Lecture 10: Continuous Functions; Exponential Function (cont.)

Key Takeaways

Okay. So last time we um we looked at power series. So you have a sequence uh am uh this is a sequence And then for each for each x you form the the new series you form the series um and the series is where a n * x ^ n and then you're summing you decide somewhere to start and then you're summing to infinitive like that. So something like this that's a power series and we looked at a particular case u where the a ends so example here is where a n is equal to 1 / n factorial uh and so we got the function e of x which is a sum from n =0 to infinity of x ^ n divided by n factorial. And then we we use this to we observed that we observed that uh e is bigger equal to zero for x bigger equal to zero. And we also observe that e of uh of zero is equal to one. And we defined little e to be e this function e of one. And then we defined another function um uh e to the q where q here is now a rational number. Right. And uh it was defined in particular E of zero was defined to be equal to one. E of one was def was uh was just defined to be E right. And uh then we also we assumed which we will prove today that e of x plus e uh e capital e of x + y is equal to e of x * e of y. So we assume that and then we saw and and then we defined right. So this little e this function based on the little e in particular e to some power m if m is an integer is just e * itself uh m times and then we defined um uh e to a power 1 / n here to be uh the positive number. This was defined to be a positive number so that uh alpha is saying so that alpha raised to the power n is equal to e. uh and and and we also defined um e to a negative integer and this was just to be defined to one / so we defined e to minus m to 1 / e to the so this way we defined uh uh we this function here the power series was defined for for all x and we defined another function little e of q uh on all rational numbers q right and then we saw we assume this here we saw that uh that these two functions so e of x is equal to e this function here is only defined on rational numbers so it's they're equal when uh for at rational numbers. So we saw that and then in in the P sets P set five I think it's for both of them you will show or maybe it's one for four and one for five you show that E is continuous [Applause] uh and uh you will also show that that E is the so it would also Show the following that if you have if f and g are two continuous functions uh and h so they are defined both of them f both f and g are defined on the real numbers. and you assume that if you restrict them to the rational numbers then the two functions are equal. So if they're both continuous and they agree on the rational numbers then actually they must agree everywhere uh on all Okay. So in other words, this one here, you know, this is a function by P sub5. It's a function that's continuous on all of R. It's agree with this function. And so this is the the only extension of this function to be continuous and defined on all of them. Okay. So now, but we still needed to prove we still needed to prove this relationship here that we assumed. So let's try to prove this one here. And I'll just do a the main case of it. Um and so we're going to prove so theorem prove that E of X * Y is equal to E of X time e to the y and I I'm going to show this proof uh in the case where where these are non negative numbers both of them the general cases is once you have this is the general case is not very difficult but I won't do the general Now how are we going to prove that? So to prove that we're going to use another theorem. So the other theorem is that so we take two suppose you have two um suppose you take two series of non each of them are of non- negative numbers. So, so the setup is so again this is going to go into the proof of that theorem over there but the setup for this is is more general. I take a series A N and I take another series BN and I'm assuming that both A and B and for all N for is are non negative so for all n so these are series of non- negative numbers and then I form another series. Um I will form a third series and that third series is going to be defined from from the two previous series and and the third series is called it CN like this. And now CN here is going to be the sum of a i b uh n minus i i i = 0 to n. So this has the effect of that that the index of this plus the index of that is always adding up to n. Right? So the sum of the two indices is is ex is exactly in right. Okay. So that's so obviously so note note that that the C ends here of course also non- negative that just sums of non- negative numbers right? So this is a new series right of non- negative numbers. And then the theorem is so the theorem is the theorem is that if if these series here this one here and this one here uh are convergent. Then this series here is also converging. And and uh we also have that the sum here that this sum here that this sum is just a product of the two sums. So it's a product of the sum here like that. So again we have these two series of non- negative numbers. From those two series you form a third series and the third third series the idea is that the third series is kind of like a product of the other two and it's uh the CN in the third series is given where you take the sum of the products of the of the elements in the other two so that the index is is at n and then if these the two series you started with are convergent then this you know product series is also convergent and the sum of the product series is is really the product of the sum of the other two series. Okay. Now so that's what we want to prove and then once we have this we're going to use it to prove this the term here. uh and so so now to prove it let me just first observe that so let me observe over here that um right so uh let me make a couple of definitions and then making an observation let me start