I limit myself in following considerations to the field $\mathbb Q_p$. As everybody interested in ultrametric calculus knows, the $p$-adic exponential function defined by series $$\exp_p(x) = \sum_{k=0}^\infty \frac{x^k}{k!}$$ converges only on a small ball $\mathcal B$ around $0$ equal to $p\mathbb Z_p$ (for $p \ge 3$) or $4 \mathbb Z_2$ (otherwise). Inspired by my previous post ,,About $\lim \left(1+\frac {x}{n}\right)^n$'' I have decided to check whether this alternative representation is also true here, in $p$-adic context.

As a side note, an analytical function $f$ from convex neighbourhood of zero satisfying $f = f'$ or $f(x+y) = f(x) f(y)$ must be identically zero (this is exercise 25.J from Schikhof's book).

Let's fix $\varepsilon > 0$ and $x \in \mathcal B$. We have $$\left|\exp_p x - \Bigl(1 + \frac xn\Bigr)^n\right| = \left(\sum_{k=n}^\infty \frac{x^k}{k!}\right) + \sum_{k=2}^n \left(\frac{x^k}{k!} - {n \choose k} \frac{x^k}{n^k}\right).$$ because $(1 +x/n)^n = 1 + x + \ldots$. For sufficiently large values of $n$, the left sum can be made smaller than $\varepsilon / 2$. Now I have tried to rewrite the summand as $$\frac{x^k}{k!} \cdot \left(1-\frac{n!}{(n-k)! \cdot n^k}\right)$$ and assumed that $n$ is of the form $p^m + p^{m-1} + \ldots + 1$, which implies $v_p(n!) =n - p^m$ by the Legendre formula and $|n|_p^k = 1$ by direct check. Now $v_p((n-k)!)$ is certainly not larger than $p^m / (p-1)$ and therefore $$v_p(n!) - v_p(k!) \ge \frac{p-2}{p-1} \cdot p^{m-1} + p^{m-2} + \ldots + 1 \to \infty.$$

It seems that $\exp_p (x) = \lim_{n \to \infty} (1 + x/n)^n$. Am I right?


2 Answers 2


No, but here is something really interesting. If you take your definition of the (real) exponential, turn it upside down, and look at it squint, you get a very unconventional way of constructing the logarithm: $$ \ln(x)=\lim_{n\to\infty}\bigl(n(\sqrt[n]x-1)\bigr)=\lim_{\varepsilon\to0} \frac{x^\varepsilon-1}{\varepsilon}, $$ in which I’ve replaced the big number $n$ by the small number $\varepsilon$, and let that go to zero instead. You can check this on a pocket calculator by just letting the $n$’s be powers of $2$, i.e. use your “$\sqrt{\>}$” button a certain number of times, subtract $1$, and them multiply by the corresponding power of $2$.

Anyhow, this idea works perfectly well if you want the logarithm of any $p$-adic number (even in algebraic extensions $K$ of $\Bbb Q_p$) as long as it’s congruent to $1$ modulo the maximal ideal of the integers of $K$. Raise your number to a tiny power, subtract $1$, and divide by that power. What’s tiny? A number of form $p^m$, of course, so the formula for the $p$-adic logarithm is $$ \ln(u)=\lim_{m\to\infty}\frac{u^{p^m}-1}{p^m}\,. $$ It’s not hard to see that this gives the same result as the formula from Calculus for $\ln(1+x)$, but that’s a story for another day.

  • $\begingroup$ I think that also the following fomula is true. Let's choose a $b \in \mathbb C_p$ such that $|b|_p = 1$, denote by $\omega$ the Teichmüller character and by $\log$ - the Iwasawa logarithm. Then $\log b = \lim_n (b^{p^{n!}} - \omega_p(b)) / (\omega_p(b) p^{n!})$. $\endgroup$
    – Santiago
    Aug 24, 2016 at 11:12
  • 2
    $\begingroup$ Yes, @Santiago, that looks right. $\endgroup$
    – Lubin
    Aug 24, 2016 at 18:21

No, the limit exists only for $x=0$

If $(a_n)$ is a sequence in $\Bbb Q_p$ and $(a_n)$ converges to anything $p$-adically, then the set $\{v_p(a_n)\}$ has a lower bound.

However, here if $x$ is nonzero and you pick an $n$ such that $v_p(n) > v_p(x)$, then $v_p(1+x/n)< 0$, and since $v_p((1+x/n)^n)$ is $n$ times that, this shows that $v_p((1+x/n)^n)$ can take arbitrarily big negative values. This prevents the sequence form converging to anything.


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