Let $p$ be a prime number and denote by $R(f)$ the radius of convergence of a power series $f(x) \in \mathbb{C}_p[[x]]$, where $\mathbb{C}_p$ is the completion of the algebraic closure of $\mathbb{Q}_p$, the field of $p$-adic numbers. Given two power series $f(x), g(x) \in \mathbb{C}_p[[x]]$, it is known that the radius of convergence of the product $h(x) = f(x)g(x)$ is at least the minimum of the radius of convergence of the two series $f(x)$ and $g(x)$. In other words, we have $$ R(h) \geq \min\{R(f), R(g)\}.$$ Keep in mind $(1-x)(1+x+x^2+\dots) = 1$ as an example for the strict inequality. Is there a way to easily predict when $R(h) > \min\{R(f), R(g)\}$ and find $R(h)$ explicitly? More specifically, I'm interested in computing the radius of convergence of power series of the form $\exp(f(x))$ for $f(x) \in x\mathbb{C}_p[[x]]$.

For example, let $f(x) = \exp(x)$ and $g(x) = \exp(x^p/p)$. Then $R(f) = R(g) = (1/p)^{1/(p-1)}$ and using the fact that the Artin-Hasse exponential series $$\text{AH}(x) = \exp(x + x^p/p + x^{p^2}/p^2 + \cdots)$$ lies in $\mathbb{Z}_p[[x]]$ (which implies $R(\text{AH}) \geq 1$), $h(x) = \exp(x + x^p/p)$ has radius of convergence $$ R(h) = R\left(\exp\left(\frac{x^{p^2}}{p^2}\right)\right) = \left(\frac{1}{p}\right)^{\frac{(2p-1)}{p^2(p-1)}}> \left(\frac{1}{p}\right)^{\frac{1}{p-1}}. $$ The importance of this example comes from the fact that if we set $\pi$ to be a root of $x+x^p/p = 0$, then $h(\pi)$ is a non-trivial $p$-th root of unity in $\mathbb{C}_p$. This provides an analytic representation of $p$-th roots of unity, exploited in particular in Dwork's proof of the rationality of zeta functions over finite fields.

More generally, using this method one can show that for any $n \geq 1$, we have $$ R(\exp(x+x^p/p+\cdots + x^{p^n}/p^n)) = R(\exp(x^{p^{n+1}}/p^{n+1})). $$ Even though I understand the details involved in this calculation, I don't know if there is some more general theory underlying these examples. It may be helpful to share similar examples that you know. For instance, is it always the case that $$ R(\exp(f(x))) > R(\exp(x)),$$ given $f(x) \in x\mathbb{C}_p[[x]]$ has a nonzero root $\alpha \in \mathbb{C}_p$ of absolute value $R(\exp(x)) = (1/p)^{1/(p-1)}$ and no non-zero roots of smaller absolute value?

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    $\begingroup$ I may be able to offer a partial answer here, but it’ll have to wait till morning at best. $\endgroup$ – Lubin Feb 7 '16 at 5:58
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    $\begingroup$ Have you looked at "Rank one solvable p-adic differential equations and finite Abelian characters via Lubin–Tate groups" by Andrea Pulita? The abstract starts with "We introduce a new class of exponentials of Artin–Hasse type, called π-exponentials". $\endgroup$ – Laurent Berger Feb 7 '16 at 12:35
  • $\begingroup$ @LaurentBerger: That looks interesting. I haven't looked at it, but I will surely do. Thank you! $\endgroup$ – Sandi Feb 7 '16 at 22:29

Here is a counterexample to your question at the end, for each $p$. Let $f_u(x) = x + ux^p/p$ for $u \in \mathbf C_p$ with $|u|_p = 1$ and $|u-1|_p = 1$. (Such $u$ can be taken in $\mathbf Z_p^\times$ if $p > 2$, but you need to go outside $\mathbf Q_p$ if $p = 2$ to an extension with residue field of size greater than $2$.) Since $|u|_p = 1$, all the nonzero roots of $f_u(x)$ in $\mathbf C_p$ have absolute value $(1/p)^{1/(p-1)} = R(\exp)$.

To find the $p$-adic radius of convergence of $\exp(f_u(x))$, write $$ \exp(f_u(x)) = \exp\left(x+\frac{x^p}{p}\right)\exp\left((u-1)\frac{x^p}{p}\right) $$ as formal power series. On the right side, the first factor has radius of convergence greater than $(1/p)^{1/(p-1)}$, as you noted.

Since $|u-1|_p = 1$, the second factor has radius of convergence equal to that of $\exp(x^p/p)$, which is $(1/p)^{1/(p-1)}$. The reciprocal $(\exp(x + x^p/p))^{-1}$ has the same radius of convergence as $\exp(x+x^p/p)$, even for $p=2$, so $\exp(f_u(x))$ has radius of convergence equal to $(1/p)^{1/(p-1)}$. This is the counterexample to your question.

