I will I'll try to speak loudly and if I'm not speaking loud enough you just let me know so as most of you know there's a book behind the set of lectures a book written by my myself and my two postdoctoral fellows problems in the theory of modular forms and we aspire to cover one chapter a day that's pretty tough so obviously you're not going to be covering every itsy-bitsy detail on such a in such a short course but the purpose of these lectures is to stimulate your interest to rush to the library right after the class and continue your study in a more detailed form that so I will try to whet your appetite - perhaps inspire you to study these things further so hopefully these lectures after the 10 days are over you will you will pick up the book perhaps and study it in more detail you will find that there are more references provided at the end in fact the references that have been provided at the end of each at the end of the chapter also has a short guide to the literature I indicate which books are accessible to the undergraduate and which books perhaps are a little more difficult the topic of modular forms is central to mathematics there's a famous aphorism or joke perhaps you might want to call it attributed to Martin Eichler that said that there are five fundamental operations in mathematics addition Multan multiplication division and modular forms so somehow modular forms are important that's just meant to be a joke but the joke is supposed to also conceal a profound truth first chapter in my book is somewhat unconventional and I did that deliberately because I want you to understand the origins of the subject of modular forms the the origin seems to be in what's called the theory of Q series perhaps beginning with the with the work of Euler is that big enough for everybody to see okay boiler studied the partition function P of n so P of n is the number of ways of writing n/a K so this is unordered it doesn't matter you know I'm not going to count 1 + 2 + 2 + 1 is different I'm gonna count 1 + 2 as just one way so representing 3 I can represent 3 as 1 + 2 but I can also represent it as 2 + 1 but I'm gonna count 1 + 2 + 2 + 1 the same if if I do that oiler realize that if you want to study this function he calculated the first couple of values but found that in order to perhaps find a closed formula for this he decided to package it in a power series so this is a power series will start with 0 and defined P of 0 just to be 1 because there's only one way of writing 0 as sum of non-negative things so these numbers of ways of writing these things with a is non-negative so if you think about it you can you can clump together you can rewrite this as 1 times J 1 + 2 times J 2 + let's say K times JK so in other words I'm grouping together how many times 1 appears and I call that j 1 x number of times 2 appears and so and so forth so in other words I'm actually trying to count the number of ways of writing m in this fashion and if i think about that that can be re-written as a product so so some eat this this this tallies how often I'm using one this is going to tell you how often I use two this is how often I use three and when you look at it in the product the coefficient of Q turns out to be exactly this so this is how Oyler thought of this problem so I think it's an important important idea in mathematics somehow when you're trying to study a sequence of numbers maybe it may it it'll be easier if you can package this sequence in either a power series or some sort of object which you can study and perhaps get a closed formula form so this is the first observation oops I should've put the product sign here this is the first observation that that Euler had of course now you can write this as a you know product because it's just a geometric series let's not worry about convergence so convergence questions will be ignored in this lecture this is a geometric series you all know how to sum that that's 1 minus 1 over 1 minus Q this is 1 minus 1 over Q squared so this is equal to 1 over 1 minus Q 1 over 1 minus Q squared 1 over 1 minus cubed dot dot so we have this nice theorem due to Euler that the partition function has a nice generating series in verse okay so this is the first observation that oil are made and using this he managed to get all sorts of interesting identities one of them is a beautiful recursion formula for the partition function which we'll derive in the course of this discussion but this is the origin of the topic of Q series so you can see now that this product makes and it makes its appearance in the context of the partition function so the way to look at this thing is the left hand side is a nice series of packaging a sequence of interesting numbers and the right hand side is a very neat crisp looking function which does not have P of n in it but I know something now about it right so I'm sure that Euler Jacobi other people of his time period what they did was they applied this method of power series to study many many problems and they were mystified by the fact that these things have remarkable identities so these identities were the surprise for many of so they they crop up in so many different ways but the origin of the topic of modular forms is really combinatorial so that's why I begin this topic of Q series so