Re: digits of a float

On 2024-12-26 19:21, Loïc Grenié wrote:
> On Thu 26 Dec, 2024, at 17:21, Ruud H.G. van Tol wrote:
>
>     [...] 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?

The code has an off-by-1 error. Corrected:

pi_digits(n, p=20)= {
   my(u=10^(n+p), 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+1));
}


With a low value of p, the rounding shows:

? pi_digits( 769, 10 )[-10..-1]
% [1, 3, 4, 9, 9, 9, 9, 9, 9, 8]  :OK

? pi_digits( 763, 6 )[-4..-1]
% [1, 3, 5, 0]  :BAD

? pi_digits( 762, 7 )[-3..-1]
% [1, 3, 5]  :BAD (also invalidates the comment about what p is needed)

-- Ruud