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 |
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2,027,888 | <h2>Exercise</h2>
<p>If $H$ is the Heaviside function, prove, using the definition below, that $\lim \limits_{t \to 0}{H(t)}$ does not exist.</p>
<hr>
<h2>Definition</h2>
<blockquote>
<p>Let $f$ be a function defined on some open interval that contains the number $a$, except possible $a$ itself. Then we say that the limit of $f(x)$ as $x$ approaches $a$ is $L$, and we write
$$\lim \limits_{x \to a}{f(x)} = L$$
if for every number $\epsilon > 0$ there is a number $\delta > 0$ such that
$$\text{if } 0 < |x - a| < \delta \text{ then } |f(x) - L| < \epsilon$$</p>
</blockquote>
<hr>
<h2>Hint</h2>
<blockquote class="spoiler">
<p> Use an indirect proof as follows. Suppose that the limit is $L$. Take $\epsilon = \frac{1}{2}$ in the definition of a limit and try to arrive at a contradiction.</p>
</blockquote>
<hr>
<h2>Attempt</h2>
<p>Let $\delta$ be any (preferably small) positive number.</p>
<p>$H(0 - \delta) = H(-\delta) = 0$</p>
<p>$H(0 + \delta) = H(\delta) = 1$</p>
<p>$H(0 - \delta) =^? H(0 + \delta) \implies 0 =^? 1 \implies 0 \neq 1 \implies H(0 - \delta) \neq H(0 + \delta)$</p>
<p>$\lim \limits_{t \to 0^-}{H(t)} \neq \lim \limits_{t \to 0^+}{H(t)} \implies \lim \limits_{t \to 0}{H(t)}$ does not exist</p>
<hr>
<h2>Request</h2>
<p>I don't even know where to begin, even with the hint.</p>
<p><strong>Can someone kickstart the proof for me?</strong>$^1$</p>
<p><strong>$^1$ Update:</strong> I've come up with an attempt. Is it valid? It seems that I don't use the hint to my advantage; so if indeed my attempt is correct, what is the alternative proof using the hint?</p>
| David | 651,991 | <p>Main idea: Set <span class="math-container">$\epsilon = 0.1$</span> and see that, for any "candidate to limit" <span class="math-container">$L$</span>, you will always find a point <span class="math-container">$X$</span> as close to <span class="math-container">$0$</span> as you want, such that <span class="math-container">$|H(X)-L| > \epsilon$</span></p>
<p>So, we've found an <span class="math-container">$\epsilon > 0$</span> such that, for any <span class="math-container">$\delta>0$</span>, there exists <span class="math-container">$x$</span> such that <span class="math-container">$|x-0| < \delta$</span> but yet <span class="math-container">$|H(X)-L| > \epsilon$</span>. But how to be precise about finding that <span class="math-container">$x$</span>?</p>
<p>If our "candidate to limit" <span class="math-container">$L \in [0.9, 1.1]$</span>, simply choose <span class="math-container">$x = - \frac{\delta}{2}$</span>, so that <span class="math-container">$H(x)=0$</span>, which is at a distance larger than <span class="math-container">$\epsilon$</span> from <span class="math-container">$L$</span>, i.e. <span class="math-container">$|f(x)-L| > \epsilon$</span></p>
<p>I think you can easily guess which <span class="math-container">$x$</span> to pick if <span class="math-container">$L \notin [0.9, 1.1]$</span></p>
|
163,585 | <p>Which (finite, undirected) graphs have this property?</p>
<p>Every vertex $v$ can be labeled with a positive integer $l(v)$.</p>
<p>Variant 1: For each vertex $v$, $l(v) \geq \Sigma_{[v,w] \in E, w \neq v} l(w)/2$.</p>
<p>Variant 2: For each vertex $v$, $l(v) > \Sigma_{[v,w] \in E, w \neq v} l(w)/2$.</p>
| Generic Human | 26,855 | <p>Here $G$ is a finite undirected graph.</p>
<p>Suppose wlog that $G$ is connected, with at least 2 vertices $1,\dots,n$ and no self-edges. There is no reason to forbid multiple edges. Define $M=A/2$ where $A$ is the adjacency matrix of $G$.</p>
<blockquote>
<p><strong>Theorem</strong>: $G$ satisfies the first condition (respectively the second condition) iff its index, that is the spectral radius of its adjacency matrix, is $\le 2$ (resp. $<2$).</p>
</blockquote>
<p><strong>Proof:</strong></p>
<ul>
<li><p>Suppose $G$ has a labeling $x\in (\mathbb N^*)^n\subset\mathbb R_+^n\setminus\{0\}$. Then using the hypothesis it's clear by induction that each component of $M^k x$ is a non-negative non-increasing function of $k$ (respectively, exponentially decaying). Because $M$ is diagonalizable, the decay of the largest component is of the form $\Theta(\lambda^k)$, where $1/2\le\lambda\le 1$ (resp. $<1$) is an eigenvalue of $M$. But $\lambda^{-k} M^k x$ converges to a positive eigenvector $y$ of $M$ associated with $\lambda$, so that by Perron-Frobenius $\lambda\le 1$ (resp. $<1$) is the spectral radius of $M$. Therefore the spectral radius of $A$ must be at most 2 (resp. $<2$).</p></li>
<li><p>Conversely, if the spectral radius of $A$ is at most 2, we can distinguish two cases. If the radius is exactly 2, then we can find an integer non-negative eigenvector of $M$ (as $\det M-I=0$ implies that $M-I$ has non-trivial kernel in $\mathbb Q^n$ thus in $\mathbb Z^n$), and therefore, because $G$ is connected and has at least one edge, a positive integer eigenvector. So there is a labeling for the first condition. If the radius is less than 2, take a positive real eigenvector $x$ of $M$ associated with $0<\lambda<1$. If $m>0$ is the smallest element of $x$, let $y = \left\lfloor\frac{2}{(1-\lambda)m}x\right\rfloor$. Then $(y-My)_i\ge 1$ so that the second condition holds.</p></li>
</ul>
<p>We know a classification of such graphs, e.g. <a href="http://math.ipm.ac.ir/tayfeh-r/papersandpreprints/graphspec.pdf" rel="nofollow">see here</a>.</p>
|
2,903,429 | <p>Show that the equation $$\frac{4}{3}{x^2} + \frac{4}{3}{y^2} + \frac{4}{3}{z^2} + \frac{4}{3}{xy} + \frac{4}{3}{xz} + \frac{4}{3}{yz} = 1$$ represents an ellipsoid. Find the position and lengths of its principal half-axes.</p>
<p>Workings
$$(x+y)^2+(y+z)^2+(z+x)^2=\frac32\,.$$
Common factor $\left(\dfrac32\right)^{\frac12}$ not sure how to find length and position half-axes??</p>
| Claude Leibovici | 82,404 | <p>$$\sum_{j=1}^n \sum_{k=j+1}^n 1=\sum_{j=1}^n (n-j)=n\sum_{j=1}^n 1-\sum_{j=1}^n j=n^2-\frac{1}{2} n (n+1)=\frac{1}{2} (n-1) n$$</p>
|
2,903,429 | <p>Show that the equation $$\frac{4}{3}{x^2} + \frac{4}{3}{y^2} + \frac{4}{3}{z^2} + \frac{4}{3}{xy} + \frac{4}{3}{xz} + \frac{4}{3}{yz} = 1$$ represents an ellipsoid. Find the position and lengths of its principal half-axes.</p>
<p>Workings
$$(x+y)^2+(y+z)^2+(z+x)^2=\frac32\,.$$
Common factor $\left(\dfrac32\right)^{\frac12}$ not sure how to find length and position half-axes??</p>
| JonHales | 413,766 | <p>How I see this is I start with $j = 1$ And so my next sum is $$\Sigma_{k = 2} ^{n} (1) = n-1$$ Now I'll add to this the value for when $j=2$, which will be $$\Sigma_{k = 3}^n (1) = n - 2$$. This pattern continues, so the nested sum you want is really just $$\Sigma_{i = 1}^n (n-i) = \Sigma_{i =1}^n(i) = \frac{1}{2} n(n-1)$$</p>
<p>Hope that helps!!</p>
|
3,358,168 | <p><span class="math-container">$x_n$</span> is bounded, <span class="math-container">$\lim_{n \rightarrow \infty}(x_{n+1}-x_n)=0$</span>, <span class="math-container">$\liminf_{n \rightarrow \infty}=l$</span>, <span class="math-container">$\limsup_{n \rightarrow \infty}=L$</span>, show that every point in <span class="math-container">$[l,L]$</span> is an accumulation point.</p>
<p>I am trying to use Stolz Cesàro theorem to show that <span class="math-container">$\lim_{n \rightarrow \infty} \frac{x_n}{n}=0$</span>, is it the right direction?</p>
| Matematleta | 138,929 | <p>Suppose <span class="math-container">$l<x<L$</span> but <span class="math-container">$x$</span> is not an accumulation point of <span class="math-container">$(x_n)$</span>. Then there are <span class="math-container">$a,b$</span> such that <span class="math-container">$l<a<x<b<L$</span> and <span class="math-container">$(a,b)\setminus \{x\}\cap (x_n)=\emptyset.$</span> Now, there is an integer <span class="math-container">$N$</span> so large that <span class="math-container">$|x_{n+1}-x_n|<\frac{b-a}{2}$</span> whenever <span class="math-container">$n>N.$</span> But there is also an integer <span class="math-container">$M>N$</span> such that <span class="math-container">$x_M>b$</span> (why?). But then, if <span class="math-container">$x_n\in \{x_M, x_{M+1},\cdots \},\ x_n>b$</span>, from which it follows that <span class="math-container">$\liminf x_n\neq l$</span>, which is a contradiction.</p>
|
2,006,870 | <p>Is the following inequality true?</p>
<p>$\left( \sum \limits_{i=1}^\infty \sum \limits_{j=1}^\infty \sum \limits_{k=1}^\infty \sum \limits_{l=1}^\infty a_{ij}\,a_{ik}\,a_{jl}\,a_{kl} \right) \leq \left( \sum \limits_{i=1}^\infty \sum \limits_{j=1}^\infty a_{ij}^2 \right)^{1/2}\left( \sum \limits_{i=1}^\infty \sum \limits_{k=1}^\infty a_{ik}^2 \right)^{1/2}\left( \sum \limits_{j=1}^\infty \sum \limits_{l=1}^\infty a_{jl}^2 \right)^{1/2}\left( \sum \limits_{k=1}^\infty \sum \limits_{l=1}^\infty a_{kl}^2 \right)^{1/2}=\left( \sum \limits_{i=1}^\infty \sum \limits_{j=1}^\infty a_{ij}^2 \right)^2$</p>
<p>where $a_{ij}$s are real numbers.</p>
| hmakholm left over Monica | 14,366 | <p>It is easy to find a collision:
$$xy+x-y = (x-1)(y+1) + 1$$
so all we need is to find some number that can be written as a product in more than one way, and then declare the two factors to be $x-1$ or $y+1$, respectively.</p>
<p>For example $12=3\cdot 4=4\cdot 3$, so your function produces $13$ both for inputs $(4,3)$ and $(5,2)$.</p>
<hr>
<p>What you want is a <a href="https://en.wikipedia.org/wiki/Pairing_function" rel="nofollow noreferrer">pairing function</a>, for which many choices exist. A fairly practical one that works for nonnegative integers, as suggested by Wikipedia, is
$$ (x,y) \mapsto \frac{(x+y+1)(x+y)}2+x $$
For theoretical purposes it is sometimes convenient to use more "wasteful" functions such as
$$ (x,y)\mapsto 2^x3^y $$
or
$$ (x,y) \mapsto 2^x(2y+1) $$</p>
|
40,532 | <p>I'm having some problems understanding the following paragraph, which I read in a analysis script (hopefully I haven't made any translation errors):</p>
<blockquote>
<p>"A map $f:U \rightarrow Y$, where $U$ is open and $X,Y$ are Banach spaces, is continuous at $x' \in U$ if $$f(x')=\lim_{x\rightarrow x'} f(x)=\lim_{h\rightarrow 0} f(x'+h),$$
where $h=x-x'$. We can decompose $h$ in a "<strong>polar</strong>" fashion in $h=ts$, where $\left\Vert h \right\Vert \geq 0$ and $s=\frac{1}{\left\Vert h \right\Vert } h$. Then
$f(x')=\lim\limits_{h\rightarrow 0} f(x'+h)$ iff $f(x'+ts)\rightarrow f(x')$ for $t \rightarrow 0^+$ <strong>uniformly with respect to</strong> $\left\Vert s \right \Vert = 1$.
No matter from which direction $s$ with $\left\Vert s \right\Vert=1$ we approach $x'$, the value of the function has to converge to $f(x')$ with a to all $s$ <strong>common "minimal speed"</strong> ".</p>
</blockquote>
<p>What I don't understand is this:</p>
<p>1) What does in means to decompose anything in a "polar" fashion ?</p>
<p>2) I thought only sequences of function can converge uniformly...and what does it mean, if something converges uniformly with respect to another thing ?</p>
<p>3) What does the author mean with "common "minimal speed""</p>
<p>Thanks in advance.</p>
| Arturo Magidin | 742 | <p>I'll only deal with (1). Like mac, I'm not sure what (3) means here.</p>
<p>(1) When working in polar coordinates, every point in the plane is written in the form $(r,\theta)$, where $r$ denotes the magnitude (distance to the origin) and $\theta$ denotes the argument (direction). Similarly, when we work with complex numbers, it is often very useful to write a complex number as if it were given by "polar coordinates" rather than "rectangular coordinates". The usual expression $z = a+bi$ with $a,b\in\mathbb{R}$ corresponds to the rectangular coordinates, with $a$ giving the $x$ coordinate and $b$ giving the $y$ coordinate (this is how Hamilton reified the complex numbers); then we can express $z$ instead in "polar coordinates", by writing $z = re^{i\theta}$, where $r$ is a nonnegative real, $\theta$ is a real number, and
$$e^{i\theta} = \cos\theta + i\sin\theta$$
gives the "direction." This is often called a "polar decomposition" of the complex number. The number $\theta$ is the argument, and the number $r$ is the magnitude (note that $\lVert z\rVert = r$).</p>
<p>By analogy, in many other circumstances where we express a quantity/object/function/transformation in terms of a "size" and a "direction", we call such an expression a "polar decomposition". This can be done for real numbers very easily: for any real number $h\neq 0$, we can write $h$ as $ts$, where $s$ is the "sign" ($s=1$ if $h\gt 0$, $s=-1$ if $h\lt 0$) and $t$ is the "magnitude" ($t=|h|$). This is the decomposition that is done here.</p>
|
65,083 | <p>Rotman's book <em>An Introduction to the Theory of Groups</em> (Fourth Edition) asks, on page 22, Exercise 2.8, to show that <span class="math-container">$S(n)$</span> cannot be embedded in <span class="math-container">$A(n+1)$</span>, where <span class="math-container">$S(n)$</span> = the symmetric group on <span class="math-container">$n$</span> elements, and <span class="math-container">$A(n)$</span> = the alternating group on <span class="math-container">$n$</span> elements. I have a proof but it uses Bertrand's Postulate, which seems a bit much for page 22 of an introductory text. Does anyone have a more appropriate (i.e., easier) proof?</p>
| Sergei Ivanov | 4,354 | <p>I think the following is sufficiently elementary: a transposition in $S_n$ is an element of order 2 commuting with at least $2(n-2)!$ elements of the group. But $A_{n+1}$ does not have such an element if $n$ is large enough. Indeed, if $\sigma\in A_{n+1}$ is of order 2, then it is a product of $k$ independent transpositions where $k$ is even and $2\le k\le(n+1)/2$. The number of elements of $A_{n+1}$ commuting with such $\sigma$ equals $2^{k-1}k!(n+1-2k)!$, and this is smaller than $2(n-2)!$ provided that $n\ge 6$.</p>
|
1,712,481 | <p><span class="math-container">$$I=\displaystyle\int\frac{dx}{(3 + 2\sin x - \cos x)}$$</span></p>
<p>If <span class="math-container">$$\tan\left(\frac{x}{2}\right)=u$$</span></p>
<p>or <span class="math-container">$$x=2\cdot\tan^{-1}(u)$$</span></p>
<p>Then,</p>
<p><span class="math-container">$$\sin{x}=\dfrac{2u}{1+u^2}$$</span></p>
<p><span class="math-container">$$\cos{x}=\dfrac{1-u^2}{1+u^2}$$</span></p>
<p><span class="math-container">$$dx=\dfrac{2}{1+u^2}$$</span></p>
<p>Substitute <span class="math-container">$$\tan\left(\dfrac{x}{2}\right)=u$$</span></p>
<p>Let us simplify the integrand before integrating</p>
<p><span class="math-container">$$\dfrac{1}{3+2\sin{x}-\cos{x}}$$</span></p>
<p><span class="math-container">$$=\dfrac{1}{3+2\frac{2u}{1+u^2}-\frac{1-u^2}{1+u^2}}$$</span></p>
<p><span class="math-container">$$=\dfrac{1}{3+\frac{4u-1+u^2}{1+u^2}}$$</span></p>
<p><span class="math-container">$$=\dfrac{1}{\frac{4u-1+u^2+3+3u^2}{1+u^2}}$$</span></p>
<p><span class="math-container">$$=\dfrac{1+u^2}{4u^2+4u+2}$$</span></p>
<p><span class="math-container">$$=\dfrac{1+u^2}{(2u+1)^2+1}$$</span></p>
<p><span class="math-container">$$I=\displaystyle\int\dfrac{1+u^2}{(2u+1)^2+1}\cdot\dfrac{2}{1+u^2}\ du$$</span></p>
<p><span class="math-container">$$=\displaystyle\int\dfrac{1}{(2u+1)^2+1}\ 2\,du$$</span></p>
<p>Now,</p>
<p>Take : <span class="math-container">$$v=2u+1$$</span></p>
<p>Therefore, <span class="math-container">$$dv=2\,du$$</span></p>
<p><span class="math-container">$$I=\displaystyle\int\dfrac{1}{v^2+1}\ dv$$</span></p>
<p><span class="math-container">$$I=\tan^{-1}(v)$$</span></p>
<p>Substitute everything back</p>
<p><span class="math-container">$$I=\tan^{-1}(2u+1)$$</span></p>
<p><span class="math-container">$$I=\tan^{-1}\left(2\tan\left(\frac{x}{2}\right)+1\right)$$</span></p>
<p><span class="math-container">$$\boxed{\displaystyle\int\frac{dx}{(3 + 2\sin x - \cos x)} = \tan^{-1}\left(2\tan\left(\frac{x}{2}\right)+1\right)+C}$$</span></p>
<p>I know that my approach is also not so difficult but still I think there must be an relatively easy approach to this integral. I have tried many different things using trigonometric identities but nothing seems to bring to the solution easily. Kindly help me out.</p>
| Noir | 324,626 | <p>Remember the following trigonometric equations, which are always very helpful to solve integrals:
$$
\begin{matrix}
2\cos(u)^2 = 1 + \cos(2u) & & & \left(\cos(u) + \sin(u)\right)^2 = 1 + \sin(2u) \\
2\sin(u)^2 = 1 - \cos(2u) & & & \left(\cos(u) - \sin(u)\right)^2 = 1 - \sin(2u) \\
\end{matrix}
$$
<em>Note that you can make a mnemonic to remember these equations.</em></p>
<hr>
<p>With them, you can solve the integral:
$$
I = \int \dfrac{dx}{3+2\sin(x)-\cos(x)}
$$
Using the following procedure:</p>
<ol>
<li><strong>Replace with $\;\;u=\frac{x}{2} \;\;\Rightarrow\;\; dx = 2 du$:</strong>
$$
\begin{align}
I &= 2 \int \dfrac{du}{3+2\sin(2u)-\cos(2u)}
\end{align}
$$</li>
<li><p><strong>Rewrite the expression to a known trigonometric expression:</strong>
$$
\begin{align}
I&= 2 \int \dfrac{du}{2\left(1+\sin(2u)\right) + \left(1-\cos(2u)\right)}
\end{align}
$$</p></li>
<li><p><strong>Replace with the corresponding trigonometric equations:</strong></p></li>
</ol>
<p>$$
\begin{align}
I&= 2 \int \dfrac{du}{2\left(\cos(u) + \sin(u)\right)^2 + 2\sin(u)^2}
\end{align}
$$</p>
<ol start="4">
<li><strong>Extract common factor by non-binomial term:</strong></li>
</ol>
<p>$$
\begin{align}
I&= 2 \int \dfrac{1}{\left(\dfrac{\cos(u)}{\sin(u)} + 1\right)^2 + 1 } \cdot\dfrac{1}{2\sin(u)^2} du = \int \dfrac{1}{\left(\cot(u) + 1\right)^2 + 1 } \cdot \csc(u)^2 du
\end{align}
$$</p>
<ol start="5">
<li><p><strong>Replace with $\;\;\omega=\cot(u) + 1 \;\;\Rightarrow\;\; d\omega = -\csc(u)^2 du$:</strong>
$$
\begin{align}
I&= \int \dfrac{-d\omega}{\omega^2 + 1 } = \mbox{arccot}(\omega) + C
\end{align}
$$</p></li>
<li><p><strong>Returning to the variable x:</strong>
$$
\omega =\cot(u) + 1 = \cot\left(\dfrac{x}{2}\right) + 1 \quad\Rightarrow\quad I = \mbox{arccot}\left(\cot\left(\dfrac{x}{2}\right) + 1\right) + C
$$</p>
<hr></li>
</ol>
<p>With a little more effort and using the same procedure, you could solve the following generic integral:
$$
I = \int \dfrac{dx}{A+B\sin(x)+C\cos(x)}
$$
The secret to this is:</p>
<ol>
<li><p><strong>In Step 2: choose the correct "known trigonometric expression" according to the signs of sine and cosine.</strong> </p></li>
<li><p><strong>In Step 4: don't forget to remove the non-binomial term to reduce it to a simple expression.</strong></p></li>
</ol>
|
1,712,481 | <p><span class="math-container">$$I=\displaystyle\int\frac{dx}{(3 + 2\sin x - \cos x)}$$</span></p>
<p>If <span class="math-container">$$\tan\left(\frac{x}{2}\right)=u$$</span></p>
<p>or <span class="math-container">$$x=2\cdot\tan^{-1}(u)$$</span></p>
<p>Then,</p>
<p><span class="math-container">$$\sin{x}=\dfrac{2u}{1+u^2}$$</span></p>
<p><span class="math-container">$$\cos{x}=\dfrac{1-u^2}{1+u^2}$$</span></p>
<p><span class="math-container">$$dx=\dfrac{2}{1+u^2}$$</span></p>
<p>Substitute <span class="math-container">$$\tan\left(\dfrac{x}{2}\right)=u$$</span></p>
<p>Let us simplify the integrand before integrating</p>
<p><span class="math-container">$$\dfrac{1}{3+2\sin{x}-\cos{x}}$$</span></p>
<p><span class="math-container">$$=\dfrac{1}{3+2\frac{2u}{1+u^2}-\frac{1-u^2}{1+u^2}}$$</span></p>
<p><span class="math-container">$$=\dfrac{1}{3+\frac{4u-1+u^2}{1+u^2}}$$</span></p>
<p><span class="math-container">$$=\dfrac{1}{\frac{4u-1+u^2+3+3u^2}{1+u^2}}$$</span></p>
<p><span class="math-container">$$=\dfrac{1+u^2}{4u^2+4u+2}$$</span></p>
<p><span class="math-container">$$=\dfrac{1+u^2}{(2u+1)^2+1}$$</span></p>
<p><span class="math-container">$$I=\displaystyle\int\dfrac{1+u^2}{(2u+1)^2+1}\cdot\dfrac{2}{1+u^2}\ du$$</span></p>
<p><span class="math-container">$$=\displaystyle\int\dfrac{1}{(2u+1)^2+1}\ 2\,du$$</span></p>
<p>Now,</p>
<p>Take : <span class="math-container">$$v=2u+1$$</span></p>
<p>Therefore, <span class="math-container">$$dv=2\,du$$</span></p>
<p><span class="math-container">$$I=\displaystyle\int\dfrac{1}{v^2+1}\ dv$$</span></p>
<p><span class="math-container">$$I=\tan^{-1}(v)$$</span></p>
<p>Substitute everything back</p>
<p><span class="math-container">$$I=\tan^{-1}(2u+1)$$</span></p>
<p><span class="math-container">$$I=\tan^{-1}\left(2\tan\left(\frac{x}{2}\right)+1\right)$$</span></p>
<p><span class="math-container">$$\boxed{\displaystyle\int\frac{dx}{(3 + 2\sin x - \cos x)} = \tan^{-1}\left(2\tan\left(\frac{x}{2}\right)+1\right)+C}$$</span></p>
<p>I know that my approach is also not so difficult but still I think there must be an relatively easy approach to this integral. I have tried many different things using trigonometric identities but nothing seems to bring to the solution easily. Kindly help me out.</p>
| Omran Kouba | 140,450 | <p>Generally, consider the real function
$$f(x)=\frac{1}{a+b\cos x+c\sin x}$$
With $a^2>b^2+c^2$ so that $f$ is defined on $\mathbb{R}$.
