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 |
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2,021,354 | <p>I have a vector <strong>x</strong> and a function that sums the elements of <strong>x</strong> like so:</p>
<p>$$f(1) = x_1$$
$$f(2) = x_1 + \sum_{i=1}^2 x_i$$
$$f(3) = x_1 + \sum_{i=1}^2 x_i + \sum_{j=1}^3 x_j$$
$$f(4) = x_1 + \sum_{i=1}^2 x_i + \sum_{j=1}^3 x_j + \sum_{k=1}^4 x_k$$</p>
<p>...and so on. How might I represent this function more compactly, please?</p>
| Dhruv Kohli | 97,188 | <p>$f(n) = \sum\limits_{i=1}^{n}\sum\limits_{j=1}^{i} x_j = \sum\limits_{i=1}^{n} (n-i+1)x_i$</p>
|
248,218 | <p>What are good job search sites, head hunter/recruiting agencies for mathematicians looking for industry jobs? </p>
<p>If this question is not appropriate for the math stack exchange, please feel free to redirect me.</p>
| Amzoti | 38,839 | <p>I think you are going to find a mixed bag of things on this topic as it is rather broad from school teachers, to professors to researchers to applied math jobs and the like.=, up to and including engineering.</p>
<p>For example, <a href="http://www.investopedia.com/financial-edge/0812/top-paying-math-related-careers.aspx"><em>Top Paying Math Careers</em></a>.</p>
<p>Here is a sampling for some ideas and sites to look at for conferences and contacts regarding math, head hunters, sites, et. al..</p>
<ul>
<li><p><a href="http://www.mathclassifieds.org/home/home.cfm?site_id=1925"><em>MAA Math Classifieds</em></a></p></li>
<li><p><a href="http://www.indeed.com/q-Mathematics-l-Charlotte,-NC-jobs.html"><em>indeed</em></a></p></li>
<li><p><a href="http://www.dwsimpson.com/"><em>DW Simpson - Actuaries</em></a></p></li>
<li><p><a href="http://www.finmathjobfair.org/"><em>National Financial Mathematics Career Fair</em></a></p></li>
<li><p><a href="http://www.phds.org/jobs/how-to-get-a-job/"><em>How to get a job - PH.D's</em></a></p></li>
<li><p><a href="http://www.math.fsu.edu/~nichols/Financial_Mathematics_Job_Resources.html"><em>Financial Mathematics Job Resources</em></a></p></li>
<li><p><a href="http://mathematics.teacher.jobs.topusajobs.com/"><em>Math Teacher Jobs</em></a></p></li>
<li><p><a href="http://www.salisbury.edu/careerservices/Students/MajorsEmployers/Math.html"><em>Math Employers</em></a></p></li>
<li><p><a href="http://www.simplyhired.com/a/jobs/list/q-applied+mathematics"><em>Simply Hired - Applied Math</em></a></p></li>
</ul>
<p>More specialized fields are also possible in research and the financial sector and the schools big in those areas can provide further support and resources in those cases.</p>
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248,218 | <p>What are good job search sites, head hunter/recruiting agencies for mathematicians looking for industry jobs? </p>
<p>If this question is not appropriate for the math stack exchange, please feel free to redirect me.</p>
| JB King | 8,950 | <p><a href="https://math.uwaterloo.ca/math/future-undergraduates/careers-mathematics" rel="nofollow">Careers in Mathematics</a> would be a page from the University of Waterloo in Canada that has a Faculty of Mathematics that covers various disciplines including Actuarial Science, Operations Research, and Computer Science in addition to conventional branches of Mathematics like Pure Math, Applied Math, Accounting and Statistics noting various career options as there can be more than a few things that may fall under the heading of a Mathematician. This isn't exactly what was asked though it is pretty close to what one could do with a Math degree which I could see as a way to know who are Mathematicians in the world.</p>
|
2,417,197 | <p>When going through with learning Grahams number, I got stuck at </p>
<p>$$3βββ3$$</p>
<p>Working it through, we have</p>
<p>$$3β3=3^3$$
$$3ββ3=3^{3^3}=3β(3β3)$$</p>
<p>As such, it would appear to me that</p>
<p>$$3βββ3=3^{3^{3^3}}=3β(3β(3β3))=3β(3ββ)$$</p>
<p>Which is incorrect; the correct answer being</p>
<p>$$3βββ3=3ββ(3ββ3)$$</p>
<p>What I'm wanting to know is where the error in the way I've worked it through, and how working $3ββ(3ββ3)$ through backwards to $3βββ3$ would look?</p>
| Simply Beautiful Art | 272,831 | <p>You wish to understand Graham's number through these arrows? If so, I'd suggest stepping back down to multiplication and building the way up.</p>
<p>Note that</p>
<p>$$a\times b=\underbrace{a+(a+(\dots+a))}_b$$</p>
<p>For example,</p>
<p>$$3\times3=3+(3+3)=3+6=9$$</p>
<p>And then exponentiation,</p>
<p>$$a^b=a\uparrow b=\underbrace{a\times(a\times(\dots\times a))}_b$$</p>
<p>For example,</p>
<p>$$3\uparrow3=3\times(3\times3)=3\times9=27$$</p>
<p>Now tetration,</p>
<p>$$a\uparrow\uparrow b=\underbrace{a\uparrow(a\uparrow(\dots\uparrow a))}_b$$</p>
<p>For example,</p>
<p>$$3\uparrow\uparrow 3=3\uparrow(3\uparrow3)=3\uparrow27=7625597484987$$</p>
<p>And beyond...</p>
<p>$$a\uparrow\uparrow\uparrow b=\underbrace{a\uparrow\uparrow(a\uparrow\uparrow(\dots\uparrow\uparrow a))}_b$$</p>
<p>$$3\uparrow\uparrow\uparrow3=3\uparrow\uparrow(3\uparrow\uparrow3)=3\uparrow\uparrow7625597484987=\underbrace{3\uparrow(3\uparrow(\dots\uparrow3))}_{7625597484987}=3^{3^{3^{3^{\dots}}}}$$</p>
|
76,600 | <p>The group of three dimensional rotations $SO(3)$ is a subgroup of the Special Euclidean Group $SE(3) = \mathbb{R}^3 \rtimes SO(3)$. The manifold of $SO(3)$ is the three dimensional real projective space $RP^3$. Does $RP^3$ cause a separation of space in the manifold of $SE(3)$? </p>
<p>(edit) Sorry about lack of clarity. My question should be worded as 'does $SO(3)$ partition any four dimensional subspace of $SE(3)$ into exactly two disjoint pieces?'</p>
<p>I am basically interested in understanding whether a generalization of the Jordan curve separation theorem works in such non Euclidean spaces. In particular, I want to know if (non) orientability of $SO(3)$ affects the generalization, especially since it is used to construct $SE(3)$ as a product space with $\mathbb{R}^3$.</p>
| John Galt | 18,078 | <p>I am not entirely sure what you mean by separation of space. But, would n't it depend on the representation of SE(3) and SO(3). For example, one can take the view that SE(3) is a dual projective space R\hat{P}^3 by using dual quaternion representation for spatial rigid body displacement.</p>
<p>I am not a mathematician, so please ignore me if what I say does not make sense.</p>
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623,810 | <p>$\omega = y dx + dz$ is a differential form in $\mathbb{R}^3$, then what is ${\rm ker}(\omega)$? Is ${\rm ker}(\omega)$ integrable? Can you teach me about this question in details? Many thanks!</p>
| Gil Bor | 118,580 | <p>$\omega$ is nowhere zero so its kernel at each point is 2 dimensional. Let us find 2 vector fields that span this kernel. If we take a vector $v=a\partial_x+b\partial_y+c\partial_z$ then $\omega(v)=ay+c.$ Hence $v\in Ker(\omega)$ iff $c=-ay$. So we can take as a basis for the kernel of $\omega$ say $v_1=\partial_x-y\partial_z$, $v_2=\partial_y$. </p>
<p>As for integrability of the distribution $Ker(\omega)$, this is equivalent to the vanishing of $\omega\wedge d\omega$. But the latter is $dx\wedge dy\wedge dz$, ie nowhere vanishing, hence the distribution is not integrable. </p>
|
2,844,902 | <p>Does $$\int_{[1,z]}\frac{1}{u}du=\log(z)$$ where $z\in\mathbb C$ ? I know that on a closed circle that contain $0$ we have $$\int_C\frac{1}{z}dz=2i\pi=\log(1),$$</p>
<p>but for $$\int_{[1,z]}\frac{1}{u}du=\log(z)$$ I don't really know to compute the integral.</p>
| md2perpe | 168,433 | <p>You can at least come to a first order differential equation by multiplying with $\dot q$ and then integration:
$$\dot q \ddot q = 4 q^{-5} \dot q$$
$$\frac{d}{dt}(\frac12 \dot q^2) = \frac{d}{dt}(-q^{-4})$$
$$\dot q = \pm 2 (C-q^{-4})$$</p>
|
4,073,757 | <p>Q: A coin is tossed untill k heads has appeared. If a mathematician knows how many heads appeared, can he figure out what is the probability that the coin was tossed <span class="math-container">$n$</span> times?</p>
<p>What I tried: The number of heads debends of the number of tosses. So I tried Bayes' theoream <span class="math-container">$P(X=k|N=n)P(N=n)/P(X=k)$</span> where <span class="math-container">$X$</span> is random variable for heads and <span class="math-container">$N$</span> for tosses. <span class="math-container">$P(X=k|N=n)$</span> is binomial distributions mass probability function, but i don't know what <span class="math-container">$P(N=n)$</span> and <span class="math-container">$P(X=k)$</span> are.</p>
<p>Because the last toss is known to be heads, we only focus on the the tosses before the last one. There is <span class="math-container">$\binom {n-1}{k-1}$</span> combinations for <span class="math-container">$k-1$</span> heads to appear before the last one. So the probability is</p>
<p><span class="math-container">$$P(N=n|X=k) = \binom{n-1}{k-1}p^k(1-p)^{n-k}$$</span></p>
<p>I believe this can be done also with Bayes Theoream. Can someone wiser show how it is done?</p>
| Karl | 279,914 | <p>Bayes' Theorem would be useful if the experiment consisted of choosing <span class="math-container">$N$</span> first, before starting to count the heads, and <em>then</em> flipping the coin exactly <span class="math-container">$N$</span> times. Then we'd have</p>
<p><span class="math-container">$$
P(N=n|X=k)
=\frac{P(X=k|N=n)P(N=n)}{P(X=k)}
=\frac{\binom nkp^k(1-p)^{n-k}P(N=n)}{\sum_{n=0}^\infty \binom nkp^k(1-p)^{n-k}P(N=n)}.
$$</span></p>
<p>In this model, the probabilities <span class="math-container">$P(N=n)$</span> need to be specified, as they describe how we make our advance choice of how many times to toss the coin.</p>
<p>The experiment you describe is different: <span class="math-container">$k$</span> is a fixed parameter and every outcome ends with a total of exactly <span class="math-container">$k$</span> heads, so treating the number of heads as a random variable ("<span class="math-container">$X$</span>") isn't useful. The random variable of interest is <span class="math-container">$N$</span>: the number of tosses needed to get exactly <span class="math-container">$k$</span> heads.</p>
<p>In order to get <span class="math-container">$N=n$</span>, we need to get exactly <span class="math-container">$k-1$</span> heads in the first <span class="math-container">$n-1$</span> tosses, in any order, followed by a head on the next toss. There are <span class="math-container">$\binom{n-1}{k-1}$</span> different <span class="math-container">$n$</span>-toss result sequences that meet these conditions, and every such sequence has probability <span class="math-container">$p^k(1-p)^{n-k}$</span>, so <span class="math-container">$P(N=n)=\binom{n-1}{k-1}p^k(1-p)^{n-k}$</span>, as you computed.</p>
<p>In other words, <span class="math-container">$P(N=n)$</span> in the given experiment is equal to <span class="math-container">$P(\sum_{i=1}^{n-1}X_i=k,X_n=1)$</span> in a <em>different</em> experiment that consists of tossing a coin <span class="math-container">$n$</span> times, and in this experiment, <span class="math-container">$X=\sum_{i=1}^{n-1}X_i$</span> is binomially distributed. But the random variables <span class="math-container">$X$</span> and <span class="math-container">$N$</span> don't exist in the same experiment, so you can't relate them with conditional probabilities.</p>
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1,088,338 | <p>There are at least a few things a person can do to contribute to the mathematics community without necessarily obtaining novel results, for example:</p>
<ul>
<li>Organizing known results into a coherent narrative in the form of lecture notes or a textbook</li>
<li>Contributing code to open-source mathematical software</li>
</ul>
<p>What are some other ways to make auxiliary contributions to mathematics?</p>
| DanielV | 97,045 | <p>There is a lot of work done towards formalizing existing proofs into logic software and proof wikis. The formalization of a proof can be very helpful economically. It can lead to more confidence in proofs and making searching for results more feasible.</p>
<p>(Mario here:) I am an undergraduate who works with the program and library <a href="http://us.metamath.org/">Metamath</a> for doing formal proofs. There is a TON of work that still needs to be done in formalizing even just the standard undergraduate curriculum, and it's all easily accessible to any reasonably bright undergraduate. It helps to be good at programming or at least thinking like a programmer, since the work you do looks a lot like programming to the uninitiated and formal math is just as unforgiving to typos as any compiler. But even though there is a lot of general problem solving involved, the path is all more or less written out by other mathematicians (as "hand proofs"), so the way forward is always clearly delineated and it might even be ungraciously called "transcription".</p>
<p>I'm going to try to keep it up in graduate school, but even just as a pastime it's good for the brain and gives you a great sense of accomplishment, in addition to making you understand the corresponding hand proof on a deeper level than pretty much any kind of study out there. It's not for the faint of heart, but for precision-minded amateur or professional mathematicians with a programming bent I could not recommend it enough.</p>
|
1,088,338 | <p>There are at least a few things a person can do to contribute to the mathematics community without necessarily obtaining novel results, for example:</p>
<ul>
<li>Organizing known results into a coherent narrative in the form of lecture notes or a textbook</li>
<li>Contributing code to open-source mathematical software</li>
</ul>
<p>What are some other ways to make auxiliary contributions to mathematics?</p>
| R K Sinha | 66,227 | <p>Many problems of theoretical physics (e.g., string theory) are related to the development of some new math. This severely requires the services of able mathematicians. In the past, Riemann, Grassman, Hilbert, Poincare, Elie Cartan, deed so for the discovery of Einstein's equation in GR. Physicist Witten was awarded Field Award for his services in Math research. </p>
|
2,965,082 | <blockquote>
<p>Suppose that <span class="math-container">$(X,\ d)$</span> and <span class="math-container">$(Y,\ \rho)$</span> are metric spaces, that
<span class="math-container">$f_n:X\to Y$</span> is continuous for each <span class="math-container">$n$</span>, and that <span class="math-container">$(f_n)$</span> convergence
pointwise to <span class="math-container">$f$</span> on <span class="math-container">$X$</span>. If there exists a sequence <span class="math-container">$(x_n)$</span> in <span class="math-container">$X$</span>
such that <span class="math-container">$x_n\to x$</span> in <span class="math-container">$X$</span> but <span class="math-container">$f_n(x_n)\not\to f(x)$</span>, show that
<span class="math-container">$(f_n)$</span> does not converge uniformly to <span class="math-container">$f$</span> on <span class="math-container">$X$</span>.</p>
</blockquote>
<p>I've managed to "prove" that the question is self-contradictory, so please find my error.</p>
<p><span class="math-container">$\forall n\in\mathbb{N}$</span>, <span class="math-container">$f_n$</span> is continuous at <span class="math-container">$x$</span>. Let <span class="math-container">$\epsilon>0$</span>. <span class="math-container">$\forall n\in\mathbb{N}$</span> <span class="math-container">$\exists\delta>0$</span> s.t. <span class="math-container">$\rho(f_n(y),\ f_n(x))<\epsilon/2$</span> for all <span class="math-container">$y\in X$</span> s.t. <span class="math-container">$d(y,\ x)<\delta$</span>. ------------ (1)</p>
<p>Since <span class="math-container">$x_m\to x$</span>, <span class="math-container">$\exists N_1\in\mathbb{N}$</span> s.t. <span class="math-container">$d(x_m,\ x)<\delta$</span> <span class="math-container">$\forall m\ge N_1$</span>. ------------- (2)</p>
<p>From (1) and (2),</p>
<p><span class="math-container">$\forall n$</span>, <span class="math-container">$\forall m\ge N_1\implies d(x_m,\ x)<\delta\implies \rho(f_n(x_m),\ f_n(x))<\epsilon/2$</span> ------------- (3)</p>
<p>Also, <span class="math-container">$f_n(y)\to f(y)$</span> for all <span class="math-container">$y\in X$</span> due to pointwise convergence. So, <span class="math-container">$\exists N_2\in\mathbb{N}$</span> s.t. <span class="math-container">$\rho(f_n(y),\ f(y))<\epsilon/2$</span> <span class="math-container">$\forall n\ge N_2$</span> and <span class="math-container">$\forall y\in X$</span>. ----------- (4)</p>
<p>Let <span class="math-container">$N_3=\max(N_1,\ N_2)$</span>. Suppose <span class="math-container">$n\ge N_3$</span>. Then <span class="math-container">$n\ge N_1$</span> and <span class="math-container">$n\ge N_2$</span>.</p>
<p><span class="math-container">$\begin{aligned}
\implies\rho(f_n(x_n), f(x))&\le\rho(f_n(x_n),\ f_n(x))+\rho(f_n(x),\ f(x)) \\
&<\epsilon/2+\epsilon/2\text{ [Using (3) and (4)]} \\
&=\epsilon
\end{aligned}$</span></p>
<p>I have thus "proved" that <span class="math-container">$f_n(x_n)\to f(x)$</span>, contradicting the question. Where have I gone wrong?</p>
| RRL | 148,510 | <p>For a correct proof (by contradiction), show that <span class="math-container">$f(x_n) \not\to f(x)$</span> is impossible if convergence is uniform using</p>
<p><span class="math-container">$$\rho(f_n(x_n), f(x)) \leqslant \rho(f_n(x_n), f(x_n)) + \rho(f(x_n), f(x))$$</span></p>
<p>Note that <span class="math-container">$f$</span> must be continuous when the sequence of continuous functions <span class="math-container">$f_n$</span> is uniformly convergent. Thus each term on the RHS is smaller than <span class="math-container">$\epsilon/2$</span> for sufficiently large <span class="math-container">$n$</span> -- the first by the uniform convergence <span class="math-container">$f_n \to f$</span> and the second by the convergence <span class="math-container">$x_n \to x$</span> and the continuity of <span class="math-container">$f$</span>. </p>
|
2,628,220 | <p>Let $(a_{n})_{n \in \mathbb N_{0}}$ be a sequence in $\mathbb Z$, defined as follows:
$a_{0}:=0,
a_{1}:=2,
a_{n+1}:= 4(a_{n}-a_{n-1}) \forall n \in \mathbb N$. </p>
<p>Required to prove: $a_{n}=n2^{n} \forall n \in \mathbb N_{0}$</p>
<p>I have gone about it in the following: </p>
<p>Induction start: $n=0$ (condition fulfilled)</p>
<p>Induction premise: $a_{n}=n2^{n}$ for a specific $n \in \mathbb N_{0}$</p>
<p>Induction step: </p>
<p>$a_{n+1}=4(a_{n}-a_{n-1})$, and here the first problem arises, since I can say that (given the premise)
$4(a_{n}-a_{n-1})=4(n2^{n}-a_{n-1})$, yet how do I get rid of the $a_{n-1}$? Surely stating the $a_{n-1}=(n-1)2^{n-1}$ is false, given that my premise is only based on an $n$ and not $n-1$. </p>
| Michael Rozenberg | 190,319 | <p>We see that $$a_{n+1}-2a_n=2(a_n-2a_{n-1}),$$
which says that $b_n=a_n-2a_{n-1}$ is geometric progression.</p>
<p>Thus, $$b_n=b_1\cdot2^{n-1}=2\cdot2^{n-1}=2^n.$$
Thus,
$$a_1-2a_0=2,$$
$$\frac{1}{2}a_2-a_1=\frac{1}{2}\cdot2^2,$$
$$\frac{1}{2^2}a_3-\frac{1}{2}a_2=\frac{1}{2^2}\cdot2^3,$$
$$.$$
$$.$$
$$.$$
$$\frac{1}{2^{n-1}}a_n-\frac{1}{2^{n-2}}a_{n-1}=\frac{2^n}{2^{n-1}},$$
which after summing sives
$$\frac{1}{2^{n-1}}a_n-2a_0=2n,$$ which gives
$$a_n=n2^n.$$</p>
|
3,541,947 | <p>How do you pronounce <span class="math-container">$\mathbb{F}_2, \mathbb{F}_2^n, \mathbb{N}^k, [n] = \{1,\ldots,n\},$</span> and <span class="math-container">$S \subseteq [n]$</span> when you're reading a text?</p>
<p>I've just started reading more advanced math textbooks and these are appearing all the time. </p>
| Randall | 464,495 | <p>I think this is a matter of taste/preference. Personally, I read them as...</p>
<p>"eff-two," "eff-two-enn," "enn-kay," and "box enn."</p>
|
892,742 | <p>Let $G$ be a finite group. How can we show that $|G/G^{'}|\leq |C_G(x)|$ for all elements $x\in G$?</p>
| Nicky Hekster | 9,605 | <p>This follows from the fact that the order of a conjugacy class $$|Cl_ G(x)| \leq |G'|$$. Namely, define a map $f : Cl_G(x) \rightarrow G'$ by $f(g^{-1}xg)=[x,g]$. This map is clearly injective. Finally, $|Cl_G(x)|=[G:C_G(x)]$.</p>
|
3,842,739 | <p>Let <span class="math-container">$H$</span> be a group and <span class="math-container">$H^m=\{ h^m \mid h\in H\}$</span>.</p>
<p>I know that this is a subgroup of <span class="math-container">$H$</span> when <span class="math-container">$H$</span> is abelian.
But I want to know that what happens if <span class="math-container">$H$</span> is not abelian.
