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) > _______________________________________________ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
