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 |
|---|---|---|---|---|
108,409 | <p>Given a commutative ring <span class="math-container">$A$</span> with unity, Grothendieck used universal polynomials to define a <em>special</em> <span class="math-container">$\lambda$</span>-ring structure on <span class="math-container">$\Lambda(A):=1+t\:A[[t]]$</span>. Suppose <span class="math-container">$A$</span> is graded, say <span class="math-container">$A=\bigoplus_{i=0}^\infty A_i$</span>. In <a href="https://doi.org/10.1007/978-1-4757-1858-4" rel="nofollow noreferrer"><strong>Riemann–Roch Algebra</strong></a>, p. 11, Fulton and Lang define <span class="math-container">$\Lambda^{\circ}(A):=\{1+a_1t+a_2t^2\dotsb\mid a_i\in A_i\}$</span>. Then on page 15 they state that since the product and <span class="math-container">$\lambda$</span> operations of <span class="math-container">$\Lambda(A)$</span> take <span class="math-container">$\Lambda^\circ(A)$</span> to itself, <span class="math-container">$\Lambda^\circ(A)$</span> becomes a <span class="math-container">$\lambda$</span>-ring (without unit). They use this <span class="math-container">$\lambda$</span>-ring structure of <span class="math-container">$\Lambda^\circ(A)$</span> in the proof of Theorem 3.1 on p. 16.</p>
<p>However, a straightforward computation shows that the product in <span class="math-container">$\Lambda(A)$</span> does <em>not</em> take <span class="math-container">$\Lambda^\circ(A)$</span> to itself. For example, if <span class="math-container">$1+a_1t+a_2t^2\dotsb$</span> and <span class="math-container">$1+b_1t+b_2t^2\dotsb$</span> are elements in <span class="math-container">$\Lambda^\circ(A)$</span>, then their product using the product of <span class="math-container">$\Lambda(A)$</span> is given by <span class="math-container">$1+P_1(a_1;b_1)t+P_2(a_1,a_2;b_1,b_2)t^2+\dotsb$</span>, where <span class="math-container">$P_1,P_2,\dotsc$</span> are certain universal polynomials. But <span class="math-container">$P_1(a_1;b_1)$</span> turns out to be <span class="math-container">$a_1b_1$</span> (<a href="https://docs.google.com/viewer?a=v&pid=sites&srcid=ZGVmYXVsdGRvbWFpbnxkYXJpamdyaW5iZXJnfGd4OjEwNzljZThlNDcwMGE5YmU" rel="nofollow noreferrer">see here</a>, p. 22) and <span class="math-container">$a_1b_1$</span> is not in <span class="math-container">$A_1$</span>, which shows the product is not in <span class="math-container">$\Lambda^\circ(A)$</span>.</p>
<p><strong>Question.</strong> Is there an error in the book? If yes, can it be fixed?</p>
<p><strong>Edit.</strong> If you know other errors in this book that one should be aware of, please share them here.</p>
| Maarten Bergvelt | 26,886 | <p>Hazewinkel in <a href="https://arxiv.org/abs/0804.3888" rel="nofollow noreferrer">Witt vectors. Part 1</a> warns about an error on page 15, second paragraph of this book. In fact he advises to "steer clear" of the book!</p>
|
1,176,615 | <p>I am invited to calculate the minimum of the following set:</p>
<p>$\big\{ \lfloor xy + \frac{1}{xy} \rfloor \,\Big|\, (x+1)(y+1)=2 ,\, 0<x,y \in \mathbb{R} \big\}$.</p>
<p>Is there any idea?</p>
<p>(The question changed because there is no maximum for the set (as proved in the following answers) and I assume that the source makes mistake)</p>
| Ryan Vitale | 95,806 | <p>Solve for one variable using the equation $(x+1)(y+1)=2$. Solving for $x$ we get $x=\frac{1-y}{1+y}$. Then rewrite the floor function as $\lfloor{xy+\frac{1}{xy}}\rfloor=\lfloor{\frac{y^2(1-y)^2+(1+y)^2}{y(1-y^2)}}\rfloor$, so there is no max, taking $y \rightarrow 1$</p>
|
1,146,759 | <p>A covering of a group $G$ a family $\{S_i\}_{i \in I}$ of subsets of $G$ such that $G = \displaystyle \bigcup _{i \in I} S_i$.</p>
<p>Why is true that: A group covered by finitely many cyclic subgroups is either cyclic or finite?</p>
<p>Remark:
Is true that by Baer (see D. Robinson, Finiteness Conditions and Generalized Soluble Groups, p.105) that:
Theorem 4.16: A group is central-by-finite if and only if it has a finite covering consisting of abelian subgroups?</p>
| Derek Holt | 2,820 | <p>Here are some hints. I denote a cyclic group of order $n$ by $C_n$.</p>
<p>Step 1. First show that $C_{\infty} \times C_n$ is not covered by finitely many cyclic groups for any $n$, including $n=\infty$.</p>
<p>Note that the condition of being covered by finitely many cyclic ubgroups is inherited by subgroups. By the Baer result, for any such group $G$, $G/Z(G)$ is finite, and $Z(G)$ is clearly finitely generated, so by Step 1, $Z(G)$ is either finite, in which case we are done, or equal to $C_\infty$.</p>
<p>Now by a result of Schur, $[G,G]$ is finite, so it has trivial intersection with $Z(G)$. If $1 \ne g \in [G,G]$ then $\langle Z(G),g \rangle = Z(G) \times \langle g \rangle \cong C_\infty \times C_n$ for some $n$, contradicting Step 1. So $[G,G]=1$ and $G=Z(G) = C_\infty$.</p>
|
879,640 | <p>Does a matrix have only one inverse matrix (like the inverse of an element in a field)? If so, does this mean that</p>
<p>$A,B \text{ have the same inverse matrix} \iff A=B$?</p>
| Michael Albanese | 39,599 | <p>Note that $GL(n, \mathbb{F})$, the set of invertible $n\times n$ matrices over the field $\mathbb{F}$, is a group. In any group, inverses are unique, so if $a^{-1} = b^{-1}$, by taking inverses it follows that $a = b$. In particular, this applies to the group $GL(n, \mathbb{F})$.</p>
|
3,845,475 | <p>Here's what I'm tasked with showing:</p>
<p>Let <span class="math-container">$(a_n)$</span> be a convergent sequence with <span class="math-container">$a_n\rightarrow a$</span> as <span class="math-container">$n\rightarrow\infty$</span>. By the Algebraic Limit Theorem, we know that <span class="math-container">$(a_n^2)\rightarrow a^2$</span>. Now prove this using the definition of convergence.</p>
<p>In doing so, I have the following:</p>
<p>Let <span class="math-container">$\epsilon>0$</span> be given. Choose some <span class="math-container">$N\in\mathbb{N}$</span> so that for all <span class="math-container">$n\geq N$</span>, <span class="math-container">$|a_n^2-a^2|<\epsilon$</span>. By algebra, <span class="math-container">$|a_n^2-a^2|=|a_n-a||a_n+a|$</span>. Consider <span class="math-container">$|a_n+a|$</span>. By the triangle inequality, <span class="math-container">$|a_n+a|\leq|a_n|+|a|$</span>, thus <span class="math-container">$|a_n|$</span> is bounded by some <span class="math-container">$M\in\mathbb{N}$</span>.</p>
<p>I know I'm trying to choose <span class="math-container">$M$</span> so that <span class="math-container">$|a_n-a|<\epsilon+M+|a|$</span>. Where should I take it from here?</p>
| fleablood | 280,126 | <blockquote>
<p>I don't see how this (sentence in italics) can be true. For example if we have S={1,2,3} the number of orderings that can be obtained are 3!=6. Following the solution's reasoning we could calculate the orderings for S by ordering the cards different form 3 and then inserting in into that ordering, that is 2! . What am I missing? perhaps the sentence in italics does not mean what I think it does?</p>
</blockquote>
<p>You are missing that there are <span class="math-container">$3$</span> positions to put the number <span class="math-container">$3$</span> is so there are <span class="math-container">$2! \times 3$</span> results.</p>
<p>Your example:</p>
<p>Shuffle the three cards and there are <span class="math-container">$3!=6$</span> options. They are <span class="math-container">$1,2,3|1,3,2|2,1,3|2,3,1|3,1,2|3,2,1$</span>.</p>
<p>Now do it the book's way.</p>
<p>Shuff the <span class="math-container">$1,2$</span>. There are <span class="math-container">$2!$</span> ways to do it. <span class="math-container">$1,2$</span> or <span class="math-container">$2,1$</span>.</p>
<p>Now put the <span class="math-container">$3$</span> into the deck. You seem to think there is only one way to do that but there are <span class="math-container">$3$</span> positions it can be placed in. If you shuffled <span class="math-container">$1,2$</span> as <span class="math-container">$a,b$</span> you can put the <span class="math-container">$3$</span> if the first position; <span class="math-container">$3,a,b$</span> or the second; <span class="math-container">$a,3,b$</span>; or the third <span class="math-container">$a,b,3$</span>.</p>
<p>So the total number of ways is <span class="math-container">$2!\times 3=6$</span>.</p>
<p>They are: if <span class="math-container">$1,2$</span> is shuffled as <span class="math-container">$1,2$</span> then <span class="math-container">$3,1,2|1,3,2|1,2,3$</span>. And if <span class="math-container">$1,2$</span> is shuffled as <span class="math-container">$2,1$</span> then <span class="math-container">$3,2,1|2,3,1|2,1,3$</span>.</p>
<blockquote>
<p>We have 3 other aces so we put AiAs, with i=c,d,h, together as one unit</p>
</blockquote>
<p>This tells allows that the <span class="math-container">$A_s$</span> con procede any other ace. BUt you did <em>not</em> specify that that ace is the first <span class="math-container">$A$</span>. If we have <span class="math-container">$A_cA_s$</span> we have only a <span class="math-container">$\frac 13$</span> chance that <span class="math-container">$A_c$</span> is the <em>first</em> ace. If <span class="math-container">$A_c$</span> is the second or third it doesn't count.</p>
<p>So the result your way would be <span class="math-container">$\frac {\frac {3\cdot 51!}3}{52!}$</span>.</p>
<p>And for the <span class="math-container">$2$</span> of clubs you'd have <span class="math-container">$\frac {\frac {4\cdot 51!}4}{52!}$</span>.</p>
|
1,859,810 | <p>Consider the function
$$
f(x)=\sum_{n=1}^{\infty}\frac{\mathrm{sign}\left(\sin(nx)\right)}{n}\, .
$$
This is a bizarre and fascinating function. A few properties of this function that SEEM to be true:</p>
<p>1) $f(x)$ is $2\pi$-periodic and odd around $\pi$.</p>
<p>2) $\lim_{x\rightarrow \pi_-} f(x) = \ln 2$. (Can be proven by letting $x = \pi-\epsilon$, expanding the sine function, and taking the limit as $\epsilon\rightarrow 0$.)</p>
<p>3) $\int_0^{\pi}dx\, f(x) = \frac{\pi^3}{8}$ (Can be "proven" by integrating each $\mathrm{sign}\left(\sin(nx)\right)$ term separately. Side question: Is such a procedure on this jumpy function even meaningful?)</p>
<p>All of this despite the fact that I can't really prove that this function converges anywhere other than when $x$ is a multiple of $\pi$!</p>
<p>A graph of this function (e.g. in Mathematica) reveals an amazing fractal shape. My question: What is the fractal dimension of the graph of this function? Does the answer depend on which definition of fractal dimension we use (box dimension, similarity dimension, ...)?</p>
<p>This question doesn't come from anywhere other than from my desire to see if an answer exists.</p>
<p>As requested, a plot of this function on the range $x\in[0,2\pi]$:</p>
<p><a href="https://i.stack.imgur.com/LHvgb.png" rel="noreferrer"><img src="https://i.stack.imgur.com/LHvgb.png" alt="Graph of fractal function"></a>
Edited to add:</p>
<p>Other, perhaps more immediate, questions about this function:</p>
<p>1) Does it converge? Conjecture: It converges whenever $x/\pi$ is irrational, but doesn't necessarily diverge if $x/\pi$ is rational. See, e.g., $x = \pi$, where it converges to zero, and apparently to $\pm \ln 2$ on either side of $x = \pi$.</p>
<p>2) I would guess that it diverges as $x\rightarrow 0_+$. How does it diverge there? If this really is a fractal function, I would suppose that the set of points where it diverges is dense. For instance, it appears to have a divergence at $x = 2\pi/3$.</p>
<p>Edit 2:</p>
<p>Another thing that's pretty straightforward to prove is that:
$$
\lim_{x\rightarrow {\frac{\pi}{2}}_-} f(x) = \frac{\pi}{4} + \frac{\ln 2}{2}
$$
and
$$
\lim_{x\rightarrow {\frac{\pi}{2}}_+} f(x) = \frac{\pi}{4} - \frac{\ln 2}{2}
$$</p>
<p>Final Edit:</p>
<p>I realize now that the initial question about this function - what is its fractal dimension - is (to use a technical term) very silly. There are much more immediate and relevant question, e.g. about convergence, etc. I've already selected one of the answers below as answering a number of these questions.</p>
<p>One final point, for anyone who stumbles on this post in the future. The term $\mathrm{sign}(\sin(nx))$ is actually a <a href="https://en.wikipedia.org/wiki/Square_wave" rel="noreferrer">square wave</a>, and so we can use the usual Fourier series of a square wave to derive an alternate way of expressing this function:
$$
f(x)=\sum_{n=1}^{\infty}\frac{\mathrm{sign}\left(\sin(nx)\right)}{n}
= \frac{4}{\pi}\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}
\frac{\sin(n(2m-1)x)}{n(2m-1)}
$$
By switching the order of the sums and doing the $n$ sum first, this could also be represented as a weighted sum of <a href="https://en.wikipedia.org/wiki/Sawtooth_wave" rel="noreferrer">sawtooth waves</a>.</p>
| John Barber | 73,626 | <p>Although I've already selected one of the excellent answers above, I thought I'd post an answer that I figured out to one of the questions I posed in the original post, namely: "I would guess that it diverges as $x\rightarrow 0_+$. How does it diverge there?"</p>
<p>So, as noted above, the term $\mathrm{sign}(\sin(n x))$ is a square wave with period $2\pi/n$. Suppose $x/\pi$ is irrational. Then $\mathrm{sign}(\sin(n x))$ is always -1 or +1, never 0. Define
$$
\phi(x,k) = \left\lfloor \frac{k\pi}{x} \right\rfloor
$$
as the largest integer such that $\phi(x,k) \, x < k \pi$. Defining $H_m$ as the $m^{\mathrm{th}}$ harmonic number, the original series can then be divided into finite positive and negative subsequences of terms of the form $\pm 1/n$ and rewritten as
\begin{align}
f(x) &=
\sum_{n = 1}^{\phi(x,1)} \frac{1}{n}
\;-\;\sum_{n = \phi(x,1)+1}^{\phi(x,2)} \frac{1}{n}
\;+\;\sum_{n = \phi(x,2)+1}^{\phi(x,3)} \frac{1}{n} \;-\; ...\\
&= H_{\phi(x,1)} - \left(H_{\phi(x,2)} - H_{\phi(x,1)}\right)+ \left(H_{\phi(x,3)} - H_{\phi(x,2)}\right) - ...\\
&= H_{\phi(x,1)} \;+\; \sum_{n = 1}^{\infty}{(-1)}^n\left(H_{\phi(x,n+1)} - H_{\phi(x,n)}\right)
\end{align}
for irrational $x/\pi$. In fact, this last expression can be used to calculate the original function. Although this harmonic number representation is ostensibly only valid for irrational $x/\pi$, it seems to overlay the original function very nicely, as shown below:</p>
<p><a href="https://i.stack.imgur.com/sRCLE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sRCLE.png" alt="Plot of original function overlaid by the function as represented by a series of harmonic number terms"></a>
Now, as to the divergence for small $x$. When $x$ is small, $\phi(x,n)$ becomes large, and we have
\begin{align}
H_{\phi(x,n)}&\approx \ln\left(\phi(x,n)\right) + \gamma\\
&= \ln\left(\left\lfloor \frac{n\pi}{x} \right\rfloor\right) + \gamma\\
&\approx \ln\left(\frac{n\pi}{x}\right) + \gamma\, .
\end{align}
The above expression for $f(x)$ then reduces to
$$
f(x) \approx \ln\left(\frac{\pi}{x}\right) + \gamma +
\sum_{n=1}^{\infty}{(-1)}^n \ln\left(1 + \frac{1}{n}\right)\, ,
$$
where
\begin{align}
\sum_{n=1}^{\infty}{(-1)}^n \ln\left(1 + \frac{1}{n}\right)
&= \ln\left(\prod_{n=1}^{\infty}\left(1 - \frac{1}{4n^2}\right)\right)\\
&= \ln\left(\frac{2}{\pi}\right)\qquad \text{By the Wallis product}\, .
\end{align}
The small-$x$ approximation then reduces to
$$
f(x) \approx -\ln x \;+\; \gamma \;+\; \ln 2\, .
$$
This is plotted below along with the original function.
<a href="https://i.stack.imgur.com/swqxI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/swqxI.png" alt="Original function along with small-x log approximation"></a>
So the answer is: A logarithmic divergence as $x\rightarrow 0_+$.</p>
|
535,757 | <p>The exercise is to give an example for two sets $M$ and $N$, and functions $f$ and $g$, for which $f \circ g = id_M$, but $g \circ f \ne id_N$.</p>
<p>My idea is a bit based on my computer programming background, where <code>(x/2)*2</code> is <code>0</code> for integers. Here it is:</p>
<p>$$M=N=\mathbb{N_0}.$$
$$f: \mathbb{N_0} \rightarrow \mathbb{N_0}, g: \mathbb{N_0} \rightarrow \mathbb{N_0}.$$
$$f(x) = 2x, g(x) = x \div 2.$$</p>
<p>Thus,</p>
<p>$$(g\circ f)(x) = g(f(x)) = g(2x) = 2x\div 2 = x = id_\mathbb{N_0}$$
$$(f\circ g)(x) = f(g(x)) = f(x\div2) = (x\div2)\cdot2 \ne id_\mathbb{N_0}$$</p>
<p>Because if $f(g(1)) = f(0) = (0\div 2)\cdot2 = 0$, and $0\ne 1$, it follows that $(f\circ g) \ne id_\mathbb{N_0}$</p>
<p>Is my example valid?</p>
| Clive Newstead | 19,542 | <p>A simpler example might be $M=[0,\infty)$ and $N = \mathbb{R}$ with $f(x) = \sqrt{x}$ and $g(x)=x^2$. You can check the details.</p>
|
2,721,836 | <p>I recently found a different method to compute prime number in $\mathcal O(\log(\log n))$ complexity. At present, that logic working fine for $300$ digits prime number, which I found on websites.I need to validate whether that logic will be working fine for a higher number of digits. At present, I have computed a prime number of $300\ 000$ digits(but I am not sure whether this would be valid),</p>
<p>My questions are:</p>
<ul>
<li>Where can I find a prime number of higher digits i.e., more than
$300\ 000$ digits?</li>
<li>Where can I validate $300\ 000$ digit prime number is valid one?</li>
</ul>
| Klangen | 186,296 | <p><a href="http://www.ellipsa.eu/public/primo/primo.html" rel="nofollow noreferrer">Primo</a> does this. It only runs on Linux, if you don't have one at home you can just create a virtual machine and run it as guest within your current OS. There are various tests that Primo can perform, please read the documentation on the website in order to adapt it to your special case.</p>
<p>Please note that it will take <em>very long</em> to certify that a $3\times10^{6}$-digit number is prime. If you want to test your method on large numbers, take a <em>much</em> smaller one to begin with, i.e., with about $10^{3}$ or $10^{4}$ digits.</p>
|
1,137,565 | <blockquote>
<p>Let <span class="math-container">$A\Delta C\subseteq A\Delta B$</span>. (<span class="math-container">$\Delta$</span> denotes symmetric difference.)</p>
<p>Prove <span class="math-container">$A\cap B \subseteq C$</span>.</p>
</blockquote>
<p>I am getting ready for a test and I could really use proof verification and any help with this.</p>
<p><em>Proof</em>: Let us look at the indicators, <span class="math-container">$x_{A\Delta C}=x_A+x_C-2x_Ax_C$</span>, <span class="math-container">$x_{A\Delta B}=x_A+x_B-2x_Ax_B$</span>, <span class="math-container">$x_{A\cap B}=x_Ax_B$</span>.</p>
<p>Let <span class="math-container">$x_{A\cap B}(a)=1$</span>. Then <span class="math-container">$x_{A\Delta B}(a)=0$</span> which means <span class="math-container">$x_{A\Delta C}(a)=0$</span>. <span class="math-container">$x_A(a)=x_B(a)=1$</span> and therefore <span class="math-container">$x_C(a)$</span> must be 1. Therefore <span class="math-container">$x_{A\cap B}(a)=1\Rightarrow x_C(a)=1$</span> <span class="math-container">$\Rightarrow A\cap B \subseteq C$</span>.</p>
| Lucian | 93,448 | <blockquote>
<p><em>Where is the $x=+1$ point ?</em></p>
</blockquote>
<p>At infinity. Just let $t\to\infty$, and evaluate the two limits.</p>
|
4,341,172 | <p><span class="math-container">$(x+1)^2y''+(x+1)y'+y=x^2+2\sin(\ln(x+1)), y(0)=\frac{1}{5},y'(0)=2$</span></p>
<p>My solution:</p>
<p><span class="math-container">$y''+\frac{y'}{(x+1)}+\frac{y}{(x+1)^2}=\frac{x^2+2\sin(\ln(x+1))}{(x+1)^2}$</span></p>
<p>First of all , I found the equation solution of <span class="math-container">$y''+\frac{y'}{(x+1)}+\frac{y}{(x+1)^2}=0$</span></p>
<p><span class="math-container">$y=c_1\cos(\ln(x+1))+c_2\sin(\ln(x+1))$</span></p>
<p>I try to solve this ode using the variation of parameters theorem</p>
<p>Get this system equation:</p>
<p>(I)<span class="math-container">$c'_1\cos(\ln(x+1))+c'_2\sin(\ln(x+1))=0$</span></p>
<p>(II)<span class="math-container">$-c'_1\sin(\ln(x+1))+c'_2\cos(\ln(x+1))=x^2+2\sin(\ln(x+1))$</span></p>
<p>Multiply (I) by <span class="math-container">$\sin(\ln(x+1))$</span> , (II) by <span class="math-container">$\frac{\cos(\ln(x+1))}{x+1}$</span>.</p>
<p>By addtion i get:</p>
<p><span class="math-container">$c'_2=\frac{\cos(\ln(x+1))[x^2+2\sin(\ln(x+1))]}{(x+1)}$</span></p>
<p>I do not know how I get <span class="math-container">$c_2$</span> by an integral ?</p>
<p>Help please</p>
<p>Thanks !</p>
| user577215664 | 475,762 | <p><span class="math-container">$$(x+1)^2y''+(x+1)y'+y=x^2+2\sin(\ln(x+1))$$</span> <span class="math-container">$$y(0)=\frac{1}{5},y'(0)=2$$</span>
Substitute <span class="math-container">$u=x+1$</span>:
<span class="math-container">$$u^2y''+uy'+y=(u-1)^2+2\sin(\ln(u))$$</span> <span class="math-container">$$y(1)=\frac{1}{5},y'(1)=2$$</span>
Substitute <span class="math-container">$u=e^t$</span>:
<span class="math-container">$$y''(t)+y(y)=(e^t-1)^2+2\sin(t)$$</span>
This is easy to solve with variation of constants method.
