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 |
|---|---|---|---|---|
759,810 | <p>I'm trying to find the last two digits of ${2012}^{2012}$. I know you can use (mod 100) to find them, but I'm not quite sure how to apply this. Can someone please explain it?</p>
| lab bhattacharjee | 33,337 | <p>Finding the last two digits of $a$ essentially $\displaystyle a\pmod{100}$ </p>
<p>Now as $\displaystyle2012\equiv12, 2012^{2012}\equiv12^{2012}\pmod{100}$</p>
<p>Again as $\displaystyle(12,100)=4$ let us find $\displaystyle12^{2012-1}\pmod{\frac{100}4}$ i.e., $\displaystyle12^{2011}\pmod{25}$</p>
<p>Now using <a href="http://en.wikipedia.org/wiki/Carmichael_function" rel="nofollow">Carmichael function</a> or Totient function $\displaystyle\lambda(25)=\phi(25)=20,2011\equiv11\pmod{20}\implies12^{2011}\equiv12^{11}\pmod{25}$</p>
<p><strong>Method</strong> $\#1:$</p>
<p>Again, $\displaystyle12^2=144\equiv-6\pmod{25}\implies12^3\equiv-6\cdot12\equiv3$</p>
<p>$\displaystyle\implies12^9\equiv3^3\equiv2\implies12^{11}\equiv2\cdot(-6)\equiv-12\pmod{25}$</p>
<p><strong>Method</strong> $\#2:$</p>
<p>$\displaystyle12^2=144=(145-1)\implies12^{10}=(-1+145)^5$
$\displaystyle=-1+\binom51\cdot145-\binom52145^2+\cdots+145^5$
$\displaystyle12^{10}\equiv-1\pmod{25}\ \ \ \ (1)$ as the rest of terms are divisible by $5^2=25$</p>
<p>As $\displaystyle a\equiv b\pmod m\implies a\cdot c\equiv b\cdot c\pmod{m\cdot c}\ \ \ \ (2)$</p>
<p>Using $(1),(2)$ with $c=12$
$\displaystyle12^{11}=12\cdot12^{10}\equiv-1\cdot12\pmod{12\cdot25}\equiv-12\pmod{300}\equiv-12\pmod{25}$</p>
<p>So, by any of the two methods $\displaystyle12^{11}\equiv-12\pmod{25}$</p>
<p>Using $(2)$ again with $c=12$,
$\displaystyle\implies12^{12}\equiv-12\cdot12\pmod{25\cdot12}\equiv-144\equiv156\pmod{300}$</p>
<p>$\displaystyle\implies12^{12}\equiv156\pmod{300}\equiv156\pmod{100}\equiv56$</p>
|
2,202,339 | <p>It's easy to prove $x^2+1$ is never divisible by $4k+3$ primes. I know a non-constructive proof for existing $x$ so that $p|x^2+1$ for $4k+1$ primes. is there any constructive one?</p>
| MR_BD | 195,683 | <p>$$ \: [(\frac{p-1}2)!]^2 \equiv -1 \:(mod \: p)$$</p>
<p>By Wilson's Theorem.</p>
|
2,005,604 | <p>Showing $\sqrt a + $$\sqrt {\cos(\sin a)} = 2$</p>
<p>I've attempted various manipulations (multiplying by one, squaring, etc.) but cannot find a way to solve for a. Anyone have an idea how I can approach this problem? Thanks. </p>
| Claude Leibovici | 82,404 | <p>For this kind of equation, there is no analytical solution (it is already the case for $x=\cos(x)$) and numerical methods should be used.</p>
<p>As Robert Israel already answered, Newton method could be the simplest to use considering $$f(x)=\sqrt{x}+\sqrt{\cos (\sin (x))}-2$$ Starting from a "reasonable" guess $x_0$, the method will updtae it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ For the present case $$f'(x)=\frac{1}{2 \sqrt{x}}-\frac{\sin (\sin (x)) \cos (x)}{2 \sqrt{\cos (\sin (x))}}$$ By inspection, we know that the root is between $0$ and $\pi$ since $f(0)=-1$ and $f(\pi)=\sqrt{\pi }-1$. So, let us start at the mid point $x_0=\frac \pi 2$.</p>
<p>Newton iterates will then be the following
$$\left(
\begin{array}{cc}
n & x_n \\
0 & 1.5707963267948966192 \\
1 & 1.5999566241894927879 \\
2 & 1.5994959773991264235 \\
3 & 1.5994958620742497549 \\
4 & 1.5994958620742425268
\end{array}
\right)$$
which the solution for twenty significant figures.</p>
|
3,020,365 | <p>Let <span class="math-container">$A=\{t\sin(\frac{1}{t})\ |\ t\in (0,\frac{2}{\pi})\}$</span>.</p>
<p>Then </p>
<ol>
<li><p><span class="math-container">$\sup (A)<\frac{2}{\pi}+\frac{1}{n\pi}$</span> for all <span class="math-container">$n\ge 1$</span>.</p></li>
<li><p><span class="math-container">$\inf (A)> \frac{-2}{3\pi}-\frac{1}{n\pi}$</span> for all <span class="math-container">$n\ge 1$</span>.</p></li>
<li><span class="math-container">$\sup (A)=1$</span></li>
<li><span class="math-container">$\inf (A)=-1$</span></li>
</ol>
<p>My answer was options <span class="math-container">$1$</span> and <span class="math-container">$2$</span> which matched with prelim answer key provided by organization, but the final answer key changed the answer to option <span class="math-container">$1$</span> only. Why is option <span class="math-container">$2$</span> not correct?</p>
<p><strong>My attempt-</strong> </p>
<p>As <span class="math-container">$-t\le t\sin(1/t)\le t$</span>. So <span class="math-container">$t\sin(1/t)$</span> will always be less than <span class="math-container">$\frac{2}{\pi}$</span> which is strictly less than <span class="math-container">$1$</span>. So option <span class="math-container">$3$</span> is false and option <span class="math-container">$1$</span> is true. Also value of <span class="math-container">$t\sin(1/t)$</span> is always greater than or equal to <span class="math-container">$-2/\pi=-0.63$</span>, so infimum cannot be <span class="math-container">$-1$</span>.</p>
<p>Now at <span class="math-container">$t=2/3\pi$</span> , function has value <span class="math-container">$\frac{-2}{3\pi}$</span>. So if option <span class="math-container">$2$</span> is false we must have some value of <span class="math-container">$t$</span> in <span class="math-container">$A$</span> for which the function attains value strictly less than <span class="math-container">$\frac{-2}{3\pi}$</span>. Right? </p>
<p>Can you please help me with option <span class="math-container">$2$</span> now. Thanks in advance. </p>
| Calvin Khor | 80,734 | <p>The graph of <span class="math-container">$\sin(1/t)$</span> turns at <span class="math-container">$t=2/3\pi$</span>, but this is not true of <span class="math-container">$t\sin(1/t)$</span>. Multiplying by <span class="math-container">$t$</span> has the effect of pushing the minimum point to the right and down slightly. </p>
<p><a href="https://i.stack.imgur.com/yg0Wo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yg0Wo.png" alt="enter image description here"></a> <a href="https://i.stack.imgur.com/yvnus.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yvnus.png" alt="enter image description here"></a>
Since <span class="math-container">$a>b-\frac{1}{\pi n}$</span> for all <span class="math-container">$n\ge 1$</span> iff <span class="math-container">$a\ge b$</span>, we are done.</p>
<p>To see this without a graph, lets start from your correct observation that for <span class="math-container">$f(t) = t\sin (1/t)$</span>, <span class="math-container">$f(2/3\pi) = -2/3\pi$</span>. The derivative at this point however is
<span class="math-container">$$ f'(3\pi/2) = \sin(3\pi/2) + 2/3\pi \cos(3\pi/2) = -1 $$</span></p>
<p>Since <span class="math-container">$f$</span> is smooth on a small neighbourhood of <span class="math-container">$2/3\pi$</span>, <span class="math-container">$f'\le -1/2<0$</span> on this neighbourhood. Thus <span class="math-container">$f$</span> is decreasing at <span class="math-container">$2/3\pi$</span>.</p>
<p>Something slightly more general is true:</p>
<blockquote>
<p>Suppose <span class="math-container">$f$</span> is smooth, <span class="math-container">$f(1)=0=f(2)$</span>, <span class="math-container">$f\le 0$</span> and <span class="math-container">$f$</span> has a unique non degenerate minimum point <span class="math-container">$x_0$</span> in <span class="math-container">$[1,2]$</span>. Then <span class="math-container">$g(x)=xf(x)$</span> has a minimum point to the right and below of <span class="math-container">$x_0$</span>.</p>
</blockquote>
<p>Proof- by a similar computation to the above, <span class="math-container">$g$</span> is decreasing at <span class="math-container">$x_0$</span>. As <span class="math-container">$g(2)=2f(2)=0$</span>, we conclude by Intermediate Value Theorem and Rolles' Theorem.</p>
|
3,020,365 | <p>Let <span class="math-container">$A=\{t\sin(\frac{1}{t})\ |\ t\in (0,\frac{2}{\pi})\}$</span>.</p>
<p>Then </p>
<ol>
<li><p><span class="math-container">$\sup (A)<\frac{2}{\pi}+\frac{1}{n\pi}$</span> for all <span class="math-container">$n\ge 1$</span>.</p></li>
<li><p><span class="math-container">$\inf (A)> \frac{-2}{3\pi}-\frac{1}{n\pi}$</span> for all <span class="math-container">$n\ge 1$</span>.</p></li>
<li><span class="math-container">$\sup (A)=1$</span></li>
<li><span class="math-container">$\inf (A)=-1$</span></li>
</ol>
<p>My answer was options <span class="math-container">$1$</span> and <span class="math-container">$2$</span> which matched with prelim answer key provided by organization, but the final answer key changed the answer to option <span class="math-container">$1$</span> only. Why is option <span class="math-container">$2$</span> not correct?</p>
<p><strong>My attempt-</strong> </p>
<p>As <span class="math-container">$-t\le t\sin(1/t)\le t$</span>. So <span class="math-container">$t\sin(1/t)$</span> will always be less than <span class="math-container">$\frac{2}{\pi}$</span> which is strictly less than <span class="math-container">$1$</span>. So option <span class="math-container">$3$</span> is false and option <span class="math-container">$1$</span> is true. Also value of <span class="math-container">$t\sin(1/t)$</span> is always greater than or equal to <span class="math-container">$-2/\pi=-0.63$</span>, so infimum cannot be <span class="math-container">$-1$</span>.</p>
<p>Now at <span class="math-container">$t=2/3\pi$</span> , function has value <span class="math-container">$\frac{-2}{3\pi}$</span>. So if option <span class="math-container">$2$</span> is false we must have some value of <span class="math-container">$t$</span> in <span class="math-container">$A$</span> for which the function attains value strictly less than <span class="math-container">$\frac{-2}{3\pi}$</span>. Right? </p>
<p>Can you please help me with option <span class="math-container">$2$</span> now. Thanks in advance. </p>
| zhw. | 228,045 | <p>Like some other answers, but written differently:</p>
<p>Let <span class="math-container">$f(t)= t\sin(1/t).$</span> Then <span class="math-container">$f'(t)= \sin(1/t) - [\cos (1/t)]/t$</span> for <span class="math-container">$t>0.$</span> Thus <span class="math-container">$f'(2/(3\pi)) = -1.$</span> Hence for <span class="math-container">$t$</span> near and to the right of <span class="math-container">$2/(3\pi),$</span> we have</p>
<p><span class="math-container">$$\frac{f(t)-f(2/(3\pi)}{t-2/(3\pi)}<-\frac{1}{2}.$$</span></p>
<p>For such <span class="math-container">$t$</span> we have <span class="math-container">$f(t)<f(2/(3\pi)=-2/(3\pi).$</span> Hence <span class="math-container">$\inf A <-2/(3\pi),$</span> which implies 2. does not hold.</p>
|
16,749 | <p>I wanted to remove the <code>Ticks</code> in my coding but i can't. Here when i try to remove the <code>Ticks</code> the number also gone. I need numbers without <code>Ticks</code>, <code>Ticks</code> and <code>GridLines</code> should be automatic and don't use<code>PlotRange</code> .</p>
<pre><code>BarChart[{{1, 2, 3}, {4, 5, 6}}, ImageSize -> 400, BarOrigin -> Left,
ChartLayout -> "Stacked", ImageSize -> {500, 300},
GridLines -> {Automatic, None}, Ticks -> {{Automatic}, None}
,
LabelStyle -> Directive[Opacity[1]],
TicksStyle -> Directive[Opacity[.3]]
,
Axes -> {True, False}, AxesStyle -> Opacity[.0],
ChartStyle -> {RGBColor[.06, .29, .66], RGBColor[.01, .56, .61],
RGBColor[1, .58, 0]},
ChartBaseStyle -> EdgeForm[GrayLevel[.6]]]
</code></pre>
<p><img src="https://i.stack.imgur.com/s28b2.png" alt="enter image description here"></p>
| Ronny | 521 | <p>You could specify the <code>Ticks</code> manually, for example by</p>
<pre><code>Ticks -> {Table[{i, i, 0}, {i, 0, 14, 2}], None},
</code></pre>
<p>The y axis stays without Ticks by the <code>None</code> the x axis gets at $i=0,2,...,14$ a Tick at $i$ with $i$ as label and zero ticks width.</p>
|
1,796,156 | <p>Let $F(n)$ denote the $n^{\text{th}}$ Fibonacci number<a href="http://mathworld.wolfram.com/FibonacciNumber.html" rel="noreferrer">$^{[1]}$</a><a href="http://en.wikipedia.org/wiki/Fibonacci_number" rel="noreferrer">$\!^{[2]}$</a><a href="http://oeis.org/A000045" rel="noreferrer">$\!^{[3]}$</a>. The Fibonacci numbers have a natural generalization to an analytic function of a complex argument:
$$F(z)=\left(\phi^z - \cos(\pi z)\,\phi^{-z}\right)/\sqrt5,\quad\text{where}\,\phi=\left(1+\sqrt5\right)/2.\tag1$$
This definition is used, for example, in <em>Mathematica</em>.<a href="http://reference.wolfram.com/language/ref/Fibonacci.html" rel="noreferrer">$^{[4]}$</a> It produces real values for $z\in\mathbb R$, and preserves the usual functional equation for Fibonacci numbers for all $z\in\mathbb C$: $$F(z)=F(z-1) + F(z-2).\tag2$$</p>
<hr>
<p>The fibonorial<a href="http://mathworld.wolfram.com/Fibonorial.html" rel="noreferrer">$^{[5]}$</a><a href="http://en.wikipedia.org/wiki/Fibonorial" rel="noreferrer">$\!^{[6]}$</a><a href="http://oeis.org/A003266" rel="noreferrer">$\!^{[7]}$</a> is usually denoted as $n!_F$, but here we prefer a different notation $\mathfrak F(n)$. It is defined for non-negative integer $n$ inductively as
$$\mathfrak F(0)=1,\quad \mathfrak F(n+1)=\mathfrak F(n)\times F(n+1).\tag3$$
In other words, the fibonorial $\mathfrak F(n)$ gives the product of the Fibonacci numbers from $F(1)$ to $F(n)$, inclusive. For example, $$\mathfrak F(5)=\prod_{m=1}^5F(m)=1\times1\times2\times3\times5=30.\tag4$$</p>
<blockquote>
<p><em>Questions:</em> Can the fibonorial be generalized in a natural way to an analytic function $\mathfrak F(z)$ of a complex (or, at least, positive real) variable, such that it preserves the functional equation $(3)$ for all arguments?</p>
<p>Is there an integral, series or continued fraction representation of $\mathfrak F(z)$, or a representation in a closed form using known special functions?</p>
<p>Is there an efficient algorithm to calculate values of $\mathfrak F(z)$ at non-integer arguments to an arbitrary precision?</p>
</blockquote>
<p>So, we can see that the fibonorial is to the Fibonacci numbers as the factorial is to natural numbers, and the analytic function $\mathfrak F(z)$ that I'm looking for is to the fibonorial as the analytic function $\Gamma(z+1)$ is to the factorial.</p>
<hr>
<p><em>Update:</em> While thinking on <a href="https://math.stackexchange.com/q/1914821/19661">this question</a> it occurred to me that perhaps we can use <a href="http://mathworld.wolfram.com/GammaFunction.html#eqn30" rel="noreferrer">the same trick</a> that is used to define the $\Gamma$-function using a limit involving factorials of integers:
$$\large\mathfrak F(z)=\phi^{\frac{z\,(z+1)}2}\cdot\lim_{n\to\infty}\left[F(n)^z\cdot\prod_{k=1}^n\frac{F(k)}{F(z+k)}\right]\tag5$$
or, equivalently,
$$\large\mathfrak F(z)=\frac{\phi^{\frac{z\,(z+1)}2}}{F(z+1)}\cdot\prod_{k=1}^\infty\frac{F(k+1)^{z+1}}{F(k)^z\,F(z+k+1)}\tag{$5'$}$$
This would give
$$\mathfrak F(1/2)\approx0.982609825013264311223774805605749109465380972489969443...\tag6$$
that appears to have a closed form in terms of the <a href="http://mathworld.wolfram.com/q-PochhammerSymbol.html" rel="noreferrer">q-Pochhammer symbol</a>:
$$\mathfrak F(1/2)=\frac{\phi^{3/8}}{\sqrt[4]{5}}\,\left(-\phi^{-2};-\phi^{-2}\right)_\infty\tag7$$
and is related to the <a href="http://mathworld.wolfram.com/FibonacciFactorialConstant.html" rel="noreferrer">Fibonacci factorial constant</a>.</p>
| Zach466920 | 219,489 | <p>(<strong>Main Result</strong>: I have a formula for what you want, its practical and it works!)</p>
<p>Since you ask multiple questions, I suppose what follows isn't a complete answer. I learned this trick the other day, so I'm more than happy to share it here,</p>
<p>If you want an integral that can generalize the recurrence,</p>
<p>$$(1) \quad \mathfrak F(0)=1,\quad \mathfrak F(n+1)=\mathfrak F(n)\times F(n+1).$$</p>
<p>You can do so by assuming the formula is of a certain form and that $k(t,n)$ satisfies a delay differential equation.</p>
<p>$$(2) \quad \mathfrak F(n)=\int_0^{\infty} e^{-t} \cdot k(t,n) \ dt$$</p>
<p>We can then demand that the eigenvalue for the differential operator is equal to $F(n+1)$.</p>
<p>$$(3) \quad \partial_t [k(t,n+1)]=F(n+1) \cdot k(t,n)$$</p>
<p>It then follows that applying integration by parts to $(2)$ yields,</p>
<p>$$\mathfrak F(n+1)=\int_0^{\infty} e^{-t} \cdot k(t,n+1) \ dt={F(n+1)} \cdot \int_0^{\infty} e^{-t} \cdot k(t,n) \ dt={F(n+1)} \cdot \mathfrak F(n)$$</p>
<p>$$\Rightarrow \mathfrak F(n+1)=\mathfrak F(n)\cdot F(n+1).$$</p>
<p>To obtain an explicit form for the integration kernel, we must solve $(3)$, which is classified as a delay differential equation. Whether or not an explicit solution exists is uncertain without further investigation. However, the kernel can certainly be found numerically.</p>
<p>However, we can certainly define the kernel that satisfies $(3)$ to be the Fibonacci kernel.</p>
<p><strong>Tentative (Approximate) Solution</strong></p>
<p>I didn't post this originally because its a bit cheap for a solution. Effectively, this solution indexes the new factorial differently. Set,</p>
<p>$$k(t,n)=t^{F(n)}$$</p>
<p>And also we have,</p>
<p>$$F(z)=(\phi^z-\cos(\pi \cdot z) \cdot \phi^{-z})/\sqrt{5}$$</p>
<p>Then we have,</p>
<p>$$\mathfrak F(n+1)=\mathfrak F(F^{-1}(F(n+1)-1)) \cdot F(n+1)$$</p>
<p>instead of,</p>
<p>$$\mathfrak F(n+1)=\mathfrak F(n) \cdot F(n+1)$$</p>
<p><strong>Asymptotics and a Formula</strong></p>
<p>I've switched notation since its the reason I came up with this. According to Wolfram Alpha, we have <a href="http://mathworld.wolfram.com/Fibonorial.html" rel="noreferrer">this</a> relation for the asymptotics of the fibonorial, </p>
<p>$$(4) \quad n!_F \sim C \cdot \cfrac{\phi^{n \cdot (n+1)/2}}{5^{n/2}}$$</p>
<p>Given this, lets hope we can find a function $g(n,t)$ such that,</p>
<p>$$(5) \quad \lim_{n \rightarrow \infty} \cfrac{n!_F \cdot g(n,t)}{(n+t)!_F}=1$$</p>
<p>I find that, and I leave you to verify, that,</p>
<p>$$(6) \quad g(n,t)=5^{-t/2} \cdot \phi^{(t^2/2+t/2)} \cdot \phi^{n \cdot t}$$</p>
<p>Now that we know what $g(n,t)$ is, multiply by both sides by $t!_F$ and simplify, assuming $t$ is a positive integer. We get,</p>
<p>$$(7) \quad t!_F=\lim_{n \rightarrow \infty} \cfrac{n!_F \cdot g(n,t)}{F(1+t) \cdot F(2+t) \ ...\ F(n+t)}$$</p>
<p>Now we formally extend $t$ from $N \gt 0$ to $t \in C$. Using this, I can construct (perhaps the first?!) table of "F-Gamma" values using $n!_F=\mathfrak F(n+1)$,</p>
<p>$$
\begin{array}{c|lcr}
n & \mathfrak F(n) & \ln(\mathfrak F(n))\\
\hline
1 & 1 & 0\\
1.1 & 0.966975 & -0.033582\\
1.2 & 0.951630 & -0.049579\\
1.3 & 0.951521 & -0.049692\\
1.4 & 0.963404 & -0.037283\\
1.5 & 0.982610 & -0.017543\\
1.6 & 1.003127 & 0.003122\\
1.7 & 1.018567 & 0.018396\\
1.8 & 1.023911 & 0.023630\\
1.9 & 1.017193 & 0.017048\\
2.0 & 1 & 0\\
\end{array}
$$</p>
<p>Here's a graph (I didn't say it'd be pretty ;))</p>
<p><a href="https://i.stack.imgur.com/pqEqF.png" rel="noreferrer"><img src="https://i.stack.imgur.com/pqEqF.png" alt="enter image description here"></a></p>
<p>Check the recurrence, defined in $(1)$. It works!!!</p>
|
269,548 | <p>I want to know how I can solve this or plot a versus b ?</p>
<pre><code>Solve[ Sqrt[a] Cosh[1.2 Log[1.65 Sqrt[1/a]]] == Sqrt[-b] Sinh[1.2 Log[1.65 Sqrt[-(1/b)]]], b]
</code></pre>
<p>Thanks</p>
| MarcoB | 27,951 | <p>Your current code mentions that <code>Solve</code> tried to rationalize your floating-point constants. If you try to do that by hand, then Solve will return <code>Solve::nsmet: This system cannot be solved with the methods available to Solve.</code></p>
<p>However, you can use <code>ContourPlot</code> to obtain a plot as a function of <span class="math-container">$a$</span> and <span class="math-container">$b$</span>:</p>
<pre><code>ContourPlot[
Sqrt[a] Cosh[12/10 Log[165/100 Sqrt[1/a]]] == Sqrt[-b] Sinh[12/10 Log[165/100 Sqrt[-(1/b)]]],
{a, 0, 8}, {b, -1/8, 0}, MaxRecursion -> 6
]
</code></pre>
<p><a href="https://i.stack.imgur.com/kknlr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kknlr.png" alt="2D contour plot of the solution of the equation in OP" /></a></p>
|
591,765 | <blockquote>
<p>What is the way to convince myself that $\left\langle(1,2),\ (1,2,3,4)\right\rangle=S_4$ but $\left\langle(1,3),\ (1,2,3,4)\right\rangle\ne S_4$?</p>
</blockquote>
<p>Let $\sigma$ be any transposition and $\tau$ be any $p-$cycle, where $p$ is a prime.
Then show that $S_p=\langle\sigma,\tau\rangle$.</p>
| Mikasa | 8,581 | <p>Besides to @Betty's points, there is another way for seeing why does this happen. We know that $S_4$ can have the following presentation:</p>
<p>$$S_4=\langle a,b\mid a^2=b^4=(ab)^3=1\rangle$$ Let's satisfy $a=(1,2),~~b=(1,2,3,4)$ in above relations. Indeed $a$ and $b$ can do that, but what will happen if we set $a=(1,3),~~b=(1,2,3,4)$? By this assumption, we see that $(ab)^3=(1,4)(2,3)$ and this happens cause of the points @Betty indicated them in detailed. Now if you are familiar to one of $D_8$'s presentation, then you'll have $$D_8=\langle a,b\mid a^2=b^4=(ab)^3=1\rangle,~~a=(1,3),~~b=(1,2,3,4)$$ instead wich is of order $8$.</p>
|
6,355 | <p>My question is located in trying to follow the argument bellow. </p>
<p>Given a normal algebraic variety $X$, and a line bundle $\mathcal{L}\rightarrow X$ which is ample, then eventually such a line bundle will have enough section to define an embedding $\phi:X\rightarrow \mathbb(H^0(X,\mathcal{L}^{\otimes d}))=\mathbb{P}^N$ (notice that $N$ depends on $d$). However, if the line bundle is NOT ample we can still say something about the existence of a certain map $\phi_d$; the so-called Iitaka fibration. I'll omit some details, but the argument of the construction goes (more or less) as follows.
