# Chapter 28. Fourier Series An Eigenvalue Problem.

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1 Chpter 28 Fourier Series Every time I close my eyes The noise inside me mplifies I cn t escpe I relive every moment of the dy Every misstep I hve mde Finds wy it cn invde My every thought And this is why I find myself wke -Filure -Tom Sher (Assemblge 23) 28. An Eigenvlue Problem. A self djoint eigenvlue problem. Consider the eigenvlue problem y + λy = 0, y() = y(π), y () = y (π). 330

2 We rewrite the eqution so the eigenvlue is on the right side. L[y] y = λy We demonstrte tht this eigenvlue problem is self djoint. v L[u] L[v] u = v u v u = [ vu ] π + v u [ v u] π v u = v(π)u (π) + v()u () + v (π)u(π) v ()u() = v(π)u (π) + v(π)u (π) + v (π)u(π) v (π)u(π) = 0 Since Green s Identity reduces to v L[u] L[v] u = 0, the problem is self djoint. This mens tht the eigenvlues re rel nd tht eigenfunctions corresponding to distinct eigenvlues re orthogonl. We compute the Ryleigh quotient for n eigenvlue λ with eigenfunction φ. We see tht the eigenvlues re non-negtive. λ = [ φφ ] π + φ φ φ φ = φ(π)φ (π) + φ()φ () + φ φ φ φ = φ(π)φ (π) + φ(π)φ (π) + φ φ φ φ = φ φ φ φ Computing the eigenvlues nd eigenfunctions. Now we find the eigenvlues nd eigenfunctions. First we consider the cse λ = 0. The generl solution of the differentil eqution is y = c + c 2 x. 33

3 The solution tht stisfies the boundry conditions is y = const. Now consider λ > 0. The generl solution of the differentil eqution is ( ) ( ) y = c cos λx + c 2 sin λx. We pply the first boundry condition. Then we pply the second boundry condition. y() = y(π) ( c cos ) ( λπ + c 2 sin ) ( ) ( ) λπ = c cos λπ + c 2 sin λπ ( ) ( ) ( ) ( ) c cos λπ c 2 sin λπ = c cos λπ + c 2 sin λπ ( ) c 2 sin λπ = 0 y () = y (π) c λ sin ( ) ( λπ + c 2 λ cos ) ( ) ( ) λπ = c λ sin λπ + c 2 λ cos λπ ( ) ( ) ( ) ( ) c sin λπ + c 2 cos λπ = c sin λπ + c 2 cos λπ ( ) c sin λπ = 0 ( λπ ) To stisify the two boundry conditions either c = c 2 = 0 or sin = 0. The former yields the trivil solution. The ltter gives us the eigenvlues λ n = n 2, n Z +. The corresponding solution is y n = c cos(nx) + c 2 sin(nx). There re two eigenfunctions for ech of the positive eigenvlues. We choose the eigenvlues nd eigenfunctions. λ 0 = 0, φ 0 = 2 λ n = n 2, φ 2n = cos(nx), φ 2n = sin(nx), for n =, 2, 3,

4 Orthogonlity of Eigenfunctions. We know tht the eigenfunctions of distinct eigenvlues re orthogonl. In ddition, the two eigenfunctions of ech positive eigenvlue re orthogonl. cos(nx) sin(nx) dx = [ ] π 2n sin2 (nx) = 0 Thus the eigenfunctions {, cos(x), sin(x), cos(2x), sin(2x)} re n orthogonl set Fourier Series. A series of the eigenfunctions φ 0 = 2, φ() n = cos(nx), φ (2) n = sin(nx), for n is ( n cos(nx) + b n sin(nx) ). This is known s Fourier series. (We choose φ 0 = so ll of the eigenfunctions hve the sme norm.) A firly generl 2 clss of functions cn be expnded in Fourier series. Let f(x) be function defined on < x < π. Assume tht f(x) cn be expnded in Fourier series f(x) ( n cos(nx) + b n sin(nx) ). (28.) Here the mens hs the Fourier series. We hve not sid if the series converges yet. For now let s ssume tht the series converges uniformly so we cn replce the with n =. 333

