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Let

$$ f(x)=\left(\sum_{n=1}^\infty \frac{\cos(nx)}{2^n}\right)^2 $$

$$\hat{f}(n) = \frac{1}{2\pi}\int_0^{2\pi} \left(\sum_{n=1}^\infty \frac{\cos(nx)}{2^n}\right)^2 e^{-inx} dx$$

Easy to see the series converges uniformly, so we can interchange the sum with the integral. What's bothering me is the power of $2$. How do I change it correctly? Is it:

$$\hat{f}(n) = \frac{1}{2\pi} \sum_{n=1}^\infty \left( \int_0^{2\pi} \frac{\cos(nx)}{2^n}\right)^2 e^{-inx} dx$$

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    $\begingroup$ If the inside sum wasn't squared, then that would work. $\endgroup$ – JimmyK4542 Dec 26 '14 at 22:07
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First, write $\displaystyle\int_{0}^{2\pi}\left(\sum_{n = 1}^{\infty}\dfrac{\cos(nx)}{2^n}\right)^2\,dx = \displaystyle\int_{0}^{2\pi}\sum_{m = 1}^{\infty}\sum_{n = 1}^{\infty}\dfrac{\cos(mx)}{2^m}\dfrac{\cos(nx)}{2^n}\,dx$.

Now, interchange the integral and the double sum to get $\displaystyle\sum_{m = 1}^{\infty}\sum_{n = 1}^{\infty}\int_{0}^{2\pi}\dfrac{\cos(mx)}{2^m}\dfrac{\cos(nx)}{2^n}\,dx$.


EDIT: Now that you have changed the question significantly, it looks like you are trying to find the Fourier series coefficients for $f(x)^2$ where $f(x) = \displaystyle\sum_{n = 1}^{\infty}\dfrac{\cos(nx)}{2^n}$. If this is indeed the case, since you already have the Fourier series coefficeints for $f(x)$, I would recommend using the fact that multiplication in the time domain corresponds to convolution in the frequency domain.

Specifically, if we know that $f(x) = \displaystyle\sum_{n = -\infty}^{\infty}a_ne^{inx}$ and $g(x) = \displaystyle\sum_{n = -\infty}^{\infty}b_ne^{inx}$, then $f(x)g(x) = \displaystyle\sum_{n = -\infty}^{\infty}c_ne^{inx}$ where $c_n = \displaystyle\sum_{k = -\infty}^{\infty}a_kb_{n-k}$.

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  • $\begingroup$ I've edited the question, but it doesn't matter... $\endgroup$ – AmitBB Dec 26 '14 at 22:11
  • $\begingroup$ question: How do you explain the change of the summation? Not all terms are positive. $\endgroup$ – AmitBB Dec 26 '14 at 22:12
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    $\begingroup$ In your question, you said that it is "Easy to see the series converges uniformly". I think that is still the case with the double summation, but I'm not sure. $\endgroup$ – JimmyK4542 Dec 26 '14 at 22:33

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