Power series $f$ such that $f(\zeta_{p^n}-1)=1/p^n$ for almost all $n \geq 0$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T04:41:03Z http://mathoverflow.net/feeds/question/35228 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35228/power-series-f-such-that-f-zeta-pn-11-pn-for-almost-all-n-geq-0 Power series $f$ such that $f(\zeta_{p^n}-1)=1/p^n$ for almost all $n \geq 0$ Tiffy 2010-08-11T13:59:59Z 2010-08-11T16:10:44Z <p>Let $p$ be a prime number and $\zeta_{p^n}$ a primitive $p^n$-th root of unity. Find $f \in \mathbf Q_p[[X]]$ fulfilling $f(\zeta_{p^n}-1)=1/p^n$ for all sufficiently large $n$.</p> http://mathoverflow.net/questions/35228/power-series-f-such-that-f-zeta-pn-11-pn-for-almost-all-n-geq-0/35245#35245 Answer by Laurent Berger for Power series $f$ such that $f(\zeta_{p^n}-1)=1/p^n$ for almost all $n \geq 0$ Laurent Berger 2010-08-11T16:10:44Z 2010-08-11T16:10:44Z <p>As Brian pointed out, the principal parts problem always has a solution on the open unit disk, and Lazard's 1962 article "Les zéros des fonctions analytiques dâ€™une variable sur un corps valué complet" gives a nice proof which is also rather explicit. </p> <p>Your problem has additional symmetries so one can be a bit more explicit, as follows.</p> <p>Let $H$ be the set of power series holomorphic on the open unit disk, let $\varphi: H \to H$ be the map defined by $\varphi(f)(X)=f((1+X)^p-1)$ and let $d$ be defined by $d(f)(X)=(1+X)df/dX$. Note that $d\varphi=p\varphi d$. You can check that if you have a function $f$ such that $\varphi(f)-pf = (1+X/2)\log(1+X)/X$, then $f(0) \neq 0$ and $f(\zeta_{p^n}-1) = p^{-n} f(0)$ so this answers your problem. </p> <p>Therefore you need to be able to solve an equation of the form $\varphi(f)-pf = g$. By taking $d$ of both sides this gives $\varphi(df)-df=dg/p$. Now you can solve an equation of the form $\varphi(a)-a=b$ if $b \in X \cdot H$ by writing $a = \sum_{n \geq 0} \varphi^n(b)$, which should converge in $H$ for its Fréchet topology. We have $d((1+X/2)\log(1+X)/X) \in X \cdot H$ (this is what the $(1+X/2)$ was put in for), and once you know $df$, you get $f$ by integrating and adjusting the constant.</p>