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vpde_lecture20 [2020/03/19 12:30]
trinh
vpde_lecture20 [2020/03/20 19:19] (current)
trinh
Line 1: Line 1:
 ====== MA20223 Lecture 20 ====== ====== MA20223 Lecture 20 ======
 +
 +<html>
 +<iframe width="560" height="315" src="https://www.youtube.com/embed/emwg9UwSn9o" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
 +</html>
  
 ==== Fourier convergence theorem ==== ==== Fourier convergence theorem ====
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 $$ $$
 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] 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
 +$$
 +b_n = \frac{2}{\pi} \int_0^\pi x^2 \sin(nx) \, \mathrm{d}x.
 +$$