Fourier Series

Any square integrable function function, $f(x)$, in the interval $\left[-\pi,\pi\right]$ can be expressed as, \begin{equation}f(x)=\sum_{n=0}^{\infty}a_{n}\cos nx\:+\sum_{n=1}^{\infty}b_{n}\sin nx\,,\end{equation} where the constants $a_{n}$ and $b_{n}$ are given by, \begin{eqnarray} a_{n}&=&\frac{1}{\pi}\,\int_{-\pi}^{+\pi}f(x)\,\cos nx\,dx\qquad n\geq0\, \\ b_{n}&=&\frac{1}{\pi}\,\int_{-\pi}^{+\pi}f(x)\,\sin nx\,dx\qquad n>0\, \end{eqnarray} This representation of function $f(x)$ is called the Fourier series.

Problem 1

Express the function \begin{equation} f(x)=\left\{ \begin{array}{cc} +1 & \left(x<0\right)\\ -1 & \left(x\geq0\right) \end{array}\right. \end{equation} In Fourier series.

The coeffients $a_{n}$ and $b_{n}$ can be determined by the above eq.
\begin{eqnarray} a_{0}&=&\frac{1}{\pi}\int_{-\pi}^{0}(-1)\,dx\,+\,\frac{1}{\pi}\int_{0}^{+\pi}(+1)\,dx=0, \\ a_{n}&=&\frac{1}{\pi}\int_{-\pi}^{0}(-\cos nx)\,dx\,+\,\frac{1}{\pi}\int_{0}^{+\pi}(+\cos nx)\,dx=0, \\ b_{n}&=&\frac{1}{\pi}\int_{-\pi}^{0}(-\sin nx)\,dx\,+\,\frac{1}{\pi}\int_{0}^{+\pi}(+\cos nx)\,dx \\ \,&=&\frac{2}{\pi}\int_{0}^{\pi}\sin nx\,dx\quad=\;\left\{ \begin{array}{cc} \frac{4}{n\pi} & \left(n=\mathrm{odd}\right)\\ 0 & \left(n=\mathrm{even}\right) \end{array}\right. \end{eqnarray} The function $f(x)$ can be written as,$$f(x)=\frac{4}{\pi}\sum_{n=\mathrm{odd}}^{\infty}\frac{1}{n}\,\sin nx $$ or $$f(x)=\frac{4}{\pi}\sum_{n=\mathrm{0}}^{\infty}\frac{1}{2n+1}\,\sin (2n+1)x $$ The plot of series by taking different number to terms are shown in figure

import numpy as np
import matplotlib.pyplot as plt
import imageio
import os
from IPython.display import Image
def f(x) :
    fx=np.ones(x.shape)
    (idx,)= np.where( x < 0 )
    fx[idx]=-1
    return fx
    
x=np.linspace(-np.pi, np.pi, 1000)
fx=f(x)


infinity=50
filenames=[]
images = []
fourierf=np.zeros(x.shape)
for n in range(infinity) :
    t1=4/np.pi * 1/(2*n+1) * np.sin((2*n+1)*x)
    fourierf+=t1
    plt.plot(x, fx, 'C0', label='f(x)')
    plt.plot(x, fourierf, 'C2',label='N='+str(n).rjust(3,'0') )
    plt.grid(alpha=0.3)
    plt.xlabel('x')
    plt.ylabel('y')
    plt.xlim([ -1.1*np.pi, 1.1*np.pi])
    plt.ylim([-1.5, 1.5])
    plt.legend()
    plt.tight_layout()
    
    fname='Fourier_'+str(n).rjust(3,'0')+'.png'
    plt.savefig(fname)
    filenames.append(fname)
    plt.close()

for filename in filenames:
            images.append(imageio.v3.imread(filename))
imageio.mimsave('Step_Fourier_Series.gif', images, duration=1)
for filename in filenames:
    os.remove(filename)
Image(url='Step_Fourier_Series.gif') 

Problem 2

Express the function $f(x)=x\qquad-\pi<x<\pi $ in terms of Fourier series. \begin{eqnarray} a_{0}&=&\frac{1}{\pi}\int_{-\pi}^{\pi}x\,dx=0 \\ a_{n}&=&\frac{1}{\pi}\int_{-\pi}^{\pi}x\,\cos nx\:dx=0 \\ b_{n}&=&\frac{1}{\pi}\int_{-\pi}^{\pi}x\,\sin nx\:dx=-\frac{2\pi}{n}\,\cos n\pi \\ \,&=&\frac{2}{n}\,\left(-1\right)^{n+1} \end{eqnarray} and the function can be written as: $$f(x)=2\sum_{n=1}^{\infty}\left(-1\right)^{n+1}\frac{\sin nx}{n}\,.$$ The polts of various terms of this series is shown in figure