Suppose in the first place that $\mathcal{L}$ is base-point base. That is to say, that there are no points (or a set) of $X$ which all the hyperplanes of $\mathbb{P}^N$ pass through. Then such a line bundle will define a linear system $|\mathcal{L}^d|$ which gives rise to a morphism $\phi_d:X\rightarrow \phi_d(X)\subset\mathbb{P}^N$ (again $N$ depends on $d$). Such a map may not be an embedding, however $\phi_d:X\rightarrow \phi(X)$ is an algebraic fiber space. My question is the following. As I increase the value $d$ the image $\phi_d(X)\subset \mathbb{P}^N$ may change, however, as a matter of fact such an image "stabilizes" as $d$ gets larger. Meaning that if $d$ is large enough, the image if $\phi_d$ is the "same" regardless $d$. </p>
<p>-What is the reason for this to happen?
-<strong>What is going on with all the sections of $\mathcal{L}^{\otimes d}$ that I am getting as I increase the value of $d$?</strong>. I'll appreciate any comment.</p>
<p>As a result, due to the fact that after a while we no longer care about the value of $d$, we can associate the space $X\rightarrow \phi(X)$ to the line bundle $\mathcal{L}$. Here, the variety $\phi(X)$ no longer depends on $d$.</p>
<p>Could someone comment further about the word "stabilizes"?.</p>
| Sándor Kovács | 10,076 | <p>Let $R(X,\mathcal L)=\oplus_d H^0(X,\mathcal L^{\otimes d})$ as graded rings, $s_0,\dots s_m\in H^0(X,\mathcal L^{\otimes d})$ and finally $R(X,\mathcal L,s_{\cdot})$ the (graded) subring generated by the $s_0,\dots s_m$ in $R(X,\mathcal L)$. </p>
<p>Facts:</p>
<ol>
<li><p>${s_0,\dots,s_m}$ define a rational map $\sigma: X\dashrightarrow \mathbb P^m$. If ${s_0,\dots,s_m}$ generate $H^0(X,\mathcal L^{\otimes d})$, then $\sigma$ agrees with $\phi_t$. </p></li>
<li><p>If $\vert\{s_0,\dots,s_m\}\vert$ is a basepoint-free linear system, then $\sigma$ is a morphism.
and $\sigma(X)\simeq {\rm Proj}\, R(X,\mathcal L, s_{\cdot})$.</p></li>
<li><p>If $R(X,\mathcal L,s_{\cdot})$ has the property that for some $a\in\mathbb N$, $R(X,\mathcal L,s_{\cdot})_{ak}=R(X,\mathcal L)_{ak}$ for all $k\in \mathbb N, k\gg 0$, then
${\rm Proj}\, R(X,\mathcal L, s_{\cdot})\simeq {\rm Proj}\, R(X,\mathcal L)$ (think of the $a$-uple embedding).</p></li>
<li><p>Since $R(X,\mathcal L)$ is finitely generated, if $s_0,\dots s_m\in H^0(X,\mathcal L^{\otimes d})$ are chosen so that they generate $H^0(X,\mathcal L^{\otimes d})$, then
for large enough $d$ the property in 3) will be satisfied, so by 1) and 2) $\phi_d(X)\simeq
{\rm Proj}\, R(X,\mathcal L)$ which is independent of $d$, hence it is stabilized.</p></li>
</ol>
|
4,413,641 | <p>I'm trying to understand a proof that given a vector space <span class="math-container">$V$</span> over the field <span class="math-container">$F$</span> and <span class="math-container">$n$</span> vectors <span class="math-container">$v_1, \ldots, v_n$</span>, <span class="math-container">$\mathrm{span}(v_1, \ldots, v_n)$</span> is the smallest subspace containing them.</p>
<p>I'm fine with most of the proof, including how it is proved in Axler, but the lecture notes I'm working through include a comment I don't fully understand.</p>
<p>I'm ok with showing that <span class="math-container">$\mathrm{span}(v_1, \ldots, v_n)$</span> is a subspace, that it contains <span class="math-container">$v_1, \ldots, v_n$</span>, and that if <span class="math-container">$S$</span> is some other subspace of <span class="math-container">$V$</span> containing <span class="math-container">$v_1, \ldots, v_n$</span>, then <span class="math-container">$\mathrm{span}(v_1, \ldots, v_n) \subset S$</span>.</p>
<p>The comment made in the lecture notes is: how do we know that this smallest subspace containing <span class="math-container">$v_1, \ldots, v_n$</span> actually exists? Axler does not provide a proof of this fact, suggesting that it isn't necessary because by definition, "smallest" means that any other subspace containing those vectors contains their span as a subset.</p>
<p>Is this something I need to show for a complete proof? If so, I cannot figure out how exactly I should go about proving this. The idea, I think, is taking the intersection of all subspaces containing <span class="math-container">$v_1, \ldots, v_n$</span>, and I know that the arbitrary intersection of subspaces is a subspace, but is it necessary to show this?</p>
| Dr. Sundar | 1,040,807 | <p>Here, <span class="math-container">$v_1, v_2, \ldots, v_n$</span> are <span class="math-container">$n$</span> vectors.</p>
<p>Let us fix some notation first.</p>
<p>We define <span class="math-container">$T = \mbox{span}\left\{ v_1, v_2, \ldots, v_n \right\}$</span>.</p>
<p>It is easy to verify that <span class="math-container">$T$</span> is a subspace of <span class="math-container">$V$</span>.</p>
<p>The question is this: How to show that <span class="math-container">$T$</span> is the smallest subspace of <span class="math-container">$V$</span> containing the <span class="math-container">$n$</span> vectors <span class="math-container">$v_1, v_2, \ldots, v_n$</span> ?</p>
<p>Indeed, let us take any subspace <span class="math-container">$S$</span> of <span class="math-container">$V$</span> that contains the <span class="math-container">$n$</span> vectors <span class="math-container">$v_1, v_2, \ldots, v_n$</span>.</p>
<p>Since <span class="math-container">$S$</span> is a subspace of <span class="math-container">$V$</span> and <span class="math-container">$v_1, v_2, \ldots, v_n \in S$</span>, <span class="math-container">$S$</span> must contain all the linear combinations of the <span class="math-container">$n$</span> vectors <span class="math-container">$v_1, v_2, \ldots, v_n$</span>.</p>
<p>In other words, for any set of <span class="math-container">$n$</span> scalars <span class="math-container">$\alpha_1, \alpha_2, \ldots,
\alpha_n$</span>,</p>
<p><span class="math-container">$$
v = \alpha_1 v_1 + \alpha_2 v_2 + \cdots + \alpha_n v_n \in S.
$$</span></p>
<p>This essentially means that <span class="math-container">$T \subset S$</span>, since <span class="math-container">$v$</span> is a general element in the subspace <span class="math-container">$T$</span>.</p>
<p>Since we have shown that <span class="math-container">$T$</span> is a subset of any subspace <span class="math-container">$S$</span> of <span class="math-container">$V$</span>
containing the vectors <span class="math-container">$v_1, v_2, \ldots, v_n$</span>, we have established that <span class="math-container">$T = \mbox{span}\{ v_1, v_2, \ldots, v_n \}$</span> is the smallest subspace of <span class="math-container">$V$</span> containing the vectors <span class="math-container">$v_1, v_2, \ldots, v_n$</span>. <span class="math-container">$ \ \ \ \ \ $$\blacksquare$</span></p>
|
3,056,121 | <p>I'm trying to find a function with infinitely many local minimum points where x <span class="math-container">$\in$</span> [0,1] and f has only 1 root. No interval should exist where the function is constant.</p>
| Felix Marin | 85,343 | <p><span class="math-container">$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\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}$</span></p>
<blockquote>
<p>With <span class="math-container">$\ds{\Re\pars{\mu} > - 1}$</span> and
<span class="math-container">$\ds{\Re\pars{\nu} > 0}$</span>:</p>
</blockquote>
<p><span class="math-container">\begin{align}
I_{\mu\nu} & \equiv
\bbox[10px,#ffd]{\int_{0}^{\infty}{x^{\mu}\ln\pars{x} \over \pars{1 + x^{2}}^{\nu}}\,\dd x}
\,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,
{1 \over 4}\int_{0}^{\infty}
{x^{\mu/2 - 1/2}\ln\pars{x} \over \pars{1 + x}^{\nu}}\,\dd x
\\[5mm] & =
\left.{1 \over 4}\,\partiald{}{\alpha}\int_{0}^{\infty}
{x^{\alpha + \mu/2 - 1/2} \over \pars{1 + x}^{\nu}}\,\dd x
\,\right\vert_{\ \alpha\ =\ 0}
\\[5mm] & \stackrel{x + 1\ \mapsto\ x}{=}\,\,\,
\left.{1 \over 4}\,\partiald{}{\alpha}\int_{1}^{\infty}
{\pars{x - 1}^{\alpha + \mu/2 - 1/2} \over x^{\nu}}\,\dd x
\,\right\vert_{\ \alpha\ =\ 0}
\\[5mm] & \stackrel{x\ \mapsto\ 1/x}{=}\,\,\,
\left.{1 \over 4}\,\partiald{}{\alpha}\int_{1}^{0}
{\pars{1/x - 1}^{\alpha + \mu/2 - 1/2} \over \pars{1/x}^{\nu}}\,
\pars{-\,{\dd x \over x^{2}}}
\right\vert_{\ \alpha\ =\ 0}
\\[5mm] & =
\left.{1 \over 4}\,\partiald{}{\alpha}\int_{0}^{1}
x^{\nu - \alpha - \mu/2 - 3/2}\pars{1 - x}^{\alpha + \mu/2 - 1/2}\,
\dd x\,\right\vert_{\ \alpha\ =\ 0}
\\[5mm] & =
{1 \over 4}\,\partiald{}{\alpha}\bracks{%
\Gamma\pars{\nu - \alpha - \mu/2 - 1/2}\Gamma\pars{\alpha + \mu/2 + 1/2} \over \Gamma\pars{\nu}}_{\ \alpha\ =\ 0}
\\[5mm] & =
\bbx{{\Gamma\pars{\mu/2 + 1/2}\Gamma\pars{\nu - \mu/2 - 1/2} \over
4\Gamma\pars{\nu}}
\bracks{H_{\mu/2 - 1/2} - H_{\nu - \mu/2 - 3/2}}}
\end{align}</span></p>
<blockquote>
<p><span class="math-container">$\ds{H_{z}}$</span> is a <a href="https://en.wikipedia.org/wiki/Harmonic_number" rel="nofollow noreferrer">Harmonic Number</a>.</p>
</blockquote>
<p><hr>
<span class="math-container">$$
\begin{array}{|c|c|}\hline
\ds{\mu \setminus \nu} & \ds{I_{\mu\nu}}
\\ \hline
\ds{0 \setminus 2} & \ds{-\,{\pi \over 4}}
\\ \hline
\ds{0 \setminus 3} & \ds{-\,{\pi \over 4}}
\\ \hline
\end{array}
$$</span></p>
<hr>
<blockquote>
<p>Note that <span class="math-container">$\ds{\bbx{I_{\mu,\mu + 1} = 0}}$</span> because, in such a case,
<span class="math-container">$\ds{H_{\mu/2 - 1/2} = H_{\nu - \mu/2 - 3/2}}$</span>.</p>
</blockquote>
|
705,945 | <p>I have this expression:
$$\sum_{\{\vec{S}\}}\prod_{i=1}^{N}e^{\beta HS_{i}}=\prod_{i=1}^{N}\sum_{S_{i}\in\{-1,1\}}e^{\beta HS_{i}} \qquad (1)$$
Where $\sum_{\{\vec{S}\}}$ means a sum over all possible vectors $\vec{S}=(S_1,...,S_N)$ with the restriction that $S_i$ can only take the values $\{-1,+1\}$, i.e. the sum is over $2^N$ different vectors: $\{\vec{S}\}$.</p>
<p>My <strong>question</strong> is: How can I be sure that (1) is right? Is there a criteria to interchange sums and products or it's always valid?</p>
| Ana S. H. | 48,197 | <p>The problem was in how to write down the sum $\sum_{\{\vec{S}\}}$. Since this sum is over all the possible vectors $\vec{S}=(S_1,...,S_N)$, (where $S_i\in\{-1,1\}$), we can rewrite this sum like</p>
<p>$$\sum_{S_{1}\in\{-1,1\}}\cdots\sum_{S_{N}\in\{-1,1\}}$$
i.e.
$$\sum_{\{\vec{S}\}}\prod_{i=1}^{N}e^{\beta HS_{i}}=\sum_{S_{1}\in\{-1,1\}}\cdots\sum_{S_{N}\in\{-1,1\}}\prod_{i=1}^{N}e^{\beta HS_{i}} \qquad(1)$$</p>
<p>Clearly this sum has $2^N$ elements of the type $\prod_{i=1}^{N}e^{\beta HS_{i}}$. Now, since</p>
<p>$$\prod_{i=1}^{N}e^{\beta HS_{i}}=e^{\beta HS_{1}}\cdots e^{\beta HS_{N}},$$
then (1) turns</p>
<p>$$\sum_{\{\vec{S}\}}\prod_{i=1}^{N}e^{\beta HS_{i}}=\sum_{S_{1}\in\{-1,1\}}\cdots\sum_{S_{N}\in\{-1,1\}}e^{\beta HS_{1}}\cdots e^{\beta HS_{N}} \qquad(2)$$
Since each $S_i$ is independent of the others, we can "factorize" the $\Sigma$'s:
$$\sum_{\{\vec{S}\}}\prod_{i=1}^{N}e^{\beta HS_{i}}=\left(\sum_{S_{1}\in\{-1,1\}}e^{\beta HS_{1}}\right)\cdots\left(\sum_{S_{N}\in\{-1,1\}}e^{\beta HS_{N}}\right)$$
And finally:
$$\sum_{\{\vec{S}\}}\prod_{i=1}^{N}e^{\beta HS_{i}}=\prod_{j=1}^{N}\sum_{S_{j}\in\{-1,1\}}e^{\beta HS_{j}} \qquad Q.E.D.$$</p>
<p>Note. This result can be generalized for vectors $\vec{S}=(S_1,...,S_N)$ with components $S_i\in\{1,...,k\}$ for some integer $k$.</p>
|
4,330,031 | <p>I want to prove that this degree sequence <span class="math-container">$(5,5,5,2,2,2,1)$</span> isn't valid to draw a graph from it, the graph needs to be simple. I am looking for a Theroem or a way to contradict the assumption that we can make a graph from it.</p>
<p>My solution was the following, for the given nodes:degrees => <span class="math-container">$(A:5; B:5; C:3; D:2; E:2; F:2; G:1)$</span></p>
<p><a href="https://i.stack.imgur.com/Z5HZ2.png" rel="nofollow noreferrer">Graph</a></p>
<p>Note that the vertex <span class="math-container">$C$</span> is the one that makes the contradiction, since we should have another 2 extra edges, but we can't add them to the previous nodes.</p>
<p>So my question is: Is there any theorem which I can use to prove this contradiction? Because I feel like my solution isn't enough.</p>
| markvs | 454,915 | <p>There are matrices <span class="math-container">$A$</span> such that <span class="math-container">$A=A^{-1}$</span>. More generally if <span class="math-container">$\det A=\pm 1$</span> then <span class="math-container">$\det(ABA)=\det B$</span> for every <span class="math-container">$B$</span> (in fact the last statement is "if and only if")..</p>
|
2,863,533 | <blockquote>
<p>Let $f(x)=x^3+ax^2+bx+c$ be a cubic polynomial with real coefficients and all real roots, also $|f(i)|=1$ where $i=\sqrt{-1}$. Prove that all three roots of $f(x)=0$ are zero. Also prove that $a+b+c=0$.</p>
</blockquote>
<hr>
<p>As $f(i)=-i-a+ib+c=1$ and $f(i)=-i-a+ib+c=-1$<br><br>
I don't know how to solve further.</p>
| farruhota | 425,072 | <p>Alternatively, note that for $z=a+bi\in \mathbb Z, \bar{z}=a-bi \in \mathbb Z$, the norm is:
$$|z|=\sqrt{z\cdot \bar{z}}=\sqrt{a^2+b^2}.$$
So:
$$|f(i)|=|c-a+(b-1)i|=\sqrt{(c-a)^2+(b-1)^2}=1 \Rightarrow (c-a)^2+b^2-2b=0 \tag{1}$$
Let $x_1,x_2,x_3$ be the roots of $f(x)=0 \iff x^3+ax^2+bx+c=0.$
By the Vieta's formulas:
$$\begin{align}\begin{cases}
x_1+x_2+x_3&=-a\\
x_1x_2+x_1x_3+x_2x_3&=b\\
x_1x_2x_3&=-c\end{cases} \tag{2}\end{align}$$
Plug $(2)$ to $(1)$:
$$(x_1+x_2+x_3-x_1x_2x_3)^2+(x_1x_2+x_1x_2+x_2x_3)^2-2(x_1x_2+x_1x_3+x_2x_3)=0 \iff \\
x_1^2+x_2^2+x_3^2+(x_1x_2)^2+(x_1x_3)^2+(x_2x_3)^2+(x_1x_2x_3)^2=0 \iff \\
x_1=x_2=x_3=0.$$</p>
|
433,639 | <p>(What follows is motivated by an answer to <a href="https://mathoverflow.net/questions/433612/fourier-optimization-problem-related-to-the-prime-number-theorem?noredirect=1#comment1116702_433612">Fourier optimization problem related to the Prime Number Theorem</a>)</p>
<p>Let <span class="math-container">$f:\mathbb{R}\to [0,\infty)$</span> be such that
<br>
(a) <span class="math-container">$\int_{\mathbb{R}} f(x) dx = 1$</span>,<br>
(b) <span class="math-container">$\widehat{f}(t)=0$</span> for all real <span class="math-container">$t$</span> with <span class="math-container">$|t|>1$</span>. <br></p>
<p>What is the choice of <span class="math-container">$f$</span> such that <span class="math-container">$$\int_{\mathbb{R}} |x| f(x) dx$$</span>is minimal? What is that minimum?</p>
<p>Remarks:</p>
<ol>
<li>It is easy to see that we can assume <span class="math-container">$f$</span> to be an even function.</li>
<li>Yes, this seems to be yet another incarnation of the uncertainty principle.</li>
</ol>
| Alexandre Eremenko | 25,510 | <p>One can prove that under your assumptions
<span class="math-container">$$A:=\int_{-\infty}^\infty|x|f(x)dx\geq 9/(4\pi),$$</span>
but estimate is not exact.</p>
<p>The proof is based on the formula
<span class="math-container">$$A=-\lim_{y\to 0+}\frac{1}{\pi}\frac{d}{dy}\left(y\int_{-1}^1\frac{\hat{f}(t)}{t^2+y^2}dt\right),$$</span>
which is easy to obtain. Now for <span class="math-container">$\hat{f}$</span> supported on <span class="math-container">$[-1,1]$</span>, Cramer obtained the estimate <span class="math-container">$\hat{f}(t)\leq 1-t^2/8$</span>,
and using this estimate and <span class="math-container">$\hat{f}(0)=1$</span>, we obtain the result.</p>
<p>Ref. H. Cramer, Random variables and probability distributions, Cambridge UP, 1970.</p>
<p>One can use a better, exact estimate for <span class="math-container">$\hat{f}$</span> in the work of</p>
<p>A. Fryntov, An extremal problem in the theory of Hermitian positive functions, Func Anal., Appl, 10, 1 (1976) 91-92;</p>
<p>this will improve the estimate, but still will not give the exact minimum of <span class="math-container">$A$</span>.</p>
<p>Comment. A somewhat simpler but similar problem would be to miminize
<span class="math-container">$$|\hat{f}''(0)|=\int_{-\infty}^\infty x^2f(x)dx.$$</span>
For this quantity, Fryntov implies the estimate <span class="math-container">$A\geq \pi^2$</span>, which is probably also not exact.</p>
|
4,515 | <p>I've been using the sentence:</p>
<blockquote>
<p>If a series converges then the limit of the sequence is zero</p>
</blockquote>
<p>as a criterion to prove that a series diverges (when $\lim \neq 0$) and I can understand the rationale behind it, but I can't find a <strong>formal proof</strong>.</p>
<p>Can you help me?</p>
| JT_NL | 1,120 | <p>Yes.</p>
<p>$$\lim_{n \to \infty} \left ( \sum_{k = 1}^{n + 1} a_k - \sum_{k = 1}^{n} a_k \right ) = \lim_{n \to \infty} a_{n + 1} $$
And both sums will converge to the same number so the limit is zero. This is by far the easiest proof I know.</p>
<p>This is the Cauchy criterion in disguise by the way, so you could use that too.</p>
|
463,239 | <p>Integrate $$\int{x^2(8x^3+27)^{2/3}}dx$$</p>
<p>I'm just wondering, what should I make $u$ equal to?</p>
<p>I tried to make $u=8x^3$, but it's not working. </p>
<p>Can I see a detailed answer?</p>
| amWhy | 9,003 | <blockquote>
<p>$$\int{x^2(8x^3+27)^{2/3}}dx$$</p>
</blockquote>
<p>It is certainly possible to work with $u = 8x^3$, but I'd suggest setting $u = 8x^3 + 27$. The key is to remember to compute and account for $\,du$. What you'll see is that for both $u = 8x^3$ and $u = 8x^3 + 27$, we have $du = 24x^2$. </p>
<p>$$u = 8x^3 + 27 \implies\,du = 24x^2 \,dx \iff \color{blue}{\bf\dfrac 1{24} \,du = x^2 \,dx}$$</p>
<p>Substituting: $$\int x^2(8x^3+27)^{2/3}dx = \int (8x^3 + 27)^{2/3} \color{blue}{\bf x^2 \,dx} = \color{blue}{\bf \dfrac1{24}} \int u^{2/3} \, \color{blue}{\bf du}$$</p>
<p>Evaluating the integral: $$\color{blue}{\bf \dfrac1{24}} \int u^{2/3} \, \color{blue}{\bf du} = \dfrac 1{24} \dfrac {u^{5/3}}{5/3} + C $$ </p>
<p>Back substituting, given $u = 8x^3 + 27$, then simplifying, gives us:</p>
<p>$$\dfrac 1{24} \dfrac {u^{5/3}}{5/3} + C = \dfrac {3}{24 \cdot 5}(8x^3 + 27)^{5/3} + C = \dfrac {1}{40}(8x^3 + 27)^{5/3} + C$$</p>
|
3,681,916 | <p>Given a matrix A, Does
<span class="math-container">$$\lim_{n \to \infty}A^n=0$$</span>
Imply that(by lower then I mean that every number in the lower matrix is closer or the same distance to 0 then it’s counterpart in the bigger one)
<span class="math-container">$$\text{if }a>b>0 \text{ then }A^a<A^b$$</span>
In R it’s true, a Number who is in between -1 and 1 is always getting closer to 0 and never goes up and back down.<br>
I’m unsure of it’s truth for matrices especially with my definition.
