Expansion of $(1+X)^b$ when $b \in \mathbb{Z}_p$ and $b \notin \mathbb{N}$

Question

For each $b \in \mathbb{N}, (1+X)^b=\sum\limits_{n=0}^b{{b}\choose {n}}X^{n}$ which is a polynomial in $X$. Do we have any expansions like that if $b \in \mathbb{Z}_p$ and $b \notin \mathbb{N}$?

Answer 1

This is certainly valid for odd $p$ as long as $X\in p\Bbb Z_p$.

Answer 2

If $\{a_n\}_{n=0}^{\infty}$ is a sequence in $\mathbb{Q}_p$ (or in any complete extension thereof) such that $a_n \rightarrow 0$, then the Mahler series

$x \mapsto \sum_{n=0}^{\infty} a_n {x \choose n}$

converges uniformly on $\mathbb{Z}_p$ and (thus) defines a continuous function. (The converse is also true: Mahler's Theorem.) See here for an exposition of Mahler series. The convergence claim is the (easy) Lemma 4.1.3.

If $a_n = X^n$ for $X \in \mathbb{Z}_p$, then $a_n \rightarrow 0$ iff $X \in p \mathbb{Z}_p$. Since $\mathbb{N}$ is dense in $\mathbb{Z}_p$, this is the unique continuous extension to $\mathbb{Z}_p$, i.e., if anything deserves to be called $(1+X)^b$, this function does.

(If $X \in \mathbb{Z}_p \setminus p\mathbb{Z}_p$, the series does not converge for $b = -1$.)