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 |
|---|---|---|---|---|
275,308 | <p>Problems with calculating </p>
<p>$$\lim_{x\rightarrow0}\frac{\ln(\cos(2x))}{x\sin x}$$</p>
<p>$$\lim_{x\rightarrow0}\frac{\ln(\cos(2x))}{x\sin x}=\lim_{x\rightarrow0}\frac{\ln(2\cos^{2}(x)-1)}{(2\cos^{2}(x)-1)}\cdot \left(\frac{\sin x}{x}\right)^{-1}\cdot\frac{(2\cos^{2}(x)-1)}{x^{2}}=0$$</p>
<p>Correct answer is -2. Please show where this time I've error. Thanks in advance!</p>
| Mikasa | 8,581 | <p>I tink if you work as follows, you will get the answer better(I hope so):</p>
<p>When $\alpha(x)$ is very small, then $\ln(1+\alpha(x))\sim\alpha(x)$ so $$\ln\left(\cos(2x)\right)=\ln\left(1+(-2\sin^2(x)\right)\sim -2\sin^2(x)$$</p>
<p>Now take your limit with this fact again. It is $-2$.</p>
|
66,801 | <p>In short, I am interested to know of the various approaches one could take to learn modern harmonic analysis in depth. However, the question deserves additional details. Currently, I am reading Loukas Grafakos' "Classical Fourier Analysis" (I have progressed to chapter 3). My intention is to read this book and then proceed to the second volume (by the same author) "Modern Fourier Analysis". I have also studied general analysis at the level of Walter Rudin's "Real and Complex Analysis" (first 15 chapters). In particular, if additional prerequisites are required for recommended references, it would be helpful if you could state them.</p>
<p>My request is to know how one should proceed after reading these two volumes and whether there are additional sources that one could use that are helpful to get a deeper understanding of the subject. Also, it would be nice to hear suggestions of some important topics in the subject of harmonic analysis that are current interests of research and references one could use to better understand these topics.</p>
<p>However, I understand that as one gets deeper into a subject such as harmonic analysis, one would need to understand several related areas in greater depth such as functional analysis, PDE's and several complex variables. Therefore, suggestions of how one can incorporate these subjects into one's learning of harmonic analysis are welcome. (Of course, since this is mainly a request for a roadmap in harmonic analysis, it might be better to keep any recommendations of references in these subjects at least a little related to harmonic analysis.)</p>
<p>In particular, I am interested in various connections between PDE's and harmonic analysis and functional analysis and harmonic analysis. It would be nice to know about references that discuss these connections. </p>
<p>Thank you very much!</p>
<p><strong>Additional Details</strong>: Thank you for suggesting Stein's books on harmonic analysis! However, I am not sure how one should read these books. For example, there seems to be overlap between Grafakos and Stein's books but Stein's "Harmonic Analysis" seems very much like a research monograph and although it is, needless to say, an excellent book, I am not very sure what prerequisites one must have to tackle it. In contrast, the other two books by Stein are more elementary but it would be nice to know of the sort of material that can be found in these two books but that cannot be found in Grafakos. </p>
| Alain Valette | 14,497 | <p>Since you have read Rudin's "Real and complex analysis", you are ready to attack Rudin's "Fourier analysis on groups", which is equally pleasant reading.</p>
<p>Still valuable for the link with Banach algebra theory, is L.H. Loomis' "An introduction to abstract harmonic analysis".</p>
<p>For connections with unitary representations: the 2nd half of Dixmier's "C*-algebras", or better R. Howe and E.C. Tan "Non-abelian harmonic analysis (applications of $SL_2(\mathbb{R})$)" (everything is in the subtitle!)</p>
<p>If you want to see connections with number theory, I recommend Weil's "Basic number theory".</p>
<p>Now you can guess my age from this list of references!</p>
|
2,135,918 | <p>Does L'Hopitals Rule hold for second derivative, third derivative, etc...? Assuming the function is differential up to the $k$th derivative, is the following true?</p>
<p>$$\lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f'(x)}{g'(x)} = \lim_{x\to c} \frac{f''(x)}{g''(x)} = \cdots = \lim_{x\to c} \frac{f^{(k)}(x)}{g^{(k)}(x)} = \lim_{x\to c} \frac{f^{(k+1)}(x)}{g^{(k+1)}(x)}$$</p>
| Michael Hardy | 11,667 | <p>When L'Hopital's rule is stated by saying that $\displaystyle \lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)},$ one of the hypotheses is that $\lim_{x\to c}f(x) = \lim_{x\to c} g(x) = 0,$ or else both limits are among the two objects $\pm\infty.$ In order to take it one more step and say $\displaystyle \lim_{x\to c} \frac{f'(x)}{g'(x)} = \lim_{x\to c}\frac{f''(x)}{g''(x)},$ you would need to know that $\lim_{x\to c} f'(x) = \lim_{x\to c} g'(x) = 0$ or else both of those are among $\pm\infty.$ And you don't need to know anything beyond L'Hopital's rule as stated at the outset in order to know that.</p>
<p>Another hypothesis of L'Hopital's rule is that the limit to the right of the "equals" sign exists. You have no guarantee that L'Hopital's rule can be applied until you know that.</p>
<p>(Where you wrote $n\to c,$ I wrote $x\to c$.)</p>
|
1,071,321 | <p>The problem is quite simple to formulate. If you have a large group of people (n > 365), and their birthdays are uniformly distributed over the year (365 days), what's the probability that every day of the year is someone's birthday?</p>
<p>I am thinking that the problem should be equivalent to finding the number of ways to place n unlabeled balls into k labeled boxes, such that all boxes are non-empty, but C((n-k)+k-1, (n-k))/C(n+k-1, n) (C(n,k) being the binomial coefficient) does not yield the correct answer.</p>
| Valentin | 310,690 | <p>Just for sake of illustration I give some numerical results:</p>
<ul>
<li>for $1000$ people we get a propability of $1.7*10^{-10}$% to exhaust all birthdays</li>
<li>for $1500$ people we get $0.2$% </li>
<li>for $2000$ people we get $22$%</li>
<li>for $2286$ people we get $50$%</li>
<li>for $2500$ people we have $68$%</li>
<li>for $3000$ people we have $91$%</li>
<li>for $4000$ people we have $99$%</li>
</ul>
|
320,704 | <p>Given an irrational <span class="math-container">$a$</span>, the sequence <span class="math-container">$b_n := na$</span> is dense and equidistributed in <span class="math-container">$\mathbb S^1$</span> where we view <span class="math-container">$\mathbb S^1$</span> as <span class="math-container">$[0, 1]$</span> with its endpoints identified. </p>
<p>Given a point <span class="math-container">$p$</span> in <span class="math-container">$\mathbb S^1$</span>, can we obtain a quantitative upper bound (that can depend on <span class="math-container">$a, p, e$</span>) on the smallest <span class="math-container">$n$</span> such that <span class="math-container">$na$</span> is in <span class="math-container">$B_e (p)$</span>?</p>
| burtonpeterj | 30,721 | <p>This is probably suboptimal, but one can use the convergents of the continued fraction expansion of <span class="math-container">$a$</span> to efficiently find <span class="math-container">$n \in \mathbb{N}$</span> such that <span class="math-container">$||na||_\mathbb{Z} \leq e$</span>, where <span class="math-container">$||\cdot||_\mathbb{Z}$</span> denotes the distance to the nearest integer. Then one can find <span class="math-container">$m \in \mathbb{N}$</span> with <span class="math-container">$m \leq ||na||_\mathbb{Z}^{-1}$</span> such that <span class="math-container">$|mna-p| \leq e$</span>.</p>
|
3,261,334 | <blockquote>
<p>Prove isomorphism of groups <span class="math-container">$\langle G, + , {}^{-1}\rangle$</span> and <span class="math-container">$\langle G, *,{}^{-1}\rangle$</span>, where <span class="math-container">$a*b=b+a$</span><br>
<span class="math-container">$\forall a,b \in G$</span></p>
</blockquote>
<p>I'm barely starting to study abstract algebra.</p>
<p>So how do I show isomorphism? I think that I should show a homomorphism somehow, but I don't know how. </p>
<p>Any thoughts/ideas would be really appreciated!</p>
| Community | -1 | <p>Let <span class="math-container">$\phi (g)=g^{-1}\,,\forall g\in G$</span>. </p>
<p><span class="math-container">$\phi$</span> is a homomorphism: <span class="math-container">$\phi (g*h)=(g*h)^{-1}=h^{-1}*g^{-1}=g^{-1}+h^{-1}=\phi(g)+\phi(h)$</span>.</p>
<p>The kernel of <span class="math-container">$\phi$</span> is trivial: <span class="math-container">$\phi(g)=e\implies g^{-1}=e\implies g=e$</span>.</p>
<p><span class="math-container">$\phi$</span> is surjective: <span class="math-container">$\forall g\in G,\phi(g^{-1})=g$</span>.</p>
<p>Thus <span class="math-container">$\phi$</span> is an isomorphism.</p>
|
320,619 | <p>Consider $\ell^\infty $ the vector space of real bounded sequences endowed with the sup norm, that is
$||x|| = \sup_n |x_n|$ where $x = (x_n)_{n \in \Bbb N}$. </p>
<p>Prove that $B'(0,1) = \{x \in l^\infty : ||x|| \le 1\} $ is not compact.</p>
<p>Now, we are given a hint that we can use the equivalence of sequential compactness and compactness without proof. </p>
<p>However, I don't understand how sequences of sequences work? Do I need to find a set of sequences and order them such that they do not converge to the same sequence? </p>
<p>Does the sequence $(y_n)$ where $y_n$ is the sequence such that $y_n = 0$ at all but the nth term where $y_n =1$ satisfy the requirement that it does not have a convergent subsequence? </p>
<p>I think I have probably just confused myself with this sequence of sequences lark. Sorry and thanks. </p>
| copper.hat | 27,978 | <p>Let $E = \{ e_n \}_n$ where $e_n$ is the vector with $1$ in the $n$th position and zero everywhere else. Clearly $E \subset \overline{B}(0,1)$, hence it is bounded. Since $\|e_n - e_m\| = 1$ whenever $n \neq m$, it is clear that each point in $E$ is isolated. Hence $E$ is closed.</p>
<p>Now let $U_n = B(e_n, \frac{1}{2}$). Then $\{ U_n\}_n$ is an open cover of $E$ that had no finite subcover. Hence $E$ is not compact.</p>
<p>If you would rather use a sequential argument, choose the sequence $x_n =e_n$. As above, we have $\|x_n -x_m\| = 1$ whenever $n \neq m$, hence no subsequence can be Cauchy. Hence no subsequence can converge, hence $E$ is not sequentially compact. </p>
|
1,652,747 | <p>Ok, so I think I'm getting the hang of this. Is this more or less on the right track?</p>
<p>$$e^{z-2}=-ie^2$$
$$e^ze^{-2}=-ie^2$$
$$e^z=-ie^4$$
$$\ln(e^z)=\ln(-ie^4)$$
$$z=\ln|-i|+iarg(-i)+2\pi ik+4$$
$$z=\frac{i\pi}{2}-\frac{i\pi}{2}+4+2\pi ik$$
$$z=4+2\pi ik$$</p>
| paradus_hex | 991,156 | <p>Continuing from line 4:
<span class="math-container">$$
z=ln\left(i^3\right)+4........(i)
$$</span>
We know,
<span class="math-container">$ln(z)=ln(r)+i(Arg(z)+2k\pi)$</span> where k=0,+-1,+-2....<br/></p>
<p>Applying this to <span class="math-container">$ln\left(-i\right)$</span>, we get:</p>
<p><span class="math-container">$$ln\left(-i\right)=ln(1)+i(-\pi/2 +2k\pi)=i(-\pi/2 +2k\pi)$$</span>
Putting this value back in equation (i):
<span class="math-container">$$z=4+i(-\pi/2 +2k\pi)$$</span></p>
|
2,186,743 | <p>I think this is a very basic question but somehow I am unable to understand the answer.</p>
<hr>
<blockquote>
<p>Find the number of zeroes of $f(x) = x^3 + x + 1$.</p>
</blockquote>
<hr>
<p>Answer in the book : </p>
<p>$f^\prime (x) = 3x^2 + 1$ since $3x^2 + 1 \ge 0$ for all $x \in \mathbb{R}$ therefore $f^\prime (x) \ge 1 $ for all $x \in \mathbb{R}$</p>
<p>Therefore by Rolle's theorem $f(x)$ has at most one zero in its domain. </p>
<p>Now $f(-1) = -1$ and $f(0) = 1$ therefore by Intermediate value theorem $f(x)$ has at least one zero in the interval $[-1, 0]$. </p>
<p>Thus $f(x)$ has one zero.</p>
<hr>
<blockquote>
<p>Rolle's theorem : A function which is continuous $[a,b]$ and differentiable on $(a,b)$ such that $f(a) = f(b) = 0$, then there exist at least a point $c \in [a,b]$ for which $f^\prime(c) = 0$ </p>
</blockquote>
<p>In the first part of the proof I am unable to understand how Rolle's theorem is applied because we don't know any $a,b$ where $f(a) = f(b) = 0$ . </p>
<p>How can conclude that $f(x)$ has at most one zero ?</p>
| Chris Leary | 2,933 | <p>Suppose the function has two zeros, then by Rolle, $f'(x)$ must be zero for some real number $c$ strictly between the two roots. This is impossible, since $f'(x) > 0$ for all real $x.$ So, the function has at most one zero.To show it has a zero, note that $f(-1) < 0$ and $f(0)>0,$ and appeal to the intermediate value theorem.</p>
|
3,743,282 | <p>One definition of complex sympletic group I have encountered is (sourced from <a href="https://en.wikipedia.org/wiki/Symplectic_group" rel="nofollow noreferrer">Wikipedia</a>):
<span class="math-container">$$Sp(2n,F)=\{M\in M_{2n\times 2n}(F):M^{\mathrm {T} }\Omega M=\Omega \}$$</span></p>
<p>What is the motivation for imposing the condition <span class="math-container">$M^{\mathrm {T} }\Omega M=\Omega$</span> instead of others such as <span class="math-container">$M^{-1}\Omega M=\Omega$</span>?</p>
| Ivo Terek | 118,056 | <p>In general, if <span class="math-container">$V$</span> is a <span class="math-container">$n$</span>-dimensional <span class="math-container">$F$</span>-vector space equipped with a bilinear form <span class="math-container">$b\colon V \times V \to F$</span> and <span class="math-container">$T\colon V \to V$</span> is an endomorphism such that <span class="math-container">$b(Tx,Ty) = b(x,y)$</span> for all <span class="math-container">$x,y \in V$</span>, one may take a basis <span class="math-container">$\mathcal{B} = (e_1,\ldots, e_n)$</span> for <span class="math-container">$V$</span>, let <span class="math-container">$b_{ij} = b(e_i,e_j)$</span> and <span class="math-container">$Te_j = \sum_{i=1}^n T^i_{~j}e_i$</span>, and compute <span class="math-container">$$b_{ij} = b(e_i,e_j) = b(Te_i,Te_j) = b\left(\sum_{k=1}^n T^k_{~i}e_k, \sum_{\ell=1}^r T^\ell_{~j}e_j\right) = \sum_{k,\ell=1}^n T^k_{~i}T^\ell_{~j}b_{k\ell}.$$</span>If <span class="math-container">$[T]_{\mathcal{B}} = (T^i_{~j})_{i,j=1}^n$</span> and <span class="math-container">$[b]_{\mathcal{B}} =(b_{ij})_{i,j=1}^n$</span>, the above identity then reads <span class="math-container">$$[b]_{\mathcal{B}} = [T]_{\mathcal{B}}^\top [b]_{\mathcal{B}}[T]_{\mathcal{B}}.$$</span>So, once a basis <span class="math-container">$\mathcal{B}$</span> (and hence the matrix <span class="math-container">$B = [b]_{\mathcal{B}}$</span>) is fixed, the isomorphism <span class="math-container">$T\mapsto [T]_{\mathcal{B}}$</span> between <span class="math-container">${\rm End}(V)$</span> and <span class="math-container">${\rm Mat}(n,F)$</span> restricts to an isomorphism <span class="math-container">$$\{T \in {\rm End}(V) \mid b(Tx,Ty) = b(x,y) \mbox{ for all }x,y \in V \}\cong \{ M \in {\rm Mat}(n,F) \mid M^\top BM = B \}.$$</span>The fact that the dimension of the space <span class="math-container">$V$</span> is even and that <span class="math-container">$\Omega$</span> is symplectic is irrelevant, this is a general mechanism regarding the relation between the matrix of a bilinear map and the matrix of its pull-back under a linear map.</p>
|
1,690,092 | <p>$a_{n} = \frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + ... + \frac{1}{\sqrt{n^2+2n-1}}$</p>
<p>and I need to check whether this sequence converges to a limit without finding the limit itself. I think about using the squeeze theorem that converges to something (I suspect '$1$').</p>
<p>But I wrote $a_{n+1}$ and $a_{n-1}$ and it doesn't get me anywhere...</p>
| Fnacool | 318,321 | <p>On the one hand, </p>
<p>$$a_n \ge \mbox{smallest summand} \times \mbox{number of summands}= \frac{1}{\sqrt{n^2+2n-1}}\times n .$$</p>
<p>To deal with the denominator, observe that </p>
<p>$$n^2+2n-1 \le n^2+2n+1=(n+1)^2.$$</p>
<p>On the other hand, </p>
<p>$$a_n \le \mbox{largest summand}\times \mbox{number of summands} = \frac{1}{\sqrt{n^2+n}}\times n.$$ </p>
<p>To deal with the denominator, observe that </p>
<p>$$n^2+n \ge n^2$$. </p>
|
3,477,795 | <p>How to use the absolute value function to translate each of the following statements into a single inequality.</p>
<p>(a) <span class="math-container">$\ x ∈ (-4,10) $</span> </p>
<p>(b) <span class="math-container">$\ x ∈ (-\infty,2] \cup[9,\infty) $</span></p>
<p>I think in the first one the absolute value of <span class="math-container">$\ x$</span> should be greater than 4 and less than 10. is that correct?
because the distance from <span class="math-container">$\ x$</span> to <span class="math-container">$\ 0$</span> should be between <span class="math-container">$\ 4$</span> and<span class="math-container">$\ 10$</span> in order for <span class="math-container">$\ x$</span> to belong in this interval.</p>
| Rachid Atmai | 99,885 | <p>Any bounded function with countably many discontinuities is Riemann Integrable. Your function has only 3 discontinuity points and is thus Riemann integrable. In addition, this function is bounded. One them computes the Riemann integral in a piecewise fashion.</p>
|
626,928 | <p>I took linear algebra course this semester (as you've probably noticed looking at my previously asked questions!). We had a session on preconditioning, what are they good for and how to construct them for matrices with special properties. It was a really short introduction for such an important research topic, so I'm not sure if I got familiar with them. I need some elementary sources on it to be able to construct preconditions for famous matrices and to make sure I fully understand the concept. Any helps would be greatly appreciated. </p>
| Yiorgos S. Smyrlis | 57,021 | <p>It is still a very hot topic of Numerical Linear Algebra - The main question is to improve the conditioning of a (usually very large and potentially ill-conditioned) matrix.</p>
<p>Take a look at the work of Dongarra: <a href="http://web.eecs.utk.edu/~dongarra/etemplates/node396.html" rel="nofollow">Preconditioning Techniques</a></p>
<p>A good textbook is:</p>
<p>H. A. van der Vorst, <em>Iterative Krylov Methods for Large Linear systems</em>,
Cambridge University Press, Cambridge, 2003.</p>
|
891,137 | <p>Count all $n$-length strings of digits $0, 1,\dots, m$ that have an equal number of $0$'s and $1$'s. Is there a closed form expression?</p>
| Mary Star | 80,708 | <p>We define $$f(x)=\log{(1+x)}-x$$</p>
<p>The domain of $f$ is $(-1, +\infty)$.</p>
<p>$$f'(x)=\frac{1}{1+x}-1=\frac{1-1-x}{1+x}=\frac{-x}{1+x}$$</p>
<p>$$f'(x)=0 \Rightarrow x=0$$</p>
<p>$$\text{ For } x \geq 0 : f'(x)\leq 0$$
$$\text{ For } -1<x\leq 0 : f'(x)\geq 0$$</p>
<p>That means that $f$ is decreasing on $[0, +\infty)$ and increasing at $(-1,0]$.</p>
<p>Therefore, $$x\geq 0 \Rightarrow f(x) \leq f(0)\Rightarrow f(x)\leq 0 \\ x\leq 0 \Rightarrow f(x)\leq f(0) \Rightarrow f(x)\leq 0$$</p>
<p>So, $f(x)\leq 0 \ \ \ \forall x \in (-1, +\infty) \Rightarrow \log{(1+x)}-x\leq 0 \ \ \ \forall x \in (-1, +\infty) \\ \Rightarrow \log{(1+x)} \leq x \ \ \ \forall x \in (-1, +\infty)$</p>
|
35,964 | <p>This is kind of an odd question, but can somebody please tell me that I am crazy with the following question, I did the math, and what I am told to prove is simply wrong:</p>
<p>Question:
Show that a ball dropped from height of <em>h</em> feet and bounces in such a way that each bounce is $\frac34$ of the height of the bounce before travels a total distance of 7 <em>h</em> feet.</p>
<p>My Work:
$$\sum_{n=0}^{\infty} h \left(\frac34\right)^n = 4h$$</p>
<p>Obviously 4 <em>h</em> does not equal 7 <em>h</em> . What does the community get?</p>
<p>I know that my calculations are correct, see Wolfram Alpha and it confirms my calculations, that only leaves my formula, or the teacher being incorrect...</p>
<p>Edit:
Thanks everyone for pointing out my flaw, it should be something like:
$$\sum_{n=0}^{\infty} -h + 2h \left(\frac34\right)^n = 7h$$</p>
<p>Thanks in advance for any help!</p>
| Bill Dubuque | 242 | <p>To get symmetry - a complete sawtooth pattern - suppose the ball is first thrown up from ground to height $h$. The total distance $x$ is that of the first tooth $= 2h$ plus the remaining sawtooth $= 3/4\ x$. Thus $x =\: 2h + 3/4\ x,\ $ so $x = 8h\:.\ $ Subtracting the intitial throw up leaves $7h.$</p>
|
40,500 | <blockquote>
<p>What are the most
fundamental/useful/interesting ways in
which the concepts of Brownian motion,
martingales and markov chains are
related?</p>
</blockquote>
<p>I'm a graduate student doing a crash course in probability and stochastic analysis. At the moment, the world of probability is a confusing blur, but I'm starting with a grounding in the basic theory of markov chains, martingales and Brownian motion. While I've done a fair amount of analysis, I have almost no experience in these other matters and while understanding the definitions on their own isn't too difficult, the big picture is a long way away.</p>
<p>I would like to <strong>gather together results and heuristics</strong>, each of which links together two or more of Brownian motion, martingales and Markov chains in some way. Answers which <strong>relate probability to real or complex analysis</strong> would also be welcome, such as "Result X about martingales is much like the basic fact Y about sequences".</p>
<p>The thread may go on to contain a Big List in which each answer is the posters' favourite as yet unspecified result of the form "This expression related to a markov chain is always a martingale because blah. It represents the intuitive idea that blah".</p>
<p>Because I know little, I can't gauge the worthiness of this question very well so apologies in advance if it is deemed untenable by the MO police.</p>
| alezok | 13,388 | <p>Let $(B_t)_{t \geq 0}$ be a Brownian motion and $\mathcal B_t:=\sigma(\{B_s : s \leq t\})$ its natural filtration. Then</p>
<ul>
<li>$B_t$ is a martingale,</li>
<li>$B_t^2-t$ is a martingale,</li>
<li>$\exp(\theta B_t - \frac{1}{2}\theta^2 t)$ is a martingale, and</li>
<li>$H_n(t,B_t)$ is a martingale for every $n$, where $H_n(t,x)$ is the Hermite polynomial defined by
$$ \exp(\theta x - \frac{1}{2}\theta^2 t) = \sum_{n=0}^{\infty} \frac{\theta^n}{n!} H_n(t,x).$$
Note that $H_1(t,x)=x$ and $H_2(t,x)=x^2-t$, so this latter statement implies the first two.</li>
</ul>
<p>When writing "is a martingale", it is obviously "with respect to the filtration $(\mathcal B_t)_{t\geq 0}$".</p>
<p>A remarkable consequence of the Levy's characterization of Brownian motion is that every continuous martingale is a time-change of Brownian motion.</p>
<p>Source: <em>L.C.G. Rogers and D. Williams, Diffusion, Markov Processes and Martingales, Vol.1 (2000)</em></p>
|
2,333,083 | <p>I want to know about when the infinite product form of the Riemann Zeta Function converges.</p>
<p>It is easy to show that $$\sum_{n=1}^\infty \frac 1{n^x}$$ converges for $\Bbb R(x)>1$.</p>
<p>When does $$\prod_{p=primes} \frac {p^x}{p^x-1}$$ converge? And how do you show that?</p>
| Dando18 | 274,085 | <p>The product converges for $\mathbb{R}(s)>1$ (from <a href="https://en.wikipedia.org/wiki/Riemann_zeta_function#Euler_product_formula" rel="nofollow noreferrer">wikipedia page</a>). This product approaches $1$ from the right hand side, which follows from the Dirichlet series for $\zeta(s)$, which converges for $\mathbb{R}(s)>1$.</p>
|
1,893,609 | <p>I am trying to show that $A=\{(x,y) \in \Bbb{R} \mid -1 < x < 1, -1< y < 1 \}$ is an open set algebraically. </p>
<p>Let $a_0 = (x_o,y_o) \in A$. Suppose that $r = \min\{1-|x_o|, 1-|y_o|\}$ then choose $a = (x,y) \in D_r(a_0)$. Then</p>
<p>Edit: I am looking for the proof of the algebraic implication that $\|a-a_0\| = \sqrt {(x-x_o)^2+(y-y_o)^2} < r \Rightarrow|x| < 1 , |y| < 1 $</p>
| Claude Leibovici | 82,404 | <p>Assuming that the points are not colinear, for a very simple procedure, consider <a href="https://en.wikipedia.org/wiki/Conic_section" rel="nofollow">the general equation of conics</a> the circle $$x^2+y^2+a x+b y +c=0$$ and build, for each data point, the equation $$f_i=x_i^2+y_i^2+a x_i+b y_i+c=0$$ This then reduce to $n$ equations $$a x_i+b y_i+c=-(x_i^2+y_i^2)$$ and you face a simple multilinear regression which will give you parameters $a,b,c$.</p>
<p>Having these numbers, at least as estimates, just complete the square $$x^2+y^2+a x+b y +c=(x+\frac a2)^2-\frac{a^2}4+(y+\frac b2)^2-\frac{b^2}4+c$$ that is to say $$(x+\frac a2)^2+(y+\frac b2)^2=\frac{a^2+b^2-4c}4$$ which makes the center located at $(-\frac a2,-\frac b2)$ with a radius equal to $\frac{\sqrt{a^2+b^2-4c}} 2$.</p>
<p>As said, this is a very simplistic solution but whatever could be the objective function you would want to minimize, it will give you very good starting values for the optimization process.</p>
<p>Quoting <a href="http://www.nlreg.com/circular.htm" rel="nofollow">NLREG documentation</a>, a more rigourous objective function would be $$F(a,b,r)=\sum_{i=1}^n \left(\sqrt{(x_i-a)^2+(y_i-b)^2}-r\right)^2$$ where $(a,b)$ are the coordinates of the circle and $r$ the radius. As you see, the problem is much more complex and would require optimization.</p>
|
1,916,297 | <p>Let $\{F_n\}_{n=1}^\infty $ be the Fibonacci sequence (defined by
$ F_n=F_{n-1}+F_{n-2}$ with $F_1=F_2=1$).</p>
<p>Is it true that for every integer number $N$ there exist positive integers
$n,m,k,l$ such that $N=F_n+F_m-F_k-F_l$ ?</p>
| Mr. Sigma. | 367,180 | <p>Edit: I'm wrong. </p>
<p>I'm getting yes!
