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 |
|---|---|---|---|---|
180,122 | <p>Could any one give me hint how to show that the zero section of any smooth vector bundle is smooth?</p>
<p>Zero section is a map $\xi:M\rightarrow E$ defined by $$\xi(p)=0\qquad\forall p\in M.$$</p>
| Mariano Suárez-Álvarez | 274 | <p>Smoothness can be checked locally on $M$ and locally $E$ is trivial. </p>
<p>Can you use these two facts to conclude what you want?</p>
|
1,113,934 | <p>How to prove that $\mathrm{Hom}(\mathbb{R})=\mathrm{Aut}(\mathbb{R})$ ?</p>
<p>(We treat it as field homomorphisms. )</p>
<p>I know that $\mathrm{Aut}(\mathbb{R})=\{\mathrm{id}\}$ and $\mathrm{Mon}(\mathbb{R})=\mathrm{Aut}(\mathbb{R})$.</p>
| Krish | 177,430 | <p><em>Hint:</em> Let $\phi: \mathbb R \to \mathbb R$ is a ring homomorphism. Prove the followings: (1). $\phi(r) = r, \forall r \in \mathbb Q.$ (2). $\phi(x) > 0, \forall x >0.$ (3). $|x-y|<\frac{1}{m} \Rightarrow |\phi(x) - \phi(y)|<\frac{1}{m}.$</p>
|
195,432 | <p>Mathematica 12 has a new function <code>FunctionCompileExportLibrary</code>, which can export a pure function as a <code>.dll</code> file like this:</p>
<pre><code>FunctionCompileExportLibrary["function.dll", Function[
Typed[arg, "MachineInteger"], arg + 1]]
</code></pre>
<blockquote>
<p><code>"...\\function.dll"</code></p>
</blockquote>
<p>But how to call it in an external program (e.g., visual studio)? I have no <code>.h</code> file, I have no <code>.lib</code> file. I even don't know the function name that I want to call... Can anyone tell me how to use it?</p>
<p>The documentation of <a href="https://reference.wolfram.com/language/ref/FunctionCompileExportLibrary.html" rel="noreferrer"><code>FunctionCompileExportLibrary</code></a> states that</p>
<blockquote>
<p>The library generated by FunctionCompileExportLibrary is suitable for linking into external programs. It can also be loaded into the Wolfram System using LibraryFunctionLoad.</p>
</blockquote>
| ilian | 145 | <p>This is a quick example of calling the exported compiled function from a standalone executable on Windows.</p>
<p>A few things to note which are currently undocumented and that can definitely be subject to change (and improvement) in the future: </p>
<ul>
<li><p>the function is named <code>Main</code> by default and the example uses an explicit prototype rather than including a header file;</p></li>
<li><p>the Wolfram Runtime Library is linked statically for convenience (of course, the dynamic WolframRTL.dll could be used instead provided the executable can find it);</p></li>
<li><p><code>CreateExecutable</code> is also used for convenience here, however the test executable could be compiled in any other way;</p></li>
<li><p>while the Windows import library (<code>function.lib</code>) is created, currently it is not placed next to the DLL. I expect this will be corrected in a paclet update.</p>
<pre><code>lib = FunctionCompileExportLibrary["function.dll",
Function[Typed[arg, "MachineInteger"], arg + 1],
"CompilerOptions" -> "CreateLibraryOptions" ->
{"SystemLibraries" -> {"WolframRTL_Static_Minimal"}}];
Needs["CCompilerDriver`"]
test = CreateExecutable["
#include <stdio.h>
#include <inttypes.h>
int64_t Main(int64_t);
int main() {
printf(\"%\" PRId64, Main(10));
}",
"test", "Libraries" -> {FileNameJoin[{$CCompilerDefaultDirectory, "function.lib"}]}];
RunProcess[test, "StandardOutput"]
(* "11" *)
</code></pre></li>
</ul>
<hr>
<p>Regarding the follow-up question in the <a href="https://mathematica.stackexchange.com/revisions/195432/8">edit</a>: I think the linker is looking for the C++ mangled name. Try declaring <code>Main</code> with C linkage, i.e. <code>extern "C"</code>. Also make sure the prototype is using the correct integer type.</p>
|
2,161,913 | <p>Consider the angle $\theta$ with $$0<\theta<\frac{\pi}{2}$$ Suppose that $$cos\theta=\frac{l}{n} \space and \space \space sin\theta=\frac{m}{n}$$
are rational numbers, with postive integers $l,m$ and $n$. Show that $l$ and $m$ cannot both odd integers.<br>
MY ATTEMPT: Assume that $l$ and $m$ are both odd integers. Then $cos^2\theta=\frac{l^2}{n^2}$ and $sin^2\theta=\frac{m^2}{n^2}$ then adding these equations and using the Pythagorean identity we obtain $$l^2+m^2=n^2$$
Notice we can conclude n is even, and we must find a contradiction. I am unsure where to go from here, any help is appreciated!</p>
| Mark Viola | 218,419 | <p>Let $\epsilon>0$ be given. Take $N$ so large that $|\log(\cos(1/i))|<\epsilon/2$ whenever $i>N$. Then, we can write for $n>N+1$</p>
<p>$$\begin{align}
\left|\log\left(\sqrt[n]{\prod_{i=1}^n\sec(1/i)}\right)\right|&=\frac1n\left|\sum_{i=1}^n\log(\cos(1/i))\right|\\\\
&\le\frac1n\sum_{i=1}^N|\log(\cos(1/i))|+\frac1n\sum_{i=N+1}^n|\log(\cos(1/i))|\\\\
&\le \frac1n\sum_{i=1}^N|\log(\cos(1/i))|+\frac\epsilon2 \left(1-\frac{N}{n}\right)\\\\
&<\epsilon
\end{align}$$</p>
<p>when $n>\max\left(N+1,\frac{2N|\log(\cos(1/N))|}{\epsilon}\right)$.</p>
<p>Therefore, we have $\lim_{n\to \infty}\log\left(\sqrt[n]{\prod_{i=1}^n\sec(1/i)}\right)=0$ and</p>
<blockquote>
<p>$$\lim_{n\to \infty}\sqrt[n]{\prod_{i=1}^n\sec(1/i)}=1 \tag 1$$</p>
</blockquote>
<hr>
<blockquote>
<p>NOTE: We could have applied the <a href="https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem" rel="nofollow noreferrer">Stolz-Cesaro Theorem</a> and obtained directly</p>
</blockquote>
<p>$$\begin{align}
\lim_{n\to \infty}\frac{\sum_{i=1}^n\log(\cos(1/i))}{n}&=\lim_{n\to \infty}\frac{\sum_{i=1}^{n+1}\log(\cos(1/i))-\sum_{i=1}^n\log(\cos(1/i))}{(n+1)-(n)}\\\\
&=\lim_{n\to \infty}\log(\cos(1/(n+1)))\\\\
&=0
\end{align}$$</p>
<p>from which we recover $(1)$ immediately.</p>
|
4,037,206 | <p><a href="https://i.stack.imgur.com/VfORl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VfORl.png" alt="enter image description here" /></a></p>
<p>I have done the following for the median <br/>
<span class="math-container">$2(x-1) = .50$</span> <br/>
results in <span class="math-container">$x = 1.25$</span> <br/></p>
<p>For the quartiles, I have <span class="math-container">$q_{1}$</span>--> <span class="math-container">$2(x-1)=.25$</span> which results in <span class="math-container">$q_{1} = 1.125$</span> <br/>
for <span class="math-container">$q_{3}$</span>--> <span class="math-container">$2(x-1)=.75$</span> which results in <span class="math-container">$q_{3}=1.375$</span> and then subtract them to get <span class="math-container">$.25$</span>
Please correct me if I am wrong! thank you in advance</p>
| heropup | 118,193 | <p>The median is the number <span class="math-container">$m$</span> for which <span class="math-container">$$\int_{x=-\infty}^m f(x) \, dx = \frac{1}{2}.$$</span> So you need to solve <span class="math-container">$$\frac{1}{2} = \int_{x=1}^m 2(x-1) \, dx = \left[x^2 - 2x\right]_{x=1}^m = m^2 - 2m - (1-2) = (m-1)^2.$$</span></p>
<p>For the IQR, you need to solve for the first and third quartiles in a similar fashion: <span class="math-container">$q_1$</span> satisfies <span class="math-container">$$\frac{1}{4} = \int_{x=1}^{q_1} f(x) \, dx,$$</span> and <span class="math-container">$q_3$</span> satisfies <span class="math-container">$$\frac{3}{4} = \int_{x=1}^{q_3} f(x) \, dx.$$</span> Then the IQR is simply <span class="math-container">$q_3 - q_1$</span>.</p>
|
733,124 | <p>I understand that reals are defined as "completing" the rationals, which (at least in ZFC) are in turn derived from the natural numbers.</p>
<p>So, if ordinal numbers are viewed as an extension of the natural numbers, are there constructs of ordinals corresponding in the same way as rationals and reals are derived from natural numbers ?</p>
<p>If there are, then does this have any serious implications for models of reality which seem to be predominantly based on the reals ?</p>
| Asaf Karagila | 622 | <p>These are called <strong>Surreal Numbers</strong>. It is a proper-class sized order field which is complete, in the sense that every subset has a least upper bound and greatest lower bound.</p>
<p>It embeds all the ordered fields in the universe [of set theory], and all the ordinals too. It does not, however, extend ordinal arithmetic since ordinal addition is not commutative.</p>
<p>There are also non-standard fields which extend the real numbers, like the <strong>hyperreal numbers</strong>, and more. But these are "mere sets". So they do not extend the ordinals.</p>
<hr>
<p>Finally, let me point out that mathematicians, in particular set theorists and model theorists don't consider their mathematics as a model of the physical universe. We keep the church and state separate from our end.</p>
<p>Moreover, if you consider the expansion of the universe and how every point look outwards thinks that it is in the center of the universe, wouldn't it be much more fitting to use some ultrametric space to model the universe (i.e. a space where every point inside an ball is its center) rather than Euclidean metric spaces? (For example the $p$-adic numbers seem like a good candidate for that.)</p>
|
733,124 | <p>I understand that reals are defined as "completing" the rationals, which (at least in ZFC) are in turn derived from the natural numbers.</p>
<p>So, if ordinal numbers are viewed as an extension of the natural numbers, are there constructs of ordinals corresponding in the same way as rationals and reals are derived from natural numbers ?</p>
<p>If there are, then does this have any serious implications for models of reality which seem to be predominantly based on the reals ?</p>
| Stella Biderman | 123,230 | <p>The complex numbers are the completion of the reals in the algebraic sense, by which I mean the set of solutions to polynomials in $\mathbb{R}[x]$ is $\mathbb{C}$.</p>
|
1,609,446 | <p>I read the definition that $f$ is in $O(x^n)$ if $|f(x)|<C|x^n|$ for some $C$.</p>
<p>I'm struggling to understand how to check this. For example, supposedly $f(x) = 5x+3x^2$ is in $O(x)$ but not $O(x^2)$?</p>
<p>If I plot $f(x) = 5x + 3x^2$ and $g(x)=x$ I see that the first goes to infinity much quicker.</p>
<p><a href="https://i.stack.imgur.com/1H5Uv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1H5Uv.png" alt="enter image description here"></a></p>
<p>If I let $g(x) = Cx$, and plot $C=1,C=10, C=20, C=100$, it looks like it overcomes $f$ for $C>10$:
<a href="https://i.stack.imgur.com/owokf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/owokf.png" alt="enter image description here"></a></p>
<p>But, if you zoom out further, you can see that's not true:</p>
<p><a href="https://i.stack.imgur.com/ZSqk5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZSqk5.png" alt="enter image description here"></a></p>
<p>So, I know it doesn't matter what $C$ is, but how can I show that there exists a $C$ to make the definition hold so that I can tell if $f$ is in $O(x^n)$?</p>
<p>If you go out far enough, and $C$ is large enough can't I make either $f$ or $g$ as close to the y-axis as I want?</p>
| Dr_Be | 242,338 | <p>The definition is not complete since the inequality is supposed to hold for all $x>k$ where $k$ is a constant. The idea behind the $\mathcal O$-notation is to provide an estimate (comparison) for large $x$. This makes life much easier for your example since $x < x^2 \ \forall\, x>1$, so that </p>
<blockquote>
<p>$0<f(x)= 5x+3x^2<5x^2+3x^2=8x^2 \ \forall \, x>1$</p>
</blockquote>
<p>So you might choose $C:=8$ and $k:=2$ here in order to see that $f\in\mathcal O(x^2)$.
By the way, this statement can be generalized to hold for polynomials of degree $n$, i.e. if $f\in \mathcal P_n$ then $f\in\mathcal O(x^n)$.</p>
|
1,335,734 | <blockquote>
<p>A library leases its photocopier. One monthly bill was 750 dollars for
12,000 copies. Another month, the bill was $862.50 for 16,500 copies.
How much does the library pay for each copy?</p>
</blockquote>
<p>How should I start set up the equation for this problem?
How exactly do I find the linear equation giving the cost in terms of the number of copies used? </p>
<p>I am sorry if this is a stupid question, I am trying to review pre-calc and I forgot some techniques used to solve the problems. </p>
| meiji163 | 201,018 | <p>Use the chain rule.</p>
<p>$$3\left({d\over{dx}}(x^2+y^2)^2\right)=3\cdot2{d\over{dx}}(x^2+y^2)(x^2+y^2)=6\left(2x+2{d\over{dx}}(y)y\right)(x^2+y^2)=6\left(2x+2y{dy\over{dx}}\right)(x^2+y^2)=12(x^2+y^2)\left(y{dy\over{dx}}+x\right)$$</p>
|
2,111,579 | <p>A question:</p>
<p>If $\displaystyle{m}^{2}={1}-{m},{\quad\text{and}\quad}{n}^{2}={1}-{n},{\quad\text{and}\quad}{n}\ne{m};$</p>
<p>Proof that $\displaystyle{m}^{7}+{n}^{7}+{30}={1}$</p>
<p>Without finding the roots of equation $\displaystyle{x}^{2}+{x}-{1}={0}$.</p>
<p>Is there such a shortcut solution?</p>
| Algebraic Pavel | 90,996 | <p>You can also use the induction along the dimension $n$ (leading to a simple but not very constructive proof though). </p>
<p>Let $A_n$ denote the matrix $A$ of dimension $n$ and $B_n$ its inverse. The statement is obviously true for $n=1$ and reduces to showing that $1=1$. Let the statement be true for $n-1$, that is, $A_{n-1}^{-1}=B_{n-1}$. If $1_{n-1}$ and $0_{n-1}$ the column vectors of ones and zeros, respectively, of dimension $n-1$, the matrix $A_n$ can be written in the block form
$$
A_n=\begin{bmatrix}A_{n-1}&0_{n-1}\\1_{n-1}^T&1\end{bmatrix}.
$$
You can easily verify that
$$
A_n^{-1}=\begin{bmatrix}A_{n-1}^{-1}&0_{n-1}\\-1_{n-1}^TA_{n-1}^{-1}&1\end{bmatrix}
$$
is the inverse of $A_n$.
By the induction assumption, $A_{n-1}^{-1}=B_{n-1}$. It is easy to check that $1_{n-1}^TA_{n-1}^{-1}=1_{n-1}^TB_{n-1}=e_{n-1}^T:=[0,\ldots,0,1]^T$ (sum of the rows of $B_{n-1}$). So
$$
A_n^{-1}=\begin{bmatrix}B_{n-1}&0_{n-1}\\-1_{n-1}^TB_{n-1}&1\end{bmatrix}=\begin{bmatrix}B_{n-1}&0_{n-1}\\-e_{n-1}^T&1\end{bmatrix}=B_n.
$$</p>
|
1,557,206 | <p>Given the series $\sum\limits_{n=1}^\infty a_n^2$ and $\sum\limits_{n=1}^\infty b_n^2$ converge. Show that the series $\sum\limits_{n=1}^\infty a_n b_n$ converges absolutely.</p>
<p>My idea so far:</p>
<ul>
<li>It's quite quite obvious that both given series converge absolutely</li>
<li>So the Cauchy-Produc tells me that $\sum\limits_{n=1}^\infty a_n^2 b_n^2 = \sum\limits_{n=1}^\infty (a_n b_n)^2$ converges absolutely</li>
</ul>
<p>I got stuck at that point. Can somehow give me a hint how to solve this ?</p>
<p>Thanks in advance!</p>
| Mirko | 188,367 | <p>$\sum |a_n b_n|\le \sum b_n^2 + \sum a_n^2$ </p>
<p>spoiler: </p>
<blockquote class="spoiler">
<p> if $|a_n|\le|b_n|$ then $|a_n\cdot b_n|\le b_n^2$ else $|a_n\cdot b_n|\le a_n^2$, so $\sum |a_n b_n|\le \sum b_n^2 + \sum a_n^2$ </p>
</blockquote>
<p>Edit: Second spoiler (complete solution with all details): </p>
<blockquote class="spoiler">
<p> if $|a_n|\le|b_n|$ then $|a_n\cdot b_n|\le b_n^2\le b_n^2+a_n^2$ else $|a_n\cdot b_n|\le a_n^2\le a_n^2+b_n^2$. In all cases $|a_n\cdot b_n|\le b_n^2+a_n^2$ so $\sum |a_n b_n|\le \sum (b_n^2 + a_n^2)=\sum b_n^2 + \sum a_n^2<\infty$. </p>
</blockquote>
|
1,557,206 | <p>Given the series $\sum\limits_{n=1}^\infty a_n^2$ and $\sum\limits_{n=1}^\infty b_n^2$ converge. Show that the series $\sum\limits_{n=1}^\infty a_n b_n$ converges absolutely.</p>
<p>My idea so far:</p>
<ul>
<li>It's quite quite obvious that both given series converge absolutely</li>
<li>So the Cauchy-Produc tells me that $\sum\limits_{n=1}^\infty a_n^2 b_n^2 = \sum\limits_{n=1}^\infty (a_n b_n)^2$ converges absolutely</li>
</ul>
<p>I got stuck at that point. Can somehow give me a hint how to solve this ?</p>
<p>Thanks in advance!</p>
| Mark Viola | 218,419 | <p>HINT:</p>
<p>From Cauchy-Schwarz we have </p>
<p>$$\sum_{n=1}^{N}\left|a_n\, b_n\right|\le\sqrt{\left(\sum_{n=1}^{N}a_n^2\right)\,\left(\sum_{n=1}^{N}b_n^2\right)}$$</p>
|
2,278,436 | <p>Does there exist integers $a, n > 1$ such that $1 + \frac{1}{1 + a} + \frac{1}{1 + 2a} + ... + \frac{1}{1 + na}$ is an integer? I have no clue how to begin. I've tried to simplify this somehow, but with no effect.</p>
| Dr. Wolfgang Hintze | 198,592 | <p><strong>Extended proof</strong></p>
<p>Here I extend the proof of my original post for $a = 2$ to arbitary integers $a > 2$.</p>
<p>Notice that I consider the proof incompete as it makes an assumption which I could not prove (but which was used by myself and others before).</p>
<p>Let</p>
<p>$$c(k)=1 + a \;k$$
$$m=\sum _{k=0}^n \frac{1}{c(k)}$$
$$x=\prod _{k=0}^n c(k)$$</p>
<p><em>Assumption</em></p>
<p>Let $p^r$ be the highest power of a prime $p$ which divides $x$.</p>
<p>In the following we <strong>assume</strong> that there is a unique maximum, i.e. we rule out the case that there are two (or more) primes in $x$ which have the same "highest" power $r$</p>
<p>Hence the numbers $p$ and $r$ are uniquely defined for each $a$ and $n$.</p>
<p>Now we extract possible powers of $p$ from the $c(k)$ writing</p>
<p>$$c(k) = p^{s(k)} q(k)$$</p>
<p>where</p>
<p>$$p\nmid q(k)\tag{1}$$</p>
<p>Hence </p>
<p>$$m=\sum _{k=0}^n \frac{1}{p^{s(k)} q(k)}\tag{2}$$</p>
<p>Here we can assume without loss of generality that the order of the summands is such that</p>
<p>$$r = s(0) < s(1) <= ... <= s(n)\tag{3}$$</p>
<p>Now multiplying $(2)$ by $p^r q(0)$ gives</p>
<p>$$m p^r q(0) = 1 + \sum _{k=1}^n \frac{p^{r-s(k)} q(0) }{q(k)}\tag{4}$$</p>
<p>Taking this relation mod $p$, observing $(3)$ and $q(k) \ne 0 \;mod(p)$ due to $(1)$, leads to the contradiction </p>
<p>$$0 = 1$$</p>
<p>which completes the proof (if the assumption is correct).</p>
<p><strong>Original post</strong></p>
<p>Proof for a = 2. </p>
<p>I adapt the beautiful proof by Bill Dubuque in <a href="https://math.stackexchange.com/questions/2746/is-there-an-elementary-proof-that-sum-limits-k-1n-frac1k-is-never-an-int/2752#2752">Is there an elementary proof that $\sum \limits_{k=1}^n \frac1k$ is never an integer?</a></p>
<p>Proof by contradiction: suppose the sum of the inverted first n odd integers is an integer m. </p>
<p>Since there is a unique denominator $\rm\:\color{#C00} {3^K}\:$ having maximal power of $3,\,$ upon multiplying all terms through by $\rm\:3^{K-1}$ one deduces the contradiction that $\rm\ 1/3\, =\, c/d \;$ with $\rm\ 3\nmid d$, $ $ e.g.</p>
<p>$$\begin{eqnarray} & &\rm\ \ \ \ \color{green}{m} &=&\ \ 1 &+& \frac{1}{3} &+& \frac{1}{5} &+& \frac{1}{7} &+&\color{#C00}{\frac{1}{9}} &+& \frac{1}{11} &+& \frac{1}{13} &+& \frac{1}{15} \\
&\Rightarrow\ &\rm\ \ \color{green}{3m} &=&\ \ 3 &+&\ 1 &+& \frac{3}{5} &+& \frac{3}{7} &+&\, \color{#C00}{\frac{1}{3}} &+& \frac{3}{11} &+& \frac{3}{13} &+& \frac{3}{15}^\phantom{M^M}\\
&\Rightarrow\ & -\color{#C00}{\frac{1}{3}}\ \ &=&\ \ 3 &+&\ 1 &+& \frac{3}{5} &-&\rm \color{green}{3m} &+& \frac{3}{11} &+& \frac{3}{13} &+& \frac{1}{5}^\phantom{M^M}
\end{eqnarray}$$</p>
<p>The sum on the r.h.s. has only denominators which have no factor 3 so reduces to a fraction with a denominator $d$ with $3\nmid d$. This contradicts the l.h.s.</p>
|
3,133,152 | <blockquote>
<p><span class="math-container">$g(x)=|\sin{x}-1|+|3-\cos{x}-\sin{x}|+2\sin{x}$</span></p>
<p>Answer: Above equality is simplified to <span class="math-container">$$1-\sin{x}+3-\cos{x}-\sin{x}+2\sin{x}=4-\cos{x}$$</span></p>
</blockquote>
<p><span class="math-container">$$-1 \le\sin{x}\le1$$</span></p>
<p>So , I know that <span class="math-container">$f(x)=|\sin{x}-1|$</span> will be equal to <span class="math-container">$1-\sin{x}$</span> when <span class="math-container">$-1 \le\sin{x}\lt1$</span>.</p>
<p>But what if <span class="math-container">$\sin{x}$</span> is exactly <span class="math-container">$1$</span>, then wouldn't the expression be <span class="math-container">$\sin{x}-1$</span>?</p>
<p>I always considered <span class="math-container">$|x|=x$</span>, whenever <span class="math-container">$x$</span> is <em>equal to</em> or greater than <span class="math-container">$0$</span>.</p>
| Carl Schildkraut | 253,966 | <p>Usually, we hear that <span class="math-container">$x=|x|$</span> is <span class="math-container">$x$</span> is "positive" and <span class="math-container">$x=-|x|$</span> if <span class="math-container">$x$</span> is "negative," and often the case of <span class="math-container">$x=0$</span> isn't really discussed. In fact, as</p>
<p><span class="math-container">$$|0|=0=-0,$$</span></p>
<p>we can include <span class="math-container">$0$</span> in <em>either</em> case above. In particular, we can say the following:</p>
<blockquote>
<p>If <span class="math-container">$x\leq 0$</span>, then <span class="math-container">$|x|=-x$</span>.</p>
</blockquote>
<p>As <span class="math-container">$\sin x\leq 1$</span>, we know <span class="math-container">$\sin x - 1 \leq 0$</span>, so...</p>
|
2,487,709 | <p>I have to proof that if for any $z$:
$$\cos(z+\omega) = \cos(z)$$</p>
<p>then $\omega = 2\pi k \quad (k = 0, \pm 1, \pm 2 \ldots)$</p>
<p>I have no idea how to prrof it strictly</p>
| Anders Beta | 464,504 | <p>This integral is in general not well defined in the theory of distributions. Note that $f(x)$ is not a test function, and it has a singularity at $x=0$. If we try to give a meaning to it one runs into difficulties. Using the step function $H(x)$ one could try to give meaning to $I(x)$ as follows:</p>
<p>$$\begin{align}
I(x)&=\int_{-\infty}^{+\infty}H(x-t)f(t)\delta'(t)\,dt=-\left[\frac{d}{dt}H(x-t)f(t)\right]_{t=0}\\
&=\left[\delta(x-t)f(t)-H(x-t)(a-b)\delta(t)\right]_{t=0}\\
&=\delta(x)f(0)-H(x)(a-b)\delta(0) = \;???\,.
