23.5. Half-Range Series. Introduction. Prerequisites. Learning Outcomes
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1 Half-Range Series 2.5 Inroducion In his Secion we address he following problem: Can we find a Fourier series expansion of a funcion defined over a finie inerval? Of course we recognise ha such a funcion could no be periodic (as periodiciy demands an infinie inerval). The answer o his quesion is yes bu we mus firs conver he given non-periodic funcion ino a periodic funcion. There are many ways of doing his. We shall concenrae on he mos useful exension o produce a so-called half-range Fourier series. Prerequisies Before saring his Secion you should... Learning Oucomes On compleion you should be able o... know how o obain a Fourier series be familiar wih odd and even funcions and heir properies have knowledge of inegraion by pars choose o expand a non-periodic funcion eiher as a series of sines or as a series of cosines 6 HELM (28): Workbook 2: Fourier Series
2 1. Half-range Fourier series So far we have shown how o represen given periodic funcions by Fourier series. We now consider a sligh variaion on his heme which will be useful in 25 on solving Parial Differenial Equaions. Suppose ha insead of specifying a periodic funcion we begin wih a funcion f() defined only over a limied range of values of, say < <. Suppose furher ha we wish o represen his funcion, over < <, by a Fourier series. (This siuaion may seem a lile arificial a his poin, bu his is precisely he siuaion ha will arise in solving differenial equaions.) To be specific, suppose we define f() = 2 < < f() 2 2 Figure 21 We shall consider he inerval < < o be half a period of a 2 periodic funcion. We mus herefore define f() for < < o complee he specificaion. Task Complee he definiion of he above funcion f() = 2, < < by defining i over < < such ha he resuling funcions will have a Fourier series conaining (a) only cosine erms, (b) only sine erms, (c) boh cosine and sine erms. Your soluion HELM (28): Secion 2.5: Half-Range Series 7
3 (a) We mus complee he definiion so as o have an even periodic funcion: f() = 2, < < f 1 () 2 (b) We mus complee he definiion so as o have an odd periodic funcion: f() = 2, < < f 2 () 2 (c) We may define f() in any way we please (oher han (a) and (b) above). For example we migh define f() = over < < : f () 2 The poin is ha all hree periodic funcions f 1 (), f 2 (), f () will give rise o a differen Fourier series bu all will represen he funcion f() = 2 over < <. Fourier series obained by exending funcions in his sor of way are ofen referred o as half-range series. Normally, in applicaions, we require eiher a Fourier Cosine series (so we would complee a definiion as in (i) above o obain an even periodic funcion) or a Fourier Sine series (for which, as in (ii) above, we need an odd periodic funcion.) The above consideraions apply equally well for a funcion defined over any inerval. 8 HELM (28): Workbook 2: Fourier Series
4 Example Obain he half range Fourier Sine series o represen f() = 2 < <. Soluion We firs exend f() as an odd periodic funcion F () of period 6: f() = 2, < < F () Figure 22 We now evaluae he Fourier series of F () by sandard echniques bu ake advanage of he symmery and pu a n =, n =, 1, 2,.... Using he resuls for he Fourier Sine coefficiens for period T from 2.2 subsecion 5, b n = 2 T ( ) 2 2n F () sin d, T T T 2 we pu T = 6 and, since he inegrand is even (a produc of 2 odd funcions), we can wrie b n = 2 ( ) 2n F () sin d = 2 ( ) n 2 sin d. 6 (Noe ha we always inegrae over he originally defined range, in his case < <.) We now have o inegrae by pars (wice!) { [ b n = 2 ( )] ( ) 2 n ( ) } n n cos + 2 cos d n { = 2 27 [ ] ( ) ( ) 6 n 6 ( ) } n cos n + sin sin d n n n n n { = 2 27 [ 18 cos n ( )] } n n n 2 2 n cos = 2 { 27 } 5 cos n + (cos n 1) n n 18 n = 2,, 6,... n = 18 n 72 n = 1,, 5,... n So he required Fourier Sine series is ( 1 F () = 18 ) ( ) sin 18 ( ) 2 2 sin ( ) sin() HELM (28): Secion 2.5: Half-Range Series 9
5 Task Obain a half-range Fourier Cosine series o represen he funcion f() = < <. f() Firs complee he definiion o obain an even periodic funcion F () of period 8. Skech F (): Your soluion F() Now formulae he inegral from which he Fourier coefficiens a n can be calculaed: Your soluion We have wih T = 8 a n = 2 ( ) 2n F () cos d 8 8 Uilising he fac ha he inegrand here is even we ge a n = 1 ( ) n ( ) cos d 2 5 HELM (28): Workbook 2: Fourier Series
6 Now inegrae by pars o obain a n and also obain a : Your soluion Using inegraion by pars we obain for n = 1, 2,,... { [ a n = 1 ( ) ( )] n 2 n sin + n = 1 ( ) ( ) [ ( )] n cos 2 n n n = 2,, 6,... i.e. a n = 16 n = 1,, 5,... n 2 2 Also a = 1 2 ( ) } n sin d = 8 [ cos(n) + 1] n 2 2 ( ) d =. So he consan erm is a 2 = 2. Now wrie down he required Fourier series: Your soluion We ge { cos ( ) + 1 ( ) 9 cos + 1 ( ) } 5 25 cos +... HELM (28): Secion 2.5: Half-Range Series 51
7 Noe ha he form of he Fourier series (a consan of 2 ogeher wih odd harmonic cosine erms) could be prediced if, in he skech of F (), we imagine raising he -axis by 2 unis i.e. wriing F () = 2 + G() 2 G() 2 Figure 2 Clearly G() possesses half-period symmery G( + ) = G() and hence is Fourier series mus conain only odd harmonics. Exercises Obain he half-range Fourier series specified for each of he following funcions: 1. f() = 1 (sine series) 2. f() = 1 (sine series). (a) f() = e 2 1 (cosine series) (b) f() = e 2 (sine series). (a) f() = sin (cosine series) (b) f() = sin (sine series) s 1. {sin + 1 sin + 15 } sin {sin 1 2 sin sin }. (a) e n 2 2 {e2 cos(n) 1} cos n (b) n=1. (a) 2 + n=1 2n + n 2 2 {1 e2 cos(n)} sin n n=2 (b) sin iself (!) { } (1 cos(1 n)) + (1 cos(1 + n)) cos n 1 n 1 + n 52 HELM (28): Workbook 2: Fourier Series
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