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 |
|---|---|---|---|---|
885,129 | <p>I want to speed up the convergence of a series involving rational expressions the expression is $$\sum _{x=1}^{\infty }\left( -1\right) ^{x}\dfrac {-x^{2}-2x+1} {x^{4}+2x^{2}+1}$$
If I have not misunderstood anything the error in the infinite sum is at most the absolute value of the last neglected term. The formula for the $n$th term is $\dfrac {-x^{2}-2x+1} {x^{4}+2x^{2}+1}$ from the definition of the series. To get the series I used Maxima the computer algebra system. I have noticed that to get 13 decimal places of the series one must wade through $312958$ terms of the series. I had to kill the computer GUI and some other system processes and run Maxima to compute the sum. I took about 5 minutes. The final sum I obtained was $0.3106137076850$. Is there any way to speed up the convergence of the sum? In general is there any way to speed up the convergence of the sum of $$\sum _{x=1}^{\infty }\left( -1\right) ^{x}\dfrac {p(x)} {q(x)}$$ </p>
<p>where both ${p(x)}$ and ${q(x)}$ are rational functions?</p>
| Claude Leibovici | 82,404 | <p>For the summation $$S=\sum _{n=1}^{\infty }\left( -1\right) ^{n}\dfrac {-n^{2}-2n+1} {n^{4}+2n^{2}+1}$$ a CAS gave as a result $$\frac{1}{8} \left(2 \pi ^2 \text{csch}^2\left(\frac{\pi }{2}\right)+(1-i) \left(\psi
^{(1)}\left(\frac{1}{2}-\frac{i}{2}\right)+i \left((-4+4 i)+\psi
^{(1)}\left(\frac{1}{2}+\frac{i}{2}\right)+\psi
^{(1)}\left(1-\frac{i}{2}\right)\right)+\psi
^{(1)}\left(1+\frac{i}{2}\right)\right)\right)$$ which is approximately $0.31061370769015654201991515962234635408157816305055$ (this could be computed for as many significant figures as required).</p>
<p>Considering Brad's result, for </p>
<p>$$T=\sum_{n=0}^\infty(-1)^n\frac{2 (n-1)}{\left(n^2+1\right)^2}$$ a CAS gave $$\left(-\frac{1}{8}+\frac{i}{8}\right) \left((2+2 i)+(1+i) \pi \left(\pi
\text{csch}^2\left(\frac{\pi }{2}\right)+2 \text{csch}(\pi )\right)+\psi
^{(1)}\left(\frac{1}{2}-\frac{i}{2}\right)+i \psi
^{(1)}\left(\frac{1}{2}+\frac{i}{2}\right)+i \psi
^{(1)}\left(1-\frac{i}{2}\right)+\psi ^{(1)}\left(1+\frac{i}{2}\right)\right)$$</p>
|
703,125 | <blockquote>
<p>Let $\{A_\alpha\}$ be a collection of connected subspaces of $X$; let
$A$ be connectted subspace of $X$. Show that if $A\cap A_\alpha \neq
\emptyset$ for all $\alpha$, then $A\cup(\cup A_\alpha)$ is connected.</p>
</blockquote>
<p>I know this theorem:<img src="https://i.stack.imgur.com/6tILz.png" alt="enter image description here"></p>
<p>And as every set in this union has a point in common with $A$, I guess I need to use this. If $\{U_\alpha \}$ was countable I could easily use this theorem with induction, but now I'm not sure if I can use such an argument.</p>
<p>I mean $A\cup A_1$ is connected, therefore $A\cup A_1 \cup A_2$ connected and so on.</p>
| dani_s | 119,524 | <p>Note that $A \cup A_{\alpha}$ is connected for all $\alpha$ and therefore $A \cup (\bigcup A_{\alpha}) = \bigcup (A \cup A_{\alpha})$ is connected by your theorem because it is the union of connected sets that have $A$ in common.</p>
|
507,454 | <p>I had a geometry class which was proctored using the Moore method, where the questions were given but not the answers, and the students were the source of all answers in the class. One of the early questions which we never solved is listed in the title.</p>
<p>In this case, use any reasonable definition of "between-ness". I believe the definition we used was "$B$ is between $A$ and $C$ if and only if $|AB|+|BC|=|AC|$". A collineation is a mapping where every line is mapped to a line. A mapping is a function that operates over the set of points within the given space and returns points in the given space.</p>
<p>When we were studying this question, we managed to get to the point that a between-ness mapping must map lines to line segments. In particular, we had managed to show that for every $A-B-C$ (read: "$B$ is between $A$ and $C$") in the mapped space by applying between-ness preserving mapping $m$, we could guarantee that pre-images $P_A, P_C$ such that $m(P_A)=A,m(P_C)=C$ implies the existence of pre-image $P_B$ such that $P_A-P_B-P_C$ and $m(P_B)=B$. I have never seen any full proof of the title statement.</p>
<p>I would like to read any hints that are known for solving this question. Feel free to completely solve it, but please hide the full solution in such a way that I can start with your hint(s) and have an opportunity to finish the solution for myself.</p>
| Zarrax | 3,035 | <p>Hint: Show $p^2 = 1 \mod 3$ and $p^2 = 1 \mod 8$ for any such $p$.</p>
|
3,739,144 | <p>I have the following proposition proved in my lectures notes, but I think there are a couple of errors and there is one think I don't get:</p>
<p>If <span class="math-container">$p_n$</span> is a Cauchy sequence in a metric space <span class="math-container">$(X,d)$</span>, the set <span class="math-container">$\{p_n| n \in \mathbb{N}\}$</span> is bounded. Moreover, if <span class="math-container">$p_n$</span> has <span class="math-container">$p_0$</span> as a limit point, then <span class="math-container">$p_n$</span> converges to <span class="math-container">$p_0$</span> <strong>----->(1)</strong></p>
<p>Proof</p>
<p>Let <span class="math-container">$\epsilon > 0$</span>. Now for each <span class="math-container">$n \geq n_{\epsilon}$</span>, <span class="math-container">$d(p_n, p_{n_{\epsilon}}) < \epsilon$</span>. Then <span class="math-container">$p_n \in B_d(p_{n_{\epsilon}}, \epsilon)$</span> for each <span class="math-container">$n\geq n_{\epsilon}$</span>, from which the thesis follows.
Moreover, if <span class="math-container">$p_0$</span> is the limit of subsequence <span class="math-container">$p_{n_k}$</span>, for <span class="math-container">$\epsilon >0$</span>, there exists <span class="math-container">$n_{\epsilon}$</span> such that <span class="math-container">$d(p_{n_k},p_0)< \epsilon/2$</span> and <span class="math-container">$d(p_{n_k},p_n)< \epsilon/2$</span> for <span class="math-container">$n\geq n_{\epsilon}$</span> and <span class="math-container">$k\geq n_{\epsilon}$</span>. <strong>----->(2)</strong></p>
<p>Then,</p>
<p><span class="math-container">$d(p_n,p_0) \leq d(p_n,p_{n_k})+d(p_{n_k},p_0) < \epsilon$</span> for <span class="math-container">$n, k \geq n_{\epsilon}$</span>
<strong>----->(3)</strong></p>
<p>from which the thesis follows</p>
<p><strong>I think there are some errors in this proof. I would like some feedback/confirmation on it</strong></p>
<p>(1) --> it should say " if <strong><span class="math-container">$p_{n_k}$</span></strong> has <span class="math-container">$p_0$</span> as a limit point" instead of " if <strong><span class="math-container">$p_{n}$</span></strong> has <span class="math-container">$p_0$</span> as a limit point"</p>
<p>(2), (3), it should be <span class="math-container">$n_k \geq n_{\epsilon}$</span>, instead of <span class="math-container">$k \geq n_{\epsilon}$</span></p>
<p><strong>And a small question that is really bothering me :</strong></p>
<p>For all <span class="math-container">$\epsilon > 0$</span>, If <span class="math-container">$p_n$</span> is a Cauchy sequence, the definition says an <span class="math-container">$n_{\epsilon}$</span> exist , such that for all <span class="math-container">$n_1, n_2 >n_{\epsilon}$</span> <span class="math-container">$d(p_{n_1},p_{n_2})<\epsilon$</span>.</p>
<p>The definition of convergence says: If <span class="math-container">$p_n$</span> converges to <span class="math-container">$p_0$</span>, then for all <span class="math-container">$\epsilon > 0$</span>, there exist and <span class="math-container">$n_{\epsilon}$</span>, such that for any <span class="math-container">$n > n_{\epsilon}$</span>, <span class="math-container">$d(p_n, p_{n_{\epsilon}})<\epsilon$</span>.</p>
<p>When they mix the definitions to prove the second part of the proposition, the one about the subsequence, they consider the <span class="math-container">$\epsilon$</span> and the <span class="math-container">$n_{\epsilon} $</span> of both definitions to be the same. I agree for the <span class="math-container">$\epsilon$</span>, because since it must be true for all <span class="math-container">$\epsilon $</span> I can choose them to be equal, but then what guarantees that the <span class="math-container">$n_{\epsilon} $</span> of the first definition is the same as the <span class="math-container">$n_{\epsilon} $</span> of the second definition ? The definitions just state some <span class="math-container">$n_{\epsilon} $</span> exists. It is assumed they are equal without saying why and I really can't figure it out.</p>
| Henno Brandsma | 4,280 | <p>Why not say : suppose <span class="math-container">$(p_n)$</span> is Cauchy. Then apply the definition of Cauchy-ness to <span class="math-container">$\varepsilon=1$</span> and we find <span class="math-container">$N \in \mathbb{N}$</span> such that</p>
<p><span class="math-container">$$\forall n,m \ge N: d(p_n, p_m) < 1$$</span></p>
<p>So we define <span class="math-container">$$M=\max\{d(p_1, p_N), \ldots, d(p_{N_1}, p_N)\}+1$$</span></p>
<p>and we have that <span class="math-container">$$\{p_n: n \in \mathbb{N}\} \subseteq B(p_N, M)$$</span></p>
<p>and so the range of the sequence is bounded.</p>
<p>If moreover <span class="math-container">$p_0$</span> is a limit point of <span class="math-container">$\{p_n: n \in \mathbb{N}\}$</span>. Then in fact <span class="math-container">$p_n \to p_0$</span> as well: let <span class="math-container">$\varepsilon >0$</span>. We find <span class="math-container">$N_1$</span> (from the Cauchy definition applied to <span class="math-container">$\frac{\varepsilon}{2}>0$</span>) such that</p>
<p><span class="math-container">$$\forall n,m \ge N_1: d(p_n, p_m) < \frac{\varepsilon}{2} .$$</span></p>
<p>As <span class="math-container">$p_0$</span> is a limit point of the range, <span class="math-container">$B(p_0, \frac{\varepsilon}{2})$</span> contains infinitely many <span class="math-container">$p_n$</span> (i.e. for infinitely many indices). So pick <span class="math-container">$p_k \in B(p_0, \frac{\varepsilon}{2})$</span> with <span class="math-container">$k \ge N_1$</span>.</p>
<p>And then for <span class="math-container">$n \ge N_1$</span> we have</p>
<p><span class="math-container">$$d(p_n,p_0) \le d(p_n, p_k) + d(p_k, p_0) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$</span></p>
<p>and as we found such <span class="math-container">$N_1$</span> for arbitrary <span class="math-container">$\varepsilon>0$</span>, <span class="math-container">$p_n \to p_0$</span> as promised.</p>
|
265 | <p>I'm a mod at Computational Science SE. Sometimes, users ask questions on Computational Science about doing something in Wolfram Alpha. As far as I can tell, Wolfram Alpha uses Mathematica for its math engine. Are those questions on topic here?</p>
<p>Note: I was made aware of <a href="https://mathematica.meta.stackexchange.com/questions/68/other-wri-product-discussion">Other WRI product discussion?</a>. I'm scoping my question specifically to uses of Wolfram Alpha that look something like <a href="https://scicomp.stackexchange.com/questions/1407/solving-system-of-equations-with-wolfram-alpha">https://scicomp.stackexchange.com/questions/1407/solving-system-of-equations-with-wolfram-alpha</a>, where the question is talking about using Wolfram Alpha for mathematics only. </p>
| rm -rf | 5 | <p>I would consider such questions explicitly <strong>off topic</strong> for this site. This site is for users of <em>Mathematica</em> (mma), not Wolfram|Alpha (W|A) and so any question that does not involve the former is out of this site's scope. I know that W|A runs on mma and it kinda sorta understands mma syntax. However, opening the door to such questions will only lower the bar and result in hit-and-run questions from folks who just want a quick result from W|A.</p>
<p>Along the same lines, questions that merely concern the <code>Free-form input (=)</code> or the <code>Wolfram|Alpha query (==)</code>, which is available from within mma from v8 onwards, without any further processing in mma should also be explicitly off-topic, as this is just a back-door for W|A questions by masking it in an mma call.</p>
<p>On the other hand, questions that involve interfacing with W|A from mma, and using the results from certain pods for further analysis/manipulation in mma should be considered on-topic. Vitaliy Kaurov's <a href="https://stackoverflow.com/a/8959477">answer here</a> on Stack Overflow is a good example of such questions (although I note that the question itself did not ask for a W|A based solution).</p>
|
3,117,459 | <p>I am interested in approximating the natural logarithm for implementation in an embedded system. I am aware of the Maclaurin series, but it has the issue of only covering numbers in the range (0; 2).</p>
<p>For my application, however, I need to be able to calculate relatively precise results for numbers in the range (0; 100]. Is there a more efficient way of doing so than decomposing each number greater than 2 into a product of factors in the (0; 2) range and summing up the results of the Maclaurin series for each factor?</p>
| Oscar Lanzi | 248,217 | <p>In theory, <a href="https://math.stackexchange.com/a/3227686/248217">a series convergent for all positive arguments exists:</a>:</p>
<p><span class="math-container">$\ln(x)=2\sum_{m=1}^\infty {\dfrac{(\frac{x-1}{x+1})^{2m-1}}{2m-1}}$</span></p>
<p>In practice, Arthur's approach of halving the argument until we get to the Taylor series convergent range (argument <span class="math-container">$<2$</span>) and adding back the appropriate number of <span class="math-container">$\ln 2$</span> terms seems more likely to gain use.</p>
<p>We may also reduce the argument per Arthur's method and then use the series above, which converges faster than the unmodified Taylor series for arguments between <span class="math-container">$1$</span> and <span class="math-container">$2$</span>.</p>
|
1,722,692 | <p>I am asked to find</p>
<p>$$\lim_{x \to 0} \frac{\sqrt{1+x \sin(5x)}-\cos(x)}{\sin^2(x)}$$</p>
<p>and I tried not to use L'Hôpital but it didn't seem to work. After using it, same thing: the fractions just gets bigger and bigger.</p>
<p>Am I missing something here?</p>
<p>The answer is $3$</p>
| carmichael561 | 314,708 | <p>I think using Taylor expansions is the best way to find the limit, but here is a more elementary approach: multiply the top and bottom by the conjugate to obtain
$$ \lim_{x\to 0}\frac{\sqrt{1+x\sin(5x)}-\cos(x)}{\sin^2(x)}=\lim_{x\to 0}\frac{(1-\cos^2x)+x\sin(5x)}{\sin^2(x)[\sqrt{1+x\sin(5x)}+\cos(x)]} $$</p>
<p>Now use $1-\cos^2x=\sin^2x$ and split the fraction into two pieces:
$$\lim_{x\to 0}\frac{(1-\cos^2x)+x\sin(5x)}{\sin^2(x)[\sqrt{1+x\sin(5x)}+\cos(x)]}$$
$$=\lim_{x\to 0}\frac{1}{\sqrt{1+x\sin(5x)}+\cos(x)}+\frac{x\sin(5x)}{\sin^2(x)[\sqrt{1+x\sin(5x)}+\cos(x)]}$$</p>
<p>The first term has limit $\frac{1}{2}$, and the second has limit $\frac{5}{2}$, using the fact that
$$ \lim_{x\to 0}\frac{\sin x}{x}=1 $$
and
$$ \lim_{x\to 0}\frac{\sin(5x)}{\sin x}=5\lim_{x\to 0}\frac{\sin(5x)}{5x}\cdot \frac{x}{\sin x}=5$$
So the result is $\frac{6}{2}=3$.</p>
|
4,000,935 | <p>How do you show that <span class="math-container">$7a^2 - 12b^2 = 8c^2$</span> has no integer solutions</p>
<hr />
<p>When <span class="math-container">$x = a/c$</span> and <span class="math-container">$y = b/c$</span> then <span class="math-container">$\gcd(a,b,c) = 1$</span> I believe.</p>
<p>If we use mod5, and neither <span class="math-container">$a,b,c$</span> is divisible by 5 then <span class="math-container">$7a^2 - 12b^2$</span> can have remainder 1 or 4 and <span class="math-container">$8c^2$</span> can only have remainder 2 or 3. But if either <span class="math-container">$a$</span> or <span class="math-container">$b$</span> is divisible by 5 then <span class="math-container">$7a^2 - 12b^2$</span> will have remainder 2 or 3.</p>
<p>I've considered using mod7 so <span class="math-container">$-12b^2$</span> would have remainder either <span class="math-container">$1,2,$</span> or <span class="math-container">$4.$</span> But <span class="math-container">$8c^2$</span> can have remainder 1,2 or 4 as well.</p>
<p>I'm not really sure how to go about this, any help would be appreciated thanks.</p>
| player3236 | 435,724 | <p>WLOG, assume that <span class="math-container">$\gcd (a,b,c) = 1$</span> as you did.</p>
<p>Taking modulo <span class="math-container">$3$</span> gives <span class="math-container">$a^2 \equiv 2c^2 \pmod 3$</span>.</p>
<p>As <span class="math-container">$n^2 \equiv 0,1 \pmod 3$</span>, we must have <span class="math-container">$a \equiv c \equiv 0 \pmod 3$</span>.</p>
<p>Hence <span class="math-container">$3 \nmid b$</span>, and <span class="math-container">$a^2 \equiv c^2 \equiv 0 \pmod 9$</span>.</p>
<p>Now taking modulo <span class="math-container">$9$</span>, we have:</p>
<p><span class="math-container">$$-3b^2 \equiv 0 \pmod 9$$</span></p>
<p>suggesting that <span class="math-container">$3 \mid b$</span>, which is a contradiction.</p>
|
1,726,416 | <p>$\displaystyle \sum_{n=1}^∞ (-1)^n\dfrac{1}{n}.\dfrac{1}{2^n}$</p>
<p>Knowing that</p>
<ol>
<li>An alternating harmonic series is always convergent</li>
<li>Riemann series are always convergent when $p>1$</li>
</ol>
<p>Is it safe to say that the product of these two is convergent (as described above)?</p>
| Jean Marie | 305,862 | <p>@Kf-Sansoo @Joanpemo</p>
<p>In fact, in my opinion, the right "product" here is the <strong>dot product</strong> $U.V$ in Hilbert space $\ell^2$, which is meaningful because </p>
<p>$$U=(-1, \dfrac{1}{2},-\dfrac{1}{3},\cdots(-1)^n \dfrac{1}{n},\cdots) \ \ and \ \ V=(1,\dfrac{1}{2},\dfrac{1}{4},\cdots\dfrac{1}{2^n}\cdots)$$</p>
<p>are square sommable with respective <em>squared</em> norm $\dfrac{\pi^2}{6}$ (classical sum) and $\dfrac{4}{3}$ (sum of a geometric series with ratio $\dfrac{1}{4}$).</p>
<p>Besides, there is another reason for which this series a sommable: it is the value for $x=1/2$ of the entire series $\sum_{n=1}^{\infty}\dfrac{(-x)^n}{n}$ which is known to be $ln(1+x)$; therefore, the value of the dot product is $ln(1+\dfrac{1}{2})=ln(\dfrac{3}{2})$.</p>
|
210,401 | <p>In other words, given a sequence $(s_n)$, how can we tell if there exist irrationals $u>1$ and $v>1$ such that </p>
<p>$$s_n = \lfloor un\rfloor + \lfloor vn\rfloor$$</p>
<p>for every positive integer $n$?</p>
<p>A few thoughts: Graham and Lin (<em>Math. Mag.</em> 51 (1978) 174-176) give a test for $(s_n)$ to be a single Beatty sequence $(\lfloor un\rfloor)$ (which they call the spectrum of $u$). Perhaps someone knows a reference for a test for sums of two or more Beatty sequences? A special case would be a test for a given sequence $(s_n)$ to be the sum of two <em>complementary</em> Beatty sequences (i.e., $1/u + 1/v = 1$).</p>
<p>In response to comments, the part of the question that says "for every positive integer n" indicates that the intended sequence is infinite. It seems to me that the question, as stated above, is okay. If it's undecidable - well, that's of interest. </p>
<p>In any case, while it may be difficult to give a test that actually finds $u$ and $v$ when such numbers exist, there are some simple tests for deciding that $(s_n)$ is <i>not</i> a sum of two Beatty sequences:</p>
<p>(1) $\lim_{n\to\infty}s(n)/n$ must exist;</p>
<p>(2) if $u$ and $v$ exist, then $u$ + $v$ = $\lim_{n\to\infty}s(n)/n$; </p>
<p>(3) $\lfloor(u+v)n\rfloor \in \{[un]+[vn],[un]+[vn]+1\}$ for every $n$;</p>
<p>(4) if $(s_n)$ is a sum of two Beatty sequences, then the difference sequence of $(s_n)$ consists of at most three terms; and if there are three, then they are consecutive integers.</p>
<p>It's easy to see how each of those generalizes to give a "negative test"; that is, a way to see that a given $(s_n)$ is not a sum of any prescribed number of Beatty sequences. I hope that someone can find more "negative tests", or even better, a "positive test", perhaps similar to Graham and Lin's result. </p>
| Gerhard Paseman | 3,206 | <p>This is the closest I can come to a positive test, but I don't know how well it will work for you. It is essentially taking the intersection of possible solution sets.</p>
<p>Let us cut down on symmetry by assuming $u \geq v \gt 1$. (I suppose if $u=v$ that would be a tipoff.) For each positive integer $i$, intersect the $u-v$ domain assumed above with the set of $(u,v)$ such that $s_i = \lfloor ui \rfloor + \lfloor vi \rfloor$. If the intersection is nonempty, then there is $(u,v)$ in the intersection which satisfies all the relations, and that gives you a positive result. It is unclear if there will be more than one pair though. It is also not clear if you can use this to predict the first few decimals of either $u$ or $v$.</p>
<p>Gerhard "Uncertain If It Is Positive" Paseman, 2015.06.29</p>
|
4,290,651 | <p>I understand the standard proof that there exists no surjection <span class="math-container">$f: X \to \mathcal{P}(X)$</span>, but I'm not able to tell whether it deals with the case that <span class="math-container">$X = \emptyset$</span> or whether I need to rule this out separately.</p>
<p>If I want to prove that <span class="math-container">$|X| < |\mathcal{P}(X)|$</span>, I need to find an injection <span class="math-container">$X \hookrightarrow \mathcal{P}(X)$</span>. In this case, I'm almost certain that I need to rule out the empty set case first. If <span class="math-container">$X = \emptyset$</span>, then the only map <span class="math-container">$X \to \mathcal{P}(X)$</span> is the empty function with codomain <span class="math-container">$\{\emptyset\}$</span>, which is vacuously injective. Otherwise, I send <span class="math-container">$x \mapsto \{x\}$</span> for each <span class="math-container">$x \in X$</span>, which is injective.</p>
<p>The proof that no surjection is tougher for me to rule out the case of the empty set.</p>
<blockquote>
<p>Suppose <span class="math-container">$f: X \to \mathcal{P}(X)$</span> is a surjection. Define <span class="math-container">$B = \{x \in X \mid x \not \in f(x)\}$</span>. As <span class="math-container">$f$</span> is surjective, <span class="math-container">$f(a) = B$</span> for some <span class="math-container">$a \in X$</span>. But then <span class="math-container">$a \in B \iff a \not \in f(a) \iff a \not \in B$</span>, which is a contradiction.</p>
</blockquote>
<p>If <span class="math-container">$X$</span> is empty, then <span class="math-container">$B$</span> is empty. I can't find an <span class="math-container">$a \in X$</span>, so that away is ruled out, but this may be a case where the statement is "vacuously" true because the definition of surjectivity starts with "for all."</p>
| Community | -1 | <ul>
<li><p>I don't see why you would need to make a special case for an injection <span class="math-container">$\emptyset\to \{\emptyset\}$</span>, because the function <span class="math-container">$f:\emptyset\to\{\emptyset\}$</span> such that <span class="math-container">$f(x)=\{x\}$</span> for all <span class="math-container">$x\in\emptyset$</span> is already the empty map.</p>
</li>
<li><p>Existence of some <span class="math-container">$a\in X$</span> such that <span class="math-container">$f(a)=\{x\in X\,:\, x\notin f(x)\}$</span> comes from the assumption of <span class="math-container">$f$</span> being surjective. In specific cases this may result in a contradiction by other means in addition to proving that <span class="math-container">$a\in f(a)\leftrightarrow a\notin f(a)$</span>, but the general way is still valid here: the same proof <span class="math-container">$$\exists f\text{ surjective}\Rightarrow \exists f\text{ surjective},\exists a\in X, (a\in f(a)\leftrightarrow a\notin f(a))$$</span> works for <span class="math-container">$X=\emptyset$</span> as well.</p>
</li>
</ul>
|
3,072,995 | <p>The only thing I know with this equation is <span class="math-container">$y=\frac{x^2+1}{x+1}=x+1-\frac{2x}{x+1}$</span>.</p>
<p>Maybe it can be solved by using inequality.</p>
| 5xum | 112,884 | <p>Since <span class="math-container">$$\lim_{x\to-\infty} \frac{x^2+1}{x+1} = -\infty,$$</span></p>
<p>the minimum does not exist.</p>
|
2,370,716 | <p>Give the postive integer $n\ge 2$,and $x_{i}\ge 0$,such $$x_{1}+x_{2}+\cdots+x_{n}=1$$
Find the maximum of the value
$$x^2_{1}+x^2_{2}+\cdots+x^2_{n}+\sqrt{x_{1}x_{2}\cdots x_{n}}$$</p>
<p>I try
$$x_{1}x_{2}\cdots x_{n}\le\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)^n=\dfrac{1}{n^n}$$</p>
| Lazy Lee | 430,040 | <p>Let $x_1\geq x_2 \geq ... \geq x_n$ without loss of generality. Notice that $$\begin{split}F(x_1,...)=\left(\sum_{i=0}^n x_i\right)^2-\sum_{i=0}^n x_i^2 - \sqrt{\prod_{i=0}^n x_i} &= 2\cdot\sum_{i<j}x_ix_j-\sqrt{\prod_{i=0}^n x_i} \\
&=\left(2\sum_{i>1}x_i\right)\cdot x_1-\sqrt{\prod_{i=2}^n x_i}\cdot\sqrt{x_1} + 2\sum_{i>j>1}x_ix_j\end{split}$$</p>
<p>This is a quadratic function of $\sqrt{x_1}$. We can then show that $F\geq0$ by showing that the quadratic function $$f(y) = \left(2\sum_{i>1}x_i\right)\cdot y^2-\sqrt{\prod_{i=2}^n x_i}\cdot y + 2\sum_{i>j>1}x_ix_j$$
satisfies $\triangle \leq 0$ for any $0\leq\sum_{i=2}^n x_i \leq 1$. This can be shown by $$\triangle = \prod_{i=2}^n x_i-16\sum_{i>1}x_i\cdot\sum_{i>j>1}x_ix_j\leq \prod_{i=2}^nx_i-16\cdot x_2^2\cdot \sum_{j>2}x_j \leq x_2\cdot\left(x_3x_4...x_n-16\cdot x_2x_3\right)\leq 0$$
Where the last inequality is because $16x_2\geq x_2\geq x_4\geq x_4x_5...$ Hence, this implies that $$F(x_1,...)\geq 0\implies x_1^2+...+x_n^2+\sqrt{x_1...x_n}\leq\left(x_1+...+x_n\right)^2 = 1$$
which occurs when $(x_1,...,x_n) = (1,0,...0)$ or some permutation of this.</p>
|
3,561,807 | <p>There is this question regarding constrained optimisation. It says, a rectangular parallelepiped has all eight vertices on the ellipsoid <span class="math-container">$x^{2}+3y^{2}+3z^{2}=1$</span>. Using the symmetry about each of the planes, write down the surface area of the parallelepiped and therefore find the maximum surface area.<br>
I know that the surface area <span class="math-container">$S(x,y,z) = 8xy+8yz+8zx$</span>. I've also defined the constraint as given <span class="math-container">$g(x) = x^{2}+3y^{2}+3z^{2}=1$</span>.
