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I want to compute the Fourier series for the following function

$$ g_n(\theta) = -2nK_{n}(\theta)\sin(n\theta)$$

where $K_n(\theta)$ is the Fejer Kernel.

I tried to compute the Fourier coefficients directly using this formula for $K_n(\theta)$

$$ K_n(\theta) = \frac{1}{n} \left(\frac{\sin^2(\frac{n\theta}{2})} {\sin^2(\frac{\theta}{2})}\right) $$

Then the coefficients are given by

\begin{align} c_k & = \frac{1}{2\pi} \int_{-\pi}^{\pi} g_n(\theta)e^{-ik\theta}d\theta \\ & = \frac{1}{2\pi} \int_{-\pi}^{\pi} \left(-2n \frac{1}{n} \left(\frac{\sin^2(\frac{n\theta}{2})} {\sin^2(\frac{\theta}{2})}\right)\sin(n\theta)\right)e^{-ik\theta}d\theta \\ & = -\frac{1}{\pi} \int_{-\pi}^{\pi} \left(\frac{\sin^2(\frac{n\theta}{2})} {\sin^2(\frac{\theta}{2})}\right)\sin(n\theta)e^{-ik\theta}d\theta \\ \end{align}

I'm not sure how to proceed after this. Can someone give me a hint as to how I should compute this Fourier series?

cdk
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  • Did you try to see if this is an odd or an even function? It seems the real part may be odd so those parts of those coefficients must vanish. – Gregory Grant Jan 27 '16 at 21:41
  • @GregoryGrant Given that $K_n$ is an even summability kernel, $g_n$ is indeed odd. I'll try computing just the sine terms. – cdk Jan 27 '16 at 21:47
  • because $K_n(\theta) = f(\theta)^2$ and that $f(\theta) = C \sum_{k \le n} e^{i k \theta}$ you get that $K_n(\theta) = C^2 \sum_{k \le n} |k| e^{i k \theta}$ so that $\sin(n \theta) K_n(\theta) = C^2 / 2i \sum_{k \le n} (|k+n|-|k-n|) e^{i k \theta}$ for some constant $C$ – reuns Jan 27 '16 at 22:03

1 Answers1

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There is a problem with the significance of index $i$.

Don't you want to compute the $k$th coefficient, which is: \begin{align} c_k & = \frac{1}{2\pi} \int_{-\pi}^{\pi} g_n(\theta)e^{-ik\theta}d\theta \ ? \end{align}

Jean Marie
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