\end{align}$$</p>
<p>If $a\ne b$ you can not make sense of this; $f(0)$ and $\delta(0)$ do not make sense. So $I(x)$ is not a well defined distribution. To give meaning to it you may want to replace $f(t)$ by some smooth version of it (physicists will call doing that "regularization"), but the result will depend on the way you do this "regularization". This is a well known problem when you try to multiply distributions with common singularities.</p>
|
3,264,655 | <p>I've found myself where I've got a bidirectional graph of roughly 5500 nodes, 14000 edges.</p>
<p>I'd like to visit ~85 of these nodes, each <em>at least</em> once, starting and ending at the same node. I'd like to get a short route (don't need explicitly shortest, though that would be nice!). What graph processing algorithm should I use in this situation?</p>
<p>I feel like this isn't that unusual of a problem, but I think I'm searching for the wrong terms, as searching around has turned up fruitless. Or maybe I just haven't practiced math in too long!</p>
<p>The cost for traveling between two connected nodes is all the same.</p>
| sebastian | 604,625 | <p>What you could do for instance is the following: </p>
<p>For every one of your <span class="math-container">$85$</span> nodes use a shortest path algorithm (like Dijkstra) to find shortest paths to all other <span class="math-container">$84$</span> nodes. Use this information to build a new complete graph with <span class="math-container">$85$</span> nodes. Now if the costs of your initial graph satisfy the triangle equality, then it will also be satisfied in the new graph. Hence you can use an approximation algorithm for the metric travelling salesman problem to find a good route in your new graph.</p>
<p>Note that every edge in the new complete graph corresponds to a path in the old graph and thus when you found a route in the new graph you can just substitute every edge on that route with the corresponding path in the old graph.</p>
<p>Note also that the following can happen: When looking for a shortest path from <span class="math-container">$v$</span> to <span class="math-container">$w$</span> in the old graph where <span class="math-container">$v, w$</span> are of the distinguished <span class="math-container">$85$</span> nodes, then it is possible that some other node <span class="math-container">$u$</span> of the distinguished nodes is on that shortest path. Hence your result might include several visits of one of the distinguished nodes.</p>
|
3,772 | <p>In <a href="https://math.stackexchange.com/questions/102764/non-polynomial-functions-on-fields-of-finite-characteristic">this</a> question, there are two answers, both of which I up-voted, but I didn't "accept" either since there didn't appear to be a reason why one stands out as clearly preferable to the other. Apparently, my lack of "acceptance" offended someone who's taking me to task for it in comments on one of my questions. Some people (or at least one person) seem to get rather passionate about acceptances. Has anyone defined codified some criteria that say when one should "accept" an answer and what that is supposed to mean?</p>
| Bill Dubuque | 242 | <p>My opinion is that it is perfectly fine not to accept any answer if one has good reasons to do so. </p>
<p>For example, during the month I was testing MathOverflow, I posted a <a href="https://mathoverflow.net/questions/31595/how-would-you-solve-this-tantalizing-halmos-problem">question</a> on a tantalizing problem of Halmos, namely: explain why the following "proof" gives the correct answer.</p>
<p>$$\rm\begin{eqnarray} (1 - ba)^{-1} &=&\rm: 1 + ba + baba + bababa + \cdots \</p>
<p>&=& \rm: 1 + b (1 + ab + abab + ababab + \cdots) a \</p>
<p>&=&\rm: 1 + b (1 - ab)^{-1} a
\end{eqnarray}$$
One easily checks that the final answer is correct in every ring, even though the derivation using (formal) power series generally cannot be interpreted in every ring. As Halmos wrote</p>
<blockquote>
<p>Why does it all this work? What goes on here? Why does it seem that the formula for the sum of an infinite geometric series is true even for an abstract ring in which convergence is meaningless? What general truth does the formula embody? I don't know the answer, but I note that the formula is applicable in other situations where it ought not to be, and I wonder whether it deserves to be called one of the (computational) elements of mathematics.</p>
</blockquote>
<p>A long time ago (perhaps sparked by a conversation with G. C. Rota), I found what I thought was a good answer to this question. But none of the current MO answers (or comments) have helped me to remember this answer. So I have not yet accepted any answer, in the hope that someday the question will get bumped into the view of MO reader who knows the method that I had in mind.</p>
|
44,310 | <p>I wanted to draw (lattice-aligned) unit cubes inside a sphere with a given radius centered on the origin, so I wrote this program:</p>
<pre><code>Cube[x_, y_, z_] := Cuboid[{x - 1, y - 1, z - 1}, {x, y, z}];
Coords[r_] := {#1, #2, Floor[Sqrt[r^2 - #1^2 - #2^2]]} & @@ # & /@
With[{t = Sqrt[r^2 - 1]},
Select[Flatten[Table[{x, y}, {x, 1, t}, {y, 1, x}], 1],
Norm[#] <= t &]];
Cubes[r_] :=
Cube @@ # & /@
Union[Flatten[Permute[#, SymmetricGroup[3]] & /@ Coords[r], 1]];
Draw[r_] :=
Graphics3D[
Union[Cubes[r], {{Green, Opacity[0.1], Sphere[{0, 0, 0}, r]}}],
PlotRange -> {{0, r}, {0, r}, {0, r}},
ViewPoint -> {r, 3 r/4, 3 r/5}];
Draw /@ Sqrt /@ {50, 100, 250, 500, 1000, 2500, 6054}
</code></pre>
<p>(Note that it shows only the cubes on the outside -- drawing the inner ones takes too long and usually Mathematica runs out of memory.) It gets the job done, but it's pretty ugly. I'm not a regular Mathematica user, and tips for improving my code?</p>
<p>That is, how can I write idiomatically-better Mathematica code?</p>
<hr>
<p>Improved code:</p>
<pre><code>Coords[r_] := {#1, #2, Floor[Sqrt[r^2 - #1^2 - #2^2]]} & @@@
With[{t = Sqrt[r^2 - 1]},
Select[Join @@ Table[{x, y}, {x, t}, {y, 1, x}], Norm[#] <= t &]];
Cubes[r_] := (Cuboid /@ (Union @@ (Permute[#, SymmetricGroup[3]] & /@
Coords[r]) - 1));
Draw[r_] :=
Graphics3D[
Union[Cubes[r], {{Green, Opacity[0.1], Sphere[{0, 0, 0}, r]}}],
PlotRange -> {{0, r}, {0, r}, {0, r}},
ViewPoint -> {r, 3 r/4, 3 r/5}];
Draw /@ Sqrt /@ {50, 100, 250}
</code></pre>
| Mr.Wizard | 121 | <p>Focusing only on syntax I think your code looks pretty good, but there are a few things I note that I think could be improved.</p>
<ul>
<li><p>The method <code>f @@ # & /@ expr</code> can be replaced with <code>f @@@ expr</code>, which is shorthand for <code>Apply[f, expr, {1}]</code>. See <a href="http://reference.wolfram.com/mathematica/ref/Apply.html" rel="nofollow"><code>Apply</code></a>.</p></li>
<li><p>You can use the listability of <code>Plus</code> to write <code>{x, y, z} - 1</code> instead of <code>{x - 1, y - 1, z - 1}</code></p></li>
<li><p><code>Union[Flatten[expr, 1]]</code> can be written <code>Union @@ expr</code></p></li>
</ul>
<p>Your first three functions with these changes (among others), and David's observation regarding <code>Cuboid</code>:</p>
<pre><code>Cube[x_, y_, z_] := Cuboid[{x, y, z} - 1];
Coords[r_] := {#1, #2, Floor[Sqrt[r^2 - #1^2 - #2^2]]} & @@@
With[{t = Sqrt[r^2 - 1]},
Select[Join @@ Table[{x, y}, {x, t}, {y, 1, x}], Norm[#] <= t &]];
Cubes[r_] := Cube @@@ Union @@ (Permute[#, SymmetricGroup[3]] & /@ Coords[r]);
</code></pre>
<p>Regarding style I think it is fairly personal whether or not something is easy to read, but here is an example of a different style which you may prefer:</p>
<pre><code>Coords[r_] :=
With[{t = Sqrt[r^2 - 1]},
Cases[
Join @@ Table[{x, y}, {x, t}, {y, 1, x}],
{x_, y_} /; Norm[{x, y}] <= t :>
{x, y, ⌊ Sqrt[r^2 - x^2 -y^2] ⌋ }
]
]
</code></pre>
<p>Fortunately <em>Mathematica</em> supports many different styles.</p>
|
330,677 | <p>I was thinking last days about the following problem - the sum of Darboux is a Darboux function?</p>
<p>Do You know a proof or counter example?</p>
| Dominic Michaelis | 62,278 | <p>No this is not true, as darboux functions are really a general class of functions, you can proof that you can write any (really any) real valued function as the sum of two Darboux functions, look <a href="http://en.wikipedia.org/wiki/Darboux%27s_theorem_%28analysis%29#Darboux_function" rel="nofollow">here</a></p>
<p>The example of such a function from the link is, with
$$\omega(x)=\limsup_{n\to \infty} \frac{1}{n} \sum_{i=1}^n a_i$$ where $x\in(0,1)$ has the dyadic expansion $x=0.a_1 a_2 \dots$
$\omega$ takes every value between $0$ and $1$ in any subinterval of $(0,1)$ and has the intermediaty value property.
Now we denote
$$g(x)=\begin{cases}
0 & \omega (x)=x\\
\omega(x) & \text{else}\\
\end{cases}$$
This function still hase the intermediate value property but $h(x)=g(x)-x$ doesn't have the intermediate value property.</p>
<p>The proof is <a href="http://books.google.de/books?id=fXfEG-F2zJUC&printsec=frontcover&hl=de#v=onepage&q&f=false" rel="nofollow">here on page 6 theorem 4.1</a></p>
|
77,466 | <p>There is a standard syllabus for a first graduate course in algebra. One teaches groups,
rings, fields, perhaps a little bit of Galois theory, perhaps a little bit of
category theory, perhaps a little bit of representation theory, all this a little bit superficially, to give an idea of the fundamental
algebraic structure to graduate students that will work in all parts of mathematics.</p>
<p>I have much more difficulties to see what to teach in a second, more advanced, course in algebra, whose student body is constituted of the grad students who like algebra, whatever they are eventually going to work in. Commutative algebra is excluded because in my department, as in many others, there is another course devoted to this specific subject.
But even so, there are so many loosely inter-related things (more category theory, more homological algebra, more representation theory, advanced theory of finite groups, study of classical groups, theory of groups defined by generators and relations, Brauer theory, etc.) one could think of that I find very difficult to
arbitrage between them. One is naturally pushed to give a course with no unity, which is not very pleasant. </p>
<p>Since the problem I experience has certainly been met by others, I'd like to know:
What did you or would you teach in such a course? What are the subjects that are
absolutely necessary to teach (if any)? How to give the course a backbone? What textbook to use?</p>
| David White | 11,540 | <p>I'm posting a separate answer because I realized my first might be too much algebraic topology. Another great second year course idea is to follow <em><a href="http://books.google.com/books/about/Lectures_on_modules_and_rings.html?id=r9VoYbk-8c4C" rel="nofollow">Lectures on Modules and Rings</a></em> by T.Y. Lam. This book is my bible for homological algebra, and I have never heard anyone claim there was a better book for this. The only "downside" is that everything is non-commutative, but Lam does a great job of telling you exactly what commutativity gets you (via corollaries to the theorems), so even students who go on to work primarily in a commutative setting will not be ill-served.</p>
<p>If you wanted to cover lots of great homological algebra in this proposed second course (without an eye towards algebraic topology), I can't think of a better book. Here are some topics:</p>
<ul>
<li>Projective, Injective, and Flat Modules</li>
<li>Semisimple, Coherent, Von Neumann Regular Rings, Cohen-Macaulay, and Gorenstein Rings</li>
<li>Homological Dimensions and Regular Local Rings</li>
<li>Localization</li>
<li>Quasi-Frobenius Rings and Algebras</li>
<li>Matrix Rings and representation theory</li>
</ul>
<p>Since this is the "second course" to his "First Course in Noncommutative Rings" one might be tempted to use that text for the first course. I'm not sure this is such a good idea. While I love Lam's writing style and the vast amount of material he covers, it seems a lot of what <em>A First Course</em> covers isn't really necessary to do algebra later on, i.e. a lot of it deals with situations which modern research avoids via standard assumptions on the rings in question.</p>
|
1,016,585 | <p>I want to solve $$\int \frac{1}{\sqrt{x^2 - c}} dx\quad\quad\text{c is a constant}$$</p>
<p>How do I do this?</p>
<p>It looks like it is close to being an $\operatorname{arcsin}$?</p>
<hr>
<p>I would have thought I could just do:
$$\int \left(\sqrt{x^2 - c}\right)^{-\frac12}\, dx=\frac{2\sqrt{c+x^2}}{2x}\text{????}$$</p>
<p>But apparently not. </p>
| Community | -1 | <p>If you try this derivative (which is not helpful in itself),
$$(\sqrt{x^2-1})'=\frac x{\sqrt{x^2-1}},$$
you can notice that by adding an $x$ term,
$$(x+\sqrt{x^2-1})'=1+\frac x{\sqrt{x^2-1}}=\frac{x+\sqrt{x^2-1}}{\sqrt{x^2-1}},$$
and
$$\frac{(x+\sqrt{x^2-1})'}{x+\sqrt{x^2-1}}=\frac1{\sqrt{x^2-1}}.$$</p>
<p>From this you can conclude (and adapt for positive or negative $c$).</p>
|
3,249,926 | <p>I’ve tried applying Vandermonde’s identity, but got stuck. Any help would be appreciated!</p>
| nonuser | 463,553 | <p>Remember that <span class="math-container">$a\leq b \iff a^2\leq b^2$</span> only if <span class="math-container">$a,b\geq 0$</span>.</p>
<p>If <span class="math-container">$x\geq 0$</span> then from <span class="math-container">$x\leq 3$</span> after squaring, we get <span class="math-container">$x^2\leq 9$</span></p>
<p>and if <span class="math-container">$x\leq 0$</span> then from <span class="math-container">$-2\leq x$</span> thus <span class="math-container">$-x\leq 2$</span> so after squaring we get <span class="math-container">$x^2\leq 4$</span>. </p>
<p>So <span class="math-container">$$0\leq x^2\leq 9$$</span></p>
|
1,782,796 | <p>Is there a particular name given to a matrix of <strong><em>m</em></strong> rows and <strong><em>n</em></strong> columns such that it must have one and only one 1 in each row and 0 elsewhere? For instance:</p>
<pre><code>0 1 0 0 0 0 0
0 0 0 0 1 0 0
0 1 0 0 0 0 0
0 0 0 0 0 0 1
0 0 1 0 0 0 0
</code></pre>
<p>Note: It may or may not be a square matrix.</p>
| Ben Grossmann | 81,360 | <p>I'm not aware of a common name for this type of matrix. However, a nice way to describe it is that the rows of your matrix are rows of the identity matrix with possible repetition.</p>
|
1,782,796 | <p>Is there a particular name given to a matrix of <strong><em>m</em></strong> rows and <strong><em>n</em></strong> columns such that it must have one and only one 1 in each row and 0 elsewhere? For instance:</p>
<pre><code>0 1 0 0 0 0 0
0 0 0 0 1 0 0
0 1 0 0 0 0 0
0 0 0 0 0 0 1
0 0 1 0 0 0 0
</code></pre>
<p>Note: It may or may not be a square matrix.</p>
| Anon | 245,264 | <p>I haven't heard of a name for the class of such matrices. If I had to suggest a name for such a matrix, I would call it a "row selection matrix", since if $A \in R^{m\times n}$ is such a matrix and $B \in R^{n\times k}$ an arbitrary matrix, then every row of $AB$ is also a row of $B$.</p>
|
403,977 | <blockquote>
<p>Let $u_n>u_{n+1}>0$ for all $n \in \Bbb N$, and suppose that $u_2+u_4+u_8+u_{16}+\dots$ diverges. Prove that $\sum_{n=1}^{\infty}\frac{u_n}{n}$diverges.</p>
</blockquote>
<p>Please provide me a hint or a full solution.</p>
| Jonas Meyer | 1,424 | <p>I suggest that you read about the <a href="http://en.wikipedia.org/wiki/Cauchy_condensation_test">Cauchy condensation test</a>.</p>
|
2,228,913 | <p>For some geometric problem that I am trying to solve, I am interested
in characterizing the cross product between a Bezier curve and its second order derivative.</p>
<p>So, let's $\mathbf{c}(t)$ be a Bezier curve, and $\ddot{\mathbf{c}}(t)$ it's derivative.</p>
<p>Is $\mathbf{w}(t) = \mathbf{c}(t) \times \ddot{\mathbf{c}}(t)$ a Bezier curve?</p>
<p>If it is the case, is it also the case in general ?</p>
<p>Thanks a lot for your time and answers. Sincerely,</p>
<p>Steve</p>
| Steve T. | 435,649 | <p>Thanks for your answers. I was in fact trying to compute the coefficients of this Bezier curve, while proving that the product is indeed a Bezier curve.</p>
<p>A friend helped me to obtain this:</p>
<p>We write:</p>
<ul>
<li>$\mathbf{c}(t) = \sum_{i=0}^n \mathbf{P}_i B^n_i(t)$</li>
<li>$\mathbf{\ddot{c}}(t) = \sum_{i=0}^{n-2} \mathbf{Q}_i B^{n-2}_i(t)$</li>
</ul>
<p>We can rewrite a Bernstein polynom as as <a href="https://en.wikipedia.org/wiki/Bernstein_polynomial#Properties" rel="nofollow noreferrer">polynom of higher degree</a> so:</p>
<ul>
<li>$\mathbf{\ddot{c}}(t) = \sum_{i=0}^{n} \mathbf{R}_i B^{n}_i(t)$</li>
</ul>
<p>Then </p>
<p>$\mathbf{w}(t) = \sum_{i=0}^{n} \sum_{j=0}^{n}( \mathbf{P}_i \times \mathbf{R}_j) B^{n}_i(t)B^{n}_j(t)$</p>
<p>$ B^{n}_i(t)B^{n}_j(t) ={{n}\choose{i}}{{n}\choose{j}} t^i(1-t)^{n-i}t^j(1-t)^{n-j}( \mathbf{P}_i \times \mathbf{R}_{k-i})$</p>
<p>We write $k = i + j$</p>
<p>$\mathbf{w}(t) = \sum_{k=0}^{2n} [\sum_{i\in[0,n]\cap[k-n,k]}^{n} {{n}\choose{i}}{{n}\choose{k-i}} ( \mathbf{P}_i \times \mathbf{R}_{k-i})\mathbf{)} t^k(1-t)^{2n-k}]$</p>
<p>$\mathbf{w}(t) = \sum_{k=0}^{2n}{{2n}\choose{k}}t^k(1-t)^{2n-k}\alpha_k) $</p>
<p>with $\alpha_k = \sum_{i\in[0,n]\cap[k-n,k]}^{n} {{n}\choose{i}}{{n}\choose{k-i}} ( \mathbf{P}_i \times \mathbf{R}_{k-i})\mathbf{)} / {{{2n}\choose{k}}}$</p>
<p>Thus </p>
<p>$\mathbf{w}(t) = \sum_{k=0}^{2n} B^{2n}_k(t)\alpha_k$</p>
<p>Does that seem right ?</p>
|
1,392,857 | <p>Evaluate the integral from: </p>
<p>$$\int_0^{\infty} \frac{x \cdot \sin(2x)}{x^2+3}dx$$</p>
<p>The way I approach this problem is </p>
<p>$$\int_0^{\infty} \frac{x \cdot \sin(2x)}{x^2+3}dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{z \cdot e^{i2z}}{(z - i\sqrt{3})(i+i\sqrt{3})}dz$$
and
$$ \text{Res}_{i\sqrt3}(f(z)) = \frac{e ^{-2\sqrt3}}{2\sqrt3}$$
Then, the integral will be:
$$\frac{1}{2}\int_{-\infty}^{\infty} \frac{z \cdot e^{i2z}}{(z - i\sqrt{3})(i+i\sqrt{3})}dz = \frac{1}{2} \cdot 2\pi i \cdot \frac{e^{-2\sqrt3}}{2\sqrt3} = \frac{\pi i e^{-2\sqrt3}}{2\sqrt3}$$</p>
<p>Is my approach correct? if not, can someone show me? Sorry because I just learn about residue theorem and don't know if my work is correct or not.</p>
| GeorgSaliba | 142,772 | <p>First of all, keep in mind that
$$\sin(2z)=\Im (e^{2iz})$$
Second, the residue is wrong, you've forgotten the $z$ in the numerator:
$$\text{Res}_{i\sqrt{3}}(f(z))=\frac{P(z).e^{i2z}}{Q'(z)}$$</p>
<p>$$\text{Res}_{i\sqrt{3}}(f(z))=\frac{z.e^{i2z}}{2z}$$
$$\text{Res}_{i\sqrt{3}}(f(z))=\frac{e^{i2z}}{2}$$</p>
<p>So the integral $\text{I}$ is:
$$\text{I}=\Im(2i\pi\text{Res}_{i\sqrt{3}}(f(z)))=\Im(2i\pi\frac{e^{-2\sqrt{3}}}{2})=\pi\frac{e^{-2\sqrt{3}}}{2}$$</p>
|
1,392,857 | <p>Evaluate the integral from: </p>
<p>$$\int_0^{\infty} \frac{x \cdot \sin(2x)}{x^2+3}dx$$</p>
<p>The way I approach this problem is </p>
<p>$$\int_0^{\infty} \frac{x \cdot \sin(2x)}{x^2+3}dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{z \cdot e^{i2z}}{(z - i\sqrt{3})(i+i\sqrt{3})}dz$$
and
$$ \text{Res}_{i\sqrt3}(f(z)) = \frac{e ^{-2\sqrt3}}{2\sqrt3}$$
Then, the integral will be:
$$\frac{1}{2}\int_{-\infty}^{\infty} \frac{z \cdot e^{i2z}}{(z - i\sqrt{3})(i+i\sqrt{3})}dz = \frac{1}{2} \cdot 2\pi i \cdot \frac{e^{-2\sqrt3}}{2\sqrt3} = \frac{\pi i e^{-2\sqrt3}}{2\sqrt3}$$</p>
<p>Is my approach correct? if not, can someone show me? Sorry because I just learn about residue theorem and don't know if my work is correct or not.</p>
| Math-fun | 195,344 | <p>Here is another approach:</p>
<p>$$f(a)=\int_0^{\infty} \frac{x \cdot \sin(ax)}{x^2+3}dx$$ take a Laplace transform with respect to $a$ to obtain
\begin{align}
\mathcal{L}(f(a))&=\int_0^{\infty} \frac{x^2}{(x^2+3)(x^2+s^2)}dx\\
&=\frac{\pi}{2\sqrt3+2s}
\end{align}
now take an inverse Laplace to obtain
$$f(a)=\frac{\pi}{2} e^{-\sqrt{3} a}$$</p>
<p>Therefore $f(2)=\frac{\pi}{2} e^{-2\sqrt{3}}$.</p>
|
159,777 | <p>In his <a href="http://www.ams.org/mathscinet-getitem?mr=44124">short paper</a> (1951) and <a href="http://www.ams.org/mathscinet-getitem?mr=74819">longer monograph</a> (1955), Postnikov introduced what I believe are now called <a href="http://en.wikipedia.org/wiki/Postnikov_system"><em>Postnikov systems</em> or <em>towers</em></a>. It is my understanding that Postnikov systems have since then been widely adopted, as a way of totally encoding the homotopy type of a topological space. Expressed algebraically, a Postnikov system of a topological space $X$ consists of its homotopy groups $\pi_i = \pi_i(X)$, where each of the higher groups $\pi_i$ ($i\ge 2$) has the structure of a $\pi_1$-module, together with a group cohomology class $[k_1] \in H^{3}(\pi_1,\pi_2)$ and higher "cohomology classes" (that are somewhat more difficult to describe algebraically) $[k_i]$ with coefficients in $\pi_{i+1}$ for each $i\ge 2$.</p>
<p>The main application of these systems, in Postnikov's original work, was to reconstruct the cohomology of the space $X$ from its homotopy invariants in a purely algebraic way. The reconstruction allowed also for cohomology with coefficients in a <a href="http://en.wikipedia.org/wiki/Local_system">local system</a>, with the local system algebraically presented as some $\pi_1$-module. In that work, this result was presented as a generalization of the earlier work (1945) of <a href="http://www.ams.org/mathscinet-getitem?mr=13312">Eilenberg and MacLane</a> (and of course others) on the algebraic reconstruction of the cohomology of spaces that are aspherical except in one degree.</p>
<p>Now my question. <strong>Is there a modern reference for the reconstruction of the cohomology (desirably with local coefficients) of a topological space from its Postnikov sequence?</strong> While Postnikov sequences themselves are treated in many places, I've not been able to find the cohomology reconstruction theorem anywhere except in Postnikov's original <a href="http://mi.mathnet.ru/tm1182">longer monograph</a> (in Russian).</p>
| Neil Strickland | 10,366 | <p>First, you say:</p>
<blockquote>
<p>It is my understanding that Postnikov systems have since then been
widely adopted, as a way of totally encoding the homotopy type of a
topological space.</p>
</blockquote>
<p>I don't think that this is really true. There are very few cases where one can actually describe the full Postnikov system explicitly, so it is not widely used, except for certain very restricted classes of spaces.</p>
<p>Next, in thinking about what is theoretically possible, you should remember the Kan-Thurston theorem. That says that for any connected space $X$ there is a group $G$ and a map $K(G,1)\to X$ that induces an isomorphism in homology. Here the space $K(G,1)$ has the simplest possible Postnikov tower, with only one layer and no $k$-invariants, but the homology can be anything you want, and does not depend in any very obvious way on $G$.</p>
|
224,474 | <p>I want to generate the following matrix for any n:</p>
<pre><code> Table[A[i, j], {i, 0, n}, {j, 0, n}];
</code></pre>
<p>Where,</p>
<pre><code> A ={{0,1,0},{0,0,2},{0,0,0}} when n=2;
A={{0,1,0,0},{0,0,2,0},{0,0,0,3},{0,0,0,0}} when n=3;
A={{0,1,0,0,0},{0,0,2,0,0},{0,0,0,3,0},{0,0,0,0,4},{0,0,0,0,0}} when n=4, and so on for any n.