Using the Lagrange multiplier <span class="math-container">$\lambda$</span> I got:<br>
<span class="math-container">$8y+8z-2\lambda x=0$</span><br>
<span class="math-container">$8x+8z-6\lambda y=0$</span><br>
<span class="math-container">$8x+8y-6\lambda z=0$</span><br>
What confuses me is what is next in the answer key. They say using symmetry, <span class="math-container">$y=z$</span>. My question is, how are we able to make that argument? I'm slightly confused. Is it because in the ellipsoid equation, the coefficients of <span class="math-container">$y$</span> and <span class="math-container">$z$</span> are equal? Does that mean for a rectangular parallelepiped inscribed in said ellipsoid, the <span class="math-container">$y$</span> and <span class="math-container">$z$</span> coordinates of its vertices will always be equal? If this is the case, can anyone explain to me why?</p>
| J. W. Tanner | 615,567 | <p>Let <span class="math-container">$n=2k+1$</span>. Then <span class="math-container">$n^2-1=(n+1)(n-1)=(2k+2)(2k)=4(k+1)k.$</span></p>
|
3,398,645 | <p>I have a doubt about value of <span class="math-container">$e^{z}$</span> at <span class="math-container">$\infty$</span> in one of my book they are mentioning that as <span class="math-container">$\lim_{z \to \infty} e^z \to \infty $</span></p>
<p>But in another book they are saying it doesn't exist.I am confused now </p>
<p>As we can see <span class="math-container">$e^{z}$</span> is entire function then as <span class="math-container">$z \to \infty $</span> then <span class="math-container">$e^{z}$</span> must go to <span class="math-container">$\infty$</span></p>
<p>Please help.</p>
| peawormsworth | 603,214 | <p>I think <span class="math-container">$e^x$</span> goes to infinity, while <span class="math-container">$e^z$</span> or <span class="math-container">$(e^i)^x$</span> spins around the unit circle at the rate of the distance around a unit circle.</p>
<p>The letter 'z' could mean Real Integer or Complex number:</p>
<p>z = integers = ( -inf, ..., -2,-1,0,1,2....inf)</p>
<p>z = complex = x + iy</p>
<p>Which are you referring to?</p>
|
3,450,908 | <p>I know how to solve problems, but I don't understand the core theory that sticks these two functions with this special relation. </p>
<p>Could you help me find the core idea that I need to understand in order to be satisfied with my understanding? </p>
| g.kov | 122,782 | <p>Just an illustration.</p>
<p>Geometrically, in 2D,
the curves <span class="math-container">$y=\exp(x)$</span> and <span class="math-container">$y=\ln(x)$</span>
are the mirror images of each other
with respect to line <span class="math-container">$y=x$</span>.</p>
<p><a href="https://i.stack.imgur.com/4CxF3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4CxF3.png" alt="enter image description here"></a></p>
|
3,370,521 | <p>I have to prove that the following series converges:
<span class="math-container">$$
\sum_{n=1}^{\infty}\frac{\sin n}{n(\sqrt{n}+\sin n)}
$$</span>
I tried to use Dirichlet's test but I was not really sure whether the denominator is a monotonically decreasing function. If it is than the problem becomes easy.</p>
| Kavi Rama Murthy | 142,385 | <p>Certainly not. Consider <span class="math-container">$\frac 1 {z-1}$</span> on the open unit disk. As <span class="math-container">$z \to -1$</span> this function tends to <span class="math-container">$-\frac 1 2 $</span>. </p>
|
68,817 | <p>I have two questions after reading the Hahn-Banach theorem from Conway's book ( I have googled to know the answer but I have not found any result yet. Also I am not sure that whether my questions have been asked here somewhere on this forum - so please feel free to delete them if they are not appropriate )</p>
<p>Here are my questions: </p>
<ol>
<li><p>We know that if $M$ is a linear subspace of $X$ and $f :M\to\mathbb{F}$ and $f$ is linear,bounded by a seminorm $p$ then $f$ can be extended onto $X$ by some functional $F$. Can $F$ be unique ? Under what condition $F$ will be an unique extension? It would be appreciate if you could give me one example that $F$ could not be unique.</p></li>
<li><p>If the above $\mathbb{F}$ is replaced a Banach space $Y$, i.e, let $M$ be a closed subspace of a Banach space $X$, and $f :M\to Y$ be a bounded linear operator, can we extend $f$ by a bounded operator $F :X\to Y$ ? if not, what condition should be put on $Y$ to have a such extension?</p></li>
</ol>
<p>thanks so much </p>
| M.González | 39,421 | <p>When $X^*$ is strictly convex and $M$ is a subspace of $X$, every norm one $f\in M^*$ admits a unique norm one extension $F\in X^*$, because given two of such extensions $F_1$ and $F_2$, $(F_1+F_2)/2$ is norm one. </p>
<p>In fact this property characterizes $X^*$ strictly convex. See Theorem 7.11 in B.V. Limaye "Functional analysis", 2nd. ed.</p>
|
2,183,809 | <p>The problem:</p>
<blockquote>
<p>If $p$ is prime and $4p^4+1$ is also prime, what can the value of $p$ be?</p>
</blockquote>
<p>I am sure this is a pretty simple question, but I just can't tackle it. I don't even known how I should begin...</p>
| kingW3 | 130,953 | <p>For any $n$ which is not divisible by $5$ you have that $n^4\equiv 1\pmod{5}$ then you must have $4p^4+1\equiv 4+1\equiv 0\pmod{5}$ but $4p^4+1$ must be prime what does that leave you with?</p>
|
2,317,166 | <p>I just pondered this question and have tried out several methods to solve it(mainly using trigonometry). However, I am not satisfied with my trigonometrical proof and is looking for better proofs.</p>
<p>Give an <em>elegant proof</em> that the diagonals are the longest lines in a square.
It would rather be nice if the proof is by <strong>"reductio ad absurdum"</strong> method. Thank you :).</p>
<p>Bonus : Also prove that the diagonals of a rectangle are the longest lines.</p>
| Archis Welankar | 275,884 | <p>Draw a circle circumscribing the square and with end points at opposite corners . Like $A,C $. This circle has diameter as $AC$ and the diameter is the longest chord of circle but this diameter is also the diagonal of the square. Thus the diagonals are the longest lines in a square.</p>
|
2,317,166 | <p>I just pondered this question and have tried out several methods to solve it(mainly using trigonometry). However, I am not satisfied with my trigonometrical proof and is looking for better proofs.</p>
<p>Give an <em>elegant proof</em> that the diagonals are the longest lines in a square.
It would rather be nice if the proof is by <strong>"reductio ad absurdum"</strong> method. Thank you :).</p>
<p>Bonus : Also prove that the diagonals of a rectangle are the longest lines.</p>
| farruhota | 425,072 | <p>It is sufficient to consider the two cases (since others are repetetive): without loss of generality, take two arbitrary points on two sides such that $0\le x\le y\le a$:</p>
<p><a href="https://i.stack.imgur.com/hUeGi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hUeGi.png" alt="enter image description here"></a></p>
<p>Case 1: From the Pythagoras: $$\sqrt{x^2+y^2}\le \sqrt{a^2+a^2}=d.$$
Case 2: From the Pythagoras: $$\sqrt{a^2+(y-x)^2}\le \sqrt{a^2+a^2}=d,$$because $y-x$ is maximum, when $y=a$ and $x=0$.</p>
|
493,600 | <p>I am presented with the following task:</p>
<p>Can you use the chain rule to find the derivatives of $|x|^4$ and $|x^4|$ in $x = 0$? Do the derivatives exist in $x = 0$? I solved the task in a rather straight-forward way, but I am worried that there's more to the task:</p>
<p>First of all, both functions is a variable to the power of an even number, so given that $x$ is a real number, we have that $|x^4| = |x|^4$. In order to force practical use of the chain rule, we write $|x|^4 = \sqrt{x^2}^4$. We are using the fact that taking a number to the power of an even number, and using the absolute value, gives us positive numbers exclusively. If we choose the chain $u = x^2$, thus $g(u) = \sqrt{u}^4$, we have that $u' = 2x$ og $g'(u) = (u^2)' = 2u$. Then we have that the derivative of the function, that I for pratical reasons will name $f(x)$, is $f'(x) = 2x^2 * 2x = 4x^3$. We see that the the general power rule applies here, seeing as we work with a variable to the power of an even number. The derivative in the point $x = 0$ is $4 * 0^3 = \underline{\underline{0}}$. Thus we can conclude that the derivative exists in $x = 0$.</p>
<p>Is this fairly logical? I'm having a hard time seeing that there is anything more to this task, but it feels like it went a bit too straightforward.</p>
| Felix Marin | 85,343 | <p>$$
{{\rm d}\left\vert x\right\vert^{4} \over {\rm d}x}
=
4\left\vert x\right\vert^{3}\,{{\rm d}\left\vert x\right\vert \over {\rm d}x}
=
4\left\vert x\right\vert^{3}\,{\rm sgn}\left(x\right)
=
4\left\vert x\right\vert^{2}
\left\lbrack\vphantom{\Large A}%
\left\vert x\right\vert\,{\rm sgn}\left(x\right)
\right\rbrack
=
4x^{2}\left\lbrack\vphantom{\Large A}x\right\rbrack
=
4x^{3}
$$
We usually define $\quad{\rm sgn}:{\mathbb R} - \left\lbrace 0\right\rbrace \to {\mathbb R}$ such that
$$
{\rm sgn}\left(x\right)
=
\left\lbrace%
\begin{array}{rl}
-1\,,\qquad & x < 0
\\[1mm]
1\,,\qquad & x > 0
\end{array}\right.
$$
However, we usually see calculations where it is assumed
"${\rm sgn}\left(0\right) = 0$" for practical purposes. The correct way is to perform the calculation for $x \not= 0$ and consider the case $x = 0$ as an independent one. It could be ( in particular cases ) that the "result $x = 0$" coincides with the "result $\not= 0$" in the limit $x \to 0^{\pm}$. </p>
<p>The above definition and "extremely care" are useful in 'practical calculations'. For example, let's solve ${\rm y}'\left(x\right) = 2\left\vert x\right\vert$
with ${\rm y}\left(-1\right) = -1$:</p>
<p>\begin{align}
{\rm y}\left(x\right) - {\rm y}\left(-1\right)
&=
{\rm y}\left(x\right) + 1
=
2\int_{-1}^{x}{\rm sgn}\left(x'\right)x'\,{\rm d}x'
=
\left.
\vphantom{\Huge A}x'^{2}{\rm sgn}\left(x'\right)
\right\vert_{-1}^{x}
-
\int_{-1}^{x}x'^{2}
\left\lbrack 2\delta\left(x'\right)\right\rbrack\,{\rm d}x'
\\[3mm]&=
x^{2}\,{\rm sgn}\left(x\right) + 1
\quad\Longrightarrow\quad
{\rm y}\left(x\right) = x\,\left\vert x\right\vert\,,\quad x \not= 0
\end{align}
We solved the differential equation without dividing the problem in two cases
( $x < 0$ and $x > 0$ ). Since
${\rm y}\left(0^{\pm}\right) = 0\left\vert 0\right\vert$, we 'adopt' as a solution ${\rm y}\left(x\right) = x\,\left\vert x\right\vert,\ {\large\forall\ x}$.</p>
<p>If you are in the Physics area, you'll find many situations like the one you addressed. </p>
|
246,384 | <p>How can I plot a hexagon inside this cylinder so that it is always in the same direction as the cylinder, no matter what the orientation is? In addition, the hexagon must always be the same size as the cylinder.</p>
<p>For example, start at the origin (10,9,8) and go to (1,2,3)?</p>
<pre><code>Graphics3D [{Cylinder [{{10, 9, 8}, {1, 2, 3}}, 0.5]}
</code></pre>
<p>I tried this:</p>
<pre><code>Table [Show [Graphics3D [Cylinder [{{0., 0., 0}, {0., 0., l}}, 1]], PolyhedronData [{"Prism", 6}]], {l, 0.1 , 3, 0.1}]
</code></pre>
| asorlik | 77,707 | <p>Summary of issue: dmHalo and dm in the original post were inner parts of a nested data structure in the format <code>{{i, {{j,k,l},{m,n,o},{...}}}}</code>. The problem is that the export formats--.csv, .dat, .txt.-- regardless of keyword--"Data","Table","List"--would convert the inner lists into strings in a way that is irreversible (at least from what I have tried i.e. using ToExpression on the resulting string).</p>
<p>SOLUTION: Using Put and Get (<a href="https://mathematica.stackexchange.com/questions/79854/mathematica-issues-storing-nested-list-on-disk">Mathematica: issues storing nested list on disk</a>) has seemed to resolve the issue and I now can recover the list format of the inner nested lists. As seen in the images, the export process converts the inner list to a string, whereas with the Put and Get commands the list format is preserved.</p>
<p><a href="https://i.stack.imgur.com/UL4ED.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UL4ED.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/ftMtY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ftMtY.png" alt="enter image description here" /></a></p>
|
1,272,647 | <p>I guess it's a simple question, but it really escaped my memory.</p>
<p>If $a + b =1$, then how can I call those $a$ and $b$ numbers? </p>
<p>$a$ is not an inversion of $b$, and it's not reciprocal of $b$.. but I'm sure that they do have a 'name'.</p>
| Chris Culter | 87,023 | <p>Grammatically, it would make sense to say that $a$ is the <strong>unit complement</strong> or <strong>unity complement</strong> of $b$. Google attests that this phrase is occasionally used, but I wouldn't call it common.</p>
|
2,798,026 | <blockquote>
<p>There are three vector $a$, $b$, $c$ in three-dimensional real vector space, and the inner product between them $a\cdot a=b\cdot b = a\cdot c= 1, a\cdot b= 0, c\cdot c= 4$. When setting $x = b\cdot c$, answer the following question: when $a, b, c$ are linearly dependent, find all possible values of $x$.
(dot here means dot product)</p>
</blockquote>
<p>For dependent condition </p>
<p>$$\begin{align}
(a×b)·c &= 0\\
a·(b×c) &= 0\\
a(bc \sin θ)&=0
\end{align}$$</p>
<p>So $\theta = 0, \pi$.</p>
<p>Then
$$\begin{align}
x&=|b||c| \cos \theta \\
x&=2 \cos \theta \\
\implies x &= 2 \cos 0, x = 2 \cos \pi \\
x &= \mp 2
\end{align}$$</p>
<p>Am I right using triple cross product? or should i find $\theta$?
