幂级数及泰勒展开习题解答

幂级数及泰勒展开习题解答公司内部档案编码：OPPTR-OPPT28-OPPTL98-OPPNN08]、求下列的收敛区间1.

xn2n(2n1)aan1aan1nnR

m 12n(22n(2n1)2(n1)(2n当x1因 1 1 1， 以 1 收2n(2n2n(nnn2敛，

n1

2n(2n1)x

(1)n2n(2n1)

绝对收敛， 收敛区间为。.

(1)nxn122n1 nmn

m 1aan1naan1n2n1 n2n n1R2当x2n1

(1)n

n为收敛的交错，nx

(1)n

2n1 n2n1 nn

发散， 收敛区间为(2,2。3.

(1)n

xn3nxnn1

2n mn

m 3aanaan1n(1)n132n1n1(1)n3n2nR1x1 1,1。3 3 334.

(1)n

(2x3)n2n1anan1annR

12n2n12n12x3

1即1

x2级数绝对。x1

()n()

1 , 1

1发散，n1

2n1 2n1n1

2n-1 2nx2

(1)n 2n1

的交错级数， 。.

ln(n1)(x1)nn1anan1annR

12)(n(nx1

1即0

x2级数绝对。x0因为1

x0，n

n1

x

x1f(x)

f(x)1xe)n

3

ln(n1)x x2

n

n1

(1)nln(n1)n1

x2因n

2 1

1 所以

发散， [0,2)。

n1

n1 2n

n1

n16.

(1)nn4nu

(x1)2n1(x(x1)2n1n4n(x1)2n1(n1

un

n

x24x24

1

x1

2x3x

(1)nn4n

(11)2n1

12

(1)n1n

x3

(1)nn4n

(31)2n1

12

(1)nn

。、求下列幂并求和函数1. x2n1(2x2n1(2nx2n1(2n

(1)n1x2n12n1

u

lim

x2n un

nx

1x

1|x1发散。x1

(1)n1

()2n1

(1)n

n1

2n1 2n1n1x1

(1)n12n1 。令S()

(1)n1x2n12n1

S(0)0S()

(1)n1x2n2

1 S()S)x

1

arctanxn1S(x)x(x

1x2

01t2.

2nx2n1

u

lim

x2n un

nx2n1(2nx2n1(2n2)x2n12n

1x

1x1。x

2n()2n1n1

2nx1

2n 区间为。令S()n1

2nx2n1S(0)0

S

x2t2n

x2x2n0 0

n1

1x2 S(x) x

2x

x.

1x2n(n1)xn

1x2aanaan1nnR

1(n(n1)(n2)n(nx1

n(n1x

n(n1)(1)n 区间为。令S()n1

n(n1)xn

S(0)0xS0

xn(n0

xn1x2n1

nxn1

x x2x2 xn

x2

1x

(1x)2 S(x)

2x

x.

(2n2nu

(1x)2x2n2

(1x)3

un

x2nx2n(2n1)2nx2n22(n1)(2n

x2x

1x

1x1。x1

2n1

1

1（通项不趋于零）2n 2nn1 n1 区间为。 2n1 令S(x

xn1x

x2n22n

S(0)2

S

x2n1t 2nt

12n11

1 2n 1

(x)(x0),S

(0)00 0 2n x

n1

2n

n1

2n x 1 1S(x)1

n1

x2n1

1

xS(x)S(0)x t 1ln(1x)21 1 01t2 21S(x)1

ln(1x2)2x0

S(x))

1 ) 2x 故

1x2

2x2 1 12),0x1S(x)1

2x21,2

x0S()

1

1x2n2

1

x2n211

1 1

1x2n n1

2n

1x2

2nn1

1x2

x2 2nn1、求下列级数的和211.

2n

12n

1

12n11

3 3n n 3333n 3 12 2333n1

n1

n1

1 3 3x 1也可以考虑利用幂级数

nxn1

xn

(|xn1

n1

1x

(1x)2

2n2

1n

2 1 33n 3

3

3 12 2n1

n1

1 3 3 ()n1

1

1 1 2.n1

(2n1)(2n

2n1

(1)n12n

12n11

(1)n1

1

(1)n1

1 (nk2 2n1 2 2n1n11

(1)n1

n11

(1)k 12 2n1 2 2k1n1

(1)n1

k21 1arctan112

n1

2n1 214 2四、利用直接展开法将下列函数展开成幂级数1.f(x)ax(aafn(xaxa)nfn(0)a)nn0

f(n)(0)xnn!

n0

(lna)nn!

xn，n

an1an1anlnan1

0R(f(n1)f(n1)(n1)!ax(lna)n1(n1)!(lna)n1(n

(x)

xn1

lim

xn1

M|x0n n

n

n

xa

a)n1

|x|

(lna)nx

(

xx n1(n1)!

n0

nn!(lna)nxax

n0

xn(n!

x。f(x)sinx21

x

1

0,

n2k,f(n)() n ,

f(n)(0)

sin

(1)k2n f(n)(0)

2 2

(1)n

2n 2

22k

n2k1 xn

x2n1，n0u

n! 22n1(2n1)!n0

un

x2nx2n3 22n1(2n22n3(2nx2n1

0R(

(x)

x2n3

|x0sin2sin2n3x2222n3(2n

n

22n3(2n|x2n3

(1)n

x2n1(

x22n3(2n3)!

n0

22n1(2n1)!xsinx

(1)n

x2n1(

x。2n0

22n1(2n1)!、使用接展开法将下列函展开幂：ex

n0

xn, (n!

