zasdfgbnm 发表于 2010-4-22 07:35:04

使用82ES实现牛顿法解方程

牛顿法解方程需要先给定一个初值x0,将该值赋值给变量x
http://www.imathas.com/cgi-bin/mimetex.cgi?x_%7B0%7D%5Crightarrow%20X
然后反复执行代码:
http://www.imathas.com/cgi-bin/mimetex.cgi?X-%5Cfrac%7Bf%28X%29%7D%7Bf%27%28X%29%7D%5Crightarrow%20X
其中f'(x)是f(x)对x的导函数,可以用手求,也可以利用
http://www.imathas.com/cgi-bin/mimetex.cgi?f%27%28x%29%5Capprox%5Cfrac%7Bf%28x+10%5E%7B-5%7D%29-f%28x%29%7D%7B10%5E%7B-5%7D%7D
打入计算器,10^-5也可以换成其他比较小的数,只要不超过计算器的精确度就不会影响最终结果。
此程序还可以精简一下:
先将x0赋值给Ans(这个都会吧~)
然后反复执行代码:
http://www.imathas.com/cgi-bin/mimetex.cgi?Ans-%5Cfrac%7Bf%28Ans%29%7D%7Bf%27%28Ans%29%7D
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牛顿法解方程效率相当高
我们可以假设根为ξ,xi为第i次执行语句后的X的值,理论分析可得
http://www.imathas.com/cgi-bin/mimetex.cgi?%5Cmid%5Cxi-x_%7Bi+1%7D%5Cmid=%5Cmu_%7Bi%7D%5Cmid%5Cxi-x_%7Bi%7D%5Cmid%5E%7B2%7D
其中,μi由方程和初值决定
例如:要解方程
http://www.imathas.com/cgi-bin/mimetex.cgi?f%28x%29=x%5E%7B3%7D-7x%5E%7B2%7D+6x+20=0
初值x0=-1.5
对于此函数,理论上可以分析出,μi<0.5对于一切正整数i成立(这点是关键,并且几乎所有方程都能得出类似结论),并且
http://www.imathas.com/cgi-bin/mimetex.cgi?%5Cmid%5Cxi-x_%7B0%7D%5Cmid%3C0.33
于是有
http://www.imathas.com/cgi-bin/mimetex.cgi?%5Cmid%5Cxi-x_%7B1%7D%5Cmid=%5Cmu_%7B0%7D%5Cmid%5Cxi-x_%7B0%7D%5Cmid%5E%7B2%7D%3C0.5%5Ctimes0.33%5E%7B2%7D%3C0.06
http://www.imathas.com/cgi-bin/mimetex.cgi?%5Cmid%5Cxi-x_%7B2%7D%5Cmid=%5Cmu_%7B1%7D%5Cmid%5Cxi-x_%7B1%7D%5Cmid%5E%7B2%7D%3C0.5%5Ctimes0.06%5E%7B2%7D%3C2%5Ctimes10%5E%7B-3%7D
http://www.imathas.com/cgi-bin/mimetex.cgi?%5Cmid%5Cxi-x_%7B3%7D%5Cmid=%5Cmu_%7B2%7D%5Cmid%5Cxi-x_%7B2%7D%5Cmid%5E%7B2%7D%3C0.5%5Ctimes%282%5Ctimes10%5E%7B-3%7D%29%5E%7B2%7D=2%5Ctimes10%5E%7B-6%7D
http://www.imathas.com/cgi-bin/mimetex.cgi?%5Cmid%5Cxi-x_%7B4%7D%5Cmid=%5Cmu_%7B3%7D%5Cmid%5Cxi-x_%7B3%7D%5Cmid%5E%7B2%7D%3C0.5%5Ctimes%282%5Ctimes10%5E%7B-6%7D%29%5E%7B2%7D=2%5Ctimes10%5E%7B-12%7D
http://www.imathas.com/cgi-bin/mimetex.cgi?%5Cmid%5Cxi-x_%7B5%7D%5Cmid%3C2%5Ctimes10%5E%7B-24%7D
http://www.imathas.com/cgi-bin/mimetex.cgi?%5Cmid%5Cxi-x_%7B6%7D%5Cmid%3C2%5Ctimes10%5E%7B-48%7D
http://www.imathas.com/cgi-bin/mimetex.cgi?%5Cmid%5Cxi-x_%7B7%7D%5Cmid%3C2%5Ctimes10%5E%7B-96%7D
http://www.imathas.com/cgi-bin/mimetex.cgi?%5Cmid%5Cxi-x_%7B8%7D%5Cmid%3C2%5Ctimes10%5E%7B-192%7D
http://www.imathas.com/cgi-bin/mimetex.cgi?%5Cmid%5Cxi-x_%7B9%7D%5Cmid%3C2%5Ctimes10%5E%7B-384%7D
http://www.imathas.com/cgi-bin/mimetex.cgi?%5Cmid%5Cxi-x_%7B10%7D%5Cmid%3C2%5Ctimes10%5E%7B-768%7D
这就是说,对此方程来说,要让精确度达到计算器的精度,只需要重复执行代码4次,如果用电脑进行高精度运算,只需执行代码10次就可以精确到小数点后至少767位(由于理论分析时进行了放大处理,所以实际精确度可达八百多位有效数字),这就是指数爆炸的威力
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想更多了解牛顿法,可以参见以下书籍(我全没读过)
Hildebrand,Introduction to Numerical Analysis,McGraw-Hill Book Co.,1956
Householder,Principles of Numerical Analysis,McGraw-Hill Book Co.,1953
Whittaker&Robinson,The Calculus of Observations,Blackie and Sons,Ltd.,1929

zasdfgbnm 发表于 2010-4-22 07:59:13

A nal也要屏蔽?

zasdfgbnm 发表于 2010-4-22 07:59:58

**ysis

hcz 发表于 2010-4-22 15:57:15

Analysis?=.=

urill 发表于 2010-4-22 19:51:58

**og....

Nero 发表于 2010-4-22 20:34:48

Analysis。。。

zasdfgbnm 发表于 2010-4-23 07:34:33

6# 991es82es
你帮忙改的么?谢谢了
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