Take k,l = 1. Then it follows.</p>
<p>N = F(n) + F(m) - 2
=> N + 2 = F(n) + F(m),Take N' = N+2. Therefore, we get.</p>
<p>=> N' = F(n) + F(m) ,N'>=2.
Which is possible since every natural number is a combination of two Fibonacci number. For N=0 & 1 also we have possible solution trivially.
Therefore your conjecture is correct.</p>
<p>Please correct me if I'm wrong since I'm newbie.
Thanks. </p>
|
2,975,665 | <p>I am asked to find the sum of the series <span class="math-container">$$\sum_{n=0}^\infty\frac{(x+1)^{n+2}}{(n+2)!}$$</span></p>
<p>For some reason (that I don't understand) I can't apply the techniques for finding the sum of the series that I usually would to this one? I think the others that I have done have been geometric series so I could just find the first few terms etc... </p>
<p>This one is really difficult for me to figure out. </p>
<p>As far as my research into this has taken me, it has something to do with writing the series out as a power series about x=a and somehow using this to find a? My teacher told me to take a few derivatives of this? Does she mean to take the derivatives of taylor series or something like that? </p>
<p>I am pretty confused as to the method I need to complete this question and online sources and class lecture information has taken me nowhere but back to the stuff I understand that applies to geometric series. What kind of series is this? How do I find its sum?</p>
| Mohammad Riazi-Kermani | 514,496 | <p><span class="math-container">$$\sum_{n=0}^\infty\frac{(x+1)^{n+2}}{(n+2)!}= \sum_{n=2}^\infty\frac{(x+1)^{n}}{(n)!} $$</span></p>
<p><span class="math-container">$$=\sum_{n=0}^\infty\frac{(x+1)^{n}}{(n)!} -1-(x+1) $$</span></p>
<p><span class="math-container">$$=e^{x+1}-x-2$$</span></p>
|
1,179,981 | <p>As the title suggests. Let $G$ be a group, and suppose the function $\phi: G \to G$ with $\phi(g)=g^3$ for $g \in G$ is a homomorphism. Show that if $3 \nmid |G|$, $G$ must be abelian.</p>
<p>By considering $\ker(\phi)$ and Lagrange's Theorem, we have $\phi$ must be an isomorphism (right?), but I'm not really sure where to go after that.</p>
<p>This is a problem from Alperin and Bell, and it is not for homework.</p>
| krirkrirk | 221,594 | <p>$\phi$ is an isomoprhism, because $g^{3} = 1 \iff g=1$ for the order of $g\in G$ must divide the order of $G$.
I guess you have to consider $\phi (gh)$ and $\phi(hg)$.
If $\forall g, h\in G, \phi(gh)=\phi(hg)$, then $ghg^{-1}h^{-1}\in ker\phi$ thus $gh=hg$ and it's won. But not really sure how to do it. </p>
|
2,931,762 | <blockquote>
<p>Given a function <span class="math-container">$f$</span> is defined for integers <span class="math-container">$m$</span> and <span class="math-container">$n$</span> as given:
<span class="math-container">$$f(mn) = f(m)\,f(n) - f(m+n) + 1001$$</span>
where either <span class="math-container">$m$</span> or <span class="math-container">$n$</span> is equal to <span class="math-container">$1$</span>, and <span class="math-container">$f(1) = 2$</span>.</p>
<p>The problem itself is to prove that <span class="math-container">$$f(x) = f(x-1) + 1001$$</span> in order to find the value of <span class="math-container">$f(9999)$</span>. </p>
</blockquote>
<p>As such, what I've already tried is:<br>
Replacing <span class="math-container">$n$</span> as <span class="math-container">$1$</span> and <span class="math-container">$m$</span> as <span class="math-container">$x$</span>, and trying to solve from there. </p>
<p><span class="math-container">$$f(x) = f(1) * f(x) - f(x+1) + 1001$$</span>
<span class="math-container">$$f(x) = 2 * f(x) - f(x+1) + 1001,$$</span></p>
<p>Although I'm not sure how I should continue from here, it did occur that the negative sign could be manipulated in some way, but I'm fairly certain that <span class="math-container">$-f(x+1)$</span>, cannot be rewritten as <span class="math-container">$f(-x-1)$</span>, or anything of the sorts.</p>
<p>So if anyone has any thoughts or insight as how I should proceed, it would be greatly appreciated.</p>
| Batominovski | 72,152 | <p><strong>Disclaimer.</strong> This is an attempt to see what happens if I drop the conditions that <span class="math-container">$f(1)=2$</span> and that at least one of <span class="math-container">$m$</span> and <span class="math-container">$n$</span> in the functional equation must be <span class="math-container">$1$</span>. That is, this answer does not answer the OP's question, but I think it is a nice relevant question. Since the work is quite long, I have to put it here as an answer. </p>
<hr>
<p>Fix <span class="math-container">$k\in\mathbb{C}$</span>. Let <span class="math-container">$f:\mathbb{Z}_{>0}\to\mathbb{C}$</span> be a function satisfying
<span class="math-container">$$f(mn)=f(m)\,f(n)-f(m+n)+k$$</span>
for all <span class="math-container">$m,n\in\mathbb{Z}_{>0}$</span>. Write <span class="math-container">$a:=f(1)$</span>. Then, plugging in <span class="math-container">$m=1$</span>, we have
<span class="math-container">$$f(n+1)=(a-1)\,f(n)+k\tag{*}$$</span>
for all <span class="math-container">$n\in\mathbb{Z}_{>0}$</span>. We have three notable scenarios: <span class="math-container">$a=1$</span>, <span class="math-container">$a=2$</span>, and <span class="math-container">$a\notin\{1,2\}$</span>. </p>
<hr>
<p>If <span class="math-container">$a=1$</span>, then <span class="math-container">$f(n)=k$</span> for all <span class="math-container">$n\in\mathbb{Z}_{>1}$</span>. That means
<span class="math-container">$$k=f(4)=f(2\cdot2)=f(2)\,f(2)-f(2+2)+k=k\cdot k-k+k\,.$$</span>
Thus, <span class="math-container">$k^2=k$</span>, or <span class="math-container">$k\in\{0,1\}$</span>. Hence, there are two possible solutions with <span class="math-container">$f(1)=1$</span>:</p>
<ul>
<li><span class="math-container">$k=0$</span>, which gives <span class="math-container">$f(1)=1$</span> and <span class="math-container">$f(n)=0$</span> for all <span class="math-container">$n\in\mathbb{Z}_{>1}$</span>;</li>
<li><span class="math-container">$k=1$</span>, which gives <span class="math-container">$f(n)=1$</span> for all <span class="math-container">$n\in\mathbb{Z}_{>0}$</span>.</li>
</ul>
<hr>
<p>Now, we suppose that <span class="math-container">$a=2$</span>. Thus, <span class="math-container">$f(n+1)=f(n)+k$</span> for all <span class="math-container">$n\in\mathbb{Z}_{>0}$</span>. This leads to <span class="math-container">$f(n)=2+k(n-1)$</span> for all <span class="math-container">$n\in\mathbb{Z}_{>0}$</span>. Consequently,
<span class="math-container">$$3k+2=f(4)=f(2\cdot 2)=f(2)\,f(2)-f(2+2)+k=(k+2)^2-(3k+2)+k\,,$$</span>
yielding
<span class="math-container">$$k^2-k=0\,.$$</span>
Thus, <span class="math-container">$k=0$</span> or <span class="math-container">$k=1$</span>. If <span class="math-container">$k=0$</span>, then <span class="math-container">$f(n)=2$</span> for all <span class="math-container">$n\in\mathbb{Z}_{>0}$</span>. If <span class="math-container">$k=1$</span>, then <span class="math-container">$f(n)=n+1$</span> for all <span class="math-container">$n\in\mathbb{Z}_{>0}$</span>.</p>
<hr>
<p>Finally, we tackle the case <span class="math-container">$a\notin \{1,2\}$</span>. Then, the solution to the recurrence relation (*) is
<span class="math-container">$$f(n)=(a-1)^{n-1}\left(a+\frac{k}{a-2}\right)-\frac{k}{a-2}\,.$$</span>
In particular, <span class="math-container">$f(2)=a(a-1)+k$</span> and <span class="math-container">$f(4)=a(a-1)^3+\left((a-1)^2+(a-1)+1\right)k$</span>. Using the same equation <span class="math-container">$2\,f(4)=\big(f(2)\big)^2+k$</span> from before, we get
<span class="math-container">$$2a(a-1)^3+2\left((a-1)^2+(a-1)+1\right)k=\big(a(a-1)+k\big)^2+k\,.$$</span>
Thus,
<span class="math-container">$$a(a-2)(a-1)^2=k(k-1)\,.$$</span>
This means <span class="math-container">$k=(a-1)^2$</span> or <span class="math-container">$k=-a(a-2)$</span>. For the subcase <span class="math-container">$k=-a(a-2)$</span>, we see that
<span class="math-container">$$f(n)=a\text{ for all }n\in\mathbb{Z}_{>0}\,.$$</span></p>
<hr>
<p>For the subcase <span class="math-container">$k=(a-1)^2$</span>, we get
<span class="math-container">$$f(n)=a(a-1)^{n-1}+(a-1)^2\left(\frac{(a-1)^{n-1}-1}{a-2}\right)\text{ for each integer }n>0\,.$$</span>
For simplicity, write <span class="math-container">$b:=a-1$</span>, so <span class="math-container">$k=b^2$</span> and <span class="math-container">$f(n)=(b+1)b^{n-1}+b^2\left(\dfrac{b^{n-1}-1}{b-1}\right)$</span>. We have
<span class="math-container">$$f(2)=(b+1)b+b^2\,,\,\,f(3)=2(b+1)b^2\,,$$</span> <span class="math-container">$$f(5)=(b+1)b^4+b^2(b^3+b^2+b+1)\,,$$</span> and <span class="math-container">$$f(6)=(b+1)b^5+b^2(b^4+b^3+b^2+b+1)\,.$$</span>
Since <span class="math-container">$f(6)=f(2\cdot3)=f(2)\,f(3)-f(2+3)+k=f(2)\,f(3)-f(5)+b^2$</span>, we obtain
<span class="math-container">$$2b^6+2b^5+b^4+b^3+b^2=(2b^2+b)(2b^3+2b^2)-(2b^5+2b^4+b^3+b^2)+b^2\,.$$</span>
That is,
<span class="math-container">$$b^2(b^2-1)(2b^2-1)=2b^6-3b^4+b^2=0\,.$$</span>
Hence, <span class="math-container">$b=0$</span>, <span class="math-container">$b=+1$</span>, <span class="math-container">$b=-1$</span>, <span class="math-container">$b=+\dfrac1{\sqrt{2}}$</span>, or <span class="math-container">$b=-\dfrac1{\sqrt{2}}$</span>. </p>
<hr>
<p>Recall that <span class="math-container">$a\notin\{1,2\}$</span>. If <span class="math-container">$b=0$</span>, then <span class="math-container">$a=1$</span>, which is a contradiction. If <span class="math-container">$b=+1$</span>, then <span class="math-container">$a=2$</span>, another contradiction. If <span class="math-container">$b=-1$</span>, then <span class="math-container">$a=0$</span>, <span class="math-container">$k=1$</span>, and
<span class="math-container">$$f(n)=\frac{1+(-1)^{n}}{2}=\left\{\begin{array}{ll}0\,,&\text{if }n\text{ is odd}\,,\\
1\,,&\text{if }n\text{ is even}\,.
\end{array}\right.$$</span></p>
<p>For <span class="math-container">$b=+\dfrac1{\sqrt{2}}$</span>, we get <span class="math-container">$a=1+\dfrac1{\sqrt{2}}$</span>, <span class="math-container">$k=\dfrac{1}{2}$</span>, and
<span class="math-container">$$f(n)=1+\frac1{\sqrt{2}}\text{ for all positive integers }n\,.$$</span>
For <span class="math-container">$b=-\dfrac1{\sqrt{2}}$</span>, we get <span class="math-container">$a=1-\dfrac1{\sqrt{2}}$</span>, <span class="math-container">$k=\dfrac12$</span> and
<span class="math-container">$$f(n)=1-\frac1{\sqrt{2}}\text{ for all positive integers }n\,.$$</span></p>
<hr>
<p>From the work above, we can now conclude our proposition. In particular, for <span class="math-container">$k=1001$</span>, the proposition states that there are two solutions:
<span class="math-container">$$f(n)=1+10\sqrt{10}\text{i}\text{ for each }n\in\mathbb{Z}_{>0}$$</span>
and
<span class="math-container">$$f(n)=1-10\sqrt{10}\text{i}\text{ for each }n\in\mathbb{Z}_{>0}\,.$$</span>
That is, if you allow <span class="math-container">$m$</span> and <span class="math-container">$n$</span> in the functional equation to be any positive integer and drop the condition that <span class="math-container">$f(1)=2$</span>, then the solutions look wildly different.</p>
<blockquote>
<p><strong>Proposition:</strong> Let <span class="math-container">$k$</span> be a complex number. Then, all solutions <span class="math-container">$f:\mathbb{Z}_{>0}\to\mathbb{C}$</span> to the functional equation
<span class="math-container">$$f(mn)=f(m)\,f(n)-f(m+n)+k\text{ for all positive integers }m\text{ and }n$$</span>
is given by the list below. <br>
(a) If <span class="math-container">$k=-a(a-2)$</span> for some <span class="math-container">$a\in\mathbb{C}$</span>, then <span class="math-container">$$f(n)=a\text{ for all }n\in\mathbb{Z}_{>0}\,.$$</span> There are two values of <span class="math-container">$a$</span> for all <span class="math-container">$k\in\mathbb{C}\setminus\{1\}$</span>, namely, <span class="math-container">$a=1-\sqrt{1-k}$</span> and <span class="math-container">$a=1+\sqrt{1-k}$</span>. For <span class="math-container">$k=1$</span>, there is only one value of <span class="math-container">$a$</span>, i.e., <span class="math-container">$a=1$</span>.<br>
(b) In the case <span class="math-container">$k=0$</span>, there is one more solution apart from two solutions covered in (a), and this extra solution is given by
<span class="math-container">$$f(n)=\left\{\begin{array}{ll}1\,,&\text{if }n=1\,,\\0\,,&\text{for }n=2,3,4,\ldots\,.\end{array}\right.$$</span><br>
(c) In the case <span class="math-container">$k=1$</span>, there are two more solutions apart from the one solution covered in (a). These solutions are the linear function
<span class="math-container">$$f(n)=n+1\text{ for all }n\in\mathbb{Z}_{>0}$$</span> and the alternating function <span class="math-container">$$f(n)=1-(n\,\text{mod}\,2)=\left\{\begin{array}{ll}0\,,&\text{if }n\text{ is odd}\,,\\
1\,,&\text{if }n\text{ is even}\,.
\end{array}\right.$$</span></p>
</blockquote>
|
230,126 | <p>A family of functions is known as <span class="math-container">$\left(\varphi_{0}, \varphi_{1}, \cdots, \varphi_{n}\right)$</span>.</p>
<p>I'd like to know how to express their inner product conveniently as follows:</p>
<p><span class="math-container">$$\left(\begin{array}{cccc}
\left(\varphi_{0}, \varphi_{0}\right) & \left(\varphi_{0}, \varphi_{1}\right) & \cdots & \left(\varphi_{0}, \varphi_{n}\right) \\
\left(\varphi_{1}, \varphi_{0}\right) & \left(\varphi_{1}, \varphi_{1}\right) & \cdots & \left(\varphi_{1}, \varphi_{n}\right) \\
\vdots & \vdots & & \vdots \\
\left(\varphi_{n}, \varphi_{0}\right) & \left(\varphi_{n}, \varphi_{1}\right) & \cdots & \left(\varphi_{n}, \varphi_{n}\right)
\end{array}\right)$$</span></p>
<p>Where <span class="math-container">$(f(x), g(x))$</span> is the inner product:
<span class="math-container">$(f(x), g(x))=\int_{a}^{b} f(x) g(x) \mathrm{d} x$</span></p>
<p>We can take <span class="math-container">$\{1,x,x^2,x^3,x^4\}$</span> and <span class="math-container">$\{a=-1,b=1\}$</span> as an example to realize the above requirements.</p>
<pre><code>Outer[Integrate[#1*#2, {x, -1, 1}] &, {1, x, x^2, x^3}, {1, x, x^2, x^3}]
</code></pre>
<p>I wonder if there are any other ways to achieve this?</p>
| A little mouse on the pampas | 42,417 | <p>The results of the two methods are the same <span class="math-container">$\color{Gray} {\text{(2001 武汉 岩石 数值分析 4)}} $</span>:</p>
<pre><code>ClearAll["Global`*"]
f[x_] := x^3
L[a_, b_, c_] := Integrate[(f[x] - (a*x^2 + b*x + c))^2, {x, 0, 1}]
StagnationPoints =
Solve[{D[L[a, b, c], a] == 0, D[L[a, b, c], b] == 0,
D[L[a, b, c], c] == 0}]
Print[Style["The best quadratic approximation polynomial is：", Red, Bold],
a*x^2 + b*x + c /. Flatten[StagnationPoints]]
LinearSolve[
Outer[Integrate[#1*#2, {x, 0, 1}] &, {1, x, x^2}, {1, x, x^2}],
Map[Integrate[#1*x^3, {x, 0, 1}] &, {1, x, x^2}]].{1, x, x^2}
</code></pre>
|
2,304,332 | <p>This is related to <a href="https://math.stackexchange.com/questions/2162294/prove-that-there-exists-f-g-mathbbr-to-mathbbr-such-that-fgx-i/2294324#2294324">Prove that there exists $f,g : \mathbb{R}$ to $\mathbb{R}$ such that $f(g(x))$ is strictly increasing and $g(f(x))$ is strictly decreasing.</a></p>
<p>But according to the proof of Ewan Delanoy, you must use a $p(x)=kx$, $q(x)=-kx$, where $k\neq1$, otherwise the iterative group is trivial, then the proof failed. I think it is still possible to construct such $f,g$ that their compositions are exactly $y=\pm x$.</p>
| robjohn | 13,854 | <p>Consider
$$
g(x)=g(f(g(x)))=-g(x)
$$</p>
|
2,547,888 | <p>A $\subset$ B, A and B closed then B-A is open</p>
<p>How can I prove this statement? Is that correct? I need it because is used to prove that a closed subset of a compact space is compact. </p>
<p>Imagine B=[0,4] and A=[1,2] contained in A, then B-A is neither open or closed, what am I doing wrong??</p>
<p>PS: maybe I need B compact in the statement, but the example works in every case.</p>
| Mr. X | 508,297 | <p>If you are considering $B$ as your ambient space, then $B-A$ would definitely be open but if your ambient space is something else then $B-A$ may or may not be open.</p>
|
2,547,888 | <p>A $\subset$ B, A and B closed then B-A is open</p>
<p>How can I prove this statement? Is that correct? I need it because is used to prove that a closed subset of a compact space is compact. </p>
<p>Imagine B=[0,4] and A=[1,2] contained in A, then B-A is neither open or closed, what am I doing wrong??</p>
<p>PS: maybe I need B compact in the statement, but the example works in every case.</p>
| Believer | 409,319 | <p>Remember this:
If A is open and B is closed,then,</p>
<p>$1.$A-B is $\color{red}{open}$</p>
<p>$2.$B-A is $\color{red}{closed}$</p>
|
176,987 | <p>Consider a random $m$ by $n$ partial circulant matrix $M$ whose entries are chosen independently and uniformly from $\{0,1\}$ and let $m < n$. Now consider a random $n$ dimensional vector $v$ whose entries are also chosen independently and uniformly from $\{0,1\}$. Let $N = Mv$ where multiplication is performed over the reals.</p>
<p>The matrices $M$ and $N$ are discrete random variables. Recall that the Shannon entropy for a discrete random variable $Z$ is $H(Z) = -\sum_z P(Z=z)\log_2{P(Z=z)}$. In the case where $P(Z=z)=0$ for some values $z$, the corresponding term in the sum is taken to be $0$.</p>
<p>We therefore know that the (base $2$) Shannon entropy $H(M) = H(v) = n$. The fact that $H(M) = n$ is a direct result of the fact that the entire matrix is defined by its first row.</p>
<p>If $m = \lfloor 10n/\ln{n} \rfloor$ then I would like to make the following conjecture.</p>
<blockquote>
<p><strong>Conjecture:</strong> For all sufficiently large $n$, $H(N) \geq n/10$.</p>
</blockquote>
<p>The value $10$ is chosen somewhat arbitrarily to be a sufficiently large constant.</p>
<p>Is this a known problem and/or can anyone see a way to approach it? Is it in fact true?</p>
| Samuel Lelièvre | 38,711 | <p>One approach is computer exploration: write a program to plot a few points, and plot of the functions you want to compare to on the same graphics.</p>
<p>I wrote a quick Sage program to compute and plot the entropy H(Z(n,m)) for all 0 <= m <= n <= 10, and I added the plots of (x=t, y=t, z=t) and (x=t, y=sqrt(t), z=t/10).
You can add a plot for (x=t, y=t/ln(t), z=t/10), and/or modify the others to find something closer to the graph.</p>
<p>You can <a href="http://sagecell.sagemath.org/?z=eJx1U02PmzAQvSPxH3yp1iQum3RvkXJq1b1WvSK0crHBVvBHjdk1_PqOgexCsvXB8ryZeTPzGGpnFKpM2_LKS6M7JJU1ziPGa9q3nsnKp0magIkEZtkpTRCcTqMzOrjFaFdGbRzSSGrEcum5e6Vtzzt8zYtH1kivzHh0pPv94yfW2dYB1PtzdO_gylvTfMO3EXqOmFHg7tbkjvveacCWXPQVKB-7JXrxQuHcaB6Zl0EFNLKwhNhacFQ3_L27oNfg_rjAlQX0O3Wed5LqX86wvvJ4h_HBkaPLSLa7EjAGkcVKYiy1zybtXqJ2UKD8kHOIUGVXY0U03KHxjEDse9ty3PUKhwLL_SX7osvdUMhyypNzgbnaZTG2JNGjlj5u-OfuC1UWY3FSZRnFP7qNnEXck4mDTXvAyqirECLODG-taIDn8ZAmHy0sWoLrKjwk5NRarhmG73HJpq9jryTTmhHghGSue8Ud9RwCxTV9moGgu4D1Jtr3ClgTRcRco47dtVT9YRSFEwqADBuk--s8DhBbj9vIxzhTmjR3BM1wh4y3SJo814BZ6qji3snqxbbGP0FvdSD1QOoxI_hAdEbgbzXu_OA4e4AmnpvPs5pAmoE0_8uKQloDawexVpBOjvz8FB2AQyN7BLxwCUDyTpg3_Cr5GwcCTysxGD2RzB54_QP6tSiG&lang=sage" rel="nofollow">find it and try it on the Sage Cell Server</a>.</p>
<p>For n > 10 plotting points from exhaustive computations becomes very slow...</p>
<p>Massive speedups are certainly possible with a better program and/or using Cython or a faster programming language,
but in any case the number of cases grows so fast that more efficient code will only get you so far.</p>
<p>The next trick is to replace exhaustive enumeration by random sampling, which might give you a rough idea of the graph in a larger range.</p>
<p>Of course, besides this exploration, a clever way to estimate H(Z(n,m)), in general or along a curve (n, m(n)), would better answer your question, but failing that, pictures are always nice.</p>
<p>In case it's any interest, Here is the Sage code I wrote.</p>
<p>from collections import defaultdict</p>
<pre><code>def h(d):
sn = 0r
sl = 0r
for n in d.itervalues():
if n:
nn = RDF(n)
sl += nn * nn.log2()
sn += nn
if sn:
return sn.log2() - sl/sn
return RDF.one()
def hh(n):
xn = xrange(n)
xnn = xrange(n+1)
cp = CartesianProduct(*((0r,1r),)*n)
dd = [defaultdict(int) for _ in xnn]
for y in cp:
for x in cp:
z = tuple(sum(x[(i+k)%n]*y[i] for i in xn) for k in xn)
for m in xnn:
dd[m][z[:m]] += 1r
return [h(d) for d in dd]
hhh = []
nmax = 10
for k in xrange(nmax):
hhh.append(hh(k))
ph = []
for n,hh in enumerate(hhh):
for m, h in enumerate(hh):
ph.append((n,m,h))
fx = lambda x: x
fy = lambda x: sqrt(x)
fz = lambda x: x/10
gx = lambda x: x
gy = lambda x: x
gz = lambda x: x
Gf = parametric_plot3d((fx,fy,fz),(0,n),color='red')
Gg = parametric_plot3d((gx,gy,gz),(0,n),color='red')
Gh = point3d(ph,size=3)
G = Gf + Gg + Gh
G.show(viewer='tachyon')
G.show()
</code></pre>
|
176,987 | <p>Consider a random $m$ by $n$ partial circulant matrix $M$ whose entries are chosen independently and uniformly from $\{0,1\}$ and let $m < n$. Now consider a random $n$ dimensional vector $v$ whose entries are also chosen independently and uniformly from $\{0,1\}$. Let $N = Mv$ where multiplication is performed over the reals.</p>
<p>The matrices $M$ and $N$ are discrete random variables. Recall that the Shannon entropy for a discrete random variable $Z$ is $H(Z) = -\sum_z P(Z=z)\log_2{P(Z=z)}$. In the case where $P(Z=z)=0$ for some values $z$, the corresponding term in the sum is taken to be $0$.</p>
<p>We therefore know that the (base $2$) Shannon entropy $H(M) = H(v) = n$. The fact that $H(M) = n$ is a direct result of the fact that the entire matrix is defined by its first row.</p>
<p>If $m = \lfloor 10n/\ln{n} \rfloor$ then I would like to make the following conjecture.</p>
<blockquote>
<p><strong>Conjecture:</strong> For all sufficiently large $n$, $H(N) \geq n/10$.</p>
</blockquote>
<p>The value $10$ is chosen somewhat arbitrarily to be a sufficiently large constant.</p>
<p>Is this a known problem and/or can anyone see a way to approach it? Is it in fact true?</p>
| domotorp | 955 | <p>This is not an answer, just a long comment.</p>
<p>As pointed out by John Mangual, this game is similar to Mastermind.