For which <span class="math-container">$n$</span>, <span class="math-container">$H^n$</span> is a subgroup of <span class="math-container">$H$</span> and for which <span class="math-container">$n$</span> it is not?</p>
<p>I tried for the nonabelian group <span class="math-container">$S_3$</span> and found that <span class="math-container">$S_3^{2k}$</span> is a subgroup of <span class="math-container">$S_3$</span> for all <span class="math-container">$k\in \mathbb{Z}$</span> and <span class="math-container">$S_3^{2k+1}$</span> is not a subgroup of <span class="math-container">$S_3$</span> for all <span class="math-container">$k\in \mathbb{Z}$</span>.</p>
<p>But I don't know whether it is true for arbitrary groups or not and I want to prove it if it is correct.</p>
<p>Any help would be appreciated.</p>
| markvs | 454,915 | <p>It is certainly true that for every finite group <span class="math-container">$H$</span> there exists <span class="math-container">$N$</span> such that whether or not <span class="math-container">$H^n$</span> is a subgroup depends only on <span class="math-container">$n\mod N$</span> and contain all multiples of <span class="math-container">$N$</span>. For this <span class="math-container">$N$</span> you can take, for example, <span class="math-container">$|H|$</span>. That is because if <span class="math-container">$m\equiv n\mod N$</span>, then <span class="math-container">$H^m=H^n$</span>, and <span class="math-container">$H^N=\{1\}$</span>.</p>
<p>For infinite groups the set of <span class="math-container">$n$</span>'s for which <span class="math-container">$H^n$</span> is a subgroup may be empty (say, the free group of rank <span class="math-container">$\ge 2$</span>).</p>
|
696,848 | <p>$\DeclareMathOperator{\rank}{rank}$
First off I'm sorry I'm still not able to make of use the built in formula expressions, I don't have time to learn it now, I'll do it before my next question.</p>
<p>I have a couple of questions regarding eigenvectors and generalized eigenvectors. To some of these questions I know the answer partially or there are some uncertainties so I will just ask in the most general form, but I can really appreciate precise answers.</p>
<p>How do I know how many eigenvectors to expect for each eigenvalue?</p>
<p>How do I know how many generalized eigenvectors to expect for each of those eigenvectors?
Consider a matrix $A$ whose eigenvalues and vectors I'd like to compute. Do basic row and column operations on either $A$ or $(A - \lambda I)$ (lambda be an eigenvalue) change any of the eigenvalues, -vectors or determinants of the two corresponding matrices?</p>
<p>Any of the following statements my be wrong and I'd appreciate it if you could point out where the errors are.</p>
<p>Consider this special case for the <a href="http://www.wolframalpha.com/input/?i=%7B%7B1,1,0,1%7D;%7B0,2,0,0%7D;%7B-1,1,2,1%7D;%7B-1,1,0,3%7D%7D" rel="nofollow">matrix A:</a></p>
<p>Its rank is $4$. The characteristic polynomial tells me there is an eigenvalue lambda with algebraic multiplicity $4$. In order to determine the geometric multiplicities to the corresponding eigenvalues (which there is just one of) I can determine the rank of $(A - \lambda I) = (A - 2I)$. Said matrix looks like <a href="http://www.wolframalpha.com/input/?i=%7B%7B-1,1,0,1%7D;%7B0,0,0,0%7D;%7B-1,1,0,1%7D;%7B-1,1,0,1%7D%7D" rel="nofollow">this</a></p>
<p>Operating with basic row and column operations on this matrix $(A - 2I)$ I can reduce down to a matrix with just one $1$ and all the other elements will be zero. Thus the rank of this matrix is $1$. In order to get the geometric multiplicity corresponding to this eigenvalue I compute
$$
\rank(A) - \rank(A-2I) = 4 - 1 = 3
$$
So the geometric multiplicity of this eigenvalue is $3$, which means I can expect $3$ eigenvectors.</p>
<p>If so far no errors have been made and no corrections have been given, consider the following:
Are the eigenvectors to this specific problem unique? Clearly I can reduce the matrix $(A - 2I)$ down to a matrix with one 1 at any element I like.</p>
<p>Let's say we picked the 1 as the first element of said matrix; are my eigenvectors just $(0,1,0,0)^T;(0,0,1,0)^T;(0,0,0,1)^T$? (the T stands for transposed)</p>
<p>How do I compute the generalized eigenvectors, which eigenvector do I pick, how do I determine which one to choose and what is it?</p>
<p>Thanks for your time!</p>
| Charles | 1,778 | <p>Probably the Riemann Hypothesis is true, in which case its falsity would not be provable (and the believability of its falsity is more a matter of psychology than mathematics), whether by computation or otherwise.</p>
|
842,365 | <blockquote>
<p>Show that a field <span class="math-container">$\mathbb{F}$</span> is finite if and only if its multiplicative group <span class="math-container">$\mathbb{F}^{\times}$</span> is finitely generated.</p>
</blockquote>
<p>The "<span class="math-container">$\Rightarrow$</span>" implication is obvious, but how to prove the otherwise?</p>
| David | 119,775 | <p>I'm not entirely clear on your proposed scheme, but it seems to me that you are doing something like this. You have "axes" representing the ten possible units digits, the ten possible first digits after the decimal point, the ten possible second digits after the decimal point, and so on; then also the ten possible digits for "tens", the ten possible digits for "hundreds" and so on. You wish to represent a real number as a "point" in this "space" of countably many dimensions. So for example you would think of
$$\pi\leftrightarrow(\ldots,0,0,3,1,4,1,5,9,\ldots)\ .$$
If I have got this right, the problem is that you talk about a countable union of countable sets, which is countable; but you are actually using a countable <strong>Cartesian product</strong> of countable sets, which is not countable, except in certain special cases.</p>
|
1,798,261 | <p>what is multilinear coefficient? I heard it a couple of times and I tried to google it, all I am getting is multiple linear regression.
I am confused at this point. </p>
| Bart W | 339,805 | <p>Just go by the definition of subspace: a subspace is a subset of the space that is also a vector space. You can easily prove that any linear combination of vectors $(a, b, c)$ for which $a+b + c = 0$, also satisfies this condition. Therefore, it is indeed a subspace.</p>
<p>I can elaborate on this if the rest of the proof is not clear to you.</p>
|
1,798,261 | <p>what is multilinear coefficient? I heard it a couple of times and I tried to google it, all I am getting is multiple linear regression.
I am confused at this point. </p>
| DonAntonio | 31,254 | <p>Prove the basic conditions a <em>subset</em> must fulfill to be a subspace: (a) check that the zero vector is in $\;W\;$ , (b) show that if two vectors are in $\;W\;$ so is their sum, and (c) if a vector $\;w\in W\; $ and $\;r\;$ is any scalar, then also $\;rw\in W\;$ .</p>
<p>As for a basis: observe that $\;a+b=0=c\;$ leaves us with only one free choice, for example $\;b\;$ . Once $\;b\;$ is chosen we already have no more choices as it must be $\;a=-b\;,\;\;c=0\;$ . This tells us that the dimension is $\;1\; $ (Remember: the ammount of free choices), so now only take your favourite number, which is $\;1\;$ (it can't be zero as then you get the zero vector, which can't be part of any basis. Why?), and $\;\{(1,-1,0)\}\;$ is a basis of $\;W\;$ .</p>
|
3,756,436 | <p>Recently I was doing a physics problem and I ended up with this quadratic in the middle of the steps:</p>
<p><span class="math-container">$$ 0= X \tan \theta - \frac{g}{2} \frac{ X^2 \sec^2 \theta }{ (110)^2 } - 105$$</span></p>
<p>I want to find <span class="math-container">$0 < \theta < \frac{\pi}2$</span> for which I can later take the largest <span class="math-container">$X$</span> value that solves this equation, i.e. optimize the implicit curve to maximize <span class="math-container">$X$</span>.</p>
<p>I tried solving this by implicit differentiation (assuming <span class="math-container">$X$</span> can be written as a function of <span class="math-container">$\theta$</span>) with respect to <span class="math-container">$\theta$</span> and then by setting <span class="math-container">$\frac{dX}{d\theta} = 0$</span>:</p>
<p><span class="math-container">\begin{align}
0 &= X \sec^2 \theta + \frac{ d X}{ d \theta} \tan \theta - \frac{g}{2} \frac{ 2 \left( X \sec \theta \right) \left[ \frac{dX}{d \theta} \sec \theta + X \sec \theta \tan \theta \right]}{ (110)^2 } - 105 \\
0 &= X \sec^2 \theta - \frac{g}{2} \frac{ 2( X \sec \theta) \left[ X \sec \theta \tan \theta\right] }{ (110)^2 } \\
0 &= 1 - \frac{ Xg \tan \theta}{(110)^2} \\
\frac{ (110)^2}{ g \tan \theta} &= X
\end{align}</span></p>
<p>This is still not an easy equation to solve. However, one of my friends told we could just take the discriminant of the quadratic in terms of <span class="math-container">$X$</span>, and solve for <span class="math-container">$\theta$</span> such that <span class="math-container">$D=0$</span>.</p>
<p>Taking discriminant and equating to 0, I get</p>
<p><span class="math-container">$$ \sin \theta = \frac{ \sqrt{2 \cdot 10 \cdot 105} }{110}$$</span></p>
<p>and, the angle from it is, 24.45 degrees</p>
<p>I tried the discriminant method, but it gave me a different answer from the implicit differentiation method. I ended up with two solutions with the same maximum value of <span class="math-container">$X$</span> but different angles: <span class="math-container">$\theta =24.45^\text{o}$</span> and <span class="math-container">$X=1123.54$</span> (from discriminant method), and
<span class="math-container">$\theta = 47^\text{o}$</span> and <span class="math-container">$X=1123.54$</span> (from implicit differentiation).</p>
<p>I later realized the original quadratic can only have solutions if <span class="math-container">$D(\theta) > 0$</span>, where <span class="math-container">$D$</span> is the discriminant. Using the discriminant, I can find a lower bound on the angle. Once I have the lower bound, if I can prove that <span class="math-container">$X$</span> decreases monotonically as a function of <span class="math-container">$\theta$</span>, then I can use the lower bound for further calculations of <span class="math-container">$\theta$</span>.</p>
<p>So then I used the implicit function theorem and got</p>
<p><span class="math-container">$$ \frac{dX}{ d \theta} =- \frac{X \sec^2 \theta -\frac{g X^2}{2 (110)^2} 2 \sec \theta ( \sec \theta \tan \theta) } {\tan \theta - \frac{g \sec^2 \theta}{2 (110)^2} 2X }$$</span></p>
<p>Now the problem here is that I can't prove this function is in monotonic in terms of <span class="math-container">$\theta$</span> as the implicit derivative is a function of both <span class="math-container">$\theta$</span> and <span class="math-container">$X$</span>.</p>
| Tryst with Freedom | 688,539 | <p>The trick is to write a quadratic in terms of <span class="math-container">$ \tan \theta $</span> and not in terms of <span class="math-container">$X$</span></p>
<p><span class="math-container">$$ 0 = X \tan \theta -\frac{g}{2} \frac{ X^2 (1 +\tan^2 \theta)}{(110)^2} - 105$$</span></p>
<p>applying the condition that tangent <span class="math-container">$ D>0$</span> for the quadratic in terms of <span class="math-container">$ \tan \theta$</span>(taking g as 10),</p>
<p>I get, this</p>
<p><span class="math-container">$$ 1> \frac{20}{(110)^2} ( 105 + \frac{5X^2}{(110)^2})$$</span></p>
<p><span class="math-container">$$ -1100 < X<1100$$</span></p>
<p>Taking the upper bound, it becomes <span class="math-container">$X=1100$</span></p>
|
493,102 | <p>I have a concern with nested quantifiers.</p>
<p>I have: $$ \forall x \exists y \forall z(x^2-y+z=0) $$ such that $$ x,y,z \in \Bbb Z^+$$ </p>
<p>My first question, can it be read like this:</p>
<p>$$ \forall x \forall z \exists y(x^2-y+z=0) $$</p>
<p>The way I did it, is I started off with $x=1, z=1 $ </p>
<p>$$ 2-y = 0 $$ $$ y =2 $$</p>
<p>Is this a good approach?</p>
| Brian M. Scott | 12,042 | <p>No, you cannot interchange the $\exists y$ and $\forall z$: doing so changes the meaning of the statement. The original statement,</p>
<p>$$\forall x\exists y\forall z\left(x^2-y+z=0\right)$$</p>
<p>says that no matter what positive integer $x$ you choose, I can find a $y\in\Bbb Z^+$ such that $x^2-y+z=0$ <strong>no matter what positive integer</strong> $z$ <strong>you choose</strong>. For instance, if you choose $x=1$, I can find some positive integer $y$ such that $1-y+z=0$ <strong>for every positive integer</strong> $z$. Since $1-y+z=0$ if and only if $z=y-1$, this is clearly false: no matter what $y$ I pick, you can choose for $z$ any positive integer except $y-1$ and ensure that $x^2-y+z$ is <strong>not</strong> $0$.</p>
<p>The modified statement</p>
<p>$$\forall x\forall z\exists y\left(x^2-y+z=0\right)\;,$$</p>
<p>on the other hand, says something very different: it says that no matter what positive integers you choose for $x$ and $z$, I can find a positive integer $y$ such that $x^2-y+z=0$. And of course I can: I just pick $y=x^2+z$.</p>
<p>Since one statement is true and the other false, they certainly cannot be equivalent.</p>
|
978,384 | <p>The following picture is constructed by connecting each corner of a square with the midpoint of a side from the square that is not adjacent to the corner. These lines create the following red octagon:</p>
<p><img src="https://i.stack.imgur.com/PZyGa.jpg" alt="enter image description here"></p>
<p>The question is, what is the ratio between the area of the octagon and the area of the square. One is supposed to find the solution without a ruler. </p>
<p>By removing some lines, I find it easy to see that the ratio between the yellow area and the square is 1/4. But I am not sure if this helps.</p>
<p><img src="https://i.stack.imgur.com/lpIZF.jpg" alt="enter image description here"></p>
| Bob | 491,960 | <p>I was frustrated by the solution being 1/6 because that is not the result if one calculates the area of a <em>regular</em> octagon with a radius that is L/4, where L is the length of large square.</p>
<p>The gridded image above makes clear that red shape is close to, but not actually, a <em>regular</em> octagon.</p>
<p>Instead, the horizontal distance from the center of the large square to a vertex is 1.5 times the length of the smallest green square. The angled distance from the center of the large square to a vertex is 'sqrt(2)' times (not 1.5 times) the length of the smallest green square. The vertices do not lie on a circle, so it is not a <em>regular</em> octagon (with equal all the included angles being equal), even though the lengths of all the sides are equal.</p>
|
2,623,324 | <p>Assume that the measure space is finite for this to make sense. Also, we know that $L^p$ spaces satisfy log convexity, that is -
$$\|f\|_r \leq \|f\|_p^\theta \|f\|_q^{1-\theta}$$
where $\frac{1}{r}=\frac{\theta}{p} +\frac{1-\theta}{q}$.
The text which I am following says 'Indeed this is trivial when $q=\infty$, and the general case then follows by convexity'. I understand that it is true when $q=\infty$, however I am unable to use that it is true for $q=\infty$ for proving the general case. I have found a proof which uses log-convexity and the fact that $\lim_{p\rightarrow 0}\|f\|_p^p=\mu(\text{supp}f)$. Is there some way that I can do this by using that it is true for $q=\infty$?</p>
| Barry Cipra | 86,747 | <p>Note that</p>
<p>$$1\le(n+1)^{1/\sqrt n}\le(2n)^{1/\sqrt n}=2^{1/\sqrt n}((\sqrt n)^{1/\sqrt n})^2$$</p>
<p>If we take $2^{1/x}$ and $x^{1/x}\to1$ as $x\to\infty$ for granted, then </p>
<p>$$2^{1/\sqrt n}((\sqrt n)^{1/\sqrt n})^2\to1\cdot1^2=1$$</p>
<p>and the Squeeze Theorem does the rest.</p>
|
2,449,581 | <p>There is a brick wall that forms a rough triangle shape and at each level, the amount of bricks used is two bricks less than the previous layer. Is there a formula we can use to calculate the amount of bricks used in the wall, given the amount of bricks at the bottom and top levels?</p>
| John Doe | 399,334 | <p>You should first work through the problem using the hints given. Here is a solution for once you are done.</p>
<blockquote class="spoiler">
<p> Let's say the lower layer has $x$ bricks, and there are a total of $n$ layers. Then the top layer has $a:=x-2(n-1)$ bricks (you can check this is correct by plugging in something like $n=2,3$). Then build an identical wall, but upside down and place it next to the first one. The lower layer of the first wall had $x$ bricks, and the lower layer of the second wall has $a$ bricks. So the lower layer of the combined wall has $x+a$ bricks, and the same is true for all $n$ layers. So the total number of bricks in this combined wall is $(x+a)n=n(x+x-2(n-1))=n(2x-2n+2)$. The total amount of bricks in the original wall is then half of this, ie $n(x-n+1).$</p>
</blockquote>
|
324,119 | <p>I've been reading about the Artin Spin operation. It's defined as taking the classical <span class="math-container">$n$</span>-knot (<span class="math-container">$S^n\hookrightarrow S^{n+2}$</span>) to an <span class="math-container">$(n+1)$</span>-knot. For the <span class="math-container">$1$</span>-knot case (in <span class="math-container">$\mathbb{R}^3$</span>), I reproduce the procedure in <a href="https://arxiv.org/pdf/math/0410606.pdf" rel="nofollow noreferrer">knot spinning</a>, p. 8, </p>
<ol>
<li><p>We manipulate a knot <span class="math-container">$K$</span> so that all but a trivial arc <span class="math-container">$a$</span> lie in the upper half space <span class="math-container">$H^3=\{(x,y,z)\>|\>z\geq 0\}$</span>. We remove the interior of <span class="math-container">$a$</span>.</p></li>
<li><p>We rotate <span class="math-container">$H^3$</span> around <span class="math-container">$\mathbb{R}^2$</span> in <span class="math-container">$\mathbb{R}^4$</span>, inducing a parameterization <span class="math-container">$(x,y,z)\mapsto (x, y, z\cos\theta, z\sin\theta)$</span></p></li>
</ol>
<p><strong>-- Question --</strong>
How is this operation similar to the suspension functor (on a topological space <span class="math-container">$X$</span>) <span class="math-container">$\Sigma X\equiv X\wedge S=\frac{X\times S}{X\vee S}$</span>?</p>
| David Corfield | 447 | <p>There's also the work of Francis Borceux and Marco Grandis, </p>
<p>Jordan-HΓΆlder, modularity and distributivity in non-commutative algebra, J. Pure Appl. Algebra 208 (2007), no. 2, 665-689, <a href="http://dx.doi.org/10.1016/j.jpaa.2006.03.004" rel="nofollow noreferrer">doi</a>.</p>
<p>There the authors prove a Jordan-HΓΆlder theorem (4.4) for all 'weakly exact' categories (1.2).</p>
|
2,867,479 | <p>From <a href="https://math.stackexchange.com/questions/2867457">ETS Major Field Test in Mathematics</a></p>
<blockquote>
<p>A student is given an exam consisting of
8 essay questions divided into 4 groups of
2 questions each. The student is required to
select a set of 6 questions to answer,
including at least 1 question from each of
the 4 groups. How many sets of questions
satisfy this requirement? </p>
</blockquote>
<p>I'm thinking $$\binom{2}{1}^4 \binom{4}{2}$$</p>
<p>because we have to pick 1 from each of the 4 groups of 2 and then from the remaining 4 questions we pick 2.</p>
| Servaes | 30,382 | <p>The student can choose two questions to omit, not both i the same group. There are $8$ options for the first question, and then there are $6$ options left for the second question. Of course the order in which the questions are chosrn doesn't matter, so we get
$$\frac{8\times6}{2}=24$$
options.</p>
<p>Alternatively, the student can choose two groups to omit a question from, and then one question from each group. This way we get
$$\binom{4}{2}\binom{2}{1}\binom{2}{1}=6\times2\times2=24$$
options.</p>
<p>Finally, in line with your own approach, we can first choose one question from each group. We can indeed do so in $16$ ways. Then we can choose two more questions, indeed in $6$ ways. But now we have overcounted; we reach the same set of questions if we had first chosen other questions in the two groups we ended up choosing both questions from. In how many ways could we have chosen questions from these two groups? A total of $4$ ways. So by this method we have coumted each set of questions $4$ times. Hence the total number of options is $\frac{96}{4}=24$.</p>
|
2,844,060 | <p>How to find the points of discontinuity of the following function $$f(x) = \lim_{n\to \infty} \sum_{r=1}^n \frac{\lfloor2rx\rfloor}{n^2}$$ </p>
| Shashi | 349,501 | <p>Let $x>0$. Notice that $$\frac{2rx - 1} {n} \leq \frac{\lfloor 2rx \rfloor} {n} \leq \frac{2rx}{n} $$
Hence $$-\frac 1 n+\frac 1 n\sum_{r=1}^n\frac{2rx } {n} \leq \frac 1 n \sum_{r=1}^n\frac{\lfloor 2rx \rfloor} {n}\leq \frac 1 n\sum_{r=1}^n\frac{2rx}{n}$$
You can see the summation on the LHS (and RHS) as a Riemann sum: $$\lim_{n\to\infty} \frac 1 n\sum_{r=1}^n\frac{2rx } {n}=x\int^1_0 2t\, dt = x$$
The squeeze theorem finishes the proof that $$\lim_{n\to\infty} \frac 1n \sum_{r=1}^n\frac{\lfloor 2rx \rfloor} {n}=x $$ A similar argument could be applied for $x<0$ yielding the same result. For $x=0$ the limit is zero. So the function in question is continuous. </p>
|
957,940 | <p>I'm "walking" through the book "A walk through combinatorics" and stumbled on an example I don't understand. </p>
<blockquote>
<p><strong>Example 3.19.</strong> A medical student has to work in a hospital for five
days in January. However, he is not allowed to work two consecutive
days in the hospital. In how many different ways can he choose the
five days he will work in the hospital?</p>
<p><strong>Solution</strong>. The difficulty here is to make sure that we do not choose two consecutive days. This can be assured by the following
trick. Let $a_1, a_2, a_3, a_4, a_5$ be the dates of the five days of
January that the student will spend in the hospital, in increasing
order. Note that the requirement that there are no two consecutive
numbers among the $a_i$, and $1 \le a_i \le 31$ for all $i$ is equivalent
to the requirement that $1 \le a_i < a_2 β 1 < a_3 β 2 < a_4 β 3 < a_5 β 4 \le 27$. In other words, there is an obvious bijection between the
set of 5-element subsets of [31] containing no two consecutive
elements and the set of 5-element subsets of [27].</p>
<p>*** Instead of choosing the numbers $a_i$, we can choose the numbers $1 \le a_i < a_2 β 1 < a_3 β 2 < a_4 β 3 < a_5 β 4 \le 27$, that is, we can
simply choose a five-element subset of [27], and we know that there
are $\binom{27}{5}$ ways to do that.</p>
</blockquote>
<p>What I don't understand here $1 \le a_i < a_2 β 1 < a_3 β 2 < a_4 β 3 < a_5 β 4 \le 27$:</p>
<ul>
<li>Why do the subtracting numbers increment with every other $a_i$? </li>
<li>Why 27?</li>
</ul>
<p>And the very last sentence (***) is unclear to me.</p>
<ul>
<li>Why is there no talk about "non-consecutive"? Why choosing 5 elements of 27 is equivalent to choosing 5 non-consecutive elements out of 31? I miss the connection. </li>
</ul>
<p>I'd be very thankful if you could help me to understand this example!</p>
| Henry | 6,460 | <p>The prohibition on consecutive days means the constraints are really $$1 \le a_1$$ $$a_1+1 \lt a_2$$ $$a_2+1 \lt a_3$$ $$a_3+1 \lt a_4$$ $$a_4+1 \lt a_5$$ $$a_5 \le 31.$$ Rewrite these so that the right hand side of each line is the same as the left hand side of the next line as $$1 \le a_1$$ $$a_1 \lt a_2-1$$ $$a_2-1 \lt a_3-2$$ $$a_3-2 \lt a_4-3$$ $$a_4-3 \lt a_5-4$$ $$a_5 -4 \le 27$$ and you can now combine these as a single line $$1 \le a_1 < a_2 β 1 < a_3 β 2 < a_4 β 3 < a_5 β 4 \le 27$$ </p>
<p><strong><em>Added for extended question:</em></strong></p>
<p>Having done that, you could let $b_1=a_1$, $b_2=a_2-1$, $b_3=a_3-2$, $b_4=a_4-3$, and $b_5=a_5-4$, so $$1 \le b_1 < b_2 < b_3 < b_4 < b_5 \le 27$$ and so the number of integer solutions for the $b_i$ is obviously the same as the number of ways of choosing $5$ distinct integers from $27$, i.e. ${27 \choose 5}$. </p>
<p>Having chosen the $b_i$ in order from smallest to largest, you can get back to the $a_i$ (the working days) by adding $0$ to the smallest, $1$ to the next, $2$ to the middle one, $3$ to the next, and $4$ to the largest. It is obvious that this will give five dates with at least ones day's gap between them. </p>
|
1,183,185 | <p>i) Show that for a particle moving with velocity $v(t), $if $ v(t)Β·vβ²(t) = 0$ for all $t$ then the speed $v$ is constant.