the first order of business today will be to prove the what's called Jacobi triple product identity it was discovered by Jacobi in 1829 and what this says is that the product and this is valid valid for X dot zero and not Q less than 1 so let's not as I said worry too much about these convergence questions but let's just write this down in this fashion No so you can see that part of what Jacobi has are terms which appear in the problem of the partition function so the term on the right hand side is a product of 1 minus 1 Q the N inverse so the product of 1 minus Q to the N is that is a thing on this side so this is already telling us something this may have some relevance to the study of the partition function ok so this is the triple product identity now it's a it when you see it for the first time you might be surprised by how could somebody come up with something like this that I will not get into you can google it and find out but what I will do is to show you another paradigm in mathematics which is appearing and and is now still in its infancy very much but seems to be a profound new paradigm which perhaps will shed light on other questions as well and this paradigm is what's called the philosophy of Q so Q so the idea is somehow that there seems to be so let me let me say a few words there seems to be a paradigm a new paradigm in mathematics and it can be labeled as the Q philosophy if there's such a word and the idea is really based on an analogy which at the moment nobody seems to really understand the analogy is that a natural number on one hand seems to have can be thought of as the limit of n q goes to one of Q the N minus 1 over Q minus 1 of course you all know that right this is because this is just limit Q goes to 1 a 1 plus Q plus Q squared to the N minus 1 that's just the geometric series which you can sum and you have this so when I plug in Q equals 1 you end up getting em okay so this basic calculus observation is the starting point of a very powerful paradigm in mathematics which the physicists are exploiting the mathematician no number theories are exploiting and commoner tourists are exploiting the li theories earth floating so something new and profound is taking place by this observation the idea is somehow there's a world of Q out there and when I take the limit as Q goes to 1 I'm actually back in the realm of natural numbers which which I so the numbers that we deal with are really some sort of a limit case of a larger world and so the Q world so that's the idea so this I mean you can't call it a theorem of mathematics you can only call it some sort of paradigm and this particular paradigm has been used to concoct and construct new functions which seem to be having remarkable consequences in much of mathematics for example some of you may have heard of the theory of the finite field of one element the theory of the finite field of one element is basically the study of FQ the finite field of Q elements and somehow try to understand the limit as Q goes to one I mean that idea of what is what do we mean by the finite field of one element well of course we don't really mean them one element thing it can't be that it's a limit case of some grandeur thing the finite fields of Q elements which you do know about and that have an important role to play so in this context people have been trying to construct Q analogues so there's a there's a kind of a philosophy or method methodology if you want to call it basically to try to construct Q Animalize of classical functions and how to construct Q analogues of these functions is not always clear but the guiding principle is this observation here so for example we are going to look at first the Q analog of the exponential function the exponential function you all know is e to the X and it's given by an infinite series X to the N over N factorial so the question then is what is the Q n log now the philosophy that says replace n by Q the N minus 1 over Q minus 1 is not a rigid philosophy it's a kind of a guiding principle so when I go to try to construct the Q analog of the exponential function whatever that may be I'm not going to change that ok but I'll change the N factorial n factorial is of course a product of N and minus 1 to 1 right that's that's your N factorial so the Q analog of that should be purely n minus 1 over Q minus 1 Q to the N minus 1 minus 1 over Q minus 1 over up to Q minus 1 over Q minus 1 I'm just going to put that in like that the way it is so the analog here that we would perhaps write down is something like that let me up to Q minus one okay so now there's a Q minus 1 to the power M here that would appear in the denominator but that can go to the top and that'll become x times Q minus 1 to the power N and that's only that's changing variables so probably this is the Q analog of the classical exponential function and I should think of the classical exponential function as some sort of limit case as Q goes to 1 of this higher dimensional incarnation of which this is just a small projection right so that's that's the idea somehow the idea is that there's a bigger world out there probably we haven't tapped and we've been fixated on natural numbers that we have not thought about queue numbers and that's why I put Q Series as the opening of this particular