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3,967,118 |
<p>How is it possible to define <span class="math-container">$\lim_{x \rightarrow a} f(x)$</span> if <span class="math-container">$f(x)$</span> is undefined at <span class="math-container">$a$</span>? There is an infinite amount of <span class="math-container">$x$</span>-values on either side of <span class="math-container">$x=a$</span>, and without a boundary or a strategically-placed, removable discontinuity, isn't the limit unknown?</p>
<p>I am using this definition of a limit [1].</p>
<blockquote>
<p>A function <span class="math-container">$f(x)$</span> approaches a limit <span class="math-container">$A$</span> as <span class="math-container">$x$</span> approaches <span class="math-container">$a$</span> if, and
only if, for each positive number <span class="math-container">$\epsilon$</span> there is another,
<span class="math-container">$\delta$</span>, such
that whenever <span class="math-container">$0 < |x-a| < \delta $</span> we have <span class="math-container">$|f(x) - A|< \epsilon$</span>.
That is, when <span class="math-container">$x$</span> is near <span class="math-container">$a$</span> (within a distance <span class="math-container">$\delta$</span>
from it), <span class="math-container">$f(x)$</span> is near <span class="math-container">$A$</span> (within a distance <span class="math-container">$\epsilon$</span> from it).
In symbols we write <span class="math-container">$\lim_{x \rightarrow a} f(x) = A$</span>.</p>
</blockquote>
<p>[1] David V. Widder. Advanced Calculus. Dover 1989.</p>
|
Ziqi Fan
| 851,072 |
<p>The limit of <span class="math-container">$f$</span> at a point <span class="math-container">$a$</span> is defined as
<span class="math-container">\begin{equation}
\lim_{x\to a}f\left(x\right) = L \longleftrightarrow \forall \epsilon > 0, \exists \delta > 0, \forall x, 0 < \lvert x-a \rvert < \delta \longrightarrow \lvert f\left(x\right) - L \rvert < \epsilon.
\end{equation}</span>
You can see that the condition <span class="math-container">$0 < \lvert x-a \rvert < \delta$</span> does not require the definition of <span class="math-container">$f$</span> at <span class="math-container">$a$</span>. Whenever you would like to tell if <span class="math-container">$L$</span> exists, you may first consult the definition. If it is not straigtforward to tell from the definition, you may use the Cauchy theorem of limit.</p>
<p>Your question is based on intuitive understanding of math, which is the type of math you learned at high school. Unfortunately, contemporary math is based on bottom-level logical systems. Using rigorous logical language is the gap between math in high school and math in college.</p>
|
261,156 |
<p>Well, I am trying to write a code that makes the number:</p>
<p><span class="math-container">$$123456\dots n\tag1$$</span></p>
<p>So, when <span class="math-container">$n=10$</span> we get:</p>
<p><span class="math-container">$$12345678910$$</span></p>
<p>And when <span class="math-container">$n=15$</span> we get:</p>
<p><span class="math-container">$$123456789101112131415$$</span></p>
<p>And when <span class="math-container">$n=4$</span> we get:</p>
<p><span class="math-container">$$1234$$</span></p>
|
murray
| 148 |
<pre><code>FromDigits@Flatten[IntegerDigits /@ Range[15]]
123456789101112131415
</code></pre>
<p>A function to do it:</p>
<pre><code>numberFromRange[n_] := FromDigits@Flatten[IntegerDigits /@ Range@n]
</code></pre>
|
261,156 |
<p>Well, I am trying to write a code that makes the number:</p>
<p><span class="math-container">$$123456\dots n\tag1$$</span></p>
<p>So, when <span class="math-container">$n=10$</span> we get:</p>
<p><span class="math-container">$$12345678910$$</span></p>
<p>And when <span class="math-container">$n=15$</span> we get:</p>
<p><span class="math-container">$$123456789101112131415$$</span></p>
<p>And when <span class="math-container">$n=4$</span> we get:</p>
<p><span class="math-container">$$1234$$</span></p>
|
flinty
| 72,682 |
<p>Without using <code>IntegerDigits</code> or string processing:</p>
<pre><code>f[x_] := Last@NestWhile[#[[1]]+{1,#[[2]]*10^IntegerLength@#[[1]]}&,{1,0},#[[1]]<=x&]
</code></pre>
<pre><code>f[105]
(* 1234567891011121314151617181920212223242526272829303132333435363738394
0414243444546474849505152535455565758596061626364656667686970717273747
5767778798081828384858687888990919293949596979899100101102103104105 *)
</code></pre>
|
261,156 |
<p>Well, I am trying to write a code that makes the number:</p>
<p><span class="math-container">$$123456\dots n\tag1$$</span></p>
<p>So, when <span class="math-container">$n=10$</span> we get:</p>
<p><span class="math-container">$$12345678910$$</span></p>
<p>And when <span class="math-container">$n=15$</span> we get:</p>
<p><span class="math-container">$$123456789101112131415$$</span></p>
<p>And when <span class="math-container">$n=4$</span> we get:</p>
<p><span class="math-container">$$1234$$</span></p>
|
user1066
| 106 |
<pre><code>ToExpression@StringJoin@Array[IntegerString,15]
(* 123456789101112131415 *)
</code></pre>
|
261,156 |
<p>Well, I am trying to write a code that makes the number:</p>
<p><span class="math-container">$$123456\dots n\tag1$$</span></p>
<p>So, when <span class="math-container">$n=10$</span> we get:</p>
<p><span class="math-container">$$12345678910$$</span></p>
<p>And when <span class="math-container">$n=15$</span> we get:</p>
<p><span class="math-container">$$123456789101112131415$$</span></p>
<p>And when <span class="math-container">$n=4$</span> we get:</p>
<p><span class="math-container">$$1234$$</span></p>
|
kglr
| 125 |
<p>Timings for all the methods (<a href="https://mathematica.stackexchange.com/a/261159/125">g1 : murray</a>, <a href="https://mathematica.stackexchange.com/a/261167/125">g2 : flinty</a>, <a href="https://mathematica.stackexchange.com/a/261188/125">g3/g4 : AccidentalFourierTransform</a>, <a href="https://mathematica.stackexchange.com/a/261162/125">g5/g6 : Syed</a>, <a href="https://mathematica.stackexchange.com/a/261176/125">g7 : David Reiss</a>, <a href="https://mathematica.stackexchange.com/a/261210/125">g8 : user1066</a>, <a href="https://mathematica.stackexchange.com/a/261158/125">g9/g10/g11 : kglr</a>) posted so far:</p>
<pre><code>ClearAll[g1, g2, g3, g4, g5, g6, g7, g8, g9, g10, g11, funcs]
g1 = FromDigits @ Flatten[IntegerDigits /@ Range[#]] &;
g2[x_] := Last @ NestWhile[#[[1]] + {1, #[[2]]*10^IntegerLength@#[[1]]} &, {1,
0}, #[[1]] <= x &]
g3[1] = 1;
g3[n_] := Block[{$RecursionLimit = 10^6}, g3[n] = g3[n - 1]*10^Floor[Log[10, 10*n]] + n]
g4 = IntegerPart[ChampernowneNumber[10]
10^((# + 1) Floor[Log[10, 10*#]] - (10^Floor[Log[10, 10*#]] - 1)/(10 - 1))] &;
g5 = FromDigits @* Flatten @* IntegerDigits @* Range @ # &;
g6 = FromDigits @ Flatten @ Last @ Reap @ Scan[Sow[IntegerDigits[#]] &] @ Range[#] &;
g7 = ToExpression[StringJoin[ToString /@ Range[#]]] &;
g8 = ToExpression @ StringJoin @ Array[IntegerString, #] &;
g9 = FromDigits @* StringJoin @* IntegerString @* Range;
g10 = Array[IntegerString, #, 1, FromDigits @* StringJoin] &;
g11 = FromDigits @ StringRiffle[Range[#], ""] &;
funcs = {g1, g2, g3, g4, g5, g6, g7, g8, g9, g10, g11};
</code></pre>
<p><strong>Timings:</strong></p>
<pre><code>ClearAll[res, timing]
Do[timing[i] = First[RepeatedTiming[
res[i] = funcs[[i]] /@ {5, 15, 55, 105, 1005, 10005}]], {i, 1, 11}]
Equal @@ (res /@ Range[11])
</code></pre>
<blockquote>
<pre><code>True
</code></pre>
</blockquote>
<pre><code>replace = {g2 -> "Last@NestWhile[#[[1]]+{1,#[[2]]*10^IntegerLength@#[[1]]}&,{1,0},#[[1]]\[LessEqual]x&]",
g3 -> "g3[1]=1;
g3[n_]:=Block[{$RecursionLimit=10^6},g3[n]=g3[n-1]*10^Floor[Log[10,10*n]]+n]"};
MapIndexed[{"g" <> ToString@#2[[1]], # /. replace, timing[#2[[1]]]} &][funcs] //
SortBy[Last] //
Prepend[{"function", "definition", "timing"}] //
Grid[#, Alignment -> {{Center, Left, "."}, Center}, Dividers -> All] &
</code></pre>
<p><a href="https://i.stack.imgur.com/bAyal.png" rel="noreferrer"><img src="https://i.stack.imgur.com/bAyal.png" alt="enter image description here" /></a></p>
|
2,801,294 |
<p>I cam across an interesting claim that:</p>
<p>$\mathcal N (\mu_f, \sigma_f^2) \; \mathcal N (\mu_g, \sigma_g^2) = \mathcal N \left(\frac{\mu_f\sigma_g^2+\mu_g\sigma_f^2}{\sigma_f^2+\sigma_g^2}, \frac {\sigma_f^2\sigma_g^2}{\sigma_f^2+\sigma_g^2}\right)$</p>
<p>In trying to understand it I consulted Bromiley:</p>
<p><a href="http://www.tina-vision.net/docs/memos/2003-003.pdf" rel="nofollow noreferrer">http://www.tina-vision.net/docs/memos/2003-003.pdf</a></p>
<p>Bromiley concludes that:</p>
<p>if </p>
<p>$f(x) = \frac{1}{\sqrt{2\pi\sigma_f^2}} e^{-\frac{(x-\mu_f)^2}{2 \sigma_f^2}}$
and
$g(x) = \frac{1}{\sqrt{2\pi\sigma_g^2}} e^{-\frac{(x-\mu_g)^2}{2 \sigma_g^2}}$</p>
<p>then:</p>
<p>$f(x)g(x) = D_{fg} \frac{1}{\sqrt{2\pi\sigma_{fg}^2}}
e^{-
\frac
{ (x - \mu_{fg})^2 }
{2 \sigma_{fg}^2 }
}$</p>
<p>where:</p>
<p>$\mu_{fg} = \frac { \sigma_g^2\mu_f + \sigma_f^2 \mu_g } {\sigma_f^2 + \sigma_g^2}$
and
$\sigma_{fg}^2 = \frac {\sigma_f^2 \sigma_g^2} {\sigma_f^2 + \sigma_g^2}$</p>
<p>$S_{fg} = \frac {1} {\sqrt{2\pi(\sigma_f^2+\sigma_g^2)}}
e^{
-\frac{(\mu_f-\mu_g)^2}{2(\sigma_f^2+\sigma_g^2)}
}$</p>
<p>Note that if $\mu_f$, $\mu_g$ , $\sigma_f$ and $\sigma_f$ are known constants then the $S_{fg}$ is a known constant too.</p>
<p>To wit, if I cast Bromiley's result in the format of the claim I'm exploring:</p>
<p>$\mathcal N (\mu_f, \sigma_f^2) \; \mathcal N (\mu_g, \sigma_g^2) = S_{fg} \; \mathcal N \left(\frac{\mu_f\sigma_g^2+\mu_g\sigma_f^2}{\sigma_f^2+\sigma_g^2}, \frac {\sigma_f^2\sigma_g^2}{\sigma_f^2+\sigma_g^2}\right)$</p>
<p>In short there is a constant scaling factor $S_{fg}$. In fact Bromiley describes the product as a scaled Gaussian. </p>
<p>Given $f(x)$ and $g(x)$ are both functions of $x$ the original claim, which reads (as a reminder):</p>
<p>$\mathcal N (\mu_f, \sigma_f^2) \; \mathcal N (\mu_g, \sigma_g^2) = \mathcal N \left(\frac{\mu_f\sigma_g^2+\mu_g\sigma_f^2}{\sigma_f^2+\sigma_g^2}, \frac {\sigma_f^2\sigma_g^2}{\sigma_f^2+\sigma_g^2}\right)$</p>
<p>implies that:</p>
<p>$\int_{-\infty}^{\infty} f(x) g(x) \;dx = 1$</p>
<p>But Bromiley's result suggests this implication is false. I presume it inetgrates to S_{fg}, or:</p>
<p>$\int_{-\infty}^{\infty} f(x) g(x) \;dx = S_{fg}$</p>
<p>My tentative conclusion is that the claim I am exploring is false, and my questions would be:</p>
<ol>
<li>Is my tentative conclusion true? (is the explored claim false?)</li>
<li>Am I right in concluding the integral would be $S_{fg}$?</li>
</ol>
<p>Those are the areas I'm a little shakey on at present and seek some review on I guess.</p>
|
Bernd Wechner
| 563,137 |
<p>We have an insight into a solution to this confusion from two observations:</p>
<ol>
<li><p>The <span class="math-container">$\mathcal N(\mu, \sigma)$</span> notation is a loose convention. I find no clear definition of it anywhere. A good reference is the Wikipedia mention here:
<a href="https://en.wikipedia.org/wiki/Normal_distribution#Notation" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Normal_distribution#Notation</a></p>
</li>
<li><p>At least one author's dive into the same apparent confusion at:
<a href="https://github.com/rlabbe/Kalman-and-Bayesian-Filters-in-Python/blob/master/04-One-Dimensional-Kalman-Filters.ipynb" rel="nofollow noreferrer">https://github.com/rlabbe/Kalman-and-Bayesian-Filters-in-Python/blob/master/04-One-Dimensional-Kalman-Filters.ipynb</a> where Roger Labbe writes:</p>
</li>
</ol>
<p><span class="math-container">$\begin{aligned}
\mathcal N(\mu, \sigma^2) &= \| prior \cdot likelihood \|\\
&=\mathcal{N}(\bar\mu, \bar\sigma^2)\cdot \mathcal{N}(\mu_z, \sigma_z^2) \\
&= \mathcal N(\frac{\bar\sigma^2 \mu_z + \sigma_z^2 \bar\mu}{\bar\sigma^2 + \sigma_z^2},\frac{\bar\sigma^2\sigma_z^2}{\bar\sigma^2 + \sigma_z^2})
\end{aligned}$</span></p>
<p>The whole confusion goes away if we table a formal definition of <span class="math-container">$\mathcal N(\mu, \sigma)$</span> as:</p>
<p><span class="math-container">$\mathcal N(x \mid \mu, \sigma^2) = \| e^{-\frac{(x-\mu)^2}{2 \sigma^2}} \|$</span></p>
<p>given:</p>
<p><span class="math-container">$\|f(x)\| \implies \frac {f(x)} {c}$</span></p>
<p>where <span class="math-container">$c$</span> is a normalisation constant, such that:</p>
<p><span class="math-container">$\int_{-\infty}^{\infty} \|f(x)\|\,dx = 1$</span>
, then in the case of a standard univariate Gaussian distribution:</p>
<p><span class="math-container">$c= \frac {1} {\sqrt{2 \pi \sigma^2}}$</span></p>
<p>In the case of the product of two Gaussian distributions <span class="math-container">$\mathcal N(x \mid \mu_f, \sigma_f) \cdot \mathcal N(x \mid \mu_g, \sigma_g)$</span>:</p>
<p><span class="math-container">$c = \frac {1} {2\pi\sigma_f\sigma_g} e^{ \frac { \mu_{fg}^2 + \mu_{fg} } {2 \sigma_{fg}^2 }}$</span> where: <span class="math-container">$\mu_{fg} = \frac { \sigma_g^2\mu_f + \sigma_f^2 \mu_g } {\sigma_f^2 + \sigma_g^2}$</span> and <span class="math-container">$\sigma_{fg}^2 = \frac {\sigma_f^2 \sigma_g^2} {\sigma_f^2 + \sigma_g^2}$</span></p>
<p>and:</p>
<p><span class="math-container">$\mathcal N(x \mid \mu_f, \sigma_f^2) \cdot \mathcal N(x \mid \mu_g, \sigma_g^2) = \mathcal N(x \mid \mu_{fg}, \sigma_{fg}^2)$</span></p>
<p>becomes a true and consistent claim (as we have embraced normalisation into the definition of the function <span class="math-container">$\mathcal N$</span>.</p>
<p>It would be prudent to note at the same time difference between a frequency distribution (Gaussian) and a Probability Density Function (PDF).</p>
<p>The Gaussian function (a frequency distribution) is:</p>
<p><span class="math-container">$f(x) = a e^{- \frac{(x-b)^2 }{ 2 c^2} }$</span> or <span class="math-container">$f(x) = a e^{- \frac{(x-\mu)^2 }{ 2 \sigma^2} }$</span></p>
<p>See: <a href="https://en.wikipedia.org/wiki/Gaussian_function" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Gaussian_function</a></p>
<p>And the Normal distribution (a probability density function, PDF) is:</p>
<p><span class="math-container">$\mathcal N(x \mid \mu, \sigma^2) = \| e^{-\frac{(x-\mu)^2}{2 \sigma^2}} \|$</span></p>
<p>or</p>
<p><span class="math-container">$\mathcal N(x \mid \mu, \sigma^2) = \frac {1} {c} e^{-\frac{(x-\mu)^2}{2 \sigma^2}}$</span></p>
<p>and <span class="math-container">$c$</span> is the normalizing constant. See: <a href="https://en.wikipedia.org/wiki/Normalizing_constant" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Normalizing_constant</a></p>
<p>that makes <span class="math-container">$\mathcal N$</span> a PDF (by ensuring its integral over its range of definition is 1, in other words, that the total probability over all possible outcomes is certain, 100% or 1 in standard algebraic contexts).</p>
<p>Thus by explicitly defining <span class="math-container">$\mathcal N$</span> the original claim is true and consistent, as is Roger Labbe's text - and this definition would seem acceptably in line with the current loose usage of the notation <span class="math-container">$\mathcal N(x \mid \mu, \sigma^2)$</span> encountered in the literature.</p>
<p>If a conflicting formal definition of <span class="math-container">$\mathcal N(x \mid \mu, \sigma^2)$</span> exists, speak up and identify it.</p>
<p>Note: Fixed minor typo in latex subscript. Canyon289 is typing this note to meet min character limit for edits</p>
|
1,341,896 |
<p>given three lines $\ell_1,\ell_2, \ell_3 $ which intersect in one point $P$. How can one construct a triangle such that the given lines become its angle bisectors?</p>
<p>So far I tried to find conditions on how the three given lines have to intersect such that the desired triangle exists. There are altogether six rays $\omega_1^1, \omega_1^2; \omega_2^1,\omega_2^2;\omega_3^1,\omega_3^2$ - any line $\ell_j$ determines the two rays $\omega_j^1,\omega_j^2$- emanating from the common point $P$.</p>
<p>Any of the three vertices $V_1,V_2,V_3$ of the desired triangle has to lie on exactly one line, i.e. $V_j\in \ell_j=\omega_j^1\cup \omega_j^2$. Hence we have either $V_j\in \omega_j^1$ or $V_j\in \omega_j^2$.</p>
<p>Therefore the triangle should exist if and only if it is possible to choose three rays $\tilde{\omega}_1\in \lbrace \omega_1^1, \omega_1^2\rbrace $, $\tilde{\omega}_2\in \lbrace \omega_1^2, \omega_1^2\rbrace$, $\tilde{\omega}_3\in \lbrace \omega_1^3, \omega_1^3\rbrace$ such that any three of the rays $\tilde{\omega_1}, \tilde{\omega_2}, \tilde{\omega_3}$ form an angle less than $\pi$.</p>
<p>But I do not know how to construct the triangle?! </p>
<p>I started to draw a circle around $P$ with any radius. Now I can choose a Point $V_1$ on the ray $\tilde{\omega_1}$ outside the circle. And now I can draw the tangents through $V_1$ to the circle. I think those tangents will meet the other rays at some points $V_2,V_3$ (Why?). Now we can draw the line $V_2,V_3$. But this line needs to be tangent to the circle and I'm not sure about that. Will this construction work, or is it done in a different way?</p>
<p>Best regards </p>
|
Wojowu
| 127,263 |
<p>Let $I$ be the point of intersection of the lines (renamed it so that it's clear that we want it to be the incenter) and let $A,B,C$ be the points we want to construct. Let $\alpha,\beta,\gamma$ be angles of the triangle, and let $\delta,\epsilon,\zeta$ be angles $BIC,CIA,AIB$ respectively (see the picture).</p>
<p><img src="https://i.stack.imgur.com/1dRmH.png" alt="enter image description here"></p>
<p>We can easily find angles $\delta,\epsilon,\zeta$ given the three lines, so now the aim is to express $\alpha,\beta,\gamma$ in terms of these. Let's try to find value of $\delta$. We have $\delta=180°-\beta/2-\gamma/2=180°-(\beta+\gamma)/2=180°-(180°-\alpha)/2=90°+\frac{\alpha}{2}$, so that $\alpha=2\delta-180°$. Similarly we can find values of $\beta,\gamma$ given $\epsilon,\zeta$. Now if we know the angles of triangle $ABC$ we can easily construct it on the lines, I'll leave it for you to do.</p>
|
3,355,215 |
<blockquote>
<p>For <span class="math-container">$x$</span> and <span class="math-container">$k$</span> real numbers, for what values of <span class="math-container">$k$</span> will the graphs of <span class="math-container">$f(x)=-2\sqrt{x+1}$</span> and <span class="math-container">$g(x)=\sqrt{x-2}+k$</span> intersect?</p>
</blockquote>
<p>I tried to make an equation of them, but I’m stuck with the two variables and I couldn’t solve it. Much appreciation.</p>
<p>We didn’t do calculus yet..</p>
|
Alessandro Cigna
| 641,678 |
<p>You want <span class="math-container">$-2\sqrt{x+1}=\sqrt{x-2}+k\; (*)$</span> has at least one <span class="math-container">$x$</span>-solution.</p>
<p>If we call <span class="math-container">$h(x)=-2\sqrt{x+1}-\sqrt{x-2}$</span> then we have that <span class="math-container">$h$</span> is defined only for <span class="math-container">$x\ge 2$</span> and <span class="math-container">$h(2)=-2\sqrt 3$</span> while <span class="math-container">$\lim_{x\to\infty}h(x)=-\infty$</span>. Then <span class="math-container">$$h'(x)=-\frac 1{\sqrt{x+1}}-\frac 1{2\sqrt{x-2}}=-\frac{2\sqrt{x-2}+\sqrt{x+1}}{2\sqrt{x+1}\sqrt{x-2}}<0\; \forall x>2$$</span></p>
<p>So <span class="math-container">$h$</span> is strictly decreasing and for intermediate values theorem it is surjective on <span class="math-container">$(-\infty,-2\sqrt 3]$</span>. We conclude that <span class="math-container">$(*)$</span> has exactly one solution for all <span class="math-container">$k\le -2\sqrt 3$</span> an no solution for other values.</p>
|
911,584 |
<p><em>Disclaimer: This thread is a record of thoughts.</em></p>
<p><strong>Discussion</strong>
Given a compact set.</p>
<blockquote>
<p>Do mere neighborhood covers admit finite subcovers?
$$C\subseteq\bigcup_{i\in I}N_i\implies C\subseteq N_1\cup\ldots N_n$$
<em>(The idea is that neighborhoods are in some sense fat.)</em></p>
</blockquote>
<p><strong>Application</strong></p>
<p>Given a locally compact space.</p>
<p>Every compact set has a compact neighborhood base:
$$C\subseteq U:\quad N\subseteq U\quad(C\subseteq N^°)$$
<em>(The above would give clues how to prove this.)</em></p>
|
gar
| 138,850 |
<p>It can be solved using ordinary generating functions:</p>
<p>Since there are three groups of numbers of our interest, take three formal variables for the ogf.</p>
<p>Let $x$ indicate 1 or 2, $z$ indicate a 6, and $y$ the other three numbers of the dice.</p>
<p>Hence, the required ogf is:
\begin{align*}
\left(2\, x + 3\, y + z\right)^5
\end{align*}
We need to consider cases with the variables, though.</p>
<p>Grouping the coefficients for which the degree of $x$ $>$ the degree of $y$, the sum turns out to be $4602$, and the
probability is
\begin{align*}
\mathbb{P} &= \frac{4602}{6^5} = \frac{767}{1296} \approx 0.5918209876543
\end{align*}</p>
|
231,317 |
<p>The task I'm faced with is to implement a poly-time algorithm that finds a nontrivial factor of a Carmichael number. Many resources on the web state that this is easy, however without further explanation why.</p>
<p>Furthermore, since Miller-Rabin exits when a nontrivial square root of 1 is found, this can be used to find a factor to the Carmichael number: $x^2 \equiv 1 = (x+1)(x-1)\equiv0 \pmod N$, where N is the Carmichael number we want to factor and $x$ the nontrivial square root of 1. Hence factors must be found using $\gcd(x+1,N)$ and $\gcd(x-1, N)$, correct? </p>
<p>Due to problems with strong liars, in some cases we will miss out on factors. Is this a major problem? Since Miller-Rabin tests only passes composites with a probability 1/4, is it correct to say that the chances of finding a factor is > 0.5?</p>
<p>Kind regards!</p>
|
Peter
| 82,961 |
<p>A Carmichael number $N$ is a probable prime to every base $a$ coprime to $N$, but there is a base $a$ with $1<a<N$ not too big and coprime to $N$ (if it is NOT coprime to $N$, $gcd(a,N)$ is a non-trivial factor), such that $N$ is not strong probable prime to base $a$. $a$ is called a witness.</p>
<p>For such an $a$, you have $\large a^{2^d\times m} \neq -1\ (\ mod\ N)$
for all $d$ with $0\le d \le k$, when $N=2^k\times m+1$ with $m$ odd. </p>
<p>But for
some $d$ with $1\le d\le k$ (choose the smallest such $d$) you have $\large a^{2^d\times m}\equiv 1\ mod \ (\ N\ )$.</p>
<p>But the congruence does not hold for the exponent $d-1$ indstead of $d$.
