As the title says, I'm trying to calculate $$\sum_{k=1}^n \sin(2^k x)$$ but I'm having some troubles since I haven't done trigonometry in a looong time. Can someone help me calculate it?
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Try a couple terms first, the answer should become clearer. The double angle formulas are $\sin{2x}=2\sin{x}cos{x}$ and $\cos{2x}=\cos^2{x}-\sin^2{x}$ – Arby Mar 01 '17 at 09:32
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2If you consider the natural frame of trigonometry is complex numbers, it's much simpler with the complex exponential. – Bernard Mar 01 '17 at 09:56
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1Check this. – Hazem Orabi Mar 01 '17 at 10:05
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There is no reason for expecting that such sum has a nice closed form for general $n$s and $x$s. It is not even clear if such sum is $o(n)$ as $n\to +\infty$. – Jack D'Aurizio Mar 01 '17 at 11:42