10.2-3: Fourier Series.

Transcription

1 10.2-3: Fourier Series.

2 10.2-3: Fourier Series. O. Costin: Fourier Series,

3 1 Fourier series are very useful in representing periodic functions. Examples of periodic functions. A function is periodic with period T if f(t + T) = f(t) for any t. The fundamental period is th smallest T with this property K x K0.4 K0.6 K0.8 K1.0 K1.2 }{{} T

4 2 The simplest periodic functions are, perhaps, sin t and cos t, with period 2π.

5 2 The simplest periodic functions are, perhaps, sin t and cos t, with period 2π. In a good sense, any periodic function can be written in terms of sin t and cos t, as a finite or infinite sum: Fourier series.

6 2 The simplest periodic functions are, perhaps, sin t and cos t, with period 2π. In a good sense, any periodic function can be written in terms of sin t and cos t, as a finite or infinite sum: Fourier series. A general Fourier series is f(x) = a m=1 a m cos mπx The period is πt/ = 2π, T = 2. + b m sin mπx

7 3 f(x) = a m=1 a m cos mπx + b m sin mπx

8 3 f(x) = a m=1 a m cos mπx The period is πt/ = 2π, T = 2. + b m sin mπx (1)

9 3 f(x) = a m=1 a m cos mπx + b m sin mπx The period is πt/ = 2π, T = 2. Example: The oscillations of the string of a string instrument is composed of the fundamental note (frequency) and multiples of it (harmonics). (1)

10 3 f(x) = a m=1 a m cos mπx + b m sin mπx The period is πt/ = 2π, T = 2. Example: The oscillations of the string of a string instrument is composed of the fundamental note (frequency) and multiples of it (harmonics). These are sines and cosines, and the full decomposition is given by an expression like (??). (1)

11 4

12 4 Given a function F, how do we find the coefficients a m, b m? f(x) = a m=1 a m cos mπx + b m sin mπx

13 4 Given a function F, how do we find the coefficients a m, b m? f(x) = a m=1 a m cos mπx + b m sin mπx This expansion is an expansion in terms of infinitely many independent functions, cos mπx, sin mπx or infinitely many linearly independent vectors.

14 5 What if we have finite expansion in terms of finitely many orthogonal vectors of magnitude one, v 1, v 2, v 3 :,

15 5 What if we have finite expansion in terms of finitely many orthogonal vectors of magnitude one, v 1, v 2, v 3 :, x = a 1 v 1 + a 2 v 2 + a 3 v 3?

16 5 What if we have finite expansion in terms of finitely many orthogonal vectors of magnitude one, v 1, v 2, v 3 :, x = a 1 v 1 + a 2 v 2 + a 3 v 3? Take scalar product with say v 2 : x v 2

17 5 What if we have finite expansion in terms of finitely many orthogonal vectors of magnitude one, v 1, v 2, v 3 :, x = a 1 v 1 + a 2 v 2 + a 3 v 3? Take scalar product with say v 2 : x v 2 = a 1 v 1 v 2 }{{} =0 +a 2 v 2 v 2 }{{} =1 +a 3 v 3 v 2 }{{} =0

18 5 What if we have finite expansion in terms of finitely many orthogonal vectors of magnitude one, v 1, v 2, v 3 :, x = a 1 v 1 + a 2 v 2 + a 3 v 3? Take scalar product with say v 2 : x v 2 = a 1 v 1 v 2 }{{} =0 +a 2 v 2 v 2 }{{} =1 +a 3 v 3 v 2 }{{} =0 = a 2!

19 5 What if we have finite expansion in terms of finitely many orthogonal vectors of magnitude one, v 1, v 2, v 3 :, x = a 1 v 1 + a 2 v 2 + a 3 v 3? Take scalar product with say v 2 : x v 2 = a 1 v 1 v 2 }{{} =0 +a 2 v 2 v 2 }{{} =1 +a 3 v 3 v 2 }{{} =0 = a 2! Is there a scalar product for functions?

