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I'm given $f\in \mathcal{C}^1[-\pi, \pi]$ with $f(-\pi)=f(\pi)$. It's fourier coefficients are given by:

$$\gamma_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-int}f(t)dt,\ n\in \mathbb{Z}$$

And now I'm asked to express the fourier coefficients of $f'$ in terms of the $\gamma_n$.

So my reasoning is that we have $f=\sum_{-\infty}^{\infty}\gamma_n e^{int}$, and so we just find that $f'=(\sum_{-\infty}^{\infty}\gamma_n e^{int})'=\sum_{-\infty}^{\infty}in\gamma_n e^{int}$. So that the coefficients of $f'$ are just $in\gamma_n$.

However I don't think this is correct. Because as far as I can tell I did not use the fact that $f(-\pi)=f(\pi)$, and it's probably not given for no reason.

So can someone tell me what I'm doing wrong here?

user2520938
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1 Answers1

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The Fourier coefficients of $f'$ are given by

$$c_n = \frac{1}{2\pi}\int_{-\pi}^\pi e^{-int}f'(t)\, dt\quad (n\in \Bbb Z).$$

By integration by parts,

$$c_n = \frac{e^{-int}f(t)}{2\pi}\bigg|_{-\pi}^\pi - \frac{1}{2\pi}\int_{-\pi}^\pi (-in e^{-int})f(t)\, dt = in\gamma_n,$$

using the boundary condition $f(-\pi) = f(\pi)$.

kobe
  • 41,901
  • Thanks for this answer, but can you tell me if there is any fault in my approach? I just treat the $\gamma_n$ as constants and use the linearity of differentiation. – user2520938 Jan 24 '15 at 13:45
  • How did you know that the series $\sum_{-\infty}^\infty in\gamma_ne^{-int}$ converges uniformly on $[-\pi,\pi]$? You would need this condition in order to differentiate the way you did. – kobe Jan 24 '15 at 13:48
  • Wouldn't I just need the uniform convergence of $\sum_{-\infty}^{\infty}\gamma_n e^{int}$? – user2520938 Jan 24 '15 at 13:54
  • @user2520938 no. However, since your $f$ is in $C^1[-\pi,\pi]$ and $f(-\pi) = f(\pi)$, there is a theorem which supports your claim that $f'(t) = \sum_{n = -\infty}^\infty in\gamma_n e^{int}$ (on $(-\pi,\pi)$). That theorem contains the condition $f(-\pi) = f(\pi)$ which you passed by in your reasoning. – kobe Jan 24 '15 at 14:26
  • Can you please tell me the name of that theorem? I have it in my book here, but they don't mention it's name and I want to look for some additional information about it. – user2520938 Jan 24 '15 at 15:45
  • @user2520938 There's no particular name for the theorem. Look into Brown and Churchill's text, "Fourier Series and Boundary Value Problems" in the sections on uniform convergence and differentiation of Fourier series. The theorem (or rather two theorems combined) I'm referring to states that if $f\in C[-\pi,\pi]$ such that $f(-\pi) = f(\pi)$ and $f'$ is piecewise continuous on $(-\pi,\pi)$, then the Fourier series of $f$ converges uniformly to $f$ on $[-\pi,\pi]$ and the differentiated Fourier series of $f'$ converges to $f'$ at every point of $(-\pi,\pi)$ where $f''$ exists. – kobe Jan 24 '15 at 15:57