FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES. To a 2π-periodic function f(x) we will associate a trigonometric series. a n cos(nx) + b n sin(nx),

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1 FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES To -periodic fuctio f() we will ssocite trigoometric series + cos() + b si(), or i terms of the epoetil e i, series of the form c e i. Z For most of the fuctios tht we will be delig with, these series re i sese equl to f. Before we do so, we eed some termiology d itroduce clss of fuctios.. Piecewise cotiuous d piecewise smooth fuctios We will be usig the followig ottio: c + mes c d > c, c mes c d < c. For fuctio f() we will use the followig f(c + ) = lim c + f() d f(c ) = lim c f() provided tht the bove limits eist d re fiite. A fuctio f is sid to be piecewise cotiuous o the closed itervl [, b] if there eit fiitely my poits such tht: c = < c < c < < c < c = b f is cotiuous i ech ope itervl (c j, c j ), for j =,, ; f( + ), f(b ) eist d for j =,,, the limits f(c j ) d f(c+ j ) eist. Hece, f is cotiuous everywhere ecept possibly t fiite umber of poits where it hs jump discotiuities. The spce of piecewise cotiuous fuctios o [, b] will be deoted by C p[, b]. Emple. Cosider the fuctio + if < < ; f() = if < < ; if < The f C p[, ]. Iside the itervl, the oly discotiuities of f re t d Dte: Februry 3, 6.

2 FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES y Figure. Piecewise cotiuous. We hve f( + ) =.5, f( ) =, f( + ) =, f( ) = f( + ) =, f( ) =. Emple. Cosider the fuctio y Figure. Not piecewise cotiuous { if < ; f() = if. The iside discotiuity of f is t =. Becuse lim f() = (ot fiite), the f / C p([, ]. A fuctio f is sid to be piecewise smooth o closed itervl [, b] if f is piecewise cotiuous o [, b] d f is lso piecewise cotiuous o [, b]. This mes tht f () eists d is cotiuous everywhere i (, b) ecept possibly t fiite umber of poits. Moreover, t ech poit where f is discotiuous the oe sided limits of f eist d re fiite. The spce of piecewise cotiuous o [, b] will be deoted by C p[, b]. Emple 3. The fuctio f() = is piecewise smooth o [, 5]. It is cotiuous o [, 5] d its derivtive eists everywhere ecept t =. At, we hve f (+) = d f ( ) =. Emple 4. The fuctio f() = { if < ; ( ) /3 if < <. is piecewise cotiuous but it is ot piecewise smooth. The fuctio is piecewise

3 FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES 3 y Figure 3. Piecewise cotiuous but ot piecewise smooth cotiuous becuse f( + ) =, f( ) =, d t =, it hs jump discotiuity (f( ) = d f( + ) = ). The derivtive f is f () = for < < d f () = 3( ) /3 for < <, The derivtive f is ot piecewise cotiuous becuse f ( ± ) re ot fiite (the fuctio f hs cusp t = ). A fuctio f is sid to be piecewise cotiuous (respectively piecewise smooth) o the whole rel lie R if f is piecewise cotiuous (resp. piecewise smooth) o ech closed itervl [, b] R. Remrk. Note tht if f C p[, b], the f is itegrble o [, b]. Tht is, b f()d is fiite umber. Ideed, with c < c < < c the jump discotiuities of f, we hve b f()d = c c f()d + c c f()d + + c c f()d d ech itegrl o the right is fiite becuse withi the limits of itegrtio, the fuctio is bouded d cotiuous.. Eve d odd fuctios A fuctio f is sid to be eve (respectively odd) fuctio if f( ) = f() (resp. f( ) = f()) i domi of f. Note tht it follows from the defiitio tht the domi of eve or odd Figure 4. Grphs of eve d odd fuctios fuctio eeds to de symmetric with respect to R. Tht is, if is i the domi, the ( ) is lso i the domi. The grph of eve fuctio is symmetric with

