qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
1,086,048 | <p>I've performed a change of variable:
$$X = \sqrt{y}$$
$$X'=\frac{1}{2}Y^{-\frac{1}{2}}$$
Thus:
$$f(\sqrt{y})*X'=f(y)=\frac{1}{2\sqrt{2\pi y}}e^{-\frac{y}{2}}$$
However the book gives:
$$f(y)=\frac{1}{\sqrt{2\pi y}}e^{-\frac{y}{2}}$$
Where did I go wrong?</p>
| Przemysław Scherwentke | 72,361 | <p>HINT: You should consider $P(-\sqrt{x}<y<\sqrt{x})$.</p>
|
1,086,048 | <p>I've performed a change of variable:
$$X = \sqrt{y}$$
$$X'=\frac{1}{2}Y^{-\frac{1}{2}}$$
Thus:
$$f(\sqrt{y})*X'=f(y)=\frac{1}{2\sqrt{2\pi y}}e^{-\frac{y}{2}}$$
However the book gives:
$$f(y)=\frac{1}{\sqrt{2\pi y}}e^{-\frac{y}{2}}$$
Where did I go wrong?</p>
| Chris | 103,950 | <p>Thanks for the hint. I computed only half the result. To complete the solution:</p>
<p>$$f(y)=\frac{1}{2\sqrt{2\pi y}}e^{-\frac{y}{2}}+\frac{1}{2\sqrt{2\pi y}}e^{-\frac{y}{2}}=\frac{1}{\sqrt{2\pi y}}e^{-\frac{y}{2}},0<y<\infty$$</p>
|
2,211,075 | <p>I don't understand the following example from Math book.</p>
<p>Solve for the equation <code>sin(theta) = -0.428</code> for <code>theta</code> in <code>radians</code> to 2 decimal places. where <code>0<= theta<= 2PI</code>.</p>
<p>And this is the answer:</p>
<p><code>theta=-0.44 + 2PI = 5.84rad and theta = PI-(0.44) = 3.58rad</code> </p>
<p>I don't understand the part why we need to add <code>2PI</code> in the first answer and add <code>PI</code> in second answer?</p>
| Clement C. | 75,808 | <p>In short: you <strong>cannot</strong> say
$$
\lim_{x\to 0} \frac{\frac{f(x)-f(0)}{x-0}}{g(x)} =
\lim_{x\to 0} \frac{f'(x)}{g(x)}
$$
which is what you used in your third step. This is simply not true in general.</p>
<hr>
<p>To see this on a simpler example, take e.g. $f$ defined by $f(x)=x^2$ and $g(x)=x$.</p>
<p>Then
$$
\lim_{x\to 0} \frac{\frac{f(x)-f(0)}{x-0}}{g(x)} =
\lim_{x\to 0} \frac{\frac{x^2}{x}}{x}
\lim_{x\to 0} 1 = 1
$$
while
$$
\lim_{x\to 0} \frac{f'(x)}{g(x)} = \lim_{x\to 0} \frac{2x}{x} = 2.
$$</p>
|
3,905,629 | <p>I need to compute a limit:</p>
<p><span class="math-container">$$\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$$</span></p>
<p>I tried to apply the L'Hôpital rule, but the emerging terms become too complicated and doesn't seem to simplify.</p>
<p><span class="math-container">$$
\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x \\
= \exp (\lim_{x \to 0+} x \ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})) \\
= \exp (\lim_{x \to 0+} \frac
{\ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})}
{\frac 1 x}) \\
= \exp \lim_{x \to 0+} \dfrac
{\dfrac {\cos \sqrt x} {x} + \dfrac {\sin \dfrac 1 x} {2 \sqrt x}
- \dfrac {\cos \dfrac 1 x} {x^{3/2}}}
{- \dfrac {1} {x^2} \left(2\sin \sqrt x + \sqrt x \sin \frac{1}{x} \right)}
$$</span></p>
<p>I've calculated several values of this function, and it seems to have a limit of <span class="math-container">$1$</span>.</p>
| Kavi Rama Murthy | 142,385 | <p>Hint: Using the fact that <span class="math-container">$\frac {\sin x} x \to 1$</span> as <span class="math-container">$x \to 0$</span> verify that <span class="math-container">$ x\ln (\frac 1 2 \sqrt x) <x\ln (2\sin\sqrt x+\sqrt x \sin (\frac 1 x)) <x \ln (3\sqrt x)$</span>. Conclude that the limit is <span class="math-container">$e^{0}=1$</span>.</p>
<p>[<span class="math-container">$x \ln x \to 0$</span> as <span class="math-container">$x \to 0+$</span>]</p>
|
3,905,629 | <p>I need to compute a limit:</p>
<p><span class="math-container">$$\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$$</span></p>
<p>I tried to apply the L'Hôpital rule, but the emerging terms become too complicated and doesn't seem to simplify.</p>
<p><span class="math-container">$$
\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x \\
= \exp (\lim_{x \to 0+} x \ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})) \\
= \exp (\lim_{x \to 0+} \frac
{\ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})}
{\frac 1 x}) \\
= \exp \lim_{x \to 0+} \dfrac
{\dfrac {\cos \sqrt x} {x} + \dfrac {\sin \dfrac 1 x} {2 \sqrt x}
- \dfrac {\cos \dfrac 1 x} {x^{3/2}}}
{- \dfrac {1} {x^2} \left(2\sin \sqrt x + \sqrt x \sin \frac{1}{x} \right)}
$$</span></p>
<p>I've calculated several values of this function, and it seems to have a limit of <span class="math-container">$1$</span>.</p>
| Adam Rubinson | 29,156 | <p><span class="math-container">$$ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... $$</span></p>
<p><span class="math-container">$$\therefore \ (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$$</span></p>
<p><span class="math-container">$$ = \left(2 \left(x^{1/2} - \frac{x^{3/2}}{3!} + \frac{x^{5/2}}{5!} -...\right) + x^{1/2} \sin \frac{1}{x}\right)^x$$</span></p>
<p><span class="math-container">$$ = \left(x^{1/2} \left(2 + \sin \frac{1}{x} - \frac{2x^{3/2}}{3!} + \frac{2x^{5/2}}{5!} -...\right) \right)^x$$</span></p>
<p><span class="math-container">$$ = \left(x^{1/2}\right)^x \left(2 + \sin \frac{1}{x} - \frac{2x}{3!} + \frac{2x^2}{5!} -...\right)^x.$$</span></p>
<p>Now <span class="math-container">$ \left(x^{1/2}\right)^x = \left(x^{x}\right)^{1/2} = (e^{x \ln x})^{1/2} = e^{\frac{1}{2} x \ln x},\ $</span> and <span class="math-container">$\ \lim\limits_{x\rightarrow 0^{+}}e^{\frac{1}{2}x\ln(x)}=e^{\lim\limits_{x\rightarrow 0^{+}}\frac{1}{2}x\ln(x)}=1. $</span></p>
<p>Lastly, for <span class="math-container">$ \left(2 + \sin \frac{1}{x} - \frac{2x}{3!} + \frac{2x^2}{5!} -...\right)^x,\ $</span> make the substitution <span class="math-container">$u = \frac{1}{x}$</span> to get:</p>
<p><span class="math-container">$\lim\limits_{u\rightarrow \infty} \left( 2 + \sin u - \frac{2}{3!u} + \frac{2}{5!u^2} - ...\right)^{\frac{1}{u}}$</span>, and the bracket oscillates between <span class="math-container">$1$</span> and <span class="math-container">$3$</span> for large <span class="math-container">$u$</span>, so this limit is <span class="math-container">$1$</span>. Thus:</p>
<p><span class="math-container">$$ \lim\limits_{x\rightarrow 0^{+}} \left(x^{1/2}\right)^x \left(2 + \sin \frac{1}{x} - \frac{2x}{3!} + \frac{2x^2}{5!} -...\right)^x = 1 \times 1 = 1.$$</span></p>
|
2,747,578 | <p>Let $S,T$ be sets with $|S|>|T|$ and $R$ some relations on $T$.<br>
Why is then $\langle S|-\rangle$ not isomorphic to $\langle T|R\rangle$ </p>
<p>This came up when I wanted to solve a different problem, <a href="https://math.stackexchange.com/q/2747383/506844">which I also asked on this site</a>. Unfortunately, the answers provided used a completely different strategy and I still wonder about how to prove this.</p>
<p>Intuitively its clear, but I look for a clear proof.</p>
| Community | -1 | <p>The group is a quotient group. The images of the elements of T generate this quotient group. If $<T|R>$ is free then any set of free generators has less than or equal to |T| generators. Thus it can't be isomorphic to $<S|->$.</p>
|
2,747,578 | <p>Let $S,T$ be sets with $|S|>|T|$ and $R$ some relations on $T$.<br>
Why is then $\langle S|-\rangle$ not isomorphic to $\langle T|R\rangle$ </p>
<p>This came up when I wanted to solve a different problem, <a href="https://math.stackexchange.com/q/2747383/506844">which I also asked on this site</a>. Unfortunately, the answers provided used a completely different strategy and I still wonder about how to prove this.</p>
<p>Intuitively its clear, but I look for a clear proof.</p>
| Giorgio Mossa | 11,888 | <p>I feel a little like cheating here, because that is basically an adaptation of the idea provided from an answer to the linked question... but again they say that mathematics is the art of finding good analogies.</p>
<p>Before I start allow me to introduce the commutator subgroup.
For every group $G$ by $G'$ I denote the subgroup of $G$ generated by the elemets of the form $xyx^{-1}y^{-1}$ where $x,y \in G$. Such $G'$ is a normal subgroup and $G/G'$ is the <em>smallest commutative quotient group of $G$</em>, for more information <a href="https://en.m.wikipedia.org/wiki/Commutator_subgroup" rel="nofollow noreferrer">take a look at this</a>.</p>
<p>Assume that you have an isomorphism
$$f \colon \langle S | - \rangle \longrightarrow \langle T | R \rangle$$
the this isomorphism should preserve the commutator subgroups, that is
$$f(\langle S| - \rangle')=\langle T|R\rangle'\ .$$</p>
<p>Because of this we get an isomorphism
$$\bar f \colon \langle S| - \rangle / \langle S| - \rangle' \longrightarrow \langle T|R\rangle/\langle T|R\rangle'$$
of abelian groups.</p>
<p>With a little effort one can easily prove that the first quotient is isomorphic to the free abelian group with a basis of cardinality $|S|$.</p>
<p>But the second group is generated by $|T|$ elements where $|T| < |S|$. This is an absurd because the cardinality of a basis of a free abelian group is the minimal number of generators for the group, so $\bar f$ cannot be an isomorphism and neither $f$ can.</p>
|
4,349,582 | <p>Its rather easy to show that <span class="math-container">$a_n=\frac{n^{1/n}}{n}$</span> ist monotonic (which means <span class="math-container">$a_{n+1}<a_n$</span> for each <span class="math-container">$n$</span>) using derivations. But how can I do it without them? Thanks.</p>
| fleablood | 280,126 | <p>First of all base 10 and base 2 are notations and not the integers themselves. And integer will have a value no matter how it is written. So if you want a <em>function</em> that converts base 10 to base two (a function being a mapping between two sets) then this is <em>not</em> a function on the integers but a function on <em>strings</em> of digits to another string of digits.</p>
<p>If you want a function from the integers themselves to a binary string (then it doesn't actually matter what notation the integer is notated in) there is no real <em>formula</em> for this but there are two recursive methods of doing this.</p>
<p>Method 1: Big to little.</p>
<p>Let <span class="math-container">$N$</span> but the natural number. Let <span class="math-container">$n=\lfloor \log_2 N \rfloor$</span> (where <span class="math-container">$\lfloor . \rfloor$</span> is the greatest integer less than or equal function). Then the string will have <span class="math-container">$n+1$</span> terms and the "leftmost/largest" term is <span class="math-container">$a_n = 1$</span>. Let <span class="math-container">$N_{n}= N$</span>.</p>
<p>Recursively for any <span class="math-container">$N_{m}$</span> and <span class="math-container">$a_m$</span> you have defined, define <span class="math-container">$N_{m-1}= N_m - a_m\cdot 2^m$</span>. If <span class="math-container">$N_{m-1}\ge 2^{m-1}$</span> define <span class="math-container">$a_{m-1}:=1$</span>. Otherwise define <span class="math-container">$a_{m-1}:=0$</span>.</p>
<p>Repeat until you have figured out <span class="math-container">$a_0$</span>.</p>
<p>For example to conver <span class="math-container">$364_{10}$</span> to binary.</p>
<p>As <span class="math-container">$2^8=256 < 364 < 2^9$</span> we know <span class="math-container">$8<\log_2 364 < 9$</span> so <span class="math-container">$n=\lfloor \log_2 364 \rfloor = 8$</span>. <span class="math-container">$N_8 = 364$</span> and <span class="math-container">$a_8 = 1$</span>.</p>
<p><span class="math-container">$N_7 = 364 - 2^8 =364-256 = 108$</span>. <span class="math-container">$108 < 128=2^7$</span> so <span class="math-container">$a_7=0$</span>.</p>
<p><span class="math-container">$N_6 = 108 - 0\cdot 2^7 = 108$</span>. <span class="math-container">$108 > 64 = 2^6$</span> so <span class="math-container">$a_6 = 1$</span>.</p>
<p><span class="math-container">$N_5 = 108 - 1\cdot 2^6=104-64=44$</span>. <span class="math-container">$44 > 32 = 2^5$</span> so <span class="math-container">$a_5 = 1$</span>.</p>
<p><span class="math-container">$N_4=40-1\cdot 2^5 = 40-32 = 12$</span>. <span class="math-container">$12 < 16 = 2^4$</span> so <span class="math-container">$a_4 = 0$</span>.</p>
<p><span class="math-container">$N_3 = 12 -0\cdot 2^4=12$</span>. <span class="math-container">$12 > 8 = 2^3$</span> so <span class="math-container">$a_3= 1$</span>.</p>
<p><span class="math-container">$N_2 = 12 - 1\cdot 2^3 =12 -8 =4$</span>. <span class="math-container">$4 \ge 4=2^2$</span> so <span class="math-container">$a_2 = 1$</span>.</p>
<p><span class="math-container">$N_1 = 4-1\cdot 2^2 =4-4=0$</span>. <span class="math-container">$0 < 2=2^1$</span> so <span class="math-container">$a_1 = 0$</span>.</p>
<p><span class="math-container">$N_0 = 0-0\cdot 2^1=0$</span>. <span class="math-container">$0 < 1 = 2^0$</span> so <span class="math-container">$a_0 = 0$</span>.</p>
<p>So in binary <span class="math-container">$364_{10} = 101101100_2$</span>.</p>
<p>Method 2: Little to big</p>
<p>Let <span class="math-container">$N = N_0$</span> be the natural number. Let <span class="math-container">$a_0$</span> be the remainder of <span class="math-container">$N_0$</span> when divided by <span class="math-container">$2$</span>.</p>
<p>For any known <span class="math-container">$N_m$</span> and <span class="math-container">$a_m$</span> let <span class="math-container">$N_{m+1} = \frac {N_m - a_m}2$</span>. And let <span class="math-container">$a_{m+1}$</span> be the remainder of <span class="math-container">$N_{m+1}$</span> divided by <span class="math-container">$2$</span>. Repeat until you come to a point where <span class="math-container">$N_m = 0$</span>.</p>
<p>Example. To convert <span class="math-container">$364_{10}$</span> to binary.</p>
<p><span class="math-container">$N_0 = 364$</span>. The remainder when we divide by <span class="math-container">$2$</span> is <span class="math-container">$0$</span>. So <span class="math-container">$a_0 = 0$</span>.</p>
<p><span class="math-container">$N_1 = \frac {364-0}2 = 182$</span>. The remainder when we divide by <span class="math-container">$2$</span> is <span class="math-container">$0$</span>. So <span class="math-container">$a_1 = 0$</span>.</p>
<p><span class="math-container">$N_2 = \frac {182-0}2 = 91$</span>. The remainder when we divide by <span class="math-container">$2$</span> is <span class="math-container">$1$</span>. So <span class="math-container">$a_2 = 1$</span>.</p>
<p><span class="math-container">$N_3 =\frac {91-1}2=45$</span>. The remainder when we divide by <span class="math-container">$2$</span> is <span class="math-container">$1$</span>. So <span class="math-container">$a_3 = 1$</span>.</p>
<p><span class="math-container">$N_4 = \frac {45-1}2= 22$</span>. The remainder when we divide by <span class="math-container">$2$</span> is <span class="math-container">$0$</span>. So <span class="math-container">$a_4 = 0$</span>.</p>
<p><span class="math-container">$N_5 = \frac {22-0}2 =11$</span>. The remainder when we divide by <span class="math-container">$2$</span> is <span class="math-container">$1$</span>. So <span class="math-container">$a_5=1$</span>.</p>
<p><span class="math-container">$N_6 = \frac {11-1}2 =5$</span>. The remainder when we divide by <span class="math-container">$2$</span> is <span class="math-container">$1$</span>. So <span class="math-container">$a_6=1$</span>.</p>
<p><span class="math-container">$N_7 = \frac {5-1}2=2$</span>. The remainder when we divide by <span class="math-container">$2$</span> is <span class="math-container">$0$</span>. So <span class="math-container">$a_7=0$</span>.</p>
<p><span class="math-container">$N_8 = \frac {2-0}2 = 1$</span>. The remainder when we divide by <span class="math-container">$2$</span> is <span class="math-container">$1$</span>. So <span class="math-container">$a_8=1$</span>.</p>
<p><span class="math-container">$N_9= \frac {1-1}2 = 0$</span>. We are done.</p>
<p>So in binary <span class="math-container">$364_{10} = 101101100_2$</span>.</p>
<p>=====</p>
<p>A third method (which can be uncoil and poke you in the eye if you are not careful) is to note that <span class="math-container">$TEN = 2^3 + 2 = 1000_2 + 10_2$</span> and to know the single digit conversion.</p>
<p>Watch</p>
<p><span class="math-container">$364_{10}= 3_{10}\times TEN^2 + 6_{10}\times TEN + 4_{10}=$</span></p>
<p><span class="math-container">$3_{10}\times(1000_2 + 10_2)^2 + 6_{10}\times (1000_2+ 10_2) + 100_2=$</span></p>
<p><span class="math-container">$(3_{10}\times 1000_2 + 3_{10}\times 10_2)(1000_2+10_2) + 110_2(1000_2 + 10_2) + 100_2=$</span></p>
<p><span class="math-container">$(11,000 + 110)(1000+10) + 110,000 + 1,100 + 100 =$</span></p>
<p><span class="math-container">$11,000,000 +110,000 + 110,000 + 1,100 + 110,000 + 1,100 + 100 =$</span></p>
<p><span class="math-container">$11,000,000 + 110,000 + 110,000 + 1,000 + 110,000 + 1,000 + 100+100 + 100=$</span></p>
<p><span class="math-container">$11,000,000 + 110,000 + 110,000 + 1,000 + 110,000 + 1,000 + 1,100=$</span></p>
<p><span class="math-container">$11,000,000 + 110,000 + 110,000 + 110,000 + 1,000 + 1,000 + 1,000 + 100=$</span></p>
<p><span class="math-container">$11,000,000 + 110,000 + 110,000 + 110,000 + 11,000 + 100=$</span></p>
<p><span class="math-container">$11,000,000 + 110,000 + 110,000 + 110,000 + 10,000 + 1,100=$</span></p>
<p><span class="math-container">$11,000,000 + 100,000 + 100,000 + 100,000 + 10,000 + 10,000 + 10,000 + 100,000 + 1,100=$</span></p>
<p><span class="math-container">$11,000,000 + 1,100,000 + 1,000,000 + 1,100=$</span></p>
<p><span class="math-container">$11,000,000 + 1,000,000 + 1,000,000 + 100,000 + 1,100=$</span></p>
<p><span class="math-container">$11,000,000 + 10,000,000 + 101,100=$</span></p>
<p><span class="math-container">$10,000,000 + 10,000,000 + 1,000,000 + 101,100=$</span></p>
<p><span class="math-container">$100,000,000 + 1,101,100=$</span></p>
<p><span class="math-container">$101,101,100_2$</span></p>
|
205,671 | <p>How would one go about showing the polar version of the Cauchy Riemann Equations are sufficient to get differentiability of a complex valued function which has continuous partial derivatives? </p>
<p>I haven't found any proof of this online.</p>
<p>One of my ideas was writing out $r$ and $\theta$ in terms of $x$ and $y$, then taking the partial derivatives with respect to $x$ and $y$ and showing the Cauchy Riemann equations in the Cartesian coordinate system are satisfied. A problem with this approach is that derivatives get messy.</p>
<p>What are some other ways to do it?</p>
| Shabir | 239,381 | <p><img src="https://i.stack.imgur.com/Ct6la.png" alt="Here is an easy approach">
Here is an easy approach to derive the Cauchy-Riemann equations in polar form.</p>
|
2,218,914 | <p>What is a boundary point when solving for a max/min using Lagrange Multipliers?
