For functions that are not periodic, the Fourier series is replaced by the Fourier transform.
The proof uses Parseval's identity, a generalization of the Pythagorean theorem. Example: Numerical Evaluation of Fourier Series for a Complicated Function //--> (f,g)=2T∫0Tf(x)g(x) dx,(f,g) = \frac{2}{T} \int_0^T f(x) g(x) \:dx,(f,g)=T2∫0Tf(x)g(x)dx, f=∑kbk(f,bk)f = \sum_k b_k (f,b_k)f=k∑bk(f,bk), The Fourier series is simply a particular way of rewriting functions, using the basis {bk}={fk}∪{gk}\{b_k\} = \{f_k\} \cup \{g_k\}{bk}={fk}∪{gk}.
As an example, look at the plot of Figure 1: Let's define a 'Fourier Series' now. Find the Fourier series of the square wave, for which the function over one period is, f(x)={1if 0≤x<12−1if 12≤x<1.f(x) = \begin{cases} 1 \quad &\text{if }~0\leq x<\frac12 \\ -1 \quad &\text{if }~\frac12 \leq x < 1. jQuery(document).ready(checkAds()); function checkAds(){if (document.getElementById('adsense')!=undefined){document.write("