 Hi, I'm Zor. Welcome to a new Zor education. I would like to talk about certain problems related to sequences. Well, problems is actually an exaggeration, I would say. These are very simple illustrative examples. So, let me just start from the first one, which is actually quite simple. And as soon as I will find this in my kindle, okay, here it is. Okay. Now, example number one. It's an illustrative example of the fact that you don't really need to remember the formula for elements of geometric or arithmetic progression. Because the common logic actually would be sufficient to come up with this formula any time you really want to. Because it's a very simple logical explanation similar to whatever I was actually talking about when I was explaining what arithmetic or geometric progression is. So, here's an example. The start was minus one. And then you increment on every step your number, your sequence element by one half. So, you add one half to minus one, you will get minus one half. You add another one half, you get zero. Then you get one half. Then you get one. Then you get one and one half or three halves, etc. So, the problem is very simple. Try to come up using very common logic with the formula which describes this particular sequence. And then I will write the formula, well, I can actually write it here, a plus m minus one d. I do remember this is the nth member of the arithmetic progression which starts with a and has a difference d. So, without this formula, let's just concentrate just on the numbers and come up with a formula based on these numbers. And then we will compare it. So, again, my purpose is to convince you that you don't really have to remember this formula. You can derive it basically based on whatever the common logic is. Well, let's think about it. So, first you start with minus one. And then I told you that the next one will be minus one plus one half, which results in minus one half. And then I will begin at one half. And that will be one and minus one zero. And then again, etc. So, look at it this way. This is element number one. This is element number two. This is element number three. This is element number four. Now, as you see, the number of times I'm adding one half is zero for element number one, once for element number two, twice for element number three, three times for element number four. So, the kind of a rule which you can imagine is that the number of times I'm adding one half is one smaller than the element number. So, if the first element number is minus one, then all I have to do is add n minus one times one half. And that would be the formula for nth element of this particular sequence. And basically, as you see, this is exactly the same. A is minus one. In this case, d is one half, because every time you step forward, we increment by one half. And obviously, the n minus one is exactly the same multiplier. So, again, don't try to remember the formula for nth element of the arithmetic progression. Divide it every time you really need it. Well, that's it. It was just an illustrative example of the fact that you don't have to remember certain formulas as long as you understand the logic this formula is derived. And the logic is basically you add the difference in what happened in this case, n minus one times for every nth element. All right. So, as I said, these are not really problems. These are illustrative examples. These are all very simple things. Figure out the formula that describes the sequence of integers starting from one and increasing in absolute value by one on each step simultaneously changing sign. All right. This is an example of a progression of a sequence which is not an arithmetic progression. So, we start with one, and then we increment absolute value by one, but change the sign at the same time. So, next one will be instead of two, minus two. Next one would be three. Next one will be minus four. Next one will be five. Next one will be minus six, et cetera. So, the signs are changing and absolute value is increasing. Now, the purpose of this exercise is, number one, to illustrate that there are sequences which are not arithmetic or geometric progressions. And by the way, during my lecture about introduction to the sequences, I was talking about Fibonacci sequence, which is also not an arithmetic or geometric progression. It's relatively simple to formulate. Like, every element is the sound of two previous ones. So, in this case, again, it's easy to explain what it is. My purpose is to come up with a formula using certain, you know, common logic. Well, let's start with common logic. Now, the common logic with this particular, in this particular case would be, let me start with this function. What does it do? Well, for number one, it gives me one. For number two, it gives me two. For number three, three, et cetera, et cetera. So, it's almost this one, but without changing the sign. So, I have to somehow modify this particular function to change the sign on every end. Well, what changes the sign for every end? Now, I'm just thinking, okay, what's my repertoire? What's changing the sign for every end? Well, the first thing which comes to my mind, at least, maybe there are some other cases, is if I will multiply minus one by itself, then once it will be minus one, second time it will be one, minus one squared, right? Then the third one, I will multiply one in this case by minus one. It will be minus one. Then if I will multiply minus one by minus one, I will get one again. So, basically, this particular sequence, which is f of m, is equal to minus one in m's degree, right? For n equals one, it's multiplied once, basically, which is minus one. For n equals two, I multiply minus one by itself two times, using minus one squared, which is one. For n equals three, I will get