Review

Periodic Functions: A function is periodic of period $T$ if $f(x+T)=f(x)$ for all $x$ in the domain of $f$.

Even Functions: A function that satisfies $f(-x)=f(x)$ for all $x$ in the domain of $f$ has a graph that is symmetric with respect to the $y$-axis. We say such a function is even. The functions $1,x^2,x^4,\dots$ are examples of even functions, as is $\cos x$.

Odd Functions: A function $f$ that satisfies $f(-x)=-f(x)$ for all $x$ in the domain of $f$ has a graph that is symmetric with respect to the origin. It is said to be an odd function. The functions $x,x^3,x^5,\dots$ are examples of odd functions, as are $\sin x$ and $\tan x$.

Knowing these properties can be useful when evaluating indefinite integrals.

If $f$ is an even piecewise continuous function on $[-a,a]$, then

$${\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx.$$

if $f$ is an odd piecewise continuous function on $[-a,a]$, then

$${\int }_{-a}^{a}f(x)dx=0.$$

Some Important Integrals

The following three integrals are crucial in Fourier series. In each, $m$ and $n$ are nonnegative integers.

The first integral:

$$\begin{array}{cc}{\int }_{-L}^L\sin \frac{m\pi x}{L}\cos \frac{n\pi x}{L}dx.& \text{(a)}\end{array}$$

We can simplify this using the trig identity for products of sines and cosines, $\sin (a)\cos (b)=\frac{1}{2}(\sin (a-b)+\sin (a+b)):$

$$\frac{1}{2}{\int }_{-L}^L\sin \frac{(m-n)\pi x}{L}+\sin \frac{(m+n)\pi x}{L}dx.$$

Since this is an integral from $-L$ to $L$ over an odd function, it, and thus $(a),$ always equals 0.

The second integral:

$$\begin{array}{cc}{\int }_{-L}^L\sin \frac{m\pi x}{L}\sin \frac{n\pi x}{L}dx.& \text{(b)}\end{array}$$

This time we use the trig identity $\sin (a)\sin (b)=\frac{1}{2}(\cos (a-b)-\cos (a+b)):$

$$\begin{array}{cc}\frac{1}{2}{\int }_{-L}^L\cos \frac{(m-n)\pi x}{L}-\cos \frac{(m+n)\pi x}{L}dx.& \text{(b.1)}\end{array}$$

Here, we have to consider two cases, $m=n$ and $m\ne n$. If $m=n$, $(m-n)=0,(m+n)=2m$; also note that this integral is over an even function from $-L$ to $L$, so (b.1)

$$\begin{array}{cc}\frac{1}{2}{\int }_{-L}^{L}1-\cos \frac{2m\pi x}{L}dx=L-{\left(\frac{L\sin \frac{2m\pi x}{L}}{2\pi m}\right)}_0^L=L-(\left(\frac{L\sin 2m\pi }{2\pi m}\right)-0)=L.& \text{(b.2)}\end{array}$$

The final $\sin$ in (b.2) is always $0$ because $\sin 2m\pi$ for integer $m$ is always $0$. So, when $m=n$, (b) evaluates to $L$.

When $m\ne n$, we can't simplify either $\cos$ in the integrand and end up with two $\sin$'s that are always 0, so when $m\ne n$, (b) evaluates to $0$.

The third integral:

$$\begin{array}{cc}{\int }_{-L}^L\cos \frac{m\pi x}{L}\cos \frac{n\pi x}{L}dx.& \text{(c)}\end{array}$$

This time we use the trig identity $\cos (a)\cos (b)=\frac{1}{2}(\cos (a-b)+\cos (a+b)):$

$$\begin{array}{cc}\frac{1}{2}{\int }_{-L}^L\cos \frac{(m-n)\pi x}{L}+\cos \frac{(m+n)\pi x}{L}dx.& \text{(c.1)}\end{array}$$

Here, we need to consider 3 cases: $m\ne n$; $m=n,m,n\ne 0$; $m=n=0$. The first two cases work out just like (b), and when $m\ne n$ we get 0, when $m=n$ but $m$ and $n$ are non-zero we get $L$.

However, when $m=n=0$, both $\cos$ in the integrand end up as $\cos 0=1$, and so the integral simplifies to:

$$\begin{array}{cc}{\int }_{-L}^{L}dx=2L.& \text{(c.2)}\end{array}$$

Thus, when $m=n=0$, (c) evaluates to 2L.