Note, that $a^m\equiv 1\ (\ mod\ N)$ is impossible, if a is a witness.</p>
<p>So, you have a congruence $x^2\equiv\ 1\ (\ mod\ N)$ , but $x\ne \pm1\ (\ mod\ N)$ and $gcd(x-1,N)$ is
a non-trivial factor of $N$ ($gcd(x+1,N)$ is a non-trivial factor as well).</p>
|
2,700,301 |
<blockquote>
<p>Suppose we throw 4 dice consecutively. What is the probability
that the sum of the dice is <span class="math-container">$24?$</span></p>
</blockquote>
<p><strong>My attempt</strong>:</p>
<p>The experiment we consider is throwing 4 dices and looking at the numbers we obtain, i.e. <span class="math-container">$\Omega := \{1,2,3,4,5,6\}^4$</span></p>
<p>Define <span class="math-container">$$X: \Omega \to \mathbb{R}: (\omega_1, \dots, \omega_4) \mapsto \sum_1^4\omega_i.$$</span></p>
<p>Then, <span class="math-container">$$\mathbb{P}(X = 24) = \mathbb{P}\left(\left\{(6,6,6,6)\right\}\right),$$</span></p>
<p>and assuming that every combination that can be thrown is as likely as the others, this is equal to <span class="math-container">$\left(\frac16\right)^4$</span>. Is this the correct way to handle such problems?</p>
|
Siong Thye Goh
| 306,553 |
<p>Yes, your approach is correct. </p>
<p>In general, for other target sum, we can use generating function to solve the problem. </p>
|
2,700,301 |
<blockquote>
<p>Suppose we throw 4 dice consecutively. What is the probability
that the sum of the dice is <span class="math-container">$24?$</span></p>
</blockquote>
<p><strong>My attempt</strong>:</p>
<p>The experiment we consider is throwing 4 dices and looking at the numbers we obtain, i.e. <span class="math-container">$\Omega := \{1,2,3,4,5,6\}^4$</span></p>
<p>Define <span class="math-container">$$X: \Omega \to \mathbb{R}: (\omega_1, \dots, \omega_4) \mapsto \sum_1^4\omega_i.$$</span></p>
<p>Then, <span class="math-container">$$\mathbb{P}(X = 24) = \mathbb{P}\left(\left\{(6,6,6,6)\right\}\right),$$</span></p>
<p>and assuming that every combination that can be thrown is as likely as the others, this is equal to <span class="math-container">$\left(\frac16\right)^4$</span>. Is this the correct way to handle such problems?</p>
|
Mohammad Riazi-Kermani
| 514,496 |
<p>Yes, your approach is correct.</p>
<p>The only way to get a sum of $24$ is having four $6$.</p>
<p>The probability of four independent event where each has the probability of $1/6$ is $ 1/6^4$</p>
|
20,802 |
<p>Look at the following example:</p>
<p>Which picture has four apples?</p>
<p>A<a href="https://i.stack.imgur.com/Tpm46.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Tpm46.png" alt="enter image description here" /></a></p>
<hr />
<p>B <a href="https://i.stack.imgur.com/AOv29.png" rel="noreferrer"><img src="https://i.stack.imgur.com/AOv29.png" alt="enter image description here" /></a></p>
<hr />
<p>C <a href="https://i.stack.imgur.com/lZNmQ.png" rel="noreferrer"><img src="https://i.stack.imgur.com/lZNmQ.png" alt="enter image description here" /></a></p>
<hr />
<p>D <a href="https://i.stack.imgur.com/BWqpH.png" rel="noreferrer"><img src="https://i.stack.imgur.com/BWqpH.png" alt="enter image description here" /></a></p>
<p>B is the expected answer but should not the correct answer be BCD? Technically if a set has <strong>exactly</strong> <span class="math-container">$m$</span> elements, then it has <span class="math-container">$k$</span> elements if <span class="math-container">$k\leq m$</span>. This is also how we talk in everyday language:</p>
<blockquote>
<p>"Do you have three dollars?"
"Yes."</p>
</blockquote>
<p>The second speaker is not indicating he has exactly three dollars. He simply indicates that he has <strong>at least</strong> three dollars.</p>
<p>So I am wondering if we are teaching children correct logic here. Shouldn't the original question be rephrased as "which picture has <strong>exactly</strong> four apples"?</p>
|
BCLC
| 5,987 |
<p>It is wrong exactly for the reasons you have pointed out. There shouldn't be an issue with including the word 'exactly' as in 'exactly 4 apples'.</p>
|
3,580,258 |
<p>Hi: The definition I'll use is this: Let <span class="math-container">$F$</span> be an abelian group and <span class="math-container">$X$</span> a subset of <span class="math-container">$F$</span>. Then <span class="math-container">$F$</span> is a free abelian group on <span class="math-container">$X$</span> if for every abelian group <span class="math-container">$G$</span> and every function <span class="math-container">$f$</span> from <span class="math-container">$X$</span> to <span class="math-container">$G$</span> there is a homomorphism <span class="math-container">$\phi$</span> from <span class="math-container">$F$</span> to <span class="math-container">$G$</span> that extends <span class="math-container">$f$</span>.</p>
<p>Let <span class="math-container">$G$</span> be a finite group and <span class="math-container">$X$</span> a subset of <span class="math-container">$G$</span>. Let <span class="math-container">$F$</span> be the free abelian group on <span class="math-container">$X$</span>. Then <span class="math-container">$F=\langle X\rangle$</span> and so <span class="math-container">$F\subseteq G$</span>. That is, every finite group has an infinite subgroup. What am I doing wrong?</p>
<p>EDIT: It will be easier to make myself clear working with free groups. I'll quote from Derek Robinson, A Course in the Theory of Groups, 2nd ed.</p>
<p><a href="https://i.stack.imgur.com/BJioC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BJioC.png" alt=" "></a></p>
<p>From this a free group is not only always free on a subset but additionally that subset generates it. If <span class="math-container">$G$</span> is a group and <span class="math-container">$X$</span> is a subset, however, indeed there will exist a free group on <span class="math-container">$X$</span> but I am unable to show it will be generated by <span class="math-container">$X$</span> based in the above quote. Which is very natural, of course. Thanks for the posts. Honestly none of the feedback, up to now, throws light in the paradox (paradox for me, of course).</p>
|
Tsemo Aristide
| 280,301 |
<p>Consider <span class="math-container">$G=\mathbb{Z}/5\mathbb{Z}$</span> and <span class="math-container">$[1]\in G$</span>, let <span class="math-container">$X=\{[1]\}$</span> <span class="math-container">$F=<X>=G$</span>, let <span class="math-container">$f:X\rightarrow\mathbb{Z}$</span> defined by <span class="math-container">$f([1])=1$</span>, <span class="math-container">$f$</span> cannot be extended to <span class="math-container">$G$</span> in <span class="math-container">$g$</span>, since we must have <span class="math-container">$g(5.[1])=5.f([1])=5$</span> and <span class="math-container">$g(5.[1])=g([0])=0$</span> since <span class="math-container">$g$</span> is a morphism of groups. </p>
|
326,600 |
<p>The intervals are: </p>
<p>$I_1=\{ x\in\mathbb{R}:a\leq x\leq b\}$ with $-1<a<b<1$.</p>
<p>$I_2=\{ x\in\mathbb{R}:-1< x< 1\}$.</p>
<p>I've shown $f_n$ is pointwise convergent to $0$ on $I_1$ and $I_2$. But what is the difference for showing uniform convergence on these two intervals? I have a few other questions that have very similar intervals and I'm not sure what I should be doing different. </p>
<p>Basically, what is different in showing uniform convergence on $I_1$ and $I_2$?</p>
|
Community
| -1 |
<p>The sequence of function $(f_n)$ is unifom convergent to $f$ in the interval $I$ if:
$$\sup_{x\in I}|f_n(x)-f(x)|=0.$$</p>
<ol>
<li><p>We have
$$ \forall x\in I_1,|f_n(x)-0|\leq \max(|a|^n,|b|^n)\to0,$$
so the sequence $(f_n)$ is uniform convergent to the zero function on $I_1$.</p></li>
<li><p>The sequence $(f_n)$ is pointwise convergent to the zero function, so if we have the uniform convergence, it must be to the zero function also.</p></li>
</ol>
<p>In another way, we have $|f_n(1-\frac{1}{n})|=(1-\frac{1}{n})^n|\sin(n-1)|,$
and this sequence has not a limit, so $(f_n)$ is not uniform convergent to the zero function on $I_2$.</p>
|
1,039,487 |
<p>Suppose $a$, $b$ are real numbers such that $a+b=12$ and both roots of the equation $x^2+ax+b=0$ are integers. </p>
<p>Determine all possible values of $a$. </p>
<p>I don't know how to go about doing this without long, messy casework. I tries $(x-s)(x-r)=x^2+ax+b$ and got $-r-s=a$ and $rs=b$, but was unable to find all solutionss based on only these and $a+b=12$. Could someone help me finish up? Thanks.</p>
|
Varun Iyer
| 118,690 |
<p>Since $a + b = 12$</p>
<p>$$-(r+s) + rs = 12$$</p>
<p>Using Simon's favorite factoring trick:</p>
<p>$$(r-1)(s-1) - 1 = 12 \iff (r-1)(s-1) = 13$$</p>
<p>Only two possible solutions arise:</p>
<p>$r -1 = 1$ and $s -1 = 13$, therefore $r = 2$ and $s = 14$</p>
<p>$r -1 = -1$ and $s -1 = -13$, therefore $r = 0$ and $s = -12$</p>
<p>Therefore, $a = -16$ and $b = 28$</p>
<p>or</p>
<p>$a = 12$ and $b = 0$</p>
|
363,881 |
<p>The problem gives the curl of a vector field, and tells us to calculate the line integral over $C$ where $C$ is the intersection of $x^2 + y^2 = 1$ and $z= y^2$. I know I should use Stokes Theorem, but how do I find $dS$? </p>
<p>I did $z = \frac{1}{2}(y^2 + 1 - x^2)$ and calculated $dS$ as $\langle-dz/dx,-dz/dy,1\rangle$ but apparently that was wrong.</p>
|
Shuhao Cao
| 7,200 |
<p>I drew a figure illustrating how these two surfaces intersecting with each other<img src="https://i.stack.imgur.com/r0IbW.png" alt="."> </p>
<p>The intersecting curve $C$ is that black line, and let's assume $C$'s direction is rotating counter-clockwise if you look from above. The surface $S$ is part of the $z=y^2$ that is inside the cylinder. Its surface normal is taking gradient of the equation $y^2 - z= 0$: $(0,2y,-1)$, normalize it we have the unit vector normal is $\mathbf{n} = (0,2y,-1)/\sqrt{4y^2+1}$</p>
<p>Suppose the vector field given is $\mathbf{F}$, then by Stokes theorem:
$$
\oint_C \mathbf{F}\cdot d\mathbf{r} = \int_S \nabla \times \mathbf{F} \cdot d\mathbf{S} = \int_S \nabla \times \mathbf{F} \cdot \mathbf{n}\,dS,
$$
where the $\mathbf{n}\,dS$ part is probably what you are looking for. Then what you need to do is just parametrizing $S$, and compute <a href="http://en.wikipedia.org/wiki/Surface_integral" rel="nofollow noreferrer">a surface integral for a scalar field</a>.</p>
|
2,194,762 |
<p>I have to find all possible values of <span class="math-container">$a$</span>, <span class="math-container">$b$</span>, <span class="math-container">$c$</span>, and <span class="math-container">$d$</span> for which the <span class="math-container">$2 \times 2$</span> matrix
<span class="math-container">$\begin{pmatrix}a & b \\ c & d \end{pmatrix}$</span> is semisimple. I know that a matrix <span class="math-container">$S$</span> is semisimple if there is a nonsingular matrix <span class="math-container">$P$</span> such that <span class="math-container">$P^{-1}SP=A$</span> is diagonal. How can I use this to find all the values <span class="math-container">$a$</span>, <span class="math-container">$b$</span>, <span class="math-container">$c$</span> and <span class="math-container">$d$</span>?</p>
|
Matthew C
| 25,959 |
<p>Since you already know semisimple is equivalent to diagonalizable (it is, in the case of complex matrices), then you can consider the possible Jordan forms for 2x2 matrices. You want to avoid the situation of a Jordan block. That means the two remaining situations are: you have a multiple of the identity matrix, or the characteristic polynomial has two roots. Both of these situations can be expressed in terms of a, b, c, and d.</p>
|
2,194,762 |
<p>I have to find all possible values of <span class="math-container">$a$</span>, <span class="math-container">$b$</span>, <span class="math-container">$c$</span>, and <span class="math-container">$d$</span> for which the <span class="math-container">$2 \times 2$</span> matrix
<span class="math-container">$\begin{pmatrix}a & b \\ c & d \end{pmatrix}$</span> is semisimple. I know that a matrix <span class="math-container">$S$</span> is semisimple if there is a nonsingular matrix <span class="math-container">$P$</span> such that <span class="math-container">$P^{-1}SP=A$</span> is diagonal. How can I use this to find all the values <span class="math-container">$a$</span>, <span class="math-container">$b$</span>, <span class="math-container">$c$</span> and <span class="math-container">$d$</span>?</p>
|
Community
| -1 |
<p>Let <span class="math-container">$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in M_n(\mathbb{C})$</span> (we work over <span class="math-container">$\mathbb{C}$</span>).</p>
<p>If <span class="math-container">$A$</span> is not diagonal, then </p>
<p><span class="math-container">$A$</span> is diagonalizable IFF <span class="math-container">$(a-d)^2+4bc\not=0$</span>.</p>
|
79,782 |
<p>Calculate $17^{14} \pmod{71}$</p>
<p>By Fermat's little theorem:<br>
$17^{70} \equiv 1 \pmod{71}$<br>
$17^{14} \equiv 17^{(70\cdot\frac{14}{70})}\pmod{71}$</p>
<p>And then I don't really know what to do from this point on. In another example, the terms were small enough that I could just simplify down to an answer, but in this example, I have no idea what to do with that $17^{(70\cdot\frac{14}{70})}$</p>
<p>What do I do from here?</p>
|
Bill Dubuque
| 242 |
<p>Here is the computation using an <a href="http://wwwhomes.uni-bielefeld.de/achim/addition_chain.html" rel="nofollow noreferrer">addition chain.</a></p>
<p>$\qquad 17^2\: \equiv\ 5$</p>
<p>$\qquad 17^3\: \equiv\ 17\cdot 17^2\:\equiv\ 5\cdot 17\ \equiv\ 14$</p>
<p>$\qquad 17^5\: \equiv\ 17^2\cdot 17^3\: \equiv\ 5\cdot 14\ \equiv\: -1$</p>
<p>$\qquad 17^7\: \equiv\ 17^2\cdot 17^5\: \equiv\ 5\:(-1)\ \equiv\: -5$</p>
<p>$\qquad 17^{14} \equiv\ 17^7\cdot 17^7\: \equiv\ (-5)^2\: \equiv\ 25$</p>
<p>See the leftmost path in the tree below.</p>
<p><img src="https://i.stack.imgur.com/mI0eb.jpg" alt="enter image description here"></p>
|
79,782 |
<p>Calculate $17^{14} \pmod{71}$</p>
<p>By Fermat's little theorem:<br>
$17^{70} \equiv 1 \pmod{71}$<br>
$17^{14} \equiv 17^{(70\cdot\frac{14}{70})}\pmod{71}$</p>
<p>And then I don't really know what to do from this point on. In another example, the terms were small enough that I could just simplify down to an answer, but in this example, I have no idea what to do with that $17^{(70\cdot\frac{14}{70})}$</p>
<p>What do I do from here?</p>
|
Angel Moreno
| 327,493 |
<p>In this curious problem, the subgroup of order 5 of <span class="math-container">$(Z_{71}^*, \cdot)$</span> is formed by these 5 numbers (5 equivalence classes modulo 71):</p>
<p><span class="math-container">$$ \{1, 5, 25, -17, -14\} $$</span></p>
<p>(When multiplying two, another one comes out of the same subgroup).</p>
<p>This group is isomorphic to the group <span class="math-container">$ (Z_5, +) $</span> and contains 4 generators.</p>
<p>The 4 different elements of 1 are generators of <span class="math-container">$\{1, 5, 25, -17, -14\}$</span></p>
<p>Using (-17) as a generator of the subgroup of order 5, gives us this order:</p>
<p><span class="math-container">$$[1, -17, 5, -14, 25]$$</span></p>
<p>So that:</p>
<p><span class="math-container">$$ 5 ^ 3 = -17 $$</span>
<span class="math-container">$$(-17) ^ 2 = 5$$</span>
<span class="math-container">$$(17) ^ {14} = (-17) ^ {14} = (-17)^{14 (mod \space 5)} = (-17) ^ 4 = 25$$</span></p>
|
4,523,363 |
<p>A set <a href="https://en.m.wikipedia.org/wiki/Lattice_(order)#:%7E:text=A%20lattice%20is%20an%20abstract,greatest%20lower%20bound%20or%20meet" rel="nofollow noreferrer">lattice</a> is a lattice where the meet is the intersection, the join is the union, and the partial order is <span class="math-container">$\subseteq$</span>. The standard example is is <span class="math-container">$P(A)$</span> where <span class="math-container">$A$</span> is a set. I’ve thought about other possible examples, yet have failed to come up with any. Are there any set/distributive lattices that aren’t isomorphic to <span class="math-container">$P(A)$</span> for some <span class="math-container">$A$</span>?</p>
|
freakish
| 340,986 |
<p>Well, not every lattice is isomorphic to <span class="math-container">$P(A)$</span>. For example the <span class="math-container">$P(A)$</span> lattice has <span class="math-container">$2^{|A|}$</span> elements, and there are lattices for any order, e.g.</p>
<p><span class="math-container">$$K_n=\big\{\{1\},\{1,2\},\ldots,\{1,2,\ldots,n-1,n\}\big\}$$</span></p>
<p>with union and intersection is a latice of exactly <span class="math-container">$n$</span> elements.</p>
<hr />
<p>So the more interesting question would be: can every lattice be embedded into <span class="math-container">$P(A)$</span> for some set <span class="math-container">$A$</span>? The answer is still no. To see that we need to look at some property of (a sublattice of) <span class="math-container">$P(A)$</span> that other lattices don't have to have. One of those properties is the <a href="https://en.wikipedia.org/wiki/Distributive_lattice" rel="noreferrer">distributivity</a> of <span class="math-container">$\vee$</span> over <span class="math-container">$\wedge$</span> (and vice versa). Union is always distributive over intersection (and vice versa), and so any subtlatice of <span class="math-container">$P(A)$</span> is distributive. But there are non-distributive lattices, e.g the <span class="math-container">$M_3$</span>:</p>
<p><span class="math-container">$$M_3=\{0,a,b,c,1\}$$</span>
<span class="math-container">$$0< a<1$$</span>
<span class="math-container">$$0< b<1$$</span>
<span class="math-container">$$0< c<1$$</span></p>
<p>In fact this is if and only if: a lattice is isomorphic to a lattice of sets (with union and intersection), and thus embeddable into some <span class="math-container">$P(A)$</span>, if and only if it is distributive. Note that this is a non-trivial theorem (see the wiki I linked earlier).</p>
|
2,203,770 |
<p>If I am asked to find all values of $z$ such that $z^3=-8i$, what is the best method to go about that?</p>
<p>I have the following formula:
$$z^{\frac{1}{n}}=r^\frac{1}{n}\left[\cos\left(\frac{\theta}{n}+\frac{2\pi k}{n}\right)+i\sin\left(\frac{\theta}{n}+\frac{2\pi k}{n}\right)\right]$$</p>
<p>for $k=0,\pm1, \pm2,...$</p>
<p>Applying this formula, I find the cubed root of $8$, which is $2$. And then when I apply it to the formula, I get the following:</p>
<p>$$z = 2\left[\cos\left(\frac{\pi}{3}+\frac{2\pi k}{3}\right)+i\sin\left(\frac{\pi}{3}+\frac{2\pi k}{3}\right)\right]$$ for $k=0,\pm1, \pm2,...$</p>
<p>I am confused, because the given solution is as follows:
$$z = 2\left[\cos\left(\frac{\pi}{2}+\frac{2\pi k}{3}\right)+i\sin\left(\frac{\pi}{2}+\frac{2\pi k}{3}\right)\right]$$ for $k=0,\pm1, \pm2,...$</p>
<p>Where did I go wrong? How would my approach changed if I was asked to find all values for $-8$, or $8i$?</p>
|
JMP
| 210,189 |
<p>You are claiming $\theta=\pi$, whereas it doesn't - actually $\theta=\dfrac{3\pi}{2}$, which gives the solution you list.</p>
<p>For $-8$ use $\theta=\pi$, and for $8i$ use $\theta=\dfrac{\pi}{2}$.</p>
|
1,232,023 |
<p>For me, intuitively, integral $2\pi y~dx$ make more sense. I know intuition can not be proof, but by far, most part of math I've learned does match with my intuition. So, I think this one should 'make sense' as well. Probably I didn't understand the way surface area is measured. It will be great if any one could tell me how 'integral $2\pi y~dx$' is wrong. (By the way, how to use mathematical symbols in texts?)</p>
|
David K
| 139,123 |
<p>Here are two ways to measure the surface area of an open cone of base radius $r$ and height $h$ (just the area of the curved surface of the cone; we are not including the area of the flat circular base of the cone).</p>
<p>The first way is, you cut the surface of the cone along a straight line from the apex to the base; then you take the surface of the cone as if it were a sheet of paper and lay it flat on a plane.</p>
<p>You can see for yourself how this works by cutting a circular disk out of a sheet of paper; then draw two radii on that disk and cut along each radius.
You will then have cut your disk into two pieces.
If you then take either piece of paper and join the two straight edges together, it will make a cone. If you let the edges separate and flatten the paper out again, it returns to its part-of-a-disk shape.</p>
<p>So when we cut and flatten the surface of the cone, it
forms part of a disk whose radius is the distance from the apex to the base circle.
That distance is $\sqrt{r^2 + h^2}.$
The length of the curved arc is the circumference of the base circle of the cone, which is $2\pi r.$
The area of this part of the disk is half the radius times the length of the arc, namely,
$$ A_1 = \frac12 \sqrt{r^2 + h^2} \left( 2\pi r \right)
= \pi r \sqrt{r^2 + h^2}.$$</p>
<p>A second way is your way: make a stack of disks inside the cone and
measure the curved surface areas on the outer edges of the disks.
The radius of a disk can be expressed as a function of its distance
from the apex; call this distance $y$, and then the disk's radius
is $ry/h.$
The curved area is $2\pi(ry/h)(\Delta y),$ where $\Delta y$
is the thickness of the disk.
If we take the limit as the thickness of the disks approaches zero,
we get the integral
$$ A_2 = \int_0^h \frac{2\pi ry}{h} dy
= \frac{2\pi r}{h} \int_0^h y\,dy
= \frac{2\pi r}{h} \left( \frac12 h^2\right) = \pi rh.
$$
As you should be able to see, $A_2 < A_1.$
(Try plugging in some numbers such as $r=h=1$ if you doubt it.)</p>
<p>To my intuition, rolling parts of paper disks into cones and unrolling them is a far more convincing way to establish the area of a cone than stacking up a bunch of very thin disks.
After all, the cone is one continuous surface, which is preserved (except the connection to itself along one cut) when we unroll it, whereas the curved parts of your disks are not connected to one another at all.
I conclude that $A_1$ is the correct area and $A_2$ is not.</p>
<p>This is the difference between the correct formula for area and your formula: the correct formula slices the solid of rotation into thin slices,
and approximates each slice by a <em>frustum</em> (a horizontal slice of a cone).
If the slices are thin enough, the curved surface of the frustum is
very, very close to of its part of the surface area of the solid of revolution,
whereas with disks we will always have grooves in the surface
(unless the solid was a cylinder to begin with).
If we only buy enough paint to paint the curved surface of each disk, it won't be enough paint to connect to the next disk below it,
even if we smooth out the grooves.
