Lecture 5: Monotone Convergence Theorem

Key Takeaways

Okay. So, uh let me just recall first that so what we were talking about so this is a very important concept uh of a sequence concept of a sequence. So sequence is is the formal definition is that you have a map from the natural number into R and then we denote so then the image of a particular natural number n it's usually denoted like a n like this you know it could be some other instead of a it could be bn potentially it could be another subscript but it's this is like the most common thing to denote it like this instead of like but it really is a function. Okay, so that's a sequence a subsequence. So subsequence is that you have a sequence and then you have another map but this map is not just any map from the integers to the integers. It has to be strictly increasing. And why does it have to be strictly increasing? Because we want for the subsequence. If you think about the numbers that come out when you plug in a when you're looking at the a then you want some of those numbers to be in the subsequence, but you want the ordering to be the same. So this map so that's why this map has to be strictly increasing right can't sort of I mean you and you can't pick the several elements many times if it wasn't picked many times in the original uh sequence right so this is so a subsequence is where you have another map from the natural number to the natural number that is strictly increasing and then you're looking at the composition of these two maps so you're looking at sorry uh this thing here right and you call this here and then that's usually denoted something like this a n k a n is the original sequence k means that it's really that you take the image of k under this strictly increasing map and then you're looking at the element in the sequence right so it's sort of n k is really you should think about it as G of K. Okay, so that's that's what a a a subsequence is. And again, it's important that it's strictly increasing because you want the elements to be in the same order than it was in the original sequence. And you don't want things you don't even want things to be picked more times than it was in original sequence. Mhm. So that that's that's you know so this this whole concept of a sequence of subunits is really important. Then the next really important concept is uh is of convergence. So if you have a sequence a n then you say that is converging to a. So so a is another real number. It's not infinity. is another real number and so a n converts. So a n converts to a if for all epsilon greater than zero there exist n an integer n you know by capital n such that if little n is bigger equal to this capital n then a n minus minus a have to be less than epsilon. Right? of those. So if you are sufficiently far out in sequence then everything in that sequence is bunch epsilon close to a okay that's what it's saying and so as if you have a sequence so a sequence that is not convergent uh is set to be divergent. Okay. Um and the next thing for sequences is [Applause] I'm just briefly reviewing this from last time. The next thing for sequences are the are the break to break uh uh laws or whatever you want to call them. And so that is that if you take a sequence, if you have a sequence a n and a n converts to a bn converts to b then if you form the sum of these two uh a n + bn. So this is a new sequence where the nth element is the sum of the nth element of the other two sequences. then this sequence here CN also converge and the limit here is the sum of the limits. Okay. And likewise if you take a if you're looking at at if you define a sequence so this was one two is that if you take a constant so this is a constant and you define cn to be c * a n then this cn also converge and the limit here is just c * a right that's the second one then uh this one I might have done it in the other order but there's no you know it's not you know there's no particular order um to have it in. Uh the third one is that if you are looking at the product. So if you define a sequence that is given by the product of the two then C and converge also and the limit is the product of the limits. And then the fourth rule is that if a n is not equal to zero for all n and the limit here is also not equal to zero. Then if you're looking at 1 / a n then those converge to one. Okay. So those are the four four the four roots. Just write it as four. And again, it's not like there's any particular convention with the ordering. And I think I might have done it. I might have exchanged the order of one and two last time. But okay. Um now, uh let's see an example of how that works. So let's define a sequence. Uh let's define a sequence uh n 2 + 1 n 2 + uh n + uh + one. Let's look at this sequence here right for each n that give us uh actually a rational number but it's only real number. Now in order to understand this I can rewrite this one here as I can factor out 1 / n well I can factor out n squ here. So I can write it like this [Applause] like that. And then of course I can get rid of the n squ in both denominator and denominator. And now I can look so so this so you may as well think about the sequence as this thing here. So now if I defined um so now I can just look at say uh I can look at uh the sequence which is one I can just first look at the sequence which is 1 / n. If I'm looking at this sequence here, right, one / n, this here converts to