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 |
|---|---|---|---|---|
3,720,272 | <p>My textbook employs a brute force method: add the number of committees that could be formed with one woman, two women, and three women in them. Then, the total number of such committees will be:
<span class="math-container">$$\left(\begin{smallmatrix} 8 \\ 1 \end{smallmatrix}\right)\cdot\left(\begin{smallmatrix} 10 \\ 2 \end{smallmatrix}\right) + \left(\begin{smallmatrix} 8 \\ 2 \end{smallmatrix}\right)\cdot\left(\begin{smallmatrix} 10 \\ 1 \end{smallmatrix}\right) + \left(\begin{smallmatrix} 8 \\ 3 \end{smallmatrix}\right)\cdot\left(\begin{smallmatrix} 10 \\ 0 \end{smallmatrix}\right) = 360 + 280 + 56 = 696$$</span>
(The number of ways of choosing one woman from eight times that of choosing two men from ten + ...)</p>
<p>I had solved the question with this reasoning: You can choose one woman from eight as a member of the committee, and for the remaining two positions, you could choose either a man or a woman. This is equivalent to choosing two people from <span class="math-container">$(10 + 8) - 1 = 17$</span>. Thus, the number of possible ways to form such a committee is:
<span class="math-container">$$\left(\begin{smallmatrix} 8 \\ 1 \end{smallmatrix}\right)\cdot\left(\begin{smallmatrix} 17 \\ 2 \end{smallmatrix}\right) = 8 \cdot 136 = 1088$$</span></p>
<p>What mistake have I made?</p>
| Anurag A | 68,092 | <p>Suppose <span class="math-container">$w_1$</span> was the first woman chosen among possible <span class="math-container">$8$</span>, then say <span class="math-container">$m_1,w_3$</span> were chosen. So your committee is <span class="math-container">$w_1,m_1,w_3$</span>.</p>
<p>In your way of counting: what if <span class="math-container">$w_3$</span> is chosen as the first woman, followed by <span class="math-container">$m_1, w_1$</span>? Then too the committee is same as before (<span class="math-container">$w_3,m_1,w_1$</span>) but you are counting this as a different committee.</p>
|
3,720,272 | <p>My textbook employs a brute force method: add the number of committees that could be formed with one woman, two women, and three women in them. Then, the total number of such committees will be:
<span class="math-container">$$\left(\begin{smallmatrix} 8 \\ 1 \end{smallmatrix}\right)\cdot\left(\begin{smallmatrix} 10 \\ 2 \end{smallmatrix}\right) + \left(\begin{smallmatrix} 8 \\ 2 \end{smallmatrix}\right)\cdot\left(\begin{smallmatrix} 10 \\ 1 \end{smallmatrix}\right) + \left(\begin{smallmatrix} 8 \\ 3 \end{smallmatrix}\right)\cdot\left(\begin{smallmatrix} 10 \\ 0 \end{smallmatrix}\right) = 360 + 280 + 56 = 696$$</span>
(The number of ways of choosing one woman from eight times that of choosing two men from ten + ...)</p>
<p>I had solved the question with this reasoning: You can choose one woman from eight as a member of the committee, and for the remaining two positions, you could choose either a man or a woman. This is equivalent to choosing two people from <span class="math-container">$(10 + 8) - 1 = 17$</span>. Thus, the number of possible ways to form such a committee is:
<span class="math-container">$$\left(\begin{smallmatrix} 8 \\ 1 \end{smallmatrix}\right)\cdot\left(\begin{smallmatrix} 17 \\ 2 \end{smallmatrix}\right) = 8 \cdot 136 = 1088$$</span></p>
<p>What mistake have I made?</p>
| richard gayle | 164,419 | <p>Anurag has put a finger on it. Your way of counting implicitly makes order of selection matter. Here's a check though. The number of committees with at least one woman should be the total number of all committees less the number that contain only men:</p>
<p><span class="math-container">$$C(18,3)-C(10,3)=696$$</span>.</p>
|
2,831,270 | <p>I am quite fascinated by the formula for the Mellin transform of the Gaussian Hypergeometric Function, which is given by:</p>
<blockquote>
<p><span class="math-container">$$\mathcal M [_2F_1(\alpha,\beta;\gamma;-x)] = \frac {B(s,\alpha-s)B(s,\beta-s)}{B(s,\gamma-s)}$$</span></p>
</blockquote>
<p><em>Source : <a href="https://authors.library.caltech.edu/43489/1/Volume%201.pdf" rel="nofollow noreferrer">Table of Integral Transforms</a> page <span class="math-container">$336$</span>, <span class="math-container">$6.9 (3)$</span></em></p>
<p>I have found this within a table of integral transforms of various functions and I would be really interested in a proof for this formula.</p>
| mrtaurho | 537,079 | <p><strong>Additum</strong></p>
<p>Recently I have come across <a href="https://en.wikipedia.org/wiki/Ramanujan%27s_master_theorem" rel="nofollow noreferrer">Ramanujan's Master Theorem</a>. This Theorem provides an elegant way to show the given relation. Therefore lets write the Gaussian Hypergeometric Function as infinite power series</p>
<p><span class="math-container">$$_2F_1(\alpha,\beta;\gamma;-x)=\sum_{k=0}^{\infty}\frac{\Gamma(\alpha+k)}{\Gamma(\alpha)}\frac{\Gamma(\beta+k)}{\Gamma(\beta)}\frac{\Gamma(\gamma)}{\Gamma(\gamma+k)}\frac{(-x)^k}{k!}=\sum_{k=0}^{\infty}\phi(k)\frac{(-x)^k}{k!}$$</span></p>
<p>For an analytic function <span class="math-container">$f(x)$</span> which is in the form of the last sum - especially with some <span class="math-container">$\phi(k)$</span> and a negative <span class="math-container">$x$</span> argument - the Mellin Transform of this function is given by</p>
<blockquote>
<p><span class="math-container">$$\int_0^{\infty}x^{s-1}f(x)dx=\Gamma(s)\phi(-s)$$</span></p>
</blockquote>
<p>From hereon by plugging in <span class="math-container">$_2F_1(\alpha,\beta;\gamma;-x)$</span> as <span class="math-container">$f(x)$</span> we get</p>
<p><span class="math-container">$$\begin{align}
\int_0^{\infty}x^{s-1}~_2F_1(\alpha,\beta;\gamma;-x)dx~&=~\Gamma(s)\phi(-s)\\
&=~\Gamma(s)\frac{\Gamma(\alpha-s)}{\Gamma(\alpha)}\frac{\Gamma(\beta-s)}{\Gamma(\beta)}\frac{\Gamma(\gamma)}{\Gamma(\gamma-s)}\\
&=~\frac{\Gamma(s)\Gamma(\alpha-s)}{\Gamma(\alpha)}\frac{\Gamma(s)\Gamma(\beta-s)}{\Gamma(\beta)}\frac{\Gamma(\gamma)}{\Gamma(s)\Gamma(\gamma-s)}\\
&=~\frac{B(s,\alpha-s)B(s,\beta-s)}{B(s,\gamma-s)}
\end{align}$$</span></p>
<blockquote>
<p><span class="math-container">$$\therefore~\mathcal M [_2F_1(\alpha,\beta;\gamma;-x)] = \frac {B(s,\alpha-s)B(s,\beta-s)}{B(s,\gamma-s)}$$</span></p>
</blockquote>
|
2,377,946 | <blockquote>
<p>The integral is:
$$\int_0^a \frac{x^4}{(x^2+a^2)^4}dx$$</p>
</blockquote>
<p>I used an approach that involved substitution of x by $a\tan\theta$. No luck :\ . Help?</p>
| Math Lover | 348,257 | <p>One way to solve this problem is to do integration by parts.<br>
$$\int{x^3 \frac{x}{(x^2+a^2)^4}dx}=-\frac{x^3}{6(x^2+a^2)^3}+\frac{1}{2}\int{x\frac{x}{(x^2+a^2)^3} dx}.$$
Continuing further
$$\int{x\frac{x}{(x^2+a^2)^3} dx} = -\frac{1}{4} \frac{x}{(x^2+a^2)^2}+\frac{1}{4}\int{\frac{1}{(x^2+a^2)^2}dx}.$$
Now you can substitute $x=a\tan{\theta}$ to get the final result.</p>
|
512,768 | <p>I am trying to intuitively understand why the solution to the following problem is $-2$. $$\lim_{x\to\infty}\sqrt{x^2-4x}-x$$
$$\lim_{x\to\infty}(\sqrt{x^2-4x}-x)\frac{\sqrt{x^2-4x}+x}{\sqrt{x^2-4x}+x}$$
$$\lim_{x\to\infty}\frac{x^2-4x-x^2}{\sqrt{x^2-4x}+x}$$
$$\lim_{x\to\infty}\frac{-4x}{\sqrt{x^2-4x}+x}$$
$$\lim_{x\to\infty}\frac{-4}{\sqrt{1-\frac{4}{x}}+1}$$
$$\frac{-4}{\sqrt{1-0}+1}$$
$$\frac{-4}{2}$$
$$-2$$
I can understand the process that results in the answer being $-2$. However, I expected the result to be $0$. I have learned that when dealing with a limit approaching $\infty$, only the highest degree term matters because the others will not be as significant. For this reason, I thought that the $4x$ would be ignored, resulting in:
$$\lim_{x\to\infty}\sqrt{x^2-4x}-x$$
$$\lim_{x\to\infty}\sqrt{x^2}-x$$
$$\lim_{x\to\infty}x-x$$
$$\lim_{x\to\infty}0$$
$$0$$
Why is the above process incorrect?</p>
| Chris Bonnell | 97,819 | <p>Intuitively, the thing you want to look at with your second attempt is that it's infinity minus infinity. Because the highest order terms are on the same order of magnitude and cancel exactly, the lower-order terms are indeed important to determine what the limit will be.</p>
|
3,956,913 | <p>For this equation :</p>
<blockquote>
<p><span class="math-container">${ (x^2 - 7x + 11)}^{x^2 - 13x +42}=1$</span></p>
</blockquote>
<p>The integer solutions of <span class="math-container">$x$</span> found by WolframAlpha using inverse (logarithmic) function are <span class="math-container">$ 2 , 5 , 6 , 7 .$</span> Why it cannot find the other solutions 3 and 4 ? Is there any full proof method to find all the solutions at once ?</p>
| player3236 | 435,724 | <p>Notice that <span class="math-container">$n^{1/n} > 1$</span> for all <span class="math-container">$n \in \mathbb N$</span>.</p>
<p>Raise both sides of <span class="math-container">$n^{1/n} \stackrel?> (n+1)^{1/(n+1)}$</span> to the <span class="math-container">$n(n+1)$</span>-th power, now we just need to show that</p>
<p><span class="math-container">$$n^{n+1} > (n+1)^n \text{ for } n \ge 3$$</span></p>
<p>which can be shown by induction quite easily.</p>
<p>See <a href="https://math.stackexchange.com/questions/77935/">Prove by induction that for all $n \geq 3$: $n^{n+1} > (n+1)^n$</a>.</p>
|
1,419,483 | <p>Can anyone please help me in solving this integration problem $\int \frac{e^x}{1+ x^2}dx \, $?</p>
<p>Actually, I am getting stuck at one point while solving this problem via integration by parts.</p>
| Will R | 254,628 | <p>This is an alternative to the other answer I have provided. I have decided to add it as another answer because I think it uses a sufficiently different approach, and nobody else seems to have hinted at it. We begin just as we begun in my other answer to this same question, and continue until we reach $$\int e^{\tan{u}}\,\mathrm{d}u.$$ From here, the two answers diverge radically, and the content of this post will be entirely concerned with evaluating this integral, which in turn answers the question.</p>
<p>In the other answer, we expanded $e^{\tan{u}}$ as a power series in $\tan{u}$. Now, instead, we shall first expand $\tan{u}$ as a power series in terms of $u$. Then $e^{\tan{u}}$ is an infinite product of exponentials, each of which can be expanded as a power series. From this sequence of expansions, we produce a power series for $e^{\tan{u}}$ which can be integrated term-by-term.</p>
<p>The <a href="https://en.wikipedia.org/wiki/Taylor_series#List_of_Maclaurin_series_of_some_common_functions" rel="nofollow">power series</a> for $\tan{u}$, valid for $|u|<\pi/2$, is $$\tan{u} = \sum_{n=1}^{\infty} {B_{2n}(-4)^{n}(1-4^{n}) \over (2n)!}u^{2n-1} = u+\frac{u^{3}}{3}+\frac{2u^{5}}{15}+\ldots.$$</p>
<p>Hence, for $|u|<\pi/2$, we have
\begin{eqnarray*}
\exp{\tan{u}} & = & \exp{\left(\sum_{n=1}^{\infty} {B_{2n}(-4)^{n}(1-4^{n}) \over (2n)!}u^{2n-1}\right)}\\
& = & \prod_{n=1}^{\infty}\exp{\left( {B_{2n}(-4)^{n}(1-4^{n}) \over (2n)!}u^{2n-1}\right)}\\
& = & \exp{u}\cdot\exp{\frac{u^{3}}{3}}\cdot\exp{\frac{2u^{5}}{15}}\cdot\ldots\\
& = & \sum_{k=1}^{\infty} \left({B_{2}^{k}(-4)^{k}(1-4)^{k} \over k!2!^{k}}u^{k}\right) \cdot \sum_{k=1}^{\infty} \left({B_{4}^{k}(-4)^{2k}(1-4^{2})^{k} \over k!4!^{k}}u^{3k}\right)\cdot\ldots\\
& = & \left( 1+\frac{u}{1!}+\frac{u^{2}}{2!}+\ldots \right)\left( 1+\frac{u^{3}}{3^{1}\cdot1!}+\frac{u^{6}}{3^{2}\cdot2!}+\ldots \right)\left( 1+\frac{2u^{5}}{15\cdot1!}+\frac{2^{2}u^{10}}{15^{2}\cdot2!}+\ldots \right)\\
& = & 1 + u + \frac{u^{2}}{2!} + \left(\frac{u^{3}}{3!}+\frac{u^{3}}{3\cdot1!}\right) + \left(\frac{u^{4}}{4!}+\frac{u}{1!}\cdot\frac{u^{3}}{3\cdot1!}\right)\\
&& \;\; + \left(\frac{u^{5}}{5!} +\frac{u^{2}}{2!}\cdot\frac{u^{3}}{3\cdot1!}+ \frac{2u^{5}}{15}\right)+\ldots\\
& = & 1 + u + \frac{u^{2}}{2} + \frac{u^{3}}{2}+\frac{3u^{4}}{8}+\frac{37u^{5}}{120}+\ldots.
\end{eqnarray*}</p>
<p>This gives us a series that I think we should be able to integrate term-by-term, to get
\begin{eqnarray*}
\int e^{\tan{u}}\,\mathrm{d}u & = & \int(1 + u + \frac{u^{2}}{2} + \frac{u^{3}}{2}+\frac{3u^{4}}{8}+\frac{37u^{5}}{120}+\ldots)\,\mathrm{d}u\\
& = & \text{constant} + u + \frac{u^{2}}{2} + \frac{u^{3}}{6}+\frac{u^{4}}{8}+\frac{3u^{5}}{40}+\frac{37u^{6}}{720}+\ldots.
\end{eqnarray*}</p>
<p>Therefore, assuming that our interval of integration is within $|\tan^{-1}{x}|<\pi/2$, we have
\begin{eqnarray*}
\int\frac{e^{x}}{1+x^{2}}\,\mathrm{d}x = \text{constant} + \tan^{-1}{x} + \frac{(\tan^{-1}{x})^{2}}{2}+ \frac{(\tan^{-1}{x})^{3}}{6}+\frac{(\tan^{-1}{x})^{4}}{8}+\frac{3(\tan^{-1}{x})^{5}}{40}+\frac{37(\tan^{-1}{x})^{6}}{720}+\ldots,
\end{eqnarray*}
and I'm pretty sure we can adjust this to other intervals of integration using the fact that $\tan{(u+\pi)}=\tan{u}$ for all $u\in\mathbb{R}$ for which $\tan{u}$ is defined.</p>
|
4,411,247 | <blockquote>
<p>If <span class="math-container">$G$</span> is finite group, how to prove that <span class="math-container">$f(g)=ag$</span>, <span class="math-container">$a \in G$</span>, is a bijection for all <span class="math-container">$g \in G$</span>? Here <span class="math-container">$ag$</span> is <span class="math-container">$a \cdot g$</span>, where <span class="math-container">$\cdot$</span> is the operator from the group <span class="math-container">$G$</span>.</p>
</blockquote>
<p>This is what I've tried so far:</p>
<p><span class="math-container">$f(g)=ag
\\
a^{-1}g(g)=g.
$</span></p>
<p>Since the inverse has been dotted so <span class="math-container">$f$</span> is a bijection, I would like to know if there's a misstep on that.</p>
| Shaun | 104,041 | <p>Fix <span class="math-container">$a\in G$</span>.</p>
<p>Define</p>
<p><span class="math-container">$$\begin{align}
\hat{f}: G&\to G,\\
x&\mapsto a^{-1}x.
\end{align}$$</span></p>
<p>Then for any <span class="math-container">$g\in G$</span>, we have</p>
<p><span class="math-container">$$\begin{align}
(f\circ \hat{f})(g)&=f(\hat{f}(g))\\
&=f(a^{-1}g)\\
&=a(a^{-1}g)\\
&=(aa^{-1})g\\
&=eg\\
&=g\\
&={\rm id}_G(g),
\end{align}$$</span></p>
<p>where</p>
<p><span class="math-container">$$\begin{align}
{\rm id}_G:G&\to G,\\
x&\mapsto x
\end{align}$$</span></p>
<p>is the identity function on <span class="math-container">$G$</span>. But <span class="math-container">$g$</span> was arbitrary. Hence</p>
<p><span class="math-container">$$f\circ \hat{f}={\rm id}_G.$$</span></p>
<p>Similarly,</p>
<p><span class="math-container">$$\hat{f}\circ f={\rm id}_G.$$</span></p>
<p>Thus <span class="math-container">$\hat{f}$</span> is the inverse of <span class="math-container">$f$</span>.</p>
<p>Hence <span class="math-container">$f$</span> is a bijection.</p>
<hr />
<p><em>Note:</em> We did not use the fact that <span class="math-container">$G$</span> is finite. It is irrelevant. Indeed, such an <span class="math-container">$f$</span> is always a bijection for any <span class="math-container">$G$</span> and any <span class="math-container">$a$</span>.</p>
|
443,475 | <p>I am reading some geometric algebra notes. They all started from some axioms. But I am still confused on the motivation to add inner product and wedge product together by defining
$$ ab = a\cdot b + a \wedge b$$ Yes, it can be done like complex numbers, but what will we lose if we deal with inner product and wedge product separately? What are some examples to show the advantage of geometric product vs other methods?</p>
| user997712 | 31,611 | <p>In addition to invertibility, as mentioned by Joe, geometric operations can be expressed in simple, co-ordinate free expressions using the geometric product.<br>
For instance, Rotation:
$$ R_{i\theta}(A) = e^{-i\theta/2}Ae^{i\theta/2} $$
rotates the blade $ A $ by an angle $ \theta $ in the plane of the bivector $ i $.<br>
Reflection:
$$ F_B(A) = (-1)^{j(k+1)}BAB^{-1} $$
reflects the $j$-blade $A$ in the $k$-blade $B$.<br>
Projection:
$$ P_B(A) = (A\cdot B)B^{-1}$$ projects the blade $ A $ onto the blade $ B $</p>
|
2,713,311 | <p>$ \lim_{x \to \infty} [\frac{x^2+1}{x+1}-ax-b]=0 \ $ then show that $ \ a=1, \ b=-1 \ $</p>
<p><strong>Answer:</strong></p>
<p>$ \lim_{x \to \infty} [\frac{x^2+1}{x+1}-ax-b]=0 \\ \Rightarrow \lim_{x \to \infty} [\frac{x^2+1-ax^2-ax-bx-b}{x+1}]=0 \\ \Rightarrow \lim_{x \to \infty} \frac{2x-2ax-a-b}{1}=0 \\ \Rightarrow 2x-2ax-a-b=0 \ \ (?) $ </p>
<p>Comparing both sides , we get </p>
<p>$ 2-2a=0 \\ a+b=0 \ $</p>
<p>Solving , we get </p>
<p>$ a=1 , \ b=-1 \ $</p>
<p>But I am not sure about the above line where question mark is there.</p>
<p><strong>Can you help me?</strong></p>
| Clayton | 43,239 | <p>What you have seems fine (though maybe overkill); another way you can approach this problem is using long division to show that $$\frac{x^2+1}{x+1}=x-1+\frac{2}{x+1}.$$ The last term goes to $0$ as $x\to\infty$ while the first two terms combine with those in the original problem to get $(1-a)x-(1+b)\to0$ as $x\to\infty$ (but this is equivalent to what you already have).</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Scott Morrison | 3 | <p>Dror Bar-Natan has begin putting <a href="http://katlas.math.toronto.edu/drorbn/index.php?title=Dbnvp">many of his lectures and talks</a> online in video format. I'm not claiming that these are the 'best' online maths videos, but they're certainly interesting, and in particular he's come up with some neat tricks to associate publicly editable annotations with particular moments in the video.</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| AgCl | 1,229 | <p>There are Stephen Boyd's lecture videos on convex optimization:</p>
<p><a href="http://www.stanford.edu/class/ee364a/videos.html">http://www.stanford.edu/class/ee364a/videos.html</a></p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Somnath Basu | 1,993 | <p>I guess all of John Conway's lectures are great. Some of those can be found here :
<a href="http://www.math.princeton.edu/facultypapers/Conway/">http://www.math.princeton.edu/facultypapers/Conway/</a></p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Daniel Pape | 11,176 | <p>A few talks under the heading "What is ..." (",,," could be "Morse Theory", for example)
given at the Freie Universität Berlin can be found here:</p>
<p><a href="http://www.scivee.tv/user/5216" rel="nofollow">http://www.scivee.tv/user/5216</a></p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Dave Novick | 13,067 | <p>The University of New South Wales in Sydney has an eLearning channel on YouTube that contains lectures on a number of topics, including Algebraic Topology, Calculus, and Linear Algebra. Some computing and engineering topics are covered as well.</p>
<p><a href="http://www.youtube.com/user/UNSWelearning#p/p" rel="nofollow">http://www.youtube.com/user/UNSWelearning#p/p</a></p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Javier Álvarez-Vizoso | 10,867 | <p>The complete <a href="http://vod.mathnet.or.kr/sub4_1.php?key_s_title=Lectures+on+Basic+Algebraic+Geometry+by+Miles+Reid+(WCU+project)" rel="nofollow noreferrer">introductory course on Algebraic Geometry by Miles Reid</a> is very interesting (28 lectures following and extending his own undergraduate book on the subject), and his other set of <a href="http://vod.mathnet.or.kr/sub4_1.php?key_s_title=Lectures+on+Algebraic+Surfaces+by+Miles+Reid+(WCU+project)" rel="nofollow noreferrer">lectures on Algebraic Surfaces</a>.</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Austin Mohr | 8,871 | <p>Timothy Gowers' "<a href="https://www.youtube.com/watch?v=BsIJN4YMZZo" rel="nofollow noreferrer">The Importance of Mathematics</a>" never fails to instill a sense of purpose in my work, even when I feel I'm doing "useless" mathematics.</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| joro | 12,481 | <p>Two recent videotaped lectures by Doron Zeilberger.</p>
<p><a href="http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/famous.html" rel="nofollow">The Joy of Dreaming to be Famous (Videotaped lecture), March 1,2012 </a></p>
<p><a href="http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/astrology.html" rel="nofollow">The Rise and Fall of Astrology and the Future Fall of the so-called Infinity (Videotaped lecture) March 29, 2012</a></p>
|
1,452,425 | <p>From what I have been told, everything in mathematics has a definition and everything is based on the rules of logic. For example, whether or not <a href="https://math.stackexchange.com/a/11155/171192">$0^0$ is $1$ is a simple matter of definition</a>.</p>
<p><strong>My question is what the definition of a set is?</strong> </p>
<p>I have noticed that many other definitions start with a set and then something. A <a href="https://en.wikipedia.org/wiki/Group_%28mathematics%29#Definition" rel="noreferrer">group is a set</a> with an operation, an equivalence relation is a set, a <a href="https://en.wikipedia.org/wiki/Function_%28mathematics%29#Definition" rel="noreferrer">function can be considered a set</a>, even the <a href="https://en.wikipedia.org/wiki/Natural_number#Constructions_based_on_set_theory" rel="noreferrer">natural numbers can be defined as sets</a> of other sets containing the empty set.</p>
<p>I understand that there is a whole area of mathematics (and philosophy?) that deals with <a href="https://en.wikipedia.org/wiki/Set_theory#Axiomatic_set_theory" rel="noreferrer">set theory</a>. I have looked at a book about this and I understand next to nothing.</p>
<p>From what little I can get, it seems a sets are "anything" that satisfies the axioms of set theory. It isn't enough to just say that a set is any collection of elements because of various paradoxes. <strong>So is it, for example, a right definition to say that a set is anything that satisfies the ZFC list of axioms?</strong></p>
| Asaf Karagila | 622 | <p>Formally speaking, sets are atomic in mathematics.<sup>1</sup> They have no definition. They are just "basic objects". You can try and define a set as an object in the universe of a theory designated as "set theory". This reduces the definition as to what <em>we</em> call "set theory", and this is not really a mathematical definition anymore.</p>
<p>In naive settings, we say that sets are mathematical objects which are collections of mathematical objects, and that there is no meaning to order and repetition of the objects <em>in</em> the collection.</p>
<p>And when we move back to formal settings, like <span class="math-container">$\sf ZF,NBG,ETCS,NF$</span><sup>2</sup> or other set theories, we try to formalize the properties <em>we</em> expect from sets to have. These may include, for example, the existence of power sets, or various comprehension schemata. But none of them is particularly canonical to the meaning of "set".</p>
<p>These are just ways to formalize, using a binary relation (or whatever you have in the language), the idea of membership, or inclusion, or whatever you think <em>should</em> be the atomic relation defining sets. But as for a <em><strong>right</strong></em> definition? In this aspect "set" is as Platonic as "chair" or "number" or "life".</p>