If we can paint the frustums, however, we have enough paint.</p>
|
1,773,303 |
<p>Let $k$ be a field and $L=k(\sqrt{a_1},\sqrt{a_2},\cdots,\sqrt{a_n})$ with $a_i\in k$. Suppose $[L:k]=2^n$. Let $a=\sqrt{a_1}+\sqrt{a_2}+\cdots\sqrt{a_n}$. Show that $L=k(a)$.</p>
|
H. Potter
| 289,192 |
<p>Since you know that being Hausdorff is an "increasing property" (that means if $\mathcal{T}'\subseteq \mathcal{T}$ and $\mathcal{T}'$ is Hausdorff, then $\mathcal{T}$ is Hausdorff), you know that the only possible topology is the discrete topology. Now your goal is just to show that any Hausdorff topology on a finite set is discrete. </p>
<p>Since $\{x\}$ is closed $\forall x\in X$, you have that $\bigcup_{x\not = x_0} \{x\} = X-\{x_0\}$ is closed since it's a finite union. Hence, $\{x_0\}$ is open. $x_0$ is an arbitrary point in $X$, so your topology has to be the discrete topology.</p>
<p>Here, you should think that, since your set $X$ is finite, there is something to do with the intersection of a lot of open sets in $X$ (or equivalently, the union of a lot of closed sets in $X$).</p>
|
3,474,847 |
<p><strong>Prop:</strong> The union of two open sets is open. </p>
<p><strong>Pf:</strong> Let <span class="math-container">$(X,d)$</span> be a metric space, and <span class="math-container">$U$</span> and <span class="math-container">$V$</span> are non empty and open sets such that <span class="math-container">$U\subset X$</span> and <span class="math-container">$V \subset X$</span>. Let <span class="math-container">$p$</span> be a point such that <span class="math-container">$d(p,q)< \epsilon_1 $</span> for some <span class="math-container">$\epsilon_1 > 0$</span> and <span class="math-container">$q \in U$</span>. And <span class="math-container">$d(p,q)<e_2$</span> for some <span class="math-container">$\epsilon_2 >0$</span> and <span class="math-container">$q \in V$</span>. Since we know that <span class="math-container">$q\in V$</span> and <span class="math-container">$q \in U$</span>, we notice <span class="math-container">$q \in( U \bigcup V)$</span>. Therefore, <span class="math-container">$q$</span> is an interior point of the union of those sets <span class="math-container">$\square$</span>.</p>
<p>I want to know if my proof is well-phrased and proves the proposition.</p>
|
Ben Grossmann
| 81,360 |
<p><em>Note: since the asker has been updating their proof since originally posting the question, this answer may no longer address the proof in its current form.</em></p>
<hr>
<p>Your proof could be substantially improved. Some criticisms below.</p>
<blockquote>
<p>Let <span class="math-container">$(X,d)$</span> be a metric space, and <span class="math-container">$U$</span> and <span class="math-container">$V$</span> are non empty sets such that <span class="math-container">$U\subset X$</span> and <span class="math-container">$V \subset X$</span>. </p>
</blockquote>
<p>You should state that <span class="math-container">$U$</span> and <span class="math-container">$V$</span> are <strong>open</strong> sets. Also note that at some point, your proof should also address the case where either <span class="math-container">$U$</span> or <span class="math-container">$V$</span> is empty.</p>
<blockquote>
<p>Then, <span class="math-container">$U$</span> and <span class="math-container">$V$</span> are also a metric space. </p>
</blockquote>
<p>I have no idea what this sentence is supposed to mean; it should probably be removed. </p>
<blockquote>
<p>We shall denote a point <span class="math-container">$p$</span> such that <span class="math-container">$d(p,q)< \epsilon_1 $</span> for some <span class="math-container">$\epsilon_1 > 0$</span>, then <span class="math-container">$q \in U$</span>. And we can find a point <span class="math-container">$p$</span> such that <span class="math-container">$d(p,q)<e_2$</span> for some <span class="math-container">$\epsilon_2 >0$</span>, then <span class="math-container">$q \in V$</span>. Since <span class="math-container">$q\in V, q \in U \implies q \in( U \bigcup V)$</span>, <span class="math-container">$q$</span> is an interior point of the union of those sets <span class="math-container">$\square$</span></p>
</blockquote>
<p>I don't understand this (which appears to be the core of your proof). At the very least, your first sentence "We shall denote a point <span class="math-container">$p$</span> such that <span class="math-container">$d(p,q)< \epsilon_1 $</span> for some <span class="math-container">$\epsilon_1 > 0$</span>, then <span class="math-container">$q \in U$</span>" makes no sense as written.</p>
<p>Because of the definition of an open set in a metric space, your proof should be structured as follows: first, define what <span class="math-container">$p$</span> is. Say something like let <span class="math-container">$p$</span> be a point ("any point" or "an arbitrary point" also work) in <span class="math-container">$U \cup V$</span>. Then, using the fact that <span class="math-container">$U$</span> and <span class="math-container">$V$</span> are open, show that there exists an <span class="math-container">$\epsilon > 0$</span> such that <span class="math-container">$q \in U \cup V$</span> whenever <span class="math-container">$d(p,q) < \epsilon$</span>. This can be divided into parts:</p>
<ul>
<li>Define <span class="math-container">$\epsilon$</span></li>
<li>State what <span class="math-container">$q$</span> is. Again, say something like "let <span class="math-container">$q$</span> be any point in <span class="math-container">$X$</span> with <span class="math-container">$d(p,q) < \epsilon$</span>". </li>
<li>Show that we have <span class="math-container">$q \in U \cup V$</span>.</li>
</ul>
|
3,606,255 |
<p>I want to ask you something about the following quadratic inequality: <span class="math-container">$$9x^2+12x+4\le 0.$$</span> Let us find the "=0" points. Here we have <span class="math-container">$9x^2+12x+4=9(x+\frac{2}{3})^2\ge0$</span> (we can factor this by calculating the discriminant <span class="math-container">$D=36-36=0$</span> and using <span class="math-container">$ax^{2}+bx+c=a(x-x_1)(x-x_2)$</span>). I am practising the wavy curve method (the method of intervals) to solve quadratic inequalities, so I would want to solve it using this way. What is the general method to approach the quadratic inequality when the discriminant is zero, and we want to use the wavy curve method?
<a href="https://i.stack.imgur.com/YjIB4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YjIB4.png" alt="enter image description here"></a></p>
|
Michael Rozenberg
| 190,319 |
<p><span class="math-container">$$(3x+2)^2\leq0$$</span> gives <span class="math-container">$$3x+2=0$$</span> or <span class="math-container">$$x=-\frac{2}{3}.$$</span>
Also, if you want to use <span class="math-container">$\Delta$</span>, so for <span class="math-container">$\Delta=0$</span> we have the following formula for the root.</p>
<p><span class="math-container">$$x_1=-\frac{b}{2a}.$$</span>
We obtain:
<span class="math-container">$$x_1=-\frac{12}{2\cdot9}=-\frac{2}{3}.$$</span></p>
<p>There are four cases:</p>
<ol>
<li><p><span class="math-container">$(x-a)^2\geq0.$</span> We obtain <span class="math-container">$\mathbb R$</span>;</p></li>
<li><p><span class="math-container">$(x-a)^2>0.$</span> we obtain <span class="math-container">$(-\infty,a)\cup(a,+\infty)$</span>;</p></li>
<li><p><span class="math-container">$(x-a)^2\leq0$</span>. We obtain <span class="math-container">$\{a\}$</span> and </p></li>
<li><p><span class="math-container">$(x-a)^2<0$</span>. We obtain <span class="math-container">$\oslash$</span></p></li>
</ol>
|
1,351,881 |
<p>This argument comes up once every while on Lambda the Ultimate. I want to know where the flaw is.</p>
<p><em>Take a countable number of TMs all generating different bitstreams. Construct a Cantor TM which runs every nth TM upto the nth bit and outputs the reversed bit.</em></p>
<p><em>Now I have established there are uncountably many TMs.</em></p>
<p>(The argument is that every bitstream is a unique real in the set [0,1]. Since every TM can be seen to be represented by that real, we can use Cantor's construction to prove that there are uncountably many TMs.)</p>
<p>Where does it go wrong?</p>
|
Asaf Karagila
| 622 |
<p>There are uncountably many Turing machines, if you allow them to use oracles (which are real numbers, really).</p>
<p>But if you disallow that, then the only way that this process gives back a Turing machine is if the enumeration itself is computable. So what you have shown is that the set of Turing machines is not computable (albeit it is recursively enumerable, but that is not enough).</p>
<p>This "liberal assumption" is the only assumption that you make which does not imply "most infinite sets related mathematics is inconsistent", by Occam razor arguments alone it shows that the assumption might be <em>a bit <strong>too</strong> liberal</em> after all.</p>
<p>As for the suggestion that we make an enumeration for enumerating the rational numbers, sure, you just get (more or less) an irrational number. Showing that <em>one</em> countable set doesn't include all the Turing machines is not a contradiction.</p>
|
1,435,665 |
<p>Matrix $\mathbf{D}$ is a full rank diagonal matrix. $\mathbf{ADA}^T=\mathbf{D}$.</p>
<p>Can we conclude that $\mathbf{A}^2=\mathbf{I}_n$? (edited from $\mathbf{A}=\pm\mathbf{I}$)</p>
<p>It's almost sure that some one has asked this question before, but I cannot find proper keywords to search for that post.</p>
|
Rory Daulton
| 161,807 |
<p>The equation $y^2+xy-3x=5$ becomes, after differentiation using the chain rule and the product rule,</p>
<p>$$2y\frac{dy}{dt}+y\frac{dx}{dt}+x\frac{dy}{dt}-3\frac{dx}{dt}=0$$</p>
<p>Can you continue from there?</p>
|
1,435,665 |
<p>Matrix $\mathbf{D}$ is a full rank diagonal matrix. $\mathbf{ADA}^T=\mathbf{D}$.</p>
<p>Can we conclude that $\mathbf{A}^2=\mathbf{I}_n$? (edited from $\mathbf{A}=\pm\mathbf{I}$)</p>
<p>It's almost sure that some one has asked this question before, but I cannot find proper keywords to search for that post.</p>
|
Brevan Ellefsen
| 269,764 |
<p>Doesn't $\frac{d}{dt} (y^2 +xy - 3x = 5) \Rightarrow 2y\frac{dy}{dt} + y\frac{dx}{dt} + x\frac{dy}{dt} - 3\frac{dx}{dt} $
??</p>
|
2,736,295 |
<p>Let $f(x,y) = xy^2$ and the domain $D = \lbrace (x,y)| x,y\geq0, x^2 + y^2 \leq 3 \rbrace$</p>
<p>$f_x(x,y) = y^2$ and $f_y(x,y) = 2xy$ </p>
<p>Therefore, the critical points should be $\lbrace (x,y)| y = 0, \sqrt{3} \geq x \geq 0 \rbrace$. </p>
<p>The determinant of the Hessian is
$$\det(Hf(x,y))=
\begin{vmatrix}
0 & 2y \\
2y & 2x
\end{vmatrix} = 0-4y^2$$</p>
<p>But this doesn't make sense to me because, if $y = 0$, then $f(x,0) = 0$. But on this interval, suppose I chose $(x,y) = (1,\sqrt{2})$, this would be larger and would be the maximum value of the system. </p>
<p>Why didn't finding the first partials and the Hessian allow me to compute the critical point? </p>
|
quasi
| 400,434 |
<p>Critical points are <em>candidates</em> for the max and min values.
<p>
Boundary points are also candidates.
<p>
But for the specified domain $D$, the only possible critical points are the points where $y=0$, and those points are on the boundary, hence are not actually critical points.
<p>
It follows that the maximum and minimum values of $xy^2$ must be on the boundary.
<p>
For the minimum, it's obvious that the minimum value is $0$, which occurs at all points on the boundary where $x=0\;$or $y=0$.
<p>
For the maximum, it's obvious that the maximum is positive, hence the only candidates are boundary points where $x,y > 0.\;$It follows that the maximum occurs on the circular part of the boundary, excluding the two endpoints.
<p>
Thus, letting $(x,y)=\bigl(\sqrt{3}\cos(t),\sqrt{3}\sin(t)\bigr)$, we want to maximize $\bigl(\sqrt{3}\cos(t)\bigr)\bigl(3\sin^2(t)\bigr)$, for $0 < t < {\large{\frac{\pi}{2}}}.\;$Then letting $g(t)=\bigl(\sqrt{3}\cos(t)\bigr)\bigl(3\sin^2(t)\bigr)$ and setting $g'(t)=0$, we get only one critical point for $0 < t < {\large{\frac{\pi}{2}}}$, namely $t=\cos^{-1}\left({\large{\frac{1}{\sqrt{3}}}}\right)$, which yields the maximum value for $f$ of $2$.
<p>
Alternatively, now that I see saulspatz's comment, since the circle has the equation $x^2+y^2=3$, maximizing $xy^2$ on the circular arc of the boundary, excluding the endpoints, is the same as maximizing the function $x(3-x^2)$ on the interval $0 < x < \sqrt{3}$. Taking the derivative and setting it to $0$, we get only one critical in the interval $0 < x < \sqrt{3}$, namely $x=1$, which yields the maximum value for $f$ of $2$.</p>
|
3,019,882 |
<p>Í already have found a proof About the statment which I did not understand, I will write it down, so maybe someone can explain me the part that I did not understand. But if there are easier ways to prove the statement I would be delighted to know. </p>
<p><span class="math-container">$\alpha:=\sqrt{2}+\sqrt{3}\notin U:= \mathbb{Q}(\sqrt{2})\cup\mathbb{Q}(\sqrt{3})\iff\alpha \notin \mathbb{Q}(\sqrt{2})$</span> and <span class="math-container">$\alpha \notin \mathbb{Q}(\sqrt{3})$</span></p>
<p>Proof: <span class="math-container">$\alpha \notin \mathbb{Q}(\sqrt{2})$</span></p>
<p>Assume <span class="math-container">$\exists_{r,s\in \mathbb{Q}}r+s\sqrt{2}=\sqrt{2}+\sqrt{3}$</span></p>
<p><span class="math-container">$\iff \sqrt{3}=r+(s-1)\sqrt2$</span></p>
<p><span class="math-container">$\iff 3=r^2+2r(s-1)\sqrt2+2(s-1)^2$</span></p>
<p>Which implicates <span class="math-container">$3=0$</span> or <span class="math-container">$\sqrt3\in \mathbb{Q}$</span> or <span class="math-container">$\sqrt{3/2}\in \mathbb{Q}$</span> or <span class="math-container">$\sqrt{2}\in \mathbb{Q}$</span></p>
<p>I don't understand the last implication, I Need some help here.</p>
<p>Many Thanks </p>
|
Joel Cohen
| 10,553 |
<p>We know that any element of <span class="math-container">$\mathbb{Q}[\sqrt{2}]$</span> can be <strong>uniquely</strong> written as <span class="math-container">$a + b \sqrt{2}$</span> with <span class="math-container">$a, b \in \mathbb{Q}$</span> (the unicity is important). So, if <span class="math-container">$\alpha \in \mathbb{Q}[\sqrt{2}]$</span>, then so does <span class="math-container">$\alpha - \sqrt{2} = \sqrt{3} \in \mathbb{Q}[\sqrt{2}]$</span>. So you can write <span class="math-container">$\sqrt{3} = r + s \sqrt{2}$</span>. Squaring that equality, we get</p>
<p><span class="math-container">$$3 = (r^2 + 2 s^2) + 2rs \sqrt{2} $$</span></p>
<p>By unicity, the number <span class="math-container">$3$</span> can be uniquely written as <span class="math-container">$3 + 0 \sqrt{2}$</span>, so in the right hand side we get <span class="math-container">$r^2 + 2 s^2 = 3$</span> and <span class="math-container">$2rs=0$</span>. From <span class="math-container">$rs = 0$</span> we get that <span class="math-container">$r=0$</span> or <span class="math-container">$s = 0$</span>. If <span class="math-container">$s=0$</span>, then <span class="math-container">$r^2 = 3$</span>, which is impossible since <span class="math-container">$r$</span> is rational. And if <span class="math-container">$r = 0$</span>, then <span class="math-container">$2s^2=3$</span> so <span class="math-container">$s^2 = \frac{3}{2}$</span>, which is also impossible since <span class="math-container">$s$</span> is rational. So in summary, <span class="math-container">$\sqrt{3}$</span> is not in <span class="math-container">$\mathbb{Q}[\sqrt{2}]$</span>.</p>
|
316,016 |
<p>Could you recommend any approachable books/papers/texts about matroids (maybe a chapter from somewhere)? The ideal reference would contain multiple examples, present some intuitions and keep formalism to a necessary minimum.</p>
<p>I would appreciate any hints or appropriate sources.</p>
|
Tara Fife
| 215,095 |
<p>There is a great introduction to matroids that is about 40 pages, free, and sounds more like what you are looking for: <em>What is a matroid?</em> by James Oxley, <a href="https://www.math.lsu.edu/~oxley/survey4.pdf" rel="nofollow">available from the author's webpage</a>.</p>
|
2,606,999 |
<p>$f: X \to Y$ prove that if $A \subseteq X$, then A is a subset of the pre-image of the image of $A$, which is shown by these symbols respectively: $f^{-1}[f[A]]$</p>
<p>This is my proof:</p>
<p>If $A \nsubseteq f^{-1}[f[A]]$, then there exists "$x$" as an element of $A$, such that $f^{-1}[f[A]] \neq x$
which is a contradiction.</p>
<p>Is this the correct way of going about doing this? I'm also not sure if we can assume that $f^{-1}[f[A]] = x$ is always true</p>
|
String
| 94,971 |
<p>It suffices to note that for any $x\in A$ we have some image $f(x)$. This implies
$$
x\in f^{-1}(f(x))\subseteq f^{-1}(f(A))
$$
To sum up, we have
$$
x\in A\implies x\in f^{-1}(f(A))
$$
which in turn means $A\subseteq f^{-1}(f(A))$.</p>
|
335,258 |
<p>Find the domain of the function:
$$f(x)= \sqrt{x^2 - 4x - 45}$$</p>
<p>I'm just guessing here; how about if I square everything and then put it in the graphing calculator?
Thanks,
Lauri</p>
|
Lauri
| 67,552 |
<p>Well, I just happened to run into a mathematician yesterday (in person) and he just pulled the polynomial out of the square root and added the "greater than or equal to"symbol.Because, a square root number can't be a negative number,but it can be zero. Then we just factored it and it factored out to (x+5)(x-9), which leaves the domain at -5 and less and also at 9 and more.I graphed it and it came out to be right.
This is the first time I used this site and I don't see where you get the math expressions to type. Maybe because I didn't join as a member?
Lauri</p>
|
167,440 |
<blockquote>
<p>Let $\Bbb F$ be a field of characteristic $p\gt 0$ and $f(x)=x^{p^n}-c \in\Bbb F[x]$ where $n$ is a positive integer. If $c \notin \{a^p:a\in \Bbb F \}$, show that $f$ is irreducible in $\Bbb F[x]$.</p>
</blockquote>
<p>I recently started studying Field theory, so I don't know How to approach this problem? Any help would be appreciated.</p>
|
Community
| -1 |
<p>Go up to a splitting field $E$ so that in there $f(x)$ splits completely. Now a root of $f(x)$ is $\sqrt[p^n]{c}$ and so we can write $f(x)$ over $E$ as </p>
<p>$$f(x) = x^{p^n} - (\sqrt[p^n]{c})^{p^n} = (x - \sqrt[p^n]{c})^{p^n}$$</p>
<p>by the schoolboy binomial theorem. Now recall $f(x) \in \Bbb{F}[x]$ so that <strong><em>each coefficient</em></strong> of $f(X)$ is <strong><em>an element</em></strong> of $\Bbb{F}$. Suppose that $f(x)$ is reducible. From the factorisation above, we see that any divisor of $f(x)$ must be of the form </p>
<p>$$(x - \sqrt[p^n]{c})^{r}$$</p>
<p>where $0 \leq r \leq p^n$. Now expand this out by the ordinary binomial theorem and look at the coefficient of the second highest power of the indeterminate $x$. Can you deduce the irreducibility of $f(x)$ from here? Hint: show that $r = 0$ or $r = p^n$.</p>
|
2,493,481 |
<p>I'm currently studying calculus of variations. I couldn't find a rigorous definition of a functional on this site.</p>
<ol>
<li>What is the general definition of a functional?</li>
<li>Why for calculus of variations in physics, I must to use for a functional <em>a convex function</em> for the space of the admissible functions?</li>
</ol>
|
Community
| -1 |
<blockquote>
<p>What is a functional?</p>
</blockquote>
<p>As it is well written in <a href="https://en.wikipedia.org/wiki/Functional_(mathematics)" rel="nofollow noreferrer">Wikipedia</a>, a functional simply refers to a mapping from a space $X$ into the real numbers. (Technically one might refer to other <a href="https://en.wikipedia.org/wiki/Field_(mathematics)" rel="nofollow noreferrer">fields</a> as well. But we can safely restrict our attention to the real numbers here.) Depending on the authors, such maps may or may not be linear. Consider the following example in the context of calculus of variations.</p>
<p>Suppose $U\subset{\bf R}^n$ is a bounded open set with smooth boundary $\partial U$ and we are given a smooth function (the Lagrangian)
$$
L:{\bf R}^n\times{\bf R}\times\overline{U}\to{\bf R}.
$$</p>
<p>We define a map $I$ with
$$
I[w]:=\int_U L(Dw(x),w(x),x)\ dx
$$
for smooth functions $w:\overline{U}\to{\bf R}$ satisfying, say, the boundary condition
$$
w=g\quad\text{on }\partial U.
$$</p>
<p>Now let $X$ be the set of smooth functions $w:\overline{U}\to{\bf R}$ such that $w=g$ on $\partial U$. Then the map $I$ is a <strong>functional</strong> mapping from $X$ to ${\bf R}$.</p>
<hr>
<blockquote>
<p>Convexity (of the Lagrangian)?</p>
</blockquote>
<p>The basic idea of calculus of variation can be summarized as follows: instead of solving a (nonlinear) PDE $A(u)=0$ directly, one associates it with some appropriate "energy" functional $I$ by writing $A$ as the "derivative" of $I$. Symbolically, we write
$$
A(\cdot)=I'[\cdot].
