4.2.3正交函數(shù)族在4.1f(x),

(x),L

(a1-1)(x (x2 (x2

f(x),

(x),L

(an-1)(x ip(mi)(x)i

f(mi)(x)

ii=1,2,L,ni

mi01L,ai-1x1x2L,

fxx1x2L,xn的插值函數(shù)

f2xLfn(x是Sf1(x),f2xLfn

nkc1c2L,cn?R使得p(x)ckfk(k下面以（4-1）中所有ai =1(i=1,2,L,n)的特殊情形為例，介紹插值函數(shù)的存在唯一性。此時插值條件為p(xi)=f(xi

i=1,2,L,n

而插值問題就是在Sp(x)ckfk(xkn

ckfk(xi)=f(xi)kc1c2L,

i=1,2,L,

f1(x),f2xLfn

LLLMMOMLD[x1,x2,L,xn]

? f1(x),f2xLfn(x)在[a,b]上滿足Haarf(x)=

-1￡x<

f(x)=

-1￡x< 0￡x￡1

0￡x￡1 1 1

x 即存在x1、x22Dx,2

=

f2(x1)=

20=2 f1(x2 f2(x2 函數(shù)值y1,y2L,yn，亦即

x1x2L,xnyi=f(xi (i=1,2,L,n)又設(shè)Sf1(x),f2xLfn(x)在[a,b]上滿足Haar

p(x)=ckfk(x)?nknp(xi)=yi

i=12L,nnnp(xi)=ckfk(xi)=yi i=1,2,L,k

fn(x1)c1

y1f(x

f2(x2 f(x)c

y2 = =

f(x f(x

f(x f(x

ynD[x,x,L,x]=f1(x2

?

故其解c1,c2cn4.1 l1(x1)=1y1, l1(xi)0yi, i2,3L,n。 L00L1l2x2=1y2, l1(xi)0yi, i=13L00L1ln(xn=1yn ln(xi0yi i=13L,n-1且l(x)=

i=,

k=1,2,L,n i=1,2,L, i? l(xL l(xL Lln(x2) MOMMMOL D[x1,x2,L,xn]l1(x),l2x),

?ML,ln(x)在[a,b]上滿足Haar一組函數(shù)l1(xl2xL,ln(x),l(x)=

i=,

k=1,2,L, i=1,2,L, i?

l1(x),l2(x),L,lnn推論4.2在定理4.1的假設(shè)下，函數(shù)p(x)=yklk 滿足插值條件p(xi)=yi i=1,2,L,n的唯一函數(shù)a￡x0<L<xn￡

y0,y1, ,ynpn(xi)=yi

i=0,1,L,

pn(x)與f(x)是相吻合的yyy=f)y=pn)ox0x1Lxxnx通常把x0<x1<L<xn-1<xn稱為插值節(jié)點,pn(x)稱為f(x)的插值多項式(函數(shù)), ,

f0(x)=1,

(x)=x

L f(x)=xnnnp

=axn

+ xn-1+L+ax+n

axn+ xn-1+L+a +

=f0a +a +

=f1axn+an

……xn-1+L+a

+

=fx x1xLx1xL1M1xMxL000

=Px-x=PPx-x?x n i>x nnxnx

i=1j 的集合）pn(xpni=fi=

i=0,1,L,f(x)在區(qū)間[a,b]上n+1a￡x0<L<xn￡

,y1, ,ynpni

f

=

i=0,1,L,

0,x1,

x0?構(gòu)造一次多項式p1(x)，滿足條件1(x0)=y0 p1(x1)=

y-y0

=x-x1-

p1(x)=y

x-x1x0-x1

x- +x-

p1(x)=l00l1 l0)=

x-x1

x)=

x-x0lx

x0-x1=

l

=x1-x1= x0- x0-l

=x1-x0=

l

=

-x0=

p1(x1)=

p1(x0)=x0,,x1,y1,x2,p2(x0)=y0 p2(x1)=y1

0+l1xy1其中l(wèi)i l1(x1)=1l2(x2)=1

l1(x1)=0

l0(x2)=0l1(x2)=0l2(x1)=下面我們l0(x)以為例來確定出：l0(x),l1(x) l2由條件l0(x1)= l0(x2)=0可知,x1,x2是l0(x)的兩個根，從l0(x)=x-1x-x2其中A為待定系數(shù)。又由l0(x0=1，

1=

-

A

-x2

l0(x)xx

(x-x1)(x-x2(x0-x1)(x0-x2 (x-x)(x-x l1(x)

x

x

l2(x)

