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?01?1????11?B?J???20?2? |?I?B???J|??2?2??????2?5??0 ?110?????1?1??????B??5J2?1 故雅克比迭代发散
（2） 高斯—塞德尔迭代矩阵
?1?1?00??1??200??1?01?1??201?1?0??22??B20????00?2??1???=?10??0?2??????G?S???2???0=??0?1??1?12?0222?12? ???00???11??000????0????42?0?1????0?2??2|?I?B|????1B1G?S????2??0 ，????G?S??2?1，故高斯—塞德尔迭代收敛
20．设矩阵A＝?aa?1112??aa?为二阶矩阵，且a21?11a12?0。证明雅克比迭代和高斯－塞德尔迭代22同时收敛或发散。
证明： 因为a11a12?0，所以a11?0,a12?0 雅克比迭代矩阵 ?1?a?0????12?Ba11a?J????11??1? |?I?BJ|??a????2?a12a21??0
?0???a??21??a?11a22?22??a22????BJ??|a21a12a| 11a22高斯－塞德尔迭代矩阵
?1??a?1B?0?a??21a0???a12?11G?S?0??11??0?a?0a?21??11??a??12a?22??00????a1??0????21?0??a? ?a?021a1211a22a?22??aa?1122?
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|?I?B?aa?0，???Baa21G?S|?????2112?a??G?S??|21|
11a22?a11a22所以，雅克比迭代和高斯－塞德尔迭代同时收敛或发散。 21．设线性代数方程组为?63??x??32??1??0?????. ??x2???1?(1) 试用最速下降法求解（取初始向量X(0)＝?0,0?T，计算到X(4)）；
(2) 试用共轭梯度法求解（取初始向量X(0)＝?0,0?T）。 解：（1）最速下降法 (k)T由p?k??b?Ax(k) t(k)??p?(pk())k?1)? 和?x(k)?t(k)p(k)
Ap(k)?T(p(k)x()K=0,1,2,3 得p(0)? ? 0??? -1? t?0?? 0.5000 x?1??? 0???? -0.5000??p(1)? ?1.500?0? 0? t?1??0.1667 x?2??? 0.25?00?? -0.50?
???00p(2)? ? 0?? 0.25?00?? -0.7500? t?2?? 0.5000 x?3????? -0.87?
?50p(3)? ?1.125?0? t?3?? 0.1667 x?4???0.437?5? 0? ???-0.87?5
?0（2）共轭梯度法 由x(k?1)?x(k)?t(k)p(k) t(k)??r(k)?T(pk())?T r?k??b?Ax(k) r?0??p(0)
Ap(k)?(p(k))Ar(k?1,)p(k)p(k?1)?r(k?1)?a(k)p(k) a(k)?????Ap(k),p(k)? K=0,1 得r?0? =?0????-1? p?0? = 0??? t?0? = 0.5 0 0 0 x?1? =? 0???-1??? -0.5000?
?r?1? =?1.5000?? 1.500?? 0? a?0? = 2.2 5 0 0 p?1? =0???? -2.25?
?00t?1?= 0.6667 x?2? =? 1???-2?，即为精确解 ?
习题四
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1.已知ln(2.0)=0.6931;ln(2.2)=0.7885,ln(2.3)=0.8329, 试用线性插值和抛物插值计算.ln2.1的值并估计误差 解：线形插值：取 x0?2.0 y0?0.693 1 x1?2.2 y1?0.788 5 x2?2.3 y2?0.832 9L1?x?x12.1?2.0x0?x1f(x0)?x?x0x1?x0f(x1)?2.1?2.32.0?2.30.6931?2.3?2.00.8329=0.7410
抛物线插值：
lx?x1)(x?x2)20?((x lx?x0)(x?x2)21?( l(x?x0)(x?x1)22?0?x1)(x0?x2) (x1?x0)(x1?x2)(x2?x0)(x2?x1)
L2?l20y0?l21y1?l22y2=0.742
2.已知x=0,2,3,5对应的函数值分别为y=1,3,2,5.试求三次多项式的插值 解：解：取x0?0 x1?2 x2?3 x3?5 l(x?x1)(x?x2)(x?x)3 l(x?x0)(x?x2)(x?x)330?(x31?0?x1)(x0?x2)(x0?x3)(x 1?x0)(x1?x2)(x1?x3)l(x?x0)(x?x1)(x?x3)32?(x l(x?x0)(x?x1)(x?x)233?2?x0)(x2?x1)(x2?x3)(x
