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 |
|---|---|---|---|---|
84,076 | <p>I think computation of the Euler characteristic of a real variety is not a problem in theory.</p>
<p>There are some nice papers like <em><a href="http://blms.oxfordjournals.org/content/22/6/547.abstract" rel="nofollow">J.W. Bruce, Euler characteristics of real varieties</a></em>.</p>
<p>But suppose we have, say, a very specific real nonsingular hypersurface, given by a polynomial, or a nice family of such hypersurfaces. What is the least cumbersome approach to computation of $\chi(V)$? One can surely count the critical points of an appropriate Morse function, but I hope it's not the only possible way.</p>
<p>(Since I am talking about dealing with specific examples, here's one:
$f (X_1,\ldots,X_n) = X_1^3 - X_1 + \cdots + X_n^3 - X_n = 0$, where $n$ is odd.)</p>
<p><strong>Update:</strong> the original motivation is the following: the well-known results by Oleĭnik, Petrovskiĭ, Milnor, and Thom give upper bounds on $\chi (V)$ or $b(V) = \sum_i b_i (V)$ that are exponential in $n$. It is easy to see that this is unavoidable, e.g. $(X_1^2 - X_1)^2 + \cdots + (X_n^2 - X_n)^2 = 0$ is an equation of degree $4$ that defines exactly $2^n$ isolated points in $\mathbb{R}^n$. I was interested in specific families of real algebraic sets with large $\chi (V)$ or $b (V)$ <em>defined by one equation of degree $3$</em>. I couldn't find an appropriate reference with such examples and it seems like a proof for such example would require some computations (unlike the case of degree $4$).</p>
| Igor Rivin | 11,142 | <p>This is quite nontrivial. See for example: </p>
<p>On Bounding the Betti Numbers and Computing the Euler Characteristic of Semialgebraic sets, by Saugata Basu (google has full text). </p>
<p>The canonical reference is a more recent book by Basu, Ricky Pollack and Marie-Francoise Roy, called "algorithms in real algebraic geometry"</p>
|
791,372 | <p>Hi I am trying to solve this double integral
$$
I:=\int_0^\infty \int_0^\infty \frac{\log x \log y}{\sqrt {xy}}\cos(x+y)\,dx\,dy=(\gamma+2\log 2)\pi^2.
$$
Thank you.</p>
<p>The constant in the result is given by $\gamma\approx .577$, and is known as the Euler-Mascheroni constant. I was thinking to write
$$
I=\Re \bigg[\int_0^\infty \int_0^\infty \frac{\log x \log y}{\sqrt{xy}}\, e^{i(x+y)}\, dx\, dy\bigg]
$$
and using Leibniz's rule for differentiation under the integral sign to write
$$
I(\eta, \xi)=\Re\bigg[ \int_0^\infty \int_0^\infty \ \frac{\log (\eta x)\log(\xi y)}{\sqrt{xy}} e^{i(x+y)}dx\,dy. \bigg]\\
$$
After taking the derivatives it became obvious that I need to try another method since the x,y constants cancel out. How can we solve this integral I? Thanks. </p>
| Zaid Alyafeai | 87,813 | <p>Using the identity </p>
<p>$$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$$</p>
<p>The integral can be written
$$
I=\int_0^\infty \int_0^\infty \frac{\log x \log y}{\sqrt {xy}}\left(\cos(x)\cos(y)-\sin(x)\sin(y)\right)\,dx\,dy $$</p>
<p>Now by splitting the integrals</p>
<p>$$\int_0^\infty \int_0^\infty \frac{\log x \log y}{\sqrt {xy}}\cos(x)\cos(y)\,dx\,dy-\int_0^\infty \int_0^\infty \frac{\log x \log y}{\sqrt {xy}}\sin(x)\sin(y)\,dx\,dy
$$</p>
<p>Notice by symmetry of the integrals we have</p>
<p>$$\left(\int^\infty_0 \frac{\log x }{\sqrt {x}}\cos(x)\,dx \right)^2-\left(\int^\infty_0 \frac{\log x }{\sqrt {x}}\sin(x)\,dx \right)^2
$$ </p>
<p>Both inegrals are solvable by using the mellin transforms </p>
<p>$$\int^\infty_0 x^{s-1}\sin(x)\,dx = \Gamma (s) \sin\left( \frac{\pi s}{2} \right)$$</p>
<p>$$\int^\infty_0 x^{s-1}\cos(x)\,dx = \Gamma (s) \cos\left( \frac{\pi s}{2} \right)$$</p>
<p>By differentiation under the integral sign and using $s=\frac{1}{2}$.</p>
<p>$$\int^\infty_0 \frac{\log x }{\sqrt {x}}\cos(x)\,dx =-\frac{1}{2} \sqrt{\frac{π}{2}} \left(2 \gamma +π+\log(16) \right) $$</p>
<p>$$\int^\infty_0 \frac{\log x }{\sqrt {x}}\sin(x)\,dx=\frac{1}{2} \sqrt{\frac{π}{2}} (-2 \gamma +π- \log(16))
$$</p>
<p>Collecting the results together we have</p>
<p>$$I=(\gamma+2\log 2)\pi^2$$</p>
|
791,372 | <p>Hi I am trying to solve this double integral
$$
I:=\int_0^\infty \int_0^\infty \frac{\log x \log y}{\sqrt {xy}}\cos(x+y)\,dx\,dy=(\gamma+2\log 2)\pi^2.
$$
Thank you.</p>
<p>The constant in the result is given by $\gamma\approx .577$, and is known as the Euler-Mascheroni constant. I was thinking to write
$$
I=\Re \bigg[\int_0^\infty \int_0^\infty \frac{\log x \log y}{\sqrt{xy}}\, e^{i(x+y)}\, dx\, dy\bigg]
$$
and using Leibniz's rule for differentiation under the integral sign to write
$$
I(\eta, \xi)=\Re\bigg[ \int_0^\infty \int_0^\infty \ \frac{\log (\eta x)\log(\xi y)}{\sqrt{xy}} e^{i(x+y)}dx\,dy. \bigg]\\
$$
After taking the derivatives it became obvious that I need to try another method since the x,y constants cancel out. How can we solve this integral I? Thanks. </p>
| Felix Marin | 85,343 | <p>$\newcommand{\+}{^{\dagger}}
\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\down}{\downarrow}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\fermi}{\,{\rm f}}
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\isdiv}{\,\left.\right\vert\,}
\newcommand{\ket}[1]{\left\vert #1\right\rangle}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large a}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
\newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{I\equiv\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{x}\ln\pars{y} \over \root{xy}}\,\cos\pars{x + y}\,\dd x\,\dd y
=\bracks{\gamma + 2\ln\pars{2}}\pi^{2}:\ {\large ?}}$</p>
<blockquote>
<p>\begin{align}
I&=\Re\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{x}\ln\pars{y} \over \root{xy}}\,\expo{\ic\pars{x + y}}\,\dd x\,\dd y
=\Re\braces{\bracks{\color{#c00000}{\int_{0}^{\infty}
{\ln\pars{x} \over \root{x}}\,\expo{\ic x}\,\dd x}}^{2}}
\end{align}</p>
</blockquote>
<p>\begin{align}
&\color{#c00000}{\int_{0}^{\infty}
{\ln\pars{x} \over \root{x}}\,\expo{\ic x}\,\dd x}
=\lim_{\mu \to -1/2}\partiald{}{\mu}\
\overbrace{\int_{0}^{\infty}x^{\mu}\expo{\ic x}\,\dd x}
^{\ds{t\ \equiv\ -\ic x\ \imp\ x\ =\ \ic t}}\
\\[3mm]&=\lim_{\mu \to -1/2}\partiald{}{\mu}
\int_{0}^{-\ic\infty}\expo{\ic\pi\mu/2}t^{\mu}\expo{-t}\,\ic\,\dd t
\\[3mm]&=\ic\lim_{\mu \to -1/2}\partiald{}{\mu}\braces{\expo{\ic\pi\mu/2}\bracks{%
\int_{0}^{\infty}t^{\mu}\expo{-t}\,\dd t
-\overbrace{\left.\lim_{R \to \infty}\int_{-\pi/2}^{0}z^{\mu}\expo{-z}\,\dd z\,
\right\vert_{z\ \equiv\ R\expo{\ic\theta}}}^{\ds{=\ 0}}}}
\\[3mm]&=\ic\lim_{\mu \to -1/2}\partiald{}{\mu}
\bracks{\expo{\ic\pi\mu/2}\Gamma\pars{\mu + 1}}
\end{align}
where $\ds{\Gamma\pars{z}}$ is the
<a href="http://people.math.sfu.ca/~cbm/aands/page_255.htm" rel="noreferrer">Gamma Function</a>
${\bf\mbox{6.1.1}}$.</p>
<blockquote>
<p>\begin{align}
I&=\color{#c00000}{\int_{0}^{\infty}{%
\ln\pars{x} \over \root{x}}\,\expo{\ic x}\,\dd x}
=\ic\lim_{\mu \to -1/2}
\bracks{\expo{\ic\pi\mu/2}\,{\ic\pi \over 2}\,\Gamma\pars{\mu + 1}
+\expo{\ic\pi\mu/2}\Gamma\pars{\mu + 1}\Psi\pars{\mu + 1}}
\end{align}
where $\ds{\Psi\pars{z}}$ is the
<a href="http://people.math.sfu.ca/~cbm/aands/page_258.htm" rel="noreferrer">Digamma Function</a> ${\bf\mbox{6.3.1}}$.</p>
</blockquote>
<p>\begin{align}
I&=\color{#c00000}{\int_{0}^{\infty}{%
\ln\pars{x} \over \root{x}}\,\expo{\ic x}\,\dd x}
=\ic\expo{-\ic\pi/4}\Gamma\pars{\half}
\bracks{{\ic\pi \over 2} + \Psi\pars{\half}}
\\[3mm]&=\root{\pi \over 2}\pars{1 + \ic}\bracks{{\ic\pi \over 2} - \gamma - 2\ln\pars{2}}
\end{align}
$\ds{\gamma}$ is the
<a href="http://people.math.sfu.ca/~cbm/aands/page_255.htm" rel="noreferrer">Euler-Mascheroni Constant</a> ${\bf\mbox{6.1.3}}$ and we used the identities
$\ds{\Gamma\pars{\half} = \root{\pi}}$ and
$\ds{\Psi\pars{\half}=-\gamma - 2\ln\pars{2}}$.</p>
<blockquote>
<p>\begin{align}
I&=\Re\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{x}\ln\pars{y} \over \root{xy}}\,\expo{\ic\pars{x + y}}\,\dd x\,\dd y
\\[3mm]&=\Re\pars{\braces{\root{\pi \over 2}\pars{1 + \ic}\bracks{{\ic\pi \over 2} - \gamma - 2\ln\pars{2}}}^{2}}
\\[3mm]&=\Re\pars{{\pi \over 2}\,2\ic\braces{\bracks{\gamma + 2\ln\pars{2}}^{2} - {\pi^{2} \over 4} - \ic\pi\bracks{\gamma + 2\ln\pars{2}}}}
\end{align}</p>
</blockquote>
<p>$$\color{#00f}{\large%
I\equiv\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{x}\ln\pars{y} \over \root{xy}}\,\cos\pars{x + y}\,\dd x\,\dd y
=\bracks{\gamma + 2\ln\pars{2}}\pi^{2}}
$$</p>
|
1,338,999 | <p><img src="https://i.stack.imgur.com/hXfn2.png" alt="sat question"></p>
<p>My friend selected option B, I did C. We're confused. Can someone please explain this for my friend?</p>
| Michael Hardy | 11,667 | <p>I got $\dfrac{(m!)^2}{(m-k)!(m+k)!}$ where you have $\left( \dfrac{m!}{(m-k)!} \right)^2$.</p>
|
2,183,390 | <p>So, I need to solve a hard problem, which reduces to this: </p>
<blockquote>
<p>Prove that $3^{\frac{1}{3}} \notin \mathbb{Q}[13^{\frac{1}{3}}]$.</p>
</blockquote>
<p>The only thing that comes into my mind is to suppose the opposite, <em>i.e.</em>, $3^{\frac{1}{3}} \in \mathbb{Q}[13^{\frac{1}{3}}]$, and then to see that $3^{\frac{1}{3}} = a+b\ 13^{\frac{1}{3}} + c\ 13^{\frac{2}{3}}$ leads to some contradiction while trying to manipulate this. But I am not sure if this would work.</p>
<blockquote>
<p>Is there a smarter solution, or a solution at all?</p>
</blockquote>
<p>(If someone really can manipulate this, I would like to see it.) </p>
| Dietrich Burde | 83,966 | <p>Suppose that $3^{\frac{1}{3}} \in \mathbb{Q}(13^{\frac{1}{3}})$. Then $\mathbb{Q}(3^{\frac{1}{3}})=\mathbb{Q}(13^{\frac{1}{3}})$, because both field have degree $3$ over $\mathbb{Q}$. This is a contradiction, since both fields have different <a href="https://en.wikipedia.org/wiki/Discriminant_of_an_algebraic_number_field" rel="nofollow noreferrer">discriminant</a>.</p>
|
1,041,226 | <p>I need to prove the following: ${n\choose m-1}+{n\choose m}={n+1\choose m}$, $1\leq m\leq n$.</p>
<p>With the definition: ${n\choose m}= \left\{
\begin{array}{ll}
\frac{n!}{m!(n-m)!} & \textrm{für \(m\leq n\)} \\
0 & \textrm{für \(m>n\)}
\end{array}
\right.$</p>
<p>and $n,m\in\mathbb{N}$.</p>
<p>I'm not really used to calculations with factorials and can't make much sense from it...</p>
| peterwhy | 89,922 | <p>$$\begin{align*}
\frac{n!}{(m-1)!(n-m+1)!} + \frac{n!}{m!(n-m)!}
&= \frac{n!}{(m-1)!(n-m)!(n-m+1)} + \frac{n!}{(m-1)!(n-m)!m}\\
&= \frac{n!}{(m-1)!(n-m)!}\left[\frac1{n-m+1}+\frac1m\right]\\
&= \frac{n!}{(m-1)!(n-m)!}\cdot\frac{n+1}{m(n-m+1)}\\
&= \frac{(n+1)!}{m!(n-m+1)!}
\end{align*}$$</p>
|
1,041,226 | <p>I need to prove the following: ${n\choose m-1}+{n\choose m}={n+1\choose m}$, $1\leq m\leq n$.</p>
<p>With the definition: ${n\choose m}= \left\{
\begin{array}{ll}
\frac{n!}{m!(n-m)!} & \textrm{für \(m\leq n\)} \\
0 & \textrm{für \(m>n\)}
\end{array}
\right.$</p>
<p>and $n,m\in\mathbb{N}$.</p>
<p>I'm not really used to calculations with factorials and can't make much sense from it...</p>
| Math-fun | 195,344 | <p>Do it intuitively: assume you have n+1 objects from which you want to choose m. Now divide your n+1 objects into two groups: one that includes n objects and one group with 1 (specific) object. Choosing m from n+1 is equivalent to choosing m out of the first group (these exclude the one specific object) PLUS choosing m-1 out of n and adding while always that one specific object to them. </p>
|
1,978,035 | <p>In the Wikipedia page about quintics, there was a list of quintics that could be solved with trigonometric roots.</p>
<p>For example:$$x^5+x^4-4x^3-3x^2+3x+1\tag1$$ has roots of the form $2\cos \frac {2k\pi}{11}$
$$x^5+x^4-16x^3+5x^2+21x-9=0\tag2$$ has roots of the form $\sum_{k=0}^{7}e^{\frac {2\pi i 3^k}{41}}$</p>
<hr>
<blockquote>
<p><strong>Question:</strong> Is there a formula or underlying structure to find the roots of the quintics?</p>
</blockquote>
<hr>
<p>Wikipedia says that the roots are the sums of the first $n$-th roots of unity with $n=10k+1$, so I'm guessing we first have to find $n$, or do we first have to find $k$?</p>
<p>Any help is appreciated.</p>
| D.Matthew | 469,027 | <p>Denoted
<span class="math-container">\begin{align*} \qquad\left\{ \begin{aligned} \varepsilon_1&=\exp\left(\dfrac{\>\pi\,\!i\>}{5}\right)=\phantom{+}\frac{1+\sqrt{5}}{4}+i\sqrt{\frac{5-\sqrt{5}}{8}}\approx\phantom{+}0.809017 + 0.587785i\qquad\\ \varepsilon_2&=\exp\left(\dfrac{2\pi\,\!i}{5}\right)=\frac{-1+\sqrt{5}}{4}+i\sqrt{\frac{5+\sqrt{5}}{8}}\approx\phantom{+}0.309017 + 0.951057i\\ \varepsilon_3&=\exp\left(\dfrac{3\pi\,\!i}{5}\right)=\phantom{+}\frac{1-\sqrt{5}}{4}+i\sqrt{\frac{5+\sqrt{5}}{8}}\approx-0.309017 + 0.951057i\\ \varepsilon_4&=\exp\left(\dfrac{4\pi\,\!i}{5}\right)=-\frac{1+\sqrt{5}}{4}+i\sqrt{\frac{5-\sqrt{5}}{8}}\approx-0.809017+0.587785i\\ \end{aligned} \right.\qquad \end{align*}</span>
and
<span class="math-container">\begin{gather*} \qquad\left\{ \begin{aligned} \alpha_1&=\sqrt[\underset{\,}{5}]{\frac{11}{4}\left(89+25\sqrt{5}+5i\sqrt{410-178\sqrt{5}}\right)}\approx3.315688+0.078842i\\ \alpha_2&=\sqrt[\underset{\,}{5}]{\frac{11}{4}\left(89-25\sqrt{5}+5i\sqrt{410+178\sqrt{5}}\right)}\approx3.197878+0.879531i\\ \alpha_3&=\sqrt[\underset{\,}{5}]{\frac{11}{4}\left(89-25\sqrt{5}-5i\sqrt{410+178\sqrt{5}}\right)}\approx3.197878-0.879531i\\ \alpha_4&=\sqrt[\underset{\,}{5}]{\frac{11}{4}\left(89+25\sqrt{5}-5i\sqrt{410-178\sqrt{5}}\right)}\approx3.315688-0.078842i \end{aligned} \right.\qquad\\ \end{gather*}</span>
Then,
<span class="math-container">\begin{align*} \bbox[#EFF,15px,border:2px solid blue]{ \left\{\begin{aligned} x_1&=\dfrac{\phantom{+}\,\alpha_1\varepsilon_1+\alpha_2\varepsilon_1-\alpha_3\varepsilon_4-\alpha_4\varepsilon_4-1}{5}\\ x_2&=\dfrac{-\,\alpha_1\varepsilon_4-\alpha_2\varepsilon_2+\alpha_3\varepsilon_3+\alpha_4\varepsilon_1-1}{5}\\ x_3&=\dfrac{-\,\phantom{\varepsilon_0}\alpha_1-\alpha_2\varepsilon_4+\alpha_3\varepsilon_1-\phantom{\varepsilon_0}\alpha_4-1}{5}\\ x_4&=\dfrac{-\,\alpha_1\varepsilon_2+\alpha_2\varepsilon_3-\alpha_3\varepsilon_2+\alpha_4\varepsilon_3-1}{5}\\ x_5&=\dfrac{\phantom{+}\,\alpha_1\varepsilon_3-\phantom{\varepsilon_0}\alpha_2-\phantom{\varepsilon_0}\alpha_3-\alpha_4\varepsilon_2-1}{5} \end{aligned}\right. } \end{align*}</span>
satisfies the equation <span class="math-container">$x^5+x^4-4x^3-3x^2+3x+1=0$</span></p>
<p><strong>References</strong></p>
<p>Dummit, D. S. "Solving Solvable Quintics."Math. Comput.57, 387-401, 1991.
