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小木虫论坛-学术科研互动平台 » 计算模拟区 » 计算模拟 » 求解高次复杂函数
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denmarkrico

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[求助] 求解高次复杂函数 已有3人参与

求解下面这个高次方程时为什么总是得到虚数解，我所用的是MATLAB2012b版

syms h c l theta11 theta21 theta31 x
h=14.5;
l=55.5;
c=19.73;
theta11=0.01/37;
theta21=0.035/37;
theta31=0.065/37;
F=((3*c + 3*h - (433*l)/250 + 2*l*cos(theta11) +3^(1/2)*c*sin(theta0) - c*cos(theta0) +l*cos(theta21) + 3^(1/2)*l*sin(theta21) + (2*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta11) - 1) - 2*l^2*(sin(theta31) - 1) + 2*c*(c + h) - 2*c*l*cos(theta11) - 2*l*cos(theta11)*(c + h) + 2*l*cos(theta31)*(c + h) + 2*c*l*cos(theta0)*cos(theta31) - 2*c*l*sin(theta0)*sin(theta31) - (4*c*l*x)/(x^2 + 1) + (2*c*(c + h)*(x^2 - 1))/(x^2 + 1)) + (3*c + 3*h + (433*l)/250 + 2*l*cos(theta11) + l*cos(theta31) - c*cos(theta0) - 3^(1/2)*c*sin(theta0) - 3^(1/2)*l*sin(theta31) + (2*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta21) - 1) - 2*l^2*(sin(theta11) - 1) - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h) - 2*l*cos(theta21)*(c + h) + (4*c*l*x)/(x^2 + 1) - (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta21))/(x^2 + 1) + (2*c*l*cos(theta21)*(x^2 - 1))/(x^2 + 1)))^2 + (((433*c)/250 + (433*h)/250 + 3*l + 2*l*sin(theta11) + 2*l*sin((2*pi)/3 + theta21) - (2*c*x)/(x^2 + 1) - (3^(1/2)*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta11) - 1) - 2*l^2*(sin(theta31) - 1) + 2*c*(c + h) - 2*c*l*cos(theta11) - 2*l*cos(theta11)*(c + h) + 2*l*cos(theta31)*(c + h) - (4*c*l*x)/(x^2 + 1) + (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta31))/(x^2 + 1) - (2*c*l*cos(theta31)*(x^2 - 1))/(x^2 + 1)) - ((433*c)/250 + (433*h)/250 - 3*l + 2*l*sin(theta11) +c*sin(theta0) + l*sin(theta31) + 3^(1/2)*l*cos(theta31) - 3^(1/2)*c*cos(theta0) + (4*c*x)/(x^2 + 1))*(2*l^2*(sin(theta21) - 1) - 2*l^2*(sin(theta11) - 1) - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h) - 2*l*cos(theta21)*(c + h) + (4*c*l*x)/(x^2 + 1) - (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta21))/(x^2 + 1) + (2*c*l*cos(theta21)*(x^2 - 1))/(x^2 + 1)))^2 + ((3*c + 3*h - (433*l)/250 + 2*l*cos(theta11) + l*cos(theta21) - c*cos(theta0) + 3^(1/2)*c*sin(theta0) + 3^(1/2)*l*sin(theta21) + (2*c*(x^2 - 1))/(x^2 + 1))*((433*c)/250 + (433*h)/250 - 3*l + 2*l*sin(theta11) +c*sin(theta0) + l*sin(theta31) + 3^(1/2)*l*cos(theta31) - 3^(1/2)*c*cos(theta0) + (4*c*x)/(x^2 + 1)) + (3*c + 3*h + (433*l)/250 + 2*l*cos(theta11) + l*cos(theta31) - c*cos(theta0) - 3^(1/2)*c*sin(theta0) - 3^(1/2)*l*sin(theta31) + (2*c*(x^2 - 1))/(x^2 + 1))*((433*c)/250 + (433*h)/250 + 3*l - 2*l*sin(theta11) + 3^(1/2)*l*cos(theta21) - l*sin(theta21) - c*sin(theta0) - 3^(1/2)*c*cos(theta0) - (4*c*x)/(x^2 + 1)))^2*(2*c*h - 2*l^2*(sin(theta11) - 1) + 2*c^2 - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h)) + ((((433*c)/250 + (433*h)/250 - 3*l + 2*l*sin(theta11) +c*sin(theta0) + l*sin(theta31) + 3^(1/2)*l*cos(theta31) - 3^(1/2)*c*cos(theta0) + (4*c*x)/(x^2 + 1))*(2*l^2*(sin(theta11) - 1) - 2*l^2*(sin(theta21) - 1) + 2*c*(c + h) - 2*c*l*cos(theta11) - 2*l*cos(theta11)*(c + h) + 2*l*cos(theta21)*(c + h) + 2*c*l*cos(theta0)*cos(theta21) - 2*c*l*sin(theta0)*sin(theta21) - (4*c*l*x)/(x^2 + 1) + (2*c*(c + h)*(x^2 - 1))/(x^2 + 1)) + ((433*c)/250 + (433*h)/250 + 3*l - 2*l*sin(theta11) + 3^(1/2)*l*cos(theta21) - l*sin(theta21) - c*sin(theta0) - 3^(1/2)*c*cos(theta0) - (4*c*x)/(x^2 + 1))*(2*l^2*(sin(theta11) - 1) - 2*l^2*(sin(theta31) - 1) + 2*c*(c + h) - 2*c*l*cos(theta11) - 2*l*cos(theta11)*(c + h) + 2*l*cos(theta31)*(c + h) + 2*c*l*cos(theta0)*cos(theta31) - 2*c*l*sin(theta0)*sin(theta31) - (4*c*l*x)/(x^2 + 1) + (2*c*(c + h)*(x^2 - 1))/(x^2 + 1)))*(2*c + 2*h + 2*l*cos(theta11) + (2*c*(x^2 - 1))/(x^2 + 1)) - ((3*c + 3*h + (433*l)/250 + 2*l*cos(theta11) + l*cos(theta31) - c*cos(theta0) - 3^(1/2)*c*sin(theta0) - 3^(1/2)*l*sin(theta31) + (2*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta11) - 1) - 2*l^2*(sin(theta21) - 1) + 2*c*(c + h) - 2*c*l*cos(theta11) - 2*l*cos(theta11)*(c + h) + 2*l*cos(theta21)*(c + h) + 2*c*l*cos(theta0)*cos(theta21) - 2*c*l*sin(theta0)*sin(theta21) - (4*c*l*x)/(x^2 + 1) + (2*c*(c + h)*(x^2 - 1))/(x^2 + 1)) - (3*c + 3*h - (433*l)/250 + 2*l*cos(theta11) +l*cos(theta21) - c*cos(theta0) + 3^(1/2)*c*sin(theta0) + 3^(1/2)*l*sin(theta21) + (2*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta11) - 1) - 2*l^2*(sin(theta31) - 1) + 2*c*(c + h) - 2*c*l*cos(theta11) - 2*l*cos(theta11)*(c + h) + 2*l*cos(theta31)*(c + h) + 2*c*l*cos(theta0)*cos(theta31) - 2*c*l*sin(theta0)*sin(theta31) - (4*c*l*x)/(x^2 + 1) + (2*c*(c + h)*(x^2 - 1))/(x^2 + 1)))*(2*l*sin(theta11) - 2*l + (4*c*x)/(x^2 + 1)))*((3*c + 3*h - (433*l)/250 + 2*l*cos(theta11) + l*cos(theta21) + 3^(1/2)*l*sin(theta21) + (3*c*(x^2 - 1))/(x^2 + 1) + (2*3^(1/2)*c*x)/(x^2 + 1))*((433*c)/250 + (433*h)/250 - 3*l + 2*l*sin(theta11) - l*sin(theta31) + 3^(1/2)*l*cos(theta31) + (2*c*x)/(x^2 + 1) - (3^(1/2)*c*(x^2 - 1))/(x^2 + 1)) + (3*c + 3*h + (433*l)/250 + 2*l*cos(theta11) + l*cos(theta31) - c*cos(theta0) - 3^(1/2)*c*sin(theta0) - 3^(1/2)*l*sin(theta31) + (2*c*(x^2 - 1))/(x^2 + 1))*((433*c)/250 + (433*h)/250 + 3*l - 2*l*sin(theta11) + 3^(1/2)*l*cos(theta21) - l*sin(theta21) - c*sin(theta0) - 3^(1/2)*c*cos(theta0) - (4*c*x)/(x^2 + 1)))；
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1楼 2013-12-13 22:28:48
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feixiaolin

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【答案】应助回帖

感谢参与，应助指数 +1
1）theta0 待求还是已知？
2）取y1=(x^2 - 1))/(x^2 + 1);
        y2=x/(x^2 + 1);
     先求y1, y2，最后求 x如何？
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denmarkrico
2楼2013-12-13 22:53:29
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ynod23