Can any of you provide hints/proves for it?<br>
Maybe counter example?<br>
Thanks in advance.</p>
| Community | -1 | <p>Consider <span class="math-container">$A=e^{u}\begin{pmatrix}\cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha\end{pmatrix}$</span>, which we'll call <span class="math-container">$A(u,\alpha)$</span>. Such matrices have the property that <span class="math-container">$(A(u,\alpha))^n=A(nu,n\alpha)$</span> and <span class="math-container">$\sqrt{ \operatorname{tr}((A(u,\alpha))^tA(u,\alpha))}=\sqrt2e^u$</span>. Select <span class="math-container">$\alpha>0$</span> small enough so that <span class="math-container">$k\alpha\in\left(0,\frac\pi2\right)$</span> for all <span class="math-container">$1\le k\le N$</span>. We have that <span class="math-container">$\lvert A(ku,k\alpha )_{2,1}\rvert=e^{ku}\lvert\sin(k\alpha)\rvert$</span>, and therefore, for all <span class="math-container">$1\le m<n\le N$</span>, <span class="math-container">$$\lvert A(nu,n\alpha)_{2,1}\rvert<\lvert A(mu,m\alpha)_{2,1}\rvert\iff u<\frac1{n-m}\ln\frac{\sin(m\alpha)}{\sin(n\alpha)}$$</span></p>
<p>Notice that the hypothesis on <span class="math-container">$\alpha$</span>, <span class="math-container">$m$</span> and <span class="math-container">$n$</span> guarantee that <span class="math-container">$\sin(m\alpha)<\sin(n\alpha)$</span> for all <span class="math-container">$1\le m<n\le N$</span>, so that the <span class="math-container">$\ln$</span> on the RHS is negative.</p>
<p>Therefore, if you fix a positive integer <span class="math-container">$N$</span>, an angle <span class="math-container">$\alpha\in\left(0,\frac\pi{2N}\right)$</span> and some <span class="math-container">$u$</span> such that <span class="math-container">$\max\limits_{1\le m<n\le N} \frac1{n-m}\ln\frac{\sin(m\alpha)}{\sin(n\alpha)}< u<0$</span>, then you'll have that <span class="math-container">$(A(u,\alpha))^m>(A(u,\alpha))^n$</span> fails for all <span class="math-container">$1\le m<n\le N$</span>.</p>
|
4,622,956 | <p>I think <span class="math-container">$\,9\!\cdot\!10^n+4\,$</span> can be a perfect square, since it is <span class="math-container">$0 \pmod 4$</span> (a quadratic residue modulo <span class="math-container">$4$</span>), and <span class="math-container">$1 \pmod 3$</span> (also a quadratic residue modulo <span class="math-container">$3$</span>).<br />
But when I tried to find if <span class="math-container">$\;9\!\cdot\!10^n+4\,$</span> is a perfect square, I didn’t succeed. Can someone help me see if <span class="math-container">$\;9\!\cdot\!10^n+4\,$</span> can be a perfect square ?</p>
| Sai Mehta | 1,137,054 | <p>Assume that <span class="math-container">$9\cdot10^n+4\equiv4$</span> is a perfect square.<br />
<span class="math-container">$9\cdot10^n+4\equiv4\pmod9$</span>, so <span class="math-container">$9\cdot10^n+4$</span> can be represented as <span class="math-container">$(9m-2)^2=81m^2-18m+4$</span> or <span class="math-container">$(9m+2)^2=81m^2+18m+4$</span>, where <span class="math-container">$m\in\mathbb{N}$</span>.<br />
If <span class="math-container">$9\cdot10^n+4=81m^2-18m+4$</span>, <span class="math-container">$$10^n=9m^2-2m=m(9m-2)$$</span>
This means that <span class="math-container">$m$</span> and <span class="math-container">$9m-2$</span> must be powers of <span class="math-container">$10$</span>.<br />
Clearly, <span class="math-container">$9m-2>m$</span> because <span class="math-container">$m>0$</span>. If <span class="math-container">$m≠1$</span>, <span class="math-container">$m\equiv0\pmod{10}$</span>, but then <span class="math-container">$9m-2\equiv8\pmod{10}$</span>, which doesn't work. If <span class="math-container">$m=1$</span>, <span class="math-container">$9m-2=7$</span>, which also doesn't work.<br />
If <span class="math-container">$9\cdot10^n+4=81m^2+18m+4$</span>, <span class="math-container">$$10^n=9m^2+2m=m(9m-2)$$</span>
This means that <span class="math-container">$m$</span> and <span class="math-container">$9m+2$</span> must be powers of <span class="math-container">$10$</span>.<br />
Clearly, <span class="math-container">$9m+2>m$</span> because <span class="math-container">$m>0$</span>. If <span class="math-container">$m≠1$</span>, <span class="math-container">$m\equiv0\pmod{10}$</span>, but then <span class="math-container">$9m+2\equiv2\pmod{10}$</span>, which doesn't work. If <span class="math-container">$m=1$</span>, <span class="math-container">$9m+2=11$</span>, which also doesn't work.<br />
Therefore, there is a contradiction and so <span class="math-container">$9\cdot10^n+4$</span> cannot be a perfect square.</p>
|
1,038,076 | <p>Solve the equation $7\times 13\times 19=a^2-ab+b^2$ for integers $a>b>0$. How many are there such solutions $(a,b)$?</p>
<p>I know that $a^2-ab+b^2$ is the norm of the Eisentein integer $z=a+b\omega$, but how can I make use of this? Thank you so much.</p>
| Rajkumar | 184,827 | <p>Plotting $7 \times 13 \times 19 = a^2 − ab + b^2$, You will get an ellipse like this: </p>
<p><img src="https://i.stack.imgur.com/H5m6B.gif" alt="enter image description here"></p>
<p>.<br>
But if we apply the condition $a$ & $b > 0$, both will be positive in only first Quadrant. And again $a > b$, we will end up with half of the ellipse in Quadrant 1 i.e. 1/8th of the total ellipse. </p>
<p><img src="https://i.stack.imgur.com/iC0TD.gif" alt="enter image description here"></p>
<p>.<br>
Set of all the points on this curve is our <strong>solution</strong>. Infinite number of Real solutions.</p>
<p>Below are the integer solutions: </p>
<p><strong>(43,3), (43,40), (45,8), (45,37), (47,15),(47,32),(48,23),(48,25)</strong></p>
|
3,060,456 | <p>Can any one explain me the intuition behind this formula ? (with permutation example)</p>
<pre><code>P(n, k) = P(n-1, k) + k* P(n-1, k-1)
</code></pre>
| Wuestenfux | 417,848 | <p>Okay, define a word <span class="math-container">$x=x_1\ldots x_k$</span> of length <span class="math-container">$k$</span> over the alphabet <span class="math-container">$[n]=\{1,\ldots,n\}$</span> to be a <span class="math-container">$k$</span>-permutation of <span class="math-container">$n$</span> if <span class="math-container">$x$</span> contains no symbol more than once. The word <span class="math-container">$x$</span> can be viewed as an injective mapping <span class="math-container">$[k]\rightarrow[n]:i\mapsto x_i$</span>. Indeed, the <span class="math-container">$k$</span>-permutations of <span class="math-container">$n$</span> correspond one-to-one with the injective mappings <span class="math-container">$[k]\rightarrow[n]$</span>.</p>
<p>Let <span class="math-container">$P(n,k)$</span> denote the number of all <span class="math-container">$k$</span>-permutations of <span class="math-container">$n$</span>. We have <span class="math-container">$P(n,1)=n$</span> and <span class="math-container">$P(n,n)=n!$</span> (bijections). For each <span class="math-container">$1<k<n$</span>, the above formula holds.</p>
<p>This can be proved by a socalled ''Pascal argument''. Let <span class="math-container">$A$</span> be the set of <span class="math-container">$k$</span>-permutations of <span class="math-container">$n$</span> which contain <span class="math-container">$n$</span>, and let <span class="math-container">$B$</span> be the set of remaining <span class="math-container">$k$</span>-permutations of <span class="math-container">$n$</span>. Then <span class="math-container">$P(n,k)= |A|+|B|$</span>.</p>
<p>In view of <span class="math-container">$A$</span>, each <span class="math-container">$k-1$</span>-permutation of <span class="math-container">$n-1$</span>, <span class="math-container">$y=y_1\ldots y_{k-1}$</span>, can be extended to a <span class="math-container">$k$</span>-permutation of <span class="math-container">$n$</span>, <span class="math-container">$y^{(i)}$</span>, such that the symbol <span class="math-container">$n$</span> is inserted into position <span class="math-container">$i$</span>. The assignment <span class="math-container">$(i,y)\mapsto y^{(i)}$</span> provides a bijection between <span class="math-container">$[k]\times \mbox{(set of $k-1$-permutations of $n-1$)}$</span> and the set <span class="math-container">$A$</span>. Thus <span class="math-container">$|A| = k\cdot P(k-1,n-1)$</span>.</p>
<p>In view of <span class="math-container">$B$</span>, the elements of <span class="math-container">$B$</span> are exactly the <span class="math-container">$k$</span>-permutations of <span class="math-container">$n-1$</span> and so <span class="math-container">$|B|=P(n-1,k)$</span>.</p>
<p>As an example, the 2-permutations of 3,
<span class="math-container">$12, 21, 13, 31, 23, 32$</span>, are decomposed into <span class="math-container">$A=\{13,31,23,32\}$</span> and <span class="math-container">$B=\{12,21\}$</span>.</p>
|
4,163,003 | <p>Let <span class="math-container">$Z \in \mathbb{R}^2$</span> be an i.i.d. Gaussian vector with mean <span class="math-container">$M$</span> where <span class="math-container">$P_{Z\mid M}$</span> is its distribution.</p>
<p>Let <span class="math-container">$g: \mathbb{R}^2 \to \mathbb{R}$</span> and consider the following equation:
<span class="math-container">$$
E[g(Z)\mid M=\mu]=0, \forall \mu \in C,
$$</span>
where <span class="math-container">$C=\{\mu: \frac{\mu_1^2}{r_1^2}+\frac{\mu_2^2}{r_2^2}=1 \}$</span> for some given <span class="math-container">$r_1,r_2 > 0$</span>. That is, <span class="math-container">$C$</span> is an ellipse.</p>
<p>It is not difficult to verify (see this <a href="https://math.stackexchange.com/questions/4159488/example-of-g-mathbbr2-to-mathbbr-s-t-egzm-mu-0-forall-mu-i/4159528?noredirect=1#comment8618967_4159528">question</a>, see also Edit 3) that a solution to this equation is given by
<span class="math-container">$$
g(x)= \frac{x_1^2}{r_1^2}+\frac{x_2^2}{r_2^2}- c,
$$</span>
where <span class="math-container">$c=\frac{1}{r_1^2}+\frac{1}{r_2^2}+1$</span>.
In fact any function <span class="math-container">$g_a(x)= a g(x)$</span> for any <span class="math-container">$a \in \mathbb{R}$</span> is a solution.</p>
<p><strong>Question:</strong> Is <span class="math-container">$g$</span> a unique solution up to a multiplicative constant?</p>
<p><strong>Edit:</strong> If we need to make an assumption on the class of allowed functions <span class="math-container">$g$</span>. Let us assume that <span class="math-container">$g$</span>'s are bounded by a quadratic monomial (i.e., for every <span class="math-container">$g$</span> there exists <span class="math-container">$a$</span> and <span class="math-container">$b$</span> such that <span class="math-container">$g(x) \le a \|x \|^2 +b$</span>).</p>
<p><strong>Edit 2:</strong> If you want to avoid expectation notation. Everything can be alternatively written as
<span class="math-container">$$
\iint g(z)\frac{1}{2 \pi} e^{-\frac{\|z-m\|^2}{2}} \, {\rm d} z=0, \, m\in C.
$$</span>
From here, one can see that this question is about a convolution.</p>
<p><strong>Edit 3:</strong> To see that <span class="math-container">$g(x)$</span> is a solution we use that the second moment of Gaussian is given by <span class="math-container">$E[Z_i^2\mid M_i=\mu_i]=1+\mu_i^2$</span>, which leads to
<span class="math-container">\begin{align}
E\left[\frac{Z_1^2}{r_1^2}+\frac{Z_2^2}{r_2^2}- c\mid M=\mu \right]&= \frac{E[Z_1^2\mid M_1=\mu_1]}{r_1^2}+\frac{ E[Z_2^2\mid M_2=\mu_2]}{r_2^2}-c\\[6pt]
&=\frac{1+\mu_1^2}{r_1^2}+\frac{ 1+\mu_1^2}{r_2^2}-c\\[6pt]
&=\frac{1}{r_1^2}+\frac 1 {r_2^2}+1-c,
\end{align}</span>
where in the last step we used that <span class="math-container">$\mu$</span> is on the ellipse.</p>
<p><strong>Edit 4:</strong> The comment below shows that the solution is not unique when an ellipse is a circle. However, I would still like to know the answer for a general ellipse.</p>
| Michael Hardy | 11,667 | <p>This will not answer the question as stated, but whoever asks such a question may be interested in what I say here.</p>
<p>First, a random variable with a "Gaussian" or "normal" distribution need not have variance <span class="math-container">$1,$</span> so I would have stated what the variance is.</p>
<p>In standard terminology, the question is asking which <b>unbiased estimators of zero</b> this parameterized family of probability distributions has (where the ellipse is the parameter space).</p>
<p>A family of probability distributions that admits no unbiased estimators of zero is called "complete", so the question is how far this family is from being complete.</p>
<p>The family of all univariate Gaussian distributions, whose parameter space may be taken to be the half-plane <span class="math-container">$\{(\mu,\sigma) : \mu\in\mathbb R,\, \sigma>0\},$</span> is complete, and that fact is essentially the same thing as the one-to-one nature of the two-sided Laplace transform.</p>
<p>A corollary of the completeness of this family of all Gaussian distributions is that with an i.i.d. sample, the sample mean has a smaller variance than every other unbiased estimator of the population mean. For example, think about how to prove that the sample median has a smaller variance than the sample mean. With this result, you get that without computing the variance of the sample median.</p>
|
3,545,548 | <p><span class="math-container">$\def\LIM{\operatorname{LIM}}$</span>
Let <span class="math-container">$(X,d)$</span> be a metric space and given any cauchy sequence <span class="math-container">$(x_n)_{n=1}^{\infty}$</span> in <span class="math-container">$X$</span> we introduce the formal limit <span class="math-container">$\LIM_{n\to \infty}x_n$</span>. We say that two formal limits <span class="math-container">$\LIM_{n\to \infty}x_n$</span> and <span class="math-container">$\LIM_{n\to \infty}y_n$</span> are equal iff <span class="math-container">$\lim_{n \to \infty}d(x_n,y_n)=0$</span>. We then define <span class="math-container">$\bar{X}$</span> to be set of all the formal limits of Cauchy sequences in <span class="math-container">$X$</span>. We define the metric <span class="math-container">$d_{\bar{X}}$</span> as follows: <span class="math-container">$$d_{\bar{X}}(\LIM_{n\to \infty}x_n,\LIM_{n\to \infty}y_n)= \lim_{n \to \infty} d(x_n,y_n)$$</span>
I have proved that <span class="math-container">$(\bar{X},d_{\bar{X}})$</span> is indeed a metric space that that the definition of metric is well defined. But I am stuck to prove that <span class="math-container">$(\bar{X},d_{\bar{X}})$</span> is a complete metric space. This problem could be resolved without taking into account topological spaces as that concept in later in the book. Any suggestion on how to go about this problem without using machinery of topology would be highly invaluable. Thanks in advance.</p>
| Arjun | 744,667 | <p>If <span class="math-container">$x,y,z$</span> are succively in AP then let each
of them be<span class="math-container">$a,a+d,a+2d$</span></p>
<p>Which means
<span class="math-container">$$3^x,3^y,3^z=3^a,3^{a+d},3^{a+2d}$$</span></p>
<p>Which when written simply is
<span class="math-container">$$3^a,3^d×3^a,3^{2d}×3^a$$</span></p>
<p>Which turns out to be the very definition of an GP.</p>
|
2,904,603 | <p>I'm working on the following question:</p>
<blockquote>
<p>Show that $G$ is a group if and only if, for every $a, b \in G$,
the equations $xa = b$ and $ay = b$ have solutions $x, y \in G$.</p>
</blockquote>
<p>I'm having trouble getting started because I'm not understanding what it means for "the equations $xa = b$ and $ay = b$ have solutions $x, y \in G$". Do they mean, there's only 1 left multiplier that takes $a$ to $b$ and one right multiplier that takes $a$ to $b$?</p>
| Sarvesh Ravichandran Iyer | 316,409 | <p>"the equations $xa = b$ and $ay = b$ have solutions $x,y \in G$", means precisely this :</p>
<blockquote>
<p>For all $a \in G$ and $b \in G$, <em>there exist</em> $x \in G$ and $y \in G$ which satisfy the equations $xa = b$ and $ay = b$ respectively.</p>
</blockquote>
<p>Since it only says that there is <em>at least one</em> solution to $xa = b$, the hypothesis does not rule out $x'a = b$ or $ay' = b$ for some <em>other</em> $x',y' \in G$. It is true that this will hold if $G$ is a group (by multiplying by the left/right inverse of $a$), but you cannot assume this from hypothesis. </p>
<p>Note : While it is not mentioned, $G$ must have a binary product on it that is associative for this to work out. (and must be non-empty, as pointed out in the comment below)</p>
|
3,292,918 | <p>Let <span class="math-container">$X$</span> be a Banach space, and denote by <span class="math-container">$B_r (x)$</span> the closed ball of radius <span class="math-container">$r > 0$</span>
around <span class="math-container">$x \in X$</span>. Furthermore, let <span class="math-container">$A \subset X$</span> be compact and <span class="math-container">$N \in \Bbb{N}$</span>.
I am interested in "optimally" covering <span class="math-container">$A$</span> by <span class="math-container">$N$</span> balls, i.e,. with as small radius as possible.
More precisely, define
<span class="math-container">$$
r_0
:= \inf \Big\{
r > 0
\colon
\exists \, x_1,\dots,x_N \in X \text{ such that } A \subset \bigcup_{i=1}^N B_r (x_i)
\Big\},
$$</span>
and assume that <span class="math-container">$r_0 > 0$</span>.</p>
<blockquote>
<p>I would like to know whether there necessarily exist <span class="math-container">$x_1,\dots,x_N \in X$</span>
such that <span class="math-container">$A \subset \bigcup_{i=1}^N B_{r_0} (x_i)$</span>.
In other words, I would like to know if the infimum above is actually a minimum.</p>
</blockquote>
<p><strong>Note</strong> I am taking the <span class="math-container">$x_i$</span> from the "surrounding" space <span class="math-container">$X$</span>, not from the compact set <span class="math-container">$A$</span>.</p>
<p>I can prove the claim in case that <span class="math-container">$X$</span> is reflexive (even only assuming that <span class="math-container">$A$</span> is bounded),
but I am not sure whether it is true for more general Banach spaces.
I will give my proof for the reflexive case below, in case one can either generalize it,
or use it to get an idea for a counterexample.</p>
<p><hr><hr></p>
<p><strong>Proof for the reflexive case:</strong>
Choose a sequence <span class="math-container">$r_n \to r_0$</span> such that for each <span class="math-container">$n$</span> there are <span class="math-container">$x_1^n,\dots,x_N^n \subset X$</span>
satisfying <span class="math-container">$A \subset \bigcup_{i=1}^N B_{r_n}(x_i^n)$</span>.
If <span class="math-container">$B_{r_n} (x_i^n) \cap A = \emptyset$</span> for some <span class="math-container">$i,n$</span>, replace <span class="math-container">$x_i^n$</span> by zero.
Note that this retains the property <span class="math-container">$A \subset \bigcup_{i=1}^N B_{r_n}(x_i^n)$</span>.</p>
<p>Since <span class="math-container">$A$</span> and the sequence <span class="math-container">$(r_n)_{n}$</span> are bounded, there is <span class="math-container">$R > 0$</span> such that <span class="math-container">$\| x \| \leq R$</span>
and <span class="math-container">$r_n \leq R$</span> for all <span class="math-container">$x \in A$</span> and <span class="math-container">$n \in \Bbb{N}$</span>:
There are now two cases for each <span class="math-container">$i,n$</span>:
1) There is some <span class="math-container">$x \in A \cap B_{r_n} (x_i^n)$</span>,
and hence <span class="math-container">$\| x_i^n \| \leq \| x_i^n - x \| + \| x \| \leq r_n + R \leq 2R$</span>.
2) There is no <span class="math-container">$x \in A \cap B_{r_n} (x_i^n)$</span>, and hence <span class="math-container">$x_i^n$</span>, whence <span class="math-container">$\| x_i^n \| \leq 2R$</span>.</p>
<p>Therefore, each of the sequences <span class="math-container">$(x_i^n)_{n \in \Bbb{N}} \subset X$</span> is bounded.
Since <span class="math-container">$X$</span> is assumed to be reflexive, we can choose a common subsequence
(which I will ignore in the notation below) such that <span class="math-container">$x_i^n \to x_i$</span> weakly for all <span class="math-container">$i = 1,\dots,N$</span>.</p>
<p>Now, let <span class="math-container">$x \in A$</span> be arbitrary.
For each <span class="math-container">$n \in \Bbb{N}$</span>, there is <span class="math-container">$i_n \in \{1,\dots,N\}$</span> satisfying <span class="math-container">$\| x - x_i^n \| \leq r_n$</span>.
Next, there is <span class="math-container">$\ell \in \{1,\dots,N\}$</span> such that <span class="math-container">$i_n = \ell$</span> for infinitely many <span class="math-container">$n \in \Bbb{N}$</span>,
say for <span class="math-container">$n = n_m$</span> with <span class="math-container">$n_m \to \infty$</span>.
Since <span class="math-container">$x - x_i^{n_m} \to x - x_i$</span> weakly and since the norm is lower semicontinuous with respect
to weak convergence, we see that
<span class="math-container">$\| x - x_\ell \| \leq \liminf_{m \to \infty} \| x - x_i^{n_m} \| \leq \liminf_{m \to \infty} r_{n_m} = r_0$</span>.
Since this holds for any <span class="math-container">$x \in A$</span>, we get <span class="math-container">$A \subset \bigcup_{\ell=1}^N B_{r_0} (x_\ell)$</span>,
as desired.</p>
| mathworker21 | 366,088 | <p>I don't feel too bad posting a "proof by references" since I worked on this problem for quite a while but failed, before realizing the <span class="math-container">$N=1$</span> case seems pretty google-able. In any event, it seems Konyagin proved that any non-reflexive Banach space admits an equivalent norm and some 3 points so that there is no smallest ball with respect to the equivalent norm containing those 3 points. A good buzzword is "Chebyshev center". Below is a link to a strengthening of Konyagin's result (and a mention of Konyagin's result). I couldn't find a free version of Konyagin's paper online.</p>
<p><a href="https://dml.cz/bitstream/handle/10338.dmlcz/119230/CommentatMathUnivCarolRetro_42-2001-1_11.pdf" rel="nofollow noreferrer">https://dml.cz/bitstream/handle/10338.dmlcz/119230/CommentatMathUnivCarolRetro_42-2001-1_11.pdf</a></p>
|
1,608,299 | <p>I am completely stuck on a question. I've done it 4 times, each times got different result, but never correct.</p>
<p>The third term of an arithmetic progression is 71 and the seventh term is 55. Find the sum of the first 45 terms.</p>
<p>Any ideas? Thanks</p>
| Archis Welankar | 275,884 | <p>$a+2d=71...1 $ while $a+6d=55$ solving you get $d=-4,a=79$ thus now sum =$\frac{45}{2}(158+(44)(-4))=-405$ thus its solved</p>
|
1,608,299 | <p>I am completely stuck on a question. I've done it 4 times, each times got different result, but never correct.</p>
<p>The third term of an arithmetic progression is 71 and the seventh term is 55. Find the sum of the first 45 terms.</p>
<p>Any ideas? Thanks</p>
| Guillemus Callelus | 361,108 | <p>We know that $a_3=71$ and $a_7=55$. Then,
$$a_3=a_1+(3-1)d\Rightarrow 71=a_1+2d$$
$$a_7=a_1+(7-1)d\Rightarrow 55=a_1+6d$$
If we subtract the two equations:
$$71-55=(a_1+2d)-(a_1+6d)\Rightarrow 16=-4d\Rightarrow d=-\dfrac{16}{4}=-4$$
Then,
$$71=a_1+2d\Rightarrow a_1=71-2d=71-2(-4)=71+8=79$$
The general term of the arithmetic progression is:
$$a_n=a_1+(n-1)d=79+(n-1)(-4)=79-4n+4=83-4n$$
$$a_{45}=83-4\cdot 45=-97$$
The formula that gives the sum of the first $n=45$ terms of an arithmetic progression is:
$$S=\dfrac{(a_1+a_{n})n}{2}=\dfrac{(a_1+a_{45})45}{2}=\dfrac{(79-97)45}{2}=\boxed{-405}$$</p>
|
228,135 | <p>I'm working to understand the Grothendieck topology version of the Zariski topology of schemes. Explained simply, it replaces the notion of "open subschemes" with "open immersions". So instead of $U\subseteq X$, we have $U\hookrightarrow X$.</p>
<p>The intersection $U\cap V$ between two open subschemes is replaced with the canonical immersion of the fiber product $(U\times_X V)\hookrightarrow X$. Is there a correspondingly simple analogue of the union, or do I have to construct it explicitly?</p>
| Community | -1 | <p>A covering of $U$ is replaced by a family of morphisms $U_i\to U$ the union of whose images is $U$. </p>
|
1,579,616 | <p>So I know it's true for $n = 5$ and assumed true for some $n = k$ where $k$ is an interger greater than or equal to $5$.</p>
<p>for $n = k + 1$ I get into a bit of a kerfuffle.</p>
<p>I get down to $(k+1)^2 + 1 < 2^k + 2^k$ or equivalently:</p>
<p>$(k + 1)^2 + 1 < 2^k * 2$.</p>
<p>A bit stuck at how to proceed at this point</p>
| Jimmy R. | 128,037 | <p>$$2^{k+1}=2\left(2^k\right)>2(k^2+1)=2k^2+2=k^2+k^2+2\overset{(*)}>k^2+2k+2=(k^2+1)+1$$ So, it remains to show that $k^2>2k$ for $k>4$. But $$k^2-2k=k(k-2)>0$$ for all $k>2$.</p>
|
164,213 | <p>Suppose I have some list with duplicates by some condition and I want to take the duplicates and apply some function to choose which duplicate to keep. Is there an efficient way to apply this transformation?</p>
<p>To clarify here is an example. Consider a list with elements with duplicate first elements:</p>
<pre><code>list={{2, 0.2}, {3, 0.}, {2, 0.1}, {4, 0.9}, {6, 0.3}, {3, 0.4}, {6, 0.3}}
</code></pre>
<p>Now I can apply, <code>DeleteDuplicatesBy[list, #[[1]] &]</code> and it will drop all the second instances of the duplicates in the list. But instead I want to do something else, say keep the duplicate with the maximum second value. This would look something like (my hypothetical function takes the arguments: list, function to find duplicates, function to act on duplicate list):</p>
<pre><code>CombineDuplicateBy[list,#[[1]]&,MaximalBy[#,#[[2]]&]]
{{2, 0.2}, {4, 0.9},{3, 0.4}, {6, 0.3}}
</code></pre>
<p>Notice my hypothetical output deleted elements <code>{3, 0.}</code> and <code>{6, 0.3}</code> since they were duplicates and had a smaller (or equal) second value. </p>
| Carl Woll | 45,431 | <p>You could use <a href="http://reference.wolfram.com/language/ref/GroupBy" rel="noreferrer"><code>GroupBy</code></a>:</p>
<pre><code>GroupBy[list, First, First@*MaximalBy[Last]]
Values @ %
</code></pre>
<blockquote>
<p><|2 -> {2, 0.2}, 3 -> {3, 0.4}, 4 -> {4, 0.9}, 6 -> {6, 0.3}|></p>
<p>{{2, 0.2}, {3, 0.4}, {4, 0.9}, {6, 0.3}}</p>
</blockquote>
|
1,056,038 | <blockquote>
<p>Each of $n$ balls is independently placed into one of $n$ boxes, with all boxes equally likely. What is the probability that exactly one box is empty? (Introduction to Probability, Blitzstein and Nwang, p.36).</p>
</blockquote>
<ul>
<li>The number of possible permutations with replacement is $n^n$</li>
<li><p>In order to have one empty box, we need a different box having $2$ balls in it. We have $\dbinom{n}{1}$ choices for the empty box, $\dbinom{n-1}{1}$ choices left for the box with $2$ balls, and $(n-2)!$ permutations to assign the remaining balls to the remaining boxes.</p></li>
<li><p>Result: $$\frac{\dbinom{n}{1} \dbinom{n-1}{1} (n-2)!}{n^n}$$</p></li>
</ul>
<p>Is this correct?</p>
| d125q | 112,944 | <p>You can think of the number of favorable arrangements in the following way: choose the empty box in $\binom{n}{1}$ ways. For each such choice, choose the box that will have at least $2$ balls (there has to be one such box) in $\binom{n - 1}{1}$ ways. And for this box, choose the balls that will go inside in $\binom{n}{2}$ ways. Now permute the remaning balls in $(n - 2)!$ ways.</p>
<p>Thus, the number of favorable arrangements is:</p>
<p>$$
\binom{n}{1} \binom{n - 1}{1} \binom{n}{2} (n - 2)! = \binom{n}{2} n!
