qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
1,264,195 | <p>So there is a way of extending the set $N$ of natural numbers with 0, equipped with ordinary multiplication, to its Grothendieck group, the group of integers with respect to addition.</p>
<p>This group, $Z$, is the <em>best</em> extension of $N$ in the sense that ...? (question)</p>
<p>So, here are my questions:</p>
<ol>
<li>I know the field of fractions $F$ of a commutative ring $R$ (without a zero divisor) is the <em>best</em> extension of $R$ into a field in the sense that given a natural inclusion map $\iota:R\to F$ (by sending $r\in R$ to $r/1\in F$), and any other field $K$ and an in injective homomorphism $\phi:R\to K$, there exists a (unique?) injective homomorphism $\overline{\phi}:F\to K$ such that the diagram commutes. So in a sense $F$, the field of fraction $F$ is the <em>smallest</em> extension of $R$.</li>
</ol>
<p>How does this idea then apply to $Z$, the Grothendieck group of $N$? Wikipedia has something like what I said, but slightly different (in that the homomorphisms are not necessarily injective, so one cannot talk about "inclusion relation").</p>
<ol start="2">
<li>Okay, the integers are not only a group with respect to addition, but it is also a ring with respect to addition and multiplication. How do you then extend a commutative monoid (such as $N$) to a commutative ring (such as $Z$)? Do we do so simply by defining one more binary operation on $(Z,+)$? If s0, how do we guarantee that it is indeed a very <em>special</em> structure (in the sense that it is the <em>smallest</em> ring extension of $N$)?</li>
</ol>
<p>Much appreciated in advance!</p>
| jgon | 90,543 | <p>Since you're interested in polynomial curves in the plane, I thought I'd write up the following.</p>
<p>Suppose your curve $C$ is the zero set in the plane $\mathbb{R}^2$ or $\mathbb{C}$ of a polynomial function $f$ in $\mathbb{R}[X,Y]$ (i.e. the set of points $x\in \mathbb{R}^2$ such that $f(x)=0$), then for any line $g(T)=(k_1T+r_1,k_2T+r_2)$, we have $f(g(t))\in \mathbb{R}[T]$, so it is either identically zero or has only finitely many zeros. I.e. for every line $L$, $C\cap L$ is finite. Note that the leading coefficient of $g(T)$ is a polynomial in $k_1$ and $k_2$ which is in fact the leading form of $f$ evaluated at $k_1$ and $k_2$, and is hence nonzero. Then if $k_1$ and $k_2$ make $g(T)$ zero then they are zeros of this form call it $F(X,Y)$. But then the projective point $[k_1:k_2]$ is a zero of the $F(X,Y)$. Since $F$ is a nonzero form, there are at most finitely many such projective points, so we can choose a point such that $g(T)\ne 0$. But this is just picking the direction of our line, so for every point, there is some direction such that the line through the point in that direction doesn't lie inside the curve.</p>
<p>Then if $x$ is a point in the plane such that $f(x)=0$, and $U$ any neighborhood of $x$, every line through $x$ intersects $U$ in an infinite set, since $U$ contains a ball of some positive radius around $x$. Then pick a line $L$ that doesn't lie inside $C$ Since $C\cap L$ is finite, while $L\cap U$ is not, there is some point of $U$ in $L$ not on $C$, which implies that every neighborhood of $x$ contains a point not on the curve for $x$ a point on a polynomial curve in the plane. Therefore $x$ is a boundary point of the curve.</p>
|
863,561 | <p>A Lambertian surface reflects or emits radiation proportional to the cosine of the angle subtended between the exiting angle and the normal to that surface. The integral of surface of the hemisphere which describes the exiting radiance is supposed to be equal to π. Is there a way I can prove that the surface of the lambertian hemisphere is equal to π?
The following is what I have tried. I assume that this hemispherical function can be described as
$$
x = \sin(\arccos(y)) = \sqrt{1-y^2}
$$
where I would attempt to use integration of a rotational surface to calculate the surface. The normal to the lambertian surface is defined by the y-axis. I want to rotate this surface about the y-axis. I thus use the following equation to calculate the rotational integration
$$
A = \int_0^1{2πx\sqrt{1+\left(\frac{dx}{dy}\right)^2}}dy
$$
I have used matlab to solve the integral symbolically and used numerical integration to try and get to the value of π, but it doesn't work. I think I am not starting with the right function to describe the surface. The following link may explain this better than above. <a href="http://fp.optics.arizona.edu/Palmer/rpfaq/rpfaq.htm#lambertian" rel="nofollow">http://fp.optics.arizona.edu/Palmer/rpfaq/rpfaq.htm#lambertian</a></p>
| Ron Gordon | 53,268 | <p>Note that <strong>intensity</strong> is proportional to $|\cos{\theta}|$, as this is a positive quantity. The integrated quantity you seek (total intensity divided by radiance) is then</p>
<p>$$\pi \int_0^{\pi} d\theta \, \sin{\theta} |\cos{\theta}| = 2 \pi \int_0^{\pi/2} d\theta \, \sin{\theta} \cos{\theta} = \pi$$</p>
|
3,996,090 | <p>I believe I have found the recurrence relation to be
<span class="math-container">$$B\left(n\right)=B\left(n-1\right)+2^{n-1}-1$$</span>
with Initial Condition B(0)=0 (I am a bit unsure about the initial condition though but I think it is correct)</p>
<p>Now I am trying to solve B(n) using iteration, this is what I have so far:
<span class="math-container">$$B\left(n\right)=B\left(n-1\right)+2^{n-1}-1$$</span>
<span class="math-container">$$B\left(n\right)=B\left(n-2\right)+2^{n-2}+2^{n-1}-\left(2\right)1$$</span>
<span class="math-container">$$B\left(n\right)=B\left(n-3\right)+2^{n-3}+2^{n-2}+2^{n-1}-\left(3\right)1$$</span>
<span class="math-container">$$=B\left(n-k\right)+2^{n-k}+2^{n-\left(k-1\right)}+2^{n-\left(k-2\right)}-\left(k\right)1$$</span>
and then I let n=k since the intial condition is B(0) but this is where I get confused I am not sure what to do from here?
<span class="math-container">$$=B\left(0\right)+2^0+2^1+2^2+...+2^{\left(?\right)}-n$$</span></p>
| Brian M. Scott | 12,042 | <p>You’re on the right track:</p>
<p><span class="math-container">$$\begin{align*}
B(n)&=B(n-1)+2^{n-1}-1\\
&=B(n-2)+2^{n-2}+2^{n-1}-2\\
&=B(n-3)+2^{n-3}+2^{n-2}+2^{n-1}-3\\
&\;\;\vdots\\
&=B(n-k)-k+\sum_{i=1}^k2^{n-i}\\
&\;\;\vdots\\
&=B(0)-n+\sum_{i=1}^n2^{n-i}\\
&=-n+\sum_{i=0}^{n-1}2^i\\
&=-n+2^n-1\\
&=2^n-n-1\,.
\end{align*}$$</span></p>
|
115,821 | <p>Here are some sample data:</p>
<pre><code>data = {{1}, {50., 53, 52, 52}, {100., 105, 104, 104}, {150., 157, 156,
156}, {200., 209, 208, 208}, {250., 261, 260, 260}, {300., 313, 312,
313}, {2}, {50., 53, 52, 51}, {100., 106, 105, 102}, {150., 158,
157, 153}, {200., 211, 210, 204}, {250., 265, 264, 256}, {300., 319,
318, 307}, {3}, {50., 53, 52, 52}, {100., 106, 105, 104}, {150.,
158, 158, 156}, {200., 211, 210, 209}, {250., 264, 263, 261}, {300.,
317, 316, 313}, {4}, {50., 51, 50, 51}, {100., 102, 101,
102}, {150., 153, 152, 152}, {200., 204, 203, 204}, {250., 256, 256,
254}, {300., 309, 309, 305}, {5}, {50., 52, 51, 52}, {100., 104,
104, 104}, {150., 156, 155, 156}, {200., 208, 208, 208}, {250., 260,
260, 260}, {300., 312, 311, 312}}
</code></pre>
<p>As we can see, there are five sets of four columns. Now we will delete the indices (1,2,3,4,5) and create the corresponding sub lists.</p>
<pre><code>d2 = SplitBy[data, Dimensions][[2 ;; ;; 2]];
</code></pre>
<p>The first column is time, while the other three are some integers (nx, ny, nz). I want to see the time-evolution of these integers. In particular, I want to <code>ListPlot</code> all five <code>nx</code> versus time in a single plot (one on top of each other). Then I suppose it would be trivial to do the same for <code>ny</code> and <code>nz</code> versus time.</p>
<p>Any suggestions?</p>
| Karsten 7. | 18,476 | <p>The easiest workaround is to use</p>
<pre><code>Table[Labeled[
ListVectorPlot[data[[i]], PlotRange -> All,
VectorColorFunction ->
Function[{x, y, vx, vy, n},
ColorData["Rainbow"][Rescale[n, {min, max}]]],
VectorColorFunctionScaling -> False, ImageSize -> 300],
BarLegend[{"Rainbow", {min, max}}, ImageSize -> 300],
{{Right, Center}}
], {i, 1, 2}]
</code></pre>
<p><a href="https://i.stack.imgur.com/MyIeg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MyIeg.png" alt="Out1"></a></p>
<hr>
<p>This is probably the correct way to specify the <code>BarLegend</code> for the given situation:</p>
<pre><code>Table[ListVectorPlot[data[[i]], PlotRange -> All,
PlotLegends ->
BarLegend[{ColorData["Rainbow"][Rescale[#, {min, max}]] &, {min, max}}],
VectorColorFunction ->
Function[{x, y, vx, vy, n}, ColorData["Rainbow"][Rescale[n, {min, max}]]],
VectorColorFunctionScaling -> False, ImageSize -> 300], {i, 1, 2}]
</code></pre>
<p><a href="https://i.stack.imgur.com/JY621.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JY621.png" alt="Out2"></a></p>
<p>Alternatively one can add the option <code>ColorFunctionScaling -> True</code> to <code>BarLegend</code>. This option is undocumented for <code>BarLegend</code> and will be highlighted in red in the FE.</p>
<pre><code>Table[ListVectorPlot[data[[i]], PlotRange -> All,
PlotLegends -> BarLegend[{"Rainbow", {min, max}}, ColorFunctionScaling -> True],
VectorColorFunction ->
Function[{x, y, vx, vy, n}, ColorData["Rainbow"][Rescale[n, {min, max}]]],
VectorColorFunctionScaling -> False, ImageSize -> 300], {i, 1, 2}]
</code></pre>
<hr>
<h3>Explanation:</h3>
<p>You specified the <code>BarLegend</code> in the format you wanted it to be displayed.<br>
However, as stated in the <a href="http://reference.wolfram.com/language/ref/BarLegend.html#1239470157" rel="nofollow noreferrer">documentation</a>, </p>
<blockquote>
<p><code>BarLegend</code> automatically picks up values of <code>ColorFunction</code> and
<code>ColorFunctionScaling</code>.</p>
</blockquote>
<p>For your specific case this means that your <code>BarLegend</code> gets replaced by </p>
<pre><code>BarLegend[{"Rainbow", {0., 12.7279}}, Legending`LegendDump`options$2394]
</code></pre>
<p>with <code>Legending`LegendDump`options$2394</code> being an internal value that picked up <code>ColorFunctionScaling -> False</code> from the rest of the input (the number at the end will be different each time the input is evaluated) and subsequently <code>BarLegend</code> becomes</p>
<pre><code>BarLegend[{"Rainbow", {0.`, 12.727922061357857`}},
Sequence[LabelStyle -> {}, LegendLayout -> "Column", LegendMarkerSize -> 300,
Charting`TickSide -> Right, ColorFunctionScaling -> False]]
</code></pre>
<p><a href="https://i.stack.imgur.com/V1lwQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/V1lwQ.png" alt="OutPNG"></a></p>
|
61,509 | <p>I'm trying to read Elias Stein's "Singular Integrals" book, and in the beginning of the second chapter, he states two results classifying bounded linear operators that commute (on $L^1$ and $L^2$ respectively).</p>
<p>The first one reads:</p>
<p>Let $T: L^1(\mathbb{R}^n) \to L^1(\mathbb{R}^n)$ be a bounded linear transformation. Then $T$ commutes with translations if and only if there exists a measure $\mu$ in the dual of $C_0(\mathbb{R}^n)$ (continuous functions vanishing at infinity), s.t. $T(f) = f \ast \mu$ for every $f \in L^1(\mathbb{R}^n)$. It is also true that $\|T\|=\|\mu\|$. </p>
<p>The second one says:</p>
<p>Let $T:L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$ be bounded and linear. Then $T$ commutes with translation if and only if there exists a bounded measurable function $m(y)$ so that $(T\hat{f})(y) = m(y) \hat{f}(y)$ for all $f \in L^2(\mathbb{R}^n)$. It is also true that $\|T\|=\|m\|_\infty$. </p>
<p>I was wondering if anyone had a reference to a proof of these two results or could explain why they are true. </p>
| robjohn | 13,854 | <p>For the first one:</p>
<p>Let $\phi_\epsilon(x)=\phi(x/\epsilon)/\epsilon^n$ where
$$
\phi(x)=\left\{\begin{array}{cl}c\;e^{|x|^2/(|x|^2-1)}&\text{if }|x|<1\\0&\text{if }|x|\ge1\end{array}\right.
$$
and $c$ is chosen so that $\int_{\mathbb{R}^n}\phi(x)\;\mathrm{d}x=1$.</p>
<p>$\|T\phi_\epsilon\|_{L^1}$ is bounded as $\epsilon\to0$, so there is a sequence $\{\epsilon_k\}$ and a measure $\mu$ so that $T\phi_{\epsilon_k}\to\mu$ weakly in $L^1$.</p>
<p>Since $T$ is continuous, linear, and commutes with translation,
$$
\begin{align}
f*\mu(x)
&=\lim_{k\to\infty}\;\int_{\mathbb{R}^n}f(y)\;T\phi_{\epsilon_k}(x-y)\;\mathrm{d}y\\
&=\lim_{k\to\infty}\;\int_{\mathbb{R}^n}f(y)\;T(\phi_{\epsilon_k}(x-y))\;\mathrm{d}y\\
&=\lim_{k\to\infty}\;\int_{\mathbb{R}^n}T(f(y)\;\phi_{\epsilon_k}(x-y))\;\mathrm{d}y\\
&=\lim_{k\to\infty}\;T\left(\int_{\mathbb{R}^n}f(y)\;\phi_{\epsilon_k}(x-y)\;\mathrm{d}y\right)\\
&=T\left(\lim_{k\to\infty}\;\int_{\mathbb{R}^n}f(y)\;\phi_{\epsilon_k}(x-y)\;\mathrm{d}y\right)\\
&=Tf(x)
\end{align}
$$
For the second one the definition should be $(Tf)^\wedge=m\hat{f}$ (otherwise $T$ is just multiplication by $m$ rather than a Fourier multiplier operator). Let $\psi(x)=e^{-\pi x^2}$ so that $\hat{\psi}=\psi$.
$$
\begin{align}
\psi(\xi)\;(Tf)^\wedge(\xi)
&=(\psi*Tf)^\wedge(\xi)\\
&=\left(\int_{\mathbb{R}^n}\psi(x-y)\;Tf(y)\;\mathrm{d}y\right)^\wedge(\xi)\\
&=\left(\int_{\mathbb{R}^n}\psi(y)\;Tf(x-y)\;\mathrm{d}y\right)^\wedge(\xi)\\
&=\left(\int_{\mathbb{R}^n}T^{\;*}\psi(y)\;f(x-y)\;\mathrm{d}y\right)^\wedge(\xi)\\
&=(T^{\;*}\psi)^\wedge(\xi)\hat{f}(\xi)
\end{align}
$$
Let $m(\xi)=(T^{\;*}\psi)^\wedge(\xi)/\psi(\xi)$, then we have
$$
(Tf)^\wedge(\xi)=m(\xi)\hat{f}(\xi)
$$
and therefore, $\|m\|_{L^\infty}=\|T\|_{L^2}$.</p>
|
39,387 | <p>It is usual to use second, third and fourth moments of a distribution to describe certain properties. Do partial moments or moments higher than fourth describe any useful properites of a distribtution.</p>
| Spencer | 4,281 | <p>The typical example I remember being given at school is that of <a href="http://en.wikipedia.org/wiki/Kurtosis" rel="nofollow">kurtosis</a>, which uses fourth moment and is easy to understand. </p>
<p>Wiki says: "Higher kurtosis means more of the variance is the result of infrequent extreme deviations, as opposed to frequent modestly sized deviations."</p>
|
1,669,096 | <p>How do I show that <span class="math-container">$\ell^{ \infty}$</span> is a normed linear space,
where <span class="math-container">$\ell^{ \infty}$</span> is define as <span class="math-container">$$\|\{a_n\}_{n=1}^{\infty}\|_{\ell^\infty}=\sup_{1 \leq k \leq \infty} |a_k|?$$</span>
There are three properties that I need to check in order for this to be a normed linear space. Nonnegativity, positive homogeneity, and the triangle inequality.</p>
<p>I am having trouble working with the supremum. Any idea will be greatly appreciated thanks</p>
| Mel | 566,704 | <p>Well begin by noticing that for <span class="math-container">$a,b\in \mathbb{R}$</span> and <span class="math-container">$(x_n),(y_n)\in \ell^\infty$</span> you have that <span class="math-container">$(ax_n+by_n)$</span> is bounded, therefore it belongs in <span class="math-container">$\ell^\infty$</span> as well.</p>
<p>For <span class="math-container">$\|\cdot\|_{\infty}$</span> to be a norm, check that it is non-negative and <span class="math-container">$\|x_n\|_{\infty}$</span> becomes zero only when <span class="math-container">$(x_n)=0$</span>.</p>
<p><span class="math-container">$\|\cdot\|_{\infty}$</span> also satisfies <span class="math-container">$\|ax_n\|_{\infty}=|a|\|x_n\|_{\infty}$</span> and use the triangle inequality for the absolute value to show that <span class="math-container">$\|x_n+y_n\|_{\infty}\leq \|x_n\|_{\infty}+\|y_n\|_{\infty}$</span>. These prove <span class="math-container">$\|\cdot\|_{\infty}$</span> is a norm on <span class="math-container">$\ell^\infty$</span>.</p>
|
2,363,840 | <blockquote>
<p>As the heading states, find all $p(x)$ for $p(x+1)=p(x)+3x(x+1)+1$ for all real $x$.</p>
</blockquote>
<p>I have no idea how to approach this. Any solution or guide how to solve these kinds of questions would be appreciated! </p>
| Ennar | 122,131 | <p>Let $g(x) = p'''(x)$. Deriving the condition $p(x+1) = p(x) + 3x(x+1)+1$ three times, we get $$p'''(x+1) = p'''(x)\implies g(x+1) = g(x).$$</p>
<p>Now, $g$ is still polynomial and the last line tells us that if $\alpha$ is a root of $g$, so is $\alpha \pm 1$ and hence, by induction, so is $\alpha + n$ for any integer $n$. Thus, either $g$ has no roots or infinitely many roots. The only polynomials that satisfy this are constant polynomials, i.e. $g\equiv const.$ Since $g(x) = p'''(x)$, it means that $p$ has degree $3$, i.e. $$p(x) = ax^3+bx^2+cx+d.$$</p>
<p>Substitute this into the starting condition and calculate $a=1,\ b = c = 0$, and thus $$p(x) = x^3+d.$$</p>
<p>Conversely, any polynomial of the form $x^3 + d$ is a solution of the problem.</p>
|
1,303,485 | <p>Evaluate the integral $$\int_{C}\frac{z^2}{z^2+9}dz$$
where C is the circle $|z|=4$</p>
<p>I know that if f is analytic in simply connected domain $D$, $C$ a simple closed positively oriented contour that lies in D and $z_o$ lies interior to $C$, then
$$\int_{C}\frac{f(z)}{z-z_o}dz=2\pi i f(z_o)$$</p>
<p>But for this problem, the circle contains both interior points which is $3i$ and $-3i$. And I found that reducing the fraction into partial fraction seems to be useless in solving the problem. So what is the $f(z)$ here?</p>
| Ron Gordon | 53,268 | <p>You can rewrite</p>
<p>$$\frac{z^2}{z^2+9} = 1 - \frac{9}{z^2+9} = 1 - \frac{9}{(z+i 3)(z-i 3)} = 1+\frac{i 3/2}{z-i 3} - \frac{i 3/2}{z+i 3}$$</p>
<p>Now apply Cauchy's integral formula.</p>
|
3,084,479 | <p><span class="math-container">$h\in \mathbb{R}$</span>, because we have defined the Trigonometric Functions only on <span class="math-container">$\mathbb{R}$</span> so far.</p>
<p>I have a look at <span class="math-container">$e^{ih}=\sum_{k=0}^{\infty}\frac{(ih)^k}{k!}=1+ih-\frac{h^2}{2}+....$</span> </p>
<p><strong>How can one describe the nth term of the sum?</strong></p>
<p>Then I look at <span class="math-container">$\frac{e^{ih}-1}{h}=\frac{(1-1)}{h}+i-\frac{h}{2}+...=i-\frac{h}{2}+....$</span> </p>
<p><strong>Again how can I describe that the nth term of the sum?</strong> </p>
<p>Because <span class="math-container">$\frac{e^{ih}-1}{h}=\sum_{k=1}^{\infty}\frac{\frac{(ih)^k}{h}}{k!}<\sum_{k=0}^{\infty}\frac{\frac{(ih)^k}{h}}{k!}=\sum_{k=1}^{\infty}\frac{(ih^{-1})^k}{k!}=e^{ih^{-1}}$</span></p>
<p>and <span class="math-container">$ih^{-1}$</span> is a complex number and the exponential-series converges absolutely for all Elements in <span class="math-container">$\mathbb{C}$</span>, I have found a convergent majorant. And I can apply the properties of Limits on <span class="math-container">$\frac{e^{ih}-1}{h}\forall, h\in \mathbb{R}$</span>.</p>
<p><strong>How can I now prove formally (i.e by chosing an explicit <span class="math-container">$\delta$</span>) that</strong> </p>
<p><span class="math-container">$$\forall_{\epsilon>0}\exists_{\delta>0}\forall_{h\in\mathbb{R}}|h-0|=|h|<\delta\Longrightarrow |(\frac{e^{ih}-1}{h}=i-\frac{h}{2}+...)-i|<\epsilon$$</span></p>
<p><strong>I am also seeking for advice how to argue in such cases more intuitively (i.e by not always ginving an explicit <span class="math-container">$\delta$</span> ).</strong></p>
| Paramanand Singh | 72,031 | <p>Here is a simple proof which I first read in Hardy's <em>A Course of Pure Mathematics</em>.</p>
<p>To reiterate the question for clarity we have <span class="math-container">$h\in\mathbb{R} $</span> and the symbol <span class="math-container">$e^{ih} $</span> is defined by the series <span class="math-container">$\sum_{n=0}^{\infty} (ih) ^n/n! $</span>. Let's assume <span class="math-container">$|h|< 1$</span> and observe that
<span class="math-container">\begin{align*}
\left|\frac {e^{ih} - 1}{h}-i\right|&=\left|\frac{i^2h}{2!}+\frac{i^3h^2}{3!}+\dots\right|\\
&\leq\frac{|h|}{2!}+\frac{|h|^2}{3!}+\dots\\
&\leq\frac{|h|} {2}+\frac{|h|^2}{2^2}+\frac{|h|^3}{2^3}+\dots\\
&=\frac{|h|}{2-|h|}\text{ (sum of infinite GP)} \\
&<|h|
\end{align*}</span>
Thus given any <span class="math-container">$\epsilon>0$</span> if we choose <span class="math-container">$\delta=\min(1,\epsilon) $</span> then we have <span class="math-container">$$\left|\frac{e^{ih} - 1}{h}-i\right|<\epsilon $$</span> whenever <span class="math-container">$0<|h|<\delta$</span>. Therefore by definition of limit we have <span class="math-container">$$\lim_{h\to 0}\dfrac{e^{ih}-1}{h}=i$$</span></p>
<hr>
<p>Hardy uses the above limit to prove the formula <span class="math-container">$$e^{ix} =\cos x+i\sin x \,\forall x\in\mathbb {R} $$</span> The idea is to show that the function <span class="math-container">$f(x) =e^{ix} $</span> is differentiable with derivative <span class="math-container">$f'(x) =if(x) $</span> and then consider the function <span class="math-container">$$g(x) =(\cos x-i\sin x) f(x) $$</span> It is easily proved that <span class="math-container">$g'(x) =0$</span> and hence <span class="math-container">$g$</span> is constant. Thus <span class="math-container">$g(x) =g(0)=1$</span> and <span class="math-container">$f(x) =\cos x+i\sin x$</span>.</p>
|
2,809,090 | <p>The stochastic vector $(X,Y)$ has a continuous distribution with pdf:
$$f(x,y) =
\begin{cases}
xe^{-x(y+1)} & \text{if $x,y>0$} \\[2ex]
0 & \text{otherwise}
\end{cases}$$<br>
Define $Z:=XY$.<br>
I would like to know what exactly $XY$ is. It seems to me that the function $f(x,y)$ has but one output, so what does $XY$ mean here? I feel like I'm missing something obvious. Thanks in advance.</p>
| egreg | 62,967 | <p>Your expression is full of red herrings. Try first removing the denominators (you can multiply by $1/6$ at the end) and writing $t$ for $x/2$. You get:
$$
\cos t\cos 3t -3\cos^2t+3\sin^2t + \sin t\sin 3t
$$
You could be tempted to expand $\cos3t$ and $\sin3t$, which is of course possible, but it would be longer than needed.</p>
<p>Instead, loot at the terms: there is just one minus sign, which makes unlikely some form of factorization. However, you can notice that
$$
\cos t\cos 3t+\sin t\sin 3t=\cos(3t-t)=\cos2t
$$
Also
$$
\cos^2t-\sin^2t=\cos2t
$$
and this accounts for the other two terms, so
$$
\cos t\cos 3t -3\cos^2t+3\sin^2t + \sin t\sin 3t
=
\cos 2t-3\cos2t=-2\cos2t
$$
Multiply by $1/6$ and go back to $2t=x$: your expression is
$$
-\frac{1}{3}\cos x
$$</p>
|
1,382,587 | <p>In $\mathbb R^3$, let $C$ be the circle in the $xy$-plane with radius $2$ and the origin as the center, i.e., </p>
<p>$$C= \Big\{ \big(x,y,z\big) \in \mathbb R^3 \mid x^2+y^2=4, \ z=0\Big\}.$$</p>
<p>Let $\Omega$ consist of all points $(x,y,z) \in \mathbb R^3$ whose distance to $C$ is at most $1$. Compute
$$\int_\Omega \left|\,x\,\right|\,dx\,dy\,dz$$</p>
<p>So, with the help of erichan (see below), I now know that the volume is a solid torus. But I am having trouble setting up the integration bounds. As erichan had suggested, we consider a union of unit-spheres, all centered on points of the radius-$2$ circle. Using spherical coordinates, I have this integral set up:</p>
<p>$$\int_0^\pi\int_0^{2\pi}\int_0^1
\Big(\big|\,\left(r\cos \theta+2\right)\sin\phi \,\big| \,r^2 \sin\phi \Big) \, dr\, d\theta \,d\phi,$$</p>
<p>Where I parameterized the solid torus as:
$$
\begin{aligned}
x &= (r\cos \theta +2)\sin \phi \\
y &= (r\sin \theta +2)\sin\phi \\
z &= r\cos \phi
\end{aligned}
$$
with Jacobian as $r^2\sin \phi$.</p>
<p>Is my setup ok? I shifted $x$ and $y$ by two units.
