nandan jha wrote:
> Hello
>
> I am trying to solve a set of non-linear equations in Mathematica 7.0
> and every time it shows a error message :
> "FindRoot::jsing: Encountered a singular Jacobian at the point
> {B1,B2,B3,Ea1,Ea2,Ea3} =
> {31.501,0.9004,38.5013,-1000.1,-1000.01,-2000.1}. Try perturbing the
> initial point(s). >> ".
>
> Sometimes it also shows this error message: "FindRoot::lstol: The line
> search decreased the step size to within tolerance specified by
> AccuracyGoal and PrecisionGoal but was unable to find a sufficient
> decrease in the merit function. You may need more than
> MachinePrecision digits of working precision to meet these tolerances.
>
>>> " .
>>>
>
> The equations that I am trying to solve are:
> ------------------------------------------------------------------------------------------------------------------------------------------------------------------
> FindRoot[{Exp[B1 + Ea1*0.000374526] + Exp[B2 + Ea2*0.000374526] +
> Exp[B3 + Ea3*0.000374526] == 0.0183,
> Exp[B1 + Ea1*0.00037925] + Exp[B2 + Ea2*0.00037925] +
> Exp[B3 + Ea3*0.00037925] == 0.00995,
> Exp[B1 + Ea1*0.00038287] + Exp[B2 + Ea2*0.00038287] +
> Exp[B3 + Ea3*0.00038287] == 0.0075,
> Exp[2*B1 + 2*Ea1*0.000374526] + Exp[2*B2 + 2*Ea2*0.000374526] +
> Exp[2*B3 + 2*Ea3*0.000374526] +
> 2*Exp[B1 + B2 + (Ea1 + Ea2)*0.000374526] +
> 2*Exp[B2 + B3 + (Ea3 + Ea2)*0.000374526] -
> 2*Exp[B1 + B3 + (Ea1 + Ea3)*0.000374526] == 0.01784*0.01784,
> Exp[2*B1 + 2*Ea1*0.00037925] + Exp[2*B2 + 2*Ea2*0.00037925] +
> Exp[2*B3 + 2*Ea3*0.00037925] +
> 2*Exp[B1 + B2 + (Ea1 + Ea2)*0.00037925] +
> 2*Exp[B2 + B3 + (Ea3 + Ea2)*0.00037925] -
> 2*Exp[B1 + B3 + (Ea1 + Ea3)*0.00037925] == 0.00983*0.00983,
> Exp[2*B1 + 2*Ea1*0.00038287] + Exp[2*B2 + 2*Ea2*0.00038287] +
> Exp[2*B3 + 2*Ea3*0.00038287] +
> 2*Exp[B1 + B2 + (Ea1 + Ea2)*0.00038287] +
> 2*Exp[B2 + B3 + (Ea3 + Ea2)*0.00038287] -
> 2*Exp[B1 + B3 + (Ea1 + Ea3)*0.00038287] == 0.00742*0.00742}, {B1,
> 40.001}, {B2, 110.9004}, {B3,
> 47.001309}, {Ea1, -1000.1}, {Ea2, -1000.01}, {Ea3, -2000.1},
> MaxIterations -> Infinity, AccuracyGoal -> Infinity]
> -------------------------------------------------------------------------------------------------------------------------------------------------------------
>
> Can someone tell me if this can be solved in sympy and shouldn't the
> Mathematica itself change the initial value and do a computation to
> get the answer.
>
> Thanks for any help in advance.
> Nandan
>
> >
>
>
You might want to look at the root finders in scipy.
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