Does $\sum_{k=0}^{\infty}\sin\left(\frac{\pi x}{2^k}\right)$ have a simple form with interesting properties?

Solution 1:

By expanding $\sin(x)$ as its Taylor series,

$$ f(x)=\sum_{k\geq 0}\sin\frac{\pi x}{2^k} = \sum_{n\geq 0}\frac{(-1)^n\pi^{2n+1}x^{2n+1}}{(2n+1)!}\sum_{k\geq 0}\frac{1}{2^{k(2n+1)}} $$ hence $$ f(x) = \sum_{n\geq 0}\frac{(-1)^n \pi^{2n+1}}{(2n+1)!\left(1-\frac{1}{2^{2n+1}}\right)}x^{2n+1}$$ but the behaviour of such function over the real line is quite chaotic:

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and it is not completely trivial to prove that such a function is bounded as it appears to be. In fact, it is not: see robjohn's comment below, proving that the EMC formula miserably fails here.