 Hi, I'm Zor. Welcome to Unizor education. This lecture is part of the advanced mathematics course for high school students presented on Unizor.com website and I do suggest you to listen to this lecture from this website, from the video which is linked to this website, because the lecture has its notes. Notes basically can be used like a textbook. Now, the topic I'm going to talk about is related to the general category of limits, which we are talking right now about. But in particular, it's also related to an exponential function with a very special base, which I denote as a letter E. Now, the E is a specific number in mathematics. It plays in calculus and mathematical analysis as important role as the number pi plays in geometry. So, I'm going to talk about this number, about its properties and about the exponential function with the base equal to E. Okay, so that's what it is about. We're talking about today. So, this lecture is heavily dependent on all the material which I have presented in the topic related to exponential functions. So, I suggest you to refresh this. Also, binomial formula will be used, which is also presented in the course as part of the mathematical concepts. The lecture was induction and specifically binomial formula in the induction category. Okay, so, first of all, what I would like to talk about is some interesting property of, well, exponential function, but basically many other functions. So, let's recall what exponential function actually is all about. That's where it is. Where A is, well, some positive number, usually. Now, the graph of this function with A greater than 1 is monotonically increasing function at point x equal to 0. A to the power of 0 is all with 1. And then, as x increasing, the function increases to infinity as x decreasing to a negative part. And you remember that negative exponent is basically 1 over corresponding positive exponent. So, like, A to the minus 2 is actually 1 over A square, right? So, the function obviously goes down to 0 when the argument goes to minus infinity. Okay, so, we will consider this function. Now, this and many other function can have at any point actually, but we will consider the point 0. We will discuss the concept which I call a steepness of this function. You see, some function can be this way, some other function can be this way. Well, obviously, the dotted line is steeper at this point 0. Now, how can it be measured, this steepness of the function? It's basically a characterization of the function. We are interested to know whether one function grows faster than another or steeper than another in this particular case. Because both of them are equal to 1 at point x equals to 0, right? So, what differentiates the behavior of this function from this function? Well, obviously, the steepness is one of the characteristics which basically reflect this behavior. Now, the way how we will try to approach this concept of a steepness, we already touched, by the way, this point when I was talking about exponential function. But now, we will talk about this in relationship to the number E and that's why I would like to spend a little bit more time. Okay, let's do it in a very large scale. Now, steepness of this function can be characterized by an angle of the line which is tangential to our line. So, this is tangential line by this angle. So, the steeper the line, actually it means that at this particular point the tangential line has a larger angle. And the angle is usually characterized by either the angle itself or by some trigonometric function of this angle or in the case which I would like to make, the angle is very much characterized by the ratio between this and this. Because this is the right triangle, obviously. So, the ratio between these two things is characterizing the angle. Obviously, the larger this ratio, the steeper the curve is. Okay, but how can I determine this particular angle and this particular ratio for a function which is given by a formula? It's not really easy thing to do, right? And here is what I suggest to do in this particular case. Let's say this is a point we would like to know what exactly this ratio is. So, I will take another point nearby and I will replace the tangential line which is somewhere here with this one. Now, obviously, if I will make the right triangle of this, well, it will not be exactly the ratio between this and this. However, if I will make this point as close as possible to this one, then this line, which is like a chord actually, this line will be closer and closer as far as its direction is concerned to the tangential line. So, if I will make this ratio going to a limit as this particular increment from this point to this point is going down to zero, then I can count on the final, the limit value of the ratio between this and this as this point comes closer and closer. It will be exactly the ratio which corresponds to a real tangential line. So, how can I accomplish this using the formula? Well, very easily, if I have a point zero, now I will increment my value of the argument by one over n. Where n is some natural number. Now, then I can calculate, well, for this particular line, I can calculate the ratio of this versus this as what? Now, this particular catechise is equal to value of the function at this point, at point one over n minus value of the function at point zero. So, this particular line is equal to a to the one n minus a to the point zero. Now, what is this? This is increment. It's between one n and zero. And the ratio is this. Now, as n increasing, my point one over n goes closer and closer and closer to zero. So, this ratio at its limit as n goes to infinity will give me exactly the value of my steepness as I have defined it for this particular function at point zero. So, this is basically an approach. Okay. So, this is a sequence and whenever my n goes to infinity, this sequence has certain limits. I mean, I hope it has certain limits. And if it does, then this limit would be exactly the steepness, the characterization