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,
|
∫π−πf(x)cosmxdx=∫π−π(a02cosmx+∞∑n=1(ancosnxcosmx+bnsinnxcosmx))dx=a02msinmx|π−π+∞∑n=1(∫π−πancosnxcosmx+∫π−πbnsinnxcosmx)dx=∞∑n=1(an∫π−πcosnxcosmxdx+bn∫π−πsinnxcosmxdx) |
But as we saw earlier,
∫π−πsinnxcosmxdx simply evaluates to zeros, whereas
∫π−πcosnxcosmxdx also evaluate to zeros except when
m=n it evaluates to
π. So,
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