4 4 FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES respect to the y-is d the grph of odd fuctio is symmetric with respect to the origi. Whe itegrtig eve or odd fuctios, it is useful to use the followig property Lemm. If f C p[ A, A] is eve fuctio, the A A f()d = If f C p[ A, A] is odd fuctio, the A A Proof. For eve fuctio f, we hve A A f()d = = = = A A A A f()d + A f()d =. A f( u)d( u) + f(u)du + f()d A f()d A f()d. f()d (substitutio u = ) f()d (use f eve) A logous rgumet c be pplied whe f is odd. Remrk. I leve it s eercise for you to prove tht the product of two eve fuctios is eve; two odd fuctios is eve; d the product of eve d odd fuctio is odd. Thtis Eve Eve = Eve Odd Odd = Eve Eve Odd = Odd 3. Periodic fuctios Recll tht fuctio f is sid to be periodic with period T (T > ) if f( + T ) = f( T ) = f() i domi of f. This implies i prticulr tht if is i the domi of f, the + kt is lso i the domi of f for every iteger k d f( + kt ) = f(). y Figure 5. Grph of periodic fuctio, T =

5 FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES 5 Emple. For emple, the fuctios si d cos re -periodic d t is -periodic. I geerl, if ω is costt, the si(ω) d cos(ω) hve period T = /ω. Emple. Deote by [ ] the gretest iteger fuctio. Thus, for R, [] the gretest iteger less or equl th. For istce, [.7] =, [ 5] =, [] = 3, y Figure 6. The step fuctio [] [.] =, [ 7] = 3, etc. Note tht [ ] stisfies the property [+] = []+ for ever R. y Figure 7. The swtooth fuctio: [] The fuctio f defied by f() = [] (the frctiol prt of ), lso clled the swtooth fuctio, is piecewise smooth i R d periodic with period T = f( + ) = ( + ) [ + ] = + ([] + ) = [] = f(), R. Emple 3. (Trigulr wve fuctio) I leve it s eercise for you to check tht the fuctio f defied by [ ] f() = + is cotiuous d periodic with period T =. A importt property of periodic fuctios is the followig. Theorem Let f be piecewise cotiuous i R d periodic with period T. The T f()d = +T f()d, R

6 6 FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES y Figure 8. The trigulr wve fuctio Proof. Cosider the fuctio H() defied for R by H() = +T prove the Theorem, we eed to show tht H is costt. We rewrite H s H() = f()d + +T f()d = +T f()d f()d. f()d. To We c compute the derivtive H () by usig the Fudmetl Theorem of Clculus d fid H () = f( + T ) f() = becuse f is T -periodic. H gives H costt. 4. Orthogolity of fuctios We defie ier product <, > i the spce Cp[, b] of piecewise cotiuous fuctios o [, b] by < f, g >= b f()g()d, for f, g C p[, b] The orm of fuctio f Cp[, b] is defied f = ( / b < f, f > = f() d). { if < < Emple. Let f() =, g() =, d h() = if < <. Note tht h() = for > d h() = for <. I Cp[, ], we hve < f, g >= < f, h >= < g, h >= ( f = ( g = ( h = f()g()d = d = ()d + / d) = d + / d) = ( )d + d = d = 3 4d) / = 5 d = + = 3

7 FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES 7 Two fuctios f, g C p[, b] re sid to be orthogol if < f, g >=. Tht is, f d g re orthogol if b f()g()d =. A set of fuctios S C p[, b] is orthoorml if the elemets of S re mutully orthogol d ech elemet of S hs orm. Tht is, < f, g >= for every f, g S, f g f = for every f S. Emple. The fuctios f() = d g() = ( ) re orthogol i Cp[, ] sice, < f, g >= ( )( ) d = 4 ( )4 = = =. The set S = {f, g} is orthogol but it is ot orthoorml becuse ( / f = ( ) d) = 3 ( / g = ( ) d) 4 = 5 However, if we replce f d g by f () = f() f = 3 f() d g () = g() g = We obti the set S = {f, g } which is orthoorml i C p[, ]. 5. The trigoometric system The followig trigoometric idetities will be used soo cos A cos B = cos(a + B) + cos(a B) cos A si B = si(a + B) si(a B) si A si B = cos(a B) cos(a + B) cos A = cos(a) + si A = cos(a) 5 g() The trigoometric system over the itervl [, ] (or over y itervl of legth ) cosists of the fuctios, cos, si, cos(), si(),, cos(k), si(k),, k Z + Lemm. The trigoometric system is orthogol over [, ]. Proof. We eed to verify tht the followig ier products re zero. <, si(k) >=<, cos(k) >= k Z + < cos(k), si(l) >= k, l Z + < cos(k), cos(l) >=< si(k), si(l) >= k, l Z +, k l We hve for k Z + <, si(k) >= si(k)d = ( cos(k) ) cos(k) cos = = k k