<a href="https://www.ams.org/journals/mcom/1991-57-195/S0025-5718-1991-1079014-X/S0025-5718-1991-1079014-X.pdf" rel="nofollow noreferrer">https://www.ams.org/journals/mcom/1991-57-195/S0025-5718-1991-1079014-X/S0025-5718-1991-1079014-X.pdf</a></p>
|
137,571 | <p>As the title, if I have a list:</p>
<pre><code>{"", "", "", "2$70", ""}
</code></pre>
<p>I will expect:</p>
<pre><code>{"", "", "", "2$70", "2$70"}
</code></pre>
<p>If I have</p>
<pre><code>{"", "", "", "3$71", "", "2$72", ""}
</code></pre>
<p>then:</p>
<pre><code>{"", "", "", "3$71", "3$71", "2$72", "2$72"}
</code></pre>
<p>And </p>
<pre><code>{"", "", "", "3$71", "","", "2$72", ""}
</code></pre>
<p>should give </p>
<pre><code>{"", "", "", "3$71", "3$71", "", "2$72", "2$72"}
</code></pre>
<p>This is my try:</p>
<pre><code>{"", "", "", "2$70", ""} /. {p : Except["", String], ""} :> {p, p}
</code></pre>
<p>But I don't know why it doesn't work. Poor ability of pattern match. Can anybody give some advice?</p>
| kglr | 125 | <p><strong>Update:</strong></p>
<pre><code>foo = # //. {a___, p : Except["", _String], Longest[b : "" ..], c___} :>
{a, p, p, ## & @@ ConstantArray["☺", Length[{b}] - 1], c} /. "☺" -> "" &;
{"", "", "", "3$71", "", "", "2$72", "", "", "", ""} // foo
</code></pre>
<blockquote>
<p><code>{"", "", "", "3$71", "3$71", "", "2$72", "2$72", "", "", ""}</code></p>
</blockquote>
<p><strong>Previous post:</strong></p>
<pre><code>rule = {a___, p : Except["", _String], Longest["" ..], b___} :> {a, p, p, b};
{"", "", "", "3$71", "", "", "2$72", "", "", "", ""} //. rule
</code></pre>
<blockquote>
<p><code>{"", "", "", "3$71", "3$71", "2$72", "2$72"}</code></p>
</blockquote>
|
51,096 | <p>Is it possible to have a countable infinite number of countable infinite sets such that no two sets share an element and their union is the positive integers?</p>
| grok_it | 14,348 | <p>For each integer $n$ put it in $A_i$ if $i$ is the smallest integer such that $n-i$ is a square. Or you can replace square with any even-degree polynomial with integer coefficients for a whole family. Or you can say n-i must be a prime.</p>
<p>Also:
$A_i$ = all integers with exactly $i$ prime-factors, counting multiplicity.</p>
|
184,361 | <p>I'm doing some exercises on Apostol's calculus, on the floor function. Now, he doesn't give an explicit definition of $[x]$, so I'm going with this one:</p>
<blockquote>
<p><strong>DEFINITION</strong> Given $x\in \Bbb R$, the integer part of $x$ is the unique $z\in \Bbb Z$ such that $$z\leq x < z+1$$ and we denote it by $[x]$.</p>
</blockquote>
<p>Now he asks to prove some basic things about it, such as: if $n\in \Bbb Z$, then $[x+n]=[x]+n$</p>
<p>So I proved it like this: Let $z=[x+n]$ and $z'=[x]$. Then we have that</p>
<p>$$z\leq x+n<z+1$$</p>
<p>$$z'\leq x<z'+1$$</p>
<p>Then $$z'+n\leq x+n<z'+n+1$$</p>
<p>But since $z'$ is an integer, so is $z'+n$. Since $z$ is unique, it must be that $z'+n=z$.</p>
<p>However, this doesn't seem to get me anywhere to prove that
$$\left[ {2x} \right] = \left[ x \right] + \left[ {x + \frac{1}{2}} \right]$$</p>
<p>in and in general that </p>
<p>$$\left[ {nx} \right] = \sum\limits_{k = 0}^{n - 1} {\left[ {x + \frac{k}{n}} \right]} $$</p>
<p>Obviously one could do an informal proof thinking about "the carries", but that's not the idea, let alone how tedious it would be. Maybe there is some easier or clearer characterization of $[x]$ in terms of $x$ to work this out.</p>
<p>Another property is
$$[-x]=\begin{cases}-[x]\text{ ; if }x\in \Bbb Z \cr-[x]-1 \text{ ; otherwise}\end{cases}$$</p>
<p>I argue: if $x\in\Bbb Z$, it is clear $[x]=x$. Then $-[x]=-x$, and $-[x]\in \Bbb Z$ so $[-[x]]=-[x]=[-x]$. For the other, I guess one could say:</p>
<p>$$n \leqslant x < n + 1 \Rightarrow - n - 1 < x \leqslant -n$$</p>
<p>and since $x$ is not an integer, this should be the same as $$ - n - 1 \leqslant -x < -n$$</p>
<p>$$ - n - 1 \leqslant -x < (-n-1)+1$$</p>
<p>So $[-x]=-[x]-1$</p>
| Aryabhata | 1,102 | <p>It is enough to prove it for $0 < x < 1$.</p>
<p>Now, let $M$ be an integer such that, $\frac{M}{n} \le x < \frac{M+1}{n}$ where $0 \le M < n$</p>
<p>Thus $[nx] = M$.</p>
<p>For $0 \le k \le n-M-1$, we have that $[x+\frac{k}{n}] = 0$. </p>
<p>For $n-1 \ge k > n-M-1$ we have that $[x + \frac{k}{n}] = 1$. </p>
<p>The result follows.</p>
<p>I don't think induction can be used here.</p>
|
1,906,146 | <p>Can the following expression be further simplified: $$a^{(\log_ab)^2}?$$</p>
<p>I know for example that $$a^{\log_ab^2}=b^2.$$</p>
| Evariste | 239,682 | <p>$$a^{(\log_ab)^2}=a^{\log_a(b)\times \log_a(b)}=b^{\log_a(b)}$$ </p>
|
2,253,645 | <p>$(1,2)$ intersection $(2,3)=\{2\}$</p>
<p>$(1,2)$ intersection $[2,3]=\{2\}$</p>
<p>$\{1,2\}$ intersection $[1,2]=[1,2]$</p>
<p>$\{1,2\}$ union $[1,2]=[1,2]$</p>
<p>$\{1,2\}$ intersection $(1,3)$ intersection $[1,3)=(1,3)$</p>
<p>$\{1,2\}$ union $(1,3)$ union $[1,3)=(1,3)$</p>
<p>Is my answer correct? I find my self confused when I have sets $\{a,b\}$ and intervals like $(a,b)$ and $[a,b]$ to find intersection and union of them.</p>
| rych | 73,934 | <p>Kincaid D., Cheney W. <em>Numerical analysis</em>, Chapter 6:</p>
<p>A <em>natural spline</em> of degree $2m+1$ is a function $s\in C^{2m}(\mathbb{R})$
that reduces to a polynomial of degree $\leq2m+1$ in each inner interval
and to a polynomial of degree <strong>at most $m$</strong> in $(-\infty,t_{1})$ and
$(t_{n},\infty)$.</p>
<p><strong>Theorem.</strong> Every natural spline can be represented using truncated power functions
as follows
$$
s(x)=\sum_{k=0}^{m}\alpha_{k}x^{k}+\sum_{i=1}^{n}\beta_{i}(x-x_{i})_{+}^{2m+1}
$$
with the coefficient conditions,
$$
\sum_{i=1}^{n}\beta_{i}x_{i}^{k}=0,\qquad k=0,1,\ldots m
$$</p>
|
756,735 | <blockquote>
<p>Let $n>0$ be a positive integer. For all $x\not=0$, prove that $f(x) = 1/x^n$ is differentiable at $x$ with $f^\prime(x) = -n/x^{n+1}$ by showing that the limit of the difference quotient exists.</p>
</blockquote>
<p>I am having trouble seeing how I can manipulate the difference quotient in order to get a limit that exists</p>
<p>so far I have</p>
<p>$$f^\prime(x)= \lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h} =\lim_{h\rightarrow 0} \frac{1/(x+h)^n - 1/x^n}{h}$$</p>
<p>All help is much appreciated.</p>
| Slade | 33,433 | <p>Collect fractions and use the binomial theorem.</p>
|
2,688,829 | <p>I was wondering how to find the growth rate of the function defined by the number of ways to partition $2^n$ as powers of 2. After a search through OEIS I came across <a href="https://oeis.org/A002577" rel="nofollow noreferrer">OEIS A002577</a> which is what I'm looking for. I can't seem to find any link to asymptotics for this function. Could someone help?</p>
| K B Dave | 534,616 | <p>Maybe this is what you're looking for: let $b(n)$ be the number of partitions of $2^n$ into powers of $2$, and assume that $2^{n+1}$ is an integer. Then de Bruijn (<a href="http://www.dwc.knaw.nl/DL/publications/PU00018536.pdf" rel="nofollow noreferrer">1948</a>) showed that there exists a function $\psi$ of period $1$ such that
$$\begin{split}\ln b(n+1)&=\tfrac{1}{2\ln 2}(n\ln 2-\ln n-\ln\ln 2)^2\\
&\quad+\left(\tfrac{1}{2}+\tfrac{1}{\ln 2}+\tfrac{\ln \ln 2}{\ln 2}\right)n\ln 2\\
&\quad-\left(1+\tfrac{\ln\ln 2}{\ln 2}\right)(\ln n+\ln\ln 2)\\
&\quad+ \psi\left(\tfrac{n\ln 2-\ln\ln n -\ln\ln2 }{\ln 2}\right)+ o(1)\text{.}\end{split}$$</p>
|
3,357,502 | <p>I'm aware that <span class="math-container">$D(n)$</span> can be calculated in O(sqrt(n)) time. Can <span class="math-container">$ D(n, k) $</span> also be calculated in O(sqrt(n)) time? What's the best algorithm?</p>
<p>For example, if <span class="math-container">$n = 8$</span> and <span class="math-container">$k = 3$</span>, then <span class="math-container">$ D(n, k) = \lfloor{\frac{8}{1}}\rfloor + \lfloor{\frac{8}{2}}\rfloor + \lfloor{\frac{8}{3}}\rfloor $</span></p>
| Vo Hoang Anh | 960,369 | <p>Yes you can, I also have a <a href="https://math.stackexchange.com/questions/4230187/faster-algorithm-for-counting-non-negative-tripplea-b-c-satisfied-a-b-c">problem</a> that require this function to be calculated fast enough.</p>
<p>You can notice that there is <span class="math-container">$O(\sqrt{n})$</span> unique values in the set S = {<span class="math-container">$\lfloor \frac{n}{1} \rfloor, \lfloor \frac{n}{2} \rfloor, \lfloor \frac{n}{3} \rfloor, \dots, \lfloor \frac{n}{n - 1} \rfloor, \lfloor \frac{n}{n} \rfloor$</span>}. Therefore you can calculate the function in <span class="math-container">$O(\sqrt{n})$</span>, even with partial divisor summatory function.</p>
<p>Even more complex but faster: using the method that Richard Sladkey described in <a href="https://arxiv.org/abs/1206.3369" rel="nofollow noreferrer">this paper</a> you can calculate the function in <span class="math-container">$O(n^{1/3})$</span></p>
|
2,188,965 | <p>Can someone explain to me how this step done? I got a different answer than what the solution said.</p>
<p>Simplify $x(y+z)(\bar{x} + y)(\bar{y} + x + z)$</p>
<p>what the solution got </p>
<p>$x(y+z)(\bar{x} + y)(\bar{y} + x + z)$ = $x(y + z\bar{x})(\bar{y} + x + z)$ (Using distrubitive)</p>
<p>What I got</p>
<p>$x(y+z)(\bar{x} + y)(\bar{y} + x + z)$ = $x(y\bar{x} + y + z\bar{x} + zy)(\bar{y} + x + z)$ (Using distrubitive)</p>
| Kanwaljit Singh | 401,635 | <p>$y\bar{x} + y + z\bar{x} + zy$</p>
<p>= $y(\bar{x} + 1) + z\bar{x} + zy$</p>
<p>= $y + z\bar{x} + zy$</p>
<p>= $y(1 + z) + z\bar{x}$</p>
<p>= $y + z\bar{x}$</p>
<p><strong>Direct rule -</strong></p>
<p>X + YZ = (X+Y)(X+Z)</p>
<p>So we have -</p>
<p>$(y + z)(y + \bar x)$</p>
<p>= $(y + z \bar x)$</p>
|
195,150 | <p>Of all the possible combinations of positive numbers that sum to 10, which has the largest multiplication?</p>
<p>I had also got a clue: it's related to <a href="http://en.wikipedia.org/wiki/E_%28mathematical_constant%29" rel="nofollow"><code>e</code></a>.</p>
<p>Please help! (I need explanation aswell)</p>
<p><strong>Notice</strong>: i said positive not natural so you can use fractions.</p>
| kahen | 1,269 | <p>Suppose we have $a_1$, $a_2$, $\dotsc$, $a_n$ all positive and summing to $10$. Then by the <a href="http://en.wikipedia.org/wiki/AM-GM_inequality" rel="nofollow">AM–GM inequality</a> their product is maximized when $a_1 = a_2 = \dotsb = a_n$. Since for $n \geq 10$ you'd have a product of numbers less than or equal to $1$, you only have to compute $(10/n)^n$ for $1 \leq n \leq 9$ and you find that the maximum occurs for $n=4$ with $2.5^4 = 39.0625$.</p>
|
195,150 | <p>Of all the possible combinations of positive numbers that sum to 10, which has the largest multiplication?</p>
<p>I had also got a clue: it's related to <a href="http://en.wikipedia.org/wiki/E_%28mathematical_constant%29" rel="nofollow"><code>e</code></a>.</p>
<p>Please help! (I need explanation aswell)</p>
<p><strong>Notice</strong>: i said positive not natural so you can use fractions.</p>
| Dan Barzilay | 40,220 | <p>You were all wrong actually... i checked with my teacher and the asnwer is <code>e</code> 3.67879441 (10/e) times so: <code>e^(10/e) = 39.5986256</code></p>
|
1,431,289 | <p>Find the average rate of change of $2x^3 - 5x$ on the interval $[1,3]$.</p>
<p>I'm really confused about this problem. I keep ending up with the answer $12$, but the answer key says otherwise. Someone please help! Thanks!</p>
| MegaboofMD | 269,606 | <p>average rate of change is the difference between the function values at two points divided by the distance between the points. $$ = \frac{f(3) - f(1) }{3-1}$$ $$ = \frac{[2*(3^3)-5*(3)] -[2*(1^3)-5*(1)]}{2}$$ $$ = \frac{[54-15]-[2-5] }{2}$$ $$ = \frac{39-(-3) }{2}$$ $$ = \frac{42 }{2}$$ $$ = \frac{21} 1$$</p>
|
4,202,490 | <p>Trying to construct an example for a Business Calculus class (meaning trig functions are not necessary for the curriculum). However, I want to touch on the limit problem involved with the <span class="math-container">$\sin(1/x)$</span> function.</p>
<p>I am sure there is a simple function, or there isn't... But would love some insight.</p>
<p>I also understand that the functions that satisfy this condition are maybe way outside the scope of the course. I'm just looking for different "flavors" of showing limits that don't exist besides just showing the limit from the left and the limit from the right does not exists.</p>
| Ethan Bolker | 72,858 | <p>As you note, this is really outside the scope of a business calculus syllabus. I might argue that anything more than a very informal discussion of limits is too.</p>
<p>In any case I think your business calculus students could profit from understanding that functions need not come from formulas. You can convey lots of the meaning and usefulness of calculus just with sketches of graphs. For this example you could sketch the graph near the origin at large magnification to show the infinitely many oscillations. If you draw the oscillations between the lines <span class="math-container">$y = \pm x$</span> you can get continuity. Between <span class="math-container">$y = \pm x^2$</span> you get differentiablity too.</p>
|
1,195,092 | <p>The book I am using for my Introduction of Topology course is Principles of Topology by Fred H. Croom.</p>
<p>I was given the following problem: </p>
<blockquote>
<p>Prove that a set $E$ is countable if and only if there is a surjection from $\mathbb{N}$ to $E$. </p>
</blockquote>
<p>I have a rough idea on how to prove this. Though the converse is tricky and I am having a hard time coming up with a proof to justify my claim. </p>
<blockquote>
<blockquote>
<p>$P \rightarrow Q$: Let $E$ be a nonempty countable set. By definition of a countable set, this implies there is a bijection between $\mathbb{N}$ and $E$. Therefore there is a surjection from $\mathbb{N}$ to $E$. </p>
<p>$Q \rightarrow P$: Let there be a surjection from $\mathbb{N}$ to $E$. Thus we have a function $f$ such that $f(\mathbb{N})=E$. Seeking to prove $E$ is countable, we have two cases: if $E$ is finite or if $E$ is infinite. </p>
</blockquote>
</blockquote>
<p>This is where I am stuck. How would I take the two cases of a finite or infinite $E$? Would I need to find (create) such a function $f$ that makes the bijection from $E$ to $\mathbb{N}$, making $E$ countable? Am I on the right track? Any suggestions?</p>
<hr>
<p>Sorry for the rather long question. If is it rather confusing, let me know so I can clarify. I sincerely thank you for taking the time to read this question. I greatly appreciate any assistance you may provide.</p>
<p><em>As a disclaimer, I put the General-Topology tag because this question was given in a topology course.</em> </p>
| owen88 | 12,981 | <p>Here is an alternative method of solution. In the following I will write $A\, \text{b.} B$ to denote the event that $A$ occurs before $B$. So you want to calculate $\textbf{P}[ 6\, \text{b.}\, 55]$.</p>
<p>Instead we work with $\textbf{P}[55 \, \text{b.} \, 6]$, and condition on the event $5\, \text{b.}\, 6$. So
\begin{align}
\textbf{P}[55 \, \text{b.} \, 6] & = \textbf{P}[55 \, \text{b.} \, 6\, | 5\, \text{b.}\, 6]\, \textbf{P}[5 \, \text{b.} \, 6]
\\
& = \frac12 \textbf{P}[55 \, \text{b.} \, 6\, | 5\, \text{b.}\, 6].
\end{align}
To evaluate the remaining probability, we condition a second time: this time conditioning on what the roll immediately after the $5$ is.
\begin{align}
\textbf{P}[55 \, \text{b.} \, 6\, | 5\, \text{b.}\, 6] & = \sum_{j=1}^6 \textbf{P}[55 \, \text{b.} \, 6\, | 5\, \text{b.}\, 6 \text{ and the next roll is $j$} ] \textbf{P}[ \text{next roll is $j$}] \\
& = \frac16 \sum_{j=1}^6 \textbf{P}[55 \, \text{b.} \, 6\, | 5\, \text{b.}\, 6 \text{ and the next roll is $j$} ] \\
& = \frac16 \textbf{P}[55 \, \text{b.} \, 6\, | 5\, \text{b.}\, 6 \text{ and the next roll is $6$} ] \\
& \qquad \qquad + \frac16 \textbf{P}[55 \, \text{b.} \, 6\, | 5\, \text{b.}\, 6 \text{ and the next roll is $5$} ] \\
& \qquad \qquad + \frac46 \textbf{P}[55 \, \text{b.} \, 6\, | 5\, \text{b.}\, 6 \text{ and the next roll is neither $5$ or $6$} ] \\
&= \Big( \frac{1}{6} \times 0 \Big)+ \Big( \frac{1}{6} \times 1 \Big) +\frac46 \textbf{P}[55 \, \text{b.} \, 6\, | 5\, \text{b.}\, 6 \text{ and the next roll is neither $5$ or $6$} ] \\
& = \frac16 + \frac46 \textbf{P}[55 \, \text{b.} \, 6],
\end{align}
in the final lines we used that
$$\textbf{P}[55 \, \text{b.} \, 6\, | 5\, \text{b.}\, 6 \text{ and the next roll is neither $5$ or $6$} ] = \textbf{P}[55 \, \text{b.} \, 6],$$
which is true since the information that we saw a $5$ but it wasn't followed by a $5$ doesn't tell us anything about the event $55 \, \text{b.} \, 6$. So substituting this into the first equation we have
\begin{align}
\textbf{P}[55 \, \text{b.} \, 6] & = \frac12 \Big( \frac16 + \frac46 \textbf{P}[55 \, \text{b.} \, 6] \Big) \\
& = \frac{1}{12} + \frac{4}{12}\textbf{P}[55 \, \text{b.} \, 6].
\end{align}
Rearranging gives: $\textbf{P}[55 \, \text{b.} \, 6] = \frac18$, and hence
$$\textbf{P}[6 \, \text{b.} \, 55] = \frac78.$$ </p>
<p><strong>Remark.</strong> In general this type of problem is more easily phrased in the context of a Markov chain, for which the probability you are wanting becomes a hitting probability. The proof above is essentially exactly the same as would be done using Markov processes, the exception being that the notation is simplified in that context.</p>
|
471,145 | <p>I'm reading Proofs from the Book, and I ran into following theorem:</p>
<p>Suppose all roots of polynomial $x^n + a_{n-1}x^{n-1} + \dots + a_0$ are real. Then the roots are contained in the interval:</p>
<p>$$ - \frac{a_{n-1}}{n} \pm \frac{n-1}{n}
\sqrt{a_{n-1}^2 - \frac{2n}{n-1} a_{n-2} } $$</p>
<p>So, if you know that all the roots of polynomial are real, you can get an interval that contains them just looking at the first two coefficients.</p>
<p>I'm interested in other theorems/tricks that let you figure out interesting things about a polynomial just by ''eyeing'' it. Especially if they are surprising!</p>
| njguliyev | 90,209 | <p>Yes.</p>
<p>Hint: Otherwise you could extend the function to larger domain using the compactness of the circle.</p>
|
3,953,674 | <p>Here is a common argument used to prove that the sum of an infinite geometric series is <span class="math-container">$\frac{a}{1-r}$</span> (where <span class="math-container">$a$</span> is the first term and <span class="math-container">$r$</span> is the common ratio):
<span class="math-container">\begin{align}
S &= a+ar+ar^2+ar^3+\cdots \\
rS &= ar+ar^2+ar^3+ar^4+\cdots \\
S-rS &= a \\
S(1-r) &= a \\
S &= \frac{a}{1-r} \, .
\end{align}</span>
I am sceptical about the validity of this argument. It feels like there is something amiss about a proof involving infinite series that makes no mention of the fact that they are typically defined as the limit of their partial sums. The third line involves 'cancelling' all of the terms other than <span class="math-container">$a$</span>. This makes it seem like an infinite series is actually an infinite string of symbols, rather than a limiting expression. If <span class="math-container">$a+ar+ar^2+ar^3+\cdots$</span> is simply a shorthand for
<span class="math-container">$$
\lim_{n \to \infty}\sum_{k=0}^{n}ar^k \, ,
$$</span>
and <span class="math-container">$ar+ar^2+ar^3+ar^4+\cdots$</span> a shorthand for
<span class="math-container">$$
\lim_{n \to \infty}\sum_{k=1}^{n}ar^k \, ,
$$</span>
then I am struggling to see how the terms actually cancel. Perhaps the third line can be written more formally as
<span class="math-container">\begin{align}
S-rS &= (a+ar+ar^2+ar^3+\cdots)-(ar+ar^2+ar^3+ar^4+\cdots) \\
&= \lim_{n \to \infty}\sum_{k=0}^{n}ar^k - \lim_{n \to \infty}\sum_{k=1}^{n}ar^k \\
&= \lim_{n \to \infty}\left(\sum_{k=0}^{n}ar^k - \sum_{k=1}^{n}ar^k\right) \\
&= \lim_{n \to \infty}\left(a +\sum_{k=1}^nar^k - \sum_{k=1}^{n}ar^k\right) \\
&= \lim_{n \to \infty}a \\
&= a \, .
\end{align}</span>
There is another concern I have. Geometric series only converge when <span class="math-container">$|r|<1$</span>. However, the argument used above seems to apply regardless of whether <span class="math-container">$|r|<1$</span>, which would yield nonsensical results such as
<span class="math-container">$$
1+2+4+8+\cdots = -1 \, .
$$</span>
So is the argument rigorous, and if so, why are my fears misplaced?</p>
| Community | -1 | <p>Let</p>
<p><span class="math-container">$$S_n=\sum_{k=0}^n ar^n.$$</span></p>
<p>Then</p>
<p><span class="math-container">$$(1-r)S_n=S_n-rS_n=a-ar^{n+1}$$</span> and</p>
<p><span class="math-container">$$S_n=a\frac{1-r^{n+1}}{1-r}.$$</span></p>
<p>This is absolutely rigorous for any <span class="math-container">$r\ne1$</span>.</p>
<p>And from this, for <span class="math-container">$|r|<1$</span>,</p>
<p><span class="math-container">$$S=\lim_{n\to\infty}S_n=\frac a{1-r}.$$</span></p>
<p>Now, for <span class="math-container">$|r|\ge1$</span>, the infinite sum diverges as does the limit of <span class="math-container">$r^{n+1}$</span>, so there is no paradox.</p>
|
127,412 | <p>How to take a 3 random given name?</p>
<p>I tried:</p>
<p><a href="https://i.stack.imgur.com/BUYIT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BUYIT.png" alt="enter image description here"></a></p>
| bill s | 1,783 | <p>One approach is to use RandomChoice[ ] to get your random names:</p>
<pre><code>list = EntityList["GivenName"];
RandomChoice[list, 3]
</code></pre>
<p><a href="https://i.stack.imgur.com/VaNss.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VaNss.png" alt="enter image description here"></a></p>
|
69,208 | <p>Consider $f:\{1,\dots,n\} \to \{1,\dots,m\}$ with $m > n$.
Let $\operatorname{Im}(f) = \{f(x)|x \in \{1,\dots,n\}\}$.</p>
<p>a.)
What is the probability that a random function will be a bijection when viewed as
$$f':\{1,\dots,n\} \to \operatorname{Im}(f)?$$</p>
<p>b.)
