??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??3.2640由此得矩阵开工的方程组为
d3??2.4300,d4?2f4???0?0??4?0M0?M4?09?2?14??32?5?3?0?7??0???M1???4.3157?2?????求解此方程组，得 M??3.26402???5???????M3???2.4300???2??M0?0,M1??1.8809M2??0.8616,M3??1.0304,M4?0又?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将M0,M1,M2,M3,M4代入得
??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)???ba?f??(x)?dx???S??(x)?dxa22b2ba?f??(x)?S??(x)?若
dx?2?S??(x)?f??(x)?S??(x)?dxab2
为插值节点，且
(2)f(xi)?S(xi)(i?0,1,?,n)，式中
xia?x0?x1???xn?b，则
?baS??(x)?f??(x)?S??(x)?dx?S??(b)?f?(b)?S?(b)??S??(a)?f?(a)?S?(a)?(1)?
?f??(x)?S??(x)?a2bab2dx2ba证明：??a??ab?f??(x)?dx???S??(x)?dx?2?f??(x)S??(x)dx?f??(x)?ba从而有
b2dx???S??(x)?dx?2?S??(x)?f??(x)?S??(x)?dxaa2b2b?
ba?f??(x)?dx???S??(x)?dxba2???f??(x)?S??(x)?2dx?2?S??(x)?f??(x)?S??(x)?dxab
第三章 函数逼近与曲线拟合
1．
f(x)?sin?2x，给出[0,1上]的伯恩斯坦多项式
B1(f,x)及
B3(f,x)。
解：?f(x)?sin其中Pk(x)??P1(x)?x?k,x?[0,1]伯恩斯坦多项式为Bn(f,x)??f()Pk(x)
n2k?0n?n?k?1?n?kx(1?x)P(x)?当时，n?10???(1?x) ?k??0??B1(f,x)?f(0)P0(x)?f(1)P1(x)?1??????(1?x)sin(?0)?xsin22?0??x当n?3时，
?1?P0(x)???(1?x)3?0??1?22P1(x)???x(1?x)?3x(1?x)?0?
?3?P2(x)???x2(1?x)?3x2(1?x)?1??3?P3(x)???x3?x3?3?k?B3(f,x)??f()Pk(x)nk?0?0?3x(1?x)2?sin?3?6?3x2(1?x)?sin?3?x3sin?23332x(1?x)2?x(1?x)?x3225?33333?623?x?x?x222?1.5x?0.402x2?0.098x3
2． 当f(x)?x时，求证Bn(f,x)?x 证明：若f(x)?x，则Bn(f,x)??f(k?0nk)Pk(x) nk?n?????xk(1?x)n?kk?0n?k?nkn(n?1)?(n?k?1)k??x(1?x)n?kk!k?0nn??(n?1)?[(n?1)?(k?1)?1]kx(1?x)n?k(k?1)!k?1nn
?n?1?kn?k???x(1?x)?k?1?k?1?n?n?1?k?1(n?1)?(k?1)?x???x(1?x)k?1?k?1??x[x?(1?x)]n?1?x
3．证明函数1,x,?,xn线性无关 证明：若a0?a1x?a2x2???anxn?0,?x?R
分别取xk(k?0,1,2,?,n)，对上式两端在[0,1]上作带权?(x)?1的内积，
得
?1??a??0?1???0n?1???a??0???????1?????此方程组的系数矩阵为希尔伯特矩????????1??????1????an??0?2n?1??n?1阵，对称正定非奇异，
?只有零解
a=0。?函数1,x,?,xn线性无关。
f?4。计算下列函数f(x)关于C[0,1]的
(1)f(x)?(x?1)3,x?[0,1]1(2)f(x)?x?,2,f1与
f2：
(3)f(x)?xm(1?x)n,m与n为正整数，
(4)f(x)?(x?1)10e?x
解：则f?x()(1)若f(x)?(x?1)3,x?[0,1]，内单调递增
f?23(?x)1?0?()(?x)1??fx3在(0,1)?maxf(x)0?x?1?max?f(0),f(1)? ?max?0,1??1f??maxf(x)0?x?1?max?f(0),f(1)? ?max?0,1??1f?(?(1?x)dx)01612211712?[(1?x)]07
?77(2)若f(x)?x?1,x??0,1?，则 2ff??maxf(x)?0?x?11121??f(x)dx011?2?1(x?)dx221?4f21
?(?f(x)dx)01212121?[?(x?)dx]2023?6
(3)若f(x)?xm(1?x)n,m
与n为正整数当x??0,1?时,f(x)?0
f?(x)?mxm?1(1?x)n?xmn(1?x)n?1(?1)?xm?1n?m(1?x)m(1?x)mn?1当x?(0,mmm)时,f?(x)?0?f(x)在(0,)内单调递减当x?(,1)n?mn?mn?m时,f?(x)?0
?f(x)在(x?(m,1)内单调递减。 n?mm,1)f?(x)?0n?mf??maxf(x)?0?x?1?m? ?max?f(0),f()?n?m??mm?nn?(m?n)m?n