class="math-container">$0 \le x \le 8$</span>. Then <span class="math-container">$$E = \left(\frac{8-x}{2}, \frac{(8-x)\sqrt{3}}{2}\right).$$</span> We then require <span class="math-container">$EF = 5$</span>, or <span class="math-container">$$25 = \left(x - \frac{8-x}{2}\right)^2 + \left(\frac{(8-x)\sqrt{3}}{2}\right)^2 = 3x^2 - 24x + 64.$$</span> Consequently <span class="math-container">$$x = 4 \pm \sqrt{3}$$</span> and depending on which root is chosen, <span class="math-container">$\angle DEC$</span> can be determined through a straightforward application of the Law of Sines.</p>
|
3,999,652 | <p>Let triangle <span class="math-container">$ABC$</span> is an equilateral triangle. Triangle <span class="math-container">$DEF$</span> is also an equilateral triangle and it is inscribed in triangle <span class="math-container">$ABC \left(D\in BC,E\in AC,F\in AB\right)$</span>. Find <span class="math-container">$\cos\measuredangle DEC$</span> if <span class="math-container">$AB:DF=8:5$</span>.</p>
<p><a href="https://i.stack.imgur.com/FZkBd.png" rel="noreferrer"><img src="https://i.stack.imgur.com/FZkBd.png" alt="enter image description here" /></a>
Firstly, I would be very grateful if someone can explain to me how I am supposed to draw the diagram. Obviously I have made it by sight.</p>
<p>Let <span class="math-container">$\measuredangle DEC=\alpha$</span>. We can note that <span class="math-container">$\triangle AEF \cong \triangle BFD \cong CDE$</span>. This is something we can always use in such configuration. So <span class="math-container">$$AE=BF=CD, $$</span> <span class="math-container">$$AF=BD=CE.$$</span> Let <span class="math-container">$AB=BC=AC=8x$</span> and <span class="math-container">$DF=DE=EF=5x$</span>. If we denote <span class="math-container">$CD=y,$</span> then <span class="math-container">$CE=AC-AE=AC-CD=8x-y$</span>. Cosine rule on <span class="math-container">$CED$</span> gives <span class="math-container">$$25x^2=(8x-y)^2+y^2-2y\cos60^\circ(8x-y)$$</span> which is a homogenous equation. I got that <span class="math-container">$\dfrac{y}{x}=4\pm\sqrt{3}.$</span> Now using the sine rule on <span class="math-container">$CED$</span> <span class="math-container">$$\dfrac{CD}{DE}=\dfrac{\sin\alpha}{\sin60^\circ}\Rightarrow \sin\alpha=\dfrac{\sqrt{3}}{10}\cdot\dfrac{y}{x}=\dfrac{4\sqrt3\pm3}{10}.$$</span> Now we can use the trig identity <span class="math-container">$\sin^2x+\cos^2x=1$</span> but it doesn't seem very rational. Can you give me a hint? I was able to find <span class="math-container">$\sin\measuredangle DEC$</span> in acceptable way, but I can't find <span class="math-container">$\cos\measuredangle DEC$</span>...</p>
| Saeed | 858,459 | <p>Here I will show how you can draw the figure. The calculations that follow are somewhat similar to other answers.
<a href="https://i.stack.imgur.com/5ciiq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5ciiq.jpg" alt="enter image description here" /></a></p>
<p>We can build triangle <span class="math-container">$DEF$</span> by scaling down triangle <span class="math-container">$ABC$</span> and rotating it around its centroid <span class="math-container">$G$</span> by some angle <span class="math-container">$\alpha$</span>.</p>
<p>Notice the hatched triangle <span class="math-container">$DHG$</span>. This is a right triangle with sides:
<span class="math-container">$$GH = \frac13 AH =\frac{4}{12} AH$$</span>
<span class="math-container">$$DG = \frac23 DY = \frac23 \times \frac58 AH = \frac{5}{12} AH $$</span>
<span class="math-container">$$HD = \sqrt{DG^2 - GH^2} = \frac{3}{12} AH$$</span>
It seems the designer of the problem intentionally used the ratio <span class="math-container">$\frac58$</span> to lead us to the <span class="math-container">$3-4-5$</span> right triangle!</p>
<p>Anyway, Now we are able to locate point <span class="math-container">$D$</span> on <span class="math-container">$BC$</span>. It is at distance <span class="math-container">$\frac14 AH$</span> from point <span class="math-container">$H$</span> , the midpoint of BC. We can similarly locate points <span class="math-container">$E$</span> and <span class="math-container">$F$</span> , hence constructing triangle <span class="math-container">$DEF$</span>.</p>
<p>Now, as for calculations, note that <span class="math-container">$\widehat{DEC}=120^o - \alpha$</span> and that <span class="math-container">$\widehat{DGH}=60^o - \alpha$</span>. So <span class="math-container">$\widehat{DEC}=60^o+\widehat{DGH}$</span> . Therefore:
<span class="math-container">$$\cos \widehat{DEC} = \cos (60^o+\widehat{DGH}) = \cos60^o \cos \widehat{DGH} - \sin60^0 \sin \widehat{DGH}$$</span>
<span class="math-container">$$= \frac12 \times \frac45 - \frac{\sqrt 3}{2} \times \frac35 = \frac{4-3\sqrt3}{10}$$</span>
Note that when locating point <span class="math-container">$D$</span> we have two options, say, above and below <span class="math-container">$H$</span>. The above calculations are for the option shown in the figure, where <span class="math-container">$D$</span> is selected <em>below</em> <span class="math-container">$H$</span>. Alternatively, we could select <span class="math-container">$D$</span> to be <em>above</em> <span class="math-container">$H$</span>, and respectively have <span class="math-container">$E$</span> <em>below</em> <span class="math-container">$K$</span>. The interested reader can verify that in that case <span class="math-container">$\widehat{DEC}$</span> would be <span class="math-container">$\alpha$</span>. And <span class="math-container">$\; \cos\alpha \;$</span> can be calculated similar to above.
<span class="math-container">$$\cos \alpha = \cos (60^o-\widehat{DGH}) = \cos60^o \cos \widehat{DGH} + \sin60^0 \sin \widehat{DGH} = \frac{4+3\sqrt3}{10}$$</span></p>
|
584,975 | <p>I'm doing an Advanced Functions course right now, and I'm wondering about something. Look at this here evaluation/simplification that I did:</p>
<p><a href="http://puu.sh/5w3XQ.png" rel="nofollow">http://puu.sh/5w3XQ.png</a></p>
<p>What I'm wondering is about the 2^4 - 2^3. I know this is is 2^3 because 2^3 is the value multiplied by 2 to get 2^4, which means I'm subtracting the same value I'm adding when going from 2^3 to 2^4 from 2^3 and thereby going back to 2^3.</p>
<p>But this is just a logical step I did it my head, and it's not something that works as a rule for all addition/subtraction. So I'm wondering if there is another way for me to solve this? I'm not comfortable with simplifying something on the premise of "I just know that's what it means", because that might mean I don't get the full score. So is there any exponent rules/laws that apply to adding or subtracting exponents with the same base which I can use here, in order to make sure I don't get less point for taking shortcuts instead of using the exponent laws?</p>
<p>Should I perhaps make it into 2^7 - 2^6 and evaluate it as 128 - 64 = 64? Or is there another way to end up with 2^6 using an exponent law? I just want to make sure I don't get any less points for doing steps in my head or simplifying something outside of the methods of exponent laws.</p>
| hhsaffar | 104,929 | <p><em>Hint</em>:</p>
<p>Consider $b_n=a_n-\frac{1}{2}$</p>
<p>Can you write a recursive equation for $b_n$?</p>
|
584,975 | <p>I'm doing an Advanced Functions course right now, and I'm wondering about something. Look at this here evaluation/simplification that I did:</p>
<p><a href="http://puu.sh/5w3XQ.png" rel="nofollow">http://puu.sh/5w3XQ.png</a></p>
<p>What I'm wondering is about the 2^4 - 2^3. I know this is is 2^3 because 2^3 is the value multiplied by 2 to get 2^4, which means I'm subtracting the same value I'm adding when going from 2^3 to 2^4 from 2^3 and thereby going back to 2^3.</p>
<p>But this is just a logical step I did it my head, and it's not something that works as a rule for all addition/subtraction. So I'm wondering if there is another way for me to solve this? I'm not comfortable with simplifying something on the premise of "I just know that's what it means", because that might mean I don't get the full score. So is there any exponent rules/laws that apply to adding or subtracting exponents with the same base which I can use here, in order to make sure I don't get less point for taking shortcuts instead of using the exponent laws?</p>
<p>Should I perhaps make it into 2^7 - 2^6 and evaluate it as 128 - 64 = 64? Or is there another way to end up with 2^6 using an exponent law? I just want to make sure I don't get any less points for doing steps in my head or simplifying something outside of the methods of exponent laws.</p>
| Felix Marin | 85,343 | <p>$\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
\newcommand{\fermi}{\,{\rm f}}%
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
\newcommand{\half}{{1 \over 2}}%
\newcommand{\ic}{{\rm i}}%
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}%
\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ket}[1]{\left\vert #1\right\rangle}%
\newcommand{\ol}[1]{\overline{#1}}%
\newcommand{\pars}[1]{\left( #1 \right)}%
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}%
\newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
\newcommand{\sech}{\,{\rm sech}}%
\newcommand{\sgn}{\,{\rm sgn}}%
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}%
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$a_{n} - 1/2 = 2^{n - 1}/4 + 3\pars{a_{n -1} - 1/2}$. Set
$a_{n} - 1/2 = 2^{n - 1}b_{n}/4\quad\iff\quad b_{n} = 2^{3 - n}\pars{a_{n} - 1/2}$
with $b_{3} = 1/2$.
$$
{1 \over 4}\,2^{n - 1}b_{n} = {1 \over 4}\,2^{n - 1}
+
3\,{1 \over 4}\,2^{n -2}b_{n - 1}\quad\imp\quad
b_{n} = 1 + {3 \over 2}\,b_{n - 1}
$$
$$
b_{n} + 2 = {3 \over 2}\pars{b_{n - 1} + 2}
= \pars{3 \over 2}^{2}\pars{b_{n - 2} + 2} = \cdots =
\pars{3 \over 2}^{n - 3}\bracks{b_{n - \pars{n - 3}} + 2}
$$
$$
a_{n} = \half + 2^{n - 3}b_{n}
= \half + 2^{n - 3}\bracks{-2 + \pars{3 \over 2}^{n - 3}\,{5 \over 2}}
= \half - 2^{n - 2}+ {5 \over 2}\,3^{n - 3}
$$
$$\color{#0000ff}{\large%
a_{n} = \half - 2^{n - 2}+ {5 \over 2}\,3^{n - 3}
}
$$</p>
|
1,507,076 | <p>I'm supposed to give a constructive proof of the following claim, but to get some hints on what to do I tried to prove the inverse and see where the proof breaks down. I have the following.</p>
<p>Claim: Define $X * Y = \{ xy \mid x \in X, y \in Y\}$. There is a ring $R$ with ideal $U$ such that $R / U$ is not closed under *.</p>
<p>What I've done:</p>
<p>Observe $S := (a + U) * (b + U) = \{ xy \mid x \in a + U, y \in b + U\}
= \{(a + u_1)(b + u_2) \mid u_1, u_2 \in U\}
= \{ ab + au_2 + u_1b + u_1u_2 \mid u_1, u_2 \in U\}$
Since $U$ is an ideal, $au_2 + u_1b + u_1u_2 \in U$. Therefore, $(a + U)*(b + U)$ could only possibly equal $ab + U$. Thus if I can find a $u \in U$ not expressible as $au_2 + u_1b + u_1u_2$, then that completes the claim, as it would make $(a + U)*(b + U)$ a strict subset of $ab + U$. But at this point I'm not sure how to proceed.</p>
| Rob Arthan | 23,171 | <p>You are being asked to produce an example, not a proof: take $R = \Bbb{Z}$ and $U = 2\Bbb{Z}$. Then $U \in R/U$, but $U*U = 4\Bbb{Z} \not\in R/U$.</p>
|
420,294 | <p>While reading Bill Thurston's <a href="http://www.ams.org/publications/journals/notices/201601/rnoti-p31.pdf" rel="noreferrer">obituary</a> in the Notices of the AMS I came across the following fascinating anecdote (pg. 32):</p>
<blockquote>
<p>Bill’s enthusiasm during the early stages of mathematical discovery was infectious. Once, while sitting in his
living room, Bill said to me, “I can do this group with
grep,” which was sort of strange to hear at first. But being
his student I knew just enough computerese to have an
inkling of what he was saying: he was able to compute
in that group with the UNIX utility for processing regular
expressions using finite deterministic automata. From
there, it was exciting to observe the quick unfolding of
the theory of automatic groups.</p>
</blockquote>
<p>Looking through David Epstein's <em>Word Processing in Groups</em> I can see that there are indeed connections between <a href="https://en.wikipedia.org/wiki/Automatic_group" rel="noreferrer">automatic groups</a> and regular expressions that should allow faster algorithms for certain computations e.g. solving the word problem.</p>
<p>But is there a concrete example of a group-theoretic computation with <code>grep</code>?</p>
| Lee Mosher | 20,787 | <p>I wrote that quote, and I'll take the <a href="https://mathoverflow.net/questions/420294/group-theory-with-grep#comment1079616_420294">hint of @SamNead</a> and try to write an answer, although the best I can do is to write a somewhat speculative extension of the story behind the quote, laced with some mathematical musings.</p>
<p>From my memories of that period (sometime in the mid-late 1980's), I only vaguely recall looking at any actual output of Bill Thurston's computer experiments using grep to compute in groups. My memory is not at all precise enough regarding any actual examples he might have produced, but I suspect that they must have been pretty simple examples, for the following reasons.</p>
<p>In the theory of regular languages and FDAs (finite deterministic automata), there are reasonably simple algorithms for going back and forth between regular expressions and finite deterministic automata, producing an expression matching exactly the words accepted by a given automaton, and producing an automaton accepting exactly the words matched by a given expression. The algorithm from regular expressions to FDAs works pretty efficiently and is the basis of the <strong>grep</strong> utility. The algorithm from FDAs to regular expressions seems to be horribly inefficient, as reported by Derek Holt in <a href="https://mathoverflow.net/questions/420294/group-theory-with-grep#comment1079534_420294">his comment</a>. So one might expect that writing an actual regular expression that matches normal forms for a certain group could be quite tedious.</p>
<p>Nonetheless if one is clever enough one can sometimes just manually come up with a regular expression to match a given language. See the exercise Derek Holt suggests in <a href="https://mathoverflow.net/questions/420294/group-theory-with-grep#comment1079534_420294">his comment</a>; carrying out that exercise will likely give you the most direct answer to your question.</p>
<p>I suspect that this is what Thurston was doing during this discovery period: noticing that certain finitely generated groups have a regular language of normal forms.</p>
<p>Thurston was also very familiar with concepts of cone types (developed by Jim Cannon around that time; Thurston and Cannon talked together a lot during that period). So at some point during this process he certainly also noticed the connections between cone types and what we now call the 2-tape "multiplier automata" which are at the heart of the theory of automatic groups. It is tempting to speculate whether Thurston's <strong>grep</strong> experiments extended to writing regular expressions for 2-tape multiplier automata, but honestly I have no idea.</p>
|
1,382,479 | <p>I would like to know why $a^p \equiv a \pmod p$ is the same as $a^{p-1} \equiv 1 \pmod p$, and also how Fermat's little theorem can be used to derive Euler's theorem, or vice versa. </p>
<p>Please keep in mind that I have little background in math, and I am trying to understand these theorems to understand the math behind RSA encryption, so I would appreciate it if the explanation could be as simple as possible (i.e. not using too much mathematical notation). I have had trouble finding resources online to explain these theorems that explain it in simple enough terms that I can understand.</p>
| Samir Khan | 183,661 | <p>The statement that $a^p\equiv a\mod p$ is the same as $a^{p-1}\equiv 1\mod p$ when $a$ and $p$ are relatively prime, because in this case we can divide both sides of the congruence by $a$, and obtain one from the other. </p>
<p>Euler's theorem says that </p>
<p>$$a^{\phi(m)}\equiv 1\mod m,$$</p>
<p>where $\phi(m)$ is the number of integers less than $m$ and relatively prime to $m$. For a prime, this is exactly $p-1$, so Fermat's Little Theorem is a special case of Euler's theorem. </p>
|
358,184 | <p>We have a lot of probabilities lower bounding as (e.g. chernoff bound, reverse markov inequality, Paley–Zygmund inequality)
<span class="math-container">$$
P( X-E(X) > a) \geq c, a > 0 \quad and \quad P(X > (1-\theta)E[X]) \geq c, 0<\theta < 1
$$</span></p>
<p>However, It would be great to know if there is any inequality bounding exactly<br>
<span class="math-container">$$
P(X > E[X]) \geq c
$$</span>
i.e., the probability that a r.v greater than its exact expected value ? (e.g., Suppose X is bounded and with bounded first and second moments)</p>
| Iosif Pinelis | 36,721 | <p>Let <span class="math-container">$Y:=X-EX$</span>. We need to obtain a lower bound on <span class="math-container">$P(Y>0)$</span>. </p>
<p>Suppose that <span class="math-container">$-a\le Y\le b$</span> for some real <span class="math-container">$a>0$</span> and <span class="math-container">$b>0$</span>, and that <span class="math-container">$EY^2\ge s^2$</span> for some real <span class="math-container">$s$</span>. Then
<span class="math-container">$$1_{Y>0}\ge\frac{aY+Y^2}{ab+b^2}.$$</span>
Taking expectations of both sides of this inequality, we get
<span class="math-container">$$P(Y>0)\ge\frac{s^2}{ab+b^2}. \tag{1}$$</span></p>
<p>In terms of <span class="math-container">$X$</span>, (1) can be rewritten as
<span class="math-container">$$P(X>EX)\ge\frac{Var\,X}{ab+b^2},$$</span>
provided that <span class="math-container">$-a\le X-EX\le b$</span>.</p>
<hr>
<p>The condition <span class="math-container">$-a\le Y\le b$</span> implies that
<span class="math-container">$$Y^2\le\frac{Y+a}{a+b}\,b^2+\frac{b-Y}{a+b}\,a^2.$$</span>
Taking expectations of both sides of this inequality, we get
<span class="math-container">$$s^2\le EY^2\le\frac{a}{a+b}\,b^2+\frac{b}{a+b}\,a^2=ab.$$</span>
So, letting now
<span class="math-container">$$p:=\frac{s^2}{(a+b)a}\quad\text{and}\quad r:=\frac{s^2}{(a+b)b}, $$</span>
we see that
<span class="math-container">$$p+r=\frac{s^2}{ab}\le1.$$</span>
Letting then <span class="math-container">$Y$</span> be a random variable taking values <span class="math-container">$-a,0,b$</span> with probabilities <span class="math-container">$p,1-p-r,r$</span> respectively, we see that <span class="math-container">$-a\le Y\le b$</span>, <span class="math-container">$EY=0$</span>, <span class="math-container">$EY^2=s^2$</span>, and
<span class="math-container">$$P(Y>0)=\frac{s^2}{ab+b^2}.$$</span>
So, the lower bound on <span class="math-container">$P(Y>0)$</span> in (1) is attained. </p>
<hr>
<p>Without the condition <span class="math-container">$EY^2\ge s^2$</span>, no nonzero lower bound on <span class="math-container">$P(Y>0)$</span> exists
even if we still assume that <span class="math-container">$-a\le Y\le b$</span> for some real <span class="math-container">$a\ge0$</span> and <span class="math-container">$b\ge0$</span> -- just let <span class="math-container">$Y$</span> be the constant <span class="math-container">$0$</span>. </p>
<p>Also, obviously, the exact lower bound <span class="math-container">$\frac{s^2}{ab+b^2}$</span> on <span class="math-container">$P(Y>0)$</span> goes to <span class="math-container">$0$</span> if either <span class="math-container">$a\to\infty$</span> or <span class="math-container">$b\to\infty$</span>. It follows that no nonzero lower bound on <span class="math-container">$P(Y>0)$</span> exists if we replace <span class="math-container">$a$</span> or <span class="math-container">$b$</span> by <span class="math-container">$\infty$</span>. </p>
<p>Thus, none of the conditions imposed on <span class="math-container">$Y$</span> can be removed if one wants to have a nonzero lower bound on <span class="math-container">$P(Y>0)$</span>. </p>
|
634,929 | <p>How can I evaluate this integral?</p>
<p>$$
\int{x^{3}\,{\rm d}x \over \left(x - 1\right)^{2}\sqrt{x^{2} + 2x + 4}}$$</p>
<p>I would be grateful for any tips.</p>
| Mike | 17,976 | <p>Let me provide an alternate method of solving the last 2 integrals from lab bhattacharjee's answer. The first step is to rewrite</p>
<p>$$\sqrt3\sin\psi-2\cos\psi\text{ as }A\cos(\psi+B)=A\cos\psi\cos B-A\sin\psi\sin B$$</p>
<p>So we have</p>
<p>$$A\cos B=-2,A\sin B=-\sqrt3$$
$$\tan B=\frac{\sqrt3}2,B=\tan^{-1}\frac{\sqrt3}2$$
$$A^2=(-2)^2+(-\sqrt3)^2=7$$
$$A=-\sqrt7$$</p>
<p>So the third integral</p>
<p>$$\int\frac{d\psi}{\sqrt3\sin\psi-2\cos\psi}=-\frac{\sqrt7}7\int\sec(\psi+\tan^{-1}\frac{\sqrt3}2)d\psi$$</p>
<p>which is a straightforward integral of a secant.</p>
<p>As for the fourth integral,</p>
<p>$$\int\frac{\cos\psi d\psi}{(\sqrt3\sin\psi-2\cos\psi)^2}=\frac17\int\frac{\cos\psi d\psi}{\cos^2(\psi+\tan^{-1}\frac{\sqrt3}2)}$$</p>
<p>$$\theta=\psi+\tan^{-1}\frac{\sqrt3}2$$</p>
<p>I'm out of time for now (I'll probably come back to it later), but if you expand the cosine in the numerator, you should end up with a $\sec\theta$ term and a $\sec\theta\tan\theta$ term, both of which can be easily integrated.</p>
|
1,842,365 | <p>I recently got acquainted with a theorem:</p>
<p>If $f(x)$ is a periodic function with period $P$, then $f(ax+b)$ is periodic with period $\dfrac{P}{a}$ , $a>0$.</p>
<p>I am having a difficulty in understanding this theorem. Does this theorem mean that $f(ax+b)=f(ax+b+ \dfrac{P}{a})$? </p>
<p>If the above meaning is true, then how can a function, initially having a single period, acquire another period by just changing the arguments?</p>
| Arthur Sinulis | 269,465 | <p>Just as another remark: a neat corollary which follows from this statement and the strong law of large numbers is that whenever $X_i$ is an iid sequence of r.v. such that the SLLN holds (e.g. $X_i \in L^1$ by Kolmogorov's SLLN), then $\mathbb{E}(X_1 \mid S_n) \to \mathbb{E}(X_1)$ $\mathbb{P}$-almost surely.</p>
|
2,178,428 | <p><strong>Goal</strong>: I need help to intuitively understand division by a fraction. </p>
<p><strong>Background</strong>: I've read <a href="https://math.stackexchange.com/questions/248385/what-is-the-physical-explanation-of-a-division-by-a-fraction">this</a> and <a href="http://www.mathsisfun.com/division.html" rel="nofollow noreferrer">this</a>. I understand that division is "even distribution" so, with integers, $\frac 6 2 = 3$ could mean that 6 friends get 2 chocolates each.</p>
<p><strong>Observation</strong>: When the denominator is >0, the solution is less than the numerator (e.g., $\frac 3 4$ and $\frac 4 3$), but when 0 < denominator < 1, the solution is larger than the numerator (e.g., $\frac {1} {\frac 1 2}$ = 2). <a href="https://www.google.com/search?hl=en&q=y%3D1%2Fx" rel="nofollow noreferrer">Graph of y=$\frac 1 x$</a>.</p>
<p><strong>Problem</strong>: "Dividing" (as understood by "into smaller parts") by a fraction doesn't seem possible: you end up multiplying, never dividing--and yes, I understand that division by a fraction is multiplication by that fraction's multiplicative inverse <a href="https://www.youtube.com/watch?v=f3ySpxX9oeM" rel="nofollow noreferrer">(Khan Academy)</a>.</p>
<p><strong>Question</strong>: Since 3--2 (three minus negative two) = 3+(-(-2))--i.e., addition not subtraction (<a href="https://math.stackexchange.com/questions/851987/subtraction-of-a-negative-number">source</a>)--is division by a fraction similar? In other words, is division by a fraction <strong>not division</strong> at all, but multiplication?</p>
| Robert Israel | 8,508 | <p>The <strong>definition</strong> of division is: $ a \div b$ is $c$ where $a = b \times c$.