After you solve the required system of equation and get the critical maxima and minima, when do you have to check for boundary points and how do you identify them?</p>
<p>e.g. Optimise (1+a)(1+b)(1+c) given constraint a+b+c=1, with a,b,c all non-negative.</p>
<p>After using the Lagrange multiplier equating the respective partial derivatives, I get (a,b,c)=(1/3, 1/3, 1/3). Clearly there must be both a maximum and minimum, and I assume this is the maximum. Where is the minimum? (0,0,1) optimises best for the minimum, and I assume using 0 is a boundary point but why? And what effect does the restriction to non-negative reals have? </p>
| Christian Blatter | 1,303 | <p>Your example serves perfectly to explain the necessary procedure.</p>
<p>You are given a function $f(x,y,z):=(1+x)(1+y)(1+z)$ in ${\mathbb R}^3$, as well as a compact set $S\subset{\mathbb R}^3$, and you are told to determine $\max f(S)$ and $\min f(S)$.</p>
<p>Differential calculus is a help in this task insofar as putting suitable derivatives to zero brings <strong>interior</strong> stationary points of $f$ in the different dimensional <strong>strata</strong> of $S$ to the fore. The given simplex $S$ is a union $S=S_0\cup S_1\cup S_2$, whereby $S_0$ consists of the three vertices, $S_1$ of the three edges (without their endpoints), and $S_2$ of the interior points of the triangle $S$.</p>
<p>If the global maximum of $f$ on $S$ happens to lie on $S_2$ it will be detected by Lagrange's method, applied with the condition $x+y+z=1$. If the maximum happens to lie on one of the edges it will be detected by using Lagrange's method with two conditions, or simpler: by a parametrization of these edges (three separate problems!). If the maximum happens to lie at one of the vertices it will be taken care of by evaluating $f$ at these vertices.</p>
<p>In all we obtain a (hopefully finite) <strong>candidate list</strong> $\{p_1,p_2,\ldots, p_N\}$. The global maximum of $f$ on the set $S$ will be the largest of the values $f(p_k)$ $(1\leq k\leq N)$. Note that we don't need to compute any second derivatives.</p>
|
109,734 | <p>I am trying to do this homework problem and I have no idea how to approach it. I have tried many methods, all resulting in failure. I went to the books website and it offers no help. I am trying to find the derivative of the function
$$y=\cot^2(\sin \theta)$$</p>
<p>I could be incorrect but a trig function squared would be the result of the trig function with the angle value and then squared. Not the angle value squared, that would give a different answer. Knowing this I also know that I can not use the table of simple trig derivatives so I know I can't just take the derivative as
$$y=\cot^2(x)$$
$$ x=\sin(\theta)$$ </p>
<p>This does not help because I can't get the derivative of cot squared. What I did try to do was rewrite it as $\frac{\cos x}{\sin x}\frac{\cos x}{\sin x}$ and then find the derivative of that but something went wrong with that and it does not produce an answer that is like the one in the book. In fact the book gets a csc squared in the answer so I know they are doing something very different.</p>
| Mike | 17,976 | <p>The way I was taught allows me to use the chain rule without really needing to think about it. This would be the way I'd write it:</p>
<p>$dy=d(\cot^2(\sin\theta))=2\cot(\sin\theta)d(\cot(\sin\theta))=2\cot(\sin\theta)(-\csc^2(\sin\theta))d(\sin\theta)=$</p>
<p>$-2\cot(\sin\theta)\csc^2(\sin\theta)\cos\theta d\theta$</p>
<p>Now what did I do to get from one step to the next? Everything inside the d() is a substitution. So my first substitution was</p>
<p>$\frac{d\cot^2(\sin\theta)}{d\theta}=\frac{du^2}{du}\frac{du}{d\theta}=2u\frac{du}{d\theta}=2\cot(\sin\theta)\frac{d\cot(\sin\theta)}{d\theta}$</p>
<p>After a while, the substitutions come without thinking. First I see I need the derivative of the square of some function, then the derivative of the cotangent of some function, etc.</p>
|
2,486,590 | <p>Let $(X,\mathcal{B},\mu,T)$ be a dynamical system and let $A \in \mathcal{B}$ such that $\mu(A)>0$ and $\forall x \in X$ we define $$L_x=\{n \in \Bbb{N}|T^nx \in A\}$$</p>
<p>Prove that $\mu(\{x:\bar{d}(L_x)>0\})>0$ where $$\bar{d}(L_x)=\limsup_n \frac{|L_x \cap \{1,2...n\}|}{n}$$.</p>
<p>My first thought is that i have to use somewhere Poincare's recurrence theorem.</p>
<p>But i cannot understand the behavior of the density function.</p>
<p>From Poincare we just know that for almost every $x \in A$ we have that $L_x$ is infinite.</p>
<p>Can some give me a hint to solve this?</p>
<p>I clearly do not want a full solution. </p>
<p>I want hint to solve this only with Poincare's Theorem and the properties of the natural density function.(Not Ergodic Theorems)</p>
<p>Any help is appreciated.</p>
<p>Thank you in advance</p>
| mathworker21 | 366,088 | <p>For almost every $x$, the limit $$\lim_{N \to \infty} \frac{1}{N}\sum_{n=1}^{N} \chi_A(T^nx) = \lim_{N \to \infty} \frac{|L_x \cap [1,N]|}{N}$$ exists. Let $f^*(x)$ be the limit. It is a fact (see Birkhoff ergodic theorem - note this does apply to non-ergodic measure preserving systems) that $\int f^*(x)d\mu = \int \chi_A d\mu$. <em>However</em>, we only know this applies if $\mu(X)$ is finite (which I am assuming it is). Finish from here.</p>
|
2,486,590 | <p>Let $(X,\mathcal{B},\mu,T)$ be a dynamical system and let $A \in \mathcal{B}$ such that $\mu(A)>0$ and $\forall x \in X$ we define $$L_x=\{n \in \Bbb{N}|T^nx \in A\}$$</p>
<p>Prove that $\mu(\{x:\bar{d}(L_x)>0\})>0$ where $$\bar{d}(L_x)=\limsup_n \frac{|L_x \cap \{1,2...n\}|}{n}$$.</p>
<p>My first thought is that i have to use somewhere Poincare's recurrence theorem.</p>
<p>But i cannot understand the behavior of the density function.</p>
<p>From Poincare we just know that for almost every $x \in A$ we have that $L_x$ is infinite.</p>
<p>Can some give me a hint to solve this?</p>
<p>I clearly do not want a full solution. </p>
<p>I want hint to solve this only with Poincare's Theorem and the properties of the natural density function.(Not Ergodic Theorems)</p>
<p>Any help is appreciated.</p>
<p>Thank you in advance</p>
| D. J. Obata | 418,537 | <p>I am also assuming $\mu$ is a probability measure. For a measurable set $A$ with positive $\mu$-measure, define $A_0$ to be the set of points $x\in A$ such that $\overline{d}(L_x)=0$. Let $\chi_{A_0}(.)$ be the characteristic function over the set $A_0$ and for any $x\in X$, consider
$$
F^+(x) = \limsup_{n\to +\infty} \frac{1}{n} \sum_{j=0}^{n-1} \chi_{A_0}\circ T^j(x).
$$</p>
<p>From the definition of $A_0$, it is easy to prove that $F^+(x) = 0$ for every $x\in X$. Also observe that
$$
F^-(x)= \liminf_{n\to + \infty} \frac{1}{n} \sum_{j=0}^{n-1} \chi_{A_0} \circ T^j(x) \geq 0.
$$</p>
<p>Thus, for every $x\in X$, it is well defined</p>
<p>$$
F(x) = \lim_{n\to + \infty} \frac{1}{n} \sum_{j=0}^{n-1} \chi_{A_0} \circ T^j(x) = 0.
$$</p>
<p>Of course for every $n\in\mathbb{N}$ and every $x\in X$, the function $ \frac{1}{n} \sum_{j=0}^{n-1} \chi_{A_0}\circ T^j(x)$ takes value between $0$ and $1$. By the dominated convergence theorem, one concludes that</p>
<p>$$ \mu(A_0) = \lim_{n\to+\infty} \int \frac{1}{n} \sum_{j=0}^{n-1} \chi_{A_0}\circ T^jd\mu= \int \lim_{n\to+\infty} \frac{1}{n} \sum_{j=0}^{n-1} \chi_{A_0}\circ T^jd\mu= \int F d\mu=0.$$</p>
<p>One concludes that $\mu$-almost every point in $A$ has a positive density.</p>
<p><strong>Remark:</strong> The same result is true if you change $\limsup$ for $\liminf$ in the definition of density, but I don't know how to prove that without using some ergodic theorem.</p>
|
4,552,723 | <p>Assume the following angles are known:
<span class="math-container">$ABD$</span>,<span class="math-container">$DBC$</span>,<span class="math-container">$BAC$</span>,<span class="math-container">$ACD$</span>.</p>
<p>Is it possible to compute <span class="math-container">$CDA$</span>?</p>
<p><a href="https://i.stack.imgur.com/nrpQL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nrpQL.png" alt="enter image description here" /></a></p>
| 冥王 Hades | 1,092,912 | <p>Generally speaking, it is possible to use the sine rule in <span class="math-container">$\triangle ABD$</span>, <span class="math-container">$\triangle DBC$</span> and <span class="math-container">$\triangle BAC$</span> to develop a system of trigonometric equations which also develops some sort of relation between the sides of the quadrilateral, and then solving the system by isolating the sine functions with the missing angle (say <span class="math-container">$\angle x$</span>) to obtain the answer. I believe such trigonometric generalizations already exist, I'd suggest looking up "Langley's adventitious angles" on Wikipedia, there, you'll find several links and more information regarding these "adventitious quadrangles" as well.</p>
<p>On the other hand, developing generalized geometric (purely euclidean) solutions is very hard if not near-impossible without further constraints, and even with that, the problem (and by extension its solution) can change significantly just by altering one or two angles. This applies to both quadrilaterals and triangles. Overall, there are several techniques and methods you can use to attack a problem using euclidean geometry, but they're very specific as I said and won't work all the time. What techniques? Well, let's take a general example. Suppose, in this scenario, we're given that <span class="math-container">$AB=AD$</span> and that <span class="math-container">$\angle ACD=30$</span>, in this case, my very first move would be to locate a point (say, point <span class="math-container">$P$</span>) outside <span class="math-container">$\triangle ADC$</span> and let it be the circumcenter for said triangle. Doing this would give us several connections such as <span class="math-container">$AB=AD=AP=DP=CP$</span> and the <span class="math-container">$\triangle APD$</span> would be equilateral. This would be a very helpful connection and help me proceed further with the solution in such a specific scenario. Alternatively, you can also reflect the <span class="math-container">$\triangle ADC$</span> about the line segment <span class="math-container">$DC$</span> and reflect point <span class="math-container">$A$</span> onto point <span class="math-container">$A'$</span> which would give you <span class="math-container">$\angle ACA'=60$</span> and thus <span class="math-container">$\triangle ACA'$</span> would be equilateral. This is yet another technique that you could use in such a specific scenario. There are definitely more, some even relying on previous results of problems I've solved (which I do at least mention whenever I use them and I don't think doing otherwise is fair), but these two would definitely things I'd try almost immediately in a scenario like this before, if this fails, proceeding to either some different techniques depending on other constraints or trying to do something completely different and being "creative" which involves staring the problem silently for 15 minutes.</p>
|
2,483,231 | <p>If $F(x)=f(g(x))$, where $f(5) = 8$, $f'(5) = 2$, $f'(−2) = 5$, $g(−2) = 5$, and
$g'(−2) = 9$, find $F'(−2)$. I'm totally lost on this problem, I'm assuming to incorporate the Chain Rule. I get $5(5) * 9 = 225$ but I am incorrect.</p>
<p>Update: Thanks guys, I see where I messed up thanks!</p>
| Donald Splutterwit | 404,247 | <p>The chain rule gives $F'(x)=g'(x)f'(g(x))$. Now just substitute the values
\begin{eqnarray*}
F'(-2)=\underbrace{g'(-2)}_{9}f'(\underbrace{g(-2)}_{5})=9\underbrace{f'(5)}_{2}=9 \times 2 =\color{red}{18}.
\end{eqnarray*}</p>
|
2,490,128 | <p>Over the domain of integers, if $(a-c)|(ab+cd)$ then $(a-c)|(ad+bc)$.</p>
<p>Note: $x|y$ means "$x$ divides $y$," i.e. $\exists k\in \mathbb{Z}. y=x\cdot k$</p>
<p>This is part of an assignment on GCD, Euclidean algorithm, and modular arithmetic.</p>
<p>My approach:</p>
<p>If $a-c$, divides a linear combination of $a$ and $c$, then $a-c$ is a common divisor of $a$ and $c$. This comes from the definition of a common divisor: that if a certain $d$ divides two integers $x$ and $y$, then $d$ divides a linear combination of $x$ and $y$. Both $ab+cd$ and $ad+bc$ are linear combinations of $a$ and $c$ so $a-c$ must divide both of them.</p>
| Cornman | 439,383 | <p>Hint:</p>
<p>$\int \frac{1}{x^2\sqrt{x^2+x+1}}\, dx=\int \frac{1}{x^2\sqrt{(x+\frac12)^2+\frac34}}\, dx$</p>
<p>Now substitute $u=x+\frac12$</p>
|
2,490,128 | <p>Over the domain of integers, if $(a-c)|(ab+cd)$ then $(a-c)|(ad+bc)$.</p>
<p>Note: $x|y$ means "$x$ divides $y$," i.e. $\exists k\in \mathbb{Z}. y=x\cdot k$</p>
<p>This is part of an assignment on GCD, Euclidean algorithm, and modular arithmetic.</p>
<p>My approach:</p>
<p>If $a-c$, divides a linear combination of $a$ and $c$, then $a-c$ is a common divisor of $a$ and $c$. This comes from the definition of a common divisor: that if a certain $d$ divides two integers $x$ and $y$, then $d$ divides a linear combination of $x$ and $y$. Both $ab+cd$ and $ad+bc$ are linear combinations of $a$ and $c$ so $a-c$ must divide both of them.</p>
| Michael Rozenberg | 190,319 | <p>$x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}$.</p>
<p>Thus, the substitution $x+\frac{1}{2}=\frac{\sqrt3}{2}\tan{t}$ must help.</p>
|
4,487,380 | <p>I was reading my calculus book wherein I came across a note, being worth of attention. It says:</p>
<blockquote>
<p>Integrals in the form of <span class="math-container">$\int P(x)e^{ax}dx$</span> have a special property. After calculating the integral, we obtain a function in the form of <span class="math-container">$Q(x)e^{ax}$</span> where <span class="math-container">$Q(x)$</span> is a polynomial of the same degree as of <span class="math-container">$P(x)$</span>. This is called method of indefinite coefficients.</p>
</blockquote>
<p>Like for example, we do <span class="math-container">$\displaystyle\int (3x^3-17)e^{2x}dx$</span>, We can do it traditionally by integration by parts but let me show you my or rather author's method :</p>
<p>Let this be equal to <span class="math-container">$(Ax^3+Bx^2+Cx+D)e^{2x}$</span></p>
<p>Now differentiating both sides we get, <span class="math-container">$$(3x^3-17)e^{2x}=2(Ax^3+Bx^2+Cx+D)e^{2x}+e^{2x}(3Ax^2+2Bx+C)$$</span>
Now we will cancel <span class="math-container">$e^{2x}$</span> on both sides and the rest is equating the coefficients.</p>
<p>I have seen that this property is applicable in every question but I don't know the mathematical proof of this. It wasn't even in the book.</p>
<p>Any help regarding the proof is greatly appreciated.</p>
| legionwhale | 685,267 | <p>The best method for this question is to use induction and integration by parts (as suggested by W. Fan), as well as the linearity of the integral. Note that if <span class="math-container">$P(x) = c_n x^n + \cdots + c_0$</span>,</p>
<p><span class="math-container">$$\int P(x) e^{ax} dx = \int (c_n x^n + \cdots + c_0) e^{ax} dx $$</span></p>
<p><span class="math-container">$$= \sum_{i=0}^n c_k\int x^ke^{ax}dx$$</span></p>
<p>Thus, it suffices to show that the result holds for <span class="math-container">$x^k$</span>. We do this by induction on <span class="math-container">$k$</span>. The basis for <span class="math-container">$k=0$</span> is simple to verify. Now consider, for <span class="math-container">$k$</span> in general, the integral:</p>
<p><span class="math-container">$$\int x^k e^{ax} dx$$</span></p>
<p><span class="math-container">$$\frac{1}{a}x^ke^{ax} -\frac{k}{a}\int x^{k-1}e^{ax} dx$$</span></p>
<p>for <span class="math-container">$a \neq 0$</span>. This latter integral is <span class="math-container">$e^{ax}$</span> multiplied by a polynomial of degree <span class="math-container">$(k-1)$</span> by the induction hypothesis, so it follows that <span class="math-container">$\int x^k e^{ax} dx$</span> is <span class="math-container">$e^{ax}$</span> multiplied by a polynomial of degree <span class="math-container">$k$</span>. <span class="math-container">$\square$</span></p>
<p><strong>Remark</strong>: note that the statement is not true for <span class="math-container">$a \neq 0$</span> (by direct verification). The statement is also missing the clause that <span class="math-container">$Q(x)e^{ax}$</span> is the antiderivative up to a constant, which is apparent when verifying the basis.</p>
<p>This argument also gives <span class="math-container">$Q$</span> explicitly with a little bit more effort. As an exercise, show that:</p>
<p><span class="math-container">$$Q(x) = \sum_{k=0}^n c_k \bigg(\frac{1}{a}x^k-\frac{k}{a^2}x^{k-1}+\frac{k(k-1)}{a^3}x^{k-2}+\cdots +{(-1)}^k\frac{k!}{a^{k+1}} \bigg)$$</span></p>
<p>where <span class="math-container">$P(x) = c_n x^n + \cdots + c_0$</span> as before.</p>
|
733,553 | <p>It's been a long time since high school, and I guess I forgot my rules of exponents. I did a web search for this rule but I could not find a rule that helps me explain this case:</p>
<p>$ 2^n + 2^n = 2^{n+1} $</p>
<p>Which rule of exponents is this?</p>
| Community | -1 | <p>$2^n + 2^n = 2^n(1+1) = 2^n(2) = 2^{n+1}$</p>
|
827,740 | <p>This is a new integral that I propose to evaluate in closed form:
$$ {\mathfrak{R}} \int_{0}^{\pi/2} \frac{x^2}{x^2+\log ^2(-2\cos x)} \:\mathrm{d}x$$
where $\Re$ denotes the real part and $\log (z)$ denotes the principal value of the logarithm defined for $z \neq 0$ by
$$ \log (z) = \ln |z| + i \mathrm{Arg}z, \quad -\pi <\mathrm{Arg} z \leq \pi.$$</p>
| pisco | 257,943 | <p>$$\color{blue}{\Re\int_{0}^{\pi/2} \frac{x^2}{x^2+\log ^2(-2\cos x)}dx = \frac{\pi }{8}\left( {1 - \ln 2 - \gamma } \right)}$$</p>
<hr>
<p>For a proof, I will continue from $$\tag{4} - 8\Re I + 2\pi J = \pi \ln 2 - \pi$$ in <a href="https://math.stackexchange.com/questions/863496">this question</a>.