οΏΌ</p>
<p>I did $(v(t))^2=|v(t)|^2=(v(t)\bullet(v(t)))$. </p>
<p>Therefore $\frac{d}{dt}(v(t))^2=2v(t)$</p>
<p>Also, $\frac{d}{dt}(v(t)\bullet(v(t))=2(v(t)Β·vβ²(t))$ </p>
<p>I'm stuck here.</p>
<p>ii) A particle of mass m with position vector r(t) at time t is acted on by a total force
F (t) = Ξ»r(t) Γ v(t),
where Ξ» is a constant and v(t) is the velocity of the particle. Show that the speed v of the particle is con- stant. (Note that Newtonβs second law of motion in its vector form is F = ma.)</p>
<p>Therefore, ma(t)=Ξ»(r(t) Γ v(t)) after which I don't know what to do.</p>
| David K | 139,123 | <p>I'll use the notation $X_{(1)}$ for the smallest $X_i$ and
$X_{(N)}$ for the largest $X_i$.</p>
<p>Suppose $X_1 \leq t_0$.
The smallest $X_i$, $X_{(1)}$, then must be either $X_1$ or some even smaller value.
In any case, $X_{(1)} \leq t_0$ and the event $A$ did not occur.
So $A$ can occur only when $X_1 > t_0$.</p>
<p>Suppose $X_1 > t_1$.
We can apply an argument to the largest $X_i$, $X_{(N)}$, following the pattern of the
argument about the smallest $X_i$, and conclude that $X_{(N)} > t_1$
and that $A$ did not occur.
So $A$ can occur only when $X_1 \leq t_1$.</p>
<p>So if $A$ then $t_0 < X_1 \leq t_1$. What is the probability of that event?</p>
<p>Now for the same reasons that $t_0 < X_1 \leq t_1$, if $A$ then we also have
$t_0 < X_2 \leq t_1$, $t_0 < X_3 \leq t_1$, and so forth, up to $t_0 < X_N \leq t_1$.
Knowing that $X_1, X_2, X_3, \ldots$ are mutually independent, what
is the probability that <em>all</em> of these inequalities are satisfied?
If all these inequalities are satisfied, is $A$ satisfied?
What can we conclude then about $\mathrm{Pr}[A]$?</p>
|
3,623,277 | <p>Show that the cycles <span class="math-container">$(1, 2, \ldots, n)$</span>, <span class="math-container">$(n, \ldots, 2, 1)$</span> are inverse permutations. </p>
| lab bhattacharjee | 33,337 | <p>As <span class="math-container">$x^3-x=(x-1)x(x+1)$</span> is a product of three consecutive integers</p>
<p><span class="math-container">$3$</span> must divide <span class="math-container">$x^3-x$</span></p>
<p>So, we need <span class="math-container">$$x^3\equiv x\pmod{5\cdot7}$$</span></p>
<p>If <span class="math-container">$(x-1)x(x+1)\equiv0\pmod 5$</span></p>
<p><span class="math-container">$\implies x\equiv0\ \ \ \ (1), x\equiv-1\ \ \ \ (2), x\equiv1\pmod5\ \ \ \ (3)$</span></p>
<p>Similarly, <span class="math-container">$x\equiv0\ \ \ \ (4), x\equiv-1\ \ \ \ (5), x\equiv1\pmod7\ \ \ \ (6)$</span></p>
<p>Now apply <a href="http://mathworld.wolfram.com/ChineseRemainderTheorem.html" rel="nofollow noreferrer">CRT</a> on <span class="math-container">$(1),(4); (1),(5);(1),(6);(2),(4); (2),(5);(2),(6)$</span></p>
|
2,359,372 | <blockquote>
<p>Given that
$$\log_a(3x-4a)+\log_a(3x)=\frac2{\log_2a}+\log_a(1-2a)$$
where $0<a<\frac12$, find $x$.</p>
</blockquote>
<p>My question is how do we find the value of $x$ but we don't know the exact value of $a$? </p>
| Robert Z | 299,698 | <p>Hint. Recall the main properties of the <a href="https://en.wikipedia.org/wiki/Logarithm#Product.2C_quotient.2C_power.2C_and_root" rel="nofollow noreferrer">logarithm</a>.
Then we have that $\frac{1}{\log_2a}=\log_a 2$.
Moreover for $x>4a/3>0$ (the argument of the logarithm should be positive),
$$\log_a((3x-4a)\cdot 3x))=\log_a(2^2\cdot(1-2a))$$
which is equivalent (logarithm is an injective function) to
$$(3x-4a)\cdot 3x=2^2\cdot(1-2a)$$
or the quadratic equation
$$9x^2 - 12ax - 4(1-2a)=0.$$
Can you take it from here?</p>
<p>P.S. We expect to find two solutions. At end remember to check the condition $x>4a/3$!</p>
|
2,881,673 | <p>I've searched all over the internet and cannot seem to factorise this polynomial.</p>
<p>$x^4 - 2x^3 + 8x^2 - 14x + 7$</p>
<p>The result should be $(x β 1)(x^3 β x^2 + 7x β 7)$</p>
<p>What are the steps to get to that result?
I've tried grouping but doesn't seem to work...</p>
| Community | -1 | <p>As $p(1)=0$, you know that $x-1$ is a factor. Now</p>
<p>$$x^4 - 2x^3 + 8x^2 - 14x + 7
\\=x^3(x-1)-x^3+8x^2-14x+7
\\=x^3(x-1)-x^2(x-1)+7x^2-14x+7
\\=x^3(x-1)-x^2(x-1)+7x(x-1)-7x+7
\\=x^3(x-1)-x^2(x-1)+7x(x-1)-7(x-1).$$</p>
|
1,855,650 | <p>Need to solve:</p>
<p>$$2^x+2^{-x} = 2$$</p>
<p>I can't use substitution in this case. Which is the best approach?</p>
<p>Event in this form I do not have any clue:</p>
<p>$$2^x+\frac{1}{2^x} = 2$$</p>
| MonK | 160,887 | <p>Let,
$2^x=y$ then the equation becomes</p>
<p>$y+\frac{1}{y}=2\\
\implies y^2+1-2y=0\\
\implies (y-1)^2=0\\
\implies y=1\\
\implies 2^x=1\\
\implies 2^x=2^0\\
\implies x=0$</p>
|
1,839,057 | <p>Where n is an integer, $n\ge1$ and $(A,B)$ just constants </p>
<blockquote>
<p>$$I=\int_{-n}^{n}{x+\tan{x}\over A
+B(x+\tan{x})^{2n}}dx=0$$</p>
</blockquote>
<p>It is obvious that</p>
<p>$$\int_{-n}^{n}x+\tan{x}dx=0$$</p>
<p>Let make a substitution for <em>I</em> $$u=x+\tan{x}\rightarrow du=1+\sec^2{x}dx$$</p>
<p>$$\int_{-n}^{n}{u\over A+Bu^{2n}}{du\over 2+\tan^2{u}}=0$$</p>
<p>I can't find a standard integral of this. I am shrugged at this point on how to continued any further, required some help please</p>
<p>Also note that </p>
<p>$$\int_{-n}^{n}{u\over A+Bu^{2n}}du=0$$</p>
<p>And</p>
<p>$$\int_{-n}^{n}{u\over A+Bu^{2n}}{du\over C+D\tan^{2k}{u}}=0$$</p>
<p>Where <em>A</em>,<em>B</em>,<em>C</em> and <em>D</em> are just constants </p>
<p>$n,k\ge1$ are both integers</p>
| lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>Use $$I=\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$$</p>
<p>$$\implies I+I=\int_a^b[f(x)+f(a+b-x)]\ dx$$</p>
<p>Here $a=?,b=?$</p>
<p>and $\tan(-x)=-\tan x$</p>
|
164,447 | <p>by default, if a number has decimal <code>.</code> after it, then Mathematica will do computation using machine Precision, which on my PC (intel hardware) running windows 7 64 bit is double Precision.</p>
<p>I'd like to get the computation also but using single and quad precision, to match the following small Fortran program output. </p>
<p>Is there a recommend way to do this? Here is the fortran program and its output</p>
<pre><code>PROGRAM foo
IMPLICIT NONE
INTEGER:: i
INTEGER, PARAMETER :: &
sp = kind(1.0), &
dp = selected_real_kind(2*precision(1.0_sp)), &
qp = selected_real_kind(2*precision(1.0_dp))
REAL(kind=sp) :: sum1,x1
REAL(kind=dp) :: sum2,x2
REAL(kind=qp) :: sum3,x3
sum1=0.0_sp; sum2=0.0_dp; sum3=0.0_qp
x1 = 0.00001_sp
x2 = 0.00001_dp
x3 = 0.00001_qp
DO i=1,10**5
sum1 = sum1 + x1
sum2 = sum2 + x2
sum3 = sum3 + x3
END DO
PRINT *,sum1; PRINT *,sum2; PRINT *,sum3;
END
</code></pre>
<p>Compiled using <code>gfortran foo.f90</code> gives</p>
<pre><code>ex1>./a.out
1.00099015 ---> SINGLE
0.99999999999808376 --> DOUBLE
0.999999999999999999999999999998395508 ----> QUAD
</code></pre>
<p>The first line is single, the second is double (which Mathematica matches exactly) and the third is quad.</p>
<p>Here is the Mathematica basic code which gives output that matches the double Precision:</p>
<pre><code>x1=0.00001;
sum1=0.0;
Do[sum1=sum1+x1,{i,1,10^5}]
InputForm[sum1]
</code></pre>
<p>Which gives</p>
<pre><code> 0.9999999999980838
</code></pre>
<p>I tried doing </p>
<pre><code> x1=0.00001`32;
sum1=0.0`32;
</code></pre>
<p>To see if will something close to the quad result, but it had not effect on final output. I still get <code>0.9999999999980838</code></p>
<p>How to change the above Mathematica code to make it give the single and quad Precision as shown by Fortran?</p>
| Bob Hanlon | 9,362 | <pre><code>$Version
x1 = 0.00001 // Rationalize // N[#, 32] &
sum1 = 0 // N[#, 32] &
Do[sum1 = sum1 + x1, {i, 1, 10^5}]
sum1 // InputForm
"11.2.0 for Mac OS X x86 (64-bit) (September 11, 2017)"
0.000010000000000000000000000000000000
0
0.999999999999999999999999999999999\
9999999999999999999979455`32.
</code></pre>
<p>Note that the zero remained exact.</p>
|
164,447 | <p>by default, if a number has decimal <code>.</code> after it, then Mathematica will do computation using machine Precision, which on my PC (intel hardware) running windows 7 64 bit is double Precision.</p>
<p>I'd like to get the computation also but using single and quad precision, to match the following small Fortran program output. </p>
<p>Is there a recommend way to do this? Here is the fortran program and its output</p>
<pre><code>PROGRAM foo
IMPLICIT NONE
INTEGER:: i
INTEGER, PARAMETER :: &
sp = kind(1.0), &
dp = selected_real_kind(2*precision(1.0_sp)), &
qp = selected_real_kind(2*precision(1.0_dp))
REAL(kind=sp) :: sum1,x1
REAL(kind=dp) :: sum2,x2
REAL(kind=qp) :: sum3,x3
sum1=0.0_sp; sum2=0.0_dp; sum3=0.0_qp
x1 = 0.00001_sp
x2 = 0.00001_dp
x3 = 0.00001_qp
DO i=1,10**5
sum1 = sum1 + x1
sum2 = sum2 + x2
sum3 = sum3 + x3
END DO
PRINT *,sum1; PRINT *,sum2; PRINT *,sum3;
END
</code></pre>
<p>Compiled using <code>gfortran foo.f90</code> gives</p>
<pre><code>ex1>./a.out
1.00099015 ---> SINGLE
0.99999999999808376 --> DOUBLE
0.999999999999999999999999999998395508 ----> QUAD
</code></pre>
<p>The first line is single, the second is double (which Mathematica matches exactly) and the third is quad.</p>
<p>Here is the Mathematica basic code which gives output that matches the double Precision:</p>
<pre><code>x1=0.00001;
sum1=0.0;
Do[sum1=sum1+x1,{i,1,10^5}]
InputForm[sum1]
</code></pre>
<p>Which gives</p>
<pre><code> 0.9999999999980838
</code></pre>
<p>I tried doing </p>
<pre><code> x1=0.00001`32;
sum1=0.0`32;
</code></pre>
<p>To see if will something close to the quad result, but it had not effect on final output. I still get <code>0.9999999999980838</code></p>
<p>How to change the above Mathematica code to make it give the single and quad Precision as shown by Fortran?</p>
| Michael E2 | 4,999 | <p>The following emulates the various precisions, by directly rounding the operation as done in floating-point. The quad seems off by 1 bit, if Fortran uses IEEE quad precision. Also, Fortran output seems to have an extra digit or two, compared to <em>Mathematica's</em> at the specified precision.</p>
<pre><code>singlebits = 23;
x1 = 0.00001 // Rationalize;
sum1 = 0;
Do[sum1 = Round[sum1 + x1, 2^Floor@Log2[N[(sum1 + x1), 9]*2^-singlebits]], {i, 1, 10^5}]
SetPrecision[sum1, 9]
(* 1.00099015 *)
(* machine precision *)
x1 = 0.00001;
sum1 = 0.0;
Do[sum1 = sum1 + x1, {i, 1, 10^5}]
SetPrecision[sum1, 17]
(* 0.99999999999808376 *)
quadbits = 112;
x1 = 0.00001 // Rationalize;
sum1 = 0;
Do[sum1 = Round[sum1 + x1, 2^Floor@Log2[N[sum1 + x1, 36]*2^-quadbits]], {i, 1, 10^5}]
SetPrecision[sum1, 36]
(* 0.999999999999999999999999999998395508 *)
</code></pre>
|
440,242 | <p>I'm pretty sure almost all mathematicians have been in a situation where they found an interesting problem; they thought of many different ideas to tackle the problem, but in all of these ideas, there was something missing- either the "middle" part of the argument or the "end" part of the argument. They were stuck and couldn't figure out what to do.</p>
<blockquote>
<ol>
<li>In such a situation what do you do?</li>
<li>Is the reason for the "missing part" the incompleteness in the theory of the topic that the problem is related to? What can be done to find the "missing part"?</li>
</ol>
</blockquote>
<p>For tenure-track/tenure professors, maybe this is not a big deal because they have "enough" time and can let the problem "stew" in the "back-burner" of their mind, but what about limited-time positions, e.g. PhD students, postdocs, etc., where the student/employee has to prove their capability to do "independent" research so that they can be hired for their next position? I think for these people it is quite a bit of a problem because they can't really afford to spend a "lot" of time thinking about the same problem.</p>
| Simon Crase | 470,900 | <p>Sometimes it's good the keep the problem in the back of your mind while you do other stuff that appears irrelevant. Here is Stanislaw Ulam's account of the invention of the Monte Carle Method--from <a href="http://www-star.st-and.ac.uk/%7Ekw25/teaching/mcrt/MC_history_3.pdf" rel="noreferrer">Los Alamos Science Special Issue 1987</a>. Anything may suggest a way to tackle your problem.</p>
<blockquote>
<p>The first thoughts and attempts I made ... were suggested by a question which occurred to me in 1946 as I was convalescing from an illness and playing solitaires. The question was what are the chances that a Canfield solitaire laid out with 52 cards will come out successfully? After spending a lot of time trying to estimate them by pure
combinatorial calculations, I wondered whether a more practical method than
βabstract thinkingβ might not be to lay it out say one hundred times and
simply observe and count the number of successful plays. This was already possible to envisage with the beginning of the new era of fast computers, and I immediately thought of problems of neutron diffusion and other questions of mathematical physics, and more generally how to change processes described by certain differential equations into an equivalent form interpretable
as a succession of random operations. Later... described the idea to John von Neumann and we began to plan actual calculations.β</p>
</blockquote>
|
1,097,134 | <p>this is something that came up when working with one of my students today and it has been bothering me since. It is more of a maths question than a pedagogical question so i figured i would ask here instead of MESE.</p>
<p>Why is $\sqrt{-1} = i$ and not $\sqrt{-1}=\pm i$?</p>
<p>With positive numbers the square root function always returns both a positive and negative number, is it different for negative numbers? </p>
| Geoff Robinson | 13,147 | <p>There are many abstract ways to construct the complex numbers, and however you do it, there is a complex number which squares to $-1$, and to which you happen to give a name, often $i$. Once you have named $i$, it becomes obvious that $-i$ also has square $-1$, and that all complex numbers have the form $a +bi$ for unique real $a$ and $b.$ It really does not matter which $i$ you choose in the first place, you could equally well have chosen $-i$ instead, and this is reflected by the fact that the mapping which sends $a + bi$ to $a-bi$ for $a,b \in \mathbb{R}$ preserves all the additive and multiplicative structure of the complex field (ie is an automorphism of the complex field fixing every element of $\mathbb{R}$). The point is that there is no mystical significance to the symbol $i$. If you make the other choice and give $-i$ a new symbol $j$, you can work equally well with expressions of the form $a + bj$ for $a,b \in \mathbb{R}$, and never know the difference.</p>
<p>When you start to represent complex numbers geometrically on the Argand diagram, you do tend to make a choice, but that is a choice of using clockwise or counterclockwise direction for rotations. It is usual to represent the complex number $\cos \theta + i\sin \theta$ by the point $(\cos \theta, \sin \theta)$ in $\mathbb{R}^{2}.$ But if we consistently used $(\cos \theta, -\sin \theta)$ instead we would have no difficulty.</p>
|
2,231,949 | <p>To find the minimal polynomial of $i\sqrt{-1+2\sqrt{3}}$, I need to prove that
$x^4-2x^2-11$ is irreducible over $\Bbb Q$. And I am stuck. Could someone please help? Thanks so much!</p>
| Arthur | 15,500 | <p>The rational root theorem (or the quadratic formula, solving for $x^2$), shows that there are no linear factors over $\Bbb Q$. That means that if the polynomial is reducible, then it reduces to two irreducible quadratic polynomials.</p>
<p>However, if that were true, then your number would be a root of one of them. The quadratic formula then says it can be written in the form $a\pm\sqrt b$ with $a,b$ rational. The number is pure imaginary, so $a=0$, which means that $b=1-2\sqrt3$, which clearly isn't rational. This is a contradiction.</p>
|
4,096,771 | <p>Given a sequence of iid random variables <span class="math-container">$(Y_i)_{i=1}^\infty$</span> on a probability space <span class="math-container">$(\Omega, \mathcal{F}, \mathbb{P})$</span> such that <span class="math-container">$\mathbb{E}|Y_i| < \infty$</span> and <span class="math-container">$\mathbb{E}Y_i = 0$</span>, consider the discrete time process given by <span class="math-container">$$X_0 := 0, \quad X_n = \sum_{i=1}^n Y_i, \quad n \in \{1,2,...\}.$$</span>
and also the filtration given by <span class="math-container">$\mathcal{F_n} = \sigma(Y_1, \dots, Y_n)$</span> and show that <span class="math-container">$(X_n)_{n=1}^\infty$</span> is a (discrete) martingale with respect to <span class="math-container">$(\mathcal{F_n})_{n=1}^\infty$</span>.</p>
<p>So far, in answering this question I believe that I have proven the first two properties of a martingale:</p>
<ul>
<li><p>We have that, since <span class="math-container">$X_n$</span> is the sum of <span class="math-container">$\mathcal{F}$</span>-measurable variables, we know that it is too <span class="math-container">$\mathcal{F}$</span>-measurable for each <span class="math-container">$n$</span>. Hence it is adapted.</p>
</li>
<li><p><span class="math-container">$\mathbb{E}|X_n| \leq \mathbb{E}(\sum_{i=1}^n| Y_i|) = \sum_{i=1}^n\mathbb{E}| Y_i| < \infty$</span>.</p>
</li>
</ul>
<p>However, I am not sure about how one could go about proving the final property of a martingale, that <span class="math-container">$\mathbb{E}(X_t | \mathcal{F_s}) = X_s$</span> for all <span class="math-container">$s \leq t$</span>.</p>
<p>Might anyone have any ways of demonstrating such a proof?</p>
| Kavi Rama Murthy | 142,385 | <p><span class="math-container">$$E(X_t|\mathcal F_s)=E(\sum\limits_{i=1}^{t}Y_i|Y_1,Y_2,..,Y_s)$$</span> <span class="math-container">$$=\sum\limits_{i=1}^{s}Y_i+E(\sum\limits_{i=s+1}^{t}Y_i|Y_1,Y_2,..,Y_s)$$</span> <span class="math-container">$$=X_s+E(\sum\limits_{i=s+1}^{t}Y_i)=X_x+0=X_s.$$</span></p>
<p>In the second equality I have used the fact that <span class="math-container">$\sum\limits_{i=1}^{s}Y_i$</span> is already measurable w.r.t <span class="math-container">$\mathcal F_s$</span>. In the next equality I have used the independence of <span class="math-container">$Y_{s+1},Y_{s+1},...,Y_t$</span> w.r.t. <span class="math-container">$\mathcal F_s$</span>.</p>
|
3,906,920 | <p>A string in <span class="math-container">$\{0, 1\}*$</span> has even parity if the symbol <span class="math-container">$1$</span> occurs in the word an even number of times; otherwise, it has odd parity.</p>
<p>(a) How many words of length <span class="math-container">$n$</span> have even parity?</p>
<p>(b) How many words of length <span class="math-container">$n$</span> have odd parity?</p>
<p>It seems to me that correct approach will be to use summations of combinations, and correct answer will yield <span class="math-container">$2^{n-1}$</span> in both cases. Is there any way to prove separately that each of them is equal to <span class="math-container">$2^{n-1}$</span>? Would be happy to know your ideas!</p>
| Mozibur Ullah | 26,254 | <blockquote>
<p>Does <em>f</em> need to be a diffeomorphism?</p>
</blockquote>
<p>It's not always possible to push-forward tangent fields on a manifold. This is why we have the restriction to diffeomorphisms, or more generally, tangent fields that are <em>f</em>-related.</p>
<p>However, it's an easy observation that when we work in the category of tangent fields <em>over</em> an auxiliary manifold, that is parametrised tangent fields, we can always do so.</p>
|
2,067,003 | <p>(Mathematics olympiad Netherlands) Let $A,B$ and $C$ denote chess players in a tournament. The winner of each match plays the next match against the oponent that did not play the current. At the end of the tournament $A$, $B$ and $C$ played $10$, $15$ and $17$ times respectively. Each match only ended up in a win. <em>Question</em>: Which player lost the second match?</p>
<p><em>UPDATE</em>: So I think I got the answer. Denote $n$ as the amount of matches between $A$ and $B$. Since $A$ plays the same amount of matches against $B$ as the otherway around, we have $15 - n = 17 - (10-n) \implies n = 4$. So $A$ plays a total of 10 matches, while there are a total of 21 matches. This is only possible if $A$ plays all the even matches <strong>and</strong> he loses that match (else contradiction to amount of matches played).</p>
| Masacroso | 173,262 | <p>Using falling factorials and <a href="https://www.cs.purdue.edu/homes/dgleich/publications/Gleich%202005%20-%20finite%20calculus.pdf" rel="nofollow noreferrer">finite calculus</a> we can write</p>
<p>$$\begin{align}\sum\frac{(-1)^{n+1}}{n(n+1)}\delta n&=\sum (-1)^{n+1}(n-1)^{\underline{-2}}\delta n\\&=(-1)^{n}(n-1)^{\underline{-1}}-\sum\frac{(-1)^{n+2}-(-1)^{n+1}}{(-1)(n+1)}\delta n\\&=\frac{(-1)^n}n+2\sum\frac{(-1)^{n}}{n+1}\delta n\end{align}$$</p>
<p>where the second step is summation by parts (check <a href="https://math.stackexchange.com/questions/1950386/abels-summation-memorizing-the-theorem/1950415#1950415">this</a>, by example), and the indefinite sum of a falling factorial is defined as</p>
<p>$$\sum n^{\underline m}\delta n=\frac{n^{\underline{m+1}}}{m+1}+ K$$</p>
<p>Now taking limits above we have</p>
<p>$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n(n+1)}=\sum\nolimits_1^\infty\frac{(-1)^{n+1}}{n(n+1)}\delta n=\frac{(-1)^n}n\bigg|_1^\infty+2\sum\nolimits_1^\infty\frac{(-1)^{n}}{n+1}\delta n=\\=1+2(\ln (2)-1)=2\ln(2)-1$$</p>
<p>because the Taylor expansion of $\ln (1+x)$ is</p>
<p>$$\ln (1+x)=\sum_{n=1}^\infty(-1)^{n+1}\frac{x^k}{k},\quad\text{whenever }x\in(-1,1]$$</p>
|
2,067,003 | <p>(Mathematics olympiad Netherlands) Let $A,B$ and $C$ denote chess players in a tournament. The winner of each match plays the next match against the oponent that did not play the current. At the end of the tournament $A$, $B$ and $C$ played $10$, $15$ and $17$ times respectively. Each match only ended up in a win. <em>Question</em>: Which player lost the second match?</p>
<p><em>UPDATE</em>: So I think I got the answer. Denote $n$ as the amount of matches between $A$ and $B$. Since $A$ plays the same amount of matches against $B$ as the otherway around, we have $15 - n = 17 - (10-n) \implies n = 4$. So $A$ plays a total of 10 matches, while there are a total of 21 matches. This is only possible if $A$ plays all the even matches <strong>and</strong> he loses that match (else contradiction to amount of matches played).</p>
| xpaul | 66,420 | <p>Let
$$ f(x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n(n+1)}x^{n+1} $$
and hence
$$ f'(x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}x^{n}, f''(x)=\sum_{n=1}^\infty(-1)^{n+1}x^{n-1}=\frac{1}{1+x}.$$
Note that
$$ f(1)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n(n+1)}, f'(0)=0.$$
So
\begin{eqnarray}
f(1)&=&\int_0^1\int_0^x\frac{1}{1+t}dtdx\\
&=&\int_0^1\int_t^1\frac{1}{1+t}dxdt\\
&=&\int_0^1\frac{1-t}{1+t}dt\\
&=&2\ln2-1.