chapter so let's try to study basic properties of the Q exponential function and out of that will pop this out this will come out so let's see how that will work so ignoring any issues about convergence let's just look at EQ of X [Music] and a small amount of reflection so let's consider this difference and I'm not actually going to do all this calculation but it's a small exercise which you can probably do in the tutorial so you write down the series for this and you write down the series for this and you just collect terms and blah blah blah and what happens you end up getting x over Q P sub Q of x over T so I leave this as a small elementary exercise for anyone to do I mean it's not gonna be done in the tutorial but I urge the students to do it and maybe in the tutorial if you have questions you can you can ask Karim Dave and and his and me if I'm floating around okay so you get something like this is a functional equation and what happens is this leads to EQ of X is 1 plus x over Q EQ of x over Q so you can now iterate this one more time so innervating you see that this is equal to product' and going from 1 to infinity 1 plus x over to the N since a limit as n goes to infinity of e sub q of x over Q the n tends to 1c when X is fixed and Q's should be thought of here by the way in this case in this thing for convergence sake this has to be you have a cubic ok so excuse the number bigger than 1 and it's shooting off the affinity here so this is going to 0 so you can see that the set constant term is 1 here therefore when the argument tends to infinity it tends to 0 the limit is 1 hmm so you end up getting sis so you get this nice neat little product so the Q exponential function that we introduced just by the play of analogy which is a very powerful method of research by the way when you see some phenomenon in one branch of mathematics and somehow mimics something else is going on there must be a deeper reason for this and so you try to explore that by building the analogy further so here we can you can call it idle curiosity but actually there's more profound reason for that the the idea of replacing n with purely n minus 1 over Q minus 1 and then moving to the X Q exponential function lo and behold you end up getting this remarkable theorem a Q exponential function equals this product so already some interesting things have have come out n equals 1 to infinity that we have a series we have an infinite series which is equal to an infinite product already something non-trivial has emerged and of course there's a small issue about the Q Q has to be bigger than 1 to get that ok that's a beautiful theorem in its own right and we got it essentially for free now if Q is bigger than 1 1 over Q is less than 1 so I'd like to actually replace Q with 1 over Q why because if I replace Q with 1 over Q this will start looking like 1 plus X Q the N is starting to look like that product that I encountered in the partition function much more natural think so we replaced Q by 1 over Q to get P sub 1 over Q of X to be product and going from 1 to infinity 1 plus X Q again so here now this is valid for mod Q 1 on the other hand if I replace Q by 1 over Q here in the series what happens by the way empty products are to be thought of as one okay so when N equals zero it's it's it's just one X to the N over 1 over Q to the N minus 1 so when you simplify all of this stuff notice that here I've got 1 minus Q the N over Q the N 1 minus Q the N minus 1 over Q the N minus 1 and so on so forth so in the denominator I've got a QD n here Q the N minus 1 here and so on so forth so the power of Q is 1 plus 2 up to n which comes out to n times n plus 1 over 2 right okay so we can simplify that and see that this is equal to and going from 0 to infinity Q the N n plus 1 over 2 X to the N over 1 minus Q that okay so so if you now if you change variables so we can do that a little bit so minor changes of variables basically what I would like to do is just this this should have started with one in terms of a but anyway what what you could do is this can be re-written as 1 plus summation and going from 1 to infinity Q the n choose 2 X to the N over 1 minus Q 1 minus Q to the N so what I'm doing is I just don't want to get stuck at this initial point initial point is always 1 then from 1 onwards I'm just going to re label the thing and you can see that this is essentially an N minus 1 I just changed variables ok so you get something like that so in other words we have the following theorem there now what's interesting is that on one hand it's this is a independent interest I mean there's nothing we don't need to tie it with anything else it's of independent interest in its own right but if we come back to the Jacoby triple product identity we see that it is slowly resembling something of that nature remember on the right hand side of the triple product identity we had Q the N squared times X to the n so in other words that was like a power series generating function for squares tops okay so now we're I think we're in a position to prove the triple product identity so let's do that so theorem for mod Q less than 1 X naught 0 this Pro infinite product is equal to summation and going from minus infinity to infinity Q the N squared next year as I said Euler and company were intrigued