A 2 To solve this recurrence relation, it is convenient to write the following expansion: F n ~ ( s ) ≃ f n − s b n. A 3 Then F n ~ ( s = 0 ) = f n yields the extinction probability.
A 4 Rearranging terms yields − s 2 b n n + s ( f n n − b n ( λ + ϕ ) ) + ( λ + ϕ ) f n = ϕ f n − 1 + λ f n + 1 − s ( ϕ b n − 1 + λ b n + 1 ).