数值分析-第五章插值与拟合

作用：由物理量离散的分布近似得到其连续的变化规律本章主要内、拉格朗日插值方、插值方、埃尔米特插值方、曲线拟问题背景插值的基本概定义：已知定义于区间[ab]上的实值函数fx)在n1个 i异节点{x [a, iyif(xi i0,1,,ax0x1xnP(x)P(xi)yi i0,1,, 作为函数y=f(x)的近似，称这样的问题为插值问题系式（5-1）的P(x)为f(x)的插值函数，f(x)为值函数[a,b]为插值区间xi}为插值节点，（5-1）式为插值条件代数代数插值：插值函数P(x)

有理插值：插值函数P(x)为有理分式函数三角插值：插值函数P(x)为三角函数P(x)fyP(x)fyf(yP(P(x)aaxax2 axn,a 0i P(xi)yif(xi i0,1, ,P(x0)

a1x0a2x2 anxny0P(x1)a0

axax2

axn P(x)

a ax2 axn

(xixj)定理5-1：满足插值条件(5-1)的不超过n次的插值多注：nf(xn+1 Rn(x)f(x) Rn(xi)f(xi)P(xi) i定理5.2f(x)

x0,x1,,[a,b]上n

次可微，则对任意x[a,b],x有关的af(n1)()Rn(x)f(x)pn(x)(n1)！n1(x) 其中

(x)(xx)(x

(xx

x) 当x=xi(i=0,1,…,n)时，Rn(xi)=f(xi)-Pn(xi)=0,xxi(i0,1,n)时g(t)f(t)Pn(t)

n1

n1反复对g’(t),g’’(t),…,g(n)(t)用罗尔定理，得到g’’(t)至少有个n零点，g’’(t)至少有个n-1零点,…,g(n+1)(t)至少有一个零点，即至少存在一点(a,b), 使得g(n1)()0.显然与所给的x有关

g(n1)(t)f(n1)(t)f(x)Pn(x)(nn1(x)f(x)

Pn(x)

f(n1)((n n1

(a,b)且与x有关拉格朗日(Lagrange)插一、线性插值(n给定两个点(x0,y0)和(x1y1),且x0 ≠x1,构造一次多项式L1(x)，使其满足条件:L1(x0)=y0, yy0(xx0 ( y0 x0

L(x)y

x

xx010101

x0

x1xL1(x)l0(x)y0l1(x)0 l(x)x 0

l(x)x x0

x1并称l0x),l1x)是x0x1的插值基函数l(x)

j

(i,j j注意li(xj 1l1l(01(n使L2x)满足L2(x0)y0 L2(x1) L2(x2)类似的，所要寻求的多项式L2x)L2(x)y0l0(x)y1l1(x)y2l2(这里lix)称为二次插值基函数，只与x0x1x2l(x)1,j (i,j0,1, jl(x) (xx1)(xx2)0 (xx)(xx0 l(x)(xx0)(xx2) l(x) (xx0)(x (xx)(xx (xx)(xx 下面我们以l0x)为例来确定出：l0(xl1(xl2(x)由条件l0(x1) l0(x2)

x1,l0(x)

l0(x)Axx1xx2l0(x01，可1AxxxxA1

x0x1x0x2从而

(xx1)(xx2(x0x1)(x0x2(xx0)(xl(x)

l(x)

(xx0)(x1 1

x2

(x2x0)(x2设x0,x1,…,xn是[a,b]上的n+1l(x)

(xx0)(x (xxi1)(xxi1) (xxn

(5-i (i

)(

(xixi1)((xixi1)(xixi1)(xixn

li(x)是一个nji(xixj

i0li(xj)ij0

i与节点有关，而与f

称为n次Lagrange插令nLn(x)l0(x)y0l1(x)y1ln(x)ynli(x)i0

(5-LLn(xi)yiin1(x)(xx0)(x

(xxn)(xxi)(xxi)(xxj d

jin1(x)(xxj)(xxi

(xxjdx ll(x)n1(x)i(xx)i(xi

jn1(xi)(xixjl(xl(x)(xx0)(x(xxi1)(xxi1)(xxni(xx)(xxi0i1(xi)(xi)(xxin注意 i这是因为 li(xj) 当in（1）n，则插值多项式不唯一n例如P(x)Ln(x)q(x)(xxi 也是一个插i多项式，其中q(x)可以是任意多项式拉格朗日插值多项式结构对称，形式简单Rn(x)f(x)Ln(x)(n1)!n1(n测试xii+1,i0,1,2,3,4,5.l2(x)的图像 1

1

10

l(x)(x1)(x2)(x4)(x5)(x 例 已知函数f(x)的如下函数值 23yif(xi 1f(x的二次Lagrange插值多项式L2(x)，并利用L2计算出f 解l(x)(x2)(x

1(x2)(x0l1(x)l2(x)

