ASKSAGE: Sage Q&A Forum - Latest question feedhttp://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Wed, 18 Apr 2012 02:38:46 -0500Weird output for differential of a non-analytic complex function.http://ask.sagemath.org/question/8899/weird-output-for-differential-of-a-non-analytic-complex-function/The following code,
var('z')
f = z*z.conjugate()
f.derivative(z)
produces
z*D[0](conjugate)(z) + conjugate(z)
Now I undestand where the problem might be, $z$ and $z^*$ are usually treated as independent variables thus. But does sage mean by this output?d3banjanWed, 18 Apr 2012 02:38:46 -0500http://ask.sagemath.org/question/8899/Complex forms and differentialshttp://ask.sagemath.org/question/8060/complex-forms-and-differentials/Can Sage handle complex differential forms and the \partial and \bar\partial operators? I tried naively defining the d-bar operator before as (in Latex) \bar d \bar u, but I got an error saying that conjugation wasn't defined for the differential form class. Is there a way of handling this other than defining the real and imaginary parts of all variables and brutally defining the \partial and \bar\partial operators?GunnarWed, 06 Apr 2011 07:34:59 -0500http://ask.sagemath.org/question/8060/Differentiating Complex Conjugated Functionshttp://ask.sagemath.org/question/7780/differentiating-complex-conjugated-functions/This is primarily a question of understanding the syntax of some output although there might be a bug hidden underneath. Consider the following code:
sage: var('x,t')
sage: q = function('q',x,t)
sage: f = q*q.conjugate()
sage: print f.derivative(x,1)
q(x, t)*D[0](conjugate)(q(x, t))*D[0](q)(x, t) + conjugate(q(x,t))*D[0](q)(x, t)
The answer is supposed to be $d/dx(q\bar{q}) = q_x \bar{q} + q \bar{q}_x$. The second term in the Sage output is correct but I'm having trouble deciphering the first term. Any thoughts?
I think I can narrow down the differences even further. Check it out:
sage: print q.conjugate().derivative(x,1)
D[0](conjugate)(q(x, t))*D[0](q)(x, t)
sage print q.derivative(x,1).conjugate()
conjugate(D[0](q)(x, t))
The independence of order isn't the issue: $q = u + iv$ means that $q_x = u_x + iv_x$, $\bar{q} = u - iv$. So $\bar{q_x} = u_x - iv_x$ and $(\bar{q})_x = (u - iv)_x = u_x - iv_x$.cswierczTue, 30 Nov 2010 09:18:04 -0600http://ask.sagemath.org/question/7780/