-x -x

p

(x-x1)(x-x2)

xxy

(x-x0)(x-y (x0-x1)(x0-x2

1

-

- x0x1,L,xn是[ab上的n+1l(x)

(x-x0)L(x-xj-1)(x-xj+1)L(x-xn

=wn+1(

j=0,1,L,j j

(x-x

)w(xj

其中wn+1(x)(xx0xx1Lxxn i= lj(xi) i,j=0,1,L, i?lj(xi)(j=0,1,L,

npn(x)=yili pn(xi)= (i=0,1,L,例1f(x)的如下函數(shù)值yi=f(xi

l(x)=(x-2)(x-

=1(x-2)(x-3) l(x)

(x-1)(x- (1-2)(1-

l(x)

(x-1)(x-

1(x-1)(x2) p2(x)=l00+l1 =1(x-2)(x-3)1)+-(x-1)(x-3)·+1(x-1)(x-2)2 22·x2222

·2

-3x+2)=

-3x+13

-3·

+1=- f(1.5)?p2(1.5)=2 f(x)[abn+1x0x1L,上的函數(shù)值f0,f1,L f0(x)fj(x)=(x-x0)(x-x1)L(x-xj-1)j=1,2,L,nnpn(x)=ajfjj

(x-xi),

=a0+a1(x-x0)+L+an(x-x0)(x-x1)L(x-xn-1)(4-a0a1L,an

p(xi)=fi

i=0,1,L,a11x=

p(x0)= p(x1)=p1(x0)=a0=f(x0p1(x1)

+a11-x0)=f(x1) f(x)-f(xp1(x)=f(x0)+ 0(x-x0x1-p2(x)=a0+a1(x-x0)+a2(x-x0)(x- p(x)=f(x)

f(x1)-f(x0)(x-x

+a2(x-x0)(x-00 x1-00p(x)=f(x)

f(x1)-f(x0)

-x)+a -x0)(x2-x1)=f(x20 0

f(x2)-f(x1)-f(x1)-f(x0 a2

x2-

x2-

p2(x)=a0f0+a1f0+a2f2=a0+a1x-x0+a2x-x0x-x1) p0 f(x2)-f(x1)-f(x1)-f(x00 p(x)

f(x)

f(x1)-f(x0

(x-x)

x2-

(x-

)(x-xxx

, j=1,L,定義4.2設(shè)函數(shù)f(x)在互異的節(jié)點x0,x1,L, f0f1L,fn

fk-f[xi,xk]

xk-

k?

f[xi,xj,xk]

f[xi,xk]-f[xi,xjxk-x

i?j?為f(x關(guān)于xixjxk的二階均差（差商）。f[x0,L,xk-2,xk]-f[x0,x1,L,xk-1]f[x0,x1,L,xk]

xk-xk-f(x)x0x1Lxkk階均差（差商） f(xj

w(x)=(x-

)L(x-x1°f[x0,x1,L,xk]=j

w+1

k

f[x0x1L,xk]

x0,x1,L,f[x0,x1,L,xk]=f[x1,x0,L,xk]=L k>f[x0,x1,L,xk]= k=

設(shè)f(x)在包含x0,x1,L, f(k)f[x0,x1,L,xk]=kminx0x1Lxkxmax(x0x1Lxk)

f(x)3x45x32x2+1f[20,21,L,24f[e0e1Le5 解:f[20,21,L,24] f(5)

-3·

=-f[e0e1L,e5 aj=fx0,x1,L,xj

j=1,L,p2(x)=f0+f[x0,x1](x-x0)+f[x0,x1,x2](x-x0)(x- p(x)

f(x)+f[x,．](x-x)

f[x,x,x](x-x)(x-x

+L+f[x0,x1,L,xn](x-x0)(x-x1)L(x-xn-1)=pn-1(x)+f[x0,x1,L,xn](x-x0)(x-x1)L(x-xn-1) 插值節(jié)點x0,x1,…,xn的均差值。 f f(x0 f(x1

f[x0,x1

f(x2 f(x3

f[x1,x2．f[x2,x3

f[x0,x1,x2f[x1,x2,x3

f[x0,x1,x2,x3pn(x)=f(x0)+f[x0,x1](x-x0)+f[x0,x1,x2](x-x0)(x-

f[x0,x1,L,xn](x-x0)(x-x1)L(x-xn-1) ，求解例1解 f

-1+1=1+1=3-

2-0=11x1)xx

-3x例2

f(0)=

f(13,f(2) f(311求f(x解 f

一階均 二階均

- -

-3-2=-1-11+6=3-

-3+5=12-17+3

10-1=3-p3(x)=2-5x+x(x-1)+3x(x-1)(x-2)=3x3-8x2+例3已知f(-2)=- f(0)= f(2)=19,f(3)=30f(xp(x) f -