with with making some definitions that is useful in the proof. So we will I'm going to define s n a just kind of like we usually do where we summing here from i = z up to n of a i right the the superscript is is come from the series the a series and then likewise I define uh S and B to be the sum here of the B eyes and then I'm defining uh S and C to be the sum of the C eyes. Oops. Up to like that. And so of course you have that we know that a i, b i and ci are all non- negative numbers. So this means that convergence so convergence of any of these series right so convergence say of this one is the same as saying that these numbers here are bound right that's by mono converter so convergence of any of these series is equivalent to that that the corresponding S n a b or c a b c is uh are bounded. Okay. So that's just an observation and again this is the monot convergence of this is by Mon. Okay. So that's a monotone convergence theorem. Uh the other thing uh see I thought of something else I wanted to to note. Um right okay so let's try to prove that so to prove it let me just try to think about it for a second here so if I'm looking at a1 say a2 plus up to a n suppose I'm looking at this here. This is like this is like SN A, right? And I have that SNA and suppose I'm now looking at the SNB. This here is the SNP. Now if you just multiply out the product of these two right then you would get and I should have started at zero sorry let's start at zero because that's what we do up there like that okay so now I multiply out and you see that you're getting a zero uh b and you're getting then you can kind of if you try to do it. So, let's try to multiply this out, but let's do it in an ordered way. Okay? And when I'm talking about ordering way, I'm not talking about just multiplying this one on that. Uh I mean, you could say that's very ordered, but it's not it's not uh it it's not related to what the the C series. So, I want to just group things so that they remind us of the C series. But if I do that, so what I starting with this uh is is this multip multiplied by this. Then I take this one here multiply by that. So a z b1 right this here is the only one where this the index the the subscripts are non- negative integers. This is the only one where the sum of the two subscript is zero. Right? If I'm looking at where the sum of the two subscript is one, this is one of them. But there's another one and that's where I take this multiply by that. So I sort of group those together. So a1 b 0. And now I want to collect the ones. So those are the only two where the sum of the subscript is one. Then if I'm looking at at the when I'm multiplying out where the sum of the subscript is two well I could start with this one but then I have to pick b2 so I take plus a z b2 then I could take a1 b1 right that's has also sum of subscript two and the last one is if I take a2 But then the B have to be the zero first. Right? And now I proceed like this. So I'm grouping it like this. Okay? Now of course there's going to be a little bit of an issue. So for a while everything is you see that this thing here this here is what we call C0. This here is C1. And this here is C2, right? And then we presumably also get C3. But then there is a point where uh where we sort of run into trouble if we want to really do it like this. So, so what we get is that. So when you do this multiplication what you will see is so you see like this a z plus a1 a n what you will see times b 0 b1 up to bn what you'll see that this pattern here it's continue for a while and you're getting that this is equal to c 0 c1 C1 plus C2 up to CN. CN in the CN, right? This is again where the sum of these indices could be this one here multiply by that the sum of the subscript will be N. Right? But if you're now looking at the next one that correspond to CN + one, you're not getting all of the CN + one because like one of the CN + one would be like a to the N +1* B 0. But you don't have that one here, right? So what you get is that you're getting all of these plus something additional. Okay. And the additional is because everything is non negative. The additional thing is a non- negative number. Right. Okay. So what we have established here is that we have so this here show that s n c this is here this here is let me just write like this. This here is sn. So you see that and this here this here was SN this here this here was S this here was um right this thing here was SN A and this here was SNB right so what you have is so this inequality just tells you that S N A time S NB is equal to S N C plus something positive not negative. So in other words, this number here is lesser equal to the product of these two. So this imply that SN c is less or equal to this here. But now you see you already have something because these here was increasing right they were increasing because the a was all non negative so these here are increasing and they are lesser equal to right there the series was supposed to be convergent so you have that sorry so I'm here I'm using that the sn a is is monot increasing up to the limit here, [Applause] right? And likewise the SN here monc increasing up to the limit, right? So this thing here so this is smaller than the limit. This is smaller than the limit of that. So this thing here is less or equal to the product of um of these guys, right? And so you see and that's that's supposed to be just a real number this here. So you see that these guys here are now bounded. So this means that because they're monotone it means that that series is also convergent. So what we have concluded here is so what we have concluded here is that uh that the series uh that this series here is convergent and we have that uh this infinite sum here is lesser equal to the product of the other ones. Right. This is what what this gave us. [Applause] Right. So this is what we have so far. Okay. Now we we want to we want to uh right we want