$$
Then the PDE $A(u)=0$ reads as $I'[u]=0$. This new formulation allows us to recognize solutions of $A(u)=0$ as being <strong>critical points</strong> of the functional $I[\cdot]$. <em>If</em> the functional $I[\cdot]$ has a minimum at $u$, then presumably $I'[u]=0$ is valid and thus $u$ is a solution of the original PDE.</p>
<p>On the other hand, one needs some conditions on the Lagrangian $L$ which ensure that the functional $I[\cdot]$ does indeed have a minimizer within an appropriate function space (the class of "admissible" functions). It turns out that one of the conditions we need is the <strong>convexity</strong> of $L$ (which together with some other assumptions imply the "weak lower semicontinuity" of the functional $I$, which allows us to take limits in a compactness argument). The convexity assumption of $L$ can be also seen by the <em>second variation</em> of $I[\cdot]$ at the function $u$. </p>
<p>For more technical mathematical details, see the first two sections in Chapter 8 of Evans's <em><a href="https://bookstore.ams.org/gsm-19-r" rel="nofollow noreferrer">Partial Differential Equations</a></em>. </p>
<hr>
<p><sup>Note: This question has been edited quite a few times. This is an answer to the <a href="https://math.stackexchange.com/revisions/2493481/24">current version</a>.</sup></p>
|
1,822,179 |
<p>Pointwise limit of the sequence of functions </p>
<p>$$h_n(x)=\begin{cases}1,&\text{if }x\ge \frac1n\\nx,&\text{if }x\in[0,\frac1n)\end{cases}$$</p>
<p>The trouble with this question is that I think that $h(0)$ is not defined but Im not completely sure. We can see that for any $n\in\Bbb N$ we have that $h_n(0)=0$. But the sequence $h_n(x)=1$ is approaching to zero as well but only reach zero in the limit cause $\lim 1/n=0$.</p>
<p>Are my assumptions correct? It is $h(0)$ undefined?</p>
|
Community
| -1 |
<p>Assuming that by $h$ you mean the pointwise limit of $(h_n)$: no, $h(0)$ is defined. In fact, $h(0) :=\lim_{n \to \infty} h_n(0) = 0$. </p>
<p>What is confusing you is that $h$ is not continuous. We have for $x > 0$, there exists $k \in \Bbb N$ such that $x > 1/k$, so $h_n(x) = 1$ for all $n \ge k$, thus $h_n(x) \to 1$. This shows that:</p>
<p>$$h(x) = \begin{cases} 0, x = 0 \\ 1, x > 0 \end{cases}$$</p>
|
33,646 |
<h1>Preamble</h1>
<p>I am a novice SE user, a toddler. In this post I want to criticize some moderator actions, which seems risky and futile, regarding unwritten policies on the meta; however, just for the record, I write this post so that unbiased readers find it helpful.</p>
<p>Please note that I am not throwing a tantrum and that I have slept enough, so it is not needed to say to me, "It is time for you to go to bed, nighty night."</p>
<h1>what's up? Why are you crying again?</h1>
<p>I, as a one being interested in contributing to the community but lacking enough math knowledge to create high-quality math content, decided to suggest grammar-fixing edits. But, many community adults ordered me not to "tinker", so I had to follow their order and stop fixing the grammar of posts.</p>
<p>After that I decided to do some <em><strong>unpaid</strong></em> janitorial task, flagging unnecessary comments, regarding that there is an option for such comments when one flags a comment and that the following excerpt from the <a href="https://meta.stackexchange.com/a/17365/502725">comment guideline</a> confirms such flagging:</p>
<blockquote>
<h3>When should I flag a comment?</h3>
<p>...</p>
<ul>
<li>It's no longer needed. <br/><br/>This comment is outdated, conversational or not relevant to the post.</li>
</ul>
</blockquote>
<p>(Also, <a href="https://meta.stackexchange.com/a/279451/502725">this community manager's answer</a> may be helpful).</p>
<p>After cleaning up many threads, the moderator team made two actions against me, which are criticized in this post:</p>
<p><strong>First</strong>, they contacted me privately sending the following message:</p>
<blockquote>
<p>... these comments are inoffensive and, while they are not strictly needed any longer, they are not causing harm to the site. On the other hand, it takes time for us to handle your flags. In the future, please think about whether or not a comment's deletion is worth the time of the moderators before you flag it.</p>
</blockquote>
<p><strong>Second</strong>, they rejected a huge number of my flags, which seems to have been deleted by the moderator(s) rejecting such flags. Since deleted comments cannot be seen by users except moderators, I cannot prove the claim to you (I think the moderator team would not deny it); however, let me put a screenshot of my declined flags here:</p>
<p><a href="https://i.stack.imgur.com/Z0Bml.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z0Bml.jpg" alt="enter image description here" /></a></p>
<p>Now, let us consider the following points:</p>
<ul>
<li>It was said, "they are not causing harm to the site." This is wrong because of the following reasons:<br/><br/><strong>First</strong>, if they were not causing harm to the site, the Stack Exchange community team would have never put the mentioned option in the list of flagging reasons and explicitly mentioned their opinion about such comments in the guideline.<br/><br/><strong>Second</strong>, they are actually causing harm to the site because many users put their conversational and unnecessary comments under posts when they see others do that, so this site would contain an uncontrollable number of such comments.</li>
<li>It was said, "it takes time for us to handle your flags." According to <a href="https://math.meta.stackexchange.com/q/33035/656359">this information</a>, the number of comment flags handled by the moderators is on average about 25 per a day. Let us suppose that I want to raise 25 comment flags per a day, which needs to be handled by the moderators (Please note that many comments are handled by the system). Since there are currently 10 moderators on the site, each moderator needs to handle 5 flags per a day, each of which should not take more than several seconds. If some moderators are not active enough on the site, I think rejecting a toddler's unpaid janitorial task is not a good solution for such a problem.</li>
<li>It was said, "think about whether or not a comment's deletion is worth the time of the moderators before you flag it." It seems that the moderators presume that I am an idle person whose only job and concern is flagging unnecessary comments on the site (?). Do you know digging up unnecessary comments takes a considerable amount of time and energy? My time and energy are worthless?</li>
<li>As you know well, I did nothing wrong about flagging such comments; I just followed the <strong>written</strong> guideline, but was treated according to an <em>unwritten</em> norm(?). According to the guideline, rejection of many flags may result in some written consequences (for example, reducing my flag daily allowance or flag ban) or <em>unwritten</em> ones (For example, a community manager disliking me may want to take it as an excuse to suspend me for 99 years).</li>
<li>If my behavior was a misdeed, why did a moderator remove the flagged comments after rejecting my flags? This reminds me of my previous story saying that a user made the same edit, which had already been rejected by the user; however, there is a difference: in this case the one who did such a thing is a moderator, not a regular user. I do not know why such things happen to a toddler like me.</li>
</ul>
<p>Finally, let me answer your question, "Why are you crying?" I am crying because I was treated like the following:</p>
<p>Suppose that you live in a city in which there is a rule asking citizens to keep it clean. You decide to keep the city clean by taking trash into a trash can <strong>for free</strong>. After you being tired, a mayor shouts at you, "Why do you waste our time? Trash cans must be emptied." Then the mayor empties a trash can onto your head.</p>
<p><strong>P.S.</strong> Although this post may seem harsh to the moderators, let me <strong>appreciate</strong> their efforts for moderating this community and handling meta issues. Please note that if I disliked the moderators, I would have never criticized them to continue making mistakes so that they would not improve themselves.</p>
|
hardmath
| 3,111 |
<p>It is reasonable and desirable that you can come here to Meta to ask about what you perceive to be "unwritten policies," and it is true (as noted in comments) that StackExchange Communities vary as to policy priorities. You seem motivated to help the Math.SE Community, and I appreciate your effort to gain understanding.</p>
<p>In this case the articulated point is that there could be better uses of Moderators' time and also of your time. Of course you are allowed to use the allotment of daily Comment flags according to your own judgement, but you have claimed in the midst of your post:</p>
<blockquote>
<p>(Please note that many comments are handled by the system).</p>
</blockquote>
<p>That was not my understanding. There are no Review Queues for Comments, but <em>as noted by quid and others below</em> the "conversational" (no longer needed) flags on Comments can under some circumstances result in an automated removal. Here's a relevant part of a <a href="https://meta.stackexchange.com/a/17365/155839">Meta StackExchange post</a>:</p>
<blockquote>
<p>If a comment is flagged by a sufficient number of users, it will be automatically deleted. There is no penalty for this. Flagged comments will be surfaced to moderators, so if you have a problem with a comment, flag it.</p>
<ul>
<li>If the comment has no upvotes and no "trigger" keywords, three flags from users will delete it.</li>
<li>If the comment has upvotes, it will require more flags to be automatically deleted: one more flag for every 3 upvotes, rounded up. (For example, a comment with 1 upvote requires 4 flags to be auto-deleted, and one with 8 upvotes requires 6 flags).</li>
<li>Comments containing certain "trigger" keywords are deleted instantly after a single flag, regardless of upvotes. The list of trigger keywords is kept secret, and may differ per site.</li>
<li>A single flag from a moderator will instantly delete the comment.</li>
</ul>
</blockquote>
<p>In the StackExchange model of Q&A posts, all comments are <a href="https://math.meta.stackexchange.com/questions/4623/any-comment-limit/4624#4624">considered ephemeral</a>. Thus ideally any important information in Comments gets transcribed either into a Question or an Answer post for the sake of posterity. But adherence to this ideal suffers both for temporal and practical circumstances.</p>
<p>Something that may have a conversational tone to you reading it years after the last activity on a Question may have served a purpose at the time that is not now obvious. But equally it may have pointed to open issues that were never fully resolved. Who would make the decision?</p>
<p>If you flag a Comment, it <em>might</em> be up to a Moderator to make the call.</p>
<p>My own experience is that I rarely have an urge to offer such flags, mainly when there has been a decisive clarification of some issue with a Question or Answer and leaving the "no longer needed" Comments up would invite confusion as to whether the resolution took place. Indeed I'm more likely to prune my own Comments than to flag those made by others.</p>
<p>A focus on current Questions and and Answers is normal, but of course you may find in searching for previous posts that there is something that is worth improving. Weigh the benefit to Math.SE Readers of flagging old conversational Comments versus researching a problem that you are interested in, and I think you'll find the latter is typically a better use of everyone's time.</p>
|
353,920 |
<p>I am studying a problem where a quadratic matrix equation emerges. The equation is as follow (all capital letters are n by n matrices)</p>
<p><span class="math-container">$(I-X^{\prime}L)X=D$</span></p>
<p>where <span class="math-container">$L$</span> and <span class="math-container">$D$</span> are both symmetric and positive definite. How much can I say about a solution <span class="math-container">$X$</span>?</p>
|
Piyush Grover
| 30,684 |
<p>A good resource is : Abou-Kandil, Hisham, Gerhard Freiling, Vlad Ionescu, and Gerhard Jank. Matrix Riccati equations in control and systems theory. Birkhäuser, 2012.</p>
|
400,715 |
<p>Consider the metric space $(\mathbb{Q},d)$ where $\mathbb{Q}$ denotes the rational numbers and $d(x,y)=|x-y|$. Let $$E:=\{x \in\mathbb{Q}:x>0, 2<x^2<3\}$$</p>
<p>Is $E$ closed and bounded in $\mathbb{Q}?$ Is it compact? Justify your answers.</p>
|
Brian M. Scott
| 12,042 |
<p>HINT: You should have no trouble deciding whether $E$ is bounded in $\Bbb Q$. To decide whether it’s closed in $\Bbb Q$, ask yourself whether $\Bbb Q\setminus E$ is open in $\Bbb Q$: can you find an open set $U$ in $\Bbb R$ such that $U\cap\Bbb Q=\Bbb Q\setminus E$? Remember, $\sqrt2$ and $\sqrt3$ are irrational.</p>
<p>For compactness, does the sequence $\langle 1.7,1.73,1.732,1.7320,1.73205,\dots\rangle$ in $E$ have a convergent subsequence?</p>
|
400,715 |
<p>Consider the metric space $(\mathbb{Q},d)$ where $\mathbb{Q}$ denotes the rational numbers and $d(x,y)=|x-y|$. Let $$E:=\{x \in\mathbb{Q}:x>0, 2<x^2<3\}$$</p>
<p>Is $E$ closed and bounded in $\mathbb{Q}?$ Is it compact? Justify your answers.</p>
|
Dimitris
| 37,229 |
<p><strong>Hint</strong></p>
<p>About boundness when do you call a set unbounded? What is your intuition about this set? Can a $x\in E$ be too much large?</p>
<p>Regarding compactness, a characterization of compact metric spaces is: </p>
<blockquote>
<p>$(X,d)$ is compact iff every sequence has a convergent subsequence</p>
</blockquote>
<p>Does it happen in the set $E$? Can you find a sequence of elements in $E$ which does not converge in $E$?</p>
|
26,192 |
<p>I have a list of rules that represents a list of parameters to be applied to a circuit model (Wolfram SystemModeler model): </p>
<pre><code>sk = {
{R1 -> 10080., R2 -> 10080., C1 -> 1.*10^-7, C2 -> 9.8419*10^-8},
{R1 -> 10820., R2 -> 4984.51, R3 -> 10000., R4 -> 10000., C1 -> 1.*10^-7,
C2 -> 1.85417*10^-7},
{R1 -> 12600., R2 -> 12600., C1 -> 1.*10^-7, C2 -> 6.29882*10^-8},
{R1 -> 16420., R2 -> 16420., C1 -> 1.*10^-7, C2 -> 3.70897*10^-8},
{R1 -> 26120., R2 -> 26120., C1 -> 1.*10^-7, C2 -> 1.46573*10^-8},
{R1 -> 76600., R2 -> 1283.61, R3 -> 10000., R4 -> 10000., C1 -> 1.*10^-7,
C2 -> 1.01704*10^-7}};
</code></pre>
<p>Before I can apply these values to the model parameters I have to rename them. The list above consists of six lists - four lists of four rules and two lists of six rules. Those that have 4 rules should be named "sallenKeyUnityGain" and those that have 6 rules should be named "sallenKey". This is what I have so far: </p>
<pre><code>Table[If[Length[sk[[i]]] > 4, cirname = "sallenKey", cirname = "sallenKeyUnityGain"];
(cirname <> ToString[i] <> "." <> ToString[sk[[i]][[All, 1]][[j]]]) -> sk[[i]][[All, 2]][[j]]
, {i, Length[sk]}, {j, Length[sk[[i]]]}]
</code></pre>
<p>And this is the output: </p>
<pre><code>{{"sallenKeyUnityGain1.R1" -> 10080., "sallenKeyUnityGain1.R2" -> 10080.,
"sallenKeyUnityGain1.C1" -> 1.*10^-7, "sallenKeyUnityGain1.C2" -> 9.8419*10^-8},
{"sallenKey2.R1" -> 10820., "sallenKey2.R2" -> 4984.51, "sallenKey2.R3" -> 10000.,
"sallenKey2.R4" -> 10000., "sallenKey2.C1" -> 1.*10^-7,
"sallenKey2.C2" -> 1.85417*10^-7},
{"sallenKeyUnityGain3.R1" -> 12600., "sallenKeyUnityGain3.R2" -> 12600.,
"sallenKeyUnityGain3.C1" -> 1.*10^-7, "sallenKeyUnityGain3.C2" -> 6.29882*10^-8},
{"sallenKeyUnityGain4.R1" -> 16420., "sallenKeyUnityGain4.R2" -> 16420.,
"sallenKeyUnityGain4.C1" -> 1.*10^-7, "sallenKeyUnityGain4.C2" -> 3.70897*10^-8},
{"sallenKeyUnityGain5.R1" -> 26120., "sallenKeyUnityGain5.R2" -> 26120.,
"sallenKeyUnityGain5.C1" -> 1.*10^-7, "sallenKeyUnityGain5.C2" -> 1.46573*10^-8},
{"sallenKey6.R1" -> 76600., "sallenKey6.R2" -> 1283.61, "sallenKey6.R3" -> 10000.,
"sallenKey6.R4" -> 10000., "sallenKey6.C1" -> 1.*10^-7,
"sallenKey6.C2" -> 1.01704*10^-7}}
</code></pre>
<p>This would work fine if all were sallenKey or if all were sallenKeyUnityGain. However, I would like the output to look like this:</p>
<pre><code> {{"sallenKeyUnityGain1.R1" -> 10080., "sallenKeyUnityGain1.R2" -> 10080.,
"sallenKeyUnityGain1.C1" -> 1.*10^-7, "sallenKeyUnityGain1.C2" -> 9.8419*10^-8},
{"sallenKey1.R1" -> 10820., "sallenKey1.R2" -> 4984.51, "sallenKey1.R3" -> 10000.,
"sallenKey1.R4" -> 10000., "sallenKey1.C1" -> 1.*10^-7,
"sallenKey1.C2" -> 1.85417*10^-7},
{"sallenKeyUnityGain2.R1" -> 12600., "sallenKeyUnityGain2.R2" -> 12600.,
"sallenKeyUnityGain2.C1" -> 1.*10^-7, "sallenKeyUnityGain2.C2" -> 6.29882*10^-8},
{"sallenKeyUnityGain3.R1" -> 16420., "sallenKeyUnityGain3.R2" -> 16420.,
"sallenKeyUnityGain3.C1" -> 1.*10^-7, "sallenKeyUnityGain3.C2" -> 3.70897*10^-8},
{"sallenKeyUnityGain4.R1" -> 26120., "sallenKeyUnityGain4.R2" -> 26120.,
"sallenKeyUnityGain4.C1" -> 1.*10^-7, "sallenKeyUnityGain4.C2" -> 1.46573*10^-8},
{"sallenKey2.R1" -> 76600., "sallenKey2.R2" -> 1283.61, "sallenKey2.R3" -> 10000.,
"sallenKey2.R4" -> 10000., "sallenKey2.C1" -> 1.*10^-7,
"sallenKey2.C2" -> 1.01704*10^-7}}
</code></pre>
<p>In other words, if I have 4 apples and 2 oranges, instead of having:<br>
apple1, orange2, apple3, apple4, apple5, orange6, I would like:<br>
apple1, orange1, apple2, apple3, apple4, orange2. How would I do this?<br>
Thank you<br>
Tatjana</p>
|
Michael E2
| 4,999 |
<p>Here's a version where the counting is done by structuring the data with <code>GatherBy</code> and using <code>MapIndexed</code>.</p>
<pre><code>lengthToNames = {6 -> "sallenKey", 4 -> "sallenKeyUnityGain"};
gatheredSk = GatherBy[sk, Length];
positionToNames = MapIndexed[
With[{basename = Length[#[[1]]] /. lengthToNames}, {First[#2],
index_, ___} :> basename <> ToString[index]] &, gatheredSk];
rewrite[rule_, name_] := ReplacePart[rule, 1 -> name <> "." <> ToString[First@rule]];
Flatten[
MapIndexed[# /. rule_Rule :> rewrite[rule, #2 /. positionToNames] &, gatheredSk, {3}],
1]
</code></pre>
<p>The key is to write a rule that maps positions in <code>gatheredSk</code> to base names.</p>
<pre><code>positionToNames
</code></pre>
<blockquote>
<pre><code>{{1, index$_, ___} :> "sallenKeyUnityGain" <> ToString[index$],
{2, index$_, ___} :> "sallenKey" <> ToString[index$]}
</code></pre>
</blockquote>
<p>In this case, it's just as easy to do it by hand. :) Except above the rule was constructed without assuming the order of elements in <code>sk</code>. (Given the problem of getting the numbering right, we probably could have assumed length 4 will always be in position 1, etc.)</p>
|
2,439,202 |
<p>i want to solve this question but i don't know what way i use :
$$\lim_{n\to \infty}a_n= \lim_{n\to\infty} \frac{\binom n 1} {n} +\frac {\binom n 2} {n^2} + \cdots + \frac {\binom n n}{n^n} $$</p>
|
Angina Seng
| 436,618 |
<p>This sum is
$$-1+\sum_{k=0}^n\binom{n}{k}\frac1{n^k}=
-1+\left(1+\frac1n\right)^n.$$
This has a well-known limit...</p>
|
262,500 |
<p>What is the Green’s function of the boundary value problem
$$
\frac{\mathrm d^2 y}{\mathrm d x^2}-\frac{1}{x}\frac{\mathrm dy}{\mathrm dx}=1,\quad y(0)=y(1)=0,
$$</p>
<p>this boundary problem is not self adjoint, so please help me how to solve it.</p>
|
Community
| -1 |
<p>Denoting $y'(x) = v(x)$, we have that $$v'(x) - \dfrac{v(x)}{x} = 1$$
Let $v(x) = x g(x)$. Then we get that $$xg'(x) + g(x) - g(x) = 1\implies g'(x) = \dfrac1x$$
Hence, we have $$g(x) = \log(x) + c_1 \implies v(x) = x \log(x) + c_1 x$$
Hence, we now need to solve for
$$y'(x) = x \log(x) + c_1 x$$
$$y(x) = \int x \log (x) + c_2 x^2 + c_3 = \dfrac{x^2}4 \left(2 \log(x) - 1 \right) + k_1 x^2 + k_2$$
$$y(0) = 0 \implies k_2 = 0$$
$$y(1) = -\dfrac14 + k_1 = 0 \implies k_1 = \dfrac14$$
Hence,
$$y(x) = \dfrac{x^2 \log(x)}2$$</p>
|
2,404,258 |
<p>I'm asked to prove that </p>
<h2>$|z-i||z+i|=2$</h2>
<p>defines an ellipse in the plane.</p>
<p>I have tried replacing $z = x+iy $ in the previous equation and brute forcing the result to no avail. Considering that $ |z-i||z+i| = |z^2+1| $ eases the algebra a bit but didn't help me that much.</p>
<p>Edit: I know it seems like there's a missing plus sign: $|z-i| + |z+i| = 2$ in the question, but that's what the exercise says. </p>
<p>In fact, the next exercise wants us to prove that $ |z-1||z+i| = 2$ defines a line in the complex plane.</p>
|
José Carlos Santos
| 446,262 |
<p>This expression <em>doesn't</em> define an ellipse. It's the set of those points of the plane such that the <em>products</em> of its distances to $i$ and to $-i$ is equal to $2$. If we were talking about the <em>sum</em> of the distances, then, yes, it would be an ellipse.</p>
<p>In order to prove that it is not an ellipse, note that four of the points of this set are $\pm1$ and $\pm\sqrt3\,i$. Furthermore, the set is symmetric with respect to both axis. There is one and only one ellipse fulfilling these conditions:$$\left\{x+yi\in\mathbb{C}\,\middle|\,x^2+\frac{y^2}3=1\right\}.$$One of the points of this ellipse is $z_0=\frac12+\frac{3i}2$. But $|z_0-i|.|z_0+i|=\frac{\sqrt{13}}2$. Therefore, your set is <em>not</em> an ellipse.</p>
|
1,810,582 |
<p>Statement 1: </p>
<blockquote>
<p>If $f :A\mapsto B$ & if $X,Y$ are subsets of $B$, then the inverse image of union of two sets $X$ & $Y$ is the union of inverse images of $X$ & $Y$.</p>
</blockquote>
<p>Statement 2:</p>
<blockquote>
<p>Let $f:A\mapsto B$ be a one-to-one correspondence from $A$ to $B$.
We define inverse of function , $(f^{-1}): B \mapsto A$ as $f(a)= b$. This inverse function also has a one-to-one correspondence.</p>
</blockquote>
<p>My question is what should i note in these two statements?</p>
<p>Second statement states that the function whose inverse is to be determined should be one-to-one correspondence. </p>
<p>Then what about the inverse in first statement?</p>
<p>Also I have doubt about these notations:</p>
<ul>
<li>A) $(f^{-1})(x)$</li>
<li>B) $(f^{-1})(\{x\})$.</li>
</ul>
<p>Can I say A) is for one one onto function & B) is for any inverse function in general?</p>
|
Jay
| 608,546 |
<p>This is what I have but I also want to add my messy work after this:</p>
<p>Let <span class="math-container">$x\in X$</span> be such that <span class="math-container">$x\in G\subset X, G$</span> open. There is a countable dense subset <span class="math-container">$E$</span> of <span class="math-container">$X$</span>. Let's construct <span class="math-container">$V_{p}$</span>. There is <span class="math-container">$r>0$</span> such that <span class="math-container">$N_{r}(x)\subset G$</span>. There is <span class="math-container">$p\in N_{r}(x)$</span> such that <span class="math-container">$p\neq x$</span>.
. Pick <span class="math-container">$r_{p}\in Q$</span> so that <span class="math-container">$0<d(p,x)<r_{p}<r$</span> and add the condition <span class="math-container">$p=q\Leftrightarrow r_{p}=r_{q}$</span>. Collect all these points and put them in a set <span class="math-container">$E_{0}$</span>. Let <span class="math-container">$Q_{0}=\left\{q\in Q^{+}:q=r_{p},p\in E_{0}\right\}$</span>.
Define <span class="math-container">$f:\left\{V_{p}\right\}_{p\in E_{0}}\longrightarrow E_{0}\times Q_{0}$</span> by <span class="math-container">$f(V_{p})=(p,r_{p})$</span> and let's prove that <span class="math-container">$f$</span> is a bijection.</p>
<p>(1) If <span class="math-container">$f(V_{p})=f(V_{q})\Rightarrow (p,r_{p})=(q,r_{q})\Rightarrow p=q$</span> and <span class="math-container">$r_{p}=r_{q}\Rightarrow V_{p}=V_{q}$</span> (same center and radius).</p>
<p>(2) Let <span class="math-container">$(p,r_{p})\in E_{0}\times Q_{0}$</span>. Then there are <span class="math-container">$x\in X$</span> and <span class="math-container">$r>0$</span> such that <span class="math-container">$x\in G\subset X, G$</span> open, and <span class="math-container">$N_{r}(x)\subset G$</span>. Then <span class="math-container">$0<d(p,x)<r_{p}<r$</span> so that <span class="math-container">$f(N_{r_{p}}(p))=(p,r_{p})$</span>.</p>
<p>Then <span class="math-container">$\left\{V_{p}\right\}_{p\in E_{0}}$</span> is a countable collection of open subsets with the property that if <span class="math-container">$x\in X\cap G$</span> and <span class="math-container">$G$</span> is open, then <span class="math-container">$x\in V_{p}$</span> for some <span class="math-container">$p\in E_{0}$</span>. </p>
<p>Messy work: Since <span class="math-container">$x$</span> is a limit point of <span class="math-container">$E$</span>, it has points <span class="math-container">$p$</span> of <span class="math-container">$E$</span>. But this does
not guarantee that the neighborhood of <span class="math-container">$p$</span> will contain <span class="math-container">$x$</span> or will "overestimate" the radius <span class="math-container">$r$</span> and not be contained in <span class="math-container">$G$</span>. So there's a problem there. This means that
the radius <span class="math-container">$r_{p}$</span> for the neighborhood of <span class="math-container">$p$</span> need to be define before mentioning that <span class="math-container">$p$</span> is in <span class="math-container">$N_{r}(x)$</span>.</p>
<p>The only thing that can work is by constructing neighborhoods with center <span class="math-container">$p\in E$</span> in such a way that its radius includes <span class="math-container">$x$</span> but does not overestimate. To guarantee uniqueness of such neighborhoods, we must impose that <span class="math-container">$r_{p}$</span> is the only radius defining <span class="math-container">$V_{p}$</span>. To guarantee that <span class="math-container">$\left\{V_{p}\right\}$</span> is countable, and to guarantee that the construction makes sense, let the radius be rational. By the Archimedean Principle, such an <span class="math-container">$r_{p}$</span> with the desired property exists. </p>
|
7,247 |
<p>Let $f(x) \in L^p(\mathbb{R})$ and $K \in C^m(\mathbb{R})$. Can I then say that $(f \ast K) (x) = \int_{\mathbb{R}} f(t) K(x-t) dt$ is in $C^m$? </p>
<p>I know that this is true if $K$ has compact support, but I was wondering if it is possible to have a stronger result (perhaps $K$ vanishing at $\infty$?). </p>
|
Willie Wong
| 1,543 |
<p>No. If $K$ does not have compact support, then $f*K$ need not even be bounded. Consider the function $K = 1 / \log( 3 + |x|)$, and $f(x) = (1 + |x|)^{-1/p - \epsilon}$. For $\epsilon > 0$. Then $f\in L^p$, but the convolution integral does not converge anywhere, even when $K$ vanishes at $\infty$. </p>
<p>What you can do is to require $K$ decays sufficiently rapidly. What you need is also <a href="http://en.wikipedia.org/wiki/Young%27s_inequality">Young's inequality for convolutions</a>, which implies that</p>
<p>$$ \sup |f * K| \leq \|f \|_p \|K\|_q $$</p>
<p>if $1/p + 1/q = 1$. So in particular if $K\in C^m(\mathbb{R}) \cap W^{m,q}(\mathbb{R})$, where $W^{m,q}$ is the Sobolev space of $m$-times weakly differentiable, $q$-integrable functions with $q = p / (p-1)$, you can conclude that the convolution is $C^m$. </p>
|
1,364,417 |
<p>Find a real number k such that the limit $$\lim_{n\to\infty}\ \left(\frac{1^4 + 2^4 + 3^4 +....+ n^4}{n^k}\right)$$ has as positive value.