3?x0)(x3?x1)(x3?x2)L?lly3362330y0?31y1?l32y2?l333=
1310x?6x2?15x?1
3.设函数f(x)在[a,b]上具有直到二阶的连续导数,且f(a)=f(b)=0, 求证:max|f(x)|?1b?a)2max|f\(x)|
a?x?b8(a?x?b解：取x0?a;x1?b，Lx?a1?a?bf(a)?x?bb?af(b)?0
''''? R1?|f(x)?L1(x)|?|f(?)(?)22(x?a)(x?b)|?|f2||(b?a)4|
''2''?|f(x)|?|L)||(b?a)f\(?)1(x)|?|f(?24|?|L1(x)|?|f(?)8||(b?a)|?|8||b?a|
4.证明n次Lagrange插值多项式基函数满足
n?xkil,xkni(x)?, 0?k?n
i?0
n解：取f(x)?x 则Ln?k?li?0nik(x)xi
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n?1 f(x)?Ln?Rn?f(()x)n(xk)(n?1)nn!?(x?xi)??(x?xi)=0
i?0n!i?0所以f(x)?Ln(x) 即证 5.证明 ln,i(x)??n(x)(x?xi)?'n(xi)
证明：、lni?(x?x0)(x?x1)?(x?xi?1)(x?xi?1)?(x?xn)(x?x i?x0)(xi?x1)?(xi?xi?1)(xi?xi?1)?(xin) ?(x?x0)(x?1x?)(?xi?x1)?(x?ix1?)?(xnx?)(xi(x
x)i?x0)(xi?1x?)(i?x?ix1)?(ix?i1x?)?(ixxn?)(xix)取 ?n?(x?x0)(x?1x?)(?x?ix1)?(xix?)(x?i1?x)?(x nx)?'n(x)?(x?1x?)(?xnx?)?(x（0x?)x2?x)?(xn?x)则 (x?x0)(x?1x?)(?xi?x1)?(x?ix1?)?(xn?x?)
?(x?x0)(x?1x?)(?xn?x1)?'n(xi)?(xi?x0)(xi?x1)?(xi?xi?1)(xi?xi?1)?(xi?xn)
所以，lni??n(x)(x?x'i)?n(x
i)
6.设f(x)?ax??an0?a1nx有n个不同的实根x1,x2,?xn.
n证明:
?xki?0,i?1f'(x???1
i)?an证明：取?(x)?xk ?n?(x?x1)?(x?xn ) 而，f(x)?an0???anx 有n个不同的实根。可以写成f(x)?an?n(x)nn?xki(xi)1n?(xi)f''?i?1(x???i)i?1an?n(xi)a?i?1(xi?x1)?(xi?xi?1)(xi?xi?1)?(xi?xn) ?1a?[x?00?k?n?21,x2,?xn]???1 n?ank?n?1
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7.分别求满足习题1和习题2 中插值条件的Newton插值 (1)
xi f[xi] f[xi?1,xi] f[xi?2,xi?1,xi] 2.0 0.6931 2.2 0.7885 0.477 2.3 0.8329 0.444 -0.11 N2(x)?f[x0]?f[x0,x1])(x?x0)?f[x0,x1,x2])(x?x0)(x?x1)
=0.693+0.477(x-2)-0.11(x-2)(x-2.2)
N2(2.1)?0.693+0.0477-0.0011=0.7419
(2) xi f[xi] f[xi?1,xi] f[xi?2,xi?1,xi] f[xi?3,xi?2,xi?1,xi] 0 1 2 3 1 3 2 -1 -2/3 5 5 3/2 5/6 3/10 N3(x)?1?x?23x(x?2)?310x(x?2)(x?3)
8.给出函数f(x)的数表如下,求四次Newton插值多项式,并由此计算f(0.596)的值
xi 0.40 0.55 0.65 0.80 0.90 1.05 f(xi) 0.41075 0.57815 0.69675 0.88811 1.02652 1.25382 解：
xi f[xi] F2 F3 F4 F5 F6 0.4 0.41075 0.55 0.57815 1.11600 0.65 0.69675 1.18600 0.28000 0.8 0.88811 1.27573 0.35893 0.19733 0.9 1.02652 1.38410 0.43347 0.18634 -0.02200 1.05 1.25382 1.51533 0.52492 0.22863 0.08846 0.16394 f(x)=0.41075+1.11600(x-0.4)+0.28(x-0.4)(x-0.55)+0.19733(x-0.4)(x-0.55)(x-0.65)