MHB Find Fourier Series for Piecewise Function

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Dustinsfl
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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?
 
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dwsmith said:
$$
b_n = \frac{1}{\pi}\int_0^{\pi}\theta\sin n\theta d\theta + \frac{1}{\pi}\int_{-\pi}^0\theta\sin n\theta d\theta + \int_{-\pi}^0\sin n\theta d\theta
$$
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.
 
So the Fourier series is
$$
\frac{\pi}{2} - 2\sum_{n = 1}^{\infty}\frac{1}{2n}\sin(2n)\theta.
$$
 
dwsmith said:
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]
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
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