to get the reverse inequality also. Right. So we want the reverse inequality of what is over there. So that we can say that those two things here are actually equal. Right? Now let's try to see how do we prove this the reverse of this inequality. The reverse we proving by examine more closely what that product was. So you see so what is this additional thing here? So again, so we have we're looking at this thing here. And what we got is that this here was equal to the sum here of C N. But then we have some additional terms right and the additional terms is the thing is that if you start looking at the additional terms so what are the additional terms you're getting a a1 bn right that's one where the sum here is n + one right and then you also get so this is coming from this uh sorry that one here multiplied by this you would also get a a2 b minus 2 you get a a3 b sorry not n minus2 n minus one b3 n minus2 right etc right so but all of these here all of these here and then right so you get that And but you don't get again you don't get the one that is missing here is a z b n + one. This one is missing, right? And the other one that is missing is this one here. But otherwise you get all of the ones that was in the CN + one and you don't get anything more. Right? And so this path continues. So like the next term would be like a2 sorry yeah then you get you would also get something where the sum of the indices was two that's like a2 bn right but here you would be missing more right you'll be missing a z bn + 2 you're missing a1 b n + one right but you don't get anything more Fine. And so the thing is like when you continue like this, the last one you get is like when you're multiplying that by that and that's a n * bn those here the sum of the indices is two two n sorry right it's 2 n but you're missing a lot now right you only have one of many many terms right but they they but it is a term that all appear and this one would appear in C 2n. So in this way what you have is that this here is less than s to the power 2 n c right they are all in there but you will be missing terms but all terms are non- negative right so this means that you have that in quarter right so you see that now we have now what we have here is that we have what we established before was the inequality that uh s n c was less than s n a * s n b. But what we have now here is that this thing here, this product is less than s 2n c. This one here is squeezed between this one here and that one here. But both these here converts to the same. Right? They both converge to this infinite sum. Right? So these here converts to the same. This means that this here the limit of this here must also be the same. Right? And so this proof this show that this here is the reverse inequality. So proof proof over there, right? Okay. So now so that's um okay. So this is this is this that if you take two series and you form the product then that also converge if the two series is non- negative numbers. So let me just make a corner of this that of course if you take a series here uh and it's a I'll just do a trivial cor but you can see how you can use this and this is kind of used in proving the general case. So we we will only yeah I mean so okay so so suppose I take two series here so this is a trivial corary of it suppose I take two series here and I know that this one here that they're both absolutely convergent If they're both absolutely convergent, then if you take this series here, CN C, sorry, CI. Well, uh, this one here I claim is also absolute converted. Again, this this sort of trivial color is is is the key ingredients in proving a more general statement that would be used say to prove this inequality here when x and y are non negative are not assumed to be non- negative but I won't get into that but I just make this observation that would get this so I claim that this thing here is also absolute convergence right so again I'm not assuming now a sign on these but I am assuming that they're absolute convergence and so if they're absolutely convergent so prove so absolutely convergent convergent. This just means that the series here, this one here and this one here are convergent, right? That's what absolute convergence mean. But now if you're looking at CI, the CI says this here is the sum here I = 0 to N of A I sorry A I B N minus I okay but this means that CI absolute value here is less than this thing here absolute value like that you can take the absolute value sign in under the summation sign so this is less than sum here from i = z to n and then a i * b n minus i like that But now you see that this here is a this here is the series. This here is the corresponding C. So that so you could call you could now define uh you could now define so you start with the sequence AI and I now define a a new sequence bi bar and I do it by just taking the absolute value like that. So I know that this new series is convergent. Likewise I defined bi bar to be the the BI absolute value and I know that this is convergent and then I defined a sequence here CI bar which is just the absolute value of the CIS and I know that this one here so I know that that this here is converging that's by assumption. This one here is convergent by assumption. And this inequality here proves that this here that the CI bar is also convergent because this one here is the one where you form the products of these two convergent sequences. So this one here is also converting. So this here is also converting by the okay anyway there was just an an observation that we we're getting this sort of statement out of it. Okay. Now this is also related. So why do we have non- negative here right or why do we talk about absolute convergence? So this is a simple fact that if you take a serious if you're trying to sum infinite numbers right we now know how to sum infinite numbers. If you're trying to do it if the numbers that you're summing are all non- negative