一階均 二階均 三階均 -2+5= -

-1+3+2=010-3=1-19-10=30-19=3-

5-30+7-5=19-7=12-11-9=1

1-1=1+ 1-1=2 1-1=3-xxx=x2 pk(x)=f(x0)+f[x0,x1](x-x0+L+f[x0,x1,L,xk](x-x0)(x-x1)L(x-xk-1)rk(x)=f(x)-pkmin(x0,x1,…,xkmax(x0,x1,…,xk)之間至少有一個x p(k)(x)=k!f[x,x,L, =

f(k)

,f(n+1)rn(x)=f(x)-pn(x) wn+1wn+1(x)(xx0(xx1L(x-

證任取x∈[a,b]當(dāng)x=x0,x1,…,xn時， 以下設(shè)x≠xi(i=0,1,…,n視x為一個節(jié)點，據(jù)一階均差的定義，f(x)=f(x0)+f[x,x0](x-x0=f[x0,x1]+f[x,x0,x1](x-

=f[x0,x1,x2]+f[x,x0M

k=12L,f[x,x0,L,xk-1]=f[x0,x1,L,xk]+f[x,x0,L,xk](x-xkf(x)=f(x0)+f[x,x0](x-x0=f(x0)+{f[x0,x1]+f[x,x0,x1](x-x1)}(x-x0=f(x0)+f[x0,x1](x-x0)+f[x,x0,x1](x-x0)(x-f(x0f(x0)+f[x0,x1](x-x0)+L+f[x0,x1,L,xn](x-x0)L(x-xn-1=

+f[x,x0,x1,L,xn](x-x0)(x-x1)L(x-xnrn(x)=f(x)-pn(x)=f[x,x0,x1,L,xn](x-x0)(x-x1)L(x-xn=f[x,x0,x1,L,xn]wn+1f(n+1)

rn(x)=(n+1)!wn+1

min(x,x0,L,xn)<x<max(x,x0,L,xnf[x0,x1]=f1

f(2)

,L

f[x,x,L,x f(n)(x = pn(x)

f(x)+fx)(x-x)+f2)(x-x)(x-x f(n)(x

(x-x0)(x-x1)L(x-xn)其中min(x0L,xnximax(x0L,xn)，i=1,2L,,p(x)

f(x)+

(x)(x-x)

(x0

(x-x)2 f(n)(x (x-x0) l”1,

lxx

=

0￡k￡in

=x)xi

nf(x)-pn(x)=1-lin

fx(n1)

nnwn+1(x)0lix”1n -n

xxk

k

x

w

”0

lxx

= n

nn

l0x 0k0 練習(xí)取節(jié)點x0=100,x1=121,x2=144，y=115?

115121115 115100115100121100

554 15· 21·== =

44·115=10.72380L

115-

(2

(115-100)(115-121)(115-115-

1·2070=￡····￡····

=THEEND 第4章插值 第4插 正交函數(shù)族在(f,g)=a

r(x)f(x)g(x)f(x)、g(x)和h(x)的內(nèi)積滿足： f,f0，當(dāng)且僅當(dāng)f0f,f0; (f,g)=g,f (lf,g)=lf,g（4）fghfh+(gh) ‖af‖=︱a︱·‖f ‖f+g‖f‖g‖；f‖為C[ab]上的一個范數(shù)。fb1=afb

f(x)dx

= a￡

fb2ar(x) f2b2ar(x) fgfg22(f,g)22

= 0￡x

x-1= 11

1

1 1 f1= -2dx=0-x-2dx+1x-2dx=f

2 2

110110x-2110x-x+42 f0(xf1(xL,fn(x)

j0(x)=f0(x)ji(x)

i=1,2,…,n(4-易 f,j= j< (i,j=0,1,2,L,

(f1,f0) (f1,f0) LL Di0,f0)0,f11,01,1 L01,iL),i=0,1,LniiLi))

0=1,00,00,0(1,0f

=f

f,f)-f

f,f

=

0=

f,f)

f,f

f,f)

