This shows you the differences between two versions of the page.
| Next revision | Previous revision | ||
|
vpde_lecture20 [2020/03/19 12:23] trinh created |
vpde_lecture20 [2020/03/20 19:19] (current) trinh |
||
|---|---|---|---|
| Line 1: | Line 1: | ||
| - | ====== | + | ====== |
| + | < | ||
| + | <iframe width=" | ||
| + | </ | ||
| + | |||
| + | ==== 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, | ||
| + | |||
| + | ==== Fourier series for even and odd- extensions ==== | ||
| + | |||
| + | We are almost done. There is one last variation to explain. Occasionally, | ||
| + | |||
| + | 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 | ||
| + | $$ | ||
| + | b_n = \frac{2}{\pi} \int_0^\pi x^2 \sin(nx) \, \mathrm{d}x. | ||
| + | $$ | ||