Discrete Mathematics | Types of Recurrence Relations – Set 2

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

Types of recurrence relations

• First order Recurrence relation :- A recurrence relation of the form : a_n = ca_n-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 :- x_n = 2x_n-1 – 1, a_n = na_n-1 + 1, etc.

  Question :- Solve the recurrence relation T(2^k) = 3T(2^k-1) + 1, T(1) = 1.
  Let T(2^k) = a_k. Therefore, a_k = 3a_k-1 + 1
  Multiplying by x^k and then taking sum,
  Σa_kx^k = 3Σa_k-1x^k + Σx^k ——> (1)
  Σa_k-1x^k = [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)^k
  Coefficient of x^k is, a_k = (3/2)3^k – (1/2)1^k
  So, a_k = [3^k+1 – 1]/2.

  
  
  
• Second order linear homogeneous Recurrence relation :- A recurrence relation of the form
  c_na_n + c_n-1a_n-1 + c_n-2a_n-2 = 0 ——> (1)
  for n>=2 where c_n, c_n-1 and c_n-2 are real constants with c_n != 0 is called a second order linear homogeneous recurrence relation with constant coefficients.
  Solution to this is in form a_n = ckⁿ where c, k!=0
  Putting this in (1)
  c_nckⁿ + c_n-1ck^n-1 + c_n-2ck^n-2 = 0
  c_nk² + c_n-1k + c_n-2 = 0 —–> (2)
  Thus, a_n = 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
  a_n = 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
  a_n = (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
  a_n = 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 a_n + a_n-1 – 6a_n-2 = 0 for n>=2 given that a₀ = -1 and a₁ = 8.
Here coefficients of a_n, a_n-1 and a_n-2 are c_n = 1, c_n-1 = 1 and c_n-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
a_n = A(-3)ⁿ + B(2)ⁿ
where 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, a_n = -2(-3)ⁿ + (2)ⁿ