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| ====== MA20223 Lecture 20 ====== | ====== MA20223 Lecture 20 ====== | ||
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| + | < | ||
| + | <iframe width=" | ||
| + | </ | ||
| ==== 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 [−π,π]. 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/π)=g(X), then this new function g(X) is 2π periodic and defined on [−π,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)=x2 === | ||
| + | |||
| + | We'll draw the odd and even periodic extension of f(x)=x2 originally defined on [0,π], and then extended in an even or odd manner to [−π,π]. 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 fe(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 | ||
| + | $$ | ||
| + | b_n = \frac{2}{\pi} \int_0^\pi x^2 \sin(nx) \, \mathrm{d}x. | ||
| + | $$ | ||