第十一章11 4b泰勒級(jí)數(shù)

1、第四節(jié)b函數(shù)的泰勒級(jí)數(shù)1.泰勒級(jí)數(shù)2.展開條件3.展開成麥克勞林級(jí)數(shù)4.展開成泰勒級(jí)數(shù)5.利用冪級(jí)數(shù)展開式求和1.泰勒公式設(shè)f(x)在(x0 - d, x0 + d)(或(x0 - d, x0 ,x0 , x0 + d)中具有n + 1階導(dǎo)數(shù),則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階泰勒公式( n+1) (x )f( xn+1其中Rn ( x)

2、 =- x0 )x0 與 x 之間).(n + 1)!( x在-拉格郎日余項(xiàng)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階麥克勞林公式泰勒級(jí)數(shù))(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為x - x0的冪級(jí)數(shù)的級(jí)數(shù)稱為泰勒級(jí)數(shù)麥克勞林級(jí)數(shù)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為x的冪級(jí)數(shù)的級(jí)數(shù)稱為麥克勞林級(jí)數(shù)2.展開條件定理1f ( x)在x0的某領(lǐng)域內(nèi)能展開成泰勒級(jí)數(shù)(1)f(x)在x0的某領(lǐng)域內(nèi)具有任意階導(dǎo)數(shù)(2)lim Rn (x) = 0n如果函數(shù) f ( x) 在 Ud ( x0 ) 內(nèi)能展開成定理 2f ( x) = a(

4、 x - x( x - x)n ,)的冪級(jí)數(shù), 即0n0n=01a=( n) ( x(n = 0,1,2,L)f)則其系數(shù)n0n!且展開式是唯一的.3.函數(shù)展開成麥克勞林級(jí)數(shù)(1)用直接法步驟求f(x)的各階導(dǎo)數(shù) 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!寫出收斂區(qū)間第四步:下面介紹某些基本初等函數(shù)的冪級(jí)數(shù)展開將f(x) = ex展開成x的冪級(jí)數(shù).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!將f ( x) = sin x展開成x的冪級(jí)數(shù).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將f ( x) = (1 + x)a (a R)展開成x的冪級(jí)數(shù).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處收斂性與a的取值有關(guān).注意:當(dāng)a = -1, 1時(shí), 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雙階乘n=0=(-1,1)xn1 - x(2)間接法通過變量代換, 四則運(yùn)算, 恒等變形, 逐項(xiàng)求導(dǎo), 逐項(xiàng)積分等方法,求展開式.1)利用冪級(jí)數(shù)的收斂性質(zhì)注意要考慮展開區(qū)間例4將下列函數(shù)展開為x的冪級(jí)數(shù)(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端點(diǎn)

9、的斂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能改變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)收斂區(qū)間不變1(3)f(x) =x2- 4x + 3將f ( x) = x - 1 在x = 1處展開成泰勒級(jí)數(shù)例64 - x(展開成x - 1的冪級(jí)數(shù))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 用逐項(xiàng)求導(dǎo)或逐項(xiàng)積分,求下列級(jí)數(shù)的和函數(shù)