David Roe wrote :
I posted a patch there that should fix it; I have to work on other
stuff, but if someone else wants to take over and write some doctests,
make sure it works in lots of cases...
David
On Mon, Oct 25, 2010 at 17:14, Burcin Erocal <bur...@erocal.org
<mailto:bur...@erocal.org>> wrote:
On Mon, 25 Oct 2010 17:00:39 -0400
David Roe <r...@math.harvard.edu <mailto:r...@math.harvard.edu>>
wrote:
> This is a good workaround, but the original problem can be traced to
> the function sage.symbolic.expression.Expression.__int__
>
> def __int__(self):
> #FIXME: can we do better?
> return int(self.n(prec=100))
>
> Presumably you could adaptively estimate to higher precision until
> your error interval included only one integer...
This is #9953 on trac:
http://trac.sagemath.org/sage_trac/ticket/9953
I prefer
return(self.n(prec=ceil(log(abs(self.n()),10)+10)))
But this answer isn't right around XXXXXXXXXXXX.999999999999 or
XXXXXXXXXXX.000000001.
And I don't see where floor(10^1234*sqrt(2)) is wrong. I test :
vt=floor(10^1234*sqrt(2)) ; sign(vt^2-2*10^2468) ; sgn((vt+1)^2-2*10^2468)
# answer =-1 and 1 for sage 4.5.3
With int the result is false (I get 1 and 1 in the both cases).
I ignore how to trace these 2 calculus.
F.
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