\end{eqnarray}</p>
|
1,492,027 | <p>Defining R to be the relationship on real numbers given by xRy iff x-y is rational, I've been asked to find the equivalence class of $\sqrt2$. My instincts say that the equivalence class of $\sqrt2$ would just be the empty set. But after a riveting conversation on a similar subject <a href="https://math.stackexchange.com/questions/1491971/subtraction-of-two-irrational-numbers-to-get-a-rational/1491978#1491978">subtraction of two irrational numbers to get a rational</a></p>
<p>I'm curious to if I am able to pick and choose x to suite my needs? say could I define x to be $\alpha+\sqrt2$ so the equivalence class would be $y=\alpha$?</p>
<hr>
<p>EDIT: This is just to see if I have the principle of equivalence classes down..</p>
<p>given the Relation R xRy s.t. x-y is rational, the following equivalence classes are..</p>
<p>[0] = {y: y is rational}</p>
<p>[1] = {y: y is rational}</p>
<p>[$\sqrt2$] = {y: y = $\alpha+\sqrt2$ s.t. $\alpha$ is rational}</p>
<p>To summarize to find the above equivalence classes on xRy iff x-y is rational</p>
<p>I would look at all of the ordered pair (0,y),(1,y),($\sqrt2$,y) fulfills the above stipulation?</p>
<p>Or as to ensure further confidence that I understand (I don't want to be spoon, fed but confirmation that I'm on the right track would be greatly appreciated.</p>
<p>Given the relation T given by (x,y)T(a,b) iff $x^2+y^2=a^2+b^2$</p>
<p>would the equivlanece class of (1,2) be the circle of radius $\sqrt5$?</p>
| Kevin Quirin | 267,868 | <p>Remember that, by definition, the equivalence class of $\sqrt 2$ is the <strong>set</strong>
$$[\sqrt 2] = \{y\in\mathbb R~|~\sqrt 2 - y \in \mathbb Q\}$$</p>
<p>Pose $A = \{\alpha + \sqrt 2~|~\alpha\in\mathbb Q\}$.</p>
<ul>
<li>Pick an element $x\in A$ : there is $\alpha \in\mathbb Q$ such that $x = \alpha + \sqrt 2$. Then $\sqrt 2 - x = -\alpha \in \mathbb Q$, thus $x\in[\sqrt 2]$.</li>
<li>Pick an element $x\in[\sqrt 2]$. Then, by definition of $[\sqrt 2]$, there is $\alpha \in \mathbb Q$ such that $\sqrt 2 - x = \alpha$, ie $x = -\alpha = \sqrt 2$. Thus $x\in A$.</li>
</ul>
<p>We just showed that $[\sqrt 2] = A$.</p>
|
1,613,185 | <p>There are five red balls and five green balls in a bag. Two balls are taken out at random. What is the probability that both the balls are of the same colour</p>
| Mithlesh Upadhyay | 234,055 | <p>Using <a href="https://en.wikipedia.org/wiki/Hypergeometric_distribution" rel="nofollow">Hypergeometric distribution</a>: Given, $5$ red and $5$ green ball. So, total number balls is $10$. </p>
<p>Two balls are taken at random, So required probability is : </p>
<p>$= \frac{^5C_2 \times ^5C_0 + ^5C_2 \times ^5C_0}{^{10}C_2}=\frac{10+10}{45}=\frac{20}{45}$</p>
|
1,569,331 | <p>Let $f$ be a continuous function on $[a,b]$. Show that </p>
<p>$$|f(x)-f(x_0)|\leq |f'(x_0)||x-x_0|.$$</p>
<p>I don't know whether differentiability of $f$ on $(a,b)$ is needed in assumption.</p>
<p>I just have seen this question in a part of a paper in the class, so I did not know exactly this is the question or something is missing.</p>
<p>Could anyone tell me what is this and how to start solving it.</p>
| Kamil Jarosz | 183,840 | <p>$$\int_1^7f\,dx=\int_1^9f\,dx+\int_9^7f\,dx=\int_1^9f\,dx-\int_7^9f\,dx=-1-5=-6$$</p>
|
1,569,331 | <p>Let $f$ be a continuous function on $[a,b]$. Show that </p>
<p>$$|f(x)-f(x_0)|\leq |f'(x_0)||x-x_0|.$$</p>
<p>I don't know whether differentiability of $f$ on $(a,b)$ is needed in assumption.</p>
<p>I just have seen this question in a part of a paper in the class, so I did not know exactly this is the question or something is missing.</p>
<p>Could anyone tell me what is this and how to start solving it.</p>
| Guilherme Thompson | 177,882 | <p>Let $a,b,c$ be 3 numbers in the real line, such $a>c>b$, and $f(x): \mathbb{R} \mapsto \mathbb{R}$ continuous in $(a,b)$, we have
$$ \int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx.$$</p>
<p>For your specific case, we have
$$ \int_1^9 f(x) dx = \int_1^7 f(x) dx + \int_7^9 f(x) dx \implies\\ \int_1^7 f(x) dx = \int_1^9 f(x) dx - \int_7^9 f(x) dx = -1 - 5 = -6.$$</p>
|
1,776,850 | <p>Given a square $ABCD$ such that the vertex $A$ is on the $x$-axis and the vertex $B$ is on the $y$-axis. The coordinates of vertex $C$ are $(u,v)$. Find the area of square in terms of $u$ and $v$ only.</p>
<p><strong>What I have done</strong></p>
<p>Let the coordinate of $A$ be $(x,0)$ and $B$ be $(0,y)$. Also let the side of the square be a units.</p>
<p>$2a^2=AC^2=(x-u)^2+v^2$</p>
<p>$a^2$ is the required area so if we write $x$ in terms of $u$ and $v$ then the job will be done. </p>
<p>Now from here I thought of two ways either using trigonometry or using rotation of axes but here none of them will work because some angles will be involved and we require just $u$ and $v$ and nothing else in the expression of the area of square.</p>
<p>So how to do it? Please help.</p>
| amd | 265,466 | <p>If the points $A(x,0)$ and $B(0,y)$ are adjacent vertices of a square, then the squareβs area is $x^2+y^2$. Assuming that the vertices are enumerated counterclockwise, $C$ has coordinates $(-y,y-x)$, i.e., $B+\operatorname{rot}(B-A,\pi/2)$, so $x=-(u+v)$, $y=-u$ and $x^2+y^2=(u+v)^2+u^2=2u^2+2uv+v^2$. If the vertices are enumerated clockwise instead, then $C$ is $(y,x+y)$, $x=v-u$, $y=u$ and the area is $2u^2-2uv+v^2$.</p>
|
255,164 | <p>$\newcommand{\al}{\alpha}$
$\newcommand{\euc}{\mathcal{e}}$
$\newcommand{\Cof}{\operatorname{Cof}}$
$\newcommand{\Det}{\operatorname{Det}}$</p>
<p>Let $M,N$ be smooth $n$-dimensional Riemannian manifolds (perhaps with smooth boundary), and let $\, f:M \to N$ be a smooth <strong>immersion</strong>. Let $\Omega^k(M,f^*TN)$ be the space of $f^*TN$-valued $k$-forms. </p>
<p>Let $d:\Omega^k(M,f^*TN) \to \Omega^{k+1}(M,f^*TN)$ be the covariant exterior derivative associated with the pullback connection of the Levi-Civita connection on $N$ (via $f$), and let $\delta$ be its adjoint.</p>
<p>Denote $\delta_{k}:\Omega^k(M,f^*TN) \to \Omega^{k-1}(M,f^*TN)$.</p>
<p><strong>Question:</strong> Is $\ker \delta_1$ is infinite dimensional?</p>
<p>Since $\delta=\star d \star$ (up to sign), $\dim(\ker \delta_1)=\dim(\ker d_{n-1})$. So, this is a question about $d$.</p>
<hr>
<p><strong>Edit:</strong></p>
<p>In the special case where $N$ is flat, $\delta_{k} \circ \delta_{k+1}=0$, so $\operatorname{Image}(\delta_{k+1}) \subseteq \ker(\delta_k)$.</p>
<p>In particulr, $\operatorname{Image}(\delta_{2}) \subseteq \ker(\delta_1)$.</p>
<p>Is $\operatorname{Image}(\delta_{2})$ is infinite-dimensional? Can we at least something when $N=\mathbb{R}^n$? or even when $M=N=\mathbb{R}^n$?</p>
<p>Let's see what happens when $M=N=\mathbb{R}^n,f=\operatorname{Id}$:</p>
<p>$$ d_{n-1}(\sigma)(e_1,...,e_n)=\sum_{j=1}^{d} (-1)^{j+1} \nabla^{T\mathbb{R}^n}_{e_j} \big( \sigma(e_1,...,\hat{e_j},...,e_d) \big)$$</p>
<p>Write $\sigma=(-1)^{j+1}a_j^idx^1 \wedge \dots \wedge \hat{dx^j} \wedge \dots \wedge dx^d \otimes e_i, a_j^i \in C^{\infty}(\mathbb{R}^n)$. </p>
<p>Then $ \sigma(e_1,...,\hat{e_j},...,e_d)=(-1)^{j+1}a_j^i e_i $, so </p>
<p>$$ \nabla^{T\mathbb{R}^n}_{e_j} \big( \sigma(e_1,...,\hat{e_j},...,e_d) \big)= \nabla^{T\mathbb{R}^n}_{e_j}((-1)^{j+1}a_j^i e_i)=(-1)^{j+1}\frac{\partial a_j^i}{\partial x_j}e_i.$$</p>
<p>Thus, $$\sigma \in \ker d_{n-1} \iff \frac{\partial a_j^i}{\partial x_j}e_i=0 $$ (in the last term there is a double summation, on $i,j$).</p>
<p>Since $e_i$ are independent, this is equivalent to
$$\sum_{j=1}^d \frac{\partial a_j^i}{\partial x_j}=0 \, \text{ for all } \, i=1,...,d $$</p>
<p>Denoting $\bar a^i=(a_1^i,...,a_d^i)$, we get that $\operatorname{div} (\bar a^i)=0$.</p>
<p>I guess it shouldn't be too hard to see now that $\dim(\ker d_{n-1}) = \infty$. (By taking the $a_j^i$ to be constants, one immediately gets $\dim(\ker d_{n-1}) \ge n^2$. Since the condition on the divergence do not touch $\frac{\partial a_j^i}{\partial x_k}$ for $k \neq j$, it seems plausible that the dimension is indeed infinite. Perhaps someone can come with a slick argument to show this.)</p>
<p>The next case we should try is $M=N=\mathbb{R}^d$, and $f$ an arbitrary mapping....</p>
<hr>
<p><strong>Comment:</strong></p>
<p>I know that $\operatorname{Cof}(df) \in \ker \delta_1$, where $\operatorname{Cof}(df)$ is the corresponding <em>cofactor map</em> of $df$:
$$ \Cof df= (-1)^{d-1} \star_{f^*TN}^{n-1} (\wedge^{n-1} df) \star_{TM}^1. $$ </p>
<p>Since $f$ is an immersion, $\operatorname{Cof}(df) \neq 0$, so $\dim (\ker \delta_1 ) \ge 1$.</p>
<p>For my purposes, It would suffice to know that $\ker \delta_1$ always contains elements which are linearly independent of $\Cof df$.</p>
| Craig | 10,749 | <p>It's clearly infinite dimensional in the flat case.</p>
<p>$\ker\delta_1 =\ker d_{n-1}$</p>
<p>And in a coordinate chart with coordinates $(x_1, \ldots, x_n)$ take various $f\in C^\infty(M)$ with support in the chart, so </p>
<p>$df\wedge dx_1 \wedge\cdots\wedge dx_{n-2} \in \ker d_{n-1}$</p>
|
405,953 | <p>Let</p>
<ul>
<li><span class="math-container">$E$</span> be the usual sobolev space <span class="math-container">$H^{1}_{0}(\Omega)$</span> on a smoothly bounded domain <span class="math-container">$\Omega$</span>,</li>
<li><span class="math-container">$E_{k}$</span> be its subspace spanned by the first <span class="math-container">$k$</span> eigenfunctions of the Laplace operator, i.e.