(x1)(x(x1)(x(31)(3

(x1)(x1(x1)(x2L2(x)l0xy0 l21(x2)(x3)1(x1)(x3)11(x1)(x2)21x25x6x24x31

3x2x23x1

3

f(1.5)

L(1.5)

3 例2

sin1,sin ,sin2 2sinx的1次、2次Lagrangesin50并估计误差。sin50n= 分别利用x0,x1以及x1,xn=

500

518

x,x

L(x)x/41x/62 /6/ /4/2sin500L5

118这 f(x)sinx,

(2)()sin,

3 f 0.01319R(5)1/*extrapolation*的实际误差x,x利 sin500.00538~518/*interpolation*sin50=n=L(n=

(x

)(x

)

(x

)(x

)

(x

)(x

) ()(

()(

2()( 2sin50

L(5)

3R(x)cos(x)(x)(x) 1cos3 3 0.00044R5218二

sin50=嘿5.3插时，全部基函数li(x)都需重新算过。Ln(x改写成x0)2(xx0)(xx1)a?n(xx0)...(xxn1)的形式，希望每加一个节f(x)n+1个互异节点x}n处的函数值fx f[x,f[x,x]f(x)f(xijijx(i xxijijf[xf[x,x,x] f[x,x]f[x,x x(ijikkff[x,x,...,x]f[x0,x1,...,xk1]f[x1,x2,...,xk01kx0kf(xk-1f(xk此外补充定义，fx0为零阶差商xxi的顺序ff[x,x,...,x]n f(x ni0(xix0 (xixi1)(xixi1(xixn即即f[x0,...,xn]nf(xiin (x其中n1xxxiinn1(xi)(xixjjjn证明：n=

f[x,x]f(x0)f(x1) f(x0) f(x x0

x0

x1n= f[x,x,x]f[x0,x1]f[x1,x2

x0f[x,x,x] f[x0,x1]f[x1,x2

x0x2

f(x1)

f(x2) f(x0)

f(x1)( x)x

x x

1

0 f(x0 (x0x1)(x

f( (x1x0)(x1

f(x2(x2x0)(x2差商具有对称性，即差商具有对称性，即f[x0x1xi的顺序无关xn]f[f[x0,x1,...,xnf(n)(n!(a,n阶差商n阶导数性质证三、插值多项

设x0x1xnn+1个互异节点，x[a,b且xxiNn.1f(x)f(x0)(xx0)f[x,12f[x,x0]f[x0,x1](xx1)f[x,x0,23f[x,x0,x1]f[x0,x1,x2](xx2)f[x,x0,x1,x23………f[x,x0,...,xn1]f[x0,...,xn](xxn)f[x,x0,...,xnff(x)f(x0)f[x0,x1](xx0)f[x0,x1,x2](xx0)(xx1)f[x0,...,xn](xx0)...(xxn1)f[x,x0,...,xn](xx0)...(xxn1)(xxnRn

f(x)Nn(x)Rn(x)Rn(x)f[x,x0,x1,,xn](xx0)(xx1)(xxnf[x,x0,x1,,xn]n1

Rn(xi)f[xi,x1,,xn]n1(xi)

Nn(xi)f(xi i由插值多项式的唯一性可知NnxLnx),aif[x0,x1,x2,...,xi i0,1,2,...,f(n1)(f[x,x0,...,xn]n1(x) n1(x)(n f[x0,...,xn]

f(n)(n

(x)N(x)f[x,x,..., k k(xx)(xx)(xx Nk(xf[x0x1xk1xx0xx1xxk) f f f

f[x0,f[x1,…

f[x0,x1,…

f … f

f[xn1,

f[xn2,xn1,

f[x0,…,xn+1f

f[xn, f[xn1,xn,

f[x1,…, f[x0,…, 已知f(0)=2,f(1)=-3,f(2)=-6,f(3)=11,求f(x)的3次插值解 f

一阶均 二阶均321 35

三阶均

633211163

21733

1013

f[0,1]5，f[0,1,2]1，f[0,1,2,3]3，N3(x)2

x(x1)3x(x1)(x 3x38x2例5：yfx)12345yi=f14786写出4次插值多项式解：xif(xi12345

N4(x)13(x1)0(x1)(x147863310236114786331023611(x1)(x2)(x3)(x4)N(x)1x49x383x233x 例 f(2)

f(1)f(2)

f(0)f(3)30,求f(x)关于上述节点组的插值多项式N(x)解 f

一阶均 二阶均 三阶均

251 53

32010311910

075972

111 112 11

230193

1193

3 f[2,1]3， f[2,1,0]1，f[2,1,0,1]0，N(x)