00

0)1

j) i,f0)i,f1

i,j

=f0(x)A0+f1(x)A1+L+fj(x)Aj+L+fi(x) i,fj=0,fj0+1,fj1+L+i,f

j0,f01,0=

0M

0,fj 1,fj

0,fj 1,fj =ii

i1,fj)

i1,fji,f0)i,f1

i,fj) i,fj對于j=i,i0,i0i1Li,fi00 01,0Li

1,1 i

1,i)= i f0(x),f1(x),L,fn(x)的線性無關(guān)性,可證Di>0因為Fi(x)

f0(x),f1(x),L,fi

j0(x),j1(x),L,jn(x)是正交函數(shù)系。 i,jj

akfk

i

0=0 k

k

k

i,ji

akfk

i+

akfk k

k i,Di-1fi)Di-1iji=Di- 0

y0(x)yi(x)

D0Di-1DiD0Di-1Di

i=1,2,L,

j(x)

Dy

=

jj)=

D

y

= ,

=0y0(x),y1(x),L,yn(x)成為標(biāo)準(zhǔn)正交函數(shù)系。三角函數(shù)系，于[-pp]上1,cosx,sinx,cos2x,sin2x,L cosnx,sinx,1,cosx,cos2x,L cosnx,1,sinx,sin2x,L sinx, 00ma令m=br(x) m=0,1,L ma

j(x)1,1),x)

,xi-1)

mi-1j(x)

x,

x,

x,xi-1)

L

i=1,2,L,

xi,1)xi,x)

xi,xi-1)

mi+1Lm2i-1

i=12L ,1),

,xi)

Di

x,x,

x,xi)= =

i

0,1,L,iiii

i,xi)

00Dy0(x)D

,y(x)

,i=1,2,L, 1Di

1 取j0(x)=m0=1dx=m=x1

=0, m

1x2dx=2 m

1x3dx=0,11 1

-1j(x)=1

1= 1= 2023 2023j(x)=m

=4x2-4=43x2-

mm

111

-122xdx=11

24

-1x=8

x

-11 1

9-=813x2dx-2=9- 11(j,j)=11

-1=8

-xx=

9-1 取j(x)=m=1

xdx=101m= 1

dx=0, m

1 x2dx=

m=

x3dx=

j(x)=

1= 1=1 1

1012 1012j2(x)=

x=1

-1=12-

例1令T0=1Tn(x)=cos(n·arccosx)，x∈[-1,1],稱Tn(x)為nTn+1(x)=2xTn(x)-Tn-

T(x)=1,T1(x)=x T2(x)=2x2-1,03T(x)=4x3-3x03

T(x)=8x

-8x2+1, 44

1Tn(x)Tm

2dx=2

n?n=m?n(x),Tm

1-1-

n=m=

T(-x)=-1n

(x =cos2k- k=1,2,L 11

Tn(x)=2n-1Tn

x)1111T2x)

T3x)例2設(shè)Ln(x)，x∈[-1,1],是以r(x)≡1為權(quán)函數(shù)的正交多Ln(x)（n=0,1,…）為n次Legendre多項式。 n n

(x2nn!

n=其具有（1）遞歸性n1Ln1(x)n+1)xLn(xnLn-1(x)L0(x)=1,

L2(x)

13x2

1,L(x)=15x3-,

L(x)=1x4-30x2+ 21

n?n(x),Lm(x)

Ln(x)Lm(x)dx=

n=2nL(-x)=nL 而n(xn

=n! n!

nnd

d

+C1(x2)n-1

n2n

n+1xn+

1yn(x恰好是n次多項式，y0(xy1(xL,yn , 定理yn(x在(a,b)內(nèi)恰有n應(yīng)保持定號。而，由于yn(x)與y0(x)≡1的正交性，可知bbyy =yxb 而x1,x2,…,xk（k＜n）為這些互異的奇數(shù)重根。fx=x-x1x-x2)Lx-xkbrxfxyx)dx 交多項式系{yn(x)}中，yn(x)與的根必相互交錯，亦即yn(x)的根為zk，k=1,2,,n，yn+1(x)的根為hk，k=1,2,…，n+1，則a<h1x1<Lxn<hn+1by0(x)=1y1=x-a0y0=x-a0

yk+1=x-akyk-bkyk-1a

y

,yk

k=0,1,2

b=yk,yk)

k=1,2 y,y

yky1(x)與y0(x)的公共根。而y0(x)≡常數(shù)≠04.5.2設(shè)S=span{f0(x),f1(x),L,fn b], f(x)?L2[a,b](n

s*(x)=a*f(x)?S, k f(x)-s*

=

f(x)-s(x)212=minbr(x)[f(x)-s(x)]2dx

s(x)?S 則稱S*(x)是f(x)在S中的最佳平方近函數(shù)，‖f(x)-S*(x)‖2稱為均方誤差顯然，求解S*(x)等價于求多元函數(shù)E(a,a,L,a)

r(x)[f(x)

af(x)]2

a k

(a*a*La*) ?E=

(i=0,1,L,

ar(x)f(x)-akfk(x)fi(x)dx=

(i=0,1,L,

k bi=r(x)f(x)f bi=

0,1,L,b akfk(x)b akfk(x)fi(x)=nnk

r(x)fk(x)fi

(i=0,1,L,ananak(fk,fi)=(f,fik

(i=0,1,L,

(f,f)a (f,f)