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  • $\begingroup$ This a great specific example. More generally, for $u \in \mathbb{C}_p$, it shows that $R(\exp(x+ux^p/p))$ depends explicitly on $|u-1|_p$ whenever $$R(\exp(x+x^p/p)) \neq R((u-1)x^p/p).$$ I wonder what happens if equality holds. Is there a chance for the radius to further increase? $\endgroup$ – Sandi Feb 7 '16 at 23:41
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    $\begingroup$ In general if $F(x)$ and $F(x)^{-1}$ have a common radius of convergence $R$, and $G(x)$ and $G(x)^{-1}$ have a common radius of convergence $S$ then the product $FG$ has radius of convergence $\min(R,S)$ if $R \not= S$. If $R=S$ then there is no simple rule for the radius of $FG$. Typically the radius for $FG$ will remain $R$, but of course there are examples where it grows, like $\exp(x)$ and $\exp(x^p/p)$, or more simply $\exp(x)$ and $\exp(-x)$. $\endgroup$ – KConrad Feb 8 '16 at 15:16

I can only give a very partial answer, and that only from my very parochial point of view.

I will use the additive valuation $v$ rather than absolute value, normalized so that $v(p)=1$, and in terms of which $R(\sum a_nx^n)=-\liminf\bigl(v(a_n)/n\bigr)$, so that when $v(z)>R(f)$, $f(z)$ is a convergent series. Examples are: if $\exp(x)=\sum_{n\ge1}x^n/n!$ and $\log(x)=\sum_{n\ge1}(-1)^{n-1}x^n/n$, then $R(\exp)=1/(p-1)$ and $R(\log)=0$. It’s for this reason that I prefer the logarithm to its inverse.

The log that I’ve named above is the logarithm of the multiplicative formal group $\hat{\mathbf G}_{\mathrm m}(x,y)=x+y+xy$, that is a formal-group homomorphism from $\hat{\mathbf G}_{\mathrm m}$ to the additive formal group $\hat{\mathbf G}_{\mathrm a}(x,y)=x+y$ with $\log'(0)=1$.

The $p$-typical logarithm $\log_{\mathrm{AH}}=x+x^p/p+x^{p^2}/p^2+\cdots$ is the logarithm of another formal group $\mathscr M$. which we might call the $p$-typical recoordinatization of $\hat{\mathbf G}_{\mathrm m}$, and the $\Bbb Z_p$-formal-group isomorphism $u:\mathscr M\to\hat{\mathbf G}_{\mathrm m}$ is exactly what’s called the Artin-Hasse Exponential. It satisfies $\log\circ u=\log_{\mathrm{AH}}$.

Now here’s the moral of my story. These two logarithms, $\log$ and $\log_{\mathrm{AH}}$, being convergent throughout the open unit disc of $\Bbb C_p$, have all sorts of interesting behavior that is not seen at all by the exponential series $\exp(x)$, except at a very far remove. In particular, they have zeros. The closest such to the origin has $v(\zeta)=1/(p-1)$, which explains immediately the radius of convergence of the exponential function.

And so I am pretty sure that your last conjecture has no chance of being correct. I tried $p=3$, $\zeta=\omega-1$, where $\omega^2+\omega+1=0$, and found that $\exp(x-x^2/\zeta)$ seemed to have as bad convergence properties as the exponential itself.


You’ve asked me to explain further why the existence of zeros of the logarithm prevents wider convergence of the exponential, and indeed, it is not quite so obvious as I was pretending.

Let $\lambda$ be a root of the above-described $\log$ with $v(\lambda)=1/(p-1)$, in fact $\lambda+1$ will be a primitive $p$-th root of unity. If $R(\exp)\ge1/(p-1)$, then the Newton polygon of $f(x)=\exp(x)-\lambda$ will have a vertex $(n,v(b_n))$ in addition to the vertex $(1,0)$. (You don’t need to know that $b_n=1/n!$.) To see this, you may look at $f(\lambda x)$ and note that its coefficients must go to zero for convergence. In particular, There will be a segment of the polygon whose negative slope is $\ge1/(p-1)$, and thus a root $\mu$ of $f$ in an algebraic extension such that $\exp(\mu)=\lambda$, impossible if $\log(\exp(x))=x$.

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  • $\begingroup$ This is great. Thank you! I was wondering if you have some references on this formal group construction of this Artin-Hasse Exponential and if this approach implies the fact that $\text{AH}(x)$ has $\mathbb{Z}_p$ coefficients and/or radius of convergence 1? What did you want to say at: "interesting behavior that is not seen at all by the exponential series $\exp(x)$, except at a very far \emph{remove}." $\endgroup$ – Sandi Feb 7 '16 at 23:27
  • $\begingroup$ Also, would you mind elaborating some more on how the root of valuation $1/(p-1)$ of the logarithm series $\log(1+x) = \sum_{n \geq 1} (-1)^{n-1}x^n/n$ explains immediately the radius of convergence of the exponential series? Is there some basic compositional inverse property I'm missing here? $\endgroup$ – Sandi Feb 7 '16 at 23:28
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    $\begingroup$ “At a very far remove” means “from very far away”. The basic reference is Hazewinkel’s big book from the 70’s. For your second question, I’ll make an addendum to my answer. $\endgroup$ – Lubin Feb 8 '16 at 3:02
  • $\begingroup$ Thank you for the EDIT. That's a nice way to argue and it seems to generalize to any two compositional inverse series where one has a bigger radius of convergence and at least one non-zero root. I looked up Hazewinkel but couldn't find if this formal group approach to the Artin-Hasse Exponential tells us $\text{AH}(x)$ has $\mathbb{Z}_p$ coefficients or if it has radius of convergence 1. $\endgroup$ – Sandi Feb 8 '16 at 21:02
  • $\begingroup$ Maybe it would be best to discuss via e-mail: I have the time, and there’s not enough space here. $\endgroup$ – Lubin Feb 9 '16 at 1:19

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