FOURIER SERIES
JIA-MING (FRANK) LIOU
1. Introduction A trigonometric polynomial is a finite sum
(1.1) t(x) = a0
2 +
n
X
k=1
{akcoskx+bksinkx},
wherea0, a1,· · · , anand b1,· · · , bnare real numbers. A trigonometric polynomial defines a continuous periodic function of period 2π onR,i.e. t(x) is a continuous function onRwith t(x+ 2π) =t(x) for allx∈R.
For each pair of integer (k, m),we set δk,m=
(1 ifk=m, 0 otherwise.
Lemma 1.1. Let k, mbe nonnegative integers. For allk, m (1)
Z 2π
0
coskxcosmxdx=δk,mπ, (2)
Z 2π
0
sinkxsinmxdx=δk,mπ, (3)
Z 2π
0
coskxsinmxdx= 0.
Proof. We leave it to the reader as an exercise.
Given a trigonometric polynomialt(x) of the form (1.1), using the fact that, Z 2π
0
coskxdx= Z 2π
0
sinkxdx= 0, k≥1, we obtain
Z 2π
0
t(x)dx= Z 2π
0
a0 2dx+
n
X
k=1
{ak Z 2π
0
coskxdx+bk Z 2π
0
sinkxdx}
=a0π.
This implies that
a0= 1 π
Z 2π
0
t(x)dx.
We can also compute Z 2π
0
t(x) cosmxdx= Z 2π
0
a0
2 cosmxdx+
n
X
k=1
{ak Z 2π
0
coskxcosmxdx+bk
Z 2π
0
sinkxcosmxdx}
=amπ
1
2 JIA-MING (FRANK) LIOU
and Z 2π
0
t(x) sinmxdx= Z 2π
0
a0
2 sinmxdx+
n
X
k=1
{ak Z 2π
0
coskxsinmxdx+bk
Z 2π
0
sinkxsinmxdx}
=bmπ.
These give us
am= 1 π
Z 2π
0
t(x) cosmxdx, bm = 1 π
Z 2π
0
t(x) sinmxdx.
Fourier thought that all the (continuous) period functions are infinite sum of trigonometric functions:
(1.2) a0
2 +
∞
X
n=1
{ancosnx+bnsinnx}.
Using the similar observation, he defined the Fourier coefficients a0, a1,· · · ,and b1, b2,· · · of a periodic functiony=f(x) by
(1) a0= 1 π
Z 2π
0
f(x)dx, (2) an= 1
π Z 2π
0
f(x) cosnxdx, (3) bn= 1
π Z 2π
0
f(x) sinnxdx.
The infinite series (1.2) is called the Fourier expansion of the functionf(x); we also use the notation
f(x)∼ a0 2 +
∞
X
n=1
{ancosnx+bnsinnx}
to say that (1.2) is the Fourier series fory=f(x).
Example 1.1. Lety=f(x) be a periodic function of period 2π so thatf(x) =xon [0,2π].
Then the Fourier series is given by
(1.3) x∼ 2π
2 +
∞
X
n=1
−2 n
sinnx.
Now, let us consider another trigonometric polynomial s(x) = α0
2 +
n
X
j=1
{αjcosjx+βjsinjx}.
We compute the integral Z 2π
0
t(x)s(x)dx= Z 2π
0
a0α0 4 dx+
n
X
j=1
a0αj 2
Z 2π
0
cosjxdx+
n
X
j=1
a0βj 2
Z 2π
0
sinjxdx
+
n
X
k=1
α0ak
2 Z 2π
0
coskxdx+
n
X
k=1
α0bk
2 Z 2π
0
sinkxdx
+
n
X
k=1 n
X
j=1
Z 2π
0
(akcoskx+bksinkx)(αjcosjx+βjsinjx)dx
FOURIER SERIES 3
= a0α0π
2 +
n
X
k=1 n
X
j=1
akαj Z 2π
0
coskxcosjxdx+
n
X
k=1 n
X
j=1
akβj Z 2π
0
coskxsinjxdx
+
n
X
k=1 n
X
j=1
bkαj Z 2π
0
sinkxcosjxdx+
n
X
k=1 n
X
j=1
bkβj Z 2π
0
sinkxsinjxdx
= a0α0π 2 +π
n
X
k=1
{akαk+bkβk}.
This shows that
1 π
Z 2π
0
t(x)s(x)dx= a0α0
2 +
n
X
k=1
{akαk+bkβk}.
When t(x) =s(x),we obtain 1 π
Z 2π
0
t(x)2dx= a20 2 +
n
X
k=1
{a2k+b2k}.
These observation motivates the following theorem.
Theorem 1.1. (Parseval Identity) Let us assume thatf(x) andg(x) are periodic (continu- ous) functions of period 2π.Suppose that the Fourier expansions off(x) andg(x) are given by
f(x)∼ a0
2 +
∞
X
n=1
{ancosnx+bnsinnx}, g(x)∼ α0
2 +
∞
X
n=1
{αncosnx+βnsinnx}.
Then
1 π
Z 2π
0
f(x)g(x)dx= a0α0
2 +
∞
X
n=1
{anαn+bnβn}.
This also leads to
Corollary 1.1. Letf(x) be as above. Then 1
π Z 2π
0
f(x)2dx= a20 2 +
∞
X
n=1
{a2n+b2n}.
Letsbe a real number and define
ζ(s) =
∞
X
n=1
1 ns.
By p-test, the infinite series ζ(s) converges when s > 1. The function ζ(s) is called the Riemann zeta function. By (1.3) and the Parseval identity, we obtain
1 π
Z 2π
0
x2dx= (2π)2
2 +
∞
X
n=1
−2 n
2
.
The identity Z 2π
0
x2dx= 8π3
3 implies that 8π2
3 = 2π2+ 4
∞
X
n=1
1 n2.
4 JIA-MING (FRANK) LIOU
This shows that
X
n=1
1 n2 = π2
6 . In other words, ζ(2) = π2
6 .
Example 1.2. The Fourier expansion off(x) =x2,0≤x≤2π, is given by x2∼
8π2 3
2 +
∞
X
n=1
− 4 n2
cosnx+
−4π n
sinnx
.
Using the Parseval identity, we obtain 1
π Z 2π
0
x4dx= 8π2
3
2
2 +
∞
X
n=1
16 n4 +
∞
X
n=1
16π2 n2 . Using the above identity and ζ(2) = π2
6 ,we obtainζ(4) :
∞
X
n=1
1 n4 = π4
90.
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