Re: digits of a float

On Thu 26 Dec, 2024, at 17:21, Ruud H.G. van Tol wrote:

>
> On 2024-12-15 12:31, Bill Allombert wrote:
> > [...] One need to round toward 0 instead of rounding to nearest
> > as usual.
> >
> > In this instance the number end by 574699992... which is (correctly)
> rounded to nearest
> > by PARI to 5747.....
> >
> > I wrote this, which increase the accuracy until there is no more
> ambiguity.
> > {
> >    my(n=19437,z);
> >    for(m=n+1,oo,
> >      localprec(m+1);
> >      z = round(Pi *10^m)/10^m;
> >      if(abs(z-Pi)>10^-(m+1),break));
> >    z*10^(n-1)\1%10
> > }
> >
> > but this need to be checked more carefully
>
> For Pi-digits, I am now using:
>
> pi_digits(n, p=20)= {
>    my(u=10^(n+p-1), f(x, u)=my(n=1, q=u\x, r=q, s=1, t);
> while(t=(q\=(x*x))\(n+=2), r+=(s=-s)*t); r*4);
>    digits((4*f(5, u) - f(239, u))\10^p);
> } \\ Machin-like, with p > the maximal number of consecutive 9-digits to
> be expected (A048940)
>

     Wouldn't floor be more adapted than round?

            Loïc