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?B?01?1???20?2??? |?I?B???11??2J?J|??2?2??????5??0
??110?????1?1??????BJ??52?1 故雅克比迭代发散 （2） 高斯—塞德尔迭代矩阵
?00??200??1?01??12??01?1???01?1???22??BG?S???220???1??00?2??110???00?21???=?0??1???1?12????000?=???????222? ??11????042??000???2?????00?1?2???2|?I?BG?S|???????1?2???0 ，???B1G?S??2?1，故高斯—塞德尔迭代收敛
20．设矩阵A＝??a11a12??aa?为二阶矩阵，且a11a12?0。证明雅克比迭代和高斯－塞德尔迭代2122?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
??10???a12?Ba11a11?J???1???? |?I?BJ|???????2?a12a21???0???a21?0a??????aa 1122?22?a22????BJ??|a21a12a| 11a22高斯－塞德尔迭代矩阵
?10????1?a12?B?a0??0?a?21?a11?G?S??11?a12a??0???a1122??01??????0?a??021???00?????a?? 21a12??a21a11a22a?022??a?11a22?
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|?I?B?aa?aaG?S|??????2112aa??0，???BG?S??|2121a|
1122?11a22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??32????x1??0??x????.
2???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 T由p?k??b?Ax(k) t(k)??p(k)?(p(k))(k?1)k)?Ap(k)?T 和(p(k))x?x(k)?t(p(k)
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???? 0.2500?? -0.5000??
p(2)? ?? 0?? -0.75?? 0 0 t?2?? 0.5000 x?3???? 0.2500?? -0.8750??
p(3)? ??1.125?0? 0?? t?3?? 0.1667 x?4????0.4375??-0.8750??
（2）共轭梯度法 (k)T(k)由x(k?1)?x(k)?t(k)p(k) t(k)??r?(p)? r?k??b?Ax(k) r?0??p(0)
Ap(k)?T(p(k))(k?1k)p(k?1)?r(k?1)?a(k)p(k) a(k)???Ar,)p(???Ap(k),p(k) K=0,1 得r?0? =??0???? p?0? =?? 0??-1?? t?0? = 0.5000 x?1?? 0?-1 =?? -0.5000??
r?1? =??1.5000?? 0?? a?0? = 2.2 5 0 0 p?1? =?? 1.5000?? -2.2500?? t?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 9L?x1x?x1?xx0?x1f(x0)?0x1?x0f(x1)?2.1?2.32.0?2.30.6931?2.1?2.02.3?2.00.8329=0.7410
抛物线插值：
l(x?x1)(x?x2)20?(x l(x?x0)(x?x2)(x?x0)(x?x1)21?0?x1)(x0?x2)(x l22?
1?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)330?(x?x?x l(x?x0)(x?x2)(x?x3)31?
01)(x0?x2)(x03)(x1?x0)(x1?x2)(x1?x3)l(x?x0)(x?x1)(x?x3)(x?x0)(32?(x?x lx?x1)(x?x)233?
20)(x2?x1)(x2?x3)(x3?x0)(x3?x1)(x3?x2)L3?l30y0?l331y1?l32y2?l33y3=
10x3?132626x?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|1a?x?bf(x)|?8(b?a)2maxa?x?b|f\(x)|
解：取xx?a0?a;x1?b，L1?a?bf(a)?x?bb?af(b)?0 ? Rf''(?)1?|f(x)?L1(x)|?|2x?a)(x?b)|?|f''(?)(b?a)2(2||4| ?|f(x)|?|Lf''(?)(b?a)2f''1(x)|?|2||4|?|L(?)8?a)|?|f\(?)1(x)|?|||(b8||b?a|
4.证明n次Lagrange插值多项式基函数满足
?nxkkiln,i(x)?x, 0?k?n
i?0
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解：取f(x)?xkn 则Ln??lni(x)xki
i?0 f(x)?Ln?Rn?f(n?1()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?x
i)?'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?1i?xn) ?(x?x0)(x?1x?)(?xi?x1)?(x?ix1?)?(xnx?)(xix)(xi?x0)(xi?1x?)(i?x?xi)?(
1ix?ix1?)?(ixnx?)(xix)取 ?n?(x?x0)(x?1x?)(?x?ix1)?(xix?)(x?i1?x)?(xn x)?'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(xi)
6.设f(x)?a0?a1x??anxn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)?nxki?1f'(x?ni?(xi)1n??(x'i)i?1an?n(xi)?i)a?i?1(xi?x1)?(xi?xi?1)(xi?xi?1)?(xi?xn) ?1a?[x?00?k?n?21,x2,?xn]????1 nank?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 N233(x)?1?x?3x(x?2)?10x(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)