<a href="https://i.stack.imgur.com/I8unM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/I8unM.png" alt="enter image description here"></a></p>
| John Douma | 69,810 | <p>Linear dependence places the vectors in the same plane. This makes the problem easier because we can focus on the angle.</p>
<p>From $c\cdot c=4$ we get that $|c|=2$.</p>
<p>So $a\cdot c=2\cos\theta=1\implies\cos\theta=\frac{1}{2}\implies\theta=60^{\circ}$.</p>
<p>From $b\cdot b=1$ and $a\cdot b=0$ we get that $b$ is a unit vector perpendicular to $a$.</p>
<p>Since the angle between $a$ and $c$ is $60^{\circ}$, the angle between $b$ and $c$ is either $30^{\circ}$ or $150^{\circ}$.</p>
<p>Therefore, $x=2\cos{30^{\circ}}$ or $x=2\cos{150^{\circ}}$ so $x=\pm \sqrt{3}$</p>
|
355,438 | <p>I want to find the geometric locus of point $M$ such that $|MA|^2 |MB|^2=a^2$ where $|AB|=2a$, Solving algebraic equation is not hard but I can't figure out the shape of this curve. Can anybody help?</p>
| MvG | 35,416 | <p>If you want to toy with these shapes, you can try this:</p>
<ol>
<li>Download and install <a href="http://de.cinderella.de/" rel="nofollow noreferrer">Cinderella</a> (free version should be enough)</li>
<li>Start it and add four free points ($A$ through $D$) to the construction</li>
<li>Press <kbd>Ctrl</kbd>+<kbd>Enter</kbd> to open a command input box</li>
<li>Check the “Permanent” checkbox next to that</li>
<li><p>Paste the following code:</p>
<pre><code>colorplot(if(dist(#,A)^2*dist(#,B)^2<(dist(A,B)/2)^2,[1,1,0],[0,0,1]),C,D,pxlres->1,startres->8);
</code></pre></li>
<li><p>Press enter</p></li>
</ol>
<p>You should see a blue plot region, its shape determined by the points $C$ and $D$, and within this one or two yellow blobs, centered around $A$ and $B$ and delimited by the curve you asked about. Something like this:</p>
<p><img src="https://i.stack.imgur.com/BtrG4.png" alt="Cinderella Screenshot"></p>
<p>The code in detail:</p>
<pre><code>colorplot( // plot pixels depending on function value
if( // function is a case distinction
dist(#,A)^2 * dist(#,B)^2 < (dist(A,B)/2)^2, // sign for your curve
[1,1,0], // yellow inside
[0,0,1] // blue outside
),
C,D, // these points control the plotting area
pxlres->1, // draw at finest resolution eventually
startres->8 // draw at coarse resolution first for smooth movements
);
</code></pre>
<p>Also see the <a href="http://doc.cinderella.de/tiki-index.php?page=CindyScript" rel="nofollow noreferrer">CindyScript reference</a> <a href="http://doc.cinderella.de/tiki-index.php?page=Function%20Plotting#colorplot" rel="nofollow noreferrer">of the <code>colorplot</code> function</a>.</p>
|
355,438 | <p>I want to find the geometric locus of point $M$ such that $|MA|^2 |MB|^2=a^2$ where $|AB|=2a$, Solving algebraic equation is not hard but I can't figure out the shape of this curve. Can anybody help?</p>
| Narasimham | 95,860 | <p>If points $A,B$ are fixed and negative distances are ruled out, the loci are quartic curves, <em>Ovals of Cassini.</em> (Incorrectly ) he assumed them at first to be planetary orbit loci.</p>
<p><a href="http://www.2dcurves.com/quartic/quarticca.html#Cassiniellipse" rel="nofollow">Ovals_Cassini1</a></p>
<p><a href="https://www.google.co.in/search?q=ovals%20of%20cassini&tbm=isch&imgil=rQTn4SpWRmIn0M%253A%253BIXGDtUv3ypcipM%253Bhttp%25253A%25252F%25252Fmathworld.wolfram.com%25252FCassiniOvals.html&source=iu&pf=m&fir=rQTn4SpWRmIn0M%253A%252CIXGDtUv3ypcipM%252C_&usg=__YHm2tK1vFEGlIjL3ZKtH1_d1zsM%3D&biw=1082&bih=474&dpr=1.25&ved=0ahUKEwjR2v-MifbKAhWVC44KHTKPCo0QyjcIMA&ei=ENS_VtGSO5WXuASynqroCA#imgrc=rQTn4SpWRmIn0M%3A" rel="nofollow">Ovals_Cassini2</a></p>
|
1,354,015 | <blockquote>
<p>If <span class="math-container">$a, b, c, d, e, f, g, h$</span> are positive numbers satisfying <span class="math-container">$\frac{a}{b}<\frac{c}{d}$</span> and <span class="math-container">$\frac{e}{f}<\frac{g}{h}$</span> and <span class="math-container">$b+f>d+h$</span>, then <span class="math-container">$\frac{a+e}{b+f} < \frac{c+g}{d+h}$</span>.</p>
</blockquote>
<p>I thought it is easy to prove. But I could not. How to prove this? Thank you.</p>
<p>The question is a part of a bigger proof I am working on. There are two strictly concave, positive valued, strictly increasing functions <span class="math-container">$f_1$</span> and <span class="math-container">$f_2$</span> (See Figure 1). Given 4 points <span class="math-container">$x_1$</span>, <span class="math-container">$x_2$</span>, <span class="math-container">$x_3$</span> and <span class="math-container">$x_4$</span> such that <span class="math-container">$x_1< x_i$</span>, <span class="math-container">$i=2, 3,4$</span> and <span class="math-container">$x_4> x_i$</span>, <span class="math-container">$i=1, 2, 3$</span>, let <span class="math-container">$d=x_2-x_1$</span>, <span class="math-container">$b=x_4-x_3$</span> <span class="math-container">$c=f_1(x_2)-f_1(x_1)$</span>, <span class="math-container">$a=f_1(x_4)-f_1(x_3)$</span>. And given 4 points <span class="math-container">$y_1$</span>, <span class="math-container">$y_2$</span>, <span class="math-container">$y_3$</span> and <span class="math-container">$y_4$</span> such that <span class="math-container">$y_1< y_i$</span>, <span class="math-container">$i=2, 3,4$</span> and <span class="math-container">$y_4> y_i$</span>, <span class="math-container">$i=1, 2, 3$</span>, let <span class="math-container">$h=y_2-y_1$</span>, <span class="math-container">$f=y_4-y_3$</span> <span class="math-container">$g=f_2(y_2)-f_2(y_1)$</span>, <span class="math-container">$e=f_2(y_4)-f_2(y_3)$</span>.</p>
<p>Since the functions are concave, we have <span class="math-container">$\frac{a}{b}<\frac{c}{d}$</span> and <span class="math-container">$\frac{e}{f}<\frac{g}{h}$</span>. And I thought in this setting, it is true that <span class="math-container">$\frac{a+e}{b+f} < \frac{c+g}{d+h}$</span> even without the restriction <span class="math-container">$b+f>d+h$</span>.</p>
<p><img src="https://i.stack.imgur.com/Onz2d.png" alt="Figure 1"></p>
| Syed Muhammad Asad | 214,964 | <p>As</p>
<p>$$\frac{a+e}{b+f} < \frac{c+g}{d+h}$$
$$(a+e)(d+h) < (c+g)(b+f)$$
$$ad+eh+ah+ed < cb+fg+cf+gb$$
here put $ah+ed = x$ and $cf+gb = y$. Now we have
$$ad+eh+x < bc+fg+y$$
Now as the given
$$\frac{a}{b}<\frac{c}{d} \implies { ad<bc}$$
and
$$\frac{e}{f}< \frac{g}{h} \implies{eh<fg}$$
so by adding both the above eqs
$$ad+eh<bc+fg$$
and by this condition we have
$$ad+eh+x<bc+fg+y$$</p>
|
275,308 | <p>Problems with calculating </p>
<p>$$\lim_{x\rightarrow0}\frac{\ln(\cos(2x))}{x\sin x}$$</p>
<p>$$\lim_{x\rightarrow0}\frac{\ln(\cos(2x))}{x\sin x}=\lim_{x\rightarrow0}\frac{\ln(2\cos^{2}(x)-1)}{(2\cos^{2}(x)-1)}\cdot \left(\frac{\sin x}{x}\right)^{-1}\cdot\frac{(2\cos^{2}(x)-1)}{x^{2}}=0$$</p>
<p>Correct answer is -2. Please show where this time I've error. Thanks in advance!</p>
| Pedro | 23,350 | <p>The known limits you might wanna use are, for $x\to 0$</p>
<p>$$\frac{\log(1+x)}x\to 1$$</p>
<p>$$\frac{\sin x }x\to 1$$</p>
<p>With them, you get</p>
<p>$$\begin{align}\lim\limits_{x\to 0}\frac{\log(\cos 2x)}{x\sin x}&=\lim\limits_{x\to 0}\frac{\log(1-2\sin ^2 x)}{-2\sin ^2 x}\frac{-2\sin ^2 x}{x\sin x}\\&=-2\lim\limits_{x\to 0}\frac{\log(1-2\sin ^2 x)}{-2\sin ^2 x}\lim\limits_{x\to 0}\frac{\sin x}{x}\\&=-2\lim\limits_{u\to 0}\frac{\log(1+u)}{u}\lim\limits_{x\to 0}\frac{\sin x}{x}\\&=-2\cdot 1 \cdot 1 \\&=-2\end{align}$$</p>
|
1,785,414 | <p>I am trying to find a closed form for the integral $$I=\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{\lfloor|\tan x|\rfloor}{|\tan x|}dx$$ So far, my reasoning is thus: write, by symmetry through $x=\pi/2$, $$I=2\sum_{n=1}^{\infty}n\int_{\arctan n}^{\arctan (n+1)}\frac{dx}{|\tan x|}=2\sum_{n=1}^{\infty}n\ln\frac{\sin\arctan(n+1)}{\sin\arctan n}$$
Using $\sin{\arctan {x}}=\frac{x}{\sqrt{1+x^{2}}}$, we get: $$I=2\sum_{n=1}^{\infty}n\ln(\frac{(n+1)\sqrt{1+n^2}}{n\sqrt{1+(n+1)^2}})=\sum_{n=1}^{\infty}n\ln\frac{(n+1)^2(1+n^2)}{n^2(1+(n+1)^2)}=\sum_{n=1}^{\infty}n\ln(1+\frac{2n+1}{n^2(n+1)^2})$$ Expanding the logarithm into an infinite series we get $$I=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{(-1)^{m+1}(2n+1)^m}{mn^{2m-1}(n+1)^{2m}}$$ Here I am a bit stuck.. Does anyone have any suggestions to go further?</p>
<p>Thank you.</p>
<p>EDIT:
keeping in mind the nice answer below, applying summation by parts to $$I_N=2\sum_{n=1}^{N}n\ln\frac{\sin\arctan(n+1)}{\sin\arctan n}=2\sum_{n=1}^{N}n(\ln\sin\arctan(n+1)-\ln\sin\arctan n)$$ gives $$I_N=2((N+1)\ln\sin\arctan(N+1)+\frac{\ln 2}{2}-\sum_{n=1}^{N}\ln\sin\arctan(n+1))$$ hence: $$I-\ln2=-\sum_{n=2}^{\infty}\ln\frac{n^2}{1+n^2}=\sum_{n=2}^{\infty}\ln\frac{1+n^2}{n^2}=\sum_{n=2}^\infty\sum_{m=1}^\infty\frac{(-1)^{m+1}}{mn^{2m}}= \sum_{m=1}^\infty\frac{(-1)^{m+1}}{m}\sum_{n=2}^\infty n^{-2m}=\sum_{m=1}^\infty\frac{(-1)^{m+1}(\zeta(2m)-1)}{m}$$<br>
Is this valid and helpful? </p>
<p>EDIT 2: Coming back to $$\sum_{n=2}^{\infty}\ln(1+\frac{1}{n^2})=\ln(\prod_{n=2}^{\infty}(1+\frac{1}{n^2}))=\ln(\prod_{n=2}^{\infty}(1-\frac{i^2}{n^2}))=\ln(\prod_{n=1}^{\infty}(1-\frac{i^2}{n^2}))-\ln2$$<br>
$$=\ln(\frac{\sin(i\pi)}{i\pi})-\ln2=\ln\frac{\sinh\pi}{\pi}-\ln2$$ hence $I=\ln\frac{\sinh\pi}{\pi}$ </p>
| David H | 55,051 | <p>$$\begin{align}
I
&=\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{\lfloor\left|\tan{\left(x\right)}\right|\rfloor}{\left|\tan{\left(x\right)}\right|}\,\mathrm{d}x\\
&=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\lfloor\left|\cot{\left(y\right)}\right|\rfloor}{\left|\cot{\left(y\right)}\right|}\,\mathrm{d}y;~~~\small{\left[x-\frac{\pi}{2}=y\right]}\\
&=2\int_{0}^{\frac{\pi}{4}}\frac{\lfloor\left|\cot{\left(y\right)}\right|\rfloor}{\left|\cot{\left(y\right)}\right|}\,\mathrm{d}y\\
&=2\int_{0}^{\frac{\pi}{4}}\frac{\lfloor\cot{\left(y\right)}\rfloor}{\cot{\left(y\right)}}\,\mathrm{d}y\\
&=2\int_{1}^{\infty}\frac{\lfloor t\rfloor}{t\left(1+t^{2}\right)}\,\mathrm{d}t;~~~\small{\left[\cot{\left(y\right)}=t\right]}\\
&=2\sum_{n=1}^{\infty}\int_{n}^{n+1}\frac{\lfloor t\rfloor}{t\left(1+t^{2}\right)}\,\mathrm{d}t\\
&=2\sum_{n=1}^{\infty}\int_{n}^{n+1}\frac{n}{t\left(1+t^{2}\right)}\,\mathrm{d}t\\
&=2\sum_{n=1}^{\infty}n\int_{n}^{n+1}\frac{\mathrm{d}t}{t\left(1+t^{2}\right)}\\
&=2\sum_{n=1}^{\infty}n\int_{n}^{n+1}\left[\frac{1}{t}-\frac{t}{1+t^{2}}\right]\,\mathrm{d}t\\
&=\sum_{n=1}^{\infty}n\left[\ln{\left(\frac{t^{2}}{1+t^{2}}\right)}\right]_{n}^{n+1}\\
&=\ln{\left(2\right)}-\sum_{n=2}^{\infty}\ln{\left(\frac{n^{2}}{1+n^{2}}\right)}\\
&=\ln{\left(2\right)}-\ln{\left(\prod_{n=2}^{\infty}\frac{n^{2}}{1+n^{2}}\right)}\\
&=\ln{\left(2\right)}-\ln{\left(2\pi\operatorname{csch}{\left(\pi\right)}\right)}\\
&=\ln{\left(\frac{\operatorname{sinh}{\left(\pi\right)}}{\pi}\right)}.\blacksquare\\
\end{align}$$</p>
|
66,801 | <p>In short, I am interested to know of the various approaches one could take to learn modern harmonic analysis in depth. However, the question deserves additional details. Currently, I am reading Loukas Grafakos' "Classical Fourier Analysis" (I have progressed to chapter 3). My intention is to read this book and then proceed to the second volume (by the same author) "Modern Fourier Analysis". I have also studied general analysis at the level of Walter Rudin's "Real and Complex Analysis" (first 15 chapters). In particular, if additional prerequisites are required for recommended references, it would be helpful if you could state them.</p>
<p>My request is to know how one should proceed after reading these two volumes and whether there are additional sources that one could use that are helpful to get a deeper understanding of the subject. Also, it would be nice to hear suggestions of some important topics in the subject of harmonic analysis that are current interests of research and references one could use to better understand these topics.</p>
<p>However, I understand that as one gets deeper into a subject such as harmonic analysis, one would need to understand several related areas in greater depth such as functional analysis, PDE's and several complex variables. Therefore, suggestions of how one can incorporate these subjects into one's learning of harmonic analysis are welcome. (Of course, since this is mainly a request for a roadmap in harmonic analysis, it might be better to keep any recommendations of references in these subjects at least a little related to harmonic analysis.)</p>
<p>In particular, I am interested in various connections between PDE's and harmonic analysis and functional analysis and harmonic analysis. It would be nice to know about references that discuss these connections. </p>
<p>Thank you very much!</p>
<p><strong>Additional Details</strong>: Thank you for suggesting Stein's books on harmonic analysis! However, I am not sure how one should read these books. For example, there seems to be overlap between Grafakos and Stein's books but Stein's "Harmonic Analysis" seems very much like a research monograph and although it is, needless to say, an excellent book, I am not very sure what prerequisites one must have to tackle it. In contrast, the other two books by Stein are more elementary but it would be nice to know of the sort of material that can be found in these two books but that cannot be found in Grafakos. </p>
| Spencer | 4,281 | <p>To my mind, the classical subject is quite different from the modern, evolved form of the subject</p>
<p>I started on the classical side with Yitzhak Katznelson's <em>An Introduction to Harmonic Analysis</em>: This is in the classical camp: Lots on Fourier Series. Very clear; very nice proofs. You will learn lots of gems about trigonometric series. In this classical camp, Zygmund's treatise <em>Trigonometric Series</em> (two volumes) deserves a mention. This is also a very beautiful book.</p>
<p>For 'harmonic analysis' as a modern field, you ought to get your hands on a copy of Stein's books as in Peter's answer. The late Tom Wolff has a very useful set of notes in this regard, available (I think, still) <a href="http://www.math.ubc.ca/~ilaba/wolff/">from Izabella Laba's homepage</a>. </p>
<p>I also second the recommendation to look at Tao's old dvi/pdf notes on his website and later on on his blog. For example, I remember finding his post on interpolating $L^p$ spaces very nice.</p>
|
66,801 | <p>In short, I am interested to know of the various approaches one could take to learn modern harmonic analysis in depth. However, the question deserves additional details. Currently, I am reading Loukas Grafakos' "Classical Fourier Analysis" (I have progressed to chapter 3). My intention is to read this book and then proceed to the second volume (by the same author) "Modern Fourier Analysis". I have also studied general analysis at the level of Walter Rudin's "Real and Complex Analysis" (first 15 chapters). In particular, if additional prerequisites are required for recommended references, it would be helpful if you could state them.</p>
<p>My request is to know how one should proceed after reading these two volumes and whether there are additional sources that one could use that are helpful to get a deeper understanding of the subject. Also, it would be nice to hear suggestions of some important topics in the subject of harmonic analysis that are current interests of research and references one could use to better understand these topics.</p>
<p>However, I understand that as one gets deeper into a subject such as harmonic analysis, one would need to understand several related areas in greater depth such as functional analysis, PDE's and several complex variables. Therefore, suggestions of how one can incorporate these subjects into one's learning of harmonic analysis are welcome. (Of course, since this is mainly a request for a roadmap in harmonic analysis, it might be better to keep any recommendations of references in these subjects at least a little related to harmonic analysis.)</p>
<p>In particular, I am interested in various connections between PDE's and harmonic analysis and functional analysis and harmonic analysis. It would be nice to know about references that discuss these connections. </p>
<p>Thank you very much!</p>
<p><strong>Additional Details</strong>: Thank you for suggesting Stein's books on harmonic analysis! However, I am not sure how one should read these books. For example, there seems to be overlap between Grafakos and Stein's books but Stein's "Harmonic Analysis" seems very much like a research monograph and although it is, needless to say, an excellent book, I am not very sure what prerequisites one must have to tackle it. In contrast, the other two books by Stein are more elementary but it would be nice to know of the sort of material that can be found in these two books but that cannot be found in Grafakos. </p>
| Martin Peters | 60,435 | <p>For the link to applications you might try Palle Jorgensen´s: <em>Analysis and Probability. Wavelets, Signals, Fractals.</em> As a special feature, this book contains a dictionary of the use of technical terms in different disciplines.</p>
|
1,025,117 | <p>Let $V$ be finite dim $K-$vector space. If w.r.t. any basis of $V$, the matrix of $f$ is a diagonal matrix, then I need to show that $f=\lambda Id$ for some $\lambda\in K$. </p>
<p>I am trying a simple approach: to show that $(f-\lambda Id)(e_i)=0$ where $(e_1,...,e_2)$ is a basis of $V$. Let the diagonal matrix be given by $diag(\lambda_1,...,\lambda_2).$ Then $$(f-\lambda Id)(e_i)=diag(\lambda_1,...,\lambda_2) (0,0,..,0,1,0,...,0)^T - (0,0,..,0,\lambda ,0,...,0)$$ $$=(0,0,...,0,\lambda_i - \lambda,0,...,0)$$ where $1,\lambda,\lambda_i-\lambda$ are in the $i^{th}$ position. I don't see how to conclude $\lambda_i=\lambda$. </p>
| DeepSea | 101,504 | <p>Let $u = \dfrac{c}{x}$, then: </p>
<p>Note that: $x\log\left(\dfrac{x+c}{x-c}\right) = c\dfrac{\log\left(\dfrac{1+u}{1-u}\right)}{u} = c\left(\dfrac{\log(1+u)}{u} - \dfrac{\log(1-u)}{u}\right) \to c(1-(-1)) = 2c$ since $x \to \infty \iff u \to 0$.</p>
|
2,186,743 | <p>I think this is a very basic question but somehow I am unable to understand the answer.</p>
<hr>
<blockquote>
<p>Find the number of zeroes of $f(x) = x^3 + x + 1$.</p>
</blockquote>
<hr>
<p>Answer in the book : </p>
<p>$f^\prime (x) = 3x^2 + 1$ since $3x^2 + 1 \ge 0$ for all $x \in \mathbb{R}$ therefore $f^\prime (x) \ge 1 $ for all $x \in \mathbb{R}$</p>
<p>Therefore by Rolle's theorem $f(x)$ has at most one zero in its domain. </p>
<p>Now $f(-1) = -1$ and $f(0) = 1$ therefore by Intermediate value theorem $f(x)$ has at least one zero in the interval $[-1, 0]$. </p>
<p>Thus $f(x)$ has one zero.</p>
<hr>
<blockquote>
<p>Rolle's theorem : A function which is continuous $[a,b]$ and differentiable on $(a,b)$ such that $f(a) = f(b) = 0$, then there exist at least a point $c \in [a,b]$ for which $f^\prime(c) = 0$ </p>
</blockquote>
<p>In the first part of the proof I am unable to understand how Rolle's theorem is applied because we don't know any $a,b$ where $f(a) = f(b) = 0$ . </p>
<p>How can conclude that $f(x)$ has at most one zero ?</p>
| Abdullah al allah | 416,261 | <p>The derivative is always positive, so the function is monotonically increasing, hence it will have one root at most if the minimum value is less than 0.</p>
|
3,298,311 | <p>How can I prove (by definition) that, if <span class="math-container">$a, b \in \mathbb{R}$</span> and <span class="math-container">$a<b$</span>, then <span class="math-container">$[a, b]$</span> is equal to the set of accumulation (limit) points?</p>
<p>Let <span class="math-container">$(E, d)$</span> a metric space and <span class="math-container">$S \subseteq E$</span>.
<span class="math-container">$x \in E$</span> is a limit point if <span class="math-container">$(B_\varepsilon(x)-\lbrace x \rbrace ) \cap S \neq \emptyset$</span> for all <span class="math-container">$\varepsilon >0$</span></p>
| Mike Pierce | 167,197 | <p>If <span class="math-container">$x \in [a,b]$</span>, the <span class="math-container">$x \in B_\varepsilon(x) \cap [a,b]$</span> for all <span class="math-container">$\varepsilon >0$</span>. If <span class="math-container">$x \notin [a,b]$</span> take <span class="math-container">$\varepsilon = \frac{1}{2}\min\{d(a,x),d(b,x)\}$</span>, and notice that <span class="math-container">$B_\varepsilon(x) \cap [a,b] = \varnothing$</span>, so <span class="math-container">$x$</span> is <em>not</em> a limit point of the set. </p>
|
445,127 | <p>I need to prove this limit:</p>
<blockquote>
<p>Given $f:(-1,1) \to \mathbb{R}\,$ and $\,f(x)>0,\,$ if $\,\lim_{x\to 0} \left(f(x) + \dfrac{1}{f(x)}\right) = 2,\,$ then $\,\lim_{x\to 0} f(x) = 1$.</p>
</blockquote>
| zuggg | 84,273 | <p>Assume $f(x)$ does not converge towards $1$ when $x\to 0$. That means there exists $\epsilon>0$ and a sequence $(x_n)_{n\in\mathbb{N}}$ such that $x_n\to 0$, and $|f(x_n)-1|>\epsilon$ for all $n$, so either $f(x_n)>1+\epsilon$ or $f(x_n)<1-\epsilon$.</p>
<p>Now, notice that the function $g:x\mapsto x+1/x$ admits a strict minimum for $x=1$. That means there exists $\sigma\in\mathbb{R}$ such that in both of the above cases, $g(f(x_n))>2+\sigma$, which is impossible, hence the result.</p>
|
445,127 | <p>I need to prove this limit:</p>
<blockquote>
<p>Given $f:(-1,1) \to \mathbb{R}\,$ and $\,f(x)>0,\,$ if $\,\lim_{x\to 0} \left(f(x) + \dfrac{1}{f(x)}\right) = 2,\,$ then $\,\lim_{x\to 0} f(x) = 1$.</p>
</blockquote>
| Hagen von Eitzen | 39,174 | <p>Let $g(x)=\max\{f(x),\frac1{f(x)}\}$ and note that $$\begin{align}h\colon [1,\infty)&\to[2,\infty)\\x&\mapsto x+\frac1x\end{align}$$ is a continous bijection with continuous inverse $$\begin{align}h^{-1}\colon [2,\infty)&\to[1,\infty)\\y&\mapsto \left(\frac{\sqrt{y-2}+\sqrt{y+2}}2\right)^2\end{align}$$ (the square roots ocurring here are $\sqrt x\pm\frac1{\sqrt x}$ because $y\pm 2=x\pm2+\frac1x$).</p>
<p>Since $h(g(x))=f(x)+\frac1{f(x)}\to 2$ as $x\to 0$, we conclude $g(x)\to h^{-1}(2)=1$.
