WEBVTT

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Part A states, Fibonacci posed the following problem.

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Suppose that rabbits live forever and that every month each pair produces a new pair which

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becomes productive at age two months old.

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So every month there's a new pair and it becomes productive after age two months they

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live forever and so on.

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And if we start with one newborn pair, how many pairs of rabbits will we have in the

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end month? And then basically it states show that the answer is fn where fn is

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the Fibonacci sequence defined in example 3c as I illustrated earlier in this

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video. And then part B states, let an equals to fn plus 1 divided by fn and show

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that an minus 1 equals to 1 plus 1 divided by an minus 2 and so on. And assuming

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that an is convergent find its limit. Alright so now let's look at part a.

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So we'll go over this rabbits once first but first let's recall the Fibonacci

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sequence defined in example 3c and it was right here. Fibonacci sequence fn is

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defined recursively by the conditions f1 equals 1, f2 equals 1 and then we have

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fn equals to fn minus 1 plus fn minus 2 for n is a positive integer greater

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than or equal to 3. So in essence we're just each term as a sum of the two

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preceding terms so f3 is going to be f1 plus f2. And the first few terms

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are 1, 1, 2, 2, 3, 5, 8, 13, 21 and again you can just add these up so 3 plus 5 is 8,

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8 plus 13 is going to be 21 and so on. And 1 plus 1 is 2. So let's look at this.

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So let an be the number of rabbit pairs in the nth month. Okay so we're asked here

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again so suppose that rabbits live forever and that every month each

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pair produces a new pair which becomes productive at age two months. So if we

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start off with one new born pair is what we start off with how many pairs are we

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left in the nth month. So let an be the number of pairs in the nth month. So

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clearly we know the first few months clearly a1 this just equals to well

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1. There's one pair and then also age I mean at month two so because rabbits

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live forever in the but they become productive at age two months. So in the

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second month the first one is going to be well it's a new born pair so then it's

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going to be equal to a2. So after two months there's still going to be one

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rabbit I mean one rabbit pair. So the first one and then it becomes productive

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and after that they're going to start producing. Yes so now in the nth month

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each pair that is two or more months old that is an minus two pairs.

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Basically all the rabbits that were around for the two months ago. So two

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months ago all the rabbits that existed then will produce new pair and to add

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to the a minus I mean a and minus one pairs. So we're going to add to the two

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months previous all the rabbits that existed all the pairs add it to all the

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pairs of last months okay or the number of those and we'll explain this as I

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illustrate this. So in other words all the rabbits that are present from two plus

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months ago multiply so we add the total number of rabbits from that time

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period to the total number of rabbits from one month ago since this number

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has no yeah I mean it has new born rabbits that haven't matured yet those

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are the ones in one month ago so you can't just take that number because some

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of them aren't going to reproduce. So thus let's just write this down to get

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this pattern. So thus we have well we have a1 equals to 1 and then a2

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equals to 1 because we still have one pair and now a3 the total number here and

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these are the third month so the first month is one second month there's two

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but then the third month what we end up having is well we take the ones in the

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first month so the number in the first month which was well a1 plus a2 now

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because we're actually a better than better than I'll write a2 plus a1

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just so it's easier to write because now we're going to do is we're going to look

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at the number of rabbits rabbit pairs in the second month which was one and then

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we're going to reproduce they're going to reproduce we'll have another one this

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is one plus one and that's because they started off at a1 so this is going to

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be well 2 and now notice how now you'll see understand this pattern so a4 so

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a4 is going to be well we take we initially have two rabbit pairs so two

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rabbit pairs a3 so we take all the ones from here now what we're going to do is

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we're going to multiply all the ones from two months previous so this is what

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this one just had one so we just add a2 equals 2 2 plus 1 equals to 3 and then

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a5 what we'll do here is we take all the number of pairs from previous so a4

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then what we're going to do is we're going to add all the ones from the previous a3

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because these are the ones are going to multiply so there's two pairs here that

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are going to be matured by the time it's a5 and these two are going to produce

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two more so that's going to be yeah so it's plus this a3 so we're going to

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have a3 plus 2 equals to 5 and I'll just write one more of these just to

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illustrate this so now a6 is going to be we take the previous pair the total

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number of them a5 and then plus then we plus the ones before those three and these

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ones are all going to multiply so these three are going to be six I'm going to

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add these three the three's three are going to produce three more here so we'll

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have a4 and that equals to 5 plus 3 this equals to 8 like that and then we

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keep going on and on then what we get is a n equals 2 well a n minus 1 plus a n

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minus 2 because you can see these a3 is a2 plus a1 a4 is a3 plus a2 and so on and

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this is well this is just the Fibonacci sequence this equals to Fibonacci

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Fibonacci sequence from sequence yeah Fibonacci sequence from example 3c earlier

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in the video so yes so thus the sequence a n equals to the sequence fn from before so

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this is the Fibonacci sequence so yeah alright so now let's look at part

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b right here so part p it says let a n equals to fn plus 1 divided by fn and

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show that a n a n minus 1 equals to 1 plus 1 over a n minus 2 and then we're

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asking assuming that a n is convergent fine it's limit very very interesting

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let's let's see what we can come up with right here so part b write this down

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part b all right so we're gonna let over here let a n equals fn plus 1 divided by

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fn so let a n equals to fn plus 1 over fn again where this is the Fibonacci

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sequence fn so and then we're also asked to show is write it down show that show

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that a n minus 1 is equal to 1 plus 1 over a n minus 2 and let me just scroll to

