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During a calculation for a physics lab, I ran into the following six non-linear equations with unknowns $x_1,x_2,x_3,x_4,x_5,x_6 \in \mathbb{R}$

Solve[{(1 + x1*0.405^2/((0.405)^2 - x2) + x3*0.405^2/((0.405)^2 - x4) + x5*0.405^2/((0.405)^2 - x6) == 2.337053326), (1 + x1*(0.635)^2/((0.635)^2 - x2) + x3*(0.635)^2/((0.635)^2 - x4) + x5*(0.635)^2/((0.635)^2 - x6) == 2.293038467), (1 + x1*(0.670)^2/((0.670)^2 - x2) + x3*(0.670)^2/((0.670)^2 - x4) + x5*(0.670)^2/((0.670)^2 - x6) == 2.289560279), (1 + x1*(0.780)^2/((0.780)^2 - x2) + x3*(0.780)^2/((0.780)^2 - x4) + x5*(0.780)^2/((0.780)^2 - x6) == 2.277436534), (1 + x1*(0.808)^2/((0.808)^2 - x2) + x3*(0.808)^2/((0.808)^2 - x4) + x5*(0.808)^2/((0.808)^2 - x6) == 2.275719732), (1 + x1*(0.850)^2/((0.850)^2 - x2) + x3*(0.850)^2/((0.850)^2 - x4) + x5*(0.850)^2/((0.850)^2 - x6) == 2.272264106), (1 + x1*(0.880)^2/((0.880)^2 - x2) + x3*(0.880)^2/((0.880)^2 - x4) + x5*(0.880)^2/((0.880)^2 - x6) == 2.268823405), (1 + x1*(0.980)^2/((0.980)^2 - x2) + x3*(0.980)^2/((0.980)^2 - x4) + x5*(0.980)^2/((0.980)^2 - x6) == 2.265389225)}, {x1, x2, x3, x4, x5, x6}]

This code does not yield any output by Mathematica, and I cannot figure out wether I'm using the wrong command or wrong syntax or something else...

Assuming that Mathematica knows methods to solving this, how do I tell the program to do so and return the values of my $x_1,\ldots,x_6$ ?

Any help is very much appreciated. ;)

EDIT: By deleting the last two equations and writing it as

Solve[ 1 + x1*0.405^2/((0.405)^2 - x2) + x3*0.405^2/((0.405)^2 - x4) + x5*0.405^2/((0.405)^2 - x6) == 2.337053326 && 1 + x1*(0.635)^2/((0.635)^2 - x2) + x3*(0.635)^2/((0.635)^2 - x4) + x5*(0.635)^2/((0.635)^2 - x6) == 2.293038467 && 1 + x1*(0.670)^2/((0.670)^2 - x2) + x3*(0.670)^2/((0.670)^2 - x4) + x5*(0.670)^2/((0.670)^2 - x6) == 2.289560279 && 1 + x1*(0.780)^2/((0.780)^2 - x2) + x3*(0.780)^2/((0.780)^2 - x4) + x5*(0.780)^2/((0.780)^2 - x6) == 2.277436534 && 1 + x1*(0.808)^2/((0.808)^2 - x2) + x3*(0.808)^2/((0.808)^2 - x4) + x5*(0.808)^2/((0.808)^2 - x6) == 2.275719732 && 1 + x1*(0.850)^2/((0.850)^2 - x2) + x3*(0.850)^2/((0.850)^2 - x4) + x5*(0.850)^2/((0.850)^2 - x6) == 2.272264106 , {x1, x2, x3, x4, x5, x6} ]

the computer returns a result, but one that I do not understand. The following output was generated by Mathematica:

{{x1 -> 1.27242, x2 -> 0.00816205, x3 -> -0.000117262, x4 -> 0.563865, x5 -> 0.0192823, x6 -> 1.70608}, {x1 -> 1.27242, x2 -> 0.00816205, x3 -> 0.0192823, x4 -> 1.70608, x5 -> -0.000117262, x6 -> 0.563865}, {x1 -> -0.000117262, x2 -> 0.563865, x3 -> 1.27242, x4 -> 0.00816205, x5 -> 0.0192823, x6 -> 1.70608}, {x1 -> -0.000117262, x2 -> 0.563865, x3 -> 0.0192823, x4 -> 1.70608, x5 -> 1.27242, x6 -> 0.00816205}, {x1 -> 0.0192823, x2 -> 1.70608, x3 -> 1.27242, x4 -> 0.00816205, x5 -> -0.000117262, x6 -> 0.563865}, {x1 -> 0.0192823, x2 -> 1.70608, x3 -> -0.000117262, x4 -> 0.563865, x5 -> 1.27242, x6 -> 0.00816205}}

Shouldn't this just be one value for each of the variables, instead of six each???

