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Question: Compute $$\int _0 ^{2\pi} \left(\sum_{n=1}^\infty \frac {\cos(nx)}{2^n}\right)^2 dx$$

Thoughts: Tried interchanging the integral and the sum, but then the integral turned out to be zero...

mathlove
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jreing
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1 Answers1

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Hint: write the squared sum as a double sum. $$ \left(\sum_{n=1}^\infty a_n\right)^2=\left(\sum_{n=1}^\infty a_n\right)\left(\sum_{m=1}^\infty a_m\right)=\sum_{m=1}^\infty\sum_{n=1}^\infty a_ma_n.$$

Tom-Tom
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    Note that if you perform the series first, its closed form is $Re\sum_{n=1}^\infty \left(\frac{\exp\mathrm{i}x}2\right)^n=\frac{2\cos x-1}{5-4\cos x}$ and you get $\int_0^{2\pi} \left(\frac{2\cos x-1}{5-4\cos x}\right)^2\mathrm{d}x=\pi/3$. – Tom-Tom Dec 20 '13 at 14:43
  • We solved it using the first hint - but now I cant seem to understand how you made the first transition in your comment (to the fraction with 2cosx-1) – jreing Dec 22 '13 at 15:31
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    You have $\cos nx=Re(\mathrm{e}^{\mathrm{i}nx})$ and then $\frac{\cos nx}{2^n}=Re\left[\left(\mathrm{e}^{\mathrm{i}x}/2\right)^n\right]$. The series is performed thanks to $\sum_{n=1}^\infty z^n=\frac{z}{1-z}$ if $|z|<1$. Using $z=\mathrm{e}^{\mathrm{i}x}/2$ and taking the real part you find the expression I have given. – Tom-Tom Dec 22 '13 at 20:56