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denmarkrico

银虫 (小有名气)

[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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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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2楼2013-12-13 22:53:29
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ynod23

金虫 (小有名气)

【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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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
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 ]
泡泡
4楼2013-12-14 01:03:00
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denmarkrico

银虫 (小有名气)

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
心 那个凉啊
6楼2013-12-14 16:07:30
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denmarkrico

银虫 (小有名气)

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
心 那个凉啊
7楼2013-12-14 16:08:44
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denmarkrico

银虫 (小有名气)

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
心 那个凉啊
8楼2013-12-14 16:09:58
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denmarkrico

银虫 (小有名气)

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;
心 那个凉啊
9楼2013-12-14 17:49:07
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denmarkrico

银虫 (小有名气)

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)
心 那个凉啊
10楼2013-12-14 17:51:45
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[考研] 材料调剂 +10 懒羊羊轻置玉臀 2026-04-02 11/550 2026-04-04 21:56 by laoshidan
[考研] 341求调剂 +3 洛多罗 2026-04-02 4/200 2026-04-04 21:36 by 智能智慧
[考研] 368求调剂 +5 今华习 2026-04-03 7/350 2026-04-04 18:47 by imissbao
[考研] 0703求调剂 +6 zizimo 2026-03-31 6/300 2026-04-04 14:16 by 无际的草原
[考研] 0710生物学336分求调剂 +6 kiyy 2026-04-01 8/400 2026-04-04 10:10 by kiyy
[考研] 调剂 +5 asdasdassda 2026-04-03 6/300 2026-04-03 20:27 by 岸上的一条鱼
[考研] 266求调剂 +18 阳阳哇塞 2026-04-01 18/900 2026-04-03 18:38 by zllcz
[考研] 一志愿华东理工大学,080500学硕,317分,求调剂 +13 s1145 2026-03-31 15/750 2026-04-03 11:44 by msi123
[考研] 材料专业求调剂 +10 月月鸟木 2026-04-01 10/500 2026-04-02 12:57 by wxiongid
[考研] 340求调剂 +4 希望如此i 2026-03-31 4/200 2026-03-31 16:40 by 690616278
[考研] 一志愿浙江大学工科动力工程370,数一121,专业课135,现在能去哪里 +3 080700调剂 2026-03-30 4/200 2026-03-31 12:00 by KLMY666
[考研] 274求调剂 +6 xiao爱同学 2026-03-30 6/300 2026-03-31 10:04 by cal0306
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