(fn,f0

(fn,fn)an (f,fn)

a*a*La* f0(x),f1(x),…,fn(x的線性無關(guān)性，可知法方程組的系數(shù)例如，特別取f0(x),f1(x),…,fn(x)為多項式系1,x,x2,…,xn，不妨設(shè)權(quán)函數(shù)r(x)=1。則1 1 xk+idx(fk,fi)=0

xdx

0 k+i

k,i=0,1,2,L, (f,f

n+1

0 1

=

n+2

n

n+

2n+1因此在實際計算中可利用S的正交基，解最佳平方近問題。j0(x),j1(x),L,jn(x)是S的正交基，則法方程組的系數(shù)矩陣

a0

(f,j0)

a= a=k k

(f,jn)(k=0,1,L,

ns*(x)

(f,jk)j

k kk

y1(x),…,yn(x)是的標(biāo)準(zhǔn)正交基，n s*(x)=(f,y)y k(xi,yi

(i=0,1,L,y?????s????y?????s?????????????mmin2oxdi=sxi-nn={f(x)|f(x)=akfk(x),"a0,L,an?km

naf(x)?af(x)? kmw(s*(x)-y

=

w(s(x)-y

s(x)?

i i則稱s*(x)為離散數(shù)據(jù)(xi,yi)(i=0,1,…,m) 數(shù)據(jù)擬合的最小二乘法（離散最小二乘近），簡稱最小二乘法， E0,a1,L,an

wi(s(xi)-yi

=

i

akfki-yi

a*a*L,a*。

k 得 得

?E=

(j=0,1,L,

wiakfkxi-yifjxi= (j=0,1,L, k wifkxijxi

=wi fjxi)(j=0,1,L,

Tk T

y=y,y,L,yT

f=fx,fx,L,fx m

m(fk,fj)=wi fki)fji

(y,fj)=wi fji

(j=0,1,L,

nk

,fj

=(y,fj

(j=0,1,L,

(f,f)a (f,f)

(fn,f0

(fn,fn)an (f,fn) mf0(x),f1(x),…,fn(x)的線性無關(guān)性，可知法方程組的系數(shù)m iimw*2i(s(x)-yii設(shè)(xiyii01L,m)s=a+

mmmin=

(s(x)-y)2i ii則稱按上述條件求s(x)的方法為離散數(shù)據(jù)i ii ((a+bx)-yE(a,b)

(s(xi)-yi

= i=0令a

a,=0,

0, m2a+bxi-yi]1=0,m即

2a+bxi-yi]xi=0,mmmm

a+-=

mm

a+bxi)xi- yi= b

mx

=m 2 mxi

+xi

=

mm

mmm

m=mm m2ii

2

y-

xi

i

xi

yia= m

x2-

xii i

mym

y-

i b

2

2 m

xi-

xi

(xiyii01L,m)s=0

+ax+a m

x2

mmiiii

x2

a ax2 x4

yi-

5 5

5=5

即

=

x2 ii

-0.15·30-5·9.8-10·-b=5·30-

=1

y=x-yyyy=x-2?2?4x???-y y

yx2.02為y= y= b ixii

-y =

和

=m

b

ln

cxb

ln i i

exi

-c yebxi=

i iy= y= lny=blnx+lnc 取lny=bx+lnc

z=lny,t=lnx,a=lncz=lny,a=lnc (i=0,1,L,或

其中ziln

ti=ln(xi,zi) (i01,L,m) zabt或

zi=lnzabx

y= + +1.50+1.75

ln

1.629+1.756+1.876+2.008+2.135=55

ln5=5

=55

bi

5xiln

即7.50

14.422·5-7.5·b5·11.875-a5·11.875-

又c e1.1223.071y= y=a+a+ yY1，y

Y=a+xX1x

y=a+b?RmEx,x,L,x

Ax-b

dAx-b

2=可得ATAxATb0ATAxAT

x y2 2

a0

y E(a,a)

