12.5 NON-PERIODIC FUNCTIONS
− 220
L
x
f(x)=x^2
Figure 12.5 f(x)=x^2 ,0<x≤2, with the range extended to give periodicity.
the range, all the coefficientsbrwill be zero. Now we apply (12.5) and (12.6) withL=4
to determine the remaining coefficients:
ar=
2
4
∫ 2
− 2
x^2 cos
(
2 πrx
4
)
dx=
4
4
∫ 2
0
x^2 cos
(πrx
2
)
dx,
where the second equality holds because the function is even inx. Thus
ar=
[
2
πr
x^2 sin
(πrx
2
)]^2
0
−
4
πr
∫ 2
0
xsin
(πrx
2
)
dx
=
8
π^2 r^2
[
xcos
(πrx
2
)] 2
0
−
8
π^2 r^2
∫ 2
0
cos
(πrx
2
)
dx
=
16
π^2 r^2
cosπr
=
16
π^2 r^2
(−1)r.
Since this expression forarhasr^2 in its denominator, to evaluatea 0 we must return to the
original definition,
ar=
2
4
∫ 2
− 2
f(x)cos
(πrx
2
)
dx.
From this we obtain
a 0 =
2
4
∫ 2
− 2
x^2 dx=
4
4
∫ 2
0
x^2 dx=
8
3
.
The final expression forf(x)isthen
x^2 =
4
3
+16
∑∞
r=1
(−1)r
π^2 r^2
cos
(πrx
2
)
for 0<x≤ 2 .
We note that in the above example we could have extended the range so as
to make the function odd. In other words we could have setf(x)=−f(−x)and
then madef(x) periodic in such a way thatf(x+4) =f(x). In this case the
resulting Fourier series would be a series of just sine terms. However, although
this will faithfully represent the function inside the required range, it does not