Re: Computing p-adic logarithm with precision two

On 09/09/2024 15:24, Georgi Guninski wrote:
> On Mon, Sep 9, 2024 at 4:13 PM Aurel Page <[email protected]> wrote:
>> By "precision 2", do you mean computing the result up to O(p^2) or
>> O(p^3)? I assume that B is assumed to be in Z_p?
>> If the former, then p*(p-a) is unnecessary, -a*p is sufficient. If the
>> latter, then you are missing one term.
>>
> Thanks.
> I mean with O(p^2) and my result is equal to pari's `log(B+O(p^2))`.
In this case, the p^2 part of p*(p-a) = p^2 - a*p is absorbed in the O(p^2).

> Does pari use the same algorithm?
Well, pari's implementation handles arbitrary precision. At precision 
O(p^2), it is not much of an algorithm, it is simply the Taylor 
expansion of log to first order. But yes, pari's implementation should 
specialise to this formula at precision O(p^2).

Cheers,
Aurel