<span class="math-container">$$y''(t)+y(t)=2 \sin t$$</span>
<span class="math-container">$$\implies y_p=At\cos t$$</span>
<span class="math-container">$$y''(t)+y(t)=e^{2t}-2e^t+1$$</span>
<span class="math-container">$$y_p=Ae^{2t}+Be^t+C$$</span></p>
<hr />
<p><strong>Edit:</strong></p>
<p>Note that:
<span class="math-container">$$\dfrac {dy}{du}=\dfrac {dy}{dt}\dfrac{dt}{du}=e^{-t} \dfrac {dy}{dt}=\dfrac 1u\dfrac{dy}{dt}$$</span></p>
<p>Then <span class="math-container">$$u\dfrac {dy}{du}=\dfrac {dy}{dt}$$</span>
Do the same for:
<span class="math-container">$$\dfrac {d^2y}{du^2}$$</span>
Then the DE is simplified and you find that :
<span class="math-container">$$y''(t)+y(t)=(e^t-1)^2+2\sin(t)$$</span></p>
|
4,580,470 | <p>Suppose <span class="math-container">$X$</span> is a Geometric random variable (with parameter <span class="math-container">$p$</span> and range <span class="math-container">$\{k\geq 1\}$</span>).</p>
<p>Let <span class="math-container">$M$</span> be a positive integer.</p>
<p>Let <span class="math-container">$Z:=\min\{X,M\}$</span>. We want to calculate the expectation <span class="math-container">$\mathbb E[Z]$</span>.</p>
<hr />
<p>My professor solved the problem by starting with <span class="math-container">$$\mathbb E[Z]=\sum_{k=1}^\infty \mathbb P[Z\geq k].$$</span></p>
<p>I can understand all other steps in his method except this starting step. Why shouldn't we have <span class="math-container">$\mathbb E[Z]=\sum_{k=1}^\infty k\mathbb P[Z= k]$</span> by the definition of expectation instead?</p>
<p>I am really confused by this starting step. Any help with understanding the step <span class="math-container">$\mathbb E[Z]=\sum_{k=1}^\infty \mathbb P[Z\geq k]$</span> will be really appreciated. Thanks!</p>
| whoisit | 1,094,230 | <p>They are actually equal:
<span class="math-container">$\sum_{k=1}^\infty k\mathbb P[Z= k] = \sum_{k=1}^\infty \mathbb P[Z\geq k] $</span></p>
<p>Consider this sum:</p>
<p><span class="math-container">$
+ \; P(1) \\
+P(2) + P(2) \\
+P(3) + P(3) + P(3) \\
\vdots
$</span></p>
<p>The rowwise grouping of this sum is the LHS, while the columnwise grouping is the RHS.</p>
|
2,757,562 | <p>I'm work through some questions relating to connection coefficents. My question is more about the summation notation being used. Why is there no starting point (or end point) for the summation here? For the i, j, k, I take these to range from 1 to 2 for a 2 dimensional surface. Does this just imply that l does likewise? </p>
<p>$$ \Gamma^i_{jk} = \sum_{l} g^{il} \left( \dfrac{\partial g_{lk}}{\partial x^j} + \dfrac{\partial g_{jl}}{\partial x^k} - \dfrac{\partial g_{jk}}{\partial x^l} \right) $$</p>
| Lee Mosher | 26,501 | <p>This formula comes from differential geometry, it takes place in the Cartesian coordinate space $\mathbb{R}^n$ for some value of the dimension $n$. Once $n$ has been specified, the indices $i,j,k,l$, by convention, take values in the index set $\{1,...,n\}$. </p>
<p>In this formula, the values $i,j,k$ are fixed and $l$ runs freely over the whole index set. So the summation symbol $\displaystyle\sum_l$, by convention, is an abbreviation for
$$\sum_{l=1}^n
$$
Whatever differential geometry text you read, there are likely to be hundreds of such summations, and there will be conventions for writing them in abbreviated forms.</p>
|
1,552,055 | <p>Why is it true that all irrational numbers are non-terminating/non-repeating decimals?</p>
<p>By definition, an irrational number is one that can't be expressed as a ratio of integers.</p>
| Hagen von Eitzen | 39,174 | <ul>
<li>If the decimal expansion of a number $x$ is terminating, with $n$ digits after the decimal point, say, then $10^nx$ is an integer $m$ (the decimal expansion is shifted by $n$ places to the left, hence has nothing after the decimal point) so that $x =\frac{m}{10^n}$ is a fraction of integers, aka. rational number.</li>
<li>If the decimal expansion of a number $x$ is eventually repeating, with a epriod of length $n\ge1$, say, then $10^nx$ has a decimal expansion that matches that of $x$ beyond some point, so that in computing $10^nx-x$ everything cancels beyond some point, i.e., $10^nx-x$ is a rational $\frac ab$ per first bullet point. Solving for $x$ we find $x=\frac{a}{(10^n-1)b}$, which is again a fraction of integers.</li>
<li>If $x$ is the ratio/fraction of integers, then <em>by definition</em> it is rational.</li>
</ul>
<p>Hence for an <em>irrational</em> number (by definition a number that is <em>not</em> a rational number) none of the three options above can occur.</p>
|
813,716 | <p>I am supposed to calculate the following as simple as possible.</p>
<p>Calcute:
$$1 + 101 + 101^2 + 101^3 + 101^4 + 101^5 + 101^6 + 101^7$$
Tip: $$ 101^8 = 10828567056280801$$</p>
<p>I have absolutely no idea how this tip is supposed to help me.<br>
Do I still have to calculate each potency?<br>
Can I somehow solve it with 101^7 * (1 + 101) = 10828567056280801?</p>
<p>As I am not allowed to use a calculator a simple technique for formulas like the one above would be welcome.</p>
| Michael Albanese | 39,599 | <p><strong>Hint:</strong> $$1 + x + x^2 + \dots + x^n = \frac{x^{n+1} - 1}{x - 1}.$$</p>
|
3,488,405 | <blockquote>
<p>Let <span class="math-container">$L(n)$</span> denote the number of positive divisors of a number <span class="math-container">$n$</span>. Prove that <span class="math-container">$\sum_{n=1}^N L(n)=\lfloor{\sqrt N}\rfloor\pmod 2$</span>.</p>
</blockquote>
<p>I wanted to prove that by induction.
For <span class="math-container">$n=1$</span> True.
Assume True for some N.
Now we have <span class="math-container">$N+1$</span>.
I observed that <span class="math-container">$L(n)=n-\phi(n)+1$</span> where phi is Euler's function.
And what now... If <span class="math-container">$N+1$</span> is a square, then <span class="math-container">$\lfloor\sqrt N\rfloor\ne\lfloor\sqrt{N+1}\rfloor\pmod 2$</span>, whereas they are the same modulo 2 if <span class="math-container">$N+1$</span> is not a square.
When we add <span class="math-container">$N+1$</span>th element to the sum, we add <span class="math-container">$N+1-\phi(N+1)+1$</span> which modulo 2 is <span class="math-container">$N-\phi(N+1)$</span>.
So, the function <span class="math-container">$k(n)=n-\phi(n+1)$</span> should be odd if <span class="math-container">$n+1$</span> is square, and even if it is not square.
But, I opened table of Euler's phi function and that function seems to be odd when n is odd and even when n is even - I looked all naturals < 500.
So, where am I wrong? </p>
<p>I would be also happy to see non-induction solution, because it is probably better way. I just cant imagine from where did that sqrt(N) came from when I think about that sum.</p>
<p>Thanks in advance for help. I will probably answer for comments tomorrow.</p>
| Will Jagy | 10,400 | <p>Your <span class="math-container">$L(n)$</span> can be odd only when <span class="math-container">$n$</span> itself is a perfect square. So, mod 2, you are just counting the squares up to the upper bound, which you called <span class="math-container">$N$</span>.</p>
<p>First, <span class="math-container">$L(1) = 1.$</span> If <span class="math-container">$n> 1,$</span> it has a prime factorization, say
<span class="math-container">$$ n = p_1^{a_1} p_2^{a_2} \cdots p_r^{a_r} $$</span>
with exponents <span class="math-container">$a_j \geq 1$</span> and <span class="math-container">$r$</span> primes with <span class="math-container">$r \geq 1.$</span> Then
<span class="math-container">$$ L(n) = (a_1+1) (a_2 + 1) \cdots (a_r + 1). $$</span>
If any of the <span class="math-container">$a_j$</span> is odd, then <span class="math-container">$a_j + 1$</span> is even, so <span class="math-container">$L(n)$</span> is even. That is the most frequent outcome. It is possible to have <span class="math-container">$L(n)$</span> odd only when all the <span class="math-container">$(a_j + 1)$</span> are odd, meaning all the <span class="math-container">$a_j$</span> are even, meaning <span class="math-container">$n$</span> is a perfect square. </p>
|
337,930 | <p>Given two polynomials</p>
<p>$$
p(x) = a_0 + a_1 x + a_2 x^2 + \ldots + a_{n-1}x^{n-1} \\
q(x) = b_0 + b_1 x + b_2 x^2 + \ldots + b_{n}x^{n}
$$</p>
<p>And the series expansion from their rational polynomial</p>
<p>$$
\frac{p(x)}{q(x)} = c_0 + c_1 x + c_2 x^2 + \ldots
$$</p>
<p>is it possible to recover the the original polynomials $a_n$, $b_n$ from only the series $c_n$ via the solution of a linear system? </p>
| Abhra Abir Kundu | 48,639 | <p>Suppose you have to select n balls from a collection of $R$ black balls and $M$ white balls.</p>
<p>Then we must select $k$ black balls and $n-k$ white balls in whatever way we do.(for $0\le k\le n$)</p>
<p>For a fixed $k\in N,0\le k\le n$ we can do this in $\binom{R}{k}\binom{M}{(n-k)}$ ways.</p>
<p>so to get the total no. of ways we must add the above for all $k:0\le k\le n$ </p>
<p>So we have the total no. of ways $=\displaystyle\sum_{k=0}^{n} \binom{R}{k}\binom{M}{(n-k)}$.</p>
<p>But if we think about it in a different way we can say that we have to select $n$ balls from a collection of $R+M$ balls and this can be done in $\displaystyle \binom{R+M}{n}$ ways.</p>
<p>So ,</p>
<p>$$\displaystyle\sum_{k=0}^{n} \binom{R}{k}\binom{M}{(n-k)}=\displaystyle \binom{R+M}{n}$$</p>
|
622,076 | <p>Continuity $\Rightarrow$ Intermediate Value Property. Why is the opposite not true? </p>
<p>It seems to me like they are equal definitions in a way. </p>
<p>Can you give me a counter-example? </p>
<p>Thanks</p>
| Betty Mock | 89,003 | <p>Let $f(x) = x^2\sin(1/x)$ on $(0,1)$ and $f(0) = 0$. Then $f$ is differentiable throughout $[0,1]$. All derivatives satisfy the intermediate value property (Darboux's Theorem); but $f'(x)$ is discontinuous at $0$.</p>
|
3,785,982 | <p>Given the following ODE,</p>
<p><span class="math-container">$$\frac{{dy}}{{dx}}=\cos ({x})-\sin ({y})+{x}^{2}; \quad {y}\left({x}_{0}=-1\right)=y_0=3$$</span></p>
<p>I have to use the Taylor Series Method to compute the value of <span class="math-container">$y(x)$</span> at <span class="math-container">$x=-0.8$</span> with a Taylor's polynomial of second-order, with <span class="math-container">$h=x-x_0=0.1$</span>.</p>
<p>Considering all this, how should this method be applied for solving this problem?</p>
<hr />
<p><strong>My attempt at a solution.</strong></p>
<p>I'm not sure if this is the correct way to apply the method, but I have written Taylor's second-order polynomial centred at <span class="math-container">$x_0=-1.0$</span>:</p>
<p><span class="math-container">$$y(x) \approx y\left(x_{0}\right)+\left(x-x_{0}\right) y^{\prime}(x_0,y_0)+\frac{1}{2}\left(x-x_{0}\right)^{2} y^{\prime \prime}(x_0,y_0)=
\\=3.0+1.39918(x+1)+0.11333(x+1)^2 $$</span></p>
<p>And I have evaluated this <span class="math-container">$y(x)$</span> at <span class="math-container">$x=-0.8$</span>, so <span class="math-container">$y(-0.8) \approx 3.28437$</span>.</p>
<p>However, this doesn't match my textbook's solution, <span class="math-container">$3.2850$</span>, neither Wolfram-Alpha's one, <span class="math-container">$3.28687$</span>.</p>
<p>Would the method be applied this way or am I missing something?</p>
| PrincessEev | 597,568 | <p>Note that, if <span class="math-container">$$x_n = \left( \frac{n+2}{n} \right)^{n^2} = \frac{(n+2)^{n^2}}{n^{n^2}}$$</span></p>
<p>then</p>
<p><span class="math-container">$$\frac{x_{n+1}}{x_n} = \underbrace{\frac{(n+3)^{(n+1)^2}}{(n+1)^{(n+1)^2}}}_{\displaystyle x_{n+1}} \cdot \underbrace{\frac{n^{n^2}}{(n+2)^{n^2}}}_{\displaystyle 1/x_n}$$</span></p>
<p>as opposed to your calculation. (Also, <span class="math-container">$r$</span> should be <span class="math-container">$1$</span> divided by the limit you calculated.)</p>
|
1,262,305 | <p>$$f(x)=\int_0^x\left(\frac{t^3-2t^2-4}{t^2+1}\right)\ dt$$
I need to find the $x$ and $y$ intercepts, and the inflection points of the function $f(x)$ (with both $x$ and $y$ coordinates). I need to find it through the calculator and explain my answer. How do I find the antiderivative?</p>
| Olivier Oloa | 118,798 | <p><strong>Hint.</strong> You may observe that, by partial fraction decomposition, you have
$$
\frac{t^3-2t^2-4}{t^2+1}=t-2-\frac{t+2}{t^2+1}
$$ giving, for the antiderivative that vanishes at $x=0$:
$$
\begin{align}
f(x)=\int_0^x\frac{t^3-2t^2-4}{t^2+1}dt&=\int_0^x\left(t-2-\frac{t+2}{t^2+1}\right)dt\\\\
&=\int_0^x\left(t-2\right)dt-\int_0^x\frac{t}{t^2+1}dt-2\int_0^x\frac{1}{t^2+1}dt\\\\
&=\frac{x^2}2-2x-\frac12\ln(x^2+1)- 2\arctan x.
\end{align}
$$
<strong>Remark.</strong> Here the constant of integration is equal to $0$ since $ \int_0^x$ is equal to $0$ when $x=0$.</p>
|
279,043 | <p>I would like to plot a complex graph of the Riemann zeta function on the Argand diagram <span class="math-container">$ς(s)$</span>, where <span class="math-container">$s = \frac{1}{2} + i t $</span>, and the value of <span class="math-container">$t$</span> is varied to get a graph in the polar form.</p>
<p>Can anyone help, please?</p>
| Nasser | 70 | <p><a href="https://i.stack.imgur.com/Pogqu.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Pogqu.gif" alt="enter image description here" /></a></p>
<pre><code>ClearAll["Global`*"]
s[t_] := 1/2 + I*t
Manipulate[
tick;
Module[{x, y},
x = Re[Zeta[s[t]]]; y = Im[Zeta[s[t]]];
AppendTo[coord, {x, y}];
ListLinePlot[coord, AxesLabel -> {"Real part", "Imaginary part"},
PlotRange -> {{-3, 3}, {-3, 3}}, GridLines -> Automatic,
GridLinesStyle -> LightGray, PlotStyle -> Red]
],
Grid[{{Button[
Text@Style["clear", 12], {coord = {}; tick = Not[tick]},
ImageSize -> {60, 40}]}}],
{{t, 0, "t"}, 0, 20, .01, Appearance -> "Labeled"},
{{coord, {}}, None},
{{tick, False}, None},
TrackedSymbols :> {t, tick}
]
</code></pre>
<blockquote>
<p>when i change the value of "t", the graph is not smooth at the
beginning for a few values of t,</p>
</blockquote>
<p>This is just how Zeta function is. You can see this more clearly by adding <code>Mesh -> All</code> to the above Plot command, which will now show the actual <span class="math-container">$x+i y$</span> points used and just the line that connects them.</p>
<p>It will show this</p>
<p><img src="https://i.stack.imgur.com/cvYP3.png" alt="Mathematica graphics" /></p>
<p>You see, that for small <span class="math-container">$t$</span>, the points are more spread than for larger <span class="math-container">$t$</span>. This is the nature of Zeta. Complain to Mr Riemann about this function :).</p>
<p>You can make the increment of <span class="math-container">$t$</span> much smaller to help reduce this problem. Say <code>0.0001</code> increment instead of <code>0.01</code> and this will make the line more smooth.</p>
<pre><code>ClearAll["Global`*"]
s[t_] := 1/2 + I*t
Manipulate[tick;
Module[{x, y}, x = Re[Zeta[s[t]]]; y = Im[Zeta[s[t]]];
AppendTo[coord, {x, y}];
ListLinePlot[coord, AxesLabel -> {"Real part", "Imaginary part"},
Mesh -> All, PlotRange -> {{-3, 3}, {-3, 3}},
GridLines -> Automatic, GridLinesStyle -> LightGray,
PlotStyle -> Red]],
Grid[{{Button[
Text@Style["clear", 12], {coord = {}; tick = Not[tick]},
ImageSize -> {60, 40}]}}], {{t, 0, "t"}, 0, 20, .0001,
Appearance -> "Labeled"}, {{coord, {}}, None}, {{tick, False},
None}, TrackedSymbols :> {t, tick}]
</code></pre>
|
84,183 | <p>I am trying to get my head around the geometric formulation of Quantum Mechanics as a projective Hilbert space (see Ashtekar, <a href="http://arxiv.org/abs/gr-qc/9706069" rel="nofollow">http://arxiv.org/abs/gr-qc/9706069</a>). So one identifies all the rays $\mathbb{C} \cdot \phi$ with the vector $\phi$ itself.</p>
<p>As the space of square-integrable functions $L^2(\Sigma, \mu)$ is the standard example of a Hilbert space I was wondering whatever there is clear characterization of its projective space. E.g. is there a good "visualization", an explicit notion of the Kähler-structure, etc?</p>
<p>A second question concerning this topic: Let $M$ be a Kähler manifold. Which further properties/conditions on $M$ are required so that $M$ can be realized as a projective Hilbert space? Is the construction than unique?</p>
<p>Thanks,
Tobi</p>
<p>P.s.: Which are the standard references regarding this topic?</p>
| Alain Valette | 14,497 | <p>Look at section 5 of this paper by Helmick and Helminck:
<a href="http://eprints.eemcs.utwente.nl/3487/01/1667.pdf" rel="nofollow">http://eprints.eemcs.utwente.nl/3487/01/1667.pdf</a></p>
|
84,183 | <p>I am trying to get my head around the geometric formulation of Quantum Mechanics as a projective Hilbert space (see Ashtekar, <a href="http://arxiv.org/abs/gr-qc/9706069" rel="nofollow">http://arxiv.org/abs/gr-qc/9706069</a>). So one identifies all the rays $\mathbb{C} \cdot \phi$ with the vector $\phi$ itself.</p>
<p>As the space of square-integrable functions $L^2(\Sigma, \mu)$ is the standard example of a Hilbert space I was wondering whatever there is clear characterization of its projective space. E.g. is there a good "visualization", an explicit notion of the Kähler-structure, etc?</p>
<p>A second question concerning this topic: Let $M$ be a Kähler manifold. Which further properties/conditions on $M$ are required so that $M$ can be realized as a projective Hilbert space? Is the construction than unique?</p>
<p>Thanks,
Tobi</p>
<p>P.s.: Which are the standard references regarding this topic?</p>
| Richard Montgomery | 2,906 | <p>Feynman, in his lecture notes, argues convincingly that you will understand the guts of Quantum Mechanics [QM] as best you can by looking at the two-slit experiment whose Hilbert space is two-dimensional. Or you could look at the Stern-Gerlach experiment, where it is three-dimensional. I recommend you stick to these cases, drop $L^2$, and see carefully what all the concepts boil down to there in terms of the Kahler geometry of projective space. Feynman does this, in physics language, in his 1957 paper with Vernon
titled ‘Geometrical Representation of the Schrodinger Equation for Solving Maser Problems’:
they solve the needed Schrodinger equation by drawing circles on the two-sphere = $CP^1$. </p>
<p>The ‘Berry Phase’ is a post-Feynman idea that is essentially the curvature of the
canonical connection for the canonical line bundle over $CP^n$. For a dictionary
from standard QM to Kahler geometry and connections you could look at ‘<a href="https://web.archive.org/web/20100630191320/http://count.ucsc.edu/~rmont/papers/aix_Heis.pdf" rel="nofollow">Heisenberg and isoholonomic inequalites</a>’.</p>
|
4,315,449 | <p>Is it correct to say that <span class="math-container">$\mathbb{R}^n \subset \mathbb{R}^n$</span> ?</p>
<p><strong>EDIT</strong>: The context of may question is that I am having a function that is defined as <span class="math-container">$f \colon D \subset \mathbb{R}^n \to \mathbb{R}^n $</span> and I am wondering if I can just generalize the definition as <span class="math-container">$f \colon \mathbb{R}^n \to \mathbb{R}^n $</span>. The symbol <span class="math-container">$\subset$</span> refers to set inclusion.</p>
<p><sup> I was just reading "J. M. Ortega, Iterative Solution of Nonlinear Equations in Several Variables". I hope I will not cause more agitation. </sup></p>
| Masacroso | 173,262 | <p>The most used convention is that the symbol <span class="math-container">$\subset $</span> applies to any subset of a set, so <span class="math-container">$A\subset A$</span> is correct for every set <span class="math-container">$A$</span>.</p>
|
1,600,063 | <p>I am trying to prove the following statement:</p>
<p>Given any two real numbers $x,y$ with $x<y$, there exists a rational number $q$ that satisfies $x<q<y$.</p>
<p>I got stuck at one point of the proof, so this is what I thought of:</p>
<p>I want to find a rational number $q$, which can be expressed as $q=\dfrac{a}{b}$, with $a$ integer and $b$ a positive integer, such that $$x<\dfrac{a}{b}<y$$</p>
<p>This is equivalent to $$(*) \space \space bx<a<by,$$</p>
<p>so if I could find an integer $a$ and a positive integer $b$ that satisfy $(*)$, then I would be done. </p>
<p>This is an exercise from Tao's Analysis I which follows right after the archimidean property, so it occurred to me to use this property:</p>
<p>Pick $x,y$ two real numbers with $x<y$, then we have $y-x$ is positive. By the archimedean property, there exists a positive integer $b$ such that $b(y-x)>1$. From here I don't know how to deduce that the integer $a$ satisfying $(*)$ exists. Any help would be appreciated.</p>
| Dave Neary | 122,612 | <p>Once you notice that <span class="math-container">$\frac{1}{y-x} < n$</span> for some positive integer <span class="math-container">$n$</span> you are home and dry. Inverting this inequality, you get <span class="math-container">$y-x>\frac{1}{n}$</span>, so you can find an integer <span class="math-container">$a$</span> such that <span class="math-container">$x<\frac{a}{n}<y$</span> - and for any integer <span class="math-container">$m>n$</span> this will work.</p>
|
858,576 | <p>Prove that the union of three subspaces of V is a subspace iff one of the subspaces contains the other two.</p>
<p>I can do this problem when I am working in only two subspaces of $V$ but I don't know how to do it with three. </p>
<p>What I tried is:
If one of the subspaces contains the other two, Then their union is obviously a subspace because the subspace that contains them is a subspace. (Is this sufficient??).</p>
<p>If the union of three subspaces is a subspace.....