How many different function f are there for which $$\sum_{i=1}^n f(i)=k?$$
Not sure where to start and have no idea what $\operatorname{Im}(f)$ means. Please help me get started on this.</p>
| Brian M. Scott | 12,042 | <p>Your algebra seems to be off:</p>
<p>$$\sin^2x + \cos^2 x \approx x^2 + \left(1-\dfrac{x^2}2\right)^2 = x^2 + 1 - x^2 + \dfrac{x^4}4 = 1 + \dfrac{x^4}4,$$ which is approximately $1$ for small values of $x$.</p>
<p><strong>Added:</strong> Note that you can’t expect to get exactly $1$, since you’re using an inexact approximation. If, as anon suggests, you were setting this equal to $1$ and solving for $x$, you weren’t vitiating the result: you were just showing that the approximation is perfect only at $x=0$.</p>
|
1,530,406 | <p>How to multiply Roman numerals? I need an algorithm of multiplication of numbers written in Roman numbers. Help me please. </p>
| 3SAT | 203,577 | <p>Make a table with two columns, and enter the two numbers to be multiplied into the first row.
Make the next row by halving the first number (discarding remainders) and doubling the second. Continue until there is nothing left to halve.
Cross out all the rows where the left number is even.
Add the remaining numbers in the second column. The result is the product of the first two numbers</p>
<p><strong>Examples</strong></p>
<p><a href="https://i.stack.imgur.com/9XF7k.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9XF7k.png" alt="enter image description here"></a></p>
<p><a href="https://rbutterworth.nfshost.com/Tables/romanmult" rel="nofollow noreferrer">Source</a></p>
|
1,903,333 | <p>Let $G$ be a group. Prove that $Z(G)$ (the center of $G$) is always nonempty.</p>
<p>Can anyone give me solution of this theoretical problem? I have just started learning group theory and I am very interested in this math branch</p>
| absolute friend | 244,073 | <p>Suppose $e$ be identity element of group $G$ then </p>
<p>$$ex=xe~~~~~~~~\forall x\in G$$</p>
<p>$$\implies e\in Z(G)\implies Z(G)\neq \emptyset$$</p>
|
3,660,101 | <p>I want to determine if the series <span class="math-container">$ \sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}}{\left(-1\right)^{n}+n} $</span> converge/diverge. the sequence in the denominator is not monotinic, so I cant use Dirichlet's or Abel's tests. My intuition is that this series converge, becuase its looks close to <span class="math-container">$ \sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}}{n} $</span> but im not sure how to prove. Any ideas will help, thanks.</p>
| Brian M. Scott | 12,042 | <p>Let </p>
<p><span class="math-container">$$s_n=\sum_{k=2}^n\frac{(-1)^k}{(-1)^k+k}=\frac13-\frac12+\frac15-\frac14+\ldots+\frac{(-1)^n}{(-1)^n+n}$$</span> </p>
<p>and </p>
<p><span class="math-container">$$s_n'=\sum_{k=2}^n\frac{(-1)^{k+1}}k=-\frac12+\frac13-\frac14+\frac15+\ldots+\frac{(-1)^{n+1}}n\;.$$</span></p>
<p>Show that <span class="math-container">$s_{2n+1}=s_{2n+1}'$</span> and <span class="math-container">$s_{2n}=s_{2n+1}'+\frac1{2n}$</span> for <span class="math-container">$n\ge 1$</span>. Use this or the fact that <span class="math-container">$s_{2n}=s_{2n}'+\frac1{2n}+\frac1{2n+1}$</span> to show that <span class="math-container">$\lim_\limits{n\to\infty}|s_n-s_n'|=0$</span>, and therefore <span class="math-container">$\lim_\limits{n\to\infty}s_n=\lim_\limits{n\to\infty}s_n'$</span>.</p>
|
3,660,101 | <p>I want to determine if the series <span class="math-container">$ \sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}}{\left(-1\right)^{n}+n} $</span> converge/diverge. the sequence in the denominator is not monotinic, so I cant use Dirichlet's or Abel's tests. My intuition is that this series converge, becuase its looks close to <span class="math-container">$ \sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}}{n} $</span> but im not sure how to prove. Any ideas will help, thanks.</p>
| CHAMSI | 758,100 | <p>Let <span class="math-container">$ n $</span> be a positive integer.</p>
<p><span class="math-container">\begin{aligned}\frac{\left(-1\right)^{n}}{n+\left(-1\right)^{n}}&=\frac{\left(-1\right)^{n}}{n}\left(\frac{n}{n+\left(-1\right)^{n}}\right)\\ &=\frac{\left(-1\right)^{n}}{n}\left(1-\frac{\left(-1\right)^{n}}{n+\left(-1\right)^{n}}\right)\\ &=\frac{\left(-1\right)^{n}}{n}+v_{n}\end{aligned}</span></p>
<p>Where <span class="math-container">$ v_{n}=-\frac{1}{n^{2}+n\left(-1\right)^{n}}=\underset{\overset{n\to +\infty}{}}{\mathcal{O}}\left(\frac{1}{n^{2}}\right) \cdot $</span></p>
<p>Since <span class="math-container">$ \sum\limits_{n\geq 1}{\frac{\left(-1\right)^{n}}{n}} $</span> converges, and <span class="math-container">$ \sum\limits_{n\geq 1}{v_{n}} $</span> converges by comparaison, we get that <span class="math-container">$ \sum\limits_{n\geq 1}{\frac{\left(-1\right)^{n}}{n+\left(-1\right)^{n}}} $</span> converges.</p>
|
93,458 | <blockquote>
<p>Let <span class="math-container">$n$</span> be a nonnegative integer. Show that <span class="math-container">$\lfloor (2+\sqrt{3})^n \rfloor $</span> is odd and that <span class="math-container">$2^{n+1}$</span> divides <span class="math-container">$\lfloor (1+\sqrt{3})^{2n} \rfloor+1 $</span>.</p>
</blockquote>
<p>My attempt:</p>
<p><span class="math-container">$$ u_{n}=(2+\sqrt{3})^n+(2-\sqrt{3})^n=\sum_{k=0}^n{n \choose k}2^{n-k}(3^{k/2}+(-1)^k3^{k/2})\in\mathbb{2N} $$</span></p>
<p><span class="math-container">$$ 0\leq (2-\sqrt{3})^n \leq1$$</span></p>
<p><span class="math-container">$$ (2+\sqrt{3})^n\leq u_{n}\leq 1+(2+\sqrt{3})^n $$</span></p>
<p><span class="math-container">$$ (2+\sqrt{3})^n-1\leq u_{n}-1\leq (2+\sqrt{3})^n $$</span></p>
<p><span class="math-container">$$ \lfloor (2+\sqrt{3})^n \rfloor=u_{n}-1\in\mathbb{2N}+1 $$</span></p>
| José Carlos Santos | 446,262 | <p>It is not hard to prove that<span class="math-container">$$(\forall n\in\mathbb N):\left(2+\sqrt3\right)^n+\left(2-\sqrt3\right)^n\in2\mathbb N.$$</span>This, together with the fact that <span class="math-container">$2-\sqrt3\in(0,1),$</span> is enough to prove that <span class="math-container">$\left\lfloor\left(2+\sqrt3\right)^n\right\rfloor$</span> is an odd integer.</p>
|
1,018,270 | <p>While I was studying about finite differences I came across an article that says "computers can't deal with limit of $\Delta x \to 0$ " in <a href="http://1drv.ms/1sB5P1B" rel="nofollow">finite differences</a>.But if computers can't deal with these equations does anybody know how they compute $ \frac {d}{dx}$ of $x^2$ and other such equations.Or whether these equations are pre-written.</p>
| Community | -1 | <p>A <a href="http://en.wikipedia.org/wiki/Differentiation_rules" rel="nofollow">table of the derivatives of primitive functions combined with differentiation rules</a> yields an algorithm that allows a computer program to <em>symbolically</em> compute the derivative of any function that is a compound of primitive functions without having to rely on the limit definition at all (computer algebra).</p>
<p>In fact, unlike symbolic integration, this algorithm is fairly easy to implement.</p>
<p><em><strong>Edit:</strong> It can only be re-emphasised that symbolic <strong>integration</strong> is a different matter entirely. In fact, while every function that is comprised of elementary functions has an elementary derivative, the same is not true for the antiderivative. This is a consequence of <a href="http://en.wikipedia.org/wiki/Liouville%27s_theorem_(differential_algebra)" rel="nofollow">Liouville's theorem</a> and finds application in the <a href="http://en.wikipedia.org/wiki/Risch_algorithm" rel="nofollow">Risch "algorithm"</a> (which is not an algorithm in the strict sense).</em></p>
|
2,790,025 | <p><strong>GIVEN:</strong></p>
<ul>
<li>$g$ is differentiable with continuous derivative on $[a,b]$.</li>
</ul>
<p><strong>WANT TO SHOW</strong></p>
<p>$$|g(x)-g(y)| \leq \bigg(\max_{x\in [a,b]} |g'(x)|\bigg)|x-y|.$$</p>
<p><strong>USING MEAN VALUE THEOREM</strong></p>
<p>Given $x<y$ in the interval $(a,b)$, the mean value theorem states that there exists a point $\lambda \in (a,b)$ for which</p>
<p>$$ \frac{g(x)-g(y)}{x-y} = g'(\lambda)$$</p>
<p>Since $g'$ is continuous on $[a,b]$, we know that it is bounded. Thus there
exists $M>0$ such that</p>
<p>$$g'(x) \leq M \space \space \forall x \in (a,b)$$ and in particular, $|g'(\lambda)| \leq M$. Hence,</p>
<p>$$\frac{|g(x)-g(y)|}{|x-y|} \leq M$$</p>
<p>$$\implies |g(x)-g(y)| \leq M|x-y|.$$</p>
<p>By the Bolzano -Weistrass Theorem there exists (since $g' \in C([a,b])$)</p>
<p>$$ M = \max_{x\in [a,b]} |g'(x)|.$$</p>
<p>Therefore, </p>
<p>$$|g(x)-g(y)| \leq \bigg(\max_{x\in [a,b]} |g'(x)|\bigg)|x-y|.$$</p>
<p>Want to prove without using mean value theorem as it is circle work.</p>
| Intelligenti pauca | 255,730 | <p>If $\angle OAD=\alpha$, then $\angle FOC=\alpha/2$ and:
$$
AO-AE=r(\csc\alpha-\cot\alpha)=r{1-\cos\alpha\over\sin\alpha}=
r\tan{\alpha\over2}=FC,
$$
where $r=OE=OF$ is the radius of the inscribed circle. In the same way one derives $BO-BF=ED$ and adding these two equalities we get the desired result.</p>
<p><a href="https://i.stack.imgur.com/3B4lj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3B4lj.png" alt="enter image description here"></a></p>
|
2,790,025 | <p><strong>GIVEN:</strong></p>
<ul>
<li>$g$ is differentiable with continuous derivative on $[a,b]$.</li>
</ul>
<p><strong>WANT TO SHOW</strong></p>
<p>$$|g(x)-g(y)| \leq \bigg(\max_{x\in [a,b]} |g'(x)|\bigg)|x-y|.$$</p>
<p><strong>USING MEAN VALUE THEOREM</strong></p>
<p>Given $x<y$ in the interval $(a,b)$, the mean value theorem states that there exists a point $\lambda \in (a,b)$ for which</p>
<p>$$ \frac{g(x)-g(y)}{x-y} = g'(\lambda)$$</p>
<p>Since $g'$ is continuous on $[a,b]$, we know that it is bounded. Thus there
exists $M>0$ such that</p>
<p>$$g'(x) \leq M \space \space \forall x \in (a,b)$$ and in particular, $|g'(\lambda)| \leq M$. Hence,</p>
<p>$$\frac{|g(x)-g(y)|}{|x-y|} \leq M$$</p>
<p>$$\implies |g(x)-g(y)| \leq M|x-y|.$$</p>
<p>By the Bolzano -Weistrass Theorem there exists (since $g' \in C([a,b])$)</p>
<p>$$ M = \max_{x\in [a,b]} |g'(x)|.$$</p>
<p>Therefore, </p>
<p>$$|g(x)-g(y)| \leq \bigg(\max_{x\in [a,b]} |g'(x)|\bigg)|x-y|.$$</p>
<p>Want to prove without using mean value theorem as it is circle work.</p>
| Blue | 409 | <p>The angle chase ---which relies upon the fact that opposite angles of an inscribed quadrilateral are supplementary--- gives us this figure:</p>
<p><a href="https://i.stack.imgur.com/ugz0i.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ugz0i.png" alt="enter image description here"></a></p>
<p>The result then follows from the identity illustrated in this trigonograph:</p>
<p><a href="https://i.stack.imgur.com/OnakW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OnakW.png" alt="enter image description here"></a></p>
<blockquote>
<p>$$\tan\theta + \cot 2\theta = \csc 2\theta$$</p>
</blockquote>
|
3,460 | <p>I asked the question "<a href="https://mathoverflow.net/questions/284824/averaging-2-omegan-over-a-region">Averaging $2^{\omega(n)}$ over a region</a>" because this is a necessary step in a research paper I am writing. The answer is detailed and does exactly what I need, and it would be convenient to directly cite the result. However, the author of the answer is anonymous... how would one deal with such a situation? I could of course very easily just reproduce the argument in my paper, but that would be academically dishonest.</p>
| Chris Godsil | 1,266 | <p>You attribute it to an anonymous author and give the link. So you are not claiming credit and your readers can check the source. </p>
<p>I would see no harm in reproducing the argument as well, especially if it’s short.</p>
|
2,928,849 | <p>I have a problem understanding the following:</p>
<p><span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are independent vatiables with
<span class="math-container">$$P(X = i) = P(Y = i) = \frac{1}{2^i}. \quad i = 1, 2, \cdots$$</span>
Now the book says<span class="math-container">$ P(X< i)= \dfrac{1}{2^i}$</span>. I really do not get that… Would be nice if you could help maybe.</p>
<p>Best<br>
KingDingeling</p>
| Travis | 568,040 | <p>I think it's a typo.</p>
<p><span class="math-container">$P(X > i) =1-\sum_{k=1}^{k=i}P(X =k)=1-\sum_{k=1}^{k=i}{P(X = k)}
= 1 - \sum_{k=1}^{k=i}{\frac{1}{2^k}}= \frac{1}{2^i}$</span></p>
|
167,904 | <p>In one the the answers to this thread " <a href="https://mathoverflow.net/questions/119850/">Can one embedd the projectivezed tangent space of CP^2 in a projective space? </a> " it was mentioned that " $\mathbb{P}(T\mathbb{P}^2)$ isomorphic to the variety of complete flags in the vector space $\mathbb{C^3}$ ". </p>
<p>I'm having a hard time understanding why this is true, and can't seem to find any references. </p>
| BlaCa | 14,514 | <p>Let $\pi:\mathbb{P}(T_{\mathbb{P}^2})\rightarrow\mathbb{P}^2$ be the projectivized tangent bundle. The point $x:=(p,[L])\in\mathbb{P}(T_{\mathbb{P}^2})$ corresponds to the point $p = \pi((p,[L]))\in\mathbb{P}^2$ and to the class of the line $L\subset\mathbb{P}^2$ passing through $p$. Now, the point $p$ is a line through the origin $V_p\subset\mathbb{C}^3$, the line $L_p$ corresponds to a plane $\Pi_p\subset\mathbb{C}^3$. Clearly, we have $V_p\subset\Pi_p\subset\mathbb{C}^3$. Now, the morphism
$$\phi:\mathbb{P}(T_{\mathbb{P}^2})\rightarrow F(1,2,3),\: x:=(p,[L])\mapsto (V_p,\Pi_p),$$
is an isomorphism because $\phi$ is injective, and $\mathbb{P}(T_{\mathbb{P}^2})$, $F(1,2,3)$ are both smooth.</p>
|
625,608 | <p>The question is to show that the function $\phi$ given by $\phi(\lambda)=\frac{\lambda}{1+|\lambda|}$ is 1-1 on the complex plane. I would be grateful for a hint on how to start.</p>
| Henry | 6,460 | <p>Hint:</p>
<p>If $\lambda = r e^{i \theta}$ and $\mu = s e^{i \psi}$ with $r=|\lambda |$ etc.,</p>
<p>then $\phi(\lambda) = \dfrac{r e^{i \theta}}{1+r} =\left(1-\frac{1}{1+r}\right) e^{i \theta}$ and $\phi(\mu) = \left(1-\frac{1}{1+s}\right) e^{i \psi}$</p>
|
111,899 | <p>Evaluate the integral using trigonometric substitutions. </p>
<p>$$\int{ x\over \sqrt{3-2x-x^2}} \,dx$$</p>
<p>I am familiar with using the right triangle diagram and theta, but I do not know which terms would go on the hypotenuse and sides in this case. If you can determine which numbers or $x$-values go on the hypotenuse, adjacent, and opposite sides, I can figure out the rest, although your final answer would help me check mine. Thanks!</p>
| Aru Ray | 13,129 | <p>$\int \frac{x}{\sqrt{4-(x+1)^2}}dx = \int \frac{2\sin\theta-1}{\sqrt{4-4\sin^2\theta}}(2\cos\theta)d\theta$
(using the substitution $x+1=2\sin\theta$)</p>
<p>$=\int\frac{2\sin\theta-1}{2\cos\theta}2\cos\theta d\theta$</p>
<p>$= \int (2\sin\theta-1) d\theta$</p>
<p>$=-2\cos\theta-\theta +C$</p>
<p>$=-2\left(\frac{\sqrt{3-2x-x^2}}{2}\right) - \sin^{-1}\left(\frac{x+1}{2}\right)+C$</p>
<p>$=-\sqrt{3-2x-x^2}- \sin^{-1}\left(\frac{x+1}{2}\right)+C$</p>
|
1,303,274 | <p>Define a sequence {$\ x_n$} recursively by</p>
<p>$$ x_{n+1} =
\sqrt{2 x_n -1}, \ and \ x_0=a \ where \ a>1
$$
Prove that {$\ x_n$} is strictly decreasing. I'm not sure where to start.</p>
| DeepSea | 101,504 | <p><strong>hint</strong>: $$s_{2n} = 0, s_{2n+1} = 1$$. </p>
|
1,303,274 | <p>Define a sequence {$\ x_n$} recursively by</p>
<p>$$ x_{n+1} =
\sqrt{2 x_n -1}, \ and \ x_0=a \ where \ a>1
$$
Prove that {$\ x_n$} is strictly decreasing. I'm not sure where to start.</p>
| mich95 | 229,072 | <p>All you need to show is that the the sequence $x_{n}=(-1)^{n}$ does not converge. Many ways to do it : First, using the theorem saying that if a sequence converges , then any subsequence converges to the same limit. $x_{2n}$ and $x_{2n+1}$ are subsequences converging obviously to different limits. Not familiar with this? No problem! Assume that $x_{n}$ converges, say to $l$. Then, given $\epsilon >0$, we can find an integer $N(\epsilon)$ such that $n \geq N(\epsilon)$ implies $|x_{n}-l| <\epsilon.$ Now, $|x_{n+1}-x_{n}| = |x_{n+1}-l+l-x_{n}| \leq |x_{n+1}-l|+|l-x_{n}| <2\epsilon$, provided $n \geq N(\epsilon)$. Now, we chose $\epsilon=1$. And I will leave it to you to see the obvious contradiction!.</p>
|
2,466,855 | <p>We were given this question for homework that the professor couldn't explain how to solve (even in class he had trouble working it out). I'm only aware that we should be using the law of large numbers but I'm not sure how to apply it as the book for the course provides no examples. The answer in class was 10 and the book gave us an 8. Any help would be appreciated.</p>
| Simply Beautiful Art | 272,831 | <p>Denote</p>
<p>$$f(z)=\sum_{n=0}^\infty\frac{z^n}{(n!)^2}$$</p>
<p>By term-wise differentiation, we find that</p>
<p>$$f'(z)+zf''(z)=f(z)$$</p>
<p>A rather simple differential equation with the general solution</p>
<p>$$f(z)=c_1I_0(2\sqrt z)+c_2K_0(2\sqrt z)$$</p>
<p>where $I_n$ is a modified Bessel function of the first kind, and $K_n$ is a modified Bessel function of the second kind. By using</p>
<p>$$f(0)=f'(0)=1$$</p>
<p>we find that</p>
<blockquote>
<p>$$f(z)=I_0(2\sqrt z)$$</p>
</blockquote>
|
4,600,131 | <blockquote>
<p>If <span class="math-container">$$f(x)=\binom{n}{1}(x-1)^2-\binom{n}{2}(x-2)^2+\cdots+(-1)^{n-1}\binom{n}{n}(x-n)^2$$</span>
Find the value of <span class="math-container">$$\int_0^1f(x)dx$$</span></p>
</blockquote>
<p>I rewrote this into a compact form.
