Chapter 11
Lecture 15 - Convergence of
Fourier Series
Example 11.1 (Completion of problem illustrating Half-range Expansions)
Periodic Extension: Assume that f (x) = x, 0 < x < 2 represents one full
period of the function so that f (x + 2) = f (x). 2L = 2 ⇒ L = 1.
a0 =
1
L
L
1
f (x) dx =
−L
2
f (x) dx =
−1
0
2
x2 =2
x dx =
2 0
since f (x + 2) = f (2).
(11.1)
(11.2)
73
Lecture 15 - Convergence of Fourier Series
n ≥ 1:
an =
1
L
L
f (x) cos
nπx −L
L
1
dx =
f (x) cos(nπx) dx
L=1
−1
2
=
x cos(nπx) dx
0
⎡
bn
⎤
2 2
x
sin(nπx)
1
−
= ⎣
sin(nπx) dx⎦
nπ
nπ
0
0
2
1 1
cos(2nπ) − 1 = 0
cos(nπx) =
=
2
2
(nπ)
(nπ)
0
L
1
nπx 1
dx = f (x) sin(nπx) dx
=
f (x) sin
L
L
−L
2
−1
⎡
cos(nπx) 2
1
=
x sin(nπx) dx = ⎣−x
+
nπ
(nπ)
0
0
−2
−2
sin(nπx) 2
=
+
=
2
nπ
(nπ)
nπ
0
2
(11.3)
⎤
cos(nπx) dx⎦
0
(11.4)
Therefore
f (x) =
∞
2 2 sin(nπx)
−
2 π
n
= 1−
2
π
n=1
∞
n=1
(11.5)
sin(nπx)
n
(11.6)
74
3
S(x)
2
1
0
−1
−4
−2
0
x
2
Figure 11.1: Full Range Expansion SN (x) = 1 −
4
2
π
N
=20
n=1
sin(nπx)
n
2
S(x)−1
1
0
−1
−2
−4
−2
0
x
2
Figure 11.2: Full Range Expansion SN (x) − 1 = − π2
4
N
=20
n=1
sin(nπx)
n
75
Lecture 15 - Convergence of Fourier Series
11.1
Convergence of Fourier Series
• What conditions do we need to impose on f to ensure that the Fourier
Series converges to f .
• We consider piecewise continuous functions:
Theorem 11.2 Let f and f be piecewise continuous functions on [−L, L]
and let f be periodic with period 2L, then f has a Fourier Series
f (x) =
a0
2
+
n=1
where
an
=
1
L
∞
L
−L
an cos
f (x) cos
nπx L
nπx L
+ bn sin
nπx L
dx and bn =
1
L
= S(x)
L
−L
f (x) sin
nπx L
(11.7)
dx.
The Fourier Series converges to f (x) at all points at which f is continuous
1
and to f (x+) + f (x−) at all points at which f is discontinuous.
2
• Thus a Fourier Series converges to the average value of the left and
right limits at a point of discontinuity of the function f (x).
76
11.2. ILLUSTRATION OF THE GIBBS PHENOMENON
11.2
Illustration of the Gibbs Phenomenon
• Near points of discontinuity truncated Fourier Series exhibit oscillations - overshoot.
1.5
SN(x) for N=5
1
0.5
0
−0.5
−1
−1.5
−2
−1
0
x/π
1
2
Figure 11.3: Fourier Series for a step function
Example 11.3 Consider the half-range sine series expansion of
f (x) = 1
f (x) = 1 =
where bn =
(11.8)
∞
bn sin(nx)
n=1
π
2
sin(nx) dx
π
0
=
Therefore f (x) =
on [0, π].
4
π
4/πn n odd
0
n even
∞
sin(nx)
n=1
n
odd
n
=
2
π
− cosnnx
π
0
=
2
πn
1 − (−1)n
(11.9)
=
4
π
∞
m=0
sin(2m+1)x
(2m+1) .
Note:
∞
4 sin (2m + 1)π/2
4
1 1
=
1 − + − · · · . There1. f (π/2) = 1 =
π
(2m + 1)
π
3 5
m=0
1 1
π
fore = 1 − + − · · · .
4
3 5
77
Lecture 15 - Convergence of Fourier Series
2. Recall the complex Fourier Series example for the function
−1 −π ≤ x < 0
f (x) =
1
0<x<π
(11.10)
which turns out to be equivalent to the odd extension of the above
function represented by the half-range sine expansion, which we can
see from the following calculation
f (x) =
=
∞
n=−∞
n
4
π
odd
∞
n=1
n
11.3
=
∞
4
π
n=1
n
sin(nx)
.
n
odd
einx −e−inx
2in
(11.11)
odd
Now consider the sum of the first N terms
SN (x) =
(x) =
SN
=
=
=
=
=
=
78
2 inx
πin e
N
N
ei(2m+1)x
4 sin(2m + 1)x
4
= Im
(11.12)
π
(2m + 1)
π
(2m + 1)
m=0
m=0
N
4
i(2m+1)x
(11.13)
ie
Im
π
m=0
N
4
m
(11.14)
ei2x
Im ieix
π
m=0
N 1 + ei2x + · · · + ei2x
4
ix
i2x
(1 − e ) (11.15)
Im ie
π
1 − ei2x
1 − ei2(N +1)x
4
ix
(11.16)
Im ie
π
1 − ei2x
4
1 − ei2(N +1)x
(11.17)
Im i
π
eix − eix
2
ei2(N +1)x − 1
Im
(11.18)
π
sin x
2 sin 2(N + 1)x
.
π
sin x
(11.19)
11.3. NOW CONSIDER THE SUM OF THE FIRST N TERMS
Therefore
t = 2(N + 1)u
SN (x) =
2
π
x
0
sin 2(N +1)u
sin u
du 2
π
2(N+1)x
0
du =
sin
t
dt
2(N +1)
(11.20)
t dt
(2/π) sin(2(N+1)x)/sin(x)
10
5
0
−5
−10
0
0.5
1
x/π
1.5
2
Figure 11.4: (2/π)sin(2(N + 1)x)/sin(x) for N = 5
2 sin 2(N + 1)x
= 0 when 2(N + 1)xN = π thus the
π
sin x
maximum value of SN (x) occurs at
Observe SN
(x) =
xN =
π
2(N + 1)
(11.21)
79
0
(2/π) ∫x sin 2(N+1)u /sin u du for N= 5
Lecture 15 - Convergence of Fourier Series
1.5
1
0.5
0
−0.5
−1
−1.5
0
0.5
1
x/π
1.5
2
Figure 11.5: Integral of (2/π)sin(2(N + 1)x)/sin(x)
80