1maxf(x)?Ih(x)?maxf??(?)hi2xi?x?xi?18a???bf(x)?x2
????f(x)?2x,f(x)?2h2?maxf(x)?Ih(x)?a?x?b419．求f(x)?x4在[a,b]上分段埃尔米特插值，并估计误差。 解：
在[a,b]区间上，x0?a,xn?b,hi?xi?1?xi,i?0,1,令h?maxhi
0?i?n?1,n?1,
f(x)?x4,f?(x)?4x3
?函数f(x)在区间[xi,xi?1]上的分段埃尔米特插值函数为
Ih(x)?(?(?(?(x?xi?12x?xi)(1?2)f(xi)xi?xi?1xi?1?xix?xi2x?xi?1)(1?2)f(xi?1)xi?1?xixi?xi?1x?xi?12)(x?xi)f?(xi)xi?xi?1x?xi2)(x?xi?1)f?(xi?1)xi?1?xi
xi4?3(x?xi?1)2(hi?2x?2xi)hixi?14?3(x?xi)2(hi?2x?2xi?1)hi?4xi(x?xi?1)2(x?xi)2hi3
4xi?13?2(x?xi)2(x?xi?1)hi误差为
f(x)?Ih(x)1(4)f(?)(x?xi)2(x?xi?1)2 4!1h?maxf(4)(?)(i)424a?x?b2?21
又
f(x)?x4
?f(4)(x)?4!?24hi4h4
?maxf(x)?Ih(x)?max?a?x?b0?i?n?1161620．给定数据表如下： Xj 0.25 0.30 Yj 0.5000 0.5477 0.39 0.6245 0.45 0.6708 0.53 0.7280 试求三次样条插值，并满足条件：
(1)S?(0.25)?1.0000,S?(0.53)?0.6868;
(2)S??(0.25)?S??(0.53)?0.解：
h0?x1?x0?0.05h1?x2?x1?0.09h2?x3?x2?0.06h3?x4?x3?0.08hj?1hj?1?hj
?j???1?,?j?hjhj?1?hj
533,?2?,?3?,?4?11457?1?924,?2?,?3?,?0?11457f(x1)?f(x0)f?x0,x1???0.9540x1?x0f?x1,x2??0.8533f?x2,x3??0.7717f?x3,x4??0.7150
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(1)S?(x0)?1.0000,S?(x4)?0.6868d0?6(f?x1,x2??f0?)??5.5200h0f?x1,x2??f?x0,x1???4.3157h0?h1d1?6d2?6d3?6d4? f?x2,x3??f?x1,x2???3.2640h1?h2f?x3,x4??f?x2,x3???2.4300h2?h36(f4??f?x3,x4?)??2.1150h3由此得矩阵形式的方程组为
2 1 M0 ?5.5200
59 2 M1 ?4.3157
141432 2 M2 ? ?3.2640
5534 2 M3 ?2.4300
77 1 2 M4 ?2.1150
求解此方程组得
M0??2.0278,M1??1.4643M2??1.0313,M3??0.8070,M4??0.6539三次样条表达式为
S(x)?Mj?(yj?(xj?1?x)36hj2?Mj?1(x?xj)36hjMj?1hj62Mjhj6)xj?1?xhj?(yj?1?)x?xjhj
(j?0,1,,n?1)?将M0,M1,M2,M3,M4代入得
23
??6.7593(0.30?x)3?4.8810(x?0.25)3?10.0169(0.30?x)?10.9662(x?0.25)??x??0.25,0.30???2.7117(0.39?x)3?1.9098(x?0.30)3?6.1075(0.39?x)?6.9544(x?0.30)???x??0.30,0.39?S(x)??33??2.8647(0.45?x)?2.2422(x?0.39)?10.4186(0.45?x)?10.9662(x?0.39)?x??0.39,0.45????1.6817(0.53?x)3?1.3623(x?0.45)3?8.3958(0.53?x)?9.1087(x?0.45)???x??0.45,0.53?(2)S??(x0)?0,S??(x4)?0d0?2f0???0,d1??4.3157,d2??3.2640d??2.4300,d
34?2f4???0?0??4?0由此得矩阵开工的方程组为
M0?M4?0??290??14??32??2???M1???4.3157??55??M???
2????3.2640?????M3?????2.4300???0372????求解此方程组，得
M0?0,M1??1.8809M2??0.8616,M3??1.0304,M4?0
又三次样条表达式为
S(x)?M(xj?1?x)3(x?x3j)j6h?Mj?1j6hjM2
?(yjhjxj?1?xj??(yMj?1h2j6)hj?1?6)x?xjjhj将M0,M1,M2,M3,M4代入得
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??6.2697(x?0.25)3?10(0.3?x)?10.9697(x?0.25)??x??0.25,0.30???3.4831(0.39?x)3?1.5956(x?0.3)3?6.1138(0.39?x)?6.9518(x?0.30)???x??0.30,0.39??S(x)??33??2.3933(0.45?x)?2.8622(x?0.39)?10.4186(0.45?x)?11.1903(x?0.39)?x??0.39,0.45????2.1467(0.53?x)3?8.3987(0.53?x)?9.1(x?0.45)???x??0.45,0.53?21．若f(x)?C2?a,b?,S(x)是三次样条函数，证明：
(1)??f??(x)?dx???S??(x)?dxaab2b2??ba?f??(x)?S??(x)?2dx?2?S??(x)?f??(x)?S??(x)?dxab2
(2)若f(xi)?S(xi)(i?0,1,,n)，式中xi为插值节点，且a?x0?x1??xn?b，
则
?baS??(x)?f??(x)?S??(x)?dx?S??(b)?f?(b)?S?(b)??S??(a)?f?(a)?S?(a)?证明：
(1)?????b?f??(x)?S??(x)?a2ba2bb2dx2ba?f??(x)?ababdx???S??(x)?dx?2?f??(x)S??(x)dx2baa
?f??(x)?dx???S??(x)?dx?2?S??(x)?f??(x)?S??(x)?dx2从而有
?a?f??(x)???
badx???S??(x)?dxa2b2?f??(x)?S??(x)?dx?2?S??(x)?f??(x)?S??(x)?dxab
第三章 函数逼近与曲线拟合
1．f(x)?sin解：
,x?[0,1]
2伯恩斯坦多项式为 f(x)?sin?2x，给出[0,1]上的伯恩斯坦多项式B1(f,x)及B3(f,x)。
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