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    $\begingroup$ Reduce returns false, which suggests that the equations are inconsistent. $\endgroup$ Commented Apr 27, 2021 at 19:41
  • $\begingroup$ You have 8 equations and only 6 unknowns. $\endgroup$ Commented Apr 27, 2021 at 19:54
  • $\begingroup$ Would deleting the last two equations yield a problem that can be solved by the program? (I care less about the actual answer than about the method of how to correctly asking Mathematica to do such a thing) $\endgroup$ Commented Apr 27, 2021 at 19:57
  • $\begingroup$ Yes it would (I know because I did that). The last two then have fairly large residuals so it seems unlikely that minor numeric tweaking would make the full set consistent. $\endgroup$ Commented Apr 27, 2021 at 20:05
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    $\begingroup$ Nonlinear equations often have multiple solutions. Mathematica finds six. $\endgroup$ Commented Apr 27, 2021 at 20:13

1 Answer 1

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One approach is to convert this to a minimisation problem. If we assume that the errors are in the values on the right hand side of each expression, and that these errors are independent and follow a Gaussian distribution the appropriate way to do it is, I think, as follows:

eqns = {(1 + x1*0.405^2/((0.405)^2 - x2) + x3*0.405^2/((0.405)^2 - x4) + x5*0.405^2/((0.405)^2 - x6) == 2.337053326), (1 + x1*(0.635)^2/((0.635)^2 - x2) + x3*(0.635)^2/((0.635)^2 - x4) + x5*(0.635)^2/((0.635)^2 - x6) == 2.293038467), (1 + x1*(0.670)^2/((0.670)^2 - x2) + x3*(0.670)^2/((0.670)^2 - x4) + x5*(0.670)^2/((0.670)^2 - x6) == 2.289560279), (1 + x1*(0.780)^2/((0.780)^2 - x2) + x3*(0.780)^2/((0.780)^2 - x4) + x5*(0.780)^2/((0.780)^2 - x6) == 2.277436534), (1 + x1*(0.808)^2/((0.808)^2 - x2) + x3*(0.808)^2/((0.808)^2 - x4) + x5*(0.808)^2/((0.808)^2 - x6) == 2.275719732), (1 + x1*(0.850)^2/((0.850)^2 - x2) + x3*(0.850)^2/((0.850)^2 - x4) + x5*(0.850)^2/((0.850)^2 - x6) == 2.272264106), (1 + x1*(0.880)^2/((0.880)^2 - x2) + x3*(0.880)^2/((0.880)^2 - x4) + x5*(0.880)^2/((0.880)^2 - x6) == 2.268823405), (1 + x1*(0.980)^2/((0.980)^2 - x2) + x3*(0.980)^2/((0.980)^2 - x4) + x5*(0.980)^2/((0.980)^2 - x6) == 2.265389225)}; vars = Union[Cases[eqns, _Symbol, ∞]];

The objective function is given by

obj = Total[eqns /. a_ == b_ -> (a - b)^2] (* (-1.33705 + (0.164025 x1)/(0.164025 - x2) + (0.164025 x3)/( 0.164025 - x4) + (0.164025 x5)/(0.164025 - x6))^2 + (-1.29304 + ( 0.403225 x1)/(0.403225 - x2) + (0.403225 x3)/(0.403225 - x4) + ( 0.403225 x5)/(0.403225 - x6))^2 + (-1.28956 + (0.4489 x1)/( 0.4489 - x2) + (0.4489 x3)/(0.4489 - x4) + (0.4489 x5)/( 0.4489 - x6))^2 + (-1.27744 + (0.6084 x1)/(0.6084 - x2) + ( 0.6084 x3)/(0.6084 - x4) + (0.6084 x5)/( 0.6084 - x6))^2 + (-1.27572 + (0.652864 x1)/(0.652864 - x2) + ( 0.652864 x3)/(0.652864 - x4) + (0.652864 x5)/( 0.652864 - x6))^2 + (-1.27226 + (0.7225 x1)/(0.7225 - x2) + ( 0.7225 x3)/(0.7225 - x4) + (0.7225 x5)/( 0.7225 - x6))^2 + (-1.26882 + (0.7744 x1)/(0.7744 - x2) + ( 0.7744 x3)/(0.7744 - x4) + (0.7744 x5)/( 0.7744 - x6))^2 + (-1.26539 + (0.9604 x1)/(0.9604 - x2) + ( 0.9604 x3)/(0.9604 - x4) + (0.9604 x5)/(0.9604 - x6))^2 *)

Minimize seeks a global minimum of this (but I haven't checked whether it succeeds in finding it)

soln = Minimize[obj, vars] (* {3.30637*10^-6, {x1 -> 1.36007, x2 -> -0.0064115, x3 -> -0.484192, x4 -> -0.0807497, x5 -> 0.361731, x6 -> -0.00424008}} *)

We can see how close the solution is to the original right hand side values

(List @@@ eqns) /. Last[soln] (* {{2.33707, 2.33705}, {2.29335, 2.29304}, {2.2889, 2.28956}, {2.27766, 2.27744}, {2.27535, 2.27572}, {2.27221, 2.27226}, {2.2702, 2.26882}, {2.26456, 2.26539}} *)

This looks like quite a good match.

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