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

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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 $x\to 0$.

• 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 $n\to \infty$, 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.

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

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)