8 8 FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES for k, l Z + d k l, we use oe of the bove trig idetities to get Whe k = l, we hve < cos(k), si(l) > = < cos(k), si(k) >= For k, l Z + d k l, we hve cos(k) si(l)d = (si(k + l) si(k l))d = ( ) cos(k + l) cos(k l) k + l k l = < si(k), si(l) > = si(k)d = si(k) si(l)d cos(4k) + cos 4k = (cos(k + l) cos(k l))d = ( ) si(k + l) si(k l) k + l k l = The verifictio of the remiig idetities is left s eercise. Lemm. The orm of the trig fuctios o [, ] re: Proof. We hve =, cos(k) = si(k) =. = d = cos(k) = cos (k)d = ( = + si(k) ) = 4k si(k) = si (k)d = ( = si(k) 4k As cosequece of the two lemms we hve ) = + cos(k) cos(k) Corollry. The followig system is orthoorml over [, ]:, si, cos si k,,, cos k, =. As we will see lter the trigoometric system forms bsis for the spce of piecewise cotiuous d -periodic fuctios.

9 FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES 9 6. Fourier series of -periodic fuctios Let f Cp(R) d -periodic, we would like to ssocite to the fuctio f series + cos() + b si() i such wy tht t ech poit where f is cotiuous the vlues of f d the series re the sme: f( ) = + cos( ) + b si( ). A immedite questio is the followig. If give fuctio f hs such represettio, how c we fid the coefficiets,, b, i terms of f? The swer is ot relly difficult if we ssume tht we c iterchge the summtio d the itegrtio. The by usig the orthogolity of the trigoometric system, we would get < f, >= Therefore, To fid d b : d + cos()d + si()d = = < f, > = f()d. < f, cos > = <, cos > + < cos(), cos > +b < si(), cos > = < cos, cos >= cos = < f, si > = <, si > + < cos(), si > +b < si(), si > Hece Similr rgumets give = b < si, si >= b si = b < f, cos > = = < f, si > b = = < f, cos k > k = cos k = < f, si k > b k = si k = We hve therefore estblished the ssocitio f() cos d ; f() si d. f() cos(k)d ; f() si kd. f() + cos() + b si()

10 FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES for -periodic fuctio f Cp(R). The ssocited series is clled the Fourier series of f. The coefficiets re give by = k = b k = f()d f() cos(k)d for k =,, f() si(k)d for k =,, Remrk. We hve worked the itegrtios over the itervl [, ]. Becuse of the -periodicity, we could hve used y itervl of legth. Hece, for y rel umber, we lso hve = k = b k = f()d f() cos(k)d for k =,, f() si(k)d for k =,, I my situtios with symmetry, it is useful to tke =. We hve the = k = b k = f()d f() cos(k)d for k =,, f() si(k)d for k =,, Remrk 3. By usig properties of eve d odd fuctios, we hve the followig. If F is eve, the the fuctios F () cos(k) re eve d the fuctios F () si(k) re odd. If G is odd, the the fuctios G() cos(k) re odd d the fuctios G() si(k) re eve. The Fourier coefficiets of eve fuctio F re = k = b k = F ()d = F () cos(k)d = F ()d F () si(k)d = for k =,, The Fourier coefficiets of odd fuctio G re = k = b k = G()d = G() cos(k)d = for k =,, G() si(k)d = F () cos(k)d for k =,, G() si(k) for k =,,

