 Hi, I'm Zor. Welcome to Unisor Education. I would like to continue solving certain problems related to exponential functions. It's not maybe so much solving problems as rather just analyzing how this function behaves in certain cases. So without further ado, I have two different problems. Actually, it's three, but the second and the third are very much related. And all of them are about steepness of exponential function. Let's talk about steepness a little bit more. We did discuss this in the previous lecture, but I would like to spend some more time. How can we determine steepness as precisely as possible? Now, if you have one function and another function which are very close in this particular point, actually they have one common five point and basically they are tangential in certain way. How to determine that this function is less steep than this function in this particular case? Well, the typical approach in this particular problem is just have a tangential line. Tangential line is actually determining the steepness. How steep tangential line is, that's exactly how steep the line is. So although this line seems to be less steep than this one, because they share the same tangent, we are talking about the same steepness in this particular case. Maybe a little further, steepness would be different, would be this or this. So in this particular point, steepness is different. But in this, where these two lines are touching each other, that's basically the steepness. That's where the steepness is the same. So steepness of the function is determined at any particular point by the tangential line. Now, we don't really know how to approach tangential lines for any kind of a curve. This is a subject of analysis in mathematics, derivatives and stuff like this. It's all related to limit theory. However, we can approximate tangential lines and therefore steepness by using a very simple approach. If you have some kind of a line, the steepness of which we would like to determine, what we can do to approximate this steepness, let's say in this particular point. We can step a little bit further from this point, let's say to the right, and have the core basically, that's what it is. Difference between increment of the function divided by the difference in increment of the argument. So how far the function grows if argument grows. So if this particular distance is relatively small, then this chord which connects these two points would probably be very close to tangential line, at least with the smooth curve. And basically having this particular ratio would give us an approximation for steepness of this particular tangential line. So I will be using this approach to analyze steepness of the exponential function in some points. So I'm going to take two points and compare what's the increment of the function as I go from one point to another relative to increment of the argument. Okay, that's the preliminary kind of introduction to whatever I'm going to talk about. Now let's consider a function and we all know that the graph looks like this. So it looks like the steepness is increasing as my argument, a greater than one, as my argument grows. How to prove it? How can I state with certainty that this steepness is increasing? Here is how. Let's just take two different, let's take three different points. This would be a natural number and this will be the next natural number and plus one and this will be the previous and minus one. And let's compare how my function increments as I go from here to here and from here to here. If for any n, I will prove that increment from n to n plus one of the function value is always greater than the increment from n minus one to n. It means that the steepness of our function is always increasing with every next natural number n because the difference between the arguments is one. So x2 minus x1 is always one. This is x2, this is x1 or this is x2 and this is x1. So the denominator is always one so I don't have to really take it into account. That's easier. So now I have to compare only the difference between increments of the function. So I'm going to prove that this piece which is a to the n plus one minus a to the n. This is n to the n plus one and this is a to the nth. I would like to prove that this is a to the nth minus a to the n minus one. So I would like to prove that a to the n plus one minus a to the n is greater than a to the n minus a to the n minus one. Now if I will be able to prove this, then what follows is that this increment is greater than this which means the function grows faster and faster and faster as we go to infinity. And that's basically the most important qualitative information or characteristic of the exponential function with base greater than one which I would like to convey. So many words for such a simple proof actually. The proof is very trivial here. Let's bring everything to the left. It would be a to the n plus one minus two to the a to the n plus a to the n minus one. I have to prove that this is greater than zero. I will factor out n minus one a to the n minus one degree. What I will have here a to the second degree rate. If I'm multiplying a to the n minus one times a to the second degree, my exponents are supposed to be added together. So n minus one plus two will be n plus one. Now in this particular case I will have minus two a. This is the first power and this is n minus one. So if I'm going to multiply a to the minus and a to the n minus one times a to the first degree, I will have a to the power of n and minus one plus one. So that's what I will have. And here I will have one. Now this is obviously true because a is greater