 Hi, I'm Zor. Welcome to a new Zor education. Today we continue solving different math problems, which are part of the course called math plus and problems. So these problems are basically not exactly the standard math problems, which are supposed to verify whether you are correctly know the theory, so to speak. So for instance you have a theorem and then you have some problems which basically illustrate that particular theorem. Now the problems which I present in this course are a little bit of a different kind. They are supposed to kind of force you to think about this problem. It's not direct implementation of theory. It's not directly related to any particular theorem as an illustration. No, you just have to find your way somehow to basically solve the problems basically. So some of them might be a little bit more difficult. Some of them are easier, but maybe present some unusual result which you kind of don't expect. For instance today I'm talking about one particular problem which I myself kind of did not expect that the result of this problem is really true. It's kind of strange looking, but anyway it is true and I'm going to prove it. Okay, so this course math plus and problems, well it's based on the previous prerequisite course which is called math proteins, presented on the same website Unisor.com. Now all the courses are totally free, there are no advertisements. If you are studying yourself you don't even have to sign in. Sign in is for supervised studying, which is also totally free. Also on the same website you have physics for teens and for some interested people even the course called relativity for all, where I present the theory of relativity in relatively not very theoretical way. Well it is actually, but anyway it may be a little bit easier than some textbook. Now what's interesting about all courses on Unisor.com, they are presented in two ways, as a lecture which you can watch and every lecture has a textual part description basically, which is like a chapter of a textbook. So I do suggest you always to watch the lecture and read the notes for this particular lecture, which are really side by side on the website. Also what is very important, problem solving is useful only if you do it yourself basically. Now here I present the problem and the solution, which means that you probably would be much better if you listen to the problem itself and then pause the video, do not really continue watching until you think about the problem yourself. Whether you come or you didn't with a solution, then you can watch whatever I'm suggesting. Maybe it's a different solution than whatever you have. Maybe you don't have a solution then it would be the only one, but in any case it's very important to think. Because the whole purpose, the whole idea behind this particular course is to develop in the students this ability to think, to analytically approach the problem. After this introduction, so let me just go straight to the problems. The first problem and as I said, I did not expect the result of this problem to be as it is presented. Okay, so here is a problem. Let's just continue with a particular example. Let's consider function which is depends on natural number n, which is actually, well, it's 1 over 1 is 1, but it's more convenient if I will put it like this. So as you see, I'm changing the sign every time. So this is 1 minus 1 half plus 1 third minus 1 fourth, et cetera. And I am ending with 2 less 1. So let me just give you an example. For n is equal to 1. So this, the end should be, oh, I'm sorry, this is minus. Signs are changing every time. So the last one will be 1 minus 1 half, so it's 1 minus 1 half. 1 minus 1 half, which is actually 1 half. For n is equal to 2, oh, okay, I did it differently. n is equal to 1 column. It will be 1, 1, okay, right. For n is equal to 2 column, 1, 1 minus 1, 2 plus 1 third minus 1 fourth, whatever it is, et cetera. With n is equal to 3 would be 1, 1 minus 1, 2 plus 1, 3 minus 1, 4 plus 1, 5 minus 1, 6. Okay, fine. That's one function. Another function is, you just have half of these numbers, the right half of these numbers, but only with a plus. g of n is equal to 1, n plus 1. So there are two n numbers. The first n, I'm cutting off. And the right side of this half starting from n plus 1 will go. And that will be with all pluses, okay. So for n is equal to 1, it will be 1 half. For n is equal to 2, it will be starting from 1 third. So it's 1 third plus 1 fourth. And for n is equal to 3, it will be 1 fourth plus 1 fifth plus 1 sixth. So what's interesting is, f of n is equal to g of n. So some of these fractions is equal to some of these fractions. And that's what we have to prove. And that's where it's good if you will pause the video and think about how to prove it. Okay, now here is my solution. Well, first, by the way, I did it for n is equal to 1, 2 and 3. And basically calculated, I added these fractions, brought them to common denominator, I mean the normal thing. And it did actually confirm that this is equal. Now, obviously it was not approved because I just checked it for 1, 2 and 3. Now we have to basically prove it for