高数下册12十二章3节

¥二、幂级数的12.3

幂级数一、函数项级数un

(x)的收敛(发散)点；n=1收敛域I；和函数S(x).¥n

n

0a

(x

-x

)收敛半径R；n=0收敛区间I

'；收敛域I.三、¥n

nn

=0a

x

的分析性质：(1)(2)(3)12.3

幂级数一、函数项级数的概念：¥若{un

(x)}是定义在区间I上的函数列：u1

(x),u2

(x),u3

(x),,un

(x),,则un

(x)

=u1(x)

+u2

(x)

+u3(x)

++un

(x)

+n=1称为区间I上的函数项级数.¥1、un

(x)在a点收敛(发散)n=1¥常数项级数un

(a)收敛(发散).n=1¥a是un

(x)的收敛(发散)点n=1¥2、un

(x)的收敛域I：n=1¥un

(x)的所有收敛点组成的集合.n=1n=1¥即：I

={x

un

(x)收敛}.¥注：同理有un

(x)的发散域.n=13、和函数：注：若记Sn

(x)=uk

(x),则k

=1nfi

¥x

˛

I时,S(x)=lim

Sn

(x).¥当x

˛

I时,un

(x)均收敛,n=1¥其和随x而变,记为S

(x),并且称S

(x)为un

(x)的和函数.n=1n二、幂级数及其敛散性：1、幂级数的定义：(1)x

=x0时，un

(x)自然收敛；n=00

¥

¥n=0nn=0n

nu

(x)

=

a

x

.(2)x

=0时，形如¥

¥n=0n=00nna

(x

-x

)的函数项级数nu

(x)

=称为幂级数.¥n

3nn解：u

(x)=n2(x

+

1

)n

\

a

=2¥¥(1)n=1nn=1nn

3(2x+1)nu

(x)

=20,

x

=-

1n

3n2nnn

2n(2x

-3)n¥

¥(2)un

(x)

=n=1

n=1例1、写出下列幂级数的an和x0：nn

2解：u

(x)=1

(x

-3

)n\

an

=

1

,

x0

=

3n

2Th1：若nx

+a处收敛,则a

(x

-x

)在

n

0

0¥n=0¥n=000|

x

-x

|<|

a

|时,nna

(x

-x

)绝对收敛；0>|

a

|时,|

x

-

x

|¥n

n

0n=0a

(x

-x

)发散.若00nx

+a处发散,则a

(x

-x

)在

n¥n=02、幂级数的敛散性：(2)I

=(x

–R)称为na

(x

-x

)的收敛区间；n

00'¥n

n

0

0a

(x

-x

)

,若I

„{x

},I

„(-¥,+¥

),Cor：对n=0则$1R

>0使：|

x

-x0

|<R时绝对收敛,|

x

-x0

|>R时发散,x

=x0

–R时可能收敛或发散.(1)R称为

an

(x

-x0

)的收敛半径；n(3)

an

(x

-x0

)的收敛域I：n(x0

–R), [

x0

–R), (

x0

–

R],[x0

–R]之一.C.R

>

3

D.R

‡

3A.R

<

3

B.R

£

3处发散,则其收敛半径练习1、若nna

x

在x

=3¥n=0R的取值范围为(

)练习2、若a

x

在xn=3处条件收敛,则其收敛

n¥n=0半径R的取值范围为(

)D.以上都不对B.R

=

3

C.R

>

3A.R

<

3¥n

=0Th

2：(1)对0nx

)

有：na

(

x

-annfi

¥

an+1R

=

lim

；(2)对缺奇(偶)次项的幂级数¥

¥n=02n-1

n=02nn

0an

(x

-x0

)

有：a

(x

-x

)

及|.anan+1R

=lim

|nfi

¥用比值审敛法可得0(x

-x

)<1时绝对收敛un(

x)lim

un+1

(

x)nfi

¥an=

lim

an+1nfi

¥annfi

¥

an+1<

lim

=

R0此时,x

-x用比值审敛法证明Th2：¥

¥(1)对幂级数n=0n=00nna

(x

-x

)

有：nu

(x)

=¥n=0¥n=02nn

0a

(x

-x

)nu

(x)

=(2)对幂级数<1时绝对收敛0(

x

-

x

)2un

(

x

)lim

un+1

(

x

)nfi

¥an=

lim

an+1nfi

¥=

R

x

-

xanan+1<

limnfi

¥0anan+10nfi

¥此时,

(x

-

x

)2

<

lim及¥n=02n-1an

(x

-x0

)

有：¥n=0nu

(x)

=3、幂级数的收敛域I的求法：¥annfi

¥

an+1R

=

lim0

I

'

=

(x

–

R),(1)对n

n

0n

=0a

(x

-x

)

有：判断x

=x0

–R时原级数的敛散性得I：(x0

–R),

[x0

–R),

(x0

–R],

[x0

–R]之一.(2)对缺奇(偶)次项的幂级数¥

¥n=02n-1

n=02nn

0an

(x

-x0

)