than one base is a minus one square. A minus one square is a square minus two a plus one. So this is positive, this is positive, so this is positive. And transformations are reversible. That's why this follows from this, this follows from this, and this is true. So this is why this is true. Therefore I have proven that as I go further towards increasing numerical value of the argument, at least if I will go by natural numbers, by integer numbers and increment them, then the steepness grows. Now does it mean that it grows for rational numbers, which are in between these natural numbers or irrational? Yes, because if you remember I did prove that the function is monotonic. So it's monotonically increasing and that's why we have always the same kind of inequality. Okay, now it's actually, what I have actually done, I've proved this theorem about increasing steepness for natural numbers which are to the right of zero. How about to the left of zero? Well, it's exactly the same thing because what is to the left of zero if you will get let's say three points here? It will be minus n minus one minus n and minus n plus one, right? Where n again is a natural number. So let's prove exactly the same inequality that a to the minus n minus one minus a to the minus n. So it's the difference between this value of the function and this one. It's greater than between the value of the function at point minus n minus n plus one. Now how can I prove this? Again, bring everything to the left, a to the n and I will just replace negative exponent with an inverted positive one. This a to the power of minus n minus one is one over a to the power of n minus one by definition. Minus two a to the n, I inverted a to the minus n as one over a to the n and plus one over a to the n plus one. I have to prove that this is greater than zero. How can I do it? Well, if I will factor out a to the power of n plus one, I will have a square here. Indeed, if you will put a square multiplied by a to the power of minus n minus n plus one in parenthesis, you will get minus n minus one plus two, which is a to the power of n minus one. Minus two a, a times a to the power of minus n plus one would be a to the power of minus n plus one. And again, this is positive and this is the same a minus one square. So everything is reversible, so this is proven as well. Again, my theorem is exactly the same. The function is exceedingly steep. As we move from left to right, we always have greater and greater increment of the function on the same unit increment of the argument. Okay, I spent a lot of time explaining this, but the proof as you saw was very easy. Now it will be a slightly different situation. The proof will be a little bit more complex and here is what I would like to talk about. Now we have different functions which have this kind of a property. This kind of a loop if you wish. Some of them are less steep, some of them are more steep. So this is two to the power of x and this is three to the power of x. Now here is my point. Steepness is different in this point. Point x is equal to zero. What's interesting is that if you draw a congenital line to this function, this is a bisector, a 45 degrees angle. So this tangential line would go below 45 angles on a positive number. And tangential line to y is equal to three to the power of x will be above. So the steepness which we have determined as the ratio between increment of the function versus increment of the argument. Now this steepness is equal to one for angle bisector, right? Because whenever you have increment of the function, you will have the same increment of the argument. Because this is 45 degrees, this is 45 degrees. So for angle bisector at 45 degrees the steepness is equal to one. Now for two to the power of x, the steepness at point zero would be less than one and for three to the power of x would be greater than one at point zero. So this is my statement that two to the power of x is below angle bisector at 45 degrees and three to the power of x is above. Okay, now let's try to prove. Here I would do a little bit more precise variation of the steepness. So what I would do is, before I was using increments of one, like n, n plus one, n plus two, etc. Here I will use a small increment. Let's say I will have an increment one over n. Now the greater n I will take, the closer this particular point will be to zero and that's why closer this increment would be to the increment of the tangential line. So if I will use this particular increment of the argument, one nth and I will prove that regardless of n, my ratio which represents the steepness is less than one for two to the power of x. For any n, it means however big, which means however close these points are, that actually will prove that even in a limit case, when these points are infinitely close to each other, I still will have the same inequality and still my tangential line would be less steep than 45 degrees angle. So the steepness would be less than one. So I will use one nth as my check point and I will basically then prove this theorem for any n, which means that the points can be however close and my evaluation would be however precise. So what does it mean for two to the power of x? Well, let's think about it. This value versus this value. So y2 minus y1 divided by x2 minus x1. So y2 would be two to the power of one n, one over n, and y1 would be two to the power of zero at this point. So this is basically the increment of the function. Now what's the increment of the argument? Well, one n, one over n minus zero. And I would like to prove that this is less than one. And I'm going to prove it for any n, regardless