of the steepness of my function at point zero. And that's how I can differentiate one line from another, one function from another. Now, my next point is to do exactly this for two functions. One is two to the power of x and another is three to the power of x. So, I will show that the steepness of this function is less than one, which means that the tangential line would have an angle less than 45 degrees. Because at 45 degrees, my ratio this to this is equal to one. But if the ratio is less than 45 degrees, less than one, then the angle is less than 45 degrees. So, this function would be the tangential to this function would make an angle less than 45 degrees. And this function greater than 45 degrees. So, the ratio would be greater than one. So, that's what I'm going to prove right now. Now, as a problem, a similar problem was discussed actually during the... It's one of the problems in the chapter for exponential functions. So, I will just use exactly the same logic. This is kind of a refreshment, if you wish. So, what I'm going to do is... So, let's consider first function two to the power of x. So, I would like to prove that two to the power of... Well, not x actually, one nth, right? Minus two to the power of zero divided by one n minus zero. I would like to prove that this is always less than one. Now, if I will prove that this is always less than one, then I can go to the limit. And the limit is also would be less than or at least equal maybe, but definitely not greater than one. That's from all these theorems about the limits which we have already learned in the previous lecture. So, that's what I'm going to do. Now, this particular inequality is obviously equivalent to this one. Minus zero I can delete. Two to the power of zero is always one. So, it's this. So, it's equivalent to two to the power of one n minus one less than one n, right? I multiply both sides of the inequality to the positive constant one over n. Or, if you wish, two to the power of one nth less than one plus one n, right? I added one to both sides. Or I can raise both sides into the power of n and this is equivalent to this. So, this particular inequality is equivalent to this one, sorry, n. And I'm going to prove this one. Now, to prove this one, what I'm going to do is I'm going to use the binomial formula, which I mentioned in the beginning. I would like you to repeat from the induction lectures in mass concepts of this course, which is a to the power of nb to the power of zero n divided by one, sorry, plus n divided by one a to the power n minus one b to the power of one plus nn minus one divided by one two nn minus two b square plus etc. The common member is nn minus one etc. n minus i plus one divided by one two three etc. i a to the power n minus i b to the power of i. And the last one would be a zero bn. Now, this is binomial formula Newton, which we have proved by induction in the lecture, which I have just referred to, and I'm going to use it for this purpose. So, a is one, b is one over n. So, let's just think about what happens in this particular case. Well, obviously, if one and one over n are a and b, all members are positive, right? That's obvious, right? So, the first member is one to the power of n, which is one, and one over n to the power of zero, which is still one. So, I have one plus one over n to the n equals two. The first member is one. The second member, n over one, which is n, times a to the power of n minus one, one in any power is one, and b to the power of one, one n to the power of one. Plus. Now, starting from the third and the fourth etc. doesn't really matter, they're all positive because I already have two, right? So, that's why one plus one over n to the power of n is greater than two. It's two plus something, and something is the whole tale of all these things. So, that's easy, right? Fine. So, I have proven this particular inequality, which means if I will go to the limit with n going to infinity, it actually resulted in the following thing, let me just repeat it. Less than one, which means that the limit of this will also be less than one, and this is tangential line two to the power of x. So, this angle is less than 45 degrees. That's about two to the power of x. Now, let's talk about three to the power of x. It's slightly more complex, but I will use exactly the same formula. So, it's three to the power of one over n minus three to the power of zero, one over n minus zero, it should be greater than one. That's what my point is right now. Alright. Or, again, minus zero we can remove. Three to the power of zero is one. So, that's equivalent to if I will multiply by one n, that's completely equivalent to this one, and this is completely equivalent to this one. And this is, I will raise to the power of n both sides equivalent to this one. So, that's what I have to prove. Again, I will use this and use the binomial formula. So, what do I have here? So, I have one over n, one plus one over n to the power of n should be less than three. Why should I, why did I have more than greater than three? Let me start from the beginning. I think I changed the sign somewhere. So, three to the power of one n minus one divided by one over n, it's supposed to be greater than one, right? So, three is supposed to be greater. Three is supposed to be greater than one plus one n to the n, right? Okay, fine. So, I have to prove that this is less than three, right? I just changed the direction. Alright, so how can I prove that? Well, again, let's just look at this particular binomial formula. Let's start from the beginning. Well, obviously the first is one and the second one is one as well, right? We already did it with this one. Now, in case of the previous one, we just completely removed all detail and obviously that resulted in this particular inequality. In this case, to prove the upper boundary, the three is actually above this, we will use all these members. And what I will do is I will actually increase every member. So, that's why I will put sign less here