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 …

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 …

@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, …

For example, $3^0$ equals 3/3, which equals $1$, but $0^0$ "equals" 0/0, which equals any number, which is why it's indeterminate. Also, 0/0 is undefined because of what I just said.

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 …

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

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 …

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 …