# MATH 461: Fourier Series and Boundary Value Problems

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 MATH 461: Fourier Series and Boundary Value Problems Chapter III: Fourier Series Greg Fasshauer Department of Applied Mathematics Illinois Institute of Technology Fall 2015 MATH 461 Chapter 3 1

2 Outline 1 Piecewise Smooth Functions and Periodic Extensions 2 Convergence of Fourier Series 3 Fourier Sine and Cosine Series 4 Term-by-Term Differentiation of Fourier Series 5 Integration of Fourier Series 6 Complex Form of Fourier Series MATH 461 Chapter 3 2

3 Piecewise Smooth Functions and Periodic Extensions Definition A function f, defined on [a, b], is piecewise continuous if it is continuous on [a, b] except at finitely many points. If both f and f are piecewise continuous, then f is called piecewise smooth. Remark This means that the graphs of f and f may have only finitely many finite jumps. MATH 461 Chapter 3 4

4 Piecewise Smooth Functions and Periodic Extensions Example The function f (x) = x defined on π < x < π is piecewise smooth since f is continuous throughout the interval, and f is discontinuous only at x = 0. Example The function { x 2, π < x < 0 f (x) = x 2 + 1, 0 x < π is piecewise smooth since both f and f are continuous except at x = 0. MATH 461 Chapter 3 5

5 Piecewise Smooth Functions and Periodic Extensions Example The function { ln(1 x), 0 x < 1 f (x) = 1, 1 x < 2 is not piecewise continuous (and therefore also not piecewise smooth) since lim x 1 x 1 f (x) = lim ln(1 x) =, i.e., f has an infinite jump at x = 1. MATH 461 Chapter 3 6

6 Piecewise Smooth Functions and Periodic Extensions Periodic Extension If f is defined on [, ], then its periodic extension, defined for all x, is given by. f (x + 2), 3 < x <, f (x), < x <, f (x) = f (x 2), < x < 3, f (x 4), 3 < x < 5,. MATH 461 Chapter 3 7

7 Piecewise Smooth Functions and Periodic Extensions Example Figure: Plot of f for f (x) = 1 x. MATH 461 Chapter 3 8

8 Piecewise Smooth Functions and Periodic Extensions Example Figure: Plot of f for f (x) = { (x + ) 2, x 0 x 2 + 1, 0 < x <. MATH 461 Chapter 3 9

9 Convergence of Fourier Series Even though we have used Fourier series to represent a given function f within our separation of variables approach, we have never made sure that these series actually converge. Moreover, even if we can assure convergence, how do we know that they converge to the function f? Remark This should not come as a total surprise, since for power series we also had to determine the interval (or radius) of convergence. MATH 461 Chapter 3 11

10 Convergence of Fourier Series Using a more precise notation, all we can say is f (x) a 0 + [ a n cos nπx + b n sin nπx ], i.e., we can associate with f this Fourier series, but not f is equal to this Fourier series. The Fourier coefficients of f, on the other hand, are never in doubt. They are given by a 0 = 1 2 a n = 1 b n = 1 f (x) dx f (x) cos nπx f (x) sin nπx dx, n = 1, 2,... dx, n = 1, 2,... MATH 461 Chapter 3 12

11 What we need is Convergence of Fourier Series Theorem (Fourier Convergence Theorem) If f is piecewise smooth on [, ], then the Fourier series of f converges. Moreover, 1 at those points x where the periodic extension f of f is continuous, the Fourier series of f converges to f (x) and 2 at jump discontinuities of the periodic extension, the Fourier series converges to 1 [ ] f (x ) + f (x+), 2 i.e., the average of the left and right limits at the jump. Remark Note that (2) actually includes (1) since 1 [ ] f (x ) + f (x+) 2 = 1 2 [ ] f (x) + f (x) = f (x) MATH 461 Chapter 3 13

12 Convergence of Fourier Series Proof. The proof of this theorem is not contained in [Haberman] and goes beyond the scope of this course. It can be found in [Pinsky, Section 1.2] or [Brown & Churchill, Section 19]. The proof requires the Dirichlet kernel D N (x) = N cos nx = sin ( N + 1 ) 2 x 2 sin x 2 as well as a careful analysis of one-sided derivatives. The calculations for Gibbs phenomenon below gives a flavor of this. MATH 461 Chapter 3 14

13 Convergence of Fourier Series Remark The theorem above is about pointwise convergence of Fourier series. In classical harmonic analysis there are also theorems about other kinds of convergence of Fourier series, such as uniform convergence or convergence in the mean. For these see, e.g., [Brown & Churchill, Pinsky]. We will talk about convergence in the mean in Chapter 5, and the Gibbs phenomenon below is evidence that uniform convergence is not guaranteed for general functions f. MATH 461 Chapter 3 15

14 Convergence of Fourier Series Example { 1, x < 0 Consider the function f (x) = 2, 0 < x [ The Fourier series of f, a 0 + a n cos nπx represented by the following graph: + b n sin nπx ], is MATH 461 Chapter 3 16

15 Convergence of Fourier Series Example (cont.) Remark Even if we know that the series converges, we have f (x) = its Fourier series only for x (, ) (and provided f is continuous at x). At all other values of x the Fourier series equals the periodic extension of f, except at jump discontinuities, where it equals the average jump. What are the Fourier coefficients for this example? MATH 461 Chapter 3 17

16 Convergence of Fourier Series Example (cont.) a 0 = 1 2 [ 0 = 1 2 f (x) dx 1 dx + = 1 2 [ + 2] = dx ] MATH 461 Chapter 3 18

17 Convergence of Fourier Series Example (cont.) a n = 1 f (x) cos nπx dx [ = 1 0 cos nπx dx + 2 [ = 1 cos nπx dx + }{{} = =0 [ sin nπx nπ ] 0 = cos nπx cos nπx dx dx ] ] MATH 461 Chapter 3 19

18 Convergence of Fourier Series Example ((cont.)) b n = 1 f (x) sin nπx dx [ = 1 0 sin nπx dx + 2 [ = 1 sin nπx dx + }{{} =0 = [ cos nπx nπ cos nπ = + cos 0 nπ nπ = 1 ( 1)n nπ = ] { 0, n even 2 nπ, n odd sin nπx sin nπx MATH 461 Chapter 3 20 dx dx ] ]

19 Convergence of Fourier Series Example (cont.) Summarizing, we have found that the function { 1, x < 0 f (x) = 2, 0 < x has Fourier series f (x) ( 1) n nπ = k=1 sin nπx 2 (2k 1)πx sin (2k 1)π MATH 461 Chapter 3 21

20 Convergence of Fourier Series Example (cont.) Figure: Plot of 10-term Fourier series approximation 10 f (x) = (2k 1)πx sin (red) together with graph of f (blue). (2k 1)π k=1 MATH 461 Chapter 3 22

21 Convergence of Fourier Series The Gibbs Phenomenon In order to understand the oscillations of the previous plots, and in particular the overshoot, we consider an almost identical function: { 1, π x < 0 f (x) = 1, 0 < x π with truncated Fourier series N 2 (1 ( 1) n ) f 2N (x) = f 2N 1 (x) = sin nx = nπ Remark N k=1 4 sin(2k 1)x π 2k 1 Compared to the previous example, the sines are simpler since = π, and we have a vertical shift by a 0 = 0 and a vertical stretching so that { b n = 2 1 ( 1)n 0, n even = nπ 4 nπ, n odd. MATH 461 Chapter 3 23

22 Convergence of Fourier Series Gibbs Phenomenon (cont.) To find the overshoot at the jump discontinuity we look at the zeros of the derivative of the truncated Fourier series (to locate its maxima), i.e., f 2N 1 (x) = 4 π N cos(2k 1)x = 4 [cos x + cos 3x cos(2n 1)x] π k=1 To find the zeros of this function we multiply both sides by sin x, i.e., sin xf 2N 1 (x) = 4 [sin x cos x + sin x cos 3x sin x cos(2n 1)x] π Using the trigonometric identity sin x cos kx = sin(k+1)x sin(k 1)x 2 we get sin xf 2N 1 (x) = 2 [(sin 2x sin 0) + (sin 4x sin 2x) (sin 2Nx sin(2n 2)x)] π = 2 sin 2Nx. π MATH 461 Chapter 3 24

23 Convergence of Fourier Series Gibbs Phenomenon (cont.) Now, sin 2Nx = 0 if 2Nx = ±π, ±2π,..., ±2Nπ. The maximum overshoot occurs at x = π 2N and its value is ( π ) f 2N 1 2N = 4 π = 2 π = 2 π N k=1 2N π N k=1 sin (2k 1)π 2N 2k 1 π N N k=1 sin (2k 1)π 2N (2k 1)π 2N sin (2k 1)π 2N 2k 1 π N. MATH 461 Chapter 3 25