by series of this form because of combinatorial questions so supposing I had knowledge of just this summation Q lien squared and I was interested in finding out how many ways can I write a natural number as a sum of two squares then I all I would need to do is take the square of this function and the coefficient of x to the n would be the number of ways of writing and as a sum of two squares if I was interested in learning how many ways can I write a natural number of the sum of K squares all I have to do is take the kate's power of this taking the K power of this is tantamount to taking the K power of this product and that should be easy in relatively speaking so there's some information that's being provided to us by this ok so that's the idea behind this thing so how do we prove this so proof of this fact it oops reason proof of this fact is let's see what place I think I'll need that so rub this off so let's call this equation star so in star replace Q by Q squared and X by QX why do you want to do that well it's for the following reason if I replace Q by Q squared the denominator that's two in the in the exponent get disappears right the denominator the exponent here is n times n minus 1 over 2 so if I put change Q to Q squared the two disappears and I get a nice NN minus 1 but that's N squared minus n to get rid of that minus n I would introduce an X a Q in this fashion to get rid of that thing so that comes up neat and clean so we get to get the following product something like that so I've replaced Q by Q squared X by QX so I'm getting that from star so it's no that's not a problem it's sphere so now the trick is to write this denominator here as a infinite product again so I'll rewrite this now don't get confused with I guess you're finished four years of undergraduate study I saw so don't get confused with these subscripts that I'm going to now put put into the calculations of Q the N squared X to the N product J going from 0 to infinity / product J going from 0 to infinity so what have I done well the this product I'm starting this product at this this numerator from n onwards so this the the exponent here starts with n onwards so what happens is all the terms before n exactly these terms will still survive so you see these are the even exponents okay and what are being cancelled by the numerator anything from 2 n plus 2 onwards is being cancelled everybody follow this okay so you have an infinite product over in so I'm taking this finite product writing it as a quotient of two infinite products that's all so when I do that what happens is the dependence on n is only in the numerator now and the dependence on n disappeared in the denominators I can move it to the other side which I will do so well maybe I write it okay thus maybe I write it here so this gives me their the point is I can do this and I'm notice that I've shifted the summation from minus infinity to infinity the original summation was from 0 to infinity less added minus infinity infinity because there's actually nothing there when n is negative there will be something that will cancel it so the product is 0 ok so it's just a convenient way of bookkeeping I can shove in and going from 0 to infinity ok now all right in again we want to use the star in star we replace Q by Q squared okay when we replace Q by Q squared and X by minus Q the 2n plus 2 so I got it I'm I'm playing on this this thing I'm keeping you know wholesale use of this identity so here I have again another infinite product that looks pretty close to that one so it doesn't it's not a big deal to see that in order to get my hands on what that what this this thing is allows you to do is put some specializations for X and in this in this calculation which is what I'm gonna do that's exactly what I'm doing so I'm gonna change X here to cure the 2n plus 2 and replace so thinking of this is a product over J hmm it'll be Q the 2j so it'll come out exactly like that we see that product J going from 0 to infinity [Applause] turns out to be equal to summation M going from viewer infinity I I will make a long story short by just writing down the answer [Applause] so so all I all I've done is basically use this identity with these with these parameters and now running on a space I have a feeling I don't need this thing again so I'm gonna rub this off okay so I think this is pretty clear you just do the obvious and you collect term the indexing is with M so finally we get we get that the sum the product in question equals product J goes from 0 to infinity 1 minus Q the 2j plus 2 inverse summation M went from 0 to infinity so there's this q squared here let's I forgot that ok so get that and it is minus 1 here the M squared plus M plus 2 m M divided by 1 minus Q squared 1 minus q 2m so that's what we ended up getting and if you if you think about it I'm going to rewrite this thing as a product J equal 0 to infinity 1 minus Q 2 J plus 2 inverse summation m going from 0 to infinity Q over X the M minus 1 to the M 1 over 1 minus Q squared 1 minus Q the 2m times to the power M plus M Squared that's it yeah so you see when I when I introduce M plus n square so the Curie M that's floating around here I've pulled to the outside I've got M squared plus 2 m n that looks like M plus n whole squared except there's that