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Does it make sense to call $\sum_{n=1}^{\infty} \frac{\sin(2^nx)}{2^n}$ continuous? For any finite n it is easy to declare that the function is continuous. Now when we let $n \to \infty$ we get a function that would seem pretty continuous, with each value very very close to its neighbors. However when we attempt to take the derivative anywhere on the function (except for a few key points) we find due the infinitesimal and ever present oscillations inherent in the function it doesn't really make sense to take the derivative.

My hunch is to say that $\sum_{n=1}^{\infty} \frac{\sin(2^nx)}{2^n}$ is continuous but not differentiable. But how can one justify that?

PS:
if you want to see an interactive, zoomable version of the function

PPS: Could the function possibly be differentiable at $f(2\pi k)$ for $k =$ some integer. At $2\pi k $ the derivative of $ \frac{\sin(2^nx)}{2^n}$ is $\frac{1}{2^n}$ for all n. Does that imply diferentiablitity at $2 \pi k$?

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    Hint: Uniform convergence theorems – 2017-01-06
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    This is basically the same as the [Weierstrass function](https://en.wikipedia.org/wiki/Weierstrass_function). I don't think it's trivial to show it's nowhere differentiable, but I'm fairly sure that, even changing $\cos$ to $\sin$, it still is. – 2017-01-06

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Claim: $f'(0)$ doesn't exist.

Proof: Because $\lim_{x\to 0} (\sin x)/x = 1,$ we can regard $(\sin x)/x$ as positive and continuous on $[0,1].$ Hence there exists $c>0$ such that $(\sin x)/x \ge c$ for $x\in [0,1],$ which gives $\sin x \ge cx$ there.

Let's see what happens to the difference quoitents of $f$ along the sequence $2^{-N}:$

$$\frac{f(2^{-N})-f(0)}{2^{-N}-0} = 2^Nf(2^{-N}) =2^N\left (\sum_{n=1}^{N} \frac{\sin (2^{n-N})}{2^n} + \sum_{n=N+1}^{\infty} \frac{\sin (2^{n-N})}{2^n}\right)$$ $$ \ge 2^N\left (cN2^{-N} - \sum_{n=N+1}^{\infty} \frac{1}{2^n}\right ).$$

That last sum adds to $1/2^{N}.$ Thus the last expression equals $cN - 1,$ which $\to \infty$ as $N\to \infty.$ This shows $f'(0)$ doesn't exist.

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    The $\frac{2}{\pi}$ comes out of nowhere... – 2017-01-06
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    Not nowhere: The inequality $\sin x \ge (2/\pi)x$ for $0\le x \le \pi/2$ is well known. If follows from the concavity of $\sin x$ on this interval, for example, which implies $\sin x$ lies above the line connecting $(0,0)$ and $(\pi/2,1).$ But I agree I sort of pulled it out of the hat, especially as I refered to the interval $[0,1]$ and not $[0,\pi/2].$ And we don't need this specific value at all, just some value. I'll edit my answer. Thanks. – 2017-01-06
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Since $$\left|\frac{\sin(2^nx)}{2^n}\right|\le\frac1{2^n}$$ and $\sum\frac1{2^n}$ converges, Weierstrass $M$-test implies that the function $$f(x)=\sum_{n=1}^\infty \frac{\sin(2^nx)}{2^n}$$ is continuous.

However, if we define $$f_n(x)=\frac{\sin(2^nx)}{2^n}$$ then $$f_n'(x)=\cos(2^nx)$$ and we can't apply the Weierstrass $M$-test anymore. This does not mean that $f$ is not differentiable, of course. (My guts say that $f$ is not differentiable at any point, but proving it seems difficult).

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    Would it make sense to say that f is differentiable at $f(2\pi k)$ for $k =$ some integer? – 2017-01-06
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    @AndoBando Yes. Differentiability is defined pointwise. Making sense and being true is not the same, though. – 2017-01-06