because I'm going to increase every member. Not this one and not this one. Starting from this member, what I'm going to do is the following. What is b? b is one over n, right? This is b. This is a. a is one, so a should be completely ignored from everywhere, right? b is one over n, so b square is one over n square. Now, notice the following thing. b square or one over n square is n and n in the denominator. In the numerator, I have n and n minus one. So, n and n minus one in the numerator and n in the denominator. Obviously, denominator is greater. So, if I will completely reduce this and this, I will only increase my value of this particular member, right? Since I'm replacing n times n minus one, I'm increasing to n times n, in which case, I will have n square here and n square here. Now, in the general member, same thing, b to the power of i is basically one over n to the power of i. Each one of these and the number of these is also i, by the way. Each of these is either equal to n or less than n. So, if I will replace them with n, I will only increase it. But if I will replace them with n, it will be n to the power of i and this is n to the power of i, right? And if I will completely eliminate them, I will increase my expression, right? So, that's why I put the sign less. So, that's one plus one plus one over one times two remaining plus one over one times two times three plus, etc. Plus one over one times two, etc., times i, etc. Plus one over n factorial, actually, from i to n. Now, the last one is obviously this, because the coefficient is one, which is actually n times n minus one, etc., etc., times one. And this is one times two times three times n. That's why I have it one. But it doesn't really matter. It doesn't really matter. The denominator one times two times, etc., times n remains. But numerator was actually replaced with n to the power of n and it reduced with n. With b to the power of n. Okay. Now, I will increase it even more. So, again, it's less than one plus one is two plus. Okay. Forget about one. It doesn't really matter. And every one of those I will replace with two, which means what? I will reduce the denominator, which means I will increase the whole number, right? So, it will be one over two plus one over two times two, which is two square, plus one over two to the third power, plus, etc. And the last one, one to the n's power, right? I will increase it again. Instead of actually ending up at number n, I will go to infinity. Because you know what this is, right? One half plus one quarter plus one eighth, etc., it's an infinite geometric progression. The sum of it is obviously is equal to one. And that's why it would be two plus one, which is three. So, that's my inequality. Finally, I proved that one plus one n's to the power of n is less than three. Because I am increasing and increasing and increasing and finally I got only three, right? So, that's why it's less than three. Okay. What does it mean graphically? Graphically, it means that the steepness of three to the power of x is greater than one. That's what it means. So, graphically, again, this is 45 degrees. Now, this is greater than 45 degrees. And that's where my three to the power of x. The tangential line at point zero is steeper than 45 degrees. That's quite interesting. You see, with exponential function, we don't need this anymore. Or we don't know for more. So, back to exponential function. So, two to the power of x has a steepness, let's call it beta, less than one. At point zero, right? Three to the power of x has steepness greater than one. Now, what does it mean? I mean, obviously, and we actually talked about it many times, the greater the base, the steepness basically is growing. And that's what's important here. It's growing from two to three from some number which is less than one to some number which is greater than one. Now, it's natural to assume that the steepness is gradually increasing from something which is less than one to something which is greater than one, as the base is increasing from two to three. Which means that somewhere in between two and three, there is some number where the steepness at point zero exactly equals to one. So, we are talking about the line which has steepness this 45 degrees. So, this base of this particular exponential function has some, this base is in between two and three. It's not an integer number. Well, obviously, there is such a number. In calculus and mathematical analysis, it's actually proven that this number is in between two and three. And it's irrational, by the way. It's interesting in geometry. Pi is irrational number, right? In analysis, in calculus, E is also irrational number. Approximately, it's equal to 2.71. Actually, there is a joke among mathematicians. Is it true that E to the power of pi is equal to pi to the power of E? Well, they are close, but they are not equal. I mean, this is completely wrong. But, you know, sometimes people are asking the questions like this. So, in any case, E is a very important point, a very important number in analysis, as I was saying. It's as important as pi in geometry. And the main characteristic, and that's how we actually introduced this particular number, is that the function E to the power of x has the steepness at point zero exactly equals 2.1. So, this is equal to this in the limit as we are moving closer and closer between these two points. Right? Now, what's the value in some... I mean, I have kind of indirectly defined this number E. It's the number which is in between 2 and 3 with this particular function E to the power of x having the steepness of 1. Can I define it a little bit more precisely? Well, yes. And here is the considerations which I can offer in this particular case. So, what I know is that E to the power of 1n minus E to the power of 0, which is 1, divided by 1n minus 0, which is okay. It's going to 1 as n goes tend to infinity, right? So, that's what I know. Because this is exactly the limit of ratio between change of the value and change of