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denmarkrico

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[求助] 求解高次復雜函數(shù) 已有3人參與

求解下面這個高次方程時為什么總是得到虛數(shù)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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心那個涼啊

1樓 2013-12-13 22:28:48

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feixiaolin

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【答案】應助回帖

感謝參與，應助指數(shù) +1

1）theta0 待求還是已知？
2）取y1=(x^2 - 1))/(x^2 + 1);
y2=x/(x^2 + 1);
先求y1, y2，最后求 x如何？

贊一下(1人)

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» 本帖已獲得的紅花（最新10朵）

denmarkrico

2樓2013-12-13 22:53:29

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ynod23

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【答案】應助回帖

感謝參與，應助指數(shù) +1

函數(shù)沒有看完，瞎了。。。
我認為，這種情況，不是查程序哪兒出問題了，而應該看方法對不對。。。。
先試試看能不能把那個大函數(shù)寫成許多函數(shù)相互運算的結果，這樣就降低了出錯的可能。如果這樣還有問題的話，也可以根據(jù)matlab的運行提示，找出是哪一行的問題。這樣把問題的規(guī)�？s小，我覺得比較靠譜。
希望能幫到你。

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denmarkrico

5樓2013-12-14 04:43:11

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普通回帖

jerkwin

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你這方程, 能解出來我覺得很奇怪

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3樓2013-12-13 23:10:14

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paopaoai11

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你這什么方程啊？

[ 發(fā)自小木蟲客戶端 ]

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泡泡

4樓2013-12-14 01:03:00

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denmarkrico

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送紅花一朵

引用回帖:

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

你這個替代法我見過

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心那個涼啊

6樓2013-12-14 16:07:30

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denmarkrico

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送紅花一朵

引用回帖:

5樓: Originally posted by ynod23 at 2013-12-14 04:43:11
函數(shù)沒有看完，瞎了。。。
我認為，這種情況，不是查程序哪兒出問題了，而應該看方法對不對。。。。
先試試看能不能把那個大函數(shù)寫成許多函數(shù)相互運算的結果，這樣就降低了出錯的可能。如果這樣還有問題的話，也可 ...

嗯謝謝你我這是機器人的正解

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心那個涼啊

7樓2013-12-14 16:08:44

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denmarkrico

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引用回帖:

2樓: Originally posted by feixiaolin at 2013-12-13 22:53:29
1）theta0 待求還是已知？
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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心那個涼啊

8樓2013-12-14 16:09:58

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denmarkrico

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引用回帖:

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

要解的方程是這樣的：

((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;

贊一下

回復此樓

心那個涼啊

9樓2013-12-14 17:49:07

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對不起大家，我用萬能公式替代后的方程式一樓的回復所示，此處的x相當于tan(theta0/2)

贊一下

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心那個涼啊

10樓2013-12-14 17:51:45

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