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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
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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
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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.

EDIT:

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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