In fact, it is even more similar to the game defined by Erdos and Renyi here:
<a href="http://www.renyi.hu/~p_erdos/1963-12.pdf" rel="nofollow">http://www.renyi.hu/~p_erdos/1963-12.pdf</a>.
In your language their game is the following:
We are given the first row of the matrix $M$, denoted by $M'$, (so $m=1$) but what we multiply it with, $X$, is not a vector, but an $n\times a$ matrix.
Our goal is to reconstruct $M$ from $N=M'X$. They show that this can be done whp if $X$ is a random matrix and $a\ge 10n/\log n$.
Of course if $M$ can be reconstructed whp, that also implies that $H(N)\ge (1-\epsilon)n$.</p>
<p>So one possible way to answer your question would be to show that their theorem holds even in the case when $X$ is a circulant matrix, as this is equivalent to $X$ being a vector and $M$ being circulant.</p>
<p><strong>new part</strong></p>
<p>So suppose we want to compute the probability that for two different random vectors, denoted by $v$ and $u$, multiplying them with the rotations of a random vector $r$ we get the same values, i.e., if the rotations of $r$ are denote by $r_1,\ldots,r_k$, then what is the chance that for all $i$ we have $vr_i=ur_i$.
For any $i$, this is like a random walk, as $v$ and $u$ are also random, so $Pr[vr_i=ur_i]\approx 1/\sqrt n$.
Now I claim that this statement is true even in the following conditional form if $k$ is small enough: $Pr[vr_i=ur_i\mid \forall j\ne i: vr_j=ur_j]\approx 1/\sqrt n.$</p>
<p>I don't think this is that hard to prove, but I cannot see a solution now.</p>
|
853,878 | <p>I'm learning calc and after learning about how to differentiate using product rule and chain rule etc. I came across marginal cost and marginal revenue. I'm pretty familiar with cost, profit and revenue. Why would I want to find the marginal revenue (aka rate of change for revenue or cost)? What does this rate of change give me? Without getting too mechanical and mathematical, can someone explain why I want to know the so called rate of change?</p>
| Kamster | 159,813 | <p>Think of it in the case of maximizing profit. Let profit be represented by $\pi$. we see $\pi$ is generally a function of our cost(C) and revenue (R) since our profit is just revenue minus cost. Letting $x$ be the goods we are selling we see revenue and cost are a function of $x$ (i.e. $R(x)$ and $C(x)$).Thus
$$\pi(C(x),R(x))=R(x)-C(x)$$
Well to find amount we should sell to maximize profit we just use calculus to find critical point
$$\frac{d}{dx}\pi(C(x),R(x))=\frac{d}{dx}\left(R(x)-C(x)\right)=\frac{dR}{dx}-\frac{dC}{dx}=0$$
Thus maximum point will be at (of course also have check is maximum with second derivative but I leave you to see that)
$$\frac{dR}{dx}=\frac{dC}{dx}$$
Well $\dfrac{dR}{dx}$ is just marginal revenue (MR) and $\dfrac{dC}{dx}$ is just marginal Cost (MC) so we maximize profit when MC=MR. Which should make sense since as long as when we increase x marginally if revenue generated is greater than cost acquired then we will keep increasing x we do this until the marginal revenue earned is equal to marginal cost (because after words the cost acquired will be greater than revenue thus resulting in negative profit)</p>
<p>This is just a glimpse why we want to look at marginal revenue. It basically comes down to that much economic and finical thinking deals with efficiency and optimization. Where these problem can be solved using calculus (thinking at the margin) </p>
|
1,303,183 | <p>If I have a vector space $V$ ( of dimension $n$ ) over real numbers such that $\{v_1,v_2...v_n\}$ is the basis for the space ( not orthogonal ). Then I can write any vector $l$ in this space as $l=\sum_i\alpha_iv_i$. Here $\alpha_1,\alpha_2...\alpha_n$ are the coefficients that define the vector $l$ according to this basis. Can another set of coefficients $\beta_1,\beta_2...\beta_n$ give the same vector $l$ ? If the basis was orthogonal the answer would be no, but I can't prove for a non orthogonal basis.</p>
| Matthias Kaul | 244,244 | <p>They are unique. Think about what happens if they aren't and you subtract two different representations from each other.</p>
|
3,259,193 | <p>I have a simple question about notation regarding limits, specifically, <span class="math-container">$$\lim_{\|x\| \rightarrow \infty}f(x).$$</span> </p>
<p><strong>Question:</strong> </p>
<p><span class="math-container">$\lim_{\|x\| \rightarrow \infty}f(x)$</span>:</p>
<p>In words what we are doing is taking the limit as the "norm" of the point <span class="math-container">$x$</span> goes to infinity. </p>
<p>My problem here is I want to make sure I am interpreting the idea behind it correctly. So with taking limits one will encounter an expression of the form: <span class="math-container">$$\lim_{x \rightarrow \infty}f(x).$$</span> Here I would visualize our value of <span class="math-container">$x$</span> just tending towards "infinity" on a graph. But I'm having trouble visualizing the behaviour in this form <span class="math-container">$$\lim_{\|x\| \rightarrow \infty}f(x).$$</span></p>
<p>What it says to me is that the "distance" of the <span class="math-container">$x$</span> value is going to infinity. So would a way to visualize it be if we had a fixed point <span class="math-container">$x_0$</span> on the number line and we kept on measuring the distance from this fixed point <span class="math-container">$x_0$</span> to some arbitrary point <span class="math-container">$x$</span> that is really far away (infinity away) from <span class="math-container">$x_0$</span>? And if this is a valid way of thinking about it, what is the benefit of writing it in this form versus the other way mentioned? </p>
| copper.hat | 27,978 | <p>It is just notation. We say <span class="math-container">$\lim_{\|x\| \to \infty} f(x) = L$</span> <strong>iff</strong>
for any <span class="math-container">$\epsilon>0$</span> there is some <span class="math-container">$B$</span> such that if <span class="math-container">$\|x\| > B$</span> then <span class="math-container">$|f(x)-L| < \epsilon$</span>.</p>
|
588,488 | <p>I know that for the harmonic series
$\lim_{n \to \infty} \frac1n = 0$ and
$\sum_{n=1}^{\infty} \frac1n = \infty$.</p>
<p>I was just wondering, is there a sequence ($a_n =\dots$) that converges "faster" (I am not entirely sure what's the exact definition here, but I think you know what I mean...) than $\frac1n$ to $0$ and its series $\sum_{n=1}^{\infty}{a_n}= \infty$?</p>
<p>If not, is there proof of that?</p>
| Andrés E. Caicedo | 462 | <p>There is no slowest divergent series. Let me take this to mean that given any sequence <span class="math-container">$a_n$</span> of positive numbers converging to zero whose series diverges, there is a sequence <span class="math-container">$b_n$</span> that converges to zero faster and the series also diverges, where "faster" means that <span class="math-container">$\lim b_n/a_n=0$</span>. In fact, given any sequences of positive numbers <span class="math-container">$(a_{1,n}), (a_{2,n}),\dots$</span> with each <span class="math-container">$\sum_n a_{i,n}=\infty$</span> and <span class="math-container">$\lim_n a_{i+1,n}/a_{i,n}=0$</span>, there is <span class="math-container">$(a_n)$</span> with <span class="math-container">$\sum a_n=\infty$</span> and <span class="math-container">$\lim_n a_n/a_{i,n}=0$</span> for all <span class="math-container">$i$</span>.</p>
<p>To see this, given <span class="math-container">$a_1,a_2,\dots$</span>, first define <span class="math-container">$b_1=a_1,b_2=a_2,\dots,b_k=a_k$</span> until <span class="math-container">$a_1+\dots+a_k>1$</span>. Second, let <span class="math-container">$b_{k+1}=a_{k+1}/2,b_{k+2}=a_{k+2}/2,\dots,b_n=a_n/2$</span> until <span class="math-container">$a_{k+1}+\dots+a_n>2$</span>, etc. That is, we proceed recursively; if we have defined <span class="math-container">$b_1,\dots,b_m$</span> and <span class="math-container">$b_m=a_m/2^r$</span>, and <span class="math-container">$b_1+\dots+b_m>r+1$</span>, let <span class="math-container">$b_{m+1}=a_{m+1}/2^{r+1},\dots,b_l=a_l/2^{r+1}$</span> until <span class="math-container">$a_{m+1}+\dots+a_l>2^{r+1}$</span>. The outcome is that <span class="math-container">$\sum b_i=\infty$</span> and <span class="math-container">$\lim b_i/a_i=0$</span>.</p>
<p>Similarly, given <span class="math-container">$(a_{1,n}),(a_{2,n}),\dots$</span>, with each <span class="math-container">$(a_{k+1,n})$</span> converging to zero faster than <span class="math-container">$(a_{k,n})$</span>, and all of them diverging, let <span class="math-container">$a_i=a_{1,i}$</span> for <span class="math-container">$i\le n_1$</span>, where <span class="math-container">$a_{1,1}+\dots+a_{1,n_1}>1$</span>, then <span class="math-container">$a_i=a_{2,i}$</span> for <span class="math-container">$n_1<i\le n_2$</span>, where we ask both that <span class="math-container">$a_{2,n_1+1}+\dots+a_{2,n_2}>1$</span> and that for any <span class="math-container">$k>n_2/2$</span> we have <span class="math-container">$a_{2,k}/a_{1,k}<1/2$</span>, etc. That is, if we have defined <span class="math-container">$n_k$</span>, we let <span class="math-container">$a_i=a_{k+1,i}$</span> for <span class="math-container">$n_k<i\le n_{k+1}$</span> where <span class="math-container">$n_{k+1}$</span> is chosen so that <span class="math-container">$a_{k+1,n_k+1}+\dots+a_{k+1,n_{k+1}}>1$</span> and for all <span class="math-container">$l>n_{k+1}/2$</span> we have <span class="math-container">$a_{k+1,l}/a_{i,l}<1/2^{k+1}$</span> for all <span class="math-container">$i<k+1$</span>. Then the series <span class="math-container">$\sum a_i$</span> diverges, and the sequence <span class="math-container">$(a_i)$</span> converges to <span class="math-container">$0$</span> faster than all the <span class="math-container">$a_{i,n}$</span>. In modern language, we would say that there are no <span class="math-container">$(\omega,0)$</span>-gaps in a certain partial order.</p>
<p>We can modify the above slightly so that given any sequences <span class="math-container">$(a_{i,n})$</span> with <span class="math-container">$\sum_n a_{i,n}<\infty$</span> and <span class="math-container">$\lim_n a_{i+1,n}/a_{i,n}=\infty$</span> for all <span class="math-container">$i$</span>, we can find <span class="math-container">$(a_n)$</span> with <span class="math-container">$\sum_n a_n<\infty$</span> and <span class="math-container">$\lim_n a_n/a_{i,n}=\infty$</span>, so there is no fastest convergent series, and not even considering a sequence of faster and faster convergent series is enough. (In modern terms, there is no <span class="math-container">$(0,\omega)$</span>-gap.)</p>
<p>What we <em>cannot do</em> in general is, given <span class="math-container">$(a_{i,n})$</span>, with all <span class="math-container">$\sum_n a_{i,n}=\infty$</span>, find <span class="math-container">$(a_n)$</span> with <span class="math-container">$\sum a_n=\infty$</span> and <span class="math-container">$a_n/a_{i,n}\to0$</span> for all <span class="math-container">$i$</span>, if the <span class="math-container">$a_{i,n}$</span> are not ordered so that <span class="math-container">$a_{i+1,n}$</span> converges to zero faster than <span class="math-container">$a_{i,n}$</span>. For example, we can have <span class="math-container">$a_n=1/n$</span> if <span class="math-container">$n$</span> is odd and <span class="math-container">$a_n=1/n^2$</span> if <span class="math-container">$n$</span> is even, and <span class="math-container">$b_n=1/n^2$</span> if <span class="math-container">$n$</span> is odd and <span class="math-container">$b_n=1/n$</span> if <span class="math-container">$n$</span> is even, and if <span class="math-container">$c_n$</span> converges to zero faster than both, then <span class="math-container">$\sum c_n$</span> converges. (These exceptions can typically be fixed by asking monotonicity of the sequences, which is how these results are usually presented in the literature.)</p>
<p>Note that the argument I gave is completely general, no matter what the series involved. For specific series, of course, nice "formulas" are possible. For example, given <span class="math-container">$a_n=1/n$</span>, we can let <span class="math-container">$b_n=1/(n\log n)$</span> for <span class="math-container">$n>1$</span>. Or <span class="math-container">$c_n=1/(n\log n\log\log n)$</span>, for <span class="math-container">$n\ge 3$</span>. Or ... And we can then "diagonalize" against all these sequences as indicated above.</p>
<p>By the way, the first person to study seriously the boundary between convergence and divergence is <a href="http://en.wikipedia.org/wiki/Paul_du_Bois-Reymond" rel="noreferrer">Paul du Bois-Reymond</a>. He proved a version of the result I just showed above, that no "decreasing" sequence of divergent series "exhausts" the divergent series in that we can always find one diverging and with terms going to zero faster than the terms of any of them. A nice account of some of his work can be found in the book <a href="http://www.gutenberg.org/ebooks/38079" rel="noreferrer"><strong>Orders of Infinity</strong></a> by Hardy. Du Bois-Reymond's work was extended by Hadamard and others. What Hadamard proved is that given <span class="math-container">$(a_i)$</span> and <span class="math-container">$(b_i)$</span> with <span class="math-container">$\sum a_i=\infty$</span>, <span class="math-container">$\sum b_i<\infty$</span>, and <span class="math-container">$b_i/a_i\to 0$</span>, we can find <span class="math-container">$(c_i),(d_i)$</span> with <span class="math-container">$c_i/a_i\to0$</span>, <span class="math-container">$b_i/d_i\to 0$</span>, <span class="math-container">$d_i/c_i\to0$</span>, <span class="math-container">$\sum c_i=\infty$</span>, <span class="math-container">$\sum d_i<\infty$</span>. More generally:</p>
<blockquote>
<p>If we have two sequences of series, <span class="math-container">$(a_{1,n}), (a_{2,n}),\dots$</span> and <span class="math-container">$(b_{1,n}),(b_{2,n}),\dots$</span>, such that</p>
<ul>
<li>Each <span class="math-container">$(a_{i+1,n})$</span> converges to zero faster than the previous one,</li>
<li>Each <span class="math-container">$(b_{i+1,n})$</span> converges to zero slower than the previous one,</li>
<li>Each <span class="math-container">$(a_{i,n})$</span> converges to zero slower than all the <span class="math-container">$(b_{j,n})$</span>,</li>
<li>Each <span class="math-container">$\sum_n a_{i,n}$</span> diverges, and</li>
<li>Each <span class="math-container">$\sum_n b_{i,n}$</span> converges,</li>
</ul>
<p>then we can find sequences <span class="math-container">$(c_n),(d_n)$</span>, "in between", with one series converging and the other diverging.</p>
</blockquote>
<p>In modern language, we say that there are no <span class="math-container">$(\omega,\omega)$</span>-gaps, and similarly, there are no <span class="math-container">$(\omega,1)$</span>- or <span class="math-container">$(1,\omega)$</span>-gaps. This line of research led to some of Hausdorff's deep results in set theory, such as the existence of so-called <a href="https://www.encyclopediaofmath.org/index.php/Hausdorff_gap" rel="noreferrer"><span class="math-container">$(\omega_1,\omega_1)$</span>- or <em>Hausdorff</em> gaps</a>. What Hausdorff proved is that this "interpolation" process, which can be iterated countably many times, cannot in general be carries out <span class="math-container">$\omega_1$</span> times, where <span class="math-container">$\omega_1$</span> is the first uncountable ordinal.</p>
|
268,635 | <p>Consider the triangle formed by randomly distributing three points on a circle. What is the probability of the center of the circle be contained within the triangle?</p>
| Charles E. Leiserson | 629,584 | <p>Here's a rigorous solution that doesn't require calculus or reasoning about distributions.</p>
<p>Let's pick the first point <span class="math-container">$A_0$</span> at random, then pick two <strong>diameters</strong> <span class="math-container">$B$</span> and <span class="math-container">$C$</span> (instead of two points) at random, and then pick which endpoint of each diameter at random by flipping a coin for each. That gives the same distribution as picking three points at random.</p>
<p>Going clockwise from <span class="math-container">$A_0$</span>, label the endpoints of the two diameters by <span class="math-container">$B_0$</span>, <span class="math-container">$C_0$</span>, <span class="math-container">$B_1$</span>, and <span class="math-container">$C_1$</span> in order. Now, there are four ways the coin flips could land to pick the endpoints of B and C, yielding the four possible triangles <span class="math-container">$(A_0, B_0, C_0), (A_0, B_0, C_1), (A_0, B_1, C_0)$</span>, or <span class="math-container">$(A_0, B_1, C_1)$</span>. Each of these outcomes is equally likely (probability <span class="math-container">$1/4$</span>) after picking <span class="math-container">$A_0$</span>. </p>
<p>Of these, the only one whose triangle contains the center is <span class="math-container">$(A_0, B_1, C_0)$</span> — and I found this interesting — regardless of the relative angles of the diameters. Triangle <span class="math-container">$(A_0, B_0, C_0)$</span> cannot contain the center, because all three points lie on the same semicircle starting at <span class="math-container">$A_0$</span> and going clockwise, which means that <span class="math-container">$B_0$</span> is an obtuse angle. Likewise, <span class="math-container">$(A_0, B_1, C_1)$</span> is out, because all three points lie on the same semicircle ending at <span class="math-container">$A_0$</span>, which means that <span class="math-container">$B_1$</span> is an obtuse angle. Finally, <span class="math-container">$(A_0, B_0, C_1)$</span> all land in the semicircle centered at <span class="math-container">$A_0$</span>, meaning that <span class="math-container">$A_0$</span> is an obtuse angle. But <span class="math-container">$(A_0, B_1, C_0)$</span> contains the center, because the minor arc from <span class="math-container">$C_0$</span> to <span class="math-container">$B_1$</span> is less than <span class="math-container">$180^{\circ}$</span>, making the angle at <span class="math-container">$A_0$</span> less than <span class="math-container">$90^{\circ}$</span>, i.e., acute. </p>
<p>Thus, the answer is <span class="math-container">$1/4$</span>.</p>
|
250,484 | <p>For a fixed set $X$ and a finite collection $E_1,E_2,\ldots,E_k\subseteq X$, define the binary relation <em>adjacency</em> as follows: $E_i,E_j$ are adjacent
if their intersection is nonempty.
We term the transitive closure of this relation by
<em>transitive adjacency</em>
and define the <em>adjacent union</em> by
$$
\tilde\cup(E_1,\ldots,E_k):=
\begin{cases}
\bigcup_{i=1}^k E_i, & \text{the $(E_i)$ are transitively adjacent}
\\
\emptyset, & \text{else}
.
\end{cases}
$$</p>
<p>Are there standard terms for <em>transitive adjacency</em> and <em>adjacent union</em>?</p>
| Bjørn Kjos-Hanssen | 4,600 | <p>In philosophy, this would be called <a href="https://en.wikipedia.org/wiki/Family_resemblance" rel="noreferrer">family resemblance</a> -- if $E_i\cap E_j\ne\emptyset$ and $E_j\cap E_k\ne\emptyset$ then $E_i$ and $E_k$ have a family resemblance.</p>
<p>That is, perhaps I have no common feature with my second cousin, but we both have common features with our common great-grandparent. Similarly, there may be no single thing that is common to all "games".</p>
<p>The idea that every-day concepts such as <em>game</em> and <em>number</em> exhibit family resemblance comes from Wittgenstein's notes from 1930, so perhaps it is prior to the notion of <strong>intersection graphs</strong> as in @ViniciusdosSantos' answer, which seems to only go back to 1945.</p>
|
2,701,658 | <p>If <span class="math-container">$1^2+2^2+3^2 + ... + 10^2=385$</span>, then what is the value of <span class="math-container">$2^2+4^2+6^2 + ... + 20^2$</span>?</p>
<p>Options are <span class="math-container">$770$</span>, <span class="math-container">$1155$</span>, <span class="math-container">$1540$</span>, <span class="math-container">$7890$</span>.</p>
<p>I haven't tried it yet and I'm writing such extra things to meet the requirements to post the question...Plz solve the above question only</p>
| Mr Pie | 477,343 | <p>You are given the equation, $$1^2 + 2^2 + 3^2 +\cdots + 10^2 = 385.$$ From this equation, you want to find the value of $x$ in the following equation: $$2^2 + 4^2 + 6^2 +\cdots + 20^2 = x.$$ I know the problem you have been given does not include a variable $x$, but we will assign the unknown expression to such a variable in order to work it out algebraically. Now let's line up both the equations: $$\begin{align} 1^2 + 2^2 + 3^2 +\cdots + 10^2 &= 385\tag1 \\ 2^2 + 4^2 + 3^2 +\cdots + 20^2 &= x.\tag2\end{align}$$ Notice that every term being squared in equation $(2)$ is twice as large as every term being squared in equation $(1)$. To make that properly clear, let's re-write equation $(2)$ as follows: $$\begin{align} 1^2 + 2^2 + 3^2 +\cdots + 10^2 &= 385\tag1 \\ (2\times 1)^2 + (2\times 2)^2 + (2\times 3)^2 +\cdots + (2\times 10)^2 &= x.\tag2\end{align}$$ Now if we divide every squared term in equation $(2)$ by $2^2$, this will definitely equal equation $(1)$, which we observe being equal to $385$. We divide by $2^2$ because each term <em>being</em> squared is twice as large as each term being squared in equation $(1)$, and since we are <em>squaring</em> them, we must divide by $2^2$. But since we are adding up all the squared terms together, then by dividing each term by $2$, we are dividing the entire sum by $2$ as well. Now we let this entire sum be equal to $x$, so now we can write that $$\begin{align} \frac{x}{2^2} &= 385 \\ \\ \frac{x}{4} &= 385.\tag{since $2^2 = 4$}\end{align}$$ Remember that whatever you do to the left hand side of an equation, you must always do the same to the right hand side as well, in order for both sides to be equal. It is a bit like balanacing a seesaw, such that if you add weight on one side, you need to add the same amount of weight on the other side for the seesaw to be levelled.</p>
<p>So how does this analogy help us?</p>
<p>Well, we want to use this analogy to solve for $x$, and we can by multiplying $4$ to both sides. Since multiplication is the opposite of division, and we are <em>dividing</em> $x$ by $4$, then to just get $x$ alone, we multiply by $4$. But remember: what we do the left hand side, we must also do to the right hand side. So, we multiply the right hand side by $4$ as well, which is equal to $385$. We should now have something that looks like this: $$\begin{align} \frac{x}{4}\times 4 &= 385\times 4 \\ x &= 385\times 4\end{align}$$ When something is alone on one side, this is known as the <em>subject</em> of the equation, and is what we want to solve for. In this case, we have assigned our subject as $x$, and we see that $x = 4\times 385$. And since $385\times 4 = 1540$, we finally conclude that $$\boxed{ \ x = 1540 \ }$$ and there is your answer!</p>
|
1,058,515 | <p>determine whether $v_1=(1,1,2)$ , $v_2=(1,0,1)$ and $v_3=(2,2,3)$ span $\mathbb{R}^3$</p>
<p>I think that it must prove that it linear Independence </p>
<p>let $α_1v_1+α_2v_2+α_3v_3=(0,0,0)$</p>
<p>so what next , is this a correct way or I'm completely wrong </p>
<p>please guide me to understand this</p>
<p>thanks</p>
| JohnD | 52,893 | <p>Put their columns in a matrix $A$. You want to know if for all vectors $\mathbf{b}=(a,b,c)$ if $A\mathbf{x}=\mathbf{b}$ has a solution. Do you know how to answer that question?</p>
|
1,058,515 | <p>determine whether $v_1=(1,1,2)$ , $v_2=(1,0,1)$ and $v_3=(2,2,3)$ span $\mathbb{R}^3$</p>
<p>I think that it must prove that it linear Independence </p>
<p>let $α_1v_1+α_2v_2+α_3v_3=(0,0,0)$</p>
<p>so what next , is this a correct way or I'm completely wrong </p>
<p>please guide me to understand this</p>
<p>thanks</p>
| Empiricist | 189,188 | <p>You are on the right way. That yields</p>
<p>$$\alpha_1 (1,1,2) + \alpha_2 (1,0,1) + \alpha_3 (2,2,3) = (0,0,0)$$</p>
<p>$$
\left( \begin{array}{ccc}
1 & 1 & 2 \\
1 & 0 & 2 \\
2 & 1 & 3 \end{array} \right)
\left( \begin{array}{c}
\alpha_1 \\
\alpha_2 \\
\alpha_3 \end{array} \right)
=
\left( \begin{array}{c}
0\\
0\\
0\end{array} \right)$$</p>
<p>which is a matrix equation. You can solve the system of linear equation or just see whether the determinant is $0$. Since this system must have a trivial solution, either it has a unique solution (hence the trivial one and thus the set is linearly independent) or infinitely many solutions (hence the set is linearly dependent).</p>
|
3,080,664 | <p>Hi I've almost completed a maths question I am stuck on I just can't seem to get to the final result. The question is:</p>
<p>Find the Maclaurin series of <span class="math-container">$g(x) = \cos(\ln(x+1))$</span> up to order 3.</p>
<p>I have used the formulas which I won't type out as I'm not great with Mathjax yet sorry. But I have obtained:</p>
<p><span class="math-container">$$ \ln(x+1) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} $$</span></p>
<p><span class="math-container">$$ \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} $$</span></p>
<p>I'm sure what I have done above is correct however I can't seem to get to the final solution shown in the answers and would like some help please. Thanks </p>
| heropup | 118,193 | <p>Since you also don't need very high orders, the straightforward calculation of derivatives is tractable, though not preferable computationally. If <span class="math-container">$f(x) = \cos \log(1+x)$</span>, then <span class="math-container">$$f'(x) = -\frac{\sin \log (1+x)}{1+x}, \quad f'(0) = 0.$$</span> Then <span class="math-container">$$f''(x) = -\frac{\cos \log (1+x)}{(1+x)^2} + \frac{\sin \log(1+x)}{(1+x)^2}, \quad f''(0) = -1.$$</span> Finally, <span class="math-container">$$f'''(x) = \frac{\sin \log(1+x)}{(1+x)^3} + \frac{2\cos \log(1+x)}{(1+x)^3} + \frac{\cos \log(1+x)}{(1+x)^3} - \frac{2 \sin \log (1+x)}{(1+x)^3}, \quad f'''(0) = 3.$$</span> Then <span class="math-container">$$f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + O(x^4) = 1 - \frac{x^2}{2} + \frac{x^3}{2} + O(x^4).$$</span>
It is worth noting that <span class="math-container">$$f^{(n)}(x) = \frac{A_n \cos \log (1+x) + B_n \sin \log (1+x)}{(1+x)^n},$$</span> for suitable constants <span class="math-container">$A_n$</span>, <span class="math-container">$B_n$</span>. A proof by induction is straightforward and yields insight into the recursion relations defining <span class="math-container">$\{(A_n, B_n)\}_{n \ge 1}$</span>: <span class="math-container">$$\begin{align*} f^{(n+1)}(x) &= -A_n \frac{\sin \log (1+x)}{(1+x)^{n+1}} - nA_n \frac{\cos \log (1+x)}{(1+x)^{n+1}} - n B_n \frac{\sin \log(1+x)}{(1+x)^{n+1}} + B_n \frac{\cos \log (1+x)}{(1+x)^{n+1}} \\
&= \frac{(B_n - nA_n) \cos \log(1+x) + (-A_n - nB_n)\sin \log(1+x)}{(1+x)^{n+1}}.