In the case of positive integers $b$ you can interpret this as even distribution of an amount $a$ into $b$ equal parts. This is because if $b$ is a positive integer, $b \times c$ is the result of adding together $b$ copies of $c$. </p>
<p>If $b$ is a fraction $m/n$, where $m$ and $n$ are positive integers, you can think of $b \times c$ as the result of adding together $m$ copies of $c/n$, i.e. you split $c$ into $n$ equal pieces, and you add together $m$ of those pieces (or pieces equal to one of those if $m$ is greater than $n$). Thus $a/b$ would be an amount $c$ that if you divide into $n$ equal pieces and add $m$ of those pieces together you get $a$.</p>
<p>However, when you go beyond fractions (e.g. to real numbers), even this doesn't make sense. Again, the definition is in terms of multiplication, the interpretation is secondary.</p>
|
2,178,428 | <p><strong>Goal</strong>: I need help to intuitively understand division by a fraction. </p>
<p><strong>Background</strong>: I've read <a href="https://math.stackexchange.com/questions/248385/what-is-the-physical-explanation-of-a-division-by-a-fraction">this</a> and <a href="http://www.mathsisfun.com/division.html" rel="nofollow noreferrer">this</a>. I understand that division is "even distribution" so, with integers, $\frac 6 2 = 3$ could mean that 6 friends get 2 chocolates each.</p>
<p><strong>Observation</strong>: When the denominator is >0, the solution is less than the numerator (e.g., $\frac 3 4$ and $\frac 4 3$), but when 0 < denominator < 1, the solution is larger than the numerator (e.g., $\frac {1} {\frac 1 2}$ = 2). <a href="https://www.google.com/search?hl=en&q=y%3D1%2Fx" rel="nofollow noreferrer">Graph of y=$\frac 1 x$</a>.</p>
<p><strong>Problem</strong>: "Dividing" (as understood by "into smaller parts") by a fraction doesn't seem possible: you end up multiplying, never dividing--and yes, I understand that division by a fraction is multiplication by that fraction's multiplicative inverse <a href="https://www.youtube.com/watch?v=f3ySpxX9oeM" rel="nofollow noreferrer">(Khan Academy)</a>.</p>
<p><strong>Question</strong>: Since 3--2 (three minus negative two) = 3+(-(-2))--i.e., addition not subtraction (<a href="https://math.stackexchange.com/questions/851987/subtraction-of-a-negative-number">source</a>)--is division by a fraction similar? In other words, is division by a fraction <strong>not division</strong> at all, but multiplication?</p>
| Timothy | 137,739 | <p>I don't have a math research job. I don't know everything with certainty. This is just my guesswork that the following is a simplifying approximation of reality.</p>
<p>There's no universal law that a binary operation whose steps include only multiplication and not division of integers cannot be called division. A rational number can be defined as an equivalence class of ordered pairs of integers <span class="math-container">$(a, b)$</span> where <span class="math-container">$b \neq 0$</span>. From now on when I talk about how operations are defined on ordered pairs of integers, I really mean that that's how it's defined on ordered pairs of integers whose second coordinate is nonzero.</p>
<p>For <span class="math-container">$(a, b)$</span> is said to be equivalent to <span class="math-container">$(c, d)$</span> when ever <span class="math-container">$ad = bc$</span>. We can do that because it can be shown that that is an equivalence relation. Addition is defined such that the sum of the equivalnce class containing <span class="math-container">$(a, b)$</span> and the equivalence class containing <span class="math-container">$(c, d)$</span> is the equivalence class containing <span class="math-container">$(ad + bc, bd)$</span> and multiplication is defined such that the product of the equivalcne class containing <span class="math-container">$(a, b)$</span> and the equivalence class containing <span class="math-container">$(c, d)$</span> is the equivalence class containing <span class="math-container">$(ac, bd)$</span>. It can also be shown that there is a way to define it that way.</p>
<p>Technically, a field is not a set but an ordered triplet of a set and operations on that set. You can denote a set with operations how ever you want and <span class="math-container">$\mathbb{Q}$</span> is just one possible variable you could use to denote it. Most mathematicians agree to use the notation <span class="math-container">$(\mathbb{Q}, +, \times)$</span> to refer to any field that is isomorphic to the desired structure of the rational numbers with those operations and call it the rational numbers with addition and multiplication. I believe the mathematical community also agreed that once you have a field and you denote the second coordinate for that field as <span class="math-container">$+$</span> and the third coordinate of it as <span class="math-container">$\times$</span>, <span class="math-container">$a - b$</span> is defined to mean <span class="math-container">$a$</span> plus the additive inverse of <span class="math-container">$b$</span> and <span class="math-container">$a \div b$</span> is defined to mean <span class="math-container">$a$</span> times the multiplicative inverse of <span class="math-container">$b$</span> if <span class="math-container">$b$</span> is not the additive identity.</p>
<p>The rational numbers the way I defined them with addition and multiplication the way I defined them can be shown to be a field. Let * denote the operation of taking the equivalence class an ordered pair of integers is in. It can be shown that when ever <span class="math-container">$a$</span> is an integer and <span class="math-container">$b$</span> is a nonzero integer, <span class="math-container">$*(a, b) = *(a, 1) \div *(b, 1)$</span>. However, this is true only when <span class="math-container">$a$</span> is an integer and <span class="math-container">$b$</span> is a nonzero integer. Since technically no rational numbers are integers according to this definition, it's in fact the case that <span class="math-container">$\forall a \in \mathbb{Q}\forall b \in \mathbb{Q}(a, b) \notin \mathbb{Q}$</span> so it is not correct to say <span class="math-container">$*(a, b) = *(a, 1) \div *(b, 1)$</span> or <span class="math-container">$*(a, b) = a \div b$</span>. Despite that, we can still define division as <span class="math-container">$*(a, b) \div *(c, d) = *(ad, bc)$</span> when ever <span class="math-container">$c$</span> is nonzero.</p>
<p>Also once <span class="math-container">$(\mathbb{Q}, +, \times)$</span> has been defined, <span class="math-container">$(\mathbb{Z}, +, \times)$</span> can be redefined in such a way that <span class="math-container">$\mathbb{Z}$</span> is a subset of <span class="math-container">$\mathbb{Q}$</span>; the operations in <span class="math-container">$(\mathbb{Z}, +, \times)$</span> are the exact same operations as the ones in <span class="math-container">$(\mathbb{Q}, +, \times)$</span> in that order; and <span class="math-container">$(\mathbb{Z}, +, \times)$</span> has the exact same structure as it was defined to have when using that to construct <span class="math-container">$(\mathbb{Q}, +, \times)$</span>. Most authors use the fractional notation synonymously with division so when <span class="math-container">$a$</span> an <span class="math-container">$b$</span> are elements of set that was originally called <span class="math-container">$\mathbb{Z}$</span> with <span class="math-container">$b$</span> nonzero, we cannot use <span class="math-container">$\frac{a}{b}$</span> to denote <span class="math-container">$*(a, b)$</span> but when <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are elements of <span class="math-container">$\mathbb{Q}$</span> with <span class="math-container">$b$</span> nonzero, we can use <span class="math-container">$\frac{a}{b}$</span> to denote <span class="math-container">$a \div b$</span>. In some books, the fractional notation is used to denote division when describing the rule for differentiating a quotient.</p>
<p>Source: <a href="https://en.wikipedia.org/wiki/Rational_number" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Rational_number</a></p>
|
549,159 | <p>How to simplify this:</p>
<p>$$(5-\sqrt{3}) \sqrt{7+\frac{5\sqrt{3}}{2}}$$</p>
<p>Dont know how to minimize to 11.</p>
<p>Thanks in advance!</p>
| DonAntonio | 31,254 | <p>Justify each step in the following:</p>
<p>$$(5-\sqrt3)\sqrt{7+\frac{5\sqrt3}2}=(5-\sqrt3)\frac{\sqrt{14+5\sqrt3}}{\sqrt2}=$$</p>
<p>$$=\frac{22}{\sqrt2}\frac{\sqrt{14+5\sqrt3}}{5+\sqrt3}=\frac{11\sqrt2}{5+\sqrt3}\sqrt{14+5\sqrt3}=$$</p>
<p>$$=\frac{11}{5+\sqrt3}\sqrt{28+10\sqrt3}=11\frac{\sqrt{(5+\sqrt3)^2}}{5+\sqrt3}=11$$</p>
|
5,253 | <p>I have an acyclic digraph that I would like to draw in a pleasing way, but I am having trouble finding a suitable algorithm that fits my special case. My problem is that I want to fix the x-coordinate of each vertex (with some vertices having the same x-coordinate), and only vary the y. My aesthetic criteria are (in order of importance):</p>
<ol>
<li>Ensure no two vertices are too close together</li>
<li>Minimize edge crossings and near misses</li>
<li>Make a reasonable use of the entire drawing space</li>
</ol>
<p>I have tried several (modified) force-directed algorithms, but they haven't met my expectations on at least one of these - usually too many edge crossings.</p>
<p>Has anyone come across a problem like this, or can you point me to some good papers that deal with restrictions like this?</p>
| David Eppstein | 440 | <p>If the x-coordinates are compatible with the acyclic structure of your DAG (that is, for an edge u->v, the x coordinate of u should always be less than that of v) then this is a standard problem in graph drawing, known as Sugiyama-style layered drawing. (Usually it is the y coordinates that are fixed but that makes no difference.) Some versions of the problem (e.g. finding the exact minimum number of edge crossings) can be NP-hard but effective heuristics are known. See e.g. chapter 9 of Di Battista, Tamassia, and Eades, "Graph Drawing: Algorithms for the Visualization of Graphs", Prentice-Hall, 1999.</p>
<p>Searching Google scholar for "layered graph drawing" should also turn up some more recent references, but be careful that some of them (including mine) which use the term to mean something unrelated.</p>
|
1,189,814 | <p>Can a finitely generated module $M$ over a commutative ring have $\operatorname{Ann}(x) \neq 0$ for all $x \in M$ while $\operatorname{Ann}(M) = 0$?</p>
<p>It's not difficult to show that there is no such module if the ring is a integral domain. For general, I guess the answer is yes. But I failed to find a desired example.</p>
| Manos | 11,921 | <p>Here is a proof that no such examples exist over integral domains:</p>
<p>Let $M$ be finitely generated over an integral domain $R$.
Let $x_1,\dots,x_n$ be generators of $M$. Take $0 \neq a_i \in \operatorname{ann}(x_i)$ for
every $i=1,\dots,n$. Define $a=a_1 \cdots a_n$. Since $R$ is an integral
domain, $a\neq 0$ and $a \in \operatorname{ann}(M)$. </p>
<p>Let me now answer the question negatively in the case where
$R$ is Noetherian and $0 \neq M= J$ is an ideal of $R$. So let us
assume that all elements of $J$ have zero annihilators. Then
all elements of $J$ are zero-divisors of $R$. Let $p_1,\dots,p_s$
be the minimal prime ideals of $R$. We know that the set of
all zero-divisors of $R$ is precisely $p_1 \cup \cdots \cup p_s$.
Consequently, $ J \subset p_1 \cup \cdots \cup p_s$. Then by
prime-avoidance we must have that $J \subset p_1$ without loss of
generality. But the minimal prime ideals have the property that
$p_i = \operatorname{ann}(x_i)$ for some $0 \neq x_i \in R, \, i=1,\dots,s$.
This implies that $J \subset \operatorname{ann}(x_1)$, thus $x_1 \, J=0$ and so
$0 \neq x_1 \in \operatorname{ann}(J)$. </p>
|
2,859,463 | <blockquote>
<p>Prove or disprove. All four vertices of every regular tetrahedron in $ \mathbb{R}^3$ have at least two irrational coordinates.</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/hYrWv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hYrWv.png" alt="enter image description here"></a></p>
<p>This question arose from my inability to construct a tetrahedron in $\mathbb{R}^3$ with all the coordinates
$$
M_x,M_y,M_z,N_x,N_y,N_z,P_x,P_y,P_z,Q_x,Q_y \;\mathrm
{and}\; Q_z,
$$
of its vertices $M$, $N$, $P$ and $Q$ being rational numbers. </p>
| John WK | 551,440 | <p>I think you want to prove that there doesn't exist a regular tetrahedron such that all its vertices have only rational coordinates. Assume the contrary, let $ABCD$ be such a tetrahedron. Consider the plane $(ABC)$ as a complex plane with origin in $A$ and let the points $B$ and $C$ have associated complex numbers $a+bi$ and $c+di$. By our assumption, we would have that $a,b,c,d\in\mathbb{Q}$. We know that the points associated to complex numbers $z_1,z_2,z_3$ form an equilateral triangle if and only if $z_1+\varepsilon z_2+\varepsilon^2z_3=0$ where $\varepsilon=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ is the root of order 3 of the unity. By plugging our numbers, we obtain
$$a+bi+\varepsilon(c+di)+\varepsilon^2\cdot 0=0\Rightarrow$$
$$\Rightarrow a+bi+\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)(c+di)=0\Rightarrow$$
$$\Rightarrow a+bi-\frac{c}{2}-\frac{d}{2}i+\frac{\sqrt{3}c}{2}i-\frac{\sqrt{3}d}{2}=0\Rightarrow$$
$$\Rightarrow a-\frac{c}{2}-\frac{\sqrt{3}d}{2}=0\Rightarrow d=0$$
also
$$b-\frac{d}{2}+\frac{\sqrt{3}c}{2}=0\Rightarrow c=0$$
thus $A=C$, so we don't actually have a triangle $ABC$ hence the conclusion.</p>
|
500,973 | <p>I am presented with the question:</p>
<blockquote>
<p>The photoresist thickness in semiconductor manufacturing has a mean of
10 micrometers and a standard deviation of 1 micrometer. Assume that
the thickness is normally distributed and that the thicknesses of
different wafers are independent.</p>
<p>(a) Determine the probability that the average thickness of 10 wafers
is either greater than 11 or less than 9 micrometers.</p>
<p>(b) Determine the number of wafers that needs to be measured such that
the probability that the average thickness exceeds 11 micrometers is
0.01.</p>
<p>(c) If the mean thickness is 10 micrometers, what should the standard
deviation of thickness equal so that the probability that the average
of 10 wafers is either greater than 11 or less than 9 micrometers is
0.001?</p>
</blockquote>
<p>I thought a) was 1-normalcdf(9,11,10,1), but that answer appears to be wrong... I'm completely stumped on the others - I don't evene know where to start. Any chance someone could explain how to go about answering this question? As always, I'm trying to understand it - not just look for answers.</p>
| copper.hat | 27,978 | <p>Suppose $t_k$ are the wafer thicknesses, each has distribution ${\cal N} (\mu, \sigma^2)$, and the $t_k$ are independent.</p>
<p>Then $t_1+\cdots +t_n$ will then have distribution ${\cal N} (n\mu, n\sigma^2)$, and so the quantity $\frac{t_1+\cdots + t_n-n \mu}{\sqrt{n} \sigma}$ will have distribution ${\cal N}(0,1)$, and has CDF $\Phi$. That is, $P(\frac{t_1+\cdots t_n-n \mu}{\sqrt{n}} \le x ) = \Phi(x)$. We note that $\Phi(x) = 1-\Phi(-x)$.</p>
<p>(a) We wish to compute the probability that $t_1+\cdots +t_n \notin [n(\mu-\sigma), n(\mu+\sigma)]$, which is the same as the probability that $\frac{t_1+\cdots + t_n-n \mu}{\sqrt{n} \sigma} \notin [-\sqrt{n}, \sqrt{n}]$. Hence the probability is $\Phi(-\sqrt{10}) + (1-\Phi(\sqrt{10})) = 2 \Phi(-\sqrt{10}) \approx 0.0015$.</p>
<p>(b) First we compute the probability that $t_1+\cdots + t_n \ge n(\mu+\sigma) $, and this is the same as the probability that $\frac{t_1+\cdots + t_n-n \mu}{\sqrt{n} \sigma} \ge \sqrt{n}$, which is $1-\Phi(\sqrt{n})$. Then we compute the minimum $n$ such that $1-\Phi(\sqrt{n}) \le 0.01$, and this gives $n=6$.</p>
<p>(c) First we compute the probability that $t_1+\cdots + t_n \notin [9n,11n]$ which is the same as the probability that $\frac{t_1+\cdots + t_n-n \mu}{\sqrt{n} \sigma} \notin [-\frac{\sqrt{n}}{\sigma}, \frac{\sqrt{n}}{\sigma}]$, which is $2 \Phi( -\frac{\sqrt{n}}{\sigma})$. Then we attempt to solve $2 \Phi( -\frac{\sqrt{10}}{\sigma}) = 0.001$, which gives $\sigma \approx 0.96102$.</p>
|
3,015,149 | <p>I’m using the formula that the number of conjugacy class is given to be <span class="math-container">$\frac{1}{|G|}\sum|C_{G}(g)|$</span>, where <span class="math-container">$C_{G}(g)=\{h \in G ; gh=hg\}$</span>, which is a special result by Burnside’s theorem.</p>
<p>I found that the number of conjugacy class in <span class="math-container">$D_8$</span> is 5, so to double check, I listed down all <span class="math-container">$C_{G}(g)=\{h \in G ; gh=hg\}$</span>.</p>
<p>Let r be a rotation counter clockwise and m be a rotation in the x axis, 1 is the identity.</p>
<p>Well, <span class="math-container">$|C_G(1)|=|D_8|=16$</span>,
<span class="math-container">$|C_G(r)|= |C_G(r^2)|= |C_G(r^3)|= |C_G(r^5)|= |C_G(r^6)|= |C_G(r^7)|=|\{1,r,r^2,r^3,r^4,r^5,r^6,r^7\}|=8$</span>, </p>
<p><span class="math-container">$|C_G(r^4)|=|\{1,r,r^2,r^3,r^4,r^5,r^6,r^7,m\}|=9$</span> since <span class="math-container">$r^4m=mr^4$</span>
Similarly, we can count for the reflections;
<span class="math-container">$|C_G(m)|=|C_G(r^4m)|=|\{1,m\}|=2$</span>, while the rest of the elements with any reflections only has one element, i.e <span class="math-container">$|C_G(r^nm)|=|\{1\}|=1,n \neq 0,4$</span></p>
<p>So the problem comes that my summation is 83, while my <span class="math-container">$|G|=16$</span>. In this case I won’t get the number of conjugacy class to be 5. Did I do something wrong here? I just merely applied the definition...</p>
| Curious student | 415,928 | <p><span class="math-container">$D_n$</span>
In general its class equation given by following </p>
<p>Case 1: n is odd</p>
<p><span class="math-container">$Z(D_n)=$</span>{<span class="math-container">$e$</span>}</p>
<p><span class="math-container">$(n-1)/2$</span> classes of {<span class="math-container">$r^i,r^-i$</span>} for <span class="math-container">$1\leq i\leq (n-1)/2$</span></p>
<p><span class="math-container">$1$</span> class of {<span class="math-container">$sr^i|0\leq i \leq n-1$</span>}</p>
<p>SO class equation <span class="math-container">$n=1+2.(n-1)/2+n$</span></p>
<p>Case 2:n is even</p>
<p><span class="math-container">$Z(D_n)=$</span>{<span class="math-container">$e,r^{n/2}$</span>}</p>
<p><span class="math-container">$(n)/2-1$</span> classes of {<span class="math-container">$r^i,r^-i$</span>} for <span class="math-container">$1\leq i\leq (n)/2-1$</span></p>
<p><span class="math-container">$1$</span> class of {<span class="math-container">$sr^{2i}|0\leq i \leq n/2-1$</span>}</p>
<p><span class="math-container">$1$</span> class of {<span class="math-container">$sr^{2i+1}|0\leq i \leq n/2-1$</span>}</p>
<p>SO class equation <span class="math-container">$n=1+1+2.(n/2-1)+n/2.1+n/2.1$</span></p>
<p>In specific:
<span class="math-container">$D_8$</span> </p>
<p><span class="math-container">$1+1+2+2+2+4+4=16 $</span>is class equation</p>
|
315,386 | <p>I think Wikipedia's polar coordinate elliptical equation isn't correct. Here is my explanation: Imagine constants <span class="math-container">$a$</span> and <span class="math-container">$b$</span> in this format -
<img src="https://upload.wikimedia.org/wikipedia/commons/thumb/6/65/Ellipse_Properties_of_Directrix_and_String_Construction.svg/800px-Ellipse_Properties_of_Directrix_and_String_Construction.svg.png" alt="image"></p>
<p>Where <span class="math-container">$2a$</span> is the total height of the ellipse and <span class="math-container">$2b$</span> being the total width.