$$</p>
|
4,615,947 | <p>Let <span class="math-container">$a,b\in\Bbb{N}^*$</span> such that <span class="math-container">$\gcd(a,b)=1$</span>. How to show that <span class="math-container">$\gcd(ab,a^2+b^2)=1$</span>?</p>
| giorgiokyn23 | 1,134,657 | <p>Complex Analysis approach: Let <span class="math-container">$$f(z)=\frac{e^{iz}}{z^2+2z+2}$$</span>Note that <span class="math-container">$Imm(f(z))=\frac{sin(z)}{z^2+2z+2}$</span>.
<span class="math-container">$f$</span> has 2 simple pole with order 1. The pole are <span class="math-container">$z_1=-1+i$</span> and <span class="math-container">$z_2=-1-i$</span> wich are the roots of <span class="math-container">$p(z)=z^2+2z+2$</span>.
So <span class="math-container">$f(z)=\frac{e^{iz}}{(z-z_1)(z-z_2)}$</span>.
Writing <span class="math-container">$f(z)=\frac{1}{z-z_1}h(z)$</span> note that <span class="math-container">$h(z_1)\not=0$</span> and <span class="math-container">$h(z)$</span> olomorphic in <span class="math-container">$z_1$</span> we have <span class="math-container">$Res(f,z_1)=h(z_1)=\frac{e^{iz_1}}{(z_1-z_2)}=\frac{e^{i(-1+i)}}{(2i)}$</span>
Analogue for <span class="math-container">$z_2$</span>.</p>
<p>Now consider <span class="math-container">$\phi(t):[-R,R]\to \Bbb C,\phi(t)=t$</span> and <span class="math-container">$\tau_R(t):[0,\pi]\to \Bbb C,\tau_R(t)=Re^{it}$</span>. Be <span class="math-container">$\Gamma=$</span>
<span class="math-container">$$ \int_{\Gamma} f(z) \ dt = \int_{\phi} f(z) \ dz + \int_{\tau_R} f(z) \ dz = 2 \pi i Res(f,z_1) $$</span>Noticing that z_1 is the only pole in <span class="math-container">$\Gamma$</span>.</p>
<p><span class="math-container">$$ \lim_{R\to \infty} \int_{\Gamma} f(z) \ dt = \lim_{R\to \infty}\int_{\phi} f(z) \ dz +\lim_{R\to \infty} \int_{\tau_R} f(z) \ dz =2 \pi i Res(f,z_1)$$</span>. Noticing that the second integral<span class="math-container">$\to 0$</span> as <span class="math-container">$R \to\infty$</span>, you consider the first integral with now is in <span class="math-container">$\Bbb R$</span> and splitting it in the real part and the imaginary part then, you conclude.</p>
|
1,747,525 | <p>Given a number $N$, how can I write down a summation of all odd numbers divisible by 5 which are also less than $N$?</p>
<p>For instance, if $N = 27$ then I am looking for a series to generate $5+15+25$.</p>
<p>Its pretty clear the series looks like </p>
<p>$$\sum_{k=0}^{???} 5(2k+1)$$</p>
<p>but I am having trouble with that upper index. It must involve flooring to the nearest integer divisible by 5 (maybe write this as $\lfloor N \rfloor_5$ perhaps there is better notation).</p>
| sayan | 312,099 | <p>We have to find the sum of all odd numbers less than N and divisible by 5</p>
<p>where N is given. For a given N you can find the number of odd numbers divisible</p>
<p>by 5 is $$\left[\frac{N}{5}\right]-\left[\frac{N}{10}\right]$$ So the reqired sum is equal to $$\sum_{i=1}^{\left[\frac{N}{5}\right]-\left[\frac{N}{10}\right]}\left(5+10(i-1)\right)$$.</p>
|
505,178 | <blockquote>
<p>Suppose $k$ is an algebraically closed field, and $f\in k[x, y]$ is
an irreducible polynomial in two variables. Furthermore, suppose that $f(u(x),
v(y))=f(x, y)$ for every $x, y\in k$, where $u\in k[x]$, $v\in k[y]$ are
polynomials of one variable. Can we conclude that either $u(x)=x$ or
$v(y)=y$?</p>
</blockquote>
<p>I think the answer is "yes". I am thinking maybe we can use some sort of degree argument, but there could be some cancellations if we try to expand $f(u(x), v(y))$. </p>
<p>This question naturally arose to me when I was reading Section 1.4 "Rational Maps" in <em>Basic Algebraic Geometry</em> by Shafarevich. But, as far as I can tell, it is not directly related to any of the results presented there. </p>
| 2'5 9'2 | 11,123 | <p>You can have $f(x,y)=x+y$, and then $u$ can shift by a constant, with $v$ shifting in reverse. </p>
|
3,736,580 | <p>Show that for <span class="math-container">$n>3$</span>, there is always a <span class="math-container">$2$</span>-regular graph on <span class="math-container">$n$</span> vertices. For what values of <span class="math-container">$n>4$</span> will there be a 3-regular graph on n vertices?</p>
<p>I think this question is slightly out of my control. Can you please help me out with this question...</p>
<p>For part two what I think is yes by handshaking I will exclude all the odd vertices as <span class="math-container">$3(2n+1)$</span> is not even number. So what should be the answer? All even number of vertices? Does that make sense?
And for part 1 it is obviously true but how can I proceed to the answer?
Thanks.</p>
| Community | -1 | <p>You can take
<span class="math-container">$n-\lfloor n(\pi-3)\rfloor$</span> times <span class="math-container">$3$</span> and <span class="math-container">$\lfloor n(\pi-3)\rfloor$</span> times <span class="math-container">$4$</span> and you get an error smaller than <span class="math-container">$\dfrac1n$</span>, because the average is</p>
<p><span class="math-container">$$3+\dfrac{\lfloor n(\pi-3)\rfloor}n=\dfrac{\lfloor n\pi\rfloor}n.$$</span></p>
<p>E.g. for <span class="math-container">$n=10000$</span>, take <span class="math-container">$8585$</span> times <span class="math-container">$3$</span> and <span class="math-container">$1415$</span> times <span class="math-container">$4$</span>, for average <span class="math-container">$3.1415$</span>.</p>
<hr />
<p>The digits <span class="math-container">$1$</span> and <span class="math-container">$2$</span> are more embarrassing than helpful.</p>
|
927,188 | <p>This question has been on my mind for a very long time, and I thought I'd finally ask it here. </p>
<p>When I was 6, my dad pulled me out of school. The classes were too easy; the professors, too dull. My father had been man of philosophy his entire life (almost got a PhD in it) and regretted not having a more quantitive background. He wanted me to have a different life and taught me math accordingly. When I was 11, I taught myself trig. When I was 12, I started taking calculus at my local university. I continued on this track, and finally got to real analysis and abstract algebra at 15. I loved every math course I ever took and found myself breezing through all that was presented to me (the university was not Princeton after all). However, around this time, I came to the conclusion that math was not for me. I decided to try a different path.</p>
<p>Why, you might ask, did I do this? The answer was simple: I didn't believe I could be a great mathematician. While I thrived taking the courses, I never turned math into a lifestyle. I didn't come home and do complex questions on a white board. I didn't read about Euler in my spare time. I also never felt I had a great intuition into problems. Once you showed me how to solve a problem, I was golden. But start from scratch on my own? It seemed like a different story entirely. To make things worse, my sister, who was at Caltech at the time, would call home with stories of all these incredible undergrads who solved the mathematical mysteries of the universe as a hobby. Whenever I mentioned math as a career, she would always issue a strong warning: you're not like these kids who spend all their time doing math. Think about doing something else. </p>
<p>Over time, I came to agree with this statement. Coincidentally, I got rejected by MIT and Princeton to continue my undergraduate studies there. This crushed me at the time; my dream of studying math at one of the great institutions had ended. Instead, I ended up at Georgia Tech (not terrible by any means, just not what I had envisioned). Being at an engineering school, I thought I'd give aerospace a shot. It had lots of math, right? Not really, or at least not enough for my taste. I went into CS. This was much better, but still didn't feel quite right. At last, as a sophomore, I felt it was time to get back on track: I'm now doubling majoring in applied math and CS. </p>
<p>My question is, how do I know I'm not making a mistake? There seems to be so many people doing math competitions, research, independent studies, etc, while I just started to take some math courses again. What should I do to test myself and see if I can really make math a career? I apologize for the long and possibly quite subjective post. I'd just really like to hear from math people who know their stuff. Thanks a bunch in advance. </p>
| ati | 175,259 | <p>You shouldn't judge yourself or your achievements by other people's standards. Do mathematics because you enjoy it, and find it interesting. Many great discoveries have come from average minds in a spirit of idle curiosity and playfulness. Likewise many brilliant minds achieve very little. In research, your approach matters as much as your intellect.</p>
<p>In short, you appear to be suffering performance anxiety because of your own or others' expectations. Get rid of those expectations and you will be free to express your potential, whatever that may be.</p>
|
927,188 | <p>This question has been on my mind for a very long time, and I thought I'd finally ask it here. </p>
<p>When I was 6, my dad pulled me out of school. The classes were too easy; the professors, too dull. My father had been man of philosophy his entire life (almost got a PhD in it) and regretted not having a more quantitive background. He wanted me to have a different life and taught me math accordingly. When I was 11, I taught myself trig. When I was 12, I started taking calculus at my local university. I continued on this track, and finally got to real analysis and abstract algebra at 15. I loved every math course I ever took and found myself breezing through all that was presented to me (the university was not Princeton after all). However, around this time, I came to the conclusion that math was not for me. I decided to try a different path.</p>
<p>Why, you might ask, did I do this? The answer was simple: I didn't believe I could be a great mathematician. While I thrived taking the courses, I never turned math into a lifestyle. I didn't come home and do complex questions on a white board. I didn't read about Euler in my spare time. I also never felt I had a great intuition into problems. Once you showed me how to solve a problem, I was golden. But start from scratch on my own? It seemed like a different story entirely. To make things worse, my sister, who was at Caltech at the time, would call home with stories of all these incredible undergrads who solved the mathematical mysteries of the universe as a hobby. Whenever I mentioned math as a career, she would always issue a strong warning: you're not like these kids who spend all their time doing math. Think about doing something else. </p>
<p>Over time, I came to agree with this statement. Coincidentally, I got rejected by MIT and Princeton to continue my undergraduate studies there. This crushed me at the time; my dream of studying math at one of the great institutions had ended. Instead, I ended up at Georgia Tech (not terrible by any means, just not what I had envisioned). Being at an engineering school, I thought I'd give aerospace a shot. It had lots of math, right? Not really, or at least not enough for my taste. I went into CS. This was much better, but still didn't feel quite right. At last, as a sophomore, I felt it was time to get back on track: I'm now doubling majoring in applied math and CS. </p>
<p>My question is, how do I know I'm not making a mistake? There seems to be so many people doing math competitions, research, independent studies, etc, while I just started to take some math courses again. What should I do to test myself and see if I can really make math a career? I apologize for the long and possibly quite subjective post. I'd just really like to hear from math people who know their stuff. Thanks a bunch in advance. </p>
| reggaemuffin | 175,350 | <p>I was a bit like you, always excelling at math in school, taking extra courses etc. but I am exactly the opposite of you. I love solving challenges but fail doing things 'like in the handbook' and so I found computer science for myself. There I can solve problems in many different ways and challenge my mind while still having enough math.</p>
<p>Like yu a looked at all the other people my age, winning prizes, doing stuff in their spare time and simply being better than me. But with some time I found what all the other good people here already told you: "There will always be someone better than you, but that shouldn't hinder you from doing wat you want to do! And makes your work no less important!"</p>
<p>While we humans always see the 1% standing out and feel bad, we forget 99% of all others struggling like us. And even being one of the 1% we will only see the 1% in any other part in life.</p>
<p>Good grades and an university with good reputations are fine. But they don't define you and what you want. Just look forward and ask yourself at every junction "where do I want to go NOW?" and never regret that decision. It is a part of you and if you come to the next junction in life you will be wiser than before.</p>
|
927,188 | <p>This question has been on my mind for a very long time, and I thought I'd finally ask it here. </p>
<p>When I was 6, my dad pulled me out of school. The classes were too easy; the professors, too dull. My father had been man of philosophy his entire life (almost got a PhD in it) and regretted not having a more quantitive background. He wanted me to have a different life and taught me math accordingly. When I was 11, I taught myself trig. When I was 12, I started taking calculus at my local university. I continued on this track, and finally got to real analysis and abstract algebra at 15. I loved every math course I ever took and found myself breezing through all that was presented to me (the university was not Princeton after all). However, around this time, I came to the conclusion that math was not for me. I decided to try a different path.</p>
<p>Why, you might ask, did I do this? The answer was simple: I didn't believe I could be a great mathematician. While I thrived taking the courses, I never turned math into a lifestyle. I didn't come home and do complex questions on a white board. I didn't read about Euler in my spare time. I also never felt I had a great intuition into problems. Once you showed me how to solve a problem, I was golden. But start from scratch on my own? It seemed like a different story entirely. To make things worse, my sister, who was at Caltech at the time, would call home with stories of all these incredible undergrads who solved the mathematical mysteries of the universe as a hobby. Whenever I mentioned math as a career, she would always issue a strong warning: you're not like these kids who spend all their time doing math. Think about doing something else. </p>
<p>Over time, I came to agree with this statement. Coincidentally, I got rejected by MIT and Princeton to continue my undergraduate studies there. This crushed me at the time; my dream of studying math at one of the great institutions had ended. Instead, I ended up at Georgia Tech (not terrible by any means, just not what I had envisioned). Being at an engineering school, I thought I'd give aerospace a shot. It had lots of math, right? Not really, or at least not enough for my taste. I went into CS. This was much better, but still didn't feel quite right. At last, as a sophomore, I felt it was time to get back on track: I'm now doubling majoring in applied math and CS. </p>
<p>My question is, how do I know I'm not making a mistake? There seems to be so many people doing math competitions, research, independent studies, etc, while I just started to take some math courses again. What should I do to test myself and see if I can really make math a career? I apologize for the long and possibly quite subjective post. I'd just really like to hear from math people who know their stuff. Thanks a bunch in advance. </p>
| pocketlizard | 173,176 | <p>It doesn't really matter if you decide your career while in undergrad. You're there to develop a strong baseline and to learn about what you don't know yet. Honestly the one piece of advise I give to most undergrads is to take more math courses, even if it's unrelated to their field of study, because math will open doors that most people won't venture down. You don't need to be a mathematician to major in math. I majored in math and marine biology and it was the best decision because it set me apart from other "pure science" undergrads when I was applying for grad school. So you don't need to know if you're going to "become a mathematician" because your goal right now should be to absorb as much information as you can so that you can apply it to the field you end up in, be it math or something else.</p>
|
2,207,848 | <p>I'm not very familiar with contraposition and so I am having some difficulties proving the statement. </p>
<blockquote>
<p>If $n$ is a positive integer such that $n \equiv 2 \pmod{4}$ or $n \equiv 3 \pmod{4}$, then $n$ is not a perfect square.</p>
</blockquote>
<p>What would be a good way to prove this?<br>
Need help please.</p>
| Bram28 | 256,001 | <p>In general, the contrapositive of a conditional:</p>
<p>'If P then Q' </p>
<p>is the statement:</p>
<p>'If not Q then not P'</p>
<p>Applied to your statement, we would thus get:</p>
<p>'If $n$ is a perfect square, then $n$ is not a positive integer such that $n \equiv 2 \pmod{4}$ or $n \equiv 3 \pmod{4}$'</p>
<p>... But somehow I doubt that's what they meant. In fact, the original statement was probably meant as:</p>
<p>'For any positive integer $n$, it holds that if $n \equiv 2 \pmod{4}$ or $n \equiv 3 \pmod{4}$, then $n$ is not a perfect square'</p>
<p>So then by taking the contrapositive of the contrapositive of the conditional that is part of that general statement about positive integers, we get:</p>
<p>'For any positive integer $n$, it holds that if $n$ is a perfect square, then it is not the case that $n \equiv 2 \pmod{4}$ or $n \equiv 3 \pmod{4}$'</p>
<p>...which makes a lot more sense.</p>
<p>Indeed, to prove this statement:</p>
<p>Take $n$ to be a positive integer and assume it is a perfect square. So, $n=k^2$ with $k$ an integer. $k$ is either even or odd. If $k$ is even, then $k=2m$ for some integer $m$, and so $n=(2m)^2=4m^2$. Hence, $n \equiv 0 \pmod{4}$. If $k$ is odd, then $k=2k+1$ for some integer $m$' and so $n=(2m+1)^2=4m^2+4m+1$, and hence $n \equiv 1 \pmod{4}$. So, it is not the case that $n \equiv 2 \pmod{4}$ or that $n \equiv 3 \pmod{4}$.</p>
|
3,383,206 | <p><strong>Question</strong>: Can <span class="math-container">$\int_0^\infty \frac{\sqrt{x}}{(1+x)^2} dx$</span> be computed with residue calculus?</p>
<p>The integral comes from computing <span class="math-container">$\mathbb{E}(\sqrt{X})$</span> where <span class="math-container">$X=U/(1-U)$</span> and <span class="math-container">$U$</span> is uniformly distributed in the unit interval. One can see that <span class="math-container">$\mathbb{E}(X)=\infty$</span> while wolfram computes the expectation of the radical as <span class="math-container">$\pi/2$</span> and I confirmed this numerically with Monte Carlo simulations in the r programming language.</p>
<p>It’s been awhile since I’ve done residue calculus so I consulted Ahlfors’ text. Ahlfors treats integrals of the form <span class="math-container">$\int_0^\infty x^\alpha R(x) dx$</span> for some rational function (i.e. ratio of polynomials) <span class="math-container">$R(x)$</span> and <span class="math-container">$0<\alpha<1$</span> which is this case with <span class="math-container">$\alpha=1/2$</span> and <span class="math-container">$R(x)=1/(1+x)^2$</span> but then states for convergence <span class="math-container">$R(x)$</span> must have a zero of at least order 2 at <span class="math-container">$\infty$</span> and at most a simple pole at the origin. But the latter is not satisfied here, there is only a zero at the origin not a pole, so this cannot be applied plus we have the pole of order 2 at <span class="math-container">$a=-1$</span> to deal with, right?</p>
<p>My idea before reviewing was to try a semi-circle of radius <span class="math-container">$R>1$</span> with indented semi-circle about <span class="math-container">$a=-1$</span>. The residue of <span class="math-container">$f$</span> at <span class="math-container">$a=-1$</span> I have computed as <span class="math-container">$-i/2$</span> and I recall indented estimation lemma resulting in the indented contour integral tending towards <span class="math-container">$i\pi Res(f,a)$</span> as the radius shrinks, which with the correct (negative) orientation would give <span class="math-container">$-\pi/2$</span> so if the entire contour integral is zero, we can add this to other side and (hopefully) show the rest vanishes except for <span class="math-container">$\int_0^\infty$</span> region but I’m obviously handwaving here. This seems problematic because it appears we’d be left with <span class="math-container">$\int_{-\infty}^{\infty}$</span> as the larger semi-circle radius grows and the smaller one about <span class="math-container">$a=-1$</span> shrinks instead of getting the integral just over the positive half line.</p>
<p>(If this can be done with real methods too, I certainly am not opposed to that answer, this is just for fun)</p>
| paw88789 | 147,810 | <p>It's not enough to say there is a neighborhood of the point that intersects the set. You need <span class="math-container">$\underline {\rm{every}}$</span> neighborhood of the point to intersect the set.</p>
|
4,036,558 | <p><span class="math-container">$f(x)=e^x(x^2+x)$</span>, derive <span class="math-container">$\dfrac{d^n\,f(x)}{dx^n}$</span></p>
<p>may use Leibniz formula but i'm not sure:(</p>
| Mathematician 42 | 155,917 | <p>Just by calculating <span class="math-container">$f',f''$</span> and <span class="math-container">$f'''$</span> by hand, you can see the following pattern:
<span class="math-container">$$\frac{\mathrm{d}^nf}{\mathrm{d}x^n}(x) = e^x(x^2+(2n+1)x+n^2).$$</span></p>
<p>Now, we can prove this by induction (the base steps are already done if you did the calculations). Assume <span class="math-container">$\frac{\mathrm{d}^nf}{\mathrm{d}x}(x) = e^x(x^2+(2n+1)x+n^2)$</span> for some <span class="math-container">$n\geq 1$</span>, then <span class="math-container">$$\frac{\mathrm{d}^{n+1}f}{\mathrm{d}x^{n+1}}(x) = e^x(x^2+(2n+1)x+n^2)+e^x(2x+2n+1)$$</span></p>
<p>The latter can be simplified as <span class="math-container">$$e^x(x^2+(2(n+1)+1)x+n^2+2n+1)=e^x(x^2+(2(n+1)+1)x+(n+1)^2).$$</span> This completes the induction step.</p>
|
4,036,558 | <p><span class="math-container">$f(x)=e^x(x^2+x)$</span>, derive <span class="math-container">$\dfrac{d^n\,f(x)}{dx^n}$</span></p>
<p>may use Leibniz formula but i'm not sure:(</p>
| zwim | 399,263 | <p>You can notice that by derivation we will get <span class="math-container">$e^xP(x)$</span> with <span class="math-container">$P$</span> polynomial, also the degree of <span class="math-container">$P$</span> stays unchanged since <span class="math-container">$(e^x)'$</span> does not bring some extra <span class="math-container">$x$</span>.</p>
<p>So we can set <span class="math-container">$\, f^{(n)}(x)=e^x(a_nx^2+b_nx+c_n)\, $</span> and try to find a recurrence relation for these sequences.</p>
<p>By derivation we get <span class="math-container">$\begin{cases}a_{n+1}=a_n&\quad a_0=1\\b_{n+1}=b_n+2a_n&\quad b_0=1\\c_{n+1}=c_n+b_n&\quad c_0=0\end{cases}$</span></p>
<p>Therefore <span class="math-container">$a_n$</span> is constant and <span class="math-container">$a_n=a_0=1$</span>.</p>
<p>The sequence <span class="math-container">$b_n$</span> is telescopic and <span class="math-container">$$b_{n}=b_0+2\sum\limits_{k=0}^{n-1} a_k=b_{n}=1+2\sum\limits_{k=0}^{n-1} 1=2n+1$$</span></p>
<p>The sequence <span class="math-container">$c_n$</span> is also telescopic and <span class="math-container">$$c_n=c_0+\sum\limits_{k=0}^{n-1} b_k=0+2\sum\limits_{k=0}^{n-1} k+\sum\limits_{k=0}^{n-1} 1=2\times\frac{(n-1)n}2+n=n^2$$</span></p>
|
2,227,280 | <p>For every positive number there exists a corresponding negative number. Would that imply that the number of positive numbers is "equal" to the number of negative numbers? (Are they incomparable because they both approach infinity?)</p>
| celtschk | 34,930 | <p>Yes, the existence of a one-to-one and onto mapping is exactly how equality of the size of sets (the technical term is "cardinality" is defined. The (cardinal) number of negative integers is the same as the cardinal number of positive integers, and the cardinal number of negative real numbers is the same as the cardinal number of positive real numbers.</p>
<p>Note however that there are more positive real numbers than positive integers, so not all infinite cardinalities are equal. The proof of that fact is done by Cantor's famous diagonal proof.</p>
|
1,590,625 | <blockquote>
<p>If $f(x)=\log \left(\cfrac{1+x}{1-x}\right)$ for $-1 < x < 1$,then
find $f \left(\cfrac{3x+x^3}{1+3x^2}\right)$ in terms of $f(x)$.</p>
</blockquote>
<p><strong>My Attempt</strong>
$$f \left(\cfrac{3x+x^3}{1+3x^2}\right)=\log\left(\cfrac{1+\cfrac{3x+x^3}{1+3x^2}}{1-\cfrac{3x+x^3}{1+3x^2}}\right)=\log \left(\cfrac{1+3x^2+3x+x^3}{1+3x^2-3x-x^3}\right)=\\\log(1+3x^2+3x+x^3)-\log(1+3x^2-3x-x^3)$$</p>
<p>Now I am kinda clueless about how to express this in terms of $f(x)$.</p>
<p>Can you guys help ?</p>
| SchrodingersCat | 278,967 | <p>$$\log(1+3x^2+3x+x^3)-\log(1+3x^2-3x-x^3)=\log(1+x)^3-\log(1-x)^3=3[\log(1+x)-\log(1-x)]=3f(x)$$</p>
|
69,378 | <p>Updated Question : How to show that in TH we never reach a state where there are no paths to the solution? ( without reversing moves, as if reversing is allowed this becomes trivial )</p>
<p>Edit : Thanks to <strong>Stéphane Gimenez</strong> for pointing out the distinction between “A deadlock would never occur” and “The problem always has a solution”, made it possible for me to state the question in a form that was the original intention.</p>
<p>Stéphane Gimenez:</p>
<blockquote>
<p>Defining deadlock as a reachable position where no more moves are
available (or alternatively as a position from which the goal cannot
be reached anymore), it's obvious that deadlocks cannot occur in the
TH game: every step along the reverse path (of a path containing valid
moves) is a valid move.</p>
</blockquote>
<p>Original Question :</p>
<blockquote>
<p>In Towers of Hanoi problem there is an implicit assumption that one
can keep moving disks, this is trivially true for 1 or 2 disks but as
obvious as it looks one can keep going with as many disks? In other
words TH with 3 sticks and n disks always has a solution?</p>
<p>The N queens problem is easyly shown will not have a solution for n>m
, where m is size of board (using Pigeon Hole ), but also it does not
have a soltion for n=m=2. But how does one show that if for some k ,
n=m=k has a solution then it will also imply for k+1 there is a
solution?</p>
</blockquote>
| davidlowryduda | 9,754 | <p>Well, one can give an exact algorithm to solve it. But I would recommend thinking of it as a recursive problem. Say we look at the 3 disk case. Well, that's really a 2 disk case on top of a third disk. So suppose we move those top 2 disks to another rod. Then we move the forgotted third to the desired end rod, and treat the 2 disks as another TH puzzle to move them on top of the third.</p>
<p>In this recursive fashion, we can see that we can always guarantee a procedure inductively. If you pay close attention, you can even give the number of moves necessary.</p>
|
2,621 | <p>Let $A$ be a commutative Banach algebra with unit.
