13 Jun
2020
13 Jun
'20
1:28 p.m.
Your example says 0^0 = binomial(0,0) = number of ways to choose zero apples from a bag with zero apples = 1.
All this of course over the integers.
It came in really handy when proving that \sum_{k \geq 0} \binom{2k}{k} x^k = \frac{1}{\sqrt{1-4x}} for | x | < 1/4, via complex calculus. I would have had to excuse the case x = 0, but with 0^0 defined as 1 it's simpler. Andres.