Our aim is to show that under reasonable assumptions a given 2π-periodic function f can be represented as convergent series

Transcription

1 8 Fourier Series Our aim is to show that uder reasoable assumptios a give -periodic fuctio f ca be represeted as coverget series f(x) = a + (a cos x + b si x). (8.) By defiitio, the covergece of the series meas that the sequece (s (x)) of partial sums, defied by s (x) = a + (a k cos kx + b k sikx), k= coverges at a give poit x to f(x), s (x) f(x). We start with some defiitios. A real valued fuctio defied o R is said to be periodic with period L, or L-periodic, is f(x + L) = f(x) for all x. I this case f is completely determied by its values o ay iterval [a,a+l). A fuctio f defied o [a,a+l) ca be exteded i a uique way to a periodic fuctio o R. Ideed, give x, there exists a uique iteger so that x belogs to [a + L,a + ( + )L) so that x L [a,a + P) ad we set f(x) = f(x L). A fuctio f o [a,b] is piecewise cotiuous if f is cotiuous except a fiite umbers of poits ad at each such poit the oe-sided limits f(x + ) = lim ε f(x + ε) ad f(x ) = lim ε f(x + ε) exist ad are fiite. We say that f is piecewise smooth (or piecewise differetiable) o [a,b] if f ad f are piecewise cotiuous o [a,b]. If f is piecewise cotiuous, the it is Riema itegrable over ay bouded iterval cotaied i its domai. I additio, if f is L-periodic, the a+l a f(x) dx = L fx) dx for every a. I what follows we focus o fuctios which are -periodic. Assume f(x) = a + (a cos x + b si x). We also assume that the covergece is well-behaved so that we ca itegrate term-by-term. We wat to compute the coefficiets a k ad b k. We use 66

2 itegral idetities, m cos mxcos x dx =, m = =, m = {, m si mxsix dx =, m = cos mxsix dx =. To compute a, we multiply both sides of (8.) by the costat fuctio ad itegrate over [,] ad get f(x) dx = a dx + = a () = a [a cos x dx + b ] si x dx To compute a m with m we we multiply both sides of (8.) by cos mx ad itegrate over [,] to get f(x)cos mx dx = a cos mx dx + [a cos xcos mx dx + b = a m cos mxcos mx dx = a m. ] si xcos mx dx Similarly, multiplyig both sides of (8.) by si mx ad itegratig over [,], we fid that f(x)si mx dx = b m. Summig up, a = b = f(x)cos x dx, f(x)si x dx,. (8.) 67

3 The coefficiets a ad b are called the Fourier coefficiets of f ad the series a + [a cos x + b si x] the Fourier series of f. We deote this fact by f(x) a + [a cos x + b si x]. The symbol should be read as f has Fourier serier. Example 8.. Let f be a -periodic fuctio give by f(x) = x for x (, ]. The itegratig by parts a = xcos x dx = [ ] xsi x si x dx = ad b = Cosequetly, xsi x dx = [ xcosx = cos ] = ( )+. f(x) ( ) + si x. + cosx dx The series coverges for every x i view of the followig Dirichlet s test. If (a ) is a decreasig sequece covergig to ad the sequece (s ) of partial sums of the series b is bouded, the the series a b coverges. I our case, a = is decreasig ad coverges to, ( ) si x = si (x + ), ad which implies that if ϑ k for all k Z. si ϑ si kϑ = si k= ( + )ϑ si kϑ si ϑ k= si ϑ Example 8.. Cosider f be a -periodic fuctio defied by f(x) = x for x [, ]. Note that f is eve. Sice the product of eve fuctio with the odd fuctio is odd, it follows that f(x)si mxdx =. 68

