Lecture 13. Fourier Series I. February 10, 2014
A Fourier series for the function f : [0, 2π] → R is an expression of the form
f (x) =
a0
+ a1 cos x + b1 sin x + a2 cos 2x + b2 sin 2x + . . .
2
How to find coefficients an , bn ? Two functions f, g : [0, 2π] → R are called orthogonal on [0, 2π]
if
Z 2π
f (x)g(x)dx = 0.
0
And it turns out that the functions
1
, cos x, sin x, cos 2x, sin 2x, . . .
2
are all orthogonal on [0, 2π]. For example,
x=2π
Z 2π
1
1
cos xdx = sin x
= 0,
2
2
0
x=0
x=2π
Z 2pi
Z 2π
1
1
cos x sin xdx =
sin 2xdx = − cos 2x = 0.
2
4
0
0
x=0
The same can be done for any two functions. So these functions form an orthogonal basis. Let
us find, for example, b1 . Multiply by sin x:
f (x) sin x =
a0
sin x + a1 cos x sin x + b1 sin2 x + a2 cos 2x sin x + b2 sin 2x sin x + . . .
2
Integrate from 0 to 2π. All terms vanish, except sin2 x. We have:
Z 2π
Z 2π
f (x) sin xdx = b1
sin2 xdx.
0
But
Z
2π
2
0
2π
Z
sin xdx =
0
0
1 − cos 2x
dx =
2
x=2π
x sin 2x −
= π.
2
4
x=0
Therefore,
1
b1 =
π
Z
2π
f (x) sin xdx.
0
Similarly,
Z
Z
1 2π
1 2π
an =
f (x) cos(nx)dx, bn =
f (x) sin(nx)dx, n ≥ 1.
π 0
π 0
Example. Let χA : R → R be a characteristic function of a subset A ⊆ R:
(
1, x ∈ A;
χA (x) =
0, else
Let us find the Fourier series for f (x) = χ(0,π) .
t=π
Z
Z
1 π
1 π
1 1
an =
f (t) cos ntdt =
cos ntdt =
sin nt
= 0, n ≥ 1,
π −π
π 0
π n
t=0
and
But
1
a0 =
π
Z
π
f (t)dt =
−π
1
π = 1.
π
t=π
1
1
sin ntdt =
− cos nt π
n
0
t=0
(
1
1
0, n = 2k,
(1 − (−1)n ) = 2
= − (cos πn − 1) =
πn
πn
, n = 2k + 1
πn
1
bn =
π
Z
π
1
f (t) sin ntdt =
π
−π
Z
π
So
1 2
f (t) = +
2 π
1
1
sin t + sin 3t + sin 5t + . . .
3
5
Example. For f (x) = cos2 x, we have: f (x) = 12 + 21 cos 2x. We decomposed it into Fourier
series simply by using trig identities, without calculating integrals.
Convergence. If f is continuous at x, then the Fourier series converges to f (x) at x. If
f is discontinuous at x, then it converges to (f (x−) + f (x+))/2, where f (x−) is the limit of
f (t) when t → x − 0 (t approaches x from the left), and f (x+) is the limit of f (t), t → x + 0).
For example, the first series: f (x) is discontinuous at x = π, and f (π−) = 1, f (π+) = 0. The
Fourier series, if you plug x = π, becones 1/2, so it converges to the half-sum of 0 and 1. (And
converges very quickly - already after the first term it is exactly 1/2!)