Zero to the power of negative numbers

Question

I lately stumbled unto a strange behaviour of my calculator (a TI-Nspire™ CX II-T CAS): $$ \begin{split} 0^{-2} = 0^{-4} &= \infty & \forall \text{ even numbers}\\ 0^{-1}, 0^{-3} &= \text{undefined} \quad& \forall \text{ odd numbers}\quad&{\scriptsize\text{(note that undefined $\ne$ undefined)}} \end{split} $$ To me, either both should result in $\infty$ ($\tilde\infty$ to be precice) or both should be $\text{undefined}$, since both involve division by zero. Somehow I cannot wrap my head around this, nor find anything about it online.

I've consultated threads like Why does zero raised to the power of negative one equal infinity?, Zero to the negative power. and some more, but they all talk about zero by the power of $-1$, nothing else.

Furthermore, I tried it on some online calculators like WolframAlpha, which directly gives me complex infinity ($\tilde \infty$), others said both would be $\text{undefined}$.

Does anyone know how this can be explained?

Answer 1

Usually such expressions are defined as limits, so for example $$0^{-n} := \lim_{x\to 0} x^{-n}.$$

You'll see that if $n$ is odd, the value of the limit depends on which side zero is approached from ($n$ odd): $$\lim_{x\to 0^-} x^{-n} = -\infty,\quad\lim_{x\to 0^+} x^{-n} = +\infty,$$ so the limit is not defined. However, for $n$ even, both one-sided limits agree, and the limit $\lim_{x\to 0} x^{-n}$ exists and is $+\infty$.

While it seems like a bug at first glance, this might be the reason why the calculator produces the observed results - they would be consistent with the above reasoning.

Answer 2

This is just a hypothesis, as I know nothing about the inner workings of the calculator or the thought process behind its design, or even its capabilities. But my immediate guess would be that the even powers $0^{-2}, 0^{-4}, \ldots$ evaluate to $+\infty$, positive infinity, while the odd powers $0^{-1}, 0^{-3}, \ldots$ evaluate to $\pm\infty$, and therefore not to a single answer, and therefore undefined.