Assignment 2 François Séguin

First of all, we will be using the following theorem proven in class about the recurrence relation of |s(n, k)| = S * (n, k). holds for any n, k, and defines a recurrence relation for S * (n, k). Let us define (t) n = t(t + 1)(t + 2) · · · (t + n − 1) to be the polynomial in t with roots 0, −1, −2,. .. , n − 1. Let us now define a(n, k) as follows (t) n = n k=0 a(n, k)t k. (1) We want to find a recurrence relation for a(n, k). We have that (t) n+1 = (t) n (t + n) = t(t) n + n(t) n and so by using equation (1) n+1 k=0 a(n + 1, k)t k = n k=0 a(n, k)t k+1 n k=0 na(n, k)t k = n+1 k=0 a(n, k − 1)t k n k=0 na(n, k)t k. Thus by equating the coefficient of t k from both sides, we have a(n + 1, k) = a(n, k − 1) + na(n, k). We notice that a(n, k) follows exactly the same recurrence relation as S * (n, k). If the initial conditions are the same, then it follows that a(n, k) = S * (n, k). The coefficient for the term t in (t) n is simply the constant term of (t) n t , which is clearly (n − 1)!. We thus have a(n, 1) = (n − 1)!.