5 We integrte Eqution 28. from to π to determine 0. f(x) dx = 2 0 f(x) dx = π 0 + dx + n cos(nx) + b n sin(nx) dx ( n cos(nx) dx + b n 0 = π f(x) dx = π 0 f(x) dx Multiplying by cos(mx) nd integrting will enble us to solve for m. f(x) cos(mx) dx = cos(mx) dx ( n cos(nx) cos(mx) dx + b n ) sin(nx) dx ) sin(nx) cos(mx) dx All but one of the terms on the right side vnishes due to the orthogonlity of the eigenfunctions. f(x) cos(mx) dx = m cos(mx) cos(mx) dx ( ) f(x) cos(mx) dx = m 2 + cos(2mx) dx m = π f(x) cos(mx) dx = π m f(x) cos(mx) dx. 334

6 Note tht this formul is vlid for m = 0,, 2,... Similrly, we cn multiply by sin(mx) nd integrte to solve for b m. The result is b m = π f(x) sin(mx) dx. n nd b n re clled Fourier coefficients. Although we will not show it, Fourier series converge for firly generl clss of functions. Let f(x ) denote the left limit of f(x) nd f(x + ) denote the right limit. Exmple For the function defined the left nd right limits t x = 0 re f(x) = { 0 for x < 0, x + for x 0, f(0 ) = 0, f(0 + ) =. Result Let f(x) be 2π-periodic function for which Fourier coefficients n = π f(x) cos(nx) dx, b n = π f(x) dx exists. Define the f(x) sin(nx) dx. If x is n interior point of n intervl on which f(x) hs limited totl fluctution, then the Fourier series of f(x) ( n cos(nx) + b n sin(nx) ), converges to 2 (f(x ) + f(x + )). If f is continuous t x, then the series converges to f(x). 335

7 Periodic Extension of Function. Let g(x) be function tht is rbitrrily defined on x < π. The Fourier series of g(x) will represent the periodic extension of g(x). The periodic extension, f(x), is defined by the two conditions: f(x) = g(x) for π x < π, f(x + 2π) = f(x). The periodic extension of g(x) = x 2 is shown in Figure Figure 28.: The Periodic Extension of g(x) = x 2. Limited Fluctution. A function tht hs limited totl fluctution cn be written f(x) = ψ + (x) ψ (x), where ψ + nd ψ re bounded, nondecresing functions. An exmple of function tht does not hve limited totl fluctution 336

8 is sin(/x), whose fluctution is unlimited t the point x = 0. Functions with Jump Discontinuities. Let f(x) be discontinuous function tht hs convergent Fourier series. Note tht the series does not necessrily converge to f(x). Insted it converges to ˆf(x) = 2 (f(x ) + f(x + )). Exmple Consider the function defined by { x for π x < 0 f(x) = π 2x for 0 x < π. The Fourier series converges to the function defined by 0 for x = x for π < x < 0 ˆf(x) = π/2 for x = 0 π 2x for 0 < x < π. The function ˆf(x) is plotted in Figure Lest Squres Fit Approximting function with Fourier series. Suppose we wnt to pproximte 2π-periodic function f(x) with finite Fourier series. f(x) N ( n cos(nx) + b n sin(nx)) Here the coefficients re computed with the fmilir formuls. Is this the best pproximtion to the function? Tht is, is it possible to choose coefficients α n nd β n such tht f(x) α N (α n cos(nx) + β n sin(nx)) 337

9 Figure 28.2: Grph of ˆf(x). would give better pproximtion? Lest squred error fit. The most common criterion for finding the best fit to function is the lest squres fit. The best pproximtion to function is defined s the one tht minimizes the integrl of the squre of the devition. Thus if f(x) is to be pproximted on the intervl x b by series N f(x) c n φ n (x), (28.2) 338

10 the best pproximtion is found by choosing vlues of c n tht minimize the error E. N 2 E f(x) c n φ n (x) dx Generlized Fourier coefficients. We consider the cse tht the φ n re orthogonl. For simplicity, we lso ssume tht the φ n re rel-vlued. Then most of the terms will vnish when we interchnge the order of integrtion nd summtion. ( b N N N ) E = f 2 2f c n φ n + c n φ n c m φ m dx E = E = E = f 2 dx 2 We complete the squre for ech term. E = f 2 dx + N c n fφ n dx + f 2 dx 2 f 2 dx + N N m= m= N c n fφ n dx + N φ 2 n dx ( c 2 n ( c n Ech term involving c n is non-negtive, nd is minimized for N c n c m φ n φ m dx N c 2 n φ 2 n dx ) φ 2 n dx 2c n fφ n dx fφ n dx φ2 n dx ) 2 ( fφ ) 2 n dx b φ2 n dx c n = fφ n dx φ2 n dx. (28.3) 339

11 We cll these the generlized Fourier coefficients. For such choice of the c n, the error is Since the error is non-negtive, we hve E = f 2 dx N c 2 n φ 2 n dx. N f 2 dx c 2 n φ 2 n dx. This is known s Bessel s Inequlity. If the series in Eqution 28.2 converges in the men to f(x), lim N E = 0, then we hve equlity s N. f 2 dx = φ 2 n dx. This is Prsevl s equlity. c 2 n Fourier coefficients. Previously we showed tht if the series, f(x) = ( n cos(nx) + b n sin(nx), converges uniformly then the coefficients in the series re the Fourier coefficients, n = f(x) cos(nx) dx, b n = π π 340 f(x) sin(nx) dx.

