三. The interpolation remainder and the error evaluationSuppose the nodes a≤x, <x, <...<x, ≤b , and f meets f eC"[a,b]J(n1) exists in [a, b].Consider truncation error R,(x)- f(x)-L,(x)(also called the remainder of interpolation polynomial)R,(x)= K(x)II(x-x)R,(x) has at least n+1i=0solutiona(t) = Rn(t) - K(x)II (t -x)Take care that here needi=0differentiate for t@(n+1)(5,)=0, 5x E(a, b)p(x)has n+2 different solutions xo ...XnxIIf(n+1)(5,)- L(n+(5,)- K(x)(n +1)! = R(n+I)(5,)- K(x) (n+1)!t(snf(n+) (5.)(x-x,)R,(x) =K(x) = l上页(n+1)!(n+1)!i=0下页返圆
上页 下页 返回 Suppose the nodes (n+1) f exists in [a , b].Consider truncation error R (x) f (x) L (x) n = − n f C [a,b] n a x x x b 0 1 n ，and f meets , 罗尔定理： 若 充分光滑， ，则 存在 使得 。 ( x) (x0 ) =(x1 ) = 0 ( , ) x0 x1 ( ) = 0 推广：若 (x0 ) =(x1 ) =(x2 ) = 0 ( , ), ( , ) 0 x0 x1 1 x1 x2 使得 ( 0 ) =( 1 ) = 0 ( , ) 0 1 使得 ( ) = 0 (x0 ) ==(xn ) = 0 存在 (a,b) 使得 ( ) 0 ( ) = n Rn (x) has at least n+1 solutions = = − n i Rn x K x x xi 0 ( ) ( ) ( ) fixed x xi (i = 0, ., n), consider = = − − n i xi t Rn t K x t 0 ( ) ( ) ( ) ( ) (x)has n+2 different solutions x0 . xn x ( ) 0, ( , ) ( 1) x x a b n = + ( ) ( ) ( 1) ! ( 1) − + + R x K x n n n Take care that here need differentiate for t. − − + = + + ( ) ( ) ( )( 1) ! ( 1) ( 1) f L x K x n n x n n ( 1)! ( ) ( ) ( 1) + = + n f K x x n = + − + = n i i x n n x x n f R x 0 ( 1) ( ) ( 1)! ( ) ( ) 三、The interpolation remainder and the error evaluation (also called the remainder of interpolation polynomial)
The remainder expression can be applied only in the presence of f(x)high order derivative.Because location can't be usually known,but,ifmax|f(n+1)(x)|= M n+1,the limit of truncation errorwe can computeasx<aprochng () is ,R(r)/ () .(n+1)!when n-l,the remainder of linear interpolation isR()==f"(E)0,(x)=f"(5)(x-xo)(x-x), e[xo,x]when n=-2,the remainder of parabolic interpolation isR(x)="(5)(x-x0)(x-)(x-x2), 5e[x0,x2]When f(x) is any order's (≤n) polynomia,R,(x)=0 ,we canknowf(n+1) (x) = (Namely,the interpolating polynamial for ≤n order isaccurate.上页下页返圆
上页 下页 返回 ' 1 ( 1) max ( ) + + = n n a x b f x M L (x) n When f(x) is any order's (≤n) polynomia, ,we can know .Namely,the interpolating polynamial for ≤n order is accurate. ( ) 0 ( 1) + f x n Rn (x) 0 The remainder expression can be applied only in the presence of f(x) high order derivative.Because location can't be usually known,but,if we can compute ,the limit of truncation error approching is . ( ) ( 1)! ( ) 1 1 x n M R x n n n + + + when n=1,the remainder of linear interpolation is ( )( )( ) [ , ] 2 1 ( ) ( ) 2 1 ( ) 1 2 0 1 0 1 R x = f x = f x − x x − x ， x x when n=2,the remainder of parabolic interpolation is ( )( )( )( ) [ , ] 6 1 ( ) 2 0 1 2 0 2 R x = f x − x x − x x − x ， x x
Example 1 considerate the interpolating polynomial through thethree points A (0, 1), B (1, 2), C(2, 3) .Solution: the value of three interpolating nodes and thecorresponding function are:Xo = 0, Jo =l; X =1, Ji = 2; xz = 2, Jz =3.based on the parabolic interpolationformula,it gets:(x-x)(x-x2)(x - xo)(x -x2)L,(x)= yo+ yi(x) -xo)(xi -x2)(x。- x))(xo - x2)(x-xo)(x-x)+y2'(x2 -xo)(x2 -x)(x-1/x-2)+2×(x-0)(x-2)=1x(0-1)(0 - 2)(1-0)(1- 2)上页(x-0)(x-1)+3x下页(2 -0)(2 -1)返圆=x+1