Hence if $\epsilon>0$. Then for $x$ sufficiently close to $1$ we have $g(x)<1+\epsilon$, i.e. $$1-\epsilon=\frac{1-\epsilon^2}{1+\epsilon}<\frac1{1+\epsilon}<f(x)<1+\epsilon.$$
This precisely says that $$\lim_{x\to1}f(x)=1.$$</p>
|
445,127 | <p>I need to prove this limit:</p>
<blockquote>
<p>Given $f:(-1,1) \to \mathbb{R}\,$ and $\,f(x)>0,\,$ if $\,\lim_{x\to 0} \left(f(x) + \dfrac{1}{f(x)}\right) = 2,\,$ then $\,\lim_{x\to 0} f(x) = 1$.</p>
</blockquote>
| N. S. | 9,176 | <p><strong>Hint</strong> Prove first that there exists some <span class="math-container">$\delta$</span> such that <span class="math-container">$f$</span> is <strong>positive and bounded</strong> on <span class="math-container">$(x- \delta, x+ \delta)$</span>.</p>
<p><strong>Hint 2:</strong>
<span class="math-container">$$\lim \limits_{x \to a}f(x)+\frac{1}{f(x)}=2 \Leftrightarrow \\
\lim \limits_{x \to a}(f(x)+\frac{1}{f(x)}-2)=0 \Rightarrow \\
\lim \limits_{x \to a}(f(x)^2-2 f(x)+1)=0
$$</span>
with the last step following from the fact that <span class="math-container">$f$</span> is positive and bounded.</p>
<p>Thus
<span class="math-container">$$\lim \limits_{x \to a}(f(x)-1)^2=0
$$</span>
Tke the square root now.</p>
|
1,690,092 | <p>$a_{n} = \frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + ... + \frac{1}{\sqrt{n^2+2n-1}}$</p>
<p>and I need to check whether this sequence converges to a limit without finding the limit itself. I think about using the squeeze theorem that converges to something (I suspect '$1$').</p>
<p>But I wrote $a_{n+1}$ and $a_{n-1}$ and it doesn't get me anywhere...</p>
| GoodDeeds | 307,825 | <p>$$\frac{1}{\sqrt{n^2+2n-1}}+\frac{1}{\sqrt{n^2+2n-1}}\cdots+\frac{1}{\sqrt{n^2+2n-1}}\le\frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + \cdots + \frac{1}{\sqrt{n^2+2n-1}}\le\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+n}}\cdots+\frac{1}{\sqrt{n^2+n}}$$
$$\frac{n}{\sqrt{n^2+2n-1}}\le\frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + \cdots + \frac{1}{\sqrt{n^2+2n-1}}\le\frac{n}{\sqrt{n^2+n}}$$
$$\lim_{n\rightarrow\infty}\frac{n}{\sqrt{n^2+2n-1}}\le\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + \cdots + \frac{1}{\sqrt{n^2+2n-1}}\le\lim_{n\rightarrow\infty}\frac{n}{\sqrt{n^2+n}}$$
$$\lim_{n\rightarrow\infty}\frac{n}{n\sqrt{1+\frac2n-\frac1{n^2}}}\le\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + \cdots + \frac{1}{\sqrt{n^2+2n-1}}\le\lim_{n\rightarrow\infty}\frac{n}{n\sqrt{1+\frac1n}}$$
$$1\le\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + \cdots + \frac{1}{\sqrt{n^2+2n-1}}\le1$$
$$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + \cdots + \frac{1}{\sqrt{n^2+2n-1}}=1$$</p>
|
626,928 | <p>I took linear algebra course this semester (as you've probably noticed looking at my previously asked questions!). We had a session on preconditioning, what are they good for and how to construct them for matrices with special properties. It was a really short introduction for such an important research topic, so I'm not sure if I got familiar with them. I need some elementary sources on it to be able to construct preconditions for famous matrices and to make sure I fully understand the concept. Any helps would be greatly appreciated. </p>
| Brian Rushton | 51,970 | <ul>
<li>Here's a famous survey paper that may be too long: <a href="http://www.mathcs.emory.edu/~benzi/Web_papers/survey.pdf" rel="nofollow">Preconditioning
Techniques for Large Linear Systems: A Survey by Michele
Benzi</a></li>
<li>Then there's this shorter introduction that is heavy on computer
programming and matlab:
<a href="http://www.asc.tuwien.ac.at/~winfried/teaching/106.079/SS2011/downloads/script-p-106-122.pdf" rel="nofollow">http://www.asc.tuwien.ac.at/~winfried/teaching/106.079/SS2011/downloads/script-p-106-122.pdf</a>
It has an intimidating wall of formulas on an early page but gets
good again later.</li>
<li>Also, this link seemed mildly interesting:
wwwhome.math.utwente.nl/~botchevma/nla/</li>
</ul>
<p>Lecture 2 has a page with simple formulas for preconditioners, and assignment 2 had a lot of work on preconditioners.</p>
<p>I'm not sure which famous matrices you would like to see (special lknear group, upper triangular, permutation matrix, sparse, etc.) but I hope this helps.</p>
|
390,771 | <p>$$\int\frac{dx}{(1+x^\frac{1}{4})x^\frac{1}{2}}$$
This is my work:
$$u^4=x$$
$$4u^3=dx$$
$$\int\frac{4u^3du}{(1+u)u^2}=\int\frac{4u^3du}{(1+u)u^2}=-4(1+x^\frac{1}{4})^{-1}+2(1+x^\frac{1}{4})^{-2}+C$$
But <a href="http://www.wolframalpha.com/input/?i=integrate%201/%28%281%2bx%5E%281/4%29%29x%5E%281/2%29%29" rel="nofollow">wolframAlpha</a> gives quite a different answer. So, have I done a mistake? Or somehow both answers are correct?</p>
| Norbert | 19,538 | <p>Check this calculations. They give the same result as wolfram alpha
$$
\int\frac{4u}{1+u}du=
4\int\left(1-\frac{1}{u+1}\right)du=
4\int du-4\int\frac{1}{u+1}du=
4u-4\ln(1+u)=\\
4x^{1/4}-4\ln(1+x^{1/4})
$$</p>
|
390,771 | <p>$$\int\frac{dx}{(1+x^\frac{1}{4})x^\frac{1}{2}}$$
This is my work:
$$u^4=x$$
$$4u^3=dx$$
$$\int\frac{4u^3du}{(1+u)u^2}=\int\frac{4u^3du}{(1+u)u^2}=-4(1+x^\frac{1}{4})^{-1}+2(1+x^\frac{1}{4})^{-2}+C$$
But <a href="http://www.wolframalpha.com/input/?i=integrate%201/%28%281%2bx%5E%281/4%29%29x%5E%281/2%29%29" rel="nofollow">wolframAlpha</a> gives quite a different answer. So, have I done a mistake? Or somehow both answers are correct?</p>
| amWhy | 9,003 | <p>You substitutions are just fine. First, we cancel the common factor of $u^2$ in the numerator and denominator.
$$\int\frac{4u^3du}{(1+u)u^2}=\int\frac{4u\,du}{1+u}$$</p>
<p>Dividing the numerator by the denominator gives us $$\int\frac{4u}{1+u}\,du = \int \left(4 - \frac 4{1+u} \right) \,du = 4\int \,du - 4\int \frac {du}{1+u}\tag{1}$$</p>
<p>Now, I'm not entirely sure how you obtained your integrated your terms, but I can only guess that you mixed up differentiating and integrating terms: <em>subtracting</em> 1 from the exponent of $(1+u)^0$ and dividing $4$ by $-1$, and <em>subtracting</em> 1 from the exponent of $(1+u)^{-1}$ and dividing $-4$ by $-2$. </p>
<p>First, when integrating a constant $\alpha$, $\int \alpha \,du$, we obtain the <em>variable with respect to which we are integrating</em>, and in this case, that's simply the variable $u$ (the $\,du\,$ indicates which variable). So we can think of $$\int \alpha\,du = \int \alpha u^0\,du = \alpha u^1 + C = \alpha u + C$$ Second, when integrating, we <em>add</em> (increment) exponents, and divide by the new exponent, <strong>except</strong> when the exponent is $-1$ (and in your second integral, we have $(1 + u)^{-1}$ as the integrand. Recall, $$\int \frac{f'(u)}{f(u)}\,du = \ln|f(u)| + C.$$</p>
<p>So back to $(1)$. Integrating, we get $$4\int \,du - 4\int \frac {du}{1+u} = 4u - 4\ln|(1+u)| + C = 4x^{1/4} - 4 \ln\left(1 + x^{1/4}\right) + C$$</p>
|
2,333,083 | <p>I want to know about when the infinite product form of the Riemann Zeta Function converges.</p>
<p>It is easy to show that $$\sum_{n=1}^\infty \frac 1{n^x}$$ converges for $\Bbb R(x)>1$.</p>
<p>When does $$\prod_{p=primes} \frac {p^x}{p^x-1}$$ converge? And how do you show that?</p>
| reuns | 276,986 | <p>Let $$A_k = \{ n \in \mathbb{N}, \text{ largest prime factor of } n \le k\}$$
Using the fundamental theorem of arithmetic, or by induction
$$\prod_{p \in \mathcal{P},p \le k} (1+p^{-s}+(p^2)^{-s}+(p^3)^{-s}+\ldots) = \sum_{n \in A_k} n^{-s}$$</p>
<p>For $s > 1$ the sequence $\{ n^{-s}\}_{n\in \mathbb{N}}$ is non-negative and summable, which means the order of summation doesn't matter, thus
$$\zeta(s) = \lim_{N \to \infty} \sum_{n \in [1,N]} n^{-s} = \lim_{k \to \infty} \sum_{n \in A_k} n^{-s}$$</p>
<p>And for $s \le 1$, since each term is positive, if $\lim_{k \to \infty} \sum_{n \in A_k} n^{-s}$ converges then $\lim_{N \to \infty} \sum_{n \in [1,N]} n^{-s}$ converges, a contradiction since we know it diverges by comparison with $\sum_{n=1}^N n^{-1} > \int_1^N x^{-1}dx = \log N$</p>
|
2,676,200 | <blockquote>
<p>A hyperbola has equation $\frac{x^2}{4}-\frac{y^2}{16}=1$. Show that every other line parallel to this asymptote, $y=2x$, intersects the hyperbola exactly once.</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/gFB6v.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gFB6v.png" alt="enter image description here"></a></p>
<p>So here's the hyperbola. The blue line represents the asymptote $y = 2x$. I am not concerned with the other asymptote that has a negative gradient, $y =-2x$, although the same principles would apply. </p>
<p>Now the problem can re-phrased as:</p>
<p>For the line $y = 2x+c , c \ne 0$</p>
<p>Prove the line intersects the hyperbola once exactly for any non-zero c.</p>
<p>The orange line represents the case where $c>0$.</p>
<p>The black line represents the case where $c<0$.</p>
<p>This is geometrically intuitive, however, I am struggling to prove this algebraically.</p>
<hr>
<p>FIRST ATTEMPT</p>
<p>First of all, substitute $y = 2x+c$ into $\frac{x^2}{4}-\frac{y^2}{16}=1$ for $y$.</p>
<p>$\frac{x^2}{4}-\frac{(2x+c)^2}{16}=1$</p>
<p>Multiply through by 16.</p>
<p>$4x^2-(2x+c)^2=16$</p>
<p>$4x^2-(4x^2+4cx+c^2)=16$</p>
<p>$-4cx-c^2=16$</p>
<p>$c^2 + 4cx + 16 = 0$</p>
<p>Take the discriminant to determine the number of times the line intersects the hyperbola.</p>
<p>$\Delta = B^2 - 4AC$ for a generic quadratic $Ax^2 + Bx + C = 0$.</p>
<p>(Using upper case $A$, $B$ and $C$ as lower case $c$ is already taken.)</p>
<p>Therefore,</p>
<p>For $0x^2 +4cx + (c^2 + 16)=0$ where the quadratic is taken in terms of $x$.</p>
<p>$\Delta = (4c)^2 -4(0)(c^2 + 16) = 16c^2$</p>
<p>As $c \ne 0$,</p>
<p>$16c^2 > 0$</p>
<p>Hence all lines parallel to $y = 2x$ must intersect the hyperbola twice.</p>
<p>I've managed to disprove what I am trying to prove. Please can someone explain where I have gone wrong.</p>
| Álvaro Serrano | 535,487 | <p>You don't have a quadratic equation, $c^2+4cx+16=0$ is a linear equation with the single solution $x=-\frac{16+c^2}{4c}$, so there is only one intersection point when $c \ne 0$ and none when $c = 0$.</p>
|
1,893,609 | <p>I am trying to show that $A=\{(x,y) \in \Bbb{R} \mid -1 < x < 1, -1< y < 1 \}$ is an open set algebraically. </p>
<p>Let $a_0 = (x_o,y_o) \in A$. Suppose that $r = \min\{1-|x_o|, 1-|y_o|\}$ then choose $a = (x,y) \in D_r(a_0)$. Then</p>
<p>Edit: I am looking for the proof of the algebraic implication that $\|a-a_0\| = \sqrt {(x-x_o)^2+(y-y_o)^2} < r \Rightarrow|x| < 1 , |y| < 1 $</p>
| Mick | 42,351 | <p>Four points constitute a quadrilateral. If these four points lie on the same circle, the quadrilateral thus formed is called cyclic quadrilateral. A cyclic quadrilateral can be but need not be a rectangle. A cyclic quadrilateral has the properties of the sum each pair of interior opposite angles is $180^0$.</p>
<p>For this problem, finding the center does not require the knowledge of whether the 4 points form a rectangle or not. The center is right at the point of intersection of the perpendicular bisectors of any two chords of that circle.</p>
<ol>
<li><p>From any two given points (say $A(x_1, y_1)$ and $B(x_2, y_2))$, find the corresponding slope ($m_{AB}$) and midpoint $(M_{AB})$.</p></li>
<li><p>Using the above info to get $(L_{AB})$, the equation of the perpendicular bisector (that passes through $M_{AB}$).</p></li>
<li><p>Repeat the above process to get $(L_{BC})$.</p></li>
<li><p>The co-ordinates of the center is given by solving the simultaneous equations $(L_{AB})$ and $(L_{BC})$.</p></li>
</ol>
<p>Note: Three (non-collinear) points uniquely determine the center of the circle passing through them. The 4th point is probably for checking the correctness of the equation of the circle formed.</p>
|
11,629 | <p>The basic logic course in school gives the impression that logic has both the syntax and the semantics aspects. Recently, I wonder whether the syntax part still plays an essential role in the current studies. Below are some of my observations, I hope the idea from the community can make them more complete.</p>
<p>Model theory: Even though model theory is stated in the language of logic, it can be viewed as the study of local isomorphism (see Poizat's "A course in Model Theory"). The syntax part is therefore a natural (though might be uncomfortable for some) way to view the theory rather than a necessity.</p>
<p>Recursion theory: The object of study is the notion of computability in different context. If we believe in Church-Turing Thesis, then these concept are independent of the formalism chosen. </p>
<p>Set theory: The intimate relationship between large cardinal and determinacy perhaps can suggest that this is a universal phenomenon. Will this phenomenon disappear if we change the language of mathematics to, for example, category theory?</p>
<p>Proof theory: I know too little to say anything.</p>
<p>If the observation is true, is it justified to demand that Turing degrees, and large cardinals receive the same mathematical status as, for example, prime numbers?</p>
| Charles Stewart | 3,154 | <p>We don't know how to abstract away from syntax in proof theory. If we say there are three main branches in proof theory:</p>
<ol>
<li>Axiomatics seem to be necessarily syntactic: formulae are what it is about;</li>
<li>Relative consistency & ordinal analysis, which are ultimately about characterising provability strength in various ways: these are maybe as syntactic as recursion theory;</li>
<li>Structural proof theory, where we care about analytic proofs: there are programs to try to abstract away from syntax, such as categorical proof theory, but too much of proof theory doesn't yield to these kinds of analysis, so we can't say that these are successful yet.</li>
</ol>
|
11,629 | <p>The basic logic course in school gives the impression that logic has both the syntax and the semantics aspects. Recently, I wonder whether the syntax part still plays an essential role in the current studies. Below are some of my observations, I hope the idea from the community can make them more complete.</p>
<p>Model theory: Even though model theory is stated in the language of logic, it can be viewed as the study of local isomorphism (see Poizat's "A course in Model Theory"). The syntax part is therefore a natural (though might be uncomfortable for some) way to view the theory rather than a necessity.</p>
<p>Recursion theory: The object of study is the notion of computability in different context. If we believe in Church-Turing Thesis, then these concept are independent of the formalism chosen. </p>
<p>Set theory: The intimate relationship between large cardinal and determinacy perhaps can suggest that this is a universal phenomenon. Will this phenomenon disappear if we change the language of mathematics to, for example, category theory?</p>
<p>Proof theory: I know too little to say anything.</p>
<p>If the observation is true, is it justified to demand that Turing degrees, and large cardinals receive the same mathematical status as, for example, prime numbers?</p>
| Justin Hilburn | 333 | <p>To elaborate slightly on Neel's post: the fact that not all proofs in intuitionist and substructural logics are identified is a strength and not a weakness when viewed through the lens of the Curry-Howard isomorphism which shows that logic has computational content.</p>
<p>Basically the Curry-Howard isomorphism states that propositions in intuitionist logic can be identified with types in a programming language and proofs can be identified with programs. In classical logic all proofs of a given proposition are identified so classical logic is too impoverished to serve as a model for computation. In other words, the fact that there are multiple proofs of the same proposition in intuitionist logic is what allows us to have multiple programs with the same type.</p>
|
1,952,562 | <p>Find an equation of the tangent line to the curve $y = sin(3x) + sin^2 (3x)$ given the point (0,0). Answer is $y = 3x$, but please explain solution steps. </p>
| Emilio Novati | 187,568 | <p>Hint:</p>
<p>Do you know that the slope of the tangent line at a point of the graph of a function is the derivative of the function at this point?</p>
<p>So, for $y = \sin(3x) + \sin^2 (3x)$ find the derivative $y'$ ( can you do?), then evaluate this derivative for $x=0$ </p>
<p>Now the line has equation $y=mx$ with $m=y'(0)$</p>
<hr>
<p>Using the chain rule the derivative is:
$$
y'=\cos(3x)\cdot(3x)'+2\sin(3x)(\sin(3x))'=3\cos(3x)+2\sin(3x)\cos(3x)(3x)'$$$$=3\cos(3x)+6\sin(3x)\cos(3x)
$$
so $y'(0)=3$.</p>
|
628,236 | <p>I am trying to give a name to this axiom in a definition: </p>
<p>$(X \bullet R) \sqcup (Y \bullet S) \equiv (X \sqcup Y) \bullet (R \sqcup S)$</p>
<p>(for all $X, Y, R, S$) where $\sqcup$ is the join of a lattice and $\bullet$ is some binary operation. It feels related to monotonicity/distributivity but I don't know a standard name for this. Any ideas? So far I have called it "full distributivity". I'd also like to have a name (possibly the same) for this axiom when $\sqcup$ isn't a lattice operation, just some binary operation.</p>
| Community | -1 | <p>Suppose that $Z^2-YZ-Y^2+X^2+2XY$ is reducible. Since this is a degree two polynomial in $Z$ with coefficients in $\Bbb C[X,Y]$ it splits into a product of two polynomials of degree one in $Z$, that is, $Z^2-YZ-Y^2+X^2+2XY=(Z+f(X,Y))(Z+g(X,Y))$. We get $f(X,Y)+g(X,Y)=-Y$ and $f(X,Y)g(X,Y)=-Y^2+X^2+2XY$. Now plug in the second relation what you get from the first and find $f(X,Y)(f(X,Y)+Y)=Y^2-2XY-X^2$. This implies that $\deg_Yf(X,Y)=1$, so $f(X,Y)=a(X)+b(X)Y$. Now just calculate what you need and get $b(X)(b(X)+1)=1$, $2a(x)b(x)+a(X)=-2X$, and $a^2(X)=-X^2$. From $b(X)(b(X)+1)=1$ we deduce that $\deg b(X)=0$ and from now on write it $b$ instead of $b(X)$. We have $b(b+1)=1$. We also get $a(X)=\frac{-2}{2b+1}X$ and plugin this into the first relation we get $4=-(2b+1)^2$, so $4=-4(b^2+b)-1$ and from $b(b+1)=1$ we get $4=-4-1$, a contradiction.</p>
|
31,261 | <p>Given a 3-manifold $M$, one can define the Kauffman bracket skein module $K_t(M)$ as the $C$-vector space with basis "links (including the empty link) in $M$ up to ambient isotopy," modulo the skein relations, which can be found in the second paragraph of section two of <a href="http://arxiv.org/abs/math/0402102" rel="nofollow">http://arxiv.org/abs/math/0402102</a>. (Side question - how can I draw these relations in Latex?)</p>
<p>If $S$ is a surface, then $K_t(S\times [0,1])$ has an algebra structure given by stacking one link on top of another. If $S$ is a boundary component of $M$, then $K_t(M)$ is a (left) $K_t(S\times [0,1])$ module, where the left module structure is given by gluing $S\times \{1\}$ to the copy of $S$ in the boundary of $M$. In this situation, we can define a left module map $K_t(A\times [0,1]) \to K_t(M)$ which is uniquely defined by "(empty link in $S\times [0,1]$) maps to (empty link in $M$)." The "peripheral ideal" is the kernel of this module map, and is a left ideal of $K_t(S\times [0,1])$.</p>
<p>The motivation for these definitions comes from knot theory - if $K$ is a knot in $S_3$, then the complement of a small tubular neighborhood of $K$ is a manifold with a torus boundary, and the algebra $K_t(T^2\times [0,1])$ and module $K_t(S^3 \setminus K)$ give information about the knot $K$.</p>
<p>Now I can ask my question: Is there a manifold $M$ with a torus boundary such that the peripheral ideal is trivial? </p>
<p>I've just recently started learning about knot theory, and I'm having a hard time trying to figure this out. One thing that I do know is that $M$ cannot be of the form $S^3 \setminus K$, because of propositions 7 and 8 in <a href="http://arxiv.org/abs/math/9812048" rel="nofollow">http://arxiv.org/abs/math/9812048</a>. I also suspect that $M$ will actually have <b>two</b> boundary components which are a torus, but I don't really have a good reason for this.</p>
<p>Also, I suspect this might be a hard question, so any hints about one might approach it would be helpful.</p>
| Bruce Westbury | 3,992 | <p>This is not an answer, more like one comment and one suggestion for an approach to this problem.</p>
<p>The comment is that this looks like a 4-dimensional TQFT. You have an algebra associated to a surface, a module associated to a 3-manifold, and a vector associated to a 4-manifold. The reason it is not usually presented this way is that the dependence on the 4-manifold is not interesting; it only depends on the signature (i.e. the cobordism class).</p>
<p>The suggestion for an approach is to look at $q=1$, the classical limit. Here the algebra associated to a surface is the coordinate ring of the character variety. The key observation is that a skein determines a function on the space of flat connections by taking traces of holonomy and the skein relation corresponds to $tr(A^{-1}B)+tr(AB)=tr(B)$. This was written up by Doug Bullock.</p>
<p>My suggestion then is for you to look at your question in this context. I don't know if this will help but it seems more likely to be a question that an expert can answer.</p>
|
2,164,946 | <p>I'm working on arc length calculation and area of surface of revolution in calculus and I'm really quite stuck on the process of how to do this. Here is a particular problem that I'm struggling with:</p>
<blockquote>
<p>Find the surface area of the surface of revolution generated by revolving the graph $$y=x^3; \qquad 0 \leq x \leq10$$ around the $x$-axis.</p>
</blockquote>
<p>I've gone through the steps that I've learned to do (listed below) and the steps for the most part seem to make sense, however I keep ending up with incorrect answers. Please help! Below I listed my general process of approaching the problem.</p>
<blockquote>
<p>$$\begin{align} &y = x^3 \\&y' = 3x^2 \\&(y')^2 = 9x^4\\&1 + (dy/dx)^2 = 1 + 9x^4\end{align}$$</p>
</blockquote>
<p>Using the formula: </p>
<blockquote>
<p>$$2\pi y \cdot\int(1 + (dy/dx)^2)^{1/2} dx$$</p>
</blockquote>
<p>Here's how I set up the integral for the problem:</p>
<blockquote>
<p>$$2\pi x^3\cdot \int_0^{10}(1+9x^4)^{1/2} dx.$$</p>
</blockquote>
<p>This came out to $$2\pi x^3\cdot(6/5)\cdot(1+9(10^4))^{3/2}\cdot x^5$$
My final answer was $$2.035785969E16.$$ </p>
<p>Please help me understand where I'm going wrong! </p>
| Mansour Alkatheeri | 443,886 | <p>$X^3$ should be placed inside the integration not outside</p>
|
195,006 | <p>I am not very familiar with mathematical proofs, or the notation involved, so if it is possible to explain in 8th grade English (or thereabouts), I would really appreciate it.</p>
<p>Since I may even be using incorrect terminology, I'll try to explain what the terms I'm using mean in my mind. Please correct my terminology if it is incorrect so that I can speak coherently about this answer, if you would.</p>
<p>Sequential infinite set: A group of ordered items that flow in a straight line, of which there are infinitely many. So, all integers from least to greatest would be an example, because they are ordered from least to greatest in a sequential line, but an infinite set of bananas would not, since they are not linearly, sequentially ordered. An infinite set of bananas that were to be eaten one-by-one would be, though, because they are iterated through (eaten) one-by-one (in linear sequence).</p>
<p>Sequential infinite subsets: Multiple sets within a sequential infinite set that naturally fall into the same order as the items of the sequential infinite set of which they are subsets. So, for example, the infinite set of all integers from least to greatest can be said to have the following two sequential infinite subsets within it: all negative integers; and all positive integers. They are sequential because the negative set comes before the positive set when ordered as stated. They are infinite because they both contain an infinite qty of items, and they are subsets because they are within the greater infinite set of all integers.</p>
<p>So I'm wondering if every (not some, but every) sequential infinite set contains within it sequential infinite subsets. The subsets (not the items within them) being sequentially ordered is extremely important. Clearly, a person could take any infinite set, remove one item, and have an infinite subset. Put the item back, remove a different item, and you have multiple infinite subsets. But I need them to be not only non-overlapping, but also sequential in order.</p>
<p>Please let me know if this does not make sense, and thank you for dumbing the answer down for me.</p>
| Brian M. Scott | 12,042 | <p>If I understand you correctly, what you mean by a <em>sequentially infinite set</em> is probably a countably infinite set with a specific ordering attached that makes it possible to specify a first element, a second element, and so on. In other words, you have what amounts to an infinite list:</p>
<p>$$a_1,a_2,a_3,\dots\;,$$</p>
<p>with an $a_n$ for every positive integer $n$. In more technical language, this is an <em>infinite sequence</em> of objects. If that is what you have in mind, just imitate what you did with the odd and even positive integers: take one set to be</p>
<p>$$\{a_2,a_4,a_6,\dots\}=\{a_n:n\text{ is even}\}\;,$$</p>
<p>and take the other set to be </p>
<p>$$\{a_1,a_3,a_5,\dots\}=\{a_n:n\text{ is odd}\}\;.$$</p>
<p>In the banana example, imagine that one banana is eaten every hour on the hour, starting at $1$ a.m. Then the first set consists of those bananas eaten at even-numbered hours, and the second set of those eaten at odd-numbered hours.</p>
<p>You can do this in general because (again, if I’m understanding you correctly) the numbering is an inherent (albeit implicit) aspect of your notion of sequentially infinite set.</p>
<p>If you want to get fancier, for any integer $m\ge 2$ you could form the $m$ subsets $A_0,A_1,\dots,A_{m-1}$, where </p>
<p>$$A_k=\{a_k,a_{k+m},a_{k+2m}\dots\}=\{a_n:n\text{ leaves a remainder of }k\text{ on division by }m\}\;.$$</p>
<p>In this way you break the original set into $m$ exhaustive but non-overlapping sequentially infinite sets.</p>
|
1,147,373 | <p>I need to prove that
<span class="math-container">$$I = \int^{\infty}_{-\infty}u(x,y) \,dy$$</span>
is independent of <span class="math-container">$x$</span> and find its value, where
<span class="math-container">$$u(x,y) = \frac{1}{2\pi}\exp\left(+x^2/2-y^2/2\right)K_0\left(\sqrt{(x-y)^2+(-x^2/2+y^2/2)^2}\right)$$</span></p>
<p>and <span class="math-container">$K_0$</span> is the modified Bessel function of the second kind with order zero.