\end{align*}.$$</span> Therefore, <span class="math-container">$$\begin{align*} A_{n+1} & = -nA_n + B_n, \\ B_{n+1} &= -A_n - nB_n, \\ A_0 &= 1, \\ B_0 &= 0. \end{align*}$$</span> In matrix form, this recurrence is equivalent to <span class="math-container">$$\begin{bmatrix}A_{n+1} \\ B_{n+1}\end{bmatrix} = \begin{bmatrix} -n & 1 \\ -1 & -n \end{bmatrix} \begin{bmatrix} A_n \\ B_n \end{bmatrix},$$</span> consequently <span class="math-container">$$\begin{bmatrix}A_n \\ B_n\end{bmatrix} = M_n \begin{bmatrix} 1 \\ 0 \end{bmatrix},$$</span> where <span class="math-container">$$M_n = \prod_{k=0}^{n-1} \begin{bmatrix} -k & 1 \\ -1 & -k \end{bmatrix}$$</span> and <span class="math-container">$f^{(n)}(0) = A_n$</span>. This lets us continue our calculation of higher orders with relative ease if we keep track of the matrix product <span class="math-container">$M_n$</span>, which is always of the form <span class="math-container">$$M_n = \begin{bmatrix} a & b \\ -b & a \end{bmatrix};$$</span> for example, <span class="math-container">$$M_3 = \begin{bmatrix} 3 & 1 \\ -1 & 3\end{bmatrix}, \quad M_4 = \begin{bmatrix} -3 & 1 \\ -1 & -3\end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 3\end{bmatrix} = \begin{bmatrix} -10 & 0 \\ 0 & -10 \end{bmatrix},$$</span> so that <span class="math-container">$A_4 = -10$</span> and the next coefficient is <span class="math-container">$-10/4! = -5/12$</span>.</p>
|
3,080,664 | <p>Hi I've almost completed a maths question I am stuck on I just can't seem to get to the final result. The question is:</p>
<p>Find the Maclaurin series of <span class="math-container">$g(x) = \cos(\ln(x+1))$</span> up to order 3.</p>
<p>I have used the formulas which I won't type out as I'm not great with Mathjax yet sorry. But I have obtained:</p>
<p><span class="math-container">$$ \ln(x+1) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} $$</span></p>
<p><span class="math-container">$$ \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} $$</span></p>
<p>I'm sure what I have done above is correct however I can't seem to get to the final solution shown in the answers and would like some help please. Thanks </p>
| K B Dave | 534,616 | <p>A heuristic way using complex numbers that's related to one of the other answers: define
<span class="math-container">$$\begin{align}
f(x)&=\cos\ln(1+x) \\
g(x)&=\sin \ln(1+x)\text{.}
\end{align}$$</span>
Then
<span class="math-container">$$\begin{align}f(0)+\mathrm{i}g(0)&=1\\
(f+\mathrm{i}g)'&=\frac{\mathrm{i}(f+\mathrm{i}g)}{1+x}
\end{align}$$</span>
so the left- and right-hand sides of
<span class="math-container">$$f(x)+\mathrm{i}g(x)=(1+x)^{\mathrm{i}}$$</span>
coincide as formal power series in <span class="math-container">$x$</span>. Thus
<span class="math-container">$$\begin{split}\cos \ln(1+x)&= 1 +\Re \mathrm{i} x+\Re \tfrac{\mathrm{i}(\mathrm{i}-1)}{2} x^2+\Re \tfrac{\mathrm{i}(\mathrm{i}-1)(\mathrm{i}-2)}{6} x^3+o(x^3)\\
&=1-\tfrac{x^2}{2}+\tfrac{x^3}{2}+o(x^3)\end{split}$$</span></p>
|
4,184,248 | <p>I'm trying to prove an inequality from an old book named "Algebra and elementary functions" by Kochetkov (ru - "Алгебра и элементарные функции" Е. С. Кочетков). This is an old book from 1970 (URSS era). The book is not copyrighted anymore, and it is freely available over the internet.</p>
<p>The inequality is <span class="math-container">$\frac{a}{\sqrt{b}} + \frac{b}{\sqrt{a}} \ge \sqrt{a} + \sqrt{b}$</span> (complex numbers are not allowed)</p>
<p>Here are my steps :</p>
<p><span class="math-container">$\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}} \ge \sqrt{a} + \sqrt{b} $</span></p>
<p><span class="math-container">$\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}} - \sqrt{a}-\sqrt{b} \ge 0$</span></p>
<p><span class="math-container">$\frac{a(\sqrt{a}-\sqrt{b})+b(\sqrt{b}-\sqrt{a}))}{\sqrt{ab}} \ge 0$</span></p>
<p>It is clear that if <span class="math-container">$a=b$</span>, then the expression <span class="math-container">$a(\sqrt{a}-\sqrt{b})+b(\sqrt{b}-\sqrt{a})$</span> is zero. Also intuitively I can see that in other cases the expression <span class="math-container">$a(\sqrt{a}-\sqrt{b})+b(\sqrt{b}-\sqrt{a})$</span> is greater than 0, but how to prove it formally ?</p>
| Huy Nguyen | 922,437 | <p>This polynomial has degree 2, therefore it can not have 3 distinct roots, hence all the coefficients must be zero</p>
|
155,373 | <p>Consider $X,Y \subseteq \mathbb{N}$. </p>
<p>We say that $X \equiv Y$ iff there exists a bijection between $X$ and $Y$.</p>
<p>We say that $X \equiv_c Y$ iff there exist a bijective computable function between $X$ and $Y$. </p>
<p>Can you show me some examples in which the two concepts disagree?</p>
| William | 13,579 | <p>Let $X = \omega$, the natural number. Let $\bar{K}$ be the complement of the $K = \{e : \phi_e(e)\downarrow\}$, the halting problem set. That is, $K$ is set of $e$ such that the $e^\text{th}$ Turing Machine on input $e$ converges. </p>
<p>Suppose there exists a computable bijection from $\omega \rightarrow \bar{K}$. Then $\bar{K}$ would be a c.e. set. </p>
<p>It is well known that $K$ is not computable, but c.e. Therefore, $\bar{K}$ is not c.e. (This follows from the fact that if a set $A$ and $\bar{A}$ are c.e., then $A$ is computable.)</p>
<p>Of course since $\bar{K}$ is an infinite subset of $\omega$, there is clearly a bijection between $\omega$ and $\bar{K}$. </p>
<hr>
<p>Also I am not sure if $\equiv_c$ as you defined it is even an equivalence relation. If $f: X \rightarrow Y$ is a computable bijection, we understand "computable" here to mean that there is some $f' : \omega \rightarrow \omega$ which is computable in the ordinary sense such that the restriction of $X$ gives the bijection $f: X \rightarrow Y$. The only problem is that do you want $f' : \omega \rightarrow \omega$ to be bijective, i.e. what is called a computable permutation. If not, I am not sure if $\equiv_c$ is symmetric. However, if you demand that $f' : \omega \rightarrow \omega$ also be bijective, then the relation is symmetric with the inverse being the witness. </p>
|
44,306 | <p>This morning I realized I have never understood a technical issue about Cauchy's theorem (homotopy form) of complex analysis. To illustrate, let me first give a definition.</p>
<p>(In what follows $\Omega$ will always denote an open subset of the complex plane.)</p>
<p><em>Definition</em> Let $\gamma, \eta\colon [0, 1]_t \to \Omega$ be two piecewise smooth closed curves (or <em>circuits</em>). We say that $\gamma, \eta$ are $\Omega$-<em>homotopic</em> if there exists a continuous mapping $H \colon [0, 1]_\lambda \times [0, 1]_t \to \Omega$ s.t. </p>
<ol>
<li>$H(0, t)=\gamma(t)$ and $H(1, t)=\eta(t), \quad \forall t \in [0, 1]$;</li>
<li>$H(\lambda, 0)=H(\lambda, 1), \quad \forall \lambda \in [0, 1]$.</li>
</ol>
<p><strong>Theorem</strong> (Cauchy) Let $f\colon \Omega \to \mathbb{C}$ be holomorphic. If $\gamma, \eta$ are $\Omega$-homotopic circuits, then </p>
<p>$$\int_{\gamma} f(z)\, dz= \int_{\eta}f(z)\, dz.$$</p>
<blockquote>
<p><strong>Problem</strong> The function $H$ above is only continuous and need not be smooth. So for $0< \lambda < 1$, the closed curve $H(\lambda, \cdot)$ may be pretty much everything (a Peano curve, for example). Does this void the validity of theorem as it is stated above? How can integration be defined over such a pathological object? </p>
</blockquote>
<p>The proof of Cauchy's theorem that I have in mind goes as follows. To begin, one observes that for a sufficiently small value of $\lambda_1$, the circuits $\gamma=H(0, \cdot)$ and $H(\lambda_1, \cdot)$ are <em>close toghether</em>; that is, they can be covered by a finite sequence of disks not leaving $\Omega$ like in the following figure:</p>
<p><img src="https://i.stack.imgur.com/E4e7r.png" alt="Proof of Cauchy's theorem"></p>
<p>Since $f$ is locally exact, its integrals over every single disk depend only on the local primitive. Playing a bit with this, one arrives at </p>
<p>$$\int_\gamma f(z)\, dz= \int_{H(\lambda_1, \cdot)} f(z)\, dz.$$</p>
<p>Then one repeats this process, yielding a $\lambda_2$ greater than $\lambda_1$ and such that</p>
<p>$$\int_{H(\lambda_1,\cdot)} f(z)\, dz= \int_{H(\lambda_2, \cdot)} f(z)\, dz.$$</p>
<p>And so on. A compactness argument finally shows that this algorithm ends in a finite number of steps. </p>
<p><em>Problem is</em>: this proof assumes implicitly that $H(\lambda_1, \cdot), H(\lambda_2, \cdot) \ldots$ are piecewise smooth, to make sense of integrals $$\int_{H(\lambda_j, \cdot)}f(z)\, dz.$$
This, however, does <em>not</em> follow from the definition if $H$ is only assumed to be continuous. Therefore this proof works only for smooth $H$. </p>
<p><strong>Is this regularity condition necessary?</strong></p>
| Sam | 3,208 | <p>As has already been pointed out by Akhil in the comments, any two smooth curves $\gamma_0, \; \gamma_1$ which are continuously homotopic are also smoothly homotopic. The point here is to approximate a continuous homotopy, rather than the curves themselves.</p>
<p>More concretely, given homotopic smooth curves $\gamma_0, \gamma_1: I \to \Omega$, let $H(s,t): I\times [0,1] \to \Omega$ be a continuous homotopy between $\gamma_0$ and $\gamma_1$. </p>
<p>By the approximation theorem of your choice (mine is due to Whitney), we can find a smooth map $G: I\times[0,1] \to \Omega$, which coincides with $H$ on $I \times \{0\} \cup I \times \{1\}$ (since $H$ is smooth there).
This means that we find a <em>smooth</em> homotopy $G$ between $\gamma_0$ and $\gamma_1$, so we are done in this special case.</p>
<p>Now, if we start out with rectifiable curves rather than smooth ones, we can find two smooth curves $\tilde \gamma_0$ and $\tilde \gamma_1$, which approximate our initial $\gamma_0$ and $\gamma_1$, respectively, and such that the linear homotopy</p>
<p>$$H_\lambda(s,t) := t\gamma_\lambda(s) + (1-t)\tilde \gamma_\lambda(s) \qquad \lambda = 0, 1$$</p>
<p>maps into $\Omega$ (choose a $\epsilon$-neighborhood of the image $\gamma_\lambda(I)$ of $I$ which is contained in $\Omega$ and take $\tilde \gamma_\lambda$ to be contained within this neighborhood).</p>
<p>It is not difficult to see that $H_\lambda$ will then be a "rectifiable" homotopy. </p>
<p>But with $\tilde \gamma_0$, $\tilde \gamma_1$ we are again in the first situation, so there is a smooth homotopy between them.
Now we can build a rectifiable homotopy between $\gamma_0$ and $\gamma_1$ in three steps</p>
<ol>
<li>Homotop $\gamma_0$ to $\tilde \gamma_0$ by the linear homotopy.</li>
<li>Use a smooth homotopy between $\tilde \gamma_0$ and $\tilde \gamma_1$</li>
<li>Go from $\tilde \gamma_1$ to $\gamma_1$ by the straight line homotopy.</li>
</ol>
<p>Thus proving that all notions of "homotopic" agree.</p>
|
164,321 | <p>I'm searching for a suitable (hopefully simple enough) solution to the following form of integral:</p>
<p>$$\int_0^\infty \mathrm{d}x~x^n J_\nu(a x) J_\nu(b x) K_\mu(c x) $$</p>
<p>Where $n$, $\nu$, and $\mu$ are all integers, and $a$, $b$, and $c$ are all real and positive.</p>
<p>If not generally, a specific case would be quite helpful:</p>
<p>$$\int_0^\infty \mathrm{d}x~x^2 J_\nu(a x) J_\nu(b x) K_1(c x) $$</p>
<p>I am aware of the following:</p>
<ul>
<li>Gradshteyn & Ryzhik eq. (6.522.3), which calculates the following integral as a relatively simple function:</li>
</ul>
<p>$$\int_0^\infty \mathrm{d}x~x K_0(ax) J_\nu(b x) J_\nu(c x)$$</p>
<ul>
<li><a href="http://dx.doi.org/10.1063/1.526062?ver=pdfcov" rel="nofollow">Gervois - integrals of three Bessel functions of the first kind</a>: by manipulating the order of the last Bessel I could change the order of the last Bessel function, but I don't know how to "replace" it by $K_\mu$.</li>
<li><a href="http://dx.doi.org/10.1002/zamm.200310059" rel="nofollow">Fabrikant - Computation of infinite integrals involving three Bessel functions
by introduction of new formalism</a>: helpful formulas, but unfortunately his eq. (21) and (22) is not valid in the case above, so all the results in his discussion are useless to me as they start from eq. (21). He does use this formula to "replace" $K_\mu$ by $J_\mu$ in eq. (24), using <a href="http://dlmf.nist.gov/10.27#E9" rel="nofollow">a property relating the two functions (or at least something close to what this links to)</a>:</li>
</ul>
<p>$$ \pi \mathrm{i} J_\nu(z) = \mathrm{e}^{-\nu\pi\mathrm{i}/2} K_\nu(-\mathrm{i} z) - \mathrm{e}^{\nu\pi\mathrm{i}/2} K_\nu(\mathrm{i} z), ~~~~|~\mathrm{arg}~z~| \leq \pi/2$$</p>
<p>I simply fail to see how this formula results in a simple change in his equations, especially because of the complex nature of the arguments introduced by the above formula, and the involved definition of $K_\mu$ for integer order.</p>
<p>Should I instead be looking for specific application of G&R eq. 6.522.17-18, which could provide the required formulas, or are there better approaches to this problem? It seems that Fabrikant in the above-linked article says the validity of these formulas is more strictly bounded than shown in G&R.</p>
<p>Any help here would be much appreciated.</p>
| rubenvb | 7,606 | <p>Although I'm sure @Zurab's answer will indeed give me a solution (and I've heard the suggestion before in other situations), I'm not very familiar with the technique.</p>
<p>Following the suggestion by @Johannes in the comments, there is a much more straightforward way that, although not fully general, fits my problem exactly (i.e. it leads to expressions for the integrands I'm coming across).</p>
<p>Starting from G&R eq. (6.522.3),
$$\int_0^\infty \mathrm{d} r~ r K_0(a r) J_\nu(b r) J_\nu(c r) = \left( \frac{r_2 - r_1}{r_2 + r_1} \right)^{|\nu|} \frac{1}{r_1 r_2}$$
where $r_1^2 = a^2 + (b-c)^2$ and $r_2^2 = a^2 + (b+c)^2$, one can use the derivative and symmetry properties of the bessel-$K$ functions to change the order from 0, incrementing the power of $r$ in the integrand at the same time. Simply calculating the derivative $c~\partial_c$ of the above righthand side gives my example from the question.</p>
|
4,330,852 | <blockquote>
<p>Let <span class="math-container">$A_1 ,\ldots ,A_m$</span> be subsets of an <span class="math-container">$n$</span>-element set such that <span class="math-container">$|A_i \cap A_j |\leq t$</span> for all <span class="math-container">$i\neq j$</span>. Prove that <span class="math-container">$$\sum_{i=1}^{m}|A_i |\leq n+t{m \choose 2}$$</span></p>
</blockquote>
<p><strong>What I've tried:</strong></p>
<p>I've tried to use the following lemma to prove the fact above:</p>
<p><em>Lemma: Let <span class="math-container">$X$</span> be a set of <span class="math-container">$n$</span> elements, and let <span class="math-container">$A_1 ,\ldots ,A_m$</span> be subsets of <span class="math-container">$X$</span> of average size at least <span class="math-container">$n/w$</span>. If <span class="math-container">$N\geq 2w^2$</span>, then there exist <span class="math-container">$i\neq j$</span> such that <span class="math-container">$|A_i \cap A_j |\geq \frac{n}{2w^2}$</span>.</em></p>
<p>Now by contrary, let <span class="math-container">$\sum_{i=1}^{m}|A_i |\geq n+t{m \choose 2}+1$</span>. Then <span class="math-container">$\frac{1}{m}\sum_{i=1}^{m}|A_i |\geq \frac{n+t{m \choose 2}+1}{m}$</span>, where <span class="math-container">$\frac{1}{m}\sum_{i=1}^{m}|A_i |$</span> is the average size of <span class="math-container">$A_1 ,\ldots ,A_m$</span>. But now I don't know what <span class="math-container">$w$</span> is here. I was wondering if someone could help me.</p>
| Youem | 468,504 | <p><em>Hint</em> Use inclusion-exclusion formula to compute the cardinality of the union of <span class="math-container">$A_i$</span> and use the fact that this cardinality is at most <span class="math-container">$n$</span></p>
|
1,897,771 | <p>Let $(a, b)$ be the open interval $\left\{z\in\mathbb{R} : a < z < b\right\}$. </p>
<p>Write the theorem "If $x,y\in(a,b)$ then $|x − y| < b − a$" in logic form, and then prove the theorem.</p>
| MarnixKlooster ReinstateMonica | 11,994 | <p>Here is an alternative form of the first answer, with some suggestions on how this proof could be designed.$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\Ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\when}{\Leftarrow}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
\newcommand{\abs}[1]{\lvert #1 \rvert}
$</p>
<p>You know something about $\;x\;$ and $\;y\;$ separately, and you have to prove something about the combination of the two. Therefore seem easiest to start with the conclusion $\;\abs{x-y} < b-a\;$, and try and split this into a part about $\;x\;$ and a part about $\;y\;$.</p>
<p>Therefore we calculate:
$$\calc
\abs{x-y} < b-a
\op\equiv\hint{basic property of $\;\abs{\cdot}\;$}
x-y < b-a \;\land\; y-x < b-a
\op\when\hints{arithmetic: $\;p<q \land r<s \;\then\; p+r<q+s\;$, twice}
\hint{-- this seems the simplest way to separate $\;x\;$ and $\;y\;$}
x < b \land -y < -a \;\land\; y < b \land -x <-a
\op\equiv\hint{arithmetic, reorder}
a < x < b \;\land\; a < y < b
\op\equiv\hint{definition of open interval}
x \in (a,b) \;\land\; y \in (a,b)
\endcalc$$</p>
|
1,856,409 | <p>In business with 80 workers, 7 of them are angry. If the business leader visits and picks 12 randomly, what is the probability of picking 12 where exactly 1 is angry? </p>
<p>(7/80)<em>(73/79)</em>(72/78)<em>(71/77)</em>(70/76)<em>(69/75)</em>(68/74)<em>(67/73)</em>(66/72)<em>(65/71)</em>(64/70)*(63/69)*12=0.4134584151106464</p>
<p>What is the probability more than 2 are angry? </p>
<p>My idea is to calculate the probability of 2,3,4,5,6, and 7 angry people just like did in the previous example and then add them altogether. </p>
<p>In the previous example I can seat the one person 12 times. In all the different 12 spots, and then times by 12. The problem I have now is, how many times can I seat 2 people in 12 spots? If I use the combinatorics formula I will get a negative factorial.</p>
<p>There must be a much easier way than this.</p>
| Joffan | 206,402 | <p>The denominators of the fractions stay constant. The total multiplication across the denominators is all the ways to pick 12 people from 80, where the order is retained:
$$ 80\cdot 79\cdot 78\cdot 77\cdot 76\cdot 75\cdot 74\cdot 73\cdot 72\cdot 71\cdot 70\cdot 69 = \frac{80!}{68!}
$$</p>
<p>We say that order is unimportant, so $\frac{80!}{68!\,12!} = {80 \choose 12}$ options</p>
<p>Then the numerators are the combination of the choices from the angry $(k)$ and non-angry $(12-k)$ groups, which are ${7 \choose k}$ and ${73 \choose 12-k}$, so overall the probability is $$\frac{{7 \choose k}{73 \choose 12-k}}{80 \choose 12}$$
and checking this against your result for $k=1$ we have
$$\frac{{7 \choose 1}{73 \choose 11}}{80 \choose 12} = \frac{68!\,12!}{80!}\frac{7!}{1!\,6!}\frac{73!}{ 62!\,11!} = \frac{12\cdot 7\cdot 68\cdot 67\cdot 66\cdot 65\cdot 64\cdot 63}{80\cdot 79\cdot 78\cdot 77\cdot 76\cdot 75\cdot 74} \approx 0.413458415
$$</p>
<p>The easiest of the possible calculations is where none of the chosen employees are angry $(k=0)$, which is
$$\frac{{7 \choose 0}{73 \choose 12}}{80 \choose 12} = \frac{68!\,12!}{80!}\cdot 1 \cdot\frac{73!}{ 61!\,12!} = \frac{68\cdot 67\cdot 66\cdot 65\cdot 64\cdot 63\cdot 62}{80\cdot 79\cdot 78\cdot 77\cdot 76\cdot 75\cdot 74} \approx 0.305171687$$</p>
<p>Then all other cases can be worked similarly. </p>
<hr>
<p>You talk about getting a negative factorial when finding the combination of how to choose $2$ seats from $12$, but that is just ${12 \choose 2}= \frac {12!}{10!\,2!} = \frac{12\cdot 11}{2} = 66$ so I don't know how you arrived at a negative factorial.</p>
|
2,372 | <p>How can I find the number of <span class="math-container">$k$</span>-permutations of <span class="math-container">$n$</span> objects, where there are <span class="math-container">$x$</span> types of objects, and <span class="math-container">$r_1, r_2, r_3, \cdots , r_x$</span> give the number of each type of object?</p>
<p>I'm still looking for the solution to this more general problem out of interest.</p>
<p>Here is an example with <span class="math-container">$n = 20, k = 15, x = 4,$</span> <span class="math-container">$ r_1 = 4 \quad r_2 = 5 \quad r_3 = 8 \quad r_4 = 3$</span>.</p>
<blockquote>
<p>I have 20 letters from the alphabet. There are some duplicates - 4 of them are <em>a</em>, 5 of them are <em>b</em>, 8 of them are <em>c</em>, and 3 are <em>d</em>. How many unique 15-letter permutations can I make?</p>
</blockquote>
<h2>Edits:</h2>
<p>I've done some more work on this problem but haven't really come up with anything useful. Intuition tells me that as Douglas suggests below there will probably not be an easy solution. However, I haven't been able to prove that for sure - does anyone else have any ideas?</p>
<p>I've now re-asked this question on <a href="https://mathoverflow.net/questions/37211/permutations-with-identical-objects">MO</a>.</p>
| robjohn | 13,854 | <p>Consider the sum defined by
<span class="math-container">$$
\begin{align}
&\sum_{k=0}^\infty s_k\frac{x^k}{k!}\\
&=\textstyle\underbrace{\left(1{+}x{+}\dots{+}\frac{x^8}{8!}\right)}_{\substack{\text{the exponent of $x$}\\\text{is the number of c's}}}\underbrace{\left(1{+}x{+}\dots{+}\frac{x^5}{5!}\right)}_{\substack{\text{the exponent of $x$}\\\text{is the number of b's}}}\underbrace{\left(1{+}x{+}\dots{+}\frac{x^4}{4!}\right)}_{\substack{\text{the exponent of $x$}\\\text{is the number of a's}}}\underbrace{\left(1{+}x{+}\dots{+}\frac{x^3}{3!}\right)}_{\substack{\text{the exponent of $x$}\\\text{is the number of d's}}}\tag1
\end{align}
$$</span>
A typical term in the product that contributes to <span class="math-container">$s_{15}\frac{x^{15}}{15!}$</span> is <span class="math-container">$\frac{x^7}{7!}\cdot\frac{x^2}{2!}\cdot\frac{x^3}{3!}\cdot\frac{x^3}{3!}$</span> which adds <span class="math-container">$\frac{15!}{7!\,2!\,3!\,3!}$</span> to <span class="math-container">$s_{15}$</span>. This accounts for all permutations with <span class="math-container">$7$</span> c's and <span class="math-container">$2$</span> b's and <span class="math-container">$3$</span> a's and <span class="math-container">$3$</span> d's. Adding up over all the collections of exponents that sum to <span class="math-container">$15$</span>, we get that <span class="math-container">$s_{15}$</span> accounts for all the permutations of <span class="math-container">$15$</span> of the <span class="math-container">$20$</span> letters.</p>
<p>That is, define the <a href="https://en.wikipedia.org/wiki/Generating_function#Exponential_generating_function" rel="nofollow noreferrer">exponential generating function</a> of the <a href="https://en.wikipedia.org/wiki/Indicator_function" rel="nofollow noreferrer">indicator function</a> for the set <span class="math-container">$\{i\in\mathbb{Z}:0\le i\le j\}$</span>:
<span class="math-container">$$
P_j(x)=\sum_{i=0}^j\frac{x_i}{i!}\tag2
$$</span>
then the number you are looking for is
<span class="math-container">$$
\underbrace{15!\left[x^{15}\right]}_\text{extract $s_{15}$}\underbrace{P_8(x)P_5(x)P_4(x)P_3(x)\vphantom{\left[x^{15}\right]}}_\text{product from $(1)$}=187957770\tag3
$$</span></p>
<hr />
<p><strong>Mathematica code</strong></p>
<p>Define</p>
<p><code>f = Sum[x^k/k!,{k,0,#}]&</code></p>
<p><code>f[j]</code> gives <span class="math-container">$P_j(x)$</span> from <span class="math-container">$(2)$</span>. Then <code>f/@{8,5,4,3}</code> gives the list of <span class="math-container">$P_j(x)$</span>'s from <span class="math-container">$(1)$</span>. <code>Times@@f/@{8,5,4,3}</code> computes their product. Thus,</p>
<p><code>15! Coefficient[Times@@f/@{8,5,4,3},x,15]</code></p>
<p>gives the result from <span class="math-container">$(3)$</span>.</p>
|
2,015,809 | <blockquote>
<p>If $$f(x)=3 f(1-x)+1$$
for all $x$, what is the value of $f(2016)$?</p>
</blockquote>
<p>I am not sure how to do this, because I see two "$f$"s.</p>
<p>All I could try is substituting,</p>
<p>$$f(x)=3(1-2016)+1\\
=-6044$$</p>
<p>Which I am pretty sure wrong.</p>
<p>How do I deal with this question? when there is $f$ around a braket?</p>
<p>Thank you</p>
| hamam_Abdallah | 369,188 | <p>We have</p>
<p>$$f(x)=3(3f(1-(1-x))+1)+1$$</p>
<p>$$=9f(x)+4$$</p>
<p>$\implies$</p>
<p>$$f(x)=-\frac12=f(2016)$$</p>
|
275,527 | <pre><code>Sqrt[Matrix[( {
{0, 1},
{-1, 0}
} )]] /. f_[Matrix[x__]] :> Matrix[MatrixFunction[f, x]]
</code></pre>
<p><code>Matrix</code> is an undefined symbol but I want to define some substitutions with it.</p>
| lericr | 84,894 | <p>I assume this is related to that other question you recently posted--maybe merge them? Anyway, I'm going to change MatrixFunction for now to hopefully get a clearer answer. Your pattern matched an expression whose body only had one argument. You need to add something to the pattern to match more arguments. Try something like this:</p>
<pre><code>Sqrt[Matrix[({{0, 1}, {-1, 0}})]] /.
