Re: sqrtint() on non-integer arguments

[Karim Belabas <Karim.Belabas@math.u-bordeaux.fr>]

* Kevin Ryde [2021-10-28 00:46]:
> Karim Belabas <Karim.Belabas@math.u-bordeaux.fr> writes:
> >
> > get instead 
> > ? sqrtint(1e40)
> > %1 = 100000000000000000000
> 
> Could keep that as precision loss error, since low bits of the root are
> essentially unknown, within selected float precision.

I have to assume that the input is correct as given otherwise the
original request is meaningless anyway: even floor(1.0) would not be
well defined if 1.0 may represent any real number within 1ulp
(the floor could be 0 or 1).

And the largest integer x such that x^2 <= 1e40 is indeed 10^20.

I chose to compute sqrtint(y) as floor(sqrt(y)) or sqrtint(floor(y))
if either is defined [if both are defined, they are equal; and the first
expression is always defined when the ѕecond is]

Cheers,

    K.B.
--
Karim Belabas, IMB (UMR 5251)  Tel: (+33) (0)5 40 00 26 17
Universite de Bordeaux         Fax: (+33) (0)5 40 00 21 23
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