數(shù)值計(jì)算答案-石瑞民

1、1習(xí)題一習(xí)題一1、取、取3.143.15，，作為作為的近似值，求各自的絕對(duì)誤差，相的近似值，求各自的絕對(duì)誤差，相722113355?對(duì)誤差和有效數(shù)字的位數(shù)。對(duì)誤差和有效數(shù)字的位數(shù)。解：14.31?x312110211021???????x?所以，有三位有效數(shù)字1x絕對(duì)誤差：，相對(duì)誤差：14.3???e??14.3??re絕對(duì)誤差限：，相對(duì)誤差限：21021????213106110321????????r?21122105.0105.0

2、1084074.000840174.015.315.3?????????????x所以，有兩位有效數(shù)字2x絕對(duì)誤差：，相對(duì)誤差：15.3???e??15.3??re絕對(duì)誤差限：，相對(duì)誤差限：11021????11061???r?31222105.0105.01012645.00012645.0722722?????????????x所以，有三位有效數(shù)字3x絕對(duì)誤差：，相對(duì)誤差：722???e??722??re絕對(duì)誤差限：，相對(duì)誤差限：2

3、1021????21061???r?1133551?x7166105.0105.01032.000000032.0113355????????????所以，有七位有效數(shù)字4x絕對(duì)誤差：，相對(duì)誤差：113355???e??113355??re36、在機(jī)器數(shù)系下、在機(jī)器數(shù)系下中取三個(gè)數(shù)中取三個(gè)數(shù)，)810(ULF41023371258.0???x，，試按，試按和兩21033678429.0??y21033677811.0???zzyx??)

4、()(zyx??種算法計(jì)算種算法計(jì)算的值，并將結(jié)果與精確結(jié)果比較。的值，并將結(jié)果與精確結(jié)果比較。zyx??解：解：3222222222241064100000.01000000641.01033677811.01033678452.01033677811.01012583367845237.01033677811.0)1033678429.01012580000002337.0(1033677811.0)1033678429.01023

5、371258.0()(????????????????????????????zyx3333342241064137126.010641371258.01006180000000.010023371258.01006180000000.01023371258.0)1033677811.01033678429.0(1023371258.0)(???????????????????????????zyx322222241064137126.

6、01012580000064137.01033677811.01033678429.01012580000002337.01033677811.01033678429.01023371258.0????????????????????zyx所以，比精確，且與相同；)(zyx??zyx??)()(zyx??zyx??因此，在做三個(gè)以上的數(shù)相加時(shí)，需要考慮相加的兩個(gè)同號(hào)數(shù)的階數(shù)盡量接近。8、對(duì)于有效數(shù)、對(duì)于有效數(shù)，，，估計(jì)下列算式的相，估計(jì)