zero. This is essentially just followed from the ardian property, right? But we already we've already seen this that 1 / n converts to zero. And the same hold for 1 / n^2, right? This here also converts to zero, right? And so if you're looking at the sequence, you could even look at the sequence. Let's call this x n. This here is yn. I'm going to run out of character very quickly. Uh set n. This is just going to be the sequence which is constant one. Right? Well, if if it's constant one, then obviously it's converging to one, right? And so now I can just think about I can think about the sequence bn to be equal to x sorry equal to set plus uh y set n plus yn right this here is a sequence here is what yn is here's what set n is right so if you add these two this is converging to one this is converging to zero by the algebraic property of limits this converts also to the sum of the limit so this here converts to one right now I can also define a sequence where I'm just adding set in and then I'm adding x in and then I'm adding y in right this here converts to one this here to zero this here to zero. So the sum here also converge to the sum of the limit but the sum of the limits is one. Now you have that one over c sorry now you have that put it up here and then then I will I'll just go over here. So now you have that um that CN converts to one. Um and let's call this here C. And so in in particular, I mean all of these numbers here, these things here are all strictly positive. So each of the CN are strictly positive. So they're definitely not zero, right? So the CN so not only do they converge to one but the CN are not equal to zero and the limit C well that's one so that's of course not equal to zero either right so you can apply the fourth rule here the fourth rule for the sequence not for the sequence CN not right so apply the fourth rule to the sequence CN and then you have that one over CN converge to one / c. But one c was one. So this is one. Right? Now you can look at at um at bn * 1 / cn. But bn * 1 / cn, right? Remember what bn was? Here's bn. And bn is really the denominator up here. Right? It's the nominator up here. when you have divided by n squ on both sides uh by not on both sides but both in the top and the denominator. So bn is the denominator and cn is the denominator. So bn divided by CN this here is a n right and now this here converge that's what we saw over there using the algebraic rule it converts to one this here also converts to one by the algebraic rules number three of the algebraic rules is that for the product it will also converge so and it converts to the product of the limits so that's one And obviously so I you know this is like when you're looking at a sequence it's often uh expressed in some form with some functions like this and you and you do this reduction and very quickly you know this would be almost automatic right I mean this is too easy almost this one I mean you know this is the first time you see it but you know in in a couple of weeks this will feel uh too easy uh then then there would be something more complicated. Okay. So that's that's the algebraic rules uh of for sequences. The next thing I want to talk about is I want to try to make more precise um you know this about square. I mean, you know, we made it pretty precise, but I want to think about uh uh I don't have a very good eraser, unfortunately. Let me just see. This is a better one. This one here will make it very dirty. Okay. So, so the next thing I want to So, we did define square of two and we talked about that quite extensively. how square root two, you know, how it really was defined and it was a real number and the real numbers followed from that the field the reals is complete. But we also we also sort of in the back of our mind we had that we already sort of knew what square root of two were, right? Because when you think about square root two, I mean at least I think about square root two. I think about it as 1.4. I mean just because you can't really express it as a fraction. So you think about it as 1.414 etc. And then I don't remember I don't remember the second years third and fourth whatever but you know you may remember just a few of the digits fine and but so how do we make so so that's sort of our intuition uh with 1.4 four with with square two. Uh but how do we make this more precise? Right? And so to make this more precise, we want to look at particular sequences. So what's called monotone sequences? So what is a mono sequence? So there's two types of monotom sequence. There's one which is monot increasing sequence and then there's similarly monotum decreasing sequence. So a monotum increasing sequence this is just means that for all n you have that the next element is larger or equal to the previous element. That's a mono increasing sequence. Right? Some people talk about strictly monotum increasing and then this have to be a strict. Uh the second thing is kind of the obvious parallel to this is monoton decreasing decreasing sequences. So that's these two types. Okay. And this just means that the next element so a n + one is less or equal to the previous one. Okay. And so now there's a theorem and and I'll prove this in a in just a minute. But I also want to and I want to come back to square root of two and then I want to look at another classical example of a monotum sequence. So um right so this is a mono sequence and now let me state the mono