</code></pre>
<p>Thanks</p>
| Alan | 19,530 | <p>The following approach is easily adapted to other cases. (But Bob Hanlon's answer is the right one for the original question.)</p>
<pre><code>makeMat[n_] := With[{
zeros = ConstantArray[0, {n, n}],
rules = (Rule[{#, 1 + #}, #] &) /@ Range[n - 1]
},
ReplacePart[zeros, rules]]
</code></pre>
|
3,926,580 | <p>I'm trying to prove with squeeze theorem that the limit of the following series equals 1:</p>
<p><span class="math-container">$$\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$$</span></p>
<p>For the left side of the inequality I did:</p>
<p><span class="math-container">$$\frac{1+\sqrt{1}+\sqrt[3]{1}+...+\sqrt[n]{1}}{n} < \frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$$</span></p>
<p>For the right side, at first I did the following:</p>
<p><span class="math-container">$$\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n} < \frac{n\sqrt[n]{n}}{n}$$</span></p>
<p>But then I realized it wasn't true and that the direction of this inequality is the opposite.</p>
<p>Do you have any idea which series with limit 1 is bigger from the original series?</p>
<p>Thanks!</p>
| AlanD | 356,933 | <p>This is not a full answer to the question, but many answers are implying that the function <span class="math-container">$n\mapsto n^{1/n}$</span> is strictly increasing. This is not the case. To see this:</p>
<p>Let <span class="math-container">$y=x^{1/x}$</span>. Then <span class="math-container">$\ln y=\frac 1x \ln x$</span> so <span class="math-container">$\frac{y'}{y}=\frac{1}{x^2}(1-\ln x)$</span>. Since <span class="math-container">$y>0$</span>, this implies that <span class="math-container">$y$</span> is increasing on <span class="math-container">$(0,e)$</span> and decreasing on <span class="math-container">$(e,\infty)$</span>.</p>
<p>Hence, <strong>do not</strong> use upper bound of <span class="math-container">$n^{1/n}$</span>.</p>
|
3,898,017 | <p><span class="math-container">$$\frac{1}{2z-1} = f(z)$$</span></p>
<p>So I don't think CR theoreum applies. So if the denominator is 0, then it's not defined. If <code>z = 1/2</code>, then it is not defined at 0. So <code>f(z)</code> it is not analytic even within C right? So there does not exist a simply connected domain D on which f is analytic and such that D contains C. So cannot apply Cauchy's Riemann equation right?</p>
<p>So we have to integrate another way.</p>
<p>So unit circle could be expressed as <span class="math-container">$e^{it} = z(t)$</span> where t is between 0 and <span class="math-container">$2\pi$</span></p>
<p>Furthermore <span class="math-container">$dz/dt = ie^{it}$</span></p>
<p><span class="math-container">$$\int_0^{2\pi} \frac{1}{2e^{it} - 1} \cdot ie^{it}$$</span></p>
<p>Am I on the right track? If so, can someone help with the integral?</p>
| Mark Viola | 218,419 | <p><strong>METHODOLOGY <span class="math-container">$1$</span>: CONTOUR DEFORMATION</strong></p>
<p>We can deform the contour <span class="math-container">$|z|=1$</span> and apply Cauchy's Integral Theorem to assert that for <span class="math-container">$0<r<1/2$</span></p>
<p><span class="math-container">$$\begin{align}
\oint_{|z|=1}\frac1{2z-1}\,dz&=\oint_{|z-1/2|=r}\frac1{2z-1}\,dz\\\\
&=\int_0^{2\pi}\frac1{2re^{it}}ire^{it}\,dt\\\\
&=i\pi
\end{align}$$</span></p>
<hr />
<p><strong>METHODOLOGY <span class="math-container">$2$</span>: PARAMETERIZATION</strong></p>
<p>Alternatively, we can parameterize the contour <span class="math-container">$|z|=1$</span> and write</p>
<p><span class="math-container">$$\begin{align}
\oint_{|z|=1}\frac1{2z-1}\,dz&=\lim_{\varepsilon\to 0^+}\int_{\varepsilon}^{2\pi-\varepsilon}\frac{ie^{it}}{2e^{it}-1}\,dt\\\\
&=\lim_{\varepsilon\to 0^+}\int_{\varepsilon}^{2\pi-\varepsilon}\frac{\sin(t)+i2(1-\cos(t))}{5-4\cos(t)}\,dt\\\\
&=\lim_{\varepsilon\to 0^+}\left.\left(\frac14\log(5-4\cos(t))+i2\left(\frac t4-\frac16 \arctan\left(3\tan(t/2)\right)\right)\right) \right|_{\varepsilon}^{2\pi -\varepsilon}\\\\
&=i\pi
\end{align}$$</span></p>
<p>as expected!</p>
<hr />
<p><strong>METHODOLOGY <span class="math-container">$3$</span>: COMPLEX LOGARITHM</strong></p>
<p>As a third approach, we cut the plane so that the natural logarithm is defined as <span class="math-container">$\log(z)=\text{Log}(|z|)+i\arg(z)$</span>, where <span class="math-container">$0\le \arg(z)<2\pi$</span> we have</p>
<p><span class="math-container">$$\oint_{|z|=1}\frac1{2z-1}\,dz=\lim_{\varepsilon\to 0^+}\left.\left(\frac12\log(2z-1)\right)\right|_{z=1+i\varepsilon}^{z=1-i\varepsilon}=i\pi$$</span></p>
|
3,898,017 | <p><span class="math-container">$$\frac{1}{2z-1} = f(z)$$</span></p>
<p>So I don't think CR theoreum applies. So if the denominator is 0, then it's not defined. If <code>z = 1/2</code>, then it is not defined at 0. So <code>f(z)</code> it is not analytic even within C right? So there does not exist a simply connected domain D on which f is analytic and such that D contains C. So cannot apply Cauchy's Riemann equation right?</p>
<p>So we have to integrate another way.</p>
<p>So unit circle could be expressed as <span class="math-container">$e^{it} = z(t)$</span> where t is between 0 and <span class="math-container">$2\pi$</span></p>
<p>Furthermore <span class="math-container">$dz/dt = ie^{it}$</span></p>
<p><span class="math-container">$$\int_0^{2\pi} \frac{1}{2e^{it} - 1} \cdot ie^{it}$$</span></p>
<p>Am I on the right track? If so, can someone help with the integral?</p>
| Raffaele | 83,382 | <p>Using <a href="https://en.wikipedia.org/wiki/Contour_integration#Evaluation_with_residue_theorem" rel="nofollow noreferrer">residue theorem</a> we easily get
<span class="math-container">$$\oint\limits_C {\frac{{dz}}{{2z - 1}} = 2\pi i}\;\text{Res}\left(\frac{1}{2 z-1},\left\{z,\frac{1}{2}\right\}\right) =\pi i$$</span></p>
<p>Because
<span class="math-container">$$\text{Res}\left(\frac{1}{2 z-1},\left\{z,\frac{1}{2}\right\}\right)=\underset{z\to \frac{1}{2}}{\text{lim}}\frac{z-\frac{1}{2}}{2 z-1}=\frac{1}{2}$$</span></p>
|
388,875 | <p>For $V$ a vector space.</p>
<p>Let $T\colon V\rightarrow V$ be a linear transformation of a finite dimensional inner product space.</p>
<p>How can I prove that $\ker(T^* T + TT^*) = $\ker(T)$\cap$$\ker(T^*)$?</p>
<p>And what is the meaning of $\ker(T^*T + TT^*)$ ?</p>
<p>Is $\ker(T^*T + TT^*)=\ker(T^*T)+\ker(TT^*)$ ?</p>
| wantToLearn | 74,557 | <p>any suggestion ?</p>
<p>I know how to prove that for every vector that is in the kernel of T and T*</p>
<p>is also at the kernel of TT*+T*T (very trivial).</p>
<p>but how to prove the other side ?</p>
|
497,609 | <p>There are 30 balls in a bag. There are 6 balls each of 5 different colors. What is the minimum number of times I have to pick a ball to pick 2 balls of each color?</p>
<ol>
<li>24</li>
<li>25</li>
<li>26</li>
<li>27</li>
<li>28</li>
</ol>
<p>Is the question correct? Someone asks me this and I'm confused.</p>
| André Nicolas | 6,312 | <p>If we want to have <strong>certainty</strong> of picking at least two balls of each colour. The worst case is if we have picked up all the balls of every colour but yellow, and only one yellow. So $25$ is not enough. </p>
<p>But $26$ is enough. For if we pick $26$, there are only $4$ balls left, so we have at least $2$ balls of each colour. </p>
<p><strong>Remark:</strong> If on the other hand we consider the simpler problem of what is the <em>least</em> number of times we need to do the picking, if we are very lucky, then the answer is $10$, since with some luck when we pick $10$ balls we will have $2$ of each colour.</p>
<p>There is some (well, more than some) ambiguity in the question. The choices provided indicate that the problem-poser did not intend the question to be interpreted in the "if we get lucky" sense. </p>
|
3,826,152 | <p>How can I verify this inequality by using induction?</p>
<p><span class="math-container">$$\frac1{2n}\le \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}$$</span></p>
<p>I understand the principle of induction, but I'm struggling with how my teacher is solving this problem. I Highlighted the area I'm stuck at In the solution - why are we able to take <span class="math-container">$1/2k$</span> and see if its less than the left side of our inducted hypothesis? (view the highlighted part of the photo)</p>
<p>Any suggestions would be great - I really don't understand how to solve this problem. If I could get a step by step walk through of this proof with side notes, I'd be very grateful</p>
<p><a href="https://i.stack.imgur.com/LU2yX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LU2yX.png" alt="enter image description here" /></a></p>
| Bernard | 202,857 | <p>Your teacher simply uses transitivity of inequalities:</p>
<p>To simplify the notations, I'll use double factorials.</p>
<p>The inductive hypothesis is that <span class="math-container">$\;\dfrac1{2(k-1)}\le \dfrac{(2k-3)!!}{(2k-2)!!}$</span>, and he/she wants to deduce that
<span class="math-container">$$\dfrac1{2k}\le \dfrac{(2k-1)!!}{(2k)!!}.$$</span></p>
<p>Now, observing that <span class="math-container">$\; \dfrac{(2k-1)!!}{(2k)!!}= \dfrac{(2k-3)!!}{(2 k-2)!!}\cdot\dfrac{2k-1}{2k}$</span>, we deduce from the inductive hypothesis that
<span class="math-container">$$\dfrac{(2k-1)!!}{(2k)!!}\ge\frac1{2k-2}\cdot \frac{2k-1}{2k}=\frac1{2k}\cdot \underbrace{\frac{2k-1}{2k-2}}_{{}>1}>\frac1{2k}.$$</span></p>
|
1,826,094 | <p>$i\cdot\bar{z} = 2+2i$</p>
<p>I know that $\bar{z} = a-bi$ so then i get $i(a-bi)=2+2i$
Then $ai+b=2+2i$ (because $i^2=-1$)</p>
<p>When 2 complex numbers are equal you usually can equal their parts
Ex: $2+2i=a+bi$ so $a=2$ and $b=2$</p>
<p>But in this case I aint got $a+bi$ I have $ai+b$</p>
<p>The answer is $z=2-2i$ but i dont understand how to get there</p>
<p>Thanks</p>
| egreg | 62,967 | <p>Suppose $z=a+bi$; then $\bar{z}=a-bi$ and $i\bar{z}=ai+b=b+ai$. Thus the real part of $i\bar{z}$ is $b$ and the imaginary part is $a$.</p>
<p>Hence, equating them gives $b=2$ and $a=2$.</p>
<p>However, you can do it more easily: since $i^{-1}=-i$, multiplying both sides by $-i$ gives
$$
\bar{z}=-i(2+2i)=2-2i
$$
and so
$$
z=\overline{2-2i}=2+2i
$$</p>
|
2,057,014 | <p>$a_n, b_n \in \mathbb R; n \in \mathbb N; x \in \mathbb R \land x > 1$ are defined as:</p>
<p>$$a_1 = 1 \land b_1 = x$$
$$a_{n+1} = \frac {2a_nb_n} {a_n+b_n} \land b_{n+1} = \frac {a_n+b_n} 2$$</p>
<p>How can I prove that $a_nb_n < a_n+b_n$ ?</p>
| ajotatxe | 132,456 | <blockquote>
<p>Your claim is true if and only if $x\le 4$.</p>
</blockquote>
<p>We will use induction to show some facts about both sequences. Namely,</p>
<p>$$a_{n+1}b_{n+1}=\frac{2a_nb_n}{a_n+b_n}\cdot\frac{a_n+b_n}2=a_nb_n=x$$</p>
<p>$$b_{n+1}-a_{n+1}=\frac{a_n+b_n}2-\frac{2a_nb_n}{a_n+b_n}=\frac{(b_n-a_n)^2}{2(a_n+b_n)}>0$$</p>
<p>Also, $a_{n+1}>a_n$:
$$a_{n+1}-a_n=\frac{2x}{a_n+b_n}-a_n>\frac{2x}{2b_n}-a_n=a_n-a_n=0$$</p>
<p>Since $a_nb_n$ is constant and $a_n$ is strictly increasing, $b_n$ is strictly decreasing.</p>
<p>All this implies that $a=\lim a_n$ and $b=\lim b_n$ exist. Also, $a_n<a\le b<b_n$ for every $n\in\Bbb N$.</p>
<p>Taking limits in the equations:
$$a=\frac{2ab}{a+b}$$
$$b=\frac{a+b}2$$
And then
$$2(a+b)^2=4ab+(a+b)^2$$
That is
$$(a-b)^2=0$$
Then $a=b$. So $a=b=\sqrt x$.</p>
<p>Avoiding formalism for the moment, for $n$ large enough, we see that $a_n+b_n$ will be near from $2\sqrt x$, which, for $x> 4$, is lesser than $x$.</p>
<p>Assuming $x\le 4$, by GM-AM inequality,
$$\frac{a_n+b_n}{a_nb_n}>\frac{2\sqrt x}x=\frac2{\sqrt x}\ge 1$$</p>
<p>But if $x>4$, for any $\epsilon>0$ there exists some $n\in\Bbb N$ such that $$\sqrt x-\epsilon<a_n<\sqrt x<b_n<\sqrt x+\epsilon$$
and then
$$a_n+b_n<2\sqrt x+\epsilon$$
Taking $\epsilon=x-2\sqrt x>0$, we see that the claim is false.</p>
|
2,057,014 | <p>$a_n, b_n \in \mathbb R; n \in \mathbb N; x \in \mathbb R \land x > 1$ are defined as:</p>
<p>$$a_1 = 1 \land b_1 = x$$
$$a_{n+1} = \frac {2a_nb_n} {a_n+b_n} \land b_{n+1} = \frac {a_n+b_n} 2$$</p>
<p>How can I prove that $a_nb_n < a_n+b_n$ ?</p>
| robjohn | 13,854 | <p>$\frac{2ab}{a+b}$ is the harmonic mean of $a$ and $b$ while $\frac{a+b}2$ is the arithmetic mean. When $a,b\gt0$, the harmonic mean is less than or equal to the geometric mean $\left(\sqrt{ab}\right)$, which is less than or equal to the arithmetic mean.</p>
<p>It is useful to note that the harmonic mean of $x$ and $\frac1x$ is $\frac{2x}{x^2+1}$ while the arithmetic mean of $x$ and $\frac1x$ is $\frac{x^2+1}{2x}$. Thus, if $a_n$ and $b_n$ are reciprocals, $a_{n+1}$ and $b_{n+1}$ are also reciprocals. In this case, $a_n$ approaches $1$ from below and $b_n$ approaches $1$ from above. Since the mean (harmonic, geometric, and arithmetic) of a multiple of two numbers is that multiple of the mean of the same two numbers, we get that
$$
a_n\nearrow\sqrt{a_1b_1}\tag{1}
$$
and
$$
b_n\searrow\sqrt{a_1b_1}\tag{2}
$$
From $(1)$, we see that $a_n\le2$ whenever $x\le4$. In this case,
$$
\frac{a_nb_n}{a_n+b_n}=\frac{a_{n+1}}{2}\le1\tag{3}
$$
If $x\gt4$, then at some point, $a_{n+1}\gt2$ and then
$$
\frac{a_nb_n}{a_n+b_n}=\frac{a_{n+1}}{2}\gt1\tag{4}
$$</p>
|
47,177 | <p>Recently I've been reading "The Wild Book" which applies semigroup theory to, among other things, chemical reactions. If I google for mathematics and chemistry together, most of the results are to do with physical chemistry: cond-mat, fluids, QM of molecules, and analysis of spectra. I'm more interested in learning about biochemistry, molecular biology, and organic chemistry — and would prefer to learn from a mathematical perspective.</p>
<p>What other books aim to teach (bio- <code>||</code> organic) chemistry specifically to those with a mathematical background?</p>
| pharmine | 15,371 | <p><strong>Organic chemistry</strong></p>
<p>S. Fujita's "<a href="http://rads.stackoverflow.com/amzn/click/0387541268">Symmetry and combinatorial enumeration in chemistry</a>" (Springer-Verlag, 1991) is one such endeavor. It mainly focuses on stereochemistry.</p>
<p><strong>Molecular biology and biochemistry</strong></p>
<p>A. Carbone and M. Gromov's "<a href="http://www-master.ufr-info-p6.jussieu.fr/2005/IMG/pdf/gazette.pdf">Mathematical slices of molecular biology</a>" is recommended, although it is not strictly a book.</p>
<p>R. Phillips, J. Kondev and J. Theriot have published "<a href="http://www.garlandscience.com/product/isbn/0815341636">Physical biology of the cell</a>", which contains biochemical topics (such as structures of hemoglobin) and is fairly accessible to mathematicians in my opinion.</p>
|
3,312,959 | <p>If the distance from the point <span class="math-container">$(K, 4)$</span> to <span class="math-container">$(2, 4)$</span> and y axis are same find k. </p>
<p>I can't understand what to calculate. According to my calculation the answer is 1, but the given answer is 0 and 4.</p>
| Fareed Abi Farraj | 584,389 | <p>I think the question is not very clear. You understood the question as "the distance from <span class="math-container">$(2,4)$</span> to <span class="math-container">$(K,4)$</span> is equal to the distance from <span class="math-container">$(K,4)$</span> to the y-axis". But actaully the question is stating that "the distance from <span class="math-container">$(2,4)$</span> to the y-axis is equal to the distance from <span class="math-container">$(2,4)$</span> to the point <span class="math-container">$(K,4)$</span>".</p>
<p>I think you know how to work it out now.</p>
<p><span class="math-container">$d_{(2,4) \to y-axis}=2=d_{(2,4) \to (K,4)}=\sqrt{(K-2)^2}=|K-2|$</span></p>
<p><span class="math-container">$\implies K-2=2 \implies K=4$</span></p>
<p>Or</p>
<p><span class="math-container">$K-2=-2 \implies K=0$</span></p>
|
2,193,306 | <p>I have to evaluate the integral of the $\sec x$ function and I do it as follows $$\int\sec xdx=\int\frac{dx}{\cos x}=\int\frac{\cos^2 x+\sin^2 x}{\cos x}dx=\sin x+\int\frac{\sin^2x}{\cos x}dx$$ Now we make a change of variables $\cos x=t$ so our integral becomes $$\sin x-\int\frac{\sqrt{1-t^2}}{t}dt$$ Now we make another change of variable $\sqrt{1-t^2}=z$ so our integral becomes $$\sin x-\int\frac{z^2-1+1}{z^2-1}dz=\sin x-\int\left(1+\frac{1}{z^2-1}\right)dz=$$ $$=\sin x-\sqrt{1-\cos^2x}+\frac12\ln\left|\frac{1+\sqrt{1-\cos^2x}{}}{1-\sqrt{1-\cos^2x}}\right|+C=\frac12\ln|\tan^2x+\sec^2x|+C$$ The <a href="http://www.integral-calculator.com/" rel="nofollow noreferrer">calculator</a> however evaluates this integral as $$\ln|\tan x+\sec x|+C$$
but I can't figure out where I made a mistake in my calculations.</p>
| imranfat | 64,546 | <p>Just another approach: <span class="math-container">$\frac{1}{\cos x}=\frac{\cos x}{\cos^ {2x}}=\frac{\cos x}{1-\sin^{2x}}$</span>.And performing a substitution <span class="math-container">$cosx=t$</span> in <span class="math-container">$\int\frac{{\cos}xdx}{1-\sin^{2x}}$</span> results in a doable integral. </p>
|
114,831 | <p>$$A_n=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$$
Try to prove $$\lim_{n \to \infty}n(\ln 2-A_n) = \frac{1}{4}$$</p>
<p>I try to decompose $\ln 2$ as $$\ln(2n)-\ln(n)=\ln\left(1+\frac{1}{2n-1}\right)+\dots+\ln\left(1+\frac{1}{n}\right)\;,$$ but I can't continue, is that right?</p>
| TMM | 11,176 | <p>Let's cheat and use one of Euler's many results:</p>
<p>$$\sum_{i=1}^{n} \frac{1}{i} = \ln n + \gamma + \frac{1}{2n} + O\left(\frac{1}{n^2}\right)$$</p>
<p>Note that:</p>
<p>$$A_n + \sum_{i=1}^{n} \frac{1}{i} = \sum_{i=1}^{2n} \frac{1}{i}$$</p>
<p>Substituting Euler's result for both summations, we get:</p>
<p>$$A_n + \ln n + \gamma + \frac{1}{2n} + O\left(\frac{1}{n^2}\right) = \ln 2n + \gamma + \frac{1}{4n} + O\left(\frac{1}{n^2}\right)$$</p>
<p>Rearranging, and using $\ln 2n = \ln 2 + \ln n$, we get</p>
<p>$$A_n = \ln 2 - \frac{1}{4n} + O\left(\frac{1}{n^2}\right)$$</p>
<p>Thus the requested limit becomes</p>
<p>$$\lim_{n \to \infty} n (\ln 2 - A_n) = \lim_{n \to \infty} n \left(\frac{1}{4n} - O\left(\frac{1}{n^2}\right)\right) = \frac{1}{4}$$</p>
|
114,831 | <p>$$A_n=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$$
Try to prove $$\lim_{n \to \infty}n(\ln 2-A_n) = \frac{1}{4}$$</p>
<p>I try to decompose $\ln 2$ as $$\ln(2n)-\ln(n)=\ln\left(1+\frac{1}{2n-1}\right)+\dots+\ln\left(1+\frac{1}{n}\right)\;,$$ but I can't continue, is that right?</p>
| Ragib Zaman | 14,657 | <p>Trying to follow the idea of the OP, write $$ \log 2 = \log (2n+2) - \log (n+1) = \log \left( 1+ \frac{1}{2n+1} \right) + \log \left( 1+ \frac{1}{2n} \right) + \cdots + \log \left( 1+ \frac{1}{n+1} \right) $$</p>
<p>so then $$ n( \log 2 - A_n) = n\log \left(1+ \frac{1}{2n+1} \right) + n\sum_{k=1}^n \left( \log \left( 1+ \frac{1}{n+k} \right) - \frac{1}{n+k} \right) .$$</p>
<p>Since near $x=0$ we have $\displaystyle \log(1+x) = x - \frac{x^2}{2} + \mathcal{O}(x^3) ,$ the first term tends to $1/2$ and the summand is $ \displaystyle \frac{-1}{2 (n+k)^2 } + \mathcal{O}(1/n^3) .$ Thus, $$n\sum_{k=1}^n \left( \log \left( 1+ \frac{1}{n+k} \right) - \frac{1}{n+k} \right) = \frac{-1}{2} \cdot \frac{1}{n} \left( \sum_{k=1}^n \frac{1}{\left(1+ \frac{k}{n} \right)^2}\right) + \mathcal{O}(1/n) $$</p>
<p>$$ \to \frac{-1}{2} \int^1_0 \frac{1}{(1+x)^2} dx= -\frac{1}{4}. $$</p>
<p>Thus, $$ n(\log 2 - A_n) \to \frac{1}{2} - \frac{1}{4} = \frac{1}{4}.$$</p>
|
812,143 | <p>City a is separated by a 2km wide river and are located as shown in Figure 1 (not drawn to scale). A road is to be built between city A to B that crosses a bridge straight across the river. </p>
<p>Use the Sign Test to determine the type of stationary point </p>
<p>Test to see if you have a minimum</p>
| Claude Leibovici | 82,404 | <p>If you want to do with algebra, $A$ being defined as the origin, let us define two points $C(x,9)$ and $D(x,11)$. So, the distance between $A$ and $C$ is given by $$d_{AC}=\sqrt {(x-0)^2+(9-0)^2}=\sqrt {x^2+81}$$ The distance between $D$ and $B$ is given by $$d_{DB}=\sqrt {(17-x)^2+(14-11)^2}=\sqrt {x^2-34x+298}$$ So taking into account the fact tht $d_{CD}=2$, the total distance from $A$ to $B$ is $$d_{AB}=\sqrt {x^2+81}+2+\sqrt {x^2-34x+298}$$ what we want to be minimum. So, compute the derivative (it will contain radicals to be removed by proper rearrangement and squaring; this will finally let you with a quadratic equation $$-72 x^2+2754 x-23409=0$$ the roots of which being $x=\frac{51}{4}$ and $x=\frac{51}{2}$. The latest must be discarded since larger than $17$. So, the solution is $x=\frac{51}{4}$; for this value of $x$, the distance between $A$ and $B$ is then $2+\sqrt{433}=22.81$ km.</p>
<p>Computing the second derivative at $x=\frac{51}{4}$ gives a value of $\frac{768}{433 \sqrt{433}}$ which is positive, then the solution corresponds to a minimum.</p>
<p>If you look at the variation of $d_{AB}$ as a function of $x$, you should notice that the largest distance would correspond to $x=0$ for which $d_{AB}=11+\sqrt{298}=28.26$ km. </p>
<p>You could also notice that, flying, the shortest distance batween $A$ and $B$ is $\sqrt {485} =22.02$ km.</p>
|
3,005,591 | <p>I wish to find the integers of <span class="math-container">$a,b,c$</span> and <span class="math-container">$d$</span> such that:
<span class="math-container">$$225a + 360b +432c +480d = 3$$</span>
which is equal to:
<span class="math-container">$$75a + 120b +144c+ 160d =1$$</span></p>
<p>I know I have to use the Euclidean algorithm. And I managed to do it for two integers <span class="math-container">$x$</span> and <span class="math-container">$y$</span>. But can't figure out, how to do it with <span class="math-container">$4$</span> integers.</p>
| user21820 | 21,820 | <p>You can <strong>systematically</strong> solve any such equation (or prove that there are no solutions) by the following:</p>
<p>Take any integers <span class="math-container">$a,b,c,d$</span>. Then the following correspond:</p>
<ul>
<li>Solutions of <span class="math-container">$75a+120b+144c+160d = 1$</span></li>
<li>Solutions of <span class="math-container">$120b+144c+160d \equiv 1 \pmod{75}$</span></li>
<li>Solutions of <span class="math-container">$45b-6c+10d \equiv 1 \pmod{75}$</span></li>
<li>Solutions of <span class="math-container">$45b-6c+10d+75p = 1$</span> where <span class="math-container">$p$</span> is an integer</li>
<li>Solutions of <span class="math-container">$45b+10d+75p \equiv 1 \pmod{6}$</span></li>
<li>Solutions of <span class="math-container">$3b+4d+3p \equiv 1 \pmod{6}$</span></li>
<li>Solutions of <span class="math-container">$3b+4d+3p+6q = 1$</span> where <span class="math-container">$q$</span> is an integer</li>
<li>Solutions of <span class="math-container">$4d \equiv 1 \pmod{3}$</span></li>
</ul>
<p>And now you simply follow the reverse correspondences.</p>
|
2,640,143 | <h2>Problem</h2>
<p>Compute following:</p>
<p>$$ \int_{1}^{2} \int_{0}^{z} \int_{0}^{y+z}\frac{1}{(z+y+x)^3}dxdydz $$</p>
<h2> Attempt to solve </h2>
<p>Now i could probably solving this by opening $(z+y+x)^3$ parenthesis and then solving 3 separate integrals involving variables $x,y,z$. Opening these parenthesis wouldn't obviously look nice and it would be a lot of work for sure. I think better way would be to substitute $u=(z+y+x)$ and then we would need to adjust the integration limits by working chainrule backwards ?</p>
<p>$$ \frac{\delta u}{\delta x}=1, \quad \delta u = \delta x $$
$$ \frac{\delta u}{\delta y}=1, \quad \delta u = \delta y $$
$$ \frac{\delta u}{\delta z}=1, \quad \delta u = \delta z $$</p>
<p>$$ \int_{1}^{2} \int_{0}^{z} \int_{0}^{y+z}\frac{1}{(z+y+x)^3}\delta x\delta y\delta z $$
$$ \int_{1}^{2} \int_{0}^{z} \int_{0}^{y+z}\frac{1}{u^3}\delta u^3$$
$$\int_{1}^{2} \int_{0}^{z} -\frac{1}{2(y+x)^2}\delta u^2$$</p>
<p>Now it is visible at this point that this cant possibly work.I don't think i have very good understanding on how i am suppose to accomplish this what i am trying to do.</p>
| Mohammad Riazi-Kermani | 514,496 | <p>While the logarithmic function is obviously the inverse of the exponential function it's main purpose is to convert large numbers to smaller numbers.</p>
<p>For example for $$x=100,000,000,$$</p>
<p>$$ log(x)= 8.$$ </p>
<p>From $$log(xy)=log (x) + log (y)$$ we get $$log(3600) = 2+log(36)=2+2(log2 +log3)$$
Applications such as Earthquake Richter's scale or Radioactive half life makes logarithms a good tool to be around.</p>
<p>Logarithmic differentiation is used to find derivative of functions such as $$f(x)= x^x$$ and$$ f(x) = x^{sin x}.$$ </p>
|
4,143,759 | <p><strong>What I understand so far :</strong></p>
<p>Intuitively this question gives me a yuuuuuge hint, it tells me at t=0, t = -4. So that let's me know that I have a phase shift of -4 immediately.</p>
<p>I understand what a period is, I understand what a phase shift is, I understand what amplitude and vertical displacement are and how they work.</p>
<p>I'm just not sure how to build a function even with that info at my disposal.</p>
<p><em><strong>My ultimate question here isn't to find an answer, but to be given a hint or a pointing in the right direction. Is there anything that you, my acquaintances at stackexchange have to offer?</strong></em></p>
| DMcMor | 155,622 | <p><strong>Hint:</strong> You know that <span class="math-container">$V = \frac{4}{3}\pi r^{3}$</span>. Use implicit differentiation to take the derivative of both sides with respect to <span class="math-container">$t$</span>. You should have <span class="math-container">$\frac{dV}{dt}$</span> on the left, and something involving <span class="math-container">$\frac{dr}{dt}$</span> on the right.</p>
|
8,741 | <p>Here is a topic in the vein of <a href="https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence" title="Describe a topic in one sentence"> Describe a topic in one sentence</a> and <a href="https://mathoverflow.net/questions/4994/fundamental-examples" > Fundamental examples </a> : imagine that you are trying to explain and justify a mathematical theory T to a skeptical mathematician who thinks T is just some sort of abstract nonsense for its own sake. The ideal solution consists
of a problem P which can be stated and understood without knowing anything about T, but which is difficult (or impossible, even better) to solve without T, and easier (or almost-trivial, even better) to solve with the help of T. What should be avoided is an example where T is "superimposed", e.g. when T is a model for some physical phenomenon, because there is always something arbitrary about the choice of a specific model. </p>
<p>A classical example is Galois theory for solving polynomial equations. </p>
<p>Any examples for homological algebra ? For Fourier analysis ? For category theory ?</p>
| Jonas Meyer | 1,119 | <p>The spectral theory of commutative Banach algebras led to an elegant proof, due to Gelfand, of the following (previously difficult) theorem of Wiener:
If $f$ is a nowhere vanishing complex valued function on the unit circle whose Fourier coefficients are absolutely summable, then the Fourier coefficients of $1/f$ are also absolutely summable.</p>
|
8,741 | <p>Here is a topic in the vein of <a href="https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence" title="Describe a topic in one sentence"> Describe a topic in one sentence</a> and <a href="https://mathoverflow.net/questions/4994/fundamental-examples" > Fundamental examples </a> : imagine that you are trying to explain and justify a mathematical theory T to a skeptical mathematician who thinks T is just some sort of abstract nonsense for its own sake. The ideal solution consists
