1 2?1(x)?a?(x?1)(x?2)?x2??1(x)??(x?1)(x?2)?x2?0(x)?a(x?1)2?x?(x?2)??0(x)??x(x?2)(x?1)2?H(x)?0??0(x)?1??1(x)?1??2(x)?0??0(x)?1??1(x)?x2(x?2)2??12x(x?3)2412x(x?1)2?(x?1)(x?2)x24
（法二）令H(x)?a0?a1x?a2x2?a3x3?a4x4 则H'(x)?a1?2a2x?3a3x2?4a4x3
?a0?a1?0?a2?02?a3?03?a4?04?0?a0?0?a0?a1?1?a2?12?a3?13?a4?14?1??234 ?a0?a1?2?a2?2?a3?2?a4?2?1?a?2a?0?3a?02?4a?03?0?a?02341?123??a1?2a2?1?3a3?1?4a4?1?1得
H(x)?923314x?x?x424
121?x(9?6x?x2)?x2(x?3)244'4、求三次多项式P3(x)，使得PPP3(0)?P3(0)?0,3(1)?1,3(2)?3
解：令P(x)?a0?a1x?a2x2?a3x3 则P'(x)?a1?2a2x?3a3x2
?a0?a1?0?a2?02?a3?03?0?2?a1?2a2?0?3a3?0?0?23?a0?a1?1?a2?1?a3?1?123??a0?a1?2?a2?2?a3?2?3a0?0a1?0 a2?a3?14a2?8a3?3
P3(x)?5211x?(?)?x3?x2(5?x) 4445?a???24?? 1?a??3??45、求一个次数?3的多项式P3(x)，使得P3(0)?1，P3(1)?2，P3(0)?P3(1)?解：令P(x)?a0?a1x?a2x2?a3x3
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''1 2则P'(x)?a1?2a2x?3a3x2
?a0?a1?0?a2?02?a3?03?1?2?a1?2a2?0?3a3?0?0.5?23?a0?a1?1?a2?1?a3?1?22?a?2a?1?3a?1?0.5123?(1)(2)(3)(4)
由（1）得a0?1 由（2）得a1?0.5 由（3）得
a0?a1?a2?a3?2 （5）
由（1）得
a1?2a2?3a3?0.5 （6）
把a0?1、a1?0.5代入（5）、（6）得
a2?1.5、a3??1
?6、给出概率积分y(x)?2P(x)?1?0.5x?1.5x2?x3
??x0e?xdx的数据表如下：
2x y(x) 0.46 0.484655 0.47 0.493745 0.48 0.502750 0.49 0.511668 试用拉格朗日插值法计算x?0.427时，该积分值等于多少？ 解：记
x1?0.46x2?0.47x3?0.48x4?0.49y4?0.511668y1?0.484655y2?0.493745y3?0.502750
将y看成x的函数y?y(x)，以x1,x2,x3,x4式:
为插值节点作y(x)的3次插值多项
L3(x)?y1?l1(x)?y2?l2(x)?y3?l3(x)?y4?l4(x)?y1?(x?x2)(x?x3)(x?x4)(x?x1)(x?x3)(x?x4)
?y2(x1?x2)(x1?x3)(x1?x4)(x2?x1)(x2?x3)(x2?x4) 17
?y3(x?x1)(x?x2)(x?x3)(x?x1)(x?x2)(x?x4) ?y4(x3?x1)(x3?x2)(x3?x4)(x4?x1)(x4?x2)(x4?x3)y(0.472)?L3(0.472)??0.02326344?0.42659568?0.108594?0.016373376 ?0.495582864?当x?0.472时，概率积分y(0.472)?
2??0.4720e?xdx
2?0.495582864
7、利用y?x在x0?100,x1?121,x2?144处函数值计算115的近似值并估计误差.
解： y?x过点（100，10）、（121，11）、（144，12），
令x0?100,y0?10,x1?121,y1?11,x2?144,y2?12, 则y?x的二次Lagrange插值多项式
L2(x)?y0l0(x)?y1l1(x)?y2l2(x)(x?121)(x?144)(x?100)(x?144)
?10??11?(100?121)(100?144)(121?100)(121?144) ?12?(x?100)(x?144)
(144?100)(144?121)?115?y(115)?L2(115)?10.722756y'''(?)|R(115)|?|(115?100)(115?121)(115?144)|3!13?5?|??2?15?6?29|3!85?13???1002?15?6?29 68?1.63125?10?3??[100,144]
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第七章 数据拟合与函数逼近
1、用最小二乘法求一个形如y?a?bx2的经验公式，使它与下列数据相拟合
xi yi 19 19.0 25 32.3 31 49.0 38 73.3 44 97.8 ?19.0?a?b?192?232.3?a?b?25??解：（法一）建立超定方程组?49.0?a?b?312
?73.3?a?b?382?2??97.8?a?b?44?1??1即：?1??1?1?192??19.0???32.3?252???a2???31?????49.0?
b??2???73.338????442??97.8????1?11???2?144???1?1?192??252?a2???31???b2???38?442???19.0??32.3??1???49.0? 442???73.3?????97.8??解
?1?192?125213121382?1?192?1112523123825327??a??271.4??5得????????53277277699??b??369321.5??a?0.972606 ??b?0.0500351?（法二）利用公式建立正规方程组
?5??1?i?1?52??xi?i?1??5?x?y??i??a??i?1i?1??????
55b????22?2x?xxy?ii???ii?i?1??i?1?2i5 19
5327??a??271.4??5?53277277699??b???369321.5? ???????a?0.972606 ???b?0.05003512、求形如y?a?ebx(a,b为常数且a?0)的经验方式，使它能和下表数据相拟合
xi yi 1.00 5.10 1.25 5.79 1.50 6.53 1.75 7.45 2.00 8.46 解：对经验方式y?a?ebx作变换，有lny?lna?bx，令
y?lny,?A?l则na,???yAbx，为了用最小二乘法求出A,b将(xi,yi)转化为
(xi,yi)
?
xi yi 1.00 1.25 1.50 1.75 2.00 1.629 1.756 1.876 2.008 2.135 （法一）建立超定方程组
?1.629?A?1.00b?1.756?A?1.25b???1.876?A?1.50b ?2.008?A?1.75b???2.135?A?2.00b即：
?1?1??1??1??11.00??1.629??1.756?1.25???A???1.50??????1.876?
??b???1.75?2.008???2.00???2.135??得正规方程组
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