10.5 Power Series. In this section, we are going to start talking about power series. A power series is a series of the form

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1 0.5 Power Series I the lst three sectios, we ve spet most of tht time tlkig bout how to determie if series is coverget or ot. Now it is time to strt lookig t some specific kids of series d we will evetully rech the poit where we c tlk bout couple of pplictios of series. I this sectio, we re goig to strt tlkig bout power series. A power series is series of the form () 3 c c0 c c c3 c 0 where is vrible d the c s re costts clled the coefficiets of the series. The first thig to otice bout power series is tht it is fuctio of. For ech fied, the series () is series of costts tht we c test for covergece or divergece. A power series my coverge for some vlues of d diverge for other vlues of. The sum of the series is fuctio f ( ) c c c c c c whose domi is the set of ll for which the series coverges. Notice tht f resembles polyomil. The oly differece is tht f hs ifiitely my terms. For emple, if we tke c for ll, the power series becomes the geometric series 0 3 which coverges whe, d diverges whe. More geerlly, series of the form () 0 c ( ) c c ( ) c ( ) c ( ) c ( ) is clled power series i ( ) or power series cetered t or power series bout. Remrk: Power series () is power series cetered t 0. For power series cetered t, series () is lwys coverget whe sice ll terms re 0. Emple : For wht vlues of is the series! coverget. 0 Solutio: We use the Rtio Test. Let deote the th term,!.

2 ( )! ( )! L lim lim lim lim( )!! It s cler tht L, whe 0, thus, the series is diverget whe 0. The series is coverget oly whe 0. Remrk: I geerl, whe we re sked to justify power series is coverget or diverget, The Rtio Test is used. Emple. For wht vlues of is the series ( ) coverget. Solutio: Let deote the th term, ( ). ( ) ( )( ) 3 ( )( ) lim lim lim lim ( ) ( 3)( ) 3 L ( ) By the Rtio Test, whe 3 5, the series is coverget. The Rtio Test provides o iformtio whe L 3, d 5, so we eed to cosider the series is coverget or diverget whe 3, d 5. Whe 3, the series lim b lim 0 d Test. Whe 5, the series ( ) b becomes ( ). is decresig, so ( ) becomes. ( ) is ltertig series, d Test. (Compre with. lim lim lim 0 d b is diverget) ( ) is coverget by ltertive series is diverget by Limit Compriso is diverget, so Thus, series ( ) is coverget whe 3 5. ( ) Emple 3. For wht vlues of is the series coverget. (!) 0

3 Solutio: Let deote the th term, ( ). (!) ( ) ( ) ( ) ( ) [( )!] ( ) (!) lim lim lim ( ) ( ) [( )!] ( ) L (!) (!) (!) lim lim 0!] [( )!] ( ) lim [( ) By the Rtio Test, the series is coverget for y vlues of, i other words, for. For the power series tht we hve looked t so fr, the set of vlues of for which the series is coverget hs lwys tured out to be itervl [ fiite itervl i Emple, the ifiite itervl (, ) i Emple 3, collpsed itervl [, ] {} i Emple ]. I geerl, we hve the followig. Theorem: For give power series 0 c ( ), there re oly three possibilities: (i) The series coverges oly whe. (ii) The series coverges for ll, i other words (, ) (iii) There is positive umber R such tht the series coverges if R d diverges if R. The umber R i cse (iii) is clled the rdius of covergece of the power series. By covetio, the rdius of covergece is R 0 i cse (i), d R i cse (ii). The itervl of covergece of power series is the itervl tht cosists of ll vlues of for which the series coverges. I cse (i) the itervl cosists of just sigle poit. I cse (ii) the itervl is (, ). I cse (iii) ote tht there re four possibilities for the itervl of covergece: ( R, R),[ R, R),[ R, R],( R, R]. We summrize here the rdius d itervl of covergece for ech of the emples lredy cosidered i this sectio. I Emple, for series 0!, the rdius of covergece is R 0, the itervl of ( ) covergece is {0} ; i Emple, for series, the rdius of covergece is R, the itervl of covergece is [3,5) ; ; i Emple 3, for series R, the itervl of covergece is (, ). 0 ( ) (!) Emple. Fid the rdius of covergece d itervl of covergece of the series, the rdius of covergece is

4 Solutio: Let 0 ( ) ( ) ( 5). ( 5) ( ) ( ) ( 5) ( ) lim lim lim ( 5) ( ) ( 5) ( 5) L ( ) lim ( 5) 5 By the Rtio Test, the series is coverget whe L 5, or Thus, the rdius of covergece is R. We kow the series coverges i the itervl (-9,-), but we must ow test for covergece t the edpoits of this itervl. Whe 9, this series becomes diverget by Divergece Test sice lim. Whe, this series becomes ( ) ( ) ( 9 5) ( ) ( 5) ( ) diverget by Divergece Test sice lim( ) does ot eist., this series is ( ) ( ), this series is Thus, the covergece of itervl is (-9,-). Emple 5. Fid the rdius of covergece d itervl of covergece of the series ( )

5 Solutio: Let ( ). ( ) lim lim lim ( ) () L ( ) lim ( ) lim ( ) 6 By the Rtio Test, the series is coverget whe L 6, or Thus, the rdius of covergece is R. 5 7 We kow the series coverges i the itervl, edpoits of this itervl. Whe 5, this series becomes, but we must ow test for covergece t the 5 ( ) ( ) ( ) Altertig Series Test sice limb lim 0 d b is decresig., d the series is coverget by Whe 7, this series becomes this series is diverget sice it is p-series with 5 7 Thus, the covergece of itervl is,. 7 ( ) ( ) p.,