Formulation of Fourier Series

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


Let us consider a series of the form

Let us consider a series of the form
$$\begin{aligned} \frac{a_0}{2}+\sum _{m=1}^\infty \left( a_m\cos \frac{m\pi x}{L}+b_m\sin \frac{m\pi x}{L}\right) . \end{aligned}$$
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.,
$$\begin{aligned} f(x):=\frac{a_0}{2}+\sum _{m=1}^\infty \left( a_m\cos \frac{m\pi x}{L}+b_m\sin \frac{m\pi x}{L}\right) . \end{aligned}$$
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 ( 1.10), ( 1.11), and ( 1.13) that the only nonzero term on the right-hand side is the one for which \(m=n\) in the first summation. Hence
$$\begin{aligned} a_n=\frac{1}{L}\int _{-L}^L f(x)\cos \frac{n\pi x}{L}\mathrm{d}x,\quad n=1,2,\ldots . \end{aligned}$$
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
$$\begin{aligned} b_n=\frac{1}{L}\int _{-L}^L f(x)\sin \frac{n\pi x}{L}\mathrm{d}x,\quad n=1,2,\ldots . \end{aligned}$$
Using ( 1.13) we can easily obtain that
$$\begin{aligned} a_0=\frac{1}{L}\int _{-L}^L f(x)\mathrm{d}x. \end{aligned}$$

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
$$\begin{aligned} f(x)\sim \frac{a_0}{2}+\sum _{m=1}^\infty \left( a_m\cos \frac{m\pi x}{L}+b_m\sin \frac{m\pi x}{L}\right) . \end{aligned}$$

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  1.5 imply that if f is even on \([-L, L]\), then the Fourier series of f has the form
$$\begin{aligned} f(x)\sim \frac{a_0}{2}+\sum _{m=1}^\infty a_m\cos \frac{m\pi x}{L}, \end{aligned}$$
and if f is odd, then
$$\begin{aligned} f(x)\sim \sum _{m=1}^\infty b_m\sin \frac{m\pi x}{L}. \end{aligned}$$
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
$$\begin{aligned} f(x)\sim \frac{a_0}{2}+\sum _{m=1}^\infty \left( a_m\cos mx+b_m\sin mx\right) , \end{aligned}$$
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
    $$\begin{aligned} a_0(\widetilde{f})&=\frac{1}{\pi }\int _{-\pi }^\pi \widetilde{f}(t)\mathrm{d}t=\frac{1}{\pi }\int _{-\pi }^\pi f\left( \frac{tL}{2\pi }+\frac{L}{2}\right) \mathrm{d}t=\frac{1}{\pi }\frac{2\pi }{L}\int _{0}^L f\left( x\right) \mathrm{d}x\\&=\frac{2}{L}\int _{0}^L f\left( x\right) \mathrm{d}x:=a_0(f), \end{aligned}$$
    $$ a_m(\widetilde{f})=(-1)^m\frac{2}{L}\int _{0}^L f(x)\cos \frac{2m\pi x}{L}\mathrm{d}x=(-1)^m a_m(f), $$
    $$ b_m(\widetilde{f})=(-1)^m\frac{2}{L}\int _{0}^L f(x)\sin \frac{2m\pi x}{L}\mathrm{d}x=(-1)^m b_m(f). $$
    $$\begin{aligned} \widetilde{f}(t)\sim \frac{a_0}{2}+\sum _{m=1}^\infty \left( a_m\cos mt+b_m\sin mt\right) , \end{aligned}$$
    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
$$\begin{aligned} f(x)\sim \sum _{n=-\infty }^\infty c_n \mathrm{e}^{Inx}, \end{aligned}$$
where the coefficients \(c_n=c_n(f)\) are equal to
$$\begin{aligned} c_n={\left\{ \begin{array}{ll} \dfrac{a_n}{2}+\dfrac{b_n}{2I},&{}n=1,2,\ldots ,\\[2ex] \dfrac{a_0}{2},&{}n=0,\\[2ex] \dfrac{a_{-n}}{2}-\dfrac{b_{-n}}{2I},&{}n=-1,-2,\ldots . \end{array}\right. }\;\text {or}\; \begin{aligned} a_n&=c_n+c_{-n},&n=1,2,\ldots ,\\ a_0&=2c_0,\\ b_n&=I(c_n-c_{-n}),&n=1,2,\ldots . \end{aligned} \end{aligned}$$
The formulas (2.3), (2.4), (2.5), and (2.11) imply that
$$\begin{aligned} c_n(f)=\frac{1}{2\pi }\int _{-\pi }^\pi f(x) \mathrm{e}^{-Inx}\mathrm{d}x \end{aligned}$$
for \(n=0,\pm 1,\pm 2 ,\ldots \). We call \(c_n(f)\) the nth Fourier coefficient of f . It can be checked that
$$\begin{aligned} c_n(f)=\overline{c_{-n}(\overline{f})}. \end{aligned}$$

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
$$ \sum _{k=1}^Na_k\cos kx. $$

Copyright information

© Springer International Publishing AG 2017

Authors and Affiliations

  1. 1.Department of Mathematical SciencesUniversity of OuluOuluFinland

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