the argument, as the argument is closer and closer to zero. That I know. Well, what does it mean? It means that as n becomes greater and greater, this is almost equal to 1, and this approximation is better and better as n increasing to infinity, right? Now, let's just change this approximation slightly. This is not as rigorous as I would prefer it to be. In higher levels of mathematics and calculus and in analysis, it's proven much more rigorously. But in this particular case, I think these intuitive manipulations are very useful and kind of natural. So, I'm using approximately equal to, with saying that the approximation is as better as the n goes to infinity. Now, if I will change this to this, right? For this, I mean this, 1 plus 1n, and then raise to the power of n, I will have this. With saying that approximation is better and better as n goes to infinity. So, what I can say is that e is the limit of 1 plus 1 over n to the power of n if n goes to infinity. This is an important formula. Now, the formula is actually exact. I did not prove it rigorously, but the formula is correct. I just illustrated how it can be derived, but not exactly rigorously, because I cannot really, with approximation, do the same as I do with the quality, obviously. I did it just for intuitive kind of purposes. So, this is the value. Now, let's just evaluate it for a couple of n's. If n is equal to 1, I have what? 1 plus 1 to the power of 1, which is 2. So, e is approximately equal to 2. If n is equal to 2, I have 1 plus 1 half, which is 3 halves, square, which is 9 fourths. So, e is approximately 9 fourths, which is 2 and 1 fourth, right? If n is equal to 3, it's 1 plus 1 third, 4 third to cube, 64, 27. 64, 27, that's 2, that's 54, so it's 10, 27, right? Now, as you see, it's growing, but we know it's limited to, it's no greater than 3. So, basically, if you will do this up to infinity, you will see that you will get some number like this. So, that's the value of e. So, we have introduced a specific value, number, actually. We have introduced the number, which we call e, and therefore we have introduced the function. This function now makes sense, because I know what e actually is. It's some real number, irrational, by the way. Now, but this function is a regular function which has a regular exponential function. So, it has the regular exponential properties, which is e to the power of 0 is equal to 1, as any other, e to the power of 0. Now, e to the power x plus y is equal to e to the power of x times e to the power of y. This is the property of all exponential functions, including this one, or something like this. So, all we know is this one. And then, I would like actually to pay attention to one specific property of the function e to the power of x, which no other function actually has. Here it is. So, we know what the property, which we call a steepness, at 0.0 is. So, we have introduced this concept. We have the increment of the function from 0, from x is equal to 0 to x is equal to 1n, and divided to the increment of the argument from 0 to 1n. And we know that this goes to 1. What I'm going to say is that very similarly, we can introduce a steepness at any point. So, here is the graph. Here is my point. I call it x0. I can have x0 plus 1n, which is this one. And I will do exactly the same thing. I divide this by this. And then, as n goes to infinity, 0.x0 plus 1n goes closer and closer to here. And this particular ratio becomes a steepness of this particular function at 0.x0, right? So, what exactly is this? Well, this is e to the power of x0 plus 1n minus e to the power of 0 divided by the difference between the arguments x0 plus 1n minus x0 equals. Now, I have just mentioned the property of exponential function that if you have some of two exponents, it means x to the power of 0 is exactly equal to this one, right? And here I can reduce by x0, so it would be 1n equals. Obviously, I can factor out e to the power of x0. And here I will have, right? What is this? This is a steepness at the point x is equal to 0 as n goes to infinity, right? So, it goes to e to the power of x0 because this steepness, as we have actually defined the number e, it's the function which has a steepness, if the function with this base has this particular steepness of equal to 1. So, this goes to 1 as n goes to infinity. That's why the steepness of function at any point x0 is equal to exactly the same function value at this point. Because e to the power of x0 is constant, right? As n goes to infinity. This goes to 1, so the product goes to e to the power of x1. Okay, this is quite a remarkable property of the function e to the power of x. It's steepness at any point equals to its value at this particular point. No other function, well, except some function which is derived from this one, like this multiplied by 2, for instance. No other, principally other function has this property. That's what's very important. It's almost like a characteristic property of the function e to the power of x. And this is the end of this lecture. That's the last property I wanted to talk about. I wanted to introduce using the limit theory and some other knowledge which we have known. I wanted to introduce the number e and the function e to the power of x. Well, that's it for today. I suggest you to read the notes to this lecture on theunizor.com. Well, I will probably use this particular function in some other very important cases. In particular, when I will talk about trigonometry and complex numbers, there is a remarkable connection between this number, the trigonometry and the theory of complex numbers which is basically, which looks like one formula, the Euler formula, which I'm going to discuss a little later in this course. It's a synergy between the number e, the complex numbers and trigonometry in one formula. It's really a beauty, quite frankly. I mean, I was always amazed how certain pieces of mathematics can be beautiful, but this is definitely. So, that's it for today. Thank you very much and good luck.