of a problem P which can be stated and understood without knowing anything about T, but which is difficult (or impossible, even better) to solve without T, and easier (or almost-trivial, even better) to solve with the help of T. What should be avoided is an example where T is "superimposed", e.g. when T is a model for some physical phenomenon, because there is always something arbitrary about the choice of a specific model. </p>
<p>A classical example is Galois theory for solving polynomial equations. </p>
<p>Any examples for homological algebra ? For Fourier analysis ? For category theory ?</p>
| Jonas Meyer | 1,119 | <p>Complex analysis and the theory of spaces of analytic functions on open subsets of $\mathbb{C}$ are used to prove the Riemann Mapping Theorem, which has the nice topological consequence that there is a unique homeomorphism class of (nonempty) simply connected open subsets of $\mathbb{C}$.</p>
|
8,741 | <p>Here is a topic in the vein of <a href="https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence" title="Describe a topic in one sentence"> Describe a topic in one sentence</a> and <a href="https://mathoverflow.net/questions/4994/fundamental-examples" > Fundamental examples </a> : imagine that you are trying to explain and justify a mathematical theory T to a skeptical mathematician who thinks T is just some sort of abstract nonsense for its own sake. The ideal solution consists
of a problem P which can be stated and understood without knowing anything about T, but which is difficult (or impossible, even better) to solve without T, and easier (or almost-trivial, even better) to solve with the help of T. What should be avoided is an example where T is "superimposed", e.g. when T is a model for some physical phenomenon, because there is always something arbitrary about the choice of a specific model. </p>
<p>A classical example is Galois theory for solving polynomial equations. </p>
<p>Any examples for homological algebra ? For Fourier analysis ? For category theory ?</p>
| José Figueroa-O'Farrill | 394 | <p>There are a number of algebraic theorems which are easier to prove using topology. The best known is probably <a href="http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra">the fundamental theorem of algebra</a>, but there are others. For example, an $n\times n$ matrix with positive entries always has an eigenvector all of whose enetries are nonnegative. (The matrix defines a continuous function from the standard simplex to itself which always has a fixed point.) I learned about this from Elmer Rees.</p>
|
8,741 | <p>Here is a topic in the vein of <a href="https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence" title="Describe a topic in one sentence"> Describe a topic in one sentence</a> and <a href="https://mathoverflow.net/questions/4994/fundamental-examples" > Fundamental examples </a> : imagine that you are trying to explain and justify a mathematical theory T to a skeptical mathematician who thinks T is just some sort of abstract nonsense for its own sake. The ideal solution consists
of a problem P which can be stated and understood without knowing anything about T, but which is difficult (or impossible, even better) to solve without T, and easier (or almost-trivial, even better) to solve with the help of T. What should be avoided is an example where T is "superimposed", e.g. when T is a model for some physical phenomenon, because there is always something arbitrary about the choice of a specific model. </p>
<p>A classical example is Galois theory for solving polynomial equations. </p>
<p>Any examples for homological algebra ? For Fourier analysis ? For category theory ?</p>
| Ilya Grigoriev | 2,467 | <p>This is not an answer as much as a request for answers. Here are some topics
for which I don't know an example of this sort, but would really like to see
one (and nobody seems to have done them yet). Feel free to edit this post if
good examples appear for some of these topics, or if you have a topic to add to
the list. They are in no particular order, </p>
<ul>
<li><p>Symplectic geometry (I'd really like to know...)</p></li>
<li><p>Algebraic geometry (there should be a ton of stuff here, it's so diverse)</p></li>
<li><p>Category theory</p></li>
<li><p>Homological algebra (is there an example simpler than some monstrous calculation?)</p></li>
<li><p>Group theory (so far, all that comes to my mind is that it applies to Galois theory,
which in turn solves equations. But I think there should be a ton of simpler and nicer applications, shouldn't there?)</p></li>
</ul>
|
942,738 | <p>First of all I must state that I am not a mathematician, so please correct me if I use wrong terminology.</p>
<p>I am building a web application which needs to calculate the rating for each entity based on both the quantity and score of the reviews for that entity.</p>
<p>In other words, I don't want to just calculate the rating based on the average score as that would make an entity with one hundred 9 score (review score can be from 0 to 10) reviews rate lower than an entity with only one 9.5 score review.</p>
<p>I need an algorithm to calculate rating and add rating "weight" to the final rating based on how many reviews the entity has, so that for instance in the above example the entity with 100 9 score reviews would get a rating that is higher than the entity with only one 9.5 score review. In other words, the final entity rating score will be based on the "relationship" between quality and quantity.</p>
<p>There is another important thing to note: an entity can not have a rating higher than 10, so the rating "weight" added by the quantity can not be linear.</p>
<p>In the algorithm we can use any data about the reviews/rating, that is individual review score, total number of reviews, sum of all reviews, number of good reviews (score 8 or higher) so far, etc, in each iteration of the rating calculation process.</p>
<p>Any kind of help or info regarding this would be appreciated.
Thank you.</p>
| Ross Millikan | 1,827 | <p>You can do all the things you speculate about. You need to decide how to weight them. You say you think 100 reviews of 9 are better than one review of 9.5, but how about 10 reviews of 9? Five? How about 90 reviews of 9 and 10 of 2-how does that compare to one review of 9.5? </p>
<p>One simple thing to do is give points for lots of reviews: no points for no reviews, 1 point for 1 review, 2 points for 2-5 reviews, and so on. You will have to scale it to your population. I am guessing that small numbers of reviews are common, which is why I started with small brackets, but you need to look at your data. Now add these points to the average score, and you have a total in the range 0 to 20. </p>
<p>Another approach is to find the average review-say it is a 7. Now just count the total number of points and subtract 7 times the number of reviews. Your single review of 9.5 is then 2.5 points above average for the number of reviews it has gotten. The 100 reviews of 9 are 200 points better, the 90 reviews of 9 and 10 reviews of 2 are $90 \cdot 2 +10 \cdot (-5)=130$ points better than average, etc.</p>
|
4,009,117 | <p>Suppose <span class="math-container">$X$</span>, <span class="math-container">$Y$</span>, and <span class="math-container">$Z$</span> are exhaustive events with <span class="math-container">$Y$</span> and <span class="math-container">$Z$</span> being mutually exclusive. Can <span class="math-container">$Y$</span> and <span class="math-container">$Z$</span> be treated as a partition? Also, as a follow-up, can the law of total probability be applied to <span class="math-container">$X$</span> this way: <span class="math-container">$P(^cX) = P(^cX \cap Y) + P(^cX\cap Z)$</span>, where <span class="math-container">$^cX$</span> is the complement of <span class="math-container">$X$</span>?</p>
| Dylan C. Beck | 390,180 | <p>By the familiar definition (i.e., the "Wikipedia" definition), for a real-valued function <span class="math-container">$f(x),$</span> we have that <span class="math-container">$\lim_{x \to \infty} f(x)$</span> is a real number <span class="math-container">$L$</span> (if it exists) such that for every real number <span class="math-container">$\varepsilon > 0,$</span> there exists a real number <span class="math-container">$M$</span> such that for all <span class="math-container">$x > M,$</span> we have that <span class="math-container">$|f(x) - L| < \varepsilon.$</span></p>
<p>Like you have mentioned, if <span class="math-container">$x$</span> is not an integer multiple of <span class="math-container">$\pi,$</span> we have that <span class="math-container">$f(x) = \frac{\sin x}{\sin x} = 1$</span>; however, for any integer <span class="math-container">$n,$</span> we have that <span class="math-container">$f(n \pi)$</span> is not defined. What can you conclude?</p>
|
3,634,490 | <p>Let <span class="math-container">$N \geq 1$</span> be an integer.</p>
<p>Let <span class="math-container">$X$</span> be a standard <span class="math-container">$\mathbb{R}^N$</span> Gaussian vector (all components are <span class="math-container">$\mathcal{N}(0, 1)$</span> and i. i. d.).</p>
<p>Let <span class="math-container">$A \in \mathcal{M}_N(\mathbb{R})$</span> be a deterministic matrix.</p>
<p>Thus, <span class="math-container">$AX$</span> is a Gaussian vector.</p>
<p>Let <span class="math-container">$\epsilon \in (0, 1)$</span>.</p>
<p>I would like to simulate Gaussian <em>vectors</em> of the form <span class="math-container">$AX$</span> that <em>mostly have a smaller infinity norm</em> than <span class="math-container">$\epsilon$</span>, with only controlling the <em>infinity norm of the matrix</em> <span class="math-container">$A$</span>.</p>
<p>So the question is: how to limit <span class="math-container">$||A||_{\infty} = \underset{i, j}{\max} |A_{i, j}|$</span> so that with high confidence / probability (let's say 95%), I get <span class="math-container">$||AX||_{\infty} \leq \epsilon$</span> ?</p>
<p>I am looking for an answer like: <strong>take <span class="math-container">$A$</span> such that <span class="math-container">$||A||_{\infty} \leq f(\epsilon)$</span></strong>. The answer is quite easy in the <span class="math-container">$1$</span>-dimensional case, I would like to generalize it, but haven't found any clear theorem addressing that.</p>
<p>Thanks a lot for your help!</p>
| Leander Tilsted Kristensen | 631,468 | <p>I doubt, that there is an exact formula to do this (that is if we want exactly 95% confidence and not just at least 95% confidence). For instance consider the two matrices
<span class="math-container">$$A=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \quad , \quad B=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$</span>
These matrices satisfy <span class="math-container">$||A||_\infty = 1 =||B||_\infty$</span>, but clearly <span class="math-container">$||Ax||_{\infty} \leq ||Bx||_\infty$</span>, thus matrices of the form <span class="math-container">$B$</span> would give larger confidence intervals.</p>
<p>The alternative is to give an upper bound for <span class="math-container">$||AX||_\infty$</span>, and then establish a rule that gives the desired result with at least 95% confidence. One possibility is to use that</p>
<p><span class="math-container">$$||AX||_\infty^2 \leq ||AX||_2^2 \leq n^2||A||_\infty^2||X||_2^2$$</span>
so we can say that
<span class="math-container">$$\mathbb{P}(||AX||_\infty < \epsilon) \geq \mathbb{P}(n^2||A||_\infty^2||X||_2^2 < \epsilon^2) = \mathbb{P}(||X||_2^2 < \frac{\epsilon^2}{n^2||A||_\infty^2})$$</span>
But since <span class="math-container">$||X||_2^2$</span> is a sum of squared <span class="math-container">$i.i.d.$</span> <span class="math-container">$N(0,1)$</span> variables it has a <span class="math-container">$\chi^2$</span> distribution with <span class="math-container">$n$</span> degrees of freedom, and thus in order to get at least 95% confidence, we could choose <span class="math-container">$\frac{\epsilon}{n^2||A||_\infty^2}$</span> to be the <span class="math-container">$0.95$</span>'th quantile of the <span class="math-container">$\chi^2(n)$</span> distribution. That is we require <span class="math-container">$||A||_\infty^2=\frac{\epsilon^2}{n^2 q}$</span>, where <span class="math-container">$q$</span> is the said quantile so you could choose your function <span class="math-container">$f$</span> as
<span class="math-container">$$f(\epsilon)= \frac{\epsilon}{n \sqrt{q}}$$</span></p>
<p>Another possibly better bound would be <span class="math-container">$||AX||_\infty \leq n ||A||_\infty ||X||_\infty$</span>, but this requires that we know the distribution of <span class="math-container">$||X||_\infty$</span>, which is a bit more complicated but possible.</p>
|
2,683,731 | <p>If a relation is not symmetric shall we say that it is anti symmetric?
In a quiz show's preliminary round i had this question looks easy but i couldn't answer.can anyone answer?</p>
| Levent | 261,831 | <p>Symmetric and anti-symmetric relations are not complements of each other. Consider $\{(a,b),(b,a),(a,c)\}$ over $\{a,b,c\}$. It is neither symmetric nor anti-symmetric.</p>
|
22,662 | <p>Suppose $V_1$ and $V_2$ are two $(g,K)$ modules of some reductive group $G$ with maximal compact $K$. Let $P$ be the minimal parabolic of $G$, $U$ its unipotent part, and $u$ its Lie algebra. Suppose the quotients $V_1/uV_1$ and $V_2/uV_2$ are isomorphic as modules for the Levi component of $P$, then what else do we need to know to conclude that $V_1$ and $V_2$ are isomorphic as $G$ modules?</p>
<p>edit:Thanks for Kevin and Emerton's comments and sorry for the confusion about the base field. Here I'm assuming REAL reductive group. </p>
| Kevin Buzzard | 1,384 | <p>EDIT: I assumed the OP was asking about reductive groups over non-arch local fields. Emerton raises the possibility that the question is actually about groups over R or C, and he's probably right. So the answer below is probably irrelevant.</p>
<p>Do correct me if I'm wrong; I'm not an expert. But I thought that if $V$ was any supercuspidal representation of, say, $GL_2(\mathbf{Q}_p)$, then $V/uV=0$. So in fact you know very little about $V$ if you only know $V/uV$. Can't $V$ basically be recovered from $V/uV$ in the $GL_2$ case when it's principal series or Steinberg, and in the supercuspidal case you have nothing? Actually, even in the Steinberg case you might have trouble distinguishing $V$ from a 1-dimensional representation...</p>
|
1,580,796 | <p>In a game players take turns saying up to 3 numbers (starting at 1 and working their way up) and who every says 21 is eliminated. So we may have a situation like the following for 3 players:</p>
<blockquote>
<p>Player 1: 1,2</p>
<p>Player 2: 3,4,5</p>
<p>Player 3: 6</p>
<p>Player 1: 7,8,9</p>
<p>Player 2:10,11</p>
<p>Player 3:13,14,15</p>
<p>Player 1: 16,17,18</p>
<p>Player 2: 19,20</p>
</blockquote>
<p>In which case Player <span class="math-container">$3$</span> would have to say <span class="math-container">$21$</span> and thus would be eliminated from the game. Player <span class="math-container">$1$</span> and <span class="math-container">$2$</span> would then face each other.</p>
<p>In the case of a two player game, the person who goes second can always win by ensuring the last number they say is a multiple of <span class="math-container">$4$</span>.</p>
<p>Let us say we are in an <span class="math-container">$N$</span> player game, and that I am the <span class="math-container">$i^{\text{th}}$</span> player to take my turn. Is there any strategy that I can take to make sure I will stay in the game? For example in the simple case of a <span class="math-container">$2$</span> player game the strategy would be 'try to end on a multiple of <span class="math-container">$4$</span> and then stay on multiples of <span class="math-container">$4$</span>'.</p>
| Antony | 517,240 | <p>You can use a strategy but only by the means on where you stand in the order of people, there is complex maths involved in probabilities but if numbers chosen ($1$-$3$) by each player are randomised then certain positions have a much lower chance of being forced into saying “$20+1$”</p>
<p>This was figured out using a simulation with that all players do not want to be landed on $21$ and therefore I used the assumption of if a player finishes on $17$, the next player will always bust the succeeding player by saying $18$, $19$, $20$ </p>
<p>Let $1^{st}$ person=$A$, $2^{nd}$=$B$, $3^{rd}$=$C$, $4^{th}$=...
Let first letter be most likely to say $21$ and last be least likely (preferable position)</p>
<p>Number Of Total Players</p>
<p>2=B,A</p>
<p>3=B,A,C</p>
<p>4=C,D,B,A</p>
<p>5=A,B,E,C,D</p>
<p>6=E,F,D,A,C,B</p>
<p>7=D,E,C,F,B,G,A</p>
<p>8=C,D,B,E,A,F,G,H</p>
<p>9=B,C,A,D,I,E,F,H,G</p>
<p>10=A,B,J,C,I,D,E,H,F,G</p>
<p>The more people there are, the more significant the benefits are of choosing a good starting position I.e. with $10$ people if you stand in position G (7th place) you have a $0.01\%$ chance of being chosen in comparison to the worst position A (1st place) with a chance of $31.88\%$</p>
<p>I don’t believe there is much of a pattern other than positions G and H are consistently good places to stand.</p>
<p>[Edit]: After more research as a basic strategy let n be number of people, c be current number before you, and b be the number you go bust/lose (I.e. 21)
Then if c+2(n-1)>b-3 then choose 3 numbers to say as this will likely cause a bust before it returns to you</p>
|
1,258,414 | <p>I want to prove that $$c = \{ (x_n)_{n\geq 0} \mid x_n \in \Bbb C \text{ and the sequence converges} \}$$ with the norm $$ \left\|(x_n)_{n\geq 0}\right\|_{\infty} =\sup_{n\geq 0} |x_n|$$ is a Banach space. I have done some work, but I am having trouble concluding.</p>
<p>So far: let $(\xi_n)_{n\geq 0} = \left( (x_k^{(n)})_{k\geq 0} \right)_{n\geq 0}$ be a $\|\cdot\|_{\infty}$-Cauchy sequence. Let $\epsilon > 0$, there exists $n_0 \in \Bbb N$ such that: $$\begin{align} \| \xi_n - \xi_m \|_{\infty} &< \epsilon, \quad \forall\, n,m > n_0 \\ \sup_{k \geq 0} |x_k^{(n)}-x_k^{(m)}| &< \epsilon, \quad \forall\,n,m > n_0 \\ |x_k^{(n)} - x_k^{(m)}| &< \epsilon, \quad \forall\, k\geq 0, \quad \forall\,n,m > n_0 \end{align}$$ So fixed $k$, $(x_k^{(n)})_{n \geq 0}$ is a $|\cdot |$-Cauchy sequence, and since $\Bbb C$ is Banach, there exists a limit $\lim_{n\to \infty} x_k^{(n)} =: x_k$. Then define $\xi = (x_k)_{k \geq 0}$. </p>
<p>Now I understand I have two things to do:</p>
<ul>
<li><p>prove that $\xi_n \stackrel{\|\cdot \|_{\infty}}{\longrightarrow} \xi $: we proceed as before. Let $\epsilon > 0$. Now, there is $n_0 \in \Bbb N$ such that: $$\begin{align} \| \xi_n - \xi_m \|_{\infty} &< \epsilon, \quad \forall\, n,m > n_0 \\ \sup_{k \geq 0} |x_k^{(n)}-x_k^{(m)}| &< \epsilon, \quad \forall\,n,m > n_0 \\ |x_k^{(n)} - x_k^{(m)}| &< \epsilon, \quad \forall\, k\geq 0, \quad \forall\,n,m > n_0 \\ \lim_{m \to \infty} |x_k^{(n)} - x_k^{(m)}| &\leq \epsilon, \quad \forall\,k\geq 0, \quad \forall\,n >n_0 \\ |x_k^{(n)} - x_k| &\leq \epsilon, \quad \forall\,k\geq 0 \quad \forall\, n>n_0 \\ \sup_{k\geq 0} |x_k^{(n)} - x_k| &\leq \epsilon,\quad \forall\, n > n_0 \\ \| \xi_n - \xi\|_{\infty} &\leq\epsilon, \quad \forall\,n>n_0, \end{align}$$ so ok.</p></li>
<li><p>prove that $\xi \in c$. I'm not sure of how to do this. I think I must use that every $(x_k^{(n)})_{k \geq 0}$ converge, because I don't seem to have used this yet. </p></li>
</ul>
<blockquote>
<p>How can I prove that $\xi \in c $?</p>
</blockquote>
<p>Thanks.</p>
| shalop | 224,467 | <p>Here is a sketch. I will use all of the same notations you are using (You should definitely check it for mistakes)</p>
<p>For each $n$, then sequence $\xi_n=(x^{(n)}_k)_{k\geq 0}$ is a convergent sequence (because it is in $c$), so let us say that $\xi_n=(x^{(n)}_k)_{k\geq 0}$ converges to $u_n \in \mathbb{C}$.</p>
<p>You can check that $|u_m-u_n|\leq ||\xi_m-\xi_n||_{\infty}$ for all $m,n \in \mathbb{N}$, so since $(\xi_n)_{n\geq 0}$ is a Cauchy sequence in $c$, it follows that $(u_n)_{n \geq 0}$ is a Cauchy sequence in $\mathbb{C}$, so there exists $u \in \mathbb{C}$ such that $\lim_{n\to \infty}u_n=u$.</p>
<p>I claim that the sequence $\xi=(x_k)_{k\geq 0}$ converges to $u$. The reason is that for any $n$ we have $$|x_k-u| \leq |x_k-x^{(n)}_k|+|x^{(n)}_k-u_n|+|u_n-u| $$ $$\;\;\;\;\;\;\;\;\;\;\;\;\;\leq ||\xi-\xi_n||_{\infty}+|x^{(n)}_k-u_n|+|u_n-u|$$Let $\epsilon>0$. Choose $n_0 \in \mathbb{N}$ such that $|u_{n_0}-u|<\frac{\epsilon}{3}$ and $||\xi-\xi_{n_0}||<\frac{\epsilon}{3}$. Choose $K \in \mathbb{N}$ such that $|x_k^{(n_0)}-u_{n_0}|<\frac{\epsilon}{3}$ whenever $k \geq K$. Then for $k \geq K$, all three terms on the right hand side of the above equation are less than $\frac{\epsilon}{3}$, and thus $|x_k-u|<\epsilon$.</p>
<p>Thus $x_k \to u$. Therefore the sequence $\xi=(x_k)_{k\geq 0}$ converges, so $\xi \in c$.</p>
<p><strong>Edit:</strong> (Alternative solution) Actually, I just realized that we don't really care about the limit of $\xi$, we only care that it converges. So it suffices to prove that it is Cauchy. To do this, you can just say that for any $n$ $$|x_k-x_l| \leq |x_k-x^{(n)}_k|+|x^{(n)}_k-x^{(n)}_l|+|x^{(n)}_l-x_l|$$ $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\leq ||\xi-\xi_n||_{\infty}+|x^{(n)}_k-x^{(n)}_l|+||\xi-\xi_n||_{\infty}$$ and then argue as before.</p>
|
215,077 | <p>This question is partly connected with the following <a href="https://mathoverflow.net/questions/214931/connection-between-stallings-end-theorem-and-seifert-van-kampen-theorem">Connection between Stalling's end theorem and Seifert-van Kampen Theorem</a>. </p>
<p>By Stalling's Theorem a group with more than one end splits over a finite subgroup, i.e. can be written as an HNN-Extension or a free product with amalgamation (over a finite subgroup).</p>
<p>However when I was working through the proofs of Dunwoody, Dunwoody & Krön and Krön, which all use Bass-Serre Theory, I was wondering if there is any way to detect if we are dealing with an HNN-Extension or a free product with amalgamation. Primarily I was thinking of some property of the Cayley graph or the action on it by the group in question. For example one could consider the induced action on the end space of the Cayley graph.</p>
<p>In my opinion this is a natural question and I would be grateful for any comment or references regarding this.</p>
<p>EDIT: I am looking for properties of the group, its Cayley graph, the action of the group on its Cayley graph and/or its end space etc., which help to distinguish if the group, supposed to split by Stalling's Theorem, is indeed an HNN-Extension <em>or</em> a free product with amalgamation.</p>
<p>For some statements one probably has to make additional assumptions such as the group is finitely generated or finitely presented etc. </p>
| Lee Mosher | 20,787 | <p>While there's no single answer to this question, there are a lot of partial results.</p>
<p>To start with, there is the still unresolved <a href="http://eprints.soton.ac.uk/46386/1/KrophollerConjecture.pdf" rel="nofollow noreferrer">Kropholler-Roller conjecture</a>, although various special cases are known, see for example <a href="https://mathoverflow.net/questions/168648/krophollers-conjecture-and-3-manifolds">this math overflow question</a>.</p>
<p>Also, there are various quasi-isometric rigidity theorems which say that if such-and-such a finitely generated group $G$ splits as an HNN-extension of a free product with amalgamations then any group quasi-isometric to $G$ also splits. Stallings theorem itself has this format, because the number of ends of a finitely generated group is a quasi-isometry invariant. For further examples of such theorems see Papasoglu's paper "Group splittings and asymptotic topology" J. Reine Angew. Math. 602 (2007), 1–16, and my paper with Whyte and Sageev, "Quasi-actions on trees II: Finite depth Bass-Serre trees", Mem. Amer. Math. Soc. 214 (2011), no. 1008</p>
<p>Of particular importance is the vast literature on JSJ decompositions, which can be thought of as the theory of splittings of one-ended groups over two-ended groups. Bowditch's paper "Cut points and canonical splittings of hyperbolic groups", Acta Math. 180 (1998), no. 2, 145–186, characterizes which hyperbolic groups $G$ have a nontrivial splitting over a two-ended group, in terms of the topology of the Gromov boundary of $G$, which gives an example of a quasi-isometric rigidity theorem in the context of JSJ decompositions.</p>
|
2,229,532 | <p>I have a circle which has a triangle inscribed in it.</p>
<p>The circle radius R = 4</p>
<p>The triangle ABC vertices divide circle into 3 arcs in 1:2:3 ratio</p>
<p>Find the perimeter and area of triangle.</p>
<p>Can you guys help me with this one? </p>
| Ángel Mario Gallegos | 67,622 | <p><strong>Hint:</strong></p>
<p>The ratio between the angles of the triangle is $1:2:3$, so the angles are $30^{\circ}$, $60^{\circ}$ and $90^{\circ}$. Now, the hypotenuse is also a diameter of the circle (why?) then the sides of the triangle are $4$, $4\cos 30^{\circ}$ and $4\sin 30^{\circ}$.</p>
|
2,229,532 | <p>I have a circle which has a triangle inscribed in it.</p>
<p>The circle radius R = 4</p>
<p>The triangle ABC vertices divide circle into 3 arcs in 1:2:3 ratio</p>
<p>Find the perimeter and area of triangle.</p>
<p>Can you guys help me with this one? </p>
| fz1795 | 435,765 | <p>Yes Angelo Mario. if you draw the triangle with these angles you could find the answer.(The longest side of the triangle goes through the center of the circle.)
the answer will be 13.856 </p>
|
2,229,532 | <p>I have a circle which has a triangle inscribed in it.</p>
<p>The circle radius R = 4</p>
<p>The triangle ABC vertices divide circle into 3 arcs in 1:2:3 ratio</p>
<p>Find the perimeter and area of triangle.</p>
<p>Can you guys help me with this one? </p>
| Alex | 293,781 | <p>The condition on the three arcs' $1:2:3$ ratio, together with the fact that the arcs add up to $360^\circ$, means that the arcs are $60^\circ$, $120^\circ$, and $180^\circ$. </p>
<p>The angles in triangle $ABC$ are inscribed angles, each being exactly half of the corresponding arc; so the angles are $30^\circ$, $60^\circ$, and $90^\circ$, i.e. $ABC$ is a right triangle. The hypotenuse of the $30$-$60$-$90$ triangle $ABC$ is the diameter of the given circle; so the hypotenuse is 8, the short leg is 4, and the long leg is $4\sqrt3$. </p>
<p>From this we find the perimeter, $12+4\sqrt3$, as well as the area $8\sqrt3$ of triangle $ABC$.</p>
|
1,401,661 | <blockquote>
<p>Dice are cubes with pips (small dots) on their sides, representing
numbers 1 through 6. Two dice are considered the same if they can be
rotated and placed in such a way that they present matching numbers on
the top, bottom, left, right, front, and back sides.</p>
<p>Below is an example of two dice that can be rotated to show that they
are the same if the 2-pip and 4-pip sides are opposite and the 3-pip
and 5-pip sides are also opposite.
