On 12/20/06, miguel lopez <miguel39123 at hotmail.com> wrote:
> The following is really horrible
>
> (%i3) tlimit(sin(sin(sin(x)^5)^3)/x^15,x,0);
> sin(sin(x)^5)^3 Assumed to be zero in `taylor'...
> (%o3) 0
> A new rule is necessary:
...
> I think that in about 1/2 - 1 hour I can write some batteries: The rule is to
> replace in products and powers f(u) with the first non zero taylor polynomial of
> f. But to be smarter follow replacing to see if last one pay the beer.
I haven't looked at the code, but the underlying problem seems to be
in the taylor routine.
ex: sin(sin(sin(x)^5)^3)/x^15$
taylor(ex,x,0,1) => "assumed to be zero" messages
taylor(ex,x,0,3) => OK
So when tlimit calls taylor, perhaps it needs to increase the number
of terms until taylor gets it right (i.e. doesn't assume anything).
Even better, of course, if taylor could get it right in the first
place.
The danger is that there may be pathological cases (essential
singularities?) where this process does not terminate -- that no
matter how many terms you ask for, taylor gives up.
-s