Wikipedia:Reference desk/Archives/Mathematics/2009 March 19

= March 19 =

Functional Convergence
In a recent thread, if I understand correctly, pma says that $$\textstyle \sum_{1\leq k\leq n}\log\cos(k^2\sqrt{n^5}s)$$ converges pointwise to $$\textstyle -s^2/10$$ as n approaces infinity. How would you prove that? Black Carrot (talk) 07:53, 19 March 2009 (UTC)
 * Won't the limit depend on what branch of log you choose for negative arguments? Algebraist 10:23, 19 March 2009 (UTC)
 * My apologies: I made a misprint there (now corrected): the change of variables was $$t= n^{-5/2}s$$, with a minus in the exponent (this is consistent with the line below, that had it). So the term $$\sqrt{n^5}$$ is at the denominator, and the argument of log goes to 1 (actually, in that computation it was always positive). Do you see how to do it now?--pma (talk) 12:40, 19 March 2009 (UTC)


 * Here it is:
 * Write the second order Taylor expansion for $$\textstyle \log \cos(x)$$ at 0, with remainder in Peano form: so, for all $$x\in [0,\pi/2)$$
 * $$\log\cos(x)=-\frac{x^2}{2}+o(x^2)$$, as $$x\to0$$.
 * For any s we only have to consider the integers $$n>4s^2/\pi^2$$. Replace $$x=\frac{k^2}{\sqrt{n^5}}s=\frac{s}{\sqrt n}\left(\frac{k}{n}\right)^2$$ in the expansion above, getting
 * $$\log\cos\left(\frac{k^2}{\sqrt{n^5}}s\right)= -\frac{s^2}{2n}\left(\frac{k}{n}\right)^4 +o(\frac{1}{n})$$, as $$n\to\infty$$, and uniformly for all $$1\leq k\leq n$$.
 * Summing over all $$1\leq k\leq n$$
 * $$\textstyle \sum_{1\leq k\leq n}\log\cos(\frac{k^2}{\sqrt{n^5}}s)= -\frac {s^2}{2n} \sum_{1\leq k\leq n}\left(\frac{k}{n}\right)^4 +o(1) $$,  as $$n\to\infty$$.
 * Then you may observe that $$\scriptstyle\frac {1}{n}\sum_{1\leq k\leq n}\left(\frac{k}{n}\right)^4$$ is the Riemann sum for the integral of $$x^4$$ on [0,1] (or use the formula for $$\scriptstyle\sum_{1\leq k\leq n}k^4$$) and conclude that the whole thing is $$\textstyle -\frac{s^2}{10}+o(1)$$.
 * Warning: I have re-edited this answer, to make it more simple and clear (hopefully) --pma (talk) 13:40, 19 March 2009 (UTC)

Differential Equation
How should one go about solving this equation.

$$y\frac{dy}{dx}=xy\frac{d^2y}{dx^2}+x\bigg(\frac{dy}{dx}\bigg)^2$$

92.9.236.44 (talk) 20:30, 19 March 2009 (UTC)


 * The right hand side is $$x\frac{d}{dx}\bigg(y\frac{dy}{dx}\bigg)$$. Does that help? — JAO • T • C 20:48, 19 March 2009 (UTC)


 * Ah yes. It seems to yeild a solution of the form $$y = \sqrt{Ax^2+B}$$ does that seem correct? —Preceding unsigned comment added by 92.9.236.44 (talk) 21:01, 19 March 2009 (UTC)
 * Check for yourself - differentiate that a couple of times, substitute everything in and see if the two sides match. If they do, you've got it right, if they don't, you haven't! --Tango (talk) 23:05, 19 March 2009 (UTC)