where $$J=\int_0^\infty {\frac{1}{{{x^2} + {\pi ^2}}}\left[ { - \frac{x}{{1 - {e^{ - x}}}} + \ln ({e^x} - 1)} \right]dx}$$
Integration by part shows
$$J = -\pi \int_0^\infty {\arctan x\frac{{{e^{\pi x}}x}}{{{{\left( {{e^{\pi x}} - 1} \right)}^2}}}dx} $$
Invoke the Binet second formula:
$$\ln\Gamma(z) = (z-\frac{1}{2})\ln z - z + \frac{\ln(2\pi)}{2}+2z\int_0^\infty \frac{\arctan t}{e^{2\pi t z} - 1} dt$$
The value of $J$ immdiately follows from differentiating with respect to $z$, and then set $z = 1/2$:
$$J = -\frac{\gamma}{2}$$</p>
<p>I hope someone can explain reminiscence of this result to <a href="https://math.stackexchange.com/questions/178790">another question</a>.</p>
|
936,138 | <p>I need help approaching a proof which deals with inequalities:</p>
<p>If p and r are the precision and recall of a test, then the F1 measure of the test is
defined to be
$$F(p, r) = \frac{2pr}{p+r}$$</p>
<p>Prove that, for all positive reals p, r, and t, if t ≥ r then F(p, t) ≥ F(p, r)</p>
<p>What's the first step to approaching this problem? Do I need to look at this with different cases? </p>
| JimmyK4542 | 155,509 | <p><strong>Hint</strong>: $F(p,r) = \dfrac{2pr}{p+r} = \dfrac{2p^2+2pr-2p^2}{p+r} = \dfrac{2p(p+r)-2p^2}{p+r} = 2p - \dfrac{2p^2}{p+r}$. </p>
<p>Can you show that this is an increasing function in $r$?</p>
|
2,572,302 | <p>I want to calculate the limit: $$ \lim_{x\to +\infty}(1+e^{-x})^{2^x \log x}$$
The limit shows itself in an $1^\infty$ Indeterminate Form. I tried to elevate $e$ at the logarithm of the function:</p>
<p>$$\lim_{x\to +\infty} \log(e^{(1+e^{-x})^{2^x \log x}}) = e^{\lim_{x\to +\infty} \log((1+e^{-x})^{2^x \log x})} = e^{\lim_{x\to +\infty} 2^x \log x \cdot \log(1+e^{-x}) }$$</p>
<p>And then rewrite the exponent as a fraction, to get an $\frac{\infty}{\infty}$
form:</p>
<p>$$= \lim_{x\to +\infty} \frac{2^x \log x}{\frac{1}{\log(1+e^{-x})}} $$</p>
<p>But I don't know how to apply an infinite comparing technique here, and even applying de l'Hôpital seems to lead to nothing...</p>
<p>Could you guys give me some help?</p>
<p>Furthermore: is there a way to calculate this limit without using series expansions or other advanced mathematic instruments?</p>
<p>Thank you very much in advance.</p>
<p>P.S. Wolfram says this limit goes to 1, but I still really want to know how.</p>
| Angina Seng | 436,618 | <p>If we call the expression $f(x)$ then
$$\ln f(x)=2^x\log x\,\ln(1+e^{-x})=O\left(\frac{2^x\log x}{e^{x}}\right)\to0$$
as $x\to\infty$ as $2/e<1$.</p>
|
2,572,302 | <p>I want to calculate the limit: $$ \lim_{x\to +\infty}(1+e^{-x})^{2^x \log x}$$
The limit shows itself in an $1^\infty$ Indeterminate Form. I tried to elevate $e$ at the logarithm of the function:</p>
<p>$$\lim_{x\to +\infty} \log(e^{(1+e^{-x})^{2^x \log x}}) = e^{\lim_{x\to +\infty} \log((1+e^{-x})^{2^x \log x})} = e^{\lim_{x\to +\infty} 2^x \log x \cdot \log(1+e^{-x}) }$$</p>
<p>And then rewrite the exponent as a fraction, to get an $\frac{\infty}{\infty}$
form:</p>
<p>$$= \lim_{x\to +\infty} \frac{2^x \log x}{\frac{1}{\log(1+e^{-x})}} $$</p>
<p>But I don't know how to apply an infinite comparing technique here, and even applying de l'Hôpital seems to lead to nothing...</p>
<p>Could you guys give me some help?</p>
<p>Furthermore: is there a way to calculate this limit without using series expansions or other advanced mathematic instruments?</p>
<p>Thank you very much in advance.</p>
<p>P.S. Wolfram says this limit goes to 1, but I still really want to know how.</p>
| Claude Leibovici | 82,404 | <p>Consider the more general case of $$y=(1+e^{-x})^{a^x\, \log(x)}$$ Tak logarithms of both sides $$\log(y)={a^x \log(x)}\log(1+e^{-x})$$ When $x$ is large $$\log(1+e^{-x})\sim e^{-x}$$ making $$\log(y)\sim \left(\frac a e\right)^x \log(x)$$ Now, consider the cases where $a<e$ and $a >e$.</p>
|
1,618,373 | <p>Prove that $S_4$ cannot be generated by $(1 3),(1234)$</p>
<p>I have checked some combinations between $(13),(1234)$ and found out that those combinations cannot generated 3-cycles.</p>
<p>Updated idea:<br>
Let $A=\{\{1,3\},\{2,4\}\}$<br>
Note that $(13)A=A,(1234)A=A$<br>
Hence, $\sigma A=A,\forall\sigma\in \langle(13),(1234)\rangle$<br>
In particular, $(12)\notin \sigma A,\forall\sigma\in \langle(13),(1234)\rangle$<br>
So we conclude that $S_4\neq\langle(13),(1234)\rangle$</p>
| Marc van Leeuwen | 18,880 | <p>The partition $\{\{1,3\},\{2,4\}\}$ is invariant under the action of the two proposed generators, but not under all of $S_4$, so they cannot generate all of $S_4$.</p>
|
1,925,867 | <p>I can't find any. For saying $H$ is a subgroup of $G$ we have notation but it seems none exists for subrings.</p>
| Daniel Buck | 293,319 | <p>As you asked for the notation, I read this as how subrings are identified in the literature so I looked up the definitions for subrings in a few well known books. I checked the books on Algebra by Artin, Dummit & Foote, Hungerford, Jacobson, Lang, van der Waerden, where notation for a subring seems to be nonexistent, i.e., "is a subring" is as good as it gets, at least in their definition of what a subring is.</p>
<p>Fraleigh in <em>A First Course in Algebra</em>, p.173 holds off <em>explicitly</em> defining any notation, but states:</p>
<blockquote>
<p>... In fact, let us say here once and for all that if we have a set, together with a certain type of <em>algebraic structure</em> (group, ring, field, integral domain, vector space, and so on), then any subset of this set, together with the natural induced algebraic structure <em>that yields an algebraic structure of the same type,</em> is a <em>substructure</em>. If $K$ and $L$ are both structures, we shall let $K\le L$ denote that $K$ is a substructure of $L$ and $K<L$ denote that $K\leq L$ but $K\neq L$.</p>
</blockquote>
<p>Then in Rotman's <em>Modern Algebra</em>, he defines a subring then gives the following on p.119:</p>
<blockquote>
<p><strong>Notation.</strong> In contrast to the usage $H \leq G$ for a subgroup, the tradition in ring theory is to write $S \subseteq R$ for a subring. We shall also write $S \subsetneq R$ to denote a <em>proper</em> subring; that is, $S \subseteq R$ and $S \neq R$.</p>
</blockquote>
<p>So it seems fair to say $S\subseteq R$, and $S\leq R$ can be safely used in place of any explicit notation reserved for subrings.</p>
|
111,795 | <p>I need to add a small graphics on top of a larger one, and the small graphics should stick very close to the large one, with their axis aligned. Here's a minimal code to work with, using some elements from this question/answer :</p>
<p><a href="https://mathematica.stackexchange.com/questions/22521/how-to-make-a-plot-on-top-of-other-plot">How to make a plot on top of other plot?</a></p>
<pre><code>Intensity[p_, q_, phi_] := Plot[
(If[p > 0, Sin[2Pi p^2 x]/(2Pi p^2 x), 1]Cos[2Pi p^2 q x + phi/2])^2,
{x, -30, 30},
PlotPoints -> 400,
MaxRecursion -> 4,
PlotRange -> All,
PlotRange -> {{-30, 30}, {0, 1}},
Axes -> False,
AspectRatio -> 1,
Frame -> True,
ImageSize -> {600, 600}
]
LumIntensity[p_, q_, phi_] := DensityPlot[
(If[p > 0, Sin[2Pi p^2 x]/(2Pi p^2 x), 1]Cos[2Pi p^2 q x + phi/2])^2,
{x, -30, 30}, {y, 0, 1},
AspectRatio -> 0.1,
PlotPoints -> {1000, 2},
Frame -> None,
ImageSize -> 600
]
GraphicsColumn[
{LumIntensity[0.25, 5, 0], Intensity[0.25, 5, 0]},
Spacings -> 0
]
</code></pre>
<p>Here's what I want to achieve (which the question/answer above don't solve) :</p>
<p><img src="https://s22.postimg.org/b3ad8su2p/interference.jpg" alt="interference"></p>
<p>Also, how can I add a black frame around the small graphics ? Using <code>Frame -> True</code> or <code>Framed[...]</code> gives an ugly output.</p>
<p>The combination would be used for a Manipulate box, since <code>p</code>, <code>q</code> and <code>phi</code> are variables.</p>
<p><strong>EDIT :</strong> Actually, it would be better if the small graphics was placed at the bottom of the large one.</p>
| ubpdqn | 1,997 | <p>I think it may perhaps be easier just to combine plots and modify (e.g. suppress unnecessary frame ticks). I post this as a motivating answer rather than definitive answer. <code>li</code> is a modified version of OP function:</p>
<pre><code>li[p_, q_, phi_, {l_, u_}] :=
DensityPlot[(If[p > 0, Sin[2 Pi p^2 x]/(2 Pi p^2 x), 1] Cos[
2 Pi p^2 q x + phi/2])^2, {x, -30, 30}, {y, l, u},
AspectRatio -> 0.1, PlotPoints -> {1000, 2}, Frame -> None,
ImageSize -> 600]
Manipulate[
Show[Intensity[0.25, 5, 0], li[0.25, 5, 0, {l, u}]], {l, -1, 0,
Appearance -> "Labeled"}, {{u, -0.5}, -1, 0,
Appearance -> "Labeled"}]
</code></pre>
<p><a href="https://i.stack.imgur.com/y7sco.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/y7sco.gif" alt="enter image description here"></a></p>
|
111,795 | <p>I need to add a small graphics on top of a larger one, and the small graphics should stick very close to the large one, with their axis aligned. Here's a minimal code to work with, using some elements from this question/answer :</p>
<p><a href="https://mathematica.stackexchange.com/questions/22521/how-to-make-a-plot-on-top-of-other-plot">How to make a plot on top of other plot?</a></p>
<pre><code>Intensity[p_, q_, phi_] := Plot[
(If[p > 0, Sin[2Pi p^2 x]/(2Pi p^2 x), 1]Cos[2Pi p^2 q x + phi/2])^2,
{x, -30, 30},
PlotPoints -> 400,
MaxRecursion -> 4,
PlotRange -> All,
PlotRange -> {{-30, 30}, {0, 1}},
Axes -> False,
AspectRatio -> 1,
Frame -> True,
ImageSize -> {600, 600}
]
LumIntensity[p_, q_, phi_] := DensityPlot[
(If[p > 0, Sin[2Pi p^2 x]/(2Pi p^2 x), 1]Cos[2Pi p^2 q x + phi/2])^2,
{x, -30, 30}, {y, 0, 1},
AspectRatio -> 0.1,
PlotPoints -> {1000, 2},
Frame -> None,
ImageSize -> 600
]
GraphicsColumn[
{LumIntensity[0.25, 5, 0], Intensity[0.25, 5, 0]},
Spacings -> 0
]
</code></pre>
<p>Here's what I want to achieve (which the question/answer above don't solve) :</p>
<p><img src="https://s22.postimg.org/b3ad8su2p/interference.jpg" alt="interference"></p>
<p>Also, how can I add a black frame around the small graphics ? Using <code>Frame -> True</code> or <code>Framed[...]</code> gives an ugly output.</p>
<p>The combination would be used for a Manipulate box, since <code>p</code>, <code>q</code> and <code>phi</code> are variables.</p>
<p><strong>EDIT :</strong> Actually, it would be better if the small graphics was placed at the bottom of the large one.</p>
| Jason B. | 9,490 | <p>Just going to throw this in to the mix. When I saw this question, it immediately seemed perfect for Jens's <a href="https://mathematica.stackexchange.com/a/6882/9490">function</a>, which I modified and used previously, and in fact I have it defined in my init.m because I use it with such regularity.</p>
<p>I have modified the original function to respect the individual aspect ratios of the constituent plots, and the definition is in the pastebin linked below. It's a reasonably large but robust function, which I've used to combine plots for publication for years now.</p>
<p>Using the functions defined in the OP, this is the plotting code,</p>
<pre><code><< "http://pastebin.com/raw/1uhTgyuJ"
plotGrid[{{LumIntensity[0.25, 5, 0]}, {Intensity[0.25, 5,
0]}}, 600, 660, "KeepAR" -> True]
</code></pre>
<p><a href="https://i.stack.imgur.com/nTNDF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nTNDF.png" alt="enter image description here"></a></p>
<p>The plot can be interactively resized with the mouse and the overall appearance remains unchanged.</p>
|
111,795 | <p>I need to add a small graphics on top of a larger one, and the small graphics should stick very close to the large one, with their axis aligned. Here's a minimal code to work with, using some elements from this question/answer :</p>
<p><a href="https://mathematica.stackexchange.com/questions/22521/how-to-make-a-plot-on-top-of-other-plot">How to make a plot on top of other plot?</a></p>
<pre><code>Intensity[p_, q_, phi_] := Plot[
(If[p > 0, Sin[2Pi p^2 x]/(2Pi p^2 x), 1]Cos[2Pi p^2 q x + phi/2])^2,
{x, -30, 30},
PlotPoints -> 400,
MaxRecursion -> 4,
PlotRange -> All,
PlotRange -> {{-30, 30}, {0, 1}},
Axes -> False,
AspectRatio -> 1,
Frame -> True,
ImageSize -> {600, 600}
]
LumIntensity[p_, q_, phi_] := DensityPlot[
(If[p > 0, Sin[2Pi p^2 x]/(2Pi p^2 x), 1]Cos[2Pi p^2 q x + phi/2])^2,
{x, -30, 30}, {y, 0, 1},
AspectRatio -> 0.1,
PlotPoints -> {1000, 2},
Frame -> None,
ImageSize -> 600
]
GraphicsColumn[
{LumIntensity[0.25, 5, 0], Intensity[0.25, 5, 0]},
Spacings -> 0
]
</code></pre>
<p>Here's what I want to achieve (which the question/answer above don't solve) :</p>
<p><img src="https://s22.postimg.org/b3ad8su2p/interference.jpg" alt="interference"></p>
<p>Also, how can I add a black frame around the small graphics ? Using <code>Frame -> True</code> or <code>Framed[...]</code> gives an ugly output.</p>
<p>The combination would be used for a Manipulate box, since <code>p</code>, <code>q</code> and <code>phi</code> are variables.</p>
<p><strong>EDIT :</strong> Actually, it would be better if the small graphics was placed at the bottom of the large one.</p>
| Edmund | 19,542 | <p>This can be achieved by matching a few layout options between the two charts. Namely, <code>ImageSize</code>, <code>ImageMargins</code>, <code>ImagePadding</code>, <code>PlotRange</code>, and <code>PlotRangePadding</code>. <code>Scaled</code> and <code>Automatic</code> can be used to simplify selection.</p>
<p>With</p>
<pre><code>f[p_, q_, phi_, x_] :=
(If[p > 0, Sin[2 Pi p^2 x]/(2 Pi p^2 x), 1] Cos[2 Pi p^2 q x + phi/2])^2
alignment =
Sequence[ImageSize -> {600, Automatic},
ImageMargins -> 0,
PlotRangePadding -> Scaled[.02],
ImagePadding -> {{Scaled[.03], Scaled[.01]}, {Automatic, Automatic}}];
</code></pre>
<p>Then</p>
<pre><code>Column[
{DensityPlot[f[.25, 5, 0, x], {x, -30, 30}, {y, 0, 1},
PlotPoints -> {1000, 2}, Frame -> None, AspectRatio -> .1,
Evaluate@alignment],
Plot[f[.25, 5, 0, x], {x, -30, 30},
PlotRange -> Full, Frame -> True,
Evaluate@alignment]},
Spacings -> Scaled[.0005]]
</code></pre>
<p><a href="https://i.stack.imgur.com/TSwZR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TSwZR.png" alt="enter image description here"></a></p>
<p>Since we want to align along the x-axis we only need the x-axis plot range, size, padding, and margin to match between the two plots. The plot range is matched in the plot calls so <code>alignment</code> only contains items for size, padding, and margin. With this the plots have the same width and amount of <em>whitespace</em> along the x-axis and so they align. </p>
<p>Hope this helps.</p>
|
833,376 | <p>I know very little in the way of math history, but I question that was bothering me recently is where the terms open and closed came from in topology. I know that it's easy to ascribe a sense of openness/closedness to said sets, but I feel like there are a lot of other, more appropriate words that could have been used. On a related note, I was wondering who first developed the idea of open/closed sets, and who first used those words to describe them.</p>
| Alexander Gruber | 12,952 | <p>This answer addresses your question on who first actually used the words "open" and "closed" to describe open and closed sets. It seems, according to <a href="http://www.sciencedirect.com/science/article/pii/S0315086008000050" rel="nofollow">this article</a>, like the first mention of this language was in René Baire's doctoral dissertation, </p>
<blockquote>
<p>Sur les fonctions de variables réelles
Annali di matematica pura ed applicata (3), 3 (1899), pp. 1–123</p>
</blockquote>
<p>which I believe first appears in page 7 of the <a href="https://archive.org/stream/surlesfonctions00bairgoog#page/n152/mode/2up" rel="nofollow">document</a> in the context of defining an "open domain." The term itself was first actually defined by Lebesgue in his dissertation, <em>Intégrale, longueur, aire</em>, (which I cannot find an online copy of) for the purpose of setting up Lebesgue measure.</p>
<p>I think the rest of your question should be addressed to your satisfaction between the linked article and related MSE/MO threads.</p>
|
2,789,002 | <p>How can I calculate the height of the tree? I am with geometric proportionality.</p>
<p><img src="https://i.stack.imgur.com/m4zMD.png"></p>
| Mr Pie | 477,343 | <p>I will denote by $a\cdot b$ the product of $a$ and $b$ (referred to as <em>dot multiplication</em>).</p>
<hr>
<p>You want to find the gradient of the dotted line. Since it is straight, it is in <em>linear form</em>, namely, $$y=mx+b\quad\text{ or }\quad y=mx+c.\tag*{$\bigg(\begin{align}&\text{depending on how} \\ &\text{you were taught}\end{align}\bigg)$}$$</p>
<hr>
<p>Firstly, you want to find the value $m$, since that is the gradient. Our unit of measurement is <em>metres</em>.</p>
<p>Now, according to the diagram: $$x_1=0\Rightarrow y_1=1.6\tag*{$\because$ $160$cm $=$ $1.6$ metres}$$ which is the height of the stick figure; and, $$x_2=16\Rightarrow y_2=17.2$$ which is the height of the smallest tree.</p>
<hr>
<p>Secondly, we want to find $y$ at $x=16+10=26$, i.e. the height of the biggest tree.</p>
<p>To do this, we use the <em>gradient formula</em>: $$m=\frac{y_2-y_1}{x_2-x_1}$$ Now, we substitute the values of $x_{1,2}$ and $y_{1,2}$ in the following way: $$\begin{align} m&=\frac{17.2-1.6}{16-0} =\frac{15.6}{16}\\ &=0.975.\end{align}$$</p>
<hr>
<p>Lastly, we find $b$ or $c$ (I will use $c$) by substituting $x=0$, since that is the $y$ intercept. It follows that, $$\begin{align}1.6&=0.975\cdot 0 + c \\ &= 0+c \\ \therefore c &= 1.6.\end{align}$$ $$\boxed{ \ \begin{align}\therefore y&=0.975\cdot 26 + 1.6 \\ &= 25.35+1.6\\ &=26.95.\end{align} \ }$$</p>
<hr>
<blockquote>
<blockquote>
<p>$$\text{Ergo, the height of the biggest tree is $26.95$ metres.}\tag*{$\bigcirc$}$$</p>
</blockquote>
</blockquote>
|
213,338 | <p>Suppose that $f$ is analytic in the unit disc D = {$z \in \mathbb{C}$ : |$z$| < 1} and $|$f($z$)$| \le 1/(1-|$z$|)$ for all $z\in D$.</p>
<p>Let $f($z$)= \sum _{n=0}^{\infty } a_nz^n$ be the power series expansion of f about $0$.</p>
<p>Prove that $$|a_n| \le (n+1)(1+1/n)^{n} < e(n+1)$$</p>
| Max Morin | 44,026 | <p>$x^x = e^{x\ln x}$, so $$\lim_{x\to 0}x^x=\lim_{x\to 0}e^{x\ln x}=e^0=1$$.</p>
|
826,011 | <p>I want to express variable from equation in WolframAlpha web. I tried several keywords but it didn't work. For example I have equation</p>
<p>$$y=x+z+k,$$</p>
<p>and I want Wolfram to rewrite it for variable $x$, $x=y-z-k$. Is it possible and how?</p>
| ronald | 155,944 | <p>You can use the <a href="http://goo.gl/NwzZDQ" rel="noreferrer">following query</a>:</p>
<pre><code>solve(y=x+z+k,x)
</code></pre>
<p><img src="https://i.stack.imgur.com/UOrXe.png" alt="enter image description here"></p>
|
826,011 | <p>I want to express variable from equation in WolframAlpha web. I tried several keywords but it didn't work. For example I have equation</p>
<p>$$y=x+z+k,$$</p>
<p>and I want Wolfram to rewrite it for variable $x$, $x=y-z-k$. Is it possible and how?</p>
| MattAllegro | 142,842 | <p>To complete my previous comment, <a href="https://www.wolframalpha.com/input/?i=solution+for+variable+x+of%3A+y%3Dx%2Bz%2Bk" rel="nofollow noreferrer">this</a> is what you may type to achieve what you asked.</p>
<p>I just copied and pasted your equation in WolphramAlpha, the remaining "syntax" was suggested to me by the usual WA output page ("solution for variable <span class="math-container">$x$</span>").</p>
|
3,320,193 | <blockquote>
<p>If given <span class="math-container">$P(B\mid A) =4/5$</span>, <span class="math-container">$P(B\mid A^\complement)= 2/5$</span> and <span class="math-container">$P(B)= 1/2$</span>, what is the probability of <span class="math-container">$A$</span>?</p>
</blockquote>
<p>I know I need to apply Bayes theorem here to figure this out, but I'm struggling a bit to understand how. </p>
<p>So far I've considered this formula:
<span class="math-container">$$P(B\mid A) = \dfrac{P (B \cap A) }{ P (B \cap A) + P(B^\complement \cap A)}$$</span></p>
<p>From this formula, I understand that <span class="math-container">$P(B \cap A) = P(A) \cdot P(B\mid A)$</span> so I plug in the given values but then only find that <span class="math-container">$P(B^\complement |A)$</span> is <span class="math-container">$2/25$</span>. But this does not get me any closer to my goal, <span class="math-container">$P(A)$</span>.</p>
<p>I imagine my understanding of this is quite backward. Any pointers would be helpful.</p>
<p>Thank you</p>
| Robert Z | 299,698 | <p>All you need is the definition of conditional probability:
<span class="math-container">$$P(X|Y)=P(X\cap Y)/P(Y).$$</span>
We have that
<span class="math-container">$$P(A^c\cap B)=P(B|A^c)P(A^c)=\frac{2}{5}(1-P(A))$$</span>
and
<span class="math-container">$$P(A \cap B)=P(B|A)P(A)=\frac{4}{5}P(A).$$</span>
Hence
<span class="math-container">$$\frac{1}{2}=P(B)=P(A^c\cap B)+P(A \cap B)=\frac{2}{5}(1-P(A))+\frac{4}{5}P(A),$$</span>
and, after solving it, we easily find that <span class="math-container">$P(A)=1/4$</span>.</p>
|
3,320,193 | <blockquote>
<p>If given <span class="math-container">$P(B\mid A) =4/5$</span>, <span class="math-container">$P(B\mid A^\complement)= 2/5$</span> and <span class="math-container">$P(B)= 1/2$</span>, what is the probability of <span class="math-container">$A$</span>?</p>
</blockquote>
<p>I know I need to apply Bayes theorem here to figure this out, but I'm struggling a bit to understand how. </p>
<p>So far I've considered this formula:
<span class="math-container">$$P(B\mid A) = \dfrac{P (B \cap A) }{ P (B \cap A) + P(B^\complement \cap A)}$$</span></p>
<p>From this formula, I understand that <span class="math-container">$P(B \cap A) = P(A) \cdot P(B\mid A)$</span> so I plug in the given values but then only find that <span class="math-container">$P(B^\complement |A)$</span> is <span class="math-container">$2/25$</span>. But this does not get me any closer to my goal, <span class="math-container">$P(A)$</span>.</p>
<p>I imagine my understanding of this is quite backward. Any pointers would be helpful.</p>
<p>Thank you</p>
| MR_BD | 195,683 | <p><a href="https://i.stack.imgur.com/hJgeo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hJgeo.png" alt="enter image description here" /></a></p>
<p>Use the Venn diagram. Let the blue side be the even that <span class="math-container">$A$</span> occurs. Also inside the ellipse be the event that <span class="math-container">$B$</span> occurs. Then the <span class="math-container">$\mathbb{P}(B|A)=\frac45$</span> means that the light blue section forms <span class="math-container">$4/5$</span> of the whole blue side. Similarly, the yellow section forms <span class="math-container">$2/5$</span> of the non-blue section.</p>
<p>Finally, let <span class="math-container">$x=P(A)$</span> and you have</p>
<p><span class="math-container">$$\frac 45 x + \frac 25 (1-x)=\frac 12,$$</span></p>
<p>which implies that <span class="math-container">$x=\frac 14$</span>.</p>
|
3,029,208 | <p>Hi I have been trying to find a way to find a combinatorial proof for <span class="math-container">${kn \choose 2}= k{n \choose 2}+n^2{k \choose 2}$</span>. </p>
| Santana Afton | 274,352 | <p>As an addendum to Daniel Robert-Nicoud’s excellent hint, I find it helpful to rewrite the equality in the following way:</p>
<p><span class="math-container">$$\binom{kn}{2} = \binom{k}{1}\binom{n}{2} + \binom{k}{2}\binom{n}{1}\binom{n}{1}.$$</span></p>
<p>By reading multiplication as “and then,” and addition as “or,” try to recover Daniel’s narrative by reading off the right-hand side.</p>
<hr>
<p>With this thought process in mind, try to come up with a formula for <span class="math-container">$\binom{nk}{3}$</span>. Can you generalize this pattern?</p>
|
2,895,284 | <blockquote>
<p>Find $\frac{d}{dx}\frac{x^3}{{(x-1)}^2}$</p>
</blockquote>
<p>I start by finding the derivative of the denominator, since I have to use the chain rule. </p>
<p>Thus, I make $u=x-1$ and $g=u^{-2}$. I find that $u'=1$ and $g'=-2u^{-3}$. I then multiply the two together and substitute $u$ in to get:</p>
<p>$$\frac{d}{dx}(x-1)^{2}=2(x-1)$$</p>
<p>After having found the derivative of the denominator I find the derivative of the numerator, which is $3x^2$. With the two derivatives found I apply the quotient rule, which states that </p>
<p>$$\frac{d}{dx}(\frac{u(x)}{v(x)})=\frac{v'u-vu'}{v^2}$$</p>
<p>and substitute in the numbers</p>
<p>$$\frac{d}{dx}\frac{x^3}{(x-1)^2}=\frac{3x^2(x-1)^2-2x^3(x-1)}{(x-1)^4}$$</p>
<p>Can I simplify this any further?Is the derivation correct?</p>
| Rhys Hughes | 487,658 | <p>It's better to use the quotient rule:
$$\frac{d(\frac fg)}{dx}=\frac{f'g-g'f}{g^2}$$
$$f=x^3\to f'=3x^2$$
$$g=(x-1)^2\to g'=2(x-1)$$
$$\to\frac{d(\frac {x^3}{(x-1)^2})}{dx}=\frac{3x^2(x-1)^2-2x^3(x-1)}{(x-1)^4}=\frac{x^2(x-3)}{(x-1)^3}$$</p>
|
4,243,344 | <blockquote>
<p><span class="math-container">${43}$</span> equally strong sportsmen take part in a ski race; 18 of
them belong to club <span class="math-container">${A}$</span>, 10 to club and 15 to club <span class="math-container">${C}$</span>. What is the
average place for (a) the best participant from club <span class="math-container">${B}$</span>; (b) the
worst participant from club <span class="math-container">${B}$</span>?</p>
</blockquote>
<hr />
<p>I've found the possible range of places the participant could get for both cases. In the case (a), the best participant from club <span class="math-container">${B}$</span> can be at any place between <span class="math-container">$1$</span> and <span class="math-container">$34$</span>. As for the case (b), the worst participant from club <span class="math-container">$B$</span> can get any place between <span class="math-container">$10$</span> and <span class="math-container">$43$</span>. To find the average place I need to compute the expected (mean) value of the this variable. But I'm not sure how to find the chances for getting each place. I suppose they should be equal, but neither <span class="math-container">$\frac{1}{33}$</span> nor <span class="math-container">$\frac{1}{43}$</span> seem to give the right answer.</p>
| user2661923 | 464,411 | <p>First see the answer of Mike Earnest, and the comments following his answer. Apparently, my computation numerically agrees with his answer.</p>
<hr />
<p>Alternative approach:</p>
<p><span class="math-container">$\underline{\text{Problem 1:}}$</span></p>
<p>For <span class="math-container">$k \in \{1,2,\cdots,34\},~$</span> let
<span class="math-container">$p(k)$</span> denote the probability that the highest member of B is in position <span class="math-container">$k$</span>.</p>
<p>Then, the expected position for the highest member of <span class="math-container">$B$</span> will be <span class="math-container">$$\sum_{k=1}^{34} \left[k \times p(k)\right].$$</span></p>
<p>Therefore, the problem reduces to computing <span class="math-container">$p(k)$</span>.</p>
<p>For <span class="math-container">$k$</span> to be the highest position of any member of B, two things have to happen:</p>
<ul>
<li>Positions <span class="math-container">$1,2,\cdots,(k-1)$</span> must be taken by someone in either A or C. If <span class="math-container">$k$</span> = 1, we can denote this event as having probability <span class="math-container">$= 1.$</span> For <span class="math-container">$k > 1$</span> then the probability of this happening is <br>
<span class="math-container">$\displaystyle \frac{33!}{[33 - (k-1)]!} \times \frac{[43 - (k-1)]!}{43!} = \frac{33!}{(34 - k)!} \times \frac{(44 - k)!}{43!}.$</span></li>
<li>Position <span class="math-container">$k$</span> must be taken by someone in B, <strong>after</strong> all of the previous positions have been taken by someone in either A or C. The probability of this happening is <br>
<span class="math-container">$\displaystyle \frac{10}{44 - k}.$</span></li>
</ul>
<p>Therefore, the expected value of the position of the highest ranking member of B is</p>
<p><span class="math-container">$$\left\{1 \times \frac{10}{43}\right\} ~+~
\left\{ ~\sum_{k=2}^{34} \left[k \times
\frac{33!}{(34 - k)!} \times \frac{(44 - k)!}{43!} \times \frac{10}{44 - k}\right]
~\right\}.
