Problem 14 - Discrete Mathematics with Applications

Question

Using generating functions, solve each LHRRWCC. $$a_{n}=a_{n-1}+2 a_{n-2}, a_{0}=3, a_{1}=0$$

Step-by-Step Solution

Answer

The generating function for the given LHRRWCC is \(A(x) = 3 + 0x + 6x^2 + 0x^3 + \cdots\), which corresponds to the sequence \(\{a_n\} = \{3, 0, 6, 0, 24, 0, 96, 0, 384, \dots\}\), where \(a_n = 2^{n}a_{n-2}\) for \(n \geq 2\).

Step 1: Define the generating function

First, we define the generating function for the sequence \(\{a_n\}\), denoted by \(A(x)\): $$ A(x) = \sum_{n=0}^{\infty} a_n x^n $$

Step 2: Translate the recurrence relation

Next, we use the recurrence relation and the initial conditions to write a functional equation for \(A(x)\): $$ a_{0}x^0 + a_{1}x^1 + a_{2}x^2 + a_{3}x^3 + \cdots = (3x^0) + (0x^1) + (a_{1} + 2a_{0})x^2 + (a_{2} + 2a_{1})x^3 + \cdots $$ Using the initial conditions, replace \(a_0\) and \(a_1\): $$ A(x) = 3 + 0x + (0 + 2(3))x^2 + (0 + 2(0))x^3 + \cdots $$

Step 3: Solve for the generating function

Now, we can simplify the equation and use the partial fraction decomposition method to write it in a simpler form: $$ A(x) = 3 + 6x^2 $$ To find the generating function, we need to rewrite this as a power series. Divide both terms by \(1 - x\) and split it into partial fractions: $$ \frac{3}{1 - x} = \frac{3}{(1 - \sqrt{x})(1 + \sqrt{x})} $$ Applying partial fraction decomposition, we obtain: $$ \frac{3}{(1 - \sqrt{x})(1 + \sqrt{x})} = \frac{A}{1 - \sqrt{x}} + \frac{B}{1 + \sqrt{x}} $$ Multiplying the denominators gives: $$ 3 = A(1 + \sqrt{x}) + B(1 - \sqrt{x}) $$ Now, choosing \(x = 1\) and \(x = -1\), we find \(A = 3\) and \(B = -3\). Therefore, the generating function becomes: $$ A(x) = 3\frac{1}{1 - \sqrt{x}} - 3\frac{1}{1 + \sqrt{x}} $$

Step 4: Convert back into the sequence

Finally, we need to convert the generating function back into the sequence. Using the geometric series expansion, we have: $$ \frac{1}{1 - \sqrt{x}} = 1 + \sqrt{x} + (\sqrt{x})^2 + (\sqrt{x})^3 + \cdots $$ $$ \frac{1}{1 + \sqrt{x}} = 1 - \sqrt{x} + (\sqrt{x})^2 - (\sqrt{x})^3 + \cdots $$ Thus, we can rewrite \(A(x)\) as the sum of two power series: $$ A(x) = 3(1 + \sqrt{x} + (\sqrt{x})^2 + (\sqrt{x})^3 + \cdots) - 3(1 - \sqrt{x} + (\sqrt{x})^2 - (\sqrt{x})^3 + \cdots) $$ $$ A(x) = 3 + 0x + 6x^2 + 0x^3 + \cdots $$ Now, we obtain the sequence by equating the coefficients of the corresponding powers of \(x\): $$ a_n = \begin{cases} 3 & \text{if } n = 0, \\ 0 & \text{if } n = 1, \\ 2^{n}a_{n-2} & \text{for } n \geq 2. \end{cases} $$ The resulting sequence is \(\{3, 0, 6, 0, 24, 0, 96, 0, 384, \dots\}\).

Key Concepts

Linear Homogeneous Recurrence Relations with Constant Coefficients (LHRRWCC)

A linear homogeneous recurrence relation with constant coefficients (LHRRWCC) is a sequence where each term is a linear combination of previous terms, and all the coefficients are constants. For the sequence to be homogeneous, each term on the right side of the equation must only include terms from the sequence itself, with no additional independent term. Consider the recurrence relation from the exercise: \[a_{n}=a_{n-1}+2 a_{n-2}\]This formula suggests that each term is calculated based on the two preceding terms. The coefficients of previous terms, here 1 and 2, remain constant throughout the sequence. Homogeneous refers to the absence of terms not containing the sequence elements. The defining characteristics of LHRRWCC are as follows:

• Each term depends linearly on a fixed number of prior terms.
• The coefficients do not change with each successive term.
• No non-homogeneous terms are on the right side of the relation.

To solve these relations, you can utilize generating functions, which transform the problem into an equivalent algebraic problem through a power series representation.

Geometric Series Expansion

The geometric series is a fundamental mathematical concept useful for simplifying the generating function into a power series. The geometric series is represented as:
\[\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots\]
This expansion is valid when the absolute value of \(x\) is less than one. In our exercise, this property helps reconstruct the generating function from partial fractions. By manipulating the coefficients through partial fraction decomposition, each part is expressed using geometric series terms.

The importance lies in converting the decomposed form into a series of powers of \(x\), allowing each term of a geometric series to correlate directly with terms in the sequence. This technique highlights the clever transformation of algebraic manipulations back into discrete sequences. When working with generating functions, implementing geometric series is crucial for simplifying results into forms that reveal the structure of the underlying sequence.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions. It simplifies the integration or comparison against a known series, like the geometric series. In the context of solving recurrence relations, after transforming your problem using generating functions, you often end up with a rational function \(\frac{P(x)}{Q(x)}\), where both \(P(x)\) and \(Q(x)\) are polynomials.

Applying partial fraction decomposition involves splitting this ratio into simpler fractions, usually linear or quadratic, that are easier to process in terms of series expansions. For example, in the solution provided, the generating function initially appears as:
\(\frac{3}{(1 - \sqrt{x})(1 + \sqrt{x})}\)
Partial fraction decomposition allows us to express this in the form:
\(\frac{A}{1 - \sqrt{x}} + \frac{B}{1 + \sqrt{x}}\)
By solving for constants \(A\) and \(B\), terms are made suitable for geometric series expansions. It's a valuable skill in algebra that makes many mathematical problems more approachable and facilitates determining each sequence's term through generating functions.