Problem 4 - Applied Discrete Structures

Question

Find closed form expressions for the generating functions of the following sequenoes: (a) \(W(n)=\left(\begin{array}{l}5 \\ n\end{array}\right) 2^{n}\) for \(0 \leq n \leq 5\) and \(W(n)=0\) for \(n>5 .\) (b) \(Q,\) where \(Q(k)+Q(k-1)-42 Q(k-2)=0\) for \(k \geq 2\), with \(Q(0)=2\) and \(Q(1)=2\). (c) \(G,\) where \(G(k+3)=G(k+2)+G(k+1)+G(k)\) for \(k \geq 0,\) with \(G(0)=G(1)=G(2)=1\)

Step-by-Step Solution

Answer

(a) Generating function: \((1 + 2x)^5\). (b) Solution: \(Q(k) = \frac{8}{13}(6^k) + \frac{18}{13}((-7)^k)\). (c) Generating function: \(G(x) = \frac{1 - 2x}{1 - x - x^2 - x^3}\).

Step 1: Understand the Binomial Generating Function

The sequence for part (a) is a binomial-type sequence. We know the binomial theorem involves summing over binomial coefficients multiplied by powers of a variable. In particular, the binomial theorem states:\[(1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k\]In this exercise, we have \(W(n) = \binom{5}{n} 2^n\) for \(0 \leq n \leq 5\). We recognize that this is equivalent to the expansion of \((1+2)^5\). Thus, the generating function is a finite polynomial:

Step 2: Write the Generating Function for Sequence W

For sequence \(W(n)\), the generating function is given by:\[W(x) = \sum_{n=0}^{5} \binom{5}{n} (2x)^n = (1 + 2x)^5\]This provides a closed form expression for the generating function of \(W(n)\).

Step 3: Setup and Solve the Characteristic Equation for Sequence Q

For part (b), the recurrence relation is \(Q(k) + Q(k-1) - 42Q(k-2) = 0\). This is a homogeneous linear recurrence relation with constant coefficients. We solve the characteristic equation:\[r^2 + r - 42 = 0\]Factoring gives: \((r - 6)(r + 7) = 0\), so the roots are \(r = 6\) and \(r = -7\).

Step 4: Formulate the General Solution and Use Initial Conditions

The general solution is:\[Q(k) = A(6^k) + B((-7)^k)\]Using initial conditions \(Q(0) = 2\) and \(Q(1) = 2\), set up equations:\[\begin{align*}A + B &= 2 \6A - 7B &= 2\end{align*}\]Solve for \(A\) and \(B\): \(A = 8/13, B = 18/13\). Thus, the solution becomes:\[Q(k) = \frac{8}{13}(6^k) + \frac{18}{13}((-7)^k)\]The generating function is built from this solution.

Step 5: Apply Recurrence Relation for Sequence G

Given \(G(k+3) = G(k+2) + G(k+1) + G(k)\) with initial conditions \(G(0) = G(1) = G(2) = 1\), we expand to find the first few terms:\[\begin{align*}G(3) &= 3 \G(4) &= 5 \G(5) &= 9\end{align*}\]Using the recurrence form, derive a generating function.

Step 6: Derive Generating Function for Sequence G

The recurrence \(G(k+3) = G(k+2) + G(k+1) + G(k)\) is solved using the generating function approach:\[G(x) = \frac{1 - 2x}{1 - x - x^2 - x^3}\]The denominator comes from transforming the relation into a third-degree polynomial term.

Key Concepts

Binomial Theorem

The Binomial Theorem is a central concept in algebra that involves expanding expressions that are raised to a power. When we say \((1 + x)^n\), it represents a series of terms involving binomial coefficients \(\binom{n}{k}\) and powers of \(x\). Each term in this expansion has the form \(\binom{n}{k} x^k\) for \(k = 0, 1, 2, \ldots, n\).
This mathematical principle becomes handy in exercises like Exercise (a), allowing us to deduce generating functions for specific sequences. For example, in the given sequence \(W(n) = \binom{5}{n} 2^n\), it mirrors the expansion of \((1 + 2x)^5\).

• This expansion means the generating function is a polynomial reflecting all terms where \(0 \leq n \leq 5\).
• The power of \(x\) aligns with different terms of our sequence.

The binomial theorem helps find closed forms and expressions for these generating functions.

Recurrence Relations

Recurrence relations are equations that define sequences recursively. Each term in the sequence is described using previous terms.
For Exercise (b), we deal with a recurrence relation: \(Q(k) + Q(k-1) - 42Q(k-2) = 0\). It means each term \(Q(k)\) depends on two prior terms.

• This reliance allows us to understand how terms in the sequence build upon each other.
• Initial conditions, like \(Q(0) = 2\) and \(Q(1) = 2\), give us starting points to solve these relations.

Recurrence relations are foundational in generating functions, forming the link between sequences and closed form expressions.

Linear Homogeneous Recurrence

Linear homogeneous recurrence relations of constant coefficients are a specific type of recurrence. They appear frequently in theoretical and applied mathematics.
For Exercise (b), the relation \(Q(k) + Q(k-1) - 42Q(k-2) = 0\) is linear (terms are raised to the first power) and homogeneous (equal to zero). The term "constant coefficients" indicates that the numbers preceding the variables don't change.

• To solve this, we use the characteristic equation, which unfolds the hidden structure of the sequence.
• Solutions to this equation, like \(r = 6\) and \(r = -7\), directly construct the form of our sequence.

The understanding of this type of recurrence helps in forming and solving characteristic equations which predict how sequences behave over time.

Characteristic Equation

Characteristic equations come into play when dealing with linear homogeneous recurrence relations like in Exercise (b). They are used to transform recurrence relations into algebraic equations.
For \(Q(k) + Q(k-1) - 42Q(k-2) = 0\), the characteristic equation is \(r^2 + r - 42 = 0\). Solving this equation gives us roots that indicate the behavior of the sequence.

• The roots of this polynomial determine the form of the general solution: \(Q(k) = A(6^k) + B((-7)^k)\).
• These roots provide the exponential factors that build the sequence solutions.

Understanding characteristic equations allows us to see how powerful transformations of recurrence relations can lead to generating accurate predictions of sequence terms.