Find the unique, closed form solution to this recurrence relation given the initial condition: an = 2an-1 + 5, a₁ = 3 ao
Find the unique, closed form solution to this recurrence relation given the initial condition: an = 2an-1 + 5, a₁ = 3 ao
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Solving a Recurrence Relation - Example
**Problem Statement:**
Find the unique, closed-form solution to the recurrence relation given the initial condition:
\[ a_n = 2a_{n-1} + 5, a_0 = 3 \]
**Explanation:**
This problem requires us to find a closed-form solution for the sequence \( \{a_n\} \) defined by the given recurrence relation, which is a standard form in mathematical sequences and series.
A recurrence relation is an equation that recursively defines a sequence where the next term is a function of the previous terms. Here, \( a_n \) is the term we want to find, and it is expressed in terms of the previous term \( a_{n-1} \) plus a constant. The initial condition provides the necessary starting value for the sequence.
**Approach:**
1. **Identify the Recurrence Relation:**
\[ a_n = 2a_{n-1} + 5 \]
2. **Initial Condition:**
\[ a_0 = 3 \]
**Step-by-Step Solution:**
To solve this recurrence relation, we can use the method of iteration or find the general solution by recognizing it's a non-homogeneous linear recurrence relation.
**Iterative Method:**
- \( a_0 = 3 \)
- \( a_1 = 2a_0 + 5 = 2(3) + 5 = 6 + 5 = 11 \)
- \( a_2 = 2a_1 + 5 = 2(11) + 5 = 22 + 5 = 27 \)
- \( a_3 = 2a_2 + 5 = 2(27) + 5 = 54 + 5 = 59 \)
**General Solution by Homogeneous and Particular Solutions:**
1. Solve the homogeneous part \( a_n = 2a_{n-1} \):
- The characteristic equation for the homogeneous part is \( r = 2 \).
- Hence, the homogeneous solution is \( a_n^{(h)} = C \cdot 2^n \) for some constant \( C \).
2. Find a particular solution \( a_n^{(p)} \):
- We look for a particular solution of the form \( a_n^{(p)} = A \), where \( A](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3eb514bb-44ba-433b-9221-45dffb06f396%2Fbc48722c-7d34-49f6-93ca-96c0b46bb754%2Fakdwqvl_processed.png&w=3840&q=75)
Transcribed Image Text:### Solving a Recurrence Relation - Example
**Problem Statement:**
Find the unique, closed-form solution to the recurrence relation given the initial condition:
\[ a_n = 2a_{n-1} + 5, a_0 = 3 \]
**Explanation:**
This problem requires us to find a closed-form solution for the sequence \( \{a_n\} \) defined by the given recurrence relation, which is a standard form in mathematical sequences and series.
A recurrence relation is an equation that recursively defines a sequence where the next term is a function of the previous terms. Here, \( a_n \) is the term we want to find, and it is expressed in terms of the previous term \( a_{n-1} \) plus a constant. The initial condition provides the necessary starting value for the sequence.
**Approach:**
1. **Identify the Recurrence Relation:**
\[ a_n = 2a_{n-1} + 5 \]
2. **Initial Condition:**
\[ a_0 = 3 \]
**Step-by-Step Solution:**
To solve this recurrence relation, we can use the method of iteration or find the general solution by recognizing it's a non-homogeneous linear recurrence relation.
**Iterative Method:**
- \( a_0 = 3 \)
- \( a_1 = 2a_0 + 5 = 2(3) + 5 = 6 + 5 = 11 \)
- \( a_2 = 2a_1 + 5 = 2(11) + 5 = 22 + 5 = 27 \)
- \( a_3 = 2a_2 + 5 = 2(27) + 5 = 54 + 5 = 59 \)
**General Solution by Homogeneous and Particular Solutions:**
1. Solve the homogeneous part \( a_n = 2a_{n-1} \):
- The characteristic equation for the homogeneous part is \( r = 2 \).
- Hence, the homogeneous solution is \( a_n^{(h)} = C \cdot 2^n \) for some constant \( C \).
2. Find a particular solution \( a_n^{(p)} \):
- We look for a particular solution of the form \( a_n^{(p)} = A \), where \( A
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