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【答案】应助回帖

感谢参与，应助指数 +1
函数没有看完，瞎了。。。
我认为，这种情况，不是查程序哪儿出问题了，而应该看方法对不对。。。。
先试试看能不能把那个大函数写成许多函数相互运算的结果，这样就降低了出错的可能。如果这样还有问题的话，也可以根据matlab的运行提示，找出是哪一行的问题。这样把问题的规模缩小，我觉得比较靠谱。
希望能帮到你。
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» 本帖已获得的红花（最新10朵）

denmarkrico
5楼2013-12-14 04:43:11
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普通回帖

jerkwin

专家顾问 (正式写手)
你这方程, 能解出来我觉得很奇怪
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3楼2013-12-13 23:10:14
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paopaoai11

木虫 (小有名气)
你这什么方程啊？

[ 发自小木虫客户端 ]
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泡泡
4楼2013-12-14 01:03:00
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denmarkrico

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送红花一朵
引用回帖:
2楼: Originally posted by feixiaolin at 2013-12-13 22:53:29
1）theta0 待求还是已知？
2）取y1=(x^2 - 1))/(x^2 + 1);
        y2=x/(x^2 + 1);
     先求y1, y2，最后求 x如何？

你这个替代法我见过
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6楼2013-12-14 16:07:30
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denmarkrico

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送红花一朵
引用回帖:
5楼: Originally posted by ynod23 at 2013-12-14 04:43:11
函数没有看完，瞎了。。。
我认为，这种情况，不是查程序哪儿出问题了，而应该看方法对不对。。。。
先试试看能不能把那个大函数写成许多函数相互运算的结果，这样就降低了出错的可能。如果这样还有问题的话，也可 ...

嗯  谢谢你 我这是机器人的正解
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7楼2013-12-14 16:08:44
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denmarkrico

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引用回帖:
2楼: Originally posted by feixiaolin at 2013-12-13 22:53:29
1）theta0 待求还是已知？
2）取y1=(x^2 - 1))/(x^2 + 1);
        y2=x/(x^2 + 1);
     先求y1, y2，最后求 x如何？

theta0 是要求的量
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8楼2013-12-14 16:09:58
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denmarkrico

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引用回帖:
2楼: Originally posted by feixiaolin at 2013-12-13 22:53:29
1）theta0 待求还是已知？
2）取y1=(x^2 - 1))/(x^2 + 1);
        y2=x/(x^2 + 1);
     先求y1, y2，最后求 x如何？

