Thank you for the opportunity to speak here. So I will begin with, OK, so now I, OK, now I managed to advance my slides. Good. So I will begin with the conjecture of Odlisklan-Punan. So this asks for the following. Let p of x be polynomial with integer coefficients. And so the leading coefficient and the constant coefficient is fixed to be 1. And the rest of the coefficients are random. And they are independent. And they can take the values 0 and 1 with equal probability. So this is the model that we are working with. And the problem is to show that such a random polynomial will be irreducible in zx with high probability approaching 1 as the degree of the polynomial goes to infinity. So this is just one instance of expressing the expectation that if you just write down polynomial randomly, then you expect it to be irreducible. So here is a result towards this conjecture. So we need to assume the extended dream on hypothesis. I will tell you about this in a minute precisely what I mean. And so then let's consider a polynomial as in the above conjecture. And then the claim is that with very high probability, so the probability that it, yeah. So this minus should not be there. So this is exponential in square root of d plus an epsilon in the exponent. So with very high probability, the polynomial can be factored as q times r, where q is irreducible. And r is a polynomial of not too big degree. So it is at most square root of d. And it is a product of cyclotomic polynomials. So it kind of says that it is kind of irreducible with very high probability. And so you can have two objections regarding the statement, why this doesn't prove the conjecture. And so one is that the conclusion is not that the polynomial is irreducible. And so in fact, this is not a big problem. So very easily from this statement, you can deduce this one that the probability that the polynomial is irreducible will be asymptotic to 1 minus square root of 2 over pi d. So some specific constant over square root of d plus an error term, which is of the order 1 over d. So the point is here that if you just look at the possibility that your polynomial might vanish at minus 1, then this will happen with roughly this probability, this constant over square root of d. And this is what contributes this term. And so this is not that too hard to work out. And so this is one instance of the cyclotomic factor that can come up. And for other low degree cyclotomic factors, it's probably with some work it's possible to work out precisely what the probability is that it arises. But if you want a statement with such a precision without the negative sign here in the exponent, then you need to allow for these cyclotomic factors. OK, so this was one objection. The other objection is that we assumed something. And so this something is the extended Riemann hypothesis. So it is the Riemann hypothesis for dedekin zeta functions, zeta sub k for all number fields, which are obtained by adjoining a root of one of these polynomials that appear in the conjecture to the rational numbers. So this is the extended Riemann hypothesis. And this is not known. So this result is conditional. And for this reason, it doesn't resolve the conjecture. So if you just want to prove the conjecture, then it is possible to weaken the hypothesis, the reliance on the Riemann hypothesis, quite significantly. So if you are not interested in this error rate, then it is enough to know that there are no zeros of these dedekin zeta functions very close to one. And how close it is log d to the 3 plus epsilon over d. And so what do we know towards this? Well, what we know is that if you draw a somewhat smaller disk around one, so with radius one over log d times d. So the log d is not in the numerator with some power, but instead it's in the denominator. Then we know that there will be at most one zero. So there could be an exceptional zero for the dedekin zeta function, which, in principle, could be very, very close to one, but then there are no more. And so in fact, there is a version of this result which allows for an exceptional zero to be quite close to one. So the exceptional zero is perhaps not the bigger issue. The bigger issue is really that the zero free region that I would like to have has the log d in the numerator. And so this bound that we have is kind of old now. And it doesn't look like that anyone has an idea how to improve on it. So it's still close to what we know. But it is probably very difficult to get this unconditionally. So final remark about this statement is that, of course, you don't necessarily have to consider this distribution that it takes the value zero and one. But instead of that, you could take an arbitrary distribution on the integers. And in fact, you can even vary this. So you can change it with the degree as the degree grows. It is subject to some mild conditions that I don't want to state now. But the point is that the method is quite general. And I will tell you about that in a minute. OK, so there are a lot of unconditional results on this problem. So let, again, P be a random polynomial as in the oldest component conjecture. So the coefficients are 0, 1, independent with the same probability, with probability 1, 1, 1, 1. So there is a result by Cognac in, which gives a lower bound on the probability that this polynomial is irreducible. And it is of the form constant over log d. So we would like to have something which is going to 1. And instead, we have something which is going to 0. But at least it goes to 0 very slowly. And very recently, this has been improved a lot by very sort of caculopoulos and Cosma. I believe one of them gave a talk on this seminar about a year ago. So they proved that, in fact, the probability can be bounded below by an absolute constant. But we still don't have that it would go to 1. So some further unconditional results are available if you change the distribution in