then actually the ordering doesn't matter. You can check you you know you can take the fifth element first and then the seventh and then the 119 and then come back and take one. It doesn't matter what order you do it in. But if the numbers are not non- negative then and the and the series is not absolutely convergent now it suddenly matters because you could end up taking all the negative numbers first and then you get really really far down to you know to minus towards 9 minus infinity then you take a few of the positive elements and then you take more of the negative and so then you would actually stay very far down towards minus infinity and and you would never be able to come up again. I mean so so this is like a basic fact that if you take a series and the series are of non- negative numbers then actually the and it's converging then you can rearrange the ordering and it will still converge and it will still converge to the same number. If you're trying to sum infinite numbers and they are not non- negative and they are don't converge absolute uh and they don't converge absolutely then the ordering does matter okay and this is related to why we assuming that all of these ai bis are non- negative numbers and okay but we won't get into this I mean you can read about it it's a sort of standard fact it's it's certainly somewhere in the book but it's not Yeah. >> Yeah. So I mean so it's related to it's related to so so I mean it's a you know again this is not part of the sort of syllabus but I think it's a useful thing to keep in mind because it also sort of explain why we're doing it in a certain way. Suppose you take the the alternating harmonic series, right? Remember that this thing here, that's a harmonic series. This here is is divergent, right? I actually proved I I proved or at least I outlined in class why this series here is convergent. And this is um okay. And it's just again the idea here is that if you take so you have something which is called the alternating series test. So this is a I'm coming back to your question just a second but but but but and this is closely related to that. So the alternating series test is that if you take a sequence a n of non- negative numbers and a end converts to zero. Let's say that a n converts to zero monot in a mono way. Right? So, so an example here of this would of course be where a in is one away right these that's a mono uh sequence right so if you take a mono sequence like this but it doesn't need to go very fast to zero this one does not convert to zero very fast right but if you have that and you're now looking at the alternating series based on This one here is convergent. This one here is convergent. And why is it convergent? It's convergent because of the following thing. And this is exactly what I did you know when I when I try to explain why this here is convergent that you're starting somewhere let's say starting at zero then you're looking at the first element right if you're summing here let's say you're summing from one to infinity. If you're summing from one, then the first element here, this here is minus one. This is non negative. So this is a negative number. So you're going down. You're starting at zero. You're going down. Right? The next element that will be positive because it's true here. So it's positive. It will go up. Right? But it's monot. So you don't go as much up as you're going down. So you're going first you're going down then you're going up but you're going less up then you're going less down then you're going less up etc. And that actually is forcing it to converge. Okay. And so now why is this related to your question about how do you reconcile these? So you see like if you're looking at something like this and you now took all of the you know if you first took many many of the of the positive elements and then you know one or two of the negative and then many of the positive element again then actually you could for this series if you rearrange the order of the things you're summing in then you could either get it to diverge or you could get it to convert to pretty much anything you want. Okay. So, so this hopefully explain why when the series is not non- negative numbers why you know the ordering really matters and and again this is related to very closely related to to this with the product of things. Why is related? Because we're getting this, you know, we're getting this remainder term, right? We're getting we got something that was kind of like the remainder and you that's the remainder that you want to be small in order to say that things are converging to the same thing and and you wouldn't have that necessary in general unless you assuming that the terms were not necessary. Okay, so let's come back to that was just a little bit of sight but I hope it makes things a bit more clear. Let's come back to this theorem here. How do we prove this? So we're going to prove it from the theorem that we did with that if you so we're going to use this theorem we just proved which is this theorem here. So we can use this theorem here to prove that theorem up there. So now it's just a matter of identifying the right series. So right so we have so this is what we want to prove. This is what we want to show. And this one here, this is the infinite sum i = 0 to infinity x + y to the power n sorry over uh uh i uh let's call it this sum from n i like that. That's what what that series is. So, and this one here, that's a sum x n factorial. And this one here is some uh y uh n factorial. Right now, so I want So, if I want cn. So the idea is that if I'm writing CN as X + Y ^ N / N factorial suppose I write CN like this and I write a n X n / n factorial and I write bn y n / n factorial. Right? Suppose I write I define them like this. I already know that this here is convergent and that here is convergent. I by by the way of course I also know that this here is convergent but but I now have that but I