convergence theorem. So suppose that so this is the following theorem. Suppose that that and there's two versions of this. There's one version for monotum increasing sequence and then there's one version for mono decreasing sequence. I just state one of them because the other one's kind of obvious parallel. So suppose that a end is a monotone increasing sequence [Applause] then it is bounded bounded. So this just means that there exists i.e. [Applause] there exist some a real number. So that's right. You have a1 this here is that is monotum right this is monotum that they cannot increase but but the claim is that now the all of the a is lesser equal to this a it's any monotum increasing sequence is clearly bounded from below right because already the first element is a lower bound right so if you have a mono increasing sequence then Um then uh that is bounded then it's convergence and uh we limit say we limit a where a is the soup of the a inch. So that is a mono convergence sequence. Okay. and and and that is the the kind of main theorem about mono convergent sequence is that if you have a monotum increasing sequence that is bounded then it's also convergence right and likewise if you have a monotum decreasing sequence that is bounded from below right then it is also convergent and so maybe maybe I will state it u but it's is the obvious parallel. So, so this is part one of the monome converting sequence. Part two of part B is that if if A is a monotone decreasing sequence sequence then uh that is bounded uh that is that is bounded from below. Then uh a is convergent with limit a where a is now the in of these numbers. Okay. Now, so you might say the first thing you might say, we'll come to square root two in just a minute, but you may see that this seems rather special, you know, like if you take a random uh sequence, obviously it's not going to be monitor. Um but but here's the thing that so uh let be any bounded sequence. So bounded again means that it's bounded from above and bounded from below. So if you take any boundless sequence then we can form two other sequences. Two other sequences I mean obviously form many other but here's two other very natural uh sequences. The first one I call BN and BN is is the soup of where you're taking where you're starting at a end and then you take the soup of all of the ones from A in and outwards in the sequence. Right? So this is one sequence right we know that that you know the set of a ends is bounded right so you can certainly take the soup and you get a real number right that's that that gives you another sequence you can also look at cn similarly where you take instead of the soup you take the in of the same that [Applause] like that. All right. So, in particular, right, just to make sure it's totally clear that that if you were looking at say BN, so what is BN? If you're looking at say B3 then this is the soup where you're starting at A3 A4 etc. And so you see of course that B23 here well obviously B3 is bigger equal to B2 right because B2 here is sorry it's no it's not bigger equal to it's lesser equal to B2 because you see in B2 the definition of B2 so let's just write B2 down so B2 is that you take the soup here of A2 and then A3 A4 etc. Right? But you see that B2 here it's both of them are defined to be soup but B2 is defined as soup over a larger set because in addition to these elements it also includes A2 right. So clearly B2 here is bigger equal to B3 right and in fact the same argument tells you that if you take B n + one then this is or bn sorry bn is bigger equal to bn + one because this here is the soup over a larger set it's soup over all the ones you take soup in here plus a n right so you have this property here right and so yeah >> what do you Yeah. If it's a monot increasing um >> yeah because monotum increasing uh then this here uh right yeah then it becomes very special. Uh so it's it's if you take a mon if the a was monoton increasing then the all of these any element here would just be the a we talked about before it would always be the limit. So so it's um it's it somehow wouldn't be that interesting a sequence. It would be a constant sequence and and it would be just >> but but but that but nevertheless it's a good example you know. I guess that's still anybody. >> Yeah. Yeah. Definitely. Right. Because it's just constant. Exactly. >> It's a it's a good example, but it's just kind of almost like too simple because you get a constant sequence, right? So, okay. So, we have for this sequence that we defined. So we have again these if you take any sequence then you get then you have two sequences B and a CN and B and here is now a decreasing sequence and in instead of writing decreasing sequence then uh it's much easier to write it like this. This is pretty standard. I mean you know it's just to uh if you just look quickly at it uh you know what this means. going to write out but let's look at CN also remember that we assume that a n we a could be any sequence but it is supposed to be bounded right so this is now a monotum decreasing sequence that is bounded so it actually has a limit right the cs what about the cs so the cns So the CN well CN was defined to be the in starting at a end like that right so it' be starting like that and so this means that if you take CN + one then you take the infom But you don't include this element. Right? So it's a infreum over fewer elements. But then this here must