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Given $$f(x) = \sum_{n=1}^\infty \frac{\cos(nx)}{n^2}$$

By plotting graph, it seems to be that this function is not differentiable at $x=0$, two sided limits are not equal with each other but I cannot show how. Also it seems that $f'(x)$ is not uniformly convergent however I cannot see how I can use that.

Hash Nuke
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3 Answers3

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$$f(x)=\sum_{n=1}^\infty\frac{\cos(nx)}{n^2}$$ $$f'(x)=-\sum_{n=1}^\infty\frac{\sin(nx)}{n}$$ so maybe the problem with $f'(0)$ is the term as ${n\to\infty}$ $$f'(0)=-\lim_{x\to 0}\sum_{n=1}^\infty\frac{\sin(nx)}{n}=-\lim_{x\to 0}\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{(-1)^k}{(2k+1)!}x^{2k+1}n^{2k}$$ this could be a way of evaluating it, but the term I think is the problem is: $$\lim_{(x,n)\to(0,\infty)}\frac{\sin(nx)}{n}$$

Henry Lee
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You can prove (see e.g. here) that $$ f(x) = \frac{x^2}{4}-\frac{\pi x}{2}+\frac{\pi^2}{6}, \qquad \forall x \in [0,2\pi]. $$ Since $f$ is $2\pi$-periodic, you can conclude that $$ f'_+(0) = -\frac{\pi}{2},\qquad f'_-(0) = \frac{\pi}{2}, $$ so $f$ is not differentiable at the origin.

Rigel
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  • that is a bit more complicated. Right now I am doing this problem from Abbott analysis. Can I show that $$ \lim_{h \to 0^+} \sum_{k=1}^\infty \frac 1 {hk^2} ( -2 \sin^2(kh ) = -\frac \pi 2 $$ – Hash Nuke Sep 21 '18 at 14:13
  • Or at least that limit $\lim_{h \to 0^+} \frac 1 h (f'(h)-f'(0)) $ is not equal to zero? – Hash Nuke Sep 21 '18 at 14:17
  • To investigate whether $f'(0)$ exists, use the definition of derivative. Since you have a formula for $f$, this should be easy. (It is possible that a derivative of a function exists everywhere but is not continuous.) – GEdgar Sep 21 '18 at 14:27
  • @HashNuke: I wonder whether there is an elementary proof of this fact. – Rigel Sep 21 '18 at 14:34
  • @GEdgar I am getting zero if I do it termwise, I get it zero. I know it is wrong to do that$$ \begin{aligned}\lim_{h \to 0} \frac{f'(h) - f'(0)}{h} &= \sum_{k=1}^\infty \frac 1 {k^2} \left ( \frac{\cos (kh) - \cos (0)}{h} \right ) \ & = \sum_{k=1}^\infty \frac 1 {k^2}\left ( \frac{-2 \sin ^2\left ( \frac{kh}2 \right ) }{h} \right ) \end{aligned} $$ I am not sure if I can use the fact that $$f ( x ) = \frac { x ^ { 2 } } { 4 } - \frac { \pi x } { 2 } + \frac { \pi ^ { 2 } } { 6 }$$ – Hash Nuke Sep 21 '18 at 14:44
  • I explain your "not sure" comment in a new answer. – GEdgar Sep 21 '18 at 20:59
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Begin with $$ f ( x ) = \frac { x ^ { 2 } } { 4 } - \frac { \pi x } { 2 } + \frac { \pi ^ { 2 } } { 6 } $$ for $x \in [0,2\pi]$, and the corresponding $$ f ( x ) = \frac { x ^ { 2 } } { 4 } + \frac { \pi x } { 2 } + \frac { \pi ^ { 2 } } { 6 } $$ for $x \in [-2\pi,0]$. So $$ \lim_{h \to 0^+}\frac{f(h)-f(0)}{h} = \lim_{h\to 0^+}\frac{h^2/4+\pi h/2}{h} = \frac{\pi}{2} $$ and $$ \lim_{h \to 0^-}\frac{f(h)-f(0)}{h} = \lim_{h\to 0^-}\frac{h^2/4-\pi h/2}{h} = -\frac{\pi}{2} $$ Therefore, $$ \lim_{h \to 0}\frac{f(h)-f(0)}{h} $$ does not exist.

Picture of the graph of $f$, showing the non-differentiable point.
graph

GEdgar
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  • what I wanted to say was I am not sure if I can assume $f ( x ) = \frac { x ^ { 2 } } { 4 } - \frac { \pi x } { 2 } + \frac { \pi ^ { 2 } } { 6 }$ because it's fairly complicated to show it. The exercise I am doing is from Abbott's Understanding Analysis. In the end, I ended up doing this. – Hash Nuke Sep 22 '18 at 08:38