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Chapter 24: Problem 617

Assume that rabbits do no reproduce during the first month of their lives but that beginning with the second month each pair of rabbits has one pair of offspring per month. Assuming that none of the rabbits die and beginning with one pair of newborn rabbits, how many pairs of rabbits are alive after \(n\) months?

Short Answer

Expert verified
The number of rabbit pairs alive after \(n\) months can be found using the Fibonacci sequence, where the rabbit population growth follows the recursive formula \(f(n) = f(n - 1) + f(n - 2)\). To find the number of rabbit pairs after \(n\) months, simply find the \(n\)-th term in the Fibonacci sequence.

Step by step solution

01

Understand the Rabbit Reproduction Pattern

The problem states that each pair of rabbits starts reproducing at the second month of their life and produce one pair of offspring every month thereafter. Therefore, we have two cases: rabbits who are not reproducing yet and rabbits who are reproducing.
02

Create a Recursive Formula

Let f(n) denote the number of rabbit pairs after n months. Since no reproduction happens during the first month, we have the base cases: f(1) = 1 (initial rabbit pair) f(2) = 1 (initial rabbit pair reproduces for the first time) For any month n > 2, the rabbit population is comprised of: 1. The pairs that were there in the previous month: f(n - 1) 2. The new pairs created by rabbits that were reproducing before the current month: f(n - 2) Thus, the recursive formula is: f(n) = f(n - 1) + f(n - 2)
03

Identify the Relation with Fibonacci Sequence

The recursive formula we derived in the previous step is similar to the Fibonacci sequence formula, which is defined as: F(n) = F(n - 1) + F(n - 2) where F(1) = F(2) = 1 From this, we can conclude that the rabbit population problem is an application of the Fibonacci sequence with f(n) = F(n).
04

Determine the Number of Rabbit Pairs after n Months

Since we established that f(n) = F(n), we just need to find the value of the n-th term in the Fibonacci sequence to get the number of rabbit pairs after n months. F(n) can be computed using various methods such as recursion, iteration, or matrix exponentiation. However, most students are already familiar with the Fibonacci sequence, so it is possible to look up Fibonacci numbers in a table or compute the numbers directly. In conclusion, to determine the number of rabbit pairs alive after n months, we just need to find the n-th term in the Fibonacci sequence.

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Key Concepts

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

Fibonacci Sequence
The Fibonacci sequence is a series of numbers where each number is the sum of the two preceding ones, usually starting with 0 and 1. It begins with the values 0, 1, 1, 2, 3, 5, 8, 13, 21, and so on. The sequence appears in many natural phenomena, from the arrangement of leaves on a stem to the branching of trees.

At its core, the Fibonacci sequence represents a model for growth patterns found in nature, making it a remarkable intersection of mathematics and the natural world. The Fibonacci sequence also has applications in computer algorithms, financial market analysis, and art.

Formula and Properties

Mathematically, the Fibonacci sequence is defined by the recursive formula: \(F(n) = F(n - 1) + F(n - 2)\), with seed values \(F(0) = 0\) and \(F(1) = 1\). This mathematical property reflects the essence of growth by accumulation, similar to the rabbit problem mentioned in the textbook exercise, where each term is built upon the foundation of the previous terms.
Recursive Formula
A recursive formula is a mathematical expression used to define the terms of a sequence with respect to the preceding terms. It is a fundamental concept in linear algebra and computer science for constructing sequences and performing calculations.

In simple terms, a recursive formula tells us how to start a sequence and how to find any term from the previous one or more known terms. Typically, a recursive formula includes the initial term(s) for the base case and an equation that defines the subsequent terms.

Understanding Recursion

To better understand the recursive formula, consider a practical example: if you have a task that can be broken down into smaller, similar tasks, the method for completing each task is effectively the recursive formula. This powerful mathematical tool not only offers a method for sequencing but also provides insights into how seemingly complex phenomena can be understood as a progression of simpler steps.
Rabbit Reproduction Problem
The rabbit reproduction problem is a classic example that introduces the concept of the Fibonacci sequence as a model for biological growth. The question posits: starting with a single pair of rabbits, if each pair produces a new pair every month from their second month, how many pairs will there be each subsequent month?

This problem mirrors natural reproduction processes, depicting how populations can expand rapidly over time. The solution to the rabbit problem relies on the recursive nature of Fibonacci's sequence; each new generation of rabbits adds to the total population in a way predicted by Fibonacci's formula.

