Recent content by JanetJanet

The problem is that I have to solve the limit as a approaches infinity of x(a)e^a. Not just x(a). That means I have to know how fast x(a) goes to 0 as a goes to infinity.

The LHS is what the generating function adds up to. s is the argument of the generating function. x(a) is just the solution to that formula at the end with s between 0 and 1.

Let x(a) be the extinction probability of a branching process whose offspring is Poisson distributed with parameter a. I need to find the limit as a approaches infinity x(a)e^a. I tried computing x(a) directly using generating functions, and I found that it's the solution to e^(a(s-1))=s, but...