How do I prove that one of the subspaces must contain the other two from here?</p>
<p>*When proving this for two I said that there is an element in one of the subspaces that is not the other and proved by contradiction that one of the subspaces must be contained in the other. How would I do this for three?</p>
| Gina | 102,040 | <p>The statement is false. Consider the following counterexample:</p>
<p>Consider the vector space $V=(\mathbb{Z}/2\mathbb{Z})^{2}$ where $F=\mathbb{Z}/2\mathbb{Z}$. Let $V_{1}$ be spanned by
$(1,0)$. Let $V_{2}$ be spanned by $(0,1)$. Let $V_{3}$ be spanned by $(1,1)$. Then we have $V=V_{1}\cup V_{2}\cup V_{3}$, but none of the $V_{1},V_{2},V_{3}$ are subspace of another.</p>
<p>You can usually count on field of characteristic $2$ to give you counterexample. There are many similar counterexample, too. In finite dimension, I think all counterexamples can be constructed this way. My intuition tells me that there are infinite dimensional counterexamples of other form, but have not checked clearly.</p>
<p>EDIT. Here is a proof of the statement with the restriction $F\not=\mathbb{Z}/2\mathbb{Z}$:</p>
<p>Without loss of generality, we can assume the whole space $V$ is in fact $V_{1}+V_{2}+V_{3}$. Easily seen that in fact we must also have $V=V_{1}\cup V_{2}\cup V_{3}$.</p>
<p>There exist $a,b\in F$ such that $a,b\not=0$ and $a-b=1$ (take $a$ to be anything except $0,1$, and take $b=a-1$).</p>
<p>Assume $V_{1}$ and $V_{2}$ neither contains another (otherwise this reduce to the 2-subspace case). For any $u\in V_{1}\setminus(V_{1}\cap V_{2})$ we take an arbitrary $w\in V_{2}\setminus(V_{1}\cap V_{2})$ (it exists due to the fact that neither $V_{1}$ nor $V_{2}$ contains another). Then $au+w$ is in neither $V_{1}$ nor $V_{2}$ (if in $V_{1}$ then since $au\in V_{1}$ we must have $w\in V_{1}$ so $w\in V_{1}\cap V_{2}$ contradiction; same for the other case but now using the fact that $a\not=0$), so $u+aw\in V_{3}$. Same argument apply to show $bu+w\in V_{3}$. Hence $u=(bu+w)-(au+w)\in V_{3}$. Hence $V_{1}\setminus(V_{1}\cap V_{2})\subset V_{3}$. Same argument apply to show $V_{2}\setminus(V_{1}\cap V_{2})\subset V_{3}$. Now for any $v\in V_{1}\cap V_{2}$ we pick a $w\in V_{2}\setminus(V_{1}\cap V_{2})\subset V_{3}$. Then $w+v\notin V_{1}\cap V_{2}$ (otherwise $w\in V_{1}\cap V_{2}$). But $w+v\in V_{2}$. Hence $w+v\in V_{2}\setminus(V_{1}\cap V_{2})\subset V_{3}$. Thus $v=(w+v)-w$ so $v\in V_{3}$. Hence $V_{1}\cap V_{2}\subset V_{3}$. Therefore $V_{1},V_{2}\subset V_{3}$.</p>
|
2,611,676 | <p>Or consider the general problem-
Find the value of n for which x^n is just greater than x!</p>
<p>I dont know even if it is possible to find the solution or not...</p>
| Salech Alhasov | 25,654 | <pre><code>import math
n=1
while 100**n < math.factorial(100):
n+=1
print(n)
</code></pre>
<p>For completeness, this python code prints the number $79$.</p>
|
446,499 | <p>I have just learned the definition of connectedness and wikipedia gives an example of a disconnected set: <span class="math-container">$(0,1)\cup \left\{ 3 \right\}$</span> (<a href="https://en.wikipedia.org/wiki/Connected_space#Examples" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Connected_space#Examples</a>). Why is it disconnected? I need a bit clarification on this. Thanks for any help!</p>
| palio | 10,463 | <p>A disconnected space $X$ is a space written as a disjoint union of open or closed sets, that is $X=Y\sqcup Z$ where $Y$ and $Z$ are both open or both closed in $X$ relatively to the topology you put on $X$. For example the disjoint union $\{0\}\sqcup (0,1]$ gives the interval $[0,1]$ which is connected, this is because $\{0\}$ is closed and $(0,1]$ is open in $[0,1]$ endowed with the subset topology from $\mathbb R$.</p>
|
2,694,740 | <p>$$\frac{2.10^{-7} - 0,4.10^{-6}}{10^{-8}} = ? $$</p>
<p>These questions are making me confused because we're dealing with the terms like $10^x$. What are your professional tips? </p>
<p><strong>My attempt:</strong></p>
<p>$$\frac{2.10^{-7} - 4.10^{-7}}{10^{-8}} \tag{1} $$
$$\frac{ -8.10^{-7}}{10^{-8}} \tag{2} $$</p>
<p>And that's where I'm stuck. </p>
| Mohammad Riazi-Kermani | 514,496 | <p>$$\frac{2.10^{-7} - 0,4.10^{-6}}{10^{-8}} = 10^8 (2.10^{-7} - 0,4.10^{-6})=20-40=-20$$</p>
|
1,656,145 | <p>Let the real function of two real variables$$u(x,y) =
\begin{cases}
x, & \quad \text{if } |y|>|x| \\
-x, & \quad \text{if } otherwise
\\ \end{cases} $$</p>
<p>Is there a sequence $\{(x_n,y_n)\}_{n \geq 0}$ which converge to $(0,0)$ such that $\lim_{n \to \infty} u(x_n,y_n) \not= u(0,0)$?</p>
<p>I tried to prove this by contrapositive of continuity, but I failed</p>
| Eric Wofsey | 86,856 | <p>If $\{(x_n,y_n)\}$ converges to $(0,0)$, then in particular $\{x_n\}$ converges to $0$. Thus for any $\epsilon>0$, there is an $N$ such that $|x_n|<\epsilon$ for all $n>N$. But $|u(x,y)|=|x|$ for all $(x,y)$, so this means $|u(x_n,y_n)|<\epsilon$ for all $n>N$. Thus $\{u(x_n,y_n)\}$ converges to $0=u(0,0)$.</p>
|
2,012,532 | <p>The following is all confirmed to be true:</p>
<p>Matrix A =
$
\begin{bmatrix}
0 & 1 & -2 \\
-1 & 2 & -1 \\
2 & -4 & 3 \\
1 & -3 & 2 \\
\end{bmatrix}
$</p>
<p>U =
$
\begin{bmatrix}
-1 & 2 & -1 \\
0 & 1 & -2 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{bmatrix}
$</p>
<p>L =
$
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
-2 & 0 & 1 & 0\\
-1 & -1 & -1 & 0\\
\end{bmatrix}
$</p>
<p>Okay so using that I need to solve the following system:</p>
<p>$
x_2 - 2x_3 = 0 \\
-x_1 + 2x_2 - x_3 = -2 \\
2x_1 -4x_2 + 3x_3 = 5 \\
x_1 - 3x_2 + 2x_3 = 1
$</p>
<p>So step one is solving $Ly = b$, where $y = Ux$</p>
<p>So that is:</p>
<p>$
y_1 = 0\\
y_2 = -2\\
-2y_1 + y_3 = 5 \\
-y_1 - y_2 -y_3 = 1 \\
$</p>
<p>How can we find $y_3$ in the last two equations? Because,</p>
<p>$
-2(0) + y_3 = 5 \\
-(0) - (-2) - y_3 = 1 \\
$</p>
<p>So in the second to last equation $y_3 = 5$, but in the last equation $y_3 = 1$. Very confused.</p>
| dantopa | 206,581 | <h2>Problem</h2>
<p>$$
\mathbf{A} =
\left[
\begin{array}{rrr}
0 & 1 & -2 \\
-1 & 2 & -1 \\
2 & -4 & 3 \\
1 & -3 & 2 \\
\end{array}
\right]
$$</p>
<h2>Associated Permutation Matrix</h2>
<p>Don't start with a $0$ pivot element. Move the first row down. The permutation matrix interchanges the first two rows.</p>
<p>$$ \mathbf{P} =
\left[
\begin{array}{cccc}
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right]
$$</p>
<h2>Input</h2>
<p>$$
\begin{align}
\mathbf{P} \mathbf{A} =
% P
\left[
\begin{array}{cccc}
0 & \boxed{1} & 0 & 0 \\
\boxed{1} & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right]
% A
\left[
\begin{array}{rrr}
0 & 1 & -2 \\
-1 & 2 & -1 \\
2 & -4 & 3 \\
1 & -3 & 2 \\
\end{array}
\right]
%
&=
% L
\left[
\begin{array}{rrr}
-1 & 2 & -1 \\
0 & 1 & -2 \\
2 & -4 & 3 \\
1 & -3 & 2 \\
\end{array}
\right]
\end{align}
$$</p>
<h2>Decomposition</h2>
<p>$$
\begin{align}
\mathbf{P} \mathbf{A} &= \mathbf{L} \mathbf{U} \\
% PA
\underbrace{\left[
\begin{array}{rrr}
-1 & 2 & -1 \\
0 & 1 & -2 \\
2 & -4 & 3 \\
1 & -3 & 2 \\
\end{array}
\right]}_{\color{blue}{m}\times \color{red}{n}}
%
&=
% L
\underbrace{\left[
\begin{array}{rrrc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
-2 & 0 & 1 & 0 \\
-1 & -1 & -1 & 1 \\
\end{array}
\right]}_{\color{blue}{m}\times \color{blue}{m}}
% U
\underbrace{\left[
\begin{array}{rrr}
-1 & 2 & -1 \\
0 & 1 & -2 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{array}
\right]}_{\color{blue}{m}\times \color{red}{n}}
\end{align}
$$</p>
|
3,995,728 | <p>For the integral
<span class="math-container">$$
\int_1^2 \frac {3^x + 2}{3^{2x} + 3^x} dx
$$</span>
choose the interval of its result: <span class="math-container">$(-\infty, 0], (0, \frac 12], (\frac 12, 1], (1, 3], (3, \infty)$</span>. According to the author of this task you do not have to compute the integral itself to determine the interval. I have managed to compute the integral but I am unsure how to determine the interval without computing it. How can I do that?</p>
| Steven Alexis Gregory | 75,410 | <p><a href="https://i.stack.imgur.com/IfSqx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IfSqx.jpg" alt="enter image description here" /></a></p>
<p>Is this what they were asking for?</p>
<p><span class="math-container">$\alpha = m \angle BAC = m \operatorname{arc} DE$</span></p>
<p><span class="math-container">$\dfrac{\pi - \alpha}{2} = m \angle ABC = \dfrac 12(m \operatorname{arc} GE)$</span></p>
<p>\end{document}</p>
|
3,512,521 | <p>Let <span class="math-container">$x_1, x_2,\dots, x_n$</span> be positive real numbers. Let <span class="math-container">$A$</span> be the <span class="math-container">$n\times n$</span> matrix whose <span class="math-container">$i,j^\text{th}$</span> entry is <span class="math-container">$$a_{ij}=\frac{1}{x_i+x_j}.$$</span></p>
<p>This is a <strong>Cauchy matrix</strong>. I am trying to show that this matrix is positive semi-definite.</p>
<p>I have been given the following hint: Consider the matrix <span class="math-container">$T=(t^{x_i+x_j})$</span> where <span class="math-container">$t>0$</span>. Use the fact that <span class="math-container">$T$</span> is positive semi-definite and that <span class="math-container">$$\frac{1}{x_i+x_j}=\int_0^1t^{x_i+x_j-1}dt.$$</span></p>
<p>I have managed to show that <span class="math-container">$T$</span> is positive semi-definite but I don't understand where to go from there or how to use the rest of the hint.</p>
<p>I would like another way to do this, preferably without involving integrals</p>
<p>Thank you.</p>
| user1551 | 1,551 | <p>There is also a proof that is similar in spirit to the hint you've mentioned. See exercise 1.6.3 (pp.24-25) of Bhatia, <em>Positive Definite Matrices</em>. The idea is that, instead of writing the Cauchy matrix as an integral of Gramians, we write it as an infinite sum of Gramians. More specifically, let <span class="math-container">$0<t<\min_ix_i$</span>. Then
<span class="math-container">$$
\frac{1}{x_i+x_j-t}
=\frac{t}{x_ix_j}\left(\frac{1}{1-\frac{(x_i-t)(x_j-t)}{x_ix_j}}\right)
=\frac{t}{x_ix_j}\sum_{k=0}^\infty\left(\frac{(x_i-t)(x_j-t)}{x_ix_j}\right)^k.
$$</span>
Therefore, by setting <span class="math-container">$\mathbf v_k=\left(\frac{(x_1-t)^k}{x_1^{k+1}},\,\frac{(x_2-t)^k}{x_2^{k+1}},\,\ldots,\,\frac{(x_n-t)^k}{x_n^{k+1}}\right)^\top$</span>, we see that <span class="math-container">$$
\left(\frac{1}{x_i+x_j-t}\right)_{i,j\in\{1,\ldots,n\}}
=t\sum_{k=0}^\infty \mathbf v_k\mathbf v_k^\top
$$</span>
is positive semidefinite. Now the result follows by passing the matrix on the LHS to the limit <span class="math-container">$t\to0$</span>.</p>
<p>A big merit of the above proof is that it can be easily extended to prove the positive semidefiniteness of the power Cauchy matrix <span class="math-container">$\left(\frac{1}{(x_i+x_j)^p}\right)_{i,j\in\{1,\ldots,n\}}$</span> for any <span class="math-container">$p>0$</span>.</p>
|
3,512,521 | <p>Let <span class="math-container">$x_1, x_2,\dots, x_n$</span> be positive real numbers. Let <span class="math-container">$A$</span> be the <span class="math-container">$n\times n$</span> matrix whose <span class="math-container">$i,j^\text{th}$</span> entry is <span class="math-container">$$a_{ij}=\frac{1}{x_i+x_j}.$$</span></p>
<p>This is a <strong>Cauchy matrix</strong>. I am trying to show that this matrix is positive semi-definite.</p>
<p>I have been given the following hint: Consider the matrix <span class="math-container">$T=(t^{x_i+x_j})$</span> where <span class="math-container">$t>0$</span>. Use the fact that <span class="math-container">$T$</span> is positive semi-definite and that <span class="math-container">$$\frac{1}{x_i+x_j}=\int_0^1t^{x_i+x_j-1}dt.$$</span></p>
<p>I have managed to show that <span class="math-container">$T$</span> is positive semi-definite but I don't understand where to go from there or how to use the rest of the hint.</p>
<p>I would like another way to do this, preferably without involving integrals</p>
<p>Thank you.</p>
| Rodrigo de Azevedo | 339,790 | <p>Complementing A.Γ.'s answer, and rephrasing a bit:</p>
<blockquote>
<p>Given <span class="math-container">$a_1, a_2, \dots, a_n > 0$</span>, we build the <span class="math-container">$n \times n$</span> symmetric Cauchy matrix <span class="math-container">$\rm C$</span> whose entries are <span class="math-container">$$c_{ij} = \frac{1}{a_i + a_j}$$</span> Show that matrix <span class="math-container">$\rm C$</span> is positive semidefinite.</p>
</blockquote>
<p>Henceforth, we shall assume that the given positive numbers are <strong>distinct</strong>, i.e., <span class="math-container">$$|\{a_1, a_2, \dots, a_n\}| = n$$</span></p>
<hr />
<p>Let <span class="math-container">${\rm A} := \mbox{diag} (a_1, a_2, \dots, a_n)$</span>. Note that <span class="math-container">$\mathrm A \succ \mathrm O_n$</span>. Consider the following matrix equation</p>
<p><span class="math-container">$${\rm A X + X A} = 1_n 1_n^\top$$</span></p>
<p>Multiplying both sides by <span class="math-container">$-1$</span>, we obtain a <a href="https://en.wikipedia.org/wiki/Lyapunov_equation" rel="nofollow noreferrer">Lyapunov matrix equation</a></p>
<p><span class="math-container">$${\rm (-A) X + X (-A)} = - 1_n 1_n^\top$$</span></p>
<p>where matrix <span class="math-container">$-\rm A$</span> is stable (or <a href="https://en.wikipedia.org/wiki/Hurwitz_matrix" rel="nofollow noreferrer">Hurwitz</a>) and the RHS is negative semidefinite. Since the pair <span class="math-container">$(-\rm A, 1_n)$</span> is <a href="https://en.wikipedia.org/wiki/Controllability" rel="nofollow noreferrer">controllable</a>, the Lyapunov equation has the following unique, symmetric, <strong>positive definite</strong> solution</p>
<p><span class="math-container">$$\rm X = \int_0^{\infty} e^{- \tau \mathrm A} 1_n 1_n^\top e^{- \tau \mathrm A} \,{\rm d} \tau = \int_0^{\infty} \begin{bmatrix} e^{- a_1 \tau}\\ e^{- a_2 \tau}\\ \vdots\\ e^{- a_n \tau}\end{bmatrix} \begin{bmatrix} e^{- a_1 \tau}\\ e^{- a_2 \tau}\\ \vdots\\ e^{- a_n \tau} \end{bmatrix}^\top \, {\rm d} \tau = \cdots = \rm C$$</span></p>
<p>because</p>
<p><span class="math-container">$$\displaystyle\int_0^{\infty} e^{-(a_i + a_j) \tau} \,{\rm d} \tau = \frac{1}{a_i + a_j}$$</span></p>
<p>Therefore, we conclude that <span class="math-container">$\rm C$</span> is positive definite.</p>
<hr />
<h3>Alternative solution</h3>
<p>Vectorizing both sides of the Lyapunov equation,</p>
<p><span class="math-container">$$\left( \mathrm I_n \otimes \mathrm A + \mathrm A \otimes \mathrm I_n \right) \mbox{vec} (\mathrm X) = 1_n \otimes 1_n$$</span></p>
<p>or,</p>
<p><span class="math-container">$$\begin{bmatrix} \mathrm A + a_1 \mathrm I_n & & & \\ & \mathrm A + a_2 \mathrm I_n & & \\ & & \ddots & \\ & & & \mathrm A + a_n \mathrm I_n\end{bmatrix} \begin{bmatrix} \mathrm x_1\\ \mathrm x_2\\ \vdots\\ \mathrm x_n\end{bmatrix} = \begin{bmatrix} 1_n\\ 1_n\\ \vdots\\ 1_n\end{bmatrix}$$</span></p>
<p>where <span class="math-container">$\mathrm x_i$</span> is the <span class="math-container">$i$</span>-th column of <span class="math-container">$\rm X$</span>. Solving for <span class="math-container">$\mathrm x_i$</span>,</p>
<p><span class="math-container">$$\mathrm x_i = \begin{bmatrix} \frac{1}{a_1 + a_i}\\ \frac{1}{a_2 + a_i}\\ \vdots\\ \frac{1}{a_n + a_i}\end{bmatrix}$$</span></p>
<p>which is the <span class="math-container">$i$</span>-th column of Cauchy matrix <span class="math-container">$\rm C$</span>. Therefore, the unique, symmetric, positive definite solution of the Lyapunov equation is <span class="math-container">$\rm C$</span>.</p>
<hr />
<h3>Addendum</h3>
<p>The <a href="https://en.wikipedia.org/wiki/Controllability" rel="nofollow noreferrer">controllability</a> matrix corresponding to the pair <span class="math-container">$(-\rm A, 1_n)$</span> is</p>
<p><span class="math-container">$$\begin{bmatrix} | & | & & |\\ 1_n & -\mathrm A 1_n & \dots & (-1)^{n-1} \mathrm A^{n-1} 1_n\\ | & | & & |\end{bmatrix}$$</span></p>
<p>which is a square <span class="math-container">$n \times n$</span> <a href="https://en.wikipedia.org/wiki/Vandermonde_matrix" rel="nofollow noreferrer">Vandermonde matrix</a> whose columns have been multiplied by <span class="math-container">$\pm 1$</span>, which does not affect its rank. Since we assumed that the given <span class="math-container">$a_1, a_2, \dots, a_n$</span> are distinct, the Vandermonde matrix has full rank and, thus, the pair <span class="math-container">$(-\rm A, 1_n)$</span> is <a href="https://en.wikipedia.org/wiki/Controllability" rel="nofollow noreferrer">controllable</a>.</p>
|
3,151,452 | <p>The context of the question is that a bakery bakes cakes and the mass of cake is demoted by <span class="math-container">$X$</span> such that <span class="math-container">$X \sim N(300, 40^2)$</span>. A sample of 12 cakes is taken and the mean of the sample is 292g. The question wants me to find the <span class="math-container">$p$</span>-value and test to see if the mean has changed at 10% significance.</p>
<p>So I know how to carry out the test as <span class="math-container">$\overline{X_{12}} \sim N(300,\frac{40^2}{12})$</span>, But what would the p-value I'm trying to calculate be? I know the p-value is 0.244.</p>
| Minus One-Twelfth | 643,882 | <p><span class="math-container">$\newcommand{\P}{\mathbb{P}}$</span>It appears that you have <span class="math-container">$X\sim N(\mu, 40^2)$</span> (known variance), and your null hypothesis is that <span class="math-container">$\mu = 300$</span>, with alternative hypothesis <span class="math-container">$\mu\ne 300$</span>. For this, we will be using a <em>two-tailed test</em>.</p>
<p>Under the null hypothesis, we have <span class="math-container">$\overline{X}_{12}\sim N\left(300, \frac{40^2}{12}\right)$</span>, or <span class="math-container">$T := \frac{\overline{X}_{12} - 300}{40/\sqrt{12}}\sim N(0,1)$</span>. The <span class="math-container">$p$</span>-value is the probability of getting a "more extreme" result for the test-statistic <span class="math-container">$T$</span> than observed under the null hypothesis (note the observed value is <span class="math-container">$\color{blue}{\frac{292-300}{40/\sqrt{12}}}$</span>), which for our two-tailed test means that the <span class="math-container">$p$</span>-value is</p>
<p><span class="math-container">$$\P\left(|T| > \left| \frac{292-300}{40/\sqrt{12}}\right| \right) \quad \text{where }T \sim N(0,1).$$</span></p>
<p>If <a href="http://wolframalpha.com/input/?i=Pr%28%7CZ%7C+>+8%2F%2840%2Fsqrt%2812%29%29%29+where+Z+~+N%280%2C1%29" rel="nofollow noreferrer">you compute this probability</a>, it seems you get double the answer you wrote. So the answers probably used a one-tailed test, which would be the case if our alternative hypothesis was <span class="math-container">$\mu < 300$</span>, in which case the <span class="math-container">$p$</span>-value would be</p>
<p><span class="math-container">$$\P\left(T < \frac{292-300}{40/\sqrt{12}} \right) \quad \text{where }T \sim N(0,1).$$</span></p>
<p>This will get you the reported answer. The reason I used a two-tailed test is because I interpreted the alternative hypothesis as being with a <span class="math-container">$\ne$</span> sign rather than <span class="math-container">$<$</span>, because the wording of the question was "mean has changed".</p>
|
3,748,739 | <p>Let <span class="math-container">$X\perp Y$</span> with <span class="math-container">$X,Y\sim N(0,1)$</span>. Let <span class="math-container">$U=\frac{(X+Y)}{\sqrt{2}}$</span> and <span class="math-container">$V=\frac{(X-Y)}{\sqrt{2}}$</span>.</p>
<ol>
<li>Find the law of <span class="math-container">$(U,V)$</span>. What is the value of <span class="math-container">$\mathbb{E}(X)$</span>, <span class="math-container">$Var(U)$</span> and <span class="math-container">$Cov(U,V)$</span>?</li>
</ol>
<p><span class="math-container">$\rightarrow \left\{\begin{matrix}
\frac{x+y}{\sqrt{2}}=u\\ \frac{x-y}{\sqrt{2}}=v
\end{matrix}\right. \Rightarrow \left\{\begin{matrix}
y=\frac{\sqrt{2}u-\sqrt{2}v}{2}\\ x=\frac{\sqrt{2}u+\sqrt{2}v}{2}
\end{matrix}\right.$</span> with <span class="math-container">$|J|=1\Rightarrow f_{UV}(u,v)=f_X(\frac{(\sqrt{2}u+\sqrt{2}v)}{2})f_Y(\frac{(\sqrt{2}u-\sqrt{2}v)}{2})|J|=\frac{1}{2\pi}e^{-\frac{(u^2+v^2)}{2}}\Rightarrow U\sim N(0,1)$</span> and <span class="math-container">$V\sim N(0,1)$</span>; <span class="math-container">$\mathbb{E}(U)=0,Var(U)=1,Cov(U,V)=0\Rightarrow U\perp V$</span></p>
<ol start="2">
<li>Calculate <span class="math-container">$\mathbb{P}(|U|<z\sqrt{2},|V|<z\sqrt{2})$</span>.</li>
</ol>
<p><span class="math-container">$\rightarrow \mathbb{P}(|U|<z\sqrt{2},|V|<z\sqrt{2})=\mathbb{P}(-z\sqrt{2}<U,V<z\sqrt{2})=$</span>
<span class="math-container">$\frac{1}{2\pi}\int_{-z\sqrt{2}}^{z\sqrt{2}}e^{-\frac{v^2}{2}}[\int_{-z\sqrt{2}}^{z\sqrt{2}}e^{-\frac{u^2}{2}}du]dv$</span></p>
<p>but now I'm stuck. How can I solve this integral?</p>
<ol start="3">
<li><span class="math-container">$\mathbb{P}(X+Y<z|X>0,Y>0)$</span>.</li>
</ol>
<hr />
<p>For point 3) I have no idea. Any suggests? Thanks in advance for any help!</p>
| Ethan Bolker | 72,858 | <p>To rule out (D) start with a nonnegative continuous function that's not differentiable - perhaps <span class="math-container">$g(x) = |x|$</span>. Then construct <span class="math-container">$f$</span> by integrating twice, so that <span class="math-container">$g$</span> is its second derivative. Then the integral of <span class="math-container">$f$</span> will be only three times differentiable.</p>
|
3,748,739 | <p>Let <span class="math-container">$X\perp Y$</span> with <span class="math-container">$X,Y\sim N(0,1)$</span>. Let <span class="math-container">$U=\frac{(X+Y)}{\sqrt{2}}$</span> and <span class="math-container">$V=\frac{(X-Y)}{\sqrt{2}}$</span>.</p>
<ol>
<li>Find the law of <span class="math-container">$(U,V)$</span>. What is the value of <span class="math-container">$\mathbb{E}(X)$</span>, <span class="math-container">$Var(U)$</span> and <span class="math-container">$Cov(U,V)$</span>?</li>
</ol>
<p><span class="math-container">$\rightarrow \left\{\begin{matrix}
\frac{x+y}{\sqrt{2}}=u\\ \frac{x-y}{\sqrt{2}}=v
\end{matrix}\right. \Rightarrow \left\{\begin{matrix}
y=\frac{\sqrt{2}u-\sqrt{2}v}{2}\\ x=\frac{\sqrt{2}u+\sqrt{2}v}{2}
\end{matrix}\right.$</span> with <span class="math-container">$|J|=1\Rightarrow f_{UV}(u,v)=f_X(\frac{(\sqrt{2}u+\sqrt{2}v)}{2})f_Y(\frac{(\sqrt{2}u-\sqrt{2}v)}{2})|J|=\frac{1}{2\pi}e^{-\frac{(u^2+v^2)}{2}}\Rightarrow U\sim N(0,1)$</span> and <span class="math-container">$V\sim N(0,1)$</span>; <span class="math-container">$\mathbb{E}(U)=0,Var(U)=1,Cov(U,V)=0\Rightarrow U\perp V$</span></p>
<ol start="2">
<li>Calculate <span class="math-container">$\mathbb{P}(|U|<z\sqrt{2},|V|<z\sqrt{2})$</span>.</li>
</ol>
<p><span class="math-container">$\rightarrow \mathbb{P}(|U|<z\sqrt{2},|V|<z\sqrt{2})=\mathbb{P}(-z\sqrt{2}<U,V<z\sqrt{2})=$</span>
<span class="math-container">$\frac{1}{2\pi}\int_{-z\sqrt{2}}^{z\sqrt{2}}e^{-\frac{v^2}{2}}[\int_{-z\sqrt{2}}^{z\sqrt{2}}e^{-\frac{u^2}{2}}du]dv$</span></p>
