Hi Richard,
Thanks for the reply. Please see my responses in context below.
On Fri, 2012-11-30 at 07:20 -0800, Richard Fateman wrote:
> You say that the form of the exponentials differ.
>
> What do the forms look like that you think should be the same?
If I run the sample code I sent (with display2d:false) and look for the
part beginning with aa[6] I get:
aa[6] = 0
bb[6] = -t[0]*xi^2*D-k[0]^2*t[3]*D-t[0]*Gam
ar = -2*k[0]^2*S(k[0])*K[0](k[0],xi)*Gam*xi^3*D^2
*%e^(t[0](-xi^2*D-Gam)-k[0]^2*t[3]*D)
-2*k[0]^4*S(k[0])*K[0](k[0],xi)*Gam*xi*D^2
*%e^(t[0](-xi^2*D-Gam)-k[0]^2*t[3]*D)...
aa[6] = 0 means that it found nothing multiplying
exp(bb[6])=exp(-t[0]*xi^2*D-k[0]^2*t[3]*D-t[0]*Gam)
ar is the remaining expression (i.e., what I started with less what has
already been parsed). The 2nd line in ar contains
%e^(t[0](-xi^2*D-Gam)-k[0]^2*t[3]*D)
which would match if the argument were expanded and rearranged.
>
> Note that radcan does not necessarily choose the same branch of
> an algebraic function that you might. It also changes exp(a+b)/exp(a).
I'm aware of that, but radcan isn't invoked while parsing the
expression, and is supposed to give zero. Also, there are no algebraic
expressions here so the branch isn't relevant.
> So far as I can tell, none of the "assume" commands should affect
> anything you are doing.
You're probably correct, but there were needed in the real problem and I
wanted to make sure that the environment used for ordering the
expressions was the same.
>
>