enter missing so I compensate for that by by the way what happened to the X there was an X missing right in this power series thing oh I replaced X X is equal to Q X exactly okay so when I do that you end up getting this no there has to be an X here see right there has to be so what happened there's an X to the N which I forgot to carry over right there's an X to the so this this this product here I'm saying is equal to this product here is I'm putting that inside here and there is an X to the N so when I calculate that and put that in it's going to be X to the M plus M because I just want to get rid of the X to the I've blood next to the M here there's an X there was an X to the N see this this product is only for this piece inside but there's an X to the end here hmm oh yeah there's a Q power N squared that's right I that's right but that's now here okay I think it all tallies there's nothing there's no you know sleight of hand here it's just oh the N on the right hand side is the end here on the left hand side hmm oh oh yes oh yeah yeah you're right I forgot to put you're right there's a some here over in thank you yeah thank you there's a sum over n yeah that that end is different than sorry I shouldn't have confused you with that there's a sum over N you're right sorry about that you're right that end has not I mean this is a this is the index here so I should I should so the M that's here should have been there's an extra summation sorry about that okay well it's you know this is all just basically you need you need sufficiently long paper in order to calculate all this stuff which is what you know people like Ramanujan Euler and they were not timid to do this type of stuff and you shouldn't be either so here we are we've got this calculation now now the question is of course you'd like to group together the these exponents and the point is that M plus M as as for a fixed so we interchange summation so let me just say this look this is not what they say it's not bream obj so you just need to do the right manipulation interchange sums and you change the m plus l squared to N squared but the idea is that if I interchange some I'm glad the M outside and I've got the N inside for a fixed M as n runs through all the integers from minus infinity to infinity so does m plus okay so that's where I can make this change of 2n squared and finally you end up getting that this is equal to this is equal to this is equal to product' J going from 0 to infinity 1 plus 0 to k plus 1 over X I'm using the theorem one more time right I'm using the the product formula so here let's let's just go this carefully so I I'm saying that this is equal to summation a product J going from 0 to infinity 1 minus Q the 2j plus 2 times summation m going from 0 to infinity Q over X to the M minus 1 to the M times the summation divided by this product 1 minus Q squared 1 minus Q the 2m and inside here I have summation and going from minus infinity to infinity Q the N squared X to the M which is the stuff that I'm after which is the stuff that I have but what are these factors well one of the factors is this guy which I can there's an inverse there was an inverse here okay which I can move now to this side and then I have to recognize what is this guy this series I go back to that Q exponential thing again and recognize that it's that infinite product with with the 1 over X okay it's Q over X so when you use that so we're using that star identity several times in the course of the calculation okay so I hope that more or less gives you a flavor of how this triple product is approved it's pretty it's it all comes from this Q exponential all right now one of the beautiful applications of this particular identity is to the study of representations of natural numbers as sums of squares so this identity here is useful because the so called theta function already appeared in mathematics at the time that people like Jacobi were studying it it already appeared in the context of the heat equation and so on and so forth so if we if we specialized a classical theta function which also was studied by Riemann in deriving his functional equation for the Riemann zeta function is where that is in the upper half plane which we will meet later on in this course consisting of complex numbers with positive imaginary parts so that's what's called the upper half plane if I'm sitting in the upper half plane the Q the absolute value of Q is less than 1 and everything converges so this is a nice function which is holomorphic in the upper half plane and so the properties of this function are central to Riemann's derivation of the analytic continuation and functional equation for the zeta function which of course is not a topic in this course but it's an important function so when when people see hey that stuff has appeared here and it's being represented as a product there's something profound going on but should be clear so if we specialize if x equals 1 and we specialize we see that we see that product summation minus infinity to infinity qdm squared can be written as a product as an infinite product which is an identity of independent interest and beautiful in its own right so that's a that's an amazing calculation now we did with we did it with bare hands nothing that's the first point I think