The computingmethod of mean difference(thetablemethod):themean-differencetab1th3th4th2th[f(x)Xkf(x)Xof[xo,x]f(x)xf[xo,X,x2]f[xo,Xi,X2,x]f[x,x]f(x2)X2f[xo,X1,",x4]f[x1,X2,x3]f[x2,xg]f[x1,x2,X3,x4]f(xg)3f[x2,X3,x4]f[x3,x]f(x4)X4上页Define function value as zero order mean difference
上页 下页 返回 ( ) ( ) ( ) ( ) ( ) ( ) 1 2 3 4 4 4 3 3 2 2 1 1 0 0 x f x x f x x f x x f x x f x x f x t h t h t h t h k k The computing method of mean difference(the table method): [ , ] 0 1 f x x [ , ] 1 2 f x x [ , ] 2 3 f x x [ , ] 3 4 f x x [ , , ] 0 1 2 f x x x [ , , ] 1 2 3 f x x x [ , , ] 2 3 4 f x x x [ , , , ] 0 1 2 3 f x x x x [ , , , ] 1 2 3 4 f x x x x [ , , , ] 0 1 4 f x x x Define function value as zero order mean difference the mean-difference table
The mean-difference table isfollowing as1thmean2thmean3thmean4thmeanf(z)differencedifferencedifferencedifferencef(zo)Zof(z,)f(zo,z1)1f(zz)12f(zi,]f(zo,i,t2)f(rs)f(tz,ts)fzo.z1,t2,ts)23f(z1.2.2s)f(z)f(zs.f(z2ts,aIf1o.zi,12.13,f(ai,22,2s,z0：**+.：...上页下页返圆
上页 下页 返回 The mean-difference table is following as
二, Newton' s interpolating formulaSuppose that interpolating polynomial isN,(x)=ao +a(x-x)+a2(x-x)(x-x)+...+a,(x-x)..(x-xn-1)satisfy the interpolating condition N,(x,) = fi , i = O,l,...,nThe undetermined coefficient isap = f.a, = f[xo,x ]a, = f[xo,Xi,x2]上页an =f[xo,xi,..,xn]下页返圆
上页 下页 返回 Suppose that interpolating polynomial is satisfy the interpolating condition Nn (xi ) = f i , i = 0,1, ,n The undetermined coefficient is 0 0 a = f [ , ] 1 x0 x1 a = f [ , , ] 2 x0 x1 x2 a = f [ , , , ] n x0 x1 xn a = f 二、Newton’s interpolating formula ( ) ( ) ( )( ) . ( ).( ) n = 0 + 1 − 0 + 2 − 0 − 1 + + n − 0 − n−1 N x a a x x a x x x x a x x x x
If we look at x ± x,,(i = o,1,..,n) as one node,then[xo,X,, x,x] = lxo,X,",x]- f[xo,X,,Xt-,x]Xk-xf[xo,x,",xk-,x] = f[xo,X,"",x]+ f[xo,X,".,x,xl(x-x)f(x)= f(x)+(x-x)f[x,xl0②f[x,x]=f[xo,x]+(x-x)f[x,xo,x]f[x,xo....,x.--]= f[xo...,x,]+(x-x.)f[x,Xo...,x.] ....n-1D +(x-x) × ②+ ... ... +(x-xo)...(x-xn-)x n-1 f(x)=f(x)+ f[xo,xl(x-x)+ f[xo,x,x,l(x-x,)(x-x)+..+ f[xo,...,x,J(x-x,)...(x-xn-)+ f[x,xo,...,x,(x-x,)...(x-xn--)(x-x,上页下页N,(x)a,= f[xo, ..., x;]返圆R,(x)
上页 下页 返回 ( ) ( ) ( ) [ , ] 0 0 0 f x = f x + x − x f x x [ , ] [ , ] ( ) [ , , ] 0 0 1 1 0 1 f x x = f x x + x − x f x x x [ , , . , ] [ , . , ] ( ) [ , , . , ] 0 n 1 0 n n 0 n f x x x = f x x + x − x f x x x − 1 2 . . . . n−1 1 + (x − x0 ) 2 + . . + (x − x0 ).(x − xn−1 ) n−1 ( ) ( ) [ , ]( ) [ , , ]( )( ) . f x = f x0 + f x0 x1 x − x0 + f x0 x1 x2 x − x0 x − x1 + [ , . , ]( ).( ) + x0 xn x − x0 x − xn−1 f [ , , . , ]( ).( )( ) x x0 xn x x0 x xn 1 x xn + f − − − − Nn (x) Rn (x) ai = f [ x0 , ., xi ] [ , , , , ] f x0 x1 xk x [ , , , , ] f x0 x1 xk−1 x x x f x x x f x x x x k k k − − = − [ , , , ] [ , , , , ] 0 1 0 1 1 [ , , , ] [ , , , , ]( ) 0 1 k x0 x1 xk x x xk = f x x x + f − If we look at as one node,then x x ,(i 0,1, ,n) i =
f(x) = N,(x)+R,(x)SOamong whichN.(r)=f(ro)+f(ro,r(r-ro)+fo..((-)+..+fro,...,(-)..(--).R(α)=f(r)-N,(a)=f(a,ro,",anJon+i(a).The polynomial N,(x) satisfies the interpolating condition,whichis N,(x)= f(x,)(j=0,,n),and the order don't exceed n.Based on theuniqueness theroem,it's the previous L,(x) ,whose coefficient isa = f[xo,", x](k=0, l,.", n)N,(x) called Newton mean difference interpolation polynomial,whoseinterpolating页amount of computing is less than Lagrange下页polynomial,and which is easier on programming.返圆
上页 下页 返回 f (x) N (x) R (x) so = n + n among which The polynomial satisfies the interpolating condition,which is ,and the order don't exceed n.Based on the uniqueness theroem,it's the previous ,whose coefficient is N (x) n N (x ) f (x ) ( j 0, ,n) n j = j ， = [ ] ( 0 1 ) 0 a f x x k n k = ，， k = ，， L (x) n N (x) n called Newton mean difference interpolation polynomial,whose amount of computing is less than Lagrange interpolating polynomial,and which is easier on programming