[Scilab-users] Re : Cauchy Integral query

[sgougeon at free.fr]

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)
>