12 Now we show tht by choosing the coefficients to minimize the squred error, we obtin the sme result. We pply Eqution 28.3 to the Fourier eigenfunctions. n = b n = 0 = f 2 dx dx = π 4 f cos(nx) dx cos2 (nx) dx = π f sin(nx) dx sin2 (nx) dx = π f(x) dx f(x) cos(nx) dx f(x) sin(nx) dx 28.4 Fourier Series for Functions Defined on Arbitrry Rnges If f(x) is defined on c d x < c + d nd f(x + 2d) = f(x), then f(x) hs Fourier series of the form Since f(x) ( ) ( ) nπ(x + c) nπ(x + c) n cos + b n sin. d d c+d c d ( ) nπ(x + c) cos 2 dx = d the Fourier coefficients re given by the formuls n = d b n = d c+d c d c+d c d c+d c d ( ) nπ(x + c) sin 2 dx = d, d ( ) nπ(x + c) f(x) cos dx d ( ) nπ(x + c) f(x) sin dx. d 34

13 Exmple Consider the function defined by x + for x < 0 f(x) = x for 0 x < 3 2x for x < 2. This function is grphed in Figure The Fourier series converges to ˆf(x) = (f(x ) + f(x + ))/2, for x = 2 x + for < x < 0 ˆf(x) = for x = 0 2 x for 0 < x < 3 2x for x < 2. ˆf(x) is lso grphed in Figure The Fourier coefficients re n = 2 ( ) 2nπ(x + /2) f(x) cos dx 3/2 3 = 2 5/2 ( ) 2nπx f(x /2) cos dx 3 /2 3 = 2 /2 ( ) 2nπx (x + /2) cos dx / /2 ( ) 2nπx (4 2x) cos dx 3 3/2 3 = ( 2nπ (nπ) sin 2 3 3/2 /2 ) [ ( nπ 2( ) n nπ + 9 sin 3 ( ) 2nπx (x /2) cos dx 3 )] 342

14 Figure 28.3: A Function Defined on the rnge x < 2 nd the Function to which the Fourier Series Converges. b n = 3/2 = 2 3 = /2 /2 /2 /2 ( ) 2nπ(x + /2) f(x) sin dx 3 ( ) 2nπx f(x /2) sin dx 3 ( ) 2nπx (x + /2) sin dx + 2 3/2 ( ) 2nπx (x /2) sin dx 3 3 / /2 ( ) 2nπx (4 2x) sin dx 3 3/2 3 ) [ ( nπ ) ( nπ )] 2( ) n nπ + 4nπ cos 3 sin 3 3 = 2 (nπ) 2 sin2 ( nπ 3 343

15 28.5 Fourier Cosine Series If f(x) is n even function, (f( x) = f(x)), then there will not be ny sine terms in the Fourier series for f(x). The Fourier sine coefficient is b n = π f(x) sin(nx) dx. Since f(x) is n even function nd sin(nx) is odd, f(x) sin(nx) is odd. b n is the integrl of n odd function from to π nd is thus zero. We cn rewrite the cosine coefficients, n = π = 2 π Exmple Consider the function defined on [0, π) by f(x) = The Fourier cosine coefficients for this function re n = 2 π = { π /2 0 0 f(x) cos(nx) dx f(x) cos(nx) dx. { x for 0 x < π/2 π x for π/2 x < π. x cos(nx) dx + 2 π π/2 for n = 0, 4 8 cos ( ) ( ) nπ πn 2 2 sin 2 nπ for n. 4 (π x) cos(nx) dx In Figure 28.4 the even periodic extension of f(x) is plotted in dshed line nd the sum of the first five nonzero terms in the Fourier cosine series re plotted in solid line. 344