It is not hard to check that the derivative of
$$F(x)=\frac{x}{d}+\frac{2}{d}\arctan\left(\frac{c\cos x-b\sin x}{d+a+b\cos x+c\sin x}\right)$$
with $d=\sqrt{a^2-b^2-c^2}$, is $f(x)$. So $\int f(x)dx=F(x)+k$. The advantage
of this expression of $F$ is that it is also defined on $\mathbb{R}$. In particular,
$$\int \frac{1}{3-\cos x+2\sin x}=\frac{x}{2}+\arctan\left(\frac{2\cos x+\sin x}{5-\cos x+2\sin x}\right)$$</p>
|
131,482 | <p>Noticed strange behaviour of <code>Export[]</code> when saving graphics to EPS. Here is an example figure I am exporting:</p>
<pre><code>fig =
Show[
ListPolarPlot[
Table[{a, .9}, {a, 0, 2 Pi, .001}],
PlotRange -> {{0, 1.1}, {0, 1.1}}, GridLines -> Automatic,
PlotStyle -> PointSize[.005]],
Graphics[
Inset[
ListPolarPlot[Table[{a, .1}, {a, 0, 2 Pi, .001}], Axes -> False,
PlotStyle -> PointSize[.05]],
{.4, .4}]
]
]
</code></pre>
<p>The exported figure does not look fine upon a close look --- horizontal grid lines seem to be rasterised:
<a href="https://i.stack.imgur.com/ZPdGn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZPdGn.png" alt="zoom into the figure 1"></a></p>
<p>If I now apply <code>GridLinesStyle -> Thick</code>, the problem disappears:
<a href="https://i.stack.imgur.com/m9PXk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/m9PXk.png" alt="zoom into the figure 2"></a> </p>
<p>Any ideas why that happens? I use Mathematica for MacOS v. 11.0.1.0.</p>
<p>Can anyone with MMA v11 reproduce the issue?</p>
| Simon Rochester | 8,253 | <p>The default grid lines style uses uses partially transparent grid lines:</p>
<pre><code>InputForm@AbsoluteOptions[fig, GridLinesStyle]
(* {GridLinesStyle -> Directive[GrayLevel[0.5, 0.4]]} *)
</code></pre>
<p>where the second argument of <code>GrayLevel</code> specifies opacity. As pointed out by george2079 in the comments, this causes the exported eps file to be partially rasterized. </p>
<p>Specifying <code>GridLinesStyle -> Thick</code> removes the default specification, fixing the problem, but changing the appearance. You can keep the original appearance with <code>GridLinesStyle -> GrayLevel[.8]</code>:</p>
<pre><code>fig = Show[ListPolarPlot[Table[{a, .9}, {a, 0, 2 Pi, .001}],
PlotRange -> {{0, 1.1}, {0, 1.1}}, GridLinesStyle -> GrayLevel[.8],
GridLines -> Automatic, PlotStyle -> PointSize[.005]],
Graphics[Inset[ListPolarPlot[Table[{a, .1}, {a, 0, 2 Pi, .001}], Axes -> False,
PlotStyle -> PointSize[.05]], {.4, .4}]]]
</code></pre>
<p><a href="https://i.stack.imgur.com/Tx6hM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Tx6hM.png" alt="enter image description here"></a></p>
|
126,270 | <p>How many seven-digit numbers divisible by 11 have the sum of their digits equal to 59?</p>
<p>I am able to get the seven-digit numbers divisible by 11 </p>
<p>and </p>
<p>I am also able to get the seven-digit numbers whose sum of their digits equal to 59.</p>
<p>But i am not able to get how i can get the count of 7 digit numbers satisying both the condition.</p>
<p>Thanks in advance.</p>
<p>Thanks in advance.</p>
| davin | 2,881 | <p>If we look at the number $a_6a_5a_4a_3a_2a_1a_0$:
$$a_6a_5a_4a_3a_2a_1a_0 = a_610^6+a_510^5+ \dots +a_010^0 \\ \equiv a_6(-1)^6 + \dots+a_0(-1)^0 \mod 11 = a_6 -a_5+a_4-a_3+a_2-a_1+a_0 \mod 11$$
So if the number will be divisible by 11 we require
$$a_6 -a_5+a_4-a_3+a_2-a_1+a_0 = 11m$$</p>
<p>We take note that $0 \leq a_i \leq 9$ so possible values of $11m$ are limited to:
$$-27 = 4\cdot0-3\cdot9\leq 11m \leq 4\cdot 9-3\cdot0 = 36$$
Which really means that $11m \in \{-22,-11,0,11,22,33\}$</p>
<p>The other requirement of the question was that $a_6 + \dots +a_0 = 59$. From this and the above equation (not sure how to number them and align them nicely in TeX) we add and subtract and get much nicer equations:</p>
<p>$$a_6 +a_4+a_2+a_0 = \frac{59+11m} 2
\\a_5+a_3+a_1 =\frac{ 59-11m}2$$</p>
<p>Of course the LHS is whole, so the right hand side must be as well, which means $m$ needs to be odd. So we reduce our options to:</p>
<p>$$11m \in \{-11,11,33\} \implies \frac{59+11m} 2 \in \{24,35,46\}$$
Of course the sum of four digits can't be $46$ from our above inequality on $a_i$, and similarly
$$\frac{59-11m} 2 \in \{35,24\}$$ But the sum of three digits can't be $35$, so we're left with
$$a_6 +a_4+a_2+a_0 = 35
\\a_5+a_3+a_1 =24$$</p>
<p>It's easy to see that the only options for the four digits is a permutation of $9998$, and the three digits must be a permutation of one of $\{699,789,888\}$.</p>
<p>Order doesn't matter, so basic combinatorics gives $4\cdot(3 + 3! + 1) = 40$ such numbers.</p>
|
307,466 | <p>Is there any simple way to construct an entiere function $f$ such that :
$$\forall p \in {\mathbb N} \quad f(2^p)=(-1)^p$$</p>
| Davide Giraudo | 9,849 | <p>By <a href="http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem" rel="nofollow">Weierstrass products</a>, for each integer $k$ we can find an entire function $f_k$ such that $f_k(2^j)=0$ if $k\neq j$ and $f_k(2^k)=(-1)^k2^k$. Define $f:=\sum_{k=0}^{+\infty}2^{-k}f_k$. Using the relation about elementary factors, we can see that the convergence of this series is uniform over compact sets. Hence this defines an entire function which does the job. </p>
|
606,656 | <p>So I heard this a long time ago and I recently started thinking about it again. So I was told that the complex function $f(z)=1/z$ maps everything inside a circle to points outside the circle (the remaining part of the complex plane). Why is this?</p>
<p>I realize we can write $$f(z)=\frac{1}{z}=\frac{\overline{z}}{|z|^2}$$ so that there is a reflection $\overline{z}$ and a dilation $1/|z|^2$. But I don't see why this then only maps to points outside the circle. Can you explain?</p>
| abiessu | 86,846 | <p>We can also write the numbers $z$ in polar form as $re^{i\theta}$, in which case $$f(z)=f(re^{i\theta})=\frac {e^{-i\theta}}r$$ So the angle is negated and the radius is inverted. This means that $r\lt1\implies \frac 1r\gt 1$, and every complex point is mirrored across the real axis, since $e^{-i\theta}$ preserves the real component (due to $\cos$ mapping a negative angle to the same value as the absolute value of that angle) and negates the imaginary component (since $\sin (-x)=-\sin x$).</p>
|
3,272,030 | <p>Let's X, Y are random variables (i.e. each one maps elements of a sample space to real numbers). In particular, let's X is such that <span class="math-container">$X:\Omega \rightarrow \mathcal{R}$</span>, where <span class="math-container">$\Omega=\{\omega_1, ..., \omega_n \}$</span> and <span class="math-container">$X(\omega_i)=i$</span></p>
<p>Then the conditional variance <span class="math-container">$Var(Y|X(\omega_i))$</span> is a random variable because it maps an element of a sample space of <span class="math-container">$\Omega$</span> to <span class="math-container">$\mathcal{R}$</span>.</p>
<p>Now consider <span class="math-container">$Var(Y|X(\omega_i)=1 \bigcup X(\omega_i)=2)$</span> This expression maps a set <span class="math-container">$\{\omega_1, \omega_2 \}$</span> to <span class="math-container">$\mathcal{R}$</span>. So, it seems no more to satisfy the definition of a random variable. Then what is it? A measure? Perhaps no.
Thank you for answering.</p>
| Paul Childs | 584,354 | <p>Actually as <span class="math-container">$\{ \omega_1, \omega_2 \} \in \Omega \oplus \Omega$</span> and that the latter is an equally valid sample space, it would <em>no less</em> satisfy the definition of a random variable.</p>
|
3,197,301 | <p>Let <span class="math-container">$OABC$</span> be a tetrahedron such that <span class="math-container">$|OA|=|OB|=|OC|$</span>. Denote by <span class="math-container">$D$</span> and <span class="math-container">$E$</span> the midpoints of segments <span class="math-container">$AB$</span> and <span class="math-container">$AC$</span> respectively. If <span class="math-container">$\alpha=\angle(DOE)$</span> and <span class="math-container">$\beta=\angle(BOC)$</span> what is the ratio <span class="math-container">$\beta/\alpha$</span>? </p>
<p>It is obvious that <span class="math-container">$|BC|=2|DE|$</span> since triangles <span class="math-container">$\Delta(ABC)$</span> and <span class="math-container">$\Delta(ADE)$</span> are similar and <span class="math-container">$D, E$</span> are midpoints by assumption. I would expect that <span class="math-container">$\beta/\alpha\geqslant 2$</span> but I haven't been able to confirm this. I tried to use the cosine law for the segments <span class="math-container">$|BC|$</span> and <span class="math-container">$|DE|$</span> but without success. </p>
| Jack D'Aurizio | 44,121 | <p>Your set is the inverse image of <span class="math-container">$(0,1)$</span> with respect to a continuous function (the distance from the origin). Since <span class="math-container">$(0,1)$</span> is an open set and the mentioned function is continuous, your set is open too.</p>
|
279,238 | <p>I want to create a regular polygon from the initial two points <span class="math-container">$A$</span>, <span class="math-container">$B$</span> and number of vertices <span class="math-container">$n$</span>,<br />
<a href="https://i.stack.imgur.com/dKr9A.png" rel="noreferrer"><img src="https://i.stack.imgur.com/dKr9A.png" alt="enter image description here" /></a></p>
<p><code>regularPolygon[{0, 0}, {1, 0}, 3]</code> gives <code>{{0, 0}, {1, 0}, {1/2, Sqrt[3]/2}}</code></p>
<p><code>regularPolygon[{x1, y1}, {x2, y2}, 4]</code> gives <code>{{x1, y1}, {x2, y2}, {x2 + y1 - y2, -x1 + x2 + y2}, {x1 + y1 - y2, -x1 + x2 + y1}}</code></p>
<p>I found a related function <a href="http://reference.wolfram.com/language/ref/CirclePoints.html" rel="noreferrer">CirclePoints</a>, it seems not suitable. Is there a simple way to implement such a function? Maybe you can use iteration.</p>
| Daniel Huber | 46,318 | <p>Corroborating cvgm interesting work, we may improve a bit. {x1,y1} is not on the polygon and the winding direction is opposite to the positive mathematical winding direction. This may be fixed by:</p>
<pre><code>{x1, y1} = {2, 3};
{x2, y2} = {5, 5};
length = Norm[{x2, y2} - {x1, y1}];
n = 5;
Graphics[{Polygon[
AnglePath[{{x2, y2}, {x2, y2} - {x1, y1}},
ConstantArray[{length, 2 Pi/n}, n]]],
Text["Point 1", {x1, y1 - 0.3}], Text["Point 2", {x2, y2 - 0.3}],
PointSize[0.02], Red, Point[{{x1, y1}, {x2, y2}}]}, Axes -> True]
</code></pre>
<p><a href="https://i.stack.imgur.com/ehlW8.png" rel="noreferrer"><img src="https://i.stack.imgur.com/ehlW8.png" alt="enter image description here" /></a></p>
|
469,521 | <p>$\theta$, $\phi$ are integrable random variables on a probability space $(\Omega,\mathcal{F},P)$ and $\mathcal{G}$ is $\sigma$-field on $\Omega$ contained in $\mathcal{F}$.
Now we want to prove $E(\theta\mid\mathcal{G})=E(\theta)$ if $\theta$ is independent of $\mathcal{G}$. The proof is, for any $B\in \mathcal{G}$, by independence, $\theta$ and $1_{B}$ are independent. And,$$\int_{B}E(\theta)dP=E(\theta)E(1_{B})=E(\theta 1_{B})=\int_{B}\theta dP,$$ and the conclusion follows. I'm really confused with the first equality $\int_{B}E(\theta)dP=E(\theta)E(1_{B})$. Can anyone explain this to me? Thanks!</p>
| Jacky Zhang | 89,915 | <p>$$\int_{B}E[\theta]dP=E[E[\theta]I_{B}]=E[I_{B}]E[\theta]$$,that's all.</p>
|
3,232,766 | <blockquote>
<p>Is <span class="math-container">$10^{100}$</span> (Googol) bigger than <span class="math-container">$100!$</span>?</p>
<p>If <span class="math-container">$10^{100}$</span> is called as Googol, does <span class="math-container">$100!$</span> have any special name to be called, apart from being called as "100 factorial"?</p>
</blockquote>
<hr />
<p>I ask this question because I get to know about the number <span class="math-container">$10^{100}$</span> on how big it is more often than <span class="math-container">$100!$</span>. If <span class="math-container">$100!$</span> is bigger than <span class="math-container">$10^{100}$</span>, then why don't we give more focus to <span class="math-container">$100!$</span> than the other number? Because for me, <span class="math-container">$100!$</span> looks simple.</p>
| Paolo Bordignon | 664,304 | <p>With simple ineqalities we have:</p>
<p><span class="math-container">$100!\geq 90^{10}\cdot 80^{10}\cdots 20^{10}\cdot 10^{10}$</span></p>
<p><span class="math-container">$100!\geq (9\cdot 8 \cdots 2 \cdot 1)^{10}\cdot 10^{90}$</span></p>
<p><span class="math-container">$100!\geq (9!)^{10}\cdot 10^{90}>10^{100}$</span></p>
|
549,245 | <p>Apparently, it is a common mistake to write $\forall x[R(x) \rightarrow P(x)]$ instead of $\forall x[R(x)\wedge P(x)]$ in some cases, however I can't seem to find the difference between the two. </p>
<p>I did some research in ProofWiki and a Discrete Mathematics book, but I couldn't find this specific difference explained.</p>
<p>Could you please explain it to me and give some examples on when to use one and when to use the other?</p>
<p>Thank you for your time. </p>
| Shaun | 104,041 | <p>Assume $[\forall x[Rx\to Px]]\leftrightarrow [\forall x[Rx\wedge Px]]$ and suppose $ \forall x[Rx\to Px] $. Then $ \forall x[Rx\wedge Px] $. Instantiate with $a$ for $x$. Now $Ra\to Pa$ and $Ra$, $Pa$. But then $\neg Ra$ is an option, a contradiction.</p>
<p>You can see these things clearly with tableaux.</p>
<p><img src="https://i.stack.imgur.com/n6Kjz.jpg" alt="Tableau"> </p>
|
1,017,320 | <p>Assume ZFC. Let $B\subseteq\mathbb R$ be a set that is not Borel-measurable. Clearly, $B$ must be uncountable, since countable sets are always Borel being a countable union of measurable singletons.</p>
<p><strong>Question:</strong> can one conclude that $B$ necessarily has the cardinality of the continuum <em>without</em> assuming either the continuum hypothesis or the negation thereof?</p>
<p>A possibly related result is that any $\sigma$-algebra that contains infinitely many <em>sets</em> must necessarily have at least the cardinality of the continuum. This result is independent of the continuum hypothesis.</p>
| Asaf Karagila | 622 | <p>No, you can't make those inferences. The Borel sets satisfy the continuum hypothesis. Namely any uncountable Borel set has necessarily the cardinality of the continuum.</p>
<p>If $B$ is not Borel measurable, in particular it is not a Borel set. All you can conclude about it is that it is uncountable. If the continuum is large, and $B$ is any cardinality between $\aleph_0$ and $2^{\aleph_0}$ then it is not Borel; and on the other hand there are plenty of non-Borel measurable (and not even Lebesgue measurable sets) of size continuum, for example Vitali sets.</p>
<hr>
<p>In fact even if you replace Borel by Lebesgue you can't say much more. Martin's axiom implies that any set of size $<2^{\aleph_0}$ has Lebesgue measure zero, and in particular each non-measurable set must have size continuum. On the other hand adding an uncountable number of Random reals to any model of $\sf ZFC+GCH$ will add a non-Lebesgue measurable set of size $\aleph_1$, while possibly blowing up the continuum to be much larger. (Note that Random reals are a technical set theoretic term, not just "arbitrary reals chosen at random".)</p>
|
1,037,632 | <p>How to find the last 2 digits of $2014^{2001}$? What about the last 2 digits of $9^{(9^{16})}$?</p>
| Anurag A | 68,092 | <p><strong>Hint:</strong></p>
<p>For finding last two digits you need to reduce this modulo $100$. That is you ne need to find
$$2014^{2001} \equiv ? \pmod{100}.$$
This is the same as asking
$$14^{2001} \equiv ? \pmod{100}.$$</p>
<p>Now in order to facilitate computation, you need to use <a href="http://en.wikipedia.org/wiki/Euler%27s_theorem" rel="nofollow">Euler's Theorem</a>. But keep in mind that $\gcd(14,100) = 2.$ So you need to adjust things a bit.</p>
|
4,056,627 | <p>Equation to minimize using Boolean Algebra Laws:
<span class="math-container">$A+\bar{A} B \bar{C}$</span></p>
<p>I have tried doing this but i am unsure of the answer:
<span class="math-container">$$
\begin{array}{l}
\text { Let } K=B \bar{C} \\
A+\bar{A} K=A+K=A+B \bar{C}
\end{array}
$$</span></p>
| kabenyuk | 528,593 | <p>If <span class="math-container">$|G:H|<9$</span>, then since <span class="math-container">$G$</span> is a simple group, it is isomorphic to a subgroup of the symmetric group <span class="math-container">$S_8$</span>. But the group <span class="math-container">$S_8$</span> has a Sylow <span class="math-container">$3$</span>-subgroup of order <span class="math-container">$9.$</span> We have a contradiction.</p>
|
597,610 | <p>can someone explain why when a radical equation is solved some solutions don't work? Is there a rule when a certain transformation is performed on the equation it gives extra solutions? thanks very much</p>
| Cameron Buie | 28,900 | <p>In general, when a transformation is <a href="http://en.wikipedia.org/wiki/One-to-one_function" rel="nofollow"><em>one-to-one</em></a> (or <em>injective</em>), then it is reversible, so results in an equivalent equation. The problem with some radical equations is (for example) that $(-1)^2=1=1^2.$ That is, squaring both sides of a false equation (like $-1=1$) may result in a true equation, so we get extraneous solutions to deal with.</p>
|
120,863 | <blockquote>
<p>A perfect dice is drawn $4$ times.What's the probability to the same number comes out at least $2$ times?</p>
</blockquote>
<p>At first, I applied to binomial law.I made all calculations.</p>
<p>I set up a random variable.The possible values for the variable were $0,1,2,3$ and $4$.
