我愛學(xué)習(xí)83冪級數(shù)應(yīng)用

naxnn

RL

(2)Llim|an1

不存在(除)不能用

|ann冪級數(shù)axn的和函數(shù)為S(x)滿足nn(1)(連續(xù)性)S(x)在其收斂域上連續(xù)即xD

limS(x)

limaxn

S(x x

x (2)(可導(dǎo)性S(x)在其收斂區(qū)間 S(x)

(axn)

naxn1 n0 n1)

0S(x)dx0

anxdx

nf

a(xx)n f(x可以作為anxx0

若可展開，如何求系數(shù)an展開式是否唯一？ 冪級數(shù)的展開與應(yīng)Taylora(xx)nn0f(x)a(xx)na(xx)nn0

x(x0R,x0討論anf(x)f(x必在(x0-Rx0R)內(nèi)任意階可導(dǎo)na(xx)n1n0f(x)na(xx)n1n0n(nn(n1)an(xx0f(x)

(xx0),f(n)(,f(n)(x)n!a0nf(

)

,f(

)

,f

)2!a2即 f(x),af(x),

f(x0),f((x0,f((x0n,f(nf(n)(x002nf(x0)

(x0)(xx0)

f(x)

(xx0)

(xx0f(xxx0處的Taylor級數(shù)若fx

an(xx0

則an

f(n)(x

nf(xx0點的冪級數(shù)展開式唯一f(xxx0可展開為冪級數(shù)，則必是其在x=x0處的Taylor級數(shù). f(n)( 0(xx0

但此級數(shù)的和函數(shù)卻未必是f 例 xf(x) x) f(0)0，f(n)(0)0，(n1,2,)f(x在x0點的Taylor級數(shù)在實數(shù)域上收斂到f(n)xn但ff(n)xn f(x)f(0)f(0)x

f(0)2!

x2f(x)的Taylor級數(shù)能否收斂于f(x)？ f(x在(x0-Rx0R)f(xx=x0點的 f(x f(n)(xf(x)f(x)f(x)(xx) 0(xx)2 0(xx)nR f(x)Sn1(x)Rn(nf(xn

f(n)(x0

(xx0

收斂于fSn1(x)f(x)(nRn(x)0(n 設(shè)f(x)在點x0的某鄰域內(nèi)任意階可導(dǎo)，則f(x)在該鄰域內(nèi)可展開為點x0處的Taylor級數(shù)limRn(x)0,Rn(x)為f(x)的 注x00時，稱注n0

f(n)

xn為f(x直接展【例1】將fx)ex展開為x的冪級數(shù)【解 f(n)(x)ex f(n)(0)林級

1x1x2 1xn n!

x

Rn(x)

(n

x

(在0x之間|

(x)

(n

xn1

e|x|

|x|n(n1)!xnn n

(c)lime|x|n

|x|n1(n1)!

limR(x)nnnex 1x1x2 1xn ，(x)2! n 1nn1 1nn1n2!【例2】將fx)sinx展開為x的冪級數(shù)【解】

(n)(x)sin

nx

f(n)(0)

n

(1)k

n2k f(x) 林級數(shù)11x31x5

x2n1

收斂域(,)3! 5!Lagrange

(2nsin(n1)

|x|nRn(x)

(n

，(0x間(n1)!

(c)

limRn(x)n n

nsinxx1x31x5 3! 5!

(2n

x2n1

，x對fx)(1x)(為常數(shù)當(dāng)為正整數(shù)時當(dāng)不是正整數(shù)時，可證明f(x) (11)內(nèi)收斂于f(x(1x)1x(1)x2 ( (n1)xn n!而在x 處視不同的值而定

x

xn 1x xn

，(1x1)1

nxn ex 1x

x2

xn n

n 2! nx2n

(x)sinx (1)n n0 (2n1)!

x2nx(1

x33! 5!

x5 (1)n (2n1)!(x)1x(1)x2 ( (n1)xn n! 1

(1)n(2n

1)!!x

(1x(1x1)1

n

2nn (1)nxn，(1x(1)nxn，(1x1

n0

(1)nln(1x)

dx(1) dx

xn 1

n0

n0

n(1)n-

(1)n1 xnx x2

xn

(1x1)n1 ln(1x)n

xnn

xnn

(1x1) x2n

x2cos

(sinx)

(1)n (1)n n

(2n1)!

n

(2n)!11x21x42! 4!

(1)n

x2 (2n)!(x)

1x2

1x2x4 (1)nx2n (1x1)

x1x

(1)nx2n12n3dx(1)nx2n12n3 x

2n

(1x1)

2(2xcos

x cos2

2n

(1)

(2n)!