You can then find the radial length, <span class="math-container">$r$</span>, at any angle <span class="math-container">$\theta$</span> to major axis as...</p>
<p><span class="math-container">$$r(\theta) = \sqrt{(b \sin(\theta))^2 + (a \cos(\theta))^2}$$</span></p>
<p>...by just following the Pythagorean theorem. Yet Wikipedia's equation for the polar coordinate ellipse is as follows:</p>
<p><span class="math-container">$$r(\theta) = \frac{ab}{\sqrt{(b \cos(\theta))^2 + (a \sin(\theta))^2}}$$</span></p>
<p>Here is the <a href="http://en.wikipedia.org/wiki/Ellipse#Polar_form_relative_to_center" rel="noreferrer">link</a> to the Wikipedia page:
Can someone explain this, please? Why divide by the hypotenuse? Why the <span class="math-container">$ab$</span>? Thank you!</p>
| Narasimham | 95,860 | <p>EDIT1:</p>
<p>What you at first proposed as ellipse looks like:</p>
<p><a href="https://i.stack.imgur.com/eOj2D.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eOj2D.png" alt="enter image description here" /></a></p>
<p>The Ellipse parametrization is done differently. To more clearly distinguish between them we should note there are two different <span class="math-container">$\theta$</span> s, viz <span class="math-container">$\theta_{deLaHire}$</span> and the standard polar coordinate <span class="math-container">$\theta_{polar}$</span> used for central conics, ellipse in this case. We are not referring to the Newton Ellipse as there is no query about it.</p>
<p>The <em><strong>first angle</strong></em> denotes by <span class="math-container">$ \theta_{deLaHire}$</span>.</p>
<p>A radial line was constructed by deLaHire originally commences at a slightly bigger angle <span class="math-container">$\theta_{\text{deLaHire}};$</span> (red lines) each point <span class="math-container">$E$</span> on ellipse in first quadrant is reached by drawing vertical and horizontal lines from points of intersection of this polar/radial line with the two circles radii <span class="math-container">$(a,b)$</span> at <span class="math-container">$(P,Q)$</span>, to meet at E as shown.</p>
<p><span class="math-container">$$ x= a \cos\theta_{deLaHire}\; ; y=b \sin\theta_{deLaHire}\;;\tag1 $$</span></p>
<p>The <em><strong>second angle</strong></em> is used in polar coordinates in the standard ellipse and is measured from center of the circles. We call it <span class="math-container">$\theta_{polar}$</span> as usual. Green radial line.</p>
<p>For an ellipse axes <span class="math-container">$(a,b)$</span> along <span class="math-container">$(x,y)$</span> coordinate axes respectively centered at origin given Wiki expression is obtained in polar coordinates thus:</p>
<p>Plug in</p>
<p><span class="math-container">$$ x=r_{polar}\cos \theta_{polar};\, y=r_{polar}\sin \theta_{polar} ; $$</span></p>
<p>casting the standard equation of an ellipse from Cartesian form:</p>
<p><span class="math-container">$$ \left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2 =1 $$</span></p>
<p>to get</p>
<p><span class="math-container">$$ OE =r_{polar}= \frac{ab}{\sqrt{(b \cos \theta_{polar})^2 + (a \sin \theta_{polar})^2}} \tag 2 $$</span></p>
<p><a href="https://i.stack.imgur.com/mWCRe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mWCRe.png" alt="deLaHire and polar EllipseCoords" /></a></p>
<p>In either case polar angles <span class="math-container">$\theta = 0$</span> and <span class="math-container">$\theta= \pi/2$</span> reach to the same points at the ends of major and minor axes respectively. The angle variations are plotted showing by comparison that starting deLaHire polar line is inclined more than (or equals to at extreme axes) the Central polar coordinate always. Can you figure out errors in the second and fourth quadrants?</p>
<p>Sketched Ellipse dimensions are <span class="math-container">$(a=5,b=3,e=0.8)$</span>.</p>
<p><a href="https://i.stack.imgur.com/ygNwI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ygNwI.png" alt="PolarCoordsStandard" /></a></p>
|
4,020,464 | <p>So <span class="math-container">$Ax = λx$</span>;</p>
<p><span class="math-container">$A(Ax) = λ(Ax) \to (A^2)x = (λ^2)x$</span></p>
<p>I kind of dont know how to get the <span class="math-container">$1$</span> and <span class="math-container">$I$</span> here...</p>
<p>Any help is appreciated</p>
| Michael Cohen | 651,310 | <p>Given independent random samples <span class="math-container">$X_{1}, X_{2}, \dots, X_{m}$</span> taken from a normal random variable with mean <span class="math-container">$\mu$</span> and variance <span class="math-container">$\sigma^2$</span>, it is a standard result that the sample mean <span class="math-container">$\bar{X}$</span> is normally distributed with mean <span class="math-container">$\mu$</span> and variance <span class="math-container">$\sigma^2/m$</span>. It follows that in your example <span class="math-container">$\bar{X}$</span> is normally distributed with mean <span class="math-container">$\mu_{1}$</span> and variance <span class="math-container">$\sigma^2_{1}/m$</span> and <span class="math-container">$\bar{Y}$</span> is normally distributed with mean <span class="math-container">$\mu_{2}$</span> and variance <span class="math-container">$\sigma^2_{2}/n$</span>. It then follows that the difference
<span class="math-container">$\bar{X} - \bar{Y}$</span> is normal with mean <span class="math-container">$(\mu_{1} -\mu_{2})$</span> and variance <span class="math-container">$(\sigma^2_{1}/m + \sigma^2_{2}/n)$</span>. As <span class="math-container">$m=n$</span> and with the given values we find that <span class="math-container">$\bar{X} - \bar{Y}$</span> has variance <span class="math-container">$(4.5/n)$</span>. After standardizing <span class="math-container">$P(-1<(\bar{X} - \bar{Y})-(\mu_{1} -\mu_{2})<1)=0.95$</span> we get <span class="math-container">$P((-1/\sqrt(4.5/n))- Z-(1/\sqrt(4.5/n)) = 0.95$</span> where <span class="math-container">$Z$</span> is the standard normal variable. The rest should be clear I hope.</p>
|
1,639,232 | <p>A really simple question, but I thought I'd ask anyway. Does $n<x^n<(n+1)$ imply $\sqrt[n] n < x < \sqrt[n] {n+1}$?</p>
<p>Thank you very much.</p>
| DanielV | 97,045 | <pre><code>Number[] a = new Number[p + 1]; // range from 1 to p
... some stuff to initialize a ...
Number[] ceps = new Number[Q + 1]; // range from 0 to Q
ceps[0] = ln(G);
for (int q = 1; q <= p; q++) {
Number sum = a[q];
for (k = 1; k <= q - 1; k++) {
sum += (k - q) / q * a[k] * ceps[q - k];
}
ceps[q] = sum;
}
for (int q = p + 1; q <= Q; q++) {
Number sum = 0;
for (int k = 1; k <= p; k++) {
sum += (k - q) / q * a[k] * ceps[q - k];
}
ceps[q] = sum;
}
</code></pre>
<p>Edit: </p>
<pre><code>(k - q) / q * a[k] * ceps[q - k]
</code></pre>
<p>should really be </p>
<pre><code>((Number)(k - q))/((Number)q) * a[k] * ceps[q - k]
</code></pre>
<p>or </p>
<pre><code>a[k] * ceps[q - k] * (k - q) / q
</code></pre>
<p>To avoid integer division.</p>
|
1,746,776 | <p>I am wondering how I could solve the integral </p>
<p>$$\iiint \frac{1-e^{-(x^2+y^2+z^2)}}{[x^2+y^2+z^2]^{2}}$$</p>
<p>over $\mathbb{R}^{3}$</p>
<p>I thought maybe I could break it up into three single integrals and multiply or something. I think it is not supposed to be difficult to solve. How should it be approached?</p>
<p>Thank you</p>
| Olivier Oloa | 118,798 | <p><strong>Hint</strong>. By the use of a symetry and by use of spherical coordinates
$$
\begin{align}
x&=r\sin\theta\cos\phi\\
y&=r\sin\theta\sin\phi\\
z&=r\cos\theta
\end{align}
$$ one gets a jacobian equal to $r^2\sin \theta \:d\phi \:d\theta\: dr$ giving
$$
\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \dfrac{1-e^{-\left(x^2+y^2+z^2\right)}}{(x^2+y^2+z^2)^2}dxdydz=8\int_0^{\pi/2}\int_0^{\pi/2}\int_0^\infty\dfrac{1-e^{-r^2}}{r^2}dr \sin \theta \:d\phi \:d\theta
$$ Integrating by parts and using the gaussian integral gives
$$
\int_0^\infty\dfrac{1-e^{-r^2}}{r^2}dr =\sqrt{\pi}
$$ leading to a conclusion.</p>
|
3,203,678 | <blockquote>
<p>Find all prime numbers <span class="math-container">$p$</span>, for which there are positive integers <span class="math-container">$m$</span> and <span class="math-container">$n$</span> such that <span class="math-container">$p=m^2+n^2$</span> and <span class="math-container">$p \mid m^3+n^3-4$</span>.</p>
</blockquote>
<p>I simplified this a little bit.</p>
<p><span class="math-container">$$m^2+n^2 \mid m^3+n^3-4 =(m+n)(m^2+n^2-mn)-4 \\ \Longrightarrow m^2+n^2 \mid (m+n)mn+4 $$</span></p>
<p>The only case that <span class="math-container">$m$</span> and <span class="math-container">$n$</span> can both be odd is <span class="math-container">$m=n=1$</span>, which leads to <span class="math-container">$p=2$</span>. If one of <span class="math-container">$m$</span> and <span class="math-container">$n$</span> is <span class="math-container">$1$</span> (For example <span class="math-container">$n=1$</span>), then <span class="math-container">$p=m^2+1$</span> and </p>
<p><span class="math-container">$$m^2+1 \mid m^2+m+4 \Longrightarrow m^2+1 \mid m+3 \Longrightarrow m=2$$</span>
Which gives <span class="math-container">$p=5$</span>.</p>
<p>For <span class="math-container">$m,n>1$</span> with different parity, I could not find a strong argument. Can you guys give it a try?</p>
| W-t-P | 181,098 | <p>There are no primes with this property other than <span class="math-container">$p=2$</span> and <span class="math-container">$p=5$</span>; here is a proof.</p>
<p>Suppose that <span class="math-container">$m,n>1$</span> and <span class="math-container">$p=m^2+n^2$</span> is a prime dividing <span class="math-container">$m^3+n^3-4$</span>. As you have observed, we have then
<span class="math-container">$$ mn(m+n) \equiv -4 \pmod p. $$</span>
We apply two operations to this congruence: squaring and multiplying by <span class="math-container">$m+n$</span>; keeping in mind that <span class="math-container">$(m+n)^2\equiv 2mn\pmod p$</span>, this gives
<span class="math-container">\begin{align*}
2m^3n^3 &\equiv 16\pmod p, \\
2m^2n^2 &\equiv -4(m+n)\pmod p;
\end{align*}</span><br>
that is,
<span class="math-container">\begin{align*}
m^3n^3 &\equiv 8\pmod p, \tag{1} \\
m^2n^2 &\equiv -2(m+n)\pmod p. \tag{2}
\end{align*}</span><br>
In view of <span class="math-container">$m^3n^3-8=(mn-2)(m^2n^2+2mn+4)$</span>, from (1) we conclude that either <span class="math-container">$mn\equiv 2\pmod p$</span>, or <span class="math-container">$m^2n^2+2mn+4\equiv 0\pmod p$</span>. In the former case (2) leads to <span class="math-container">$p\mid m+n+2$</span>, and as a result to <span class="math-container">$p=m^2+n^2\le m+n+2$</span>, which is easily seen to be possible only if <span class="math-container">$m=2$</span> and <span class="math-container">$n=1$</span>, or vice versa, contradicting the assumption <span class="math-container">$m,n>1$</span>. Suppose thus that
<span class="math-container">$$ m^2n^2+2mn+4\equiv 0\pmod p. $$</span>
Combined with (2), this gives
<span class="math-container">$$ m+n \equiv mn+2\pmod p, $$</span>
whence
<span class="math-container">$$ p = m^2+n^2 \le mn-m-n+2. $$</span>
But, recalling that <span class="math-container">$m,n>1$</span>,
<span class="math-container">$$ m^2+n^2 \ge 2mn = mn - m - n + 2 + ((m+1)(n+1) - 3) > mn -m -n + 2, $$</span>
a contradiction.</p>
|
1,557,353 | <p>Context: I'm taking calc based physics, and we are supposed to be able to integrate moment of inertia for a cylinder. I referenced a mit vid, and though I have no education on multiple integrals, I got all but one thing. <a href="https://www.youtube.com/watch?v=iYFogDTPlRo" rel="nofollow">https://www.youtube.com/watch?v=iYFogDTPlRo</a></p>
<p>My progress is:</p>
<p>1) $\int r^2 dm$</p>
<p>2) $ \int r^2 \delta dV $ because $ \delta = \frac{dm}{dV} $ -> $dm = \delta dV$ where $\delta$ is density</p>
<p>3) $\delta \int_0^{2\pi} \int_0^r \int_0^h r^2 * r dzdrd\theta$ <- I dont fully understand this step.</p>
<p>Can someone explain why $dV = r dz dr d\theta$. Can it be justified in a similar manner as step 2, perhaps in multiple parts even? I get the idea I believe, but can't justify it.</p>
<p>EDIT:
Thank you everyone. I don't have enough rep to upvote, but the answers given by David, Hamed, B. Pasternak, and root were all helpful. I have a pretty good understanding now.</p>
| David | 119,775 | <p>Here is an intuitive (though not rigorous) geometrical argument.</p>
<p>Consider the volume element you get if you start at a point $(r,\theta,z)$ and increase $r$ by $dr$ and increase $\theta$ by $d\theta$ and increase $z$ by $dz$. You get a "curved box" where one side is a line with length $dr$, one is a line with length $dz$ and one is an arc of a circle with radius $r$, angle $d\theta$ and therefore length $r\,d\theta$. If this were an actual rectangular box (not curved), its volume would be
$$dV=r\,dr\,d\theta\,dz\ ;$$
but as $dr$, $d\theta$ and $dz$ become vanishingly small, this will become a better and better approximation to the volume of the "curved box".</p>
<p>Another point worth considering: the "obvious" answer $dV=dr\,d\theta\,dz$ cannot be right as it is dimensionally incorrect: the right hand side is an area, not a volume.</p>
|
2,812,314 | <p>For integers $n\geq 1$ we denote the square-free kernel as $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p,$$ that is the product of distinct primes dividing an integer $n>1$ with the definition $\operatorname{rad}(1)=1$ (the Wikipedia's article dedicated to this multiplicative function is <a href="https://en.wikipedia.org/wiki/Radical_of_an_integer" rel="nofollow noreferrer"><em>Radical of an integer</em></a>).</p>
<p>An integer $n\geq 1$ <strong>is said to be an odd perfect number</strong> if $n$ is an odd integer and satisfies $$\sigma(n)=2n,$$
where $$\sigma(m)=\sum_{d\mid m}d=\prod_{\substack{p\mid m\\p\text{ prime}}}\frac{p^{e_p+1}-1}{p-1}$$ is the sum of the positive divisors of $m=\prod_{\substack{p\mid m\\p\text{ prime}}}p^{e_p}$. </p>
<p>It is well-known that if there exists an odd perfect number (it is an open problem) then has a specific form, I mean mainly the Euler's theorem for odd perfect numbers but many results are known about odd perfect numbers, for example Touchard's theorem or theorems about their prime factorizations. In particular are known upper bounds for $$\operatorname{rad}(n)\leq \text{Upper bound}\tag{1}$$
where here in $(1)$ with $\text{Upper bound}=\text{Upper bound}(n)$ we denote a function of $n$. See Proposition 1 of [1], or [2] (on assumption of certain conditions that satisfy the odd perfect numbers).</p>
<blockquote>
<p><strong>Question.</strong> I would like to know if it is possible to deduce some statement of the form $$\text{Lower bound}\leq \operatorname{rad}(n),\tag{2}$$
where here in $(2)$ with $\text{Lower bound}=\text{Lower bound}(n)$ we denote a function of $n$ an odd perfect number.</p>
<p>Please if you know literature about it provide the references answering my question as a reference request and I try to find and read those propositions. In case that isn't in the literature, can work can be done about $(2)$ unconditionally or on assumption that our odd perfect number $n$ does satisfy any conditions? <strong>Many thanks.</strong></p>
</blockquote>
<h2>References:</h2>
<p>[1] Florian Luca and Carl Pomerance, <em>On the radical of a perfect number</em>, The New York Journal of Mathematics, Volume: 16 (2010), page 23-30.</p>
<p>[2] Ph. Ellia, <em>A remark on the radical of a perfect numbers</em>, The Fibonacci Quarterly VOLUME 50, NUMBER 3 (AUGUST 2012).</p>
| José Carlos Santos | 446,262 | <p>You could note that<span class="math-container">$$\sum_{n=1}^\infty\frac{n-2n\sqrt n+n^2}{n^3}=\sum_{n=1}^\infty\frac{1-2\sqrt n+n}{n^2}$$</span>and that<span class="math-container">$$\lim_{n\to\infty}\frac{\frac{1-2\sqrt n+n}{n^2}}{\frac1n}=\lim_{n\to\infty}\frac{1-2\sqrt n+n}n=1.$$</span>Therefore, your series diveres, by the comparison test.</p>
|
720,282 | <p>I was trying to come up with reasons, why we naturally consider the topology of uniform convergence on compact sets as the appropriate framework for spaces of holomorphic functions such as e.g. $H(\mathbb{C}^n)$ (which is the space of entire functions on $\mathbb{C}^n$).</p>
<p>I understand that e.g. by Weierstrass' theorem the above space is closed, when $f_n$ converges compact to f. Moreover, that we can make such spaces into Fréchet spaces in case the domain behaves nicely enough (e.g. is the union of countable many compact sets, which clearly is the case for $\mathbb{C}^n$), which provides some nice topological properties.</p>
<p>But I would like to hear some more motivation or reasons, why to consider this topology naturally, which stem (purely) from complex analysis.</p>
<p>Thanks in advance.</p>
<p>PS: this is my first question I ask here, so please don't be to hard on me, if it doesn't belong here.</p>
| fgp | 42,986 | <p>I'd say being closed alone is sufficient reason. Usually the <em>main</em> reason for imbueing a topology onto a space is to be able to reason about convergence. Non-closed spaces are burdensome then, because there will be sequences which <em>look</em> like they converge (for metrizable spaces, think cauchy sequences), but don't because the limit element is "missing".</p>
|
1,182,432 | <p>Is it possible, that everyone is a pseudo-winner in a tournament with 25 people?
(pseudo-winner means that either he won against everyone, or if he lost against someone, then he beated someone else, who beated the one who he lost to).</p>
<p>In the "language" of graph theory: Is it possible in the directed K(25) (all edges drawn with 25 vertices), that from all vertices, every vertices can be reached with a maximum of 2-long route.</p>
<p>I thought about this, and my answer is no. I tried to do the proving, starting with K(3), which is a triangle. Triangle works, but I don't know how to continue this until 25 vertices.</p>
<p>Maybe any easier solutions?</p>
<p>Thanks!</p>
| Leader47 | 214,065 | <p>We can prove that this is possible for any odd integer $2n+1$ by induction. </p>
<p>For $n=1$ it is possible. </p>
<p>If it is possible for $n\ge 1$ Let $G$ have $2(n+1)+1$ vertices. Let $u,v$ be some $2$ vertices and the others are $v_1,v_2,...,v_m$ where $m=2n+1$ Connect $v_1,...,v_m$ by induction. so that from ever vertex there is path of length at most $2$ to any other. Now draw directed edges from $u->v_i$ and $v_i->v$ for $i=1,...,m$ and $v->u$. This graph has all pseudo-winners and has $2(n+1)+1$ vertices thus induction is complete. </p>
<p>Thus for $25$ it is possible aswell.</p>
|
2,010,434 | <p>Say we are given n piles of stones.