<p>but now I'm stuck. How can I solve this integral?</p>
<ol start="3">
<li><span class="math-container">$\mathbb{P}(X+Y<z|X>0,Y>0)$</span>.</li>
</ol>
<hr />
<p>For point 3) I have no idea. Any suggests? Thanks in advance for any help!</p>
| Robert Israel | 8,508 | <p>Hint for (C): if <span class="math-container">$f'' \ge 0$</span> and <span class="math-container">$f'(x) > 0$</span>, then <span class="math-container">$f'(t) \ge f'(x)$</span> for all <span class="math-container">$t \ge x$</span>, so <span class="math-container">$f(t) \ge f(x) + (t-x) f'(x)$</span> for <span class="math-container">$t \ge x$</span>. Similarly in the opposite direction if <span class="math-container">$f'(x) < 0$</span>.</p>
|
1,222,909 | <p>I was thinking to convert to cartesian coordinates and then find when the slope of the tangent line is $1$, but I get a messy equation $2\cos^2\theta -2\sin^2\theta=4\sin^2\theta\cos\theta$
I was wondering if there was an easy way as it is hard to get values from this.</p>
<p>Edit: The equation ends up simplifying to $\tan(2\theta) = 1$, but for future reference is this the best method?</p>
| Poppy | 41,695 | <p>The curve is $\alpha(\theta)=(2sin{\theta}\cos{\theta},2\sin{\theta}\sin{\theta})$. The velocity vector is $\alpha'(\theta)=(2\cos{2\theta},2\sin{2\theta})$. Now we have to find all points so that this vector is proportional to $(1,1)$.</p>
<p>$\theta=\dfrac{\pi}{8}$ or $\theta=\dfrac{5\pi}{8}$</p>
|
272,057 | <p>Let $\mu$ and $\nu$ be two probability measures on $\mathbb R^n$ with finite first moment. Denote by $d:=W_1(\mu,\nu)$, where $W_1(\cdot,\cdot)$ stands for the Wasserstein distance of order $1$. </p>
<p>My question is the following: Let $X$ be a random variable defined on some probability space (rich enough) with law $\mu$, could we find <strong>a measurable function $f:\mathbb R^n\times \mathbb R^n\to\mathbb R^n$ and a random variable $G$ independent of $X$</strong> s.t.</p>
<p>$$Y:=f(X,G)~\sim~\nu~~~~~~ \mbox{ and }~~~~~~ \mathbb E[|X-Y|]~\le ~2d~?$$</p>
<p><strong>Thought 1:</strong> Let $d_0:=\rho(\mu,\nu)$, where $\rho(\cdot,\cdot)$ denotes the Prokhorov distance. Then we have a measurable function $f_0:\mathbb R^n\times \mathbb R^n\to\mathbb R^n$ and a random variable $G_0$ independent of $X$ s.t.</p>
<p>$$Y_0:=f_0(X,G)~\sim~\nu~~~~~~ \mbox{ and }~~~~~~ \mathbb P[\{|X-Y_0|\ge2d_0\}]~\le ~2d_0.$$</p>
<p>The above construction is from the paper <em>On a representation of random variables</em> by Skorokhod, but I can't find this paper. </p>
<p><strong>Thought 2:</strong> Let $\pi(dx,dy)$ be the optimal transport plan, i.e. $\pi(A\times\mathbb R^n)=\mu(A)$ and $\pi(\mathbb R^n\times A)=\nu(A)$ for all measurable $A\subset\mathbb R^n$. Disintegration w.r.t. the first coordinate $x$, one has $\pi(dx,dy)=\mu(dx)\otimes \lambda_x(dy)$, where $(\lambda_x)_{x\in\mathbb R^n}$ denotes the r.c.p.d. (regular conditional probability distribution). But I've no idea how to recover the function $f$ using $\lambda_x$.</p>
<p>Any answer, help or comment is highly appreciated. Thanks a lot!</p>
| Benoît Kloeckner | 4,961 | <p>Yes, you can even ensure $\mathbb{E}(|X-Y|)=d$ and all you need for $G$ is that it has an atomless law, $\rho$ say.</p>
<p>Take the disintegration $(\lambda_x)$ of the optimal coupling $\Pi$ with respect to $\mu$ (note your formula has a $dx$ where one should read a $dy$). What you want is that $f(x,\cdot)$ sends the law $\rho$ of $G$ to $\lambda_x$. As soon as $\rho$ has no atom, this can be done. Then you only have to check that you can collect the $f(x,\cdot)$ into a measurable map $f$ (you need to be a bit careful in the construction for this to hold; in dimension $1$ using distribution functions should help, and in higher dimension you can use the fact that all standard space are isomorphic; alternatively you can possibly take advantage of the extra bit of margin you took in the question).</p>
|
4,363,409 | <blockquote>
<p>Define <span class="math-container">$X_0=\alpha\in(0,1)$</span> the initial capital and <span class="math-container">$X_n$</span> as the remaining capital after each game.
A player bets <span class="math-container">$1-X_n$</span> if <span class="math-container">$X_n>1/2$</span> and <span class="math-container">$X_n$</span> if <span class="math-container">$X_n\leq 1/2$</span> such that each game is a Bernoulli<span class="math-container">$(1/2)$</span>.
Define <span class="math-container">$A_n=\{X_n\in(0,1)\}$</span>, the event that the player neither wins everything or reachs ruin. Show by induction that <span class="math-container">$P(A_n)\leq 2^{-n}$</span>.</p>
</blockquote>
<p>If <span class="math-container">$\alpha< 1/2$</span> then the player either goes broken with probability <span class="math-container">$1/2$</span> or owns <span class="math-container">$2X_n$</span> in the next game. If <span class="math-container">$\alpha >1/2$</span> then the player owns every available resource with probability <span class="math-container">$1/2$</span> or owns <span class="math-container">$2X_n -1$</span> in the next game. Should I use total law of probability and try to work with the conditionals upon the last fortune as in</p>
<p><span class="math-container">$$P(A_n) = \sum_{k=1}^m P(A_n|B_m)P(B_m)$$</span>
where <span class="math-container">$\bigcup B_m= \Omega$</span>.</p>
<p>Also if this is the right path, should I split three-ways with a <span class="math-container">$X_n=1/2$</span> case where the player reach either <span class="math-container">$0$</span> or <span class="math-container">$1$</span>?</p>
<p>I'm having a little trouble working this problem out.</p>
| David Quinn | 187,299 | <p>Imagine that the surface of the water is part of a large flat mirror that extends through the earth and up to the horizon.</p>
<p>For each object you want to reflect, such as the large tree in the foreground, decide how high the base of it is above the level of the water, and imagine where the level of the mirror would be if it was underneath it.</p>
<p>Then the object is reflected in that mirror.</p>
|
4,363,409 | <blockquote>
<p>Define <span class="math-container">$X_0=\alpha\in(0,1)$</span> the initial capital and <span class="math-container">$X_n$</span> as the remaining capital after each game.
A player bets <span class="math-container">$1-X_n$</span> if <span class="math-container">$X_n>1/2$</span> and <span class="math-container">$X_n$</span> if <span class="math-container">$X_n\leq 1/2$</span> such that each game is a Bernoulli<span class="math-container">$(1/2)$</span>.
Define <span class="math-container">$A_n=\{X_n\in(0,1)\}$</span>, the event that the player neither wins everything or reachs ruin. Show by induction that <span class="math-container">$P(A_n)\leq 2^{-n}$</span>.</p>
</blockquote>
<p>If <span class="math-container">$\alpha< 1/2$</span> then the player either goes broken with probability <span class="math-container">$1/2$</span> or owns <span class="math-container">$2X_n$</span> in the next game. If <span class="math-container">$\alpha >1/2$</span> then the player owns every available resource with probability <span class="math-container">$1/2$</span> or owns <span class="math-container">$2X_n -1$</span> in the next game. Should I use total law of probability and try to work with the conditionals upon the last fortune as in</p>
<p><span class="math-container">$$P(A_n) = \sum_{k=1}^m P(A_n|B_m)P(B_m)$$</span>
where <span class="math-container">$\bigcup B_m= \Omega$</span>.</p>
<p>Also if this is the right path, should I split three-ways with a <span class="math-container">$X_n=1/2$</span> case where the player reach either <span class="math-container">$0$</span> or <span class="math-container">$1$</span>?</p>
<p>I'm having a little trouble working this problem out.</p>
| Michael Thwaites | 627,654 | <p>All previous comments taken into account, I should add that reflections of objects often appear taller than the actual image of the object seen directly. When I worked north of Lake Merritt in Oakland, the reflections of the buildings across the lake were noticeably taller than the buildings. I concluded this is do to the partial reflections on the tips of small wave crests on the lake between me and the buildings.</p>
|
126,549 | <p>For a quadratic form $q(\mathbf{v})$, when you change the basis do you <em>always</em> change the quadratic form? Can you have the same quadratic form with respect to different basis? Or is the quadratic form unique to the basis. </p>
<p>Also, if you're given a quadratic form say $q(\mathbf{v}) = 3x^2 + y^2 - 2z^2 + 4xy - 2xz$, $\:$ and you can clearly deduce the matrix from this
$$\begin{pmatrix}3&2&-1\\2&1&0\\-1&0&-2\end{pmatrix}$$
what is the basis for this quadratic form? Is it unique? Can you deduce it from this? Perhaps with a different basis is there a different 'method' of deducing the matrix?</p>
<p>Thanks! </p>
| Sunni | 10,800 | <p>The basis for your quadratic form is the natural basis, i.e., $(1, 0,\cdots, 0), \cdots, (0, \cdots, 0, 1)$. The matrix of quadratic form depends on the basis. Canonical form of a symmetric matrix is a diagonal matrix. You may compare these two forms...</p>
|
126,549 | <p>For a quadratic form $q(\mathbf{v})$, when you change the basis do you <em>always</em> change the quadratic form? Can you have the same quadratic form with respect to different basis? Or is the quadratic form unique to the basis. </p>
<p>Also, if you're given a quadratic form say $q(\mathbf{v}) = 3x^2 + y^2 - 2z^2 + 4xy - 2xz$, $\:$ and you can clearly deduce the matrix from this
$$\begin{pmatrix}3&2&-1\\2&1&0\\-1&0&-2\end{pmatrix}$$
what is the basis for this quadratic form? Is it unique? Can you deduce it from this? Perhaps with a different basis is there a different 'method' of deducing the matrix?</p>
<p>Thanks! </p>
| Sonu Lamba | 573,334 | <p>This <a href="https://drive.google.com/file/d/1cNxG-Mf6Uq4W48Wi186p9e6mOJ7fvwKt/view?usp=drivesdk" rel="nofollow noreferrer">link</a> may help you. See <strong><em>proposition 2.3</em></strong>. I think that implies that matrix of quad form changes on changing basis. That is there are many other basis too (other than that of std basis).</p>
|
2,208,755 | <p>I got stuck on this question: find all solutions $x$ for $a\in R$:</p>
<p>$$\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{(a^2-a+1)^3}{a^2(a-1)^2}$$</p>
<p>I see that if we simplify we get:</p>
<p>$$\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{[(x-{\frac 12})^2+{\frac 34}]^3}{[(x-{\frac 12})^2-{\frac 14}]^2}$$</p>
<p>From the expression $(x-{\frac 12})^2$, I see that if $x=x_1$ is a solution, then $x=1-x_1$ is also a solution. But in the solution to this exercise, it was stated that $x=\frac{1}{x_1}$ must also be a solution, and I don't see how.</p>
<p>[EDIT]</p>
<p>Ok, thx for the help guys. What do you think of this solution (doesn't involve any above precalculus math, and needs no long calculations)?</p>
<p>From the above we know that if $x_1=a$ is a solution, then $x_2=1-a$ is also a solution.</p>
<p>Also, from here:</p>
<p>$$\require{cancel}\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{\cancel{x^3}(x+{\frac 1x}-1)^3}{\cancel{x^3}(x+{\frac 1x}-2)}$$</p>
<p>in the expression $x+{\frac 1x}$ we see that if $x=x_1$ is a solution, then $x=\frac{1}{x_1}$ is also a solution, so $x_3=\frac{1}{a}$.</p>
<p>With these two rules we can now keep generating roots until we have 6 total.</p>
<p>If $x=x_2$ is a solution, then $x=\frac{1}{x_2}$ is also a solution, so $x_4=\frac{1}{1-a}$.</p>
<p>If $x=x_3$ is a solution, then $x=1-x_3$ is also a solution, so $x_5=\frac{a-1}{a}$.</p>
<p>Finally, if $x=x_5$ is a solution, then $x=\frac{1}{x_5}$ is also a solution, so $x_6=\frac{a}{a-1}$</p>
<p>The 6 obtained values are distinct, so they cover all the roots.</p>
<p>[EDIT2]</p>
<p>I guess this is answered. No sure whose particular answer to actually select as the right one since they're all correct, so I'll just leave it like this.</p>
| Jean Marie | 305,862 | <p>Let <span class="math-container">$$f(x)=\frac{(x^2-x+1)^3}{x^2(x-1)^2}\tag{1}$$</span></p>
<p>(representative curve in Fig. 1).</p>
<p>Your question can be reformulated in the following way : </p>
<p><span class="math-container">$$\text{for a given} \ a, \ \ \text{find all} \ x \ \text{such that} \ \ f(x)=f(a) \tag{2}$$</span></p>
<p>This answer will have 3 parts : A) Solving the problem itself, B) Understanding a little the associated group, C) Considering function <span class="math-container">$f$</span> in a larger scope.</p>
<p><a href="https://i.stack.imgur.com/Zjepo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Zjepo.jpg" alt="enter image description here"></a></p>
<p><em>Fig. 1. The red dots correspond to:
<span class="math-container">$f(-3)=f(-1/3)=f(1/4)=f(3/4)=f(4/3)=f(4)=2197/144$</span> where the values <span class="math-container">$-3,-1/3,\cdots 4$</span> are the values <span class="math-container">$\varphi_k(-3)$</span> (the first one corresponding to identity function with notations found below). The vertical asymptotes and the asymptotic parabola (with equation <span class="math-container">$y=x^2-x+3$</span>) are also featured.</em></p>
<p>A) <strong>Solving the problem itself</strong></p>
<p>First of all, the variations of <span class="math-container">$f$</span> are easily established once we remark that </p>
<p><span class="math-container">$$\tag{3}f'(x)=\underbrace{(x^2-x+1)}_{>0 \ \text{for all} \ x}\dfrac{(x-2)(2x-1)(x+1)}{x^3(x-1)^3}.$$</span></p>
<p>We deduce from (3) the existence of 3 minima in <span class="math-container">$(-1,a), (0.5,a)$</span> and <span class="math-container">$(2,a)$</span> where <span class="math-container">$a=27/4=6.75.$</span> </p>
<p>We can observe (by computations) the following facts, with the introduction of spécific notations</p>
<ul>
<li><p><span class="math-container">$f(x)=f(\underbrace{1-x}_{\varphi_2(x)})$</span>. </p></li>
<li><p><span class="math-container">$f(x)=f(\underbrace{1/x}_{\varphi_3(x)})$</span>.</p></li>
</ul>
<p>In a natural way the composition of these two invariant transforms, <span class="math-container">$\varphi_3\circ\varphi_2$</span> and <span class="math-container">$\varphi_2\circ\varphi_3$</span>, gives rise to a new invariance property :</p>
<ul>
<li><p><span class="math-container">$f(x)=f(1-1/x)$</span> meaning that if <span class="math-container">$x$</span> is solution, <span class="math-container">$a=1-1/x$</span> is also a solution ; in the same vein :</p></li>
<li><p><span class="math-container">$f(x)=f(1/(1-x))$</span> meaning that if <span class="math-container">$x$</span> is solution, <span class="math-container">$a=1/(1-x)$</span> is also a solution.</p></li>
</ul>
<p>A last invariant transformation :</p>
<ul>
<li><span class="math-container">$f(x)=f(x/(x-1))$</span> meaning that if <span class="math-container">$x$</span> is solution, <span class="math-container">$a=x/(x-1)$</span> is also a solution.</li>
</ul>
<p>We can summarize all this in the following manner: </p>
<p><span class="math-container">$$\tag{4}f(x)=f(\varphi_k(x))$$</span></p>
<p>for all "homographic" functions belonging to the so-called (6 elements) anharmonic group (in the sense of group theory):</p>
<p><span class="math-container">$$\phi_1(x)=x, \ \phi_2(x)=1-x, \ \phi_3(x)=\tfrac{1}{x}, \ \phi_4(x)=1-\tfrac{1}{x}, \ \phi_5(x)=\tfrac{1}{1-x}, \ \phi_6(x)=\tfrac{x}{x-1}.$$</span></p>
<blockquote>
<p>Conclusion : Equation (2) has </p>
<ul>
<li><p>6 solutions if <span class="math-container">$a\neq -1,1/2,2$</span> : <span class="math-container">$ \ x=a, \ 1-a, \ \tfrac{1}{a}, \ 1-\tfrac{1}{a}, \ \tfrac{1}{1-a}, \ \tfrac{a}{a-1} $</span></p></li>
<li><p>3 solutions if <span class="math-container">$a=-1,1/2,2$</span> which are... <span class="math-container">$x=-1,1/2,2$</span></p></li>
</ul>
</blockquote>
<p>Are there other transformations that we haven't considered ? No, because for a given <span class="math-container">$a$</span>, i.e, for a given <span class="math-container">$b=f(a)$</span>, there are 5 other points of the curve with a common ordinate <span class="math-container">$b$</span>, not less not more (set apart the special case of the minimal points). But in fact the fundamental interplay between function <span class="math-container">$f$</span> and the group just described can only be understood through "higher algebra" considerations that we will see in part C.</p>
<p>Remark : we could have taken profit of the fact that the curve has a symmetry with respect to a vertical axis situated at <span class="math-container">$x=\dfrac12$</span>. Indeed, setting <span class="math-container">$x=X+\dfrac12$</span>, we get a new (even) equation simpler than the initial one :</p>
<p><span class="math-container">$$Y=\dfrac{(4X^2+3)^2}{4 (4X^2-1)^2}$$</span></p>
<p>B) <strong>Understanding a little the associated anharmonic group</strong></p>
<p>(see (<a href="https://en.wikipedia.org/wiki/Cross-ratio" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Cross-ratio</a>))</p>
<p>It can be considered as a (finite) subgroup of <span class="math-container">$PGL(2,\mathbb{Z})$</span> through a classical linear representation (in short : representation by matrices) :</p>
<p><span class="math-container">$$\dfrac{az+b}{cz+d} \ \ \ \leftrightarrow \ \ \ \begin{pmatrix}a&b\\c&d\end{pmatrix}$$</span></p>
<p>It is isomorphic to <span class="math-container">$S_3$</span>, the group of permutation on 3 objects. </p>
<p>It is generated by <span class="math-container">$\varphi_2$</span> and <span class="math-container">$\varphi_3$</span>.</p>
<p>It has a nice vizualization in terms of "fundamental regions":</p>
<p><a href="https://i.stack.imgur.com/PgFNO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PgFNO.jpg" alt="enter image description here"></a></p>
<p><em>Fig. 2 : if <span class="math-container">$z$</span> belongs to the green "triangular" region, then <span class="math-container">$1-z$</span> belongs to... etc. Please note that straight line <span class="math-container">$x=\tfrac12$</span> is an axis of symmetry like for Fig. 1.</em></p>
<p>C) <strong>Considering function <span class="math-container">$f$</span> in a larger scope</strong></p>
<p>Why is this working so well ? Because this is the tip of an iceberg. Indeed, function <span class="math-container">$f$</span> is the "Klein j-invariant" (see (<a href="https://arxiv.org/pdf/1810.08742.pdf)(https://math.stackexchange.com/q/3418400)" rel="nofollow noreferrer">https://arxiv.org/pdf/1810.08742.pdf)(https://math.stackexchange.com/q/3418400)</a> (<a href="http://www.cis.upenn.edu/~jean/gma-v2-chap5.pdf" rel="nofollow noreferrer">http://www.cis.upenn.edu/~jean/gma-v2-chap5.pdf</a>) exercice 5.14) associated with anharmonic group in projective geometry ; it would be too long to explain it here. See for that the Wikipedia article (<a href="https://en.wikipedia.org/wiki/Cross-ratio" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Cross-ratio</a>). This invariant is associated with elliptic curves. </p>
<p>Remarks : other connections : <a href="https://math.stackexchange.com/q/3277840">How is $E = S(x)$, where $E = k(x), k$ a field, and S= k(I) of all rationals functions of $I = I(x) = \frac{(x^2 -x+1)^3}{x^2(x-1)^2}$?</a> ; about Lüroth theorem <a href="https://upload.wikimedia.org/wikipedia/commons/3/3e/Another_elementary_proof_of_Luroth%27s_theorem-06.2004.pdf" rel="nofollow noreferrer">https://upload.wikimedia.org/wikipedia/commons/3/3e/Another_elementary_proof_of_Luroth%27s_theorem-06.2004.pdf</a> ; <a href="https://math.stackexchange.com/q/956473">Klein's invariant and negative discriminant</a>} ; an interesting inequality :<a href="https://artofproblemsolving.com/community/c6h1447345p8277143" rel="nofollow noreferrer">https://artofproblemsolving.com/community/c6h1447345p8277143</a></p>
|
2,947,953 | <p>Given the two functions <span class="math-container">$$f(x) = \ln\left(\frac{x+1}{x-1}\right)$$</span> and <span class="math-container">$$g(x) = \ln(x+1)-\ln(x-1)$$</span>
I can justify independently why <span class="math-container">$\text{dom}(f) = (-\infty, -1) \cup (1,\infty)$</span>, and <span class="math-container">$\text{dom}(g)= (1,\infty)$</span>, but I'm not sure why these two functions have different domains. Can anyone enlighten me on what I'm missing? </p>
<p>EDIT: I plan on trying to explain this to freshman students taking their first calculus course. How can I justify that these domains are different? Is it true that <span class="math-container">$f(x)=g(x)$</span>? </p>
| Anik Bhowmick | 354,636 | <p>For <span class="math-container">$f(x)$</span>, you are considering about the domain of <span class="math-container">$ln$</span> and also of <span class="math-container">$\frac{x+1}{x-1}$</span>. But for <span class="math-container">$g(x)$</span>, you are considering about the domain of <span class="math-container">$ln(x+1)$</span> and <span class="math-container">$ln(x-1)$</span> separately.</p>
|
2,947,953 | <p>Given the two functions <span class="math-container">$$f(x) = \ln\left(\frac{x+1}{x-1}\right)$$</span> and <span class="math-container">$$g(x) = \ln(x+1)-\ln(x-1)$$</span>
I can justify independently why <span class="math-container">$\text{dom}(f) = (-\infty, -1) \cup (1,\infty)$</span>, and <span class="math-container">$\text{dom}(g)= (1,\infty)$</span>, but I'm not sure why these two functions have different domains. Can anyone enlighten me on what I'm missing? </p>
<p>EDIT: I plan on trying to explain this to freshman students taking their first calculus course. How can I justify that these domains are different? Is it true that <span class="math-container">$f(x)=g(x)$</span>? </p>
| Jean-Luc Bouchot | 24,153 | <p>Going back to the definition, a mapping <span class="math-container">$f: \begin{array}{ccc} D & \to & E \\ x &\mapsto &f(x)\end{array}$</span>is characterized by three elements: </p>
<ol>
<li>A domain <span class="math-container">$D$</span> where the mapping is defined (i.e. an <em>input</em> space)</li>
<li>A co-domain <span class="math-container">$E$</span> where the function may land (i.e. an <em>output</em> space)</li>
<li>A mechanism or black-box <span class="math-container">$f$</span> that <em>computes</em> (or gives) an output to a given output (for every input <span class="math-container">$x \in D$</span>, <span class="math-container">$f$</span> spits out the output <span class="math-container">$f(x)$</span>, hopefully in <span class="math-container">$E$</span>)</li>
</ol>
<p>Two functions can only be equal if these three elements are the same.</p>
<p>Even though the mechanisms (i.e. the black-box affecting output values to potential inputs) are the same, the domains, are not, as the way the black box is wired is not the same. </p>
<p>More precisely about your example... </p>
<p><span class="math-container">$log$</span> is only defined for inputs <span class="math-container">$> 0$</span>. So, considering <span class="math-container">$f$</span>, you need what is <em>inside</em> the <span class="math-container">$\log$</span> to be strictly positive, i.e.
