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Let $f(x)= \displaystyle \sum \limits_{n=1}^\infty \frac{\sin(nx)}{n^3}.$ Show that $f(x)$ is differentiable and that the derivative f'(x) is continuous.

In class we solved a similar problem, and I think we had to show that both $f(x)$ and f'(x) converge uniformly, but I am not really sure why that is what we have to show. I think we also showed that the partial sums are continuous.

I would really appreciate it if someone could solve this problem and maybe explain a little the theory behind it.

Thanks!

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    Just as an aside, this is a Clausen function: http://mathworld.wolfram.com/ClausenFunction.html2010-11-22

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To show uniform convergence of the original series and the series of derivatives you can use the facts that $\sin$ and $\cos$ are bounded and that $\sum_{n=1}^\infty \frac{1}{n^3}$ and $\sum_{n=1}^\infty \frac{1}{n^2}$ converge.

The uniform limit of continuous functions is continuous. To see that $f$ is differentiable with derivative equal to the series of derivatives, you can apply an integral convergence theorem. E.g., uniform convergence allows you to pass limits through integrals, and (using the fundamental theorem of calculus) $f(x)=\sum_{n=1}^\infty \int_0^x \frac{\cos(nt)}{n^2}dt$. You can use the fundamental theorem of calculus again to finish, also using the fact that the series of derivatives represents a continuous function by the first sentence of this paragraph.

(I'm intentionally leaving some details for you to fill in.)

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    @George: [clarification of uneditable comment I'm deleting.] Yes, continuity allows the integrability of the series of derivatives to follow, so that the FTC can be applied, and without knowing this a priori it takes considerably more work. (And in general, the derivative need not even be Riemann integrable. E.g., one could consider a constant sequence consisting of an everywhere differentiable function without integrable derivative, as seen e.g. in http://mathoverflow.net/questions/6711/integrability-of-derivatives/6716#6716.) Thanks for the comment.2010-11-22
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You had to show that both $(f_n)$ and (f_n') converge uniformly because you want to apply a theorem like this:

Theorem 10.7 (from "Real Analysis" by N.L. Carothers):

Suppose that $(f_n)$ is a sequence of real-valuled functions, each having a continuous derivative on $[a,b]$, and suppose that the sequence of derivatives (f_n') converges uniformly to a function $g$ on $[a,b]$. If $(f_n(x_0))$ converges at any point $x_0$ in $[a,b]$, then in fact, $(f_n)$ converges uniformly to a differentiable function $f$ on $[a,b]$. Moreover f' = g. That is, (f_n') converges uniformly to f' on $[a,b]$.

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Well, to see that $f(x)$ *converges uniformly* please apply the Weierstrass M-Test.

Now, i am aware of the fact that term by term differentiation and integration is possible if the series is uniformly convergent. But i don't know much about this, i don't want to answer that, but i am sure someone here will help you out regarding this.

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    @jonas: yes u are right!2010-11-22