24 Convergence of Fourier Series Gibbs Phenomenon (cont.) If we interpret N k=1 sin (2k 1)π 2N (2k 1)π 2N as a partial Riemann sum with x = π N and midpoints x = π 2N, 3π 2N,..., (2N 1)π 2N for the partition 0, π N, 2π N,..., (N 1)π N, π of [0, π] then ( π ) f 2N 1 2 π sin x dx for N. 2N π x This integral can be evaluated numerically to get ( π ) f 2N N 0 π N MATH 461 Chapter 3 26

25 Convergence of Fourier Series Gibbs Phenomenon (cont.) Since the actual size of the jump discontinuity is 2, we have an approximately 9% overshoot. This is true in general [Pinsky, p. 60]: Theorem If f is piecewise smooth on ( π, π) then the overshoot of the truncated Fourier series of f at a discontinuity x 0 (the Gibbs phenomenon) is approximately 9% of the jump, i.e., Remark 0.09 [f (x 0 +) f (x 0 )]. The Gibbs phenomenon was actually discovered by Henry Wilbraham in Gibbs was just more famous, published in a better journal (50 years later), and built in some mistakes perhaps drawing more attention to his work (for further discussion see [Trefethen, Chapter 9]). MATH 461 Chapter 3 27

26 Fourier Sine and Cosine Series We begin by reviewing the concepts of odd and even functions: Definition f is an odd function if f ( x) = f (x) for all x in the domain of f. Remark The graph of an odd function is symmetric about the origin. For an odd function we have f (x) dx = 0. f (x) dx = 0 = = 0 0 f (x) }{{} u = x du = dx f ( u) }{{} = f (u) dx + du + f (u) du f (x) dx f (x) dx f (x) dx = 0 MATH 461 Chapter 3 29

27 Fourier Sine and Cosine Series Definition f is an even function if f ( x) = f (x) for all x in the domain of f. Remark The graph of an even function is symmetric about the y-axis. For an even function we have f (x) dx = 2 0 f (x) dx, which can be shown similarly to the analogous property for odd functions. MATH 461 Chapter 3 30

28 Fourier Sine and Cosine Series et s consider the Fourier series of an odd function with Therefore, a 0 = 1 2 b n = 1 f (x) a 0 + f (x) dx = 0, }{{} odd f (x) }{{} odd sin nπx } {{ } odd } {{ } even f (x) [ a n cos nπx dx = 2 a n = b n sin nπx ] B n sin nπx, f (x) }{{} odd cos nπx } {{ } even } {{ } odd f (x) sin nπx dx(= B n) i.e., the Fourier series is automatically a Fourier sine series. dx = 0 MATH 461 Chapter 3 31

29 Fourier Sine and Cosine Series What does the Fourier sine series of f converge to? Theorem If f is piecewise smooth on [0, ], then the Fourier sine series of f converges. Moreover, 1 at those points x where the odd periodic extension of f is continuous, the Fourier sine series converges to the odd periodic extension and 2 at jump discontinuities of the odd periodic extension, the Fourier sine series converges to the average of the left and right limits at the jump. MATH 461 Chapter 3 32

30 Fourier Sine and Cosine Series Example Figure: Plot of odd periodic extension of f (x) = 1 x. MATH 461 Chapter 3 33

31 Fourier Sine and Cosine Series Remark Even if f is not an odd function, it may still be necessary to represent it by a Fourier sine series. Example The heat equation problem u t (x, t) = k 2 u (x, t), x 2 0 < x <, t > 0 u(0, t) = u(, t) = 0 u(x, 0) = cos πx has sines as eigenfunctions, so we need to find the Fourier sine series expansion of f (x) = cos πx. MATH 461 Chapter 3 34

32 Fourier Sine and Cosine Series Example (cont.) We know u(x, t) = ϕ(x)g(t), with eigenvalues λ n = ( ) nπ 2, n = 1, 2, 3,... and eigenfunctions as well as G n (t) = e λnkt. Therefore and u(x, t) = u(x, 0) = ϕ n (x) = sin nπx B n sin nπx nπ e k( ) 2 t B n sin nπx! = cos πx MATH 461 Chapter 3 35

33 Fourier Sine and Cosine Series Example (cont.) In HW 3.3.2a you should show a B n = 2 0 cos πx { nπx 0, n odd sin dx = n even 4n π(n 2 1), Therefore, letting n = 2k (even), we have cos πx = 8k 2kπx π(4k 2 sin 1) k=1 and the equality is true for 0 < x < (since the cosine equals its odd periodic extension there). For x = 0 and x = the series is zero (which is equal to the average jump of the cosine function there). a Remember that we established the orthogonality of sine and cosine only on [, ], not on [0, ] MATH 461 Chapter 3 36

34 Fourier Sine and Cosine Series Consider the Fourier series of an even function [ f (x) a 0 + a n cos nπx + b n sin nπx ] with Therefore, a 0 = 1 2 a n = 1 b n = 1 f (x) dx = 1 }{{} even f (x) cos nπx }{{ } even f (x) sin nπx }{{ } odd f (x) A dx = 2 dx = 0 f (x) dx(= A 0 ) 0 A n cos nπx, f (x) cos nπx dx(= A n) i.e., the Fourier series is automatically a Fourier cosine series. MATH 461 Chapter 3 37

35 Fourier Sine and Cosine Series Theorem If f is piecewise smooth on [0, ], then the Fourier cosine series of f converges. Moreover, 1 at those points x where the even periodic extension of f is continuous, the Fourier cosine series converges to the even periodic extension and 2 at jump discontinuities of the even periodic extension, the Fourier cosine series converges to the average of the left and right limits at the jump. Remark Note that jump discontinuities are possible only for 0 < x <, i.e., if f itself had jump discontinuities. The even periodic extension cannot have any jumps at x = 0 or x = ±. MATH 461 Chapter 3 38

36 Fourier Sine and Cosine Series Example Figure: Plot of even periodic extension of f (x) = x MATH 461 Chapter 3 39

37 Fourier Sine and Cosine Series Example Find the Fourier cosine series expansion of the function { 1, 0 x < f (x) = 2 0, 2 x and sketch its graph. nπx MATH 461 Chapter 3 40

38 Fourier Sine and Cosine Series Example (cont.) et s compute the Fourier cosine coefficients: A 0 = 1 f (x) dx 0 [ = 1 /2 ] 1 dx + 0 dx 0 /2 = 1 2 = 1 2 MATH 461 Chapter 3 41

39 Fourier Sine and Cosine Series Example (cont.) Therefore A n = 2 f (x) cos nπx 0 dx [ = 2 /2 cos nπx ] 0 dx + 0 cos nπx /2 dx = 2 [ sin nπx ] /2 nπ 0 { = 2 nπ sin nπ 0, n = 2k even 2 = 2 (2k 1)π ( 1)k+1, n = 2k 1 odd f (x) k=1 2( 1) k+1 (2k 1)πx cos (2k 1)π Note that equals = for all x [0, ] except x = 2. MATH 461 Chapter 3 42

40 Fourier Sine and Cosine Series Even and odd parts of functions Theorem Any function f can be written as the sum of an even and an odd function: Proof. Indeed, f e is even f (x) = 1 [f (x) + f ( x)] + 1 [f (x) f ( x)] } 2 {{}} 2 {{} :=f e(x) :=f o(x) and f o is odd f e ( x) = 1 2 [f ( x) + f (x)] = f e(x), f o ( x) = 1 2 [f ( x) f (x)] = 1 2 [f (x) f ( x)] = f o(x). MATH 461 Chapter 3 43

41 Fourier Sine and Cosine Series Now, the Fourier series for an arbitrary function f is given by f (x) a 0 + a n cos nπx + b n sin nπx with Fourier coefficients a 0 = 1 2 f (x)dx, a n = 1 On the other hand, } {{ } even } {{ } odd f (x) cos nπx dx, b n = 1 f (x) sin nπx dx. b n sin nπx is a Fourier sine series. However, it is the Fourier sine series of f o, not of f! Note that the Fourier sine series of f has coefficients B n = 2 0 f (x) sin nπx MATH 461 Chapter 3 44 dx.