any integer, positive integer n. Okay. Usually, if you have something like this to prove for any natural number n, the good way is to approach it using mathematical induction. So what is mathematical induction? You check it for some initial value of n. For example, for n is equal to 1. And by the way, for n is equal to 1, we have already checked it because this will be 1, 1 minus 1, 2, which is 1, 2. And this for n is equal to 1 will be 1, 2 only, which is equal. So we checked it for some initial value of n. Then we assume that it's true for some number n, just abstract number n, assume. And then if using this, we will be able to prove that the next one also would be true. That would be a sufficient proof. Why? Because if I have this, from this follows this, and I checked it from n is equal to 1 in this particular case, then I can say that, okay. If this formula is true for n, from this follows that it's true for n plus 1, now I checked it already for n is equal to 1. So for n is equal to 1, it's true. And therefore, it would be for n plus 1, which is 2. Okay, so now basically using this particular logic, I can say that if it's true for n is equal to 1, then it's true for n is equal to 2. Great, I've proven it for 2. Now, but since I have already proven it for 2, I can say then, using the same logic, it's true for 3, etc. So basically for any n, I can sequentially make this type of logical connection and go to million or billion or whatever number of n, which basically means that we have proven it for any n. Okay, so how can we prove that from this follows this? So for n is equal to 1, we have already gotten, right? So let's just try to prove this particular logic. Now, what is the difference between fn and f of n and f of n plus 1? Well, I'm adding two more numbers, f of n plus 1 is equal to f of n, and then two more numbers, which is plus 1 over 2n plus 1 minus 1 over 2n plus 2. So if I will put n plus 1 into this, I have to finish with 2n plus 2, right? So the next number will be plus 2n plus 1, and the next one will be 2n plus 2 with a minus n. Okay, fine. Now, the g of n plus 1 will be g of n plus 1 over 2n plus 1 plus 1 over 2n plus 2, but we have to get rid of this because it's supposed to be only half of these numbers, right? So if I start from 2n plus 2, then, I mean, if I end 2n plus 2, I should cut one of them because it's only half. Half of this would be n plus 1, so n plus 1 I should cut off. So starting from n plus 2. So I should really minus 1 over n plus 1. Okay, now, if I assume that these two are equal, how can I prove that these two are equal? Well, they just compare whatever we have extra here and extra here. If I will prove that 1 over 2n plus 1 minus 1 over 2 plus 2 is equal to this, then basically my theorem is proven. From the equality of this, I have proven the equality of these two, but just by basically comparing the extras, right? So this extra we can call f lowercase f plus 1 and this one we have n g of n plus 1. Okay, so we have to basically prove that f of n plus 1 is equal to g of n plus 1. Now, how can this type of thing will be, how can prove this? Well, very simply, just get the common denominator, multiply them one by one by one with whatever the necessary is in the numerator. Let's do the calculations basically and see if it's actually equal. Now, I've done this calculation in the texture part of this lecture. I don't want to spend, it's trivial kind of thing, so I don't want to spend any time. You just have to trust me or you have to check it yourself. Now, I do recommend you to check it yourself that some of these, this plus this minus this, if you will get to common denominator and have the numerator completely open all the parentheses, et cetera, you will have exactly the same as this one. Because here we have three different multipliers in the denominator, 2n plus 1, 2n plus 2 and n plus 1. Here we have only two, but whenever we will just bring whatever is necessary to numerator and add them up, you will have in numerator n plus 1. So n plus 1 would cancel this and you will have exactly the same as this one. So I'm not going to do it, you do it yourself and you can check your calculations. I mean, it is equal. So if you do not get the equal, which is then something is wrong with your calculations. So yes, they are equal and that's why we can conclude that from this follows this. And again, using this mathematical induction logic, since I have already checked it from n equal n is equal to 1, then it follows that this formula is true for n is equal to 2, from 2 is equal to 3 is following, et cetera, et cetera. So we have proven it for every n. So it's a good illustration of method of mathematical induction, obviously. But again, it's kind of unusual result. I never expected quite frankly myself that this would be equal to this. Well, it is what it is. Next problem. Next problem is also kind of illustration of something which you know about quadratic functions. The illustration is a little bit more vivid. It's kind of physical, so to speak. Let's consider you have two perpendicular roads and you have two cars approaching the intersection. Now, at