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vpde_lecture20 [2020/03/19 12:24]
trinh 		vpde_lecture20 [2020/03/20 19:19] (current)
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 ====== 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 ====
  
-We will first start by reviewing Lecture 19 and our introduction of the Fourier convergence theorem. This involves **Theorem 12.5** in the notes. We'll draw some pictures of how to visualiser the 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 2π-periodic function with f and f′ continuous on the interval (−π,π), 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 2π-periodic functions on [−π,π] 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 [−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, 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)=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
 +$$
 +b_n = \frac{2}{\pi} \int_0^\pi x^2 \sin(nx) \, \mathrm{d}x.
 +$$
Last modified: 2020/03/19 12:24