42 Fourier Sine and Cosine Series We can make a similar observation for cosine. Therefore, [ a 0 + a n cos nπx + b n sin nπx ] }{{} Fourier series of f = a 0 + a n cos nπx }{{} cosine series of f e + b n sin nπx }{{} sine series of f o MATH 461 Chapter 3 45

43 Fourier Sine and Cosine Series Summary: Convergence of Fourier series et f be piecewise smooth. Then The Fourier series of f is continuous for all x f is continuous on [, ] (i.e., no jumps inside) and f ( ) = f () (i.e., no jumps at end). MATH 461 Chapter 3 46

44 Fourier Sine and Cosine Series The Fourier cosine series of f is continuous for all x f is continuous on [0, ]. MATH 461 Chapter 3 47

45 Fourier Sine and Cosine Series The Fourier sine series of f is continuous for all x f is continuous on [0, ] (i.e., no jumps inside) and f (0) = f () = 0 (i.e., no jumps at end). MATH 461 Chapter 3 48

46 Term-by-Term Differentiation of Fourier Series Recall that in HW 2.5.5c we had to deal with the boundary condition where u(r, θ) = u (1, θ) = f (θ), r B n r 2n sin 2nθ. In order to determine the coefficients b n we needed to differentiate the infinite series, i.e., find u r (r, θ) = 2nB n r 2n 1 sin 2nθ. Was this justified? Does this new series converge? If so, does it converge to u r (r, θ)? MATH 461 Chapter 3 50

47 Term-by-Term Differentiation of Fourier Series Example Consider the function f (x) = x, and find its Fourier sine series. Then, compare the termwise derivative of the series with the correct derivative f (x) = 1. We know x B n sin nπx with B n = 2 0 x sin nπx dx. MATH 461 Chapter 3 51

48 Term-by-Term Differentiation of Fourier Series Example (cont.) Using integration by parts we have B n = 2 [ = 2 = 2 = 2 0 x sin nπx x nπ [ 2 dx cos nπx cos nπ + 2 nπ ] [ 2 cos nπ nπ 0 (nπ) + cos nπx nπ 0 nπx ] sin 2 0 = 2 nπ ( 1)n dx ] Therefore, x 2 π ( 1) n+1 n sin nπx for which we know that equals = for 0 x <. MATH 461 Chapter 3 52 (1)

49 Term-by-Term Differentiation of Fourier Series Example (cont.) Now we consider the termwise derivative of the Fourier sine series i.e., x 2 π 2 π ( 1) n+1 n ( 1) n+1 = 2 n nπ sin nπx, cos nπx ( 1) n+1 cos nπx. Remark Note that this is a divergent series since the terms in the sequence do not approach zero for n, and therefore the series diverges by the standard test for divergence from calculus. MATH 461 Chapter 3 53

50 Term-by-Term Differentiation of Fourier Series Example (cont.) Obviously, we must conclude that 1 2 ( 1) n+1 cos nπx. In fact, the Fourier cosine series of f (x) = 1 is given by f (x) = 1 1, i.e., a 0 = 1, a n = 0 for n 1. Obviously, we could replace by =. MATH 461 Chapter 3 54

51 Term-by-Term Differentiation of Fourier Series Example (cont.) Why did this not work? What caused the trouble? Figure: Plot of f (x) = x for 0 < x <. MATH 461 Chapter 3 55

52 Term-by-Term Differentiation of Fourier Series Example (cont.) Figure: Plot of odd extension of f (x) = x. MATH 461 Chapter 3 56

53 Term-by-Term Differentiation of Fourier Series Example (cont.) Figure: Plot of odd periodic extension (actually, Fourier sine series) of f (x) = x. The jumps in the Fourier sine series at odd multiples of prevent the series from being differentiable. MATH 461 Chapter 3 57

54 Term-by-Term Differentiation of Fourier Series Theorem (Differentiation of Fourier Series) A continuous Fourier series can be differentiated term-by-term provided f is piecewise smooth. Remark In other words, the Fourier series of a continuous function f which satisfies f ( ) = f () can be differentiated term-by-term provided f is piecewise smooth. Piecewise smoothness of f ensures that its Fourier series converges. MATH 461 Chapter 3 58

55 Proof Term-by-Term Differentiation of Fourier Series The Fourier series of f is with a 0 = 1 2 f (x) a 0 + f (x)dx, a n = 1 [ a n cos nπx + b n sin nπx ] (2) f (x) cos nπx dx, b n = 1 f (x) sin nπx dx. Since f is piecewise smooth, it has a convergent Fourier series of the form [ f (x) A 0 + A n cos nπx + B n sin nπx ] (3) with A 0 = 1 2 f (x)dx, A n = 1 f (x) cos nπx dx, B n = 1 f (x) sin nπx dx. MATH 461 Chapter 3 59

56 Term-by-Term Differentiation of Fourier Series If allowed, term-by-term differentiation of the Fourier series (2), i.e., would yield f (x) a 0 + f (x) [ a n cos nπx [ nπ a n sin nπx + b n sin nπx ] + nπ b n cos nπx ]. Therefore, comparing with (3), we need to show that A 0 = 0, A n = nπ b n, B n = nπ a n. MATH 461 Chapter 3 60

57 Term-by-Term Differentiation of Fourier Series et s actually compute the Fourier coefficients of f based on the information we have: A 0 = 1 f (x) dx 2 = 1 2 [f (x)] = 1 [f () f ( )] 2 Since we assumed that the Fourier series of f is continuous, i.e., in particular, that f () = f ( ), we have A 0 = 0. MATH 461 Chapter 3 61

58 Term-by-Term Differentiation of Fourier Series A n = 1 [ = 1 [ parts u = cos nπx dx = f (x) cos nπx f (x) cos nπx + f (x) nπ = 1 [ f () cos nπ f ( ) cos( nπ) }{{}, du = nπ dv = f (x)dx, v = f (x) ] nπx sin dx }{{} nπb n ] +nπb n =cos nπ = 1 [( ) ] f () f ( ) cos nπ + nπb }{{} n =0, since F.S. of f cont. = nπ b n. nπx sin dx ] B n is treated similarly. MATH 461 Chapter 3 62

59 Term-by-Term Differentiation of Fourier Series If the Fourier series of f is not continuous, i.e., if f ( ) f (), then the proof above shows us that A 0 = 1 [f () f ( )], 2 A n = 1 ( 1)n [f () f ( )] + nπ b n, B n = nπ a n. Therefore, even if the Fourier series of f itself is not continuous, the Fourier series of the derivative of a continuous function f is given by f (x) 1 2 [f () f ( )] + ( ( 1) n nπ a n sin nπx. [f () f ( )] + nπ b n ) cos nπx MATH 461 Chapter 3 63

60 Term-by-Term Differentiation of Fourier Series Theorem (Differentiation of Fourier Cosine Series) A continuous Fourier cosine series can be differentiated term-by-term provided f is piecewise smooth. Remark In other words, the Fourier cosine series of a continuous function f can be differentiated term-by-term provided f is piecewise smooth. No additional end conditions are required for f since f ( ) = f () is automatically satisfied due to even extension. Proof. HW 3.4.4b MATH 461 Chapter 3 64

61 Term-by-Term Differentiation of Fourier Series Example Consider again the function f (x) = x, but now find its Fourier cosine series. Can we apply term-by-term differentiation? If so, what is the derivative? We know x A 0 + A n cos nπx with A 0 = 1 A n = x dx = = 2, x cos nπx dx. MATH 461 Chapter 3 65

62 Term-by-Term Differentiation of Fourier Series Example (cont.) A n = 2 [ parts = 2 = = 0 x cos nπx x nπ 2 nπ nπ dx nπx sin nπx cos 0 2 (nπ) 2 [( 1)n 1] = 0 nπ 0 sin nπx dx ] { 0, for n even 4 (nπ) 2, for n odd Therefore, with n = 2k 1, x (2k 1)πx π 2 cos (2k 1) 2 k=1 (4) MATH 461 Chapter 3 66

63 Term-by-Term Differentiation of Fourier Series Example (cont.) Figure: Plot of odd periodic extension (i.e., Fourier cosine series) of f (x) = x. From the plots it is clear that equals = in (4) for 0 x. MATH 461 Chapter 3 67

64 Term-by-Term Differentiation of Fourier Series Example (cont.) Since f and its Fourier cosine series are continuous we can now perform the term-by-term derivative of the Fourier cosine series from (4) i.e., Remark x 2 4 π π 2 = 4 π k=1 k=1 k=1 1 (2k 1)πx cos (2k 1) 2 π (2k 1)πx sin (2k 1) 1 (2k 1) (2k 1)πx sin. (5) Note that this is the Fourier sine series of f (x) = 1, for 0 < x <., MATH 461 Chapter 3 68

