The context for this question is basic curiosity. I know that there are formulas for calculating $\displaystyle \sum_{k=0}^{n}\sin({kx})$ and $\displaystyle \sum_{k=0}^{n}\sin({2kx})$. Is there a formula/method/identity that could be used to find $\displaystyle \sum_{k=0}^{n}\sin({2^kx})$ ?
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Possible route to a solution: write $\sin(kx) = \frac{e^{ikx}-e^{-ikx}}{2i}$, then as there are finite sums you can split $\sum_i A_i-B_i = \sum_i A_i-\sum_i B_i$. Then you just need to be able to evaluate $\sum_{i = 1}^n e^{i2^k x}$. If it were geometric (like for $\sin(kx)$, it'd be trivial). – Mark Schultz-Wu Nov 03 '16 at 21:00
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4Don't remember a formula for $\sum_{k=0}^n z^{2^k}$ offhand. – dxiv Nov 03 '16 at 21:01
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1Short answer: no. Otherwise, this question (http://math.stackexchange.com/questions/1976417/a-community-project-prove-or-disprove-that-sum-n-geq-1-frac-sin2nn) would be simple to deal with. – Jack D'Aurizio Nov 03 '16 at 21:06
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To prove an inequality like $$\sum_{k=0}^{N}\sin(2^k x) \ll\frac{N}{\log\log N},$$ that is just a little stronger than the trivial bound $\ll N$, just for some values of $x$, would have deep consequences. – Jack D'Aurizio Nov 03 '16 at 22:18
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@Jack D'Aurizio - I take it your statement excludes obvious values of x? There are infinite numbers for which $\sum_{k=0}^{N}\sin(2^k x)$ converges. Namely any number that can be represented as ${\frac{a\pi}{2^b}}$ | a and b are any integers. The convergent value being strictly less than $\log(2b) = b$. Thus $\sum_{k=0}^{N}\sin(2^k x) \ll b$ for these values of x. – C Shreve Nov 08 '16 at 18:58
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@JackD'Aurizio - In addition, any number that cannot be expressed as ${\frac{a\pi}{2^b}}$ can be approximated (to an arbitrary accuracy) by a number that can be expressed as ${\frac{a\pi}{2^b}}$. For example, ${\frac{\pi}{3}}$ can be approximated as ${\frac{21,845\pi}{2^{16}}}$. These are the same number to the ten thousandths decimal place. Hope this is helpful in some way. – C Shreve Nov 08 '16 at 21:38