Fourier Series and Partial Differential Equations Lecture Notes

Chapter 2. Fourier series 15
for every n. Also, f(x)cos(nx) is even, so
an = 1
π
f(x)cos(nx)dx,
π −π
= 2
π
f(x)cos(nx)dx,
π 0
= 2
π
xcos(nx)dx,
π 0
= 2
π π xsin(nx) sin(nx)
−
π n 0 0 n
= − 2
−cos(nx)
π n2 π ,
0
= 2
cos(nπ)−cos(0)
π n2
,
= 2
π
dx ,
[(−1) n −1]
n 2 . (2.37)
Note that this is not valid for n = 0. In fact, a0 = π. If n is even, n = 2m say, we have
If n is odd, n = 2m+1 say, we obtain
2.2.2 Sine and cosine series
Let f be 2π-periodic. If f is odd then
where
a2m = 2((−1)2m −1)
π(2m) 2 = 0. (2.38)
a2m+1 = 2(−1−1)
=
π(2m+1) 2
f(x) ∼
−4
π(2m+1) 2.
(2.39)
∞
bnsin(nx), (2.40)
n=1
bn = 1
π
f(s)sin(ns)ds =
π −π
2
π
f(s)sin(ns)ds, (2.41)
π 0
i.e. f has a Fourier sine series. In this case an = 0 because f(x)cos(nx) is odd. This is
also true if f(x) = −f(−x) for x = nπ, n ∈ Z, i.e. f is odd apart from the end points and
zero.
If f is even then
f(x) ∼ 1
2 a0
∞
+ ancos(nx), (2.42)
n=1
where
an = 2
π
f(s)cos(ns)ds,
π 0
(2.43)
i.e. f has a Fourier cosine series.