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As every number theorist learns, the radius of convergence of $exp(x)$, defined by the usual power series in a neighborhood of zero, is $$\rho = p^{-1/(p-1)}.$$ This is typically proven by computing the $p$-adic absolute value of $n!$.

I imagine that this might also be proven by using the fact that the differential equation $Df - f = 0$ (to which $f(x) = exp(x)$ is a solution) has an irregular singular point at $\infty$. My intuition in the $p$-adic case, from looking at a paper of Bombieri and Dwork a long time ago, is that this irregular singular point "pushes" convergence away from infinity -- hence the relatively small radius of convergence of $exp(x)$ $p$-adically, near zero.

Is there a somewhat general statement along these lines -- that an irregular singular point of a differential equation at a point $s$ (on a smooth projective curve, let's say) will cause the (generic?) radius of convergence of a series solution to the differential equation at a point $t$ to be less than something (involving $s$, $t$, and invariants of irregular singular points, etc.)? Something that, when you "plug in" the differential equation $Df - f = 0$, and the points $s = \infty$ and $t = 0$, will output the radius of convergence of $exp(x)$?

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Marty: your radius of convergence is wrong. The series exp(x) doesn't converge at 1 but it's less than your radius. Take a reciprocal. – KConrad Jan 27 '11 at 10:11
Whoops. Forgot a minus sign. It's corrected now. – Marty Jan 27 '11 at 16:46
@Marty: it's good that you did say " every number theorist learns ..." :) (I am not confident about my ability to write down the formula for the radius of convergence from memory either.) Anyway, very interesting question! – Pete L. Clark Jan 27 '11 at 19:41
It will indeed be interesting if this approach can predict the radius of convergence; but I like to look at the exponential as a secondary thing, the inverse of the $p$-adically much more fundamental logarithm. This is defined on the whole open unit disk, with derivative a unit, so $\exp$ can be defined only on an open disk of the same size as the one in which $\log$ is one-to-one, in other words on the disk $z\colon v(z)>1/(p-1)$. No need for memory here: sketch the Newton polygon of the log and see what (nonunit) homothetic conjugation needs to be applied to make the result integral. – Lubin Jan 27 '11 at 21:44
@Marty: Right you are. It's considerations of exactly that sort that made me delay my comment. But the exponential has, in some sense, no properties, at least in comparison to the log, which has loads of structure (I'm thinking of how the roots are placed.) Maybe, like Abel, you'd get a handle on things by looking at the inverse of any function you got as a solution of a DE? – Lubin Jan 28 '11 at 0:14

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