<hr />
<p>Footnotes:</p>
<ol>
<li><p>This assumes that you take a foundational approach based on set theory. There are other approaches to mathematics, e.g. type theory, in which the notion of "type" is primitive, and sets are just a certain type of objects.</p>
<p>Sufficiently strong set theories can interpret these foundations as well, reducing them to sets if you choose to, or not if you choose not to.</p>
</li>
<li><p>These are Zermelo-Fraenkel, von Neumann-Goedel-Bernays, Elementary Theory of Category of Sets, and New Foundations. These are not the only set theories, of course. And the point of the answer is that these just offer formal frameworks for the notion of "set" as a primitive object (in one way or another).</p>
</li>
</ol>
|
1,452,425 | <p>From what I have been told, everything in mathematics has a definition and everything is based on the rules of logic. For example, whether or not <a href="https://math.stackexchange.com/a/11155/171192">$0^0$ is $1$ is a simple matter of definition</a>.</p>
<p><strong>My question is what the definition of a set is?</strong> </p>
<p>I have noticed that many other definitions start with a set and then something. A <a href="https://en.wikipedia.org/wiki/Group_%28mathematics%29#Definition" rel="noreferrer">group is a set</a> with an operation, an equivalence relation is a set, a <a href="https://en.wikipedia.org/wiki/Function_%28mathematics%29#Definition" rel="noreferrer">function can be considered a set</a>, even the <a href="https://en.wikipedia.org/wiki/Natural_number#Constructions_based_on_set_theory" rel="noreferrer">natural numbers can be defined as sets</a> of other sets containing the empty set.</p>
<p>I understand that there is a whole area of mathematics (and philosophy?) that deals with <a href="https://en.wikipedia.org/wiki/Set_theory#Axiomatic_set_theory" rel="noreferrer">set theory</a>. I have looked at a book about this and I understand next to nothing.</p>
<p>From what little I can get, it seems a sets are "anything" that satisfies the axioms of set theory. It isn't enough to just say that a set is any collection of elements because of various paradoxes. <strong>So is it, for example, a right definition to say that a set is anything that satisfies the ZFC list of axioms?</strong></p>
| user247327 | 247,327 | <p>One of the reasons mathematics is so useful is that it is applicable to so many different fields. And the reason for that is that its logical structure starts with "undefined terms" that can then be given definitions appropriate to the field of application. "Set" is one of the most basic "undefined terms". Rather than defining it a-priori, we develop theories based on generic "properties" and allow those applying it to a particular field (another field of mathematics, or physics or chemistry, etc.) to use a definition appropriate to that field (as long as it has those "properties" of course).</p>
|
1,452,425 | <p>From what I have been told, everything in mathematics has a definition and everything is based on the rules of logic. For example, whether or not <a href="https://math.stackexchange.com/a/11155/171192">$0^0$ is $1$ is a simple matter of definition</a>.</p>
<p><strong>My question is what the definition of a set is?</strong> </p>
<p>I have noticed that many other definitions start with a set and then something. A <a href="https://en.wikipedia.org/wiki/Group_%28mathematics%29#Definition" rel="noreferrer">group is a set</a> with an operation, an equivalence relation is a set, a <a href="https://en.wikipedia.org/wiki/Function_%28mathematics%29#Definition" rel="noreferrer">function can be considered a set</a>, even the <a href="https://en.wikipedia.org/wiki/Natural_number#Constructions_based_on_set_theory" rel="noreferrer">natural numbers can be defined as sets</a> of other sets containing the empty set.</p>
<p>I understand that there is a whole area of mathematics (and philosophy?) that deals with <a href="https://en.wikipedia.org/wiki/Set_theory#Axiomatic_set_theory" rel="noreferrer">set theory</a>. I have looked at a book about this and I understand next to nothing.</p>
<p>From what little I can get, it seems a sets are "anything" that satisfies the axioms of set theory. It isn't enough to just say that a set is any collection of elements because of various paradoxes. <strong>So is it, for example, a right definition to say that a set is anything that satisfies the ZFC list of axioms?</strong></p>
| epi163sqrt | 132,007 | <p><strong>Note:</strong> According to OPs formulation in the bounty text this answer is aimed to give at least a glimpse of some aspects around the definition of set and set theories with focus on the axiom of choice. The top voted answers already contain the essential information.</p>
<p><strong>On the definition of sets:</strong></p>
<blockquote>
<p>In modern times sets (and all other mathematical objects) are defined not in order to specify what they <em>are</em> but instead what we want to do with them. This led to different set theories which emerged essentially due to three different strands and common to all of them was the goal to develop a substantial mathematical theory.</p>
<ul>
<li><p>Set theory as a tool in understanding the <em>infinite</em> leading to the theory of cardinals and ordinals. We owe Georg Cantor the fundamentals of this theory who did it more or less single-handed and against all odds.</p>
</li>
<li><p>Set theory as <em>foundation</em> of supplying the subject matter of mathematics. This claim reflects the mainstream and in many books we can find something like <em>set theory is the foundation of mathematics</em>.</p>
</li>
<li><p>Set theory as supplier of a <em>common mode of reasoning</em> for diverse areas of mathematics. This is strongly related with the second strand and the <em>axiom of choice</em> is a famous set-theoretical principle of this type.</p>
</li>
</ul>
</blockquote>
<p>This reasoning from M. Potter's <em><a href="https://rads.stackoverflow.com/amzn/click/com/0199270414" rel="nofollow noreferrer" rel="nofollow noreferrer">Set Theory and It's Philosophy</a></em> is followed by</p>
<blockquote>
<p>The history of such theories is now a century old ... - and yet there is, even now, <em>no consensus</em> in the literature about the form they should take.</p>
</blockquote>
<p><strong>Conclusion:</strong> There is no widely agreed <em>single</em> set theory which is favored by the mathematicians from which we could derive a <em>right</em> definition of a set. Instead depending on the area of research and the richness of results within these areas <em>different set theories</em> like ZF, ZFC, etc. are taken as their basis.</p>
<hr />
<p><strong>Set theory today:</strong></p>
<p>Presumably most of the daily work is explicitly or implicitely based upon <strong>ZF</strong>, <strong>ZFC</strong> or a somewhat weaker version in between. But we should be aware that each of these set theories has benefits as well as drawbacks.</p>
<blockquote>
<p>U. Felgner writes in <em><a href="https://rads.stackoverflow.com/amzn/click/com/3540055916" rel="nofollow noreferrer" rel="nofollow noreferrer">Models of ZF-Set theory</a></em></p>
<p>We believe that the <strong>ZF</strong>-axioms describe in a correct way our intuitive contemplations concerning the notion of sets. The axiom of choice (<strong>AC</strong>) is intuitively not so clear as the other <strong>ZF</strong>-axioms are, but we have learned to use it because it seems to be indispensible in proving mathematical theorems. On the other hand (<strong>AC</strong>) has <em>strange</em> consequences, such as <em>every set can be well-ordered</em> and we are unable to <em>imagine</em> a well-ordering of the set of real numbers.</p>
</blockquote>
<p>Besides the well-order theorem (WOT) there are many other equivalents to the AC which are then also to accept.</p>
<p><strong>Historical aspects around AC:</strong></p>
<p>Some mathematicians had difficulties deciding for or against this axiom e.g. van der Waerden:</p>
<blockquote>
<p>In 1930, van der Waerden published his <em>Modern Algebra</em>, detailing the exciting new applications of the axiom. The book was very influential, providing Zorn und Teichmüller with a proving ground for their versions of choice, but van der Waerden's Dutch colleagues persuaded him to abondon the axiom in the second edition of 1937. He did so, but the resulting limited version of abstract algebra brought such a strong protest from his fellow algebraists that he was moved to reinstate the axiom and all its consequences in the third edition of 1950. (<a href="http://www.cs.umd.edu/%7Egasarch/BLOGPAPERS/belaxioms1.pdf" rel="nofollow noreferrer">P. Maddy, 1988</a>)</p>
</blockquote>
<p>H. Herrlich summarizes the historical development in his <em><a href="https://rads.stackoverflow.com/amzn/click/com/3540309896" rel="nofollow noreferrer" rel="nofollow noreferrer">Axiom of Choice</a></em></p>
<blockquote>
<p>After Gödel (1938) proved the relative consistency of the Axiom of Choice by constructing within a given model of <strong>ZF</strong> a model of <strong>ZFC</strong>, the proponents of <strong>AC</strong> gained ground. Most modern textbooks take <strong>AC</strong> for granted and the vast majority of methematicians use <strong>AC</strong> freely.</p>
<p>However, after Cohen (1963) proved the relative consistency of the negation of <strong>AC</strong> and, moreover, provided a method, called <em>forcing</em>, for producing a plethora of models of <strong>ZF</strong> that have or fail to have a wide range of specified properties, a growing number of mathematicians started to investigate the <strong>ZF</strong> world by substituting <strong>AC</strong> by a variety of possible alternatives, sometimes just by weaking <strong>AC</strong> and sometimes by replacing <strong>AC</strong> by axioms that contradict it.</p>
</blockquote>
<p>And with respect to a <em>true</em> definition of sets he continues</p>
<blockquote>
<p>All this work demonstrates how useful or convenient such axioms as <strong>AC</strong> and its possible alternatives are. But the question of the truth of <strong>AC</strong> is not touched, and Hilbert's First Problem remains unanswered. It is conceivable, even likely, that it will never be solved, despite Hilbert's optimistic slogan expressed in his Paris lecture: <em>in mathematics there is no ignorabimus</em>.</p>
</blockquote>
<hr />
<p><strong>Pros and Cons of AC:</strong></p>
<p>Herrlich's book is an interesting source of information around AC. He presents many equivalents of AC and some related concepts to AC. The main part are the chapters <em>Disasters without Choice</em> consisting of 11 sections organised by mathematical disciplines and the chapter <em>Disaster with Choice</em> consisting of 2 sections. To get a glimpse about such consequences I pick out two, three easy understandable examples:</p>
<blockquote>
<p><strong>From Section 4: Disasters without Choice</strong></p>
<p>Section 4.4: Disasters in Algebra I: Vector Spaces</p>
<p>In <strong>ZFC</strong> every vector space is uniquely determined, up to isomorphism, by a single cardinal number, its dimension. Each of the two fundamental results which together enable us to associate dimension with a given vector space fail badly in <strong>ZF</strong>.</p>
<p>Disaster 4.42: The following can happen:</p>
<ol>
<li>Vector spaces may have no bases</li>
<li>Vector spaces may have two bases with different cardinalities.</li>
</ol>
<p>Theorem 4.44: Equivalent are:</p>
<ol>
<li>Every vector space has a basis</li>
<li>AC</li>
</ol>
<hr />
<p>Section 4.6: Disasters in Elementary Analysis: The Reals and Continuity</p>
<p>Disaster 4.53: The following can happen</p>
<ol>
<li><span class="math-container">$\mathbb{R}$</span> may fail to be Fréchet, i.e., not every accumulation point <span class="math-container">$x$</span> of a subset A may be reachable by a sequence <span class="math-container">$(a_n)$</span> in <span class="math-container">$A$</span>.</li>
</ol>
<p>... (9 more to follow)</p>
</blockquote>
<p>Though the Axiom of Choice is responsible for many beautiful results, it is equally responsible for the existence of several dreadful monstrosities - unwelcome and unneeded.</p>
<blockquote>
<p><strong>From Section 5: Disasters with Choice</strong></p>
<p>Section 5.1: Disasters in Elementary Analysis:</p>
<p>Definition 5.1: The equation <span class="math-container">$f(x+y)=f(x)+f(y)$</span> is called the <em>Cauchy-equation</em></p>
<p>Consider a function <span class="math-container">$f:\mathbb{R}\rightarrow\mathbb{R}$</span> that satisfies the Cauchy-equation for all real <span class="math-container">$x$</span> and <span class="math-container">$y$</span>. Then it is easily seen that</p>
<ul>
<li><span class="math-container">$f(r\cdot x)=r\cdot f(x)$</span> for all rational <span class="math-container">$r$</span> and real <span class="math-container">$x$</span>, i.e. <span class="math-container">$f$</span> is <span class="math-container">$\mathbb{Q}$</span>-linear.</li>
</ul>
<p>In particular:</p>
<ul>
<li><span class="math-container">$f(r)=f(1)\cdot r$</span> for all rational <span class="math-container">$r$</span>.</li>
</ul>
<p>And continuity of <span class="math-container">$f$</span> would imply</p>
<ul>
<li><span class="math-container">$f(x)=f(1)\cdot x$</span> for all <span class="math-container">$x\in \mathbb{R}$</span></li>
</ul>
<p>Are there solutions of the Cauchy-equation that fail to be continuous? None has ever been constructed and in <strong>ZF</strong> none will ever be. However the Axiom of Choice guarantees the existence of such monsters; even worse, under <strong>AC</strong> there are far mor undesirable solutions of the Cauchy-equation than there are desirable ones:</p>
<p>Disaster 5.2: In <strong>ZFC</strong> there are</p>
<ol>
<li><p><span class="math-container">$2^{\aleph_0}$</span> continuous solutions <span class="math-container">$f:\mathbb{R}\rightarrow\mathbb{R}$</span> and</p>
</li>
<li><p><span class="math-container">$2^{(2^{\aleph_0})}$</span> non-continuous solutions <span class="math-container">$f:\mathbb{R}\rightarrow\mathbb{R}$</span> of the Cauchy-equation.</p>
</li>
</ol>
</blockquote>
<p><strong>Conclusion:</strong> It is good to be aware that there are benefits and drawbacks in ZF as well as ZFC and it's plausible that this is also the case for other set theories. So, there is no <em>true</em> definition of a set theory as framework for a true definition of a set.</p>
<p><strong>Two hints:</strong></p>
<p>A classic source to read and think about sets is <em><a href="https://rads.stackoverflow.com/amzn/click/com/1614271313" rel="nofollow noreferrer" rel="nofollow noreferrer">Naive Set Theory</a></em> by P. Halmos. It presumably covers most of the aspects around sets you might need for daily work.</p>
<p>On the other hand if you are curious, to see which functions in real analysis can be defined or not be defined according to underlying set theories, you may want to have a short look into <em><a href="https://rads.stackoverflow.com/amzn/click/com/B00UVB40QS" rel="nofollow noreferrer" rel="nofollow noreferrer">Strange Functions in Real Analysis</a></em>.</p>
|
1,186,517 | <p>I do some ex for preparing discrete mathematics exam, i get stuck in one problem, anyone could help me?</p>
<blockquote>
<p>How many ways we can partition set {1,2,...,9} into subsets of size 2
and 5?</p>
</blockquote>
<p>anyway, some tutorials for solving such a question...</p>
<p>Edit: Like always Scott is Right...</p>
| Michael E | 221,334 | <p>Since there are an odd number of elements in the set you must have a subset with 5 elements and therefore you must have two subsets of 2 elements.</p>
<p>First Choose the set with five elements. There are $9 \choose 5$ ways to do this. For <em>each</em> of these choices there are then 4 elements left and you have to pick two. Finally the last subset is determined by your choices of the first two. So the total number of ways should be.</p>
<p>$$
{9 \choose 5} {4 \choose 2}
$$</p>
<p>...</p>
<p>Might as well leave my mistake in as it helps make a good point at the expense of only slight embarrassment. The final answer is of course half this amount. Thanks to Brian for pointing this out. I am choosing two numbers from 4 but I'm putting 2 numbers in set A and 2 numbers in set B. However A, B is the same partition as B,A. </p>
|
87,538 | <p>My problem is the following:</p>
<p>I have a finite surjective morphism $f: X\rightarrow Y$ of noetherian schemes and know that $Y$ is a regular scheme.
(Indeed, in my situation, the two schemes are topologically the same and the arrow is topologically the identity.)</p>
<p>I don't know if $f$ is étale or smooth. But I know that $f$ has a section.</p>
<p>I want to conclude somehow that $X$ is also a regular scheme. What further minimal conditions (on $f$ or $X$) would imply this?</p>
<p>I would be glad about just some hints of what one could do here.</p>
| Sándor Kovács | 10,076 | <p>I may be misunderstanding the question, but it seems rather straightforward to me. </p>
<p>a) As stated, without assuming, that $X$ is, say, reduced, it is certainly false:
Let $X=\mathrm{Spec}k[\varepsilon]=k[x]/(x^2)$, $Y=\mathrm{Spec} k$ and $f:X\to Y$ the structure map of $X$ as a $Y$-scheme. This is obviously a homeomorphism topologically and it has a section that maps $Y$ isomorphically to $X_{\mathrm{red}}$. </p>
<p>b) It seems to me that pretty much this is the only thing that can go wrong: If $f$ has a section, then by definition of a section, $Y$ is isomorphic to the image of the section. </p>
<p>As $Y$ is regular, it is necessarily reduced, so if $f$ is one-to-one, then $f$ and the section induce an isomorphism between $Y$ and $X_{\mathrm{red}}$. So, if $X$ is reduced, the statement is true, if $X$ is not reduced, then the question is whether there exists a morphism $X\to X_{\mathrm{red}}$ that's an isomorphism on $X_{\mathrm{red}}$. The above example shows that this can happen, so some assumption is required to rule out this possibility.</p>
<p>If $f$ is not one-to-one, something similar would still happen. The section would still induce an isomorphism between $Y$ and the image of the section which would have to be an irreducible component of $X_{\mathrm{red}}$. On the other hand, in this case you would probably want to assume that $X$ and $Y$ are irreducible, since otherwise you can just add an additional component that screws things up. However, then (assuming that $X$ is irreduciblke) this implies that $Y\simeq X_{\mathrm{red}}$ and you end up in the previous case again.</p>
|
971,160 | <p>So, this is actually 2 questions in 1. I apologize if that is bad practice, but I didn't want to write 2 questions when they're a word different. So, I have</p>
<ol>
<li>Prove or disprove that if $a|(sb+tc), \forall s,t \in\mathbb{Z}$, then $a|b$, and $a|c$.</li>
</ol>
<p>and then,</p>
<ol start="2">
<li>Prove or disprove that if $a|(sb+tc)$ for some $s,t \in\mathbb{Z}$, then $a|b$, and $a|c$.</li>
</ol>
<p>I know how to prove if $a|b$ and $a|c$, then $a|(sb+tc), \forall s,t \in\mathbb{Z}$, but I'm certain I can't just write the proof backwards, and show that it works. So, how would I tackle these two problems, especially the second one?</p>
| DeepSea | 101,504 | <ol>
<li>True. </li>
</ol>
<p>Choose $s = 0, t = 1$, then $a|c$, and choose $s = 1, t = 0$, then $a|b$</p>
<ol start="2">
<li>False.</li>
</ol>
<p>Let $a = 2, b = 3, c = 5$, and $s = t = 2$.</p>
|
3,215,381 | <p>When we write <span class="math-container">$\mathbb{Z}_3$</span>, does it mean <span class="math-container">$\mathbb{Z}/3\mathbb{Z}$</span>? Also, does <span class="math-container">$3\mathbb{Z}_3$</span> mean <span class="math-container">$0 \pmod 3$</span>?</p>
| Xander Henderson | 468,350 | <p><em>Many</em> notations in mathematics are overloaded. Whenever you are working in mathematics, it is very important to keep track of the <em>context</em> in which you are working. For example, in the expression
<span class="math-container">$$ 3^s = 2 \implies s \in \left\{ \log_{3}(2) + i\frac{2\pi}{\log(3)} \right\},$$</span>
they symbol <span class="math-container">$i$</span> (clearly?) represents the imaginary unit, i.e. the complex number which, when squared, gives <span class="math-container">$-1$</span>. On the other hand, in the notation
<span class="math-container">$$ \sum_{i=0}^{\infty} \alpha^i = \frac{1}{1-\alpha}, $$</span>
the variable <span class="math-container">$i$</span> is the index of summation. In both cases, the notation should be unambiguous, as the context makes it clear what is going on.</p>
<p>Similarly, the notation <span class="math-container">$\mathbb{Z}_3$</span> is potentially ambiguous, but is typically clear with the correct context. In the context of anything <span class="math-container">$p$</span>-adic, the notation <span class="math-container">$\mathbb{Z}_3$</span> unambiguously represents the <span class="math-container">$3$</span>-adic integers. If you are working in a <span class="math-container">$p$</span>-adic setting (or, more generally, if the words "local field" appear in your work), then it is safe to assume that <span class="math-container">$\mathbb{Z}_3$</span> is the set of <span class="math-container">$3$</span>-adic integers. On the other hand, it is quite common to use <span class="math-container">$\mathbb{Z}_3$</span> to denote the quotient group <span class="math-container">$\mathbb{Z}/3\mathbb{Z}$</span> in other contexts. If you are not explicitly working in a setting where the <span class="math-container">$3$</span>-adic integers might pop up, it is safe to assume that <span class="math-container">$\mathbb{Z}_3 = \mathbb{Z}/3\mathbb{Z}$</span>. If you fear that there is any <em>potential</em> for ambiguity, then make it clear—unambiguously define your notation.</p>
<p>With respect to the last part of your question, <span class="math-container">$3\mathbb{Z}_3$</span> almost certainly denotes the set
<span class="math-container">$$ 3\mathbb{Z}_3 = \{ 3x \mid x \in \mathbb{Z}_3 \}, $$</span>
which is the ball of radius <span class="math-container">$\frac{1}{3}$</span> centered at zero in the space of <span class="math-container">$3$</span>-adic numbers.</p>
|
3,002,325 | <p>Proof <span class="math-container">$t$</span> is irrational <span class="math-container">$ t = a-bs $</span> , Given <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are rational numbers, <span class="math-container">$b \neq 0$</span> and <span class="math-container">$s$</span> is irrational.