don't have I haven't proved so but I claim now that this CN is really kind of form in this way. Okay. So let's see. So CN here if I define CN to be this in here by the binomial formula. So x + y if you take this here to the power n this thing here is the sum here of x i * y n - i and then the coefficient is sorry x i * y to n - y i = 0 to n and the coefficient is n choose i. That's a binomial formula. Okay. Now what is what is n choose i? Well, n choose i. n choose i. This is just n * n - 1 down to n - i + one. Right? This here gives you n elements and then divided by i factorial. Right? But you can write this in here as n factorial. If you do that then you included too many. So you have to divide by n minus i factorial and then of course you still have the i factorial in the denominator. So you can write it like that. Okay. So this means that the CN here our CN is so we define CN like this X to the power + Y to ^ N / N factorial and if we use the binomial formula then it was equal to that. So this here was equal to 1 / n factorial sum i = 0 to n x i y n minus i and then n choose i like that. Now I write in this little this little equation here and so I get that 1 / n factorial summing from i = 0 to n and then n factorial n - i factorial * i factorial x to the i * y - Okay. And now you see that the n factorials here cancel with the n factorials. So you're left with i = 0 to n. And now I can group it. I can group it. I can write it as x i / i factorial * y / n - i factorial and this is to the power n minus i like that. But now you see that this here is exactly this here is exactly when we defined uh a n like this and bn like this. This here is exactly the sum from i = z to n of a i b n minus i. Right? And so you see that what we have is that this thing here that so so e of x + y with our definition of the c's this was just defined to be equal to this limit but but it's also the cn is also given by the product of these two other series but this means that it must be the product of the two like that. It's product of the two sums. Infinite sums like that, right? But this thing here is of course just e x and this is just e y, right? So this proved it, right? Okay. And again this here I was just assuming that the X and Y are non negative but given what I said also this colorary it's not very hard to prove the general case from that okay so now um but I won't do that but let's so let's um let's just talk about a little bit more about continuous functions so let me one more time just remind you about it and then let me give you a sort of wild example of of um of a function. Um so so so first let me remind you with continuous so continuous right is a really important concept also in in this class. So I have a function and it doesn't need to be defined. It's defined say on some subset of R into R and then we say that uh and then we have some point X0 in the domain and we say that f here is continuous at X0. If the following hold if the following holds that for all epsilon for all epsilon greater than zero there exist delta greater than zero such that If um x is in the domain and x - x0 is less than delta then the images deviate at most by epsilon. Okay. So this is again what it's say to be continuous and here unlike the previous couple of times when I define it here it's really defined you see it makes total sense on any subset of R it doesn't need to be a interval even right doesn't need to be all of R but it doesn't need to be an interval um yep now so this is what it means means uh for a function to be continuous. Let me give you an example of a sort of somewhat wild function. Uh and so continuous is a very nice property. It's continuous if it's continuous at all points. You know, usually continuous is a nice property to have. Often it's sort of the minimum bare minimum. It's often not enough. you know, you want something more than continuous, but but of course other times you you know, you may have much less. So here's an example. So this is an example. So it's a function. So it's a function defined on all of R. Example of a function defined on all of R that is is nowhere continuous by nowhere continuous I mean that it that's not a single point where it's continuous Okay. And so this is the function f ofx and the function uh is uh if x here is uh is a rational point uh then f of x is sure and if uh if x here is a irrational point then I setting it equal to one. Okay, just a pretty simple function. And so I claim that this is not continuous at any point. And so the thing is that see if you take any points, let's take a rational points first. So suppose here suppose you take a rational point here. This here is x0 is rational then you have that if you are looking at square root of two we know that this here is this here is not rational. We proved that right. We showed that this here is not rational. And so in fact by if you take if you take uh one over sorry square<unk> two divided by n where n here is sufficiently large then you can make this thing here very close to zero. Right? So you could look at and then you could add this thing here. So you could look at x0 and you could look at square root of two over n for n really large. If n is really large, you can make this thing here. So this is<unk> 2 plus x0, right? So you can make this really really close to x0. This thing here would be irrational and so the value of the function here would be one. Okay, let's make this precise a little bit more precise, right? And so let me maybe say just before I make it more precise. So let let's just draw the graph here. So we have here x0. We have here this this here is this irrational number which which you could think about but it doesn't need to be this number. It could be anything that is irrational. This one here is irrational. And by making n super large right this here is super close to x0. The value of the function here is zero. The value of the function here is is one. Right? And