Question

While tutoring a student recently, I have come across the situation of explain logarithms by first introducing functions of the form $$f(x)= a^x$$ where $a \ge 0,x\in \mathbb{R}$. My student then asked me a seemingly innocent question, what if $a < 0.$ I explained to her, that it suffices to consider the basic case $$g(x)= (-1)^x$$ however I told her that the answer to that is beyond the scope of the material. I did tell her the answer though, namely that $$g(x) = (-1)^x = (e^{i \pi})^x = e^{i x \pi} $$ which is the unit circle in the complex plane. Of course this opens up a new can of worms.

This got me to thinking, in the course of history the most probable chronological sequence was that functions of the form of $f(x)$ were considered before complex numbers were formalized. Then did mathematicians of the past simply state that functions with $a<0$ were simply undefined? Or was it the case that complex numbers were considered first?

Answer 1

Short answer: Most likely undefined.

Long answer: The "naive" definition of $f(x) = a^x$ where $a, x \in \mathbb{R}$ and $a > 0$ is as follows. You know how to define $f(n)$ where $n$ is an integer. You can define $a^{p/q}$ to be the unique positive real number satisfying $(a^{p/q})^q = a^p$, so you know how to define $f(n)$ where $n$ is rational. When $a > 0$, it turns out that $f$ is continuous on $\mathbb{Q}$, so one can sensibly define $f(x)$ for all real $x$ by finding a Cauchy sequence of rationals converging to $x$ and taking the limit.

When $a < 0$, this reasoning falls apart at the second step: there is no positive real number satisfying $(a^{p/q})^q = a^p$ if $p$ is odd and $q$ is even. Even at the other rational numbers, the resulting function is highly discontinuous. So the only sensible thing to do, from a real-variable perspective, is to ignore this case.