I'm not so sure about the parametrization of $z$ (should I leave it as it is normally?).
And should I change the Jacobian factor?</p>
<p>I welcome any answers to this problem – I had previously requested just hints.
I am wondering whether there is a simpler way to compute this integral, using symmetry of the torus.</p>
<p>Thanks,</p>
| zzchan | 252,453 | <p>Since you only want hints...</p>
<p>For each point on the circle of radius $2$ consider a solid sphere of radius $1$. Now consider the union of all these spheres. What do you get?</p>
|
1,392,257 | <p><strong>The definition of a conjugate element</strong> </p>
<p>We say that $x$ is conjugate to $y$ in $G$ if $y = g^{-1}xg $ for some $g \in G$</p>
<p>Now for the group $G=Q_8$ , we have the group presentation $$Q_8 = \big<a,b: a^4 =1,b^2 = a^2, b^{-1}ab = a^{-1} \big>$$</p>
<p>Now the elements of $Q_8$ are $\{1,a,a^2,a^3,ab,a^2b,a^3b,b\}$ and after some calculation we would get $5$ different conjugacy classes, namely $a^G = \{a,a^3\}$ where $a^G$ denotes the conjugacy class of $a$ in $G = Q_8$,</p>
<p>also we have </p>
<p>$1^G = \{1 \}$, ${a^2}^G = \{ a^2 \}$, ${(a^2b)}^G = \{a^2b,b \}$ and ${(ab)}^G = \{ab,a^3b\}$</p>
<p>Of course , there is no surpise that for every element $x \in G$ we have $x \in x^G$ because $x = 1^{-1}x1$. However, we see that all the conjugacy classes for $Q_8$ contain the element and it's inverse. Like $a^{-1} = a^3$, ${(a^2)}^{-1} = a^2$, ${(a^2b)}^{-1} = b$ and so on.</p>
<p>My question is does this hold true for all groups ?</p>
<p>More formally , Is it true that for an element $x \in G$ then $x,x^{-1} \in x^G$ ?</p>
| Geoff Robinson | 13,147 | <p>In a finite group $G$ of odd order $2n+1$, only the identity is conjugate to its inverse. For if $gxg^{-1} = x^{-1}$, then it is easy to see that $g^{2}xg^{-2} =x$, so that $g^{2}$ commutes with $x$. But $g = (g^{2})^{n+1}$ is a power of $g^{2}$, so that $g$ commutes with $x$. Hence $x = x^{-1}$, so that $x^{2} = 1_{G}$. But then $x = (x^{2})^{n+1} = 1_{G}$.</p>
|
306,788 | <p>I understand determinants but I cannot understand the following question, can someone explain it to me ?</p>
<p>Suppose that a $4 x 4$ matrix with rows $v_1,v_2,v_3$ and $v_4$ has determinant det A = -1. Find the following determinants: </p>
<p>$$det \begin{bmatrix}v_1\\6v_2\\v_3\\v_4 \end{bmatrix}=$$
$$det \begin{bmatrix}v_2\\v_1\\v_4\\v_3 \end{bmatrix}=$$
$$det \begin{bmatrix}v_1\\v_2+7v_1\\v_3\\v_4 \end{bmatrix}=$$</p>
<p>I know how to find the determinate of a matrix but do not understand how this works.</p>
| Clayton | 43,239 | <p>Hint: For the first one, use the property that $$\det(A\cdot B)=\det(A)\cdot\det(B).$$See if you can change the identity matrix just right to get the 'new' matrix, where the second row is multiplied by $6$.</p>
<p>For the second one, use a similar idea, but this time the rows of the identity matrix get jumbled up a little bit. Determine which row has a $1$ in the first column, which row has a $1$ in the second column, etc. Now calculate this determinant and multiply it by the determinant of your original matrix.</p>
<p>For the third one, this builds on the previous exercise. Now you can begin with the identity matrix and change one of the zero entries so that when you multiply it to your original matrix, you get $v_2+7v_1$.</p>
|
3,430,008 | <p>Currently stuck on the last part of 15.2.8(e) of this problem:</p>
<p><a href="https://i.stack.imgur.com/UvGJK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UvGJK.png" alt="enter image description here"></a></p>
<p>I don't know how to apply Fubini's theorem since one index relies on the other.</p>
<p>Having slept on it, I've almost got it figured out except for one thing.
Part of Fubini's theorem states (in my book) that if</p>
<p><span class="math-container">$$\sum_{(n,m) \in \mathbb{N} \times \mathbb{N}} f(n,m)$$</span> </p>
<p>converges absolutely to some limit <span class="math-container">$L$</span>, then </p>
<p><span class="math-container">$$\sum_{n=0}^{\infty} \left ( \sum_{m=0}^{\infty} f(n, m)\right )$$</span></p>
<p>also converges absolutely to <span class="math-container">$L$</span>.</p>
<p>What I'm trying to figure out is if the converse of this statement is true.
Because it seems that in order to apply Fubini's theorem to this problem it needs to be.</p>
| Matematleta | 138,929 | <p>I think Rudin's approach is very clear. Following him then, without loss of generality <span class="math-container">$a=0,$</span> so <span class="math-container">$(b-s,b+s)\subseteq (-r,r).$</span> Then, by the binomial theorem,</p>
<p><span class="math-container">$\sum^{\infty}_{n=0}c_nx^n=\sum^{\infty}_{n=0}c_n((x-b)+b)^n=\sum^{\infty}_{n=0}c_n\left[\sum^{n}_{m=0}\binom{n}{m}b^{n-m}(x-b)^m\right].$</span></p>
<p>In this double sum, we are summing the (infinite number of) rows of a lower triangular matrix. We want to change this to a sum over the columns, so that the above becomes </p>
<p><span class="math-container">$\sum^{\infty}_{m=0}\left[\sum^{\infty}_{n=m}\binom{n}{m}c_nb^{n-m}\right](x-b)^m$</span>, and then we may set the bracketed term equal to <span class="math-container">$d_n.$</span></p>
<p>So all that remains is to see that </p>
<p><span class="math-container">$\sum^{\infty}_{n=0}\left|c_n\left[\sum^{n}_{m=0}\binom{n}{m}b^{n-m}(x-b)^m\right]\right|=\sum^{\infty}_{n=0}\sum^{n}_{m=0}\left|c_n\binom{n}{m}b^{n-m}(x-b)^m\right|=$</span></p>
<p><span class="math-container">$\sum^{\infty}_{n=0}|c_n|\cdot (|(x-b)|+|b|)^n$</span> converges by the comparison test, once <span class="math-container">$|(x-b)|+|b|<r.$</span></p>
<p>But if <span class="math-container">$x\in (b-s,b+s),$</span> then <span class="math-container">$|x-b|<s<r-|b|\Rightarrow |(x-b)|+|b|<r,$</span> which completes the proof. </p>
|
4,435,088 | <p>In an <span class="math-container">$8×8$</span> table one of the square is colored black and all the others are white . Prove that one cannot make all the boxes white by recoloring the rows and columns . "Recoloring" is the operation of changing the color of all boxes in a row or in a column .</p>
<p>This is a problem taken from Mathematical Circles by Dimitri Fomin , Sergey Genkin and IIa Itenberg.</p>
<p>My solution goes like this:</p>
<blockquote>
<p>We have a square colored black while all other squares are white. So, in order to make it white we must first recolor its row or its column in which the black colored square is present . Now , if we recolor the row or the column the number black squares will be increased and it's parity will still be odd as we now have <span class="math-container">$7$</span> black squares in that particular row or column . Now, if we again recolor that row or column it will again get reverted back to its initial configuration. So, we must now color each of the row or column in which we have one of those <span class="math-container">$7$</span> black sqaures . But this will also result in an odd number of black squares as we have then <span class="math-container">$7-1+7=13$</span> black squares in total. So, after any transformations we are left with an odd number of black squares. If all the squares are colored black then we have <span class="math-container">$0$</span> black squares . This has an even parity. So, this is not possible to have all squares colored white.</p>
</blockquote>
<p>I want to verify my proof whether it is alright or not? Is this valid? Of course there is a duplicate link about this question but that asks for a different thing. I am asking whether thus proof is valid or not...that link asks probably for a verification of a different proof ...but I want to know whether this proof is valid or not?....</p>
<p>Also if we have a <span class="math-container">$3×3$</span> square like the one given in the figure can we do the do the same thing.<a href="https://i.stack.imgur.com/yfXmxm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yfXmxm.jpg" alt="enter image description here" /></a></p>
<p>Can we solve this using the same above reasoning?</p>
| JMP | 210,189 | <p>The only moves that affect the corner cells are the top/bottom rows and left/right columns.</p>
<p>Neither of these alter the total parity of the corner cells, and as the initial parity is <span class="math-container">$1$</span>, parity <span class="math-container">$0$</span> is unachievable.</p>
|
3,742,456 | <blockquote>
<p>If <span class="math-container">$ABC$</span> is a right angled triangle at <span class="math-container">$C$</span> then prove that <span class="math-container">$a^n+b^n<c^n$</span> for all <span class="math-container">$n>2$</span></p>
</blockquote>
<p>This is an olympiad book problem. I know that <span class="math-container">$a+b>c$</span> and <span class="math-container">$a^2+b^2=c^2$</span>, <span class="math-container">$a^n+b^n≠c^n$</span> for all <span class="math-container">$n>2$</span>, and <span class="math-container">$(a+b)^n>c^n$</span> but I cant find a way to put these together and prove the above, may be I am not hitting up the right approach .......Need help.</p>
| atul ganju | 788,520 | <p>Well, we know <span class="math-container">$a^2+b^2 = c^2$</span> and we know <span class="math-container">$a\lt c$</span> and <span class="math-container">$b\lt c$</span> so we can write this:</p>
<p><span class="math-container">$$a^n+b^n = a^2a^{n-2}+b^2b^{n-2} \leq a^2(\max\{a, b\})^{n-2}+b^2(\max\{a, b\})^{n-2} \lt (a^2+b^2)c^{n-2} = c^n$$</span></p>
|
118,074 | <p>Just based on some reading, I know that every Möbius transformation is a bijection from the Riemann sphere to itself. </p>
<p>I'm curious about the converse. For any holomorphic bijection on the sphere, why is it necessarily a Möbius transformation? Is there a proof or reference of why this converse is true? Thanks.</p>
<p>(I would appreciate an explanation at the level of someone whose just self-studying complex analysis for the first time.)</p>
| M.C. | 752,637 | <p>Here is another proof using results from the theory of Riemann surfaces.</p>
<p>Any holomorphic automorphism <span class="math-container">$F\colon\mathbb{P}^1\rightarrow \mathbb{P}^1$</span> of the Riemann sphere <span class="math-container">$\mathbb{P}^1$</span> can be viewed as a meromorphic function <span class="math-container">$f\colon\mathbb{P}^1\rightarrow \mathbb{C}$</span> in the obvious way. This works since, by the Identity Theorem, the fiber <span class="math-container">$F^{-1}(\{\infty\})$</span> is discrete. (Since <span class="math-container">$\mathbb{P}^1$</span> is compact, the fiber is even finite.)</p>
<p>Now, every meromorphic function on <span class="math-container">$\mathbb{P}^1$</span> is rational, i.e. there are complex univariate polynomials <span class="math-container">$g,h$</span> with <span class="math-container">$f=\frac{g}{h}$</span>. A proof can be found after Corollary 2.9. of Forster's <em>Lectures on Riemann Surfaces</em>. For completeness, let me recall Forster's proof:</p>
<blockquote>
<p>Note that, by the Identity Theorem, <span class="math-container">$f$</span> has only finitely many poles and zeroes. Without loss of generality, assume that <span class="math-container">$\infty$</span> is not a pole of <span class="math-container">$f$</span>. Otherwise, consider <span class="math-container">$\frac{1}{f}$</span>. Let <span class="math-container">$a_1,\ldots,a_n\in\mathbb{C}$</span> denote the poles of <span class="math-container">$f$</span>. For each <span class="math-container">$\nu=1,\ldots,n$</span>, consider
<span class="math-container">$$h_\nu(z)=\sum_{j=-k_\nu}^{-1}c_{\nu j}(z-a_\nu)^j$$</span>
the principal part of the Laurent expansion of <span class="math-container">$f$</span> around the pole <span class="math-container">$a_\nu$</span>. Then the function <span class="math-container">$f-(h_1+\ldots +h_n)$</span> is holomorphic on <span class="math-container">$\mathbb{P}^1$</span>. Now, assume for a contradiction that the function <span class="math-container">$f-(h_1+\ldots +h_n)$</span> were non-constant. Then it would be open by the Open Mapping Theorem. Since <span class="math-container">$\mathbb{P}^1$</span> is compact, the function would hence be surjective. Consequently, <span class="math-container">$\mathbb{C}$</span> would be compact which is absurd. It follows that <span class="math-container">$f$</span> is rational.</p>
</blockquote>
<p>Since a rational function <span class="math-container">$\frac{g}{h}$</span> is bijective only if <span class="math-container">$g$</span> and <span class="math-container">$h$</span> are linear (see <a href="https://math.stackexchange.com/questions/3389265/prove-that-injective-rational-function-on-complex-numbers-are-mobius-transformat?noredirect=1&lq=1">here</a>), we have that <span class="math-container">$f$</span>, and thus <span class="math-container">$F$</span>, is of the form <span class="math-container">$z \mapsto \frac{az+b}{cz+d}$</span> with <span class="math-container">$a,b,c,d\in \mathbb{C}$</span>. Such a mapping is bijective if and only if <span class="math-container">$ad-bc\neq 0$</span>.</p>
|
2,732,562 | <p>I'm trying to figure out how to find the number of ternary strings of length $n$ that have 3 or more consecutive 2's.
So far I've been able to establish that there is $n(2^{n-1})$ with a single 2.
