Week 6 Lectures, Math 6451, Tanveer

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1 Fourier Series Week 6 Lectures, Math 645, Tanveer In the context of separation of variabe to find soutions of PDEs, we encountered or and in other cases f(x = f(x = a 0 + f(x = a 0 + b n sin nπx { a n cos nπx a n cos nπx for x (0, ( + b n sin nπx } for x (0, ( for x (, (3 The genera representation (3 is caed the Fourier representation of f(x in the (, interva, whie ( and ( are the Fourier sine and cosine representations of f(x in the interva (0,. Our discussions revove around some basic questions. When are represenations (-(3 vaid and in what sense?. How do we determine coefficients a n, b n in terms of f(x. 3. What conditions aow term by term differentiation of the Fourier-Series. Genera L theory We need enough generaity to be abe to ay the framework for discussion of more genera series representations of L, i.e. square integrabe functions than (-(3, since they arise in other PDE probems. In the space L (a, b of generay compex vaued functions in the interva (a, b, we introduce the L inner-product: (f, g = We note that the L norm is reated through b a f(xḡ(xdx (4 [ / b f = (f, f / = f (xdx] (5 This may be generaized to any number of dimension, with x repaced by x and integration over the interva (a, b, repaced by integration over appropriate n-dimensiona rectange. a

2 Definition A sequence {X n } L (a, b is orthogona if (X n, X m = 0 iff m n (6 This sequence is said to be orthonorma, if in addition (X n, X n = X n =. Theorem Let {X n } L (a, b be a orthogona set of functions. Let f <. Let N be a fixed positive integer. The choice of A n that minimizes mean square error E N = f N A nx n is given by A n = (f, X n X n (7 Further, we have f (f, X n X n Besse inequaity Proof. Define the square of the ( E N = f A n X n = f A n X n, f A n X n Expanding the above using properties of inner product and the orthogonaity of X n, we get E N = (f, f A n (X n, f n A n(f, X n + A n A n(x n, X n (8 We minimize E n as a function of N rea variabes, {(c n, d n } N, where A n = c n + id n. On taking partia derivatives, we get Aso, E N c n = (X n, f (f, X n + c n (X n, X n = 0 impying c n = R {(X n, f} X n E N = i(x n, f + i(f, X n + d n (X n, X n = 0 impying d n = I {(X n, f} d n X n Therefore A n = c n + id n is given by (7. Again with A n given by (7, the corresponding E N becomes E N = f A n X n 0 n N Taking the imit of N, we obtain Besse inequaity. Theorem 3 (Parseva s equaity The mean square error E N, defined in Theorem, 0 as N, if and ony if (f, X n X n = f ; Parseva equaity (9 When condition (9 hods for any function f L (a, b, the orthogona sequence {X n } is compete and forms a basis in L (a, b.

3 Proof. From Theorem, im E N = f N (f, X n X n Therefore, it is zero iff and ony if Parseva s equaity hods. Lemma 4 The sequence is an orthogona sequence in L (, π {, cosx, sin x, cosx, sin x,...} {X n } Proof. This invoves a simpe cacuation. First we note that for any n, (, sinnx = sinnxdx = 0 and aso (, cosnx = 0. Further, for m n, whie for m n, (sin mx, cosnx = [sin(m + nx + sin(m nx] dx = 0 (cosmx,cos nx = On the other hand for m = n, [cos(m nx + cos(m + nx] dx = 0 and Whie (sin mx, cosnx = (sinnx, cosnx = (sin nx, sin nx = (cosnx, cos nx = sin nxdx = cos nxdx = π sin nxdx = 0 ( cosnxdx = π ( + cosnxdx = π Coroary 5 The sequence {, cos πx is an orthogona sequence in L (,., sin πx, cos πx, sin πx },.. Proof. This foows simpy by introducing rescaed independent variabe xπ to (, π. We can then use the previous Lemma., which maps (, Remark By using a shift and a scaing, one can get a simiar Fourier representations for square integrabe functions in (a, b. 3

4 Lemma 6 For a L function f in (, interva, the choice of a m and b m that minimizes the mean-square error ( f(x a M 0 a m cos mπx M b m sin mπx dx for any positive integer M is given by a m = b m = m= m= f(xcos mπx dx for 0 m M f(xsin mπx dx for 0 m M These are referred to as the Fourier-Coefficients of f(x. Further, ( a 0 + a m + b m f(x dx (0 m= m= Proof. This foows simpy by using the formua in Theorem by identifying (a, b = (,, X =, X = cosx, X 3 = sin x, X 4 = cosx, X 5 = sin x, etc and using vaues of integras (X n, X n. The inequaity (0 is simpy a restatement of Besse inequaity for this case. Coroary 7 For odd functions f L (,, the cosine coefficients a m = 0, whie for even square integrabe functions, the sine coefficients b m = 0. Proof. Proof foows simpy by noting that if f is odd, then the formua that a m invoves integra of an odd function over (,, which is 0. When f is even, the same happens to the formua for b m. Remark The above coroary impies that we do not need a theory for Fourier Sine and Fourier Cosine Series in ( and (, separate from the fu Fourier Series (3, because we can do an odd or even extension of the function f(x to the interva (,. Then (3 wi reduce to either ( or ( depending on whether the extension was odd or even. 3 Pointwise convergence of Fourier Series The primary purpose of this section is to prove the foowing Theorem: Theorem 8 (Pointwise Convergence of Fourier Series Assume f (x ± exists at each point x (, and that f(x is bounded in [, ]. Then,. The Fourier Series ( for f converges pointwise on (, provided f C 0 [, ].. More generay, is f PC 0 [, ], i.e. continuous in the interva [, ] except for a finite set of points, then the Fourier series converges at every point x (,. The sum is [f(x+ + f(x ] for x (, and [f ext(x + + f ext (x ] for x (,, where f ext (x is the -periodic extension of f(x. 4