65 Term-by-Term Differentiation of Fourier Series Example (cont.) Figure: Plot of Fourier sine series of f (x) = 1. Note that due to the jumps in the graph of the Fourier sine series equals = in (5) only for 0 < x <. MATH 461 Chapter 3 69

66 Term-by-Term Differentiation of Fourier Series Theorem (Differentiation of Fourier Sine Series) A continuous Fourier sine series can be differentiated term-by-term provided f is piecewise smooth. Remark In other words, the Fourier sine series of a continuous function f which satisfies f (0) = f () = 0 can be differentiated term-by-term provided f is piecewise smooth. Proof. See the textbook on pages MATH 461 Chapter 3 70

67 Term-by-Term Differentiation of Fourier Series From the proof of the theorem we get that if f is continuous, but does not satisfy f (0) = f () = 0, with Fourier sine series f (x) B n sin nπx then, provided f is piecewise smooth, we get the Fourier cosine series f (x) 1 [f () f (0)] + ( nπ B n + 2 ) [( 1)n f () f (0)] cos nπx. (6) MATH 461 Chapter 3 71

68 Term-by-Term Differentiation of Fourier Series Example We saw earlier that for the function f (x) = x we have x = 2 π ( 1) n+1 n sin nπx, 0 < x <. Since f (0) = 0 f () = we can t expect term-by-term differentiation to yield f (x) (and we observed this in the earlier example). However, (6) does provide the expected (and correct) answer f (x) 1 [f () f (0)] + = 1 [ 0] + = 1 + [ ( nπ ( nπ B n + 2 [( 1)n f () f (0)] 2( 1) n+1 nπ 2( 1) n+1 + 2( 1) n }{{} =0 ) cos nπx + 2 ) [( 1)n 0] cos nπx ] cos nπx = 1 MATH 461 Chapter 3 72

69 Term-by-Term Differentiation of Fourier Series Another ook at Separation of Variables: The Eigenfunction Perspective By starting our discussion of the solution of the heat equation with an eigenfunction expansion we are able to obtain a justification for why the separation of variables approach works. Remark The main advantage of taking this different point of view is that it can be applied to nonhomogeneous problems as well (see HW and ). MATH 461 Chapter 3 73

70 Term-by-Term Differentiation of Fourier Series Example et s once more solve the 1D heat equation u t = k 2 u x 2, 0 < x <, t > 0 u(0, t) = u(, t) = 0 u(x, 0) = f (x). We know that the eigenfunctions for this problem are { sin πx } 2πx 3πx, sin, sin,... and therefore we make the Ansatz u(x, t) = B n (t) sin nπx. Note the time-dependence of the Fourier sine coefficients. MATH 461 Chapter 3 74

71 Term-by-Term Differentiation of Fourier Series Example (cont.) The first thing to do is to enforce the initial condition u(x, 0) = f (x), i.e., f (x) B n (0) sin nπx with B n (0) = 2 0 f (x) sin nπx dx, n = 1, 2, 3,... Remark Note that now we are approaching the problem from a different angle, and so we don t know yet whether u satisfies the heat equation. MATH 461 Chapter 3 75

72 Term-by-Term Differentiation of Fourier Series Example (cont.) To check whether u satisfies the heat equation we compute all the required partial derivatives using term-by-term differentiation: u(x, t) = u x (x, t) B n (t) sin nπx = 2 u x 2 (x, t) ( nπ and nπ B n(t) cos nπx u t (x, t) B n(t) sin nπx ) 2 Bn (t) sin nπx (7) (8) (9) MATH 461 Chapter 3 76

73 Term-by-Term Differentiation of Fourier Series Example (cont.) Were all of these differentiations justified? (7) was OK since we differentiated the sine series of a continuous function (for fixed t) which satisfies u(0, t) = u(, t) = 0. (8) was OK since we differentiated the cosine series of a continuous function (for fixed t). (9) was questionable. So far we have no theorem covering this case see below. MATH 461 Chapter 3 77

74 Term-by-Term Differentiation of Fourier Series Example (cont.) Using (8) and (9), u satisfies the heat equation if B n(t) sin nπx = k [ ( nπ ) 2 Bn (t) sin nπx ]. Comparing coefficients of sines of like frequencies we get an ODE for the coefficients B n : ( nπ ) 2 B n(t) = k Bn (t), n = 1, 2, 3,... This ODE is easily solved and yields B n (t) = B n (0)e k( nπ ) 2 t which is the same answer we had earlier using separation of variables. MATH 461 Chapter 3 78

75 Term-by-Term Differentiation of Fourier Series We close the section with the theorem that justifies the derivation of (9) above. Theorem If u = u(x, t) is a continuous function of t with time-dependent Fourier series u(x, t) = a 0 (t) + [ a n (t) cos nπx + b n(t) sin nπx ] then u t (x, t) = a 0 (t) + [ a n(t) cos nπx provided u t is piecewise smooth. + b n(t) sin nπ ] MATH 461 Chapter 3 79

76 Integration of Fourier Series Theorem The Fourier series of a piecewise smooth function f can always be integrated term-by-term. Moreover, the result is a continuous infinite series (but not necessarily a Fourier series) which converges to the integral of f on the interval [, ], i.e., if f has the Fourier series f (x) a 0 + [ a n cos nπx then, for all x [, ], we have + b n sin nπx ], x x [ an f (t)dt = a 0 (x +)+ nπ sin nπx + b n nπ ( cos nπ cos nπx ) ]. (10) MATH 461 Chapter 3 81

77 Integration of Fourier Series Remark The integrals in the theorem need not be from to x. They can also be from a to b, a, b [, ], since we can always write b a... = a b a b = , and the latter two integrals are covered by the formula in the theorem. MATH 461 Chapter 3 82

78 Integration of Fourier Series Before we prove the theorem we note the following facts: x x x a 0 dt = a 0 t x = a 0(x + ) cos nπt dt = nπt x sin nπ = nπx sin nπ sin nπt dt = nπt x cos nπ = ( cos nπ cos nπx ) nπ This shows that the coefficients in formula (10) indeed are likely candidates for term-by-term integration of the Fourier series. MATH 461 Chapter 3 83

79 Proof. Integration of Fourier Series We begin by defining the function F as F(x) = x f (t) dt. Since we assumed f to be piecewise smooth, F is continuous. Its Fourier series is continuous if and only if F( ) = F(). However, F( ) = F() = f (t) dt = 0 f (t) dt = 2a 0, where a 0 is one of the Fourier coefficients of f. These are in general not the same. Therefore, the Fourier series of F is not continuous in general, and we cannot assume that F (x) equals its Fourier series for x. MATH 461 Chapter 3 84

80 Proof (cont.) In addition to F(x) = x now also define and Integration of Fourier Series H(x) = a 0 (x + ) G(x) = F(x) H(x). f (t) dt we Clearly, H denotes the line passing through (, 0) and (, 2a 0 ). As a consequence we have G( ) = G() = 0 (since F( ) = 0 and F() = 2a 0 ), G is continuous (since F and H are) so that G(x) equals its Fourier series on [, ]. MATH 461 Chapter 3 85

81 Proof (cont.) Integration of Fourier Series et s write the Fourier series of G in the form G(x) = A 0 + [ A n cos nπx + B n sin nπx ] with (remember that G(x) = F(x) H(x) = F(x) a 0 (x + )) A 0 = 1 2 A n = 1 B n = 1 and let s compute A 0, A n and B n. [F(x) a 0 (x + )] dx [F(x) a 0 (x + )] cos nπx [F(x) a 0 (x + )] sin nπx dx dx MATH 461 Chapter 3 86

82 Proof (cont.) Integration of Fourier Series A n = 1 = 1 = 1 [ = nπ b n [F(x) a 0 (x + )] cos nπx [F(x) a 0 ] }{{} =u, du=f (x)dx (F(x) a 0 ) nπ cos nπx }{{ } =dv, v= nπx sin nπ sin nπx } {{ } 0 dx dx 1 a 0 x cos nπx dx }{{} nπ B n = nπ a n is computed similarly (see HW 3.5.5) =0, odd ] f (x) sin nπx dx }{{} =b n MATH 461 Chapter 3 87

83 Proof (cont.) Integration of Fourier Series To compute A 0 we note that Therefore 0 = G() = A 0 + A 0 = [ A n cos nπ A n cos nπ = Putting everything together we get ] + B n sin nπ }{{ } =0 nπ b n cos nπ. F(x) = H(x) + G(x) = a 0 (x + ) + A 0 + A n cos nπx + B n sin nπx which matches the claim of the theorem if we use the representations of A 0, A n and B n. MATH 461 Chapter 3 88