Sizes are $s_{1}, s_{2}, .. , s_{n}$, they can be any positive integer numbers.
The game is played by two players, they alternate their moves.
The allowed moves are:
1. Take exactly 1 stone from 1 pile.
2. Take all stones from 1 pile.</p>
<p>Wins the player who mades the last move.
Both play optimally.</p>
<p>Is it possible to find the winner for some state?
I was thinking about Grundy numbers, but stuck with definitions and do not know can it be applied here. </p>
| DLIN | 355,583 | <p>We can consider the positive case, i.e. $a,b,c>0$.</p>
<p>We can categorize into the 3 cases(from simple to copmlex):</p>
<p>1): $a=b=c$. Then $\frac1a+\frac1b+\frac1c=\frac3a\in\mathbb Z$, so $a|3$, hence $a=1,3$. </p>
<p>2): $0<a<b<c$. Then $1\leq\frac1a+\frac1b+\frac1c<\frac3a$, $1\leq a<3$, So $a=1,2$.</p>
<p>If $a=1$, then $\frac1b+\frac1c<\frac22=1$, impossible, so $a=2.$</p>
<p>Then $b\geq3$(since $b>a$), and
$$1\leq\frac1a+\frac1b+\frac1c<\frac12+\frac2b\leq\frac12+\frac23<2,$$
thus the sum equals to $1$. Moreover, if $b\geq4$, then $\frac1b+\frac1c<\frac24$, which means the sum is not an integer.
So $a=2,~b=3~,\Rightarrow c=6.$</p>
<p>3): $a=b<c$. By the method of 2), we have $1\leq\frac2a+\frac1c<\frac3a,~\Rightarrow a<3$.</p>
<p>If $a=1$, then $\frac1c<1$, impossible. </p>
<p>If $a=2$, $\frac1c<1$, still impossible. </p>
<p>Hence, in this case, there is no result.</p>
<p>4): $a=b>c$. By the above argument, $1\leq\frac2a+\frac1c<\frac3c$.
$c=1,2$.</p>
<p>If $c=1$, then $a=2$.</p>
<p>If $c=2$, $\frac2a\in \mathbb Z+\frac12$, and $a\geq3$. So $a=4$.</p>
<p><strong>PS</strong>: For the non-positive case, there are infinite solutions. </p>
<p>e.g. $a=b=2n,~c=-n$, where $n\in \mathbb N^+.$</p>
|
28,104 | <p>It occured to me that the Sieve of Eratosthenes eventually generates the same prime numbers, independently of the ones chosen at the beginning. (We generally start with 2, but we could chose any finite set of integers >= 2, and it would still end up generating the same primes, after a "recalibrating" phase).</p>
<p>If I take 3 and 4 as my first primes, starting at 5:</p>
<ul>
<li>5 is prime,</li>
<li>6 is not (twice 3),</li>
<li>7 is prime,</li>
<li>8 is not,</li>
<li>9 is not,</li>
<li>10 is not,</li>
<li>etc.</li>
</ul>
<p>Eventually, I will find all the primes as if I had started with 2 only.</p>
<p>To me, it means that we can generate big prime numbers without any knowledge of their predecessors (until a certain point). If I want primes higher than 1000000, then I can generate them without any knowledge of primes under, say, 1000. (It may not be as effective computationnally, but I find this philosophically interesting.)</p>
<p>Is this result already known ? </p>
<p>Are there any known implications ?</p>
<p><strong>Edit</strong></p>
<p>The number from which we are assured to get the right primes again is after the square of the last missing natural prime.</p>
<p>That is, if I start with a seed set containing only the number 12, the highest missing natural prime is 11, so I'll end up having 121 as my last not-natural prime.
Non-natural primes found are 12,14,15,16,18,20,21,22,25,27,33,35,49,55,77 and 121.</p>
<p>This is a bit better than what I thought at first (namely, as stated below, somewhere under the square of the highest seed).</p>
| Theo Johnson-Freyd | 78 | <p>So, let's see if I can precisify your claim. We start with a finite "seed set" that we assert to be Ghi-Om-prime (the seed set must not contain 1). Numbers smaller than the largest seed we completely ignore. Now for numbers larger than any seed prime, we run the Seive. You claim that there is some cut-off, depending of course on the seed set but hopefully not growing too quickly, so that after that cut-off, the primes match the primes in the seed set.</p>
<p>Recall how the Seive works. A number is GO-prime iff it is not divisible by any smaller GO-prime. Conversely, the GO-composites are precisely those numbers that are divisible by some smaller number that is either in your seed set or larger than the largest number in the seed set. In particular, every GO-composite is a true composite. So suppose there is a true composite that is not GO-composite. Then all its proper divisors are less than the largest seed. But every non-prime has a proper divisor that is at least as large as its square root. So the largest true composite that is not GO-composite is strictly less than the square of the largest seed.</p>
<p>This proves your claim, with the bound you suggested in the comment to Noldorin.</p>
<p>In any case, some remarks. First, if we state the rules for the Sieve correctly, and in particular disallow 1 as either prime or composite, then the true primes are precisely the GO primes where the seed set is empty.</p>
<p>Second, although I haven't seen this particular variation of the Sieve before, as I said in the comments to the question, you should check out <a href="http://www.inference.phy.cam.ac.uk/sanjoy/thesis/thesis-letter.pdf" rel="nofollow">Sanjoy Mahajan's thesis (PDF)</a>. In the last chapter, he proposes the following probabilistic variation of the Sieve. Namely, let's define the probability that $p$ SM-divides $n$ to be $1/p$. Then the probability that $n$ is SM-prime is
$$ \prod_{p < \sqrt n} (\text{probability that $p$ is SM-prime})\times (\text{probability that $p$ does not SM-divide $n$})$$
The point is that SM-primality is nicely smooth, rather than exhibiting the strange jumps in the true prime spectrum, but still approximates the spectrum well for certain types of estimates. Sanjoy also considers a few other probabilistic models (whether to stop the product at $\sqrt n$ or $n$, for example), including some where he seeds the model with some early assertions of SM-primality. Sanjoy is a physicist, and so does not work at mathematicians' level of rigor, and says as much: the point of his model is to lead to new conjectures about distribution of true primes.</p>
|
2,138,241 | <p>I tried to prove $$\lim_{x\to \infty}\frac 1x = 0$$
I started as thus
$$\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac x{x^2}$$
Applying <a href="https://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule" rel="nofollow noreferrer">L'Hospital's Rule</a>
$$\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac x{x^2}=\lim_{x\to \infty}\frac 1{2x}=\frac12\lim_{x\to \infty}\frac 1x$$
Thus,
$$\frac12\lim_{x\to \infty}\frac 1x=\lim_{x\to \infty}\frac 1x$$
which therefore implies
$$\lim_{x\to \infty}\frac 1x = 0$$
QED.</p>
| Ethan Bolker | 72,858 | <p>Not an answer - essentially a comment and too long for a comment that I don't want lost in the flurry of existing comments.</p>
<p>Many students try L'Hopital unthinkingly when faced with the limit of an indeterminate form like $0/0$. Often the application is incorrect. Even when it works it's often not the easiest method, and it's rarely the most illuminating. You learn much more thinking about simple order of magnitude inequalities or the first few terms of Taylor series expansions.</p>
<p>There are many answers on this site that illustrate that. Here are some; other answerers should feel free to edit this answer to link to more. </p>
<p><a href="https://math.stackexchange.com/questions/1461433/lhopitals-displaystyle-lim-x-rightarrow-infty-ln-x3-x">lHopitals $ \displaystyle \lim_{x\rightarrow \infty} \; (\ln x)^{3 x} $?</a></p>
<p><a href="https://math.stackexchange.com/questions/98533/finding-the-limit-of-a-function-with-a-trigonometric-exponent">Finding the limit of a function with a trigonometric exponent</a></p>
<p><a href="https://math.stackexchange.com/questions/2068098/computing-lim-x-to0-frac-8-x8-left-1-cos-fracx2-2-cos-fracx">Computing $\lim_{x\to0} \frac 8 {x^8} \left[ 1 - \cos\frac{x^2} 2 - \cos\frac{x^2}4 + \cos\frac{x^2}2\cos\frac{x^2}4 \right]$ without using L'Hospital</a></p>
<p><a href="https://math.stackexchange.com/questions/2128621/find-the-limit-lim-x-to0-frac-arcsin-x-xx2">Find the limit $\lim_{x\to0}\frac{\arcsin x -x}{x^2}$</a></p>
|
3,700,814 | <p>Consider the vector space <span class="math-container">$V=C(\mathbb{R},\mathbb{R})$</span> and <span class="math-container">$V\ni U=\{f\in C(\mathbb{R},\mathbb{R}) |f(0)=0\}$</span>. I want to find a complement of <span class="math-container">$U$</span>, such that <span class="math-container">$V=U\oplus W$</span>. This condition is the same as finding a set <span class="math-container">$W$</span> that satisfies <span class="math-container">$V=U+W$</span> and <span class="math-container">$U\cap W={f_0}$</span> where <span class="math-container">$f_0$</span> is the null element. At a lecture, the professor defined <span class="math-container">$W=Span({\mathbb{\mathbf{1}}})$</span>, where <span class="math-container">$\mathbf{1}$</span> is the function defined by <span class="math-container">$\mathbf{1}(x)=1\ \forall x\in \mathbb{R}$</span> and proved that this is a complement. He did not, however, give an explanation of how intuitively to find this complement without knowing beforehand that it indeed does satisfy the conditions. </p>
<p>Looking at the definition of <span class="math-container">$U$</span>, I would rather define <span class="math-container">$W$</span> as:
<span class="math-container">$$
W=\{f \in C(\mathbb{R}, \mathbb{R}): f(0) \neq 0\} \cup\left\{f_{0}\right\}
$$</span>
But I would imagine that if it were that simple, the professor would have done the same. Is there something wrong with my definition?</p>
| Rick | 601,449 | <p>The issue is that the condition <span class="math-container">$V=U\oplus W$</span> requires <span class="math-container">$V$</span> and <span class="math-container">$W$</span> to be <em>vector subspaces</em>, as oposed to just <em>sets</em> as you say; in particular they have to be closed under vector addition. Having this in mind, if you define <span class="math-container">$$W=\{f \in C(\mathbb{R}, \mathbb{R}): f(0) \neq 0\} \cup\left\{f_{0}\right\},$$</span>
then it's clear that this is not a vector space. To see this consider <span class="math-container">$f(x) = x^2+1$</span> and <span class="math-container">$g(x) = -1$</span> for all <span class="math-container">$x \in \mathbb R$</span>; then <span class="math-container">$f,g \in W$</span>, and if <span class="math-container">$W$</span> were indeed a vector space one would have <span class="math-container">$f + g \in W$</span>, yet <span class="math-container">$$(f+g)(x) = (x^2+1)+ (-1)=x^2,$$</span> which is not in <span class="math-container">$W$</span>. </p>
|
2,784,784 | <p>Let $f(x)=$$x-1 |x \in \mathbb{Q} \brace 5-x| x \in \mathbb{Q}^c$</p>
<p>Show that $\lim_{x \to a}f(x)$ does not exists for any $a \not= 3$</p>
<p>I first showed that $lim_{x \to 3}f(x)=2$. </p>
<p>I don't know how to approach this part. Can anyone please guide? I was thinking of using density theorem at first to show that there will exist sequences of rationals and irrationals that will approach a but their limits would not equal but that is problematic as I don't really know the limit.</p>
<p>thank you</p>
| Fred | 380,717 | <p>Let $a \ne 3$.</p>
<p>If $(x_n)$ is a sequence in $ \mathbb Q$ with $x_n \to a$, then $f(x_n)=x_n-1 \to a-1$.</p>
<p>If $(y_n)$ is a sequence in $ \mathbb Q^c$ with $y_n \to a$, then $f(x_n)=5-y_n \to 5-a$.</p>
<p>Since $a \ne 3$ we have $a-1 \ne 5-a$, hence $\lim_{x \to a}f(x)$ does not exist.</p>
|
2,495,555 | <p>I want to use Konig's theorem to show that the pictured graph $G$ has no perfect matching. By this theorem it suffices to find a vertex cover of size $|G|/2-1= 20$, but so far I have only been able to find vertex covers of size 21. I'm just doing this by inspection as opposed to using any algorithms, and it's not immediately obvious to me how to find a cover of size 20.</p>
<p><a href="https://i.stack.imgur.com/3gsoV.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/3gsoV.jpg" alt="enter image description here"></a></p>
| Misha Lavrov | 383,078 | <p>If you don't want to just stumble around blindly until you get there, here's how you find the solution.</p>
<p>The solution you found with $21$ vertices probably looks something like this:</p>
<p><a href="https://i.stack.imgur.com/DNNc5.png" rel="noreferrer"><img src="https://i.stack.imgur.com/DNNc5.png" alt="Naive solution"></a></p>
<p>If this is the optimal solution, then there is a perfect matching in which every edge has one endpoint among the chosen vertices. But in fact, if we try to find such a matching, we get irreparably stuck:</p>
<p><a href="https://i.stack.imgur.com/4e4Gu.png" rel="noreferrer"><img src="https://i.stack.imgur.com/4e4Gu.png" alt="Naive solution gets stuck"></a></p>
<p>We started by giving a bunch of vertices an edge, but the purple vertex has no edge to an unused neighbor. There's no way to fix this, because the $11$ vertices we've looked at so far (the $10$ we've given an edge, plus the purple vertex) have only $10$ neighbors... but that's exactly what we want to improve the vertex cover. Just replace these $11$ vertices by their $10$ neighbors, and you get a solution of size $20$:</p>
<p><a href="https://i.stack.imgur.com/d0W2w.png" rel="noreferrer"><img src="https://i.stack.imgur.com/d0W2w.png" alt="Improved solution"></a></p>
|
1,016,884 | <p>Four friends, Andrew, Bob, Chris and David, all have different heights. The sum of their heights is 670 cm.
Andrew is 8cm taller than Chris and Bob is 4cm shorter than David.
The sum of the heights of the tallest and shortest of the friends is 2cm more than the sum of the heights of the other two.
Find the height of each friend.</p>
| Jimmy R. | 128,037 | <p>You are given that $$\begin{cases}A=C+8\\B=D-4\\A+B+C+D=670\\A\neq B\neq C\neq D\end{cases}$$ Substituting the first two in the third you obtain $$C+D=333$$ therefore
$$\begin{cases}A=C+8\\B=329-C\\D=333-C\\A\neq B\neq C\neq D\end{cases}$$</p>
<p>Now compare the sums by pairs to check which pairs satisfy the condition in the second to last sentence (i.e. that sums differ by 2): $$A+B=337 \qquad \text{ vs } \qquad C+D=333 \qquad 337-333=4\neq2\text{ no }$$ $$A+D=341 \qquad \text{ vs } \qquad B+C=329 \qquad 341-329=12\neq2\text{ no }$$ So, since $A>C$ the only possibility is that $A$ is the tallest and $D$ the shortest $$A+C=2C+8 \qquad \text{ vs } \qquad B+D=662-2C \qquad \text{ yes }$$ which gives $$2C+8=662-2C+2 \implies 4C=656 \implies C=164$$</p>
|
2,570,008 | <p>For what value of a will this equation have only one real root: </p>
<p>$$(2a−5)x^2−2(a−1)x+3=0$$</p>
<p>Note: $x$ is a variable</p>
<p>If found that $a=4$ works, but there seems to be another solution. Any help?</p>
| Andrei | 331,661 | <p>You can have only one solution if the discriminant is $0$, or if the coefficient of the $x^2$ term is $0$. $a=5/2$</p>
|
3,748,392 | <blockquote>
<p>Let <span class="math-container">$G=\mathbb{Z}_{7}\rtimes_{\rho}\mathbb{Z}_{6}$</span> with <span class="math-container">$|\ker\rho| = 2$</span>. How many <span class="math-container">$3$</span>-Sylow subgroups are there in <span class="math-container">$G$</span>?</p>
</blockquote>
<p>I know that the number is <span class="math-container">$1$</span> or <span class="math-container">$7$</span>, but I'm stuck.</p>
| markvs | 454,915 | <p>If the number of Sylow <span class="math-container">$3$</span>-subgroups is <span class="math-container">$1$</span>, then the Sylow <span class="math-container">$3$</span>-subgroup is normal and trivially intersects the normal Sylow <span class="math-container">$7$</span>-subgroup, whence it commutes with the Sylow <span class="math-container">$7$</span>-subgroup element-wise. So it is contained in the kernel of <span class="math-container">$\rho$</span>, a contradiction. So the number is <span class="math-container">$7$</span>.</p>
|
3,254,290 | <p>Consider two random variables <span class="math-container">$X,Y$</span>. <span class="math-container">$Y$</span> has support <span class="math-container">$\mathcal{Y}\equiv \mathcal{Y}_1\cup \mathcal{Y}_2$</span>. <span class="math-container">$X$</span> has support <span class="math-container">$\mathcal{X}$</span>. <span class="math-container">$\mathcal{X}$</span> and <span class="math-container">$\mathcal{Y}$</span> are finite sets.</p>
<p>Take a function <span class="math-container">$f(X,Y)$</span> and a parameter <span class="math-container">$\beta$</span>. Suppose that
<span class="math-container">$$
E(f(X,y)-\beta)\geq 0 \text{ }\forall y\in \mathcal{Y}_1
$$</span></p>
<p>The textbook asks me to show that this implies
<span class="math-container">$$
E(f(X,Y)-\beta| Y\in \mathcal{Y}_1)\geq 0
$$</span></p>
<p>I would like your help to understand </p>
<p>1) What is <span class="math-container">$E(f(X,Y)| Y\in \mathcal{Y}_1)$</span>? I know that <span class="math-container">$\forall y \in \mathcal{Y}_1$</span>
<span class="math-container">$$
E(f(X,Y)| Y=y)\equiv \sum_{x\in \mathcal{X}} f(x,y)P(X=x| Y=y)
$$</span>
but I don't understand how <span class="math-container">$E(f(X,Y)| Y\in \mathcal{Y}_1)$</span> is defined.</p>
<p>2) How to get that implication.</p>
| Gautam Shenoy | 35,983 | <p>Hint: When faced with a random variable <span class="math-container">$X$</span> and an event <span class="math-container">$A$</span>, the following may help:
<span class="math-container">$$E[X|A] = \frac{E[X1_{A}]}{P(A)}$$</span>
where <span class="math-container">$1_{()}$</span> is the indicator function.
In your case replace <span class="math-container">$X$</span> with <span class="math-container">$f(X,Y)$</span> and <span class="math-container">$A$</span> with <span class="math-container">$\{y \in \mathcal{Y}\}$</span>.</p>
<p>Let me know if you need more help.</p>
|
405,648 | <p>Is there a sensible characterization of groups <span class="math-container">$G$</span> with the following property?</p>
<blockquote>
<p>Every extension of groups <span class="math-container">$1\to G\to H\to K\to 1$</span> is split.</p>
</blockquote>
<p>A complete group <span class="math-container">$G$</span> has that property and in fact such a group has a <em>normal</em> complement in every group that contains it as a normal subgroup (moreover, completeness is characterized by this)</p>
<p>For comparison, a group <span class="math-container">$K$</span> has the property that every extension <span class="math-container">$1\to G\to H\to K\to 1$</span> is split iff it is free (there is such an extension with <span class="math-container">$H$</span> free, and the splitting map <span class="math-container">$K\to H$</span> is injective, so gives an isomorphism of <span class="math-container">$K$</span> with a subgroup of <span class="math-container">$H$</span>, which is free by the Nielsen–Schreier theorem)</p>
<p><strong>NB:</strong> the title does use the old meaning of «extension of a group»...</p>
| Brauer Suzuki | 332,108 | <p>Since my late comment to YCor's answer is easily overlooked, I allow myself to repeat it here: The question was answered in [J. S. Rose, Splitting properties of group extensions, Proc. London Math. Soc. (3) 22 (1971), 1–23] with exactly the same outcome as in YCor's answer.</p>
|
2,936,608 | <blockquote>
<p>Let <span class="math-container">$a_n$</span> where <span class="math-container">$n \in \mathbb {N}$</span> be a sequence of rational numbers converging to <span class="math-container">$a$</span>. Suppose <span class="math-container">$a \neq 0$</span>, for <span class="math-container">$k = 1, 2, ...$</span> let
<span class="math-container">$$b_k=\begin{cases} 0 & \text{if}\;a_k=0\\\\\frac{1}{a_k} &\text{if}\;a_k \neq 0\end{cases}$$</span>
Prove that <span class="math-container">$b_n$</span> converges to <span class="math-container">$\frac{1}{a}$</span>.</p>
</blockquote>
<p>I was studying real analysis and got stuck on this problem.
