Dear Allin,

Thanks for mentioning fdjac2.c:

It helped me to find quickly

weak spot:

Near zero the step h will be

inadequately small

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@wfu.edu>:

> On Tue, 29 Dec 2015, oleg_komashko@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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