The radius of convergence of the p-adic exponential function. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T06:38:52Z http://mathoverflow.net/feeds/question/53456 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53456/the-radius-of-convergence-of-the-p-adic-exponential-function The radius of convergence of the p-adic exponential function. Marty 2011-01-27T06:08:17Z 2011-01-29T06:50:55Z <p>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 $n!$.</p> <p>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.</p> <p>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 $exp(x)$?</p>