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- Thread starter dwsmith
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- Feb 7, 2012

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I think the only way to do this would be if you already happen to know the Fourier series for the functions $$ f(\theta) = \theta, \qquad g(\theta) = \begin{cases} 1 & (0 < \theta < \pi) \\ 0 & (\theta = 0, \pm\pi) \\ -1 & (-\pi < \theta < 0) \end{cases}.$$ Then you can use the fact that $h(\theta) = \frac12f(\theta) + \frac\pi2g(\theta)$ to write down the answer.Let

$$

h(\theta) = \begin{cases}

\frac{1}{2}(\theta + \pi), & 0 < \theta < \pi\\

0, & \theta = 0, \pm\pi\\

\frac{1}{2}(\theta - \pi), & -\pi < \theta < 0

\end{cases}

$$

How can I find the fourier series without doing any integration?

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ThenI think the only way to do this would be if you already happen to know the Fourier series for the functions $$ f(\theta) = \theta, \qquad g(\theta) = \begin{cases} 1 & (0 < \theta < \pi) \\ 0 & (\theta = 0, \pm\pi) \\ -1 & (-\pi < \theta < 0) \end{cases}.$$ Then you can use the fact that $h(\theta) = \frac12f(\theta) + \frac\pi2g(\theta)$ to write down the answer.

$$

h(\theta) = \sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\frac{\sin n\theta}{n} + \frac{1}{2}\sum\limits_{n = 1}^{\infty}\frac{\sin(2n - 1)\theta}{2n - 1}.

$$

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