Part b) Find the Fourier Series for the functions f(x)=x^2, x^3, and x^4 on the interval [0,1]
Part b) Find the Fourier Series for the functions f(x)=x^2, x^3, and x^4 on the interval [0,1]f[x_] := x^2
In[43]:=
a[0] := (1/1) Integrate[f[x], {x,0,1}]
In[44]:=
a[0]
Out[44]=
1 - 3
This is the value of the constant term, a0 (the average of the function).
In[45]:=
a[m_] := (2/1) Integrate[f[x]*Cos[2 m Pi x/1], {x,0,1}]
In[46]:=
a[m]
Out[46]=
2 2
2 m Pi Cos[2 m Pi] - Sin[2 m Pi] + 2 m Pi Sin[2 m Pi]
-------------------------------------------------------
3 3
2 m Pi
In[47]:=
% /. TrigId
Out[47]=
2 m
(-1)
-------
2 2
m Pi
I will introduce another simplifying function, goodId.
In[48]:=
goodId = {(-1)^(2*n_) -> 1}
Out[48]=
2 (n_)
{(-1) -> 1}
In[49]:=
a[m] /. TrigId /. goodId
Out[49]=
1
------
2 2
m Pi
These are the coefficients am--the coefficients of the Cosine terms.
In[50]:=
b[n_] := (2/1) Integrate[f[x]*Sin[2 n Pi x/1], {x,0,1}]
In[51]:=
Simplify[b[n] /. TrigId] /. goodId
Out[51]=
1
-(----)
n Pi
These are the coefficients bn for the Sine terms.
In[52]:=
Clear[FullSeries]
Here is the F Series for x^2 on [0,1].
In[53]:=
FullSeries[x_,N_] := 1/3 + \
Sum[(1/((m^2)(Pi^2)))Cos[2 Pi m x/(1)],{m,1,N}] + \
Sum[(-1/(n Pi))Sin[2 Pi n x/1],{n,1,N}]
In[54]:=
Plot[{FullSeries[x,2], f[x]}, {x,0,1}]
Out[55]=
-Graphics-
In[56]:=
Plot[{FullSeries[x,8], f[x]}, {x,0,1}]
Out[57]=
-Graphics-
Now, let's find the Fourier Series for x^3 over the interval [0,1].
In[58]:=
f[x_] := x^3
In[59]:=
a[0] := (1/1) Integrate[f[x], {x,0,1}]
In[60]:=
a[0]
Out[60]=
1 - 4
This is the value of the constant term, a0.
In[61]:=
a[m_] := (2/1) Integrate[f[x]*Cos[2 m Pi x/1], {x,0,1}]
In[62]:=
Simplify[a[m] /. TrigId /. goodId]
Out[62]=
3
--------
2 2
2 m Pi
These are the coefficients am--the coefficients of the Cosine terms.
In[63]:=
b[n_] := (2/1) Integrate[f[x]*Sin[2 n Pi x/1], {x,0,1}]
In[64]:=
Simplify[b[n] /. TrigId /. goodId]
Out[64]=
2 2
3 - 2 n Pi
------------
3 3
2 n Pi
These are the coefficients bn for the Sine terms.
In[65]:=
Clear[FullSeries]
Here is the F Series for x^3 on [0,1].
In[66]:=
FullSeries[x_,N_] := 1/4 + \
Sum[(3/(2(m^2)(Pi^2)))Cos[2 Pi m x/(1)],{m,1,N}] + \
Sum[((3 - 2*n^2*Pi^2)/(2*n^3*Pi^3))Sin[2 Pi n x/1],{n,1,N}]
Now, let's find the Fourier Series for x^4 over the interval [0,1].
In[67]:=
f[x_] := x^4
In[68]:=
a[0] := (1/1) Integrate[f[x], {x,0,1}]
In[69]:=
a[0]
Out[69]=
1 - 5
This is the value of the constant term, a0.
In[70]:=
a[m_] := (2/1) Integrate[f[x]*Cos[2 m Pi x/1], {x,0,1}]
In[71]:=
Simplify[a[m] /. TrigId /. goodId]
Out[71]=
2 2
-3 + 2 m Pi
-------------
4 4
m Pi
These are the coefficients am--the coefficients of the Cosine terms.
In[72]:=
b[n_] := (2/1) Integrate[f[x]*Sin[2 n Pi x/1], {x,0,1}]
In[73]:=
Simplify[b[n] /. TrigId /. goodId]
Out[73]=
2 2
3 - n Pi
----------
3 3
n Pi
These are the coefficients bn for the Sine terms.
In[74]:=
Clear[FullSeries]
Here is the F Series for x^4 on [0,1].
In[75]:=
FullSeries[x_,N_] := 1/5 + \
Sum[((-3 + 2 m^2 Pi^2)/((m^4)(Pi^4)))Cos[2 Pi m x/(1)],{m,1,N}] + \
Sum[((3 - n^2*Pi^2)/(n^3*Pi^3))Sin[2 Pi n x/1],{n,1,N}]
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