 Hello, I'm Zor. Welcome to Unisor Education. Today's topic will be induction, a problem in the category of induction problem number five. And in this particular case, it's a real problem. It's much more difficult than those little exercises we did before. So let me go straight to the point. This is actually a theorem which was attributed to a French mathematician Cauchy. Here it goes. First of all, there are two definitions. The arithmetic average and geometric average. Now, arithmetic average of n numbers is, as you know, their sound divided by their number. The geometric average of the same numbers is an nth root of their product. Well, intuitively, it is kind of obvious that both averages are really average. They are somewhere in between these numbers. However, they are quite different and there is a famous inequality between these two averages. Arithmetic average is always greater or equal than geometric average of n positive numbers. We'll talk about positive numbers just because we don't want to have something like a square root of negative numbers or whatever. So all numbers are positive and arithmetic is greater or equal than geometric average. The problem number one is let's prove it. Well, first of all, let's start from a simple case. Only two variables and we will try to prove it for this case when n is equal to 2. So we have to prove that x1 plus x2 divided by 2, that's average of two numbers, is greater than square root of their product. So this is arithmetic average. This is the geometric average. Well, this is actually quite simple because if you will take this obvious inequality, since this is the square root of x1 minus square root of x2, now this is obvious because this is the square. So every number positive or negative being squared will give a non-negative result. But if you will open this, you will see that this is square of the first, which is x1 minus 2 product plus square of the second, which is x2 and that's greater than equal to 0, which basically gives us this inequality. This goes to the right, so we'll have x1 plus x2 greater than 2. Now, the multiplication is product of two square root. The root is obviously a square root of their multiplication. That's the property of the power and this is the same as this, that just divide by 2. So in case n is equal to 2, it's really very simple thing, straightforward check and there is no problem. Now, obviously since I started the whole problem in the category of induction, we will try to prove the whole thing using the method of mathematical induction. So we have proved for one particular beginning value of n and a typical way of proving something using the method of mathematical induction is, well, first you check it for some beginning number like for instance n equals to 2 and then you prove a theory assuming that the formula or the quality or whatever is true for n is equal to k, assuming that that is true, prove that it will be true for n is equal to k plus 1. Then, using the standard logic from n equals to 2, using this theory, we go to n equal to 3 from 3 to 4, etc. And that's how we prove it for any finite number. In this particular case, it's not that easy because if you will go to 3 for instance and you have to prove this, the third root of n x1 x2 x3. Now, it's not really obvious how from this we can go to this. However, let's make another observation. If I will use n is equal to 4 instead. So I have four numbers and I have to prove that this is the fourth root of their product. This actually is quite easily deducible from the previous one. How? Well, very simply. Just divided by two pairs. So it will be x1 plus x2 divided by 2, x3 plus x4 divided by 2 divided by 2. Now, obviously, this left part is the same as this one, right? The common denominator 2. So it's the sum of these 4 divided by 2 and divided by 2 more, so it's divided by 4. Now, but this, let's consider this as a one number and this another number. And we know that their arithmetic average, just two numbers, we will use this thing, is greater or equal than square root of their product, right? So these are individual numbers and I'm using my theorem which I have actually proved for n equals 2. In this case, it's one number, two number. Two numbers, their average greater than their arithmetic average is greater than their geometric average. Now, in turn, I can say that this is greater or equal than. Now, let's consider just these two arithmetic averages and use this same theorem. So it was the square root of x1, x2 times square root of x3 times x4. Now, obviously, square root of square root is the fourth root, right? So this will be fourth root of x1, x2, x3, x4. And that's exactly what we had to prove from the very beginning. So we have switched from 2 to 4 elements from n equal to 2 to n equal to 4 very easily, which is divided by