Re: floor function not doing what it should

[Bill Allombert <Bill.Allombert@math.u-bordeaux1.fr>]

On Sun, Mar 14, 2010 at 06:54:54AM +0100, ewan.Delanoy@math.unicaen.fr wrote:
> 
>    Dear PARI-GP users,
> 
>   my problem can be summed up in a one-line command :

Hello Ewan,

> parisize = 4000000, primelimit = 500000
> ? floor(6400^(3/2))
> %1 = 511999

floor is not at fault:
? 6400^(3/2)
%1 = 512000.0000000000000000000000
? \p100
   realprecision = 105 significant digits (100 digits displayed)
? %1
%2 = 511999.99999999999999999999999
? 6400^(3/2)==(512000*(1-2^-95))
%3 = 1

So actually %1 is slightly less than 512000.

6400^(3/2) is computed using the formula exp(3/2*log(6400))
which incurrs some rounding errors.
Using the formula sqrt(6400)^3 give a better result:
? sqrt(6400)^3
%2 = 512000.00000000000000000000000000000000
or even
? sqrtint(6400)^3
%3 = 512000

To convert real to i$ntegers, I would suggest you to use round:

? round(6400^(3/2))
%4 = 512000

Cheers,
Bill.