要解的方程是这样的：

((3*c + 3*h - (433*l)/250 + 2*l*cos(theta11) + l*cos(theta21) + 3^(1/2)*l*sin(theta21) + (3*c*(x^2 - 1))/(x^2 + 1) + (2*3^(1/2)*c*x)/(x^2 + 1))*((433*c)/250 + (433*h)/250 - 3*l + 2*l*sin(theta11) + l*sin(theta31) + 3^(1/2)*l*cos(theta31) + (6*c*x)/(x^2 + 1) + (3^(1/2)*c*(x^2 - 1))/(x^2 + 1)) + (3*c + 3*h + (433*l)/250 + 2*l*cos(theta11) + l*cos(theta31) - 3^(1/2)*l*sin(theta31) + (3*c*(x^2 - 1))/(x^2 + 1) - (2*3^(1/2)*c*x)/(x^2 + 1))*((433*c)/250 + (433*h)/250 + 3*l - 2*l*sin(theta11) - l*sin(theta21) + 3^(1/2)*l*cos(theta21) - (6*c*x)/(x^2 + 1) + (3^(1/2)*c*(x^2 - 1))/(x^2 + 1)))^2*(2*c*h - 2*l^2*(sin(theta11) - 1) + 2*c^2 - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h)) + ((3*c + 3*h - (433*l)/250 + 2*l*cos(theta11) + l*cos(theta21) + 3^(1/2)*l*sin(theta21) + (3*c*(x^2 - 1))/(x^2 + 1) + (2*3^(1/2)*c*x)/(x^2 + 1))*((433*c)/250 + (433*h)/250 - 3*l + 2*l*sin(theta11) - l*sin(theta31) + 3^(1/2)*l*cos(theta31) + (2*c*x)/(x^2 + 1) - (3^(1/2)*c*(x^2 - 1))/(x^2 + 1)) + (3*c + 3*h + (433*l)/250 + 2*l*cos(theta11) + l*cos(theta31) - 3^(1/2)*l*sin(theta31) + (3*c*(x^2 - 1))/(x^2 + 1) - (2*3^(1/2)*c*x)/(x^2 + 1))*((433*c)/250 + (433*h)/250 + 3*l - 2*l*sin(theta11) - l*sin(theta21) + 3^(1/2)*l*cos(theta21) - (6*c*x)/(x^2 + 1) + (3^(1/2)*c*(x^2 - 1))/(x^2 + 1)))*(((3*c + 3*h + (433*l)/250 + 2*l*cos(theta11) + l*cos(theta31) - 3^(1/2)*l*sin(theta31) + (3*c*(x^2 - 1))/(x^2 + 1) - (2*3^(1/2)*c*x)/(x^2 + 1))*(2*l^2*(sin(theta21) - 1) - 2*l^2*(sin(theta11) - 1) - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h) - 2*l*cos(theta21)*(c + h) + (4*c*l*x)/(x^2 + 1) - (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta21))/(x^2 + 1) + (2*c*l*cos(theta21)*(x^2 - 1))/(x^2 + 1)) - (3*c + 3*h - (433*l)/250 + 2*l*cos(theta11) + l*cos(theta21) + 3^(1/2)*l*sin(theta21) + (3*c*(x^2 - 1))/(x^2 + 1) + (2*3^(1/2)*c*x)/(x^2 + 1))*(2*l^2*(sin(theta31) - 1) - 2*l^2*(sin(theta11) - 1) - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h) - 2*l*cos(theta31)*(c + h) + (4*c*l*x)/(x^2 + 1) - (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta31))/(x^2 + 1) + (2*c*l*cos(theta31)*(x^2 - 1))/(x^2 + 1)))*(2*l*sin(theta11) - 2*l + (4*c*x)/(x^2 + 1)) - (((433*c)/250 + (433*h)/250 - 3*l + 2*l*sin(theta11) + l*sin(theta31) + 3^(1/2)*l*cos(theta31) + (6*c*x)/(x^2 + 1) + (3^(1/2)*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta21) - 1) - 2*l^2*(sin(theta11) - 1) - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h) - 2*l*cos(theta21)*(c + h) + (4*c*l*x)/(x^2 + 1) - (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta21))/(x^2 + 1) + (2*c*l*cos(theta21)*(x^2 - 1))/(x^2 + 1)) + ((433*c)/250 + (433*h)/250 + 3*l - 2*l*sin(theta11) - l*sin(theta21) + 3^(1/2)*l*cos(theta21) - (6*c*x)/(x^2 + 1) + (3^(1/2)*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta31) - 1) - 2*l^2*(sin(theta11) - 1) - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h) - 2*l*cos(theta31)*(c + h) + (4*c*l*x)/(x^2 + 1) - (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta31))/(x^2 + 1) + (2*c*l*cos(theta31)*(x^2 - 1))/(x^2 + 1)))*(2*c + 2*h + 2*l*cos(theta11) + (2*c*(x^2 - 1))/(x^2 + 1))) + ((3*c + 3*h + (433*l)/250 + 2*l*cos(theta11) + l*cos(theta31) - 3^(1/2)*l*sin(theta31) + (3*c*(x^2 - 1))/(x^2 + 1) - (2*3^(1/2)*c*x)/(x^2 + 1))*(2*l^2*(sin(theta21) - 1) - 2*l^2*(sin(theta11) - 1) - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h) - 2*l*cos(theta21)*(c + h) + (4*c*l*x)/(x^2 + 1) - (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta21))/(x^2 + 1) + (2*c*l*cos(theta21)*(x^2 - 1))/(x^2 + 1)) - (3*c + 3*h - (433*l)/250 + 2*l*cos(theta11) + l*cos(theta21) + 3^(1/2)*l*sin(theta21) + (3*c*(x^2 - 1))/(x^2 + 1) + (2*3^(1/2)*c*x)/(x^2 + 1))*(2*l^2*(sin(theta31) - 1) - 2*l^2*(sin(theta11) - 1) - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h) - 2*l*cos(theta31)*(c + h) + (4*c*l*x)/(x^2 + 1) - (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta31))/(x^2 + 1) + (2*c*l*cos(theta31)*(x^2 - 1))/(x^2 + 1)))^2 + (((433*c)/250 + (433*h)/250 + 3*l + 2*l*sin(theta11) + 2*l*sin((2*pi)/3 + theta21) - (2*c*x)/(x^2 + 1) - (3^(1/2)*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta11) - 1) - 2*l^2*(sin(theta31) - 1) + 2*c*(c + h) - 2*c*l*cos(theta11) - 2*l*cos(theta11)*(c + h) + 2*l*cos(theta31)*(c + h) - (4*c*l*x)/(x^2 + 1) + (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta31))/(x^2 + 1) - (2*c*l*cos(theta31)*(x^2 - 1))/(x^2 + 1)) - ((433*c)/250 + (433*h)/250 - 3*l + 2*l*sin(theta11) + l*sin(theta31) + 3^(1/2)*l*cos(theta31) + (6*c*x)/(x^2 + 1) + (3^(1/2)*c*(x^2 - 1))/(x^2 + 1))*(2*l^2*(sin(theta21) - 1) - 2*l^2*(sin(theta11) - 1) - 2*c*(c + h) + 2*c*l*cos(theta11) + 2*l*cos(theta11)*(c + h) - 2*l*cos(theta21)*(c + h) + (4*c*l*x)/(x^2 + 1) - (2*c*(c + h)*(x^2 - 1))/(x^2 + 1) + (4*c*l*x*sin(theta21))/(x^2 + 1) + (2*c*l*cos(theta21)*(x^2 - 1))/(x^2 + 1)))^2;
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9楼2013-12-14 17:49:07
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denmarkrico

