c pHabala 2007
MA3 lecture 13
MA3: Lectures content, week 13
27. Fourier series
Remark: If f is T -periodic, then also its derivative f 0 is T -periodic, but this is not true for its
Rt
RT
antiderivative F (t) = f (u) du. This one is T -periodic if f (u) du = 0, i.e. a0 = 0.
0
0
Theorem.
Let f be a T -periodic function that is piecewise continuous on [0, T ) and it has a piecewise
∞ P
ak cos(kωt) + bk sin(kωt) .
continuous derivative on [0, T ). Let f ∼ a20 +
a0
2 +
(i) Then f 0 ∼
(ii) If
RT
∞ P
k=1
−ak sin(kωt)kω+bk cos(kωt)kω =
k=1
∞ P
a0
+
b
kω
cos(kωt)−a
kω
sin(kωt)
.
k
k
2
k=1
f (u) du = 0 (that is, a0 = 0), then
0
F (t) =
Rt
f (u) du ∼
∞ ∞ P
P
−bk
1
1
− bk cos(kωt) kω
=
ak sin(kωt) kω
kω cos(kωt) +
k=1
0
k=1
ak
kω
sin(kωt) .
Remark: If we know that Fourier series converges to f on some interval [a, b], then we unfortunately cannot claim that the convergence is uniform there. At the ends of the interval we have
the so-called Gibbs problem.
Definition.
∞ p
P
ak cos(kωt) + bk sin(kωt) . Denote Ak = a2k + b2k , find ϕk so that
Let f ∼ a20 +
k=1
bk = Ak cos(ϕk ) and ak = Ak sin(ϕk ). Then f ∼
a0
2
+
∞
P
Ak sin(kωt + ϕk ).
k=1
This series is called the Fourier series in amplitude-phase form.
Denote c0 =
a0
2
a ck = 21 (ak − j bk ), c−k = 12 (ak + j bk ) for k ∈ IN . Then f ∼
∞
P
ck ejkωt .
k=−∞
This series is called the Fourier series in complex form.
Remark ϕk = arctg abkk , or ϕk = arccotg
coordinates to polar.
Fact.
We have ck =
1
T
RT
bk
ak
, or some shifts, see transformation of Cartesian
f (t)e−jkωt dt.
0
28. Application of series
Example.
1, t ∈ [2k, 2k + 1);
00
y +y =
0, t ∈ [2k − 1, 2k).
Expand the right hand-side f =
1
2
+
∞
P
k=1
1−(−1)k
kπ
sin(kπt).
We assume that a solution can be found of the form y =
We substitute into the equation and obtain
∞
P
a0
+
[ak (1 − k 2 π 2 ) cos(kπt) + bk (1 − k 2 π 2 ) sin(kπt)] =
2
k=1
1
a0
2
+
1
2
+
∞
P
[ak cos(kπt) + bk sin(kπt)].
k=1
∞
P
k=1
1−(−1)k
kπ
sin(kπt).
c pHabala 2007
MA3 lecture 13
k
1−(−1)
Comparing both sides we get a0 = 1, ak = 0 for k ≥ 1 and bk = kπ(1−k
2 π2 )
∞
∞
k
P
P
1−(−1)
2
1
and thus y = 12 +
kπ(1−k2 π 2 ) sin(kπt)] = 2 +
(2k+1)π(1−(2k+1)2 π 2 ) sin((2k + 1)πt)].
k=1
k=0
Example.
y 00 − x3 y = 24x2 around x0 = 0.
On the right we have a power series with x0 = 0, we will try to find a solution of the form
∞
P
y=
ak xk .
k=0
∞
P
We substitute into the equation and get
∞
P
2a2 + 6a3 x1 2a4 x2 +
∞
P
ak k(k − 1)xk−2 −
k=2
ak xk+3 = 24x2 , hence
k=0
[ak+2 (k + 1)(k + 2) − ak−3 ]xk = 24x2 .
k=3
Comparing both sides we get a2 = 0, a3 = 0, a4 = 2, and equations 3·5a5 −a0 = 0, 5·6a6 −a1 = 0,
6 · 7a7 − a2 = 0, 7 · 8a8 − a3 = 0, 8 · 9a9 − a4 = 0, 9 · 10a10 − a5 = 0, 10 · 11a11 − a6 = 0, etc.
b
1
a
b
a
We choose a0 = a, a1 = b, then a5 = 20
, a6 = 30
, a7 = a8 = 0, a9 = 36
, a10 = 1800
, a11 = 3300
,
a12 = a13 = 0, etc.
a 5
b 6
1 9
a
b
and thus y(x) = a + bx + 2x4 + 20
x + 30
x + 36
x + 1800
x10 + 3300
x11 + . . .
Example.
1−cos(x) x2
x→0
lim
2
Example.
R 1
R
sin(x)
dx
=
x
=
4
1−[1− x2 + x4! −... ] x2
x→0
= lim
∞
P
k=0
1
x
∞
P
k=0
(−1)k
2k+1
(2k+1)! x
(−1)k
2k+1
(2k+1)!(2k+1) x
= lim
dx =
x2
2
x→0
∞
P
k=0
4
− x4! +... x2
(−1)k
(2k+1)!
+ C.
2
R
1
x→0 2
= lim
x2k dx =
−
∞
P
k=0
x2
4!
+ . . . = 21 .
(−1)k x2k+1
(2k+1)! 2k+1
+C