<span class="math-container">$$\sum_{k=1}^n\binom{n}{k}(x-k)^2(-1)^{k-1}$$</span>
Now,
<span class="math-container">$$\int_0^1\sum_{k=1}^n\binom{n}{k}(x-k)^2(-1)^{k-1}dx$$</span>
<span class="math-container">$$=\sum_{k=1}^n\binom{n}{k}\frac{(1-k)^3}{3}(-1)^{k-1}-\sum_{k=1}^n\binom{n}{k}\frac{(-k)^3}{3}(-1)^{k-1}$$</span>
<span class="math-container">$$=\sum_{k=1}^n\binom{n}{k}\frac{(1-k)^3}{3}(-1)^{k-1}+\sum_{k=1}^n\binom{n}{k}\frac{k^3}{3}(-1)^{k-1}$$</span>
After this, I took <span class="math-container">$\dfrac13$</span> common and did some simplifications but nothing useful came out.</p>
<p>Any help is greatly appreciated.</p>
| K.defaoite | 553,081 | <p>A devious little problem indeed! I am interested in where you found it. We in fact have a very nice formula:</p>
<p><span class="math-container">$$\sum_{k=1}^n (-1)^{k-1} \binom{n}{k} (x-k)^2 =x^2$$</span></p>
<p>It follows from:</p>
<p><span class="math-container">$$F(x)=\sum_{k=1}^n (-1)^{k-1} \binom{n}{k} (x-k)^2 \\ =\sum_{k=1}^n (-1)^{k-1}\binom{n}{k}\big(x^2-2kx+k^2\big) \\ =x^2\color{red}{\sum_{k=1}^n (-1)^{k-1}\binom{n}{k}}-2x\color{green}{\sum_{k=1}^n (-1)^{k-1}\binom{n}{k}k}+\color{blue}{\sum_{k=1}^n (-1)^{k-1}\binom{n}{k}k^2}$$</span>
Now, we need to somehow show</p>
<p><span class="math-container">$$\color{red}{\blacksquare}=1 \\ \color{green}{\blacksquare}=\color{blue}{\blacksquare}=0$$</span></p>
<p>These sums have been studied before:</p>
<p><span class="math-container">$\color{red}{\blacksquare}$</span> : <a href="https://math.stackexchange.com/questions/3797962/alternating-sum-of-binomial-coefficients-equal-to-1?noredirect=1">Alternating sum of binomial coefficients equal to $1$</a></p>
<p>Follows from doing a binomial expansion of <span class="math-container">$(1-1)^n$</span>.</p>
<p><span class="math-container">$\color{green}{\blacksquare}$</span> : <a href="https://math.stackexchange.com/questions/399464/binomial-coefficient-series-sum-limits-k-1n-1k1-k-binom-nk-0?noredirect=1">Binomial coefficient series $\sum\limits_{k=1}^n (-1)^{k+1} k \binom nk=0$</a></p>
<p>Follows from the recurrence <span class="math-container">$k\binom{n}{k}=n\binom{n-1}{k-1}$</span>.</p>
<p><span class="math-container">$\color{blue}{\blacksquare}$</span> : This is the hard one. We proceed as follows:</p>
<p><span class="math-container">$$\sum_{k=1}^n (-1)^{k-1}k^2\binom{n}{k}=n\sum_{k=1}^n (-1)^{k-1}k \binom{n-1}{k-1} \tag{1}$$</span> <span class="math-container">$$ =n\sum_{l=0}^{n-1}(-1)^l(l+1)\binom{n-1}{l} \\ =n\sum_{l=0}^{n-1}(-1)^l l\binom{n-1}{l}+n\underbrace{\sum_{l=0}^{n-1}(-1)^l \binom{n-1}{l}}_{=(1-1)^{n-1}=0} $$</span>
Finally, since <span class="math-container">$\binom{n-1}{n}=0$</span>, and since the <span class="math-container">$l=0$</span> summand is zero, we can remove the <span class="math-container">$l=0$</span> index and add a <span class="math-container">$l=n$</span> index, and then rename the index back to <span class="math-container">$k$</span>:
<span class="math-container">$$n\sum_{l=0}^{n-1}(-1)^l l\binom{n-1}{l}=n\sum_{k=1}^n(-1)^k k\binom{n-1}{k}$$</span>
Now, using the recurrence relation for the binomials,
<span class="math-container">$$n\sum_{k=1}^n(-1)^k k\binom{n-1}{k}=n\underbrace{\sum_{k=1}^n k(-1)^k \binom{n}{k}}_{=\color{green}{\blacksquare}=0}-n\sum_{k=1}^n (-1)^kk \binom{n-1}{k-1}$$</span>
Again, <span class="math-container">$k\binom{n-1}{k-1}=k^2\binom{n}{k}$</span> and hence
<span class="math-container">$$-n\sum_{k=1}^n (-1)^kk \binom{n-1}{k-1}=-n\sum_{k=1}^n (-1)^kk^2 \binom{n}{k}=n\sum_{k=1}^n (-1)^{k-1}k^2 \binom{n}{k}\tag{2}$$</span>
But, retracing our steps from <span class="math-container">$(1)$</span> to <span class="math-container">$(2)$</span>, we have just proved that
<span class="math-container">$$\sum_{k=1}^n (-1)^{k-1}k^2\binom{n}{k}=~\boldsymbol{n}~\sum_{k=1}^n (-1)^{k-1}k^2\binom{n}{k}$$</span></p>
<p>This can <strong>only</strong> be true for general <span class="math-container">$n$</span> if the sum is zero. Hence,</p>
<p><span class="math-container">$$\boxed{\sum_{k=1}^n (-1)^{k-1} \binom{n}{k} (x-k)^2=x^2\color{red}{\blacksquare}-2x\color{green}{\blacksquare}+\color{blue}{\blacksquare} \\ =x^2}$$</span></p>
<p>QED!!</p>
|
3,044,318 | <p><span class="math-container">$$\frac{e^{z^2}}{z^{2n+1}}$$</span>
Am I right that limit as z approaches infinity does not exist? So its residue at infinity is equal to <span class="math-container">$c_{-1}$</span> of Laurent series. How am I supposed to get Laurent series of this function? Where is it centered? What range?</p>
<p>So <span class="math-container">$$e^{z^2}z^{-2n-1}$$</span> should somehow get to <span class="math-container">$$\sum_{k=-\infty}^{+\infty}{A_k(z-z_0)^k}$$</span></p>
| Angina Seng | 436,618 | <p>Strictly speaking, one finds residues of differentials, not of functions.
Here the differential is
<span class="math-container">$$\alpha=\frac{\exp(z^2)}{z^{2n+1}}\,dz$$</span></p>
<p>To find the residue at <span class="math-container">$\infty$</span>, set <span class="math-container">$z=1/w$</span> and expand in powers of <span class="math-container">$w$</span>. The coefficient of <span class="math-container">$w^{-1}\,dw$</span> is the residue.</p>
<p>Applying this to <span class="math-container">$\alpha$</span> gives
<span class="math-container">$$\alpha=w^{2n+1}\exp(1/w^2)\left(-\frac{dw}{w^2}\right)
=-\left(\sum_{k=0}^\infty\frac{w^{2n-1}}{k!w^{2k}}\right)\,dw.$$</span>
The coefficient of <span class="math-container">$w^{-1}\,dw$</span> is <span class="math-container">$-1/n!$</span>, and that is the residue
of the differential <span class="math-container">$\alpha$</span> at <span class="math-container">$\infty$</span>.</p>
|
250,496 | <p>Please help me out.. Is there some appropriate method to draw Hasse diagram</p>
<blockquote>
<p>My question is $L=\{1,2,3,4,5,6,10,12,15,30,60\}$</p>
</blockquote>
<p>Please explain me by step by step solution... Thanks for help..</p>
| Brian M. Scott | 12,042 | <p>I would start at the bottom. Clearly $1$ divides every member of $L$, so $1$ is the minimum element in the divisibility order on $L$. Now what is the next layer up of the Hasse diagram? It must contain those members of $L$ that are divisible by $1$ (and of course by themselves) but not by any other element of $L$. Those elements are $2,3$, and $5$: every other element of $L$ is divisible by at least one of these three numbers. Thus, your partial Hasse diagram now looks like this:</p>
<pre><code> 2 3 5
\ | /
\|/
1
</code></pre>
<p>What numbers will make up the third row? I claim that the third row contains $4,6,10$, and $15$. First, why not $12,30$, or $60$? $12$ is out because $6$ must be below $12$, and $6$ isn’t in the second row; $30$ is out because $15$ must be below $30$, and $15$ isn’t in the second row; and $60$ is out because every other element of $L$ must be below it. A bit of thought should convince you that no member of $L$ must come below $4,6,10$, or $15$ but above the second row, so that $4,6,10$, and $15$ really do belong in the third row. Specifically, $4$ must lie above $2$; $6$ must lie above $2$ and $3$; $10$ must lie above $2$ and $5$; and $15$ must lie above $3$ and $5$.</p>
<p>You can probably finish it up pretty easily from there: $12$ and $30$ go in the fourth row, with $12$ above $4$ and $3$, and $30$ above $6$ and $5$. Finally, $60$ will be the top element, sitting above $12$ and $30$. As a quick cross-check that $12$ and $30$ are the only members of $L$ that should be immediately below $60$, note that every other member of $L$ (besides $12,30$, and $60$) is a divisor of either $12$ or $30$ and therefore goes somewhere below at least one of these two numbers.</p>
|
21,569 | <p>Recently I commented on a question which was about tiling a 100 by 100 grid with a 1 x 8 square. (Actually to prove this was impossible.) One of our users posted a link to a very interesting looking paper on tiling problems as a comment, and I also commented, but now looking through my comments I can't find the post. I assume it was deleted. However, I really want to find this paper! I had planned on giving a talk about it next week. Is there any way for me to locate the question? </p>
| hardmath | 3,111 | <p>If a Question is self-deleted by the poster, it will <em>not</em> show under the 10K User Tools report of <a href="https://math.stackexchange.com/tools">recently deleted items</a>. Also the items shown are limited to deletion actions in the past 30 days, if the date range selector in the upper right corner there is used.</p>
<p>Regardless, if one had the foresight or luck to "favorite" the item, the Deleted Question would still appear <a href="https://meta.stackexchange.com/a/97110/155839">in ones personal Favorites</a>. Will Jagy <a href="https://math.meta.stackexchange.com/a/12301/3111">once described this</a> as a strategy for certain "very bad questions".</p>
<p>If the Question ID (number following "questions/" in the SE URL) is known (e.g. from ones browser history), the item can still be viewed by 10K users. Some ideas for users < 10K <a href="https://math.meta.stackexchange.com/questions/13328/are-there-some-possibillities-for-users-below-10k-to-see-deleted-posts-of-other">were discussed here</a>.</p>
<p>There was a post not long ago here on meta.Math.SE about using a query feature for deleted items in SE Data Explorer's schema (apparently novel). However I cannot find that post (which I commented on) and suspect it has itself been deleted.</p>
|
1,699,627 | <p>Why is the following equation true?$$\int_0^\infty e^{-nx}x^{s-1}dx = \frac {\Gamma(s)}{n^s}$$
I know what the Gamma function is, but why does dividing by $n^s$ turn the $e^{-x}$ in the integrand into $e^{-nx}$? I tried writing out both sides in their integral forms but $n^{-s}$ and $e^{-x}$ don't mix into $e^{-nx}$. I tried using the function's property $\Gamma (s+1)=s\Gamma (s)$ but I still don't know how to turn it into the above equation. What properties do I need?</p>
| Graham Kemp | 135,106 | <p>The probability that you will have <em>at most 3</em> kings is the probability that you will have <em>less than 4</em>.</p>
<p>$$\mathsf P(K\leq 3) = 1 -\mathsf P(K=4)$$</p>
<p>The probability that you will have exactly all four kings is the count of ways to select 4 kings and 1 other card divided by the count of ways to select any 5 cards.</p>
<p>$$\mathsf P(K=4)~=~\dfrac{{^{4}\mathrm C_{4}}\cdot{^{(52-4)}\mathrm C_{(5-4)}}}{^{52}\mathrm C_5}$$</p>
<p>Put it together.</p>
<p>$$\mathsf P(K\leq 3) ~=~ 1 -\dfrac{{^{4}\mathrm C_{4}}\cdot{^{(52-4)}\mathrm C_{(5-4)}}}{^{52}\mathrm C_5}$$</p>
|
1,699,627 | <p>Why is the following equation true?$$\int_0^\infty e^{-nx}x^{s-1}dx = \frac {\Gamma(s)}{n^s}$$
I know what the Gamma function is, but why does dividing by $n^s$ turn the $e^{-x}$ in the integrand into $e^{-nx}$? I tried writing out both sides in their integral forms but $n^{-s}$ and $e^{-x}$ don't mix into $e^{-nx}$. I tried using the function's property $\Gamma (s+1)=s\Gamma (s)$ but I still don't know how to turn it into the above equation. What properties do I need?</p>
| user247327 | 247,327 | <p>Yet another way: there are 52 cards in the deck, 4 of which are kings. The probability the first card is a king is 4/52= 1/13. There are then 51 cards left, 3 of them kings. The probability the second card is a king is 3/51= 1/17. There are then 50 cards left, 2 of them kings. The probability the third card is a king is 2/50= 1/25. There are then 49 cards left, 1 of them a king. The probability the fourth card is a king is 1/49. There are no longer any kings in the deck so the probability the fifth card is <strong>not</strong> a king is 1.</p>
<p>The probability of getting "four kings and one non-king" <strong>in that order</strong> is (1/13)(1/17)(1/25)(1/49). There are 5 ways to order "four kings and one non- king" (the non-king being in any of the 5 places) so the probability of "four kings and one non-king" in any order is 5(1/13)(1/17)(1/25)(1/49).</p>
<p>Finally, the probability of get "at most three kings in a five card hand" is 1- 5(1/13)(1/17)(1/25)(1/49).</p>
|
3,168,130 | <p><a href="https://math.stackexchange.com/questions/2690416/mathematical-proof-of-uniform-circular-motion">Here</a> is a mathematical proof that any force <span class="math-container">$F(t)$</span>, which affects a body, so that <span class="math-container">$\vec{F(t)} \cdot \vec{v(t)} = 0$</span>, where <span class="math-container">$v(t)$</span> is its velocity cannot change the amount of this velocity.</p>
<p>Further, there is stated that <span class="math-container">$\vec{v(t)}$</span> itself cannot change, what I think is nonsense. Since:</p>
<p><span class="math-container">$$\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \times t = \begin{pmatrix} 1 \\ t \\ 0 \end{pmatrix} $$</span></p>
<p>But maybe I am just wrong.
Now, I am further wondering in how far <span class="math-container">$\vec{F(t)} \cdot \vec{v(t)} = 0$</span> leads to a circular motion and how to proof this using Netwons laws and calculus? Can you explain why a circle is created, if you let the time tick?</p>
| nonuser | 463,553 | <p>Since <span class="math-container">$$ \vec{F} = m{d\vec{v}\over dt} =m\cdot \vec{v}'$$</span></p>
<p>so <span class="math-container">$$\vec{v}'\cdot \vec{v} =0\implies (\vec{v}^2)' = 0 \implies \vec{v}^2 = constant$$</span></p>
<p>So <span class="math-container">$$|\vec{v}|^2 = constant\implies |\vec{v}| = constant_2$$</span> </p>
<p>So the magnitude of velocity is constant, but not the velocity it self.</p>
|
249,047 | <p>I have the following matrix: $$A=
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1 \\
\end{bmatrix}
$$
What is the norm of $A$? I need to show the steps, should not use Matlab...<br>
I know that the answer is $\sqrt{\sqrt{5}/2+3/2}$. I am using the simple version to calculate the norm but getting different answer: $\sum_{i=0}^3\sum_{j=0}^3(a_{ij})^2=\sqrt{1+1+1+1}=2$
Maybe this is some different kind of norm, not sure.</p>
<p>This might help - i need to get a condition number of $A$, which is $k(A)=\|A\|\|A^{-1}\|$...that is why i need to calculate the norm of $A$. </p>
| marty cohen | 13,079 | <p>$1/(n-1)-1/(n+1) = 2/(n^2-1)
$, so
$$\sum_{n=4}^{\infty} \frac{1}{n^2-1}
= (1/2)\sum_{n=4}^{\infty} \left(\frac{1}{n-1} - \frac{1}{n+1}\right)
= \frac{1}{3} + \frac{1}{4}
$$
since all the later terms are cancelled out.</p>
<p>Whoops - as pointed out by Limitless, this should be
$\frac{1}{2}\left(\frac{1}{3} + \frac{1}{4}\right)$.</p>
<p>Note that this also allows you to get an explicit expression for
$\sum_{n=a}^b \frac{1}{n^2-1}$
for any integers $a$ and $b$.</p>
|
962,573 | <p>I have just learned Fermat's little theorem.</p>
<p>That is,</p>
<blockquote>
<p>If $p$ is a prime and $\gcd(a,p)=1$, then $a^{p-1} \equiv 1 \mod p$</p>
</blockquote>
<p>Well, there's nothing more explanation on this theorem in my book.</p>
<p>And there are exercises of this kind</p>
<blockquote>
<p>If $\gcd(a,42)=1$, then show that $a^6\equiv 1 \mod 168$</p>
</blockquote>
<p>I don't have any idea how to approach this problem.</p>
<p>The only thing I know is that $168=3\cdot 7\cdot 8$.</p>
<p>Hence, I get $[a^2\equiv 1 \mod 3]$ and $[a^6\equiv \mod 7]$ and $[a\equiv 1\mod2]$.</p>
<p>That's it.</p>
<p>I don't know what to do next.</p>
<p>Please help.</p>
| Bruce Zheng | 181,584 | <p>If $x \equiv 1 \bmod y$ and $x \equiv 1 \bmod z$, then $x \equiv 1 \bmod yz$ if $y$ and $z$ are coprime.</p>
<p>It also turns out that $a \equiv 1 \bmod 2 \Rightarrow a^2 \equiv 1 \bmod 2^3$. This is because if $a = 2k + 1$, $a^2 = 4(k^2 + k) + 1$. Since $k^2 \equiv k \bmod 2$, $k^2 + k$ is even. Thus $8 \mid 4(k^2+k)$.</p>
<p>Then $a^6 \equiv 1 \bmod 3$, $a^6 \equiv 1 \bmod 7$, and $a^6 \equiv 1 \bmod 8$. Hence $a^6 \equiv 1 \bmod (3 \cdot 7 \cdot 8)$, or $a^6 \equiv 1 \bmod 168$.</p>
|
609,770 | <p>We have an empty container and $n$ cups of water and $m$ empty cups. Suppose we want to find out how many ways we can add the cups of water to the bucket and remove them with the empty cups. You can use each cup once but the cups are unique. </p>
<p>The question: In how many ways can you perform this operation.</p>
<p>Example: Let's take $n = 3$ and $m = 2$.</p>
<p>For the first step we can only add water to the bucket so we have 3 choices.
For the second step we can both add another cup or remove a cup of water.
So for the first 2 steps we have $3\times(5-1) = 12$ possibilities.
For the third step it gets more difficult because this step depends on the previous step.
There are two scenarios after the second step. 1: The bucket either contains 2 cups of water or 2: The bucket contains no water at all.</p>
<p>1) We can both add or subtract a cup of water
2) We have to add a cup of water</p>
<p>So after step 3 we have $3\times(2\cdot3 + 2\cdot2) = 30$ combinations.</p>
<p>etc.</p>
<p>I hope I stated this question clearly enough since this is my first post. This is not a homework assignment just personal curiosity. </p>
| CoffeeIsLife | 97,554 | <p>The method in proving that $f(x)=\frac{1}{x}$ is not uniformly continuous is similar to the method in proving that $f(x)=\frac{1}{x^2}$. I will address $f(x)=\frac{1}{x^2}$ , and you can infer from it how to prove $f(x)=\frac{1}{x}$.</p>
<p>Assume that $f(x)=\frac{1}{x^2}$ is not uniformly continuous. This means that there exists $\varepsilon>0$ and $\delta>0$ such that $$\vert x-y\vert<\delta\Rightarrow\vert f(x)-f(y)\vert\geq\varepsilon \tag 1$$ (This is the reverse of the definition of uniform continuity)</p>
<p>As Michael Hardy pointed out, to prove that $f(x)$ is not uniformly continuous, we simply need to show that (1) is true for some value of $\varepsilon$. This is because the definition of uniform continuity is for all $\varepsilon$. Hence, we only need to show that uniform continuity fails (or that (1) holds) for one value of $\varepsilon$. Now, let $\varepsilon=1$ and let $y=x+\frac{\delta}{2}$ (the value of $\varepsilon$ and $y$ were chosen arbitrarily...we do this to change the equation to one variable which simplifies things). Thus we have</p>
<p>$$\vert f(x)-f(x+\frac{\delta}{2})\vert\geq 1$$</p>
<p>Thus we have that $$\left| \frac{1}{x}-\frac{1}{x+\frac{1}{\delta}}\right|\geq 1$$</p>
<p>We can rearrange the expression in the absolute values by using the fact that $f(x)-f(y)=\frac{(y-x)(y+x)}{x^2y^2}$ (let $y=x+\frac{\delta}{2}$) </p>
<p>Thus, we have that $$1\leq\frac{(x+\frac{\delta}{2}-x)(x+\frac{\delta}{2}+x)}{x^2(x+\frac{\delta}{2})^2}$$</p>
<p>Arbitrarily let $x=\delta$, then we have </p>
<p>$$\frac{5\delta^2}{9\delta^4}\geq 1$$. Note that we can pick a delta that satisfies such that the above inequality holds. For example, we could let $\delta=1/3$. Therefore, there exists $\delta$ such that (1) holds.</p>
|
609,770 | <p>We have an empty container and $n$ cups of water and $m$ empty cups. Suppose we want to find out how many ways we can add the cups of water to the bucket and remove them with the empty cups. You can use each cup once but the cups are unique. </p>
<p>The question: In how many ways can you perform this operation.</p>
<p>Example: Let's take $n = 3$ and $m = 2$.</p>
<p>For the first step we can only add water to the bucket so we have 3 choices.
For the second step we can both add another cup or remove a cup of water.
So for the first 2 steps we have $3\times(5-1) = 12$ possibilities.
For the third step it gets more difficult because this step depends on the previous step.
