Fourier Series

Posted on May 07, 2019 (Update: Jun 23, 2019)
Tags: Fourier

A trigonometric polynomial (三角多项式) is a finite sum of the form

where $a_0,a_1,\ldots,a_N, b_1,\ldots,b_n$ are complex numbers but $x$ is real. It can be written in the form

\begin{equation}\label{60} f(x) = \sum_{-N}^Nc_ne^{inx}\,. \end{equation}

It is clear that every trigonometric polynomial is periodic, with period $2\pi$.

If $n$ is a nonzero integer, $e^{inx}$ is the derivative of $e^{inx}/in$, which also has period $2\pi$. Hence

Multiply \eqref{60} by $e^{-imx}$, where $m$ is an integer, if we integrate the product, then we have

\begin{equation}\label{62} c_m = \frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-imx}dx \end{equation}

for $\vert m\vert\le N$. If $\vert m\vert >N$, the integral is 0.

In agreement with \eqref{60}, define a trigonometric series to be a series of the form

\begin{equation}\label{63} \sum_{-\infty}^\infty c_ne^{inx}\quad (x \text{ real})\,. \end{equation}

If $f$ is an integrable function on $[-\pi,\pi]$, the numbers $c_m$ defined by \eqref{62} for all integers $m$ are called the Fourier coefficients of $f$, and the series \eqref{63} formed with these coefficients is called the Fourier series of $f$.

Natural question: whether the Fourier series of $f$ converges to $f$, or more generally, whether $f$ is determined by its Fourier series.

Let $\{\phi_n\}(n=1,2,3,\ldots)$ be a sequence of complex functions on $[a, b]$, such that

Then $\{\phi_n\}$ is said to be an orthogonal system of functions on $[a,b]$. If, in addition,

for all $n$, $\{\phi_n(x)\}$ is said to be orthonormal.

For example, the function $(2\pi)^{-1/2}e^{inx}$ form an orthonormal system on $[-\pi,\pi]$. So do the real functions

If $\{\phi_n\}$ is orthonormal on $[a,b]$ and if

we call $c_n$ be the $n$th Fourier coefficient of $f$ relative to $\{\phi_n\}$. We write

and call this series the Fourier series of $f$ (relative to $\{\phi_n\}$).

Let $\\{\phi_n\\}$ be orthonormal on $[a,b]$. Let $$s_n(x) = \sum_{m=1}^nc_m\phi_m(x)$$ be the $n$th partial sum of the Fourier series of $f$, and suppose $$t_n(x) = \sum_{m=1}^n\gamma_m\phi_m(x)\,.$$ Then $$\int_a^b \vert f-s_n\vert^2dx\le \int_a^b\vert f-t_n\vert^2dx\,,$$ and equality holds iff $$\gamma_m=c_m\quad (m=1,\ldots,n).$$ That is to say, among all functions $t_n$, $s_n$ gives the best possible mean square approximation to $f$.

Let

be the $N$th partial sum of the Fourier series of $f$.

(Parseval's theorem) Suppose $f$ and $g$ are Riemann-integrable functions with period $2\pi$, and $$f(x)\sim \sum_{-\infty}^\infty c_ne^{inx},\qquad g(x) \sim \sum_{-\infty}^\infty \gamma_ne^{inx}\,.$$ Then \begin{align*} \lim_{N\rightarrow\infty}\frac{1}{2N}\int_{-\pi}^\pi \vert f(x) - s_N(f;x)\vert^2dx &=0\\ \frac{1}{2\pi}\int_{-\pi}^\pi f(x)\overline{g(x)} dx &= \sum_{-\infty}^\infty c_n\bar\gamma_n\\ \frac{1}{2\pi}\int_{-\pi}^\pi \vert f(x)\vert^2dx =\sum_{-\infty}^{\infty}\vert c_n\vert^2\,. \end{align*}

$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$

Apply Parseval’s identity to the function $f(x)=x$,

where

for $n\neq 0$, and $c_0=0$. Thus,

Therefore,

Published in categories Note