Math 21b Fourier Series 2 √ Recap. The functions 1/ 2, cos(x), sin(x), cos(2x), sin(2x),... are an orthonormal basis for the the space of 2π-periodic functions (or equivalently the space of functions f with domain [−π, π] satisfying f(−π) = f(π)).

The Fourier series of f is ∞ ∞ a0 X X √ + ak cos(kx) + bk sin(kx) 2 k=1 k=1 where the Fourier coefficients are 1 Z π 1 1 Z π 1 Z π a0 = √ f(t)dt ak = cos(kt)f(t)dt bk = sin(kt)f(t)dt π −π 2 π −π π −π A function f is equal to its Fourier series at the points on (−π, π) where f is continuous.

Computation Tricks. A function f is even if f(−x) = f(x) (symmetric about the y-axis) and is odd if f(−x) = −f(x) (symmetric about the origin). 1. Even and odd functions multiply like even and odd integers add: the product of two even functions is even, the product of two odd functions is even, and the product of an even and an odd function is odd.

2. If f is even then all of the Fourier coefficients bk are 0.

3. If f is odd then all of the Fourier coefficients a0, ak are 0.

4.A trigonometric polynomial is a function f which can be written in the form f(x) = a0 + a1 cos x + b1 sin x + ··· + an cos(nx) + bn sin(nx). Any trigonometric polynomial is its own Fourier series.

n 2 Example. Suppose ~u1, . . . , ~un is an orthonormal basis for R . If ~v = c1~u1 + ··· + cn~un then what is k~vk ? Parseval’s Identity. The length of f is related to it’s Fourier coefficients by

∞ ∞ 2 2 X 2 X 2 kfk = a0 + ak + bk k=1 k=1

∞ X 2(−1)k+1 Example. Last time we saw that the Fourier series for f(x) = x is given by sin(kx) k k=1 1. What does Parseval’s identity say here?

π 2. Use the Fourier series of to find a formula for 2 . P∞ 1 Example. The function ζ(s) = k=1 ks is called the Riemann zeta function and is important in number 1 2 π4 theory. Use Parseval’s identity and the Fourier series for x to show ζ(4) = 90 .

1The Riemann zeta function is related to prime numbers by the infinite product 1 1 1 1 ζ(s) = · · ··· ··· (p prime) 1 − 2−1 1 − 3−1 1 − 5−1 1 − p−1 A big unsolved problem is the Riemann Hypothesis, which conjectures that ζ has zeroes only at negative even integers and complex numbers with real part 1/2. The problem is so notoriously hard to solve that there is currently a $1,000,000 prize for a valid solution. Fourier Series in General. In this course we are only talking about Fourier series for functions defined on [−π, π], but the same idea works for functions defined on any interval [a, b]. Set L = b − a.

1. The rule 2 Z b hf, gi = f(t)g(t)dt L a is an inner product on C[a, b] the space of (piecewise continuous) functions with domain [a, b]. 2. The functions 1 2πmx 2πmx √ , cos , sin (m = 1, 2,... ) 2 L L form an orthonormal basis for C[a, b].

3. Any function f :[a, b] → R equals its Fourier series ∞   ∞   a0 X 2πmx X 2πmx f(x) = √ + a cos + b sin k L k L 2 k=1 k=1

at the points in (a, b) where f is continuous.

Fourier Series and DEs. We saw before that the functions eλt (where λ is any real number) gave an eigen- basis for the differential operator D on C∞. This let us solve differential equations by writing solutions as linear combinations of these eigenfunctions. √ The functions 1/ 2, cos(kt) and sin(kt) (where k is an integer) are the eigenfunctions for the operator D2 ∞ on Cper, the periodic functions. A Fourier series is just expressing a function in terms of an eigenbasis for the differential operator D2.

A Table of Analogies. The following table summarizes how functions and Fourier series are analogous to vectors and orthonormal bases. Vectors Functions Dot product Inner product 1 R π ~v · ~w = v1w1 + ··· + vnwn hf, gi = π −π f(t)g(t)dt

Orthonormal basis√ Fourier basis ~u1, . . . , ~un 1/ 2, cos(kt), sin(kt) for k ≥ 1

Basis representation√ Fourier series P∞ ~v = c1~u1 + ··· + cn~un f(x) = a0/ 2 + k=1 (ak cos(kx) + bk sin(kx))

Coordinates√ Fourier coefficients ci = ~v · ~ui a0 = hf, 1/ 2i, ak = hf, cos(kt)i, bk = hf, sin(kt)i

Pythagorean Theorem Parseval’s identity 2 2 2 2 2 P∞ 2 2
k~vk = c1 + ··· + cn kfk = a0 + k=1 ak + bk