Part a) Find the Fourier Series for the functions f(x)=x^2, x^3, and x^4 on the interval [-Pi,Pi]
Part a) Find the Fourier Series for the functions f(x)=x^2, x^3, and x^4 on the interval [-Pi,Pi]f[x_] := x^2
In[2]:=
a[0] := (1/(2*Pi)) Integrate[f[x], {x,-Pi,Pi}]
In[3]:=
a[0]
Out[3]=
2
Pi
---
3
This is the value of the constant term, a0.
In[4]:=
TrigId = {Cos[Pi n_] -> (-1)^n, Sin[Pi n_] -> 0}
Out[4]=
n
{Cos[Pi (n_)] -> (-1) , Sin[Pi (n_)] -> 0}
In[5]:=
a[m_] := (2/(2*Pi)) Integrate[f[x]*Cos[2 m Pi x/(2*Pi)], {x,-Pi,Pi}]
In[6]:=
a[m] /. TrigId
Out[6]=
m
4 (-1)
-------
2
m
These are the coefficients am--the coefficients of the Cosine terms.
In[7]:=
b[n_] := (2/(2*Pi)) Integrate[f[x]*Sin[2 n Pi x/(2*Pi)], {x,-Pi,Pi}]
In[8]:=
b[n]
Out[8]=
0
These coefficients are all zero because x^2 is even and Sin(x) is odd.
In[9]:=
Clear[FullSeries]
Here is the F Series for x^2 on [-Pi,Pi].
In[10]:=
FullSeries[x_,N_] := (Pi^2/3) + \
Sum[((4*(-1)^m)/(m^2))Cos[2 Pi m x/(2*Pi)],{m,1,N}]
In[11]:=
Plot[{FullSeries[x,1], f[x]}, {x,-Pi, Pi}]
Out[12]=
-Graphics-
In[13]:=
Plot[{FullSeries[x,3], f[x]}, {x,-Pi, Pi}]
Out[14]=
-Graphics-
In[15]:=
Plot[{FullSeries[x,10], f[x]}, {x,-Pi, Pi}]
Out[16]=
-Graphics-
Now, let's find the Fourier Series for x^3 over the interval [-Pi, Pi].
In[17]:=
f[x_] := x^3
In[18]:=
a[0] := (1/(2*Pi)) Integrate[f[x], {x,-Pi,Pi}]
In[19]:=
a[0]
Out[19]=
0
The constant term a0 is 0 because 1 is even and x^3 is odd.
In[20]:=
a[m_] := (2/(2*Pi)) Integrate[f[x]*Cos[2 m Pi x/(2*Pi)], {x,-Pi,Pi}]
In[21]:=
a[m]
Out[21]=
0
All of the Cosine terms are zero because Cosine(x) is even and x^3 is odd.
In[22]:=
b[n_] := (2/(2*Pi)) Integrate[f[x]*Sin[2 n Pi x/(2*Pi)], {x,-Pi,Pi}]
In[23]:=
b[n] /. TrigId
Out[23]=
n n 3 3 n n 3 3
6 (-1) n Pi - (-1) n Pi -6 (-1) n Pi + (-1) n Pi
--------------------------- - ----------------------------
4 4
n n
----------------------------------------------------------
Pi
In[24]:=
Simplify[%]
Out[24]=
n 2 2
2 (-1) (6 - n Pi )
--------------------
3
n
The full F series will consist entirely of Sine terms with the coefficients bn as above.
In[25]:=
Clear[FullSeries]
Here is the F Series for x^3 on [-Pi,Pi].
In[26]:=
FullSeries[x_,N_] :=
Sum[((2(-1)^n)(6-n^2*Pi^2)/(n^3))Sin[2 Pi n x/(2*Pi)],{n,1,N}]
In[27]:=
Plot[{FullSeries[x,1], f[x]}, {x,-Pi, Pi}]
Out[28]=
-Graphics-
In[29]:=
Plot[{FullSeries[x,3], f[x]}, {x,-Pi, Pi}]
Out[30]=
-Graphics-
In[31]:=
Plot[{FullSeries[x,10], f[x]}, {x,-Pi, Pi}]
Out[32]=
-Graphics-
Now, let's find the Fourier Series for x^4 over the interval [-Pi, Pi].
In[33]:=
f[x_] := x^4
In[34]:=
a[0] := (1/(2*Pi)) Integrate[f[x], {x,-Pi,Pi}]
In[35]:=
a[0]
Out[35]=
4
Pi
---
5
This is the value of the constant term, a0.
In[36]:=
a[m_] := (2/(2*Pi)) Integrate[f[x]*Cos[2 m Pi x/(2*Pi)], {x,-Pi,Pi}]
In[37]:=
Simplify[a[m] /. TrigId]
Out[37]=
m 2 2
8 (-1) (-6 + m Pi )
---------------------
4
m
These are the coefficients am--the coefficients of the Cosine terms.
In[38]:=
b[n_] := (2/(2*Pi)) Integrate[f[x]*Sin[2 n Pi x/(2*Pi)], {x,-Pi,Pi}]
In[39]:=
b[n]
Out[39]=
0
These coefficients are all zero because x^4 is even and Sin(x) is odd.
In[40]:=
Clear[FullSeries]
Here is the F Series for x^4 on [-Pi,Pi].
In[41]:=
FullSeries[x_,N_] := (Pi^4/5) + \
Sum[((8*(-1)^m)(-6+m^2 Pi^2)/(m^4))Cos[2 Pi m x/(2*Pi)],{m,1,N}]
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