高數(shù)下冊(cè)十二章4節(jié)

復(fù)習(xí)：(1)Taylor公式：若f

(x)有直到(n

+1)階導(dǎo)數(shù),則：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：展開點(diǎn)、展開階數(shù)、余項(xiàng)的形式.(2)Maclaurin公式：若f

(x)有直到(n

+1)階導(dǎo)數(shù),則：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、泰勒級(jí)數(shù)：若f

(x)在U

(x0

)內(nèi)能展開成x

-x0的冪級(jí)數(shù),則：所得必為Taylor展開式：12.4

函數(shù)展開成冪級(jí)數(shù)(1)式右邊稱為f

(x)在x0處的Taylor

級(jí)數(shù).nf

(n)

(0)n!￥n=0f

(x)

=

x

(2)注：當(dāng)x0

=

0時(shí),

有：若f

(x)在U

(0)內(nèi)能展開成x的冪級(jí)數(shù),則：所得必為Maclaurin展開式：(2)式右邊稱為f

(x)的Maclaurin

級(jí)數(shù).二、函數(shù)f

(x)展開成x的冪級(jí)數(shù)：1、前提：函數(shù)存在各階導(dǎo)數(shù).2、展式形式：若存在則唯一即M.展式.3、方法：直接法、間接法.Th、若f

(x)在U

(x0

)內(nèi)有各階導(dǎo)數(shù),則：

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、函數(shù)f

(x)展開成x的冪級(jí)數(shù)的方法：1、直接法：(1)求f(x)的各階導(dǎo)數(shù)在x=0處的值f

(n

)(0),n

=0,1,2,f

(n)

(0)n=0n!xn及其I

',I；￥(2)寫出f

(

n

+1)

(x)'n

+1(3)驗(yàn)證x

?

I

時(shí)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、由直接法及相關(guān)性質(zhì)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!(二項(xiàng)展開式)=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、間接法：采取變量替換、初等運(yùn)算、分析運(yùn)算等利用公式求冪級(jí)數(shù).1

+

2

x21例2、將f

(x)=展開成x的冪級(jí)數(shù).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、將f(x)=展成x

-2的冪級(jí)數(shù).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、將f

(x)展開成x的冪級(jí)數(shù)：(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)函數(shù)y

=ln(1+x)+ln(1-2x)；(2)記A

=ln(1+x),B

=ln(1-2x)；(3)得冪級(jí)數(shù)A,B及收斂域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、前提條件：有各階導(dǎo)數(shù)2、展式的形式：T.展式或M.展式3、展開方法：直接法、間接法4、常用公式(1)--(5)及其應(yīng)用作業(yè)12--4：285頁2(2,4)小結(jié)：函數(shù)展開成冪級(jí)數(shù)一、主要知識(shí)：1、級(jí)數(shù)的概念、性質(zhì)和常用結(jié)論.2、級(jí)數(shù)審斂法：正項(xiàng)級(jí)數(shù)、交錯(cuò)級(jí)數(shù)、任意項(xiàng)級(jí)數(shù).3、冪級(jí)數(shù)收斂半徑、收斂域的求法.4、冪級(jí)數(shù)求和及應(yīng)用.5、展開函數(shù)成冪級(jí)數(shù).第十二章 總結(jié)1、級(jí)數(shù)的概念、性質(zhì)和常用結(jié)論的記用.2、判斷任意項(xiàng)級(jí)數(shù)的斂散性：絕對(duì)收斂、條件收斂、發(fā)散；3、求冪級(jí)數(shù)的收斂半徑R、收斂域I；4、求冪級(jí)數(shù)的和函數(shù)并解決相關(guān)問題；5、展開函數(shù)成冪級(jí)數(shù).322--323頁總習(xí)題十二：3,5(1,2),7(3,4)二、常見題型：：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記A(x)=n

-1n

+1x

-xn+1￥n=2￥n=2￥n=2

1

.(n2

-1)2n1、求A

=三、思考與練習(xí)：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,則當(dāng)x

?

(-1,1)時(shí)02、設(shè)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、將f

(x)=1

ln

1

+x

+1

arctan

x

-x4 1

-

x

2展成x的冪級(jí)數(shù).解：

f

(0)=0

且當(dāng)|

x

|<1時(shí)an

-an+1

?0an

?an+1\{an

}單減有下界nnnfi

￥=

a

￡

a

liman+1證(1)an=

1

a

+

1

?12

a

n

n2

a

nfi

￥(1)極限lim

an存在；anna￥n=1u

=￥n

1

1

4、設(shè)a1

=

2,

an+1

=

a

+

,求證：

n=1n+1

-1收斂.(2)級(jí)數(shù)(akn\

Sn

￡a1k=11ak+11a1-

1

)

=

a

(

1

-

1

)11ak+1-

)

= -1a1

1

a

a1

a￡

a

(\{Sn

}單增有上界-