f_[Matrix[mat_], args___] :> Matrix[MatrixFunction[f[#, args] &, mat]]
</code></pre>
<p><code>Matrix[MatrixFunction[Sqrt[#1] & , {{0, 1}, {-1, 0}}]]</code></p>
<p>(Thanks BobHanlon)</p>
|
1,384,203 | <p>The additive character of a finite field $\mathbb F_q$ is obtained by using the trace function to base field $F_p$.</p>
<p>Can we write some of them into a middle field. Using trace function from $\mathbb F_{q^n}$ to $\mathbb F_q$? If yes, how?</p>
| Jyrki Lahtonen | 11,619 | <p>Morgan gave you already the negative answer. For the sake of completeness...</p>
<p>There are plenty of trace maps around here. There is the relative trace
$$
tr^{q^n}_q:\Bbb{F}_{q^n}\to\Bbb{F}_q, x\mapsto x+x^q+x^{q^2}+\cdots+x^{q^{n-1}}.
$$
Then we have the two (absolute) trace maps of both these fields all the way down to the base field (assuming $q=p^m$ so $q^n=p^{mn}$:
$$
tr^q_p:\Bbb{F}_q\to\Bbb{F}_p,x\mapsto x+x^p+x^{p^2}+\cdots+x^{p^{m-1}}
$$
and
$$
tr^{q^n}_p:\Bbb{F}_q\to\Bbb{F}_p,x\mapsto x+x^p+x^{p^2}+\cdots+x^{p^{mn-1}}.
$$
It is a useful exercise to show that
$$
tr^{q^n}_p=tr^q_p\circ tr^{q^n}_q,
$$
i.e. composing the relative trace with the absolute trace of its codomain gives the absolute trace of the domain.</p>
<p>ALL of the additive characters of the finite field $\Bbb{F}_q$ can be gotten by the following recipe. The characters are parametrized by an element $a\in\Bbb{F}_q$, and are given by the formula
$$
\chi_a(x)=\zeta^{tr^q_p(ax)},
$$
where $\zeta=e^{2\pi i/p}$ is a primitive complex $p$th root of unity.</p>
<p>Proving this takes a while, but is easy if you already know that the number of additive characters must be equal to $q$. This is because it is easy to show that different choices of $a$ lead to different functions.</p>
<p>The choice $a=0$ gives the trivial character (=constant function taking the value $1$ everywhere). The character gotten with the choice $a=1$ is relatively often called the canonical character, but this term is not in universal use.</p>
|
1,804,360 | <p>The following is a classically valid deduction for any propositions <span class="math-container">$A,B,C$</span>.
<span class="math-container">$\def\imp{\rightarrow}$</span></p>
<blockquote>
<p><span class="math-container">$A \imp B \lor C \vdash ( A \imp B ) \lor ( A \imp C )$</span>.</p>
</blockquote>
<p>But I'm quite sure it isn't intuitionistically valid, although I don't know how to prove it, which is my first question.</p>
<p>If my conjecture is true, my next question is what happens if we add this rule to intuitionistic logic. I do not think we will get classical logic. Is my guess right?</p>
<p>[Edit: The user who came to close and downvote this 4 years after I asked this presumably did not see my comment: "The BHK interpretation suggests to me that it isn't intuitionistically valid, but that's just my intuition...". If you are not familiar with Kripke frames (and I was not at that time), good luck trying to figure out what to try!]</p>
| Hanno | 81,567 | <p>Are you familiar with Kripke semantic of intuitionistic logic (see <em>Semantics of intuitionistic logic</em> <a href="https://en.wikipedia.org/wiki/Kripke_semantics" rel="noreferrer">here</a>)?<span class="math-container">$\def\imp{\rightarrow}$</span></p>
<p><em>Intuitively</em>, a proposition is not interpreted as globally truth or false, but rather as the set of 'knowledge states' in which it is 'known'. Knowledge states may evolve, and the only restriction to the interpretation of propositions is that knowledge cannot be revoked. </p>
<p><em>Formally</em>, you pick a <em>Kripke frame</em>, i.e. a set <span class="math-container">$W$</span> equipped with a preorder <span class="math-container">$\leq$</span>, and interpret any propositional variable as a subset of <span class="math-container">$W$</span> that is closed under passing to <span class="math-container">$\leq$</span>-greater elements. With the interpretations of the propositional variables fixed, you interpret <span class="math-container">$\lor$</span> and <span class="math-container">$\land$</span> as union and intersection, respectively. Also, you interpret <span class="math-container">$\bot$</span> as none of <span class="math-container">$W$</span> (the empty set) and <span class="math-container">$\top$</span> as all of <span class="math-container">$W$</span>. The interesting part is the interpretation of implication: If <span class="math-container">$\psi$</span> and <span class="math-container">$\phi$</span> are interpreted as <span class="math-container">$A,B\subset W$</span>, then <span class="math-container">$\psi\imp\phi$</span> is interpreted as the set of states <span class="math-container">$w\in W$</span> such that for any <span class="math-container">$v\in W$</span> with <span class="math-container">$w\leq v$</span> and <span class="math-container">$v\in A$</span>, you also have <span class="math-container">$v\in B$</span>. Appealing to the intuition again, it means that however knowledge evolves, whenever <span class="math-container">$\psi$</span> is known then so is <span class="math-container">$\phi$</span>. For example, <span class="math-container">$\neg\neg\phi \equiv ( \phi \imp \bot ) \imp \bot$</span> is known at <span class="math-container">$w$</span> precisely if for any <span class="math-container">$v\in W$</span> with <span class="math-container">$w\leq v$</span> there is <span class="math-container">$z\in W$</span> with <span class="math-container">$w\leq z$</span> such that <span class="math-container">$\phi$</span> holds at <span class="math-container">$z$</span> - in other words, you can never refute <span class="math-container">$\phi$</span>. This is much weaker than asserting that <span class="math-container">$\phi$</span> holds for all of <span class="math-container">$W$</span>, as you can see e.g. from the example at the end of the post, where <span class="math-container">$W = \{0\to 1\}$</span>.</p>
<p>Anyway, the Kripke semantics is sound for intuitionistic logic, so to prove that a formula is <em>not</em> an intuitionistic tautology, it suffices to provide <em>some</em> Kripke frame with an interpretation of the formulas' propositional variables such that the associated interpretation is not true in the whole Kripke frame. </p>
<p>Let's do this for <span class="math-container">$(A\imp (B\lor C))\imp ((A\imp B)\lor (A\imp C))$</span>: Consider the frame <span class="math-container">$1\leftarrow 0\rightarrow 2$</span> and interpret <span class="math-container">$A$</span> as known only at <span class="math-container">$1,2$</span>, <span class="math-container">$B$</span> as known only at <span class="math-container">$1$</span> and <span class="math-container">$C$</span> as known only at <span class="math-container">$2$</span>. Then <span class="math-container">$A\imp B$</span> and <span class="math-container">$A\imp C$</span> are both not known at <span class="math-container">$0$</span> since there is an evolution which validates <span class="math-container">$A$</span> but does not validate <span class="math-container">$B$</span> (similarly <span class="math-container">$C$</span>). However, <span class="math-container">$A\imp (B\lor C)$</span> is valid at <span class="math-container">$0$</span>, since for any evolution from <span class="math-container">$0$</span> to either <span class="math-container">$1$</span> or <span class="math-container">$2$</span>, either <span class="math-container">$B$</span> or <span class="math-container">$C$</span> is valid in the new state.</p>
<p>Concerning your second question: An example like the above can only exist if the frame <span class="math-container">$W$</span> under consideration has branches; if, in contrast, evolution is unique - meaning formally that for any three elements <span class="math-container">$u,v,w$</span> in the frame satisfying <span class="math-container">$u\leq v$</span> and <span class="math-container">$u\leq w$</span>, we have either <span class="math-container">$v\leq w$</span> or <span class="math-container">$w\leq v$</span> - the formula <span class="math-container">$(A\imp (B\lor C))\imp ((A\imp B)\lor (A\imp C))$</span> does indeed hold under any interpretation. </p>
<p>Therefore, if you add <span class="math-container">$(A\imp (B\lor C))\imp ((A\imp B)\lor (A\imp C))$</span> as an axiom to intuitionistic logic, you will still have sound Kripke semantics when restricting to frames with unique evolution. Hence, to check that a formula is not a tautology with respect to this stronger calculus, it suffices to find an interpretation in such a uniquely evolving frame which does not validate the formula. For <span class="math-container">$\neg \neg A$</span>, you can pick the frame <span class="math-container">$W = \{0\to 1\}$</span> and interpret <span class="math-container">$A$</span> as known at <span class="math-container">$1$</span> but unknown at <span class="math-container">$0$</span>: Then <span class="math-container">$\neg\neg A$</span> is globally known (see above), but <span class="math-container">$A$</span> is not. Hence, <span class="math-container">$\neg \neg A \imp A$</span> is not a tautology in the enhanced calculus, and therefore the enhanced calculus is still strictly contained in classical propositional logic.</p>
|
2,247,445 | <p>Find isomorphism between <span class="math-container">$\mathbb F_2[x]/(x^3+x+1)$</span> and <span class="math-container">$\mathbb F_2[x]/(x^3+x^2+1)$</span>.</p>
<hr />
<p>It is easy to construct an injection <span class="math-container">$f$</span> satisfying <span class="math-container">$f(a+b)=f(a)+f(b)$</span> and <span class="math-container">$f(ab)=f(a)f(b)$</span>. However, I am stuck how to construct such a mapping that is bijective.</p>
<p>Thank you for help!</p>
| Kanwaljit Singh | 401,635 | <p>Hint -</p>
<p>Make a right angle triangle. We know that $\tan \theta = \frac{\text{perpendicular}} {\text{base}}$.</p>
<p>Comparing it with $\tan \alpha$ you get perpendicular = 1 and base = 3.</p>
<p>Now find 3rd side of triangle using phythagorous theorem. Then you can find value of $\cos \alpha$ and $\sin \alpha$.</p>
|
1,297,023 | <p>I have a question mainly in functional analysis. </p>
<p>Suppose that $H^1(\mathbb R^n)$ is the standard Sobolev space that we all know. My question is as follows:</p>
<p>Does $a_n \in H^1(\mathbb R^n)$, $|a_n| <1$, $a_nb_n \in L^1(\mathbb R^n)$, and $b_n \rightharpoonup 0$ weakly in $H^1(\mathbb R^n)$ imply $\int_{\mathbb R^n} a_n b_n dx \to 0$?</p>
<p>Thank you in advance.</p>
| daw | 136,544 | <p>No.</p>
<p>Let $b$ be a smooth function with support of $b$ included in the unit ball. Assume that $\|b\|_{H^1}=1$. Then define $b_n$ to be translates of $b$:
$$
b_n(x) = b(x + 2n e_1).
$$
Then $b_n\rightharpoonup 0$ in $H^1$, but $b_n$ does not converge stronlgy to zero in $H^1$ or $L^2$. Set $a_n:=b_n$. Then $\|a_n\|_{H^1}\le 1$, $a_nb_n\in L^1$, but $\int_{\mathbb R^n} a_nb_n = \|b\|_{L^2}^2$ converges not to zero.</p>
|
1,297,023 | <p>I have a question mainly in functional analysis. </p>
<p>Suppose that $H^1(\mathbb R^n)$ is the standard Sobolev space that we all know. My question is as follows:</p>
<p>Does $a_n \in H^1(\mathbb R^n)$, $|a_n| <1$, $a_nb_n \in L^1(\mathbb R^n)$, and $b_n \rightharpoonup 0$ weakly in $H^1(\mathbb R^n)$ imply $\int_{\mathbb R^n} a_n b_n dx \to 0$?</p>
<p>Thank you in advance.</p>
| Peter | 111,826 | <p>I think it is not true. The following should work.
Take $g_n:=a_n = b_n \in C_c^\infty$ with $\text{supp} g_n \in [n,n+1]$ and $0\leq g_n \leq 1$ and $g_n = 1$ on $[n+\frac 14, n+\frac 34]$. </p>
|
3,817,760 | <p>In geometry class, it is usually first shown that the medians of a triangle intersect at a single point. Then is is explained that this point is called the centroid and that it is the balance point and center of mass of the triangle. Why is that the case?</p>
<p>This is the best explanation I could think of. I hope someone can come up with something better.</p>
<p>Choose one of the sides of the triangle. Construct a thin rectangle with one side coinciding with the side of the triangle and extending into it. The center of mass of this rectangle is near the midpoint of the side of the triangle. Continue constructing thin rectangles, with each one on top of the previous one and having having the lower side meet the two other sides of the triangle. In each case the centroid of the rectangle is near a point on the median. Making the rectangles thinner, in the limit all the centroids are on the median, and therefore the center of mass of the triangle must lie on the median. This follows because the center of mass of the combination of two regions lies on the segment joining the centroids of the two regions.</p>
| CiaPan | 152,299 | <p>Split a triangle into narrow stripes, parallel to one side.</p>
<p><a href="https://i.stack.imgur.com/OKf4c.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OKf4c.png" alt="enter image description here" /></a></p>
<p>A center of mass of each stripe is close to the middle of its length, which in turn is on the median. In a limit of the stripes' width approaching zero, centers of mass of all stripes get exactly on the median. Their weighted mean, i.e. the center of mass of the triangle, is on the same line.</p>
|
4,046,265 | <p>I'm evaluating a line integral of the function <span class="math-container">$T= x^2 + 4xy + 2yz^3$</span> from <span class="math-container">$a = (0,0,0)$</span> to <span class="math-container">$b=(1,1,1)$</span> on the path <span class="math-container">$z = x^2$</span>, and <span class="math-container">$y = x$</span> without using the fundamental theorem.</p>
<p>My question is how to factor in the boundaries of the integral when I parameterize the path in terms of <span class="math-container">$t$</span></p>
<p>So far I have:<br />
Let <span class="math-container">$x = t$</span><br />
so <span class="math-container">$r= \langle t,t,t^2\rangle$</span> and <span class="math-container">$dr = \langle 1,1,2t\rangle$</span></p>
<p>how do I factor in the boundaries <span class="math-container">$a=(0,0,0)$</span> and <span class="math-container">$b=(1,1,1)$</span> for my integral? After I have the boundaries, solving the line integral is not a problem</p>
<p>Thank you</p>
| Andreas Cap | 202,204 | <p>It seems to me that you got the computation of the derivative wrong. As you have written (with a small simplification as observed in the comment by @guidoar and using the chain rule), you get
<span class="math-container">$$
\Phi\circ\Psi^{-1}(x,v)=((\phi\circ\psi^{-1})(x), D(\phi\circ\psi^{-1})(x)(v)).
$$</span>
(I have changed the notion to emphasize that <span class="math-container">$v$</span> is a tangent vector, while <span class="math-container">$x$</span> is a point in <span class="math-container">$M$</span>.) But when differentiating this, you have to distinguish wheter you differentiate in <span class="math-container">$x$</span>-directions or in <span class="math-container">$v$</span>-directions: Differentiating the second component in <span class="math-container">$x$</span> directions, you get a second derivative, whereas the fact that the derivative is linear in <span class="math-container">$v$</span> implies that you just get back the first derivative when differentiating in <span class="math-container">$v$</span>-directions. More formally this means that
<span class="math-container">$$
D(\Phi\circ\Psi^{-1})(x,v)(w_1,w_2)=(D(\phi\circ\psi^{-1})(x)(w_1),D^2(\phi\circ\psi^{-1})(x)(w_1,v)+D (\phi\circ\psi^{-1})(x)(w_2)).
$$</span>
In matrix notation, this leads to a block form as
<span class="math-container">$$
\begin{pmatrix} D(\phi\circ\psi^{-1})(x) & 0\\
D^2(\phi\circ\psi^{-1})(x)(\_,v) & D(\phi\circ\psi^{-1})(x) \end{pmatrix}
$$</span>
Taking the determinant, the off-diagonal block does not play a role, so you conclude that <span class="math-container">$\mathcal J(\Phi\circ\Psi^{-1})=\mathcal J(\phi\circ\psi^{-1})^2>0$</span>, which implies the claim.</p>
|
158,469 | <p>This is another exercise from Golan's book.</p>
<p><strong>Problem:</strong> Let $V$ be an inner product space over $\mathbb{C}$ and let $\alpha$ be an endomorphism of $V$ satisfying $\alpha^*=-\alpha$, where $\alpha^*$ denotes the adjoint. Show that every eigenvalue of $\alpha$ is purely imaginary.</p>
<p>My proposed solution is below.</p>
| Martin Argerami | 22,857 | <p>Let me show another argument which applies to a more general setting: if $\alpha$ is a linear operator on a Hilbert space satisfying $\alpha^*=-\alpha$, then the spectrum of $\alpha$ is purely imaginary (i.e. real part equal zero). </p>
<p>Indeed, one simply needs to notice that $\alpha-\lambda\,\text{id}$ is invertible if and only if $(\alpha-\lambda\,\text{id})^*$ is invertible. As $$(\alpha-\lambda\,\text{id})^*=\alpha^*-\overline\lambda\,\text{id}=-\alpha-\overline\lambda\,\text{id}=-(\alpha+\overline\lambda\,\text{id}),$$ we conclude that any $\lambda$ in the spectrum of $\alpha$ satisfies $\overline\lambda=-\lambda$.</p>
|
3,586,001 | <p>Let <span class="math-container">$k$</span> be a positive integer. Is it true that there are infinitely many primes of the form <span class="math-container">$p=6r-1$</span> such that <span class="math-container">$r$</span> and <span class="math-container">$k$</span> are coprime? Feel free to assume any well known (even if hard to prove) theorem such as Dirichlet's theorem for arithmetic progressions.</p>
<p>Update: can we do it only with assuming the version that for any coprime <span class="math-container">$a$</span> and <span class="math-container">$d$</span> there are infinitely many primes <span class="math-container">$p$</span> with <span class="math-container">$p\equiv a \pmod d$</span>?</p>
<p>Any help appreciated!</p>
| Sungjin Kim | 67,070 | <p>Denote <span class="math-container">$(a,b)$</span> by the greatest common divisor of <span class="math-container">$a$</span> and <span class="math-container">$b$</span>. Denote also <span class="math-container">$q^a||k$</span> when <span class="math-container">$q^a$</span> is the largest power of a prime <span class="math-container">$q$</span> that divides <span class="math-container">$k$</span>. </p>
<p>We apply Chinese Remainder Theorem. Find a solution to the system of congruences, </p>
<p><span class="math-container">$$
x\equiv 1 \ (\textrm{mod} \ q^a) \ \textrm{for primes powers }q^a ||k, \ q\neq 5 \ \ (1),
$$</span>
<span class="math-container">$$
x\equiv 2 \ (\textrm{mod} \ 5^b) \ \textrm{ if } 5^b ||k \ \ (2).
$$</span></p>
<p>This system is solvable, we have <span class="math-container">$x\equiv x_0$</span> mod <span class="math-container">$k$</span>. Note that if <span class="math-container">$5\nmid k$</span> then we do not use (2) and just have <span class="math-container">$x\equiv 1$</span> mod <span class="math-container">$k$</span> with <span class="math-container">$x_0=1$</span>. </p>
<p>Then we see that <span class="math-container">$(x_0,k)=1$</span>, so <span class="math-container">$(k,kx+x_0)=1$</span> and we take <span class="math-container">$r=kx+x_0$</span> so that <span class="math-container">$p=6(kx+x_0)-1=6kx+6x_0-1$</span>.</p>
<p>By the above construction, we have <span class="math-container">$(6k, 6x_0-1)=1$</span>. Thus, by Dirichlet's theorem, there are infinitely many <span class="math-container">$x$</span> such that <span class="math-container">$6kx+6x_0-1$</span> is prime. </p>
|
4,642,772 | <p>I am trying to understand this proof:
<a href="https://math.stackexchange.com/questions/3201004/prove-that-varphipk-pk-pk-1-for-prime-p">Prove that $\varphi(p^k)=p^k-p^{k-1}$ for prime $p$</a></p>
<p>At some point it is stated that the number of multiples of p in range <span class="math-container">$[1,p^k)$</span> is <span class="math-container">$p^{k-1}$</span>.<br />
I am struggling to get why. I tried this:<br />
The multiples of p are :
<span class="math-container">$$p, 2p, .., (p-1)p, p^2, 2p^2, .., (p-1)p^2, .., p^{k-1}, 2p^{k-1}, .., (p-1)p^{k-1}$$</span></p>
<p>Now I can think in terms of powers of p : Each power <span class="math-container">$p^i$</span> contributes <span class="math-container">$p-1$</span>
mutliples : <span class="math-container">$$p^i, 2p^i, .., (p-1)p^i$$</span></p>
<p>And we have <span class="math-container">$k-1$</span> distinct powers so <span class="math-container">$(k-1)\times (p-1)$</span> multiples of p in total.
Of course that's really different from the actual result.
I cannot see what I am missing here.</p>
| cigar | 1,070,376 | <p>Anything less than or equal to <span class="math-container">$p^{k-1}$</span> can be multiplied by <span class="math-container">$p$</span> and not go over <span class="math-container">$p^k.$</span></p>
<p>So there's exactly <span class="math-container">$p^{k-1}$</span> non-totatives.</p>
<p>Subtract those out to get <span class="math-container">$p^k-p^{k-1}$</span> totatives.</p>
|
975 | <p>The usual <code>Partition[]</code> function is a very handy little thing:</p>
<pre><code>Partition[Range[12], 4]
{{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}}
Partition[Range[13], 4, 3]
{{1, 2, 3, 4}, {4, 5, 6, 7}, {7, 8, 9, 10}, {10, 11, 12, 13}}
</code></pre>
<p>One application I'm working on required me to write a particular generalization of <code>Partition[]</code>'s functionality, which allowed the generation of sublists of unequal lengths, for as long as the lengths were appropriately commensurate. (Let's assume for the purposes of this question that the list lengths being commensurate is guaranteed, but you're welcome to generalize further to the incommensurate case.) Here's my generalization in action:</p>
<pre><code>multisegment[lst_List, scts_List] := Block[{acc},
acc = Prepend[Accumulate[PadRight[scts, Length[lst]/Mean[scts], scts]], 0];
Inner[Take[lst, {#1, #2}] &, Most[acc] + 1, Rest[acc], List]]
multisegment[CharacterRange["a", "x"], {3, 1, 2}]
{{"a", "b", "c"}, {"d"}, {"e", "f"}, {"g", "h", "i"}, {"j"}, {"k", "l"},
{"m", "n", "o"}, {"p"}, {"q", "r"}, {"s", "t", "u"}, {"v"}, {"w", "x"}}
</code></pre>
<p>(Thanks to halirutan for optimization help with <code>multisegment[]</code>.)</p>
<p>The problem I've hit into is that I wanted <code>multisegment[]</code> to also support offsets, just like in <code>Partition[]</code>. I want to be able to do something like the following:</p>
<pre><code>multisegment[Range[14], {4, 3}, {3, 1}]
{{1, 2, 3, 4}, {4, 5, 6}, {5, 6, 7, 8},
{8, 9, 10}, {9, 10, 11, 12}, {12, 13, 14}}
</code></pre>
<p>How might a version of <code>multisegment[]</code> with offsets be accomplished?</p>
| kglr | 125 | <p>Also related to Mike's solution:</p>
<pre><code>ClearAll[irregPartition];
irregPartition[list_List, sizes_List, offsets_List] :=
Module[{offsetlist, sizelist},
offsetlist = (NestWhile[Join[offsets, #] &,offsets, (Tr@# < Length@list) &] //
NestWhile[Most, #, (Tr@# >= Length@list) &] & // Accumulate //
Prepend[#, 0] & ) + 1;
sizelist = NestWhile[Join[sizes, #] &,sizes,(Length@# <= Length@offsetlist) &] //
NestWhile[Most, #, Length@# > Length@offsetlist &] &;
{offsetlist, sizelist} =Transpose@TakeWhile[
Partition[Riffle[offsetlist, sizelist], 2], #[[1]] + #[[2]] <= Length@list &];
MapThread[Take[#1,Min[#2, Length@#1]] &, {Drop[list,
Min[#, Length@list]] & /@ (offsetlist - 1), sizelist}] //
If[Length@Last@# < Last@sizelist, Most@#, #] &];
irregPartition[list_List, sizes_List] := irregPartition[list, sizes, sizes];
</code></pre>
<p>Example:</p>
<pre><code>list = Range[10];
sizes = {5, 4};
offsets = {2, 3};
irregPartition[list, sizes, offsets]
(*
==> {{1,2,3,4,5},{3,4,5,6},{6,7,8,9,10}}
*)
irregPartition[list, sizes]
(*
==> {{1,2,3,4,5},{6,7,8,9}
*)
</code></pre>
<p>Another example:</p>
<pre><code>list = CharacterRange["a", "x"];
sizes = {4, 5, 3, 4};
offsets = {2, 7, 3, 5};
irregPartition[list, sizes, offsets]
(*
==> {{"a","b","c","d"},{"c","d","e","f","g"},{"j","k","l"},{"m","n","o","p"},{"r","s","t","u"},{"t","u","v","w","x"}}
*)
irregPartition[list, sizes]
(*
==> {{"a","b","c","d"},{"e","f","g","h","i"},{"j","k","l"},{"m","n","o","p"},{"q","r","s","t"}}
*)
</code></pre>
|
855,463 | <p>I have the following question regarding support vector machines: So we are given a set of
training points $\{x_i\}$ and a set of binary labels $\{y_i\}$.</p>
<p>Now usually the hyperplane classifying the points is defined as:</p>
<p>$ w \cdot x + b = 0 $</p>
<p><strong>First question</strong>: Here $x$ does not denote the points of the training set, but the points on the separating (hyper)plane, right?</p>
<p>In a next step, the lecture notes state that the function:</p>
<p>$f(x) = \mathrm{sign}\, (w \cdot x + b)$</p>
<p>correctly classify the training data.<br>
<strong>Second question</strong>: Now I don't understand that, since it was stated earlier that $ w \cdot x + b = 0 $, so how can something which is defined to be zero have a sign?</p>
<p>Two additonal questions:</p>
<p>(1) You have added that a slack variable might be introduced for non-linearly separable data - how do we know that the data is not linearly separable, as far as I understand the mapping via kernel has as a purpose to map the data into a vector space where in the optimal case it would be linearly separable (and why not using a non-linear discriminant function altogether instead of introducing a slack variable?)</p>
<p>(2) I've seen that for the optimization only one of the inequalities $w \cdot x = b \geqslant 1$ is being used as a linear constrained - why?</p>
| Joel Bosveld | 108,482 | <p>For the first equation, $w\cdot x+b=0$, $w$ is the direction normal (orthogonal/perpendicular) to the hyperplane. You are correct that the $x$ (satisfying this equation) are the points on the hyperplane.</p>
<p>In the second equation, the $x$ are the training (or test) data. These should not lie on the hyperplane, where $f(x)=0$, but should lie on either side of the hyperplane - since, by construction, the hyperplane separates the two classes.</p>
<p>Points that lie on the hyperplane, where $f(x)=0$, will not be able to be classified.</p>
|
297,251 | <p>In his well-known <a href="http://matwbn.icm.edu.pl/ksiazki/fm/fm101/fm101110.pdf" rel="nofollow noreferrer">paper</a> Bellamy constructs an indecomposable continua with exactly two composants. The setup is as follows:</p>
<p>We have an inverse-system $\{X(\alpha); f^\alpha_\beta: \beta,\alpha < \omega_1\}$ of metric indecomposable continua and retractions. For each $X(\alpha)$ a composant $C(\alpha) \subset X(\alpha)$ is specified and each $C(\alpha)$ maps into $C(\beta) $. </p>
<p>The inverse limit $X$ has exactly two composants. The first is the union
$\bigcup\{X(\beta): \beta < \omega_1\}$ where we identify $X(\beta)$ with the set of sequences $(x_\alpha) \in X$ with $x_\beta = x_{\beta+1} = x_{\beta+2} = \ldots . $ The second composant is the inverse limit $\{C(\alpha); f^\alpha_\beta: \beta,\alpha < \omega_1\}$.
