Real Analysis

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 18 July 2014

Lecture 22

Fourier series and $\zeta \left(2\right)$

Taylor series at $x=0$ are about expanding functions $f\left(x\right)$ in powers of $x\text{.}$

Find the Taylor series of $f\left(x\right)={\text{cos}}^{2}x$ at $x=0\text{.}$

Since $$\begin{array}{ccc}{\displaystyle \text{cos}\hspace{0.17em}2x}& =& {\displaystyle {\text{cos}}^{2}x-{\text{sin}}^{2}x}\\ & =& {\displaystyle {\text{cos}}^{2}x-(1-{\text{cos}}^{2}x)}\\ & =& {\displaystyle 2{\text{cos}}^{2}x-1,}\end{array}$$ $$\begin{array}{ccc}{\displaystyle {\text{cos}}^{2}x}& =& {\displaystyle \frac{\text{cos}\hspace{0.17em}2x+1}{2}=\frac{1}{2}+\frac{1}{2}\text{cos}\hspace{0.17em}2x}\\ & =& {\displaystyle \frac{1}{2}+\frac{1}{2}(1-\frac{{\left(2x\right)}^2}{2!}+\frac{{\left(2x\right)}^4}{4!}-\frac{{\left(2x\right)}^6}{6!}+\cdots )}\\ & =& {\displaystyle \frac{3}{2}-\frac{2x^2}{2!}+\frac{2^3{x}^4}{4!}-\frac{2^5{x}^6}{6!}+\cdots }\end{array}$$

Taylor series at $x=3$ are about expanding functions $f\left(x\right)$ in powers of $x-3\text{.}$

Find the Taylor series of $f\left(x\right)=x^3$ at $x=3\text{.}$

$$\begin{array}{ccc}{\displaystyle x^3}& =& {\displaystyle {(x-3+3)}^3={(x-3)}^3+3{(x-3)}^2·3+3(x-3)·3^2+3^3}\\ & =& {\displaystyle 27+27(x-3)+9{(x-3)}^2+{(x-3)}^3\text{.}}\end{array}$$

Fourier series are about expanding functions $f\left(x\right)$ in powers of $e^{ix}\text{.}$

Since $$\begin{array}{c}{\displaystyle \text{sin}\hspace{0.17em}kx=\frac{e^{ikx}-e^{-ikx}}{2i}\text{and}\text{cos}\hspace{0.17em}kx=\frac{e_i+e^{-ix}}{2}}\\ {\displaystyle e^{ikx}=\text{cos}\hspace{0.17em}kx+i\text{sin}\hspace{0.17em}kx\text{and}{e}^{-ikx}=\text{cos}\hspace{0.17em}kx-i\text{sin}\hspace{0.17em}kx\text{.}}\end{array}$$

Fourier series are about expanding functions $f\left(x\right)$ in terms of $\text{cos}\hspace{0.17em}kx$ and $\text{sin}\hspace{0.17em}kx\text{.}$

Let $k\in \mathbb{Z}\text{.}$

(a)	${\displaystyle \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{e}^{ikx}dx}=\{\begin{array}{cc}0,& \text{if}\hspace{0.17em}k\ne 0,\\ 1,& \text{if}\hspace{0.17em}k=0\text{.}\end{array}$
(b)	${\displaystyle \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{e}^{ikx}-e^{-i\ell x}dx}={\delta }_{kl},$ where ${\delta }_{k\ell }=\{\begin{array}{cc}0,& \text{if}\hspace{0.17em}k\ne \ell ,\\ 1,& \text{if}\hspace{0.17em}k=\ell \text{.}\end{array}$
(c)	If $$f\left(x\right)=c_0+c_1{e}^{ix}+c_1{e}^{-ix}+c_2{e}^{2ix}+c_{-2}{e}^{-i2x}+\cdots$$ then $$c_k=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left(x\right)e^{i(-kx)}dx\text{.}$$

