History of $p$-adic logarithm of integers

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2026-08-16 19:57 bmatschke the constraint said k = 1 mod p and 702 of the 856 entries are not; they are the integers coprime to p, and the logarithm reaches them through log(k^(p-1))/(p-1) current
2026-08-14 21:30 bmatschke how well the digits are known: proven (p-adic arithmetic (Qp(k).log()); precision tracked by Qp)
2026-08-13 22:16 bmatschke how well the digits are known: proven (p-adic arithmetic (Qp(k).log()); precision tracked by Qp)
2026-08-13 22:05 bmatschke how well the digits are known: heuristic (a fixed-precision value wrapped in an interval field)
2026-08-09 09:12 flattening entries rewritten as records with named parameters
2026-08-09 08:34 data-repository import the current state of the data repository reviewed
2021-04-30 12:12 bmatschke from the data repository, 698e7e1f

What changed between 2026-08-14 21:30 and 2026-08-16 19:57

from line 8 (5 lines) @@ -8,5 +8,5 @@
   k:     type: Z-    constraints: $k = 1 \mod p$+    constraints: $p \nmid k$ Comments:   comment-extension-of-log: 'The $p$-adic logarithm is defined by its Taylor series  for
from line 18 (8 lines, 3 more than before) @@ -18,5 +18,8 @@
     Similarly, we can extend $\log_p$ to any $x\in \mathbb{Q}_p^\times$ by prescribing     an arbitrary value for $\log_p p$, such as $\log_p p = 0$.'-Formulas: {}+Formulas:+  formula-extension: $\log_p(k) = \frac{1}{p-1}\log_p(k^{p-1})$. The series $\log_p(1+u)+    = \sum_{n\geq1} (-1)^{n+1} u^n/n$ converges only for $k \equiv 1 \bmod p$; this+    extends it to every unit, since $k^{p-1} \equiv 1 \bmod p$ whatever $k$ was. Programs: {} References: {} 

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