Fourier series: Introduction

The theory of the Fourier series is based on a wild assumption: any "well-behaved" periodic function can be represented as a linear combination of sine and cosine functions, and that expression is unique for the given function. This theory (eventually) provided much of the foundations of modern mathematics. But it is also of tremendous practical importance.

Trigonometric functions are periodic. But what is a periodic function in general?

Definition (Periodic function)

A function $f(x)$ is said to be a periodic function if there exists a real number $T>0$ such that

$$f(x+T)=f(x).$$

In this case, $T$ is called a period of the function. Note that if $T$ is a period of $f(x)$, then $2T$, $3T$, $\cdots ,nT$ ($n\in \mathbb{N}$) is are also periods of $f(x)$. In fact, 

$$f(x+nT)=f(x+(n-1)T+T)=f(x+(n-1)T)=\cdots =f(x).$$

The smallest (non-zero) period is called the fundamental period.

Remark. If we say simply a period, we usually mean the fundamental period. □

Example. For any $n\in \mathbb{N}$, $\cos (nx)$ and $\sin (nx)$ have a period of $2\pi$, their fundamental periods are $\frac{2\pi }{n}$ for each $n$. □

Definition (Fourier series)

A Fourier series is a series of the form

$$\frac{1}{2}{a}_0+\sum _{n=1}^{\mathrm{\infty }}\{a_n\cos (nx)+b_n\sin (nx)\}$$

where $a_n$ ($n=0,1,\cdots$) and $b_n$ ($n=1,2,\cdots$) are (usually) real constants. 

The factor $\frac{1}{2}$ in the constant term ($\frac{1}{2}{a}_0$) is by convention as well as for convenience. Each term of the series has a period of $2\pi$, so the domain of the above function of $x$ may be $\mathbb{R}$ or $[-\pi ,\pi ]$ or $[0,2\pi ]$.

Definition (Fourier expansion of a function)

Let $f(x)$ be a function on $\mathbb{R}$ that has a period of $2\pi$. If

$$\begin{array}{}\text{(eq:FF)}& f(x)=\frac{1}{2}{a}_0+\sum _{n=1}^{\mathrm{\infty }}\{a_n\cos (nx)+b_n\sin (nx)\}\end{array}$$

holds for all except for finitely many $x\in \mathbb{R}$, the right-hand side is said to be 

a Fourier expansion or Fourier series expansion of the function $f$. 

If term-wise integration is allowed, the coefficients $a_n$ and $b_n$ are readily determined. Note the following formulae: For any $m,n\in \mathbb{N}$, 

$$\begin{array}{rcl}{\int }_0^{2\pi }\cos (nx)\cos (mx)dx& =& \pi {\delta }_{m,n},\\ {\int }_0^{2\pi }\sin (nx)\sin (mx)dx& =& \pi {\delta }_{m,n},\\ {\int }_0^{2\pi }\sin (nx)\cos (mx)dx& =& 0\end{array}$$

where ${\delta }_{m,n}$ is Kronecker's delta. That is, the functions $\cos x,\cos (2x),\cdots ,\sin x,\sin (2x),\cdots$ are orthogonal. By multiplying (eq:FF) by $\cos (mx)$ or $\sin (mx)$ and then integrating, we have

$$\begin{array}{rcl}{\int }_0^{2\pi }f(x)\cos (mx)dx& =& \pi a_m(m=0,1,\cdots ),\\ {\int }_0^{2\pi }f(x)\sin (mx)dx& =& \pi b_m(m=1,2,\cdots ).\end{array}$$

Thus, we have

$$\begin{array}{}\text{(eq:fab)}& \begin{array}{cccc}{a}_m& =& \frac{1}{\pi }{\int }_0^{2\pi }f(x)\cos (mx)dx& (m=0,1,\cdots ),\\ b_m& =& \frac{1}{\pi }{\int }_0^{2\pi }f(x)\sin (mx)dx& (m=1,2,\cdots ).\end{array}\end{array}$$

Since we are assuming the period of $2\pi$ for $f$, the range of integration can be $[-\pi ,\pi ]$ instead of $[0,2\pi ]$. 

The sequences $\{a_m\}$ and $\{b_m\}$ defined by (eq:fab) are called the Fourier coefficients of $f$. The Fourier coefficients of $f$ can be determined if $f$ is integrable on $[0,2\pi ]$. Given the Fourier coefficients of $f$, we can formally define the following series, which is called the Fourier series of $f$, denoted by $S[f]$:

$$\begin{array}{}\text{(eq:Sf)}& S[f]=\frac{1}{2}{a}_0+\sum _{n=1}^{\mathrm{\infty }}\{a_n\cos (nx)+b_n\sin (nx)\}.\end{array}$$

In this case, we also write

$$\begin{array}{}\text{(eq:Sfsim)}& f\sim \frac{1}{2}{a}_0+\sum _{n=1}^{\mathrm{\infty }}\{a_n\cos (nx)+b_n\sin (nx)\}.\end{array}$$

Note that, in this case, $S[f]$ may not be the same function as $f$. In fact, whether and/or when $f=S[f]$ (note it's "$=$", not "$\sim$") is a fundamental question in the theory of Fourier series.

The fundamental problems in the theory of the Fourier series are

• Under what conditions on $f$ does $S[f]$ converge?
• What does the sum $S[f]$ represent?

The continuity of $f$ alone is known to be insufficient. To state the sufficient condition, we need the language of Lebesgue integral, or the measure theory, which is far beyond the scope of this post. We give it below anyway without proof.

Theorem (Carlson (1966))

Let $f$ be a function that is measurable on $[0,2\pi ]$ and $L^2$-integrable, i.e.,

$${\int }_0^{2\pi }|f(x){|}^{2}dx<+\mathrm{\infty },$$

then the Fourier series $S[f]$ converges to $f$ almost everywhere.

Remark. The technical terms such as measurable, $L^2$-integrable, and almost everywhere come from the theory of Lebesgue integral. The Lebesgue integral is a generalization of the Riemann integral. Much of modern mathematics depends on the Lebesgue integral. □