of n. Well, okay, let's do it. First of all, we can simplify it. Two to the power of zero is one. So it would be two to the power of one over n minus one. And I will multiply, now this is one over n. So I will multiply both sides of the inequality by one over n. So I will have to prove this. This is invariant transformation. So this is what we have to prove. Okay, let me make it even better. I will put one on the right side. So it would be two to the power of one n should be less than one plus one over n. And now what is two to the power of one over n? It's nth root of two. So I will raise both sides to the nth degree, nth power of n. I will have two on the left and one plus one tenths to the nth degree on the right. So this is the final inequality which I would like to prove. And this is completely equivalent to this one because I did not make any non-invariant transformation. All transformations are invariant from here to here to here to here. So I'm going to prove this. Okay, now what's interesting is that if we know something important, it's very easy to prove. And what is this important thing? Some time ago when I was lecturing about mathematical induction, I introduced the binomial formula which was derived by Newton first. So I'm going to use it. Now for those who don't remember, I do suggest you to go to the corresponding lecture on induction. It's in mass concepts chapter of this course. And just refresh your memory. Anyway, I'm just going to use it here. Now it's a to the power of n plus n divided by one a to the power of n minus one b. So I will increase the number of multipliers here and the power is being reduced by one on every new member for a and increased by one on every new member for b. Now the common element of this particular case, let's say element is n multiplied by n minus one, etc. n minus k plus one divided by one times two, etc. times k, right? Should be k multipliers on the top starting from n down, on the bottom starting from one up, k multipliers. And the number of multipliers represents the same as powers, n minus k for a and b. And the last member would be n, n minus one, blah, blah, blah, up to one from one to up to n b. Well actually I can put a to the zero and b to the n. The same way as here I can put for commonality b to the power of zero. So this is my binomial formula. Again, derived by Newton and you can refresh your memory by looking at the lecture. I'm going to use it for this particular expression and it's quite obvious that right now if I will drop all the members of this particular formula, which are all positive by the way, except the first two, so I will leave in this formula only the first two members which is one to the power of n. So we don't need this, we don't need this. So one plus one over n to the power of n equals. So a is one, b is one over n and n is n. Now one to the power of n which is one, b to the power of zero plus n which is the exponent times a which is one to the power of n minus one which is one and times b. Plus doesn't matter what, I already have two. You see this is one plus one which is two. So without this tail I have two. With this tail I will have greater than two. That's it. So if you know the formula it's very easy to derive this particular inequality. Okay, so we have proven that two to the power of x lies below 45 degrees angle at a point zero. Now let's talk about three to the power of x. I will prove that it will be above. Now very similarly, y2 minus y1 divided by x2 minus x1 if y is equal to three to the power of x. Now my x1 is equal to zero, x2 is equal to one over n, right? So that's my graph. This is zero, this is one over n. This is my 45 degrees and I am claiming that, well, I have to draw it not through point zero, obviously, but to the point one. So I'm claiming that it goes this way, above this particular line. Alright, so y2 minus y1 would be at the point where x is equal to one, one ends. It would be three to the power of one over n. This point is x1 is zero, so it would be three to the power of zero divided by one over n minus minus zero. And I would like to prove that in this particular case it would be greater than one. Okay, so this is inequality which I would like to prove and I will do exactly the same as before. So this is one, this is zero, right? So it's three to the power of one over n minus one should be greater than one ends or three to the power of one ends should be greater than one plus one ends or if I will raise both into the power of n, it would be three should be greater than one plus one over n to the ends degree. So that's what I'm going to do. That's what I'm going to prove using the same binomial formula. So let me just write it down on the left. One plus one over n to the ends degree should be less than three. I reversed it, right? It would be three greater than this. I wrote it as one plus one ends over n to the power of n less than three. Okay, how can I prove that using exactly the same formula? Now it's a little bit more difficult. However, not impossible. Let me do it this way. So a is one and b is one nth. So let me rewrite it in the following fashion. One plus one n to the ends degree is equal and I will use this formula. So wherever I have a, I can just completely drop it because a is equal to one and this is multiplied everywhere. So I will use only b which is one ends. Now in this particular case, it would be b to the power of zero which is one. Then I will have n divided by one times, drop the a, b is one ends, plus n, n minus one divided by one times two, drop the a, it would be one over n squared, plus, etc. What's my common member? Common member is