<span class="math-container">$$
\frac{x+1}{x+1} > 0 \Leftrightarrow x+1 \text{ and } x-1 \text{ have the same sign.}
$$</span>
This is happening if (<span class="math-container">$x+1 > 0$</span> and <span class="math-container">$x-1 > 0$</span>) or (<span class="math-container">$x+1 < 0$</span> and <span class="math-container">$x- 1< 0$</span>), which again is true if <span class="math-container">$x \in (1,+\infty)$</span> or <span class="math-container">$x \in (-\infty, -1)$</span>, i.e. <span class="math-container">$x \in (-\infty,-1) \cup (1,\infty)$</span>.</p>
<p>Now, looking at the <span class="math-container">$g$</span>, you see that you have two <span class="math-container">$\log$</span>s that need to exist, i.e. you need <span class="math-container">$x+1 > 0$</span> AND <span class="math-container">$x-1 > 0$</span> which gives you the domain <span class="math-container">$x \in (1,\infty)$</span>. </p>
<p>Another way to look at this, is to think in terms of what you compute when. </p>
<p><strong>Case <span class="math-container">$f$</span>:</strong> Given a <span class="math-container">$x$</span> in some domain, you start by computing <span class="math-container">$x+1$</span> and keep the output in a stack on the side. Then you compute <span class="math-container">$x-1$</span> and save somewhere (so far, we have no issue with the domain). Then you unpile your two computed values and compute the ration <span class="math-container">$\frac{x+1}{x-1}$</span>, here you red-flag <span class="math-container">$x$</span> as not being equal to <span class="math-container">$1$</span> for this ratio to exist. Only now do you apply the <span class="math-container">$\log$</span> function, where you need the previous ration to be <span class="math-container">$>0$</span>. </p>
<p><strong>Case <span class="math-container">$g$</span>:</strong> Given a <span class="math-container">$x$</span> in some domain, you start by computing <span class="math-container">$x+1$</span>, and then throw the result into the <span class="math-container">$\log$</span> machine. At this moment, you read-flag <span class="math-container">$x+1 > 0$</span>. You leave the result on the side. You know deal with the second part of the addition; you compute <span class="math-container">$x-1$</span> and again throw the result into the <span class="math-container">$\log$</span> machine, flagging that <span class="math-container">$x-1 > 0$</span> and leave the result on the side. As a last step, you subtract the second result from the first one. </p>
<p>Looking at this problem from an algorithmic point of view shows you that you indeed have two different domains for the two different functions, even if both have the same output (on the elements both of them can be used). </p>
<p>Think about <span class="math-container">$f(x) = \sqrt{x^2}$</span> and <span class="math-container">$g(x) = (\sqrt{x})^2$</span>, it's probably easier to understand. </p>
|
99,799 | <p>I have a <code>Solve</code> similar to the following:</p>
<pre><code>Solve[e^2 - c^2 == -15, {e, c}, Integers]
(* {{e -> -7, c -> -8}, {e -> -7, c -> 8}, {e -> -1, c -> -4},
{e -> -1, c -> 4}, {e -> 1, c -> -4}, {e -> 1, c -> 4},
{e -> 7, c -> -8}, {e -> 7, c -> 8}} *)
</code></pre>
<p>I need to add a region constraint to get the solution I want from the unconstrained list of solutions. I tried the following:</p>
<pre><code>Solve[e^2 - c^2 == -15 ∧ {e, c} ∈ Interval[{0, 4}], {e, c}, Integers]
(* {{e -> {1}, c -> {4}}} *)
</code></pre>
<p>However, when I do this it wraps the variable's solutions in <code>List</code>. Is there a way to turn this off so I just get <code>{{e -> 1, c -> 4}}</code> or <code>{e -> 1, c -> 4}</code> as the result? The current result is a pain as I have to massage it for use with <code>Replace</code>. Also, can any explain why it is doing this when I constrain the variables? </p>
| Adam Strzebonski | 6,258 | <p>Interval is a 1D region, so <code>Element[e, Interval[...]]</code> makes e a 1D vector
not a scalar. If you want <code>e</code> to be a scalar use <code>Element[{e}, Interval[...]]</code>.</p>
<pre><code>In[1]:= Solve[e^2 - c^2 == -15 && Element[{e}|{c}, Interval[{0, 4}]], {e, c}, Integers]
Out[1]= {{e -> 1, c -> 4}}
</code></pre>
<p>Compare to:</p>
<pre><code>In[2]:= Solve[Element[e, Disk[]], e, Integers]
Out[2]= {{e -> {-1, 0}}, {e -> {0, -1}}, {e -> {0, 0}}, {e -> {0, 1}},
> {e -> {1, 0}}}
In[3]:= Solve[Element[{x, y}, Disk[]], {x, y}, Integers]
Out[3]= {{x -> -1, y -> 0}, {x -> 0, y -> -1}, {x -> 0, y -> 0},
> {x -> 0, y -> 1}, {x -> 1, y -> 0}}
</code></pre>
|
78,641 | <p>I am interested in the relation between the property of countable chain condition (ccc) and the property of separable. Could someone recommend some papers or books about this to me? thanks in advance.</p>
| Stefan Geschke | 7,743 | <p>Nathan's answer seems to indicate that there are some strange spaces that are ccc but not separable. But in fact, such spaces are rather common:</p>
<p>All products of separable Hausdorff spaces are ccc, but if the spaces have at least two different points, then products with more than $2^{\aleph_0}$ factors are not separable anymore. You can find this is basic books on topology, such as Engelking's book.</p>
<p>Also, in set-theoretic forcing, partial orders with the ccc are very common.
If you take the completion of a ccc po, you get a ccc Boolean algebra whose Stone space is ccc. Separability of the Stone space translates into $\sigma$-centeredness of p.o.s.
And there are a lot of partial orders used in forcing that are ccc but not $\sigma$-centered. </p>
|
834,228 | <p>$$u_{1}=2, \quad u_{n+1}=\frac{1}{3-u_n}$$
Prove it is decreasing and convergent and calculate its limit.
Is it possible to define $u_{n}$ in terms of $n$?</p>
<p>In order to prove it is decreasing, I calculated some terms but I would like to know how to do it in a more "elaborated" way.</p>
| AnnieOK | 130,682 | <p>Assuming the claim is true for $n$: $u_n < u_{n-1}$ </p>
<p>We'll show the claim is also right for $n+1$. Indeed:<br>
$$u_{n+1} = \frac{1}{3-u_n} < \frac{1}{3-u_{n-1}} = u_n$$</p>
<p>The inequality is of course based on the assumption.
For a full proof you should add the base case as well.</p>
|
781,776 | <blockquote>
<p>A red die, a blue die, and a yellow die (all six sided) are rolled. Given that no two of the dice land on the same number, what is the conditional probability that blue is less than yellow which is less than red?</p>
</blockquote>
<p>The Answer is a sixth. I have absolutely no idea how to do this though.</p>
| Caleb Stanford | 68,107 | <p>Your answer seems to be mixing computations where the computers are distinguishable, and computations where the computers are indistinguishable. In fact the problem specifies every computer is the same, so refer to David's answer for the correct solution.</p>
<p>If the computers <em>were</em> distinguishable, the start of your answer would be right. Then the entire formula would be
$$
\sum_{i = 0}^{4}(C(15,i) \left[ \sum_{j=0}^{15-i} C(15 - i, j) \right]
= \sum_{i=0}^4 C(15,i) 2^{15-i}.
$$</p>
|
448 | <p>Let's say, I have 4 yellow and 5 blue balls. How do I calculate in how many different orders I can place them? And what if I also have 3 red balls?</p>
| mau | 89 | <p>The case of two colors is simple: if you have m yellow balls and n blue ones you only need to choose m positions among (m+n) possibilities, that is (m+n)!/(m!·n!). The other balls' positions are automatically set up.</p>
|
3,419,276 | <p>I'm reading about the directional derivative:</p>
<blockquote>
<p>Let <span class="math-container">$(E,\|\cdot\|)$</span> and <span class="math-container">$(F,\|\cdot\|)$</span> be Banach spaces over the field <span class="math-container">$\mathbb{K}$</span>, and <span class="math-container">$X$</span> an open subset of <span class="math-container">$E$</span>. A function <span class="math-container">$f: X \rightarrow F$</span> is differentiable at <span class="math-container">$a \in X$</span> if there is <span class="math-container">$A \in \mathcal{L}(E, F)$</span> such that <span class="math-container">$$f(x)=f\left(a\right)+A\left(x-a\right)+o\left(\left\|x-a\right\|\right) \quad\left(x \rightarrow a\right)$$</span></p>
</blockquote>
<p>The author continues to define the directional derivative:</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/pmF7C.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pmF7C.png" alt="enter image description here"></a></p>
</blockquote>
<p>and prove a proposition:</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/xWPS6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xWPS6.png" alt="enter image description here"></a></p>
</blockquote>
<p>We define <span class="math-container">$g:\mathbb K \rightarrow F, \quad t \mapsto f\left(x_{0}+t v\right)$</span>. Our goal is to find the derivative of <span class="math-container">$g$</span> at <span class="math-container">$t=0$</span>.</p>
<p>Because <span class="math-container">$f$</span> is differentiable at <span class="math-container">$x_0$</span>, <span class="math-container">$$ f(x)=f\left(x_{0}\right)+\partial f\left(x_{0}\right)\left(x-x_{0}\right) + o(\left\|x-x_{0}\right\|) \quad (x \to x_0)$$</span> for all <span class="math-container">$x \in X$</span>. It follows that <span class="math-container">$$\begin{aligned}g(t) &= f(x_0+tv) \\ &=f\left(x_{0}\right)+\partial f\left(x_{0}\right)\left((x_0+tv)-x_{0}\right)+o(\left\|(x_0+tv)-x_{0}\right\|) \quad ((x_0+tv) \to x_0)\\ &= g\left(0 \right)+\partial f\left(x_{0}\right)\left(tv\right)+o(\left\|tv\right\|) \quad (t \to 0) \\&= g\left(0\right)+ \partial f\left(x_{0}\right)\left(v\right) \cdot (t-0) + o(\left\|t\right\|) \quad (t \to 0) \end{aligned}$$</span></p>
<p>Let <span class="math-container">$\partial g(0) \in \mathcal{L}(\mathbb K, F)$</span> be the derivative of <span class="math-container">$g$</span> at <span class="math-container">$t=0$</span>. It follows that <span class="math-container">$$g(t)= g(0) + \partial g(0)(t-0) + o(\left\|t\right\|) \quad (t \to 0)$$</span></p>
<p>To sum up, we have <span class="math-container">$$\begin{aligned}g(t) &=g\left(0\right)+ \partial f\left(x_{0}\right)\left(v\right) \cdot (t-0) + o(\left\|t\right\|) \quad (t \to 0)\\ &= g(0) + \partial g(0)(t-0) + o(\left\|t\right\|) \quad (t \to 0)\end{aligned}$$</span></p>
<p>Hence <span class="math-container">$\partial f\left(x_{0}\right)\left(v\right) \cdot (t-0) = \partial g(0)(t-0)$</span> and thus <span class="math-container">$\partial f\left(x_{0}\right)\left(v\right) \cdot t = \partial g(0)(t)$</span>. Because <span class="math-container">$(t)$</span> on the right hand is just the input of the function <span class="math-container">$\partial g(0)$</span>, we have <span class="math-container">$\partial f\left(x_{0}\right)\left(v\right) \cdot t = \partial g(0)$</span>.</p>
<p>In my understanding, <span class="math-container">$\partial g(0)(t)$</span> denote the value of the function <span class="math-container">$\partial g(0)$</span> at <span class="math-container">$t$</span>. On the contrary, <span class="math-container">$\partial f\left(x_{0}\right)\left(v\right) \cdot t$</span> denoted the product of <span class="math-container">$\partial f\left(x_{0}\right)\left(v\right)$</span> and <span class="math-container">$t$</span>.</p>
<p><strong>My question:</strong></p>
<p>I could not understand why the answer given in my textbook is just <span class="math-container">$\partial f\left(x_{0}\right)\left(v\right)$</span>, which is lack of <span class="math-container">$t$</span>.</p>
<p>Could you please elaborate on this point?</p>
| fleablood | 280,126 | <p>Assuming <span class="math-container">$b >0; b\ne 1$</span> then <span class="math-container">$\log_b u$</span> is, by definition, <span class="math-container">$\text{"whatever power we must raise b to in order to get u"}$</span>.</p>
<p>So <span class="math-container">$b^{\log_b u} = b^{\text{"whatever power we must raise b to in order to get u"}}=\text{b raised to a power that gives us u as a result}= \text{u as a result} = u$</span></p>
<p>In other words.... It's just the definition of logarithm.</p>
|
2,990,580 | <p>I am doing my maths A-level*. Often when I am at home I get questions about why we solve certain problem types in a certain way. One example is "why does completing the square work?"</p>
<p>Is there a website which collects explanations like these together for me to read? <strong><em>Preferably one that is aimed at A-level students.</em></strong></p>
<p>*Roughly equivalent to American AP classes, to give an idea of the level needed.</p>
| M. Damon | 573,313 | <p>Would this work instead, </p>
<p><span class="math-container">$$x \in A \rightarrow x \in A\cup C \rightarrow x\in B\cup C\\\text{so we either have that:}\\(1)\;x\in B , x\notin C\\(2)\;x\in C, x\notin B \\(3)\;x\in B, x\in C\\ \text{For (1) we have that $x \in B$. For (3) we have that $x \in B$. So we would just need to worry about (2).} \\\text{(2) States $x \in C$. Therefore, $x\in C$ and $x\in A$. So $x\in A\cap C.$ But then $x \in B\cap C$, so $x \in B$. But $x\notin B$.}\\\text{So only (1) and (3) could hold. In which case, $x \in B$ and $A \subseteq B.$}$$</span></p>
<p>And it would be the same type of logic for <span class="math-container">$B \subseteq A.$</span> Would this work instead? </p>
|
2,007,173 | <p>I've managed to severely confuse myself in my attempts to simplify this seemingly straightforward expression:
$$
\arctan(\cot(\alpha)),\quad\text{with $0<\alpha\leq\pi$.}
$$
It seems like maybe there are some issues with domain, as $\cot$ and $\tan$ are defined on $(0,\pi)$ and $(-\frac{\pi}{2},\frac{\pi}{2})$, respectively. However, $\arctan$ <em>is</em> able to "deal" with the fact that $\cot(\pi)=-\infty$, since its limit exists at negative infinity.</p>
<p>What method should I be using to simplify this? I believe that I should be getting $\frac{\pi}{2}-\alpha$, but I'm not sure how to show it symbolically.</p>
| Stefano | 387,021 | <p>One way to see this is to calculate the derivative of $\arctan(\cot(\alpha))$, which is identically equal to $-1$. Therefore, since you are restricting yourself to the interval $(0, \pi)$, you can conclude that your expression is equal, on that interval, to $-\alpha + c$ for some $c \in \mathbb R$. Finally, setting $\alpha = \pi/2$ gives $c = \pi/2$.</p>
<p>p.s. Even if the limit of $\arctan(\cot(\alpha))$ as $\alpha \to \pi$ is finite, you still can't evaluate that expression at $\alpha = \pi$. Same thing happens at zero, by the way!</p>
|
1,200,919 | <p>Let $x$ be the solution of the equation $x^x=2$. Is $x$ irrational? How to prove this?</p>
| Bart Michels | 43,288 | <p>Suppose $x$ is rational. Clearly $x<2$ and $x>1$, so $x$ is not an integer. Now from this question we obtain that $x^x=2$ is irrational, a contradiction: <a href="https://math.stackexchange.com/questions/978174">If $x$ is a positive rational but not an integer, is $x^x$ irrational?</a></p>
|
85,309 | <p>Let A and B be positive semidefinite matrices. It is not hard to see that $(A-B)^2 \leq 2A^2 + 2B^2$. In fact, $2A^2 + 2B^2 - (A-B)^2 = (A+B)^2$ is positive semidefinite. </p>
<p>My question is: Is there a constant c (independent of A and B and the dimension) such that</p>
<p>$$(A-B)^2 \leq c (A+B)^2?$$</p>
<p>Thanks.</p>
| Noam D. Elkies | 14,830 | <p>There is no such $c$ even if we use only $2 \times 2$ matrices.
For any $c \geq 1$ let $A,B$ be the positive-semidefinite matrices
$$
A = \left( \begin{array}{lc} c^2 & c \cr c & 1 \end{array} \right),
\phantom\infty
B = \left( \begin{array}{cc} 1 & 0 \cr 0 & 0 \end{array} \right).
$$
of rank $1$. Then we calculate that the difference
$$
D := c(A+B)^2 - (A-B)^2
$$
has determinant $(c-1)^2 - 4c^3 < 0$, and is thus not positive semidefinite.</p>
<p>In fact this counterexample works for all $c \in \bf R$:
looking around $\ker A = {\rm span} \lbrace(-1,c)\rbrace$
we find the negative vector $v = (-c, c^2+1)$.
To verify that $\langle v, Dv \rangle < 0$, recall that
for any vector $x$ and any symmetric matrix $M$ of the same order we have
$$
\langle x, M^2 x \rangle = \langle Mx, Mx \rangle = |Mx|^2.
$$
Here we compute $v(A+B) = (0,1)$ and $v(A-B) = (2c,1)$, so
$$
\langle v, Dv \rangle = c \phantom. |(0,1)|^2 - |(2c,1)|^2 = -4c^2 + c - 1,
$$
which is negative for all real $c$. If $c$ is small enough that
$\det(D) > 0$ then $D$ is <em>negative</em> definite.</p>
|
356,306 | <p>If $f:X_1 \rightarrow X_2$ and $g:X_2 \rightarrow X_3$ are homomorphisms.
If $g \circ f =0$ does it imply that $Im f \subseteq ker g$? and how to show that? do you have an example?
thanks :)</p>
| Federica Maggioni | 49,358 | <p>Take an element $y\in Im f$. Then $y$ is of the form $y=f(x)$, for some $x\in X_1$. Now apply $g$ to $y$. You get $g(y)=g(f(x))=(g\circ f)(x)$, which is zero by assumption. As an example, take $f$ arbitrary and $g(y)=0$ for every $y\in X_2$.</p>
|
1,234,471 | <p>Given two sequences $(a_k),(b_k)$ with $a_k\geq0,b_k>0$ such that the power series $\sum_{k=0}^\infty a_k b_kr^{k}$ and $\sum_{k=0}^\infty a_kr^k$ converge for each $r>0$. My question now is: Does there exist a constant $c$ (depending only on $(b_k))$ such that
\begin{align*}
\sum_{k=0}^\infty a_kb_kr^{k}\geq c\sum_{k=0}^\infty a_kr^{k}
\end{align*}
for all large $r>0$? This is obvious for if $(b_k)$ is bounded away from zero, but what if $\liminf b_k=0?$
In my case, $b_k=1/k!$, and I have no idea how to approach this problem...
