MATH 414 Lecture 8

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Sine and Cosine Series

A regular Fourier series expresses a function in terms of both sines and cosines. For example, on the interval $[-\pi ,\pi ]$,

$\begin{array}{rl}f(x)& =a_0+\sum _{k=1}^{\mathrm{\infty }}{a}_k\cos (kx)+b_k\sin (kx)\\ a_0& =\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)\mathrm{d}x\\ a_n& =\frac{1}{\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)\cos (nx)\mathrm{d}x\\ b_n& =\frac{1}{\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)\sin (nx)\mathrm{d}x\end{array}$

Sine Series

The Fourier Sine Series of a function $f(x)$ on an interval $[0,\pi ]$ is a fourier series that produces an odd extension of the function $f(x)$ (i.e. for $-\pi <x<0$, we have $-f(-x)$.)

$\begin{array}{r}f(x)=\sum _{k=1}^{\mathrm{\infty }}{b}_k\sin (kx)b_n=\frac{2}{\pi }\underset{0}{\overset{\pi }{\int }}f(x)\sin (nx)\mathrm{d}x\end{array}$

Cosine Series

The Fourier Cosine Series of a function $f(x)$ on an interval $[0,\pi ]$ is a fourier series that produces an even extension of the function $f(x)$ (i.e. for $-\pi <x<0$, we have $f(-x)$.

$\begin{array}{rl}f(x)& =a_0+\sum _{k=1}^{\mathrm{\infty }}{a}_k\cos (kx)\\ a_0=\frac{1}{\pi }\underset{0}{\overset{\pi }{\int }}f(x)\mathrm{d}x\\ a_n=\frac{2}{\pi }\underset{0}{\overset{\pi }{\int }}f(x)\cos (nx)\mathrm{d}x\end{array}$

Complex Fourier Series

Recall

$\begin{array}{rl}{\mathrm{e}}^{inx}& =\cos (nx)+i\sin (nx)\\ {\mathrm{e}}^{-inx}& =\cos (nx)-i\sin (nx)\end{array}$

Let's add and divide by 2:

$\begin{array}{rl}\cos (nx)& =\frac{1}{2}({\mathrm{e}}^{inx}+{\mathrm{e}}^{-inx})\\ \sin (nx)& =\frac{1}{2i}({\mathrm{e}}^{inx}+{\mathrm{e}}^{-inx})\end{array}$

Substituting this into the fourier series definition and simplifying gives

$f(x)=a_0+\sum _{k=1}^{\mathrm{\infty }}\left(\frac{a_k-i{b}_k}{2}\right){\mathrm{e}}^{inx}+\sum _{k=1}^{\mathrm{\infty }}\left(\frac{a_k+i{b}_n}{2}{\mathrm{e}}^{-inx}\right)$

We'll rewrite this formula a different way. Let $c_0:=a_0$, $c_k:=\frac{a_k-i{b}_k}{2}$. If we change the index on our sum by negating, we get

$f(x)=\sum _{n=0}^{\mathrm{\infty }}{c}_n{\mathrm{e}}^{inx}+\sum _{n=-\mathrm{\infty }}^{-}1\left(\frac{a_{-n}+i{b}_{-n}}{2}\right){\mathrm{e}}^{inx}$

Observe that $\left(\frac{a_{-n}+i{b}_{-n}}{2}\right)$ is the same as $c_n$, so we're left with

$\begin{array}{rl}f(x)& =\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_n{\mathrm{e}}^{inx}\\ c_0& =\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x){\mathrm{e}}^{-0x}\mathrm{d}x=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)\mathrm{d}x\\ c_n& =\frac{1}{2\pi }\int {-\pi }^{\pi }f(x){\mathrm{e}}^{-inx}\mathrm{d}x\end{array}$