Fourier series

A Fourier series is a mathematical tool that represents a function, which is periodic on a specific interval, as a linear combination of cosine and sine waves.

Consider the trigonometric functions ${g(x)=cos\biggr$\frac{2\pi nx}{L}\biggr$}$ and ${h(x)=sin\biggr$\frac{2\pi nx}{L}\biggr$}$ that are periodic on the interval $0\leq x\leq L$ :

$g(x)=cos\biggr\[\frac{2\pi n(x+L)}{L}\biggr\]=cos\biggr$\frac{2\pi nx}{L}+2\pi n\biggr$=cos\biggr$\frac{2\pi nx}{L}\biggr$$

$h(x)=sin\biggr\[\frac{2\pi n(x+L)}{L}\biggr\]=sin\biggr$\frac{2\pi nx}{L}+2\pi n\biggr$=sin\biggr$\frac{2\pi nx}{L}\biggr$$

Question

i) What is the period of $h(x)=sin(bx)$ ?

ii) Show that the sum of two periodic functions produces another periodic function if the ratio of the periods of the component functions is a rational number.

Answer

i) The period is the interval after which the range of the function repeats itself. The sine wave completes one full cycle when $bx=2\pi$ , which happens when ${x=\frac{2\pi}{b}}$ . Hence, the period of $h(x)$ is ${\frac{2\pi}{b}}$ .

ii) Let $p(x)=cos(ax)$ and $q(x)=sin(bx)$ with periods ${\frac{2\pi}{a}}$ and ${\frac{2\pi}{b}}$ , respectively, where $a,b\neq 0$ . For all $x$ , the output of the function $f(x)=Acos(ax)+Bsin(bx)$ repeats itself when

$\begin{align}f(x)&=Acos\biggr\[a\biggr$x+m\frac{2\pi }{a}\biggr$\biggr\]+Bsin\biggr\[b\biggr$x+n\frac{2\pi }{b}\biggr$\biggr\]\\&=Acos(ax+m2\pi)+Bsin(bx+n2\pi)\\&=Acos(ax)+Bsin(bx)\end{align}$

where $m,n\in \mathbb{Z}$ .

In other words, the period of $f(x)$ is ${T=m\frac{2\pi}{a}=n\frac{2\pi}{b}}$ , which implies that ${\frac{m}{n}=\frac{a}{b}}$ . Therefore, ${\frac{a}{b}}$ and hence ${\frac{2\pi}{b}/\frac{2\pi}{a}}$ must be a rational number.

The linear combination ${f(x)=Ag(x)+Bh(x)=Acos\biggr$\frac{2\pi nx}{L}\biggr$+Bsin\biggr$\frac{2\pi nx}{L}\biggr$}$ is a periodic function since the ratio of the periods of the cosine and sine functions is a rational number. It follows that ${f(x)+Ccos\biggr$\frac{2\pi nx}{L}\biggr$}$ is also a periodic function. Therefore, we can express a function that is periodic on the interval $0\leq x\leq L$ as an infinite sum of cosine and sine functions:

$\begin{align}f(x)&=\sum_{n=0}^{\infty}\biggr\[a_ncos\biggr$\frac{2\pi nx}{L}\biggr$+b_nsin\biggr$\frac{2\pi nx}{L}\biggr$\biggr\]\\&=a_0+\sum_{n=1}^{\infty}\biggr\[a_ncos\biggr$\frac{2\pi nx}{L}\biggr$+b_nsin\biggr$\frac{2\pi nx}{L}\biggr$\biggr\]\;\;\;\;\;\;\;\;105\end{align}$

Eq105 is known as a Fourier series.

To find $a_0$ , we integrate eq105 with respect to $x$ over the interval $L$ :

$\int_0^Lf(x)dx=\int_0^L\biggr\{a_0+\sum_{n=1}^{\infty}\biggr\[a_ncos\biggr$\frac{2\pi nx}{L}\biggr$+b_nsin\biggr$\frac{2\pi nx}{L}\biggr$\biggr\]\biggr\}dx$

The cosine and sine functions have the period of $L/n$ . As we are integrating over $n$ multiples of $L/n$ , we have ${a_n\int_0^Lcos\biggr$\frac{2\pi nx}{L}\biggr$dx=b_n\int_0^Lsin\biggr$\frac{2\pi nx}{L}\biggr$dx=0}$ . Hence, ${\int_0^Lf(x)dx=a_0L}$ or