20 5 What if we have finite expansion in terms of finitely many orthogonal vectors of magnitude one, v 1, v 2, v 3 :, x = a 1 v 1 + a 2 v 2 + a 3 v 3? Take scalar product with say v 2 : x v 2 = a 1 v 1 v 2 }{{} =0 +a 2 v 2 v 2 }{{} =1 +a 3 v 3 v 2 }{{} =0 = a 2! Is there a scalar product for functions? Yes. A function can be thought of as a vector with infinitely many components, f(x) = f x ; x R

21 5 What if we have finite expansion in terms of finitely many orthogonal vectors of magnitude one, v 1, v 2, v 3 :, x = a 1 v 1 + a 2 v 2 + a 3 v 3? Take scalar product with say v 2 : x v 2 = a 1 v 1 v 2 }{{} =0 +a 2 v 2 v 2 }{{} =1 +a 3 v 3 v 2 }{{} =0 = a 2! Is there a scalar product for functions? Yes. A function can be thought of as a vector with infinitely many components, f(x) = f x ; x R. The scalar product x y = (x 1, x 2, x 3 ) (y 1, y 2, y 3 ) =

22 5 What if we have finite expansion in terms of finitely many orthogonal vectors of magnitude one, v 1, v 2, v 3 :, x = a 1 v 1 + a 2 v 2 + a 3 v 3? Take scalar product with say v 2 : x v 2 = a 1 v 1 v 2 }{{} =0 +a 2 v 2 v 2 }{{} =1 +a 3 v 3 v 2 }{{} =0 = a 2! Is there a scalar product for functions? Yes. A function can be thought of as a vector with infinitely many components, f(x) = f x ; x R. The scalar product x y = (x 1, x 2, x 3 ) (y 1, y 2, y 3 ) = x 1 y 1 + x 2 y 2 + x 3 y 3

23 6 (x, y) = x y = (x 1, x 2, x 3 ) (y 1, y 2, y 3 ) =

24 6 (x, y) = x y = (x 1, x 2, x 3 ) (y 1, y 2, y 3 ) = x 1 y 1 + x 2 y 2 + x 3 y 3 ikewise, (f(x), g(x) = x f xg x

25 6 (x, y) = x y = (x 1, x 2, x 3 ) (y 1, y 2, y 3 ) = x 1 y 1 + x 2 y 2 + x 3 y 3 ikewise, (f(x), g(x) = x f xg x What should be?

26 6 (x, y) = x y = (x 1, x 2, x 3 ) (y 1, y 2, y 3 ) = x 1 y 1 + x 2 y 2 + x 3 y 3 ikewise, (f(x), g(x) = x f xg x What should be?, of course. Definition (u, v) = β α u(x)v(x)dx

27 7 (u, v) = β α u(x)v(x)dx

28 8 Proposition 1. If α = and β =, then we have (sin mπx nπx, cos ) = 0 m, n

29 8 Proposition 1. If α = and β =, then we have (sin mπx, cos nπx ) = 0 m, n (2) (sin mπx nπx, sin ) = 0 m n

30 8 Proposition 1. If α = and β =, then we have (sin mπx (sin mπx (cos mπx, cos nπx ) = 0 m, n (2), sin nπx ) = 0 m n (3) nπx, cos ) = 0 m n

31 8 Proposition 1. If α = and β =, then we have (sin mπx (sin mπx, cos nπx ) = 0 m, n (2), sin nπx ) = 0 m n (3), cos nπx ) = 0 m n (4) (cos mπx (sin nπx nπx, sin ) =

32 8 Proposition 1. If α = and β =, then we have (sin mπx, cos nπx ) = 0 m, n (2) (sin mπx, sin nπx ) = 0 m n (3), cos nπx ) = 0 m n (4) (sin nπx nπx, sin ) = (5) (cos mπx (cos nπx nπx, cos ) =

33 8 Proposition 1. If α = and β =, then we have (sin mπx, cos nπx ) = 0 m, n (2) (sin mπx, sin nπx ) = 0 m n (3) (cos mπx, cos nπx ) = 0 m n (4) (sin nπx nπx, sin ) = (5) (cos nπx nπx, cos ) = (6) These vectors are orthogonal, (7) any two distinct are!! (8)

34 9 Any two distinct sin nπx and/or cos nπx are orthogonal.

