DISCRETE MATHEMATICS/Student Engineering Notebook
Page 08·notes
COMBINATORICS & RECURRENCE RELATIONS—~34 MIN READ

Recurrence Relations & Ordinary Generating Functions

“Linear homogeneous recurrence with distinct/repeated roots, non-homogeneous particular solutions, and OGF power series.”

1.Solving Linear Homogeneous Recurrences

A recurrence relation of the form a_n + c_1 a_{n-1} + c_2 a_{n-2} + ... + c_k a_{n-k} = 0 has constant coefficients.

Method of Characteristic Roots:

Substitute trial solution a_n = r^n into the recurrence to obtain the characteristic equation:

r^k + c_1 r^{k-1} + ... + c_k = 0.

1. Distinct Real Roots r_1, r_2, ..., r_k:

a_n = A_1 r_1^n + A_2 r_2^n + ... + A_k r_k^n.

2. Repeated Roots (Root r with multiplicity m):

a_n = (A_0 + A_1 n + A_2 n^2 + ... + A_{m-1} n^{m-1}) r^n.

Constants A_i are determined from initial boundary conditions.

2.The Ordinary Generating Function (OGF) Method

An Ordinary Generating Function encapsulates an infinite numerical sequence {a_0, a_1, a_2, ...} as coefficients of a formal power series:

G(x) = sum_{n=0}^infty a_n x^n = a_0 + a_1 x + a_2 x^2 + ...

Standard Power Series Expansions:

- frac{1}{1 - x} = sum_{n=0}^infty x^n = 1 + x + x^2 + ...

- frac{1}{1 - a x} = sum_{n=0}^infty a^n x^n

- frac{1}{(1 - x)^2} = sum_{n=0}^infty (n + 1) x^n

- frac{1}{(1 - x)^k} = sum_{n=0}^infty binom{n + k - 1}{n} x^n

WORKED EXAM ANALYSIS · STEP-BY-STEP DERIVATIONOriginal Practice Problem · Modeled on GATE Pattern (2 Marks)

3. Worked Problem: Solving Second-Order Recurrence Relation

⚠️ Pitfall Warning:

Ensure initial conditions a0 and a1 are substituted correctly into the general solution to avoid algebraic sign flips.

Problem: Solve the recurrence relation a_n = 5a_{n-1} - 6a_{n-2} for n >= 2, with initial conditions a_0 = 1, a_1 = 4.

Step 1: Write characteristic equation:

r^2 - 5r + 6 = 0

(r - 2)(r - 3) = 0 => Distinct roots r1 = 2, r2 = 3.

Step 2: Set up general homogeneous solution:

a_n = c1 (2^n) + c2 (3^n).

Step 3: Solve for constants using initial conditions:

For n = 0: a_0 = c1(2^0) + c2(3^0) = c1 + c2 = 1.

For n = 1: a_1 = c1(2^1) + c2(3^1) = 2c1 + 3c2 = 4.

Step 4: Solve linear 2x2 system:

From eq 1: c1 = 1 - c2.

Substitute: 2(1 - c2) + 3*c2 = 4 => 2 + c2 = 4 => c2 = 2.

c1 = 1 - 2 = -1.

Step 5: Write final explicit closed-form expression:

a_n = 2 * (3^n) - (2^n).

Check for n = 2: a_2 = 2*(9) - 4 = 18 - 4 = 14.

From recurrence: a_2 = 5(4) - 6(1) = 20 - 6 = 14. Perfect match!

SOURCE TRANSPARENCY & CITATIONS

Primary Academic References & Exam Sources

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Discrete Mathematics and Its Applications (8th Edition) ·
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