)sin

(1)n

x2n1

, (

)s

n0n0

(1)n

(2n1)!x2n, ((2n)!

)41x

1n0

nxn, (1n!

1)(5)

11

n0

xn, (1x11

n0

(1)nxn, (1

1)11x

n0

(1)nx2n, (1

1)61x)

(1)n1

xn, (1xn7n

(1)n

x

(1)n1

x

, (1

x1)n0

2n1 2n1n1.。1．f(x)ex2tn解由

n0

(t令n!

x2得ex2

n0

(1)n

x2nn!

)。f(x)sin2xsin

n0

(1)n

t2n1(2n1)!

(t

2x得sin2x

n0

(1)n

(2x)2n1(2n1)!

(

)。f(x)sin2x

t2n

1 costn0

(1)n

(t及sin2(2n)!

12x2

t2x得sin2x

1 1

(2t)2n(2n)!

(2t)2n( 2(2n

(

)。n0

)!n1f(x)xf(x)

11x

n0

(1)nx2n(1

1)arctan

x 1 t

(1)n

t

(1)n

x

(1

1)01t2

n0

n0

2n1x时均为收敛的交错级数。5. f(x)

152x

11

n0

tn(|t

f(x)

152x

1 15125

t x得2521

2n 2 5f(x)

52x

5 n5

xn,

x 15

)21t6. f(x)x 1x2)1t1

1

13 (2n(1)n tn

1

(1)n

(2n1)!!tn(1t，n1得

24 (2n) (2n)!!n11xx1x211xx1x21x2

11x1x2

(2n1)!!x2n(2n)!!

|1)1x21x2

1 dt

() () tn n1t2x1t2

0(2n

x2n1

n1

(2n)!! 0n11x7. f(x)1x

(2n1)(2n)!!f(x)

21x

2n0

x2n(1

1)1

x 2 t

x2n1

(1

。1x

01t

n0

2n1、在指定点处将下列函数展开成幂级数f(x)x

在x处 tnt

n1

(1)n1

(1t1)及n

ln(2

2)ln21x

221

2，令t

x2得 2 2 2 x2n 2

x2nx2

2

x4)。n1f(xex,在

n n2nn1e

eex

(n0

1)nn!

)。

f(x)

x2xx0n(n2)f(x)

x2

()k1x

k(1)kk

xk2k0

k kk0f(n)(x)

kn2

(k2)(k1) (k3n)(1)kxk2nkf(n)(0)

n!n

1)n3n(n1)(n。1、设有两条抛物线ynx2 和y(n1)x21n

1n1

，记它们的交点横坐标的绝对值为an求a的表达式n求这两条抛物线所围成的图形的面积Snn求级数S 的和nann(n1)解：（1）a n(n1)n

(n1)x2 1 a34 1 ; 4n 4n

n n1 3 n33n(nn(n 1

mn

1 m

1

1

1

1，得n(n

k(k1)

k k1

n(n1)n1

k14 2

4

k11 4

n n a .an

3 n1

3n1

n(n1) 3幂级数部分习题课常用幂级数展式：ex

n0

xn, (n!

x)inx

(1)n

x2n1

(1)n1

x2n1

, (

x)n0

(2n1)! (2n1)!n1x2ncosx

n0

(1)n

, ((2n)!

x)（4）

1n0

) nxn, (1n!

x(5)

11

n0

xn, (1x11

n0

(1)nxn, (1

1)11x

n0

(1)nx2n, (1

1)61x)

(1)n1

xn, (1n

1)7n

(1)n

x

(1)n1

x

, (1

1)n0

2n1 2n1n1.、 把下列成x1. f(x)

x9x2x x 1

x

x2n

x2n1解f(x)

(1)n

,(3

。9x2 9

x2 9

3

32n231 31x1x2

n0

n0f(x)

由

f(x)

x 1x2

1 2x1x22 1x22 1x2f

0

n0

2n1

,(1

1)f(x)xft)0

n0

(1)n

x2n2(2n1)(2n

,(1x

。3. f(x)

114 1

1xx2

f(x)1 1

1 1 1

1

1 1

x4n,(1

1) f

41x 1x0

21x

1x

n1f(x)

f

x

,(1x

。0

4n1f(x)xx2x3x4)f(x)1

x2x3x4)

1x1x

x5)x)(x

1)而1x5)

(x5)n

x5n

,(1

1)n1

n nn11x)

(x)n

xn,(1

1)n1故

n nn1 x5n

xn

1x4n)f(x) n n n

xn,(1

。n1

n1

n1、 把下列函数在指定点展成幂级数1. f(xxx1处

(x1)nf(x)n

(x

1)]

n1

(1)n1

,(0xn

2)2. f(x)

x23x

2x1处f(x)(x

11)(

2)

1 1x1 x21 1 1 1 1

x1n

(x1)n x1

(1)n

(1)n

,(1

3)x1 2(x

212

2n0

2

n0

2n11 1 1 1 1

x1n

(x1)n x1

(1)n

(1)n

,(2

x4)x2 3(x1) 313

3n0

3

n0

3n1故f()

(1)n

1 1

(x1)n,(1

x n0

2n1 3n1f(x)

dexex1处 dx

x1

(x1)n

exe

(x)n1e

eex1e

n0

n! x1

n!n1f(x)

dexee

(x1)n2(x。dxx1

n(n2)! n24. f(xxx处4xx4 4 2x 4 4及x 4 42 cosx4x4 2n1

x4x4

(1)n

(2n

(

x)n0

2n

x4x4

(1)n

(2n)!

(

x