# Formulation of Fourier Series

Chapter
Part of the Applied Mathematical Sciences book series (AMS, volume 197)

## Abstract

Let us consider a series of the form

Let us consider a series of the form
\begin{aligned} \frac{a_0}{2}+\sum _{m=1}^\infty \left( a_m\cos \frac{m\pi x}{L}+b_m\sin \frac{m\pi x}{L}\right) . \end{aligned}
(2.1)
This series consists of 2L-periodic functions. Thus, if the series (2.1) converges for all x, then the function to which it converges will also be 2L-periodic. Let us denote this limiting function by f(x), i.e.,
\begin{aligned} f(x):=\frac{a_0}{2}+\sum _{m=1}^\infty \left( a_m\cos \frac{m\pi x}{L}+b_m\sin \frac{m\pi x}{L}\right) . \end{aligned}
(2.2)
To determine $$a_m$$ and $$b_m$$ we proceed as follows: assuming that the integration can be legitimately carried out term by term (it will be, for example, if $$\sum _{m=1}^\infty (|a_m|+|b_m|)<\infty$$), we obtain
\begin{aligned} \int _{-L}^L f(x)\cos \frac{n\pi x}{L}\mathrm{d}x&=\frac{a_0}{2}\int _{-L}^L \cos \frac{n\pi x}{L}\mathrm{d}x +\sum _{m=1}^\infty a_m\int _{-L}^L \cos \frac{m\pi x}{L}\cos \frac{n\pi x}{L}\mathrm{d}x\\&+\sum _{m=1}^\infty b_m\int _{-L}^L \sin \frac{m\pi x}{L}\cos \frac{n\pi x}{L}\mathrm{d}x \end{aligned}
for each fixed $$n=1,2,\ldots$$. It follows from the orthogonality relations (), (), and () that the only nonzero term on the right-hand side is the one for which $$m=n$$ in the first summation. Hence
\begin{aligned} a_n=\frac{1}{L}\int _{-L}^L f(x)\cos \frac{n\pi x}{L}\mathrm{d}x,\quad n=1,2,\ldots . \end{aligned}
(2.3)
A similar expression for $$b_n$$ is obtained by multiplying (2.2) by $$\sin \frac{n\pi x}{L}$$ and integrating termwise from $$-L$$ to L. The result is
\begin{aligned} b_n=\frac{1}{L}\int _{-L}^L f(x)\sin \frac{n\pi x}{L}\mathrm{d}x,\quad n=1,2,\ldots . \end{aligned}
(2.4)
Using () we can easily obtain that
\begin{aligned} a_0=\frac{1}{L}\int _{-L}^L f(x)\mathrm{d}x. \end{aligned}
(2.5)

## Definition 2.1.

Let f be integrable (not necessarily periodic) on the interval $$[-L, L]$$. The Fourier series of f is the trigonometric series (2.1), where the coefficients $$a_0, a_m$$ and $$b_m$$ are given by (2.5), (2.3), and (2.4), respectively. In that case, we write
\begin{aligned} f(x)\sim \frac{a_0}{2}+\sum _{m=1}^\infty \left( a_m\cos \frac{m\pi x}{L}+b_m\sin \frac{m\pi x}{L}\right) . \end{aligned}
(2.6)

## Remark 2.2.

This definition does not imply that the series (2.6) converges to f or that f is periodic.

Definition 2.1 and Lemma imply that if f is even on $$[-L, L]$$, then the Fourier series of f has the form
\begin{aligned} f(x)\sim \frac{a_0}{2}+\sum _{m=1}^\infty a_m\cos \frac{m\pi x}{L}, \end{aligned}
(2.7)
and if f is odd, then
\begin{aligned} f(x)\sim \sum _{m=1}^\infty b_m\sin \frac{m\pi x}{L}. \end{aligned}
(2.8)
The series (2.7) and (2.8) are called the Fourier cosine series and Fourier sine series , respectively.
If $$L=\pi$$, then the Fourier series (2.6) ((2.7) and (2.8)) transforms to
\begin{aligned} f(x)\sim \frac{a_0}{2}+\sum _{m=1}^\infty \left( a_m\cos mx+b_m\sin mx\right) , \end{aligned}
(2.9)
where the coefficients $$a_0$$, $$a_m$$, and $$b_m$$ are given by (2.5), (2.3), and (2.4) with $$L=\pi$$.