<span class="math-container">$$E_{k}:=\text{span}\{\varphi_{j}\in E: -\Delta\varphi_{j}=\lambda_{j}\varphi_{j},~j=1,2\dots,k \},$$</span></li>
<li><span class="math-container">$P$</span> be the positive cone in <span class="math-container">$E$</span>, i.e,
<span class="math-container">$$P:=\{u \in E~; u \geq 0 ~a.e.\},$$</span></li>
</ul>
<p>Now set <span class="math-container">$P_{k}=P\cap E_{k}~~u^{-}=min\{u,0\}$</span>.</p>
<p><strong>My question</strong>: does it exist <span class="math-container">$C_{k}>0$</span>, such that
<span class="math-container">$$\text{dist}_{L^{2}}(u,P_{k}) \leq C_{k} \text{dist}_{L^{2}}(u,P)~\text{ holds }~\forall u\in E_{k}\; ?$$</span>
or
<span class="math-container">$$\text{dist}_{E}(u,P_{k}) \leq C_{k} \text{dist}_{E}(u,P)~\text{ holds }~\forall u\in E_{k}\; ?$$</span>
or <span class="math-container">$$\text{dist}_{E}(u,P_{k}) \leq C_{k} \|u^{-}\|_{E}~\text{ holds }~\forall u\in E_{k}\; ?$$</span>
If not, could you please show me a counterexample? Thanks.</p>
<p>Moreover, this problem is raised from the proof of lemma 3.6 in "Infinitely many solutions to perturbed elliptic equations",<a href="https://pdf.sciencedirectassets.com/272601/1-s2.0-S0022123605X05468/1-s2.0-S0022123605002582/main.pdf?X-Amz-Security-Token=IQoJb3JpZ2luX2VjEIf%2F%2F%2F%2F%2F%2F%2F%2F%2F%2FwEaCXVzLWVhc3QtMSJIMEYCIQCA3DmdTW7ToN5U0cdmubTTGzQJCXX38jpSYRu%2FaTQqOgIhANqGdebL%2FSXPmzKTqmmN0Xr%2F8itx2hkp7WGf6i72BrqTKvoDCBAQBBoMMDU5MDAzNTQ2ODY1Igz5FlE6QYbbcb7YH8kq1wMEn%2BjofTAgOGQ%2B7YALRJ5jJO286FtV2mHjiOBbpm1uIdXmNcC7GQJyghCxd4oRxfBUUIvPrbOwfdDH09ESAULHmDMXJNKlFyTF%2F33Jn2DSngmTTTq%2F8JFpuR2Bqe7p4k%2FDYnmnBm%2BMIsCoevLmtmJQ%2BNbL0bx%2FnvwCnvW%2BI45ClJddGgxdHCYBLrLxg0AOtNgy%2FIEagUE8CxkIXDro7QeNKau9lwQfEWrVa9lakzcMfXC70%2BnlVk2VzXAIdX2cKYDsf%2Fl8wl7Dkcs1%2FlqbbXNxz95n8CuAg7ifFqc0dfnetoIf02pLxKaw4hMx40lKKaWRdMLyKnT9C5fFToz5n%2Ff1liqy%2BDPvphnNV18lN7fZ%2Bbw5hUYj8q8rCxCIc8DnSeExbF78GjTXwR%2FsHHuHUZoXjIah2vK8SJUeVspjWfFK3CDxHokr%2BibxlkNhX2pjXIrZeAHqwMavPCT0bkvyU90XbZuI1vOc3h0FxnHfFXhyReh%2Bz6NjMl7OUv2YiaQgoxQEFRp4g2nNLau0RaPfxLTG77edF5nlL6b3bmSyu6KV%2FzWE2dOGEu9vk9cG%2F4ID3R1jUl8sSwEDiCMHZAaAMxjAld0l%2BtLwvKGq0046wxg5Uy9klhYCFbUwit2UiwY6pAHeFguTRVuilMp%2BddCPI10eVsY1ugttXA0kflCcFX73U8SfGzOTjmoheDKj%2FCNTQrzgGp8p4yhBjLvODQ8croHzoNyV6PoOl92rAARM5%2Fa9NOlTDvT2RTMpQzOI0sfSCvXEuUClp%2BrdylgeZJSLSMIT%2FQ6lyj12iGeJWZpoygEIvc3oTNdx8YjZSwbV4cv8smABlGArjB4p3eRWS0R5gFzpL8SZgA%3D%3D&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20211012T080454Z&X-Amz-SignedHeaders=host&X-Amz-Expires=300&X-Amz-Credential=ASIAQ3PHCVTY4Y4CPKBJ%2F20211012%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=e9f461d9634a663aa1a87583e2578ad501f687454a69dff0e960c653c345a3f2&hash=ebaa2e2d930797073bf18eae43724b9df1cc260ef6f19e04102732392edbc536&host=68042c943591013ac2b2430a89b270f6af2c76d8dfd086a07176afe7c76c2c61&pii=S0022123605002582&tid=spdf-ca8bb609-0d0a-4a63-8c85-58d7098f6e09&sid=e27151265f16754ed3986a79985943117767gxrqa&type=client" rel="nofollow noreferrer">doi:10.1016/j.jfa.2005.06.014</a>
<a href="https://i.stack.imgur.com/ruowq.png" rel="nofollow noreferrer">enter image description here</a></p>
<p>also see lemma 5.4 in "On finding sign-changing solutions",<a href="https://pdf.sciencedirectassets.com/272601/1-s2.0-S0022123606X05591/1-s2.0-S0022123605003319/main.pdf?X-Amz-Security-Token=IQoJb3JpZ2luX2VjEJ7%2F%2F%2F%2F%2F%2F%2F%2F%2F%2FwEaCXVzLWVhc3QtMSJIMEYCIQDyfP9hsP1tHkBhbzcqwuQOpM51AhMSmUw8lQjARVyomQIhAL%2BkqkTXKKLwXuf08PPLGidPv2ZzyX2LYEMUTEyKhGbgKvoDCCcQBBoMMDU5MDAzNTQ2ODY1Igy%2F9ma67gic8nnsmhIq1wMLrIT%2FDxI9G6FQavBW3P21mCaVAw3de%2B4W15bgV7bTepA4sSaYvFGgm84PGZscT2JoxYpbJXPpHh%2FHncPeCKdSwMB6KisZ3qj3TUi5UVctygkh7PsrPp48LPms%2B98r%2FmEHdrrGD3PejQlZKrHT6eZ7S48OpXV7qtNt3FBqPGj12kZU3npDDdO4y3eOdo4vTyBkab7tyPAovwCr4frZiAooONzBayGkUEmKZ0bAnYgkElIXAgUbkGawBjFzPsfoiXmPsgX0GwiSIIV%2BBj4BEW4VglGabDqRtAO7xjGPegxoVM7chmabrwHwcOIuzT6pZhD4rqOeTjTtq88dBlRgUWVm3Ko6j%2BWDcfNVQxsbpgZZ3FjDKlJCt2Ki%2FKhBoET%2BC2CNNWMww8XJvT789JwPTI5MR7icHX5g54pxcbZV7iSgTrFHCUjE%2B%2BYyGg3cRNz73Wv65FvL1YnebwAeAGinMYhKSGPA5Vujv4f%2B3UrZM7TWOEK76t8ASgklluYS2B4hNdfNxMuhElgannyqfO8Mao2JVUcag5mBM1dJHcPc6xGbRgJQJkKG9AZ01CQ6trBMe9gCDy3ZSwRZAsgSu9xf0gr6%2BjjH8WzmMqoyE%2B7KLyqOS5i2ecsyQcgw%2F%2BaZiwY6pAEBkjdSZRAyQgflJe%2FR0CGWu1dHufRVHsGeRCeSEwzyfaBkii6sCmJv5IjB7QLnMiHrZYogDf4FdczVdD5ij%2Bp%2BL5Kd7q7FWkUAhLTx9Gm32EiH%2B4Nn8UO7Iczh0aMSuqPXdyec3P%2BRiF50K5bXrOKwo2fWF9nfUbGZMTFAh4UjeuIU1WD9OKgaIcMDNlMNzFMWJCUN4bS5AVZ1pV8UsBE5IpoaVQ%3D%3D&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20211013T062711Z&X-Amz-SignedHeaders=host&X-Amz-Expires=300&X-Amz-Credential=ASIAQ3PHCVTYY3F5SIHU%2F20211013%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=74d463a3fdfe47d9fe039c0ce0d0c8d8647cae0dbe5669c7c1dd2f6fe8e6ddac&hash=4ab29c5d1d48cc6b4b9a75d0b843958d134747e1605579583d6db363c5a6e3b2&host=68042c943591013ac2b2430a89b270f6af2c76d8dfd086a07176afe7c76c2c61&pii=S0022123605003319&tid=spdf-9e5fb13a-6657-456c-a6e4-cfaa6bcc0bf1&sid=8b90046339b6054a78486f6-8787b72cf34agxrqa&type=client" rel="nofollow noreferrer">doi:10.1016/j.jfa.2005.09.004</a>
<a href="https://i.stack.imgur.com/CLUMD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CLUMD.png" alt="enter image description here" /></a></p>
| Giorgio Metafune | 150,653 | <p>The first is not true, and probably also the others.</p>
<p>Take <span class="math-container">$L^2(0, \pi)$</span> and <span class="math-container">$u_1=\sin x$</span>, <span class="math-container">$u_2=\sin (2x)$</span>, so that <span class="math-container">$E_2=\{u=a\sin x+b \sin (2x)\}$</span> and <span class="math-container">$u \geq 0$</span> iff <span class="math-container">$a \geq 0$</span> and <span class="math-container">$2|b| \leq a$</span>. If <span class="math-container">$v=\alpha sin x+\beta \sin (2x)$</span>, then <span class="math-container">$\|u-v\|_2^2=\frac{\pi}{2} \left((a-\alpha)^2+(b-\beta)^2\right)$</span> and, if <span class="math-container">$v_\epsilon=\sin x-\frac{1+\epsilon}{2}\sin (2x)$</span>, the closest positive <span class="math-container">$u \in E_2$</span> is <span class="math-container">$u=\sin x-\frac{1}{2}\sin (2x)$</span> and <span class="math-container">$\|v_\epsilon -u\|_2 \approx \epsilon$</span>. On the other hand, <span class="math-container">$v_\epsilon$</span> is negative in an interval starting from 0 of length <span class="math-container">$\approx \sqrt \epsilon$</span> where the function is of order <span class="math-container">$\epsilon^{3/2}$</span> and <span class="math-container">$\|v^-\|_2 \leq C\epsilon^{7/4}$</span>.</p>
|
3,632,576 | <p>Considering that input <span class="math-container">$x$</span> is a scalar, the data generation process works as follows:</p>
<ul>
<li>First, a target t is sampled from {0, 1} with equal probability.</li>
<li>If t = 0, x is sampled from a uniform distribution over the interval
[0, 1]. </li>
<li>If t = 1, x is sampled from a uniform distribution over the
interval [0, 2].</li>
</ul>
<p>I'm trying to find the formula for <span class="math-container">$P(t=1)$</span>, <span class="math-container">$P(t=0)$</span>, <span class="math-container">$P(x|t=1)$</span> and <span class="math-container">$P(x|t=0)$</span> and then find the posterior probability <span class="math-container">$P(t = 0|x)$</span> as a function of <span class="math-container">$x$</span>.</p>
<p>So far I have that <span class="math-container">$P(t=1)$</span> and <span class="math-container">$P(t=0)$</span> <span class="math-container">$=$</span> <span class="math-container">$\frac12$</span> but I wasn't sure how to find <span class="math-container">$P(x|t=1)$</span> and <span class="math-container">$P(x|t=0)$</span>.</p>
<p>I know from there we can just use <span class="math-container">$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$</span> to compute the probability <span class="math-container">$P(t = 0|x)$</span> as a function of <span class="math-container">$x$</span>. Is that correct?</p>
| Paramanand Singh | 72,031 | <p>Using <span class="math-container">$g(x) =f(x) - a$</span> we can reduce the problem to case when <span class="math-container">$a=0$</span>. So lets prove the result when <span class="math-container">$a=0$</span>.</p>
<p>Let <span class="math-container">$\epsilon >0$</span> and we have a number <span class="math-container">$M>0$</span> such that <span class="math-container">$$-\epsilon<f(x) <\epsilon$$</span> whenever <span class="math-container">$x>M$</span>. Integrating this on interval <span class="math-container">$[M, x] $</span> we get <span class="math-container">$$-\epsilon(x-M) <\int_{M}^{x} f(t) \, dt<\epsilon(x-M) $$</span> or <span class="math-container">$$-\epsilon\left(1-\frac{M}{x}\right)+\frac{1}{x}\int_{0}^{M}f(t)\,dt<\frac{1}{x}\int_{0}^{x}f(t)\,dt<\epsilon\left(1-\frac{M}{x}\right)+\frac{1}{x}\int_{0}^{M}f(t)\,dt$$</span> Letting <span class="math-container">$x\to \infty $</span> we get <span class="math-container">$$-\epsilon\leq\liminf_{x\to\infty} \frac{1}{x}\int_{0}^{x}f(t)\,dt\leq\limsup_{x\to\infty} \frac{1}{x}\int_{0}^{x}f(t)\,dt\leq\epsilon $$</span> Since <span class="math-container">$\epsilon$</span> is arbitrary our proof is complete. </p>
|
2,528,716 | <p>How can I prove </p>
<blockquote>
<p>$$x^2+y^2-x-y-xy+1β₯0$$</p>
</blockquote>
<p>I tried $(x+y)^2-3xy-(x+y)+1β₯0 \rightarrow(x+y-1)(x-y)-3xy+1β₯0$ I can not continue</p>
| Zaharyas | 457,607 | <blockquote>
<p>$$x^2+y^2-x-y-xy+1β₯0 \Rightarrow \frac 12( 2x^2+2y^2-2x-2y-2xy+2)β₯0 \Rightarrow \frac 12( x^2-2x+1+y^2-2y+1+x^2+y^2-2xy)β₯0 \Rightarrow \frac 12((x-1)^2+(y-1)^2+(x-y)^2)β₯0$$</p>
</blockquote>
|
2,471,633 | <p>Let $B$ an open ball in $\mathbb{R}^{n}$, and $(K_{j})_{j}$ be an increasing sequence of compact subsets of $B$ whose union equals $B$. For each $j$, let $\rho_{j}$ be a cut-off function in $C_{c}^{\infty}(B)$ that equals 1 on a neighborhood of $K_{j}$ and whose support is in $K_{j+1}$. Finally, let $\theta$ be a smooth function whose laplacian $\Delta\theta\equiv1$, and set
$$\phi_{j}:=\rho_{j}\theta.$$
My question: can we conclude that
$$\sup_{(x,j)\in B\times\mathbb{N}}|\Delta\phi_{j}(x)|$$
is bounded?</p>
| Rigel | 11,776 | <p>I think you cannot conclude that the $\sup$ is finite, because $\nabla\rho_j$ is not bounded.</p>
<p>Consider the following one-dimensional example.
Let $B = (-1,1)$, and let $K_j := [-1+1/j, 1-1/j]$.
In a first approximation you can think $\rho_j$ a Lipschitz function (piecewise affine) such that $\rho_j = 1$ on $K_j$, $\rho_j = 0$ outside $K_{j+1}$, and $\rho_j$ affine in the two intervals $I_j := (1-1/j, 1-1/(j+1))$, $I_j' :=
(-1+1/(j+1), -1+1/j)$.
(Your "real" $\rho_j$ will be a smooth approximation of this Lipschitz function.)</p>
<p>On these intervals you have $\rho_j' = j(j+1)$.
If you compute
$$
\phi_j'' = \theta \rho_j'' + \rho_j \theta'' + 2 \rho_j' \theta'
$$
you see that you cannot bound the last term.
(Probably also the first term cannot be bounded.)</p>
|
487,749 | <p>An opinion poll in a certain city indicated that 69 people in a random sample of 120 said that they would vote for Mr. Jones, while in a second random sample of 160, 93 said that they would vote for Mr. Jones. Find an unbiased estimate of the proportion of people in the city who will vote for Mr. Jones.</p>
| Caleb Stanford | 68,107 | <p><strong>Hint:</strong> By "unbiased", they probably mean that <em>every person is counted equally in the estimate</em>.</p>
<p>The first random sample had $69/120 = 0.575$ in favor of Mr. Jones. The second random sample had $93/160 = 0.58125$ in favor of Jones. But <em>we can't just take the average of these two numbers</em>, because the second number included more people than the first number, so <em>we would not be representing all people equally</em>.</p>
<p>Instead, try pretending that there was only one poll, instead of two, and that that poll included all of the people in both of these random samples combined.</p>
|
487,749 | <p>An opinion poll in a certain city indicated that 69 people in a random sample of 120 said that they would vote for Mr. Jones, while in a second random sample of 160, 93 said that they would vote for Mr. Jones. Find an unbiased estimate of the proportion of people in the city who will vote for Mr. Jones.</p>
| AlexR | 86,940 | <p>Another approach (yielding the same result as Goos) is using a weighted average, taking the number of people asked as the weight:
$$\bar{X}_\omega = \frac{\sum_{i=1}^n \omega_i X_i}{\sum_{i=1}^n \omega_i}$$
Where in this case $n=1, \omega = (120, 160), X = (69,93)$</p>
|
487,749 | <p>An opinion poll in a certain city indicated that 69 people in a random sample of 120 said that they would vote for Mr. Jones, while in a second random sample of 160, 93 said that they would vote for Mr. Jones. Find an unbiased estimate of the proportion of people in the city who will vote for Mr. Jones.</p>
| Felix Marin | 85,343 | <p>\begin{align}
{\cal F}\left(p\right)
&=
{\left(120 p - 69\right)^{2} + \left(160 p - 93\right)^{2} \over 2}
\\[3mm]
{\cal F}\,'\left(p\right)
&=
120\left(120 p - 69\right) + 160\left(160 p - 93\right)
\\[5mm]&\mbox{}
\end{align}</p>
<p>$$
{\cal F}\,'\left(p\right) = 0
\qquad\Longrightarrow\qquad
p
=
{120\times 69 + 160\times 93\over 120^{2} + 160^{2}}
=
\color{#ff0000}{\large 0.579}
$$</p>
|
1,611,506 | <blockquote>
<p>$$\int (2x^2+1)e^{x^2} \, dx$$</p>
</blockquote>
<p>It's part of my homework, and I have tried a few things but it seems to lead to more difficult integrals. I'd appreciate a hint more than an answer but all help is valued.</p>
| user84413 | 84,413 | <p>$\textbf{Hint:}$ Write the integral as $\int2x^2e^{x^2}dx+\int e^{x^2}dx$.</p>
<p>Then use integration by parts on the first integral, with $dv=2xe^{x^2}dx$</p>
|
422,143 | <p>f differentiable function in R. $f(x)= e^{f'(x)}$
$f(0)=1$</p>
<p>I have proved that $f(x)=1$ for every $x\lt0$. im stuck for $x\gt0
$</p>
| Christian Blatter | 1,303 | <p>In the domain $H:=\{(x,y)\ |\ y>0\}\subset{\mathbb R^2}$ your differential equation is equivalent to</p>
<p>$$y'=\log y=:\psi(x,y)\ .\tag{1}$$</p>
<p>As $\psi$ is a continuously differentiable function in $H$ it is locally Lipschitz with respect to $y$ in $H$. It follows that for any $(x_0,y_0)\in H$ the initial value problem
$$y'=\log y,\qquad y(x_0)=y_0\tag{2}$$
has a unique solution $x\mapsto y=\phi(x)$ whose graph will leave any given compact subset $K\subset H$. Given the initial point $(x_0,y_0):=(0,1)$ it is obvious that $y=\phi(x)\equiv1$ solves the IVP $(2)$ for $-\infty<x<\infty$, so there is not more to it.</p>
<p>The right side $\psi(x,y)$ in $(1)$ being independent of $x$ it follows that in order to obtain a picture of all solutions of $(1)$ it is enough to consider initial points $(x_0,y_0):=(0,b)$ for $b>0$. Separation of variables will lead to solutions that can be expressed in terms of the logarithmic integral function ${\rm Li}$.</p>
|
3,455,009 | <p>In the proof of the expectation of the binomial distribution,</p>
<p><span class="math-container">$$E[X]=\sum_{k=0}^{n}k \binom{n}{k}p^kq^{n-k}=p\frac{d}{dp}(p+q)^n=pn(p+q)^{n-1}=np$$</span></p>
<p>Why is <span class="math-container">$\sum_{k=0}^{n}k \binom{n}{k}p^kq^{n-k}= p\frac{d}{dp}(p+q)^n$</span>?</p>
<p>I know that by the binomial theorem <span class="math-container">$(p+q)^n=\sum_{k=0}^{n}\binom{n}{k}p^kq^{n-k}$</span>. </p>
| Community | -1 | <p>Plugging your <span class="math-container">$z_1$</span> in the equation, we get</p>
<p><span class="math-container">$$\left(\frac{1+i}2\right)^3=\left(\frac{-1+i}2\right)^3,$$</span> which cannot be true.</p>
<hr>
<p>The solutions of </p>
<p><span class="math-container">$$3z^2+3z+1=0$$</span> are</p>
<p><span class="math-container">$$\frac{-3\pm\sqrt{9-12}}6=\frac{-3\pm i\sqrt3}{6}.$$</span></p>
<hr>
<p>For a more "trigonometric" solution,</p>
<p><span class="math-container">$$\left(1+\frac1z\right)^3=1=e^{i2k\pi}$$</span></p>
<p>so that</p>
<p><span class="math-container">$$z=\frac1{e^{i2k\pi/3}-1},$$</span></p>
<p>for <span class="math-container">$k=1,2$</span>.</p>
|
1,973,686 | <p>I am stuck on two questions :</p>
<ol>
<li>If $f,g\in C[0,1]$ where $C[0,1]$ is the set of all continuous functions in $[0,1]$ then is the mapping $id:(C[0,1],d_2)\to (C[0,1],d_1)$ continuous ? where $id$ denotes the identity mapping.</li>
</ol>
<p>where $d_2(f,g)=(\int _0^1 |f(t)-g(t)|^2dt )^{\frac{1}{2}} $ and $d_1(f,g)=(\int _0^1 |f(t)-g(t)|dt )$
?</p>
<p>2.If $f\in L^2(\Bbb R)$ does it imply that $f\in L^1(\Bbb R)$.</p>
<p><strong>My try</strong>:</p>
<p>1.The first question reduces to proving the fact that if $(\int _0^1 |f(t)-g(t)|^2dt )^{\frac{1}{2}} <\infty$ does it imply that $(\int _0^1 |f(t)-g(t)|dt )<\infty$ which I am unable to prove.</p>
<p>I am also unable to conclude anything for the 2nd one.</p>
<p>Please give some hints</p>
| H. H. Rugh | 355,946 | <p>The answer to the first is yes, by the Cauchy-Schwarz inequality (you simply have $d_1(f,g)\leq d_2(f,g)$). But it only works because the interval $[0,1]$ is finite. On ${\Bbb R}$ it does not hold (as already mentioned in other posts).</p>
|
2,663,537 | <p>Suppose G is a group with x and y as elements. Show that $(xy)^2 = x^2 y^2$ if and only if x and y commute.</p>
<p>My very basic thought is that we expand such that $xxyy = xxyy$, then multiply each side by $x^{-1}$ and $y^{-1}$, such that $x^{-1} y^{-1} xxyy = xxyy x^{-1}$ , and therefore $xy=xy$.</p>
<p>I realize that this looks like a disproportionate amount of work for such a simple step, but that is what past instruction has looked like and that is perhaps why I am confused. Moreover, "if and only if" clauses have always been tricky for me since I took Foundations of Math years ago, but if I remember correctly, the goal here should be to basically do the proof from right to left and then left to right, so to speak. Anyhow, I think that I am overthinking this problem.</p>
| Matthew Leingang | 2,785 | <p>You're right that you should do the proof from right to left and then left to right. But make sure you know what βleftβ and βrightβ are.</p>
<p>βleftβ is the equation $(xy)^2=x^2y^2$, while βrightβ is the statement β$x$ and $y$ commute.β You can write the last statement as the equation $xy=yx$. </p>
<p>This means you want to show:</p>
<ol>
<li>$(xy)^2 = x^2y^2 \implies xy =yx$</li>
<li>$xy=yx \implies (xy)^2 = x^2y^2$</li>
</ol>
<p>For the first, you can expand <em>so that</em> (better word choice than <em>such that</em>, in my opinion) $xyxy = xxyy$. Multiply on the left by $x^{-1}$ and on the right by $y^{-1}$, so that $yx=xy$.</p>
<p>For the second, $xy=yx$ means that you can swap $x$ and $y$ in $xxyy$. So $x^2y^2=xyxy=(xy)^2$.</p>
|
481,086 | <blockquote>
<p>Find a formula (provide your answer in terms of $f$ and its derivatives) for the curvature of a curve in $\mathbb{R}^3$ given by
$\{(x,y,z)\ | \ x=y, f(x)=z\}$.</p>
</blockquote>
<p>How will I be able to do this problem? </p>
<p>I know that a regular parametrization of a curve then the curvature at $\mu(t)$ is given
by: $$\kappa(t) = \left|\left|\left(\frac{\mu'(t)}{||\mu'(t)||}\right)'\frac{1}{||\mu'(t)||}\right|\right|.$$</p>
<p>But since $f$ is not a parametrization I don't know how to continue this problem. </p>
<p>Since $x=y$ and there exists a function $f$ such that $f(x)=z$ and $f(y)=z$ is it safe to assume that it is a one to one function? </p>
| bubba | 31,744 | <p>You were given a parametric curve, but it's slightly disguised. Let $\mu(x) = (x,x,f(x))$ and then apply the formula you gave in your question, but with $t=x$. To get you started, note that $\mu'(x) = (1,1,f'(x))$. Can you take it from there?</p>
|
3,700,367 | <p><strong>What is the <em>average</em> distance from any point on a unit square's perimeter to its center?</strong></p>
<p>The distance from a square's corner to its center is <span class="math-container">$\dfrac{\sqrt{2}}{2}$</span> and from a point in the middle of a square's side length is <span class="math-container">$\dfrac{1}{2}$</span>. <a href="https://i.stack.imgur.com/exKkJ.png" rel="nofollow noreferrer">A visual explanation of what I'm trying to explain</a></p>
<p>So, what would the <em>average</em> distance be, accounting for all the points along a square's perimeter?</p>
<p>Also if possible, a general formula for finding the average distance from center to edge of any <span class="math-container">$n$</span>-sided regular polygon would be super awesome. </p>
| paulinho | 474,578 | <p>The forward direction is quite straightforward. Let us prove the contrapositive of this statement. If <span class="math-container">$M^{n-1}$</span> were zero, then the nontrivial linear combination composed of only <span class="math-container">$M^{n-1}$</span> yields zero, so the elements cannot be independent.</p>
<p>For the backward direction, we will also prove the contrapositive: if <span class="math-container">$I, M, \cdots , M^{n-1}$</span> are dependent, then <span class="math-container">$M^{n-1} = 0$</span>. Suppose there is a nontrivial linear combination of these elements yielding zero, and let <span class="math-container">$M^k$</span> be the smallest power of <span class="math-container">$M$</span> that has a nonzero coefficient in this linear combination. Multiplying on both sides of the linear combination by <span class="math-container">$M^{n - k - 1}$</span>, we obtain the equation <span class="math-container">$M^{n-1} = 0$</span> as desired, since by assumption all terms with lower powers of <span class="math-container">$M$</span> are zero and all terms with higher powers have power at least <span class="math-container">$n$</span> and are hence annihilated.</p>
|
1,988,420 | <p>An Ant is on a vertex of a triangle. Each second, it moves randomly to an adjacent vertex. What is the expected number of seconds before it arrives back at the original vertex?</p>
<p>My solution: I dont know how to use markov chains yet, but Im guessing that could be a way to do this. I was wondering if there was an intuitive way to solve this problem. I would have guessed 3 seconds as an answer. </p>
<p>I'm assuming that if it is at Vertex A, there is a 1/2 chance of going to Vertex B or C. So minimum number of seconds is 2 seconds. Max number could be infinite if it keeps bouncing back between B and C without returning to A.</p>
<p>I'm still not sure how to do this puzzle.</p>
| bof | 111,012 | <p>I assume the ant starts at vertex $A.$ We want to find the expected of value the random variable $X$ which is the number of vertex-to-vertex steps (i.e. the number of seconds) the ant takes. Now
$$X=X_{A,B}+X_{B,A}+X_{B,C}+X_{C,B}+X_{C,A}+X_{A,C}$$
where $X_{A,B}$ is the number of times the ant walks from $A$ to $B,\ $ $X_{B,A}$ the number of times it walks from $B$ to $A,\ $ $X_{B,C}$ the number of times it walks from $B$ to $C,$ and so on. By the linearity of expectation,
$$E(X)=E(X_{A,B})+E(X_{B,A})+E(X_{B,C})+E(X_{C,B})+E(X_{C,A})+E(X_{A,C}).$$
By symmetry we have
$$E(X_{A,B})=E(X_{A,C})\text{ and }E(X_{B,A})=E(X_{B,C})=E(X_{C,B})=E(X_{C,A}).$$
Also
$$X_{A,B}+X_{A,C}=1\text{ and }X_{B,A}+X_{C,A}=1$$
because the vertex leaves the vertex $A$ exactly once and enters it exactly once. It follows that
$$E(X_{A,B})=E(X_{B,A})=E(X_{B,C})=E(X_{C,B})=E(X_{C,A})=E(X_{A,C})=\frac12$$
and so
$$E(X)=6\cdot\frac12=\boxed3.$$
This is a special case of a general theorem about random walks on finite <a href="https://en.wikipedia.org/wiki/Graph_theory" rel="nofollow noreferrer">graphs</a>. Namely, if an ant goes on a random walk on a finite connected undirected graph, which ends the first time it returns to the starting vertex, then the expected length of the walk is $\frac{2e}d,$ where $e$ is the number of edges in the graph and $d$ is the degree of the starting vertex. In your case the graph is $K_3,$ the complete graph on three vertices, so $e=3$ and $d=2.$</p>
|
389,750 | <p>Given A(1,4) and B (3,-5) use the dot product to find point C so that triangle ABC is a right angle triangle.</p>
| rurouniwallace | 35,878 | <p>Find the vector connecting between the two:</p>
<p>$$\vec{c}=<1-3,4+5>=<-2,9>$$</p>
<p>You can use the dot product between $\vec{c}$ and a unit vector $\hat{b}=<-1,0>$:
$$\cos{\theta}=\frac{\vec{c}\cdot\hat{b}}{||\vec{c}||\space||\hat{b}||}$$</p>
<p>The length of the opposing two angles are $||\vec{c}||\sin{\theta}$ and $||\vec{c}||\cos{\theta}$.</p>
|
1,464,522 | <blockquote>
<p>Let <span class="math-container">$O_n(\mathbb Z)$</span> be the group of orthogonal matrices (matrices <span class="math-container">$B$</span> s.t. <span class="math-container">$BB^T=I$</span>) with entries in <span class="math-container">$\mathbb Z$</span>.<br>
1) How do I show that <span class="math-container">$O_n(\mathbb Z)$</span> is a finite group and find its order?<br>
2) I need to show also that symmetric group <span class="math-container">$S_n$</span> is a subgroup of <span class="math-container">$O_n(\mathbb Z)$</span>.</p>
</blockquote>
<p>So it needs to satisfy associativity/identity/inverse.</p>
<p>It is easy to see that every orthogonal matrix <span class="math-container">$A \in O(\mathbb Z)$</span> has an inverse, namely <span class="math-container">$A^T$</span>. Moreover, the product of two orthogonal matrices is orthogonal since <span class="math-container">$(AB)^T = B^T A^T$</span>. If <span class="math-container">$A, B \in O_n(\mathbb Z)$</span> then <span class="math-container">$(AB)^T(AB) = B^T A^T AB = BIB^T = BB^T = I$</span>, hence <span class="math-container">$O_n(\mathbb Z)$</span> is closed under multiplication, since <span class="math-container">$I \in O_n(\mathbb Z)$</span>.</p>
| Pablo Herrera | 135,689 | <p>Let <span class="math-container">$A$</span> be an orthogonal <span class="math-container">$n \times n$</span> matrix with integer entries. First of all, we know that <span class="math-container">$\det(A)=\pm 1$</span>. This means that <span class="math-container">$A$</span> must be an invertible matrix.</p>
<p>Secondly, let <span class="math-container">$A_1,A_2,\dots,A_n \in \mathbb{Z}^n$</span> be the columns vectors of <span class="math-container">$A$</span>. <span class="math-container">$A$</span> is orthogonal so <span class="math-container">$||A_i||=1$</span> for <span class="math-container">$i=1,2,...,n$</span>. Then <span class="math-container">$a_{i1}^2+\cdots+a_{in}^2=1$</span>, <span class="math-container">$A_i \in \mathbb{Z}^n$</span>, hence <span class="math-container">$A_i$</span> is a canonical vector for all <span class="math-container">$i$</span>. The vectors <span class="math-container">$A_i$</span> must be an independent set.
