P(x) should meet the interpolating conditionP(x)= f ,i=0,1,..,nhasP(x)= f =aoa= f.fi-fa =P(x)=fi =ao +a,(x -x)X-xoP(x2)= f, =ao +a(x2 -xo)+az(x -xo)(xz -x)f-f。fi-fxz-xo Xi-xoa,=Xz -XiIf it goes on,the form of the undetermined coefficient will bemore complex.上页so,introduce the concept of difference quotient(mean difference)下页返圆
上页 下页 返回 0 0 0 has P(x ) = f = a ( ) ( ) 1 1 a0 a1 x1 x0 P x = f = + − 0 0 a = f 1 0 1 0 1 x x f f a − − = ( ) ( ) ( )( ) 2 2 a0 a1 x2 x0 a2 x2 x0 x2 x1 P x = f = + − + − − 2 1 1 0 1 0 2 0 2 0 2 x x x x f f x x f f a − − − − − − = If it goes on,the form of the undetermined coefficient will be more complex. so,introduce the concept of difference quotient(mean difference) P(xi ) = f i , i = 0,1, ,n P(x) should meet the interpolating condition
-,Themean difference andits propertyDefinitionlSuppose that the function's value of f(x) on thedifferent nodes is f ,i=o,l,...,ncallfi-J(i±j)f[x,x,]=x,-x;for 1th mean difference of f(x) on the nodes x;,xjf[x,x,]-f[x,,x,](ij+k)f[x,x,x,]=Xk-Xjis 2th mean difference of f(x)on the nodes x,,xj,xk上页The rest may be deduced by analogy下页返圆
上页 下页 返回 Definition1. call [ , ] (i j) x x f f f x x i j i j i j − − = ( ) [ , ] [ , ] [ , , ] i j k x x f x x f x x f x x x k j i k i j i j k − − = The rest may be deduced by analogy. 一、The mean difference and its property Suppose that the function's value of on the different nodes is f (x) f i ,i = 0,1, ,n for 1th mean difference of on the nodes f (x) i j x , x is 2th mean difference of on the nodes f (x) xi x j xk ,
f[xi,,X,,..",Xi-]-f[xi,,X,,".,Xi,,Xikf[Xi,X,",Xi-,,X;,]=Xir-- -Xixis f(x)'s k order difference quotient on the nodes x,X,,xixiObviouslyf[xo,Xi,..",Xk-1]-f[xo,Xi,..",Xk-2,x,]f[xo,Xi,"",Xk-1,x,l=Xk-1 -Xk上页下页返圆
上页 下页 返回 [ , , , , ] 0 1 k 1 k xi xi xi xi f − [ , , , , ] x0 x1 xk 1 xk f − Obviously k k k k k i i i i i i i i i x x f x x x f x x x x − − = − − − 1 0 1 1 0 1 2 [ , ,, ] [ , ,, , ] k k k k k x x f x x x f x x x x − − = − − − 1 0 1 1 0 1 2 [ , ,, ] [ , ,, , ] is k order difference quotient on the nodes f x s ' ( ) k k i i i i x , x , , x , x 0 1 −1
Themeandifferencehas thefollowing property(1)kth mean difference f[xo,Xi,...,xk-,x] of f(x) can beexpressed by linearly conbination of f(xo), f(x),.:.,f(x)f[xo,Xi,...,Xk-1,Xh]-2a - f(x,)(2)The difference quotient has the symmetry.That's tosay,whatever the order of nodes is,the value ofdifference quotient won't be changed.For examplef[Xo,Xi,x2] = f[xo,X2,xi]= f[x2,Xi,x]上页下页返圆
上页 下页 返回 The mean difference has the following property: [ , , , , ] x0 x1 xk 1 xk f − = − − − − + − = k i i i i i i i k i x x x x x x x x f x 0 0 1 1 ( ) ( )( ) ( ) ( ) (2) The difference quotient has the symmetry.That's to say,whatever the order of nodes is,the value of difference quotient won't be changed. [ , , ] x0 x1 x2 f [ , , ] x0 x2 x1 = f [ , , ] x2 x1 x0 = f For example (1)kth mean difference of can be expressed by linearly conbination of f[x0 , x1 , , xk−1 , xk ] f (x) ( ), ( ), , ( ) 0 1 k f x f x f x
Based on property 1 and property 2, we can get:[x,x,x-+x] = x,,+-fx,x,x]X-Xk(3)when f()(x) exists on the interval which includes thenodesxo,Xi..",x.There must be a point which isamong xo,xi,-,xto makeProved by theequal of theremainderI[xo,x, x] - [(5)k!上页下页返圆
上页 下页 返回 [ , , , ] x0 x1 xk f Proved by the e qual of the remainder ! ( ) ( ) k f k = [ , , , , ] x0 x1 xk 1 xk f − Based on property 1 and property 2，we can get： k k k x x f x x x f x x x − − = − 0 0 1 1 1 2 [ , ,, ] [ , ,, ] (3)when exists on the interval which includes the nodes ( ) ) f x （k x x xk , , , 0 1 There must be a point which is among to make x x xk , , , 0 1