import numpy as np
import matplotlib.pyplot as plt
import imageio
import os
from IPython.display import Image
def linear(x) :
    fx=np.copy(x)
    return fx
    
x=np.linspace(-np.pi, np.pi, 1000)
fx=linear(x)


infinity=50
filenames=[]
images = []
fourierf=np.zeros(x.shape)
for n in range(1,infinity) :
    t1=2 * 1/n * np.sin(n*x) *(-1)**(n+1)
    fourierf+=t1
    plt.plot(x, fx, 'C0', label='f(x)')
    plt.plot(x, fourierf, 'C2',label='N='+str(n).rjust(3,'0') )
    plt.grid(alpha=0.3)
    plt.xlabel('x')
    plt.ylabel('y')
    plt.xlim([ -1.1*np.pi, 1.1*np.pi])
    plt.ylim([-4, 4])
    plt.legend()
    plt.tight_layout()
    
    fname='Fourier_'+str(n).rjust(3,'0')+'.png'
    plt.savefig(fname)
    filenames.append(fname)
    plt.close()

for filename in filenames:
            images.append(imageio.v3.imread(filename))
imageio.mimsave('Linear_Fourier_Series.gif', images, duration=1)
for filename in filenames:
    os.remove(filename)
Image(url='Linear_Fourier_Series.gif')

Extended domain

The Fourier series can be extended to artitrary interval $\left[-\pi,\,\pi\right]$ to $\left[-L,\,L\right]$ by scaling. Any $L^{2}$ function in ter interval $\left[-L,\,L\right]$ can be expressed in terms of Fourier series as given in eq above \begin{equation} f(x)=\sum_{n=0}^{\infty}a_{n}\cos nx\,+\,\sum_{n=1}^{\infty}b_{n}\sin nx\,. \end{equation} where the coeffients $a_{n}$ and $b_{n}$ are given by, \begin{eqnarray} a_{n}&=&\frac{1}{L}\int_{-L}^{+L}\:f(x)\:\cos\left(\frac{n\pi x}{L}\right)\,dx\,,\\ b_{n}&=&\frac{1}{L}\int_{-L}^{+L}\:f(x)\:\sin\left(\frac{n\pi x}{L}\right)\,dx\, \end{eqnarray} An even function $f(x)$ is such that, $f(x)=f(-x)\,.$ For example $\cos x $ is an even function. An odd function is such that $f(x)=-f(x)\,.$The function $\sin x$ is an example for odd function. An even function can be expaned in Fourier cosine series and coeffients $b_{n}$ will be zero. An odd function can be expanded in terms of Fourier sine seires and the coeffients $a_{n}$ will be identically zero.

Problem

Express the function $$f(x)=\left\{ \begin{array}{cc} x & 0<x<L\\ -x & -L<x<0 \end{array}\right.$$ Since the function is an even function, the coeffients $b_{n}$ are identically zero, after simple computation, we get: \begin{equation} f(x)=\frac{L}{2}\,-\,\frac{4}{L}\,\sum_{n=\mathrm{odd}}^{\infty}\:\frac{\cos\left(\frac{n\pi x}{L}\right)}{(\pi n)^{2}}\,, \end{equation} and the plots are given in the figure

import numpy as np
import matplotlib.pyplot as plt
import imageio
import os
from IPython.display import Image
def absx(x) :
    fx=np.abs(x)
    return fx
L=1.0 
x=np.linspace(-L, L, 1000)
fx=absx(x)


infinity=50
filenames=[]
images = []
fourierf=np.ones(x.shape)*L/2
for n in range(1,infinity,2) :
    t1=-4/L*np.cos(n*np.pi*x/L)/n/n/np.pi/np.pi
    fourierf+=t1
    plt.plot(x, fx, 'C0', label='f(x)')
    plt.plot(x, fourierf, 'C2',label='N='+str(n).rjust(3,'0') )
    plt.grid(alpha=0.3)
    plt.xlabel('x')
    plt.ylabel('y')
    plt.xlim([ -1.1*L, 1.1*L])
    plt.ylim([-.5, 1.5])
    plt.legend()
    plt.tight_layout()
    
    fname='Fourier_'+str(n).rjust(3,'0')+'.png'
    plt.savefig(fname)
    filenames.append(fname)
    plt.close()

for filename in filenames:
            images.append(imageio.v3.imread(filename))
imageio.mimsave('Abs_Fourier_Series.gif', images, duration=1)
for filename in filenames:
    os.remove(filename)
Image(url='Abs_Fourier_Series.gif')

