Fourier made the astounding claim that any function can be represented by an infinite sum of sines and consines. That is,

\[ \begin{aligned} f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty (\color{blue}{a_n} \cos{nx} + \color{blue}{b_n} \sin{nx}) & x \in (-\pi, \pi] \end{aligned} \]

Suppose that is true, how do we figure out the coefficients \(a_n\) and \(b_n\) ? Multiply and integrate! For example,

\[ \begin{aligned} \int_{-\pi}^{\pi} f(x) \cos{mx}\,dx &= \int_{-\pi}^{\pi} \left(\frac{a_0}{2}\cos{mx} + \sum_{n=1}^\infty (a_n \cos{nx}\cos{mx} + b_n \sin{nx}\cos{mx})\right)\,dx \\ &= \frac{a_0}{2m}\sin{mx} \big|_{-\pi}^{\pi} + \sum_{n=1}^\infty \left( \int_{-\pi}^{\pi} a_n \cos{nx}\cos{mx} + \int_{-\pi}^{\pi} b_n \sin{nx}\cos{mx} \right)\,dx \\ &= \sum_{n=1}^\infty \left( a_n \int_{-\pi}^{\pi} \cos{nx}\cos{mx}\,dx + b_n \int_{-\pi}^{\pi} \sin{nx}\cos{mx}\,dx \right) \\ \end{aligned} \]

But as we saw earlier, \(\displaystyle \int_{-\pi}^{\pi} \sin{nx}\cos{mx}\,dx\) simply evaluates to zeros, whereas \(\displaystyle \int_{-\pi}^{\pi} \cos{nx}\cos{mx}\,dx\) also evaluate to zeros except when \(m = n\) it evaluates to \(\pi\). So,

\[ \begin{aligned} \int_{-\pi}^{\pi} f(x) \cos{nx}\,dx = a_n \pi \\ \therefore \color{blue}{a_n = {1 \over \pi} \int_{-\pi}^{\pi} f(x) \cos{nx}\,dx} \\ \end{aligned} \]

Similarly, we can multiply both sides with \(\sin{mx}\) and integrate, and come up with:

\[ \begin{aligned} \color{blue}{b_n} &\color{blue}{= {1 \over \pi} \int_{-\pi}^{\pi} f(x) \sin{nx}\,dx} \\ \end{aligned} \]

Furthermore, similar methods can be applied to the complex plane. Suppose any complex function can be represented as:

\[ \begin{aligned} f(x) &= \sum_{-\infty}^{\infty} \color{blue}{c_n} e^{in \pi x/L} & x \in [-L, L] \\ \end{aligned} \]

To find \(c_n\), we can multiply both sides with \(\displaystyle e^{-im \pi x/L}\) and then integrate from \(-L\) to \(L\). The integral will evaluate to zeros except when \(n=m\), the value becomes \(2L\). So

\[ \begin{aligned} \color{blue}{c_n} &\color{blue}{= {1 \over 2L} \int_{-L}^{L} f(x) e^{-in \pi x/L}\,dx} \\ \end{aligned} \]

\(\Box\)

Source: Computational Methods for Data Analysis.

# posted by rot13(Unafba Pune) @ 7:29 AM