We will first start by reviewing Lecture 19 and our introduction of the Fourier convergence theorem. This involves **Theorem 12.5** in the notes. In particular, the theorem states that if $f$ is a $2\pi$-periodic function with $f$ and $f'$ continuous on the interval $(-\pi, \pi)$, then the Fourier series of $f$ at $x$ converges to the average of the left- and right-limits. Thus
$$
\frac{1}{2}[f(x_-) + f(x_+)] = \frac{a_0}{2} + \sum_{1}^\infty [a_n \cos(nx) + b_n \sin(nx)]
$$

We'll draw some pictures of how to visualise the theorem. We will also discuss again this notion of pointwise convergence vs. uniform convergence and the notion of the Gibbs' Phenomenon.

Next, we want to simply note that the Fourier series you've derived for $2\pi$-periodic functions on $[-\pi, \pi]$ can be easily extended to functions defined on $[-L, L]$. The truth is that we should really just have done the derivation like this from the get-go! This leads to:

**Theorem 12.7:** (Fourier coefficients for $2L$-periodic function) Let $f$ be a periodic function with period $2L$. Then
$$
f(x) \sim \frac{a_0}{2} + \sum_{n=1}^\infty \left[ a_n \cos\left(\frac{n\pi x}{L}\right) + b_n \sin\left(\frac{n\pi x}{L}\right)\right]
$$
where now the coefficients are calculated from
$$
a_n = \frac{1}{L} \int_{-L}^L f(x) \cos\left(\frac{n\pi x}{L}\right) \, \mathrm{d}{x}
$$
$$
b_n = \frac{1}{L} \int_{-L}^L f(x) \sin\left(\frac{n\pi x}{L}\right) \, \mathrm{d}{x}
$$

There is a simple way to prove this based on what we already know. Let's take the function we have on $[-L. L]$ and simply transform the domain so that it is now between $[-\pi, \pi]$. We can do that via $$ X = \frac{\pi x}{L}. $$ Go ahead and verify that this works as indicated. You can verify that if $f(x) = f(LX/\pi) = g(X)$, then this new function $g(X)$ is $2\pi$ periodic and defined on $[-\pi, pi]$. So now we have the Fourier series for $g(X)$. Go and write that down. After you have done that, you'll notice that you get the above formulae.

**Remark 12.8.** There is an important note here. Because of the $2L$-periodicity, suppose we asked you to determine the Fourier series for a function on $[0, 2L]$ instead of $[-L, L]$. Then this is equivalent to using all the same formulae, but instead you integrate from $0$ to $2L$.

We are almost done. There is one last variation to explain. Occasionally, we want to derive the Fourier series of an even or odd- extension of a function that is defined on $[0, L]$. In other words, you would take that original function on $[0, L]$, plot the even/odd extension to $[-L, 0]$, and then at that point you have a full function from $[-L, L]$ so you can apply the usual Fourier series.

It is a lot easier to explain how this is done via a picture.

We'll draw the odd and even periodic extension of $f(x) = x^2$ originally defined on $[0, \pi]$, and then extended in an even or odd manner to $[-\pi, \pi]$. So for example $$ f_e(x) = \begin{cases} x^2 & x\in[0, \pi] \\ x^2 & x\in[-\pi, 0] \end{cases} $$ is the even extension, while $$ f_o(x) = \begin{cases} x^2 & x\in[0, \pi] \\ -x^2 & x\in[-\pi, 0] \end{cases} $$ is the odd extension.

Once you have drawn (or derived) the extension, then it becomes a simple matter to calculate the Fourier series. For example, since $f_e(x)$ is an even function then only cosines are present, and $$ f_e(x) \sim \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos(nx), $$ where $$ a_n = \frac{2}{\pi} \int_0^\pi x^2 \cos(nx) \, \mathrm{d}x. $$ Above, we have used the even property to double up the integral over the positive $x$ values. Similarly, the Fourier series for the odd extension is $$ f_o(x) \sim \sum_{n=1}^\infty b_n \sin(nx), $$ where $$ b_n = \frac{2}{\pi} \int_0^\pi x^2 \sin(nx) \, \mathrm{d}x. $$

Last modified: 2020/03/20 19:19