0 otherwise. sin( nx)sin( kx) 0 otherwise. cos( nx) sin( kx) dx 0 for all integers n, k.

Transcription

1 . Computtio of Fourier Series I this sectio, we compute the Fourier coefficiets, f ( x) cos( x) b si( x) d b, i the Fourier series To do this, we eed the followig result o the orthogolity of the trigoometric fuctios. I d lie you to prove the followig. Questio: The followig itegrl reltios hold: if cos( x)cos( x) if otherwise if si( x)si( x) otherwise cos( x) si( x) for ll itegers,. A equivlet wy of sttig this theorem is tht the collectio cos( x) cos( x) si( x) si( x),,,,,, is orthoorml set of fuctios i L (, ) Proof. First, we proof the legth of ech fuctio i L (, ) is. whe is ozero iteger, cos( x) cos ( x) ( cos( x) ( / si( x)) si( x) si ( x) ( cos( x) ( / si( x)) Now, prove tht y two of them the ier product is. I fct, if we hve the followig, the we hve proved the result.

2 For ll egtive itegers, si(( ) x) si(( ) x cos( x)cos( x) cos(( ) x) cos(( ) x) si(( ) x) si(( ) x si( x)si( x) cos(( ) x) cos(( ) x) For ll itegers,, cos(( ) x) cos(( ) x si( x)cos( x) si(( ) x) si(( ) x) cos( x) cos( x) si( x) si( x) Now, we use the orthogolity reltios:,,,,,, compute the Fourier coefficiets. We strt with the equtio () f ( x) cos( x) bsi( x) to To fid,,, we multiply both sides by cos( x) bove:, d itegrte, the use the properties i the f ( x)cos( x) cos( x) bsi( x) cos( x) = cos( x) cos( x)cos( x) bsi( x)cos( x) = To fid, we itegrte both sides of (), we hve f ( x) cos( x) b si( x) To fid b, we multiply both sides of () by si( x) bove:, d itegrte, the use the properties i the f ( x)si( x) cos( x) bsi( x) si( x) = si( x) cos( x)si( x) b si( x)si( x) = b

3 Theorem : If f ( x) cos( x) bsi( x) o the itervl x, the The f ( x) f ( x)cos( x) for b f ( x)si( x) d b re clled the Fourier coefficiets of the fuctio f. Now, we cosider the geerl itervl x. Suppose F is fuctio defied o the itervl [, ]. The substitutio x t / leds to the followig chge of vribles formul: t F( x) F( ) dt By usig this chge of vribles, the followig theorem c be derived from the bove theorem. Theorem : If f ( x) cos( x / ) bsi( x / ) o the itervl x, the Exmple : Let f () t dt f ( t)cos( t / ) dt for b f ( t)si( t / ) dt if x f( x) otherwise Plese compute the Fourier series for f vlid o the itervl x. Solutio: With i the bove Theorem, the Fourier cosie coefficiets re for, ( ) ( ) f t dt f t dt dt si( / ) f ( t)cos( t / ) dt f ( t)cos( t / ) dt cos( t / ) dt

4 Whe is eve, these coefficiets re zero. Whe is odd, the si( / ) ( ). Therefore, ( ), ( ) Similrly, b f ( t)si( t / ) dt si( t / ) dt (cos( / ) ) whe 4 j, b whe 4 j, b (4 j ) whe 4 j, b ( j ) whe 4 j 3, b (4 j 3) Thus, the Fourier series for f is F( x) cos( x / ) b si( x / ) where, b give s bove. Lter, we will cosider the questio, whether or ot the Fourier series Fx ( ) for f equls f( x ) itself. Cosie d Sie Expsios Eve d Odd Fuctios Defiitio: Let f : R R be fuctio; f is eve if f ( x) f ( x); f is odd if f ( x) f ( x).