Evaluating the integral numerically with Mathematica for different values of <span class="math-container">$x$</span> gives the result of <span class="math-container">$2.38$</span>, but I want to know if it is possible to show analytically.</p>
<p>Increasing <span class="math-container">$x$</span> results in an increase of the exponential term on the left, but it also then strongly increases the argument of modified Bessel function, thus reducing its value.</p>
<p>To show that integral is independant of <span class="math-container">$x$</span>, it is sufficient to show that <span class="math-container">$\int^{\infty}_{-\infty}\frac{\, d}{\, dx}u(x,y) = 0$</span> but any differentiation looks more and more ugly.</p>
<p><strong>EDIT</strong>
Mathematica test:</p>
<pre><code> x = 100
NIntegrate[
(1/(2 Pi))* Exp[x*x/2 - y*y/2] BesselK[0,
Sqrt[(x - y)*(x - y) + (x*x/2 - y*y/2)*(x*x/2 -
y*y/2 )]], {y, -Infinity, x, Infinity}, MaxRecursion -> 22]
</code></pre>
<p>This gives an answer of <span class="math-container">$0.378936$</span> independent of the choice of <span class="math-container">$x$</span>. In the earlier calculation I missed the factor <span class="math-container">$\frac{1}{2\pi}$</span>.</p>
| Upax | 157,068 | <p>I think there is a typo somewhere. I tell you why. First of allI write your equation as folllow
<span class="math-container">\begin{equation}
u(x,y)=g(x,y) K_0(\rho(x,y))
\end{equation}</span>
where
<span class="math-container">\begin{equation}
g(x,y) =\frac{1}{2\pi}\exp\left(+x^2/2-y^2/2\right)
\end{equation}</span>
And
<span class="math-container">\begin{equation}
\rho(x,y) =\sqrt{(x-y)^2+(-x^2/2+y^2/2)^2}
\end{equation}</span>
Now please note that when differentiating <span class="math-container">$g(x,y)$</span> with respect to <span class="math-container">$x$</span> we have
<span class="math-container">\begin{equation}
g_x(x,y) =x g(x,y)
\end{equation}</span>
Then
<span class="math-container">\begin{equation}
\frac{d}{dx} u(x,y)=g(x,y) \left(x K_0(\rho(x,y))-K_1(\rho(x,y)) \rho_x(x,y)\right)=0
\end{equation}</span>
Differentiating <span class="math-container">$u(x,y)$</span> a second time we have:
<span class="math-container">\begin{equation}
\frac{d}{dx} u_x(x,y)=g_x(x,y) \left(x K_0(\rho(x,y))-K_1(\rho(x,y)) \right)+g(x,y) \left(K_0(\rho(x,y))-x K_1(\rho(x,y)) \rho_x(x,y)- K_1(\rho(x,y)) \rho_{xx}(x,y) +\frac{\rho^2_x(x,y)}{2} \left(K_0(\rho(x,y))+K_2(\rho(x,y)) \right)\right )=0
\end{equation}</span>
Setting the first and second derivative to 0, using the equation <span class="math-container">$g_x(x,y)= x g(x,y)$</span> and the property of the modified Bessel function of second kind:
<span class="math-container">\begin{equation}
\rho(x,y) \left(K_2(\rho(x,y))-K_0(\rho(x,y)) \right)=2 K_1(\rho(x,y))
\end{equation}</span>
It is possible to write the following expression:
<span class="math-container">\begin{equation}
\rho(x,y) \rho_x^3(x,y)+x \rho_x^2(x,y)+ \rho_x(x,y) \rho(x,y)(1-x^2)-\rho_{xx}\rho(x,y)=0
\end{equation}</span>
This equation must be satisfied by <span class="math-container">$\rho(x,y)$</span> defined above. If I dind't do any error (please check because I filled few pages and I'm not completely sure) the <span class="math-container">$\rho(x,y)$</span> as define above doesn't satisfy the differential equation for <span class="math-container">$\rho$</span>.</p>
|
3,513,581 | <p>I'm referring to Sakasegawa's forumla for calculating average line length in a queuing system.</p>
<p><a href="https://i.stack.imgur.com/Ydg3Y.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ydg3Y.png" alt="enter image description here"></a></p>
<p>I don't understand the result intuitively. For example, our server is able to serve 10 customers per hour, and we have 8 arriving every hour. So we have 1 queue and 1 server, and a utilization of 0.8:</p>
<ul>
<li>I think to myself, "we're underutilized, and not running at full capacity, so there should be no line". I.e. average line length = 0</li>
<li>But the actual answer from the formula is 3.2</li>
</ul>
<p>How can we have 3 people lined up if we are able to process them faster than they come in?? I just don't see it intuitively.</p>
| shortmanikos | 729,993 | <p>Disclaimer: I first heard of queueing theory today :)</p>
<p>I think the key is that 8 customers per hour do not arrive at a steady rate of 1 every 7.5 minutes but are randomly distributed. Taking an extreme example say you have 16 people appear at the start of the day. Nobody else arrives for the next two hours. </p>
<p>You obviously average 8 customers per hour for the first two hours and you have at least 6 people standing in line for 1 full hour. So for the first two hours your average queue is bigger than 3 people.</p>
<p>The key intuition is that there are times where your server sits around doing nothing.</p>
|
3,201,996 | <p>I have an 8-digit number and you have an 8-digit number - I want to see if our numbers are the same without either of us passing the other our actual number. Hashing the numbers is the obvious solution. However, if you send me your hashed number and I do not have it - it is very easy to hash all the permutations of an 8-digit number and see your number.</p>
<p>I am looking for a way to increase the complexity of the 8-digit number while maintaining uniqueness and a universal process (i.e. we need to be able to apply the same process on both ends.) Squaring the number or something like that will not work because there are the same number of unique squares of an 8 digit number as there are unique 8 digit permutations. Salting will not work for the same reason.</p>
<p>Is there anything I can do to the number to make brute-forcing all permutations not viable?</p>
| kelalaka | 338,051 | <p>In Cryptography, There are two interesting problems;</p>
<ol>
<li><a href="https://en.wikipedia.org/wiki/Yao%27s_Millionaires%27_Problem" rel="nofollow noreferrer">Yao's Millionaires Problem</a>: In which two millionaires want to know who is the reacher without revealing their riches to the other.</li>
<li><a href="https://en.wikipedia.org/wiki/Socialist_millionaires" rel="nofollow noreferrer">Socialist Millionaires</a> : In which two millionaires want to know their wealth is equal or not without revealing their riches to the other.</li>
</ol>
<p>You can think the first one as a secure <span class="math-container">$>$</span> and the second one as secure <span class="math-container">$=$</span> </p>
<p>Therefore, the second one can compare the values without revealing the value. </p>
<p><strong>Note 1:</strong> these problems are part of <a href="https://en.wikipedia.org/wiki/Secure_multi-party_computation" rel="nofollow noreferrer">secure multi-party computation</a>.</p>
<p><strong>Note 2:</strong> You can also solve this by using SPAKE2. I've given a complete answer in <a href="https://crypto.stackexchange.com/a/72740/18298">Cryptography</a></p>
|
2,547,888 | <p>A $\subset$ B, A and B closed then B-A is open</p>
<p>How can I prove this statement? Is that correct? I need it because is used to prove that a closed subset of a compact space is compact. </p>
<p>Imagine B=[0,4] and A=[1,2] contained in A, then B-A is neither open or closed, what am I doing wrong??</p>
<p>PS: maybe I need B compact in the statement, but the example works in every case.</p>
| hmakholm left over Monica | 14,366 | <p>What you're missing is that for "closed subset of a compact space is compact", $A$ is not only compact and closed: $A$ <em>is the entire space</em> and therefore <em>open</em>. Thus $A\setminus B = A\cap \overline B$ is open too.</p>
|
176,987 | <p>Consider a random $m$ by $n$ partial circulant matrix $M$ whose entries are chosen independently and uniformly from $\{0,1\}$ and let $m < n$. Now consider a random $n$ dimensional vector $v$ whose entries are also chosen independently and uniformly from $\{0,1\}$. Let $N = Mv$ where multiplication is performed over the reals.</p>
<p>The matrices $M$ and $N$ are discrete random variables. Recall that the Shannon entropy for a discrete random variable $Z$ is $H(Z) = -\sum_z P(Z=z)\log_2{P(Z=z)}$. In the case where $P(Z=z)=0$ for some values $z$, the corresponding term in the sum is taken to be $0$.</p>
<p>We therefore know that the (base $2$) Shannon entropy $H(M) = H(v) = n$. The fact that $H(M) = n$ is a direct result of the fact that the entire matrix is defined by its first row.</p>
<p>If $m = \lfloor 10n/\ln{n} \rfloor$ then I would like to make the following conjecture.</p>
<blockquote>
<p><strong>Conjecture:</strong> For all sufficiently large $n$, $H(N) \geq n/10$.</p>
</blockquote>
<p>The value $10$ is chosen somewhat arbitrarily to be a sufficiently large constant.</p>
<p>Is this a known problem and/or can anyone see a way to approach it? Is it in fact true?</p>
| john mangual | 1,358 | <p>This reminds me a bit of the game of <a href="https://en.wikipedia.org/wiki/Mastermind_(board_game)" rel="nofollow noreferrer">mastermind</a>.</p>
<p><img src="https://cimsec.org/wp-content/uploads/2013/04/mastermind13.jpg" width="200"></p>
<hr>
<p>Since your matrix is <a href="https://en.wikipedia.org/wiki/Circulant_matrix" rel="nofollow noreferrer">circulant</a>, this entire matrix is determined by the first row ${\bf m}$, the vector $X$ and the shift map taking $T:(m_1, \dots, m_n) \mapsto (m_2, \dots, m_n, m_1)$</p>
<p>You are given $m$ numbers (please excuse the notation clash): $T^i({\bf m})\cdot {\bf X}$ for $i = 1, \dots, m$ so there are at most $m \log n$ bits of information.</p>
<p>The bound $ \log n \cdot \tfrac{10n}{\log n} \geq H(N) \geq \tfrac{n}{10}$ is consistent with your problem, at least.</p>
|
2,018,071 | <p>The matrix is banded and symmetric, and is positive definite. It's very similar to a Laplacian matrix, though it's slightly modified, signs are flipped and every value is divided by the same constant. I need to find the smallest two eigenvalues in MATLAB, without using the eig() function. </p>
<p>Since I know it's positive definite and symmetric, I coded Rayleigh Quotient Iteration with an initial guess for the smallest eigenvalue as 0. So, I have the smallest eigenvalue and it's respective eigenvector. However, I'm not sure what to do next.</p>
<p>I can't transform matrix to send the smallest eigenvalue to exactly 0, since the value for the smallest eigenvalue isn't exact. The matrix is also extremely large and sparse, so I'd like to avoid taking the inverse of the matrix since that would be expensive.</p>
<p>Any tips would be appreciated. Thanks!</p>
| Max | 130,322 | <p>i might be wrong, because i think my idea sounds really naive: let $\lambda_{M}$ be the largest eigenvalue of $M$. (It should be possible to get a good approximation for the greatest eigenvalue of any symmetric positive definite matrix at least by an iterative method) Let $C\geq \lambda_M$ be any constant (to compensate the fact that we have an approximation for the greatest eigenvalue only). Let $I$ be the identity matrix. Consider $N:=CI-M$ and let $\nu_M$ be the greatest eigenvalue of $N$. Let $P=N-I\nu_M$ and $\rho_M$ the greatest eigenvalue of $P$. Now $C-\rho_M$ should be the second smallest eigenvalue of $M$ and $C-\nu_M$ should be the smallest.</p>
|
2,018,071 | <p>The matrix is banded and symmetric, and is positive definite. It's very similar to a Laplacian matrix, though it's slightly modified, signs are flipped and every value is divided by the same constant. I need to find the smallest two eigenvalues in MATLAB, without using the eig() function. </p>
<p>Since I know it's positive definite and symmetric, I coded Rayleigh Quotient Iteration with an initial guess for the smallest eigenvalue as 0. So, I have the smallest eigenvalue and it's respective eigenvector. However, I'm not sure what to do next.</p>
<p>I can't transform matrix to send the smallest eigenvalue to exactly 0, since the value for the smallest eigenvalue isn't exact. The matrix is also extremely large and sparse, so I'd like to avoid taking the inverse of the matrix since that would be expensive.</p>
<p>Any tips would be appreciated. Thanks!</p>
| pixel | 27,824 | <p>Matlab has <code>eigs</code> to give the largest or smallest few eigenvalues of a matrix. See the <a href="http://www.mathworks.com/help/matlab/ref/eigs.html" rel="nofollow noreferrer">documentation</a>. The <code>sm</code> option takes the matrix inverse, but the <code>sa</code> option does not, which might be more efficient in your case.</p>
|
1,590,262 | <p>Let $ABC$ be of triangle with $\angle BAC = 60^\circ$
. Let $P$ be a point in its interior so that $PA=1, PB=2$ and
$PC=3$. Find the maximum area of triangle $ABC$.</p>
<p>I took reflection of point $P$ about the three sides of triangle and joined them to vertices of triangle. Thus I got a hexagon having area double of triangle, having one angle $120$ and sides $1,1,2,2,3,3$. We have to maximize area of this hexagon. For that, I used some trigonometry but it went very complicated and I couldn't get the solution.</p>
| nikola | 213,926 | <p>Let E be the point outside triangle such that AEPC is a parallelogram , let F be point outside the triangle such that CPBF is a parallelogram, and let D be a point outside ABC such that APBC is paralellogram. Now, you have hexagon AECFBD with sides EC=1,CF=2,FB=3,BD=1, AD=2, EA=3, and draw line EB. Then, ADBE and EBFC are the same(congruent) and now surface of a hexagon is double of a surface of ADEB. Now, in that, you have three sides which are determined, plus AD is parallel to EB. Now consider ADBE. There, ED=1,DA=2,AB=3. Let M be perpendicular to EB from D and let N be perpendicular to EB from A. Now let EM be x, and NB be y. Let DM be h, as well. We have to maximise the surface of ADEB. We can see that surface is sum of surfaces of NMDA and EMD and ANB. So, it is $2h+\frac{hx}{2}+\frac{hy}{2}$. Relations that we have imply from triangles EMD and ANB, Now, those are $1=h^2+x^2, 9 = h^2+y^2$ so $y^2-x^2=8$ ,and $h^2=1-x^2$.</p>
|
1,977,577 | <p>if a function is lebesgue integrable, does it imply that it is measurable?
(without any other assumption)</p>
<p>The reason why I ask this is because royden, in his book, kind of imply about a measurable function when assuming the function to be lebesgue integrable</p>
| Adam | 82,101 | <p>By definition a function $f$ is called Lebesgue integrable if $f$ is measurable and $$\int |f(x)| \, \mu(dx) < \infty.$$</p>
<p>This has to be included in the definition because otherwise the term $$ \int |f(x)| \, \mu(dx)$$ is not defined.</p>
|
3,283,600 | <p>"Find all the complex roots of the following polynomials</p>
<p>A) <span class="math-container">$S(x)=135x^4 -324x^3 +234x^2 -68x+7$</span>, knowing that all its real roots belong to the interval <span class="math-container">$(0.25;1.75)$</span></p>
<p>B)<span class="math-container">$M(x)=(x^3 -1+i)(5x^3 +27x^2 -28x+6)$</span>
"</p>
<p>Well, in A) I don't know how to use the given information about real roots. I mean, I know that I can apply Bolzano but I don't think that's very useful. To find the complex roots I should have some information about a complex root in particular so that I could use Ruffini, but this is not the case.</p>
<p>And in B) I know that <span class="math-container">$(x^3 -1+i)$</span> is giving me some information related to a complex root, but that "^3" bothers me. If it wasn't there, I would know that <span class="math-container">$1-i$</span> is a root...</p>
| Lubin | 17,760 | <p>I’ll deal with (B) only. You have probably seen that <span class="math-container">$\frac35$</span> is a root, so that <span class="math-container">$x-\frac35$</span> is a factor, and when you divide <span class="math-container">$5x^3 +27x^2-28x+6$</span> by that, you get a <span class="math-container">$\Bbb Q$</span>-irreducible quadratic, which you’ll have to use the Quadratic Formula on.</p>
<p>The factor <span class="math-container">$x^3 - (1-i)$</span> actually is much easier. You’re looking for the three cube roots of <span class="math-container">$1-i=\sqrt2\bigl(\cos(-45^\circ)+i\sin(-45^\circ)\bigr)$</span>, since the point <span class="math-container">$(1,-1)$</span> is <span class="math-container">$\sqrt2$</span> units from the origin and on the ray <span class="math-container">$45^\circ$</span> clockwise from the <span class="math-container">$x$</span>-axis. You need the cube root of <span class="math-container">$\sqrt2$</span>, which you can write <span class="math-container">$2^{1/6}$</span> or <span class="math-container">$\sqrt[6]2$</span>. The angles that triple to <span class="math-container">$-45^\circ$</span> are <span class="math-container">$-15^\circ$</span>, <span class="math-container">$105^\circ$</span>, and <span class="math-container">$225^\circ$</span>. So your three linear factors of <span class="math-container">$x^3-(1-i)$</span> are <span class="math-container">$x-\bigl(\sqrt[6]2(\cos\theta+i\sin\theta)\bigr)$</span>, for <span class="math-container">$\theta$</span> taking each of the explicit angle values I mention above.</p>
|
1,303,183 | <p>If I have a vector space $V$ ( of dimension $n$ ) over real numbers such that $\{v_1,v_2...v_n\}$ is the basis for the space ( not orthogonal ). Then I can write any vector $l$ in this space as $l=\sum_i\alpha_iv_i$. Here $\alpha_1,\alpha_2...\alpha_n$ are the coefficients that define the vector $l$ according to this basis. Can another set of coefficients $\beta_1,\beta_2...\beta_n$ give the same vector $l$ ? If the basis was orthogonal the answer would be no, but I can't prove for a non orthogonal basis.</p>
| muaddib | 242,554 | <p>They cannot. Suppose so. Then $\sum \alpha_i v_i = \sum \beta_i v_i$. This implies:
$$0 = \sum (\alpha_i - \beta_i) v_i$$
which violates the basis being a set of linearly independent vectors.</p>
|
258,215 | <p>How can we show that if $f:V\to V$
Then for each $m\in \mathbb {N}$ $$\operatorname{im}(f^{m+1})\subset \operatorname{im}(f^m)$$
Please help,I am stuck on this.</p>
| Chris Eagle | 5,203 | <p>No, of course not. For example, let $f$ be a constant function, $f(x)=c$. Then for any $B\subset Y$, $f^{-1}(B)$ is either $X$ (if $c \in B$) or $\varnothing$ (if $c \not \in B$). So if $X$ has more than one element, then $f^{-1}$ is not surjective.</p>
|
1,549,490 | <p>$f(x) = 3x - \frac{1}{x^2}$</p>
<p>I am finding this problem to be very tricky:</p>
<p><a href="https://i.stack.imgur.com/7RjYG.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7RjYG.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/jwUHp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jwUHp.jpg" alt="enter image description here"></a></p>
| Dr. Sonnhard Graubner | 175,066 | <p>we have $$\frac{f(x+h)-f(x)}{h}=\frac{3(x+h)-\frac{1}{(x+h)^2}-3x+\frac{1}{x^2}}{h}$$ for the numerator of the right-hand side we obtain
$$\frac{h \left(3 h^2 x^2+6 h x^3+h+3 x^4+2 x\right)}{x^2 (h+x)^2}$$ thus we have $$\frac{f(x+h)-f(x)}{h}=\frac{3h^2x^2+6hx^3+h+3x^4+2x}{x^2(h+x)^2}$$
now you can calculate the limit</p>
|
905,672 | <p>Let $S$ be a subset of a group $G$ that contains the identity element $1$ and such that the left cosets $aS$ with $a$ in $G$, partition $G$.Prove that $S$ a is a subgroup of $G$.</p>
<p>My try:</p>
<p>For $h$ in $S$, If I show that $hS=S$, then that would imply that $S$ is closed. </p>
<p>Now $hS$ is a partiton of $S$ and contains $h$ since $1$ is in $S$. Also $h$ is in $S$. Hence $h \in S\cap hS$. Moreover both of these are partitions and two partitions are either disjoint or equal. Hence $S=hS$ which says that $S$ is closed. </p>
<p>Does this seem alright??</p>
<p>Thanks!!</p>
| Murtuza Vadharia | 132,945 | <p>It was difficult to follow your argument, but I think you got it right.
My try:
We will prove that S is closed under multiplication. Let a,b∈S be arbitrary. Since ab∈aS, it will be sufficient to prove that aS=S. Since e∈S, we have a=ae∈aS. By assumption, we also have a∈S=eS. Since aS and eS are both cells of the partition, they are either equal or disjoint. So the fact that they both contain aimplies that aS=eS=S.</p>
|
2,494,069 | <p>All the calculators I've tried overflow at 2^1024. I'm not sure how to go about calculating this number and potentially even larger ones.</p>
| James Garrett | 457,432 | <p>Just use the property $\ln(2^{2048})=2048\cdot \ln(2)$. Hence the answer is approximately $2048\cdot 0,693=1419.264$.</p>
|
268,635 | <p>Consider the triangle formed by randomly distributing three points on a circle. What is the probability of the center of the circle be contained within the triangle?</p>
| Joffan | 206,402 | <p>Take a circle with circumference $2$, and place the first point arbitrarily. Then the second point which will fall a distance $a$ away from that first point, where $a$ is somewhere between $0$ and $1$, with equal probability.</p>
<p>Then in order for the triangle to contain the centre of the circle, the third point must fall in a section of the circumference of length $a$ - which happens with probability $a/2$ - opposite the first two points, as illustrated:</p>
<p><a href="https://i.stack.imgur.com/Niv6M.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Niv6M.png" alt="enter image description here"></a></p>
<p>Note that the third point has to be on the other side of the diameter from the first point than the second point - so at least half the circle is always excluded.</p>
<p>Thus the probability can be evaluated as:</p>
<p>$$\int_0^1 \frac a2 \ da = \left[ \frac{a^2}{4} \right]_0^1 = \frac 14$$</p>
|
268,635 | <p>Consider the triangle formed by randomly distributing three points on a circle. What is the probability of the center of the circle be contained within the triangle?</p>
| Isaak | 560,931 | <p>Okay so not a math whiz or anything, so don't expect an elegant answer, but I did this question earlier today because I saw it online, and this is basically what I got:</p>
<p>You're placing 3 points (A, B and C) randomly around the circumference of a circle. What's the probability that a triangle drawn between those points will contain the center of the circle within its area?</p>
<p>The answer I got was 1/4. Here's why.