35 9 Any two distinct sin nπx and/or cos nπx are orthogonal. Proof: Just a calculation of trig integrals. For example, we have the formula: sin (ax) sin (bx) dx = sin ((a b) ) a b sin ((a + b) ) a + b

36 9 Any two distinct sin nπx and/or cos nπx are orthogonal. Proof: Just a calculation of trig integrals. For example, we have the formula: sin (ax) sin (bx) dx = sin ((a b) ) a b sin ((a + b) ) a + b where a = nπ, b = mπ, and thus the integral vanishes provided a b!. x v 2

37 9 Any two distinct sin nπx and/or cos nπx are orthogonal. Proof: Just a calculation of trig integrals. For example, we have the formula: sin (ax) sin (bx) dx = sin ((a b) ) a b sin ((a + b) ) a + b where a = nπ, b = mπ, and thus the integral vanishes provided a b!. x v 2 = a 2 Thus, if f(x) = a m=1 a m cos mπx + b m sin mπx

38 9 Any two distinct sin nπx and/or cos nπx are orthogonal. Proof: Just a calculation of trig integrals. For example, we have the formula: sin (ax) sin (bx) dx = sin ((a b) ) a b sin ((a + b) ) a + b where a = nπ, b = mπ, and thus the integral vanishes provided a b!. x v 2 = a 2 Thus, if f(x) = a m=1 a m cos mπx + b m sin mπx a m = (f, cos mπx )

39 10 What are the Fourier coefficients? f(x) = a m=1 a m cos mπx + b m sin mπx

40 10 What are the Fourier coefficients? f(x) = a m=1 a m cos mπx + b m sin mπx a m = (f, cos mπx )

41 10 What are the Fourier coefficients? f(x) = a m=1 a m cos mπx + b m sin mπx a m = (f, cos mπx ) 1. Find, so that the period is [, ].

42 10 What are the Fourier coefficients? f(x) = a m=1 a m cos mπx + b m sin mπx a m = (f, cos mπx ) 1. Find, so that the period is [, ]. 2. Then, a n = nπx f(x) cos dx

43 10 What are the Fourier coefficients? f(x) = a m=1 a m cos mπx + b m sin mπx a m = (f, cos mπx ) 1. Find, so that the period is [, ]. 2. Then, a n = 3. b n = nπx f(x) cos dx nπx f(x) sin dx

44 10 What are the Fourier coefficients? f(x) = a m=1 a m cos mπx + b m sin mπx a m = (f, cos mπx ) 1. Find, so that the period is [, ]. 2. Then, a n = 3. b n = nπx f(x) cos dx nπx f(x) sin dx

45 11 Examples. Consider the function 0 1 < x and x < 1/3 G(x) = 1 1/3 < x and x < 1/3 0 1/3 < x and x < 1

46 11 Examples. Consider the function 0 1 < x and x < 1/3 G(x) = 1 1/3 < x and x < 1/3 0 1/3 < x and x < 1 (extended periodically.)

47 11 Examples. Consider the function 0 1 < x and x < 1/3 G(x) = 1 1/3 < x and x < 1/3 0 1/3 < x and x < 1 (extended periodically.) K1.5 K1.0 K x

48 12 1. Find, so that the period is [, ]. Here, = 1

49 12 1. Find, so that the period is [, ]. Here, = 1 2. Then, a n = nπx f(x) cos dx a n = 1/3 1/3 cos nπxdx = 2 sin (nπ/3) ; a 0 = 2/3 nπ 3. b n = 1/3 1/3 cos nπxdx = 0

50 12 1. Find, so that the period is [, ]. Here, = 1 2. Then, a n = nπx f(x) cos dx a n = 1/3 1/3 cos nπxdx = 2 sin (nπ/3) ; a 0 = 2/3 nπ 3. b n = 1/3 1/3 cos nπxdx = 0 Maple 11 > plot(2/3 + sum(2 sin(n Pi/3)/n/Pi cos(n Pi x), n = 1..30), x = 1..1);

51 K1.0 K x K1.0 K x K1.0 K x