 Hi, I'm Zor. Welcome to a new Zor education. We have spent some time explaining Euler's formula for trigonometric representation of the complex exponent. Just by itself, this formula looks quite amazing. Well, at least for me. At the same time, let's not forget that this represents a polar representation of complex numbers. And the polar representation is very convenient in a way that you can have something like this. So raising a complex number into a certain power actually is equivalent to rotation. You are increasing the face, the argument of the complex number in polar representation by this factor which is actually equal to the power you are raising the complex number to. So we were talking about this formula. Basically, we have proved it for n equals to 2. You just open parentheses and everything seems to be very obvious. For any n, it can be proven by induction, also basically using the trigonometric formula for the sum of two angles. Now, what's interesting is that this combination of trigonometry and exponent, which represents basically the complex numbers, allows us to solve certain problems in certain unusual way, I would say. So the problem which I would like to present today is the following. You have to summarize this series and this one. Same thing for sine. That's true. Now, there are certain non-related to exponential representation methods to solve this problem. And I will address them later on because they actually require certain lucky guess, if you wish. And unless you know this particular trick, you will not be able to come up with it. However, if you know the complex numbers and their representation in this form using the Euler's formula, you can actually use this type of relationship to calculate these two series. And here is how. Let's just consider that you would like to represent the following number. Pi Pn plus iRn. What is this? Well, that's cosine x plus i sine x plus cosine 2x plus i sine 2x plus etc. Plus the last member cosine nx plus i sine nx, right? So this is multiplied by i added to this one and I will group together the corresponding members, which have the same argument for trigonometric functions. So x was x, 2x was 2x, etc. Now, look at this formula. It obviously means that instead of doing this, you can write it in a way, this is e to the power of ix, this is e to the power of i, well let's call it 2ix plus etc. The last member is e to the power n ix. Now, why is it better? For very obvious reason, this is a geometric progression, right? Since e to the power of k ix is actually e to the power of ix to the k's degree, remember this is the property of any exponential function. So I can write it in a way that pn plus irn equals e to the power of ix plus e to the power of ix square plus etc. plus e to the power of ix to the n's degree. Now, it's obvious that this is a geometric progression, right? Now, we know how to sum the geometric progression, right? So we will sum it up, we will replace e to the power of ix in that final formula with cos ix plus ix, and basically bring it into some technically acceptable form. So that's basically the whole idea. And to tell the truth, I would love to finish this on this particular note. I don't want to go through the calculations, summing the geometric progression. I mean, we all know how to do it. What's the most important in this case? The most important is to organize two series which we have into this particular way to use the property of exponents of the complex argument. Thereby reducing it to something no. It's basically like, here is a comparison, a philosophical comparison. If you have two points A and B, and you know the road from A to B and it's maybe long, it's tedious, et cetera, but you know the road, there is basically no doubts that you can actually go from A to B. Now, but if you don't know the road from A to B, even if they are very close together but you don't know the road, well, that's actually very difficult to get from A to B, right? So, you have certain known, technically, already went through procedure, like summarizing the geometric progression, converting complex exponent in trigonometric format, et cetera. So, you know all this stuff. What is the most important idea to use this thing? This is an inspiration which really should come to you because everything else is just technicality. Now, on my website, Unizord.com, where this lecture is represented, it also has all the required calculations for summing the geometric progression, converting back to trigonometric formula, and finally, after certain manipulations, I've got the final result. And here is the final result. Let me just write it for you. I don't want to go through this right now. You can go to my website and find out what it is. The final result for this is Pn, which is the sum of cosines, is equal to sin 2n plus 1 over 2x minus sin x over 2 divided by 2 sin x over 2. And Rn is equal to, well, let's wait for Rn. We'll talk about this later on. Now, once I have basically discovered this formula using whatever the techniques we have, summing the geometric progression, converting in the trigonometric form, et cetera, now what seems to me is interesting is that just looking at this formula, I can come up with certain, well, a trick, if you wish. Let's just think about it. So if this sum is equal to this, then if I will multiply this sum by 2 sin x over 2, I might actually get, well, some simplification using some trigonometry to come up with the difference, and that would prove from a completely different aspect of this formula. So let's try to do it. So what I will do is I will multiply this by the sin of x, and I will use the formula of sin A times cos B. Well, let's use Greek letters since we are talking about angles, sin of alpha cos sin beta. Now, what is it equal? Let's just think about it. Sin of alpha plus beta is sin cos