And I think (but am not certain) that this can be extrapolated to the number of strings with a single group of 2's of length $x$ by:
$$\bigl(n-(x-1)\bigr)(2^{n-x})$$
What I'm getting caught on is the 'or more part', any help would be greatly appreciated.</p>
| Christian Blatter | 1,303 | <p>Denote by $c_n$ the number of ternary words containing no $222$. Then $c_1=3$, $c_2=9$, $c_3=26$, and we have the recursion
$$c_n=2c_{n-1}+2c_{n-2}+2c_{n-3}\qquad(n\geq4)\ .\tag{1}$$
<em>Proof.</em> An admissible word $w$ can begin with $Z$, $2Z$, or $22Z$ with $Z\in\{0,1\}$, followed by an admissible word $w'$.</p>
<p>Unfortunately the characteristic equation of $(1)$ has no rational roots, hence it remains to list the first few values:
$$(c_n)_{n\geq1}=(3, 9, 26, 76, 222, 648, 1892, 5524, 16128, 47088, 137480, 401392,\ldots)\ .$$
The values $a_n=3^n-c_n$ required by the OP then are
$$(a_n)_{n\geq1}=(0, 0, 1, 5, 21, 81, 295, 1037, 3555, 11961, 39667, 130049,\ldots)\ .$$</p>
|
2,011,003 | <p>I stumbled upon this logic question in a math class recently. </p>
<p>My teacher told us that a statement that is not tested/is empty is true. For example, that if I stated that: "if the team A wins the game, I am gonna buy you a coke", and then team B goes on and wins the game, the statement would be true, independent of me buying a coke. Could anybody elaborate how this can be the case, and why?</p>
<p>It came up as an explanation to why the the empty-set is both an open and a closed set. </p>
| PMar | 388,671 | <p>The short answer is, this is true because we have adopted that resolution by convention, because, for all its faults, doing so comes closest to resembling the intuitive notion of 'if-then' while having a definite logical value in all mathematical situations. To see what I am getting at, consider the following 'truth table':</p>
<pre><code>A | B | A-->B
T | T | T
T | F | F
F | T | *
F | F | *
</code></pre>
<p>This illustrates the truth about our intuitive notion of 'if-then' - When A is true, B has to be true, <em>and when A is false we don't care</em>. But mathematical logic doesn't allow 'we don't care', so we have to replace the '*'-s with definite truth-values. Then the question becomes: how do we do so and still match our intuition? There are clearly four possible resolutions:</p>
<pre><code>A | B | 1 | 2 | 3 | 4
T | T | T | T | T | T
T | F | F | F | F | F
F | T | F | F | T | T
F | F | F | T | F | T
</code></pre>
<p>If we choose (1) to mean 'if-then', we are saying 'if A is true then B is true' means 'A and B are both true', and we are also implying 'if B is true then A is true'. But this is clearly not what we want: When we have both 'A-->B' and 'C-->B', this would be saying 'A, B, and C' are all true together'; this doesn't work because sometimes we need A and C to be mutually-exclusive <em>(typical example: a number-theory proof-by-cases, where A means 'n is even' and C means 'n is odd')</em>. A similar problem occurs if we choose (2), which says 'A and B are both true -or- both false'. If we choose (3) to mean 'if-then' we are saying 'if A is true then B is true' means merely 'B is true' - REGARDLESS of A. Clearly that also doesn't work - generally we want B to depend on A <em>somehow</em>. So all that is left is (4).</p>
|
1,095,621 | <p>I am looking for a way to integrate $$\int \sqrt{x^2-4}\ dx $$ using trigonometric substitutions. </p>
<p>All my attempts so far lead to complicated solutions that were uncomputable.</p>
| Git Gud | 55,235 | <p>I'll focus on $\displaystyle \int \sqrt{x^2-1}\,\mathrm dx$ for simplicity. This is how I think about this sort of problem when I don't remember the appropriate substitution. </p>
<p>If you suspect that it's useful to use a trigonometric substitution, since the fundamental property $$\forall \theta \in\mathbb R\left((\sin(\theta))^2+(\cos(\theta))^2=1\right)$$ involves only squared numbers, the substitution $x=\varphi(\theta)$ has to be such that $(\varphi(\theta))^2-1$ is a square as opposed to the negative of a square (because of the square root).</p>
<p>So now the goal is to obtain something that looks like $(\varphi(\theta))^2-1$ from $(\sin(\theta))^2+(\cos(\theta))^2=1$.</p>
<p>Clearly something like $(\sin(\theta))^2-1=-(\cos(\theta))^2$ won't help because the RHS is the negative of a square. </p>
<p>The next usual trick with this trigonometric identities is to divide by $\cos$ yielding $(\tan(\theta))^2+1=(\sec(\theta))^2$ from where one gets $(\tan(\theta))^2=(\sec(\theta))^2-1$ which satisfies the criteria above.</p>
<p>Consider the substitution $x=\sec(\theta)$. It will result, after a few calculations, in known antiderivatives.</p>
|
1,095,621 | <p>I am looking for a way to integrate $$\int \sqrt{x^2-4}\ dx $$ using trigonometric substitutions. </p>
<p>All my attempts so far lead to complicated solutions that were uncomputable.</p>
| Dylan | 135,643 | <p>Using the substitution $x = \sec \theta$, we end up with</p>
<p>$$ 4\int \sec \theta \tan^2 \theta \,d\theta $$</p>
<p>This can be done using integration by parts</p>
<p>$$ \begin{align} \int \sec \theta \tan^2 \theta \,d\theta &= \int \tan \theta \,d(\sec \theta)
\\&= \tan \theta \sec \theta - \int \sec\theta \,d(\tan\theta)
\\&= \tan \theta \sec \theta - \int \sec\theta \,(\tan^2 \theta + 1) \,d\theta
\\&= \tan \theta \sec \theta - \int \sec \theta \tan^2 \theta \,d\theta - \int \sec\theta \,d\theta
\\ 2\int \sec \theta \tan^2 \theta \,d\theta &= \tan \theta \sec \theta - \int \sec\theta\,d\theta
\end{align}$$
For reference: <a href="http://en.wikipedia.org/wiki/Integral_of_secant_cubed" rel="nofollow">Integral of secant cubed</a></p>
|
3,054,898 | <h3>Problem</h3>
<p>Evaluate <span class="math-container">$$\int_0^{2\pi}(t-\sin t)(1-\cos t)^2{\rm d}t.$$</span></p>
<h3>Comment</h3>
<p>It's very complicated to compute the integral applying normal method. I obtain the result resorting to the skillful formula</p>
<blockquote>
<p><span class="math-container">$$\int_0^{2\pi}xf(\cos x){\rm d}x=\pi\int_0^{2\pi}f(\sin x){\rm d}x,$$</span>
where <span class="math-container">$f(x) \in C[-1,1].$</span></p>
</blockquote>
<p><span class="math-container">\begin{align*}
\require{begingroup}
\begingroup
\newcommand{\dd}{\;{\rm d}}\int_0^{2\pi} (t-\sin t)(1-\cos t)^2 \dd t
&= \int_0^{2\pi} t(1-\cos t)^2 \dd t - \int_0^{2\pi} \sin t(1-\cos t)^2 \dd t \\
&= \pi\int_0^{2\pi} (1-\sin t)^2 \dd t - \int_0^{2\pi} (1-\cos t)^2 \dd (1-\cos t) \\
&= \pi\int_0^{2\pi} \left(\frac32-\frac12\cos2t-2\sin t\right) \dd t - \left[\frac13(1-\cos t)^3\right]_0^{2\pi}\\
&= \pi\left[\frac32t-\frac14\sin2t+2\cos t\right]_0^{2\pi}\\
&= 3\pi^2
\endgroup
\end{align*}</span></p>
<p><strong>But any other solution?</strong></p>
| mechanodroid | 144,766 | <p>Notice that your integral is in fact an area integral of the function <span class="math-container">$(x,y) \mapsto x$</span> over one arch of a cycloid given by
<span class="math-container">$$\begin{cases} x = t - \sin t \\ y = 1-\cos t\end{cases}$$</span>
Indeed, if we assume that the cycloid is explicitly given by <span class="math-container">$y = f(x)$</span>, we have
<span class="math-container">\begin{align}
\int_{\text{cycloid}} x \,dA &= \int_0^{2\pi} \int_0^{f(x)} x\,dy\,dx\\
&= \int_0^{2\pi} xf(x)\,dx \\
&= \begin{bmatrix} x = t-\sin t \\ f(x) = 1-\cos t \\ dx = (1-\cos t)\,dt\end{bmatrix}\\
&= \int_0^{2\pi} (t-\sin t)(1-\cos t)^2\,dt
\end{align}</span></p>
<p>Now recall the formula for the <span class="math-container">$x$</span>-coordinate of the centroid:
<span class="math-container">$$\frac1A \int_{\text{cycloid}} x \,dA = x\text{-coordinate of the centroid} = \pi$$</span></p>
<p>since centroid is clearly at <span class="math-container">$x = \pi$</span> by symmetry.</p>
<p>Using the same metrod as above area of the cycloid is <span class="math-container">$$A = \int_{\text{cycloid}}dA = \int_0^{2\pi} (1-\cos t)^2 \,dt = 3\pi$$</span></p>
<p>this integral being a lot easier than the original one.</p>
<p>It follows <span class="math-container">$$\int_{\text{cycloid}} x \,dA = A\pi = 3\pi^2$$</span></p>
|
3,054,898 | <h3>Problem</h3>
<p>Evaluate <span class="math-container">$$\int_0^{2\pi}(t-\sin t)(1-\cos t)^2{\rm d}t.$$</span></p>
<h3>Comment</h3>
<p>It's very complicated to compute the integral applying normal method. I obtain the result resorting to the skillful formula</p>
<blockquote>
<p><span class="math-container">$$\int_0^{2\pi}xf(\cos x){\rm d}x=\pi\int_0^{2\pi}f(\sin x){\rm d}x,$$</span>
where <span class="math-container">$f(x) \in C[-1,1].$</span></p>
</blockquote>
<p><span class="math-container">\begin{align*}
\require{begingroup}
\begingroup
\newcommand{\dd}{\;{\rm d}}\int_0^{2\pi} (t-\sin t)(1-\cos t)^2 \dd t
&= \int_0^{2\pi} t(1-\cos t)^2 \dd t - \int_0^{2\pi} \sin t(1-\cos t)^2 \dd t \\
&= \pi\int_0^{2\pi} (1-\sin t)^2 \dd t - \int_0^{2\pi} (1-\cos t)^2 \dd (1-\cos t) \\
&= \pi\int_0^{2\pi} \left(\frac32-\frac12\cos2t-2\sin t\right) \dd t - \left[\frac13(1-\cos t)^3\right]_0^{2\pi}\\
&= \pi\left[\frac32t-\frac14\sin2t+2\cos t\right]_0^{2\pi}\\
&= 3\pi^2
\endgroup
\end{align*}</span></p>
<p><strong>But any other solution?</strong></p>
| Nosrati | 108,128 | <p>With substitution <span class="math-container">$t=\pi+x$</span> we see that
<span class="math-container">$$I=\int_0^{2\pi}(t-\sin t)(1-\cos t)^2{\rm d}t=\int_{-\pi}^{\pi}(\pi+x+\sin x)(1+\cos x)^2{\rm d}x.$$</span>
the part <span class="math-container">$(x+\sin x)(1+\cos x)^2$</span> is an odd function so it's integral over <span class="math-container">$[-\pi,\pi]$</span> is zero, then
<span class="math-container">$$I=\pi\int_{-\pi}^{\pi}(1+\cos x)^2{\rm d}x=\pi\int_{-\pi}^{\pi}\dfrac32+2\cos x+\dfrac12\cos2x{\rm d}x=\color{blue}{3\pi^2}$$</span></p>
|
2,130,807 | <p>How to Prove $G$ is connected, if $G$ is an acyclic graph on $n \ge 1$ vertices containing exactly $n − 1$ edges?</p>
| Smylic | 100,361 | <p>Induction on $n$ is a good option. Hint:</p>
<blockquote class="spoiler">
<p> If $|G| \ge 2$ and $G$ is acyclic and $G$ has at least one edge then $G$ has a vertex of degree 1.</p>
</blockquote>
|
3,267,499 | <p>Let <span class="math-container">$k$</span> be a field.
<span class="math-container">$k[x,y]$</span> is a UFD by the following known argument taken from <a href="https://en.wikipedia.org/wiki/Unique_factorization_domain" rel="nofollow noreferrer">wikipedia</a>:
"If <span class="math-container">$R$</span> is a UFD, then so is <span class="math-container">$R[X]$</span>, the ring of polynomials with coefficients in <span class="math-container">$R$</span>. Unless <span class="math-container">$R$</span> is a field, <span class="math-container">$R[X]$</span> is not a principal ideal domain. By induction, a polynomial ring in any number of variables over any UFD (and in particular over a field) is a UFD". </p>
<p><strong>(1)</strong> It seems that <span class="math-container">$k[x,x^{-1},y]$</span> is a UFD, isn't it?
Is the proof based on the result that <span class="math-container">$k[x,y]$</span> is a UFD? (probably yes?)</p>
<p><strong>(2)</strong> Can one find all irreducible=prime elements of <span class="math-container">$k[x,x^{-1},y]$</span>?
("In UFDs, every irreducible element is prime").</p>
<p>Thank you very much! </p>
| Con | 682,304 | <p>1) Localization of a UFD is again a UFD (at least if you are not inverting <span class="math-container">$0$</span>). Maybe you can try to prove that. </p>
<p>2) Is there any condition on <span class="math-container">$k$</span>?</p>
|
4,650,979 | <blockquote>
<p>Solve the following recurrence equation: <span class="math-container">$T(n) = T(n-2)+n^2$</span>, having <span class="math-container">$T(0)=1$</span>, <span class="math-container">$T(1)=5$</span>.</p>
</blockquote>
<p>I need to solve this equation but when I get to the particular solution with <span class="math-container">$n^2$</span> some of the terms I need cancel out and it's kind of impossible to find the constants at that point.</p>
<p>Here is the way I'm doing it, using <span class="math-container">$an^2+bn+c$</span> to replace <span class="math-container">$T(n)$</span>;
<span class="math-container">$$\begin{split}
0&=an^2+bn+c - a(n-2)^2-b(n-2)-c-n^2\\
&=an^2+bn- a(n-2)^2-b(n-2)+n^2\\
&=an^2+bn- a(n^2-4n+4)-b(n-2)-n^2\\
&=an^2+bn- an^2+4an-4a-bn-2b-n^2\\ &=4an-4a-2b-n^2\\
&=n(4a-n)+(-4a+2b)\\
\end{split}$$</span>
But beyond that, I cannot find any way to find the constants.</p>
| Alexander Burstein | 499,816 | <p>Since <span class="math-container">$n^2=2\binom{n}{2}+\binom{n}{1}$</span>, it follows from the hockey-stick identity that
<span class="math-container">$$
\begin{split}
\sum_{k=0}^{n}{k^2}&=2\binom{n+1}{3}+\binom{n+1}{2}=\frac{(n+1)n(n-1)}{3}+\frac{(n+1)n}{2}\\
&=\frac{(n+1)n}{6}(2(n-1)+3)=\frac{n(n+1)(2n+1)}{6}.
\end{split}
$$</span>
This identity lets us find the values <span class="math-container">$T(n)$</span> separately for even <span class="math-container">$n$</span> and for odd <span class="math-container">$n$</span>.</p>
<p>For even <span class="math-container">$n$</span>:
<span class="math-container">$$
\begin{split}
T(2n)&=T(0)+\sum_{k=1}^{n}(2k)^2=1+4\sum_{k=1}^{n}k^2=1+\frac{4n(n+1)(2n+1)}{6}\\
&=\frac{4n^3+6n^2+2n+3}{3}=\frac{(2n+3)(2n^2+1)}{3}.
\end{split}
$$</span>
For odd <span class="math-container">$n$</span>:
<span class="math-container">$$
\begin{split}
T(2n+1)&=T(1)+\sum_{k=1}^{n}(2k+1)^2=5+\sum_{k=1}^{2n+1}k^2-1^2-\sum_{k=1}^{n}(2k)^2=\\
&=4+\frac{(2n+1)(2n+2)(4n+3)}{6}-\frac{4n(n+1)(2n+1)}{6}=\\
&=4+\frac{(2n+1)(n+1)}{3}((4n+3)-2n)=4+\frac{(n+1)(2n+1)(2n+3)}{3}\\
&=\frac{4n^3+12n^2+11n+15}{3}=\frac{(2n+5)(2n^2+n+3)}{3}.
\end{split}
$$</span></p>
|
1,664,081 | <p>Solve the equation $2^x - 3^{x-1}=-(x+2)^2$</p>
<p>How I got this question? I created this question so I know the answer. The answer is 5. But I have no idea how to solve it. Take note that I cannot do logarithm, guess and check and modulus. Does anybody know how to solve this? I have no idea how to start.</p>
<p>$2^x - 3^{x-1}$ is infactorisable. Even if I did $2^x - (2+2^0)^{x-1}$, it is STILL infactorisable. I was hoping to solve by hand.</p>
<p>So, I do I solve the equation?</p>
| SS_C4 | 242,290 | <p>The only solution is $x=5$.</p>
<p>You can see this by graphing the LHS and RHS separately and seeing that there is only one point of intersection.
<a href="https://i.stack.imgur.com/1JJWv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1JJWv.png" alt="Graph"></a></p>
|
1,664,081 | <p>Solve the equation $2^x - 3^{x-1}=-(x+2)^2$</p>
<p>How I got this question? I created this question so I know the answer. The answer is 5. But I have no idea how to solve it. Take note that I cannot do logarithm, guess and check and modulus. Does anybody know how to solve this? I have no idea how to start.</p>
<p>$2^x - 3^{x-1}$ is infactorisable. Even if I did $2^x - (2+2^0)^{x-1}$, it is STILL infactorisable. I was hoping to solve by hand.</p>
<p>So, I do I solve the equation?</p>
| Piquito | 219,998 | <p>HINT.-You have $$3^{x-1}-(x+2)^2=2^x$$ Searching for an eventual rational solution one has $$(3^{\frac{x-1}{2}}+x+2)(3^{\frac{x-1}{2}}-x-2)=2^x$$
We do now $$3^{\frac{x-1}{2}}+x+2=2^{x-h}$$ $$3^{\frac{x-1}{2}}-x-2=2^h$$ Hence
$$2x+4=2^{x-h}-2^h\iff x+2=2^{x-h-1}-2^{h-1}\qquad (*)$$ Trying with $h=1$ we find out the easy $$x+2=2^{x-2}-1$$ where one can see the integer solution $x=5$.</p>
<p>For values $h\gt 1$ we have in $(*)$ that $x$ is necessarily even excepting when $x=h+1$. In the first case this contradicts the original equation. I stop here. I wanted just to show the value $x=5$ can be deduced.</p>
|
222,555 | <p>I would like to find a simple equivalent of:</p>
<p>$$ u_{n}=\frac{1}{n!}\int_0^1 (\arcsin x)^n \mathrm dx $$</p>
<p>We have:</p>
<p>$$ 0\leq u_{n}\leq \frac{1}{n!}\left(\frac{\pi}{2}\right)^n \rightarrow0$$</p>
<p>So $$ u_{n} \rightarrow 0$$</p>
<p>Clearly:</p>
<p>$$ u_{n} \sim \frac{1}{n!} \int_{\sin(1)}^1 (\arcsin x)^n \mathrm dx $$</p>
<p>But is there a simpler equivalent for $u_{n}$?</p>
<p>Using integration by part:</p>
<p>$$ \int_0^1 (\arcsin x)^n \mathrm dx = \left(\frac{\pi}{2}\right)^n - n\int_0^1 \frac{x(\arcsin x)^{n-1}}{\sqrt{1-x^2}} \mathrm dx$$</p>
<p>But the relation </p>
<p>$$ u_{n} \sim \frac{1}{n!} \left(\frac{\pi}{2}\right)^n$$</p>
<p>seems to be wrong...</p>
| WimC | 25,313 | <p>Substitute $x = \sin(t)$ to get</p>
<p>$$
u_n = \frac{1}{n!}\int_0^{\pi/2} t^n \cos(t) dt.
$$</p>
<p>The following equalities hold for all integers $m \geq 0$ (which can be checked by partial integration):</p>
<p>$$
\frac{1}{n!}\int_0^{\pi/2} t^n \left(\tfrac{\pi}{2} - t \right)^m dt = \frac{m!}{(n + m + 1)!}\left(\frac{\pi}{2}\right)^{n + m + 1}.
$$</p>
<p>Then it follows from the Taylor expansion of $\cos(t)$ around $t = \pi/2$</p>
<p>$$
\cos(t) = \sin \left(\tfrac{\pi}{2} - t \right) = \sum_{m=0}^{\infty}\frac{(-1)^m}{(2m + 1)!}\left(\tfrac{\pi}{2} - t\right)^{2m+1}
$$
that</p>
<p>$$
u_n = \sum_{m=0}^{\infty} \frac{(-1)^m}{(n+2m+2)!}\left(\frac{\pi}{2}\right)^{n + 2m + 2}.
$$
The growth in $n$ is determined by the first term ($m=0$) so
$$
u_n \sim \frac{1}{(n+2)!}\left(\frac{\pi}{2}\right)^{n + 2}.
$$</p>
|
2,698,121 | <p>Came across a question about CDF and PDF in my homework:</p>
<p><a href="https://i.stack.imgur.com/bYqvr.jpg" rel="nofollow noreferrer">Click here for the question</a></p>
<p><a href="https://i.stack.imgur.com/bYqvr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bYqvr.jpg" alt="enter image description here"></a></p>
<p>a. What is the cumulative distribution function?</p>
<p>Knowing that pdf is the derivative of cdf, I integrated the piecewise function to: 0 if x < 0, (3/4)x if 0 ≤ x ≤ 1, 3/4 + 1/4(x-3) if 3 ≤ x ≤ 4, and 1 for 4 < x. However, I am given that the answer to 1 ≤ x ≤ 3 is 3/4, but I do not know how to find it. </p>
<p>b. What is P(X>1)?</p>
<p>I know that for P(X < u) I will sub in the value of u into the cdf function and find the answer. However, I don't think the same will work for P(X > u). Do I need to integrate the pdf from 0 to 2? </p>
<p>c. What is E(X)?</p>
<p>I know that E(X) is the integral of xf(x) from negative infinity to infinity, but since this is a piecewise function, do I need to take the integral of each interval and find their sum?</p>
<p>Sorry for asking so many questions at once, I am really lost in the whole CDF and PDF idea.</p>
<p>Thanks!</p>
| Dave | 334,366 | <p>I'll get you going on the CDF. You're correct that the derivative of the CDF is the PDF. By definition, the CDF (denote it $F(x)$) is $F(x):=P(X<x)=\int_{-\infty}^xf(x)~dx$, where $f(x)$ is the PDF. The PDF has a piecewise definition, and the support is cut up into sections. To compute the CDF we need to examine these sections in turn. </p>
<p>The first section is $x<0$: in this case, $f(x)=0$ so $F(x)=0$. </p>
<p>Now we consider $0\leq x\leq 1$: in this case, $f(x)=\frac{3}{4}$, so $$F(x)=\int_0^x\frac{3}{4}~dx$$</p>
<p>Now we consider $1<x<3$: in this case, $f(x)=0$, but we have to account for the stuff we already integrated over, so $$F(x)=\int_0^1\frac{3}{4}~dx+\int_1^x0~dx=\int_0^1\frac{3}{4}~dx$$</p>
<p>Now we consider $3\leq x\leq 4$: in this case, $f(x)=\frac{1}{4}$, so accounting for what we've already integrated over we get $$F(x)=\int_0^1\frac{3}{4}~dx+\int_1^30~dx+\int_3^x\frac{1}{4}~dx=\int_0^1\frac{3}{4}~dx+\int_3^x\frac{1}{4}~dx$$</p>
<p>Lastly, for $x>4$ we have integrated out the whole support of $f(x)$, so $F(x)=1$.</p>
<hr>
<p>I will add some suggestions for the other parts.</p>
<p>For (b), to find $P(x>1)$, this is equal to $1-P(x\leq 1)=1-F(1)$. Do you see why?</p>
<p>For (c), you will need to do two integrals essentially for each segment of the support. So yes, you can integrate $xf(x)$ for $0\leq x\leq 1$ and add this to the integral of $xf(x)$ for $3\leq x\leq 4$ to obtain $E(X)$.</p>
|
1,089,635 | <blockquote>
<p>Is it possible to display a result in Mathcad as a function of $\pi$? </p>
</blockquote>
<p>I'm studying physics and I need to show exact results at the exams. I know I can set Mathcad to give me the result in decimals or in fractions, but none of them are good enough. </p>
<p><strong>Example: calculating 400/200 $\pi$</strong></p>
<p>What mathcad can give me: $(400/200) \cdot \pi = 6.283$ or $(400/200) \cdot \pi = 10838702/1725033 $</p>
<p>What I want: $(400/200) \cdot \pi = 2 \pi$</p>
| Antonio Velazquez Bustamante | 250,223 | <p>If you want to see results as an expression of something else, then put that at the end of the result (in this case, after 6.283). In that way you can express results in different units or if you want to see it in PIs, then inset a pi symbol (p and alt+g) at the end. If the result is not shown properly, then highlight pi and set it as a constant. In that way it will turn green and the result will be accurate.</p>
<p><img src="https://i.stack.imgur.com/m1lii.jpg" alt="enter image description here"></p>
|
1,299,266 | <p>How many zeros are there in the number $50!$?</p>
<p>My attempt:</p>
<p>The zeros in every number come from the 10s that make up the number. The 10s are, in turn, made up of 2s and 5s.</p>
<p>So: $\frac{50}{5*2} = 5$ zeros?</p>
| David Quinn | 187,299 | <p>In answer to your specific question there are not only 12 zeros at the end of the number but an additional 7 zeros occurring amongst the other non-zero digits</p>
|
1,379,513 | <p>A hot dog stand has 12 different toppings available. How many different kinds of hot dogs can be made, assuming the order of the toppings does not make a difference. I believe the correct answer is 882050, with the maximum varieties per number of toppings selected being 665280 when there six toppings. I am also not sure about how to create a formula that would arrive at this result.</p>
| user176717 | 257,988 | <p>It's not necessarily a yes or no question on each topping.</p>
<p>For each topping, I would make 4 different 'states'</p>
<ul>
<li>none</li>
<li>regular</li>
<li>extra</li>
<li>easy</li>
</ul>
<p>$4^{12}=16,777,216$ individual hot dogs, which would definitely not fit in the stand. You could refine it more to have different standardized volumes of topping, but then you'd just get ridiculous. You could further confuse the issue by counting the bun as being a topping, because you wouldn't really have easy or extra bun, hun.</p>
|
3,191,402 | <p>I have tried to answer by taking change the variable <span class="math-container">$\theta$</span> to <span class="math-container">$\theta/2$</span>, so the integration is now over unit circle, then I have taken <span class="math-container">$z=e^{i\theta}$</span>. Now I tried to use residue formula for integration, but I failed.</p>
| Peter Foreman | 631,494 | <p>This method solves the integral when <span class="math-container">$n\in\mathbb{N}$</span> and <span class="math-container">$a\in\mathbb{C}$</span>.</p>
<p>By using binomial expansion the integral becomes
<span class="math-container">$$\int_0^\pi\sum_{k=0}^n\left( \binom{n}{k} a^{n-k} \cos^k{(\theta)} \right)d\theta=\sum_{k=0}^n\left( \binom{n}{k} a^{n-k}\int_0^\pi \cos^k{(\theta)} d\theta\right)$$</span>
Now we can use the <a href="https://en.wikipedia.org/wiki/Beta_function" rel="nofollow noreferrer">Beta function</a> in order to find the latter integral
<span class="math-container">$$B(x,y)=\frac{(x-1)!(y-1)!}{(x+y-1)!}=2\int_0^\frac\pi2 \sin^{2x-1}{(\theta)}\cos^{2y-1}{(\theta)}d\theta$$</span>
<span class="math-container">$$\int_0^\pi \cos^k{(\theta)} d\theta=\begin{cases}
0&k\,\text{ odd}\\
B\left(\frac12,\frac{k+1}2\right)&\text{otherwise}
\end{cases}$$</span>
Now we just have to simplify
<span class="math-container">$$\begin{align}
B\left(\frac12,\frac{k+1}2\right)
&=\frac{(-\frac12)!(\frac{k}{2}-\frac12)!}{(\frac{k}2)!}\\
&=\frac{(-\frac12)!(\frac{k}{2}-\frac12)(\frac{k}{2}-\frac32)\dots(\frac32)(\frac12)(-\frac12)!}{(\frac{k}2)!}\\
&=\frac{((-\frac12)!)^2(k-1)(k-3)\dots(3)(1)}{2^{\frac{k}2}(\frac{k}2)!}\\
&=\frac{((-\frac12)!)^2(k-1)(k-3)\dots(3)(1)}{2^{\frac{k}2}(\frac{k}2)(\frac{k}2-1)\dots(2)(1)}\\
&=\frac{((-\frac12)!)^2(k-1)(k-3)\dots(3)(1)}{(k)(k-2)\dots(4)(2)}\\
&=\frac{((-\frac12)!)^2(k-1)!!}{k!!}\\
\end{align}$$</span>
Where <span class="math-container">$k!!$</span> denotes the <a href="https://en.wikipedia.org/wiki/Double_factorial" rel="nofollow noreferrer">Double factorial</a>. The value of <span class="math-container">$(\frac12)!$</span> is well known as
<span class="math-container">$$\left(\frac12\right)!=\frac12\left(-\frac12\right)!=\frac{\sqrt{\pi}}2\implies\left(-\frac12\right)!=\sqrt{\pi}$$</span>
Hence our integral is finally given by
<span class="math-container">$$\int_0^\pi \cos^k{(\theta)} d\theta=\begin{cases}
0&k\,\text{ odd}\\
\frac{(k-1)!!}{k!!}\pi&\text{otherwise}
\end{cases}$$</span>
This gives the solution to the original integral as
<span class="math-container">$$\boxed{\large{\int_{0}^{\pi}(a+\cos{\theta})^nd\theta=\pi a^n\sum_{k=0}^{\lfloor\frac{n}2\rfloor} \left(\binom{n}{2k} \frac{(2k-1)!!}{a^{2k}(2k)!!}\right)}}$$</span></p>
|
4,315,283 | <p>Let <span class="math-container">$X$</span> be a continuous random variable, having pdf <span class="math-container">$f(x)$</span> and let be <span class="math-container">$Y=f(X)$</span>.
