是否有可能在Matlab中得到以下方程的显式解?

时间:2015-04-19 19:37:19

标签: matlab equation-solving equations

注意,问题已被编辑。

我按照建议设置限制并收到以下结果:

syms f1 f2 M b1 b2
assume (M > 1 & b1 > 0 & b2 > 0)
S = solve ( [(f1^M)/(f2^M) - M/(b1*f1) + M + 1 == 0, ...
(f2^M)/(f1^M) - M/(b2*f2) + M + 1 == 0], [f1,f2] )

Warning: The solutions are parameterized by the symbols: z, z1. To include parameters and conditions in the solution, specify the 'ReturnConditions' option. 
> In solve>warnIfParams at 511
  In solve at 361 


 Warning: The solutions are valid under the following conditions: M + z^M/z1^M - M/(b1*z) + 1 == 0 & M + z1^M/z^M - M/(b2*z1) + 1 == 0 & z < 0 & z1 < 0. To include parameters and conditions in the solution, specify the 'ReturnConditions' option. 


> In solve>warnIfParams at 518
  In solve at 361 

S = 

f1: [1x1 sym]
f2: [1x1 sym]

即使我将 f1 f2 的值限制为负值,我也会收到与上面相同的输出。

根据上述等式,我试图复制出现在Hirshleifer J.(1995),“无政府主义及其崩溃”中的反应曲线。 学家政治经济学(如果您需要更多信息)

2 个答案:

答案 0 :(得分:1)

很可能你没有得到一个明确的,封闭的解决方案,因为没有任何一个适用于任何真实Mb1b2的一般情况。

特别是对于案例M = 0,两个方程都简化为

  (...)^0 + 0 / (...) + 0 + 1   =   1 + 0 + 0 + 1   =   2    =    0

显然根本没有解决方案。您如何期待Matlab处理这个问题?

如果您对Mb有一些额外限制,则应尝试添加它们。

答案 1 :(得分:-1)

@mbschenkel,我按照建议设置限制并收到以下结果:

syms f1 f2 M b1 b2
assume (M > 1 & b1 > 0 & b2 > 0)
S = solve ( [(f1^M)/(f2^M) - M/(b1*f1) + M + 1 == 0, ...
(f2^M)/(f1^M) - M/(b2*f2) + M + 1 == 0], [f1,f2] )

Warning: The solutions are parameterized by the symbols: z, z1. To include parameters and conditions in the solution, specify the 'ReturnConditions' option. 
> In solve>warnIfParams at 511
  In solve at 361 


Warning: The solutions are valid under the following conditions: M + z^M/z1^M - M/(b1*z) + 1 == 0 & M + z1^M/z^M - M/(b2*z1) + 1 == 0 & z < 0 & z1 < 0. To include parameters and conditions in the solution, specify the 'ReturnConditions' option. 


> In solve>warnIfParams at 518
  In solve at 361 

S = 

f1: [1x1 sym]
f2: [1x1 sym] 

即使我将 f1 f2 的值限制为负值,我也会收到与上面相同的输出。

根据上述等式,我试图复制出现在Hirshleifer J.(1995),“无政府主义及其崩溃”中的反应曲线。 学家政治经济学(如果您需要更多信息)