有：a

(x

-x

)

及anan+1R

=lim

|nfi

¥0|

I'

=(x

–R),判断x

=x0

–R时原级数的敛散性得I：(x0

–R),

[x0

–R),

(x0

–R],

[x0

–R]之一.例2、求幂级数的收敛域I：2n+1(-1)n

xnn

n=1

n=1¥

¥(1)u

(x)

=[分析]：(1)由通项un

(x)确定an

,x0；(4)得收敛域I.(3)判断x

=x0

–R时原级数的敛散性；(2)求R

=limnfi

¥0,

得I

'

=

(x

–

R)；anan+1详解n解：u

(x)

=n2n+1(-1)

xn\

I

=(-1,1]收敛当x

=1时,原级数为¥n=1(-1)n2

n+1(Leibniz准则)发散¥12n+1当x

=-1时,原级数为n=1(同阶法+调和级数)=

lim

2n+1

(

-1)n+12(

n+1)+1(

-1)nnfi

¥nnfi

¥

an+1R

=

lim

a

0,

x

=0(-1)2n+1\

a

=nn'=1

\

I

=(-1,1)nn!解：u

(x)

=

1

xn\

I

=(-¥,+¥).nfi

¥

an+1\

R

=

lim

n

an!\

an

=

1

,

x0

=0¥¥(2)n!xn=1

n

n=1nu

(x)

==

¥=

lim

1

n!

1

(

n+1)!nfi

¥¥n=0¥(3)n=0nn!xnu

(x)

=n解：u

(x)

=

n!xn\

an

=

n!,

x0

=

0nfi

¥

an+1\

R

=

lim

an(n+1)!=

lim

n!

=

0nfi

¥\

I

={0}\

I

=[-1

–

3)

=[-2,1).2

22=

32n+1(n+1)

3nnfi

¥

n

3nnfi

¥

an+1R

=lim

an

=lim

2n2

21

3n

3nn解：u

(x)=n2(x

+

1

)n

\

a

=2

312

2当x

=-当x

=-+时,-时,原级数为(Leibniz准则)(调和级数)¥¥(4)n=1nn=1nn

3(2x+1)nu

(x)

=20,

x

=-

1n

3n2nn收敛¥n(-1)n发散原级数为n=1¥n=11n\

I

'

=(-1

–

3)2

2\

I

=[3

–1)

=[1

,

5

).2

2

2(n

+1)

=1=lim

1nfi

¥

an+1R

=lim

annfi

¥

n2

3当x

=+1nn

2解：u

(x)=1

(x

-3

)n2

3当x

=-1时,原级数为(Leibniz准则)(调和级数)n

2n(2x-3)n¥

¥(5)un

(x)

=n=1

n=1\

an

=

1

,

x0

=

3n

2收敛¥时,原级数为n(-1)n发散n=1¥n=11n2\

I'

=(3

–1)¥当x

=–1时,原级数为n=1\

I

=[-1,1].=1

\

I

'

=(-1,1)nlim(-1)2(n+1)-1(-1)n+1nfi

¥

an+1R

=

lim

an

=nfi

¥

2n-1¥n=12n-1(-1)n+1均收敛(-1),2n-1n(-2n-11)

x2n-1n解：u

(x)

=n(Leibniz准则)(6)¥¥2n+1n=1n=1n

xu

(x)

=(-1)n2n-10,x

=0且缺偶次项2n-1n=

(-1)\

an=

2

n+1

4-(

n+1)nnfi

¥nfi

¥

an+1R

=

lim

an

=

lim

4-n¥n=1当x

=1–2时,原级数为\

I

=

(1–

2)

=(-1,3).1n均发散(调和级数)(x

-1)2nnn解：u

(x)

=4-n2nn(x

-1)4(7)u

(x)

=n

¥n=1¥n=1-n0,x

=1且缺奇次项\

an-n=

4n\

I

'

=

(1–

2)(1)x

˛

I时,S

(x)连续.=S(x),则：'设x

˛

I

=

(-R,

R)时a

xnn¥n=0¥n-1=nan

xn=1'¥n

na

x

n=0

(2)x

˛

I

'时,S

'(x)=n

)'¥n=0=n(a

x三、的分析性质：¥n

=0nna

x注：分析性质结合

x

1-xnx

=¥n=1x

˛

(-1,1)时¥n=1n-1nx可求形如¥n=0xn+1

的和函数,1n+1及S(t)dt=(3)x

˛

I时,xnx¥n=0a

tn

)dt00(¥n=0xn+1n+1=an¥n=0

0=xnna

t

dtnu

(x)

=2nx2n-1

=(x2n

)''

¥¥

n=1

n=1\

S(x)

=(x2n)'

=x2n

；(4)得S(x),x˛

I.¥例3、求和函数S

(x)：(1)S(x)=2nx2n-1n=1[分析]：(1)求R,得I

'=(-R,R)；(2)x

˛

I

'时,求S

(x)：详解(3)判定x

=–R时原级数敛散性得I；(1-x2)2\

I

'

