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LESK樂木蟲 (小有名氣)
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[求助]
用matlab解非線性方程組
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如題,方程組在下面,x1 x2 x3 x4的初值非別為0.2, 0.6,0.75,1.45,希望大神幫忙解答 2*(cos(x1)-cos(x2)+cos(x3)-cos(x4))-1=π/4(等號(hào)右邊是4分之pai) 2*(cos(3*x1)-cos(3*x2)+cos(3*x3)-cos(3*x4))-1=0 2*(cos(9*x1)-cos(9*x2)+cos(9*x3)-cos(9*x4))-1=0 2*(cos(15*x1)-cos(15*x2)+cos(15*x3)-cos(15*x4)))-1=0 [ Last edited by LESK樂 on 2013-10-21 at 14:24 ] |
木蟲 (小有名氣)
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謝謝,不過為什么我用計(jì)算器把答案帶進(jìn)去算出來的方程答案不對(duì)啊,http://www.ilovematlab.cn/thread-263705-1-1.html這是我之前發(fā)的帖子,里面的結(jié)果跟你的一樣 |
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編寫xfsolve.m文件: CODE: function xfsolve clear all;clc x0=[0.2,0.6,0.75,1.45]; x1=fsolve(@abc,x0); function f=abc(x) f(1)=2*(cos(x(1))-cos(x(2))+cos(x(3))-cos(x(4)))-1-pi/4; f(2)=2*(cos(3*x(1))-cos(3*x(2))+cos(3*x(3))-cos(3*x(4)))-1; f(3)=2*(cos(9*x(1))-cos(9*x(2))+cos(9*x(3))-cos(9*x(4)))-1; f(4)=2*(cos(15*x(1))-cos(15*x(2))+cos(15*x(3))-cos(15*x(4)))-1; |
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