5 Remark 3 The hypothesis for f can be weakened even further, though it wi not be necessary for the appications we have in mind. The proof of Theorem 8 wi have to await some preiminary Lemmas. We wi ony prove it for the specia case = π, since the genera case woud foow merey by rescaing variabe x, as seen before. Definition 9 For f L [, π] we define S N (x = a 0 + N where a m and b m are determined from Lemma 0 Further, a m = b m = S N (x = π m= π m= (a m cosmx + b m sin mx ( f(xcos mπx dx for 0 m M ( f(xsin mπx dx for 0 m M (3 K N (x yf(ydy where K N (θ = sin[( N + sin θ 0 K N (θdθ = 0 K N (θdθ = π Proof. Using the formua for Fourier Coefficients, we get S N (x = [ ] π + (cosmy cosmx + sin my sin mx f(ydy = K(x yf(ydy π π Therefore, K N (θ = + cosnθ = n= N θ ] e inθ = e inθ e i(n+θ e iθ = sin [( ] N + θ sin θ Equation (5 foows simpy by using the cosine series expansion of K N (θ in (6 and integrating term by term. A terms are zero except the first one, which gives a haf. Lemma Assume the same conditions on f as stated in Theorem 8. For x (,, define (4 (5 (6 g ± (θ = f(x + θ f(x± sin θ (7 Then, 0 g +(θ < and 0 g (θ <. 5

6 Proof. A sufficient condition for each of these integras to exist is for im θ 0 + g + (θ and im θ 0 g (θ to exist, since for other vaues of θ conditions on f(x make g ± square integrabe over the intervas (0, π, (, 0 respectivey. However, it is easy to see im g f(x + θ f(x + +(θ = im θ 0 + θ 0 + θ θ sin θ = f (x + and each of which exists. f(x + θ f(x im g (θ = im θ 0 θ 0 θ θ sin θ = f (x Proof of Theorem 8 We wi assume = π since otherwise, we can rescae x. Note we ony need to prove the second part, as the first part is a specia case. S N (x [ f(x + + f(x ] K N (θ [ = f(x + θ f(x + ] 0 K N (θ [ dθ+ f(x + θ f(x ] dθ 0 π π {( = g + (θsin N + } 0 {( θ dθ + g + (θsin N + } θ dθ (8 0 It is easy to check that { sin ( } N + θ is an is an orthogona set of functions either in the N= interva (, 0 or (0, π. Therefore, from appying Besse s inequaity, it foows that each of the integras on the right of (8 tend to zero N, since Lemma shows that g + and g are square integrabe in (0, π and (, 0 respectivey. Further, for x outside the interva (, π, it is cear that the Fourier Series is periodic and therefore converges to the periodicay extended function [f ext(x + + f ext (x ]. Lemma If f C 0 (, and is periodic, and f (x ± exists at every point x (,, then S N (x converges to f(x pointwise. Proof. We note that outside the (,, the extended function f ext (x = f(x itsef. Since the function is continuous every where, [f ext(x + + f ext (x ] = f(x. From Theorem (8, the coroary foows. Coroary 3 If conditions of Theorem (8 hod, then the Fourier series ( converges at the end points ± to f(± if and ony if f( = f(. Proof. We simpy note that when the condition f( = f(, then the extended function f ext C 0 (, and its derivatives exists at every point in (,. Appying previous emma, the concuson foows. 4 Uniform Convergence We wi now seek stronger conditions on f so that S N (x is uniformy convergent to f for f C 0 (,. 6

7 Lemma 4 Assume f exists at each point in [, ] and that f L (,, and f( = f(. Then, S N (x converges to f(x uniformy in [, ], i.e. im f S N 0 N Proof. We wi assume = π since otherwise we can introduce appropriate rescaed variabe. The conditions on f guarantees that and a m π b m π f (xcos mx dx f (xsin mx dx exists, and from Besse s inequaity [ π (a { 0 + (a m + (b m }] which impies that m= im N m=n+ { (a m + (b m } 0 On integration by parts, the Fourier coefficients of f itsef becomes: f(x S N (x a m = π b m = π f (x < f(xcos mxdx = f (xsin mxdx = b m πm m f(xsin mxdx = f (xcos mxdx = a m πm m a m cosmx + b m sin mx N+ ( Therefore im N f S N = 0. N+ ( a m + b m N+ N+ m ( a m + b m / { } / m (a m + (b m 0 as N (9 N+ Remark 4 If f C 0 (, and periodic (even without any condition on derivatives it can be uniformy approximated in the maximum norm sense by a trigonometric poynomia, which is defined to be T(x = c 0 + N m= ( c n cos nπx + d n sin nπx In order to show this, we first prove that such an f can be approximated we by a function -periodic g C (,, which wi aow us to use Lemma 4. 7

