[Scilab-users] Cauchy Integral query

[Lester Anderson]

Many thanks Samuel.
Happy New Year!

Lester

On Wed, 5 Jan 2022 at 13:32, <sgougeon at free.fr> wrote:

> Hello Lester,
>
> The integrand is y = exp((z^2))/(z-2), not y = exp((z^2)).
> Then, provided that the (undocumented) absolute tolerance is increased wrt
> the default one,
> we get the expected result:
>
> --> function y=f(z)
>   >   y = exp((z.^2))./(z-2)
>   > endfunction
>
> --> fz=intl(0, 2*%pi, 2, 1, f,1e-10) // gives round-off error
>  fz  =
>    4.199D-13 + 343.05029i
>
> --> 2*%pi*%i*%e^4
>  ans  =
>    0. + 343.05029i
>
> Regards
> Samuel
>
> > ----- Mail d'origine -----
> > De: Lester Anderson
> > Envoyé: Wed, 05 Jan 2022 09:46:47 +0100 (CET)
> >
> > Hello,
> >
> > I am trying to understand how to work the Cauchy integral inputs and
> > replicate the results of a published example:
> >
> > .e.g. Compute the integral of e^(z^2) / (z-2) assumes C is closed
> > (anticlockwise) and z=2 is inside C (a simple circle). The solution
> should
> > be 2*pi*i*f(2) = 2*pi*i*e^4
> >
> > In Scilab, the solution is defined from the Cauchy Integral (intl):
> > y = intl(a, b, z0, r, f)
> > a and b are real and z complex
> >
> > function y=f(z)
> >   y = exp((z^2)) // solution uses f(z) =  e^(z^2)
> > endfunction
> >
> > fz=intl(0, 2*%pi, 2+0*%i, 1, f) // gives round-off error
> > // z position +2(real z), 0(imaginary z)
> >
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