Euler noticed that you could write down other interesting things Jacobi also noticed more things and so let me kind of quickly tell you about those things in the time remaining so so here a bunch of exercises that we could we could kind of do after we have okay so let me let me let me just give you the some of the corollaries of this thing and you could you can try to do them you know it's just a matter of plugging and specializing values of X and Q and so forth right so here are corollary exercises corollaries what's called the Euler pentagonal hero pentagonal number theorem okay so from here we want to get to that kind of relation so it's just a matter of giving two minutes of thought to how you would do that and you can immediately see that you want to replace these guys by something like 1 minus Q to the N and so there's a little trick so hint replace so you can probably do this at lunchtime replace Q by Q by Q to the three-halves and X by minus Q the minus 1/2 in the Jacobi triple product identity and twiddle a little bit and lo and behold you get this ok so you just have to do that it's corollary one corollary to Oh before I move to corollary to let me just point out you will remember at the very beginning of the lecture I thought I did the partition function problem so summation P of n to the N and going from 0 to infinity was the product and going from 1 to infinity 1 minus Q the n inverts right we got this at the very beginning well now we have a formula for this so what coiler sees is immediately that if this is the inverse of that we see that immediately K going from minus infinity to infinity minus 1 to the K Q to the K 3 K minus 1 over 2 times this partition function equals 1 which means that there are the only coefficient is the one coffee everybody else is zero so this allows you to get a recursion relation for the partition function and this is how boiler studied the partition function calculated all these values of the partition function two huge numbers and you may wonder how did it how in the world did he do it did he actually write down all the ways of writing and you know 10 million as a bunch of cements no he used this recursion so this is an important step and this is the interest of Euler purely combinatorial you catch it and by the way why these are called pentagonal numbers is if you try to stack oranges in some sort of Pentagon just you know put put oranges in in a pentagonal shape the the sequence of numbers is like this these are there you know these are the only things that you get so that's why it's called pentagonal number second corollary is Jacobi's formula which is also going to be useful later on so this is Jacobi and how do you get this put replace X with - X and and factor 1 minus X cube out and take limits and take limit as Q goes to as limit as X goes to one working so there's a little tricky and maybe this can be covered in the tutorial in the afternoon you just need to do a little bit else cool and whatnot okay so that's we'll leave it at that okay so these are our two very classical theorems in well this falls under the common rhetoric chapter and this actually falls into the number theory chapter it'll be useful later on for other purposes so the jacobi coral reefs italy give it a little bit more difficult than the euler corollary oops and so let me just give one more exercise which will be useful in a moment and these exercises you can do in the afternoon tutorial yeah so I'll leave that for you to do okay so now let me in the half an hour remaining let me give you a taste of two applications the first I will do in detail now and second will be done by Karim Dave in the afternoon and this is the classical question of how many ways can I write a number as sum of two squares and the corresponding question how many ways can I write it as sum of four squares so the four squares will be handled by Karim leaving the afternoon it's using exactly similar method but I will now show you in detail how to answer this question this question can be approached in other ways I mean this is not the way to classically the the way of representing a number as sum of two squares the classical approaches through gaussians gaussian integers ring the gaussian integers and the ring of gaussian integers for example you consist of all integers of the form a plus bi with NB integers and the norm of a Gaussian integer is a squared plus b squared so you recognize you change the problem of how how many ways can i write a number as a sum of two squares as asking how many Gaussian integers are there with the norm n okay and then that takes you into the algebraic number theory root and you can answer the question using the arithmetic of the ring of Gaussian integers which probably some of you have already seen in some basic course in algebra here we will take a combinatorial mode we'll take the root that we just figured out so Jacobi 2 squared theorem let our to again with a number of ways number a B with a and B integers then are to M is equal to four times D one of M minus D three of em where D one of M di of M is the number of divisors of n with D condom - I mod 4 so not only is Jacobi going to answer the question he's going to give you a formula for the ways of number of such representations and the same thing will happen