Finally those matrices have the following form:
<span class="math-container">$$ A=\Bigg[ \pm e_{\sigma(1)} \pm e_{\sigma(2)}\dots\pm e_{\sigma(n)} \Bigg], $$</span>
where <span class="math-container">$\sigma(i)$</span> is a permutation of <span class="math-container">$I_n$</span>.</p>
<p>When <span class="math-container">$A$</span> has only positive coordinates, <span class="math-container">$A$</span> is a permutation matrix, and it is not difficult to see that those matrices are isomorphic to <span class="math-container">$S_n$</span>. </p>
|
2,632,696 | <p>I have this equation: $x^2y'+y^2-1=0$. It's an equation with separable variable. When I calculate the solution do I have to consider the absolute value for the argument of the log? </p>
| Community | -1 | <p>$$\frac1{1-y^2}$$ is defined for all $y\ne\pm1$ and its antiderivative can be expressed as</p>
<p>$$\frac12(\log|y+1|-\log|y-1|)=\log\sqrt{\left|\frac{y+1}{y-1}\right|}.$$</p>
<p>Depending on the range of $y$, this function is one of $\text{artanh(y)}$ or $\text{arcoth(y)}$. Hence the solution to the ODE is one of</p>
<p>$$y=\tanh\left(C-\frac1x\right),\\
y=\coth\left(C-\frac1x\right)=\dfrac1{\tanh\left(C-\dfrac1x\right)}.$$</p>
<p>Notice that due to the singularities, the solution is not allowed to cross the values $y=\pm1$.</p>
<p>Also, these solutions are not defined at $x=0$. But then the original equation degenerates to $y=\pm1$ and the solution has a jump there. The cotangent solution is also undefined at $Cx=1$.</p>
|
537,965 | <p><span class="math-container">$X_0:\Omega\rightarrow I$</span> is a random variable where <span class="math-container">$I$</span> is countable. Also <span class="math-container">$Y_1,Y_2,\dots$</span> are i.i.d. <span class="math-container">$\text{Unif}[0,1]$</span> random variables. </p>
<p>Define a sequence <span class="math-container">$(X_n)$</span> inductively as <span class="math-container">$X_{n+1}=G(X_n,Y_{n+1})$</span>, where <span class="math-container">$G:I\times[0,1] \rightarrow I$</span>. Show that <span class="math-container">$(X_n)$</span> is a Markov chain and determine the transition matrix in terms of G? </p>
| Shuchang | 91,982 | <p>Since $X_{n+1}=G(X_n,Y_{n+1})$ where $Y_{n+1}$ is independent with $X_i$ for all $i$, of course we have
$$\begin{align}P(X_{n+1}|X_n,...,X_0)&=P(G(X_n,Y_{n+1})|X_n,...,X_0)\\&=P(G(X_n,Y_{n+1})|X_n)\\&=P(X_{n+1}|X_n)\end{align}$$
which indicates $\{X_n\}$ is a Markov chain.</p>
|
177,209 | <p>I found the following problem while working through Richard Stanley's <a href="http://www-math.mit.edu/~rstan/bij.pdf">Bijective Proof Problems</a> (Page 5, Problem 16). It asks for a combinatorial proof of the following:
$$ \sum_{i+j+k=n} \binom{i+j}{i}\binom{j+k}{j}\binom{k+i}{k} = \sum_{r=0}^{n} \binom{2r}{r}$$
where $n \ge 0$, and $i,j,k \in \mathbb{N}$, though any proof would work for me.</p>
<p>I also found a similar identity in Concrete Mathematics, which was equivalent to this one, but I could not see how the identity follows from the hint provided in the exercises.</p>
<p>My initial observation was to note that the ordinary generating function of the right hand side is $\displaystyle \frac {1}{1-x} \frac{1}{\sqrt{1-4x}}$, but couldn't think of any way to establish the same generating function for the left hand side.</p>
| Grigory M | 152 | <p>Turns out, this indeed follows (relatively) directly from Strehl's identity.</p>
<p>Let's rewrite LHS in terms of <span class="math-container">$i$</span>, <span class="math-container">$k$</span> and <span class="math-container">$l=i+j$</span>:
<span class="math-container">$$
\text{LHS}=
\sum_{l+k=n}\sum_{i=0}^l\binom li\binom{k+l-i}{l-i}\binom{k+i}i=
\sum_{l+k=n}\sum_{i=0}^l(-1)^l\binom li\binom{-k-1}{l-i}\binom{-k-1}i;
$$</span>
application of <a href="https://math.stackexchange.com/q/586138/">Strehl's identity</a> to the internal sum (in that post's notation, for <span class="math-container">$n=-k-1$</span>, <span class="math-container">$m=l$</span>) yields
<span class="math-container">$$
\sum_{l+k=n}\sum_i(-1)^l\binom{-k-1}i\binom li\binom{2i}l=
\sum(-1)^{l-i}\binom{k+i}i\binom li\binom{2i}l=
\sum(-1)^{l-i}\binom{k+i}i\binom i{l-i}\binom{2i}i;
$$</span>
the <span class="math-container">$\binom{2i}i$</span> terms suggests that we're almost there β and indeed what we've got is
<span class="math-container">$$
\sum_{i=0}^n\binom{2i}i\sum_j(-1)^j\binom{n-j}i\binom ij
$$</span>
and the internal sum is (always) equal to 1 by Vandermonde convolution. QED.</p>
<hr>
<p><em>Remark.</em> In principle, this is not that far from being bijective, since we have a bijective proof of Strehl's identity β but that proof works only for positive parameters...</p>
|
1,089,078 | <p>Suppose we have a deck of cards, shuffled in a random configuration. We would like to find a $k$-bit code in which we explain the current order of the cards. This would be easy to do for $k=51 \cdot 6=306$, since we could encode our deck card-by-card, using $2$ bits for the coloring and $4$ bits for the number on each card.</p>
<p>We would like to optimalise this code. There exist $52!$ ways to arrange our deck, and so $k$ will have to be at least: $\lceil\log(52!)\rceil=226$. I'm asked to find a code for a $k$-value halfway between $306$ and $226$.</p>
<p>I understand that my code cannot work card-by-card, since there exist $52$ different cards and $\lceil \log(52)\rceil=6$. Therefore any card-by-card codation will lead to $k\geq306$.</p>
<p>Therefore I concluded my strategy should encode blocks of cards, another idea I had was to encode the colouring first and the numbers second.</p>
<p>Could anyone give me a hint about where to go from here?</p>
| Seyhmus GΓΌngΓΆren | 29,940 | <p>According to Hoffman, the optimal lossless coding needs at least $H(s)$ bits on average for each code, where $h(s)$ is the entropy of the source $s$.</p>
<p>In your case there is no redundancy in the numbers that you want to encode. They are simly numbers starting from $1$ to $52!$. I could suggest using <a href="http://en.wikipedia.org/wiki/Variable-length_code" rel="nofollow">variable length coding</a>. One needs to be careful about the code construction since it needs to be decodable and not all codes are decodable. It is explained in the wiki article how to do it.</p>
|
1,684,741 | <p>I'm able to show it isn't absolutely convergent as the sequence $\{1^n\}$ clearly doesn't converge to $0$ as it is just an infinite sequence of $1$'s. How do I prove the series isn't conditionally convergent to prove divergence!</p>
| Gottfried Helms | 1,714 | <p>The following is -in principle-still "searching" but structures the space to be searched into simpler subspaces:
$$ \begin{array}{} &4 &= y^4 \pmod 7 \\
& y^4 - 4 &\equiv 0 \pmod 7 \\
&(y^2 - 2)(y^2+2) &\equiv 0 \pmod 7 \\
&& \text{giving two factors}\\
&y^2 - 2 &\equiv 0 \pmod 7 \\
\text{ or } & y^2 + 2 &\equiv 0 \pmod 7 \\
& y^2 &\equiv k \cdot 7 +2 & \to k=1,y^2=9 \text{ or } k=2,y^2=16 \text{ or ...}\\
\text{ or } & y^2 &\equiv j \cdot 7- 2 &\to k=?? \\
\end{array}$$</p>
|
8 | <p>Contexts have backticks, which conflict with the normal way to enter inline code. How do I enter an inline context, since the initial approach:</p>
<pre><code>`System``
</code></pre>
<p>doesn't work ( `System`` ).</p>
| David Z | 79 | <p>According to <a href="https://meta.stackexchange.com/questions/12694/escaping-backticks-fails">this MSO question</a>, you can use double backticks set off by spaces to surround a code snippet:</p>
<pre><code>`` System` ``
</code></pre>
<p>produces <code>System`</code>.</p>
|
2,408,223 | <p>Compute $\int_0^2 \lfloor x^2 \rfloor\,dx$.</p>
<p>The challenging part isn't the problem itself, but the notation around the x^2. I don't know what it is. If someone could clarify, that would be great!</p>
<p>Edit: Clarified that it represents the floor function, can anyone give me a hint on how to start working on the problem?</p>
| John Wayland Bales | 246,513 | <p>The following diagram shows the effect that the greatest integer function has on the graph of $y=x^2$.
<a href="https://i.stack.imgur.com/jltzM.png" rel="noreferrer"><img src="https://i.stack.imgur.com/jltzM.png" alt="Greatest integer function of x square"></a></p>
<p>The answer will equal the area of the three rectangles on the right. Note that the endpoints of the three intervals on the $x$ axis will be $1,\sqrt{2},\sqrt{3},2$.</p>
<p>You should be able to take it from here.</p>
|
43,611 | <p>I posted this on Stack Exchange and got a lot of interest, but no answer.</p>
<p>A recent <a href="http://people.missouristate.edu/lesreid/POW12_0910.html" rel="nofollow">Missouri State problem</a> stated that it is easy to decompose the plane into half-open intervals and asked us to do so with intervals pointing in every direction. That got me trying to decompose the plane into closed or open intervals. The best I could do was to make a square with two sides missing (which you can do out of either type) and form a checkerboard with the white squares missing the top and bottom and the black squares missing the left and right. That gets the whole plane except the lattice points. This seems like it must be a standard problem, but I couldn't find it on the web. <strong>Question:</strong> So can the plane be decomposed into unit open intervals? closed intervals?</p>
| Fedor Petrov | 4,312 | <p>Let me answer for closed intervals. (It is well-known, but I do have only Russian references in mind.) We may decompose closed rectangle (easy). Then, if we have rectangle $R_1=a\times b$ already decomposed, then we double it, make $R_2=2R_1=a\times 2b$ and cover $R_2\setminus R_1$. So, doubling in different directions initial rectangle, we get whole plane.</p>
|
4,008,420 | <p>Suppose we had a differentiable curve <span class="math-container">$C$</span> in <span class="math-container">$\mathbb{R}^2$</span> that serves as our "light container". Light is shining in from all directions, so the space of incoming light-beams is <span class="math-container">$\mathbb{R} \times S^1$</span> where the <span class="math-container">$S^1$</span> parametrises the angle that the light is coming in from, and the <span class="math-container">$\mathbb{R}$</span> parametrises the offset from the origin. When the light hits <span class="math-container">$C$</span> it bounces as you would expect it to off a mirror. Some of the incoming light beams will be trapped inside <span class="math-container">$C$</span>, and keep bouncing forever without escaping, however most will just bounce off.</p>
<ul>
<li>Is there such a finite curve <span class="math-container">$C$</span> such that it actually captures a non-zero amount of light? (take the natural measure on <span class="math-container">$\mathbb{R} \times S^1$</span>, is there a curve <span class="math-container">$C$</span> that captures a non-null subset of the light?) My intuition seems to be telling me that there isn't (because I can't seem to write down any convincing examples), but I can't really think of a reason why not.</li>
<li>If so, what's the curve <span class="math-container">$C$</span> that captures the most light? (for a given length <span class="math-container">$L$</span>. Or, I guess any collection of curves <span class="math-container">$C_i$</span> whose total length is <span class="math-container">$L$</span>).</li>
</ul>
<hr />
<p><a href="https://i.stack.imgur.com/Cbiow.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/Cbiow.jpg" alt="enter image description here" /></a>
Sorry about the confusion, I mean that the light is shining in from infinity. Here for example, the black curve C is our jar, and the blue lines represent incoming light. Most just reflect off, some may get trapped forever. We can draw such C that captures a discrete amount of light, but can we draw C that captures a non-trivial amount of light?</p>
| Joce NoToPutinsWarInUkraine | 138,627 | <p>The isosceles right triangle with an opening in one of the edges close to a vertex away from the right angle has this property.</p>
<p>First let us consider a case where only a null subset of the light.</p>
<p>Consider <span class="math-container">$ABC$</span> an isosceles triangle with right angle at <span class="math-container">$C$</span>. Let's <span class="math-container">$\mathcal{C}= (AB] \cup [BC] \cup [CA)$</span>. Then according to Lemma 3.1 of Tokarsky (1995), light beams entering through <span class="math-container">$A$</span>, i.e. in <span class="math-container">$\{0\} \times (0,\frac{\pi}{4})$</span>, never exit.</p>
<p>G. W. Tokarsky, Polygonal Rooms Not Illuminable from Every Point, The American Mathematical Monthly, Vol. 102, No. 10 (Dec., 1995), pp. 867-879. <a href="https://doi.org/10.2307/2975263" rel="nofollow noreferrer">https://doi.org/10.2307/2975263</a></p>
<p><a href="https://i.stack.imgur.com/PO5MC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PO5MC.png" alt="enter image description here" /></a></p>
<p>His proof is quite intuitive, thanks to the fact that you can easily represent multiple reflections of the triangle by its faces as a lattice of <span class="math-container">$\mathbb{R}^2$</span> :</p>
<p><a href="https://i.stack.imgur.com/ywOEH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ywOEH.png" alt="enter image description here" /></a></p>
<p>Then, it is easy to see that to exit a ray has to cross some reflection of <span class="math-container">$A$</span>, and that before doing so it will always encounter first the reflection of another vertex. This corresponds to absorption with infinite reflections.</p>
<p>From there, I believe you can extend this to any curve <span class="math-container">$\mathcal{C}'$</span>, <span class="math-container">$[BC] \varsubsetneq \mathcal{C}' \subset \mathcal{C}$</span>. Indeed, what you want is to find elements of <span class="math-container">$S^1$</span> for offsets corresponding to points in <span class="math-container">$\mathcal{C}'\setminus\mathcal{C}$</span> such that reflections of <span class="math-container">$B$</span> or <span class="math-container">$C$</span> are encountered at shorter distances than <span class="math-container">$A$</span>.</p>
<p>Let us take such a curve <span class="math-container">$\mathcal{C}' = (AB] \cup [BC] \cup [CD]$</span>, with an "opening" <span class="math-container">$AD = \varepsilon > 0$</span>.</p>
<p>Let us consider any ray passing through <span class="math-container">$(s,0)$</span> with <span class="math-container">$0 < s < \varepsilon$</span>, then all rays with slope <span class="math-container">$(1+2j)/(2k+2j-s)$</span> with <span class="math-container">$(j,k) \in \mathbb{N}\times\mathbb{N}^*$</span> cross reflections of point <span class="math-container">$B$</span> and are thus trapped. The set of corresponding angles is dense in <span class="math-container">$(0,\pi/4)$</span>, so we have a non-zero amount of light trapped.</p>
|
3,358,449 | <blockquote>
<p>I have 8 variables; <span class="math-container">$A$</span>, <span class="math-container">$B$</span>, <span class="math-container">$C$</span>, <span class="math-container">$D$</span>, <span class="math-container">$E$</span>, <span class="math-container">$F$</span>, <span class="math-container">$G$</span>, and <span class="math-container">$H$</span>.</p>
<p>What are they all equal to?</p>
<p>Rules:</p>
<ol>
<li><p>All the variables are equal to integer values between one and eight.</p>
</li>
<li><p>None of the variables are equal to each other.</p>
</li>
<li><p><span class="math-container">$D + E = F$</span></p>
</li>
<li><p><span class="math-container">$B + C = D$</span></p>
</li>
<li><p><span class="math-container">$B + H = A$</span></p>
</li>
<li><p><span class="math-container">$C + G = F$</span></p>
</li>
<li><p><span class="math-container">$E + G = H$</span></p>
</li>
<li><p><span class="math-container">$B + E = G$</span></p>
</li>
</ol>
</blockquote>
<p>I know that <span class="math-container">$1$</span> and <span class="math-container">$2$</span> both must be either <span class="math-container">$B$</span>, <span class="math-container">$C$</span>, or <span class="math-container">$E$</span> since there is no <span class="math-container">$0$</span> so <span class="math-container">$1$</span> must be a letter that does not have an equation answer. For <span class="math-container">$2$</span> I know that there is no equation that is the a variable plus itself.</p>
| Community | -1 | <p>As given in the post <span class="math-container">${B,C,E}$</span> contains the numbers <span class="math-container">$1$</span> and <span class="math-container">$2$</span>. Similarly <span class="math-container">${A,F}$</span> contains 8.</p>
<p>The equations give <span class="math-container">$A=2(B+E),H=B+2E,G=B+E$</span>. <span class="math-container">$A$</span> must therefore be either <span class="math-container">$6$</span> or <span class="math-container">$8$</span>.</p>
<p>If <span class="math-container">$A=6$</span>, then <span class="math-container">$B=1,E=2, F=8,G=3,H=5$</span> or <span class="math-container">$B=2,E=1,G=3,F=8,H=4$</span>.</p>
<p>If <span class="math-container">$A=8$</span>, then <span class="math-container">$B=1,C=2,E=3,G=4,H=7$</span> or <span class="math-container">$B=3,C=2,E=1,G=4,H=5$</span>.</p>
<blockquote>
<p>The actual solution is <span class="math-container">$A=6,B=2,C=5,D=7,E=1,F=8,G=3,H=4$</span>.</p>
</blockquote>
|
1,579,528 | <p>You decide to play a holiday drinking game. You start with 100 containers of eggnog in a row. The 1st container contains 1 liter of eggnog, the 2nd contains 2 liters, all the way until the 100th, which contains 100 liters. You select a container uniformly at random and take a one liter sip from it. If the container is empty after taking this sip, you remove it from the row and select only from the remaining bottles. You continue this process until there is only 1 bottle remaining. What is the expected number of liters of eggnog in this last bottle? What is this as this as a function of n, the number of starting bottles?</p>
<p>I came up with this problem myself recently, and I'm not really sure how to approach it. I can find the conditional expectation of a bottle given that it is the last one remaining using linearity of expectations, but it's not clear to me how to use this to get the overall expectation. </p>
| gar | 138,850 | <p>Finding the exact answer may not be feasible for 100 containers, I think. I managed to compute up to 5 containers using recurrence and a computer. The following python code generates the recurrence for 5 containers with the boundary conditions:</p>
<pre><code>def g(n):
bac = 'f'+str(n)+'('+','.join(['x'+str(i) for i in xrange(1,n+1)])+')'
if n == 2:
print 'f2(0,x2)==x2'
print 'f2(x1,0)==x1'
print 'f2(x1,x2)==(f2(x1-1,x2)+f2(x1,x2-1))/2'
return
a = []
for i in xrange(1,n+1):
print bac.replace('x'+str(i), '0')+ '=='+bac.replace('x'+str(i), '').replace(',,', ',').replace('(,', '(').replace(',)',')').replace('f'+str(n),'f'+str(n-1))
a = []
for i in xrange(1,n+1):
a.append(bac.replace('x'+str(i), 'x'+str(i)+'-1'))
print bac+'==('+'+'.join(a)+')/'+str(n)
return g(n-1)
g(5)
</code></pre>
<p>which gives the recurrence</p>
<pre><code>f5(0,x2,x3,x4,x5)==f4(x2,x3,x4,x5)
f5(x1,0,x3,x4,x5)==f4(x1,x3,x4,x5)
f5(x1,x2,0,x4,x5)==f4(x1,x2,x4,x5)
f5(x1,x2,x3,0,x5)==f4(x1,x2,x3,x5)
f5(x1,x2,x3,x4,0)==f4(x1,x2,x3,x4)
f5(x1,x2,x3,x4,x5)==(f5(x1-1,x2,x3,x4,x5)+f5(x1,x2-1,x3,x4,x5)+f5(x1,x2,x3-1,x4,x5)+f5(x1,x2,x3,x4-1,x5)+f5(x1,x2,x3,x4,x5-1))/5
f4(0,x2,x3,x4)==f3(x2,x3,x4)
f4(x1,0,x3,x4)==f3(x1,x3,x4)
f4(x1,x2,0,x4)==f3(x1,x2,x4)
f4(x1,x2,x3,0)==f3(x1,x2,x3)
f4(x1,x2,x3,x4)==(f4(x1-1,x2,x3,x4)+f4(x1,x2-1,x3,x4)+f4(x1,x2,x3-1,x4)+f4(x1,x2,x3,x4-1))/4
f3(0,x2,x3)==f2(x2,x3)
f3(x1,0,x3)==f2(x1,x3)
