高数下册十二章4节

复习：(1)Taylor公式：若f

(x)有直到(n

+1)阶导数,则：0f

(x)

=nk

=0kf

(k

)

(x

)

0

k!(x

-

x

)n+

R

(x)f

(n+1)

(x)0,n+1(x

-

x

)(n

+1)!n其中R

(x)=0x˛(x

,

x)注意：展开点、展开阶数、余项的形式.(2)Maclaurin公式：若f

(x)有直到(n

+1)阶导数,则：nkf

(k

)

(0)k

!x

+

Rn

(x)f

(x)

=

k

=0f

(n+1)

(qx)n

1(n

+1)!n其中R

(x)

=x+

,

(0<q<1)

0

0nf

(n)

(x

)n!¥n=0f

(x)

=(x

-x

)

(1)一、泰勒级数：若f

(x)在U

(x0

)内能展开成x

-x0的幂级数,则：所得必为Taylor展开式：12.4

函数展开成幂级数(1)式右边称为f

(x)在x0处的Taylor

级数.nf

(n)

(0)n!¥n=0f

(x)

=

x

(2)注：当x0

=

0时,

有：若f

(x)在U

(0)内能展开成x的幂级数,则：所得必为Maclaurin展开式：(2)式右边称为f

(x)的Maclaurin

级数.二、函数f

(x)展开成x的幂级数：1、前提：函数存在各阶导数.2、展式形式：若存在则唯一即M.展式.3、方法：直接法、间接法.Th、若f

(x)在U

(x0

)内有各阶导数,则：

0

0f

(x)

=nf

(n)

(x

)n!¥n=0(x

-x

)0n

1f

(n+1)

(x)(x-x

)

+

=0nnfi

¥limR

(x)

=limnfi

¥

(n+1)!三、函数f

(x)展开成x的幂级数的方法：1、直接法：(1)求f(x)的各阶导数在x=0处的值f

(n

)(0),n

=0,1,2,f

(n)

(0)n=0n!xn及其I

',I；¥(2)写出f

(

n

+1)

(x)'n

+1(3)验证x

˛

I

时limn

fi

¥x

=

0(n

+1)!nf

(n)

(0)n!¥x

,

x˛

I(4)得f

(x)=n=0(1)ex

=n!1

xn

,

(x

˛

R)¥n=0

1

2n-1(2)sin

x

=(

1)-

n+1¥n=1x

,(x˛

R)(2n-1)!例1、由直接法及相关性质得常用公式：(教材P281---285)x2nn=01(

1)n,(x˛

R)(2n)!¥cos

x

=

-n=0n!¥

n

ax

=

(ln

a)

xn

,(a

>

0,

a

„1,

x

˛

R)¥(3)

1

=xn,(|

x|<1)1-xn=01(3.1)1+x¥n=0=(-1)n

xn,(|

x|<1)(-1)n

x2n,(|

x

|<1)1+x2

=¥n=0

1

(3.2)n+1

1

n

+1(4)ln(1+

x)

=x

,(-1<

x

£1)n(-1)¥n=02n+1¥n=0

1

2n

+1(5)

arctan

x

=x

,(-1£

x

£1)n(-1)n=0An¥*(6)

(1+

x)m

=

m

xn

,

(|

x

|<1)n!(二项展开式)=1-

1

x

+

13

x2

-

13

5

x3

+

13

5

7

x4

-,(-1<

x

£1)2 2

4 2

4

6 2

4

6

8n

(2n-1)!!

n(2n)!!=1+(-1)x

,

(-1

<

x

£1)¥n=11(6.1) 1+

x

=

(1+

x)2=1+

1

x

-

1

x2

+

13

x3

-

135

x4

+,(|

x

|£1)2 2

4 2

4

6 2

4

6

8-12=

(1+

x)(6.2)

1

1+x2、间接法：采取变量替换、初等运算、分析运算等利用公式求幂级数.1

+

2

x21例2、将f

(x)=展开成x的幂级数.n=01n¥1-x解：

=x

,(|

x

|<1)¥n=0=

(-2x2

)n

,(|

-2x2

|<1)即：f

(x)=¥n=0n

2n(-2)

x

,2

22

2x˛(-

,

)2

1\

f

(x)

=1-(-2x

)1例3、将f(x)=展成x

-2的幂级数.x

-

531=

-

1x

-

5

3

1-(

x-2

)1\

f

(x)

=,

x˛(-1,5)3¥1

x

-

2

n3=

-

n=0

1n(x

-2)¥n+1n=0

-3\

f

(x)

=3,

x-2

<1n=01n¥1-x解：

=x

,(|

x

|<1)1\

f

(x)

=n(xln

2)

,n!¥n=0例4、将f

(x)展开成x的幂级数：(1)

f

(x)

