Fourier sine series, Fourier cosine series

When a function is an odd function, its Fourier series contains only the sine functions. Such a Fourier series is called a Fourier sine series. When a function is an even function, its Fourier series contains only the cosine functions (and possibly a constant). Such a Fourier series is called a Fourier cosine series.

Example. Consider the function $f\in {\mathcal{P}}_{2\pi }$ defined by

$$f(x)=x(-\pi <x<\pi ).$$

Note that $f$ is an odd function (i.e., $f(-x)=-f(x)$). The Fourier coefficients are obtained as (exercise!)

$$\begin{array}{rcl}{a}_n& =& \frac{1}{\pi }{\int }_0^{2\pi }f(x)\cos (nx)dx=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f(x)\cos (nx)dx\\ & =& 0\\ b_n& =& \frac{1}{\pi }{\int }_0^{2\pi }f(x)\sin (nx)dx=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f(x)\sin (nx)dx.\\ & =& (-1{)}^{n+1}\frac{2}{n}.\end{array}$$

Thus, we have

$$x\sim 2[\frac{\sin x}{1}-\frac{\sin (2x)}{2}+\frac{\sin (3x)}{3}-\cdots +(-1{)}^{n+1}\frac{\sin (nx)}{n}+\cdots ]$$

for $-\pi <x<\pi$. □

More generally, if $f\in {\mathcal{R}}_{2\pi }^1$ is an odd function, the coefficients of $\cos (nx)$ are all 0 so $S[f]$ consists of only the $\sin (nx)$ terms,

$$f\sim S[f]=\sum _{n=1}^{\mathrm{\infty }}{b}_n\sin (nx).$$

Such a Fourier series is called a Fourier sine series.

In contrast, if $f\in {\mathcal{R}}_{2\pi }^1$ is an even function, the coefficients of $\sin (nx)$ are all 0 so $S[f]$ consists of only the $\cos (nx)$ terms,

$$f\sim S[f]=\frac{1}{2}{a}_0+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cos (nx).$$

Such a Fourier series is called a Fourier cosine series.

Example (evenfunc). Consider the function $f\in {\mathcal{P}}_{2\pi }$ defined by

$$f(x)=1-\frac{|x|}{\pi }(|x|\le \pi ).$$

$f$ is a continuous, even, and periodic function. Clearly, $b_n=0$ (verify!). It is easy to see $a_0=1$. For $n=1,2,\cdots$,

$$\begin{array}{rcl}\pi a_n& =& {\int }_{-\pi }^{\pi }f(x)\cos (nx)dx=2{\int }_0^{\pi }(1-\frac{x}{\pi })\cos (nx)dx\\ & =& \frac{2}{n^2\pi }\{1-(-1{)}^n\}(n=1,2,\cdots ).\end{array}$$

That is,

$$a_n=\{\begin{array}{cc}1& (n=0),\\ \frac{4}{{\pi }^2{n}^2}& (n=1,3,5,\cdots ),\\ 0& (n=2,4,6,\cdots ).\end{array}$$

Therefore,

$$1-\frac{|x|}{\pi }\sim \frac{1}{2}+\frac{4}{{\pi }^2}(\frac{\cos x}{1^2}+\frac{\cos (3x)}{3^2}+\frac{\cos (5x)}{5^2}+\cdots )(-\pi <x<\pi ).$$

□

Now, suppose we have a function $g(x)$ on $I_0=(0,\pi )$ that is integrable and continuous on $I_0$ except for finitely many points. Based on this function, let us define the following odd function $f_o$:

$$\begin{array}{rcl}{f}_o(x)& =& g(x)\text{ for }x\in (0,\pi ),\\ f_o(x)& =& -g(-x)\text{ for }x\in (-\pi ,0),\\ f_o(x+2n\pi )& =& f_o(x)\text{ for }x\in (-\pi ,\pi )\setminus \{0\},n\in \mathbb{N}.\end{array}$$

Similarly, we can define the following even function $f_e$:

$$\begin{array}{rcl}{f}_e(x)& =& g(x)\text{ for }x\in (0,\pi ),\\ f_e(x)& =& g(-x)\text{ for }x\in (-\pi ,0),\\ f_e(x+2n\pi )& =& f_e(x)\text{ for }x\in (-\pi ,\pi )\setminus \{0\},n\in \mathbb{N}.\end{array}$$

The corresponding Fourier series are of the forms

$$\begin{array}{rcl}{f}_o& \sim & \sum _{n=1}^{\mathrm{\infty }}{b}_n\sin (nx),\\ f_e& \sim & \frac{1}{2}{a}_0+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cos (nx),\end{array}$$

where

$$\begin{array}{rcl}{a}_n& =& \frac{1}{\pi }{\int }_{-\pi }^{\pi }{f}_e(x)\cos (nx)dx=\frac{2}{\pi }{\int }_0^{\pi }g(x)\cos (nx)dx,\\ b_n& =& \frac{1}{\pi }{\int }_{-\pi }^{\pi }{f}_o(x)\sin (nx)dx=\frac{2}{\pi }{\int }_0^{\pi }g(x)\sin (nx)dx.\end{array}$$

Using these coefficients, $g(x)$ on $I_0=(0,\pi )$ can be expressed in two ways:

$$\begin{array}{rcl}g(x)& \sim & \sum _{n=1}^{\mathrm{\infty }}{b}_n\sin (nx),\\ g(x)& \sim & \frac{1}{2}{a}_0+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cos (nx).\end{array}$$