My thought was that any number have a probability of $\frac{1}{6}$ to come out.</p>
<p>Then I started to think.In the $4$ launchs two different numbers can come out $2$ times, each one.So my previous thoughts were wrong.</p>
<p>Can you give me an idea on how to solve this problem?</p>
| Michael Hardy | 11,667 | <p>If you find the probability that a $1$ appears at least two times, and similarly a $2$, and a $3$, etc., then you've got the complication that those six events are not mutually exclusive: a $1$ could appear twice <b>and</b> a $2$ twice.</p>
<p>So the simpler way is to find the probability that the event you're looking for does <b>not</b> occur, i.e. no number appears more than once.</p>
<p>The probability that the second number differs from the first is $\dfrac 5 6$.</p>
<p>The probability that the third number differs from the first two is $\dfrac 4 6$.</p>
<p>The probability that the fourth number differs from the first three is $\dfrac 3 6$.</p>
<p>Multiply those: $\dfrac{5\cdot4\cdot3}{6^3} = \dfrac{5}{18}$.</p>
<p>So what you're looking for is $\dfrac{13}{18}$.</p>
|
3,270,504 | <blockquote>
<p>Prove that <span class="math-container">$\log|e^z-z|\leq |z|+1$</span> where <span class="math-container">$z\in\mathbb{C}$</span> with <span class="math-container">$|z|\geq e$</span>.</p>
</blockquote>
<p><strong>Background:</strong></p>
<p>This is from a proof that <span class="math-container">$e^z-z$</span> has infinitely many zeroes. The present stage is that we assumed in contradiction that <span class="math-container">$e^z-z$</span> hasn't any zero.</p>
<p><strong>My attampt:</strong></p>
<p>I assume that the meaning of <span class="math-container">$\log$</span> here is the principal branch of <span class="math-container">$\log$</span>.</p>
<p>We know that <span class="math-container">$|w|\in\mathbb{R} ,\ \forall w\in\mathbb{C}$</span>. Because <span class="math-container">$\log$</span> is increasing in <span class="math-container">$\mathbb{R}^+$</span> and according to the triangle inequality we get <span class="math-container">$$\log|e^z-z|\leq\log(|e^z|+|z|)$$</span> But I'm not sure how to proceed. Thanks.</p>
| Chrystomath | 84,081 | <p><span class="math-container">\begin{align*}
|e^z-z|&\le|e^z|+|z|\\
&\le e^{|z|}+|z|\quad\textrm{from series expansion}\\
&\le e^{|z|+1}\quad\textrm{again from series expansion}
\end{align*}</span></p>
|
3,281,965 | <p>It is known that a way to check whether a number <span class="math-container">$n$</span> is prime, is to check for divisors of <span class="math-container">$n$</span> from <span class="math-container">$2$</span> to <span class="math-container">$\lfloor\sqrt{n}\rfloor$</span>. If we find any divisor, then <span class="math-container">$n$</span> is not prime. If we don't, then we don't need to check for divisors bigger than <span class="math-container">$\sqrt{n}$</span> (and <span class="math-container">$n$</span> is prime).</p>
<p>An "approximation" of this method would be to check for divisors the same way but from <span class="math-container">$2$</span> to <span class="math-container">$log_2n$</span>. If we find any divisor we declare that <span class="math-container">$n$</span> is not prime. If we don't find any divisor we declare that <span class="math-container">$n$</span> is prime. Of course this method will not always give the correct results. My question is:</p>
<p>If the second algorithm declares a number to be prime, what is the probability that this number is actually prime?</p>
| Robert Israel | 8,508 | <p>Suppose <span class="math-container">$n$</span> is a random positive integer from <span class="math-container">$1$</span> to <span class="math-container">$N$</span>. By the Prime Number Theorem, the probability that it is prime is on the order of <span class="math-container">$1/\log N$</span>. On the other hand, for each
prime <span class="math-container">$p < \log N$</span> the probability that it is not divisible by <span class="math-container">$p$</span> is <span class="math-container">$1 - 1/p$</span>, so the probability that it has no prime factors <span class="math-container">$< \log N$</span> is on the order of
<span class="math-container">$$ \prod_{p \le \log N} (1 - 1/p) \approx \exp \left( - \sum_{k=2}^{\log N} 1/(k \log k) \right) \approx \exp(- \log\log\log N) = \frac{1}{\log \log N}$$</span>
Thus the probability that a number that passes the test is prime is on the order of
<span class="math-container">$$ \frac{\log \log N}{\log N} $$</span></p>
|
1,647,442 | <p>Prof using the binomial theorem: for all integers $n ≥0$ and for all nonnegative real numbers $x$, $1+nx ≤(1+x)^n$. </p>
<p>Don't have a idea to start this one. I don't know how to use math induction yet, so I need a answer without that.</p>
| Thomas | 26,188 | <p>To take $10\%$ of a quantity $X$, you multiply $X$ by $\frac{1}{10}$. So, for example, $10\%$ of $20$ is $\frac{1}{10}\cdot 20 = 2$.</p>
|
2,859,411 | <p>Let $y(t)$ be a real valued function defined on the real line such that
$y'= y(1 − y)$, with $y(0) \in [0, 1]$. Then $\lim_{t\to\infty} y(t) = 1$.</p>
<p>The solution is given as false .But i have no idea about that. I try some counterexample but it won't work.</p>
<p>How can i find solution with 3 minutes?</p>
| mechanodroid | 144,766 | <p>$$\frac{dy}{dx} = y(1-y) \implies dx = \frac{dy}{y(1-y)}$$</p>
<p>so $$x+C = \ln\frac{y}{1-y} \implies Ce^x = \frac{y}{1-y}$$</p>
<p>and therefore $$y = \frac{Ce^x}{1+Ce^x}$$</p>
<p>The initial condition is</p>
<p>$$y(0) = \frac{C}{1+C} \implies C = \frac{y(0)}{1-y(0)}$$</p>
<p>so $$y = \frac{\left(\frac{y(0)}{1-y(0)}\right)e^x}{1+\left(\frac{y(0)}{1-y(0)}\right)e^x}$$</p>
<p>If we pick $y(0) = 0$ then $y = 0$ so $\lim_{t\to\infty} y(t) = 0 \ne 1$.</p>
|
161,766 | <p>What is the best way to write a quick function which takes in a vertex (of some undirected graph), and <strong>gives the value of the k-Core's in which that vertex is in?</strong></p>
<p>The obvious modification of <code>kCoreComponents[Graph,k]</code> using <code>Intersection</code> is either taking ages, since it needs to re-check the core composition for each vertex (I have over 7000), or spends ages "loading" after the script is complete.</p>
| Jason B. | 9,490 | <p>I can somewhat reproduce this error. Except for me, if I try to export a graphic with no display set, the kernel just hangs. The answer is to use an <a href="https://en.wikipedia.org/wiki/Xvfb" rel="noreferrer">Xvfb display server</a>, as described <a href="http://elementalselenium.com/tips/38-headless" rel="noreferrer">here</a>.</p>
<p>I have the file, <code>exportSVG.m</code>, with the contents</p>
<pre><code>plot = Plot[x, {x, -3, 3}];
Export["/home/jasonb/some.svg", plot];
</code></pre>
<p>Then another file, <code>exportSVG.sh</code>, </p>
<pre><code>#!/usr/bin/env bash
xvfb-run /usr/local/bin/wolframscript -script /home/jasonb/exportSVG.m
</code></pre>
<p>If I add this to my <code>crontab</code> with out the <code>xvfb-run</code>, it hangs. If I add in that command, it works. You'll need to ensure you have it installed, check via <code>which xvfb-run</code>.</p>
|
3,173,125 | <p>Let <span class="math-container">$f(x) = x^3 + a x^2 + b x + c$</span> and <span class="math-container">$g(x) = x^3 + b x^2 + c x + a\,$</span> where <span class="math-container">$a, b, c$</span> are integers and <span class="math-container">$c\neq 0\,$</span>. Suppose that the following conditions hold:</p>
<ol>
<li><span class="math-container">$f(1)=0$</span> </li>
<li>The roots of <span class="math-container">$g(x)$</span> are squares of the roots of <span class="math-container">$f(x)$</span>.</li>
</ol>
<p>I'd like to find <span class="math-container">$a, b$</span> and <span class="math-container">$c$</span>.</p>
<p>I tried solving equations made using condition 1. and relation between the roots, but couldn't solve.
The equation which I got in <span class="math-container">$c$</span> is <span class="math-container">$c^4 + c^2 +3 c-1=0$</span> (edit: eqn is wrong).
Also I was able to express <span class="math-container">$a$</span> and <span class="math-container">$b$</span> in terms of <span class="math-container">$c$</span>. But the equation isn't solvable by hand. </p>
| lhf | 589 | <p><em>Hint:</em> Condition 2 can be expressed as <span class="math-container">$f(x)$</span> divides <span class="math-container">$g(x^2)$</span>.</p>
|
1,614,111 | <p>Is there a sequence of positive real numbers $x_1,\ldots,x_n$ for which
$$
\sum_{1\leq i,j\leq n}\left[\frac{\sqrt{ij}}{2}-\min(i,j)\right]x_ix_j> 0?
$$</p>
| Jason | 195,308 | <p>The answer is "yes". Consider the matrix <span class="math-container">$A_n=[a_{ij}]_{1\le i,j\le n}$</span>, defined by</p>
<p><span class="math-container">$$ a_{ij} = \min(i,j) - \frac{\sqrt{ij}}2. $$</span></p>
<p>As mentioned in a comment, this is positive definite for small <span class="math-container">$n$</span>, but it turns out not to be the case always. Since the <span class="math-container">$k^\text{th}$</span> minor of <span class="math-container">$A_n$</span> is simply <span class="math-container">$A_k$</span>, <a href="https://en.wikipedia.org/wiki/Sylvester%27s_criterion" rel="nofollow noreferrer">Sylvester's criterion</a> implies that <span class="math-container">$A_n$</span> being positive definite is equivalent to <span class="math-container">$\det(A_k) > 0$</span> for <span class="math-container">$k\le n$</span>. By implementing the following Python code:</p>
<pre><code>import numpy as np
def special_matrix(n):
'''Define the matrix A_n'''
matrix = np.identity(n)
for i in range(n):
for j in range(n):
matrix[i,j] = min(i+1,j+1) - np.sqrt((i+1)*(j+1)) / 2
# Note we use i+1, j+1 since Python indexes starting at 0
return matrix
for n in range(2,100):
if np.linalg.det(special_matrix(n)) < 0:
print(n)
break
>>> 56
</code></pre>
<p>we see that <span class="math-container">$\det(A_n) \ge 0$</span> for <span class="math-container">$n < N:= 56$</span>, and <span class="math-container">$\det(A_N) < 0$</span>. This means <span class="math-container">$A_N$</span> is <em>not</em> positive definite. In particular, <span class="math-container">$A_N$</span> has a negative eigenvalue. If this eigenvalue has a corresponding positive eigenvector <span class="math-container">$\mathbf x = (x_1,\ldots,x_N)$</span>, then clearly this will provide the desired sequence. We then implement the following code:</p>
<pre><code>N = 56
Matrix = special_matrix(N)
evals, evecs = np.linalg.eig(Matrix)
for i in range(N):
if evals[i] <= 0:
evec = -evecs[:,i]
# Negate the eigenvector, since by default it returns an eigenvector
# with all negative entries
print(evec)
break
>>> [0.97135177 0.17110863 0.08503891 0.05411931 0.03839335 0.02907463
0.02301033 0.018801 0.01573758 0.01342565 0.0116299 0.01020206
0.00904451 0.00809062 0.00729348 0.00661925 0.00604291 0.00554566
0.00511308 0.00473398 0.00439953 0.0041027 0.00383783 0.00360028
0.00338627 0.00319267 0.00301683 0.00285657 0.00271002 0.00257558
0.0024519 0.00233782 0.00223232 0.00213452 0.00204366 0.00195907
0.00188016 0.00180641 0.00173736 0.0016726 0.00161176 0.00155453
0.0015006 0.00144973 0.00140166 0.00135619 0.00131313 0.0012723
0.00123355 0.00119672 0.00116169 0.00112834 0.00109669 0.00107966
0.00226057 0.11233599]
</code></pre>
<p>We have found a positive eigenvector <span class="math-container">$\mathbf x$</span>, so it provides the desired sequence. Let us double check that this really does make the desired sum positive:</p>
<pre><code>desired_sum = 0
for i in range(N):
for j in range(N):
desired_sum += (np.sqrt((i+1)*(j+1)) / 2 - min(i+1,j+1))*evec[i]*evec[j]
# Note again the use of i+1 and j+1
print(desired_sum)
>>> 0.011233898661515196
</code></pre>
<p>confirming our answer.</p>
<p>Note that, since we also incidentally demonstrated that such a sequence cannot have length shorter than <span class="math-container">$56$</span>, it seems unlikely that there will be a nice analytic (i.e. non-numerical) solution to this problem.</p>
|
2,880,271 | <p>Let a, b, c and d be real numbers that are not all zero. Let
ax + by = p
cx + dy = q
be a pair of equations in the variables x and y with p, q ∈ R. </p>
<p>Show this system of equations has a unique solution if and only if ab − cd != 0.</p>
<hr>
<p>From Determinant of coefficient matrix, I know (ad -bc) =0 => no unique solution.
Have tried substitution of one equation into another and replacement.</p>
<p>=> ab = cd....show that solution is unique</p>
<p><= solution is unique ....show that ab - cd != 0</p>
<p>Pointers?</p>
| mfl | 148,513 | <p>Assume $ad-bc\ne 0.$ Then </p>
<p>$$\begin{array}{l}ax+by=p\\cx+dy=q\end{array}\implies \begin{array}{l}adx+bdy=dp\\bcx+bdy=bq\end{array}\implies x=\dfrac{dp-bq}{ad-bc}.$$ In a similar way you can get $y:$</p>
<p>$$\begin{array}{l}ax+by=p\\cx+dy=q\end{array}\implies \begin{array}{l}acx+bcy=cp\\acx+ady=aq\end{array}\implies y=\dfrac{aq-cp}{ad-bc}.$$</p>
<p>Conversely, suppose that the system has a solution $(x,y).$ If $ab-cd=0$ then the solution is not unique because $(x+d,y-c)$ and $(x+b,y-a)$ are also solutions. So, if the solution is unique it must be $ab-cd\ne 0.$</p>
|
1,549,506 | <p>To Prove $$\lim_{x \to 0}\frac{x^2+2\cos x-2}{x\sin^3 x}=\frac{1}{12}$$
I tried with L'Hospital rule but in vain.</p>
| Idris Addou | 192,045 | <p>We can use the basic limits
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\sin x}{x}=1,\ \ \ \ \ \ \ \ \
\lim\limits_{x\rightarrow 0}\dfrac{\cos x-1+\frac{1}{2}x^{2}}{x^{4}}=\dfrac{1%
}{24}
\end{equation*}
as follows
\begin{equation*}
\lim\limits_{x\rightarrow 0}\frac{x^{2}+2\cos x-2}{x\sin ^{3}x}%
=\lim\limits_{x\rightarrow 0}\left( \frac{x}{\sin x}\right)
^{3}\lim\limits_{x\rightarrow 0}\left( \frac{\cos x-1+\frac{1}{2}x^{2}}{x^{4}%
}\right) \left( 2\right) =1^{3}\cdot \frac{1}{24}\cdot 2=\frac{1}{12}.
\end{equation*}</p>
<p>The basic limits can be computed by repeated use of L'Hospital's rule. </p>
|
423,793 | <p>A group of 3141 students gather together. Some of them have 13 friends in this group, some have 33 friends, and the rest has 37 friends. Prove using graph theory that this group does not exist. Assume that if A is friends with B, then B is friends with A.</p>
| Nick Peterson | 81,839 | <p>This seems like homework; so, let me just give you a hint.</p>
<p>Say that you have $a$ people with 13 friends, $b$ with 33, and the remaining $3141-a-b$ have 37. In the graph, this corresponds to having $a$ vertices of degree 13, $b$ vertices of degree $33$, and the rest of degree 37.</p>
<p>So, the total degree of the graph is
$$
13a+33b+37(3141-a-b).