16 Figure 28.4: Fourier Cosine Series Fourier Sine Series If f(x) is n odd function, (f( x) = f(x)), then there will not be ny cosine terms in the Fourier series. Since f(x) cos(nx) is n odd function, the cosine coefficients will be zero. Since f(x) sin(nx) is n even function,we cn rewrite the sine coefficients b n = 2 π 0 f(x) sin(nx) dx. 345

17 Exmple Consider the function defined on [0, π) by { x for 0 x < π/2 f(x) = π x for π/2 x < π. The Fourier sine coefficients for this function re /2 b n = 2 x sin(nx) dx + 2 π 0 π = 6 ( nπ ) ( nπ ) πn cos sin π/2 (π x) sin(nx) dx In Figure 28.5 the odd periodic extension of f(x) is plotted in dshed line nd the sum of the first five nonzero terms in the Fourier sine series re plotted in solid line Complex Fourier Series nd Prsevl s Theorem By writing sin(nx) nd cos(nx) in terms of e ınx nd e ınx we cn obtin the complex form for Fourier series ( n cos(nx) + b n sin(nx) ) = ( ) n 2 (eınx + e ınx ) + b n ı2 (eınx e ınx ) = ( 2 ( n ıb n ) e ınx + ) 2 ( n + ıb n ) e ınx where = n= c n e ınx ( 2 n ıb n ) for n c n = 0 2 for n = 0 ( 2 n + ıb n ) for n. 346

18 Figure 28.5: Fourier Sine Series. The functions {...,e ıx,, e ıx, e ı2x,...}, stisfy the reltion e ınx e ımx dx = e ı(n m)x dx { 2π for n = m = 0 for n m. Strting with the complex form of the Fourier series of function f(x), f(x) c n e ınx, 347

19 we multiply by e ımx nd integrte from to π to obtin If f(x) is rel-vlued then f(x) e ımx dx = c n e ınx e ımx dx c m = f(x) e ımx dx 2π c m = f(x) e ımx dx = f(x)(e 2π 2π ımx ) dx = c m where z denotes the complex conjugte of z. Assume tht f(x) hs uniformly convergent Fourier series. f 2 (x) dx = = 2π = 2π = 2π ( m= n= ( n= c n c n ( c m e ımx )( n= c n e ınx ) dx [ ] 4 ( n + ıb n )( n ıb n ) ) ( 2 n + b 2 n) [ ] ) 4 ( n ıb n )( n + ıb n ) This yields result known s Prsevl s theorem which holds even when the Fourier series of f(x) is not uniformly convergent. 348

20 Result Prsevl s Theorem. If f(x) hs the Fourier series then f(x) ( n cos(nx) + b n sin(nx)), f 2 (x) dx = π π ( 2 n + b 2 n) Behvior of Fourier Coefficients Before we jump hip-deep into the grunge involved in determining the behvior of the Fourier coefficients, let s tke step bck nd get some perspective on wht we should be looking for. One of the importnt questions is whether the Fourier series converges uniformly. From Result 2.2. we know tht uniformly convergent series represents continuous function. Thus we know tht the Fourier series of discontinuous function cnnot be uniformly convergent. From Section 2.2 we know tht series is uniformly convergent if it cn be bounded by series of positive terms. If the Fourier coefficients, n nd b n, re O(/n α ) where α > then the series cn be bounded by (const) /nα nd will thus be uniformly convergent. Let f(x) be function tht meets the conditions for hving Fourier series nd in ddition is bounded. Let (, p ), (p, p 2 ), (p 2, p 3 ),...,(p m, π) be prtition into finite number of intervls of the domin, (, π) such tht on ech intervl f(x) nd ll it s derivtives re continuous. Let f(p ) denote the left limit of f(p) nd f(p + ) denote the right limit. f(p ) = lim f(p ǫ), f(p+ ǫ 0 + ) = lim f(p + ǫ) ǫ 0 + Exmple The function shown in Figure 28.6 would be prtitioned into the intervls ( 2, ), (, 0), (0, ), (, 2). 349

21 Figure 28.6: A Function tht cn be Prtitioned. Suppose f(x) hs the Fourier series f(x) n cos(nx) + b n sin(nx). We cn use the integrl formul to find the n s. n = π = π ( p f(x) cos(nx) dx f(x) cos(nx) dx + p2 ) f(x) cos(nx) dx + + f(x) cos(nx) dx p p m 350