question. So now I take for the coefficients from which are not the leading coefficient and the constant coefficient. So they are still independent. But now they take the values between 1 and capital N, where capital N is some fixed number, which should be a parameter. So there is a result by Berry-Sorokan and Cosma, which proves the analog of the conjecture in this setting. If capital N is an integer, which is divisible by at least four distinct primes. So the smallest example is 210. And then in the paper that I have already mentioned, this is improved to any integer, which is at least certified. But somehow the probability that they get is not as good as what is available under the Riemann hypothesis. So their error is polynomial N, which is D, I guess. So this N should be D. Sorry about that. And of course, this work is also, this method will also work for more general distributions. But they need a little bit more spread out distribution than something supported on just two points. But if something is sufficiently spread out, then it has some nice Fourier transform, then it will work. OK, so another comment is about the Galois group in question. So if you know that the polynomial is irreducible, you might also be interested in what the Galois group of this polynomial is. And if you ever tried to find examples in a Galois theory course for a problem sheet, then you will know that finding polynomials where the Galois group is the symmetry group is very easy, because most of the things that you would write down will have the symmetry group as the Galois group. But that's something that we can't quite prove. So what can be done is that in all of these results, the statement that the polynomial is irreducible, it can be exchanged for the statement that the Galois group contains the alternating group. And it is difficult to distinguish between the alternating group and the symmetry group. And so this is actually a very frustrating problem. So when I was first working on this, then I made some computer experiments and I tested all polynomials that have been in the odalis copoon and conjecture up to degree 31. So that's a little bit more than a billion polynomials. And none of them has Galois group which is equal to the alternating group. So there will be some which have Galois group contained in the alternating group, but exactly equal to the alternating group that are now. And still, we are unable to prove even that most of them is not the alternating group. Peter, we have a question in the chat by Jacob Strypoop. Maybe Jacob, you want to unmute and ask away. No, how good my microphone is. I hear you fine. OK, the constant term changed from 1 to a 0 in the statement of the OP conjecture to this unconditional result. It's a 0 significant. Say again? The constant term in the polynomial on the previous term. Yes, so if you allow 0 to be the constant term, then so maybe if you would say that the constant term is also 0, 1 with probability 1, 1.5, then the constant term will be 0 with probability 1, 1.5. And if the constant term is 0, then the polynomial is not irreducible because you can factor out the factor x. But in the original conjecture, it was always 1, right? But now it's about to vary? I might be missing something. What do you mean? Wasn't the leading coefficient under comes? Oh, you mean in this? Oh, OK, so in this here, right here, right? So the way I formulated it is that the values will be between 1 and n, 1 and capital N. So 0 is excluded. So basically, that's the only thing that I was careful to exclude that the constant term can't be 0. So they have a more general version of the result, which maybe would allow an arbitrary interval. And but in that case, they modify, they model to exclude 0 for the possible values of a 0. So that's the only thing that you need to be careful about. OK, thank you. Thank you for the question. We have a further question by Jordan Elmerk. Jordan, maybe. Yes, please. Oh, yeah, so I was trying to do this in my head, and I couldn't. I mean, the discriminant of a polynomial of degree D with 0, 1 coefficients, I expect is pretty big. And someone could heuristically say, OK, if an integer of size n is a square, 1 over root n of a time, in other words, that's what distinguishes whether you're AD or SD. So I mean, does one just expect that naive heuristic grounds that because the discriminants are so large, there should be only one. Yes, so the discriminant is expected to be roughly of the size d to the d with some constants. Oh, yeah. Around the, I mean, theoretically, it could be smaller for some polynomials, but somehow it shouldn't happen. And then the probability that such a thing would hit a square is just extremely low. So it is not surprising. Agreed. So it will be occasionally square. So it could happen that, so most of the time this happens, you are in a situation when the Galois group is not primitive. So it could happen that your polynomial is a polynomial in x square or x cube or something like that. And in that case, in that case, your Galois group will be primitive and sometimes then the discriminant will be square. But basically, those are kind of the situations that I found. But some are still proving. You see the problem is that the discriminant is some sort of complicated, several variables polynomial. And you are plugging in only 0 and 1 in the variables. So handling this with analytic number theory methods is really frightening. But anyhow, it's a problem that I really like. And so far, I was not able to solve it. Other questions? OK, so please stop me if there is something. So let me just remind you what the theorem that we proved with Emanuel was. So we assume the extended Riemann hypothesis. And we take random polynomial as in the oldest component conjecture. So the coefficients that you see here are independent and 0, 1 with probability half, half. And then the claim is that the polynomial can