Is $0^x$ not analytic? Is $0^x$ not analytic?
Assume that we define it as follows:
$$f(x)=0\ (x>0)$$
$$f^{(k)}(x)=0\ (x>0)$$
If it were analytic, I could extend its domain by taylor's theorem:
$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n$$
If we have $a>0$, then this reduces down to
$$f(x)=0$$
So why can't we define $0^x=0$ like this for all $x$?

For clarification, I am not asking what $0^{-1}$ or $0^0$ is, I am asking why we can't have analytic continuation of $0^x$ to any $x$ value.
 A: You are mistaking "function" for "formula of a function". Keep in mind that one and the same function may have different formulae on different domains. Let me explain this by an analogy.
Consider the (convergent) series $\sum \limits _{n \ge 1} \frac 1 {n^x}$. This is denoted as $\zeta (x)$ and I believe that you know this. $\zeta$ may be extended by analyticity to the whole of $\Bbb C \setminus \{1\}$, but the formula $\sum \limits _{n \ge 1} \frac 1 {n^x}$ may not.
Another, easier, example: consider $\log : (0, \infty) \to \Bbb R$. Remember that on $(0,2]$
$$\log x = \sum \limits _{n \ge 1} (-1)^{n-1} \frac {(x-1)^n} n .$$
$\log$ is analytic on the whole domain of definition, but that series representation only holds on $(0,2]$.
This is exactly what happens in your example: $0^x$, initially defined only on $(0, \infty)$ may be extended by analyticity to the whole $\Bbb R$ (or $\Bbb C$, depending on your context), but the formula $0^x$ no longer holds on this new, larger domain.
A: Sure, you can redefine $0^x$ however you want, and you can obviously construct an analytic continuation of the constant zero function to wherever you like. There's nothing stopping you, this is mathematics after all.
However before you do that, maybe you want to consider the nice properties of real exponentiation that you'll break because of that. For example, if you have $x \leq 0$, then your $0^x$ is no longer the limit of $y^x$ for $y \searrow 0$.
Finally, why do you even want to do that? Is there any particular phenomenon you want to describe, or any mathematical statement which becomes easier to formulate when you use your modified exponentiation? I know of no such phenomenon or statement.
A: You can have an analytic continuation. Indeed, the constant $0$ function is the analytic continuation. However, to denote it by $0^{x}$ is rather pointless and designed to cause confusion. 
For a quite popular related situation you can have an analytic continuation of $$\sum_{n\ge 1} n^{-s}$$  from $\Re(s)>1$ to most of the complex plane. The value of this continuation at $-1$ happens to be $-1/12$ and some are fond of  writing this as 
$$\sum_{n\ge 1} n = -\frac{1}{12}$$
Yet a more transparent way to do this is to say that $\zeta(s)$ is defined by $\zeta(s)= \sum_{n\ge 1} n^{-s}$  for $\Re(s)>1$ and analytic continuation to most of the rest of the complex plane. And $\zeta(-1)=-1/12$. 
