MA20223 Lecture 21

We finished the example from lecture 18, which was to calculate the Fourier series of the $2\pi$-periodic extension of the function $f(x) = x^2$ defined on $[0, 2\pi]$. The Fourier series is given by $$f(x) \sim \frac{a_0}{2} + \sum_{n=1}^\infty [a_n \cos(nx) + b_n\sin(nx)]$$ where we showed that $$a_0 = \frac{1}{\pi}\frac{(2\pi)^3}{3}, \qquad a_n = \frac{4}{n^2}, \ n \geq 1$$ and $$b_n = -\frac{4\pi}{n}.$$ We then used a Matlab code to plot it.

x = linspace(-pi, 3*pi, 5000);
Nvals = [1, 5, 10, 50];
 
figure(1); hold all
for j = 1:length(Nvals)
    N = Nvals(j);
    a0 = 1/pi*(2*pi)^3/3;
    S = a0/2;
    for n = 1:N
        S = S + 4/n^2*cos(n*x) - 4*pi/n*sin(n*x);
    end
    plot(x, S, 'LineWidth', 2)    
end
xlabel('x');
ylabel('f');
title('Fourier series for x^2 on [0, 2*pi]');
xx = linspace(0, 2*pi, 50);
plot(xx, xx.^2, 'k', 'LineWidth', 2);
grid on

We then tackled the following problem: calculate the $2\pi$-periodic odd extension of the function $f(x) = x^2$ on $[0, \pi]$.

We showed that to do this, first construct the odd extension to $[-\pi, \pi]$, and then construct the $2\pi$ extension from there. Then the Fourier series is $$f(x) \sim \sum_{n=1}^\infty b_n\sin(nx)$$ where we showed that $$b_n = \frac{2}{\pi}\left[-\frac{\pi^2 (-1)^n}{n} + \frac{2}{n^2} [(-1)^n - 1)]\right].$$ We then used the following code to plot it.

x = linspace(-2*pi, 2*pi, 5000);
Nvals = [1, 5, 10, 50];
 
b = @(n) 2/pi*(-pi^2*(-1).^n./n + 2./n.^2.*((-1).^n - 1));
 
figure(1); hold all
for j = 1:length(Nvals)
    N = Nvals(j);
    S = 0;
    for n = 1:N
        S = S + b(n)*sin(n*x);
    end
    plot(x, S, 'LineWidth', 2)    
end
xlabel('x');
ylabel('f');
title('Fourier series for the odd extension of x^2');
xx = linspace(0, pi, 30);
plot(xx, xx.^2, 'k', 'LineWidth', 2);
grid on