11 FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES The Fourier series of eve fuctio F d odd fuctio G re: F () + cos(), G() b si(), = b = 7. Emples We give emples of Fourier series of some simple fuctios F () cos()d ; G() si()d. Emple. Let f() be the -periodic fuctio defied o [, ] by f() =. Hece the grph of f is the trigulr wve. Note tht f is eve fuctio d Figure 9. Trigulr wve so its Fourier series cotis oly the cosie terms (b = for every Z + ). The coefficiets of the cosies re: Hece, = d = ; = cos()d = ( ) cos() = = () =, d = (() ) = ( si() ) = cos() si() d if = k (eve) 4 (k + ) if = k + (odd) We hve obtied the ssocited Fourier series of f o [, ] 4 j= cos(j + ) (j + ). Emple. Let f() be the -periodic fuctio defied o [, ] by { if < < ; f() = if < <. Hece the grph of f is the rectgulr wve. Note tht f is odd fuctio d

12 FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES Figure. Rectgulr wve so its Fourier series cotis oly the sie terms ( = d = for every Z + ). The coefficiets of the sies re: ( ) cos() b = si()d = = () Hece, b = ( () ) = if = k (eve) 4 (k + ) if = k + (odd) We hve obtied the ssocited Fourier series of f o [, ] f() 4 si(j + ). (j + ) Emple 3. Let f() be the -periodic fuctio defied o [, ] by { if < < ; f() = if < <. j= The grph of f is the give below. The fuctio f is either eve or odd d so Figure. Grph of f its Fourier series cotis sies d cosies terms. Its Fourier coefficiets re: We hve = = b = f()d = f() cos()d = f() si()d = f() 4 j= d = ; cos(j + ) (j + ) + cos()d = () si()d = () j= () j si j j..

13 FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES 3 8. The comple epoetil form of Fourier series Recll tht the cosie d sie fuctio c be epressed i terms of the epoetil s { e iθ = cos θ + i si θ e iθ = cos θ i si θ Now we c rewrite the trigoometric series cos θ = eiθ + e iθ si θ = eiθ e iθ i + cos + b si, = i eiθ e iθ with rel coefficiets i terms of these epoetils. For this, we use cos + b si e i + e i e i e i = ib = ib e i + + ib e i = c e i + c e i where c deotes the comple cojugte of c For -periodic, R-vlued fuctio fuctio f C p(r), we hve with c f() c + c e i + c e i = ib = f() (cos i si ) d = f()e i d This comple form of the Fourier is equivlet to the oe give with cosies d sies. I my pplictios it is esier d more coveiet to use the comple form. Note tht the bove ssocitio tke the form f() c + Re ( c e i) where Re(C) deotes the rel prt of the comple umber C. Emple. Let f be the -periodic fuctio defied over the itervl (, ) by { if < < f() = if < <

14 4 FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES To compute the Fourier coefficiets, we use itegrtio by prts. e i d = e i e i i i d = e i i e i e i ( i) + ( i) d = e i i = i e i We fid the the Fourier coefficiets c = d for, c = = (i e i = e i ( i) + e i f()e i d = + e i + e i ( i) 3 + C d = 6 i e i 3 e i d i e i 3 It gives ( c = () () + i + ) () 3 The Fourier series of f is therefore 6 + Re [ c e i] ) + C We c rewrite the Fourier series i terms of cosie d sie. First ( c e i = () () cos + ) () 3 si + ( ( () () +i si + ) ) () 3 cos The Fourier series of f is: 6 + () ( () cos + ) () 3 si Emple. Let f be the -periodic fuctio defied o [, ] s f() = e. The -th Fourier coefficiet of f is c = = e e i d = ( ) e ( i) = e ( i) d i = = (e ) i = (e ) + i + Hece (fter clcultio), we fid [ ] Re(c e i ) = (e ) cos si + +

15 FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES 5 The Fourier series of f is therefore, e c + Re(c e i ) [ ] (e ) cos si Eercises I ech eercise, fid the fourier series of the -periodic fuctio f tht is give by Eercise. f() = for < <. Eercise. f() = for < <. Eercise 3. f() = cos { if < < Eercise 4. f() = if < < Eercise 5. f() = cos Eercise 6. f() = si { if < < Eercise 7. f() = si if < < { /(d) if < d Eercise 8. f() = if d < < with d positive costt. Eercise 9. f() = e d for < <, where d positive costt Eercise. f() = cosh for < <. Eercise. f() = cosh for < <. Eercise. f() = cosh for < <. { ( )/(d) if < d Eercise 3. f() = with d positive costt. if d < <