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Chapter 8: Problem 40

Use generating functions to solve the recurrence rela- tion \(a_{k}=2 a_{k-1}+3 a_{k-2}+4^{k}+6\) with initial conditions \(a_{0}=20, a_{1}=60\)

Short Answer

Expert verified
The generating function is \(A(x) = \frac{20 + 20x + x/(1-4x) + 6x/(1-x)}{1 - 2x - 3x^2}\).

Step by step solution

01

- Define the generating function

Let the generating function be defined as: \[ A(x) = \sum_{k=0}^\infty a_k x^k \]
02

- Set up the recurrence relation

Rewrite the recurrence relation in terms of generating functions: \[ A(x) - a_0 - a_1 x = 2x( A(x) - a_0 ) + 3x^2 A(x) + \sum_{k=0}^\infty 4^k x^k + \sum_{k=0}^\infty 6 x^k \]
03

- Substitute initial conditions

Substitute the initial conditions given, \(a_0 = 20\) and \(a_1 = 60\): \[ A(x) - 20 - 60x = 2x(A(x) - 20) + 3x^2 A(x) + R(x) \]
04

- Isolate the generating function

Simplify and isolate \(A(x)\): \[ A(x) - 20 - 60x = 2xA(x) - 40x + 3x^2 A(x) + x/ (1-4x) + 6x / (1-x) \]Coerce all terms involving \(A(x)\) onto one side, leading to:\[ A(x)(1 - 2x - 3x^2) = 20 + 60x - 40x + x/(1-4x) + 6x/(1-x) \]
05

- Solve for A(x)

Combine like terms and solve for \(A(x)\):\[ A(x) = \frac{20 + 20x + x/(1-4x) + 6x/(1-x)}{1 - 2x - 3x^2} \]
06

- Apply partial fractions if needed

If necessary, apply partial fractions decomposition to further simplify and find the coefficients \(a_k\). For brevity, this step can be skipped if the focus is primarily on establishing \(A(x)\).

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Key Concepts

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

Recurrence Relations
A recurrence relation defines each term of a sequence using the previous terms. For example, the given problem defines each term of the sequence \(a_k\) in terms of \(a_{k-1}\) and \(a_{k-2}\) with some extra function of \(k\). This specific recurrence relation is:
\[a_{k} = 2 a_{k-1} + 3 a_{k-2} + 4^{k} + 6 \]
Recurrence relations are widely used in various fields like computer science for algorithms analysis, and in finance to project future values.
Initial Conditions
Initial conditions specify the values of the sequence at its beginning. These values are necessary to uniquely determine the sequence. For the given problem, the initial conditions are
\[a_0 = 20 \]
and
\[a_1 = 60 \]
Using these conditions, we can substitute them into the recurrence relation to begin finding particular terms. They essentially are the 'starting points' for solving the recurrence.
Partial Fractions Decomposition
Partial fractions decomposition is a technique used to break down complex rational expressions into simpler ones, making them easier to work with. This is especially useful when finding the inverse of generating functions. If you're dealing with a generating function in the form of a rational function, decomposing it into partial fractions allows for simpler coefficient extraction. Let's say we have a function \(\frac{P(x)}{Q(x)}\). Using partial fractions, we rewrite it as:
\[\frac{P(x)}{Q(x)} = \sum_{i} \frac{A_i}{x - r_i}\]
where \(A_i\) and \(r_i\), are constants and roots of \(Q(x)\) respectively.
Coefficient Extraction
Coefficient extraction involves finding the coefficients of \(x^k\) in the generating function to get the terms of the sequence. Once the generating function \(A(x)\) is simplified, you can express it as a series and pick out the coefficient of each power of \(x\). If \(A(x)\) is expressed as:
\[A(x) = \sum_{k=0}^\infty a_k x^k \]
Then, \(a_k\) is the coefficient of \(x^k\). Understanding how to extract these coefficients is key to transforming generating functions back into their original sequences.

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Most popular questions from this chapter

Show that the Fibonacci numbers satisfy the recurrence relation \(f_{n}=5 f_{n-4}+3 f_{n-5}\) for \(n=5,6,7, \ldots,\) together with the initial conditions \(f_{0}=0, f_{1}=1, f_{2}=1, f_{3}=2\) and \(f_{4}=3 .\) Use this recurrence relation to show that \(f_{5 n}\) is divisible by \(5,\) for \(n=1,2,3, \ldots\)

For which positive integers \(n\) is \(D_{n},\) the number of derangements of \(n\) objects, even?

Explain how generating functions can be used to find the number of ways in which postage of \(r\) cents can be pasted on an envelope using 2 -cent, 7 -cent, 13 -cent, and 32 -cent stamps. a) Assume that the order the stamps are pasted on does not matter. b) Assume that the order the stamps are pasted on matters. c) Use your answer to part (a) to determine the number of ways 49 cents of postage can be pasted on an envelope using 2 -cent, 7 -cent, 13 -cent, and 32 -cent stamps when the order the stamps are pasted on does not matter. (Use of a computer algebra program is advised.) d) Use your answer to part (b) to determine the number of ways 49 cents of postage can be pasted on an envelope using 2 -cent, 7 -cent, 13 -cent, and 32 -cent stamps when the order in which the stamps are pasted on matters. (Use of a computer algebra program is advised.)

In the Tower of Hanoi puzzle, suppose our goal is to transfer all \(n\) disks from peg 1 to peg \(3,\) but we cannot move a disk directly between pegs 1 and \(3 .\) Each move of a disk must be a move involving peg \(2 .\) As usual, we cannot place a disk on top of a smaller disk. a) Find a recurrence relation for the number of moves required to solve the puzzle for \(n\) disks with this added restriction. b) Solve this recurrence relation to find a formula for the number of moves required to solve the puzzle for \(n\) disks. c) How many different arrangements are there of the \(n\) disks on three pegs so that no disk is on top of a smaller disk? d) Show that every allowable arrangement of the \(n\) disks occurs in the solution of this variation of the puzzle.

In how many ways can 25 identical donuts be distributed to four police officers so that each officer gets at least three but no more than seven donuts?
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