There are two scenarios after the second step. 1: The bucket either contains 2 cups of water or 2: The bucket contains no water at all.</p>
<p>1) We can both add or subtract a cup of water
2) We have to add a cup of water</p>
<p>So after step 3 we have $3\times(2\cdot3 + 2\cdot2) = 30$ combinations.</p>
<p>etc.</p>
<p>I hope I stated this question clearly enough since this is my first post. This is not a homework assignment just personal curiosity. </p>
| CoffeeIsLife | 97,554 | <p>Or you could use the theorem of uniform continuity regarding Cauchy sequences. A theorem states that if $f$ is uniformly continuous on a set $S$ and $(s_n)$ is a Cauchy sequence in $S$, then $f(s_n)$ is a Cauchy sequence. </p>
<p>Choose $(s_n)=\frac{1}{n}$ (here $n\in\mathbb{N}$), which is a Cauchy sequence. However, $f(s_n)=n$. Since $f((s_n))=n$ is not Cauchy, $f$ is not uniformly continuous.</p>
<p>Both proofs that I have given are from Kenneth Ross's "Elementary Analysis".</p>
|
2,699,753 | <p>$a,b$ and $c$ are all natural numbers, and function $f(x)$ always returns a natural number. If$$
\sum_{n=b}^{a} f(n) = c,$$
in terms of $b,c$ and $f$, how would you solve for $a$? Do I require more information to solve for $a$?</p>
<p>EDIT: If $x$ increases $f(x)$ increases</p>
| saulspatz | 235,128 | <p>"The general solution to an inhomogeneous linear recurrence relation is the general solution to the associated homogeneous recurrence, plus any particular solution of the inhomogeneous recurrence."</p>
<p>Quite a mouthful, but what is means is this:</p>
<ol>
<li><p>Solve the homogeneous recurrence you get by eliminating the right-hand side: $f(n)=9f(n-1).$ We know the answer is $f(n) = c9^n,$ for some constant $c$.</p></li>
<li><p>Find any solution at all to the inhomogeneous recurrence. In this case we can try $f(n) \equiv k.$ That leads to $k=9k-14\implies k = 7/4.$</p></li>
<li><p>Now the general solution to the inhomogeneous recurrence is the sum of the results in the two previous steps: $f(n)= c9^n - 7/4$</p></li>
<li><p>Finally, use the initial condition to solve for $c$. I leave that to you.</p></li>
</ol>
|
2,699,753 | <p>$a,b$ and $c$ are all natural numbers, and function $f(x)$ always returns a natural number. If$$
\sum_{n=b}^{a} f(n) = c,$$
in terms of $b,c$ and $f$, how would you solve for $a$? Do I require more information to solve for $a$?</p>
<p>EDIT: If $x$ increases $f(x)$ increases</p>
| Math1000 | 38,584 | <p>Let $A(z) = \sum_{n=0}^\infty f(n)z^n$. Multiplying both sides of the recurrence by $z^n$ and summing over $n$, we have
$$
\sum_{n=1}^\infty f(n)z^n = \sum_{n=1}^\infty 9f(n-1)z^n - \sum_{n=1}^\infty 14z^n.
$$
Writing the above in terms of $A(z)$,
$$
A(z) - 3 = 9zf(z) -\frac{14z}{1-z}.
$$
Solving for $A(z)$, we have
$$
A(z) = \frac3{1-9z} - \frac{14z}{(1-z)(1-9z)}.
$$
Partial fraction decomposition yields
$$
A(z) = \frac74\cdot\frac1{1-z} + \frac54\cdot\frac1{1-9z},\\
$$
with power series representation
$$
A(z) = \sum_{n=0}^\infty \left(\frac 74 + \frac 54\cdot 9^n\right)z^n.
$$
It follows that
$$
f(n) = \frac 74 + \frac 54\cdot 9^n, \; n\geqslant0.
$$</p>
|
2,699,753 | <p>$a,b$ and $c$ are all natural numbers, and function $f(x)$ always returns a natural number. If$$
\sum_{n=b}^{a} f(n) = c,$$
in terms of $b,c$ and $f$, how would you solve for $a$? Do I require more information to solve for $a$?</p>
<p>EDIT: If $x$ increases $f(x)$ increases</p>
| farruhota | 425,072 | <p>You can find several terms and see pattern:
$$\begin{align} &f(1)=9f(0)-14; \\
&f(2)=9f(1)-14=9(9f(0)-14)-14=9^2f(0)-9\cdot 14-14; \\
&f(3)=9f(2)-14=9(9^2f(0)-9\cdot 14-14)-14=9^3f(0)-9^2\cdot 14-9\cdot 14-14; \\
&\cdots \\
&f(n)=9^nf(0)-(9^{n-1}+9^{n-2}+\cdots+9+1)\cdot 14=9^nf(0)-\frac{9^n-1}{9-1}\cdot 14=9^n\left(f(0)-\frac74\right)+\frac74.\end{align}$$
Now use the initial value:
$$f(n)=9^n\left(3-\frac74\right)+\frac74=\frac54\cdot 9^n+\frac74.$$
The general formula:</p>
<blockquote>
<p>$$f(n+1)=af(n)+b, f(0): \\
f(n)=\left(f(0)-\frac{b}{1-a}\right)\cdot a^n+\frac{b}{1-a}.$$</p>
</blockquote>
<p>Applying the formula:
$$f(n)=\left(3-\frac{-14}{1-9}\right)+\frac{-14}{1-9}=\frac54\cdot 9^n+\frac74.$$</p>
|
1,690,854 | <p>Solve the equation
$$-x^3 + x + 2 =\sqrt{3x^2 + 4x + 5.}$$
I tried. The equation equavalent to
$$\sqrt{3x^2 + 4x + 5} - 2 + x^3 - x=0.$$
$$\dfrac{3x^2+4x+1}{\sqrt{3x^2 + 4x + 5} + 2}+x^3 - x=0.$$
$$\dfrac{(x+1)(3x+1)}{\sqrt{3x^2 + 4x + 5} + 2}+ (x+1) x (x-1)=0.$$
$$(x+1)\left [\dfrac{3x+1}{\sqrt{3x^2 + 4x + 5} + 2}+ x (x-1)=0\right ]=0.$$
How can I prove the equation
$$\dfrac{3x+1}{\sqrt{3x^2 + 4x + 5} + 2}+ x (x-1)=0$$
has no solution?</p>
| John_dydx | 82,134 | <p>I would start by squaring both sides of the equation:</p>
<p>$$ (-x^2 + x +2)^2 = 3x^2 +4x +5$$</p>
<p>$$x^6 -2x^4-4x^3-2x^2-1 =0$$</p>
<p>As suggested by Deepak, $x = -1$ is a solution.
You can factorise fully by dividing the above polynomial by $x+1$ to obtain other factors and solutions (if any).</p>
|
2,915,735 | <p>So I just got done showing explicitly that an isomorphism exists between these two rings if the $\gcd(m, n) = 1$, and I did not have much trouble with that. For some reason I'm having a lot harder of a time showing that the result <em>is not</em> true if $m$ and $n$ are not relatively prime. Can somebody help me out here? Thanks.</p>
| N. S. | 9,176 | <p><strong>Hint</strong> Show that $\mbox{lcm} (m,n) \cdot (x,y) =(0,0)$ for all elements $(x,y) \in \mathbb Z_m \times \mathbb Z_n$. </p>
|
9,416 | <p>Say I pass 512 samples into my FFT</p>
<p>My microphone spits out data at 10KHz, so this represents 1/20s.</p>
<p>(So the lowest frequency FFT would pick up would be 40Hz).</p>
<p>The FFT will return an array of 512 frequency bins
- bin 0: [0 - 40Hz)
- bin 1: [40 - 80Hz)
etc</p>
<p>So if my original sound contained energy at say 115Hz, how can I accurately retrieve this frequency?</p>
<p>That is going to lie in bin #2, but very close to bin #3. so I would expect both bins to contain something nonzero.</p>
<p>Question: how about the bins either side of this? Would they be guaranteed to be zero if there are no other frequencies close in the original signal?</p>
<p>Main question: is there some algorithm for deciphering the original frequency given the relative bin strengths?</p>
| P i | 3,096 | <p>In the end I derived a formula that calculates the exact frequency from the rate at which each bin rotates.</p>
<p>My result can be seen here: <a href="https://stackoverflow.com/questions/4633203/extracting-precise-frequencies-from-fft-bins-using-phase-change-between-frames">https://stackoverflow.com/questions/4633203/extracting-precise-frequencies-from-fft-bins-using-phase-change-between-frames</a></p>
<p>To understand <strong>bin rotation</strong>, just look at a single bin. So let's consider theta = 2pi/12, so exp(i.n.theta) for n in {0, ..., 11} will give us our twelve 12th-roots of unity; if you look on the complex plane it will make a 12 pointed star.</p>
<p>Now if the signal matches that frequency, Ie 12 samples makes one wave, then repeatedly projecting that signal onto that star ( proj[k] = sig[k]*star[k%12] ) will stretch out one spine (the one corresponding to the maximum positive amplitude of sig[k], the peak of the sine wave if you like) and will squash or negatively stretch (depending on whether the sine wave is vertically centred, i.e. DC offset = 0) the diametrically opposite spine.</p>
<p>If you keep a running vector sum of the last 12 projections, the result will come out in this direction, by symmetry. And the angle of the result gives the bin phase, and length gives the bin magnitude.</p>
<p>And the phase remains constant. So no bin rotation.</p>
<p>But now if you change the input frequency to {13 samples = one wave}, you can see that this vector sum is now going to rotate 1/12 of a cycle
If a peak corresponds to an angle of 0, the next peak will no longer line up, it will correspond to 1/12 of a revolution</p>
<p>And this is the bin rotation.</p>
<p>While I wouldn't expect the reader to derive the answer from this basis, it will hopefully give some intuitive understanding of what is going on.</p>
|
55,232 | <p>I'm looking for a concise way to show this:
$$\sum_{n=1}^{\infty}\frac{n}{10^n} = \sum_{n=1}^{\infty}\left(\left(\sum_{m=0}^{\infty}{10^{-m}}\right){10^{-n}}\right)$$
With this goal in mind:
$$\sum_{n=1}^{\infty}\left(\left(\sum_{m=0}^{\infty}{10^{-m}}\right){10^{-n}}\right) =
\sum_{n=1}^{\infty}\left(\left(\frac{10}{9}\right){10^{-n}}\right) = \frac{10}{81}$$</p>
<p>So far I've been looking at it by replacing $n$ in the LHS with $(\sum_{m=1}^{n}1)$ like this:
$$\sum_{n=1}^{\infty}\left(\left(\sum_{m=1}^{n}{1}\right){10^{-n}}\right) = \sum_{n=1}^{\infty}\left(\left(\sum_{m=0}^{\infty}{10^{-m}}\right){10^{-n}}\right)$$</p>
<p>And here I hit a particularly uncreative brick wall. This equation is obvious to me in a common sense way - I could easily demonstrate it by writing out the RHS as a huge addition problem and showing that the LHS just has the digit columns added ahead of time - but I don't know what to do in between for a proof.</p>
| Braindead | 2,499 | <p>Are you allowed to use derivatives? If so, then...</p>
<p>$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$</p>
<p>$\frac{1}{(1-x)^2} = \sum_{n=1}^\infty n x^{n-1}$</p>
<p>$\frac{x}{(1-x)^2} = \sum_{n=1}^\infty n x^n$</p>
<p>Letting $x=\frac{1}{10}$, you get</p>
<p>$\frac{10}{81} = \sum_{n=1}^\infty \frac{n}{10^n}$</p>
|
4,536,320 | <p>Let <span class="math-container">$R$</span> be a ring and <span class="math-container">$A \subseteq R$</span> be finite, say <span class="math-container">$A = \{a\}$</span>. The set <span class="math-container">$$RaR = \{ras\:\: : r,s\in R\}$$</span> Why is this not necessarily closed under addition?</p>
<p>Take <span class="math-container">$r_1as_1$</span> and <span class="math-container">$r_2as_2$</span>. Is it because the ring must be commutative in order to guarantee that we can add these two elements and still stay within <span class="math-container">$RaR$</span>?</p>
<p>Even though I see why commutativity will make this possible, I'm still a bit confused because the very definition of <span class="math-container">$RaR$</span> includes all possible combinations of elements of the form <span class="math-container">$ras$</span>.</p>
| user2661923 | 464,411 | <p>For any given <span class="math-container">$k \in \Bbb{Z}$</span> such that <span class="math-container">$1 \leq k \leq 10$</span>, <span class="math-container">$m$</span> must be some element in <span class="math-container">$\{0,1,2,\cdots,2k\}$</span>. Therefore, for each such <span class="math-container">$k$</span>, the number of possible ordered pairs <span class="math-container">$(m,a)$</span> is <span class="math-container">$(2k+1)$</span>.</p>
<p>For any given <span class="math-container">$k \in \Bbb{Z}$</span> such that <span class="math-container">$11 \leq k \leq 20$</span>, <span class="math-container">$m$</span> must be some element in <span class="math-container">$\{2k - 20, 2k - 19, \cdots, 20.\}$</span>. Therefore, for each such <span class="math-container">$k$</span>, the number of possible ordered pairs <span class="math-container">$(m,a)$</span> is <span class="math-container">$[(20 + 1) - (2k-20)] = 41 - 2k$</span>.</p>
<p>Here, if <span class="math-container">$k = 2$</span>, I am distinguishing between (for example) <span class="math-container">$(m,a) = (3,1)$</span> and <span class="math-container">$(m,a) = (1,3).$</span></p>
|
3,651,287 | <p>I would be very grateful if you could help me, I have a question about the Cauchy sequences, they have given me the definition that a Cauchy sequence if:</p>
<p>A sequence <span class="math-container">$(r_{n})_{n\in \mathbb{N}}$</span> is of Cauchy if:</p>
<p><span class="math-container">$\forall \epsilon>0,$</span> <span class="math-container">$\exists N \in \mathbb{N}$</span> such that if <span class="math-container">$n,m>N$</span>, then <span class="math-container">$|r_{n}-r_{m}|<\epsilon$</span></p>
<p>My question is, I found a different definition in another book and I would like to know how they are equivalent since the second definition I see as only limits the difference in the sequences. But I think I am wrong and I would like to know how they are equivalent definitions.</p>
<p>The other definition I found is:</p>
<p>A secuence <span class="math-container">$(r_{n})_{n\in \mathbb{N}}$</span> is Cauchy if given <span class="math-container">$\epsilon>0$</span>, there exists a positive integer <span class="math-container">$m$</span> such that:</p>
<p><span class="math-container">$|r_{n}-r_{m}|<\epsilon\,,$</span> <span class="math-container">$[n>m$</span></p>
<p>first of all, Thanks!</p>
| Kavi Rama Murthy | 142,385 | <p>They are equivalent. If the second condition is satisfied then <span class="math-container">$|r_k-r_j| \leq |r_k-r_m|+|r_m-r_j| <2 \epsilon$</span> for all <span class="math-container">$j,k >m$</span>. Do you now see the equivalence of the two definitions?</p>
|
1,043 | <p>Hi all,</p>
<p>The short-time fourier transform decomposes a signal window into a sin/cosine series.</p>
<p>How would one approximate a signal in the same way, but using a set of arbitrary basis functions instead of sin/cos? These arbitrary basis functions are likely in my case to be very small discrete chunks of a 1-dimensional non-periodic waveform. </p>
<p>Is this something wavelets are used for? </p>
<p>Please excuse my tag, 'signal-analysis' does not exist and I can not create it. </p>
| Francisco Pereira | 704 | <p>You can use an arbitrary set of basis functions that can be placed or scaled at any point along your signal, if I understand what you want. The problem you'd be solving would be to find placement locations (another set of parameters) and scales (another set), to minimize squared error relative to the original signal (or some other measure that is easy to minimize). I think the main issue will be ensuring that you get a single solution (if there are many) or prevent situations where some basis functions cancel out the others. One way to do this would be to use something like wavelets, indeed, though there you don't have an arbitrary set. Another would be to enforce some sort of regularization on the parameters (e.g. a penalty that would make most of them 0). I don't know too much about this myself but the thing to look for is "overcomplete basis".</p>
|
1,043 | <p>Hi all,</p>
<p>The short-time fourier transform decomposes a signal window into a sin/cosine series.</p>
<p>How would one approximate a signal in the same way, but using a set of arbitrary basis functions instead of sin/cos? These arbitrary basis functions are likely in my case to be very small discrete chunks of a 1-dimensional non-periodic waveform. </p>
<p>Is this something wavelets are used for? </p>
<p>Please excuse my tag, 'signal-analysis' does not exist and I can not create it. </p>
| Vasile Moșoi | 1,093 | <p>If the signal is not periodic we talk about the integral transform of the signal (in case of Fourier transform) and it have a continuum spectrum but not discreet.
In case of polynomial decomposition the desired accuracy of approximation is valid in the finite interval in which the basis functions are orthogonal. Outside, the error can be significant. </p>
|
256,373 | <p>I've not been able to find a package which will deal with Geometric Algebra. Perhaps somebody can help?</p>
| nsap | 81,217 | <p>You might want to use FindInstance in these case:</p>
<pre><code>Table[{n,
FindInstance[x^3 + y^3 + z^3 == n && 0 < n < 100, {x, y, z},
Integers]}, {n, 1, 10}]
</code></pre>
|
4,609,236 | <p>When finding the derivative of <span class="math-container">$f(x) = \sqrt x$</span> via the limit definition, one gets</p>
<p><span class="math-container">$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt x}{h}$$</span></p>
<p>For this, I could get the answer from applying L'Hopital's rule... but to me, this line of reasoning is a bit circular. I'm computing a derivative from first principles while needing to use derivative rules (from L'Hopital's) to compute this derivative.</p>
<p>Can I use L'Hopital's rule when finding a derivative using the limit definition?</p>
| PrincessEev | 597,568 | <p>The line of reasoning is in fact circular; you need to know the derivative of the function to apply L'Hopital's rule in the first place.</p>
<p>One can (infamously) likewise apply this to</p>
<p><span class="math-container">$$\lim_{x \to 0} \frac{\sin(x)}{x}$$</span></p>
<p>but</p>
<p><span class="math-container">$$\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x} \sin(x) &= \lim_{h\to 0}\dfrac{\sin(x+h)-\sin x}{h} \\
&= \lim_{h\to 0}\dfrac{\sin x\cos h+\cos x\sin h-\sin x}{h} \\
&= \cos x\cdot\left(\lim_{h\to 0}\dfrac{\sin h}{h}\right)+ \sin x\cdot\left(\lim_{h\to 0}\dfrac{\cos h -1}{h}\right)
\end{align*}$$</span></p>
<p>which requires knowledge of the limit we originally sought! And if you don't know the derivative, how can you apply L'Hopital's rule to our original limit, which uses that definition?</p>
<p>That is, to find the original limit this way is to assume - without basis - what the derivative of sine is. <em>(Of course, who's to say this is the only way you can find the derivative? If you can justify that the derivative is cosine some other way, that doesn't make use of this limit, then there is no issue.)</em></p>
<p>To find the derivative in your case with <span class="math-container">$f(x) = \sqrt x$</span>, using L'Hopital's rule, is to assume what the derivative is before you've even found it.</p>
<p>Some discussion is on this MSE post <a href="https://math.stackexchange.com/questions/2118581/lhopitals-rule-and-frac-sin-xx">here</a> for the <span class="math-container">$\sin(x)/x$</span> limit.</p>
|
4,609,236 | <p>When finding the derivative of <span class="math-container">$f(x) = \sqrt x$</span> via the limit definition, one gets</p>
<p><span class="math-container">$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt x}{h}$$</span></p>
<p>For this, I could get the answer from applying L'Hopital's rule... but to me, this line of reasoning is a bit circular. I'm computing a derivative from first principles while needing to use derivative rules (from L'Hopital's) to compute this derivative.</p>
<p>Can I use L'Hopital's rule when finding a derivative using the limit definition?</p>
| PierreCarre | 639,238 | <p>The circular use of the knowledge of <span class="math-container">$f'(x)$</span> in its computation using the definition is not reasonable and proves nothing. Your actual conclusion would just be that "If <span class="math-container">$f'(x) = \frac{1}{2 \sqrt{x}}$</span> then <span class="math-container">$f'(x) = \frac{1}{2 \sqrt{x}}$</span>". The way to go in this case is just multiplying and dividing by the conjugate expression:
<span class="math-container">$$
\lim_{h\to 0}\dfrac{\sqrt{x+h}-\sqrt{x}}{h} = \lim_{h \to 0}\dfrac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})} = \dfrac{1}{\sqrt{x}+\sqrt{x}}=\frac{1}{2\sqrt{x}}.
$$</span></p>
<hr />
<p>Just for laughs, suppose that you <strong>wrongly</strong> assume <span class="math-container">$(\sqrt{x})' = \frac 1x$</span>. Now using the definition + L'Hôpital's rule would give you
<span class="math-container">$$
(\sqrt{x})' = \lim_{h\to 0}\dfrac{\sqrt{x+h}-\sqrt{x}}{h} = \lim_{h\to 0} \dfrac{\frac{1}{x+h}}{1} = \frac 1x ...
$$</span></p>
<p>Voila, confirmed! See the problem here?</p>
|
928,826 | <p>I have a function </p>
<p>$$f(x)=\frac{2x^2 - x - 1}{x^2 + 3x + 2}$$</p>
<p>from the interval $[0,\infty)$</p>
<p>The limit of this function is $2$. Is the range then simply from $f(0)$ to $2$, and if yes, would I write it as $[f(0],2]$ or $[f(0),2)$, i.e open brackets or closed? </p>
<p>Also, would i first need to argue that it's a monotonous - growing - function, thus $f(0)$ has to be the lower end of the range? THX</p>
| Pierre Alvarez | 175,318 | <p>If you can prove that your function is monotonous, then you need to write this argument first, and then you can deduce that the range is $[f(0),2[$. </p>
|
928,826 | <p>I have a function </p>
<p>$$f(x)=\frac{2x^2 - x - 1}{x^2 + 3x + 2}$$</p>
<p>from the interval $[0,\infty)$</p>
<p>The limit of this function is $2$. Is the range then simply from $f(0)$ to $2$, and if yes, would I write it as $[f(0],2]$ or $[f(0),2)$, i.e open brackets or closed? </p>
<p>Also, would i first need to argue that it's a monotonous - growing - function, thus $f(0)$ has to be the lower end of the range? THX</p>
| Jean-Claude Arbaut | 43,608 | <p>Yes, you have to prove your function is increasing. No too difficult, since the derivative is, after simplification,</p>
<p>$$f'(x)=\frac{7x^2+10x+1}{(x^2+3x+2)^2}>0$$</p>
<p>Hence the range is $[f(0), \lim_{x\to+\infty}f(x)[=[-1/2,2[$, <em>open</em> on the right.</p>
<hr>
<p>I'm not sure what you mean in your comment, anyway here is a plot of $f(x)$ (blue) and $f'(x)$ (red). When the curve of $f$ is flattening, the derivative tends to zero, which is rather expected.</p>
<p><img src="https://i.stack.imgur.com/OzPBi.png" alt="enter image description here"></p>
|
1,439,920 | <blockquote>
<p>So, the question is:<br>
Calculate the probability that 10 dice give more than 2 6s.</p>
</blockquote>
<p>I've calculated that the probability for throwing 3 6s is 1/216.</p>
<p>And by that logic: 1/216 + 1/216 + .. + 1/216 = 10/216.</p>
<p>But I've been told that this isn't the proper way set it up.</p>
<p>Anyone having a good way to calculate this?</p>
| Lutz Lehmann | 115,115 | <p>The Newton polygon tells us that the dominant binomials are </p>
<ul>
<li>$x^5-1102x^4$ for large roots, resulting in a root close to $1102$ and</li>
<li>$-1102x^4-2015x$ for small roots, resulting in roots close to $0$ and the three third roots of $-\frac{2015}{1102}\approx (-1.22282549628)^3$.</li>
</ul>
<p>Single real roots stay real under small perturbations, thus giving exactly 3 real roots and a pair of complex conjugate roots. Indeed the numerical approximations confirm this, they are (thanks to <a href="http://www.akiti.ca/PolyRootRe.html" rel="nofollow">http://www.akiti.ca/PolyRootRe.html</a>):</p>
<pre><code> 0
0.6111860961336238 + 1.0593896464200445 i
0.6111860961336238 - 1.0593896464200445 i
-1.2223736979388697
1102.0000015056714
</code></pre>
|
4,171,907 | <blockquote>
<p>If <span class="math-container">$3\sin x +5\cos x=5$</span> then prove that <span class="math-container">$5\sin x-3\cos x=3$</span></p>
</blockquote>
<p>What my teacher did in solution was as follows</p>
<p><span class="math-container">$$3\sin x +5\cos x=5 \tag1$$</span></p>
<p><span class="math-container">$$3\sin x =5(1-\cos x) \tag2$$</span></p>
<p><span class="math-container">$$3=\frac{5(1-\cos x)}{\sin x} \tag3$$</span></p>
<p><span class="math-container">$$3=\frac{5\sin x}{(1+\cos x)} \tag4$$</span></p>
<p><span class="math-container">$$5\sin x-3\cos x=3 \tag5$$</span></p>
<p>However this should not be true when <span class="math-container">$\sin x=0$</span>, as division by zero is not defined; and also, if <span class="math-container">$\sin x=0$</span>, then the expression we have to prove evaluates to <span class="math-container">$-3$</span>. In other words question is incomplete but my teacher denied it.</p>
<p>Am I correct in my reasoning?</p>
| José Carlos Santos | 446,262 | <p>If <span class="math-container">$c=\cos x$</span> and <span class="math-container">$s=\sin x$</span>, then you know that<span class="math-container">$$\left\{\begin{array}{l}3s+5c=5\\c^2+s^2=1.\end{array}\right.$$</span>This system is easy to solve. One of the solutions is <span class="math-container">$(c,s)=(1,0)$</span> and the other one is <span class="math-container">$(c,s)=\left(\frac 8{17},\frac{15}{17}\right)$</span>. In the first case (which is the case that you get when <span class="math-container">$x=0$</span>), <span class="math-container">$5s-3c=-3$</span>; in the second case, <span class="math-container">$5s-3c=3$</span>. So, you are right; some hypothesis is missing. But this also shows that, other than the case in which <span class="math-container">$x$</span> is an integer multiple of <span class="math-container">$2\pi$</span>, the statement is true.</p>
|
1,868,263 | <p>A Relation R on the set N of Natural numbers be defined as (x,y) $\in$R if and only if $x^2-4xy+3y^2=0$ for allx,y $\in$N then show that the relation is reflexive,transitive but not SYMMETRIC.</p>
<p>i got how this relation is reflexive or transitive but i am not able to think of any reason of why this relation is not symmetric.</p>
| Robert Z | 299,698 | <p>R is not symmetric because $(3,1)\in R$ but $(1,3)\not\in R$.
