Math 21b Fourier Series 2 p 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 1 1 a0 X X p + 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 = p 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 1 1 2 2 X 2 X 2 kfk = a0 + ak + bk k=1 k=1 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 . P1 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 p ; 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 1 1 a0 X 2πmx X 2πmx f(x) = p + 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 C1. This let us solve differential equations by writing solutions as linear combinations of these eigenfunctions. p The functions 1= 2, cos(kt) and sin(kt) (where k is an integer) are the eigenfunctions for the operator D2 1 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 basisp Fourier basis ~u1; : : : ; ~un 1= 2, cos(kt), sin(kt) for k ≥ 1 Basis representationp Fourier series P1 ~v = c1~u1 + ··· + cn~un f(x) = a0= 2 + k=1 (ak cos(kx) + bk sin(kx)) Coordinatesp 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 P1 2 2 k~vk = c1 + ··· + cn kfk = a0 + k=1 ak + bk.