Observe there is no reason <em>a priori</em> for the second composant to be nonempty. However I do not believe an example is know.</p>
<p>My question is an easier one. Can you think of an example of a metric continuum $M$ and a $\omega_1$-indexed decreasing nest of dense semicontinua <strong>with empty intersection</strong>? We call the set $S \subset M$ a <em>semicontinuum</em> to mean for each $x,y \in S$ there exists a continuum $K$ with $\{x,y\} \subset K \subset S$.</p>
<p>If the second composant was empty the family $\{f^\alpha_0(C(\alpha)): \alpha < \omega_1\}$ would be such a nest for $M = X_0$.</p>
<p>If we index by $\omega$ instead an example is easy to come by. Let $M$ be the unit disc and $Q = \{q_1,q_2, \ldots\}$ and enumeration of the rational points on the boundary. Let $S(n)$ be formed by drawing the straight line segment from each element of $\{q_n,q_{n+1}, \ldots\}$ to each rational point of $(0,1/n) \times \{0\}$. Then add in $(0,1/n) \times \{0\}$ itself to make the space a semicontinuum.</p>
<p>Indexing by $\omega_1$ must somehow get around the fact that any $\omega_1$ decreasing nest of closed subsets of a metric continuum eventually stabilizes.</p>
<p>It feels like this would be easier if we assume the Continuum Hypothesis.</p>
| Vít Tuček | 6,818 | <p>The <a href="http://inspirehep.net/record/1591349" rel="nofollow noreferrer">article</a> that was linked by David G. Stork actually contains information about how one can get these pictures. Namely, the <a href="http://doc.sagemath.org/html/en/reference/discrete_geometry/sage/geometry/lattice_polytope.html#sage.geometry.lattice_polytope.ReflexivePolytope" rel="nofollow noreferrer">SageMath</a> computer algebra system contain a database of all reflexive polytopes and the figure from the article includes their position in this database (off by one). You can do much more with these objects in Sage than just getting their polar dual. Thus I can give you easily more than just a picture, I can give you a 3D model that you can rotate and zoom in your browser. Just paste the following code into <a href="http://sagecell.sagemath.org/?z=eJxVjc0KgzAQhO-BvMNCLxaimLU_6aHvIL2KFFvXKsRENC3t23cjvfQwLN_O7M4G7n3jHgShJ3DP8UYzdLMfQUPwoI0UDs7rnLz9BD_RcrXDEnhZaQV7BQcFRwVGATIgEzJi5JOCImdxziCrYO1YMafZRMxrKaQo-dmFOkvv4UXlrybh9H9l5VJdp3orRRsvyoztZk5WzibrQ9FG-ALekjeY&lang=sage" rel="nofollow noreferrer">SageMathCell</a> and hit <em>Evaluate</em>. </p>
<pre><code># change the number from 1 to 18
n = 3
polytopes_list = [1, 5, 6, 7, 8, 25, 26, 27, 28, 29, 30, 31, 82, 83, 84, 85, 219, 220]
P = ReflexivePolytope(3, polytopes_list[n-1]-1)
dP = P.polar()
dP.plot3d()
</code></pre>
<p><a href="https://i.stack.imgur.com/8kwR7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8kwR7.png" alt="enter image description here"></a></p>
|
3,361,224 | <blockquote>
<p>Show that a dense subspace <span class="math-container">$Y$</span> of a first countable separable topological space <span class="math-container">$X$</span> is separable.</p>
</blockquote>
<p><em>Proof:</em></p>
<p><span class="math-container">$X$</span> is separable.
Let <span class="math-container">$S=\{x_n \in X | n \in \mathbb{N}\}$</span> be a countable dense subset of <span class="math-container">$X$</span>.</p>
<p><span class="math-container">$Y$</span> is also dense in <span class="math-container">$X$</span>.</p>
<p>Because <span class="math-container">$X$</span> is first-countable, thus for each <span class="math-container">$x_n$</span> where <span class="math-container">$n \in \mathbb{N}$</span> there exists a countable local-basis around <span class="math-container">$x_n$</span>.
Let the countable local-basis around <span class="math-container">$x_n$</span> be <span class="math-container">$S_n=\{\text{ }B_n^k \text{ } | \text{ }k \in \mathbb{N} \}$</span></p>
<p>Because <span class="math-container">$Y$</span> is dense in <span class="math-container">$X$</span> thus for each <span class="math-container">$x_n$</span> where <span class="math-container">$ n=1,2,3 \dots $</span> and for each <span class="math-container">$B_n^k$</span> where <span class="math-container">$k=1,2,3,4 \dots$</span>, we have <span class="math-container">$Y \cap B_n^k \neq \phi$</span>. </p>
<p>Say <span class="math-container">$y_n^k \in Y \cap B_n^k \neq \phi$</span></p>
<p>Denote <span class="math-container">$Z=\{ y_n^k \in Y \text{ } | \text{ } n,k \in \mathbb{N} \}$</span></p>
<p>Claim: <span class="math-container">$Z$</span> is a countable dense set of <span class="math-container">$Y$</span>.</p>
<p>Choose <span class="math-container">$y \in Y$</span> and any open set <span class="math-container">$V$</span> in <span class="math-container">$Y$</span> containing y.
<span class="math-container">$V$</span> is open in <span class="math-container">$Y$</span> implies that <span class="math-container">$V=U \cap Y$</span> where <span class="math-container">$U$</span> is an open set in <span class="math-container">$X$</span>.</p>
<p>Thus <span class="math-container">$y \in U \in \tau$</span> and <span class="math-container">$y \in Y$</span></p>
<p><span class="math-container">$y \in U$</span> and <span class="math-container">$U$</span> is open in X. Because <span class="math-container">$S$</span> is dense in X, we have that <span class="math-container">$U \cap S \neq \phi $</span>.</p>
<p>Let <span class="math-container">$x_n \in U \cap S$</span>, Thus <span class="math-container">$x_n \in U$</span> and <span class="math-container">$U$</span> is open in <span class="math-container">$X$</span>.</p>
<p>Considering that <span class="math-container">$S_n$</span> is a countable local-basis around <span class="math-container">$x_n$</span> we have an element <span class="math-container">$B_n^{k_0}$</span> such that <span class="math-container">$x_n \in B_n^{k_0} \subset U$</span>. choose the corresponding <span class="math-container">$y_n^{k_0}$</span> as done in the construction above. Then we have <span class="math-container">$y_n^{k_0} \in B_n^{k_0} \subset U$</span>. Thus <span class="math-container">$y_n^{k_0} \in U \cap Y = V$</span> and hence <span class="math-container">$V \cap Z \neq \phi$</span> as it contains <span class="math-container">$y_n^{k_0}$</span>.</p>
<p>Hence <span class="math-container">$Y$</span> has a countable dense subset. <span class="math-container">$Y$</span> is separable.</p>
<p>Hence proved!</p>
<p>Please check my solution. I need to correct my mistakes and learn.
Thank You.</p>
| bof | 111,012 | <p>Looks good to me. Here's a slightly different way to look at it. Consider the following properties of a topological space <span class="math-container">$X$</span>:</p>
<p>(1) <span class="math-container">$X$</span> is separable and first countable;</p>
<p>(2) <span class="math-container">$X$</span> has a countable <span class="math-container">$\pi$</span>-base, i.e., a countable collection <span class="math-container">$\mathcal B$</span> of nonempty open sets such that every nonempty open set contains a member of <span class="math-container">$\mathcal B$</span> as a subset;</p>
<p>(3) <span class="math-container">$X$</span> is separable.</p>
<p>You showed that a dense subspace of a space with property (1) has property (3). With the same ideas you can show that <span class="math-container">$(1)\implies(2)\implies(3)$</span>, and that a dense subspace of a space with property (2) has property (2).</p>
<p><br><span class="math-container">$(1)\implies(2)$</span>: Suppose <span class="math-container">$X$</span> is separable and first countable. Let <span class="math-container">$S$</span> be a countable dense subset of <span class="math-container">$X$</span>, for for each <span class="math-container">$x\in S$</span> let <span class="math-container">$\mathcal B_x$</span> be a countable local base at <span class="math-container">$x$</span>. Then <span class="math-container">$\bigcup_{x\in S}\mathcal B_x$</span> is a countable <span class="math-container">$\pi$</span>-base for <span class="math-container">$X$</span>.</p>
<p><br><span class="math-container">$(2)\implies(3)$</span>: Suppose <span class="math-container">$\mathcal B$</span> is a countable <span class="math-container">$\pi$</span>-base for <span class="math-container">$X$</span>. By choosing one point from each member of <span class="math-container">$\mathcal B$</span>, we get a countable dense subset of <span class="math-container">$X$</span>.</p>
<p><br>Finally, suppose <span class="math-container">$X$</span> has property (2) and <span class="math-container">$Y$</span> is a dense supspace of <span class="math-container">$X$</span>. Let <span class="math-container">$\mathcal B$</span> be a countable <span class="math-container">$\pi$</span>-base for <span class="math-container">$X$</span>; then <span class="math-container">$\mathcal B_Y=\{B\cap Y:B\in\mathcal B\}$</span> is a countable <span class="math-container">$\pi$</span>-base for <span class="math-container">$X$</span>.</p>
|
989,118 | <p>The sum is $$\sum_{n>0} \mathrm{i}^n\frac{J_n(x)}{n}\sin\frac{2\pi n}{3}\stackrel{?}{=}0$$ and $$\sum_{n>0} (\mathrm{-i})^n\frac{J_n(x)}{n}\sin\frac{2\pi n}{3}\stackrel{?}{=}0$$</p>
<p>I suspect they are zero because I am working on a project which from symmetry consideration they should be zero. </p>
<p><strong>If anyone can point out they cannot be zero, or they cannot identically be zero, that will be great, too.</strong></p>
<p>I've asked a question about Bessel function before, see <a href="https://math.stackexchange.com/questions/985566/does-this-infinite-summation-of-bessel-function-has-a-closed-form">Does this Infinite summation of Bessel function has a closed form?</a> I hope the reference it contains may do some help.</p>
| Saketh Malyala | 250,220 | <p>So we know that $i^2$ = -1, and that squared is $i^4$ which is 1. As aforementioned, $i^2 = -i^4$. Because, $i^4=1$, $i^2=i^6=i^{10}...i^{38}$, and $i^4=i^8..i^{36}$. Therefore, canceling out terms leaves us with $(i^{38})^2$, which evaluates to $(-1)^2$, which we know to be one. </p>
<p>==> $(((i^2-i^4)+(i^6-i^8)...(i^{34}-i^{36})+i^{38})^2$ = $(0+0+0+..+(-1))^2$ = (-1)^2 = 1</p>
|
51,903 | <p>Does anyone know of a tool which</p>
<ol>
<li>Can display formulas neatly, preferably like this website without hassle. (Unlike wikipedia with <code>:<math></code>) </li>
<li>Has a wiki like structure: i.e categories of pages, individual articles with hyperlinked sections, subsections etc. </li>
<li>Preferrably can be used online and does not require installation of some software.</li>
<li>Comes with a free host, i.e for people with little money and no university server.</li>
</ol>
<p>So basically I need a notebook on steroids :)</p>
<p><em>Update</em>: Edited this question to remove the essay I wrote on blogs. You may refer to the revision history if you're interested. I am keeping it open in case anyone knows of further alternatives than have already been mentioned. At present I have settled on and am fairly content with Drupal.</p>
| Ilmari Karonen | 9,602 | <p>It <a href="http://michaelnielsen.org/polymath1/index.php?title=Special%3aVersion" rel="nofollow">looks like</a> the polymath wiki you mention is just a plain old no-frills MediaWiki installation. Getting MediaWiki installed on a webhost isn't actually very hard, although admittedly getting the <a href="http://www.mediawiki.org/wiki/Math" rel="nofollow">LaTeX support</a> working and customizing the wiki to your needs may take some effort. (If you don't like the standard texvc based LaTeX renderer, you could also install the <a href="http://www.mediawiki.org/wiki/Extension%3aMathJax" rel="nofollow">MathJax extension for MediaWiki</a>.)</p>
<p>If you want a free wiki host, try <a href="http://www.google.com/search?q=free+wiki+hosting" rel="nofollow">Googling for one</a>, or start <a href="http://en.wikipedia.org/wiki/Comparison_of_wiki_farms" rel="nofollow">here</a>. However, keep in mind that <em>there ain't no such thing as a free lunch</em>. At best, you'll probably have to live with third-party ads on your wiki; at worst, you may find that the fine print says that your wiki doesn't really belong to you, and that the host can just take it over and do what they want with it if <em>you</em> try to do something they don't like (like, say, closing it or moving it to another host).</p>
<p>Most MediaWiki hosts will probably support the standard texvc based <math> syntax, but I doubt they'll have the MathJax extension installed out of the box. However, even if you can't install the extension, you might still be able to add rudimentary MathJax support to your wiki via custom <a href="http://www.mediawiki.org/wiki/Manual%3aInterface/JavaScript" rel="nofollow">site JS</a>.</p>
|
180,121 | <p>Let $f:X\rightarrow Y$ be a morphism of schemes and let $\mathcal{F}$ be a sheaf on $Y$. Then there is a natural map $$\Psi:\mathcal{F} \rightarrow f_{\ast}f^{\ast}(\mathcal{F})$$ and localizing $$\Psi_p:\mathcal{F}_p \rightarrow f_{\ast}f^{\ast}(\mathcal{F})_p=f_{p,\ast}f_p^{\ast}(\mathcal{F}_p)$$</p>
<p>I'm interested in learning under what conditions on $X,Y,f,\mathcal{F},p$ is $\Psi_p$ an isomorphism. E.g. is it an isomorphism if $\mathcal{F}$ is quasi-coherent and $f$ is étale?</p>
<p>Thanks in advance for any insight.</p>
| Sasha | 4,428 | <p>If $f_*O_X=O_Y$ then this is true for any locally free (or flat) sheaf (and if $f$ is flat then for any quasicoherent sheaf) by projection formula. </p>
|
2,228,635 | <p>I have tried to start with $\epsilon> 0$, and with $f$ uniformly continuous at $(a,b)$, so then there exists a $\delta>0$ such that for $x,u \in (a,b)$ with $|x - u|<δ$ then $|f(x)-f(u)|<\epsilon$. Since $x,u < b$ then $b-x >0$ so we have $0 < b - x <\delta$ then $|f(x)-f(b)|<\epsilon$. I'm not sure with this idea because $b$ is not in the interval $(a,b)$.</p>
| skyking | 265,767 | <p>As said, you can always prove things like this with Venn diagrams, truth tables or Karnaugh diagram. In this case I'd say that would be the easiest way. </p>
<p>Otherwise you just use the distributive law together with absorbtion:</p>
<p>$$(A\cup B)\cap(B\cup C)\cap(C\cup A)
\\= ((A\cap B)\cup (A\cap C)\cup (B\cap B)\cup (B\cap C))\cap(C\cup A)
\\= ((A\cap B)\cup (A\cap C)\cup B \cup (B\cap C))\cap(C\cup A)
\\= (A\cap B\cap C)\cup (A\cap C)\cup (B\cap C) \cup (B\cap C) \cup
(A\cap B)\cup (A\cap C)\cup (A\cap B) \cup (A\cap B\cap C)$$</p>
<p>Now you use absorbtion $(A\cap B)\cup (A\cap B\cap C)= A\cap B$ and so you get rid of the $A\cap B\cap C$ terms and also remove duplicate terms (since $X\cup X=X$) and arrive at</p>
<p>$$(A\cup B)\cap(B\cup C)\cap(C\cup A)
= (A\cap C)\cup (B\cap C) \cup
(A\cap B) $$</p>
|
2,228,635 | <p>I have tried to start with $\epsilon> 0$, and with $f$ uniformly continuous at $(a,b)$, so then there exists a $\delta>0$ such that for $x,u \in (a,b)$ with $|x - u|<δ$ then $|f(x)-f(u)|<\epsilon$. Since $x,u < b$ then $b-x >0$ so we have $0 < b - x <\delta$ then $|f(x)-f(b)|<\epsilon$. I'm not sure with this idea because $b$ is not in the interval $(a,b)$.</p>
| Henno Brandsma | 4,280 | <p>I managed to convince myself of this equality as follows: $x$ is in the left hand side, iff $x$ is in at least two of the sets $A,B,C$ (by definition of union, and then intersection ). In that case it will be in all unions $A \cup B, B \cup C, A \cup C$, because to avoid that you have to miss at least two distinct sets. And if you're in all of the unions, you cannot avoid any pair of sets out of the three, so you're in at least 2 of them.</p>
|
412,944 | <p>Is it mathematically acceptable to use <a href="https://math.stackexchange.com/questions/403346/prove-if-n2-is-even-then-n-is-even">Prove if $n^2$ is even, then $n$ is even.</a> to conclude since 2 is even then $\sqrt 2$ is even? Further more using that result to also conclude that $\sqrt [n]{2}$ is even for all n?</p>
<p>Similar argument for odd numbers should give $\sqrt[n]{k}$ is even or odd when k is even odd.</p>
<p>My question is does any of above has been considered under a more formal subject or it is a correct/nonsensical observation ? </p>
| Key Ideas | 78,535 | <p>Yes, essentially analogous arguments work to extend the notion of <strong>parity</strong> from the ring of integers to many rings of algebraic integers such as $\,\Bbb Z[\sqrt[n]{k}].\ $ </p>
<p>The key ideas are: one can apply parity arguments in any ring that has $\ \mathbb Z/2\ $ as an image, e.g. the ring of all rationals with odd denominator, or the Gaussian integers $\rm\:\mathbb Z[{\it i}\,],\:$ where the image $\rm\ \mathbb Z[{\it i}\,]/(2,{\it i}-\!1) \cong \mathbb Z/2\ $ yields the natural parity definition that $\rm\ a\!+\!b\,{\it i}\ $ is even iff $\rm\ a\equiv b\ \ (mod\ 2),\ $ i.e. $ $ if $\rm\ a+b\,{\it i}\ $ maps to $\:0\:$ via the above isomorphism, which maps $\rm\ 2\to 0,\ i\to 1\:$. </p>
<p>Generally, it is easy to show that if $\rm\:2\nmid f(x)\in \Bbb Z[x]\setminus \Bbb Z\:$ then the number of ways to define parity in the ring $\rm\ \mathbb Z[w] \cong \mathbb Z[x]/(f(x))\ $ is given by the number of roots of $\rm\: f(x)\: $ modulo $2\:.\ $ For suppose there exists a homomorphism $\rm\ h\, :\, \mathbb Z[w]\to \mathbb Z/2.\:$ Then $\rm\:w\:$ must map to a root of $\rm\:f(x)\:$ in $\rm\ \mathbb Z/2.\ $ Thus if $\rm\ f(0)\equiv 0\ (mod\ 2)\ $ then $\rm\: \mathbb Z[w]/(2,w) \cong \mathbb Z[x]/(2,x,f(x)) \cong \mathbb Z/2\ $ by $\rm\: x\mid f(x)\ (mod\ 2),\, $ and $\rm\, \!f(1)\equiv 0\ (mod\ 2) $ $\Rightarrow$ $\rm \mathbb Z[w]/(2,w\!-\!1) \cong \mathbb Z[x]/(2,x\!-\!1,f(x)) \cong \mathbb Z/2\, $ by $\rm\, x\!-\!1\,|\, f(x)\ (mod\ 2). $ </p>
<p>Let's consider some simple examples. Since $\rm\ x^2\!+1\ $ has the unique root $\rm\ x\equiv 1\ (mod\ 2),\:$ the Gaussian integers $\rm\ \mathbb Z[{\it i}\,]\cong \mathbb Z[x]/(x^2\!+1)\ $ have a unique definition of parity, with $\:i\:$ being odd. Since $\rm\ x^2\!+x+1\ $ has no roots modulo $\: 2,\: $ there is no way to define parity for the Eisenstein integers $\rm\ \mathbb Z[w] \cong \mathbb Z[x]/(x^2\!+x+1).\, $ Indeed since $\rm\ w^3 = 1\ $ we infer that $\rm\: w \equiv 1\ (mod\ 2)\ $ contra $\rm\ w^2\!+w+1 = 0.\ $ On the other hand $\rm\ \mathbb Z[w] \cong \mathbb Z[x]/(x^2\!+x+2)\ $ has two parity structures since both $\:0\:$ and $\rm\:1\:$ are roots of $\rm\ x^2\! + x + 2\ $ modul0 $\rm\:2,\:$ so we can define $\rm\:w\:$ to be either even or odd.</p>
|
1,413,212 | <blockquote>
<p>Given $x$ the number of circles of radius $r$,
what is a good approximate size of the radius of a bigger circle which they fit in?</p>
</blockquote>
<p>To explain in actual problem terms. I want to move units in a video games which take up a certain amount of area around, I don't want those units overlapping so I want the points there told to move to to be spaced far enough apart to make sure they aren't. I know how to make sure each point is far enough apart but i'm picking points randomly in a circle and then checking to make sure they are far enough apart. </p>
<p>The problem is if a circle is too small in size there simply isn't enough valid movement points for all the units and you get an infinite loop endlessly trying to find an answer that doesn't exist.
If the circle is too large you end up with units spaced out over vast areas way out of proportion. </p>
<p>I do not need a perfect minimum radius answer I just need to be certain the amount will fit while not being too large. To show that I did think about it my rough guess was.</p>
<p>the square root of (amount needed) * (radius of each circle * 2)</p>
| Community | -1 | <p>By illegal calculations, we get $\lim_{x \to 1}f(x) = 4$. To prove that now, use the $\epsilon - \delta$ definition of limits, with $\delta = \epsilon/10$. </p>
|
1,413,212 | <blockquote>
<p>Given $x$ the number of circles of radius $r$,
what is a good approximate size of the radius of a bigger circle which they fit in?</p>
</blockquote>
<p>To explain in actual problem terms. I want to move units in a video games which take up a certain amount of area around, I don't want those units overlapping so I want the points there told to move to to be spaced far enough apart to make sure they aren't. I know how to make sure each point is far enough apart but i'm picking points randomly in a circle and then checking to make sure they are far enough apart. </p>
<p>The problem is if a circle is too small in size there simply isn't enough valid movement points for all the units and you get an infinite loop endlessly trying to find an answer that doesn't exist.
