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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 $[-\pi, \pi]$. 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 | ||
+ | $$ | ||
+ | b_n = \frac{2}{\pi} \int_0^\pi x^2 \sin(nx) \, \mathrm{d}x. | ||
+ | $$ |