I have already proved these two following results:</p>
<ol>
<li>If <span class="math-container">$f(x)$</span> is strictly increasing, <span class="math-container">$Y$</span> has the same distribution as <span class="math-container">$X$</span> if and only if</li>
</ol>
<p><span class="math-container">$$
f(x)=
\begin{cases}
x, \qquad x \in \,[0, \sqrt{2}] \\
0, \qquad otherwise
\end{cases}
$$</span></p>
<ol start="2">
<li>If <span class="math-container">$f(x)$</span> is strictly decreasing, <span class="math-container">$Y$</span> has the same distribution as <span class="math-container">$X$</span> if and only if</li>
</ol>
<p><span class="math-container">$$
f(x)=
\begin{cases}
\frac{a^{2}}{x}, \qquad x \in \,[ae^{-\frac{1}{2a^2}}, ae^{\frac{1}{2a^2}}] \\
0, \qquad otherwise
\end{cases}
$$</span></p>
<p>Now, I should investigate about the random variable that has pdf partially increasing and partially decreasing:</p>
<p><span class="math-container">$$
f(x)=
\begin{cases}
x, \qquad x \in [0,1] \\
\frac{1}{x}, \qquad x \in \,[1, e^{\frac{1}{2}}]
\end{cases}
$$</span></p>
<p>What can I say about the distribution of the random variable having aforementioned pdf? Do the last pdf have some simmetry (<span class="math-container">$\frac{1}{x}$</span> is the reciprocal of <span class="math-container">$x$</span>)?</p>
| Siong Thye Goh | 306,553 | <p><span class="math-container">$Y\in [0,1]$</span>,</p>
<p><span class="math-container">$$Y=\begin{cases} X, & X \in [0,1] \\ \frac1X, & X \in [1, e^\frac12]\end{cases}$$</span></p>
<p>If <span class="math-container">$y \in [0, e^{-\frac12}],$</span></p>
<p><span class="math-container">\begin{align}
P(Y \le y) &= P(X \le y)\\
&= \int_0^y x \, dx
\end{align}</span></p>
<p><span class="math-container">$$f_Y(y) = y$$</span>
If <span class="math-container">$y \in [e^{-\frac12}, 1]$</span>,</p>
<p><span class="math-container">\begin{align}
P(Y\le y) &= \int_0^y f(x) \, dx + \int_{\frac1y}^{e^\frac12} f(x) \, dx\\
&=\int_0^yx \, dx + \int_{\frac1y}^{e^{\frac12}} \frac1x \, dx
\end{align}</span></p>
<p><span class="math-container">$$f_Y(y) = y-y\left( -\frac1{y^2}\right)=y+\frac1y$$</span></p>
<p>In summary,</p>
<p><span class="math-container">$$f_Y(y) = \begin{cases} y, & y\in [0, e^{-0.5}]\\ y + \frac1y , & y \in [e^{-0.5},1] \\ 0, & \text{otherwise} \end{cases}$$</span></p>
|
2,725,317 | <blockquote>
<p>For which $p,q\in \mathbb R$ is the following system stable?$$\frac{\mathrm dx}{\mathrm dt} = \begin{bmatrix} p & -q \\ q & p \end{bmatrix}x(t)$$ </p>
</blockquote>
<p>If I'm correct about this, isn't it just when the eigenvalues are $< 1?$ Or is there something more fancy to it? Any and all help appreciated.</p>
| Lutz Lehmann | 115,115 | <p>No, the condition you think of is for discrete-time dynamical systems. </p>
<p>Here in the continuous-time case you need that the eigenvalues $p\pm i\,q$ have non-positive real part. You need an extra test for the purely imaginary case, as that can not in general be decided from the eigenvalues alone.</p>
|
1,156,907 | <p>I don't know anything about measure theory, I'm studying real analysis and this showed up in the book I'm reading as a way to characterize integrable functions. The author defined that a subset $X \subset \mathbb{R}$ has measure zero if for each $\epsilon > 0$ we can find infinitely countable open intervals $I_n$ such that $X \subset \bigcup_{n=1}^{\infty}I_n$ and $\sum_{n=1}^{\infty} |I_n| < \epsilon $ where $|I|$ is the size of $I$, as in, if $I = (a,b)$, then $|I| = b - a$.</p>
<p>Now, the author gives the following proof that the countable union of measure-zero sets has measure zero:</p>
<p>"Let $Y =\bigcup_{i=1}^{\infty} X_i $, where each $X_i$ has measure zero. Now, given $\epsilon > 0 $ we can, for each $n$, write $X_n \subset \bigcup_{i=1}^{\infty} I_{n_i}$ where each $I_{n_i}$ is an open subset and $\sum_{i=1}^{\infty}|I_{n_i}| < \epsilon / 2^n$. Therefore, $Y \subset \bigcup_{n,j=1}^{\infty} I_{n_J}$ where $\sum_n \sum_j |I_{n_J}| < \sum_{i=1}^{\infty} \epsilon /2^n = \epsilon$. Therefore, $m(Y) = 0$"</p>
<p>I'm really confused about the ending. It is very intuitive, but it's not rigorous enough for me, I wanna see formally why this holds:</p>
<p>$\sum_n \sum_j |I_{n_J}| < \sum_{i=1}^{\infty} \epsilon /2^n$</p>
<p>Like maybe looking at the definition of a series, the limit of the sequence of partial sums. Any help?</p>
| user2566092 | 87,313 | <p>Equalities and inequalities with series of series are preserved under exchanging order of summation indices and also taking inner limits before outer limits, vs. summing all the terms in any order you want, so long as the series are either all positive or all negative terms. Thus you don't have to worry about the double summation and how it affects the inequality.</p>
|
4,036,049 | <p>In order for <span class="math-container">$n^5 - n$</span> divisible by <span class="math-container">$5$</span>, <span class="math-container">$n^5 - n = 5 x + 0$</span> (for some <span class="math-container">$x$</span>, <span class="math-container">$x$</span> is a natural number)
I simplified the <span class="math-container">$(n^5 - n) = n(n^2+1)(n+1)(n-1)$</span> and I do not know what to do next.</p>
<p>And I tried something else like <span class="math-container">$(n^5 - n) = n (n^4 - 1)$</span>, but now I need to show <span class="math-container">$(n^4 - 1)$</span> is divisible by <span class="math-container">$5$</span>, how do I do that? Thanks in advance.</p>
| Paul Frost | 349,785 | <p>You have to solve the cubic equation <span class="math-container">$-3t^3+45t^2+600t = 3750$</span> which is equivalent to
<span class="math-container">$$t^3 -15t^2 -200t +1250 = 0. \tag{1}$$</span></p>
<p>In my opinion the best strategy is to verify first whether it has a rational root. Here all coefficients are integers and it is well-known that a rational root (if it exists) must be an integer which divides <span class="math-container">$1250$</span>. We have <span class="math-container">$1250 = 2 \cdot 5^4$</span> and a simple trial shows that <span class="math-container">$t = 5$</span> is a root. For the other roots you need to solve
<span class="math-container">$$t^2 - 10t -250 = 0$$</span>
which gives <span class="math-container">$t = 5 \pm 5\sqrt{11}$</span>. Among these three roots <span class="math-container">$t = 5$</span> is the only one between <span class="math-container">$0$</span> and <span class="math-container">$10$</span>.</p>
<p>Cubic equations can also be solved via the <a href="https://math.stackexchange.com/q/2838797">Cardano formula</a>. However, we know that <span class="math-container">$(1)$</span> has three real roots, thus we are in the so-called <em>casus irreducibilis</em>. This means that the roots of <span class="math-container">$(1)$</span> are represented as sums of two complex numbers - and this is not very nice.</p>
|
1,793,854 | <p>I am messed up on solving this question. What should I do first in order to get the answer ?</p>
<p><a href="https://i.stack.imgur.com/hE4rG.png" rel="nofollow noreferrer">This is the trigonometric function</a></p>
<p>$$ \lim \limits_{x \rightarrow 0} \frac{(a+x)\sec(a+x) - a \sec(a)}{x} $$</p>
| user5713492 | 316,404 | <p>Method 1: This is the definition of
$$\frac d{da}(a\sec a)=\sec a+a\sec a\tan a$$
Method 2: Change those secants to cosines
$$
\begin{align}
\lim_{x\rightarrow0}\frac{(a+x)\cos a-a\cos(a+x)}{x\cos a\cos(a+x)}&=\lim_{x\rightarrow0}\frac{a\cos a+x\cos a-a\cos a\cos x+a\sin a\sin x}{x\cos a\cos(a+x)}\\
&=\lim_{x\rightarrow0}\frac{a\cos a(1-\cos x)+x\cos a+a\sin a\sin x}{x\cos a\cos(a+x)}\\
&=\lim_{x\rightarrow0}\left[\frac{2a\sin^2(x/2)}{x\cos(a+x)}+\frac1{\cos(a+x)}+\frac{a\sin a\sin x}{x\cos a\cos(a+x)}\right]\\
&=0+\sec a+a\tan a\sec a
\end{align}
$$
In the last line we have used the limits
$$\lim_{x\rightarrow0}\frac{\sin x}{x}=\lim_{x\rightarrow0}\frac{\sin(x/2)}{(x/2)}=1
$$
$$\lim_{x\rightarrow0}\sin(x/2)=0
$$</p>
|
1,738,968 | <blockquote>
<p>Let $V$ be a vector space and let $T \in \operatorname{End}(V)$. If $\operatorname{rank}(T)$ and $\operatorname{null}(T)$ are finite, prove that $\dim(V)$ is finite.</p>
</blockquote>
<p>I cannot use the Rank-Nullity Theorem as it only applies to finite
dimensional vector space and I don't know whether $V$ is finite or infinite dimensional. </p>
| mathlove | 78,967 | <p>We have two lines :
$$L_1 : a_1x+b_1y+c_1=0,\quad L_2 : a_2x+b_2y+c_2=0$$</p>
<p>and the angle bisectors :
$$L_{\pm} : \frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$$</p>
<p>If we let $\theta$ be the (smaller) angle between $L_+$ and $L_1$, then we have
$$\cos\theta=\frac{\left|a_1\left(\frac{a_1}{\sqrt{a_1^2+b_1^2}}-\frac{a_2}{\sqrt{a_2^2+b_2^2}}\right)+b_1\left(\frac{b_1}{\sqrt{a_1^2+b_1^2}}-\frac{b_2}{\sqrt{a_2^2+b_2^2}}\right)\right|}{\sqrt{a_1^2+b_1^2}\sqrt{\left(\frac{a_1}{\sqrt{a_1^2+b_1^2}}-\frac{a_2}{\sqrt{a_2^2+b_2^2}}\right)^2+\left(\frac{b_1}{\sqrt{a_1^2+b_1^2}}-\frac{b_2}{\sqrt{a_2^2+b_2^2}}\right)^2}}$$</p>
<p>$$=\frac{\left|\sqrt{a_1^2+b_1^2}-\frac{a_1a_2+b_1b_2}{\sqrt{a_2^2+b_2^2}}\right|}{\sqrt{a_1^2+b_1^2}\sqrt{2-2\frac{a_1a_2+b_1b_2}{\sqrt{(a_1^2+b_1^2)(a_2^2+b_2^2)}}}}\times\frac{2\frac{1}{\sqrt{a_1^2+b_1^2}}}{2\frac{1}{\sqrt{a_1^2+b_1^2}}}=\sqrt{\frac{1-\frac{a_1a_2+b_1b_2}{\sqrt{(a_1^2+b_1^2)(a_2^2+b_2^2)}}}{2}}$$</p>
<p>Hence, we can see that
$$\begin{align}a_1a_2+b_1b_2\gt 0&\iff\cos\theta\lt 1/\sqrt 2\\&\iff \theta\gt 45^\circ\\&\iff \text{$L_+$ is the obtuse angle bisector}\end{align}$$
as desired.</p>
<p>(Note that "$c_1,c_2$ both are of same sign" is irrelevant.)</p>
|
751,053 | <p>T : Rn → Rm is a linear transformation where n,m>= 2</p>
<p>Let V be a subspace of Rn
and let W ={T(v ) | v ∈ V}
. Prove completely that W
is a subspace of Rm. </p>
<p>For this question how do I show that the subspace is non empty, holds under scaler addition and multiplication! I have never proved subspaces with transformations before!</p>
<p>I think I know this subspace is non empty when v is the 0 vector so is T(v) right?</p>
<p>b)Assume now that nullity(T) = 0 and that A =
{
a1; a2; . . . ; ap
}
is a basis for V as
in part (a). Prove completely that the set C =
{
T(a1); T(a2); . . . . ; T(ap)
}
is a basis
for W as in part (a) </p>
<p>Overhere Im lost!
since the nullity is 0 we know the matrix representation of the transformation is invertible. but then what</p>
| Vincent Boelens | 94,696 | <p>$W$ is nonempty since $0=T(0)\in W$. Now, suppose $x,y\in W$. Then there are $u,v\in V$ such that $Tu=x$ and $Tv=y$. Use this to show $x+y\in W$. A similar argument can be used to prove $ax\in W$ for all scalars $a$.
For b), use the fact that :$$0=\sum_{i=1}^n\lambda_iT(a_i)=T\left(\sum_{i=1}^n\lambda_ia_i\right) \iff \sum_{i=1}^n\lambda_ia_i=0$$</p>
|
3,903,016 | <p>Why is
<span class="math-container">$$
\sum_{n=1}^{\infty} \left( \frac{1}{(10n-9)^2} + \frac{1}{(10n-1)^2} \right) = \frac{\pi^2}{50} \frac{1}{1- \cos \frac{\pi}{5}}
$$</span>
?</p>
<p>Each form of
<span class="math-container">$$
\sum_{n=1}^{\infty} \frac{1}{(10n-9)^2}
$$</span>
and
<span class="math-container">$$
\sum_{n=1}^{\infty} \frac{1}{(10n-1)^2}
$$</span></p>
<p>couldn't be calculated but the sum of them has a closed form</p>
| xpaul | 66,420 | <p>Using
<span class="math-container">$$\sum_{n\in\mathbb{Z}}\frac{1}{(n-a)^2+b^2}=\frac{\pi\sinh2\pi b}{b\left(\cosh2\pi b-\cos2\pi a\right)}$$</span>
from <a href="https://math.stackexchange.com/questions/386667/closed-form-for-sum-n-infty-infty-frac1n-a2b2">this</a>, one has
<span class="math-container">\begin{eqnarray}
&&\sum_{n=1}^{\infty} \left( \frac{1}{(10n-9)^2} + \frac{1}{(10n-1)^2} \right)\\
&=&\sum_{n\in\mathbb{Z}}\frac{1}{(10n-1)^2}=\frac{1}{10^2}\sum_{n\in\mathbb{Z}}\frac{1}{(n-\frac1{10})^2}\\
&=&\frac1{10}\lim_{b\to0}\frac{\pi\sinh(2\pi b)}{b\left(\cosh(2\pi b)-\cos(\frac{2\pi}{10})\right)}\\
&=&\frac{\pi^2}{50} \frac{1}{1- \cos \frac{\pi}{5}}.
\end{eqnarray}</span></p>
|
3,631,587 | <p>Assume that our samples are high dimensional points (i.e., d is large) and we use
PCA to reduce it to k = 10 dimensions. After this step, we found that all the 10 new dimensions have
continuous values (e.g., in other words, each feature in the transformed dimension is not from discrete
domain, but rather, continuous domain). Describe in detail, how we can now use parametric method to
train our model to do classification. In particular, discuss how we can compute the correlation matrix
estimation, and when a new point arrive, what procedure we need to do so to make a classification
prediction (assume in general, we have K > 2 classes).</p>
| Aderinsola Joshua | 395,530 | <p><span class="math-container">$circumference^2 = 4×\pi×area$</span></p>
|
1,760,242 | <p>Can anybody tell me where can I find some REAL problems (i.e. form real life) that can be solved using a 3x3 system of linear equations? Or, can anybody give me an example? A solution could be a circuit in electrical engineering, but this is not very interesting, and it doesn't seem to be so real.
Thank you!</p>
| Robert Israel | 8,508 | <p>A factory employs three classifications of worker, each type having a different wage rate: \$10/hour for $A$'s, \$15/hour for $B$'s, \$20/hour for $C$'s. Each $A$ worker works $20$ hours per week, while $B$'s and $C$'s each work $40$ hours per week. Unfortunately, some will have to be laid off. Management has decided to</p>
<ol>
<li>Reduce the total number of workers by $100$.</li>
<li>Reduce the total number of hours worked per week by $3200$.</li>
<li>Reduce the total wages per week by $\$50000$.</li>
</ol>
<p>How many A's, B's and C's will be laid off?</p>
|
1,760,242 | <p>Can anybody tell me where can I find some REAL problems (i.e. form real life) that can be solved using a 3x3 system of linear equations? Or, can anybody give me an example? A solution could be a circuit in electrical engineering, but this is not very interesting, and it doesn't seem to be so real.
Thank you!</p>
| Doug M | 317,162 | <p>One which I deal with in on a somewhat regular basis. Treasury notes have 3 primary risk factors (called shift, twist and butterfly). These are in fact all dimension of interest rate risk. However, once you have determined the factor loading for any given security, it behaves as a linear factor.</p>
<p>You and your strategy team work out an ideal risk profile based on your outlook for the market and have balanced your portfolio accordingly. The next day a client gives you $10 million to be put into the fund. Pick 3 treasury notes, calculate their factor loading, and determine how much of each you need to buy to bring your exposures back to target. </p>
|
88,156 | <p>I understand that <code>Round</code> give the nearest even integer for cases where the number is between two integers, i.e. <code>Round[2.5] = 2</code> and <code>Round[3.5] = 4</code> (see this <a href="https://mathematica.stackexchange.com/questions/29122/why-do-numberform-and-round-apparently-use-different-tie-breaking-methods/33939#33939">question</a>). This carries over to rounding to non-integers that are not orders of magnitude as well:</p>
<pre><code>Round[1.3, 0.2] = 1.2
Round[1.5, 0.2] = 1.6
Round[Range[1, 2, 0.1], 0.2] = {1., 1.2, 1.2, 1.2, 1.4, 1.6, 1.6, 1.6, 1.8, 1.8, 2.}
</code></pre>
<p>However, the <code>Floor</code> function seems to behave oddly in a similar situation:</p>
<pre><code>Floor[Range[1, 2, 0.1], 0.2] = {1., 1., 1., 1.2, 1.2, 1.4, 1.6, 1.6, 1.8, 1.8, 2.}
</code></pre>
<p>but <code>Ceiling</code> does what I would expect it to do:</p>
<pre><code>Ceiling[Range[1, 2, 0.1], 0.2] = {1., 1.2, 1.2, 1.4, 1.4, 1.6, 1.6, 1.8, 1.8, 2., 2.}
</code></pre>
<p>My question therefore is why does <code>Floor[1.2, 0.2] = 1.</code>? I understand the behavior of <code>Round</code> for this test set, but I do not understand <code>Floor</code>. Then, with <code>Floor</code> giving something I don't expect, I am very surprised that <code>Ceiling</code> gives exactly what I would expect.</p>
<hr>
<h2>Background for my Problem</h2>
<p>I have a regular array of 2D data (<code>{x, y, probability}</code>) that I want to bin by summing, so <code>Round</code> doesn't work because different numbers of elements are included in consecutive bins because it rounds to evens, but I expected <code>Floor</code> to work.</p>
<pre><code>gathered = GatherBy[predictedProbabilityDist,
{ Round[#[[1]], newGridSpacing],
Round[#[[2]], newGridSpacing] } &];
</code></pre>
<p>versus</p>
<pre><code>gathered = GatherBy[predictedProbabilityDist,
{ Floor[#[[1]], newGridSpacing],
Floor[#[[2]], newGridSpacing] } &];
</code></pre>
<p>or the nearly equivalent expression with <code>Ceiling</code>, followed by something like</p>
<pre><code>predictedProbDist = Map[
{ Min[#[[All, 1]]],
Min[#[[All, 2]]],
Total[#[[All, 3]]] } &,
gathered];
</code></pre>
<p>Based on the testing with the simple <code>Range</code> above, it looks like I need to use <code>Ceiling</code>, but I don't know why for sure that is the case.</p>
| Sjoerd C. de Vries | 57 | <p>Welcome in the amazing world of machine precision arithmetic!</p>
<p>If you examine the binary representation of both numbers you see the following:</p>
<pre><code>RealDigits[1.2, 2]
(* {{1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1,
1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0,
0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1}, 1} *)
RealDigits[.2, 2]
(* {{1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0,
1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1,
0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0}, -2} *)
</code></pre>
<p>The .2 part is represented by an infinite series of 0,0,1,1,.. that necessarily has to be truncated to fit into the allocated space. Since the fractional part has slightly more space in the representation of 0.2 than in 1.2 (because of the missing integer part) it is therefore slightly larger than the fractional part of 1.2 and therefore 1.2 is seen as smaller than a multiple of 0.2. Compare also</p>
<pre><code>Floor[1.20000000000000001, .2]
(* 1. *)
</code></pre>
<p>and</p>
<pre><code>Floor[1.200000000000001, .2]
(* 1.2 *).