n, n minus one, etc. n minus k plus one divided by one times two, etc. k, drop a one over n to the power of k, plus, etc. Forget about the last one. Now from here, it's quite obvious now that this is one, this is one. Now this is greater than, this is greater than one over two. I mean, let's, sorry, it's less. Why? Here is why. This is n times n minus one in the numerator and denominator is greater, it's n times n. So n times n minus one will always be less than n squared. And two remains as it is. Now, as far as my common member is less than, I will do exactly the same separately for numerator and denominator. Now, numerator n times n minus one, etc., etc., divided by n to the power of k. This is less than one because numerator is less than denominator in this case. And these I will replace as two times two, etc., times two. So I will decrease this part, denominator, since I decreased denominator from one times two times three times four times k, I put only one times two times two times two times two. So my denominator is diminished by this because every multiplier is now either the same or smaller. And, not even that, I'm increasing this particular fraction by replacing this piece with one. So I'm increasing this fraction by twice actually, by decreasing the denominator and increasing whatever is left of it. So when I'm replacing n times n minus one, etc., divided by n to the power of k, this is the same number of multipliers as here. k different numbers here and k here. These numbers are smaller than these numbers, right? This is n which is the same, next one n minus one and this is n. So every one of them on the top is smaller than this one. So if I will replace it with one, I'm only increasing the whole element. And I'm also increasing by replacing whole numbers in the denominator with twos. So this will be my common number. And what actually I am ending up with? I'm ending up with two, plus one, two, one second, plus one quarter, plus one eighths, right? It's number, it's two multiplied by itself, plus, etc., plus, etc. Now, obviously from progressives, from progressions which are addressed in another lecture, this is obviously less than one, so it's less than two plus one, which is equal to three. Well, this is actually quite easy to, even if you don't know progressions, it's quite easy to see because what is this? Let's draw a square, one half of the square, which is this, plus one quarter of the square, which is this, plus one eighths of the square, which is this, plus one sixteenths, which is this, etc. So we are gradually filling up whatever it is, but no more than the whole square. So even if you don't know progressions, they are addressed a little bit further maybe down in the course, but you can definitely see that this piece is less than one, so the whole thing is less than three. So I'm using certain pieces of knowledge which might be a little bit more advanced, however, they are addressed in the lecture, so you're welcome to refresh your memory or learn it separately. So the binomial formula of Newton and some of the progression are used in this particular case, but if you have it in your repertoire, the whole thing is actually simple, because all I did is I just reduced each number, each member of this sequence to get whatever I need. So that basically completes the proof that the graph of three to the power of x is above four to five degrees. And here is just one small side comment about this. Let's just consider again the graph. Okay, this is point one, zero one, x, y. So this is 45 degrees angle. So one of my functions which is, well actually I can make it a little longer here, one of my functions which is three to the power of x goes this way, it's above, and another function, they have many different colors, goes below this, like this. This is two to the power of x. I also know, and I did prove it before, that the greater the base, the steeper at every point, the steeper the curve of the exponential function. So if I will increase the base from two to three, somewhere between two and three, there is a number where my tangential line is exactly 45 degrees. I mean that's kind of intuitively understandable statement. So if two to the power of x has a tangential line which is below 45 degrees, and three to the power of x is above 45 degrees, it should be something in between which is exactly equal. And actually it does exist. And that's one of the, well, reasonable definition for a new number, which in mathematics is designated by a letter e, which is irrational number, and approximately it's 2.71, but this is just an approximation, it's an internet number of decimal digits. So the number e plays extremely important role in the analysis. Strangely, however strangely it sounds, this irrational number is probably one of the most important numbers in analysis. It's like pi in geometry, which is the ratio between the circumference of the circle and diameter. It's irrational number pi, and we're probably used to think that nice numbers like 1, 2, 3 are the most important. Actually, no. In geometry, pi is probably the most important number, and in analysis e is probably the most important number, and those are irrational. Anyway, that's it for today. I do encourage you to try to reconstruct the proof again. By yourself, if you can't just listen to the lecture again, it's very important for you to understand how these things are logically derived from some previous knowledge which you might have, which is the binomials and progressions. And it's interesting what's the origin of this particular number, why is it so important? That's it. Thank you very much and good luck.