Problem

Give Fourier series expansion for the function$$f(x)=\left\{ \begin{array}{cc} \,\frac{1}{2}\left(L-x\right) & \left(0<x<L\right)\\ -\frac{1}{2}\left(L+x\right) & -L<x<0 \end{array}\right.$$ The Fourier series expansion is given by \begin{equation} f(x)=\sum_{n=1}^{\infty}\frac{1}{\pi n}\,\sin\left(\frac{n\pi x}{L}\right) \end{equation} The plots are given in the figure

import numpy as np
import matplotlib.pyplot as plt
import imageio
import os
from IPython.display import Image
def halflx(x, L) :
    fx=0.5*(L-x)
    (idx,)=np.where(x<0);
    fx[idx]=-0.5*(L+x[idx])
    return fx
    
    
L=1.0 
x=np.linspace(-L, L, 1000)
fx=halflx(x,L)


infinity=50
filenames=[]
images = []
fourierf=np.zeros(x.shape)
for n in range(1,infinity) :
    t1= 1/n/np.pi * np.sin(n*np.pi*x/L)
    fourierf+=t1
    plt.plot(x, fx, 'C0', label='f(x)')
    plt.plot(x, fourierf, 'C2',label='N='+str(n).rjust(3,'0') )
    plt.grid(alpha=0.3)
    plt.xlabel('x')
    plt.ylabel('y')
    plt.xlim([ -1.1*L, 1.1*L])
    plt.ylim([-.6, .6])
    plt.legend()
    plt.tight_layout()
    
    fname='Fourier_'+str(n).rjust(3,'0')+'.png'
    plt.savefig(fname)
    filenames.append(fname)
    plt.close()

for filename in filenames:
            images.append(imageio.v3.imread(filename))
imageio.mimsave('Halflx_Fourier_Series.gif', images, duration=1)
for filename in filenames:
    os.remove(filename)
Image(url='Halflx_Fourier_Series.gif')

Dirac $\delta$

The Dirac delta function $\delta\left(x-x'\right)$ is such that, \begin{equation} f(x')=\int_{a}^{b}f(x)\,\delta(x-x')\,dx=\left\{ \begin{array}{cc} f(x') & x'\in\left[a,b\right]\\ 0 & x'\notin\left[a,b\right] \end{array}\right. \end{equation} can be expressed in terms of Fourier seires as follows \begin{eqnarray} \delta\left(x-x'\right)&=& S+ C \\ S& = &\left(\frac{1}{L}\right)\sum_{n=1}^{\infty}\sin\left(\frac{n\pi x}{L}\right)\sin\left(\frac{n\pi x'}{L}\right)\,, \nonumber \\ C & = & \left(\frac{1}{L}\right)\sum_{n=0}^{\infty}\cos\left(\frac{n\pi x}{L}\right)\cos\left(\frac{n\pi x'}{L}\right)\,, \end{eqnarray} where $L=b-a$. The coeffients $\left\{ a_{n},\,b_{n}\right\}$ can easily determined

import numpy as np
import matplotlib.pyplot as plt
import imageio
import os
from IPython.display import Image
L=1.0 
x=np.linspace(-L, L, 1000)
xp=0.5
infinity=50
filenames=[]
images = []
deltaS=np.zeros(x.shape)
deltaC=np.zeros(x.shape)
for n in range(1,infinity) :
    s1= 1/L * np.sin(n*np.pi*x/L)*np.sin(n*np.pi*xp/L)
    c1= 1/L * np.cos(n*np.pi*x/L)*np.cos(n*np.pi*xp/L)
    deltaS+=s1
    deltaC+=c1
    
    plt.plot(x, deltaS, 'C1--',label='Sin Series N='+str(n).rjust(3,'0') )
    plt.plot(x, deltaC, 'C2--',label='Cos Series N='+str(n).rjust(3,'0') )
    plt.plot(x, deltaS+deltaC, 'C3',label='Sin+Cos Series N='+str(n).rjust(3,'0') )
    plt.grid(alpha=0.3)
    plt.xlabel('x')
    plt.ylabel('y')
    plt.xlim([ -1.1*L, 1.1*L])
    plt.ylim([-50,50])
    plt.legend()
    plt.tight_layout()
    
    fname='Fourier_'+str(n).rjust(3,'0')+'.png'
    plt.savefig(fname)
    filenames.append(fname)
    plt.close()

for filename in filenames:
            images.append(imageio.v3.imread(filename))
imageio.mimsave('Delta_Fourier_Series.gif', images, duration=1)
for filename in filenames:
    os.remove(filename)
Image(url='Delta_Fourier_Series.gif')