Hence show that <span class="math-container">$(\sqrt3-1)/(\sqrt3+1)$</span> is irrational</p>
| Ethan Bolker | 72,858 | <p>For the first part: in a comment you have solved for <span class="math-container">$s$</span> in terms of <span class="math-container">$a$</span>, <span class="math-container">$b$</span> and <span class="math-container">$t$</span>. You know the first two are rational. What could you say about <span class="math-container">$s$</span> if <span class="math-container">$t$</span> were rational too?</p>
<p>You have started the second part correctly. Now it's in a form where you can apply the first part.</p>
|
3,009,362 | <p>I need to find
<span class="math-container">$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}$$</span></p>
<p>Looking at the graph, I know the answer should be <span class="math-container">$\frac{20}{17}$</span>, but when I tried solving it, I reached <span class="math-container">$0$</span>.</p>
<p>Here are the two ways I approached this:</p>
<p>WAY I:</p>
<p><span class="math-container">$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35} =
\lim_{x\rightarrow -5} \frac{\require{cancel} \cancel{x^2}(2- \frac{50}{x^2})} {\require{cancel} \cancel{x^2}(2+ \frac{3}{x}-\frac{35}{x^2})}
=\frac{2-2}{\frac {42}{5}}=0
$$</span></p>
<p>WAY II:
<span class="math-container">$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35} =
\lim_{x\rightarrow -5} \frac{\require{cancel} \cancel{2}(x^2- 25)} {\require{cancel} \cancel{2}(x^2+ \frac{3}{2}x-\frac{35}{2})}
=\lim_{x\rightarrow -5} \frac{{\require{cancel} \cancel{(x-5)}}(x+5)}{{\require{cancel} \cancel{(x-5)}}(x+3.5)}= \frac{-5+5}{-5+3.5}=0
$$</span></p>
<p>What am I doing wrong here?</p>
<p>Thanks!</p>
| user | 505,767 | <p>As an alternative by <span class="math-container">$y=x+5 \to 0$</span></p>
<p><span class="math-container">$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}=\lim_{y\rightarrow 0} \frac{2(y-5)^2-50}{2(y-5)^2+3(y-5)-35}=\lim_{y\rightarrow 0} \frac{2y^2-20y}{2y^2-17y}=\lim_{y\rightarrow 0} \frac{2y-20}{2y-17}$$</span></p>
|
2,138,448 | <p>Survival game: Consider $3$ players, $A, B$ and $ C$, taking turns shooting at each other. Any player can shoot at only one opponent at a time (and each of them has to make a shot whenever it is his/her turn). </p>
<p>Each shot of $A$ is successful with probability $1/3$, each shot of $B$ is successful with probability $1$, and each shot of C is successful with probability $1/2$ (with all the outcomes being independent). </p>
<p>$A$ goes first, then $B$,
then $C$, then $A$, and so on, until one of them dies. Then, the remaining two will be shooting at each other, so that nobody ever makes two shots in a row: e.g. if $A$ gets $C$ shot, then $B$ goes next. </p>
<p>The game continues until only one player is left. Assume that every player is trying to find a strategy that maximises his/her probability of survival. Assume also that every player acts optimally and knows that the other players will act optimally too. </p>
<p>Who should player A shoot at first? What is the probability of survival of $ A$ (assuming he/she acts optimally)?</p>
<p>*Hint. A strategy is a sequence of decisions on who to shoot at at any given turn, given who is still left in the game. It is clear that, after $B$ shoots for the first time, there will be at most two players left, and, hence, for the remaining players, there will be no need to make any choices. Therefore, it is convenient to solve the problem recursively, starting from the decision of $B$, and assuming that all players are alive by the time $B$ shoots (otherwise, again, there are no decisions for $B$ to make). </p>
<p>It is clear that,
given a choice between $A$ and $C$, $B$ will shoot at $C$, because playing against $A$ only
will give $B$ a higher probability of survival than playing against $C$ only (i.e. $2/3$
vs. $1/2$). Knowing this, A needs to choose whether it is optimal to shoot at $B$ or at
$C$. Considering the possible outcomes produced by each of the two choices, you will notice that, in one case, the survival probability can be computed by hand, and, in the other case, it can be reduced to the computation of an exit probability of a simple Markov chain.</p>
<p>My trial:</p>
<ol>
<li><p>if $A$ shoots $C$: $P$($A$ hits $C$ *$B$ misses $ A$ * $A$ hits $B$) + $P$ ($A$ misses $C$ * $B$ hits $ C$ * $A$ hits $B ) = 0 +1/3*1*1/3 = 1/9$</p></li>
<li><p>if $A$ shoots $B$: $P$($A$ hits $B$ *$C$ misses $ A$ * $A$ hits $C$) + $P$ ($A$ misses $B$ * $B$ hits $ C$ * $A$ hits $B ) = 1/3*1/2*1/3 +2/3*1*1/3 = 5/18$</p></li>
</ol>
<p>So A should shoot B, is it correct? </p>
| Anatoly | 90,997 | <p>It has already correctly been noted that the best strategy for $A $ in this game - assuming no possibility of shooting into the air - is to begin by shooting at $B $. I would add some details on this. </p>
<p>The case in which $A $ decides to shoot at $C $ is relatively simple, since the only possibility for him to survive is to fail the first shot to $C $, to wait for $B $ killing $C $, and to have success in his second shot to $B $. Because $B $ does not fail, in this case there are no other possibilities for $A $ to survive: this directly leads to a probability of $2/3 \cdot 1 \cdot 1/3 =2/9 \,\,\,\, $ (note that for this case there is a typo in the OP, which reports a probability of $1/9 \,$). </p>
<p>The case in which $A $ decides to shoot at $B $ is a bit more complex. Since the game continues until only one player is left, in the first subcase, that where $A $ kills $B $, the possibilities for $A $ to survive in the subsequent shots between $C $ and him include any successive sequence of failures that ends with a successful shot by $A $. Calling $P_n $ the probability of occurrence for a sequence of this type that ends with a successful $n^{th} $ shot by $A $, we have</p>
<p>$\begin{align*}
P_1= & \frac {1}{2} \cdot \frac {1}{3}\\
P_2= & \frac {1}{2}\cdot \left( \frac {2}{3} \cdot \frac {1}{2} \right) \cdot \frac {1}{3}\\
P_3= &\frac {1}{2}\cdot \left( \frac {2}{3} \cdot \frac {1}{2} \right) \cdot \left ( \frac {2}{3} \cdot \frac {1}{2} \right) \cdot \frac {1}{3}
\end{align*}$</p>
<p>and so on (here $n $ does not include the initial shot of $A $ to $B $). So we get the sum </p>
<p>$$\sum_{k=0}^\infty \, \frac {1}{2} \cdot \left(\frac {2}{3} \cdot \frac {1}{2} \right)^k \cdot \frac {1}{3} \\ = \sum_{k=0}^\infty \, \frac {1}{6} \cdot \left(\frac {1}{3} \right)^k =\frac {1}{4}$$</p>
<p>Multiplying this to $\frac {1}{3} $ (probability that in the initial shot $A $ kills $B $), we get for this first subcase a survival probability of $\frac {1}{12} $.</p>
<p>The second subcase, that where $A $ shoots at $B $ but misses him, is again rather simple. The only possibility of $A $ to survive in this subcase is to wait for $B $ killing $C $, and then to kill $B $ in the successive shot. This can occur with probability $2/3 \cdot 1 \cdot 1/3=2/9\,\,\,\,$. </p>
<p>Collecting these results we have that, in the case in which $A $ decides to begin by shooting at $B $, his survival probability is $$\frac {1}{12} + \frac {2}{9} = \frac {11}{36} $$</p>
|
1,081,383 | <p>Assume that $A$ is an $n\times n$ symmetric positive-definite matrix.</p>
<p>Prove that:</p>
<blockquote>
<p>the element of $A$ with maximum magnitude must lie on the diagonal. </p>
</blockquote>
| Algebraic Pavel | 90,996 | <p>For an $n\times n$ Hermitian positive semidefinite matrix $A=(a_{ij})$ and $1\leq i,j\leq n$, $$|a_{ij}|^2\leq a_{ii}a_{jj}\leq\max\limits_{k}a_{kk}^2 \quad \Rightarrow \quad|a_{ij}|\leq\max_k a_{kk}.$$</p>
|
2,572,802 | <p>We were asked to find the number of five digit numbers $N=d_1d_2d_3d_4d_5$, where $d_i$ is the $i$th digit of the number and $d_1 < d_2 < d_3 < d_4 < d_5 $. The solution was trivial as for a given selection of five random distinct digits, there is only one way to arrange them in strict increasing order. Now to raise the difficulty level of the question, our teacher changed all the "less than" signs into "less than or equal to" signs, thus yielding the new constraint $d_1 \leq d_2 \leq d_3 \leq d_4 \leq d_5$.</p>
<p>I proceeded as follows .....</p>
<p>Let $x_i = d_{i+1} - d_i$. So, $$ x_1 + x_2 + x_3 + x_4 = d_5 - d_1,$$ or $$d_1 + x_1 + x_2 + x_3 + x_4 = d_5 \leq 9.$$ But $d_1 \geq 1, x_i \geq 0$ for all $i $ belonging to the set ${1,2,3,4} $. So, the problem reduces to find the number of non negative integral solutions of the inequality $$d_1 + x_1 + x _2 + x_3 + x_4 \leq 8,\quad
(*)$$ which comes out to be $^{13} C_ {5}$. </p>
<p>But looking at the simplicity of the answer, I am bound to think that there must exist some elegant approach to this problem. In Combinatorics, every one thinks differently, which is why I ask this question. I just want to explore the different ways one can approach a problem. Please note that I am just a beginner in Combinatorics. I am familiar with Permutations and Combinations, but I do not know a single thing in Graph Theory and all those "advanced" concepts. So please answer accordingly.</p>
| DanielWainfleet | 254,665 | <p>Let $F(n)$ be the number of $n$-digit numbers whose digits are strictly increasing from L to R. The answer to your Q is $$(F(5)+4F(4)+3F(3)+2F(2)+F(1))+(3F(3)+2F(2))=$$ $$=F(5)+4F(4)+6F(3)+4F(2)+F(1).$$ </p>
<p>Explanation: </p>
<p>(i). $F(5)$ ways with $5$ different digits. </p>
<p>(ii). Given $4$ strictly increasing digits, there are $4$ different numbers obtained by inserting a $5$th digit equal to and adjacent to one of those $4$ digits, giving $4F(4) .$</p>
<p>(iii). Given $3$ strictly increasing digits $d_1d_2d_3,$ there are $3$ numbers obtained by inserting $2$ identical digits, equal to and adjacent to one of those three (i.e. $d_1d_1d_1d_2d_2d_3$ or $d_1d_2d_2d_2d_3$ or $d_1d_2 d_3d_3d_3$), giving $3F(3).$ </p>
<p>(iv). Given increasing $d_1d_2$ there are $2$ numbers obtained by inserting $3$ digits adjacent to and equal to $d_1$ or to $d_2,$ giving $2F(2).$ </p>
<p>(v). Given one digit, adding $4$ more equal to it gives $F(1)$ different numbers of the form $ddddd.$ </p>
<p>(vi). Given $3$ increasing digits there are $3$ numbers obtained by repeating one of those $3$ digits and repeating another of those $3$ digits, giving $3F(3).$</p>
<p>(vii). Given increasing $d_1d_2$ there are two numbers obtained by inserting $2$ copies of $d_1$ and one copy of $d_2$ or vice-versa (i.e. $d_1d_1d_1d_2d_2$ and $d_1d_1d_2d_2d_2),$ giving $2F(2).$</p>
<p>The expression $F(5)+4F(4)+6F(3)+4F(2)+F(1)$ is suggestive of the formula $(x+1)^4=x^4+4x^3+6x^2+4x+1,$ so there may be a slicker generalization involving generating polynomials. If it were not 4 a.m. here, I might think of one (if there is one). </p>
|
2,572,802 | <p>We were asked to find the number of five digit numbers $N=d_1d_2d_3d_4d_5$, where $d_i$ is the $i$th digit of the number and $d_1 < d_2 < d_3 < d_4 < d_5 $. The solution was trivial as for a given selection of five random distinct digits, there is only one way to arrange them in strict increasing order. Now to raise the difficulty level of the question, our teacher changed all the "less than" signs into "less than or equal to" signs, thus yielding the new constraint $d_1 \leq d_2 \leq d_3 \leq d_4 \leq d_5$.</p>
<p>I proceeded as follows .....</p>
<p>Let $x_i = d_{i+1} - d_i$. So, $$ x_1 + x_2 + x_3 + x_4 = d_5 - d_1,$$ or $$d_1 + x_1 + x_2 + x_3 + x_4 = d_5 \leq 9.$$ But $d_1 \geq 1, x_i \geq 0$ for all $i $ belonging to the set ${1,2,3,4} $. So, the problem reduces to find the number of non negative integral solutions of the inequality $$d_1 + x_1 + x _2 + x_3 + x_4 \leq 8,\quad
(*)$$ which comes out to be $^{13} C_ {5}$. </p>
<p>But looking at the simplicity of the answer, I am bound to think that there must exist some elegant approach to this problem. In Combinatorics, every one thinks differently, which is why I ask this question. I just want to explore the different ways one can approach a problem. Please note that I am just a beginner in Combinatorics. I am familiar with Permutations and Combinations, but I do not know a single thing in Graph Theory and all those "advanced" concepts. So please answer accordingly.</p>
| Christian Blatter | 1,303 | <p>Your approach can maybe streamlined somewhat, but it is the textbook solution for this problem. </p>
<p>I'd put it this way: Look at weakly ascending seven letter words $\ 1\ d_1\ d_2\ d_3\ d_4\ d_5\ 9\ $. There are six slots for an increment $x_k\geq0$, and the sum of the increments should be $9-1=8$. By "stars and bars" you can choose these increments in ${8+5\choose 5}=1287$ ways.</p>
|
1,730,352 | <p>I am really struggling to understand what modular forms are and how I should think of them. Unfortunately I often see others being in the same shoes as me when it comes to modular forms, I imagine because the amount of background knowledge needed to fully appreciate and grasp the constructions and methods is rather large, so hopefully with this post some clarity can be offered, also for future readers.. The usual definitions one comes across are often of the form: (here taken from <a href="https://en.wikipedia.org/wiki/Modular_form">wikipedia</a>)</p>
<blockquote>
<p>A modular form is a (complex) analytic function on the upper
half-plane satisfying a certain kind of functional equation with
respect to the group action of the modular group, and also satisfying
a growth condition.</p>
<p>A modular form of weight $k$ for the modular group $$
\text{SL}(2,\mathbb{Z})=\left\{\begin{pmatrix} a & b \\ c & d
\end{pmatrix}| a,b,c,d \in \mathbb{Z} , ad-bc = 1 \right\} $$ is a
complex-valued function $f$ on the upper half-plane $\mathbf{H}=\{z
\in \mathbb{C},\text{Im}(z)>0 \}$, satisfying the following three
conditions:</p>
<ol>
<li>$f$ is a holomorphic function on $\mathbf{H}.$</li>
<li>For any $z \in \mathbf{H}$ and any matrix in $\text{SL}(2,\mathbb{Z})$ as above, we have: $$
f\left(\frac{az+b}{cz+d}\right)=(cz+d)^k f(z) $$</li>
<li>$f$ is required to be holomorphic as $z\to i\infty.$</li>
</ol>
</blockquote>
<p><strong>Questions</strong>:</p>
<ul>
<li>(a): I guess what I'm having least familiarity with is the <em>modular group</em> part. My interpretation of $\text{SL}(2,\mathbb{Z}):$ <em>The set of all $2$ by $2$ matrices, with integer components, having their determinant equal to $1.$</em> But where does the name come from, as in why do we call this set a group and what modular entails?</li>
<li>(b): If I understand correctly, the group operation here is function composition, of type: $\begin{pmatrix}a & b \\ c & d\end{pmatrix}z = \frac{az+b}{cz+d}$ which is also called a linear fractional transformation. How should one interpret the condition $2.$ that $f$ has to satisfy? My observation is that, as a result of the group operation of $\text{SL}$ on a given integer $z,$ the corresponding image is multiplied by a polynomial of order $k$ (which is the weight of the modular form). </li>
<li>(c) The condition $3.$ I interpret as: $f$ should not exhibit any poles in the upper half plane, not even at infinity. About right? </li>
<li>(d) A more general question: Given the definition above, it is tempting to see modular forms as particular classes of functions, much like the Schwartz class of functions, or $L^p$ functions and so on. Is this an acceptable assessment of modular forms?</li>
<li>(e) Last question: It is often said that modular forms have interesting Fourier transforms, as in their Fourier coefficients are often interesting (or known) sequences. Is there an intuitive way of seeing, from the definition of modular forms, the above expectation of their Fourier transforms?</li>
</ul>
| Mathmo123 | 154,802 | <p>The definition of a modular form seems extremely unmotivated, and as @AndreaMori has pointed out, whilst the complex analytic approach gives us the quickest route to a definition, it also clouds some of what is really going on.</p>
<p>A good place to start is with the theory of <em>elliptic curves</em>, which have long been objects of geometric and arithmetic interest. One definition of an elliptic curve (over $\mathbb C$) is a quotient of $\mathbb C$ by a lattice $\Lambda = \mathbb Z\tau_1\oplus\mathbb Z\tau_2$, where $\tau_1,\tau_2\in\mathbb C$ are linearly independent over $\mathbb R$ ($\mathbb C$ and $\Lambda$ are viewed as additive groups): i.e.
$$E\cong \mathbb C/\Lambda.$$</p>
<p>In this viewpoint, one can study elliptic curves by studying lattices $\Lambda\subset\mathbb C$. Modular forms will correspond to certain functions of lattices, and by extension, to certain functions of elliptic curves.</p>
<blockquote>
<p>Why the upper half plane?</p>
</blockquote>
<p>For simplicity, since $\mathbb Z\tau_1 = \mathbb Z(-\tau_1)$, there's no harm in assuming that $\frac{\tau_1}{\tau_2}\in \mathbb H$. </p>
<blockquote>
<p>What about $\mathrm{SL}_2(\mathbb Z)$?</p>
</blockquote>
<p>When do $(\tau_1,\tau_2)$ and $(\tau_1',\tau_2')$ define the same lattice? Exactly when
$$(\tau_1',\tau_2')=(a\tau_1+b\tau_2,c\tau_1+d\tau_2)$$where $\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\mathrm{SL}_2(\mathbb Z)$. Hence, if we want to consider functions on lattices, they had better be invariant under $\mathrm{SL}_2(\mathbb Z)$.</p>
<blockquote>
<p>Functions on lattices:</p>
</blockquote>
<p>Suppose we have a function $$F:\{\text{Lattices}\}\to\mathbb C.$$ First observe that multiplying a lattice by a non-zero scalar (i.e. $\lambda\Lambda$ for $\lambda\in\mathbb C^\times$) amounts to rotating and rescaling the lattice. So our function shouldn't do anything crazy to rescaled lattices.</p>
<p>In fact, since we really care about elliptic curves, and $\mathbb C/\Lambda\cong\mathbb C/\lambda\Lambda$ under the isomorphism $z\mapsto \lambda z$, $F$ should be completely invariant under such rescalings - i.e. we should insist that </p>
<p>$$F(\lambda \Lambda) = F(\Lambda).$$
However, if we define $F$ like this, we are forced to insist that $F$ has no poles. This is needlessly restrictive. So what we do instead is require that
$$F(\lambda\Lambda) = \lambda^{-k}F(\Lambda)$$
for some integer $k$; the quotient $F/G$ of two weight $k$ functions gives a fully invariant function, this time with poles allowed.</p>
<blockquote>
<p>Where do modular forms come in?</p>
</blockquote>
<p>If $\Lambda = \mathbb Z\tau\oplus\mathbb Z$ with $\tau\in\mathbb H$, define a function $f:\mathbb H\to\mathbb C$ by $f(\tau)=F(\Lambda)$. For a general lattice, we have</p>
<p>$$\begin{align}F(\mathbb Z\tau_1\oplus\mathbb Z\tau_2)&=F\left(\tau_2(\mathbb Z({\tau_1}/{\tau_2})\oplus\mathbb Z)\right)\\
&=\tau_2^{-k}f({\tau_1}/{\tau_2})
\end{align}$$
and in particular,
$$\begin{align}f(\tau) &= F(\mathbb Z\tau\oplus\mathbb Z) \\&=F(\mathbb Z(a\tau+b)\oplus\mathbb Z(c\tau+d)) &\text{by }\mathrm{SL}_2(\mathbb Z)\text{ invariance}\\&= (c\tau+d)^{-k} f\left(\frac{a\tau+b}{c\tau+d}\right).\end{align}$$</p>
<p>This answers your first two questions. </p>
<p>At this point, there's no reason to assume that condition (3) holds, and one can study such functions without assuming condition (3). However, imposing cusp conditions is a useful thing to do, as it ensures that the space of weight $k$ modular forms is <strong>finite dimensional</strong>.</p>
<p>To answer your fourth question, yes, and this is exactly the viewpoint taken in most research done on modular forms and their generalisations, where one considers <em>automorphic representations</em>.</p>
|
3,156,570 | <p>I need to evaluate the following limit:
<span class="math-container">$$
\lim_{x\downarrow 0} \dfrac{(1 - e^x)^{-1}}{x^c}
$$</span>
for different values of the constant <span class="math-container">$c$</span>.</p>
<p><em>What I've tried thus far:</em></p>
<p>We have that
<span class="math-container">$$
\lim_{x\downarrow 0} \dfrac{(1 - e^x)^{-1}}{x^c} = \lim_{x\downarrow 0}\dfrac{1}{x^c(1 - e^x)}
$$</span>
Now I know that <span class="math-container">$1 - e^x \to 0$</span> as <span class="math-container">$x\to 0$</span>. If <span class="math-container">$c\geq 0$</span> then <span class="math-container">$x^c\to 0$</span> as <span class="math-container">$x\to 0$</span> so we have that
<span class="math-container">$$
\lim_{x\downarrow 0}\dfrac{1}{x^c(1 - e^x)} = \infty
$$</span>
What I'm not so sure about is what happens when <span class="math-container">$c<0$</span>. If <span class="math-container">$c< 0$</span> then we can write <span class="math-container">$x^c$</span> as <span class="math-container">$x^{-\gamma}$</span> where <span class="math-container">$\gamma = \left|c\right|$</span>. I know that <span class="math-container">$\lim_{x\downarrow 0}x^{-\gamma}\to \infty$</span>, but I'm not sure whether <span class="math-container">$x^{-\gamma}\to \infty$</span> quicker than <span class="math-container">$1 - e^x\to 0$</span> as <span class="math-container">$x\to 0$</span>. I need to know because right now I'm not sure what the term <span class="math-container">$x^c(1 - e^x)$</span> does as <span class="math-container">$x\downarrow 0$</span> when <span class="math-container">$c <0$</span>.</p>
<p><strong>Question:</strong> </p>
<p>How do I evaluate
<span class="math-container">$$
\lim_{x\downarrow 0}\dfrac{1}{x^c(1 - e^x)}
$$</span>
when <span class="math-container">$c< 0$</span>?</p>
| little o | 543,867 | <p>If <span class="math-container">$-1< c < 0$</span> then <span class="math-container">$\frac {1} {x^c(1-e^x)} \rightarrow -\infty$</span> as <span class="math-container">$x \downarrow 0.$</span> If <span class="math-container">$c=-1$</span> then <span class="math-container">$\frac {1} {x^c(1-e^x)} \rightarrow -1$</span> as <span class="math-container">$x \downarrow 0.$</span> If <span class="math-container">$c<-1$</span> then <span class="math-container">$\frac {1} {x^c(1-e^x)} \rightarrow 0$</span> as <span class="math-container">$x \downarrow 0.$</span></p>
<p>For <span class="math-container">$c=-1$</span> Kavi Rama Murthy sir has already explained below. For <span class="math-container">$-1<c<0$</span> write <span class="math-container">$d=-c.$</span> Then <span class="math-container">$0<d<1.$</span> Therefore <span class="math-container">$\frac {1} {x^c(1-e^x)} = \frac {x^d} {1-e^x}$</span> which is a <span class="math-container">$\frac 0 0$</span> form as <span class="math-container">$x \downarrow 0.$</span> Applying L'Hospital rule we get <span class="math-container">$\lim\limits_{x \downarrow 0} \frac {dx^{d-1}} {-e^{x}}.$</span> Since <span class="math-container">$d-1 < 0$</span> so <span class="math-container">$x^{d-1} \rightarrow \infty$</span> and <span class="math-container">$-e^{x} \rightarrow -1,$</span> as <span class="math-container">$x \downarrow 0.$</span> Hence <span class="math-container">$$\lim\limits_{x \downarrow 0} \frac {dx^{d-1}} {-e^{x}} = -\infty$$</span> since <span class="math-container">$d>0.$</span></p>
<p>Can you do it similarly for <span class="math-container">$c<-1$</span>?</p>
|
2,398,215 | <p>If $f$ is continuous on $\mathbb{R}$ any of the following conditions are satisfied then $f$ must be a constant.</p>
<p>(1).$f(x)=f(mx),\forall x\in \mathbb{R},|m|≠1,m\in \mathbb{R}$</p>
<p>(2).$f(x)=f(2x+1),\forall x\in \mathbb{R}$</p>
<p>(3).$f(x)=f(x^2),\forall x\in \mathbb{R}$.</p>
<p>Suppose $f$ satisfy (1). I assumed, $f$ is not a constant, since f is not defined in the closed and bounded interval I am not able to use the major theorems of continuity.</p>
<p>Similarly I tried to apply the same theorems to (2) and (3). I am not able to proceed. Please help me to solve this question.</p>
| Tsemo Aristide | 280,301 | <ol>
<li>Suppose $|m|>1$ $f(x)=f(mx)$ implies that $f(x/m)=f(m(m/x))=f(x)$. We deduce that for every integer $n$, $f({x\over m^n})=f(x)$, since $lim_n{x\over m^n}=0$ and $f$ is continue, we deduce that $f(x)=lim_{n\rightarrow +\infty}f({x\over m^n})=f(0)$.</li>
</ol>
<p>The case $m<1$ consider the sequence $f(m^nx)$.</p>
|
2,185,489 | <blockquote>
<p>Let $\xi \in \mathcal{L}^2(\Omega,P)$ be a random variable with finite variance. Show that $$(E(\xi))^2 \leq E(\xi^2)$$ </p>
</blockquote>
<p>Since $$\operatorname{Var}(\xi) := E(\xi^2) - (E(\xi))^2$$ this boils down to showing $$\operatorname{Var}(\xi) \geq 0$$ which is quite restrictive. Since the lecture is not really good, I have not much of an idea what I should use. Has anyone a hint for me?</p>
| Scientifica | 164,983 | <p>Notice that $$\text{Var}(\xi)=E((\xi-E(\xi))^2)$$</p>
<p>is the definition of variance (at least in a number of books).</p>
|
3,931,807 | <p>I need to find max and min of <span class="math-container">$f(x,y)=x^3 + y^3 -3x -3y$</span> with the following restriction: <span class="math-container">$x + 2y = 3$</span>.</p>
<p>I used the multiplier's Lagrange theorem and found <span class="math-container">$(1,1)$</span> is the minima of <span class="math-container">$f$</span>. Apparently, the maxima is <span class="math-container">$(-13/7, 17/7)$</span> but I could not find it via Lagrange's theorem.</p>
<p>Here's what I did:</p>
<p>I put up the linear system:</p>
<p><span class="math-container">$\nabla f(x,y) = \lambda \, \nabla g(x,y)$</span></p>
<p><span class="math-container">$g(x,y) = 0$</span></p>
<p>then,</p>
<p><span class="math-container">$(3x^2 -3, 3y^2 -3) = \lambda (1,2)$</span></p>
<p><span class="math-container">$x + 2y -3 = 0$</span></p>
<p>Solving for <span class="math-container">$\lambda$</span>, I got <span class="math-container">$\lambda = 0$</span>, which gave me <span class="math-container">$x = 1$</span> and <span class="math-container">$y = 1$</span>.</p>
<p>How can I find the maxima if lambda only gives one value which is <span class="math-container">$0$</span>?</p>
| Théophile | 26,091 | <p>As others have said, you don't necessarily need to use Lagrange multipliers. But since you've set up the system, we can see what happens:</p>
<p><span class="math-container">$$
3x^2-3=\lambda\\
3y^2-3=2\lambda\\
x+2y-3=0
$$</span></p>
<p>From the first two equations, we have <span class="math-container">$3y^2-3=2(3x^2-3)$</span>, which simplifies to <span class="math-container">$y^2-1=2(x^2-1)$</span>. Rearranging the linear constraint, we have <span class="math-container">$x=3-2y$</span>. Putting this information together leads to
<span class="math-container">$$7y^2-24y+17=0.$$</span></p>
<p>You could solve this using the quadratic formula, but it is quicker to observe that <span class="math-container">$-24 = -7-17$</span>:</p>
<p><span class="math-container">$$7y^2-7y-17y+17=0$$</span></p>
<p>and so <span class="math-container">$(7y-17)(y-1)=0$</span>.</p>
<p>This will give you the two local extrema.</p>
|
287,129 | <p>The standard definition of computability, for a sequence $s\in\{0,1\}^\omega$, is that there is a Turing machine outputting $s[i]$ on input $i$.</p>
<p>I'm looking for strengthenings of this notion; for example, in the above definition it's not decidable whether there is a $1$ in $s$; or, given $i$, whether there is a $1$ in $s$ at position $\ge i$. I would be happy to be shown a "natural" definition of computability that makes these predicates computable.</p>
<p>To the above: if there were an algorithm that, from the Turing machine producing $s$, tells us whether $s$ contains a $1$ then I could do the following: from any Turing machine $M$, program a Turing machine outputting $s[i]=1$ if $M$ stops after $\le i$ steps. This sequence $s$ is obviously computable --- I said how to compute it --- but an algorithm determining if the sequence contains a $1$ would solve the Halting problem.</p>
<p>A search through the literature didn't show anything, so links and references are most welcome!</p>
| Bjørn Kjos-Hanssen | 4,600 | <p>You can look at <a href="http://www.math.cornell.edu/~minnes/LCsurvey.pdf" rel="noreferrer">automatic structures</a>, replacing Turing machines by finite automata. In that setting, the whole first order theory is decidable (in particular you can add existential quantifiers like you describe), by a Theorem of Hodgson.</p>
<p><em>Hodgson, Bernard R.</em>, <a href="http://dx.doi.org/10.1016/0304-3975(82)90042-1" rel="noreferrer"><strong>On direct products of automaton decidable theories</strong></a>, Theor. Comput. Sci. 19, 331-335 (1982). <a href="https://zbmath.org/?q=an:0493.03002" rel="noreferrer">ZBL0493.03002</a>.</p>
|
2,536,163 | <p>How to integrate using contour integration?
$$\int_1^{\infty}\frac{\sqrt{x-1}}{(1+x)^2}dx$$
I was putting $y = x-1$ then $\frac{dy}{dx}$= $1-0$, ${dy}={dx} $ then i get $$\int_1^{\infty}\frac{\sqrt{y}}{(2+y)^2}dy$$</p>
<p>I don't know how to take it from here. I would appreciate if someone could help me and give me some hints or the solution. Thanks.</p>
| Aditya Narayan Sharma | 335,483 | <p>I won't prefer a contour here, making the substitution $y=x-1$ we have,
$$ \displaystyle \int_1^\infty \dfrac{\sqrt{x-1}}{(1+x)^2}\; dx = \int_0^\infty \dfrac{\sqrt{y}}{(2+y)^2}\; dy$$
Next substitute $y=u^2$ ,
$$\displaystyle I = 2\int_0^\infty \dfrac{u^2}{(u^2+2)^2}\; du $$
Using partial fractions this will be equal to,
$$ \displaystyle \begin{align} I &= 2\int_0^\infty \dfrac{du}{u^2+2}-\int_0^\infty \dfrac{4\; du}{(u^2+2)^2} \\ &= \dfrac{\pi}{\sqrt{2}} - \dfrac{1}{\sqrt{2}}\int_0^{\pi/2} (1+\cos 2\theta)\; d\theta \\ &=\dfrac{\pi}{\sqrt{2}}-\dfrac{\pi}{2\sqrt{2}} \\ &= \dfrac{\pi}{2\sqrt{2}} \end{align}$$</p>
<p><strong>Footnotes :</strong></p>
<p>1) $\displaystyle \int \dfrac{dx}{x^2+a^2} = \dfrac{1}{a} \tan^{-1}\dfrac{x}{a}$</p>
<p>If you integrate it then,</p>
<p>$\displaystyle \int_0^{+\infty} \dfrac{dx}{x^2+a^2} = \dfrac{1}{a}\times \dfrac{\pi}{2}$</p>
<p>2) $\displaystyle \int_0^{+\infty} \dfrac{4\; du}{(u^2+2)^2}$</p>
<p>We make the substitution $u=\sqrt{2}\tan \theta$ , this means $du=\sqrt{2}\sec^2 \theta d\theta$ .The new limits for $\theta$ would be according to
$ \displaystyle \theta = \tan^{-1}\dfrac{u}{\sqrt{2}}$ .The new limits are thus from $0\to\dfrac{\pi}{2}$ since $\tan^{-1}\dfrac{+\infty}{\sqrt{2}}=\dfrac{\pi}{2}$ & $\tan^{-1}\dfrac{0}{\sqrt{2}}=0$ . The integral now reads :
$$\displaystyle \begin{align}\int_0^{+\infty} \dfrac{4\; du}{(u^2+2)^2} &= 4\int_0^{\pi/2} \dfrac{\sqrt{2}\sec^2 \theta}{(2(\tan^2 \theta+1))^2} \; d\theta \\ &= 4\sqrt{2}\int_0^{\pi/2} \dfrac{\sec^2 \theta}{4(\sec^2 \theta)^2}\; d\theta \\ &= \sqrt{2} \int_0^{\pi/2} \dfrac{\sec^2 \theta}{\sec^4 \theta} \; d\theta \\ &= \sqrt{2} \int_0^{\pi/2} \cos^2 \theta \; d\theta \\ &= \dfrac{\sqrt{2}}{2}\int_0^{\pi/2} 2\cos^2 \theta \; d\theta \\ &= \dfrac{1}{\sqrt{2}}\int_0^{\pi/2} (1+\cos 2\theta)\; d\theta \\ &= \dfrac{1}{\sqrt{2}} \int_0^{\pi/2} d\theta + \dfrac{1}{\sqrt{2}}\int_0^{\pi/2} \cos 2\theta \; d\theta \\ &= \dfrac{\pi}{2\sqrt{2}} - \dfrac{1}{2\sqrt{2}}\left[\sin 2\theta\right]_0^{\pi/2} \\ &= \dfrac{\pi}{2\sqrt{2}} - \dfrac{1}{2\sqrt{2}}(\sin \pi - \sin 0) \\ &= \dfrac{\pi}{2\sqrt{2}} \end{align} $$</p>
|
2,536,163 | <p>How to integrate using contour integration?