minus one cubed, which is, again, minus one times minus one times minus one, which is minus one. For n equals four, it will be minus one to the fourth degree, which is one, again. So, this is another function, and this function gives me changing of the sign, if you wish. Now, all I have to do is combine these two functions to get my signals. So, what I do is I write this function. So, first, I take care of the sign, and then I take care of the value. So, this is the formula which describes this. Well, sorry, actually, I made a small mistake. In this case, I start with n equals one, I start with minus one, and then I'm changing the sign. Actually, I have to start with plus one. So, I have to subtract one here. So, I will change the sign again. So, it will be, for n equals one, I will have minus one to the zero's degree, which is one. So, it's one. For n equals two, I will have the absolute value two, and the minus one will be to the first degree, which is minus one. So, the result will be minus two. And then again, all the signs will be interchangeable. So, basically, that's how I gradually, using nothing but common sense, come up with this particular formula. I start with n, which basically gives me values one, two, three, four, five. Then I talk about what is my formula for changing sign, and then slightly adjust it to have the sign exactly what I want. Let's introduce a function for any real argument, floor, okay? Floor of x. First of all, I would like to explain the function here. Y equals to floor of x. Now, what is floor of x? Floor of x is, for any x, is the largest integer number, which does not exceed x. So, for, let's say, one-half is the largest integer, which does not exceed one-half, which is zero, right? One already exceeds, so it's zero. If I will have floor of 35.7, the largest integer which does not exceed this is 35. Last example, floor of minus one-and-a-half. What's the largest integer which does not exceed? Well, minus one is greater than minus one-and-a-half, right? Minus one, minus two, and this is minus one-and-a-half. So, what's the largest number, which is smaller, means to the left, actually, from minus one to half, with minus two? As long as this is understood, then I can formulate next problem. And the next problem is figure out the formula by using this function to express sequence of repeating natural numbers, okay? The sequence is one, one, two, two, three, three, et cetera. So, every number is repeated once. And why did I talk about the function floor before? It's because I can actually write the formula for this particular sequence using this function. And let me start from something simple. Let's put f of n equals floor of n divided by two. Let's examine this. Let's start with this function. Now, what happens? For n equals one, it's one-and-a-half. The floor is zero. For n equals one, for n equals two, it's two over two, it's one. So, what's the floor of one? What's the largest integer which does not exceed one? Well, this is one, obviously. Next is n equals three. Three is one-and-a-half, right? If you divide by two. The largest integer which does not exceed one-and-a-half is one. Okay? Sounds promising, right? One, one. For four, n equals four, I have four divided by two and the floor is two. For five, it's five over two, which is two-and-a-half, and the floor is again two. And as you see, I will have this sequence. It's almost the same as I need. Now, how can I convert this into this? Well, very simple. For n equals one, instead of zero, I have to have the next one, which means my real function should be four of n plus one over two. In this case, if n equals one, it will be one plus one is two divided by two, it will be one. For n equals two, it will be two plus one, which is three divided by two, it's one-and-a-half, and the floor is still one. And then you can just make sure that everything will be exactly like in this particular case. So this is the function. How did I come up with this? Well, the bright idea was to use the floor, and then all I was just doing was kind of trying to manipulate with this function, which is almost what I want to get exactly what I want by just changing from n to n plus one. It would be probably quite illustrative if I will draw a graph of the function. What would be the graph? Let's just think about the three, four minus one, minus two, minus three, et cetera. So let's start from zero, let's say. The floor of zero is zero, that's correct? Now, Sx is increasing from zero to one. The floor, which is the largest integer not exceeding X, and X is somewhere here, and the largest integer which does not exceed is still zero, which means my graph would coincide here, except at the point one, when X is equal to one, Y is equal to one. So my graph jumps to this point, which can be actually shown on the graph. I will put a little arrow here, which means the graph goes to the point one but does not include the point one in this particular case and includes from this side. So starting from X equals one, my largest integer which does not exceed X during changing of the X from one to two would be always one. Again, not including the last point one. Starting from two, my graph will be on the level two, up to three. And from three, it will be three up to four. Now, what's on this side? Well, let's think about it. If X is here between minus one and zero, the largest integer which does not exceed the X would be the one which is on the left of it, which is minus one. So for all axes in this particular interval, it will be minus one. Well, except zero, it will be zero. So again, I put a little arrow here at the graph. And here I will have exactly the same thing. From minus two to minus one, it will be minus two, including this point but not including minus one, because minus