Re: Why is precision(sqrt(0)) < precision(sqrt(1)) ??

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

* Jacques Gélinas [2016-02-29 22:12]:
> Now, I can understand lowering the precision by half for sqrt(0.), 
> but not for sqrt(0)

This is the unfortunate result of a generic convention that (most of the
time) an exact input is converted to floating point before a
"transcendental" function is applied, so sqrt(0) really calls sqrt(0.),
which really means sqrt(eps) for some 0 <= eps < 2^-bitprecision.
The result is then some eps' such that 0 <= eps' < 2^(-bitprecision/2)

This is arguably a bug since *some* functions (e.g. cos / sin) avoid
the  preliminary conversion that when it would lead to a loss of accuracy.
Compare sin(10^38) and sin(1e38).

I'll fix that.

> and then why is is not also done for sqrt(1)  ?

Because there is no loss of accuracy there : (1+eps)^(1/2) ~ 1 + eps/2,
we even gain 1 bit :-)

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 69 50
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