I'm trying to solve an admittedly gnarly system of equations:

h11 = 2;
NN = 2;
A = Array[a, NN];
Q = Array[q, {h11, NN}];
\[Rho] = Array[r, h11];
eqs = Table[Sum[A[[i]] Q[[nn, i]] Exp[-Sum[Q[[j, i]] \[Rho][[j]], {j,1,h11}]], {i, 1, NN}] == 0, {nn, 1, h11}];
Reduce[eqs, \[Rho]]

The question comes in two parts:

1) For h11=2, NN=2, it can solve this, but for any higher values it says it can't do it. Is the system really too complicated for Mathematica?

2) For h11=2, NN=2, it solves the equations, but it seems to put a condition on the Q's. It gives a solution for r[2] in terms of r[1], but it also specifies that $$q[2,2] = \frac{q[1,2] q[2,1]}{q[1,1]}$$. Am I really to believe that the only way to get a solution to this equation is if the q's satisfy this relationship? How can I be sure that Mathematica is giving me all the solutions?

  • 1
    $\begingroup$ Mathematica (11.0.1 Windows ) Solve[eqs, \[Rho] ] (*{}*) states there is no solution! $\endgroup$ – Ulrich Neumann Nov 17 '18 at 16:45
  • $\begingroup$ For 2x2 it looks like your system simplifies to {exponential1 == -exponential2, exponential3 == -exponential4} and you can take logs of both sides reducing it to (roughly) a pair of linear equations, with a couple of complexes thrown in. All your q variables become coefficients and seem to have to have particular values for the system to have a solution. If you don't want any conditions on the q's then what else is left to form a solution? For 3x3 or greater logs of sums don't simplify unless they are very special. I'd guess that is why there is no simple closed form solution. $\endgroup$ – Bill Nov 18 '18 at 4:32

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