Fourier Series and Partial Differential Equations Lecture Notes

Chapter 2. Fourier series 16
2.2.3 Question 2
Recall Question 2: with these an, bn, when, if ever, is equation (2.19) true? Consider what
happens in the following example.
Example 2.2 Consider the Fourier series of the function f which is periodic with period
2π and such that
1 0 < x ≤ π,
f(x) =
−1 −π < x ≤ 0.
(2.44)
−π
y
1
Note that f is odd, so we can conclude that f(x)cos(nx) is odd, giving an = 0 without
computation. On the other hand, f(x)sin(nx) is even, so
i.e.
and hence
−1
bn = 1
π
f(x)sin(nx)dx =
π −π
2
π
π 0
b2m = 0, b2m+1 =
f(x) ∼ 4
π
∞
m=0
Consider Question 2 for this case: when is
Recall that
means limn→∞sn(x) where
f(x) = 1
2 a0 +
4
π
π
sin(nx)dx = − 2[(−1)n −1]
, (2.45)
nπ
4
, (2.46)
(2m+1)π
1
sin[(2m+1)x]. (2.47)
2m+1
∞
[ancos(nx)+bnsin(nx)]? (2.48)
n=1
∞
m=0
sn(x) = 4
π
sin[(2m+1)x]
, (2.49)
2m+1
n
m=0
sin[(2m+1)x]
. (2.50)
2m+1
The question is therefore, does sn(x) converge for each x? If it does, is the limit f(x)?
Some partial sums, sn, are plotted in Figure 2.1.
x