The product of 0 and anything is $0$, and seems like it would be reasonable to assume that $0! = 0$. I'm perplexed as to why I have to account for this condition in my factorial function (Trying …

Inclusion of $0$ in the natural numbers is a definition for them that first occurred in the 19th century. The Peano Axioms for natural numbers take $0$ to be one though, so if you are …

@Arturo: I heartily disagree with your first sentence. Here's why: There's the binomial theorem (which you find too weak), and there's power series and polynomials (see also Gadi's answer). …

Jan 12, 2015 · It is possible to interpret such expressions in many ways that can make sense. The question is, what properties do we want such an interpretation to have? $0^i = 0$ is a good …

Your title says something else than "infinity times zero". It says "infinity to the zeroth power". It is also an indefinite form because $$\infty^0 = \exp (0\log \infty) $$ but $\log\infty=\infty$, so the …

Oct 9, 2013 · Is a constant raised to the power of infinity indeterminate? I am just curious. Say, for instance, is $0^\\infty$ indeterminate? Or is it only 1 raised to the infinity that is?

Oct 28, 2019 · In the context of limits, $0/0$ is an indeterminate form (limit could be anything) while $1/0$ is not (limit either doesn't exist or is $\pm\infty$). This is a pretty reasonable way to …

@Swivel But 0 does equal -0. Even under IEEE-754. The only reason IEEE-754 makes a distinction between +0 and -0 at all is because of underflow, and for +/- ∞, overflow. The …

My daughter is stuck on the concept that $$2^0 = 1,$$ having the intuitive expectation that it be equal to zero. I have tried explaining it, but I guess not well enough. How would you explain the

Feb 4, 2018 · So there is a sense in which the only "really geometrically meaningful" integral of $0$ is $0$ itself. But your friend is still wrong, since the term "integral" in this context means …