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[SOLVED] piecewise fourier

dwsmith

Well-known member
Feb 1, 2012
1,673
Find the Fourier series for
$$
f(\theta) = \begin{cases}
\theta, & 0\leq \theta \leq\pi\\
\pi + \theta, & -\pi\leq \theta < 0
\end{cases}.
$$

$$
a_0 = \frac{1}{\pi}\int_0^{\pi}\theta d\theta + \frac{1}{\pi}\int_{-\pi}^0\theta d\theta + \int_{-\pi}^0 d\theta
$$
The first and second integral together are 0 so the $a_0 = \pi$
$$
a_n = \frac{1}{\pi}\int_0^{\pi}\theta\cos n\theta d\theta + \frac{1}{\pi}\int_{-\pi}^0\theta\cos n\theta d\theta + \int_{-\pi}^0 \cos n\theta d\theta
$$
The first and second integral together are 0 so the $a_n = 0$
$$
b_n = \frac{1}{\pi}\int_0^{\pi}\theta\sin n\theta d\theta + \frac{1}{\pi}\int_{-\pi}^0\theta\sin n\theta d\theta + \int_{-\pi}^0\sin n\theta d\theta
$$
The first and second integral together are 0 so the $b_n = \begin{cases}\frac{-2}{n}, & \text{if n is odd}\\0, & \text{if n is even}\end{cases}$

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

Correct?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
$$
b_n = \frac{1}{\pi}\int_0^{\pi}\theta\sin n\theta d\theta + \frac{1}{\pi}\int_{-\pi}^0\theta\sin n\theta d\theta + \int_{-\pi}^0\sin n\theta d\theta
$$
The first and second integral together are 0 so the $b_n = \begin{cases}\frac{-2}{n}, & \text{if n is odd}\\0, & \text{if n is even}\end{cases}$

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

Correct?
Hi dwsmith, :)

I think you have made a little error. As you can see \(g(\theta)=\theta\sin n\theta=g(-\theta)\). Therefore \(g\) is an even function. Hence,

\[b_n = \frac{1}{\pi}\int_0^{\pi}\theta\sin n\theta d\theta + \frac{1}{\pi}\int_{-\pi}^0\theta\sin n\theta d\theta + \int_{-\pi}^0\sin n\theta d\theta= \frac{2}{\pi}\int_0^{\pi}\theta\sin n\theta d\theta + \int_{-\pi}^0\sin n\theta d\theta\]

Kind Regards,
Sudharaka.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
So the Fourier series is
$$
\frac{\pi}{2} - 2\sum_{n = 1}^{\infty}\frac{1}{2n}\sin(2n)\theta.
$$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,715
So the Fourier series is
$$
\frac{\pi}{2} - 2\sum_{n = 1}^{\infty}\frac{1}{2n}\sin(2n)\theta.
$$
Correct! You can check that this formula works by using the excellent Desmos graphing application that now comes for free with MathHelpBoards. See the graph below. In fact, click on the graph and you will see the formula used to generate it. Use the slider to see the sum of up to 20 terms of this Fourier series, and notice how it converges to the midpoint of each jump in the graph.

[graph]ymrd1ovgce[/graph]