moment t is equal to 0. One car is on a distance a from intersection and goes towards. Another car is on the distance b and goes towards intersection. And let's just assume that a is not equal to b for a simple reason. We don't want any collisions because they are moving with the same speed b. Okay. So this is the condition. Now, obviously, when they are approaching, this distance is changing. Now, at certain time, it all depends on a and b and the speed, obviously. This distance should have some minimum value. Because they are not at the same time at the intersection. So the question is when exactly this minimum value will be? Well, to do this, we will just have to calculate the distance between them as a function of time. Now, what's the distance, let's call it a of t at the moment t? a of t is equal to a minus speed times time, right? If this is speed, this is the initial distance and every second we are reducing distance by v. Now, the b of time would be b minus vt. Speed is the same. Now, what's the distance between them? Well, if I know these two, this is a of t, this is b of t, this is hypotenuse. So the distance square of t is equal to a square of t plus b square of t. Which is equal to, let me just get another. So this is equal to a minus vt square plus b minus vt square. This is square of a distance. We have to minimize it. Now, well, we have to minimize the distance, not the square of a distance, but they are making actually this minimum value at the same time. So whenever the distance is minimized, the square of a distance is minimized as well, obviously. So when this is minimized, we need to find time, which I call t minimum, when this thing, this d square of t, whenever it takes a minimum value. Well, this is a quadratic function. So it's basically a parabola, right? So we know how to calculate the minimum value. So when it's minimum and what exactly the minimum is itself. Okay. In case you have a parabola, something like px square plus qx plus r, y is equal. Now, when this parabola takes the minimum value, when x is equal to minus q divided by 2p, this is definitely a known fact. If you forgot about this, you can check the course which is prerequisite for this one, mass proteins, whenever the quadratic function is explained. It's very easy to prove. Basically, it's supposed to be in between two roots of this, two intersections with x right in between. So that's what you will have. And the value, minimum value, would be if you substitute instead of x into this formula, you substitute this one and you will have the minimum value. Alright, so we will do the same thing, but we have to really bring it into this type of form. So what is coefficient at t square? Now, coefficient of t square will be v square and another v square, when we will open the parenthesis, right? So it will be 2v square t square. This is my d square of t. Now, what is the coefficient at v? Minus 2avt and minus b, 2bvat. So it would be minus 2a plus bvt. And if we remember, the r would be a square plus b square. From which we conclude that this particular thing, so instead of x, we have t. Instead of q, we have this. This is q. And instead of p, we have this. This is p. So minus q, which is 2a plus bd divided by 2pi. So it's 2p, so it's 2 4v square. So we have this 2, this, and we have a plus b divided by 2v. And this is t minimum. So at this particular moment in time, the distance between them would be minimum. Okay, so we find the time. How about the value? Well, to find the value, we just have to substitute this thing into this and see what will be. And take the square root, obviously. So let's see what happens. So v times t is t minimum. Is a plus b over 2, okay? So if we will substitute with the a minus a plus b over 2 square plus b minus a plus b minus a divided by square. So that's my d square minimum. Equals. So it's no longer a function of time of t. It's just plain constant, which depends only on initial position in d. So what will that be? It's 2a minus a minus b, so it's a minus b square divided by 2 plus 2b minus. So it would be b minus a square divided by 2, which is equal to a minus b square and a minus b and b minus a. When it's square, it's the same thing. So we can say a minus b square divided by 2. So my d minimum equals square root of this, which is a minus b absolute value divided by square root of 2. And that's the result. So this would be the minimum distance. As they are moving, the minimum distance would be this. So by the way, if a is equal to b, they will reach the same, at the same point, intersection. Well, there would be a collision, but that's when we will have distance between them equal to zero. Otherwise, the distance would not be equal, minimum distance would not be equal to zero. Okay, that's it. I suggest you to read the notes for this lecture. And again, if you can, don't look at the answer or solution whatever is in the textual part of this lecture. Do it yourself first. If you can, great. If you cannot, well, then read the solution and basically learn from it. So next time it might be easier for you. That's it for today. Thank you very much and good luck.