If I am not mistaken every even $k$ can be the answer. But the answer is 5.</p>
|
GohP.iHan
| 151,481 |
<p>Do you know Riemann Sum?</p>
<p>$\begin{eqnarray}
&& \lim_{n\to\infty} \left( \frac{1^4+2^4+3^4+\ldots+n^4 }{n^k } \right) \\
&=& \lim_{n\to\infty}n^{4-k} \left( \left(\frac1n\right)^4 +\left(\frac2n\right)^4 + \left(\frac3n\right)^4 + \ldots + \left(\frac nn\right)^4 \right) \\
\end{eqnarray} $</p>
<p>Should it converge to a finite non-zero value, it should be in the form of $\displaystyle \lim_{n\to\infty} \frac1n \displaystyle\sum_{k=1}^n f\left( \frac kn \right) $</p>
<p>So $4-k = -1 \Rightarrow k = 5 $.</p>
<p>If you want to evaluate the limit: it becomes $ \displaystyle \int_0^1 x^4 \, dx = \frac15 $.</p>
<hr>
<p>Alternatively, you can apply Stolz Cesaro Theorem: $a_n= 1^4 + 2^4 + \ldots+n^4 $, $ b_n = n^k $.</p>
<p>Show that $ \displaystyle \lim_{n\to\infty} = \frac{a_{n+1} - a_n}{b_{n+1} - b_n } = \lim_{n\to\infty} \frac{(n+1)^4}{(k+1)n^{k-1}} $, then we have $k-1 =4 $ again.</p>
|
2,354,383 |
<p>Why doesn't a previous event affect the probability of (say) a coin showing tails?</p>
<p>Let's say I have a <strong>fair</strong> and <strong>unbiased</strong> coin with two sides, <em>heads</em> and <em>tails</em>.</p>
<p>For the first time I toss it up the probabilities of both events are equal to $\frac{1}{2}$.
This much is <em>intuitive and clear</em> to me.</p>
<p>Now suppose that I toss it up $1000000000$ times and the scores are,</p>
<p>$501000000$ Heads</p>
<p>$499000000$ Tails</p>
<p>Now, for the $1001000000^{th}$ toss, shouldn't the probability of a tail coming up be <strong>greater</strong> than that of heads showing up?</p>
<p>I have seen many books which say that even for the $1001000000^{th}$ toss, the probabilities of both events are equal to $\frac{1}{2}$.</p>
<p>This <strong>seems wrong to me</strong> since the same books affirm that <strong>if a coin is tossed a large number of times, the quantity $\frac{heads}{tails}$ will approach $1$.</strong></p>
<p>I know this is very elementary and naive, yet I had only superficially studied probability and I hope you all will bear with me.</p>
<p><strong>My Objections with some of the top-voted answers</strong></p>
<blockquote>
<p>It isn't that future flips compensate for the imbalance, it is that there are so many of them it doesn't matter. </p>
</blockquote>
<p>I don't get this statement. What exactly does the second sentence mean? Moreover, if what you said is true then, the following comment by a user should be wrong,</p>
<blockquote>
<p>Law of large numbers</p>
</blockquote>
<p>So these are contradicting each other I feel. Please bear with my lack of knowledge.</p>
|
paw88789
| 147,810 |
<p>Your question mentions a 'fair and unbiased coin.' That very concept means a coin that on any given flip has a $0.5$ probability of heads and a $0.5$ of tails on any given flip, $\underline{\rm{independently}}$ of what has happened on any other flip.</p>
<p>To understand the statement 'It isn't that future flips compensate for the imbalance, it is that there are so many of them it doesn't matter,' let's think about an example: You want to make a $50-50$ mixture of oil and vinegar. But accidentally you start with $1$ cup of oil and $2$ cups of vinegar. Fortunately you want a large amount of the mixture, so you add $1000$ cups of oil and $1000$ cups of vinegar. You haven't fixed the imbalance of $1$ extra cup of vinegar; but you have added so much of the proper mixture, that you hardly notice the imbalance. (Your mixture went from about $66.67\%$ vinegar to about $50.02\%$ vinegar, even though there is still one full extra cup of vinegar.)</p>
<p>Things aren't quite as clear with the coin-flipping situation because there is no certainty that you will get exactly $1000$ heads and $1000$ tails in the next $2000$ flips; but it is spectacularly likely that the ratio of $2000$ flips will be a lot closer to $50-50$ than the first $3$ flips (that's where the law of large numbers comes in).</p>
|
316,374 |
<p>If <span class="math-container">$a_n$</span> satisfies the linear recurrence relation <span class="math-container">$a_n = \sum_{i=1}^k c_i a_{n-i}$</span> for some constants <span class="math-container">$c_i$</span>, then is there an easy way to find a linear recurrence relation for <span class="math-container">$b_n = a_n^2$</span> ?</p>
<p>For example, if <span class="math-container">$a_n = a_{n-1} + a_{n-3}$</span>, then <span class="math-container">$b_n=a_n^2$</span> seems to satisfy <span class="math-container">$b_n=b_{n-1}+b_{n-2}+3b_{n-3}+b_{n-4}-b_{n-5}-b_{n-6}$</span>.</p>
|
Qiaochu Yuan
| 290 |
<p>Yes. Take the <a href="https://en.wikipedia.org/wiki/Companion_matrix" rel="noreferrer">companion matrix</a> <span class="math-container">$M$</span> of the <a href="https://en.wikipedia.org/wiki/Characteristic_polynomial" rel="noreferrer">characteristic polynomial</a> of your original recurrence. Then the squared recurrence satisfies a recurrence with the characteristic polynomial of the <a href="https://en.wikipedia.org/wiki/Symmetric_algebra" rel="noreferrer">symmetric square</a> <span class="math-container">$S^2(M)$</span> of <span class="math-container">$M$</span>. If the original recurrence has order <span class="math-container">$n$</span> this new recurrence has order <span class="math-container">${n+1 \choose 2}$</span>, and its coefficients are polynomials (depending only on <span class="math-container">$n$</span>) in the coefficients of the old recurrence. </p>
<p>To prove this it suffices to consider the case where the characteristic polynomial has distinct roots, by a density argument. Then <span class="math-container">$a_n$</span> is a linear combination of the sequences <span class="math-container">$\lambda_i^n$</span> where <span class="math-container">$\lambda_i$</span> are the roots of the characteristic polynomial. So <span class="math-container">$a_n^2$</span> is a linear combination of the sequences <span class="math-container">$(\lambda_i \lambda_j)^n$</span> (and we may have <span class="math-container">$i = j$</span>), and <span class="math-container">$S^2(M)$</span> has the <span class="math-container">$\lambda_i \lambda_j$</span> as its eigenvalues. In general these are all also distinct, so the characteristic polynomial of <span class="math-container">$S^2(M)$</span> is minimal with this property and it's not possible to reduce the order further than this in general. </p>
<p>The same argument shows that for <span class="math-container">$k^{th}$</span> powers we can use the symmetric powers <span class="math-container">$S^k(M)$</span>, and that for general products <span class="math-container">$a_n b_n$</span> of sequences satisfying linear recurrences we can use tensor / Kronecker products of the companion matrices of their characteristic polynomials. </p>
|
3,102,218 |
<p>Given a fraction:</p>
<p><span class="math-container">$$\frac{a}{b}$$</span></p>
<p>I now add a number <span class="math-container">$n$</span> to both numerator and denominator in the following fashion:</p>
<p><span class="math-container">$$\frac{a+n}{b+n}$$</span></p>
<p>The basic property is that the second fraction is suppose to closer to <span class="math-container">$1$</span> than the first one. My question is how can we prove that?</p>
<p>What I have tried:</p>
<p>I know <span class="math-container">$\frac{n}{n} = 1$</span> so now adding numbers <span class="math-container">$a$</span> and <span class="math-container">$b$</span> to it would actually "move it away" from <span class="math-container">$1$</span>. But I cannot understand why <span class="math-container">$\frac{a}{b}$</span> is actually farther away from <span class="math-container">$1$</span> than <span class="math-container">$\frac{a+n}{b+n}$</span>.</p>
<p>Why is that? What does it mean to add a number to both the numerator and denominator?</p>
|
fleablood
| 280,126 |
<p>Intuition?</p>
<p>For me the intuition is this: The absolute difference in size becomes less significant when we are comparing big things than when we are comparing small thing. e.g. If one person ways <span class="math-container">$100$</span> lbs more than another that is significant. If one elephant is <span class="math-container">$100$</span> lbs heavy then another that's noticeable if you look really close but not significant. If a building is <span class="math-container">$100$</span> lbs heavier than another it is ludicrous to even attempt to point that out (and darn near impossible to actually measure accurately). If a mastiff is <span class="math-container">$100$</span> lbs heavier than a rabbit... well, that shows they are entirely different things.</p>
<p>Adding a positive <span class="math-container">$n$</span> to both terms of a fraction "pushes" them both to a large frame of reference where the <em>actual</em> difference between them <span class="math-container">$(a-b)$</span> is less significant. <span class="math-container">$(a-b) = 2$</span> is a big part of <span class="math-container">$a = 3$</span> (<span class="math-container">$67\%$</span>) and a big part of <span class="math-container">$b = 5$</span> (<span class="math-container">$40\%$</span>) when it comes to comparing <span class="math-container">$a$</span> to <span class="math-container">$b$</span> the fact that they are not equal but apart by <span class="math-container">$2$</span> is going to make a big difference. But <span class="math-container">$(a-b) =2 $</span> not such a big deal when <span class="math-container">$a = 10$</span> (then <span class="math-container">$2$</span> is only <span class="math-container">$20$</span> percent) and <span class="math-container">$b = 12$</span> (then <span class="math-container">$2$</span> is only <span class="math-container">$17\%$</span>) then the fact that they are not equal isn't that important because there are only <span class="math-container">$2$</span> which is a small proportion of either.</p>
<p>But that's just intuition. A proof needs to be done algebraicly and that's.... straightforward.</p>
<blockquote>
<p>What does it mean to add a number to both the numerator and denominator?</p>
</blockquote>
<p>Well, nothing mysterious. You are comparing the proportion of two numbers and adding <span class="math-container">$n$</span> to both means you are a different pair of numbers-- a pair where each term is <span class="math-container">$n$</span> more.</p>
<p>I guess a proof that is focused on this idea might be: if we assume <span class="math-container">$a - b =m$</span> (<span class="math-container">$m \ne 0$</span> but <span class="math-container">$m < 0$</span> is possible if <span class="math-container">$b < a$</span>) then:</p>
<p><span class="math-container">$\frac ab = \frac {b+m}b = 1 + \frac mb$</span>. Whereas <span class="math-container">$\frac {a+n}{b+n} = \frac {b+m + n}{b+n} = 1 + \frac m{b+n}$</span> </p>
<p>And <span class="math-container">$|\frac m{b+n}| < |\frac m{b+n}|$</span> so <span class="math-container">$1 + \frac m{b+n}$</span> is closer to <span class="math-container">$1$</span> than <span class="math-container">$1 + \frac m{b}$</span> is.</p>
<p>... or in other words...</p>
<p>If we notice that <span class="math-container">$\frac ab = 1 \pm \delta$</span> then <span class="math-container">$\delta = \frac {|numerator - denominator|}{denominator}$</span>, then as the denominator becomes larger but the difference between the numerator and the denominator stay the same, <span class="math-container">$\delta$</span> becomes smaller and less significant.</p>
<p>.... or in my opinion best yet.....</p>
<p>Distance between <span class="math-container">$1$</span> and <span class="math-container">$\frac ab=|1 - \frac ab| = |\frac {b-a}b|$</span>.</p>
<p>Distance between <span class="math-container">$1$</span> and <span class="math-container">$\frac {a+n}{b+n} =|1 - \frac {a+n}{b+n}| = |\frac {(b+n) - (a+n)}{b+n}| = |\frac {b-a}{b+n}|$</span>.</p>
<p>An <span class="math-container">$|\frac {b-a}{b+n}| < |\frac{b-a}{b+n}|$</span>.</p>
|
299,795 |
<p>I seem to have completely lost my bearing with implicit differentiation. Just a quick question:</p>
<p>Given $y = y(x)$ what is $$\frac{d}{dx} (e^x(x^2 + y^2))$$</p>
<p>I think its the $\frac d{dx}$ confusing me, I don't what effect it has compared to $\frac{dy}{dx}$. Any help will be greatly appreciated.</p>
|
amWhy
| 9,003 |
<p>Note:
$$x^9 + x^8 + x^7 = 0 \;\; \iff \;\;x^7(x^2+x+1)=0.$$</p>
<p>Simply solve the equations $x^7=0$ and $x^2+x+1=0.$</p>
<p>No need for WolframAlpha: simply use the quadratic formula to obtain solutions to the second equation.</p>
<p>You'll get one real and two complex (non-real) solutions.</p>
|
1,721,345 |
<p>Found a game related to knight's tour <a href="http://www.emathhelp.net/math-games-and-logic-puzzles/knight-score/" rel="nofollow" title="Knight's Score">Knight's Score</a></p>
<p>The basic idea is to gather as much points as possible, avoiding some cells.</p>
<p>What is the math of this?
Can algorithm be constructed to visit all cells except certain ones (to avoid negative values in the game)?</p>
<p>Is the problem always solvable?</p>
|
Hagen von Eitzen
| 39,174 |
<p>Yes, the game is solvable in the sense that an optimal move sequence can be found. This follows from the fact that there are only finitely many move sequences.</p>
|
897,955 |
<p>For a function $f:X\rightarrow Y$ we have $f(U)\subset V\iff U\subset f^{-1}(V)$ proof</p>
<p>I'm following a book and it just uses this, it doesn't say anything about the function, so I've not assumed it's one-to-one. I've spent to much time on this (as in it should be 4 lines for both ways at most) so I hope you can help me.</p>
<p><strong>What have I tried?</strong></p>
<p>For $\implies$</p>
<p>$x\in U\implies f(x)\in f(U)$ then by hypothesis $f(x)\in V$<br>
Here's where I get stuck, because $f^{-1}(\{f(x)\})\subset f^{-1}(V)$ isn't true.</p>
<p>Wait, It is because $f(x)\in V$ everything that maps to $f(x)$ will be in $f^{-1}(V)$, writing this really does help!</p>
<p>I then say $x\in f^{-1}(\{f(x)\})$ thus $x\in f^{-1}(V)$ - is that right?</p>
<p>For $\impliedby$</p>
<p>$y\in f(U)\implies \exists x\in U:f(x)=y$ thus $x\in f^{-1}(V)$ by hypothesis, so $f(x)\in f(f^{-1}(V))$ which is the same as $y\in V$</p>
<p>I don't like these sort of proves largely because they're glossed over in the first chapter of all books, and the term "function" is used very ambiguously (is it one-to-one implicitly for example?)</p>
<p>Are these proofs okay?</p>
|
drhab
| 75,923 |
<p>First of all note that by definition of the preimage of $V\subset Y$ under $f:X\rightarrow Y$ we have: $$x\in f^{-1}\left(V\right)\iff f\left(x\right)\in V$$</p>
<hr>
<p>$\Rightarrow$</p>
<p>$x\in U\Rightarrow f\left(x\right)\in f\left(U\right)\subset V\Rightarrow x\in f^{-1}\left(V\right)$. Application of definition</p>
<p>$\Leftarrow$</p>
<p>$y\in f\left(U\right)\iff\exists x\in U\; f\left(x\right)=y$.
Here $x\in U\subset f^{-1}\left(V\right)$ so that $y=f\left(x\right)\in V$ again applying definition.</p>
|
1,109,706 |
<p>I have just started going over abstract algebra.</p>
<p>One of the question is </p>
<p>$*$ is defined on $\mathbb C$ such that $a*b=|ab|$</p>
<p>I tried to check three axioms : 1) Associativity 2) identity 3) inverse</p>
<p>I found out with long computation that Associativity works.</p>
<p>For identity, $a*e=e*a=a$ , where $a*e=|ae|=a$</p>
<p>I am quite confused how to find out the value of e, since this is a complex number.</p>
<p>Can anyone show me why this is a group or not a group?</p>
|
k.stm
| 42,242 |
<p>It’s <em>not</em> a group as it fails to have an identity.</p>
<p><em>Hint</em>. Look at $a = -1$ and the range of the defined operation “$*$”.</p>
|
1,109,706 |
<p>I have just started going over abstract algebra.</p>
<p>One of the question is </p>
<p>$*$ is defined on $\mathbb C$ such that $a*b=|ab|$</p>
<p>I tried to check three axioms : 1) Associativity 2) identity 3) inverse</p>
<p>I found out with long computation that Associativity works.</p>
<p>For identity, $a*e=e*a=a$ , where $a*e=|ae|=a$</p>
<p>I am quite confused how to find out the value of e, since this is a complex number.</p>
<p>Can anyone show me why this is a group or not a group?</p>
|
MJD
| 25,554 |
<p>There is a theorem that if $\langle G, \ast\rangle$ is a group, then each function $f_a$ defined by $$f_a(x) = a\ast x$$ must be a bijection. (That is, the left coset $aG$ must be equal to $G$.) For example, the function $f_3(x) = 3 + x$ is a bijection of the real numbers.</p>
<p>This is not hard to show. To see that $f_a$ is onto $G$, just note that each $y\in G$ is in the range of $f$ because $f_a(a^{-1}*y) = y$. To see that $f_a$ is one-to-one, suppose we have $f_a(x) = f_a(x')$. Then $a*x=a*x'$, and multiplying by $a^{-1}$ on the left we obtain $x=x'$.</p>
<p>Here your function $f_a(x) = |ax|$ is evidently not a bijection of the complex numbers because its value is always a positive real number. So the operation is not a group operation.</p>
|
489,528 |
<p>I need to check for convergence here. Why is every example completely different and nothing which I've learned from the examples before helps me in the next? Hate it!</p>
<p>$$ \sum_{n\geq1} \frac{n^{n-1}}{n!} $$</p>
<p>so I tried the ratio test and I got</p>
<p>$$ \frac{1}{1+\frac{1}{n}} $$
which converges to $1$ which isn't $>$ or $<$ then $1,$ so that doesn't work.</p>
<p>which test will work and how can I transform the series?</p>
|
marty cohen
| 13,079 |
<p>Let's look at the more general function
$\sum_{n\geq1} x^n\frac{n^{n-1}}{n!}$</p>
<p>The ratio of consecutive terms is</p>
<p>$\begin{align}
x\frac{\frac{(n+1)^n}{(n+1)!}}{\frac{n^{n-1}}{n!}}
&=x\frac{(n+1)^n}{(n+1)!}\frac{n!}{n^{n-1}}\\
&=x\frac{(n+1)^n}{(n+1)n^{n-1}}\\
&=x\frac{(n+1)^{n-1}}{n^{n-1}}\\
&= x(1+\frac1{n})^{n-1}\\
&\to xe\\
\end{align}
$</p>
<p>For this to converge,
it is necessary that
$|xe| \le 1$,
or
$|x| \le \frac1{e}$.</p>
<p>Since $x = 1$,
the sum diverges.</p>
|
2,260,975 |
<p>Problem: Let $\mu : \mathcal{B}(\mathbb{R}^k) \rightarrow \mathbb{R}$ be a nonnegative and finitely additive set function with $\mu(\mathbb{R}^k) < \infty $.
Suppose that
\begin{equation}
\mu(A) = \sup \{ \mu(K) : K \subset A, K \text{ compact}\}
\end{equation}
for each $A \in \mathcal{B}(\mathbb{R}^k)$.
Then $\mu$ is a finite measure.</p>
<p>I'm stuck showing that $\mu$ is countably additive.
First, let $A_{i}$ disjoint, and $K_i \subset A_i$ compact such that $\mu(A_i) \leq \mu(K_i)+\epsilon/2^i$ for $\epsilon > 0$.
Then
\begin{equation}
\mu\left(\bigcup_{i=1}^{\infty} A_i \right)
\geq \mu\left(\bigcup_{i=1}^{n} A_i \right)
\geq \mu\left(\bigcup_{i=1}^{n} K_i \right)
= \sum_{i=1}^n \mu(K_i)
\geq \sum_{i=1}^n \mu(A_i) + \frac{\epsilon}{2^i} ,
\end{equation}
which follows from monotonicity (obvious) and finite additivity (on the class of compact sets).
Hence, letting $n \rightarrow \infty$, and since $\epsilon$ was arbitrary,
\begin{equation}
\mu\left(\bigcup_{i=1}^{\infty} A_i \right)
\geq \sum_{i=1}^\infty \mu(A_i) .
\end{equation}
Next, I want to establish the reverse inequality.
This is where I'm stuck. Any ideas?</p>
<p>Thanks in advance!</p>
<p>Chris</p>
|
Community
| -1 |
<p>First, $\mu$ is $\sigma$-additive iff it is continuous at $\emptyset$. Suppose that $\mu$ is not $\sigma$-additive, i.e. there exists a decreasing sequence of sets $\{A_n\}$ s.t. $\cap_{n\ge 1}A_n=\emptyset$ but $\inf_n\mu(A_n)=\epsilon>0$. Let $K_n\subset A_n$ compact s.t. $\mu(K_n)\le \mu(A_n)+\epsilon2^{-(n+1)}$. Then, letting $B_n=\cap_{k=1}^nK_n$, we get
$$
\mu(A_n\setminus B_n)\le \sum_{k=1}^n\mu(A_k\setminus K_k)<\frac{\epsilon}{2}.
$$
Thus, $\mu(B_n)>0$ so that $\{B_n\}$ is a decreasing sequence of nonempty compact sets. Hence, $\cap_{n\ge 1}K_n\ne \emptyset$, a contradiction.</p>
|
2,900,014 |
<p>How would solve for $a$ in this equation without using an approximation ?