<a href="https://www.dropbox.com/s/6q56njm11hu3f36/Screenshot%202015-08-18%2012.02.11.png?dl=0" rel="nofollow">https://www.dropbox.com/s/6q56njm11hu3f36/Screenshot%202015-08-18%2012.02.11.png?dl=0</a></p>
<p>How many different dice exist? That is, how many ways can you make
distinct dice that cannot be rotated to show they are the same? Note:
This problem does not involve rolling the dice or the probability of
roll outcomes.</p>
</blockquote>
<p>I'm having trouble understanding exactly what is being asked in this question. I understand that I have to find how many different ways the dice can be placed to show that they are the same, but saying they cannot be rotated confuses me.</p>
<p>Could somebody make an attempt at rewording this? Or walking me through how to solve this?</p>
| Maverick Meerkat | 342,736 | <p>In the case there are <strong>no limitation</strong> on the way we sort the different values (i.e. it's not a normal dice which opposite sides sum up to 7), and there's also <strong>no rotation</strong> allowed (the cube is fixed) - then the answer would be 6! = <strong>720</strong> options.</p>
<p>(n-Factorial is the number of different unique combinations you can arrange all the numbers 1,...,n in a series)</p>
<p>Now let's consider <strong>1-axis rotation</strong> is allowed.
You can choose 6 values for the upside face, then you would be left with 5 values for the downside face. Sorting out the 4 faces of the sides would normally be 4! but since you're allowing rotation, this needs to be divided by 4, since each no-rotation combination is equivalent to 4 with-rotation combinations (1234 = 2341 = 3412 = 4123).
So the answer would be 6!/4 = 6*5*4!/4 = <strong>180</strong>. </p>
<p>(Note that it does matter here which value you choose for the upper and for the lower, since if you turn them around, the dices will not be the same - but will switch between clockwise, or counter-clockwise, is. a dice with 6 on the top, 5 on the bottom, and 1234 on the sides, is different than a dice with 5 on the top, 6 on the bottom and 1234 on the sides)</p>
<p>Now let's consider <strong>2-axis rotation</strong> is allowed.
We would need to take our last answer, and then divide it to take into account that now we're allowed to "play" even more with the dice. We now find out that each "unique" combination in the 1-axis version is equivalent to 6 different combinations in the 2-axis version. 6,1234,5 = 5,4321,6 = 1,5264,3 = 3,4625,1 = 2,5361,4 = 4,1635,2 (the notation I'm using is upper,4-sides,lower).</p>
<p>So our answer will be 6!/4/6 = 6*5*(4!/4)/6 = <strong>30</strong>.</p>
<p>Now what would be the answer in the case we limit ourselves to <strong>normal dices</strong>, i.e. dices that follow the pattern that each opposite sides sum must be equal to 7? </p>
<p>This means that we can now only freely choose 3 faces of the dice, corresponding to its 3 dimensions space, since choosing one face number to be equal to a value, immediately sets the value of the opposite side. </p>
<p>So for <strong>no rotation</strong> - we could choose 6 values for the upper side, but then immediately the lower side is set, so you can choose 4 value for a another face, but then the opposite face is again set, and finally you can choose 2 values for the next face, which will immediately set the final face. I.e. the answer is 6*4*2 = <strong>48</strong>. </p>
<p>For <strong>1-axis rotation</strong> - we saw that we need to divide by 4: 48/4 = <strong>12</strong>. </p>
<p>For <strong>2-axis rotation</strong> - we need to further divide by 6: 12/6 = <strong>2</strong>. </p>
<p>And that actually corresponds with the two normal 6-sided dices that exist in the world, called: clockwise (Also - left handed), and counter-clockwise (Also - right handed). </p>
|
1,401,661 | <blockquote>
<p>Dice are cubes with pips (small dots) on their sides, representing
numbers 1 through 6. Two dice are considered the same if they can be
rotated and placed in such a way that they present matching numbers on
the top, bottom, left, right, front, and back sides.</p>
<p>Below is an example of two dice that can be rotated to show that they
are the same if the 2-pip and 4-pip sides are opposite and the 3-pip
and 5-pip sides are also opposite.
<a href="https://www.dropbox.com/s/6q56njm11hu3f36/Screenshot%202015-08-18%2012.02.11.png?dl=0" rel="nofollow">https://www.dropbox.com/s/6q56njm11hu3f36/Screenshot%202015-08-18%2012.02.11.png?dl=0</a></p>
<p>How many different dice exist? That is, how many ways can you make
distinct dice that cannot be rotated to show they are the same? Note:
This problem does not involve rolling the dice or the probability of
roll outcomes.</p>
</blockquote>
<p>I'm having trouble understanding exactly what is being asked in this question. I understand that I have to find how many different ways the dice can be placed to show that they are the same, but saying they cannot be rotated confuses me.</p>
<p>Could somebody make an attempt at rewording this? Or walking me through how to solve this?</p>
| Eric Padmore | 784,379 | <p>Assuming that the question refers to "normal" die, where the opposite sides must total seven, (why would you make any other kind?) I believe this is a spatial problem rather than a mathematical one.</p>
<p>Lets begin with the 4 side facing up, it is symmetrical around both the horizontal and vertical axis so it doesn't matter how you set it down. The addition rule means that the 3 side is on the bottom but we can't see that so we ignore it for now. </p>
<p>Next we arbitrarily choose to place the 1 side on the vertical face to the right of the 4. It is also symmetrical so this is trivial. The addition rule dictates that the 6 side must then be placed on the vertical face to the left of the 4. At this point there are two options, the dots on the 6 form two parallel tracks of three which can either be turned as if they were a continuation of the dots on the 4 face, or as if they were 'perpendicular' to the 4 face.</p>
<p>We then place the 5 on the front face of the die as we have set it down. It is symmetrical so it doesn't matter how it goes on. This forces the 2 onto the back face of the die in one of only two possible configurations with the dots placed on the face top-right bottom-left or, top-left bottom-right.</p>
<p>If we now flip the die over and place the 3 on top, we see we have the same options as the 2, only with a dot in the center.</p>
<p>So each of the choices for the 6, 3 and 2 faces gives us eight ways (2x2x2)to build our die. The arbitrary choice to place the 1 on the right and the 5 on the front is akin to chirality so that there are mirror images of the eight choices given above. Thus there are a total of 16 ways you can build a 'normal' die.</p>
|
200,414 | <p>Some people may carelessly say that you need calculus to find such a thing as a local maximum of $f(x) = x^3 - 20x^2 + 96x$. Certainly calculus is <em>sufficient</em>, but whether it's <em>necessary</em> is another question.</p>
<p>There's a global maximum if you restrict the domain to $[0,8]$, and $f$ is $0$ at the endpoints and positive between them. Say the maximum is at $x_0$. One would have
$$
\frac{f(x)-f(x_0)}{x-x_0}\begin{cases} >0 & \text{if }x<x_0, \\ <0 & \text{if }x>x_0. \end{cases}
$$
This difference quotient is undefined when $x=x_0$, but mere algebra tells us that the numerator factors and we get
$$
\frac{(x-x_0)g(x)}{x-x_0} = g(x)
$$
where $g(x)$ is a polynomial whose coefficients depend on $x_0$. Then of course one seeks its zeros since it should change signs at $x_0$.</p>
<p>Have we tacitly used the intermediate value theorem, or the extreme value theorem? To what extent can those be avoided? Must one say that <em>if</em> there is a maximum point, <em>then</em> it is at a zero of $g(x)$? And can we say that without the intermediate value theorem? (At least in the case of this function, I think we stop short of needing the so-called fundamental theorem of algebra to tell us some zeros of $g$ exist!)</p>
| Community | -1 | <p>There is a kind of algebraic structure called a <em>real closed field</em>. It's a structure with constants $0$, $1$, and has operations $+$ and $\cdot$ and a relation $<$.</p>
<p>The theory of real closed fields is interesting, because any (first-order) logical statement you can make using just $0,1,+,\cdot,<$ is true in a single real closed field if and only if it is true in all real closed fields. Similarly, the ones that are true can be <em>proven</em> using just first-order logic and the axioms of real closed fields -- i.e. there is a purely algebraic proof of it.</p>
<p>For example, for any particular algebraic function, the statement that a particular value is a local maximum can be expressed in first-order logic. Thus, if you can find a local maximum with calculus, you can find it with pure algebra too.</p>
<p>Of course, the axioms of real closed field typically include some variant of the intermediate value theorem: e.g.</p>
<blockquote>
<p>every polynomial of odd degree has a root</p>
</blockquote>
<p>This is the IVT, since an odd polynomial must be positive somewhere and negative somewhere. Of course, this can also be seen as a variant of the fundamental theorem of algebra too. Ultimately, in the purely algebraic setting, there isn't much difference between the two.</p>
|
17,398 | <p>I am teaching my daughter, who is currently about <span class="math-container">$46$</span> months old, additions. She is very curious and asks a lot of <em>good</em> questions. For example, when I told her that <span class="math-container">$2+6=8$</span> and <span class="math-container">$4+4=8$</span>, she asked me the following question:</p>
<blockquote>
<p>Why are they the same?</p>
</blockquote>
<p>Surely I know the logical answer this question using <a href="https://en.wikipedia.org/wiki/Peano_axioms" rel="noreferrer">Peano Axioms</a> and the definition of natural numbers. But she has not learnt the Peano Axiom at this age.</p>
<p>So how to answer my three-year-old daughter's question that </p>
<blockquote>
<p>Why <span class="math-container">$2+6$</span> is the same as <span class="math-container">$4+4$</span>?</p>
</blockquote>
| Jamie Watts | 13,034 | <p>Start with two equations that are the same in every way, like this:</p>
<p>2 + 2 = 4 and 2 + 2 = 4</p>
<p>Now, add "4" more to each of the two equations, one at a time, so you'll get 8 in each equation. But do it different ways in each of the two equations.</p>
<p>1.) Add all of the "4" more to the second "2" in the first equation: 2 + 6 = 8 (and 2 + 2 = 4)</p>
<p>2.) Now split up the "4" you're adding to the second equation into two "2's" and add one "2" to each "2" in the second equation: (2 + 6 = 8 and) 4 + 4 = 8</p>
<p>I think this is similar to the first answer someone gave above, but without using fingers. I explained it like this (a similar question - not this identical question) years ago to my five year old son and again to my four year old niece.</p>
|
933,515 | <p>Let $T$ be a bounded operator acting on a Banach space $X$. The point spectrum $\sigma_p(T)$ is of $T$ is defined to be $$\sigma_p(T):=\{\lambda\in\mathbb C~|~T-\lambda\text{ has nonempty kernel}\}$$ and the residual spectrum $\sigma_r(T)$ is defined to be $$\sigma_r(T):=\{\lambda\in\mathbb C~|~T-\lambda\text{ has empty kernel but Ran}(T-\lambda)\text{ is not even dense in }X\}.$$
For the continuous spectrum there seem to exist two non-equivalent definitions. For example, Dunford-Schwartz defines
$$\sigma_c(T):=\{\lambda\in\mathbb C~|~T-\lambda\text{ has empty kernel but Ran}(T-\lambda)\text{ is a dense proper subset of } X\}$$ whereas for example Kato or Schmüdgen define
\begin{align*}\sigma_c'(T)&:=\{\lambda\in\mathbb C~|~\text{Ran}(T-\lambda)\text{ is not a closed subset of } X\}\\
\sigma_r'(T)&:=\{\lambda\in\mathbb C~|~T-\lambda\text{ has a bounded inverse defined only on a proper subset of } X\}.
\end{align*}</p>
<p>Does anyone have insights on the difference of these definitions? What are the advantages? The former one gives a partition of the spectrum, so that's kind of nice, but why is the second definition (more?) useful?</p>
<hr>
<p><strong>Edit:</strong> I just created a table that categorizes parts of the spectrum in terms of the questions:</p>
<ul>
<li>Is $K_\lambda:=$Ker$(T-\lambda)$ zero or nonzero?</li>
<li>Is $R_\lambda:=$Ran$(T-\lambda)$ closed or not?</li>
<li>Is $R_\lambda:=$Ran$(T-\lambda)$ dense or not?</li>
</ul>
<p><img src="https://i.stack.imgur.com/PeFKz.png" alt="Table of spectral Decomposition"></p>
<p>According to this table the point spectrum can be subdivided into four subtypes and the residual spectrum can be subdivided into two subtypes. The alternative definitions would correspond to \begin{align*}\sigma_c'(T)&=\sigma_{c}(T)\cup\sigma_{r,1}(T)\cup\sigma_{p,2}(T)\cup\sigma_{p,3}(T)\\
\sigma_r'(T)&=\sigma_{r,1}(T)
\end{align*} </p>
<p>The approximate point spectrum and compression spectrum on the other hand are given by
\begin{align*}
\sigma_{ap}(T)&=\sigma_{c}(T)\cup\sigma_{r,1}(T)\cup\sigma_{p,1}(T)\cup\sigma_{p,2}(T)\cup\sigma_{p,3}(T)\cup\sigma_{p,4}(T)\\
\sigma_{com}(T)&=\sigma_{r,1}(T)\cup\sigma_{r,2}(T)\cup\sigma_{p,3}(T)\cup\sigma_{p,4}(T).
\end{align*} </p>
| Jonas Dahlbæk | 161,825 | <p>I think the property of having closed range is simply more useful in certain applications, so you want to single out the 'worst case scenario' by considering those $\lambda$ for which $(T-\lambda)$ fails to have closed range. Note for instance that in the Hilbert space setting, you have the very useful direct sum decomposition
$$
\mathcal H= \mathcal R(T-\lambda)\oplus \mathcal N (T^*-\bar \lambda)
$$
if and only if $\lambda\notin \sigma '_c(T)$.</p>
<p>On the other hand, the set $\sigma_c(T)$ enjoys the nice property that $(T-\lambda)$ has a densely defined inverse when $\lambda\in \sigma_c$. In a sense, $\sigma_c'(T)$ singles out a poorly-behaved subset of the spectrum, whereas $\sigma_c(T)$ singles out a well-behaved subset of $\sigma_c'(T)$.</p>
|
510,672 | <p>My prof taught us that during Gaussian Elimination, we can perform three elementary operations to transform the matrix:</p>
<p>1) Multiple both sides of a row by a non-zero constant
2) Add or subtract rows
3) Interchanging rows</p>
<p>In addition to those, why isn't removing zero rows an elementary operation? It doesn't affect the system in any way. Define zero rows to be a row with no leading variables.</p>
<p>For example isn't $\begin{bmatrix}a & b & k\\c & d & m\end{bmatrix} \rightarrow
\begin{bmatrix}a & b & k\\c & d & m\\0 & 0 & 0\end{bmatrix}$</p>
| Trevor Wilson | 39,378 | <p>Because we want the dimensions of the matrix to remain fixed. I can think of two reasons for this.</p>
<ol>
<li><p>So that performing an elementary row operation will correspond to multiplication on the left by an elementary matrix.</p></li>
<li><p>Changing the dimensions of the matrix by adding or removing zero rows does not help us solve the system, so why allow it? The (other) row operations do not change the size of the matrix, so let's keep things simple and fix it once and for all.</p></li>
</ol>
|
510,672 | <p>My prof taught us that during Gaussian Elimination, we can perform three elementary operations to transform the matrix:</p>
<p>1) Multiple both sides of a row by a non-zero constant
2) Add or subtract rows
3) Interchanging rows</p>
<p>In addition to those, why isn't removing zero rows an elementary operation? It doesn't affect the system in any way. Define zero rows to be a row with no leading variables.</p>
<p>For example isn't $\begin{bmatrix}a & b & k\\c & d & m\end{bmatrix} \rightarrow
\begin{bmatrix}a & b & k\\c & d & m\\0 & 0 & 0\end{bmatrix}$</p>
| Marc van Leeuwen | 18,880 | <p>Whether you want addition or removal of rows with all entries zero to be allowed or not depends on the precise application one has for using row operations. If row operations are intended to model the action of the general linear group by left multiplication, then addition or removal of rows should be forbidden because the mentioned action never does that. However if the purpose is to define a normal form for subspaces defined by systems of linear equations, then removal of rows is <em>needed</em> to obtain such a normal form (equations with all coefficients zero have no effect on the solution space, and there must be some means to bring equivalent systems with a different number of equations to a common normal form). The normal form would be row reduced echelon form without any zero rows. Addition of zero rows is then also needed to ensure one can get back from the normal form to any system equivalent to it. </p>
|
510,672 | <p>My prof taught us that during Gaussian Elimination, we can perform three elementary operations to transform the matrix:</p>
<p>1) Multiple both sides of a row by a non-zero constant
2) Add or subtract rows
3) Interchanging rows</p>
<p>In addition to those, why isn't removing zero rows an elementary operation? It doesn't affect the system in any way. Define zero rows to be a row with no leading variables.</p>
<p>For example isn't $\begin{bmatrix}a & b & k\\c & d & m\end{bmatrix} \rightarrow
\begin{bmatrix}a & b & k\\c & d & m\\0 & 0 & 0\end{bmatrix}$</p>
| Community | -1 | <p>Here is a slightly more long-range answer: a matrix corresponds to a linear operator $T:V\to W$ where $V$ and $W$ are vector spaces with some chosen bases. The elementary row operations (or elementary column operations) then correspond to changing the basis of $W$ or of $V$ to give an equivalent matrix: one which represents the same linear operator but with the bases changed around. Under this correspondence you can get all the possible matrices corresponding to the linear operator $T$ by doing elementary row and column operations.</p>
<p>Adding or removing a row of zeroes will not give you a matrix corresponding to the linear operator $T$.</p>
|
13,553 | <p>(This issue has been discussed before, specifically <a href="https://math.meta.stackexchange.com/questions/8485/how-civil-should-we-be/8493">here</a>. A different kind of censorship has been discussed <a href="https://math.meta.stackexchange.com/questions/8306/defacement-censorship-of-peoples-answers">here</a>.)</p>
<p>There are two reasons I am bringing this up again. Firstly, the previous discussion on this issue did not seem conclusive. In particular, the <a href="https://math.meta.stackexchange.com/a/8487/68107">top answer</a> from Mad Scientist states that this is "a matter of professionalism" and that "if the swear words are not necessary for the post, I see no problem in removing them." Conversely, the <a href="https://math.meta.stackexchange.com/a/8520/68107">second-to-top answer</a> (which I agree with much more) states that "I know this is supposed to be a professional environment, and that is precisely why we need to allow every form of expression possible."</p>
<p>Secondly, a moderator has recently edited <a href="https://math.stackexchange.com/a/775883/68107">this post</a> to remove a swear word. Perhaps this was a response to a flag. Instead of rolling back and starting a potential edit war, I am bringing it up here to solicit the broader community opinion.</p>
<p>With regards to this specific post, I am of the strong opinion that the swear word should not have been replaced, because the way it was used was effective and clear. The replacement--"horrible open access journals"--simply doesn't get the point across. I think this censorship should be reversed.</p>
<p>More generally, I disagree with removing swear words merely because they aren't necessary, and think they should only be removed if they <em>take away</em> from a post. I believe it is unprofessional and pointless to censor any form of effective communication. I feel we should tolerate swearing to the extent that it is used well, and I see no need to favor the small percentage of us who are offended by it.</p>
<p>What do others think?</p>
| user642796 | 8,348 | <p>Answering just to provide some of my thoughts (as the ♦-mod in question):</p>
<ol>
<li><p>The answer was flagged. I don't go out of my way (and I don't think anyone goes out of their way) to find the odd swear word on the site. They're easy enough to find, however, for anyone who wants to.</p></li>
<li><p>I did feel that the original word choice was unnecessary and was used as a slur against a particular group. Furthermore, I felt that the underlying idea could be expressed in a manner much less likely the elicit future flags. At the same time...</p></li>
<li><p>... in hindsight I do agree that "horrible" was a... err... horrible choice. There are much better adjectives out there to express one's low opinion (<em>e.g.</em>, wretched, contemptible, atrocious, abhorrent). At least some of which likely have an even more negative connotation that the original word.</p></li>
<li><p>As everyone likely knows, users are in general free to edit/suggest edits to posts. Were someone to edit the answer again it is fairly unlikely that I would ever know about it (until it is flagged, and even then only if I'm around to handle it). Heck, I haven't even checked that the answer hasn't already been altered, and am unlikely to revisit it any time soon. (Of course, the existence of <em>this thread</em> makes it much more likely that for a period of time users <em>will</em> be paying greater attention to <em>that</em> particular answer.)</p></li>
</ol>
|
4,014,144 | <p>Let <span class="math-container">$f(x) = \max(c_1^Tx, c_2^Tx, \dots, c_k^Tx)$</span>.
where <span class="math-container">$x, c_1, c_2, \dots, c_k \in \mathbb R^n$</span>.
What fast iterative methods are available for finding the (approximate) min of <span class="math-container">$f$</span> with the constraint <span class="math-container">$\lVert x \rVert_2 = 1$</span>?</p>
<p>Notes:</p>
<ol>
<li><span class="math-container">$f$</span> is convex and and non negative, that is, <span class="math-container">$c_i$</span> positively span <span class="math-container">$\mathbb R^n$</span></li>
<li>For my use case <span class="math-container">$k \gg n$</span> and (rougly) <span class="math-container">$3 \le n \le 50$</span> and <span class="math-container">$100 \le k \le 10000$</span>.
I tried <a href="https://en.wikipedia.org/wiki/Subgradient_method" rel="nofollow noreferrer">projected subgradients</a> (projected to the sphere) but it can be slow to converge and improving the initial guess doesn't seem to accelerate the method. From the literature, this seems to be equivalent to Riemannian manifold subgradient methods which use the exponential map. I haven't tried bundle methods yet but I am investigating them now. They seem like they might be too slow.</li>
</ol>
<p>I've also tried approximating <span class="math-container">$f$</span> with a smooth maximum <a href="https://en.wikipedia.org/wiki/LogSumExp" rel="nofollow noreferrer">LogSumExp</a> and doing gradient descent with projection. I also tried unconstrained gradient descent with a <a href="https://en.wikipedia.org/wiki/Penalty_method" rel="nofollow noreferrer">quadratic penalty</a>. The approximation is too inaccurate and the evaluation of exp is too slow for my use case. I'm not interested in <a href="https://en.wikipedia.org/wiki/Sequential_quadratic_programming" rel="nofollow noreferrer">SQP</a> because I suspect whatever method used to smooth (e.g. LogSumExp, hyperbolic) will be too costly/inaccurate to evaluate.</p>
<p>Edit:
I believe this problem is equivalent to a linear programming problem with quadratic <strong>equality</strong> constraints. I haven't been able to to find much literature on this type of problem. Perhaps I can use a <a href="https://en.wikipedia.org/wiki/Semidefinite_programming" rel="nofollow noreferrer">SDP</a> method but I'm not sure.</p>
| River Li | 584,414 | <p>We need to solve the following optimization problem (P<span class="math-container">$0$</span>):
<span class="math-container">\begin{align}
(\mathrm{P0})\qquad &\min_{x} \max\{c_1^\mathsf{T}x, \cdots, c_k^\mathsf{T}x\}\\
&\mathrm{s.t.}\quad \|x\|_2 = 1.