Note that $x^2-4xy+3y^2=(x-y)(x-3y)$.</p>
|
1,386,307 | <p>If you consider that you have a coin, head or tails, and let's say tails equals winning the lottery. If I participate in one such event, I may not get tails. It's roughly 50%. But if a hundred people are standing with a coin and I or them get to flip it, my chances of having gotten a tail after these ten attempts, is higher, is it not? Way higher than 50% though I'm not sure how to calculate it.</p>
<p>So why is it different for lotteries? Or is it? I was once told that in a certain lottery, I had a one in 12 million chance of winning. And like the coin toss, each lottery is different with different odds, but would the accumulated odds be way higher if I participate, be it in this same lottery over a thousand times, or this lottery and thousand other lotteries around country, thereby increasing my chances of getting a win, a tail? </p>
<p>I appreciate a response, especially at level of high school or first year university (did not do math past first year university). Thank you. </p>
| David | 119,775 | <p>There are a few similar but different problems here, perhaps that is what is causing confusion.</p>
<ul>
<li>If you are aiming to win the lottery <em>at least once</em>, then the more times you enter, the better your chance of success. It's the same as the coin problem you described.</li>
<li>If you are aiming to win the lottery <em>every time you enter</em>, then the more times you enter, the worse your chance of success.</li>
<li>If you are aiming to win <em>overall from the financial point of view</em> - that is, you want your winnings to be more than the amount you paid for lottery tickets - then the more times you enter, the worse your chance of success. At least this is true for a "normal" lottery, but if you imagine an "altruistic" lottery where the organisers pay out more than they receive in entry fees, then the reverse would be true. If you hear of an "altruistic" lottery like this, please let me know in the comments ;-)</li>
</ul>
|
2,109,347 | <p>My statistics note states that the variance of the empirical distribution is
$v= \sum_{i=1}^{n}(x_i-\bar x )^2\frac {1} {n}$ which the author then re-writes as
$v= \sum_{i=1}^{n}x_i^2 (\frac {1} {n}) - \bar x^2$. How is this achieved?</p>
| Joda | 392,273 | <p>It's just algebra</p>
<p>$$
\begin{aligned}
v&=\frac{1}{n}\sum_{i=1}^n(x_i-\bar{x})^2=\frac{1}{n}\sum_{i=1}^n\left(x_i^2-2x_i\bar{x}+\bar{x}^2\right)\\
&=\frac{1}{n}\sum_{i=1}^nx_i^2-2\bar{x}\frac{1}{n}\sum_{i=1}^nx_i+\bar{x}^2\frac{1}{n}\sum_{i=1}^n1\\
&=\frac{1}{n}\sum_{i=1}^nx_i^2-2\bar{x}^2+\bar{x}^2\\
&=\frac{1}{n}\sum_{i=1}^nx_i^2-\bar{x}^2
\end{aligned}
$$</p>
|
770,544 | <p>It's clear that a system of two quadratic equations can have none, one or two solutions. </p>
<p>For example: $y = x^2 + 2$ and $y = - x^2 + 1$ have none. $y = x^2$, $2x^2 - 8x + 8$ and $y = - x^2 + 8x - 8$ have $4$ as common solution. And $2x^2 - 8x + 8 = x^2 - 4x + 5$ have $1$ and $3$ as solutions.</p>
<p>Is it also possible to have more solutions? Intuitively I'd say that two is the max number of solutions, but where is the proof of this?</p>
<p>Note (I found it after asking the question, and receiving answers): the question <a href="https://math.stackexchange.com/questions/186569/on-the-number-of-possible-solutions-for-a-quadratic-equation">On the number of possible solutions for a quadratic equation.</a> is strongly related to this question.</p>
| Mark Bennet | 2,906 | <p>It depends what you mean by a system of quadratics.</p>
<p>If you have $y=ax^2+by+c$ and $y=px^2+qx+r$ then you must have $(a-p)x^2+(b-q)x+(c-r)=0$. This equation has at most two solutions for $x$ (unless $a=p, b=q, c=r$) and each solution for $x$ gives a value for $y$.</p>
<p>However there is another way of thinking about quadratic curves in the plane, which would include circles (like x^2+y^2=r^2), hyperbolae (for example, xy=c), and ellipses - and rotated versions too. A parabola can cut a circle in four points, for example.</p>
|
1,176,938 | <p>How do you show that $-1+(x-4)(x-3)(x-2)(x-1)$ is irreducible in $\mathbb{Q}$?</p>
<p>I don't think you can use the eisenstein criterion here</p>
| Lubin | 17,760 | <p>Another argument, special to this polynomial.</p>
<p>Any number of arguments (including multiplying the thing out) show that the polynomial is $\equiv X^4-2\pmod5$. Thus in characteristic $5$, any root is a $16$-th root of unity. But the smallest field of characteristic $5$ whose cardinality is $\equiv1\pmod{16}$ is $\Bbb F_{625}$. Thus the polynomial is already irreducible over $\Bbb F_5$, thus over $\Bbb Z$, thus over $\Bbb Q$.</p>
|
3,483,260 | <p>Given the set <span class="math-container">$\{1,2,3,4,5,6,7\}$</span>.</p>
<p>We would like to create a string of size 8 so that each of the elements of the set appears at least once in the result. How many ways are there to create such a set?</p>
<p>I think that the answer should be: order 7 elements <span class="math-container">$7!$</span> and chose 1 number out of 7 to reappear and chose a position out of 8 available.</p>
<p>Am I correct? Or am I missing something?</p>
| Clement Yung | 620,517 | <p>For a non-negative, increasing, convex function, we have <span class="math-container">$\lim_{x \to -\infty}(f(x) - f'(x)) \geq 0$</span> exists. Consider:
<span class="math-container">$$
g(x) = (f(x) - f'(x))e^x
$$</span>
Differentiating yields:
<span class="math-container">$$
g'(x) = (f(x) - f''(x))e^x \geq 0
$$</span>
We have that <span class="math-container">$\lim_{x \to -\infty} g(x) \geq 0$</span>. Since <span class="math-container">$g'(x) \geq 0$</span>, <span class="math-container">$g$</span> is increasing so this implies <span class="math-container">$g(x) \geq 0$</span> <span class="math-container">$\forall x \in \mathbb{R}$</span>. Since <span class="math-container">$e^x > 0$</span>, we have <span class="math-container">$f(x) \geq f'(x)$</span>.</p>
<hr>
<p><strong>EDIT</strong>: To prove that <span class="math-container">$\lim_{x \to -\infty}(f(x) - f'(x)) \geq 0$</span>, we shall show that <span class="math-container">$\lim_{x \to -\infty} f(x) \geq 0$</span> and <span class="math-container">$\lim_{x \to -\infty} f'(x) = 0$</span>.</p>
<p><span class="math-container">$\lim_{x \to -\infty} f(x)$</span> exists because of monotone convergence theorem. Take any <span class="math-container">$(x_n)_{n \in \mathbb{Z}^+}$</span> such that <span class="math-container">$x_n \to -\infty$</span>, and we may assume WLOG that <span class="math-container">$x_n$</span> is monotone (as otherwise we can remove the terms which violate monotonicity, and the asymptotic behavior remains unchanged). Since <span class="math-container">$x_n \geq 0$</span> <span class="math-container">$\forall n$</span>, we have that <span class="math-container">$x_n$</span> converges. Since <span class="math-container">$f(x) \geq 0$</span>, the limit must also be non-negative.</p>
<p>Similarly, we can prove that <span class="math-container">$\lim_{x \to -\infty} f'(x) \geq 0$</span> as we have <span class="math-container">$f'(x) \geq 0$</span> and <span class="math-container">$f''(x) \geq 0$</span> so <span class="math-container">$f'$</span> is a monotonically increasing function bounded from below. We show further that <span class="math-container">$\lim_{x \to -\infty} f'(x) = 0$</span> by contradiction, in which if <span class="math-container">$\lim_{x \to -\infty} f'(x) = \epsilon > 0$</span>, then for some <span class="math-container">$M > 0$</span> we have <span class="math-container">$f'(x) > \frac{\epsilon}{2}$</span> for all <span class="math-container">$x < -M$</span>. This would violate the assumption that <span class="math-container">$f(x) \geq 0$</span>.</p>
|
3,249,809 | <p>What’s the explicit rule for for this number sequence?</p>
<p><span class="math-container">$$\displaystyle{1 \over 100},\ -{3 \over 95},\ {5 \over 90},\
-{7 \over 85},\ {9 \over 80}$$</span></p>
<p>The numerator changes to negative every other term, while the denominator subtracts <span class="math-container">$5$</span> every term. </p>
| beelal | 674,630 | <p>with convention we start from <span class="math-container">$n= 0$</span>, <span class="math-container">$x_n = \frac{(-1)^n(2n+1)}{100-5n}$</span> for <span class="math-container">$n=0,1, \ldots, 19$</span></p>
|
2,231,487 | <p>In [Mathematical Logic] by Chiswell and Hodges, within the context of natural deduction and the language of propositions LP (basically like <a href="http://www.cs.cornell.edu/courses/cs3110/2011sp/lectures/lec13-logic/logic.htm" rel="nofollow noreferrer">here</a>) it is asked to show, by counter-example that a certain 'sequent rule' is 'unacceptable'.</p>
<p>I suppose the proof should follow an example a few pages earlier that shows that the sequent rule $$(p_0 \to p_1 \vdash p_1)$$ is unacceptable due to the following counter-example: let both $p_0$ and $p_1$ mean $(2=3)$. The book argues that indeed if $(2=3)$ then $(2=3)$, so the left side is true, but the right side: $(2=3)$ is false. The conclusion is that we found a counter-examples and the sequent rule is unacceptable.</p>
<p>It is now asked to prove that this sequent rule is unacceptable: </p>
<p>If $$(\Gamma \vdash (\phi \lor \psi))$$ is correct (i.e has a derivation), then at least one of $$(\Gamma \vdash (\phi))$$ and $$(\Gamma \vdash (\psi))$$ is also correct. </p>
<p>The book hints that one should first try to give a counter-example 'for both sequents $(\vdash p_0)$ and $(\vdash \lnot p_0)$'. What could such a $p_0$ be? Does the book mean using as $p_0$ something like 'red is square'? Or some liar kind of sentence 'this sentence is false'? Or some known undecidable statement (which I doubt due to the level of this book)? Or am I totally off and misunderstand something?</p>
| Daniel Schepler | 337,888 | <p>My approach would be a bit different: if $\Gamma = \{ p \vee q \}$ where $p$ and $q$ are atomic formulae, then $\Gamma \vdash p \vee q$ so the rule would imply that either $\{ p \vee q \} \vdash p$ or $\{ p \vee q \} \vdash q$. But in the first case, assigning $p := (2 = 3)$ and $q := (3 = 3)$ would disprove that; and in the second case, assigning $p := (3 = 3)$ and $q := (2 = 3)$ would disprove that.</p>
<p>(It is interesting to observe, however, that if $\Gamma = \emptyset$ and you're restricted to intuitionistic logic, then the statement is actually true: if $\phi \vee \psi$ is an intuitionistic tautology then either $\phi$ is an intuitionistic tautology or $\psi$ is an intuitionistic tautology. Again, that's obviously false if you extend to classical logic.)</p>
|
19,285 | <p>Is anyone aware of Mathematica use/implementation of <a href="http://en.wikipedia.org/wiki/Random_forest">Random Forest</a> algorithm?</p>
| Daniel Lichtblau | 51 | <p>Disclaimer: This is not an implementation of the Random Forest Algorithm. Also, while I have on occasion used random florists, until today I had not heard of the Random Forest Algorithm.</p>
<p>I poked around a bit on the Net and learned that these take subsamples of data, subsampling the variables as well, and form decision trees for the subsetted subsamples. These then can be used to classify data points. One tests against each subsamples (tree) and returns the modal value.</p>
<p>Okay there is more that can be done with these, and I probably don't have even the above part correct. Whatever.</p>
<p>For some types of problem one can regard the "space" as having a vector of either continuous, discrete, or perhaps mixed values, with each datum then evaluating to some new value. That is to say, we pretend we have an unknown function <code>f</code> with <code>f(x_1,...,x_n) = y</code> The classifier goal is to find <code>` given the vector</code>(x_1,...,x_n<code>). The idea I have is to use</code>NearestFunction<code>as a surrogate for a tree. One creates each tree using a random subset of data points, and a random subset of dimensions. Classification of data can then be done by a voting system:. For a particular input, see what values the forest (set of</code>NearestFunction`s) gives, and use the most common as the end result. or can use the several most common if a viable set of candidate values is desired.</p>
<p>I will assume the input data is of the form <code>{vec1->val1,...,vecn->valn}</code>. For purposes of terminology to describe the functions below:</p>
<p><code>rules</code> is the master set
<code>ntrees</code> is how many trees to create (size of forest)
<code>subdim</code> is the dimension used for each tree
<code>sizes</code> is the number of rules to use in each subtree</p>
<pre><code>makeForest[rules_, ntrees_, subdim_, sizes_] := Module[
{len = Length[rules], dim = Length[rules[[1, 1]]],
slots, leaves, subleaves, nf},
Table[
leaves = rules[[RandomSample[Range[len], sizes]]];
slots = Sort[RandomSample[Range[dim], subdim]];
subleaves = Map[#[[1, slots]] -> #[[2]] &, leaves];
nf = Nearest[subleaves];
{slots, nf}
, {ntrees}]
]
classify[vector_, forest_, n_] := Module[
{vals, tree, subvec},
vals = Flatten[Table[
tree = forest[[j]];
tree[[2]][vector[[tree[[1]]]]]
, {j, Length[forest]}]];
Reverse[SortBy[Tally[vals], Last][[-n ;; -1]]]
]
</code></pre>
<p>That's it.</p>
<p>Here is an example. I'll take a million points in the 8-cube {-1,1}^8. We give them ordinal values based on orthant (use base 2...) but before evaluating we "pollute" them first with Gaussian noise.</p>
<pre><code>dim = 8;
size = 6;
pts = RandomReal[{-1, 1}, {10^size, dim}];
ruls = Map[# ->
FromDigits[(Sign[# +
RandomReal[NormalDistribution[0, .4], Length[#]]] + 1)/2,
2] &, pts];
</code></pre>
<p>Now we'll create a "forest" (I really should use a different term since I doubt this is what an actual random forest is, but...). I use 300 trees, each taking 4 (of the 8) vector positions, and each using 1000 elements from the full set, chosen at random.</p>
<pre><code>Timing[rForest = makeForest[ruls, 300, 4, 1000];]
(* Out[80]= {4.740000, Null} *)
</code></pre>
<p>Now i create an random 8-vector with coordinates in <code>{-1,1}</code>, and also evaluate it according to our actual function.</p>
<pre><code>SeedRandom[11111];
randvec = RandomReal[{-1, 1}, dim]
FromDigits[(Sign[randvec] + 1)/2, 2]
(* Out[82]= {-0.472529240338, 0.534376840246, 0.286266627232, \
0.690979660248, -0.663034535046, 0.715836737332, -0.389931777484, \
0.66884997164} *)
(* Out[83]= 117 *)
</code></pre>
<p>We run it through <code>classify[]</code>.</p>
<pre><code>Timing[candidates = classify[randvec, rForest, 5]]
(* Out[84]= {1.370000, {{117, 13}, {245, 10}, {116, 10}, {101, 10}, {53,
10}}} *)
</code></pre>
<p>While I doubt it will always give the "expected" result at the top of the list, it is encouraging to see that it might do so. Also the next ones do tend to share bits with 117.</p>
<p>Of course it might be unrealistic to use half of the full set of dimensions in the trees, so I really do now know how well this might scale to "real world" problems of high dimension. But I thought it might be worth showing this approach since it is simple in terms of code and others might have ideas on tweaking it for better accuracy/performance.</p>
|
3,053,386 | <p>This might be a very basic question for some of you. Indeed in <span class="math-container">$\textbf Z$</span>, it's very easy. For example, <span class="math-container">$\textbf Z / \langle 2 \rangle$</span> consists of <span class="math-container">$\langle 2 \rangle$</span> and <span class="math-container">$\langle 2 \rangle + 1$</span>. Obviously just two elements. In general, if <span class="math-container">$p$</span> is a positive prime in <span class="math-container">$\textbf Z$</span>, then <span class="math-container">$\textbf Z / \langle p \rangle$</span> consists of one principal ideal and <span class="math-container">$p - 1$</span> cosets.</p>
<p>I guess it's also easy in imaginary quadratic integer rings, since we can visualize them in the complex plane, e.g., <span class="math-container">$\textbf Z[i] / \langle 1 + i \rangle$</span> consists of <span class="math-container">$\langle 1 + i \rangle$</span>, <span class="math-container">$\langle 1 + i \rangle + 1$</span> and <span class="math-container">$\langle 1 + i \rangle + i$</span>... wait a minute, three elements? I'm not sure that's quite right.</p>
<p>And I really have no idea how to go about, say, <span class="math-container">$\textbf Z[\sqrt{14}] / \langle 4 + \sqrt{14} \rangle$</span>. To say nothing of something like <span class="math-container">$\textbf Z[\sqrt{10}] / \langle 2, \sqrt{10} \rangle$</span>.</p>
<p>Given a ring <span class="math-container">$R$</span> of algebraic integers of degree <span class="math-container">$2$</span>, and a prime ideal <span class="math-container">$\mathfrak P$</span>, how do you determine how many elements there are in <span class="math-container">$R / \mathfrak P$</span>?</p>
| Kenny Wong | 301,805 | <p>For principal ideals, it's easy: if <span class="math-container">$L$</span> is a number field with ring of integers <span class="math-container">$\mathcal O_L$</span>, and if <span class="math-container">$(a)$</span> is an ideal in <span class="math-container">$\mathcal O_L$</span>, then the number of elements in <span class="math-container">$\mathcal O_L / (a)$</span> is <span class="math-container">$|N_{L / \mathbb Q} (a)|$</span>, where <span class="math-container">$N_{L / \mathbb Q}(a)$</span> denotes the norm of the element <span class="math-container">$a$</span>.</p>
<p>For a quadratic number field <span class="math-container">$\mathbb Q(\sqrt{d})$</span>, the norm of an element <span class="math-container">$x + y \sqrt{d}$</span> is simply <span class="math-container">$x^2 - dy^2$</span>.</p>
<p>So the number of elements in <span class="math-container">$\mathbb Z[\sqrt{14}] / (4 + \sqrt{14})$</span> is <span class="math-container">$| 4^2 - 14 \times 1^2 | = 2$</span>.</p>
<p>Similarly, the number of elements in <span class="math-container">$\mathbb Z[i] / (1 + i)$</span> is <span class="math-container">$|1^2 + 1 \times 1^2 | = 2$</span>. (You double counted: <span class="math-container">$1 + (1 + i)$</span> and <span class="math-container">$i + (1 + i)$</span> are the same thing, since <span class="math-container">$1 - i \in (1 + i)$</span> by virtue of the fact that <span class="math-container">$-i(1 + i) = 1 - i$</span>.)</p>
<hr>
<p>As for your final example, <span class="math-container">$\mathbb Z[\sqrt{10}] / (2, \sqrt{10})$</span>, it might be easier to use an integral basis.</p>
<p>It is well known that <span class="math-container">$\{ 1, \sqrt{10} \}$</span> is an integral basis for <span class="math-container">$\mathbb Z[\sqrt{10}]$</span>, which is to say that every element in <span class="math-container">$\mathbb Z[\sqrt{10}]$</span> can be written in the form <span class="math-container">$a + b\sqrt{10}$</span> for a unique choice of <span class="math-container">$a, b \in \mathbb Z$</span>. Therefore, the elements in <span class="math-container">$(2, \sqrt{10})$</span> are precisely the elements of the form</p>
<p><span class="math-container">$$ (u + v\sqrt{10})\times 2 +(w + z\sqrt{10})\times \sqrt{10} = (2u + 10z) + (2v + w) \sqrt{10}$$</span>
where <span class="math-container">$u, v, w, z \in \mathbb Z$</span>. And if you think about it, this includes precisely those elements <span class="math-container">$a + b \sqrt{10}$</span> in <span class="math-container">$ \mathbb Z[\sqrt{10}]$</span> where <span class="math-container">$a$</span> is <em>even</em>. Hence <span class="math-container">$\mathbb Z[\sqrt{10}] / (2, \sqrt{10})$</span> contains two cosets: <span class="math-container">$(2, \sqrt{10})$</span> and <span class="math-container">$1 + (2, \sqrt{10})$</span>.</p>
<p>I hope these examples provide you with enough techniques to be getting on with....</p>
|
3,817,104 | <p>For <span class="math-container">$a,b,c \in \Big[\dfrac{1}{3},3\Big].$</span> Prove<span class="math-container">$:$</span></p>
<p><span class="math-container">$$(a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) \leqslant 25.$$</span></p>
<p>Assume <span class="math-container">$a\equiv \text{mid}\{a,b,c\},$</span> we have<span class="math-container">$:$</span></p>
<p><span class="math-container">$$25-(a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) =\dfrac{2}{bc} (10bc-b^2-c^2) +\dfrac{c+b}{abc} (a-b)(c-a)\geqslant 0.$$</span></p>
<p>I wish to find a proof with <span class="math-container">$a:\neq {\rm mid}\left \{ a, b, c \right \},$</span> or another proof<span class="math-container">$?$</span></p>
<p>Actually<span class="math-container">$,$</span> I also found a proof true for all <span class="math-container">$a,b,c \in \Big[\dfrac{1}{3},3\Big],$</span> but very ugly.</p>
<p>After clearing the denominators<span class="math-container">$,$</span> need to prove<span class="math-container">$:$</span></p>
<p><span class="math-container">$$f:=22abc-a^2c-a^2b-b^2c-ab^2-bc^2-ac^2\geqslant 0$$</span></p>
<p>but we have<span class="math-container">$:$</span></p>
<p><span class="math-container">$$f=\dfrac{1}{32} \left( 3-a \right) \left( 3-b \right) \Big( c-\dfrac{1}{3} \Big) +
\left( 3-a \right) \left( a-\dfrac{1}{3} \right) \left( b-\dfrac{1}{3} \right) +\\+{
\frac {703}{32}}\, \left( a-\dfrac{1}{3} \right) \left( b-\dfrac{1}{3} \right) \left(
c-\dfrac{1}{3} \right) +{\frac {9}{32}} \left( 3-a \right) \left( 3-c
\right) \left( a-\dfrac{1}{3} \right) +\dfrac{1}{4} \left( 3-b \right) \left( 3-c
\right) \left( c-\dfrac{1}{3} \right) +\dfrac{5}{4} \left( 3-c \right) \left( c-\dfrac{1}{3}
\right) \left( a-\dfrac{1}{3} \right) +{\frac {49}{32}} \left( 3-c \right)
\left( b-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) + \left( 3-b \right)
\left( b-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) +\\+{\frac {21}{16}}\,
\left( 3-b \right) \left( a-\dfrac{1}{3} \right) \left( b-\dfrac{1}{3} \right) \\+\dfrac{5}{4}\,
\left( 3-a \right) \left( c-\dfrac{1}{3} \right) \left( a-\dfrac{1}{3} \right) +\dfrac{1}{32}
\, \left( 3-a \right) ^{2} \left( 3-c \right) +\dfrac{1}{4}\, \left( 3-b
\right) \left( b-\dfrac{1}{3} \right) ^{2}+\dfrac{1}{32} \left( 3-b \right) ^{2}
\left( a-\dfrac{1}{3} \right) +{\frac {9}{32}} \left( a-\dfrac{1}{3} \right) \left(
b-\dfrac{1}{3} \right) ^{2}+\dfrac{1}{4} \left( a-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) ^{
2}+\dfrac{1}{4} \left( b-\dfrac{1}{3} \right) \left( 3-b \right) ^{2}+{\frac {9}{32}}
\, \left( b-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) ^{2}$$</span></p>
<p>So we are done.</p>
<p>If you want to check my decomposition<span class="math-container">$,$</span> please see the text <a href="https://github.com/tthnew/Text/blob/master/79133" rel="noreferrer">here</a>.</p>
| nguyenhuyenag | 410,198 | <p>I found a better estimation
<span class="math-container">$$ (a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) \leqslant \frac{209}{9}.$$</span>
Equality occur when <span class="math-container">$a=b=3,\,c=\frac 13$</span> or <span class="math-container">$a=b=\frac 13,\,c=3.$</span></p>