The following table (from "Differential Equations and Boundary Value Problems", Nagle, et al, 2017, p574) summarizes these integrals nicely:

Fourier Series

Definition: Let $f$ be a piecewise continuous function on the interval $[-L,L]$. The Fourier Series of $f$ is the trigonometric series

$$\begin{array}{cc}f(x)\sim \frac{a_0}{2}+\sum _{n=1}^{\mathrm{\infty }}\{a_n\cos \frac{n\pi x}{L}+b_n\sin \frac{n\pi x}{L}\},& \text{(8)}\end{array}$$

where the $a_n$'s and $b_n$'s are given by the formulas:

$$\begin{array}{cc}{a}_n=\frac{1}{L}{\int }_{-L}^{L}f(x)\cos \frac{n\pi x}{L}dx,n=0,1,2,\dots ,& \text{(9)}\end{array}$$

$$\begin{array}{cc}{b}_n=\frac{1}{L}{\int }_{-L}^{L}f(x)\sin \frac{n\pi x}{L}dx,n=0,1,2,\dots ,& \text{(10)}\end{array}$$

Formulas (9) and (10) are called the Euler-Fourier formulas. The $\sim$ symbols in (8) means that the series is associated with $f(x)$ but may not converge to $f(x)$.

Example 1

Find the Fourier series for $f(x)=\Vert 3x\Vert$, .

Here, $L$ = $pi$. We get lucky here because $\Vert 3x\Vert$ is even, so all $b_n$ terms are 0 (if it were odd, all $a_n$ terms would be 0). First, we'll find $a_0$ using (8). Since $f(x)$ is even, we simplify and integrate over just the positive half and double the result:

$$a_0=\frac{1}{\pi }={\int }_{-\pi }^{\pi }|3x|dx=\frac{6}{\pi }{\int }_0^{\pi }xdx=3\pi .$$

Now, we find $a_n$ using (8), again taking advantage of our integrand being even, and then using integration by parts:

$$a_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }|3x|\cos (nx)dx=\frac{6}{\pi }{\int }_0^{\pi }x\cos nxdx=\frac{6}{\pi n^2}((-1{)}^n-1).$$

Now, using (9) we can write our Fourier series:

$$|3x|\sim \frac{3\pi }{2}+\sum _{n=1}^{\mathrm{\infty }}(\frac{6}{\pi n^2}((-1{)}^n-1)\cos (nx))$$

The first few terms are:

$$|3x|\sim \frac{3\pi }{2}-\frac{12}{\pi }\cos x-\frac{12}{9\pi }\cos 3x-\frac{12}{25\pi }\cos 5x-\cdots$$

Extensions

If we have a function $f(x)$ that is defined on , and we recall that an odd function defined on $[-L,L]$ has only sine terms, we might wish to construct an artificial extension of $f(x)$ onto $(-L,L)$ in such a way that the extended function is odd, so we could construct a Fourier series for the function that consists only of sine terms. We can accomplish this by defining the function

and extending $f_o(x)$ to all $x$ using $2L$-periodicity (well, all $x$ other than integer multiples of $L$). Since $f_o(x)$ is odd, its Fourier series will contain only sine terms, and since $f_o(x)=f(x)$ on $(0,L),$ $f_o(x)$ is an extension of $f(x)$. Specifically, it's called the odd 2L-periodic extension of $f(x).$ The resulting Fourier series expansion is called a half-range expansion for $f(x),$ since it represents the function $f(x)$ on $(0,L,)$ which is half the interval $(-L,L)$ where it represents $f_o(x).$

Similarly, we can define the even 2L-periodic extension of $f(x)$ as

Its Fourier series would contain only cosines.

Fourier Cosine and Sine Series

Using the notions of expansions above, we have the following definitions.

Let $f(x)$ be a piecewise continuous function on the interval $[0,L].$ The Fourer cosine series of $f(x)$ on $[0,L]$ is

$$\begin{array}{cc}\frac{a_0}{2}+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cos \frac{n\pi x}{L},& \text{(4)}\end{array}$$

where

$$a_n=\frac{2}{L}{\int }_0^{L}f(x)\cos \frac{n\pi x}{L}dx,n=0,1,\dots .$$

The Fourer sine series of $f(x)$ on $[0,L]$ is

$$\begin{array}{cc}\sum _{n=1}^{\mathrm{\infty }}{b}_n\sin \frac{n\pi x}{L},& \text{(6)}\end{array}$$

where

$$b_n=\frac{2}{L}{\int }_0^{L}f(x)\sin \frac{n\pi x}{L}dx,n=1,2,\dots .$$

The series in (4) is the Fourier series for $f_e(x)$, and that in (6) is the series for $f_o(x)$. These are called half-range expansions for $f(x)$.