Discrete Mathematics | Types of Recurrence Relations Set 2 (original) (raw)

Last Updated : 11 Jul, 2025

Prerequisite - Solving Recurrences, Different types of recurrence relations and their solutions, Practice Set for Recurrence RelationsThe sequence which is defined by indicating a relation connecting its general term an with an-1, an-2, etc is called a recurrence relation for the sequence.

Types of recurrence relations

• First order Recurrence relation :- A recurrence relation of the form : an = can-1 + f(n) for n>=1 where c is a constant and f(n) is a known function is called linear recurrence relation of first order with constant coefficient. If f(n) = 0, the relation is homogeneous otherwise non-homogeneous.Example :- xn = 2xn-1 - 1, an = nan-1 + 1, etc.Question :- Solve the recurrence relation T(2k) = 3T(2k-1) + 1, T(1) = 1. Let T(2k) = ak. Therefore, ak = 3ak-1 + 1 Multiplying by xk and then taking sum, Σakxk = 3Σak-1xk + Σxk ------> (1) Σak-1xk = [a0x + a1x2 + ......] = x[a0 + a1x + ......] = x[G(x)] (1) becomes G(x) - 3xG(x) - x/(1-x) = 0 G(x)(1-3x) - x/(1-x) = 0 G(x) = x/[(1-x)(1-3x)] = A/(1-x) + B/(1-3x) --> A = -1/2 and B = 3/2 G(x) = (3/2)Σ(3x)k - (1/2)Σ(x)kCoefficient of xk is, ak = (3/2)3k - (1/2)1kSo, ak = [3k+1 - 1]/2.
• Second order linear homogeneous Recurrence relation :- A recurrence relation of the form cnan + cn-1an-1 + cn-2an-2 = 0 ------> (1) for n>=2 where cn, cn-1 and cn-2 are real constants with cn != 0 is called a second order linear homogeneous recurrence relation with constant coefficients. Solution to this is in form an = ckn where c, k!=0 Putting this in (1) cnckn + cn-1ckn-1 + cn-2ckn-2 = 0 cnk2 + cn-1k + cn-2 = 0 -----> (2) Thus, an = ckn is solution of (1) if k satisfies quadratic equation (2). This equation is called characteristic equation for relation (1). Now three cases arises,Case 1 : If the two roots k1, k2 of equation are real and distinct then, we take an = A(k1)n + B(k2)n as general solution of (1) where A and B are arbitrary real constants. Case 2 : If the two roots k1, k2 of equation are real and equal, with k as common value then, we take an = (A + Bn)kn as general solution of (1) where A and B are arbitrary real constants. Case 3 : If the two roots k1 and k2 of equation are complex then, k1 and k2 are complex conjugate of each other i.e k1 = p + iq and k2 = p - iq and we take an = rn(Acosnθ + Bsinnθ) as general solution of (1) where A and B are arbitrary complex constants, r = |k1| = |k2| = √p2 + q2 and θ = tan-1(q/p).

Question :- Solve the recurrence relation an + an-1 - 6an-2 = 0 for n>=2 given that a0 = -1 and a1 = 8. Here coefficients of an, an-1 and an-2 are cn = 1, cn-1 = 1 and cn-2 = -6 respectively. Hence, characteristic equation is k2 + k - 6 or (k + 3)(k - 2) = 0 ------> (1) The roots of (1) are k1 = -3 and k2 = 2 which are real and distinct. Therefore, general solution is an = A(-3)n + B(2)nwhere A and B are arbitrary constants. From above we get, a0 = A + B and a1 = -3A + 2B A + B = -1 -3A + 2B = 8 Solving these we get A = -2 and B = 1 Therefore, an = -2(-3)n + (2)n