f3(x1,x2,0)==f2(x1,x2)
f3(x1,x2,x3)==(f3(x1-1,x2,x3)+f3(x1,x2-1,x3)+f3(x1,x2,x3-1))/3
f2(0,x2)==x2
f2(x1,0)==x1
f2(x1,x2)==(f2(x1-1,x2)+f2(x1,x2-1))/2
</code></pre>
<p>which can be input to friCAS and we can compute values like f5(1,2,3,4,5).</p>
<p>Here are the answers for number of containers being 2,3,4,5:</p>
<p>\begin{align*}
\frac{3}{2}, \frac{125}{72}, \frac{157885}{82944}, \frac{685466694095183}{335923200000000}
\end{align*}</p>
<p>And for 100 containers, a monte-carlo simulation gives an answer close to $5.6$</p>
|
754,583 | <p>Write <span class="math-container">$$\phi_n\stackrel{(1)}{=}n+\cfrac{n}{n+\cfrac{n}{\ddots}}$$</span> so that <span class="math-container">$\phi_n=n+\frac{n}{\phi_n},$</span> which gives <span class="math-container">$\phi_n=\frac{n\pm\sqrt{n^2+4n}}{2}.$</span> We know <span class="math-container">$\phi_1=\phi$</span>, the <a href="http://en.wikipedia.org/wiki/Golden_ratio" rel="nofollow noreferrer">Golden Ratio</a>, so let's take <span class="math-container">$\phi_n\stackrel{(2)}{=}\frac{n+\sqrt{n(n+4)}}{2}$</span>. (Is that justified?)</p>
<p><a href="http://m.wolframalpha.com/input/?i=%28n%2B%E2%88%9A%28n%28n%2B4%29%29%29%2F2&x=0&y=0" rel="nofollow noreferrer">Wolfram Alpha</a> states that, with <span class="math-container">$(2)$</span>, <span class="math-container">$$\lim\limits_{n\to -\infty}\phi_n=-1.$$</span> Why? Can I infer that this is true for <span class="math-container">$(1)$</span> and, if so, <em>why</em>?</p>
<p><strong>I wonder what happens in <span class="math-container">$(1)$</span> for <span class="math-container">$n\in\mathbb{C}\backslash\mathbb{Z}$</span> too</strong>. I got something horrendous looking in <span class="math-container">$(2)$</span> for <span class="math-container">$n=i$</span>.</p>
<hr>
<p><em>Clarification:</em> I'm trying to <strong>find <span class="math-container">$\phi_n$</span> in terms of <span class="math-container">$n$</span></strong>. See the comments below.</p>
| Lutz Lehmann | 115,115 | <p>The usual trick is to apply the third binomial formula, so that
$$
\frac{n+\sqrt{n(n+4)}}{2}
=\frac{n^2-(n^2+4n)}{2(n-\sqrt{n(n+4)})}
=-\frac{2 |n| }{|n|+\sqrt{|n|(|n|-4)}}
$$
Now standard limit procedurs for fractions apply, cancel $|n|$ in numerator and denominator, move the limit inside the square root in the denominator, ..., to get the limit $-1$.</p>
<hr>
<p><s>To the general question, I do not think that the convergence of this continued fraction is well-defined. Depending on the order of the convergent, one can cancel the $n$ to either get
$$
n+\frac{n}{1+\frac{1}{1+\frac{1}{1+\frac{1}{\vdots+1}}}}
$$
or
$$
n+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{\vdots+1}}}},
$$
so that these even and odd subsequences of convergents converge to $n\cdot\phi_1$ and $n-1+\phi_1$. </p>
<blockquote>
<p>The convergents in the tradional sense form, for $n\ne 1$, an oscillating sequence.</s></p>
</blockquote>
<p>Cancellation leaves some more of the $n$ in place
$$
n+\frac{n}{n+\frac{1}{1+\frac{1}{n+\frac{1}{\vdots+1}}}}
$$
or
$$
n+\frac{1}{1+\frac{1}{n+\frac{1}{1+\frac{1}{\vdots+1}}}},
$$
so that after cancellation one obtains a classical continued fraction $[n;1,n,1,n,...]$ or $n\cdot[1;n,1,n,1,...]$, which do indeed converge for integer $n\ge 1$ in the classical sense.</p>
<p>The resulting recursion
$$
a_{k+1}=F(a_k)=n+\frac1{1+\frac1{a_k}}
$$
is contractive on $a_k\in (-\infty,-2)\cup(0,\infty)$. For $n\in\Bbb R\setminus[-4,0]$ the fixed point
$$
a_*=\frac n2\left(1+\sqrt{1+\frac 4n}\right)
$$
falls inside this contraction region and is thus the value of the given continued fraction.</p>
|
39,762 | <p>Happy new year mathematica gurus of stack exchange!</p>
<p>As I see it one of the major obstacles in getting decent at programming mathematica is that, not only do you need to learn how certain commands work, but rather that you mainly need to understand how to write your syntax. This is a typical such situation, I was hoping that someone might shine some light on how to do it.</p>
<p>I run this piece of code:</p>
<pre><code>For[i = 1, i <= samples, i++,
AppendTo[fList,
f1[randomSeeds[[1 + (n - 1) (i - 1)]][[1]],
randomSeeds[[1 + (n - 1) (i - 1)]][[2]], 0, 0, 0, 0, 0, 0]];
]
For[i = 1, i <= samples, i++,
AppendTo[fList,
f1[randomSeeds[[1 + (n - 1) (i - 1)]][[1]],
randomSeeds[[1 + (n - 1) (i - 1)]][[2]],
randomSeeds[[2 + (n - 1) (i - 1)]][[1]],
randomSeeds[[2 + (n - 1) (i - 1)]][[2]], 0, 0, 0, 0]];
]
For[i = 1, i <= samples, i++,
AppendTo[fList,
f1[randomSeeds[[1 + (n - 1) (i - 1)]][[1]],
randomSeeds[[1 + (n - 1) (i - 1)]][[2]],
randomSeeds[[2 + (n - 1) (i - 1)]][[1]],
randomSeeds[[2 + (n - 1) (i - 1)]][[2]],
randomSeeds[[3 + (n - 1) (i - 1)]][[1]],
randomSeeds[[3 + (n - 1) (i - 1)]][[2]], 0, 0]];
]
For[i = 1, i <= samples, i++,
AppendTo[fList,
f1[randomSeeds[[1 + (n - 1) (i - 1)]][[1]],
randomSeeds[[1 + (n - 1) (i - 1)]][[2]],
randomSeeds[[2 + (n - 1) (i - 1)]][[1]],
randomSeeds[[2 + (n - 1) (i - 1)]][[2]],
randomSeeds[[3 + (n - 1) (i - 1)]][[1]],
randomSeeds[[3 + (n - 1) (i - 1)]][[2]],
randomSeeds[[4 + (n - 1) (i - 1)]][[1]],
randomSeeds[[4 + (n - 1) (i - 1)]][[2]]]];
]
</code></pre>
<p>As you can see most of the stuff is identical in the for loops, it's just the number of zeroes that varies in the end. In fact, I only wish to run one of these for loops at the time. Just above these for loops I specify the number of dimensions I'm working in (n in the code), for n=2 I want to run the first loop, for n=3 I want to run the second etc. At the moment I comment an uncomment the undesired parts of the code, but that seems like a very ugly solution to me. </p>
<p>So my question is this: is there a simple way of reducing these for copies of code into one copy? It seems to me as though this is a quite ineffective way of doing things.</p>
<p>Edit: Small clarification: Ideally I want something like this: I give the program "2" as input and it chooses the first for loop above etc. </p>
<p>Cheers,
David</p>
| C. E. | 731 | <p>I certainly agree with everything Mr. Wizard says in his answer. Taking the question at face value you can give each loop a symbol so that <code>f[1] := For[... f[2] := For[ ...</code> and then use <code>Switch</code>:</p>
<pre><code>Switch[n,1,f[1],2,f[2]...]
</code></pre>
<p>Or as Kuba suggests:</p>
<pre><code>f[n]
</code></pre>
<p>But that is only valid if you use <code>f[1]...f[n]...</code> as your symbols.</p>
|
134,407 | <p>Some shapes, such as the disk or the <a href="http://en.wikipedia.org/wiki/Reuleaux_triangle" rel="nofollow noreferrer">Releaux triangle</a> can be used as manholes,
that is, it is a curve of constant width.
(The width between two parallel tangents to the curve are independent of the orientation of the curve.)</p>
<p><strong>(1) Is it possible to tile the plane with such shapes?</strong></p>
<p>The shapes should be simply connected, and all must have an area greater than some fixed uniform $\epsilon.$
Otherwise, we'd just tile the plane with disks of various diameters where some are arbitrarily small, similar to the Apollonian circle.
By scaling the tiling, we can take $\epsilon=1$.</p>
<p>As a note,
<a href="https://mathoverflow.net/questions/31810/">an earlier question here on MO gave a positive answer on the existence of non-convex simply connected manholes.</a></p>
<p><em>Some clarifications:</em> By tiling, we mean that all manholes used are closed sets,
and there is no open ball that is simultaneously in the interior of two different manholes. Thus, boundaries of manholes may intersect.
Note, we may use several different manholes in a tiling (otherwise, we are essentially asking for a solution for the open einstein problem).</p>
<p>If the answer is negative, a more general question is the following.
Define the roundness of an object as the minimum width divided by the maximum width of the shape (width is the distance between two non-equal parallel tangents).
The roundness of a circle or a Releaux triangle is 1,
and the square has roundness $1/\sqrt{2}.$ Define the roundness of a tiling as the minimum of the roundness of all shapes in the tiling. </p>
<p>For non-convex shapes, roundness can be defined as follows: It is the factor I need to re-size the hole with, so that the original shape cannot fall through that hole.
For example, a square with sides $1$ cannot fall through a square hole with sides $1/\sqrt{2}.$</p>
<p><strong>(2) What is the best possible roundness $R$ a tiling of the plane can have?</strong></p>
<p>Trivially, $\sqrt{3}/2 \leq R \leq 1$ since we can tile the plane with equilateral triangles, and a positive answer to question 1 gives $R=1.$</p>
<p>The applications are evident: This would be a nice way to tile a ceiling,
instead of using regular square tiles, that sometimes falls down.</p>
| WΕodzimierz HolsztyΕski | 8,385 | <p>I believe that <strong>@Per</strong> meant to ask (for $n=2$) about the following (for arbitrary $\ n>1$):</p>
<p><strong>CONJECTURE A</strong> There exists real $\delta_n > 0\ $ such that for every family $\ F\ $ of bounded constant width convex bodies $\ B\in F\ $ such that each such $\ B\ $ contains a ball of radius $\ 1\ $ there exists an open ball $\ K\ $ of radius $\ \delta_n\ $ such that $\ K\subseteq \mathbb R^n\setminus \bigcup F$.</p>
<p>Each convex body $\ B\in F\ $can have its own diameter (but it has to be $\ \ge 2\ $ due to the assumption about a ball). The constant should be just one constant $\ \delta_n\ $, the same for all said families F.</p>
<p>This seems to be an easy conjecture. The <strong>true challenge</strong> seems to compute the maximal possible $\ \delta_n.\ $ In a sophisticated case (not too realistic?) the $\ \sup \delta_n\ $ may be NOT attained.</p>
<hr>
<blockquote>
<p><strong>COMMENTS</strong> The original <strong><em>Question</em></strong> considered convex and non-convex manholes, etc. Already squares give a perfect tiling. Thus I thought that it is still sufficiently interesting to concentrate on a restriction to the manholes which are convex bodies of constant width.</p>
</blockquote>
|
2,292,713 | <blockquote>
<p><strong>Definition.</strong> Let <span class="math-container">$E$</span> be a nonempty subset of <span class="math-container">$X$</span>, and let <span class="math-container">$S$</span> be the set of all real numbers of the form <span class="math-container">$d(p, q)$</span>, with <span class="math-container">$p,q\in E$</span>. The sup of <span class="math-container">$S$</span> is called the diameter of <span class="math-container">$E$</span>.</p>
<p><strong>Theorem 3.10.</strong> If <span class="math-container">$\overline{E}$</span> is the closure of a set <span class="math-container">$E$</span> in a metric space <span class="math-container">$X$</span>, then <span class="math-container">$$\text{diam }\overline{E} = \text{diam }E.$$</span></p>
</blockquote>
<p>Proof: Fix <span class="math-container">$\varepsilon>0$</span>, and choose <span class="math-container">$p, q \in \overline{E}$</span>. By the definition of <span class="math-container">$\overline{E}$</span>, there are points <span class="math-container">$p',q' \in E$</span> such that <span class="math-container">$d(p,p') < \varepsilon$</span> and <span class="math-container">$d(q,q') < \varepsilon$</span>. Hence <span class="math-container">$$d(p, q) \le d(p,p') + d(p', q') + d(q', q) < 2\varepsilon + d(p', q') \le 2\varepsilon + \text{diam }E.$$</span></p>
<p>Ok until here. But then they use the inequality above to come up with <span class="math-container">$$\text{diam }\overline{E} \le 2\varepsilon + \text{diam }E$$</span></p>
<p>Where I can only see the strict inequality because of the strict inequality relation made in the inequalities above.</p>
| James Shapiro | 148,829 | <p>Another subtlety to this proof, just in case anyone missed it. The reason that we know that there are <span class="math-container">$p, q \in E$</span> such that <span class="math-container">$d(p, p') < \epsilon$</span>, <span class="math-container">$d(q, q') < \epsilon$</span> is the following:</p>
<p>Every point in <span class="math-container">$\overline{E}$</span> is either</p>
<p>(1.) a limit point of <span class="math-container">$E$</span></p>
<p>(2.) a point in <span class="math-container">$E$</span></p>
<p>If <span class="math-container">$p$</span> is a limit point of <span class="math-container">$E$</span>, then we can find a point <span class="math-container">$p' \in N_{\epsilon}(p)$</span> such that <span class="math-container">$p' \not = p.$</span> (By the definition of a limit point). And since <span class="math-container">$p'$</span> is in that neighborhood with radius <span class="math-container">$\epsilon$</span>, <span class="math-container">$d(p, p') < \epsilon$</span>.</p>
<p>If <span class="math-container">$p \in E$</span>, then we set <span class="math-container">$p' = p$</span> and we have that <span class="math-container">$d(p,p') = 0 < \epsilon$</span>.</p>
|
3,841,542 | <p>I am trying to show that <span class="math-container">$\sqrt{\sqrt{2}+5}$</span> is constructible through a diagram.</p>
<p>I know how to show something of the form <span class="math-container">$\sqrt[n]{a}$</span> is constructible through a diagram, but I am really having a difficult time with this one.</p>
<p>Any tips? Thanks.</p>
| Aadhaar Murty | 826,105 | <p>As suggested by @Abhi, differentiating w.r.t. <span class="math-container">$k$</span> will give you a direct answer. We have,</p>
<p><span class="math-container">$$\frac {d}{dk} \left(\frac {k}{a^{2}+ k^{2}}\right) = \int_{0}^{\infty} e^{-ax} \cdot \partial_{k}\sin(kx) dx = \int_{0}^{\infty}xe^{-ax} \cos(kx) dx = \boxed{\frac {a^{2}-k^{2}}{(a^{2}+k^{2})^{2}}}$$</span></p>
<p>Differentiating with respect to <span class="math-container">$a$</span> gives the value of the other integral -</p>
<p><span class="math-container">$$-\frac {d}{da}\left(\frac {k}{a^{2}+ k^{2}}\right) = -\int_{0}^{\infty} \sin(kx) \cdot \partial_{a}(e^{-ax})dx = \int_{0}^{\infty} xe^{-ax} \sin(kx) dx = \boxed {\frac{2ak}{(a^{2}+k^{2})^{2}}}$$</span></p>
|
2,040,961 | <blockquote>
<p>Written with <a href="https://stackedit.io/" rel="noreferrer">StackEdit</a>. </p>
<p>Suppose $(a_i)$ is a sequence in $\Bbb R$ such that $\sum\limits_{i=1}^{ \infty} |a_i||x_i| < \infty$ whenever $\sum\limits_{i=1}^{\infty} |x_i| < \infty$. Then is $(a_i)$ a bounded sequence?</p>
</blockquote>
<p>Look at the end of the question for the right answer. </p>
<p>If the statement '$(a_i)$ is a properly divergent sequence implies that there exists some $k \in \Bbb N$ such that $\sum\limits_{i=1}^{\infty} {1/{a_i}}^k$ is convergent' was true, we could have easily proven $(a_i)$ is bounded by using sub-sequences but since that is dis-proven by $ln(n)$, can we use something around it? Like can all the functions which do not satisfy the 'statement' I mentioned be considered as a special case of functions? </p>
<p>Correct Answer - Yes, $(a_i)$ is bounded.</p>
<p>Source - Tata Institute of Fundamental Research Graduate Studies 2013</p>
| zhw. | 228,045 | <p>If $a_n$ is unbounded, then there exist integers $0 < n_1 < n_2 < \cdots \to \infty$ such that $|a_{n_k}| > k^2.$ Define $x_n$ as follows: $x_{n_k} = 1/k^2, k = 1,2, \dots,$ $x_n=0$ for all other $n.$ Then $\sum |x_n| < \infty,$ while $\sum |a_n||x_n|$ has infinitely many terms $> 1,$ hence diverges, contradiction.</p>
|
2,040,961 | <blockquote>
<p>Written with <a href="https://stackedit.io/" rel="noreferrer">StackEdit</a>. </p>
<p>Suppose $(a_i)$ is a sequence in $\Bbb R$ such that $\sum\limits_{i=1}^{ \infty} |a_i||x_i| < \infty$ whenever $\sum\limits_{i=1}^{\infty} |x_i| < \infty$. Then is $(a_i)$ a bounded sequence?</p>
</blockquote>
<p>Look at the end of the question for the right answer. </p>
<p>If the statement '$(a_i)$ is a properly divergent sequence implies that there exists some $k \in \Bbb N$ such that $\sum\limits_{i=1}^{\infty} {1/{a_i}}^k$ is convergent' was true, we could have easily proven $(a_i)$ is bounded by using sub-sequences but since that is dis-proven by $ln(n)$, can we use something around it? Like can all the functions which do not satisfy the 'statement' I mentioned be considered as a special case of functions? </p>
<p>Correct Answer - Yes, $(a_i)$ is bounded.</p>
<p>Source - Tata Institute of Fundamental Research Graduate Studies 2013</p>
| Tacet | 186,012 | <p><strong>Hint</strong>: Look at this simple fact: <a href="https://math.stackexchange.com/questions/388898/if-the-positive-series-sum-a-n-diverges-and-s-n-sum-limits-k-leqslant-na">If the positive series $\sum a_n$ diverges and $s_n=\sum\limits_{k\leqslant n}a_k$ then $\sum \frac{a_n}{s_n}$ diverges as well</a>.</p>
<p><strong>Hint2</strong>:</p>
<p>Consider $x_n = \dfrac{1}{\sum\limits_{i=1}^{n}a_i}$</p>
|
445,816 | <p>I have to show that</p>
<blockquote>
<blockquote>
<p>$\mathbb{C}=\overline{\mathbb{C}\setminus\left\{0\right\}}$,</p>
</blockquote>
</blockquote>
<p>what is very probably an easy task; nevertheless I have some problems.</p>
<p>In words this means: $\mathbb{C}$ is the smallest closed superset of $\mathbb{C}\setminus\left\{0\right\}$.</p>
| Community | -1 | <p>$\partial(\mathbf{C}^\ast)$ consists of $0$ alone, use $\overline{\mathbf{C}^\ast}=\partial(\mathbf{C}^\ast)\cup\mathbf{C}^\ast$.</p>
|
997,463 | <p>For example, a complex number like $z=1$ can be written as $z=1+0i=|z|e^{i Arg z}=1e^{0i} = e^{i(0+2\pi k)}$.</p>
<p>$f(z) = \cos z$ has period $2\pi$ and $\cosh z$ has period $2\pi i$.</p>
<p>Given a complex function, how can we tell if it is periodic or not, and further, how would we calculate the period? For example, how do we find the period of $f(z)=\tan z$?</p>
| Anastasiya-Romanova η§ | 133,248 | <p><strong>Hint:</strong></p>
<p>Put $x = \frac{3}{2}\tan\theta \Rightarrow dx = \frac{3}{2} \sec^2 \theta \ d\theta$, we have
\begin{align}
\int\frac{x^3}{\left(\sqrt{4x^2+9}\right)^3}\,dx&=\frac{3}{16}\int\frac{\tan^3\theta}{\sec^3\theta}\cdot\sec^2\theta\,\,d\theta\\
&=\frac{3}{16}\int\frac{\sin^3\theta}{\cos^3\theta}\cdot\cos\theta\,\,d\theta\quad\Rightarrow\quad\tan\theta=\frac{\sin\theta}{\cos\theta}\,\,\mbox{and}\,\,\sec\theta=\frac{1}{\cos\theta}\\
&=\frac{3}{16}\int\frac{\sin^2\theta}{\cos^2\theta}\cdot\sin\theta\,\,d\theta\\
&=\frac{3}{16}\int\frac{1-\cos^2\theta}{\cos^2\theta}\cdot\sin\theta\,\,d\theta\quad\Rightarrow\quad\mbox{set}\,\,t=\cos\theta \Rightarrow dt=-\sin\theta\,\,d\theta\\
&=\frac{3}{16}\int\frac{t^2-1}{t^2}\,dt
\end{align}
Can you take it from here?</p>
|
165,489 | <p>I have problem solving this equation, smallest n such that $1355297$ divides $10^{6n+5}-54n-46$. I tried everything using my scientific calculator, but I never got the correct results(!).and finally I gave up!. Could you help me find the first 2 solutions for this equation ? (thanks.)</p>
| Sjoerd Smit | 43,522 | <p>How about this?</p>
<pre><code>cf = Compile[{{max, _Integer}, {p, _Integer}, {j, _Integer}},
Module[{list = ConstantArray[0, 0]},
Do[
If[
PowerMod[10, 6 n + 5, p] - Mod[54 n + 46, p] == 0,
AppendTo[list, n];
If[Length[list] >= j, Break[]]
],
{n, Range[1, max]}
];
list
]
];
In[20]:= result = cf[10^8, 1355297, 2]
Out[20]= {2331259, 3776127}
</code></pre>
<p>Check the result:</p>
<pre><code>In[21]:= Table[Mod[10^(6 n + 5) - 54 n - 46, 1355297], {n, result}]
Out[21]= {0, 0}
</code></pre>
|
1,476,847 | <p>I am a little puzzled by some notations in optimization community. Is there anyone who can explain why $f_1:\mathbb{R}^n\rightarrow\mathbb{R}$ is a finite valueed but $f_2:\mathbb{R}^n\rightarrow\mathbb{R}\cup\{\infty\}$ is not?? I have never have this kind of notations. For function $f_1$ I always calculated limit when $x\rightarrow \infty$ and nobody said you can't as infinity is not a member of real values. </p>
<p>Can anyone clarify this for me?
