 Hi, I'm Zor. Welcome to Unizor Education. We continue the course of advanced mathematics for high school students. It's presented on Unizor.com and that's the website which I'm recommending you to listen to this lecture because it contains the notes for every lecture including this one and it helps you to basically follow the logic maybe much much much better. Now Today we will talk about the number E, which I have introduced in the previous lecture as a function actually as a base of a the basis of the function which has a very important property. If you recall these exponential functions have a shape something like this with the base greater than one. Now I have introduced the concept of a steepness at point zero in this particular case which is basically a characterization of this angle. Angle of the tangential line at the point zero. Apparently this angle is less than 45 degrees with exponential function with a base is equal to A and it's greater than 45 when exponential function has a base equal to three. So the three goes something like this much steeper and its angle will be greater than 45 angle. So I presume that there is some number in between two and three some real number which if used as a base of exponential function it will have this tangential line exactly at 45 degrees and I call this number E and as a consequence of this as a consequence of this I have come up with the following thing. How can I determine the value of this angle? Well, basically it's very simple. Let me just make it in a larger scale. Okay, here is my larger scale. Now this is zero. I will step from zero by one nth where n is some number. All right, and then I will take a chord basically. Now if I will divide this piece which is basically A to the power of one n minus one because at zero all the exponential function are equal to zero. So this is this piece and divide by this piece which is equal to one n. Now, what does it mean that my tangential line would be at angle of 45 degrees? Well, it means that this particular ratio will be almost one because it's 45 degrees. These are two cateches of right triangle. With the 45 degrees, cateches are equal to each other, right? So basically the the steepness can be characterized by this ratio. Now if I will increase n so my point comes closer and closer to zero, my chord will be closer and closer to the tangential line. So if this ratio tends to one as my n goes to infinity, that means that my tangential line would be at angle 45 degrees. So basically I have determined that there is some kind of a number which is if used as e, then this would be tending to one. That's how I define basically the number e as a base of an exponential function which has this type of a tendency, right? Okay, now I kind of presume that this particular approximation exists and the approximation is is better and better as n increasing to infinity. From which I have derived that e to the one n approximately equals to one n plus one, right? So one plus one n over n and then I raise both sides into n's degrees. So e is approximately one plus one n to the n's. Now these manipulations with approximation is not exactly rigorous mathematically. So today I'm going to prove the following thing that this sequence does have a limit. And then you can actually say that this limit is e, this number e. All right, so this is the most important goal of my today's lecture to prove that this sequence has a limit as n goes to infinity and this limit, since I know it exists, I call it e. Now I definitely know that this limit is in between two and three because in the previous lecture I actually proved that this expression for any n is in between two and three. It's greater than two and less than e, than less than three. So there is some number in between two and three, some real number, which I can call basically the letter e. All right. So how can I prove that this thing has a limit a little bit more rigorously than what I did in the previous lecture? The previous lecture was kind of an intuitive approach. Now I will make it a little bit more rigorous. Now there was a theory in which I have proved in one of the previous lectures dedicated to limits that if I have a sequence which is monotonically increasing and bound from above, let me just do it graphically. So this is my bounder a and my sequence is monotonically increasing and always less than eight. So my points are here. So the theorem says that if my sequence is monotonically increasing and always bound it on the top by some constant a, then the Riesli limit and this limit is actually the least upper bound for this particular sequence, which is called supremum. So what my point right now is that I will use this theorem which states that if it's bounded and monotonically increasing then there is a limit and that's what actually I have to prove here that there is a limit which I can call the number e. So if I will prove that this is monotonically increasing sequence, I already know it's bounded from above by the number 3, that would be sufficient to prove that there is a limit and if there is a limit I can call it anything I want including this beautiful letter e which mathematicians like very much. All right, so all I have to prove again this. That's what monotonically increasing means, right? That the next one is greater than the previous one. Now, there are actually many different methods to approach this including some tricky manipulations with powers, etc. What I have decided is to do it relatively straightforward, although maybe a little lengthier than some other methods, I will just use the binomial formula in both cases. Now, the binomial formula again it was discussed before in one of the lectures before and you know that this is the following. It's sum of the common member. Common member is coefficient is n factorial divided by n minus a factorial and i factorial, i from 0 to n from 0 times a to the power n minus i b to the power of i. All right, that's the general binomial formula by Newton. So, I will use this formula for this and for this case and I will basically prove that every element of the sum on the left would be greater than the corresponding element of the sum on the right. That's what I'm going to prove. Now, one more interesting thing. The number of elements in this sum, in this case it's n, in this case it's n minus 1. So, the number of elements is greater than here. So, from 0 to n and here from 0 to n minus 1. So, what I will get to do is that if every member on the left is greater than corresponding member on the right plus in addition this member also has one extra member that would make it even better. All right, so all I have to do is have i-th member of this sum. I will compare with i-th member of that sum and I will show that one is greater than another. All right, for every i from 0 to n minus 1. Now, what's my i-th member on the left? Well, a is 1 so I don't need this and b is 1 over n so this is 1 over n to the power of i. Now, that's my member on the left. i-th member where i is 0, 1, 2, 3, etc. up to n. On the on the right it's correspondingly n minus 1 factorial divided by n minus i minus 1 factorial i factorial and 1 over n minus 1 factorial. So, I would like to prove that this is greater than this. Okay, here's how I can do it. First of