April 27th
Today I learned that some self-contained calculus can be done over the $p$-adics, from here . Of course, Tate's thesis is all about doing volume integrals over local fields (and by extension, the adeles), but here we're talking about trying to treat $\QQ_p$ for $p \lt \infty$ prime like $\RR$ and doing the analysis.
We define the formal derivative of a power series instead of the normal derivative.
Definition. Suppose $f\in\QQ_p[[x]]$ takes the form $f(x)=\sum_{k=0}^\infty a_kx^k$ for $\{a_k\}_{k\in\NN}\subseteq\QQ_p.$ Then we define the formal derivative by \[f'(x):=\sum_{k=0}^\infty(k+1)a_{k+1}x^k.\]
We would like the formal derivative to be defined at least where $f(x)$ is defined. We can do this directly because $p$-adic analysis is so nice.
Lemma. Suppose $f\in\QQ_p[[x]]$ takes the form $f(x)=\sum_{k=0}^\infty a_kx^k$ for $\{a_k\}_{k\in\NN}\subseteq\QQ_p.$ Then, for fixed $x\in\QQ_p,$ if $f(x)$ converges, then $f'(x)$ converges as well.
The fact that $f(x)$ converges is equivalent to the fact that the terms of the sum approach $0,$ or\[\left|a_kx^k\right|_p\to0.\]To show that $f'(x)$ converges, it is again sufficient to show that the terms $\left|(k+1)a_{k+1}x^k\right|_p\to0.$ However, this is\[\left|(k+1)a_{k+1}x^k\right|_p\le\left|a_{k+1}x^k\right|=\frac1{|x|_p}\left|a_{k+1}x^{k+1}\right|,\]which goes to $0$ by hypothesis. So we are done here. $\blacksquare$
We also say that we can talk a little more generally about convergence of power series; namely, they have a radius of convergence just like power series in $\RR$ do.
Proposition. Suppose $f\in\QQ_p[[x]]$ takes the form $f(x)=\sum_{k=0}^\infty a_kx^k$ for $\{a_k\}_{k\in\NN}\subseteq\QQ_p.$ Then the radius of convergence of $f$ is \[\rho:=\left(\limsup_{k\to\infty}\sqrt[k]{|a_k|_p}\right)^{-1}.\] We note that $\rho=\infty$ is permitted.
This pretty much follows from the same fact that a series converges if and only if its terms vanish. Namely, fixing $x\in\QQ_p,$ we know $f(x)$ will converge if and only if\[\left|a_kx^k\right|_p=|a_k|_p\cdot|x|_p^k\to0.\]Now, if $|x|_p \lt \rho,$ then\[\limsup_{k\to\infty}\sqrt[k]{|a_k|_p}\cdot|x|_p \lt 1.\]
As an aside, we remark that $\limsup_{n\to\infty}\sqrt[n]{|n|_p}=1,$ so the extra linear factor in the coefficients of the formal derivative does not affect the radius of convergence, which is what we expected from the lemma. We can be more precise about the equality cases for the radius of convergence here, but we don't bother.
The main attraction is as follows: the formal derivative for power series behaves like theusual derivative.
Proposition. Fix $f\in\QQ_p[[x]].$ Then for any $x$ where $f(x)$ converges, \[f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h.\]
We remark that the limit is done with respect to $|\bullet|_p,$ as it should be; we will write $|\bullet|$ for $|\bullet|_p$ in what follows, for brevity. Anyways, this proof is roughly the same proof done for polynomials. Let $f(x)=\sum_{k=0}^\infty a_kx^k.$ Very quickly, note that for $h$ sufficiently small, we have $|h| \lt |x|,$ so\[\left|a_k\left(x+h\right)^k\right|=|a_k|\cdot\left|x+k\right|^k=|a_k|\cdot|x|^k=\left|a_kx^k\right|.\]In particular, this goes to $0$ as $k\to\infty$ where $f(x)$ converges. The point of saying this is so that we have $f(x+h)$ converges where $f(x)$ converges, for $h$ sufficiently small.
Now we begin the computation. Observe that\[f(x+h)-f(x)=\left(\sum_{k=0}^\infty a_k(x+h)^k\right)-\left(\sum_{k=0}^\infty a_kx^k\right)=\sum_{k=0}^\infty a_k\left((x+h)^k-x^k\right).\]The sum still converges because $\left|(x+h)^k-x^k\right|\le\max\left\{\left|(x+h)^k\right|,\left|-x^k\right|\right\}=|x|^k$ for sufficiently small $h.$ Anyways, we use the binomial theorem now to conclude\[f(x+h)-f(x)=\sum_{k=0}^\infty\left(\sum_{\ell=1}^k\binom k\ell h^\ell x^{k-\ell}\right)a_k.\]Dividing by $h$ and rearranging, we see\[\frac{f(x+h)-f(x)}h=\underbrace{\sum_{k=1}^\infty ka_kx^{k-1}}_{f'(x)}+\underbrace{\sum_{k=2}^\infty\left(\sum_{\ell=2}^k\binom k\ell h^{\ell-1}x^{k-\ell}\right)a_k}_{\text{error}}.\]We already know that $f'(x)$ converges.
It remains to talk about the error sum, which we must show converges to $0$ as $h\to0.$ Factoring out an $h,$ it suffices to show that\[\sum_{k=2}^\infty\left(\sum_{\ell=2}^k\binom k\ell h^{\ell-2}x^{k-\ell}\right)a_k\]converges at all, for we are interested in what happens as $h\to0.$ So we need to know that the summands above go to $0.$ Fixing a $k,$ we note that the summand is bounded by\[\left|\left(\sum_{\ell=2}^k\binom k\ell h^{\ell-2}x^{k-\ell}\right)a_k\right|\le|a_k|\max_{2\le\ell\le k}\left|\binom k\ell h^{\ell-2}x^{k-\ell}\right|\le|a_k|\max_{2\le\ell\le k}\left|h^{\ell-2}x^{k-\ell}\right|.\]So for sufficiently small $h,$ say $|h|\le|x|,$ we can bound the term above by $|a_k|\cdot\left|x^{k-2}\right|.$ Now, $f(x)$ converges, so $\left|a_kx^k\right|$ goes to $0,$ so for $x\ne0$ ($x=0$ gives convergence for free), we may divide by $x^2$ to conclude $\left|a_kx^{k-2}\right|$ goes to $0$ as well. This finishes. $\blacksquare$
I find this statement at least amusing because it reconnects what are typically purely formal derivatives with the usual geometric view. I suppose this is the natural way to define differentiability in $\QQ_p$ and then for power series in $\QQ_p[[x]],$ but the fact it all works properly is nice.