84 Integration of Fourier Series Example Integrate the following Fourier cosine series (see (4)) from 0 to x: x = 2 4 π 2 k=1 1 (2k 1)πx cos (2k 1) 2 Solution We immediately have and x 0 x 0 t dt = x dt = x 2 MATH 461 Chapter 3 89

85 Integration of Fourier Series Solution (cont.) The remaining part becomes 4 π 2 k=1 1 (2k 1) 2 x 0 cos (2k 1)πt dt = 42 π 3 = 42 π 3 Putting all three parts together we have x 2 = x 82 π 3 k=1 k=1 1 (2k 1)πt sin (2k 1) 3 k=1 1 (2k 1)πx sin (2k 1) 3 1 (2k 1)πx sin (2k 1) 3 Note that this is in agreement with the statement of the theorem. Due to the presence of the linear term x this is not a Fourier (sine) series.. x 0 MATH 461 Chapter 3 90

86 Integration of Fourier Series Solution (cont.) We can interpret x 2 = x 82 π 3 k=1 1 (2k 1)πx sin (2k 1) 3 differently. Namely, we do have the following two sine series: x x 2 = 82 π 3 k=1 1 (2k 1)πx sin (2k 1) 3 and, using the Fourier sine series of f (x) = x (see (1)), x 2 = 2 π ( 1) n+1 k sin nπx 82 π 3 k=1 1 (2k 1)πx sin (2k 1) 3 MATH 461 Chapter 3 91

87 Integration of Fourier Series Example Use the fact established earlier (see (5)) that to show that 1 = 4 π k=1 k=1 1 2k 1 (2k 1)πx sin, 0 < x < 1 (2k 1) 2 = = π2 8. MATH 461 Chapter 3 92

88 Integration of Fourier Series Solution The main idea is to integrate the given identity from 0 to x. For the left-hand side we have while the right-hand side is x 0 4 π k=1 x 1 (2k 1)πt sin dt = 4 2k 1 π = 4 π dt = x, = 4 π 2 1 x (2k 1)πt sin dt 2k 1 0 x 1 (2k 1)πt cos (2k 1) 2 k=1 k=1 [ ] 1 (2k 1) 2 1 (2k 1)πx cos (2k 1) 2 k=1 0 MATH 461 Chapter 3 93

89 Integration of Fourier Series Solution (cont.) Therefore, splitting into two series, x = 4 π 2 k=1 1 (2k 1) 2 4 π 2 k=1 This is a cosine series with constant term A 0 = 4 π 2 k=1 and we can now conclude that k=1 1 (2k 1) 2 = 1 1 (2k 1)πx cos (2k 1) 2 1 (2k 1) 2 = π2 4 2 = π x dx = 2,. MATH 461 Chapter 3 94

90 Integration of Fourier Series Solution (cont.) Alternatively, we could have evaluated the series expansion x = 4 1 π 2 (2k 1) (2k 1)πx π 2 cos (2k 1) 2 k=1 k=1 for some special value of x. For example, for x = we get = 4 π 2 so that = 4 π 2 = 2 4 π 2 k=1 k=1 1 (2k 1) 2 4 π 2 1 (2k 1) 2 4 π 2 k=1 k=1 k=1 1 (2k 1) 2 1 (2k 1)π cos (2k 1) 2 1 cos(2k 1)π (2k 1) 2 }{{} = 1 k=1 1 (2k 1) 2 = π2 8. MATH 461 Chapter 3 95

91 Integration of Fourier Series Solution (cont.) For x = 2 we get 2 = 4 π 2 = 4 π 2 k=1 k=1 1 (2k 1) 2 4 π 2 1 (2k 1) 2 4 π 2 k=1 k=1 1 (2k 1) 2 cos (2k 1)π 2 1 (2k 1) 2 cos(2k 1)π 2 }{{} =0 so that 2 = 4 π 2 k=1 1 (2k 1) 2 k=1 1 (2k 1) 2 = π2 8. MATH 461 Chapter 3 96

92 Integration of Fourier Series Example The Basel problem, first proved by eonhard Euler in 1735: 1 n 2 = π2 6 One can prove this as we did for the squares of odd integers above. Here we evaluate the Fourier series of f (x) = x 2, x 2 = π 2 at x =. ( 1) n See [Proofs from THE BOOK] for three different proofs. n 2 cos nπx, MATH 461 Chapter 3 97

93 Complex Form of Fourier Series Fourier series are often expressed in terms of complex exponentials instead of sines and cosines. The main ingredient for understanding this translation in notation is Euler s formula e iθ = cos θ + i sin θ. This, of course, implies and so e iθ = cos θ i sin θ, cos θ = eiθ + e iθ 2 sin θ = eiθ e iθ. 2i MATH 461 Chapter 3 99

94 Complex Form of Fourier Series We can therefore rewrite the Fourier series as f (x) a 0 + f (x) a 0 + = a [a n e i nπx [ a n cos nπx [( a n + b n i nπx i + e 2 + b n sin nπx ] e i nπx + b n ) ( e i nπx + a n b n i ] nπx i e 2i ) e nπx i ] MATH 461 Chapter 3 100

95 Complex Form of Fourier Series We break this into two series and use 1 i = i to arrive at f (x) a (a n ib n ) e i nπx nπx i (a n + ib n ) e Now we perform an index transformation, n n, on the first series to get f (x) a n= 1 nπx i (a n ib n ) e Note that, using the symmetries of cosine and sine, a n = 1 b n = 1 f (x) cos ( n)πx f (x) sin ( n)πx nπx i (a n + ib n ) e dx = a n dx = b n MATH 461 Chapter 3 101

96 We can therefore rewrite as f (x) a f (x) a Complex Form of Fourier Series n= 1 n= 1 If we introduce new coefficients nπx i (a n ib n ) e nπx i (a n + ib n ) e c 0 = a 0 and c n = a n + ib n 2 then we get the exponential form of the Fourier series nπx i f (x) c n e with Fourier coefficients c 0 = 1 2 n= f (x) dx nπx i (a n + ib n ) e nπx i (a n + ib n ) e MATH 461 Chapter 3 102

97 Complex Form of Fourier Series and c n = a n + ib n [ 2 = 1 2 = 1 2 = 1 2 f (x) cos nπx dx + i f (x) f (x)e i nπx [ cos nπx dx nπx + i sin f (x) sin nπx ] dx dx ] Note that this formula also gives the correct value for c 0. Remark Sometimes the formula for the Fourier coefficients c n is referred to as the finite Fourier transform of f. MATH 461 Chapter 3 103

98 References I Appendix References Aigner, Martin, Günter M. Ziegler, and Karl H. Hofmann. Proofs from THE BOOK (4th Ed.). Springer, J. W. Brown and R. V. Churchill. Fourier Series and Boundary Value Problems. McGraw-Hill (5th ed.), New York, R. Haberman. Applied Partial Differential Equations. Pearson (5th ed.), Upper Saddle River, NJ, M. A. Pinsky. Partial Differential Equations and Boundary-value Problems with Applications. American Mathematical Society (reprint of 3rd ed.), Providence, RI, N. Trefethen. Approximation Theory and Approximation Practice. Society of Industrial and Applied Mathematics, Philadelphia, PA, MATH 461 Chapter 3 104

### Fourier Series Chapter 3 of Coleman

Fourier Series Chapter 3 of Coleman Dr. Doreen De eon Math 18, Spring 14 1 Introduction Section 3.1 of Coleman The Fourier series takes its name from Joseph Fourier (1768-183), who made important contributions

### Maths 361 Fourier Series Notes 2

Today s topics: Even and odd functions Real trigonometric Fourier series Section 1. : Odd and even functions Consider a function f : [, ] R. Maths 361 Fourier Series Notes f is odd if f( x) = f(x) for

### Fourier Series. Chapter Some Properties of Functions Goal Preliminary Remarks

Chapter 3 Fourier Series 3.1 Some Properties of Functions 3.1.1 Goal We review some results about functions which play an important role in the development of the theory of Fourier series. These results

Week 5 Vladimir Dobrushkin 5. Fourier Series In this section, we present a remarkable result that gives the matehmatical explanation for how cellular phones work: Fourier series. It was discovered in the

### 16 Convergence of Fourier Series

16 Convergence of Fourier Series 16.1 Pointwise convergence of Fourier series Definition: Piecewise smooth functions For f defined on interval [a, b], f is piecewise smooth on [a, b] if there is a partition

### Lecture VI. Review of even and odd functions Definition 1 A function f(x) is called an even function if. f( x) = f(x)

ecture VI Abstract Before learning to solve partial differential equations, it is necessary to know how to approximate arbitrary functions by infinite series, using special families of functions This process

### Fourier series. Jan Philip Solovej. English summary of notes for Analysis 1. May 8, 2012

Fourier series Jan Philip Solovej English summary of notes for Analysis 1 May 8, 2012 1 JPS, Fourier series 2 Contents 1 Introduction 2 2 Fourier series 3 2.1 Periodic functions, trigonometric polynomials

### Fourier Series. Some Properties of Functions. Philippe B. Laval. Today KSU. Philippe B. Laval (KSU) Fourier Series Today 1 / 19

Fourier Series Some Properties of Functions Philippe B. Laval KSU Today Philippe B. Laval (KSU) Fourier Series Today 1 / 19 Introduction We review some results about functions which play an important role

### Fourier Series. A Fourier series is an infinite series of the form. a + b n cos(nωx) +

Fourier Series A Fourier series is an infinite series of the form a b n cos(nωx) c n sin(nωx). Virtually any periodic function that arises in applications can be represented as the sum of a Fourier series.

