> restart; assume 0 ! n ;
Fourier series for the example we did in class:
f(x) = x2
on the interval x∈ (-π..π)
[To do any other function, just change the definition here.]
> f d x/x2 ;
f := x/x2
(1)
> plot f x , x =K2 p ..2 p ;
30
20
10
K2 p
3p
2
K
Kp
0
p
2
K
p
2
p
3p
2
2p
x
Of course, Maple may be used to calculate the coefficients:
> a d n/
1
$ int f x $cos n$x , x =Kp ..p ;
p
p
f x cos n x dx
a := n/
Kp
p
(2)
> b d n/
1
$ int f x $sin n$x , x =Kp ..p ;
p
p
f x sin n x dx
b := n/
Kp
p
(3)
The b(n) are all zero, as f(x) is even:
> eval b n
;
0
(4)
Lets do a0 separately, to be sure not to divide by zero:
> a0 d
1
$int f x , x =Kp ..p ;
p
a0 :=
2 2
p
3
Now calculate the Nth partial sum, that is the sum of the Fourier series up to the Nth term:
>
a0
> ps d N, x /
C sum 'a n $cos n$x C b n $sin n$x ', 'n = 1 ..N' ;
2
1
ps := N, x / a0 C sum 'a n cos n x C b n sin n x ', 'n = 1 ..N'
2
(5)
(6)
Here are the 4th and 5th partial sums of the series:
> ps 4, x ;
1 2
4
1
p K 4 cos x C cos 2 x K
cos 3 x C
cos 4 x
3
9
4
> ps 5, x
1 2
4
1
4
p K 4 cos x C cos 2 x K
cos 3 x C
cos 4 x K
cos 5 x
3
9
4
25
And now the graphs; we look at partial sums with N getting larger.
>
> plot
ps 1, x , f x , x =K2 p ..2 p, y =K1 ..10 ;
(7)
(8)
10
8
6
y
4
2
K2 p
3p
2
K
Kp
p
2
K
0
p
2
p
x
> plot
ps 2, x , f x , x =K2 p ..2 p, y =K1 ..10
3p
2
2p
10
8
6
y
4
2
K2 p
3p
2
K
Kp
p
2
K
0
p
2
p
x
> plot
ps 3, x , f x , x =K2 p ..2 p, y =K1 ..10
3p
2
2p
10
8
6
y
4
2
K2 p
3p
2
K
Kp
p
2
K
0
p
2
p
x
> plot
ps 4, x , f x , x =K2 p ..2 p, y =K1 ..10
3p
2
2p
10
8
6
y
4
2
K2 p
3p
2
K
Kp
p
2
K
0
p
2
p
x
> plot
ps 8, x , f x , x =K2 p ..2 p, y =K1 ..10
3p
2
2p
10
8
6
y
4
2
K2 p
3p
2
K
Kp
p
2
K
0
p
2
p
x
> plot
ps 12, x , f x , x =K2 p ..2 p, y =K1 ..10
3p
2
2p
10
8
6
y
4
2
K2 p
3p
2
K
Kp
p
2
K
0
p
2
p
x
> plot
ps 20, x , f x , x =K2 p ..2 p, y =K1 ..10
3p
2
2p
10
8
6
y
4
2
K2 p
3p
2
K
Kp
p
2
K
0
p
2
p
3p
2
2p
x
By the Convergence Theorem, the series converges to f(x) at points of continuity in the interval x∈(-π,π).
The series is periodic with period 2π. As there are no discontinuities, there are no jumps (and the
convergence is rapid!)
> plot
ps 50, x , x =K5 p ..5 p, scaling = constrained
9
7
5
3
1
K5 p K4 p K3 p K2 p
0
Kp
p
2p
3p
4p
5p
x
Now let's do the Half-Range Sine Series for f(x)=x2
2
> B d n/ $ int f x $sin n$x , x = 0 ..p ;
p
p
2
B := n/
f x sin n x dx
0
(9)
p
> pss d N, x / sum 'B n $sin n$x ', 'n = 1 ..N' ;
pss := N, x /sum 'B n sin n x ', 'n = 1 ..N'
> eval B n
;
2
2 2 K 2 cos p n~ C n~2 cos p n~ p K 2 n~ sin p n~ p
K
> plot
(10)
p n~3
pss 25, x , pss 15, x , f x , x =K2 p ..2 p, y =K12 ..12 ;
(11)
10
y
5
K2 p
3p
2
K
Kp
0
p
2
K
p
2
p
x
K5
K10
3p
2
2p