案例高数课件

y=y=exy=1+oy=y=ln(1+of(x)»f(x0)+f(x0)(x-x0 1、精确度不高；2、误差不能估计。 问题:寻找函数Px)fx)P误差Rx)fxPx)fx)在含有x0的开区间(ab)内具有直到(n1)Px)为多项式函数P(x)=a+a(x-x)+a(x-x)2+L+a(x-x 误差Rnx)fxPnPn(x)=a0+a(x-x)+a(x-x)2+L+a(x-x fx)在(ab)内具有n1阶导数，误差Rnx)=fxPn二、Pn和Rn好

yoyo(x0((x0fx0)L L

f(xPn(x)=a0+a1(x-x0)+a2(x-x0)2+L+an(x-xfx)在(ab)内具有n1Rnx)=fxPn 1:Pn(x)在x0处的函数值和它的直 n阶导数值与f(x P(k)(x)＝f(k)(x)k Pnx)与fx)在x0附近有较好吻合，即当有:fx)－Px)＝oxx)n)

xfix0‹柯西中值定理

Pnx)－fx

‹用f(n+1x)给出因此，将Pnx)对x求导，直至n阶，并令xx0a0=f(x0),1a=

(x),2!a=

(x)L

n!a=

(n)(x 10得ak =0

f(k)(x (k=0,1,2,L,

代入Pnx)中得f¢(x

f(n)(xP(x)=f(x)+f(x)(x-x)+ 0(x-x

+L+ 0(x-x

fx)在x0处所对应的n阶泰勒多项式。泰勒(Taylor)中值定理如果函数fx在含有x0的某个开区间(ab)内具有直到(n1阶的导数,则当x在(ab)内时,fx可以表示为xx0的一个nf(x)=f(x)+f(x)(x-x)+f(x0)(x-x)20+L

f(n)(x (x-x0)n

f(n+1) n其中Rnx)

(n+

(x-x0

(x在x0与xf(n+1) n

(n+

(x-x0

(x在x0与xRn(x)=f(x)-

(

P(x)=f(x)+f'(x)(x-x)+L+f(n)(x)(x-x 证明:Rnx)在(ab)内具有直到(n1)R(x)=R(x)=R(x)= =R(n)(x)= 令G(x)(x

x0

则Gx0Gx0Gx0L

(n)(x)=0两函数Rn(x)及G（x）在以x0及x为端点的区间上满 Rn(

=Rn(x)-Rn(x0)

R1

(x在x与x之间 G(x)-G(x G'(x 再对Rn'(x)及G’(x)在以x0 Rn'(x1)

Rn'(x1)-Rn'(x0

R'(x2G'(x G'(x)-G'(x

G''(x

R(

(n)(x

R(n)(x)-

(n)(x

R(

在x与x =L = = = ( G(n)(x)-G(n)(x

G(n+1)

即在x与xn又QGx)xx0

\G(n+1)(x)=(n+

R(n+1)(x)=f(n+1)( R(n+1)

f(n+1)

则由上式得Rn(x)= G(x)G(n+1)

(x-x)n+1

0P(x)

f(k)(x (x-x f(k)(x f(x)= (x-x) +R( nf(x)=n

f(k)(x0

(x-x)k

Rn(

fx)按xx0)nk 0+f(n+1) n0+Rn(x)=n+!(x-x0

Rn(x)

f(n+1)

(x-

(x-

)n+1n+

n+

Rn( =

Rx)oxx)n0xfix0(x-x0

n\f(x)=

f(k)(x

(x-x

+o[(x-x)n0k 0

f(x) f(k)f(x) =

(x-x)k

fn+1)(xxx)n+1 落在x与x之间。 0k0n

f(k)(x

(n+f(n+1)(

+q(x-xf(x)

(x-x)

+ (x-x)n+1

0<q<0k 0

(n+ :1.当n=1时 变成拉氏中(x)( 2.取x0=0,x在0与x之间,令x= (0<qf(n+1)则余项Rnx)

(n+

xn+1 f f(n)

f(n+1)(0)

x2+L

xn

(0<q< (n+f f(n) f(x)=f(0)+

(0) +L

+o(x

f(x)=f(0)+

(0)

f(n)+L

f(n+1)n+n

(0<q< (n+例1求f(x)=ex的n阶麦克劳 解 f(x)=f(x)=L=f(n)(x)=ex\f(0)=f)=f(0)=L=f(n)(0)=

x xn

,e =1+x

2!+L+n!x xn

(n+

xn+1

ex»1+x(x

2!+L+ Rn(x)=(n+1)!<(n+1)!

(0<q取x1e111L

(n+

< (n+ f(k)(x

f(n+1)(

+q(x-x0f(x)= 0(x-x0

+ (x-x)n+1

k

(n+ 例2求f(x)=sinx的n 解：fxsinxsinxp

f''(x)=sin(x

f(n)(x)=sin(x+np故：f(0)0,f'(0)

1,f''(0)

=0,f'''(0)=

1,

依次循环取代 （n＝2m），有

x x5 x

sinx=x

1 + (2m- (2m+0<q<1,m=

x2m取m＝1，则sinxx，误差为

= 2x

£1x6类似地取m3则可得sinx的3次和5次近似式。 f(k)(x f(x)=k

0(x-x) +R(x) 求函数fx)x33x22x4在x01 解f(-1)f(x)=3x2+6x-f(x)=6x+f(x)=

f(1)=-(f(1)=f(x)=8-f(x)=8-5(x+1)+(x+1)3+R(3其中R3x0.(因f(4x sinx=x -L+(-1)n +o(x2n+2) x

(2n+xcosx=1 +L+(-1)n

ln(1+x)=x-+-L+(-1)n +o(xn+1) 1

=1+x+x2+L+xn+o(xn(1+x)m=1+mx+m(m-1)x2+L+m(m-1)L(m-n+1) ：m=-1,m=1/2,m=-1/2

+o(xn

x

xex=1+x +L

cosx=1 +L+(-1)n 2

xfi

2ex+2cosx-3ln(1+x4)

.

解 ln(1+x4)~

x x4 e =1+x22

x4+o(x4

cosx=1

++o(x5) \e +2cosx-3= + )x4+o(x47原式lim

+o(x4 =xfi x4 解：

x2 x3 ex »1+xex

2!

+...

R

<n+1 ，则e11e3 e3误差Rn(1)< 要使Rn

n+1 n+1<10－6，只需取n＝9即可 9+1

<绝对值10－6的e的近似值为：e思考题利 x2 x3

xfi

exsinx-x(1+x)x3x3解Qex=1+x ++o(x3

sinx=x

exsinx-x(1+\ xfi x3

3 1+x +o(x)x +o(x)-x(1+limxfi

=

-

+o(x31=1xfi x3 y=y