22 (1)n x2

(x) 2n

(2n)!ln1 ln(1x)ln(1x1(1)n

xn1

xn1

x2kn0

n n0n

k02k2x2x3 3

2n

x2n

(1x1)將fx)ln(x24x3)(x+2)的冪級數(shù)f(x)ln[1(x2)2 n1 [(x n1 (1)n(x2)2

(x2)4(x2)6

(x2)2nn(3x1)將fx)

x24x

x=1展開成冪級數(shù)f(x)

(x1)(x

2(x 2(x3)1 4 x1

x11

21

(1)n(x 2n

(1)n(x1) 22n01

n(1)n1 (x1)n

(1x3) 2

22n1n 1x11

x11

1x3x

1x二.二.Taylor①直接法②拆項法③遞推法

的和Sarctan arctan1arctan1

1 11

arctan3 Sarctan1arctan 假設(shè) arctankk arctank1

k arctan1

(n n 故

4①恒等變②冪級數(shù)的運算性質(zhì)(四則運算、分析運算

n

ln2，

4

n

的收斂域、和函數(shù)S(x)，并計

(n1)nn1(n

n

n

(n1 n1

S(x)

x x2exn1

n1(n

n0(n1)

n

n11S(2)

2n n0 14e2(e21)e22

xn1

14916

1t，設(shè)S(tx

n2tn1，t(1,tt tt則

0

n2tn1dt

nt

tntn1 t

t 1

1

(1S(t)

1(1

(1t S1

x2(x1) x

(x (|x| 14916

S122

n2

1)2n

的和n【解

S(x)n2

n21x

(1x1

1 xn

1 xn則S(x) n22n

n1

2

2n2n2 xn

nn1 n

12x

(1xn利 nn1

ln(1x 1x1) S(x)xln(1x)2

1ln(1x)x2x

x2(|x|1,x2 2 S153ln2．2

(n21)2n 1

x2S(x)

ln(12

2xln(1x)x2 2 1 ln(1 2x

(|x|1,x0)2 1 S(0)limS(x)lim ln(1

x0

2

2x 1 S(1)limS(

limx1

ln(1x)2x 2

1

S(1)limS(

lim ln(1

x1 2x 構(gòu)造一個冪級數(shù)

axn

axn nn求an 的收斂域nnn若x0 ，則 發(fā)n

x0

axn

S(x)unS(x0 nanxan的處理方法常不唯一n

n1

(1)n(2n1)!

的和【解】構(gòu)

(1)n

x2n1n1(2nlimn

x22n(2n3)

(1)n1(1)n1(n1)x2n(2n(2n∴收斂域為,設(shè)冪級數(shù)

(1)n

x2n1

n1

n1(2n(1)nS(1)(2n1)!法1Sx)n1

(1)n(2n

x2n1

(1)n(2n1

x2n2n1

(2n1)!1

(1)n

x2n1

(1)n

x2n12n1

(2n)!

n1

(2n1)! 1

(1)n

x2

(1)n

x2n12 n

(2n)!

n

(2n1)! 1x(cosx1)(sinx21(xcosxsinx)

x(,法2Sx)n1

(1)n(2n

x2nx2 (1)n(2n) x2n n1

(2n1)!x2

(1) 2n1

(2n

x2nx x2

(1) '

x2n12

n

(2n1)! x2

' 2

sinx1

(xcosxsinx)S(0)limS(x)x0S(x)1(xcosxsinx)2

x(,)法3Sx)n1

(1)n(2n

x2n x (1)n

x2ndx0n1

(2n)! x (1)n

x2ndx 1

n1(2n1)! x(xsinx)dx 1(xcosxsinx)

x(,) (1)n

S(1)

(cos1sinn1(2n1)! 本題也可假設(shè)Sx)n1

(1)n(2n

x2n S(x)n1

(1)n(2n

x2n

f(n)nf(x)n

an(x

an

f(n)(a)n!a【例】若fx)

x41

，求f(n 【解】fx)x4x3n(1)nx3n

(1xn0當(dāng)n3k4

f(n)(0) f(3k4)(0)(3k4)! k0,1, f(0) f(7)(0)f(314)(0)7!(1)17!f(14)(0) xex2dx,

sin

dx

x……

lnxex2x

x

(1)n

x2ndx

(1)n

x2n0xsinx

n0x

n!(1)n

n0(2n1)n!x(,)x2ndx

n0(2n1)!(1)n

x2nn0(2n1)(2n

x(,)

1sin

dx的近似值，精確到10-4

sinx11x21x41x6

I

1sin1

3!dx1

7! 1 1 33! 55! 77!誤

|Rn|un1(2n3)(2nn要|R|104n

(2n3)(2n

104

104

104

1043 5

77!

294120I1

3 5

f(x

f(n)(xf(x)f(x)f(x)(xx) 0(xx)2 0(xx)n

近似

f(x)

|R(x)|

f(n1)(x (xx)n1n兩類問題

(n1)! 給出精度,確定項數(shù) 方法1：若是Leibniz型級數(shù),可用|Rn(x)|un1來估計;方法2.：|Rn(x)||un1(x)un2(x) |un1(x)||un2(x)|

【例1】計算e的近似值，精確到10-4exex1x1x22!1xnn!,1n!212!1n!212!n!2!

|R| (n1)!

(n2)!

1 (n1)! n 1

(n1)! n (n

n 11

nn要|R|104n7!e2117! 3!

nn

1042.7183

n7【例2】計算ln2的近似值，精確到10-4【分析】ln(1xx1x2

(1)n11xn ,(1x (1)n11n,ln211(1)n11n, 誤 |R|

要|R|104 須n= n f(x)ln1xln(1x)ln(1x)1x2x2x3 3

n

x2n 3