is it possible?</p>
<p>where $x>0$ and $0<a<\infty$</p>
<p>$x=\Sigma _{i=1}^{n} i^a$</p>
<p>for example $120=\Sigma _{i=1}^{6} i^a$ what is $a$ in this equation?</p>
|
user
| 505,767 |
<p>There is not a general closed form but we can use integral estimation</p>
<p>$$\sum_{i=1}^{6} i^a\approx \int_{0.5}^{6.5}x^adx=\left[\frac{x^{a+1}}{a+1}\right]_{0.5}^{6.5}=\frac{6.5^{a+1}}{a+1}-\frac{0.5^{a+1}}{a+1}=120 \implies a\approx 2.175$$</p>
<p>where the value for $a$ needs to be evaluated by numerical methods and by a direct check we obtain</p>
<p>$$\sum_{i=1}^{6} i^{2.175}\approx 119.21$$</p>
|
1,400,876 |
<p>Given a measurable space $(\Omega, \cal{F})$, $f:(\Omega, \cal F) \to (\Bbb{B},\cal B)$, where $\cal B$ is the Borel $\sigma$-algebra of $\Bbb R$, is said to be $\cal {F}$-measurable if $f^{-1}(B)\in \cal F $ for any $B\in \cal B$.</p>
<p>Any function $f:\Omega\to\Bbb R$ is trivially $2^\Omega$ measurable. A constant function on $\Omega$ is $\{\emptyset, \Omega\}$ measurable. If $\Omega$ is Lebesgue measurable and $\cal F$ is the set of all Lebesgue measurable subsets of $\Omega$, then all Lebesgue measurable functions are $\cal F$-measurable.</p>
<p><strong>My question is</strong> can anyone help provide an example of $\cal F$-measurable function besides the above three examples? Thank you!</p>
|
layman
| 131,740 |
<ol>
<li><p>For any $\sigma$-algebra $\mathcal{F}$ on $\Omega$, and any $A \in \mathcal{F}$, the indicator function $\Bbb 1_{A}(x) : = \begin{cases} 1 & x \in A \\ 0 & x \not \in A \end{cases}$ is $\mathcal{F}$-measurable. It's a good (and easy) exercise to prove this.</p></li>
<li><p>Also, if $\Omega$ is a topological space, and $\mathcal{F}$ contains $\mathcal{B}(\Omega)$, the Borel $\sigma$-algebra on $\Omega$, then every continuous function (defined to be a function where the pre-image of each open set is open) is $\mathcal{F}$-measurable. This requires a tiny bit of work to prove, but it's not hard. You will have to use the fact that $\{ A \subseteq \Bbb R \mid f^{-1}(A) \in \mathcal{B}(\Omega) \}$ is a $\sigma$-algebra (which contains the open sets if $f$ is continuous...thus it contains $\mathcal{B}(\Bbb R)$...which is good because?).</p></li>
</ol>
|
620,370 |
<p>I am tackling a problem which asks:</p>
<p>Find the sum of all the multiples of 3 or 5 below 1000.</p>
<p>My reasoning is that since
Since $\left\lfloor\frac{1000}{3}\right\rfloor = 333$ and $\left\lfloor\frac{1000}{5}\right\rfloor = 200$</p>
<p>This sum can be denoted as:
\begin{equation}
\sum\limits_{n=1}^{333}3n + \sum\limits_{m=1}^{200} 5m = 3\sum\limits_{n=1}^{333}n + 5\sum\limits_{m=1}^{200} m
\end{equation}
Using the identity
$$
\sum\limits_{i=1}^{n} i = \frac{n(n+1)}{2}
$$
It can be solved analytically as
$$
3\sum\limits_{n=1}^{333}n + 5\sum\limits_{m=1}^{200} m = \frac{3}{2}(333(333+1)) + \frac{5}{2}(200(200+1)) = 267333
$$</p>
<p>But apparently this is incorrect? Where did I go wrong in my reasoning?</p>
<h2>Edit</h2>
<p>I tried to correct the double multiples with:
$$3\sum\limits_{i=1}^{333}i + 5\sum\limits_{j=1}^{200}j - 15\sum\limits_{k=1}^{66} k = 234168$$</p>
<p>But this seems to be wrong as well...</p>
<h2>Edit 2:</h2>
<p>GOT IT!</p>
<p>Up to and <strong>not</strong> including.</p>
<p>$$3\sum\limits_{i=1}^{333}i + 5\sum\limits_{j=1}^{199}j - 15\sum\limits_{k=1}^{66} k = 233168$$</p>
<p>Or, more ugly:
$$3\sum\limits_{i=1}^{\left\lfloor\frac{999}{3}\right\rfloor}i + 5\sum\limits_{j=1}^{\left\lfloor\frac{999}{5}\right\rfloor}j - 15\sum\limits_{k=1}^{\left\lfloor\frac{999}{3\cdot 5}\right\rfloor} k = 233168$$</p>
|
Jeppe Stig Nielsen
| 70,134 |
<p>Some numbers, such as $15$ and $30$, are multiples of <em>both</em> $3$ and $5$. These should be counted only once.</p>
<p><strong>Edit:</strong> After comments to the question (not to this answer), I see that another issue is that you include $1000$ as the last term, but $1000$ is not below $1000$.</p>
|
4,295,212 |
<p>I have a statics problem in which my approach differs from the suggested solution, so I was wondering if anyone can point out my flawed logic. Below is the question.</p>
<p>Two identical uniform rods <span class="math-container">$ab$</span> and <span class="math-container">$bc$</span>, each of length <span class="math-container">$2l$</span> and weight <span class="math-container">$W$</span>, are smoothly hinged at <span class="math-container">$b$</span>. Ends <span class="math-container">$a$</span> and <span class="math-container">$c$</span> rest on a smooth horizontal floor. The system is kept in
equilibrium in a vertical plane by a light inextensible string joining <span class="math-container">$a$</span> to the midpoint of <span class="math-container">$bc$</span>. the and between <span class="math-container">$ab$</span> and <span class="math-container">$bc$</span> is <span class="math-container">$2\alpha$</span>. Show the tension of the string is <span class="math-container">$T = \frac{W}{4} \sqrt{9\tan^2\alpha + 1}$</span>.</p>
<p>So below is the force diagram I have created.</p>
<p><a href="https://i.stack.imgur.com/rzxc9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rzxc9.jpg" alt="enter image description here" /></a></p>
<p>The suggested solution for this problem resolves the <span class="math-container">$\vec{j}$</span> components of the system as:</p>
<p><span class="math-container">$$ R_a + R_c = 2W $$</span></p>
<p>without taking into consideration the resolved tension force, <span class="math-container">$T$</span>, of the light inextensible string.</p>
<p>Also taking moments about the point <span class="math-container">$a$</span> of the system abc the solution is:</p>
<p>$R_c \cdot 4l \sin \alpha = l \sin\alpha \cdot W + 3l\sin \alpha \cdot W $$</p>
<p>Once again no mention of the tension.</p>
<p>So I am I wrong to include the tension force for the <span class="math-container">$\vec{j}$</span> components of the system and the moments about <span class="math-container">$a$</span>.</p>
<p>Any help appreciated.</p>
|
Cesareo
| 397,348 |
<p>Calling the forces indexed by point of application</p>
<p><span class="math-container">$$
\cases{
T_a = T(\cos\beta,\sin\beta)\\
F_a = (0,R_a)\\
F_b = (0,R_b)\\
F_c = (H_c,V_c)\\
F_{g_1} = (0,-W)\\
F_{g_2} = (0,-W)\\
a = (0,0)\\
b = 2l(\sin\alpha,0)\\
c = 2l(\sin\alpha,\cos\alpha)\\
g_1 = \frac 12 (a+c)\\
g_2 = \frac 12(b+c)\\
}
$$</span>
with <span class="math-container">$\tan\beta = \frac 13\cot\alpha$</span> we have the equilibrium equations</p>
<p><span class="math-container">$$
\cases{
F_a+T_a+F_{g_1}+F_b = 0\\
F_{g_1}\times (g_1-a)+F_c\times(c-a) = 0\\
}
$$</span></p>
<p>and</p>
<p><span class="math-container">$$
\cases{
-F_b+F_{g_2}-T_a + F_c = 0\\
(-F_c)\times(b-c)+(F_{g_2}-T_a)\times(g_2-c) = 0
}
$$</span></p>
<p>Solving the corresponding linear system we got</p>
<p><span class="math-container">$$
\cases{
R_a = \frac{W (\cos (\alpha -\beta )+6 \cos (\alpha +\beta ))}{6 \cos (\alpha ) \cos (\beta )-2 \sin (\alpha ) \sin (\beta )} \\
R_b = \frac{W (\sin (\alpha ) \sin (\beta )+5 \cos (\alpha ) \cos (\beta ))}{6 \cos (\alpha ) \cos (\beta )-2 \sin (\alpha ) \sin (\beta )} \\
H_c = \frac{2 W \sin (\alpha ) \cos (\beta )}{\sin (\alpha ) \sin (\beta )-3 \cos (\alpha ) \cos (\beta )} \\
V_c = -\frac{W \cos (\alpha -\beta )}{6 \cos (\alpha ) \cos (\beta )-2 \sin (\alpha ) \sin (\beta )} \\
T = \frac{2 W \sin (\alpha )}{3 \cos (\alpha ) \cos (\beta )-\sin (\alpha ) \sin (\beta )} \\
}
$$</span></p>
<p>and using the relationship <span class="math-container">$\tan\beta = \frac 13\cot\alpha$</span> we have finally</p>
<p><span class="math-container">$$
T = \frac{3}{4} W \tan \alpha \sqrt{\frac{\cot^2\alpha}{9}+1} = \frac W4\sqrt{9\tan^2\alpha+1}
$$</span></p>
|
1,990,105 |
<p>Presume that $(x_n)$ is a sequence s. t.</p>
<p>$|x_n-x_{n+1}| \le 2^{-n}$ for all $n \in \mathbb N$</p>
<p>Prove that $x_n$ converges.</p>
<p>What I've tried to think: since $2^{-n}$ converges to 0, and the difference between the terms $x_n$ and $x_{n+1}$ is smaller or equal to it, then $x_n$ must be a cauchy sequence and converge. How do I make this an actual proof?</p>
|
ILoveMath
| 42,344 |
<h3>Hint:</h3>
<p>For <span class="math-container">$m>n$</span>, write</p>
<p><span class="math-container">$$|x_n - x_m| = |x_n + x_{n+1} - x_{n+1} + x_{n+2} - x_{n+2} + ... + -x_{m-n-1} + x_{m-n-1} - x_m|$$</span></p>
<p>now use triangle inequality</p>
|
310,513 |
<p>I haven't found this yet and I'm somehow not sure if my idea is correct.</p>
<p>The Problem: Let $X$ be a separable Banach-Space, let $x_k\to x$ weakly and such that for every $\lambda_k \to \lambda$ weakly-* there holds $\lambda_k(x_k)\to \lambda(x)$. Then $x_k\to x$ strongly (in the norm of $X$).</p>
<p>My idea was to give a proof with contradiction. Hence assume there holds for some $\epsilon >0$ and a subsequence of $x_k$ denoted again by $k$:
$$ \epsilon <||x_k-x||=|\lambda^*_k(x_k-x)| $$
for a functional provided by Hahn-Banach theorem with norm 1. From that and the separabilty we conclude that there is a further subsequence such that $\lambda_{k_l}^*\to \lambda^*$ weakly-* . Since each subsequence of $x_k$ also converges weakly to $x$ we use the "weakly-*" assumption to receive a contradiction since for the previous subsequence
$$|\lambda^*_{k_l}(x_{k_l}-x)|=|\lambda^*_{k_l}(x_{k_l})-\lambda^*_{k_l}(x)|\to 0$$</p>
<p>Somehow this seems to easy and I feel like I'm not using especially the weak convergence in the right way. </p>
|
A Blumenthal
| 54,337 |
<p>I think the reason for your confusion is the condition made on the convergence $x_k \rightarrow x$. The condition that for any $\lambda_k \rightarrow \lambda$ weak* we have $|\lambda_k(x_k) - \lambda(x)| \rightarrow 0$ implies $x_k \rightarrow x$ weakly, by taking $\lambda_k$ to be the constant sequence. You certainly use this condition, so there's nothing to worry about.</p>
|
2,672,497 |
<p>$$\lim _{n\to \infty }\sum _{k=1}^n\frac{1}{n+k+\frac{k}{n^2}}$$ I unsuccessfully tried to find two different Riemann Sums converging to the same value close to the given sum so I could use the Squeeze Theorem. Is there any other way to solve this?</p>
|
RRL
| 148,510 |
<p>Writing this as</p>
<p>$$\lim_{n \to \infty}S_{nn} = \lim_{n \to \infty}\frac{1}{n}\sum _{k=1}^n\frac{1}{1+\frac{k}{n}+\frac{k}{n}\frac{1}{n^2}}
$$</p>
<p>a technique that often works is to evaluate the double limit</p>
<p>$$\lim_{m \to \infty} \lim_{n \to \infty} S_{mn} = \lim_{m \to \infty} \lim_{n \to \infty} \frac{1}{n}\sum _{k=1}^n\frac{1}{1+\frac{k}{n}+\frac{k}{n}\frac{1}{m^2}} = \lim_{m \to \infty}\int_0^1 \frac{dx}{1 +x + x/m^2} = \int_0^1 \frac{dx}{1 +x}
$$</p>
<p>where the last step is justified by DCT.</p>
<p>We can justify $ \lim_{n \to \infty} S_{nn} = \lim_{m \to \infty} \lim_{n \to \infty} S_{mn} = \log 2$ by showing one of the iterated limits exhibits uniform convergence.</p>
<p>An example with more details is given <a href="https://math.stackexchange.com/a/2259535/148510">here</a>. The argument for justifying the use of the double limit in this case will be similar.</p>
|
1,018,672 |
<blockquote>
<p><span class="math-container">$$\int_0^{\infty} \frac{1}{x^3-1}dx$$</span></p>
</blockquote>
<p>What I did:</p>
<p><span class="math-container">$$\lim_{\epsilon\to0}\int_0^{1-\epsilon} \frac{1}{x^3-1}dx+\lim_{\epsilon\to0}\int_{1+\epsilon}^{\infty} \frac{1}{x^3-1}dx$$</span></p>
<hr />
<p><span class="math-container">$$\lim_{\epsilon\to0}\int_0^{1-\epsilon}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2(x^2+x+1)}dx+\lim_{\epsilon\to0}\int_{1+\epsilon}^{\infty}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2(x^2+x+1)}dx$$</span></p>
<hr />
<p><span class="math-container">$$\lim_{\epsilon\to0}\int_0^{1-\epsilon}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2[(x+\frac{1}{2})^2+\frac{3}{4}]}dx+\lim_{\epsilon\to0}\int_{1+\epsilon}^{\infty}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2[(x+\frac{1}{2})^2+\frac{3}{4}]}dx$$</span></p>
<hr />
<p><span class="math-container">$$[\frac{1}{3}ln(x-1)-\frac{1}{6}ln(x^2+x+1)-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{3}ln(x-1)-\frac{1}{6}ln(x^2+x+1)-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$</span></p>
<hr />
<p><span class="math-container">$$[\frac{1}{6}(2ln(x-1)-ln(x^2+x+1))-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{6}ln{2(x-1})-\frac{1}{6}ln(x^2+x+1)-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$</span></p>
<hr />
<p><span class="math-container">$$[\frac{1}{6}ln(\frac{(x-1)^2}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{6}ln(\frac{(x-1)^2}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$</span></p>
<hr />
<p><span class="math-container">$$[\frac{1}{6}ln(\frac{x^2-2x+1}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{6}ln(\frac{x^2-2x+1}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$</span></p>
<hr />
<p><span class="math-container">$$\lim_{\epsilon\to0}[\frac{1}{6}ln(\frac{(1-\epsilon)^2-2(1-\epsilon)+1}{(1-\epsilon)^2+1-\epsilon+1})-\frac{1}{\sqrt3}\arctan(\frac{2(1-\epsilon)+1}{\sqrt3})+\frac{1}{\sqrt3}\arctan(\frac{1}{\sqrt3})]+\lim_{\epsilon\to 0} [ \frac{1}{6}ln(\frac{(\infty)^2-2(\infty)+1}{(\infty)^2+(\infty)+1})+\cdots]$$</span></p>
<hr />
<blockquote>
<p>This is where my problem is, what is :</p>
<p><span class="math-container">$$ \frac{1}{6}ln(\frac{(\infty)^2-2(\infty)+1}{(\infty)^2+(\infty)+1})$$</span></p>
</blockquote>
<p>^^^ If I know past this, I know how to proceed. The only thing stopping me is this ^^^. Please help.</p>
|
Lucian
| 93,448 |
<ul>
<li>Subdivide $(0,\infty)$ into $(0,1)$ and $(1,\infty)$.</li>
<li>On the latter, let $t=\dfrac1x$.</li>
<li>$a^3-b^3=\big(a-b\big)\Big(a^2+ab+b^2\Big)$.</li>
<li>Complete the square in the denominator.</li>
</ul>
|
1,018,672 |
<blockquote>
<p><span class="math-container">$$\int_0^{\infty} \frac{1}{x^3-1}dx$$</span></p>
</blockquote>
<p>What I did:</p>
<p><span class="math-container">$$\lim_{\epsilon\to0}\int_0^{1-\epsilon} \frac{1}{x^3-1}dx+\lim_{\epsilon\to0}\int_{1+\epsilon}^{\infty} \frac{1}{x^3-1}dx$$</span></p>
<hr />
<p><span class="math-container">$$\lim_{\epsilon\to0}\int_0^{1-\epsilon}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2(x^2+x+1)}dx+\lim_{\epsilon\to0}\int_{1+\epsilon}^{\infty}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2(x^2+x+1)}dx$$</span></p>
<hr />
<p><span class="math-container">$$\lim_{\epsilon\to0}\int_0^{1-\epsilon}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2[(x+\frac{1}{2})^2+\frac{3}{4}]}dx+\lim_{\epsilon\to0}\int_{1+\epsilon}^{\infty}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2[(x+\frac{1}{2})^2+\frac{3}{4}]}dx$$</span></p>
<hr />
<p><span class="math-container">$$[\frac{1}{3}ln(x-1)-\frac{1}{6}ln(x^2+x+1)-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{3}ln(x-1)-\frac{1}{6}ln(x^2+x+1)-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$</span></p>
<hr />
<p><span class="math-container">$$[\frac{1}{6}(2ln(x-1)-ln(x^2+x+1))-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{6}ln{2(x-1})-\frac{1}{6}ln(x^2+x+1)-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$</span></p>
<hr />
<p><span class="math-container">$$[\frac{1}{6}ln(\frac{(x-1)^2}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{6}ln(\frac{(x-1)^2}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$</span></p>
<hr />
<p><span class="math-container">$$[\frac{1}{6}ln(\frac{x^2-2x+1}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{6}ln(\frac{x^2-2x+1}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$</span></p>
<hr />
<p><span class="math-container">$$\lim_{\epsilon\to0}[\frac{1}{6}ln(\frac{(1-\epsilon)^2-2(1-\epsilon)+1}{(1-\epsilon)^2+1-\epsilon+1})-\frac{1}{\sqrt3}\arctan(\frac{2(1-\epsilon)+1}{\sqrt3})+\frac{1}{\sqrt3}\arctan(\frac{1}{\sqrt3})]+\lim_{\epsilon\to 0} [ \frac{1}{6}ln(\frac{(\infty)^2-2(\infty)+1}{(\infty)^2+(\infty)+1})+\cdots]$$</span></p>
<hr />
<blockquote>
<p>This is where my problem is, what is :</p>
<p><span class="math-container">$$ \frac{1}{6}ln(\frac{(\infty)^2-2(\infty)+1}{(\infty)^2+(\infty)+1})$$</span></p>
</blockquote>
<p>^^^ If I know past this, I know how to proceed. The only thing stopping me is this ^^^. Please help.</p>
|
robjohn
| 13,854 |
<p><strong>Cauchy Principal Value</strong></p>
<p>The integral, as written, diverges. In the case of an improper integral such as this,
$$
\int_0^\infty\frac1{x^3-1}\mathrm{d}x
=\int_0^1\frac1{x^3-1}\mathrm{d}x+\int_1^\infty\frac1{x^3-1}\mathrm{d}x\tag{1}
$$
however, neither of the integrals on the right converge.</p>
<p>On the other hand, if what we want is the <a href="http://en.wikipedia.org/wiki/Cauchy_principal_value" rel="nofollow">Cauchy Principal Value</a>, then we are asking for
$$
\lim_{\epsilon\to0^+}\left(\int_0^{1-\epsilon}\frac1{x^3-1}\mathrm{d}x+\int_{1+\epsilon}^\infty\frac1{x^3-1}\mathrm{d}x\right)\tag{2}
$$
In many cases, this will exist, even when the actual integral fails to converge.</p>
<hr>
<p><strong>A Real Approach to Computing the Cauchy Principal Value</strong></p>
<p>Substituting $x\mapsto\frac1x$, we get
$$
\begin{align}
\int_{1+\epsilon}^\infty\frac1{x^3-1}\mathrm{d}x
&=\int_0^{\frac1{1+\epsilon}}\frac{x}{1-x^3}\mathrm{d}x\\
&=\int_0^{1-\epsilon}\frac{x}{1-x^3}\mathrm{d}x+\color{#C00000}{\int_{1-\epsilon}^{\frac1{1+\epsilon}}\frac{x}{1-x^3}\mathrm{d}x}\tag{3}
\end{align}
$$
Since $\frac1{1+\epsilon}-(1-\epsilon)=\frac{\epsilon^2}{1+\epsilon}$, we have that
$$
\begin{align}
\color{#C00000}{\int_{1-\epsilon}^{\frac1{1+\epsilon}}\frac{x}{1-x^3}\mathrm{d}x}
&\le\frac{\epsilon^2}{1+\epsilon}\frac{\frac1{1+\epsilon}}{1-\left(\frac1{1+\epsilon}\right)^3}\\
&=\frac{\epsilon(1+\epsilon)}{3+3\epsilon+\epsilon^2}\\[9pt]
&\stackrel{\epsilon\to0^+}{\to}0\tag{4}
\end{align}
$$
Therefore, using $(3)$, $(4)$, and $x+\frac12=\frac{\sqrt3}2\tan(\theta)$, we can conclude
$$
\begin{align}
&\mathrm{PV}\int_0^\infty\frac1{x^3-1}\mathrm{d}x\\
&=\lim_{\epsilon\to0^+}\left(\int_0^{1-\epsilon}\frac1{x^3-1}\mathrm{d}x
+\int_{1+\epsilon}^\infty\frac1{x^3-1}\mathrm{d}x\right)\\
&=\lim_{\epsilon\to0^+}\left(\int_0^{1-\epsilon}\frac1{x^3-1}\mathrm{d}x
-\int_0^{1-\epsilon}\frac{x}{x^3-1}\mathrm{d}x
+\color{#C00000}{\int_{1-\epsilon}^{\frac1{1+\epsilon}}\frac{x}{1-x^3}\mathrm{d}x}\right)\\
&=\lim_{\epsilon\to0^+}\left(-\int_0^{1-\epsilon}\frac{x-1}{x^3-1}\mathrm{d}x\right)
+\color{#C00000}{0}\\
&=-\int_0^1\frac1{x^2+x+1}\mathrm{d}x\\
&=-\int_0^1\frac1{(x+\frac12)^2+\frac34}\mathrm{d}x\\
&=-\frac2{\sqrt3}\int_{\pi/6}^{\pi/3}1\mathrm{d}\theta\\[9pt]
&=-\frac\pi{3\sqrt3}\tag{5}
\end{align}
$$</p>
|
1,018,672 |
<blockquote>
<p><span class="math-container">$$\int_0^{\infty} \frac{1}{x^3-1}dx$$</span></p>
</blockquote>
<p>What I did:</p>
<p><span class="math-container">$$\lim_{\epsilon\to0}\int_0^{1-\epsilon} \frac{1}{x^3-1}dx+\lim_{\epsilon\to0}\int_{1+\epsilon}^{\infty} \frac{1}{x^3-1}dx$$</span></p>
<hr />
<p><span class="math-container">$$\lim_{\epsilon\to0}\int_0^{1-\epsilon}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2(x^2+x+1)}dx+\lim_{\epsilon\to0}\int_{1+\epsilon}^{\infty}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2(x^2+x+1)}dx$$</span></p>
<hr />
<p><span class="math-container">$$\lim_{\epsilon\to0}\int_0^{1-\epsilon}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2[(x+\frac{1}{2})^2+\frac{3}{4}]}dx+\lim_{\epsilon\to0}\int_{1+\epsilon}^{\infty}\frac{1}{3(x-1)}-\frac{2x+1}{6(x^2+x+1)}-\frac{1}{2[(x+\frac{1}{2})^2+\frac{3}{4}]}dx$$</span></p>
<hr />
<p><span class="math-container">$$[\frac{1}{3}ln(x-1)-\frac{1}{6}ln(x^2+x+1)-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{3}ln(x-1)-\frac{1}{6}ln(x^2+x+1)-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$</span></p>
<hr />
<p><span class="math-container">$$[\frac{1}{6}(2ln(x-1)-ln(x^2+x+1))-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{6}ln{2(x-1})-\frac{1}{6}ln(x^2+x+1)-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$</span></p>
<hr />
<p><span class="math-container">$$[\frac{1}{6}ln(\frac{(x-1)^2}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{6}ln(\frac{(x-1)^2}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$</span></p>
<hr />
<p><span class="math-container">$$[\frac{1}{6}ln(\frac{x^2-2x+1}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{0}^{1-\epsilon}+[\frac{1}{6}ln(\frac{x^2-2x+1}{x^2+x+1})-\frac{1}{\sqrt3}\arctan(\frac{2x+1}{\sqrt3})]_{1+\epsilon}^{\infty}$$</span></p>
<hr />
<p><span class="math-container">$$\lim_{\epsilon\to0}[\frac{1}{6}ln(\frac{(1-\epsilon)^2-2(1-\epsilon)+1}{(1-\epsilon)^2+1-\epsilon+1})-\frac{1}{\sqrt3}\arctan(\frac{2(1-\epsilon)+1}{\sqrt3})+\frac{1}{\sqrt3}\arctan(\frac{1}{\sqrt3})]+\lim_{\epsilon\to 0} [ \frac{1}{6}ln(\frac{(\infty)^2-2(\infty)+1}{(\infty)^2+(\infty)+1})+\cdots]$$</span></p>
<hr />
<blockquote>
<p>This is where my problem is, what is :</p>
<p><span class="math-container">$$ \frac{1}{6}ln(\frac{(\infty)^2-2(\infty)+1}{(\infty)^2+(\infty)+1})$$</span></p>
</blockquote>
<p>^^^ If I know past this, I know how to proceed. The only thing stopping me is this ^^^. Please help.</p>
|
Felix Marin
| 85,343 |
<p><span class="math-container">$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\on}[1]{\operatorname{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$</span>
<span class="math-container">\begin{align}
&\bbox[5px,#ffd]{\mrm{P.V.}\int_{0}^{\infty}
{\dd x \over x^{3} - 1}} =
\Re\int_{0}^{\infty}{\dd x \over
\pars{x - 1 + \ic 0^{+}}\pars{x^{2} + x + 1}}
\\[5mm] = &\
-\Re\int_{\infty}^{0}{\ic\,\dd y \over
\pars{\ic y - 1 + \ic 0^{+}}\pars{-y^{2} + \ic y + 1}} =
\Im\int_{0}^{\infty}{\dd y \over 1 + \ic y^{3}}
\\[5mm] = &\
{1 \over 3}\,\Im\int_{0}^{\infty}{y^{\color{red}{1/3} - 1} \over 1 + \ic y}\,\dd y
\end{align}</span>
Note that
<span class="math-container">$\ds{{1 \over 1 + \ic y} = \sum_{k = 0}^{\infty}
\color{red}{\Gamma\pars{1 + k}\expo{\ic\pi k/2}}
\,\,{\pars{-y}^{k} \over k!}}$</span>
<span class="math-container">\begin{align}
&\bbox[5px,#ffd]{\mrm{P.V.}\int_{0}^{\infty}
{\dd x \over x^{3} - 1}} =
{1 \over 3}\,\Im\bracks{%
\Gamma\pars{\color{red}{1 \over 3}}
\Gamma\pars{1 - \color{red}{1 \over 3}}
\expo{\ic\pi\pars{\color{red}{-1/3}}/2}\,}
\\[5mm] = &\
{1 \over 3}\bracks{-\sin\pars{\pi \over 6}}
{\pi \over \sin\pars{\pi/3}} =
{1 \over 3}\pars{-\,{1 \over 2}}\,{\pi \over \root{3}/2}
\\[5mm] = &\
\bbx{-\,{\root{3} \over 9}\,\pi}
\approx -0.6046 \\ &
\end{align}</span></p>
|
1,056,045 |
<blockquote>
<p>Show that </p>
<p>$2^{3n}-1$ is divisble by $7$ for all $n$ $\in \mathbb N$</p>
</blockquote>
<p>I'm not really sure how to get started on this problem, but here is what I have done so far:</p>
<p>Base case $n(1)$:</p>
<p>$\frac{2^{3(1)}-1}{7} = \frac{8-1}{7} = \frac{7}{7}$</p>
<p>But not sure where to go from here. Tips?</p>
|
Crostul
| 160,300 |
<p>$$2^{3n}-1 = 8^n-1 = (8-1)(8^{n-1} + 8^{n-2} + \dots + 8^2 +8+1) = 7 \cdot (\mbox{something})$$</p>
|
1,056,045 |
<blockquote>
<p>Show that </p>
<p>$2^{3n}-1$ is divisble by $7$ for all $n$ $\in \mathbb N$</p>
</blockquote>
<p>I'm not really sure how to get started on this problem, but here is what I have done so far:</p>
<p>Base case $n(1)$:</p>
<p>$\frac{2^{3(1)}-1}{7} = \frac{8-1}{7} = \frac{7}{7}$</p>
<p>But not sure where to go from here. Tips?</p>
|
Josh B.
