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Chapter 8: Problem 21

What is the general form of the solutions of a linear homogeneous recurrence relation if its characteristic equation has roots \(1,1,1,1,-2,-2,-2,3,-4 ?\)

Short Answer

Expert verified
\[a_n = (c_1 + c_2 n + c_3 n^2 + c_4 n^3) \cdot 1^n + (c_5 + c_6 n + c_7 n^2) \cdot (-2)^n + c_8 \cdot 3^n + c_9 \cdot (-4)^n\]

Step by step solution

01

Identify the characteristic polynomial roots

In the given problem, the characteristic equation roots are: 1, 1, 1, 1, -2, -2, -2, 3, -4.
02

Determine the multiplicity of each root

Count the number of times each root appears: - Root 1 has multiplicity 4 - Root -2 has multiplicity 3 - Root 3 has multiplicity 1 - Root -4 has multiplicity 1
03

Write the general form of the solution terms for each root

For each root, the general solution includes terms based on its multiplicity: - For root 1 (multiplicity 4): The terms are \(c_1 \cdot 1^n, c_2 \cdot n \cdot 1^n, c_3 \cdot n^2 \cdot 1^n, c_4 \cdot n^3 \cdot 1^n\) - For root -2 (multiplicity 3): The terms are \(c_5 \cdot (-2)^n, c_6 \cdot n \cdot (-2)^n, c_7 \cdot n^2 \cdot (-2)^n\) - For root 3 (multiplicity 1): The term is \(c_8 \cdot 3^n\) - For root -4 (multiplicity 1): The term is \(c_9 \cdot (-4)^n\)
04

Combine all terms to form the general solution

Sum all the terms obtained from each root: \[a_n = (c_1 + c_2 n + c_3 n^2 + c_4 n^3) \cdot 1^n + (c_5 + c_6 n + c_7 n^2) \cdot (-2)^n + c_8 \cdot 3^n + c_9 \cdot (-4)^n \]

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Key Concepts

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

Characteristic Equation
In the study of linear homogeneous recurrence relations, the characteristic equation is a fundamental tool. It's derived by assuming a solution of the form \(a_n = r^n\). This is substituted into the recurrence relation itself.
The characteristic equation's roots play a crucial role in determining the general solution. For a recurrence relation of the form:
\(\rof = c_1 a_{n-1} + c_2 a_{n-2} + \text{...} + c_k a_{n-k}\)\
the characteristic equation becomes:
\(r^k - c_1 r^{k-1} - c_2 r^{k-2} - \text{...} - c_k = 0\)
Here are some key points about the characteristic equation:
* The \r\ values are found by factoring it, which gives you the 'roots'.
* The roots tell us about the patterns in the sequence of the recurrence relation.
Identifying the roots accurately is crucial because they determine the structure of the solution.
Multiplicity
Multiplicity refers to the number of times a particular root appears in the characteristic equation. This concept is essential to formulating the solutions correctly.

For example:
* If root \r\ appears \(m\) times, its multiplicity is \(m\).
For our characteristic equation roots given in the example:
* Root \(1 \) has a multiplicity of 4
* Root \(-2\) has a multiplicity of 3
* Root \(3\) has a multiplicity of 1
* Root \(-4\) has a multiplicity of 1
* Each root and its multiplicity contribute to the form of the solution in certain intuitive ways.
If a root \r\ has a multiplicity \(m\), the terms involving \r\ in the general solution will include:
* \(c_1 r^n\)
* \(c_2 n r^n\)
* \(c_3 n^2 r^n\)
* So on... up to terms involving \(n^{m-1} r^n\).
General Solution
Combining all solutions derived from each root's multiplicity gives us the general solution of the linear homogeneous recurrence relation.

In our given example, the characteristic equation roots led to specific multiplicities:

1. Root \(1\) (Multiplicity 4):
* Terms: \(c_1 \times 1^n\), \(c_2 \times n \times 1^n\), \(c_3 \times n^2 \times 1^n\), \(c_4 \times n^3 \times 1^n\)

2. Root \(-2 \) (Multiplicity 3):
* Terms: \(c_5 \times (-2)^n\), \(c_6 \times n \times (-2)^n\), \(c_7 \times n^2 \times (-2)^n\)

3. Root \(3\) (Multiplicity 1):
* Term: \(c_8 \times 3^n\)

4. Root \(-4\) (Multiplicity 1):
* Term: \(c_9 \times (-4)^n\)

Summing all these terms, the general solution for our example is:
\[\rof = (c_1 + c_2 n + c_3 n^2 + c_4 n^3) 1^n + (c_5 + c_6 n + c_7 n^2) (-2)^n + c_8 3^n + c_9 (-4)^n\]

This general solution encompasses all possible sequence solutions to the given recurrence relation.

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Most popular questions from this chapter

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