|
3,817,104 | <p>For <span class="math-container">$a,b,c \in \Big[\dfrac{1}{3},3\Big].$</span> Prove<span class="math-container">$:$</span></p>
<p><span class="math-container">$$(a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) \leqslant 25.$$</span></p>
<p>Assume <span class="math-container">$a\equiv \text{mid}\{a,b,c\},$</span> we have<span class="math-container">$:$</span></p>
<p><span class="math-container">$$25-(a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) =\dfrac{2}{bc} (10bc-b^2-c^2) +\dfrac{c+b}{abc} (a-b)(c-a)\geqslant 0.$$</span></p>
<p>I wish to find a proof with <span class="math-container">$a:\neq {\rm mid}\left \{ a, b, c \right \},$</span> or another proof<span class="math-container">$?$</span></p>
<p>Actually<span class="math-container">$,$</span> I also found a proof true for all <span class="math-container">$a,b,c \in \Big[\dfrac{1}{3},3\Big],$</span> but very ugly.</p>
<p>After clearing the denominators<span class="math-container">$,$</span> need to prove<span class="math-container">$:$</span></p>
<p><span class="math-container">$$f:=22abc-a^2c-a^2b-b^2c-ab^2-bc^2-ac^2\geqslant 0$$</span></p>
<p>but we have<span class="math-container">$:$</span></p>
<p><span class="math-container">$$f=\dfrac{1}{32} \left( 3-a \right) \left( 3-b \right) \Big( c-\dfrac{1}{3} \Big) +
\left( 3-a \right) \left( a-\dfrac{1}{3} \right) \left( b-\dfrac{1}{3} \right) +\\+{
\frac {703}{32}}\, \left( a-\dfrac{1}{3} \right) \left( b-\dfrac{1}{3} \right) \left(
c-\dfrac{1}{3} \right) +{\frac {9}{32}} \left( 3-a \right) \left( 3-c
\right) \left( a-\dfrac{1}{3} \right) +\dfrac{1}{4} \left( 3-b \right) \left( 3-c
\right) \left( c-\dfrac{1}{3} \right) +\dfrac{5}{4} \left( 3-c \right) \left( c-\dfrac{1}{3}
\right) \left( a-\dfrac{1}{3} \right) +{\frac {49}{32}} \left( 3-c \right)
\left( b-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) + \left( 3-b \right)
\left( b-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) +\\+{\frac {21}{16}}\,
\left( 3-b \right) \left( a-\dfrac{1}{3} \right) \left( b-\dfrac{1}{3} \right) \\+\dfrac{5}{4}\,
\left( 3-a \right) \left( c-\dfrac{1}{3} \right) \left( a-\dfrac{1}{3} \right) +\dfrac{1}{32}
\, \left( 3-a \right) ^{2} \left( 3-c \right) +\dfrac{1}{4}\, \left( 3-b
\right) \left( b-\dfrac{1}{3} \right) ^{2}+\dfrac{1}{32} \left( 3-b \right) ^{2}
\left( a-\dfrac{1}{3} \right) +{\frac {9}{32}} \left( a-\dfrac{1}{3} \right) \left(
b-\dfrac{1}{3} \right) ^{2}+\dfrac{1}{4} \left( a-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) ^{
2}+\dfrac{1}{4} \left( b-\dfrac{1}{3} \right) \left( 3-b \right) ^{2}+{\frac {9}{32}}
\, \left( b-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) ^{2}$$</span></p>
<p>So we are done.</p>
<p>If you want to check my decomposition<span class="math-container">$,$</span> please see the text <a href="https://github.com/tthnew/Text/blob/master/79133" rel="noreferrer">here</a>.</p>
| richrow | 633,714 | <p>Let <span class="math-container">$f(a,b,c)=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$</span>. Note that <span class="math-container">$f$</span> is concave of each variable (if other variables are fixed). Hence, since concave on <span class="math-container">$I$</span> fucntion attains its maximum at endpoint of <span class="math-container">$I$</span> (here <span class="math-container">$I=[m,M]=\left[\frac{1}{3},3\right]$</span>)
<span class="math-container">$$
\max_{(a,b,c)\in I^3} f=\max_{(a,b,c)\in\{m,M\}^3} f.
$$</span>
Thus, we need only to compute these 8 values and choose the maximal one.</p>
<p><strong>Details:</strong> consider any point <span class="math-container">$(a,b,c)$</span>, fix <span class="math-container">$b$</span> and <span class="math-container">$c$</span> and consider <span class="math-container">$f$</span> as a function of <span class="math-container">$a$</span>. We obtain
<span class="math-container">$$
f(a,b,c)\leq\max\{f(m,b,c),f(M,b,c)\},
$$</span>
so we can assume that <span class="math-container">$a\in\{m,M\}$</span>. Now repeat this argument for <span class="math-container">$b$</span> and <span class="math-container">$c$</span>.</p>
|
1,287,225 | <p>So I have come across a question asked by my peers. </p>
<p>Define: $$g:=\sqrt{E[|y_r(t)|^2]}$$</p>
<p>Given that $$y_r(t)=\sqrt{t}\cdot h+k,$$ where $h$ and $k$ are independent random variables with variance $\sigma_h$ and $\sigma_k$ respectively. </p>
<p>So what is $g$ in this case? The answer key gives: $g=\sqrt{t \sigma_h + \sigma_k}$
.</p>
<hr>
<p><strong>Well, I think this answer make sense. But my friend ask me, which random variables does it apply the expectation?</strong></p>
<p>I am confuse to his question, I thought when you take the expectation, you take it over all the random variable. But he says this doesn't make sense to him. Now I am confused. </p>
| Michael Hardy | 11,667 | <p>$\newcommand{\E}{\operatorname{E}}\newcommand{\var}{\operatorname{var}}$
It follows from independence that $\E(hk)=\E(h)\E(k)$.</p>
<p>If $\E(h)=\E(k) = 0$ then $\E(h^2)= \var(h)$ and $\E(k^2)=\var(k)$ and we have
\begin{align}
\E\left((\sqrt{t}\, h + k)^2\right) = \E(th^2 + 2\sqrt t\,hk + k^2) & = t\E(h^2) + 2\sqrt t\E(h)\E(k) + \E(k^2) \\[10pt]
& = t\var(h) + 2\cdot0\cdot0 + \var(k).
\end{align}
Therefore the answer key is right provided $\E(h)=\E(k)=0$.</p>
|
1,287,225 | <p>So I have come across a question asked by my peers. </p>
<p>Define: $$g:=\sqrt{E[|y_r(t)|^2]}$$</p>
<p>Given that $$y_r(t)=\sqrt{t}\cdot h+k,$$ where $h$ and $k$ are independent random variables with variance $\sigma_h$ and $\sigma_k$ respectively. </p>
<p>So what is $g$ in this case? The answer key gives: $g=\sqrt{t \sigma_h + \sigma_k}$
.</p>
<hr>
<p><strong>Well, I think this answer make sense. But my friend ask me, which random variables does it apply the expectation?</strong></p>
<p>I am confuse to his question, I thought when you take the expectation, you take it over all the random variable. But he says this doesn't make sense to him. Now I am confused. </p>
| ronno | 32,766 | <p>Building on Michael Hardy's answer. I use the notation $\bar h = E(h)$ and $\bar k = E(k)$ for readability. Since $h,k$ are independent, $E(hk) = \bar h \bar k$.</p>
<p>By definition, $E(h^2) = \sigma_h + \bar h^2$ and $E(k^2) = \sigma_k + \bar k^2$. We have,
$$\begin{align*}E((\sqrt{t}h+k)^2)& =E(th^2+2\sqrt{t}hk + k^2)\\ &= tE(h^2)+2\sqrt{t}E(hk)+E(k^2)\\ &= t(\sigma_h + \bar h^2)+2\sqrt{t}\bar h \bar k + \sigma_k + \bar k^2\\ &= (t\sigma_h + \sigma_k) + (\sqrt{t} \bar h + \bar k)^2 \,\text{.}\end{align*}$$
Therefore the answer key is right iff $\sqrt{t}E(h)+E(k)=E(y_r(t)) = 0$.</p>
<p>The expectation is with respect to the product variable $(h,k)$, the distribution of which is determined by those of $h$ and $k$, since they are independent.</p>
|
540,227 | <p>I was try to understand the following theorem:-</p>
<p><strong>Let $X,Y$ be two path connected spaces which are of the same homotopy type.Then their fundamental groups are isomorphic.</strong></p>
<p><strong>Proof:</strong> The fundamental groups of both the spaces $X$ and $Y$ are independent on the base points since they are path connected. Since $X$ and $Y$ are of the same homotopy type, there exist continuous maps $f:X\to Y $ and $g:Y\to X$ such that $g\circ f\sim I_X$ by a homotopy, say, $F$ and $f\circ g \sim I_Y$ by some homotopy, say $G$. Let $x_0\in X$ be a base point. Let<br>
$$f_\#:\pi_1(X,x_0)\to \pi_1(Y,f(x_0))$$ and $$g_\#:\pi_1(Y,f(x_0))\to \pi_1(X,g(f(x_0)))$$ be the induced homomorphisms. Let $\sigma$ be the path joining $x_0$ to $gf(x_0)$ defined by the homotopy $F$.</p>
<p>After that the author says that $\sigma_\#$ is a isomorphism. obviously $\sigma_\#$ is a homomorphism but I could not understand how it becomes a isomorphism.</p>
<p>Can someone explain me please. thanks for your kind help and time.</p>
| Edoardo Lanari | 77,181 | <p>$f_{\#}:\pi_1(X,x_0) \to \pi_1(Y,f(x_0))$ is an isomorphism since $f$ is an homotopy equivalence; moreover $\pi_1(Y,f(x_0)) \simeq \pi_1(Y,y_0)$ since $Y$ is path-connected. </p>
|
1,494,409 | <blockquote>
<p>Let <span class="math-container">$\Bbb R^+$</span> denote the real numbers. Suppose <span class="math-container">$\phi:\Bbb R^+\to\Bbb R^+$</span> is an automorphism of the group <span class="math-container">$\Bbb R^+$</span> under multiplication with <span class="math-container">$\phi(4)=7$</span>.</p>
<p>Find <span class="math-container">$\phi(2)$</span>, <span class="math-container">$\phi(8)$</span>, and <span class="math-container">$\phi(1/4).$</span></p>
</blockquote>
<p>I've done this problem in groups of integers modulo <span class="math-container">$n$</span> but I am not sure how to start here. Please explain what <span class="math-container">$\phi(4)=7$</span> is.</p>
<p>Thank you!</p>
| CPM | 119,124 | <p>If $\phi(4)=7$, then $7=\phi(4)=\phi(2\cdot 2)=\phi(2)\phi(2) = \phi(2)^2$. Can you take it from there?</p>
|
37,804 | <p>I'm trying to gain some intuition for the usefullness of the spectral theory for bounded self adjoint operators. I work in PDE and any interesting applications/examples I've ever encountered are concerning <em>compact operators</em> and <em>unbounded operators</em>. Here I have the examples of $-\Delta$, the laplacian and $(-\Delta)^{-1}$, the latter being compact.</p>
<p>The most common example I see of a bounded non-compact operator is the shift map on $l_2$ given by $T(u_1,u_2,\cdots) = (u_2,u_3,\cdots)$. While this nicely illustrates the different kind of spectra, I don't see why this is useful or where this may come up in practice.<em>Why does knowing things about the spectrum of the shift operator help you in any practical way?</em></p>
<p>Secondly, concerning the spectral theorem for bounded, <em>self adjoint</em> operators. All useful applications I have encountered concern <em>compact or unbounded operators</em>. Is there an example arising in PDE (preferably) or some other applied field where knowing the spectral representation for a bounded, non-compact operator is useful? I have yet to encounter one that didn't just reduce to the compact case. Any insight/suggestions are appreciated.</p>
<p>Best,
dorian</p>
| Paul Siegel | 4,362 | <p>I think Helge's answer cuts to the historical heart of the matter: solution operators for various differential equations tend to be bounded, non-compact operators (obtained in many cases from an unbounded differential operator via the functional calculus), and it is often quite useful from that point of view to know something about their spectra. This is one reason why the theory of elliptic operators over non-compact spaces is more complicated than the corresponding theory for compact spaces: one has to deal with the fact that the eigenvalues of the solution operators can accumulate at 0.</p>
<p>What I find more persuasive, however, are the ways in which spectral theory mediates the relationship between functional analysis and geometry. In many cases you will miss this relationship unless you ignore compact operators entirely. The celebrated Atiyah-Singer Index Theorem provides a particularly dramatic example of this phenomenon, but I'll focus on more digestible examples (A-S is more about spectral theory for unbounded operators anyway).</p>
<p>First, consider the classical Toeplitz operators. Given a complex valued function $g \in C(S^1)$, the Toeplitz operator $T_g$ with symbol $g$ is defined as follows. Form the Hardy space $H^2(S^1)$ by considering the $L^2$-closure of the space of polynomial functions on $S^1$, regarding $S^1$ as a subspace of $\mathbb{C}$, and let $P$ denote the orthogonal projection from $L^2(S^1)$ to $H^2(S^1)$. Then define $T_g: H^2(S^1) \to H^2(S^1)$ to be $T_g(f) = T(fg)$. This is a bounded operator, and the classical Toeplitz index theorem asserts that its Fredholm index (the dimension of its kernel minus the dimension of its cokernel) is precisely the winding number of $g$. Thus an analytic invariant of the Toeplitz operator with symbol $g$ calculates a topological invariant of $g$.</p>
<p>That result by itself isn't heavy on the spectral theory. The connection with spectral theory is revealed by a more refined statement. Recall that the essential spectrum of a bounded operator $T$ on a Hilbert space $H$ is the spectrum of the image of $T$ in the Calkin algebra $Q(H)$ (which is the space of bounded operators on $H$ modulo the space of compact operators). A consequence of the Toeplitz index theorem (and its proof) is the fact that the essential spectrum of the Toeplitz operator $T_g$ is precisely the range of $g$. While this statement alone is scant evidence, this suggests a deep relationship between the essential spectrum of a bounded operator and geometry. This line of thinking culminates in the Brown-Douglas-Fillmore theorem, which makes the following startling assertion. Let $X$ be a nonempty subset of the complex plane, and define $Ext(X)$ to be the space of essential unitary equivalence classes of essentially normal operators with essential spectrum $X$ (here "essential" always means "modulo compact operators"). Direct sum of operators gives $Ext(X)$ the structure of a commutative semigroup, and the BDF theorem asserts that $Ext(X)$ is naturally isomorphic to the space $Hom(\pi^1(X), \mathbb{Z})$ of group homomorphisms between the first cohomotopy group of $X$ and $\mathbb{Z}$. (Note: it is not even obvious that $Ext(X)$ has a zero element!) Thus spectral theory helps to classify certain kinds of bounded operators mod compacts in a particularly beautiful way (via algebraic topology).</p>
<p>As mentioned above, there are also fruitful interactions between functional analysis and geometry - mediated by spectral theory - which flow from analysis to geometry. Aside from Atiyah-Singer, there are fruitful generalizations of the Toeplitz index theorem along these lines. But let me give a different sort of example in the theory of hyperbolic diffeomorphisms.</p>
<p>Informally, a diffeomorphism $f: M \to M$ on a smooth manifold $M$ is said to be Anosov (or uniformly hyperbolic) if $M$ admits transverse stable and unstable foliations for the action of $f$. Prototypical examples of Anosov diffeomorphisms on the 2-torus can be obtained by considering $2 \times 2$ matrices with integer entries and irrational eigenvalues. It turns out that spectral theory has a great deal to say about smooth dynamical systems in general and Anosov diffeomorphisms in particular.</p>
<p>Given any diffeomorphism $f: M \to M$, consider the bounded operator $f_*$ on the Banach space $\Gamma^0(TM)$ of continuous vector fields on $M$ defined by </p>
<p>$(f_*v)(x) = df(v)(f^{-1}(x))$ </p>
<p>If the non-periodic orbits of $f$ are dense in $M$, then a theorem of Mather asserts that the spectrum of $f_*$ is a disjoint union of finitely many annuli centered at the origin. If $H_i$ is the invariant subspace for $f_*$ corresponding to the $i$th annulus then the subspaces $E_i(x)$ of $T_x M$ consisting of the vectors $v(x)$ for $v \in H_i$ form a $df$ invariant continuous distribution on $M$, and the direct sum of the $E_i(x)$'s gives the whole tangent space $T_x M$. So the Mather spectral theory of $f$ is very closely related to its dynamics. Indeed, one can characterize the Anosov diffeomorphisms as precisely those $f$ for which $1$ is not in the spectrum of $f_*$. Pesin used this idea to prove that Anosov diffeomorphisms are structurally stable, meaning they form an open subset of the full diffeomorphism group of $M$ (so that a small perturbation of an Anosov diffeomorphism is still Anosov). The same strategy also works for partially hyperbolic dynamical systems, which have a slightly different spectral characterization.</p>
|
37,804 | <p>I'm trying to gain some intuition for the usefullness of the spectral theory for bounded self adjoint operators. I work in PDE and any interesting applications/examples I've ever encountered are concerning <em>compact operators</em> and <em>unbounded operators</em>. Here I have the examples of $-\Delta$, the laplacian and $(-\Delta)^{-1}$, the latter being compact.</p>
<p>The most common example I see of a bounded non-compact operator is the shift map on $l_2$ given by $T(u_1,u_2,\cdots) = (u_2,u_3,\cdots)$. While this nicely illustrates the different kind of spectra, I don't see why this is useful or where this may come up in practice.<em>Why does knowing things about the spectrum of the shift operator help you in any practical way?</em></p>
<p>Secondly, concerning the spectral theorem for bounded, <em>self adjoint</em> operators. All useful applications I have encountered concern <em>compact or unbounded operators</em>. Is there an example arising in PDE (preferably) or some other applied field where knowing the spectral representation for a bounded, non-compact operator is useful? I have yet to encounter one that didn't just reduce to the compact case. Any insight/suggestions are appreciated.</p>
<p>Best,
dorian</p>
| Pietro Majer | 6,101 | <p>In Ergodic theory unitary operators naturally arise as $U_T:f\mapsto f\circ T$ where $T$ is a measure preserving transformation on a probability space. The spectral theory of the operator $U_T$ carries some information on the dynamics of $T$. </p>
|
2,201,085 | <p>Let
$$x_{1},x_{2},x_{3},x_{5},x_{6}\ge 0$$ such that
$$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=1$$
Find the maximum of the value of
$$\sum_{i=1}^{6}x_{i}\;x_{i+1}\;x_{i+2}\;x_{i+3}$$
where
$$x_{7}=x_{1},\quad x_{8}=x_{2},\quad x_{9}=x_{3}\,.$$</p>
| Empy2 | 81,790 | <p>Let $B=(x_2+x_6)/2,C=(x_3+x_5)/2,tR=(x_2-x_6)/2,tS=(x_3-x_5)/2$.<br>
The numbers are now $A,B(1+tR),C(1+tS),D,C(1-tS),B(1-tR)$. The sum is
$$2AB^2C+2BC^2D+2ABCD-2t^2(AB^2CR^2+BC^2DS^2-ABCDRS)$$
Let $t$ vary. The maximum is either when $t=0$ or when $t$ is as large as possible. </p>
<p>$t$ is at its largest when $B(1-tR)$ or $C(1-tS)$ has reached zero. It is simple to calculate the sum's maximum when one value is zero; that maximum is $1/256$. </p>
<p>If the maximum is when $t=0$, then the numbers are $A,B,C,D,C,B$. That is, $x_2=x_6$ and $x_3=x_5$.<br>
A similar argument shows that any maximum other than $1/256$ has $x_2=x_4$ and $x_1=x_5$.<br>
So the numbers are $C,B,C,B,C,B$ and the sum is $6B^2C^2$. That is maximum when $B=C=1/6$, and the sum is $1/216$. </p>
|
2,121,583 | <p>Question: Let $f,g: X \rightarrow \mathbb{R}$ continous (over $X$, and $X$ is a metric space). If $\overline{Y}\subset X $, and $f(y)=g(y)$ for every $y\in Y $, prove that $\left.f\right|_\overline{Y}= \left.g\right|_\overline{Y}$.</p>
<p>Attempt:</p>
<p>Since $\overline{Y} \subset X$, it follows that $\left.f\right|_\overline{Y}$ and $\left.g\right|_\overline{Y}$ are continuous functions. Given $\epsilon>0$, let $a\in \overline{Y}$. Since $\overline{Y}=Y\cup Y'$, assume that $a \in Y'$. </p>
<p>$\left.f\right|_\overline{Y}$ and $\left.g\right|_\overline{Y}$ are continuous, so there exists $\delta_1,\delta_2>0$ such that for $x\in \overline{Y}$, $|x-a|<\delta_1$ and $|x-a|<\delta_2$ we have:</p>
<p>$|f(x)-f(a)|<\epsilon/2$ and $|g(x)-g(a)|<\epsilon/2$, respectively.</p>
<p>Since we are supposing that $a\in Y'$, the continuity of $\left.f\right|_\overline{Y}$ and $\left.g\right|_\overline{Y}$ also gives us that $\lim_{x \to a} f(x) = f(a)$ and $\lim_{x \to a} g(x) = g(a)$. Using the hyphothesis, we conclude that, in particular, $f(a)=g(a)$.</p>
<p>Hence, taking $\delta_0$=min$\{\delta_1,\delta_2\}$, $x \in \overline{Y}$ and $x\in(a-\delta_0,a+\delta_0)$ implies that :</p>
<p>$|f(x)-g(x)|\leq |f(x)-f(a)|+|f(a)-g(x)|<\epsilon/2+\epsilon/2=\epsilon.$</p>
<p>Since $\epsilon$ is arbitrary, it proves that $f(x)=g(x)$ for every $x\in \overline{Y}$.</p>
<p>Doing the same, supposing that $a\in Y$, we still get that $f(x)=g(x)$ for every $x\in \overline{Y}$.</p>
<p>In either case, we coclude that $\left.f\right|_\overline{Y}= \left.g\right|_\overline{Y}$.</p>
<p>Is it correct?</p>
| Momo | 384,029 | <p>HINT: Write each point of $\bar{Y}-Y$ as a limit of a sequence in $Y$, then use continuity.</p>
|
2,992,416 | <blockquote>
<p>A pendulum of length <span class="math-container">$1$</span> m and mass <span class="math-container">$100$</span> g attached to the end. Another 100 g mass move horizontally with speed 2 m/s. When collision happens this ball sticks with the pendulum and move together. Find the initial linear speed of the block.</p>
</blockquote>
<p>conservation of energy :-</p>
<pre><code>E(initially) = .5*m*v*v = .5*.1*2*2 = .2 joule
E(initially) = E(final) = .5*(.2)*v*v = .2 joule
v=sqrt(2)
</code></pre>
<p>conservation of momentum :-</p>
<pre><code>m1 v1 + m2 v1 = m1 v2 + m2 v2
.1*0 + .1*2 = .1 (2v)`
v=1 m/s
</code></pre>
<p>main question: why they are different <code>1 m/s</code> and <code>sqrt(2)</code></p>
| David Lui | 445,002 | <p>Consider the bijection <span class="math-container">$\{-1, 1\}^{\mathbb{Z}^2}$</span> to <span class="math-container">$P(\mathbb{Z}^2)$</span> by sending a function <span class="math-container">$f$</span> to the set <span class="math-container">$\{x \in \mathbb{Z}^2 : f(x) = 1\}$</span>. The inverse is <span class="math-container">$S \rightarrow f$</span> where <span class="math-container">$f(x) = \begin{cases} 1 \text{ if x $\in S$} \\ -1 \text{ otherwise} \end{cases}$</span> </p>
<p>The power set of any infinite set is uncountable by Cantor's theorem.</p>
|
2,992,416 | <blockquote>
<p>A pendulum of length <span class="math-container">$1$</span> m and mass <span class="math-container">$100$</span> g attached to the end. Another 100 g mass move horizontally with speed 2 m/s. When collision happens this ball sticks with the pendulum and move together. Find the initial linear speed of the block.</p>
</blockquote>
<p>conservation of energy :-</p>
<pre><code>E(initially) = .5*m*v*v = .5*.1*2*2 = .2 joule
E(initially) = E(final) = .5*(.2)*v*v = .2 joule
v=sqrt(2)
</code></pre>
<p>conservation of momentum :-</p>
<pre><code>m1 v1 + m2 v1 = m1 v2 + m2 v2
.1*0 + .1*2 = .1 (2v)`
v=1 m/s
</code></pre>
<p>main question: why they are different <code>1 m/s</code> and <code>sqrt(2)</code></p>
| Robert Z | 299,698 | <p>Hint. Assume that <span class="math-container">$\{f_n\}_{n\in\mathbb{N}}$</span> is a countable list of ALL such functions from <span class="math-container">$\mathbb{Z}^2\to \{-1,1\}$</span>. Now define a new function <span class="math-container">$g:\mathbb{Z}^2\to \{-1,1\}$</span> such that for any <span class="math-container">$(n,m)\in \mathbb{Z}^2$</span>, <span class="math-container">$g(n,m):=-f_{|n|}(n,m)$</span>. Does the function <span class="math-container">$g$</span> belong to the list?</p>
|
1,031,559 | <p>I'm getting ahead in my differential equations textbook (<em>Fundamentals of Differential Equations</em> by Nagle et. al) and in the chapter of Laplace Transforms it states that the rectangular window function $\Pi_{a,b}\left(t\right)$ is given by
\begin{align}
\Pi_{a,b}\left(t\right):=u\left(t-a\right)-u\left(t-b\right)=\begin{cases}
0, & t<a, \\
1, & a<t<b, \\
0, & b <t.