5 The grph of eve fuctio is symmetric bout the y xis, s illustrted i Figure. f ( x) x, f ( x) cos x re eve fuctios. The grph of odd fuctio is symmetric bout the origi, s illustrted i Figure 3. f ( x) x, f ( x) si x re odd fuctios. The followig properties follow from the defiitio. Eve X Eve = Eve Eve X Odd = Odd Odd X Odd = Eve Aother importt property of eve d odd fuctios is give i the ext Theorem. Theorem: () If F is eve fuctio, the F( x) F( x) () If F is odd fuctio, the F ( x ) This theorem follows esily from Figure d 3. If F is eve, the the itegrl over the left hlf itervl [,] is the sme s the itegrl over the right hlf itervl [,]. Thus, the itegrl over [, ] is twice the itegrl over [, ]. If F is odd, the the itegrl over the left hlf itervl [,] ccels with the itegrl over the right hlf itervl [,]. I this cse, the itegrl over [, ] is. If the Fourier series of fuctio oly ivolves the cosie terms, the it must be eve fuctio (sice cosie is eve). Liewise, Fourier series tht oly ivolves sies must be odd. The coverse of this fct is lso true, which is the cotet of the ext Theorem. Theorem 3: () If f( x ) is eve fuctio, the its Fourier series o the itervl [, ] will oly ivolve cosies. Tht is, f ( x) cos( x / ) with f ( x) f ( x)cos( x / ) for (). If f( x ) is odd fuctio, the its Fourier series o the itervl [, ] will oly ivolve sies. Tht is, f ( x) b si( x / ) with b f ( x)si( x / ) Fourier Cosie d Sie Series o Hlf Itervl. Suppose f is defied o the itervl [, ]. By cosiderig eve or odd extesios of f, we c expd f s cosie or sie series. To express f s cosie series, we cosider the eve extesio of f :

6 The fuctio Fourier expsio: e f ( x) if x fe( x) f ( x) if x f is eve fuctio defied o [, ]. Therefore, oly cosie terms pper i its () fe( x) cos( x / ) x where is give i Theorem 3. Sice f ( x) f ( x) for x, the itegrl formuls i Theorem 3 oly ivolve f( x ) rther th f ( x ) d so () becomes with e e f ( x) cos( x / ) x f ( x) f ( x)cos( x / ) for Liewise, to express f s sie series, the we cosider the odd extesio of f : f ( x) if x f( x) f ( x) if x The odd fuctio, f ( x ) hs oly sie terms i its Fourier expsio. Sice f ( x) f ( x) for x, we obti the followig sie expsio for f : where b is give i Theorem 3: f ( x) b si( x / ) x b f ( x)si( x / ) Let f be fuctio d let Fx ( ) be its Fourier series o [, ] : F( x) cos( x) b si( x) N lim cos( x) b si( x) N

7 where, b re the Fourier coefficiets of f. We sy tht the Fourier series coverges if the precedig limit exists (s N ). Theorem d Theorem oly compute the Fourier series of give fuctio. We hve yet show tht give Fourier series coverges (or wht it coverges to). Lter, we will show tht uder mild hypothesis o the differetibility of f, the followig priciple holds: Let f be periodic fuctio. If f is cotiuous t poit x, the its Fourier series, Fx ( ), coverges d F( x) f ( x). If f is ot cotiuous t poit x, the its Fourier series, Fx ( ), coverges the verge of the left d right limits of f t x, tht is F ( x ) lim f (t) lim f (t). tx tx The secod sttemet icludes the first becuse if f is cotiuous t x, the the left d right limits of f t x re both equl to f( x ) d so i this cse, F( x) f ( x). We will prove the bove results lter. I the followig, we preset severl exmples to gi isight ito the computtio of Fourier series d the rte t which Fourier series coverge. Exmple : Cosider the fuctio f ( x) x o x. Fid its Fourier series. Solutio: This fuctio is odd d so oly the sie coefficiets re ozero. Its Fourier coefficiets re b f ( x)si( x) xsi( x) d so its Fourier series for the itervl x is ( ) F( x) si( x) ( ) (usig itegrtio by prts) Discussios: The fuctio f ( x) x is ot periodic. Its periodic extesio, f, is give i Figure 4.