be larger equal right to that. So you see that the sequence CN is monotum. This here is monotum increasing again is bounded and and and just copying from over there. The BN was decreasing. >> Okay. This is like I mean it's also like I mean this is an important thing. It's also something that you would see in lot of different context where you define something some sort of sub and then you have these property and so now the thing is that if you're looking at so aure it might not be so clear to compare a random element in this sequence to a random element in that but what is clear is that if you try to compare BN with CN And right it's the same set. It's the same set. And one you take the soup, the other you take the in. So of course BN is bigger equal to CN, right? So you have this, right? And so now you see that if you take like a random element here, if it was like further out, if it was N, sorry, if it was C N further out, I just write it as something further out. So you adding something else to it. So this would be like further out. We know that this sequence here. Yeah. Anyway, so you get Sorry, now you're getting this. I'm getting this here. like that and right so you have this so what you should what I should have done is that this sequence here so you have so let let me rather draw it like this so I have that this that so here is maybe C1 here is B1 and now the next one the next element the sequence C n is increasing and the sequence bn is decreasing. So then but you always have this here. So c2 have to lie to the right of this but it has to lie to the left of b2. So you're getting this here. So, so you see that this is a way to kind of you know this gives you a way to compare right that that the CN will always lie to the left of the BN. So you have so so now what we have done here is we took any boundary sequence and we formed two mono sequences. They're both mono And we have so so this means that now from the monot convergence that we haven't quite proven yet but from the mono convergence uh convergence theorem because these are bounded we have that bm converge converging they are going uh they're going down right so they're converging so one often would write it like this more suggestively that is converging downward to a limit B and the CN is converging upward to a limit C. Right? And so what you get is because of all of these elements, all the C ends lie to the left of the B ends. This means that and we we'll prove this later but it seems probably pretty obvious that the limit here C is less or equal to B and this so we'll come back to this also next time um when we talked I if I have time I'll talk a little bit about cushy sequence but I won't really prove too much about it and and and but this here will play a important role uh when we prove the the basic theorem about cushy sequences. Okay. So so the takeaway here at the moment is just that uh so you have you have this concept of a monot uh either increasing or decreasing sequence. If such a sequence is bounded then it's convergent. That's a mono convergence theorem. And again from any sequence any boundary sequence you can act that's there that's a very natural mono increasing sequence and a very natural mono decreasing sequence. Now let's try to prove the monotom convergence theorem and then let's get back to how to think about square root of two in a more intuitive way using this. [Applause] Okay. So now the proof of the monome. So we So now let me just remind you what we have. So I'm only going to prove one of the two versions of it where the sequence is monotum increasing because the other one is just proof is is just the same with obvious modification. So I'm assuming so this is the assumption. So this is the assumption. So, so this is just that the next element is larger equal to the previous one and they're all bounded by this a right now I take I we set little a equal to the soup uh of the a Okay. And I want to show we want to show that a converts to a. Now to do that what we need is we need this stuff about that for a given epsilon there exist a capital n. Okay. So now let me first make the observation that that the that the soup right so a here since a here is the soup it is right. Remember that the soup of a set is the smallest upper bound right so but since it is so it's partic so a is in particular an upper bound so you have that a n is less than a for all n since a is an upper bound. It's actually the smallest upper bound. But for this you just need it's a upper bound. Okay. So I'm going to look at a n sorry I'm going to look at a minus a n. And what I want to prove is that for a given epsilon if I'm sufficiently far out if I'm bigger than some capital if little n is bigger than some capital n then this here is less than epsilon right that's what I want to prove but this thing here you see this here I don't need the absolute value because I already know that this is bigger than that so I don't need the absolute value so I can just so this is what I want to prove that I want to prove that for given given epsilon greater than zero there exist. So maybe let me just write it a little larger here. So what I want to prove is that given epsilon greater than zero there exist a capital n such that if little n is bigger equal to this capital n then a minus a n which I know is positive I just need to prove that this is less than epsilon. Okay. Now, so now how do you do that? Well, I know. So, at the