$$\int_1^{\infty}\frac{\sqrt{x-1}}{(1+x)^2}dx$$
I was putting $y = x-1$ then $\frac{dy}{dx}$= $1-0$, ${dy}={dx} $ then i get $$\int_1^{\infty}\frac{\sqrt{y}}{(2+y)^2}dy$$</p>
<p>I don't know how to take it from here. I would appreciate if someone could help me and give me some hints or the solution. Thanks.</p>
| Jack D'Aurizio | 44,121 | <p>$$\begin{eqnarray*}\int_{1}^{+\infty}\frac{\sqrt{x-1}}{(x+1)^2}\,dx &\stackrel{x\mapsto z+1}{=}& \int_{0}^{+\infty}\frac{\sqrt{z}}{(2+z)^2}\,dz\stackrel{z\mapsto 2u}{=}\frac{1}{\sqrt{2}}\int_{0}^{+\infty}\frac{\sqrt{u}}{(1+u)^2}\,du\\&\stackrel{u\mapsto v^2}{=}&\frac{1}{\sqrt{2}}\int_{0}^{+\infty}\frac{2v}{(1+v^2)^2}\cdot v\,dv\\&\stackrel{\text{IBP}}{=}&\frac{1}{\sqrt{2}}\int_{0}^{+\infty}\frac{dv}{1+v^2}=\frac{1}{2\sqrt{2}}\int_{\mathbb{R}}\frac{dv}{1+v^2}\\&=&\frac{\pi i}{\sqrt{2}}\cdot \operatorname*{Res}_{v=i}\left(\frac{1}{v^2+1}\right)=\frac{\pi i}{\sqrt{2}}\cdot\frac{1}{2i}=\color{blue}{\frac{\pi}{2\sqrt{2}}}. \end{eqnarray*}$$</p>
|
680,319 | <p>Let's restate this question in using mathematical notation. Let $n,k \in \mathbb{N}$. Let $f(n)=\left\lfloor{\frac{n}{k}}\right\rfloor$. Is it possible to rewrite this using the addition, multiplication, and exponentiation operators? I know it's possible for the case where $k=1$. Quite simply, note that $f(n)=\left\lfloor{\frac{n}{1}}\right\rfloor=\lfloor{n}\rfloor=n$. I know it's possible when $k=2$. Let $f(n)=\left(\frac{n}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}\right)(-1)^n$. The trick there was to start halfway between $\frac{n}{2}-\frac{1}{2}$ and $\frac{n}{2}$ and using the $(-1)^n$ term to figure out the rest. </p>
<p>I'm wondering if it's possible if $k=3$? I think it's possible with the use of infinite sums and trigonometric functions (though I haven't thought about that yet). However, I'd like to avoid those if necessary. I feel they may not be necessary -- though I may be wrong.</p>
<p>Even a slight discussion about the topic would be great.</p>
<p>EDIT: </p>
<p>So, I like the @user44197's solution. It gives rise to the solution I was looking for -- which was simpler than I thought. Let $f(n)=\left\lfloor{\frac{n}{k}}\right\rfloor$, where $n$ is a variable and $k$ is a constant (for simplicity, let $k$ be some constant in the naturals). Why not treat it like a recursion? Let $R(n) = R(n-k) + 1$. To solve this, we need to solve the characteristic polynomial $\lambda^k-1=0$. Therefore, $R_{Homogeneous}(n)=\sum_{i=0}^{k-1}n^{i}c_{i}(1)^n=\sum_{i=0}^{k-1}n^{i}c_{i}$ where $c_i$ is the coefficient that will solve the recursion. Now you just need to find the non-homogeneous part, solve for the coefficients, and you're done.</p>
| JJacquelin | 108,514 | <p>The floor function on form of Fourrier series :</p>
<p><img src="https://i.stack.imgur.com/YQ0ag.jpg" alt="enter image description here"></p>
|
3,394,050 | <p>I'm having trouble with this problem.</p>
<blockquote>
<p>Using logical equivalencies prove that <span class="math-container">$(p \land q)\implies (p \lor q)$</span> is a tautology.</p>
</blockquote>
| Mikhail D | 398,197 | <p>A tautology is a statement that is always 'True'.</p>
<p>In your case you have two variables, each may take the value of either 'true' or 'false'.</p>
<p>You may simply plug these four cases in your expression and get 'True' in each case you try. Note that your logical expression is a composition of three operations, i.e. 'and', 'or' and 'implies'. Please, check with Wikipedia for their truth tables.</p>
<p><a href="https://en.wikipedia.org/wiki/Modus_ponens" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Modus_ponens</a> </p>
|
4,488,991 | <p>Let <span class="math-container">$M,M'$</span> be oriented connected compact smooth manifolds of the same dimension, let <span class="math-container">$S$</span> be a smooth manifold,
and let <span class="math-container">$\nu : S\times M\rightarrow M'$</span> be some smooth map.
Let <span class="math-container">$\nu_s : M\rightarrow M'$</span> denote <span class="math-container">$\nu_s(\cdot) = \nu(s,\cdot)$</span> for <span class="math-container">$s\in S$</span>.</p>
<p>If I'm not mistaken, we can formalize the claim that the per-<span class="math-container">$s$</span> degree map <span class="math-container">$s \mapsto \mathrm{degree}(M\xrightarrow{\nu_s} M')$</span>,
as a map <span class="math-container">$S \rightarrow \mathbb{Z}$</span>, should depend continuously on <span class="math-container">$s \in S$</span> (hence be locally constant) by using e.g. de Rham theory.</p>
<p>(Details: Fix some top-degree differential form <span class="math-container">$\omega$</span> on <span class="math-container">$M'$</span> which is non-exact; then <span class="math-container">$\mathrm{deg}(\nu_s) = (\int_M \nu_s^*\omega)/(\int_{M'}\omega)$</span>, which we can probably argue depends continuously on <span class="math-container">$s$</span>.)</p>
<hr />
<p>More generally, if <span class="math-container">$M,M'$</span> are Poincaré duality spaces (with chosen fundamental classes), and <span class="math-container">$S$</span> is a topological space and "smooth" everywhere above is replaced by "continuous", does the same claim hold? How can the proof be formalized?</p>
| Adayah | 149,178 | <p>Here is an abstract way to construct a bunch of rectangular triangles satisfying the property. They don't need to have integer side lengths, though.</p>
<p>Take any rectangular triangle <span class="math-container">$T$</span>. Let <span class="math-container">$x$</span> denote the product of the lengths of the two shorter line segments from a corner to the center of the inscribed circle, and let <span class="math-container">$y$</span> denote the length of the longest one. We want <span class="math-container">$x = y$</span>.</p>
<p>Now let <span class="math-container">$\alpha > 0$</span> and consider a triangle <span class="math-container">$S$</span> which results from scaling <span class="math-container">$T$</span> by the factor <span class="math-container">$\alpha$</span>. Then the corresponding values <span class="math-container">$x'$</span>, <span class="math-container">$y'$</span> in the new triangle satisfy</p>
<p><span class="math-container">$$x' = \alpha^2 x, \qquad y' = \alpha y.$$</span></p>
<p>Therefore it's possible to find <span class="math-container">$\alpha$</span> such that <span class="math-container">$x' = y'$</span>, so the property holds for the triangle <span class="math-container">$S$</span>.</p>
<p>This actually shows that for every rectangular triangle we can find a (unique, actually) triangle similar to it that has the desired property.</p>
|
2,600,776 | <blockquote>
<p>A continuos random variable $X$ has the density
$$
f(x) = 2\phi(x)\Phi(x), ~x\in\mathbb{R}
$$
then</p>
<p>(<em>A</em>) $E(X) > 0$</p>
<p>(<em>B</em>) $E(X) < 0$</p>
<p>(<em>C</em>) $P(X\leq 0) > 0.5$</p>
<p>(<em>D</em>) $P(X\ge0) < 0.25$</p>
<p>\begin{eqnarray}
\Phi(x) &=& \text{Cumulative distribution function of } N(0,1)\\
\phi(x) &=& \text{Density function of } N(0, 1)
\end{eqnarray}</p>
</blockquote>
<p>I don't have a slightest clue where to start with. Can someone give me a little push. I saw some answers on same question like this but I didn't understand how should I integrate it when calculating expectation. </p>
| drhab | 75,923 | <p>Hint:</p>
<p>If $x>0$ then $f(-x)<f(x)$</p>
|
303,933 | <p>How can I prove that
$$ \lim_{n\to\infty} \frac{(\ln(n))^a}{n^b} = 0 \;\forall a,b > 0 $$
? Intuitively it is clear to me because of the behavior of the functions. Thanks for all.</p>
<p><strong>Edit</strong> I'm not able to use L'Hopital rule. Sorry.</p>
| Ishan Banerjee | 52,488 | <p>$lim_{n\to\infty} \frac {log(n)}{n^a}=0$ $ a>0$
(This can be proved with L'hopital's)</p>
<p>Use this to prove your result.</p>
|
1,831,243 | <p>Kronecker "delta" function is generally defined as
$\delta(i,j)=1$ if $i$ is equal to $ j$, otherwise $0$.</p>
<p>How about if $j$ is not an integer? I mean let $j$ is a half open interval defined as $j=(0,1]$ and $ i$ has any value on interval $[0,1]$,
then can we use Kronecker delta to find if $i$ belongs to $j$? In other words, can we define like: $\delta(i,j)=1$ if $i$ is in $j$, $0$ otherwise?</p>
| Marc van Leeuwen | 18,880 | <p>No, the convention is that the Kronecker delta tests for <em>equality</em> and not for some other relation such as set membership. Also the elements tested for equality are most often numbers, and even integers, but in a stretch you could use it to test equality of objects of other types. So using the Kronecker delta with one index a number and the other an interval goes against the original definition of the symbol. (This is compounded by the confusion caused by having the letter $j$ denote an <em>interval</em> rather than a number.)</p>
<p>However there is an easy notation generalising the Kronecker $\delta$ that allows you to test all kinds of relations, including set membership, which is the <a href="https://en.wikipedia.org/wiki/Iverson_bracket" rel="nofollow">Iverson bracket</a>: the convention that by enclosing any condition in brackets one obtains an expression that denotes $1$ if the condition holds, and $0$ if it fails. So instead of $\delta_{i,j}$ you can then write $[i=j]$ which makes the equality test explicit, but you can equally well use $[i\leq j]$ for an "upper triangular" relation, or if you want to test membership of an interval $I$ such as $(0,1]$ (or indeed any set) write $[i\in I]$. Personally I find this much more readable than using one of the many notations $\mathbf1_I(i)$, $\mathbf I_I(i)$, $\chi_I(i)$ for an indicator function that are in use, and also avoids monopolising yet another symbol for a limited special purpose.</p>
<p>To my regret the use of Iverson bracket is not yet very common in mathematical texts, but it certainly deserves to be used where it can (combinatorics for instance provides many potential uses). Since the notation introduces a logical relation in the middle of a formula, it takes a little getting used to. But this my be facilitated by typographically making the brackets stand out a bit with respect to other uses of brackets and parentheses, for instance making them bold and/or giving them additional spacing; however even without it does not really cause any conflict with other notations.</p>
|
3,293,082 | <p>I know this question was asked on this site, but I didn't understand the answer. Could someone give me the simplest explanation of this? (High school level explanation)</p>
| Justin Benfield | 297,916 | <p><span class="math-container">$\arccos(x)$</span> is the inverse function of <span class="math-container">$\cos(x)$</span> (restricted to the interval <span class="math-container">$[0,\pi]$</span>)</p>
<p><span class="math-container">$\sec(x)$</span> is the reciprocal, <span class="math-container">$\dfrac{1}{\cos(x)}$</span>.</p>
<p>in otherwords, it is the difference between <span class="math-container">$f^{-1}(x)$</span> and <span class="math-container">$\dfrac{1}{f(x)}$</span>.</p>
|
3,293,082 | <p>I know this question was asked on this site, but I didn't understand the answer. Could someone give me the simplest explanation of this? (High school level explanation)</p>
| J.G. | 56,861 | <p><span class="math-container">$\arccos x$</span> is a value whose cosine is <span class="math-container">$x$</span>, whereas <span class="math-container">$\sec x=\frac{1}{\cos x}$</span>. The two kinds of "inverse" trigonometric functions can feel a bit esoteric when you're new to them, but let's pretend <span class="math-container">$\cos x=7x+12$</span>. Then <span class="math-container">$\arccos x=\frac{x-12}{7}$</span> while <span class="math-container">$\sec x=\frac{1}{7x+12}$</span>, which is clearly completely different.</p>
|
2,000,940 | <p>Suppose $A \subseteq \mathbb{R}$ is measurable and $f\colon A \to \mathbb{R}$ is Lipschitz on the set $A$, i.e there is some $K\ge 0$ such that $\lvert f(x)-f(y)\rvert \le K \lvert x-y\rvert$ for $x,y \in A$. </p>
<p>I'm trying to prove that
$$
m^\ast(f(E)) \le K\,m^\ast(E)\textrm{ for every set }E \subseteq A.
$$</p>
<p>I've tried covering the set $E$ by intervals, yet the function $f$ may not have been defined outside the set $A$. Also, approaching the set $E$ from inside can encounter problem when $E$ is non-measurable.</p>
| Daniel Fischer | 83,702 | <blockquote>
<p>I've tried covering the set $E$ by intervals, yet the function $f$ may not have been defined outside the set $A$.</p>
</blockquote>
<p>Good that you noticed that problem.</p>
<p>But that problem can be dealt with, we can extend $f$ to a Lipschitz continuous $F \colon \mathbb{R} \to \mathbb{R}$ with the same Lipschitz constant $K$:</p>
<p>First, since Lipschitz functions are uniformly continuous, $f$ has a unique continuous extension $\overline{f} \colon \overline{A} \to \mathbb{R}$. This extension is also Lipschitz continuous with Lipschitz constant $K$: for $x,y \in \overline{A}$, choose sequences $(x_n),\, (y_n)$ in $A$ with $x_n \to x,\, y_n \to y$. Then</p>
<p>$$\lvert \overline{f}(x) - \overline{f}(y)\rvert = \lim_{n\to\infty} \lvert f(x_n) - f(y_n)\rvert \leqslant \lim_{n\to\infty} K\lvert x_n - y_n\rvert = K\lvert x-y\rvert.$$</p>
<p>Then we can extend $\overline{f}$ to all of $\mathbb{R}$. $\mathbb{R}\setminus \overline{A}$ is a disjoint union of countably many open intervals. For a bounded interval $(a,b)$ in the complement of $\overline{A}$, we have $a,b \in \overline{A}$ and can define the extension by linear interpolation,</p>
<p>$$F(x) := \frac{x-a}{b-a} \overline{f}(b) + \frac{b-x}{b-a} \overline{f}(a).$$</p>
<p>If $\overline{A}$ is bounded above, and $a = \max \overline{A}$, define $F(x) = \overline{f}(a)$ for $x > a$. If $\overline{A}$ is bounded below and $b = \min \overline{A}$, define $F(x) = \overline{f}(b)$ for $x < b$. (And if $A = \varnothing$, there's nothing to do, but we can set $F(x) = 0$ for all $x$.)</p>
<p>Verify that the extension $F$ thus defined is a globally Lipschitz function with $\lvert F(x) - F(y)\rvert \leqslant K\lvert x-y\rvert$ for all $x,y\in \mathbb{R}$.</p>
<p>Then covering $E$ by intervals and looking at the corresponding cover of $f(E) = F(E)$ works.</p>
|
125,592 | <p>I'm finding in trouble trying to resolve this exercise. I have to calculate the convolution of two signals:</p>
<p>$$y(t)=e^{-kt}u(t)*\frac{\sin\left(\frac{\pi t}{10}\right)}{(\pi t)} $$</p>
<p>where $u(t)$ is Heavside function</p>
<p>well I applied the formula that says that the convolution of this two signal is equal to</p>
<p>$$Y(f)=X(f)W(f)$$</p>
<p>where $X(f)$ is the fourier transform of the first signal and $W(f)$ is the fourier transform of second signal</p>
<p>well fourier transform of $e^{-kt}u(t)$ is $X(f)=\frac{1}{k+j2\pi f}$. I have to make second signal as equals as possible to $\operatorname{sinc}\left(\frac{\pi t}{10}\right)$ so I do this operation:
$\frac{\sin\left(\frac{\pi t}{10}\right)}{\left(\frac{\pi t}{10}\right)}{\left(\frac{1}{10}\right)}$. this is equal ${\left(\frac{1}{10}\right)}\operatorname{sinc}\left(\frac{\pi t}{10}\right)$</p>
<p>right or not?</p>
<p>Edit</p>
<p>If something is not clear please advice me</p>
| dayar | 24,144 | <p>It's possible to do this integral in a couples of lines using the residue theorem from complex analysis.</p>
<p>Details: The usual trick to do definite integrals going from $0$ to $2\pi$ is to let $\cos x = \dfrac {z^2 + 1} {2z}$ where $z = {\rm e} ^{{\rm i} x}$. This substitution also implies that ${\rm d} x = \dfrac {{\rm d} z} {{\rm i} z}$. Then this is reduced to the contour integral of $\left( \dfrac {z^2 + 1} {2z} \right) ^n \dfrac {{\rm d} z} {{\rm i} z}$ where the contour is the unit circle in the complex plane. Then you can expand this using the binomial theorem and get the coefficient of $\dfrac 1 z$ and apply the residue theorem to get the answer.</p>
|
125,592 | <p>I'm finding in trouble trying to resolve this exercise. I have to calculate the convolution of two signals:</p>
<p>$$y(t)=e^{-kt}u(t)*\frac{\sin\left(\frac{\pi t}{10}\right)}{(\pi t)} $$</p>
<p>where $u(t)$ is Heavside function</p>
<p>well I applied the formula that says that the convolution of this two signal is equal to</p>
<p>$$Y(f)=X(f)W(f)$$</p>
<p>where $X(f)$ is the fourier transform of the first signal and $W(f)$ is the fourier transform of second signal</p>
<p>well fourier transform of $e^{-kt}u(t)$ is $X(f)=\frac{1}{k+j2\pi f}$. I have to make second signal as equals as possible to $\operatorname{sinc}\left(\frac{\pi t}{10}\right)$ so I do this operation:
$\frac{\sin\left(\frac{\pi t}{10}\right)}{\left(\frac{\pi t}{10}\right)}{\left(\frac{1}{10}\right)}$. this is equal ${\left(\frac{1}{10}\right)}\operatorname{sinc}\left(\frac{\pi t}{10}\right)$</p>
<p>right or not?</p>
<p>Edit</p>
<p>If something is not clear please advice me</p>
| Christian Blatter | 1,303 | <p>Qiaochu Yuan's hint seems to be the simplest approach: By the binomial theorem for any $n\geq0$ one has
$$2^n\cos^n x=(e^{ix}+e^{-ix})^n=\sum_{k=0}^n {n\choose k} (e^{ix})^k\ (e^{-ix})^{n-k}=\sum_{k=0}^n {n\choose k} e^{(2k-n)ix}\ .\qquad(*)$$
Since
$$\int_0^{2\pi}e^{i\ell x}\ dx=\cases{2\pi&$\quad(\ell=0)$\cr 0&$\quad(\ell\ne0)$\cr}$$
at most one term on the right of $(*)$ contributes to the integral $J_n:=\int_0^{2\pi}\cos^n x\ dx$. When $n$ is odd then $2k-n\ne0$ for all $k$ in $(*)$, therefore $J_n=0$ in this case. When $n$ is even then $k=n/2$ gives the only contribution to the integral, and we get
$$\int_0^{2\pi} \cos^n x\ dx={2\pi\over 2^n}{n\choose n/2}\ .$$</p>
|
125,592 | <p>I'm finding in trouble trying to resolve this exercise. I have to calculate the convolution of two signals:</p>
<p>$$y(t)=e^{-kt}u(t)*\frac{\sin\left(\frac{\pi t}{10}\right)}{(\pi t)} $$</p>
<p>where $u(t)$ is Heavside function</p>
<p>well I applied the formula that says that the convolution of this two signal is equal to</p>
<p>$$Y(f)=X(f)W(f)$$</p>
<p>where $X(f)$ is the fourier transform of the first signal and $W(f)$ is the fourier transform of second signal</p>
<p>well fourier transform of $e^{-kt}u(t)$ is $X(f)=\frac{1}{k+j2\pi f}$. I have to make second signal as equals as possible to $\operatorname{sinc}\left(\frac{\pi t}{10}\right)$ so I do this operation:
$\frac{\sin\left(\frac{\pi t}{10}\right)}{\left(\frac{\pi t}{10}\right)}{\left(\frac{1}{10}\right)}$. this is equal ${\left(\frac{1}{10}\right)}\operatorname{sinc}\left(\frac{\pi t}{10}\right)$</p>
<p>right or not?</p>
<p>Edit</p>
<p>If something is not clear please advice me</p>
| dohmatob | 168,758 | <p>Let <span class="math-container">$X$</span> be a standard normal random variable. Then, your integral <span class="math-container">$I$</span> can be computed as</p>
<p><span class="math-container">$$
I = 2\pi\cdot \mathbb E[X^n] = \begin{cases}2\pi\cdot (n-1)!! = \frac{2\pi}{2^n}{n\choose n/2},&\mbox{ if }n\text{ is even},\\
0,&\mbox{ else.}
\end{cases}
$$</span></p>
|
1,608,645 | <p>Is there supposed to be a fast way to compute recurrences like these?</p>
<p>$T(1) = 1$</p>
<p>$T(n) = 2T(n - 1) + n$</p>
<p>The solution is $T(n) = 2^{n+1} - n - 2$. </p>
<p>I can solve it with:</p>
<ol>
<li><p>Generating functions.</p></li>
<li><p>Subtracting successive terms until it becomes a pure linear recurrence $T(n) = 4T(n-1) - 5T(n-2) + 2T(n-3)$ and then solving it using the powers-of-roots approach. </p></li>
<li><p>Repeated substitution, which gives a few simple closed-forms but one messy sum $\sum_{k=1}^{n-2} 2^k k$ which to me is not easy to do quickly.</p></li>
</ol>
<p>Each one of these approaches takes me several minutes to flesh out, but I feel like this is supposed to be one of those questions I should be able to answer in a few seconds and move on. What am I missing? Is there some quick trick to doing these recurrences?</p>
| Kaynex | 296,320 | <p>Solve the homogeneous equation by removing any non-functional terms. In this case, simply remove the n.</p>
<p>Solve the specific solution by guessing. In this case, it's not difficult. Try plugging in Ax + b into the functional equation, and see if you can solve for the coefficients.</p>
<p>Then, add both solutions together for the general solution.</p>
|
2,396,073 | <p>Let $\omega_1$ be the first uncountable ordinal. In some book, the set $\Omega_0:=[1,\omega_1)=[1,\omega_1]\backslash\{\omega_1\}$ is called the set of countable ordinals. Why? It is obvious that it is an uncountable set, because $[1,\omega_1]$ is uncountable. The most possible reason I think is that for any $x\prec \omega_1$, the set $[1,x)$ is countable. </p>
| Noah Schweber | 28,111 | <p>In fact, $\Omega_0=\omega_1$ (EDIT: with $0$ removed): remember that each ordinal is the set of all smaller ordinals. So in particular, since $\omega_1$ is the smallest uncountable ordinal, it is also the <em>set</em> of all <em>countable</em> ordinals (since each countable ordinal is smaller than $\omega_1$).</p>
|
484,117 | <p>What is the simplest $\Bbb{R}\to\Bbb{R}$ function with two peaks and a valley?</p>
<p>I have a set of points in $\Bbb{R^2}$ and I would like to fit a curve to the points, the points approximately lie on a curve like the one depicted in the following figure:</p>
<p><img src="https://i.stack.imgur.com/8tVLu.png" alt="A $\Bbb{R}\to\Bbb{R}$ function with two peaks and a valley."></p>
<p>My points are such that $a$ (the width of the valley) is almost constant while $b$ (the height of the peaks with respect to the valley) can change.</p>
<p>My informal definition of "simplest" is based on the following requirements:</p>
<ol>
<li>the function should be sufficient smooth</li>
<li>the fit should be easy to do with some off-the-shelf algorithm</li>