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see if we got the right one there so yes a n minus 1 equals 1 plus 1 over a n

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minus 2 so let's go ahead for this first of all let's look at this a n minus 1

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let's put an arrow across there so we have a n minus 1 this is equal to well

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this is gonna be minus 1 here so the top is going to be set up fn plus 1 so fn

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plus 1 minus 1 fn the bottom is going to be fn minus 1 so we just place n with n

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minus 1 like that and now fn recall fn is just equal to well that says the

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Fibonacci sequence and scroll up so much sequence again is just fn equals fn

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minus 1 plus fn minus 2 or n is greater than or equal to 3 so we have fn

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minus 1 plus fn minus 2 over fn minus 1 yeah so that's just the Fibonacci

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sequence on top so now this equals 2 well divide this out this is going to be

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fn minus 1 divided by fn minus 1 plus fn minus 2 divided by fn minus 1 so this

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becomes 1 so what we get is equals to 1 plus fn minus 2 over fn minus 1 and what

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I'll do here now is I'm going to multiply the top and bottom by well this power

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right here 1 over fn minus 2 because remember we want to show it it's like that

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so then we're going to try to get rid of this make it a 1 so get rid of the

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numerator by dividing by it on both sides the top and bottom fn minus 2

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so then what we end up having is 1 plus this one cancels we get a 1

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so 1 plus 1 over and we have this fn minus 1 divided by fn minus 2 and now this fn minus 1

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divided by fn minus 2 well we know that fn plus 1 divided by fn is equal to an

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so yeah then we subtract one more we get fn over fn minus 1 but this is fn minus 1 over

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fn minus 2 so in other words this just becomes fn minus yeah let me just

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this just becomes an minus 2 so add another one so fn minus 1 fn minus 2

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so this becomes an minus 1 I mean minus 2 an minus 2 so thus yeah so thus

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an minus 1 equals to 1 plus 1 over so 1 plus 1 over well an minus 2

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so there is it we've proven this so far yeah and that's exactly what we've proven over there

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as I written above here yeah so we just prove that now we're assuming that an is convergent

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find its limit so let's find the limit now all right so now if you want to get the limit of

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this one so we we know that the limit exists so we are given that the limit yeah the limit

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exists in other words is convergent so what we'll do is write

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so if if a limit so limit as n approaches infinity of an is equal to l yeah then

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then by the earlier theorem then basically this limit then limit as n approaches infinity

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of an minus 1 this also is going to equal l and this also equals to limit as n approaches infinity

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of an minus 2 and this is a from the theorem that I just went over what was that was a theorem

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theorem 2 I believe school above actually no theorem 3 actually no I'm not a theorem one or two

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I mean and from exercise 1 remember this one here limit as n approaches infinity of an plus 1

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equals to limit as n approaches infinity of an and again you could just see this one here

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instead of having a plus one we could have a minus one but in the exactly the same thing

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we're just going to have m would be n plus 1 so on so it's exactly the same thing as

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matter what it is it they're still approaching infinity we're approaching the limit so the

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limit is being approached regardless where your initial starting point is yeah so from exercise 1

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let's put a bracket from exercise x yeah from exercise 1 and also just using a

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bit of a deduction yeah so that's what we have as well we take this an from above so thus

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limit as an minus 1 I mean a limit as n approaches 0 my bad and approaches infinity

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limit as n approaches infinity of a n minus 1 this equals to l which equals to over there

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this entire limit limit as n approaches infinity of 1 plus 1 over a n minus 2

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so in other words what we end up having is this is becomes limit of this one using limit laws

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and then we just get the we get the equation l equals to 1 plus 1 over l and now multiply

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both sides by l and what we end up having is multiplied by l on both sides so that nothing

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changes we end up getting is l squared equals to um yeah l this one cancels now plus 1

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so just move this onto the option move the other ones to the other side

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like that what we end up having is um we'll put the zero on the other side or I should put this

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with zero equals to l squared minus l minus 1 or just put the other side for convenience

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so we have it like that so l is going to be on the other side I mean now zero is going to

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be on the other side l squared and move everything here minus that we have this

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so yeah and so now what we do is we'll recall the quadratic formula again just copy and paste

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from the one I've written above so what we have is l is going to equal 2 over here

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where now b I mean yeah b is negative 1 so we're going to have a negative of a negative 1

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plus or minus square root of a negative 1 squared minus 4 and then a which is 1

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then times the by c which is negative 1 like that all divide this by 2 times a which is 1

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now this equals to well negative 1 times negative 1 is plus 1 plus or minus square root now we have

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negative 1 squared is 1 minus 4 times 1 times negative 1 this becomes positive

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and that becomes well plus 4 like that plus 4 over 2 and again this just becomes 5

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same time and there is our answer this equals to the limit actually now it doesn't need a limit

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because we have the plus or minus yeah so it's a plus or minus so thus l has to be positive so we

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take this one here thus l equals to 1 plus square root 5 over 2 and put in a bracket l just put here

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since l must be positive like that number it's a number of yeah it's a number of of rabbit pairs

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that all is positive again of a negative rabbit pair and again a school up here so yeah this is

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basically part b which is giving us a easy way to find the limit by showing that it's you could

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write it in this form and use the exercise one theorem that I covered so now yeah the actual

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numerical answer to this limit is actually very interesting and very important in mathematics

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l equals to put this in our my built-in one-note calculator 1 plus square root 5 old brackets

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closed brackets divided by 2 equals 1.618033988749895