  • Month 1: 1 pair of newborn rabbits
  • Month 2: 1 pair of mature rabbits
  • Month 3: 2 pairs (the original pair and their offspring)
  • Month 4: 3 pairs (the first offspring pair matures and the original pair reproduces again)
As with the Fibonacci sequence, the number of rabbit pairs forms a pattern that precisely follows the sequence's recursive formula, illustrating the sequence's practical application in understanding natural phenomena.

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Most popular questions from this chapter

There are three types of grocery stores in a given community. Within this community (with a fixed population) there always exists a shift of customers from one grocery store to another. On January \(1,1 / 4\) shopped at store \(\mathrm{I}, 1 / 3\) at store II and \(5 / 12\) at store III. Each month store I retains \(90 \%\) of its customers and loses \(10 \%\) of them to store II. Store II retains \(5 \%\) of its customers and loses \(85 \%\) of them the store I and \(10 \%\) of them to store III. Store III retains \(40 \%\) of its customers and loses \(50 \%\) of them to store I and \(10 \%\) to store II. a) Find the transition matrix. b) What proportion of customers will each store retain by Feb. 1 and Mar. 1 ? c) Assuming the same pattern continues, what will be the long-run distribution of customers among the three stores?

Recent archeological investigations in Asia Minor have unearthed clay tablets that give details of transactions between merchants from 1900 B.C. Suppose the following hypothetical communication matrix represents tablets between members of a group of merchants. Show how this matrix can help to identify contemporary merchants.$$ \begin{array}{lllllllllll} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ 2 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 3 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\ 5 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 6 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 7 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 8 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 9 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 10 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} $$

A camera store stocks a particular model camera that can be ordered weekly. Let \(D_{1}, D_{2}, \ldots\), represent the demand for this camera during the first week, second week, \(\ldots\), respectively. It is assumed that the \(D_{i}\) are independent and identically distributed random variables having a known probability distribution. Let \(\mathrm{X}_{0}\) represent the number of cameras on hand at the outset, \(\mathrm{X}_{1}\) the number of cameras on hand at the end of week one, \(\mathrm{X}_{2}\) the number of cameras on hand at the end of week two, and so on. Assume that \(\mathrm{X}_{0}=3 .\) On Saturday, night, the store places an order that is delivered in time for the opening of the store on Monday. The store uses the following \((\mathrm{s}, \mathrm{S})\) ordering policy. If the number of cameras on hand at the end of the week is less than \(\mathrm{s}=1\) (no cameras in stock), the store orders (up to) \(\mathrm{S}=3\). Otherwise, the store does not order (if there are any cameras in stock, no order is placed). It is assumed that sales are lost when demand exceeds the inventory on hand. Assuming further that \(\mathrm{X}_{\mathrm{t}}\) is the number of cameras in stock at the end of the \(\mathrm{t}^{\text {th }}\) week (before an order is received, and that each \(D_{t}\) has a Poisson distribution with parameter \(\lambda=1\), determine a) the one-step transition probabilities b) Given that there are two cameras left in stock at the end of a week, what is the probability that there will be three cameras in stock two weeks and four weeks later. c) Compute the expected time until the cameras are out of stock, assuming the process is started when there are three cameras available; i.e., the expected first passage time, \(\mu_{30}\), is to be obtained. d) Determine the long-run steady-state probabilities.

\(A\) and \(B\) play the following game. Each tosses a die in turn. If a six is tossed the game is over and whoever tossed the six wins the bets. If a 4 or a 5 is tossed the player tosses again. If a 1,2 or 3 is tossed the die passes to the other player. Is the game biased in favor of the starting player?

Three species of bacteria coexist in a test tube and feed on three resources. Suppose that a bacterium of the ith species consumes on average an amount \(c_{i j}\) of the jth resource per day. Define \(c_{i}=\left[c_{i 1} c_{i 2} c_{i 3}\right]\) to the consumption vector for the ith species. Suppose that \(c_{1}=[1,1,1], c_{2}=[1,2,3]\) and \(c_{3}=[1,3,5]\) and suppose that there are 15,000 units of the first resource supplied each day to the test tube, 30,000 units of the second resource and 45,000 units of the third resource. Assuming that all the resources are consumed, what are the populations of the three species that can coexist in this environment?
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