</code></pre>
<p>Using arbitrary precison arithmetic you can make the representation of the bit series in 1.2 longer than in 0.2 and the result is now:</p>
<pre><code>Floor[1.2`200, .2`10]
(* 1.200000000 *)
</code></pre>
|
60,152 | <p>What is the motivation behind topology?</p>
<p>For instance, in real analysis, we are interested in rigorously studying about limits so that we can use them appropriately. Similarly, in number theory, we are interested in patterns and structure possessed by algebraic integers and algebraic prime numbers.</p>
<p>Some googling and wiki-ing gave me that topology studies about deformation of objects in some space i.e. how an object in some space behaves under a continuous map. However, when I started reading the subject it starts of by defining what a topology is i.e. a set of subsets of a set with certain properties. I fail to see the connection immediately. I would appreciate if someone could give a short bird's eye view of topology.</p>
<p>Thanks,
Adhvaitha</p>
| Michael Hardy | 11,667 | <p>This will answer only the question in the second paragraph.</p>
<p>Imagine the half-open interval $[0,1)$ with the usual open sets. In particular, an open neighborhood of $0$ contains $0$ itself and all positive numbers sufficiently close to $0$.</p>
<p>Now alter the definition of "open set", so that every open neighborhood of $0$ contains not only $0$ and all positive numbers close enough to $0$, but also all numbers close enough to $1$. I.e. $[0,\varepsilon)\cup(1-\eta,1)$.</p>
<p>Can you see how that alters the way in which the whole space is connected together?</p>
<p>Thus: how the space is connected together, is simply a matter of which sets are open. That's the connection between the two things.</p>
|
60,152 | <p>What is the motivation behind topology?</p>
<p>For instance, in real analysis, we are interested in rigorously studying about limits so that we can use them appropriately. Similarly, in number theory, we are interested in patterns and structure possessed by algebraic integers and algebraic prime numbers.</p>
<p>Some googling and wiki-ing gave me that topology studies about deformation of objects in some space i.e. how an object in some space behaves under a continuous map. However, when I started reading the subject it starts of by defining what a topology is i.e. a set of subsets of a set with certain properties. I fail to see the connection immediately. I would appreciate if someone could give a short bird's eye view of topology.</p>
<p>Thanks,
Adhvaitha</p>
| Tilo Wiklund | 13,504 | <p>This may not be quite the answer you are looking for but I think a "birds-eye" view of topology is somewhat hard. One example which I found somewhat enlightening though, was when I first thought about the topology of Baire space.</p>
<p>The topology is defined on the set of infinite sequences over the natural numbers. The topology is then defined to be such that each open set is given by the set of all sequences which coincide on some finite set of points. This can be made somewhat simpler by saying that the topology is generated by a basis of open sets given by finite sequences, to each finite sequence you associate the set of sequences beginning with this finite sequence, and then let the topology be given by all unions and intersections of such sets.</p>
<p>A continuous function from this Baire space into the the real numbers (with the metric topology, that is to say the topology generated by open intervals) now correspond, intuitively speaking, to those functions such that given "enough" information about he sequence (that is to say, a large enough initial segment) you get a "good enough" approximation of where the value of the whole sequence will be (i.e. a small interval within which the value will be). </p>
<p>Of course you can choose the codomain to be, again, Baire space. In which case the continuous functions correspond to those where an arbitrary large initial segment of the "output" can be generated by giving a large enough initial segment of the "input". Such continuous functions are, thus, in a sense, the computable functions, since they do not "get stuck".</p>
<p>I rather liked this example since it indicates a way in which topology gives a sense of "information content" and things behaving well with respect to approximation. I guess this language is more close to topology as its studied in areas as domain theory, but I figured it might help to consider it from a somewhat non-geometric perspective. Also it suggest, as so often seems to be the case with mathematics, that the interesting things happen when you start consider structure preserving maps, and that the definition of a topology is, perhaps, more interesting as a way to be able to describe the notion of continuity of a function.</p>
<p>(As a remark that I am aware that the Baire topology is a metric topology, but I feel that the definition in terms of a topology is much more natural than in terms of the ultrametric distance)</p>
|
2,567,607 | <p>$$\arctan 2x +\arctan 3x = \left(\frac{\pi}{4}\right)$$
$$\arctan \left(\frac{2x+3x}{1-2x*3x}\right)=\frac {\pi}{4}$$
$$\frac {5x}{1-6x^2}=\tan \frac{\pi}{4}=1$$
$$6x^2 + 5x -1 = 0$$
$$(6x-1)(x+1)=0$$
$$x=-1, \frac{1}{6}$$</p>
<p>The answer however rejects the solution $x=-1$ saying that it makes the L.H.S of the equation negative. I don't understand this, I don't see how $x=-1$ makes the L.H.S. negative.</p>
| Michael Rozenberg | 190,319 | <p>Because
$$\arctan(-2)+\arctan(-3)=-135^{\circ}<0$$
By the way, $\tan(-135^{\circ})=1.$</p>
|
398,010 | <p>1.17 THEOREM</p>
<p>Let $f: X \to [0, \infty)$ be measurable. There exist simple measurable functions $s_n$ on $X$ such that</p>
<p>(a) $0 \le s_1 \le s_2 \le \dots \le f.$</p>
<p>(b) $s_n(x) \to f(x)$ as $n \to \infty \forall x$</p>
<p>PROOF Put $\delta_n = 2^{-n}$. To each positive integer $n$ and each real number $t$ corresponds a unique integer $k = k_n(t)$ that satisfies $k \delta_n \le t < (k + 1) \delta_n$. Define</p>
<p>$
\phi_n(t)=
\begin{cases}
k_n(t) \delta_n \space \space \text{if} \space 0 \le t <n \\
n \space \space \text{if} \space n \le t < \infty \\
\end{cases}
$</p>
<p>Each $\phi_n$ is then a Borel function on $[0,\infty)$, </p>
<p>$t~\delta_n< \phi_n(t)\le t \space \space \space \space \text{if} \space \space 0 \le t <n$</p>
<p>$0 \le \phi_1(t) \le \phi_2(t) \le \dots \le t$ and $\phi_n(t) \to t$ as $n \to \infty$ It follows that the functions</p>
<p>$s_n = \phi_n (f)$</p>
<p>satisfy (a) and (b); they are measurable, by Theorem 1.12(d). $\square$</p>
<p>$\textbf{Question}$: Why can't we set $\phi_n(t)= n$ ? What would we lose? </p>
<p>$0 \le \phi_1(t) \le \phi_2(t) \le \dots \le t$ is only true for $t<n$. I just don't see what we gain by having $k_n(t) \delta_n \space \space \text{if} \space 0 \le t <n $</p>
| Sargera | 30,988 | <p>The construction partitions a subset of the range of $f$, i.e. the values of $f$ that fall in the interval $[0,n]$. The definition of $\phi_{n}$ ensures that the interval goes to $[0,\infty)$ (i.e. covers the entire range of $f$) while the requirement $\phi$ be equal to $k_{n}(t)\delta_{n}$ ensures that the range is partitioned finer and finer (in particular, at a "faster rate" than the interval is being extended). What you lose by just setting $\phi$ equal to $n$ is the approximation of $f$!</p>
|
398,010 | <p>1.17 THEOREM</p>
<p>Let $f: X \to [0, \infty)$ be measurable. There exist simple measurable functions $s_n$ on $X$ such that</p>
<p>(a) $0 \le s_1 \le s_2 \le \dots \le f.$</p>
<p>(b) $s_n(x) \to f(x)$ as $n \to \infty \forall x$</p>
<p>PROOF Put $\delta_n = 2^{-n}$. To each positive integer $n$ and each real number $t$ corresponds a unique integer $k = k_n(t)$ that satisfies $k \delta_n \le t < (k + 1) \delta_n$. Define</p>
<p>$
\phi_n(t)=
\begin{cases}
k_n(t) \delta_n \space \space \text{if} \space 0 \le t <n \\
n \space \space \text{if} \space n \le t < \infty \\
\end{cases}
$</p>
<p>Each $\phi_n$ is then a Borel function on $[0,\infty)$, </p>
<p>$t~\delta_n< \phi_n(t)\le t \space \space \space \space \text{if} \space \space 0 \le t <n$</p>
<p>$0 \le \phi_1(t) \le \phi_2(t) \le \dots \le t$ and $\phi_n(t) \to t$ as $n \to \infty$ It follows that the functions</p>
<p>$s_n = \phi_n (f)$</p>
<p>satisfy (a) and (b); they are measurable, by Theorem 1.12(d). $\square$</p>
<p>$\textbf{Question}$: Why can't we set $\phi_n(t)= n$ ? What would we lose? </p>
<p>$0 \le \phi_1(t) \le \phi_2(t) \le \dots \le t$ is only true for $t<n$. I just don't see what we gain by having $k_n(t) \delta_n \space \space \text{if} \space 0 \le t <n $</p>
| AlexM | 72,471 | <p>What would we lose? Well, we need to keep our eyes on the prize: approximating $f$ by simple functions. The $k_n(t)\delta_n$ part is what is actually doing it. </p>
<p>Imagine filling up a glass with some ice in the bottom, ripe for filling. As you pour water in, ignoring displacement there will be a level it would go up to, but in some spots it will be "stopped" by the ice. </p>
<p>That is whats going on here essentially, you are trying to fill up to level "n" but the function gets in the way (the $k_n(t)\delta_n$ part). But, as you fill up more and more, the degree to which you are approximating $f$ gets better. </p>
<p>A good picture of what is going on can be found in Folland, chapter 2. </p>
|
1,307,269 | <p>In my textbook, before introducing the epsilon delta definition, they gave a working definition of what a limit is. The definition sounded something like this "$\lim \limits_{x \to a}f(x) = L$, if when $x$ gets closer to $a$, $f(x)$ gets closer to $L$" </p>
<hr>
<p>But is that always the case with limits? What if $f(x) = 4,$ then we have $\lim \limits_{x \to 2}f(x) = 4$, but it is never true that when x gets closer to 2, f(x) gets closer to 4. Maybe instead we should say: "$\lim \limits_{x \to a}f(x) = L$, if when $x$ gets closer to $a$, $ f(x)$ gets closer to or equals $L$".</p>
<hr>
<p>Please correct me if I'm wrong. I'm pretty new to this stuff. Btw, i understand that the epsilon delta definition has the constant function limit case covered, but I'm more interested in the working definition.</p>
| Ittay Weiss | 30,953 | <p>You are correct. You present a nuance of the 'working definition', a special case where "getting closer" is misleading. You should interpret "getting closer" as "getting as close as you like". This is a more accurate 'working definition' in any case. Then the constant function scenario works just fine. You can get as close as you like to the constant, quite trivially. </p>
<p>The "getting closer" definition makes it sounds as if a limit is somehow an indefinite thing, moving around, getting closer to things. This is misleading. A better intuition is "the limit at $x_0$ of $f$ is $L$" is "the value $f(x)$ can be made as close as you like to $L$ for all $x$ that is sufficiently close to but not equal to $x_0$". </p>
|
457,557 | <p>Use a triple integral to find the volume of the solid: The solid enclosed by the cylinder $$x^2+y^2=9$$ and the planes $$y+z=5$$ and $$z=1$$<br>
This is how I started solving the problem, but the way I was solving it lead me to 0, which is incorrect. $$\int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\int_{1}^{5-y}dzdxdy=\int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\left(4-y\right)dxdy=\int_{-3}^3\left[4x-xy\right]_{-\sqrt{9-y^2}}^\sqrt{9-y^2}dy= {8\int_{-3}^3{\sqrt{9-y^2}}dy}-2\int_{-3}^3y{\sqrt{9-y^2}}dy$$<br>
If this is wrong, then that would explain why I'm stuck. If this is correct so far, that's good news, but the bad news is that I'm still stuck. If someone could help me out, that would be wonderful, thanks!</p>
| Christian Blatter | 1,303 | <p>This was an exercice on triple integrals; but at the end we are inclined to check our result against geometric reasoning. Such reasoning tells us that the given solid is half a cylinder of radius $3$ and height $2\cdot(5-1)=8$. Therefore its volume is $36\pi$.</p>
|
915,631 | <blockquote>
<p>Let <span class="math-container">$V$</span> be the vector space of <span class="math-container">$n\times n$</span> matrices over <span class="math-container">$K$</span> under addition and let the linear operator <span class="math-container">$f$</span> be given by <span class="math-container">$f(A)=A^{T}$</span>, where <span class="math-container">$A^T$</span> denotes the transpose of matrix <span class="math-container">$A$</span>.</p>
<ol>
<li><p>Show that <span class="math-container">$\pm 1$</span> are the only eigenvalues of <span class="math-container">$f$</span>.</p>
</li>
<li><p>Suppose the characteristic of <span class="math-container">$K$</span> is not <span class="math-container">$2$</span>. What are the dimensions of the eigenspaces <span class="math-container">$E(1)$</span> and <span class="math-container">$E(-1)$</span>?</p>
</li>
</ol>
</blockquote>
<p>Eigenvalues are found from equation: <span class="math-container">$f(v)=\lambda v$</span>, therefore we have <span class="math-container">$v^T=\lambda v$</span>, I don't understand how this shows <span class="math-container">$\lambda =\pm 1$</span>.</p>
<p>I know that <span class="math-container">$\dim E(1)+\dim E(2)=n$</span> would this suffice for the second part?</p>
| Hamou | 165,000 | <p>$f(A)=\lambda A$, hence $f(f(A))=\lambda f(A)=\lambda^2 A$, but $f(f(A))=A$, then
$A=\lambda^2 A$, so $\lambda^2=1$.</p>
|
1,740,458 | <p>I saw somewhere on Math Stack that there was a way of finding integrals in the form $$\int \frac{dx}{a+b \cos x}$$ <strong>without</strong> using Weierstrass substitution, which is the usual technique. </p>
<p>When $a,b=1$ we can just multiply the numerator and denominator by $1-\cos x$ and that solves the problem nicely. </p>
<p>But I remember that the technique I saw was a nice way of evaluating these even when $a,b\neq 1$.</p>
| omegadot | 128,913 | <p>Another way to get to the same point as C. Dubussy got to is the following:
\begin{align*}
\frac{1}{a + b \cos x} &= \frac{1}{a \left (\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} \right ) + b \left (\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} \right )}\\
&= \frac{1}{(a - b) \sin^2 \frac{x}{2} + (a + b) \cos^2 \frac{x}{2}}\\
&= \frac{\sec^2 \frac{x}{2}}{(a + b) + (a - b) \tan^2 \frac{x}{2}},
\end{align*}
and then make the substitution of $t = \tan \frac{x}{2}$ in the integral.</p>
|
3,046,979 | <p>Recently I've been trying to tackle the integral <span class="math-container">$\int_0^{\frac{\pi}{2}} \ln(\sin(\theta))d\theta$</span> using the Beta function
<span class="math-container">$$\frac{B(\frac{x}{2},\frac{1}{2})}{2}=\int_0^{\frac{\pi}{2}} \sin^{x-1}(\theta)d\theta=\frac{\sqrt{\pi}}{2}\left(\Gamma\left(\frac{x+1}{2}\right)\right)^{-1}$$</span>
Differentiating both sides
<span class="math-container">$$\int_0^{\frac{\pi}{2}} \ln(\sin(\theta))\sin^{x-1}(\theta)d\theta=-\frac{\sqrt{\pi}}{4}\frac{\psi(\frac{x+1}{2})}{\Gamma(\frac{x+1}{2})}$$</span>
However, at <span class="math-container">$x=1$</span> <span class="math-container">$$\int_0^{\frac{\pi}{2}} \ln(\sin(\theta))d\theta\ne\frac{\gamma\sqrt{\pi}}{4}$$</span></p>
<p>Where did I go wrong?</p>
| J.G. | 56,861 | <p>You forgot the <span class="math-container">$\Gamma(x/2)$</span> factor.</p>
|
3,046,979 | <p>Recently I've been trying to tackle the integral <span class="math-container">$\int_0^{\frac{\pi}{2}} \ln(\sin(\theta))d\theta$</span> using the Beta function
<span class="math-container">$$\frac{B(\frac{x}{2},\frac{1}{2})}{2}=\int_0^{\frac{\pi}{2}} \sin^{x-1}(\theta)d\theta=\frac{\sqrt{\pi}}{2}\left(\Gamma\left(\frac{x+1}{2}\right)\right)^{-1}$$</span>
Differentiating both sides
<span class="math-container">$$\int_0^{\frac{\pi}{2}} \ln(\sin(\theta))\sin^{x-1}(\theta)d\theta=-\frac{\sqrt{\pi}}{4}\frac{\psi(\frac{x+1}{2})}{\Gamma(\frac{x+1}{2})}$$</span>
However, at <span class="math-container">$x=1$</span> <span class="math-container">$$\int_0^{\frac{\pi}{2}} \ln(\sin(\theta))d\theta\ne\frac{\gamma\sqrt{\pi}}{4}$$</span></p>
<p>Where did I go wrong?</p>
| Ininterrompue | 622,553 | <p>Another way that <em>only</em> requires knowledge of the beta function (and expansion of the gamma function) is given as follows. Consider the integral</p>
<p><span class="math-container">$$\begin{aligned}\int_{0}^{\pi/2}\sin^{\epsilon}x\,\mathrm{d}x &= \int_{0}^{\pi/2}e^{\epsilon\ln\sin x}\,\mathrm{d}x = \sum_{n=0}^{\infty}\frac{\epsilon^{n}}{n!}\int_{0}^{\infty}\ln^{n}\sin x\,\mathrm{d}x \\
&= \frac{\Gamma(1/2)\Gamma(1/2+\epsilon/2)}{2\,\Gamma(1+\epsilon/2)}. \end{aligned}$$</span> </p>
<p>We have to match the appropriate coefficient of <span class="math-container">$\epsilon$</span> found by expanding the latter expression to the coefficient of the series in the first line. The integral we are interested in is the <span class="math-container">$n=1$</span> term. <span class="math-container">$\epsilon^{2}$</span> and higher order terms can then be neglected. Due to the trigonometric form of the beta function, integrals of this type often require the use of Legendre's duplication formula</p>
<p><span class="math-container">$$ \frac{\Gamma(1+\epsilon)}{\Gamma(1/2+\epsilon)} = \frac{2^{2\epsilon}}{\sqrt{\pi}}\frac{\Gamma^{2}(1+\epsilon)}{\Gamma(1+2\epsilon)}, $$</span></p>
<p>derived also by using the beta function. The expansion of the gamma function around <span class="math-container">$1$</span> is given as</p>
<p><span class="math-container">$$ \ln\Gamma(1+\epsilon) = -\gamma\epsilon + \sum_{k=2}^{\infty}\frac{(-1)^{k}\zeta(k)}{k}\epsilon^{k} $$</span></p>
<p>which is highly useful. Here <span class="math-container">$\gamma$</span> is the Euler-Mascheroni constant and <span class="math-container">$\zeta(k)$</span> is the Riemann zeta function. Note that the form of the expansion of the gamma function and the ratio of gamma functions <span class="math-container">$\Gamma(1+\epsilon)/\Gamma^{2}(1+\epsilon/2)$</span> means that the first order term of this ratio cannot contribute (it is just <span class="math-container">$1$</span>), so up to first order,</p>
<p><span class="math-container">$$\begin{aligned} \frac{\Gamma(1/2)\Gamma(1/2+\epsilon/2)}{2\,\Gamma(1+\epsilon/2)} &= \frac{\sqrt{\pi}}{2}\frac{\sqrt{\pi}}{2^{\epsilon}}\frac{\Gamma(1+\epsilon)}{\Gamma^{2}(1+\epsilon/2)} \\
&\approx \frac{\pi}{2}\frac{1}{e^{(\ln 2)\epsilon}} \approx \frac{\pi}{2}(1 - (\ln 2)\epsilon). \end{aligned}$$</span></p>
<p>In general, if we look for the coefficient of the <span class="math-container">$n$</span>th term, we have to multiply by <span class="math-container">$n!$</span> to account for the factorial in the original expansion. First order term is trivial: <span class="math-container">$1! = 1$</span>, so the first order coefficient is the answer</p>
<p><span class="math-container">$$ \int_{0}^{\pi/2}\ln\sin x\,\mathrm{d}x = -\frac{\pi}{2}\ln 2. $$</span></p>
|
2,650,182 | <ol>
<li>Between every two distinct real numbers, there is a rational number </li>
</ol>
<p>Answer: There is no rational numbers between two non-distinct real numbers.</p>
<ol start="2">
<li>For all natural numbers $n ∈ N, \sqrt n$ is either a natural number or an
irrational number</li>
</ol>
<p>Answer: For all natural numbers $n$, $\sqrt n$ is either not a natural number or not a irrational number. </p>
<ol start="3">
<li>Given any real number $x ∈ R$, there exists $n ∈ N$ satisfying $n>x$.</li>
</ol>
<p>Answer: ??</p>
<p>Can someone tell me what is the general way to look at these things. </p>
| Ove Ahlman | 222,450 | <p>The general rule is that "For all" statements should be changed to "There exists" and "There exists" should be changed to "For all". Then apply this inward in the statement. It is often easier to rewrite statements in the syntactic logical way to get a better formal grip. (i.e. using $\exists x, \forall y $ etc.)</p>
<ol>
<li>Say that "for all distinct ..." so we change this to a "there exists." This leads to the statement
<blockquote>
<p>There exists two distinct real numbers such that there is no rational number between these.</p>
</blockquote></li>
</ol>
<p>Now it is not clear exactly how far one should push the negation, currently I let the negation stand on the "There exists" part, because this makes the formulation clear, but we could reformulate it as </p>
<blockquote>
<p>There exists two distinct real numbers such that for any rational number, this number is not between the two distinct real numbers.</p>
</blockquote>
<p>which pushes the negation as far as it can go. On the other hand a negation of the sentence is also </p>
<blockquote>
<p>It is not the case that between every two distinct real numbers, there is a rational number</p>
</blockquote>
<p>But this might not be what anyone wanted, still it is very clearly a negation of the sentence.</p>
<p>For sentence 2 and 3 there are similar discussions regarding the quantifiers which you need to change. Thus when the sentence starts with "For any real number x there exists a natural number n such that $n>x$ we will need to swap quantifiers to get</p>
<blockquote>
<p>There is a real number $x$ such that for any natural number $n, n\not > x$</p>
</blockquote>
|
3,179,505 | <p>Help me please , I am not able to solve this problem.I have tried in many ways to figure out such as Ration test , Integral test , Comparison test , Limit Comparison Test , Root Test but i can't find the way out . This is my first question and i'm not good at English. If there is something wrong or you are not comfortable with my language usage I'm so sorry.</p>
| Mark | 470,733 | <p>If you let <span class="math-container">$d$</span> be <strong>any</strong> metric on <span class="math-container">$X$</span> then of course the answer is no. Let's say we take <span class="math-container">$\mathbb{R}$</span> with the usual metric and <span class="math-container">$\mathbb{R^2}$</span> with the discrete metric. (i.e <span class="math-container">$d(x,y)=\delta_{xy})$</span>.Then the sequence <span class="math-container">$(\frac{1}{n},\frac{1}{n})$</span> is not Cauchy in <span class="math-container">$\mathbb{R^2}$</span> but in both coordinates you have Cauchy sequences. </p>
|
1,915,782 | <p>I'm attempting to teach myself some vector calculus before starting university next month in hope of getting my head around some of the concepts as I can foresee this being a weak topic for me.</p>
<p>I have been 'learning' from some online lecture notes related to my course. The notes talk about line integrals but as far as I understand say little on how to evaluate them and only gives one quick example in the form below that I didn't find terribly useful. As a result I'm not entirely sure how to evaluate the line integral below and so I would ask that someone answer the below question, but if possible perhaps give more detail than would usually be necessary, talking through each step with a specific emphasis on the difference between evaluating (i) and (ii), thank you.</p>
<blockquote>
<p>Evaluate explicitly the line integral $\int(y$ $dx+x$ $dy+dz)$ along
(i) the straight path from the origin to $x=y=z=1$ and (ii) the
parabolic path given parametrically by $x = t,y = t,z = t^2$ from $t=0$ to
$t=1$.</p>
</blockquote>
<p>Any help is appreciated.</p>
<p>Thank you.</p>
| Bernard | 202,857 | <p>You have to make a substitution with the parameter. I'll denote paths under the generic name $\gamma$.</p>