| 4,308 |
<p>Assume $2^{3n}-1$ is divisible by 7. Look at $2^{3(n+1)}-1$ and show that it is divisible by 7.</p>
<p>$2^{3(n+1)}-1 = 2^{3n+3}-1 = 8*2^{3n}-1 = (7+1)*2^{3n}-1 = (7*2^{3n} + 2^{3n})-1 = 7*2^{3n} + (2^{3n}-1)$ which is divisble by 7.</p>
|
1,394,550 |
<p>I have tried to prove the following inequality:</p>
<p>$$
\left(1+\frac{\log n}{n}\right)^n \gt\frac{n+1}{2}, \mbox{for}\;n\in\{2,3,\ldots\}
$$</p>
<p>which seems to be correct (confirmed by numerical result).</p>
<p>Can anyone give me some help or hint? Thanks a lot.</p>
|
k170
| 161,538 |
<p>$$
\left(1+\frac{\log n}{n}\right)^n\gt\frac{n+1}{2}
$$
Let $n=2$, then
$$
\left(1+\frac{\log 2}{2}\right)^2\gt\frac{2+1}{2}
$$
$$
1+\log 2+\frac14(\log 2)^2\gt1+\frac12
$$
Let $n=k+1$, then
$$\left(1+\frac{\log(k+1)}{k+1}\right)^{k+1}\gt\left(1+\frac{\log k}{k}\right)^k+\frac12$$
Applying the induction hypothesis, we have
$$\left(1+\frac{\log k}{k}\right)^k+\frac12\gt\frac{k+1}{2}+\frac12$$
Therefore
$$\left(1+\frac{\log(k+1)}{k+1}\right)^{k+1}\gt\frac{(k+1)+1}{2}$$
Which proves the hypothesis for $n\geq 2$.</p>
|
3,324,803 |
<p><strong>definition.</strong></p>
<p>The phase flow of the differential equation <span class="math-container">$\dot{x}=\vec{v}\ (x)$</span> is the one-parameter diffeomorphism group for which <span class="math-container">$\vec{v}$</span> is the phase velocity vector field, namely,
<span class="math-container">$$
\vec{v}=\frac{d}{dt} \Big|_{t=0} (g^tx)
$$</span></p>
<p>In the book, for the problem to find the phase flow of <span class="math-container">$\dot{x}=x-1$</span>, the provided answer, <span class="math-container">$g^tx=(x-1)e^t+1$</span>, is easy to verify. However, I have not idea to solve this problem. Any helps?</p>
|
sirous
| 346,566 |
<p><span class="math-container">$$\log_3(x+1)+\log_2 x=5$$</span></p>
<p><span class="math-container">$$\log_3(x+1)=5-\log_2 x$$</span></p>
<p><span class="math-container">$$x+1=3^{5-\log_2 x}$$</span></p>
<p>For integer solutions for x we must have:</p>
<p><span class="math-container">$$5-\log_2 x>0$$</span></p>
<p>⇒ <span class="math-container">$$\log_2 x<5$$</span></p>
<p>Therefore we must check numbers 4, 3,2, 1 which gives:</p>
<p><span class="math-container">$\log_2 x= 1, 2, 3, 4$</span></p>
<p>⇒<span class="math-container">$x=2, 4, 8, 16$</span></p>
<p>These solution must also satisfy the initial equation; corresponding values are:</p>
<p><span class="math-container">$\log_3 (x+1)=5-\log_2x=5-1=4,5-2= 3,5-3= 2,5-4= 1$</span></p>
<p>⇒ <span class="math-container">$x+1= 3^4=81, 3^3=27, 3^2=9, 3^1=3$</span></p>
<p>⇒<span class="math-container">$x= 80, 26, 8, 2$</span></p>
<p>The only common solution is 8. </p>
|
2,564,256 |
<p>The definition of general complex log for any non-zero complex number $z$ is
$$Log(z)=\log|z|+i[\arg(z)+2m\pi], m\in \mathbb{Z}$$</p>
<p>With this, if $n\in \mathbb{N}$ then
$Log(z^{1/n})=\frac{1}{n} Log(z)$ holds for all non-zero complex number $z$. </p>
<p>I verified this for $Log(i^{1/2})=\frac12 Log(i)$ successfully but could not make up with the following: $$Log(i^{1/3})=\frac13 Log(i)$$</p>
<p>Let me show what I have done and where I got stuck. </p>
<p>Since $i=\cos(2n\pi+\frac{\pi}{2})+i \sin(2n\pi+\frac{\pi}{2}), n\in \mathbb{Z}$ then by De-Moivre' s theorem, we have
\begin{align}
i^{1/3}= & \cos\left(\frac{2n\pi+\frac{\pi}{2}}{3}\right)+i\sin\left(\frac{2n\pi+\frac{\pi}{2}}{3}\right), n=0,1,2 \\
=& \cos\left(\frac{(4n+1)\pi}{6}\right)+i\sin\left(\frac{(4n+1)\pi}{6}\right), n=0,1,2\\
= &\begin{cases}
\cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6}) \\
\cos(\frac{5\pi}{6})+i\sin(\frac{5\pi}{6}) \\
\cos(\frac{9\pi}{6})+i\sin(\frac{9\pi}{6})
\end{cases}\\
= &\begin{cases}
\cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6}) \\
\cos(\pi-\frac{\pi}{6})+i\sin(\pi-\frac{5\pi}{6}) \\
-\cos(\frac{\pi}{2})-i\sin(\frac{\pi}{2})
\end{cases}\\
= &\begin{cases}
\frac{\sqrt{3}}{2}+i\frac{1}{2} \\
-\frac{\sqrt{3}}{2}+i\frac{1}{2} \\
0-i
\end{cases}
\end{align}</p>
<p>Now LHS:
\begin{align}
\frac13 Log(i)=&\frac13[\log|i|+i\{\arg(i)+2n_1 \pi\}], n_1\in \mathbb{Z}\\
=&\frac13(2n_1\pi+\frac{\pi}{2}), n_1\in \mathbb{Z}\\
=&(4n_1+1)\frac{\pi i}{6}, n_1\in \mathbb{Z}
\end{align}</p>
<p>whereas we see that
\begin{align}
&Log(i^{1/3})\\
=&\begin{cases}
\log|\frac{\sqrt{3}}{2}+i\frac{1}{2}|+i[\arg(\frac{\sqrt{3}}{2}+i\frac{1}{2})+2m_1\pi]\\
\log|-\frac{\sqrt{3}}{2}+i\frac{1}{2}|+i[\arg(-\frac{\sqrt{3}}{2}+i\frac{1}{2})+2m_2\pi]\\
\log|-i|+i[\arg(-i)+2m_3\pi]
\end{cases}\\
=&\begin{cases}
i[\frac{\pi}{6}+2m_1\pi]\\
i[\frac{5\pi}{6}+2m_2\pi]\\
i[\frac{-\pi}{2}+2m_3\pi]
\end{cases}, m_1, m_2, m_3\in \mathbb{Z}
\end{align}</p>
<p>And here I got stuck. I don't know how to finish. Any help will be appreciated. </p>
|
Dylan
| 135,643 |
<p>You have </p>
<p>$$ i = \exp\left(i\frac{\pi}{2} + i2n\pi\right) $$</p>
<p>Taking the third power yields</p>
<p>$$ i^{1/3} = \exp\left(i\frac{\pi}{6} + i \frac{2n\pi}{3} i\right) $$</p>
<p>Since $|i| = |i^{/3}| = 1$, taking the log yields</p>
<p>$$ \log (i^{1/3}) = i\left(\frac{\pi}{6} + \frac{2n\pi}{3} \right) $$</p>
<p>On the other hand, we have
$$ \frac{1}{3}\log (i) = \frac{i}{3} \left(\frac{\pi}{2} + 2n\pi \right) = i \left( \frac{\pi}{6} + \frac{2n\pi}{3} \right) $$</p>
<p>Does this help?</p>
<p><strong>EDIT:</strong> Continuing from your work, you can show that the 3 arguments for $\log(i^{1/3})$ are evenly spaced by an angle of $2\pi/3$. Therefore they can all be combined as
$$ \left\{\begin{aligned} i\left(-\frac{\pi}{6} + 2m_1\pi\right) \\ i\left(\frac{5\pi}{6} + 2m_2\pi\right) \\ i\left(-\frac{\pi}{2} + 2m_3\pi\right) \end{aligned}\right. = i\left(\frac{\pi}{6} + \frac{2n\pi}{3}\right) = i(4n+1)\frac{\pi}{6} $$</p>
<p>where $n$ is mapped by every alternating triplet of $(m_1,m_2,m_3)$. For example $m_3 = 0, m_1 = 0, m_2 = 0 \to n = -1,0,1$, etc</p>
|
2,051,785 |
<p>$PQRS$ is a convex quadrilateral. The intersection of the two diagonals is $O$. If the areas of triangles $PQS$, $QRP$, and $SPR$ are $1, 2,$ and $3$, respectively. What is the area of triangle $PQO$?</p>
|
Jean Marie
| 305,862 |
<p>With the excellent hint given by @Triskele, one finds $y=2/5$ under the assumption that <strong>such a quadrilateral exists</strong>.</p>
<p>This existence can be established. Moreover, there is no unicity: up to a rotation, there is one degree of freedom. More precisely, taking line PR as the $x$ axis and $O$ as the origin on this axis, we can take:</p>
<p>$$P\left(\frac45,0\right), \ \ Q\left(a,1\right), \ \ R\left(-\frac{16}{5},0\right), \ \ S\left(-\frac32a,-\frac32\right)$$
as illustrated on figure below with the simplest choice $a=0$.</p>
<p><a href="https://i.stack.imgur.com/pjzmb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pjzmb.jpg" alt="enter image description here"></a></p>
|
4,005,738 |
<p>I am interested in a method of how to compute the expected number of steps for absorpion in a Markov Chain with only one absorption node and given the starting and pre-final node (before absorption).</p>
<p>So, if the transition probability matrix is given, for example by:</p>
<p><span class="math-container">$P = \begin{bmatrix}
0 & 1/5 & 2/5 & 1/5 & 1/5\\
1/5 & 0 & 1/5 & 1/5 & 2/5\\
2/5 & 0 & 1/5 & 1/5 & 1/5\\
1/5 & 1/5 & 0 & 1/5 & 2/5\\
0 & 0 & 0 &0 & 1
\end{bmatrix}$</span></p>
<p>what will be the expected number of steps for being absorbed <strong>given that we start in node 1 (first line) and the last state before absorption is node 2 (second line)</strong> ?</p>
<p>So what I tried so far... I know the very known technique of using the fundamental matrix <span class="math-container">$N$</span> and then computing <span class="math-container">$N = (I-Q)^{-1}$</span> where <span class="math-container">$Q$</span> is the recurrent probabilities matrix (from the canonical form of <span class="math-container">$P$</span>) and the expected number of steps would be the first entry of the vector <span class="math-container">$(I-Q)^{-1}1$</span>, where <span class="math-container">$1$</span> is the five dimensional vector with all entries equal to <span class="math-container">$1$</span>, in the case where we dont know the information about where we are absorbed.</p>
<p>Then, as we know that we are absorbed from node two, I tried to replace all the probabilties of the last columns by <span class="math-container">$0$</span> (execept for line <span class="math-container">$2$</span> and for the aborption node that I kept <span class="math-container">$2/5$</span> and <span class="math-container">$1$</span>, respectively), then I am not pretty sure of how re-normalizing all the others...</p>
<p>Thank you very much!</p>
|
Jeroen van der Meer
| 874,176 |
<p>I'm slightly correcting my comments, and turning it into an answer. Let <span class="math-container">$P$</span> be the Markov chain you're interested in. Write <span class="math-container">$j_i$</span> for the initial state, <span class="math-container">$j_f$</span> for the last state before absorption, and <span class="math-container">$j_a$</span> for the absorption state. Write <span class="math-container">$P'$</span> for the sub-Markov chain consisting of all states except <span class="math-container">$j_a$</span>. If <span class="math-container">$i$</span> and <span class="math-container">$j$</span> are states in a Markov chain, write <span class="math-container">$\mu(i,j)$</span> for the expected amount of steps needed to reach <span class="math-container">$j$</span> given that you start at <span class="math-container">$i$</span>.</p>
<p>To find the expected amount of time to reach absorption given that you're in state <span class="math-container">$j_f$</span> right before that, what you'll want to do is make a summation over the amount of times that you've passed through <span class="math-container">$j_f$</span> before landing in <span class="math-container">$j_a$</span>. With probability <span class="math-container">$P_{j_f,j_a}$</span>, you'll need to pass through <span class="math-container">$j_f$</span> only once. With probability <span class="math-container">$(1 - P_{j_f,j_a}) P_{j_f,j_a}$</span>, you'll need to pass through <span class="math-container">$j_f$</span> twice. And so on. Given that you pass through <span class="math-container">$j_f$</span> precisely <span class="math-container">$n+1$</span> times, the expected amount of time that you'll need to reach <span class="math-container">$j_a$</span> is
<span class="math-container">$$\mu(j_i,j_f) + n \cdot\mu(j_f,j_f) + 1,$$</span>
where <span class="math-container">$\mu$</span> is computed in <span class="math-container">$P'$</span> (not in <span class="math-container">$P$</span>; this is because we assume that on our path from <span class="math-container">$j_f$</span> back to itself, we don't get absorbed along the way). Now it's a matter of appropriate summation. Modulo errors I think the answer will be
<span class="math-container">$$E(T) = \sum_{n = 0}^\infty P_{j_f,j_a} (1 - P_{j_f,j_a})^{n} \big(\mu(j_i,j_f) + n \cdot \mu(j_f,j_f) + 1\big)$$</span>
which can be simplified to
<span class="math-container">$$E(T) = \mu(j_i,j_f) + 1 + \sum_{n = 0}^\infty n P_{j_f,j_a} (1-P_{j_f,j_a})^{n} \mu(j_f,j_f),$$</span>
and yet further to
<span class="math-container">$$E(T) = \mu(j_i,j_f) + \frac{1-P_{j_f,j_a} }{P_{j_f,j_a}}\mu(j_f,j_f) + 1.$$</span></p>
<p>Here's an approach for finding <span class="math-container">$\mu$</span>. In a general Markov chain <span class="math-container">$Q$</span>, the expected amount of time needed to pass from state <span class="math-container">$i$</span> to state <span class="math-container">$j$</span> satisfies the recurrence relation
<span class="math-container">$$\mu(i,j) = 1 + \sum_{k \neq j} Q_{i,k} \mu(k,j),$$</span>
where the sum is taken over all states <span class="math-container">$k$</span> not equal to <span class="math-container">$j$</span>. This formula should be enough to find <span class="math-container">$\mu(j_i,j_f)$</span> and <span class="math-container">$\mu(j_f,j_f)$</span> in <span class="math-container">$P'$</span> numerically.</p>
|
3,668,702 |
<p>As the title says, how should I go about finding the shortest distance between all pairs of nodes (Each node has x and y co-odrinates associated with it) on a graph?</p>
<p>A brute force method is to run shortest path finding algorithms between all the pairs of the points. Is there a better way to approach this problem. (The reason I need this is I am trying to solve Traveling Salesman Problem using Ant Colony Optimziation which requires the cost matrix between each pair of nodes)</p>
|
peter.petrov
| 116,591 |
<p>There is a standard algorithm just for this. </p>
<p><a href="https://en.wikipedia.org/wiki/Floyd%E2%80%93Warshall_algorithm" rel="nofollow noreferrer">Floyd-Warshall's algorithm</a> </p>
<p>Its execution time is <span class="math-container">$O(n^3)$</span> where n is the number of vertices.<br>
It is easy to implement and often used in competitive programming. </p>
|
362 |
<p>The Math Stack Exchange chat room has had some heated arguments/suspensions in the past and I was wondering if there were any plans for the newly opened Math Overflow chat room to have its own moderators to provide the required special attention to prevent similar problem situations?</p>
|
Tim Post
| 35,352 |
<p><a href="https://mathoverflow.net/users?tab=moderators">Moderators of Math Overflow</a> <em>are also</em> moderators in chat. That said ...</p>
<p>Math Overflow can certainly have an election like any other site, and the same procedure would apply as on any other site. </p>
<p>Once a year, we check in with the moderation teams to ask how they're doing and if they feel they need any additional help - and we generally just go with whatever the existing team says. If they indicate that they need two more moderators, and that seems sane, then we schedule an election for them. </p>
<p>Once in a while we may prod more frequently if we notice that flags are taking an exceptional amount of time to clear - but we'll usually have heard cries for help when this is consistently the case.</p>
<p>If and when <a href="https://mathoverflow.net/users?tab=moderators">the existing Math Overflow moderation team</a> reaches out to us to ask for more help, we'll gladly schedule an election for them using the same STV system that Stack Overflow and the rest of the sites use. We like leaving this up to the moderation teams themselves, because they generally have the best feel for when one might be needed.</p>
<p>Regarding chat - Yes, other moderators on the network have moderation abilities in chat. When Math Overflow transitioned to the Stack Exchange 2.0 platform, some of our other community moderators from other academic sites joined chat to make themselves available to anyone that had questions or needed guidance. This was a very nice thing for them to do, they're genuinely helpful people and wanted to help make everyone feel welcome.</p>
<p>Still, your existing mod team are the ones that are in charge of making sure that the site runs smoothly.</p>
|
1,770,656 |
<p>In lecture I was given the following reasoning:</p>
<p>There are 7 options for the first character, 6 options for the last character, and 9! combinations for the 9 characters in between. Then you divide out the repeats -- 4 I's, 4 S's, and 2 P's. So the answer is 7 * 6 * 9! / (4! * 4! * 2!) = 13230</p>
<p>My own reasoning went like this:</p>
<p>There are 11! / (4! * 4! * 2!) combinations if you ignore the restriction on the I's. Then you just subtract out the combinations where I starts and ends the word, which is the same thing as finding out how many combinations are of the form I _ _ _ _ _ _ _ _ _ I, with the possible letters being 1 M, 2 I's (since we used two at the beginning and end), 4 S's, and 2 P's => 9! / (4! * 2! * 2!).</p>
<p>11! / (4! * 4! * 2!) - 9! / (4! * 2! * 2!) = 30870</p>
<p>Where does my reasoning go wrong?</p>
|
Graham Kemp
| 135,106 |
<p>You were wanting to count words that <em>neither</em> begin <em>nor</em> end with I. However you have found the count of words that do not <em>both</em> begin <em>and</em> end in I.</p>
<p>Let $\Omega$ be the set of all words formed with those letters (distinct permutations of the string), $B$ be the set of those that begin with I, and $E$ those that end with I. You are after $\lvert B^\complement\cup E^\complement\rvert$</p>
<p>$$\lvert B^\complement\cup E^\complement\rvert~=~\lvert\Omega\rvert-\lvert B\rvert-\lvert E\rvert + \lvert B\cap E\rvert$$</p>
<p>You found instead: $\lvert (B\cap E)^\complement\rvert = \lvert \Omega\rvert -\lvert B\cap E\rvert$</p>
|
1,771,279 |
<p>Let $B=\{x \in \ell_2, ||x|| \leqslant 1\}$ - Hilbert ball $X \subset B$ - open convex connected set in Hilbert ball, $\bar{X}$ - closure of $X$. $F: \bar{X} \to \bar{X}$ - continuous map that holomorphic into each point of $X$. Is it true in general, that $F$ has fixed point?</p>
|
Chill2Macht
| 327,486 |
<p>At first ignoring the holomorphic assumption, in general, no, since the Hilbert Ball is not compact for an infinite-dimensional Hilbert space.</p>
<p>Moreover, just because $X$ is open and connected does not mean that it is convex, and even though $\bar{X}$ is closed, since it is not a priori a subset of a compact set, we cannot assume that it is compact either.</p>
<p>So in terms of topological fixed point theorems, even the powerful Brouwer fixed point theorem will not apply in general. <a href="https://en.wikipedia.org/wiki/Brouwer_fixed-point_theorem" rel="nofollow">https://en.wikipedia.org/wiki/Brouwer_fixed-point_theorem</a></p>
<p>As regards whether the holomorphy properties of $F$ are sufficient to remedy this situation, I cannot say -- I am not sufficiently familiar with complex analysis on Hilbert spaces. I apologize.</p>
|
4,101,068 |
<p>Can two random variables <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> have the covariance matrix</p>
<p><span class="math-container">$\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}? $</span></p>
<p>The matrix is positive definite if <span class="math-container">$xy<0$</span>. Does this imply that it could be the convariance matrix for some <span class="math-container">$X$</span> and <span class="math-container">$Y$</span>?</p>
|
Sandipan Dey
| 370,330 |
<p>In order to show that the matrix <span class="math-container">$A=\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$</span> can't be a covariance matrix, we need to show that it's not a PSD matrix, i.e., it's sufficient to show an example vector (<span class="math-container">$\exists \mathbb{x} \in \mathbb{R^2}$</span>), such that <span class="math-container">$x^TAx<0$</span>.</p>
<p>Let <span class="math-container">$x=[1,1]^T$</span>, then we have</p>
<p><span class="math-container">$x^TAx=\begin{pmatrix} 1 & 1 \end{pmatrix}. \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}.\begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} -1 & -1 \end{pmatrix}.\begin{pmatrix} 1 \\ 1 \end{pmatrix}=-2 < 0$</span>.</p>
<p>Hence, <span class="math-container">$A$</span> is not a PSD matrix and can't be a covariance matrix.</p>
|
1,609,945 |
<p>Prove that the set of points for which $\arg(z-1)=\arg(z+1) +\pi/4$ is part of the set of points for which $|z-j|=\sqrt 2$ and show which part clearly in a diagram.</p>
<p>Intuition for the geometrical interpretation of this would be really appreciated. </p>
<p>And I was wondering greatest and lowest value for $|z+k|$ and why does this represent a circle ?
The question might be too basic but I just needed a intuition behind this. Concept of vectors is quite hard to follow.
Thanks in advance .</p>
|
Rafael
| 303,887 |
<p>We will prove that $X=\{|s-t|:s\in S, t\in T\}$ is closed.</p>
<p>Let $(|s_{n}-t_{n}|)$ be a sequence in $X$ such that $lim|s_{n}-t_{n}|=x$. Then $(s_{n})$ is a sequence in $S$ and $(t_{n})$ is a sequence in $T$.
Since $S$ is limited, there is a subsequence $(s_{n_{k}})$ of $(s_{n})$ such that $lim\ s_{n_{k}}=s$; since $S$ is closed, $s\in S$.</p>
<p>Now, note that $$|t_{n_{k}}|=|t_{n_{k}}-s_{n_{k}}+s_{n_{k}}|\leq|t_{n_{k}}-s_{n_{k}}|+|s_{n_{k}}|$$
and, since $|t_{n_{k}}-s_{n_{k}}|$ and $|s_{n_{k}}|$ converges, both are limited. So $|t_{n_{k}}|$ is limited.</p>
<p>Let $(t_{n_{k_{l}}})$ be a subsequence of $|t_{n_{k}}|$ such that $lim\ t_{n_{k_{l}}}=t$. Since $T$ is closed then $t\in T$. We conclude that $$x=lim|s_{n_{k_{l}}}-t_{n_{k_{l}}}|=|s-t|\in X$$
so $X$ is closed.</p>
<p>Take $\delta=inf(X)$. Since $X$ is closed, $\delta\in X$. If $\delta=0$, there is $s\in S,t\in T$ such that $|s-t|=0$, i.e., $s=t$. But $S\cap T=\emptyset$; therefore $|s-t|>\delta$ for all $s\in S, t\in T$.</p>
|
1,956,855 |
<p>I'm doing some math work involving proofs, and one of the definitions is:</p>
<p>|a| = -a when a < 0</p>
<p>Isn't the absolute value of a, positive a no matter what a is in the beginning? Am I looking at this wrong? Could use an explanation.</p>
|
adjan
| 219,722 |
<p>$|a|$ is always positive, but if $a$ is negative at the beginning, the absolute value is <em>equal</em> to $-a$, because negating a negative number results in a positive. This is merely a a property of $a$, not of $|a|$.</p>
|
1,956,855 |
<p>I'm doing some math work involving proofs, and one of the definitions is:</p>
<p>|a| = -a when a < 0</p>
<p>Isn't the absolute value of a, positive a no matter what a is in the beginning? Am I looking at this wrong? Could use an explanation.</p>
|
Mikhail Tikhonov
| 272,803 |
<p>If $a<0$ you can consider it as $-|a|$ since $|a|$ is absolute value(like unsigned value).</p>
<p>So $a=-|a|$, and $|a|=-a$</p>
<p>Example: $a=-1$, $|a|=1=-(-1)=-a$</p>
|
4,580,478 |
<p>So I'm currently reading about signed measures in "Real analysis" by Folland. In it, he defines a signed measure as follows.</p>
<p><em>Let <span class="math-container">$(X,\mathcal{M})$</span> be a measurable space. Then a signed measure <span class="math-container">$\nu$</span> is a function from <span class="math-container">$\mathcal{M}$</span> to <span class="math-container">$[-\infty,\infty]$</span> satisfying<br />
(i) <span class="math-container">$\nu(\emptyset) = 0$</span><br />
(ii) At most one of the values <span class="math-container">$\pm\infty$</span> are assumed<br />
(iii) If <span class="math-container">$A_1, A_2, \dots$</span> are (pairwise) disjoint elements of <span class="math-container">$\mathcal{M}$</span>, then
<span class="math-container">$\nu(\cup_iA_i) = \sum_i\nu(A_i)$</span>
where the sum on the RHS converges absolutely if the LHS is finite.</em></p>
<p>Now suppose we have a sequence <span class="math-container">$(A_i)$</span> of sets in <span class="math-container">$\mathcal{M}$</span>. If <span class="math-container">$\nu(\cup_iA_i) = \infty$</span>, it is not obvious to me that, under these hypotheses, for any permutation <span class="math-container">$\sigma$</span>, one automatically has <span class="math-container">$\sum_i\nu(A_{\sigma(i)}) = \infty$</span>.</p>
<p>Is this in fact automatically true or is it an additional assumption that we bake into the definition?</p>
|
GEdgar
| 442 |
<p>Answer for the original question.</p>
<p>Let <span class="math-container">$a_n \in [-\infty,\infty]$</span>, for <span class="math-container">$n=1,2,3,\dots$</span>. Write <span class="math-container">$\mathbb N = \{1,2,3,\dots\}$</span>. Let <span class="math-container">$\mathfrak S$</span> be the set of all bijections <span class="math-container">$\sigma : \mathbb N \to \mathbb N$</span>.