pairs. Now, if I will have instead of 4, if I will have 8 numbers, how that would be? Well, it's kind of obvious that I will use exactly the same technique for 8. I will divide, so for n is equal to 8, what I will have is, I will have x1 plus, etc. plus x8 divided by 8. That's my beginning on the left, right? All 8 numbers. I don't want to write each one of them. But then I can pair them and I will have x1 plus x2 divided by 2 plus x3 plus x4 divided by 2, etc. x7 plus x8 divided by 2. And then, divide by 4. Well, probably if let me just write down all 4 components would be nicer. So next one will be x5 plus x6 divided by 2. And next will be x7 plus x8 divided by 2. So obviously, the arithmetic average of 8 numbers can be represented as average of 4 numbers where each one is a pair average of the previous 2. Okay, now, but this is the same as this because I can just consider these as separate numbers. And if I will use this previous theorem for n is equal to 4, I can say that this is greater or equal than fourth root of their product, right? x1 plus x2 divided by 2. x3 plus x4 divided by 2. x5 plus x6 divided by 2. And x7 plus x8 divided by 2. So here we're using this one. Now, again, let's use the same theorem for a pair. So instead of x1 plus x2, I can reduce it even further. So it will be square root of x1 x2. Same here, here, and here. So basically, it will be greater or equal than fourth root of 2 roots, which means it's the 8 roots of their product. And that's exactly what we had to prove. So as you see, we are relatively easily going from 2 to 4, from 4 to 8. So my first step in the proof of the original theorem for any number n would be not a transformation from n equals k to n equals k plus 1. I really don't know how to do it easily. Instead, I will transform it from n is equal to 2 to the k's power to 2 to the k plus 1's power. So from 2 to 4, from 4 to 8, 2 to the third power, from 8 to 16, 2 to the fourth power, etc. Now, I will prove this as a separate step in the whole theorem, in the whole proof of the whole theorem. Now, so I'll have, instead of having every number, I have 2, 4, 8, 16, etc. Now, how about 3? How about 5, 6, and 7? How about 9, 10, etc.? Well, this will be the second step of the theorem, of the proof. Instead of going forward now, since I have any big number, basically, at my disposal, where the theorem is already proved in the first step, I will go backwards. Apparently, to go backwards is easy. So my second proof will be, second step in the proof will be, if the first step is equal for n is equal to k, then it's equal for n is equal to k minus 1. So, again, it's a very important difference. From the traditional one step proof, from k to k plus 1, I will do it in 2 steps. First, for n is equal to 2 in the k's degree to k plus 1, that's how I will get my big numbers. And in the second step, I will prove that from any number, I can go backwards. And that's how I will cover all the steps, how, for instance, I can get to number 9. Well, very simply, the theorem was proven for 2. Using this, I can therefore say that it's true for 4 and for 8 and for 16. Then, if it's true for 16, using the second step, I will go backwards to 15, 14, 13, etc. to 9. So that's why the combination of these two steps basically completes the proof for any number. Alright, so we have the plan. We know more or less how to prove this one. I'll just repeat the same thing I did, like 2, 4 and 8. And then this will be a separate step. So step number 1. Right, before whatever I was talking, before was just a preliminary kind of explanation of why I have chosen to do it this way. And you're obviously welcome to try to do it in a classical mode, like from k to k plus 1. Personally, I found it extremely difficult. I just wasn't able to prove it that way. So maybe there is no good proof. So anyway, you're welcome to try if you want to. But in any case, if you would like to follow this particular plan, which I definitely recommend this one, this is a good point in time to place the pause button, try to do it yourself. And obviously, this one you have already basically seen in a couple of examples how to do that. And as far as the next step, I can give you a hint. To transform from k to k minus 1 is relatively easy. If you know that for k it's true, but you have to prove that it's true for k minus 1, you just have to add extra number to these k minus 1. And so you will get the k numbers. And I will not tell you which number to add. That's kind of up to you. But this is just a little hint. In any case, I will do both steps right now without any interruptions. So go ahead, press the pause, think about it and then turn it back upon again and we'll continue with the proof. Okay. Now, let's consider the theorem is true for n is equal to 2 to the k's degree. And we have to prove that it is true