8 Accordig to the precedig priciple, Fx ( ) coverges to f( x ), t poits where f is cotiuous. At poits of discotiuity of f, ( x,,, ), Fx ( ) will coverge to the verge of the left d right limits of f( x ). For exmple, F( ) (sice si ), which is the verge of the left d right limit of f t x. To see how fst the prtil sums of this Fourier series coverges to f( x ), we grph the prtil sum For vrious vlues of N. The grph of N ( ) SN ( x) si( x) ( ) S( x) si( x) is give i Figure 5 together with the grph of f( x ) (the squiggly curve is the grph of S ). First, otice tht the ccurcy of the pproximtio of f( x) by S ( ) x gets worse the closer x is to poit of discotiuity. For exmple, er x, the grph of S ( ) x must trvel from bout y to y i very short itervl, resultig i slow rte of covergece er x.

9 Secod, otice the blips i the grph of the Fourier series just before d just fter the poits of discotiuity of f( x ). This effect is clled the Gibbs pheomeo. Figure 6. Gibbs pheomeo for S 5 Exmple 3. Cosider the swtooth wve illustrted i Figure 7. The formul for f o the itervl x give by x if x / f() t x if / x d exteds to the itervl x s eve fuctio. Compute its Fourier series. Sice f is eve fuctio, oly the cosie terms re ozero. By Theorem 3, we hve Tht is, F( x) cos( x) with f ( x) 4

10 For f ( x)cos( x) / = x cos( x) ( x)cos( x) / 4cos( / ) cos( ) = Oly the 4. These coefficiets simplify to ( ) 4 So, the Fourier series for the swtooth wve is F( x) cos((4 ) x) 4 ( ) The swtooth wve is lredy periodic d it is cotiuous. Thus its Fourier series Fx ( ) equls the swtooth wve, f( x ), for every x by the priciple stted t the begiig of this sectio. The covergece rte is much fster th for the Fourier series i Exmple. To illustrte the rte of covergece, with plot the sum of the first two terms of its Fourier series, cos( x) S( x) I Figure 8. The sum of just two terms of this Fourier series gives more ccurte pproximtio of the swtooth wve th or 5 terms of the Fourier series i the previous (discotiuous) exmple. Ideed, the grph of the first terms of this Fourier series (give i Figure 9) is lmost idistiguishble form the origil fuctio.

11 Exmple 4. Fid the Fourier series of f ( x) si but you c get the Fourier series usig trigoometric idetity: f x x x ( ) si cos (, ) x. You c use the formuls to get the Fourier series, Questio: Fid the Fourier sie series for the fuctio f x vlid o the itervl x. ( ) x Aswer: we first exted f s odd fuctio: f ( x) o x x x x for for The Fourier coefficiets for f o is ( )si( ) ( )si( ) b f x x x x ( ) ( ) 3 3

12 Whe, b Whe, b ( ) Thus, the Fourier si series for f x o the itervl x is ( ) x 3 4 F( x) si si( ) x 3 3 () Questio: Solve the het equtio: ut ( x, t) uxx( x, t) t, x u( x,) f ( x) x u(, t) u(, t) where x if x / f( x) x if / x Setch of the swer: From the discussio i., the solutio to this problem is (3) u( x, t) b si( x) e t Let t i (3), we hve f ( x) u( x,) b si( x). Therefore, the b must be chose s the Fourier sie coefficiets of f( x ), which by Thorem 3, re b f ( t)si( t) dt j 4( ) After do the itegrl, we hve. Whe j, b ; whe j, bj ( j ) j 4( ) u( x, t) si(( j ) x) e ( j ) ( j) t