\end{align}</span></p>
<p>Assume that <span class="math-container">$\{c_1, c_2, \cdots, c_k\}$</span> satisfies
<span class="math-container">$\max\{c_1^\mathsf{T}x, \cdots, c_k^\mathsf{T}x\} > 0$</span> on <span class="math-container">$\mathbb{R}^n\backslash \{0\}$</span>.</p>
<p>(P<span class="math-container">$0$</span>) is equivalent to <span class="math-container">$\min_{x} \frac{\max\{c_1^\mathsf{T}x, \cdots, c_k^\mathsf{T}x\}}{\|x\|_2}$</span>
and <span class="math-container">$\max_{x} \frac{\|x\|_2}{\max\{c_1^\mathsf{T}x, \cdots, c_k^\mathsf{T}x\}}$</span> and
<span class="math-container">\begin{align}
&\max_{x} \|x\|_2^2\\
&\mathrm{s.t.}\quad \max\{c_1^\mathsf{T}x, \cdots, c_k^\mathsf{T}x\} = 1
\end{align}</span>
and
<span class="math-container">\begin{align}
(\mathrm{P1})\qquad &\max_{x} \|x\|_2^2\\
&\mathrm{s.t.}\quad c_i^\mathsf{T}x \le 1, \ i=1, 2, \cdots, k.
\end{align}</span>
(P<span class="math-container">$1$</span>) is a Quadratic Programming (QP) problem (<a href="https://en.wikipedia.org/wiki/Quadratic_programming" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Quadratic_programming</a>).</p>
<p>Some online and software Resources for QP are given in [1]. Some quadratic programming codes are given in [2].</p>
<p><em>References</em></p>
<p>[1] <a href="https://neos-guide.org/content/quadratic-programming" rel="nofollow noreferrer">https://neos-guide.org/content/quadratic-programming</a></p>
<p>[2] <a href="http://www.numerical.rl.ac.uk/people/nimg/qp/qp.html" rel="nofollow noreferrer">http://www.numerical.rl.ac.uk/people/nimg/qp/qp.html</a></p>
|
1,794 | <p>I have always loved the beauty of mathematics and physics. However I'm severely dyslexic and find it hard to keep numbers in my head, any more than 4 numbers at a time and they melt together and lose their meaning. (You know that feeling when you wake up after a lucid dream but as you think about it, it just evaporates? Well that's what it's like.) However, I'm pretty good at understanding concepts like Cantor Sets, mathematical logic, etc.</p>
<p>I have degrees in the philosophy of science and have studied epistemology, so I'm comfortable with abstract concepts BUT, I want to do some mathematics, I want to try algebra and trigonometry, to play with equations but I just don't know how to start.</p>
<p>Are there any books designed for the likes of me? Can someone who finds it hard to add two two-digit numbers together in his head, and even takes a second to decipher a 4-digit number on paper be taught to do advanced math? Is it like teaching a 1-legged man to run or a pig to fly?</p>
<p>I'm interested in hearing the opinions of people who love math, who can do math and maybe understand what a lot of my friends don't - namely why I at least want to try. Thanks.</p>
<hr>
<p>Thanks for your comments,</p>
<p>Let me rephrase my original question if it helps.</p>
<ol>
<li><p>Being dyslexic. I have a very poor working memory (like a computer with very little ram if you like) the processing power required to recognise the shape of numbers written down as characters (i.e. 1, 5, 9, 3, 2, 8, etc) and then understand what they represent requires all my processing power. That means if I want to calculate two numbers in my head, 7+5 for example it takes a second but I can do it. However, if I want to add 14+63 my brain will spin and crash. Ive not been able to move past what most 7 year olds can do in math class.</p></li>
<li><p>That said, in common with many dyslexics I’m very good at understanding complex and abstract theories and ideas (spatial reasoning). I have degrees (BA and MA form Durham) in subjects that can broadly be described as philosophy. I have been fascinated with subjects like mathematics and physics since I was a child and I have always read popular books on the subject. I can understand, for example, ‘set theory’ .. at a conceptual level, mathematic proof, .. at a conceptual level. </p></li>
</ol>
<p>What you have written is encouraging, but math - even at a basic level is a foreign land to me. I would love to do some sort of higher math (no, I don’t really know what that means). but I would love to try to understand equations. But can I learn to do algebra or trigonometry and beyond when I can’t recite the 6x table or add 34 and 58 in my head? If I can, and I think you are saying that I might just be able to, where do I start? </p>
| Joseph O'Rourke | 511 | <p>I look forward to the responses to this interesting question.
Instead of answering directly,
permit me to make an analogy to art.</p>
<p><a href="http://en.wikipedia.org/wiki/Frank_Stella" rel="nofollow noreferrer">Frank Stella</a> is a renowned artist whose influence and compositional talent cannot be questioned.
Nevertheless, I happen to know from reliable second-hand reports, that
he cannot "draw" in the sense that, say, Leonardo, or even a contemporary
cartoonist, can "draw." This is roughly equivalent to limited mental arithmetic
abilities.</p>
<p>Just as the essence of art is the compositional inventiveness with shapes and colors
that resonate with viewers—not drawing talent, similarly
the essence of mathematics is patterns of interrelated, fertile concepts—not calculation abilities.
<hr />
<img src="https://i.stack.imgur.com/A0JzN.jpg" alt="FrankStellaWikipedia"></p>
|
235,425 | <p>Ok, here is what I have for the proof of this conjecture. Let me know if I'm on the right path? all input appreciated.</p>
<p>There exist integers $j$, $k$, and $m$, such that,
$b = aj $ and $ c = ajk.$ Then $c = ajk $ (substituting $aj$ for $b$)
let $m = jk$, then $c = ma, => a|c.$ </p>
| The Substitute | 17,955 | <p>Yes, $b=aj$ and $c=bk=ajk=(jk)a$. Since $jk \in \mathbb{Z}$, we have $a|c$.</p>
|
223,521 | <p>Suppose that $a^2-b^2 =x$ where $a,b,x$ are natural numbers.</p>
<p>Suppose $x$ is fixed. If there is one $(a,b)$ found, can there be another $(a,b)$?</p>
<p>Also, would there be a way to know how many such $(a,b)$ exists?</p>
| Mark Bennet | 2,906 | <p>Note that $a^2-b^2=(a+b)(a-b)$ and that $(a+b)=c$ and $(a-b)=d$ are either both odd or both even.</p>
<p>If $x=cd$ where $c>d$ and $c$ and $d$ have the same parity then $a=\frac {c+d}2, b=\frac {c-d}2$ are natural numbers which work for $x$.</p>
<p>It is always possible to express an odd number $x$ in this way, using $c=x, d=1$. If $x$ is prime there is no other factorisation.</p>
<p>If $x$ is even, it must be divisible by 4, and if $x=4y$ then $c=2y, d=2$ will do provided $y>1$.</p>
<p>The number of solutions depends on the number of factorisations into factors of like parity.</p>
|
223,521 | <p>Suppose that $a^2-b^2 =x$ where $a,b,x$ are natural numbers.</p>
<p>Suppose $x$ is fixed. If there is one $(a,b)$ found, can there be another $(a,b)$?</p>
<p>Also, would there be a way to know how many such $(a,b)$ exists?</p>
| ShreevatsaR | 205 | <p>You want $x = a^2 - b^2 = (a-b)(a+b)$. Let $m = a-b$ and $n = a+b$, then note that $a = (m+n)/2$ and $b = (n-m)/2$. For these to be natural numbers, you want both $m$ and $n$ to be of the same parity (i.e., both odd or both even), and $m \le n$. For any factorization $x = mn$ satisfying these properties, $a = (m+n)/2$ and $b = (n-m)/2$ will be a solution.</p>
<p>The answer to your question of how many such $(a,b)$ exist is therefore the same as how many ways there are of writing $x = mn$ with both factors of the same parity and $m \le n$. Let $d(x)$ denote the number of divisors of $x$. (For instance, $d(12) = 6$ as $12$ has $6$ factors $1, 2, 3, 4, 6, 12$.)</p>
<ul>
<li><p>If $x$ is odd, note that for any divisor $m$ of $x$, the factorization $x = mn$ (where $n = x/m$) has both factors odd. Also, out of any two factorizations $(m,n)$ and $(n,m)$, one of them will have $m < n$ and the other will have $m > n$, so the number of "good" factorizations of $x$ is $d(x)/2$. In the case where $x$ is a perfect square, this means either $\lceil d(x)/2 \rceil$ or $\lfloor d(x)/2 \rfloor$ depending on whether or not you want to allow the solution $a = \sqrt{x}$, $b = 0$.</p></li>
<li><p>If $x$ is even, then $m$ and $n$ can't be both odd so they must be both even, so $x$ must be divisible by $4$. Say $x = 2^k \cdot l$, where $k \ge 2$. How many factorizations $x = mn$ are there with both $m$ and $n$ being even? Well, there are $d(x)$ factorisations of $x$ as $x = mn$. Of these, the factorisations in which all the powers of $2$ go on one side can be got by taking any of the $d(l)$ factorisations of $l$, and then putting the powers of two entirely on one of the $2$ sides. So the number of representations $x = mn$ with $m$ and $n$ both being even is $d(x) - 2d(l)$. Again, the number where $m \le n$ is half that, namely $(d(x) - 2d(l))/2$, where in the case $x$ is a perfect square, you mean either $\lceil (d(x) - 2d(l))/2 \rceil$ or $\lfloor (d(x) - 2d(l))/2 \rfloor$ depending on whether you want to allow the $b = 0$ solution or not.</p></li>
</ul>
<hr>
<p>Here is a program using <a href="http://sagemath.org/" rel="nofollow">Sage</a> that can print all solutions $(a,b)$ for any given $x$.</p>
<pre><code>#!/path/to/sage
print "Hello"
def mn(x):
'''Returns all (m,n) such that x = mn, m <= n, and both odd/even'''
for m in divisors(x):
n = x/m
if m <= n and (m % 2 == n % 2): yield (m,n)
elif m > n: break
def ab(x):
'''Returns all (a,b) such that a^2 - b^2 = x'''
return [((m+n)/2, (n-m)/2) for (m,n) in mn(x)]
def num_ab(x):
'''The number of (a,b) such that a^2 - b^2 = x'''
dx = number_of_divisors(x)
if x % 2: return ceil(dx / 2)
l = odd_part(x)
dl = number_of_divisors(l)
return ceil((dx - 2*dl) / 2)
# Do it in two ways to check that we have things right
for x in range(1,1000): assert num_ab(x) == len(ab(x))
# Some examples
print ab(12)
print ab(100)
print ab(23)
print ab(42)
print ab(999)
print ab(100000001)
</code></pre>
<p>It prints:</p>
<pre><code>Hello
[(4, 2)]
[(26, 24), (10, 0)]
[(12, 11)]
[]
[(500, 499), (168, 165), (60, 51), (32, 5)]
[(50000001, 50000000), (2941185, 2941168)]
</code></pre>
<p>and you can verify that, for instance, $168^2 -165^2 = 999$.</p>
|
223,521 | <p>Suppose that $a^2-b^2 =x$ where $a,b,x$ are natural numbers.</p>
<p>Suppose $x$ is fixed. If there is one $(a,b)$ found, can there be another $(a,b)$?</p>
<p>Also, would there be a way to know how many such $(a,b)$ exists?</p>
| Adrián Barquero | 900 | <p>You need to exploit the fact that the right hand side of your equation can be factored. For example for part (1) of the exercise, if $x$ is odd, say $x = 2n+1$ for some integer $n$, then</p>
<p>$$
x = 2n + 1 = y^2 - z^2 = (y - z)(y + z)
$$ </p>
<p>Now try to consider a trivial factorization of $2n+1$ like $2n+1 = 1 \cdot (2n+1)$ and compare the two factorizations to get a system of equations</p>
<p>$$
\begin{align}
y - z &= 1\\
y + z &= 2n + 1
\end{align}
$$</p>
<p>I think you can take it from here, but feel free to ask if you get stuck.</p>
|
2,797,039 | <p><strong>1)</strong> Let $\{f_n\}$ be a sequence of nonnegative measurable functions of $\mathbb R$ that converges pointwise on $\mathbb R$ to $f$ integrable. Show that</p>
<p>$$\int_{\mathbb R} f = \lim_{n\to \infty}\int_{\mathbb R}f_n \Rightarrow \int_{E} f = \lim_{n\to \infty}\int_{E}f_n $$</p>
<p>for any measurable set $E$</p>
<p>I know that $\int_{\mathbb R} f = \int_{\mathbb R \setminus E} f + \int_{E} f$ and $\int_{\mathbb R \setminus E} f \le \liminf_{n\to {\infty}}\biggr(\int_{\mathbb R \setminus E} f_n \biggr)$ from Fatau's Lemma.</p>
<p>I couldn't obtain $\int_{E} f = \liminf_{n\to \infty}\int_{E}f_n = \limsup_{n\to \infty}\int_{E}f_n$ and I have seen that inequality below for obtaining it but I couldn't understand. Could someone explain me please?</p>
<blockquote>
<p>$$\liminf_{n\to \infty}\int_{\mathbb R \setminus E}f_n = \int_{\mathbb R}f-\limsup_{n\to \infty}\int_{E}f_n$$</p>
</blockquote>
<p><strong>2)</strong> It has been written "since $\int_Ef_n \le \int_Ef$ (this inequality from monotonicity I have understood) thus </p>
<blockquote>
<p>$$\limsup\int_Ef_n \le \int_Ef$$</p>
</blockquote>
<p>in proof of The Monotone Convergence Theorem in Royden's Real Analysis. I couldn't see why that inequality obtains.</p>
<p>Thanks for any help</p>
<p>Regards</p>
| Green.H | 456,457 | <p>Suppose the die shows $n$ (where $n \geq 2$). Then, the probability that among $n$ randomly pocked cards, the number of Jacks is equal to $2$ is </p>
<p>$$\frac{\binom{4}{2}\binom{48}{n-2}}{\binom{52}{n}}.$$</p>
<p>Namely, there are $\binom{4}{2}$ events that we choose $2$ out of $4$ Jacks and $\binom{48}{n-2}$ events that we choose other $n-2$ cards. Thus we have the numerator of the expression. It is easy to see that there are $\binom{52}{n}$ possibilities of choosing $n$ cards randomly, and this is the denominator. </p>
<p>Next, we are interested in all $n$ numbers that is at least $2$, or $n = \{2,3,4,5,6\}$. The probability that each of those number show is $\frac{1}{6}$. Then, we sum up over all $n$ to obtain the final answer:</p>
<p>$$\frac{1}{6}\sum_{n=2}^6 \frac{\binom{4}{2}\binom{48}{n-2}}{\binom{52}{n}}.$$</p>
|
2,365,120 | <p>I'm doing some revision on indices and surds. How do you simplify </p>
<blockquote>
<p>$(xy^2)^p\sqrt{x^q} $</p>
</blockquote>
<p>Bit confused because my textbook says the answer is </p>
<blockquote>
<p>$x^{p+q/2}y^{2p}$</p>
</blockquote>
<p>I can understand simplifying but only when it's the same base. I'm confused with this specific question - step-by-step help would be much appreciated!</p>
<p>[[Edit: changed $x^{p+1/2q}$ to $x^{p + q/2}$.]]</p>
| Surb | 154,545 | <p>$$(x\cdot y^2)^p\cdot\sqrt{x^q}=x^p\cdot(y^2)^p \cdot(x^q)^{1/2}=x^p\cdot y^{2p}\cdot x^{q/2}\\=x^p\cdot x^{q/2}\cdot y^{2p}=(x^p\cdot x^{q/2})\cdot y^{2p}=x^{p+q/2}\cdot y^{2p}$$</p>
|
2,866,399 | <p>I was studying Ring and modules and now I am studying Group theory. I feel very confuse regarding generating set .
In Ring Theory if we have an Ideal $ \ I = < x,y > $
Then x doesn't contained in that Ideal, however in Group theory if we have $ \ G = < x, y >$ then x is in the group.
Can any one please show how can we write x as a product of the generators.</p>
| Rob Arthan | 23,171 | <p>The right way to think of the subgroup $\langle X \rangle$ of a group $G$ generated by a subset $X$ of $G$ is that it is the smallest subgroup of $G$ that contains $X$. Or more formally, it is the intersection of all subgroups of $G$ that contain $X$. From that point of view, it is clear that $X \subseteq \langle X \rangle$ (because there is at least one subgroup of $G$, namely $G$ itself, that does contain $X$). Given this definition of $\langle X \rangle$, we can show that $\langle X \rangle$ comprises all elements of $G$ that can be written as products of elements of $X$ or their inverses, where, if $x \in X$, we view $x$ itself as a product of 1 element, namely $x$ itself.</p>
<p>The above description generalises to other situations with "products of elements" replaced by the appropriate algebraic operations: e.g., the ideal generated by a subset $X$ of a ring $R$ is the smallest ideal of $R$ containing $X$ and comprises all the elements of $R$ you can obtain from elements of $X$ by addition and by multiplication by arbitrary elements of $R$.</p>
|
903,831 | <p>One may be curious why one wishes to convert a polynomial ring to a numerical ring. But as one of the most natural number system is integers, and many properties of rings can be easily understood in parallel to ring of integers, I think converting a polynomial ring to a numerical (i.e. integer) ring is useful.</p>
<p>What I mean by converting to a numerical ring is: in the standard ring of integers, $+$ and $\cdot$ are defined as in usual arithmetic. But is there universal way of converting any polynomial/monomial rings such that each object in the ring gets converted to an integer, and $+$ and $\cdot$ can be defined differently from standard integer $+$ and $\cdot$? This definition would be based on integer arithmetic, though. </p>
| Bill Dubuque | 242 | <p>If $\,R\,$ is a ring (or any algebraic structure) then one can <a href="https://math.stackexchange.com/a/190326/242">transport the structure</a> of $\,R\,$ to <em>any</em> set $\,S\,$ of the same cardinality, by push/pulling the algebra operations along any bijection of their underlying sets. When $\,S\,$ = $\,\Bbb N\,$ or $\,\Bbb Z,\,$ this can be viewed simply as indexing (or coding) the elements of $\,R\,$ (e.g. in computer representations of $\,R\,$ where indices are memory addresses).</p>
<p>For example, to add $\,m,n\in\Bbb Z\,$ we first unindex them to $\,i^{-1}(m),\,i^{-1}(n)\in R,\,$ then perform the addition in $\,R,\,$ then index the result, i.e. the transported addition $\,\oplus\,$ in $\,S=\Bbb Z\,$ is</p>
<p>$$ m \oplus n\, =\, i\,(i^{-1}(m)+i^{-1}(n))$$</p>
<p>and analogously for all other operations of $\,R.\,$ This implies that the unindex map $\,i^{-1}$ is a ring homomorphism, i.e. $\, i^{-1}(m\oplus n) = i^{-1}(m)+i^{-1}(n),\,$ and similarly for other operations, yielding a ring isomorphism $\, R\cong \Bbb Z\,$ with said transported operations.</p>
|
2,873,309 | <p>Let's say each month a utility bill comes and it needs to be split among a variable number of roommates. Let's say each person occupied the house for a certain percentage of the month: person 1 100%, person 2 100%, person 3, 75%, person 4, 50%. How can I calculate how much each person has to pay? (The number of people could be greater than number of rooms because of people moving in and out during the month.) Let's say the utility bill is $150 and length of month is 28 days.</p>
<p>My algebraic and correct solution is to calculate utility per day: $5.357. Then I go through each day of the month and check number of people that lived in the house. I do 5.357/number of people in house at day d. Then I add up for each for person for each day.</p>
<pre><code><first 7 days>: 2 people occupied
<second 7 days>: 3 people occupied
<last 14 days>: 4 people occupied
This gives me a correct answer of:
-person 1: 50 = 18.75 + 12.5 + 18.75
-person 2: 50 = 18.75 + 12.5 + 18.75
-person 3: 31.25 = 12.5 + 18.75
-person 4: 18.75 = 18.75
</code></pre>
<p>How can I make this calculation without having to divide up the month day by day? My intuition is that there is a calculus solution to this. </p>
| Micah | 30,836 | <p>If the goal is to get exactly the same result as the solution you've presented, there's pretty much no hope except doing it day-by-day. Just knowing the number of days each roommate occupied isn't enough information, because it's more expensive to live in the house when there are fewer other people living there.</p>
<p>For example, consider two possible scenarios:</p>
<ul>
<li><p>In scenario $A$, person 1 always lives in the house. Person 2 lives there for the first $2$ weeks, and then is replaced by person 3 for the last $2$ weeks.</p></li>
<li><p>In scenario $B$, person 1 lives alone for the first $2$ weeks, then is joined by persons 2 and 3 for the last $2$ weeks.</p></li>
</ul>
<p>Note that in both these scenarios, person 1 lives in the house for $4$ weeks and the other two people live there for $2$ weeks each. However:</p>
<ul>
<li>In scenario $A$, person 1 is always splitting the utility bill with exactly one other person. So person 1 pays $\$75$, while each of the other two pay $\$37.50$.</li>
<li>In scenario $B$, person 1 owes the entire $\$75$ utility bill for the first half of the month, and then the bill is split three ways for the last half. So person 1 pays $\$100$, while the other two pay $\$25$ each.</li>
</ul>
<p>That is, the amount each person owes is different even though their occupancy times are the same.</p>
<p>That being said, utility bills are likely higher when more people live in the house. So splitting it up by shares as suggested by the other answer might be reasonable.</p>
|
2,661,718 | <p>My basic understanding is that each time
I have 30% chance of winning the prize
so between 3 and 4 tries I should win it</p>
<p>cause .3 +.3 +.3 = 90% as I need to win only once with 3 try
.3+.3+.3+.3 = 120% with 4 try</p>
<p>I do remember a formula saying 1-(0.7)^3 = 65.7% chance but I dont remember what that number mean exactly</p>
<p>thanks for answering this noob question my math are way behind me now</p>
| Dr. Sonnhard Graubner | 175,066 | <p>HINT: write $$\left(\frac{x}{\sqrt{x+1}}\right)^2\geq \left(\log(x+1)\right)^2$$
and split into two cases
$$\left(\frac{x}{\sqrt{x+1}}-\log(x+1)\right)\left(\frac{x}{\sqrt{x+1}}+\log(x+1)\right)\geq 0$$</p>
|
294,383 | <p>Evaluate :
$$\int_{0}^{+\infty }{\left( \frac{x}{{{\text{e}}^{x}}-{{\text{e}}^{-x}}}-\frac{1}{2} \right)\frac{1}{{{x}^{2}}}\text{d}x}$$</p>
| Mhenni Benghorbal | 35,472 | <p><a href="https://math.stackexchange.com/questions/513413/show-that-int-0-infty-frac-cosat-cosbtt-ln-fracba">Related technique</a>. Here is a closed form solution of the integral</p>
<p>$$\int_{0}^{+\infty }{\left( \frac{x}{{{\text{e}}^{x}}-{{\text{e}}^{-x}}}-\frac{1}{2} \right)\frac{1}{{{x}^{2}}}\text{d}x} = -\frac{\ln(2)}{2}. $$</p>
<p>Here is the technique, consider the integral</p>
<p>$$ F(s) = \int_{0}^{+\infty }{e^{-sx}\left( \frac{x}{{{\text{e}}^{x}}-{{\text{e}}^{-x}}}-\frac{1}{2} \right)\frac{1}{{{x}^{2}}}\text{d}x}, $$</p>
<p>which implies</p>
<p>$$ F''(s) = \int_{0}^{+\infty }{e^{-sx}\left( \frac{x}{{{\text{e}}^{x}}-{{\text{e}}^{-x}}}-\frac{1}{2} \right)\text{d}x}. $$</p>
<p>The last integral is the Laplace transform of the function </p>
<p>$$ \frac{x}{{{\text{e}}^{x}}-{{\text{e}}^{-x}}}-\frac{1}{2} $$ </p>
<p>and equals</p>
<p>$$ F''(s) = \frac{1}{4}\,\psi' \left( \frac{1}{2}+\frac{1}{2}\,s \right) -\frac{1}{2s}. $$</p>
<p>Now, you need to integrate the last equation twice and determine the two constants of integrations, then take the limit as $s\to 0$ to get the result. </p>
|
1,219,851 | <p>If $n$ is a positive integer such that $2^n+n^2$ is a prime number , then is it true that $6|n-3$ ? Trivially $n$ cannot be even , so this leaves us only with the possibilities $n \equiv1,3,5( \mod 6) $ , but I cannot reduce the cases . Please help , Thanks in advance . </p>
| lab bhattacharjee | 33,337 | <p>$$2^{6a+1}+(6a+1)^2=64^a\cdot2+(6a+1)^2\equiv0\pmod3$$ as $64\equiv1\pmod3\implies64^a\equiv1^a$</p>
<p>Similarly,
$$2^{6a+5}+(6a+5)^2=64^a\cdot32+36a^2+60a+25\equiv0\pmod3$$ </p>
<hr>
<p>Alternatively, if $(n,3)=1,n\equiv\pm1\pmod3\implies n^2\equiv1$</p>
<p>$2^n+n^2\equiv2^n+1\pmod3\iff(-1)^n+1$</p>
<p>Now $(-1)^n+1\equiv0\pmod3$ which is true if $n$ is odd</p>
<p>We need $n\equiv1\pmod2$ and $n\equiv\pm1\pmod3\implies n\equiv1,5\pmod6$</p>
|
2,148,979 | <p>I want to find $A$ such that $$A\sum \limits_{n=m}^{\infty}1/n^{3}=1$$</p>
<p>for any natural value of $m$.</p>
| Robert Israel | 8,508 | <p>If you're interested in approximations,</p>
<p>$$A = 2m^2 - 2m + 1 - \frac{1}{6 m^2} - \frac{1}{6 m^3} + \frac{1}{12 m^4} + \frac{1}{3m^5} + O\left(\frac{1}{m^6}\right) \ \text{as}\ m \to \infty $$</p>
|
2,061,655 | <p>Could we compute the limits
$$\lim_{x\rightarrow 0}\frac{\sin (x)-x+x^3}{x^3} \\ \lim_{x\rightarrow 0}\frac{e^x-\sin (x)-1}{x^2}$$
without using the l'Hospital rule and the Taylor expansion?</p>
| Robert Israel | 8,508 | <p>Sure: use <a href="http://www.wolframalpha.com/input/?i=limit%20of%20(exp(x)-sin(x)-1)%2Fx%5E2%20as%20x%20-%3E%200" rel="nofollow noreferrer">Wolfram Alpha</a> or Maple or ...</p>
|
1,647,399 | <p>Suppose that $(X,d)$ is a compact metric space and $(E_n)$ is any sequence of nonempty closed subsets of $X$ with $E_{n+1}\subset E_n$ for all $n\in\mathbb{N}$. Show that $\cap_{n=1}^\infty E_n$ is nonempty.</p>
<p><strong>What I know:</strong> Since $X$ is compact, we need to use the fact that every sequence in $X$ has subsequence which converges in $X$. $E_n$ is nonempty and closed, that means it contains all its limit points. And $E_n$ is also compact right? </p>
<p><strong>My doubts are:</strong> How can we construct a sequence <strong>that has a subsequence that has a limit in $X$</strong>? I don't think we should use prove by contradiction here. My guess is to show the limit that we have from the fact that $X$ is compact is in the intersection. But I don't even have the sequence yet. </p>
<p>Can anyone please give some clue on the line of reasoning of the proof? Many thanks!</p>
| Martín-Blas Pérez Pinilla | 98,199 | <p>Alternative idea:
$$\bigcap_{n=1}^\infty E_n = \emptyset\implies \bigcup_{n=1}^\infty(X\setminus E_n) = X$$</p>
|
1,327,755 | <p>Prove that the antipodal mapping $A: S^n \to S^n$ given by $A(p)=-p$ is an isometry. </p>
<p>I know that in order to prove that a map $f$ is an isometry of a smooth manifold $M$ it must hold true that
$$\langle v,u\rangle_p = \langle d_f v, d_fu \rangle_{f(p)}$$</p>
<p>Now I found online <a href="http://image.sciencenet.cn/olddata/kexue.com.cn/upload/blog/file/2009/9/2009941352126256.pdf" rel="noreferrer">this document</a> that proves this exercise (Exercise 1.1.). I cannot follow it though. It basically says that in order to prove this we first need to claim that $T_{p}S^n = T_{A(p)}S^n$ and for this it is enough to prove that $T_{p}S^n \subset T_{A(p)}S^n$. <strong>Why is this true?</strong> In order to show this they claim that $T_{A(p)}S^n \subset \subset T_{A \circ A(p)}S^n = T_pS^n$. <strong>What does the $\subset \subset$ mean</strong> and <strong>why the last relation holds true, i.e. why $T_{A(p)S^n}=T_pS^n$ since these tangent spaces are defined in different points?</strong></p>
