Suppose we start (at the end of month 0 ) with 10 pairs of...

Suppose we start (at the end of month 0 ) with 10 pairs of baby rabbits, and that after baby rabbits mature for one month they begin to reproduce, with each pair producing two new pairs at the end of each month afterwards. Suppose further that over the time we observe the rabbits, none die. Let $a_{n}$ be the number of rabbits we have at the end of month $n$. Show that $a_{n}=a_{n-1}+2 a_{n-2}$. This is an example of a second order linear recurrence with constant coefficients. Using a method similar to that of Problem 211 , show that $$ \sum_{i=0}^{\infty} a_{i} x^{i}=\frac{10}{1-x-2 x^{2}} $$ This gives us the generating function for the sequence $a_{i}$ giving the population in month $i$; shortly we shall see a method for converting this to a solution to the recurrence.

Suppose we start (at the end of month 0 ) with 10 pairs of baby rabbits, and that after baby rabbits mature for one month they begin to reproduce, with each pair producing two new pairs at the end of each month afterwards. Suppose further that over the time we observe the rabbits, none die. Let $a_{n}$ be the number of rabbits we have at the end of month $n$. Show that $a_{n}=a_{n-1}+2 a_{n-2}$. This is an example of a second order linear recurrence with constant coefficients. Using a method similar to that of Problem 211 , show that
$$
\sum_{i=0}^{\infty} a_{i} x^{i}=\frac{10}{1-x-2 x^{2}}
$$
This gives us the generating function for the sequence $a_{i}$ giving the population in month $i$; shortly we shall see a method for converting this to a solution to the recurrence.

Key Concepts

Generating Functions

A generating function is a formal power series in which the coefficients correspond to the terms of a sequence. This tool converts problems about sequences into problems about functions, making it easier to manipulate and solve recurrence relations algebraically through power series techniques.

Recurrence to Generating Function Conversion

This technique involves multiplying each term of the recurrence by a power of x and summing over the index to transform the recurrence relation into an algebraic equation in terms of the generating function. The approach provides a systematic way to derive the generating function that encapsulates the entire sequence, which can then sometimes be used to deduce closed form solutions.

Second Order Linear Recurrence

A second order linear recurrence is a sequence where each term is defined as a linear combination of the previous two terms with fixed coefficients. Problems of this type often involve finding closed-form expressions, analyzing long-term behavior, and performing stability analysis using characteristic equations.

Constant Coefficients

In the context of recurrence relations, constant coefficients mean that the multipliers of the previous terms do not depend on the index. This feature simplifies the analysis and allows the use of characteristic equations to seek solutions of the form r^n, ultimately aiding in the derivation of closed-form expressions.

Transcript

00:01 On an island after n months.

00:01 And so first, if a .n.

00:04 Is months in the future, uh, the first thing is that any rabbits that are just a month old are not reproducing yet, right? so those rabbits will be the exact same number in this month as they were last month.

00:26 And so that gives us a m minus one is staying the same.

00:31 Uh, then for rabbits that are two, months old in this case these are each reproducing two pairs of rabbits so we need to take the number of rabbits that we have in the previous month subtract the number of ones that we had two months ago because that will tell us all the ones that are two months old and then these are all reproducing right so we're gonna multiply that by two and then the other condition here is that rabbits that are older than two months are having six babies right and so in this case, the rabbits that were born two months ago or n minus two earlier, each had six pairs of a babies.

01:33 So this is going to be six times a m minus two.

01:44 Okay.

01:45 And so if you combine all this and simplify it a little bit, combining the like terms.

01:50 So we have a .n minus one plus two, a .n minus one for three times a .n.

01:57 Minus one.

01:58 And then we're going to add the 6 a .m.

02:01 Minus 2, but minus the 2, a .n.

02:03 Minus 2, a .n.

02:05 Minus 2.

02:08 Okay, and this is our recurrence relation for the bunnies.

02:15 So that would be part a.

02:17 And so now we want to solve it.