Section 25.2 Fourier series
Using the period \(T=1\text{,}\) we find that the functions \(f_k(x) = \exp (2\pi i k x)\) are orthonormal with respect to the inner product
(For other periods they are orthogonal but not orthonormal.) This makes it easy to compute the Fourier coefficients of a given function \(f\) in the basis \(\{f_k\colon k\in \mathbb Z\}\text{:}\)
For sufficiently nice periodic functions (having a continuous derivative is enough), the Fourier series \(\sum c_k f_k\) converges to function \(f\text{.}\)
Example 25.2.1. Compute Fourier coefficients.
Use the built-in command integral
to compute the Fourier coefficients \(c_k\) for \(k=-3, \dots, 3\) for the complex-valued function
Then plot both the function and its partial Fourier sum \(\sum_{|k|\le 3} c_k f_k\) for comparison.
The syntax of integral
is integral(@(x) ..., a, b)
where \(a, b\) are the limits of integration and the function is expressed in a form that allows vectorized evaluation. Here is the computation of Fourier coefficient
f = @(x) sqrt(1 + cos(2*pi*x)) + 1i*log(2 + sin(2*pi*x)); n = 3; c = zeros(1, 2*n+1); for k = -n:n c(k+n+1) = integral(@(x) f(x).*exp(-2*pi*1i*k*x), 0, 1); end disp(c)
Then we combine both complex plots using different colors.
t = linspace(0, 1, 1000); k = -n:n; p = c*exp(2*pi*1i*k'*t); hold on plot(f(t)) plot(p, 'r') hold off
Note that the lack of smoothness at one point slows down the convergence of Fourier series. If the function had sqrt(1.1 + cos(2*pi*x))
instead, the convergence would be much better.