$p$-adic exponential function

Question

Simple question about $p$-adic numbers $\mathbb{Q}_p$. How do I evaluate $$\exp(x)=\sum_{n=0}^\infty \frac{x^n}{n!}$$ where $x\in {\mathbb Q}_p$. I'm a bit confused, because as far as I understand it, each term is added in the same way as in $\mathbb R$, so why should the result be different? As a start let's evaluate $x=1$ in $p=3$.

Answer 1

If you want to try $\exp(3)$ in $p=3$, here are the first few partial sums: $$ \eqalign{1 &= 1\cr 1 + 3 &= 1 + 3\cr 1 + 3 + \frac{9}{2} &= 1+3+2 \cdot 3^2+ 3^3+3^4+\ldots\cr 1 + 3 + \frac{9}{2} + \frac{9}{2} &= 1 + 3 + 3^2\cr 1+ 3 + \frac{9}{2} + \frac{9}{2} + \dfrac{27}{8} &= 1 + 3 + 3^2 + 2 \cdot 3^3 + 2 \cdot 3^4 + \ldots\cr 1 +3 + \frac{9}{2} + \frac{9}{2} + \dfrac{27}{8} + \dfrac{81}{40} &= 1 + 3 + 3^2 + 2 \cdot 3^3 + 0 \cdot 3^4 + \ldots\cr 1 + 3 + \frac{9}{2} + \frac{9}{2} + \dfrac{27}{8} + \dfrac{81}{40} + \frac{81}{80} &= 1 + 3 + 3^2 + 2 \cdot 3^3 + 2 \cdot 3^4 + \ldots\cr}$$ and it turns out all later terms have numerators with higher powers of $3$ than $3^4$, so the start of the $3$-adic expansion of $\exp(3)$ is $ 1 + 3 + 3^2 + 2 \cdot 3^3 + 2 \cdot 3^4 + \ldots$. As was mentioned, this is a $3$-adic number, not the same as the real number $\exp(3)$, nor any other real number.

Answer 2

The reals (and complexes) are truly different from the various $p$-adic fields $\Bbb Q_p$. One of the glaring differences is the domain of convergence of exponential and logarithm.

For $\Bbb R$ and $\Bbb C$, the exponential series is convergent for no matter what $z$ you plug into it, while the logarithmic series is convergent only for $|z|<1$.

For any $p$-adic field (including the complete extensions of $\Bbb Q_p$), the logarithmic series is still convergent for $z$ with $|z|<1$, but for convergence of the exponential series, you need the very restrictive condition $|z|<\varepsilon^{\frac1{p-1}}$, where $\varepsilon=|p|$ must be a positive real number $<1$, and $\varepsilon=1/p$ is the conventional choice. The above restriction means that you can calculate the $3$-adic value of $\exp(3)$, as @RobertIsrael has done in his answer, but you can not calculate the $2$-adic exponential of $2$.

Now that I’m on my hobby-horse, I must ride it a bit further. The real logarithm extends from its domain of convergence to a homomorphism from the group $\Bbb R^{>0}$ onto $\Bbb R$, multiplicative group to additive group. The complex logarithm has no such extension: there is no group containing $1+\{z\in\Bbb C:\,|z|<1\}$ on which the logarithm can be defined as a homomorphism.

The $p$-adic logarithm is much more wonderful (and useful) than the complex, for if $k$ is a complete $p$-adic field, the set $1+\{z\in k:\,|z|<1\}$ is a multiplicative group, and the logarithmic series defines a homomorphism from it to the additive group $k$. If you have a $p$-adic number $z$, you could try to define $\exp(z)$ as a number $\zeta$ such that $\log(\zeta)=z$, and this equation is always solvable, for $\zeta$ perhaps in a finite field extension of $\Bbb Q_p$. Only problem is that there are infinitely many such $\zeta$, and in most cases, there is little reason to choose one of these over the others. In my opinion, this is the deep reason why you can’t define the exponential of most $p$-adic numbers.