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!)

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