$$
Simplify this, and see if you can prove that no matter how you choose $a$ and $b$, the resulting total degree will always be odd. Why is this impossible in a graph?</p>
|
229,748 | <p>I have an integral given by <code>intS[zs,c]</code> and I want to plot the expression <code>functionS</code> (which contains <code>intS[zs,c]</code>) with respect to another integral function <code>inta[zs,c]</code> and <code>zs</code>. However, the values of <code>inta[zs,c]</code> are extremely small compared to <code>functionS</code> so in the <code>ParametricPlot3D</code> I got a badly looking plot that looks smashed.</p>
<p>My code is as follows,</p>
<pre><code>d = 4;(*dimensions*)
inta[zs_?NumericQ, c_?NumericQ] := NIntegrate[((c zs^(d + 1))/(2 d)) (y^((1 - d)/(2 d)))/(1 - c^2 zs^(2 d) y)^(1/2), {y, 0, 1}]
intS[zs_?NumericQ, c_?NumericQ] := NIntegrate[((c^2 zs^(2 d))/(d - 1)) (y^d)/(1 - c^2 zs^(2 d) y^(2 d))^(1/2), {y, 0, 1}]
functionS = ((-(1 - c^2 zs^(2 d))^(1/2))/(d - 1) - intS[zs, c] + 1/zs^(d - 1));
function = Log[10, functionS];
ParametricPlot3D[{zs, inta[zs, c], function}, {zs, 0, 10}, {c, 0, 10}, PlotStyle -> {LightBlue}, PlotRange -> Full, LabelStyle -> Directive[Bold, Medium], ImageSize -> Large] // Quiet
</code></pre>
<p><a href="https://i.stack.imgur.com/hUP27.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hUP27.jpg" alt="Image" /></a></p>
| Sumit | 8,070 | <p>You need to fix <code>BoxRatios</code> for 3d plots like this. I am going to show with a simple example</p>
<pre><code>ParametricPlot3D[{zs, zs^2, zs + c}, {zs, 0, 10}, {c, 0, 10},
PlotStyle -> {LightBlue}, PlotRange -> Full,
LabelStyle -> Directive[Bold, Medium], ImageSize -> 300]
</code></pre>
<p><a href="https://i.stack.imgur.com/kIsGq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kIsGq.png" alt="enter image description here" /></a></p>
<p>Now adjust equal length for each axis with <code>BoxRatios->{1,1,1}</code></p>
<pre><code>ParametricPlot3D[{zs, zs^2, zs + c}, {zs, 0, 10}, {c, 0, 10},
PlotStyle -> {LightBlue}, PlotRange -> Full,
LabelStyle -> Directive[Bold, Medium], ImageSize -> 300,
BoxRatios -> {1, 1, 1}]
</code></pre>
<p><a href="https://i.stack.imgur.com/6Lv4o.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6Lv4o.png" alt="enter image description here" /></a></p>
|
1,656,963 | <p>If $(a_n) \to 0$, then applying the algebraic limit theorem, what is $ \lim_\limits{n\to\infty} \frac {1+2a_n}{1+3a_n - 4a^2_n}$.</p>
<p>Would just be able to do : $\lim_\limits{n\to\infty} \frac {1+2(0)}{1+3(0) - 4(0)}$. </p>
<p>$\lim \frac {1}{1} = 1$.</p>
| Tsemo Aristide | 280,301 | <p>Hint: The function $f(x) = {{1+2x}\over{1+3x-4x^2}}$ is continuous, so $f(a_n)$ converges towards $f(a)$.</p>
|
3,436,804 | <p>Does the following series converge? If yes, what is its value in simplest form?</p>
<p><span class="math-container">$$\left( \frac{1}{1} \right)^2+\left( \frac{1}{2}+\frac{1}{3} \right)^2+\left( \frac{1}{4}+\frac{1}{5}+\frac{1}{6} \right)^2+\left( \frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10} \right)^2+\dots$$</span></p>
<p>I have no idea how to start. Any hint would be really appreciated. THANKS!</p>
| Jam | 161,490 | <h3>Suggestion.</h3>
<p>This is merely a suggestion that is too large to fit in a comment. Euler's integral formula for the harmonic numbers, <span class="math-container">$H_n=\int_0^1\frac{1-x^n}{1-x}\ \mathrm{d}x$</span>, gives us a formula for the series as <span class="math-container">$a\to1^-$</span>:</p>
<p><span class="math-container">$$S=4\sum_{n=1}^{\infty}\left[f_a(n(n+1))-f_a(n(n-1))\right]^2$$</span></p>
<p>where <span class="math-container">$f_a(x)=\int_{0}^{a}\frac{t^{x+1}}{1-t^{2}}\ \mathrm{d}t$</span>. The function, <span class="math-container">$f_a$</span>, can also be given in terms of the hypergeometric function, as <span class="math-container">$\frac{a^{x+2}}{x+2}\cdot{_2F}_1\left(1,\frac{x}2+1;\frac{x}2+2;a^{2}\right)$</span>.</p>
<p>Thus, it could potentially aid in finding an analytic solution to <span class="math-container">$S$</span> (or showing whether that would be impossible) by finding a closed form for the square of the difference of hypergeometric functions, <span class="math-container">$\left[f_a(n(n+1))-f_a(n(n-1))\right]^2$</span>. However, no such solution has leapt out to me so far.</p>
|
1,105,787 | <p>I was given an excercise in my calculus class that i don't really understand, the problem says : Find the area limited by the curves $$ y = \frac{x+4}{x^2+1} ,\space x = -2 ,\space x = 3,\space y = 0 $$</p>
<p>I don't really know what approach to follow here, my guess would be to solve it using riemann sums or maybe definite integrals and using $ x = -2 $ and $ x = 3 $ as the interval but i'm totally lost.</p>
| Tim Raczkowski | 192,581 | <p>Yes, this is a definite integral problem. You can solve it by evaluating</p>
<p>$$\int_{-2}^3{x+4\over x^2+1}\,dx.$$</p>
<p>Hint: Split it into</p>
<p>$$\int_{-2}^3\frac x{x^2+1}\,dx+4\int_{-2}^3\frac1{x^2+1}\,dx,$$</p>
<p>and recall that the derivative of $\tan^{-1}x$ is $\frac1{x^2+1}$.</p>
|
2,265,368 | <p>Let $R$ be a commutative ring with unit and $N$ its nilradical. If $R$ is connected then $R/N$ is also connected; the quotient map induces a homeomorphism between the spectra. I'd like to see a more hands-on proof, but I'm unable to make it work. That is to say, I'd like to show that if $e^2-e\in N$ for some $e\in R$, then $e\in N$ or $e-1\in N$. After scribbling full quite a couple of pages going around in circles, any idea is welcome.</p>
| egreg | 62,967 | <p>See <a href="http://www.math.hawaii.edu/~lee/algebra/idempotent.pdf" rel="nofollow noreferrer">http://www.math.hawaii.edu/~lee/algebra/idempotent.pdf</a>, but I'll repeat the argument here.</p>
<p>Suppose $e-e^2\in N$ and set $f=1-e$. Then $e^kf^k=0$, for some $k\ge1$.</p>
<p>Note that $e^k+N=e+N$ and $f^k+N=f+N$, so we can replace $e$ and $f$ with $e^k$ and $f^k$, so now $ef=0$.</p>
<p>Set $x=1-e-f\in N$; then $x^l=0$, for some $l\ge1$. Hence $1-x$ as the inverse $u=1+x+x^2+\dots+x^{l-1}$. Also $u-1\in N$ and
$$
ue+N=e+N,\qquad uf+N=f+N
$$
so we can replace $e$ and $f$ by $ue$ and $uf$.</p>
<p>Now $e+f=1$. So
$$
e=e(e+f)=e^2+ef=e^2
$$</p>
|
492,031 | <p>I am currently reading through Hatcher's Algebraic Topology book. I am having some trouble understanding the difference between a deformation retraction and just a retraction. Hatcher defines them as follows:</p>
<p>A <strong>deformation retraction</strong> of a space $X$ onto a subspace $A$ is a family of maps $f_t:X \to X$, $t \in I$, such that $f_0=\mathbb{1}$ (the identity map), $f_1(X)=A$, and $f_t|A=\mathbb{1}$ for all $t$.</p>
<p>A <strong>retraction</strong> of $X$ onto $A$ is a map $r:X \to X$ such that $r(X)=A$ and $r|A=\mathbb{1}$.</p>
<p>Is the notion of time the important characteristic that sets the two ideas apart? (It seems that the definition of deformation retraction utilizes time in its definition, whereas retraction seems to not.)</p>
<p>Any insight is appreciated. Also, if anyone have additional suggested reading material to help with concepts in Algebraic topology, that would be much appreciated.</p>
| Thomas Belulovich | 831 | <p>The difference between a retraction and a deformation retraction does have to do with the "notion of time" as you suggest. </p>
<p>Here's a strong difference between the two:</p>
<p>1) For any $x_0 \in X$, $\{x_0\} \subset X$ has a retract. Choose $r : X \to \{x_0\}$ to be the unique map to the one-point set. Then, certainly, $r(x_0) = x_0$.</p>
<p>2) However, $\{x_0\} \subset X$ only has a deformation retraction if $X$ is contractible. To see, why, notice there has to be a family of maps $f_t : X \to X$ such that $f_0(x) = x$, $f_1(x) = x_0$, and $f_{t}(x_0) = x_0$ for every $t$. This gives a homotopy from $id_X$ to the constant map at $x_0$, which makes $X$ contractible.</p>
<p>In fact, showing a deformation retract from $X$ onto a subspace $A$ always exhibits that $A$ and $X$ are homotopy equivalent, whereas $A$ being a retract of $X$ is weaker. (But often, still useful! Two spaces being homotopy equivalent is very strong indeed!)</p>
|
3,974,768 | <p>Suppose <span class="math-container">$a_n \rightarrow a$</span>, <span class="math-container">$b_n \rightarrow b$</span> and <span class="math-container">$ a < b$</span>.
Is it true that there is <span class="math-container">$ N \in \mathbb{N} $</span> such that for any <span class="math-container">$ m > N $</span>, <span class="math-container">$a_m < b_m $</span>?</p>
<p>I think it's trivial but couldn't prove it.</p>
| heropup | 118,193 | <p>Use <a href="https://en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem" rel="nofollow noreferrer">Rouché's theorem</a> with the choice <span class="math-container">$f(z) = z^7 + 1$</span> and <span class="math-container">$g(z) = 7z^4 + 4z$</span>. Then you need to show that for <span class="math-container">$z = 2e^{i\theta}$</span>; i.e., on the boundary of the disk of radius <span class="math-container">$2$</span> at the origin, <span class="math-container">$|g(z)| < |f(z)|$</span>. Then <span class="math-container">$f+g$</span> and <span class="math-container">$f$</span> have the same number of zeroes in the disk. Then because <span class="math-container">$f$</span> has all seven zeroes on the unit circle, the result follows.</p>
<p>(I edited the previous version to make it a bit easier to show <span class="math-container">$g$</span> satisfies the condition.)</p>
|
4,041,595 | <p>I've been having lectures in group theory with Hungerford's book. We were presented with the following theorem:</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/pkL0f.png" rel="noreferrer"><img src="https://i.stack.imgur.com/pkL0f.png" alt="enter image description here" /></a></p>
</blockquote>
<p>And then with:</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/ZP9eR.png" rel="noreferrer"><img src="https://i.stack.imgur.com/ZP9eR.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/jpiHf.png" rel="noreferrer"><img src="https://i.stack.imgur.com/jpiHf.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/jYsqM.png" rel="noreferrer"><img src="https://i.stack.imgur.com/jYsqM.png" alt="enter image description here" /></a></p>
</blockquote>
<p>Previous to the lectures about them, I was understanding most of the stuff, that is: I kinda could figure out the motivation for things. But when I came into these theorems, things got pretty weird for me. I don't understand the motivation for them. I have the following guesses:</p>
<ul>
<li><p>Theorem 5.6 is used to prove Corollary 5.7 but I think it has a utility on its own: If given an homomorphism <span class="math-container">$f: G \to H$</span> with <span class="math-container">$N\lhd G$</span>, we have a unique homomorphism <span class="math-container">$\overline{f} : G/N\to H$</span>, then we could use this to know what homomorphisms could exist betwen <span class="math-container">$G$</span> and <span class="math-container">$H$</span> and this depends only on the normal subgroups of <span class="math-container">$G$</span>. Does that make sense?</p>
</li>
<li><p>For the isomorphism theorems, I noticed they are corollaries so they must be very important. One of the reasons I conjectured for them is that we can construct very interesting isomorphisms. But why are those isomorphisms interesting? It's not clear to me what interesting stuff one could do with them.</p>
</li>
</ul>
<p>I know my guesses are maybe too obvious or wrong, but I can't figure out on my own the answer to that question. Can you help me?</p>
| SSA | 863,540 | <p>Not sure, how much below information can help you. But may help others and I too can get more intuitive insight of this very important theorem, by some experts.</p>
<p>The way I understand the group is a kind of Symmetries of an object. On which one can perform some actions.
Now when it comes to comparing two groups or symmetries, what mathematical approach we can take, to find it if they are 'same' or 'similar' to some extent.
we understand these by a function which maps elements(vertices,edges etc..) of symmetries to other objects symmetries, and then apply action on elements and compare
how do they appear on other symmetries.</p>
<p>As you know isomorphism and homomorphism are the tool to compare two symmetries. Most of the time we would like to know when two symmetries doesn't result
same kind of o/p , then what extent they behave same? i.e., if we give up bijection constraint and try to find what we loose and still behave somewhat similar
we use a tool called homomorphism.</p>
<p>Now, one can ask , that what are the properties that is lost when we move from isomorphism to homomorphism, and what is still preserved?
and two of the important properties are <span class="math-container">${Ker(\phi)}$</span> and <span class="math-container">${Im(\phi)}$</span> , So image of <span class="math-container">${Im(\phi)}$</span>, which help us to find out what part
of the other symmetry H is going to hit by this function. <span class="math-container">${Ker(\phi)}$</span> : If we sacrifice one-to-oneness, then in our structure preserving function
more than one elements will hit to identity of the other symmetry. i.e, some non identity element from the domain symmetry is going to be mapped on the identity of the other symmetry.
And it is the most important property which help us to find out how many homomorphism is possible between two group.</p>
<p>so mathematically any function which follow <span class="math-container">${\phi (a \cdot b)= \phi(a) \cdot \phi(b)}$</span>, has enough power to tell us something about these two groups even if it is not bijection.
for ex. <span class="math-container">${D_4 \rightarrow Z_4}$</span> are two different order group, but still we can find homomorphism between them.</p>
<p>Continuing same example of <span class="math-container">${D_4 \rightarrow Z_4}$</span> if we partition D4 into rotation and reflection and send rotation to 0 <span class="math-container">${(e_H)}$</span> in <span class="math-container">${Z_4}$</span>
and reflection to 2 in <span class="math-container">${Z_4}$</span>, These partition acts almost similar to sub group {0,2} of <span class="math-container">${Z_4}$</span> (is it normal subgroup?)</p>
<p>Having these two piece of information <span class="math-container">${Im(\phi)}$</span> and <span class="math-container">${Ker(\phi)}$</span> we can compare narrower part of symmetries and gain some information about the structure of other group.</p>
<p>First Isomorphism Theorem: Here we use homomorphism to tell us something about of the structure of that symmetry via this theorem.</p>
<p>which tells that image of that homomorphism is isomorphic to the factor group of a domain by the kernel of it.(Kernel is a property of homomorphism and not of a group)
This also, tell us, that every homomorphism is hiding isomomorphism inside them.</p>
<p>for ex. <span class="math-container">${Z_{12} \rightarrow Z_3}$</span> by mapping <span class="math-container">${\phi(x)=x mod3}$</span>. <span class="math-container">${Ker(\phi)= \{0,3,6,9\} \cong Z_4}$</span>, using this we can get more information
i.e, factoring out <span class="math-container">${Z_{12}}$</span> with kernel, there are still 3 cosets are hiding under the domain group can be achieved via <span class="math-container">${Im(\phi)}$</span> which is isomorphic to <span class="math-container">${Z_3}$</span>
these 3 coset form a factor group which is isomorphic to cyclic group <span class="math-container">${Z_3}$</span></p>
<p>Hence, if you give me the homomorphism between two group, I look at the Kernel of that homomorphism, that is the normal subgroup of domain group and they are exactly
the building blocks of symmetries(groups).</p>
<ol>
<li>Use homomorphism to discover normal subgroups and to determine factor group. 2. building a cosets out of this and which will be in one to one correspondence with other group that you get via <span class="math-container">${Im(\phi)}$</span>.</li>
</ol>
|
2,288,358 | <p>Is there a straightforward way to prove the following inequality:
$$|1 + k\big(\exp(it)-1\big)|\leq 1 $$
where $k\in(0,1)$ and $t \in \mathbb{R}$ (correction, see dxiv answer) with $|t| \leq 1$, other than writing the quantity into its real and imaginary parts and checking that they satisfy the required inequalities (which is long and seems inelegant) ? </p>
| user407691 | 407,691 | <p>Since the region of integration in u-v coordinates is still a triangle, it can be converted to polar coordinates. We can look at the bounds for r and <span class="math-container">$\theta$</span> separately.</p>
<p>In general with polar coordinates, it's easier when the inner integral is evaluated with respect to r (<span class="math-container">$\theta$</span> held constant), so we'll do that here.</p>
<p>What are the bounds for <span class="math-container">$\theta$</span>? Looking at the following diagram, we can see that <span class="math-container">$\theta$</span> goes from 0 to approximately <span class="math-container">$\frac{\pi}{4}$</span>.</p>
<p><a href="https://i.stack.imgur.com/t3Rvj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/t3Rvj.png" alt="u-v region" /></a></p>
<p>To find <span class="math-container">$\theta$</span> exactly, we can use basic trigonometry. The triangle has legs of length <span class="math-container">$\sqrt{3}$</span> and 1. Therefore, <span class="math-container">$\tan(\theta) = \frac{\sqrt{3}}{1}$</span>, from which we can take the inverse to find <span class="math-container">$\theta=\tan^{-1}(\sqrt 3)$</span>. So, with the bounds of the outer integral, we're effectively doing a circular "scan" from <span class="math-container">$\theta = 0$</span> to <span class="math-container">$\theta = \tan^{-1}(\sqrt 3)$</span>.</p>
<p>To find the bounds of the inner integral for r, imagine fixing a particular value of <span class="math-container">$\theta$</span> between 0 and <span class="math-container">$\tan^{-1}(\sqrt 3)$</span>. Now, we need to figure out the line that r will traverse, i.e., where does r start and where does it stop, for a particular fixed value of <span class="math-container">$\theta$</span>.</p>
<p>Such a line will look like the orange dotted lines in the following diagram (each orange line corresponds to a separate fixed <span class="math-container">$\theta$</span>).</p>
<p><a href="https://i.stack.imgur.com/RVVsS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RVVsS.png" alt="r bounds" /></a></p>
<p>So then, r starts at r = 0, and ends at the line u = 1. The standard formulas for polar coordinates are <span class="math-container">$x = r\,\cos(\theta)$</span> and <span class="math-container">$y=r\,\sin(\theta)$</span>, except that we are using u and v instead of x and y. Therefore, u = 1 implies that <span class="math-container">$r\,\cos(\theta) = 1$</span>, which implies that <span class="math-container">$r = \frac{1}{\cos(\theta)} = \sec(\theta)$</span>.</p>
<p>The final integral should look like this:
<span class="math-container">$$\frac{1}{\sqrt 3}\int_{0}^{\tan^{-1}(\sqrt 3)}\int_0^{\sec(\theta)}r^3\,\cos(\theta)\,dr\,d\theta$$</span></p>
|
3,126,080 | <p>Is there any particular equation which doesn't work on the real plane of numbers but works on other planes?</p>
| FormerMath | 174,432 | <p><span class="math-container">$x^2+1=0$</span> doesn't admit roots on <span class="math-container">$\mathbb{R} $</span>, but it does on <span class="math-container">$\mathbb{C}$</span></p>
|
1,827,612 | <p>I have to simplify $\sum_{i,j}\left[{n}\atop{i+j}\right]\binom{i+j}{i}$. I looks like we have $n$ children and we have to answer how many times we can arrange them into circles and color some of circles red.</p>
| Matematleta | 138,929 | <p>Let $\epsilon >0$ and note that there are partitions $P, Q$ such that </p>
<p>$\tag1 \underline{\int_{a}^{b}}{f(x)dx}-\epsilon<L(P,f)$ </p>
<p>$\tag2\underline{\int_{a}^{b}}{g(x)dx}-\epsilon <L(g,Q)$</p>
<p>These inequalities persist if we pass to a common refinement, $T$, so </p>
<p>$\tag 3\underline{\int_{a}^{b}}{f(x)dx}+\underline{\int_{a}^{b}}{g(x)dx}-2\epsilon<L(T,f)+L(T,g)$. </p>
<p>But also, \begin{equation*}L(T,f) + L(T,g)\leq L(T,f+g)\leq\underline{\int_{a}^{b}}{(f+g)(x)dx} \tag{4}\end{equation*}</p>
<p>from which we get</p>
<p>$\tag5\underline{\int_{a}^{b}}{f(x)dx}+\underline{\int_{a}^{b}}{g(x)dx}-2\epsilon<\underline{\int_{a}^{b}}{(f+g)(x)dx}$</p>
<p>which is what we want.</p>
<p>For the second part, here is a hint: observe that if $P$ is any partition of [a,b], we may assume without loss of generality, that $c\in P$. Then $P'=P\cap[a,c]$ and $P''=P\cap[c,b]$ are parititions of $[a,c]$ and $[c,b]$,resp. and $P=P\cup P''$.</p>
|
1,307,460 | <p>Need to integrate this function. Need help with my assignment. Thanks</p>
| lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>Use $I=\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$</p>
<p>$\implies I+I=\int_a^bf(x)\ dx+\int_a^bf(a+b-x)\ dx=\int_a^b[f(x)+f(a+b-x)]\ dx$</p>
<p>Then</p>
<p><strong>Method</strong> $\#1:A\cos x+B\sin x=\sqrt{a^2+b^2}\cos\left(x-\arctan \dfrac BA\right)$</p>
<p><strong>Method</strong> $\#2:$ <a href="http://en.wikibooks.org/wiki/Calculus/Integration_techniques/Tangent_Half_Angle" rel="nofollow">Weierstrass substitution</a></p>
|
4,644,847 | <p>Is there a formal fallacy to describe lack of an observation is not proof that it does not exist, or lack of an occurrence is not proof that it can never happen?</p>
| Prem | 464,087 | <p>In most general terms , we can state it Mathematically / logically / formally like this :</p>
<p>Let X be the Universe (eg Integers or real numbers or .... )<br />
Let Y be the non-empty subset which DOES NOT have Property P.</p>
<p><span class="math-container">$ Y = \{ x \in X : \lnot P(x) \} $</span></p>
<p>Now , <span class="math-container">$ \forall y \in Y , \lnot P(y) \implies \forall x \in X , \lnot P(x) $</span> is the fallacy.<br />
We are not able to observer Property P in Y , hence we think (over generalizing) that it must be true every where , in X.<br />
We must not take Examples from Y to over generalize to X.</p>
<p>Let Z be some arbitrary non-empty subset of Y.<br />
Here too , <span class="math-container">$ \forall z \in Z , \lnot P(z) \implies \forall x \in X , \lnot P(x) $</span> is the fallacy.<br />
We are not able to observer Property P in some small Collection Z , hence we think (over generalizing) that it must be true every where , in X.<br />
We must not take Examples from Z to over generalize to X.</p>
<p><strong>Example 1 (lack of observation) :</strong><br />
Let a Continent have many countries , where the Presidents are all over the age of 70.<br />
We may think , that in all continents , in all countries , the Presidents are all over the age of 70.<br />
We then check in the nearest Continent , & observe the same , hence , with these Examples , we conclude on this "fact".<br />
When we then check more Continents , we see that there are Countries with young Presidents , hence the fallacy gets exposed.</p>
<p><strong>Example 2 (wrong observation) :</strong><br />
We have checked Polynomial Equations like <span class="math-container">$Ax+B=0$</span> & Determined the Solution.<br />
We have checked Polynomial Equations like <span class="math-container">$Ax^2+Bx+C=0$</span> & <a href="https://en.wikipedia.org/wiki/Quadratic_formula" rel="nofollow noreferrer">Determined the Solution</a>.<br />
We may think , we can always get the Solution in "Closed-form" radicals.<br />
We then check Polynomial Equations like <span class="math-container">$Ax^3+Bx^2+Cx+D=0$</span> & <a href="https://en.wikipedia.org/wiki/Cubic_formula" rel="nofollow noreferrer">Determined the Solution</a>.<br />
Hence , with these Examples , we "conclude" that all Polynomials Equations ( including <a href="https://en.wikipedia.org/wiki/Quartic_formula" rel="nofollow noreferrer">Degree 4</a> ) will have the Solution in "Closed-form" radicals.<br />
Then <a href="https://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem" rel="nofollow noreferrer">Galois (along with Abel & Ruffini)</a> shows that it is Impossible , there by exposing the fallacy.</p>
<p><strong>In Summary , we can not observe some small set to then take these Examples to over generalize to the universal Set.</strong></p>
<p>More here :<br />
<a href="https://en.wikipedia.org/wiki/Proof_by_example" rel="nofollow noreferrer">Proof By Example</a></p>
|
2,651,486 | <p>I have an exercise here that is asking me to write the MacLaurin formula of orders II, III, IV for a multivariable function. Example: $ f(x,y)=\cos x \cos y$</p>