<strong>(I have written a summed up version at the bottom for people who don't like reading my boring writing, which is everyone)</strong>.</p>
<p>Point A can be placed anywhere. It doesn't matter because the other two points can all be measured relative to point A. I placed Point A at the top of the circle because that's visually easier for me, but it really doesn't matter. (Though I will be using language as if it were at the top of the circle henceforth).</p>
<p>Point B can also be placed anywhere, though there's a bit more here. If we divide down from point A into two even semi-circles, we can see that the probability of point B falling on either half must be 1/2.
However, because A will always be at the top, you can afford to disregard one whole half of the circle as candidates for point B. In my drawn diagram I chose to place point B on the anti-clockwise half of point A.
What I mean by that is even if point B does fall on the clockwise side of point A (It's randomly distributed after all, all points are possible.), then I could just flip the circle, effectively moving the point to the opposite side of the circle, without altering the relative distances between any of the points (A, B or the center itself).
To put it simply, we can assume that Point B will fall on the anti-clockwise side of the circle.</p>
<p>But can we gather more information about where B will fall? Well, not for any individual instance. But for the probability we can take averages. I don't know how many of you use dice, but when you work out the average of a dice you add the maximum and minimum values then divide by two (or add all values and divide by number of values. It's the same as how you work out regular averages).
In this case the lowest possible value of point B is 0, in that it falls in the exact same place as point A, there are 0 degrees of separation. </p>
<p>(By the way, I measure distances here in 'degrees of separation', which is basically just the angle formed by lines from the two points meeting in the center of the circle. I do this because there are no set distances, so angles and ratios are the only constants in this scenario.)</p>
<p>The most separation possible is 180 degrees. There's a completely even chance that point B could fall on or between any value between 0 and 180, which means the average value, if you selected 1000 different point B's at random, would be (0 + 180)/2 = 180/2 = 90. So we can say that on average there will be about 90 degrees of separation between points A and B.
Obviously, the majority of cases will not actually have exactly 90 degrees of separation, but for every case with more separation, there will (statistically speaking) be a case with equally less separation, so in the end it averages out to 90.</p>
<p>My next step was to draw lines from point A (at the top) and point B (at 90 degrees anticlockwise of point A) through the center of the circle and to the opposite side of the circle. The result is the circle divided into four quarters vertically and horizontally. For the triangle to contain the center of the circle, point C must fall within the opposite quarter to the one between Points A and B.
In other words, Point C must fall on or within the points which are directly opposite points A and B, in order to contain the center of the circle within the area of triangle ABC.
For a while I pondered the placing of point B, because the farther from point A point B is placed, the more likely point C will be within the desired area, but as before, for every more likely scenario, there's an equally likely scenario which is as equally unlikely for point C to fall within the desired area.
What this boils down to is that we can assume Point B is at its average, because that will ultimately give us the average probability.</p>
<p>Having decided this, we can clearly see that the zone of the circle within point C must fall to contain the center of the circle is one quarter of the entire circumference of the circle.</p>
<p>Therefore, for every point that point C could create a 'correct' solution there are 3 that do not.</p>
<p>There is a 1/4 chance that point C will fall in place to create a triangle which contains the center of the circle.</p>
<p><strong>If I were to try to sum everything up simply, I would say;</strong>
- Point A can be chosen arbitrarily, because no matter where it falls there's an equally likely chance of the other points falling in a 'correct' position.</p>
<ul>
<li><p>Point B will be, on average, separated by 90 degrees from point A. Therefore we can set point B as 90 degrees from point A to find the average zone within point C must fall.</p></li>
<li><p>Point C will have to, on average, fall within a zone which is equal to 1/4 of the entire circumference. This means that, on average, there's a 1/4 chance of point C falling in a place which creates a triangle containing the circle's center.</p></li>
</ul>
<p>Sorry for the painful read, but frankly I don't really have the mathematical vocabulary to simplify this all down into a compact equation.</p>
|
2,701,658 | <p>If <span class="math-container">$1^2+2^2+3^2 + ... + 10^2=385$</span>, then what is the value of <span class="math-container">$2^2+4^2+6^2 + ... + 20^2$</span>?</p>
<p>Options are <span class="math-container">$770$</span>, <span class="math-container">$1155$</span>, <span class="math-container">$1540$</span>, <span class="math-container">$7890$</span>.</p>
<p>I haven't tried it yet and I'm writing such extra things to meet the requirements to post the question...Plz solve the above question only</p>
| user061703 | 515,578 | <p>$1^2+2^2+3^2+...+10^2=385$</p>
<p>$\Rightarrow 4\times(1^2+2^2+3^2+...+10^2)=1540$</p>
<p>$\Rightarrow 1^2\times 4+2^2\times 4+3^2 \times 4+...+10^2 \times 4=1540$</p>
<p>$\Rightarrow 1^2\times 2^2+2^2 \times 2^2+3^2\times 2^2+...+10^2\times 2^2=1540$</p>
<p>$\Rightarrow (1\times2)^2+(2\times2)^2+(3\times2)^2+(4\times2)^2+...+(10 \times 2)^2=1540$</p>
<p>$\Rightarrow 2^2+4^2+6^2+8^2+...+20^2=1540$.</p>
<p>You can just reverse the steps above to find your answer.</p>
|
3,080,664 | <p>Hi I've almost completed a maths question I am stuck on I just can't seem to get to the final result. The question is:</p>
<p>Find the Maclaurin series of <span class="math-container">$g(x) = \cos(\ln(x+1))$</span> up to order 3.</p>
<p>I have used the formulas which I won't type out as I'm not great with Mathjax yet sorry. But I have obtained:</p>
<p><span class="math-container">$$ \ln(x+1) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} $$</span></p>
<p><span class="math-container">$$ \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} $$</span></p>
<p>I'm sure what I have done above is correct however I can't seem to get to the final solution shown in the answers and would like some help please. Thanks </p>
| Keen-ameteur | 421,273 | <p><span class="math-container">$f(x)=o(x^k)$</span> if:</p>
<p><span class="math-container">$\underset{x\rightarrow 0}{\lim} \frac{f(x)}{x^k}=0$</span></p>
<p>Since smaller powers of <span class="math-container">$x$</span> matter more when tending to <span class="math-container">$0$</span>, it essentially means that something is negligible relative to <span class="math-container">$x^k$</span> when tending to <span class="math-container">$0$</span>. Moreover by simple calculus one can see that if <span class="math-container">$f=o(x^m)$</span> and <span class="math-container">$g=o(x^\ell)$</span>, then:</p>
<p><span class="math-container">$f(x)+g(x)=o(x^{\min \{m,\ell \} })$</span></p>
<p>And for <span class="math-container">$k<\ell$</span> we have that:</p>
<p><span class="math-container">$a\cdot x^\ell = o(x^k)$</span></p>
<p>Therefore by uniqueness of the Taylor series, you want to write your function of the form:</p>
<p><span class="math-container">$\cos \big( \ln(1+x) \big)= p_3(x)+o(x^3)$</span></p>
<p>Where <span class="math-container">$p_3$</span> is a polynomial of <span class="math-container">$x$</span> with degree at most <span class="math-container">$3$</span>. If we denote by <span class="math-container">$T_m^f(x)$</span> the Taylor polynomial up to order <span class="math-container">$m$</span> at <span class="math-container">$0$</span>. Then by your calculations:</p>
<p><span class="math-container">$T_4^{\ln(x+1)}(x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}\quad$</span> and <span class="math-container">$\quad T_6^{\cos(x)}(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}$</span></p>
<p>Then the function <span class="math-container">$T_6^{\cos(x)}\circ T_4^{\ln(x+1)}(x) $</span> is a polynomial, and if you erase the powers negligible to <span class="math-container">$x^3$</span> in the final polynomial, then you obtain the Taylor polynomial at <span class="math-container">$0$</span> of order <span class="math-container">$3$</span>. In particular:</p>
<p><span class="math-container">$\cos(\ln(1+x))= 1 - \Big( T_4^{\ln(x+1)}(x) \Big)/2 + \Big( T_4^{\ln(x+1)}(x) \Big)^4/24+o(x^4)=$</span> </p>
<p><span class="math-container">$=1- \frac{1}{2} \Big( x- \frac{1}{2} x^2+ +\frac{x^3}{3} +o(x^3) \Big)^2 +o(x^4)=$</span></p>
<p><span class="math-container">$=1-\frac{1}{2} \Big( x^2- x^3 +o(x^3) \Big)+ o(x^4)= 1-\frac{x^2}{2}+\frac{x^3}{2}+o(x^3)$</span></p>
<p>Where I neglected things that I saw which were going to be of powers strictly greater than <span class="math-container">$3$</span>.</p>
<p>I think it is helpful to go by this algorithm when unsure.</p>
|
4,184,248 | <p>I'm trying to prove an inequality from an old book named "Algebra and elementary functions" by Kochetkov (ru - "Алгебра и элементарные функции" Е. С. Кочетков). This is an old book from 1970 (URSS era). The book is not copyrighted anymore, and it is freely available over the internet.</p>
<p>The inequality is <span class="math-container">$\frac{a}{\sqrt{b}} + \frac{b}{\sqrt{a}} \ge \sqrt{a} + \sqrt{b}$</span> (complex numbers are not allowed)</p>
<p>Here are my steps :</p>
<p><span class="math-container">$\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}} \ge \sqrt{a} + \sqrt{b} $</span></p>
<p><span class="math-container">$\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}} - \sqrt{a}-\sqrt{b} \ge 0$</span></p>
<p><span class="math-container">$\frac{a(\sqrt{a}-\sqrt{b})+b(\sqrt{b}-\sqrt{a}))}{\sqrt{ab}} \ge 0$</span></p>
<p>It is clear that if <span class="math-container">$a=b$</span>, then the expression <span class="math-container">$a(\sqrt{a}-\sqrt{b})+b(\sqrt{b}-\sqrt{a})$</span> is zero. Also intuitively I can see that in other cases the expression <span class="math-container">$a(\sqrt{a}-\sqrt{b})+b(\sqrt{b}-\sqrt{a})$</span> is greater than 0, but how to prove it formally ?</p>
| cansomeonehelpmeout | 413,677 | <p>Assume there are 3 distinct roots <span class="math-container">$r_1,r_2,r_3$</span>. Then we have the system of equations <span class="math-container">$$lr_1^2+mr_1+n=0\\lr_2^2+mr_2+n=0\\lr_2^2+mr_2+n=0\\$$</span> as <span class="math-container">$r_1\neq r_2\neq r_3$</span> we can add and subtract lines to get <span class="math-container">$$l(r_1+r_2)+m=0\\l(r_2+r_3)+m=0\\lr_2^2+mr_2+n=0$$</span> but then <span class="math-container">$l(r_1+r_2)=l(r_2+r_3)$</span> which is onlye possible if <span class="math-container">$l=0$</span>. If <span class="math-container">$l=0$</span> then <span class="math-container">$mr_1=mr_2$</span> which again yields that <span class="math-container">$m=0$</span>. And so we also have <span class="math-container">$n=0$</span>.</p>
|
1,220,002 | <p>I am performing the ratio test on series and I come across many situations where I have any of those 3 in the numerator and denominator and I was wondering what I could cancel as n approaches infinity.</p>
| Elaqqad | 204,937 | <p>We have the following orders we have $n!<<n^n$ and $(n+1)^n\sim e n^n$:</p>
<ul>
<li>When we are dealing with fractions with theses terms the best approsimation for $n!$ is given by Stirling's formula:
$$n\sim \left (\frac{n}{e} \right)^n\sqrt{2\pi n} $$</li>
<li>Note also that:
$$\left (1+\frac{x}{n}\right)^n=e^x $$</li>
</ul>
<p>so for example $(n+x)^{(n+y)}\sim e^{x}n^n$</p>
|
1,220,002 | <p>I am performing the ratio test on series and I come across many situations where I have any of those 3 in the numerator and denominator and I was wondering what I could cancel as n approaches infinity.</p>
| hunter | 108,129 | <p>No, neither of these are true. First,
$$
\lim_{n \to \infty} \frac{(n+1)^n}{n^n} = e.
$$
This is because
$$
\frac{(n+1)^n}{n^n} = \left(\frac{n+1}{n}\right)^n = \left(1 + \frac{1}{n}\right)^n
$$
The limit of the last expression is the definition of $e$.</p>
<p>On the other hand
$$
\lim_{n \to \infty} \frac{n!}{n^n} = 0.
$$
That's because
$$
\frac{n!}{n^n} = \frac{n \cdot (n-1) \cdot \ldots \cdot 1}{n \cdot n \cdot \ldots \cdot n} = \frac{n}{n} \cdot \frac{n}{n-1} \cdot \ldots \cdot \frac{1}{n},
$$
and that last product clearly goes to $0$ as $n \to \infty$.</p>
|
1,176,381 | <p>I have many sets containing three values like $\{1, -2, 5\}$.
I am want to write in mathematical form to filter set where exist one value with different sign. (and for sure, none of them should be zero)</p>
<p>I am not sure about tag. (please correct the tag if it is not correct)/ </p>
| Community | -1 | <p>How about ($\mathcal M$ is your family of sets) $\mathcal M \cap \left(F_1 \cup F_2 \cup F_3\right)$, where:
$$F_1 = \{\{a,b,c\} : a,b \in \mathbb R_+, c \in \mathbb R_-\} \\
F_2 = \{\{b,c,a\} : a,b \in \mathbb R_+, c \in \mathbb R_-\} \\
F_3 = \{\{c,a,b\} : a,b \in \mathbb R_+, c \in \mathbb R_-\}. \\
$$</p>
|
2,472,313 | <blockquote>
<p>Suppose $x$ and $y$ are real numbers and $x^2+9y^2-4x+6y+4=0$. Then find the maximum value of $\left(\frac{4x-9y}{2}\right)$</p>
</blockquote>
| user8277998 | 516,330 | <p>Use parameterisation, </p>
<p>$$x^2+9y^2-4x+6y+4=(x-2)^2+(3y+1)^2- 1 = 0$$</p>
<p>Let $x - 2 = \cos t$ and $3y + 1 = \sin t$, then</p>
<p>$$f(t) := \dfrac{4x - 9y}{2} = \dfrac{4(\cos t + 2) - 3(\sin t - 1)}{2} = \dfrac{4\cos t - 3\sin t + 11}{2}$$</p>
<p>Equating first derivative to zero we get, $f^\prime(t) = 0$ if $x = 2n \pi + 2 \tan^{-1}(3)$ or $x = 2n \pi - 2 \tan^{-1}\left(\dfrac 13\right)$ where $n \in \Bbb Z$.</p>
<p>Using second derivative test the maxima is given at $x = 2n \pi - 2 \tan^{-1}\left(\dfrac 13\right)$ and the maxima
is $f\left(-\tan^{-1}\left(\dfrac13\right)\right) = 8$.</p>
|
1,972,002 | <p>Show an open set $A \subseteq \mathbb{R}$ contains no isolated points.</p>
<p>My attempt:
well if we can show an arbitrary element $a \in A$ is a limit point then we will be done. So since $A$ is open, there is an $\varepsilon > 0$ such that $B_\varepsilon(a) = (a - \varepsilon, a + \varepsilon) \subseteq A$. To show $a$ is a limit point, we need to construct a sequence $(a_n)$ in $A$ with $a_n \neq a$ for every $n \in \mathbb{N}$ and such that $(a_n) \rightarrow a$. I'm confused on how to construct such a sequence. Or is it easier to prove this statement by contradiction? Thanks.</p>
| Mark Joshi | 106,024 | <p>let $a_n = a + \epsilon/(2n)$ . This is a sequence in the set that converges to $a.$</p>
|
4,132,439 | <p>I know that <span class="math-container">$\int_1^\infty \frac1xdx$</span> diverges. I can probe that <span class="math-container">$\int_1^\infty \frac1{\ln(x)}dx$</span> diverges, as <span class="math-container">$\forall x>1:\frac1x<\frac1{\ln(x)}$</span></p>
<p>I also know that <span class="math-container">$\int_1^\infty \frac1{x^2}dx$</span> converges <span class="math-container">$ \therefore \int_1^\infty \frac1{x^7}dx$</span> also converges.</p>
<p>I know that <span class="math-container">$\forall x>1:x>\ln(x) \rightarrow x^7>\ln^7(x)\rightarrow \frac1{x^7}<\frac1{\ln^7(x)}$</span> but this doesn't help me much.</p>
| Yuta73 | 435,516 | <p>Nevermind, I made variable subtitution of <span class="math-container">$t=ln(x) \rightarrow e^tdt=dx$</span></p>
<p>So the integral ends up as
<span class="math-container">$$\int_0^\infty \frac{e^t}{t^7}dt$$</span>
But the limit of the integrand goes to infinity
<span class="math-container">$$\lim_{t\to+\infty} \frac{e^t}{t^7}=\lim_{t\to+\infty} \frac{e^t}{7t^6}=...=\lim_{t\to+\infty} \frac{e^t}{7!}\to+\infty$$</span>
<span class="math-container">$\therefore \int_0^\infty \frac{e^t}{t^7}dt$</span> diverges so <span class="math-container">$\int_1^\infty \frac1{ln^7(x)}dx$</span> diverges</p>
|
4,132,439 | <p>I know that <span class="math-container">$\int_1^\infty \frac1xdx$</span> diverges. I can probe that <span class="math-container">$\int_1^\infty \frac1{\ln(x)}dx$</span> diverges, as <span class="math-container">$\forall x>1:\frac1x<\frac1{\ln(x)}$</span></p>
<p>I also know that <span class="math-container">$\int_1^\infty \frac1{x^2}dx$</span> converges <span class="math-container">$ \therefore \int_1^\infty \frac1{x^7}dx$</span> also converges.</p>
<p>I know that <span class="math-container">$\forall x>1:x>\ln(x) \rightarrow x^7>\ln^7(x)\rightarrow \frac1{x^7}<\frac1{\ln^7(x)}$</span> but this doesn't help me much.</p>
| Sanjoy Kundu | 245,640 | <p>Take <span class="math-container">$x = e^u$</span>. Your integral transforms to <span class="math-container">$\int_0^\infty \frac{e^u}{u^7}du$</span>. You can now compare this with <span class="math-container">$\int_0^\infty \frac{1}{u^7}du$</span>, since <span class="math-container">$1 < e^u$</span> for all <span class="math-container">$u > 0$</span>. Since <span class="math-container">$\int_0^\infty \frac{1}{u^7}du$</span> diverges, it follows that <span class="math-container">$\int_0^\infty \frac{e^u}{u^7}du$</span> must diverge via Direct Comparison.</p>
|
557,468 | <p>Let $f$ be a continuous map from $[0,1]$ to $[0,1].$ Show that there exists $x$ with $f(x)=x. $</p>
<p>I have $f$ being a continuous map from $[0,1]$ to $[0,1]$ thus $f: [0,1]\to [0,1]$. Then I know from the intermediate value theorem there exists an $x$ with $f(x)=x$ but I don't know how to formally prove it? </p>
<p>Is there another way of proving this besides using $g(x) = f(x) - x$? </p>
| Community | -1 | <p>If it helps, you can think of this problem graphically as saying that for any function $f:[0,1]\to[0,1]$, $f$ must cross the diagonal of the square with vertices at $(0, 0),\ (1,1).$</p>
<p>Here's a picture:
<img src="https://i.stack.imgur.com/6ji5w.png" alt="enter image description here"></p>
<p>Hopefully it's clear from this picture that $f$ is going to have to cross this diagonal, since $f$ starts somewhere on the $y$-axis and and ends up somewhere on the rightmost side of the square.</p>
<p>Now you can use your idea with the function $g(x)=f(x)-x$. Note that $g(0)=f(0) \geq 0$, and $g(1)=f(1)-1 \leq 0$. Can you finish the rest? </p>
|
557,468 | <p>Let $f$ be a continuous map from $[0,1]$ to $[0,1].$ Show that there exists $x$ with $f(x)=x. $</p>
<p>I have $f$ being a continuous map from $[0,1]$ to $[0,1]$ thus $f: [0,1]\to [0,1]$. Then I know from the intermediate value theorem there exists an $x$ with $f(x)=x$ but I don't know how to formally prove it? </p>
<p>Is there another way of proving this besides using $g(x) = f(x) - x$? </p>
| user104235 | 104,235 | <p>Can anyone check my answer to see if I understand this?</p>
<p>We have three cases: When $f(x) > x$ for every $x \in [0,1]$. When $f(x)< x$ for every $x \in [0,1]$ And finally, when $f(x)<x$ at some points of $x$ and $f(x)>x$ at other points of $x$. </p>
<p>If $f(x)> x$ at some points of $x\in [0,1]$ and $f(x)<x$ at other points of $x\in [0,1]$ then we can use the IVT and we are done. </p>
<p>Now, consider when $f(x) >x$ for all $x\in [0,1].$ Now, to map $[0,1] \to [0,1]$ we must have that $f(0)$ must exist and since $f$ is a continuous function such that $f: [0,1]\to [0,1]$ we have that $f(0) = 0$ which contradicts $f(x) < x$ for all $x\in [0,1].$</p>
<p>Now consider when $f(x) < x$ then in order to map $f: [0,1]\to [0,1]$ we must have the function $f(1)$ exists but this is not that case since $f(x)< x$ for all $x$ which contradicts the choice of $x\in [0,1]$. Hence contradiction. </p>
<p>Is this correct?</p>
|
2,555,811 | <p>Is there a name for the following type of ordering on some set $S$ {$a,b,c$} that includes only $>$ and $=$ for example: </p>
<p>$$a>b>c$$
$$a>b=c$$
$$a=b>c$$
$$a=b=c$$</p>
<p>Is there some name for these orderings? </p>
<hr>
<p>I know that all these <em>satisfy</em> a <em>total preorder</em> on $S$, since a preoder on $S$ is simply one in which the elements are ordered by the $\geq$ relation. But is there a name for these particular orderings?</p>
<p>Are my examples all instances of total orderings, since all members are comparable?</p>
<p>Is it okay to call these simply various "orderings" on $S$?</p>
| robjohn | 13,854 | <p>Since $e^t=1+t+O\!\left(t^2\right)$, set $t=\frac{\log(k)}x$
$$
\begin{align}
\lim_{x\to\infty}\left(\frac1n\sum_{k=1}^nk^{1/x}\right)^{nx}
&=\lim_{x\to\infty}\left(1+\frac1n\sum_{k=1}^n\frac{\log(k)}x+O\!\left(\frac1{x^2}\right)\right)^{nx}\\
&=\lim_{x\to\infty}\left(1+\frac1n\sum_{k=1}^n\frac{\log(k)}x\right)^{nx}\lim_{x\to\infty}\left(1+O\!\left(\frac1{x^2}\right)\right)^{nx}\\[3pt]
&=\lim_{x\to\infty}\left(1+\frac1n\frac{\log(n!)}x\right)^{nx}\cdot1\\[9pt]
&=e^{\log(n!)}\\[15pt]
&=n!
