91Ó°ÊÓ

Suppose that there are two goats on an island initially. The number of goats on the island doubles every year by natural reproduction, and some goats are either added or removed each year. a) Construct a recurrence relation for the number of goats on the island at the start of the \(n\) th year, assuming that during each year an extra 100 goats are put on the island. b) Solve the recurrence relation from part (a) to find the number of goats on the island at the start of the \(n\) th year. c) Construct a recurrence relation for the number of goats on the island at the start of the \(n\) th year, assuming that \(n\) goats are removed during the \(n\) th year for each \(n \geq 3\) . d) Solve the recurrence relation in part (c) for the number of goats on the island at the start of the \(n\) th year.

Step by step solution

01

- Construct Recurrence Relation for Part (a)

Define the number of goats at the start of the n-th year as \( G_n \). Initially, there are 2 goats, so \( G_0 = 2 \). Since the number of goats doubles each year and an extra 100 goats are added each year, the recurrence relation is \( G_{n+1} = 2G_n + 100 \) for \( n \geq 0 \).

02

- Solve the Recurrence Relation for Part (a)

To solve the recurrence relation, start with the homogeneous part: \( G_{n+1} = 2G_n \). Solving this gives \( G_n^{(h)} = A \times 2^n \), where \( A \) is a constant. For the particular solution, assume a constant solution \( G_n^{(p)} = B \), then: \[ B = 2B + 100 \] This simplifies to \( B = -100 \). The general solution is: \[ G_n = A \times 2^n - 100 \]. Using \( G_0 = 2 \), we find \( A = 102 \), so the solution becomes: \[ G_n = 102 \times 2^n - 100 \]

03

- Construct Recurrence Relation for Part (c)

For the recurrence relation with goats removed, we start with \( G_{n+1} = 2G_n \) and account for removing \( n \) goats during the \( n \)th year: \[ G_{n+1} = 2G_n - n, \text{ for } n \geq 2 \] For \( n=2 \) and before, it remains as \( G_{n+1} = 2G_n \).

04

- Solve the Recurrence Relation for Part (c)

Solve the relation by breaking into initial conditions and for \( n \geq 3 \): Initial: \( G_0 = 2 \), \( G_1 = 2 \times 2 = 4 \), \( G_2 = 2 \times 4 = 8 \).For \( n \geq 2 \), use iteration:\[ G_3 = 2G_2 - 2 = 16 - 2 = 14 \]\[ G_4 = 2G_3 - 3 = 28 - 3 = 25 \]Continuing this, we generally have:\[ G_{n+1} = 2G_n - n \].

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Key Concepts

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

Population Dynamics

Population dynamics studies how populations change over time. In our problem, we analyze how the number of goats on an island evolves.
We start with a set number of goats and then take into account how they reproduce and are managed.
This specific scenario explores both natural growth and external factors, such as adding or removing goats.
Understanding population dynamics helps in predicting future population sizes and managing wildlife effectively.
With recurrence relations, we get a mathematical way to describe these changes over discrete time intervals.

Homogeneous Recurrence Relation

A recurrence relation is a way to define sequences based on previous terms in the sequence.
A homogeneous recurrence relation has the form where each term is a multiple of previous terms, without any additional constants.
In our case, the homogeneous part of the recurrence relation is: \[ G_{n+1} = 2G_n \]
This shows that each new term is simply twice the previous term.
Such relations are easy to solve but real-world scenarios often involve more complexities, which is where non-homogeneous elements come into play.

Particular Solution

To solve a non-homogeneous recurrence relation, we need to find both the homogeneous solution and a particular solution.
The homogeneous solution, as discussed, follows the simple form without extra terms.
The particular solution deals with constant or variable additional factors.
For instance, in the relation \[ G_{n+1} = 2G_n + 100 \], 100 is the non-homogeneous part.
By assuming a constant particular solution, we set \[ B = 2B + 100 \], solving this gives \[ B = -100 \].
The complete solution combines both parts into: \[ G_n = A \times 2^n - 100 \].

Initial Conditions

Initial conditions are the starting values of your sequence.
They are essential to find specific solutions to recurrence relations.
In our problem, the initial condition is given by: \[ G_0 = 2 \].
This value helps determine the final form of the equation.
Using this, we solve for the constant in the complete solution: \[ 102 \times 2^0 - 100 = 2 \], which simplifies to \[ A = 102 \].
So, our specific solution becomes: \[ G_n = 102 \times 2^n - 100 \], valid for all n.

Iteration Method

The iteration method is a step-by-step approach to solve recurrence relations.
It involves calculating each term sequentially based on initial conditions.
For instance, if goats are removed, we modify the relation to: \[ G_{n+1} = 2G_n - n \].
We start with initial values: \[ G_0 = 2 \] and \[ G_1 = 4 \].
Then, iteratively compute: \[ G_2 = 8 \], \[ G_3 = 14 \], and so forth.
This method is straightforward and helps verify the accuracy of our general solution.