Can you help me solve this problem or give me some hints?</p>
<p>Thanks</p>
<p>edit: is it possible to solve this in terms of Cauchy Sequence?</p>
| TheSilverDoe | 594,484 | <p>It is not necessary to use Cauchy sequences.</p>
<p>Let <span class="math-container">$\varepsilon = \frac{|a|}{2} > 0$</span> (because <span class="math-container">$a \neq 0)$</span>. The sequence <span class="math-container">$(a_k)$</span> converges to <span class="math-container">$a$</span>, so there exists <span class="math-container">$N \in \mathbb{N}$</span>, such that for all <span class="math-container">$k \geq N$</span>, <span class="math-container">$a-\varepsilon < a_k < a +\varepsilon$</span>. By definition of <span class="math-container">$\varepsilon$</span>, you get <span class="math-container">$a_k \neq 0$</span> for all <span class="math-container">$k \geq N$</span>. </p>
<p>So, for all <span class="math-container">$k \geq N$</span>, <span class="math-container">$b_k = \frac{1}{a_k}$</span>. Taking the limit, you get immediately that <span class="math-container">$(b_k)$</span> converges to <span class="math-container">$\frac{1}{a}$</span>.</p>
|
3,855,736 | <p>Mathematics is not my primary discipline, but I know enough about both it and academics in general to know that many to most mathematical researchers do what they do because they enjoy doing it. This would seem to make "recreational mathematics" a rhetorical tautology, yet the term is used as if it were a subdiscipline in its own right. Universities offer courses on recreational mathematics. There are academic journals on recreational mathematics. There's a tag right here on SE Mathematics that reads <code>recreational-mathematics</code>. It's defined as "Mathematics done just for fun, often disjoint from typical school mathematics curriculum." Yet it does seem to be applied to a certain specific, albeit rather eclectic category of mathematical problems.</p>
<p>A lot of math that is called recreational falls quite neatly under some other category of math, often logic or combinatorics. In some cases, a kind of recreational math seems to be characterized as such only because some other formulation of the same ideas "got there first": it's relaxing to draw shapes with a compass and straightedge in much the same way that folding paper is relaxing, and the mathematics behind origami apprehend most if not all of the same mathematical truths that Euclidean geometry does, but mathematical origami is considered recreational while Euclidean geometry generally isn't.</p>
<p>With that in mind, what <em>is</em> recreational math? "Math that's done for fun" doesn't seem to make sense, because, again, that applies in some capacity to all math. Is it math that's done casually, with less concern for rigorous proof? Is it math that has not (yet) found an application in engineering or the empirical sciences? Is it the mathematical counterpart to popular science? What <em>is</em> it?</p>
<p>It's good math, but I don't understand this label attached to it. What makes a particular bit of math recreational?</p>
| J.G. | 56,861 | <p>The history of mathematics is replete with examples of something novel being done to address a problem at the time. It might have been "applied" in the sense of helping with a technological innovation, or it might have been "pure", e.g. mathematicians trying to make sense of something unrigorous or paradoxical in their present formulations of mathematics. Recreational mathematics is different again. Let's contrast three examples:</p>
<p><strong>Applied</strong>: how can we encrypt data so eavesdroppers can't decrypt it fast enough to crack our security before we update it? <a href="https://en.wikipedia.org/wiki/RSA_encryption" rel="nofollow noreferrer">At least one main approach</a> relies on number theory.</p>
<p><strong>Pure</strong>: asymptotically, how many prime numbers are there below <span class="math-container">$n$</span>? (Answer <a href="https://en.wikipedia.org/wiki/Prime_number_theorem" rel="nofollow noreferrer">here</a>.) More to the point, what kinds of mathematics do we need to prove the answer? Complex numbers came up in early proofs, but we eventually found we didn't need them. There are all sorts of interesting stories there.</p>
<p><strong>Recreational</strong>: when is a perfect power plus one also a perfect power? (<a href="https://en.wikipedia.org/wiki/Catalan%27s_conjecture" rel="nofollow noreferrer">We know now</a>.)</p>
<p>Recreational mathematics involves solving a lot of <a href="https://en.wikipedia.org/wiki/Diophantine_equation" rel="nofollow noreferrer">Diophantine equations</a> that don't enrich our understanding of mathematics as a whole.</p>
<p>Now, the applied/pure distinction gets foggy at times, especially when something that seemed pure at first becomes "useful" later, or when something invented for applications has "pure" spin-offs. But distinguishing these two from recreational is easier, as recreational problems aren't expected to enrich even our mathematical insight in general.</p>
|
3,003,672 | <p>Say I have an infinte 2D grid (ex. a procedurally generated world) and I want to get a unique number for each integer coordinate pair. How would I accomplish this?</p>
<p>My idea is to use a square spiral, but I cant find a way to make a formula for the unique number other than an algorythm that just goes in a square spiral and stops at the wanted coords.</p>
<p>The application for this converstion could be for example a way to save an n dimensional shape to a file where each line represents a chunk of the shape (by using <span class="math-container">$u(x, y, z) = u(u(x, y), u(y, z))$</span> ), or have a very unique random seed for each integer point (ex. a way to hash an integer vector to a data point in an n dimensional array)</p>
| Ross Millikan | 1,827 | <p>You need the Cantor <a href="https://en.wikipedia.org/wiki/Pairing_function" rel="nofollow noreferrer">pairing function</a>, tuned up to accept integers instead of naturals. The basic function takes a pair of naturals (including zero) <span class="math-container">$x,y$</span> and returns a natural <span class="math-container">$\pi(x,y)=\frac 12(x+y)(x+y+1)+y$</span>. It is invertible, so given <span class="math-container">$\pi(x,y)$</span> you can recover <span class="math-container">$x$</span> and <span class="math-container">$y$</span>. Now just take your integers to naturals by <span class="math-container">$$f(z)=\begin {cases} 2z&z\ge 0\\-2z-1& z \lt 0\end {cases}$$</span> pair them and you have your result. </p>
<p>You can do your idea with the Ulam spiral:
<a href="https://i.stack.imgur.com/L6zmy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/L6zmy.png" alt="enter image description here"></a>
The cell numbered <span class="math-container">$1$</span> is the origin. Note the odd squares follow the downward right diagonal and the even squares start just above <span class="math-container">$1$</span> and follow an upward left diagonal. If we are given a coordinate <span class="math-container">$(x,y)$</span> we first find the direction the side is going. If <span class="math-container">$x \gt 0, x \gt |y|$</span>, the cell is on an upward side. The corner at the bottom is <span class="math-container">$(x,-x)$</span> and has number <span class="math-container">$(2x-1)^2$</span> The number in our cell is <span class="math-container">$y+x$</span> cells above, so the number is <span class="math-container">$(2x-1)^2+y+x$</span>. You can go through the other three sides similarly. To go the other direction, given a cell number find the perfect square below it. Find the location of the square on the diagonal, then count the number of squares from there as needed.</p>
|
4,128,510 | <p>Let <span class="math-container">$(V, ||.||)$</span> be a normed space and define <span class="math-container">$d(x,y) = ||x|| + ||y|| $</span> if <span class="math-container">$ x \neq y$</span> and as <span class="math-container">$0$</span> if <span class="math-container">$x=y$</span>. Describe all convergent sequences in <span class="math-container">$(V,d)$</span>.
I'm sure that all eventually sequences are convergent sequences as in this case <span class="math-container">$d(x_n,x) \to 0$</span>.
I'm not really sure how to prove these are the only convergent sequences in this space / whether these are in fact all the convergent sequences in this space.</p>
<p>Also, what is the general approach for finding all convergent sequences in a metric space?</p>
<p>How do I go about proving this is complete?</p>
<p>Thanks!</p>
| alphaomega | 775,794 | <p><span class="math-container">$d(x_n, x)\to 0$</span> iff <span class="math-container">$\|x_n\| +\|x\|\to 0$</span> iff <span class="math-container">$\|x_n\|\to 0 \ \& \ \|x\|\to 0$</span> iff <span class="math-container">$x_n\overset{\tiny\| \cdot \|}{\to} 0 \ \& \ x=0$</span></p>
|
4,128,510 | <p>Let <span class="math-container">$(V, ||.||)$</span> be a normed space and define <span class="math-container">$d(x,y) = ||x|| + ||y|| $</span> if <span class="math-container">$ x \neq y$</span> and as <span class="math-container">$0$</span> if <span class="math-container">$x=y$</span>. Describe all convergent sequences in <span class="math-container">$(V,d)$</span>.
I'm sure that all eventually sequences are convergent sequences as in this case <span class="math-container">$d(x_n,x) \to 0$</span>.
I'm not really sure how to prove these are the only convergent sequences in this space / whether these are in fact all the convergent sequences in this space.</p>
<p>Also, what is the general approach for finding all convergent sequences in a metric space?</p>
<p>How do I go about proving this is complete?</p>
<p>Thanks!</p>
| Alessandro | 671,329 | <p>Let <span class="math-container">$x_n \to x$</span>, by definition: <span class="math-container">$\forall \varepsilon > 0 \ \exists \ N \in \mathbb{N}: \ n \geqslant N \Rightarrow ||x||+||x_n||=d(x_n,x)<\varepsilon.$</span></p>
<p>Note that <span class="math-container">$|| x || \leqslant ||x|| + ||x_n||< \varepsilon. \ $</span> Since <span class="math-container">$\varepsilon$</span> is arbitrary, we must have <span class="math-container">$||x||=0 \Rightarrow x=0.$</span> If we substitute <span class="math-container">$||x||=0$</span> for the property <span class="math-container">$x_n \to x$</span>, we have:
<span class="math-container">$\forall \varepsilon>0 \ \exists \ N \in \mathbb{N}: \ n \geqslant N \Rightarrow ||x_n||<\varepsilon $</span>. Since <span class="math-container">$||x_n||=||x_n-0||$</span>, it's easy to see that convergence wrt <span class="math-container">$d$</span> implies convergence to <span class="math-container">$0$</span> with the metric <span class="math-container">$d'(x,y)=||x-y|| $</span>, the standard metric derived from norm.</p>
<p>On the other hand any sequence converging to <span class="math-container">$0$</span> wrt the metric coming from the norm, clearly converges with respect to <span class="math-container">$d$</span>, it's just a matter of checking the definitions.</p>
<p>Completeness:</p>
<p>if <span class="math-container">$x_n$</span> is Cauchy, <span class="math-container">$\forall \varepsilon > 0, \ \exists \ N \in \mathbb{N}: n,m > N \Rightarrow d(x_n,x_m)<\varepsilon$</span>.</p>
<p>Note that <span class="math-container">$||x_n|| \leqslant d(x_n,x_m) = ||x_n||+||x_m|| < \varepsilon$</span>. Thus <span class="math-container">$x_n \to 0$</span> with respect to the metric coming from <span class="math-container">$|| \cdot ||$</span>.</p>
<p>Any sequence <span class="math-container">$x_n \to 0$</span> wrt to <span class="math-container">$||\cdot ||$</span>, is Cauchy wtr to <span class="math-container">$d$</span>.</p>
|
4,128,510 | <p>Let <span class="math-container">$(V, ||.||)$</span> be a normed space and define <span class="math-container">$d(x,y) = ||x|| + ||y|| $</span> if <span class="math-container">$ x \neq y$</span> and as <span class="math-container">$0$</span> if <span class="math-container">$x=y$</span>. Describe all convergent sequences in <span class="math-container">$(V,d)$</span>.
I'm sure that all eventually sequences are convergent sequences as in this case <span class="math-container">$d(x_n,x) \to 0$</span>.
I'm not really sure how to prove these are the only convergent sequences in this space / whether these are in fact all the convergent sequences in this space.</p>
<p>Also, what is the general approach for finding all convergent sequences in a metric space?</p>
<p>How do I go about proving this is complete?</p>
<p>Thanks!</p>
| Henno Brandsma | 4,280 | <p>If <span class="math-container">$x_n \to x$</span> in <span class="math-container">$d$</span>, there are two cases:</p>
<p><span class="math-container">$x=0$</span>. Then <span class="math-container">$d(x_n,x)= \|x_n\| + \|x\| = \|x_n\|\to 0$</span> and so <span class="math-container">$x_n \to x$</span> under <span class="math-container">$\|.\|$</span>. So <span class="math-container">$(V,\|.\|)$</span> and <span class="math-container">$(V,d)$</span> have the same convergent sequences with limit <span class="math-container">$0$</span>.</p>
<p><span class="math-container">$x \neq 0$</span>. Then also <span class="math-container">$d(x_n,x) \to 0$</span> but <span class="math-container">$d(x_n,x) \ge \|x\| >0$</span> so this can never happen unless <span class="math-container">$x_n = x$</span> for a tail of the sequence (because then the other clause of the definition of <span class="math-container">$d$</span> kicks in). So a sequence can <em>only</em> converge in <span class="math-container">$d$</span> to a <span class="math-container">$x \neq 0$</span> if it's eventually constant with value <span class="math-container">$x$</span>.</p>
<p>This completely characterises the convergence in <span class="math-container">$(V,d)$</span>: if a given sequence is not eventually constant (then it has the obvious limit), it converges (to <span class="math-container">$0$</span>) iff it already did under the norm.</p>
<p>Similar considerations show that <span class="math-container">$(V,d)$</span> is complete: if <span class="math-container">$(x_n)$</span> is Cauchy <span class="math-container">$d(x_n, x_m) = \|x_n\| + \x_m\|$</span> must tend to <span class="math-container">$0$</span> for both <span class="math-container">$n,m \to \infty$</span>. So if all <span class="math-container">$x_n$</span> are bounded away from <span class="math-container">$0$</span>, an injective sequence cannot be Cauchy, so <span class="math-container">$x_n \to 0$</span> follows by necessity.</p>
|
1,585,772 | <p>I am finding this problem confusing :</p>
<blockquote>
<p>If,for all $x$,$f(x)=f(2a)^x$ and $f(x+2)=27f(x)$,then find $a$.</p>
</blockquote>
<p>When $x=1$ I have that $f(1)=f(2a)$ using the first identity.</p>
<p>Then when $x=2a$ I have by the second identity that $f(2a+2)=27f(2a)$,after that I simple stare at the problem without having a clue of how to proceed.</p>
<p>What's the trick the problem is calling for ?</p>
<p>I've thought of finding the inverse of the function $f(x)$ but It's not really clear to me how to apply this idea as I don't have linear functions .</p>
<p>Can you guys give me a hint ?</p>
| Steven Stadnicki | 785 | <p>In fact, you barely even need the second relation. Hint: first set $f(2a)=c$. Now you know that $f(x)$ is an exponential function, $f(x)=c^x$ (and the second relation implies that $c\neq 1$ — this is all it does; 27 could as easily be 0.27 or $10^{27}$ and it wouldn't change the answer); this function is one-to-one over its domain, and you've already figured out that $f(1)=f(2a)$. Apply $f^{-1}$ to both sides.</p>
|
327,291 | <p>I am struggling to find the English translation of Malcev's paper "On a class of homogenous spaces" providing foundational material for nil-manifolds. To be precise this paper: <i>Malcev, A. I. On a class of homogeneous spaces. Amer. Math. Soc. Translation 1951, (1951). no. 39, 33 pp.</i> (<a href="https://mathscinet.ams.org/mathscinet-getitem?mr=39734" rel="noreferrer">mathscinet link</a>) . It would be really important, for a project I am doing, to find this paper and I did not succeed neither on the website of the AMS nor by standard googling, which gives tons of papers referring to it. </p>
<p>Can anyone provide a reference to a place where to download the paper? I am at an institution with free access virtually everywhere, I just need a place with the actual paper in English (yeah in Russian I could find it). </p>
| Ben McKay | 13,268 | <p><a href="http://www.mathnet.ru/links/f4bbcad8dff13663fdc1e07b7d350dd1/im3161.pdf" rel="nofollow noreferrer">The Russian original</a>:</p>
<p>You can use <a href="https://www.deepl.com/en/translator" rel="nofollow noreferrer">DeepL</a> to translate it, for free.
Here is a DeepL translation of the title and the abstract, to show how well it works:</p>
<p><em>Izvestiya ACADEMY Nauke USSR Matematicheskaya Series 13 (1949), 9-32 A. I. MALTSEV ON ONE CLASS OF ONE-STANDING SPACES (Edited by Academician A. N. Kolmogorov) I. MALCEV ON ONE CLASS OF ONE-LINE SPACE (Edited by Academician A. N. Kolmogorov) We consider a manifold with a transitive nilpotent topological group acting on it. These manifolds can be dissolved into the topological product of Euclidean manifold and compact manifold" having the same property. Compact manifolds on which nilpotent groups act transitive are uniquely defined by their fundamental groups, and an abstract group % is the fundamental group of a compact manifold with a transitive nilpotent group if and only if Ъ is a nilpotent group without elements of finite order with a finite number of generators</em></p>
<p>Translated with <a href="http://www.DeepL.com/Translator" rel="nofollow noreferrer">www.DeepL.com/Translator</a> (free version)</p>
<p>Obviously you need to work a little on it, but I think you can read it.</p>
|
654,198 | <p>$6x^3 -11x^2 + 6x + 5 \equiv (Ax-1)(Bx - 1)(x - 1) + c$</p>
<p>Find the value of A, B and C.</p>
<p>I started it like this: </p>
<p>$6x^3 -11x^2 + 6x + 5 \equiv (Ax-1)(Bx - 1)(x - 1) + c$</p>
<p>Solving the right hand side:</p>
<p>$ (ABx^2 - Ax - Bx + 1)(x - 1) + C$</p>
<p>$ ABx^3 - ABx^2 - Ax^2 + Ax - Bx^2 + Bx + x - 1 + C$</p>
<p>$ABx^3 - (AB + A + B)x^2 + (A + B + 1)x - 1 + C$</p>
<p>Comparing the coefficients: </p>
<p>$AB = 6$</p>
<p>$A = \frac6 B$</p>
<p>$AB + A + B = 11$</p>
<p>Then substitute the value of A in the above equation...is this right? Is there any error?</p>
| Devgeet Patel | 116,131 | <p>Equating x coefficients you will get A+B=5.You already have AB=6.Solving you will get A=2 or 3.This will be the easier approach.</p>
|
1,503,958 | <p>In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions?</p>
<p>I tried shifting the second term to the rhs and squaring.Even after that i'm left with square roots.No idea how to proceed.Help!</p>
| Ángel Mario Gallegos | 67,622 | <p><strong>Hint:</strong></p>
<p>Notice that $$x+3-4\sqrt{x-1}=x-1-4\sqrt{x-1}+4=(\sqrt{x-1}-2)^2$$</p>
<p>and</p>
<p>$$x+8-6\sqrt{x-1}=x-1-6\sqrt{x-1}+9=(\sqrt{x-1}-3)^2=(3-\sqrt{x-1})^2$$</p>
<p>After, you can try by cases.</p>
|
1,503,958 | <p>In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions?</p>
<p>I tried shifting the second term to the rhs and squaring.Even after that i'm left with square roots.No idea how to proceed.Help!</p>
| Zach466920 | 219,489 | <p><a href="https://i.stack.imgur.com/qVT0Y.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qVT0Y.png" alt="enter image description here"></a></p>
<p>I found that to be odd. So the solutions are $x \in [5,10]$. I guess it wouldn't be too hard to formally prove that. </p>
<blockquote>
<p><em>Hint: the imaginary parts cancel on this interval.</em></p>
</blockquote>
|
1,503,958 | <p>In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions?</p>
<p>I tried shifting the second term to the rhs and squaring.Even after that i'm left with square roots.No idea how to proceed.Help!</p>
| Ennar | 122,131 | <p>\begin{align}
&\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}} = 1\\ \implies &\sqrt{(x-1)-4\sqrt{x-1} + 4}+\sqrt{(x-1)-6\sqrt{x-1}+9}=1\\ \implies &\sqrt{(\sqrt{x-1}-2)^2}+\sqrt{(\sqrt{x-1}-3)^2}=1\\ \implies &|\sqrt{x-1}-2| + |\sqrt{x-1}-3| = 1\tag{1}
\end{align}</p>
<p>This calls for casework:</p>
<hr>
<p><strong>1.</strong> $\quad\sqrt{x-1}\geq 3$</p>
<p>$(1)\implies \sqrt{x-1}-2 + \sqrt{x-1}-3 = 1\implies \sqrt{x-1} = 3\implies x=10$</p>
<hr>
<p><strong>2.</strong> $\quad 2\leq\sqrt{x-1} < 3$</p>
<p>$(1)\implies \sqrt{x-1}-2 - \sqrt{x-1}+3 = 1\implies 1 = 1$ and thus all $x$ such that $2\leq\sqrt{x-1} < 3$, i.e. $x\in [5,10\rangle$</p>
<hr>
<p><strong>3.</strong> $\sqrt{x-1} < 2$</p>
<p>$(1)\implies -\sqrt{x-1}+2 - \sqrt{x-1}+3 = 1\implies \sqrt{x-1} = 2 \implies x = 5$, but this $x$ doesn't satisfy our condition 3 (although it does satisfy condition 2, and is already included as a solution)</p>
<hr>
<p>Taking union, we conclude that any $x\in[5,10]$ is a solution to the equation.</p>
|
746,750 | <p>This is a consequence of the exponential rule, but how do I actually prove it to be true?</p>
| Ant | 66,711 | <p>How do you define $e^x$?</p>
<p>The most common way to define it is</p>
<p>$$e^x = \sum_{n = 0}^{\infty} \frac{x^n}{n!}$$</p>
<p>If you take this definition, then if $x > 0$, $e^x$ is the sum of an infinite number of positive terms. </p>
<p>One should show that the series converges, but if you take that for granted, then of course if converges to a positive number.</p>
<p>If $x < 0$ then $e^{-x} > 0$ ; and since $$ e^x \cdot e^{-x} = 1 > 0 \Rightarrow e^x > 0$$ since $e^{-x} > 0$</p>
|
1,179,195 | <p>Good day everyone. </p>
<p>I need to know automata theory. Can you advice me the best way to study math?
