湖南大学出版社《数值计算方法》课后题答案
8
1
2
3
4
5
6
0.54030,
0.87758,
0.94496,
0.96891,
0.98007,
0.98614,
0.98981.
x
x
x
x
x
x
x
=
=
=
=
=
=
=
解:由埃特金加速公式
k
k
k
k x
x
x
x
x
x
x
+
-
-
-
=+
2
1
2
)
(
~计算,结果列下表:
6.令初值
1
x=
,分别用牛顿迭代法,双点弦割法和单点弦割法求解方程2
()60
f x x
=-=的解。
解:牛顿迭代法
2
)1('>
=
f,0
2
)2(''>
=
f,满足0
)1(''
)1('≥
f
f,由牛顿迭代法的收敛条件知当取初值为0
1
x=时迭代法收敛。
牛顿迭代格式为:k
k
k
k
k
k x
x
x
x
x
x
f
x
f
x
x
3
2
2
6
)
('
)
(2
1
+
=
-
-
=
-
=
+