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(a) Let $ a_1 =a, a_2 = f(a), a_3 = f(a_2) = f( f…

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Problem 83 Hard Difficulty

(a) Fibonacci posed the following : Suppose that rabbits live forever and that every month each pair produces a new pair which becomes productive at age 2 months. If we start with one newborn pair, how many pairs or rabbits will we have in the $ n $th month? Show that the answer is $ f_n $ where $ \{ f_n \} $ is the Fibonacci sequence defined in Example 3(c).
(b) Let $ a_n = f_{n + 1} / f_n $ and show that $ a_{n - 1} = 1 + 1/a_{n - 2}. $ Assuming that $ \{ a_n \} $ is convergent, find its limit.

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Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 1

Sequences

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Video Transcript

for this problem will let fn to note the number of pairs of rabbits at the beginning of the end of month Here were given that in the beginning of the first month we're starting with one newborn Pierre. So this means f of one is one. Now, in the beginning of the second month, this first peer of rabbits takes two months to reproduce. So in the beginning of the second month, this first pier is still not able to reproduce. So a month later, we still have on ly one pair the same pair that we started with. However, if in the beginning of month three now it's been two months. So that means not only do we have this original rabbit here, but they've produce offspring a peer of their own. And that leaves us with a total of two piers now going in someone for it. This pair that was born in one three needs two months to reproduce, so they won't be having any offspring yet. However, this original pair that we started with keeps producing after the second month, so they'LL produce another pair giving us a total of three pairs. Okay, so in General, we have the following. This is for and bigger than three. We've already given some values over here. Now, here's a general formula for the number of rapids at the end of month, and the reason this is true is this is where this first expression here this is the number of pairs at the beginning of month and minus one and now before and minus one, so a month. And we know that the on ly rabbits that can reproduce are the ones that have been alive for at least two months. So this means that we have to ignore month end in month and minus one if we want to consider offspring. So this explains the second term Over here. This is the number of newborn piers. So this is just based on the fact that it takes two months to become reproductive. So any rabbit born honor before month and minus two will have offspring in month end. And so this agrees with the Fibonacci sequence that was given in the mentioned example, and that also completes part aid. So now we'LL go on to part B. So for party, let's just observe the following formula Before we go on to the next age, we're given a formula for a N and we want to drive a formula for a N and then find the limit of AM So I'LL go ahead and go on to the next page here. I'm running out of space. Okay, so he was just a reminder of what we have I want to show. So this is what we're trying to prove what we want to show its true. Okay, so we have Let's start with the right hand side of the formula we're trying to verify that's this expression over here Now, using this given formula for a n just replace and with and minus two And also you see here that the A and minus two is the reciprocal. So we'LL go ahead and flip this fraction. So this is just by using the definition of am Okay, The next step here is let's use the fact that by part, eh, we have the following sn f n minus one plus f n minus two. So go ahead and solve that for F N minus two and then go ahead and plug this back into the numerator two so we can go ahead and simplify this a little bit. So let's see here F n minus two. So I made a typo here, let me go back and correct this. This should be a f and minus f n minus one. Sorry about that. Now go ahead and just split up that fraction. Cancel that A ones are. Excuse me. Cancel the ones. And then by definition of a n, we're left with a and minus one. And that's exactly what we wanted to end up with. So now let's go ahead to the next stage. But to find the limit of an come on. So we have so first, go ahead and recall the Fibonacci from party. And then let's go ahead and divide both sides by F n Plus one. Now we're allowed to assume that this limit exists and party. So let's just go ahead and call this limit. L And so also note that what we have here, by definition of a n, this is the equivalent to a M plus one equals one plus one over a end. Now go ahead and take the limit on both sides. Here limit as n goes to infinity. Now, if you add once hen, that's still going to infinity. So this expression now becomes l equals one plus one over l. Now let's go ahead and simplify this right? As the quadratic said it. Zero use the contract IQ formula. However you see, you get two answers here once positive one's negative. But in our problem, we know that and should be a positive number by definition of an so recall. And is this expression over here? So this is the definition, and we know FN is positive because it's a number of rabbit pears. So in our case, we should go ahead and just take the positive square room here. So that's the limit of AM

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(a) Let $ a_1 =a, a_2 = f(a), a_3 = f(a_2) = f( f…

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Caleb Elmore

Kristen Karbon

Samuel Hannah

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Caleb Elmore

Kristen Karbon

Samuel Hannah

James StewartCalculus: Early Transcendentals

Anna Marie Vagnozzi

Caleb Elmore

Kristen Karbon

Samuel Hannah

James Stewart

Calculus: Early Transcendentals