Power Series: A power series about the center, x = 0, is a function of x of the form

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1 You are familiar with polyomial fuctios, polyomial that has ifiitely may terms. 2 p ( ) a0 a a 2 a. A power series is just a Power Series: A power series about the ceter, = 0, is a fuctio of of the form c c0 c c2 c3 c 0 A power series about the ceter, = a, is a series of the form c ( a) c0 c( a) c2( a) c3( a) c ( a) 0 k Eample: Let f( ) ( 2). k! Evaluate f (): k Does the series, f () coverge? Or diverge? I fact, if we the fuctio f () oly makes sese as a fuctio if it coverges to a real umber. Therefore, it is importat to ask for what values does the fuctio, or power series, coverge? This would be the domai of our fuctio f ad would be the iterval of covergece of the power series. Eample: Of which series does the series below remid you? 2 ( 2) ( 2) Does't rig a bell? Try substitutig u. Now, the series becomes 2 uu u Oh! It's the geometric series, which coverges to, if - < u <. That must mea that u if 2. Or more simply, if 0 4, 2 2 ( 2) ( 2) ( 2) ( 2) 2 4 8

2 Does the series coverge at = 0 or = 4? We plug i these values to check 2 ( 2) ( 2) 2 = 0, diverget (2) (2) ( 2) = 4, does ot coverge We say the power series is cetered at 2 radius of covergece is 4 the iterval of covergece is (0,4) The Covergece Theorem for Power Series: If the power series < c. a 0 coverges for ozero c, = c, the it coverges absolutely for all If the power series > d. a 0 diverges for = d, the it coverges absolutely for all Practice Problems: For each of the followig power series, idetify the ceter, radius of covergece, ad covergece iterval. ) ( ) Ceter: Radius of Covergece: Covergece Iterval: 2) 0! Ceter: Radius of Covergece: Covergece Iterval:

3 MacLauri Series Cosider the polyomial f( ) Note that the coefficiets are a0 a a2 a3 a4 WITHOUT SIMPLIFYING, compute the followig: f (0) a0 ( f '(0) a ( f ''(0) a2 ( f '''( 0) a3 ( ) ) ) ) (4) f (0) a ( ) 4 This meas we ca solve for the coefficiets of the polyomial. a0 a a2 a3 a 4 So if a fuctio f is ifiitely differetiable i a ope iterval aroud 0, we ca write the fuctio as a power series fuctio for some iterval cotaiig 0. f ( ) f(0) 0 f '(0) f (0) 2 f (0) 3 f (0) 0!! 2! 3!! k f (0) k k 0 k! Eample: Compute the MacLauri series for f ( ) cos( ). th term th derivative th derivative evaluated at 0 0 f ( ) cos( ) f (0) cos(0) f '( ) si( ) f '(0) si(0) 0 2 f ''( ) cos( ) f ''(0) cos(0) 3 f '''( ) si( ) f '''(0) si(0) 0 4 f (4) ( ) cos( ) 5 f (5) ( ) si( ) 6 f (6) ( ) cos( ) th f (7) ( ) si( ) f (4) (0) cos(0) f (5) (0) si(0) 0 f (6) (0) cos(0) f (7) (0) si(0) ( ) 2 cos( ) ! 2! 4! 6! (2)! ( ) 2)! 2 4 2! 4! ( 2

4 Notice the more terms that are added to the MacLauri Series of cos(), the closer the graph is to the actual graph of cos(). If you remember the MacLauri Series for the fuctios below, write them dow. If ot, compute them usig the formula: f( ) k 0 f ( k ) (0) k! k ) f ( ) si( ) 2) f ( ) e 3) f( )

