Re: [Gretl-devel] fdjac: fdjac(x=10^-9,exp(x))=0 instead of 1!
Dear Allin, Thanks for mentioning fdjac2.c: It helped me to find quickly weak spot: Near
zero the step h will be inadequately small
set fdjac_quality 2
x={10^-6}
eval fdjac(x,exp(x))
eval exp(10^-6)
x={10^-7}
eval fdjac(x,exp(x))
eval exp(10^-7)
x={10^-8}
eval fdjac(x,exp(x))
eval exp(10^-8)
x={10^-9}
eval fdjac(x,exp(x))
eval exp(10^-9)
x={10^-12}
eval fdjac(x,exp(x))
eval exp(10^-12)
29 грудня 2015, 03:39:57, від "Allin Cottrell" <cottrell(a)wfu.edu>:
On Tue, 29 Dec 2015, oleg_komashko(a)ukr.net wrote:
> Dear Riccardo, My simple script uses exactly Richardson
> extrapolation but it uses a simplistic method for defining the
> step. 3 digits better is by far too large. Clearly, fdjac uses a
> more advanced way to find h. It looks, as if it has a buglet
> inside.
I don't think there's any bug in Jack's Richardson code. In fact I
believe the only reason you're getting more accurate results in this
case is that you got lucky with your "simplistic method" for setting
h. If I just change the code in our minpack-derived fdjac2.c to use
your choice of
h = 1000 * sqrt(DBL_EPSILON)
in place of
h = fabs(x[j]) * sqrt(DBL_EPSILON)
I get a value for sin'(0.1) which is good to 12 decimal places!
I don't suppose this would be consistent across arbitrary functions,
but it may be worth experimenting with a bigger h, or a different
formula for h, in the "quality = 2" case.
Allin
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