<li>I have just the points lying on the curve in figure, so I think that the function should smoothly go to zero to the left of the left peak and to the right of the right peak.</li>
</ol>
<p>My goal is to estimate $b$.</p>
| doraemonpaul | 30,938 | <p>Let $u=\sqrt{f(t)-y}$ ,</p>
<p>Then $y=f(t)-u^2$</p>
<p>$\dfrac{dy}{dt}=\dfrac{df(t)}{dt}-2u\dfrac{du}{dt}$</p>
<p>$\therefore\dfrac{df(t)}{dt}-2u\dfrac{du}{dt}=u$</p>
<p>$2u\dfrac{du}{dt}+u=\dfrac{df(t)}{dt}$</p>
<p>This belongs to an Abel equation of the second kind.</p>
<p>Let $u=-\dfrac{v}{2}$ ,</p>
<p>Then $\dfrac{du}{dt}=-\dfrac{1}{2}\dfrac{dv}{dt}$</p>
<p>$\therefore\dfrac{v}{2}\dfrac{dv}{dt}-\dfrac{v}{2}=\dfrac{df(t)}{dt}$</p>
<p>$v\dfrac{dv}{dt}-v=2\dfrac{df(t)}{dt}$</p>
<p>This belongs to an Abel equation of the second kind in the canonical form.</p>
<p>Please follow the method in <a href="https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf" rel="nofollow">https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf</a> or in <a href="http://www.iaeng.org/IJAM/issues_v43/issue_3/IJAM_43_3_01.pdf" rel="nofollow">http://www.iaeng.org/IJAM/issues_v43/issue_3/IJAM_43_3_01.pdf</a></p>
|
2,706,165 | <p>So if $y=\log(3-x) = \log(-x+3)$ then you reflect $\log(x)$ in the $y$ axis to get $\log(-x)$.</p>
<p>Then because it is $+3$ inside brackets you then shift to the left by $3$ giving an asymptote of $x=-3$ and the graph crossing the $x$ axis at $(-4,0)$. </p>
<p>However this does not work. The answer shows the $+3$ in the bracket shifting the curve to the right by $3$ giving an asymptote of $x=3$ and the curve crossing the $x$ axis at $(2,0)$. </p>
<p>Why does it do this? Can anyone please explain?</p>
| user | 505,767 | <p>Note that</p>
<ul>
<li>$3-x$ put the origin in x=3 (right translation) and reverse axis direction</li>
<li>thus the vertical asympthote is at $x=3$ and the root is at $x=2$</li>
</ul>
|
4,644,904 | <p>Find the greatest and the least values of the function <span class="math-container">$f(x)=\sin x\sin2x$</span> on the interval <span class="math-container">$(-\infty,\infty)$</span>.</p>
<p>The solution presented is as follows:</p>
<blockquote>
<p>Represent the function <span class="math-container">$y=f(x)=\sin x\sin 2x$</span> in the form <span class="math-container">$y=\frac{\cos x-\cos 3x}{2},$</span> whence it is seen that the function is an even one and has a period <span class="math-container">$2\pi.$</span> Hence, it is sufficient to seek the greatest and the least values among the extrema on the interval <span class="math-container">$[0, \pi].$</span> Find the derivative <span class="math-container">$y'=\frac12(3\sin {3x}-\sin x)$</span> In <span class="math-container">$[0,\pi]$</span> the derivative vanishes at the points <span class="math-container">$x_1=0,x_2=\arccos \frac {1}{\sqrt{3}},x_3=\arccos(-\frac{1}{\sqrt{3}}),x_4=\pi.$</span> Compute the values of the function at these points: <span class="math-container">$y(0)=y(\pi)=0,y[\arccos(\pm\frac{1}{\sqrt{3}})]=\pm\frac {4}{3\sqrt{3}}.$</span> Hence, the least value of the function in the interval <span class="math-container">$(-\infty,\infty)$</span> is equal to <span class="math-container">$-\frac{4}{3\sqrt 3}$</span> and <span class="math-container">$\frac{4}{3\sqrt 3}.$</span></p>
</blockquote>
<p>However, in the above solution, I dont get, how do they assert "Hence, it is sufficient to seek the greatest and the least values among the extrema on the interval <span class="math-container">$[0, \pi].$</span>" ? The function is periodic and even but that only leads us to investigate it in the interval <span class="math-container">$[0, 2\pi].$</span> I don't get why investigating this function at <span class="math-container">$[0, \pi]$</span> is sufficient. I am not quite getting it.</p>
| Community | -1 | <p><strong>Alternative resolution:</strong></p>
<p>By direct differentiation, we have to solve</p>
<p><span class="math-container">$$\sin(2x)\cos(x)+2\sin(x)\cos(2x)=0$$</span></p>
<p>or</p>
<p><span class="math-container">$$2\sin(x)(3\cos^2(x)-1)=0.$$</span></p>
<p>When <span class="math-container">$\sin(x)=0$</span>, the function equals <span class="math-container">$0$</span>. When <span class="math-container">$3\cos^2(x)-1=0$</span>, the function <span class="math-container">$\sin(x)\sin(2x)=2(1-\cos^2(x))\cos(x)$</span> equals one of <span class="math-container">$\pm\dfrac4{3\sqrt3}$</span>.</p>
|
35,151 | <p>Many complexity theorists assume that $P\ne NP.$ If this is proved, how would it impact quantum computing and quantum algorithms? Would the proof immediately disallow quantum algorithms from ever solving NP-Complete problems in Quantum Polynomial time?</p>
<p><a href="http://en.wikipedia.org/wiki/QMA" rel="nofollow">According to Wikipedia</a>, quantum complexity classes BQP and QMA are the bounded-error quantum analogues of P and NP. Is it likely that a proof that $P\ne NP$ can be adapted to the quantum setting to show that $BQP \ne QMA?$</p>
| Artem Kaznatcheev | 8,239 | <p>To your second question. It is unlikely that the current under-review proof of P != NP will allow you to seperate BQP and QMA (or BQP and P, or BQP and NP, or even BPP and NP...). Deolalikar's proof uses descriptive complexity, in particular it uses a correspondence between statements expressible in certain logics and the complexity classes P and NP. As far as I know there is no such nice one-to-one correspondences between BPP, BQP, or QMA and nice logics.</p>
|
259,308 | <p>The output of <code>ListPointPlot3D</code> is shown below:
<a href="https://i.stack.imgur.com/ypt73.png" rel="noreferrer"><img src="https://i.stack.imgur.com/ypt73.png" alt="enter image description here" /></a>
I only want to connect the dots in such a way that it forms a ring-like mesh. However, when I use <code>ListPlot3D</code>, the output is like this:
<a href="https://i.stack.imgur.com/zmaI1.png" rel="noreferrer"><img src="https://i.stack.imgur.com/zmaI1.png" alt="enter image description here" /></a>
Which is not what I want.</p>
<p>May I know how I should modify the code to achieve the intended result? Thank you.</p>
<p>Here is the data I used (with fewer data points):</p>
<pre><code>{{15.,-4.33154,0.015},{15.4114,-3.89839,0.0154114},{15.7792,-3.46523,0.0157792},{16.1037,-3.03208,0.0161037},{16.3847,-2.59893,0.0163847},{16.6224,-2.16577,0.0166224},{16.8169,-1.73262,0.0168169},{16.968,-1.29946,0.0169681},{17.076,-0.866309,0.017076},{17.1408,-0.433154,0.0171408},{17.1624,-6.01122*10^-16,0.0171624},{17.1408,0.433154,0.0171408},{17.076,0.866309,0.017076},{16.968,1.29946,0.0169681},{16.8169,1.73262,0.0168169},{16.6224,2.16577,0.0166224},{16.3847,2.59893,0.0163847},{16.1037,3.03208,0.0161037},{15.7792,3.46523,0.0157792},{15.4114,3.89839,0.0154114},{14.698,-4.79395,2.99474},{15.1194,-4.31455,3.08061},{15.4564,-3.83516,3.14927},{15.7246,-3.35576,3.2039},{15.9361,-2.87637,3.24701},{16.1009,-2.39697,3.28058},{16.2264,-1.91758,3.30616},{16.3185,-1.43818,3.32492},{16.3815,-0.95879,3.33775},{16.4181,-0.479395,3.34521},{16.4301,-6.65294*10^-16,3.34766},{16.4181,0.479395,3.34521},{16.3815,0.95879,3.33775},{16.3185,1.43818,3.32492},{16.2264,1.91758,3.30616},{16.1009,2.39697,3.28058},{15.9361,2.87637,3.24701},{15.7246,3.35576,3.2039},{15.4564,3.83516,3.14927},{15.1194,4.31455,3.08061},{13.8101,-5.25915,5.85509},{14.2225,-4.73323,6.02996},{14.5189,-4.20732,6.15563},{14.7316,-3.6814,6.24579},{14.8837,-3.15549,6.31026},{14.9917,-2.62957,6.35607},{15.0675,-2.10366,6.3882},{15.1193,-1.57774,6.41015},{15.1527,-1.05183,6.42432},{15.1714,-0.525915,6.43225},{15.1774,-7.29853*10^-16,6.4348},{15.1714,0.525915,6.43225},{15.1527,1.05183,6.42432},{15.1193,1.57774,6.41015},{15.0675,2.10366,6.3882},{14.9917,2.62957,6.35607},{14.8837,3.15549,6.31026},{14.7316,3.6814,6.24579},{14.5189,4.20732,6.15563},{14.2225,4.73323,6.02996},{12.3716,-5.68073,8.48201},{12.7528,-5.11265,8.74341},{13.0014,-4.54458,8.91387},{13.1635,-3.97651,9.02496},{13.2689,-3.40844,9.09726},{13.3374,-2.84036,9.14419},{13.3815,-2.27229,9.17442},{13.4094,-1.70422,9.19355},{13.4263,-1.13615,9.20515},{13.4353,-0.568073,9.21134},{13.4382,-7.88359*10^-16,9.21329},{13.4353,0.568073,9.21134},{13.4263,1.13615,9.20515},{13.4094,1.70422,9.19355},{13.3815,2.27229,9.17442},{13.3374,2.84036,9.14419},{13.2689,3.40844,9.09726},{13.1635,3.97651,9.02496},{13.0014,4.54458,8.91387},{12.7528,5.11265,8.74341},{10.4398,-6.03187,10.7708},{10.7685,-5.42868,11.1099},{10.9656,-4.8255,11.3133},{11.0838,-4.22231,11.4352},{11.1547,-3.61912,11.5083},{11.1971,-3.01594,11.552},{11.2223,-2.41275,11.5781},{11.2372,-1.80956,11.5935},{11.2457,-1.20637,11.6022},{11.25,-0.603187,11.6066},{11.2513,-8.3709*10^-16,11.608},{11.25,0.603187,11.6066},{11.2457,1.20637,11.6022},{11.2372,1.80956,11.5935},{11.2223,2.41275,11.5781},{11.1971,3.01594,11.552},{11.1547,3.61912,11.5083},{11.0838,4.22231,11.4352},{10.9656,4.8255,11.3133},{10.7685,5.42868,11.1099},{8.09191,-6.3005,12.6302},{8.35003,-5.67045,13.033},{8.49481,-5.0404,13.259},{8.57602,-4.41035,13.3858},{8.62155,-3.7803,13.4569},{8.64707,-3.15025,13.4967},{8.66132,-2.5202,13.5189},{8.66922,-1.89015,13.5313},{8.67348,-1.2601,13.5379},{8.67557,-0.63005,13.5412},{8.67619,-8.74371*10^-16,13.5421},{8.67557,0.63005,13.5412},{8.67348,1.2601,13.5379},{8.66922,1.89015,13.5313},{8.66132,2.5202,13.5189},{8.64707,3.15025,13.4967},{8.62155,3.7803,13.4569},{8.57602,4.41035,13.3858},{8.49481,5.0404,13.259},{8.35003,5.67045,13.033},{5.42138,-6.48148,13.986},{5.5955,-5.83333,14.4352},{5.68862,-5.18519,14.6754},{5.73842,-4.53704,14.8039},{5.76505,-3.88889,14.8726},{5.77928,-3.24074,14.9093},{5.78687,-2.59259,14.9289},{5.79089,-1.94444,14.9393},{5.79297,-1.2963,14.9446},{5.79396,-0.648148,14.9472},{5.79425,-8.99486*10^-16,14.9479},{5.79396,0.648148,14.9472},{5.79297,1.2963,14.9446},{5.79089,1.94444,14.9393},{5.78687,2.59259,14.9289},{5.77928,3.24074,14.9093},{5.76505,3.88889,14.8726},{5.73842,4.53704,14.8039},{5.68862,5.18519,14.6754},{5.5955,5.83333,14.4352},{2.53472,-6.57254,14.7843},{2.6163,-5.91529,15.2601},{2.65878,-5.25803,15.5079},{2.6809,-4.60078,15.6369},{2.69241,-3.94352,15.704},{2.69841,-3.28627,15.739},{2.70152,-2.62902,15.7572},{2.70313,-1.97176,15.7666},{2.70395,-1.31451,15.7713},{2.70433,-0.657254,15.7735},{2.70444,-9.12123*10^-16,15.7742},{2.70433,0.657254,15.7735},{2.70395,1.31451,15.7713},{2.70313,1.97176,15.7666},{2.70152,2.62902,15.7572},{2.69841,3.28627,15.739},{2.69241,3.94352,15.704},{2.6809,4.60078,15.6369},{2.65878,5.25803,15.5079},{2.6163,5.91529,15.2601},{-0.452986,-6.5727,14.9932},{-0.467535,-5.91543,15.4747},{-0.475076,-5.25816,15.7243},{-0.478985,-4.60089,15.8537},{-0.481011,-3.94362,15.9207},{-0.48206,-3.28635,15.9555},{-0.482603,-2.62908,15.9734},{-0.482883,-1.97181,15.9827},{-0.483024,-1.31454,15.9873},{-0.483089,-0.65727,15.9895},{-0.483108,-9.12146*10^-16,15.9901},{-0.483089,0.65727,15.9895},{-0.483024,1.31454,15.9873},{-0.482883,1.97181,15.9827},{-0.482603,2.62908,15.9734},{-0.48206,3.28635,15.9555},{-0.481011,3.94362,15.9207},{-0.478985,4.60089,15.8537},{-0.475076,5.25816,15.7243},{-0.467535,5.91543,15.4747},{-3.42264,-6.48177,14.6043},{-3.5319,-5.8336,15.0705},{-3.58953,-5.18542,15.3164},{-3.61993,-4.53724,15.4462},{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</code></pre>
| Michael E2 | 4,999 | <p>Using <code>Partition</code>:</p>
<pre><code>Graphics3D[
Polygon[
Flatten[
Partition[Partition[dat, 20], {2, 2}, 1, 1][[All, 1 ;; -2]],
{{1, 2}, {3, 4}}][[All, {1, 2, 4, 3}]]
]
]
</code></pre>
<p><a href="https://i.stack.imgur.com/UIgpv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UIgpv.png" alt="enter image description here" /></a></p>
<p>Or as a <code>GraphicsComplex</code>:</p>
<pre><code>Graphics3D[
GraphicsComplex[
dat,
Polygon[
Flatten[
Partition[Partition[Range@Length@dat, 20], {2, 2}, 1, 1][[All, 1 ;; -2]],
{{1, 2}, {3, 4}}][[All, {1, 2, 4, 3}]]
]
]
]
</code></pre>
|
138,520 | <p>I am attempting to show that the series $y(x)\sum_{n=0}^{\infty} a_{n}x^n$ is a solution to the differential equation $(1-x)^2y''-2y=0$ provided that $(n+2)a_{n+2}-2na_{n+1}+(n-2)a_n=0$</p>
<p>So i have:
$$y=\sum_{n=0}^{\infty} a_{n}x^n$$
$$y'=\sum_{n=0}^{\infty}na_{n}x^{n-1}$$
$$y''=\sum_{n=0}^{\infty}a_{n}n(n-1)x^{n-2}$$</p>
<p>then substituting these into the differential equation I get:</p>
<p>$$(1-2x+x^2)\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-2}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$</p>
<p>$$\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-2}-2\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n-1}+\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$</p>
<p>relabeling the indexes: </p>
<p>$$\sum_{n=-2}^{\infty}(n+2)(n+1)a_{n+2}x^{n}-2\sum_{n=-1}^{\infty}n(n+1)a_{n+1}x^{n}+\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$</p>
<p>and then cancelling the $n=-2$ and $n=-1$ terms:</p>
<p>$$\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}-2\sum_{n=0}^{\infty}n(n+1)a_{n+1}x^{n}+\sum_{n=0}^{\infty}n(n-1)a_{n}x^{n}-2\sum_{n=0}^{\infty} a_{n}x^n=0$$</p>
<p>but this doesn't give me what I want (I don't think) as I have $n^2$ terms as I would need</p>
<p>$(n^2+3n+2)a_{n+2}-(2n^2+n)a_{n+1}+(n^2-n-2)a_{n}=0$</p>
<p>I'm not sure where I have gone wrong?</p>
<p>Thanks very much for any help</p>
| agt | 6,752 | <p>You are correct.<br>
Only you need to go on and observe that the lhs of your last equation factorizes as: $$(n+1)[(n+2)a_{n+2}-2n a_{n+1}+(n-2)a_n]$$</p>
|
3,715,715 | <p>Let’s say I have a set <span class="math-container">$X$</span> and a set <span class="math-container">$Y$</span>, and <span class="math-container">$X \subseteq Y$</span>. Is it possible to state that <span class="math-container">$|X| \leq |Y|$</span> (<span class="math-container">$|X|$</span> cardinality of <span class="math-container">$X$</span>)? How can I demonstrate that?</p>
| Brian M. Scott | 12,042 | <p>Yes, it is true. The function <span class="math-container">$\iota:X\to Y:x\mapsto x$</span> is an injection from <span class="math-container">$X$</span> to <span class="math-container">$Y$</span>, and the existence of such an injection by definition means that <span class="math-container">$|X|\le|Y|$</span>.</p>
<p>If <span class="math-container">$Y$</span> is a finite set, you can go further: if <span class="math-container">$X\subsetneqq Y$</span>, then <span class="math-container">$|X|<|Y|$</span>. If <span class="math-container">$Y$</span> is infinite, however, it has proper subsets of the same cardinality as itself. E.g., the set of even integers has the same cardinality as the set of integers.</p>
|
1,846,168 | <p>Let <span class="math-container">$R$</span> be a commutative ring, <span class="math-container">$a \in R$</span>, and <span class="math-container">$\forall i = 1, ...,r \ \ f_i(x) \in R[x]$</span>.</p>
<p>Prove the equality of ideals</p>
<p><span class="math-container">$(f_1(x), ..., f_r(x), x-a ) = (f_1(a), ...f_r(a), x-a)$</span>.</p>
<hr />
<p>We want to show that <span class="math-container">$$\forall g_1(x), ..., g_{r+1}(x) \in R[x] \ \ \exists y_1(x), ..., y_{r+1}(x) \in R[x]: \ \ g_1(x)f_1(x) + ... + g_r(x)f_r(x) + g_{r+1}(x)(x-a) = y_1(x)f_1(a) + ... + y_r(x)f_r(a) + y_{r+1}(x)(x-a)$$</span>
and vice versa.</p>
| m.idaya | 337,282 | <p>as the coefficient dominant of $x-a$ is 1 ivertible in R the euclidien division is ok for any polynomial by $x-a$. so $f_i(x)=g(x)+b $ for some constante $b$ the equality with x=a come $f(a)=b$ and then as reauired the ideal $(f_1(x),...,f_r(x),(x-a))$ is the ideal $(f_1(a),...,f_r(x),(x-a))$</p>
|
750,417 | <p><strong>Question:</strong></p>
<blockquote>
<p>Initially we have a list of numbers $1,2,3,\cdots,2013$.an operation is defined that taking two numbers $a, b$ out from the list, but add $a+b$ into it instead, what is the minimum number of operations required that the sums of any number of numbers in the list can never be $2014$</p>
</blockquote>
<p><strong>This is Math competition in jiangxi province in 2014 the first problem</strong></p>
<p>my idea: if we change this digital: such this:</p>
<p>Initially we have a list of numbers $1,2,3,4,5$.an operation is defined that taking two numbers $a,b$ out from the list, but add $a+b$ into it instead, what is the minimum number of operations required that the sums of any number of numbers in the list can never be $6$</p>
<p>if we an operation is defined that taking two numbers $1,2$then add $1+2=3$,so must operation is defined that taking two numbers $3,3$ out from the list.and $3+3$.so we at lest operation two.</p>
<p><img src="https://i.stack.imgur.com/kQQBs.jpg" alt="enter image description here">
It is said this reslut is $1007$
Thank you for you help </p>
| user139388 | 139,388 | <p><strong>EDIT:</strong></p>
<p>I originally had the interpretation that after taking two pieces of paper with numbers written on them, we add the two together and put a piece of paper with that new number on it into the bag. This allows for duplicates (e.g. two copies of "$3$"). The following solution is for the situation where duplicates are not allowed:</p>
<p>Leave $a=1$ and $b=2013$ unoperated on and perform the following operations:</p>
<p>Step $1$ -> take $2012$ and $2011$ and add them together to get $s_1 = 4023$
Step $n$ -> take $s_{n-1}$ and $2012-n$, add them to get $s_n = s_{n-1}+2012-n$</p>
<p>This can be done for $n=1,\ldots,2011$, at which point the remaining elements of the set are $1,2013$, and $t_{2011} = \sum_{i=2}^{2012} i$. It is necessary to perform one more operation at this point, so the number is at least $2012$.</p>
<p>On the other hand suppose that $2012$ operations have been performed. At this point at most $2$ numbers can remain in the set (every operation reduces the number of elements by at least $1$), but it isn't possible that both are less than $2014$, so the number is exactly $2012$.</p>
|
679,904 | <p>The question is let $a \in \mathbb{R} $ does not contain 0. Prove that $|a+\frac{1}{a}| \ge 2$. I have no idea how to start this problem and any help on it would be greatly appreciated.</p>
| Community | -1 | <p>We have</p>
<p>$$\left|a+\frac 1 a\right|\ge2\iff a^2+1\ge2|a|\iff (|a|-1)^2\ge0\;\text{which's true}$$</p>
|
1,158,642 | <p>Let $E$ be a vector bundle of rank $r$ and let $\phi:E\rightarrow \mathbb C_p$ non vanishing map to the skyscraper sheaf.
consider the kernel $F$ of this sheaf which is a sub-bundle of $E$, every fiber of $F$ has a rank $r$, just that over $p$ which has rank $r-1$.
So why we say that $F$ has a rank $r$??
thanks </p>
| Roland | 113,969 | <p>If you know that $F$ is a vector bundle, its rank may only be the rank of $F_{|U}$ where $U$ is the open set $U=X\setminus p$. Hence it has rank $r$, even at $p$ !! Here is an example of this situation.</p>
<p>Consider $X=\mathbb{P}^1$ and let $p$ be any point. Let $E=X\times\mathbb{C}$ be the trivial vector bundle so that its sheaf of section is $\mathcal{O}_X$ and $\mathcal{O}_X\rightarrow\mathbb{C}_p$ is the evaluation map $f\mapsto f(p)$. It is a surjective map of sheaves and its kernel $I_p$ is isomorphic to $\mathcal{O}_X(-1)$, and in particular is locally free of rank 1.</p>
<p>While this is true that the sequence of stalks (not fibers)
$$ 0 \rightarrow (I_p)_p\rightarrow \mathcal{O}_{X,p}\rightarrow \mathbb{C}\rightarrow 0$$
is exact, it does not mean that the sequence of fibers is. That is the fiber of $I_p$ at $p$ is not the kernel of the map between the fibers. Hence, the fiber of $I_p$ is not of dimension $r-1$ ($r=1$ in our example).</p>
<p>Recall that if $f:E\rightarrow X$ is a vector bundle over $X$ with sheaf of section $\mathcal{E}$, the fiber at $p$ is $f^{-1}(p)$ and this is not the same thing as the stalk $\mathcal{E}_p$. Algebraically, you get the fiber by considering the tensor product $\mathcal{E}\otimes_{\mathcal{O}_X}\mathbb{C}_p$. The algebraic reason why fibers do not commute with kernel is that $\mathbb{C}_p$ is not a flat $\mathcal{O}_X$-module.</p>
|
792,356 | <p>There is a dark night and there is a very old bridge above a canyon. The bridge is very weak and only 2 men can stand on it at the same time. Also they need an oil lamp to see holes in the bridge to avoid falling into the canyon.</p>
<p>Six man try to go through that bridge. They need 1,3,4,6,8,9(first man, second man etc.) minutes to pass the bridge.</p>
<p>What is the fastest way for those six men to pass this bridge?</p>
| Soumyadipto | 149,724 | <p>The correct answer is 31 minutes. This is a very common puzzle question which has been included in various books on brain-teasers, and a few mini games are also based on this puzzle.</p>
<p>The basic trick here is that 1, being the fastest, should ideally cross the bridge with each of the others, so that bringing back the lamp will take the least time. However, the important <strong>catch</strong> here is that the two slowest people(8 and 9) should always cross together, with one of the faster people waiting on the other side to bring the lamp back. This is due to the fact that if they go separately, it will take 8+9 = 17 minutes to cross, plus additional time to bring the lamp back. If they go together, however, it will take only 9 minutes, and even if the next slowest person(6) brings the lamp back, it will only take 9+6 = 15 minutes.</p>
<p>The algorithm for the solution is as follows (One can make minor variations to this without changing the outcome) :
1> 1 and 3 cross, 3 brings back the lamp : 3+3 = 6 minutes.
2> 8 and 9 cross, 1 brings back the lamp : 9+1 = 10 minutes.
3> 1 and 3 cross, 1 brings back the lamp : 3+1 = 4 minutes.
4> 1 and 4 cross, 1 brings back the lamp : 4+1 = 5 minutes.