3x2(x1)xx26xxiaih(i0,..., hban

fif(xi

fifi1

称为在点xi处的i f(i

fi

向后差分

fi

称为在点xi处的mi f(mi

fi

性质 kf[x0,x1 ,xk

] 0 (k1,2, ,n)k!hk xkx0kh, k0,1,2,...,k

f[x,x]f1

设kj时结论成立当k=j+1

f[

,

,

]f[x0,

,xj]f[x0xj

,xj1]f[

,

,

]f[x0,

,xj]f[x0xj

,xj1] j j k1(j1)hj!hj j!hj k j1k(j1)!hj1

hm(m0(),(x,x),0nf(n)(xkx0kh,fkf(xk),k0,1,2,...,f[x,x

mx]

f[x0,

f(m)(, ], ]m1m

mNn(x)f(x0)f[x0,x1](xx0)...f[x0,...,xn](xx0)...(x向前插值公当插值点x位于插值区间左端点x0 xx0 0tff[x0,x1,x]kk k!hkN(x)

xx0 (xx0)(xx1)2 (xx0)(xn!hn(xxn

2!h2

0n0xxk(x0th)(x0kh)(tk)h

k N(xth) 0 (ti) k!hkk1 i nkf0

0t(t

(tkk kCkt(Ckt(tt(tkk!

Ckkt0kt0(tf(n1)(t

Rn[f]

(n

hn1t(ttf(n1)( t向后插值公当插值点xxn令xxn 1tNn(x)f(xn)f[xn,xn1](xxn)...f[xn,...,x0](xxn)...(xff[xk,xk1,x1,x0]kkkxxni(t

k N(xth)f n (t kk1 i k

nt(t

k!nf

Rn[f]

f(n1)((n

hn1t(t(t(tt

f(n1)(注：注：xx0xn例4ysinx)yi=f(xi分别利用前插和后插公式计算sin(0.42351)的近似值精确值精确值 精确值0.08521 txx20.423510.6 5.4埃尔米特插不仅要求函数值重合，而且要求若干阶导数(xi)f(xi(xi)f(xi(xi)f(xi

问题：已知函数fx在互异节点

i

f(x

f(x) i

,2n+1的多项式H2n1(x) 满足如下的2n+2个条H2n1(xi)f(xiH

(x)f(

i0,1,2, ,) 2n1 设满足前述2n+2个条件的插值多项式为 H2n1(x)f(xi)i(x)f(xi)i(x)ix)，i(x)

xj)(x)

(x)

ii

,i,j0,1,,ix)和ix)i(x和i(x)均为2n+1次多项式，且有n个二重x0

,xi1,xi1

,令x)(axb)l2x) (xixi1(xixi1)(xixi1(xixni

(xx0)(x (xxi1)(xxi1) (xxn(xix0)(xi(x)al2(x)(axb)l2( l2i(x)2l(x)ll2i(x)2l(x)l(ii

i(xi)ai2(aixibi)li(xi)

nnkk

n1k0xi

1ai

12xik(x(xi)(xi)(xx(x)(x)n(xx

xiil(x)i

(x

)(x (

)(

iiiiii nn

xi

(x

xi1)(x

(x

x(x)[1(x)[12(xx)iikkn x]l(x), i0,1,2i,ik(x)i,j,ij

c

(x)(x)(xx)l2(i H2n1(x)f(xi)i(x)f(xi)i( n ii(x)[12(xiiki

xi

]l2((x)(xx)l2(

i0,1,2,,

(2n

(x)f(x)

(2n

如n=1时，埃尔米特插值多项式H3(x) xxxx xxxxH(x)f(x)12 1f(x)12 0 x x x x 1 1 0 0xx xx f(x)(xx) 1f(x)(xx) H9(H9(x)yyf(x)yH9(x)

例1yfx)

i(i0,1,

012f(xi123f(xi-01 H5(x)f(xi)i(x)f(xi)i(i i2(x)[12(xx) ]l2(x)i ikki

xixk (x)(xx)l2

i0,1,(x)[12(x 0 0 1(13x)(x1)2(x2)2(x)[12(x

1x2(x

1 1 (x)[12(x

2 2 1(73x)x2(x4(x)(x0)l2(x)1x(x1)2(x (x)(x1)l2(x)(x1)x2(x (x)(x2)l2(x)1(x2)x2(x H5(x)f(xi)i(x)f(xi)i(i i1x25x25x31x41442二、导数值不完全的Hermite f(x)y,i 和f'(x)

H3

)

,i

和Hx f(x)y,i f'(x)y' f''(x) H(x)y

(x)y

H'(x)y'

H''(x)

f i 第1步：由3个函数值f0,f1,f2,构造二次插值多项式P2(x)N2(x)f(x0)f[x0,x1](xx0)f[x0,x1,x2](xx0)(x02xx(x1)x2 H3(x)N2(x)(xx0)(xx1)(xx2其中λ为待定常量.显然有H3(xi)=f(xi),i=0,1,2.H'(x)f'(x) H'(x)2x1(3x

来确定6x

H'(x