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 …

Feb 6, 2021 · $$ 0! = \Gamma (1) = \int_0^ {\infty} e^ {-x} dx = 1 $$ If you are starting from the "usual" definition of the factorial, in my opinion it is best to take the statement $0! = 1$ as a part of the …

@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, …

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 …

Jul 20, 2010 · If you take the more general case of lim x^y as x,y -> 0 then the result depends on exactly how x and y both -> 0. Defining 0^0 as lim x^x is an arbitrary choice. There are unavoidable …

Is there a consensus in the mathematical community, or some accepted authority, to determine whether zero should be classified as a natural number? It seems as though formerly $0$ was considered i...

Aug 10, 2023 · A value of "0" doesn't tell the reader that we actually do know that the value is < 0.1. Would we not want to report it as 0.00? And if so, why wouldn't we also say that it has 2 significant …

This definition of the "0-norm" isn't very useful because (1) it doesn't satisfy the properties of a norm and (2) $0^ {0}$ is conventionally defined to be 1.

Nov 8, 2013 · That $0$ is a multiple of any number by $0$ is already a flawless, perfectly satisfactory answer to why we do not define $0/0$ to be anything, so this question (which is eternally recurring it …

Apr 6, 2017 · Since $0\in\mathbb R,$ there is no dispute that $0$ is a real number. So, by symmetry, it makes sense to also consider $0$ an imaginary number. (Wolfram Alpha agrees.) Hence, it makes …