sin plus cos sin, right? So we would like to reduce, so let me just again, sin alpha plus beta equals sin alpha cos sin beta plus cos sin alpha sin beta, right? And we want to get rid of this. If I have minus here, I will have minus here, right? So if I will summarize them plus sin of alpha minus beta, I will get plus and minus, so this would reduce. And the only thing is I have double this, so I have to put this, right? So that's the formula. I will use this formula, and I will multiply this sum Pn by 2 sin of x over 2. So alpha is x over 2 and beta is x, 2x, etc. and x. Now, what happens in this case? Let's just think about it. Let's temporarily wipe out this. We don't really need this. So Pn times 2 sin x over 2 equals 2. So 2 sin over 2 times cos x. So this is alpha, this is beta. This is alpha, this is beta. So it's sin of their sum, which is x over 2 plus x, which is 3x over 2, plus sin of their difference. Now the difference is x over 2 minus x, so it's minus x over 2, right? And since sin is an odd function, I have minus here. Okay, next. Beta is equal to 2x. Now, it would be plus sin of their sum, which is x over 2 plus 2x. It's 5x over 2. Minus difference x over 2 minus 2x is minus 3x over 2. Now, I think you already see that this is plus, this is minus. Then this plus next would be minus. So let me just go to the very last one. The last one would be x over 2 plus nx. That would be sin of 2n plus 1 over 2x. And minus would be 2n minus 1 with a minus sign. Minus sin of 2n minus 1 over 2x. And what happens? Every second one is reduced with a previous. This reduced with this, this reduced with something else, and this reduced. And what's remaining is from the first member I have this, and from the last member I have this. So it's sin of 2n plus 1 over 2x minus sin x. Which is exactly this. So you see, I probably would not be able to guess that the way to solve this problem might be just to multiply this by sin of x over 2. Unless I'm told that this is the way to do it. Yes, if I will do this, I will come up with the same solution. I multiply pn by this. And I got the difference between these, which is this. So that's the formula. And very similarly I can do exactly the same with sum of sines. Let me do it now. Again, my final formula, which I have received, basically brute force method. Using the brute force method of summarizing geometric progression. Converting into complex format, etc, etc. So my formula for rn, which is the sum of sines is cos sin x over 2 minus cos sin 2n plus 1 over 2x. 2x divided by 2 sin of x over 2. So again, if I kind of know this, I can come up with an easier way to basically do this formula. How? Well, I will multiply rn, which is sin x plus sin 2x plus etc plus sin of nx. I can multiply it by sin of x over 2, 2 sin of x over 2. And what do I get? Now let's think about it. It's sin times sin. How can I get sin times sin? Here it is. Cos sin of alpha plus beta is equal to cos sin alpha cos sin beta minus sin alpha sin beta, right? And cos sin of alpha minus beta is equal to cos sin alpha cos sin beta plus sin alpha sin beta. So if I will subtract from this, I will subtract this. My cos will go out and I will have 2 sin times sin, right? So 2 sin times sin is a cos of their difference plus, no, minus cos of their sum, right? Alright, so 2 sin x times sin x over 2 cos of their difference is cos sin of x over 2, right? x minus x over 2 minus cos of their sum, which is 3x over 2. Next, 2x, their difference is 3x over 2 minus their sum, which is 5x cos sin 5x over 2, etc. The very last one, the nx, the difference is cos sin 2n minus 1 over 2x minus their sum 2n plus 1 over x. And what do we see? These are all reduced. And what's left? I have this one and this one, which is this minus this. Same basically approach. Once I know this, I can come up with this trick. If I didn't know that there is a trick of such, such a thing, then probably I'm at loss. Well, so what helps in this particular case? Well, first of all, the more problems you solve, the more tricks you have in your repertoire, of course. So this is one of the things which you might actually remember. And next time you would think about what if I will have a sum of all odd, for instance, or all even multipliers, factors. Maybe there is something which I can use, which looks like this trick. Maybe I can come up with some multiplier, which would convert this into plus minus plus minus, if everything will be reduced. But if not, then you can actually use the heavy artillery, because you always know that these type of problems are very easily converted into sum of geometric progression if you will use the Euler's formula. This is my kind of purpose today to, on one hand, to show that a completely different theory seemingly unrelated to the problem at hand to summarize these series. What I mean actually is a representation of the complex exponent in the trigonometric form. The Euler's formula can help you in this particular case, if you don't know some special tricks, etc. So the more theoretical background you have, the wider are your options to resolve the problems. And sometimes, as I'm saying, if you cannot really come up with a trick like this one, well use the heavy artillery, which is a theory. A theory of representation of complex numbers in trigonometric form in this particular case. It's always helpful to go outside of this box, wherever you are, like trigonometric box, to go outside of the box and think about this from a different perspective. And that's how many problems actually have been solved. You have to go completely, you have to forget about all the ways how the problem was addressed before and go completely outside of the box. And in this case, the Euler's formula is actually completely outside of the trigonometric box, which you have here. Well, that's it. Thank you very much and good luck.