5 Taylor Series MacLauri Series of f : If a fuctio, f, is ifiitely differetiable aroud = 0, the f has a power series epasio cetered about = 0, give by f ( ) 0 ( ) f (0)! Suppose g has ifiitely may derivatives aroud = c. The g () c eists for ay. Just as we did with power series, we ca chage the variable, u = - c, or = u + c Now, let us defie, a ew fuctio f ( u) g( u c) Use Chai Rule to compute the derivative of d( g( u c)) d Sice f satisfies the properties to have a Mac Lauri Series, we kow there eists a power series epasio ( ) f (0) ( ) ( ) f u u 0! ( ) g () c ( ) c 0! gu ( + c) g ( ) Taylor Series Epasio of f aroud = c: Let g be a ifiitely differetiable fucito i a iterval aroud = c, the g has a power series epasio defied by ( ) g () c g ( ) ( c )! 0 Eample: Fid the Taylor Series Epasio of f( ) l( ) about =. f() 0 f '( ) f ''() f '''() 4 l( ) ( ) ( ) ( ) ( ) ( ) ( )

6 Radius ad Iterval of Covergece I order to fid the radius of covergece, we must look at the Taylor Series Epasio ad fid at which it coverges. We might be able to use the Ratio or Root Test to this purpose. Eample: Suppose f ( ) ( )! (3) for all, ad f (3) =. Fid the radius of covergece. 3 Solutio: By Taylor Series Epasio we kow that ( ) f (3) ( ) ( 3) f 0! By the Ratio Test a ( )! 3! lim lim 3 a 3 ( )! 3 ( )! 3 lim 3( ) 3 lim The series will oly coverge by the ratio test if Therefore the radius of covergece is 3. To fid the iterval of covergece may be (0, 6), but we have to check the edpoits. If = 0, the the series is give by ( 3) ( ) f (0) which is kow to coverge because it is a alteratig harmoic series. Cosequetly, at = 6, the series is: (you do) Geeralized Mea Value Theorem: Let f be a differetiable fuctio o a ope iterval ( a, ) ad cotiuous o a closed iterval [ a, ]. The there eists a poit c( a, ) such that f '( c) f( ) f ( a) a

7 If we let ( k ) f ( a) P ( ) ( a) k! k0 k be the th degree Taylor Polyomial evaluated aroud = a, the Taylor s theorem says there eists a c( a, ) such that ( ) f () c f ( ) P ( ) ( )! ( a) The Origial Mea Value Theorem is the case where =0. Are the statemets the same for =0? The idea is to proceed by iductio, but as you ca imagie, it gets a little complicated. So we wo't prove it here. Taylor's Theorem: If f is -differetiable o a iterval cotaiig a, (a, a ),the there eists a umber c ( a, b) such that ( ) f( a) 2 f ( a) f( ) f( a) f '( a)( a) ( a) ( a) R ( ) 2!! P ( ) R ( ) We call ( ) f () c R ( ) ( a) ( )! the remaider of the th degree Taylor Polyomial, P ( ). The purpose of Taylor s Theorem

8 l() is differetiable at = The Taylor Polyomial of degree 5 is a good estimatio, but the error icreases as gets further away from. The error at 2 is give by The error at = 2, is How ca we estimate the error at each poit? LaGrage s Error Boud

9 The idea behid this uses Taylor s Theorem, but the statemet is as follows: ( ) If f( ) has derivatives ad f ( c) M for all c ( a, ), a the the remaider R ( ) M ( )! Practice Problem: [999 BC 2] The fuctio f has derivatives of all orders for all real umbers. Assume f(2) 3, f '(2) 5, f ''(2) 3, ad f '''(2) 8. a) Write the third degree Taylor Polyomial for f about = 2 ad use it to approimate f (.5). b) The fourth derivative of f satisfies the iequality f (4) ( ) 3 for all i the closed iterval [.5, 2]. Use the Lagrage Error Boud o the approimatio to f (.5) foud i part (a ) to eplai why f (.5) 5. c) Wrote the fourth degree Taylor Polyomial, P ( ) for g 2 ( ) f( 2) about = 0.