上页 下页 返回 Example 1 considerate the interpolating polynomial through the three points A (0, 1), B (1, 2), C (2, 3) . Solution：the value of three interpolating nodes and the corresponding function are: 0 1 1 2 2 3. x0 = ，y0 = ；x1 = ，y1 = ；x2 = ，y2 = ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( ) 2 0 2 1 0 1 2 1 0 1 2 0 2 1 0 1 0 2 1 2 2 0 x x x x x x x x y x x x x x x x x y x x x x x x x x L x y − − − − + + − − − − + − − − − = (2 0)(2 1) ( 0)( 1) 3 (1 0)(1 2) ( 0)( 2) 2 (0 1)(0 2) ( 1)( 2) 1 − − − − + + − − − − + − − − − = x x x x x x = x + 1 based on the parabolic interpolation formula,it gets:
Example 2 Consider the following interpolation problem:compute 2thpolymial P(x) ,satisfy P(xo)= yo ，P'(x)=y，P(x2)=y2amongwhich Xo+x2Yo、 y y2are the given datas,and give thecondision to make the problem's solution exist and unique.Solution:suppose that P(x)= ax? +bx+c ,so P'(x)= 2ax +b .Accordingto the known condision,get :ax +bxo+c=yoax2+bx2+c=y22axi+b=y'so,the only solvability of original problems are reduced to the only oneof the above equations.The sufficient conditions of the onlysolvability for the latter isIx?1xox1￥0X22x101$*+(: xo -x, 0)That is to say: 2x(xo -x2)-(x -x2)+0.This上页,So2下页is the condision of existing P(x) and being unique返圆
上页 下页 返回 P(x) 0 0 P(x ) = y 1 1 P(x ) = y Example 2 Consider the following interpolation problem:compute 2th polymial ,satisfy , , among which , are the given datas,and give the condision to make the problem's solution exist and unique. 2 2 P(x ) = y 0 2 x x 0 1 2 y 、y 、y Solution:suppose that ,so .According to the known condision,get : P x = ax +bx + c 2 ( ) P(x) = 2ax + b + = + + = + + = 1 1 2 2 2 2 0 0 2 0 2ax b y ax bx c y ax bx c y so,the only solvability of original problems are reduced to the only one of the above equations.The sufficient conditions of the only solvability for the latter is 0 2 1 0 1 1 1 2 2 2 0 2 0 x x x x x That is to say: ,so .This is the condision of existing and being unique. 2 ( ) ( ) 0 2 2 2 1 0 2 0 x x − x − x − x ( 0) 2 0 2 0 2 1 − + x x x x x P(x)
Example sn-1，sin-方 sing-学Know:use the lagrange interpolationsinx's 1th、 2th1 to5元compute sin 50°,and estimate the error50°18usexo,Xiandxj,X,tocomputesolutionn=1x.X2XT元1$use4.×1+×=元/6L;(x)=x-元 /4X4V2元/6元/4～2～元/4—元/6sin 50°~ L,(5元) ~ 0.77614这里f(x)=sinx, f(2)()=-sinx,e(,)6'3(2(5)(x-)(x-)Interpolation is better than46extrapolation.If Select the xinterval'sendpoint,the面sin50=0.7660444...00762interpolationeffectisbetterthe actual erextrapolation~-0.01001元$use5元七元sin 50°~0.76008, 0.00538<R<0.006604，=318上页theactualerroe of interpolation~0.00596下页返圆
上页 下页 返回 Example3： 2 3 3 , sin 2 1 4 , sin 2 1 6 sin = = = Know:use the lagrange interpolationsinx's 1th、2th to compute sin 50,and estimate the error solution： 0 x x1 x2 18 5 500 = n = 1 use x0 , x1 and x1 , x2 to compute 4 , 6 0 1 x = x = use 2 1 / 4 / 6 / 6 2 1 / 6 / 4 / 4 ( ) 1 − − + − − = x x L x 这里 ) 3 , 6 ( ) sin , ( ) sin , ( (2) f x = x f x = − x x 而 ) 4 )( 6 ( 2 ! ( ) , ( ) 2 3 sin 2 1 (2) 1 = x − x − f R x x x ) 0.00762 18 5 0.01319 ( − 1 − R sin 50 = 0.7660444. ) 18 5 sin 50 (1 0 L 0.77614 the actual error of extrapolation −0.01001 3 , 4 1 2 use x = x = sin 50 0.76008, 0.00660 18 ~ 5 0.00538 1 R the actual erroe of interpolation 0.00596 Inte r pol atio n i s bet te r t ha n extrapolation.If Select the x i n t e r v a l ' s e n d p o i n t , t h e interpolation effect is better