《数值计算方法》课后题答案详解吉 林 大 学第一章 习 题 答 案1. 已知(1)2,(1)1,(2)1f f f −===，求()f x 的Lagrange 插值多项式。

解：由题意知：()01201212001020211012012202121,1,2;2,1,1()()(1)(2)()()6()()(1)(2)()()2()()(1)(1)()()3(1)(2)(1)(2)()2162nj j j x x x y y y x x x x x x l x x x x x x x x x x l x x x x x x x x x x l x x x x x x x x L x y l x ==−=====−−−−==−−−−+−==−−−−−+−==−−−−+−==×+×−∴∑()2(1)(1)131386x x x x +−+×=−+2. 取节点01210,1,,2x x x ===对x y e −=建立Lagrange 型二次插值函数，并估计差。

解11201201210,1,;1,,2x x x y y e y e −−======1)由题意知：则根据二次Lagrange插值公式得：02011201201021012202110.510.520.51()()()()()()()()()()()()()2(1)(0.5)2(0.5)4(1)(224)(43)1x x x x x x x x x x x x L x y y y x x x x x x x x x x x x x x x x e x x e e e x e e x −−−−−−−−−−−−=++−−−−−−=−−+−−−=+−+−−+22)Lagrange 根据余项定理，其误差为(3)2210122()1|()||()||(1)(0.5)|3!61max |(1)(0.5)|,(0,1)6()(1)(0.5),()330.5030.2113()61()0.2113(0.21131)(0.21130.5)0.008026x f R x x e x x x x x x t x x x x t x x x x t x R x ξξωξ−+≤≤==−−≤−−∈′=−−=−+=−==≤××−×−=∴取 并令 可知当时，有极大值3. 已知函数y =在4, 6.25,9x x x ===处的函数值，试通过一个二次插值函数求的近似值，并估计其误差。

解：0120124, 6.25,9;2, 2.5,3y x x x y y y =======由题意 （1） 采用Lagrange 插值多项式220()()j j j y L x l x y ==≈=∑27020112012010*********()|()()()()()()()()()()()()(7 6.25)(79)(74)(79)(74)(7 6.25)2 2.532.255 2.25 2.75 2.7552.6484848x y L x x x x x x x x x x x x x y y y x x x x x x x x x x x x ==≈−−−−−−=++−−−−−−−−−−−−=×+×+××−××= 其误差为(3)25(3)25(3)2[4,9]2()(7)(74)(7 6.25)(79)3!3()83max |()|40.0117281|(7)|(4.5)(0.01172)0.008796f R f x x f x R ξ−−=−−−==<∴<=又则（2）采用Newton插值多项式2()y N x =≈224(7)2(74)()(74)(7 6.25) 2.64848489495N =+×−+−×−×−≈ 4. 设()()0,1,...,k f x x k n ==，试列出()f x 关于互异节点()0,1,...,i x i n =的Lagrange 插值多项式。

注意到：若1n +个节点()0,1,...,i x i n =互异，则对任意次数n ≤的多项式()f x ，它关于节点()0,1,...,i x i n =满足条件(),0,1,...,i i P x y i n ==的插值多项式()P x 就是它本身。

可见，当k n ≤时幂函数()(0,1,...,)kf x x k n ==关于1n +个节点()0,1,...,i x i n =的插值多项式就是它本身，故依Lagrange 公式有()00(),0,1,...,nn n k kk i j j j j j i j ii jx x x l x x x k n x x ===≠−=≡=−∑∑∏特别地，当0k =时，有()0001nn n ij j j i j ii jx x l x x x ===≠−=≡−∑∑∏而当1k =时有()000n nni j j j j j i j i i j x x x l x x x x x ===≠⎛⎞−⎜⎟=≡⎜⎟−⎜⎟⎝⎠∑∑∏ 5. 依据下列函数表分别建立次数不超过3的Lagrange 插值多项式和Newton 插值多项式，并验证插值多项式的唯一性。