$$</span></p>
<hr />
<p><span class="math-container">$\underline{\text{Problem 2:}}$</span></p>
<p>Since most of the groundwork has already been done in Problem 1, the easiest approach for Problem 2 is to consider it to be the <strong>mirror</strong> problem to Problem 1. That is, the math is all exactly the same, but for each event that would be multiplied by the scalar <span class="math-container">$k$</span>, you instead multiply it by the scalar <span class="math-container">$(44 - k)$</span>.</p>
<p>Therefore, there is no need to repeat the analysis from Problem 1. The expected value of the position of the lowest ranking member of B is</p>
<p><span class="math-container">$$\left\{43 \times \frac{10}{43}\right\} ~+~
\left\{ ~\sum_{k=2}^{34} \left[(44 - k) \times
\frac{33!}{(34 - k)!} \times \frac{(44 - k)!}{43!} \times \frac{10}{44 - k}\right]
~\right\}.
$$</span></p>
|
244,433 | <p>I have a list:</p>
<pre><code>data = {{2.*10^-9, 0.0025}, {4.*10^-9, 0.0025}, {6.*10^-9, 0.0025}, {8.*10^-9, 0.0025}, {1.*10^-8, 0.0025}, {7.*10^-9, 0.0023}, {3.*10^-9, 0.0025},...}
</code></pre>
<p>And I wanted to remove every third pair and get</p>
<pre><code> newdata = {{2.*10^-9, 0.0025}, {4.*10^-9, 0.0025}, {8.*10^-9, 0.0025}, {1.*10^-8, 0.0025}, {3.*10^-9, 0.0025},...}
</code></pre>
| Chris Degnen | 363 | <p>Using <code>Partition</code>, padded with <code>Nothing</code></p>
<pre><code>Flatten[Take[#, UpTo[2]] & /@ Partition[data, 3, 3, {1, 1}, Nothing], 1]
</code></pre>
|
1,400,352 | <p>Confusion with the eccentricity of ellipse. On <a href="https://en.wikipedia.org/wiki/Ellipse#Directrix" rel="nofollow noreferrer">wikipedia</a> I got the following in the directrix section of ellipse.</p>
<blockquote>
<p>Each focus F of the ellipse is associated with a line parallel to the minor axis called a directrix. Refer to the illustration on the right, in which the ellipse is centered at the origin. The distance from any point P on the ellipse to the focus F is a constant fraction of that point's perpendicular distance to the directrix, resulting in the equality e = PF/PD. The ratio of these two distances is the eccentricity of the ellipse. This property (which can be proved using the Dandelin spheres) can be taken as another definition of the ellipse.
Besides the well-known ratio e = f/a, where f is the distance from the center to the focus and a is the distance from the center to the farthest vertices (most sharply curved points of the ellipse), it is also true that e = a/d, where d is the distance from the center to the directrix.</p>
</blockquote>
<p>It is given that $e=\frac fa=\frac ad$</p>
<p><a href="https://i.stack.imgur.com/pPLFX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pPLFX.png" alt="enter image description here"></a></p>
<p>In my book it was only given that $e=f/a$ (in my book there is nothing given about directrix of an ellipse).</p>
<p><strong>My question</strong></p>
<p>Knowing that $e=f/a$ how can I get $e=a/d$ and $e=PF/PD$?</p>
| Harish Chandra Rajpoot | 210,295 | <p><strong>Hint</strong>: If the eccentricity $e$ & the major axis $2a$ of an ellipse are known then we have the following </p>
<ol>
<li>Distance of each focus from the center of ellipse
$$=\text{(semi-major axis)}\times \text{(eccentricity of ellipse)}=\color{red}{ae}$$</li>
<li>Distance of each directrix from the center of ellipse
$$=\frac{\text{semi-major axis} }{\text{eccentricity of ellipse}}=\color{red}{\frac ae}$$</li>
</ol>
|
1,400,352 | <p>Confusion with the eccentricity of ellipse. On <a href="https://en.wikipedia.org/wiki/Ellipse#Directrix" rel="nofollow noreferrer">wikipedia</a> I got the following in the directrix section of ellipse.</p>
<blockquote>
<p>Each focus F of the ellipse is associated with a line parallel to the minor axis called a directrix. Refer to the illustration on the right, in which the ellipse is centered at the origin. The distance from any point P on the ellipse to the focus F is a constant fraction of that point's perpendicular distance to the directrix, resulting in the equality e = PF/PD. The ratio of these two distances is the eccentricity of the ellipse. This property (which can be proved using the Dandelin spheres) can be taken as another definition of the ellipse.
Besides the well-known ratio e = f/a, where f is the distance from the center to the focus and a is the distance from the center to the farthest vertices (most sharply curved points of the ellipse), it is also true that e = a/d, where d is the distance from the center to the directrix.</p>
</blockquote>
<p>It is given that $e=\frac fa=\frac ad$</p>
<p><a href="https://i.stack.imgur.com/pPLFX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pPLFX.png" alt="enter image description here"></a></p>
<p>In my book it was only given that $e=f/a$ (in my book there is nothing given about directrix of an ellipse).</p>
<p><strong>My question</strong></p>
<p>Knowing that $e=f/a$ how can I get $e=a/d$ and $e=PF/PD$?</p>
| wltrup | 232,040 | <p>If I understood your question correctly, you're essentially asking how one can find the equation for the directrix if one only has the equation for an ellipse with a given eccentricity.</p>
<p>You start with the equation below
$$
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
\qquad(1)
$$
where $a$ and $b$ are positive real-valued constants.</p>
<p>If you then <em>define</em> two points on the $x$ axis, $F$ and $F'$, by their coordinates,
$$
F \equiv (\epsilon a, 0)
\qquad(2)
$$
$$
F' \equiv (-\epsilon a, 0)
\qquad(3)
$$
for some as-yet undetermined real value $\epsilon \ge 0$, and then <em>require</em> that the sum of the distances from $F$ to $P$ and from $P$ to $F'$ be equal to $2a$ - for any arbitrary point $P$ on the ellipse - you'll find out that that is only possible if you choose $\epsilon$ to satisfy
$$
\epsilon = \sqrt{1 - \frac{b^2}{a^2}}
\qquad(4)
$$</p>
<p>So far, the above has nothing to do with the directrix. It's just a way to construct the foci and find the eccentricity, starting from some equation and then requiring that the curve described by that equation must have the fundamental property of an ellipse (namely, that the sum of the distances from any point on the ellipse to the two foci is a constant).</p>
<p>Now, let's see how we can prove the directrix property. Imagine that there is a vertical line at $x=d$ for some as-yet-undetermined real value $d \ge 0$ and let's see if we can find a value of $d$ such that the directrix property is satisfied. The focus for positive $x$ is $F$ (see its coordinates above), and an arbitrary point $D$ in the would-be directrix has coordinates $D = (d, y)$. The directrix property mandates that
$$
\epsilon = \frac{\overline{PF}}{\overline{PD}}
\qquad(5)
$$</p>
<p>Now, let's look at the square of the distances involved, since that gets rid of the square roots:
$$
\overline{PF}^{\,2} = (x - \epsilon a)^2 + (y - 0)^2 = (x - \epsilon a)^2 + y^2
$$</p>
<p>But $P$ lies on the ellipse so $(x,y)$ satisfies $(1)$. Therefore, eliminating $x^2$ in favour of $y^2$, but leaving the $x$ term alone, we have
$$
\overline{PF}^{\,2} =
a^2\,(1 + \epsilon^2) + y^2\,(1 - \frac{a^2}{b^2}) - 2\epsilon ax
$$</p>
<p>How about $\overline{PD}^{\,2}$? Note that $D$ and $P$ have the same $y$ coordinate so
$$
\overline{PD}^{\,2} = (x - d)^2 + (0)^2 = (x - d\,)^2
$$</p>
<p>Here's the crux now. <em>Can we find a value of $d$ such that $(5)$ is true?</em>
Imposing
$$
\frac{\overline{PF}^{\,2}}{\overline{PD}^{\,2}} =
\frac{a^2\,(1 + \epsilon^2) + y^2\,(1 - \frac{a^2}{b^2}) - 2\epsilon ax}{(x - d\,)^2} = \epsilon^2
$$
we get
$$
a^2\,(1 + \epsilon^2) + y^2\,(1 - \frac{a^2}{b^2}) - 2\epsilon ax = \epsilon^2\,(x - d\,)^2
$$</p>
<p>Note that the complicated expression above is something like this:
$$
(\mbox{terms independent of $x$ and $y$}) + (\mbox{terms involving $y^2$}) - 2\epsilon ax + 2\epsilon^2 xd = 0
$$
because the $x^2$ terms can be removed using $(1)$. Since the above has to be true for all $x$, the only hope for $(5)$ to be possible is if we choose $d$ such that
$$
2\epsilon ax - 2\epsilon^2 xd = 0
\qquad\Rightarrow\qquad
d = \frac{a}{\epsilon}
$$</p>
<p>With this choice, it's then not hard to show that the terms not written above also vanish so, indeed,
$$
\epsilon = \frac{\overline{PF}}{\overline{PD}}
$$
is true, provided we make that choice for $d$.</p>
<blockquote>
<p>So, what's the conclusion? The conclusion is that an ellipse has <em>two equivalent properties</em>, namely, (A) the sum of the distances from the foci to any arbitrary point on the ellipse is a constant, and (B) the ratio of (the distance from an arbitrary point on the ellipse to one of the foci) to (the distance from that point to a fixed vertical line parallel to the semi-minor axis) is equal to the eccentricity of the ellipse.</p>
<p>And we also proved that $\epsilon = a/d$.</p>
<p>(Technically, I only proved that (A) above implies (B). We'd also have to prove that (B) implies (A) for the equivalence I referred to be proven. That's an exercise I leave for the reader...)</p>
</blockquote>
|
1,876,133 | <p>Let x be a object that is not a set</p>
<p>Let S be a set</p>
<p>Would the following statement:</p>
<p>x ⊆ S</p>
<p>evaluate to False, or considered not a well formed statement (as x is not even a set).</p>
| Ted Shifrin | 71,348 | <p><strong>HINT</strong>: Mean Value Theorem. (Consider $x\le x_0$ and $x\ge x_0$ separately.)</p>
|
1,876,133 | <p>Let x be a object that is not a set</p>
<p>Let S be a set</p>
<p>Would the following statement:</p>
<p>x ⊆ S</p>
<p>evaluate to False, or considered not a well formed statement (as x is not even a set).</p>
| Asinomás | 33,907 | <p>Prove the following with the mean value theorem, given that $x<y$:</p>
<ul>
<li>$f(x)\leq f(y)$</li>
<li>if $f(x)=f(y)$ then $x_0\in(x,y)$</li>
<li>$f(x)<f(\frac{x+y}{2})\leq f(y)$ or $f(x)\leq f(\frac{x+y}{2})<f(y)$</li>
</ul>
|
39,684 | <p>In order to have a good view of the whole mathematical landscape one might want to know a deep theorem from the main subjects (I think my view is too narrow so I want to extend it).</p>
<p>For example in <em>natural number theory</em> it is good to know quadratic reciprocity and in <em>linear algebra</em> it's good to know the Cayley-Hamilton theorem (to give two examples).</p>
<p>So, what is one (<strong>per post</strong>) deep and representative theorem of each subject that one can spend a couple of months or so to learn about? (In Combinatorics, Graph theory, Real Analysis, Logic, Differential Geometry, etc.)</p>
| Eric Naslund | 6,075 | <p>In Analytic Number Theory, the <a href="http://en.wikipedia.org/wiki/Prime_number_theorem" rel="nofollow">Prime Number Theorem</a> is a great result. </p>
<p>Riemann's 1859 paper, which outlined a possible approach to proving the PNT, is credited with motivating a large amount of the research done in Complex Analysis in the 19th century. </p>
|
39,684 | <p>In order to have a good view of the whole mathematical landscape one might want to know a deep theorem from the main subjects (I think my view is too narrow so I want to extend it).</p>
<p>For example in <em>natural number theory</em> it is good to know quadratic reciprocity and in <em>linear algebra</em> it's good to know the Cayley-Hamilton theorem (to give two examples).</p>
<p>So, what is one (<strong>per post</strong>) deep and representative theorem of each subject that one can spend a couple of months or so to learn about? (In Combinatorics, Graph theory, Real Analysis, Logic, Differential Geometry, etc.)</p>
| Community | -1 | <ul>
<li>In Group theory i would say "Structure theorem for finitely generated abelian groups"</li>
</ul>
|
145,303 | <p>Another question about the convergence notes by Dr. Pete Clark:</p>
<p><a href="http://alpha.math.uga.edu/%7Epete/convergence.pdf" rel="nofollow noreferrer">http://alpha.math.uga.edu/~pete/convergence.pdf</a></p>
<p>(I'm almost at the filters chapter! Getting very excited now!)</p>
<p>On page 15, Proposition 4.6 states that for the following three properties of a topological space <span class="math-container">$X$</span>,</p>
<p><span class="math-container">$(i)$</span> <span class="math-container">$X$</span> has a countable base.</p>
<p><span class="math-container">$(ii)$</span> <span class="math-container">$X$</span> is separable.</p>
<p><span class="math-container">$(iii)$</span> <span class="math-container">$X$</span> is Lindelof (every open cover admits a countable subcover).</p>
<p>we always have <span class="math-container">$(i)\Rightarrow (ii)$</span> and <span class="math-container">$(i)\Rightarrow (iii)$</span>.</p>
<p>Also, we if <span class="math-container">$X$</span> is metrizable, we have <span class="math-container">$(iii)\Rightarrow (i)$</span>, and <em>thus all three are equivalent</em>.</p>
<p>This last part confuses me. We establish all the implications claimed in the proof, but there seems to be a missing link in the claim that all three are equivalent: namely <span class="math-container">$(ii)\Rightarrow (iii)$</span>.</p>
| Carl | 18,702 | <p>Note that a separable metric space has a countable basis. Specifically, we take a countable dense subset $S$ and take the set of balls centered at $s$ with radius $1/n$ for each $n \in N$, $s \in S$. This can be checked to be a basis. So then $(ii) \Rightarrow (i)$ is proven, which is the missing link.</p>
|
3,583,778 | <p>This problem came to me when I was solving another binomial coefficient summation problem in this site.</p>
<p>I want to prove that <span class="math-container">$\sum_{i=0}^{p}{\sum_{j=0}^{q+i}{\sum_{k=0}^{r+j}{\binom{p}{i}\binom{q+i}{j}\binom{r+j}{k}}}}=4^{p}3^{q}2^{r}$</span></p>
| Rezha Adrian Tanuharja | 751,970 | <p><span class="math-container">$$
\begin{aligned}
(3+x)^{p}(2+x)^{q}(1+x)^{r}&=(1+(2+x))^{p}(2+x)^{q}(1+x)^{r}\\
&=\sum_{i=0}^{p}{\binom{p}{i}(2+x)^{i}(2+x)^{q}(1+x)^{r}}\\
&=\sum_{i=0}^{p}{\binom{p}{i}(2+x)^{q+i}(1+x)^{r}}\\
&=\sum_{i=0}^{p}{\binom{p}{i}(1+(1+x))^{q+i}(1+x)^{r}}\\
&=\sum_{i=0}^{p}{\sum_{j=0}^{q+i}{\binom{p}{i}\binom{q+i}{j}(1+x)^{j}(1+x)^{r}}}\\
&=\sum_{i=0}^{p}{\sum_{j=0}^{q+i}{\binom{p}{i}\binom{q+i}{j}(1+x)^{r+j}}}\\
&=\sum_{i=0}^{p}{\sum_{j=0}^{q+i}{\sum_{k=0}^{r+j}{\binom{p}{i}\binom{q+i}{j}\binom{r+j}{k}x^{k}}}}\\
\end{aligned}
$$</span></p>
<p>Now substitute <span class="math-container">$x=1$</span></p>
|
1,840,352 | <blockquote>
<p>For every $A \subset \mathbb R^3$ we define $\mathcal{P}(A)\subset \mathbb R^2$ by
$$ \mathcal{P}(A) := \{ (x,y) \mid \exists_{z \in \mathbb R}:(x,y,z) \in A \} \,. $$
Prove or disprove that $A$ closed $\implies$ $\mathcal{P}(A)$ closed and $A$ compact $\implies$ $\mathcal{P}(A)$ compact.</p>
</blockquote>
<p>This $\mathcal{P}(A)$ seems to me the projection on the $xy$-plane, so intuitively both make sense to me.</p>
<p><strong>My try</strong></p>
<p>For the first one, to prove something is closed seems easiest to do with sequences, so for all sequences $(x^{(n)})$ in $A$ with $x^{(n)} \to x$ we have that $x \in A$. But I do not know how to progress further in a useful direction. For the second one I don't even know where to start.</p>
<p><strong>Update</strong></p>
<p>For the first one I tried to argue that for any sequence going to $x \in A$, a projection of that sequence on the $xy$-plane will also go to the projection of $x$, so $x \in \mathcal{P}(A)$, so hence $\mathcal{P}(A)$ must be closed. </p>
| Shashi | 349,501 | <p><strong>Hint:</strong></p>
<p>The first one is not true.</p>
<p>For the second one try this:</p>
<p>I assume you know Heine Borel's theorem about compact sets.</p>
<p>Heine Borel says $A$ is bounded. So for all components of $(x_1,x_2,x_3)\in A $ you can find an $M>0$: $|x_i|<M$. That means $|x_i|^2<M^2$. Now we have:
\begin{align} ||(x_1,x_2)||^2=|x_1|^2+|x_2|^2<2M^2\end{align}</p>
<p>Can you finish now?</p>
<p>Heine Borel says $A$ is also closed. Choose a convergent sequence $x^n \rightarrow x$ in $\mathcal{P}(A)$. There is a sequence $x^n_3$ with $(x_1^n,x_2^n,x_3^n)\in A$. You know $x_3^n$ is boundend. By Bolzano Weierstrass theorem for sequence we have now there is $x^{n_k}_3 \rightarrow x_3$. What can you say about the original sequence in $\mathcal{P}(A)$ now?</p>
<p>Combine them both and by Heine Borel you have $\mathcal{P}(A)$ is compact.</p>
|
1,485,310 | <blockquote>
<p>Write a formula/formulae for the following sequence:</p>
<p>b). 1,3,6,10,15,...</p>
</blockquote>
<p>I am not getting any pattern here, from which to derive a formula.