### 1. Periodic Fourier series. The Fourier expansion of a 2π-periodic function f is:

CONVERGENCE OF FOURIER SERIES 1. Periodic Fourier series. The Fourier expansion of a 2π-periodic function f is: with coefficients given by: a n = 1 π f(x) a 0 2 + a n cos(nx) + b n sin(nx), n 1 f(x) cos(nx)dx

### Recap on Fourier series

Civil Engineering Mathematics Autumn 11 J. Mestel, M. Ottobre, A. Walton Recap on Fourier series A function f(x is called -periodic if f(x = f(x + for all x. A continuous periodic function can be represented

### Sine and Cosine Series; Odd and Even Functions

Sine and Cosine Series; Odd and Even Functions A sine series on the interval [, ] is a trigonometric series of the form k = 1 b k sin πkx. All of the terms in a series of this type have values vanishing

### Fourier Series and Sturm-Liouville Eigenvalue Problems

Fourier Series and Sturm-Liouville Eigenvalue Problems 2009 Outline Functions Fourier Series Representation Half-range Expansion Convergence of Fourier Series Parseval s Theorem and Mean Square Error Complex

### 1. the function must be periodic; 3. it must have only a finite number of maxima and minima within one periodic;

Fourier Series 1 Dirichlet conditions The particular conditions that a function f(x must fulfil in order that it may be expanded as a Fourier series are known as the Dirichlet conditions, and may be summarized

### Fourier Series Representations

Fourier Series Representations Introduction Before we discuss the technical aspects of Fourier series representations, it might be well to discuss the broader question of why they are needed We ll begin

### M344 - ADVANCED ENGINEERING MATHEMATICS Lecture 9: Orthogonal Functions and Trigonometric Fourier Series

M344 - ADVANCED ENGINEERING MATHEMATICS ecture 9: Orthogonal Functions and Trigonometric Fourier Series Before learning to solve partial differential equations, it is necessary to know how to approximate

### CHAPTER 2 FOURIER SERIES

CHAPTER 2 FOURIER SERIES PERIODIC FUNCTIONS A function is said to have a period T if for all x,, where T is a positive constant. The least value of T>0 is called the period of. EXAMPLES We know that =

### Lecture 3: Fourier Series: pointwise and uniform convergence.

Lecture 3: Fourier Series: pointwise and uniform convergence. 1. Introduction. At the end of the second lecture we saw that we had for each function f L ([, π]) a Fourier series f a + (a k cos kx + b k

### Introduction to Fourier Series

Introduction to Fourier Series A function f(x) is called periodic with period T if f(x+t)=f(x) for all numbers x. The most familiar examples of periodic functions are the trigonometric functions sin and

### Introduction to Fourier Series

Introduction to Fourier Series MA 16021 October 15, 2014 Even and odd functions Definition A function f(x) is said to be even if f( x) = f(x). The function f(x) is said to be odd if f( x) = f(x). Graphically,

### Taylor Polynomials and Taylor Series Math 126

Taylor Polynomials and Taylor Series Math 26 In many problems in science and engineering we have a function f(x) which is too complicated to answer the questions we d like to ask. In this chapter, we will

### 3 Trigonometric Fourier Series

3 Trigonometric Fourier Series Ordinary language is totally unsuited for epressing what physics really asserts, since the words of everyday life are not sufficiently abstract. Only mathematics and mathematical

### Fourier Series & Fourier Transforms

Fourier Series & Fourier Transforms nicholas.harrison@imperial.ac.uk 19th October 003 Synopsis ecture 1 : Review of trigonometric identities Fourier Series Analysing the square wave ecture : The Fourier

### An Introduction to Separation of Variables with Fourier Series Math 391w, Spring 2010 Tim McCrossen Professor Haessig

An Introduction to Separation of Variables with Fourier Series Math 391w, Spring 2010 Tim McCrossen Professor Haessig Abstract: This paper aims to give students who have not yet taken a course in partial

### Fourier Series, Integrals, and Transforms

Chap. Sec.. Fourier Series, Integrals, and Transforms Fourier Series Content: Fourier series (5) and their coefficients (6) Calculation of Fourier coefficients by integration (Example ) Reason hy (6) gives

### Engineering Mathematics II

PSUT Engineering Mathematics II Fourier Series and Transforms Dr. Mohammad Sababheh 4/14/2009 11.1 Fourier Series 2 Fourier Series and Transforms Contents 11.1 Fourier Series... 3 Periodic Functions...

### Math 201 Lecture 30: Fourier Cosine and Sine Series

Math ecture 3: Fourier Cosine and Sine Series Mar. 3, Many examples here are taken from the textbook. The first number in ( refers to the problem number in the UA Custom edition, the second number in (

### ) + ˆf (n) sin( 2πnt. = 2 u x 2, t > 0, 0 < x < 1. u(0, t) = u(1, t) = 0, t 0. (x, 0) = 0 0 < x < 1.

Introduction to Fourier analysis This semester, we re going to study various aspects of Fourier analysis. In particular, we ll spend some time reviewing and strengthening the results from Math 425 on Fourier

### ME231 Measurements Laboratory Spring Fourier Series. Edmundo Corona c

ME23 Measurements Laboratory Spring 23 Fourier Series Edmundo Corona c If you listen to music you may have noticed that you can tell what instruments are used in a given song or symphony. In some cases,

### Math 241: More heat equation/laplace equation

Math 241: More heat equation/aplace equation D. DeTurck University of Pennsylvania September 27, 2012 D. DeTurck Math 241 002 2012C: Heat/aplace equations 1 / 13 Another example Another heat equation problem:

### THE DIFFUSION EQUATION

THE DIFFUSION EQUATION R. E. SHOWALTER 1. Heat Conduction in an Interval We shall describe the diffusion of heat energy through a long thin rod G with uniform cross section S. As before, we identify G

### , < x < Using separation of variables, u(x, t) = Φ(x)h(t) (4) we obtain the differential equations. d 2 Φ = λφ (6) Φ(± ) < (7)

Chapter1: Fourier Transform Solutions of PDEs In this chapter we show how the method of separation of variables may be extended to solve PDEs defined on an infinite or semi-infinite spatial domain. Several

### Fourier Series Expansion

Fourier Series Expansion Deepesh K P There are many types of series expansions for functions. The Maclaurin series, Taylor series, Laurent series are some such expansions. But these expansions become valid

### CHAPTER 3. Fourier Series

`A SERIES OF CLASS NOTES FOR 2005-2006 TO INTRODUCE LINEAR AND NONLINEAR PROBLEMS TO ENGINEERS, SCIENTISTS, AND APPLIED MATHEMATICIANS DE CLASS NOTES 4 A COLLECTION OF HANDOUTS ON PARTIAL DIFFERENTIAL

### Fourier Series. 1. Introduction. 2. Square Wave

531 Fourier Series Tools Used in Lab 31 Fourier Series: Square Wave Fourier Series: Triangle Wave Fourier Series: Gibbs Effect Fourier Series: Coefficients How does a Fourier series converge to a function?