<p>Can anyone tell me what the formula looks like for a multivariable function and maybe guide me through this example? Would be much appreciated!</p>
| Christian Blatter | 1,303 | <p>Your example is special in so far as the function is "separated" into two functions of one variable. In order to prove the formula given by Wikipedia consider for <em>fixed</em> $x$ and $y$ the auxiliary function
$$g(t):= f(t\,x,t\,y)$$
of one variable $t$, and compute its value at $t=1$, using the Taylor expansion of a function of one variable:
$$f(x,y)=g(1)=\sum_{p=0}^n {1\over p!} g^{(p)}(0)+R_n\ ,$$
and compute the higher derivatives $g^{(p)}(0)$ using repeatedly the chain rule. Collecting equal terms you obtain
$$g^{(p)}(0)=\sum_{k=0}^p{p\choose k}f_{x^{p-k} y^k}(0,0)\> x^{p-k}\,y^k\ .$$</p>
|
3,019,989 | <p>I'm watching the Limits series on Khan Academy. In many videos Sal repeatedly says that although some people say that functions that tend to infinity have a limit infinity. (For example, in <a href="https://www.khanacademy.org/math/calculus-1/cs1-limits-and-continuity/modal/v/infinite-limits-and-asymptotes" rel="nofollow noreferrer">this</a> video, the says that the function <span class="math-container">$ y = \frac {2}{x-1} $</span> (here's a link to the <a href="https://www.desmos.com/calculator/d1ioaqbnnn" rel="nofollow noreferrer">graph</a>) is unbounded as x approaches 1 from the left side, although "some" people would say that the function is tending towards infinity (i.e., the <span class="math-container">$ \lim_{x \to c} f(x) = \infty $</span>. </p>
<p>Over a series of videos, I caught hold of that argument and always said that function don't tend towards infinity; they're just unbounded. But, in a different video (<a href="https://www.khanacademy.org/math/calculus-1/cs1-limits-and-continuity/modal/v/introduction-to-infinite-limits" rel="nofollow noreferrer">here</a>), Sal essentially says:</p>
<p><span class="math-container">$$ \lim_{x \to 0} \frac 1{x^2} = \infty $$</span>
<span class="math-container">$$ \lim_{x \to 0^+} \frac 1{x} = \infty $$</span>
<span class="math-container">$$ \lim_{x \to 0^-} \frac 1{x} = -\infty $$</span></p>
<p>There is an answer to <a href="https://math.stackexchange.com/questions/1887018/unbounded-and-sequences-tending-to-infinity">this question</a> that says that unbounded sequences don't always tend to infinity. So it seems that we can't say that the limit of unbounded functions as x tends to some number is infinity -- what I'm confused about, and my question is: When do we ever say that the limit of some function is infinity? Specifically, are the limits of the above functions <span class="math-container">$ \frac 1x $</span> and <span class="math-container">$$ \frac 1{x^2} $$</span> correct? Or are they unbounded? If we can say that these limits are correct, why can't we say this: <span class="math-container">$$ \lim_{x \to 1^-} \frac 2{x-1} = - \infty $$</span></p>
| YiFan | 496,634 | <p>Like Arthur said in the comments, functions being "bounded" and "tending to infinity" mean different things. If <span class="math-container">$\lim_{x\to c}f(x)=\infty$</span>, then it tends to infinity and gets arbitrarily large in the process. Hence it is not bounded, so there does not exist real numbers <span class="math-container">$a,b$</span> so that <span class="math-container">$f(x)\in[a,b]$</span> for all <span class="math-container">$x$</span> in the domain. However, if a function is unbounded it does not necessarily tend to infinity. In fact, the limit need not even exist. Just consider <span class="math-container">$f(x)=\sin(1/x)/x$</span>.</p>
|
3,019,989 | <p>I'm watching the Limits series on Khan Academy. In many videos Sal repeatedly says that although some people say that functions that tend to infinity have a limit infinity. (For example, in <a href="https://www.khanacademy.org/math/calculus-1/cs1-limits-and-continuity/modal/v/infinite-limits-and-asymptotes" rel="nofollow noreferrer">this</a> video, the says that the function <span class="math-container">$ y = \frac {2}{x-1} $</span> (here's a link to the <a href="https://www.desmos.com/calculator/d1ioaqbnnn" rel="nofollow noreferrer">graph</a>) is unbounded as x approaches 1 from the left side, although "some" people would say that the function is tending towards infinity (i.e., the <span class="math-container">$ \lim_{x \to c} f(x) = \infty $</span>. </p>
<p>Over a series of videos, I caught hold of that argument and always said that function don't tend towards infinity; they're just unbounded. But, in a different video (<a href="https://www.khanacademy.org/math/calculus-1/cs1-limits-and-continuity/modal/v/introduction-to-infinite-limits" rel="nofollow noreferrer">here</a>), Sal essentially says:</p>
<p><span class="math-container">$$ \lim_{x \to 0} \frac 1{x^2} = \infty $$</span>
<span class="math-container">$$ \lim_{x \to 0^+} \frac 1{x} = \infty $$</span>
<span class="math-container">$$ \lim_{x \to 0^-} \frac 1{x} = -\infty $$</span></p>
<p>There is an answer to <a href="https://math.stackexchange.com/questions/1887018/unbounded-and-sequences-tending-to-infinity">this question</a> that says that unbounded sequences don't always tend to infinity. So it seems that we can't say that the limit of unbounded functions as x tends to some number is infinity -- what I'm confused about, and my question is: When do we ever say that the limit of some function is infinity? Specifically, are the limits of the above functions <span class="math-container">$ \frac 1x $</span> and <span class="math-container">$$ \frac 1{x^2} $$</span> correct? Or are they unbounded? If we can say that these limits are correct, why can't we say this: <span class="math-container">$$ \lim_{x \to 1^-} \frac 2{x-1} = - \infty $$</span></p>
| egreg | 62,967 | <p>A <em>necessary</em> condition for a function to have infinite limit is that it is unbounded.</p>
<p>More precisely, if <span class="math-container">$\lim_{x\to c}f(x)=\infty$</span>, then <span class="math-container">$f$</span> is upper unbounded; if <span class="math-container">$\lim_{x\to c}f(x)=-\infty$</span>, then <span class="math-container">$f$</span> is lower unbounded. (<em>The limit in the previous statements can also be one-sided.</em>)</p>
<p>One could be more precise and say that <span class="math-container">$f$</span> must be (upper/lower) unbounded in <em>every</em> punctured neighborhood of <span class="math-container">$c$</span> (or punctured right/left neighborhood of <span class="math-container">$c$</span>).</p>
<p>This essentially follows from the definition.</p>
<p>However, this condition is by no means sufficient. Consider
<span class="math-container">$$
f(x)=\frac{1}{x}\sin\frac{1}{x}
$$</span>
Then this function is unbounded in every punctured neighborhood of <span class="math-container">$0$</span>, but its limit is neither <span class="math-container">$\infty$</span> nor <span class="math-container">$-\infty$</span>.</p>
<p>Restricting to the limit from the right or the left doesn't improve the situation.</p>
<p>Confusing “unbounded” with “has infinite limit” may be very dangerous.</p>
|
3,648,315 | <p><a href="https://i.stack.imgur.com/IKnlD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IKnlD.png" alt="enter image description here"></a></p>
<p>I have an item, let's say sword. </p>
<p>As described above, <br/>
If my sword is at state 1, I can try upgrading it. There are 3 possibilities. <br/>
It can be upgraded with prob = 0.3, remain still with prob = 0.68, can be destroyed with prob = 0.02.</p>
<p>If my sword is at state 2, I still can try to upgrade it. <br/>
It can be upgraded with prob = 0.3, can be downgraded to state 1 with prob = 0.68, can be destroyed with prob = 0.02.</p>
<p>Once my sword destroyed, there is no turning back. <br/>
Once my sword reached at state 3, no need to do something else. I'm done.</p>
<p>I know it's a Markov chain problem. <br/>
I can express this situation with matrix, and if I multiply it over and over, it can reach equilibrium state.</p>
<pre><code>p2 = matrix(c(1, rep(0, 3),
0.02, 0.68, 0.3, 0,
0.02, 0.68, 0, 0.3,
rep(0, 3), 1), 4, byrow = T)
p2
## [,1] [,2] [,3] [,4]
## [1,] 1.00 0.00 0.0 0.0
## [2,] 0.02 0.68 0.3 0.0
## [3,] 0.02 0.68 0.0 0.3
## [4,] 0.00 0.00 0.0 1.0
matrix.power <- function(A, n) { # For matrix multiplication
e <- eigen(A)
M <- e<span class="math-container">$vectors
d <- e$</span>values
return(M %*% diag(d^n) %*% solve(M))
}
round(matrix.power(p2, 1000), 3)
## [,1] [,2] [,3] [,4]
## [1,] 1.000 0 0 0.000
## [2,] 0.224 0 0 0.776
## [3,] 0.172 0 0 0.828
## [4,] 0.000 0 0 1.000
</code></pre>
<p>But how can I get the <code>Pr(Reach state 3 without destroyed | currently at state 2)</code> using Markov chain?</p>
<p>I could get <code>Pr(Reach state 2 without destroyed | currently at state 1)</code> by using sum of geometric series.</p>
<p>Thank you.</p>
| Ekaveera Gouribhatla | 31,458 | <p>Since our aim is to find <span class="math-container">$x+y$</span>, Let <span class="math-container">$x+y=k$</span>, where <span class="math-container">$k \in \mathbb{Z}$</span></p>
<p>We have <span class="math-container">$$k+2x(k-x)=83$$</span></p>
<p>So</p>
<p><span class="math-container">$$2x^2-2kx+83-k=0 $$</span></p>
<p>The roots are <span class="math-container">$$x_1,x_2=\frac{k}{2}\pm\frac{1}{2}\sqrt{(k+1)^2-167}$$</span></p>
<p>So <span class="math-container">$$(k+1)^2-167=r^2$$</span> and <span class="math-container">$167$</span> being Prime we get:</p>
<p><span class="math-container">$$k+1+r=1$$</span>
<span class="math-container">$$k+1-r=167$$</span> OR <span class="math-container">$$k+1+r=-1$$</span>
<span class="math-container">$$k+1-r=-167$$</span></p>
<p>Giving <span class="math-container">$k=83$</span> and <span class="math-container">$k=-85$</span></p>
|
1,639,521 | <p>Let <span class="math-container">$f:[0,\infty)\to\mathbb{R}$</span> differentiable and suppose that <span class="math-container">$$\lim_{x\to\infty}f'(x)=L.$$</span> How can I prove that <span class="math-container">$$\lim_{x\to\infty}\frac{f(x)}{x} = L\;?$$</span></p>
<p>I have solved some similar problems using the Mean Value Theorem, and I am trying to use it again in this one, but nothing works. For example, I tried to apply the MVT in <span class="math-container">$[x, 2x]$</span> but it does not work. Some hint?</p>
| Joe | 107,639 | <p>Maybe you're searching for a direct computation.</p>
<p>Let us check the case $L\neq0$.</p>
<p>\begin{align*}
L=\lim_{x\to\infty}f'(x)
&=\lim_{x\to\infty}\lim_{y\to x}\frac{f(y)-f(x)}{y-x}\\
&=\lim_{x\to\infty}\lim_{y\to x}\frac{f(x)-f(y)}{x-y}\\
&=\lim_{x\to\infty}\lim_{y\to x}\frac{f(x)\left[1-\frac{f(y)}{f(x)}\right]}{x\left[1-\frac yx\right]}\\
\end{align*}</p>
<p>Provided the limit interchange, this last one equals to
$$
\lim_{y\to x}\lim_{x\to\infty}\frac{f(x)\left[1-\frac{f(y)}{f(x)}\right]}{x\left[1-\frac yx\right]}
$$</p>
<p>Now $L\neq0$ by hypotesis, then $f(x)\stackrel{x\to\infty}{\to}{\infty}$, thus
$$
\left[1-\frac{f(y)}{f(x)}\right]\to1
$$
as well as $1-\frac yx\to1$ as $x\to \infty$. Hence
$$
\lim_{y\to x}\lim_{x\to\infty}\frac{f(x)\left[1-\frac{f(y)}{f(x)}\right]}{x\left[1-\frac yx\right]}
=\lim_{y\to x}\lim_{x\to\infty}\frac{f(x)}{x}
=\lim_{x\to\infty}\frac{f(x)}{x}
$$
which concludes the case $L\neq 0$.</p>
<p>$L=0$ is even simpler.</p>
|
440,528 | <p>My question is about group theory:</p>
<blockquote>
<p>How many subgroups does a non-cyclic group contain whose order is 25?</p>
</blockquote>
<p>How can i answer that question?</p>
<p>Can you generalize the answer?</p>
<p>Thanks for your help.</p>
| DonAntonio | 31,254 | <p>As a power of a prime, a non-cyclic group $\;G\;$ of order $\;25\;$ is isomorphic to $\,C_5\times C_5\;$ , with $\,C_5\cong\Bbb Z/5\Bbb Z\;$ is the cyclic group of order $\;5\;$</p>
<p>Now, some hints for you to answer your question:</p>
<p>== A group of the form $\;G=\underbrace{C_p\times C_p\times\ldots\times C_p}_{n\;\text{times}}\;$ is a vector space of dimension $\;n\;$ over the field $\,\Bbb Z/p\Bbb Z\;$ and is therefore isomorphic to $\;\left(\Bbb Z/p\Bbb Z\right)^n\;$</p>
<p>== The subgroups of $\,G\,$ as above are in $\,1-1\,$ correspondence with the vector subspaces of $\,\left(\Bbb Z/p\Bbb Z\right)^n\;$</p>
|
35,375 | <p>Good morning,
today I have read that "number theory is nothing but the study of $\mathrm{Gal}(\mathbb{\bar{Q}}/\mathbb{Q})$", here <a href="http://www.math.uconn.edu/~alozano/elliptic/finding%20points.pdf" rel="nofollow">http://www.math.uconn.edu/~alozano/elliptic/finding%20points.pdf</a>
can anyone give a very naive layman definition of what it actually means?</p>
<p>Furthermore, I got this doubt that $\bar {\mathbb{Q}}$ is the algebraic closure of $\mathbb{Q}$, and the thing that confuses me is
the field of rational numbers $\mathbb{Q}$ is not a algebraically closed as there exists a polynomial with $a_{1},a_{2},\dotsc,a_{n}\in \mathbb{Q}$ and $(x-a_{1})(x-a_{2})\cdots(x-a_{n})+1$ has no zero in $\mathbb{Q}$.</p>
<p>Then why are we considering the field extension of $\bar {\mathbb{Q}}/\mathbb{Q}$ when $\mathbb{Q}$ is not algebraically closed, won't it contradict the definition of algebraic closure?</p>
<p>But I am not getting an answer i was looking for ,i want what are the things going on behind the $\mathrm{Gal}(\mathbb{\bar{Q}}/\mathbb{Q})$ like what is the thing we get if we take the $\mathbb{\bar{Q}/\mathbb{Q}}$ and what does taking the $\mathrm{Gal}(\mathbb{\bar{Q}}/\mathbb{Q})$ give someone,</p>
<p>Thank you</p>
| Keenan Kidwell | 628 | <p>I would say it's not unreasonable to (loosely) describe <em>algebraic</em> number theory as the study of number fields (number fields are the objects you get when you adjoin roots of polynomials like $X^2+1$ to $\mathbb{Q}$). This is admittedly an incredibly vague description and could be taken in any number of ways, but, if you allow it, then, via Galois theory (for infinite degree extensions of $\mathbb{Q}$, like $\bar{\mathbb{Q}}$), Galois number fields are (roughly) in bijection with finite quotients of the Galois group of $\bar{\mathbb{Q}}$ over $\mathbb{Q}$. (I say roughly because really the bijection is with normal subgroups of finite index of the Galois group that are closed for a certain topology.) For instance, a sort of natural question in algebraic number theory would be "what finite groups occur as Galois groups of Galois field extensions of $\mathbb{Q}$?" This is equivalent to the question "what are the finite quotients of the Galois group of $\bar{\mathbb{Q}}$ over $\mathbb{Q}$?"</p>
|
128,533 | <p>For the description, I have a simplified problem like the following:</p>
<pre><code>MapAt[f[1, #1], {a, b, c, d}, #2] & @@@ {{1, 2}, {3, 4}}
</code></pre>
<p>will give</p>
<blockquote>
<pre><code>{{a, f[1, 1][b], c, d}, {a, b, c, f[1, 3][d]}}
</code></pre>
</blockquote>
<p>But actually <code>{a, f[1, 1][b], c, f[1, 3][d]}</code> is what I expected. What happened? How to adjust the code?</p>
<hr>
<p><strong>Update:</strong></p>
<p>My real case is</p>
<pre><code>bigList = Range @ 10;
veryBigList = {{1, 3}, {1, 4}, {2, 7}, {2, 9}, {4, 10}};
Function[{binLevel, place},
MapAt[BitSet[#, binLevel] &, bigList, place]] @@@ veryBigList
</code></pre>
<blockquote>
<p>{{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, {1, 2, 3, 6, 5, 6, 7, 8, 9, 10}, {1,
2, 3, 4, 5, 6, 7, 8, 9, 10}, {1, 2, 3, 4, 5, 6, 7, 8, 13, 10}, {1,<br>
2, 3, 4, 5, 6, 7, 8, 9, 26}}</p>
</blockquote>
<p>In my case, the list size is very huge. If I use <code>Fold</code>, it will act on a very big result list every calculation, for the RAM so I want to avoid using <code>Fold</code>.</p>
| Kuba | 5,478 | <pre><code>Fold[
Function[{data, spec}, MapAt[f[1, #1], data, #2] & @@ spec],
{a, b, c, d},
{{1, 2}, {3, 4}}
]
</code></pre>
<blockquote>
<pre><code>{a, f[1, 1][b], c, f[1, 3][d]}
</code></pre>
</blockquote>
|
3,113,850 | <blockquote>
<p><span class="math-container">$$f(x)=\sum_{n=1}^{\infty}{\frac{x^{n-1}}{n}},$$</span></p>
<p>Prove that <span class="math-container">$f(x)+f(1-x)+\log(x)\log(1-x)=\frac{{\pi}^2}{6}$</span></p>
</blockquote>
<p>In my mind though,I think that this is related to Basel problem<span class="math-container">$\left(\sum\limits_{n=1}^{\infty}{\frac{1}{n^2}}\right)$</span>,but I don't know how to solve this.</p>
<p>Any help would be greatly appreciated :-)</p>
<p>Edit:</p>
<p>My attempt:</p>
<p>I cannot use latex expertly,so I post image.The circled part<a href="https://i.stack.imgur.com/gFDdf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gFDdf.jpg" alt="enter image description here " /></a></p>
| Mirko | 248,138 | <p>The matrix</p>
<p><span class="math-container">$$A=\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix} $$</span></p>
<p>satisfies <span class="math-container">$adj(A)=0$</span> and <span class="math-container">$\det(A)=0$</span>. The inverse does not exist if the determinent is zero.</p>
<p>This answers your question!</p>
|
2,459,651 | <p>What additional properties must an operation have besides commutativity so that commutativity along with other properties implies associativity?</p>
<p>Where can I read about such structures?</p>
| paul garrett | 12,291 | <p>This is a reasonable question, but a reasonable answer (in my opinion) basically is negative: commutativity and associativity are not reasonably related, in most natural contexts.</p>
<p>The most disturbing-to-me example is that of Jordan algebras, which are commutative but not associative, despite being reasonably natural. A simple case is that of (real or complex) matrices of some fixed size, where the "Jordan product" is $a*b=ab+ba$. </p>
<p>For that matter, Lie algebras' anti-commutativity do suggest that associativity is a quite different thing ...</p>
|
1,429,937 | <p>From <a href="http://www.gottfriedville.net/mathprob/comb-subrect.html" rel="nofollow">this</a> page it can be shown that the number of possible rectangles in and m*n grid can be found by first choosing $2$ lines from a possible $m+1$ to form the vertical sides of the rectangle, and then $2$ from the $n+1$ horizontal sides, to give ${{m+1} \choose {2}}*{{n+1} \choose {2}}$ rectangles. </p>
<p>I want to extend this proof to cover only squares</p>
<p>I have tried choosing $2$ from ${m+1}$ for one set of sides, as a square can be formed with any two lines, but I am having trouble on how many choices there are for the remaining sides. If the first two sides are beside each other, then the last two can be any two sides that are beside each other, so there are so n choices, but after that I get lost. </p>
<p>The formula given on the page using a different approach is $$\frac {{n}{(2n+1)}{(n+1)}}{6}$$ or the sum of squares when m = n</p>
| Clive Newstead | 19,542 | <p>If you restrict yourself to the real numbers, then it is not possible in general.</p>
<p>For example, take $a=1$ and $b=2$. Given any $x \in \mathbb{R}$ we have
$$\sin(x+\pi)=-\sin x$$
but since $\sin 2(x+\pi) = \sin 2x$ and $\cos 2(x+\pi) = \cos 2x$, no function of $\sin 2x$ and $\cos 2x$ can have this property.</p>
|
2,713,555 | <p>I have a question stating</p>
<blockquote>
<p>Let $A= \{ x_1 , x_2 , x_3 , \ldots , x_n \}$ be a set consisting of $n$ distinct elements. How many subsets does it have? How many proper subsets?</p>
</blockquote>
<p>My thought is that there would be subsets with $1$ element, $2$ elements, $3$ element and so on, up to $n$ elements. The number of subsets of each size would be:</p>
<p>$$\begin{array}{c|c}
\text{Subset size} & \text{no. of subsets} \\
\hline
1 & n \\ 2 & n-1 \\ 3 & n-2 \\ \vdots & \vdots \\ n-1 &n-(n-2) \\ n & n- (n-1)
\end{array}$$
From this it seems the number of subsets would be $\displaystyle \sum_{k=0}^{n-1} (n-k) $. And for proper subsets, I would just not include the subsets of size $n$ , so $\displaystyle \sum_{k=0}^{n-2} (n-k) $. Is this correct?</p>
| gt6989b | 16,192 | <p>I disagree. For example, the number of subsets of size 2 is $$ \binom{n}{2} = \frac{n(n-1)}{2} \ne n-1 $$ and in general you are looking at
$$
\sum_{k=0}^n \binom{n}{k} = 2^n,
$$
which is also easy to see from fundamentals by a simple counting argument -- each of the $n$ elements can be either included or excluded, independently of others. So you have $2$ choices $n$ times, a total of $2^n$.</p>
|
4,419,360 | <p>In chapter 9 of Spivak's <em>Calculus</em>, on derivatives, he mentions the "Leibnizian Notation" for the derivative of a function <span class="math-container">$f$</span>, <span class="math-container">$\frac{df(x)}{dx}$</span>. In a footnote on page 155, he writes</p>
<blockquote>
<p>Leibniz was led to this symbol by his intuitive notion of the
derivative, which he considered to be, not the limit of quotients
<span class="math-container">$\frac{f(x+h)-f(x)}{h}$</span>, but the "value" of this quotient when <span class="math-container">$h$</span> is
an "infinitely small" number. This "infinitely small" quantity was
denoted <span class="math-container">$dx$</span> and the corresponding "infinitely small" difference
<span class="math-container">$f(x+dx)-f(x)$</span> by <span class="math-container">$df(x)$</span>. <strong>Although this point of view is impossible
to reconcile with properties (P1)-(P13) of the real numbers</strong>, some
people find this notion of the derivative congenial.</p>
</blockquote>
<p>The bold section has been highlighted by me. What does he mean with that?</p>
| José Carlos Santos | 446,262 | <p>Within that axiomatic, there are no infinitesimal numbers, that is, there is no number <span class="math-container">$\mu>0$</span> such that <span class="math-container">$(\forall n\in\Bbb N):\mu<\frac1n$</span>. That's so because the Archimedean property follows from those axioms. And that property states that <span class="math-container">$\Bbb N$</span> has no upper bound. But if such a number <span class="math-container">$\mu$</span> existed, we would have <span class="math-container">$(\forall n\in\Bbb N):n<\frac1\mu$</span>.</p>
|
301,106 | <p>In the game connect four with a $7 \times 6$ grid like in the image below, how many game situations can occur?</p>
<p><strong>Rules</strong>:</p>
<blockquote>
<p>Connect Four [...] is a two-player game in which the players first
choose a color and then take turns dropping colored discs from the top
into a seven-column, six-row vertically-suspended grid. The pieces
fall straight down, occupying the next available space within the
column. The object of the game is to connect four of one's own discs
of the same color next to each other vertically, horizontally, or
diagonally before your opponent.</p>
</blockquote>
<p><sup>Source: <a href="http://en.wikipedia.org/wiki/Connect_four" rel="noreferrer">Wikipedia</a></sup></p>
<p><img src="https://i.stack.imgur.com/pMjBU.gif" alt="Source: Wikipedia commons, File:Connect_Four.gif"></p>
<p><sup>Image source: <a href="http://commons.wikimedia.org/wiki/File:Connect_Four.gif" rel="noreferrer">http://commons.wikimedia.org/wiki/File:Connect_Four.gif</a></sup></p>
<p><strong>Lower bound</strong>:</p>
<p>$7 \cdot 6 = 42$, as it is possible to make the grid full without winning</p>
<p><strong>Upper bound</strong>:</p>
<p>Every field of the grid can have three states: Empty, red or yellow disc. Hence, we can have $3^{7 \cdot 6} = 3^{42} = 109418989131512359209 < 1.1 \cdot 10^{20}$ game situations at maximum. </p>
<p>There are not that much less than that, because you can't have four yellows in a row at the bottom, which makes $3^{7 \cdot 6 - 4} = 1350851717672992089$ situations impossible. This means a better upper bound is $108068137413839367120$</p>
<p>How many situations are there?</p>
<p>I think it might be possible to calculate this with the approach to subtract all impossible combinations. So I could try to find all possible combinations to place four in a row / column / vertically. But I guess there would be many combinations more than once.</p>
| Damien Merle | 920,944 | <p>So I've worked on it, but I'm not really qualified to do so. First I wanted to calculate every possibility like 7 then 49 and then it was complicated to avoid redundancies and I was facing another problem, the fact that after 6 pieces played, there was a possibility of having a full columns. Too hard for me. So I decided to calculate every possible full board of 21 0 and 21 1. I found 538257874440 possibility, but, first of all I wasn't taking in account the wining board. And the fact that it's not possible to have the two first line full of one kind of piece. So I figured I wasn't taking the good path since it was only the full board, useless.