What themes will I need to know to understand automata theory. What a sequence of study? What level will I need to study intermediate themes? Maybe can you say something yet, what can help me quickly learn automata theory?</p>
| John Gowers | 26,267 | <p>You seem to be a bit confused about how to think of a contraction as a line on a graph. Let's assume that we are looking for a contraction on $[0,1]$, though we could use any complete subset of the reals (or even the real line itself.</p>
<p>Intuitively, we have:</p>
<ul>
<li>The line $y=x$ plotted on $[0,1]$ (so $x,y\in[0,1]$). </li>
</ul>
<p>Then a curve corresponding to a <em>contraction</em> is one such that:</p>
<ul>
<li>The curve corresponds to a function $y=f(x)$ for $0\le x\le1$. </li>
<li>The curve's $x$ and $y$ values never leave the square $[0,1]\times[0,1]$ - in other words, we always have $0\le x\le1,0\le y\le1$. </li>
<li>The slope of the curve is bounded above by some fixed constant $\lambda<1$.</li>
</ul>
<p>Then the curve has a unique intersection with $y=x$; i.e., there is a unique point such that $x=f(x)$. For example:</p>
<p><img src="https://i.stack.imgur.com/EzcwP.gif" alt="enter image description here"></p>
<p>Of course, we don't require the curve to be differentiable in general, so it might not always make sense to talk about the 'slope' of the curve, but that's what you should have in mind. </p>
<p>Your curves don't count as contraction mappings - the first one has slope greater than $1$ at some points, and the second leaves the square given by $0\le x\le1,0\le y\le1$, which is not allowed </p>
|
123,587 | <p>I always wonder how many inbuilt functions does Mathematica have (of course you can google for it) and how they are connected with each other! So I tried this (v10.1).</p>
<pre><code>SetDirectory[$InstallationDirectory<>"/Documentation/English/System/ReferencePages/Symbols"]
comms = FileNames[];
ncomms = Length[comms]
</code></pre>
<blockquote>
<p>4613</p>
</blockquote>
<p>I believe this is the total number of documented functions. Now the second part where I am stuck in. <strong>How to find the connections among them?</strong></p>
<p>What I am thinking is to get the list of functions from the <code>See Also</code> section of an example. Here I do one manually. Let's say I start with <code>Plot</code>. <code>See Also</code> in <code>Plot.nb</code> contains ({DiscretePlot, ListLinePlot, ParametricPlot, PolarPlot, Plot3D, ContourPlot, Graphics, Show}). So I store them.</p>
<pre><code>link["Plot"] = {"DiscretePlot", "ListLinePlot", "ParametricPlot" ,
"PolarPlot", "Plot3D", "ContourPlot", "Graphics", "Show"};
</code></pre>
<p>Now I want to scan each element of the list. For example, I take <code>Plot3D</code> and <code>Graphics</code></p>
<pre><code>link["Plot3D"] = {"ListPlot3D", "ContourPlot", "DensityPlot", "ParametricPlot3D",
"Graphics3D", "ListSurfacePlot3D", "Plot", "Show"};
link["Graphics"] = {"Plot", "ListPlot", "ListLinePlot", "ParametricPlot", "DensityPlot",
"ArrayPlot", "RegionPlot", "ContourPlot", "Show", "Graphics3D",
"Image", "Import", "Sound"};
</code></pre>
<p>Now combine and map them</p>
<pre><code>map = Join[DirectedEdge[num["Plot"], num[#]] & /@ link["Plot"],
DirectedEdge[num["Plot3D"], num[#]] & /@ link["Plot3D"],
DirectedEdge[num["Graphics"], num[#]] & /@ link["Graphics"]];
Graph[map, VertexLabels -> Table[n -> sets[[n]], {n, nsets}], ImagePadding -> 10]
</code></pre>
<p><a href="https://i.stack.imgur.com/zgODy.png" rel="noreferrer"><img src="https://i.stack.imgur.com/zgODy.png" alt="enter image description here"></a></p>
<p><strong>How can I put this whole thing in a Mathematica code?</strong></p>
| Szabolcs | 12 | <p>Something like this:</p>
<pre><code>g = SimpleGraph@Graph[
Catenate[
Thread /@
Normal@DeleteMissing@
WolframLanguageData[WolframLanguageData[], "RelatedSymbols", "EntityAssociation"]
],
DirectedEdges -> False
];
</code></pre>
<p>It's slow, like most new <code>*Data</code> functions, but it works.</p>
|
123,587 | <p>I always wonder how many inbuilt functions does Mathematica have (of course you can google for it) and how they are connected with each other! So I tried this (v10.1).</p>
<pre><code>SetDirectory[$InstallationDirectory<>"/Documentation/English/System/ReferencePages/Symbols"]
comms = FileNames[];
ncomms = Length[comms]
</code></pre>
<blockquote>
<p>4613</p>
</blockquote>
<p>I believe this is the total number of documented functions. Now the second part where I am stuck in. <strong>How to find the connections among them?</strong></p>
<p>What I am thinking is to get the list of functions from the <code>See Also</code> section of an example. Here I do one manually. Let's say I start with <code>Plot</code>. <code>See Also</code> in <code>Plot.nb</code> contains ({DiscretePlot, ListLinePlot, ParametricPlot, PolarPlot, Plot3D, ContourPlot, Graphics, Show}). So I store them.</p>
<pre><code>link["Plot"] = {"DiscretePlot", "ListLinePlot", "ParametricPlot" ,
"PolarPlot", "Plot3D", "ContourPlot", "Graphics", "Show"};
</code></pre>
<p>Now I want to scan each element of the list. For example, I take <code>Plot3D</code> and <code>Graphics</code></p>
<pre><code>link["Plot3D"] = {"ListPlot3D", "ContourPlot", "DensityPlot", "ParametricPlot3D",
"Graphics3D", "ListSurfacePlot3D", "Plot", "Show"};
link["Graphics"] = {"Plot", "ListPlot", "ListLinePlot", "ParametricPlot", "DensityPlot",
"ArrayPlot", "RegionPlot", "ContourPlot", "Show", "Graphics3D",
"Image", "Import", "Sound"};
</code></pre>
<p>Now combine and map them</p>
<pre><code>map = Join[DirectedEdge[num["Plot"], num[#]] & /@ link["Plot"],
DirectedEdge[num["Plot3D"], num[#]] & /@ link["Plot3D"],
DirectedEdge[num["Graphics"], num[#]] & /@ link["Graphics"]];
Graph[map, VertexLabels -> Table[n -> sets[[n]], {n, nsets}], ImagePadding -> 10]
</code></pre>
<p><a href="https://i.stack.imgur.com/zgODy.png" rel="noreferrer"><img src="https://i.stack.imgur.com/zgODy.png" alt="enter image description here"></a></p>
<p><strong>How can I put this whole thing in a Mathematica code?</strong></p>
| BoLe | 6,555 | <pre><code>(* docs on my system (10.3, Windows) *)
base = FileNameJoin[{$InstallationDirectory, "Documentation",
"English", "System", "ReferencePages", "Symbols"}];
FileNames[FileNameJoin[{base, "*"}]] // Length
</code></pre>
<blockquote>
<p>4811</p>
</blockquote>
<pre><code>(* symbols in See Also section *)
also[name_String] :=
Module[{import},
import = Import[
FileNameJoin[{base, name <> ".nb"}], {"Cells", "SeeAlso"}];
Cases[import, TextData[s_String] :> s, Infinity]]
also["I"]
</code></pre>
<blockquote>
<p>{"Complex", "Re", "Im", "ComplexExpand", "GaussianIntegers"}</p>
</blockquote>
<pre><code>edges[start_String, n_Integer] :=
Reap[Module[{visited = {}},
Nest[
Function[visiting,
Module[{temp},
temp = Join @@ Table[
Sow[DirectedEdge[v, #] & /@ also[v]], {v, visiting}];
visited = Join[visited, visiting];
Complement[Last /@ temp, visited]]],
{start}, n]]][[2, 1]] // Flatten
graph[edges : {__DirectedEdge}, start_String, opts___] :=
Graph[edges, opts,
VertexLabels -> "Name",
VertexStyle -> {start -> Red}]
</code></pre>
<h2>Example 1</h2>
<pre><code>With[{x = "I"}, graph[edges[x, 2], x]]
</code></pre>
<p><a href="https://i.stack.imgur.com/E84rA.png" rel="noreferrer"><img src="https://i.stack.imgur.com/E84rA.png" alt="enter image description here"></a></p>
<h2>Example 2</h2>
<pre><code>(* this took me 40 seconds *)
e5 = edges["Plot", 5];
g = graph[e5, "Plot",
VertexLabels -> None,
EdgeStyle -> Directive[Opacity[.2]],
EdgeShapeFunction -> ({Arrowheads[.01], Arrow[#]} &)];
Through[{VertexCount, EdgeCount}@g]
</code></pre>
<blockquote>
<p>{1463, 4192}</p>
</blockquote>
<p>Majority of vertices are not reachable from a general source (dark red area). Note that: a) graph is directed, and b) crawler stopped after five steps.</p>
<pre><code>dm = GraphDistanceMatrix[g];
dm // MatrixPlot
</code></pre>
<p><a href="https://i.stack.imgur.com/boxHk.png" rel="noreferrer"><img src="https://i.stack.imgur.com/boxHk.png" alt="enter image description here"></a></p>
<pre><code>max = Max[dm /. Infinity -> 0]
</code></pre>
<blockquote>
<p>13</p>
</blockquote>
<pre><code>pos = Position[dm, max];
pairs = VertexList[g][[#]] & /@ pos;
</code></pre>
<p>Example of a longest shortest path (via <code>pairs</code>). Starting in Help with <code>Repeated</code>, 13 clicks are needed at least to get to <code>Arrow</code>.</p>
<pre><code>sp = FindShortestPath[g, "Repeated", "Arrow"];
(* {"Repeated", "BlankSequence", ... "Arrow"} *)
HighlightGraph[g, PathGraph[sp, DirectedEdges -> True]]
</code></pre>
<p><a href="https://i.stack.imgur.com/k3Pj3.png" rel="noreferrer"><img src="https://i.stack.imgur.com/k3Pj3.png" alt="enter image description here"></a></p>
|
1,349,654 | <p>Is there some one show me how do i evaluate this integral :$$ \int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx $$</p>
<p>Note :By mathematica,the result is :
$\frac{Gamma\left(\frac1 4\right)Gamma\left(\frac5 4\right)}{\sqrt{\pi}}-\sqrt{2} Hypergeometric2F1\left(\frac1 4,\frac3 4,\frac5 4,\frac1 4\right).$
and i think it elliptic integral .</p>
<p>Thank you for any kind of help </p>
| Claude Leibovici | 82,404 | <p><em>This is not an answer but it is too long for a comment.</em></p>
<p>As I wrote in comment, there is something wrong somewhere since $$\int\frac{\sqrt{\tan x}}{\sin x}dx=-2 \sqrt{\cos (x)} \, _2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};\cos
^2(x)\right)$$ and $$I=\int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx=\frac{\Gamma \left(\frac{1}{4}\right) \Gamma \left(\frac{5}{4}\right)}{\sqrt{\pi
}}-\sqrt{2} \, _2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};\frac{1}{4}\right)\approx 0.379133$$ which does not match with Mercy's result $$\ln\left(\frac{1+\sqrt2}{\sqrt3}\right) \approx 0.332067$$</p>
<p>Reusing all steps done for Mercy, I thing that in fact, we should arrive to $$I=\int_1^{\sqrt{3}}\frac{1}{\sqrt t \,\sqrt{1+t^2}}\,dt=\frac{4 \Gamma \left(\frac{5}{4}\right)^2}{\sqrt{\pi }}-\frac{2 \,
_2F_1\left(\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{1}{3}\right)}{\sqrt[4]{3}}\approx 0.379133$$ which is still a nightmare ( the <strong>I</strong>nverse <strong>S</strong>ymbolic <strong>C</strong>alculator did not find any solution for this number).</p>
<p>Totally unable to find any closed form derivative, I came back to the original problem and expanded the integrand as a Taylor series built at $x=\frac \pi 4$. This gave $$\frac{\sqrt{\tan x}}{\sin x}=\sqrt{2}+\sqrt{2} \left(x-\frac{\pi }{4}\right)^2+\frac{7 \left(x-\frac{\pi
}{4}\right)^4}{3 \sqrt{2}}+\frac{139 \left(x-\frac{\pi }{4}\right)^6}{45
\sqrt{2}}+\frac{5473 \left(x-\frac{\pi }{4}\right)^8}{1260
\sqrt{2}}+\cdots$$ Integrating between the given bounds leads to $$I\approx \frac{\pi }{6 \sqrt{2}}+\frac{\pi ^3}{2592 \sqrt{2}}+\frac{7 \pi ^5}{3732480
\sqrt{2}}+\frac{139 \pi ^7}{11287019520 \sqrt{2}}+\frac{5473 \pi
^9}{58511909191680 \sqrt{2}}+\cdots$$ which gives the correct value for six significant figures.</p>
<p><strong>Edit</strong></p>
<p>For sure, the exact solution exits but it involves elliptic integrals. Using the tangent half-angle substitution $t=\tan(\frac x2)$ and the fiven bounds, the result is $$I=-2 i \sqrt{2} \left(F\left(\left.\sin
^{-1}\left(\sqrt[4]{3}\right)\right|-1\right)-F\left(\left.\sin
^{-1}\left(\sqrt{1+\sqrt{2}}\right)\right|-1\right)\right)$$ where appear elliptic integrals of the first kind.</p>
<p><strong>Edit</strong></p>
<p>If we change variable $x=y+\frac \pi 4$, expanding the trigonometric functions and simplifying, we can arrive at $$I=\int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx=\sqrt{2}\int_{0}^{\pi /12}\frac{dy}{\sqrt{\cos (2 y)}}=\sqrt{2}\, F\left(\left.\frac{\pi }{12}\right|2\right)$$</p>
|
387,202 | <p>What is the smallest 3-regular graph to have a unique perfect matching?</p>
<p>With a large enough number of nodes, it is possible for a 3-regular graph to have no perfect matching (example can be seen in this question <a href="https://mathoverflow.net/questions/98385/cubic-graphs-without-a-perfect-matching-and-a-vertex-incident-to-three-bridges">Cubic graphs without a perfect matching and a vertex incident to three bridges</a> ). So I believe 3-regular graphs with a unique matching likely exists, but I am unsure how to go about constructing and proving what the smallest one is. Likely there is no better answer than to brute force check all the possibilities, so I am hoping someone happens to know what this graph looks like.</p>
<p>Even better: Does anyone know of an online searchable graph database that allows searching for small graphs with certain properties?</p>
| Matthieu Latapy | 158,328 | <p>Regarding your second <em>even better</em> question, I warmly suggest the <a href="https://users.cecs.anu.edu.au/%7Ebdm/data/" rel="nofollow noreferrer">Brendan McKay page on combinatorial objects</a>, that gives many kinds of graph examples.</p>
|
55,965 | <p>I'm a games programmer with an interest in the following areas:</p>
<ul>
<li>Calculus</li>
<li>Matrices</li>
<li>Graph theory</li>
<li>Probability theory</li>
<li>Combinatorics</li>
<li>Statistics</li>
<li>More linguistic related fields of logic such as natural language processing, generative grammars</li>
</ul>
<p>Here are some examples of topics I've come across in the last 18 months in my design/development work, that have been of interest in solving certain problems. I grasp the outlines of these topics enough to know how they would help me to solve certain problems in my designs, but I don't even scratch the surface in understanding how to apply the math involved.</p>
<ul>
<li>Matrix math for spatial transformations</li>
<li>Minkowski sums for spatial expansion</li>
<li><a href="http://www.cs.brown.edu/~rt/gdhandbook/chapters/planarity.pdf" rel="nofollow">Planarity testing and embedding to convert logical non-planar to planar graphs</a></li>
<li>Generative grammars and natural language for narrative generation (linguistics / logic)</li>
</ul>
<p>My maths ability is sorely lacking. I know enough to get by for the relatively simple games I write. My logical and analytical skills are generally good, being a programmer. I enjoyed math in high school, but college was a different story -- my lecturer was terrible, and I didn't get any individual tutoring as I did before that. Anything that was in my head has long since departed. I would need to relearn what I learnt, which in mostly centred around "the calculus".</p>
<p>Bearing in mind that I need to balance my time between improving as a game designer, developer and mathematician/logician, what is the best way for me to tackle these gaping holes in my knowledge, enough to work in-depth mathematical descriptions into working algorithms?</p>
| Community | -1 | <ul>
<li><span class="math-container">$\textbf{Theorem.}$</span> Let <span class="math-container">$|G| = 2^{n} \cdot m$</span> where <span class="math-container">$2 \nmid m$</span>. If <span class="math-container">$G$</span> has a cyclic <span class="math-container">$2$</span>- Sylow subgroup, then <span class="math-container">$G$</span> has a normal subgroup of order <span class="math-container">$m$</span>.</li>
</ul>
<p>Your question is just a corollary to this theorem. Please see <span class="math-container">$\textbf{Theorem 6.9}$</span> in Prof. Keith Conrad blurb here:</p>
<ul>
<li><a href="http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/sylowapp.pdf" rel="nofollow noreferrer">http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/sylowapp.pdf</a></li>
</ul>
|
3,019,506 | <p>I am stuck on this problem during my review for my stats test. </p>
<p>I know I have to use the convolution formula, and I understand that:</p>
<p><span class="math-container">$f_{U_1}(U_1) = 1$</span> for <span class="math-container">$0≤U_1≤1$</span> </p>
<p><span class="math-container">$f_{U_2}(U_2) = 1$</span> for <span class="math-container">$0≤U_2≤1$</span></p>
<p>but I do not know how to continue on from there. How do I use the convolution formula in this question? Thanks</p>
| Zeno Rogue | 265,219 | <p>You would need a surface of negative curvature.</p>
<p><a href="https://i.stack.imgur.com/m7Erq.png" rel="noreferrer"><img src="https://i.stack.imgur.com/m7Erq.png" alt="illustration of a hexagon in the hyperbolic plane"></a></p>
<p>It is best to use a hyperbolic plane for this, where you can easily fit any regular n-gon with given angles as long as the sum of its external angles is greater than 360 degrees. The problem is that the hyperbolic plane does not fit in Euclidean space.</p>
<p>The pseudosphere is a small fragment of the hyperbolic plane. You can draw a right-angled hexagon on the pseudosphere only if you allow it to wrap over itself. (Edit: actually I am not completely sure about this; <a href="http://roguetemple.com/z/hyper/patterns/ps-coverage-6666.png" rel="noreferrer">see here</a>, you get a pseudosphere by cutting the part covered with white dots; it appears that a hexagon is slightly larger than the area covered by the pseudosphere, but I am not sure. Should be possible to prove.)</p>
<p>You can also draw it on a Dini's surface -- that is basically an unrolled pseudosphere where you have several layers, and thus you avoid the intersection problem. But it would be hard to see anything because it is rolled very tightly. See <a href="http://roguetemple.com/z/hyper/patterns/dini-6666.png" rel="noreferrer">here</a>.</p>
<p>Less smooth, but probably the best way would be to use something similar to a hyperbolic crochet. See our <a href="http://www.roguetemple.com/z/hyper/online2.php?c=-rugon+-geo+9+-7+-canvas+r+-noscr+-noplayer+-each+10+-msg0+-nohud" rel="noreferrer">computer simulation</a> (arrow keys to rotate).</p>
|
93,099 | <p>Consider $n$ points generated randomly and uniformly on a unit square. What is the expected value of the area (as a function of $n$) enclosed by the convex hull of the set of points?</p>
| kaleidoscop | 16,934 | <p>For any convex set $K$ in dimension $d$ with volume $V(K)$, it is aymptotically $$V(K)-\frac{T(K)}{(d+1)^{d-1}(d-1)!}n^{-1}ln(n)^{d-1}+O(n^{-1}ln(n)^{d-2}ln(ln(n)))$$ (see {New perspectives in stochastic geometry} by W. Kendall and I. Molchanov, p. 49) where $T(K)$ is the number of "flags", i.e. of sequences $f_0\subset f_1 \subset ... \subset f_{d-1}$ where $f_i$ is an $i$-dimensional facet. There is an abundant literature for random convex hulls and if you're interested there might be an exact closed formula for the square.</p>
|
1,998,391 | <p>To define a <strong>real</strong> exponential function
$$f(x)=a^x=e^{x \,\mathrm{lg} a}$$</p>
<p>It is strictly necessary that $a>0$.</p>
<p>But is the same true if the exponent is a <strong>natural</strong> number?</p>
<hr>
<p>In function series and moreover in power series I see things like
$$\sum_{n\geq0} f(n) x^n, \,\,\,\, x\in \mathbb{R}$$
So there is not the resctriction $x>0$.</p>
<p>It makes sense because natural number in the exponent do not create problems like even roots of negative numbers, so I think I'm ok with this.</p>
<p>Nevertheless if I rewrite $x^n$ using the definition of logaritm (for example this passage is useful while evaluating limits of such expressions), I get</p>
<p>$$\sum_{n\geq0} f(n) \,\, e^{n \, \mathrm{lg}x}$$</p>
<p>So suddenly I get a $\mathrm{lg}x$ and I need the condition $x>0$. </p>
<p>On the one hand this passage should be valid because it is just the definition of logaritm, but on the other hand it looks like I cannot perform it without meet a restriction on $x$ that was not present in the form $x^n$.</p>
<hr>
<p>So what is the point here? Is the use of the definition of logaritm not allowed in these cases, or is it wrong to avoid setting the condition $x>0$ in $x^n$?</p>
| user90369 | 332,823 | <p>Of course if you use $e^{x\ln a}$ instead of $a^x$ then $a>0$ because otherwise the exponent $\ln a$ is not real any more. </p>
<p>(1) It's $e^{x\ln a}=a^x$ for $a>0$ ;</p>
<p>(2) it's only $a^x$ (and not $e^{x\ln a}$) valid for $a\in\mathbb{R}\setminus\{0\}$ with $x\in\mathbb{Z}$. </p>
<p>$e^{x\ln a}$ and $a^x$ are two different kind of constructions for evaluations and the value ranges have to be considered.</p>
|
2,262,661 | <p>The question is in how many ways can we select 20 different items from the empty set?</p>
<h3>ans:</h3>
<p>Obviously in 0 ways since the empty set has no items. I mean, this seem obvious, but maybe there is a trick to this question.</p>
| JMJ | 295,405 | <p>If you wanted to sound more mathematical you could make an induction argument:</p>
<p><em>There are exactly zero ways to select $n>0$ items from the empty set.</em></p>
<p><em>Proof:</em> Take $n =1$. Since the empty set has no members it is impossible to select a member of the empty set, thus zero ways to select a single member. Assuming that there are zero ways to select $n$ members, there also must be zero ways to select $n+1$ members, for if there were more than zero ways to select $n+1$ members but zero ways to select $n$ members, it would imply there are more than zero ways to select a single member, contradicting the base case.</p>
<p>It doesn't change the answer but the style is somewhat more formal...</p>
|
1,476,982 | <p>I'm trying to understand why the volume of a parallelepiped whos sides are $s,u,w$ is $ V = s \cdot(u \times w)$.</p>
<p>Even the units of measurement don't add up. The length of the vectors $s,u,w$ is measured in centimeters, the volume is measured in cubic cm.</p>
<p>$u\times w$ is a vector. It is a vector that is orthogonal to $u$ and $w$, but still a vector, so its length is again measure in cms. So overall $V=s \cdot(u \times w)$ means that $V$ is equal to the product of $2$ vectors, so the unit of measurement for $V$ is squared centimeters, not cubed.</p>
<p>I'm struggling to understand how can $|u\times w|$ be equal to the area of a parallelogram. That is equivalent to saying "The time it takes for me to solve a problem is the distance between New York and London."</p>
| vadim123 | 73,324 | <p>I think the apparent paradox in this question derives from the mistaken belief that a vector has associated units of length. This is incorrect. In three-dimensional space, the vector $(1,2,3)$ could be thought of as having units of ordered triples of centimeters (not the same as centimeters cubed). We could also have the coordinates be centimeters, seconds, liters, but it is still an ordered triple.</p>
<p>Now, the length of a vector $|(1,2,3)|=\sqrt{1^2+2^2+3^2}$ is indeed in centimeters, because each of $1,2,3$ is in centimeters, so when squared is in square centimeters, and then the sum has square root applied, which takes us back to centimeters.</p>
<p>If the units were centimeters, seconds, liters, then the length of the vector no longer has any meaningful units anymore.</p>
|
4,173 | <p>I asked this question on mathoverflow, but it was deemed too simple, so I'm posting here instead -- </p>
<p>Is there a nice way to characterize an orthonormal basis of eigenvectors of the following $d\times d$ matrix?</p>
<p>$$\mathbf{I}-\frac{1}{d} \mathbf{v}\mathbf{v}'$$</p>
<p>Where $\mathbf{v}$ is a $d\times 1$ vector of 1's. This is similar to the <a href="http://en.wikipedia.org/wiki/Householder_transformation" rel="nofollow">Householder matrix</a>, except the $v's$ are not normalized. One eigenvector is $\mathbf{v}$ with corresponding eigenvalue 0, remaining eigenvalues should be 1. I'm looking for an expression in terms of unknown d.</p>
<p>Motivation: this is covariance matrix of uniform multinomial distribution, so expression for orthonormal basis produces a linear transformation that will make variables uncorrelated for large n</p>
<p>Example: below are 5 orthonormal eigenvectors vectors I get from Gram-Schmidt for d=5...what is the expression for general d? An even bigger example -- columns of <a href="http://yaroslavvb.com/upload/bigmatrix.png" rel="nofollow">this</a> form orthonormal basis for d=20</p>
<p>$$-\frac{1}{\sqrt{2}},0,0,0,\frac{1}{\sqrt{2}}$$</p>
<p>$$-\frac{1}{\sqrt{6}},0,0,\sqrt{\frac{2}{3}},-\frac{1}{\sqrt{6}}$$</p>
<p>$$-\frac{1}{2 \sqrt{3}},0,\frac{\sqrt{3}}{2},-\frac{1}{2 \sqrt{3}},-\frac{1}{2 \sqrt{3}}$$</p>
<p>$$-\frac{1}{2 \sqrt{5}},\frac{2}{\sqrt{5}},-\frac{1}{2 \sqrt{5}},-\frac{1}{2 \sqrt{5}},-\frac{1}{2 \sqrt{5}}$$</p>
<p>$$\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}}$$</p>
<p><b>Update 09/08</b>
I came across another interesting characterization, when d=2^k, for some k, then Walsh Functions form orthogonal basis for this matrix. In particular, let {$\mathbf{x_i}$} represent the list of vectors of binary expansion of integers 1 to d, ie {(0,0,0),(0,0,1),(0,1,0)...}. Then, rows (and columns) of $M$ define the orthonormal basis of matrix in question, where</p>
<p>$$M_{ij}=(-1)^{x_i \cdot x_j}$$</p>
| Community | -1 | <p>One way is to find the householder matrix Q that maps v to a multiple of e_1 (first coordinate basis vector). Then (since Q is symmtric and orthogonal) w_2=Q*e_2 ... will be a basis of the orthogonal complement of v.