5> 1 and 6 cross - everyone has reachedthe other side : 6 minutes
Total time = 6 + 10 + 4 + 5 + 6 = <strong>31 minutes</strong></p>
<p>So the basic trick is to make the two slowest people cross only once, together. Cheers!</p>
|
2,497,216 | <p>I have this exercise, but I feel like something is wrong. As far as I know, if $m$ is a maximal ideal, then $m \subsetneq A$. But with this hypothesis, I think to take $I = \lbrace 1_A \rbrace$, so the only ideal that contains $I$ is $A$, so $I$ is not maximal. </p>
<p>EDIT:</p>
<p>I wrote something wrong. I didn't meant that I was the ideal, but that if $m$ is an ideal such that $I \subseteq m$, then $m = A$, so m is not maximal</p>
| Andres Mejia | 297,998 | <p>The ideal $(1)$ generates the whole ring, so we usually do not consider it an ideal. A different approach is via Zorn's lemma. If you take the collection of ideals that contain $I$, and any chain in it, we just have to show that the chain has an upper bound (as in an ideal that contains $I$.)</p>
<p>This amounts to checking that</p>
<p>$$M:=\bigcup_{i \in \mathcal{I}} I_i$$
is a proper ideal since it clearly contains $I$. It is proper since $1 \notin I_i$ for any $i \in \mathcal{I}$.</p>
<p>To see that it is an ideal ,note that it is closed under multiplication by any $r \in A$, since for any $a\in M$, we have that $a \in I_i$ for some $i$, so $ra \in I_i$. This method also shows the existence of additive inverses by taking $r=-1$.</p>
<p>Closure under addition is the part that uses the ordering. WLOG, suppose that $I_i \subset I_j$. Then $i+j \subset I_j$, since $i,j \in I_j$.</p>
<p>Hence, $M$ is an ideal that bounds the chain, so Zorn's lemma applies. </p>
|
582,478 | <p>Please simplify this logic expression for me with helping boolean algebra :</p>
<p>A'C'D + A'BD + BCD + ABC + ACD'</p>
<p>I know that must use consensus theorem .</p>
<p>my solve :</p>
<p>STEP 1 : Terms 1 & 3 ---eliminate---> Term 2</p>
<p>STEP 2 : Terms 3 & 5 ---eliminate---> Term 4</p>
<p>STEP 3 : Terms 2 & 4 ---eliminate---> Term 3</p>
<p>But truth table said step 3 is incorrect . but why ?</p>
<p>please tell me why step 3 is Wrong ?</p>
<p>and tell me What is the simplest form of it ?</p>
| sai kiran grandhi | 93,701 | <p>I think you are confusing during the application of minimization by using Boolean algebra formula</p>
<p>First you apply consensus theorem for terms 1,2,3 by taking D as a common factor. This helps in removing the term 2. So the minimized expression is A'C'D + BCD + ABC + ACD'</p>
<p>Now combining the last three terms similarly leads to elimination of the term ABC</p>
<p>So the minimized expression is A'C'D + BCD + ACD'. </p>
<p>Here you don't have any terms for which consensus theorem can be applied i.e., so you question describes there were no more 2 and 4 terms to eliminate the term 3.</p>
<p>So the minimized expression is A'C'D + BCD + ACD'</p>
|
221,351 | <p>I asked the following question (<a href="https://math.stackexchange.com/questions/1487961/reference-for-every-finite-subgroup-of-operatornamegl-n-mathbbq-is-con">https://math.stackexchange.com/questions/1487961/reference-for-every-finite-subgroup-of-operatornamegl-n-mathbbq-is-con</a>) on math.stackexchange.com and received no answers, so I thought I would ask it here. I've asked several people in my department who were all stumped by the question.</p>
<p>The question is: why is every finite subgroup of $\operatorname{GL}_n(\mathbb{Q})$ conjugate to a finite subgroup of $\operatorname{GL}_n(\mathbb{Z})$?</p>
<p>Note that at least for $n=2$ the question of isomorphism is much easier, since one can (with some effort) work out exactly which finite groups can be subgroups of $\operatorname{GL}_2(\mathbb{Q})$. Further, there are isomorphic finite subgroups of $\operatorname{GL}_2(\mathbb{Q})$ that are not conjugate to each other. For example, the group generated by $-I_{2 \times 2}$ and $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ are both isomorphic to $C_2$, but they cannot be conjugate to each other because the eigenvalues of the two generators are different.</p>
<p>If there is a relatively simple proof, that would be ideal, but a reference with a potentially long proof is fine as well.</p>
<p>Thanks for any assistance.</p>
| John Binder | 30,726 | <p>I think this should work: let $S$ be the set of primes occurring in the denominators of the entries of the elements of $G$, and let $\mathbb{Z}[S^{-1}]$ be the subring of $\mathbb{Q}$ generated by adjoining $1/p$ for $p\in S$.</p>
<p>Let $\mathbb{Q}_S = \prod_{p\in S} \mathbb{Q}_p$ and let $\mathbb{Z}_S = \prod_{p\in S} \mathbb{Z}_p$. We'll show that we can conjugate $G$ into $GL_n(\mathbb{Z}_S)$ by an element of $x$ $GL_n(\mathbb{Z}[S^{-1}])$. This is enough, since it shows then the $p$-adic valuation of all entries of elements of $x^{-1}Gx$ is nonnegative, and of course we still have $x^{-1}Gx\in GL_n(\mathbb{Q})$.</p>
<p>To this end, because $G$ is finite, it compact, so given a $p$ there is some $x_p\in GL_n(\mathbb{Q}_p)$ such that $x_p^{-1}Gx_p \leq GL_n(\mathbb{Z}_p)$. In fact, the set of such $x_p$ is open in $GL_n(\mathbb{Q}_p)$ since $G$ is finite and the map $x_p \mapsto x_p^{-1}gx_p$ is continuous. Because $GL_n(\mathbb{Z}[S^{-1}])$ is dense in $GL_n(\mathbb{Q}_S)$, there is in fact an $x\in GL_n(\mathbb{Z}[S^{-1}])$ that conjugates $G$ into $GL_n(\mathbb{Z}_S)$, completing the proof.</p>
|
2,416,424 | <p>It is known that the collection of finite mixtures of Gaussian Distributions over $\mathbb{R}$ is dense in $\mathcal{P}(\mathbb{R})$ (the space of probability distributions) under convergence in distribution metric.</p>
<p>I'm interested to know the following:</p>
<p>Let $P_X$ be a random variable with finite $p$ th moment i.e. $\mathbb{E}_{P_X}[|X|^p]<\infty$, and $P_{X_n}\stackrel{d}{\to} P_X$ where $P_{X_n}$ are mixtures of Gaussian distributions. Then, suppose $X_n \sim P_{X_n}$ and $X\sim P_X$, does it follow that
$$\mathbb{E}[|X_n|^p] \to \mathbb{E}[|X|^p]$$? $\quad (*)$</p>
<p><strong>My attempt:</strong></p>
<p>I've been able to show that
$$\liminf_{n \to \infty} \mathbb{E}[|X_n|^p] \geq \mathbb{E}[|X|^p]$$
In fact this didn't even use the mixtures part. However I'm having difficulty showing the upper bound. Here are some steps:
\begin{eqnarray}
\mathbb{E}[|X_n|^p] &=& \mathbb{E}[|X_n|^p1_{\{|X_n|\leq A\}}] + \mathbb{E}[|X_n|^p1_{\{|X_n|>A\}}] \\
&=& \mathbb{E}[|X_n|^p1_{\{|X_n|\leq A\}}] + A^pPr[|X_n|\geq A] + \mathbb{E}[(|X_n|^p-A^p)1_{\{|X_n|>A\}}]
\end{eqnarray}
Let $f(x)=x1_{0\leq x\le A} + A1_{x\geq A}$ which is a continuous and bounded function. Then the first two terms of RHS are equal to $\mathbb{E}[f(|X_n|^p)]$, which by definition of weak convergence, will converge to $\mathbb{E}[f(|X|^p)]$ as $n \to \infty$. Now we may use MCT as $A\to \infty$ to get $\mathbb{E}[|X|^p]$. Hence in a nutshell, we need to show:
\begin{equation}
\limsup_{A\to\infty} \limsup_{n \to \infty} \mathbb{E}[(|X_n|^p-A^p)1_{\{|X_n|>A\}}] \leq 0
\end{equation}
However I do not know how to proceed from here.</p>
<p>I'd like to point out that I don't know the answer to the question I asked in $(*)$ but my guess is that this is true. The reason is because I read a similar result in Soren Asmussen's book. The result states that for distributions over non-negative reals, we have phase type distributions not only being weak dense but also the moments converging. But in the provided proof, they said "It can be easily shown that the moments converge".</p>
<p><strong>Update:</strong>
Sincerest apologies for I forgot one very essential condition. The mixtures are not any mixtures but specific ones. Namely, I'm looking at distributions $P_X$ such that $P_{X_n}$ has the form $\sum_{k=1}^n \alpha_k \mathcal{N}(\mu_k;P_k)$ ($\mathcal{N}(\mu_k;P_k)$ stands for Gaussian distribution with mean $\mu_k$ and variance $P_k$.) where given $P<\infty$, $\sum_{k=1}^n \alpha_k =1$, $\alpha_k >0$ and $\sum_{k=1}^n\alpha_k P_k = P$. </p>
<p><strong>Update 2:</strong> The reference I mentioned earlier is Soren Asmussen "Applied probability and queues", 2nd edition, page 84.</p>
<p><strong>Update 3:</strong> Looks like I misinterpreted Asmussen. What he wanted to say was that for any distribution $P_X$ with finite $p$th moment, <strong>there exists</strong> a sequence of phase type distributions $P_X^k$ such that $P_X^k\stackrel{d}{\to}P$ and $E_{P_X^K}[|X|^p] \to E_{P_X}[|X|^p]$. This doesn't mean any weak converging mixture will have moment convergence as the answer points out.</p>
| Dr. Sonnhard Graubner | 175,066 | <p>HINT: write $$\frac{2^x}{2^x-8}-2=\frac{2^x-2^{x+1}+8}{2^x-8}>0$$ and do case work
solving this we get $$3<x<4$$</p>
|
3,449,589 | <p>In Example 1.4 of <em>Lee's Introduction to Smooth Manifolds</em>, which is showing that the <span class="math-container">$n$</span>-sphere, <span class="math-container">$\mathbb{S}^n$</span> is a topological <span class="math-container">$n$</span>-manifold, the following is stated.</p>
<p>In the part where the author shows that <span class="math-container">$\mathbb{S}^n$</span> is locally Euclidean, he does the following. For <span class="math-container">$1\leq i \leq n+1$</span> he let's <span class="math-container">$$U_i^+ = \{(x^1, \cdots, x^{n+1}) \in \mathbb{R}^{n+1} \ | \ x^i > 0 \}$$</span> and <span class="math-container">$$U_i^{-} = \{(x^1, \cdots, x^{n+1}) \in \mathbb{R}^{n+1} \ | \ x^i < 0 \}.$$</span> Then he defines <span class="math-container">$f : \mathbb{B}^n \to \mathbb{R}$</span> by <span class="math-container">$$f(u) = \sqrt{1-|u|^2}$$</span> and claims that <span class="math-container">$U_i^+ \cap \mathbb{S}^n$</span> is the graph of the function <span class="math-container">$$x^i = f(x^1, \dots, \widehat{x^i}, \dots, x^{n+1})$$</span> and <span class="math-container">$U_i^- \cap \mathbb{S}^n$</span> is the graph of the function <span class="math-container">$$x^i = -f(x^1, \dots, \widehat{x^i}, \dots, x^{n+1})$$</span> and where the hat indicates omission of the <span class="math-container">$x^i$</span>.</p>
<p>Now here is where my confusion comes about. Let me call the first function that <span class="math-container">$U_i^+ \cap \mathbb{S}^n$</span> is supposedly the graph of to be <span class="math-container">$g_i$</span> and let me label <span class="math-container">$h_i$</span> as the function which <span class="math-container">$U_i^- \cap \mathbb{S}^n$</span> is the graph of. So now <span class="math-container">$$g_i = f(x^1, \dots, \widehat{x^i}, \dots, x^{n+1})$$</span> and <span class="math-container">$$h_i = -f(x^1, \dots, \widehat{x^i}, \dots, x^{n+1}).$$</span> The problem that I'm having is that if I want to write down the domain and codomain of <span class="math-container">$g_i$</span> and <span class="math-container">$h_i$</span> set-theoretically, they are both going to be maps from <span class="math-container">$\mathbb{B}^{n+1} \to \mathbb{R}$</span>, because surely one would need both <span class="math-container">$g_i$</span> and <span class="math-container">$h_i$</span> to take inputs of the form <span class="math-container">$(x^1, \dots, x^i, \dots, x^{n+1})$</span> to even begin talking about removing the <span class="math-container">$x^i$</span>, so we must have <span class="math-container">$g_i, h_i : \mathbb{B}^{n+1} \to \mathbb{R}$</span>, but the graph of these functions would then be a subset of <span class="math-container">$\mathbb{R}^{n+2}$</span> whereas <span class="math-container">$U_i^- \cap \mathbb{S}^n$</span> is a subset of <span class="math-container">$\mathbb{R}^{n+1}$</span>.</p>
<p>So my question is how can we rigorously (in terms of set theory) define these functions <span class="math-container">$g_i$</span> and <span class="math-container">$h_i$</span> so that <span class="math-container">$U_i^+ \cap \mathbb{S}^n$</span> is the graph of <span class="math-container">$g_i$</span> and <span class="math-container">$U_i^- \cap \mathbb{S}^n$</span> is the graph of <span class="math-container">$h_i$</span>?</p>
| Berci | 41,488 | <p>No, <span class="math-container">$g_i$</span> and <span class="math-container">$h_i$</span> <em>ignore</em> the <span class="math-container">$i$</span>th coordinate in the input, and is defined on <span class="math-container">$\Bbb B^n$</span>, just as <span class="math-container">$f$</span>. <br>
The two possible values for <span class="math-container">$x_i$</span> in the presence of the other coordinates will be obtained by <span class="math-container">$g_i$</span> and <span class="math-container">$h_i$</span>. </p>
|
1,789,373 | <p>I'm trying to figure out why the following is true:</p>
<p>Let $ \kappa $ be an uncountable, regular cardinal. Suppose we turn it into a group (i.e. there are operations $ (\cdot, ^{-1}, e) $ with which $ \kappa $ is a group. My aim is to prove that the set</p>
<p>$$ \{ \alpha \in \kappa : \alpha \text{ is a subgroup of } \kappa\}$$</p>
<p>is a club. It's rather obvious that it is closed, however I'm not quite sure why it would be unbounded. Suppose we have an $ \alpha \in \kappa $ which is a subgroup. Somehow we need to enlarge $ \alpha $ so that it will still be both a group and an ordinal.</p>
<p>We can close $ \alpha + 1 $ under the group operations, but we have no warranty that it will be an ordinal (it wont be equal to $ \kappa $ since $ \kappa $ is regular).</p>
<p>I don't see any way to approach this. I would appreciate a hints</p>
| Jack D'Aurizio | 44,121 | <p>By the Cauchy-Schwarz inequality,</p>
<p>$$ (8-e)^2 = (a+b+c+d)^2 \leq 4(a^2+b^2+c^2+d^2) = 4(16-e^2) $$
from which it follows that $e\leq \color{red}{\frac{16}{5}}$. Now it is enough to show that the inequality holds as an equality for some $(a,b,c,d,e)\in\mathbb{R}^5$, pretty easy. In the same way you may also show that $e\geq 0$.</p>
|
1,808,258 | <p>I was reading about orthogonal matricies and noticed that the $2 \times 2$ matrix
$$\begin{pmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{pmatrix} $$
is orthogonal for every value of $\theta$ and that every $2\times 2$ orthogonal matrix can be expressed in this form. I then wondered if this can be generalized to any smooth paramtrization of the unit circle. More precisely, let $x(t)$ and $y(t)$ be a smooth parametrization of the unit circle for all $t$ in some interval $I \subseteq \Bbb R$ such that $|\langle x(t),y(t) \rangle| = 1$ for all $t \in I$. Is the matrix
$$A= \begin{pmatrix}x(t)&y(t)\\x'(t) & y'(t) \end{pmatrix} $$
necessarily orthogonal?
At first I thought yes, but I'm having trouble proving it. Letting $v = \langle x(t), y(t) \rangle$, it suffices to show three things: 1) $|v| = 1$, 2) $v \cdot v' = 0$, and 3) $|v'| = 1$. (1) follows straight from how $x(t)$ and $y(t)$ were defined. (2) can be obtained by differentiating the equation $x(t)^2 + y(t)^2 = 1$:
\begin{align*}
&\frac{d}{dt} \Big[ x(t)^2 + y(t)^2 \Big]= 0 \\
&\implies 2x(t)x'(t) + 2y(t)y'(t)= 0 \\
&\implies v \cdot v' = 0.
\end{align*}
But I couldn't find a way to prove (3). Now I am unsure whether (3) is true at all. Is $A$ even orthogonal in the first place? Any hints would be much appreciated.</p>
| Andrew D. Hwang | 86,418 | <p>First, not every orthogonal matrix is of the stated form, only <em>rotation</em> matrices. There are orthogonal <em>reflection</em> matrices, with determinant $-1$:
$$
\left[
\begin{array}{@{}rr@{}}
\cos\theta & \sin\theta \\
\sin\theta & -\cos\theta
\end{array}\right].
$$</p>
<p>Second, your conjecture is not true: Only <em>unit speed</em> parametrizations have the stated property, because the rows of an orthogonal matrix form an orthonormal set (in the obvious sense), and particularly must be unit vectors.</p>
|
1,808,258 | <p>I was reading about orthogonal matricies and noticed that the $2 \times 2$ matrix
$$\begin{pmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{pmatrix} $$
is orthogonal for every value of $\theta$ and that every $2\times 2$ orthogonal matrix can be expressed in this form. I then wondered if this can be generalized to any smooth paramtrization of the unit circle. More precisely, let $x(t)$ and $y(t)$ be a smooth parametrization of the unit circle for all $t$ in some interval $I \subseteq \Bbb R$ such that $|\langle x(t),y(t) \rangle| = 1$ for all $t \in I$. Is the matrix
$$A= \begin{pmatrix}x(t)&y(t)\\x'(t) & y'(t) \end{pmatrix} $$
necessarily orthogonal?
At first I thought yes, but I'm having trouble proving it. Letting $v = \langle x(t), y(t) \rangle$, it suffices to show three things: 1) $|v| = 1$, 2) $v \cdot v' = 0$, and 3) $|v'| = 1$. (1) follows straight from how $x(t)$ and $y(t)$ were defined. (2) can be obtained by differentiating the equation $x(t)^2 + y(t)^2 = 1$:
\begin{align*}
&\frac{d}{dt} \Big[ x(t)^2 + y(t)^2 \Big]= 0 \\
&\implies 2x(t)x'(t) + 2y(t)y'(t)= 0 \\
&\implies v \cdot v' = 0.
\end{align*}
But I couldn't find a way to prove (3). Now I am unsure whether (3) is true at all. Is $A$ even orthogonal in the first place? Any hints would be much appreciated.</p>
| hmakholm left over Monica | 14,366 | <p>No, that is not the case. Ad you show, the rows in your $A$ will necessarily be orthogonal to each other, and the first row has norm $1$ -- but the norm of the <em>second</em> row is not necessarily $1$.</p>
<p>If you scale each of $x'(t)$ and $y'(t)$ in the second row by $\dfrac{1}{\sqrt{x'(t)^2+y'(t)^2}}$, you do get all of $SO(2)$, assuming that your $x'(t)$ and $y'(t)$ never vanish at the same time.</p>
|
4,506,026 | <p>Consider the set of equations:
<span class="math-container">$$
\begin{cases}
x^2 &= -4y-10\\y^2 &= 6z-6\\z^2 &= 2x+2\\
\end{cases}$$</span></p>
<p>With <span class="math-container">$x,y,z$</span> being real numbers.</p>
<p>By adding the three equations, after simple manipulations, we easily obtain
<span class="math-container">$$
(x-1)^2+(y+2)^2+(z-3)^2=0
$$</span>
Which yields
<span class="math-container">$$
\begin{cases}
x&=1\\y&=-2\\z&=3\\
\end{cases}
$$</span>
However, plugging that in we find that the solution is invalid. How can one reason about this? In other words, <em>why</em> is it invalid? What causes it to be invalid?</p>
<p><a href="https://i.stack.imgur.com/EQMj6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EQMj6.jpg" alt="enter image description here" /></a></p>
| fleablood | 280,126 | <p>A solution to three equations will be a solution to the sum of the equations. But the solution to the sum need not be a solution to all three.</p>
<p>Consider <span class="math-container">$x^2 = 6z-6; y^2 = 2x + 2; z^2 = -4y-10$</span>. Those are <em>different</em> equations and will have different solutions but if you add them you get the same sums.</p>
<p>Even worse we could have <span class="math-container">$x^2 = -4y+10; y^2 =6z-10; z^2 = 2x - 14$</span> and the sum will be the same but <em>obvious</em> can't have the same solution.</p>
<p>You are basically losing information when you sum them as you are no longer requiring that <span class="math-container">$x^2$</span> <em>must</em> equal <span class="math-container">$-4y -10$</span> (etc) but can be some other combination.</p>
<p>====</p>
<p>An even simpler example could be <span class="math-container">$2x + 3y + z = 17; x-y+2z =7; x+z = 6$</span>. Three equations, three unknowns. One solution <span class="math-container">$x=2;y=3;z=4$</span></p>
<p>But if we add them: <span class="math-container">$4x + 2y +4z = 30$</span>. One equation, three unknowns. Infinite number of solutions of which <span class="math-container">$x=2;y=3;z=4$</span> is just one of them.</p>
|
1,088,734 | <p>It's possible the integral bellow. What way I must to use for solve it.</p>
<p>$$\int \sin(x)x^2dx$$</p>
| Henry | 130,750 | <p>Using Gauss Lemma (also called Rational Root Theorem), the integer solutions can only be factors of $26$, in this case positive factors. This leaves us with $1$, $2$, $13$ or $26$. Also, since the roots are distinct and by Viete's formula, their product must be $26$, the roots are thus $1$, $2$ and $13$. Thus, again by Viete's formula, $a=-16$ and $b=41$.</p>
|
1,640,217 | <p><em>Use the dot/scalar product to solve the problem</em></p>
<p>Line 1 has vector equation $(2\mathrm{i}-\mathrm{j}) + \lambda(3\mathrm{i} + 2\mathrm{j})$
Find the vector equation of the line perpendicular to Line 1 and passing through the point with position vector $(4\mathrm{i} + 3\mathrm{j})$.</p>
<p>I can solve this problem by converting Line 1 into cartesian equation, but I dont know how to use the dot/scalar product to solve it.</p>
| Frentos | 298,557 | <p>Keeping it abstract, you have a line $L_1$ given by $\vec P + \lambda \vec v$, and you want to find a line $L_2$ perpendicular to $L_1$ and passing through an external point $\vec E$.</p>
<p>We want to find the point $\vec C$ on $L_1$ closest to $\vec E$, because then $L_2$ will be the line through $\vec C$ and $\vec E$.
The vector $\vec{CE}$ will be perpendicular to $L_1$, i.e. perpendicular to $\vec v$.</p>
<p>Finding $\lambda_C$ for $\vec C$:
$$\vec{CE}\cdot\vec v= \vec 0\\
(\vec E-(\vec P+\lambda_C \vec v))\cdot \vec v = \vec 0\\
(\vec E-\vec P-\lambda_C \vec v)\cdot \vec v = \vec 0\\
(\vec E-\vec P)\cdot\vec v=\lambda_C \vec v\cdot \vec v\\
\lambda_C = \dfrac{(\vec E-\vec P)\cdot\vec v}{\vec v\cdot \vec v}\\
$$</p>
<p>In your case, $\lambda_C = \dfrac{((4\mathrm i+3\mathrm j)-(2\mathrm i -\mathrm j))\cdot(3\mathrm i+2\mathrm j)}{(3\mathrm i+2\mathrm j)\cdot(3\mathrm i+2\mathrm j)}=\dfrac{(2\mathrm i+4\mathrm j)\cdot(3\mathrm i+2\mathrm j)}{9+4}=\dfrac{6+8}{13}=\dfrac{14}{13}$</p>
<p>Sub $\lambda_C$ into the equation for $L_1$ to get $\vec C$. Then $L_2$ is the line through $\vec C$ and $\vec E$.</p>
|
2,436,419 | <p>So I get the general sense of all of this and calculated all of the partial derivatives but I am unsure what exactly it is asking me for. Any advice or explanation would be greatly appreciated. Thanks!</p>
<p>Let <span class="math-container">$F: R^{3} \to R^{2}$</span> be defined by <span class="math-container">$F(x,y,z) = (y(z+6)^{1/3}, z\sin(3x-y))$</span>. Find <span class="math-container">$DF(1,3,2)$</span>. Using the first part approximate <span class="math-container">$F(.99, 3.01, 2.02)$</span> by <span class="math-container">$F(1,3,2)$</span>. </p>
| David | 136,138 | <p>Calculating the partial derivatives basically tells you the slope of the function in the $x,y,z$ direction. When you calculate $\partial F(1,3,2)$ with respect to $x,y$ and $z$, you get the slope values at the point (1,3,2). What it is asking you is to approximate a point close to the point whose slope you calculated. To do this is</p>
<p>$$ F(0.99, 3.01, 2.02) \approx F(1,3,2) - .01F_x(1,3,2) + .01F_y(1,3,2) + .02F_z(1,3,2) $$</p>
<p>where $F_x, F_y, F_z$ are the partial derivatives of $F$. Another way to write this is</p>
<p>$$ F(a + \alpha, b + \beta, c + \gamma) \approx F(a,b,c) + (\alpha, \beta, \gamma) \cdot\nabla F(a,b,c) $$</p>
<p>where $(\alpha, \beta, \gamma)$ are small perturbations to the point $(a,b,c)$ and $\nabla$ is the gradient.</p>
|
2,436,419 | <p>So I get the general sense of all of this and calculated all of the partial derivatives but I am unsure what exactly it is asking me for. Any advice or explanation would be greatly appreciated. Thanks!</p>
<p>Let <span class="math-container">$F: R^{3} \to R^{2}$</span> be defined by <span class="math-container">$F(x,y,z) = (y(z+6)^{1/3}, z\sin(3x-y))$</span>. Find <span class="math-container">$DF(1,3,2)$</span>. Using the first part approximate <span class="math-container">$F(.99, 3.01, 2.02)$</span> by <span class="math-container">$F(1,3,2)$</span>. </p>
| amd | 265,466 | <p>Go back to the definition of $DF$: it’s the best <em>linear</em> approximation to the change in $F$ at a point. Symbolically, $$F(\mathbf x+\Delta\mathbf x)=F(\mathbf x)+DF(\mathbf x)\Delta\mathbf x+o(\Delta\mathbf x).$$ That is, for a small displacement $\Delta\mathbf x$ from $\mathbf x$, the value of $F(\mathbf x+\Delta\mathbf x)$ is approximately $F(\mathbf x)+DF(\mathbf x)\Delta\mathbf x$. </p>
<p>In your problem, $\mathbf x=(1,3,2)^T$ and $\Delta\mathbf x=(0.99,3.01,2.02)^T-(1,3,2)^T=(-0.01,0.01,0.02)^T$. $DF$ is the Jacobian matrix of $F$: the $2\times 3$ matrix of the partial derivatives at $\mathbf x$, which you’ve already computed. The rest is a matter of plugging in these values.</p>
|
3,029,446 | <p>If </p>
<ul>
<li><p><span class="math-container">$A$</span> is <span class="math-container">$m \times n$</span> (<span class="math-container">$m<n$</span>), and its rows are independent</p></li>
<li><p><span class="math-container">$B$</span> is <span class="math-container">$n \times p$</span> (<span class="math-container">$p<n$</span>), and its columns are independent</p></li>
<li><p>We also know <span class="math-container">$m\ge n$</span>.</p></li>
</ul>
<p>does <span class="math-container">$AB$</span> have linearly independent columns?</p>
<p>Or what additional requirements are needed for <span class="math-container">$AB$</span> to have linearly independent columns?</p>
| Robert Israel | 8,508 | <p>Since <span class="math-container">$A$</span> has rank <span class="math-container">$m$</span> and <span class="math-container">$B$</span> has rank <span class="math-container">$p$</span>, <span class="math-container">$AB$</span> has rank at most <span class="math-container">$\min(m,p)$</span>.