x-y2=

a+ax-y2=

- 1

=Ax-

M Ma

i

i

1

xm ym 0

x=a0

aA= 2a

1

b=-0.1 1.1 4

1.9dE(a,a

Ax-b = =0, ATAx-ATb=0ATAx=AT 0 1

1

AA=

4

2= 3

1

ATb=

-0.1= 41.1

9.81.9

10a -

ATAxATb, 得

0= y=x-

30a1

9.82x+4y=3x-5y=x+2y=2x+y=

4

x

3A z ,b 2 1

y

67 4

2

-5x

23

= 1 2y -3x=51

167 7

46y

x3.04029y=1.24176THEEND(JosephLouisLagrange，1736～1813）是參議員，伯爵，并被授予大十字勛章。1755年拉格朗日第一篇“極大和極小的方法研究”，發(fā)展了歐拉所開創(chuàng)魯士通訊。1783年，被任命為都靈名譽院長。出任法國米制大的努力。1791年，拉格朗日被選為英國學(xué)會會員，又先后在巴黎高等師范學(xué)院 。拉格朗日科學(xué)涉及的領(lǐng)域極其廣泛。他在數(shù)學(xué)上最突出的貢獻是使數(shù)學(xué)拉格朗日在代數(shù)方程和方程的解法上，作出了有價值的貢獻，推動了代數(shù) 第4章插值 第44.2.4Hermite

f(x),f(x),L f(a1-1)(x f(x),f(x),L f(a2-1) f(x),f(x),L f(as-1)

為正整數(shù)，記a1

Lasn+1,p(mi)(x)=f(mi)(x)=y(mi)

i=1,2L,s mi0,1,L,ai-1

Li,k(x),i=1,2,L,s;k=0,1,L,ai-p(x)

y(k) (x)

i=1ks=ys

(x)+ (x)+L+y(ai-1)

L(h)(x)= L(h)(x)

m?h?k

h=0,1,L,am-1;h,k=0,1,L,a-

s

h=k

pn(x)

y (x)+ (x)+L+y(ai-1)

sp(x)=

y (x)+ (x)+L+y(ai-1) (x

=L+y

(x)+

(x)+L+y(am-1)

(x+ =ymLm,0(xm)=yms

p(xm

yL￠(x)+yL￠(x)+L+y(ai-1)

(x

=L+yL￠(x)+y￠L￠(x)+L+y(am-1)L￠(x)+ 1 以下構(gòu)造Li,k(x1(x-x

vA(x) (xx)av，v(ai-k- li,k(x)

(x-x)ai(ai-k-

(xiai

(x-xi)

(xi

-k-

(x-x)ai

(x-x)ai

1(x-x

x-xi

2 2

x-xi)+i i

ili,k

(x)=1

+a(x-

)+L+

(x-x)ai-k- ?k

i

}ai-k-i (x)i

(x-x)a1L(x- )ai-1(x-x)k(x- )ai+1L(x-x)as

(x)

(x

(x-x

=

i(x-x i

(xi

(x-x

(x-x

Li,k(x)

(x-xi ks

i (xivA(x) (xx)avvi=1,2,L,s;k=0,1,L,ai-a1+L+ai-1+k+ai+1+L+as+ai-k-=a1+L+ai+L+as-1=s例3a1=a2=L=as =1的情形。p(x)=yiLi,0sx-x i A(x)(xi

xi1L

L

x-xLi,0(x)=(x-x1)L(x-xi-1)(x-xi+1)L(x-xs)1 iA(x)(x A(xix

x1x2L,

ps-1(x)=

(x-

A(xi

例4s=1,x1=a的情形 p(x)

=k

y(k) 1記A(x)=(x-x =(x-a)a，1(x

-a(a-k-

(x-L1,k(x)= (a

x-

i

(i

(a)L1,k(x)i

f(i)

(x-a)i例5a1=a =L=as =s

p(x)=

+記A(xs2(x)k01

s(x)=(x-x1)(x-x2)L(x-xss(x)

2 Li,0

=x-x

1 s2

ii2s

x

+2 i

x-x =(x-x

(xi

s(x)s x-x

x-x +2

x-x

s(x) i

i i

s(x) s

2 j

x-

jx- j

x-xis(x i

s

x-x

x=

2

j

x-x

x-xi ix-xs(x

s(x) 2

(xx=

x-xis(xi) (xi s(x)

-x

Li,1(

-x

x

xi

ii2s

= x-

x-xi)s(x)