There are different approaches if the function f is defined on an asymmetric interval [0, L] with arbitrary $$L>0$$.

1. (1)
Even extension . Define a function g(x) on the interval $$[-L, L]$$ as
\begin{aligned} g(x)={\left\{ \begin{array}{ll} f(x),&{} 0\le x\le L,\\ f(-x),&{}-L\le x <0. \end{array}\right. } \end{aligned}
Then g(x) is even and its Fourier (cosine) series (2.7) represents f on [0, L].

2. (2)
Odd extension . Define a function h(x) on the interval $$[-L, L]$$ as
\begin{aligned} h(x)={\left\{ \begin{array}{ll} f(x),&{} 0\le x\le L,\\ -f(-x),&{}-L\le x <0. \end{array}\right. } \end{aligned}
Then h(x) is odd, and its Fourier (sine) series (2.8) represents f on [0, L].

3. (3)
Define a function $$\widetilde{f}(t)$$ on the interval $$[-\pi ,\pi ]$$ as
\begin{aligned} \widetilde{f}(t)=f\left( \frac{tL}{2\pi }+\frac{L}{2}\right) . \end{aligned}
If $$f(0)=f(L)$$, then we may extend f to be periodic with period L. Then
\begin{aligned} a_0(\widetilde{f})&=\frac{1}{\pi }\int _{-\pi }^\pi \widetilde{f}(t)\mathrm{d}t=\frac{1}{\pi }\int _{-\pi }^\pi f\left( \frac{tL}{2\pi }+\frac{L}{2}\right) \mathrm{d}t=\frac{1}{\pi }\frac{2\pi }{L}\int _{0}^L f\left( x\right) \mathrm{d}x\\&=\frac{2}{L}\int _{0}^L f\left( x\right) \mathrm{d}x:=a_0(f), \end{aligned}
$$a_m(\widetilde{f})=(-1)^m\frac{2}{L}\int _{0}^L f(x)\cos \frac{2m\pi x}{L}\mathrm{d}x=(-1)^m a_m(f),$$
and
$$b_m(\widetilde{f})=(-1)^m\frac{2}{L}\int _{0}^L f(x)\sin \frac{2m\pi x}{L}\mathrm{d}x=(-1)^m b_m(f).$$
Hence,
\begin{aligned} \widetilde{f}(t)\sim \frac{a_0}{2}+\sum _{m=1}^\infty \left( a_m\cos mt+b_m\sin mt\right) , \end{aligned}
and at the same time,
\begin{aligned} f(x)\sim \frac{a_0}{2}+\sum _{m=1}^\infty (-1)^m\left( a_m\cos \frac{2m\pi x}{L}+b_m\sin \frac{2m\pi x}{L}\right) , \end{aligned}
where $$a_0$$, $$a_m$$, and $$b_m$$ are the same and $$\displaystyle x=\frac{tL}{2\pi }+\frac{L}{2}$$.