moment, I just use that A was an upper bound. Now, I'm going to use that is the least upper bound. So since a is the least upper bound then uh a minus epsilon is not an upper bound. Right? A minus epsilon is not an upper bound. So this means that there must exist so there exist capital n so that a n is bigger or equal. So a capital n just this one here is bigger than a minus epsilon. Right? because otherwise it would have been an upper box but it's actually strictly but maybe I should strictly because I'll put it up. Okay, but I already know that this thing here is less than a so but and I also know in fact I know more than this. I know that this thing here is less or equal to a little n as long as little n is bigger than capital n. That was just because the sequence was increasing and all of them is smaller than this upper bound. So now you see that so here you have a all the elements a in lie to the left and if you're looking at just the one from a capital N outwards then here you have a minus epsilon and then they lie between this one here and that in particular they lie to the right of this and to the left of that. So this means that they're squeezed in this little interval and it's so this difference here this difference from here to here is certainly bounded by the distance to here which was epsilon right? So this proved that this in here is less than. Okay. And with obvious modification uh the proof of that if you take so this was an increasing sequence. The proof uh that if you have a decreasing sequence that is bounded right that that that the limit of that is the in proof with obvious changes is is know the same. Okay. So now let's come back to square root of two. Right. And so we have this idea that square<unk> two in a way the way you know to just get an idea about how large it is you think about it as uh you know you think about it as one and then a number of digits afterwards right that's a kind of approximation so so let's try to make this more precise Um [Applause] and so I'm going to look at a I'm going to define uh two sequence uh two sequences uh a n and bn sorry two. So we'll define two sequences a n and bn and a n will really be a n where where a n is going to be bn / 10 to ^ nus one. Remember that anything to you know if it's 10 to the power zero by definition that's one. Okay. So this here in particular means that a a1 should be equal to b1. And we're going to define it like this. And we are going to define the BN. So and the sequences here the the so there would be some properties here is that the sequence uh sequence uh a is monotone increasing a just real numbers that it's monot increasing and um and uh and the a um is going to be defined to right so and and and a is defined like that sorry um and the bn and the bn are just natural numbers Okay. So, so a n is defined from bn. It's really a n that we are interested in, but bn is going to be a useful tool here. And so, and it's going to be and the bn to some extent is very simple because they're just natural numbers. So, now um so now we define bn. So bn is the largest natural number such that uh b² is bigger and less or equal to 2 uh * uh 10 to the power 2 n - 2. So that's how right so this here so this so the sets of natural number that is bounded by that right that's like a finite set so you just take the max of soup or whatever the max of that's it right it's just the max of that's it so that's a natural number okay and now and now again we right and then we define. So now we have the bs and then again a ends the a n are defined to be bn divided by 10 to the n minus one. Right? Right. And so again you see that and so right. So now the first thing you observe is that if n is equal to one if n is equal to one then this thing here is zero so this means this has one so it's the largest natural number so that the square is less than two so this means that bn so n= to one bn is equal to one and because this thing here is also zero. So this the denominator is is one. So a n a1 sorry a1 is one. b1 is one and a1 has to be the same as b1. And so that's also one. Okay. So that's how it starts. And now what we want to prove is that um we want to prove that the sequence a n is monotm increasing right we want to prove that a end is monotum increasing so let's just try to understand that right and so again for mono increasing you just need to compare two consecutive elements right so so we want to show so so this here remember that there's a kind of you know there's ways to write a formal proof and then there's like things you have on the scrap paper to get the idea. So this is would be what we're going on may maybe the first statement is so we want to show this you could certainly put as part of the proof want to show that a n is less or equal to a n + one and so this is equivalent to that right? So a n since a n here is equal to bn / 10 ^ n minus 2. You want to prove that this thing here is less than a n + 1 which is b n + 1 over 10 to the n + 1 - one. So this here is b n + 1 over 10 uh to the power n. That's what you want to prove. So we want to show so so you see that we want to show so this is what we want to prove that this thing here is less than this. If you multiply by 10 to the^ n on both sides then you get over here and after you reduce it you get 10. So en so need to show needs to show that 10 bn is less or equal to bn + one right that's what we need to prove right and again this just comes from