=x-

s(

x-xi i

2 (x p2s-1(x)=

f(xi)1- (xi(xi

(xi s 2 s(x (x)= 1- i(x-x)is(x)(x-x (xi Li,1(x)=s(x)(x-x)

(xxi) 例6s=2且a1=a2=2p(x)=y1 此時A(x(xx)2xx 22

(x)

=x-x2

21-x21

) ￠=x-x

x-x22 x-x22

1x=x1

x-x2)

2x-2=x-x22

3x- 2 2x-x3= 2 1x1-x2 1-x2

x-x

=x

x x-x

= x-x

-x2

x-x 11

x-x2 2x-xL2,0(x)= 2x2-x1 2x- L2,1(x)= 1x-x2x2-x1 x- x-

x- 2 H(x)=f(x)1-2

2

(x1 2 x1-x2x1-x2

-x2 x-x x-x2 x- +f(x2)1-

1+

(x)(x-x) x2-x1x2-x1 x2-x1 H3(xk(xkk=1, a￡x1<x2<L<xs￡b

f(2s)f(x)-p2s(x)

(x)]2, x?[a,

(2s)!min(x1x2Lxsxmax(x1x2Lxs)若xx1x2…xs中的某一個，則（4-24）顯然成立。x≠xi(i=1,2,…,s)，由插值條件，可設(shè)f(x)-p2s-

(x)=(x-x)2L(x-x)2K

K(x), f(t)=f(t)- (t)-K(x)(t-x)2L(t-x)22s- f(t)=f(t)- (t)-K(x)(t-x)2L(t-x2s- f(xi

i=1,2,L,

f(x)0min(x,x1,Lxs)<x<max(x,x1,L,xs f(2s)(x)=f(2s)(x)-0-K(x)(2s)!=0

K(x)

f(2s)我們還可以用基函數(shù)法來構(gòu)造三次多項式H3(x)。(x1)(x2其中L1,0(x),L2,0(x),L1,1(x),L2,1(x),為插值基函數(shù)。 0(x10(x1) 1(x2L2,0(x2)=1L2,0(x1)

,0(x1)=,0(x2)=1(x21(x1)

0(x2 (x)=x-x2ax+ L￠(x)=2x-xax+b+ax-x L1,0x11,0(x10

-x2

+=

ax+b= x-x2x+b+ax-x=0a=- 123x-12

x-x

(x) x-x

L1,1(x)= 2x

x-

x-x

x1-x2 2 (x)=

2- 1-

x-x2

x- L2,1(x)= 1x-x2

x2-x1

x2-x1

x2-x1

x?p,p 2

fp=sinp 24 424 4

fp=cosp 24 4 24 4 fp=sin

fp=cosp=0,

p

p

p

p p

x-x-

px-2

-x- x

41-2

p

p+f4x-4

p+f21-2

ppp

- -

-

- - 2 2

2

4 4 p

p

p+8x-4x-

4x-

p-8x-24x-42 4 2+2x-p 2+ 22

2 4

定理4.3’f(x)∈C3[a,b]，在（a,b）內(nèi)4階可導(dǎo)，f(4) f(x)-p3(x)

)x x?[a,min(x1x2xmax(x1x2利用插值法構(gòu)造近似函數(shù)時，為了提高近精度，經(jīng)常 f

1+

510 k01,L,n n=6、n=8n=10作出插值多項式pn(x)近f(xyx)x)

x)

f= 1+x2- - - - max

hmaxf-n=maxInf,x-Inf,x￡ li

fi-

h=max

a￡x￡b n

n其中

a￡x￡b

i(

In(f,x=lixfi

f=x)fix0,x1xn滿足a≤x0<x1<…<xn≤b在每一個區(qū)x- x-hL(k)(x)=h

xk-xk

+yk+1

-

令Lh(x)

L(0)hh(x),hhM

x?[x0,x?[x1,x2M

h

(x),

Lh(xi)=yi(i=0,1,…,n),Lh(x)f(x)在[a,b]上的分段y=Lh(x)的圖形是平面上連接點(x0,y0)、yL(k)hyL(k)h1+x2- - - - R(x)=f(x)-L(k

