[Scilab-users] Cauchy Integral query

[Heinz Nabielek]

interesting...


> On 05.01.2022, at 17:39, Lester Anderson <arctica1963 at gmail.com> wrote:
> 
> I just checked my version and result:
> 
> 2.154D-13 + 343.05029i
> 
> --> [version, options] = getversion()
>  version  =  "scilab-6.1.1"
>  options  =  "VC++"  "x64"  "tk"  "release"  "Jul 15 2021"  "15:32:10"
> 
> Windows 10
> 
> Lester
> 
> 
> 
> On Wed, 5 Jan 2022 at 13:45, Heinz Nabielek <heinznabielek at me.com> wrote:
> Is there a reason that my round-off error is 2.831D-13 ?
> 
> Is there a way to get SciLab to print always 2.831e-13 so that I can copy numbers over to EXCEL?
> 
> Heinz
> 
> 
> --> 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  =    2.831D-13 + 343.05029i
> 
> --> 2*%pi*%i*%e^4
>  ans  =    0. + 343.05029i
> 
> Scilab Version: 6.1.1.988271013
> macOS Catalina Version 10.15.7 
> 
> 
> ______________-
> 
> > On 05.01.2022, at 14:31, 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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