Explicitly, I get
w_k = e_k - u
where u = (v+sqrt(d)*e_1)/(d+sqrt(d))</p>
|
2,109,197 | <p><strong>Update:</strong><br>
(because of the length of the question, I put an update at the top)<br>
I appreciate recommendations regarding the alternative proofs. However, the main emphasis of my question is about the correctness of the reasoning in the 8th case of the provided proof (with a diagram).</p>
<p><strong>Original question:</strong> </p>
<p>I would like to know, whether the following proof, is a valid way to prove that $a^2 + b^2 \neq 3c^2$ for all $a, b, c \in Z$ (except the trivial case, when $a=b=c=0$). More formally, we have to prove the correctness of the following statement:</p>
<p>$$P: (\forall a,b,c \in Z, a^2 + b^2 \neq 3c^2 \lor (a=b=c=0))$$</p>
<p><strong>Proof.</strong> <em>(by contradiction)</em><br>
For the sake of contradiction let's assume, that there exist such $a, b, c \in Z$, that $a^2 + b^2 = 3c^2$ (and the combination of $a,b,c$ is not a trivial case). More, formally, let's assume that $\neg P$ is <em>true</em>:</p>
<p>$$\neg P: (\exists a,b,c \in Z, a^2 + b^2 = 3c^2 \land \neg (a=b=c=0))$$</p>
<p>There are $2^3$ possible combinations of different parities of $a,b,c$ (8 disjoint cases, which cover entire $Z^3$). So, in order to prove the original statement, we have to consider each case, and show that the <em>true</em>-ness of the $\neg P$ always leads to some sort of contradiction.</p>
<p>Let's consider 8 possible cases (7 of which are simple, <strong>whereas the 8th case looks a bit intricate, and I am not sure regarding its correctness</strong>):</p>
<p><strong>Case 1)</strong> $a$ is odd, $b$ is odd, $c$ is odd<br>
Thus:<br>
$a = (2x + 1)$, $b = (2y + 1)$, $c = (2z + 1)$ for some $x, y, z \in Z$<br>
So:<br>
$$
a^2 + b^2 = 3c^2 \\
\implies (2x + 1) ^2 + (2y + 1)^2 = 3 \cdot (2z + 1)^2 \\
\implies 2 \cdot (2x^2 + 2x + 2y^2 + 2y + 1) = 2 \cdot (6z^2 + 6z + 1) + 1 \\
\implies even\ number = odd\ number \\
$$</p>
<p>However, the derived result contradicts to the fact that <em>odd numbers</em> and <em>even numbers</em> can't be equal.
Hence: $(even\ number = odd\ number) \land (even\ number \neq odd\ number)$, or equivalently: $(even\ number = odd\ number) \land \neg (even\ number = odd\ number)$. Contradiction.</p>
<p><strong>Case 2)</strong> $a$ is odd, $b$ is odd, $c$ is even<br>
Thus:<br>
$a = (2x + 1)$, $b = (2y + 1)$, $c = 2z$ for some $x, y, z \in Z$<br>
So:<br>
$$
a^2 + b^2 = 3c^2 \\
\implies (2x + 1) ^2 + (2y + 1)^2 = 3 \cdot (2z)^2 \\
\implies 2 \cdot (2x^2 + 2x + 2y^2 + 2y + 1) = 12z^2 \\
\implies 2 \cdot (x^2 + x + y^2 + y) + 1 = 6z^2 \\
\implies odd\ number = even\ number
$$
Contradiction.</p>
<p><strong>Case 3)</strong> $a$ is odd, $b$ is even, $c$ is odd<br>
Thus:<br>
$a = (2x + 1)$, $b = 2y$, $c = (2z + 1)$ for some $x, y, z \in Z$<br>
So:<br>
$$
a^2 + b^2 = 3c^2 \\
\implies (2x + 1) ^2 + (2y)^2 = 3 \cdot (2z + 1)^2 \\
\implies 4x^2 + 4x + 1 + 4y^2 = 12z^2 + 12z + 3 \\
\implies 4\cdot(x^2 + x + y^2) = 2 \cdot (6z^2 + 6z + 1) \\
\implies 2\cdot(x^2 + x + y^2) = 6z^2 + 6z + 1 \\
\implies even\ number = odd\ number
$$
Contradiction.</p>
<p><strong>Case 4)</strong> $a$ is odd, $b$ is even, $c$ is even<br>
The square of an odd number is odd (so, $a^2$ is odd).<br>
The square of an even number is even (so, $b^2$ and $3c^2$ are even).<br>
Fact: <em>the sum of an even number and an odd number is odd</em>.<br>
However, equality: $a^2 + b^2 = 3c^2$ leads to the conclusion, that: $odd\ number + even\ number = even\ number$<br>
Contradiction.</p>
<p><strong>Case 5)</strong> $a$ is even, $b$ is odd, $c$ is odd<br>
Symmetric to the Case 3 (because $a$ and $b$ are mutually exchangeable), which shows the contradiction.</p>
<p><strong>Case 6)</strong> $a$ is even, $b$ is odd, $c$ is even<br>
Symmetric to the Case 4, which shows the contradiction.</p>
<p><strong>Case 7)</strong> $a$ is even, $b$ is even, $c$ is odd<br>
Thus:<br>
$a = 2x$, $b = 2y$, $c = (2z + 1)$ for some $x, y, z \in Z$<br>
So:<br>
$$
a^2 + b^2 = 3c^2 \\
\implies 4x^2 + 4y^2 = 12z^2 + 12z + 3 \\
\implies even\ number = odd\ number
$$
Contradiction.</p>
<p><strong>Case 8)</strong> $a$ is even, $b$ is even, $c$ is even<br>
Thus:<br>
$a = 2x$, $b = 2y$, $c = 2z$ for some $x, y, z \in Z$<br>
So:
$$
a^2 + b^2 = 3c^2 \\
\implies 4x^2 + 4y^2 = 3 \cdot 4z^2 \\
\implies x^2 + y^2 = 3z^2
$$</p>
<p>Now, we are faced with the similar instance of the problem, <strong>however, the size of the problem is strictly smaller</strong> ($x = {a \over 2}$, $y = {b \over 2}$, $z = {c \over 2}$).<br>
At first glance, it seems that we have to consider again the eight possible parities of $x, y, z$. However, if we analyze all dependencies between the cases of the problem, we will notice that the only possible outcomes are either <em>contradiction</em> or <em>the trivial case</em>:</p>
<p><a href="https://i.stack.imgur.com/KVWUL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KVWUL.png" alt="enter image description here"></a></p>
<p>We have shown the contradiction in all cases, hence we have subsequently proved the original statement.
$\blacksquare$</p>
<p>So, I would like to know, if there is any problem with reasoning in the 8th case?</p>
| Community | -1 | <p>We know that $z_1,z_2,z_3$ are collinear iff there exists some $t\in\mathbb{R}$ such that $$t(z_1-z_3)=z_1-z_2.$$
The problem becomes trivial if $z_1=z_3$ so we can suppose this is not the case, and then it is legal to write
$$
t=\frac{z_1-z_2}{z_1-z_3}.
$$
Multiplying both sides of this equation by $|z_1-z_3|^2=(\overline{z_1-z_3})(z_1-z_3)$ we get
\begin{align}
t|z_1-z_3|^2=(\overline{z_1-z_3})(z_1-z_2)=&|z_1|^2-\bar z_1z_2 -\bar z_3z_1+\bar z_3z_2\\
=&|z_1|^2-\bar z_1z_2 -\bar z_3z_1\underbrace{-\bar z_2z_3+\bar z_2z_3}
_{=0}+\bar z_3z_2.\\
\end{align}
Expressing this as
$$
t|z_1-z_3|^2-|z_1|^2-(\overline{\bar z_3z_2}+\bar z_3z_2)=-(\bar z_1z_2 +\bar z_2z_3+\bar z_3z_1)
$$
and noticing that the LHS is a real number leads to the desired conclusion.</p>
|
2,679,153 | <p>Let $A\in\mathbb{R}^{n\times n}$ be a generic <em>lower triangular</em> matrix and let $P\in\mathbb{R}^{n\times n}$ be a symmetric <em>positive definite</em> matrix.</p>
<blockquote>
<p><strong>True or false.</strong> Does $AP + PA^\top=0$ imply $AP=0$? </p>
</blockquote>
| user | 505,767 | <p>Let $a^2+b^2=c^2$ then</p>
<ul>
<li>$\frac12 ab = 51.2$</li>
<li>$\frac{a}b=\frac{1}{1.6}$</li>
</ul>
<p>$$\implies b=1.6 a \implies0.8a^2=51.2\implies a=8 \quad b=12.8 \quad c=\sqrt{8^2+12.8^2}$$</p>
|
1,025,671 | <p>When we have something in this form</p>
<p>$$\sqrt{x + a} = \sqrt{y + b},$$</p>
<p>a common technique to solve is to square both side so that:</p>
<p>$$(\sqrt{x + a})^2 = (\sqrt{y + b})^2 \implies x + a = y + b.$$</p>
<p>I'm an engineer and not a mathematician. As I understand it engineers do lots of things that are mathematically frowned upon. However, while those things do make intuitive sense to me this technique makes no sense to me. Specifically what if we consider the alternative road.</p>
<p>\begin{align*}
\sqrt{x + a} &= \sqrt{y + b}\\
\sqrt{x + a} \cdot \sqrt{y + b} &= y + b\\
&\,\, \vdots
\end{align*}</p>
<p>It is unclear to me how we can conclude that $(\sqrt{x + a})^2 = (\sqrt{y + b})^2 \implies x + a = y + b$ is true.</p>
| Ojas | 154,392 | <p>This is true because square root is not a Many$\to$1 function. i.e., 2 numbers can't possibly have the same square root. Thus, if square root of 2 numbers is the same, it means that the 2 numbers must be equal.</p>
|
866,808 | <p>In my lecture notes:</p>
<p>Let $m,n\in \mathbb{N}$ be relatively prime. The fundamental theorem of arithmetic implies that each divisor of $mn$ is the product of two unique positive relatively prime integers $d_1|m$ and $d_2|n$.</p>
<p>Please could someone help me understand how this is implied? I have no idea</p>
<p>thanks!</p>
| Adam Hughes | 58,831 | <p>Simply write</p>
<blockquote>
<p>$$m=\prod_{i=1}^rp_i^{e_i},\quad n=\prod_{j=1}^sq_j^{f_j}$$</p>
</blockquote>
<p>Since they are coprime, no $p_i=q_j$.</p>
<p>A divisor of $mn$ is determined by taking each $p_i$ to some power $0\le n_{p_i}\le e_i$ and $q_j$ to power $0\le n_{q_j}\le f_j$. Grouping the primes $\{p:p|m\}$ and $\{q: q|n\}$ together, we get</p>
<blockquote>
<p>$$d_1=\prod_{i=1}^rp_i^{n_{p_i}},\quad d_2=\prod_{j=1}^sq_i^{n_{q_j}}$$</p>
</blockquote>
|
16,733 | <p>This is a variant on the question <a href="https://matheducators.stackexchange.com/questions/14492/small-real-numbers">small real numbers</a>.</p>
<p>I have a disagreement with someone about the meaning of "bigger" real numbers.</p>
<p>Say we have the real number <span class="math-container">$-1.$</span> Is <span class="math-container">$0$</span> "bigger" or "smaller" than <span class="math-container">$-1$</span>? In other words, should we interpret "bigger" to be a synonym of "greater than"?</p>
<p>This may be considered an "opinion-based" question, but is there a consensus about how to assign meaning to "bigger" or "smaller" (real) numbers?</p>
<p>My interpretation is that "bigger" real numbers have larger magnitude, irrespective of sign. For example, I consider that <span class="math-container">$-2$</span> is "bigger" than <span class="math-container">$-1.$</span> To avoid ambiguity (we are mathematicians!), I think it better to say/write "<span class="math-container">$-2$</span> has a larger magnitude/size than <span class="math-container">$1$</span>".</p>
<p>Further, I try to avoid using "bigger" or "smaller" with respect to comparing numbers unless it is clear we are comparing their magnitude (e.g. vectors or complex numbers), where we are only comparing non-negative values. I would stick to "greater / less than" or "closer to / further from zero".</p>
<p>Regards,</p>
| Tommi | 2,083 | <p>Since "bigger" and "smaller" are ambiguous, it is best to avoid them, as you mention. The methods you mention seem reasonable, though I am not native English speaker.</p>
<p>Someone may be able to refer to a credible source or official standard, but regardless, you can only know how someone else understands the terms by asking them. So you might as well do that and dispel the confusion.</p>
<p>A mathematician should know that definitions are matters of convenience and communication. As such, a mathematician can easily adapt to the terminology the other person prefers, for the purposes of a given discussion. Most people have less practice with working with explicit and sometimes arbitrary definitions than mathematicians (and philosophers) do.</p>
|
1,030,274 | <p>$A,B,X,Y$ are four invertible matrices. If $AYB=XY$, can I express matrix $X$ in the terms of $A$ and $B$?</p>
| user26857 | 121,097 | <blockquote>
<p>The ideal $I=(2,X)$ of $\mathbb Z[X]$ is not a direct sum of (non-zero) cyclic $\mathbb Z[X]$-modules.</p>
</blockquote>
<p>Let's suppose that $I=(2,X)$ is a direct sum of (non-zero) cyclic $\mathbb Z[X]$-modules. Then there exists a family $(N_{\alpha})_{\alpha\in A}$ of (non-zero) cyclic submodules of $I$ such that $I=\sum_{\alpha\in A}N_{\alpha}$, and $N_{\beta}\cap\sum_{\alpha\ne\beta}N_{\alpha}=0$ for all $\beta\in A$. In particular, $N_{\alpha}$ are principal ideals in $\mathbb Z[X]$. But $N_{\alpha}\cap N_{\beta}\ne 0$ for $\alpha\ne\beta$ (if $x_{\alpha}\in N_{\alpha}$ and $x_{\beta}\in N_{\beta}$, then $x_{\alpha}x_{\beta}\in N_{\alpha}\cap N_{\beta}$). This leads us to the conclusion that $|A|=1$, that is, $I$ is principal, <a href="https://math.stackexchange.com/questions/36169/show-that-langle-2-x-rangle-is-not-a-principal-ideal-in-mathbb-z-x">a contradiction</a>.</p>
|
267,051 | <p>Games appear in pure mathematics, for example, <a href="https://en.wikipedia.org/wiki/Ehrenfeucht%E2%80%93Fra%C3%AFss%C3%A9_game" rel="noreferrer">Ehrenfeucht–Fraïssé game</a> (in mathematical logic) and <a href="https://en.wikipedia.org/wiki/Banach%E2%80%93Mazur_game" rel="noreferrer">Banach–Mazur game</a> (in topology).</p>
<p>But the Game Theory behind those applications is not so deep, and we don't need to know some fundamental theorems for them. Maybe except Zarmelo Theorem.</p>
<p>Are there applications of <strong>advanced</strong> (anything behind the basic definitions) game theory ideas in pure mathematics?</p>
<p>Thanks!</p>
| Michael Greinecker | 35,357 | <p>The purity of math is in the eye of the beholder, but maybe some of the following examples qualify:</p>
<p>There is a proof of Kolmogorov's strong law of large numbers in the book <a href="http://www.probabilityandfinance.com/" rel="noreferrer">Probability and Finance</a> by Shafer and Vovk that uses the determinacy of quasi-Borel games. The book generally shows how one can use game theory to prove probabilistic results.</p>
<p>Versions of the minimax theorem can be used to prove results in convex analysis. There is even a journal called <a href="http://www.heldermann.de/MTA/mtacover.htm" rel="noreferrer">Minimax Theory and its Applications</a>. It is probably the most useful mathematical tool that came out of game theory.</p>
<p>Konrad Podczeck and I have a <a href="https://link.springer.com/article/10.1007/s00199-014-0815-1" rel="noreferrer">purification theorem</a> for measure-valued maps whose proof is at least heavily based on game theoretic intuitions.</p>
<p>I would say the most useful applications of game theory to other areas of mathematics are based on zero-sum games, which are of least interest from the perspective of game theory as a tool of social sciences. </p>
|
52,802 | <p>This is partly a programming and partly a combinatorics question.</p>
<p>I'm working in a language that unfortunately doesn't support array structures. I've run into a problem where I need to sort my variables in increasing order.</p>
<p>Since the language has functions for the minimum and maximum of two inputs (but the language does not allow me to nest them, e.g. <code>min(a, min(b, c))</code> is disallowed), I thought this might be one way towards my problem.</p>
<p>If, for instance, I have two variables $a$ and $b$, I only need one temporary variable so that $a$ ends up being less than or equal to $b$:</p>
<pre><code>t = min(a, b);
b = max(a, b);
a = t;
</code></pre>
<p>for three variables $a,b,c$, the situation is a little more complicated, but only one temporary variable still suffices so that $a \leq b \leq c$:</p>
<pre><code>a = min(a, b);
t = max(a, b);
c = max(t, c);
t = min(t, c);
b = max(a, t);
a = min(a, t);
</code></pre>
<p>Not having a strong combinatorics background, however, I don't know how to generalize the above constructions if I have $n$ variables in general. In particular, is there a way to figure out how many temporary variables I would need to sort out $n$ variables, and to figure out what is the minimum number of assignment statements needed for sorting?</p>
<p>Thanks in advance!</p>
| Gerry Myerson | 8,269 | <p>Patrick's got it. Here's the same, in a somewhat different presentation. </p>
<p>The symmetric difference of two sets is the stuff that's in exactly one of the two sets. If the symmetric difference of $A$ and $B$ is $A$, then $B$ can't have any elements that are in $A$ (since those elements of $A$ wouldn't be in the symmetric difference), and $B$ can't have any elements that aren't in $A$ (since those non-elements of $A$ would be in the symmetric difference), so $B$ can't have any elements. Thus if $B$ is to be the identity, $B$ must be empty; moreover, if $B$ is empty, it works as an identity element for symmetric difference. </p>
|
1,315,199 | <p>From the wikipedia article on sine waves:</p>
<blockquote>
<p>The sine wave is important in physics because it retains its wave
shape when added to another sine wave of the same frequency and
arbitrary phase and magnitude. It is the only periodic waveform that
has this property. This property leads to its importance in Fourier
analysis and makes it acoustically unique.</p>
</blockquote>
<p>I don't follow these statements. If you add two sine waves together with identical phase, isn't the amplitude doubled? Sure, it's still a sine wave, but isn't the same true of a square wave? If I add two square waves with equal phase, it's still a square wave, just with doubled amplitude. It has retained its shape. What am I missing?</p>
| Kishan | 247,422 | <p>Well you already gave the answer that when you add 2 sinusoids you get a sinusoid again.</p>
<p>You thought you gave a counter example by saying when 2 square wave get added it produces a square wave. But what you missed out is that a square wave can be represented by a summation of many sinusoids, which is the Fourier Series.</p>
<p>So in general "any" signal can be represented in terms of sinusoids by using Fourier Series and thus we only talk any theory in terms of sinusoids.</p>
<p>Hope you understood the crux of this discussion!</p>
|
1,225,359 | <p>Q: The sum of all the coefficients of the terms in the expansion of $(x+y+z+w)^{6}$ which contain $x$ but not $y$ is:</p>
<p>What I tried to do was make pairs of two terms and the expand it as a binomial expression and then again expand the binomial in the resulting series which gave me an expression with lot of unknowns and I got stuck. </p>
<p>Any help would be appreciated.