Finally I had a good idea. Taking a columns alone with 0 1 and 2, so it made 3^6 possibility then I just had to substract the one with 0 on the right of another number (gravity issue) and it wasn't that difficult :
0000000, 0000001, 0000002, 0000011, 0000012 ,000021 ,0000022 ,0000111 ,0000112, 0000121, 0000122, 0000211, 0000212, 0000221, 0000222
If you don't see a patern, it's 1 then 2 then 4 then 8... , which give us 2^0 + 2^1 ... + 2^6 = 2 ^7 - 1 = 127 possibility for a columns, wich make 127^7 = 532875860165503 possible board. Then I figured there was still the same problem with the two lines at the begining and the wining board. So I gave up.</p>
|
283,473 | <p>I'd like to be able to construct polynomials $p$ whose graphs look like this:</p>
<p><img src="https://i.stack.imgur.com/8Rk3a.jpg" alt="enter image description here"></p>
<p>We can assume that the interval of interest is $[-1, 1]$. The requirements on $p$ are:</p>
<p>(1) Equi-oscillation (or roughly equal, anyway) between two extremes. A variation of 10% or so in the values of the extrema would be OK.</p>
<p>(2) Zero values and derivatives at the ends of the interval, i.e. $p(-1) = p(1) =p'(-1) = p'(1) = 0$</p>
<p>I want to do this for degrees up to around 30 or so. Just even degrees would be OK.</p>
<p>If it helps, these things are a bit like Chebyshev polynomials (but different at the ends).</p>
<p>The one in the picture has equation $0.00086992073067855669451 -
0.056750328789339152999 t^2 +
0.60002383910750621904 t^4 -
2.3217878459074773378 t^6 +
4.0661558859963998471 t^8 -
3.288511471137768132 t^{10} + t^{12}$</p>
<p>I got this through brute-force numerical methods (solving a system of non-linear equations, after first doing a lot of work to find good starting points for iteration). I'm looking for an approach that's more intelligent and easier to implement in code.</p>
<p>Here is one idea that might work. Suppose we want a polynomial of degree $n$. Start with the Chebyshev polynomial $T_{n-2}(x)$. Let $Q(x) = T_{n-2}(sx)$, where the scale factor $s$ is chosen so that $Q(-1) = Q(1) = 0$. Then let $R(x) = (1-x^2)Q(x)$. This satisfies all the requirements except that its oscillations are too uneven -- they're very small near $\pm1$ and too large near zero. Redistribute the roots of $R$ a bit (somehow??) to level out the oscillations. </p>
<p><strong>Comments on answers</strong></p>
<p>Using the technique suggested by achille hui in an answer below, we can very easily construct a polynomial with the desired shape. Here is one:</p>
<p><img src="https://i.stack.imgur.com/b0Es2.jpg" alt="achille hui solution"></p>
<p>The only problem is that I was hoping for a polynomial of degree 12, and this one has degree 30.</p>
<p>Also, I was expecting the solution to grow monotonically outside the interval $[-1,1]$, and this one doesn't, as you can see here:</p>
<p><img src="https://i.stack.imgur.com/eIKkN.jpg" alt="behaviour beyond unit interval"></p>
| achille hui | 59,379 | <p>Starting with any Chebyshev polynomial of $1^{st}$ kind $T_n(x), n > 2$,
and any positive root $a$ of it. Compose $T_n(ax)$ with any polynomial $q(x)$
which is monotonic over $[-1,1]$ that satisfies $q(-1) = -1, q(1) = 1$ and
$q'(-1) = q'(1) = 0$ , then $T_n(a q(x))$ is something you want. For example,
$$
T_n\left(\cos\left(\frac{\pi}{2n}\right) \frac{x(3-x^2)}{2}\right)
$$</p>
|
1,691,685 | <p>The answer for it is $$3 + \sum_{k=1}^n (3+k(k-1)2^{k-2})\frac{(-1)^k}{k!} x^k + o(x^n)$$
Well, I've tried to change every $e^x$ to $1 + x + \frac{x}{2!} + ... + o(x^n)$ and got nothing useful. I know how to get Maclaurin series for polynomials but I don't really know what should I do with fractions and how get that kind of result.</p>
| Nick Peterson | 81,839 | <p><strong>Hint:</strong> I would start by simplifying:
$$
\frac{x^2+3e^x}{e^{2x}}=x^2e^{-2x}+3e^{-x}.
$$</p>
|
2,019,711 | <p>Correction: For what values of the real number $a$, can
$$
a(x_1^2+x_2^2+x_3^2)+2(x_1x_2+x_1x_3+x_2x_3)
$$
be expressed as sum of the form
$$
\alpha x^2+\beta y^2+\gamma z^2
$$
where $\alpha,\beta,\gamma$ are real numbers? </p>
| Community | -1 | <p>The deep reason is that Newton is a "single point" approach which is <em>unaware of the existence of a root</em> ! It can keep iterating even if there is none, and widly wander.</p>
<p>This contrasts with "two points" methods (such as dichotomy, regula falis or the more sophisticated Brent method) that keep track of a <em>change of sign</em> and narrow it down.</p>
<p>Newton is extremely efficient when the first order Taylor approximation holds</p>
<p>$$f(x)=f(y)+f'(y)(x-y)+R(x,y)$$ with a small remainder, i.e. "close" to a root, and very poor otherwise.</p>
|
2,755,785 | <p>I recently came across a problem in which I had to show that a discrete random variable $X$ has two modes $m_1$ and $m_2$. The information given was that $$\frac{P(X=n)}{P(X=n-1)} = \frac{0.9(n-1)}{n-3}$$</p>
<p>I need to show that this distribution has two modes, $m_1$ and $m_2$. My initial thoughts were to try and find the probability distribution, but could not do so from the data above. I don't really know how to approach this problem so any inputs would be appreciated.</p>
| lab bhattacharjee | 33,337 | <p>Hint:</p>
<p>$$\tan(\tan^{-1}x+\tan^{-1}y+\tan^{-1}z)=\dfrac{x+y+z-xyz}{1-(xy+yz+zx)}$$</p>
<p>Now we have $$x+y+z=xyz=-\dfrac p1$$</p>
<p>and $xy+yz+zx=\dfrac q1$</p>
<p>But $\dfrac{x+y+z-xyz}{1-(xy+yz+zx)}$ will be undefined if $xy+yz+zx=q=1$</p>
|
3,440,171 | <p>How does one write the Taylor expansion of <span class="math-container">$\log\big(\frac{\sin(x)}x\big)$</span>? It is not defined on <span class="math-container">$x=0$</span>.</p>
| user76284 | 76,284 | <p>According to <a href="http://oeis.org/A046989" rel="nofollow noreferrer">OEIS</a>,</p>
<p><span class="math-container">$$ \log \frac{x}{\sin x} = \sum_{n > 0} x^{2n} \frac{2^{2n-1} (-1)^{n+1} B(2n)}{n (2n)!} $$</span></p>
<p>where <span class="math-container">$B$</span> is the <a href="https://en.wikipedia.org/wiki/Bernoulli_number" rel="nofollow noreferrer">Bernoulli number</a>. Note that the singularity at <span class="math-container">$x=0$</span> is <a href="https://en.wikipedia.org/wiki/Removable_singularity" rel="nofollow noreferrer">removable</a>.</p>
|
1,046,792 | <p>Given:
$$\sin(x+iy)=\cos\theta+i\sin\theta$$
To prove:
$$x=\arccos (\sqrt{\sin\theta})$$
How I tried:
$$\begin{align*}
\sin x \cosh y &= \cos\theta \\
\cos x \sinh y &= \sin\theta
\end{align*}$$
Then tried to use logarithm of hyperbolic complex number.</p>
<p>Also various trignometric form manipulation but I can't get the answer.</p>
| M. Wind | 30,735 | <p>We start off with two real numbers $x$ and $y$, and consider the complex function $\sin(x+iy)$. As has been pointed out by others in this thread, this is a complex number that can written as $z = \sin(x + iy) = \sin(x)\cosh(y)+i\cos(x)\sinh(y)$. </p>
<p>We are then asked to express this result in terms of a new variable $\theta$ by means of the relation $z = e^{i\theta} = \cos(\theta) + i\sin(\theta)$. In general this is only possible if $\theta$ itself is a complex number, and equally so for the functions in which it appears. For example $\sin(\theta)$ can be written as $\sin(\theta) = \frac{(z - 1/z)}{2i}$.</p>
<p>However, the OP has now stated that $\theta$ is a <strong>real</strong>. This implies that $z$ lies on the unit circle. Using the fact that the norm of $z$ is equal to one, we see $x$ and $y$ are related by: $\sinh(y) = \pm\cos(x)$. Note that this equation can only have a solution when $\sinh(y)$ is in the interval $(-1, 1)$, hence values of $y$ are restricted to the interval $(-0.881377, +0.881377)$. </p>
<p>The two branches of the solution are described by the following equations.</p>
<p>A) $y = +arc\sinh(\cos(x))$. Also $\sin(\theta) = \cos^2(x)$ and $\cos(\theta) = \sin(x)\sqrt{1+cos^2(x)}$</p>
<p>B) $y = -arc\sinh(\cos(x))$. Also $\sin(\theta) = -\cos^2(x)$ and $\cos(\theta) = \sin(x)\sqrt{1+cos^2(x)}$</p>
<p>We see that in both branches $y$ and $\theta$ are periodic functions of $x$. If we combine the two branches in a single plot, $y$ and $\theta$ oscillate perfectly in phase, each having period $\pi$. The amplitude of $y$ is $arc\sinh(1) = 0.881377$ and the amplitude of $\theta$ is $\pi/2$. Note however that this interpretation does not apply to the separate branches, because then the sign of $\theta$ remains the same.</p>
|
199,235 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/9505/xy-yx-for-integers-x-and-y">$x^y = y^x$ for integers $x$ and $y$</a> </p>
</blockquote>
<p>Determine the number of solutions of the equation
$n^m = m^n$
where both m and n are integers.</p>
| robjohn | 13,854 | <p><strong>Hint:</strong></p>
<p>Since $m^n=n^m$, take logs and separate the variables:
$$
\frac{\log(m)}{m}=\frac{\log(n)}{n}
$$
This suggests considering the function $f(x)=\frac{\log(x)}{x}$.</p>
<p>$\hspace{2cm}$<img src="https://i.stack.imgur.com/NFeRg.png" alt="enter image description here"></p>
<p><strong>Another Approach:</strong></p>
<p>Start by comparing $n^{n+1}$ vs $(n+1)^n$. Divide both by $n^n$, to get $n$ vs $\left(1+\frac1n\right)^n$. We can use the binomial theorem to get
$$
\begin{align}
\left(1+\frac1n\right)^n
&=\sum_{k=0}^\infty\binom{n}{k}\frac1{n^k}\\
&=\sum_{k=0}^\infty\frac1{k!}\frac{n}{n}\frac{n-1}{n}\cdots\frac{n-k+1}{n}\\
&<\sum_{k=0}^\infty\frac1{k!}\\
&<1+\sum_{k=1}^\infty\frac1{2^{k-1}}\\
&=3
\end{align}
$$
Thus, for $n\ge3$, we have
$$
n\ge3>\left(1+\frac1n\right)^n
$$
Multiplying both sides by $n^n$ yields that for $n\ge3$
$$
n^{n+1}>(n+1)^n
$$
Taking the $n(n+1)$ root of both sides gives
$$
n^{1/n}>(n+1)^{1/(n+1)}
$$
So we have determined that $n^{1/n}$ is monotonically decreasing for $n\ge3$. What does that say about $m^n$ and $n^m$ when $n>m\ge3$?</p>
<p><strong>Simpler Proof by Induction</strong></p>
<p>I just noted that $n^{n+1}>(n+1)^n$ for $n\ge3$ can be also proven pretty simply by induction.</p>
<p>Note that $3^4=81>64=4^3$.</p>
<p>Suppose that $n^{n+1}>(n+1)^n$. Divide through by $n^n$ to get
$$
n>\left(1+\frac1n\right)^n
$$
Multiply through by $1+\frac1n$ to get
$$
n+1>\left(1+\frac1n\right)^{n+1}
$$
Since $1+\frac1n>1+\frac1{n+1}$ we get
$$
n+1>\left(1+\frac1{n+1}\right)^{n+1}
$$
Multiply through by $(n+1)^{n+1}$ to get
$$
(n+1)^{n+2}>(n+2)^{n+1}
$$
This finishes the induction.</p>
|
1,567,473 | <p>$-\frac12·\frac{iz-2z}{z^2}+\frac{-1-i}{2i-2z}=
\frac{\frac{-3}4-\frac12i}{z}$</p>
<p>if $z=a+bi$,
How to find $a$ and $b$?
Thank you.</p>
| Martin R | 42,969 | <p>In your case $x_0 = y_0 = 0$. For any $a > 0, b > 0$, the function
$f(x, y) = y^2 + \cos^2 x$ is defined and Lipschitz continous on
the rectangle $R=\{|x|\leq a, |y|\leq b\}$.</p>
<p>On this rectangle, $$|f(x, y)| \le |y|^2 + |\cos x|^2 \le b^2 + 1$$
with equality for $x = 0$ and $y = b$, so
the supremum is $M = b^2 + 1$ and therefore
$$h= \min\{a,\frac{b}{b^2 + 1}\}$$</p>
<p>The Picard existence theorem states that the IVP has a (unique)
solution on the interval $[-h, h]$, or – if you restrict the problem
to $x \ge 0$ – on $[0, h]$.</p>
<p>The task is now to choose $a$ and $b$ such that $h$ becomes as
large as possible. From the AM-GM inequality is follows that
$$
b = \sqrt{1 \cdot b^2} \le \frac{b^2 + 1}2 \Longrightarrow \frac{b}{b^2 + 1} \le \frac 12
$$
with equality for $b=1$.</p>
<p>It follows that $h \le \frac 12$ for <em>any</em> choice of $a, b$, and
$h = \frac 12$ for $a = \frac 12, b = 1$. So
$h = \frac 12$ is the largest value that can be obtained by this
method.</p>
|
2,334,305 | <blockquote>
<p>Let $F$ be the distribution function of the continuous-type random variable $X$, and assume that $F(x)=0$ for $x \le 0$ and $0 \lt F(x) \lt1$ for $0 \lt x$. Prove that if $P(X \gt x+y |X \gt x)=P(X \gt y),$ then $F(x)=1-e^{- \lambda x}, 0 \lt x.$</p>
</blockquote>
<p>My attempt: </p>
<p>Let $F(x)=P(X \le x)$. Then </p>
<p>$ \frac {P(X \gt x+y)}{P(X \gt x)}=P(X \gt y)$</p>
<p>$\Rightarrow \frac {1-P(X\le x+y)}{1-P(X \le x)}=1-p(X \le y)$</p>
<p>$ \Rightarrow \frac {1-F(x+y)}{1-F(x)}=1-F(y)$</p>
<p>Then I have no idea where to go from here. Any help is appreciated. </p>
| Jonathan Davidson | 457,443 | <p>Let $F(s) = P(X>s)$ Using conditional probability manipulations, $$F(s) = P(X>s) = P(X > s+t\mid X >t) = \frac{P(X>s+t)}{P(X>t)} = \frac{F(s+t)}{f(t)}$$ yielding the functional equation $$F(s+t) = F(s)F(t)$$ Next we consider the difference quotient $$\frac{F(x+h)-F(x)}{h} = F(x)\left(\frac{F(h)+1}{h}\right)$$ Taking limits of both sides and applying the definition of the derivative, we obtain the differential equation $$F'(x) = -\lambda F(x)$$ where $\lambda = -F'(0)$. The solution of this differential equation is the exponential function, $F(x) = e^{-\lambda x}$. The cumulative distribution function is then $$P(X\le x) = 1-P(X>x) = 1-e^{-\lambda x}$$</p>
|
115,657 | <p>In topology the spheres <span class="math-container">$S^n$</span> are the "simplest" closed manifolds, and they are like "Dirac's delta at <span class="math-container">$n$</span>" for (reduced) cohomology groups. Furthermore they are boundaries of the simplest compact manifolds-with-boundary, i.e. the disks <span class="math-container">$D^{n+1}$</span>, which are contractible. And <span class="math-container">$S^{n}$</span> is obtained by glueing two copies of <span class="math-container">$D^{n}$</span> along their boundary <span class="math-container">$S^{n-1}$</span>.