### Some Notes on Taylor Polynomials and Taylor Series

Some Notes on Taylor Polynomials and Taylor Series Mark MacLean October 3, 27 UBC s courses MATH /8 and MATH introduce students to the ideas of Taylor polynomials and Taylor series in a fairly limited

### Learning Objectives for Math 165

Learning Objectives for Math 165 Chapter 2 Limits Section 2.1: Average Rate of Change. State the definition of average rate of change Describe what the rate of change does and does not tell us in a given

### Math Spring Review Material - Exam 3

Math 85 - Spring 01 - Review Material - Exam 3 Section 9. - Fourier Series and Convergence State the definition of a Piecewise Continuous function. Answer: f is Piecewise Continuous if the following to

### Problem 1 (10 pts) Find the radius of convergence and interval of convergence of the series

1 Problem 1 (10 pts) Find the radius of convergence and interval of convergence of the series a n n=1 n(x + 2) n 5 n 1. n(x + 2)n Solution: Do the ratio test for the absolute convergence. Let a n =. Then,

### Chapter 11 Fourier Analysis

Chapter 11 Fourier Analysis Advanced Engineering Mathematics Wei-Ta Chu National Chung Cheng University wtchu@cs.ccu.edu.tw 1 2 11.1 Fourier Series Fourier Series Fourier series are infinite series that

### 1 Review of complex numbers

1 Review of complex numbers 1.1 Complex numbers: algebra The set C of complex numbers is formed by adding a square root i of 1 to the set of real numbers: i = 1. Every complex number can be written uniquely

### f x a 0 n 1 a 0 a 1 cos x a 2 cos 2x a 3 cos 3x b 1 sin x b 2 sin 2x b 3 sin 3x a n cos nx b n sin nx n 1 f x dx y

Fourier Series When the French mathematician Joseph Fourier (768 83) was tring to solve a problem in heat conduction, he needed to epress a function f as an infinite series of sine and cosine functions:

### Solutions for homework assignment #4

Solutions for homework assignment #4 Problem 1. Solve aplace s equation inside a rectangle x, y, with the following boundary conditions:, y) =,, y) =, ux, ) =, ux, ) = fx). y ux, y) = b + b n sinh nπ )

### FOURIER SERIES. 1. Periodic Functions. Recall that f has period T if f(x + T ) = f(x) for all x. If f and g are periodic

FOURIER SERIES. Periodic Functions Reca that f has period T if f(x + T ) f(x) for a x. If f and g are periodic with period T, then if h(x) af(x) + bg(x), a and b are constants, and thus h aso has period

### SECTION 1.5: PIECEWISE-DEFINED FUNCTIONS; LIMITS AND CONTINUITY IN CALCULUS

(Section.5: Piecewise-Defined Functions; Limits and Continuity in Calculus).5. SECTION.5: PIECEWISE-DEFINED FUNCTIONS; LIMITS AND CONTINUITY IN CALCULUS LEARNING OBJECTIVES Know how to evaluate and graph

### Chapt.12: Orthogonal Functions and Fourier series

Chat.12: Orthogonal Functions and Fourier series J.-P. Gabardo gabardo@mcmaster.ca Deartment of Mathematics & Statistics McMaster University Hamilton, ON, Canada Lecture: January 10, 2011. 1/3 12.1:Orthogonal

### Lecture Notes for Math 251: ODE and PDE. Lecture 33: 10.4 Even and Odd Functions

Lecture Notes for Math 51: ODE and PDE. Lecture : 1.4 Even and Odd Functions Shawn D. Ryan Spring 1 Last Time: We studied what a given Fourier Series converges to if at a. 1 Even and Odd Functions Before

### The Dirichlet Problem on a Rectangle

The Dirichlet Problem on a Rectangle R. C. Trinity University Partial Differential Equations March 18, 014 The D heat equation Steady state solutions to the -D heat equation Laplace s equation Recall:

### Fourier cosine and sine series even odd

Fourier cosine and sine series Evaluation of the coefficients in the Fourier series of a function f is considerably simler is the function is even or odd. A function is even if f ( x) = f (x) Examle: x

### Infinite series, improper integrals, and Taylor series

Chapter Infinite series, improper integrals, and Taylor series. Introduction This chapter has several important and challenging goals. The first of these is to understand how concepts that were discussed

### ANALYTICAL MATHEMATICS FOR APPLICATIONS 2016 LECTURE NOTES Series

ANALYTICAL MATHEMATICS FOR APPLICATIONS 206 LECTURE NOTES 8 ISSUED 24 APRIL 206 A series is a formal sum. Series a + a 2 + a 3 + + + where { } is a sequence of real numbers. Here formal means that we don

### Fourier Series and Integrals

Fourier Series and Integrals Fourier Series et f(x) be a piecewise linear function on [,] (This means that f(x) may possess a finite number of finite discontinuities on the interval). Then f(x) can be

### Analysis for Simple Fourier Series in Sine and Co-sine Form

Analysis for Simple Fourier Series in Sine and Co-sine Form Chol Yoon August 19, 13 Abstract From the history of Fourier series to the application of Fourier series will be introduced and analyzed to understand

### The Fourier Transform

The Fourier Transorm Fourier Series Fourier Transorm The Basic Theorems and Applications Sampling Bracewell, R. The Fourier Transorm and Its Applications, 3rd ed. New York: McGraw-Hill, 2. Eric W. Weisstein.

### Course MA2C02, Hilary Term 2012 Section 8: Periodic Functions and Fourier Series

Course MAC, Hiary Term Section 8: Periodic Functions and Fourier Series David R. Wikins Copyright c David R. Wikins Contents 8 Periodic Functions and Fourier Series 37 8. Fourier Series of Even and Odd

### Fourier Series. 1. Full-range Fourier Series. ) + b n sin L. [ a n cos L )

Fourier Series These summary notes should be used in conjunction with, and should not be a replacement for, your lecture notes. You should be familiar with the following definitions. A function f is periodic

### Finding Antiderivatives and Evaluating Integrals

Chapter 5 Finding Antiderivatives and Evaluating Integrals 5. Constructing Accurate Graphs of Antiderivatives Motivating Questions In this section, we strive to understand the ideas generated by the following

### Lecture5. Fourier Series

Lecture5. Fourier Series In 1807 the French mathematician Joseph Fourier (1768-1830) submitted a paper to the Academy of Sciences in Paris. In it he presented a mathematical treatment of problems involving

### Problems for Quiz 14

Problems for Quiz 14 Math 3. Spring, 7. 1. Consider the initial value problem (IVP defined by the partial differential equation (PDE u t = u xx u x + u, < x < 1, t > (1 with boundary conditions and initial

### Advanced Engineering Mathematics Prof. Jitendra Kumar Department of Mathematics Indian Institute of Technology, Kharagpur

Advanced Engineering Mathematics Prof. Jitendra Kumar Department of Mathematics Indian Institute of Technology, Kharagpur Lecture No. # 28 Fourier Series (Contd.) Welcome back to the lecture on Fourier

### The Alternating Series Test

The Alternating Series Test So far we have considered mostly series all of whose terms are positive. If the signs of the terms alternate, then testing convergence is a much simpler matter. On the other

### So here s the next version of Homework Help!!!

HOMEWORK HELP FOR MATH 52 So here s the next version of Homework Help!!! I am going to assume that no one had any great difficulties with the problems assigned this quarter from 4.3 and 4.4. However, if

### Fourier Series. 5.1 Introduction

5 Fourier Series 5. Introduction In this chapter we will look at trigonometric series. Previously, we saw that such series epansion occurred naturally in the solution of the heat equation and other boundary

### VI. Transcendental Functions. x = ln y. In general, two functions f, g are said to be inverse to each other when the

VI Transcendental Functions 6 Inverse Functions The functions e x and ln x are inverses to each other in the sense that the two statements y = e x, x = ln y are equivalent statements In general, two functions

### Applications of Fourier series

Chapter Applications of Fourier series One of the applications of Fourier series is the evaluation of certain infinite sums. For example, n= n,, are computed in Chapter (see for example, Remark.4.). n=

### From Fourier Series to Fourier Integral

From Fourier Series to Fourier Integral Fourier series for periodic functions Consider the space of doubly differentiable functions of one variable x defined within the interval x [ L/2, L/2]. In this

### Sequences of Functions

Sequences of Functions Uniform convergence 9. Assume that f n f uniformly on S and that each f n is bounded on S. Prove that {f n } is uniformly bounded on S. Proof: Since f n f uniformly on S, then given

### x + 5 x 2 + x 2 dx Answer: ln x ln x 1 + c

. Evaluate the given integral (a) 3xe x2 dx 3 2 e x2 + c (b) 3 x ln xdx 2x 3/2 ln x 4 3 x3/2 + c (c) x + 5 x 2 + x 2 dx ln x + 2 + 2 ln x + c (d) x sin (πx) dx x π cos (πx) + sin (πx) + c π2 (e) 3x ( +

### The Geometric Series

The Geometric Series Professor Jeff Stuart Pacific Lutheran University c 8 The Geometric Series Finite Number of Summands The geometric series is a sum in which each summand is obtained as a common multiple

