2) f(x)?xn?a, f?(x)?nxn?1, f??(x)?n(n?1)xn?2,
n?1?f(x)?nx???f?(x)?xn?a??1，
f??(x)f?(x)?n(n?1)xnxn?1n?2?n?1x
Newton 迭代格式 xk?1n?1?nxk??xk???nx?a?k??1
1nan ?xk?nxk?an?1nxkn?(1?)xk?xk1?n
limk??a?xk?1na?xkaxn??1n?12na?1?n2?na
3) f(x)?1?, f?(x)?anx?(n?1), f??(x)??an(n?1)x?(n?2)
axn
f(x)f?(x)f??(x)f?(x)1??anx?(n?1)?xn?1an?nx,
n?1x??an(n?1)xanx?(n?2)?(n?1)??
Newton 迭代格式
xk?1n?1n?1?xk?xk?xk??xk???nxk??(1?n)xk?f?(xk)an?an?f(xk)
nlimk??a?xk?12(a?xk)n??*f??(x)2f?(x)*?n?12?na
12.试写出求方程?c?0(其中c 为已知正常数)的Newton迭代格式，并证
x1明当初值 x0满足0?x0?算？
解：记 f(x)?c?1x2c时迭代格式收敛。该迭代格式中是否含有除法运
,则求等价于求方程f(x)?0的根.
c126
f?(x)?1x2, f??(x)??2x3
Newton迭代格式为
c??xk?1xk2c21xk?xk(2?cxk)， k?0,1,2,?
xk?1?xk?f(xk)f?(xk)对任意 x0?(0,)，存在充分小的δ(δ<),(δ<1)使得
cx0?[δ
1, ?δ] 现在考虑区间[a,b]=[δ, ?δ]
cc1?1(c??1)?0
221o f(a)?f(?)?c?f(b)?f(2c????)?c?12c???c?c2?c??2c?c??c2?c?2?c(1?c?)2?c??0
f(a)f(b)?0
23o 当 x?[a,b]时 当 x?[a,b]时 ??f(?)f?(?)f?(x)?0
of??(x)?0
4o??(2?c?)?2c??
?(2?c?)?2c?? }
{(c??1)(c??2)?0 c?(2?c?)?2?c?f(???f?(2c2c??)?(??)
2c22???)2?c(???cc?)?? ? ?1c?c?(2?c?)??
27
因而 当x20?(0,c)时，Newton迭代格式收敛。
直接证明
xk?1?xk(2?cxk)
1?cx2k?1?cxk(2?cxk)?(1?cxk)
1?cx2?kk?(1?cxk?1)??(1?cx20)
limx12k?0k??k?c?lim(1?cxk??k)?0?lim(1?cxk??0)
?1?cx20?1?x0?(0,c)
用劈因子法解方程x3?3x2?x?9?0(取?20(x)?x?4x?6，算至
r0?0.005，r1?0.005)
解：f(x)?x3?3x2?x?9
取 ?0(x)?x2?4x?6
1+1 1
1?4?6 1?3?1?9 1?4?61?1?0
1?4?6 1?4?6 1?7?9 ?5?6 1?4?6 ?3?3
5?u??v??3 ?u??0.545455 ?6?u??v?3 ?v??0.272727
于是得到 ?1(x)??0(x)??ux??v?x2?4.54546x?5.72727 6.09092?u??v?0.29756
?5.72727?u?1.54546?v?0.14873 ?u?0.0205500，?v?0.172392
?2(x)?x2?4.5249x1?5.89966 228
18.
f(x)??2(x)(x?1.5249)1?0.0004x2?0.0035 5
??2(x)(x?1.52491)
x1,2?2.26246?0.883718i
x3??1.52491
19.用适当的迭代法求下列方程组的根，精确至4位有效数：
?1?x?sin?y?
?2??1x?cos??312?x? ?解： xk?1?sin(yk) yk?1?cos(13xk)
k?0,1,2,?
k 0 1 2 3 4 5 6 xk yk 0 0 0.479426 0.479426 0.473825 0.473825 0.473955 0 1 1 0.987258 0.987258 0.987553 0.987553
k 7 8 9 xk yk 0.473955 0.473952 0.473952 0.987546 0.987546 0.987546
?
x?0.4740*
y*?0.987 529
习题3
1. 设L为单位下三角阵，试写出解方程组的算法。
?1?l?21?l31?????ln11l32?ln21?ln3???x1??d1??????xd??2??2???x3???d3? ????????????1????xn????dn??解：
x1?d1
i?1?lijxjj?1?xi?di，i?2,3,?,n
x1?d1
i?1xi?di??lijxj，i?2,3,?,n
j?12. 为阶的上三角阵，试计算用回代算法解上三角方程组所需的乘除法运算次数。
?r11??解：?????r12r22r13r23????rn?1,n?1r1n??x1??c1??r2n?x2??c2?????????????c3??????rn?1,n?xn?1?????????rn,n???xn??cn?
rn,nxn?cn
n?rijxjj?i?ci，i?n?1,n?2,?,1
xn?cnrn,n
nxi?(ci?n?rijxj)j?i?1rii ， i?n?1,n?2,?,1
12n(n?1)
乘除法运算次数 ??(n?i?1)?1?i?12???n?3. 试用Gauss消去法解下列方程组，计算过程按5位小数进行：
30