for 2 to the k plus 1. Well, very easy. Since now I have x1 plus x2 plus et cetera plus x2 to the k plus 1, I told by relatively big numbers, big letters, so you will see. And we divide by their numbers. And obviously, to prove that this is greater than their geometric average, I will separate pairs of these numbers. Now, if total number of my numbers is 2 to the power of k plus 1, and if I pair them, how many pairs do I have? Well, obviously, I have to divide the total number of numbers by 2, since every two elements will give me a pair. Now, if I divide 2 to the k plus 1 by 2, I guess I will have 2 to the k's degree, 2 to the power of k. So, if I will do it like this, divide it by 2 plus x3 plus x4, 2 to the 2 plus et cetera plus. My two last elements will be x with an index 2 to the power of k plus 1 minus 1 plus x with an index of 2 to the power of k plus 1. That's my last pair. It's a little bit difficult on indexing here, so it's x with an index of 2 to the power of k plus 1, and then minus 1, because this is the previous member. And the last member is x with an index of 2 to the power of k plus 1. So, how many pairs, as I said, to the power of k? So, if I divide it 2 to the power of k, that's what we will have. Now, in this particular case, so far, this is not greater or equal, this is just reticle. So, all I did, I separated into pairs, divided each pair by 2. So, now I have to divide by 2 to the case, and all together will be, this 2 will go down, so 2 times 2 to the case, it will be 2 to the k plus 1, exactly what we have in the beginning. So, that's how we reduce the number of elements from 2 to the power of k plus 1 to the number of elements 2 to the power of k. Now, since we have assumed that our theorem is true for number of elements equal to 2 to the power of k, I will use it here right now, and I can say that this is greater or equal than their geometric average. Now, the geometric average is, the number of elements is 2 to the power of k root, same number as here. So, this is 2 to the power of k. And under this root, I have their product, which is x1 plus x2 divided by 2 times x3 plus x4 divided by 2, etc. And the last number is, again this kind of a difficulty to write, x with an index 2 to the power of k plus 1 minus 1 plus x with an index of 2 to the power of k plus 1 and divided by 2, the whole thing. Okay. So, that's what we have. And obviously, now we put even greater than, greater than even more. Now, since our theorem is proven for n is equal to 2, each arithmetic average of a pair, I will replace with their geometric average, so I will have the same 2 to the power of k root. And here I have a square root of x1 times x2 times again, square root of x3 times x4, etc. And the last root, square root will be product of x with an index of 2 to the power of k plus 1 minus 1 times k with an index of 2 to the power of k plus 1. Okay, that's my last square root, and this is my 2 to the power of k root. Well, and now it's basically obvious. If you multiply all these square roots, you can have only one square root from everything, and if you will have a square root and then root of a degree with a degree of 2 to the power of k, obviously the result will be 2 to the power of k plus 1, root of their product. And the last one is x with an index of 2 to the power of k plus 1, which is exactly what we needed to prove. So, from k, 2 to the power of k, we can always step forward to the next element, the element 2 to the power of k plus 1. So, that's basically the first step of the proof. Since it's true for n is equal to 2, it's true for 4, for 8, for 16, for 32, for 64, etc., etc., we double and double the number of elements, and that's how we can move forward in bigger and bigger steps to get the points where we have already proven our theory. Now, let's talk about the second step, how to move backwards, and from let's say 16, we can go backwards to 15, or from 32, we can move backwards to 31 and then 30 and 29, let's say. And that would complete the whole proof for any number. So, let's assume that the theorem is true for n is equal to k and we have to prove it from k to switch to k minus 1. Okay, so for n is equal to k, we assume it's true and we have to prove it for n is equal to k minus 1. Alright, so let's have k minus 1 elements divided by k minus 1. Alright, here is an interesting thing. It's kind of obvious that if you have an average of certain number of elements, arithmetic average, and then you have one extra number which is exactly equal to their average, the whole combination shouldn't really be different as far as the average is concerned from the one before. So, what I'm saying is that if I have an average of k minus 1, let's call it y. And now let's have an average of k elements which are original k minus 1 plus their average y. And I have an average of k