$a_0=\frac{1}{L}\int_0^Lf(x)dx\;\;\;\;\;\;\;\;106$

To find $a_n$ , we multiply eq105 by ${cos\biggr$\frac{2\pi mx}{L}\biggr$}$ and integrate with respect to $x$ over the period $L$ :

$\int_0^Lf(x)cos\biggr$\frac{2\pi mx}{L}\biggr$dx=\int_0^L\biggr\{a_0+\sum_{n=1}^{\infty}\biggr\[a_ncos\biggr$\frac{2\pi nx}{L}\biggr$+b_nsin\biggr$\frac{2\pi nx}{L}\biggr$\biggr\]\biggr\}cos\biggr$\frac{2\pi mx}{L}\biggr$dx$

Since $m$ is an integer, ${a_0\int_0^Lcos\biggr$\frac{2\pi mx}{L}\biggr$dx=\frac{a_0L}{2\pi m}sin2\pi m=0}$ . Using the trigonometric identity $2sinAcosB=sin(A+B)+sin(A-B)$ , we have ${b_n\int_0^Lsin\biggr$\frac{2\pi nx}{L}\biggr$cos\biggr$\frac{2\pi mx}{L}\biggr$dx=\frac{b_n}{2}\int_0^L\biggr\{sin\biggr\[(n+m)\frac{2\pi x}{L}\biggr\]+sin\biggr\[(n-m)\frac{2\pi x}{L}\biggr\]\biggr\}dx=0}$ because we are again integrating over an integral number of complete periods for either sine function. Therefore,

$\int_0^Lf(x)cos\biggr$\frac{2\pi mx}{L}\biggr$dx=\sum_{n=1}^{\infty}a_n\int_0^Lcos\biggr$\frac{2\pi nx}{L}\biggr$cos\biggr$\frac{2\pi mx}{L}\biggr$dx$

Using the trigonometric identity of $2cosAcosB=cos(A+B)+cos(A-B)$ ,

$\int_0^Lf(x)cos\biggr$\frac{2\pi mx}{L}\biggr$dx=\sum_{n=1}^{\infty}\frac{a_n}{2}\int_0^L\biggr\{cos\biggr\[(n+m)\frac{2\pi x}{L}\biggr\]+cos\biggr\[(n-m)\frac{2\pi x}{L}\biggr\]\biggr\}dx\;\;\;\;\;\;\;\;107$

Since we are again integrating over an integral number of complete periods for either cosine function, all the terms in the summation equals zero except when $n=m$ . We have ${\int_0^Lf(x)cos\biggr$\frac{2\pi mx}{L}\biggr$dx=\frac{a_m}{2}\int_0^Ldx=\frac{a_m}{2}L}$ or

$a_n=\frac{2}{L}\int_0^Lf(x)cos\biggr$\frac{2\pi nx}{L}\biggr$dx\;\;\;\;\;\;n=1,2,\cdots\;\;\;\;\;\;\;\;108$

where we have relabelled the index from $m$ to $n$ .

To find $b_n$ , we multiply eq105 by ${sin\biggr$\frac{2\pi mx}{L}\biggr$}$ and integrate with respect to $x$ over the period $L$ . Using the same logic as the derivation of eq108 and the trigonometric identity $2sinAsinB=cos(A-B)-cos(A+B)$ , we have

$b_n=\frac{2}{L}\int_0^Lf(x)sin\biggr$\frac{2\pi nx}{L}\biggr$dx\;\;\;\;\;\;n=1,2,\cdots\;\;\;\;\;\;\;\;109$

Question

i) Can a Fourier series be defined for a function that is periodic on the interval ${-\frac{L}{2}\leq x\leq \frac{L}{2}}$ instead of $0\leq x\leq L$ ?

ii) Can we represent a periodic function of period $L$ with a Fourier series on the interval $0\leq x\leq 2L$ instead of $0\leq x\leq L$ ?

Answer

i) Yes, the representation is the same as eq105 but has slightly different formulae for the coefficients. Repeating the above steps in determining the coefficients, we have

$a_0=\frac{1}{L}\int_{-L/2}^{L/2}f(x)dx\;\;\;\;\;\;\;\;109a$

$a_n=\frac{2}{L}\int_{-L/2}^{L/2}f(x)cos\biggr$\frac{2\pi nx}{L}\biggr$dx\;\;\;\;\;\;\;\;109b$