4 Hece b = for all. To compute b ote that the product of a eve fuctio with a eve fuctio is eve so that a = If =, the f(x)cosx dx = a = f(x)cos xdx = x dx =, ad if, the itegratig by parts, oe fids that xcos x dx = [ ] xsi x = [ cosx Hece a = if is eve ad a = 4 f(x) 4 ] si x dx = [( ) ]. whe is odd, ad hece cos( )x ( ). xcosx dx. Sice the series coverges, the M-Weierstarss test implies that the above series coverges. I the above examples we have used the followig idetities, { L L g(x) dx = g(x) dx if g is eve if g is odd. L Usig that the sum of two eve fuctios is eve ad the sum of two odd fuctios is odd, that the product of two eve fuctios or two odd fuctios is eve, ad the product of a eve fuctio ad a odd fuctio is odd, we have the followig propositio. Propositio 8.3. (i) If f is eve, the its Fourier sie coefficiets b are equal to ad f is represeted by Fourier cosie series f(x) a + a cos x where a = f(x)cos x dx for. 69

5 (ii) If f is odd, the its Fourier cosie coefficiets a are equal to ad f is represeted by Fourier sie series f(x) a si x where b = f(x)si x dx for. 8.. Covergece Theorem Theorem 8.4. Assume that f is a -periodic fuctio which is piecewise smooth. The [ ] f(x ) + f(x + ) = a + [a k cos kx + b k si kx] where f(x ± ) = lim y x ± f(y). I particular, if f is cotiuous at x, the k= f(x) = a + [a k cos kx + b k si kx]. k= Example 8.5. Cosider the fuctio f from Example 8.. The f is smooth at all poits except poits m where m is odd. The oe sided limits at these poits are f(m ) = lim x m f(x) = ad f(m + ) = lim x m + f(x) = provided that m is odd. Cosequetly, the Fourier series coverges of f coverges to f at every poit except poits m with m odd. At these poits the Fourier series coverges to [f(m ) + f(m + )] =. I particular, x = ( ) + si x, for x (, ). Takig x = /, we fid that 4 = ( ). 7

6 Example 8.6. Cosider the fuctio -periodic fuctio f give by f(x) = x for x (, ). (See Example 8..) The f is cotiuous at every poit ad it is smooth except poits m with m odd. Hece the Fourier series of f coverges to f at every poit. I particular, x = 4 Substitutig x =, we fid that Set S =. The we ote that ad hece cos( )x ( ) for x [, ]. 8 = ( + ). S 4 = S = S S 4 = From this we coclude that = S = Itegratio of Fourier series = () () = = 6. = ( + ) = 8. We start with the followig observatio. Assume that f is -periodic, piecewise cotiuous, ad let F(x) = x f(y) dy. The F is periodic if ad oly if f(y)dy =. Ideed, we have F(x + ) F(x) = x+ x f(y)dy = f(y)dy. This meas that the costat term i the Fourier series of f is equal to. Sice a = f(y) dy, the umber a / is the mea of the fuctio f over the iterval [.]. Theorem 8.7. Assume that f is -periodic ad piecewise cotiuous ad its mea is equal to. The its Fourier series f(x) [a cos x + b si x] 7

7 ca be itegrated term by term ad produce the Fourier series x F(x) = f(y) dy C + [ b cos x + a ] si x where the costat C = F(y) dy. Example 8.8. Cosider a -periodic fuctio f give by f(x) = x o (, ]. (See Example 8..) Sice f is odd, its mea value over [, ] is equal to. Its Fourier series is give by x = ( ) si x, Itegratig term by term we fid that for x (, ]. x = 6 ( ) cosx for x (, ] where /6 = (x /) dx. Differetiatio of Fourier series Propositio 8.9. Assume that f is -periodic, cotiuous, ad piecewise smooth. Abbreviate by a,b the Fourier coefficiets of f ad by a,b the Fourier coefficiets of f. The Proof. Itegratig by parts, a = Similarly, b = a. f (x)cos x dx = As a cosequece, we get: a = b ad b = a. [ ] f(x)cosx + f(x)si x dx = b. Theorem 8.. Let f be -periodic, cotiuous, ad piecewise smooth. I additio, assume that f is piecewise smooth. If the f(x) = a + [a cos x + b si x], [ ] f (x ) + f (x + ) = [b cos x a si x]. I particular, at poits x at which f (x) exists, the series coverges to f (x). 7

