《高数双语》课件section 7-4.pptx

1、Expansion of Functions in Power Series12Taylor and Maclaurin SeriesIf a function f(x)can be expressed by a convergent power series in a neighborhood of a,that is0()(),(,),nnnf xcxaxaR aR (,).aR aRthen this series is called an expansion in power series of the function f,and we also say the function f

2、 can be expanded in power series at the point a in the interval 3Constructing a Series(1)If a function has derivatives of all orders on an interval ,can it be expressed as a power series on?(2)If it can,what will its coefficients be?()f xIIWe known that within its interval of convergence,the sum of

3、a power series is a continuous function with derivatives of all orders,but what the other way around?we need to solve the following two problems:4Constructing a SeriesIf we assume that is the sum of a power series()f x0()()nnnf xcxa 2012()()()nncc xacxacxawith a positive radius of convergence.211232

4、2342534()2()3()()()1 22 3()3 4()()1 2 32 3 4()3 4 5()nnfxccxacxancxafxccxacxafxccxacxa By repeated term-by-term differentiation within the interval of convergence I,we obtain5These formulas reveal a marvelous pattern in the coefficients of any power series that converges to the value of f on I(“repr

5、esents f on I”,we say).0()nnncxa Constructing a SeriesWith the nth derivative,for all n,being()()!a sum of terms with()as a factor.nnfxn cxaSince these equations all hold at ,we havexa 0123(),(),()2,()3!f acfacfacfacand,in general,()()!.nnfan c If there is such a series(still an open question),then

6、there is only one such series and its nth coefficient is()().!nnfacn 6Constructing a SeriesIf f has a series representation,then the series must be()2()()()()()()()2!nnfafaf af axaxaxan(2)7Let f be a function with derivatives of all orders throughout some interval containing a as an interior point.T

7、hen the Taylor series generated by f at x=a is()20()()()()()()()()!2!()()!kkknnfafaxaf af axaxakfaxan Taylor and Maclaurin Series()()0(0)(0)(0)(0)(0)!2!knknkfffxffxxxkn The Maclaurin series generated by f isthe Taylor series generated by f at x=0.8Finding a Taylor Series Find the Taylor series gener

8、ated by at Where,if anywhere,does the series converge to?1()f xx 2.a 1xSolution We need to find .(2),(2),(2),.fffwe get1122()()(1)11(),(2)221(),(2)2(2)(1)()(1)!,.!2nnnnnnf xxffxxfffxn xn Taking derivatives,9Finding a Taylor SeriesThe Taylor series is()22231(2)(2)(2)(2)(2)(2)(2)2!1(2)(2)(2)(1).2222nn

9、nnnffffxxxnxxx This is a geometric series with first term and ratio .1222xr 1 211.1(2)22(2)xxxIt converges absolutely for ,and its sum is|2|2x In this example,the Taylor series generated by at 1()f xx 2a converges to for or .|2|2x 1x04xFinish.10A Function f whose Taylor Series Converges at every x b

10、ut converges to f(x)only at x=0It can be shown(although not easily)that has derivatives of all orders at x=0 and that for all n.()(0)0nf 11A Function f whose Taylor Series Converges at every x but converges to f(x)only at x=0Hence,the Taylor series generated by f at x=0 is()22(0)(0)(0)(0)2!0000000.n

11、nnffffxxxnxxx The series converges for every x(its sum is 0)but converges to f(x)only at x=0.Then the question is:For what values of x can wenormally expect a Taylor seriesto converge to its generatingfunction?12Remainder of a Taylor PolynomialWe have already known the Taylor Theorem and its remaind

12、er,which can be used to estimate the difference between function f and its Taylor Polynomial.()()()nnf xP xRxThe absolute value is called the error associated with the approximation .|()|()()|nnRxf xP x()nP xLet f be a function with derivatives of all orders throughout some interval I containing a.T

13、he power series converges to f,iff lim()0,(,),nnRxxaR aR In this case,we say the Taylor series(2)is the Taylor expansion of the function f.13The Maclaurin Series for ex Show that the Taylor series generated by at converges to for every real value of x.()xf xe 0 x ()f xSolutionThe function has deriva

14、tives of all orders throughout the interval and we have(,)I 21(),2!nxnxxexRxnwhere1()for some between 0 and.(1)!nneRxxxn Since is an increasing function of x,lies between and .e xe01e xe When x is zero,and1xe ()0.nRx When x is negative,so is ,and .1e When x is positive,so is ,and .xee 14The Maclauri

15、n Series for exSolution(continued)1|()|when 0,(1)!nxnxRxexn and1|()|when 0,(1)!nnxRxxn Finally,because1lim0for every,(1)!nnxxn we have ,and the series converges to for every x.lim()0nnRx xe201,(,).!2!nnxnxxxexxnn Expansion ofxeThus Finish.Show that the Taylor series generated by at converges to for

16、every real value of x.()xf xe 0 x ()f x15Estimating the Remainderl It and Taylor Theorem can be used together to settle questions of convergence.l It can be used to determine the accuracy with which a function is approximated by one of its Taylor polynomials.Let f be a function with derivatives of a

17、ll orders throughout some interval I,if and ,such that0KNN ()|()|,nfxKnNxI then the power series converges to f in the interval I and1|()|.(1)!nnxaRxKn 16The Maclaurin Series for sin x Show that the Maclaurin Series for sin x converges to sin x for all x.SolutionThe function and its derivatives are(