These three alternatives allow us to consider (for simplicity) only the case of a symmetric interval $$[-\pi ,\pi ]$$ such that the Fourier series will be of the form (2.9) i.e.
\begin{aligned} f(x)\sim \frac{a_0}{2}+\sum _{m=1}^\infty \left( a_m\cos mx+b_m\sin mx\right) . \end{aligned}
Using Euler’s formula, we will rewrite this series in the complex form
\begin{aligned} f(x)\sim \sum _{n=-\infty }^\infty c_n \mathrm{e}^{Inx}, \end{aligned}
(2.10)
where the coefficients $$c_n=c_n(f)$$ are equal to
\begin{aligned} c_n={\left\{ \begin{array}{ll} \dfrac{a_n}{2}+\dfrac{b_n}{2I},&{}n=1,2,\ldots ,\\[2ex] \dfrac{a_0}{2},&{}n=0,\\[2ex] \dfrac{a_{-n}}{2}-\dfrac{b_{-n}}{2I},&{}n=-1,-2,\ldots . \end{array}\right. }\;\text {or}\; \begin{aligned} a_n&=c_n+c_{-n},&n=1,2,\ldots ,\\ a_0&=2c_0,\\ b_n&=I(c_n-c_{-n}),&n=1,2,\ldots . \end{aligned} \end{aligned}
(2.11)
The formulas (2.3), (2.4), (2.5), and (2.11) imply that
\begin{aligned} c_n(f)=\frac{1}{2\pi }\int _{-\pi }^\pi f(x) \mathrm{e}^{-Inx}\mathrm{d}x \end{aligned}
(2.12)
for $$n=0,\pm 1,\pm 2 ,\ldots$$. We call $$c_n(f)$$ the nth Fourier coefficient of f . It can be checked that
\begin{aligned} c_n(f)=\overline{c_{-n}(\overline{f})}. \end{aligned}
(2.13)

## Exercise 2.1.

Prove formulas (2.10), (2.11), (2.12), and (2.13).

## Exercise 2.2.

Find the Fourier series of
1. (1)

$${{\mathrm{sgn}}}(x)={\left\{ \begin{array}{ll} -1,&{}-\pi \le x<0,\\ 0,&{}x=0,\\ 1,&{}0<x\le \pi . \end{array}\right. }$$

2. (2)

$$|x|, -1\le x\le 1$$.

3. (3)

$$x, -1\le x\le 1$$.

4. (4)

$$f(x)={\left\{ \begin{array}{ll} 0,&{}-L\le x\le 0,\\ L,&{}0< x\le L. \end{array}\right. }$$

5. (5)

$$f(x)=\sin x,|x|\le 2$$.

## Exercise 2.3.

Prove, using Part (2) of Exercise 2.2, that
$$\frac{\pi ^2}{8}=\sum _{k=1}^\infty \frac{1}{(2k-1)^2}\quad \text {and}\quad \frac{\pi ^2}{6}=\sum _{k=1}^\infty \frac{1}{k^2}.$$

## Exercise 2.4.

Suppose that
$$f(x)={\left\{ \begin{array}{ll} 1-x,&{}0\le x\le 1,\\ 0,&{}1< x\le 2. \end{array}\right. }$$
Find the Fourier cosine and sine series of f(x).

## Exercise 2.5.

Find the Fourier series of $$f(x)=\cos (x/2)$$, $$|x|\le \pi$$. Using this series, show that
1. (1)

$$\displaystyle \pi = 2+\sum _{k=1}^\infty \frac{(-1)^{k+1}}{k^2-1/4}$$;

2. (2)

$$\displaystyle \frac{\pi }{4} =\sum _{k=1}^\infty \frac{(-1)^{k+1}}{2k-1}$$;

3. (3)

$$\displaystyle \frac{1}{2} =\sum _{k=1}^\infty \frac{1}{4k^2-1}$$.

## Exercise 2.6.

Show that if N is odd, then $$\sin ^N x$$ can be written as a finite sum of the form
$$\sum _{k=1}^Na_k\sin kx,$$
which means that this finite sum is the Fourier series of $$\sin ^N x$$ and the coefficients $$a_k$$ (which are real) are the Fourier coefficients of $$\sin ^N x$$.

## Exercise 2.7.

Show that if N is odd, then $$\cos ^N x$$ can be written as a finite sum of the form
$$\sum _{k=1}^Na_k\cos kx.$$