Differentiability of a Special Function 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.
 A: 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? :)
A: 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.
