数学分析6.3泰勒公式(练习)

11n2n1)!!,113x35xx35211123xx335+3xx6xxxx21x33)=11n2n1)!!,113x35xx35211123xx335+3xx6xxxx21x33)=第章分值理其用泰公练题1、式√1x

;到含x

5

;到x5

项解2

3

,f”(x)=

34

5

,,f(n)2n√(1x)2n1

∴f

(n)

n

2n1)!!2n

√1x

()x+x24·2

2

+…+(-1)

2n1)2nn!

n+(xn).(2)∵f’(x)=(1+x2)-1

,f2-2

,f”’(x)=-2(1+x2)-2

2

2

)-3

,f(4)

2)-3

3

(1+x2),f(x)=24(1+x)-32(1+x2)-4(1+x2)∴fff”(4)

f

++o(x535

(3)∵f’(x)=secx,ff”’(x)=4secxtanx+2sec

x,f

(4)

(x)=8secxtanx+16sec4f

(5)

(x)=16secxtan

x+88secxtanx+16sec6∴fff”(4)

f

(5)

∴

x3

2x15

+o(x2、限(1)limx

e

x

sinxx(1x)x

;lim[xxx∞

ln(1

;(3)(ctanx).xxxx解e++(x326

x6

(x2+

312

+o(x3),36x

e

x

sinxx(1x)x

=limx

35631236x

𝑜(x

3

)

=lim(x312

x36

𝑜(xx

)13

112111xx1(3)lim(−ctanx)=xxx−𝑜xx𝑜x=lim=lim(+)1−1))x1xx−3255πx3xsin(θx+)45112111xx1(3)lim(−ctanx)=xxx−𝑜xx𝑜x=lim=lim(+)1−1))x1xx−3255πx3xsin(θx+)455π|x|11,|x|xxx−233x21(2)∵ln(1+xx2x

+o()，xlim[x−xx

(1)][x−(x−)xx∞

𝑜()212

]=211sinx−xcosxxxxxxsinx+𝑜x)x1𝑜(x)13sinx3x3xsinxx

=limx

323362xsinx

]3、式(1)f(x)=x3+4x2x=1;,在x=0处.1+x解’(x)=3x2fff(n)(n≥∴fff”f(n)∴f(x)=10+11(x-1)+7(x-1)2+(x-1)3.

(n≥

−11+x2

,f”(x)=

21+x3

,…,f(n)1+x

,f(n+1)(x)=

(n+1)!1+xn+2∴ff,f(n)

nn!,f

(n+1)(0)=(-1)n+1

∴2+…nx+θxn+2

xn+1,(0<4、(1)sinx≈x,x|≤(2)1+x=1+628解(1)sinx=x−+2,(0<6120

,当∴公：(x)|=|

xsin(θx+)52=1201203840(2)1++28161+θx5

,(0<∴公：(x)|=|16θx5

|,16

xxex2nθxn+1θ11enθ<11xxxx234xxex2nθxn+1θ11enθ<11xxxx234523+4555、数到10-9

;-5.e++2n!(n+1)!

,e=1+1++…+,(0<2n!(n+1)!当|R(x)|=

e(n+1)!(n+1)!

-9

n≥，e≈+…2

=2.718281828.12!(2)∵+−+24θx+1)5

,