Thanks</p>
| 5xum | 112,884 | <p>You cannot. Basically, you want some constants $a,b,c$, which are possibly dependent on $n$, such that $$\det(A+B)\leq a\det (A) + b\det (B) + c .$$</p>
<p>However, take $A=\begin{bmatrix}1 & 0 &\dots &0\\
0&0&\dots &0\\
\vdots &\vdots &\ddots &\vdots\\
0&0&\dots&0\end{bmatrix}$
and $B=I_n - A$.</p>
<p>Then, $\det(\alpha (A+B)) = \alpha$, and $\det(A)=\det(B)=0$, which means that for every real value $\alpha$, you have $\alpha \leq c$. Obviously, no such $c$ exists.</p>
|
3,238,563 | <p>I have a question about a proof I saw in a book about basic algeba rules. The rule to prove is:
<span class="math-container">\begin{eqnarray*}
\frac{1}{\frac{1}{a}} = a, \quad a \in \mathbb{R}_{\ne 0}
\end{eqnarray*}</span></p>
<p>And the proof: </p>
<p><span class="math-container">\begin{eqnarray*}
1 = a \frac{1}{a} \Longrightarrow 1 = \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow a = a \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow \frac{1}{\frac{1}{a}} = a
\end{eqnarray*}</span></p>
<p>Why is it allowed to just replace <span class="math-container">$a$</span> with <span class="math-container">$1/a$</span>? What's the explanation behind it? </p>
| auscrypt | 675,509 | <p>Let <span class="math-container">$x=\frac{1}{a}$</span>. Then:</p>
<p><span class="math-container">\begin{eqnarray*}
1 = x \frac{1}{x} \Longrightarrow 1 = \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow a = a \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow \frac{1}{\frac{1}{a}} = a
\end{eqnarray*}</span></p>
<p>Better? You're right, replacing <span class="math-container">$a\to\frac{1}{a}$</span> is a minor abuse of notation, because they're implicitly changing the variable without telling you. But you can fix that by just letting <span class="math-container">$x=\frac{1}{a}$</span> at the start.</p>
|
3,238,563 | <p>I have a question about a proof I saw in a book about basic algeba rules. The rule to prove is:
<span class="math-container">\begin{eqnarray*}
\frac{1}{\frac{1}{a}} = a, \quad a \in \mathbb{R}_{\ne 0}
\end{eqnarray*}</span></p>
<p>And the proof: </p>
<p><span class="math-container">\begin{eqnarray*}
1 = a \frac{1}{a} \Longrightarrow 1 = \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow a = a \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow \frac{1}{\frac{1}{a}} = a
\end{eqnarray*}</span></p>
<p>Why is it allowed to just replace <span class="math-container">$a$</span> with <span class="math-container">$1/a$</span>? What's the explanation behind it? </p>
| Allawonder | 145,126 | <p>I proceed in steps:</p>
<p>The first implication follows because the reciprocal of <span class="math-container">$1$</span> is <span class="math-container">$1.$</span> Thus, they got <span class="math-container">$$1=\frac 1a\frac{1}{\frac 1a}$$</span> by taking reciprocals of both sides of <span class="math-container">$$1=a\frac 1a.$$</span></p>
<p>The second implication follows since they only multiplied both sides by <span class="math-container">$a.$</span> This gives the last equation, which is what was to be proved.</p>
<hr>
<p>A shorter way is to first define <span class="math-container">$1/a=a^{-1}.$</span> Then it is almost trivial to see that <span class="math-container">$$a\frac 1a=1\implies a\frac 1a\left(\frac 1a\right)^{-1}=\left(\frac 1a\right)^{-1}\implies a=\frac{1}{\frac 1a}.$$</span></p>
|
2,929,094 | <p>Differentiation of
<span class="math-container">$\int_{a(x)}^{b(x)} f(x,t)\,\text{d}t$</span> is done by Leibniz's integral rule:
<span class="math-container">$$\frac{\text{d}}{\text{d}x} \left (\int_{a(x)}^{b(x)} f(x,t)\,\text{d}t \right )= f\big(x,b(x)\big)\cdot \frac{\text{d}}{\text{d}x} b(x) - f\big(x,a(x)\big)\cdot \frac{\text{d}}{\text{d}x} a(x) + \int_{a(x)}^{b(x)}\frac{\partial}{\partial x} f(x,t) \,\text{d}t,$$</span></p>
<p>if <span class="math-container">$-\infty<a(x),b(x)<\infty$</span>.</p>
<p>Can we say anything in general about the derivative of the powers of the <span class="math-container">$\int_{a(x)}^{b(x)} f(x,t)\,\text{d}t$</span> in a similar fashion? So for example
<span class="math-container">$$\frac{\text{d}}{\text{d}x} \left(\left(\int_{a(x)}^{b(x)} f(x,t)\,\text{d}t\right)^2\right)=~?$$</span></p>
| Hw Chu | 507,264 | <p><a href="https://i.stack.imgur.com/ChL5h.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ChL5h.png" alt="enter image description here"></a></p>
<p>Denote the intersection of <span class="math-container">$\overline{AC}$</span> and <span class="math-container">$\overline{BD}$</span> by <span class="math-container">$O$</span>. Since <span class="math-container">$\overline{AD}||\overline{BE}$</span>,</p>
<p><span class="math-container">$$
\overline{DO}:\overline{OB} = \overline{AO}:\overline{OE}.
$$</span></p>
<p>Since <span class="math-container">$\overline{BC}||\overline{AF}$</span>,</p>
<p><span class="math-container">$$
\overline{BO}:\overline{OF} = \overline{CO}:\overline{OA}.
$$</span></p>
<p>Multiplying the ratios,
<span class="math-container">$$
\overline{DO}:\overline{OF} = \overline{CO}:\overline{OE}.
$$</span></p>
|
187,432 | <p>Can we evaluate the integral using <a href="http://en.wikipedia.org/wiki/Jordan%27s_lemma#Application_of_Jordan.27s_lemma">Jordan lemma</a>?
$$ \int_{-\infty}^{\infty} {\sin ^2 (x) \over x^2 (x^2 + 1)}\:dx$$</p>
<p>What de we do if removeable singularity occurs at the path of integration?</p>
| Mhenni Benghorbal | 35,472 | <p>Hint, note that $ \cos(2x)=1-2\sin(x)^2 $, this suggest to consider the integral</p>
<p>$$ \int_{C} \frac{ {\rm e}^{2 i z} - 1 }{ z^2 (z^2 + 1)} dz \,.$$ </p>
|
3,700,440 | <p>As stated in the title, it is requested to define a linear transformation <span class="math-container">$T:\Bbb R^3 \to \Bbb R^3$</span> such that the null space of <span class="math-container">$T$</span> is the <span class="math-container">$z$</span>-axis, and the range of <span class="math-container">$T$</span> is the plane: <span class="math-container">$x+y+z=0$</span> </p>
<hr>
<p>I don't really know how to begin with the solution of the exercise, I think that I should try to get a matrix using the standard base, but after that, I don't have any concrete ideas.</p>
| Rodrigo Dias | 375,952 | <p>Choose a basis <span class="math-container">$\mathcal{B}=\{u,v\}$</span> for <span class="math-container">$x+y+z=0$</span> (for instance, it could be <span class="math-container">$\{(2,-1,-1), (1,1,-2)\}$</span>).</p>
<p>Letting <span class="math-container">$T$</span> be the linear extension of
<span class="math-container">$$\left\{\begin{array}{c} e_1\mapsto u \\ e_2 \mapsto v \\ e_3\mapsto 0\end{array}\right.$$</span>
should work.</p>
|
499,840 | <p>I have three points $A=(2,3), B=(6,4)$ and $C=(6,6).$
Given $\vec{AB}=\vec v$ and $\vec{BC}={0 \choose 2}$. I have also that for every $t\in [0,1]$ there is a point $D$ given as $\vec{AD}=t\vec{v}.$ </p>
<p>My question is determine $t$ such that the area of triangle $ADC$ equals area of the triangle $DBC$.</p>
<p>My suggestions is Can I say that $D$ is on line $AB$ dividing the area of $ABC$ in to two equal parts, namely $ADC$ and $DBC$? If this is true then why is it true?</p>
<p>Thanks a lot.</p>
| Γmer | 55,199 | <p>$$\overrightarrow{AD}=t\overrightarrow{v}$$ means that $AD$ and $v$ are linearly dependent. Also we can say $D\in[AB]$ because of $t\in[0,1]$. For $ADC$ and $DBC$ have same areas $t$ must be $1/2$ and hence $D=(4,7/2)$. </p>
|
1,652,846 | <p>Let $s$ be any complex number, $t = e^s$ and $z = t^{1/t}$. Define the sequence $(a_n)_{n\in\mathbb{N}}$ by $a_0 = z $ and $a_{n+1} = z^{a_n} $ for $n \geq 0$, that is to say $a_n$ is the sequence $z$, $z^z$, $z^{z^z}$, $z^{z^{z^{z}}}$ and so on.</p>
<p>I want to show that the sequence $(a_n)_{n\in\mathbb{N}}$ converges to $t$ <em>if and only if</em> $s$ lies in the unit disk. I know that when the sequence converges the limit is $\frac{W(-\ln(z))}{-\ln(z)}$ where $W$ is the Lambert W function.</p>
<p>I have verified the above statements numerically for several thousand values of $z$ but I have no idea how to actually prove it.</p>
<p>I graphed the natural logs of the limits on my computer. I got what appeared to be the unit disk. </p>
<p>If we take the log of the limit we get $ln(\frac{W(βln(z)}{βln(z)})=-ln(βln(z))βW(βln(z))+ln(βln(z))=βW(βln(z))$</p>
<p>So what I want to prove is equivalent to showing the sequence $a_n$ is convergent if and only if $|W(βln(z))| \leq 1$ </p>
<p>I was reading the Wikipedia article on the Lambert W Function and I found this proof that the limit $c$, when it exists, is $c= \frac{W(-ln(z))}{-ln(z)}$</p>
<p>$z^c = c\implies z = c^{1/c} \implies z^{-1} = c^{-1/c} \implies 1/z = (1/c)^{1/c} \implies -ln(z) = \frac{ln(1/c)}{c} \implies -ln(z) = e^{ln(1/c)}ln(1/c) \implies ln(1/c) = W(-ln(z)) \implies 1/c = e^{W(-ln(z))} \implies \frac{1}{c} = \frac{-ln(z)}{W(-ln(z)} \implies c = \frac{W(-ln(z)}{-ln(z)}$</p>
<p>I can only assume that at least 1 step is not justified when $|W(-ln(z)| > 1$ though I am not sure which one. I think that part of the problem is the equation $z^c=c$ has a solution for every non-zero complex number $c$ while the sequence $a_n$ only converges for certain special values of z. In other words the convergence of the $a_n$ is a sufficient but not necessary condition for the existence of a solution to the equation.</p>
| reuns | 276,986 | <p>$$a_0 = z\qquad \qquad \qquad a_{n+1} = z^{a_n}$$ </p>
<p>let $b_n = \ln a_n$ so $a_n = e^{b_n}$ and $$b_{n+1} = \ln \left(z^{e^{b_n}}\right) = e^{b_n} \ln z$$ </p>
<p>if $b_n$ converges to $b$ then $b = e^b \ln z = - e^{b} (-\ln z)$ so </p>
<p>$$b = e^b \ln z = W(-\ln z)$$</p>
<p>where $W$ is (<strong>one of the branches of</strong> ?) the <a href="https://fr.wikipedia.org/wiki/Fonction_W_de_Lambert#Comme_limite_d.27une_suite" rel="nofollow">Lambert function</a>.</p>
<p>let $c_n = b_n - b$ so $$c_{n+1} = e^{(b + c_n)} \ln z - b= (e^b \ln z) e^{c_n} - b = b e^{c_n} - b = b (e^{c_n} - 1)$$ </p>
<blockquote>
<p>suppose $|b| > 1$ and $\ln z \ne 0$. then suppose $b_n \to b$ thus $c_n \to 0$ so $e^{c_n}-1 \sim c_n$ so $c_{n+1} \sim b c_n $ which clearly cannot converge except if $c_0 = 0$ which would imply $c_n = 0$ for every $n$ which is not the case because $b_0 = \ln z$ so $b_1 = e^{\ln z} \ln z \ne b_0$ so $c_1 \ne c_0$ $\implies$ a contradiction.</p>
</blockquote>
<p>hence if $\ln z \ne 0$ $$|b| = \left|W(-\ln z)\right| \le 1$$ is a necessary condition for the convergence of $(b_n)$ and $(a_n)$</p>
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25,284 | <p>I recently worked my way through Walter Warwick Sawyer's book, <em>Mathematician's Delight</em>, which has opened my eyes to Maths. I used to fear maths, feeling I was incapable. Sawyer (among other authors) has a gift for teaching the subject. I now feel much more confident tackling Maths problems, I have a better intuitive understanding of Maths and a renewed interest in it.</p>
<p>There's a nice summary of Sawyer's life and work here: <a href="https://plus.maths.org/content/os/latestnews/may-aug08/sawyer/index" rel="nofollow noreferrer">https://plus.maths.org/content/os/latestnews/may-aug08/sawyer/index</a></p>
<p>Have you had a similar experience after encountering Sawyer's work?</p>
<p>Stephen</p>
| paul garrett | 63 | <p>Yes, I remember vividly my chance encounter at the library with that book of his! Yes, it had a big impact on me. The idea that mathematics was a real thing in its own right, like music, and not just a school subject, and not just a device to filter people out.</p>
|
3,039,040 | <p>In the equation <span class="math-container">$3^x=2y^2-1$</span>,
<span class="math-container">$x$</span>, <span class="math-container">$y$</span> are natural numbers.
I found <span class="math-container">$x=1$</span> or <span class="math-container">$2$</span> (mod <span class="math-container">$4$</span>), and <span class="math-container">$y^2=1$</span> or <span class="math-container">$4$</span> (mod <span class="math-container">$120$</span>)
but I even don't know if the number of solutions is infinite.
Is there a way to find the solution of this indeterminate equation?</p>
| Community | -1 | <p>You can even solve this question using Mathematical Induction.
However, answer is also correct and easy.</p>
<p>And its solution is x=0 and y=1.</p>
|
2,755,213 | <p><strong>Question.</strong> Find, with proof, the possible values of a rational number $q$ for which $q+\sqrt{2}$ is a reduced quadratic irrational.</p>
<p>So, by definition a <em>quadratic irrational</em> is one of the form $u+v\sqrt{d}$ where $u,v\in\Bbb Q, v\neq 0$ and $d$ being square-free. Then, it is said to be <em>reduced</em> if it exceeds $1$ and has conjugate in the interval $(-1, 0)$.</p>
<p>For the question at hand; this corresponds to having $-1<q-\sqrt{2}<0$ and $q+\sqrt{2}>1$.</p>
<p>But I'm not quite sure how to proceed from here?</p>
| Arthur | 15,500 | <p>$-1<q-\sqrt{2}<0$ implies that
$$
-1+\sqrt2<q<\sqrt2
$$</p>
|
2,083,460 | <p>While trying to answer <a href="https://stackoverflow.com/questions/41464753/generate-random-numbers-from-lognormal-distribution-in-python/41465013#41465013">this SO question</a> I got stuck on a messy bit of algebra: given</p>
<p>$$
\log m = \log n + \frac32 \, \log \biggl( 1 + \frac{v}{m^2} \biggr)
$$</p>
<p>I need to solve for $m$. I no longer remember enough logarithmic identities to attempt to do this by hand. Maxima canβt do it at all, and Wolfram Alpha <a href="http://www.wolframalpha.com/input/?i=solve+[log+m+%3D+log+n+%2B+%283%2F2%29*log%281+%2B+v%2F%28m%5E2%29%29]+for+m" rel="nofollow noreferrer">coughs up a hairball</a> that appears to be the zeroes of a quartic, with no obvious relationship to the original equation.</p>
<p>Is there a short, tidy solution? Failing that, an explanation of how WA managed to turn this into a quartic, and the quartic itself, would be ok.</p>
| M. Chen | 403,559 | <p>$logm$=$logn$+$3/2log(1+v/m^2)$<br>
$logm$=$logn$+$log((m^2+v)/m^2)$<br>
$logm$=$logn$+$log(m^2+v)^(1.5))$<br>
$logm$=$logn$+$log(m^2+v)^3/2$-$logm^3$<br>
$logm^3$+$logm$-$logn$=$log(m^2+v)^3/2$<br>
$log(m^4/n)$=$log(m^2+v)$<br>
$m^4/n$=$(m^2+v)^(3/2)$ From here think you gotta use the cubic expension and square out the root
It is my first answer so plz dont judge me...</p>
|
2,593,627 | <p>I struggle to find the language to express what I am trying to do. So I made a diagram.</p>
<p><a href="https://i.stack.imgur.com/faHgE.png" rel="noreferrer"><img src="https://i.stack.imgur.com/faHgE.png" alt="Graph3parallelLines"></a></p>
<p>So my original line is the red line. From (2.5,2.5) to (7.5,7.5).</p>
<p>I want to shift the line away from itself a certain distance but maintaining the original angle of the line(so move it a certain distance away at a 90 degree angle). So after the shifting the line by the distance of +1 it would become the blue line or by -1 it would be come the yellow line.</p>
<p>I don't know a lot of the maths terminology so if anybody could manage an explanation in layman terms it would be appreciated.</p>
<p>Thanks,
C</p>
| Nominal Animal | 318,422 | <p>Let's use basic vector algebra. (This is extremely useful for any kind of graphics programmers, so if you are not familiar with the basics yet, I warmly recommend you look up some tutorials on the net first. Basics of linear algebra, namely matrices, and vector-matrix and matrix-matrix multiplication, is of tremendous help with transformations (rotations et cetera). If you add unit quaternions (also known as versors) describing rotations, and descriptive geometry basics (how to do projections), you've got a very powerful mathematical toolbox to work with all kinds of 2D and 3D graphics.)</p>
<p>Let's say you have a line segment between points $\vec{p}_1 = (x_1 , y_1)$ and $\vec{p}_2 = (x_2 , y_2)$, and you wish to do a parallel translation by $d$. (If $d$ is positive, when we look from $\vec{p}_1$ towards $\vec{p}_2$, the translation is to the left, if $x$ axis increases right and $y$ axis up; if $d$ is negative, to the right. This choice is arbitrary, but you'll see this choice made often in vector algebra tutorials, so it'll feel and look familiar to many.)</p>
<hr>
<p>Step 1.</p>
<p>Let's define $\hat{n}$ as the unit direction vector from $\vec{p}_1$ towards $\vec{p}_2$. (Unit vector means its length is 1, i.e. $\left\lVert\hat{n}\right\rVert = 1$. The hat mark, $\hat{\cdot}$, is often (but not always) used to distinguish <em>unit</em> vectors from all other vectors, $\vec{\cdot}$; I use that convention here just to make that point, to convey the math clearer.)</p>
<p>We calculate $\hat{n}$ by
$$\hat{n} = \frac{\vec{p}_2 - \vec{p}_1}{\left\lVert \vec{p}_2 - \vec{p}_1 \right\rVert} \tag{1a}\label{NA1a}$$
If we use $\hat{n} = ( x_n , y_n )$, then in coordinate form,
$$\begin{cases}
x_n = \frac{x_2 - x_1}{\sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 }} \\
y_n = \frac{y_2 - y_1}{\sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 }} \end{cases} \tag{1b}\label{NA1b}$$</p>
<hr>
<p>Step 2.</p>
<p>Rotating $\hat{n} = ( x_n , y_n )$ 90Β° counterclockwise yields $\hat{v} = ( -y_n , x_n )$.</p>
<p>Explanation:</p>
<p>In two dimensions, counterclockwise rotation (this is the choice I made above) by angle $\varphi$ is described by matrix $\mathbf{R}$,
$$\mathbf{R} = \left [ \begin{array}{cc} \cos\varphi & -\sin\varphi \\ \sin\varphi & \cos\varphi \end{array} \right ]$$
so that the rotation of a vector $\vec{p} = ( x , y )$ by $\varphi$ yields $\vec{p}^{,}$:
$$\vec{p}^{,} = \mathbf{R}\vec{p} \iff \begin{cases}
x^{,} = x \cos(\varphi) - y \sin(\varphi) \\
y^{,} = x \sin(\varphi) + y \cos(\varphi) \end{cases}$$
Note that a common "bug" in first implementations is to not realize that you must use the old coordinates for the entire calculation; you cannot do say $x$ first, and use the new $x^{,}$ when calculating $y^{,}$.</p>
<p>In our current particular situation, $\varphi = 90Β°$, and therefore
$$\mathbf{R} = \left [ \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right ]$$
or, in coordinate form, $$\begin{cases} x^{,} = x 0 - y 1 = - y\\ y^{,} = x 1 + y 0 = x \end{cases}$$</p>
<hr>
<p>Step 3.</p>
<p>Move the line segment along the rotated unit vector $\hat{v} = ( -y_n , x_n )$ by the desired distance $d$.</p>
<p>If we use $\vec{p}_3 = ( x_3 , y_3 )$ and $\vec{p}_4 = ( x_4 , y_4 )$ for the translated line segment endpoints $\vec{p}_1$ and $\vec{p}_2$, respectively, then</p>
<p>$$\begin{cases} \vec{p}_3 = \vec{p}_1 + d \hat{v} \\
\vec{p}_4 = \vec{p}_2 + d \hat{v} \end{cases} \tag{2}\label{NA2}$$</p>
<hr>
<p>Summary:</p>
<p>In coordinate form, and summarizing all you need to calculate to do this programmatically:
$$\begin{array}{l}
r = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \\
x_\Delta = \frac{d}{r} ( y_1 - y_2 ) \\
y_\Delta = \frac{d}{r} ( x_2 - x_1 ) \\
x_3 = x_1 + x_\Delta \\
y_3 = y_1 + y_\Delta \\
x_4 = x_2 + x_\Delta \\
y_4 = y_2 + y_\Delta \end{array}\tag{3}\label{NA3}$$
with the translated line segment being from $( x_3 , y_3 )$ to $( x_4 , y_4 )$.</p>
|
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