22 Using integrtion by prts, = ( [ ] p [ ] p2 [ ] ) π f(x) sin(nx) + f(x) sin(nx) + + f(x) sin(nx) nπ p p m ( p p2 ) f (x) sin(nx) dx + f (x) sin(nx) dx + f (x) sin(nx) dx nπ p p m = { [f(p ) f(p + ) ] sin(np ) + + [ f(p nπ m) f(p + m) ] } sin(np m ) π f (x) sin(nx) dx n π = n A n n b n where A n = π m sin(np j ) [ f(p j ) f(p+ j )] j= nd the b n re the sine coefficients of f (x). Since f(x) is bounded, A n = O(). Since f (x) is bounded, b n = π f (x) sin(nx) dx = O(). Thus n = O(/n) s n. (Actully, from the Riemnn-Lebesgue Lemm, b n = O(/n).) 35

23 Now we repet this nlysis for the sine coefficients. b n = π = π = nπ f(x) sin(nx) dx p2 ) f(x) sin(nx) dx + f(x) sin(nx) dx + + f(x) sin(nx) dx p p m { [f(x) ] p cos(nx) + [ f(x) cos(nx) ] p 2 p + + [ f(x) cos(nx) ] } π p m + ( p p2 f (x) cos(nx) dx + f (x) cos(nx) dx + nπ p ( p = n B n + n n ) f (x) cos(nx) dx p m where B n = ( )n [ ] f() f(π) π π m cos(np j ) [ f(p j ) f(p+ j )] j= nd the n re the cosine coefficients of f (x). Since f(x) nd f (x) re bounded, B n, n = O() nd thus b n = O(/n) s n. With integrtion by prts on the Fourier coefficients of f (x) we could find tht n = n A n n b n where A n = π m j= sin(np j)[f (p j ) f (p + j )] nd the b n re the sine coefficients of f (x), nd b n = n B n + n n where B n = ( )n π [f () f (π)] π m j= cos(np j)[f (p j ) f (p + j )] nd the n re the cosine coefficients of f (x). 352

24 Now we cn rewrite n nd b n s n = n A n + n 2B n n 2 n b n = n B n + n 2A n n 2b n. Continuing this process we could define A (j) n nd B (j) n so tht n = n A n + n 2B n n 3A n n 4B n + b n = n B n + n 2A n + n 3B n n 4A n. For ny bounded function, the Fourier coefficients stisfy n, b n = O(/n) s n. If A n nd B n re zero then the Fourier coefficients will be O(/n 2 ). A sufficient condition for this is tht the periodic extension of f(x) is continuous. We see tht if the periodic extension of f (x) is continuous then A n nd B n will be zero nd the Fourier coefficients will be O(/n 3 ). Result Let f(x) be bounded function for which there is prtition of the rnge (, π) into finite number of intervls such tht f(x) nd ll it s derivtives re continuous on ech of the intervls. If f(x) is not continuous then the Fourier coefficients re O(/n). If f(x), f (x),...,f (k 2) (x) re continuous then the Fourier coefficients re O(/n k ). If the periodic extension of f(x) is continuous, then the Fourier coefficients will be O(/n 2 ). The series n cos(nx) cn be bounded by M /n2 where M = mx ( n + b n ). Thus the Fourier series converges to f(x) uniformly. n Result If the periodic extension of f(x) is continuous then the Fourier series of f(x) will converge uniformly for ll x. If the periodic extension of f(x) is not continuous, we hve the following result. 353

25 Result If f(x) is continuous in the intervl c < x < d, then the Fourier series is uniformly convergent in the intervl c + δ x d δ for ny δ > 0. Exmple Different Rtes of Convergence. A Discontinuous Function. Consider the function defined by f (x) = { for < x < 0, for 0 < x <. This function hs jump discontinuities, so we know tht the Fourier coefficients re O(/n). Since this function is odd, there will only be sine terms in it s Fourier expnsion. Furthermore, since the function is symmetric bout x = /2, there will be only odd sine terms. Computing these terms, b n = 2 sin(nπx) dx 0 [ ] = 2 nπ cos(nπx) 0 = 2 ( ( )n nπ ) nπ = { 4 nπ for odd n 0 for even n. The function nd the sum of the first three terms in the expnsion re plotted, in dshed nd solid lines respectively, in Figure Although the three term sum follows the generl shpe of the function, it is clerly not good pproximtion. 354

26 Figure 28.7: Three Term Approximtion for Function with Jump Discontinuities nd Continuous Function. A Continuous Function. Consider the function defined by x for < x < /2 f 2 (x) = x for /2 < x < /2 x + for /2 < x <. 355