8 Lemma 5 If f C 0 (, and periodic. For any ǫ > 0, there exists a function g C (, and periodic such that f g < ǫ. Proof. For each δ > 0. Let F δ (x δ x+δ x δ f(ydy = δ δ δ f(x + tdt It is easiy verified that F δ is -periodic and F δ C (,. Since f(x C 0 (, and periodic, it is uniformy continuous. Given ǫ > 0, there exists δ > 0 independent of x, so that f(x + t f(x < ǫ for t < δ. Hence, if we choose such a δ and define g(x = F δ (x, g(x f(x δ δ This is true for a x, hence Lemma foows. δ(f(x + t f(xdt δ δ δ f(x + t f(x dt ǫ Theorem 6 (Weirstrass Approximation Theorem If f C 0 (, and periodic. For any ǫ > 0, there exists a trigonometric poynomia T(x so that f T < ǫ. Proof. According to previous Lemma, for given ǫ, there exists -periodic g C (, so that f g < ǫ. Further, for such a g, Lemma 4 impies there exists T(x, namey its Fourier series truncated to N terms for a sufficienty arge N, so that g T < ǫ. Hence f T f g + g T < ǫ Theorem 7 (Competeness of Fourier Series The Fourier Series of any function f L [, ] converges to f in the L norm, i.e. in the mean-square sense and the Parseva s equaity hods. Proof. It is known that L space is the competion of the set of C 0 [, ] with the L norm. This means that that any ǫ > 0, there exists f c C 0 [, ] so that f f c < ǫ, where. denotes the L norm. We can aso arrange f c ( = f c (. Now we appy periodic extension to f c, so that f c C 0 (,. Appy the Weirstrass approximation theorem, there exists trigonometric poynomia T so that T f c < ǫ. Therefore, over the finite interva [, ], Therefore, [ / T f c = T(x F c (x dx] ǫ T f T f c + f f c ǫ Now, we know that for any trignometric poynomia with N terms the best approximation in the. sense is through Fourier Series trucated to N term. If we denote this Fourier Poynomia by T f, we get the resut that T f f T f ǫ This is true for any ǫ. Hence Fourier Series converges to f in the mean-square sense. 8

9 5 Non-Uniform Convergence near a point of discontinuity Remark 5 For piecewise continuous functions, near points of discontinuity, S N (x does not approach f(x uniformy, even though it does so pointwise, except at the discontinuity itsef. This is caed the Gibbs phenomena. Theorem 8 (Gibb s pheneomena Assume f PC 0 [, ] and f (x ± exists at each x [, ]. If there is a point of point of discontinuity x d (,, then im sup N x (x d, im sup N x (,x d S N (x f(x 0 S N (x f(x 0 Proof. We wi ony show the first part, since the proof of the second part proof is very simiar. Aso, for simpicity of notation, we take = π, x d = 0. We first consider the piecewise continuous function h(x defined as h(x = for π > x > 0 and h(x = for π < x < 0 We can cacuate its Fourier Coefficient and find its Fourier Series to be sin nx nπ We cacuate ( S N (x = 0,3,5.. 0 K N (x y dy ( 0 sin[(n + /(x y] 4π = 0 sin [(x y/] Define M = N +. In the first integra we take θ = M(x y, and in the second integra, take θ = M(y x. Then S N (x = ( Mx Mx M(x M(x+π Mπ Mx Mx Mπ+Mx ( Mx sin θ dθ M sin [θ/(m] π ( sin θ dθ Mx M sin[θ/(m] π = Mx Mπ+Mx Mπ Mx Using cacuus, we find that the first integra is maximized at x = π M. ( ( π Mπ+π sin θ dθ S N = M M sin(θ/(m π Mπ dy 4π sin θ dθ M sin [θ/(m] π The second integra tends to 0 as M, since the denominator in the integrand is bounded beow by M. Therefore, as M, ( π sin θ dθ S N M θ π = 0.59 (0 9

10 There is an overshoot. We eave the detais for a more genera piecewise continuous function as an exercise, except to note that if f(x has a jump at 0, then if we decompose f(x = [f(0+ f(0 ]h(x + ( f(x [f(0+ f(0 ]h(x then the term within (. is continuous at x = 0, aowing use of previous theorems. Remark 6 In the foowing theorem, we consider a usefu condition that aows a Fourier series to be differentiated term by term. Theorem 9 (Differentiabiity of Fourier Series If the Fourier series has the property that f(x = a 0 + m= (a m cosmx + b m sinmx m( a m + b m < m Then the Fourier Series can be differentiated term by term and the differentiated series converges uniformy to f. Proof. Leave it as an exercise. 0