Question

1. Consider the difference equation (1) Sn+2 - -8n+1Sn with initial conditions so = 0, 81 = 1 Let (x) n-0 denote the generating function of (sn) (a) Derive the fact that I++ = (x)f to show that (b) Use the generating function (x) r2+a+1 0 if n is a multiple of 3 1 if n is of the form 3j + 1 . -1 if n is of the form 3j 2 There are a number of ways to go about this. One way is to factor 2+x1 over the complex numbers, perform a partial fractions decomposition, and expand the resulting two series as geometric series. If you wish to avoid complex numbers another possibility is to expand a/(1+x+a2) = x/(1- (--a2)) as a geometric series with initial term x and common ratio x - x2. No complex numbers wil be involved, but facility with the binomial theorem and interchanging order of summation in a double series may be necessary

Answer 1

We have the differential equation s_n + 2 = -s_n + 1 - s_n therefore s_n + 2 + s_n + 1 + s_n = 0 - 1 Also we have s_0 = 0, s_1 = 1 Let f(x) = sigma^infinity_n = 0 s_n x^n - 2 From (1) we get sigma^infinity_n = 0 s_n + 2 x^n + 2 + sigma^infinity_n = 0 s_n + 1 x^n + 2 + sigma^infinity_n = 0 s_n x^n + 2 = 0 therefore s(x) - s_1 . x - s_0 + x . (s(x) - s_0) + x^2 . s(x) = 0 therefore s(x) . [x^2 + x + 1] = s_1 . x + s_0 - s_0 . x = x hence s_1 = 1, s_0 = 0 therefore s(zx) = x/x^2 + x + 1, is the generating function. we have x^2 + x + 1 = (x - w)(x - w^2) where w = -1/2 + I squareroot 3/2, w = -1/2 - I squareroot 3/2 therefore 1/x^2 +