Differences

This shows you the differences between two versions of the page.

--- vpde_lecture20 [2020/03/19 12:30]
trinh
+++ vpde_lecture20 [2020/03/19 15:59]
trinh [Fourier series for even and odd- extensions]
@@ Line 17: / Line 17: @@
 $$
 f(x) \sim \frac{a_0}{2} + \sum_{n=1}^\infty \left[ a_n \cos\left(\frac{n\pi x}{L}\right) + b_n \sin\left(\frac{n\pi x}{L}\right)\right]
+$$
+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
+$$
+b_n = \frac{2}{\pi} \int_0^\pi x^2 \sin(nx) \, \mathrm{d}x.
+$$

Trinh @ Bath

Page Tools

Site Tools

User Tools

Differences