Any help is highly appreciated! Thanks in advance.</p>
| String | 94,971 | <p>Let $1_n$ be the set of such strings of length $n$ ending in $1$. Similarly let $0_n$ be the set of such strings ending in $0$. Then we have
$$
\begin{align}
1_{n+1}&=\{x\parallel 1\mid x\in 1_n\cup 0_n\}\\
0_{n+1}&=\{x\parallel 0\mid x\in 1_n\}
\end{align}
$$</p>
<hr>
<p>Define $A_n=|1_n|$ and $B_n=|0_n|$. Then we have from above
$$
\begin{align}
A_{n+1}&=A_n+B_n\\
B_{n+1}&=A_n
\end{align}
$$
with $A_1=B_1=1$ and thus $A_2=2$. Combining those we get
$$
A_{n+1}=A_n+A_{n-1}
$$
also known as the Fibonacci recurrence relation. Hence
$$
A_n=F_n=\frac{(1+\sqrt 5)^2+(1-\sqrt 5)^n}{2^n\sqrt 5}
$$
and the number you seek is
$$
A_{10}+B_{10}=F_{10}+F_{9}=89+55=144
$$</p>
|
1,384,053 | <p>Which number is bigger? $1.01^{101}$ or $2$? and how about $e^{\pi}$ or $\pi^e$?</p>
<p>Tried some algebraic manipulations to no end, so would love some suggestions or some different ways to approach those kind of problem</p>
| Ben Grossmann | 81,360 | <p><strong>Hint for the first:</strong> With $x = 0.01$ and $n = 101$, note that
$$
(1 + x)^n = 1 + n\,x + \binom n2 x^2 + \cdots \geq 1 + n\,x
$$</p>
<p><strong>Hint for the second:</strong> It is equivalent to show that $e^{1/e} > \pi^{1/\pi}$. In order to do so, consider the function $f(x) = x^{1/x}$ and use calculus.</p>
|
1,384,053 | <p>Which number is bigger? $1.01^{101}$ or $2$? and how about $e^{\pi}$ or $\pi^e$?</p>
<p>Tried some algebraic manipulations to no end, so would love some suggestions or some different ways to approach those kind of problem</p>
| Claude Leibovici | 82,404 | <p>Consider $$A=1.01^{101}$$ So $$\log(A)=101\log(1.01)$$ Now use the very fast convergent expansion $$\log\Big(\frac{1+x}{1-x}\Big)=2\,\Big(\frac {x}{1}+\frac {x^3}{3}+\frac {x^5}{5}+\cdots\Big)$$ and make $\frac{1+x}{1-x}=\frac{101}{100}$ that is to say $x=\frac{1}{201}$. The first term (to which only positive terms will be added) is already $$\log(A)=101\log(1.01)=2\times 101\times \frac{1}{201}=\frac{202}{201}>1$$ and then $A>e>2$</p>
|
3,450,581 | <p>I know convergence-preserving functions have been discussed a fair amount in the past; however, I was a looking at <a href="https://math.stackexchange.com/questions/1337042/sum-a-n-converges-iff-sum-fa-n-converges/1337057#1337057">another post</a>, and I saw the following result: if <span class="math-container">$f$</span> is Lipschitz and <span class="math-container">$f(0)=0$</span>, then <span class="math-container">$\sum |a_n| <\infty \Rightarrow \sum |f(a_n)|<\infty$</span>. How exactly would I go about proving this statement? I'm also guessing from other posts that this is a sufficient but not necessary condition?</p>
<p>Edit: made it absolute convergence instead.</p>
| Cye Waldman | 424,641 | <p>I think I understand your dilemma now. Say that <span class="math-container">$m=a+b$</span> is fixed and we let <span class="math-container">$b=1/n$</span>, where <span class="math-container">$n$</span> is an integer, so that we satisfy the condition that <span class="math-container">$m/b$</span> is indeed an integer. Now, what happens when we let <span class="math-container">$n\to \infty$</span>? We can never actually achieve <span class="math-container">$b=0$</span>. It's sort of like Zeno's paradox; we just can't get there. I demonstrated this numerically, i.e., for any value of <span class="math-container">$n$</span> the arc length is <span class="math-container">$L=8m$</span>. If you plot this for large <span class="math-container">$n$</span> it <em>looks</em> like a circle, but it isn't. There are a large number of cusps and adding up the total length always gives the same result. If you try to <em>force</em> <span class="math-container">$b=0$</span> then you are confronted with the problem that <span class="math-container">$m/b$</span> is not really defined. At the same time, we can look at the area enclosed, which I believe to be <span class="math-container">$A=\pi m(m-b)$</span>, we see that area of the epicycloid approaches that of the circle of larger radius for large <span class="math-container">$n$</span>.</p>
|
632,891 | <p>I'm trying to solve this limit, for which I already know the solution thanks to Wolfram|Alpha to be $\sqrt[3]{abc}$:</p>
<p>$$\lim_{n\rightarrow\infty}\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n:\forall a,b,c\in\mathbb{R}^+$$</p>
<p>As this limit is an indeterminate form of the type $1^\infty$, I've been trying to approach it by doing:</p>
<p>$$\lim_{n\rightarrow\infty}\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{1}{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{1}{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}}\right)^{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}\cdot\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n}=e^{\lim_{n\rightarrow\infty}\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n}$$</p>
<p>But now when I approach that top limit this is what I get:</p>
<p>$$\lim_{n\rightarrow\infty}\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n=\lim_{n\rightarrow\infty}\frac{n\cdot a^{\frac{1}{n}}}{3}+\frac{n\cdot b^{\frac{1}{n}}}{3}+\frac{n\cdot c^{\frac{1}{n}}}{3}-n=\lim_{n\rightarrow\infty}\frac{n\cdot a^0}{3}+\frac{n\cdot b^0}{3}+\frac{n\cdot c^0}{3}-n=\lim_{n\rightarrow\infty}\frac{n}{3}+\frac{n}{3}+\frac{n}{3}-n=0$$</p>
<p>And hence the final limit should be $e^0=1$ which is clearly wrong but I honestly don't know what I did wrong, so what do you suggest me to solve this limit?</p>
| Yiorgos S. Smyrlis | 57,021 | <p>Fact I.
$$
\lim_{n\to\infty}\left(1+\frac{a}{n}+\frac{b}{n^2}\right)^{\!n}=\mathrm{e}^a.
$$</p>
<p>Fact II. For every $a>0$, there exists a $b>0$, such that
$$
1+\frac{\ln a}{n}\le a^{1/n} \le 1+\frac{\ln a}{n}+\frac{b}{n^2}.
$$</p>
<p>Using the two facts:
$$
1+\frac{\ln a+\ln b+\ln c}{3n}\le\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\le
1+\frac{\ln a+\ln b+\ln c}{3n}+\frac{k}{n^2}
$$
and hence
$$
\left(1+\frac{\ln\sqrt[3]{abc}}{n}\right)^{\!n}\le
\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^{\!n}
\le\left(1+\frac{\ln(\sqrt[3]{abc})}{n}+\frac{k}{n}\right)^{\!n}
$$
and as the both the left and right hand side tend to $\sqrt[3]{abc}$, so does the middle one.</p>
|
51,390 | <p>Say $g$ is a matrix which is given as, $g = g_0 + xg_2 + x^2 g_4 .. +x^{d/2 -1}g_{d-2}+ x^{d/2}(g_d + h_d(\log (x)))$ where $d$ is an even number and each $g_i$ is a matrix (same dimension as $g$) and $h_d$ is another matrix. </p>
<ul>
<li>For such a set of arbitrary matrices, how can one power-series expand $\sqrt {\det(g)}$ in $x$? </li>
</ul>
| Carl Woll | 45,431 | <p>You can use my function <a href="https://mathematica.stackexchange.com/a/162063/45431">MatrixD</a> from the question <a href="https://mathematica.stackexchange.com/q/161918/45431">det simplification</a>:</p>
<pre><code>MatrixD[expr_, x__] := With[
{old = OptionValue[SystemOptions[], "DifferentiationOptions"->"ExcludedFunctions"]},
Internal`WithLocalSettings[
SetSystemOptions["DifferentiationOptions"->"ExcludedFunctions"->Join[old, {Det, Inverse, Tr}]];
Unprotect[D];
(* handle list derivatives *)
D[h:((Det|Tr|Inverse)[m_]), {z_, n_Integer}] := Nest[D[#, Replace[z, _List :> {z}]]&, h, n];
D[h:((Det|Tr|Inverse)[m_]), {z_List}] := D[h, #]& /@ z;
D[h:((Det|Tr|Inverse)[m_]), z_, y___] := D[D[h, z], y];
(* define derivatives for Det, Tr, and Inverse *)
D[Det[m_], z:Except[_List]] := Det[m] Tr[Inverse[m] . D[m,z]];
D[Tr[m_], z:Except[_List]] := Tr[D[m,z]];
D[Inverse[m_], z:Except[_List]] := -Inverse[m] . D[m, z] . Inverse[m],
D[expr, x],
SetSystemOptions["DifferentiationOptions"->"ExcludedFunctions"->old];
Clear[D];
Protect[D]
]
]
</code></pre>
<p>For your example:</p>
<pre><code>g[x_] := g0 + x g2 + x^2 g4 + x^3 g6 + x^3 hd Log[x];
</code></pre>
<p>Then:</p>
<pre><code>coeffs = Table[
MatrixD[Sqrt[Det[g[x]]] /. Log[x]->u, {x, n}]/n! /. {x->0, u->Log[x]},
{n, 0, 3}
];
Print @* TeXForm /@ coeffs;
</code></pre>
<blockquote>
<p>$\sqrt{\left| \operatorname{g0}\right| }$</p>
<p>$\frac{1}{2} \sqrt{\left| \operatorname{g0}\right| } \operatorname{Tr}\left[\operatorname{g0}^{-1}.\operatorname{g2}\right]$</p>
<p>$\frac{1}{2} \left(\frac{1}{2} \sqrt{\left| \operatorname{g0}\right| } \operatorname{Tr}\left[\left(-\operatorname{g0}^{-1}.\operatorname{g2}.\operatorname{g0}^{-1}\right).\operatorname{g2}+\operatorname{g0}^{-1}.(2 \operatorname{g4})\right]+\frac{1}{4} \sqrt{\left| \operatorname{g0}\right| } \operatorname{Tr}\left[\operatorname{g0}^{-1}.\operatorname{g2}\right]^2\right)$</p>
<p>$\frac{1}{6} \left(\frac{1}{2} \sqrt{\left| \operatorname{g0}\right| } \operatorname{Tr}\left[2 \left(-\operatorname{g0}^{-1}.\operatorname{g2}.\operatorname{g0}^{-1}\right).(2 \operatorname{g4})+\left(-\left(-\operatorname{g0}^{-1}.\operatorname{g2}.\operatorname{g0}^{-1}\right).\operatorname{g2}.\operatorname{g0}^{-1}-\operatorname{g0}^{-1}.\operatorname{g2}.\left(-\operatorname{g0}^{-1}.\operatorname{g2}.\operatorname{g0}^{-1}\right)-\operatorname{g0}^{-1}.(2 \operatorname{g4}).\operatorname{g0}^{-1}\right).\operatorname{g2}+\operatorname{g0}^{-1}.(6 \operatorname{g6}+6 \operatorname{hd} \log (x))\right]+\frac{3}{4} \sqrt{\left| \operatorname{g0}\right| } \operatorname{Tr}\left[\operatorname{g0}^{-1}.\operatorname{g2}\right] \operatorname{Tr}\left[\left(-\operatorname{g0}^{-1}.\operatorname{g2}.\operatorname{g0}^{-1}\right).\operatorname{g2}+\operatorname{g0}^{-1}.(2 \operatorname{g4})\right]+\frac{1}{8} \sqrt{\left| \operatorname{g0}\right| } \operatorname{Tr}\left[\operatorname{g0}^{-1}.\operatorname{g2}\right]^3\right)$</p>
</blockquote>
<p>Let's evaluate the first few terms of the series using an explicit set of matrices:</p>
<pre><code>SeedRandom[2];
rules = Thread[{g0, g2, g4, g6, hd} -> RandomReal[1,{5,3,3}]]
</code></pre>
<blockquote>
<p><code>{g0 -> {{0.72224, 0.109449, 0.470703}, {0.535582, 0.583178,
0.293942}, {0.165154, 0.601258, 0.754218}},
g2 -> {{0.771123, 0.778574, 0.0236104}, {0.922757, 0.992454,
0.350409}, {0.0450047, 0.501359, 0.633756}},
g4 -> {{0.642208, 0.389875, 0.664971}, {0.843882, 0.56904,
0.398212}, {0.238652, 0.673513, 0.419507}},
g6 -> {{0.587398, 0.00833523, 0.942441}, {0.771263, 0.147503,
0.964774}, {0.898747, 0.332963, 0.204548}},
hd -> {{0.839035, 0.250388, 0.238638}, {0.616616, 0.879303,
0.404504}, {0.402517, 0.516192, 0.292009}}}</code></p>
</blockquote>
<p>And the comparison:</p>
<pre><code>N @ CoefficientList[Series[Sqrt[Det[g[x]]] /. rules, {x, 0, 3}], x]
coeffs /. rules //Expand
</code></pre>
<blockquote>
<p>{0.507317, 0.663253, 0.436586, 0.0826776 + 0.572028 Log[x]}</p>
<p>{0.507317, 0.663253, 0.436586, 0.0826776 + 0.572028 Log[x]}</p>
</blockquote>
|
3,082,080 | <p><a href="https://i.stack.imgur.com/Dcf40.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Dcf40.png" alt="enter image description here"></a></p>
<blockquote>
<p>The sides <span class="math-container">$AB, BC, CD$</span> of trapezoid <span class="math-container">$ABCD$</span> touches the circle with
center <span class="math-container">$O$</span> and they are equal. <span class="math-container">$AD$</span>, goes through the point O. If
diameter is <strong>2</strong>, then the area of the trapezoid is <span class="math-container">$√a$</span> . What is
the value of a?</p>
</blockquote>
<p>Source: <a href="https://www.matholympiad.org.bd/" rel="nofollow noreferrer">Bangladesh Math Olympiad 2016 Junior Catagory</a></p>
<p>I tried but I could not find any possible solution. How am I supposed to get <span class="math-container">$BC + AD$</span>?</p>
| dfnu | 480,425 | <p><strong>Hint</strong>. </p>
<p>Once you showed that <span class="math-container">$ABO$</span>, <span class="math-container">$OBC$</span> and <span class="math-container">$COD$</span> are equilater triangles, note that the altitude of such triangles is equal to <span class="math-container">$r=1$</span>. By Pythagorean theorem on half of the triangle determine the side, i.e. <span class="math-container">$\frac{2\sqrt{3}}{3}$</span>. The total area of the trapezoid is equal to three times the area of the triangle, i.e. <span class="math-container">$\sqrt{3}$</span></p>
|
4,347,308 | <p><em><strong>Definition:</strong></em></p>
<p>Let <span class="math-container">$(X,\mathscr{A},\mu)$</span> be a measurable space, an atom of the measure <span class="math-container">$\mu$</span> is a set <span class="math-container">$A \in\mathscr{A}$</span> with the property that
<span class="math-container">$\mu(A) > 0$</span> and for any <span class="math-container">$B\in \sigma (A)$</span> either <span class="math-container">$\mu(B) = 0$</span>, or <span class="math-container">$\mu(A \setminus B) = 0$</span>. If a measure has
atoms it is called atomic; in the opposite case, the measure is called non-atomic (or
atomeless). A measure is called purely atomic if <span class="math-container">$X$</span> can be written as the union of a
finite or countable number of atoms.</p>
<p>From the definition of atoms, we get the following corollary:</p>
<p><em><strong>Corollary:</strong></em></p>
<p><em>Every purely atomic measure is an atomic measure.</em></p>
<p>I am trying to find an example of an atomic measure that is not purely atomic, can anyone help me?</p>
| Paul | 202,111 | <p><span class="math-container">$$\frac{1+x^3}{1+x^2} =x+\frac{1-x}{1+x^2}=x+\frac{1}{1+x^2}-\frac{x}{1+x^2}$$</span>
So you can use the Maclaurin series for <span class="math-container">$\frac{1}{1+x^2}$</span> to get the answer.</p>
|
3,009,477 | <p>I want to determine <span class="math-container">$f(x) = x+\sin x$</span> is homeomorphic or not on <span class="math-container">$\mathbb{R}$</span>?</p>
<p>A bijective continuous function is homeomorphic if its inverse is also continuous.
I know that <span class="math-container">$f$</span> is bijective. Also <span class="math-container">$f$</span> is continuous being the sum of two continuous functions.</p>
<p>How to look for the continuity of <span class="math-container">$f^{-1}$</span>.</p>
| egreg | 62,967 | <p>Since <span class="math-container">$H$</span> is commutative, <span class="math-container">$x\circ y=xy+yx=2xy$</span>. This new multiplication is obviously commutative, but need not have a neutral element. For instance, if <span class="math-container">$H$</span> has characteristic <span class="math-container">$2$</span>, we have <span class="math-container">$x\circ y=0$</span> for every <span class="math-container">$x,y\in H$</span>.</p>
<p>Is the operation associative? <span class="math-container">$(x\circ y)\circ z=(2xy)\circ z=4xyz$</span> and <span class="math-container">$x\circ(y\circ z)=2x(y\circ z)=2x(2yz)=4xyz$</span>.</p>
<p>Distributivity is obvious.</p>
<p><strong>Note.</strong> If <span class="math-container">$n$</span> is an integer and <span class="math-container">$x\in H$</span>, <span class="math-container">$nx$</span> is the standard multiple.</p>
<p>Under what condition does the circle operation have a neutral element <span class="math-container">$e$</span>? We need it satisfies <span class="math-container">$2ex=x$</span>, for every <span class="math-container">$x$</span>, in particular for <span class="math-container">$\mathbf{1}$</span> (the neutral element of <span class="math-container">$H$</span>), so <span class="math-container">$2e=\mathbf{1}$</span>. This condition is also easily seen to be sufficient: indeed, if <span class="math-container">$2e=\mathbf{1}$</span>, then <span class="math-container">$e\circ x=2ex=\mathbf{1}x=x$</span>.</p>
<p>Thus the ring <span class="math-container">$(H,+\circ)$</span> is unital if and only if <span class="math-container">$2\mathbf{1}=\mathbf{1}+\mathbf{1}$</span> is invertible in <span class="math-container">$(H,+,\cdot)$</span>. This is granted if the characteristic of <span class="math-container">$H$</span> is finite and coprime with <span class="math-container">$2$</span>. It is false whenever the characteristic is divisible by <span class="math-container">$2$</span>: if the characteristic is <span class="math-container">$2k$</span>, then <span class="math-container">$k\mathbf{1}=(k\mathbf{1})\circ e=2ke=0$</span> gives a contradiction.</p>
<p>If the characteristic is <span class="math-container">$0$</span>, it may happen or not: the identity doesn't exist for <span class="math-container">$H=\mathbb{Z}$</span>, it obviously exists for <span class="math-container">$H=\mathbb{Q}$</span>.</p>
|
550,188 | <p>Okay so I have an equation in my book which is as follows..
$$
\frac {a}{s(s+a)}
$$
it says "using partial fractions this can be expanded to
$$
\frac {1}{s} + \frac {-1}{s+a}
$$</p>
<p>My usual method would be to cross multiply and do something like this
$$
\frac {a}{s(s+a)} = \frac {A(s+a)}{s(s+a)} + \frac {B(s)}{s(s+a)}
$$</p>
<p>Then cancel off the denominators and solve..</p>
<p>$$
a = A(s+a) + B(s)
$$</p>
<p>usually though the a would be some constant but here I have no values to play around with.. how has he done it in the book?</p>
| Martin Argerami | 22,857 | <p>The equality you have is
$$
a=(A-B)s+aA.
