Finding a Non-Recursive Formula
Date: 06/10/99 at 13:32:16 From: Gabriel Alume Subject: Method of iteration I cannot for the life of me see the pattern in this problem based on iteration. I understand that once you find a pattern in the solutions to the recurrence relation, you guess a formula to solve. Am I not seeing something? Because I can't find a pattern. Please help. recurrence relation: s_n = - [s_(n-1)] - n^2 initial condition: s_0 = 3 s_1 = -4 s_2 = 0 s_3 = -9 s_4 = -7 s_5 = -18
Date: 06/10/99 at 17:21:25 From: Doctor Anthony Subject: Re: Method of iteration The recurrence relation is u(n) + u(n-1) = -n^2 Let u(r) = v(r) + w(r) where v(r) is solution of v(r) + v(r-1) = 0 So we solve x + 1 = 0 x= -1 and v(r) = A.(-1)^r and we use a trial solution w(r) = a.r^2 + b.r Substituting into the equation a.r^2 + b.r + a(r-1)^2 + b(r-1) = -r^2 a.r^2 + b.r + a(r^2 - 2r + 1) + b(r-1) = -r^2 2a.r^2 + r(2b-2a) + a-b = -r^2 2a.r^2 + 2r(b-a) + a-b = -r^2 comparing coefficients of r on both sides we require a = -1/2, b = a = -1/2 and so w(r) = -(1/2)[r^2 + r] Therefore the general solution is u(n) = A(-1)^n - (1/2)[n^2 + n] and the initial condition is u(0) = 3 so A = 3 We get u(n) = 3(-1)^n - (1/2)[n^2 + n] Check with n = 5 u(5) = -3 - (1/2)[25 + 5] = -3 - (1/2)(30) = -3 - 15 = -18 which checks. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/
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