It is well known that if the Gelfand transform $\hat{x}$ of $x\in A$ is non-zero, then $x$ is invertible in $A$ (the so called Wiener Lemma in the case when $A$ is the Banach algebra of absolutely convergent Fourier series).</p>
<p>As a converse of the above, let $B$ be a Banach space contained in $A$ and suppose $B$ is closed under inversion - i.e.: If $x\in B$ and $x^{-1}\in A$ then $x^{-1}\in B$.</p>
<p>(1) Prove that $B$ is a Banach algebra.</p>
<p>(2) Must $A$ and $B$ have the same norm? If not are the norms similar?</p>
<p>(3) Do $A$ and $B$ have the same maximal ideal space?</p>
| John D. Cook | 247 | <p>Monte Carlo methods are very useful in numerically evaluating high-dimensional integrals. With traditional integration methods, the number of integrand evaluations required to maintain accuracy grows quickly as dimension increases. With Monte Carlo integration, the number of integrand evaluations needed is <em>independent</em> of dimension. For many high-dimensional integrals, Monte Carlo methods are the only practical choice.</p>
|
700,012 | <p>Ok, so the question is to prove by induction that:</p>
<p>$${n \choose k} \le n^k$$</p>
<p>Where $N$ and $k$ are integers, $k \le n$;</p>
<p>How do I approach this? Do i choose a $n$ and a $k$ to form my base case?</p>
| RghtHndSd | 86,816 | <p>While there are various ways to approach this, I would recommend fixing (an arbitrary) $n$ and inducting on $k$. So your base case would then be to show that $\binom{n}{0} \leq n^0$. To do the inductive step, figure out what you would need to multiply $\binom{n}{k-1}$ by to get $\binom{n}{k}$.</p>
|
2,436,336 | <p>I am a bit puzzled. Trying to solve this system of equations: </p>
<p>\begin{align*}
-x + 2y + z=0\\
x+2y+3z=0\\
\end{align*}</p>
<p>The solution should be \begin{align*}
x=-z\\
y=-z\\
\end{align*} </p>
<p>I just don't get the same solution. Please advice.</p>
| Dr. Sonnhard Graubner | 175,066 | <p>adding both equations we get
$$y=-z$$ plugging this in the first equation we get $$-x-2z+z=0$$ tgherefore $$x=-z$$</p>
|
1,344,284 | <p>if $Z=X+iy$ then determine the locus of the equation $\left | 2Z-1 \right | = \left | Z-2 \right |$.I can tell that it a circle equation and it is $x^2 + y^2 = 1$.There are a lot of equation in my book such as $\left | Z-8 \right | +\left | Z+8 \right |=20$,$\left | Z-2 \right | = \left | Z-3i \right |$,$\left | 2Z+3 \right |= \left | Z+6 \right |$.Every time I have to do a long calculation.<br>
Is there any short way to find out that if the given equation is circle,ellipse, parabola, hyperbola or straight line.This is needed for my MCQ exam.</p>
| user251257 | 251,257 | <p>It is true.
Let $R$ be the symmetric root of $A$, which exists positive for semidefinite matrices. That is, $R$ is symmetric, positive semidefinite, and $R^2=A$. </p>
<p>For $x\in\mathbb R^n$ we have
$$ x^T A x - x^T A e e^T A x \ge 0 $$
if and only if
$$ \|Rx\|^2 \underbrace{\|Re\|^2}_{\langle A,J \rangle=1} = x^T A x \ge x^T A e e^T A x = ((Rx)^T (Re))^2, $$
which is true by Cauchy-Schwarz.</p>
|
4,213,207 | <p>I am trying to find all homomorphisms from <span class="math-container">$\mathbb{Z}_{20}$</span> to <span class="math-container">$\mathbb{Z}_8$</span>. I understand how to do it - one completely determines any homomorphism, say <span class="math-container">$\phi$</span>, by computing multiples of <span class="math-container">$\phi(1)$</span> or more generally <span class="math-container">$\phi(g)$</span> where <span class="math-container">$g$</span> generates <span class="math-container">$\mathbb{Z}_{20}$</span>. Call <span class="math-container">$\phi_i$</span> the mapping taking <span class="math-container">$1$</span> to <span class="math-container">$i$</span> in <span class="math-container">$\mathbb{Z}_8$</span>. I don’t know how else to determine whether each <span class="math-container">$\phi_i$</span> is a homomorphism besides going through the painstaking process of calculating each <span class="math-container">$\phi_i(x+y)$</span> and comparing it to <span class="math-container">$\phi_i(x)+\phi_i(y)$</span>.</p>
<p>Is there a way to easily rule out certain <span class="math-container">$\phi_i$</span>s?</p>
<p>Any help would be greatly appreciated.</p>
| Community | -1 | <p>By the First Homomorphism Theorem, the quotient <span class="math-container">$\Bbb Z_{20}/\operatorname{ker}\phi$</span> is isomorphic to the image of <span class="math-container">$\phi$</span>. Since <span class="math-container">$|\operatorname{im}\phi|$</span> must be a common divisor of <span class="math-container">$20$</span> and <span class="math-container">$8$</span>, the only possible options for the image of <span class="math-container">$\phi$</span> are:</p>
<ol>
<li><span class="math-container">$\operatorname{im}\phi=\{0\}$</span>;</li>
<li><span class="math-container">$\operatorname{im}\phi\cong \Bbb Z_2$</span>;</li>
<li><span class="math-container">$\operatorname{im}\phi\cong \Bbb Z_4$</span>.</li>
</ol>
<p>Option 1 corresponds to the trivial homomorphism: all the elements of <span class="math-container">$\Bbb Z_{20}$</span> are sent to <span class="math-container">$0$</span>.</p>
<p>Option 2 would correspond to <span class="math-container">$\operatorname{ker}\phi\cong \Bbb Z_{10}$</span>: <span class="math-container">$\operatorname{ker}\phi$</span> is sent to <span class="math-container">$0$</span>, and <span class="math-container">$1+\operatorname{ker}\phi$</span> (which has order <span class="math-container">$2$</span> in the quotient) is sent to <span class="math-container">$4$</span>. You can check that this is indeed a homomorphism.</p>
<p>Option 3 would correspond to <span class="math-container">$\operatorname{ker}\phi\cong \Bbb Z_{5}$</span>:</p>
<ul>
<li><span class="math-container">$\operatorname{ker}\phi$</span> is sent to <span class="math-container">$0$</span>;</li>
<li><span class="math-container">$1+\operatorname{ker}\phi$</span> (which has order <span class="math-container">$4$</span> in the quotient) is sent to either <span class="math-container">$2$</span> or <span class="math-container">$6$</span>;</li>
<li><span class="math-container">$2+\operatorname{ker}\phi$</span> (which has order <span class="math-container">$2$</span> in the quotient) is sent to <span class="math-container">$4$</span>;</li>
<li><span class="math-container">$3+\operatorname{ker}\phi$</span> (which has order <span class="math-container">$4$</span> in the quotient) is sent to either <span class="math-container">$6$</span> or <span class="math-container">$2$</span>, according to where <span class="math-container">$1+\operatorname{ker}\phi$</span> was sent to before.</li>
</ul>
<p>You can check that both these are indeed homomorphisms.</p>
|
2,894,126 | <blockquote>
<p>$$\int \sin^{-1}\sqrt{ \frac{x}{a+x}} dx$$</p>
</blockquote>
<p>We can substitute it as $x=a\tan^2 (\theta)$ . Then:</p>
<p>$$2a\int \theta \tan (\theta)\sec^2 (\theta) d\theta$$</p>
<p>Using integration by parts will be enough here. But I wanted to know if this particular problem can be solved by any other method. Because the above method is quite lengthy. </p>
| Community | -1 | <p>I don't see anything lengthy here.</p>
<p>$$\int \theta\frac{\sin\theta}{\cos^3\theta}d\theta=-\frac\theta{2\cos^2\theta}+\frac12\int\frac{d\theta}{\cos^2\theta}$$ and the last integral is immediate.</p>
|
1,729,308 | <p>The sum of the first $n$ $(n>1)$ terms of the A.P. is $153$ and the common difference is $2$. If the first term is an integer , then number of possible values of $n$ is </p>
<p>$a)$ $3$</p>
<p>$b)$ $4$</p>
<p>$c)$ $5$</p>
<p>$d)$ $6$</p>
<p>My approach : I used the formula for the first $n$ terms of an A.P. to arrive at the following quadratic equation
$n^2 + n(a-1) -153 = 0 $</p>
<p>Next up I realised that since we are talking about the number of terms , thus the
possible values which n can take must be whole numbers. That is the discriminant of the above quadratic should yield a whole number in other words </p>
<p>$ (a-1)^2 + 612 = y^2 $ for some y . </p>
<p>However I am stuck at this point , as from here I am unable to figure out the number of such a's ( i.e. the initial terms of an AP ) which will complete the required pythagorean triplet
The answer mentioned is $5$</p>
<p>Please let me know , if I am doing a step wrong somewhere . Or If you have a better solution , that will be welcomed too. </p>
| lulu | 252,071 | <p>To summarize the (extensive) discussion in the comments:</p>
<p>The OP's method is sound and nearly complete. To finish it off we look at the relation $$612=y^2-(a-1)^2=(y+(a-1))(y-(a+1)$$ To solve that (over the integers) we simply need to factor $612=cd$ where the factors must have the same parity. There are three possible such factorings: $$\{18,34\},\;\{2,306\},\:\{6,102\}$$</p>
<p>Each of these gives rise to two possible starting points for our progressions. We get $$a\in \{-151,-47,-7,9,49,153\}$$ </p>
<p>We reject the "degenerate" case $a=153$ as that progression just has a single term (and the OP specified $n>1$). Thus we have $5$ solutions.</p>
|
1,729,308 | <p>The sum of the first $n$ $(n>1)$ terms of the A.P. is $153$ and the common difference is $2$. If the first term is an integer , then number of possible values of $n$ is </p>
<p>$a)$ $3$</p>
<p>$b)$ $4$</p>
<p>$c)$ $5$</p>
<p>$d)$ $6$</p>
<p>My approach : I used the formula for the first $n$ terms of an A.P. to arrive at the following quadratic equation
$n^2 + n(a-1) -153 = 0 $</p>
<p>Next up I realised that since we are talking about the number of terms , thus the
possible values which n can take must be whole numbers. That is the discriminant of the above quadratic should yield a whole number in other words </p>
<p>$ (a-1)^2 + 612 = y^2 $ for some y . </p>
<p>However I am stuck at this point , as from here I am unable to figure out the number of such a's ( i.e. the initial terms of an AP ) which will complete the required pythagorean triplet
The answer mentioned is $5$</p>
<p>Please let me know , if I am doing a step wrong somewhere . Or If you have a better solution , that will be welcomed too. </p>
| Hypergeometricx | 168,053 | <p>Interesting question (+$1$). </p>
<p>As the common difference is $2$, the series is either one of odd numbers or even numbers only. As the sum is an odd number, it must be a series of odd numbers with an odd number of terms. This narrows it down to $3$ terms or $5$ terms from the choices given. </p>
<p>Since $153\div 3=51$, a quick check shows that $49+51+53=153$ hence the answer is $3.\;\blacksquare$</p>
<p>NB: $153\div 5\approx 30$. The four closest odd numbers to $30$ add up to $120$, and the next one is either $25$ or $35$, none of which would result in $153$.</p>
|
4,013,065 | <p>I need help interpreting the answer to a question about the base and dimension of a subspace within linear algebra. I have a subspace W of <span class="math-container">$R^5$</span> that is spanned by the vectors:</p>
<p><span class="math-container">$${v_1}=\begin{pmatrix} 1 \\ 2 \\ 3 \\ -1 \\ 1 \end{pmatrix} , {v_2}=\begin{pmatrix} 0 \\ 1 \\ 1 \\ -1 \\ 2 \end{pmatrix} , {v_3}=\begin{pmatrix} -1 \\ -1 \\ -2 \\ 1 \\ 2 \end{pmatrix}
\text{and} \,{v_4}= \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \\ -4 \end{pmatrix}$$</span></p>
<p>and I will determine the dimension of W and also specify the base of W.</p>
<p>To do this I put the vectors as columns in a matrix and simplified this using gauss:</p>
<p><span class="math-container">$$\begin{pmatrix}1 & 0 & -1 & 1 \\ 2 & 1 & -1 & 0 \\ 3 & 1 & -2 & 1 \\ -1 & -1 & 1 & 0 \\ 1 & 2 & 2 & -4\end{pmatrix}\sim\begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}.\\$$</span></p>
<p>Here I concluded that the dimension max can be equal to three as we have three rows of pivot elements, then I also concluded that you can express v4 = -v2-v3 (I have also tested this and it is true) and the two vectors v2 and v3 must then be linearly independent. But what I have a hard time interpreting is if this means that v2 and v3 are the basis of W (and not v1,v2 and v3 as my first thought was)? And if the dimension in that case becomes two and not three (as the maximum can be)?</p>
<p>Thanks in advance!</p>
| Timur Bakiev | 855,963 | <p>The dimension is exactly three since you have three linearly independent vectors <span class="math-container">$v_1, v_2, v_3$</span> in the generating system of <span class="math-container">$W$</span> and <span class="math-container">$v_4 = -v_2-v_3$</span>.</p>
|
20,982 | <p>Let E be an elliptic curve over a finite field k (char(k) is not 2) be given by y^2 = (x-a)(x-b)(x-c) where a,b and c are distinct and are in k. Then why is (c,0) is in [2]E(k) iff c-a and c-b is a square in k-{0}? </p>
| Bjorn Poonen | 2,757 | <p>Here is a conceptual explanation that applies to any $E/k$ with $\operatorname{char} k \ne 2$ and $E[2] \subseteq E(k)$, and that explains why $x-a$ and $x-b$ are the relevant rational functions (the field $k$ need not be finite).</p>
<p>The map $[2]\colon E \to E$ makes the function field $k(E)$ a finite extension of itself, say $L/K$, and $L$ is the unique unramified $(\mathbb{Z}/2\mathbb{Z})^2$-extension of $K$ in which the point at infinity splits completely (here we use that $E[2]$ is rational). The field $K(\sqrt{x-a},\sqrt{x-b})$ satisfies these conditions, so it is $L$. To say that a point $(x_0,y_0) \in E(k)$ lies in $2 E(k)$ means that it splits in $L/K$. When $x_0 \notin \{a,b,\infty\}$, this means simply that $x_0-a$ and $x_0-b$ should be squares in $k$.</p>
|
1,873,596 | <p>Near the end of <a href="http://www.maa.org/sites/default/files/pdf/upload_library/2/Rice-2013.pdf" rel="nofollow noreferrer">this MAA piece about elliptic curves</a>, the author explains why the complex domain of the cosine function is a sphere: since it's periodic, its domain can be taken as a cylinder, wrapping up the real axis. And because cosine of $\theta\pm i\infty$ is $\infty$, the two ends of the cylinder can be identified with a single point $\infty$. Ok, great, but this sounds to me like a pinched torus. Can I have a clearer explanation why this is a sphere?</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/LwmN0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LwmN0.png" alt="enter image description here"></a></p>
</blockquote>
| Max | 2,633 | <p>Let me start by saying you are right in a sense, and I think the article is at best being unclear, but the larger point that the equation of ellipse cuts out a sphere is also right. There are a few things going on, hence a long answer below.</p>
<p>What one is trying to describe is the topology of a set defined by a quadratic equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Of course it is crucially important to explain what values $x$ and $y$ are allowed to take. In the context to the article, that means whether one allows only real values or complex ones as well (this is the question of choosing the field of coefficients), and whether one allows infinite values and in what sense (this is a choice of compactification). Both of these choices affect the topology of the solution set. If only finite real values are allowed the result is the familiar ellipse in the plane (which has topology of a circle); now one has various options about how to treat adding infinite values of $x$ and $y$. However the (real) ellipse does not go to infinity in any way (it stays in finite part of the plane; or, in other words, it is already compact so needs no compactification; algebraically, no real vector with large modulus can solve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$). Thus for this equation over real numbers the second choice is irrelevant. However (spoiler!) it is not irrelevant over complex numbers.</p>
<p>To better see what's going on let's stick to the real variables for now, but consider the equation $uw=1$ instead. The solutions to this in $\mathbb{R}^2$ define a hyperbola and this has topology of 2 lines. Now if we compactify $\mathbb{R}^2$ to the sphere - treating all points at infinity as one - we get one extra point, and the topology of solution set is a wedge of two circles. If we treat $u$-infinities and $v$-infinities as different (compactify each $\mathbb{R}$ to $S^1$, so $\mathbb{R}^2$ is compactified to $S^1\times S^1$), then we get 2 extra points ("$(\infty, 0)$" and "$(0, \infty)$") and a topology of a single circle. We can also compactify $\mathbb{R}^2$ to $\mathbb{R}P^2$ (and still get a circle), or to $\mathbb{D}^2$ (and get two closed segments) or any other number of things, some more natural than others. To finish with the discussion over $\mathbb{R}$, note that if we consider $(u-v)(u+v)=1$ instead of $uv=1$, the two curves in $\mathbb{R}^2$ are isomorphic, but in the compactification $S^1\times S^1$ we get now a wedge of two circles instead of a single circle. This is because the coordinate change of $\mathbb{R}^2$ given by $u_n=u-v$, $v_n=u+v$ taking curve $(u-v)(u+v)=1$ to $u_n v_n=1$ does not extend to these compactifications. This never happens for $\mathbb{R}P^2$, which is one big reason why this is usually the prefered choice of compactification.</p>
<p>Now over complex numbers, the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ has a solution set in $\mathbb{C}^2$ which is a cylinder (aka tangent bundle of $S^1$). Then we can compactify by adding one point for "all infinity" and get a "pinched torus" with both infinities of the cylinder filled in by that one extra point; or by adding a point at infinity to each coordinate separately - this is what is done in the article - and still get a "pinched torus", since $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ forces $x$ to have infinite modulus as soon as $y$ does and vice versa, so the only added point is $(\infty, \infty)$. However, if we compactify to $\mathbb{C}P^2$ we get two extra points ($[a:ib:0]$ and $[a:-ib:0]$), and topology of a sphere. This is in a sense the "right" compactification of the image, sitting inside the "right compactification" $\mathbb{C}P^2$.</p>
<p>Finally, we should note that this is not so much about the domain of $\cos$ but more about the image of $z \to (a\cos z, b\sin z)$ in $\mathbb{C}^2$ and its compactifications. Whether we can extend $\cos z$ or $\sin z$ or $(\cos z, \sin z)$ to a map of some compactification of $\mathbb{C}$ to somewhere depends on where (which extension/compactification of $\mathbb{C}$ or $\mathbb{C}^2$) we are mapping to, and does not really have so much intrinsic meaning.</p>
|
166,460 | <p>I work with linear combinations of graphs,
$$c_1 G_1 + c_2 G_2 + \dotsc,$$
and I want to represent them in my Mathematica code. I represent graphs as adjacency matrices, e.g.</p>
<pre><code>{{0,1},{1,0}}
</code></pre>
<p>The next step is to write down linear combinations of these matrices. However, I want to implement formal linear combinations of the kind</p>
<pre><code>lin = 5 * AdjMtx[{{0,1},{1,0}}] + 3 * AdjMtx[{{1,0},{0,1}}]
</code></pre>
<p>with an unevaluated "function" or type <code>AdjMtx</code>. The reason I don't want to write</p>
<pre><code>5 * {{0,1},{1,0}} + 3 * {{1,0},{0,1}}
</code></pre>
<p>is that Mathematica then treats the adjacency matrices as normal matrices which admit multiplication my scalars etc.</p>
<p><code>AdjMtx</code> should allow other functions to access the contents of adjacency matrices as usual. For example, a function should be able to search through a linear combination of adjacency matrices and read out their respective matrix elements.</p>
<p>I guess that moving through a linear combination, e.g. <code>lin</code> from above, is just done using <code>Part</code>. Then the question is how to do implement something like</p>
<pre><code>AdjMtx[...][[3,4]]
</code></pre>
| Albert Retey | 169 | <p>I am not sure whether I understand correclty what you are after, but I think that using <code>UpValues</code> should help you solve your problem, e.g. this definition:</p>
<pre><code>Part[AdjMtx[data_], idcs___] ^:= Part[data, idcs]
</code></pre>
<p>will make it possible to access the matrix elements just as desired:</p>
<pre><code>a = AdjMtx[{{1,0},{0,1}}];
a[[1,1]]
</code></pre>
|
1,776,260 | <p>After understanding the Cardano's formula for solving the depressed cubic (of the form $x^3+mx=n$, of course), I tried to find the solution of the equation $$x^3+6x=20.$$
After plugging into the formula
$$x=(n/2+\sqrt{ \frac{n^2}{4}+ \frac{m^3}{27} })^{1/3}+(-n/2+\sqrt{ \frac{n^2}{4}+ \frac{m^3}{27} })^{1/3}$$
where $m=6$ and $n=20$, we get
$$x=(10+ \sqrt{108})^{1/3}-(-10+ \sqrt{108})^{1/3}.$$
However, we notice that, without using Cardano's formula, that $x=2$ is the solution for the equation $x^3+6x=20.$
My question is: how does the equation $$x=(10+ \sqrt{108})^{1/3}-(-10+ \sqrt{108})^{1/3}$$ get simplified to $x=2$?</p>
<p>P.S. I understand that it was Niccolo Fontana who first figured out how to solve depressed cubic, to give one the proper credit.</p>
| Siong Thye Goh | 306,553 | <p>\begin{align*}
&\left(10+\sqrt{108}\right)^{1/3}-\left(-10+\sqrt{108}\right)^{1/3}\\
&=\left(10+6\sqrt{3}\right)^{1/3}-\left(-10+6\sqrt{3}\right)^{1/3}\\
&=\left((1+\sqrt{3})^3\right)^{1/3}-\left((\sqrt{3}-1)^3\right)^{1/3}\\
&=(1+\sqrt{3})-(\sqrt{3}-1)\\
&=2
\end{align*}</p>
<p>The formula:
$$(A+B\sqrt{3})^3=A^3+3\sqrt{3}A^2B+9B^2A+3\sqrt{3}B^3$$ is useful.</p>
|
4,050,831 | <p>Suppose 40% of all seniors have a computer at home and a sample of 64 is taken. What is the probability that more than 30 of those in the sample have a computer at home?"</p>
<p>My attempt:</p>
<p>n=64</p>
<p>0.4x64=25.6</p>
<p>p=?</p>
<p>x=??</p>
<p>A>30=??</p>
<p>Don't have an idea of what equation would be appropriate to determine proportion so as to use s=root p(1−p)/n</p>
| BruceET | 221,800 | <p>The number <span class="math-container">$X$</span> of seniors in a random sample of size <span class="math-container">$n = 64,$</span> who will have computers at home has <span class="math-container">$X \sim \mathsf{Binom}(n = 64, p = 0.4).$</span> You seek
<span class="math-container">$P(X > 30) = 1 - P(X\le 30).$</span> You may be expected to
solve this problem using software to get an exact answer
or using a normal approximation to get a useful approximate answer.</p>
<p><strong>Using R statistical software,</strong> where <code>pbinom</code> is a binomial DCF and <code>dbinom</code> is a binomial PDF, you could obtain the answer <span class="math-container">$0.1062776$</span> in either of two ways, as shown below. [The second method is a software implementation of the sum displayed in @user's (+1) Answer.]</p>
<pre><code>1 - pbinom(30, 64, 0.4)
[1] 0.1062776
sum(dbinom(31:64, 64, 0.4))
[1] 0.1062776
</code></pre>
<p><strong>Normal approximation</strong> begins by finding <span class="math-container">$ \mu = E(X) = np = 64(.4) = 25.6$</span> and <span class="math-container">$\sigma = \sqrt{64(.4)(.6)} = 3.9192.$</span> Then <span class="math-container">$X \stackrel{aprx}{\sim}
\mathsf{Norm}(\mu=25.6, \sigma = 3.9192).$</span>
Using a continuity correction, you would find
<span class="math-container">$P(X > 30.5) = 1 - P(X < 30.5) \approx 0.1056.$</span> This could be evaluated in R, where <code>pnorm</code> is a normal CDF:</p>
<pre><code>1 - pnorm(30.5, 25.6, 3.9192)
[1] 0.1056032
</code></pre>
<p>Often, a normal approximation of a binomial probability
is accurate to two decimal places, provided <span class="math-container">$n$</span> is large
enough that <span class="math-container">$np$</span> and <span class="math-container">$n(1-p)$</span> both exceed <span class="math-container">$5.$</span> (Roughly speaking, this ensures that there is very little normal
probability below <span class="math-container">$9$</span> or above <span class="math-container">$n.)$</span></p>
<p><strong>Standardizing and using a printed standard normal table</strong>
is done as follows:</p>
<p><span class="math-container">$$P(X > 30.5) =
P\left(\frac{X=\mu}{\sigma} > \frac{30.5 - 25.6}{3.9192}\right)\\
\approx P(Z > 1.25) = 0.1056,$$</span>
where <span class="math-container">$Z$</span> is standard normal and the final answer
is obtained from a printed table. Sometimes this method
is a little less precise because printed table give
z-scores only to two decimal places.</p>
<p>The plot below shows the good fit of the normal distribution to this particular binomial distribution.</p>
<pre><code>x = 10:40; PDF= dbinom(x, 64, .4)
hdr = "BINOM(64, 0.4) with Normal Approx."