$$
This suggests taking $A=B=1$. </p>
|
3,024,169 | <p>First, my apologies. This question may have been asked many times before but I do not know the correct terms to search on..... and my school trigonometry is many years ago. Pointing me to an appropriate already-answered question would be an ideal solution for me.</p>
<p>I am writing a program to do 3D view from an observer. I want to calculate motion based upon the direction the observer is looking. I currently define this as angle from directly ahead in two planes. Rotation about the x (left-right) axis gives me elevation (elevationRadians), and rotation about y (up-down) axis gives me left-right. Rotation about z never happens.</p>
<p>I need to calculate the change in cartesian coordinates caused by moving D units in the direction of view.</p>
<pre><code>dx = D * cos(elevationRadians)
dy = D * sin(deflectionRadians)
</code></pre>
<p>But I now get two components for dz</p>
<pre><code>dz = D * cos(deflectionRadians)
dz = D * sin(elevationRadians)
</code></pre>
<p>How should I combine these terms to give realistic movement? Should I add them, average them, or something else? Part of me thinks that simply adding them will give too large a dz. Could someone confirm, deny, or point me to a good resource for this please.......</p>
<p>Edit: rotation about y gives 'deflection.. </p>
<p>My axes: x: left -> right, y: down -> up, z: behind -> in front.</p>
<p>Thanks to Andei's amswer which almost worked (I think our axes differed), I wound up with this:</p>
<p>dx = D cos(elevationRadians) cos(deflectionRadians)
dy = D sin(elevationRadians) sin(deflectionRadians)
dz = D sin(elevationRadians)</p>
<p>which <em>seems</em> to work a lot better than what I had.</p>
<p>What is the mathematical term for what I am trying to do? I need to read some theory preferably at the simple end of the scale.</p>
| Andrei | 331,661 | <p>I think you don't define your axes correctly. You have rotation around <span class="math-container">$z$</span>, and in the next sentence you say it never happens. </p>
<p>I think you should use polar coordinates. If you call <span class="math-container">$\theta$</span> the angle from the vertical direction <span class="math-container">$z$</span>, <span class="math-container">$\theta=\pi/2-\text{elevationRadians}$</span>, and <span class="math-container">$\phi=\text{deflectionRadians}$</span> is the angle that the projection in the horizontal plane makes with the <span class="math-container">$x$</span> axis, you have:<span class="math-container">$$\begin{align}x&=D\sin\theta\cos\phi\\y&=D\sin\theta\sin\phi\\z&=D\cos\theta\end{align}$$</span></p>
|
3,845,570 | <p>Premises: <span class="math-container">$\neg(A \to B)\ ,\ \neg B \to C$</span> .</p>
<p>Conclusion: <span class="math-container">$C$</span></p>
<p>My intuition is that I should do a sub-derivation where I prove <span class="math-container">$\neg C$</span> is an absurdity. However, I soon run into issues. If I could prove that <span class="math-container">$B$</span> is an absurdity, that would work also, but I'm not sure how to do so using the first premise.</p>
| DanielV | 97,045 | <p>Start by assuming that B is true.</p>
<blockquote class="spoiler">
<p> You can prove a contradiction from that, to establish <span class="math-container">$\lnot B$</span>.</p>
</blockquote>
<blockquote class="spoiler">
<p> Within the assumption you can prove <span class="math-container">$A \to B$</span></p>
</blockquote>
|
2,352,721 | <h2>Question</h2>
<blockquote>
<p>Four fair six-sided dice are rolled. The probability that the sum of the results being <span class="math-container">$22$</span> is <span class="math-container">$$\frac{X}{1296}.$$</span> What is the value of <span class="math-container">$X$</span>?</p>
</blockquote>
<h2>My Approach</h2>
<p>I simplified it to the equation of the form:</p>
<blockquote>
<p><span class="math-container">$x_{1}+x_{2}+x_{3}+x_{4}=22, 1\,\,\leq x_{i} \,\,\leq 6,\,\,1\,\,\leq i \,\,\leq 4 $</span></p>
</blockquote>
<p>Solving this equation results in:</p>
<p><span class="math-container">$x_{1}+x_{2}+x_{3}+x_{4}=22$</span></p>
<p>I removed restriction of <span class="math-container">$x_{i} \geq 1$</span> first as follows-:</p>
<p><span class="math-container">$\Rightarrow x_{1}^{'}+1+x_{2}^{'}+1+x_{3}^{'}+1+x_{4}^{'}+1=22$</span></p>
<p><span class="math-container">$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$</span></p>
<p><span class="math-container">$\Rightarrow \binom{18+4-1}{18}=1330$</span></p>
<p>Now i removed restriction for <span class="math-container">$x_{i} \leq 6$</span> , by calculating the number of <strong>bad cases</strong> and then subtracting it from <span class="math-container">$1330$</span>:</p>
<p>calculating <strong>bad combination</strong> i.e <span class="math-container">$x_{i} \geq 7$</span></p>
<p><span class="math-container">$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$</span></p>
<p>We can distribute <span class="math-container">$7$</span> to <span class="math-container">$2$</span> of <span class="math-container">$x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$</span> i.e<span class="math-container">$\binom{4}{2}$</span></p>
<p>We can distribute <span class="math-container">$7$</span> to <span class="math-container">$1$</span> of <span class="math-container">$x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$</span> i.e<span class="math-container">$\binom{4}{1}$</span> and then among all others .</p>
<p>i.e</p>
<p><span class="math-container">$$\binom{4}{1} \binom{14}{11}$$</span></p>
<p>Therefore, the number of bad combinations equals <span class="math-container">$$\binom{4}{1} \binom{14}{11} - \binom{4}{2}$$</span></p>
<p>Therefore, the solution should be:</p>
<p><span class="math-container">$$1330-\left( \binom{4}{1} \binom{14}{11} - \binom{4}{2}\right)$$</span></p>
<p>However, I am getting a negative value. What am I doing wrong?</p>
<p><strong>EDIT</strong></p>
<p>I am asking for my approach, because if the question is for a larger number of dice and if the sum is higher, then predicting the value of dice will not work.</p>
| Toby Mak | 285,313 | <p>This is a method similar to @GTonyJacobs and @adhg's answers, but in this answer I use a slightly different way of thinking: </p>
<p>The maximum number possible is a set of six $4$'s $(6, 6, 6, 6)$, which totals $6 * 4 = 24$. To reach $22$, we need to take off $2$ in total from the set. </p>
<p>There are $2$ ways to do this: </p>
<p>– take off $1$ from $2$ numbers out of $4$.</p>
<p>– take off $2$ from $1$ number out of $4$.</p>
<p>Following this, the first method can be done in $4 \choose 2$ ways, while the second can be done in $4 \choose 1$ ways, which is $10$ ways in total.</p>
<p>Therefore, the probability of $4$ dice totalling $22$ is $\frac{10}{6^4}$, which is approximately $0.0077$.</p>
|
271,824 | <p>I have a list= {4, 8, 10, 11, 12, 14, 16, 7, 9}</p>
<p>How can i partition the list by group of Arithmetic Progression with common difference 1 :</p>
<p>{{4}, {8}, { 10, 11, 12}, {14}, {16}, {7}, {9}}</p>
| Lukas Lang | 36,508 | <p>The reason <code>Except</code> is not working for you is because the entire list <code>li1</code> also matches that pattern, so everything is simply replaced by <code>x</code>. To fix it, you need to make sure <code>Except</code> matches only what you want. The easiest is to simply use <a href="https://reference.wolfram.com/language/ref/Replace.html" rel="noreferrer"><code>Replace</code></a> with a level specification of 1:</p>
<pre><code>Replace[li1, Except[Alternatives @@ li2] -> x, 1]
(* {x, {2, 9}, x, {4, 7}, {5, 6}, x, x, x, {9, 2}, x} *)
</code></pre>
<p>An alternative is to use the second argument of <a href="https://reference.wolfram.com/language/ref/Except.html" rel="noreferrer"><code>Except</code></a> to force it to only match pairs of things (note that this will fail when <code>li1</code> is itself two elements long):</p>
<pre><code>li1 /. Except[Alternatives @@ li2, {_, _}] -> x
(* {x, {2, 9}, x, {4, 7}, {5, 6}, x, x, x, {9, 2}, x} *)
</code></pre>
|
4,041,842 | <p>I want to solve for <span class="math-container">$t \in \mathbb{R}, u'(t)=-u(t)\ln \lvert u(t) \rvert$</span>.</p>
<p>I defined two cases: <span class="math-container">$\mathbb{R^*_+}$</span> and <span class="math-container">$\mathbb{R^*_-}$</span>.</p>
<p>For <span class="math-container">$\mathbb{R^*_+}$</span>:</p>
<p><span class="math-container">$$\frac{du}{u}=-\ln(u(t))dt$$</span></p>
<p>And by integrating, it follows that:</p>
<p><span class="math-container">$$\ln\lvert u \rvert = -u\ln(u) + u$$</span></p>
<p>But I'm stuck here, I don't know what to do next.</p>
| JJacquelin | 108,514 | <p><span class="math-container">$$\frac{du}{u}=-\ln(u(t))dt$$</span>
You cannot integrate <span class="math-container">$\int-\ln(u(t))dt$</span> directly.</p>
<p>The mistake is : You integrate <span class="math-container">$\quad \int-\ln(u(t))du= -u\ln(u)+u\quad$</span> but this is not <span class="math-container">$\quad \int -\ln(u(t))dt$</span>.</p>
<p>Back to the original ODE :
<span class="math-container">$$\frac{du}{dt}=u\ln(u)$$</span>
This is a separable ODE thus
<span class="math-container">$$\frac{du}{u\ln(u)}=dt$$</span>
<span class="math-container">$$\int\frac{du}{u\ln(u)}=\int dt$$</span>
<span class="math-container">$$\ln(\ln(u))=t+c$$</span>
<span class="math-container">$$u=\exp(\exp(t+c))$$</span></p>
|
578,487 | <p>Maybe this is a well know result, however, I could not find it. Before stating it, let me write here a well know result (at least for me)</p>
<blockquote>
<p>Assume that $\Omega\subset\mathbb{R}^N$ is a open domain and $f:\Omega\to\mathbb{R}$. If there is constants $L>0$ and $\alpha>1$ such that $$|f(x)-f(y)|\leq L |x-y|^\alpha,\ \forall\ x,y\in\Omega$$
then, $f$ is constant in each connected componente of $\Omega$.</p>
</blockquote>
<p>The above result can be proved, for example, by showing that $\nabla f=0$ and then we join points in the same connected component by a continuous curve.</p>
<p>Now my question is: </p>
<blockquote>
<p>Assume that $\Omega\subset\mathbb{R}^N$ and $f:\Omega\to\mathbb{R}$. Suppose that there is constants $L>0$ and $\alpha>1$ such that $$|f(x)-f(y)|\leq L |x-y|^\alpha,\ \forall\ x,y\in\Omega$$
Can we conclude that $f$ is constant in each connected componente of $\Omega$?</p>
</blockquote>
<p>Maybe it is necessary to add the hypothesis that each connected component of $\Omega$ is pathwise connected? </p>
<p>Remark: Note that in the question, $\Omega$ does not need to be a open set. It is now any set.</p>
| Disintegrating By Parts | 112,478 | <p>Claim: For $f$ as stated, and $p > 1$, assume that the region $\Omega$ is path connected by continuous paths of finite total variation. Then $f$ is constant on $\Omega$.</p>
<p>To see this, let $x$, $y$ be given, and choose a path $v(t) : [0,1]\rightarrow \Omega$ of finite arc length $l(v)$ that connects $x$ to $y$. Without loss of generality assume $l(v)\ne 0$ (otherwise $f(x)=f(y)$.) Let $\epsilon > 0$ be given. Find a partition
$$
{0=t_{0} < t_{1} < \cdots t_{n}=1}
$$
refined enough so that $|v(t_{j-1})-v(t_{j})| < (\epsilon/2l(v))^{1/(p-1)}$ for all $j$, which is possible because $x$ is continuous on $[0,1]$ and, hence, also uniformly continuous on $[0,1]$. Then
$$
\begin{align}
|f(x)-f(y)| & \le \sum_{j=1}^{n}|f(v(t_{j-1}))-f(v(t_{j}))| \\
& \le L\sum_{j=1}^{n}|v(t_{j-1})-v(t_{j})|^{p-1+1} \\
& \le \frac{\epsilon}{2l(v)}\sum_{j=1}^{n}|v(t_{j})-v(t_{j-1})|\le \frac{\epsilon}{2} < \epsilon.
\end{align}
$$
Because $\epsilon$ was arbitrary, then $f(x)=f(y)$. This is true for all points $y$ connected to $x$ by such paths, which is everything in this case. So $f$ is constant on $\Omega$.</p>
|
1,111,952 | <p><strong>My Try:</strong> </p>
<p>We substitute $y = x^{2/3}$. Therefore, $x = y^{3/2}$ and $\frac{dx}{dy} = \frac{2}{3}\frac{dy}{y^{1/3}}$</p>
<p>Hence, the integral after substitution is: </p>
<p>$$ \frac{3}{2} \int_0^\infty \sin(y)\sqrt{y} dy$$</p>
<p>Let's look at:</p>
<p>$$\int_0^\infty \left|\sin(y)\sqrt{y} \right| dy = \sum_{n=0}^\infty \int_{n\pi}^{(n+1)\pi}\left|\sin(y)\right| \sqrt{y} dy \ge \sum_{n=0}^\infty \sqrt{n\pi} \int_{n\pi}^{(n+1)\pi}\left|\sin(y)\right| dy \\= \sum_{n=1}^\infty \sqrt{n\pi} \int_{n\pi}^{(n+1)\pi}\sqrt{\sin(y)^2}$$</p>
| abel | 9,252 | <p>the convergence of $\int_0^\infty \sin(x^{2/3}$ at the lower limit $x = 0$ is not a problem. the trouble is at the upper limit $x = \infty$</p>
<p>to handle the upper limit, i will make a change of variable $x = t^{3/2}, dx = 3/2 t^{1/2} dt.$
then
$\int_0^\infty \sin x^{2/3} dx = \frac{3}{2} \int_0^\infty t^{1/2} \sin t \ dt$</p>
<p>taking the idea from GEdgar's answer<br>
$$ \int t^{1/2} \sin t \ dt = -t^{1/2}\cos t + \dfrac{1}{2}t^{-1/2}\sin t - \dfrac{1}{4}\int t^{-3/2}\sin t \ dt \tag 1$$</p>
<p>the last two terms are alright at $t = \infty,$ and the first term has no limit at $t = \infty.$</p>
|
4,547,918 | <p>Given the torus and given the point p <span class="math-container">$\in$</span> M corresponding to the parameters <span class="math-container">$s=\frac{\pi }{4}$</span> and <span class="math-container">$t=\frac{\pi }{3}$</span>.
Determine the cartesian equation of the tangent plane to M in p.</p>
<p><span class="math-container">$\begin{cases} x=\left(3+\sqrt{2}cos\left(s\right)\right)cos\left(t\right) \\ y=\left(3+\sqrt{2}cos\left(s\right)\right)sin\left(t\right) \\ z=\sqrt{2}sin\left(s\right) \end{cases}$</span></p>
<p>Could someone give me a hint or help me? I'm not sure if I firstly should go from the given parametric equation to a cartesian equation.</p>
| Z Ahmed | 671,540 | <p>Use <span class="math-container">$\cos^2x=1-\sin^2x$</span>. Then <span class="math-container">$I=\int (1-\sin x) dx$</span></p>
|
3,154,244 | <p>I tried to ask this in a different way and did not correctly explain myself.</p>
<p>I am ok integrating the line <span class="math-container">$y = x$</span> , let us say from <span class="math-container">$0$</span> to <span class="math-container">$2$</span> using calculus.
If I want to get the square I can easily multiply by two and using calculus the dimensions work. The area is a square and when we integrate we have a square.</p>
<p>Here is my question. If I use a straight line above the x axis my equation becomes <span class="math-container">$y$</span> = ( some constant) . Now I use calculus and integrate from 0 to 2 , I also get the correct answer BUT I have to imagine it is a rectangle because when you ingtegrate you have <em>one</em> <span class="math-container">$x$</span> term and it is not a <em>square.</em> The answers match OK it's the dimensions that bother me. </p>
<p>I did not do a good job explaining this on my previous question. Sorry </p>
| Paras Khosla | 478,779 | <p>The dimensions do work correctly in case of a rectangle as well. Let us say the height of the rectangle is <span class="math-container">$k$</span> and width is <span class="math-container">$b-a$</span>.</p>
<p>Recall that integration is basically summing up infinitely many small rectangles with infinitesimally small widths.</p>
<p><span class="math-container">$$\begin{aligned}\text{Area }&=\lim_{n\to \infty}\sum_{i=1}^{n}\underbrace{k}_{\text{height}}\cdot\underbrace{\left(\dfrac{b-a}{n}\right)}_{\text{width of each infinitesimal rectangle}}\\&=\int_{a}^{b}\underbrace{k}_{\text{units}}\underbrace{\mathrm dx}_{\text{units}}\\&=kx\biggr|_{a}^{b}=k(b-a) \text{ sq. units}\end{aligned}$$</span></p>
|
2,757,469 | <p>The set $A=\{1, \frac{1}{2}, \frac{1}{4}, \dots \}$ is obviously not closed in $\mathbb{R}$ with the Euclidean metric, as the sequence $1, \frac{1}{2}, \frac{1}{4} \dots$ converges to $0 \notin A$.</p>
<p>But if we look at the complement, $A'$ and let $p\in A'$ then there exist some $a_i$ and $a_{i+1}$ such that $a_{i+1}<p<a_i$. Letting $\epsilon=\frac{a_i-a_{i+1}}{2}$ we see that the open ball $B_\epsilon(p) \cap A = \emptyset$ so $B\epsilon(p) \subset A'$ thus proving that $A'$ is open and $A$ closed.</p>
<p>What am I doing wrong?</p>
| Tsemo Aristide | 280,301 | <p>Since $A$ is not closed, its complement cannot be open, $0$ is in the complement of $A$ and every ball which contains $0$ meets $A$.</p>
|
2,757,469 | <p>The set $A=\{1, \frac{1}{2}, \frac{1}{4}, \dots \}$ is obviously not closed in $\mathbb{R}$ with the Euclidean metric, as the sequence $1, \frac{1}{2}, \frac{1}{4} \dots$ converges to $0 \notin A$.</p>
<p>But if we look at the complement, $A'$ and let $p\in A'$ then there exist some $a_i$ and $a_{i+1}$ such that $a_{i+1}<p<a_i$. Letting $\epsilon=\frac{a_i-a_{i+1}}{2}$ we see that the open ball $B_\epsilon(p) \cap A = \emptyset$ so $B\epsilon(p) \subset A'$ thus proving that $A'$ is open and $A$ closed.</p>
<p>What am I doing wrong?</p>
| dyf | 556,562 | <p>Your complement $A'$ includes $0$ but you cannot find an open ball around it. Your mistake is tacitly assuming $p \in A'$ is in $(0,1)$ as well.</p>
|
1,158,956 | <p>To show that orthogonal complement of a set A is closed.</p>
<p>My try: I first show that the inner product is a continuous map. Let $X$ be an inner product space. For all $x_1,x_2,y_1,y_2 \in X$, by Cauchy-Schwarz inequality we get,
$$|\langle x_1,y_1\rangle - \langle x_2,y_2\rangle| = |\langle x_1- x_2,y_1\rangle + \langle x_2, y_1-y_2\rangle| $$
$$\leq \|x_1- x_2\|\cdot\|y_1\| +\|x_2\|\cdot\| y_1-y_2\|$$</p>
<p>This implies continuity of inner products.</p>
<p>Let $A \subset X$ and $y \in A^\perp$. To show that $ A^\perp$ is closed, we have to show that if $(y_n)$ is convergent sequence in $ A^\perp$, then the limit $y$ also belong to $ A^\perp$.</p>
<p>Let $x \in A$, then using that inner product is a continuous map,
$$\langle x,y\rangle = \langle x, \lim_{n\to \infty} (y_n)\rangle = \lim_{n\to \infty} \langle x, y_n\rangle = 0.$$</p>
<p>Since $\langle x, y_n\rangle = 0$ for all $x \in A$ and $y_n \in A^\perp$. Hence $y \in A^\perp$.</p>
<p>Is the approach\the proof correct??</p>
<p>Thank You!!</p>
| MUH | 214,956 | <p>Let's denote the orthogonal complement of <span class="math-container">$A$</span> by <span class="math-container">$A^{\perp}$</span>. Also, we denote the scalar product <span class="math-container">$\langle \cdot, y \rangle : V \rightarrow \mathbb{F}$</span> as the function <span class="math-container">$\varphi_y$</span>. So, all the orthogonal elements corresponding to <span class="math-container">$y$</span> is nothing but <span class="math-container">$Ker \; \varphi_y$</span>.
Hence,</p>
<p><span class="math-container">$$
A^{\perp} = \cap_{y \in A} \varphi_y
$$</span></p>
<p>Since <span class="math-container">$Ker \; \varphi_y$</span> is closed subset of <span class="math-container">$V$</span> (why?) and arbitrary intersection of closed set is again a closed set, we have the result.</p>
|
2,166,075 | <blockquote>
<p>Prove $a_1+\cdots+a_n=\dfrac{(a_1+a_n)n}{2}$ inductively.</p>
</blockquote>
<p>Where $a_i=a_{i+1}-r$.</p>
<p>I tried to start proving it inductively, but any try lead to a bad conclusion, so I ended up proving it by making $a_n$ depend on $a_i$.</p>
<p>But I didn't know how to prove it inductively, so there is the problem.</p>
<p><strong>EDIT:</strong></p>
<p>I'm looking for a valid induction steps to reach the conclusion.</p>
| Siong Thye Goh | 306,553 | <p>If $n=1$, the result is trivial.</p>
<p>Suppose $$\sum_{i=1}^k a_i = \frac{(a_1+a_k)k}{2}$$</p>
<p>\begin{align}\sum_{i=1}^{k+1} a_i &= \frac{(a_1+a_k)k}{2}+a_{k+1}\\&=\frac{a_1k+a_{k+1}(k+1)-rk+a_{k+1}}{2}\\
&= \frac{a_1k+a_{k+1}(k+1)+a_1}{2}\\
&=\frac{(a_1+a_{k+1})(k+1)}{2}\end{align}</p>
|
4,047,601 | <p>I did a question <span class="math-container">$\int_{0}^{1}\frac{1}{x^{\frac{1}{2}}}\,dx$</span>, and evaluating this is divergent integral yes? Then as a general form <span class="math-container">$\int_{0}^{1} \frac{1}{x^p}\,dx$</span>, <span class="math-container">$p \in \mathbb{R}$</span>, what values of <span class="math-container">$p$</span> can give me <span class="math-container">$\int_{0}^{1} \frac{1}{x^p}\,dx = \frac{4}{3}$</span>? This is a easy integral to calculate, make it <span class="math-container">$\int_{0}^{1}x^{-p}dx$</span> and calculate, etc. Then how do I get this to solve <span class="math-container">$p$</span>? I am using the fundamental theorem of calculus and confused here.</p>
| DMcMor | 155,622 | <p>Start with the antiderivative, assuming that <span class="math-container">$p\ne 1$</span> <span class="math-container">$$\int x^{-p}\,dx = \frac{1}{1-p}x^{1-p} + C.$$</span> Then it must be the case that <span class="math-container">$$\frac{1}{1-p}x^{1-p}\bigg|_{0}^{1} = \frac{1}{1-p}\left(1-\lim_{x\to 0}x^{1-p}\right).$$</span> Now, in order for <span class="math-container">$\lim_{x\to 0}x^{1-p}$</span> to exist we need <span class="math-container">$1-p>0,$</span> in which case the limit is zero. So, we need two conditions:
<span class="math-container">\begin{align}
p&<1\\[5pt]
\frac{1}{1-p} &= \frac{4}{3}.
\end{align}</span></p>
<p>We can calculate <span class="math-container">$p$</span> from the second condition to get <span class="math-container">$p = 1/4$</span>. This also satisfies the condition that <span class="math-container">$p<1$</span>. To confirm that we have the correct result:
<span class="math-container">$$\int_{0}^{1}x^{-1/4}\,dx = \frac{4}{3}x^{3/4}\bigg|_{0}^{1} = \frac{4}{3} - 0 = \frac{4}{3}.$$</span></p>
|
3,375,459 | <p>I am trying to study differential geometry.</p>
<p>I am confused with regards to the following function for finding the length of
a curve <span class="math-container">$\gamma$</span> connecting two points <span class="math-container">$p, q ∈ S^2$</span></p>
<p><span class="math-container">$$L(γ) = \int^1_0|\dot{γ}(t)| dt,γ(0) = p, γ(1) = q$$</span></p>
<p>Where <span class="math-container">$S^2$</span> is a 2-dimensional sphere sitting in the three dimensional Euclidean space <span class="math-container">$R^3$</span></p>
<p>I am unfamiliar with the "dot above function" notation (dot above <span class="math-container">$\gamma$</span>), what does it mean? And from where is this function derived or what is it called?</p>
| Surb | 154,545 | <p><strong>Hint</strong></p>
<p>If <span class="math-container">$X\sim \mathcal N(0,1)$</span>,</p>
<p><span class="math-container">$$\mathbb E[f(X)]=\int_{\mathbb R}f(x)e^{-\frac{x^2}{2}}\,\mathrm d x.$$</span></p>
<p>Moreover, <span class="math-container">$$\int_{\mathbb R}e^{-x^2}\,\mathrm d x=\pi.$$</span></p>
<p>All these formulas will allow you to conclude.</p>
|
103,397 | <p>Is there functionality in <em>Mathematica</em> to expand a function into a series with Chebyshev polynomials? </p>
<p>The <code>Series</code> function only approximates with Taylor series.</p>
| J. M.'s persistent exhaustion | 50 | <p>One slick way to derive the analytic Chebyshev series of a function is to use the relationship between the Chebyshev polynomials and the cosine, and then use the built-in <code>FourierCosSeries[]</code>. As an example:</p>
<pre><code>f[x_] := Exp[x];
n = 5; (* degree of approximation *)
approx[x_] = FourierCosSeries[f[Cos[t]], t, n] /. Cos[k_. t] :> ChebyshevT[k, x]
</code></pre>
<p>(Note that the result of that evaluation contains modified Bessel functions of the first kind, which arise as the coefficients.)</p>
<pre><code>{Plot[{f[x], approx[x]}, {x, -1, 1}],
Plot[approx[x] - f[x], {x, -1, 1}, PlotStyle -> ColorData[97, 3]]} // GraphicsRow
</code></pre>
<p><img src="https://i.stack.imgur.com/FsR7o.png" alt="plot of function and Chebyshev series, and the approximation error"></p>
<p>See how good the approximation is? Note the equiripple behavior of the error at the right.</p>
<hr>
<p>Of course, not every function will admit a closed form Chebyshev series representation, since the Fourier integrals involved won't necessarily have a closed form known to <em>Mathematica</em>. In that case, you can of course use <code>NIntegrate[]</code> instead. In fact, <em>Mathematica</em> does provide a package for numerically evaluating those integrals. Thus,</p>
<pre><code>Needs["FourierSeries`"]
f[x_] := 3 x^2 Exp[2 x] Sin[2 π x];
n = 12;
cof = Table[If[k == 0, 1/2, 1] NFourierCosCoefficient[f[Cos[t]], t, k, Method -> "LevinRule"],
{k, 0, n}];
(* Clenshaw recurrence for a Chebyshev series *)
chebval[c_?VectorQ, x_] := Module[{n = Length[c], u, v, w},
u = c[[n - 1]] + 2 x (v = c[[n]]);
Do[w = v; v = u; u = c[[k]] + 2 x v - w, {k, n - 2, 2, -1}];
c[[1]] + x u - v]
approx[x_] = chebval[cof, x];
{Plot[{f[x], approx[x]}, {x, -1, 1}],
Plot[approx[x] - f[x], {x, -1, 1}, PlotStyle -> ColorData[97, 3]]} // GraphicsRow
</code></pre>
<p><img src="https://i.stack.imgur.com/BEBEw.png" alt="not too good"></p>
<p>though the approximation in this case is not too good.</p>
|
1,232,363 | <p>I have to solve a probability problem and it says that we take a random sample of size 10. But I don´t understand the concept (I´m on my first course on probability). </p>
<p>Suppose that we have a box with 100 balls and I take a random sample of size 10</p>
<p>Is a random sample of size 10 if</p>
<ul>
<li>I take AT THE SAME TIME 10 balls? or</li>
<li>I take one ball, then return it and repeat this process ten times?</li>
</ul>
<p>Thanks for your help.</p>
| Mark Viola | 218,419 | <p>Let's use the "brute force" approach here. We have the matrix equation $Av=\lambda v$. Thus, the eigenvalue equation becomes</p>
<p>$$(A-\lambda I)v=0$$</p>
<p>which implies that the determinant of $A-\lambda I$ is zero. Thus, we have </p>
<p>$$(\lambda -a)(\lambda -d)-bc=0$$</p>
<p>which implies that </p>
<p>$$\begin{align}
\lambda &=\frac{(d+a)\pm \sqrt{(d+a)^2-4(ad-bc)}}{2}\\
&=\frac{(d+a)\pm \sqrt{(d-a)^2+4bc}}{2}
\end{align}$$</p>
<p>Note that </p>
<p>$$y=\frac{\lambda -a}{b}x=\left(\frac{(d-a)\pm \sqrt{(d-a)^2+4bc}}{2}\right)x$$</p>
<p>Thus, for the positive-valued eigenvalue, $y$ is of the same sign as $x$, which means that there is one eigenvector for which $x>0$ and $y>0$.</p>
|
870,030 | <p>Q: Prove that the relation given by $a\sim b\Leftrightarrow a-b\in\mathbb{Z}$ is a congruence relation on the additive group $\mathbb{Q}$.</p>
<p>A: Maybe...
