Yes, of course, but it does not do what I want:
why using an approximation which in some cases
could produce a wrong result?
For instance when the approximated value
is smaller to 3 by a value smaller than
machine precision, the result could be wrong.
I think it would be nice to
have floor(log(1331)/log(11)) evaluated to 3
but maybe it is complicated to implement/correct.
For this specific problem it would be
enough to have logs in any base.
This would solve the problem of extimating
the log in base 11 because:
floor(log(1331)) does indeed work.
Fabrizio
On Wed, 13 Dec 2006 sen1 at math.msu.edu wrote:
> On Wed, 13 Dec 2006, Fabrizio Caruso wrote:
>
> > Hi
> >
> > Should floor work on log?
> > In particular should
> > floor(log(1331)/log(11))
> > output 3?
> floor(float(log(1331)/log(11)))
>
> works.
>
> -sen
> >
> > Fabrizio
> >
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> >
>
>