## Part a) Find the Fourier Series for the functions f(x)=x^2, x^3, and x^4 on the interval [-Pi,Pi]

In[1]:=
`  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}] ```