Thanks for the help Vinzent.
I am actually not looking for infinite accuracy but I need some answer
rather then just an error message that Perturb your initial value.
I know it is not trivial but i think Mathematica somehow should have
an option for perturbing the initial point itself.
Thanks again,
Nandan.
On Jun 27, 1:55 pm, Vinzent Steinberg
<vinzent.steinb...@googlemail.com> wrote:
> On Jun 26, 5:44 pm, nandan jha <nandanjh...@gmail.com> 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]
>
> ^^^^^^^^
> I'm not familiar with Mathematica, but does this mean you want
> infinite accuracy? This seems hard to achieve. :)
>
> > -------------------------------------------------------------------------------------------------------------------------------------------------------------
>
> > Can someone tell me if this can be solved in sympy
>
> In sympy use nsolve(), but this uses simply Newton's method and
> requires a very good starting point (so rather try scipy). Mathematica
> is more advanced I think.
>
> > and shouldn't the
> > Mathematica itself change the initial value and do a computation to
> > get the answer.
>
> That is sadly not trivial.
>
> Vinzent
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