one equals minus one. And from minus three to minus two, it will be minus three. So the graph looks like a staircase. This is the graph of the function floor of X. Again, just for illustrative purposes, there's nothing to do with sequences, but we were using this function to come up with a formula describing repeating integer numbers, repeating natural numbers as a sequence. Okay, next. Figure out formula that describes a sequence of all integer numbers in the order of their absolute value with changing signs. So all integer numbers in the order of absolute value, what does it mean? Well, what's the minimum absolute value of any integer number? Zero. So zero is the first integer number with the smallest absolute value. What's the next number which has a greater absolute value? One. Now, don't forget that minus one has exactly the same absolute value. So if I'm trying to put my integer numbers in the sequence of absolute value, after one, I should have minus one. Then would be two, which has a greater absolute value than one, which is two, and obviously I have to do minus two as well. So in a way, it's a combination of something which I did before. Before I was having an example of the sequence one, one, two, two, three, three, etc. That was one example. Another example was when I was changing the sign, one minus two, three minus four, etc. So this is basically a combination of these two. So I have to really think about combining the previous two approaches to this one. So number one is how can I get repeating numbers? One, one, two, two, etc. Well, if you remember, this is floor of n over two. So let's start with this one. Now, what it will give me? For n equals one, it's zero. For n equals two, it's one. For n equals three, it's one again. And then two, and two again, etc. That's how I will get it, right? But that's not what I want. I want changing signs. Now, as you remember, changing signs can be accomplished by multiplying minus one to some degree, like nth degree, let's say. So, and in case I had the wrong sign, I just had to use n plus one or n minus one. So let's just check this particular formula that I will get. For n equals one, I will have, one half is zero. Doesn't matter what this is, it will be zero. For n equals two, floor of two over two, which is one, is one, times minus one to the second degree, which is one, so I will have one. Now, for n equals three, floor of three seconds, three halves would be, it's one and a half, which means one. But minus one to the power of three would be minus one. So I will get this. Next, four. Four over two is two. Floor of two is two. And minus one to the fourth degree is one, so I will have two. For five, obviously I will have minus two. So this is exactly what I need. In the case I basically combined my two previous problems, the problems which I have already solved, one gives me repeating integer numbers and the second one gives me interchanging signs. So I combine them together into one formula and basically can check how it works and it looks like it works. But what's interesting is that the more problems you solve, the more problems you will be able to solve because, like in this particular case, I solved two problems before and it allowed me basically without much efforts to come up with a solution for the third problem. That's the purpose of solving problems. If you will solve enough problems, you will have a whole repertoire of tools which you can use to solve something new. Because there is no really new thing under the sun. Everything, whatever it was before will happen. So basically it's all kind of combination in different order of something which you already know if you know enough. Obviously if you don't know enough, you will have plenty of new information. But the more you know, the less it will be for you to move forward to use the tools which you have already learned. Every once in a while I'm going into this philosophical mood to explain why we are solving problems. Alright. Next. Imagine a segment of a length L. Okay. Imagine a segment of a length L. What you do, you cut one third from this segment and you get two segments, each one having lengths of L divided by 3. But you have two segments. So the total length which remains is two thirds of L. Now I will repeat exactly the same operation with every segment which is left. Now I have two segments left. So each one I will divide in three pieces and wipe out the middle one. So now I have four segments, but the lengths will be each one, will be one third of the previous. And then I'm repeating and repeating this process again and again. On each step, I will cut the middle third from every segment which is left. My question is, what's the total length after the end step, which is remaining? Now, how can I approach it? Well, again, we have to think about this from the position of segments. What happens on each step? If I cut out the one third of each segment, it means two thirds are left, right? Which means on every step from whatever lengths before I had, remaining lengths would be two thirds of it. So next would be two thirds squared L. Next would be two thirds cubed L. So on every step, I am shortening the total lengths of remaining segments by one third of each one, which means two thirds remains. So my total lengths on each step would be less than the previous and the proportional quotient would be two thirds. So basically, we're talking about geometric progression. So all I'm saying right now is that if I will have two thirds times N, sorry, to the power of N, this is the lengths of old segments left after the end step. Let's check again. After the first step, I have two thirds of L. And indeed, if I cut one third, two thirds are left. On the second step, I will have two thirds squared because I already had