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March 19

Functional Convergence

In a recent thread, if I understand correctly, pma says that $\sum _{1\le k\le n}\log \cos (k^2\sqrt{n^5}s)$ converges pointwise to $-s^2/10$ 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 $\log \cos (x)$ 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 $x\to 0$.

• 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 $n\to \mathrm{\infty }$, and uniformly for all $1\le k\le n$.

• Summing over all $1\le k\le n$

$\sum _{1\le k\le n}\log \cos (\frac{k^2}{\sqrt{n^5}}s)=-\frac{s^2}{2n}\sum _{1\le k\le n}{\left(\frac{k}{n}\right)}^4+o(1)$, as $n\to \mathrm{\infty }$.

• Then you may observe that $\frac{1}{n}\sum _{1\le k\le n}{\left(\frac{k}{n}\right)}^4$ is the Riemann sum for the integral of $x^4$ on [0,1] (or use the formula for $\sum _{1\le k\le n}{k}^4$) and conclude that the whole thing is $-\frac{s^2}{10}+o(1)$.

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.

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

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

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

Ah yes. It seems to yeild a solution of the form $y=\sqrt{A{x}^2+B}$ 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)