[SOLVED]Fourier coefficient

dwsmith

Well-known member
$f(\theta) = \theta$ 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
$$2\pi\sum_{n =1}^{\infty}\frac{1}{2n-1}\sin(2n-1)\theta - \frac{1}{2n}\sin 2n\theta\mbox{?}$$

Last edited:

Opalg

MHB Oldtimer
Staff member
$f(\theta) = \theta$ 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
$$2\pi\sum_{n =1}^{\infty}\frac{1}{2n-1}\sin(2n-1)\theta - \frac{1}{2n}\sin 2n\theta\mbox{?}$$
Correct except that the $\dfrac1\pi$ in front of the integral has got lost somewhere along the line, so there should not be a $\pi$ in the final formula. Also, it would look better if you used some parentheses to indicate that everything to the right of the $\sum$ is meant to be included in the summation: $$2\sum_{n =1}^{\infty}\Bigl(\frac{1}{2n-1}\sin(2n-1)\theta - \frac{1}{2n}\sin 2n\theta\Bigr).$$