two thirds of L. And now I cut one third of each piece, which means two thirds of each piece remains. So it's two thirds of the previous value. The previous value was two thirds of L. So next one would be two thirds squared of L, etc. Then the cube, etc. So every time I'm multiplying my initial lengths by two thirds to get to the next step. And that's basically the formula of geometric progression. I didn't really use the formula. Well, the formula is A, Q to the nth degree where Q is the quotient and A is the first element. And yes, indeed, that's really the first element is L because I started with L. Well, in this case, actually, L is an initial thing. So the first one would be two thirds of N after the first step. And the second one, two thirds, squared, etc. So this is the formula. I came up with this formula using the common logic, not the formula. And that's, by the way, one of the points of this particular exercise. I mean, it's a simple formula. You can obviously remember it, but you don't have to. Common sense is sufficient in this case to come up with the formula for lengths and whatever the remaining segments is after N step. Now, what I wanted actually to show here is that sequences, they can be everywhere. In this case, we have some kind of a, I don't know, segments, it's a geometry, right? So it's a geometrical problem in a way. But we're talking about lengths, so that makes it a numerical problem. And that's where the sequences really come very handy. All right, next. All right, next is a very practical example, which for whatever reason, I don't know, people like to talk about bank operations. All right, so you have a certain sum of money. It's called S. You deposited it to the bank. Bank for simplicity, annually, pays you P% interest on this sum. So the question is, how much money will you have after N years? So again, this is kind of a theoretical problem because in practice, banks is calculating this on a monthly basis, or sometimes even on a daily basis, but forget about this. This is a simple case. Once a year, on December 31st, your account is increasing by P%. So you have deposited on January 1st of a certain year. Question is, after a certain number of years and years, how much money will you have? Well, let's think about it. Let's talk about the first year only. What will be at the end of the year? At the end of the year, you will have, first of all, you will have your old amount, S, plus interest. Now, what is interest? Interest is P% on S. P% is P divided by 100. So that's what you will have at the end of the first year. Now, I can express it differently. I can express it as S times E over 100. That will be at the end of the first year. So let's call it S first. Now, what will be at the end of the second year? S second. It will be S first. So whatever was in the beginning of the second year, which is the result of the first year, right? Plus this amount would be multiplied by P%. So that would be at the end of your second year, which is S1 times 1 plus P over 100, which is, now, what's S1? S1 is S1 plus P over 100. And 1 plus P over 100 again. So it will be 1 plus P over 100 squared, right? So we multiply S by 1 plus P over 100 to get S1, and now we multiply again. So your combined interest, well, your combined amount, combined amount after the second year would be this. Now, let's continue this process. Now, I'm sure you can guess the result. Your S3 would be S2 plus S2, 50%, right? Which is S2 1 plus P over 100, which is, now, S2 is this, right? It's S times 1 over P 100 squared, and again, which is S times 1 plus P over 100. So you see basically the rule here, the amount after year number three would be, this is the search, right? Next will be four, et cetera. So I can write that this is an amount after the year number N. Now, I basically came up with this formula. How can it be really proven? Well, the proof is very simple. You can actually do it by induction. Considering that this is, let's first check. Okay, for N equals 1, it would be obviously what I did already. Let's assume that the formula is correct for N years, and now to check it for N plus first year, we obviously put S N plus first is equal to S N times 1 plus P over 100, and that would be, again, S 1 plus P over 100 to the N plus first week. Because S N is this one. This is basically the proof, which is the same formula for N equals N plus 1. So that's something which is definitely a geometric sequence, as you see, right? This is the formula for geometric sequence. What is the first member for N equals 1? It's S times 1 plus P over 100. That's the first element of this. How much you have after one year, and this is 8, and quotient is equal to 1 plus P over 100. This is the multiplier which is used to come up with the next element of the progression, geometric progression from the previous one. So as you see, banking operations for accumulating certain interests is basically geometric progression, which is a known thing, and it can be calculated very simply. So all you have to do is just do a little arithmetic using this formula to find out how much your total amount will be after a certain number of years. Well, that's it for today. These are, as I was saying, very illustrative examples. They're not real problems. Just to show what's the place of sequences in mathematics and what kind of sequences can occur. That's it. Thank you very much for listening to me, and I do recommend you to register to Unisor.com and make sure you take exams. And one more thing. I mean, obviously I'm repeating this all the time. All the problems which are presented on the website, it's very much recommended for you to, number one, try to solve them yourself before you listen to the lecture, and then again after the lecture is finished. Thank you very much.