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对不起大家，我用万能公式替代后的方程式一楼的回复所示，此处的x相当于tan(theta0/2)
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10楼2013-12-14 17:51:45
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[基金申请] 今天放榜没戏了吧 +6 yuleib84 2026-08-19 7/350 2026-08-21 09:24 by zxg7980
[基金申请] 基金啊基金 +4 longfie172 2026-08-20 4/200 2026-08-21 08:58 by mark mao
[基金申请] 让我中一个面上吧！ +9 大萍1987 2026-08-20 11/550 2026-08-21 08:19 by littlezl
[基金申请] 时间戳今天，20号变了 +4 archvillain 2026-08-20 4/200 2026-08-21 08:10 by gatelove
[基金申请] 科研孤儿太难了 +14 我4大白菜 2026-08-20 15/750 2026-08-20 22:33 by 袁向阳007
[基金申请] 时间戳变了，能看出什么问题？ +18 基诺咪客 2026-08-17 23/1150 2026-08-20 17:19 by Godzela
[基金申请] 重要消息，中午系统在维护 +11 yuleib84 2026-08-18 12/600 2026-08-20 11:09 by xskun
[基金申请] filecode，4个jtjc了 +11 ziyangfang 2026-08-19 13/650 2026-08-19 21:28 by aasahr
[教师之家] 为什么余额宝的年化利率越来越低？主要原因有哪些？ +5 瞬息宇宙 2026-08-15 5/250 2026-08-19 20:23 by super2002521
[基金申请] 今天放榜吗？ +14 布布和一二 2026-08-19 15/750 2026-08-19 18:07 by gltch
[基金申请] filecode=后面第一个是大写字母 +10 wangze12014 2026-08-14 12/600 2026-08-19 16:56 by 苦难博士
[基金申请] 2027广东省杰青 +4 奶牛小黑 2026-08-15 10/500 2026-08-19 11:02 by wanfengnew
[基金申请] 朋友圈看到的 +6 wangzilk 2026-08-18 8/400 2026-08-19 10:55 by Haru815
[基金申请] 什么时候开奖？ +6 CrisMessi 2026-08-18 6/300 2026-08-19 08:06 by Equinoxhua
[基金申请] 明天放榜？ +5 Shxjjxjkx 2026-08-18 5/250 2026-08-18 18:14 by -大大大大大-
[论文投稿] 投稿咨询 +4 wwm09 2026-08-17 6/300 2026-08-18 15:36 by wwm09
[基金申请] 今天维护系统维护 祝所有人 高中 +8 gjjjzhong 2026-08-18 9/450 2026-08-18 13:01 by 家与远方
[基金申请] 93BebMhtakh前后11位开头都是大写 +4 且听虎啸 2026-08-17 5/250 2026-08-18 00:49 by 蔡棒棒菂
[基金申请] 感觉是下周放榜了 +6 angus9576 2026-08-17 11/550 2026-08-17 23:57 by angus9576
[基金申请] 各位道友，我要去昆明玩几天，回来见。 +7 Tide man 2026-08-14 8/400 2026-08-15 01:11 by arzu_hma
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