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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

01

Define the Generating Function

Define the generating function for the sequence \(\big\{a_k\big\}\): \[A(x) = \sum_{k=0}^{\infty} a_k x^k.\]
02

Substitute the Recurrence Relation

Substitute the recurrence relation \(\big\{a_k = 3a_{k-1} + 2\big\}\) into the generating function: \[A(x) = a_0 + \sum_{k=1}^{\infty} (3a_{k-1} + 2) x^k.\]
03

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 \(\sum_{k=1}^{\infty} x^k = \frac{x}{1-x}\big\}\).
04

Simplify the Expression

Simplify the expression using the previous step: \[A(x) = 1 + 3xA(x) + 2\frac{x}{1-x}\]
05

Solve for A(x)

Rearrange and solve for \(A(x)\): \[A(x) - 3xA(x) = 1 + \frac{2x}{1-x}\]\[A(x)(1-3x) = 1 + \frac{2x}{1-x}\]\[A(x) = \frac{1 + \frac{2x}{1-x}}{1-3x}\]
06

Simplify Further

Combine the fractions in the numerator: \[A(x) = \frac{1 - x + 2x}{(1-x)(1-3x)}\]\[A(x) = \frac{1 + x}{(1-x)(1-3x)}\]
07

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.
08

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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Most popular questions from this chapter

Find a closed form for the exponential generating function for the sequence \(\left\\{a_{n}\right\\},\) where \(\begin{array}{ll}{\text { a) } a_{n}=2 .} & {\text { b) } a_{n}=(-1)^{n}} \\\ {\text { c) } a_{n}=3^{n} .} & {\text { d) } a_{n}=n+1} \\ {\text { e) } a_{n}=1 /(n+1)}\end{array}\)

Exercises 33鈥�37 deal with a variation of the Josephus problem described by Graham, Knuth, and Patashnik in [GrKnPa94]. This problem is based on an account by the historian Flavius Josephus, who was part of a band of 41 Jewish rebels trapped in a cave by the Romans during the Jewish Roman war of the first century. The rebels preferred suicide to capture; they decided to form a circle and to repeatedly count off around the circle, killing every third rebel left alive. However, Josephus and another rebel did not want to be killed this way; they determined the positions where they should stand to be the last two rebels remaining alive. The variation we consider begins with n people, numbered 1 to n, standing around a circle. In each stage, every second person still left alive is eliminated until only one survives. We denote the number of the survivor by J(n). Determine \(J(100), J(1000),\) and \(J(10,000)\) from your formula for \(J(n) .\)

Find \(f(n)\) when \(n=2^{k},\) where \(f\) satisfies the recurrence relation \(f(n)=8 f(n / 2)+n^{2}\) with \(f(1)=1\)

In Exercises \(29-33,\) assume that \(f\) is an increasing function satisfying the recurrence relation \(f(n)=a f(n / b)+c n^{d},\) where \(a \geq 1, b\) is an integer greater than \(1,\) and \(c\) and \(d\) are positive real numbers. These exercises supply a proof of Theorem \(2 .\) Show that if \(a=b^{d}\) and \(n\) is a power of \(b,\) then \(f(n)=\) \(f(1) n^{d}+c n^{d} \log _{b} n .\)

Use generating functions to find the number of ways to choose a dozen bagels from three varieties鈥攅gg, salty, and plain鈥攊f at least two bagels of each kind but no more than three salty bagels are chosen.
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