18、2)(21)()sin,()cos,()sin,()cos,()(1)sin,()(1)cos,kkkkf xxfxxfxxfxxfxxfxx so(2)(21)(0)0and(0)(1).kkkff The series has only odd-powered terms and,for ,we have21nk352121(1)sin().3!5!(21)!kkkxxxxxRxk ()sinsin()2nxxn 17The Maclaurin Series for sin xSolution(continued)All the derivatives of sin x have abso

19、lute values less that or equal to 1,so we can apply the corollary of Remainder Estimation to obtain2221|1.(22)!kkxRk Since as ,whatever the value of x,22|0(22)!kxk k 21()0,kRx and the Maclaurin series for sin x converges to sin x for every x.Finish.()sinsin()2nxxn Show that the Maclaurin Series for

20、sin x converges to sin x for all x.18The Maclaurin Series for cos xSolution We add the remainder term to the Taylor polynomial for cos x with :2nk 2422cos1(1)().2!4!(2)!kkkxxxxRxk Because the derivatives of the cosine have absolute value less that or equal to 1,we have212|1.(21)!kkxRk For every valu

21、e of x,as .Therefore,the series converges to cosx for every value of x.20kRk Finish.352121024220sin(1)(1)(all real)3!5!(21)!(21)!cos1(1)(1)(all real)2!4!(2)!(2)!nnnnnnnnnnxxxxxxxnnxxxxxxnn Expansion of sinx and cosx Show that the Maclaurin Series for cos x converges to cos x for all x.19Combining Ta

22、ylor SeriesOn the intersection of their intervals of convergence,Taylor series can be added,subtracted,and multiplied by constants and powers of x,and the results are once again Taylor expansion.Then we obtain a new method to find some other Taylor expansion.This method is sometimes called the indir

23、ect expansion method(间接展开法间接展开法).20Finding a Maclaurin Expansion by SubstitutionSolution Find the Maclaurin expansion for .cos2xfor in the Maclaurin expansion for :cos x2xxWe can find the Maclaurin expansion for by substitutioncos2x20(2)cos2(1)(2)!nnnxxn 246(2)(2)(2)12!4!6!xxx 22446622212!4!6!xxx 22

24、02(1).(2)!nnnnxn that it holds for so the newly created series converges for all x.2,x Since the Taylor series generated by holds for implyingcos x,x Finish.21Finding a Maclaurin Series by MultiplicationSolution Find the Maclaurin expansion forsin.xx We can find the Maclaurin expansion for by multip

25、lyingsinxx3574682sin3!5!7!3!5!7!xxxxxx xxxxxthe Maclaurin expansion for by x:sin xThe new series converges for all x because the series for sin x converges for all x.Finish.22Finding a Maclaurin Expansion by AdditionSolution Find the Maclaurin expansion for ,and find .21()12f xxx Notice that21112()1

26、23 112f xxxxxand 2011 1124(2)2,.122 2nnnnxxxxxx then2011()(1),(1,1)1nnnnxxxxxx Thus 100012(1)21 1()(1)2,.3332 2nnnnnnnnnnf xxxxx Finish.nf()(0)nnnfn1()(1)2(0)!.3 and23A Series for arctan x and arcsin x by Term-by-Term IntegrationSolution Since 242211()(1,1),1nxxxxx we have211()(1,1),1nxxxxx Integrat

27、ion from 0 to x on both sides of the above series,we obtain3521arctan(1),(1,1).3521nnxxxxxxn Similarly,to find the Maclaurin Series for we can substitutearcsin x2x for to the series for and then integration both sides of x11x the series.Find the Maclaurin expansion for and arctanxarcsin.xFinish.Macl

28、aurin Expansion of Some functions2020242220231113511(|1)111()(1)(|1)111()(1)(|1)1ln(1)(1)(1)(11)23arctan(1)35nnnnnnnnnnnnnnnnnxxxxxxxxxxxxxxxxxxxxxxxxxnnxxxx 21210312112102(1)(|1)212111 3(2)1 3(2)arcsin(|1)232!212!21(1)(1)(1)(1)1,(11)2!nnnnnnnnnnnnxxxnnxxxxxxnnnnnxxxxxn 25Applications of Power Serie

29、s For example,the Maclaurin series for converges to for all x,but we still need to decide how many terms to use to approximate to a given degree of accuracy.We get this information from the Remainder Estimation Theorem.xexexe Calculating the Number e.1111(1),2!neRn Solution Since with ,we have1x 201

30、2!nnxnxxxexnn and1(1)for some between 0 and 1.(1)!nRen 26Applications of Power SeriesSolution(continued)For the purposes of this example,we assume that we know that .3e We are certain that13(1)(1)!(1)!nRnnbecause for .13e 01 By experiment,we find that ,where .61109!631010!Thus,weshould take to be at

31、 least 10 or n to be at least 9 with an error(1)n less than ,610 111112.718282.23!9!e LFinish.Calculating the Number e.27Applications of Power Series Find the approximate value of the integral 210 xedx such that the error is not great than 10-4.Solution Since 2222220()()1()(1),(,),2!nnxnnxxxexxnn af

32、ter integration term by term we obtain 2211000111135721100000(1)!(1)31042(21)!nxnnnnxedxdxnxxxxxnn 11111(1).31042(21)!nnn 28Applications of Power SeriesSolution(continued)By the theory of alternating series,we require 4110,(21)!nn and this requires only n 6.22771100001(1)(1)!(21)!0.74684.nxnnnnxedxdxnnn We take n=7 and we have Finish.Find the approximate value of the integral 210 xedx such that the error is not great than 10-4.