<ul>
<li>$a\sim a\Leftrightarrow a-a=0\in \mathbb{Z}$ ✓
<li>$a\sim b\Leftrightarrow a-b\in \mathbb{Z}$. $a\in \mathbb{Z}\Rightarrow -a\in \mathbb{Z}$ and $-b\in\mathbb{Z}\Rightarrow b\in \mathbb{Z}$, yielding $a-b+(-2a+2b)\in\mathbb{Z}\Leftrightarrow -a+b\in \mathbb{Z}\Leftrightarrow b\sim a.$ ✓
<li>$a\sim b$ and $b\sim c$ so $a-b\in \mathbb{Z}$ and $b-c\in \mathbb{Z}$ so $a-b+b-c\in\mathbb{Z}\Leftrightarrow a-c\in\mathbb{Z}\Leftrightarrow a\sim c.$ ✓
<li>$a_1\sim a_2$ and $b_1\sim b_2$. So $(a_1-a_2)(b_1-b_2)\in \mathbb{Z}\Leftrightarrow a_1b_1-a_1b_2-a_2b_1+a_2b_2\in\mathbb{Z}\Leftrightarrow a_1b_1+a_2b_2\in\mathbb{Z}$<br/>$\Leftrightarrow a_1b_1\sim a_2b_2$ ✓
</ul>
</p>
<p><p>The last bullet shows that the equivalence relation is a congruence relation on $\mathbb{Q}$.</p>
| Lee Mosher | 26,501 | <p>The second bullet is wrong. It should be $a \sim b \iff a-b \in \mathbb{Z} \iff b-a \in \mathbb{Z} \iff b \sim a$.</p>
|
2,619,638 | <p>I know a function which is not equal a.e to a continuous function is the step function or the characteristic of any interval and I also know the Dirichlet function is not an a.e continuous function but I want an example of a function with both properties.</p>
| Dr. Sonnhard Graubner | 175,066 | <p>The first derivative of $\arccos$ is given by $$-\frac{1}{\sqrt{1-x^2}}$$, so the first derivative of $$\arccos\left(\frac{z}{B}\right)$$ is given by $$-\frac{1}{B \sqrt{1-\frac{z^2}{B^2}}}$$</p>
|
7,130 | <p>I'm looking for an explanation on how reducing the Hamiltonian cycle problem to the Hamiltonian path's one (to proof that also the latter is NP-complete). I couldn't find any on the web, can someone help me here? (linking a source is also good).</p>
<p>Thank you.</p>
| richard | 395,150 | <p>Just remove one edge from G, this will construct a G'. There's a Hamiltonian path in G' iff There's a Hamiltonian cycle in G.</p>
|
712,736 | <p>For any continuous function $f(z)$ of period $1$. Show that $\varphi'=2\pi \varphi+f(t)$ has a unique solution of period $1$.</p>
<p>Is this problem wrong with the counter example $\varphi(t)=e^{2\pi t}$. Shall we change it into $\varphi'=2\pi i\varphi+f(t)$</p>
| user127096 | 127,096 | <p>What you want is a solution of the boundary value problem on $[0,1]$ with periodic boundary condition: $\varphi(1)=\varphi(0)$. It helps to focus on the map $\mathbb R\to\mathbb R$ defined by $T: \varphi(0) \mapsto \varphi(1) $. You should check that: </p>
<ol>
<li>$T$ is injective. This follows from the uniqueness of the IVP with initial condition specified at $1$. </li>
<li>$T$ is an affine map: $Tx=ax+b$ where $a,b$ are constant. Indeed, $b$ is what you get from $\varphi(0)=0$. The linearity follows from the equation being linear in $\varphi$. The constant $a$ has nothing to do with $f$, it comes from the homogeneous equation $\varphi'=2\pi \varphi$. You can and should write it down explicitly. </li>
<li>Key point: $a\ne 1$.</li>
</ol>
<p>Conclude that there is a unique $x\in\mathbb R$ such that $Tx=x$. </p>
|
77,744 | <p>Hopefully a simple one for you (or at least seemingly)!</p>
<p>I import a .txt file, from which i make a ListLinePlot. I simply want to read in the name of the file, store it in a variable so I can use it to tag my plots later.</p>
<pre><code> Data = Import["C:\\Users\\Name\\Folder\\test2.txt", "CSV"];
ListLinePlot[Transpose[{Data[[All, 1]], Data[[All, 2]]}],
Frame -> True,
PlotRange -> {{Data[[All, 1]][[
First@Flatten[Position[Data, {_, Max[Data[[All, 2]]]}]]]]
- 0.0003,
Data[[All, 1]][[
First@Flatten[Position[Data, {_, Max[Data[[All, 2]]]}]]]]
+ 0.0003}}, FrameTicks -> Automatic, Axes -> False,
PlotStyle -> Directive[Black, Thin],
FrameLabel -> {"SIMULATION TIME STEP [\[Mu]s]",
"PARTICLE NUMBER [#]"}]
</code></pre>
<p>Here is my code I'm using to plot, the filename is "test2", and it is that I would like to extract.</p>
<p>Thanks in advance,</p>
<p>QP</p>
| Basheer Algohi | 13,548 | <p>I don't know if I understand you correctly. but here is what I got for you:</p>
<pre><code>FileNameTake["C:\\Users\\Name\\Folder\\test2.txt"]
(*"test2.txt"*)
</code></pre>
<p>If you want without extension, then:</p>
<pre><code>FileBaseName["C:\\Users\\Name\\Folder\\test2.txt"]
(*test2*)
</code></pre>
|
2,825,652 | <p>I have a magnetometer sensor on each vertices of an isosceles triangle. I also have a magnet that can be anywhere on the triangle (inside, on edges, etc). I have the magnitude reading from each sensor (essentially giving me the distance the magnet is from each vertices of the triangle). I'd like to calculate the x,y coordinates from these three distances. </p>
<p>The magnet can be assumed to always be inside the triangle (or on it's borders).</p>
<p>How would I calculate this?</p>
| amd | 265,466 | <p>In theory, you just compute the common intersection of three circles. In practice, noisy real-world data make it likely that the resulting system of equations is inconsistent: the three measured circles don’t have a common intersection point. You will in all likelihood need to find an approximate solution. </p>
<p>A fairly inexpensive approximation uses a method by which the intersection might be computed in the ideal case, anyway: the pairwise radical axes of any set of three circles are guaranteed to be coincident. This point is known as the <em>radical center</em> of the configuration. When two circles intersect, their radical axis is the secant line through the intersection points, an equation for which can be found by subtracting one circle equation from the other. Thus, you compute the intersection of three lines with equations of the form $$2(x_j-x_i)x+2(y_j-y_i)y+(x_i^2+y_i^2-r_i^2)-(x_j^2+y_j^2-r_j^2)=0.$$ Since you know that the three lines are coincident, you can simply compute the intersection of any two of them, which can be done in homogeneous coordinates with a cross product. Once again, truncation and other errors can make the full three-equation system inconsistent, so if you’re being extra paranoid, compute a least-squares solution to it. Indeed, if you’re going to be doing a lot of measurements relative to static sensor configuration, you can precompute most of the latter solution so that computing each location requires at most squaring the three distances, a matrix-vector multiplication and a vector addition.</p>
|
1,328,909 | <p>I know how to find for which $n$ $\phi(n)=n/2$ or $\phi(n)=n/3$, my method for finding those was simply to find primes $p$ that satisfy $\Pi_p$$_|$$_n$$1-1/p$ $ = 1/2$ or $1/3$.</p>
<p>However, I don't know how to find $\Pi_p$$_|$$_n$$1-1/p = n/6$. Intuitively it seems that if I combine results for both $\phi(n) = n/2$ and $\phi(n) = n/3$ I'll get $\phi(n) = n/6$ but it does not work, cause I get number of the form $2^a3^b$ which gives $\phi(n) = n/3$ again.</p>
<p>Is there a way to find $n$ for which $\phi(n)=n/6$? Or do such numbers exist at all?</p>
| Ian | 83,396 | <p>In general, if <span class="math-container">$X$</span> is a topological space, <span class="math-container">$Y$</span> is a metric space, and <span class="math-container">$f : X \to Y$</span> is a function, then the set of continuity points of <span class="math-container">$f$</span> must be a <span class="math-container">$G_\delta$</span> set in <span class="math-container">$X$</span>, meaning a countable intersection of open sets. If <span class="math-container">$Y=\mathbb{R}$</span> and <span class="math-container">$X$</span> is a metric space with no isolated points (for instance, <span class="math-container">$\mathbb{R}$</span>) then the converse is true, i.e. for any <span class="math-container">$G_\delta$</span> set <span class="math-container">$A \subset X$</span>, there exists <span class="math-container">$f : X \to \mathbb{R}$</span> such that the set of continuity points of <span class="math-container">$X$</span> is <span class="math-container">$A$</span>. A proof of this converse is given here: <a href="http://alpha.math.uga.edu/%7Epete/Kim99.pdf" rel="nofollow noreferrer">http://alpha.math.uga.edu/~pete/Kim99.pdf</a> The construction is very similar to Nescio's comment; it is based on the construction of a dense set whose complement is dense, which in <span class="math-container">$\mathbb{R}$</span> is naturally given by <span class="math-container">$\mathbb{Q}$</span>.</p>
<p>Your result follows from this, because an open set is certainly a <span class="math-container">$G_\delta$</span> set.</p>
<p>The proof of the first result is not difficult, but it is a little bit tricky. You first introduce the set <span class="math-container">$A_n$</span> of "<span class="math-container">$1/n$</span>-continuity points", i.e. the points <span class="math-container">$x$</span> where there exists a neighborhood <span class="math-container">$U_x$</span> such that if <span class="math-container">$y \in U_x$</span> then <span class="math-container">$d(f(x),f(y))<1/n$</span>. Then you prove that <span class="math-container">$A_n$</span> is open, and that the set of continuity points of <span class="math-container">$f$</span> is <span class="math-container">$\bigcap_{n=1}^\infty A_n$</span>. In a somewhat less general context, this is an exercise in Royden and Fitzpatrick.</p>
<p>The "dual" problem, asking whether there is a function continuous on any given <em>closed</em> set, has the same answer. This follows from the second result above in concert with this: <a href="https://math.stackexchange.com/questions/317479/a-closed-set-in-a-metric-space-is-g-delta">A closed set in a metric space is $G_\delta$</a> But the trend ends here, because there are <span class="math-container">$F_\sigma$</span> sets such as <span class="math-container">$\mathbb{Q}$</span> which cannot be a set of continuity points. You can prove this by assuming that <span class="math-container">$\mathbb{Q}$</span> is a <span class="math-container">$G_\delta$</span> and then using the Baire category theorem to conclude that <span class="math-container">$\mathbb{Q} \cap \mathbb{I} = \emptyset$</span> is dense in <span class="math-container">$\mathbb{R}$</span>, which is an obvious contradiction.</p>
|
313,437 | <p>I have to find out the convergence of the next integral:
$$\int^{\pi/2}_0{\frac{\ln(\sin(x))}{\sqrt{x}}}dx$$
Any help? Thanks</p>
| Julien | 38,053 | <p>The only problem is at $0$.</p>
<p>We have, as $x$ approaches $0$,
$$
\sin x\sim x
$$
so
$$
\frac{\log(\sin x)}{\sqrt{x}}\sim\frac{\log x}{\sqrt{x}}
$$
and the integral of the latter converges at $0$.</p>
<p>If you did not know that, compare for instance with $\frac{1}{x^{2/3}}$.
Since
$$
x^{2/3}\frac{\log x}{\sqrt{x}}\longrightarrow 0
$$
we have
$$
\lvert \frac{\log x}{\sqrt{x}}\rvert\leq \frac{C}{x^{2/3}}.
$$
Now it is easily seen that the integral of $1/x^{2/3}$ converges at $0$.</p>
<p>So your integral converges.</p>
|
1,617,890 | <blockquote>
<p>Question: Solve $\sin(3x)=\cos(2x)$ for $0≤x≤2\pi$.</p>
</blockquote>
<p>My knowledge on the subject; I know the general identities, compound angle formulas and double angle formulas so I can only apply those.</p>
<p>With that in mind</p>
<p>\begin{align}
\cos(2x)=&~ \sin(3x)\\
\cos(2x)=&~ \sin(2x+x) \\
\cos(2x)=&~ \sin(2x)\cos(x) + \cos(2x)\sin(x)\\
\cos(2x)=&~ 2\sin(x)\cos(x)\cos(x) + \big(1-2\sin^2(x)\big)\sin(x)\\
\cos(2x)=&~ 2\sin(x)\cos^2(x) + \sin(x) - 2\sin^2(x)\\
\cos(2x)=&~ 2\sin(x)\big(1-\sin^2(x)\big)+\sin(x)-2\sin^2(x)\\
\cos(2x)=&~ 2\sin(x) - 2\sin^3(x) + \sin(x)- 2 \sin^2(x)\\
\end{align}
<strong>edit</strong> </p>
<p>\begin{gather}
2\sin(x) - 2\sin^3(x) + \sin(x)- 2 \sin^2(x) = 1-2\sin^2(x) \\
2\sin^3(x) - 3\sin(x) + 1 = 0
\end{gather} </p>
<p>This is a cubic right? </p>
<p>So $u = \sin(x)$,</p>
<p>\begin{gather} 2u^3 - 3u + 1 = 0 \\
(2u^2 + 2u - 1)(u-1) = 0
\end{gather}</p>
<p>Am I on the right track?<br>
This is where I am stuck what should I do now?</p>
| lab bhattacharjee | 33,337 | <p>$$\cos2x=\sin3x=\cos\left(\dfrac\pi2-3x\right)$$</p>
<p>$$\iff2x=2m\pi\pm\left(\dfrac\pi2-3x\right)$$ where $m$ is any integer</p>
<p>Alternatively, $$\sin3x=\cos2x=\sin\left(\dfrac\pi2-2x\right)$$</p>
<p>$$3x=n\pi+(-1)^n\left(\dfrac\pi2-2x\right)$$ where $n$ is any integer</p>
|
3,391,280 | <p>Prove by Induction on n that <span class="math-container">$\exists x,y,z \in Z$</span> s.t. <span class="math-container">$x\ge 2, y\ge 2, z\ge 2$</span> satisfies <span class="math-container">$x^2+y^2=z^{2n+1}$</span> </p>
<p>I'm a lot more comfortable with proving induction with <span class="math-container">$\forall$</span> I haven't really seen one of this format yet where there's an <span class="math-container">$\exists$</span>. Since this is obviously not true for all <span class="math-container">$x,y,z\in Z$</span> it's harder for me to figure out how to solve it.</p>
| lhf | 589 | <p><em>Hint:</em> The <a href="https://en.wikipedia.org/wiki/Brahmagupta%E2%80%93Fibonacci_identity" rel="nofollow noreferrer">Brahmagupta–Fibonacci identity</a> implies that if <span class="math-container">$z_1$</span> and <span class="math-container">$z_2$</span> are sums of squares, then so is their product <span class="math-container">$z_1 z_2$</span>.</p>
|
700,004 | <p>I have been working on this proof for a few hours and I can not make it work out.</p>
<p>$$\sum_{i=1}^{n}\frac{1}{i(i+1)}=1-\frac{1}{(n+1)}$$</p>
<p>i need to get to $1-\frac{1}{k+2}$</p>
<p>I get as far as $$1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}$$
then I have tried $1-\frac{(k+2)+1}{(k+1)(k+2)}$ which got me exactly nowhere. </p>
| ljfa | 127,600 | <p>Hint: Write $\frac 1{i(i+1)}$ as $\frac 1i - \frac 1{i+1}$.
Then you have a telescoping sum.</p>
|
2,705,794 | <p>I ran across this problem on a practice Putnam worksheet. Completely stumped.</p>
<p>Is $$\large \frac{m^{6} + 3m^{4} + 12m^{3} + 8m^{2}}{24}$$ an integer for all $m \in \mathbb{N}$?</p>
<p>I suspect it is an integer for any $m$. It checks out for small cases.</p>
<p>Any hints for proving the general case?</p>
| Raghib | 462,749 | <p>It is always an integer. A standard "brute force" approach is simply to show that each factor of 24 divides the expression in the numerator. This is equivalent to showing that 3 and 8 both divide it in all cases.</p>
<p>The case of the factor 3 splits into two sub cases: one where m is divisible by 3 and one where it is not. If m is divisible by 3 then clearly the numerator is also divisible by 3 and we are done. If m is not divisible by 3, we use the fact that $m^2\equiv 1 \mod 3$ to get $m^6+3m^4+12m^3+8m^2\equiv 1+8\equiv 9\equiv 0\mod 3$ and we are done.</p>
<p>Now we need to prove that 8 divides the expression. We factor it as $m^2(m^4+3m^2+12m+8)$. Clearly there are two subcases: m is either even or odd. If m is even the two factors in the previous factorized form are even, and further, the first factor is a square meaning divisible by 4, hence the two factors together are divisible by 8 and we are done. If m is odd, we can forget the $m^2$ factor since it is odd. We focus on showing that $m^4+3m^2+12m+8\equiv m^4+3m^2+4m\mod 8$ is in fact $0\mod 8$. Here I simply did case work on $m\equiv \{1,3,5\}\mod 8$ (since m is odd) since it isn't too much casework.</p>
|
2,705,794 | <p>I ran across this problem on a practice Putnam worksheet. Completely stumped.</p>
<p>Is $$\large \frac{m^{6} + 3m^{4} + 12m^{3} + 8m^{2}}{24}$$ an integer for all $m \in \mathbb{N}$?</p>
<p>I suspect it is an integer for any $m$. It checks out for small cases.</p>
<p>Any hints for proving the general case?</p>
| robjohn | 13,854 | <p>By noting the values at $m=0$, $m=1$, ... $m=6$, it is easy to compute
$$
\tfrac{m^6+3m^4+12m^3+8m^2}{24}
=\textstyle30\binom{m}{6}+75\binom{m}{5}+68\binom{m}{4}+30\binom{m}{3}+8\binom{m}{2}+\binom{m}{1}
$$</p>
|
1,875,351 | <p>I saw this problem in a book (not homework),</p>
<p>Assuming $L(n) = F(n)$ for$ n = 1,2,\cdots, k$
where $L(n)$ is Lucas Number and $F(n)$ is Fibonacci number.</p>
<p>$$\qquad L(k+1) = L(k) + L(k-1) \qquad \tag{by definiton}$$</p>
<p>$$ \qquad\qquad= F(k) + F(k-1) \tag{assumption}$$</p>
<p>$$ \ =F(k+1)$$</p>
<p>Hence, by principle of Mathematical induction, $F(n) = L(n)$ for each positive number $n$.</p>
<p>Is the reason why the proof is incorrect is because you can't assume that $L(n) = F(n)$ because that is not the relationship / definition of Lucas numbers in terms of Fibonacci numbers.</p>
| Bill Dubuque | 242 | <p>Your induction step is well-defined only for $\,k\ge 2\,$ since for $\,k\le 1\,$ the term $\,L(k-1)\,$ is not defined. The recurrence lifts equality at the prior two indices to the current index. It cannot be used to deduce equality at the <em>initial</em> indices $\,n = 1,2$ (the "initial conditions" = base cases).</p>
<p>Generally, two solutions of the recurrence are equal iff they have equal $\rm\color{#c00}{initial\ conditions}$. Below is a proof of the uniqueness theorem for second order recurrences</p>
<p><strong>Theorem</strong> $\ $ If $\,f',f\,$ are both solutions of the recurrence $\,f_{n+2} = a f_{n+1} + b f_n\,$ for all $\,n\ge 0\,$ then $\,f_n = f'_n\,$ for all $\,n\ge 0\iff \color{#c00}{f'_0 = f_0}\,$ and $\,\color{#c00}{f'_1 = f_1}$</p>
<p><strong>Proof</strong> $\ (\Rightarrow)\ $ Clear. $\ (\Leftarrow)\ $ $\,f'_n = f_n\,$ for $\,n = 0,1\,$ by hypothesis. If $\,n\ge 2\,$ then suppose for induction they are equal at all indices $< n.\,$ Applying the recurrence and induction we have </p>
<p>$$\begin{align} f'_n\ =&\ \ a\ f'_{n-1}\! + b\, f'_{n-2}\\
=&\ \ a\,f_{n-1}\! + b\, f_{n-2}\ \ \text{ by induction hypothesis}\\
=&\ \ f_n\end{align}$$</p>
<p><strong>Remark</strong> $\ $ The recurrence enables us to inductively lift the equality of the sequences at any two consecutive integer indices (initial conditions) to equality at all larger indices. The initial conditions provide the basis (foundation) of the lifting. Exactly the same proof works for a $k\,$th order recurrences, except you need equality at $\,k\,$ consecutive integers to serve as the base of the induction.</p>
|
129,788 | <blockquote>
<p>Let be A and B two events from the same sample set. If $\space P(A)+P(B)=1$, can one say that they are opposite events?</p>
</blockquote>
<p>In my thought:</p>
<p>$\space P(A)+P(B)=1$</p>
<p>$\space P(A)=1-P(B)$</p>
<p>So they are opposite events. But my book says no! It says that is not necessary true.</p>
<p>Can you explain me, why not?</p>
<p>Thanks</p>
| Michael Hardy | 11,667 | <p>Doesn't work. Throw a die. The sample space is $\{1,2,3,4,5,6\}$.</p>
<p>Let $A=\{1,2,3,4\}$ and $B=\{1,2\}$.</p>
<p>Then $P(A)+P(B)=1$ but $A$ and $B$ are not complementary events.</p>
|
3,995,119 | <p>I've difficulties calculating the following integral</p>
<p><span class="math-container">$$\int_z^\infty\mu\mathrm e^{-\mu y}(\mathrm e^{-\lambda z}-\mathrm e^{-\lambda y})\ \mathrm{d}y$$</span></p>
<p>I'm gonna use her to find a joint distribution of two random variables. I've try to apply the following substitution <span class="math-container">$u=e^{\lambda y}$</span> but I couldn't do much after that.
I appreciate any help. Thank you in advance. Sorry for my bad English.</p>
| Community | -1 | <p><strong>Hint:</strong></p>
<p><span class="math-container">$$\int_u^\infty e^{-at}dt=-\left.\frac{e^{-at}}a\right|_u^\infty=\frac{e^{-au}}a.$$</span></p>
<p>Your integral is</p>
<p><span class="math-container">$$\mu e^{-\lambda z} \int_z^\infty e^{-\mu y}dy-\mu\int_z^\infty e^{-(\mu+\lambda) y}dy.$$</span></p>
|
3,059,571 | <p><span class="math-container">$$\lim_{x\to \frac\pi2} \frac{(1-\tan(\frac x2))(1-\sin(x))}{(1+\tan(\frac x2))(\pi-2x)^3}$$</span></p>
<p>I only know of L'hopital method but that is very long. Is there a shorter method to solve this?</p>
| klirk | 385,702 | <p>Another approach is to use that <span class="math-container">$\tan x=\frac{\sin x}{\cos x}$</span>.</p>
<p>Then <span class="math-container">$$\frac{1-\tan(x/2)}{1+\tan(x/2)}=\frac{\cos(x/2)-\sin(x/2)}{\cos(x/2)+\sin(x/2)}=\frac{(\cos(x/2)-\sin(x/2))^2}{\cos^2(x/2)-\sin^2(x/2)}=\frac{1-\sin(x)}{\cos(x)} .$$</span></p>
<blockquote>
<p>So the original limit reduces to</p>
<p><span class="math-container">$$\lim_{x\to \frac\pi 2} \frac{(1-\sin (x))^2}{\cos(x)(\pi-2x)^3}.$$</span></p>
</blockquote>
<p>This is the product of the two (finite) limits
<span class="math-container">$$\lim_{x\to \frac\pi 2} \frac{1-\sin (x)}{\cos(x)(\pi-2x)} ~~~~~~\text{ and } ~~~~~\lim_{x\to \frac\pi 2} \frac{(1-\sin (x))}{(\pi-2x)^2}.$$</span></p>
<p>Now these limits can be calculated using l'Hospital (in second order).</p>
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.