		Proof.
	(a) Case 1: $k=0$ $$\begin{array}{ccc}{\displaystyle \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{e}^{i0x}dx}& =& {\displaystyle \frac{1}{2\pi }{\int }_{-\pi }^{\pi }dx=\frac{1}{2}\left(x{|}_{x=-\pi }^{x=\pi }\right)=\frac{1}{2\pi }(\pi -(-\pi ))}\\ & =& {\displaystyle \frac{1}{2\pi }·2\pi =1\text{.}}\end{array}$$ Case 2: If $k\ne 0$ then $$\begin{array}{ccc}{\displaystyle \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{e}^{ikx}dx}& =& {\displaystyle \frac{1}{2\pi }\left(\frac{e^{ikx}}{ik}{|}_{x=-\pi }^{x=\pi }\right)}\\ & =& {\displaystyle \frac{1}{2\pi ik}(e^{i\pi k}-e^{-i\pi k})}\\ & =& {\displaystyle \frac{1}{2\pi ik}({(-1)}^k-{(-1)}^{-k})=0\text{.}}\end{array}$$ (b) To show: $\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{e}^{ikx}{e}^{-i\ell x}dx={\delta }_{k\ell }\text{.}$ $$\begin{array}{ccc}{\displaystyle \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{e}^{ikx}-e^{-i\ell x}dx}& =& {\displaystyle \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{e}^{i(k-\ell )x}dx}\\ & =& {\displaystyle \{\begin{array}{cc}0,& \text{if}\hspace{0.17em}k-\ell \ne 0,\\ 1,& \text{if}\hspace{0.17em}k-\ell =0,\end{array}=\{\begin{array}{cc}0,& \text{if}\hspace{0.17em}k\ne \ell ,\\ 1,& \text{if}\hspace{0.17em}k=\ell ,\end{array}={\delta }_{k\ell }\text{.}}\end{array}$$ (c) Assume $f\left(x\right)=c_0+c_1{e}^{ix}+c_{-1}{e}^{-ix}+c_2{e}^{i2x}+c_{-2}{e}^{-i2x}+\cdots \text{.}$ Then $$\begin{array}{ccc}{\displaystyle \frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left(x\right)e^{-ikx}dx}& =& {\displaystyle \frac{1}{2\pi }{\int }_{-\pi }^{\pi }(c_0+c_1{e}^{ix}+c_{-1}{e}^{-ix}+\cdots )e^{-ikx}dx}\\ & =& {\displaystyle c_0\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{e}^{-ikx}dx+c_1\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{e}^{ix}{e}^{-ikx}dx+c_{-1}\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{e}^{-ix}-e^{-ikx}dx+\cdots }\\ & =& {\displaystyle 0+0+\cdots +0+c_k·1+0+0+\cdots =c_k\text{.}}\end{array}$$ So $$c_k=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left(x\right)e^{-ikx}dx\text{.}$$ $\square$