<span class="math-container">$AB$</span> is <span class="math-container">$m\times p$</span>, so it could have linearly independent columns if <span class="math-container">$m \ge p$</span>,
but not if <span class="math-container">$m < p$</span>.</p>
|
23,953 | <p>I cited the diagonal proof of the uncountability of the reals as an example of a <a href="https://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/23708#23708">`common false belief'</a> in mathematics, not because there is anything wrong with the proof but because it is commonly believed to be Cantor's second proof. The stated purpose of <a href="http://resolver.sub.uni-goettingen.de/purl?GDZPPN002113910" rel="nofollow noreferrer">the paper where Cantor published the diagonal argument </a> is to prove the existence of uncountable infinities, avoiding the theory of irrational numbers. I have no problem believing that Cantor himself realized that a diagonal proof of the uncountability of <strong>R</strong> was possible but I have not even found an allusion to this in his collected works. The earliest appearance in print that I know is on page 43 of <em>The theory of sets of points</em> by W. H. Young and Grace Chisholm Young (1906). I would be very grateful for any reference to some scrap of paper where Cantor himself mentions the possibility of using the diagonal method to prove the set of reals uncountable. </p>
| Harald Hanche-Olsen | 802 | <p><strong>Edit:</strong> A closer look reveals that the proof in the reference below is not the <em>diagonal</em> proof. I am leaving the answer up since the reference might be of interest anyhow.</p>
<p>Cantor had a paper in Crelle's Journal <strong>77</strong> (1874) 258–262. In Christopher P. Grant's translation, the title of the paper is <em>On a Property of the Class of all Real Algebraic Numbers</em>. In §2, we can read</p>
<blockquote>
<p>If an infinite sequence of distinct real numerical quantities $$\omega_1, \omega_2,\ldots,\omega_\nu,\ldots\qquad(4)$$
(obtained according to whatever rule) is given, then in each prespecified interval $(\alpha\ldots\beta)$ a number $\eta$ (and consequently infinitely many such numbers) can be specified, which does not occur in the sequence (4); this will now be proven.</p>
</blockquote>
<p>I am not sure where I found the translation. Sloppy of me.</p>
<p><strong>2nd Edit:</strong> here is a very brief outline of Cantor's non-diagonal proof. By induction on $k$, find $\alpha_k$, $\beta_k$ as early as possible from the given sequence with $$\alpha<\alpha_1<\alpha_2<\cdots<\beta_2<\beta_1<\beta$$ and note that any number the the closed interval $[\alpha_\infty,\beta_\infty]$ is not in the given sequence. If the induction fails after step $k$, the interval $(\alpha_k,\beta_k)$ contains at most one point from the given sequence.</p>
|
23,953 | <p>I cited the diagonal proof of the uncountability of the reals as an example of a <a href="https://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/23708#23708">`common false belief'</a> in mathematics, not because there is anything wrong with the proof but because it is commonly believed to be Cantor's second proof. The stated purpose of <a href="http://resolver.sub.uni-goettingen.de/purl?GDZPPN002113910" rel="nofollow noreferrer">the paper where Cantor published the diagonal argument </a> is to prove the existence of uncountable infinities, avoiding the theory of irrational numbers. I have no problem believing that Cantor himself realized that a diagonal proof of the uncountability of <strong>R</strong> was possible but I have not even found an allusion to this in his collected works. The earliest appearance in print that I know is on page 43 of <em>The theory of sets of points</em> by W. H. Young and Grace Chisholm Young (1906). I would be very grateful for any reference to some scrap of paper where Cantor himself mentions the possibility of using the diagonal method to prove the set of reals uncountable. </p>
| Hilbert7 | 112,109 | <p>The evidence that you look for can already be found in the second paragraph of <a href="http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=PPN237853094&DMDID=DMDLOG_0033" rel="nofollow">Cantor's original paper</a>. There he states "Es läßt sich aber von <em>jenem Satze</em> ein viel einfacherer Beweis liefern, der unabhängig von der Betrachtung der Irrationalzahlen ist.", meaning that his diagonal argument supplies a much simpler proof of <em>the theorem proved in his first paper</em> on the uncountability of the real numbers. This does not only mention but declare that his method can be applied to real numbers.</p>
<p>Probably in order to emphasize its independence of numbers, Cantor did not use numerals 0 and 1, but m and w which (and this answers your last remark) in German are abbreviations of male and female. So he had a good substitution for 0 and 1 or up and down or yes and no - and he deliberately or unconsciously circumvented the problem that this proof without reservations, as stated in his original text, would fail on binary sequences.</p>
|
2,473,780 | <p>So I have the limit $$\lim_{x\rightarrow \infty}\left(\frac{1}{2-\frac{3\ln{x}}{\sqrt{x}}}\right)=\frac{1}2,$$ I now want to motivate why $(3\ln{x}/\sqrt{x})\rightarrow0$ as $x\rightarrow\infty.$ I cam up with two possibilites:</p>
<ol>
<li><p>Algebraically it follows that $$\frac{3\ln{x}}{\sqrt{x}}=\frac{3\ln{x}}{\frac{x}{\sqrt{x}}}=\frac{3\sqrt{x}\ln{x}}{x}=3\sqrt{x}\cdot\frac{\ln{x}}{x},$$
Now since the last factor is a standard limit equal to zero as $x$ approaches infinity, the limit of the entire thing should be $0$. However, isn't it a problem because $\sqrt{x}\rightarrow\infty$ as $x\rightarrow \infty$ gives us the indeterminate value $\infty\cdot 0$?</p></li>
<li><p>One can, without having to do the arithmeticabove, directly motivate that the function $f_1:x\rightarrow \sqrt{x}$ increases faster than the function $f_2:x\rightarrow\ln{x}.$ Is this motivation sufficient? And, is the proof below correct?</p></li>
</ol>
<p>We have that $D(f_1)=\frac{1}{2\sqrt{x}}$ and $D(f_2)=\frac{1}{x}$. In order to compare these two derivateives, we have to look at the interval $(0,\infty).$ Since $D(f_1)\geq D(f_2)$ for $x\geq4$, it follows that $f_1>f_2, \ x>4.$</p>
| marty cohen | 13,079 | <p>For any $a > c > 0, x > 1$,
we have</p>
<p>$\begin{array}\\
\ln(x)
&=\int_1^x \frac{dt}{t}\\
&\lt\int_1^x \frac{dt}{t^{1-c}}\\
&=\int_1^x t^{c-1}dt\\
&=\dfrac{t^c}{c}\big|_1^x\\
&=\dfrac{x^c-1}{c}\\
&<\dfrac{x^c}{c}\\
\text{so}\\
\dfrac{\ln(x)}{x^a}
&<\dfrac1{x^a}\dfrac{x^c}{c}\\
&=\dfrac{x^{c-a}}{c}\\
&\to 0
\qquad\text{as } x \to \infty
\text{ since } a > c\\
\end{array}
$</p>
<p>For your case,
where $a = \frac12$,
we can choose
$c = \frac14$
to get
$\dfrac{\ln(x)}{x^{1/2}}
\lt\dfrac{x^{\frac14-\frac12}}{\frac14}
=4x^{-\frac14}
\to 0
$.</p>
<p>In general
we can choose
$c = a/2$
to get
$\dfrac{\ln(x)}{x^{a}}
\lt\dfrac{x^{a/2-a}}{a/2}
=\dfrac{2}{ax^{a/2}}
\to 0
$.</p>
|
362,881 | <p>I am going to try to explain this as easily as possible. I am working on a computer program that takes input from a joystick and controls a servo direction and speed. I have the direction working just fine now I am working on speed. To control the speed of rotation on the servo I need to send it so many pulses per second using PWM. The servo that I am using takes arguments for speed between 120-150. 120 is %100 speed and 150 is %0 or stopped. 135 is %50 speed. How would I convert percentage from 0-100 into a number between 120-150 including 1/10ths? I hope this makes sense if you need me to explain further please let me know. I really don't know what tag this falls under either.</p>
| A Blumenthal | 54,337 | <p>Hint: $\delta(x)$ is a linear functional on some space of suitably nice (in particular, continuous) functions of $\mathbb{R}$. Given a suitably nice function $f$, what does $\delta(x)[f]$ equal? </p>
<p>When you say $e^x \delta(x) = \delta(x)$, what you're saying is that the left hand side and the right hand side act on continuous functions in the same way. Is that true?</p>
|
2,087,235 | <p>I have a question about this question. Find all complex numbers $z$ such that the equation
$$t^2 + [(z+\overline z)-i(z-\overline z)]t + 2z\overline z\ =\ 0$$
has a real solution $t$.</p>
<p><strong>Attempt at a solution</strong></p>
<p>The discriminant is</p>
<p>$[(z+\overline z) - i(z-\overline z)]^2 - 4(2z\overline z)$<br>
$=\ (z+\overline z)^2 - 2i(z+\overline z)(z-\overline z) + [i(z-\overline z)]^2 -8z\overline z$<br>
$=\ (z^2+2z\overline z+\overline z^2) -2i(z^2-\overline z^2) - (z^2-2z\overline z+\overline z^2)-8z\overline z$<br>
$=\ -4z\overline z - 2iz^2 + 2i\overline z^2$</p>
<p>For real solutions, the discriminant must be non-negative. But $z$ is a complex number; how can complex numbers be positive or negative? This is what I don't understand.</p>
<p>Would appreciate any help. Thanks.</p>
| Bernard | 202,857 | <p>Just
set $z=u+iv$. The discriminant becomes
$$\Delta=4[(u+v)^2-2(u^2+v^2)]=-4(u-v)^2.$$
Hence the condition is $\;u=v,\;$ or $\;\arg z\equiv\dfrac\pi4\mod\pi$.</p>
|
148,257 | <p>Let $f:\mathbb R\to\mathbb R$ be continuous. Suppose $(x_n)_n$ and $(y_n)_n$ are sequences in $\mathbb R$ such that the sequence $(x_n-y_n)_n$ converges to $0$. Does this mean that the sequence $(f(x_n)-f(y_n))_n$ converges to $0$?</p>
<p>I feel like it is true, since the definition of continuity states that $f$ preserves limits of convergent sequences, but I do not know how to prove it.</p>
| Davide Giraudo | 9,849 | <ul>
<li>The result is true is $f$ is uniformly continuous on $\Bbb R$. </li>
<li>But if it's not the case, we can find $\{x_n\}$ and $\{y_n\}$ which contradict this fact, using the definition of uniform continuity. </li>
</ul>
|
176,488 | <p><strong>Summary:</strong> My question, in a nutshell, is how we should intuitively imagine a generic real number (as opposed to a random one), and whether we can construct numbers which empirically behave like generic numbers in the same way that $e$ or $\pi$ behave empirically like random ones. I hope this is not too vague, informal or philosophical for MO. Let me explain what I'm asking in greater detail.</p>
<p><strong>Background:</strong> First recall the classical "duality" between (Lebesgue) measure and (topological) category: a subset of $\mathbb{R}$ is said to be "negligible" iff it is of Lebesgue measure zero, and "meager" iff it is contained in a countable union of nowhere dense closed sets (i.e., closed sets with empty interior). By Lebesgue measure theory, resp. by the Baire category theorem, a negligible, resp. meager set has empty interior. Both are "small" in a certain sense, but in an incompatible way since $\mathbb{R}$ is the union of a negligible and a meager set (as a simple example is given below). There is also a classical theorem by Erdős (refining an earlier result by Sierpiński) showing that, under the Continuum Hypothesis, there is an involution of $\mathbb{R}$ to itself which takes negligible subsets to meager ones and conversely.</p>
<p>[What follows is badly written: jump to "edit/clarification" below for an attempt at saying things more clearly.]</p>
<p>There are various meanings of the word "random", but the general flavor is that a real number is random iff it does not belong to a negligible Borel set which can somehow be described or coded in a simple way (e.g., random over a transitive model of set theory means that it does not belong to a negligible Borel set coded by a sequence in that model; but there are some weaker meanings of "random" where we forbid belonging only to negligible Borel sets of simpler description, e.g., those which can be described by a code computable by a Turing machine). For example, a random number will be normal in every base, because the set of real numbers which are not normal in every base is a simply described Borel set which is negligible.</p>
<p>Now a real number like $e$ or $\pi$ is not random even in the weakest sense, because it is, well, equal to $e$ or $\pi$, and that is not random (it belongs to the Borel set $\{e\}$ or $\{\pi\}$ which is negligible and certainly computable). It does seem to be, however, "empirically random", in a sense that we (or at least, I) don't know how to make precise, but the idea being that it won't belong to any simply defined negligible Borel set which hasn't been specifically constructed to contain it. For example, it is conjectured that $e$ and $\pi$ are normal in every base: we expect their decimals to pass statistical tests of randomness. The same holds for a huge number of "naturally defined" real numbers (and not just real numbers: see <a href="https://mathoverflow.net/questions/156301/distribution-of-digits-of-pq-adic-idempotents-aka-automorphic-numbers">this question</a> for another case). Philosophically, it is also generally expected that the real number whose binary expansion is obtained by flipping an unbiased coin (or, better, taking some physical source of randomness) will be random in a strong sense. So, anyway, we have a good intuition of what a random number feels like.</p>
<p>The dual notion of a "generic" number, however, is more obscure: a real number is generic iff it does not belong to a meager Borel set which can be somehow described or coded in a simple way (e.g., belonging to a transitive model of set theory, or Turing-computable, or something like this).</p>
<p>Here is an example of something we can say about generic numbers: call a real number (between $0$ and $1$, say) an "oft-repeater in base $b$" iff its expansion in base $b$ repeats an infinite number of times <em>all</em> the digits up to that point. In other words, there exist arbitrarily large $n$ such that the digits $c_n$ to $c_{2n-1}$ are equal to $c_0$ to $c_{n-1}$. It is easy to see that the set of oft-repeaters in base $b$ (and therefore, in any base) is negligible but comeager (=contained in a countable intersection of open dense subsets): so a random real number is <em>not</em> an oft-repeater in any base, but a generic real number <em>is</em> an oft-repeater in every base.</p>
<p>We don't know this either way, but I don't think anyone would seriously conjecture that $e$ or $\pi$ is an oft-repeater in any base: clearly we expect them to be empirically random and <em>not</em> empirically generic. Or in other words, we expect measure theory to be a better predictor of what $e$ and $\pi$ behave like than category. An "empirically generic" real number, however, would be (among many other things) an oft-repeater in every base (and it would <em>not</em> be normal in any base: for example, in a generic number, there are arbitrarily large $n$ such that all digits $c_n$ to $c_{n^2}$ are zero — the set of normal reals is meager).</p>
<p><strong>Questions:</strong> So, my questions are something like this:</p>
<ul>
<li><p>Are there "naturally defined" real numbers which are "empirically generic" rather than "empirically random"? Or at least, can we give some examples of (non-"naturally defined") such numbers?</p></li>
<li><p>Is there some kind of process (physical or idealized), analog to throwing a coin, that would produce a (somewhat!) generic real number?</p></li>
<li><p>Is there a philosophical argument explaining why measure theory predicts better than category how the numbers naturally encountered in mathematics behave? Why should we expect $e$ and $\pi$ to behave more randomly than generically (when, in fact, they are neither)?</p></li>
<li><p>How can one intuitively visualize a generic random number? (I think I can picture a random one, and the idea of it being normal makes sense, but the fact that a generic random number has infinitely large $n$ such that all digits $c_n$ to $c_{n^2}$ are zero seems very difficult to imagine.)</p></li>
<li><p>How would one even test empirically if a given real number is generic? (Assume you have a true generic oracle and a fake one: how would you proceed to detect which is the true one? Can we have a "genericity test" like we have randomness tests?)</p></li>
</ul>
<p>Perhaps the idea that genericity should behave symmetrically to randomness is naïve: please don't hesitate to tell me why this is naïve!</p>
<hr>
<p><strong>Edit/Clarification:</strong> The above discussion was probably too messy or informal. Let me try to give a clearer restatement:</p>
<p><strong>Definition:</strong> If $\mathscr{T}$ is a set of Turing degrees, a real number is said to be $\mathscr{T}$-random, resp. $\mathscr{T}$-generic, iff it does not belong to any negligible, resp. meager, Borel set which can be coded by a sequence whose Turing degree is in $\mathscr{T}$ ("coding" of Borel sets being done, say, as in Jech's <em>Set Theory</em>).</p>
<p>If $\mathscr{T}$ is the set of degrees belonging to some transitive model $\mathfrak{M}$ of ZFC, the reals in question are said to be random over $\mathfrak{M}$ resp. Cohen/generic over $\mathfrak{M}$ (cf. Jech, definition 26.3 and lemma 26.4 in the Third Millennium edition). If $\mathscr{T}$ is simply the degree $\mathbf{0}$ of Turing computability, I think we get a definition equivalent to Martin-Löf random numbers, and something analogous for "generic": of course, this is a much weaker property than being random, resp. generic, over a model of ZFC. We could perhaps define even weaker versions of "random", resp. "generic", by replacing $\mathscr{T}$ by a set of finer degrees, maybe primitive recursive degrees (but if the degrees are too fine, then the definition will become too sensitive on how Borel sets are coded and probably not the right way to proceed).</p>
<p><strong>Question number 1:</strong> While I think I have an intuitive grasp of how a random number real behaves (irrespective of what $\mathscr{T}$ is), e.g., by imagining an coin being tossed an infinite number of times, the corresponding "generic" notion is much more obscure. Is there some way to picture it intuitively?</p>
<p>Now there is the matter of numbers like $e$ and $\pi$. Of course <strong>these numbers are not random (nor are they generic)</strong> in the sense of the above definition, or even of any reasonable weakening I can imagine.</p>
<p>Nevertheless, $e$ and $\pi$ behave <em>in certain ways</em> like random real numbers, and I claim that they behave "more like random reals than like generic reals". For example, if we are to make a conjecture as to the lim.sup. and lim.inf. of the sequence $\frac{1}{n}\sum_{k=0}^{n-1} c_k$ where $c_k$ denotes the $k$-th binary digit of $\pi$, and if Pr. Eugsebel predicts "I conjecture that the limit is $\frac{1}{2}$, because the set of real numbers for which this is the case is of full measure", while Pr. Eriab predicts "I conjecture that the lim.inf. is $0$ and the lim.sup. is $1$, because the set of real numbers for which this is the case is comeager", then experimentally, it appears that Pr. Eugsebel is right and Pr. Eriab is wrong: measure theory seems to predict the <em>empirical</em> behavior of the decimals of $\pi$ correctly, and category does not; alternatively, $\pi$ behaves <em>empirically</em> like a random number (even though it is not at all random!), in this limited respect, and it <em>does not</em> behave empirically like a generic number.</p>
<p><strong>Informal definition</strong> (which probably cannot be made rigorous): Say that a real number is "empirically random" when it behaves like a random real number for this kind of simple tests. (Perhaps "pseudorandom" would be a better term for this.) For example, an "empirically random" real number should, at least, be normal in any base (note that the set of real numbers that are normal in any base is of full measure). Analogously, we want to define a number to be "empirically generic" (or "pseudogeneric") when it behaves like a generic real number. For example, an "empirically generic" real number should at least be an "oft-repeater" in any base (meaning that there exist arbitrarily large $n$ such that the digits $c_n$ to $c_{2n-1}$ are equal to $c_0$ to $c_{n-1}$); also, the lim.sup. and lim.inf. of the sequence $\frac{1}{n}\sum_{k=0}^{n-1} c_k$ where $c_k$ denotes the $k$-th binary digit of the number should be $0$ and $1$ (note that the set of real numbers satisfying these criteria comeager).</p>
<p><strong>Question number 2:</strong> Whereas $e$, $\pi$ and many others can reasonably be conjectured to be "empirically random", is there, dually, any real number that has been explicitly defined in mathematics that one can reasonably expect to be "empirically generic"? Or could one be defined?</p>
<p>(By "explicitly", I mean to forbid something like "take some number outside of the union of all meager Borel sets with a computable code": this would indeed define a generic real number, not just an "empirically generic" one, but this is not explicit by any means.)</p>
<p>The underlying philosophical question is something like this: "How come is it that randomness appears to be a much more natural notion than genericity?" But I don't really expect anyone to have an answer to that.</p>
| Bjørn Kjos-Hanssen | 4,600 | <p>It sounds like you are talking about what in computability theory and set theory are known as <strong>Cohen generic</strong> reals (the lowest level of which in computability theory is <strong>1-generic</strong>, then 2-generic and so on).</p>
<p>I don't know any really natural example of a 1-generic real, but there is a fairly simple construction of one, see e.g. the book <em>Lerman, Degrees of Unsolvability, 1983</em>.</p>
<p>To visualize you can imagine that we put down more and more digits in our number, and every now and then we stop and say "what kind of digits could we possibly put down?", and then we put down some digits like that. For example, you could put down 100 times as many 0s as the number of digits you've put down so far. The important thing is that you eventually cover all kinds like that. So something like:</p>
<p>02345234</p>
<p>Now let's add 23 zeroes:</p>
<p>0234523400000000000000000000000</p>
<p>Now let's add what we already have two more times:</p>
<p>023452340000000000000000000000002345234000000000000000000000000234523400000000000000000000000</p>
<p>... and so on.
But we could also have done it in a different order, so say we have</p>
<p>78345786345</p>
<p>and then add what we already have two more times:</p>
<p>783457863457834578634578345786345</p>
<p>and then add 23 zeroes</p>
<p>78345786345783457863457834578634500000000000000000000000</p>
|
4,206,205 | <p>I came across a theorem in algebraic number theory:</p>
<blockquote>
<p><strong>Theorem</strong> Let <span class="math-container">$A$</span> be a Dedekind ring and <span class="math-container">$M, N$</span> two modules over <span class="math-container">$A$</span>. If <span class="math-container">$M_\mathfrak{p} \subset N_\mathfrak{p}$</span> for all prime ideals <span class="math-container">$\mathfrak{p} \subset A$</span>, then <span class="math-container">$M \subset N$</span>.</p>
</blockquote>
<p><em>Proof.</em> Let <span class="math-container">$a \in M$</span>. For each <span class="math-container">$\mathfrak{p}$</span> we can find <span class="math-container">$x_\mathfrak{p} \in N$</span> and <span class="math-container">$s_\mathfrak{p} \in A \setminus \mathfrak{p}$</span> such that <span class="math-container">$a/1=x_\mathfrak{p}/s_\mathfrak{p}$</span>. Let <span class="math-container">$\mathfrak{b}$</span> be the prime ideal generated by <em>the</em> <span class="math-container">$s_\mathfrak{p}$</span>. Then <span class="math-container">$\mathfrak{b}$</span> is the unit ideal, and we can write
\[
1=\sum y_\mathfrak{p}s_\mathfrak{p}
\]
with elements <span class="math-container">$y_\mathfrak{p} \in A$</span> all but a finite number of which are <span class="math-container">$0$</span>. This yields
\[
a = \sum y_\mathfrak{p}s_\mathfrak{p}a = \sum y_\mathfrak{p}x_\mathfrak{p} \in N
\]
as desired. <span class="math-container">$\square$</span></p>
<hr />
<p>Here <span class="math-container">$M_\mathfrak{p}$</span> means the localisation of <span class="math-container">$M$</span> at <span class="math-container">$\mathfrak{p}$</span>.</p>
<p>My question is not about the theorem per se, but I guess there should be some counterexamples. Chances are there are modules <span class="math-container">$M,N$</span> such that</p>
<p><span class="math-container">$$
M_\mathfrak{p} \subset N_\mathfrak{p} \text{ does not imply } M \subset N.
$$</span></p>
<p>But I find it pretty hard to construct an example. I try to breakdown this question into pieces:</p>
<ol>
<li>Does this coincide with local property? It's pretty close:</li>
</ol>
<blockquote>
<p>Let <span class="math-container">$\phi:M \to N$</span> be an <span class="math-container">$A$</span>-modue homomorphism (no futher property assumed on <span class="math-container">$A$</span> at this point), then the followings are equivalent:</p>
<ol>
<li><p><span class="math-container">$\phi$</span> is injective.</p>
</li>
<li><p><span class="math-container">$\phi_\mathfrak{p}:M_\mathfrak{p} \to N_\mathfrak{p}$</span> is injective for each prime ideal <span class="math-container">$\mathfrak{p}$</span>.</p>
</li>
<li><p><span class="math-container">$\phi_\mathfrak{m}:M_\mathfrak{m} \to N_\mathfrak{m}$</span> is injective for each maximal ideal <span class="math-container">$\mathfrak{m}$</span>.</p>
</li>
</ol>
</blockquote>
<p>It looks like we are talking about 2 implies 1. But the inclusion map <span class="math-container">$M \to N$</span> is not given in this theorem. Also in the proof of the theorem, Serge Lang used properties of Dedekind ring (fractional ideals form a group) explicitly. If local property is the case, why would he ignore it?</p>
<ol start="2">
<li><p>Is it possible to get a desired counterexample given by <span class="math-container">$M_\mathfrak{p} \not\subset N_\mathfrak{p}$</span> for at least one prime ideal?</p>
</li>
<li><p>Is it possible to get a desired counterexample given by <span class="math-container">$A$</span> not being Dedekind, or even wilder, <span class="math-container">$A$</span> is not Noetherian? Noetherian counterexamples will certainly be more interesting.</p>
</li>
</ol>
<p>Thanks in advance!</p>
<hr />
<p>Edit 1: I added proof of the theorem to avoid unnecessary confusion.</p>
| hm2020 | 858,083 | <p><strong>Question:</strong> "My question is not about the theorem per se, but I guess there should be some counterexamples. Chances are there are modules <span class="math-container">$M,N$</span> such that <span class="math-container">$M_p⊂N_p$</span> does not imply <span class="math-container">$M⊂N$</span>."</p>
<p><strong>Answer:</strong> Let <span class="math-container">$A$</span> be any commutative unital ring and let <span class="math-container">$L,L'$</span> be invertible <span class="math-container">$A$</span>-modules with</p>
<p><span class="math-container">$$L_{\mathfrak{p}}\cong L_{\mathfrak{p}}' \cong A_{\mathfrak{p}}$$</span></p>
<p>for all primes. Hence there is an abstract inclusion <span class="math-container">$L_{\mathfrak{p}}\subseteq L'_{\mathfrak{p}}$</span> for any prime ideal.This does not imply there is a "global" inclusion <span class="math-container">$L \subseteq L'$</span>.</p>
<p><strong>Example:</strong> Let <span class="math-container">$K$</span> be any number field with ring of integers <span class="math-container">$A$</span>, and let <span class="math-container">$I,J$</span> be two invertible <span class="math-container">$A$</span>-modules that are not contained in each other. This gives a counter example: The <span class="math-container">$A$</span>-modules <span class="math-container">$I,J$</span> have isomorphic stalks since they both have rank one.</p>
|
69,050 | <p>The basic concept of Quotient Group is often a confusing thing for me,I mean can any one tell the intuitive concept and the necessity of the Quotient group,
I thought that it would be nice to ask as any basic undergraduate can learn the intuition seeing the question.