\end{align}
$$</p>
|
2,372 | <p>How can I find the number of <span class="math-container">$k$</span>-permutations of <span class="math-container">$n$</span> objects, where there are <span class="math-container">$x$</span> types of objects, and <span class="math-container">$r_1, r_2, r_3, \cdots , r_x$</span> give the number of each type of object?</p>
<p>I'm still looking for the solution to this more general problem out of interest.</p>
<p>Here is an example with <span class="math-container">$n = 20, k = 15, x = 4,$</span> <span class="math-container">$ r_1 = 4 \quad r_2 = 5 \quad r_3 = 8 \quad r_4 = 3$</span>.</p>
<blockquote>
<p>I have 20 letters from the alphabet. There are some duplicates - 4 of them are <em>a</em>, 5 of them are <em>b</em>, 8 of them are <em>c</em>, and 3 are <em>d</em>. How many unique 15-letter permutations can I make?</p>
</blockquote>
<h2>Edits:</h2>
<p>I've done some more work on this problem but haven't really come up with anything useful. Intuition tells me that as Douglas suggests below there will probably not be an easy solution. However, I haven't been able to prove that for sure - does anyone else have any ideas?</p>
<p>I've now re-asked this question on <a href="https://mathoverflow.net/questions/37211/permutations-with-identical-objects">MO</a>.</p>
| Douglas S. Stones | 139 | <p>There's probably not going to be an easy way to do this... Consider two different examples of 15-letter "permutations". Then the number of permutations with that multiset of digits depends on the proportions of the chosen digits.</p>
<p>If you really want to do this, you can sum $k!/(s(1)! s(2)! \cdots s(x)!)$ (called the <a href="http://en.wikipedia.org/wiki/Multinomial_theorem" rel="nofollow noreferrer">mulitnomial coefficient</a>) over all partitions of $s(1)+s(2)+ \cdots +s(x)=k$ [in this partition s(i) is allowed to be zero and order is important] such that $s(i) \leq r(i)$ for all i. The part s(i) says you have s(i) copies of the i-th letter. The number of permutations with s(i) i-th letters is given as above, by the <a href="http://en.wikipedia.org/wiki/Group_action" rel="nofollow noreferrer">Orbit Stabiliser Theorem</a>.</p>
<p>Although, this is only one step better than the caveman's counting formula: sum_P 1 where P is the set of permutations you want to count. I.e. just count them one-by-one.</p>
<p>EDIT: While I'm making a few touch-ups, here's GAP code that implements the above formula.</p>
<pre><code>NrPermIdent:=function(k,T)
local PSet,x;
x:=Size(T);
PSet:=Filtered(OrderedPartitions(k+x,x)-1,p->ForAll([1..x],i->p[i]<=T[i]));
return Sum(PSet,p->Factorial(k)/Product(p,i->Factorial(i)));
end;;
</code></pre>
<p>where T is a list of bounds and k is the number of terms in the partition.</p>
<p>For example:</p>
<pre><code>gap> NrPermIdent(15,[4,5,8,3]);
187957770
</code></pre>
<p>As another indication that finding a simple formula for these numbers is not going to be easy, observe that NrPermIdent(n,[n,k]) is equal to $\sum_{0 \leq i \leq k} {n \choose k}$ (which is considered a difficult sum to find -- see: <a href="https://mathoverflow.net/questions/17202/">https://mathoverflow.net/questions/17202/</a>). I remember reading somewhere (most likely in <a href="http://www.math.upenn.edu/~wilf/AeqB.html" rel="nofollow noreferrer">A=B</a>) that you can prove there is no "closed-form" solution for this.</p>
|
1,849,003 | <p>The following question was asked in an theoretical computer science entrance exam in India:</p>
<blockquote>
<p>A spider is at the bottom of a cliff, and is n inches from the top.
Every step it takes brings it one inch closer to the top with
probability $1/3$, and one inch away from the top with probability
$2/3$, unless it is at the bottom in which case, it always gets one
inch closer. What is the expected number of steps for the spider to
reach the top as a function of $n$?</p>
</blockquote>
<p>How do we solve this problem? The recurrence for the expectation seems tricky because of the boundary conditions :(</p>
| Jaydeep Pawar | 761,596 | <p>The recurrence relation can be thought as,</p>
<p>T<sub>n</sub> = (1/3)<em>(T<sub>n-1</sub> + 1) + (2/3)</em>(T<sub>n-1</sub> + 1 + T<sub>n</sub> - T<sub>n-2</sub>)</p>
<p>On further simplification</p>
<p>T<sub>n</sub> = 3* T<sub>n-1</sub> - 2*T<sub>n-2</sub> + 3</p>
<p>The intuition is: To find expected number of steps to reach n<sup>th</sup> inch we first need to reach (n-1)<sup>th</sup> inch now after reaching (n-1)<sup>th</sup> inch we need 1 more step with probability 1/3 to complete n<sup>th</sup> inch this "move forward" step incurred additional cost of 1 otherwise with probability 2/3 we have to move 1 inch behind from (n-1)<sup>th</sup> inch so this "move back" step incurred additional cost 1 so we are now at (n-2)<sup>th</sup> inch, expected number of steps to complete n inches is T<sub>n</sub> and expected number of steps to complete n-2 inches is T<sub>n-2</sub> so after reaching (n-2)<sup>th</sup> inch we require T<sub>n</sub>-T<sub>n-2</sub> additional cost to complete n inches.
Hope this helps</p>
|
602,248 | <p>This is the system of equations:
$$\sqrt { x } +y=7$$
$$\sqrt { y } +x=11$$</p>
<p>Its pretty visible that the solution is $(x,y)=(9,4)$</p>
<p>For this, I put $x={ p }^{ 2 }$ and $y={ q }^{ 2 }$.
Then I subtracted one equation from the another such that I got $4$ on RHS and factorized LHS to get two factors in terms of $p$ and $q$.</p>
<p>Then $4$ can be represented as $2*2$, $4*1$ or $1*4$. Comparing the two factors on both sides, I got the solution.</p>
<p>As you can see, the major drawback here is that I assumed this system has only integral solutions and then went further. Is there any way I can prove that this system indeed has only integral solutions or is there any other elegant way to solve this question?</p>
| egreg | 62,967 | <p>You want
\begin{cases}
x=(7-y)^2\\
y=(11-x)^2\\
0\le x\le 11\\
0\le y\le 7
\end{cases}</p>
<p>The equation becomes
$$
x=49-14(121-22x+x^2)+(121-22x+x^2)^2
$$
which reduces to
$$
(x-9)(x^3-35x^2+397x-1444)=0
$$
(courtesy of WolframAlpha). The polynomial $f(x)=x^3-35x^2+397x-1444$ has at least one real root. It has indeed three. One of them satisfies the condition $0\le x\le 11$ and it's approximately $7.87$. With this value of $x$ we get $y\approx 9.79$ that doesn't satisfy $0\le y\le 7$.</p>
<p>One can be more precise: call $\alpha$ the least root of $f$. Then $7<\alpha<8$, so that $3<11-\alpha<4$ and so $9<(11-\alpha)^2<16$, which shows that the limitation on $y$ is not fulfilled.</p>
<p>How can you know this? Compute</p>
<ul>
<li>$f(7)=-37$</li>
<li>$f(8)=4$</li>
<li>$f(12)=8$</li>
<li>$f(13)=-1$</li>
<li>$f(15)=11$</li>
</ul>
<p>Thus you know that the three roots of $f$ are $7<\alpha<8$, $12<\beta<13$ and $13<\gamma<15$.</p>
|
4,461,482 | <p>In <span class="math-container">$ΔABC$</span>, if <span class="math-container">$(a+b+c)(a−b+c)=3ac$</span>, then which of the following is <span class="math-container">$\color{green}{\text{True}}$</span>?</p>
<ul>
<li><span class="math-container">$\angle B=60^\circ $</span></li>
<li><span class="math-container">$\angle B=30^\circ $</span></li>
<li><span class="math-container">$\angle C=60^\circ $</span></li>
<li><span class="math-container">$\angle A + \angle B=120^\circ $</span></li>
</ul>
| Wang YeFei | 955,668 | <p>You must show your attempts here first before any help can be given to you. As <strong>a hint</strong>: expand the left side and use the identity: <span class="math-container">$b^2=a^2+c^2-2ac\cos B$</span> to solve for <span class="math-container">$\cos B$</span> and then for <span class="math-container">$B$</span>.</p>
|
255,902 | <p>I have a double lattice sum and I was wondering how I could calculate this with Mathematica. In particular, I have a function <span class="math-container">$F:\mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}$</span> which takes as arguments a pair of points <span class="math-container">$x,y\in\mathbb{R}^2$</span>, and I want to evaluate the sum</p>
<p><span class="math-container">$$
\sum_{x\in L}\sum_{y\in L} F(x,y)
$$</span></p>
<p>where <span class="math-container">$L\subset\mathbb{R}^2$</span> is some finite lattice, i.e. a collection of points in <span class="math-container">$\mathbb{R}^2$</span>.</p>
<p>Now, I have tried using <code>Outer</code> for this via</p>
<p><code>Total[Flatten@Outer[F[#1,#2] &, L, L]]</code></p>
<p>but with this, I obtain a sum of <span class="math-container">$F$</span> evaluated at real pairs <span class="math-container">$s,t\in\mathbb{R}$</span></p>
<p><code>F[s,t] + ...</code></p>
<p>which doesn't make sense, instead of <span class="math-container">$F$</span> evaluated at real vector pairs <span class="math-container">$x=(x_1,x_2),y=(y_1,y_2)\in\mathbb{R}^2$</span></p>
<p><code>F[{x_1,x_2},{y_1,y_2}] + ...</code></p>
<p>as it should be. I'm not sure how to remedy this, but any help with this would be much appreciated.</p>
<p>I am also open to recommendations for any alternative/better ways to perform such a double sum.</p>
<p><strong>Extra Information</strong>
My function <code>F</code> is defined as a conditional, i.e.</p>
<p><code>F[{x_,y_},{s_,t_}] = If[Abs[g[{x, y}]] < eps || Abs[g[{s, t}]] < eps, 0, W[{x - s, y - t}] * g[{x, y}] * g[{s, y}]]</code></p>
<p>for some other lattice functions <code>g</code> and <code>W</code>.</p>
<p>Thanks to answer below, I am able to evaluate the summation, but I obtain as an answer</p>
<p><code>If[Abs[\[Piecewise] 1 {-(27/2),(7 Sqrt[3])/2}==0&&0<={-27,7 Sqrt[3]}<=2 Sqrt[3]&&{-(27/2)+8 Sqrt[3],Sqrt[3]/2}==0 0 True ]<2.22045*10^-14,0,W[{{-(27/2),(7 Sqrt[3])/2},{-(27/2),(7 Sqrt[3])/2}}] g[{{-(27/2),(7 Sqrt[3])/2},{-(27/2),(7 Sqrt[3])/2}}]] +...</code></p>
<p>It seems that the double summation has not been executed entirely.</p>
<p><strong>Edit</strong></p>
<p>Thank you for all the solutions and comments, they have been helpful and I shall surely use those in the future. Sadly, due to the nature of my functions, they didn't work perfectly, but that was just because of how my functions were defined.</p>
<p>Although, I have found a solution that seems to work: To evaluate the double sum, I used the Map function twice</p>
<p><code>Total[Flatten@Map[Function[y, Map[Function[x, F[x,y]], L, L]]]]</code></p>
<p>I found this solution in another post given by <a href="https://mathematica.stackexchange.com/questions/15480/how-do-i-designate-arguments-in-a-nested-map">@halirutan</a></p>
| Domen | 75,628 | <p>You can use <code>Sum</code> to iterate twice through the set of lattice points.</p>
<pre><code>lattice = Catenate@Table[{x, y}, {x, 0, 3}, {y, 0, 3}]
(* {{0, 0}, {0, 1}, {0, 2}, {0, 3}, {1, 0}, {1, 1}, {1, 2}, {1, 3}, {2,
0}, {2, 1}, {2, 2}, {2, 3}, {3, 0}, {3, 1}, {3, 2}, {3, 3}} *)
Sum[F[p1, p2], {p1, lattice}, {p2, lattice}]]
(* F[{0, 0}, {0, 0}] + F[{0, 0}, {0, 1}] + F[{0, 0}, {0, 2}] +
F[{0, 0}, {0, 3}] + ... *)
</code></pre>
<hr />
<p>Edit:
<em>The following old answer can be significantly less memory efficient when performed on a large set of lattice points, as kindly warned by Henrik.</em></p>
<pre><code>Total[Flatten@Table[F[p1, p2], {p1, lattice}, {p2, lattice}]]
</code></pre>
|
1,969,479 | <blockquote>
<p>Prove that the symmetric point of orthocenter $H$ of a triangle with respect to the midpoint of any side resides on the triangle's circumcircle.
<a href="https://i.stack.imgur.com/tkh4c.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tkh4c.png" alt="enter image description here"></a></p>
</blockquote>
<p>In the above figure,it's sufficient to prove that triangle $BHM$ is equal to triangle $CKM$, but how? I tried to make use of the symmetric point of $H$ with respect to $BC$ which resides on circumcircle but failed...</p>
| Community | -1 | <p>Use that BHCK is a quadrilateral whose diagonals intersect in the middle. So is a parallelogram. Then show that BHCK is inscribed.</p>
|
172,157 | <p>Today I was asked if you can determine the divergence of $$\int_0^\infty \frac{e^x}{x}dx$$ using the limit comparison test.</p>
<p>I've tried things like $e^x$, $\frac{1}{x}$, I even tried changing bounds by picking $x=\ln u$, then $dx=\frac{1}{u}du$. Then the integral, with bounds changed becomes $\int_1^\infty \frac{1}{\ln u}du$ This didn't help either.</p>
<p>This problem intrigued me, so any helpful pointers would be greatly appreciated.</p>
| Robert Israel | 8,508 | <p>Comparison to $e^{x/2}$ should work (taking $x \to \infty$). So should $1/x$ (either as $x \to 0+$ or as $x \to \infty$). </p>
|
1,563,004 | <p>Assume we that we calculate the expected value of some measurements $x=\dfrac {x_1 + x_2 + x_3 + x_4} 4$. what if we dont include $x_3$ and $x_4$, but instead we use $x_2$ as $x_3$ and $x_4$. Then We get the following expression $v=\dfrac {x_1 + x_2 + x_2 + x_2} 4$.</p>
<p>How do I know if $v$ is a unbiased estimation of $x$?</p>
<p>I am not sure how to approach this problem, any ideas are appreciated!</p>
| Olivier Oloa | 118,798 | <p><strong>Hint</strong>. You may use
$$
(u \times v)'=u'v+uv'
$$ in the form
$$
\left(10^x\log_{10}(x)\right)'=\left(e^{x \ln 10} \times \frac{ \ln x}{\ln 10}\right)'.
$$ Can you take it from there?</p>
|
971,333 | <p>Given $\int \int dxdy$, I want to find the area bounded by $y=\ln\left(x\right)$ and $y=e+1-x$, and the $x$ axis. </p>
<p>I think the limits of integral in $y$ axis are from $y=\ln\left(x\right)$ to $y=e+1-x$ so $\int \limits_{y=\ln\left(x\right)}^{e+1-x}dy\int \:dx$, but I don't know how to find the limits of the integral in $x$ axis since the area bounded by $x$ axis? Please give me a clue to solve this problem.</p>
| g.kov | 122,782 | <p>In this case it is convenient to use inverse representation of the curves,
$x=\mathrm{e}^y,\ x=\mathrm{e}+1-y$:</p>
<p><img src="https://i.stack.imgur.com/AFeBR.png" alt="enter image description here"></p>
<p>$$ \int_0^1 \int_{\mathrm{e}^y}^{\mathrm{e}+1-y} dx dy=\frac{3}{2}$$.</p>
|
17,285 | <p>I have problems in understanding few concepts of elementary set theory. I've choosen a couple of problems (from my problems set) which would help me understand this concepts. To be clear: it's not a homework, I'm just trying to understand elementary set's theory concepts by reading solutions.
<hr/>
<strong>Problem 1</strong></p>
<p>(I don't understand this; I mean - not at all)</p>
<p>Let $f: A \to B$, where $A,B$ are non-empty sets, and let $R$ be an equivalence relation in set $B$. We define equivalence relation $S$ in $A$ set by condition:</p>
<p>$aSb \Leftrightarrow f(a)R f(b)$
Determine, which inclusion is always true:</p>
<p>(a) $f([a]_S) \subseteq [f(a)]_R$</p>
<p>(b) $[f(a)]_R \subseteq f([a]_S)$</p>
<p>Notes:</p>
<p>$[a]_S$ is an equivalence class
<hr/>
<strong>Problem 2</strong></p>
<p>(I suppose, that (a) is true & (b) is false)</p>
<p>Which statement is true, and which is false (+ proof):</p>
<p>(a) If $f: A \xrightarrow{1-1} B$ and $f(A) \not= B$ then $|A| < |B|$</p>
<p>(b) If $|A| < |B|$ and $C \not= \emptyset$ then $|A \times C| < |B \times C|$
<hr/>
<strong>Problem 3</strong></p>
<p>(I don't know, how to think about $\mathbb{Q}^{\mathbb{N}}$ and $\{0,1\}^∗$.)</p>
<p>Which sets have the same cardinality:</p>
<p>$P(\mathbb{Q}), \mathbb{R}^{\mathbb{N}},\mathbb{Z}, \mathbb{Q}^{\mathbb{N}}, \mathbb{R} \times \mathbb{R}, \{ 0,1 \}^*, \{ 0,1 \}^{\mathbb{N}},P(\mathbb{R})$</p>
<p>where $\{ 0,1 \}^*$ means all finite sequences/words that contains $1$ and $0$, for example $000101000100$ or $1010101010101$ etc. $P(A)$ is a Power Set.
<hr/>
<strong>Problem 4</strong></p>
<p>(I don't understand this; I mean - not at all)</p>
<p>What are: maximum/minimum/greatest/lowest elements in set:</p>
<p>$\{\{2; 3; 3; 5; 2\}; \{1; 2; 3; 4; 6\}; \{3\}; \{2; 1; 2; 1\}; \{1; 2; 3; 4; 5\}; \{3; 4; 2; 4; 1\}; \{2; 1; 2; 2; 1\}\}$</p>
<p>ordered via subset inclusion
<hr/>
<strong>Problem 5</strong></p>
<p>How many equivalence relations there are in $\mathbb{N}$ which also are partial order?
<hr/>
These are simple problems, but I really need to understand, how to solve this kind of problems. I would apreciate Your help.</p>
| Jason DeVito | 331 | <p>I'll do 1 and 5.</p>
<p>For one, the first inclusion is true. For let $x\in f([a]_b)$. We want to show that $x\in [f(a)]_r$, i.e., that $x$ $r$ $f(a)$. Since $x\in f([a]_s)$, there is a $b\in A$ such that $b$ $s$ $a$ and f(b) = x. Since $b$ $s$ $a$, we know $f(b)$ $r$ $f(a)$. Thus, $x$ $r$ $f(a)$ so $x\in [f(a)]_r$.</p>
<p>The reverse inclusion is false. Suppose $A = \{a\}$ and $B = \{b,c\}$ with $b$ and $c$ distinct. Use the relation $r = B\times B$ - all elements are $r$-related to all elements. Use the function $f:A\rightarrow B$ given by $f(a) = b$. Then it's not too hard to show that $c\in [f(a)]_r$ but $c\notin f([a]_s)$.</p>
<p>For 5, notice that the only difference between a partial order and an equivalence relation is in the symmetry/antisymmetry. In other words, if $r$ is an equivalence relation, then $a$ $r$ $b$ implies $b$ $r$ $a$. However, if $r$ is a partial order, then $a$ $r$ $b$ and $b$ $r$ $a$ implies $a = b$.</p>
<p>(To be clearer, I think there are several competing definitions of "partial order" - some, drop the antisymmetry condition. If that's the case, ignore my answer to number 5).</p>
<p>Now, suppose $r$ is an equivalence relation and a partial order. Given $a$, what are all the elements it can be related to?</p>
<p>Well, suppose $a$ $r$ $b$. Then we conclude that $b$ $r$ $a$ since $r$ is an equivalence relation. Since $r$ is a partial order, we conclue from this that $a = b$. Said another way, if $a\neq b$, then $a$ $r$ $b$ cannot be true. Thus, the only possibility to consider is whether or not $a$ $r$ $a$. But in an equivalence relation, this must be the case.</p>
<p>Hence, our relation $r$ is simply the set of all ordered pairs $(a,a)$ - there is a unique relation on $\mathbb{N}$ (or any set!) which is both an equivalence relation and a partial order.</p>
|
4,642,772 | <p>I am trying to understand this proof:
<a href="https://math.stackexchange.com/questions/3201004/prove-that-varphipk-pk-pk-1-for-prime-p">Prove that $\varphi(p^k)=p^k-p^{k-1}$ for prime $p$</a></p>
<p>At some point it is stated that the number of multiples of p in range <span class="math-container">$[1,p^k)$</span> is <span class="math-container">$p^{k-1}$</span>.<br />
I am struggling to get why. I tried this:<br />
The multiples of p are :
<span class="math-container">$$p, 2p, .., (p-1)p, p^2, 2p^2, .., (p-1)p^2, .., p^{k-1}, 2p^{k-1}, .., (p-1)p^{k-1}$$</span></p>
<p>Now I can think in terms of powers of p : Each power <span class="math-container">$p^i$</span> contributes <span class="math-container">$p-1$</span>
mutliples : <span class="math-container">$$p^i, 2p^i, .., (p-1)p^i$$</span></p>
<p>And we have <span class="math-container">$k-1$</span> distinct powers so <span class="math-container">$(k-1)\times (p-1)$</span> multiples of p in total.
Of course that's really different from the actual result.
