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 …

@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.

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 …

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 …

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 …

Nov 17, 2014 · I began by assuming that $\dfrac00$ does equal $1$ and then was eventually able to deduce that, based upon my assumption (which as we know was false) $0=1$. As this is clearly false …

@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 …

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 argument of …

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.