第11章無窮-11-5函數(shù)展開成冪級數(shù)

第五一、主要內(nèi)三、同步練四、同步練習一、主要(一)函數(shù)的冪級數(shù)展泰勒Taylor展開函數(shù)定義若f(x)an(xx0)n xI(I為區(qū)間n則稱fx在I上可以展開成xx0的冪級數(shù)問題 (1)如果能展開 an是什么定理(展開式的唯一性)若在鄰域U(x0,Rf(x能展成冪級數(shù)f(x)an(xx0)nn

xU(x0,R

f(n)(x0

(n0,1,2,L)定義(泰勒級數(shù))設(shè)fx)在x0f(x)~

f(n)(x0)(x

x0)n

f(

) f(x0)(xx0)f(x0)(xx)22!

L

(xx0)nf(n)(x0為f(n)(x0麥克勞林級數(shù)(x0=f(0)f(0)xf(0)x2L

n 泰勒構(gòu)造

f(n)(x0)

(xx0)nI內(nèi)，級數(shù)是否一定f(x f(n)(x0 f(x)n0

(xx0)，x答：不一定.(二)函數(shù)展開成冪級數(shù)的充分定理 設(shè)f(x)在區(qū)間I上具有各階導數(shù)f(xIf(x)

f(n)(x0

(xx0)n, x limRn(x)

(xI其中Rnx)

f(n1)((n1)!(xx0

(ξ在x，x0之間(三)函數(shù)展開成冪級數(shù)的方直接展開法展開間接展開法1of(n)(x),f(n)(0),n012···2o

n

f(n)

xn

R3o判 limRn(x)

根據(jù)展開式的唯一性,利用常見展開式,通過變量代換四則運算恒等變形逐項求導,逐項積分等方法,求展開式.二、典型例1將f(x)e 展開成x的冪級數(shù)

f(n)(x)ex f(n)(0)1(n0,1,L),2oexex~f(0)f(0)x

f

x2L

f(n)

xn 1x1x21x3L1xn 3!1

收斂半徑R

n!(n

(,) ex=S(x)1x1x2L

1xnx1xn x

eξx(nx(n

xn1

(介于x0之間n 當n時eex1x1x21x3L1xnL,x(,)例2 f(x)sinx展開成x的冪級數(shù)解 f(n)(x)sin(xn

f(n)(0)sinn f(2k)(0)f(2k1)(0)(1)k

(k0,1,2,L)

n

(k0,1,2,L) (0)

n2k sin

~(1)k

x2k(2k

收斂半徑Rk

f(n1)((n3°f(n1)((n

sin[(n1)Rn(

xxsinxxsinxx1x31x5L1x2n1)

xn1n

2xn1(n0例 將f(x)(1x)m展開成x的冪級(m任意常數(shù)解

f(0)1,f(0)m, f(0)m(m1)f(n)(0)m(m1)(m2)L(mn1),21mxm(m1)x2Lm(m1)L(mn1)xn R

n

x

m3Fx),下證Fx(1x)m

1x(m1)L(mn)(m1)L(mn1) (n F'(x)m1m1xL(m1)L(mn1)xn1L (nxF(x)mxm1x2 (m1)L(mn1)xnL (n(1x)F(x)m1mxm(m1)x 2m(m1)L(mn1)xnLmF( n xm(1x)F(x)mF(x),F(0)1xm

xxF(x)dxx F(

1

dFF(x)(1x)m，x(1,1)lnF(x)lnF(0)mln(1二項展開式(1x)m1mx

m(m1)x2Lm(m1)L(mn1)xnL(1x1)1°在x1處收斂性與m的取值有關(guān)2m為正整數(shù)時得二項式定理(1x)m1mxm(m1)x2Lx(1(1mn(1x1)3°m1,

1x11x1x213x3135x4 2468(1x1) 11x13x2135x31357x4

2

24

246(1x1)例4將cosxx的冪級數(shù)

解sin

)cosx(sin

x2n1

L

)例5 f(x)ln(1x)展開成x的冪級數(shù)解[ln(1

1

(1)nxn

(1x1

ln(1x)

dx01

0 dx, xnn

1x1ln(1x)

nn

，1x注取x1得,ln2111L

n4例6將sinx展成x4

sinxsinπ(xπ sinπcos(xπ)cosπsin(xπ 1cos(xπ)sin(xπ

(x 2111(xπ)2 1(xπ)2

2 4

x

)1(xπ

（x

2k4

3

(2k 2 1(xπ) 1(x2 2

π)21(xπ)3L 3 (x例

x23x

展開成x4的冪級數(shù) x23x

x1x2

x x3(1x3

2(1x42 x4n1

x4

x4 3n0

2n0 )x4x23x

n

2n1 3n1

(6x2)1x例8將arcsinx展開 1xarcsinx

1(

2

m12

1 13 1x

21

2x 2 2 31x3!!x25!!x3 2n1!!1nx

2n!!1n

1n2n1!!x2n

1x121x121xn1x1x

x2

3 L

42n1!!x2n1

n

2n1!!x22n!!11xx 11xx0arcsinxxn

2n12n

x2n2n1

x

1

(2n (2n)!! 2n若0a則aa若0a則aabb

2n

3(2n

(2n1)!!, 13L2n124

1 12n 2 312n

2nnv2n

2n

vn unvn

2n

而3 而33(2n 3

(2n un收 故1

收 2n當x1

[1

(2n 2n x arcsinxxn

2n12n

x2n2n1

x例 將fx

32

x1的冪級數(shù)拆展fxlnxln32拆展ln[1x1]ln[12x

n1x1

2x11n n 1 1n12

x1

x11n

2x1

1n12n

1

x3時發(fā)散

2

收斂 n x[1,3

三、同步f(xln(2x3x2在x=0處展為冪級數(shù)1將x24x 展成x－1的冪級數(shù)x的冪級數(shù)f(x)arctan11將fxxarctanx

1xx的冪級數(shù)四、同步練習f(xln(2x3x2在x=0處展為冪級數(shù)解fxln(1xln2ln(132ln(1x) xn (1x

2x3(1x)(2322

3x)

(2(2

(1)n1(3

(2x f(x)ln2

xnn

n

3x)nf(x)ln2

1[1(3)n]xn(2x2n1

1將x24x 展成x－1的冪級數(shù) x24x

(x1)(x

2(1 2(3 241x12

481x142

(x12n 1

x1(x

L(1)n(x

L42n 42n1

x1(x1)2

n(x 8 8

(

L(1)

2n2 22n3

(x

(1x3f(x)的f(x)的 f(x)arctan11f(x)

x2n x(1,1)1 1f(x)f(0)

(1)n

xx2ndx

(1)n

2nnn

2n1f(x)π4nf(x)π4n02nx2n14x[1,1]1x將fxxarctan1x解fx(arctanx

1x

arctan1x 1fx