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

1. Does this sum converge to a simpler looking function? $$\sum\limits_{k=0}^{\infty}\sin\left(\frac{\pi x}{2^k}\right)$$

2. 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.

3. 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).

• The title is misleading, since it suggests that the question is about the convergence of the given series, which is an easy exercise. Maybe change it? – Alex M. Sep 23 '16 at 15:40
• I accepted an edit actually. My original question was just the formula (which I admit is not great either). Please suggest/edit. Thanks – Srini Sep 23 '16 at 15:45

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.

• i think that a bound might be optained by euler mac laurin formula – tired Sep 23 '16 at 15:32
• @tired: I agree. With that approach, the problem boils down to showing that the SineIntegral function is bounded, that is true for sure. – Jack D'Aurizio Sep 23 '16 at 15:34
• nice jack... (+1) – tired Sep 23 '16 at 15:37
• Note that $$\sin\left(\frac{2\pi}7\right)+\sin\left(\frac{4\pi}7\right) +\sin\left(\frac{8\pi}7\right) =1.3228756555322952953$$ If $x=\frac{8^n}7$, then $$\sum_{k=0}^{3n-1}\sin\left(\frac{\pi x}{2^k}\right)=1.3228756555322952953\,n$$ and $$\left|\,\sum_{k=3n}^\infty\sin\left(\frac{\pi x}{2^k}\right)\,\right|\le\frac{2\pi}{7}$$ Thus, $$\sum_{k=0}^\infty\sin\left(\frac{\pi x}{2^k}\right)\ge1.322875\,n-\frac{2\pi}7$$ Therefore, I don't believe that $f$ is bounded. – robjohn Sep 23 '16 at 21:32
• It is entire and bounded by $C |x| e^{ \pi |x|}$ right ? – reuns Sep 23 '16 at 22:38