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 …

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.

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...

@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, …

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 …

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?

@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 …

Oct 28, 2019 · That means that 1/0, the multiplicative inverse of 0 does not exist. 0 multiplied by the multiplicative inverse of 0 does not make any sense and is undefined. Therefore both 1/0 and 0/0 are …

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