Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 27 / 2 / ii / Solution

Let $c_n$ be the Mahler coefficients of $f$. Define its discrete antiderivative byBecause $c_n\to 0$ and the binomial polynomials have field absolute value at most one on ${\mathbb{Z}}_p$, this series converges uniformly to a continuous ${\mathbb{Z}}_p$-valued function. Every summand vanishes at zero. Pascal's identity gives $\left(\genfrac{}{}{0ex}{}{x+1}{n+1}\right)-\left(\genfrac{}{}{0ex}{}{x}{n+1}\right)=\left(\genfrac{}{}{0ex}{}{x}{n}\right)$, so taking differences through the uniformly convergent series proves $g(x+1)-g(x)=f(x)$. For nonnegative integers, induction from $g(0)=0$ gives precisely $g(m)={\sum }_{j=0}^{m-1}f(j)$. Thus this is the required continuous extension, unique because the nonnegative integers are dense in ${\mathbb{Z}}_p$.

The Mahler expansion of $g$ has coefficient zero in degree zero and coefficient $c_{m-1}$ in degree $m\ge 1$. This discrete antidifferentiation on the p-adic integers is the mechanism behind the next part.