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Sample

Fourier Series

The Fourier Series is a representation of a function as an infinite sum of sinusoids.

The Fourier Series

History

Jean-Baptiste 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 eigen-solutions that are sinusoids.

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
FresnelS
2
π
x
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

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]=
-
1
8
π
1/4
(-1)
-

4


2

Erf
1
2
1/4
(-1)
(1-2π)-Erf
1
2
1/4
(-1)
(1+2π)-Erfi
1
2
1/4
(-1)
(-1-2π)+Erfi
1
2
1/4
(-1)
(-1+2π)
◼
Express it numerically:
In[79]:=
N[%]
Out[79]=
0.096227-1.38778×
-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]=
-
1
8
π
1/4
(-1)
-

4


2

Erf
1
2
1/4
(-1)
(1-2π)-Erf
1
2
1/4
(-1)
(1+2π)-Erfi
1
2
1/4
(-1)
(-1-2π)+Erfi
1
2
1/4
(-1)
(-1+2π),-
1
8
π
1/4
(-1)
-


2

Erf
1/4
(-1)
(1-π)-Erf
1/4
(-1)
(1+π)+Erfi
1/4
(-1)
(-1+π)+Erfi
1/4
(-1)
(1+π),-
1
8
π
1/4
(-1)
-
9
4

9
2

Erf
1
2
1/4
(-1)
(3-2π)-Erf
1
2
1/4
(-1)
(3+2π)-Erfi
1
2
1/4
(-1)
(-3-2π)+Erfi
1
2
1/4
(-1)
(-3+2π)
In[81]:=
N[%]
Out[81]=
{0.096227-1.38778×
-17
10
,-0.00837573+6.48353×
-17
10
,-0.340173+2.77556×
-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]=
-
1
8
π
1/4
(-1)
-

4
+x


2

Erf
1
2
1/4
(-1)
(1-2π)-Erf
1
2
1/4
(-1)
(1+2π)-Erfi
1
2
1/4
(-1)
(-1-2π)+Erfi
1
2
1/4
(-1)
(-1+2π)-
1
8
π
1/4
(-1)
-

4
-x


2

Erf
1
2
1/4
(-1)
(1-2π)-Erf
1
2
1/4
(-1)
(1+2π)-Erfi
1
2
1/4
(-1)
(1-2π)+Erfi
1
2
1/4
(-1)
(1+2π)+
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.-0.25)-(0.+1.)x
2.71828
+(0.0932356+0.0238069)
(0.-0.25)+(0.+1.)x
2.71828
-(0.00452542+0.00704793)
(0.-1.)-(0.+2.)x
2.71828
-(0.00452542+0.00704793)
(0.-1.)+(0.+2.)x
2.71828
+(0.213688-0.264679)
(0.-2.25)-(0.+3.)x
2.71828
+(0.213688-0.264679)
(0.-2.25)+(0.+3.)x
2.71828
and we can plot said function with Sin(
2
x
) to compare.
◼
Plot Sin(
2
x
) and our Fourier Series approximation on the same graph:
In[4]:=
Plot[{Sin[x^2],fourierSeries3},{x,-4,4},PlotLegends"Expressions"]
Out[4]=
sin(
2
x
)
fourierSeries3

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]=
sin(
2
x
)
fourierSeries1
◼
Create the 4th order Fourier Series approximation:
In[13]:=
fourierSeries4=N[FourierSeries[Sin[x^2],x,4]];
◼
and plot:
In[14]:=
Plot[{Sin[x^2],fourierSeries4},{x,-4,4},PlotLegends"Expressions"]
Out[14]=
sin(
2
x
)
fourierSeries4
◼
Create the 7th order Fourier Series approximation:
◼
and plot:

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

Consider the differential equation: y’’ = - kx, k > 0 with boundary condition y(0) = 0, y’(0) = 0.
We will solve this using Fourier series.
◼
First find the Fourier Series of the right-hand side of the equation.
We show the first few terms here:
◼
To begin, we compute the second derivative on the left-hand side of the equation for each term:
◼
We then solve for the unknown coefficients, after generalizing the coefficients on the right-hand side.
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