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[SOLVED] Fourier series

dwsmith

Well-known member
Feb 1, 2012
1,673
Calculate the Fourier series of the function $f$ defined on the interval $[\pi, -\pi]$ by
$$
f(\theta) =
\begin{cases} 1 & \text{if} \ 0\leq\theta\leq\pi\\
-1 & \text{if} \ -\pi < \theta < 0
\end{cases}.
$$
$f$ is periodic with period $2\pi$ and odd since $f$ is symmetric about the origin.
So $f(-\theta) = -f(\theta)$.
Let $f(\theta) = \sum\limits_{n = -\infty}^{\infty}a_ne^{in\theta}$.
Then $f(-\theta) = \sum\limits_{n = -\infty}^{\infty}a_ne^{-in\theta} = \cdots + a_{-2}e^{2i\theta} + a_{-1}e^{i\theta} + a_0 + a_{1}e^{-i\theta} + a_{2}e^{-2i\theta}+\cdots$
$-f(\theta) = \sum\limits_{n = -\infty}^{\infty}a_ne^{-in\theta} = \cdots - a_{-2}e^{-2i\theta} - a_{-1}e^{-i\theta} - a_0 - a_{1}e^{i\theta} - a_{2}e^{2i\theta}-\cdots$

$a_0 = -a_0 = 0$

I have solved many Fourier coefficients but I can't think today.

What do I need to do next?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
a_n = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(\theta)e^{-im\theta}d\theta
$$
Since my function is defined piecewise, would I write it as
$$
a_n = \frac{1}{2\pi}\left[\int_0^{\pi}e^{-im\theta}d\theta - \int_{-\pi}^0e^{-im\theta}d\theta\right]
$$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,705
$$
a_n = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(\theta)e^{-im\theta}d\theta
$$
Since my function is defined piecewise, would I write it as
$$
a_n = \frac{1}{2\pi}\left[\int_0^{\pi}e^{-im\theta}d\theta - \int_{-\pi}^0e^{-im\theta}d\theta\right]
$$
Yes. (Yes)

(But since this is an odd function, you might find it easier to use the real rather than the complex Fourier series. The cosine terms will all be zero and you will only have to deal with the sine terms. To evaluate them, do just what you are doing with the complex terms, writing them as the difference between the integrals on the intervals [0,1] and [-1,0].)
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Yes. (Yes)

(But since this is an odd function, you might find it easier to use the real rather than the complex Fourier series. The cosine terms will all be zero and you will only have to deal with the sine terms. To evaluate them, do just what you are doing with the complex terms, writing them as the difference between the integrals on the intervals [0,1] and [-1,0].)

When I solve, I have
$$
-\frac{1}{2\pi}\int_0^{\pi}\sin m\theta d\theta = \frac{1}{\pi m}
$$
and
$$
-\frac{1}{2\pi}\int_{-\pi}^0\sin m\theta d\theta = -\frac{1}{\pi m}
$$

I think the integral has to be
$$
-\frac{1}{\pi}\int_0^{\pi}\sin m\theta d\theta = \frac{2}{\pi m}
$$

Now what?
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,705
When I solve, I have
$$
-\frac{1}{2\pi}\int_0^{\pi}\sin m\theta d\theta = \frac{1}{\pi m}
$$
and
$$
-\frac{1}{2\pi}\int_{-\pi}^0\sin m\theta d\theta = -\frac{1}{\pi m}
$$

I think the integral has to be
$$
-\frac{1}{\pi}\int_0^{\pi}\sin m\theta d\theta = \frac{2}{\pi m}
$$

Now what?
Try doing those integrals again. $$\begin{aligned}\int_0^\pi\sin m\theta\,d\theta &=\Bigl[-\tfrac1m\cos m\theta\Bigr]_0^\pi \\ &= -\tfrac1m\bigl(\cos m\pi - \cos 0\bigr) \\ &= -\tfrac1m\bigl((-1)^m - 1\bigr) \\ &= \begin{cases}0 &\text{ (if $m$ is even),} \\2/m &\text{ (if $m$ is odd).}\end{cases} \end{aligned}$$

The integral from -1 to 0 is the same but with a minus sign. You should then find that the Fourier series for $f(\theta)$ is $$\sum_{k=0}^\infty\frac4{(2k+1)\pi}\sin(2k+1) \theta.$$