with 4 squares you're going to get out of this method you're going to get closer formulas alright so how do you do this we do this by using the triple product identity you see people keep writing Jacobi triple product identity and not shorten it to Jacobi identity because there's a Jacobi identity in the theory of Lee algebras you don't want to get that confused so Jacobi triple product identity we will rot it so that's that's your identity will keep feeling this the product here goes from zero to 30 okay so now how do we solve this two square theorem well so the proof of this theorem proof of the Jacobi 2-squared theorem begins by it's this is due to michael hirsh form so in the triple product identity in dagger put in the for q 4x q squared for q to get I'll tell you in a second what the motivation is for this but let's just understand the steps to start with so if I change you know in this X I'm putting into the 4q right so I'm going to get and and I'm changing Q to Q squared so by changing Q to Q square I'm going to get 4n plus 2 and the extra Q that's coming from X will become make it for n plus 3 and here it'll be the other way around so I'm getting numbers convert 3 mod 4 numbers come to 1 mod 4 this way what apply this process and then the this term here of course is of when I change Q to Q squared they'll all be multiples of 4 by changing the index of summation you know you get that okay so that's basically what how that you get this so this is not a big deal so now multiply both sides by by a and differentiate with respect to a so a is being thought of as a parameter so if I multiply this by a and differentiate with respect to a well what do I get I get I have a to the 4n plus 1 and I'm differentiating a to the 4n plus 1 that's going to be 4 n plus 1 times a to the 4 n right so I end up getting summation for n plus 1 a to the 4n q the 2n squared plus N and on the right hand side I'm supposed to multiply by a and take the derivative okay well a times some function I have to differentiate well use the product rule so it's easy to differentiate a that's just 1 so you end up getting this product it's like that and then I have to keep the a and differentiate the product wait a minute that's the product rule now when you want to differentiate some product it's useful to do the following make the following observation if P of a is product and going from 1 to infinity some fnm a then by taking logarithms and differentiating logarithmic derivative it's called you end up getting P prime of a over P of a is equal to summation F n prime of a so that the derivative that I want to now look at can be written as a product of P of a time's the sum because the the P of a now would be nice this factors there so I can kind of move it out so it should be clear then we get this way we obtain summation from minus infinity to infinity for n plus 1 e to the 4 n to the 2n squared plus n equals this product times 1 plus well there was that a there in front of that thing and then differentiating it in the differentiating turns out to be a nice sum it's n going from 1 to infinity for a cubed here the 4n minus 1 divided by 1 plus a 2 for 0 for n minus 1 minus 4 a to the minus 5 to the 4 n minus 3 divided by 1 plus e to the minus 4 Q the for n minus 3 hmm oh there's a - yeah this is a - here here oh is it one - well maybe I I copied it incorrectly yes you could be right what do I have here I actually copied this thing wrong so when I multiply when I change variables you know make this this should have been a plus sign here you should have been a plus sign here first from where oh did I make a typo there oh the further you mean the four yeah the first term let's write the at this one yeah sorry about that okay so when you when you work it out you end up getting this I mean there's nothing I mean there's nothing as I said this is not bromo Vidya to manipulate these power series so it should be pretty straightforward and the final punch line here is that we put a equals one to get [Applause] I think there's a typo in the book I think the the sign you know this should be so this so what's happened this has got to be minus sign there's a small typo in the on page eight there should be a there's I think it should be a minus sign because what will happen now is I expand so let's try and understand this I expand to two things one important point I think I rubbed up one of the earlier exercises in the earlier exercise I gave you that the left hand side is can be you can use the triple product identity again so the Jacobi identity rather the Corolla ring the corollary remember the corollary that we had so using Jacobi's corollary the corollary to that I wrote down Jacobi's you know the the cube of cube of 1 minus Q the N cubed was the was the thing write that one using that and exercise one point two point four everybody has a copy of the book right it's a part of the package everybody's got a copy the book okay so exercise one point two point four which I wrote down earlier if you use that you end up showing that the left hand side shows that the left hand side equals product and going from one to infinity 1 minus Q the NQ and so and when you use the which which can be rewritten I mean I'm just separating out the odd numbers and the even numbers and this is a one