Consider $f:[-\pi ,\pi ]\to ℝ$ given by $f\left(x\right)=x^2\text{.}$ Find the expansion $$f\left(x\right)=c_0+c_1{e}^{ix}+c_{-1}{e}^{-ix}+c_2{e}^{2ix}+c_{-2}{e}^{-2ix}+\cdots \text{.}$$ First find $c_0\text{:}$ $$\begin{array}{ccc}{\displaystyle c_0}& =& {\displaystyle \frac{1}{2\pi }{\int }_{-\pi }^{\pi }f\left(x\right)dx=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{x}^{2}dx=\frac{1}{2\pi }\left(\frac{x^3}{3}{|}_{x=-\pi }^{x=\pi }\right)}\\ & =& {\displaystyle \frac{1}{2\pi }(\frac{{\pi }^3}{3}-\frac{{(-\pi )}^3}{3})=\frac{1}{2\pi }\frac{2{\pi }^3}{3}=\frac{{\pi }^2}{3}\text{.}}\end{array}$$ To find $c_k=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{x}^2{e}^{-ikx}dx,$ first do the integral $$\begin{array}{ccc}{\displaystyle \int x^2{e}^{-ikx}dx}& =& {\displaystyle x^2\frac{e^{-ikx}}{-ik}-\int 2x\frac{e^{-ikx}}{-ik}dx\text{(by integration by parts)}}\\ & =& {\displaystyle \frac{i{x}^2}{k}{e}^{-ikx}(2x\frac{e^{-ikx}}{{(-ik)}^2}-\int 2\frac{e^{-ikx}}{{(-ik)}^2}dx)}\\ & =& {\displaystyle \frac{i{x}^2}{k}{e}^{-ikx}+\frac{2x}{k^2}{e}^{-ikx}+\frac{2e^{-ikx}}{{(-ik)}^3}\text{.}}\end{array}$$ So $$\begin{array}{ccc}{\displaystyle c_k}& =& {\displaystyle \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{x}^2{e}^{-ikx}=\frac{1}{2\pi }(\frac{i{x}^2}{k}{e}^{-ikx}+\frac{2x}{k^2}{e}^{-ikx}-\frac{2i}{k^3}{e}^{-ikx}){|}_{x=-\pi }^{x=\pi }}\\ & =& {\displaystyle \frac{1}{2\pi }(\frac{i{\pi }^2}{k}+\frac{2\pi }{k^2}-\frac{2i}{k^3})e^{-i\pi k}-\frac{1}{2\pi }(\frac{i{\pi }^2}{k}-\frac{2\pi }{k^2}-\frac{2i}{k^3})e^{i\pi k}}\\ & =& {\displaystyle \frac{1}{2\pi }(\frac{i{\pi }^2}{k}+\frac{2\pi }{k^2}-\frac{2i}{k^3}){(-1)}^k-\frac{1}{2\pi }(\frac{i{\pi }^2}{k}-\frac{2\pi }{k^2}-\frac{2i}{k^3}){(-1)}^k\text{.}}\end{array}$$ So $$c_k=\frac{1}{2\pi }{(-1)}^k(\frac{2\pi }{k^2}+\frac{2\pi }{k^2})={(-1)}^k\frac{4\pi }{2\pi k^2}={(-1)}^k\frac{2}{k^2}\text{.}$$ So $$\begin{array}{ccc}{\displaystyle x^2}& =& {\displaystyle c_0+\sum _{k=1}^{\infty }{c}_k{e}^{ikx}+c_{-k}{e}^{-ikx}}\\ & =& {\displaystyle \frac{{\pi }^2}{3}+\sum _{k=1}^{\infty }{(-1)}^k\frac{2}{k^2}{e}^{ikx}+{(-1)}^{-k}\frac{2}{{(-k)}^2}{e}^{-ikx}}\\ & =& {\displaystyle \frac{{\pi }^2}{3}+\sum _{k=1}^{\infty }{(-1)}^k\frac{2}{k^2}(e^{ikx}+e^{-ikx})}\\ & =& {\displaystyle \frac{{\pi }^2}{3}+\sum _{k=1}^{\infty }{(-1)}^k\frac{2}{k^2}·2\text{cos}\hspace{0.17em}kx}\\ & =& {\displaystyle \frac{{\pi }^2}{3}+\sum _{k=1}^{\infty }{(-1)}^k\frac{1}{k^2}4\text{cos}\hspace{0.17em}kx}\\ & =& {\displaystyle \frac{{\pi }^2}{3}-4\text{cos}\hspace{0.17em}kx+\frac{1}{4}4\text{cos}\hspace{0.17em}2x-\frac{1}{9}4\text{cos}\hspace{0.17em}3x+\frac{1}{16}4\text{cos}\hspace{0.17em}4x+\cdots \text{.}}\end{array}$$ If $x=\pi$ then $$\begin{array}{ccc}{\displaystyle {\pi }^2}& =& {\displaystyle \frac{{\pi }^2}{3}+\sum _{k=1}^{\infty }{(-1)}^k\frac{1}{k^2}4\text{cos}\hspace{0.17em}k\pi }\\ & =& {\displaystyle \frac{{\pi }^2}{3}+\sum _{k=1}^{\infty }{(-1)}^k\frac{4}{k^2}{(-1)}^k}\\ & =& {\displaystyle \frac{{\pi }^2}{3}+\sum _{k=1}^{\infty }\frac{4}{k^2}\text{.}}\end{array}$$ So $$\sum _{k=1}^{\infty }\frac{1}{k^2}=\frac{1}{4}({\pi }^2-\frac{{\pi }^2}{3})=\frac{1}{4}\frac{2{\pi }^2}{3}=\frac{{\pi }^2}{6}\text{.}$$

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100430Lect22.pdf and was given on 30 April 2010.

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