=f

(x-x)(x-

|(x-x)(x- )|￡M2h2

M2=max|f(x)|,h=maxhka￡

hk=xk+1-h易證，當(dāng)f(x)∈C[a,b]時，limLh(x)=f h

L(k)

h

hL(n-1)h

（1）當(dāng)x∈[xkxk+1]，且x偏向xk時，選擇xk-1xkxk+1 當(dāng)x∈[xkxk+1]，且x偏向xk+1時，選擇xkxk+1xk+2當(dāng)x∈(xn-1xn]，或x＞xn時，選擇xn-2xn-1xnTHEEH法國數(shù)學(xué)家。年月日生于法國洛林，法國。巴黎綜合工科學(xué)校和巴黎理學(xué)院教授。還有許多國家的科學(xué)的 科目還是──數(shù)學(xué)。 Hermite中 。Hermit上 "

工科學(xué)系，Hermite只 確定的──

應(yīng)用數(shù)學(xué)》《五。第44．3樣條函數(shù)是一個重要的近工具，在插值、數(shù)值微分、曲D:-￥<x1<x2<L<xn<+￥樣條節(jié)點，以x1,x2,…,xn為節(jié)點的m次樣條函數(shù)的全體記為： (x-a)m x?+(x-a)m+

x<

+易見(x-a)m 是Cm-1(-∞,+∞)（表示(-∞,+∞)上m-1次連續(xù)++a +an s(x)=p(x)+c(x- j

-￥<x< n s(x)=p(x)+c(x-x j

-￥<x+￥上的表達式只相差一項，c(xx)m即 S(x)

+Q+

x-x

x?x j-1

j+1n S(x)=P Cx-x1,x,L,xm

(xx)mxx)mL,(xx)m

x=x= x xj記h(x)=q(x)-p(x)∈Cm-1[x h(k)(x)=p(k)(x)-q(k)(x) k=0,1,2,L,m-

p(k)(x)=q(k)(x)

k=0,1,2,L,m-h(k)(x)=mjm

k=0,1,2,L,m-mxm

q(xp+Cjxxj)S(x)=

1-28+25x+9x2+26+19x+3x2-26+19x+

-1￡x<0￡定理驗證。因為

(28+25x+9x2+x3)-(1-2x)=(x+3)3(26+19x+3x2-x3)-(28+25x+9x2+x3)=-2(x+1)3(26+19x+3x2)-(26+19x+3x2-x3)=x3S(x)

+

0￡x<1

+bx2+cx-

1￡x￡ 解：1）由ax3+bx2+cx-1-3+x2=x-3 a-3+b-2+cx-1=x3-3x2+3x-a-1=1a=2b-1=-3b=-2,c33bx2cx-13x2x

xa+b+c-1=13+12= a+b+c=3bx2cx-￠3x2 x 2+2bx+x

=2+x3a+2b+c=3+ 3a+2b+c=3+bx2+cx-x

3x2 即xx+x

=x+

6a+2b=6+ 6a+2b=

a+b+c= 3a+2b+c=a2,b2,c3

6a+2b= i01L,n。s(x?S3(x1x2Lxn-使s(xi)=yi i=0,1,L,

分段Hermite三次多項式插值問題，只有值函數(shù)在所有

S(x)=ax3+bx2+cx+

x?

,

]i=0,1,L,n-插值條件n+1S(xi

i=0,1,L,

-0)=S(xi+0)i=0,1,L,n-S(xiS(xi i=0,1,L,n-S(xiS(xi i=0,1,L,n-S(x(x0(x0(x0

S(xn(xn(xn(xn

(x0(xn

S(x0-0)=S(xn+0) S(x0S(xn(x0f(x0)=f(xn

+0)0000 s(x)=

+bx2+

x+

S(x)=

s(x)=ax3+bx2+cx+

x?,s(x)=3ax2+2bx+c

s(x)=

x+ s(x)=3ax2+2bx+c

s)=

(0))(0)s1(0)10s10-a0+b0-c0+d0=1,-6a0+2b0=0

d0=

6a1+2b1=0,

c0-c1=0, d1=a=-a=1,b=b=3

c0c1b0b10 S(x)=

1x3+3 -1x3+3

x?,階線性方程組，而且力學(xué)意義明顯。設(shè)s(xk)=mk(k=0,1,L, hk=xk+1- (k=0,1,L,n- x-x

x-

x- 2ks(x)=s(x)1-k

+mk(x-xk)

k xk-xk+1xk-xk+1 xk-xk+1 x-x

x- 2

x- 2 )1-

k

(x- k

xk+1-xkxk+1-xk

k k

- k -即s(x)=

+2(x-xk)(x- )2y+hk-2(x-xk+1)(x-x)2 k

k (x-x)(x-

mk

(x- )(x-x

因此，求s(x)的關(guān)鍵在于確定n+1個常數(shù)m0m1mn。s(x)=6