plot(x, PDF, type="h", lwd=2, main=hdr)
curve(dnorm(x, 25.5, 3.9192), add=T, col="red", lwd=2)
abline(v=30.5, col="blue", lty="dotted", lwd=2)
abline(h=0, col="green2")
</code></pre>
<p><a href="https://i.stack.imgur.com/gGtcK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gGtcK.png" alt="enter image description here" /></a></p>
|
929,502 | <p>Here are two succinct statements of the 'same' question:</p>
<p><strong>Statement 1:</strong>
Take $a>0$ and $S \subseteq \mathbb{R}^N; S=\{(x_1,\dots,x_N)| \frac{1}{N}\sum_i x_i = a; x_i>0\}$. Define a 'product function' $f:S\rightarrow \mathbb{R}^N; f(x_1,\dots,x_i)=\prod_ix_i$. There are many proofs that the max of $f$ over all of $S$ occurs at $\vec{v}_a=(a,a,\dots,a)$. But here is the 'extension': </p>
<blockquote>
<p>Consider two points $c,d \in S$ with $||\vec{c}-\vec{v}_a||_1\leq||\vec{d}-\vec{v}_a||_1$ ('taxicab' $\ell_1$ norm). Why does $f(\vec{d})<f(\vec{c})<f(\vec{v}_a)$?</p>
</blockquote>
<p><strong>Statement 2:</strong>
Take two sets of real numbers $C=\{c_1,\dots, c_N\}, D=\{d_1,\dots,d_N\}$ with $\sum_i c_i=\sum_i d_i = 0$ and $\sum_i|c_i|<\sum_i |d_i|$. Take some $a>0$ great enough so that $a+x_i>0$ for any $x_i\in C \cup D$. </p>
<blockquote>
<p>Why is $\prod_i (a+d_i) < \prod_i(a+c_i)$?</p>
</blockquote>
<p>Apparently it is a known result, but after a day of grief on my own I haven't been able to make any useful progress. My most promising approaches seemed to be:</p>
<ul>
<li>Fiddling with the AM-GM inequality</li>
<li>Halve $C,D$ about their means (or centers?) to get more sets. Then try and somehow use the fact that for $a>b$, $(a-b)(a+b)<a^2$. Iterate recursively until you have the desired product.</li>
<li>Show that $f$'s derivative is largest in the $(1,1,\dots, 1)$ direction, $\vec{c},\vec{d}$. Then show that in $S$ you are 'closest' to this direction the smaller $\ell_1$-distance you are from $\vec{v}_a$. Note that $S$ is a positive hyperplane, but it's shifted so it's not a subspace.</li>
<li>Not-so-promising was expanding the polynomial product into a sum of $2^N$ $N^{\mathrm{th}}$-degree terms.</li>
</ul>
| Тимофей Ломоносов | 54,117 | <p>$\ln x$ is an increasing function. therefore $\arg \max\prod\limits_{i=1}^{x_n}x_i = \arg \max \ln \prod\limits_{i=1}^N x = \arg\max\sum\limits_{i=1}^n\ln x_i$. Once you've noticed it, it should be easier.</p>
|
2,249,020 | <p><a href="https://i.stack.imgur.com/L7PXf.jpg" rel="nofollow noreferrer">The Math Problem</a></p>
<p>I have issues with finding the Local Max and Min, and Abs Max and Min, after I find the Critical Point. How do I do this problem in its entirety? </p>
| Zain Patel | 161,779 | <p>Draw two position vectors, $\mathbf{v}_1$ and $\mathbf{v}_2$ with unit magnitude and at angles $\alpha, \beta$ to the positive $x$-axis. Then the angle between the two is $\alpha - \beta$ (assuming $\alpha > \beta$ w.l.o.g). But $\mathbf{v}_1 \cdot \mathbf{v}_2$ is the cosine of the angle between them. So $\cos (\alpha - \beta) = \mathbf{v}_1 \cdot \mathbf{v}_2$. </p>
<p>But remember that the two vectors lie on the unit circle and have components $\mathbf{v}_1 = \cos \alpha \mathbf{i} + \sin \alpha \mathbf{j}$ and $\mathbf{v}_2 = \cos \beta \mathbf{i} + \sin \beta \mathbf{j}$. By the definition of the dot product $$\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta.$$</p>
<hr>
<p>To solve the given problem: note that $\beta \mapsto -\beta$ gives $$\cos (\alpha - (-\beta)) = \cos \alpha \cos (-\beta) + \sin \alpha \sin (-\beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta = \cos(\alpha + \beta)$$ using the oddity of $\sin$ and even-ness of $\cos$.</p>
|
3,988,808 | <p>I recently got into set theory and i was wondering what is the cardinality of a set of all finite sequences of natural numbers?
I know that it is N for natural numbers and 2^N is for real numbers but how can i prove it?</p>
| Ethan Bolker | 72,858 | <blockquote>
<p>Theorem: The set of all finite-length sequences of natural numbers is
countable.</p>
</blockquote>
<p>More is true: that follows from</p>
<blockquote>
<p>Theorem: (Assuming the axiom of countable choice) The union of
countably many countable sets is countable.</p>
</blockquote>
<p>You can find proofs on this wikipedia page:
<a href="https://en.wikipedia.org/wiki/Countable_set" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Countable_set</a></p>
|
3,988,808 | <p>I recently got into set theory and i was wondering what is the cardinality of a set of all finite sequences of natural numbers?
I know that it is N for natural numbers and 2^N is for real numbers but how can i prove it?</p>
| R.V.N. | 730,220 | <p>You might notice that your set is <span class="math-container">$\bigcup_{n\in\mathbb{N}}\mathbb{N}^n$</span>. I will assume you know that for every <span class="math-container">$n\in\mathbb{N}$</span>, <span class="math-container">$|\mathbb{N}^n|=|\mathbb{N}|$</span>. Then, for each <span class="math-container">$n\in\mathbb{N}$</span> there exists a bijection <span class="math-container">$h_n:\mathbb{N}^n\longrightarrow\mathbb{N}$</span>. The function:
<span class="math-container">$$\Psi:(x_1,\ldots,x_k)\in\bigcup_{n\in\mathbb{N}}\mathbb{N}^n\longrightarrow\Psi(x_1,\ldots,x_k):=(k,h_k(x_1,\ldots,x_k))\in\mathbb{N}\times\mathbb{N}$$</span>
is bijective. To check it's injective, if <span class="math-container">$\Psi(x_1,\ldots,x_k)=\Psi(y_1,\ldots,y_p)$</span> then <span class="math-container">$(k,h_k(x_1,\ldots,x_k))=(p,h_p(y_1,\ldots,y_p))$</span>, so <span class="math-container">$k=p$</span> and <span class="math-container">$h_k=h_p$</span>. As <span class="math-container">$h_k$</span> is a bijection <span class="math-container">$h_k(x_1,\ldots,x_k)=h_k(y_1,\ldots,y_k)\implies(x_1,\ldots,x_k)=(y_1,\ldots,y_k)$</span>. To check it's surjective, given <span class="math-container">$(m,n)\in\mathbb{N}\times\mathbb{N}$</span>, <span class="math-container">$\Psi(h_m^{-1}(n))=(m,h_m(h_m^{-1}(n))=(m,n)$</span>.
Then, <span class="math-container">$|\bigcup_{n\in\mathbb{N}}\mathbb{N}^n|=|\mathbb{N}\times\mathbb{N}|=|\mathbb{N}|$</span>.</p>
|
148,127 | <p>I am trying to get an analytic expression for this integral:</p>
<pre><code> Integrate[Sign[Cos[q]]/(q + 1), {q, 0, x}, Assumptions -> x > 0]
</code></pre>
<p>Mathematica gives the answer:</p>
<pre><code> Abs[Cos[x]] Log[1 + x] Sec[x]
</code></pre>
<p>However, when I compare it to the numerical integration, I see that the answer is (incorrectly) discontinuous:</p>
<pre><code> Plot[{Abs[Cos[x]] Log[1 + x] Sec[x],
NIntegrate[Sign[Cos[q]]/(q + 1), {q, 0, x}]}, {x, 0, 4 Pi}]
</code></pre>
<p><a href="https://i.stack.imgur.com/ruX8o.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ruX8o.png" alt="enter image description here"></a></p>
<p>Is there a way to solve this issue and get the correct answer? Thank you</p>
| zhk | 8,538 | <p>The functionality you are looking for is <code>Epilog</code>,</p>
<pre><code>c1 = ContourPlot3D[
x z + y z - x y z == 0, {x, 0, 2}, {y, 0, 2}, {z, 0, 2},
Epilog -> {Text[Style["c=0", 22], Scaled[{0.1, 0.0}]]}];
c2 = ContourPlot3D[
x z + y z - x y z == 0.2, {x, 0, 2}, {y, 0, 2}, {z, 0, 2},
Epilog -> {Text[Style["c=0.2", 22], Scaled[{0.1, 0.0}]]}];
c3 = ContourPlot3D[
x z + y z - x y z == 0.4, {x, 0, 2}, {y, 0, 2}, {z, 0, 2},
Epilog -> {Text[Style["c=0.4", 22], Scaled[{0.1, 0.0}]]}];
Grid[{{c1, c2, c3}}]
</code></pre>
<p><a href="https://i.stack.imgur.com/n1wHv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/n1wHv.png" alt="enter image description here"></a></p>
|
110,373 | <p>Are there classes of infinite groups that admit Sylow subgroups and where the Sylow theorems are valid?</p>
<p>More precisely, I'm looking for classes of groups <span class="math-container">$\mathcal{C}$</span> with the following properties:</p>
<ul>
<li><span class="math-container">$\mathcal{C}$</span> includes the finite groups</li>
<li>in <span class="math-container">$\mathcal{C}$</span> there is a notion of Sylow subgroups that coincides with the usual one when restricted to finite groups</li>
<li>Sylow's theorems (or part of them) are valid in <span class="math-container">$\mathcal{C}$</span></li>
</ul>
<p>An example of such a class <span class="math-container">$\mathcal{C}$</span> is given by the class of profinite groups.</p>
| Igor Rivin | 11,142 | <p>Well, the Wikipedia gives <a href="https://en.wikipedia.org/wiki/Sylow_theorems#Sylow_theorems_for_infinite_groups" rel="nofollow noreferrer">an example of a Sylow theorem</a>, and there is more on this in notes <a href="https://people.brandeis.edu/%7Eigusa/Math131b/Sylow.pdf" rel="nofollow noreferrer">Sylow theorems</a> by Igusa. There is also the following paper of Baer:</p>
<p><a href="http://projecteuclid.org/euclid.dmj/1077492065" rel="nofollow noreferrer">Sylow theorems for infinite groups</a></p>
<p>Reinhold Baer</p>
<p>Source: Duke Math. J. Volume 6, Number 3 (1940), 598-614.</p>
|
110,373 | <p>Are there classes of infinite groups that admit Sylow subgroups and where the Sylow theorems are valid?</p>
<p>More precisely, I'm looking for classes of groups <span class="math-container">$\mathcal{C}$</span> with the following properties:</p>
<ul>
<li><span class="math-container">$\mathcal{C}$</span> includes the finite groups</li>
<li>in <span class="math-container">$\mathcal{C}$</span> there is a notion of Sylow subgroups that coincides with the usual one when restricted to finite groups</li>
<li>Sylow's theorems (or part of them) are valid in <span class="math-container">$\mathcal{C}$</span></li>
</ul>
<p>An example of such a class <span class="math-container">$\mathcal{C}$</span> is given by the class of profinite groups.</p>
| Sh.M1972 | 44,949 | <p>The best reference for this subject is the book of Martyn Dixon: Locally finite groups and Sylow theory.</p>
|
1,613,863 | <p>How to express $\log_3(2^x)$ using $\log_{10}$? And how to evaluate $4^{\log_4y}$? </p>
| Jimmy R. | 128,037 | <p>If $n\ge 2$ (to be sure that $n+2\ge 4$ and $n\ge 2$ in the given binomial coefficients, otherwise see the comment below), you can write the LHS as $$\dbinom{n+2}{4}=\frac{(n+2)!}{4!(n+2-4)!}=\frac{(n+2)(n+1)n(n-1)(n-2)!}{4!(n-2)!}=\frac{(n+2)(n+1)n(n-1)}{24}$$ and the RHS as $$6\dbinom{n}{2}=6\frac{n!}{2(n-2)!}=6\frac{n(n-1)(n-2)!}{2(n-2)!}=3n(n-1)$$ Equating these two you get $$\frac{(n+2)(n+1)n(n-1)}{24}=3n(n-1) \implies (n+2)(n+1)=72$$ which gives $n=7$ (or $n=-10$ but this should be rejected).</p>
|
1,177,721 | <p>A fair $6$-sided die is rolled $6$ times independently. For any outcome, this is the set of numbers that showed up at least once in the different rolls. For example, the outcome is $(2,3,3,3,5,5)$, the element set is $\{2,3,5\}$. What is the probability the element set has exactly $2$ elements? how about $3$ elements? </p>
<p>I know the sample space is $6^6$. The counting for $2$ elements is $6C2 \cdot 6!$? I would really appreciate the help! :)</p>
| N. F. Taussig | 173,070 | <p>Exactly two numbers appear:</p>
<p>There are $\binom{6}{2}$ ways for two of the six numbers to appear. On each of the six rolls of the die, there are two possible outcomes, giving $\binom{6}{2} \cdot 2^6$ possible outcomes in which two of the six numbers appear. However, of these $2^6$ sequences, there are two in which the same number is rolled six times. Thus, the probability of exactly two numbers appearing in six rolls of the die is
$$\frac{\binom{6}{2} \cdot (2^6 - 2)}{6^6}$$</p>
<p>Exactly three numbers appear:</p>
<p>There are $\binom{6}{3}$ ways for three of the six numbers to appear. On each of the six rolls of the die, there are three possible outcomes, giving $3^6$ sequences containing these three numbers. However, we have counted the sequences in which not all three of the numbers appear. There are $\binom{3}{2}$ ways for two of the numbers to appear. For each such pair, there are $2^6$ sequences, of which two consist of sequences in which one number is rolled all six times. Thus, there are $\binom{3}{2}(2^6 - 2)$ sequences in which exactly two of the three numbers appear. The number of sequences in which exactly one of the three numbers appears is $3$. Hence, the probability that exactly three of the numbers appear is
$$\frac{\binom{6}{3}[3^6 - \binom{3}{2}(2^6 - 2) - 3]}{6^6}$$ </p>
<p>My thanks to @bof for clarifying my thinking about the first problem. Any errors that remain are entirely my responsibility.</p>
|
11,994 | <p>Now that we get to see the SE-network wide list of "hot" questions, I am just shaking my head in disbelief. At the time I am writing this, the two hot questions from Math.SE are titled (get a barf-bag, quick)</p>
<ul>
<li><a href="https://math.stackexchange.com/q/599520/8348">https://math.stackexchange.com/q/599520/8348</a></li>
<li><a href="https://math.stackexchange.com/q/600373/8348">$1/i=i$. I must be wrong but why?</a></li>
</ul>
<p>Who gets to select these questions? How? Irrespective of how this is done, this is ridiculous, as neither question has any even remotely serious content (the latter one is more or less a common <em>fake-proof</em>). </p>
<p>My proposal:</p>
<blockquote>
<p>The representatives of Math.SE on this list should be based only on the votes of
people who are active on Math.SE. Not just all voting members (suspecting/pointing
finger at SOers, who get the right to vote from association bonus alone).</p>
</blockquote>
<p>The exact rep limit (if any) is open to debate, may be 1000? Probably shouldn't put the bar too high, for that would introduce different kind of problems. But something that ensures a valued history of contributions on this site - not elsewhere on the SE network.</p>
| user642796 | 8,348 | <p>These lists are automatically generated by a formula as described <a href="https://meta.stackexchange.com/a/61343/214632">here</a>.</p>
<p>According to <a href="https://meta.stackexchange.com/a/11604/214632">this</a> accepted MSO answer, in the past there was no intent to change the formula to account for the reputation of voters.<sup>1</sup> And who's to say that "active" math.SE users are not upvoting our poetry question? Five answers were given by $\geq 1000$ reputation users. Note, also, that the current formula does account for the number of answers given, and many "serious" mathematical questions get only very few answers.</p>
<p><sup>1</sup><sub>It's an old answer, but maybe MadScientist's feature request will yield a new answer.</sub></p>
|
3,260,911 | <p>I am currently struggling with the following exercise:</p>
<blockquote>
<p>Let <span class="math-container">$B$</span> be a Banach space and <span class="math-container">$C, D \subset B$</span> closed subspaces of <span class="math-container">$B$</span>.<br>
There is a <span class="math-container">$M \in ]0, \infty[$</span> such that <span class="math-container">$\forall x \in D : \operatorname{dist}(x, C \cap D) \leq M \cdot \operatorname{dist}(x, C)$</span> holds.</p>
<p>Show that <span class="math-container">$C + D$</span> is closed.</p>
</blockquote>
| D. A. | 671,115 | <p>I have proven it as follows:</p>
<p><span class="math-container">$B$</span> is a Banach space and <span class="math-container">$D, C \subseteq B$</span> are closed.
Thus <span class="math-container">$D$</span> and <span class="math-container">$C$</span> are themselves Banach spaces.
<span class="math-container">$C \cap D$</span> is also a closed subspace of B and thus <span class="math-container">$D / C \cap D$</span> is a Banach space.</p>
<p>The second isomorphism theorem states that <span class="math-container">$D / C \cap D \simeq (D + C)/D$</span> given by the isomorphism of vector spaces <span class="math-container">$\varphi : D / C \cap D \to (D + C)/C, D + C \cap D \mapsto D + C$</span>.</p>
<p>We now show that <span class="math-container">$\varphi$</span> and <span class="math-container">$\varphi^{-1}$</span> are continuous:</p>
<p>We know that <span class="math-container">$C \cap D \subseteq D \implies \forall x \in D : ||[x]||_{(D+C)/C} = \operatorname{dist}(x, C) \leq \operatorname{dist}(x, C \cap D) = ||[x]||_{D/C \cap D}$</span>.
Thus <span class="math-container">$\varphi$</span> is continuous.</p>
<p>Now show that <span class="math-container">$\varphi^{-1} : (D+C)/C \to D / C \cap D, [D+C] \mapsto [D]$</span> is continuous:
The inequation given results in: <span class="math-container">$\forall D \in D, C \in C : ||[D]||_{D / C \cap D} = \operatorname{dist}(D, C \cap D) \leq M \cdot \operatorname{dist}(D, C) = M \cdot ||[D]||_{(D+C)/C}$</span> which proves that <span class="math-container">$\varphi^{-1}$</span> is continuous.</p>
<p>Using <span class="math-container">$\varphi$</span> we see that <span class="math-container">$(D+C)/C$</span> is a Banach space, since <span class="math-container">$D / C \cap D$</span> is one.</p>
<p>Let <span class="math-container">$(x_n)_{n \in \mathbb N}$</span> be a Cauchy sequence.
<span class="math-container">$(x_n)$</span> converges to a <span class="math-container">$z \in B$</span>, since <span class="math-container">$B$</span> is complete.
Now look at the induced Cauchy sequence <span class="math-container">$([x_n])_{n \in \mathbb N}$</span>. It converges in the Banach space <span class="math-container">$(D + C)/C$</span> to <span class="math-container">$[z] \in (D + C)/C$</span>.</p>
<p>Thus we have <span class="math-container">$||[z] - [x_n]||_{(D+C)/C} = \operatorname{dist}(z - x_n, C) \overset{n \to \infty}{\longrightarrow} 0$</span>.
