Fourier Series
Fourier Series
The Fourier Series is a representation of a function as an infinite sum of sinusoids.
The Fourier Series
The Fourier Series
History
History
JeanBaptiste Joseph Fourier introduced the Fourier Series as a way of solving the heat equation in a metal plate. In turn, he concluded that any arbitrary continuous function can be represented by a trigonometric series based on the set of Sin(n·x) and Cos(n·x) functions. In general, the techniques are quite useful when applied to differential equations involving eigensolutions that are sinusoids.
Motivation
Motivation
Consider a function that does not have a closed form integral formula (i.e. cannot be expressed without using numeric methods):
◼
Plot said function: Sin()
2
x
In[16]:=
Plot[Sin[x^2],{x,4,4}]
Out[16]=
We attempt to the take the integral...
◼
Integrate the function:
In[17]:=
Integrate[Sin[x^2],x]
Out[17]=
π
2
2
π
and we can plot the function numerically 
◼
Plot the integral formula:
In[18]:=
Plot[%,{x,4,4}]
Out[18]=
However there does not exist a closed form solution to such an integral equation.
Theory
Theory
We can define a series of sinusoids that converge to the original function, with the correct constants.
◼
Find the first coefficient in the Fourier Series:
In[78]:=
FourierCoefficient[Sin[x^2],x,1]
Out[78]=
Erf(12π)Erf(1+2π)Erfi(12π)+Erfi(1+2π)
1
8
π
1/4
(1)

4
2
1
2
1/4
(1)
1
2
1/4
(1)
1
2
1/4
(1)
1
2
1/4
(1)
◼
Express it numerically:
In[79]:=
N[%]
Out[79]=
0.0962271.38778×
17
10
◼
Create a list of the first 3 coefficients in the series:
In[80]:=
Table[FourierCoefficient[Sin[x^2],x,i],{i,1,3}]
Out[80]=
Erf(12π)Erf(1+2π)Erfi(12π)+Erfi(1+2π),Erf(1π)Erf(1+π)+Erfi(1+π)+Erfi(1+π),Erf(32π)Erf(3+2π)Erfi(32π)+Erfi(3+2π)
1
8
π
1/4
(1)

4
2
1
2
1/4
(1)
1
2
1/4
(1)
1
2
1/4
(1)
1
2
1/4
(1)
1
8
π
1/4
(1)

2
1/4
(1)
1/4
(1)
1/4
(1)
1/4
(1)
1
8
π
1/4
(1)

9
4
9
2
1
2
1/4
(1)
1
2
1/4
(1)
1
2
1/4
(1)
1
2
1/4
(1)
In[81]:=
N[%]
Out[81]=
{0.0962271.38778×,0.00837573+6.48353×,0.340173+2.77556×}
17
10
17
10
17
10
We attach these constants to the corresponding functions and obtain an approximation.
◼
Create the 1st order Fourier Series of the function:
In[82]:=
FourierSeries[Sin[x^2],x,1]
Out[82]=
Erf(12π)Erf(1+2π)Erfi(12π)+Erfi(1+2π)Erf(12π)Erf(1+2π)Erfi(12π)+Erfi(1+2π)+
1
8
π
1/4
(1)
+x
4
2
1
2
1/4
(1)
1
2
1/4
(1)
1
2
1/4
(1)
1
2
1/4
(1)
1
8
π
1/4
(1)
x
4
2
1
2
1/4
(1)
1
2
1/4
(1)
1
2
1/4
(1)
1
2
1/4
(1)
FresnelS[
2π
]2π
◼
Create the 3rd order Fourier Series of the function:
In[3]:=
fourierSeries3=FourierSeries[Sin[x^2],x,3]//N
Out[3]=
0.245943+(0.0932356+0.0238069)+(0.0932356+0.0238069)(0.00452542+0.00704793)(0.00452542+0.00704793)+(0.2136880.264679)+(0.2136880.264679)
(0.0.25)(0.+1.)x
2.71828
(0.0.25)+(0.+1.)x
2.71828
(0.1.)(0.+2.)x
2.71828
(0.1.)+(0.+2.)x
2.71828
(0.2.25)(0.+3.)x
2.71828
(0.2.25)+(0.+3.)x
2.71828
and we can plot said function with Sin() to compare.
2
x
◼
Plot Sin() and our Fourier Series approximation on the same graph:
2
x
In[4]:=
Plot[{Sin[x^2],fourierSeries3},{x,4,4},PlotLegends"Expressions"]
Out[4]=
Mathematical Exploration
Mathematical Exploration
While we can see that the functions do not overlap exactly, we will explore the convergence of the series.
◼
Create the 1st order Fourier Series approximation:
In[11]:=
fourierSeries1=N[FourierSeries[Sin[x^2],x,1]];
◼
and plot:
In[12]:=
Plot[{Sin[x^2],fourierSeries1},{x,4,4},PlotLegends"Expressions"]
Out[12]=
◼
Create the 4th order Fourier Series approximation:
◼
and plot:
◼
Create the 7th order Fourier Series approximation:
◼
and plot:
Application: Numerical Analysis
Application: Numerical Analysis
We can then compare the numerical integral to the integral of the Fourier Series (of which all the terms are integrable).
◼
Compute a list of differences between the integrals as the order of the series increases and plot:
and we can see that the difference is tending towards 0. Thus taking the integral of the Fourier Series of a function is one way to approximate the area under a curve.
Another Application: Solving Differential Equations
Another Application: Solving Differential Equations
Consider the differential equation: y’’ =  kx, k > 0 with boundary condition y(0) = 0, y’(0) = 0.
We will solve this using Fourier series.
We will solve this using Fourier series.
◼
First find the Fourier Series of the righthand side of the equation.
We show the first few terms here:
We show the first few terms here:
◼
To begin, we compute the second derivative on the lefthand side of the equation for each term:
◼
We then solve for the unknown coefficients, after generalizing the coefficients on the righthand side.
For the even terms:
For the even terms:
◼
For the odd terms:
◼
Sum up the terms, set k = 1, and plot the solution:
Further Explorations
Explore the Taylor Series Expansion of a Function
Explore Other Sets of Basis Functions That Can Approximate a Function
Authorship information
Michael Dobbs
21.06.17
dobbsm@sonoma.edu