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Chapter 5: Problem 11

(a) Solve the recurrence relation \(a_{n+1}=-8 a_{n}-\) \(16 a_{n-1}, n \geq 1\), given \(a_{0}=5, a_{1}=17\) (b) Solve the recurrence relation \(a_{n+1}=-8 a_{n}-\) \(16 a_{n-1}+5, n \geq 1\), given \(a_{0}=2, a_{1}=-1\)

Short Answer

Expert verified
(a) Solution: \( a_n = (5 - n)(-4)^n \); (b) Solution: \( a_n = (\frac{51}{23} - \frac{14}{23}n)(-4)^n - \frac{5}{23} \).

Step by step solution

01

Understand the recurrence relation for part (a)

The given recurrence is homogeneous: \( a_{n+1} = -8a_n - 16a_{n-1} \). It does not involve any additional terms, making it easier to solve using characteristic equations.
02

Set up the characteristic equation for part (a)

To find the characteristic equation for the homogeneous recurrence relation, use \( r^2 = -8r - 16 \). Rearrange it to get \( r^2 + 8r + 16 = 0 \).
03

Solve the characteristic equation for part (a)

The characteristic equation \( r^2 + 8r + 16 = 0 \) is a quadratic equation. Solve it using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), where \( a = 1, b = 8, c = 16 \). The discriminant \( b^2 - 4ac = 0 \), giving a repeated root \( r = -4 \).
04

Write the general solution for part (a)

Since the root is repeated \( r = -4 \), the solution will be \( a_n = (C_1 + C_2 n)(-4)^n \).
05

Apply initial conditions for part (a)

Use the initial conditions to solve for \( C_1 \) and \( C_2 \). Given \( a_0 = 5 \), substitute \( n = 0 \) to find \( C_1 = 5 \). For \( a_1 = 17 \), substitute \( n = 1 \) into the general solution to find \( C_2 \).
06

Solve for C_2 using initial conditions for part (a)

Using \( a_1 = 17 \), set \( 17 = (5 + C_2)(-4) \). Solve this equation to find \( C_2 = -1 \).
07

Final general solution for part (a)

Substitute \( C_1 \) and \( C_2 \) back into the general solution: \( a_n = (5 - n)(-4)^n \).
08

Understand the recurrence relation for part (b)

The given recurrence is non-homogeneous: \( a_{n+1} = -8a_n - 16a_{n-1} + 5 \). This requires finding a particular solution in addition to solving the homogeneous part.
09

Solve the homogeneous part for part (b)

The homogeneous part is the same as in part (a): \( a_{n+1} = -8a_n - 16a_{n-1} \), with the general solution \( a_n^{(h)} = (C_1 + C_2 n)(-4)^n \).
10

Find a particular solution for the non-homogeneous part (b)

Guess a constant solution for the recurrence. Let \( a_n^{(p)} = A \). Substitute it into the original equation: \( A = -8A - 16A + 5 \), solve for \( A \) to find \( A = -\frac{5}{23} \).
11

Form the general solution including particular solution for part (b)

The general solution will be \( a_n = a_n^{(h)} + a_n^{(p)} = (C_1 + C_2 n)(-4)^n - \frac{5}{23} \).
12

Apply initial conditions for part (b)

Use \( a_0 = 2 \) and \( a_1 = -1 \) to find \( C_1 \) and \( C_2 \). Substitute \( n = 0 \) into the general solution to find \( 2 = C_1 - \frac{5}{23} \).
13

Solve for C_1 for part (b)

Add \( \frac{5}{23} \) to both sides of \( 2 = C_1 - \frac{5}{23} \) to find \( C_1 = \frac{51}{23} \).
14

Use second initial condition to find C_2 for part (b)

Substitute \( n = 1 \) into the general solution and solve \( -1 = (\frac{51}{23} + C_2)(-4) - \frac{5}{23} \) to find \( C_2 = -\frac{14}{23} \).
15

Final general solution for part (b)

Substitute \( C_1 = \frac{51}{23} \) and \( C_2 = -\frac{14}{23} \) back into the general solution: \( a_n = (\frac{51}{23} - \frac{14}{23}n)(-4)^n - \frac{5}{23} \).

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Key Concepts

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

Homogeneous Recurrence
A homogeneous recurrence relation is a type of equation involving sequences where each term is a linear combination of a fixed number of preceding terms. There are no additional or external terms added. This makes them somewhat easier to handle, especially for stability and pattern recognition.
In our context from the problem, the recurrence equation given in part (a) is
  • Homogeneous because it can be written as: \( a_{n+1} = -8a_n - 16a_{n-1} \)
  • Note that there aren't any additional terms influencing the sequence, such as constants or other functions of \(n\)
The task of solving homogeneous recurrence relations often starts with looking for solutions of a characteristic equation, which allows us to express the sequence in a closed form. This closed form has parameters determined by initial conditions.
Non-homogeneous Recurrence
In contrast to homogeneous recurrence relations, non-homogeneous recurrence relations do include additional terms that are not simply linear combinations of previous terms. This makes their solution slightly more involved because it requires us to find a particular solution alongside the homogeneous one.
For part (b) of our exercise, we see a non-homogeneous recurrence relation:
  • It starts similarly to the homogeneous case, \( a_{n+1} = -8a_n - 16a_{n-1} \)
  • The additional \(+5\) term makes it non-homogeneous.
When solving, we first focus on the associated homogeneous equation. Then, we find a particular solution that fits the non-homogeneous form. The complete solution is the sum of these two solutions.
Characteristic Equation
The characteristic equation is a key tool for solving linear recurrence relations, especially when dealing with homogeneous cases. Through it, we find the roots that indicate the nature of the sequence's solutions.
For example, in part (a) of our given problem, the characteristic equation is derived as follows:
  • Starting from \( a_{n+1} = -8a_n - 16a_{n-1} \)
  • We replace terms with a trial solution \( a_n = r^n \), leading to \( r^{n+1} = -8r^n - 16r^{n-1} \)
  • By simplifying, we obtain the characteristic equation: \( r^2 + 8r + 16 = 0 \)
Solving this quadratic gives us the roots, which are used to form the general solution. The nature of the roots (real, repeated, complex) significantly impacts the type of sequence solution.
Initial Conditions
Initial conditions are essential for determining the specific solutions of recurrence relations, especially because they allow us to solve for any constants that appear in the general solution.
In both parts of the exercise, initial conditions are provided:
  • For part (a), they are \( a_0 = 5 \) and \( a_1 = 17 \)
  • For part (b), they are \( a_0 = 2 \) and \( a_1 = -1 \)
These initial conditions are used to:
  • Substitute into the general solution equation
  • Solve for any constants that were introduced during solving the characteristic equation
Thus, while the general solution gives the framework, the initial conditions tailor it to fit the specific sequence exactly.

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Most popular questions from this chapter

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