8 Proof. By Propositio 8.9, f (x) [b cosx a si x], ad sice f is piecewise smooth, the theorem follows from Theorem 8.4. Example 8.. Recall the fuctio -periodic fuctio f give by f(x) = x for x (, ). (See Example 8. ad Example 8.6.) The fuctio f is cotiuous ad piecewise smooth, ad x = 4 The derivative of f is give by I view of Theorem 8., 4 d dx x = si( )x ( ) cos( )x ( ), x [, ]. {, x >,, x <. {, < x <, =, < x <. At x =, the series coverges to [f (x ) + f (x + )] = [ + ] =. 8.. Absolute ad Uiform Covergece Theorem 8.. Let f be -periodic, cotiuous, ad piecewise smooth. The its Fourier series coverges absolutely ad uiformly. Proof. We prove the result uder additioal assumptio that f is piecewise smooth. This meas that f is piecewise cotiuous. Abbreviate by a, b the Fourier coefficiets of f ad by a, b the Fourier coefficiets of f. Sice f is cotiuous, f(x) = a + [a cosx + b si x]. To prove the absolute covergece it suffices to show that a, b M/ for some costat M idepedet of. The, by the compariso test, a ad b coverge which imply the absolute covergece of the Fourier series. Also, the Weierstrass M-test implies the uiform covergece. By Propositio 8.9, a = b /. That is, a = f(x)cos x dx = f (x)si x dx. 73

9 By assumptio, f is piecewise cotiuous ad so f is cotiuous except a fiite umber of poit at which it has a jump. If (a, b) is a iterval o which f is cotiuous, the f (x)si x dx = = [ ] b f (x)cos x [ f(b )cosx f(a + )cosx ] a + b a b a f (x)cos x dx f (x)cos x dx. Sice f (x), f (x) K for some costat K ad all x (at poits x where f ad f has a jump discotiuity, f (x) oe has to take left- ad had-side derivatives). So, f (x)si x dx K( + (b a) K( + 4K. Sice f has a fiite umber of discotiuities, the itegral f (x)si x dx ca be writte as the fiite umber, say m, of itegrals b a f (x)si x dx over itervals o which f is cotiuous. The a = f (x)si x dx 4Km. Similarly, b 4Km. Theorem 8.3. Let f be -periodic. Assume that f is of class C k ad that f (k ) is piecewise smooth. If the Fourier coefficiets satisfy a C k+α ad b C k+α for some α > ad C >, the f is of class C k. Proof. Let a (j) ad b (k) are coefficiets of the Fourier series of f (j). The usig Propositio 8.9 a (j) = j a if j is eve ad a (j) = k b if j is odd. Similarly, for b (k). Observe, usig the assumptio, that if j k, the j a C k j+α C α for j k. By the Weierstrass M-test, the series j a coverges absolutely ad uiformly for j k. Similarly, the series j b coverges absolutely ad uiformly for j k Hece, f (j) = [aj) cosx + b j) si x] for j k ad f (k) is cotiuous. 74

10 8. Chagig a scale So far we have cosidered -periodic fuctios ad their Fourier series over the iterval [,]. Now, give a fuctio f defied o [ L,L], we defie F(y) = f ( L y). The F is defied o [,] ad its Fourier series is give by where a = F(y) a + [ a cos y + b si y ] F(y)cos y dy ad b = Sice f(x) = F( Lx) for x [ L,L], we deduce that f(x) a + [ a cos L x + b si L x] F(y)si y dy. (8.3) The coefficiets a ad b ca be computed by itegratig (8.3) by substitutio x = L y. We get a = F(y)cos y dy = ( ) L f y cos y dy = L f(x)cos x (8.4) L L dx ad similarly L b = F(y)si y dy = L L L f(x)cos x L dx. (8.5) 75