[NOISE] The last part of this lecture will be devoted to Euler's Pentagonal Theorem. We will study the generating function for partitions, the generating function which we have written down in the beginning of the lecture in more detail. Let's take P(q) and let's take the inverse to the series. Well, we have a presentation of P(q) as a, as an infinite product of fractions 1 over 1 minus q to the power k. So, the inverse is just the denominator of this end to end product. Namely it is (1-q) (1-q squared) (1-q cubed) etc. So this is the infinite product of the factors of the formula 1-q to the power k, where k goes from one to infinity. Okay, and let us denote this expression by Q (q). Okay, now let us compute the first serial terms of these expression. Let us take a couple of brackets of the form and multiply them. So if we do this, well I propose to you to compute it for instance, five or six or seven terms by hand or if you want you can log them into your computer. You will get something like Q(q) = 1-q-q squared + q to the fifth. There will be no terms with the third and fourth power, + q to the seventh- q to the power of 12th- q to the power of 15th + q to the 22nd + q to the 26th -- etc. And so, we get such a series and this looks very interesting. Well, there are, we can make several observations, namely In the first, strange thing is that all coefficients of the series, which is retained as product of this brackets, are either 1 or negative 1 or 0. Let's write this down. And note that the Entries with the 1 or negative 1 becomes more and more sparse. Okay, one more interesting thing is that we have 1 in the beginning, then we have two negative ones, two ones. Then again, two negative ones and two ones. So we have 1, -1, -1, 1, 1, -1, -1. So 1s and negative 1s comes in pairs. Okay, what else can be said about this series? Another observation is that in each pair, the difference between exponents is first 1, 2 minus 1. Then 2, 7 minus 5, then are three, 15 minus 12, then four, 26 minus 22 And so on. This little contain the dots, so the differences of exponents. Are equal to, 1, 2, 3, etc. Okay, and one more question is, what are the exponents occurring in each pair? What are these numbers 1, 5, 12, 22, etc. Euler knew this sequence very well. These are the so-called pentagonal numbers Okay, probably the sequence was already known to ancient Greeks. They considered triangle numbers, squares etc. So, what are the sequences? Suppose you have marbles and you arrange these marbles in forms of some geometric shapes. You can arrange them in forms of triangles both sides, one, two, we have a triangle formed by three marbles of length, three. On this case, we have six marbles. A triangle with a side of four. Then, the fourth triangle number is ten, and solve. Or alternatively, you can form these marbles into squares. Obtaining one, four, nine, 16 and so on. 1, 4, 9, 16, so these are squares of natural numbers, and that's why they're called squares. Or otherwise, you can form them into pentagons. So you can take a pentagon with, or create one formed only by one marble, or you can take a pentagon with side of length one. It will be formed by five marbles. Then, we can take a bigger one, with side two. It will look like this. It will be formed by one, two, three, four, five, six, seven, eight, nine, ten, 11, 12 marbles. You can take a bigger one. With A side of length three and in this case, we'll have 22 marbles. And so on. You can easily show that the nth pentagonal number is nothing but n(3n-1) divided by 2. And these are exactly the exponents occurring in this series, 1, 5, 12, 22, etc. And what are all the second numbers in each pair? These are just n(3n+1) over 2 + n, as we have observed here. And this is equal to n(3n+1) divided by 2. Okay, so, this allows us to conjecture. What does this series look like? So, this theory was conjectured by Euler and proved by him some ten year later. It took him a very long time to complete the proof. So the theorem is, This theorem is called Euler's pentagonal theorem. This theorem says that, This function, Q(q) = 1 + the following sum. It should take -1 to the power n multiplied by q to the power n(3n-1) divided by 2 + q to the power n(3n+1) divided by 2. So this is the nth integral number, and this is the nth integral number plus n. Now, we are going to prove this theorem. [SOUND] [MUSIC]