### Math 241: Laplace equation in polar coordinates; consequences and properties

Math 241: Laplace equation in polar coordinates; consequences and properties D. DeTurck University of Pennsylvania October 6, 2012 D. DeTurck Math 241 002 2012C: Laplace in polar coords 1 / 16 Laplace

### The Heat Equation. Lectures INF2320 p. 1/88

The Heat Equation Lectures INF232 p. 1/88 Lectures INF232 p. 2/88 The Heat Equation We study the heat equation: u t = u xx for x (,1), t >, (1) u(,t) = u(1,t) = for t >, (2) u(x,) = f(x) for x (,1), (3)

### 1 if 1 x 0 1 if 0 x 1

Chapter 3 Continuity In this chapter we begin by defining the fundamental notion of continuity for real valued functions of a single real variable. When trying to decide whether a given function is or

### An Introduction to Partial Differential Equations in the Undergraduate Curriculum

An Introduction to Partial Differential Equations in the Undergraduate Curriculum J. Tolosa & M. Vajiac LECTURE 11 Laplace s Equation in a Disk 11.1. Outline of Lecture The Laplacian in Polar Coordinates

### 1 The Dirichlet Problem. 2 The Poisson kernel. Math 857 Fall 2015

Math 857 Fall 2015 1 The Dirichlet Problem Before continuing to Fourier integrals, we consider first an application of Fourier series. Let Ω R 2 be open and connected (region). Recall from complex analysis

### Examination paper for Solutions to Matematikk 4M and 4N

Department of Mathematical Sciences Examination paper for Solutions to Matematikk 4M and 4N Academic contact during examination: Trygve K. Karper Phone: 99 63 9 5 Examination date:. mai 04 Examination

### 1 Mathematical Induction

Extra Credit Homework Problems Note: these problems are of varying difficulty, so you might want to assign different point values for the different problems. I have suggested the point values each problem

### Lecture 5 : Continuous Functions Definition 1 We say the function f is continuous at a number a if

Lecture 5 : Continuous Functions Definition We say the function f is continuous at a number a if f(x) = f(a). (i.e. we can make the value of f(x) as close as we like to f(a) by taking x sufficiently close

### Fourier Analysis. Be able to follow derivations of Fourier Series coefficients, relations between Fourier Series and Fourier Integrals etc.

Self Learn Module Fourier Analysis Contents Fourier Series for functions of period 2π Fourier Series for functions of period L Fourier Integrals Complex Fourier Integrals Fourier analysis of standing wave

### MATH31011/MATH41011/MATH61011: FOURIER ANALYSIS AND LEBESGUE INTEGRATION. Chapter 4: Fourier Series and L 2 ([ π, π], µ) ( 1 π

MATH31011/MATH41011/MATH61011: FOURIER ANALYSIS AND LEBESGUE INTEGRATION Chapter 4: Fourier Series and L ([, π], µ) Square Integrable Functions Definition. Let f : [, π] R be measurable. We say that f

### Solutions to Sample Midterm 2 Math 121, Fall 2004

Solutions to Sample Midterm Math, Fall 4. Use Fourier series to find the solution u(x, y) of the following boundary value problem for Laplace s equation in the semi-infinite strip < x : u x + u

### CALCULUS 2. 0 Repetition. tutorials 2015/ Find limits of the following sequences or prove that they are divergent.

CALCULUS tutorials 5/6 Repetition. Find limits of the following sequences or prove that they are divergent. a n = n( ) n, a n = n 3 7 n 5 n +, a n = ( n n 4n + 7 ), a n = n3 5n + 3 4n 7 3n, 3 ( ) 3n 6n

### Fourier Series and Fejér s Theorem

wwu@ocf.berkeley.edu June 4 Introduction : Background and Motivation A Fourier series can be understood as the decomposition of a periodic function into its projections onto an orthonormal basis. More

### 5 Indefinite integral

5 Indefinite integral The most of the mathematical operations have inverse operations: the inverse operation of addition is subtraction, the inverse operation of multiplication is division, the inverse

### Chapter 14: Fourier Transforms and Boundary Value Problems in an Unbounded Region

Chapter 14: Fourier Transforms and Boundary Value Problems in an Unbounded Region 王奕翔 Department of Electrical Engineering National Taiwan University ihwang@ntu.edu.tw December 25, 213 1 / 27 王奕翔 DE Lecture

### Limit processes are the basis of calculus. For example, the derivative. f f (x + h) f (x)

SEC. 4.1 TAYLOR SERIES AND CALCULATION OF FUNCTIONS 187 Taylor Series 4.1 Taylor Series and Calculation of Functions Limit processes are the basis of calculus. For example, the derivative f f (x + h) f

### Limits at Infinity Limits at Infinity for Polynomials Limits at Infinity for the Exponential Function Function Dominance More on Asymptotes

Lecture 5 Limits at Infinity and Asymptotes Limits at Infinity Horizontal Asymptotes Limits at Infinity for Polynomials Limit of a Reciprocal Power The End Behavior of a Polynomial Evaluating the Limit

### FOURIER SERIES, GENERALIZED FUNCTIONS, LAPLACE TRANSFORM. 1. A (generalized) function f(t) of period 2L has a Fourier series of the form

FOURIER SERIES, GENERAIZED FUNCTIONS, APACE TRANSFORM 1 Fourier Series 1.1 General facts 1. A (generalized) function f(t) of period has a Fourier series of the form or f(t) a + a 1 cos πt + a cos πt +

### 11.2. Using a table of derivatives. Introduction. Prerequisites. Learning Style. Learning Outcomes

Using a table of derivatives 11.2 Introduction In Block 1 you were introduced to the idea of a derivative and you calculated some derivatives from first principles. Rather than always calculate the derivative

### Fourier series and transforms

Chapter 5 Fourier series and transforms Physical wavefields are often constructed from superpositions of complex exponential traveling waves, e i(kx ω(k)t). (5.) Here the wavenumber k ranges over a set

### Chapter Three. Functions. In this section, we study what is undoubtedly the most fundamental type of relation used in mathematics.

Chapter Three Functions 3.1 INTRODUCTION In this section, we study what is undoubtedly the most fundamental type of relation used in mathematics. Definition 3.1: Given sets X and Y, a function from X to

### 11.2. Using a Table of Derivatives. Introduction. Prerequisites. Learning Outcomes

Using a Table of Derivatives 11.2 Introduction In Section 11.1 you were introduced to the idea of a derivative and you calculated some derivatives from first principles. Rather than calculating the derivative

### The Derivative and the Tangent Line Problem. The Tangent Line Problem

The Derivative and the Tangent Line Problem Calculus grew out of four major problems that European mathematicians were working on during the seventeenth century. 1. The tangent line problem 2. The velocity

### nπt a n cos y(x) satisfies d 2 y dx 2 + P EI y = 0 If the column is constrained at both ends then we can apply the boundary conditions

Chapter 6: Page 6- Fourier Series 6. INTRODUCTION Fourier Anaysis breaks down functions into their frequency components. In particuar, a function can be written as a series invoving trigonometric functions.

### Solvability of Laplace s Equation Kurt Bryan MA 436

1 Introduction Solvability of Laplace s Equation Kurt Bryan MA 436 Let D be a bounded region in lr n, with x = (x 1,..., x n ). We see a function u(x) which satisfies u = in D, (1) u n u = h on D (2) OR

### An introduction to generalized vector spaces and Fourier analysis. by M. Croft

1 An introduction to generalized vector spaces and Fourier analysis. by M. Croft FOURIER ANALYSIS : Introduction Reading: Brophy p. 58-63 This lab is u lab on Fourier analysis and consists of VI parts.

### Solutions to Problems for 2D & 3D Heat and Wave Equations

Solutions to Problems for D & 3D Heat and Wave Equations 18.303 Linear Partial Differential Equations Matthew J. Hancock Fall 006 1 Problem 1 A rectangular metal plate with sides of lengths L, H and insulated

### Section 1.1 Real numbers. Set Builder notation. Interval notation

Section 1.1 Real numbers Set Builder notation Interval notation Functions a function is the set of all possible points y that are mapped to a single point x. If when x=5 y=4,5 then it is not a function

### Distributions, the Fourier Transform and Applications. Teodor Alfson, Martin Andersson, Lars Moberg

Distributions, the Fourier Transform and Applications Teodor Alfson, Martin Andersson, Lars Moberg 1 Contents 1 Distributions 2 1.1 Basic Definitions......................... 2 1.2 Derivatives of Distributions...................