第十一章11 4b泰勒级数

1、第四节b函数的泰勒级数1.泰勒级数2.展开条件3.展开成麦克劳林级数4.展开成泰勒级数5.利用幂级数展开式求和1.泰勒公式设f(x)在(x0 - d, x0 + d)(或(x0 - d, x0 ,x0 , x0 + d)中具有n + 1阶导数,则x (x0 - d, x0 + d)(或(x0 - d, x0 ,x0 , x0+ d)(x - x) + f (x0 )(x - x有f(x) = f (x) + f (x)200002!f (n)(x)0+ L +(x - x0 )+ Rn (x)nn!- -f(x)在x0的n + 1阶泰勒公式( n+1) (x )f( xn+1其中Rn ( x)

2、 =- x0 )x0 与 x 之间).(n + 1)!( x在-拉格郎日余项f ( x 0 ) ( x -f ( x) + f ( x在 f ( x ) =)( x -x) +) 2x00002!( n ) ( x)0f+L +( x -+x 0=nx 0 )Rn ( x )中令0 ,n!得 :f ( x) =f (0)( n) (0)ff (0) +f (0)x + L +x 2xn2!n!( n+1) (qx)f(0 q 1)xn+1+(n + 1)!- -f(x)的n + 1阶麦克劳林公式泰勒级数)(x - x) + f (x0 )(x - x形如: f (x) + f (x+ L)20

3、0002!(n)+ f(x0 )(x - x+ L = f (x) + f (n) (x)0(x - x)n)n000n!n!n=1为x - x0的幂级数的级数称为泰勒级数麦克劳林级数f (0)f (n) (0)形如: f (0) + f (0)x +2+ L +x+ Lnx2!n!= f (0) + f (n) (0)nxn!n=1为x的幂级数的级数称为麦克劳林级数2.展开条件定理1f ( x)在x0的某领域内能展开成泰勒级数(1)f(x)在x0的某领域内具有任意阶导数(2)lim Rn (x) = 0n如果函数 f ( x) 在 Ud ( x0 ) 内能展开成定理 2f ( x) = a(

4、 x - x( x - x)n ,)的幂级数, 即0n0n=01a=( n) ( x(n = 0,1,2,L)f)则其系数n0n!且展开式是唯一的.3.函数展开成麦克劳林级数(1)用直接法步骤求f(x)的各阶导数 f (x),f (x),Lf (n)(x)L求 f (0),f (0),Lf (n)(0)L第一步:第二步:f (0)f (n) (0)f (x) = f (0) + f (0)x +x+ L +x+ L2n第三步:2!n!写出收敛区间第四步:下面介绍某些基本初等函数的幂级数展开将f(x) = ex展开成x的幂级数.例1f (n) ( x) = e x ,f (n)(0) = 1.(

5、n = 0,1,2,L)解( n) (0)f1 an =n!n!f (0)f (n) (0)f (x) = e= f (0) + f (0)x +xx+ L +2n+ Lx2!1 xnn!= 1 + x +1 x2 + L + Lx (-,+)2!n!= n=0xnn!将f ( x) = sin x展开成x的幂级数.例2f (n) (0) = sin np ,f (n) ( x) = sin( x + np),解22(n = 0,1,2,L) f( 2n) (0) = 0,f (2n+1)(0) = (-1)n ,f (0)f (n) (0)f (x) = sin x = f (0) + f

6、(0)x +x+ L +x+ L2n2!nn!x2n+111= x -x+35-L + (-1)+ Lx(2n + 1)!3!5!(-1)n x2n+1(-1)n-1 x2n-1x (-,+)= n=0或=(2n + 1)!(2n - 1)!n=1将f ( x) = (1 + x)a (a R)展开成x的幂级数.例3Q f (n)( x) = a(a - 1)L(a - n + 1)(1 + x)a-n ,解(n = 1,2,L)f (n)(0) = a(a - 1)L(a - n + 1),f (0)f (n) (0)af (x) = (1 + x)= f (0) + f (0)x +x+

7、L +2n+ Lx2!n!= 1 + ax + a(a - 1) x2+ L + a(a - 1)L(a - n + 1) xn+ L2!n!= 1 + a(a - 1)(a - 2)L(a - n + 1) xnx (-1,1)n!n=1在x = 1处收敛性与a的取值有关.注意:当a = -1, 1时, 有2= 1 - x + x2 - x31+ L + (-1)n xn + L(-1,1)1 + x1 3(2n - 3)! xn1 + x = 1 + 1 x -1+L+ (-1)n-1x2+x3+L2 42 4 62(2n)!-1,1= 1 - 1 x + 1 3 x2- 1 3 5 x3

8、(2n - 1)! xn1+ L + (-1)n+ L1 + x2 42 4 62(2n)!(-1,11双阶乘n=0=(-1,1)xn1 - x(2)间接法通过变量代换, 四则运算, 恒等变形, 逐项求导, 逐项积分等方法,求展开式.1)利用幂级数的收敛性质注意要考虑展开区间例4将下列函数展开为x的幂级数(1)f(x)=cosx解:由cos x = (sin x)x2n+111Qsin x = x -+x5 -L + (-1)n + Lx3(2n + 1)!3!15!x2n1cos x = 1 -+- L + (-1)n + Lx2x42!4!(2n)!x (-,+):(2)arctanx端点

9、的敛1dxx(arctanx) =arctanx =解:由1 + x21 + x20散1= (-1)x( - 1 x 1)性可nnxn=0Q=(-1)xdxn2n1 + x0n=0 11 + x2=(-1)x能改变n2n2n+1n=0 1 -1 x 1)x= (-1)n=0n(-1 x22n + 1x2n+12n + 1- 1+ 1x -1,1= x35- L + (-1)n+ Lxx35(3) f(x)=ln(1+x)1解:由 (ln(1 + x) =1 + xQ 1= dx1 + xxln(1 + x) =(-1)n xn( - 1 x 1)1 + x0n=0x=(-1)n xndx0n=

10、0xn+1= (-1)nn + 1+ x3n=0xn11n-1= x -x2- L + (-1)+ L23x (-1,1n2)套公式(2)f(x) = ln(5 + x)收敛区间不变1(3)f(x) =x2- 4x + 3将f ( x) = x - 1 在x = 1处展开成泰勒级数例64 - x(展开成x - 1的幂级数)并求f ( n) (1).111解=,Q3(1 - x - 1)4 - x3 - ( x - 1)3= 11 + x - 1 + ( x - 1)2+ L + ( x - 1)n+ L3333x - 1 3 x - 1 = ( x - 1) 14 - x4 - x( x - 1)2( x - 1)3( x - 1)n= 1( x - 1) +3+ L + L32333nx - 1 0 , b 0).4、a+ bnn二、利 用逐项求导或逐项积分,求下列级数的和函数