====== MA20223 Lecture 20 ======
==== Fourier convergence theorem ====
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.
==== Fourier series on any interval ====
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 $2L$.
==== Fourier series for even and odd- extensions ====
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.
=== Odd- and even extensions of $f(x) = x^2$ ===
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
$$
b_n = \frac{2}{\pi} \int_0^\pi x^2 \sin(nx) \, \mathrm{d}x.
$$