My question is:</p>
<blockquote>
<p>Are there some objects of algebraic geometric nature that somehow reproduce the same pattern, or that are considerable as the equivalent of spheres from topology?</p>
<p>More generally, are there "homology spheres" for some homology theory like -say- Chow groups? What about an "algebraic Poincaré conjecture"?</p>
</blockquote>
<p>If they do exist, I don't expect them to be standard varieties or schemes, otherwise they probably would have made their appearence "classically".</p>
| Piotr Achinger | 3,847 | <p>Probably not, at least treating the question naively, since algebraic geometry focuses on smooth projective varieties, and these have nonzero cohomology (for basically any cohomology theory) in degrees $0, 2, \ldots 2\cdot \dim$: the class of the hyperplane section in $H^2$ has nonzero top cup product with itself. </p>
|
4,587,608 | <p>The following is the definition of a change-of-coordinate matrix from a textbook I'm using:</p>
<p>Let <span class="math-container">$\beta$</span> and <span class="math-container">$\beta'$</span> be two ordered bases for a finite-dimensional vector space V, and let <span class="math-container">$Q=[I_v]_{\beta}^{\beta'}$</span>. Then <span class="math-container">$Q$</span> is called a change of coordinate matrix.</p>
<p>There is, however, no explanation of what <span class="math-container">$[I_v]$</span> means, and I can't understand it.</p>
<p>Presumably, it should mean the identity matrix of the size appropriate for the vector space <span class="math-container">$V$</span>. It appears that the identity matrix is the same independent of the basis (1s on the diagonal), but in that case, how could it possible take us from one basis to another?</p>
<p>Any help with what <span class="math-container">$[I_v]$</span> signifies would be greatly appreciated.</p>
| azif00 | 680,927 | <p>Let <span class="math-container">$u = t$</span> and <span class="math-container">$dv = \frac{-t}{\sqrt{1-t^2}}dt$</span>. Then <span class="math-container">$du=dt$</span> and <span class="math-container">$v = \sqrt{1-t^2}$</span>; so
<span class="math-container">\begin{align}
\int \frac{-t^2}{\sqrt{1-t^2}}dt = \int udv
&= uv - \int vdu \\
&= \textstyle t \sqrt{1-t^2} - \displaystyle \int \textstyle \sqrt{1-t^2} dt.
\end{align}</span>
Can you take it from here?</p>
|
696,370 | <p>Is it possible that $R/I$ is a field when $R$ is non-commutative ring with unit and $I$ is a maximal left ideal of $R$? If it is not, can anyone give an example of such $R$ and $I$? Thanks.</p>
| DonAntonio | 31,254 | <p>By Cauchy's Theorem, the additive group of $\;\Bbb F\;$ has an element of order two, meaning</p>
<p>$$a+a=a(1+1)=0\implies 1+1=0$$</p>
<p>and clearly (why?) this is the minimal possible integer with this characteristic...</p>
|
997,116 | <p>If one root of the equation $x^2 + ax + b = 0$ and $x^2 + bx + a = 0$ is common and $a \ne b$ then:</p>
<p>The options are as follows:
$$\begin{array}{ll}
(A)\quad& a + b = 0\\
(B)& a + b = -1\\
(C)& a - b = 1\\
(D)& a + b = 1
\end{array}$$</p>
<p>Idk how to solve this, please help me.</p>
| Macavity | 58,320 | <p>Hint: Substitute the common root and subtract. </p>
|
1,330,078 | <p>Given a quadrilateral $MNPQ$ for which $MN=26$, $NP=30$, $PQ=17$, $QM=25$ and $MP=28$ how do I find the length of $NQ$?</p>
| Community | -1 | <p>Use the cosine law to find the two partial angles at $M$. Then you know the total angle and you can find $NQ$.</p>
<p>$$\mu_0=\arccos\left(\frac{MQ^2+MP^2-PQ^2}{2MQ\cdot MP}\right)\\
\mu_1=\arccos\left(\frac{MP^2+MN^2-NP^2}{2MP\cdot MN}\right)\\
NQ=\sqrt{MQ^2+MN^2-2MQ\cdot MN\cos\left(\mu_0+\mu_1\right)}.$$</p>
|
18,691 | <p>Let's say I have a folowing set of data:</p>
<ul>
<li>k = 1 : list of values </li>
<li>k = 3 : list of values </li>
<li>k = 10 : list of values</li>
</ul>
<p>I know that to make a <code>BoxWhiskerChart</code> I have to give it a list of listen of these values as data and ks as labels. </p>
<p>How do I force the offset between the boxes for different ks to be proportional to the values of ks?</p>
<p>This is like combining <code>ListPlot</code> and <code>BoxWhiskerChart</code> - list plot gives appropriate position of boxes relative to the x-axis.</p>
| Jens | 245 | <p>Since the <code>BoxWhiskerChart</code> doesn't label its x-axis coordinate, I decided to make use of the functionality that already exists in <code>ErrorListPlot</code>. The latter is part of the <code>"ErrorBarPlots"</code> package which has to be loaded explicitly.</p>
<p>First I generate the data, then I extract a list of coordinates for <code>ErrorListPlot</code>, then I specify a function that renders the "error bar" as a <code>BoxWhiskerChart</code> for each data set individually. The scale has to be common to all plots, and that's achieved with a combination of fixed <code>PlotRange</code> and <code>AspectRatio -> Full</code> for the inset whisker plots.</p>
<pre><code>data = {{1, RandomInteger[{-10, 10}, 10]}, {3,
RandomInteger[{-10, 10}, 10]}, {10, RandomInteger[{-10, 10}, 10]}};
data = {{1, {7, -9, 5, 1, 3, 8, 10, 10, -3,
0}}, {3, {-8, -6, -6, 7, -8, -8, 2, -8, -2, -4}}, {10, {-9, -3,
9, 5, -1, 0, 9, 2, 3, -1}}};
dataValues = data[[All, 2]];
</code></pre>
<p>The original chart is here:</p>
<pre><code>BoxWhiskerChart[dataValues]
</code></pre>
<p><img src="https://i.stack.imgur.com/2phkS.png" alt="chart"></p>
<p>Now for the modifications to <code>ErrorListPlot</code>: </p>
<pre><code>{min, max} = {Min[#], Max[#]} &@Flatten[dataValues]
(* ==> {-9, 10} *)
kValues = MapIndexed[Flatten[{#, min, #2}] &, data[[All, 1]]];
whiskers = Map[
BoxWhiskerChart[#, Frame -> None, PlotRange -> {min, max},
AspectRatio -> Full, PlotRangePadding -> 0,
PlotRangeClipping -> False] &,
dataValues];
Needs["ErrorBarPlots`"]
errorfunction[{x_, y_}, err_] := Inset[
whiskers[[Last[Flatten[List @@ err]]]],
{x, y}, {Center, Bottom}, Scaled[{.1, 1}]]
ErrorListPlot[kValues, ErrorBarFunction -> errorfunction,
PlotMarkers -> "", PlotRange -> {min, max}]
</code></pre>
<p><img src="https://i.stack.imgur.com/qj0Mw.png" alt="error plot"></p>
<p>The tooltips of the original chart still work in this plot, too.</p>
<p><strong>Edit</strong></p>
<p>In response to the additional question about logarithmic x axis, we can work with the same data but can't use <code>ErrorListPlot</code>. So instead one could do this:</p>
<pre><code>ListLogLinearPlot[
List /@ kValues[[All, 1 ;; 2]],
PlotMarkers -> Map[{#, 1} &, whiskers],
PlotRange -> {min, max}] /.
Inset[i_, pos_, align_, scl_] :>
Inset[i, pos, {Center, Bottom}, Scaled[{.1, 1}]]
</code></pre>
<p>The idea is the same as above: <em>"inject"</em> the <code>BoxWhiskerChart</code> at the correct location by means of an <code>Inset</code>. The positioning can be taken care of by the plot function, but the <em>scaling</em> is not under my control in the plotting process. So I have to fix that by <em>post-processing</em> using the replacement <code>/.</code> with a rule that looks for <code>Inset</code> in the finished plot and aligns its bottom with the plot point coordinate. The latter is always equal to the minimum data value (<code>min</code>), so that by setting <code>PlotRange</code> and using <code>Scaled[{.1, 1}]</code> for the fourth <code>Inset</code> argument we get the same result as in <code>ErrorListPlot</code>, only with a logarithmic axis:</p>
<p><img src="https://i.stack.imgur.com/ZJSYi.png" alt="log axis"></p>
<p>This last approach would of course also work with <code>ListPlot</code> and could therefore be a replacement for my my first solution.</p>
<p>A limitation of both approaches is that the vertical <code>PlotRange</code> must correspond exactly to the one in the <code>BoxWhiskerChart</code>, so that the scaling works properly by assuming the <code>Inset</code> should span <code>100%</code> of the vertical space. This means you can't add <code>PlotRangePadding</code> in the vertical direction. You can add it in the <em>horizontal</em> direction, though. Here is an illustration of an explicit horizontal <code>PlotRange</code> to keep the bars away from the edges:</p>
<pre><code>ListLogLinearPlot[
List /@ kValues[[All, 1 ;; 2]],
PlotMarkers -> Map[{#, 1} &, whiskers],
AxesOrigin -> {.9, 0},
PlotRange -> {{.9, 11}, {min, max}}] /.
Inset[i_, pos_, align_, scl_] :>
Inset[i, pos, {Center, Bottom}, Scaled[{.1, 1}]]
</code></pre>
<p><img src="https://i.stack.imgur.com/pDxxx.png" alt="plotrange example"></p>
|
2,758,338 | <p>I need to find the laurent series and the residue of the following complex function
$$f(z)=(z+1)^2e^{3/z^2}$$
at $z=0$.</p>
<p>Since $e^z=\sum z^n/n!$, then
$$e^{3/z^2}=\sum_{n=0}^\infty \frac{3^n/n!}{z^{2n}}$$
thus
$$f(z)=(z^2+2z+1)\sum_{n=0}^\infty \frac{3^n/n!}{z^{2n}}=\sum_{n=0}^\infty \frac{3^n/n!}{z^{2(n-1)}}+\sum_{n=0}^\infty \frac{2\cdot3^n/n!}{z^{2n-1}}+\sum_{n=0}^\infty \frac{3^n/n!}{z^{2n}}$$
which, with a shift of index and expansion of positive powers, can be expressed as
$$f(z)=z^2+3+\sum_{n=1}^\infty\left(\frac{3}{n+1}+1\right)\frac{3^n/n!}{z^{2n}}+\sum_{n=1}^\infty\frac{2\cdot3^n/n!}{z^{2n-1}}$$
so the residue is given by evaluating the numerator of the second series at $n=1$, so its value is $6$. I tried using WolframAlpha and Mathematica to check my answer, but both would not return a value. Would this be correct? Also, is there a way to put the two sums together (one gives the coefficients of even powers, while the other of the odd) so I can have the principal part of the laurent series expressed only with one sum?</p>
| Ashwin Trisal | 343,481 | <p>No, this isn't true. Consider $S_5$ (not minimal but easy). Then consider the groups $\langle (1\,2)\rangle$ and $\langle (3\, 4)\rangle$. Both are isomorphic to $\mathbb{Z}/2\mathbb{Z}$, but they're not equal.</p>
|
300,154 | <p>I am considering an optimization problem of the form:
\begin{equation}
\begin{split}
f(s) &= \min_{X} \mathrm{tr}(C(s)X) \\
&\;\;\;\;\;\;\;\;\;\;\; X \ge 0, \\
&\;\;\;\;\;\;\;\;\;\;\; \mathrm{tr}(A_iX) = a_i, \;\; 1 \le i \le M,
\end{split}
\end{equation}
where the minimization is over $n\times n$ Hermitian matrices $X$. Further, $A_i$ for $1 \le i \le M$ denote some $n\times n$ Hermitian matrices which together with $a_i \in \mathbb{R}$ determine linear constraints on $X$. Finally, the matrix-valued function $C(s)$ is of the block form:
\begin{equation}
C(s) = \left( \begin{array}{cc} C_{1}(s) & 0 \\ 0 & 0\end{array} \right),
\end{equation}
where the upper left block $C_1(s)$ is of size $(n_1 + 1) \times (n_1 + 1)$ for some $n_1 < n$, and is given by:
\begin{equation}
C_1(s) = \left( \begin{array}{ccccc} I_{n_1\times n_1} & -ic \mathbb{I}_{n_1\times n_1} & \cdot & \cdot \\ i c \mathbb{I}_{n_1\times n_1} & \cdot & -i \frac{s}{2} \mathbb{I}_{n_1\times n_1} & \cdot \\ \cdot & i \frac{s}{2} \mathbb{I}_{n_1\times n_1} & \cdot & \cdot \\ \cdot & \cdot & \cdot & s^2\end{array} \right).
\end{equation}
Here, $c \in \mathbb{R}$, $I_{n_1\times n_1}$ is the $n_1 \times n_1$ matrix of ones and $\mathbb{I}_{n_1\times n_1}$ denotes the $n_1 \times n_1$ identity matrix (whereas all entries indicated by $\cdot$ vanish).</p>
<p>Can it be shown that $f(s)$ is convex?</p>
<p>If not, which further requirements has the optimization to fulfill in order to guarantee convexity of $f(s)$?</p>
| DSM | 155,380 | <p>Consider the following argument for a slightly changed problem (with <span class="math-container">$s^2$</span> in <span class="math-container">$C(s)$</span> replaced by <span class="math-container">$-s^2$</span>) . Not sure if this would be of any help, but writing it anyway. Note that due to concavity of <span class="math-container">$-s^2$</span> in C(s) and the rest of the terms being either constant or linear in <span class="math-container">$s$</span> (in <span class="math-container">$C(s)$</span>), we have:
<span class="math-container">$$
C(\lambda s_1 + (1-\lambda)s_2) \succeq \lambda C(s_1) + (1-\lambda)C(s_2).
$$</span>
Therefore,
<span class="math-container">$$
f(\lambda s_1 + (1-\lambda)s_2) \geq \min_{X\in \Gamma} \left\{ \lambda \mbox{Tr}(C(s_1)X) + (1-\lambda)\mbox{Tr}(C(s_2)X) \right\} \geq \lambda \min_{Y\in \Gamma} \left\{ \mbox{Tr}(C(s_1)Y)\right\} + (1-\lambda) \min_{Z\in \Gamma} \left\{ \mbox{Tr}(C(s_2)Z)\right\} = \lambda f(s_1) + (1-\lambda)f(s_2).
$$</span>
And hence <span class="math-container">$f(.)$</span> is concave in <span class="math-container">$s$</span>.</p>
|
4,337,887 | <p>I am solving an exercise:</p>
<p>Let <span class="math-container">$T: V \rightarrow W$</span> be a linear transformation.
<span class="math-container">$V$</span> and <span class="math-container">$W$</span> are finite-dimensional inner product spaces. Prove <span class="math-container">$T^*T$</span> and <span class="math-container">$TT^*$</span> are <strong>semidefinite</strong>.</p>
<p>This is a solution that I don't understand: <br>
<span class="math-container">$T^*T$</span> and <span class="math-container">$TT^*$</span> are self-adjoint, then we have <span class="math-container">$T^*T(x) = \lambda x$</span>. Hence: <br>
<span class="math-container">$$\lambda = ⟨T^*T(x),x⟩ = ⟨T(x),T(x)⟩ ≥ 0.$$</span>
<span class="math-container">$\lambda$</span> is <span class="math-container">$≥ 0$</span>, hence <span class="math-container">$T^*T$</span> is semidefinite.</p>
<p>I don't understand why the eigenvalue is equal to <span class="math-container">$⟨T^*T(x),x⟩$</span>.
Thank you for any kind of help!</p>
| paw88789 | 147,810 | <p>The number of divisors function, sometimes denoted <span class="math-container">$\tau$</span> (tau) includes <span class="math-container">$1$</span> and <span class="math-container">$n$</span> as divisors also. So for instance <span class="math-container">$\tau(12)=6$</span> (divisors are <span class="math-container">$1, 2, 3, 4, 6, 12$</span>.)</p>
|
2,006,927 | <p>I want to prove that $\cap_{i=1}^{\infty}\left(A_i \cap \left(\cup_{j=1}^{\infty}B_j\right)\right)=\cup_{j=1}^{\infty}\left(\left(\cap_{i=1}^{\infty}A_i\right) \cap B_j\right)$ for sets $A_i,B_j$ and natural numbers $i,j$. If an element $x$ belongs to the left hand side, then $x\in A_1$ and $x\in$ some of $B_j$ and $x\in A_2$ and $x\in$ some of $B_j$ and so forth. Then $x\in A_1$, $x\in A_2$, $x\in A_3$ etc so $x \in \cap_{i=1}^{\infty}A_i$ but I don't see how I can proceed with the B:s and get to $x \in B_1$ and $x\in$ all $A_i$ or $x \in B_2$ and $x\in$ all $A_i$ or $x \in B_3$ and $x\in$ all $A_i$ and so forth.</p>
| DanielWainfleet | 254,665 | <p>For any sets $I,J$ let $A=\cap_{i\in I}A_i,\; B=\cup_{j\in J}B_j,\;C=\cap_{i\in I}(A_i\cap B),\; D=\cup_{j\in J}(A\cap B_j).$</p>
<p>If $I\ne \emptyset$ then for any $x$ we have $$x\in C\iff \forall i\in I\;(x\in A_i\land x\in B)\iff$$ $$\exists j\in J\; (\;x\in B_j\land (\forall i\in I\;(x\in A_i))\;)\iff$$ $$ \exists j\in J\;(x\in B_j\land x\in A)\iff$$ $$ \exists j\in J\;(x\in A \cap B_j)\iff$$ $$ x\in \cup_{j\in J}(A\cap B_j)=D.$$</p>
<p>If $I=\emptyset$ then by def'n, $A= C=\emptyset$ and also $A\cap B_j=\emptyset$ for every $j\in J,$ so $D=\cup_{j\in J}\emptyset=\emptyset =C.$ </p>
<p>Note: We cannot write $x\in \cap_{f\in F}\iff \forall f\in F\;(x\in f)$ when $F$ is empty, as then we would have $x\not \in \cap_{f\in F}\iff \exists f\in F\; (x\not \in f),$ but since there does not exist any $f\in F,$ this would imply that NO $x$ fails to belong to $\cap_{f\in F}f,$ which would be inconvenient. So we define $\cap_{f\in \emptyset}f=\emptyset.$</p>
|
1,000,938 | <p>$\lim_{X \to \infty} \int_0^Xf(x)^2 > 2(\lim_{X \to \infty}\int_0^Xf(x))^2$</p>
| Christian Blatter | 1,303 | <p>Since you just need one example, consider
$$f(t):=e^{-4t}\qquad(t\geq0)\ .$$
Then
$$f(x):=\int_0^x f^2(t)\>dt={1\over8}\bigl(1-e^{-8x}\bigr)$$
and
$$g(x):=2\left(\int_0^x f(t)\>dt\right)^2={1\over 8}\left(1-e^{-4x}\right)^2\ .$$
It follows that
$${f(x)\over g(x)}=\coth(2x)>1\qquad(x>0)\ .$$</p>
|
1,115,793 | <p>Let's consider a number of linear operators, defined on a finite dimensional complex vector space, which two by two commutes with each other. (the amount of them can be infinite). How to prove that that will have a common eigenvector?</p>
<p>The finite case can be done by induction:
1) $n=2$, $AB=BA$, then let $x$ be an eigenvector of $A$ (it does exist, because we are working over a $\mathbb{C}$) and $\alpha$ - an eigenvalue. Then, $A(x)=\alpha \cdot x, B(A(x))=A(B(x))=B(\alpha x)=\alpha B(x)$, so $B(x)$ is also an eigenvector of $A$, associated with $\alpha$ eigenvalue.