that you we want to prove forget about the a now you can just think about that we need to prove that this in here is less than this multiply over by 10 to the n on both sides then there's still 10 * bn left and here it's just b okay so we need to show this one here right and so if I was to write it down I would probably have some of this on scrap paper and then I would instead sort of establish this here first and then say therefore you know and then I'll conclude why you So you but but of course that's not the way you're thinking about it right I mean you first try to analyze what you have to prove and then once you know that then you try to write it down nicely okay so so we need to prove this here and so now uh right so now the thing is that we have we have that so The definition of bn which is right here that's the definition of bn but let me copy it over here. So we know that bn squared is less or equal to it's the largest integer. So that you have this in here, right? But this means that if you're looking at 10 bn, that's still an integer. And if you square this here, right? Then this is the same as 100 times bn squared. And so that's that actually has to be less than 100 * this. So this kills those uh I mean so it just becomes less or equal to two 10 to the 2n right like that and this here you can also write as 2 to the 10 * 10 to ^ 2 n + 1 - 2 right but now you see that bn so this so this so 10 bn is an integer and it has this property but bn + one was supposed to be the largest integer so that the square here is less than 2 * 10 to to 2 n + 1 - 2 you see that bn + one is a is the largest one with this property so you conclude that so since bn + one is the largest such and both are integers and and bn is is an integer then uh bn + one must be bigger equal to bn uh must be bigger than 10 * bn right and that's exactly what we want to prove right that that uh bn + one was bigger than bn. So we now have that the sequence a defined this slightly cryptical way but really it's not I mean it sounds maybe cryptical but it's just to instead of using somehow digital numbers you just multiply you're just thinking about you want this and that many uh by so uh like the n is referring to the number of digits but including the digit that is for the the point right so it's actually not very cryptical but it maybe sounds a little formal here okay so we have a n is a increasing sequence okay now why is a n bounded right so but the a is defined from the bn so in order to show that a n is bounded you have to figure out you have to figure out what is a bound for the BN right but you have this here bn is the largest such number if I take if I take two * 10 ^ n minus one suppose I take this number here right and I square it this is a definitely a integer I square it then I'm getting 4 * 10 ^ 2 n - 2 right so this thing here This here is the same as over here but it's now four. So this number here would not be allowed as a bn. In fact it would be larger than bn. Right? So this here so bn is less than 2 to the^ 10 to the n minus one. Right? So this means that a n which is equal to bn over 10 ^ n minus one but this is now you see less or equal to two right so the sequence a n that we have defined is uh as you would have expected they are they're increasing because they really define the way they're really if you think about it is really defined to take you know the number of digits sort of it's going to be like the number of digits in square roo including the one the starting one right so and so um and so they all are going to be less than two right okay now let's try to Okay. Right. So now um we have so we have that a n is an increasing sequence and is bounded. So a end here uh is convergent by the monitor uh convergence here. Right? So the a is convergent by the monotone convergence. But now if I'm looking at a n squ a n squ well a n squ it's the same as a end obviously times a n right but um but um but uh a n by definition was bn divided by 10 uh uh 10 to the power uh n - one right and so I can write like this 10 ^ n minus one and so you see that you can write this as bn^ squ and then these here combines to 2 n minus 2 right and by assumption if you divide by this here this thing here is lesser equal to two so you have that. So you have that a n converts to a. And if you want to think about it, you think about this sequence again. And so now you know that a n by this uh by the product rule for sequences a n times a n the sequence times itself is converging to a n square a square. Right? And this thing here because they are because each of these squares here each of them is less or equal to two then the limit must be lesser equal to two. What I want to prove now is that this a that we get as a limit that the square is in fact equal to two. So I want to show that it's the square is so I have so I have so you know when we're doing it this this is one more proof of that square root of two is a real number but this is sort of in a way how you think about it in terms of the uh decimals right and so we have this sequence a n and we have proven that it's converging to some a and we've shown that this a here has a property that is lesser a squ is lesser equal to true. So I want to prove so I want to show that a and that sorry that a squ is actually equal to not one but two right that's what I want to prove right and but I already proven that a squ is less or equal to two so enough to show