My Question is :</p>
<ol>
<li>Why is the name Quotient Group kept,normally in the case of division, let us take the example of $\large \frac{16}{4}$ the Quotient of the Division is '$4$' which means that there are four '$4$'s in $16$, I mean we can find only $4$ elements with value $4$
<blockquote>
<p>So how can we apply the same logic in the case of Quotient Groups,like consider the Group $A$ and normal subgroup $B$ of $A$, So if $A/B$ refers to "Quotient group", then does it mean:</p>
<blockquote>
<p>Are we finding how many copies of $B$ are present in $A$??, like in the case of normal division, or is it something different ??</p>
</blockquote>
</blockquote></li>
</ol>
<p>I understood the Notion of Cosets and Quotient Groups,b ut I want a different Perspective to add Color to the concept. Can anyone tell me the necessity and background for the invention of Quotient Groups?</p>
<p>Note: I tried my level best in formatting and typing with proper protocol,if in case, any errors still persist, I beg everyone to explain the reason of their downvote (if any), so that I can rectify myself, Thank you.</p>
| Jack Schmidt | 583 | <blockquote>
<p>Quotient groups $A/B$ count cosets of $B$ inside $A$. The counting even works well with addition.</p>
</blockquote>
<p>The Cartesian plane forms a group A, and a line through the origin is a subgroup B. The cosets of B inside A are all the parallel lines. How many are there?</p>
<p>Suppose <em>B</em> is the line:
$$ B = \{ (x,y) : y = 2x \}$$
or just <em>B</em> is <em>y</em> = 2<i>x</i> for short. Parallel lines are parameterised by their "intercept" <em>b</em>, the coset <em>b</em> + <em>B</em> is the line with slope 2 and intercept <em>b</em>.
$$b+B = \{(x,y) : y = 2x+b\}$$
This means that there is exactly one coset for every real number. In some sense we have counted the parallel lines.</p>
<p>The way we count them even keeps track of addition. If I took a point on the line <em>B</em>, say (2,4), and added it to a point on 7 + <em>B</em>, say (3,13), then I get the point (5,17), which is on the line $$(0 + B) + (7 + B) = 7 + B.$$
If I add the point (3,7) on 1 + <em>B</em> to the point (8,20) on 4 + <em>B</em>, then I get the point (11,27) on $$(1+B) + (4+B) = 5 + B$$
This is just because if x = 11, then 2<i>x</i> is 22, and 27 is 2<i>x</i>+5.</p>
<p>If one wanted to be more precise, I suppose one should say (0,5) + <em>B</em>, since it should be an element of $A$ plus $B$, but just like $A/B \cong \mathbb{R}$, we only need one number here too.</p>
|
69,050 | <p>The basic concept of Quotient Group is often a confusing thing for me,I mean can any one tell the intuitive concept and the necessity of the Quotient group,
I thought that it would be nice to ask as any basic undergraduate can learn the intuition seeing the question.
My Question is :</p>
<ol>
<li>Why is the name Quotient Group kept,normally in the case of division, let us take the example of $\large \frac{16}{4}$ the Quotient of the Division is '$4$' which means that there are four '$4$'s in $16$, I mean we can find only $4$ elements with value $4$
<blockquote>
<p>So how can we apply the same logic in the case of Quotient Groups,like consider the Group $A$ and normal subgroup $B$ of $A$, So if $A/B$ refers to "Quotient group", then does it mean:</p>
<blockquote>
<p>Are we finding how many copies of $B$ are present in $A$??, like in the case of normal division, or is it something different ??</p>
</blockquote>
</blockquote></li>
</ol>
<p>I understood the Notion of Cosets and Quotient Groups,b ut I want a different Perspective to add Color to the concept. Can anyone tell me the necessity and background for the invention of Quotient Groups?</p>
<p>Note: I tried my level best in formatting and typing with proper protocol,if in case, any errors still persist, I beg everyone to explain the reason of their downvote (if any), so that I can rectify myself, Thank you.</p>
| Beginner | 15,847 | <p>Two of the most basic concepts in mathematics, namely "Sets" and "Relations" are much useful to create new things from old. For example, "integers" can be constructed from natural numbers by putting an equivalence relation on the set $\mathbb{N}\times \mathbb{N}$ (see Concrete Abstract Algebra: Niels Lauritzen), rationals can be constructed from integers by putting an equivalence relation on $\mathbb{Z}\times \mathbb{Z}$; in some way reals, complex numbers can also be constructed. The only things we have used are "<em>Sets and Equivalence Relation</em>".</p>
<p>In this way, we can think of quotient group of a group $G$ as a set of equivalence classes of group $G$ with a natural binary operation, induced from that of $G$:</p>
<p>If $H$ is a subgroup of $G$, define an equivalence relation on $G$ by $x\sim y$ iff $x^{-1}y\in H$, denote equivalence class of $a$ by $[a]$. If $K$ is the set of equivalence classes of $G$, a natural way to define a binary operation on $K$ is to define $*\colon K \times K\rightarrow K$ by $[a]*[b]=[ab]$. </p>
<p>(So we have constructed a new set using equivalence relation and trying to put a binary on it) The <em>necessary (and sufficient)</em> for this map to be well defined (hence binary operation) is that $H$ should be normal subgroup of $G$. Once we convince for it, it is easy to verify other conditions in the definition of group. The new group is called as quotient group of $G$ by $H$ (denoted $G/H$)</p>
<p>Therefore, the quotient group of a group by <em>normal subgroup</em> is a set of equivalence of classes of $G$ with a natural binary operation, induced from $G$</p>
|
69,050 | <p>The basic concept of Quotient Group is often a confusing thing for me,I mean can any one tell the intuitive concept and the necessity of the Quotient group,
I thought that it would be nice to ask as any basic undergraduate can learn the intuition seeing the question.
My Question is :</p>
<ol>
<li>Why is the name Quotient Group kept,normally in the case of division, let us take the example of $\large \frac{16}{4}$ the Quotient of the Division is '$4$' which means that there are four '$4$'s in $16$, I mean we can find only $4$ elements with value $4$
<blockquote>
<p>So how can we apply the same logic in the case of Quotient Groups,like consider the Group $A$ and normal subgroup $B$ of $A$, So if $A/B$ refers to "Quotient group", then does it mean:</p>
<blockquote>
<p>Are we finding how many copies of $B$ are present in $A$??, like in the case of normal division, or is it something different ??</p>
</blockquote>
</blockquote></li>
</ol>
<p>I understood the Notion of Cosets and Quotient Groups,b ut I want a different Perspective to add Color to the concept. Can anyone tell me the necessity and background for the invention of Quotient Groups?</p>
<p>Note: I tried my level best in formatting and typing with proper protocol,if in case, any errors still persist, I beg everyone to explain the reason of their downvote (if any), so that I can rectify myself, Thank you.</p>
| Community | -1 | <p>I modify Jack Schmidt's exquisite answer for more details.</p>
<blockquote>
<p>Quotient groups $A/B$ count cosets of $B$ inside $A$. The counting even works well with addition.</p>
</blockquote>
<p>The Cartesian plane forms a group A, and a line through the origin is a subgroup B. The cosets of B inside A are all the parallel lines. How many are there?</p>
<p>Suppose <em>B</em> is the line: $ B = \{ (x,y) : y = 2x \}$
or just $y = 2x$ for short. Parallel lines are parameterised by their "intercept" <em>b</em>. Hence the coset <em>b</em> + <em>B</em> is the line with slope 2 and intercept <em>b</em>.
$$b+B = \{(x,y) : y = 2x+b\}$$
This means there is exactly one coset for every real number $x$. In some sense we've counted the parallel lines.</p>
<p>The way we count them even keeps track of addition.<br>
Take a point on the line $\color{blue}{B = 0 + B}$, say (2,4).<br>
Add it to a point on $\color{blue}{7 + B}$, say (3,13).<br>
Then I get the point $(2,4) + (3,13) = (5,17)$,<br>
which is on the line $\color{blue}{(0 + B) + (7 + B)} = 7 + B.$</p>
<p>If I add the point (3,7) on 1 + <em>B</em> to the point (8,20) on 4 + <em>B</em>, then I get the point (11,27) on $$(1+B) + (4+B) = 5 + B$$
This is just because if x = 11, then 2<i>x</i> is 22, and 27 is 2<i>x</i>+5. To boot, $y = 2(11) + 5$. </p>
<p>If one wanted to be more precise, one should say (0,5) + <em>B</em>, since a coset should be an element of $A$ plus $B$. But just like $A/B \cong \mathbb{R}$, we only need one number here too.</p>
|
8,382 | <h3>Context</h3>
<p>I'm writing a function that look something like:</p>
<pre><code>triDiagonalQ[mat_] := MapIndexed[ #1 == 0 || Abs[#2[[1]]-#2[[2]]] <= 1 &, mat, {2}] //
Flatten // And @@ # &
</code></pre>
<p>Now, things like <code>#2[[1]]</code> and <code>#2[[2]]</code> are somewhat hard to read. I'd prefer to do something like:</p>
<pre><code>triDiagonalQ[mat_] := MapIndexed[ #1 == 0 || Abs[i-j] <= 1 &, mat, {2}] //
Flatten // And @@ # & (* with a {i, j} <- #2 somewhere *)
</code></pre>
<h3>Question</h3>
<p>Is there someway to do something like "destructuring" in <em>Mathematica</em>?</p>
<hr>
<p>The following links convey what I mean by "destructuring":</p>
<ul>
<li><a href="http://clojure.org/special_forms">http://clojure.org/special_forms</a></li>
<li><a href="http://java.dzone.com/articles/clojure-destructuring">http://java.dzone.com/articles/clojure-destructuring</a></li>
<li><a href="http://blog.jayfields.com/2010/07/clojure-destructuring.html">http://blog.jayfields.com/2010/07/clojure-destructuring.html</a></li>
</ul>
<p>(These have nothing to do with <em>Mathematica</em>; they're posted mainly to demonstrate what is meant by "destructuring")</p>
| Mr.Wizard | 121 | <p>My first answer explains that <em>Mathematica</em>'s replacement rules perform destructuring. This answer is intended to complement jVincent's method, which I see is appreciated.</p>
<p>My aim is to provide <code>Attributes</code> for the pattern-based function. This requires that the head evaluate therefore <code>SubValues</code> may not be used. Here are two separate approaches.</p>
<h2>Module</h2>
<p>A direct approach is to automate the creation of a new <code>DownValue</code> function using <code>Module</code>.</p>
<pre><code>SetAttributes[pFunc1, HoldAll]
pFunc1[pattern_, body_, attrib_: {}] :=
Module[{pf},
SetAttributes[pf, attrib];
pf[pattern] := body;
pf
]
pFunc1[_[__, b_, c_], Binomial[b, c], HoldAllComplete][9 + 7 + 6 + 3]
pFunc1[PatternSequence[_[__, b_, c_], x_], Binomial[b, c]/x, HoldAllComplete][9 + 7 + 6 + 3, 4]
</code></pre>
<blockquote>
<pre><code>20
5
</code></pre>
</blockquote>
<h2>ReplaceAll</h2>
<p>The first method creates a new symbol <code>pf$xxx</code> for each use which could be seen as undesirable. A method that does not is to simply embed a <code>ReplaceAll</code> operation in the body of <code>Function</code> like this:</p>
<pre><code>SetAttributes[pFunc2, HoldAll]
pFunc2[pattern_, body_, attrib_: {}] :=
Function[
Null,
Unevaluated@{##} /. pattern :> body,
attrib
]
</code></pre>
<p>This method returns an actual <code>Function</code> expression rather than a <code>pf$xxx</code> symbol.<br>
It also allows mixing <code>Slot</code> (<code>#</code>) and pattern references in <code>body</code> should that be desired.</p>
<p>It however does not return an unevaluated function expression when the arguments do not match the pattern. (Though one could add a <code>{__} :> $Failed</code> type of rule to <code>/.</code>, if that is all that is desired.)</p>
|
17,115 | <p>I am pretty sure that the following statement is true. I would appreciate any references (or a proof if you know one).</p>
<p>Let $f(z)$ be a polynomial in one variable with complex coefficients. Then there is the following dichotomy. Either we can write $f(z)=g(z^k)$ for some other polynomial $g$ and some integer $k>1$, or the restriction of $f(z)$ to the unit circle is a loop with only finitely many self-intersections. (Which means, more concretely, that there are only finitely many pairs $(z,w)$ such that $|z|=1=|w|$, $z\neq w$ and $f(z)=f(w)$.)</p>
<p>EDIT. Here are a couple reasons why I believe the statement is correct.</p>
<p>1) The statement is equivalent to the following assertion. Consider the set of all ratios $z/w$, where $|z|=1=|w|$ and $f(z)=f(w)$ (here we allow $z=w$). If $f$ is a nonconstant polynomial, then this set is finite.</p>
<p>[[ Here is a proof that the latter assertion implies the original statement. Suppose that there are infinitely many pairs $(z,w)$ such that $|z|=1=|w|$, $z\neq w$ and $f(z)=f(w)$. Then some number $c\neq 1$ must occur infinitely often as the corresponding ratio $z/w$. However, this would imply that $f(cz)=f(z)$ (as polynomials). It is easy to check that this forces $c$ to be a root of unity, and if $k$ is the order of $c$, then $f(z)=g(z^k)$ for some polynomial $g(z)$. ]]</p>
<p>Going back to the latter assertion, note that the set of all such ratios is a compact subset of the unit circle, and it is not hard to see that 1 must be an isolated point of this set. So it is plausible that the whole set is discrete (which would mean that it is finite).</p>
<p>2) If I am not mistaken, experiments with polynomials that involve a small number of nonzero monomials (such as 2 or 3) also confirm the original conjecture.</p>
| Jonas Meyer | 1,119 | <p>You're right. Quine proved in "<a href="http://www.jstor.org/stable/2039005">On the self-intersections of the image of the unit circle under a polynomial mapping</a>" that if the degree is $n$ and $f(z)\neq g(z^k)$ with $k>1$, then the number of points with at least 2 distinct preimage points is at most $(n-1)^2$. An example shows that this is sharp. Here's the <a href="http://www.ams.org/mathscinet-getitem?mr=313485">review</a> in MR.</p>
<p>After the proof, there is a remark: </p>
<blockquote>
<p>As a simple consequence of this theorem we note that a polynomial $p$
cannot map $|z| < 1$ conformally onto a domain with a slit, for in this case
$p(e^{i\phi})$ would have an infinite number of vertices.</p>
</blockquote>
|
17,115 | <p>I am pretty sure that the following statement is true. I would appreciate any references (or a proof if you know one).</p>
<p>Let $f(z)$ be a polynomial in one variable with complex coefficients. Then there is the following dichotomy. Either we can write $f(z)=g(z^k)$ for some other polynomial $g$ and some integer $k>1$, or the restriction of $f(z)$ to the unit circle is a loop with only finitely many self-intersections. (Which means, more concretely, that there are only finitely many pairs $(z,w)$ such that $|z|=1=|w|$, $z\neq w$ and $f(z)=f(w)$.)</p>
<p>EDIT. Here are a couple reasons why I believe the statement is correct.</p>
<p>1) The statement is equivalent to the following assertion. Consider the set of all ratios $z/w$, where $|z|=1=|w|$ and $f(z)=f(w)$ (here we allow $z=w$). If $f$ is a nonconstant polynomial, then this set is finite.</p>
<p>[[ Here is a proof that the latter assertion implies the original statement. Suppose that there are infinitely many pairs $(z,w)$ such that $|z|=1=|w|$, $z\neq w$ and $f(z)=f(w)$. Then some number $c\neq 1$ must occur infinitely often as the corresponding ratio $z/w$. However, this would imply that $f(cz)=f(z)$ (as polynomials). It is easy to check that this forces $c$ to be a root of unity, and if $k$ is the order of $c$, then $f(z)=g(z^k)$ for some polynomial $g(z)$. ]]</p>
<p>Going back to the latter assertion, note that the set of all such ratios is a compact subset of the unit circle, and it is not hard to see that 1 must be an isolated point of this set. So it is plausible that the whole set is discrete (which would mean that it is finite).</p>
<p>2) If I am not mistaken, experiments with polynomials that involve a small number of nonzero monomials (such as 2 or 3) also confirm the original conjecture.</p>
| Ian Agol | 1,345 | <p>The image of the unit circle is a real-algebraic curve, so the number of self-intersections should be finite. </p>
<p>Addendum:
I'm not sure how to complete the argument, but here's a heuristic (following Speyer's suggestion). The curve $x^2+y^2=1$ is a rational curve (i.e., birational to $\mathbb{CP}^1$, the Riemann sphere), so it's image is also a rational curve (here, I'm thinking of the map $p:\mathbb{C}\to \mathbb{C}$ as a polynomial map $\mathbb{R}^2 \to \mathbb{R}^2$, and its extension to $\mathbb{C}^2\to \mathbb{C}^2$ and $\mathbb{CP}^2\to \mathbb{CP}^2$). Complex conjugation gives an antiholomorphic involution of $x^2+y^2=1$, fixing the circle (on the Riemann sphere, this must be conjugate to complex conjugation). The image (in $\mathbb{CP}^2$) is a singular sphere, and again complex conjugation should be an anti-holomorphic involution fixing the image of the unit circle. So the map should be a composition of a polynomial map with a map of the Riemann sphere sending the circle to the circle. All such maps of the Riemann sphere are products of Mobius transformations of the form $\frac{z-\varphi}{1-\overline{\varphi}z}$ (this is an exercise in Ahlfors, making use of the Schwarz lemma). If the composition is to be a polynomial map, then $\varphi$ must $=0$ in each factor, and the map is of the form $z^k$. My algebraic geometry is quite weak, so I'm not sure if this argument can be made rigorous (and I probably shouldn't have posted an answer in the first place!). </p>
|
875,458 | <p>Is it possible to draw a triangle, if the length of its medians $(m_1, m_2, m_3)$ are given only?</p>
<p>Someone asked me this question, but I can not see it. Is it really possible?</p>
<p><strong>UPDATE</strong></p>
<p>Apart from the algebraic solution given by <em>JimmyK4542</em>, can anyone give me a direct construction? I mean, it should sound like:</p>
<blockquote>
<p>Draw a line segment sufficiently long. Cut the length of $m_1$ from it. Then $\ldots$</p>
</blockquote>
| Christian Blatter | 1,303 | <p>From JimmyK4542's formulas it follows that
$a={2\over3}\sqrt{2s^2-m_a^2}\>$, where $s:=\sqrt{m_b^2+m_c^2}$. From this one derives the following construction of ${3\over2}a$: </p>
<p><img src="https://i.stack.imgur.com/kskHT.jpg" alt="enter image description here"></p>
<p>Construct $s$ as hypotenuse of a right triangle with legs $m_b$, $m_c$; then draw a square with side $s$ and find $\sqrt{2}\>s$ as length of the diagonal $d$. Draw a Thales semicircle with diameter $d$; then construct a right triangle with hypotenuse $d$ and one leg $m_a$. The other leg then is ${3\over2}a$. The rest should be easy.</p>
|
1,957,304 | <p>I'm proving the compact-to-Hausdorff lemma (probably not a universal name for it) which is stated as:</p>
<blockquote>
<p>If $X$ is compact, $Y$ Hausdorff, $f:X \rightarrow Y$ a continuous bijection, then $f$ is a homeomorphism.</p>
</blockquote>
<p>However, the following line has popped up in a proof of it:</p>
<blockquote>
<p>If $f: X \rightarrow Y$ is a bijection, then $f(U) = Y\setminus f(X \setminus U)$ (where $U$ is open in $X$)($\star$).</p>
</blockquote>
<p>I know that if $f$ is a bijection, then $f(X\setminus U) = f(X) \setminus f(U)$. Using this, I've tried to draw a little picture to try to see that $f(U) = Y\setminus f(X \setminus U)$, but it hasn't actually helped.</p>
<p>What's a proof of ($\star$)?</p>
| carmichael561 | 314,708 | <p>If $y\in f(U)$, then there exists $x\in U$ such that $f(x)=y$. Since $f$ is injective, there cannot be an element $z\in X\setminus U$ such that $f(z)=y$. Therefore $y\not\in f(X\setminus U)$, i.e. $y\in Y\setminus f(X\setminus U)$.</p>
<p>On the other hand, suppose that $y\in Y\setminus f(X\setminus U)$. There exists some $x\in X$ such that $f(x)=y$ because $f$ is surjective, and $x\not\in X\setminus U$ because $y\not\in f(X\setminus U)$, so $x\in U$. Therefore $y\in f(U)$.</p>
|
1,957,304 | <p>I'm proving the compact-to-Hausdorff lemma (probably not a universal name for it) which is stated as:</p>
<blockquote>
<p>If $X$ is compact, $Y$ Hausdorff, $f:X \rightarrow Y$ a continuous bijection, then $f$ is a homeomorphism.</p>
</blockquote>
<p>However, the following line has popped up in a proof of it:</p>
<blockquote>
<p>If $f: X \rightarrow Y$ is a bijection, then $f(U) = Y\setminus f(X \setminus U)$ (where $U$ is open in $X$)($\star$).</p>
</blockquote>
<p>I know that if $f$ is a bijection, then $f(X\setminus U) = f(X) \setminus f(U)$. Using this, I've tried to draw a little picture to try to see that $f(U) = Y\setminus f(X \setminus U)$, but it hasn't actually helped.</p>
<p>What's a proof of ($\star$)?</p>
| egreg | 62,967 | <p>The property you want to prove is the same as
$$
Y\setminus f(U)=f(X\setminus U)
$$
Suppose $y\notin f(U)$; then $f^{-1}(y)\notin U$ and therefore $f^{-1}(y)\in X\setminus U$. Therefore $y\in f(X\setminus U)$.</p>
<p>Similarly for the converse inclusion.</p>
<hr>
<p>Note that, in general, when you have a function $g\colon A\to B$, then for every subset $C$ of $B$ we have
$$
g^{-1}(B\setminus C)=A\setminus g^{-1}(C)
$$
(where $g$ need not be bijective: $g^{-1}(C)=\{a\in A:g(a)\in C\}$; in order to avoid confusions, I usually write $g^{\gets}(C)$, instead, but $g^{-1}(C)$ is commonly used).</p>
<p>Your case is exactly this one, with $g=f^{-1}$.</p>
|
882,590 | <p><img src="https://i.stack.imgur.com/BMapu.png" alt="enter image description here"></p>
<p>long method: Determine an equation for each and solve using average value formula</p>
<p>alternative methods? </p>
<p>How could you prove the average value to be C over an interval [a,b] if you are given a graph.... looking for most efficient/unique methods. </p>
| Varun Iyer | 118,690 | <p>Your equation for this question is:</p>
<p>$$\frac{1}{6}*\int_0^6 h(x) = 2$$</p>
<p>Therefore,</p>
<p>$$\int_0^6 h(x) = 12$$</p>
<p>For answers $c$ and $d$ you can use the area of a triangle,</p>
<p>If you try answer choice $c$,</p>
<p>We get that:</p>
<p>$$A = \frac{1}{2}*4.5*6 - \frac{1}{2}*1.5*(-2) = 12$$</p>
<p>Therefore our answer is C.</p>
|
3,802,806 | <p>I and a friend are trying to find all endomorphisms <span class="math-container">$f$</span> of <span class="math-container">$\mathcal{M}_n(\mathbb{R})$</span> such that <span class="math-container">$f({}^t M)={}^t f(M)$</span> for all <span class="math-container">$M$</span>. We believe they are of the form <span class="math-container">$M\mapsto\lambda M+\mu {}^t M$</span> for a fixed <span class="math-container">$(\lambda,\mu)\in\mathbb{R}^2$</span>. Any help is appreciated, thank you.</p>
| Nick | 27,349 | <p>First, consider a basis of the space <span class="math-container">$\mathcal{M}_n$</span>. For example, the <span class="math-container">$n^2$</span> elementary matrices <span class="math-container">$e_{ij}$</span> which are all zero's except a <span class="math-container">$1$</span> in the <span class="math-container">$(i,j)$</span> spot. An element <span class="math-container">$f \in \mathrm{End}(\mathcal{M}_n)$</span> is determined by where it sends each <span class="math-container">$e_{ij}$</span>. Call these <span class="math-container">$M_{ij}$</span>, so that <span class="math-container">$f(e_{ij}) = M_{ij}$</span>. Then for an arbitrary matrix <span class="math-container">$A$</span> with entries <span class="math-container">$a_{ij}$</span>, you have
<span class="math-container">$$ f(A) = \sum_{i,j} a_{ij} M_{ij} $$</span></p>
<p>Now, you want to consider when <span class="math-container">$f$</span> commutes with the transpose operation. Consider this in the special case of the basis elements <span class="math-container">$e_{ij}$</span>. Notice that <span class="math-container">${}^te_{ij} = e_{ji}$</span>, so the condition becomes :</p>
<p><span class="math-container">$$
\begin {align*}
f({}^te_{ij}) &= {}^tf(e_{ij}) \\
f(e_{ji}) &= {}^tM_{ij} \\
M_{ji} &= {}^tM_{ij}
\end {align*}
$$</span></p>
|
130,465 | <p>i just started university so im pretty new to all this new math. My problem is to solve this <code>recursive sequence</code>: $a_{n+1} = a_{n}^3$ with: $a_{0} = \frac{1}{2}$ and: $n \in N$</p>
<p>I've to analyse convergence and if its convergent i've to get the limit of this sequence.</p>
<p>I dont know how to start and this <code>to the power of three</code> confuses me.</p>
<p>thanks in advance!</p>
| Riemann | 27,899 | <p>Due to $a_{n+1}=a_n^{3}=(a_{n-1}^3)^3=a_{n-1}^{3^2}=……=a_{0}^{3^{n+1}}=(\frac{1}{2})^{3^{n+1}}$. if you know this ,you can do this question.</p>
|
3,753,060 | <blockquote>
<p>If <span class="math-container">$\int f(x)dx =g(x)$</span> then <span class="math-container">$\int f^{-1}(x)dx $</span> is equal to</p>
<p>(1) <span class="math-container">$g^{-1}(x)$</span></p>
<p>(2) <span class="math-container">$xf^{-1}(x)-g(f^{-1}(x))$</span></p>
<p>(3) <span class="math-container">$xf^{-1}(x)-g^{-1}(x)$</span></p>
<p>(4) <span class="math-container">$f^{-1}(x)$</span></p>
</blockquote>
<p>My approach is as follows:
Let <span class="math-container">$f(x)=y$</span>, therefore <span class="math-container">$f^{-1}(y)=x$</span>, <span class="math-container">$\int f^{-1}(f(x))dx =g(f(x))$</span></p>
<p>On differentiating we get <span class="math-container">$x=g'(f(x))f'(x)$</span></p>
<p>After this step, I am not able to proceed.</p>
| Knight wants Loong back | 569,595 | <p>All the other answers have already given you the finest methods of solving the problem. I want to say that as you’re given the options for the integral <span class="math-container">$\int f^{-1} (x) dx$</span>, it would be a nice thing to differentiate the options to see if we get <span class="math-container">$f^{-1}(x)$</span> (it should follow from <a href="https://en.m.wikipedia.org/wiki/Fundamental_theorem_of_calculus" rel="nofollow noreferrer">The Fundamental Theorem of Calculus</a> ).</p>
<p>Let’s try the first option:
<span class="math-container">$$
If ~~~~ \int f^{-1}(x) dx = g^{-1} (x) \\
then~~~~~ f^{-1} (x)= \frac{d}{dx} g^{-1} (x) \\
f^{-1} (x)= \left(
\frac{d~g(x)}{dx} \right)^{-1}\\
f^{-1} (x)= \frac{1}{f(x)}$$</span> (in the third step I once again used the FTC for the function<span class="math-container">$f(x)$</span>) the last equality is not true in general, therefore this option is not valid.</p>
<p>Let’s try the second option:
<span class="math-container">$$
If~~~~ \int f^{-1}(x) dx = xf^{-1} (x) - g\left( f^{-1}(x)\right) \\
then~~~~ f^{-1}(x)= \frac{d}{dx} \left[xf^{-1} (x) - g\left( f^{-1}(x)\right) \right] \\
f^{-1} (x)= f^{-1}(x) + x \frac{d~f^{-1}(x)}{dx} - f\left(f^{-1} (x)\right) \frac{d~f^{-1}(x)}{dx}\\
f^{-1}(x) = f^{-1}(x) $$</span>
Hence, the second option is correct.</p>
<p>Hope it adds something to this thread!</p>
|
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