<p>Finally, <strong>why they later claim that $dA_pv = -v$?</strong> I do not understand why this also holds true.</p>
| Lee Mosher | 26,501 | <p>To address the remaining question, the proof of the equation of tangent planes $T_p S^2 = T_{A(p)}S^2$ is obtained by tracing through the definitions and doing some multivariable calculus computations.</p>
<p>First, $S^2$ is the solution set of the equation
$$x^2 + y^2 + z^2-1=0
$$
Second, the gradient of the left hand side of this equation at the point $p=(a,b,c)$ is computed to be
$$\nabla_p(f) = \langle 2a,2b,2c\rangle
$$
and the tangent plane $T_p S^2$ is the perpendicular plane to $\nabla_p(f)$ which is computed to be the set vectors
$$T_p S^2 = \{\langle x,y,z \rangle \bigm| ax+by+cz=0 \}
$$
Third, the gradient at the point $A(p)=(-a,-b,-c)$ is computed to be
$$\nabla_{A(p)}(f)=\langle-2a,-2b,-2c\rangle
$$
and $T_{A(p)}S^2$ is the perpendicular plane to $\nabla_{A(p)}(f)$ which is computed to be the exact same set of vectors
$$T_{A(p)}S^2 = \{\langle x,y,z \rangle \bigm| ax+by+cz=0 \}
$$</p>
|
3,208,003 | <p>How do you calculate the sum of this series?</p>
<p><span class="math-container">$$ \sum_{n=1}^\infty 0.5n \left(\frac{2}{3}\right)^n $$</span></p>
| denklo | 562,687 | <p>There is a simple trick, for <span class="math-container">$a < 1$</span>:
<span class="math-container">$$\sum\limits_{n=0}^\infty n a^n = a\partial_a\sum\limits_{n=0}^\infty a^n = a\partial_a \frac{1}{1-a} = \frac{a}{(1-a)^2}$$</span></p>
|
3,208,003 | <p>How do you calculate the sum of this series?</p>
<p><span class="math-container">$$ \sum_{n=1}^\infty 0.5n \left(\frac{2}{3}\right)^n $$</span></p>
| Community | -1 | <p><strong>Hint:</strong></p>
<p><span class="math-container">$$\color{green}{\sum_{n=1}^\infty na^n}=\sum_{n=0}^\infty(n+1)a^{n+1}=a\color{green}{\sum_{n=1}^\infty na^n}+\sum_{n=0}^\infty a^{n+1}.$$</span></p>
<blockquote class="spoiler">
<p> <span class="math-container">$$|a|<1\implies(1-a)S=\frac a{1-a}.$$</span></p>
</blockquote>
|
3,208,003 | <p>How do you calculate the sum of this series?</p>
<p><span class="math-container">$$ \sum_{n=1}^\infty 0.5n \left(\frac{2}{3}\right)^n $$</span></p>
| PrincessEev | 597,568 | <p>Note that we can pull out the <span class="math-container">$0.5$</span> from the summation, and focus on </p>
<p><span class="math-container">$$\sum_{n=1}^\infty n \left( \frac 2 3 \right)^n$$</span></p>
<p>This bears resemblance to a geometric series. We know that, for a geometric series with ratio <span class="math-container">$r \in (-1,1)$</span>,</p>
<p><span class="math-container">$$\sum_{n=0}^\infty r^n = \frac{1}{1-r}$$</span></p>
<p>We want to manipulate this expression into the desired form. First, we see the derivative is given by </p>
<p><span class="math-container">$$\sum_{n=0}^\infty nr^{n-1} = \frac{1}{(1-r)^2}$$</span></p>
<p>and multiplication by <span class="math-container">$r$</span> gives</p>
<p><span class="math-container">$$\sum_{n=0}^\infty nr^{n} = \frac{r}{(1-r)^2}$$</span></p>
<p>Since the first summand of the sum (<span class="math-container">$n=0$</span>) is itself <span class="math-container">$0$</span>, we can start the summation at <span class="math-container">$n=1$</span> if we choose, which would match the form of the sum we choose to focus on. Take <span class="math-container">$r=2/3$</span> and we have the "important" sum; once you have the result, multiply by <span class="math-container">$1/2$</span> again to account for the pulling out of it at the start, and we're done!</p>
|
4,537,390 | <p>Let <span class="math-container">$C$</span> be a circle centred at <span class="math-container">$O$</span> and <span class="math-container">$P$</span> be a point inside the circle.</p>
<p>The line <span class="math-container">$OP$</span> intersects the circle <span class="math-container">$C$</span> at <span class="math-container">$Q_1$</span> and <span class="math-container">$Q_2$</span>.</p>
<p>Let <span class="math-container">$Q$</span> be a point on the circle <span class="math-container">$C$</span>.</p>
<p>Then the length of <span class="math-container">$PQ$</span> is in between the lengths of <span class="math-container">$PQ_1$</span> and <span class="math-container">$PQ_2$</span>.</p>
<p><a href="https://i.stack.imgur.com/bVoba.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bVoba.png" alt="enter image description here" /></a></p>
<p>I have observed that if <span class="math-container">$P = O$</span>, then we have <span class="math-container">$PQ_1 = PQ_2 = PQ$</span>.</p>
<p>How to prove the general case. Any hints will be useful.</p>
<hr />
<p>Source: Homework Assignment.</p>
| Eugen Ionascu | 83,478 | <p>Let's extend PQ and have the intersection with the circle denoted by R. It is known that <span class="math-container">$$PQ\cdot QR=PQ_1\cdot PQ_2,$$</span> (similarity of triangles) so if we denote by <span class="math-container">$s=PQ+QR=PR$</span> and <span class="math-container">$p=PQ_1\cdot PQ_2$</span> we see that the segments <span class="math-container">$PQ$</span> and <span class="math-container">$PR$</span> are the solutions of the quadratic <span class="math-container">$x^2-sx+p=0$</span>. Hence, <span class="math-container">$$PQ=(s\pm \sqrt{s^2-4p})/2$$</span> and these two functions of <span class="math-container">$s$</span> are monotone with the max/min when <span class="math-container">$s$</span> is the length of the diameter.</p>
|
1,764,162 | <p>I have $$f(x) = x \left| x - 1 \right|$$</p>
<p>Here my given value for $x$ is 1</p>
<p>And I need to test if the function is "continuous" when $x$ is $1$ and also when reaching
$$ f(1)$$</p>
<p>$$ \lim\limits_{x \to 1} f(x)$$</p>
<p>So I plug in 1 to the equation to find out the value of $f(1)$</p>
<p>$$ f(1) = 1|1-1| = 0 $$
Then I check the $lim$ to see if I get $0$ as well.</p>
<p>$$ \lim\limits_{x \to 1} f(x) = x|x-1| = 1|1-1| = 0$$</p>
<p>What I am not understanding is why I need to even check for $lim$ even from the first place because the process looks identical. But I am taught to do it anyway.</p>
<p>Could an equation not have the same value for $f(AnyValue)$ and $\lim\limits_{x \to AnyValue} f(x)$ ?</p>
| Siddharth Bhat | 261,373 | <p>I think you are confused between the definition of any old function and a continuous function.</p>
<p>As you said, it is <em>only</em> for continuous functions that the fact </p>
<p>$$\lim_{x \to a} f(x) = f(a)$$</p>
<p>Any discontinuous function <strong>will not</strong> exhibit this behaviour.</p>
<p>As an example, consider $f(x) = \frac{1}{x}$ as $\lim{x \to 0+}$ (that is, approaching from the right hand side) versus $\lim{x \to 0-}$. I have put up the function here:</p>
<p><a href="https://i.stack.imgur.com/T7s1P.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/T7s1P.gif" alt="image of y = 1/x"></a></p>
<p>If you look at the graph of $\frac{1}{x}$, the function is discontinuous at $x = 0$. Morever, if we approach it from the left ($\lim{x \to 0-}$), then we get the values as "$-\infty$" (this is only conventional shorthand to say that diverges off to negative infinity. It does not "reach" infinity).</p>
<p>However, if we look at the graph from the right hand as $\lim x \to 0+$, then we get "$+ \infty$" (which, vice versa, means that the function diverges off to positive infinity)</p>
<p>Now, the graph of your function looks like this: <a href="https://www.wolframalpha.com/input/?i=graph+of+x+*+%7C+x+-+1%7C" rel="nofollow noreferrer">WolframAlpha plot</a></p>
<p>So, in your question, you are asked to look at $$\lim_{x \to 1} f(x) = x |x - 1|$$</p>
<p>As you have verified, the limit $$\lim_{x \to 1} f(x) = 1 * |1 - 1| = 0$$ is equal to the value at $$f(1) = 0$$</p>
<p>This means that the function is continuous.</p>
<p>To get a better understanding of why this is a criteria for continuity, think of what is happening. The condition is basically saying that in the smallest of regions, the value <strong>at the point $a$ of the function $f$</strong> should be same <strong>in the smallest region around $a$</strong>. This does not give the function a lot of wiggle room, and is hence forced to be "close to each other" at small intervals, thereby ensuring what we see as continuity.</p>
|
26,736 | <p>I am still a newbie to <em>Mathematica</em>, so I am still getting used to the syntax and logic of the program - but I've been trying to do the following 3D plot:</p>
<pre><code>g[x_, t_] := y /. FindRoot[y == Tanh[(2*y + x)/t], {y, x, t}];
p1 = ContourPlot3D[{z ==
Sech[(2 g[x, t] + x)/t]^2/(t - 2 Sech[(2 g[x, t] + x)/t]^2)}, {t,
0.001, 3}, {x, -1.5, 1.5}, {z, 0, 2.5}, Mesh -> None,
ContourStyle -> Opacity[0.25],
ColorFunction -> ColorData["ThermometerColors"]];
Rasterize[
Show[p1, PlotRange -> {{0.001, 3}, {-1.5, 1.5}, {0, 2.5}},
ViewPoint -> {0.9*Pi, -0.45*Pi, 1.05*Pi}], ImageResolution -> 1200]
</code></pre>
<p>The <code>ContourPlot3D</code> of the transcendental equation <code>y == Tanh[(2*y + x)/t]</code> itself comes out perfect - but after some 10-15 mins of evaluating z, I seem to be getting the right 3D surface from the 2D solutions I've done (cuts through the 3D surface I am trying to plot), but the plot is filled with artifacts. I am the tinkering type, so I tend to need to see things work right a few times, before things like the syntax and logic of <em>Mathematica</em> start to stick in my memory - and I can't figure out this one. What am I doing wrong?</p>
| Michael E2 | 4,999 | <h2>Edit: Introduction</h2>
<p>In my original answer, I had noticed a few things about this problem but hadn't pursued them. Then every few days something would occur to me and I would poke around little, be done, and then another thing would occur to me. This problem has some features that I think make an interesting case study. They are mainly mathematical features, but ones which invite <em>Mathematica</em> solutions.</p>
<ul>
<li><p>The formula to be graphed is a composition of explicit function and an implicitly defined function that has a fold in it that yields three branches over part of the domain; so the composition itself is discontinuous along the branch curve. Further, the equation for the implicit function is discontinuous at <code>t == 0</code>, and the three branches collapse into one.</p>
<p><img src="https://i.stack.imgur.com/gzBqR.png" width="250" alt="constraint surface"></p></li>
<li><p>Root finding over the branched domain is unstable if the initial points are far from the sought-after roots.</p></li>
<li><p>Relatively speaking, the equations are simple in some aspects and this specific problem can be improved via algebra and calculus.</p></li>
</ul>
<p>Two things can be improved, the accuracy of the plot and the speed. Accuracy is probably more important. Nevertheless there are some principles here that can be used generally to speed things up.</p>
<ul>
<li><p><strong>Pick good initial points</strong> for <code>FindRoot</code> to solve for implicit function values. The original answer gave a zeroth order approach using <code>ContourPlot3D</code> that might be applied to a range of such problems. In this case it is possible to improve this.</p></li>
<li><p><strong>Separate</strong> everything to do with the <strong>branches</strong> in code to forestall switching to the wrong branch. We can solve for the equation of the branch curve, so this is easy.</p></li>
<li><p><strong>Parametrize the domain</strong> so that the branch curve, where the plotted function is discontinuous, corresponds to a parameter being constant (such as <code>u == 0</code>). This helps in a couple of ways:</p>
<ul>
<li>One can treat the discontinuity, either with the <code>Exclusions</code> option or by dividing the graph cleanly in two (for instance, into <code>u < 0</code> and <code>u > 0</code>). Dividing the domain works better in this case.</li>
<li>One can now prevent <code>FindRoot</code> jumping to the wrong branch. (In a sense, the parametrization completes the separation of the branches.)</li>
</ul></li>
<li><p><strong><a href="http://reference.wolfram.com/mathematica/tutorial/FunctionsThatRememberValuesTheyHaveFound.html" rel="nofollow noreferrer">Memoize</a></strong> the results of <code>FindRoot</code>, i.e. the function <code>g</code>, which is called several times on the same input.</p></li>
<li><p>Use the <strong><code>NormalsFunction</code></strong> option. With numeric functions, the plot routines have to figure out the surface normals. Using calculus to supply a function for the surface normals speeds plotting up by a factor of three.</p></li>
<li><p>A couple of these steps are implemented in a "<strong>vectorized</strong>" fashion that speeds up the step.</p></li>
</ul>
<p>Finally, I add one more thing. Silvia's answer made me wonder whether a plot of the whole surface was desired. One can solve explicitly for <code>x</code> in terms of <code>y</code> and <code>t</code>, instead of <code>y</code> in terms of <code>t</code> and <code>x</code>. There are still some issues in using the resulting parametrization, but one no longer has to deal with <code>FindRoot</code>. Parametrizing in terms of <code>y</code> and <code>t</code> makes quick work of plotting the whole surface. To trim the overlap to get graphs of functions would require <code>FindRoot</code> again to get a good image, which would slow things down. I omit that, since we already have other methods for plotting the functions.</p>
<h3>Overview</h3>
<p>Here is the progression of plots. The numbers and plot labels indicate the <code>Timing</code>s.</p>
<p>The OP's code but the same initial value <code>y == 0.5</code> for <code>FindRoot</code>. It makes a nice octopus, but the discontinuities are chaotic. (Different initial values yield different chaotic plots.)</p>
<p><img src="https://i.stack.imgur.com/vigXh.png" width="250" alt="naive plot - octopus"></p>
<p>My initial answer: the three functions resulting from the three branches.</p>
<p><img src="https://i.stack.imgur.com/rg0Kp.png" alt="First answer - with almost no erratic behavior at discontinuities"></p>
<p>This <strong>updated answer</strong> - an improvement along the same lines.</p>
<p><img src="https://i.stack.imgur.com/T36LQ.png" alt="Updated answer - no erratic behavior"></p>
<p>The whole surface, parametrized in terms of <code>t</code>, <code>y</code>.</p>
<p><img src="https://i.stack.imgur.com/Mp1gk.png" width="250" alt="entire self-intersecting surface"></p>
<h2>Simplification of the problem</h2>
<p>Since <code>y == g[x, t]</code>, the original system is equivalent to</p>
<pre><code>{ y == Tanh[(2*y + x)/t],
z == Sech[(2 y + x)/t]^2 / (t - 2 Sech[(2 y + x) / t]^2) }
</code></pre>
<p>This can be rewritten</p>
<pre><code>{ -x - 2 y + t ArcTanh[y] == 0,
(t - 2 (1 - y^2)) z == (1 - y^2) }
</code></pre>
<p>since <code>Sech[(2 y + x)/t]^2</code> equals <code>1 - Tanh[(2 y + x)/t]^2</code> or <code>1 - y^2</code>. We will let
<code>F = -x - 2 y + t ArcTanh[y]</code>
so that the first equation is <code>F == 0</code>. The branch curve is where the partial derivative of <code>F</code> with respect to <code>y</code> is zero. From this <code>y</code> can be solved in terms of <code>t</code> and then <code>x</code> in terms of <code>t</code>.</p>
<pre><code>F = -x - 2 y + t ArcTanh[y]
D[F, y] (* --> -2 + t/(1 - y^2) *)
branchY = Solve[D[F, y] == 0, y]
branchX = Simplify @ First @ Solve[#, x] & /@ (F == 0 /. branchY)
(* {{y -> -(Sqrt[2 - t]/Sqrt[2])},
{y -> Sqrt[2 - t]/Sqrt[2]}} *)
(* {{x -> Sqrt[4 - 2 t] - t ArcTanh[Sqrt[1 - t/2]]},
{x -> -Sqrt[4 - 2 t] + t ArcTanh[Sqrt[1 - t/2]]}} *)
</code></pre>
<p>The branch curve divides into two segments that meet at a point where <code>t == 2</code>. The segments are mirror images, one where <code>y</code> (resp. <code>x</code>) is positive (resp. negative), and vice versa.</p>
<p>Another property of the constraint <code>F == 0</code> that is easily seen is that for each fixed <code>y</code>, the solution set is a line and treating <code>x</code> or <code>t</code> as functions of <code>y</code> and the other coordinate is easy. This will be applied to get a plot of the full surface.</p>
<p>Finally, we can derive a formula for the surface <strong>normals</strong>. Since our plotted function will be calculated numerically, the plot routines cannot analyze it symbolically. We can help (and speed up plotting) by deriving a formula for the surface normal of our plot. First get the partial derivatives of <code>y</code> with respect to <code>t</code> and <code>x</code>. Then substituting <code>y -> g[t, x]</code>, we find the gradient of surface (2nd equation above), which is normal to the surface. Since the direction of the normal is all that is used, we can multiply it by a scalar (to clear denominators). We do have to check later (visually at least) that we get a consistent orientation.</p>
<pre><code>dydt = -D[F, t]/D[F, y]; (* -(ArcTanh[y]/(-2 + t/(1 - y^2))) *)
dydx = -D[F, x]/D[F, y]; (* 1/(-2 + t/(1 - y^2)) *)
(-2 + t + 2 y^2) D[(t - 2 (1 - y^2)) z - (1 - y^2) /. y -> g[t, x], {{t, x, z}}] /.
{Derivative[1, 0][g][t, x] -> dydt,
Derivative[0, 1][g][t, x] -> dydx} /. g[t, x] -> y // Simplify
(* {(-2 + t + 2 y^2) z + 2 y (-1 + y^2) (1 + 2 z) ArcTanh[y],
-2 (-1 + y^2) (y + 2 y z),
(-2 + t + 2 y^2)^2} *)
</code></pre>
<p>We will use this formula for the normals later.</p>
<h2>Plotting the functions</h2>
<p>Some of the calculations are specific to the region being plotted. Therefore we make the following definitions. Because of the nature of the problem, there are limitations on the range of the domain (noted in the comments).</p>
<pre><code>{xMin, xMax} = {-1.5, 1.5}; (* implicitly assumed be symmetric and between -2, 2 *)
{tMin, tMax} = {0, 4}; (* tMin is implicitly assumed to be 0 in what follows *)
domain = {{tMin, tMax}, {xMin, xMax}};
</code></pre>
<p>The starting <code>t</code> value of the branch curve at the boundary of the <code>x</code> domain. This will show up in the parametrization of the domain.</p>
<pre><code>tBr0[x2_] := tBr0[x2] = t /. FindRoot[First[x /. branchX] == x2, {t, 0.1}];
tBr0[xMax]
(* 0.218768 *)
</code></pre>
<h2>Parametrizing the domain</h2>
<p>Of the two parameters <code>u0</code> runs from <code>tBr0[xMax]</code> to <code>4 - tBr0[xMax]</code>, and <code>v0</code> runs roughly from <code>-1</code> to <code>1</code>. The parameter <code>u0</code> is substituted for <code>t</code> in <code>branchX</code>, and so it determines the <code>x</code> coordinate from the corresponding point on the branch curve. The parameter <code>v0</code> linearly interpolates between the boundaries, <code>t == tMin</code> (<code>0</code>) and <code>t == tMax</code> (<code>4</code>). For <code>-1 < v0 < 0</code>, the parametrization stays inside the region where there are three solutions to <code>F == 0</code>; for <code>0 < v0 <= 1</code>, it stays in the one-solution region. At <code>v0 == -1</code> and <code>v0 == 0</code>, there are discontinuities in <code>y</code> and <code>z</code>, so those values are avoided in plotting <code>z</code>. (The point symmetry for <code>u0</code>, namely <code>u0 == 2</code>, seemed convenient at the time; the corresponding reflection <code>4 - u0</code> appears in the code.)</p>
<pre><code>Block[{u0, x}, (* we Block these because of Evaluate - just in case *)
myx[u0_?NumericQ] /; 0. <= u0 <= 2. := Evaluate[First[x /. branchX] /. t -> u0];
myx[u0_?NumericQ] /; 2. < u0 <= 4. := Evaluate[Last[x /. branchX] /. t -> (4 - u0)];
myt[u0_?NumericQ, v0_?NumericQ, {t1_, t2_}] /; 0. <= u0 <= 2. && v0 <= 0. :=
Rescale[v0, {-1, 0}, {t1, u0}];
myt[u0_?NumericQ, v0_?NumericQ, {t1_, t2_}] /; 0. <= u0 <= 2. && v0 > 0. :=
Rescale[v0, {0, 1}, {u0, t2}] ;
myt[u0_?NumericQ, v0_?NumericQ, {t1_, t2_}] /; 2. <= u0 <= 4. && v0 <= 0. :=
Rescale[v0, {-1, 0}, {t1, 4 - u0}];
myt[u0_?NumericQ, v0_?NumericQ, {t1_, t2_}] /; 2. <= u0 <= 4. && v0 > 0. :=
Rescale[v0, {0, 1}, {4 - u0, t2}];
]
</code></pre>
<p>The following shows how the parametrization works.</p>
<pre><code>ParametricPlot[{myt[u, v, {tMin, tMax}], myx[u]},
{u, tBr0[xMax], 4. - tBr0[xMax]}, {v, -1, 1},
MeshFunctions -> {#3 &, #4 &, #4 &},
Mesh -> {Range[0, 4, 1/5], {0}, DeleteCases[Range[-1, 1, 1/8], 0]},
MeshStyle -> {Directive[Thin, Gray], Red}, FrameLabel -> {t, x}]
</code></pre>
<p><img src="https://i.stack.imgur.com/x1JTY.png" alt="Parametrization of domain, showing mesh, branch curve"></p>
<h2>Solving the implicit equation</h2>
<p>The OP used <code>FindRoot</code> to solve <code>y == Tanh[(2*y + x) / t</code> for <code>y</code> for given <code>x</code>, <code>t</code>. This was called from <code>ContourPlot3D</code> twice each time the contour function was evaluated. (Sometimes, apparently, <code>ContourPlot</code> will call the function more than once on the same input.) There are a few things to do to improve the accuracy and speed.</p>
<ul>
<li><p>good initial values for <code>y</code></p></li>
<li><p>memoizing <code>g</code> so that <code>FindRoot</code> is called only once on each unique input</p></li>
<li><p>making sure <code>y</code> stays on the right branch, even if <code>FindRoot</code> overshoots.</p></li>
</ul>
<h3>Good initial values</h3>
<p>There are four regions of <code>F ==0</code>, three where there are three solutions and one where there is only one. Near the boundaries of these regions (the branch curve), and where <code>t == 0</code> where there is another discontinuity, we seed special initial points for setting up <code>FindRoot</code> later.</p>
<p>Note: We will need to identify these regions in some way. The 3-solution branches will be identified by <code>branch</code> being equal to <code>-1.</code>, <code>0.</code>, <code>1.</code>; the top and bottom branches correspond to <code>1.</code> and <code>-1.</code> resp., because that is the limit <code>y</code> approaches, a fact we will take advantage of; <code>0.</code> is between and has no other significance. The 1-solution region is simply identified by "OneSolution". Two functions test which type of region a point <code>{x, t}</code> lies, one with one solution to <code>F == 0</code> or one with three solutions (<code>oneSolutionQ</code>, <code>threeSolutionsQ</code> resp.). We make vectorized versions for speed; a pointwise version of only one is needed. For the vectorized versions, parallel lists of <code>t</code> and <code>x</code> coordinates are passed as arguments, as is typical for such functions. (The case where there are two solutions, the branch curve, is where <code>z</code> is discontinuous, so it is omitted.)</p>
<p>We use the second partial derivative of <code>y</code> with respect to <code>t</code> to estimate <code>y</code> near the branch curve where the first partial derivative vanishes, with the function <code>d2y</code>. The points for initial values near the boundaries are stored in <code>iv0</code>, <code>iv3a</code>, <code>iv3b</code>.</p>
<p>We store some points across the whole constraint <code>F == 0</code> in <code>ivGen</code> using <code>ContourPlot3D</code> as before. One problem is that <code>ContourPlot3D</code> is approximate and for some of the <code>y</code> coordinates, <code>Abs[y] >= 1</code>, outside the domain of <code>F</code>. Therefore we clean up the <code>y</code> coordinates a little.</p>
<p>Next we separate the initial value points <code>ivGen</code> into <code>ivpoints[branch]</code> according to which region they approximate. We use these to define separate <code>NearestFunction</code>s for finding the initial value for <code>y</code> for the closest point on the constraint <code>F == 0</code> for a given <code>t</code>, <code>x</code> on a given <code>branch</code>. They is an <code>nXT[branch]</code> for each of the four regions <code>branch</code>. The function <code>initPt</code> returns the <code>y</code> for point on the contstraint nearest <code>t</code>, <code>x</code> on <code>branch</code>. (Here <code>branch</code> will have only one of the three values <code>-1.</code>, <code>0.</code>, or <code>1.</code>. The fourth region <code>"OneSolution"</code> is common to all three of the functions for <code>z</code>.) </p>
<pre><code> (* auxiliary functions *)
oneSolutionQ[t_?NumericQ, x_?NumericQ] :=
With[{sqrt = Sqrt[4 - 2 t],
log = Chop[1/2 t Log[(2 - Sqrt[4 - 2 t])/(2 + Sqrt[4 - 2 t])]]},
t > 2. || x > sqrt + log || x < -sqrt - log];
oneSolutionQ[t_?VectorQ, x_?VectorQ] :=
With[{sqrt = Chop[Sqrt[4 - 2 t]],
log = Chop[1/2 t Log[(2 - Sqrt[4 - 2 t])/(2 + Sqrt[4 - 2 t])]]},
MapThread[Or, Positive[{t - 2., x - (sqrt + log), (-sqrt - log) - x}]]];
threeSolutionsQ[t_?VectorQ, x_?VectorQ] :=
With[{sqrt = Chop[Sqrt[4 - 2 t]],
log = Chop[1/2 t Log[(2 - Sqrt[4 - 2 t])/(2 + Sqrt[4 - 2 t])]]},
MapThread[And, Positive[{2. - t, (sqrt + log) - x, x - (-sqrt - log)}]]];
d2y[t0_, Δt_] := Evaluate[Sqrt[-2 Δt D[F, t]/D[F, {y, 2}]] /.