<p>For the first integral, the parametrisation of the straight line is clearly $\;\begin{cases}x=t\\y=t\\z=t\end{cases}\enspace(0\le t\le 1)$, so $\;\mathrm d\mkern1mux=\mathrm d\mkern1mut=\mathrm d\mkern1muy=\mathrm d\mkern1muz $, and
$$\int_\gamma(y\,\mathrm d\mkern1mu x+x\,\mathrm d\mkern1mu y+\mathrm d\mkern1muz)=
\int_0^1(t\,\mathrm d\mkern1mu t+t\,\mathrm d\mkern1mu t+\mathrm d\mkern1mu t)=\int_0^1(2t+1)\,\mathrm d\mkern1mu t= t^2+t\,\bigg\vert_0^1=2.$$</p>
|
3,970,959 | <blockquote>
<p><span class="math-container">$S = \frac{1}{1001} + \frac{1}{1002}+ \frac{1}{1003}+ \dots+\frac{1}{3001}$</span>.</p>
</blockquote>
<blockquote>
<p>Prove that <span class="math-container">$\dfrac{29}{27}<S<\dfrac{7}{6}$</span>.<br></p>
</blockquote>
<p>My Attempt:<br>
<span class="math-container">$S<\dfrac{500}{1000} + \dfrac{500}{1500}+ \dfrac{500}{2000}+ \dfrac{500}{2500}+\dfrac{1}{3000} =\dfrac{3851}{3000}$</span></p>
<p>(Taking 250 terms together involves many fractions and is difficult to calculate by hand.)</p>
<p>Using AM-HM inequality gave me <span class="math-container">$S > 1$</span>, but the bounds are weak.</p>
<p><a href="https://math.stackexchange.com/questions/688432">Prove that $1<\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\dots+\frac{1}{3001}<\frac43$</a> <br><br>
<a href="https://math.stackexchange.com/questions/605111">Inequality with sum of inverses of consecutive numbers</a> <br><br>
The answers to these questions are nice, but the bounds are weak.</p>
<p>Any help without calculus and without calculations involving calculators would be appreciated.</p>
<p>(I encountered this question when I was preparing for a contest which neither allows calculators nor calculus(Only high-school mathematics.))</p>
| WA Don | 542,712 | <p>Dividing the sum into groups as you have done will obtain the limits if you take enough buckets (around 10).</p>
<p>A similar approach is to add <span class="math-container">$1/1000$</span> and then pair the terms, with two terms left over added to the end
<span class="math-container">\begin{align}
S&=\frac{1}{1000} + \frac{1}{3000} + \frac{1}{1001} + \frac{1}{2999} + \cdots +
\frac{1}{1999} + \frac{1}{2001} + \frac{1}{3001} - \frac{1}{1000} \\
&= \frac{4000}{1000\cdot 3000} + \frac{4000}{1001\cdot 2999} + \cdots +\frac{4000}{1999\cdot 2001} + \frac{1}{3001}-\frac{1}{1000}
\end{align}</span></p>
<p>Then try grouping these into <span class="math-container">$m$</span> buckets, each with <span class="math-container">$n$</span> terms. The lower bound is
<span class="math-container">\begin{align}
\frac{4000n}{(1000+n-1)(3000-n+1)}+\cdots+\frac{4000n}{(1000+mn-1)(3000-mn+1)} \\ \quad+\frac{1}{3001}-\frac{1}{1000}
\end{align}</span>
and the upper bound,
<span class="math-container">\begin{align}
\frac{4000n}{(1000)(3000)}+\cdots+\frac{4000n}{(1000+(m-1)n)(3000-(m-1)n)}\\ \quad+\frac{1}{3001}-\frac{1}{1000}
\end{align}</span>
Using a calculator, and (still) choosing <span class="math-container">$m=4, n=250$</span> can create bounds comparable to the ones you have, but it is still cumbersome.</p>
<hr />
<p>With <span class="math-container">$n=250$</span>, the upper bound is
<span class="math-container">\begin{align}
U &= \frac{1}{3}+\frac{16}{55}+\frac{4}{15}+\frac{16}{63}+\frac{1}{3001}-\frac{1}{1000} \\
&\leqslant \frac{1}{3}+\frac{16}{55}+\frac{4}{15}+\frac{16}{63} \\
&\leqslant \frac{20 + 18+16+16}{60} \\
&= \frac{7}{6}
\end{align}</span></p>
|
254,030 | <p>I am following a course in basic algebra, and we have covered rings & groups in class, but I am having trouble visualising them. Are there applications of group &/or ring theory that can be more easily visualized than the abstract object? For instance, are there objects, or properties of objects, that behave as elements of a group in physics, chemistry, or other fields?</p>
| 000 | 22,144 | <p>This post hinges on whether or not you consider a polygon an abstract object. That is a highly debatable philosophical claim. As a result, I'm submitting this answer on good faith.</p>
<p>The most basic thing that comes to mind for me is the group of rotations of a polygon about its center point $P_k$. For instance,</p>
<p>$$(P_{3},r):=\{r(n): n \in \mathbb{Z}\},$$</p>
<p>the group of rotations of a triangle about its center where every element is a rotation of $n$ radians. We have a fundamental identity:
$$r(0)=r(2\pi)=r\left(\frac{2\pi}{3}\right).$$
This is of interest because it directly relates to a triangle's three lines of symmetry. In fact, . . .</p>
<p>If we define $(P_k,r)$ to be the group of rotations of a $k$-gon about its center, we have the following identity for all values of $k$:
$$r(0)=r(2\pi)=r\left(\frac{2\pi}{k}\right).$$</p>
<p>This is because every polygon has as many lines of symmetry as it has sides. As a result, you can inscribe a polygon in a circle and 'cut' the polygon $k$ ways via its lines of symmetry. This shows us that every rotation is the angle $\dfrac{2\pi}{k}$ radians because we cut the <em>circle</em> into $k$ equal parts.</p>
<hr>
<p>For you to explore: Consider $(P_k,r,r')$ where $r'$ is the set of reflections across a line through the center. Is it a ring? If so, what properties arise?</p>
|
1,128,623 | <p>Question: Let S be the set of sequences of $0$s and $1$s. For $x = (x_1, x_2, x_3, ...)$ and $y = (y_1, y_2, y_3, ...)$. Define</p>
<p>$d(x,y)=\sum_{i=1}^\infty \dfrac{|x_i - y_i|}{2^i}$ </p>
<p>Proof the infinite sum in the definition of $d(x,y)$ converges for all $x$ and $y$. </p>
<p>Incomplete answer: Since the max of $d(x,y)$ happens when all elements of one of $x$ and $y$ is $1$ and the other is $0$, and the min of $d(x,y)$ happens when $x_n=y_n$ for all indices $n$, then </p>
<p>$0=\sum_{i=1}^\infty \dfrac{0}{2^i}\leq d(x,y)=\sum_{i=1}^\infty \dfrac{|x_i - y_i|}{2^i}\leq \sum_{i=1}^\infty \dfrac{1}{2^i}=1 $</p>
<p>But how to prove that $d(x,y)$ converges to some point when $x$ and $y$ have fixed arbitrary elements?</p>
<p>Thank you. </p>
| Community | -1 | <p>The sequence $a_n = \sum_{k=1}^n \frac{|x_k-y_k|}{2^k}$ is monotonic and bounded from above, thus converges.</p>
<p>$$a_n \le \sum_{k=1}^n \frac{1}{2^k} \le \sum_{k=1}^\infty \frac{1}{2^k} = 1$$</p>
|
3,680,658 | <p>In how many ways are we able to arrange <span class="math-container">$k$</span> identical non-overlapping dominoes on a circle of <span class="math-container">$2n$</span> labelled vertices?</p>
<hr>
<p>The problem can be reduced to the number of ways to choose <span class="math-container">$k$</span> non-consecutive vertices from the <span class="math-container">$2n$</span> labelled vertices,and now there are two cases,assuming we are arranging these <span class="math-container">$k$</span> identical dominoes counterclockwise:</p>
<ul>
<li><p>If vertex <span class="math-container">$1$</span> in included,then the adjacent vertex (from the left) cannot be chosen,since the dominoes are non-overlopping,so we are left with the other <span class="math-container">$2n-2$</span> vertices and we want to choose <span class="math-container">$k-1$</span> nonconsecutive vertices ,this can be done in <span class="math-container">$\binom{2n-2-(k-1)+1}{k-1}=\binom{2n-k}{k-1}$</span> ways.</p></li>
<li><p>If vertex <span class="math-container">$1$</span> in not included,so we are left with the other <span class="math-container">$2n-1$</span> vertices and we want to choose <span class="math-container">$k$</span> nonconsecutive vertices ,this can be done in <span class="math-container">$\binom{2n-1-k+1}{k}=\binom{2n-k}{k}$</span> ways.</p></li>
</ul>
<p>Now summing these two cases gives the answer.</p>
<hr>
<p>I'm not sure about the proof,besides dos it make difference if we do the process clockwise?</p>
| Will Orrick | 3,736 | <p>Let's use <span class="math-container">$d_k$</span> to denote the number non-overlapping arrangements of <span class="math-container">$k$</span> dominoes on a circle of <span class="math-container">$2n$</span> vertices.</p>
<p>Your calculation contains an error. If vertex <span class="math-container">$1$</span> is included, then <em>three</em> vertices must be excluded, not two. Since the domino will cover vertices <span class="math-container">$1$</span> and <span class="math-container">$2$</span>, neither of these may be included. In addition, as you point out, vertex <span class="math-container">$2n$</span> must be excluded. So the correct result is
<span class="math-container">$$
d_k=\binom{2n-k-1}{k-1}+\binom{2n-k}{k}.
$$</span></p>
<p>Here's an alternative method. Let's place a token on each vertex that is not covered by a domino. So every vertex is now covered either by a token or by a domino and the number of dominoes and tokens taken together is <span class="math-container">$k+2(n-k)=2n-k$</span>.</p>
<p>Let's now make a special mark on one object, either domino or token. Since there are <span class="math-container">$2n-k$</span> objects we could choose to mark, the number of marked arrangements is <span class="math-container">$(2n-k)d_k$</span>.</p>
<p>Marked arrangements can be counted in a different way, however, and equating the two answers will allow us to find <span class="math-container">$d_k$</span>. This second method is to lay the <span class="math-container">$k$</span> dominoes and <span class="math-container">$2(n-k)$</span> tokens out in a line rather than a circle. There are <span class="math-container">$\binom{2n-k}{k}$</span> ways to do this. The first object in the line will be the marked one. Now we wrap the arrangement on the circle. Let the vertices be numbered <span class="math-container">$1$</span> to <span class="math-container">$2n$</span>. We just need to pick one of these vertices to be the starting point. So there are <span class="math-container">$2n\binom{2n-k}{k}$</span> marked arrangements.</p>
<p>Equating this to our previous answer shows that
<span class="math-container">$$
d_k=\frac{2n}{2n-k}\binom{2n-k}{k}.
$$</span>
Since <span class="math-container">$\binom{2n-k-1}{k-1}=\frac{k}{2n-k}\binom{2n-k}{k}$</span>, the two methods agree.</p>
|
3,680,658 | <p>In how many ways are we able to arrange <span class="math-container">$k$</span> identical non-overlapping dominoes on a circle of <span class="math-container">$2n$</span> labelled vertices?</p>
<hr>
<p>The problem can be reduced to the number of ways to choose <span class="math-container">$k$</span> non-consecutive vertices from the <span class="math-container">$2n$</span> labelled vertices,and now there are two cases,assuming we are arranging these <span class="math-container">$k$</span> identical dominoes counterclockwise:</p>
<ul>
<li><p>If vertex <span class="math-container">$1$</span> in included,then the adjacent vertex (from the left) cannot be chosen,since the dominoes are non-overlopping,so we are left with the other <span class="math-container">$2n-2$</span> vertices and we want to choose <span class="math-container">$k-1$</span> nonconsecutive vertices ,this can be done in <span class="math-container">$\binom{2n-2-(k-1)+1}{k-1}=\binom{2n-k}{k-1}$</span> ways.</p></li>
<li><p>If vertex <span class="math-container">$1$</span> in not included,so we are left with the other <span class="math-container">$2n-1$</span> vertices and we want to choose <span class="math-container">$k$</span> nonconsecutive vertices ,this can be done in <span class="math-container">$\binom{2n-1-k+1}{k}=\binom{2n-k}{k}$</span> ways.</p></li>
</ul>
<p>Now summing these two cases gives the answer.</p>
<hr>
<p>I'm not sure about the proof,besides dos it make difference if we do the process clockwise?</p>
| Brian M. Scott | 12,042 | <p>You appear to be off a bit: in your first case <span class="math-container">$3$</span> vertices are unavailable, not <span class="math-container">$2$</span>.</p>
<p>I’ve numbered the vertices from <span class="math-container">$1$</span> through <span class="math-container">$2n$</span>. For my first case I put a domino on vertices <span class="math-container">$1$</span> and <span class="math-container">$2$</span>. Now I need to choose <span class="math-container">$k-1$</span> of the <span class="math-container">$2n-3$</span> vertices <span class="math-container">$3,4,\ldots,2n-1$</span>, ensuring that no two chosen vertices are adjacent. This can be done in</p>
<p><span class="math-container">$$\binom{(2n-3)-(k-2)}{k-1}=\binom{2n-1-k}{k-1}$$</span></p>
<p>ways.</p>
<p>For my second case I put a domino on vertices <span class="math-container">$2n$</span> and <span class="math-container">$1$</span>; the analysis is the same, so we get another <span class="math-container">$\binom{2n-1-k}{k-1}$</span> arrangements.</p>
<p>Any other arrangement must avoid vertex <span class="math-container">$1$</span> entirely. In that case we need to choose <span class="math-container">$k$</span> of the <span class="math-container">$2n-2$</span> vertices <span class="math-container">$2,3,\ldots,2n-1$</span>, ensuring that no two chosen vertices are adjacent. This can be done in</p>
<p><span class="math-container">$$\binom{(2n-2)-(k-1)}k=\binom{2n-1-k}k$$</span></p>
<p>ways. I get a total of</p>
<p><span class="math-container">$$2\binom{2n-1-k}{k-1}+\binom{2n-1-k}k=\binom{2n-1-k}{k-1}+\binom{2n-k}k$$</span></p>
<p>arrangements. I’ve checked this by hand with <span class="math-container">$n=4$</span> and <span class="math-container">$k=3$</span>.</p>
|
4,508,796 | <p>How to find the integral
<span class="math-container">$$\int_0^1 x\sqrt{\frac{1-x}{1+x}}dx$$</span></p>
<p>I tried by substituting <span class="math-container">$x=\cos a$</span>. But it's leading to a form <span class="math-container">$\sin2a\cdot\tan a/2$</span> which I can't integrate further.</p>
| Ucchash Hore | 878,812 | <p>Well I tried this problem yesterday so here is my solution .</p>
<p>Let <span class="math-container">$x= \cos (2a)$</span>; hence ,<span class="math-container">$\mathrm dx=2\sin(2a) \mathrm da$</span>
putting this value you will get<br />
<span class="math-container">$x\cdot \sqrt{\frac{1-x}{1+x}}\\=\cos (2a)\cdot \tan a\cdot 2\sin (2a)\\
=\cos (2a)\cdot \tan a\cdot 4\sin a \cdot \cos a
\\=\cos (2a)\cdot 4(\sin a)^2 \\=2\cos (2a)\cdot (1-\cos (2a))
\\=2\cos (2a) -2(\cos 2a)^2 \\=2\cos 2a -(1+\cos (4a))$</span></p>
<p>I think you can do the rest.</p>
|
615,375 | <p>I would appreciate if somebody could help me with the following problem</p>
<p>Q: Let $f:[0,1]\longrightarrow \Bbb R$ be a continuously function such that
$$m\leq f(x)\leq M, m+M=1 (m:\text{minimum of} f(x), M:\text{maximum of} f(x) )$$
Prove that for every $x \in[0,1]$, there exists $c\in[0,1]$ such that $f(c)=1-f(x)$ </p>
| copper.hat | 27,978 | <p>We have $f([0,1]) = [m,M]$, and $m+M=1$. Hence if we let $\phi(x) = 1-f(x)$, we have $\phi([0,1]) = [1-M,1-m] = [m,M]$.</p>
<p>Hence $f([0,1]) = \phi([0,1])$.</p>
<p>It should be straightforward to finish from here.</p>
|
773,324 | <p>I know Dijkstra's algorithm to find the shortest way between 2 nodes, but is there a way to find the shortest path between 3 nodes among $n$ nodes? Here are the details:</p>
<p>I have $n$ nodes, some of which are connected directly and some of which are connected indirectly, and I need to find the shortest path between 3 of them.</p>
<p>For example, given $n = 6$ nodes labelled A through F, and the following graph:</p>
<pre><code>A-->B-->C
A-->D-->E
D-->F
</code></pre>
<p>How can I find the shortest path between the three nodes (A,E,F)?</p>
<p>I am looking for a solution similar to Dijkstra's shortest path algorithm, but for 3 nodes instead of 2.
<br/>
Please Note : <br/>
1- The Starting Node is A <br/>
2- The Sequential is not important just the path needs to cover all these Nodes <br/>
3- Their is no return back to A <br/>
Please find the diagram Image
<img src="https://i.stack.imgur.com/M1wxF.png" alt="enter image description here">
Regards & Thanks<br />
Nahed</p>
| Lily Chung | 32,765 | <p>For the case of a start node S and two target nodes X and Y, one could use the following algorithm:</p>
<p>Use Dijkstra's shortest-path algorithm to find the shortest path from S to X and the shortest path from S to Y. If path from S to X is shorter, use Dijkstra's shortest-path algorithm to find the shortest path from X to Y, and follow the paths found from S to X and then from X to Y. Else (if the second path is shorter), find the shortest path from Y to X and follow the paths found from S to Y and then from Y to X.</p>
<p>Since this always uses Dijkstra's algorithm exactly 3 times, it is asymptotically just as efficient as Dijkstra's algorithm.</p>
<hr>
<p>Note that, as Tyler Olsen and ml0105 point out, if there are in fact a variable number of nodes you need to pass through instead of only 3, this problem is NP-Hard.</p>
|
2,304,448 | <blockquote>
<p>Does any group homomorphism $\Bbb Z \to \Bbb Z/n$ have kernel isomorphic to $\Bbb Z$?</p>
</blockquote>
<p>Here $n$ is any natural number.</p>
| Tomasz Kania | 17,929 | <p>I am confused because what is required here is almost the version of the open mapping theorem that I was taught as an undergraduate:</p>
<p><em>Suppose that $X$ and $Y$ are Banach spaces and $T\colon X\to Y$ is a bounded linear operator. If the range of $T$ is not meagre in $Y$, then $T$ is surjective.</em></p>
<p>See <a href="https://math.stackexchange.com/questions/177530/range-of-bounded-operator-is-of-first-category/177586#177586">here</a> for the proof. In your case, you simply apply contraposition of this statement as certainly in your case $T$ is not surjective, having non-closed range.</p>
|
2,580,822 | <p>first I will describe my problem (1), then solutions I found (2) and then pose the question (3).</p>
<p>1) I am a student making simple program for statistical analysis of data for biological laboratory. The situation is as follows: They have a protein and measure some property three times (to eliminate measurement error). Then they mix the same batch of protein with some chemical and measure the same property again twice. In the end they want to know, if the measurements are different enough. (I know the number of measurements is low, but it is the maximum they can do).</p>
<p>2) If we assume, that the samples are independent I would choose Welch's t-test, because we cannot assume the same variance. However they say that, due to the fact that we measure the same batch of protein three times, and then two times with some chemical, the before and after measurements are related samples. But for related samples I only found paired t-tests, which cannot be used here, because there are no explicit pairs. The only thing that comes to my mind is to pair each measurement from each set. </p>
<p>3) How to measure statistically significant difference of a property, when we have 3 measurements of batch of protein before applying a chemical and 2 measurements after applying a chemical. Measurements are repeated to eliminate error in measurement device.</p>
| Archis Welankar | 275,884 | <p>A short hint. Since the number is low and same batch of proteins is used to do two different tasks you can choose t-test $t=\frac {d\bar-\mu }{\frac {s}{\sqrt {n}}} $ where $d\bar$ difference between results of two tests and $\mu,s$ are the mean and standard deviations of the $d\bar $. n is the total number of samples tested. Hope it helps!</p>
|
2,580,822 | <p>first I will describe my problem (1), then solutions I found (2) and then pose the question (3).</p>
<p>1) I am a student making simple program for statistical analysis of data for biological laboratory. The situation is as follows: They have a protein and measure some property three times (to eliminate measurement error). Then they mix the same batch of protein with some chemical and measure the same property again twice. In the end they want to know, if the measurements are different enough. (I know the number of measurements is low, but it is the maximum they can do).</p>
<p>2) If we assume, that the samples are independent I would choose Welch's t-test, because we cannot assume the same variance. However they say that, due to the fact that we measure the same batch of protein three times, and then two times with some chemical, the before and after measurements are related samples. But for related samples I only found paired t-tests, which cannot be used here, because there are no explicit pairs. The only thing that comes to my mind is to pair each measurement from each set. </p>
<p>3) How to measure statistically significant difference of a property, when we have 3 measurements of batch of protein before applying a chemical and 2 measurements after applying a chemical. Measurements are repeated to eliminate error in measurement device.</p>
| BruceET | 221,800 | <p>This semi-answer used to be a Comment, but too abbreviated to make sense:</p>
<p>Everything "they say" is not necessarily helpful. I think a two-sample t test is the right thing to use. The two groups in any two-sample test have <em>something</em> in common. Details are sparse, but the argument that you don't have two
independent samples (as presented) seems pathetic.</p>
<p>However, Welch 2-sample t will probably not find a difference with three and two replications per group. Even with a pooled 2-sample t test, you're down to
three degrees of freedom, so the absolute value of the t statistic would
have to be large to get a significant result. A Welch t test will characteristically have even fewer degrees of freedom.</p>
<p>If all of the before measurements are smaller than any of the after measurements (or the reverse), then you'd have a P-value of only $1/{5 \choose 3} = 1/10,$ based on relative sizes of the measurements. So no kind of rank-based test (or permutation test) is going to find a significant result at the 5% level. </p>
<p>Recommendation: Cross fingers; use 2-sample pooled t test; hope for the best. The two samples are based on testing the same medium, with the same method and
with the same chemistry students students using the technology. Maybe OK to hope for equal population variances.</p>
<hr>
<p>Here are the Welch and the Pooled two-sample t tests done in R statistical
software with some "data" I made up--with the idea of maybe being able
to show a significant difference. Welch not significant at the 5% level;
Pooled is significant.</p>
<pre><code>x = c(12.173, 13.531, 11.044); y = c(17.681, 21.321) # My fake data
t.test(x,y) # Default in R is Welch
Welch Two Sample t-test
data: x and y
t = -3.7058, df = 1.32, p-value = 0.1188 # P-val > .05: Not Signif
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-21.552561 7.049228 # Includes 0: Not signif
sample estimates:
mean of x mean of y
12.24933 19.50100
t.test(x,y, var.eq=T) # Parameter 'var.eq=T' forces Pooled test
Two Sample t-test
data: x and y
t = -4.4118, df = 3, p-value = 0.02161 # P-value < 5%: Significant
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-12.482607 -2.020727 $ Does NOT include 0: Significant
sample estimates:
mean of x mean of y
12.24933 19.50100
qt(.975, c(1.32, 3))
## 7.311460 3.182446 # 5% critical values for Welch and Pooled 2-sided test ...
# ... |T| > crit. val implies Reject Null Hypothesis
</code></pre>
|
4,059,426 | <ul>
<li>In a longer derivation I ran into the following quantity:
<span class="math-container">$$
\nabla\left[\nabla\cdot\left(%
{\bf r}_{0}\,{\rm e}^{{\rm i}{\bf k} \cdot {\bf r}}\,\right)
\right]
$$</span>
( i.e., the gradient of the divergence ) where <span class="math-container">${\bf k}$</span> is a vector of constants and <span class="math-container">${\bf r}$</span> is a position vector.</li>
<li>Can someone help explaining how to calculate this?