$b_n=\frac{2}{L}\int_{L/2}^{L/2}f(x)sin\biggr$\frac{2\pi nx}{L}\biggr$dx\;\;\;\;\;\;\;\;109c$

ii) Yes. However, the Fourier series representation is $f(x)=a_0+\sum_{n=1}^{\infty}\biggr\[a_ncos\biggr$\frac{\pi nx}{L}\biggr$+b_nsin\biggr$\frac{\pi nx}{L}\biggr$\biggr\]$ , which is slightly different from eq105. The corresponding coefficients are ${a_0=\frac{1}{2L}\int_0^{2L}f(x)dx}$ , ${a_n=\frac{1}{L}\int_0^{2L}f(x)cos\biggr$\frac{\pi nx}{L}\biggr$dx}$ and ${b_n=\frac{1}{L}\int_0^{2L}f(x)sin\biggr$\frac{\pi nx}{L}\biggr$dx}$ . The function in this representation now repeats every $2L$ instead of every $L$ .

The fourier series can also be expressed in the complex form. Substituting the identities ${cosy=\frac{e^{iy}+e^{-iy}}{2}}$ and ${siny=\frac{e^{iy}-e^{-iy}}{2i}}$ , where ${y=\frac{2\pi nx}{L}}$ in eq105, we have

$f(x)=a_0+\sum_{n=1}^{\infty}\biggr\[\biggr$\frac{a_n-ib_n}{2}\biggr$e^{iy}+\biggr$\frac{a_n+ib_n}{2}\biggr$e^{-iy}\biggr\]\;\;\;\;\;\;\;\;110$

Let ${c_n=\frac{a_n-ib_n}{2}}$ and ${c_{-n}=\frac{a_{-n}-ib_{-n}}{2}}$ , where $n=1,2,\cdots$ . From eq108, $a_{-n}=\frac{2}{L}\int_0^Lf(x)cos(-y)dx=\frac{2}{L}\int_0^Lf(x)cos(y)dx=a_n$ . Similarly, from eq109, $b_{-n}=-b_n$ . Therefore, ${c_{-n}=\frac{a_{n}+ib_{n}}{2}}$ and eq110 becomes

$f(x)=c_0+\sum_{n=1}^{\infty}c_ne^{i\frac{2\pi nx}{L}}+\sum_{n=1}^{\infty}c_{-n}e^{-i\frac{2\pi nx}{L}}\;\;\;\;\;\;\;\;111$

where we have relabelled the constant $a_0$ as $c_0$ without loss of generality (see Q&A below).

If we let the dummy index $n=-k$ for the second summation and subsequently relabel $k$ as $n$ , we have

$f(x)=c_0+\sum_{n=1}^{\infty}c_ne^{i\frac{2\pi nx}{L}}+\sum_{k=-1}^{-\infty}c_{k}e^{i\frac{2\pi kx}{L}}=c_0+\sum_{n=1}^{\infty}c_ne^{i\frac{2\pi nx}{L}}+\sum_{n=-1}^{-\infty}c_{n}e^{i\frac{2\pi nx}{L}}$

Therefore, $f(x)$ can be expressed as

$f(x)=\sum_{n=-\infty}^{\infty}c_{n}e^{i\frac{2\pi nx}{L}}\;\;\;\;\;\;\;\;112$

To find $c_n$ , we substitute eq108 and eq109 in ${c_n=\frac{a_n-ib_n}{2}}$ to give

$c_n=\frac{1}{L}\int_0^Lf(x)e^{-i\frac{2\pi nx}{L}}dx\;\;\;\;\;\;\;\;113$

Question

i) Show that $c_0=a_0$ .

ii) What is the corresponding equation for eq113 if $f(x)$ is periodic on the interval ${-\frac{L}{2}\leq x\leq \frac{L}{2}}$ instead of $0\leq x\leq L$ ?

Answer

i) From eq113, $c_0=\frac{1}{L}\int_0^Lf(x)dx$ . Comparing this equation with eq106, $c_0=a_0$ . This implies that eq113 is valid for all $n$ , including $n=0$ even though the derivation involves ${c_n=\frac{a_n-ib_n}{2}}$ , which is for $n=1,2,\cdots$ .

ii) Repeating the above steps for the derivation of the complex form of a Fourier series using eq109a, eq109b and eq109c, we have

$c_n=\frac{1}{L}\int_{-L/2}^{L/2}f(x)e^{-i\frac{2\pi nx}{L}}dx\;\;\;\;\;\;\;\;114$

Question

Answer

Question

Answer

Question

Answer

Next article: Fourier transform

Previous article: Vibrational selection rules for molecules

Content page of vibrational spectroscopy

Content page of advanced chemistry

Leave a Reply Cancel reply