Fourier series

Transcription

1 Fourier series Yurii Lyubarskii, NTNU September 5, 2016

2 Periodic functions Function f defined on the whole real axis has period p if Properties f (t) = f (t + p) for all t R If f and g have period p then so does their linear combinations af (x + p) + bg(x + p) = af (x) + bg(x). f (t + np) = f (t) for all integer n Basic period i the smallest possible period for a given function f (t) has period p g(t) = f (ct) has period p/c. Therefore it suffices to study the functions of some fixed period. Then we can make scaling. We will deal mainly with functions of period 2π. A periodic function with period p can be reconstructed from its values on a segment of length p

3 Examples In nature - Pendulum - Wave motion - Cristal structures - Sound waves. In mathematics - 1, sin nt, cos nt n = 1, 2, 3,... - trigonometric system - e int, n = 0, 1, 3,... - exponential system They are related by the Euler formulas or e iα = cos α + i sin α sin α = eiα e iα, cos α = eiα + e iα 2i 2 Refresh your knowledge on complex numbers!!

4 Basic fact Each 2π-periodic function f can be represented as a series in trigonometrical or in exponential system Trigonometrical Fourier series: f (t) = a 0 + ( an cos nt + b n sin nt ) n=1 Exponential (complex) Fourier series: f (t) = c n e int.

5 Basic question Q1: How does one find the coefficients a n, b n or c n? Q2: When and in what sense does the Fourier series converge to the function f? We answer these question starting from the trigonometrical Fourier series

6 Q1: Coefficients of the trigonometrical Fourier series Basic formulas: Let Then f (t) = a 0 + ( an cos nt + b n sin nt ) n=1 a 0 = 1 2π a n = 1 π b n = 1 π Idea of the proof: ORTHOGONALITY. f (t)dt f (t) cos ntdt f (t) sin ntdt

7 Orthogonality Model: x R 3, x = x 1 e 1 + x 2 e 2 + x 3 e 3. Then x j = x, e j. This is because { 0, l j e j, e l = 1, l = j. Ortogonality relations for Fourier series: cos nx cos mx dx = 0, m m; sin nx cos mx dx = 0; sin nx sin mx dx = 0, m m; cos 2 nx dx = 0 = sin 2 nxdx = π. Now it is easy to obtain expressions for the coefficients...

8 Fourier series Given a 2π-periodic function f, consider its Fourier series ( S f (x) = a 0 + an cos nx + b n sin nx ) n=1 where a 0 = 1 2π a n = 1 π b n = 1 π f (x)dx f (x) cos nxdx f (x) sin nxdx

9 Rectangular wave Example Let f (x) = { 1, < x < 0 1, 0 < x < π.

10 Rectangular wave Example Let f (x) = { 1, < x < 0 1, 0 < x < π Then S f (x) = 4 (sin x + 13 π sin 3x + 15 ) sin 5x +... = 4 1 sin(2n 1)x π 2n 1. n=1

11 Q2: Convergence Precise statement: Theorem Let f be a piece-wise continuous function on [, π] (it can be extended 2π-periodically). Suppose that f has left and right derivatives at each point. Then the Fourier series S f converges at each point and S f (x) = f (x 0) + f (x + 0), where f (x ± 0) = lim h 0+ f (x ± h). Informal statement: If f is not very bad continuous 2π periodic function, than its Fourier series convergent. Attention: we need continuity on the whole real axis.

12 Further examples Example Let f (x) = x, x π and f is extended to a 2π-periodic function. Find the Fourier series of f.

13 Further examples Example Let f (x) = x, x π and f is extended to a 2π-periodic function. Find the Fourier series of f. a 0 = 1 2π x dx = 1 π 0 x dx = 1 π π 2 2 = π 2

14 Further examples Example Let f (x) = x, x π and f is extended to a 2π-periodic function. Find the Fourier series of f. a n = 1 π a 0 = 1 2π x dx = 1 π x cos nx dx = 2 π 0 0 x dx = 1 π π 2 2 = π 2 x cos nx dx =... = 2 cos nx n 2 π x=π x=0

15 Further examples Example Let f (x) = x, x π and f is extended to a 2π-periodic function. Find the Fourier series of f. a n = 1 π a 0 = 1 2π x dx = 1 π 0 x dx = 1 π π 2 2 = π 2 x cos nx dx = 2 π 0 x cos nx dx =... = { a n = 4 n 2 π, n odd 0, n even, b n = 0 S f (x) = π 2 4 π n=1 cos(2n 1)x (2n 1) 2 2 cos nx n 2 π x=π x=0

16 Further examples Example f (x) = { t + π, < t < 0, t π, 0 < t < π. Attention: you may evaluate Fourier coefficients along any interval whose length is the period. Sometimes It may be convenient to chose appropriate interval. Dirty exam trick: Find the Fourier series for f (x) = (cos x + sin x) 2

17 Odd and even functions f is odd if f (x) = f ( x). Fact: f is odd all a n = 0 (cos coefficients) f is even if f (x) = f ( x). Fact: f is even all b n = 0 (sin coefficients) Given a function f on (0, π) we can prolongate it either as odd or as even 2π periodic function expand either as sin or cos Fourier series. Which of them is true? Half range expansions.

18 Further exercises Exercises: 1. Produce formulas for Fourier coefficients for half range expansions 2. Write basic formulas for the Fourier series for functions with period with period 2L.