March 19

In a recent thread, if I understand correctly, pma says that ${\displaystyle \textstyle \sum _{1\leq k\leq n}\log \cos(k^{2}{\sqrt {n^{5))}s)}$ converges pointwise to ${\displaystyle \textstyle -s^{2}/10}$ as n approaces infinity. How would you prove that? Black Carrot (talk) 07:53, 19 March 2009 (UTC)[reply]

Won't the limit depend on what branch of log you choose for negative arguments? Algebraist 10:23, 19 March 2009 (UTC)[reply]

My apologies: I made a misprint there (now corrected): the change of variables was ${\displaystyle t=n^{-5/2}s}$, with a minus in the exponent (this is consistent with the line below, that had it). So the term ${\displaystyle {\sqrt {n^{5))))$ 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)[reply]

Here it is:

• Write the second order Taylor expansion for ${\displaystyle \textstyle \log \cos(x)}$ at 0, with remainder in Peano form: so, for all ${\displaystyle x\in [0,\pi /2)}$

${\displaystyle \log \cos(x)=-{\frac {x^{2)){2))+o(x^{2})}$, as ${\displaystyle x\to 0}$.

• For any s we only have to consider the integers ${\displaystyle n>4s^{2}/\pi ^{2))$. Replace ${\displaystyle x={\frac {k^{2)){\sqrt {n^{5))))s={\frac {s}{\sqrt {n))}\left({\frac {k}{n))\right)^{2))$ in the expansion above, getting

${\displaystyle \log \cos \left({\frac {k^{2)){\sqrt {n^{5))))s\right)=-{\frac {s^{2)){2n))\left({\frac {k}{n))\right)^{4}+o({\frac {1}{n)))}$, as ${\displaystyle n\to \infty }$, and uniformly for all ${\displaystyle 1\leq k\leq n}$.

• Summing over all ${\displaystyle 1\leq k\leq n}$

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

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

Warning: I have re-edited this answer, to make it more simple and clear (hopefully) --pma (talk) 13:40, 19 March 2009 (UTC)[reply]

How should one go about solving this equation.

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

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

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

Ah yes. It seems to yeild a solution of the form ${\displaystyle y={\sqrt {Ax^{2}+B))}$ does that seem correct? —Preceding unsigned comment added by 92.9.236.44 (talk) 21:01, 19 March 2009 (UTC)[reply]

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)[reply]