=

(-1,1)nfi

¥

2(n+1)解：

R=

lim

2n

=1¥n=12n

'n=1¥S(x)

=(x

)

=(x

)¥且x

=–1时原级数为–2n发散(性质5)n=1(1-x2

)2\

S(x)

=

2x

,

x˛

(-1,1))'=

2x

1-x222n

'

=(

x当x

˛

(-1,1)时,[分析]：(1)求R,得I

'=(-R,R)；(2)x

˛

I

'时,求S

(x)：(3)判定x

=–R时原级数敛散性得I；(4)得S(x),x˛

I.详解¥xn+11n=1n+1(2)S(x)

=n+1(

1

xn+1)'

=

xn¥\

S'

(x)

=xnn=1S

(t)dt

S(x)

=x0'当x

˛

(-1,1)时,1'¥n+1

'n=1n+1S

(x)

=[x

]

=xn+1)'(¥n=11n+1n¥x

==x1-xn=1¥且x

=-1时,原级数为n=1(-1)n+1n+1

收敛(L准则)=-x

-ln(1-x)

,

x˛[-1,1)dt\

S(x)

=x

t

1-t0\

I

'

=

(-1,1)n+1nfi

¥解：

R

=

lim

n+2

=1¥n=1n+1x

=1时,原级数为1

发散(同阶法+调和级数)小结：幂级数1、定义2、收敛半径R和收敛域I

的求法3、和函数的求法及应用作业12--3：277页1(5,6,8),2(1,2)求收敛域例3(3--6)、例4求幂级数的收敛半径R及收敛域I：由通项un

(x)确定an

,x0以及是否缺项；求收敛半径R：(3)判断x

=x0

–R时原级数的敛散性；(4)得收敛域I：(x0

–R),

[x0

–R),

(x0

–R],

[x0

–R]之一.；anan+1；anan+1不缺项：R

=limnfi

¥缺奇(偶)次项：R

=limnfi

¥¥例3、求和函数S

(x)：(3)S(x)=nxnn=1[分析]：I

=(-1,1)n

n

n(1)un

(x)

=

nx

=

(n

+1)x

-

x¥

¥

S(x)

=(n

+1)xn

-xnn=1n=1n

n-1(2)un

(x)

=

nx

=

xnx¥

S(x)

=

xnxn-1n=1解1解2n=1

n=1¥且x

=–1时原级数为–

n发散(性质5)n=1n=1

n=1解1：易得R

=1,当x

˛

(-1,1)时¥

¥

¥

¥S(x)

=(n

+1)xn

-xn

=(xn+1)'

-xn¥

¥n=1

n=1n+1

')

-x=(x(1-x)2\

S(x)

=

x

,

x˛

(-1,1)1-x

1-x2n

=(

x

)'

-

x(1-x)2x=¥

¥S(x)

=

2nxn

+xnn=1

n=1¥x

=–1时原级数为–(2n

+1)n=1\

I

=

(-1,1)¥(4)S(x)

=

(2n

+1)xnn=1[分析]：

I

'=(-1,1)¥

¥S(x)

=

2(n

+1)xn

-xnn=1

n=1解1解2解1：易得R

=1,当x

˛

(-1,1)时¥

¥n=1n=1(n

+1)xn

-S(x)

=

2¥且x

=–1时原级数为–(2n

+1)发散(性质5)n=1(1-x)22\

S(x)

=

3x-x

,

x˛

(-1,1)¥(4)S(x)

=

(2n

+1)xnn=11-x1-xxn

=(2x2

)'

-(1-x)22x

=

3x-x解2：易得R

=1,当x

˛

(-1,1)时

¥

¥n=1

n=1nxn

+S(x)

=

2¥且x

=–1时原级数为–(2n

+1)发散(性质5)n=1(1-x)22\

S(x)

=

3x-x

,

x˛

(-1,1)¥(4)S(x)

=

(2n

+1)xnn=1xxn

=

+1-x2x(1-x)2(1-x)22=

3x-x

¥

¥1n+11xn=1

n=1nx

-S(x)

=¥n=1解：易得R

=1,且x

=0时S

(x)=0当x

˛

(-1,0)(0,1)时

n

n+1(5)S(x)

=nx+1

ln(1-x)xn+1

=

11-x

x时原级数–且x

=–1¥n=1n+1n发散(x

=0x˛(-1,0)(0,1)+1

ln(1-x)\

S(x)

=

11-x

x0性质5)解：易得R

=1,且x

=0时S

(x)=0当x

˛

(-1,0)(0,1)时0

1

ln(1-x)1\

S

(x)

=dt

=-1-tx¥1nn(n+1)(6)S(x)

=1

211xxxn+1

=S

(x)-1

S

(x)n1nx

-S(x)

=n=1n+1¥

¥n=11nxn

]'1¥n=1S

'

(x)

=[¥n=1n

'(

x

)=¥=n

11-xn=1xn-1

=

1

¥

¥x

=1n=1

n=1

n=1xnn