This sequence does not look like the examples I could solve: like</p>
<blockquote>
<p>a) 1,0,1,0,1...</p>
</blockquote>
<p>where I got that <span class="math-container">$S_n =1 $</span> (for <span class="math-container">$n=1,3,5,7,...$</span>) and <span class="math-container">$S_n=0$</span> (for <span class="math-container">$n=2,4,6,8,...$</span>)</p>
<p>or</p>
<blockquote>
<p>c.) 1,1,1,2,1,3,1,4,1,...</p>
</blockquote>
<p>where <span class="math-container">$S_n=1$</span> for <span class="math-container">$n=1,2,3,5,7,...$</span> and <span class="math-container">$S_n= n/2$</span> for <span class="math-container">$n=4,6,8,.. $</span></p>
| PTDS | 277,299 | <p>Mike Pierce has already given the answer. </p>
<p>Still let me mention that the $n$-th term of the sequence is the sum of the first $n$ natural numbers. </p>
<p>Hence $$s_n = \frac{n(n+1)}{2}$$</p>
<p>It is also possible to derive the above by solving the recurrence relation given by Amey Deshpande.</p>
|
4,345,671 | <p>I have a series of cubic polynomials that are being used to create a trajectory. Where some constraints can be applied to each polynomial, such that these 4 parameters are satisfied.
-Initial Position
-final Position
-Initial Velocity
-final Velocity</p>
<p>The polynomials are pieced together such that the ends of one polynomial are identical to the beginnings of the next to preserve continuity.</p>
<p>I instead want to represent these polynomials as cubic Bézier curves.</p>
<p><strong>How would I find the x,y position of each control point for the cubic Bézier curves, such that it matches the curvature of the cubic polynomial.</strong></p>
<p>Here is what I have so far, made in desmos.</p>
<p><a href="https://www.desmos.com/calculator/agsywptfno" rel="nofollow noreferrer">https://www.desmos.com/calculator/agsywptfno</a></p>
<p>Currently the bezier curve is defined as a binomial, with a polynomial for X and or Y
e.g. Bezier = (X(t), Y(t))</p>
| Andrew D. Hwang | 86,418 | <p>As noted in the comments and Karl's (+1) answer, the issue is in flattening the "<a href="https://en.wikipedia.org/wiki/Gore_(fabrics)" rel="nofollow noreferrer">gores</a>," which does not preserve area.</p>
<p>One correct reckoning is instead to use a quasi-triangle with two right angles at the base, angle <span class="math-container">$d\theta$</span> at the apex, and width <span class="math-container">$r\sin\varphi\, r\, d\theta$</span> at height <span class="math-container">$\varphi$</span> (below, left), as noted by md2perpe and bubba in the comments. The area of the flat region is correct,
<span class="math-container">$$
\int_{0}^{\pi/2} r\sin\varphi\, d\varphi\, (r\, d\theta) = r^{2}\, d\theta,
$$</span>
though this region <a href="https://en.wikipedia.org/wiki/Theorema_Egregium" rel="nofollow noreferrer">does not wrap without distortion</a> (i.e., not <a href="https://en.wikipedia.org/wiki/Differential_geometry_of_surfaces#Isometries" rel="nofollow noreferrer">isometrically</a>) onto a gore, but only in an area-preserving way.</p>
<p>A related correct accounting (below, right, which shows a unit sphere; for radius <span class="math-container">$r$</span>, multiply all lengths by <span class="math-container">$r$</span>) slices the quasi-triangle by latitudes. The portion of a sphere between longitude <span class="math-container">$\theta$</span> and <span class="math-container">$\theta + d\theta$</span>, and between colatitude <span class="math-container">$\varphi$</span> and <span class="math-container">$\varphi + d\varphi$</span>, is an infinitesimal trapezoid. Project <em>axially</em> (away from the axis through the poles) onto a circumscribed cylinder. Up to an error infinitesimal compared to <span class="math-container">$d\theta\, d\varphi$</span>, the trapezoid is a rectangle with width <span class="math-container">$\sin\varphi\, d\theta$</span> and height (measured along the sphere) <span class="math-container">$d\varphi$</span>. The image of this quadrilateral under axial projection is a curved (but <a href="https://en.wikipedia.org/wiki/Developable_surface" rel="nofollow noreferrer">developable</a>) rectangle of width <span class="math-container">$d\theta$</span> and height (along the cylinder) <span class="math-container">$\sin\varphi\, d\varphi$</span>. Consequently, <a href="https://mathworld.wolfram.com/ArchimedesHat-BoxTheorem.html" rel="nofollow noreferrer"><em>axial projection from a sphere to a circumscribed cylinder is area-preserving</em></a>. (!!) Particularly, the area of a sphere is the area of a circumscribed cylinder, <span class="math-container">$(2\pi r)(2r) = 4\pi r^{2}$</span>.</p>
<p><a href="https://i.stack.imgur.com/jHAZQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jHAZQ.png" alt="Triangular and quasi-triangular gores for a sphere" /></a><span class="math-container">$\rule{1in}{0pt}$</span>
<a href="https://i.stack.imgur.com/MqmnD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MqmnD.png" alt="Axial projection from a sphere to a circumscribed cylinder preserves area" /></a></p>
|
764,905 | <p>Calculate $$\int_{D}(x-2y)^2\sin(x+2y)\,dx\,dy$$ where $D$ is a triangle with vertices in $(0,0), (2\pi,0),(0,\pi)$.</p>
<p>I've tried using the substitution $g(u,v)=(2\pi u, \pi v)$ to make it a BIT simpler but honestly, it doesn't help much.</p>
<p>What are the patterns I need to look for in these problems so I can get an integral that's viable to calculate? Everything I try always leads to integrating a huge function and that's extremely error prone.</p>
<p>I mean, I can obviously see the $x-2y$ and $x+2y$ but I don't know how to use it to my advantage. Also, when I do my substitution, I get $\sin(2\pi(u+v))$ and in the triangle I have, $u+v$ goes from 0 to 1, so the $\sin$ goes full circle. Again, no idea if that helps me.</p>
<p>Any help appreciated.</p>
| Santosh Linkha | 2,199 | <p>Here is how I calculate those types of integral on plane.
$$\int_D 1\,dy dx = \mathrm{Area \;of \; triangle}$$</p>
<p>Now, area of triangle is just area under the straight line.
$$\int_0^{2\pi}\left ( \pi - \frac x2\right ) dx$$
Now I change single integral into double integral,
$$\int_0^{2\pi}\left(\int_0^{ \pi - \frac x2 } \,dy\right)\, dx$$
Finally I substitute that thing inside it and evaluate it.
$$\int_0^{2\pi}\int_0^{ \pi - \frac x2 } (x-2y)^2\sin(x+2y)dydx$$</p>
<p>Somehow <a href="http://www.wolframalpha.com/input/?i=Integrate%5B+%28x%E2%88%922y%29%5E2sin%28x%2B2y%29%2C+%7Bx%2C+0%2C+2pi%7D%2C+%7By%2C+0%2C+pi-x%2F2%7D%5D" rel="nofollow">wolf</a> calculates it to be $2 \pi -\frac{4 \pi ^3}{3}$.</p>
|
90,656 | <p>In the introduction to 'A convenient setting for Global Analysis', Michor & Kriegl make this claim: "The study of Banach manifolds per se is not very interesting, since they turnout to be open subsets of the modeling space for many modeling spaces."</p>
<p>But finite-dimensional manifolds are found to be interesting even though they can be embedded in some Euclidean space (of larger dimension). (Actually this seems to me, to make the above claim intuitively plausible, so that claim should be no more than we should expect).</p>
<p>But they do go on to say that "Banach manifolds are not suitable for many questions of Global Analysis, as ... a Banach Lie group acting effectively on a finite dimensional smooth manifold it must be finite dimensional itself.", which does seem a rather strong limitation. </p>
| Liviu Nicolaescu | 20,302 | <p>I am of two minds on this topic. It is much easier to work on Banach manifolds because the implicit function theorem on such spaces has a simple formulation. On the other hand, as the examples of gauge theory or the theory of pseudo-holomorphic curves show, in these contexts one works not with one Banach manifold, but with several, determined by stronger and stronger Sobolev norms. One important part of the game is to conclude that objects with a priori weaker Sobolev regularity are in fact smooth. This feels very much like we are implicitly working on a Frechet manifold.</p>
<p>One draw back of Banach spaces is that they do not have many smooth functions on them, and the notion of real analycity on such spaces is problematic. Let's take the example of Seiberg-Witten equations. These are quadratic equations in its variables, so intuitively they ought to be real analytic, though I do not know how to formulate this rigorously in a Sobolev context.</p>
<p>Why do I care about real analycity? In the real analytic context one can formulate an intersection theory involving not necessarily smooth objects. For example, the point $0\in\mathbb{R}$ is a solution of the quadratic equation $x^2=0$. It is a degenerate zero, and from the point of view of intersection theory it has multiplicity $0$. My hope is that this real analytic point of view would allow one to deal with mildly degenerate solutions of the the Seiberg-Witten equations, and assign multiplicities to such solutions. </p>
|
679,544 | <p>How to prove this for positive real numbers?
$$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$$</p>
<p>I tried AM-GM, CS inequality but all failed.</p>
| r9m | 129,017 | <p>Using Cauchy-Schwarz Inequality twice:</p>
<p>$a^4 + b^4 +c^4 \geq a^2b^2 +b^2c^2 +c^2a^2 \geq ab^2c +ba^2c +ac^2b = abc(a+b+c)$</p>
|
679,544 | <p>How to prove this for positive real numbers?
$$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$$</p>
<p>I tried AM-GM, CS inequality but all failed.</p>
| zhangwfjh | 115,670 | <p>I have come up with an answer with myself. Using CS inequality
$$(a^4+b^4+c^4)(1+1+1)\geq(a^2+b^2+c^2)^2$$
$$(a^2+b^2+c^2)(1+1+1)\geq(a+b+c)^2$$
Hence we have
$$a^4+b^4+c^4\geq\frac{(a+b+c)^4}{27}=(a+b+c)\left(\frac{a+b+c}{3}\right)^3\geq abc(a+b+c)$$</p>
|
525,611 | <p>Let $f: [0,1] \to \mathbb{R}$ be defined by letting $f = 0 $ on $\mathcal{C}$, the Cantor set and $f(x) = k $ for every $x$ in each interval of lenght $\frac{1}{3^k}$ which has been removed from $[0,1]$. We want to calculate $\int\limits_{[0,1]} f dm $.</p>
<p>How CAn I express $f$ as a simple function?</p>
<p>So, if $$f(x) = \sum_{k}^{\infty} \sum_{j = 1}^{2^{k-1} } k 1_{A_{kj}}$$, then </p>
<p>$$\int\limits_{[0,1]} f dm = \sum_{k}^{\infty} \sum_{j = 1}^{2^{k-1} } k m(A_{kj}) = \sum_{k}^{\infty} \sum_{j = 1}^{2^{k-1} } \frac{k}{3^k} = \sum_{k}^{\infty} \frac{2^{k-1}k}{3^k} = 3 $$</p>
| Dietrich Burde | 83,966 | <p>Euclid's formula generates all primitive triples, and this can be modified to give all triples: set $a = k\cdot(m^2 - n^2) ,\ \, b = k\cdot(2mn) ,\ \, c = k\cdot(m^2 + n^2)$, where $m, n$, and $k$ are positive integers with $m > n, m − n$ odd, and with $m$ and $n$ coprime. Then $a^2+b^2=c^2$, and $c$ is the given hypotenuse.</p>
|
525,611 | <p>Let $f: [0,1] \to \mathbb{R}$ be defined by letting $f = 0 $ on $\mathcal{C}$, the Cantor set and $f(x) = k $ for every $x$ in each interval of lenght $\frac{1}{3^k}$ which has been removed from $[0,1]$. We want to calculate $\int\limits_{[0,1]} f dm $.</p>
<p>How CAn I express $f$ as a simple function?</p>
<p>So, if $$f(x) = \sum_{k}^{\infty} \sum_{j = 1}^{2^{k-1} } k 1_{A_{kj}}$$, then </p>
<p>$$\int\limits_{[0,1]} f dm = \sum_{k}^{\infty} \sum_{j = 1}^{2^{k-1} } k m(A_{kj}) = \sum_{k}^{\infty} \sum_{j = 1}^{2^{k-1} } \frac{k}{3^k} = \sum_{k}^{\infty} \frac{2^{k-1}k}{3^k} = 3 $$</p>
| poetasis | 546,655 | <p>You can use equations I created for a paper I am writing. To find a triplet with a matching <span class="math-container">$sideC$</span> for any odd number (<span class="math-container">$C_1$</span>), set <span class="math-container">$$k_c=\frac{-(2n-1)+\sqrt{2C_1-(2n-1)^2}}{2}$$</span></p>
<p>Now try values of <span class="math-container">$n$</span> from <span class="math-container">$1$</span> up to <span class="math-container">$(m-1)$</span> where <span class="math-container">$4m^2+1>C_1$</span>. Then, for any value of <span class="math-container">$n$</span> that yields a positive integer for <span class="math-container">$k$</span>, you have a triplet with a matching <span class="math-container">$sideC$</span>. If you do not find such an integer for <span class="math-container">$k$</span> using values of <span class="math-container">$n$</span> between <span class="math-container">$1$</span> and <span class="math-container">$(m-1)$</span>, then there is no matching side.</p>
<p>If you do find a positive integer for <span class="math-container">$k$</span> for some value of <span class="math-container">$n$</span>, then you can find the triplet sides A,B,C using the following functions:</p>
<p><span class="math-container">$$A(n,k)=(2n-1)^2+2(2n-1)k$$</span>
<span class="math-container">$$B(n,k)=2(2n-1)k+2k^2$$</span>
<span class="math-container">$$C(n,k)=(2n-1)^2+2(2n-1)k+2k^2$$</span></p>
<p>Examples:
If you want to find a triplet where <span class="math-container">$C_1=125$</span>, you will find for <span class="math-container">$k(n)$</span> that <span class="math-container">$k(3)=5$</span> and <span class="math-container">$k(5)=2$</span> and that means that plugging <span class="math-container">$(3,5)$</span> and <span class="math-container">$(5,2)$</span> into the A.B.C equations will yield triplets with <span class="math-container">$sideC=125$</span>, i.e. <span class="math-container">$75,100,125$</span> and <span class="math-container">$117,44,125$</span>, respectively. Here are more examples I found in the spreadsheet I used for my study of matching values of <span class="math-container">$C$</span> where <span class="math-container">$f(n,k)$</span> is the triplet generated using the functions above:
<span class="math-container">$$f(1,11)=(23,264,265), f(7,3)=(247,96,265) $$</span>
<span class="math-container">$$f(1,13)=(21,220,221), f(5,5)=(171,140,221) $$</span>
<span class="math-container">$$f(1,21)=(43,924,925), f(7,14)=(533,756,925) $$</span>
<span class="math-container">$$f(1,28)=(57,1624,1625), f(8,20)=(825,1400,1625)$$</span>
<span class="math-container">$$f(2,19)=(123,836,845), f(7,13)=(507,676,845) $$</span>
<span class="math-container">$$f(2,26)=(165,1508,1517), f(8,19)=(795,1292,1517)$$</span>
<span class="math-container">$$f(3,16=(185,672,697), F(7,11)=(455,528,697) $$</span>
<span class="math-container">$$f(3,23)=(255,1288,1313), F(8,17)=(735,1088,1313)$$</span>
<span class="math-container">$$f(4,19)=(315,988,1037), f(8,14)=(645,812,1037)$$</span>
<span class="math-container">$$f(5,14)=(333,644,725), f(8,10)=(525,500,725)$$</span>
If you go into the higher sets you will find more and more matching triplets in the lower sets
<span class="math-container">$$f(1,23)=(47,1105,1105), f(10,12)=(817,744,1105) , f(15,4)=(1073,264,1105)$$</span>
<span class="math-container">$$f(1,36)=(73,2664,2665), f(9,27)=(1207,2376,2665), f(15,19)=(1943,1824,2665)$$</span>
<span class="math-container">$$f(7,23)=(767,1656,1825), f(13,15)=(1375,1200,1825), f(15,12)=(1537,904,1825)$$</span></p>
|
3,200,330 | <p>Suppose <span class="math-container">$\Phi: A\to A$</span> is a transformation of the set <span class="math-container">$A$</span>. I want to understand what it means for a subset <span class="math-container">$B\subseteq A$</span> to be invariant under <span class="math-container">$\Phi$</span>. </p>
<p><a href="https://press.princeton.edu/titles/3098.html" rel="nofollow noreferrer">Halmos</a> states that this means <span class="math-container">$\forall b\in B (\Phi(b)\in B)$</span>. Later on, in the same book, he characterizes an invariant subset <span class="math-container">$B$</span> as one satisfying <span class="math-container">$\Phi(B)\subset B$</span>. I want to understand if these two are equivalent.</p>
<p>It seems to me that the statement <span class="math-container">$\forall b\in B (\Phi(b)\in B)$</span> is equivalent to <span class="math-container">$B\subset \Phi^{-1}(B)$</span>, (<span class="math-container">$\Phi^{-1}$</span> here denotes the inverse image, which is always defined; not the inverse of the function) and not <span class="math-container">$\Phi(B)\subset B$</span>. So my question is this: </p>
<blockquote>
<p>What does it really mean for a subset <span class="math-container">$B$</span> to be invariant under <span class="math-container">$\Phi$</span>? is it <span class="math-container">$B\subset \Phi^{-1}(B)$</span> or <span class="math-container">$\Phi(B)\subset B$</span>; or are the above characterizations equivalent for an arbitrary <span class="math-container">$\Phi$</span>?</p>
</blockquote>
<p>A tiny playing around with the definitions tells me that <span class="math-container">$A\subset B \Rightarrow \Phi(A)\subset \Phi(B)$</span>, but in general the arrow doesn't go the other way around, i.e., <span class="math-container">$\Phi(A)\subset \Phi(B)$</span> does not, in general, imply that <span class="math-container">$A\subset B$</span>. So the above definitions of invariant subset are not equivalent.</p>
| José Carlos Santos | 446,262 | <p>Yes, they are equivalent. Asserting that <span class="math-container">$A\subset B$</span> is equivalent to asserting that <span class="math-container">$(\forall a\in A):a\in B$</span>. And asserting that <span class="math-container">$\Phi(B)\subset B$</span>, in particular, is equivalent to <span class="math-container">$(\forall b\in B):\Phi(b)\in B$</span>.</p>
|
2,315,647 | <p>Compute the gravitational attraction on a unit mass at the origin due to the mass (of constant density) occupying the volume inside the sphere $r = 2a$ and above the plane $z=a$. Use spherical coordinates.</p>
<p>So I know the function should be
$$(G/r^2) dM$$
What are the limits of integration? What should the integral look like?</p>
| Rafa Budría | 362,604 | <p>$0\leq\theta\leq\arccos(a/2a)$ or $0\leq\theta\leq\pi/3$; $0\leq\phi\leq 2\pi$ and $a/\cos\theta\leq r\leq 2a$</p>
<p>$$F=\int_0^{\pi/3}\int_0^{2\pi}\int_{a/\cos\theta}^{2a}\dfrac{\hat r\rho}{r^2}r^2\sin\theta d\phi d\theta dr$$</p>
<p>With $\hat r=\cos\phi\sin\theta\hat x+\sin\phi\sin\theta\hat y+\cos\theta\hat z$</p>
<p>$$F=\rho\int_0^{\pi/3}\int_0^{2\pi}\int_{a/\cos\theta}^{2a}(\cos\phi\sin\theta\hat x+\sin\phi\sin\theta\hat y+\cos\theta\hat z)\sin\theta d\phi d\theta dr$$</p>
<p>The first two terms go to zero as we are integrating $\sin\phi$ and $\cos\phi$ over a complete period.</p>
<p>$$F=2\pi\rho\hat z\int_0^{\pi/3}\left(2a-a/\cos\theta\right)\cos\theta\sin\theta d\theta=\dfrac{1}{2}\pi a\rho\hat z$$</p>
<p>$$V=\int_0^{\pi/3}\int_0^{2\pi}\int_{a/\cos\theta}^{2a}r^2\sin\theta d\phi d\theta dr=\dfrac{5}{3}\pi a^3$$</p>
<p>$$F=\dfrac{3}{10}\dfrac{M}{a^2}\hat z$$</p>
<p>I let $G=1$, t's not an unusual way to procceed, but it had to be said...</p>
|
11,519 | <p>Hello,</p>
<p>If $f:\mathbb{R} \to \mathbb{R}$ a differentiable function, it is very easy to find its Lipschitz constant. Is there any way to extend this to functions $f: \mathbb{R} \to \mathbb{R}^n$ (or similar)?</p>
| Johannes Hahn | 3,041 | <p>In fact a statement similar to what was described by Pete Clark is true for all normed vector spaces:
Let $X$ and $Y$ be normed vector spaces. A (total) differentiable function $f:X\to Y$ is Lipschitz iff its derivative is bounded. Every upper bound for the differential is a Lipschitz constant.</p>
<p>One direction follows from the mean value theorem:
$\|f(x)-f(y)\|\leq \|Df(\xi)\|\cdot\|x-y\|$ for some $\xi$ on the straight line from $x$ to $y$.</p>
<p>The other follows immediately from
$Df(x)=\lim_{h\to 0}\frac{f(x)-f(x+h)}{\|h\|}$</p>
|
2,208,113 | <p>Let $x$ and $y \in \mathbb{R}^{n}$ be non-zero column vectors, from the matrix $A=xy^{T}$, where $y^{T}$ is the transpose of $y$. Then the rank of $A$ is ?</p>
<hr>
<p>I am getting $1$, but need confirmation .</p>
| dxiv | 291,201 | <p>Hint: $\;x,y,z \ge \frac{1}{4}$ for the square roots to be defined. Assume WLOG that $x \ge y \ge z\,$, then $\sqrt{4z-1}=x+y \ge z+z = 2z \implies 4z-1 \ge 4z^2\iff (2z-1)^2 \le 0 \,$.</p>
|
752,045 | <p>Write the area $D$ as the union of regions. Then, calculate $$\int\int_Rxy\textrm{d}A.$$</p>
<p>First of all I do not get a lot of parameters because they are not defined explicitly (like what is $A$? what is $R$?).</p>
<p>Here is what I did for the first question:</p>
<p>The area $D$ can be written as:</p>
<p>$$D=A_1\cup A_2\cup A_3\cup A_4\cup A_5.$$</p>
<p>Where: </p>
<p>$$A_1=\{(x, y)\in\mathbb{R}^2: x\geq-1\}.$$
$$A_2=\{(x, y)\in\mathbb{R}^2: y\geq-1\}.$$
$$A_3=\{(x, y)\in\mathbb{R}^2: x\leq1\}.$$
$$A_4=\{(x, y)\in\mathbb{R}^2: x\leq y^2\}.$$
$$A_5=\{(x, y)\in\mathbb{R}^2: y\leq1+x^2\}.$$</p>
<p>First, for me I see that $D$ is the intersection of these regions and not the union. Am I wrong?</p>
<p><img src="https://i.stack.imgur.com/frEJp.jpg" alt="enter image description here"></p>
<p>P.S. This is a homework.</p>
| fgp | 42,986 | <p>The easiest way to do this is to write $A$ as $$
A = Q \setminus \bigl(
\underbrace{\{(x,y) \in Q \mid y > 1+x^2\}}_{:=B_1} \cup
\underbrace{\{(x,y) \in Q \mid x > y^2\}}_{:=B_2}
\big)
$$
where $Q = [-1,1]\times[-1,2]$ (i.e. a rectangle), $B_1$ is the missing part at the top and $B_2$ the missing part on the right. Note that, per your picture, $B_1$ and $B_2$ are disjoint.</p>
<p>You can further use that the area of $B_1$ is the same as the area under the curve $f(x) = 1 - x^2$ (why?) and that the area of $B_2$ is twice the area under the curve $g(x) = \sqrt{x}$ (again, why?).</p>
|
752,045 | <p>Write the area $D$ as the union of regions. Then, calculate $$\int\int_Rxy\textrm{d}A.$$</p>
<p>First of all I do not get a lot of parameters because they are not defined explicitly (like what is $A$? what is $R$?).</p>
<p>Here is what I did for the first question:</p>
<p>The area $D$ can be written as:</p>
<p>$$D=A_1\cup A_2\cup A_3\cup A_4\cup A_5.$$</p>
<p>Where: </p>
<p>$$A_1=\{(x, y)\in\mathbb{R}^2: x\geq-1\}.$$
$$A_2=\{(x, y)\in\mathbb{R}^2: y\geq-1\}.$$
$$A_3=\{(x, y)\in\mathbb{R}^2: x\leq1\}.$$
$$A_4=\{(x, y)\in\mathbb{R}^2: x\leq y^2\}.$$
$$A_5=\{(x, y)\in\mathbb{R}^2: y\leq1+x^2\}.$$</p>
<p>First, for me I see that $D$ is the intersection of these regions and not the union. Am I wrong?</p>
<p><img src="https://i.stack.imgur.com/frEJp.jpg" alt="enter image description here"></p>
<p>P.S. This is a homework.</p>
| robjohn | 13,854 | <p><strong>Hint:</strong> the area inside each of the parabolic indentations is
$$
\int_{-1}^1(1-x^2)\,\mathrm{d}x
$$</p>
|
1,556,298 | <p>If we have $p\implies q$, then the only case the logical value of this implication is false is when $p$ is true, but $q$ false.</p>
<p>So suppose I have a broken soda machine - it will never give me any can of coke, no matter if I insert some coins in it or not.</p>
<p>Let $p$ be 'I insert a coin', and $q$ - 'I get a can of coke'.</p>
<p>So even though the logical value of $p \implies q$ is true (when $p$ and $q$ are false), it doesn't mean the implication itself is true, right? As I said, $p \implies q$ has a logical value $1$, but implication is true when it matches the truth table of implication. And in this case, it won't, because $p \implies q$ is <strong>false</strong> for true $p$ (the machine doesn't work).</p>
<p>That's why I think it's not right to say the implication is true based on only one row of the truth table. Does it make sense?</p>
| kviiri | 187,461 | <p>Implication is just like any other logical connective, and in fact, $a \to b$ can also be written using negation and disjunction as $\neg a \lor b$. Just as with logical disjunction and conjunction, it is very possible to know the value of the entire operation by observing just one operand - if $a = F$ we know $a \to b$ for all $b$, just as we'd know $a = T$ would result in $(a \lor b)$ being true for all $a$.</p>
<p>In your example, the implication is true if and only if the machine gives a drink (at least) when a coin is inserted. It can also spew out drinks freely at other times too ($p = F,\ q = T$) and the implication would remain true. If the machine is broken in a manner that $p = T,\ q = F$, then the implication is false, just like $(p \land q)$ would be false for the same values.</p>
|
166,925 | <p>I have a function <code>u[y]</code> and I want to find the limit of integration that integration is equal zero.</p>
<pre><code>Λ = -30;
u[η_] := (2*η - 2*η^3 + η^4) + Λ/6*(η - 3*η^2 + 3*η^3 - η^4);
θ = Integrate[u[η]*(1 - u[η]), {, 0, 1}] // N;
δ = 1/θ;
u[y_] := Piecewise[{{1,y > δ}}, (2*y/δ - 2*(y/δ)^3 + (y/δ)^4) + Λ/6*
((y/δ) - 3*(y/δ)^2 + 3*(y/δ)^3 - (y/δ)^4)];
FindRoot[Integrate[u[y], {y, 0, yd}] , {yd, 5}]
</code></pre>
<p>I have the following error:
"Unable to prove that integration limits {0,yd} are real. Adding assumptions may help."</p>
| chuy | 237 | <p>You can do something like this (assuming of course that the <code>InterpolatingFunction</code> <code>f1</code> and <code>f2</code> are generating using the same mesh):</p>
<pre><code>f12 = ElementMeshInterpolation[{mesh}, f1["ValuesOnGrid"] + f2["ValuesOnGrid"]]
</code></pre>
<p>Where <code>ElementMeshInterpolation</code> is from the </p>
<pre><code>NDSolve`FEM`
</code></pre>
<p>package.</p>
<pre><code>f1[1, 1] + f2[1, 1]
(* -1.84159 *)
f12[1, 1]
(* -1.84159 *)
Do[f1[0, 0], 100000] // AbsoluteTiming
(* {0.223145, Null} *)
Do[f12[0, 0], 100000] // AbsoluteTiming
(* {0.213769, Null} *)
</code></pre>
|
887,200 | <p>So I have the permutations:
$$\pi=\left( \begin{array}{ccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
2 & 3 & 7 & 1 & 6 & 5 & 4 & 9 & 8
\end{array} \right)$$
$$\sigma=\left( \begin{array}{ccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
9 & 5 & 6 & 8 & 7 & 1 & 2 & 4 & 3
\end{array} \right)$$</p>
<p>I found $\pi\sigma$ and $\sigma\pi$ to be</p>
<p>$$\pi\sigma=\left( \begin{array}{ccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
5 & 6 & 2 & 9 & 1 & 7 & 8 & 3 & 4
\end{array} \right)$$</p>
<p>$$\sigma\pi=\left( \begin{array}{ccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
8 & 6 & 5 & 9 & 4 & 2 & 3 & 1 & 7
\end{array} \right)$$.</p>
<p>So my question is: did I do these correctly or did I mix them up (i.e, the permutation matrix I have for $\pi\sigma$ is actually $\sigma\pi$)?</p>
| evinda | 75,843 | <p>You mixed them up.If you look at the permutation matrix $\pi \sigma $ you have to look ,firstly at $\sigma$ and then at $\pi$.</p>
<p>For example,for the first column of $\pi \sigma$,it is:</p>
<p>$$1 \to 9, 9 \to 8$$</p>
|
539,448 | <p>$a,b,c,d,e>o$. Show that</p>
<p>$$ a^{b+c+d+e}+ b^{c+d+e+a}+c^{ d+e+a+b}+ d^{e+a+b+c}+e^{a+b+c+d}>1$$ </p>
| KeithS | 23,565 | <p>Consider cases in which $a=b=c=d=e$. In such cases, the inequality (as I think you wanted to write it) simplifies to $5x^{4x} > 1$. To figure out if this is true, we need to consider the left side as a function of $x$, and find the vertex of this function. To do that, you need to take the derivative, $f'(x)$, and figure out where the derivative equals zero (and show that this vertex is a local minimum by demonstrating that $f'(x) < f(x\pm\delta)$ for small deviations $\delta$ from $x$), which makes this a calculus problem. Hint: for $y=x^x$, try taking the natural log of both sides first: $\ln(y) = \ln(x^x) = x \ln(x)$. Now, what's the derivative of a log?</p>
<p>I simply cheated and plugged it into a graphing calculator, and found that the vertex occurs at around $f(.36)=1.15$, so the inequality holds. For an exact answer, I leave the calculation of the derivative to you.</p>
<p>As per Calvin's comment, for cases where they are not all equal, any $x,y>1$ will mean that $x^y>1$, and any $x,y<1$ will mean $x,y<x^y<1$. The only way you can raise a base to a power and get a number smaller than either, is if $x<1<y$ or $y<1<x$, so in this inequality, you need five numbers < 1 but for which any four sum to > 1, thus making the result of each term smaller than each base and giving you the greatest chance for success.</p>
<p>You can show with a few examples that the higher the base, or the lesser the exponent, the larger the result, and vice-versa. Since every number will be a base and exponent at least once each, trying to minimize the base or maximize the exponent of any one term will backfire when the values you are manipulating swap places (for instance, $0.1^{10}=.0000000001$, but $10^{0.1}=1.2589$). </p>
<p>The cases where all numbers are equal, eliminating any delta between terms, is therefore the ideal, and as I've shown above, there's no single value that would make the inequality false.</p>
|
2,536,553 | <p>I know that is possible to apply the spectral decomposition (diagonalization) to a matrix when the sum of the dimensions of its eigenspaces is equal to the size of the matrix.</p>
<p>The spectral decomposition is:</p>
<p>$$
F=P\Lambda P^{-1}
$$</p>
<p>where $\Lambda$ is the diagonal matrix of eigenvalues and $P$ is the matrix of eigenvectors. </p>
<p>I have the following matrix:</p>
<p>$$
F=\begin{pmatrix}\phi_{1} & \phi_{2} & \phi_{3} & \cdots & \phi_{p-1} & \phi_{p}\\
1 & 0 & 0 & \cdots & 0 & 0\\
0 & 1 & 0 & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & 0 & 0\\
0 & 0 & 0 & \cdots & 1 & 0
\end{pmatrix}
$$</p>
<p>where $\phi_{p}\neq0$ and all $\phi_{i}$ are real valued. How can I be sure that its possible to apply the spectral decomposition to $F$?</p>
| Nightgap | 506,645 | <p>According to JayTuma you get (w.l.o.g. $i=1$):</p>
<p>$q_1\ldots q_m=p_1\ldots p_n=u_1q_1p_2\ldots p_n$</p>
<p>so $q_2\ldots q_m=u_1p_2\ldots p_n$ since every PID is an integral domain. Now substitute $p_2$ by (w.l.o.g.) $u_2q_2$ to get $q_3\ldots q_m=u_1u_2p_3\ldots p_n$ and so on. From this you easily deduce that $n$ has to be equal to $m$.</p>
|
2,337,583 | <p>I cannot understand the inductive dimension properly. I read something on Google but mostly there only are conditions or properties. Not a definition. I got to know about it from the book “ The fractal geometry of nature”. ( I am a 12 grader.)</p>
| Henno Brandsma | 4,280 | <p>The usual definition of the small inductive dimension <span class="math-container">$\operatorname{ind}(X)$</span> is as follows and defines <span class="math-container">$\operatorname{ind}(X) \le n$</span> by recursion: <span class="math-container">$\operatorname{ind}(X)$</span> is a function from topological spaces <span class="math-container">$X$</span> to <span class="math-container">$\{-1,0,1, 2 \ldots, \infty\}$</span> defined as follows</p>
<p>(i) <span class="math-container">$\operatorname{ind}(\emptyset) =-1$</span></p>
<p>Suppose we know already what <span class="math-container">$\operatorname{ind}(Y) \le n$</span> means for all spaces <span class="math-container">$Y$</span> and some <span class="math-container">$n \in \{-1, 0, \ldots\}$</span>, next we define this for <span class="math-container">$n+1$</span>:</p>
<p>(ii) <span class="math-container">$\operatorname{ind}(X) \le n+1$</span> iff for every <span class="math-container">$x \in X$</span>, and for all open sets <span class="math-container">$U$</span> that contain <span class="math-container">$x$</span>, there is an open set <span class="math-container">$V$</span> that contains <span class="math-container">$x$</span> and such that <span class="math-container">$\overline{V} \subseteq U$</span> and <span class="math-container">$\operatorname{ind}(\partial V) \le n$</span>.</p>
<p>A space <span class="math-container">$X$</span> has <span class="math-container">$\operatorname{ind}(X) = n$</span> iff <span class="math-container">$\operatorname{ind}(X) \le n$</span> and <em>not</em> <span class="math-container">$\operatorname{ind}(X) \le n-1$</span>.</p>
<p>A space has <span class="math-container">$\operatorname{ind}(X) = \infty$</span> iff for no <span class="math-container">$n \in \{-1, 0, 1,2, \ldots\}$</span> we have <span class="math-container">$\operatorname{ind}(X) \le n$</span>.</p>
<p>Sometimes we see the <span class="math-container">$n=0$</span> defined as the base case (<span class="math-container">$X$</span> has a clopen base, which means a base of sets with empty boundary, as <span class="math-container">$\partial V = \emptyset$</span> iff <span class="math-container">$\operatorname{int}(V) = \overline{V} = V$</span>. The <span class="math-container">$\operatorname{ind}(\emptyset) = -1$</span> is a sort of trick to make this original base case the first inductive stage. The intuition behind this definition is that the boundary of a set "should" have a dimension one lower than the set we start with. This seems to hold for all normal subsets of the plane, e.g.</p>
<p>Sets <span class="math-container">$X$</span> with <span class="math-container">$\operatorname{ind}(X) = 0$</span> include all discrete spaces (all sets are clopen), spaces like <span class="math-container">$\mathbb{Q}$</span> (as sets of the form <span class="math-container">$(a,b) \cap \mathbb{Q}$</span> form a base for it, and these are clopen iff <span class="math-container">$a,b$</span> are both irrational.</p>
<p>We have <span class="math-container">$\operatorname{ind}(\mathbb{R}) \le 1$</span> as open intervals have discrete boundary sets. It does not have <span class="math-container">$\operatorname{ind}(\mathbb{R}) \le 0$</span> as this would imply that <span class="math-container">$\mathbb{R}$</span> is disconnected. So <span class="math-container">$\operatorname{ind}(\mathbb{R}) = 1$</span>, as it should be.</p>
<p>Next, Brouwer was the first to show (using his fixed point theorem) that <span class="math-container">$\operatorname{ind}(\mathbb{R}^n) = n$</span> for all <span class="math-container">$n$</span>. This is a deep theorem (to get the exact value, <span class="math-container">$\le n$</span> is easier. As homeomorphic spaces have the same dimension (everything only depends on the topology), this was the first proof that different <span class="math-container">$n$</span> the spaces <span class="math-container">$\mathbb{R}^n$</span> were not homeomorphic. This was an open problem for a while (and space filling curves has been found, so people did not trust their intuitions about dimension very much).</p>
|
1,767,880 | <p>Let $A$ be a square matrix in the form $A=B+O(h^2)$, where $B$ is a fixed matrix, and $O(h^2)$ is a matrix with very small elements. Assume that: $$(I-A)^{-1}=I+A+A^2+A^3+...$$
How can I esimate the right hand side of the equality above ? Is that right $(I-B)^{-1}+O(h^2) ?$</p>
| Josh Hunt | 282,747 | <p><strong>Edit:</strong> <em>the below answer is incorrect - the binomial theorem doesn't apply because $O(h^2)$ doesn't necessarily commute past $B$.</em></p>
<p>A rather slapdash (<em>i.e.</em> non-rigourous) way of doing it is just to substitute $A = B + O(h^2)$ in the right-hand side, and notice that by the binomial theorem we have $$(B+O(h^2))^n = B^n +nO(h^2)B^{n-1} = B^n + O(h^2)$$</p>
<p>This gives you $I + B + B^2 + \ldots + O(h^2) = (I-B)^{-1} + O(h^2)$, as you say.</p>
|
1,767,880 | <p>Let $A$ be a square matrix in the form $A=B+O(h^2)$, where $B$ is a fixed matrix, and $O(h^2)$ is a matrix with very small elements. Assume that: $$(I-A)^{-1}=I+A+A^2+A^3+...$$
How can I esimate the right hand side of the equality above ? Is that right $(I-B)^{-1}+O(h^2) ?$</p>
| Community | -1 | <p>Let $A=B-K$, where $K$ is a small variable matrix. Then, by the Taylor's formula $(I-A)^{-1}=(I-B+K)^{-1}=(I-B)^{-1}-(I-B)^{-1}K(I-B)^{-1}+O(||K||^2)$. If $||K||=O(h^2)$, then $||(I-B)^{-1}K(I-B)^{-1}||=O(h^2)$ and we are done.</p>
|
43,172 | <p>I am trying to solve $\frac{dx}{dt} + \alpha x = 1$, $x(0) = 2$, $\alpha > 0$ where $\alpha$ is a constant. </p>
<p>[some very badly done mathematics deleted]</p>
<p>Continuing with Gerry's suggestion:</p>
<p>$\log|1-\alpha x | = -t\alpha + \log|1-2\alpha|$</p>
<p>$1-\alpha x = e^{-t\alpha}(1-2\alpha)$</p>
<p>$x(t) = \frac{1 - e^{-t\alpha}(1-2\alpha)}{\alpha}$</p>
<p>Then the asymptotic behaviour of $x(t)$ as $t$ goes infinity would be $e^{-t\alpha}$ approaching zero, therefore the overall $x(t)$ would approach $\frac{1}{\alpha}$. </p>
| joriki | 6,622 | <p>This is an inhomogeneous first-order linear ordinary differential equation. The standard way to solve such an equation is to find all solutions of the corresponding homogeneous equation and then add any particular solution of the inhomogeneous equation.</p>
<p>The inhomogeneous equation is solved by a constant:</p>
<p>$$x(t)=x_0 \rightarrow \alpha x_0=1 \rightarrow x_0=\frac{1}{\alpha}\;.$$</p>
<p>The standard way to find all solutions of the homogeneous equation</p>
<p>$$\frac{\mathrm dx}{\mathrm dt}+\alpha x=0$$</p>
<p>is through the <em>ansatz</em></p>
<p>$$x(t)=c\mathrm e^{\lambda x}\;,$$</p>
<p>which leads to the characteristic equation</p>
<p>$$\lambda+\alpha=0\;,$$</p>
<p>and hence $\lambda=-\alpha$ and</p>
<p>$$x(t)=c\mathrm e^{-\alpha t}\;.$$</p>
<p>So the general solution of the inhomogeneous equation is</p>
<p>$$x(t)=c\mathrm e^{-\alpha t}+\frac{1}{\alpha}\;.$$</p>
<p>Substituting $x(0)=2$ yields</p>
<p>$$c+\frac{1}{\alpha}=2\;,$$</p>
<p>$$c=2-\frac{1}{\alpha}\;,$$</p>
<p>and thus</p>
<p>$$x(t)=\left(2-\frac{1}{\alpha}\right)e^{-\alpha t}+\frac{1}{\alpha}\;.$$</p>
|
3,710,710 | <p>Suppose <span class="math-container">$A(t,x)$</span> is a <span class="math-container">$n\times n$</span> matrix that depends on a parameter <span class="math-container">$t$</span> and a variable <span class="math-container">$x$</span>, and let <span class="math-container">$f(t,x)$</span> be such that <span class="math-container">$f(t,\cdot)\colon \mathbb{R}^n \to \mathbb{R}^n$</span>.</p>
<p>Is there a chain rule for <span class="math-container">$$\frac{d}{dt} A(t,f(t,x))?$$</span></p>
<p>It should be something like <span class="math-container">$A_t(t,f(t,x)) + ....$</span>, what is the other term?</p>
| Jacky Chong | 369,395 | <p>Let <span class="math-container">$y=(y_1, \ldots, y_n) = (f_1(t, x), \ldots, f_n(t, x))$</span> Consider the <span class="math-container">$a_{ij}$</span> entry, then we see that
<span class="math-container">\begin{align}
\frac{d}{dt}a_{ij}(t, f(t,x)) = \partial_ta_{ij}(t, f(t, x))+\sum^n_{k=1} \underbrace{a_{ij, k}(t, f(t, x)}_{\partial_k a_{ij}(t, f(t, x))}\partial_t f_k(t, x).
