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Chapter 8: Problem 34

Find all solutions of the recurrence relation \(a_{n}=\) \(7 a_{n-1}-16 a_{n-2}+12 a_{n-3}+n 4^{n} \quad\) with \(\quad a_{0}=-2\) \(a_{1}=0,\) and \(a_{2}=5\)

Short Answer

Expert verified
Combine homogeneous and non-homogeneous parts; solve with initial conditions.

Step by step solution

01

- Identify the Homogeneous Part

First, solve the homogeneous part of the recurrence relation: a_{n} = 7a_{n-1} - 16a_{n-2} + 12a_{n-3}.
02

- Find the Characteristic Equation

Determine the characteristic equation of the homogeneous recurrence relation: r^3 - 7r^2 + 16r - 12 = 0.
03

- Solve the Characteristic Equation

Find the roots of the characteristic equation. By factoring, solve for:(r-2)(r-3)(r-2) = 0,which gives r = 2 (twice) and r = 3.
04

- Form the General Solution of the Homogeneous Equation

Use the roots to form the general solution of the homogeneous equation:a_{n,h} = c_{1}2^{n} + c_{2}n2^{n} + c_{3}3^{n}.
05

- Solve the Non-Homogeneous Part

For the non-homogeneous term treat it as term = n4^{n} and propose a particular solution of the form a_{n,p} = n^m d_{n}4^{n}+ ... .
06

- Combine Solutions

Combine the homogeneous and non-homogeneous solutions to get the full recurrence relation solution:a_{n} = c_{1}2^{n} + c_{2}n2^{n} + c_{3}3^{n} + n f(n) 4^{n}.
07

- Apply Initial Conditions

Use the initial conditions a_{0} = -2, a_{1} = 0, and a_{2} = 5 to find the values of c_{1}, c_{2}, c_{3}, and other coefficients.

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Key Concepts

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

Homogeneous Recurrence Relation
A recurrence relation expresses each term of a sequence as a function of its preceding terms. When all terms on the right side depend solely on preceding terms and there is no external input, it is called a homogeneous recurrence relation. For example, in the exercise given, the homogeneous part is:

  • \[a_{n} = 7a_{n-1} - 16a_{n-2} + 12a_{n-3}\]
This equation has no additional terms and depends only on the previous terms of the sequence. To tackle such a recurrence relation, we first decompose it into simpler parts that can be solved individually, namely the characteristic equation.
Characteristic Equation
The characteristic equation helps find the general form of a solution to a homogeneous linear recurrence relation. To derive the characteristic equation, we replace each occurrence of \(a_{n-k}\) with \(r^k\). For the given exercise, the characteristic equation associated with the recurrence

  • \[a_{n} = 7a_{n-1} - 16a_{n-2} + 12a_{n-3}\]
If we replace \(a_{n}\) by \(r^{n}\), \(a_{n-1}\) by \(r^{n-1}\), and so on, we obtain:
  • \[r^3 - 7r^2 + 16r - 12 = 0\]
Solving this polynomial equation provides the roots, which are essential for constructing the general solution of the recurrence.
Particular Solution
For non-homogeneous recurrence relations, we need to find a particular solution that accounts for the non-homogeneous term. In our exercise, the non-homogeneous term is \(n4^{n}\). Hence, we guess a particular solution of the form:

  • \[a_{n,p} = n f(n) 4^{n}\]
Here, \(f(n)\) is a function that depends on \(n\). Plugging this guess into the non-homogeneous recurrence and equating terms appropriately, we can solve for unknown coefficients to get a specific solution.
Initial Conditions
Initial conditions are crucial because they allow us to pinpoint a specific solution out of the infinitely many ones provided by the general form. For the given exercise, the initial conditions are:

  • \(a_{0} = -2\)
  • \(a_{1} = 0\)
  • \(a_{2} = 5\)
Using these initial values, we can create a system of equations. By substituting these into the general solution, we solve for the constants \(c_{1}\), \(c_{2}\), and \(c_{3}\). Thus, we obtain a unique and complete solution to the original recurrence relation.
General Solution
The general solution of a recurrence relation combines both the homogeneous and particular solutions. Using our previously solved components, we achieve:

  • Homogeneous solution: \(a_{n, h} = c_{1}2^{n} + c_{2}n2^{n} + c_{3}3^{n}\)
  • Particular solution: \(a_{n,p} = n f(n) 4^{n}\)
Combining both, we get the overall solution:

\[ a_{n} = c_{1}2^{n} + c_{2}n2^{n} + c_{3}3^{n} + n f(n) 4^{n} \]Substituting the values of \(c_{1}\), \(c_{2}\), and \(c_{3}\) determined by the initial conditions, we acquire the unique solution for the given problem. This final solution captures all behaviors of the sequence defined by the recurrence relation.

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Most popular questions from this chapter

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.

Exercises \(38-45\) involve the Reve's puzzle, the variation of the Tower of Hanoi puzzle with four pegs and \(n\) disks. Before presenting these exercises, we describe the Frame-Stewart algorithm for moving the disks from peg 1 to peg 4 so that no disk is ever on top of a smaller one. This algorithm, given the number of disks \(n\) as input, depends on a choice of an integer \(k\) with \(1 \leq k \leq n .\) When there is only one disk, move it from peg 1 to peg 4 and stop. For \(n>1\) , the algorithm proceeds recursively, using these three steps. Recursively move the stack of the \(n-k\) smallest disks from peg 1 to peg \(2,\) using all four pegs. Next move the stack of the \(k\) largest disks from peg 1 to peg \(4,\) using the three-peg algorithm from the Tower of Hanoi puzzle without using the peg holding the \(n-k\) smallest disks. Finally, recursively move the smallest \(n-k\) disks to peg \(4,\) using all four pegs. Frame and Stewart showed that to produce the fewest moves using their algorithm, \(k\) should be chosen to be the smallest integer such that \(n\) does not exceed \(t_{k}=k(k+1) / 2,\) the \(k\) th triangular number, that is, \(t_{k-1}

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

a) Find a recurrence relation for the number of ternary strings of length n that do not contain two consecutive 0s or two consecutive 1s. b) What are the initial conditions? c) How many ternary strings of length six do not contain two consecutive 0s or two consecutive 1s?

A coding system encodes messages using strings of base 4 digits (that is, digits from the set \(\\{0,1,2,3\\} )\) . A codeword is valid if and only if it contains an even number of 0 \(\mathrm{s}\) an even number of 1 \(\mathrm{s}\) . Let \(a_{n}\) equal the number of valid codewords of length \(n .\) Furthermore, let \(b_{n}, c_{n},\) and \(d_{n}\) equal the number of strings of base 4 digits of length \(n\) with an even number of 0 \(\mathrm{s}\) and an odd number of \(1 \mathrm{s},\) with an odd number of 0 \(\mathrm{s}\) and an even number of \(1 \mathrm{s},\) and with an odd number of 0 \(\mathrm{s}\) and an odd number of 1 \(\mathrm{s}\) , respectively. a) Show that \(d_{n}=4^{n}-a_{n}-b_{n}-c_{n}\) . Use this to show that \(a_{n+1}=2 a_{n}+b_{n}+c_{n}, b_{n+1}=b_{n}-c_{n}+4^{n},\) and \(c_{n+1}=c_{n}-b_{n}+4^{n} .\) b) What are \(a_{1}, b_{1}, c_{1},\) and \(d_{1} ?\) c) Use parts (a) and (b) to find \(a_{3}, b_{3}, c_{3},\) and \(d_{3}\) d) Use the recurrence relations in part (a), together with the initial conditions in part (b), to set up three equa- tions relating the generating functions \(A(x), B(x),\) and \(C(x)\) for the sequences \(\left\\{a_{n}\right\\},\left\\{b_{n}\right\\},\) and \(\left\\{c_{n}\right\\},\) respec- tively. e) Solve the system of equations from part (d) to get ex- plicit formulae for \(A(x), B(x),\) and \(C(x)\) and use these to get explicit formulae for \(a_{n}, b_{n}, c_{n},\) and \(d_{n}\)
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