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 to learn …

Jan 12, 2015 · In the context of natural numbers and finite combinatorics it is generally safe to adopt a convention that $0^0=1$. Extending this to a complex arithmetic context is fraught with risks, as is …

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 working with these …

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?

@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). For all this, …

@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 intention is if you have …

Show that ∇· (∇ x F) = 0 for any vector field [duplicate] Ask Question Asked 9 years, 7 months ago Modified 9 years, 7 months ago

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 think about …

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 antiderivative …

Nov 17, 2014 · The reason $0/0$ is undefined is that it is impossible to define it to be equal to any real number while obeying the familiar algebraic properties of the reals. It is perfectly reasonable to …