Hi Troels,

For large negative values, the exponential value should be zero.  So
returning zero is to be expected.  Negative values usually only occur
at the end of the exponential curve once it has decayed to statical
zero (i.e. it's value is less than the noise).  Then a return value of
zero from func() is perfectly normal, and minfx will handle this.

Regards,

Edward




On 31 August 2014 14:13, Troels E. Linnet
<[email protected]> wrote:
> Update of bug #22552 (project relax):
>
>                 Priority:           9 - Immediate => 5 - Normal
>                 Severity:           4 - Important => 2 - Minor
>
>     _______________________________________________________
>
> Follow-up Comment #1:
>
> Corrections to this statement.
>
> The C-code was not initialized with function values.
> Therefore a return with 0.0
>
> from target_functions.relax_fit import jacobian, jacobian_chi2, setup
> from specific_analyses.relax_fit.optimisation import func_wrapper,
> dfunc_wrapper, d2func_wrapper
> func = func_wrapper
>
> times = array([ 0.7,  1. ,  0.8,  0.4,  0.9])
> I = array([ 476.76174875,  372.43328777,  454.20339981, 656.87936253,
> 419.16726341])
> errors = array([  9.48032653,  11.34093541,   9.35149017,  10.84867928,
> 12.17590736])
>
> scaling_list = [1.0, 1.0]
> setup(num_params=2, num_times=len(times), values=I, sd=errors,
> relax_times=times, scaling_matrix=scaling_list)
>
> R = - 500.
> I0 = 1000.
> params = [R, I0]
>
> chi2 = func(params)
>
> The question is, if minfx can handle inf values?
>
>
>     _______________________________________________________
>
> Reply to this item at:
>
>   <http://gna.org/bugs/?22552>
>
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