that a squ is bigger = to okay so let's try to do that and I'm going to use this is going to be another illustration of the algebraic rules for sequences so I I haven't really I mean I I here now I really you know to prove that is less or equal to two. I didn't use fully the definition of the A because I didn't use I didn't really use that it was the largest such that A was defined in terms of BN where BN was the largest such uh such um such number and so if I'm looking at suppose now I'm looking at BN and I'm adding one and I'm squaring it. Right? Well, this here cannot also be smaller than this here because then then then the BN should have been this thing here, right? So, this here must actually be strictly larger than 2 10 ^ 2nus 2. Okay. And so now you write, you know, you just divide by by this here on both sides. So you have b n + 1 squared and you divide it by 10 to the power 2 n - 2. So you know that this thing is bigger than two. So now but you can write this in here. This denominator 10 ^ 2 nus 2 you can write as 10 ^ n -1 * 10 ^ n minus one. So in this way you can split this up and you can write it as bn + 1 over uh 10 ^ n minus one and then times itself. This is is just to make sure that it's not it's BN and then to the BN you're adding one. So like that you write like that. But you see this thing here, this is this here, this thing here. You can split the fraction up. You can write it as bn uh over 10 ^ n minus one. And then you can add 1 / 10 ^ nus1 and then you can multiply that by itself. Right? You can write it like that. Right? But now now you see now I can write this in here as so now I can think about it in a slightly different way. So I have that. So so I can write this thing here. This thing here is just that is the definition of a n. Right? So I can say so what I have here is that a n + 1 / 10 ^ n minus one times itself but I'm I would just write it nicely out like this that this thing here remember the way the it go that this is equal to that which is bigger than one two so this here is bigger than two right Now 10 1 / 10 to the n you can think about that as a sequence right you can think about this as a sequence that sequence go to zero this here goes to a the sum of these two sequences by the algebraic rules go to the sum of the limits so this is a plus zero so that's a so this thing here goes to a and and this is the same so this goes to a so you see that the limit here then goes to by the product rule for the limits. The limit here is so you get that a squared here and since it holds for each element then it also holds for the limit right except you had to be a little bit careful it's may not be strict inequality but just big or equal to and of course in fact it is by equal to because now you have established the other inequality that a^2 was bigger equal to 2. We had already proven that a^2 was less or equal to two. So you conclude that a^2 is equal to two. If you kind of just think about it, this is really how it sounds maybe a little bit convoluted, but it's really kind of writing down precisely what you think about when you're singing about a square root of two because you think about the number of digits. And so if you take, you know, including the first the one in 1.4 four B including that one. If you think about the number of digits like the A N then the A end is really defined if you were to make it precise you would define it in terms of this other sequence BN. Okay. Okay. So now the next thing is so this was just like a little illustration of how the mono convergence theorem worked. and and and and then and we this will you know play an important role next time we Yeah. >> So uh the reason why at the top we have like strictly greater than two and then at the bottom we have greater or equal to because the a and the a limits whereas on the top we're considering each element. >> So so this um the a this you were talking about this here right? >> Yeah. >> Yeah. So the a n converts to a right this here you I think about this here. So think about a sequence plus another sequence, right? This sequence here converts to zero, right? So the sum of these two sequences converts to zero. And likewise for this and then I'm thinking about it as a product of this sequence with that sequence. And so but but it's like each element is in that in this in the product sequence is strictly bigger than two. But it doesn't necessarily mean that the limit is strictly but the limit has to be bigger than two, right? Because they could kind of approach two and that's exactly what happened of course. Yeah. Uh okay. So and and so I also just want to say that this with lee soup and le mint well we didn't actually define it but I defined you know and this is it is how you defining but but we haven't talked about that concept yet but this I talked about with that if you take a boundary sequence then you could form two mono sequences this this will play a really important role for several things this is how one define the soup and the min haven't talked about those yet. It's also going to be used to prove an important theorem really important theorem and that's what and that I will talk about that theorem next but we won't prove it before next time. So what is what is that stuff going to be used and and and how does the monotone in what is another application and in fact there'll be several other applications of the monotone convergence theorem but so here is um here's