If the circle is too large you end up with units spaced out over vast areas way out of proportion. </p>
<p>I do not need a perfect minimum radius answer I just need to be certain the amount will fit while not being too large. To show that I did think about it my rough guess was.</p>
<p>the square root of (amount needed) * (radius of each circle * 2)</p>
| Paramanand Singh | 72,031 | <p>Use the algebra of limits to get $$\lim_{x \to 1}f(x) = \lim_{x \to 1}f(x) - 4 + 4 = \lim_{x \to 1}\frac{f(x) - 4}{x - 1}\cdot (x - 1) + 4 = 10\cdot 0 + 4 = 4$$</p>
|
2,086,598 | <p>Let's start with a well known limit:
$$\lim_{n \to \infty} \left(1 + {1\over n}\right)^n=e$$</p>
<p>As $n$ approaches infinity, the expression evaluates Euler's number $e\approx 2.7182$. Why that number? What properties does the limit have that leads it to 2.71828$\dots$ and why is this number important? Where can we find $e$, in what branch of Mathematics?</p>
| Travis | 404,488 | <p>$e$ is used in recursive interest.</p>
<p>It's related to the fact that 5% interest ever 6 months is better than 10% every 12 months</p>
<p>Euler's number is the limit of intrest, ie: if you made an infinitesimal amount of interest constantly (at an infinitesimal interval, in the end, you would have $x*e$ money, x being the money you started with.</p>
<p>A popular forumula containing ie is: $e^{\pi i} = -1$</p>
|
2,086,598 | <p>Let's start with a well known limit:
$$\lim_{n \to \infty} \left(1 + {1\over n}\right)^n=e$$</p>
<p>As $n$ approaches infinity, the expression evaluates Euler's number $e\approx 2.7182$. Why that number? What properties does the limit have that leads it to 2.71828$\dots$ and why is this number important? Where can we find $e$, in what branch of Mathematics?</p>
| Doug M | 317,162 | <p>It is not so important that it equals $2.718\cdots$ but that it converges to some number. </p>
<p>When Napier,Jacob Bernoulli, et. al. were looking at it, they were thinking about the implications of the compound interest calculation. And the effect of increasing the compounding frequency has an upper bound.</p>
<p>Euler, a student of Bernoulli's, was working with the calculus of exponential functions and found $e^x$ to be unchanged when subjected to differentiation. This makes it an incredibly important function to the calculus.</p>
<p>He then went on to show how the exponential function becomes a trigonometric function when when taken to an "imaginary" power. That is $e^{ix} = \cos x + i\sin x$ or $e^{\pi i} = -1.$ This opens up the field of complex analysis.</p>
|
3,476,181 | <p>I'm trying to solve this group theory problem, and I'm really not sure how to approach this. The question is:</p>
<p>Show that <span class="math-container">$\mathbb{F}_7[\sqrt{-1}] := \{a+b\sqrt{-1}$</span> | <span class="math-container">$a,b \in \mathbb{F}_7\}$</span> is a ring.</p>
<p>I have been stuck on this problem for hours and I really cannot figure it out. </p>
<p>This is my progress so far:</p>
<p><span class="math-container">$(a+b\sqrt{-1}) + (c+d\sqrt{-1}) = (a+c) + (b+d)\sqrt{-1}$</span></p>
<p>I'm not sure what I'm doing, I'll appreciate it if anyone can help me out. Thanks in advance!</p>
| bof | 111,012 | <p>Your intuition is correct. Quoting Wikipedia on <a href="https://en.wikipedia.org/wiki/Higman%27s_lemma" rel="noreferrer">Higman's lemma</a>:</p>
<blockquote>
<p>Higman's lemma states that the set of finite sequences over a finite alphabet, as partially ordered by the subsequence relation, is <a href="https://en.wikipedia.org/wiki/Well-quasi-ordering" rel="noreferrer">well-quasi-ordered</a>. That is, if <span class="math-container">$w_1,w_2,\dots$</span> is an infinite sequence of words over some fixed finite alphabet, then there exist indices <span class="math-container">$i\lt j$</span> such that <span class="math-container">$w_i$</span> can be obtained from <span class="math-container">$w_j$</span> by deleting some (possibly none) symbols. More generally this remains true when the alphabet is not necessarily finite, but is itself well-quasi-ordered, and the subsequence relation allows the replacement of symbols by earlier symbols in the well-quasi-ordering of labels. This is a special case of the later <a href="https://en.wikipedia.org/wiki/Kruskal%27s_tree_theorem" rel="noreferrer">Kruskal's tree theorem</a>. It is named after <a href="https://en.wikipedia.org/wiki/Graham_Higman" rel="noreferrer">Graham Higman</a>, who published it in 1952. </p>
</blockquote>
|
3,684,158 | <p>You are given two circles:</p>
<p>Circle G: <span class="math-container">$(x-3)^2 + y^2 = 9$</span></p>
<p>Circle H: <span class="math-container">$(x+3)^2 + y^2 = 9$</span></p>
<p>Two lines that are tangents to the circles at point <span class="math-container">$A$</span> and <span class="math-container">$B$</span> respectively intersect at a point <span class="math-container">$P$</span> such that <span class="math-container">$AP + BP = 10$</span></p>
<p>Find the locus of all points <span class="math-container">$P$</span>.</p>
<p><a href="https://i.stack.imgur.com/zKl67.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zKl67.png" alt="enter image description here"></a></p>
<hr>
<p>This problem is solvable if we set point <span class="math-container">$P = (x,y)$</span> and solve the equation <span class="math-container">$AP + BP = 10$</span>. After substituting <span class="math-container">$GP^2 = AP^2 + 3^2$</span> and <span class="math-container">$HP^2 = BP^2 + 3^2$</span> and getting the following equation for an ellipse </p>
<p><span class="math-container">$16x^2 +25y^2 = 625$</span></p>
<p>That's a lot of math and algebra to do, so my question is: What is the geometric reasoning behind why is the locus an ellipse (without using analytical geometry) or is there any other elegant proofs that lack heavy calculations?</p>
| Ng Chung Tak | 299,599 | <p>Not answering the question but giving further observation,</p>
<p><span class="math-container">\begin{align}
\sqrt{(x-r)^2+y^2-r^2} \pm \sqrt{(x+r)^2+y^2-r^2} &= 2s \\
\sqrt{x^2-2rx+y^2} \pm \sqrt{x^2+2rx+y^2} &= 2s \\
2(x^2+y^2) \pm 2\sqrt{(x^2+y^2)^2-4r^2x^2} &= 4s^2 \\
(x^2+y^2)^2-4r^2x^2 &= 4s^4-4s^2(x^2+y^2)+(x^2+y^2)^2 \\
(s^2-r^2)x^2+s^2y^2 &= s^4 \\
\end{align}</span></p>
<ul>
<li><p>Positive sign is taken when <span class="math-container">$s^2>r^2$</span> giving an ellipse.</p></li>
<li><p>Negative sign is taken for constant difference instead, the locus can be two horizontal lines <span class="math-container">$(s^2=r^2)$</span>, a hyperbola <span class="math-container">$(s^2<r^2)$</span> or a vertical line <span class="math-container">$(s=0)$</span>.</p></li>
<li><p>The loci always pass through the point <span class="math-container">$(0,\pm s)$</span>.</p></li>
<li><p>The loci don't have contact with two circles when <span class="math-container">$s^2>2r^2$</span>.</p></li>
<li><p>The eccentricity is given by <span class="math-container">$e=\dfrac{r}{|s|}$</span>.</p></li>
</ul>
|
2,796,854 | <p>what happens if both the second and first derivatives at a certain point are $0$? Is it an infliction point, an extremum point, or neither? Can we say anything at all about a point in such a case? </p>
| tien lee | 557,074 | <p>• if $\frac{df}{dx}(p) = 0$, then $x = p$ is called a critical point of $f(x)$, and we do not know anything new about the behavior of $f( x) $ at $x = p$.</p>
<p>• if $\frac{d^2f}{dx^2} (p) = 0$ at $x = p$, then we do not know anything new about the behavior of $f(x)$ at $x = p$.</p>
<p>As @arugula mentioned $f(x)=\begin{cases} x^3\sin(1/x)&x\neq 0\\ 0&x=0\end{cases}$</p>
|
3,407,174 | <p>I'm dealing with a specific polynomial function. The first derivative of it is displayed below. As you can see, it has the shape of an asymetric function. But what does this tells us about the initial function in general and in terms of finding a maximum or a minimum for it? I know it is a general question, but I find hardly any documentation on this and I think it's good if someone can give general information when the first derivative takes this form.</p>
<p>Any information is appreciated.</p>
<p><a href="https://i.stack.imgur.com/ECkgk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ECkgk.png" alt="enter image description here"></a></p>
<p>edit: when I have different inputs in the function, the first derivative looks like this:</p>
<p><a href="https://i.stack.imgur.com/Cfv3Z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Cfv3Z.png" alt="enter image description here"></a></p>
| ling | 670,949 | <p>Cauchy inequality
<span class="math-container">$$(a_1+a_2+\cdots +a_m)^2\leq m(a_1^2+a_2^2+\cdots +a_m^2).$$</span>
Hence
<span class="math-container">$$(1+\frac{1}{\gamma})(x_1^2+\cdots +x_m^2)-\frac{(x_1+\cdots +x_m)^2}{m}\geq \frac{1}{\gamma}|x|^2>0.$$</span>
That is positive definite.</p>
|
1,638,490 | <p>Yes, I know the title is bizarre. I was urinating and forgot to lift the seat up. That made me wonder: assuming I maintain my current position, is it possible for the toilet seat (assume it is a closed, but otherwise freely deformable curve) to be moved/deformed such that stream does not pass through the hole anymore without it intersecting the curve (in other words, spraying urine everywhere!)?</p>
| Ross Millikan | 1,827 | <p>If it is deformable, pass it under your feet, up your back, and over your head. It takes a bunch of deformation to make that work.</p>
|
55,071 | <p><strong>Bug introduced in 10.0.0 and persisting through 10.3.0 or later</strong></p>
<hr>
<p>I've upgraded my home installation of <em>Mathematica</em> from version 9 to 10 today on a Windows 8.1 machine, and I'm getting a weird font issue - the fonts are not anti-aliased, and look unbalanced and weird. Just look:</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/cnuXE.png" alt="HALP"></p>
</blockquote>
<p>For comparison, here what it looks on Linux with <em>Mathematica</em> V10</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/Qf6kO.png" alt="Mathematica graphics"></p>
</blockquote>
<p>At this point you may object, as these issues are too minor to get worked up about. But neither I nor my otherwise benign OCD can work like this. Any ideas?</p>
<p><strong>EDIT</strong> I've just had an idea, that maybe I need to manually remove old fonts leftover from the old <em>Mathematica</em> 9 installation. I've read somewhere that they were not going to pollute the main font catalog with symbol fonts in the next release, and maybe the new <em>Mathematica</em> is using old fonts for some reason. I can't test it right now myself, unfortunately.</p>
| Karsten 7. | 18,476 | <p>I analyzed which fonts are loaded when starting Mma v10 compared to v9.
The problem could be tracked down to the loading of the fonts in the Folder </p>
<p><code>$InstallationDirectory\SystemFiles\Fonts\TrueType</code></p>
<p>If you open <code>Mathematica-Bold.ttf</code> or <code>MathematicaMono-Bold.ttf</code>, you will see how ugly these are.</p>
<p>To prevent loading these fonts one should make a backup copy of this <code>TrueType</code> folder and than delete all <code>.ttf</code> files that are in it. A simple renaming of the folder increased the Mma startup time and is therefore not a good idea. </p>
<p><strong>Edit/Warning:</strong><br>
AutoReplacements will not work after this customization!</p>
|
3,885,025 | <p>Let be <span class="math-container">$(A,M)$</span> a local ring (if it needs noetherian ring), <span class="math-container">$P_1$</span> and <span class="math-container">$P_2$</span> two prime ideals with <span class="math-container">$P_1\subseteq P_2$</span>, <span class="math-container">$\ a_1\in M\setminus P_2$</span>. I don't know if it suffices to conclude this sentence:</p>
<p>If <span class="math-container">$P_1+(a)=P_2+(a)$</span> so <span class="math-container">$P_1=P_2$</span>.</p>
<p>Someone can help me?
Maybe in a more general setting it is possible to say that if <span class="math-container">$P+I=Q+I$</span> so <span class="math-container">$P=Q$</span>.</p>
<p>Thank you.</p>
| Eric Wofsey | 86,856 | <p>Here is an example to show that this need not be true if <span class="math-container">$A$</span> is not Noetherian. Let <span class="math-container">$A$</span> be a valuation ring with value group <span class="math-container">$\mathbb{Z}^2$</span> with the lexicographic order. Let <span class="math-container">$P_1=0$</span> and let <span class="math-container">$P_2$</span> be the prime ideal consisting of elements whose valuation has positive first coordinate. Let <span class="math-container">$a\in A$</span> be any element with valuation <span class="math-container">$(0,1)$</span>. Then <span class="math-container">$P_1\subset P_2\subset (a)$</span> so <span class="math-container">$P_1+(a)=P_2+(a)=(a)$</span> but <span class="math-container">$P_1\neq P_2$</span>.</p>
|
3,097,640 | <p>So I have the following problem: $a + b = c + c.
I want to prove that the equation has infinitely many relatively prime integer solutions.</p>
<p>What I did first was factor the right side to get:
(</p>
| Mark Bennet | 2,906 | <p>Suppose <span class="math-container">$c=u^2+v^2$</span> then you have <span class="math-container">$$(uc^2+v)^2+(vc^2-u)^2=c(c^4+1)$$</span> and you want <span class="math-container">$uc^2+v$</span> and <span class="math-container">$vc^2-u$</span> to be coprime.</p>
<p>Now just choose <span class="math-container">$u=1$</span> so that the expressions are <span class="math-container">$a=c^2+v$</span> and <span class="math-container">$b=vc^2-1$</span> with <span class="math-container">$c=v^2+1$</span></p>
<p>Note then that that a common factor of <span class="math-container">$a$</span> and <span class="math-container">$b$</span> is also a factor of <span class="math-container">$va-b=v^2+1=c$</span> and <span class="math-container">$ac^2-b=c^4+1$</span>, but <span class="math-container">$c$</span> and <span class="math-container">$c^4+1$</span> are coprime.</p>
<hr>
<p>So you get a family of solutions with </p>
<p><span class="math-container">$c=v^2+1$</span> and </p>
<p><span class="math-container">$a=c^2+v=v^4+2v^2+v+1$</span> and </p>
<p><span class="math-container">$b=vc^2-1=v^5+2v^3+v-1$</span></p>
<p>[Which, I notice is just Sam's parametric solution with <span class="math-container">$u=1$</span>. It should be obvious what to do to generalise.]</p>
|
2,219,266 | <blockquote>
<p>Compute integral $\displaystyle \int_{-\infty}^{\infty}e^{-k^2t+ikx}\, dk$.</p>
</blockquote>
<p>Hint: Complete the square in the Exponent.</p>
<p>Okay, for the Exponent, we have
$$
-k^2t+ikx=-t\cdot\left(k-\frac{ix}{2t}\right)^2-\frac{x^2}{4t}.
$$</p>
<p>Now, is it easier to compute
$$
\int_{-\infty}^{\infty}\exp\left(-t\cdot\left(k-\frac{ix}{2t}\right)^2-\frac{x^2}{4t}\right)\, dk?
$$
Don't see the trick.</p>
| Bram28 | 256,001 | <p>The answer depends on <em>how</em> "you know that at least one of them is male".</p>
<p>Here are two different types of scenarios:</p>
<ol>
<li><p>Someone one day just told you that this woman had a triplet and that 'at least one of them is a boy'</p></li>
<li><p>You met this woman some day before on the street (or store, or ..) and she had 1 child with her that was a boy, and she told you this was one of a triplet that she had</p></li>
</ol>
<p>Notice that in both scenarios you know that she had a triplet and that at least one of them is male.</p>
<p>In scenario 1 (which I would guess is what the book/instructor is looking for), you use the calculation used by ConnorHarris, since out of the 8 possible and initially equally likely triplets (MMM, MMF, MFM, FMM, MFF, FMF, FFM, FFF), you can rule out the FFF one, so you have a $\frac{3}{7}$ chance of one male and two females, a $\frac{3}{7}$ chance of two male and one female, and a $\frac{1}{7}$ chance of all three being male, and so the probability of the one child being with her today being male is:</p>
<p>$\left(\dfrac{1}{3} \times \dfrac{3}{7}\right) + \left(\dfrac{2}{3} \times \dfrac{3}{7}\right) + \left( \dfrac{3}{3} \times \dfrac{1}{7} \right) = \dfrac{4}{7}.$</p>
<p>In scenario 2 (which is actually a far more likely kind of scenario to happen in the real world; indeed, note that you are meeting her <em>today</em> in that exact kind of situation ...), there is a $\frac{1}{4}$ chance that both other children are female, a $\frac{1}{2}$ chance that there is one more male and one more female, and a $\frac{1}{4}$ chance that both other children are male. In other words, there is a $\frac{1}{4}$ chance of one male and two females, a $\frac{1}{2}$ chance of two male and one female, and a $\frac{1}{4}$ chance of all three being male.</p>
<p>So now you get that the probability of the one child with her today being male is:</p>
<p>$\left(\dfrac{1}{3} \times \dfrac{1}{4}\right) + \left(\dfrac{2}{3} \times \dfrac{1}{2}\right) + \left( \dfrac{3}{3} \times \dfrac{1}{4} \right) = \dfrac{2}{3}.$</p>
<p>In sum: the question is actually ambiguous, and even though the book is most likely looking for the answer compatible with scenario 1, if a situation like this ever comes up in real life, you are more likely to deal with a scenario 2 type situation!</p>
|
903,117 | <p>I am trying to evaluate
$$\int_{-\infty}^{\infty} \frac{\sin(x)^2}{x^2} dx $$
Would a contour work? I have tried using a contour but had no success.
Thanks.</p>
<p>Edit: About 5 minutes after posting this question I suddenly realised how to solve it. Therefore, sorry about that. But thanks for all the answers anyways.</p>
| SuperAbound | 140,590 | <p>This is an approach using contour integration.
\begin{align}
\int_{\mathbb{R}}\frac{\sin^2{x}}{x^2}{\rm d}x
&=-\frac{1}{4}\lim_{\epsilon \to 0}\int_{\mathbb{R}}\frac{e^{2ix}-2+e^{-2ix}}{(x-i\epsilon)^2}{\rm d}x\tag1\\
&=-\frac{1}{4}\lim_{\epsilon \to 0}2\pi i\lim_{z \to i\epsilon}\frac{{\rm d}}{{\rm d}z}(e^{2iz}-2)\tag2\\
&=\frac{1}{4}\lim_{\epsilon \to 0}4\pi e^{-2\epsilon}\\
&=\pi
\end{align}
Explanation:<br/>
$(1)$: Expand $\sin^2{x}$ in terms of complex exponentials and shift the pole upwards. <br/>
$(2)$: Split the integral into $2$. Integrate the first along a semicircle in the uhp, and the second along a semicircle in the lhp. The second integral $=0$ since it encloses no poles. <br/><br/>
Alternatively, one may use the fact that
$$\int^\infty_0t^{n-1}e^{-xt}{\rm d}t=\frac{\Gamma(n)}{x^n}$$
It follows that
\begin{align}
2\int^\infty_0\frac{\sin^2{x}}{x^2}{\rm d}x
&=2\int^\infty_0t\int^\infty_0e^{-xt}\sin^2{x} \ {\rm d}x \ {\rm d}t\\
&=2\int^\infty_0\int^\infty_0e^{-xt}\sin{2x} \ {\rm d}x \ {\rm d}t \tag{Integrated by parts}\\
&=2\int^\infty_0\frac{2}{t^2+4}{\rm d}t\\
&=\pi
\end{align}</p>
|
119,532 | <p>I have a <code>ListAnimate</code> where the frames consist of graphics made of lines and points. I want to make sure that, when the animation is paused and then resumed, all frames appear normal again, so if you stop it to rotate the plot (it's 3D), after a loop it looks normal again.</p>
<p>My code is the following:</p>
<pre><code>ListAnimate[
{Labeled[Show[Graphics3D[g0, ViewPoint -> Front], ImageSize -> Full], S0, Top],
Labeled[Show[Graphics3D[g1, ViewPoint -> Front], ImageSize -> Full], S1, Top],
...,
Labeled[Show[Graphics3D[g51, ViewPoint -> Front], ImageSize -> Full], S51, Top]},
4, AnimationRunning -> False]
</code></pre>
<p>where g1, ..., g51 are the images and S1, ..., S51 are timestamps that I print on the top.</p>
<p>I read in another answer that setting <code>PreserveImageOptions -> False</code> should take care of this. However, the option doesn't do anything, I tried all possible placing of it as well as wrapping the whole thing in a <code>Manipulate</code> expression.</p>
| MarcoB | 27,951 | <p><code>PlusMinus</code> formats nicely, but it does not have a built-in meaning. You may work around that:</p>
<pre><code>h = 0.08; t = 0.13;
Plot[
Evaluate[
Sqrt[(1/4) + (1/64 h^2) - x^2] - (1/8 h) + {-1, 1} ((3/8) t (1 - 2 x) Sqrt[1 - (2 x)^2])
], {x, -.5, .5}
]
</code></pre>
<p><img src="https://i.stack.imgur.com/EtaiY.png" alt="Mathematica graphics"></p>
<hr>
<p>If need be, you could also define your own meaning for <code>PlusMinus</code>:</p>
<pre><code>Clear[PlusMinus]
PlusMinus[a__] := {-1, 1} (a)
3 + PlusMinus[a]
(* Out: {3 - a, 3 + a} *)
</code></pre>
<hr>
<p>As @JM mentioned, however, I suspect that your parametrization may not be the most advantageous; here is a complex parametrization of the Joukowsky transformation (see <a href="https://en.wikipedia.org/wiki/Joukowsky_transform" rel="noreferrer">Wikipedia</a> as well).</p>
<p>The Joukowsky transform is the following transformation of complex numbers:</p>
<p>$$z=\zeta+\frac{1}{\zeta}$$</p>
<p>We can obtain a complex parametrization of the results of the transform $z$ as follows:</p>
<pre><code>Block[{$Assumptions = {chi > 0, eta > 0}, x, y},
zeta = chi + I eta;
{x, y} = Together@Simplify@ReIm@ComplexExpand[zeta + 1/zeta]
]
(* {(chi*(1 + chi^2 + eta^2))/(chi^2 + eta^2),
(eta*(-1 + chi^2 + eta^2))/(chi^2 + eta^2)}
*)
</code></pre>
<p><img src="https://i.stack.imgur.com/QJshl.png" alt="Mathematica graphics"></p>
<p>The interesting property of the Joukowski transform is the fact that the transform of a circle passing through $\zeta=1$ and containing in its interior the point $\zeta=-1$ describes the profile of an airfoil. Even more interesting for practical purposes is the fact that the streamlines of the original circular profile (typically easy to calculate) transform to the streamlines of the airfoil profile (otherwise hard to calculate) (see e.g. <a href="http://mathfaculty.fullerton.edu/mathews/c2003/JoukowskiTransMod.html" rel="noreferrer">here</a>).</p>
<p>Here is a quick manipulator to visualize those conditions and find some reasonable starting values for such circles:</p>
<pre><code>Manipulate[
Graphics[{
PointSize[0.02], Point[{{1, 0}, {-1, 0}}],
Circle[center, radius],
Inset["Center=" <> ToString@N@center, {-1, 0.5}],
Inset["Radius=" <> ToString@InputForm@radius, {0.6, 0.8}]
}, Axes -> True
],
{{center, {-1/2, 2/10}}, Locator},
{{radius, 1}, 0, 2, 1/10}
]
</code></pre>
<p><img src="https://i.stack.imgur.com/E3r3T.png" alt="Mathematica graphics"></p>
<p>A good set of values is, for instance, a circle centered at $(-1/8,3\ \sqrt{31}/40)$ with radius $1.2$. We can use a simple parametrization of a circle (<code>{r Cos[t] + xc, r Sin[t] + yc}</code>) toghether with the transformation results obtained above, and use ParametricPlot to generate a plot of the Joukowski-transformed circle:</p>
<pre><code>ParametricPlot[
{chi (chi^2 + eta^2 + 1)/(chi^2 + eta^2),
eta (chi^2 + eta^2 - 1)/(chi^2 + eta^2)
} /. {chi -> r Cos[t] - 1/8, eta -> r Sin[t] + 3 Sqrt@31/41} /. r -> 12/10,
{t, 0, 2 Pi},
PlotRange -> All, PlotRangePadding -> {Scaled[0.05], Scaled[0.25]},
Frame -> True, Axes -> None,
AspectRatio -> 0.4
]
</code></pre>
<p><img src="https://i.stack.imgur.com/2yMhs.png" alt="Mathematica graphics"></p>
|
3,777,049 | <p>Suppose that <span class="math-container">$(H_i)_{i \in I}$</span> is a collection of closed <strong>orthogonal</strong> subspaces of the Hilbert space <span class="math-container">$H$</span>. Suppose that <span class="math-container">$\sum_{i \in I} \Vert x_i \Vert^2 < \infty$</span>. Prove that <span class="math-container">$\sum_{i \in I} x_i$</span> converges in <span class="math-container">$H$</span>.</p>
<p>Here <span class="math-container">$\sum_{i \in I} x_i$</span> is the norm-limit of the net <span class="math-container">$(\sum_{i \in J} x_i)$</span> where <span class="math-container">$J$</span> ranges over all finite subsets of <span class="math-container">$I$</span>, ordered by inclusion.</p>
<p><strong>Attempt</strong>:</p>
<p>It suffices to check that <span class="math-container">$(\sum_{i \in J} x_i)_J$</span> is a Cauchy net in <span class="math-container">$H$</span>. So, let <span class="math-container">$\epsilon > 0$</span>. Since <span class="math-container">$\sum_{i \in I} \Vert x_i \Vert^2 < \infty$</span>, we have that <span class="math-container">$(\sum_{i \in J} \Vert x_i \Vert^2)_J$</span> is a Cauchy net. Thus, there is a finite subset <span class="math-container">$J_0 \subseteq I$</span> such that if <span class="math-container">$K,L$</span> are finite subsets of <span class="math-container">$I$</span> containing <span class="math-container">$J_0$</span>, then <span class="math-container">$$\sum_{K \triangle L} \Vert x_i \Vert^2 = |\sum_K \Vert x_i \Vert^2 - \sum_L \Vert x_i\Vert^2 | < \epsilon$$</span></p>
<p>Here <span class="math-container">$K \triangle L = (K \setminus L) \cup (L \setminus K)$</span> is the symmetric difference.</p>
<p>Consequently, for <span class="math-container">$K,L$</span> as above
<span class="math-container">$$\Vert \sum_K x_i - \sum_L x_i \Vert ^2 = \Vert \sum_{K\triangle L} x_i \Vert ^2 = \sum_{K \triangle L}
\Vert x_i \Vert^2 < \epsilon$$</span></p>
<p>Hence <span class="math-container">$(\sum_{i \in J} x_i)_J$</span> is a Cauchy net in <span class="math-container">$H$</span> and we are done.</p>
<p>Is this correct? I think the step with the <span class="math-container">$\triangle $</span> might be flawed.</p>
| Paul Frost | 349,785 | <p>I guess you forgot to mention that <span class="math-container">$x_i \in H_i$</span>. It would also be sufficient to assume that the <span class="math-container">$x_i$</span> are pairwise orthogonal. This implies that for finite <span class="math-container">$A \subset I$</span> and <span class="math-container">$\sigma(i) = \pm 1$</span>
<span class="math-container">$$\Vert \sum_A (-1)^{\sigma(i)}x_i \Vert^2 = \left\langle \sum_A (-1)^{\sigma(i)}x_i ,\sum_A (-1)^{\sigma(i)}x_i \right\rangle = \sum_{i \in A, j \in A} (-1)^{\sigma(i)}(-1)^{\sigma(j)}\langle x_i,x_j\rangle\\ = \sum_{A} (-1)^{\sigma(i)}(-1)^{\sigma(i)}\lVert x_i \rVert^2 = \sum_{A} \lVert x_i \rVert^2 . $$</span>
It is in general not true that
<span class="math-container">$$\sum_{K \triangle L} \Vert x_i \Vert^2 = \left|\sum_K \Vert x_i \Vert^2 - \sum_L \Vert x_i\Vert^2 \right| .$$</span>
Anyway, we do not need it. Your equation
<span class="math-container">$$\Vert \sum_K x_i - \sum_L x_i \Vert ^2 = \Vert \sum_{K\triangle L} x_i \Vert ^2 = \sum_{K \triangle L} \Vert x_i \Vert^2 \tag{1}$$</span>
is true, but requires a proof which you have not given and moreover needs an explanation what it has to with <span class="math-container">$(\sum_J x_i)$</span> being a Cauchy net.</p>
<p>Let us therefore start at the beginning. What we know is that
<span class="math-container">$$\left|\sum_A \Vert x_i \Vert^2 - \sum_B \Vert x_i\Vert^2 \right| < \epsilon $$</span>
for finite <span class="math-container">$A, B \supset J_0$</span>. Thus in particular for any finite set <span class="math-container">$F \subset I$</span> such that <span class="math-container">$F \cap J_0 = \emptyset$</span> we may take <span class="math-container">$A = J_0 \cup F$</span> and <span class="math-container">$B = J_0$</span> and obtain
<span class="math-container">$$\sum_{F}\Vert x_i \Vert^2 < \epsilon . \tag{2}$$</span></p>
<p>For finite <span class="math-container">$K, L \supset J_0$</span> we therefore get
<span class="math-container">$$\Vert \sum_K x_i - \sum_L x_i \Vert ^2 = \Vert \sum_{K \triangle L} (-1)^{\sigma(i)}x_i \Vert ^2 = \sum_{K \triangle L} \Vert x_i \Vert^2 < \epsilon$$</span>
because <span class="math-container">$K \triangle L$</span> is disjoint from <span class="math-container">$J_0$</span>. Note that in <span class="math-container">$(1)$</span> the middle term <span class="math-container">$\Vert \sum_{K\triangle L} x_i \Vert ^2$</span> is irrelevant.</p>
|
4,320,209 | <p><em>I am stuck on the following two questions on matrix Algebra:</em></p>
<p>Let <span class="math-container">$x$</span> be a vector. Find the following:</p>
<ul>
<li><span class="math-container">$\frac{\partial}{\partial x}||x\otimes x||^2$</span>, here ||.|| denotes Euclidean norm of a vector and <span class="math-container">$\otimes$</span> denotes the kroneckar product.</li>
<li><span class="math-container">$\frac{\partial}{\partial x^T}(\frac{x}{||x||})$</span>.</li>
</ul>
<p><strong>My try</strong>:
<strong>(i)</strong>
Let us take <span class="math-container">$x=(x_1,x_2,\dots,x_n)$</span>.