all, i factorial present in here and here. So, if I will compare it, I don't really need this. I will prove that this is greater than this. Both are containing this multiplier i factorial so it doesn't really change the relationship between them. So, how can I prove that this is greater than this? Let me open all these factorials in this particular case. What is n factorial? It's n times n minus 1, etc, etc, up to 1. n minus i factorial is starting from n minus 1 to 1. So, if I will divide one by another, let me just write it again what happens here. This is n, n minus 1, etc, 1 and n minus i, n minus i minus 1, etc, 1. So, somewhere along the line starting from n minus i, I can basically reduce it a certain number and what will be left here. On the top, I will have n minus, etc, n minus i plus 1. Starting from next one, n minus i, it will be reduced with the denominator, right? And this is divided by n to the i's degree. Similarly, this one would be n starting from n minus 1. I will have a certain number of multipliers and the last one would be n minus i because starting from the next one, n minus i minus 1, it would be reduced with the denominator. So, it would be n minus 1, n minus 2, etc, n minus i, that would be my last member. Divide it by n minus 1. I did get the term, no, that's wrong, n minus 1. n minus i, sorry, n minus i to the i's degree, n minus 1 to the i's degree. Okay, now how can we compare these? Well, this is actually quite easy because let's spread this n to the i's degree as n times n times n, etc, n times n, etc. Same thing here, n minus 1 times n minus 1, etc, n minus 1. It's i multipliers on the top and i multipliers on the bottom. Same thing here. So, what I'm going to do is I'm going to show that any member of any result of a division of the corresponding member on the top divided by bottom, which is n minus k divided by n. That's the common member, right? Where k is 0, 1, etc, i minus 1, right? That's my last k. Here, the common member is also n minus k minus 1 divided by n minus 1, where k is 0, 1, etc, i minus 1, right? Because if k is i minus 1, so it's i minus 1 plus 1 and minus 1 it will be minus i, right? So, this is a product of these and this is the product of these. But look, every this one member of this product is greater than this one. k is exactly the same. Now, n minus k divided by n is greater than n minus k minus 1 divided by n minus 1. Now, why is this obvious? Well, because this is equal to 1 minus k over n, right? And this equals to 1 minus k divided by n minus 1, right? So, from 1, we subtract this and from 1, we subtract this. This is less than this because the numerators are the same denominator. This is bigger denominator. So, the ratio is smaller. So, we are subtracting a smaller number. So, the result would be bigger than this one. So, that's why this is bigger. And that's why this is bigger than this one and this one. And that's why the product of these is greater than the product and everything is going back to original inequality. So, that's how I have proven that my sequence is monotonically increasing. Now, since it's monotonically increasing and it's bounded from the off by number 3 and it's actually bounded on the bottom by number 2, it means that my limit exists. This is a valid formula, a valid expression. There is such a limit and I call it E. Now, E happened to be irrational number. The approximate value is 2.71 but then it goes to infinity. Obviously, it's irrational number and it plays a very significant role in the analysis as I was basically telling you many times. All right. Now, the only one thing more I would like to add to this particular expression. What I will do now, I will prove this. So, from this, I will prove this. So, if I will have E to the power of X, then it's the limit of instead of 1, I will have X here. Now, how can it be proved? Well, intuitively, again, it's obvious and here is Y. Now, n is an infinitely increasing number, right? X is just a fixed number, just some real number X, whatever it is. So, X over n is decreasing, obviously, right? Now, in this particular case, we have the same thing, the denominator here and the power there and we know it goes to a constant. So, what I'm going to do is I'm going to consider this. Let's consider X is not equal to 0, right? I multiply by X and divide it by X. Now, when I multiply something in the power, as you remember, A to the power of B to the power of C is equal to A to the power of B C, right? So, this is equal to 1 plus X over n n over X and then over there, I have another power of X, right? That's what it is. It's the same thing. This is n. n is n times X divided by X. So, n divided by X, I'm using here and X is there. Now, let me ask you, what's the difference between this inside the square brackets? What's the difference between this and this? n is increasing to infinity, right? Now, that means n divided by X where X is some kind of a constant. Well, let's assume right now it's not equal to 0. It's a positive constant. n is also increasing to infinity. I mean n over X is also increasing to infinity in exactly the same fashion. So, it doesn't really matter whether it's n or n divided by 2 or n divided by 3 and 7 quarters. It doesn't really matter what kind of a constant is here. The limit of this inside the square brackets is the same thing, e, and that's why the whole thing is e to the power of X. So, without actually going into more detailed analytical calculations with which goes to a definition of limit, which means that for every distance d we have to find number n, which makes our sequence closer than d to this limit, etc. I don't want to go into all these details. They are trivial in this case. But intuitively, you perfectly understand that that's exactly how it's supposed to be done. So, since this is true and n is going to infinity, then this is exactly the same kind of limit, whether it's X divided by n or 1 divided by n. Doesn't matter because X is a constant and it's exactly the same in the power and here. All right. So, that concludes my last statement which I wanted to make, namely this. Now, it's very important actually to understand that there are many different approaches to the same thing and the more different approaches you found which lead you to the same thing, the more natural it becomes and if you wish even beautiful the theory is, if you can approach the same formula from different sides, right? So, in this particular case, this is one of the solutions which I'm offering. And there are some others as well. My purpose was to introduce you to this particular limit. That's the purpose of this particular lecture. All right. Thanks very much. I suggest you to read the notes in Unizord.com to this lecture or listen to it again. I mean it's always beneficiary. And so, for those people who really want to get a little bit more involved in taking exams for instance and getting some grades, you can sign as a student to the website to Unizord.com. And then you will probably have to have a supervisor who will assign you some topic to learn and at the end of this topic you will pass exam. So, that will give you a structural process, educational process in advanced mathematics. I certainly suggest you to do this. Very beneficial. All right. Thanks very much and good luck.