Analogically, we do it for $n>2$.</p>
<p>But, what can i do, while working with an infinite number of operators( induction doesn't work here, actually).</p>
<p>Any help would be appreciated. </p>
| voldemort | 118,052 | <p>Let's say your vector space is $\mathbb{C^n}$. Then $M_n(\mathbb{C})$, i.e. all $n$ by $n$ matrices are the bounded operators on $\mathbb{C^n}$.</p>
<p>Now, $M_n(\mathbb{C})$ is finite dimensional. So, even if you have infinitely many operators, say $A_1,A_2,\cdots$, there will exist $i_1,i_2,\cdots,i_k$ such that $A_{i_1},\cdots,A_{i_k}$ will span the rest of the matrices. So, your case for finitely many operators will work.</p>
|
36,272 | <p>Is there a characterisation for which $x\in\mathbb{R}$ the value $\arctan(x)$ is a rational multiple of $\pi$? </p>
<p>Or reformulated: What is the "structure" of the subset $A\subseteq\mathbb{R}$ which fulfils
$$ \arctan(x) \in \pi\mathbb{Q} \Leftrightarrow x\in A$$
for all $x\in\mathbb{R}$?</p>
| Philip Thomas | 114,952 | <p>It is easy to show that for
$$\frac{\pi}{2^{n+1}}=\arctan x,$$
where $n=\left\{0,1,2,3,\ldots\right\}$, the argument $x$ is always an irrational number. Therefore $\arctan x$ cannot be a rational multiple of $\pi$ at $x\in\mathbb{R}$ for this specific case.</p>
|
4,494,199 | <p>This is a problem from a past qualifying exam in complex analysis. I'm working through these to study for my own upcoming qual. For this question, I think my proof is fairly straightforward, but I'd like to know whether or not it is correct and complete. I'm also interested in other ways of answer the question. Thanks!</p>
<p><strong>Problem:</strong></p>
<p>Find how many solutions (counting multiplicity) the equation <span class="math-container">$\sin z = ez^4$</span> has on the unit disk <span class="math-container">$|z|<1$</span>. Justify your answer.</p>
<p><strong>My Solution:</strong></p>
<p>Let <span class="math-container">$g(z) = \sin(z)$</span> and <span class="math-container">$f(z) = -ez^4$</span> and consider these functions on the unit circle <span class="math-container">$|z|=1$</span>. We will show by Rouche's Theorem that since <span class="math-container">$|g(z)|<|f(z)|$</span>, then <span class="math-container">$f(z)+g(z)$</span> has the same number of zeros inside the unit circle as <span class="math-container">$f(z)$</span> counting multiplicities, thus there are four solutions to the given equation.</p>
<p>First, we need to show that <span class="math-container">$|\sin(z)|<|e|$</span>. We have
<span class="math-container">$$
\begin{align*}
|\sin(z)| &= \left\vert \frac{e^{iz} - e^{-iz}}{2i}\right\vert \\
&= \frac{1}{2} |e^{i(x+iy)} - e^{-i(x+iy)}| \\
&\leq \frac{1}{2}(|e^{i(x+iy)}| + |- e^{-i(x+iy)}|)\\
&= \frac{1}{2}(e^{-y} + e^y)
\end{align*}
$$</span></p>
<p>On <span class="math-container">$|z|=1$</span>, we have <span class="math-container">$|y|\leq 1$</span>, thus
<span class="math-container">$$
|\sin(z)| \leq \frac{1}{2}(e^{-y} + e^y) \leq \frac{1}{2}(e^{1}+e^{-1})< \frac{1}{2}(2e) = e.
$$</span></p>
<p>Now, we have that when <span class="math-container">$|z|=1$</span>
<span class="math-container">$$
|\sin(z)|<e = e|z|^4 = |-ez^4|,
$$</span></p>
<p>thus <span class="math-container">$|g(z)|<|f(z)|$</span>, and by Rouche's Theorem, <span class="math-container">$f(z)+g(z)$</span> has the same number of zeros inside the unit circle as <span class="math-container">$f(z)$</span>. Since <span class="math-container">$f(z) = -ez^4$</span> is a polynomial, we know by the fundamental theorem of algebra that it has exactly four roots counting multiplicity. Thus,
<span class="math-container">$$
f(z) + g(z) = \sin(z) - ez^4
$$</span></p>
<p>has exactly four roots and <span class="math-container">$\sin(z) = ez^4$</span> has exactly four solutions. <span class="math-container">$\blacksquare$</span></p>
| Arctic Char | 629,362 | <p>I guess this is a typo, they meant to say</p>
<p><span class="math-container">$$ 1-a \ge b-a,$$</span></p>
<p>which is true from</p>
<p><span class="math-container">$$ (1-a)(1+a) \ge b(b-a)$$</span>
since <span class="math-container">$1+a \ge 1 \ge b$</span>.</p>
<p>(Of course, <span class="math-container">$1-a\ge b-a$</span> is the same as <span class="math-container">$1\ge b$</span>)</p>
|
158,978 | <p><a href="https://i.stack.imgur.com/5nKVV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5nKVV.jpg" alt="Explore the Fourier Series for a square wave"></a></p>
<pre><code>f (t) = (4/pi) Sum[(1/n) sin (2 pi (f (t))), {n, 1, Infinity}]
</code></pre>
| m_goldberg | 3,066 | <p>First you need to learn the proper syntax for writing Mathematica expressions. When the sum is written properly as</p>
<pre><code>f[t_] = (4/Pi) Sum[(1/n) Sin[2 Pi n t], {n, 1, ∞, 2}]
</code></pre>
<blockquote>
<p><code>(2 I (ArcTanh[E^(-2 I π t)] - ArcTanh[E^(2 I π t)]))/π</code></p>
</blockquote>
<p>then the plot is simple:</p>
<pre><code>Plot[f[t], {t, 0, 3}]
</code></pre>
<p>plot<a href="https://i.stack.imgur.com/AhDas.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AhDas.png" alt="enter image description here"></a></p>
|
414,400 | <p>Let $ \mathbb{Z}[i]$ denote the ring of the Gaussian intergers. For which of the following value of n is the quotient ring $ \mathbb{Z}[i]/n\mathbb{Z}[i]$ an integral domain?</p>
<p>$ a. 2$</p>
<p>$ b. 13$</p>
<p>$ c. 19$</p>
<p>$ d. 7$</p>
<p>I'm doubtful with the following attempt I made.</p>
<ul>
<li><p><strong>I think all 4 options are correct:</strong> It suffices to show $n\mathbb Z[i]$ is a prime ideal of $\mathbb Z[i]$ if $n$ is prime. Now $(n)=n\mathbb Z[i].$ So $n$ is prime element of $n\mathbb Z[i]\implies(n)$ is a prime ideal of $\mathbb Z[i].$</p>
<p>Let $n$ be a prime integer. Of course then $n$ is non zero and non unit. Let $n|(a+ib)(c+id).$ That's $n|(ac-bd)+i(ad+bc)\\\implies\dfrac{ac-bd}{n},\dfrac{ad+bc}{n}\in\mathbb Z\\\implies n|ac,bd,ad,bc\\\implies n|\{a~or~c\}~and~\{b~or~d\}~and~\{a~or~d\}~and~\{b~or~c\}\\\implies n\text{ divides at least $3$ of }a,b,c,d.$</p>
<p>WLG let $n|a,b\implies n|a+ib.$</p></li>
</ul>
<p><strong>Is my attempt correct?</strong></p>
| zacarias | 35,464 | <p>The following theorem is well known (see a book on algebraic number theory)</p>
<p>Theorem: Let $p$ be a rational prime (that is, a prime in $\mathbb Z$). Then, $p$ is a prime in the Gaussian integers $\mathbb Z[i]$ if and only if $p\equiv 3 \pmod 4$.</p>
<p>Since $\mathbb Z[i]/n \mathbb Z[i]$ is an integral domain if and only if $n$ is prime in $\mathbb Z[i]$, it follows that $(3)$ and $(4)$ are correct. </p>
|
874,607 | <blockquote>
<p>There were 10 questions on a test. A student gets 5 points for every correct answer and 3 points for every partially correct answer. If the student got 19 points, how many correct and partial answers did they have?</p>
</blockquote>
<p>To solve the problem I express the total number of points as the sum of multiples of 5 and 3.</p>
<blockquote>
<p>5x+3y=19</p>
</blockquote>
<p>After that, the only thing I can do is find the solution by brute forcing it. Is there a more mathematical way of finding it?</p>
| DeepSea | 101,504 | <p>$5x + 3y = 19$. So $5x \leq 19$, thus $x \leq 3$. Check $x = 2$, and $y = 3$ is the only integer values solution. </p>
|
318,299 | <blockquote>
<p>Let <span class="math-container">$U$</span> be an open set in <span class="math-container">$\mathbb R$</span>. Then <span class="math-container">$U$</span> is a countable union of disjoint intervals. </p>
</blockquote>
<p>This question has probably been asked. However, I am not interested in just getting the answer to it. Rather, I am interested in collecting as many different proofs of it which are as diverse as possible. A professor told me that there are many. So, I invite everyone who has seen proofs of this fact to share them with the community. I think it is a result worth knowing how to prove in many different ways and having a post that combines as many of them as possible will, no doubt, be quite useful. After two days, I will place a bounty on this question to attract as many people as possible. Of course, any comments, corrections, suggestions, links to papers/notes etc. are more than welcome.</p>
| Guldam | 37,753 | <p>$\mathbb{R}$ with standard topology is second-countable space.</p>
<p>For a second-countable space with a (not necessarily countable) base, any open set can be written as a countable union of basic open set.</p>
<p><a href="https://math.stackexchange.com/questions/1395714/open-set-in-a-second-countable-space-which-is-not-a-countable-union-of-basic-ope/1395740#1395740">Given any base for a second countable space, is every open set the countable union of basic open sets?</a></p>
<p>Clearly, collection of open intervals is a base for the standard topology.
Hence any open set in $\mathbb{R}$ can be written as countable union of open intervals.</p>
<p>If any two of exploited open intervals overlap, merge them. Then we have disjoint union of open intervals, which is still countable. </p>
|
318,299 | <blockquote>
<p>Let <span class="math-container">$U$</span> be an open set in <span class="math-container">$\mathbb R$</span>. Then <span class="math-container">$U$</span> is a countable union of disjoint intervals. </p>
</blockquote>
<p>This question has probably been asked. However, I am not interested in just getting the answer to it. Rather, I am interested in collecting as many different proofs of it which are as diverse as possible. A professor told me that there are many. So, I invite everyone who has seen proofs of this fact to share them with the community. I think it is a result worth knowing how to prove in many different ways and having a post that combines as many of them as possible will, no doubt, be quite useful. After two days, I will place a bounty on this question to attract as many people as possible. Of course, any comments, corrections, suggestions, links to papers/notes etc. are more than welcome.</p>
| Asinomás | 33,907 | <p>The balls with radii $\frac{1}{n}$ and center at a rational number form a basis for the euclidean topology. This family is countable as it is a countable union of countable sets. We have found a countable basis, so we are done.</p>
|
318,299 | <blockquote>
<p>Let <span class="math-container">$U$</span> be an open set in <span class="math-container">$\mathbb R$</span>. Then <span class="math-container">$U$</span> is a countable union of disjoint intervals. </p>
</blockquote>
<p>This question has probably been asked. However, I am not interested in just getting the answer to it. Rather, I am interested in collecting as many different proofs of it which are as diverse as possible. A professor told me that there are many. So, I invite everyone who has seen proofs of this fact to share them with the community. I think it is a result worth knowing how to prove in many different ways and having a post that combines as many of them as possible will, no doubt, be quite useful. After two days, I will place a bounty on this question to attract as many people as possible. Of course, any comments, corrections, suggestions, links to papers/notes etc. are more than welcome.</p>
| CopyPasteIt | 432,081 | <p>The balls with radii $\frac{1}{n}$ and center at a rational number form a basis for the euclidean topology. This family is countable since $\mathbb N \times \mathbb Q \equiv \mathbb N$ and we have a countable basis $(B_\lambda)_{ \, \lambda \in \mathbb N \times \mathbb Q}$ of open intervals for $\mathbb R$.</p>
<p>Let $U$ be a nonempty open set in $\mathbb R$; we can express it as countable union of open balls from $(B_\lambda)$.
Also,
$\tag 1 \text{ }$
$\quad$ If two open intervals have a nonempty intersection, then their union is also an open interval.</p>
<p>So if $U$ is a finite union of the $B_\lambda$, it is an easy matter to combine the $B_\lambda$, if necessary, and writing $U$ as a disjoint union of a finite number of open intervals. So assume, WLOG, that</p>
<p>$\tag 2 U = \bigcup_{\, n \in \mathbb N \,} B_n$. </p>
<p>We define a relation on our (new) index set $\mathbb N$ with $m\sim n$ if $B_m \cap B_n \ne \emptyset$ or there is is a finite 'nonempty intersection $B\text{-}$chain' connecting $B_m$ with $B_n$. It is easy to see that this partitions $\mathbb N$ and that taking the corresponding unions of the $B_n$ over an index $\lambda \text{-}$ block gives a partition of $U$. Also, using (1), we can show that we have also expressed $U$ as a countable union of disjoint open intervals.</p>
|
183,768 | <p>Prove convergence\divergence of the series:
$$\sum_{n=1}^{\infty}\dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)}$$</p>
<p>Here is what I have at the moment:</p>
<p><strong>Method I</strong></p>
<p>My first way uses a result that is related to <strong><a href="http://en.wikipedia.org/wiki/Wallis_product" rel="nofollow noreferrer">Wallis product</a></strong> that we'll denote by $W_{n}$. Also,<br>
we may denote $\dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)}$ by $P_{n}$. Having noted these and taking a large
value of $n$<br>
we get:
$$(P_{n})^2 =\frac{1}{W_{n} \cdot (2n+1)}\approx\frac{2}{\pi}\cdot \frac{1}{2n+1}$$
$$P_{n}\approx \sqrt {\frac{2}{\pi}} \cdot \frac{1}{\sqrt{2n+1}}$$ </p>
<p>Further we have that:
$$\lim_{n\to\infty}\sqrt {\frac{2}{\pi}} \cdot \frac{n}{\sqrt{2n+1}} \le \sum_{n=1}^{\infty} P_{n}$$
that obviously shows us that the series diverges.</p>
<p><strong>Method II</strong></p>
<p>The second way is to resort to the powerful <strong><a href="http://mathworld.wolfram.com/KummersTest.html" rel="nofollow noreferrer">Kummer's Test</a></strong> and firstly proceed with the ratio test:
$$\lim_{n\to\infty} \frac{P_{n+1}}{P_{n}}=\frac{2n+1}{2n+2}=1$$
and according to the result, the ratio test is inconclusive.</p>
<p>Now, we apply Kummer's test and get:
$$\lim_{n\to\infty} \frac{P_{n}}{P_{n+1}}n-(n+1)=\lim_{n\to\infty} -\frac{n+1}{2n+1}=-\frac{1}{2} \le 0$$
Since
$$\sum_{n=1}^{\infty} \frac{1}{n} \longrightarrow \infty$$
our series diverges and we're done.</p>
<p>On the site I've also found <a href="https://math.stackexchange.com/questions/118383/convergence-of-sum-limits-n-1-infty-left-dfrac-1-cdot3-cdots-2n-1?rq=1">a related question</a> with answers that can be applied for my question.
Since I've already have some answers for my question you may regard it as a recreational one and if you have a nice proof to share I'd be glad to receive it. I like this question very much and want to make up a collection with nice proofs for it. Thanks. </p>
| robjohn | 13,854 | <p>$$
\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}=\frac{(2n)!}{2^{2n}n!^2}\tag{1}
$$
Using Stirling's Formula, we get that
$$
\frac{(2n)!}{2^{2n}n!^2}\sim\frac1{\sqrt{\pi n}}\tag{2}
$$
By the $p$-test,
$$
\sum_{n=1}^\infty \frac1{n^p}\tag{3}
$$
diverges for $p\le1$,
$$
\sum_{n=1}^\infty\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\tag{4}
$$
diverges.</p>
<p><strong>Derivation of (1):</strong></p>
<p>$$
\begin{align}
\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}
&=\frac{1\cdot\color{#C00000}{2}\cdot3\cdot\color{#C00000}{4}\cdot5\cdot\color{#C00000}{6}\cdots(2n-1)\cdot\color{#C00000}{(2n)}}{2\cdot4\cdot6\cdots(2n)\color{#C00000}{2\cdot4\cdot6\cdots(2n)}}\\
&=\frac{(2n)!}{(2^nn!)^2}
\end{align}
$$</p>
|
2,056,209 | <p>We know that the union of countably many countable sets is countable.
What can se say about the union of infinitely many countable sets and its cardinality?
Thanks :)</p>
| BrianO | 277,043 | <p>The cardinality of a union of an infinite collection $\mathscr{S}$ of countably infinite sets is infinite, and is at most equal to the cardinality $\lvert\mathscr{S}\rvert$ of that collection. </p>
<p>That is: </p>
<blockquote>
<p>Suppose $\mathscr{S}$ is an infinite set of countably infinite sets. Then
$$
\aleph_0 \le \left\lvert\bigcup \mathscr{S}\right\rvert \le \left\lvert \mathscr{S}\right\rvert.
$$</p>
</blockquote>
<p>The first inequality is clear. For the second, note that
$$\begin{align}
\left\lvert \mathscr{S}\right\rvert &= \max(\left\lvert \mathscr{S}\right\rvert, \aleph_0) \\
&= \left\lvert \mathscr{S} \right\rvert \left\lvert \omega \right\rvert \tag{cardinal multiplication} \\
&= \left\lvert \mathscr{S} \times \omega \right\rvert.
\end{align}$$
Let $D$ be the (or, <em>a</em>) disjoint union of the sets in $\mathscr{S}$:
$$
D = \{(S, x) \mid x\in S \in \mathscr{S}\}.
$$
Then $\left\lvert D \right\rvert = \left\lvert \mathscr{S} \times \omega \right\rvert = \left\lvert \mathscr{S}\right\rvert$. From this, the inequality follows, because the usual function
$$
(S, x)\mapsto x \colon D \to \bigcup \mathscr{S}
$$
is a surjection.</p>
|
768,925 | <p><img src="https://i.stack.imgur.com/SHRqU.png" alt="enter image description here"></p>
<p>This is quite a tricky question for me, but this is how far I got:<img src="https://i.stack.imgur.com/UV0ou.jpg" alt="enter image description here"></p>
<p>My drawing may not be precise, but I do know the points of tangency. I am a little stuck now, and I would appreciate it if someone can guide me through the answer. Thanks.</p>
| Justin Lindberg | 212,486 | <p>The figure is self-similar; it contains a smaller version of itself. The ratio of the radius of middle circle to that of the largest circle is the same as ratio of the radius of the smallest circle to that of the middle circle. That is, if we let $r$ be the radius of the middle circle, then
$$\frac r{99}=\frac{19}r.$$
Multiplying both sides by $99r$ gives us
$$r^2=19\times 99=1881,$$
so $$r=\sqrt{1881}.$$</p>
|
1,824,638 | <p>The figure shows a piece of string tied to a circle with a radius of one unit. The string is just long enough to reach the opposite side of the circle. Find the area of the region, not including the circle itself that is traced out when the string is unwound counterclockwise and continues counterclockwise until it reaches the opposite side again.</p>
<p><a href="https://i.stack.imgur.com/KzJRQ.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KzJRQ.gif" alt="enter image description here"></a></p>
| John Molokach | 90,422 | <p>I tried graphing the equations given in the accepted answer by Doug M, but they do not seem to trace out the path of the end of the string as it is being unwound (although his method does gives the correct answer). The way I approached the problem was to copy the equations (5) and (6) from the page <a href="http://mathworld.wolfram.com/GoatProblem.html" rel="nofollow noreferrer">http://mathworld.wolfram.com/GoatProblem.html</a>, using $a=1$ and then use polar coordinates. I translated everything to the right one unit so the area of part 1 of the calculation looks like this: </p>
<p><a href="https://i.stack.imgur.com/O0Kb0.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O0Kb0.gif" alt="Silo Problem"></a></p>
<p>... (where the orange point on the $y-$axis is located at $(0,\pi)$) so that the equations needed are
$$x=\cos\theta+\theta\sin\theta+1\quad\text{and}\quad y=\sin\theta-\theta\cos\theta.$$
Now using $r^2=x^2+y^2$ eventually gives us
$$r^2=\theta^2+2\theta\sin\theta+\cos\theta+2$$ and subtracting the area of the upper half of circle $r=2\cos\theta$, we have the area between the curves as
$$A=\left[\frac12\int_0^\pi(\theta^2+2\theta\sin\theta+\cos\theta+2)\,d\theta\right]-2\pi=\boxed{\frac{\pi^3}6}.$$</p>
<p>Doubling this value and adding the semicircle from "part 2" described by Doug M gives the final result of $\boxed{\dfrac{5\pi^3}6}$.</p>
|
64,905 | <p>Let's see if we could use MO to put some pressure on certain publishers...</p>
<p>Although it is wonderful that it has been put
<a href="http://www.jmilne.org/math/Books/DMOS.pdf" rel="nofollow">online</a>, I think it would make an even greater read if "Hodge Cycles, Motives and Shimura Varieties" by Deligne, Milne, Ogus and Shih would be (re)written in the latex typesetting (well, if I could understand its content..).</p>
<p>But enough about my opinion, what do you think? Which book(s) would you like to see "texified"? As customary in a CW question, one book per answer please.</p>
| Paolo Aceto | 5,001 | <p>Rolfsen - Knots and Links</p>
|
64,905 | <p>Let's see if we could use MO to put some pressure on certain publishers...</p>
<p>Although it is wonderful that it has been put
<a href="http://www.jmilne.org/math/Books/DMOS.pdf" rel="nofollow">online</a>, I think it would make an even greater read if "Hodge Cycles, Motives and Shimura Varieties" by Deligne, Milne, Ogus and Shih would be (re)written in the latex typesetting (well, if I could understand its content..).</p>
<p>But enough about my opinion, what do you think? Which book(s) would you like to see "texified"? As customary in a CW question, one book per answer please.</p>
| Spencer | 4,281 | <p>Leon Simon - Lectures on Geometric Measure Theory</p>
|
64,905 | <p>Let's see if we could use MO to put some pressure on certain publishers...</p>
<p>Although it is wonderful that it has been put
<a href="http://www.jmilne.org/math/Books/DMOS.pdf" rel="nofollow">online</a>, I think it would make an even greater read if "Hodge Cycles, Motives and Shimura Varieties" by Deligne, Milne, Ogus and Shih would be (re)written in the latex typesetting (well, if I could understand its content..).</p>
<p>But enough about my opinion, what do you think? Which book(s) would you like to see "texified"? As customary in a CW question, one book per answer please.</p>
| hce | 2,781 | <p>Adams - Lectures on Lie Groups</p>
|
2,031,842 | <p>If 3 is the remainder when dividing $P(x)$ with $(x-3)$, and $5$ is the remainder when dividing $P(x)$ with $(x-4)$, what is the remainder when dividing $P(x)$ with $(x-3)(x-4)$?</p>
<p>I'm completely puzzled by this, I'm not sure where to start...</p>
<p>Any hint would be much appreciated. </p>
| Bill Dubuque | 242 | <p>Notice $\ p\, = 3+(x\!-\!3)q\,\ $ by $\,\ p(3) = 3$</p>
<p>$5 = p(4) = 3+q(4)\,\Rightarrow\, {\color{#c00}2}\, =\, q(\color{#0a0}4) \, $ </p>
<p>Hence $\, p = 3 + (x\!-\!3)(\underbrace{\color{#c00}2+(x\!-\!\color{#0a0}4)r}_{\Large q}\!)\, =\, 2x\!-\!3\, +\, (x\!-\!3)(x\!-\!4)r$</p>
|
1,314,992 | <p>What operations do I need to perform the following conversion?</p>
<p>$$
\frac{\partial ^2y}{\partial x\partial z} \mapsto \frac{\partial ^2y}{\partial x\partial t}
$$</p>
| Intrepid Traveller | 244,431 | <p>Hint: Use the chain rule by changing your z variable to t</p>
|
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