Then <span class="math-container">$x\otimes x=\begin{pmatrix}x_1^2 &x_1x_2& x_1x_3 &\dots & x_1x_n
\\
x_2x_1 &x_2^2& x_2x_3 &\dots & x_2x_n
\\
\dots & \dots& \dots & \dots & \dots\\
\dots & \dots& \dots & \dots & \dots\\
x_nx_1 &x_nx_2& x_nx_3 &\dots & x_n^2
\end{pmatrix}$</span></p>
<p>I am confused about how do I take <span class="math-container">$||x\otimes x||$</span> and find its derivative in terms of <span class="math-container">$x$</span>.</p>
<p><strong>(II)</strong> Since <span class="math-container">$||x||$</span> is a scalar, so we can write <span class="math-container">$\frac{\partial}{\partial x^T}(\frac{x}{||x||})=\frac{1}{||x||}\frac{\partial}{\partial x^T}(x)$</span>.
Here I am stuck how to find the derivative of <span class="math-container">$x$</span> with respect to <span class="math-container">$x^T$</span>.</p>
<p>Can someone please help me to clear my doubts and complete the above problems?</p>
| greg | 357,854 | <p><span class="math-container">$
\def\a{\alpha}\def\b{\beta}\def\l{\lambda}\def\o{{\tt1}}
\def\B{\Big}\def\L{\left}\def\R{\right}
\def\LR#1{\L(#1\R)}
\def\BR#1{\B(#1\B)}
\def\trace#1{\operatorname{Tr}\LR{#1}}
\def\v#1{\operatorname{vec}\LR{#1}}
\def\qiq{\quad\implies\quad}
\def\t{\otimes}
\def\p{\partial}
\def\grad#1#2{\frac{\p #1}{\p #2}}
\def\c#1{\color{red}{#1}}
$</span>The Frobenius product is a concise notation for the trace
<span class="math-container">$$\eqalign{
A:B &= \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} \;=\; \trace{A^TB} \\
A:A &= \big\|A\big\|^2_F \\
}$$</span>
This is also called the double-dot or double contraction product.
<br>When applied to vectors <span class="math-container">$(n=\o)$</span> it reduces to the standard dot product.
<br>The distributive rule for a mixed Kronecker-Frobenius product is
<span class="math-container">$$\eqalign{
(A\t B):(X\t Y) = (A:X)\t(B:Y) \\
}$$</span></p>
<p>First calculate the gradient of the simple function
<span class="math-container">$$\eqalign{
\a &= \|x\|^2 = x:x \qiq \c{d\a = 2x:dx} \\
}$$</span>
Then calculate the differential (and gradient)
for the first function of interest <span class="math-container">$(\b)$</span>.
<span class="math-container">$$\eqalign{
\b &= \|x\t x\|^2 \\&= (x\t x):(x\t x) \\&= (x:x)\t(x:x) \\&= \a^2 \\
d\b &= 2\a\;\c{d\a} \\&= 2\a\,\LR{2x\,dx} \\&= 4\|x\|^2\,x\,dx \\
\grad{\b}{x} &= 4\|x\|^2\,x \\
}$$</span>
The second function of interest <span class="math-container">$(y)$</span>
can also make use of the simple <span class="math-container">$\a$</span> function.
<span class="math-container">$$\eqalign{
\l &= \|x\| \\
\l^2 &= \a \qiq 2\l\,d\l = d\a = 2x^Tdx \\
d\l &= \l^{-1}x^Tdx \\
\\
y &= \frac{x}{\|x\|} = \l^{-1}x \\
dy &= \l^{-1}dx-x\l^{-2}\,\c{d\l} \\
&= \l^{-1}I\,dx-x\l^{-2}\,\c{\l^{-1}x^Tdx} \\
&= \l^{-1}\LR{I-yy^T}\,dx \\
\grad{y}{x} &= \l^{-1}\LR{I-yy^T} \\\\
}$$</span></p>
<hr />
<p><strong>NB:</strong> Your expansion of <span class="math-container">$(x\t x)$</span> is incorrect. What you have
written is actually the matrix <span class="math-container">$xx^T$</span> which you must
<a href="https://en.wikipedia.org/wiki/Vectorization_(mathematics)#Compatibility_with_Kronecker_products" rel="nofollow noreferrer">vectorize</a> to recover the desired Kronecker product, i.e.
<span class="math-container">$$(x\t x) = \v{xx^T}$$</span>
However, there is another property of the Frobenius product
<span class="math-container">$$\v{A}:\v{A} = A:A$$</span>
which allows you to use your matrix without vectorizing it
<span class="math-container">$$\eqalign{
\b &= \v{xx^T}:\v{xx^T} \\&= \LR{xx^T}:\LR{xx^T} \\
&= \trace{xx^Txx^T} \\&= \a^2
}$$</span></p>
|
3,745,097 | <p>In my general topology textbook there is the following exercise:</p>
<blockquote>
<p>If <span class="math-container">$F$</span> is a non-empty countable subset of <span class="math-container">$\mathbb R$</span>, prove that <span class="math-container">$F$</span> is not an open set, but that <span class="math-container">$F$</span> may or may not but a closed set depending on the choice of <span class="math-container">$F$</span>.</p>
</blockquote>
<p>I already proved that <span class="math-container">$F$</span> is not opened in the euclidean topology, but why is the second part true?</p>
<p>If <span class="math-container">$F$</span> is countable then <span class="math-container">$F \sim \mathbb N$</span>. This means that we can list the elements of <span class="math-container">$F$</span>, so we can write: <span class="math-container">$F=\{f_1,...,f_k,...\}$</span></p>
<p><span class="math-container">$\mathbb R \setminus F= (-\infty, f_1) \cup \bigcup \limits _{i=1}^{\infty}(f_i,f_{i + 1})$</span></p>
<p>We have that <span class="math-container">$(-\infty, f_1) \in \tau$</span> and that every <span class="math-container">$(f_i,f_{i + 1}) \in \tau$</span>. Because the union of elements of <span class="math-container">$\tau$</span> is also a element of <span class="math-container">$\tau$</span>, we have that <span class="math-container">$(-\infty, f_1) \cup \bigcup \limits _{i=1}^{\infty}(f_i,f_{i + 1}) \in \tau$</span>, then <span class="math-container">$F$</span> is closed.</p>
<p>Is this correct, because the statement says that "may or may not but a closed set depending on the choice of <span class="math-container">$F$</span>"?</p>
| atul ganju | 788,520 | <p>A countably large set consisting of the points in this sequence <span class="math-container">$a_n = \frac{1}{2^n}$</span> is not closed. It does not contain the limit point <span class="math-container">$0$</span>.</p>
|
3,962,058 | <p>Let <span class="math-container">$A$</span> be a commutative ring with identity, and <span class="math-container">$P$</span> a prime ideal of <span class="math-container">$A$</span>. I want to show that <span class="math-container">$\lim_{f \notin P} A_{f} = A_{P}$</span>. <span class="math-container">$A_f$</span> is the localization of <span class="math-container">$A$</span> at <span class="math-container">$\{1\} \cup \{f^{n}: n \in \mathbb N\}$</span>, and <span class="math-container">$A_P$</span> is the localization of <span class="math-container">$A$</span> at <span class="math-container">$A \setminus P$</span>. There is a natural map from <span class="math-container">$A_{f} \to A_P$</span> if <span class="math-container">$f$</span> is not in <span class="math-container">$P$</span>, so by the universal property of the direct limit, we get a map from <span class="math-container">$\Phi : \lim_{f \notin P} A_{f} \to A_{p}$</span>. Then it remains to show that this canonical map is an isomorphism. Surjectivity can be proven just using the universal property without knowing what <span class="math-container">$\Phi$</span> looks like. However I cannot prove the injectivity using the universal property. It seems that one has to use the explicit form of <span class="math-container">$\Phi$</span> in order to achieve that.</p>
| Karl Kroningfeld | 67,848 | <p>Take an element <span class="math-container">$\xi$</span> of the kernel of <span class="math-container">$\Phi$</span>, represented by <span class="math-container">$\frac a{f^n}$</span> for some <span class="math-container">$f\not\in P$</span>. We know that <span class="math-container">$$\frac a{f^n}=0\quad\text{in the ring $A_P$}$$</span>whence there exists <span class="math-container">$g\not\in P$</span> such that <span class="math-container">$ga=0$</span>, so that <span class="math-container">$$\frac a{f^n}=\frac {ag^n}{(fg)^n} = 0\quad\text{in $A_{fg}$}$$</span>This means that <span class="math-container">$\xi=0$</span> (in <span class="math-container">$\varinjlim A_f$</span>).</p>
|
2,102,356 | <blockquote>
<p>Under the assumption that a pair of dice is fair, the probability is
0.88 that the number of 9's appearing in 288 throws of the dice will lie within 32 $\pm$ K. What is K?</p>
</blockquote>
<p>I have found the standard deviation using </p>
<p>$\sqrt{mean\:\cdot \:trials\:\cdot \:possibility\:of\:pair\:of\:fair\:dice\:show\:9}$</p>
<p>but I cannot go any further from that.</p>
| M.Diggerson | 401,465 | <p><strong>Hint</strong>
$\Pr$(pair of fair dice show 9) = $4/36 = 1/9$</p>
<p>$(1/9 + 8/9)^{288}$ models the outcomes of 288 trials</p>
|
2,102,356 | <blockquote>
<p>Under the assumption that a pair of dice is fair, the probability is
0.88 that the number of 9's appearing in 288 throws of the dice will lie within 32 $\pm$ K. What is K?</p>
</blockquote>
<p>I have found the standard deviation using </p>
<p>$\sqrt{mean\:\cdot \:trials\:\cdot \:possibility\:of\:pair\:of\:fair\:dice\:show\:9}$</p>
<p>but I cannot go any further from that.</p>
| lulu | 252,071 | <p>We note that this is a standard Binomial distribution with probability $p=\frac 19$. </p>
<p>If we like, we can approximate this with a normal distribution..with mean $\mu=32$ and $\sigma =\sqrt {288\times \frac 19\times 89}=\frac {16}3$ </p>
<p>We then seek $k$ such that $$\Phi(32+K,\mu,\sigma)-\Phi(32-K,\mu,\sigma)=.88$$</p>
<p>Here $\Phi$ denotes the cumulative distribution function for the associated normal distribution. That's easy enough to solve numerically as it is, but if you want to use standard functions it helps to remark that $$\Phi(32+K,\mu,\sigma)+\Phi(32-K,\mu,\sigma)=1$$ From which we conclude that $$2\Phi(32+K,\mu,\sigma)-1=.88\quad \implies \quad \Phi(32+K,\mu,\sigma)=.94$$</p>
<p>Which gives $K=8.292125838$. </p>
<p>Of course, $K$ is meant to be an integer for the true (discrete) situation so one is lead to suspect that $K=8$.</p>
<p>Of course, it is not difficult to do this exactly using the Binomial Distribution. Let $p_n$ denote the probability of getting exactly $n$ $9's$. We have $$p_n=\binom {288}n \left(\frac 19\right)^n\left(\frac 89\right)^{288-n}$$</p>
<p>It's not hard to compute these with mechanical assistance. Inspired by the normal approximation we start with $n=24$ and compute $$p_{24}=0.025039361,p_{25}=0.033051957,\cdots,p_{39}=0.030441785,p_{40}=0.023687514$$ We add these up to get $0.89005738$. So, I'd have liked $.89$ instead of $.88$ in the question, but $K=8$ appears to get the job done. if you only add from $p_{25}$ to $p_{39}$ you get $0.841330505$ which is too low</p>
|
2,010,329 | <p>The function under consideration is:</p>
<p>$$y = \int_{x^2}^{\sin x}\cos(t^2)\mathrm d t$$</p>
<p>Question asks to find the derivative of the following function. I let $u=\sin(x)$ and then $\tfrac{\mathrm d u}{\mathrm d x}=\cos(x)$. Solved accordingly but only to get the answer as
$$ \cos(x)\cos(\sin^2(x))-\cos(x)\cos(x^4) $$
but the answer is given as:
$$ \cos(x)\cos(\sin^2(x))-2x\cdot\cos(x^4) $$</p>
<p>May I know where I went wrong? Is my substitution wrong in the first place?</p>
| Hugo | 41,494 | <p>Well, when i to solve this kind of problem i usually do the following.</p>
<p>1) I define a two variable function like:</p>
<p>$$F(u,v) = \int_{v}^{u} \cos(t^2)dt.$$</p>
<p>2) What we want to calculate is $$\frac{dy}{dx}(x)$$ where $y(x) = F(\sin(x),x^2)$. Then we have:</p>
<p>$$\frac{dy}{dx}(x) = \frac{d}{dx}\{F(\sin(x),x^2)\} = \frac{\partial F}{\partial u}(\sin(x),x^2)\frac{d}{dx}\{\sin(x)\}+\frac{\partial F}{\partial v}(\sin(x),x^2)\frac{d}{dx}\{ x^2\}.$$</p>
<p>Since $$\frac{\partial F}{\partial u} = \cos(u^2),$$
and $$\frac{\partial F}{\partial v} = -\cos(v^2),$$
we obtain
$$\frac{dy}{dx}(x) = \cos(\sin^2(x))\cos(x)-2x\cos(x^4).$$</p>
|
215,061 | <p>Is it possible to do a 2D plot taking the point coordinates from separate lists of x and y values? These lists were produced by calculations within the same notebook. For a small number of points it is easy to enter x & y values by hand but it gets tedious as the number of points grows (I've done it.). If this is possible, I cannot find it in the documentation. Presently, I have over 40 points to plot and I expect the lists to grow. Any advice would be appreciated.</p>
| kglr | 125 | <p>Make each element of <code>replace</code> a function by wrapping it with <code>Function</code>. </p>
<pre><code>functions = Function /@ replace
</code></pre>
<blockquote>
<p>{(αu #1)/(ku + #1) &, Du #1 &, (αv #1)/(kv + #1) &}</p>
</blockquote>
<p>Use <code>functions</code> with <code>Through</code> to apply all functions to the same argument:</p>
<pre><code>Through[functions@1]
</code></pre>
<blockquote>
<p>{αu/(1 + ku), Du, αv/(1 + kv)}</p>
</blockquote>
<p>Use <code>Map</code> to apply this function to a list of arguments:</p>
<pre><code>Through[functions@#] & /@ {1, 2}
</code></pre>
<blockquote>
<p>{{αu/(1 + ku), Du, αv/(1 + kv)},<br>
{(2 αu)/(2 + ku), 2 Du, (2 αv)/(2 + kv)}}</p>
</blockquote>
<p>Alternatively, map <code>Function</code> on <code>reactions</code> before you create <code>replace</code>:</p>
<pre><code>reactions2 = Function /@ reactions;
replace2 = {a, c, d} /. reactions2;
Through[replace2@#] & /@ {1, 2}
</code></pre>
<blockquote>
<p>same result</p>
</blockquote>
|
1,350,062 | <p>In almost all of the physics textbooks I have ever read, the author will write the oscillating function as</p>
<blockquote>
<p>$$x(t)=\cos\left(\omega t+\phi\right)$$</p>
</blockquote>
<p>My question is that, is there any practical or historical reason why we should prefer $\cos$ to $\sin$ here? One possible explanation I can think of is that, to trigger a harmonic oscillation movement, we usually push the mass (to the maximum displacement) from the balance point at the initial moment, for which the cosine function will be neater to use than sine ($\phi=0$). But is it really the case?</p>
| Sergei M. | 264,641 | <p>Direct current (I = const, U = const, etc.) can be assumed as "alternating current with ω = 0 and ϕ = 0" only when using cosine, but not sine. This reason went from the electrical engineering.</p>
|
189,308 | <p>I have this problem: Find integration limits and compute the following integral.</p>
<p>$$\iiint_D(1-z)\,dx\,dy\,dz \\ D = \{\;(x, y, z) \in R^3\;\ |\;\; x^2 + y^2 + z^2 \le a^2, z\gt0\;\}$$</p>
<p>I can compute this as an indefinite integral but finding integration limits beats me. As indefinite integral the result looks like this (hopefully without any careless mistakes):</p>
<p>$$\iiint(1-z)\,dx\, dy\, dz \\ = \iint(x(1-z) + C_x)\,dy\, dz\\ = \iint (x - xz + C_x)\,dy\, dz \\ = \int (xy - xyz + yC_x + C_y)\,dz\\ = xyz - \frac{xyz^2}{2} + yzC_x + zC_y + C_z$$</p>
| DonAntonio | 31,254 | <p>Another approach, with cylindrical coordinates:
$$x=r\cos\theta\,\,,\,\,y=r\sin\theta\,\,,\,\,z=z\Longrightarrow |J|=r\,\, (\,\text{Jacobian})$$
Since we want to integrate on the uper half sphere $\,x^2+y^2+z^2\leq a^2\,\,,\,\,z\geq 0$ , we get:
$$0\leq r\leq a\,\,,\,\,0\leq\theta\leq 2\pi\,\,,\,\,0\leq z\leq \sqrt {a^2-x^2-y^2}=\sqrt {a^2-r^2}$$
so our integral becomes
$$\int_0^adr\int_0^{2\pi}d\theta\int_0^\sqrt{a^2-r^2}r(1-z)dz=2\pi\int_0^ar\,dr\left.\left(z-\frac{1}{2}z^2\right)\right|_0^\sqrt{a^2-r^2}=$$
$$=-\pi\int_0^a(-2r\,dr)\sqrt{a^2-r^2}-\pi\int_0^a\left(a^2r-r^3\right)dr=$$
$$=\left.-\frac{2\pi}{3}\left(a^2-r^2\right)^{3/2}\right|_0^a-\left.\pi\left(\frac{a^2r^2}{2}-\frac{r^4}{4}\right)\right|_0^a=\frac{2\pi}{3}a^3-\frac{\pi}{4}a^4$$</p>
|
832,206 | <p>I want to approximate the derivative of f(x)</p>
<h2>Finite difference</h2>
<p>$f'(x) \approx \frac{f(x+h)-f(x)}{h}$</p>
<p>I was taught that the error from the subtraction is blown up for small h. This I can verify with MATLAB.</p>
<h2>Smartass method</h2>
<p>So I though maybe the following would fix the problem:</p>
<p>$f'(x) \approx \frac{\log{\frac{\exp(f(x+h))}{\exp(f(x))}}}{h}$</p>
<p>However, I get the same relative errors. (Which also start increasing for h smaller than approx. $\epsilon_{mach}/2$</p>
<p>Why is this?</p>
| Community | -1 | <p>This is proven in these 3 textbooks. 1. Saber Elaydi, <em>An Introduction to Difference Equations</em> (2005 3 ed). Pages 66, 126.</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/JWuLj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JWuLj.jpg" alt="enter image description here" /></a></p>
</blockquote>
<blockquote>
<p><a href="https://i.stack.imgur.com/fvlLm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fvlLm.jpg" alt="enter image description here" /></a></p>
</blockquote>
<ol start="2">
<li>Walter G. Kelley, <em>Difference equations an introduction with applications</em> by (2001 2 ed), p 50.</li>
</ol>
<p><a href="https://i.stack.imgur.com/zNFAB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zNFAB.jpg" alt="enter image description here" /></a></p>
<ol start="3">
<li>Mickens, Ronald E, <em>Difference Equations Theory, Applications and Advanced Topics</em> (2015 3 ed), pp 84-85.</li>
</ol>
<p><a href="https://i.stack.imgur.com/TeCTo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TeCTo.jpg" alt="enter image description here" /></a></p>
|
827,072 | <p>How to find the equation of a circle which passes through these points $(5,10), (-5,0),(9,-6)$
using the formula
$(x-q)^2 + (y-p)^2 = r^2$.</p>
<p>I know i need to use that formula but have no idea how to start, I have tried to start but don't think my answer is right.</p>
| AMANE otti | 155,953 | <p>if the problem has a solution (the three points are on a circle) then you only need to calculate the equations of the two mediators and the intersection should be the center and the distance between one of the points and this center gives you r.</p>
|
2,322,481 | <p>Look at this limit. I think, this equality is true.But I'm not sure.</p>
<p>$$\lim_{k\to\infty}\frac{\sum_{n=1}^{k} 2^{2\times3^{n}}}{2^{2\times3^{k}}}=1$$
For example, $k=3$, the ratio is $1.000000000014$</p>
<blockquote>
<p>Is this limit <strong>mathematically correct</strong>?</p>
</blockquote>
| robjohn | 13,854 | <p>$$
\begin{align}
\lim_{k\to\infty}\frac{\sum_{n=1}^k2^{2\cdot3^n}}{2^{2\cdot3^k}}
&=\lim_{k\to\infty}\sum_{n=1}^k2^{-2\left(3^k-3^n\right)}\\
&=1+\lim_{k\to\infty}\sum_{n=1}^{k-1}2^{-2\left(3^k-3^n\right)}\\
&\le1+\lim_{k\to\infty}(k-1)2^{-2\left(3^k-3^{k-1}\right)}\\[6pt]
&=1+\lim_{k\to\infty}(k-1)2^{-4\cdot3^{k-1}}\\[9pt]
&=1
\end{align}
$$</p>
|
3,407,474 | <p>I can understand that The set <span class="math-container">$M_2(2\mathbb{Z})$</span> of <span class="math-container">$2 \times 2$</span>
matrices with even integer entries is an infinite non commutative ring. But why It doesn't have any unity? I think <span class="math-container">$I(2\times2)$</span> is a unity for this ring.
Where is my misunderstanding in subject of <span class="math-container">$I(2\times2)$</span> being a unity?</p>
| José Carlos Santos | 446,262 | <p>If <span class="math-container">$I(2\times2)=\left[\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right]$</span>, then <span class="math-container">$I(2\times2)\notin M_2(2\mathbb Z)$</span> since <span class="math-container">$1$</span> is odd.</p>
|
1,975,199 | <blockquote>
<p>In this problem, construct a bijection to show the identity. Use the definitions of these quantities, not formulas for them.</p>
<p>For <span class="math-container">$0≤j≤n$</span>
<span class="math-container">$$\binom{n}{j}=\binom{n}{n-j}$$</span></p>
</blockquote>
<p>I know how to prove this by induction, but I have no idea what "construct a bijection" stands for. Could anyone explain this or (better) show me an example? Thanks.</p>
| Graham Kemp | 135,106 | <p>A bijection is a one-to-one correspondence between sets. If there is a bijection between two sets they have the same cardinal size (which, for finite sets that is basically: "count of elements").</p>
<p>In this case, $\binom n j$ is the count of ways to select $j$ elements from a set of size $n$.</p>
<p>Meanwhile, $\binom n {n-j}$ is the count of ways to select $n{-}j$ elements from a set of size $n$.</p>
<p>Let us call $N$ the set of $n$ items. Let $J$ be the set of sets of size $j$ that can be selected from $N$. Let $K$ be the set of sets of size $n-j$ that can be selected from $N$.</p>
<p>Show that there is a bijection from $J\mapsto K$. If the sets have the same size then the above counts have the same value.</p>
<hr>
<p>Well, for any set in $J$ there is a single $N$-relative complement of size $n-j$. ...</p>
|
2,649,756 | <p>I was given the equation</p>
<p>$x=y^4$</p>
<p>and asked to find the points where the curvature was the biggest and the smallest. I know the curvature equation:</p>
<p>$κ(x)= \frac{|y″|}{\left(1+\left(y′\right)^2\right)^{\frac{3}{2}}}$</p>
<p>and I know $f(x)=y=\sqrt[4]{x}$ </p>
<p>but I am not sure how to minimize or maximize the curvature. Any help would be great!</p>
| Lukas Heger | 348,926 | <p>First, some obvervations: For two subgroups $A, B$ of an abelian group, the set $A+B:= \{a+b \mid a \in A, b \in B\}$ is again a subgroup</p>
<p>We will prove by induction on $n$, that for any $n$ rational numbers $r_1, \dots, r_n$, the set $\{\sum_{i=1}^nn_ir_i|n_i\in\mathbb{Z}\}$ is a cyclic group.</p>
<p>The case $n=1$ is trivial and $n=2$ is treated in the linked question, so I won't repeat the argument here.<br>
In the induction step, we have $$\{\sum_{i=1}^{n+1}n_ir_i|n_i\in\mathbb{Z}\}= \{n_{n+1}r_{n+1} \mid n_{n+1} \in \Bbb Z\}+\{\sum_{i=1}^nn_ir_i|n_i\in\mathbb{Z}\}$$</p>
<p>By the induction hypothesis, $\{\sum_{i=1}^nn_ir_i|n_i\in\mathbb{Z}\}$ is cyclic, so there exists a $s \in \Bbb Q$ such that $\{\sum_{i=1}^nn_ir_i|n_i\in\mathbb{Z}\} = \{ks \mid k \in \Bbb Z\}$, thus</p>
<p>$$\{n_{n+1}r_{n+1} \mid n_{n+1} \in \Bbb Z\}+\{\sum_{i=1}^nn_ir_i|n_i\in\mathbb{Z}\}=\{n_{n+1}r_{n+1} \mid n_{n+1} \in \Bbb Z\}+\{ks \mid k \in \Bbb Z\}= \{n_{n+1}r_{n+1}+ks \mid n_{n+1}, k \in \Bbb Z\}$$</p>
<p>This group is cyclic by the case $n=2$.</p>
|
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