I am hoping it gives:
<span class="math-container">$$
\nabla\left[\nabla\cdot\left(%
{\bf r}_{0}\,{\rm e}^{{\rm i}{\bf k} \cdot {\bf r}}\,\right)
\right] =
-{\bf k}\left({\bf k}\cdot{\bf r}_{0}\right)
$$</span>
( because then the rest of my equations add up ).</li>
</ul>
| 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]{\nabla\bracks{%
\nabla\cdot\pars{{\bf r}_{0}\expo{\ic{\bf k}\cdot{\bf r}}}}} =
\nabla\bracks{%
\overbrace{\pars{\nabla\cdot{\bf r}_{0}}}^{\ds{=\ 0}}\
\expo{\ic{\bf k}\cdot{\bf r}} +
{\bf r}_{0}\cdot\nabla\pars{\expo{\ic{\bf k}\cdot{\bf r}}}} =
\ic\nabla\pars{{\bf r}_{0}\cdot{\bf k}\,\expo{\ic{\bf k}\cdot{\bf r}}}
\\[5mm] = &\
\ic\,{\bf r}_{0}\cdot{\bf k}\,\nabla\expo{\ic{\bf k}\cdot{\bf r}} =
\ic\,{\bf r}_{0}\cdot{\bf k}\,\pars{\expo{\ic{\bf k}\cdot{\bf r}}
\,\ic{\bf k}} =
\bbx{-{\bf k}\pars{{\bf k}\cdot{\bf r}_{0}}\expo{\ic{\bf k}\cdot{\bf r}}}
\\ &
\end{align}</span>
There are a <a href="https://dlmf.nist.gov/1.6" rel="nofollow noreferrer">lot of vectorial identities over here</a>.</p>
|
3,445,768 | <p>I am trying to solve a nonlinear differential equation of the first order that comes from a geometric problem ; <span class="math-container">$$x(2x-1)y'^2-(2x-1)(2y-1)y'+y(2y-1)=0.$$</span></p>
<p>edit1 <strong><em>I am looking for human methods to solve the equation</em></strong> </p>
<p>edit2 the geometric problem was discussed on this french forum <a href="http://www.les-mathematiques.net/phorum/read.php?8,1779080,1779080" rel="nofollow noreferrer">http://www.les-mathematiques.net/phorum/read.php?8,1779080,1779080</a> </p>
<p>We can see the differential equation here <a href="http://www.les-mathematiques.net/phorum/read.php?8,1779080,1780782#msg-1780782" rel="nofollow noreferrer">http://www.les-mathematiques.net/phorum/read.php?8,1779080,1780782#msg-1780782</a></p>
<p>edit 3 I do not trust formal computer programs: look at Wolfram's answer when asked to calculate the cubic root of -1 <a href="https://www.wolframalpha.com/input/?i=%7B%5Csqrt%5B3%5D%7B-1%7D%7D%29+" rel="nofollow noreferrer">https://www.wolframalpha.com/input/?i=%7B%5Csqrt%5B3%5D%7B-1%7D%7D%29+</a>.</p>
| Francois Rideau | 728,345 | <p>This problem is about <a href="http://mathworld.wolfram.com/IsotomicTransversal.html" rel="nofollow noreferrer">isotomic transversals</a>: </p>
<p>In general, each point of the plane is on a pair of isotomic transversals with respect to some given triangle <span class="math-container">$ABC$</span>.
So is defined a field of directions in the plane.
Write down the differential equation linked to this field and integrate it.</p>
|
409,626 | <p>I'm studying elementary group theory, and just seeing the ways in which groups break apart into simpler groups, specifically, a group can be broken up as the sort of product of any of its normal subgroups with the quotient group of that subgroup. So I wondered how you could do the inverse of that operation:</p>
<ol>
<li>Given two groups $A$ and $B$, construct a group $G$ which admits a normal subgroup $H$ isomorphic to $A$, such that $G/H$ is isomorphic to $B$.</li>
</ol>
<p>I think I have a proof that the cartesian product $A \times B$ (with the usual component-wise operation) verifies (1), but since I'm just starting out I'm not totally confident in my construction. Furthermore, if I'm right, is this the <em>only</em> group up to isomorphism satisfying (1)?</p>
<p>Edit: I just noticed <a href="https://math.stackexchange.com/questions/220487/proving-the-direct-product-d-of-two-groups-g-h-has-a-normal-subgroup-n-such-th?rq=1">Proving the direct product D of two groups G & H has a normal subgroup N such that N isomorphic to G and D/N isomorphic to H</a>, which seems to positively answer my question. In that case I'd like to draw attention to the follow up question above (uniqueness up to isomorphism).</p>
| Zev Chonoles | 264 | <p>You're correct that $A\times B$ <em>satisfies</em> (<a href="http://www.jmilne.org/math/words.html" rel="nofollow">see here</a>) the stated property, but in general it will not be the only such group. </p>
<p>The simplest example is with $A=B=\mathbb{Z}/2\mathbb{Z}$, in which case $\mathbb{Z}/4\mathbb{Z}$ also has the desired property.</p>
|
409,626 | <p>I'm studying elementary group theory, and just seeing the ways in which groups break apart into simpler groups, specifically, a group can be broken up as the sort of product of any of its normal subgroups with the quotient group of that subgroup. So I wondered how you could do the inverse of that operation:</p>
<ol>
<li>Given two groups $A$ and $B$, construct a group $G$ which admits a normal subgroup $H$ isomorphic to $A$, such that $G/H$ is isomorphic to $B$.</li>
</ol>
<p>I think I have a proof that the cartesian product $A \times B$ (with the usual component-wise operation) verifies (1), but since I'm just starting out I'm not totally confident in my construction. Furthermore, if I'm right, is this the <em>only</em> group up to isomorphism satisfying (1)?</p>
<p>Edit: I just noticed <a href="https://math.stackexchange.com/questions/220487/proving-the-direct-product-d-of-two-groups-g-h-has-a-normal-subgroup-n-such-th?rq=1">Proving the direct product D of two groups G & H has a normal subgroup N such that N isomorphic to G and D/N isomorphic to H</a>, which seems to positively answer my question. In that case I'd like to draw attention to the follow up question above (uniqueness up to isomorphism).</p>
| Dan Rust | 29,059 | <p>Your idea about the direct product working is true. The general idea you are looking for however is that of a <a href="http://en.wikipedia.org/wiki/Semidirect_product" rel="nofollow">semidirect product</a> which refutes your claim of uniqueness in the general setting.</p>
<p>It's interesting to ask <em>when</em> the direct product is the only possibility and that can actually be answered by finding the group $\text{Ext}(B,A)$. This is know as an <a href="http://en.wikipedia.org/wiki/Extension_problem#Extension_problem" rel="nofollow">extension problem</a>. We have that if $\text{Ext}(B,A)=1$ then $A\times B$ is the unique extension of $B$ by $A$.</p>
<p>I believe the converse is false however, but I can't think of an example at the moment. That is, there exist pairs $(B,A)$ such that $\text{Ext}(B,A)\neq 1$ but $B\times A$ is the unique (up to group isomorphism, not extension equivalence) extension of $B$ by $A$.</p>
|
3,245,945 | <p>Suppose <span class="math-container">$A$</span> is a linearly ordered set without maximum or minimum and every closed interval is a finite set. I want to show <span class="math-container">$A$</span> is isomorphic to the set of integers with the usual order.</p>
<p>I know that if <span class="math-container">$A$</span> is countable then I can use induction to construct partial isomorphisms and hence an isomorphism.</p>
<p>Any help is appreciated.</p>
| Community | -1 | <p>Consider <span class="math-container">$x_0\in A$</span> and the function <span class="math-container">\begin{align}f:A&\to \Bbb Z\\ f(x)&=\lvert[x_0,x)\rvert-\lvert (x,x_0]\rvert\end{align}</span></p>
<p>This function is injective (which is sufficient for the specific passage you need). It's also surjective and an isomorphism of orders, however.</p>
|
972,530 | <blockquote>
<p>If $S_1 = \sqrt{2}$, and</p>
<p>$S_{n+1} = \sqrt{2 + \sqrt{S_n}}$ (n = 1,2,3....),</p>
<p>prove that $\{S_n\}$ converges, and that $S_n < 2$ for all $n \in \Bbb{N}$</p>
<p>This is one the questions from Principles of Mathematical Analysis by Rudin. I am not sure how to proceed with the solution.</p>
</blockquote>
| Kevin Sheng | 150,297 | <p>Well if we prove that $S_n<2$ for $n=1,2,3,\ldots$ then the first part comes easily as $S_n$ is a sequence in $\mathbb{R}$ which is complete. Since $S_n$ is a monotone increasing sequence with an upper bound (namely 2), it converges. We can prove $S_n<2$ for $n \ge 1$ via induction as follows:</p>
<p>$\textbf{I:}$ $ S_1 < 2 $, i.e. $\sqrt{2} < 2$, can be checked by direct computation.</p>
<p>$\textbf{II:}$ Suppose $S_{n-1}<2$ but that $S_n \ge 2$. But then we have $\sqrt{2 + \sqrt{S_{n-1}}} \ge 2 \Rightarrow \sqrt{S_{n-1}} \ge 2$, a contradiction. Thus $S_n < 2$.</p>
<p>$\textbf{III:}$ Thus $S_n<2$ for all $n = 1,2,3,\ldots$.</p>
|
381,067 | <p>I am asked this question:</p>
<blockquote>
<p>Prove that $x^5 - 5x + 1$ has no double roots in $\mathbb C[x]$. </p>
</blockquote>
<p>Now here's what I said:</p>
<blockquote>
<p>$p(x) \in \mathbb C[x]$ has no double roots if and only if $gcd(p,p') = 1$. Now we need to prove that. So:
$p(x) = x^5 - 5x + 1$ and $p'(x) = 5x^4 - 5$ and I start doing the Euclidean algorithm and I got to a point where I have to do $\frac{-4x+1}{ \frac{5}{4}x^3-5 }$, which we obviously can not do. </p>
</blockquote>
<p>What do we do in such a situation? What does that mean that while doing the Euclidean algorithm we can not continue dividing?</p>
| Ma Ming | 16,340 | <p>You are almost finished. $\gcd(-4x+1, \frac{5}{4} x^3-5)=1$</p>
|
449,413 | <p>I'm trying to construct a norm on the space $\mathcal{D}(\Omega) := \{ f \in C^\infty(\Omega) | f $ is compactly supported on $ \Omega \}$ where $\Omega$ is an open subset of $\mathbb{R}$. I want this norm to include, somehow, the $L^\infty$-norms of <strong>all</strong> the derivatives of the smooth function to which it is applied. Specifically, I want to be able to encapsulate the statement</p>
<p>"<em>$f_n$$^{(m)} \rightarrow f^{(m)} ($as $n \rightarrow \infty)$ uniformly for all non-negative values of $m$.</em>"</p>
<p>as the statement</p>
<p>"$\|f_n - f\| \rightarrow 0$ $(n \rightarrow \infty)$.",</p>
<p>where $\|\cdot\|$ denotes my desired norm.</p>
<p>So far I've considered trying to write it as something along the lines of</p>
<p>\begin{align}\|f\| := \sum\limits_{m \in \mathbb{N}} \frac{1}{m!} \|f^{(m)}\|_\infty, \end{align}
where $\|\cdot\|_\infty$ denotes the usual $L^\infty$-norm. I am very unsure of the validity of this sort of "definition", as I can't see how to prove that this (or a related/similar) series converges. Any help on this would be greatly appreciated! Thanks in advance.</p>
| AD - Stop Putin - | 1,154 | <p>If $\Omega$ is open then there is no norm on $C^\infty(\Omega)$, it is a classical Montel space. </p>
<p>In addition, for compact $K\subset\Omega$ we may put $\sup_{x\in K}|f^{(k)}(x)|$ is finite, but $\sup_K\sup_{x\in K}|f^{(k)}(x)|$ may be infinite.</p>
|
2,919,561 | <p>What is the following limit ? How do I solve it ?
$$\lim \limits_{x\to0}\frac{6\sin x}{x-3\tan x}$$</p>
| user | 505,767 | <p><strong>HINT</strong></p>
<p>We don't need l'Hopital indeed we have</p>
<p>$$\lim \limits_{x\to0}\frac{6\sin x}{x-3\tan x}=\lim \limits_{x\to0}\frac{6\frac{\sin x}x}{1-3\frac{\tan x}x}$$</p>
<p>then refer to standard limits.</p>
|
439,745 | <blockquote>
<p>Prove:$|x-1|+|x-2|+|x-3|+\cdots+|x-n|\geq n-1$</p>
</blockquote>
<p>example1: $|x-1|+|x-2|\geq 1$</p>
<p>my solution：(substitution)</p>
<p>$x-1=t,x-2=t-1,|t|+|t-1|\geq 1,|t-1|\geq 1-|t|,$</p>
<p>square,</p>
<p>$t^2-2t+1\geq 1-2|t|+t^2,\text{Since} -t\leq -|t|,$</p>
<p>so proved.</p>
<p><em>question1</em> : Is my proof right? Alternatives?</p>
<p>one reference answer: </p>
<p>$1-|x-1|\leq |1-(x-1)|=|1-x+1|=|x-2|$</p>
<p><em>question2</em> : prove:</p>
<p>$|x-1|+|x-2|+|x-3|\geq 2$</p>
<p>So I guess:( I think there is a name about this, what's that? wiki item?)</p>
<p>$|x-1|+|x-2|+|x-3|+\cdots+|x-n|\geq n-1$</p>
<p>How to prove this? This is <em>question3.</em> I doubt whether the two methods I used above may suit for this general case.</p>
<p>Of course, welcome any interesting answers and good comments.</p>
| Thomas Andrews | 7,933 | <p>Actually, $|x-n| + |x-1| = |n-x| + |x-1| \geq |n-1|=n-1$. Don't need the other terms.</p>
<p>Your first proof is correct. You have to take some care moving from $U^2\geq V^2$ to $U\geq V$, but in this case, $U=|t-1|$ is non-negative.</p>
|
3,484,136 | <blockquote>
<p>Show that <span class="math-container">$n^2+n$</span> is even for all <span class="math-container">$n\in\mathbb{N}$</span> by contradiction.</p>
</blockquote>
<p>My attempt: assume that <span class="math-container">$n^2+n$</span> is odd, then <span class="math-container">$n^2+n=2k+1$</span> for all <span class="math-container">$k\in\mathbb{N}$</span>.</p>
<p>We have that
<span class="math-container">$$n^2+n=2k+1 \Leftrightarrow \left(n+\frac{1}{2}\right)^2-\frac{1}{4}=2k+1\Leftrightarrow n=\sqrt{2k+\frac{5}{4}}-\frac{1}{2}.$$</span></p>
<p>Choosing <span class="math-container">$k=1$</span>, we have that <span class="math-container">$n=\sqrt{2+\frac{5}{4}}-\frac{1}{2}\notin\mathbb{N}$</span>, so we have a contradiction because <span class="math-container">$n^2+n=2k+1$</span> should be verified for all <span class="math-container">$n\in\mathbb{N}$</span> and for all <span class="math-container">$k\in\mathbb{N}$</span>.</p>
<p>Is this correct or wrong? If wrong, can you tell me where and why? Thanks.</p>
| ViHdzP | 718,671 | <p>Other users have already pointed out your big mistake: you can’t say that <span class="math-container">$n^2+n=2k+1$</span> <strong>for all <span class="math-container">$\mathbf k$</span></strong>, since being odd only means that this is true for a <strong>specific <span class="math-container">$\mathbf k$</span></strong>. We can complete your proof as follows. If <span class="math-container">$n^2+n=2k+1$</span>, by your same calculation, <span class="math-container">$$n=\sqrt{2k+\frac54}-\frac12=\frac12\left(\sqrt{8k+5}-1\right).$$</span> However, <span class="math-container">$8k+5$</span> isn’t a square <strong>for any <span class="math-container">$\mathbf k$</span></strong> (as a square must give a remainder of <span class="math-container">$0$</span>, <span class="math-container">$1$</span> or <span class="math-container">$4$</span> when divided by <span class="math-container">$8$</span>), so that, even when assuming the original statement for a single <span class="math-container">$k$</span>, we arrive at the contradiction that <span class="math-container">$n$</span> is irrational. <span class="math-container">$\blacksquare$</span></p>
|
1,130,855 | <p>I've searched this website and while there are a few questions similar to mine, I couldn't find what I was looking for/a specific method for what I want to do.</p>
<p>I want to understand how one would prove that the remainder of $5^{336}$ by $23$ is $8$, or in other words, $$5^{336}\equiv 8 \pmod{23}.$$</p>
<p>Can anyone help me understand how one would approach a problem like this?</p>
| AlexR | 86,940 | <p>$23$ is prime, so $\phi(23) = 22$. You can thus <em>always</em> take the exponent modulo $22$ (by <a href="http://en.wikipedia.org/wiki/Euler%27s_theorem" rel="nofollow">Euler's theorem</a>, or in this special case <a href="http://en.wikipedia.org/wiki/Fermat%27s_little_theorem" rel="nofollow">Fermat's little theorem</a>).
$$5^{336} \equiv 5^6 \equiv 2^3 = 8 \pmod{23}$$
The last congruence is because $5^2 = 25 \equiv 2$</p>
<p><strong>Remark on Brute-Forcing</strong><br>
We actually have $\mathrm{ord}_{23}(5) = 22$ so you'd have to do many computations until you arrive there. This is highly impractical.</p>
<p><strong>Remark on modulus in exponents</strong><br>
We know that
$$a^{\mathrm{ord}_n(a)}\equiv 1 \pmod n$$
by definition of the order of an element. Also we know that the order of an element is a divisor of the group order, alas $\mathrm{ord}_n(a) | \phi(n)$. This allows us to make two statements:
$$a^k \equiv a^{k\bmod \mathrm{ord}_n(a)} \pmod n$$
and as a corollary of that, $a^k \equiv a^{k\bmod \phi(n)} \pmod n$.</p>
|
1,250,132 | <p>Below is part of a solution to a critical points question. I'm just not sure how the equation on the left becomes the equation on the right. Could someone please show me the steps in-between? Thanks.</p>
<blockquote>
<p>$$\frac{-1}{x^2}+2x=0 \implies 2x^3-1=0$$</p>
</blockquote>
| Jeel Shah | 38,296 | <p>The solution has multiplied both left and right side by $x^2$</p>
<p>Observe:
\begin{align}
&-\frac{1}{x^2} * x^2 + 2x(x^2) = 0 * x^2 \\
&\implies -\frac{x^2}{x^2} + 2x^3 = 0 \\
&\implies -1 + 2x^3 = 0
\end{align}</p>
|
1,014,476 | <p>I pick 6 cards from a set of 13 (ace-king). If ace = 1 and jack,queen,king = 10 what is the probability of the sum of the cards being a multiple of 6? </p>
<p><strong>Tried so far:</strong>
I split the numbers into sets with values:
6n, 6n+1, 6n+2, 6n+3
like so:</p>
<p>{6}{1,7}{2,8}{3,9}{4,10,j,q,k}{5}</p>
<p>and then grouped the combinations that added to a multiple of 6:</p>
<p>(5c4)(1c1)(2c1) + (2c2)(2c2)(2c2) + (5c4)(2c2) + (5c2)(1c1)(1c1)(2c1)(2c1)
/ (13c6)</p>
<p>= 10/1716</p>
<p>I am almost certain I am missing combinations but am having trouble finding out which.</p>
| David K | 139,123 | <p>It will help to explicitly list the values of cards in each combination, as other answers have done, <em>then</em> count the number of ways you can make each combination.
If doing this by hand rather than by computer, you were wise to group the cards by
their remainders after division by $6$; you can use just the remainders as the terms
of your sum (as already shown in two answers).</p>
<p>For example, you can make a multiple of $6$ by taking $4+4+4+4+5+3=24.$
There are exactly $\binom54\binom11\binom21$ ways to draw cards without replacement
whose remainders would add up in exactly this way.