elements now, original k minus 1 plus their arithmetic average, divided by k. I would like to say that it's very obvious that we should really think about this combination as really not being different as far as its average is concerned from the original one. And how to prove it? Well, let's just open up this thing and check if we will go and we will be reduced to the same formula as before. So, what this is is x1 plus x2 plus x3 plus et cetera plus xk minus 1 plus. Now, y is again x1 plus x2 plus et cetera a k minus 1 divided by k minus 1 and divided by k, right? That's what it is. Okay. Now, we will use k minus 1 divided by each of those guys to bring them with the common denominator k minus 1. So, what is this? Well, it's x1 will be k minus 1. Okay. We will put k minus 1 down and this k. Now, x1 will be with k minus 1 plus k minus 1 I will multiply for x2 plus et cetera plus k minus 1 we will multiply by xk minus 1 plus all these guys so plus x1 plus x2 plus et cetera xk minus 1, right? So, x1 was k minus 1 here and 1x1 is here from here. So, all together will be k times x1. Same thing for x2. We have k minus 1 times x2 and plus another x2. So, it's k again, kx2. As you understand, it will be exactly the same thing for everyone up to k minus 1. And we divide by k minus 1, k. And obviously, you reduce it by k and you have a arithmetic average of k minus 1. So, this is a proof that adding arithmetic average of k minus 1 numbers does not change an average of a set. If it was y, it will be exactly the same y at the end. So, adding average to a set of numbers does not change the average of the combined set. This is a very important project and we will definitely use this in this particular case. Since we have just proven that average doesn't average of k minus 1 numbers previous k minus 1 plus their average is exactly the same thing, then we will do the following. I have to prove that this, a arithmetic average of k minus 1 is greater than their geometric average. But since I know that this thing is the same as x1 plus x2 plus xk minus 1 plus their average, average by k, since I have just proved that, now I will work with this. But now this is an average of k elements, right? And I can use an assumption which I have. Because we started from let's assume that it's true for n equals k and then let's just move to k minus 1. I can actually represent as an average of k numbers original k minus 1 plus their average. And this we can use our assumption for n is equal to k and say that this is greater than k root, case root of x1, x2, xk minus 1 and what is this? This is k minus 1 root. Well, let me do it in two steps. x1 plus x2, et cetera, xk minus 1 divided by k. Okay. This is the product of from 1 to k minus 1 and the kth component is, sorry, k minus 1 is their arithmetic average. It would be a little easier for me. If I will use, instead of roots, I will use the powers. You know that ath root of b is basically b to the power of 1 over 8. It's just easier to work with. So I will have it as x1 times x2 times x1 times x2 times, et cetera, times 1 plus x2 plus xk minus 1 to k minus 1 and the whole thing is in the power of 1 over k. Okay. Now, you see, here is a very important thing. This component is exactly equal to this component. So what I will do, I will just transfer this piece into this link, basically, the regular transformation. Because this is equal to, since this is the power and this is the product, I can separate the components. So it will be x1, x2, et cetera, xk minus 1 to the power 1 over k times x1 plus x2 plus, et cetera, plus xk minus 1, to the power 1 over k. So, this is inequality which I have. This piece is greater or equal than this piece. Let me just write it again so it would be a little easier to understand what I have done here. Greater or equal than this. I will just rewrite x1 times x2 minus xk minus 1 1 over k x1 plus x2 plus x k minus 1 divided by k minus 1 to the power 1 over k. So that's what I can prove immediately. So, again, what I'm saying is that this member you see, looks like this member. So, what can we do about this? Well, first of all, we will raise those left and right to the power of k. What will be is x1 plus x2 plus, et cetera, plus xk minus 1 that's plus to k minus 1 to the power of k. Greater or equal. Now, if I will raise to the power of k these guys, the powers are multiplying. So that will be x1, x2, xk minus 1, 1 over k to the power of k will be 1. So it will be just this. And same thing here. 1 over k to the power of k will be the same thing. And now I can obviously reduce by this number. I will put this and k minus 1 here, right? That's the reduction. And this is basically the end of the proof because now we will just have the k minus 1's root from both sides of the equation of the equation and we will get what we want. Greater or equal