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$f(\theta) = \theta$ for $-\pi < \theta \leq \pi$

$f$ is odd so $\sum\limits_{n = 1}^{\infty}a_n\sin n\theta$.

$$

\frac{1}{\pi}\int_{-\pi}^{\pi}\theta\sin n\theta d\theta

$$

So I have that

$$

a_n = \begin{cases}

\frac{-2\pi}{n}, & \text{if n is even}\\

\frac{2\pi}{n}, & \text{if n is odd}

\end{cases}

$$

So the solution would be

$$

2\pi\sum_{n =1}^{\infty}\frac{1}{2n-1}\sin(2n-1)\theta - \frac{1}{2n}\sin 2n\theta\mbox{?}

$$

$f$ is odd so $\sum\limits_{n = 1}^{\infty}a_n\sin n\theta$.

$$

\frac{1}{\pi}\int_{-\pi}^{\pi}\theta\sin n\theta d\theta

$$

So I have that

$$

a_n = \begin{cases}

\frac{-2\pi}{n}, & \text{if n is even}\\

\frac{2\pi}{n}, & \text{if n is odd}

\end{cases}

$$

So the solution would be

$$

2\pi\sum_{n =1}^{\infty}\frac{1}{2n-1}\sin(2n-1)\theta - \frac{1}{2n}\sin 2n\theta\mbox{?}

$$

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