27 Figure 28.8: Three Term Approximtion for Function with Continuous First Derivtive nd Comprison of the Rtes of Convergence. Since this function is continuous, the Fourier coefficients will be O(/n 2 ). Also we see tht there will only be odd sine terms in the expnsion. b n = = 2 /2 /2 0 ( x ) sin(nπx) dx + /2 /2 x sin(nπx) dx + x sin(nπx) dx + 2 ( x) sin(nπx) dx /2 = 4 (nπ) sin(nπ/2) { 2 4 ( ) (n )/2 for odd n (nπ) = 2 0 for even n. 356 /2 ( x + ) sin(nπx) dx

28 The function nd the sum of the first three terms in the expnsion re plotted, in dshed nd solid lines respectively, in Figure We see tht the convergence is much better thn for the function with jump discontinuities. A Function with Continuous First Derivtive. Consider the function defined by f 3 (x) = { x( + x) for < x < 0 x( x) for 0 < x <. Since the periodic extension of this function is continuous nd hs continuous first derivtive, the Fourier coefficients will be O(/n 3 ). We see tht the Fourier expnsion will contin only odd sine terms. b n = = x( + x) sin(nπx) dx + x( x) sin(nπx) dx = 4( ( )n ) (nπ) 3 = { 4 (nπ) 3 for odd n 0 for even n. 0 x( x) sin(nπx) dx The function nd the sum of the first three terms in the expnsion re plotted in Figure We see tht the first three terms give very good pproximtion to the function. The plots of the function, (in dshed line), nd the three term pproximtion, (in solid line), re lmost indistinguishble. In Figure 28.8 the convergence of the of the first three terms to f (x), f 2 (x), nd f 3 (x) re compred. In the lst grph we see closeup of f 3 (x) nd it s Fourier expnsion to show the error. 357

29 28.9 Gibb s Phenomenon The Fourier expnsion of f(x) = { for 0 x < for x < 0 is f(x) 4 π n sin(nπx). For ny fixed x, the series converges to 2 (f(x ) + f(x + )). For ny δ > 0, the convergence is uniform in the intervls + δ x δ nd δ x δ. How will the nonuniform convergence t integrl vlues of x ffect the Fourier series? Finite Fourier series re plotted in Figure 28.9 for 5, 0, 50 nd 00 terms. (The plot for 00 terms is closeup of the behvior ner x = 0.) Note tht t ech discontinuous point there is series of overshoots nd undershoots tht re pushed closer to the discontinuity by incresing the number of terms, but do not seem to decrese in height. In fct, s the number of terms goes to infinity, the height of the overshoots nd undershoots does not vnish. This is known s Gibb s phenomenon Integrting nd Differentiting Fourier Series Integrting Fourier Series. Since integrtion is smoothing opertion, ny convergent Fourier series cn be integrted term by term to yield nother convergent Fourier series. Exmple Consider the step function f(x) = { π for 0 x < π for π x <

30 Figure 28.9: Since this is n odd function, there re no cosine terms in the Fourier series. b n = 2 π sin(nx) dx π 0 [ = 2 ] π n cos(nx) 0 = 2 n ( ( )n ) = { 4 n for odd n 0 for even n. 359

31 Integrting this reltion, f(x) oddn 4 sin nx n Since this series converges uniformly, 4 oddn x n 4 2 oddn f(t) dt F(x) = = = 4 x oddn x oddn oddn oddn 4 n oddn 4 n sin(nt)dt sin(nt) dt [ 4 ] x n n cos(nt) 4 n 2( cos(nx) + ( )n ) n 4 2 oddn cos(nx) n 2 { cos(nx) x π for π x < 0 = F(x) = n 2 x π for 0 x < π. The vlue of the constnt term is 4 oddn n = 2 F(x) dx = 2 π 0 π. 360

32 Thus π 4 oddn { cos(nx) x π for π x < 0 = n 2 x π for 0 x < π. Differentiting Fourier Series. Recll tht in generl, series cn only be differentited if it is uniformly convergent. The necessry nd sufficient condition tht Fourier series be uniformly convergent is tht the periodic extension of the function is continuous. Result The Fourier series of function f(x) cn be differentited only if the periodic extension of f(x) is continuous. Exmple Consider the function defined by { π for 0 x < π f(x) = for π x < 0. f(x) hs the Fourier series f(x) oddn 4 sin nx. n The function hs derivtive except t the points x = nπ. Differentiting the Fourier series yields f (x) 4 cos(nx). oddn For x nπ, this implies 0 = 4 cos(nx), oddn 36

33 which is flse. The series does not converge. This is s we expected since the Fourier series for f(x) is not uniformly convergent. 362

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