数值分析习题1

13. 解：

?f(x)?ln(x?x2?1),?f(30)?ln(30?899)设u?899,y?f(30)则u*?29.98331*??(u)??10?42故1*?(y)???(u)*30?u1?(u*)0.0617?3?10?3?若改用等价公式ln(x?x2?1)??ln(x?x2?1)则f(30)?-ln(30?899)此时1*?(y)?-?(u)*30?u1??(u*)59.9833?7?8?10

14. 解：

*

由p(x)?3x5?2x3?x?7得a0?3,a1?0,a2??2,a3?0,a4?1,a5?7当x*?3时，bi?bi?1x*?ai?b0?3b1?9,b2?25,b3?75,b4?226,b5?685*?p(x)?b5?685

15. 解：

?xk?1??x1?x3?x5?1,x0?11?xk1112?,x2??,1?x021?x131315?,x4??,1?x251?x3818??0.61538461951?x413?1?5又?x*??0.61803398821*得x5-x??10-22?x5有两位有效数字 16. 解： (1)

(2)复合梯形公式：

?120exdx?en1x20?e?e0?1.64872?11?0.648721

1211b?a?h?,a?0,b?,xi?a?ih,h?42nb?a?n??2n

nI?f????f?xi?1??f?xi???Tni?12

11x0?0,x1?,x2?42hh?I?f???f?x0??f?x1????f?x1??f?x2??22

?

111f?x0??f?x1??f?x2?848

（3）利用T1及T2的松弛法： 公式为：

110111?e?e4?e2848?0.652096

b?a?f?a??f?b??T1?2b?aa?b?f?a??2f?c??f?b??,C?T2?4241S1?T2?T133b?a?f?a??4f?c??f?b???61??1??1??2??f?0??4f???f???6??4??2??11?1?042??e?4e?e???12???0.0833?303.4280?008.137393 17.解

?0.648735

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