∫C e^z (z-1)(z-2)^2 / sin^2 pi z
where C is the circle │z│= 2.5
residues at the other points as I still don't have agreement with my numerical
integral (he said giving away the secret of why he was not happy with his previous
But this is the problem I have. At 0 for instance:
is analytic so the residue is (2 pi i times) the coefficient of x in the taylor series
of this about , which of course now that I compute it I find it non zero
(where before it seemed to be 0 )
So now I am with you.
RonLCode:This is EULER, Version 2.3 RL-06. Type help(Return) for help. Enter command: (20971520 Bytes free.) Processing configuration file. Done. > > >dthe=0.001*pi; > >the=dthe/2:dthe:2*pi; >rc=2.5; >z=rc*exp(I*the); > >fz=exp(z)*(z-1)*(z-2)^2/(sin(pi*z))^2; > > >..numerical integral: > >II=sum(fz*rc*I*exp(I*the))*dthe 7.26249e-014+4.29033i > > >e=exp(1) 2.71828 > >..TD!'s residues > >(-8/e^2/pi^2+3/e/pi^2+4/pi^2+e/pi^2)*2*pi*I 0+4.29033i >
their agreement is a sanity check.
It is always important to be able to have two or more independent
techniques available. The ability of the mind to see what it wants is
always with us. An idependent check helps prevent us being carried away
(resolution disagreements also helps develop technique and understanding
of the limitations of our methods).
Finally, it is also instructive to set about computing the integral numericaly
a few things becom clear that otherwise would remain hidden.