\end{align}</span>
Hence it follows
<span class="math-container">\begin{align}
\frac{d}{dt}A(t, f(t, x))= \frac{\partial A}{\partial t}+\nabla A\cdot \partial_t f.
\end{align}</span>
Here <span class="math-container">$\nabla A$</span> is the gradient of a rank <span class="math-container">$2$</span> tensor which is a rank <span class="math-container">$3$</span> tensor (I am assuming everything is in Euclidean space). In general, <span class="math-container">$\nabla A$</span> is called the covariant derivative of <span class="math-container">$A$</span>. Also, note that <span class="math-container">$\nabla A\cdot \partial_t f$</span> is the directional derivative of <span class="math-container">$A$</span> in the direction <span class="math-container">$\partial_t f$</span>. </p>
|
172,894 | <p>Suppose $A$ is an integral domain with integral closure $\overline{A}$ (inside its fraction field), $\mathfrak{q}$ is a prime ideal of $A$, and $\mathfrak{P}_1,\ldots,\mathfrak{P}_k$ are the prime ideals of $\overline{A}$ lying over $\mathfrak{q}$. Show that $\overline{A_\mathfrak{q}} = \bigcap\overline{A}_\mathfrak{P_i}$ (note that the LHS is the integral closure of a localization, whereas the RHS is the intersection of localizations of integral closures of $A$).</p>
<p>If it would help, I suppose we could assume that $A$ has dimension 1, so that $\overline{A}$ is Dedekind, though I don't think that assumption is required.</p>
<p>(Geometrically, we're comparing the integral closure of a local ring at a singular point Q of some variety with the intersection of local rings at points in the normalization mapping to Q).</p>
<p>Thanks.</p>
| Community | -1 | <p>Consider $\dim A=1$. One can assume that $A$ is a local Noetherian domain with maximal ideal $m$ and have to prove that $\overline{A}=\bigcap_{i=1}^n\overline{A}_{P_i}$, where $P_1,\dots,P_n$ are all the prime ideals of $\overline{A}$ lying over $m$. (There are only finitely many primes in $\overline{A}$ lying over $m$ because $\overline{A}$ is Dedekind!) Now one can use the well known fact that an integral domain (in our case $\overline{A}$) is the intersection of all its localizations at maximal ideals. Note that a prime ideal $P$ of $\overline{A}$ is lying over $m$ or over $(0)$. The prime ideals $P$ lying over $(0)$ give "large" localizations, i.e. $\overline{A}_P=K$ when $P\cap A=(0)$, where $K$ is the field of fractions of $A$. In particular these primes can be removed from the intersection finally remaining only the primes $P$ lying over $m$.</p>
|
3,425,415 | <p>I need to define a bijection <span class="math-container">$f:\mathbb Q\to\mathbb Q$</span> such that <span class="math-container">$f(0) = 0$</span> and <span class="math-container">$f(1) = 1$</span> while also preserving order (i.e. if <span class="math-container">$a < b$</span>, then <span class="math-container">$f(a) < f(b)$</span>). Also, <span class="math-container">$f$</span> cannot be the identity function. </p>
<p>Whenever I try to come up with a function, it either becomes not injective, not surjective, or it does not preserve order. Any help would be appreciated. </p>
| Rushabh Mehta | 537,349 | <p>There are plenty of examples. The following function <span class="math-container">$f:\mathbb Q\to\mathbb Q$</span> is one example: <span class="math-container">$$f(x) = \begin{cases}2x&x<0\\x&x\geq0\end{cases}$$</span>Now, if you also require that the bijection preserves addition/multiplication, then such a map doesn't exist.</p>
|
899,230 | <p>It seems that both isometric and unitary operators on a Hilbert space have the following property:</p>
<p><span class="math-container">$U^*U = I$</span> (<span class="math-container">$U$</span> is an operator and <span class="math-container">$I$</span> is an identity operator, <span class="math-container">$^*$</span> is a binary operation.) </p>
<p>What is the difference between <strong>isometry</strong> and <strong>unitary</strong>? Which one is more general,
or are they the same? Are they isomorphic?</p>
| glS | 173,147 | <p>In finite dimensions, there is a straightforward characterisation of isometries and unitaries in terms of their matrix representations.</p>
<p>The basic observation is that <span class="math-container">$U^*U=I$</span> means that the columns of (the matrix representation of) <span class="math-container">$U$</span> are orthonormal, while <span class="math-container">$UU^*=I$</span> means that the <em>rows</em> of <span class="math-container">$U$</span> are orthonormal.</p>
<p>A unitary <span class="math-container">$U$</span> is a matrix whose columns (equivalently, rows) form an orthonormal <em>basis</em>. This is equivalent to <em>both</em> its columns and rows being orthonormal.
An isometry, on the other hand, only requires that the columns are orthonormal, but not that they form a basis.</p>
<p>In summary, the distinction between the two objects can be stated as follows: <em>an isometry is a matrix whose columns are orthonormal, while a unitary is <strong>squared</strong> matrix whose columns are orthonormal</em>.</p>
<p>For completeness: one can also talk about <em>partial isometries</em>, which are matrices such that <span class="math-container">$U^* U$</span> is an orthogonal <em>projection</em>, rather than being the identity. This is equivalent to asking for <span class="math-container">$U$</span> to be an isometry on the orthogonal complement of its ker.</p>
|
405,866 | <p>Original question:
For compact metric spaces, plenty of subtly different definitions converge to the same concept. Overtness can be viewed as a property dual to compactness. So is there a similar story for overt metric spaces?</p>
<p>Edit: Since overtness is trivially true assuming the Law of the Excluded Middle, clearly the question is primarily interesting when we do not assume the LEM.</p>
<p>Edit 2: It looks like it is extremely difficult for a metric space to not be overt even in constructive settings. So editing the question to ask if there is ANY model where metric spaces are not overt.</p>
<p>Edit 3: For these reasons I changed the question again, from "Is there any model of mathematics where there exists a metric space that is not overt?". PT</p>
| Ingo Blechschmidt | 31,233 | <p>The spectrum of a commutative ring, defined as the classifying locale of its prime filters, is overt if and only if any element is nilpotent or not nilpotent (Proposition 12.51 in <a href="https://rawgit.com/iblech/internal-methods/master/notes.pdf#page=130" rel="noreferrer">these notes of mine</a>).</p>
<p>Hence for almost any nontrivial base scheme <span class="math-container">$X$</span>, inside the topos of sheaves over <span class="math-container">$X$</span>, the spectrum of <span class="math-container">$\mathcal{O}_X$</span> will be an example for a locale which is not overt.</p>
<p>In impredicative (constructive or classical) mathematics, you will not find a topological space which would not be overt.</p>
<p>(Note to people familiar with algebraic geometry's relative spectrum construction who would expect <span class="math-container">$\operatorname{Spec}(\mathcal{O}_X)$</span> to be the one-point locale (so that it externalizes to <span class="math-container">$X$</span> again): The usual localic spectrum construction, when carried out internally in the topos of the base scheme, does <em>not</em> coincide with the relative spectrum construction. But a variant does. Details are in Section 12 of the linked notes.)</p>
|
4,463,373 | <p>If <span class="math-container">$\frac{\partial f}{\partial x}(0,0) = \frac{\partial f}{\partial y}(0,0) = 0$</span>, then <span class="math-container">$f'((0,0);d)=0$</span> (directional derivative) for every direction <span class="math-container">$d \in \mathbb{R}^n$</span>.</p>
<p>Is this true? I'm trying to find a counterexample to prove it false, but nothing comes to mind.</p>
| Fred | 380,717 | <p>For <span class="math-container">$x>0$</span> we have <span class="math-container">$f(x)=x^4$</span>, hence <span class="math-container">$f'(x)=4x^3$</span>.</p>
<p>For <span class="math-container">$x<0$</span> we have <span class="math-container">$f(x)=-x^4$</span>, hence <span class="math-container">$f'(x)=-4x^3$</span>.</p>
<p>For <span class="math-container">$x=0$</span> we use the definition:</p>
<p><span class="math-container">$$\frac{f(h)-f(0)}{h}=|h^3|=|h|^3 \to 0$$</span></p>
<p>as <span class="math-container">$h \to 0.$</span> Hence <span class="math-container">$f$</span> is differentiable at <span class="math-container">$0$</span> and <span class="math-container">$f'(0)=0.$</span></p>
|
55,076 | <p>Let $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3 $ be the linear operator </p>
<p>$$
A = \begin{pmatrix}
1 & 0 & 0 \\
1 & 2 & 0 \\
0 & 0 & 3 \end{pmatrix}
$$</p>
<p>The problem I am trying to understand is the following. </p>
<p>True or False? If $W$ is a $T$-invariant subspace of $\mathbb{R}^3$ $\exists T$-invariant subspace $W'$ of $\mathbb{R}^3$ such that $W \oplus W' = \mathbb{R}^3$. I think the answer is true and I will list my ideas below but I think there must be an easier way to do approach this.</p>
<p>Since $A$ has distinct eigenvalues $1,2,3$ we see the minimal polynomial is $m_A(x) = (x-1)(x-2)(x-3)$</p>
<p>and therefore using the fundamental structure theorem for PID's we have $\mathbb{R}^3 = \mathbb{R}/(x-1)\oplus\mathbb{R}/(x-2)\oplus \mathbb{R}/(x-3) = \mathbb{R}\oplus\mathbb{R}\oplus\mathbb{R} $</p>
<p>From this calculation does it follow that $W' = \mathbb{R}\oplus\mathbb{R}$ is the invariant subspace we are looking for?</p>
<p>Is there a way to do this problem without using the fundamental structure theorem for modules over a PID?</p>
| Geoff Robinson | 13,147 | <p>I am not used to your notation, but I think I understand what you mean. If a linear transformation $T$ on an $n$-dimensional vectors space $V$ has $n$ distinct eigenvalues,
say $\{\lambda_1, \lambda_2,\ldots, \lambda_n \}$, then $V$ has a basis consisting of eigenvectors of $T$. Most proofs of this fact make (at least) implicit use of the fact that the $n$-polynomials $ \{p_i(x): 1 \leq i \leq n \}$, where $p_i(x)= \prod_{j \neq i}(x- \lambda_j)$ have greatest common divisor one. Probably the way to disguise this use as far as possible is to say that the minimum polynomial divides the characteristic polynomial, while whenever $\lambda$ is an eigenvalue of a linear transformation,
$x - \lambda$ is a factor of its minimum polynomial. Hence when there are $n$ distinct eigenvalues, as here, the minimum polynomial must be $\prod_{i=1}^{n}(x-\lambda_i)$.
This means that, with the earlier notation, $p_i(T) \neq 0$ for $1 \leq i \leq n.$
On the other hand, $(T- \lambda_i)p_i(T)= [0]$ for each $i$. Thus for each $i$,
$p_i(T)V$ is a non-zero subspace of $V$ consisting of eigenvectors for $T$ associated to eigenvalue $\lambda_i$. Since eigenvectors associated to different eigenvalues are linearly independent, this produces a basis of eigenvectors of $T$. Let $V_i$ be the
$1$-dimensional $\lambda_i$-eigenspace for $T$ on $V$. Then we have $V = \oplus_{i=1}^{n}V_i$.</p>
<p>If $W$ is a $T$-invariant subspace of $V$, then for some subset $I$ of $\{1,2,\ldots, n \}$, the minimum polynomial of $T$ on $W$ is $\prod_{i \in I}(x-\lambda_i)$ (it must divide the minimum polynomial on $V$. Then for each $i \in I$, arguing as above, $W$
must contain the $\lambda_i$-eigenspace of $T$, as $W$ does contain an eigector
with that eigenvalue, but the eigenspace in $V$ is $1$-dimensional. Hence $W = \oplus_{i \in I}V_i$. Thus $V = W \oplus W'$, where $W' = \oplus_{i \in I'} V_i$ and
$I' = \{1,2,\ldots, n\} \backslash I$. </p>
|
1,275,461 | <p>I am trying to proof that $-1$ is a square in $\mathbb{F}_p$ for $p = 1 \mod{4}$. Of course, this is really easy if one uses the Legendre Symbol and Euler's criterion. However, I do not want to use those. In fact, I want to prove this using as little assumption as possible.</p>
<p>What I tried so far is not really helpful:</p>
<p>We can easily show that $\mathbb{F}_p^*/(\mathbb{F}_p^*)^2 = \{ 1\cdot (\mathbb{F}_p^*)^2, a\cdot (\mathbb{F}_p^*)^2 \}$ where $a$ is not a square (this $a$ exists because the map $x \mapsto x^2$ is not surjective). Now $-1 = 4\cdot k = 2^2 \cdot k$ for some $k\in \mathbb{F}_p$.</p>
<p>From here I am trying to find some relation between $p =1 \mod{4}$ and $-1$ not being a multiple of a square and a non-square.</p>
| Daniel Fischer | 83,702 | <p>On $\mathbb{F}_p^\ast$, define the equivalence relation</p>
<p>$$x\sim y :\!\!\iff (x = y) \lor (x = -y) \lor (xy = 1) \lor (xy = -1).$$</p>
<p>Generally, the equivalence class of $x$ has four elements, $\{ x, -x, x^{-1}, -x^{-1}\}$. Since we always have $x \neq -x$, every class has at least two elements, and if $x = x^{-1}$ or $x = -x^{-1}$, then the equivalence class of $x$ has precisely two elements. The case $x = x^{-1}$ (or $x^2 = 1$) yields the class $\{1,-1\}$, so there is always at least one class with two elements.</p>
<p>If $p \equiv 1 \pmod{4}$, the number of elements of $\mathbb{F}_p^\ast$ is divisible by $4$, therefore the number of two-element classes must be even, hence $2$, and that means there must be an $x$ with $x = -x^{-1}$, or $x^2 = -1$.</p>
|
2,312,016 | <p>Prove limit of three variables using (ε, δ)-definition.</p>
<p>$$\lim_{(x, y, z)\to (0, 1, 2)} (3x+3y-z)=1$$</p>
<p>I have no idea how to do this with three variables.</p>
| SBM | 422,110 | <p>We need to show that for every $\varepsilon > 0$, there is a value $\delta \gt 0$ such that $\left|x - 0\right| \lt \delta, \left|y-1\right| \lt \delta, \text{and } \left|z-2\right| \lt \delta$ imply that $\left|3x + 3y - z - 1 \right| \lt \varepsilon$</p>
|
158,662 | <p>I know to prove a language is regular, drawing NFA/DFA that satisfies it is a decent way. But what to do in cases like</p>
<p>$$
L=\{ww \mid w \text{ belongs to } \{a,b\}*\}
$$</p>
<p>where we need to find it it is regular or not. Pumping lemma can be used for irregularity but how to justify in a case where it can be regular?</p>
| hmakholm left over Monica | 14,366 | <p>Suppose the language was regular and had a DFA.</p>
<p>After reading, for example, "$\underbrace{aa\ldots a}_nb$" the DFA is in some state, and the identity of this state determines <em>completly</em> what the rest of an input that the machine accepts can be.</p>
<p>But if $n\ne m$, then the possible tails that can follow $\underbrace{aa\ldots a}_nb$ and $\underbrace{aa\ldots a}_m b$ are <em>different sets</em>. That means that the machine must be in <em>different states</em> after having read them.</p>
<p>Now, remember what the F in DFA stands for?</p>
|
2,764,381 | <p>one thing I don't understand is what is sin(0) and sin(1) exactly? I am alright with the concept of radian (pi) but don't understand 0 and 1. What does it mean?</p>
| INDIAN | 556,781 | <p>If $1$ is considered as $1$ radian then $1$ less than $\pi /2$ so $sin x$ is a strictly increasing function on that interval.</p>
|
3,505,397 | <p>well, I have to find the Taylor polynomial of <span class="math-container">$f(x,y)=\sin(x)\sin(y)$</span> at <span class="math-container">$(0,\pi/4)$</span>. I found:</p>
<p>Is <span class="math-container">$T_3(x,y)=-\frac{1}{12}\sqrt{2}x(16x^2+48y^2-24\pi+3\pi^2)$</span> correct?</p>
| Ryan Shesler | 585,375 | <p><span class="math-container">$\sin(x) = \cos(\frac{\pi}{2} - x)$</span>.</p>
<p>Using this you get <span class="math-container">$\cos^{-1} (\cos(\frac{\pi}{2} - \frac{16\pi}{7}))$</span>. Then you can finish it from here (but keep in mind the domain of <span class="math-container">$\cos^{-1}(x)$</span>)</p>
|
3,505,397 | <p>well, I have to find the Taylor polynomial of <span class="math-container">$f(x,y)=\sin(x)\sin(y)$</span> at <span class="math-container">$(0,\pi/4)$</span>. I found:</p>
<p>Is <span class="math-container">$T_3(x,y)=-\frac{1}{12}\sqrt{2}x(16x^2+48y^2-24\pi+3\pi^2)$</span> correct?</p>
| lab bhattacharjee | 33,337 | <p><span class="math-container">$$f(x)=\sin\dfrac{16\pi}7=\sin\dfrac{2\pi}7=\cos(\dfrac\pi2-\dfrac{2\pi}7)=\cos\dfrac{\pi(7-4)}{14}$$</span></p>
<p>Now <span class="math-container">$\cos^{-1}f(x)=2n\pi\pm\dfrac{3\pi}{14}$</span> where <span class="math-container">$n$</span> is an integer such that <span class="math-container">$0\le\cos ^{-1}f(x)\le\pi$</span></p>
|
578,337 | <p>For $n=1,2,3,\dots,$ and $|x| < 1$ I need to prove that $\frac{x}{1+nx^2}$ converges uniformly to zero function. How ?. For $|x| > 1$ it is easy. </p>
| Community | -1 | <p>Let $f_n(x)=\frac{x}{1+nx^2}$ then we have
$$f'_n(x)=\frac{1-nx^2}{(1+nx^2)^2}=0\iff x=n^{-1/2}:=x_n$$
hence
$$||f_n||_\infty=f_n(x_n)=\frac{1}{2}x_n\to0$$
so we have the uniform convergence to $0$.</p>
|
136,086 | <p>I've been given the following problem as homework:</p>
<blockquote>
<p>Q: <strong>Compute the number of subgraphs of <span class="math-container">$K_{15}$</span> isomorphic to <span class="math-container">$C_{15}$</span></strong>.</p>
<p><span class="math-container">$K_{15}$</span> means complete graph with 15 vertices. <span class="math-container">$C_{15}$</span> means cyclic graph, where the whole graph is a cycle, with 15 vertices. For example, <span class="math-container">$C_3$</span> is a triangle, <span class="math-container">$C_4$</span> is a square, and <span class="math-container">$C_5$</span> is a pentagon.</p>
</blockquote>
<p>My efforts: In order to try to figure out a general formula for <span class="math-container">$C_n$</span>, I tried doing this problem with <span class="math-container">$C_5$</span>. After a huge amount of trial and error, it looks like the formula is something like <span class="math-container">$$\binom{15}{n}\binom{n-1}{2}(n-3)!$$</span><br />
However, I can't seem to come up with a good reason for this formula or verify whether it's correct. I'm doubting it is correct.</p>
<p>This is homework, so I'm NOT looking for solutions. Could you give some tips for figuring this out?</p>
| Yuval Filmus | 1,277 | <p>Let's start with the case $n = 3$. A triangle in $K_{15}$ is given by any three vertices. So the number of triangles is $\binom{15}{3}$.</p>
<p>Let's proceed to the case $n = 4$. A rectangle in $K_{15}$ is given by any four vertices. But the vertices $a,b,c,d$ define three different rectangles. Indeed, suppose that $a$ is connected to $b,c$, and let $d$ be the remaining vertex. Then the rectangle must be
$$ab,bd,dc,ca$$
So the number of different rectangles is $\binom{15}{4} \cdot 3$.</p>
<p>When $n = 15$, there is only one way of choosing $15$ vertices. The earlier examples don't seem to help. Instead, I suggest trying to count how many copies of $C_n$ there are in $K_n$ for small $n$, and then generalize.</p>
|
3,168,119 | <p>How do I solve for n?</p>
<p><span class="math-container">$125 = x * 2^n$</span></p>
<p>This is what I have so far:</p>
<p><span class="math-container">$5^3 = x * 2^n$</span></p>
<p>I do remember that according to the exponential rules,
that the powers should be the same if the equation is like this:</p>
<p><span class="math-container">$8 = 2^n$</span></p>
<p><span class="math-container">$2^3 = 2^n \iff 2^3 = 2^3$</span></p>
<p>I am not sure if this rule can be used in the equation above.</p>
| YiFan | 496,634 | <p>You cannot solve for <span class="math-container">$n$</span>, unless you want your answer to be in terms of <span class="math-container">$x$</span>. In that case, <span class="math-container">$125/x=2^n$</span> tells you <span class="math-container">$n=\log_2(125/x)=3\log_2(5)-\log_2(x)$</span>.</p>
<p>The reason it's not possible to solve for <span class="math-container">$n$</span> is because, no matter what <span class="math-container">$n$</span> you choose, there's always an appropriate choice of <span class="math-container">$x$</span> so that the equation is satisfied. (Why? Can you find an expression for such an <span class="math-container">$x$</span>?) On the other hand, if <span class="math-container">$x$</span> is restricted to an <strong>integer</strong>, then you want to consider prime factorizations of both sides. I'll leave this to you.</p>
|
2,240,405 | <p>The question asks me to find the Laurent series of $$f(z) = {5z+2e^{3z}\over(z-i)^6}\,\,at\,\,z=i$$I know the following $$e^z=\sum_{n=0}^\infty {z^n\over n!}$$ What I want to know, is if I can do this: $$={1\over (z-i)^6}\sum_{n=0}^\infty ({2(3z)^n\over n!}+5z)$$ $$=\sum_{n=0}^{n=6}(z-i)^{n-6}({2(3z)^n\over n!}+5z)+\sum_{n=7}^{\infty}(z-i)^{n-6}({2(3z)^n\over n!}+5z)$$
This is all I could think of possibly doing and I'm not even sure if it is even something that I can even do, and I don't really know how to continue. I also need to find the radius of convergence but I feel once I have the series it will be easy to apply whatever test necessary to acquire it.</p>
| Angina Seng | 436,618 | <p>For this kind of problem I recommend changing the variable to push the pole to zero. Here I'd set $w=z-i$. The problem becomes:
find the Laurent series of
$$g(w)=\frac{5(w+i)+2e^{3w+3i}}{w^6}$$
at $w=0$. The numerator is
$$5i+5w+2e^{3i}\left(1+3w+\frac{3^2w^2}{2}+\frac{3^3w^3}{3!}+\cdots\right)$$
etc.</p>
|
1,009,503 | <p>Theorem 15 in Chapter 15 of Peter Lax's functional analysis book says</p>
<p>$X$ is a Banach space, $Y$ and $Z$ are closed subspaces of $X$ that complement each other $X = Y \oplus Z$, in the sense that every $x\in X$ can be decomposed uniquely as $x = y+z$ where $y\in Y$, $z\in Z$. Denote the two complements of $y$ and $z$ of $x$ by $y = P_Y x$, $z = P_Z x$. Then</p>
<p>1) $P_Y$ and $P_Z$ are linear maps of $X$ on $Y$ and $Z$ respectively.</p>
<p>2) $P_Y^2 = P_Y$, $P_Z^2 = P_Z$, $P_YP_Z = 0$.</p>
<p>3) $P_Y$ and $P_Z$ are continuous.</p>
<p>However, I would like to say that for all $\hat y\in Y$,
$\|y-x\| \leq \|\hat y - x\|$.</p>
<p>which is a property of projections we already know, but doesn't seem to be built from the definition of projections given by Lax. I'm not too familiar with Banach spaces, so I don't really want to use this property until I know for sure it is implied by the previous properties. </p>
<p>Can anyone give me some intuition if 1) this property is implied by the definitions, 2) this property may not hold in some special case (Not sure if $X$ is reflexive), or 3) this is implicit in the definition of projection and I'm just reading the text wrong? </p>
<p>Thanks!</p>
<p>Edit: Just to add clarification, I'm considering any complemented Banach space (infinite dimensional). Not necessarily a Hilbert space, and no indication that it is reflexive.</p>
| MJD | 25,554 | <p>If the probability of getting tails is $q = 1-p$, then you expect to get, on average, $q$ tails per throw, because that is exactly what a probability is: the average number of tails per throw.</p>
<p>Expectations are additive, so if you get $q$ tails per throw it requires $\frac nq$ throws in order to expect $n$ tails, and $\frac1q$ throws in order to expect to have one tail. That means that the number of heads you expect to get just before the first tail is $\frac 1q - 1$.</p>
<p>If the coin is fair, $q=\frac12$, so the expected number of heads in a row is $1$. If the coin throws heads with probability $0.8$, then $q=0.2$, so the expected number of heads in a row is $4$. In general, if the probability of throwing heads is $p$, the expected number of heads in a row before the first tail is $$\frac1{1-p} - 1 = \frac p{1-p}.$$</p>
<p>Note that although the infinite summation gets the same answer, it is not needed to solve this problem.</p>
|
1,369,990 | <p>I came across a quesion -
<a href="https://www.hackerrank.com/contests/ode-to-code-finals/challenges/pingu-and-pinglings" rel="nofollow">https://www.hackerrank.com/contests/ode-to-code-finals/challenges/pingu-and-pinglings</a></p>
<p>The question basically asks to generate all combinations of size k and sum up the product of numbers of all combinations..Is there a general formula to calculate the same,as it is quite tough to generate all the possible combinations and operate on them..
