1、已知 f （x ）=ln x 的数值表如下，分别用线性及二次 Lagrange 插值法计算f (0.54) 的近似值，并估计误差。
解：
（1）线性插值法：
因为()ln f x x = 时递增函数，所以取0.5和0.6为插值节点。
则线性插值多项式为：
0011()()()()f x F x f l x f l x ==+
0.540.60.540.5
(0.54)(0.54)(0.5)
(0.6)0.50.60.60.50.6930.6(0.510)0.40.6198f F f f --≈=+
-- =-⨯+-⨯=-
截断误差：101011
()''()()(),(,)2
R x f x x x x x x ξξ=
--∈ 12
1
(0.54)0.0012
R ξ=
11(0.5,0.6)
110.0012(0.54)0.00120.360.25
,0.0032(0.54)0.0048
R R ξ∈∴⨯
<<⨯<<即 （2）二次lagrange 插值法：
A ：若取0.4，0.5和0.6为插值点，0120.916,0.693,0.510f f f =-=-=-
012(0.540.5)(0.540.6)
0.12
(0.40.5)(0.40.6)(0.540.4)(0.540.6)0.84(0.50.4)(0.50.6)(0.540.4)(0.540.5)0.28
(0.60.4)(0.60.5)l l l --=
=-----==----==--
001122(0.54)(0.54)(0.54)(0.54)(0.54)0.615f F f l f l f l ≈=++=-
截断误差：20121
()'''()()()()3!
R x f x x x x x x ξ=
---
则
23
1
(0.54)0.000112
R ξ
=
333
233
[0.4,0.6],()111,0.60.411
0.000112(0.54)0.0001120.40.6
f x R ξξ∈∴
<<-⨯<<-⨯递增；
即
20.00175(0.54)0.000519R -<<-
B ：若取0.5，0.6和0.7为插值点，0120.693,0.510,0.357f f f =-=-=-
012(0.540.6)(0.540.7)
0.48
(0.50.6)(0.50.7)(0.540.5)(0.540.7)0.64(0.60.5)(0.60.7)(0.540.5)(0.540.6)0.12
(0.70.5)(0.70.6)l l l --=
=----==----==---
001122(0.54)(0.54)(0.54)(0.54)(0.54)0.6162f F f l f l f l ≈=++=-
截断误差：20121
()'''()()()()3!
R x f x x x x x x ξ=--- 则
23
1
(0.54)0.000128
R ξ=
333
233
[0.5,0.7],()111,0.70.5
11
0.000128(0.54)0.0001280.70.5
f x R ξξ∈∴
<<⨯<<⨯递增；
即
20.000373(0.54)0.001024R <<
2、已知f(x)=e -x 的一组数据见下表，用抛物插值法计算e -2.1的近似值。
解：
001122()()()()F x f l x f l x f l x =++ 001122(2.12)(2.13)
(2.1)0.045,0.3679;
(12)(13)(2.11)(2.13)
(2.1)0.99,0.1353;(21)(23)(2.11)(2.12)
(2.1)0.055,0.0498;
(31)(32)
l f l f l f --=
=-=----===----===--
(2.1)(2.1)0.1201f F ≈= 截断误差：20121
()'''()()()()3!
R x f x x x x x x ξ=--- 故
32(2.1)= -0.0165e R - ⨯
3131
[1,3],()e ;
0.01650.01650.0165x f x e e e e e e ξξξ-------∈=∴<<∴⨯<⨯<⨯递减；
3、根据如下数表，构造不超过三次的 Lagrange 插值多项式。
解：
四点三次拉格朗日插值公式：
32(1)(2)(4)(0)(2)(4)
()19(01)(02)(04)(10)(12)(14)(0)(1)(4)(0)(1)(2)
233(20)(21)(24)(40)(41)(42)11451
1
442x x x x x x F x x x x x x x x x x ------=⨯
+⨯
------------ +⨯+⨯
------ =-+-+ 三点二次多项式：
取0120,1,2x x x ===，
2(1)(2)(0)(2)(0)(1)
()1923(01)(02)(10)(12)(20)(21)351
x x x x x x F x x x ------=⨯
+⨯+⨯
------ =++
取0120,1,4x x x ===，
2(1)(4)(0)(4)(0)(1)
()1233(01)(04)(10)(14)(40)(41)
521
1
22x x x x x x F x x x ------=⨯
+⨯+⨯
------ =-++
取0120,2,4x x x ===，
2(2)(4)(0)(4)(0)(2)
()1233(02)(04)(20)(24)(40)(42)
2143
1
42x x x x x x F x x x ------=⨯
+⨯+⨯
------ =-++
取0121,2,4x x x ===，
2(2)(4)(0)(4)(0)(2)
()9233(02)(04)(20)(24)(40)(42)83821
x x x x x x F x x x ------=⨯
+⨯+⨯
------ =-+-
线性插值多项式： 取010,1x x ==，10
()191010110x x F x x --=⨯+⨯=--- 取010,2x x ==，20
()1231110220x x F x x --=⨯
+⨯=+-- 取010,4x x ==，401
()13104402x x F x x --=⨯
+⨯=+-- 取011,2x x ==，21
()9231451221x x F x x --=⨯
+⨯=--- 取011,4x x ==，41
()932111441x x F x x --=⨯
+⨯=-+-- 取0
12,4x x ==，42
()23310432442
x x F x x --=⨯
+⨯=-+--。