Since <span class="math-container">$C$</span> is closed, we also have <span class="math-container">$\operatorname{dist}(z - x_n, C) \overset{n \to \infty}{\longrightarrow} \operatorname{dist}(z - x, C)$</span>.</p>
<p>Thus it follows that <span class="math-container">$\operatorname{dist}(z - x, C) = 0 \overset{C~\text{closed}}{\implies} z -x \in C \implies z - x \in C \implies x \in z + C \subseteq D + C$</span>.</p>
<p>Thus we have <span class="math-container">$\overline{D+C} \subseteq D + C$</span> which shows that <span class="math-container">$D + C$</span> is closed. qed.</p>
|
1,482,104 | <p>Let $X$ have a uniform distribution with p.d.f. $f(x) = 1$, $x$ is in $(0, 1)$, zero elsewhere.
Find the p.d.f. of $Y = -2 \ln X$.</p>
<p>I don't think this is a very difficult question, I just don't really understand what it is asking or where to start. Any help would be very much appreciated. Thank you! </p>
<p>Update: I did
$$F(y)= P(Y \le y) = P(-2\ln x \le y) = P(\ln x \ge -y/2) = P(x \ge e^{-y/2}).$$
Then $ x=e^{-y/2}$ and $dx/dy =-1/2e^{-y/2}$
Is this all I need to do? Also, I'm not 100% sure why I am using inequalities here- can someone give me a quick explanation? </p>
| UnknownW | 78,627 | <p>Let $f(x)=1_{(0,1)}(x)=F_{X}'(x)$. Note that
$$F_{Y}(y)=P(X\geq e^{-y/2})=1-P(X\leq e^{-y/2})=1-F_{X}(e^{-y/2}).$$
Then the density is
\begin{align}
F_{Y}'(y)&=(1-F_{X}(e^{-y/2}))'\\
&=-F_{X}'(e^{-y/2})\frac{\mathrm{d} }{\mathrm{d} y}(e^{-y/2})\\
&=-1_{(0,1)}(e^{-y/2})\cdot \left ( -\frac{1}{2}e^{-y/2} \right )\\
&=1_{(0,\infty)}(y)\frac{1}{2}e^{-y/2}.
\end{align}</p>
|
1,482,104 | <p>Let $X$ have a uniform distribution with p.d.f. $f(x) = 1$, $x$ is in $(0, 1)$, zero elsewhere.
Find the p.d.f. of $Y = -2 \ln X$.</p>
<p>I don't think this is a very difficult question, I just don't really understand what it is asking or where to start. Any help would be very much appreciated. Thank you! </p>
<p>Update: I did
$$F(y)= P(Y \le y) = P(-2\ln x \le y) = P(\ln x \ge -y/2) = P(x \ge e^{-y/2}).$$
Then $ x=e^{-y/2}$ and $dx/dy =-1/2e^{-y/2}$
Is this all I need to do? Also, I'm not 100% sure why I am using inequalities here- can someone give me a quick explanation? </p>
| mathreadler | 213,607 | <p>It was such a long time ago I did this, but this seems to work too. At least in my computer simulations. <strong>Maybe it is just a coincidence</strong> it works for this particular function, if that is the case, please show me why.</p>
<p>$Y = -2 \ln(X)$ means that if a sample of X is x, then the corresponding sample of Y is $y = -2\ln(x)$</p>
<p>$y = \ln(x^{-2})$ and then $e^y = x^{-2}$ and $e^{-y/2} = x$. We see that since x is between 0 and 1, y will be between 0 and infinity. Now we just need to normalize, i.e. find $k$ so that $\int_0^\infty ke^{-y/2} dy = 1$ and that is an easy exercise in calculus.</p>
|
1,482,104 | <p>Let $X$ have a uniform distribution with p.d.f. $f(x) = 1$, $x$ is in $(0, 1)$, zero elsewhere.
Find the p.d.f. of $Y = -2 \ln X$.</p>
<p>I don't think this is a very difficult question, I just don't really understand what it is asking or where to start. Any help would be very much appreciated. Thank you! </p>
<p>Update: I did
$$F(y)= P(Y \le y) = P(-2\ln x \le y) = P(\ln x \ge -y/2) = P(x \ge e^{-y/2}).$$
Then $ x=e^{-y/2}$ and $dx/dy =-1/2e^{-y/2}$
Is this all I need to do? Also, I'm not 100% sure why I am using inequalities here- can someone give me a quick explanation? </p>
| Upstart | 312,594 | <p>$ P(-2logX\le x)=P(logX\ge -x/2)=P(X\ge e^{\frac{-x}{2}})=\int_{e^{\frac{-x}{2}}}^1dt=1-e^{\frac{-x}{2}}$</p>
<p>$F_X(x)=1-e^{\frac{-x}{2}}$</p>
<p>hence $f_X(x)=\frac{1}{2}e^{\frac{-x}{2}}$<br>
which is the pdf of exponential with parameter $\frac{1}{2}$</p>
|
1,492,660 | <p>I'm teaching a course on discrete math and came across <a href="http://ac.els-cdn.com/0097316573900204/1-s2.0-0097316573900204-main.pdf?_tid=700c69c2-78e3-11e5-9825-00000aacb35e&acdnat=1445535573_08d35be15f0f7d7d939fc2800d9be60b" rel="nofollow">a paper related to the Hadwiger-Nelson problem</a>. The question asks how many colors are needed to color every point in $\mathbb{Q}^2$ such that no two points at distance one are the same color.</p>
<p>The proof given in the paper works as follows. First, it defines an equivalence relation over $\mathbb{Q}^2$ where two points are related if the differences of their coordinates can be written with odd denominators. Next, it shows how to color the equivalence class containing the origin with two colors so that no two points at distance one are the same color. Finally, it argues that this coloring can be translated to the other equivalence classes, giving a 2-coloring.</p>
<p>My question is whether this last step - arguing that the coloring can be translated to the other equivalence classes - requires the axiom of choice. I suspect that it does because translating the coloring would require some way to identify which equivalence class a particular point is in and then using that to define the color, but I'm not entirely sure.</p>
<p>Does this result rely on choice? If so, if we reject the axiom of choice, is the chromatic number still two? Or does it change?</p>
| Asaf Karagila | 622 | <p>There is no actual use of the axiom of choice. Since $\Bbb Q^2$ is a countable set, we can enumerate it and identify each equivalence class with the least-indexed point in that class.</p>
<p>Generally if $X$ can be well-ordered, then there is no need to use the axiom of choice in order to choose from equivalence classes defined on $X$.</p>
|
1,992,143 | <p>I'm trying to determine if $\sum \limits_{n=1}^{\infty} \sin(n\pi + \frac{1}{2n})$ absolutely converges or not.</p>
<p>Help me check it. I don't know how to do it. Advance thanks. :)</p>
| Jacky Chong | 369,395 | <p>Observe
\begin{align}
\sin\left(n\pi +\frac{1}{2n}\right) = (-1)^n\sin\frac{1}{2n}
\end{align}
then apply alternating series test. </p>
<p>It should be noted for large enough $n$ the terms $\sin\frac{1}{2n}$ decreases monotonically to zero. </p>
<p>Hence the series converges conditionally.</p>
<p>Edit: Let us consider the case of absolute convergence. Observe
\begin{align}
\sin x = x-\frac{x^3}{3!} +\mathcal{O}(x^5)
\end{align}
which means
\begin{align}
\sin x \geq x-\frac{x^3}{3!}
\end{align}
for a neigbhorhood close to $0$. Hence it follows
\begin{align}
\sum^\infty_{n=N} \sin \frac{1}{2n} \geq \sum^\infty_{n=N} \frac{1}{2n} -\frac{1}{3!}\sum^\infty_{n=N} \frac{1}{8 n^3}
\end{align}
which shows that the series diverges absolutely because $\sum^\infty_{n=N} \frac{1}{n} = \infty$. </p>
|
749,714 | <p>Does anyone know how to show this preferable <strong>without</strong> using modular</p>
<p>For any prime $p>3$ show that 3 divides $2p^2+1$ </p>
| Andreas Caranti | 58,401 | <p>If you want to avoid modular arithmetic, you can easily do the explicit calculations, distinguishing the two cases</p>
<ul>
<li>$p = 1 + 3 k$, then $$2 p^{2} + 1 = 2 (1 + 6k + 9 k^{2}) + 1 = 3 (1 + 4k + 6 k^{2})$$</li>
<li>$p = 2 + 3 k$, then $$2 p^{2} + 1 = 2 (4 + 12k + 9 k^{2}) + 1 = 3 (3 + 8k + 6 k^{2})$$</li>
</ul>
|
4,498,498 | <p>Lets say you have two sequences of non negative integers each of length <span class="math-container">$n$</span>.</p>
<p>ie <span class="math-container">$(a_1,a_2,...,a_n)$</span> and <span class="math-container">$(b_1,b_2,...,b_n)$</span> such that
<span class="math-container">$\max(a_i) < k$</span>
and
<span class="math-container">$\max(b_i) < k$</span></p>
<p><strong>Game rule:</strong></p>
<p>You can edit both sequence with <span class="math-container">$\mathrm{swap}(a_i, b_i)$</span> for <span class="math-container">$1 ≤ i≤ n$</span>,</p>
<p><strong>Goal:</strong></p>
<p><span class="math-container">$a_i ≤ a_j$</span> for all <span class="math-container">$i ≤ j$</span>
and
<span class="math-container">$b_i ≤b_j$</span> for all <span class="math-container">$i ≤ j$</span></p>
<p>But not all initial sequence <span class="math-container">$a$</span> and <span class="math-container">$b$</span> can be solved. For example <span class="math-container">$(2,0)$</span> and <span class="math-container">$(0,1)$</span> is a pair of sequence that can't be solved.</p>
<p>Now given <span class="math-container">$n$</span> and <span class="math-container">$k$</span>, count number of different pair of initial sequence <span class="math-container">$(a,b)$</span> that can be solved with game described above.</p>
<p><strong>Example:</strong></p>
<p>for <span class="math-container">$n=1$</span>,<span class="math-container">$k=2$</span>:
These are the cases: <span class="math-container">${[(0),(0)],[(0),(1)],[(1),(0)],[(1),(1)]}$</span>.
Hence answer would be <span class="math-container">$4$</span>.</p>
| AlvinL | 229,673 | <p>Not an answer but an elaboration on P. Quinton's remark.</p>
<p>A pair of sequences <span class="math-container">$a$</span> and <span class="math-container">$b$</span> is solvable if and only if the pair of sequences
<span class="math-container">$$(\min (a_1,b_1),\ldots, \min(a_n,b_n))\quad\mbox{and}\quad (\max (a_1,b_1),\ldots,\max(a_n,b_n)) $$</span>
is solved.</p>
<p>Proof sketch. <span class="math-container">$\Rightarrow$</span> Say <span class="math-container">$\min (a_1,b_1) > \min (a_2,b_2)$</span>, then <span class="math-container">$a_1,b_1 > \min (a_2,b_2)$</span> contradicting solvability. Similar argument for the maximums. <span class="math-container">$\Leftarrow$</span> Winning strategy is clear.</p>
<p>Call a pair of sequences a <em>pair of min/max sequences</em> if it is invariant with respect to the min/max strategy. A pair of min/max sequences can be scrambled in <span class="math-container">$2^n$</span> different ways (for each index either swap or no swap). So the number of all solvable pairs of sequences is bounded above by <span class="math-container">$2^nP(n,k)$</span>, where <span class="math-container">$P(n,k)$</span> is the number of pairs of min/max sequences of length <span class="math-container">$n$</span> and bound <span class="math-container">$k$</span>.</p>
<p>There is a caveat. A pair of constant and equal sequences, for instance, is invariant with respect to scrambling. I don't know how to efficiently account for the <span class="math-container">$a_i=b_i$</span> cases that will otherwise lead to considerable overestimation.</p>
|
2,258,697 | <p>I recently encountered this question and have been stuck for a while. Any help would be appreciated!</p>
<p>Q: Given that
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{5} \tag{1} \label{eq:1}$$
$$abc = 5 \tag{2} \label{eq:2}$$
Find $a^3 + b^3 + c^3$. It wasn't specified in the question but I think it can be assumed that $a, b, c$ are real numbers.</p>
<p>My approach:
$$ ab + ac + bc = \frac{1}{5} abc = 1 $$
$$ a^3 + b^3 + c^3 = (a+b+c)^3 - 3[(a + b + c)(ab + ac + bc) - abc] $$
$$ a^3 + b^3 + c^3 = (a+b+c)^3 - 3(a+b+c) + 15 $$
From there, I'm not sure how to go about solving for $a + b + c$.
Something else I tried was letting $x = \frac{1}{a}, y = \frac{1}{b}, z = \frac{1}{c}$, so we get $$ xyz = x + y + z = \frac{1}{5} $$Similarly, I'm not sure how to continue from there. </p>
| copper.hat | 27,978 | <p>By inspection we see that $A (1,1,1,1)^T = 2 (1,1,1,1)^T $.</p>
<p>Similarly, $A (1,0,-1,0)^T = 0 $, $A (0,1,0,-1)^T = 0 $ and
$A(1,-1,1,-1)^T = 2 ((1,-1,1,-1)^T$.</p>
<p>All four vectors are linearly independent, hence constitute a
basis of eigenvectors.</p>
|
2,258,697 | <p>I recently encountered this question and have been stuck for a while. Any help would be appreciated!</p>
<p>Q: Given that
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{5} \tag{1} \label{eq:1}$$
$$abc = 5 \tag{2} \label{eq:2}$$
Find $a^3 + b^3 + c^3$. It wasn't specified in the question but I think it can be assumed that $a, b, c$ are real numbers.</p>
<p>My approach:
$$ ab + ac + bc = \frac{1}{5} abc = 1 $$
$$ a^3 + b^3 + c^3 = (a+b+c)^3 - 3[(a + b + c)(ab + ac + bc) - abc] $$
$$ a^3 + b^3 + c^3 = (a+b+c)^3 - 3(a+b+c) + 15 $$
From there, I'm not sure how to go about solving for $a + b + c$.
Something else I tried was letting $x = \frac{1}{a}, y = \frac{1}{b}, z = \frac{1}{c}$, so we get $$ xyz = x + y + z = \frac{1}{5} $$Similarly, I'm not sure how to continue from there. </p>
| Community | -1 | <p>Developing $|A-\lambda I|$, on the first row you get</p>
<p>$$
(1-\lambda)\begin{vmatrix}
1-\lambda & 0 & 1\\
0 & 1-\lambda & 0\\
1 & 0 & 1-\lambda\\
\end{vmatrix}+
\begin{vmatrix}
0 & 1-\lambda & 1\\
1 & 0 & 0\\
0 & 1 & 1-\lambda\\
\end{vmatrix}.
$$</p>
<p>Then developing again,</p>
<p>$$(1-\lambda)((1-\lambda)^3+\lambda-1)-(1-\lambda)^2+1=l^4-4l^3+4l^2.$$</p>
|
248,706 | <p>Let $X$ be a compact connected manifold. Since $\mathbb T^1$ is an Eilenberg-MacLane space $K(\mathbb Z,1)$, it follows that for every morphism $\varphi\colon\pi_1(X)\to\pi_1(\mathbb T^1)$ there is a continuous map $f\colon X\to\mathbb T^1$ such that $\varphi$ coincides with the induced morphism $f_*$.</p>
<p>Now assume that $G$ is a compact connected Lie group and $\varphi\colon\pi_1(G)\to\pi_1(\mathbb T^1)$ is a morphism. Does there exist a <em>morphism of Lie groups</em> $h\colon G\to\mathbb T^1$ with $h_*=\varphi$? If not, are there any necessary or sufficient conditions for this to be the case? What about non-compact Lie groups $G$?</p>
| Qiaochu Yuan | 290 | <p>There is no need to assume that $X$ is compact, connected, or a manifold in your first claim. Anyway, here is some context in which to put this question. A morphism $G \to S^1$ of Lie groups gives a map $BG \to BS^1 \cong B^2 \mathbb{Z}$ of classifying spaces, and hence a cohomology class in $H^2(BG, \mathbb{Z})$. If $G$ is connected, $BG$ is simply connected, and hence</p>
<p>$$\pi_1(G) \cong \pi_2(BG) \cong H_2(BG, \mathbb{Z}).$$</p>
<p>Now, by universal coefficients, we have an isomorphism</p>
<p>$$H^2(BG, \mathbb{Z}) \cong \text{Hom}(H_2(BG, \mathbb{Z}), \mathbb{Z})$$</p>
<p>from which it follows that every morphism $\pi_1(G) \to \mathbb{Z}$ corresponds to a map $BG \to B^2 \mathbb{Z}$ of classifying spaces, and hence to a map $G \to S^1$ of $\infty$-groups (meaning, loosely, a group homomorphism up to coherent homotopy). The remaining question is when we can strictify a map $G \to S^1$ of $\infty$-groups to a genuine Lie group homomorphism. </p>
<p>As Anton says in the comments, this is not possible if $G$ is noncompact, and $SL_2(\mathbb{R})$ is an explicit counterexample (the point being that the inclusion $SO(2) \to SL_2(\mathbb{R})$ of the maximal compact is a homotopy equivalence). It's less clear to me what happens when $G$ is compact. Certainly if $G$ is itself a torus it's always possible. On the other hand, if $G$ is semisimple then $\pi_1(G)$ is finite, so admits no nontrivial homomorphisms to $\mathbb{Z}$. So it might be possible to construct a more exotic counterexample by taking something like $SU(2) \times S^1$ mod $(-1, -1)$? I haven't thought much about these groups. </p>
|
15,205 | <p>I'm a young math student. And I live with the effort of always wanting to understand everything I study, in mathematics. This means that for every thing I face I must always understand every single demonstration, studying the basics every time if I don't remember them. And this makes it impossible for me to prepare the exams, because I can't go on, I fix myself on wanting to derive by myself a theorem and I lose days in it. And so I ask mathematicians if it is always necessary to be able to prove everything, or we must accept what the theorems say and give it for good. If possible I also ask you some advices to help me study, knowing my problem.</p>
| guest | 11,881 | <p>I am going to assume that you are teaching a calculus "helper" versus the entire physics class. Your initial statements don't match that. But then all your content described is math, not physics. And also 50 minutes per week sounds rather light for a whole class. [If the converse is the case, I would spend your time on...physics.]</p>
<p>With that in mind, my advice is NOT to look for some interesting activities. Instead use this as an opportunity for you to learn about the math required in stereotypical physics 101 classes. And give the kids a brushup related to that. Look at the homework and see what are the key math steps in them. And give a little help on them. Don't teach the physics of it (no word problems, they get that in real class). But abstract it down just to the math itself.</p>
<p>I would try to parallel whatever is being done in physics class. If vectors are important this week, spend some time on vectors this week. If integrals are important a different week, spend time on that. Now, it may not match perfect match, and some remediation may be needed of issues in previous semester. But let that be a general guide.</p>
<p>In terms of format, I would try to do something interactive, using a handout each week. Some lecture and some practice, using the sheet.</p>
|
2,231,003 | <p>I am trying to prove that the function $f_a(x) = e^{-\frac{1}{(x-a)}}$ is differentiable for all $x>a$. However, I do not know how to show $|\frac{e^{-\frac{1}{(x-a)}}-e^{-\frac{1}{(p-a)}}}{x-p} - (-\frac{e^{-\frac{1}{(p-a)}}}{(p-a)^2})| < \epsilon$ if $0<|x-p|<\delta$. </p>
<p>My next idea was to apply the natural logarithmic function to $e^{-\frac{1}{(x-a)}}$ to get $\hat{f_a}(x)=-\frac{1}{(x-a)}$. From here, I can prove that $\hat{f_a}(x)$ is differentiable for any $x>a$, but I do not know if this implies that $f_a(x)$ is differentiable for any $x>a$. </p>
<p>I know that $f_a(x):(-\infty,\infty) \rightarrow (0,\infty)$ and $\hat{f_a}(x):(-\infty,\infty) \rightarrow (-\infty,\infty)$. </p>
<p>Can you please help me understand how I can prove $f_a(x)$ is differentiable for all $x>a$? Thank you for your time and help. </p>
| Yes | 155,328 | <p>You were almost there! Note that $f_{a}: x \overset{g}{\mapsto} \frac{-1}{x-a} =: y \overset{h}{\mapsto} e^{y}$. You also know that both $g$ and $h$ are differentiable. So by the chain rule $f_{a} = h \circ g$ is also differentiable (its proof is contained as a special case in the proof of chain rule). </p>
<p>The comments above already gave some great examples for your reference. Let me supplement something that may help you conceptually. A theorem of the form "if $P$ then $Q$" is <strong>not applicable</strong> when $P$ does not hold. </p>
|
628,682 | <p>As both a programmer and a math student, I am trying to come up with a fool-proof way to handle errors from subtractive cancellation caused by trying to evaluate $x-y$, where x,y are extended (long double) precision floating-point numbers. (Obviously, if x is very close to y, this causes problem.) I found two equivalent forms, ${x^2-y^2\over x+y}$ and $(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})$. I was trying to evaluate regions where either form would work better, as well as the regions where either form might produce the same result, and/or a worse result in comparison to a straight subtraction.</p>
<hr>
<p>I have tried to do the error-analysis as such:</p>
<p>Let the long double c be defined such that $x-y=c$.Then, ${x^2-y^2\over x+y}=x-y=c$, which rewrites as ${x^2-y^2\over x-y}={x^2-y^2\over c}=x+y$, which means that, if ${x^2-y^2\over x+y}$ does better than x-y, then $|x+y|>1$.