so you see that you have these points as close to this point here as you want but the value of the function is very different. Right? So so again just to kind of spell it out um you know given an epsilon you see that if epsilon here If epsilon here was was smaller than one then you would have that uh f ofx minus f of x0 this here is uh a rational point. So this here is this is zero. So this is just f of x and if this here is irrational. So if x here is irrational then this here is one. So you would have that you would have that you can choose this irrational point. So for any delta for any delta we can choose uh x which is irrational and so that x - x0 is smaller than delta. But we have that f of x here is actually equal to one. And so this but this difference here was supposed to be smaller than delta. That's a that it's not the case. Right? If I if if epsilon smaller than epsilon if epsilon is bigger smaller than one it's definitely not. It's it's one. Right? And so similarly if you take so this was proving so this showed right. So this proof this proof prove that f is discontinuous at all uh rational points. Now let's prove that it's discontinuous at all irrational points. So so if x here zero is is irrational then we have of course that f of x0 is equal to one. But we but but we also know that arbitrary close. So for any for any uh delta greater than zero there exist there exist a uh rational point a rational number x uh so that x - X0 is smaller than delta. Right? But now again we have that f of x minus f of x0. This thing here is one. This thing here is zero. So this difference here is one. And so you couldn't make it strictly less than one right by having this delta sufficient small. So this pro sort of same thing. It proves that the function is also discontinuous at all um at all irrational points. Okay. Now let's talk a little bit more. Right? So this is uh okay. So let's talk a little bit more about some of the algebraic rules that we we we discussed last time. and I wanted to single out two of them. So, so we had so if we had the following if f and g are continuous functions and c is a constant. Then last time we talked about uh the following. We talked about that f plus g the sum of these two functions is also continuous. We talked about that if you take a function and you multiply by constant this is also continuous. We talked about that if you're looking at the product of these functions that's also continuous. The ones we mentioned but we didn't actually I mean you know these here was so elementary. This one here I sort of outline and let's talk about the last two properties. One is close to what we had for sequences and that was that if f here does not vanish is nowhere vanishing then one / f here is also continuous. And the last one which is sort of different from anything we had for four sequences is that um that if you take if you have now a function so if uh so you have these two functions f and g but now I'm thinking about f here as going from say some subset of r into r right but it may starts right by into some set D. This is a subset of R. This is another subset of R. And then I have G here goes from this set into R. Right? It needs to be right. So you need this here to be defined on the images of F. If you have that and these two functions are continuous then if you form the if you form the composition. So this is g of f ofx this new function h here is also continuous. Okay. And so I want to in a few remaining minutes uh I want to talk about uh these two properties and I I will start with the last one because that's somewhat different from stuff we tal about last time. I mean for for for uh for sequences So let's try to prove that this composition here is continuous. So proof of five. So given epsilon given epsilon right well we know that the function so you sort of have to start backwards given epsilon you know right sorry uh let's before I take given let me just say that I want to prove it let's show that um that h which is a composition is continuous at x0. Okay. And and uh so I want to prove that and for for for convenience I'm just going to call the image of x0 under f and we'll call that y. Okay. So now given epsilon given epsilon bigger than zero. There uh since since uh g is continuous at y zeote that this is this is y z. So since G is continuous at Y Z there exist at delta let's call it delta G. So that if y - y0 if this is less than delta g then there images sorry on the g f but on the g. So then their images g of y minus g of y0 the images under g is smaller than epsilon. Okay. So this is just because g is continuous and I'm using that it's continuous at y z. Now I kn need to use that f is continuous. So now so now I'm thinking about this delta G as my epsilon that in F right so from from delta G and the continuity Okay. Uh there exist there exist uh there exist um delta of f such that if x - x0 is less than delta of f then the images f ofx minus f of x0 the images is less than delta g right but now you see now I have this thing here f ofx this thing here this is f ofx uh and this thing here f of x0 was y z right so this is the same as this but now you see that so that I have these two points that they are and where one of them is y zero and then I have this other point and their difference is bounded by delta g so I'm in this iteration So it means that the images of these two. So this imply that the images of this thing here under G and this here on G. So G of F of X minus G of F sorry G of Y zero. This thing here is less than epsilon, right? This is just because I that was how delta G was chosen, right? And you see that one of them one of these two points have to be y zero. But that's exactly what it is over there. So I have this but you see this thing here but but this thing here is just But you see that f g of f ofx this is just the composition. So this is what we call h of x g of y 0. Well y z was defined to be f of x0. Right? So