[sage-support] Re: why is this contour integral wrong?

2011-12-06 Thread emil
Applying expand(ratsimp( )); to your %o15 one can obtain Dan's result. The problem is that the integral should not depend on the center of the circle containing the pole. It looks like maxima bug (?) Sometimes ago there was an example of failing complex calculations on ask.sagemath.org

[sage-support] Re: why is this contour integral wrong?

2011-12-06 Thread kcrisman
The problem is that the integral should not depend on the center of the circle containing the pole. It looks like maxima bug (?) I've reported this at https://sourceforge.net/tracker/?group_id=4933atid=104933 Dan, if you want to open a ticket, just be sure to refer to that. -- To post to

[sage-support] Re: why is this contour integral wrong?

2011-12-06 Thread achrzesz
On Dec 6, 3:26 pm, kcrisman kcris...@gmail.com wrote: The problem is that the integral should not depend on the center of the circle containing the pole. It looks like maxima bug (?) I've reported this athttps://sourceforge.net/tracker/?group_id=4933atid=104933 Dan, if you want to

[sage-support] Re: why is this contour integral wrong?

2011-12-05 Thread achrzesz
On Dec 5, 5:31 am, Dan Drake dr...@kaist.edu wrote: I keep wondering whether Sage is making a mistake, or I'm not understanding complex analysis. I'm a little afraid to learn the answer. :) Take f(z) = (z-I)*(z-1)^2/(z-(-1/2-I/3)). It's analytic everywhere except at -1/2-I/3, where it has

[sage-support] Re: why is this contour integral wrong?

2011-12-05 Thread achrzesz
On Dec 5, 10:04 am, achrzesz achrz...@wp.pl wrote: On Dec 5, 5:31 am, Dan Drake dr...@kaist.edu wrote: I keep wondering whether Sage is making a mistake, or I'm not understanding complex analysis. I'm a little afraid to learn the answer. :) Take f(z) = (z-I)*(z-1)^2/(z-(-1/2-I/3)).

[sage-support] Re: why is this contour integral wrong?

2011-12-05 Thread kcrisman
So it sounds like you should file a ticket, Dan. Maybe we're just sending it to Maxima wrong. (%i9) f(z):=(z-%i)*(z-1)^2/(z-(-1/2-%i/3)); 2 (z - %i) (z - 1) (%o9) f(z) := -

[sage-support] Re: why is this contour integral wrong?

2011-12-05 Thread achrzesz
On Dec 5, 4:56 pm, kcrisman kcris...@gmail.com wrote: So it sounds like you should file a ticket, Dan.  Maybe we're just sending it to Maxima wrong. (%i9) f(z):=(z-%i)*(z-1)^2/(z-(-1/2-%i/3));                                                    2                                    (z - %i)

[sage-support] Re: why is this contour integral wrong?

2011-12-04 Thread achrzesz
On Dec 5, 5:31 am, Dan Drake dr...@kaist.edu wrote: I keep wondering whether Sage is making a mistake, or I'm not understanding complex analysis. I'm a little afraid to learn the answer. :) Take f(z) = (z-I)*(z-1)^2/(z-(-1/2-I/3)). It's analytic everywhere except at -1/2-I/3, where it has