Thanks.</p>
| RE60K | 67,609 | <blockquote>
<p>Q: The sum of all the coefficients of the terms in the expansion of $(x+y+z+w)^{6}$ which contain $x$ but not $y$ is:</p>
</blockquote>
<p>Sum of terms with no y : $3^6$ (y=0 rest all 1)<br>
Sum of terms with no y and no x: $2^6$ (x,y=0 rest all 1)<br>
Sum of terms with no y but x: $3^6-2^6=665$ (subtract the above)</p>
|
1,683,414 | <p>$$\begin{cases}
x^2 = yz + 1 \\
y^2 = xz + 2 \\
z^2 = xy + 4
\end{cases}
$$</p>
<p>How to solve above system of equations in real numbers? I have multiplied all the equations by 2 and added them, then got $(x - y)^2 + (y - z)^2 + (x - z)^2 = 14$, but it leads to nowhere.</p>
| André Nicolas | 6,312 | <p>Multiplying both sides of the first equation by $z$, of the second by $x$, and of the third by $y$, we get
$$x^2z=z^2y+z, \quad y^2x=x^2z+2x,\quad z^2y=y^2x+4y.$$
Adding up and cancelling, we get<br>
$$2x+4y+z=0.$$
Similarly,
$$x^2y=y^2z+y, \quad y^2z=z^2x+2z, \quad z^2x=x^2y+4x,$$
giving<br>
$$4x+y+2z=0.$$ </p>
<p>To finish, use the linear equations to express $y$ and $z$ in terms of $x$, and substitute in $x^2=yz+1$.</p>
|
2,482,868 | <p>I am trying to find</p>
<p>$$ \lim\limits_{x\to \infty }[( 1+x^{p+1})^{\frac1{p+1}}-(1+x^p)^{\frac1p}],$$</p>
<p>where $p>0$. I have tried to factor out as</p>
<p>$$(1+x^{p+1})^{\frac1{p+1}}- \left( 1+x^{p}\right)^{\frac{1}{p}} =x\left(1+\frac{1}{x^{p+1}}\right)^{\frac{1}{p+1}}- x\left(1+\frac{1}{x^{p}}\right)^{\frac{1}{p}},$$
but still was not able to make progress. Any other approach to this is welcome.</p>
| Jack D'Aurizio | 44,121 | <p>For any $x>0$ and $\alpha>0$ we clearly have $(1+x^\alpha)^{\frac{1}{\alpha}}\geq x$. If $\alpha=p\in\mathbb{N}^+$,</p>
<p>$$ (1+x^p)^{\frac{1}{p}}=\text{GM}\left[\underbrace{x,\ldots,x}_{p-1\text{ times}},x+x^{1-p}\right]\leq \text{AM}\left[\underbrace{x,\ldots,x}_{p-1\text{ times}},x+x^{1-p}\right]=x+\frac{1}{px^{p-1}}$$
and $\left(x+\frac{1}{px^{p-1}}\right)^p\geq x^p+1$, for any $p>1$, can also be seen as a consequence of Bernoulli's inequality. It follows that the difference between $ (1+x^p)^{\frac{1}{p}}$ and $ (1+x^{p+1})^{\frac{1}{p+1}}$ goes to zero as $x\to +\infty$.<br></p>
<p>It just remains to study the case $p\in(0,1]$.</p>
|
685,642 | <p>I am trying to find a basis for the set of all $n \times n$ matrices with trace $0$. I know that part of that basis will be matrices with $1$ in only one entry and $0$ for all others for entries outside the diagonal, as they are not relevant.</p>
<p>I don't understand though how to generalize for the entries on the diagonal. Maybe just one matrix with $1$ in the $(1, 1)$ position and a $-1$ in all other $n - 1$ positions?</p>
| Ben Grossmann | 81,360 | <p>In order to finish constructing your basis, you could add the set of matrices for which the $(1,1)$ entry is $1$, the $(i,i)$ entry is $-1$ for some $i \neq 1$, and all other entries are zero.</p>
<p>Note that the space of $n\times n$ matrices with trace $0$ is $n^2 - 1$ dimensional, so you should have this many elements in your basis in total.</p>
|
685,642 | <p>I am trying to find a basis for the set of all $n \times n$ matrices with trace $0$. I know that part of that basis will be matrices with $1$ in only one entry and $0$ for all others for entries outside the diagonal, as they are not relevant.</p>
<p>I don't understand though how to generalize for the entries on the diagonal. Maybe just one matrix with $1$ in the $(1, 1)$ position and a $-1$ in all other $n - 1$ positions?</p>
| Yulia Alexandr | 337,050 | <p>Since you have to find the dimension of the subspace of all matrices whose trace is $0$, having a linear transformation T: $M(n×n)→ ℝ$, all it really comes down to is finding the size of <em>ker(T)</em>. </p>
<p>In order to do so, notice that the standard matrix for the given transformation will have the dimension $1$ × $n^2$. Hence, applying Gauss Jordan, we get only one pivot variable and $n^2-1$ free variables. Since the number of free variables gives us the nullity of T (dimension of the kernel), and it is equal to the dimension of the subspace we're looking for, dimension of all <em>n</em> × <em>n</em> matrices with zero trace is simply $n^2-1$.</p>
<p>Then you can figure out what the actual basis could look like.</p>
|
184,940 | <p>Gauss has proven in his famous Theorema Egregium, that it is possible, to calculate the gaussian curvature from measuring angles and distances on the surface, irrespective of how the surface is embedded into space.</p>
<blockquote>
<p><strong>Question:</strong></p>
<p>is it also possible, to calculate the euclidean distance of two points on the surface, also from distance- and angle-measurements on the surface, alone?</p>
</blockquote>
| Thomas Richard | 8,887 | <p>The answer is no.
The fact that you can bend without stretching a piece of paper should convince you of that. Consider for instance a piece of a plane and piece of a cylinder.</p>
<p>The question you ask can be rephrased in the following way: take a 2 dimensional Riemannian manifold $(M^2,g)$. Under which condition is there a unique isometric embedding $f:M^2\to\mathbb{R}^3$ which is unique up to rigid motion ? </p>
<p>The only case where I know this is true is the case of convex surfaces: assume $M$ is diffeomorphic to $\mathbb{S}^2$ and has nonnegative Gauss curvature. Then Cohn-Vossen has proved that there exist a unique convex body $K\subset \mathbb{R}^3$ (up to rigid motion) such that $(M^2,g)$ is isometric to the boundary of $K$ with the induced Riemannian metric. This shows that, on a theoretical level the intrinsic metric on $M^2$ determines the distance of between two point in the extrinsic distance given by the embedding. </p>
<p>However, from what I know of the proof building the embedding from the metric is not easy at all and won't give any practical way to compute the extrinsic distance between two points in terms of the intrinsic geometry of the surface.</p>
<p><strong>EDIT:</strong> I'll come back later to add some references.</p>
|
1,437,073 | <p>Ok guys, I have to solve this ODE</p>
<p>$$
\frac{d^2y}{dx^2}=f(x), \quad
x>0,\quad y\left(0\right) = 0, \quad
\left.\frac{dy}{dx}\right\lvert_{x=0}=0
$$
The solution I should get is in the form of
$$y\left(x\right)=\int_0^x k\left(t\right)\, dt $$
Moreover, I should tell what the function $\,k\left(t\right)\,$ is in a simple form. I have tried by substitution, with $\,u=y',\,$ but I have to integrate two times $\,f\left(x\right)\,$, and this seems not at all like a simple form... The function $\,k\left(t\right)\,$ is undoubtedly an exponential, but in which form? Hope that somebody can help!</p>
| Blazej | 155,834 | <p>You are given a form that will do the trick. The thing you need to do now is to differentiate it twice. By comparing result with your equation you will get condition on function $k$ that will make it work. This might involve double integration but you can try to simplify it by changing the order of integrals. This way you can make it a single integral (over slightly modified function)</p>
|
509,928 | <p>There is a square cake. It contains N toppings - N disjoint axis-aligned rectangles. The toppings may have different widths and heights, and they do not necessarily cover the entire cake.</p>
<p>I want to divide the cake into 2 non-empty rectangular pieces, by either a horizontal or a vertical cut, such that the number of toppings I destroy (i.e. cross in the interior) is minimized.</p>
<p>What is the number of toppings I will have to destroy, in the worst case, as a function of N?</p>
<p>CURRENT BOUNDS:</p>
<p><strong>Upper bound $N/2$</strong>:</p>
<p>Take any horizontal cut. If it crosses no more than N/2 toppings, then we are done. Otherwise, make a vertical cut between two of the crossed rectangles. This vertical cut does not cross any rectangle crossed by the horizontal cut, therefore it crosses at most N/2 toppings.</p>
<p><strong>Lower bound $N/4$</strong>:</p>
<p>In the following cake, with 4 toppings, every cut must cross at least 1 topping:</p>
<pre><code>aaaaaaaa bb
aaaaaaaa bb
cc ..... bb
cc ..... bb
cc ..... bb
cc dddddddd
cc dddddddd
</code></pre>
<p>As MvG suggested, it is possible to cut each rectangle into $N/4$ parallel strips, forcing a cut to destroy at least $⌊N/4⌋$ toppings.</p>
<p>NOTE: I just found out that this problem is related to the topic of <a href="https://en.wikipedia.org/wiki/Geometric_separator" rel="nofollow">Geometric separators</a>. The lower bound example and the upper bound proofs are given in Section 4 of <a href="http://users.monash.edu.au/~nwormald/papers/geomsep.ps.gz" rel="nofollow">Smith and Wormald (1998)</a>. There is still a gap between the lower and upper bound.</p>
| san | 229,191 | <p>The lower bound $r=\lfloor \frac N4\rfloor$ is also an upper bound: </p>
<p>Assume that in a square with toppings one must cut at least $r>1$ toppings with every cut.
Number the edges of the square with $E_1,E_2,E_3,E_4$ clockwise, and set $\varepsilon_i=min\{d(E_i,R_j)\ :\ d(E_i,R_j)>0\}$,
where $R_j$ is any of the edges of a rectangle of the toppings which is parallel to $E_i$. Note that $\varepsilon_i$ can be the distance of a rectangle to the edge $E_i$ or the width of that rectangle (and in that case the rectangle has an edge in $E_i$). Choose one of the rectangles which define $\varepsilon_i$ and name it $R_i$. </p>
<p>Set $T_i$ to be the line parallel to $E_i$ at a distance $\varepsilon_i$. By our assumption, $T_i$ must cut at least $r$ rectangles, denote the set of these rectangles with $S_i$. Note that, by the minimality of $\varepsilon_i$, all rectangles in $S_i$ have an edge contained in $E_i$. </p>
<p>Clearly $R_i\notin S_i$, and it is also easy to check that $R_i\notin S_{i+2}$, the subindices taken $i\mod 4$, since then there would be a cut along a side of $R_i$ which cuts at most 1 topping. We also have $S_1\cap S_3=\emptyset$, since a rectangle $R$ in $S_1\cap S_3$ allows us to cut through the middle of $R$, parallel to $E_2$, cutting through only one rectangle (topping). Similarly $S_2\cap S_4=\emptyset$. Now we assume that for a given number $r$ (of minimal number of toppings destroyed by a cut) the number $N$ of rectangles is minimal. We can assume that in this configuration we can choose the rectangles $R_i$ to have an edge contained in $E_i$. In fact, if no rectangle with an edge in $T_i$ touches $E_i$, then we can extend one of the rectangles with distance
$\varepsilon_i$ so that it touches $E_i$. The new configuration has the same number of rectangles, and the number of toppings destroyed by each cut is either equal or has increased by one.</p>
<p>Let $r_1$ be the new minimal number of toppings destroyed by a cut, and so $r_1\ge r$. Note that eventually the $\varepsilon_i$ have changed. But now for each edge $E_i$ there are $r_1+1$ rectangles with an edge contained in $E_i$, the $r_1$ rectangles in (the new) $S_1$ and the rectangle $R_i$.
None of these rectangles can have two edges contained in opposite edges $E_i,E_{i+2}$, and there can be at most one rectangle with a corner in $E_i,E_{i+1}$. Hence there are at least $|S_1|+|S_2|+|S_3|+|S_4|+4-4$ distinct rectangles. We arrive at
$$
N\ge |S_1|+|S_2|+|S_3|+|S_4|\ge 4r_1\ge 4r\quad\Rightarrow \quad 4r\le N \quad\Rightarrow \quad r\le \left\lfloor \frac N4\right\rfloor.
$$ </p>
|
463,619 | <p>let us consider following problem:</p>
<p>Roger sold a watch at a profit of $10$%. If he had bought it at $10\%$ less and sold it for $13$ dollar less,then he would have made a profit of $15$%. What is the cost price of the watch?</p>
<p>suppose that price of watch is $x$ dollar, $profit=sell -cost $</p>
<p>so let us denote price of watch by $x$,then if he would buy it $10$ less,then cost of this would be $x-0.1*x=0.9*x$, and if it sold $13$ dollar less,then sold price would be $x-13$, profit is $15$%,then it would be $0.15*x$ right? or we have</p>
<p>$(x-13)-0.9*x=0.15*x$</p>
<p>but it makes negative equation like $13=-0.05*x$,so what is my mistake,how can i used information like Ronger sold watch at a profit of $10\%$? Thanks in advance</p>
| lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>Let Roger has bought the watch at $100x$ dollar</p>
<p>So, the current selling price $=110x$ dollar</p>
<p>If he had bought it at $10$% less, the buying price would have been $=90x$ dollar</p>
<p>If he had sold it for $13$ dollar less, selling price $=110x-13$ dollar</p>
<p>So, $$90x \left(1+\frac{15}{100}\right)=110x-13$$</p>
|
4,154,205 | <p>I study maths as a hobby. I am stuck on this question:</p>
<p>Find the values of c for which the line <span class="math-container">$2x-3y = c$</span> is a tangent to the curve <span class="math-container">$x^2+2y^2=2$</span> and find the equation of the line joining the points of contact.</p>
<p>I have established that <span class="math-container">$c=\pm \sqrt17$</span>.</p>
<p>To get the line joining the points of contact I thought one way would be to start by finding the points of contact.</p>
<p>For <span class="math-container">$c =\sqrt 17$</span> the equation of the line is y = <span class="math-container">$\frac{2x-\sqrt 17}{3}$</span> and the equation of the curve is <span class="math-container">$y = \sqrt \frac{2-x^2}{2}$</span></p>
<p>At the tangent</p>
<p><span class="math-container">$\frac{2x-\sqrt 17}{3} = \sqrt \frac{2-x^2}{2}$</span></p>
<p><span class="math-container">$\rightarrow \frac{4x^2-4x\sqrt17+17}{9} = \frac{2-x^2}{2}$</span></p>
<p><span class="math-container">$\rightarrow 17x^2-8x\sqrt17+16=0$</span></p>
<p><span class="math-container">$\rightarrow x = \frac{8\sqrt17\pm\sqrt1088-1088}{34}=\frac{8\sqrt17}{34}$</span> but proceeding in this fashion seems very messy and I don't see how it will lead to the answer given in the book as <span class="math-container">$3x+4y=0$</span></p>
| Eric Wofsey | 86,856 | <blockquote>
<p>I can really only understand this through examples, though.</p>
</blockquote>
<p>There's nothing wrong with this! "Without loss of generality" is an imprecise term and can be used in a wide variety of contexts, so there's no fixed rules on exactly how it can be used or what exactly it means. Ultimately, it is intentionally introducing an easy-to-fill gap in a proof and relying on the reader to fill the gap, giving them a hint that it can be done by some sort of symmetry argument or easy reduction to a special case. Your understanding of the meaning seems pretty much correct.</p>
<p>As for the specific example you're asking about, to justify the "without loss of generality" you have to actually fill in the gap in the proof that it introduces. That is, you have to prove the statement in the general case (where some <span class="math-container">$X_i$</span> may be empty) using the specific case (where you assume they are all nonempty). You should never use "without loss of generality" unless you yourself know how to fill in the gap in the proof and trust that your reader can do so as well. There are various ways to do that in this case; here is one. Let <span class="math-container">$Y_i=X_i\cup\{0\}$</span> for each <span class="math-container">$i$</span>. Then each <span class="math-container">$Y_i$</span> is nonempty, and is still countable. So now we are in the specific case where none of our sets are empty and we can deduce that <span class="math-container">$\bigcup_{i\in\mathbb{N}}Y_i$</span> is countable. But <span class="math-container">$\bigcup_{i\in\mathbb{N}}Y_i=\bigcup_{i\in\mathbb{N}}X_i\cup\{0\}$</span>, so <span class="math-container">$\bigcup_{i\in\mathbb{N}}X_i$</span> is a subset of a countable set and thus countable.</p>
<p>Here, the essence of the "without loss of generality" is that if you <em>changed</em> the setup in some simple way to make the assumption true (e.g. added an element to the <span class="math-container">$X_i$</span>'s so they can't be empty), then this would still give a conclusion that would be strong enough to deduce what you originally wanted (since you would only enlarge the union, and a subset of a countable set is still countable).</p>
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3,892,583 | <p>I'm trying to find an example of a function <span class="math-container">$f: A \to B$</span> and <span class="math-container">$X \subset A$</span> so that <span class="math-container">$f^{-1}(f(X)) \ne X$</span>, and similarly where <span class="math-container">$Y \subset B$</span> so that <span class="math-container">$f(f^{-1}(Y)) \ne Y$</span>.</p>
<p>I thought to have <span class="math-container">$f = x^2$</span>, which has no inverse, thus making it vacuously true that <span class="math-container">$f^{-1}(f(X)) \ne X$</span> and <span class="math-container">$f(f^{-1}(Y)) \ne Y$</span>, but that seems like a copout. Any help is greatly appreciated.</p>
| peter.petrov | 116,591 | <p>You should apply the substitution: <span class="math-container">$x = t^{14}$</span></p>
<p>This will get you a rational function of <span class="math-container">$t$</span>.</p>
<p>And then... as we know all rational functions can be integrated,<br />
there is a well-known procedure for that.</p>
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3,892,583 | <p>I'm trying to find an example of a function <span class="math-container">$f: A \to B$</span> and <span class="math-container">$X \subset A$</span> so that <span class="math-container">$f^{-1}(f(X)) \ne X$</span>, and similarly where <span class="math-container">$Y \subset B$</span> so that <span class="math-container">$f(f^{-1}(Y)) \ne Y$</span>.</p>
<p>I thought to have <span class="math-container">$f = x^2$</span>, which has no inverse, thus making it vacuously true that <span class="math-container">$f^{-1}(f(X)) \ne X$</span> and <span class="math-container">$f(f^{-1}(Y)) \ne Y$</span>, but that seems like a copout. Any help is greatly appreciated.</p>
| lab bhattacharjee | 33,337 | <p>Hint:</p>
<p>As <span class="math-container">$[2,7,14]=14,$</span></p>
<p>Let <span class="math-container">$x^{1/14}=y, x=y^{14}, dx=14y^{13}dy$</span></p>
<p><span class="math-container">$$\int\dfrac{y^2+y^7}{y^{16}+y}\cdot14y^{13}dy$$</span></p>
<p><span class="math-container">$$=14\int\dfrac{y^{15}(1+y^5)}{y(1+y^{15})}dy$$</span></p>
<p><span class="math-container">$$=14\int\dfrac{y^{10}}{1-y^5+y^{10}}\cdot y^4dy$$</span></p>
<p>Set <span class="math-container">$y^5=z$</span></p>
<p>The rest is left as an exercise!</p>
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