a really important concept and so sometimes it's hard to you know in in if it was a monotom sequence then you could kind of write down explicitly what the limits were, right? And in all of the examples we saw so far, it was easy to write down the limits, but in general, you might not be able to write down the limits, but you want to prove that the limit exists. And so, this is the following notion that is then becomes really handy, and that's a notion of a cushy sequence. So what is a cushy sequence? So a cushi sequence. So it's a sequence a sequence with the property that for all epsilon greater than zero there exist n there exist a natural number n such that if n and m are bigger than this natural number then the two elements a n and and a m are less than epsilon apart. So like the so this is a different notion than convergence. Convergence say that if you that a sequence is convergent if everything kind of far out accumulates around the limit. Right? This here say that that and and of course it will. So this has say that if you that a cushie sequence that's a sequence so that if you are far out then syncs a bunch together basically right now of course we will prove and so we prove for the on r we will prove that the cy sequence so we prove that so we prove the following um put for that that's on R but everything is on R at the moment uh that any that a sequence is a cushy sequence sequence if and only if only if it is convergent. Okay, I won't prove this now but but I'll prove this next time and and again the proof of this involve this stuff with we define these two sequences from a bounded sequence. So it it would be useful there. Um uh now let me give you some examples. Um so one uh one uh one um example and the thing is that that and we'll get to that later but as at the moment we talk everything is in R and then we will later generalize this right and so to generalize this a lot of these concepts work much more general and and that's extremely useful and Um and it can be used to find solutions to differential equations but uh and but then you have to abstract these concepts. So uh to where sequences are not anymore real numbers but they are of some other space and then you have to talk about convergence etc. And so you can use this to to define um solutions to def to show that various differential equations uh have solutions. Okay. And so that often goes over using what's called the contracting mapping the mapping. and and the contracting mappings theorem, you know, have all kind of forms and uh the but they're all, you know, it's just a matter of what space you're looking at and and so on. Uh but so we're looking at R here in this case. And so the contracting mapping theorem, that's a map is just a it's just a map here now from R to R. And that the map is contracting means that if you take two elements uh in uh so it means that sorry T is a contracting map is a contracting map [Music] if they exist. [Applause] And this is really important. C is is positive. I mean that's not important. It sort of had to be positive. So that's but the the the important point is that C has to be strictly less than one. That that's a crucial point. Um and then it then it has a property that when you apply when you take X and Y in this in R and you're looking at the how close X and Y are together. So the how close that the image uh under t of x and y how close they are together then those have to be some factor and this is the factor the c is the factor and that has so they have to come closer. This is why c have to be strictly less than one because they have to come closer. If they come closer by a factor than the original one was then then so the contracting mapping theorem say so this is a contracting map sorry so a contracting map is a map from R to R so that there exists some constant strictly less than one so that for all X and Y you have that the difference between the images is this factor that is strictly less than on with what they were originally and so now you get the contracting mapping theorem and we'll talk about the proof next time and and again this you know you know once you have know the more abstract concepts then it's quite easy I mean the proof is going to be the same etc etc and will give you existence of solutions to a lot of differential equation so this is the contracting mapping term So it say the following that if uh t from r to r is a contracting Then then C has a fixed point. It has a fixed point. And this means that t of x is equal to x. And when you apply this theorem, the contract. So this is the contracting mapping. And again, it's crucial that this C is less than one. That's what makes everything go. If it's if it was equal to one, you wouldn't get anything. Okay? It has to be strictly less than one. And this thing here, if you set it up more generally, you know, if the spaces are more general than just the real numbers, then then this here will be that you have a solution. It it could be it can be applied to many other things, but it could be that you have a solution to a differential equation. This statement here, you know, is like a standard application to get solutions to differential equations and you know, you can do use for all kind of stuff. Okay. And so uh so we'll talk about that next time. Uh and not maybe not the the the I talk about the contracting mapping the won't talk quite yet about the applications. Any questions? Okay. Great.