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My friend asks me a question what value is the infinite series $$\sum\sin\frac{1}{2^i}$$

It is obviously a convergence series, however I have no idea to compute the value. So is there a value of it? Thank you.

gaoxinge
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    Since $\sin(x)\sim x$ near $0$ and $x > \sin(x)$ for positive $x$, it has a somewhat accurate upper bound of $\sum_{i\ge1}\frac 1 {2^i} = 1$ – Regret Nov 18 '14 at 08:36

3 Answers3

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It's not a very nice series, I don't think you will get a closed-form analytic expression. However, you can compute its value to any desired accuracy numerically. You can also bound it from above:

$$\sum_{n=0}^\infty \sin 2^{-n}<\sum_{n=0}^\infty 2^{-n}=2$$ This also means that you can compute it much more accurately than just summing the first few terms: you can approximate the rest with a geometric sum.

Example:

$$\sum_{n=0}^\infty \sin 2^{-n}<\sin 1+\sin \frac12 + \sum_{n=2}^\infty 2^{-n}\approx$$ $$0.841471+0.479426+\frac{1}{2}=1.8209$$ The accurate value for 5 decimal places (Wolfram Alpha) is $1.81793...$.

orion
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  • Why we can't calculate the certain value of this series? Thank you. – gaoxinge Nov 18 '14 at 08:39
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    Very few sums (the same is true for integrals) have an analytical solution. There are way too many functions that can be expressed as a sum to be expressible with that very few functions that we limit ourselves to (usually just trigonometry, exponentials, logarithms, powers and division/multiplication). But that doesn't mean that when you see a particularly nice integral or a sum, you can't just name it and use it as another function from an extended set. There are many special functions defined either as integrals, sums, or solutions of transcendental equations. But yours isn't one of them :) – orion Nov 18 '14 at 08:43
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Since $$\sin x = \sum_{j=0}^{+\infty}\frac{(-1)^j}{(2j+1)!}x^{2j+1}$$ we have that: $$\sum_{i\geq 0}\sin\frac{1}{2^i}=\sum_{i\geq 0}\sum_{j\geq 0}\frac{(-1)^j}{(2j+1)!}\cdot\frac{1}{2^{i(2j+1)}}=\sum_{j\geq 0}\frac{(-1)^j}{(2j+1)!(1-2^{-(2j+1)})}=1.81792872\ldots$$ where the last series is very fast converging. However, I would not bet on "nicer" closed forms.

Jack D'Aurizio
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One possible way is to start with the expansion of $\sin(x)$ for small values of $x$ $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\frac{x^9}{362880}-\frac{x^{11}}{399 16800}+O\left(x^{13}\right)$$ and replace $x$ by $\frac{1}{2^i}$. So, $$\sin(\frac{1}{2^i})=2^{-i}-\frac{2^{-3 i-1}}{3} +\frac{2^{-5 i-3}}{15} -\frac{2^{-7 i-4}}{315} +\frac{2^{-9 i-7}}{2835}-\frac{2^{-11 i-8}}{155925}+\cdots$$ and now the summation corresponds to the sum of geometric series. Limiting to the above terms, we arrive to $$\sum_{i=0}^{\infty}\sin\frac{1}{2^i}=\frac{1167338371295774}{642125490254325}\approx 1.817928721$$

If we use $n$ terms for the Taylor expansion, we can find $$S_1=2$$ $$S_2=\frac{38}{21}\approx 1.809523810$$ $$S_3=\frac{5918}{3255}\approx 1.818125960$$ $$S_4=\frac{450902}{248031}\approx 1.817925985$$ $$S_5=\frac{10368507238}{5703472845}\approx 1.817928746$$ $$S_6=\frac{1167338371295774}{642125490254325}\approx 1.817928721$$ $$S_7=\frac{372905075405009059958}{205126345736253866925}\approx 1.817928721$$

Claude Leibovici
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