For example for n=3(no of elements) and k=2
and the given 3 numbers are 4 2 1,then the answer will be 14 as
For k=2 the combinations are {4,2},{4,1},{2,1} so answer is (4×2)+(4×1)+(2×1)=8+4+2=14.
I hope i am clear in asking my question.</p>
| Shuaib Lateef | 880,979 | <p>I guess this is a general case for size 2:
<span class="math-container">$$\sum_{i=1}^{n-1}\sum_{j=i+1}^na_ia_j=\frac{(\sum_{i=1}^na_i)^2-\sum_{i=1}^na_i^2}{2}$$</span></p>
|
1,353,498 | <p>The problem is to prove or disprove that there is a noncyclic abelian group of order $51$. </p>
<p>I don't think such a group exists. Here is a brief outline of my proof:</p>
<p>Assume for a contradiction that there exists a noncyclic abelian group of order $51$.</p>
<p>We know that every element (except the identity) has order $3$ or $17$. Assume that $|a|=3$ and $|b|=17$. Then I managed to prove that the subgroups generated by $a$ and $b$ only intersect at the identity element, from which we can show that $ab$ is a generator of the whole group, so it is cyclic. Contradiction.</p>
<p>So every element (except the identity) has the same order $p$, where $p$ is either $3$ or $17$. </p>
<p>If $p=17$, take $a$ not equal to the identity, and take $b$ not in the subgroup generated by $a$. Then we can prove that $a^kb^l$ where $k,l$ are integers between $0$ and $16$ inclusive are distinct, hence the group has more than $51$ elements, contradiction.</p>
<p>If $p=3$, take $a$ not equal to the identity and take $b$ not in the subgroup generated by $a$. Then we can prove that $a^kb^l$ where $k,l$ are integers betwen $0$ and $2$ inclusive are distinct. This subgroup has $9$ elements so we can find $c$ that's not of the form $a^kb^l$. Then we can prove that $a^kb^lc^m$ where $k,l,m$ are integers betwen $0$ and $2$ inclusive are distinct. Then this subgroup has $27$ elements so we can find $d$ that's not of the form $a^kb^lc^m$. Then we prove that $a^kb^lc^md^n$ where $k,l,m,n$ are integers between $0$ and $2$ inclusive are distinct, this being $81$ elements. Contradiction.</p>
| coldnumber | 251,386 | <p>Using the Sylow theorems would shorten your proof considerably, because from the first Sylow theorem it follows that if a prime divides the order of a group, then the group contains an element of that order. (This eliminates the need to check cases where all elements have order 3 or all have order 17.)</p>
<p>Hence, if $\left|G\right|=51$, then $G$ has an element $a$ of order 3 and an element $b$ of order 17. </p>
<p>Then, as you said, $ab$ has order 51; you can show this directly, without using the proposition for product groups:</p>
<p>Since $\left|G\right|=51$, by Lagrange's theorem, the order of $ab$ divides 51.</p>
<p>Now, since $G$ is abelian, $(ab)^3=a^3b^3=b^3 \neq 1$ because $b$ has order 17, and $(ab)^{17}=a^{17}b^{17}=a^2\neq 1$, because $a$ has order 3.</p>
<p>Hence, $ab$ generates the group, so $G$ is cyclic.</p>
|
3,118,462 | <p>cars arrives according to a Poisson process with rate=2 per hour and trucks arrives according to a Poisson process with rate=1 per hour. They are independent. </p>
<p>What is the probability that <strong>at least</strong> 3 cars arrive before a truck arrives? </p>
<p>My thoughts:
Interarrival of cars A ~ Exp(2 per hour), Interarrival of trucks B ~ Exp(1 per hour). </p>
<p>Probability that <strong>at least</strong> 3 cars arrive before a truck arrives</p>
<p><span class="math-container">$= 1- Pr(B<A) - Pr(A<B)Pr(B<A) - Pr(A<B)Pr(A<B)Pr(B<A)
\\= 1 - (\frac{1}{3})-(\frac{2}{3}\cdot\frac{1}{3})-(\frac{2}{3}\cdot\frac{2}{3}\cdot\frac{1}{3})\\=\frac{8}{27}.$</span> </p>
<p>Is this correct?</p>
| Ross Millikan | 1,827 | <p>You can use the <a href="https://en.wikipedia.org/wiki/Quadratic_formula" rel="nofollow noreferrer">quadratic formula</a>. Any quadratic <span class="math-container">$ax^2+bx+c$</span> factors as <span class="math-container">$a\left(x+ \frac{b+\sqrt {b^2-4ac}}{2a}\right)\left(x+ \frac{b-\sqrt {b^2-4ac}}{2a}\right)$</span></p>
|
9,462 | <p>In my question I ask for practical tips for the mathematical research practice, if not personal,I look for some articles/websites/books/guides/faq related, or if was already asked on Math.SE, the link to the question.</p>
<p><a href="https://math.stackexchange.com/questions/386520/practical-tips-research-and-discoveries">Practical Tips: Mathematical research and discoveries</a></p>
<p>the motivation of the closure was this.</p>
<blockquote>
<p>"As it currently stands, this question is not a good fit for our Q&A
format. We expect answers to be supported by facts (1), references, or
specific expertise, but this question will likely solicit debate (2),
arguments, polling, or extended discussion(3). If you feel that this
question can be improved and possibly reopened"</p>
</blockquote>
<p>(1) Why is not supported by facts? Is a pratical problem, and I'm asking for tips.</p>
<p>(2) I'm asking for a list of links/books not for opinions.</p>
<p>(3) I don't want do discuss the answers.</p>
<p>Another reason was:" Please see the FAQ about what questions are appropriate to ask here"</p>
<p>and in the faqs I read:</p>
<blockquote>
<p>You should only ask <strong>practical</strong>, <strong>answerable</strong> questions based on <strong>actual problems that you face.</strong></p>
</blockquote>
<p>The question is about <strong>mathematical practice.</strong>
Well, If someone do mathematical research or know some books/guides/related questions, than he can <strong>answer</strong> this question. And this is an <strong>actual problem that I face.</strong></p>
<p>So where is the problem?</p>
<p>Thanks in advance</p>
<p><strong>QUESTION IS RECLOSED AGAIN!?</strong></p>
| Douglas S. Stones | 139 | <p>I cast the fifth re-open vote (and thus has been re-opened). Seems like a reasonable enough question to me. There seems to be one answer in the comments already.</p>
<p>Note that, in order to close a question, merely 5 close votes are required. From this, we cannot deduce that there is a consensus for closing. We can only deduce that from the community (with a diverse range of opinions of what's good and what's not for the site) there are 5 people who have the ability to vote to chose, and voted for this particular question.</p>
<p>The reasons for closure can be varied, and the real reason can be quite dissimilar from the reason the software makes us choose (and each voter might have their own distinct reason, etc.).</p>
|
1,383,380 | <p>On page 12 of Stein, Shakarchi textbook 'Complex analysis', the authors state that the <em>Cauchy-Riemann equations link complex and real analysis</em>. I have completed courses on real and complex analysis, but I feel that this is somewhat of an over-statement. But perhaps it is just me which doesnt have a good enough overview.</p>
<p>$$
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}
$$</p>
<p>If anyone with a clear insight is able to concisely explain how one could justify writing something like this -- then that insight would be most valuable.</p>
| Brian Tung | 224,454 | <p>There does not <em>have</em> to be a fixed point $x$ such that $f(x) = x$. Here's a counterexample: the constant function $f(x) = 2$. Obviously, there is no $x \in [0, 1]$ for which $x = 2$.</p>
<p>Given a function $f(x)$ as described in the problem, consider the function $g(x) = f(x)-2x$. We have $g(0) \in [0, 2]$, and $g(1) \in [-2, 0]$. Since $g(x)$, like $f(x)$, is continuous, there must—by way of the intermediate value theorem—exist a value $x \in [0, 1]$ such that $g(x) = 0$. For that value of $x$, we must have $f(x) = 2x$.</p>
<p>More broadly, the reason why there must be an $x \in [0, 1]$ such that $f(x) = 2x$ is that the line $y = 2x$ is a diagonal of the rectangle $[0, 1] \times [0, 2]$. You start off on the left side of the rectangle, and end up on the right side, and in so doing, must cross the diagonal. The other diagonal also works this way: One might just as effectively say that there must be an $x \in [0, 1]$ such that $f(x) = 2-2x$. Hopefully, you can now see why that must be so.</p>
|
464,489 | <p>We are given $H = \{(1),(13),(24),(12)(34),(13)(24),(14)(23),(1234),(1432)\}$ is a subgroup of $S_4$. Also assume $K = \{(1),(13)(24)\}$ is a normal subgroup of $H$. Show $H/K$ isomorphic to $Z_2\oplus Z_2$. </p>
<p>This is just a practice question (not assignment). So I have tried finding $H/K$ explicitly.</p>
<p>$H/K = \{\{(1),(13)(24)\},\{(13),(24)\},\{(14)(23),(12)(34)\},\{(1234),(1423)\}\}$. We know there are only $2$ groups of order $4$. One of the elements in $H/K$ we see is $(1234)K$, doesn't this element have a order of $4$, making $H/K$ cyclic and hence not isomorphic to $Z_2\oplus Z_2$? </p>
| user66733 | 66,733 | <p>I think your doubt about these two being isomorphic has been solved but just in case you still don't see how to define a precise isomorphism between them note that $\mathbb{Z}_2 \bigoplus \mathbb{Z}_2 $ has two generators, namley $(1,0)$ and $(0,1)$ also $(1,0) + (0,1) = (1,1)$. Therefore I think it's very easy for you to know what precisely your isomorphism should be and which coset should be send to which generator. You can also use the fact that there are only two groups of order 4 up to isomorphism as you already pointed out.</p>
|
1,017,965 | <p>Just as the title, my question is what is the matrix representation of Radon transform (Radon projection matrix)? I want to have an exact matrix for the Radon transformation. </p>
<p>(I want to implement some electron tomography algorithms by myself so I need to use the matrix representation of the Radon transformation.)</p>
<p>(An introduction of the Radon transformation: <a href="http://en.wikipedia.org/wiki/Radon_transform" rel="nofollow noreferrer">click here</a>.)</p>
| Yi-Chun | 642,317 | <p>Currently I am solving the same problem as you guys.
I found a way to find the matrix.</p>
<p>First, we may view image(MxN) as a vector by certain reshape.
Second, set the ONLY one entry to 1, and others to 0.
Third, input this image to the function 'radon' then you way get the transformed image(LxP)</p>
<p>Loop over these three steps for every entry of image, which may causes MxN times.
cascade the result image(LxP) MxN times, then you will get the answer.</p>
<pre><code>I.e. for 2x3-->3x4
[1 0 0 0 0 0]-->X_1 (reshape to a row vector)
[0 1 0 0 0 0]-->X_2 (reshape to a row vector)
[0 0 1 0 0 0]-->X_3 (reshape to a row vector)
[0 0 0 1 0 0]-->X_4 (reshape to a row vector)
[0 0 0 0 1 0]-->X_5 (reshape to a row vector)
[0 0 0 0 0 1]-->X_6 (reshape to a row vector)
Finally, stack result row vectors
[X_1, X_2, X_3, X_4, X_5, X_6].T
</code></pre>
|
3,345,329 | <p>In Bourbaki Lie Groups and Lie Algebras chapter 4-6 the term displacement is used a lot. For example groups generated by displacements. But I can not find a definition of the term displacement given anywhere. I also looked at Humphreys Reflection Groups and Coxeter groups book but I could not find it. Can someone provide a definiton of displacements in the context of reflection groups and root systems? </p>
| Jake Levinson | 43,565 | <p>This is not an answer, but it's too long for a comment.</p>
<p>Clearly, the answer must scale proportionally to <span class="math-container">$\sigma^2$</span>, so let's just assume <span class="math-container">$\sigma=1$</span>.</p>
<p>Sampling <span class="math-container">$x_1, \ldots, x_n$</span> i.i.d. from a standard Gaussian is equivalent to sampling a random Gaussian vector <span class="math-container">$\vec{x} \in \mathbb{R}^n$</span> with mean <span class="math-container">$\vec{0}$</span> and identity covariance matrix. The median coordinate <span class="math-container">$\mathrm{Med}(\vec{x})$</span> satisfies <span class="math-container">$$\mathrm{Med}(\vec{x}) = ||\vec{x}|| \mathrm{Med}(\hat{x})$$</span> where <span class="math-container">$\hat{x} = \frac{\vec{x}}{||\vec{x}||}$</span> is the corresponding unit vector.</p>
<p>For Gaussians, the radius and unit vector are independent, so the calculation factors:
<span class="math-container">$$\mathbb{E}_{\vec{x} \sim N(\vec{0}, I)}[\mathrm{Med}(\vec{x})^2] = \mathbb{E}_{r,\ \hat{x} \sim \mathrm{unif}(S^{n-1})}[r^2 \mathrm{Med}(\hat{x})] = \mathbb{E}_r[r^2] \mathbb{E}_{\hat{x} \sim \mathrm{unif}(S^{n-1})}[\mathrm{Med}(\hat{x})^2].$$</span></p>
<p>Here <span class="math-container">$r$</span> is just the norm of <span class="math-container">$\vec{x}$</span> so <span class="math-container">$\mathbb{E}_r[r^2] = \mathbb{E}_{\vec{x}}[||\vec{x}||^2] = n$</span>. The distribution of <span class="math-container">$\hat{x}$</span> is uniform on the sphere by symmetry.</p>
<p>I don't know how to calculate this integral over the sphere though. You can do obvious things like dividing by <span class="math-container">$n!$</span> and restricting to the sector <span class="math-container">$U_0 = \{x_1 \leq x_2 \leq \cdots \leq x_n\} \subset S^{n-1}$</span>:
<span class="math-container">$$\mathbb{E}_{\hat{x} \sim \mathrm{unif}(S^{n-1})}[\mathrm{Med}(\hat{x})^2] = n! \int_{\hat{x} \in U_0} (x_{(n+1)/2})^2 d\mu(\hat{x}),$$</span>
but already for <span class="math-container">$n=5$</span> Mathematica doesn't manage to compute the integral.</p>
|
2,307,785 | <p>Things I understand:</p>
<p>Shannon entropy- </p>
<ul>
<li>is the expected amount of information in an event from that
distribution. </li>
<li>In game of 20 questions to guess an item, it is the
lower bound on the number questions one could ask.</li>
</ul>
<p>Doubt:</p>
<p>It gives the lower bound on the number of bits needed on average to encode symbols drawn from a distribution.</p>
<p>I don't understand the why it mentions on <strong>average</strong>. Isn't it just a lower bound.
Also, please elaborate on the lower bound if possible.</p>
| spaceisdarkgreen | 397,125 | <p>To see why there needs to be a notion of lower bound, consider that you need to specify a code. You could choose an arbitrarily inefficient code that takes a ton of bits to encode anything. So this is a statement about how well you <em>can</em> do.</p>
<p>To see why it must be an average, note that in general the length of the encoded sequence of symbols will depend on what symbols actually occur (since each symbol in general will have a different number of bits for its code word.) Since the symbols are randomly drawn, the length of the code will be random.</p>
|
3,725,007 | <p>Consider the rings, <span class="math-container">$Z_2[x]/(1 + x^2)$</span> and <span class="math-container">$ Z_2[x]/(1 + x + x^2)$</span>, despite having different polynomial as divisor, I have been told that -</p>
<p><span class="math-container">$$Z_2[x]/(1 + x^2) = \{0, 1, x, 1 + x\}$$</span></p>
<p>and</p>
<p><span class="math-container">$$Z_2[x]/(1 + x + x^2)= \{0, 1, x, 1 + x\}$$</span></p>
<p>Then what is the difference between <span class="math-container">$Z_2[x]/(1 + x^2)$</span> and <span class="math-container">$ Z_2[x]/(1 + x + x^2)$</span>?</p>
| Oliver Díaz | 121,671 | <p>Let <span class="math-container">$G(x)=\int^x_0\frac{\sin t}{t}\,dt$</span>. If One proves that <span class="math-container">$|G(x)|\leq M$</span> for some <span class="math-container">$M>0$</span> and all <span class="math-container">$x\geq0$</span>, then</p>
<p><span class="math-container">$$|G(b)-G(a)|\leq |G(b)|+|G(a)|\leq 2M$$</span></p>
<p>Hint: <span class="math-container">$\lim_{x\rightarrow+\infty}G(x)$</span> exists (it is <span class="math-container">$\frac{\pi}{2}$</span> but the value for this exercise is not that important)</p>
<p>If you know this, then the problem is almost done. If you don't, then you may try to write <span class="math-container">$G(x)$</span> as sum of decreasing alternating terms just to show that indeed <span class="math-container">$\lim_{x\rightarrow\infty}G(x)$</span> exists. For instance
for <span class="math-container">$n\pi\leq x<(n+1)\pi$</span>,</p>
<p><span class="math-container">$$G(x)=\int^x_{n\pi}\frac{\sin t}{t}\,dt +\sum^n_{k=1}\int^{n\pi}_{(n-1)\pi}\frac{\sin t}{t}\,dt
$$</span></p>
<p>The first term is bounded by <span class="math-container">$\pi$</span>, the second is a sum of terms <span class="math-container">$\sum^n_{k=1}a_k$</span> which alternate in sign but that in absolute value <span class="math-container">$|a_n|\xrightarrow{n\rightarrow\infty}0$</span>.</p>
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2,853,278 | <p><a href="https://math.stackexchange.com/a/203701/312406">This answer</a> suggests the idea, that a local ring $(R, \mathfrak{m})$ whose maximal ideal is nilpotent is in fact an Artinian ring.</p>
<p>Is this true? If so, how is it proven?</p>
| Louis | 75,278 | <p>You need that $R$ is noetherian, else there are counterexamples. </p>
<p>E.g., take $R = K[x_i]_{i \in \mathbb{N}}/(x_i | i \in \mathbb{N})^2$.</p>
<p>If $R$ is Noetherian, this is a special case of <a href="https://stacks.math.columbia.edu/tag/00KD" rel="noreferrer">Lemma 10.59.4 here</a> which states that a Noetherian ring of dimension $0$ is Artinian.</p>
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