\end{cases}
\end{align}
However, in another textbook I'm reading about Fourier Transforms (which I ATM know very little about, just the basics since I just got it, <em>Fourier Transforms: an introduction for Engineers</em> by Gray et. al.) they've stated that the rectangle function, a variation of the box function, <em>considered by Bracewell</em> (in <em>The Fourier Transform and its applications</em>, 1965) is given by
\begin{align}\Pi\left(t\right)=
\begin{cases}
1, & \left|t\right|<\frac{1}{2}, \\
\frac{1}{2}, & \left|t\right|=\frac{1}{2}, \\
0, & \text{otherwise}.
\end{cases}
\end{align}
Why are they so different yet have such <em>similar</em> names?</p>
| AnonSubmitter85 | 33,383 | <p>Other than the value of $1/2$ on the edges, they're not different, just different notation. I don't know the motivation for the $1/2$'s, since it's still discontinuous, but they presumably explain why in the text. If we use the second notation, then the first definition (ignoring the $1/2$'s) is just
$$
\Pi \left( { t - (a+b)/2) } \over { a-b } \right).
$$</p>
|
1,031,559 | <p>I'm getting ahead in my differential equations textbook (<em>Fundamentals of Differential Equations</em> by Nagle et. al) and in the chapter of Laplace Transforms it states that the rectangular window function $\Pi_{a,b}\left(t\right)$ is given by
\begin{align}
\Pi_{a,b}\left(t\right):=u\left(t-a\right)-u\left(t-b\right)=\begin{cases}
0, & t<a, \\
1, & a<t<b, \\
0, & b <t.
\end{cases}
\end{align}
However, in another textbook I'm reading about Fourier Transforms (which I ATM know very little about, just the basics since I just got it, <em>Fourier Transforms: an introduction for Engineers</em> by Gray et. al.) they've stated that the rectangle function, a variation of the box function, <em>considered by Bracewell</em> (in <em>The Fourier Transform and its applications</em>, 1965) is given by
\begin{align}\Pi\left(t\right)=
\begin{cases}
1, & \left|t\right|<\frac{1}{2}, \\
\frac{1}{2}, & \left|t\right|=\frac{1}{2}, \\
0, & \text{otherwise}.
\end{cases}
\end{align}
Why are they so different yet have such <em>similar</em> names?</p>
| Olli Niemitalo | 230,294 | <p>With Fourier transforms and sampled band-limited signals the concept of a <a href="https://en.wikipedia.org/wiki/Dirac_delta_function" rel="nofollow noreferrer">Dirac delta function</a> and the <a href="https://en.wikipedia.org/wiki/Dirac_comb" rel="nofollow noreferrer">Dirac comb</a> sometimes comes up. They are zero everywhere else but at certain points. The integral over each of such special point equals one. Sometimes you need to multiply a Dirac comb by a rectangular function. If either of the edges coincides with a Dirac delta, then it matters what the value of the rectangular function is exactly at the edge. The multiplication results in scaling of the Dirac delta (and its integral) by that value. Setting the edge values to $\frac{1}{2}$ may give the most meaningful result, see: <a href="https://dsp.stackexchange.com/a/38294/15347">Whittaker-Shannon ($\operatorname{sinc}$) interpolation for a finite number of samples</a>.</p>
<p>Where the rectangle ends is a matter of convention. It is better to define the rectangle function for the reader if you ever happen to use it.</p>
|
132,591 | <p>Let $f(x)$ be a positive function on $[0,\infty)$ such that $f(x) \leq 100 x^2$. I want to bound $f(x) - f(x-1)$ from above. Of course, we have $$f(x) - f(x-1) \leq f(x) \leq 100 x^2.$$ This is not good for me though. I need a bound which is linear (or at worst linear-times-root) in $x$.</p>
<p>Is there an inequality of the form $f(x) - f(x-1) \leq f^\prime (x)=200 x$?</p>
| André Nicolas | 6,312 | <p>For $2^n \le x<2^{n+1}$, let $f(x)=100(2^{n})^2$. There is an enormous jump from $f(2^{n+1}-1)$ to $f(2^{n+1})$. So even if we assume that $f$ is non-decreasing, we can have jumps of size comparable to $100x^2$. At the cost of complicating the description, we can modify the above $f(x)$ to make it strictly increasing. </p>
|
3,037,291 | <p><a href="https://i.stack.imgur.com/5t6LN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5t6LN.png" alt="enter image description here"></a></p>
<p>I am trying to build a <a href="http://mathworld.wolfram.com/SnubCube.html" rel="nofollow noreferrer">snub cube</a>. I have made <span class="math-container">$6$</span> squares and <span class="math-container">$32$</span> equilateral triangles (out of perler beads if you're curious).
I am trying to figure out the angles at which I adjoin the squares to the triangles, and the triangles to other triangles. </p>
<p>I have found a few formulas, but I think I am a bit overwhelmed by the vocabulary used and do not understand what the listed variables are. </p>
<p><a href="https://www.academia.edu/11784745/Table_of_solid_angles_subtended_by_Archimedean_solids_uniform_polyhedrons_at_their_vertices_by_HCR" rel="nofollow noreferrer">H. Rajpoot</a> says "There is a general expression of the solid angle subtended by the snub cube at any of its <span class="math-container">$24$</span> vertices is given by the general expression
<span class="math-container">\begin{align}\Omega&=2\sin^{-1}\left(\frac{(1-\sqrt{1-K^2})-\sqrt{2K^2-1}}{K^2\sqrt{2}}\right)+8\sin^{-1}\left(\frac{(1-\sqrt{1-K^2})-\sqrt{4K^2-1}}{2K^2\sqrt{3}}\right)\\&\approx 3.589629551 \space sr,\end{align}</span>
where <span class="math-container">$K\approx 0.928191378"$</span>.</p>
<p>and <a href="https://math.stackexchange.com/questions/500913/formula-for-angle-given-three-3d-coordinates">Felix Marin</a> says that the formula to find the angles is <span class="math-container">$$
\cos\left(\vphantom{\Large A}\angle{\rm ABC}\right)
=
{\left(\vec{A} - \vec{B}\right)\cdot\left(\vec{C} - \vec{B}\right)
\over
\left\vert\vec{A} - \vec{B}\right\vert\;\left\vert\vec{C} - \vec{B}\right\vert}
$$</span> where <span class="math-container">$A$</span>, <span class="math-container">$B$</span>, and <span class="math-container">$C$</span> are are vectors <span class="math-container">$A:[x_1,y_1,z_1]$</span>, <span class="math-container">$B:[x_2,y_2,z_2]$</span>, and <span class="math-container">$C:[x_3,y_3,z_3]$</span>.</p>
<p><strong>I suppose, I am completely overwhelmed.
I have a sight feeling that finding the 'subtended angle' is not the same as the angle I am trying to find. Is that true?
What is <span class="math-container">$s$</span>? <span class="math-container">$r$</span>?
Why are <span class="math-container">$A$</span>, <span class="math-container">$B$</span>, & <span class="math-container">$C$</span> vectors and how do I know which vectors to use?</strong></p>
<p>I saw <a href="https://wikivisually.com/wiki/Snub_cube#Cartesian_coordinates" rel="nofollow noreferrer">online, here</a> that the the coordinates for the vertices of a snub cube are all the even permutations of <span class="math-container">$(±1, ±1/t, ±t)$</span> with an even number of plus signs, along with all the odd permutations with an odd number of plus signs, where <span class="math-container">$t ≈ 1.83929$</span> is the tribonacci constant.</p>
<p><strong>Are these the values I am supposed to use to find the vectors to use the second equation?
Is there an easier way to do this?</strong> I fell like I have way over-complicated this.</p>
<p>edit: okay, I found <a href="http://www.coolmath.com/reference/polyhedra-snub-cuboctahedron" rel="nofollow noreferrer">this website</a> that says the square-triangle angle is <span class="math-container">$142$</span> degrees, <span class="math-container">$59$</span> minutes and the triangle-triangle angle is <span class="math-container">$153$</span> degrees, <span class="math-container">$14$</span> minutes. Would still be stoked to know how on earth to figure this out on my own. thanks!</p>
| Dr. Richard Klitzing | 518,676 | <p>For the mere requested values you might want to have a look <a href="https://bendwavy.org/klitzing/incmats/snic.htm" rel="nofollow noreferrer">here</a>.</p>
<p>A more descriptive way on the derivation of these values might be found already in the old German book of Max Brückner, "Vielecke und Vielflache, Theorie und Geschichte", Leipzig, Teubner Verlag (1900), at page 139.</p>
<p>--- rk</p>
|
3,037,291 | <p><a href="https://i.stack.imgur.com/5t6LN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5t6LN.png" alt="enter image description here"></a></p>
<p>I am trying to build a <a href="http://mathworld.wolfram.com/SnubCube.html" rel="nofollow noreferrer">snub cube</a>. I have made <span class="math-container">$6$</span> squares and <span class="math-container">$32$</span> equilateral triangles (out of perler beads if you're curious).
I am trying to figure out the angles at which I adjoin the squares to the triangles, and the triangles to other triangles. </p>
<p>I have found a few formulas, but I think I am a bit overwhelmed by the vocabulary used and do not understand what the listed variables are. </p>
<p><a href="https://www.academia.edu/11784745/Table_of_solid_angles_subtended_by_Archimedean_solids_uniform_polyhedrons_at_their_vertices_by_HCR" rel="nofollow noreferrer">H. Rajpoot</a> says "There is a general expression of the solid angle subtended by the snub cube at any of its <span class="math-container">$24$</span> vertices is given by the general expression
<span class="math-container">\begin{align}\Omega&=2\sin^{-1}\left(\frac{(1-\sqrt{1-K^2})-\sqrt{2K^2-1}}{K^2\sqrt{2}}\right)+8\sin^{-1}\left(\frac{(1-\sqrt{1-K^2})-\sqrt{4K^2-1}}{2K^2\sqrt{3}}\right)\\&\approx 3.589629551 \space sr,\end{align}</span>
where <span class="math-container">$K\approx 0.928191378"$</span>.</p>
<p>and <a href="https://math.stackexchange.com/questions/500913/formula-for-angle-given-three-3d-coordinates">Felix Marin</a> says that the formula to find the angles is <span class="math-container">$$
\cos\left(\vphantom{\Large A}\angle{\rm ABC}\right)
=
{\left(\vec{A} - \vec{B}\right)\cdot\left(\vec{C} - \vec{B}\right)
\over
\left\vert\vec{A} - \vec{B}\right\vert\;\left\vert\vec{C} - \vec{B}\right\vert}
$$</span> where <span class="math-container">$A$</span>, <span class="math-container">$B$</span>, and <span class="math-container">$C$</span> are are vectors <span class="math-container">$A:[x_1,y_1,z_1]$</span>, <span class="math-container">$B:[x_2,y_2,z_2]$</span>, and <span class="math-container">$C:[x_3,y_3,z_3]$</span>.</p>
<p><strong>I suppose, I am completely overwhelmed.
I have a sight feeling that finding the 'subtended angle' is not the same as the angle I am trying to find. Is that true?
What is <span class="math-container">$s$</span>? <span class="math-container">$r$</span>?
Why are <span class="math-container">$A$</span>, <span class="math-container">$B$</span>, & <span class="math-container">$C$</span> vectors and how do I know which vectors to use?</strong></p>
<p>I saw <a href="https://wikivisually.com/wiki/Snub_cube#Cartesian_coordinates" rel="nofollow noreferrer">online, here</a> that the the coordinates for the vertices of a snub cube are all the even permutations of <span class="math-container">$(±1, ±1/t, ±t)$</span> with an even number of plus signs, along with all the odd permutations with an odd number of plus signs, where <span class="math-container">$t ≈ 1.83929$</span> is the tribonacci constant.</p>
<p><strong>Are these the values I am supposed to use to find the vectors to use the second equation?
Is there an easier way to do this?</strong> I fell like I have way over-complicated this.</p>
<p>edit: okay, I found <a href="http://www.coolmath.com/reference/polyhedra-snub-cuboctahedron" rel="nofollow noreferrer">this website</a> that says the square-triangle angle is <span class="math-container">$142$</span> degrees, <span class="math-container">$59$</span> minutes and the triangle-triangle angle is <span class="math-container">$153$</span> degrees, <span class="math-container">$14$</span> minutes. Would still be stoked to know how on earth to figure this out on my own. thanks!</p>
| Maxim | 491,644 | <p>A way to obtain the coordinates of the vertices is given <a href="http://paulscottinfo.ipage.com/polyhedra/semiregular/snub-cube.html" rel="nofollow noreferrer">here</a>. To find the coordinates of <span class="math-container">$B$</span>, rotate <span class="math-container">$A = (1, v, w)$</span> by <span class="math-container">$\pi/2$</span> around the <span class="math-container">$z$</span>-axis to map the blue face to the yellow face, then by <span class="math-container">$\pi/2$</span> around the <span class="math-container">$x$</span>-axis, then by <span class="math-container">$\pi/2$</span> around the <span class="math-container">$y$</span>-axis:
<span class="math-container">$$(1, v, w) \to (-v, 1, w) \to (-v, -w, 1) \to (1, -w, v) = B.$$</span>
To find the coordinates of <span class="math-container">$C$</span>, repeat the first two steps above and rotate by <span class="math-container">$\pi$</span> around the <span class="math-container">$z$</span>-axis to obtain
<span class="math-container">$$(-v, -w, 1) \to (v, w, 1) = C.$$</span>
Then an outward normal to the triangular face is
<span class="math-container">$$\mathbf n = ((1, -w, v) - (v, w, 1)) \times ((1, v, w) - (v, w, 1))$$</span>
and the dihedral angle between a square and an adjacent triangular face is
<span class="math-container">$$\phi_1 = \arccos \frac {\mathbf n \cdot (-1, 0, 0)} {|\mathbf n|} =
\pi - \arcsin \sqrt {\frac {n_y^2 + n_z^2} {n_x^2 + n_y^2 + n_z^2}}.$$</span>
The rational function under the square root simplifies to at most a quadratic polynomial in <span class="math-container">$v$</span> since <span class="math-container">$v$</span> is a root of a cubic polynomial, giving
<span class="math-container">$$\phi_1 = \pi - \arcsin \sqrt {\frac {2 v} 3}.$$</span>
Similarly, the angle between two adjacent triangular faces is
<span class="math-container">$$\phi_2 = \pi - \arcsin \frac {2\sqrt {1 - v \,}} 3.$$</span></p>
|
395,618 | <p>If n squares are randomly removed from a $p \ \cdot \ q$ chessboard, what will be the expected number of pieces the chessboard is divided into? </p>
<p>Can anybody please provide how can I approach the problem? There are numerous cases and when I go through case consideration it becomes extremely complex.</p>
| Jon Claus | 77,104 | <p>I am posting this as an answer to make it more visible. This is currently a Brilliant.org problem, in which the specific case $ 2 \times 500 $ is asked for. While it is an interesting problem meriting discussion, that should wait until after 8:00 PM EST 5/19 when the problem set is closed.</p>
|
2,419,057 | <p>Trying to assimilate the meaning of the differential I have looked for different examples of functions which:</p>
<ol>
<li>Admits all directional derivatives but are not continuous $(f: \mathbb{R}^2 \rightarrow \mathbb{R} \quad ,\quad (x,y) \mapsto
\begin{cases}
0 & \text{for } (x,y)=(0,0) \\
\frac{xy^4}{x^4+y^8} & \text{for } (x,y) \neq (0,0)
\end{cases})$</li>
<li>Admits all directional derivatives and are continuous but not differentiable:<a href="https://math.stackexchange.com/questions/372070/f-not-differentiable-at-0-0-but-all-directional-derivatives-exist?rq=1">$f$ not differentiable at $(0,0)$ but all directional derivatives exist</a></li>
</ol>
<p>For what I am looking for now I will use:</p>
<p>Suppose we have a function $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$</p>
<p>that in a point, let say $x_0$, admits all direccional derivatives. Then, we can consider the map</p>
<p>$\phi_{x_{0}}: \mathbb{R}^n \rightarrow \mathbb{R}^m \quad$ $v \mapsto D_{v}f(x_{0}) \quad$ , which sends a vector to its directional derivative in $x_0$.</p>
<p>If I am not mistaken, if this map is linear, it will be the differential of $f$.
What I am searching now is:</p>
<blockquote>
<p>A function $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$, continuos at $x_0$, which admits all directional derivatives at this point, but in which $\phi_{x_{0}}$ is not continuous.</p>
</blockquote>
<p>I don´t know if this would be possible.
In the example 2, the function $\phi_{(0,0)}$ is smooth (continuous, although non linear) and I can´t imagine how you could "break" the smoothness without breaking the continuity of $f$ too (that happens in the example 1).
I would appreciate if the example is in 2 dimensions ($f: \mathbb{R}^2 \rightarrow \mathbb{R}$) for it to be possible to visualize.</p>
<p>Thanks in advance</p>
| zhw. | 228,045 | <p>The function $f(x,y)= \text { sgn }(x)\sqrt {x^2+y^2}$ is continuous at $(0,0),$ but $D_u f(0,0) = 1$ for all unit vectors $u$ except for $u =(0,1),(0,-1),$ in which case $D_u f(0,0) = 0.$</p>
|
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