=

2x解：

f

(x)=2x

=(eln2

)x

=exln2ex

=

1

xn

,(x

R)¥˛n=0

n!(ln

2)nn!nx

,

(x

˛

R)¥即：f

(x)=n=0(xln2˛

R)(2)

f

(x)

=1+x

-2x2x1

,3

1-

x1-(-2x)

解：

f

(x)

=

1

1

-11-xnx

,(|

x|¥n=0=<1),33nn1

¥

1

¥n=0n=0\

f

(x)

=x

-(-2x)

,

(|

x|<1且|

-2x|<1)3nx

,¥1-(-2)n即：f

(x)=n=01 1

x˛-

,2

233nn

n1

¥

1

¥n=0=x

-(-2)

x

,

n=033nn1

¥

1

¥n=0n=0\

f

(x)

=x

-(-2x)

,

(3)y

=

f

(x)

=ln(1-

x

-2x2

)分析：(1)函数y

=ln(1+x)+ln(1-2x)；(2)记A

=ln(1+x),B

=ln(1-2x)；(3)得幂级数A,B及收敛域I1

,I

2；(4)所求y

=A

+B且I

=I1

I2

.解：

y

=ln(1+x)+ln(1-2x)；且ln(1+x)=,n+1n

xn+1¥n=0(-1)n=0n

(-2x)n+1ln(1-2x)

=(-1),n+1¥n

xn+1¥n=0\

y

=(-1)¥n=0n

(-2x)n+1+(-1)n+1,n+12-1

£x

<

12(-1<

x

£1)(-1<-2x

£1)xn+1,¥(-1)n

-2n+1n+1即：y

=n=0

1

1x˛

-

,

2

21、前提条件：有各阶导数2、展式的形式：T.展式或M.展式3、展开方法：直接法、间接法4、常用公式(1)--(5)及其应用作业12--4：285页2(2,4)小结：函数展开成幂级数一、主要知识：1、级数的概念、性质和常用结论.2、级数审敛法：正项级数、交错级数、任意项级数.3、幂级数收敛半径、收敛域的求法.4、幂级数求和及应用.5、展开函数成幂级数.第十二章 总结1、级数的概念、性质和常用结论的记用.2、判断任意项级数的敛散性：绝对收敛、条件收敛、发散；3、求幂级数的收敛半径R、收敛域I；4、求幂级数的和函数并解决相关问题；5、展开函数成幂级数.322--323页总习题十二：3,5(1,2),7(3,4)二、常见题型：：1

24=

1

S

(x)-S

(x)

,

x

˛

(-1,1)1114¥¥n+1

1

n-1

1

n+1-n-1

2

2

n=2n=2解A

=n-11411记A(x)=n

-1n

+1x

-xn+1¥n=2¥n=2¥n=2

1

.(n2

-1)2n1、求A

=三、思考与练习：1n=21

,=1-x¥xn-2S'

(x)

=x2'2S

(x)

=n¥x

=1-xn=210dt

=-ln(1-x)1-tx\

S

(x)

=t222dt

=-1

x

-x-ln(1-x)01-txS2(x)

=1214S

(x)\

A(x)

=

S

(x)-

\所求A

=A

1

1

1

38

2

41

2

=25

38

4+

+

ln1-

=-

ln221

324=

x

+x+

ln(1-x)

1

n+112¥¥n=0I

=n+1

n

1xn=0¥n+1n=0n+1令S(x)=n=01-xS'

(x)

=xn

=

1

¥S(x)

=-ln(1-x)n=0¥

1

=ln(2

+

2)2

\所求In

=S

解：易得,则当x

˛

(-1,1)时02、设In

=p

4¥n=0nIn.sin

x

cosxdx,求1+x2f

'(x)

=

1

(

1

+

1 )

+14

1+x

1-x

2¥

¥=-1+x4n

=x4n

1

-1=

1

-1n=0

n=1(|x|<1)1-x44n0xx

dx¥x

¥4n

dx

=

(

n=1

n=1\

f

(x)

=0

x1x4n+1¥=4n+1n=13、将f

(x)=1

ln

1

+x

+1

arctan

x

-x4 1

-

x

2展成x的幂级数.解：

f

(0)=0

且当|

x

|<1时an

-an+1

‡0an

‡an+1\{an

}单减有下界nnnfi

¥=

a

£

a

liman+1证(1)an=

1

a

+

1

‡12

a

n

n2

a

nfi

¥(1)极限lim

an存在；anna¥n=1u

=¥n

1

1

4、设a1

=

2,

an+1

=

a

+

,求证：

n=1n+1

-1收敛.(2)级数(akn\

Sn

£a1k=11ak+11a1-

1

)

=

a

(

1

-

1

)11ak+1-

)

= -1a1

1

a

a1

a£

a

(\{Sn

}单增有上界-