# [SOLVED]Fourier series without integration

#### dwsmith

##### Well-known member
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?

#### Opalg

##### MHB Oldtimer
Staff member
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?
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.

#### dwsmith

##### Well-known member
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.
Then
$$h(\theta) = \sum\limits_{n = 1}^{\infty}(-1)^{n + 1}\frac{\sin n\theta}{n} + \frac{1}{2}\sum\limits_{n = 1}^{\infty}\frac{\sin(2n - 1)\theta}{2n - 1}.$$

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