I need advice on how to use radcan



Dear Mr. Fateman,

>> Obviously, the mathematically correct answer would read abs(b-a) or
>> abs(a-b).
> Actually, this is not obvious.
of course you are right; I should not write emails concerned with 
mathematics late in the evening... sorry!

The conceptual difficulty I had was how to tell maxima/radcan that the 
argument to sqrt in fact is a squared distance and therefore, a 
non-negative, real result for sqrt is required in the following 
calculations in order to yield meaningful results. I now proceed by 
substituting the argument by a helper variable d^2. The calculations now 
require a lot of subst()-commands, but I'm not frightened anymore.  The 
final result turned out to be identical to that obtained using radcan.

> Richard Fateman (who first wrote radcan in 1969 or so)
I wasn't even born at that time! I think I wrote my first Fortran code 
in 1994.

Regards
Dragan Stoikovitch