On 9/8/2011 9:25 AM, Raymond Toy wrote:
> Setting radexpand:false makes f(x) and g(x) be different.
>
> It seems that maxima has simplified g(x) incorrectly by making
> sqrt(x^2) be x. This shouldn't happen since radexpand is true by
> default. So at most maxima should make g(x) be abs(x)/sqrt(1+sqrt(x)).
>
> Ray
>
>
> On Tue, Sep 6, 2011 at 1:25 PM, Richard Fateman
> <fateman at eecs.berkeley.edu <mailto:fateman at eecs.berkeley.edu>> wrote:
>
> f(x):=x/sqrt(1+sqrt(x));
> g(x):=sqrt(x^2/(1+sqrt(x)));
>
> f(x)-g(x) returns 0
>
> f(3)-g(3) returns 0
>
> f(-3) - g(-3) is NOT zero.
>
> indeed, f(x)-g(x) is not zero if x<0.... is there a remedy?
>
> RJF
>
>
>
> _______________________________________________
> Maxima mailing list
> Maxima at math.utexas.edu <mailto:Maxima at math.utexas.edu>
> http://www.math.utexas.edu/mailman/listinfo/maxima
>
>
oddly enough, this came up in a Mathematica mailing list, but there it
had to do with taylor series.
For maxima,
taylor(f(x),x,0,3) and ...g(x)... give the same expression, regardless
of radexpand.
This is, I think, correct. The series are identical but with different
radii of convergence?
This perhaps opens up another option for taylor series... that is, on
the off chance that someone can made some useful observation about the
function represented here,
namely something about convergence, maybe add that info to the series data.
There is a literature in numerical analysis that attempts to "compute"
from some number of terms of a series its radius of convergence. This
is obviously heuristic in nature, and presumably makes assumptions about
how nasty the components are of a function being expanded. I haven't
looked at it in many years, but perhaps an interested person could look
at it.
RJF