Fourier Series and Partial Differential Equations Lecture Notes

Chapter 2. Fourier series 22
because f is continuous on [0,L). If we put x = 0 we calculate
0 = L
4 +
∞
m=0
−2L
(2m+1) 2 π 2,
which proves (2.69). If we set x = L in equation (2.73) we obtain
giving
f(L+)+f(L−)
2
0+L
2
which gives equation (2.69) again.
2.4.1 Sine and cosine series
= L
4 +
∞
m=0
= L
4 +
(2.75)
−2L
(2m+1) 2 cos[(2m+1)π], (2.76)
π2 ∞
m=0
−2L
(2m+1) 2 π 2,
(2.77)
Given a function f defined on [0,L] we require an expansion with only cosine terms or
only sine terms. This will be done by extending f to be even (for only cosine terms) or
odd (for only sine terms) on (−L,L] and then extending to a 2L-period function. The
series obtained will then be valid on (0,L).
Definition If f is definedon [0,L], theeven extension for f, denoted by fe, is the periodic
extension of
f(x) x ∈ [0,L],
fe(x) =
f(−x) x ∈ (−L,0),
(2.78)
so that we have fe(x) = fe(−x) for all x. Thus:
where
an = 1
L
fe(x)cos
L −L
fe(x) ∼ a0
2 +
is called the Fourier cosine series of f.
∞
nπx
ancos , (2.79)
L
n=1
nπx
dx =
L
2
L
L
nπx
f(x)cos dx, (2.80)
L
Definition The odd extension for f, denoted by fo, is the periodic extension of
f(x) x ∈ [0,L],
fo(x) =
−f(−x) x ∈ (−L,0),
so that fo(x) = −fo(−x) for all x = nL. Similarly,
where
fo(x) ∼
is called the Fourier sine series for f.
0
n=1
0
(2.81)
∞
nπx
bnsin , (2.82)
L
bn = 2
L
nπx
f(x)sin dx,
L L