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Prove that $f(x)=\sum_{n=1}^\infty 1/2^n(\cos3^nx)$ is continuous but nowhere differentiable on $\mathbb{R}$.

I have proved the continuity part, but unable to do the second one. Thanks for any help.

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    His method won't work directly, but the concept behind it will still apply, mainly due to the self-similarity issue this curve will have concerning differentiation. More specifically the curve can at most be differentiated at a countable number of points (and in this case that will not happen due to the lack of smoothness). For an exploration into fractal differentiation see: http://arxiv.org/pdf/1010.0881.pdf2012-12-10

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I am not a proper mathematician, so I'll try to give a heuristic argument, why the function is not differentiable at all $x$.

We should start by applying the definition of the derivative to the function: $f'(x) = \lim_{\delta \to 0} \frac{f(x + \delta) - f(x)}{\delta}$

Since the limit and the sum are commutable, we can get the expression: $f'(x) = \sum_{n=1}^{\infty} 2^{-n} \lim_{\delta \to 0} {\left( \cos(3^nx) ( \cos(3^n\delta) - 1 ) - \sin(3^nx)\sin(3^n\delta) \right) \over \delta}$

One can see that no proper limit can be found for this sum, as for large $n$ the value of some of the terms becomes indeterminate when the inequality $3^n\delta \ll 1$ is no longer satisfied. Since some of the terms in the sum don't have a proper limit, the whole sum doesn't have a proper limit.

This is by no means a proper, mathematically rigorous answer, but maybe it helps.

EDIT:

Now let's consider the 'last' terms in the series: $\lim_{n \to \infty} 2^{-n} \lim_{\delta \to 0} {\left( \cos(3^nx) ( \cos(3^n\delta) - 1 ) - \sin(3^nx)\sin(3^n\delta) \right) \over \delta}$

or rather

$\lim_{n \to \infty} 2^{-n} n{\left( \cos(3^nx) ( \cos(3^n/n) - 1 ) - \sin(3^nx)\sin(3^n/n) \right)}$

which does not have a proper limit (I hope that the L'Hoptal's theorem can be applied here, but I suspect that is not the case)... Is this not sufficient for mathematicians? :)

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    Amending the above after further consideration: upon looking more closely there is an issue with the derivative, but the sequence of partial sums could still converge even if some of the tail-end points in the sequence don't alone.2012-12-10
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Think about the nature of symmetry of the curve here, mainly given an interval $(-a,a)$ (or a similar location about another point than 0) what does the function look like compared to the interval $(\frac{-a}{2}, \frac{a}{2})$, and what would that mean for intervals of the form $(\frac{-a}{2^n}, \frac{a}{2^n})$ as n gets larger?

Another point would be to direct you to look at the Weirstrass function, but that will likely lead to a give away of the answer, so depending on what approach you want to take you can either look there or take the hint I gave and run with it.