Problem 35 Use generating functions to solv... [FREE SOLUTION]

Chapter 8: Problem 35

Use generating functions to solve the recurrence relation \(a_{k}=3 a_{k-1}+2\) with the initial condition \(a_{0}=1\)

Short Answer

Expert verified

The generating function method gives a sequence generated by \(1+x\)Terms in the summation \(\frac{1+x}{(1-x)(1-3x)} = (1+x) .\sum_{n=0}^{\infty} (\sum_{i+j=n} (3x)^i j\).\}

Step by step solution

Define the Generating Function

Define the generating function for the sequence \(\big\{a_k\big\}\): \[A(x) = \sum_{k=0}^{\infty} a_k x^k.\]

Substitute the Recurrence Relation

Substitute the recurrence relation \(\big\{a_k = 3a_{k-1} + 2\big\}\) into the generating function: \[A(x) = a_0 + \sum_{k=1}^{\infty} (3a_{k-1} + 2) x^k.\]

Separate the Summation Parts

Separate the summation into two parts based on the recurrence relation: \[A(x) = a_0 + 3\sum_{k=1}^{\infty} a_{k-1} x^k + 2\sum_{k=1}^{\infty} x^k\]Notice that \(\big\{\sum_{k=1}^{\infty} a_{k-1} x^k = x \sum_{k=0}^{\infty} a_k x^k = xA(x)\big\}\) and \(\sum_{k=1}^{\infty} x^k = \frac{x}{1-x}\big\}\).

Simplify the Expression

Simplify the expression using the previous step: \[A(x) = 1 + 3xA(x) + 2\frac{x}{1-x}\]

Solve for A(x)

Rearrange and solve for \(A(x)\): \[A(x) - 3xA(x) = 1 + \frac{2x}{1-x}\]\[A(x)(1-3x) = 1 + \frac{2x}{1-x}\]\[A(x) = \frac{1 + \frac{2x}{1-x}}{1-3x}\]

Simplify Further

Combine the fractions in the numerator: \[A(x) = \frac{1 - x + 2x}{(1-x)(1-3x)}\]\[A(x) = \frac{1 + x}{(1-x)(1-3x)}\]

Expand A(x) into a Series

Expand \(A(x)\) into a power series: \[A(x) = \frac{1+x}{(1-x)(1-3x)} = (1+x) \sum_{n=0}^{\infty} (\sum_{i+j=n} x^i (3x)^j)\](using binomial theorem)Notice that \(1+x\) can be distributed.

Coefficients Identification

By comparing with the original series \(A(x) = \sum_{k=0}^{\infty} a_k x^k\), the coefficients of \(x^k\) gives \(a_k\) values.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

recurrence relation

A recurrence relation is an equation that recursively defines a sequence. Each term of the sequence is expressed as a function of one or more previous terms. In the given exercise, the recurrence relation is:
\[ a_k = 3a_{k-1} + 2 \]
This means that each term is three times the previous term plus two. Understanding this relation is crucial for defining the sequence and solving it.

generating function

A generating function transforms a sequence into a power series. This function makes it easier to handle and manipulate the sequence. For the sequence \(\{a_k\}\), the generating function \(A(x)\) is defined as:
\[ A(x) = \sum_{k=0}^{\infty} a_k x^k \]
By substituting the recurrence relation into this generating function, we can derive an expression that represents the entire sequence in a compact form.

series expansion

Series expansion involves expressing a function as a sum of terms in a sequence, usually involving powers of a variable. In our example, when we substitute the recurrence relation into the generating function, we get:
\[ A(x) = 1 + 3xA(x) + 2\sum_{k=1}^{\infty} x^k \]
Separating and simplifying the summations, we eventually expand \(A(x)\) into a power series. This allows us to find the individual terms \(a_k\) easily.

binomial theorem

The Binomial Theorem provides a way to expand expressions that are raised to a power. In this problem, we use it to expand the generating function. Specifically, we get:
\[ (1+x) \sum_{n=0}^{\infty} \sum_{i+j=n} x^i (3x)^j \]
This step simplifies the manipulation of our generating function, turning it into a form where series and individual coefficients can be easily identified.

coefficient identification

Coefficient identification is the process of finding the coefficients of the powers of x in the expanded series. These coefficients correspond to the terms in the original sequence. By comparing the expanded generating function to:
\[ A(x) = \sum_{k=0}^{\infty} a_k x^k \]
We determine the values for each \(a_k\). Each coefficient directly represents each term in the sequence.

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