Let <span class="math-container">$\mathfrak S_0$</span> be the set of all <span class="math-container">$\sigma \in \mathfrak S$</span> such that
<span class="math-container">$\lim_{n \in \mathbb N}\sum_{j=1}^n a_{\sigma(j)}$</span> exists in <span class="math-container">$[-\infty,\infty]$</span>. For <span class="math-container">$\sigma \in \mathfrak S_0$</span>, write
<span class="math-container">$$
A(\sigma) = \lim_{n \in \mathbb N}\sum_{j=1}^n a_{\sigma(j)} .
$$</span>
Write <span class="math-container">$A_+ = \sum_{j : a_j>0} a_j$</span> and <span class="math-container">$A_- = \sum_{j : a_j<0} a_j$</span>.</p>
<p>Assume
<span class="math-container">$$
\text{there exists $\sigma_1, \sigma_2 \in \mathfrak S_0$ with $A(\sigma_1) \ne A(\sigma_2)$}.
\tag1$$</span>
We claim <span class="math-container">$A_+ = +\infty$</span>
and <span class="math-container">$A_- = -\infty$</span>.</p>
<p>First, we eliminate some trivial cases.</p>
<p><span class="math-container">$\bullet$</span> If there exist <span class="math-container">$j_1, j_2 \in \mathbb N$</span> with <span class="math-container">$a_{j_1} = +\infty$</span> and <span class="math-container">$a_{j_2} = -\infty$</span>, then <span class="math-container">$\mathfrak S_0 = \varnothing$</span>. This contradicts <span class="math-container">$(1)$</span>.</p>
<p><span class="math-container">$\bullet$</span> If there exists <span class="math-container">$j_1 \in \mathbb N$</span> with <span class="math-container">$a_{j_1} = +\infty$</span>, but there is no <span class="math-container">$j_2$</span> with <span class="math-container">$a_{j_2}=-\infty$</span>, then <span class="math-container">$A(\sigma) = +\infty$</span> for all <span class="math-container">$\sigma$</span>. This contradicts <span class="math-container">$(1)$</span>.</p>
<p><span class="math-container">$\bullet$</span> If there exists <span class="math-container">$j_2 \in \mathbb N$</span> with <span class="math-container">$a_{j_2} = -\infty$</span>, but there is no <span class="math-container">$j_1$</span> with <span class="math-container">$a_{j_1}=+\infty$</span>, then <span class="math-container">$A(\sigma) = -\infty$</span> for all <span class="math-container">$\sigma$</span>. This contradicts <span class="math-container">$(1)$</span>.</p>
<p>So, we assume from now on that <span class="math-container">$a_j \in \mathbb R$</span> for all <span class="math-container">$j$</span>.</p>
<p><span class="math-container">$\bullet$</span> If <span class="math-container">$A_+ = +\infty$</span> and <span class="math-container">$A_- \ne -\infty$</span>, then <span class="math-container">$A(\sigma)=+\infty$</span> for all <span class="math-container">$\sigma$</span>. This contradicts <span class="math-container">$(1)$</span>.</p>
<p><span class="math-container">$\bullet$</span> If <span class="math-container">$A_+ \ne +\infty$</span> and <span class="math-container">$A_- = -\infty$</span>, then <span class="math-container">$A(\sigma)=-\infty$</span> for all <span class="math-container">$\sigma$</span>. This contradicts <span class="math-container">$(1)$</span>.</p>
<p><span class="math-container">$\bullet$</span> If <span class="math-container">$A_+ \ne +\infty$</span> and <span class="math-container">$A_- \ne -\infty$</span>, then the series converges absolutely. <strong>THEOREM</strong> For an absolutely convergetn series <span class="math-container">$\sum a_j$</span>, we have <span class="math-container">$A(\sigma_1) = A(\sigma_2)$</span> for all <span class="math-container">$\sigma_1,\sigma_2 \in \mathfrak S$</span>. This contradicts <span class="math-container">$(1)$</span>.</p>
<p>The only case remaining is: <span class="math-container">$A_+=+\infty$</span> and <span class="math-container">$A_-=-\infty$</span>.</p>
<hr />
<p>So we see that the only nontrivial case is the <strong>THEOREM</strong>, which is in every calculus text. We can consider this result an easy restatement of the <strong>THEOREM</strong>.</p>
<hr />
<p>For the the revised version of the question, it seems to me there is nothing to do. <span class="math-container">$$\bigcup_{j=1}^\infty A_j = \bigcup_{j=1}^\infty A_{\sigma(j)}$$</span> for any permutation, so if the sets are pairwise disjiont, then by (iii) twice we have
<span class="math-container">$$
\sum_{j=1}^\infty \nu(A_j)
=\nu\left(\bigcup_{j=1}^\infty A_j\right) =
\nu\left(\bigcup_{j=1}^\infty A_{\sigma(j)}\right)
=\sum_{j=1}^\infty \nu(A_\sigma(j))
$$</span>
even if the series do not converge absolutely.</p>
|
2,796,618 |
<p>I am trying to</p>
<p>i) determine the infimum</p>
<p>ii) show that there's a function for which $\int_{0}^{1} {f'(x)}^2 dx$ is the infimum</p>
<p>iii) show if such function is unique.</p>
<p>I tried out several functions that suit the given condition, but couldn't see how $\int_{0}^{1} {f'(x)}^2 dx$ changes as $f(x)$ changes. How could we solve this problem?</p>
|
rapier
| 521,292 |
<p>You can solve this problem with variational calculus.
Let $y=f(x)$ and $g(x,y,y')=(y')^2=(f'(x))^2$. Then the function $f(x)$ that gives infimum satisfies
\begin{equation}
\frac{\partial g}{\partial y}-\frac{d}{dx}\left(\frac{\partial g}{\partial y'}\right)=0,
\end{equation}
if we use the Euler–Lagrange equation.
As $\cfrac{\partial g}{\partial y}=0$,
\begin{align}
&\frac{d}{dx}\left(2f'(x)\right)=0\\
&f(x)=ax+b
\end{align}
where $a,b$ are constant. Using the condition that $f(0)=0$ and $f(1)=1$,
\begin{equation}
f(x)=x\ .
\end{equation}</p>
|
2,185,434 |
<p>I recently read this problem</p>
<blockquote>
<p>A bag contains $500$ beads, each of the same size, but in $5$ different colors. Suppose there are $100$ beads of each color and I am blindfolded. What is the least number of beads I must pick before I can be sure there are $5$ beads of the same color among the beads I have picked blindfolded? <br/></p>
</blockquote>
<p>My intuitive answer was $401$, because the worst case would be picking $100$ beads of $4$ of the colors, then picking $1$ of the last remaning color. However, the given answer was $21$. Could someone explain how this answer was achieved?</p>
|
vrugtehagel
| 304,329 |
<p>You probably read the question wrong. You need $5$ beads of <em>the same</em> color. This means you can get $4$ beads of each color maximum to not-have $5$ beads that are the same color (which is $20$ beads) and then the next one makes sure you have $5$ beads of the same color (that was the $21$st bead).</p>
|
2,185,434 |
<p>I recently read this problem</p>
<blockquote>
<p>A bag contains $500$ beads, each of the same size, but in $5$ different colors. Suppose there are $100$ beads of each color and I am blindfolded. What is the least number of beads I must pick before I can be sure there are $5$ beads of the same color among the beads I have picked blindfolded? <br/></p>
</blockquote>
<p>My intuitive answer was $401$, because the worst case would be picking $100$ beads of $4$ of the colors, then picking $1$ of the last remaning color. However, the given answer was $21$. Could someone explain how this answer was achieved?</p>
|
The Count
| 348,072 |
<p>You have $5$ colors, and the worst possible scenario is picking $4$ of each color before you pick a fifth of one color. So you can get $20$ beads before you get the fifth of at least one of the colors. Does this make sense?</p>
|
1,985,849 |
<p>I am really stumped by this equation.
$$e^x=x$$
I need to prove that this equation has no real roots. But I have no idea how to start.</p>
<p>If you look at the graphs of $y=e^x$ and $y=x$, you can see that those graphs do not meet anywhere. But I am trying to find an algebraic and rigorous proof. Any help is appreciated. ( $e^\infty=\infty$ looks like a solution but is it?)</p>
|
Barry Cipra
| 86,747 |
<p>We have $e^x=x$ if and only if $x=\ln x$, which means, for one thing that there can be no solution with $x\le0$. For $x\gt0$, we have</p>
<p>$$\ln x=\int_1^x{dt\over t}\le\int_1^xdt=x-1\lt x$$</p>
<p>so there is no solution for $x\gt0$ either.</p>
<p>Remark: The key inequality between the two integerals may look puzzling for $x\lt1$, since ${1\over t}\gt1$ in that case; but it's still correct because the integrals are both <em>negative</em>.</p>
|
90,548 |
<p>Suppose we are handed an algebra $A$ over a field $k$. What should we look at if we want to determine whether $A$ can or cannot be equipped with structure maps to make it a Hopf algebra?</p>
<p>I guess in order to narrow it down a bit, I'll phrase it like this: what are some necessary conditions on an algebra for it to be a Hopf algebra?</p>
<p>Thoughts so far:</p>
<p>The first obvious condition is that $A$ must be augmented, i.e. there must be a nontrivial character $\varepsilon : A \to k$. Since this is generally not that hard to determine if we are given the algebra in some fairly concrete way, let's suppose that $A$ is given to us with an augmentation map.</p>
<p>If $A$ is finite-dimensional, then $A$ must be a Frobenius algebra. But not every finite-dimensional Frobenius algebra is a Hopf algebra, e.g. $\Lambda^\bullet(k^2)$ is not a Hopf algebra if the characteristic of $k$ is not 2. And generally I am more interested in the infinite-dimensional case.</p>
<p>All I can come up with is this: the category of finite-dimensional $A$-modules must be a (left) rigid monoidal category. But I don't know if that is a helpful observation: given a category with a forgetful functor to finite-dimensional vector spaces over some field, how can one prove that it can't be given the structure of a <s>braided</s> rigid monoidal category?</p>
<p>And perhaps there are some homological invariants that one can look at?</p>
<p>To sum up, the question is:</p>
<h3>Question</h3>
<p>Given a $k$-algebra $A$ and a nonzero character $\varepsilon : A \to k$, are there invariants we can look at in order to show that $A$ cannot be given the structure of a Hopf algebra?</p>
|
James
| 49,213 |
<p>In the book <em>Hopf algebras</em> by Eiichi Abe (1980), there is a theorem on page 57-58 that reads
that </p>
<blockquote>
<p>Given a $k$-linear space $H$, suppose that there are $k$-linear maps $\mu:H\otimes H\rightarrow H$, $\eta:k\rightarrow H$, $\Delta:H\rightarrow H\otimes H$, $\varepsilon:H\rightarrow k$ such that $(H,\mu,\eta)$ is a $k$-algebra and $(H,\Delta,\varepsilon)$ is a $k$-coalgebra. Then the following conditions are equivalent. (i) $\mu,\eta$ are $k$-coalgebra morphisms. (ii) $\Delta,\varepsilon$ are $k$-algebra morphisms ...</p>
</blockquote>
<p>and that </p>
<blockquote>
<p>When a $k$-linear space $H$ together with $k$-linear maps $\mu,\eta,\Delta,\varepsilon$ satisfy one of the equivalent conditions of Theorem 2.1.1. (quote above) then $(H,\mu,\eta,\Delta,\varepsilon)$ or simply $H$ is called a $k$-bialgebra.</p>
</blockquote>
<p>Since a Hopf algebra is a bialgebra with an antipodal map, if you have an algebra such that $\mu$ and $\eta$ are not coalgebra morphisms, or $\Delta,\varepsilon$ are not algebra morphisms then it would fail to be a bialgebra and thus a Hopf algebra.</p>
<p>It won't tell you for every algebra since there are bialgebras that are not Hopf algebras, but it is a start.</p>
<p>After looking into it further I found in the book <em>Hopf Algebras: An Introduction</em> by Sorin Dascalescu, et al. (2001), there is a remark on page 151 that reads</p>
<blockquote>
<p>In a Hopf algebra, the antipode is unique, being the inverse of the element I in the algebra $Hom(H^c,H^a)$. The fact that $S:H\rightarrow H$ is the antipode is written as $S*I=I*S=\eta\varepsilon$, and using the sigma notation $\sum S(h_1)h_2=\sum h_1S(h_2)=\varepsilon(h)1$ </p>
</blockquote>
<p>[Note: $H^c$ is the bialgebra looking at the coalgebra structure on it, or $(H,\Delta,\varepsilon)$ from above. $H^a$ is the bialgebra looking at the algebra structure on it, or $(H,\mu,\eta)$ from above. Also I changed the names of the multiplication, comultiplication, unit, and counit maps to correspond with what I had above.]</p>
<p>So if you cannot find an antipodal map for your bialgebra, it fails to be a Hopf algebra. Again, this doesn't completely answer the question because I do not know how you would go about figuring this out, but it should help you take a step in the right direction.</p>
|
782,929 |
<p>prove or disprove following matrix is
positive definite matrix ?</p>
<p>$$\begin{bmatrix}
\dfrac{\sin(a_1-a_1)}{a_1-a_1}&\cdots&\dfrac{\sin(a_1-a_n)}{a_1-a_n}\\
\vdots& &\vdots\\
\dfrac{\sin(a_n-a_1)}{a_n-a_1}&\cdots&\dfrac{\sin(a_n-a_n)}{a_n-a_n}
\end{bmatrix}_{n\times n}$$
and where we defind $$\dfrac{\sin 0}{0}=1$$</p>
<p>Maybe this problem can use Integral to solve it.
since
$$\frac{1}{x} = \int_0^\infty e^{-tx} t \, dt$$?
Thank you</p>
|
Adam
| 89,966 |
<p>Are there any other assumptions you forgot to add? A counter-example which comes to mind is to take $n = 2$ and $a_1 = a_2 = 1$ so that the matrix is just
$$ \left( \begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix} \right), $$
which is positive semidefinite, but not positive definite since one of its eigenvalues is $0$.</p>
|
1,595,243 |
<p>Let $f:\mathbb R \to \mathbb R$ be a function such that $|f(x)|\le x^2$, for all $x\in \mathbb R$. Then, at $x=0$, is $f$ both continuous and differentiable ?</p>
<p>No idea how to begin. Can someone help?</p>
|
user295959
| 295,959 |
<p>Note $|f(x)|\leq x^2$ implies $f(0)=0$. Then $$\lim_{x\rightarrow 0}\frac{|f(x)-f(0)|}{|x|}\leq \lim_{x\rightarrow 0}|x|=0.$$</p>
|
2,666,640 |
<p>Assume I have a monoid $M$ and I am not guaranteed that all elements have an inverse.</p>
<p>Say I have the property that:</p>
<p>$a^m = a^{m+n}$</p>
<p>Can I claim than it must be that $i=a^n$?</p>
<p>Why or why not?</p>
|
vadim123
| 73,324 |
<p>The answer is yes, if the monoid is a <a href="https://en.wikipedia.org/wiki/Cancellation_property" rel="nofollow noreferrer">cancellative</a> monoid. </p>
<p>Here's an example of a non-cancellative monoid under multiplication: $$\{[1],[4]\}\subseteq \mathbb{Z}/6\mathbb{Z}$$</p>
<p>$[4][4]=[4]$, but $[4]\neq [1]$</p>
|
2,666,640 |
<p>Assume I have a monoid $M$ and I am not guaranteed that all elements have an inverse.</p>
<p>Say I have the property that:</p>
<p>$a^m = a^{m+n}$</p>
<p>Can I claim than it must be that $i=a^n$?</p>
<p>Why or why not?</p>
|
lisyarus
| 135,314 |
<p>In general you cannot simplify expressions in monoids this way. Consider a monoid $M = \{i,x\}$, where $i$ is the neutral element and $x$ satisfies $x^2=x$. Then, $x^m=x^{n+m}$ for $n,m \geq 1$, but $i \neq x^n$.</p>
<p>However, in some monoids you can do this (for example, in groups and submonoids of groups).</p>
|
1,760,148 |
<p>If I have a connected metric space $X$, is any ball around a point $x\in X$ also connected?</p>
|
Community
| -1 |
<p>It is easy to see the answer is no. Take the unit circle in the plane, with the metric inherited from the Euclidian metric in the plane. Then remove the "top" point (coordinates $(0,1)$). Then a "ball" (small disk) around a point close to "the top", restricted to the "circle with the top removed," will be disconnected. Here the metric space is the circle <strong>with the top point removed</strong>.</p>
|
483,131 |
<p>How can you prove that if $t$ is less than or equal to $1$, that the probability of the sum of a sequence of uniform random variables being less than or equal to $t$ equals $t^k/k!$ ?</p>
<p>In other words:</p>
<p>Prove if $t \leq 1$, $$P(U_1+U_2 + \dots +U_k \leq t)=\frac{t^k}{k!}$$</p>
<p>My thought is that integration is needed <em>CDF</em>, but I can't figure out how to analytically solve this. </p>
<p>Thanks!</p>
|
André Nicolas
| 6,312 |
<p>We are told that $a-b$ is divisible by all of the $p_i$, where $i$ ranges from $1$ to $n$. We want to show that $a-b$ is divisible by $p_1p_2\cdots p_n$. </p>
<p>We <strong>must</strong> assume that the $p_i$ are <strong>distinct</strong>. If they are not, the result does not necessarily hold. </p>
<p>So we want to show that for any $c$, if all the $p_i$ divide $c$, then $p_1p_2\cdots p_n$ divides $c$.</p>
<p>From the fact that $p_1$ divides $c$, we find that $c=p_1c_1$ for some $c_1$.</p>
<p>But $p_2$ divides $c$, so $p_2$ divides $p_1c$. But $p_2$ is a prime, and does not divide $p_1$, so $p_2$ divides $c_1$, say $c_1=p_2c_2$.</p>
<p>It follows that $c=p_1p_2c_2$.</p>
<p>But $p_3$ divides $c$. Since $p_3$ is prime, and $p_3$ does not divide $p_1$ and $p_2$, it follows that $p_3$ divides $c_2$. Thus $c_2=p_3c_3$ for some $c_3$.</p>
<p>It follows that $c=p_1p_2p_3 c_3$.</p>
<p>Continue. After a while, we find that $c=p_1p_2\cdots p_n c_n$, which is the required result.</p>
<p><strong>Remark:</strong> You can make the argument a little nicer by doing a formal induction.
The result is trivially true for $n=1$. </p>
<p>Suppose the result is true for $n=k$. We want to show that the result holds for $n=k+1$.</p>
<p>So we are told that $c$ is divisible by $p_1,p_2,\dots,p_k,p_{k+1}$. By the induction assumption, $c$ is divisible by $p_1p_2\cdots p_k$, so $c=(p_1p_2\cdots p_k)q$ for some $q$. From the fact that $p_{k+1}$ divides $(p_1p_2\cdots p_k)q$, and that $p_{k+1}$ does not divide any of the $p_i$, $i=1$ to $k$, we conclude that $p_{k+1}$ divides $q$. This essentially completes the proof.</p>
|
1,009,868 |
<p>I had the question of $0^{-1}$ on a math test and I naturally assumed that this evaluates to zero, but from what I have seen from various sources it is equal to infinity which I do not quite understand. I would sooner believe that this it is just undefined.</p>
|
k170
| 161,538 |
<p>$$ 0^{-1}=\frac{1}{0}=\mbox{undefined} $$
because
$$ \lnot\exists x\in\mathbb{R}:1=0\times x $$</p>
|
1,009,868 |
<p>I had the question of $0^{-1}$ on a math test and I naturally assumed that this evaluates to zero, but from what I have seen from various sources it is equal to infinity which I do not quite understand. I would sooner believe that this it is just undefined.</p>
|
Graham Kemp
| 135,106 |
<p>Strictly speaking we say <span class="math-container">$0^{-1}$</span> is undefined.</p>
<p><span class="math-container">$x^{-1}$</span> is the multiplicative inverse of the number <span class="math-container">$x$</span>. By definition of the multiplicative inverse, the product of a number and its multiplicative inverse equals one. So we would have <span class="math-container">$0\times 0^{-1} = 1$</span>.</p>
<p>However, by definition of zero and multiplication, the product of zero and <em>any number</em> equals zero. So <span class="math-container">$0\times 0^{-1} = 0$</span>.</p>
<p>So, unless <span class="math-container">$0=1$</span>, these definitions conflict! Hence the multiplicative inverse <em>of zero</em> is undefined.</p>
<hr />
<p>Another way.</p>
<p>We can examine the behaviour of the function <span class="math-container">$f(x)=x^{-1}$</span> as <span class="math-container">$x$</span> approaches zero. Plot the curve <span class="math-container">$y=1/x$</span> to visualise what is happening. (It is a hyperbola.)</p>
<p>On the right side, the limit of <span class="math-container">$x^{-1}$</span> tends towards positive infinitude as <span class="math-container">$x$</span> tends downwards to zero.</p>
<p><span class="math-container">$$\lim_{0<x\to 0} \frac 1 x = +\infty$$</span></p>
<p>On the left side, the limit of <span class="math-container">$x^{-1}$</span> tends towards negative infinitude as <span class="math-container">$x$</span> tends upwards towards zero.</p>
<p><span class="math-container">$$\lim_{0>x\to 0} \frac 1 x = -\infty$$</span></p>
<p>So there is a discontinuity in the function <span class="math-container">$f(x)= x^{-1}$</span> at <span class="math-container">$x=0$</span>. Thus the quantity of <span class="math-container">$0^{-1}$</span> is indefinite.</p>
|
4,373,216 |
<p>Find power series solution for <span class="math-container">$y''-xy=0$</span> close to <span class="math-container">$x=0$</span>.</p>
<p><span class="math-container">$\sum_{n=2}^\infty n(n-1)a_nx^{n-2}-x\sum_{n=0}^\infty a_nx^n=0$</span></p>
<p>Then,
<span class="math-container">$a_{n+2}(n+2)(n+1)-a_{n-1}=0 \implies a_{n+2}(n^2+3n+2)-a_{n-1}=0 \implies a_{n+3}=a_n\cdot \frac{1}{(n+3)(n+2)}$</span></p>
<p>I got two problems:</p>
<p><span class="math-container">$(1)$</span> The answer is <span class="math-container">$a_{n+3}=(n+3)(n+2)a_n$</span> , where am I wrong ?</p>
<p><span class="math-container">$(2)$</span> How can I find the general power series solution ?</p>
|
2'5 9'2
| 11,123 |
<p>Use this property of integrals:</p>
<p><span class="math-container">$$\int_{-\infty}^{\infty}x^2f(x)\,dx=\int_{-\infty}^{0}x^2f(x)\,dx+\int_{0}^{1}x^2f(x)\,dx+\int_{1}^{2}x^2f(x)\,dx+\int_{2}^{\infty}x^2f(x)\,dx$$</span></p>
|
1,918,674 |
<p>For what value $k$ is the following function continuous at $x=2$?
$$f(x) = \begin{cases}
\frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2} & x \neq 2 \\
k & x = 2
\end{cases}$$</p>
<p>I was thinking about multiplying the numerator by it's conjugate, but that makes the denominator very messy, so I don't rlly know what to do.</p>
|
Mark
| 147,256 |
<p>You actually <em>should</em> multiply by the conjugate, in this case. For $x \neq 2$, we have
$$\begin{aligned}
f(x) &= \frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2} \cdot \frac{\sqrt{2x+5}+\sqrt{x+7}}{\sqrt{2x+5}+\sqrt{x+7}}\\
&= \frac{(2x+5)-(x+7)}{(x-2)(\sqrt{2x+5}+\sqrt{x+7})}\\
&= \frac{1}{\sqrt{2x+5}+\sqrt{x+7}}
\end{aligned}$$
You can now use this expression to determine the value of $k$.</p>
|
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