this thing here is g of f of x0 but that's just a composition at x0. So you see what we have now is that so so therefore we have that if x - x0 if this thing here is less than this delta f then we have we have this thing here right but this thing here is h of x minus h of x0 Z is less than epsilon. Right? And so this prove that the composition of these two function is continuous. Right? And so again when you have when you have two function and you look at the composition, it's really important to first check that the second one is defined. So the second one should be defined where you end up under the first map. Right? So okay. So now um so now let me just say a few words about the other one number four. But that's the number four is closer to things we had for for the sequences. So you so we are now pro of four. So what we have is we have this function. It's never zero and we have some x0 in the domain. And we know that f here is continuous. We only need it to be continuous at x0. And then I want to show that 1 / f ofx is continuous. And so I'm looking at um so I want to estimate this difference here and I want to say that I if I pick an epsilon then I can pick X as close if if I pick X to be within delta but delta sufficiently small of x0 then this difference here can be smaller than epsilon. So now just let me just estimate this thing here and so this thing here is just by so now I just put it on the common denominator. So I have f of x0 minus f ofx divided by f of x0 uh f of x right and now it's sort of the usual problem but but now we we maybe not so worried about that usual problem the usual problem is this s x x0 is fixed so this is just some number. So that doesn't really matter too much. This one here is not fixed, right? But so that's a usual problem. I mean this is problem we've seen before. We saw it in three up there and we saw it in the similar things for sequences. But now the thing is that that since since uh f of x0 is not equal to zero then you can just take so then now you can first take so we're going to uh right so now we just going to we we're going to pick the delta in uh in uh the delta we want in in two in two steps the first step is since this here is not zero then we know that they exist at delta 1 greater than zero. So that if x - x0 is smaller than delta 1, then the image between f of x and f of x0. I can make that smaller than f of x0. over two. Right? You see I'm just using this one here first as my epsilon. Right? Once I have that, but this here implies of course that f ofx see I have f ofx in the denominator and I want to bound this here from above. Right? Right. I want to bound it from above. So I just need to get a lower bound for f ofx. Right? So f ofx here. So I have that f ofx uh plus um I can write sorry I can write um using this I can write f of x0 as uh f of x minus uh sorry f of x0 like this minus f ofx plus f of x right obviously I can write like that and so this means that f of x0 taking slabbing absolute value sign on it this here is less than f of x0 - f ofx plus f ofx, right? And so now I can move this one here over on the other side. So I have that. So I have that. So if I move this one here over on the other side, I get that f ofx absolute value is bigger or equal to f of x0 absolute value minus this difference here like that. Right? But if I make x and x0 delta one close to each other, then this difference here is at, you know, it's actually less than this thing here. So there's still another half. You see this one here must be bigger equal to 12 of f of x0, right? Because this here takes up strictly less than half of this thing here. So there's still another half left. Okay. And so this give us a lower bound for this. And now we're out of time, but this give us a lower bound for this as long as we make sure that X and X0 is delta one close. And once you have then you can just treat this here as a constant. And so now it's clear that you then make another choice of of delta possibly even smaller than the one this one here. And then for this smallest of these two numbers then you can now conclude that this fraction here is smaller than epsilon. Okay. Any questions? >> Yeah. Um, I have a question going back to what we're talking about. >> I was a little bit confused because like we're defining it to be like non continuous >> but like by definition of continuity like if you had if you picked some um like two x's near each other that were both rational. >> Yeah. >> Like and f of x is both zero then wouldn't that be continuous? >> Right. So, so the first of all, let me say that if you were to picture this function, right? I mean, of course, it wouldn't be very easy to picture it, but but you would sort of if you were to draw it, right? Then actually, it would look it wouldn't look like a function anymore because like you would have it on all the rational numbers here, it would be zero. On all the irrational numbers here, it would be one. So, if you picture it, it would look like two lines. But of course it wouldn't be two lines because it's right but the points are just so close together that you couldn't really draw it right. And so the thing is that but it also means that if you this picture here illustrate what you're saying that for if you take here if you take a rational numbers well all the rational numbers that is really you know any rational number whether it's close or not it will all be zero on those. So that's not where the problem with continuity occur. The problem occur because you also have irrational numbers close to it and it has to be for all points close to this that it the value had to be almost the same. Right? So it's really for the irrational numbers that the problem occur with continuity in a rational numbers and vice versa. for an irrational number. If you're looking at continuity around an irrational number, the problem is not from other irrational numbers, but it's from the rational numbers.