(You could also write this expression $(5C4)(1C1)(2C1)$ as you did in the question.)</p>
<p>But you can also make a multiple of $6$ by taking $4+4+4+4+2+0=24.$
There are <em>also</em> exactly $\binom54\binom11\binom21$ ways to draw such a set of cards
(or you could write $\binom54\binom21\binom11$ so that the terms you are multiplying
are in the same order as the corresponding terms of the sum, for clarity),
and these sets are all different from the ones that are counted in the previous paragraph.</p>
<p>You might make a table of sums and the number of ways to draw each sum:</p>
<p>$$\begin{array}{lcrcr}
4+4+4+4+5+3=24 && \binom54\binom11\binom21&=&10 \\
4+4+4+4+2+0=24 && \binom54\binom21\binom11&=&10 \\
4+4+4+4+1+1=24 && \binom54\binom22&=&5 \\
4+4+4+5+1+0=18 && \binom53\binom11\binom21\binom11&=&20
\end{array}$$</p>
<p>Continue this column for several more rows until you have listed all the sums that
are possible to draw that are divisible by $6.$
<em>Then</em> you can add up the values in the right-hand column.</p>
|
1,221,914 | <blockquote>
<p>Let <span class="math-container">$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 ,\ n\ge 1$</span> have <span class="math-container">$n$</span> roots <span class="math-container">$x_1,x_2,\ldots,x_n \le -1$</span> and <span class="math-container">$a_0^2+a_1a_n=a_n^2+a_0a_{n-1}$</span>.
Find all such <span class="math-container">$P(x)$</span>. (Poland 1990).</p>
</blockquote>
<p>I used Viete Theorem and get: <span class="math-container">$$(x_1x_2\cdots x_n)^2 \pm \sum_{j=1}^n\prod_{i \not=j} x_i=1\pm(x_1x_2 \cdots x_n)(x_1+x_2+\cdots+x_n)$$</span> but not succeeded.</p>
| Karanko | 227,830 | <p>Let $s_i$ denote the <a href="https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial#Definition" rel="nofollow">elementary symmetric polynomials</a> in $x_1,x_2,...,x_n$.</p>
<p>Then $a_0/a_n=(-1)^ns_n$, $a_1/a_n=(-1)^{n-1}s_{n-1}=(-1)^{n-1}s_n\left(\sum_{k=1}^{n}\frac{1}{x_k}\right)$, and $a_{n-1}/a_n=-s_1$.</p>
<p>We can re-interpret $a_0^2+a_1a_n=a_n^2+a_0a_{n-1}$ as</p>
<p>$$s_n^2+(-1)^{n-1}s_{n-1}=1-s_ns_{n-1}$$</p>
<p>$$s_n(s_n-s_{n-1})+(-1)^{n-1}s_{n-1}=1$$</p>
<p>Assume $n$ is even. Then, $s_n$ is positive, $-s_{n-1}$ is positive, $(-1)^{n-1}s_{n-1}$ is positive. If at least one of the $x_k<-1$ then $(-1)^{n-1}s_{n-1}>1$ and the equality can't happen. Therefore, all $x_k=-1$.</p>
<p>Assume now that $n$ is odd. Similarly as before, $s_n(s_n-s_{n-1})$ is positive, $(-1)^{n-1}s_{n-1}$ is positive again. If at least one of the $x_k<-1$ then the sum on the left is $>1$. Therefore all $x_k=-1$.</p>
<p>Hence $P(x)=a_n(x+1)^{n}$, or forgetting (dividing out) about the $a_n$, $P(x)=(x+1)^{m}$. This would give us $a_0=1$, $a_1=n$, $a_{n-1}=n$. We need $$1^2+n=1^2+n$$ which holds.</p>
|
4,572,336 | <p>This is the
<a href="https://i.stack.imgur.com/BRjsf.jpg" rel="nofollow noreferrer">question</a></p>
<p>I have managed to find the formula for part a) which is n(n+1) for all natural numbers n, but I'm not sure how to prove it directly. Can I just say that the summation of 2i is the same as the sums of the summation of i plus the summation of i as the proof?</p>
<p>I have managed to do part b) which uses induction to prove that the summation of even numbers can be found using the formula n(n+1).</p>
| Scene | 723,326 | <p>We have
<span class="math-container">$$
2+4+\dots+2n =2(1+2+\dots+n) =2\left (\frac{n(n+1)}{2}\right)=n(n+1)
$$</span></p>
|
4,572,336 | <p>This is the
<a href="https://i.stack.imgur.com/BRjsf.jpg" rel="nofollow noreferrer">question</a></p>
<p>I have managed to find the formula for part a) which is n(n+1) for all natural numbers n, but I'm not sure how to prove it directly. Can I just say that the summation of 2i is the same as the sums of the summation of i plus the summation of i as the proof?</p>
<p>I have managed to do part b) which uses induction to prove that the summation of even numbers can be found using the formula n(n+1).</p>
| Dean Menezes | 452,340 | <p><span class="math-container">$$\begin{align} 2 + 4 + \dots + 2n &= (1 + \dots + n) + (1 + \dots + n) \cr & = (1 + 2 + \dots + n) + (n + (n-1) + \dots + 1) \cr & = (n+1) + \dots + (n+1) = n(n+1)\end{align}$$</span></p>
|
4,572,336 | <p>This is the
<a href="https://i.stack.imgur.com/BRjsf.jpg" rel="nofollow noreferrer">question</a></p>
<p>I have managed to find the formula for part a) which is n(n+1) for all natural numbers n, but I'm not sure how to prove it directly. Can I just say that the summation of 2i is the same as the sums of the summation of i plus the summation of i as the proof?</p>
<p>I have managed to do part b) which uses induction to prove that the summation of even numbers can be found using the formula n(n+1).</p>
| Nayjer | 1,117,302 | <p>Yes, that should be enough. You can also say:</p>
<p><span class="math-container">$\sum_{i=1}^{n}2i = 2 \cdot \sum_{i=1}^{n}i = 2 \cdot \frac{n(n+1)}{2}$</span>.</p>
|
4,572,336 | <p>This is the
<a href="https://i.stack.imgur.com/BRjsf.jpg" rel="nofollow noreferrer">question</a></p>
<p>I have managed to find the formula for part a) which is n(n+1) for all natural numbers n, but I'm not sure how to prove it directly. Can I just say that the summation of 2i is the same as the sums of the summation of i plus the summation of i as the proof?</p>
<p>I have managed to do part b) which uses induction to prove that the summation of even numbers can be found using the formula n(n+1).</p>
| user | 293,846 | <p>Hint for even <span class="math-container">$n$</span>:
<span class="math-container">$$\left.\begin{align}
&2+2n&=2(n+1)\\
&4+(2n-2)&=2(n+1)\\
&6+(2n-4)&=2(n+1)\\
&\dots\\
&n+(n+2)&=2(n+1)
\end{align}\right \}\frac n2\text{ rows }$$</span></p>
|
14,735 | <p>This question is somewhat related to <a href="https://mathematica.stackexchange.com/questions/4576/changing-the-order-of-elements-in-a-chart-legend">this</a> one.</p>
<p>Let's say this is the <code>BarChart</code> i want to make:</p>
<pre><code>BarChart[Range[5], ChartStyle -> "Rainbow", BarOrigin -> Left,
ChartLegends -> Map[ToString, Range[5]]]
</code></pre>
<p><img src="https://i.stack.imgur.com/dQQw3.jpg" alt="BarChart reflected"></p>
<p>The <code>BarChart</code> runs from the bottom and the legend starts at the top. The related question (linked above) gives a solution to reverse the legend, but i'd rather have the axis (and thus the purple bar) on top. It should look like this (made with <code>ImageReflect[]</code>):</p>
<p><img src="https://i.stack.imgur.com/KU79U.jpg" alt="BarChart reflected"></p>
<p>I tried <code>AxesOrigin</code>, but that doesn't work. My question seems pretty trivial, but i haven´t found a solution for it yet.</p>
<p><strong>Question: is there a way to specify the origin of a <code>BarChart</code>?</strong></p>
| J. M.'s persistent exhaustion | 50 | <p>This might be what you were expecting:</p>
<pre><code>In[]:=
{{-1/4, 1/4 + 1/Sqrt[2], -1/2 + 1/(2 Sqrt[2])},
{1/4 - 1/Sqrt[2], -1/4, -1/2 - 1/(2 Sqrt[2])},
{-1/2 - 1/(2 Sqrt[2]), -1/2 + 1/(2 Sqrt[2]), 1/2}} // Eigenvalues // ExpToTrig
Out[]=
{-1/2 + (I Sqrt[3])/2, -1/2 - (I Sqrt[3])/2, 1}
</code></pre>
<p>where <code>ExpToTrig[]</code> converts the exponentials like $(-1)^{2/3}$ accordingly.</p>
|
1,068,719 | <p>The problem is to prove that the quintic
$$x^5+10x^4+15x^3+15x^2-10x+1$$
is irreducible in the rationals. </p>
<p>I don't have much knowledge in group theory, and certainly not in Galois theory, and I'm pretty sure this problem can be solved without those tools. </p>
<p>I know about Eisenstein's criterion, but it cannot be applied to this particular quintic because $5$ does not divide the constant term. If we somehow manipulate the polynomial so that $5$ divides the constant term, we still have to make sure that $25$ doesn't. </p>
<p>So is there any other easy ("elementary") way to solve this?</p>
| Will Jagy | 10,400 | <p>Major simplification: result of Gauss that one may factor a polynomial with integer coefficients and get the same answer as with rational numbers, <a href="http://en.wikipedia.org/wiki/Gauss%27s_lemma_%28polynomial%29" rel="nofollow">http://en.wikipedia.org/wiki/Gauss%27s_lemma_%28polynomial%29</a> </p>
<p>This is Theorem 3.10.1 on page 160 of <em>Topics in Algebra</em> by Herstein.</p>
<p>So, the choices to finish are
$$ (x^2 + a x + 1)(x^3 + b x^2 + c x + 1) = x^5 + 10 x^4 + 15 x^3 + 15 x^2 - 10 x + 1 $$
and
$$ (x^2 + a x - 1)(x^3 + b x^2 + c x - 1) = x^5 + 10 x^4 + 15 x^3 + 15 x^2 - 10 x + 1 $$</p>
<p>It is really the same thing to point out the rational roots theorem, the only possible roots (and linear factors, therefore) are $\pm 1.$</p>
|
135,911 | <p>Liouville's theorem gives such a proof for antiderivatives of functions like <span class="math-container">$e^x/x$</span> or <span class="math-container">$e^{x^2}$</span>, and differential Galois theory extends that to Bessel functions, say. But what tools exist for implicit functions like Lambert's W?</p>
| Igor Khavkine | 2,622 | <p>It seems that the non-linear differential equation satisfied by the Lambert <span class="math-container">$W$</span> function is simple enough for this question to have already been answered. This paper proves that the Lambert <span class="math-container">$W$</span> is non-elementary by appealing to a result of Rosenlicht (1969):</p>
<blockquote>
<p>Bronstein, M., Corless, R. M., Davenport, J. H. and Jeffrey, D. J.
(2008) Algebraic properties of the Lambert W Function from a
result of Rosenlicht and of Liouville. <em>Integral Transforms and
Special Functions</em>, <strong>19</strong> (10). pp. 709-712.
<a href="https://dx.doi.org/10.1080/10652460802332342" rel="noreferrer">https://dx.doi.org/10.1080/10652460802332342</a></p>
</blockquote>
<p>The argument runs roughly as follows. First note that <span class="math-container">$W(x)$</span> satisfies a two-variable algebraic-differential system, where the differential part has a special form:</p>
<p><span class="math-container">\begin{equation*}
W e^W = x \implies W'\!/W + W' = 1/x \implies \{Y = x/W,\; Y'\!/Y = W'\} .
\end{equation*}</span></p>
<p>Now, consider the differential field <span class="math-container">$\mathbb{C}(x)(Y,W)$</span> where <span class="math-container">$x'=1$</span>, <span class="math-container">$Y'/Y = W'$</span> and <span class="math-container">$f(x,Y,W)=0$</span>, with <span class="math-container">$f$</span> polynomial in <span class="math-container">$Y$</span> and <span class="math-container">$W$</span> over <span class="math-container">$\mathbb{C}(x)$</span>. According to a more general result of Rosenlicht, the field <span class="math-container">$\mathbb{C}(x)(Y,W)$</span> is Liouvillian only if <span class="math-container">$Y$</span> and <span class="math-container">$W$</span> are algebraic over <span class="math-container">$\mathbb{C}(x)$</span>. So, if <span class="math-container">$W(x)$</span> were elementary (elementary is a special case of Liouvillian) then it would be algebraic.</p>
<p>It remains to check that <span class="math-container">$W(x)$</span> is not algebraic. The above article shows that as well. I'll just leave it at saying that it follows from the fact that it satisfies a transcendental equation.</p>
|
164,309 | <p>I came across this recurrence function:</p>
<blockquote>
<p>$$F(n) = a \times F(n-1) + b$$</p>
</blockquote>
<p>where $F(0) =1$. We have to solve for $F(n) \pmod {m}$</p>
<p>But for very large $n$, solving it with computer is also taking time. Is there anyway to simplify this. I think the values will be repeated again after some '$n$' based upon the values of $a,b$ and $m$. But I am unable to figure out how to solve it.</p>
| pritam | 33,736 | <p>$F(n)=a.F(n-1)+b=a(a.F(n-2)+b)=a^2F(n-2)+ab=a^2(a.F((n-3)+b)+ab=a^3F(n-3)+a^2b+ab=\ldots=a^nF(0)+a^{n-1}b+\cdots +ab$</p>
<p>Now if $a\mod m\ne 1$ $$F(n)\mod m=(a\mod m)^n+(b\mod m)\frac{1-(a\mod m)^n}{1-a\mod m}$$</p>
<p>If $a\mod m=1$ then $a^i\mod m=1$ for all $i$, hence $F(n)\mod m=1+(n-1)(b\mod m)$</p>
|
164,309 | <p>I came across this recurrence function:</p>
<blockquote>
<p>$$F(n) = a \times F(n-1) + b$$</p>
</blockquote>
<p>where $F(0) =1$. We have to solve for $F(n) \pmod {m}$</p>
<p>But for very large $n$, solving it with computer is also taking time. Is there anyway to simplify this. I think the values will be repeated again after some '$n$' based upon the values of $a,b$ and $m$. But I am unable to figure out how to solve it.</p>
| robjohn | 13,854 | <p>If $m$ is small enough, compute when the sequence
$$
1,1+a,1+a+a^2,1+a+a^2+a^3,\dots\tag{1}
$$
becomes $0\bmod{m}$. If the $k^{\text{th}}$ term of $(1)$ is $0\bmod{m}$, then
$$
\frac{a^k-1}{a-1}\equiv0\pmod{m}\quad\quad\text{and}\quad\quad a^k\equiv1\pmod{m}\tag{2}
$$
Reduce $n'\equiv n\pmod{k}$ and you have $F(n)\equiv F(n')\pmod{m}$. If $m$ is much smaller than $n$, this will reduce the number of iterations.</p>
<hr>
<p>If $m$ is large, then you can still compute (as shown by Marvis, et al)
$$
F(n)=a^nF(0)+b\frac{a^n-1}{a-1}\pmod{m}\tag{3}
$$
Although $a^n$ and $\frac{a^n-1}{a-1}$ may get very big, $(3)$ takes very little time.</p>
|
194,664 | <p>How to generate a list of fixpoint free permutations of n elements in mathematica?</p>
| Carl Woll | 45,431 | <p>You can use <a href="http://reference.wolfram.com/language/ref/Solve" rel="noreferrer"><code>Solve</code></a> to find the roots. Here is a function that finds the roots between <span class="math-container">$-\pi$</span> and <span class="math-container">$\pi$</span>:</p>
<pre><code>f[x_] := a /. Solve[Sin[a]/a==x && -Pi < a < Pi, a, Reals]
</code></pre>
<p>Examples:</p>
<pre><code>f[1/2]
f[1/10]
</code></pre>
<blockquote>
<p>{Root[{-2 Sin[#1] + #1 &, -1.89549426703398094714}],
Root[{-2 Sin[#1] + #1 &, 1.89549426703398094714}]}</p>
<p>{Root[{-10 Sin[#1] + #1 &, -2.8523418944500916483}],
Root[{-10 Sin[#1] + #1 &, 2.8523418944500916483}]}</p>
</blockquote>
<p>You can use <a href="http://reference.wolfram.com/language/ref/N" rel="noreferrer"><code>N</code></a> to obtain approximate answers:</p>
<pre><code>N[f[1/100], 40]
</code></pre>
<blockquote>
<p>{-3.110482807621505335413758229364966288043,
3.110482807621505335413758229364966288043}</p>
</blockquote>
|
2,061,363 | <p>I have the complex power series $ \sum_{k=1}^{\infty}(\frac{z^4}{4} - \frac{\pi}{7})^k$. </p>
<p>Through algebraic manipulation I obtain $ \sum_{k=1}^{\infty}(\frac{1}{4})^k(z^4 - \frac{4}{7}\pi)^k$. I now argue that this is a power series around $\frac{4}{7}\pi$ with radius of convergence R = 4, using the euler root test, asserting all the while that the power 4 in the argument z doesn't affect those two quantities. However, I'm not sure how to prove or even heuristically show that last bit. In fact, I'm not even sure I'm right in asserting that the power doesn't matter, it's really more of a hunch. Does anyone know how to handle this scenario? </p>
| zhw. | 228,045 | <p>The series in this problem is not a power series, so the concept of "radius of convergence" doesn't really apply. Because of basic results for geometric series, the series convergeges iff $|z^4/4 - \pi/7|<1.$ The set of such $z$ is not a disc, so what does "radius" mean? This set actually looks kind of complicated.</p>
|
3,547,384 | <p>I saw this equation<span class="math-container">$$S(q)=\int_a^bL(t,q(t),\dot q(t))dt$$</span>
in <a href="https://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation" rel="nofollow noreferrer">wikipedia</a>.</p>
<p>So I would think that <span class="math-container">$f(x,y)$</span> must be equal to <span class="math-container">$f(x,y,t)$</span> if <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are the function of <span class="math-container">$t$</span>. So let's take an experiment.</p>
<p>Firstly, just let <span class="math-container">$f = f(x,y)$</span>, where <span class="math-container">$x = x(t)$</span>, <span class="math-container">$y=y(t)$</span>, so <span class="math-container">$f=f\left(x(t),y(t)\right)$</span></p>
<p><span class="math-container">$$\cfrac{df}{dt}=\cfrac{\partial f}{\partial x}\cfrac{dx}{dt}+\cfrac{\partial f}{\partial y}\cfrac{dy}{dt}\qquad (1)$$</span></p>
<p>Secondly, since you all see that <span class="math-container">$f$</span> is actually also a function of <span class="math-container">$t$</span>, so we have</p>
<p><span class="math-container">$$f = f(x,y)=f(x,y,t)\qquad (2)$$</span></p>
<p>Now, </p>
<p><span class="math-container">$$\cfrac{df}{dt}=\cfrac{\partial f}{\partial x}\cfrac{dx}{dt}+\cfrac{\partial f}{\partial y}\cfrac{dy}{dt}+\cfrac{\partial f}{\partial t}\cfrac{dt}{dt}\qquad (3)$$</span></p>
<p><span class="math-container">$$\cfrac{df}{dt}=\cfrac{\partial f}{\partial x}\cfrac{dx}{dt}+\cfrac{\partial f}{\partial y}\cfrac{dy}{dt}+\cfrac{\partial f}{\partial t}\qquad (4)$$</span></p>
<p>Because it is always correct that <span class="math-container">$\cfrac{df}{dt}=\cfrac{\partial f}{\partial t}$</span>,</p>
<p><span class="math-container">$$\cfrac{\partial f}{\partial x}\cfrac{dx}{dt}+\cfrac{\partial f}{\partial y}\cfrac{dy}{dt}=0\qquad (5)$$</span></p>
<p>Substitute (5) into (1),</p>
<p><span class="math-container">$$\cfrac{df}{dt}=0\qquad (6)$$</span></p>
<p>This is not a correct outcome, since (6) are not always equal to zero for all cases.</p>
<p>So what's wrong???</p>
| Quiver | 564,698 | <p>If we write <span class="math-container">$f(x,y,t)$</span> we mean an <em>explicit</em> dependence of time. This makes the assertion you make </p>
<blockquote>
<p>Because it is always correct that <span class="math-container">$$\frac{df}{dt} = \frac{\partial f}{\partial t} $$</span></p>
</blockquote>
<p><strong>incorrect</strong>. Whenever there's an explicit dependence on time, the partial derivative takes it into account. Let's say for example that you have the following lagrangian </p>
<p><span class="math-container">$$ L(x(t), y(t), t) = \frac{1}{2} m \left(\dot{x}^2+\dot{y}^2\right) - At$$</span> you can compute the total derivative of the lagrangian and see by yourself that indeed your assertion is wrong.</p>
<p>Physcally speaking we put the explicit dependence on time in the lagrangian because if there's some <em>dissipative</em> terms, they usually depende directly on time while, as well, the position depends on time. So the explicit dependence is made clear for this reasons, not because the position depends on time. That's an <em>implicit</em> dependence on time.</p>
|
1,241,695 | <p>I was reading A First Course in Probability by Sheldon Ross. I think I quite understood the below problem but I still feel fuzzy.</p>
<blockquote>
<p><strong>Problem</strong>: In answering on a multiple choice test, a student either know the answer or guesses. Let p be the probability that the students knows the answer and 1-p be the probability that the student guesses. Assume that a student who guesses at the answer will be correct with probability 1/m, where m is the number of multiple choice alternatives. What is the conditional probability that a student knew the answer to a question when he or she answered it correctly?</p>
<p><strong>Solution as given in the book</strong>:
Let C be the event that the student answered the question correctly. And
Let K be the event that the student actually knows the answer.
<span class="math-container">$$P(K|C) = \frac{P(KC)}{P(C)}$$</span>
<span class="math-container">$$ = \frac{P(C|K)P(K)}{P(C|K)P(K)+P(C|K^C)P(K^C)}$$</span>
<span class="math-container">$$ = \frac{p}{p+(1/m)(1-p)}$$</span></p>
</blockquote>
<p>Now this seems reasonable, only confusing is that how is</p>
<ol>
<li>Probability that the student knows the answer <span class="math-container">$= P(KC) = p$</span> but not <span class="math-container">$P(C|K)$</span></li>
<li>While on the other hand the probability that a student who guesses the answer will be correct <span class="math-container">$= P(C|K^C) = 1/m$</span> but this time not <span class="math-container">$P(CK^C)$</span> or <span class="math-container">$P(K^CC)$</span>.</li>
</ol>
<p>I think its just that I am finding it difficult to determine the probabilities relation from the sentence formations.</p>
<ol start="3">
<li>Is there any other simpler, non fuzzy approach to such problems?</li>
</ol>
| elDin0 | 195,164 | <p>The way I work through these is by thinking "the probability of <strong>B GIVEN A</strong> is equal to the probability of <strong>A AND B</strong> divided by the probability of <strong>just A</strong>. </p>
<p>So in this case, the probability that the student knew the answer given he answered it right is the probability he knew it and got it right divided by the probability he answered it right. </p>
<p>So probability is $\frac{\mathbb{P}(\text{Knew it and got it right}) [= \mathbb{P}(\text{knew it})]}{\mathbb{P}(\text{got it right})} = \frac{p}{p + \frac{1}{m}(1-p)}$ since our denominator is the probability he got it right since he knew it, added to the probability he got it right through guessing. Note the top equality comes since if he knows it, he clearly gets it right, too, as you noted with point 1. </p>
<p>I feel thinking through the problem logically like this is much easier than trying to brute force the way through with formulae like your textbook does.</p>
|
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