First@branchY /. First@branchX /. t -> t0]
(* seed points for the initial values for y - for use with Nearest *)
iv0 = Table[{0.001, x, -x/2}, {x, xMin, xMax, (xMax - xMin)/100}]; (* near t = 0 *)
ivGen = First @ Cases[
ContourPlot3D[y == Tanh[(2*y + x)/t],
{t, tMin + 0.01, tMax}, {x, xMin, xMax}, {y, -1.03, 1.03},
PlotPoints -> 50, MaxRecursion -> 0, Mesh -> None],
GraphicsComplex[pts_, rest___] :> pts, Infinity];
ivGen = ivGen /. {t_, x_, y_} /; Abs[y] >= 1 :> {t, x, Clip[y, {-1. + 10^-10, 1. - 10^-10}]};
ivpoints["OneSolution"] = Pick[ivGen,
MapThread[And,
{Positive[Times @@ Transpose@ivGen[[All, 2 ;; 3]]],
oneSolutionQ @@ Transpose@ivGen[[All, 1 ;; 2]]}]];
ivpoints[0.] = Pick[#, MapThread[And,
{Negative[Abs[#[[All, 3]]] - (y /. branchY[[-1, 1]] /. t -> tBr0[xMax])],
Negative[D[F, y] /. {t -> #1, x -> #2, y -> #3} & @@ Transpose@#],
threeSolutionsQ @@ Transpose@#[[All, 1 ;; 2]]}]] &@
Join[ivGen, iv0];
sign[1.] := Positive; sign[-1.] := Negative;
ivpoints[branch : 1. | -1.] := ivpoints[branch] =
Pick[ivGen, MapThread[And,
{sign[branch][ivGen[[All, 3]]],
MapThread[Or, {Positive[D[F, y] /. {t -> #1, x -> #2, y -> #3} & @@
Transpose @ ivGen]}],
threeSolutionsQ @@ Transpose @ ivGen[[All, 1 ;; 2]]}]];
(* defines the NearestFunction for each region *)
Set @@@ ({nXT[#], Nearest[Most /@ # -> #] &@
Join[ivpoints[#]]} & /@ {"OneSolution", -1., 0., 1.})
(* interface for use with FindRoot *)
initPt[t_, x_, branch_] /; t >= 2. := nXT["OneSolution"][{t, x}][[1, -1]]; (* for speed *)
initPt[t_, x_, branch_] /; oneSolutionQ[t, x] := nXT["OneSolution"][{t, x}][[1, -1]];
initPt[t_, x_, branch_] := nXT[branch][{t, x}][[1, -1]];
</code></pre>
<p>Here is what the point sets for the regions look like.</p>
<pre><code>Graphics3D[MapIndexed[{Hue[First@#2/4], Point@#} &,
Cases[#, {_Real ..}] & /@ ivpoints /@ {"OneSolution", 0., 1., -1.}]
Axes -> True, AxesLabel -> {t, x, y}]
</code></pre>
<p><img src="https://i.stack.imgur.com/kYwLr.png" alt="Initial points, colored by region"></p>
<h2>Getting accurate graphs</h2>
<h3>Improving the use of FindRoot</h3>
<p>There are two improvements, memoizing and correcting <code>y</code> if <code>FindRoot</code> overshoots in the 3-solution domain. As before, we also define <code>g</code> to evaluate only on numeric arguments. We divide the definition of <code>g</code> into two parts, one for the 3-solution domain (<code>v < 0</code>) and one for the 1-solution domain (<code>v > 0</code>). Over the 3-solution domain, for a given <code>t</code>, the <code>y</code> coordinate of the branch curve divides the <code>y</code> coordinates of the branches. We can use it and <code>Clip</code> to ensure that the return value of <code>g</code> is on the desired <code>branch</code>. We fudge a little (scaling the extreme values of <code>y</code> by <code>0.9999</code> and <code>0.999999</code> to use in <code>Clip</code>) to keep <code>y</code> strictly on the correct side.</p>
<pre><code>g[x0_?NumericQ, t0_?NumericQ, v_?Negative, branch_] := g[x0, t0, v, branch] =
Clip[y /.
With[{y0 = initPt[t0, x0, branch]}, FindRoot[y == Tanh[(2*y + x0)/t0], {y, y0}]],
If[branch == 0.,
0.9999 y /. branchY /. t -> t0,
Sort[{branch y /. Last@branchY /. t -> t0, 0.999999 branch}]]];
g[x0_?NumericQ, t0_?NumericQ, v_?Positive, branch_] := g[x0, t0, v, branch] =
y /. With[{y0 = initPt[t0, x0, branch]}, FindRoot[y == Tanh[(2*y + x0)/t0], {y, y0}]];
(* surface normals *)
normals = Function[{t0, x0, z0, u0, v0, y0},
{(-2 + t0 + 2 y0^2) z0 + 2 y0 (-1 + y0^2) (1 + 2 z0) ArcTanh[y0],
-2 (-1 + y0^2) (y0 + 2 y0 z0),
(-2 + t0 + 2 y0^2)^2}];
(* parametrization *)
param[t_?NumericQ, x_?NumericQ, y0_?NumericQ] := {t, x, (1 - y0^2)/(t - 2 (1 - y0^2))};
(* plotting function *)
plot2[branch_, opts___] :=
Show[
ParametricPlot3D[
param[
myt[u, v, {tMin, tMax}],
myx[u],
g[myx[u], myt[u, v, {tMin, tMax}], v, branch]],
{u, tBr0[xMax], 4. - tBr0[xMax]}, Evaluate@{v, Sequence @@ #},
NormalsFunction -> (normals[##, g[#2, #1, #5, branch]] &), opts] & /@
If[branch == 0., {{-0.98, -0.001}, {0.001, 1}}, {{-0.98, 0}, {0, 1}}]
]
</code></pre>
<p>The plots in the Overview were made with the following command:</p>
<pre><code>Timing[plot2[#,
Mesh -> None, PlotStyle -> Opacity[0.7],
BoxRatios -> {1, 1, 0.5}, ColorFunctionScaling -> False,
ColorFunction -> (ColorData["ThermometerColors"][Rescale[#3, {0, 1.5}]] &),
PlotRange -> {All, All, {-2.5, 2.5}}, AxesLabel -> {t, x, z},
ImageSize -> 200]] & /@ {-1., 0., 1.} // Transpose // Grid
</code></pre>
<h2>Plotting the surface, parametrized by y</h2>
<p>The code below plots the whole surface. We parametrize the surface in terms of <code>y</code> and <code>t</code>, where <code>x</code> lies on a <strong>line</strong>, since for a fixed <code>y</code>, the constraint <code>F == 0</code> is a linear equation in <code>x</code> and <code>t</code>. This line is passes through the branch curve at a given <code>y</code>, so this parametrization turns out to be convenient for separating the components of the surface. As <code>y</code> changes, this line slides along the branch curve and rotates through a half-turn.</p>
<p>One issue to deal with is to cut off the line at the boundary of the plot range. <em>Mathematica</em> can do it automatically, but it turns out to produce a rather ragged looking plot. So we explicitly solve for the end points and rescale.</p>
<p>Another issue is that as <code>y</code> approaches an end point <code>1</code> or <code>-1</code>, the constraint becomes very flat and the rate at which the line rotates approaches infinity. To get enough plot points near the end points, we reparametrize <code>y -> Sin[(Pi/2)v]</code>. A "perfect" solution might have the angle of the line rotate uniformly. This does not but it produces a satisfactory graph nonetheless.</p>
<p>Some utility functions:</p>
<ul>
<li><code>xOFty</code>: returns <code>x</code> for a given <code>t</code>, <code>y</code>;</li>
<li><code>tOFxy</code>: returns <code>t</code> for a given <code>x</code>, <code>y</code>;</li>
<li><code>xBrOFy</code>: returns the <code>x</code> coordinate on the branch curve for a given <code>y</code>.</li>
</ul>
<p>The functions <code>tP</code>, <code>xP</code>, <code>zP</code> parametrize the <code>t</code>, <code>x</code>, <code>z</code> coordinates of the surface in terms of <code>y</code> and a parameter <code>u</code>. The surface normals function <code>normalsY</code> is updated to work with the parametrization in terms of <code>y</code> and <code>t</code>. This is the fastest way, but it does not produce a graph of <code>z</code> as a function of <code>t</code> and <code>x</code>. </p>
<pre><code>xOFty[t_, y_] := Evaluate[x /. Simplify @ First @ Solve[F == 0, x]];
tOFxy[x_, y_] := Evaluate[t /. Simplify @ First @ Solve[F == 0, t]];
xBrOFy[y_] :=
Evaluate[x /. Simplify @ First @ Solve[F == 0 /. First@Solve[D[F, y] == 0, t], x]];
xP[u_, y_, {{t1_, t2_}, {x1_, x2_}}] /; y == 0 := 0;
xP[u_, y_, {{t1_, t2_}, {x1_, x2_}}] /; u >= 0 && y == 1 := Rescale[u, {0, 1}, {x1, x2}];
xP[u_, y_, {{t1_, t2_}, {x1_, x2_}}] /; u >= 0 && y == -1 := Rescale[u, {0, 1}, {x2, x1}];
xP[u_, y_, {{t1_, t2_}, {x1_, x2_}}] /; u >= 0 && y > 0 :=
Rescale[u, {0, 1}, {xBrOFy[y], Min[x2, xOFty[t2, y]]}];
xP[u_, y_, {{t1_, t2_}, {x1_, x2_}}] /; u >= 0 && y < 0 :=
Rescale[u, {0, 1}, {xBrOFy[y], Max[x1, xOFty[t2, y]]}];
xP[u_, y_, {{t1_, t2_}, {x1_, x2_}}] /; u < 0 && y != 0 :=
Rescale[u, {-1, 0}, {-2 y, xBrOFy[y]}];
tP[u_, y_, {{t1_, t2_}, {x1_, x2_}}] /; y == 1 || y == -1 := 0;
tP[u_, y_, {{t1_, t2_}, {x1_, x2_}}] /; u >= 0 && y == 0 := Rescale[u, {0, 1}, {2, t2}];
tP[u_, y_, {{t1_, t2_}, {x1_, x2_}}] /; u < 0 && y == 0 := Rescale[u, {-1, 0}, {t1, 2}];
tP[u_, y_, {{t1_, t2_}, {x1_, x2_}}] /; -1 < y < 1 :=
tOFxy[xP[u, y, {{t1, t2}, {x1, x2}}], y];
zP[u_, y_, {{t1_, t2_}, {x1_, x2_}}] /; y == 1 || y == -1 := 0;(* limits *)
zP[u_, y_, {{t1_, t2_}, {x1_, x2_}}] :=
(1 - y^2)/(tP[u, y, {{t1, t2}, {x1, x2}}] - 2 (1 - y^2));
(* normals function*)
normalsY = Compile[{t0, x0, z0, u0, y0},
Sign[z0] {(-2 + t0 + 2 y0^2) z0 + 2 y0 (-1 + y0^2) (1 + 2 z0) ArcTanh[y0],
-2 (-1 + y0^2) (y0 + 2 y0 z0),
(-2 + t0 + 2 y0^2)^2}]
(* parametrization *)
domain = {{tMin, tMax}, {-2, 2}};
param[u_?NumericQ, y_?NumericQ] := {tP[u, y, domain], xP[u, y, domain], zP[u, y, domain]};
(* the plot *)
Show[ParametricPlot3D[param[u, y] /. y -> Sin[(Pi/2) v],
Evaluate@{u, Sequence @@ #}, {v, -1, 1},
PlotPoints -> {15, 27},
NormalsFunction -> (normalsY[#1, #2, #3, #4, Sin[(Pi/2) #5]] &),
Mesh -> None, PlotStyle -> Opacity[0.5], BoxRatios -> {1, 1, 0.5},
ColorFunction -> (ColorData["ThermometerColors"][Rescale[#3, {0, 1.5}]] &),
ColorFunctionScaling -> False, AxesLabel -> {t, x, z},
PlotRange -> {All, All, {-2.5, 2.5}}, ImageSize -> 300] & /@
{{-1, 0}, {0, 1}}
] // Timing
</code></pre>
<p>The output is shown in the Overview above.</p>
<hr>
<h2>Original Answer</h2>
<p><em>[Slightly modified]</em></p>
<p>First, in general, it's a good idea to use <code>_?NumericQ</code> on the arguments of <code>g</code>, as you can discover <a href="http://support.wolfram.com/kb/3820" rel="nofollow noreferrer">here</a>, as well as <a href="https://mathematica.meta.stackexchange.com/questions/436/candidate-topics-posts-for-canonical-questions/438#438">here</a> and <a href="https://mathematica.stackexchange.com/questions/18393/what-are-the-most-common-pitfalls-awaiting-new-users/26037#26037">here</a>.</p>
<p>Second, the <code>ContourPlot3D</code> should really be a <code>Plot3D</code>, since the equation is already in the form <code>z = f[t, x]</code>. <code>ContourPlot3D</code> will generally evaluate the function far more often and blindly (needlessly) than <code>Plot3D</code>.</p>
<p>Third, the equation has three branches (solutions) in part of the region in which you are plotting. Each is discontinuous along a branch line.</p>
<p>Fourth, <code>FindRoot</code> can be sped up by giving a good starting point for the search. (It can also be sped up by setting <code>AccuracyGoal</code> and/or <code>PrecisionGoal</code> lower, when as in plotting some accuracy can be spared. That doesn't matter too much here because the number of function evaluation is not extremely great -- but it can be if you increase <code>PlotPoints</code> or <code>MaxRecursion</code>.)</p>
<h3>Finding good starting points</h3>
<p>One can see the folds in the equation defining <code>g[x, t]</code> above.
In the region below <code>t == 2</code>, there are three solutions for <code>y</code> for a given <code>t</code>, <code>x</code>. Root-finding algorithms generally have trouble consistently finding the same branch unless the initial value is close to the desired root. This will be difficult near the boundary line, since two branches approach one another.</p>
<p>One can get good starting points from <code>ContourPlot3D</code> by extracting the points from the <code>GraphicsComplex</code> it constructs for the surface:</p>
<pre><code>yPts = First @ Cases[
ContourPlot3D[y == Tanh[(2*y + x)/t],
{t, 0.001, 3}, {x, -1.5, 1.5}, {y, -1.5, 1.5},
PlotPoints -> 200, MaxRecursion -> 0, Mesh -> None],
GraphicsComplex[pts_, rest___] :> pts, Infinity]
</code></pre>
<p>We can use <code>Nearest</code> to construct a function that returns the <code>y</code> coordinates for the points nearest a given <code>t</code>, <code>x</code>:</p>
<pre><code>nearestY = Nearest[Most /@ yPts -> Last /@ yPts]
</code></pre>
<p>Among the value(s), return we can select the one closest to the branch, where <code>branch</code> is a number close to the desired branch:</p>
<pre><code>Nearest[nearestY[{t, x}, 15], branch][[1]] (* why 15? *)
</code></pre>
<p>Why pick the 15 closest points? The closest point might not correspond to a value of <code>y</code> on the desired branch. This turns out to be true sometimes even for the 3 nearest values. As the number values grows large, and the number of initial values on the desired branch grows, and more likely it is that the chosen initial value leads <code>FindRoot</code> astray to the wrong branch. In this case 5-20 work well; that number depends on the number of <code>yPts</code>, too.</p>
<h3>Plotting the graph</h3>
<p>As I said above, one should use <code>Plot3D</code> instead of <code>ContourPlot3D</code> in this case. Here I define a function to plot a branch. A branch will be defined by <code>branch</code> being equal to <code>-1.</code>, <code>0</code>, or <code>1</code>. The line of code with <code>Sow</code> in it sows the initial points for the points <code>{t, x}</code> at which <code>Plot3D</code> calls <code>g[x, t]</code> -- this may be omitted, but I used it to examine the relation between the initial values and the discontinuities in the plot (<code>Reap</code> will harvest the sown points).</p>
<pre><code>plot[branch_] := Module[{g},
g[x_?NumericQ, t_?NumericQ] :=
y /. FindRoot[y == Tanh[(2*y + x)/t],
{y,
(Sow[{t, x, #}]; #) &@ (* this line can be removed *)
Nearest[nearestY[{t, x}, 15], branch][[1]]}]; (* why 15 *)
Plot3D[{Sech[(2 g[x, t] + x)/t]^2/(t - 2 Sech[(2 g[x, t] + x)/t]^2)},
{t, 0.001, 3}, {x, -1.5, 1.5},
Mesh -> None,
PlotStyle -> Opacity[0.5], ColorFunction -> (ColorData["ThermometerColors"][2 #3 - 1] &),
PlotRange -> {-2.5, 2.5}, ClippingStyle -> None, AxesLabel -> {t, x, z}]
]
</code></pre>
<h3>Initial evaluation</h3>
<p>I plotted the branches together with the sample points and initial <code>y</code> values. Here's is the branch "0". It is clearly much quicker than 10 minutes!</p>
<pre><code>{p1, yp} = Reap@plot[-0.]; // Timing
</code></pre>
<blockquote>
<pre><code>{1.754386, Null}
</code></pre>
</blockquote>
<pre><code>Show[p1, Graphics3D[{Red, Point@yp}]]
</code></pre>
<p><img src="https://i.stack.imgur.com/kG5LV.png" alt="Branch 0 & sample points"></p>
<p>One can see that <code>FindRoot</code> does a good (and quick) job finding the right branch, except close to the discontinuity along the branch line. I think the spikes are due to <code>FindRoot</code> finding the wrong branch (and the granularity of plotting) -- I'll leave others to think through that.</p>
<h3>All three branches</h3>
<pre><code>Show[#, ImageSize -> 200] & /@ plot /@ {-1., 0., 1.} // Row
</code></pre>
<p><img src="https://i.stack.imgur.com/0XYtm.png" alt="All three branches"></p>
|
3,592,368 | <p>I managed to solve this question using trigonometry. But I wondered if there'd be anyway of doing it using only synthetic geometry. Here it is.</p>
<blockquote>
<p>Let <span class="math-container">$ABC$</span> be a right isosceles triangle of hypotenuse <span class="math-container">$AB$</span>. Let also <span class="math-container">$\Gamma$</span> be the semicircle whose diameter is the line segment <span class="math-container">$AC$</span> such that <span class="math-container">$\Gamma\cap\overline{AB} = \{A\}$</span>. Consider <span class="math-container">$P\in\Gamma$</span> with <span class="math-container">$PC = k$</span>, with <span class="math-container">$k \leq AC$</span>. Find the area of triangle <span class="math-container">$PBC$</span>.</p>
</blockquote>
<p>Here is my interpretation of the picture:
<a href="https://i.stack.imgur.com/Vz0ck.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Vz0ck.png" alt="enter image description here" /></a></p>
<p>I managed to get the solution via trigonometry as below.</p>
<p><a href="https://i.stack.imgur.com/dQ8hU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dQ8hU.png" alt="enter image description here" /></a></p>
<p>Then, the area <span class="math-container">$S$</span> requested is:</p>
<p><span class="math-container">$$\begin{align} S &= \displaystyle\frac{PC\cdot BC\cdot \sin(90^\circ + \beta)}{2}\\
&= \displaystyle\frac{k\cdot d\cdot \cos\beta}{2}\\
&= \displaystyle\frac{k\cdot d\cdot \frac{k}{d}}{2}\\
&= \displaystyle\frac{k^2}{2}.\\
\end{align}$$</span></p>
| Michael Rozenberg | 190,319 | <p>Let <span class="math-container">$PD$</span> be a perpendicular from <span class="math-container">$P$</span> to the line <span class="math-container">$BC$</span>.</p>
<p>Thus, since a line <span class="math-container">$BC$</span> is a tangent to the circle and <span class="math-container">$\measuredangle DCP=\measuredangle PAC$</span>, we see that <span class="math-container">$\Delta DCP\sim\Delta PAC,$</span> </p>
<p>which gives:
<span class="math-container">$$\frac{PD}{k}=\frac{k}{AC}$$</span> or
<span class="math-container">$$PD=\frac{k^2}{AC}.$$</span>
Id est, <span class="math-container">$$S_{\Delta PBC}=\frac{1}{2}BC\cdot PD=\frac{1}{2}BC\cdot\frac{k^2}{AC}=\frac{k^2}{2}.$$</span></p>
|
701,677 | <ol>
<li><p>If I know something about the representation theory of the general linear group $\mathrm{GL}_n(\mathbb C)$, what can I say about the representation theory of the unitary group $\mathrm U_n(\mathbb C)$? E.g. branching rules</p></li>
<li><p>Similarly for the groups $\mathrm{SO}_n(\mathbb C) \subset \mathrm O_n(\mathbb C)$?</p></li>
<li><p><strike>Is module over ${\mathbb C\mathrm{GL}_n(\mathbb C)}$ that is not semi-simple necessarily infinite-dimensional? How does it typically look like?</strike> See <a href="https://math.stackexchange.com/questions/707911/how-do-non-semisimple-modules-over-mathbb-c-mathrmgl-n-mathbb-c-look-like">new Question here</a></p></li>
<li><p><strike>If I have information about the restriction of representations of the general linear group, can I make any statements about the induction (by Frobenius reciprocity)? E.g. I know $$\mathrm{res}^{\mathrm{GL}_n}_{\mathrm{GL}_k\times \mathrm{GL}_{n-k}} V(\lambda)_n \cong \bigoplus_{\alpha, \beta} c_{\alpha, \beta}^\lambda V(\alpha)_k \otimes V(\beta)_{n-k}$$
where $V(\lambda)_n$ is the irreducible polynomial representations corresponding to a partition (or Young diagram) $\lambda$ of $\mathrm{GL}_n(\mathbb C)$ and $c^\lambda_{\alpha,\beta}$ are the Littlewood-Richardson numbers. Is it true that
$$ \mathrm{ind}_{\mathrm{GL}_k\times \mathrm{GL}_{n-k}}^{\mathrm{GL}_n} V(\alpha)_k \otimes V(\beta)_{n-k} \cong \bigoplus_\lambda c_{\alpha,\beta}^\lambda V(\lambda)_n ?$$
(I know it is not but true but it should be true up to being semi-simple.)</strike> See <a href="https://math.stackexchange.com/questions/707929/reciprocity-for-branching-rules-of-mathrmgl-n-mathbb-c">new question here</a></p></li>
</ol>
| goblin GONE | 42,339 | <p>Assume for a contradiction that $k \in A \cap B$ and that $H$ is surjective. Then $H(K) = \langle \{k\}_A,\emptyset_B\rangle$ for some $K : A \cup B \rightarrow 2.$ Fix any such $K$.</p>
<p>Case 1. Assume $k \in K$. Then $k \in [H(K)]_0$ and $k \in [H(K)]_1$, thus $H(K) \neq \langle \{k\}_A,\emptyset_B\rangle,$ a contradiction.</p>
<p>Case 2. Assume $k \notin K$. Then $k \notin [H(K)]_0$ and $k \notin [H(K)]_1$, thus $H(K) \neq \langle \{k\}_A,\emptyset_B\rangle,$ a contradiction.</p>
<p><strong>In conclusion:</strong> if $H$ is surjective, then $A \cap B$ must be empty.</p>
|
1,218,912 | <p>My problem is following:
$$\binom{n}{r} + \binom{n+1}{r+1} + \binom{n+2}{r+2} + \dots + \binom{n+M}{r+M}$$</p>
<p>how can we reduce it to a more short solution</p>
<p>Here $\dbinom{n}{r} = \dfrac{n!}{r! (n-r)!}$ and thus same as regular.Please help me in solving the above expression</p>
| Brian M. Scott | 12,042 | <p>$$\begin{align*}
\sum_{k=0}^m\binom{n+k}{r+k}&=\sum_{k=0}^m\binom{n+k}{n-r}\\
&=\sum_{\ell=r}^{m+r}\binom{n-r+\ell}{n-r}\\
&=\sum_{\ell=0}^{m+r}\binom{n-r+\ell}{n-r}-\sum_{\ell=0}^{r-1}\binom{n-r+\ell}{n-r}\\
&=\binom{n+m+1}{n-r+1}-\binom{n}{n-r+1}\\
\end{align*}$$</p>
<p>by a standard identity.</p>
|
78,461 | <p>I'm trying to solve the following problem:</p>
<p>For a vector $v$ of length $c$, $\min \frac{\sum v[i]^4}{(\sum v[i]^2)^2}$ subject to $\sum v[i] = N$.</p>
<p>I can solve this numerically for a given $c$ using the following command: </p>
<pre><code>Minimize[{Total[z^4]/Total[z^2]^2, Total[z] == N}, z]
</code></pre>
<p>Is it possible to solve this problem symbolically leaving $c$ as a free variable? </p>
| David G. Stork | 9,735 | <p>Isn't this is a problem of mathematics more than programming? Under the stated conditions, by permutation symmetry of the indexes of $v$ all entries of $v$ should be the same (up to absolute value, because in the problem statement all elements are squared). Thus one answer is $v$ is an array of $c$ entries each of which is $n/c$. This solution is confirmed through code for $n = 1, 2$ and $3$. </p>
<p>According to @2012rcampion, these are other solutions for odd $c$:</p>
<pre><code>otherSolutions[c_?OddQ] :=
Permutations[Join[Table[n, {c/2 + 1/2}], Table[-n, {c/2 - 1/2}]]]
</code></pre>
|
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