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Tuesday, January 07, 2014

 

Fourier Series

Fourier made the astounding claim that any function can be represented by an infinite sum of sines and consines. That is,
  f(x)=a02+n=1(ancosnx+bnsinnx)x(π,π]
Suppose that is true, how do we figure out the coefficients an and bn ? Multiply and integrate! For example,
  ππf(x)cosmxdx=ππ(a02cosmx+n=1(ancosnxcosmx+bnsinnxcosmx))dx=a02msinmx|ππ+n=1(ππancosnxcosmx+ππbnsinnxcosmx)dx=n=1(anππcosnxcosmxdx+bnππsinnxcosmxdx)
But as we saw earlier, ππsinnxcosmxdx simply evaluates to zeros, whereas ππcosnxcosmxdx also evaluate to zeros except when m=n it evaluates to π. So,
  ππf(x)cosnxdx=anπ
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.


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