Wikipedia:Reference desk/Archives/Mathematics/2009 March 19#Functional Convergence

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= March 19 =

== Functional Convergence ==

In a recent thread, if I understand correctly, pma says that $\backslash textstyle\; \backslash sum\_\{1\backslash leq\; k\backslash leq\; n\}\backslash log\backslash cos(k^2\backslash sqrt\{n^5\}s)$ converges pointwise to $\backslash textstyle\; -s^2/10$ 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 $\backslash sqrt\{n^5\}$ 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 $\backslash textstyle\; \backslash log\; \backslash cos(x)$ at 0, with remainder in Peano form: so, for all $x\backslash in\; [0,\backslash pi/2)$

::$\backslash log\backslash cos(x)=-\backslash frac\{x^2\}\{2\}+o(x^2)$, as $x\backslash to0$.

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

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

:*Summing over all $1\backslash leq\; k\backslash leq\; n$

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

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

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