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Solve the simultaneous recurrence relations $$ \begin{aligned} a_{n} &=3 a_{n-1}+2 b_{n-1} \\ b_{n} &=a_{n-1}+2 b_{n-1} \end{aligned} $$ with \(a_{0}=1\) and \(b_{0}=2\)

Short Answer

Expert verified
First terms: \(a_{0}=1\), \(a_{1}=7\), and \(a_{2}=31\). First terms: \(b_{0}=2\), \(b_{1}=5\), and \(b_{2}=17\).

Step by step solution

01

- Identify Initial Conditions

Given the initial conditions: \(a_{0} = 1\) and \(b_{0}=2\).
02

- Find the First Term

Using the initial conditions in the recurrence relations, find \(a_{1}\) and \(b_{1}\):\[\begin{aligned} a_{1} &= 3a_{0} + 2b_{0} = 3(1) + 2(2) = 3 + 4 = 7 \ b_{1} &= a_{0} + 2b_{0} = 1 + 2(2) = 1 + 4 = 5\end{aligned}\]
03

- Find the Second Term

Using \(a_{1} = 7\) and \(b_{1} = 5\) in the recurrence relations, find \(a_{2}\) and \(b_{2}\):\[\begin{aligned} a_{2} &= 3a_{1} + 2b_{1} = 3(7) + 2(5) = 21 + 10 = 31 \ b_{2} &= a_{1} + 2b_{1} = 7 + 2(5) = 7 + 10 = 17\end{aligned}\]
04

- General Term Approach

To find a general solution, express the recurrence relations in matrix form and solve for the eigenvalues and eigenvectors. However, for simpler calculations, the process can be repeated for as many terms as needed.

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Key Concepts

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

Initial Conditions
Initial conditions are the starting values of a sequence or system. In our problem, the initial conditions provided are:
  • \( a_{0} = 1 \)
  • \( b_{0} = 2 \)

These values are crucial as they allow us to initiate the sequence using recurrence relations. Essentially, they set the foundation for calculating subsequent terms. When we're given initial conditions, we can substitute these values into the recurrence relations to find the first few terms of the sequences.

In this case, we used the given initial conditions to compute \( a_{1} \) and \( b_{1} \). By knowing the beginning point, we systematically build up the sequence by repeating the process.
Recurrence Relations
Recurrence relations are equations that express each term of a sequence as a function of the preceding terms. The given relations in the exercise encapsulate how each new term is generated from the previous ones:
  • \( a_{n} = 3a_{n-1} + 2b_{n-1} \)
  • \( b_{n} = a_{n-1} + 2b_{n-1} \)

These formulas specify a step-by-step process for moving from one term to the next.
For example, to find \( a_{1} \), you use the values \( a_{0} \) and \( b_{0} \). Similarly, to compute \( b_{1} \), the same initial values are used. Through this iterative process, more terms can be generated to build the sequence.

Recurrence relations can be found in many fields including computer science, engineering, and natural sciences, as they model the way systems evolve over time.
Matrix Form Solution
Converting the system of recurrence relations into matrix form is a powerful method to solve it. The given relations can be expressed in matrix form as:

\begin{aligned}\begin\br [ a_{n} \b_{n} ] &= [ 3 2 \ 1 2 ] [ a_{n-1} \b_{n-1} ] \end{aligned}\begin{aligned}\begin\br [ a_{n} \b_{n} ] &= A [ a_{n-1} \b_{n-1} ] \end\br \begin{aligned}\end{aligned}

where matrix \( A \) is:

\begin{bmatrix} 3 & 2 1 & 2 \begin{aligned}\end{bmatrix}\begin{aligned}\end{br \begin{aligned}\end{matrix}\end{aligned}\end{bmatrix}\end{matrix}\end{aligned}\begin{aligned}\end{noun}\begin{aligned}\begin{aligned}A^{n} [ a_{0} \b_{0} ] <\br>Using this matrix form, we systematically apply the transformation to compute subsequent terms. Matrix representation abstracts the relations, simplifying complex systems, and is widely used in more advanced mathematical modeling.
Eigenvalues and Eigenvectors
To solve recurrence relations expressed in matrix form, eigenvalues and eigenvectors of the transformation matrix play a crucial role. Eigenvalues \( \lambda \) and eigenvectors \(v\) satisfy:

\[\begin{equation}A v = \lambda v \end{equation}\]<\br>For our matrix A: \begin{bmatrix}3 & 2 \1 & 2 \end{br}numerical methods or characteristic equation solving yields eigenvalues. Once we find the eigenvalues, corresponding eigenvectors are computed. These eigenvectors help in expressing the sequence terms as linear combinations, facilitating easier computation.

This approach provides deep insights into the system’s behavior, revealing patterns. Eigenvalues indicate growth rates of the sequences, while eigenvectors help in defining the direction of these growth patterns.

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Most popular questions from this chapter

Exercises 33–37 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 the value of \(J(n)\) for each integer \(n\) with \(1 \leq\) \(n \leq 16 .\)

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}

a) Find the recurrence relation satisfied by \(S_{n},\) where \(S_{n}\) is the number of regions into which three-dimensional space is divided by \(n\) planes if every three of the planes meet in one point, but no four of the planes go through the same point, b) Find \(S_{n}\) using iteration.

Generating functions are useful in studying the number of different types of partitions of an integer \(n .\) A partition of a positive integer is a way to write this integer as the sum of positive integers where repetition is allowed and the or- der of the integers in the sum does not matter. For exam- ple, the partitions of 5 (with no restrictions) are \(1+1+1+1+\) \(1+1,1+1+1+2,1+1+3,1+2+2,1+4,2+3,\) and \(5 .\) Exercises \(53-58\) illustrate some of these uses. Show that the coefficient \(p_{d}(n)\) of \(x^{n}\) in the formal power series expansion of \((1+x)\left(1+x^{2}\right)\left(1+x^{3}\right) \cdots\) equals the number of partitions of \(n\) into distinct parts, that is, the number of ways to write \(n\) as the sum of positive inte- gers, where the order does not matter but no repetitions are allowed.

a) Find a recurrence relation for the number of bit strings of length n that contain three consecutive 0s. b) What are the initial conditions? c) How many bit strings of length seven contain three consecutive 0s?

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