I cannot see what I am missing here.</p>
| Parcly Taxel | 357,390 | <p>Your powers-based counting only counts those numbers that are "pure" powers of <span class="math-container">$p$</span> – those numbers that when written in base <span class="math-container">$p$</span> have only one nonzero digit. Your counting omits numbers like <span class="math-container">$p+p^2$</span>.</p>
<hr />
<p>Ellipses can also deceive you. Did you notice when writing
<span class="math-container">$$p, 2p, .., (p-1)p, p^2, 2p^2, .., (p-1)p^2, .., p^{k-1}, 2p^{k-1}, .., (p-1)p^{k-1}$$</span>
that the separation between, say, <span class="math-container">$p^2$</span> and <span class="math-container">$2p^2$</span> is greater than <span class="math-container">$p$</span>? This means there are more multiples of <span class="math-container">$p$</span> in between them.</p>
|
975 | <p>The usual <code>Partition[]</code> function is a very handy little thing:</p>
<pre><code>Partition[Range[12], 4]
{{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}}
Partition[Range[13], 4, 3]
{{1, 2, 3, 4}, {4, 5, 6, 7}, {7, 8, 9, 10}, {10, 11, 12, 13}}
</code></pre>
<p>One application I'm working on required me to write a particular generalization of <code>Partition[]</code>'s functionality, which allowed the generation of sublists of unequal lengths, for as long as the lengths were appropriately commensurate. (Let's assume for the purposes of this question that the list lengths being commensurate is guaranteed, but you're welcome to generalize further to the incommensurate case.) Here's my generalization in action:</p>
<pre><code>multisegment[lst_List, scts_List] := Block[{acc},
acc = Prepend[Accumulate[PadRight[scts, Length[lst]/Mean[scts], scts]], 0];
Inner[Take[lst, {#1, #2}] &, Most[acc] + 1, Rest[acc], List]]
multisegment[CharacterRange["a", "x"], {3, 1, 2}]
{{"a", "b", "c"}, {"d"}, {"e", "f"}, {"g", "h", "i"}, {"j"}, {"k", "l"},
{"m", "n", "o"}, {"p"}, {"q", "r"}, {"s", "t", "u"}, {"v"}, {"w", "x"}}
</code></pre>
<p>(Thanks to halirutan for optimization help with <code>multisegment[]</code>.)</p>
<p>The problem I've hit into is that I wanted <code>multisegment[]</code> to also support offsets, just like in <code>Partition[]</code>. I want to be able to do something like the following:</p>
<pre><code>multisegment[Range[14], {4, 3}, {3, 1}]
{{1, 2, 3, 4}, {4, 5, 6}, {5, 6, 7, 8},
{8, 9, 10}, {9, 10, 11, 12}, {12, 13, 14}}
</code></pre>
<p>How might a version of <code>multisegment[]</code> with offsets be accomplished?</p>
| Mike Honeychurch | 77 | <p><strong>This is a complete re-write</strong></p>
<p>This is the original solution which was done in haste but i will leave here. It works in limited cases:</p>
<pre><code>multisegment[lst_List, scts_List, offset_List] :=
Module[{acc, offs},
offs = 1+Prepend[Accumulate[PadRight[offset,
1 + Ceiling[Length[lst]/Total[offset]], offset]], 0];
acc = PadRight[scts, Length[offs],scts];
acc = acc + offs - 1;
Inner[Take[lst, {#1, #2}] &, offs, acc, List]
]
multisegment[Range[14], {4, 3}, {3, 1}]
{{1, 2, 3, 4}, {4, 5, 6}, {5, 6, 7, 8}, {8, 9, 10}, {9, 10, 11,
12}, {12, 13, 14}}
</code></pre>
<p>To solve this you note that the starting position (for <code>Part</code> or <code>Take</code>) of the list depends solely on the offset list:</p>
<pre><code>{1,4,5,8,9,12}
</code></pre>
<p>The "span to" position is determined by adding the partition list</p>
<pre><code>{4,3,4,3,4,3}
</code></pre>
<p>to the offset list (minus 1) to give</p>
<pre><code>{4,6,8,10,12,14}
</code></pre>
<p>From there, proceed as before with <code>Inner</code> and use either <code>Take</code> or <code>Part</code>. So this becomes an exercise in generating the correct offset list. As earlier failed attempts have shown, this is dependent on both the total of the offsets and the length of the offsets (list).</p>
<p>But also you do not want your <code>Take</code> or "span to" range exceeding the length of your target list. I have taken the easy way out here but using <code>DeleteCases</code>. A more exact and possibly elegant, but maybe not faster (?), approach is to actually work this out based on the partition list.</p>
<pre><code>multisegment[lst_List, scts_List, offset_List] :=
Module[{fin, offs, len = Length[lst], tot = Total[offset], len2 = Length[offset]},
offs = 1 + Prepend[Accumulate[
PadRight[offset, Ceiling[len2*len/tot], offset]], 0];
fin = PadRight[scts, Length[offs], scts] + offs - 1;
fin = DeleteCases[Transpose[{offs, fin}], {_, x_ /; x > len}];
Take[lst, #] & /@ fin]
(* case for no offsets *)
multisegment[lst_List, scts_List] := multisegment[lst, scts, scts]
</code></pre>
<p>I prefer to layout the code in steps rather than combine multiple steps into a one (or two) liner. Feel free to do that if you wish but I think this way makes it easier for people to check out what is happening.</p>
<p>Also a qualifier: checks and/or conditions should be added. you cannot have {0} for your partition or offset. Must be integers etc. as per Simon's comments.</p>
<p>Usage. First the base case of an uneven partition with no offset</p>
<pre><code>multisegment[Range[14], {3, 4}]
{{1, 2, 3}, {4, 5, 6, 7}, {8, 9, 10}, {11, 12, 13, 14}}
</code></pre>
<p>now add an offset</p>
<pre><code>multisegment[Range[14], {3, 4}, {1, 2}]
{{1, 2, 3}, {2, 3, 4, 5}, {4, 5, 6}, {5, 6, 7, 8}, {7, 8, 9}, {8, 9,
10, 11}, {10, 11, 12}, {11, 12, 13, 14}}
</code></pre>
<p>Examples that previously failed:</p>
<pre><code>multisegment[Range[10], {5, 4}, {2, 3}]
{{1, 2, 3, 4, 5}, {3, 4, 5, 6}, {6, 7, 8, 9, 10}}
multisegment[Range[100], {5, 4}, {2, 3}]
{{1, 2, 3, 4, 5}, {3, 4, 5, 6}, {6, 7, 8, 9, 10}, {8, 9, 10, 11}, {11,
12, 13, 14, 15}, {13, 14, 15, 16}, {16, 17, 18, 19, 20}, {18, 19,
20, 21}, {21, 22, 23, 24, 25}, {23, 24, 25, 26}, {26, 27, 28, 29,
30}, {28, 29, 30, 31}, {31, 32, 33, 34, 35}, {33, 34, 35, 36}, {36,
37, 38, 39, 40}, {38, 39, 40, 41}, {41, 42, 43, 44, 45}, {43, 44,
45, 46}, {46, 47, 48, 49, 50}, {48, 49, 50, 51}, {51, 52, 53, 54,
55}, {53, 54, 55, 56}, {56, 57, 58, 59, 60}, {58, 59, 60, 61}, {61,
62, 63, 64, 65}, {63, 64, 65, 66}, {66, 67, 68, 69, 70}, {68, 69,
70, 71}, {71, 72, 73, 74, 75}, {73, 74, 75, 76}, {76, 77, 78, 79,
80}, {78, 79, 80, 81}, {81, 82, 83, 84, 85}, {83, 84, 85, 86}, {86,
87, 88, 89, 90}, {88, 89, 90, 91}, {91, 92, 93, 94, 95}, {93, 94,
95, 96}, {96, 97, 98, 99, 100}}
</code></pre>
<p>Example showing it working with increasing offset list length</p>
<pre><code>multisegment[Range[44], {3, 4}, {1, 3, 2}]
{{1, 2, 3}, {2, 3, 4, 5}, {5, 6, 7}, {7, 8, 9, 10}, {8, 9, 10}, {11,
12, 13, 14}, {13, 14, 15}, {14, 15, 16, 17}, {17, 18, 19}, {19, 20,
21, 22}, {20, 21, 22}, {23, 24, 25, 26}, {25, 26, 27}, {26, 27, 28,
29}, {29, 30, 31}, {31, 32, 33, 34}, {32, 33, 34}, {35, 36, 37,
38}, {37, 38, 39}, {38, 39, 40, 41}, {41, 42, 43}}
multisegment[Range[44], {3, 4}, {1, 3, 2, 4}]
{{1, 2, 3}, {2, 3, 4, 5}, {5, 6, 7}, {7, 8, 9, 10}, {11, 12, 13}, {12,
13, 14, 15}, {15, 16, 17}, {17, 18, 19, 20}, {21, 22, 23}, {22, 23,
24, 25}, {25, 26, 27}, {27, 28, 29, 30}, {31, 32, 33}, {32, 33, 34,
35}, {35, 36, 37}, {37, 38, 39, 40}, {41, 42, 43}}
</code></pre>
<p>and so on, and so forth.</p>
|
975 | <p>The usual <code>Partition[]</code> function is a very handy little thing:</p>
<pre><code>Partition[Range[12], 4]
{{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}}
Partition[Range[13], 4, 3]
{{1, 2, 3, 4}, {4, 5, 6, 7}, {7, 8, 9, 10}, {10, 11, 12, 13}}
</code></pre>
<p>One application I'm working on required me to write a particular generalization of <code>Partition[]</code>'s functionality, which allowed the generation of sublists of unequal lengths, for as long as the lengths were appropriately commensurate. (Let's assume for the purposes of this question that the list lengths being commensurate is guaranteed, but you're welcome to generalize further to the incommensurate case.) Here's my generalization in action:</p>
<pre><code>multisegment[lst_List, scts_List] := Block[{acc},
acc = Prepend[Accumulate[PadRight[scts, Length[lst]/Mean[scts], scts]], 0];
Inner[Take[lst, {#1, #2}] &, Most[acc] + 1, Rest[acc], List]]
multisegment[CharacterRange["a", "x"], {3, 1, 2}]
{{"a", "b", "c"}, {"d"}, {"e", "f"}, {"g", "h", "i"}, {"j"}, {"k", "l"},
{"m", "n", "o"}, {"p"}, {"q", "r"}, {"s", "t", "u"}, {"v"}, {"w", "x"}}
</code></pre>
<p>(Thanks to halirutan for optimization help with <code>multisegment[]</code>.)</p>
<p>The problem I've hit into is that I wanted <code>multisegment[]</code> to also support offsets, just like in <code>Partition[]</code>. I want to be able to do something like the following:</p>
<pre><code>multisegment[Range[14], {4, 3}, {3, 1}]
{{1, 2, 3, 4}, {4, 5, 6}, {5, 6, 7, 8},
{8, 9, 10}, {9, 10, 11, 12}, {12, 13, 14}}
</code></pre>
<p>How might a version of <code>multisegment[]</code> with offsets be accomplished?</p>
| Mr.Wizard | 121 | <p>My first method was related to Mike's, yet unrefined. This is another method designed for speed. As written it only returns the portion of the list which can be partitioned into a complete set of partitions. This behavior can be changed with 4th+ argument of <code>Partition</code> and/or filtering.</p>
<pre><code>dpCyclic[l_List, p : {__Integer?Positive}, {os__Integer?Positive}] :=
Module[{ranges, blocks},
ranges = {# + 1, # + p} & @ Most @ Accumulate @ {0, os};
blocks = Partition[l, Max @ Last @ ranges, +os];
MapThread[blocks[[All, # ;; #2]] &, ranges] ~Flatten~ {2, 1}
]
</code></pre>
<p>As an example of different behavior (with implicit <code>Null</code>s):</p>
<pre><code>dpCyclic[l_List, p : {__Integer?Positive}, {os__Integer?Positive}] :=
Module[{ranges, blocks},
ranges = {# + 1, # + p} & @ Most @ Accumulate @ {0, os};
blocks = Partition[l, Max @ Last @ ranges, +os, 1,];
MapThread[blocks[[All, # ;; #2]] &, ranges] ~Flatten~ {2, 1} ~DeleteCases~ {___,}
]
</code></pre>
|
3,361,224 | <blockquote>
<p>Show that a dense subspace <span class="math-container">$Y$</span> of a first countable separable topological space <span class="math-container">$X$</span> is separable.</p>
</blockquote>
<p><em>Proof:</em></p>
<p><span class="math-container">$X$</span> is separable.
Let <span class="math-container">$S=\{x_n \in X | n \in \mathbb{N}\}$</span> be a countable dense subset of <span class="math-container">$X$</span>.</p>
<p><span class="math-container">$Y$</span> is also dense in <span class="math-container">$X$</span>.</p>
<p>Because <span class="math-container">$X$</span> is first-countable, thus for each <span class="math-container">$x_n$</span> where <span class="math-container">$n \in \mathbb{N}$</span> there exists a countable local-basis around <span class="math-container">$x_n$</span>.
Let the countable local-basis around <span class="math-container">$x_n$</span> be <span class="math-container">$S_n=\{\text{ }B_n^k \text{ } | \text{ }k \in \mathbb{N} \}$</span></p>
<p>Because <span class="math-container">$Y$</span> is dense in <span class="math-container">$X$</span> thus for each <span class="math-container">$x_n$</span> where <span class="math-container">$ n=1,2,3 \dots $</span> and for each <span class="math-container">$B_n^k$</span> where <span class="math-container">$k=1,2,3,4 \dots$</span>, we have <span class="math-container">$Y \cap B_n^k \neq \phi$</span>. </p>
<p>Say <span class="math-container">$y_n^k \in Y \cap B_n^k \neq \phi$</span></p>
<p>Denote <span class="math-container">$Z=\{ y_n^k \in Y \text{ } | \text{ } n,k \in \mathbb{N} \}$</span></p>
<p>Claim: <span class="math-container">$Z$</span> is a countable dense set of <span class="math-container">$Y$</span>.</p>
<p>Choose <span class="math-container">$y \in Y$</span> and any open set <span class="math-container">$V$</span> in <span class="math-container">$Y$</span> containing y.
<span class="math-container">$V$</span> is open in <span class="math-container">$Y$</span> implies that <span class="math-container">$V=U \cap Y$</span> where <span class="math-container">$U$</span> is an open set in <span class="math-container">$X$</span>.</p>
<p>Thus <span class="math-container">$y \in U \in \tau$</span> and <span class="math-container">$y \in Y$</span></p>
<p><span class="math-container">$y \in U$</span> and <span class="math-container">$U$</span> is open in X. Because <span class="math-container">$S$</span> is dense in X, we have that <span class="math-container">$U \cap S \neq \phi $</span>.</p>
<p>Let <span class="math-container">$x_n \in U \cap S$</span>, Thus <span class="math-container">$x_n \in U$</span> and <span class="math-container">$U$</span> is open in <span class="math-container">$X$</span>.</p>
<p>Considering that <span class="math-container">$S_n$</span> is a countable local-basis around <span class="math-container">$x_n$</span> we have an element <span class="math-container">$B_n^{k_0}$</span> such that <span class="math-container">$x_n \in B_n^{k_0} \subset U$</span>. choose the corresponding <span class="math-container">$y_n^{k_0}$</span> as done in the construction above. Then we have <span class="math-container">$y_n^{k_0} \in B_n^{k_0} \subset U$</span>. Thus <span class="math-container">$y_n^{k_0} \in U \cap Y = V$</span> and hence <span class="math-container">$V \cap Z \neq \phi$</span> as it contains <span class="math-container">$y_n^{k_0}$</span>.</p>
<p>Hence <span class="math-container">$Y$</span> has a countable dense subset. <span class="math-container">$Y$</span> is separable.</p>
<p>Hence proved!</p>
<p>Please check my solution. I need to correct my mistakes and learn.
Thank You.</p>
| Henno Brandsma | 4,280 | <p>This proof looks fine. Quite detailed. See Daniel's comment for an alternative faster proof.</p>
<p>To see that you need the first countable assumption on <span class="math-container">$X$</span>: if <span class="math-container">$X=[0,1]^\mathbb{R}$</span>, then <span class="math-container">$X$</span> is separable (but not first countable), and <span class="math-container">$Y=\Sigma_0[0,1]^\mathbb{R} := |\{f \in X: |\{x: f(x) \neq 0\}| \le \aleph_0 \}$</span> is dense in <span class="math-container">$X$</span> and not separable. Think about it.</p>
|
412,751 | <p>Let <span class="math-container">$G\subseteq \mathbb{C}^n$</span> be a bounded domain. Consider the Carathéodory metric <span class="math-container">$C_G$</span> on <span class="math-container">$G$</span>. If <span class="math-container">$G=\mathbb{D}^n$</span> (unit polydisc), then <span class="math-container">$C_G(a,z)=\max_{1\leq j\leq n}p(a_j,z_j)$</span>, where <span class="math-container">$p$</span> denotes the Poincaré metric on <span class="math-container">$\mathbb{D}$</span>.</p>
<p>My question: Is there a similar formula in case <span class="math-container">$D=\mathbb{B}^2\times \mathbb{D}\subseteq \mathbb{C}^3$</span>, where <span class="math-container">$\mathbb{B}^2$</span> denotes the unit ball in <span class="math-container">$\mathbb{C}^2$</span>? More generally, is there a formula (or some estimate) for the Carathéodory metric of the product domain, in terms of the Carathéodory metric on the components?</p>
| Jackson Morrow | 56,667 | <p>For a general formula, you should check out Kobayashi's book <em>Hyperbolic Complex Spaces</em>. In Proposition 3.1.11 of that text, he proves the following:</p>
<blockquote>
<p>Let <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> be complex spaces. For <span class="math-container">$(x,y), (x',y') \in X\times Y$</span>, we have
<span class="math-container">$$
\text{max}\{ c_X(x,x'),c_Y(y,y')\} \leq c_{X\times Y}((x,y),(x',y')) \leq c_X(x,x') + c_Y(y,y').
$$</span></p>
</blockquote>
|
412,751 | <p>Let <span class="math-container">$G\subseteq \mathbb{C}^n$</span> be a bounded domain. Consider the Carathéodory metric <span class="math-container">$C_G$</span> on <span class="math-container">$G$</span>. If <span class="math-container">$G=\mathbb{D}^n$</span> (unit polydisc), then <span class="math-container">$C_G(a,z)=\max_{1\leq j\leq n}p(a_j,z_j)$</span>, where <span class="math-container">$p$</span> denotes the Poincaré metric on <span class="math-container">$\mathbb{D}$</span>.</p>
<p>My question: Is there a similar formula in case <span class="math-container">$D=\mathbb{B}^2\times \mathbb{D}\subseteq \mathbb{C}^3$</span>, where <span class="math-container">$\mathbb{B}^2$</span> denotes the unit ball in <span class="math-container">$\mathbb{C}^2$</span>? More generally, is there a formula (or some estimate) for the Carathéodory metric of the product domain, in terms of the Carathéodory metric on the components?</p>
| Peter Pflug | 343,739 | <p>In fact, there is the product property <span class="math-container">$\max\{c_X(x,x′),c_Y(y,y′)\}=c_{X×Y}((x,y),(x′,y′))$</span> for arbitrary domains <span class="math-container">$X, Y$</span>. See chapter 18 in "Invariant distances and metrics in complex analysis" de Gruyter (2013).</p>
|
50,479 | <p>Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies?</p>
<p>x,y are integral points on an Elliptic Curve and are easy to find using enumeration of divisors of n (assuming n can be factored).</p>
<p>If yes, will large number of solutions give moderate rank EC?</p>
<p>If one drops $-1$ i.e. $xy(x-y)=n$ the number of solutions can be unbounded via multiples of rational point(s) and then multiplying by a cube. (Explanation): Another unbounded case for varying $a , n$ is $xy(x-y-a)=n$. If $(x,y)$ is on the curve then $(d x,d y)$ is on $xy(x-y-a d)=n d^3$. Find many rational points and multiply by a suitable $d$. Not using the group law seems quite tricky for me. The constant $-1$ was included on purpose in the initial post.</p>
<p>I would be interested in this computational experiment: find $n$ that gives a lot of solutions, say $100$ (I can't do it), check which points are linearly independent and this is a lower bound on the rank.</p>
<p>What I find intriguing is that <strong>all integral points</strong> in this model come from factorization/divisors only.</p>
<p><strike>
Current record is n=<strong>179071200</strong> with 22 solutions with positive x,y. Due to Matthew Conroy.</p>
<p>Current record is n=<strong>391287046550400</strong> with 26 solutions with positive x,y. Due to Aaron Meyerowitz</p>
<p>Current record is n=<strong>8659883232000</strong> with 28 solutions with positive x,y. Found by Tapio Rajala.
</strike></p>
<p>Current record is n=<strong>2597882099904000</strong> with 36 solutions with positive x,y. Found by Tapio Rajala.</p>
<p>EDIT: $ab(a+b+9)=195643523275200$ has 48 positive integer points. – Aaron Meyerowitz (<em>note this is a different curve and 7 <= rank <= 13</em>)</p>
<p>A variation: $(x^2-x-17)^2 - y^2 = n$ appears to be eligible for the same question. The quartic model is a difference of two squares and checking if the first square is of the form $x^2-x-17$ is easy.</p>
<p>Is it possible some relation in the primes or primes or divisors of certain form to produce records: Someone is trying in $\mathbb{Z}[t]$ <a href="https://mathoverflow.net/questions/51193/can-the-number-of-solutions-xyx-y-1n-for-x-y-n-in-zt-be-unbounded-as-n">Can the number of solutions xy(x−y−1)=n for x,y,n∈Z[t] be unbounded as n varies?</a> ? Read an article I didn't quite understand about maximizing the Selmer rank by chosing the primes carefully.</p>
<p>EDIT: The curve was chosen at random just to give a clear computational challenge.</p>
<p>EDIT: On second thought, can a symbolic approach work? Set $n=d_1 d_2 ... d_k$ where d_i are variables. Pick, well, ?some 100? ($d_i$, $y_i$) for ($x$,$y$) (or a product of $d_i$ for $x$). The result is a nonlinear system (last time I tried this I failed to make it work in practice).</p>
<p>EDIT: Related search seems <strong>"thue mahler" equation'</strong></p>
<p>Related: <a href="https://mathoverflow.net/questions/50661/unboundedness-of-number-of-integral-points-on-elliptic-curves">unboundedness of number of integral points on elliptic curves?</a></p>
<p>Crossposted on MATH.SE: <a href="https://math.stackexchange.com/questions/14932/can-the-number-of-solutions-xyx-y-1-n-for-x-y-n-in-z-be-unbounded-as-n">https://math.stackexchange.com/questions/14932/can-the-number-of-solutions-xyx-y-1-n-for-x-y-n-in-z-be-unbounded-as-n</a></p>
| Tapio Rajala | 11,716 | <p><em>Although this is not a mathematical answer I will put the results of my brute force search as an aswer as requested by jerr18. I didn't get anywhere with the thinking part.</em></p>
<h2>Code</h2>
<p>You can find the (non-optimal) C-code I wrote under <a href="http://users.jyu.fi/~tamaraja/temp/solu.c">my webpages</a>. The biggest limitation with the program is that it uses 64-bit integers. Feel free to run, test, tweak and/or mutilate the code as you wish.</p>
<p>The program constructs first $n$ with a recursion and then $y$ with a recursion (this way I avoid considering values of $y$ that don't divide $n$). Finally it checks if the positive solution $x$ to the equation $$xy(x-y-1) = n$$ is an integer.</p>
<h2>Results</h2>
<p>Here are some values found using this program (in roughly 4 hours).</p>
<h3>36 positive solutions</h3>
<p>$$n = 2597882099904000 = 2^9 · 3^3 · 5^3 · 7 · 13 · 17 · 23 · 29 · 31 · 47$$</p>
<h3>30 positive solutions</h3>
<p>$$ n = 34747990981704000 = 2^6 · 3^4 · 5^3 · 7^2 · 11 · 13 · 17 · 19^2 · 29 · 43 $$</p>
<h3>28 positive solutions</h3>
<p>$$n = 105140926800 = 2^4 · 3^3 · 5^2 · 7^2 · 13 · 17 · 29 · 31 $$
$$n = 8659883232000 = 2^8 · 3^3 · 5^3 · 7 · 11 · 13 · 17 · 19 · 31 $$
$$n = 3783439308448800 = 2^5 · 3^4 · 5^2 · 7^3 · 11 · 13 · 19 · 31 · 43 · 47$$
$$n = 9928464968822400 = 2^7 · 3^4 · 5^2 · 7^2 · 11^2 · 13 · 17 · 23 · 31 · 41$$
$$n = 18680310941292000 = 2^5 · 3^4 · 5^3 · 7 · 11^2 · 13^2 · 17 · 19 · 29 · 43$$
$$n = 88550619849291600 = 2^4 · 3^5 · 5^2 · 7^2 · 11^2 · 13 · 17 · 19 · 23 · 37 · 43$$</p>
<p><em>Note that I did not check the results after my program handed them to me...</em></p>
<p><strong>Edit:</strong> Just out of curiosity I tried also with $+1$ instead of $-1$. For example the equation
$$ xy(x-y+1) = 388778796252000 = 2^5 · 3^3 · 5^3 · 7^2 · 11 · 17 · 19 · 23 · 29 · 31$$
has 38 positive solutions.</p>
|
989,118 | <p>The sum is $$\sum_{n>0} \mathrm{i}^n\frac{J_n(x)}{n}\sin\frac{2\pi n}{3}\stackrel{?}{=}0$$ and $$\sum_{n>0} (\mathrm{-i})^n\frac{J_n(x)}{n}\sin\frac{2\pi n}{3}\stackrel{?}{=}0$$</p>
<p>I suspect they are zero because I am working on a project which from symmetry consideration they should be zero. </p>
<p><strong>If anyone can point out they cannot be zero, or they cannot identically be zero, that will be great, too.</strong></p>
<p>I've asked a question about Bessel function before, see <a href="https://math.stackexchange.com/questions/985566/does-this-infinite-summation-of-bessel-function-has-a-closed-form">Does this Infinite summation of Bessel function has a closed form?</a> I hope the reference it contains may do some help.</p>
| agha | 118,032 | <p>$\textbf{Hint:}$ Note that $i^{4n}-i^{4n+2}=2$ for $n \in \mathbb{N}$.</p>
|
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