解：（1） Lagrange 插值多项式330()()j j j L x l x y ==∑ 30,()j iii j i jx x l x x x =≠−=−∏3120010203124()010204x x x x x x x x x l x x x x x x x −−−−−−=••=••−−−−−−=3271488x x x −+−−0321101213024()101214x x x x x x x x x l x x x x x x x −−−−−−=••=••−−−−−−=32683x x x−+0312202123014()202124x x x x x x x x x l x x x x x x x −−−−−−=••=••−−−−−−=32544x x x−+−0123303132012()404142x x x x x x x x x l x x x x x x x −−−−−−=••=••−−−−−−=323224x x x−+()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()32222321240241901020410121401401223320212440414212313243685432848114511442x x x x x x L x x x x x x x x x x x x x x x x x x x x x x −−−−−−=×+×+−−−−−−−−−−−−×+×−−−−−−=−−+−+−+−−++−+=−+−+（2） Newton 插值多项式kk x ()k f x一阶差商二阶差商三阶差商0 0 1 1 1 9 8 2 2 23 14 3 3 4 3 -10 -8-11/43001001201()()(,)()(,,)()()N x f x f x x x x f x x x x x x x =+−+−−0123012(,,,)()()()f x x x x x x x x x x +−−−1118(0)3(0)(1)(0)(1)(2)4x x x x x x =+−+−−−−−−32114511442x x x =−+−+由求解结果可知：33()()L x N x =说明插值问题的解存在且唯一。

6. 已知由数据1(0,0),(0.5,),(1,3)(2,2)y 和构造出的Lagrange 插值多项式()3L x 的最高次项系数是6，试确定1y 。

解：31200102030.512()00.50102x x x x x x x x x l x x x x x x x −−−−−−=××=××−−−−−−=3277122x x x −+−+0321101213012()0.500.510.52x x x x x x x x x l x x x x x x x −−−−−−=××=××−−−−−− =328(32)3x x x −+x 0 1 2 4 ()f x1 9 23 30312********.52()1010.512x x x x x x x x x l x x x x x x x −−−−−−=××=××−−−−−−=32252x x x −+−012330313200.51()2020.521x x x x x x x x x l x x x x x x x −−−−−−=××=××−−−−−−=32111326x x x −+3()L x 中最高次项系数为：1810(1)(2)32633y ×−++−×+×=⇒1174y =7. 设()4f x x =，试利用Lagrange 余项定理给出()f x 以1,0,1,2−为节点的插值多项式()3L x 。

解：由Lagrange 余项定理(1)1()()()()()(1)!n n n n f R x f x L x x n ξ++=−=+ [,]a b ξ∈可知：当3n =时，(1)(4)()()4!n x f f x ξξ+===301234!()()()()()()(31)!L x f x x x x x x x x x =−−−−−+4(1)(0)(1)(2)x x x x x =−+−−− 3222x x x =+−8. 设[]2(),f x C a b ∈且()()0f a f b ==，求证21max ()()max ()8a xb a x b f x b a f x ≤≤≤≤′′≤−证明：以,a b 为节点进行线性插值，得()()()x b x aL x f a f b a b b a−−=+−−1由于()()0f a f b ==,故1()0L x =。

于是由''1()()()()(),2!f f x L x x a x b ξ−=−− a b ξ<<有''()()()()2f f x x a x b ξ=−−，令()()() [,]t x x a x b x a b =−−∈()2()0()2t x x a b a bx t x ′=−+=+=∴时有极大值21max ()=max ()max ()()21max ()()()2221=()max ()8a xb a x ba xb a x b a x b f x f x x a x b a b a b f x a b b a f x ≤≤≤≤≤≤≤≤≤≤′′•−−++′′=•−−′′−∴9. 求作()1n f x x +=关于节点()0,1,,i x i n =L 的Lagrange 插值多项式，并利用插值余项定理证明()()1001nnnn i ii i i xl x +===−∑∏式中()i l x 为关于节点()0,1,,i x i n =L 的Lagrange 插值基函数。