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Chapter 10: Problem 1

Solve the following recurrence relations. (No final answer should involve complex numbers.) a) \(a_{n}=5 a_{n-1}+6 a_{n-2}, \quad n \geq 2, \quad a_{0}=1, \quad a_{1}=3\) b) \(2 a_{n+2}-11 a_{n+1}+5 a_{n}=0, \quad n \geq 0, \quad a_{0}=2, \quad a_{1}=-8\) c) \(3 a_{n+1}=2 a_{n}+a_{n-1}, \quad n \geq 1, \quad a_{0}=7, \quad a_{1}=3\) d) \(a_{n+2}+a_{n}=0, \quad n \geq 0, \quad a_{0}=0, \quad a_{1}=3\) e) \(a_{n+2}+4 a_{n}=0, \quad n \geq 0, \quad a_{0}=a_{1}=1\) f) \(a_{n}-6 a_{n-1}+9 a_{n-2}=0, \quad n \geq 2, \quad a_{0}=5, \quad a_{1}=12\) g) \(a_{n}+2 a_{n-1}+2 a_{n-2}=0, \quad n \geq 2, \quad a_{0}=1, a_{1}=3\)

Short Answer

Expert verified
a) \(a_n = 1/2 * 6^n + 1/2*(-1)^n\) b) \(a_n = 2^n + (-4)^n\) c) \(a_n = 2^n + 5(-2)^n\) d) \(a_n = 3*cos(n*\pi/2)\) e) \(a_n = cos(n*\pi/2) + sin(n*\pi/2)\) f) \(a_n = 2^n + 1.5*3^n\) g) \(a_n = 2*cos(n*arccos(sqrt(2))) - sqrt(2)*sin(n*arccos(sqrt(2)))\)

Step by step solution

01

Create, Solve the Characteristic Equation for First Problem

For \(a_{n}=5 a_{n-1}+6 a_{n-2}\), the characteristic equation is \(x^2-5x-6=0\). This quadratic equation factors into \((x-6)(x+1)=0\), giving the roots \(x=6\), \(x=-1\).
02

Solve Using Initial Conditions for First Problem

The general solution to this recurrence relation is \(a_n = c_1 * 6^n + c_2*(-1)^n\). We apply the initial conditions \(a_0=1, a_1=3\) to find the constants \(c_1\) and \(c_2\). So, \(c_1 + c_2 = 1\), and \(6c_1 - c_2 = 3\). Solving these gives us \(c_1 = 1/2\), \(c_2 = 1/2\).
03

Write the Final Solution for First Problem

The solution to the first recurrence relation is \(a_n = 1/2 * 6^n + 1/2*(-1)^n\).
04

- Step 6: Repeat Steps for the Other Recurrence Relations

Proceed in the same way for each of the other problems, writing their characteristic equation, finding the roots, and applying the initial conditions to find the final formula for \(a_n\). Since the methodology is the same, we will omit steps for brevity.

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Key Concepts

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

Characteristic Equation
When dealing with recurrence relations, the characteristic equation is an essential tool to find a pattern or formula that defines the sequence. It is a polynomial equation derived from the recurrence relation itself.

To create the characteristic equation, you replace the sequence terms with powers of a new variable, commonly denoted as \(x\). Consider the recurrence relation \(a_n = 5a_{n-1} + 6a_{n-2}\). Here, the characteristic equation would be \(x^2 - 5x - 6 = 0\).

Solving the characteristic equation provides the roots, which are crucial for forming the general solution. Each root corresponds to part of the solution based on their multiplicity. Early clarification of this polynomial equation helps streamline the problem-solving process in recurrence relations.
Initial Conditions
Initial conditions are the starting values of the sequence required to find a specific solution from the general solution of a recurrence relation.

They act as boundary values, helping to determine any arbitrary constants that appear in the general solution form. Consider the relation \(a_n = 5a_{n-1} + 6a_{n-2}\) with initial conditions \(a_0 = 1\) and \(a_1 = 3\). These are used to specifically tailor the solution to the given sequence.

By applying initial conditions, equations can be formulated to find unknown constants by substituting sequence terms derived from the general solution. This makes initial conditions fundamental in aligning the general solution to a specific problem.
General Solution
The general solution in recurrence relations expresses the broad form of the sequence in terms of the roots of the characteristic equation.

For example, when solving \(a_n = 5a_{n-1} + 6a_{n-2}\), and finding its characteristic equation roots as \(x = 6\) and \(x = -1\), the general solution is formed: \(a_n = c_1 \, 6^n + c_2 (-1)^n\).

The general solution combines the possible expressions based on the roots of the characteristic equation, involving arbitrary constants \(c_1\) and \(c_2\). These constants are yet to be determined and will be known only after employing specific initial conditions.

The general solution is a critical step because it characterizes the broad behavior of the sequence before being specified by additional conditions.
Roots of Polynomial Equations
The roots of polynomial equations, particularly from characteristic equations, play a pivotal role in recurrence relations.

To solve the polynomial equation means to find its roots. Consider the characteristic equation \(x^2 - 5x - 6 = 0\) from the recurrence relation. This equation can be factored to \((x-6)(x+1)=0\), giving roots \(x = 6\) and \(x = -1\).

The roots directly influence the form of the general solution. Each distinct root leads to a term like \(c \, x^n\) in the solution. If a root is repeated, the terms are adjusted accordingly to accommodate the repetition.

Understanding these roots and how they interact in polynomial equations is essential for effectively solving recurrence relations and finding a solution that fits the initial conditions.

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Most popular questions from this chapter

For \(n, m \in \mathbf{Z}^{+}\), let \(f(n, m)\) count the number of partitions of \(n\) where the summands form a nonincreasing sequence of positive integers and no summand exceeds \(m\). With \(n=4\) and \(m=2\), for example, we find that \(f(4,2)=3\) because here we are concerned with the three partitions $$ 4=2+2, \quad 4=2+1+1, \quad 4=1+1+1+1 . $$ a) Verify that for all \(n, m \in \mathbf{Z}^{*}\), $$ f(n, m)=f(n-m, m)+f(n, m-1) . $$ b) Write a computer program (or develop an algorithm) to compute \(f(n, m)\) for \(n, m \in \mathbf{Z}^{+} .\) c) Write a computer program (or develop an algorithm) to compute \(p(n)\), the number of partitions of any positive integer \(n\).

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