第五章 插值方法

第五章 插值方法1. 填空(1) 若已知f (X )二y , i二0,1,2,n，且x ,x，,x互异，则作n次插值多项式P (x)，使其满足i i01 nnP (x ) = y (i二0,1,2,n ),这就是所谓的拉格朗日插值.n ii(2) 过下列三点(2,4)，(3,9)，(5,25)，用抛物插值计算x二4处的函数值y二16(3 )设f (x) = x4，试用插值余项定理写出以-1, 0,1, 2为节点的三次插值多项式为P (x) = 2x 3 + x 2 2x f()忆)n!3f(472)Q 即4 雹汇47)：需爲)x 04846555+ 注-。购 1(.472-凹 x 0.4937452(0.47 - 0.46) x (0.47 - 0.48)+ 回2 一 46)X(川2 47)x 0.5027498 =0.4955529. (0.48 - 0.46) x (0.47 - 0.47)4-给定节点x0 = 7二二h 3二3, 3二4，试分别对下列函数导出拉格朗日插值余项:(1) f (x) = 4x3 - 3x + 2解 R3( x)=伴4( x) = 0(2) f (x)二 x4 - 2x3解 R (x) = f &)(x)=里-(x +1)(x -1)(x - 3)(x - 4) = x4 - 7x3 + 11x2 +17x -1234!44!5.证明：对于次数不超过n的多项式f (x)，其n次拉格朗日多项式P (x)二f (x). n证明：设 / (x)二 C0 + 7 + 叮，由拉格朗日余项定理知R (x) = f (x) - P (x)=nnf (n+1)点)(x) = 0，P (x)二 f (x) (n + 1)!n+1n丿6. 已知函数表x0143f (x)0-785求三次牛顿插值多项式，并计算f (2.5)的值，若增加一个节点(6,14)，求f (0,1, 4, 3, 6)及四次牛顿插 值多项式.解 列差商表xif (x )i1阶差商2阶差商 3阶差商4阶差商00-7P (x)二 0 + (7) - x + 3x(x -1) + (4 / 3)x(x -1)(x - 4)二(4 /3)x3 + (29 /3)x2 + (46/ 3)x 3f (2.5)沁 P (2.5)二 1.25, f (0,1,4,3,6)二 23/903P (x) = P (x) + (23 / 90)x(x -1)(x - 4)(x - 3) 43=(23/90)x4 - (152/45)x3 + (1307/90)x2 - (92/5)x7. (1)选择插值节点0.5 ,0.7； (2)略8. (1) f (x)二1在节点x ,x，,x的n次Lagrange插值多项式为01 nP (x)=工 l (x) f (x )=工 l (x)nkkkk =0k=0由 Lagrange 插值余项定理有= f(x)-P (x)=气n+1诞)3(x) = 0,(n +1)!n+1P (x) = f (x), Hl (x) = 1 nkk=0(2) f (x) = xj (j = 1,2,n)在节点x ,x，,x的n次Lagrange插值多项式为01 nP ( x ) = H l ( x ) f ( x ) = H l ( x ) x jnkkk kk =0k =0由 Lagrange 插值余项定理知R = f ( x ) - P ( x ) =nnf (+1诞)3(n +1)!n+1(x) = 0, P (x) = f(x),Hl (x)x j = x jnk kk=0(3)k=0Hn (x -x)jl(x)=HnHj Clxl(-x)j-ll (x) =Hj (Hn xll (x)Cl (-x) j-lkj kkk kjk =0 l=0l=0 k=0=HClxl(-x)j-l =(x-x)j =0jl=09.因为 R = f (x) P (x) = -(x x )(x x )1 1 2! 1 2所以 If (x) - P( x) 1=f 代)max |f(x)|-(X - x )(x - x ) x0-xX1I (x - x )(x - x ) I2!i22oimax(x)()2x + x、/x + x x , (x 一 x )2-xoxxI ( o i 一 x )(oi x ) I= i o max (x)。220218x-x-x0110. -(x) = xn+1关于节点x ,x，,x的n次Lagrange插值多项式为o inP (x)=工 l (x) (x )=工 xn+1l (x)nkkk kk=0k =0由 Lagrange 插值余项定理知R = (x)-P (x) = +弋) (x) = & (：)弋 肝(x-x ) =H(x-x )nn(n +1)! n+1(n +1)!kkk=ok =o所以工 xn+1l (x) = xn+1 - (x - x )(x - x )(x - x )k ko1nk=o工 xn+1l (0) = 0 n+1 - (0 - x )(0 - x )(0 - x ) = (-1) nx x xk ko1no 1 n(4)设f (x) = xn-2，则f (x)的n次插值多项式为P (x) = xn-2.n差商具有对称性，差商与导数的关系是f (x0, x1，,xn )=(6)设 f (x) = x7 + 5x* * 3 +1，则差商 f (20,21) = 162， f (20,21 *,22) = 2702， f (20,21, ,27)=丄,f (20,21, ,28) = 0.2拉格朗日插值多项式P(x)逼近f (x) = x3，要求:取节点x0 =-1，x1 = 1作线性插值;1 (1)解 P1( x) = 1+F -(x+1) =x取节点x0 = j, x1 = 0, x2 = 1做抛物插值;吕 x (-1“ 十 X 0 + 譽 X1 = x(3)取节点x = 1, x = 0, x = 1, x = 2作三次插值.0123f (4) (E)解由拉格朗日插值余项定理知R (x) = f (x) P (x)=(x)334!4由于 f(4)(x)=(x3)(4) =0，所以 f(x)P(x)=0， P(x)=x3. 333.给出概率积分的数据表，用抛物插值计算当x = 0.472时该积分值等于多少？解 由于 10.46 0.472 11 0.49 0.472 I，所以选择插值节点x = 0.46,x = 0.47,x = 0.48,0 1 2