minus qbn cube so I can partition those into odds and evens and do it right rewrite it like this and then you do some twit rewrite this a little bit in such a fashion that you end up getting that this can be rephrased arya formulated as being n equals 1 to infinity of 1 minus Q the 2 n minus 2 n minus 1 to the power 4 and 1 minus Q the 2n squared and this if I if I had two here and one here two here and one here notice they're just triple product again it's summation QT and it's if this essentially it's this ok so this is the square therefore the left hand side is nothing but summation to the N squared squared the left-hand side is exactly what you are interested in the number of ways of writing a number and sum of two squares so that's that's actually calculating the number of ways of writing sum of two squares and so the right-hand side though is non-trivial so when you when you when you so after having pulled it all out you end up getting that this is equal to there is still a sign error isn't it there's it maybe there should be a plus anyway let's just let's just look at these these terms here Q to the 4 n minus 1 over 1 minus Q the 4 n minus 1 can be written as a power series so there are multiples of 4 and minus 1 and the other ones are oh they're maybe the sign was right maybe that was the right one because 4 and minus 1 is a 3 mod 4 business right and then this one is going to be 4 and minus 3 4 and minus 3 is essentially 4 k plus 1 okay so minus 3 is plus 1 so so you get that the this is actually Kelly this is actually rewritten as a power series with powers of Q but you're factoring the number the coefficient of Q the M will be the number of ways of writing m as a product of something conquering to 1 mod 4 or 3 mod 4 depending on the side so when you piece this all together along with this other part of the thing you get that explicit formula that Jacobi had so as I I'm going to be running out of time pretty soon I just wanted to make some concluding remarks but I hope it's clear that this remarkable application comes practically for free the idea of introducing this parameter a and then differentiating with respect to that a is a standard trick that goes back at least to the 16th 17th century because when when you meet integrals if you want to evaluate them a standard trick was to introduce one more variable and then different see if it satisfies a differential equation solve the differential equation and therefore solve the integral I mean that kind of trick was pretty much in the subconscious mind of many people like Jacobi and others so it's not a surprise to do that kind of trick here as well so that's about the only new idea that's needed in order to get the 2 squared theorem and a similar maybe more detailed careful derivation the afternoon will give you some four squares so for those things you get these beautiful applications finally before we stop for the day in the book Jacobi foursquare theorem is proved by you a technique of Ramanujan which is you know in a sense more complicated than this but Karim Dave's presentation this afternoon will be will show you that the same idea can lead to similar results so in the few minutes remaining I just want to say that the theory of modular forms in many ways starts with the 1916 paper of Ramanujan although modular forms were being studied much earlier somehow Ramanujan in his 1916 paper was intrigued by this product which is now familiar to us you see we met 1 minus Q the M cubed was Jacobi's identity ok 1 minus Q Leanne was the Euler identity so the thing to the power one was the Euler pentagonal theorem to the power 3 was Jacobi's thing so you may ask are there other powers for which you can get some nice results that's a very loaded question and the answer surprisingly is in Lee Theory has something to do with a fine root systems but Ramanujan noticed that just as we were able to get nice formulas for the number of ways of writing and a sum of two squares and four squares in the afternoon is it possible to do other powers of K squares and it turns out that you can and you can get beautiful formulas all the way up to 22 and the problem arises when it's 24 squares and when 24 squares this is the function that kind of enters the picture to mess up the calculation you don't get an exact formula like the kind that we just wrote down you get an exact part of it is an exact formula but part of it is a problem term and when you expand this in an infinite series there's a queue expansion you get some coefficients and the Ramanujan call these the tau tau n so this is called the Ramanujan tall function and in his 1916 paper made fundamental conjectures about the tau function firstly that it was multiplicative tava MN is tau of M times tau of n whenever m and n are relatively prime then he made some conjectures about various estimates and so on so forth that was the paper that somehow spawned the development in the theory of modular forms and it was only in the 1930s that Hecker came up with the proper theory in order to understand what's going on here so that's taken up in these subsequent lectures and subsequent chapters but that's basically where we're going and I hope I have stimulated your interest in trying to study products on this shape which will appear throughout this course