k minus 1's root of their product. And that's what we have to prove. So, again, what we have done here, how to switch from assumption that it's true for n equals k to the proof that that would be true for n is equal to k minus 1. To k minus 1, original elements, I added the k's element, which is their arithmetic average. And since it doesn't change the whole arithmetic average, now we have k elements, original k minus 1 plus their arithmetic average. And I have applied the formula of the inequality for n equals k, which we have assumed is true. And then it's just pure arithmetic reduction power here, power there. That's just very simple. So that completes the proof of the whole theorem. Again, it's attributed to French mathematician Cauchy and it was actually very interesting theorem about this comparison between arithmetic and geometric averages of two sets of numbers. Now, there are many averages and by the way, in finance, if you're interested, all these averages are used quite substantially. Let me just talk a little bit about something which is called moving average. Now, moving average average is a concept very often used in the finance when you are analyzing stock market quotes or something like that. Here it is. If you have, let's say the value of Donald Johnson industrial for at the end of every day for a few years history. So you have date number one industrial average had the value of x1. Day number two it has the value of x2, etc. Day, I don't know, 3,000, that's it. And industrial had x3,000. What is moving average? Moving average is the following. If this is today, date number 3,000. What we can say is okay, what's the average of Donald Johnson industrial during the last 100 days? Well, you should really add together x3,000 plus 2,999, 2,998, etc. So you go backwards by whatever number of days I decided to have my average, let's say it's by 100 days. So this would be 2,9001. So x2,901. So we are averaging these numbers. And this constitute my average, let's call it y3,000. This is an average of the last 100 days as of today. Okay, great. What was it yesterday? Well yesterday I had to add from x2,900 to x2,999. Yesterday it was 2,999 and 100 days before it was this. So this constitute y2,999 of 100 days. So for every day for today, which is 3,000, or yesterday, which is 2,999, or the day before, which was 2,999, 8, we can have the average of the previous 100 days. Now these numbers constitute another sequence. It's the sequence of averages. So if x were sequence of values, y is sequence of their moving averages, where average is calculated always for the last 100 days. So that's basically something which again financial specialists are using every day to analyze how certain indices or certain stocks or certain markets, whatever, currencies for instance, exchange rates, how they behave. What do they accomplish? Well obviously every individual day can represent certain hectic movement, up, down, etc. But if you're averaging throughout let's say 100 days or something like that, you are smoothing the curve. So if originally a daily graph would be something like this, your moving average graph would be much more smoother something like this. And it represents bigger ways, so to speak. If these are daily ways then averaging by the last whatever, 5, 10, or 100 days you are smoothing out all these daily fluctuations and only the bigger waves are actually staying in this particular graph. So just one of the examples of how averages are used in finance obviously they're used everywhere for statistical purposes, etc. But again what's interesting is that even in finance if you are saying which average, arithmetic average, or geometric average, and there are other averages like quadratic average for instance which we might actually have another problem about. So all these are important and as you see if you use arithmetic average it's always greater or equal than geometric average. Okay that completes for this problem for today. There is one interesting detail that I would like to mention for those who are interested in these arithmetic averages or geometric averages. This inequality which we have, that arithmetic average is always greater or equal than geometric average, I said it's always greater or equal. Now the question is when is it greater or when is it equal? Well, interesting point is that if you take only two now obviously the equal sign will be if they are equal to itself. If both are the same x plus x divided by 2 will be x, x times x will be x square and the root will be again x. So if x1 is equal to x2 this is equality and the interesting thing is that for any number of elements, equality will only be in case when they are all equal to each other. And I'm not going to prove that try to prove it yourself that's an interesting problem. Thank you very much, enjoy it.