Series function not differentiable at a point

Consider the function

$\displaystyle f(x) = \sum_{n=1}^\infty 3^{-n/2} \cos (3^n x)$

How to show that it is not differentiable at x=0?

It is absolutely convergent but the derivative of partial sum is not, so I know it does not make sense to just differentiate and put x=0 in, but what should I be looking at instead for non-differentiability?

Consider the function

$\displaystyle f(x) = \sum_{n=1}^\infty 3^{-n/2} \cos (3^n x)$

How to show that it is not differentiable at x=0?

It is absolutely convergent but the derivative of partial sum is not, so I know it does not make sense to just differentiate and put x=0 in, but what should I be looking at instead for non-differentiability?

What happens as x->0+ and x->0-?

Cosine is a smooth even function so the behaviour of f should be the same from either side? We get f(0) which is convergent

you have written the function twice.... the first version looks better :

Consider the function

$\displaystyle f(x) = \sum_{n=1}^\infty 3^{-n/2} \cos (3^n x)$

How to show that it is not differentiable at x=0?

It is absolutely convergent but the derivative of partial sum is not, so I know it does not make sense to just differentiate and put x=0 in, but what should I be looking at instead for non-differentiability?

For the derivative? So its -sin() with coeffs 3^(n/2) so ...

I might be wrong, but I think you have to actually use the definition of the derivative at x=0 here; just because differentiating term by term gives something that doesn't converge as x->0 doesn't necessarily mean the derivative doesn't exist.

C.f. f(x) = x^2 sin(1/x), which is differentiable at x=0 but f'(x) doesn't converge to a limit as x -> 0.

The 3^(n/2) coeffs mean that its rough everywhere, so asking about x=0 isnt that special for this question as you say, though I guess it makes it a bit more precise. I was just imagining that you pick a sequence for x positive (so the peaks of sin(3^n x) and argue it -> inf, then you must get the converse for x negative?

Moved to a computer instead of trying to answer on my phone.

Actually, I think the x = 0 case is not too bad. The key advantage is that each cos(3^n x) term is 1 at x =0. So for x=/=0, if we're arguing that f(x) is close to f(0), we can always replace terms with 1, and know we're not making it worse. So take $x_k = \pi / 3^k$; split the sum into $\sum_0^{k-1} 3^{-n/2} \cos(3^n x) + \sum_k^\infty 3^{-n/2} \cos(3^n x)$, then for the first part just replace cos(3^n x) by 1, and explicltly sum the 2nd part and it should be easy to show (f(x_k)-f(0))/x_k is unbounded.

Edit: it would be nastier for the x=/=0 case because you could possibly have the difference caused by the "2nd part" of the sum cancelling with the diffference caused by the 1st part. But when x = 0 you know all the differences have to be in the same direction (i.e. whatever h is, cos(3^n h) is always <=1).

I see, so converting from $x\to 0$ to observing specific points $x_k = \dfrac{\pi}{3^k}$ as $k \to \infty$ should help here.

Let me check if I got your reasoning right;

$\displaystyle \lim_{x \to 0} \dfrac{f(x) - f(0)}{x} = \lim_{k \to \infty} \dfrac{f(x_k) - f(0)}{x_k}$

$\displaystyle = \lim_{k \to \infty} \dfrac{ \sum_{n=1}^{k-1} 3^{-n/2} \cos(3^{n-k} \pi) + \sum_{n=k}^{\infty} 3^{-n/2} \underbrace{\cos(3^{n-k} \pi)}_{=-1} - \dfrac{1}{3^{1/2} - 1}}{x_k}$

$\displaystyle = \lim_{k \to \infty} \dfrac{ \sum_{n=1}^{k-1} 3^{-n/2} \cos(3^{n-k} \pi) - \dfrac{3^{-k/2 + 1/2}}{3^{1/2}-1} - \dfrac{1}{3^{1/2} - 1}}{x_k}$

I don't understand why we are replacing $\cos(3^{n-k} \pi)$ by 1 in the first sum next? Surely this is only a good approximation for $n \ll k$ but when $n\approx k$ we have $\cos(\pi/3^2) \approx 0.94$ or even $\cos(\pi/3) \approx 0.5$.