Now, we take a look at $(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})=x-y=c$. A rewriting of this yields $\sqrt{x}-\sqrt{y}={c\over \sqrt{x}+\sqrt{y}}$, which means that $\sqrt{x}+\sqrt{y}<1$ for this decomposition to work. (This also poses the additional condition that x,y>0.) It would then follow that $x+y=x-y+2y>1$, or, in other terms, $x-y>1-2y$. Also, $\sqrt{x}+\sqrt{y}<1$ can be rewritten as $\sqrt{x}<1-\sqrt{y}$, or after squaring both sides, $x-y<1-2\sqrt{y}$. This means that $1-2y<1-2\sqrt{y}$, which simplifies to $y>\sqrt{y}$ which returns $y>1$ (this is the only place where it is true....)</p>
<p>So, if y > 1, then the decomposition works better, right? Well, not according to my program, which can be found here: <a href="http://ideone.com/amwv9H" rel="nofollow">http://ideone.com/amwv9H</a> ; // Disclaimer: It is written in C++</p>
| Mike Warren | 78,453 | <p>I think my error lies in getting from $x<1-\sqrt{y}$ to $x-y<1-2\sqrt{y}$. Since I have already imposed the condition that x,y>0, it would then follow that $x,y\in [0,1]$. I then went from $x-y<1-2\sqrt{y}$ to $y>1$, <strong>which contradicts the previous statement</strong>.</p>
|
3,141,618 | <p>The exercise is:</p>
<blockquote>
<p>Show that if <span class="math-container">$A \subset \mathbb{R} $</span> is bounded and <span class="math-container">$ A \neq \varnothing $</span> then <span class="math-container">$sup(A)=max(\overline{A} ).$</span></p>
</blockquote>
<p>Now, I wanted to ask you <strong>whether my proof is watertight</strong>:</p>
<hr>
<blockquote>
<p>Let <span class="math-container">$A \subset \mathbb{R}$</span> be a non-empty and bounded set.</p>
<p>Then <span class="math-container">$A$</span> has a finite supremum <span class="math-container">$sup(A) \equiv \widetilde{x}$</span>, which is
the least upper bound on <span class="math-container">$A$</span>.</p>
<p>Further define <span class="math-container">$\overline{x} \equiv max(\overline{A} )$</span>.</p>
<p>Assume, for the sake of contradiction, that <span class="math-container">$\overline{x} \neq
\widetilde{x}$</span>, which implies that there exists a distance <span class="math-container">$
d(\overline{x}, \widetilde{x} ) \equiv \epsilon > 0$</span>. Given that
<span class="math-container">$\widetilde{x} \geq x, \forall x \in A,$</span> we have that <span class="math-container">$B_{\epsilon /2}
(\overline{x}) \cap A = \varnothing $</span>. </p>
<p>This is a contradiction of the definition of closure. </p>
<p>Therefore, <span class="math-container">$sup(A)=max(\overline{A} )$</span>.</p>
</blockquote>
| Henno Brandsma | 4,280 | <p>First show that <span class="math-container">$\sup(A)$</span> (which exists by boundedness of <span class="math-container">$A$</span> and completeness of <span class="math-container">$\mathbb{R}$</span>) is in <span class="math-container">$\overline{A}$</span>. This follows from the definition of sup and the fact that we have the order topology on the reals. Then at least <span class="math-container">$\sup(A)$</span> is a candidate for being <span class="math-container">$\max(\overline{A})$</span>. </p>
|
977,446 | <p>Prove that $A\cap B = \emptyset$ iff $A\subset B^C$. I figured I could start by letting $x$ be an element of the universe and that $x$ is an element of $A$ and not an element of $B$. </p>
| AlexR | 86,940 | <p>Recall some definitions: Let $\Omega$ denote the universe (At least $A\cup B$).
$$B^C = \{ x \in\Omega | x\notin B\} \\
M\subset C :\Leftrightarrow \forall m\in M: m\in C \\
A\cap B := \{x\in\Omega | x\in A \wedge x\in B\}$$
Now plug this in: $A\subset B^C \Rightarrow A\cap B = \emptyset$ starts with any $x\in \Omega$ we know that $x\in A \Rightarrow x\notin B$ so $\nexists x\in \Omega : x\in A \wedge x\in B \Rightarrow A\cap B = \emptyset$. Now try the other direction.</p>
|
3,489,347 | <p><strong>Is there a simple way to characterize the functions in <span class="math-container">$C^\infty((0,1])\cap L^2((0,1])$</span>?</strong></p>
<p>That is, given a function <span class="math-container">$f(t)\in C^\infty((0,1])$</span>, is there a necessary/sufficient condition I can check to see if it's square integrable? An example of such a function is <span class="math-container">$f(t)=t^{-1/3}$</span>, which diverges as <span class="math-container">$t\to0$</span> but satisfies <span class="math-container">$\int_0^1 f(t)^2\,dt=3<\infty.$</span></p>
<hr>
<p><strong>Notes:</strong> I was hoping to prove something to the effect that <span class="math-container">$f(t)$</span> is square integrable if and only if <span class="math-container">$$\lim_{t\to0} \frac{f(t)^2}{t^p}=L < \infty$$</span> for some <span class="math-container">$p>-1$</span>. This is certainily a sufficient condition by the "<a href="https://services.math.duke.edu/~cbray/Stanford/2003-2004/Math%2042/limitcomp.pdf" rel="nofollow noreferrer">limit comparison test</a>" for improper integrals, but I'm not sure if it's necessary. (But, I also couldn't find a simple counterexample!)</p>
| ZAF | 609,023 | <p><span class="math-container">$N \in \mathbb{N}$</span> ?</p>
<p><span class="math-container">$10N\log(N)> 2N^2$</span> </p>
<p>If and only if</p>
<p><span class="math-container">$5 \log(N)> N$</span> </p>
<p>If and only if </p>
<p><span class="math-container">$e^{5\log(N)} > e^{N}$</span></p>
<p>If and only if</p>
<p><span class="math-container">$N^5 > e^{N}$</span></p>
|
3,893,440 | <p>Suppose we have <span class="math-container">$4$</span> books on Math, <span class="math-container">$5$</span> books on English and <span class="math-container">$6$</span> books on History. In how many ways you can put them on your bookshelf if you want :- <br/>
<span class="math-container">$1)$</span> The first book is a math book. <br/>
<span class="math-container">$2)$</span> All math books are at the beginning. <br/>
<span class="math-container">$3)$</span> Math and English books will stay together. <br/>
<span class="math-container">$4)$</span> The first book and the last book are both Math books.</p>
<p>(I'm completely lost with this problem. If someone can give me an explanation that would be great.)</p>
<p>I know that we have <span class="math-container">$3$</span> subjects so there are <span class="math-container">$3$</span>! Or <span class="math-container">$6$</span> possibilities and for <span class="math-container">$4$</span> math books we have <span class="math-container">$4!$</span> or <span class="math-container">$24$</span> ways to order. When it is <span class="math-container">$5$</span> English books we have <span class="math-container">$5!$</span> or <span class="math-container">$120$</span> ways to order. If it is <span class="math-container">$6$</span> history books we have <span class="math-container">$6!$</span> Or <span class="math-container">$720$</span> ways to order. So I think the answer is <span class="math-container">$6 \times 24 \times 120 \times 720=12,441,600$</span> ways to order the books .</p>
| Scott Hahn | 714,928 | <p>These are <span class="math-container">$4$</span> separate homework problems, right? In total we have <span class="math-container">$15$</span> books.</p>
<p><span class="math-container">$1)$</span> <span class="math-container">$4 * 14!$</span> (First choose any math book, then for each choice, we can order the remaining 14 however we want)</p>
<p><span class="math-container">$2)$</span> <span class="math-container">$4! * 11!$</span> (First, order the math books 4! number of ways, then for each ordering, we can order the remaining 11 however we want)</p>
<p><span class="math-container">$3)$</span> Unclear/left as an exercise for the reader.</p>
<p><span class="math-container">$4)$</span> <span class="math-container">$4 * 3 * 13!$</span> (First, choose the first book to be a math book, then choose the last book to be a math book, then order the remaining 13 in the middle however we want)</p>
|
3,893,440 | <p>Suppose we have <span class="math-container">$4$</span> books on Math, <span class="math-container">$5$</span> books on English and <span class="math-container">$6$</span> books on History. In how many ways you can put them on your bookshelf if you want :- <br/>
<span class="math-container">$1)$</span> The first book is a math book. <br/>
<span class="math-container">$2)$</span> All math books are at the beginning. <br/>
<span class="math-container">$3)$</span> Math and English books will stay together. <br/>
<span class="math-container">$4)$</span> The first book and the last book are both Math books.</p>
<p>(I'm completely lost with this problem. If someone can give me an explanation that would be great.)</p>
<p>I know that we have <span class="math-container">$3$</span> subjects so there are <span class="math-container">$3$</span>! Or <span class="math-container">$6$</span> possibilities and for <span class="math-container">$4$</span> math books we have <span class="math-container">$4!$</span> or <span class="math-container">$24$</span> ways to order. When it is <span class="math-container">$5$</span> English books we have <span class="math-container">$5!$</span> or <span class="math-container">$120$</span> ways to order. If it is <span class="math-container">$6$</span> history books we have <span class="math-container">$6!$</span> Or <span class="math-container">$720$</span> ways to order. So I think the answer is <span class="math-container">$6 \times 24 \times 120 \times 720=12,441,600$</span> ways to order the books .</p>
| Anonymous | 772,237 | <p>Here is the solution for the <span class="math-container">$3$</span>rd part .</p>
<p>Assume the <span class="math-container">$4$</span> Math and <span class="math-container">$5$</span> English books to be in a line, as in each arrangement they have to be consecutively placed. Take this to be a block of <span class="math-container">$9$</span> books.</p>
<p>In this block the number of ways to arrange the <span class="math-container">$4$</span> Math and <span class="math-container">$5$</span> English books would be :- <span class="math-container">$$\frac{9!}{4!*5!} = 126$$</span></p>
<p>Now this whole block can be placed with <span class="math-container">$6$</span> History books in this number of ways :-
<span class="math-container">$$\frac{7!}{6! * 1!} = 7$$</span></p>
<p>Add both of these up to get total number of ways in which Math and English books stay together :- <span class="math-container">$126 + 7 = 133$</span> .</p>
|
1,067,051 | <p>How can I find the point of intersection of <span class="math-container">$y=e^{-x}$</span> and <span class="math-container">$y=x$</span> ?</p>
<p><a href="https://i.stack.imgur.com/VoX32.png" rel="nofollow noreferrer">Here's the graph</a></p>
| Joel | 201,006 | <p>First:</p>
<p>There is no closed form for the solution. You can get an arbitrarily good approximation with root finding.</p>
<p>Have you learned about cobweb diagrams? That's one of the simplest versions of root finding. Otherwise, perhaps Newton's method? I don't want to give too much away (looks like a homework problem), so I'm being vague right now.</p>
|
1,365,489 | <p>What is the value of the following expression?</p>
<p>$$\sqrt[3]{\ 17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38}$$</p>
| No_way | 244,997 | <p>It is $2\sqrt{5}$. Note that $(17\sqrt{5}+38)^{\frac{1}{3}}=2+\sqrt{5}$. </p>
|
126,739 | <p><strong>I changed the title and added revisions and left the original untouched</strong> </p>
<p>For this post, $k$ is defined to be the square root of some $n\geq k^{2}$. Out of curiousity, I took the sum of one of the factorials in the denominator of the binomial theorem;
$$\sum _{k=1}^{\infty } \frac{1}{k!} \equiv e-1$$
<a href="http://oeis.org/A091131" rel="nofollow">OEIS A091131</a></p>
<p>Because I need to show that only the contiguous non-overlapping sequences of size $k$ up to $k^{2}+2k$ are valid for my purpose, I took the same sum with the denominator multiplied by $k+2$:
$$\sum _{k=1}^{\infty } \frac{1}{(k+m) k!} \equiv \frac{1}{2}\text{ for $m=2$ }$$
<a href="http://oeis.org/A020761" rel="nofollow">OEIS A020761</a></p>
<p>This is not a sum that I expected.</p>
<p>When $m\neq2$ the convergence returns alternating values like $\frac{1}{k}(-x+y e)$ and $\frac{1}{k}(x^{\prime}-y^{\prime} e)$, so $\frac{1}{2}$ seems to be the only value constructed out of integers.</p>
<p>Two questions:</p>
<p>$1)$ Is there a proof technique that can use this specific convergence to show that $k+2$ is the natural limit to my sequences? And that those specific non-overlapping sequences are the only ones that apply?</p>
<p>$2)$ Is this convergence interesting enough to put into OEIS?</p>
<p>I need some hints for my next step.</p>
<p><strong>Edit</strong><br>
Q1 is answered. I have enough info to keep me going for a few months.<br>
Q2: if you look at the OEIS entries for constants like $\pi$ and $e$, you will see dozens of identities. The entry for $\frac{1}{2}$ has only two identities. I feel it should have many more. But, just because I find this series interesting, doesn't mean others do, therefore, the question. </p>
<p>My motivation is to prove <a href="http://en.wikipedia.org/wiki/Oppermann%27s_conjecture" rel="nofollow">Oppermann's conjecture</a>. Thanks for the great answers and comments, and your patience.</p>
<p><strong>Revised</strong></p>
<p>Original post revised to use $k=0$ as starting index. And we show an example of the underlying pattern. </p>
<p>$ e= \sum_{k=0}^{\infty} 1/k!\textit{ Revised }$ </p>
<p>$ e-1= \sum_{k=0}^{\infty} 1/((k+m)k!)\text{ for }m=1$ </p>
<p>$ 1= \sum_{k=0}^{\infty} 1/((k+m)k!)\text{ for }m=2$ </p>
<p>$\sum_{k=0}^{\infty} 1/((k+m)k!)\not \in \textbf{Q} \text{ for }m>2$ </p>
<p>Example of underlying pattern for (say) $k=3$: </p>
<p>$(1, 2, 3), (4, 5, 6), (7, 8, 9), (10, 11, 12), (13, 14, 15)$<br>
$(1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3), (1, 2, 3)$<br>
$(1, 2, 3), (2, 1, 2), (1, 2, 3), (2, 1, 2), (1, 2, 3)$ </p>
<p>Top: Number line partitioned into $k+2$ non-overlapping ordered lists<br>
Middle: Equivalence classes $n-1 \mod k +1$<br>
Bottom: Least divisors. $1= p_{x}$ </p>
<p>What is it about these patterns that causes the convergence result for $m=2$ to be $\in \textbf{Q}$?</p>
<p><strong>Coda</strong></p>
<p>Removed the identities as not quite in step. Below I show the summand of my function on left, the summand of an 'instep' identity, and a variation of the identity.</p>
<p>$$\frac{1}{(k+2)k!} \equiv \frac{1}{(k+1)!+k!} \equiv \frac{1}{\Gamma(k+2)+k!}$$ </p>
<p>So, $\frac{1}{(k+2)k!}$ sums two consecutive factorials. Why? </p>
<p><strong>New</strong> This ratio equals $(e-1)^{-1}$ as shown <a href="http://mathworld.wolfram.com/ContinuedFraction.html" rel="nofollow">here</a>,</p>
<p>$$
\frac{\sum _{k=0}^{\infty } \frac{1}{(k+2) k!}}{\sum _{m=0}^{\infty } \left(\sum _{k=m}^{\infty } \frac{1}{(k+2) k!}\right)}=\frac{1}{1+\frac{2}{2+\frac{3}{3+\frac{4}{4+\frac{5}{5+\frac{6}{6+\frac{7}{7+\frac{8}{8+\frac{9}{9+\frac{10}{10+11}}}}}}}}}}
$$</p>
<p><strong>Another interesting pattern for the series:</strong><br>
$$
11_2,22_3,33_4,44_5,55_6,66_7,77_8,88_9,99_{10},\text{AA}_{11},\text{BB}_{12},\text{CC}_{13}{}{}{}
$$</p>
| Douglas Zare | 2,954 | <p>In the comments, quid showed that $\sum_{k=0}^\infty \frac{1}{(k+m)k!} = (-1)^m\bigg((m-1)! - !(m-1)e \bigg)$ where $!a = a! \sum_{k=0}^a \frac{(-1)^k}{k!}$ is the number of <a href="http://oeis.org/wiki/Number_of_derangements" rel="nofollow">derangements</a> in the symmetric group on $a$ objects. For example, $\sum_{k=0}^\infty \frac{1}{(k+10)k!} = 9! - !9 e = 362880-133496e$.</p>
<p>This explains why $m=2$ is the only case which is rational. The magnitude of the coefficient of $e$ is the number of derangements in $S_{m-1}$, and the only case where the number of derangements is $0$ is in $S_1$. While $S_0 \cong S_1$, a crucial difference between the trivial group $S_0$ and the trivial group $S_1$ is that the identity in $S_0$ <em>is</em> a derangement, and the corresponding sum for $m=1$ is $-1+e$.</p>
|
1,288,584 | <p>If <span class="math-container">$f(x)=f(x_0)+f'(x_0)(x-x_0)+\ldots+\frac{f^{(n-1)}(x_0)}{(n-1)!}(x-x_0)^{n-1}+\frac{f^{(n)}(\xi(x))}{n!}(x-x_0)^n,$</span> prove that <span class="math-container">$x \rightarrow f^{(n)}(\xi(x)) $</span> is continuous on <span class="math-container">$[x_0-\beta, x_0+\beta]$</span>, if <span class="math-container">$f\in C^n[x_0-\beta,x_0+\beta]$</span>.</p>
<p>How should I approach this problem?</p>
| DeepSea | 101,504 | <p>Let $g(x) = f^{(n)}\left(\xi(x)\right)$, and let $t \in[x_0-\beta,x_0+\beta]$, we show $g$ is continuous at $x = t$, i.e. $\displaystyle \lim_{x \to t} g(x) =g(t)$. We have: $x_0 < \xi(t) < t$, and $x_0 < \xi(x) < x$. We prove: $\displaystyle \lim_{x\to t} \xi(x) = \xi(t)$. To this end, assume WLOG:
$x_0 < \xi(t) < t < \xi(x) < x$. Then $\xi(x) - \xi(t) < x-x_0= (x-t)+(t-x_0)$. Now by the Archimedian principle, $\exists n \in \mathbb{N}: t-x_0 < n(x-t) \Rightarrow 0<\xi(x)-\xi(t) < (x-t)+n(x-t) = (n+1)(x-t)$. Thus by squeeze theorem: $\xi(x) \to \xi(t)$ when $x \to t$. Thus: $f^{(n)}(\xi(x)) \to f^{(n)}(\xi(t))$, or $g(x) \to g(t)$, and $g$ is continous at $x = t$. QED.</p>
|
1,975,018 | <p>I have an equation: </p>
<p>$$y = \Bigl(x + f(x^2 - 1)\Bigr)^{1/2},$$
(so a square root function.) I am asked to find $dy/dx$ when $x = 3$, given that $f(8) = 0$, and $f'(8) = 3$.</p>
<p>I apply the chain rule as I usually do for this problem, derivative of the inside times derivative of the outside. I get:</p>
<p>$$1/2\Bigl(x + f(x^2-1)\Bigr)^{-1/2} \Bigl(1 + f'(x^2-1)\Bigr).$$
However, when my teacher did this example, she multiplied the $f'(x^2-1)$ part by $2x$. Why did she do so?</p>
<p>Thank you! </p>
| Claude Leibovici | 82,404 | <p>Consider $$y = \Bigl(x + f[t(x)]\Bigr)^{1/2}$$ An easy way could be logarithmic differentiation $$y = \Bigl(x + f[t(x)]\Bigr)^{1/2}\implies \log(y)=\frac 12 \log\Bigl(x + f[t(x)]\Bigr)$$ $$\frac {y'}y=\frac 12 \frac{1+\frac{df}{dt}\times \frac{dt}{dx}}{x + f[t(x)]}\implies y'= \frac{1+\frac{df}{dt}\times \frac{dt}{dx}}{2\Bigl(x + f[t(x)]\Bigr)^{1/2}}$$</p>
|
1,975,018 | <p>I have an equation: </p>
<p>$$y = \Bigl(x + f(x^2 - 1)\Bigr)^{1/2},$$
(so a square root function.) I am asked to find $dy/dx$ when $x = 3$, given that $f(8) = 0$, and $f'(8) = 3$.</p>
<p>I apply the chain rule as I usually do for this problem, derivative of the inside times derivative of the outside. I get:</p>
<p>$$1/2\Bigl(x + f(x^2-1)\Bigr)^{-1/2} \Bigl(1 + f'(x^2-1)\Bigr).$$
However, when my teacher did this example, she multiplied the $f'(x^2-1)$ part by $2x$. Why did she do so?</p>
<p>Thank you! </p>
| operatorerror | 210,391 | <p>Hint: Rewrite $f(x^2-1)$ as $(f\circ g )(x)$ where $g(x)=x^2-1$, and then take the derivative of the composition using the chain rule. </p>
|
1,866,931 | <p>I would like to see a proof to this fact.</p>
<blockquote>
<p>If $A$ is an invertible matrix and $B \in \mathcal{L}(\mathbb{R}^n,\mathbb{R}^n)$, that is an bounded linear opertor in $\mathbb{R}^n$. Then, if there holds
$$
\|B-A\| \|A^{-1}\| <1,
$$
we have that B is invertible.</p>
</blockquote>
<p>Moreover, if possible, how to use it to prove that the aplication $A \rightarrow A^{-1}$ is continuous?</p>
| Anon | 245,264 | <p>Your formulation is incorrect. You're doing the right steps, but you're writing them down wrong, the logical meaning of what you write is different from what you mean. Here's proper formulation:</p>
<p>Let $R$ be a relation on $A$ such that $R \circ R \subseteq R$.</p>
<p>Let $x, y, z \in A$ such that $(x, y), (y, z) \in R$. By definition of $R \circ R$, it follows that $(x, z) \in R \circ R$.</p>
<p>Now $R\circ R \subseteq R$, so $(x, z) \in R$. This shows transitivity.</p>
|
755,571 | <p>$$a_n=3a_{n-1}+1; a_0=1$$</p>
<p>The book has the answer as: $$\frac{3^{n+1}-1}{2}$$</p>
<p>However, I have the answer as: $$\frac{3^{n}-1}{2}$$</p>
<p>Based on:</p>
<p><img src="https://i.stack.imgur.com/4vJrQ.png" alt="enter image description here"></p>
<p>Which one is correct?</p>
<p>Using backwards substitution iteration, the end of this will be
$$3^{n-1}a_0+3^{n-2}+3^{n-3}+...+3+1$$</p>
<p>which is $$=3^{n-1}+3^{n-2}+3^{n-3}+...+3+1=\sum_{i=0}^{n-1}3^i$$</p>
<p>Which according to the theorem should be $$\frac{3^{(n-1)+1}-1}{(3-1)}=\frac{3^{n}-1}{2}$$</p>
| AsdrubalBeltran | 62,547 | <p>Ok your answer is correct, but you have the formula for the $(n-1)$-term, if you need the formula for the $n$-term, then note that, according to recurrence</p>
<p>$$a_n=3a_{n-1}+1$$</p>
<p>You have that $$a_{n-1}=\frac{3^n-1}{2}$$ then:
$$a_n=3\cdot\frac{3^n-1}{2}+1=\frac{3^{n+1}-1}{2}$$</p>
<p>God bless</p>
|
1,217,175 | <p><strong>Here's the question:</strong></p>
<p>Is the following true or false?</p>
<p>There is a function $f: \mathbb R \to \mathbb R$ that satisfies the following condition:</p>
<p>For every $a \in \mathbb R $ and $ \epsilon \gt 0 $ there is $\delta \gt 0$ such that $\left| f(x)-f(a) \right| \lt \epsilon \implies \left| x-a \right| \lt \delta $.</p>
<p><strong>My initial response:</strong></p>
<p>I said that this is true for any constant function, e.g. $f(x) = 0$. In this case, $\left| f(x)-f(a) \right| = \left|0-0 \right|=0 \lt \epsilon \text{ and} \left| x-a \right|=\left|0-0 \right|=0 \lt \delta $. I know that this doesn't work because in the $\left| x-a \right|$ case, $x$ may not be $0$, obviously. However, this was the last question on a test, so I kind of just guessed because I was running out of time. Now that I've had time to think about it, though, I can't seem to figure it out (i.e., whether or not it's true or false). Any and all help here is appreciated, as always. Thanks. </p>
| Ali Caglayan | 87,191 | <p>Let $x=a+bi$ then $\sqrt{x}=\sqrt{a+bi}=-1$. Then square both sides to get $a+bi=1$. Thus $x$ must be a 'real' number and not complex. Now if $a=1$, taking the <em>positive</em> square root, which is what $\sqrt{.}$ means, will give you that $\sqrt{1}=-1$. Which is not strictly true when using the radical sign.</p>
<p>You can however consider the solutions to the equation $x^2-1=0$ and define the square root(s) of a number $c$ to be the solutions to $x^2-c=0$.</p>
<p>When working with the real numbers, we had the convenience of not always having to consider the negative solutions for a result to make sense. This is not strictly true in the complex numbers, therefore operations like taking square roots must be more precisely defined and hence it must be taken into account that such an operation can have multiple results.</p>
|
3,848,929 | <p>The question is entirely in the title. I need this result as part of a much bigger question. If this is possible, please give me an example of such <span class="math-container">$u$</span> and <span class="math-container">$v$</span>.</p>
<p><strong>My progress:</strong>
Let <span class="math-container">$u=2k+1$</span> and <span class="math-container">$v=2l+1$</span> such that <span class="math-container">$k>l>0$</span>. Then what we get is
<span class="math-container">\begin{equation}
k-l\mid 2k+1\qquad\&\qquad k-l\mid 2l+1
\end{equation}</span>
But I can't go any further.</p>
| metamorphy | 543,769 | <p>Let <span class="math-container">$u,v$</span> be coprime and <span class="math-container">$u+v=(u-v)d$</span>. Then <span class="math-container">$(d-1)u=(d+1)v$</span>.</p>
<p>Thus, <span class="math-container">$u\mid(d+1)v$</span>, hence <span class="math-container">$u\mid(d+1)$</span> and <span class="math-container">$d+1=uw$</span> for some <span class="math-container">$w$</span>.</p>
<p>Similarly, we get <span class="math-container">$d-1=vw$</span> (with <em>the same</em> <span class="math-container">$w$</span>!), and <span class="math-container">$2=(u-v)w$</span>.</p>
<p>This means that <span class="math-container">$u-v$</span> <em>cannot</em> be greater than <span class="math-container">$2$</span>.</p>
|
3,848,929 | <p>The question is entirely in the title. I need this result as part of a much bigger question. If this is possible, please give me an example of such <span class="math-container">$u$</span> and <span class="math-container">$v$</span>.</p>
<p><strong>My progress:</strong>
Let <span class="math-container">$u=2k+1$</span> and <span class="math-container">$v=2l+1$</span> such that <span class="math-container">$k>l>0$</span>. Then what we get is
<span class="math-container">\begin{equation}
k-l\mid 2k+1\qquad\&\qquad k-l\mid 2l+1
\end{equation}</span>
But I can't go any further.</p>
| Haran | 438,557 | <p>Assume <span class="math-container">$u-v>2$</span>. Let <span class="math-container">$p$</span> be an odd prime divisor of <span class="math-container">$u-v$</span>. Then, <span class="math-container">$p \mid (u+v)$</span> as well. Adding and subtracting, we can see that <span class="math-container">$p$</span> divides <span class="math-container">$2u$</span> and <span class="math-container">$2v$</span>. Since <span class="math-container">$p$</span> is an odd prime, this would imply that <span class="math-container">$p$</span> divides both <span class="math-container">$u$</span> and <span class="math-container">$v$</span>, which is a contradiction, since they are relatively prime. Thus, there are no odd prime divisors of <span class="math-container">$u-v$</span>, i.e. it is a power of <span class="math-container">$2$</span>.</p>
<p>Now, since <span class="math-container">$u-v>2$</span> and is a power of <span class="math-container">$2$</span>, we can see that <span class="math-container">$4 \mid (u-v)$</span>. Then, we also need <span class="math-container">$4 \mid (u+v)$</span>. However, adding gives <span class="math-container">$4 \mid 2u$</span> which would give <span class="math-container">$2 \mid u$</span> and this is a contradiction since <span class="math-container">$u$</span> is odd. Thus, we cannot have <span class="math-container">$u-v>2$</span> as required.</p>
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