Does this sum converge to a simpler looking function? $$\sum\limits_{k=0}^{\infty}\sin\left(\frac{\pi x}{2^k}\right)$$
If there is no known closed form expression for this, what interesting properties does this sum have? I know this is not a concrete question, so I am just asking for opinions.
Are there general series expansions for functions in terms of sinusoids with decreasing frequencies (like my example above)?
What I have done so far:
I tried plotting this in wolfram for a few terms(as much as the text box would allow me). I also wrote a program to numerically compute the sum up to a thousand terms. I could not find any recognizable patterns, but one interesting thing I observed (in answering my Q2 above) is that the curve is always positive on the right of y-axis and always negative on the left of y-axis. If I use positive integers for $k$ (i.e. $\frac{\pi x}{k}$) instead of using decreasing powers of two (i.e $\frac{\pi x}{2^k}$) for frequencies, I see that the curve crosses the x-axis (zero-crossing) at multiple places in both left and right planes.
My thoughts on Q3. Since the highest frequency sinusoid has a frequency of $\frac{1}{2}$ in my expression, I don't expect that any function can be arbitrarily expressed using decreasing frequency sinusoids. I am merely wondering if there is a small subset of functions that may be expressed this way (with a scaling factor for each sinusoid of course).
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:
$\hspace1in$
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.