be factored in the form of an irreducible polynomial plus something of a small degree which only contains cyclotomic factors. And this happens with very high probability. So I want to tell you very briefly the main scheme of the proof. And so the main reason why I want to tell you this is because it's quite a general method. And later on in the talk, I will tell you about some newer developments where this method was used. So the proof is based on the following idea, which is a consequence of the Cheboterev density theorem. Or in this case, actually, it's just a special case of it which is called the prime ideal theorem, which says the following. So now let's take a fixed polynomial with integer coefficients. And I want to determine whether it is irreducible or not. And here is one way of doing this. So if the polynomial is irreducible, then this polynomial will have, on average, one root in the finite field Fp if you are averaging over the primes. So this is a fact. And so what you can do with this is that if you take a general polynomial now and now you solve this polynomial in finite fields with various primes, and for each prime you count how many solutions, how many roots you have. And the average number of the roots will tell you how many distinct irreducible factors of your polynomial have in Z of x. Because each such factor will contribute one average. So this simple fact is the main idea behind the proof. And so here is how it is used. So I take a random polynomial as in the conjecture. And here is the quantity that I'm interested in. It is the expected number of distinct irreducible factors of my polynomial. Then using this observation on the previous slide, I can write this as the expectation over the random polynomial of the expectation for choosing a random prime in some interval, capital N to N. And I'm averaging the number of roots of the polynomial in Fp. So if I'm just looking at what's inside the outer expectation, then this will be the number of distinct irreducible factors of P. And then I'm just averaging over the polynomial. And of course, then I have some error term that I will tell you in a minute. And OK, so now comes the big idea, which is always the same. So you have two expectations, then you exchange them. So what you are going to do is that you are first averaging over the prime, and then you are averaging over the polynomial. And if you fix the prime, then you can take this step further. And what you can do is that you can take all elements in the finite field, and you can sum up the probability that that particular element is a root of the polynomial. So that's by definition the expected number of roots. OK, so I have this expression. So basically, the hope is that I will be able to say something about what the probability is that the random polynomial is 0 evaluated at a given A for a given Fp. And so basically, the hope is that when I plug in that into this formula, then I will get the result that the average number of distinct irreducible factors of p will be exactly 1. And then I will be happy. So now, of course, one thing that I'm worried about is this error. So I want to know something about this. And this is where the zeros of the data kinds of functions come in. So if you want to prove this fact on the previous slide about an irreducible polynomial having 1 root on average in Fp, then the proof of this will be very similar to the usual proof of the prime number theorem. And the difference is that you need to plug in the data kinzata function of some number field instead of the Riemann zeta function. And then what this error is, of course, it depends on capital N. So how far up you are going with the primes. And it depends on where the zeros of the data kinzata function is. And so if you assume the Riemann hypothesis, then this error will be small as soon as you take primes which are larger than some polynomial of the degree, maybe d squared. I don't exactly remember. But basically, the point is that you only need to go up with the prime up to some power of the degree. And this is something that we will care about later very much. So the goal is to show that the probability that given element of a finite field is a 0 of a polynomial is roughly 1 over p. This is what we want to show. And in fact, what we are going to show is that if you fix a prime and you fix an element in the finite field, then if you evaluate your random polynomial on this given element, then you get roughly a uniform distribution on Fp. So not only that, it will take the value 0 with a roughly probability 1 over p. But in fact, it will take any values with roughly 1 over p. OK. So it turns out that you can think about this as some kind of random walk on the finite field Fp. So this is the random walk that you need to consider. So the elements are x0, x1, et cetera. And so you start the walk from 1, x0 is equal to 1. And then this is how you propagate it. So at each step, you multiply the previous step by a. And then you add something random. So it will be either 0 or 1. And the steps are independent. So this is a nice Markov chain on Fp. And so it turns out that if you make formally the substitution that you change the bj to ad minus 1, then the distribution of p of a will be the d step of the random walk, except that in the last step, you are forced to move by plus 1. And plus 0 is not allowed. This is because the leading coefficient and the constant term were fixed to b1. OK, so basically we got a nice Markov chain on Fp. And so basically what Markov chains do is that the distribution of the j step of this Markov chain will converge to the uniform distribution on the finite field. And this is precisely what we want to show. And now the question is that how fast does it converge to the uniform distribution? So remember that we have to take somewhat large primes compared to the degree so that the error term in the Chebotary density theorem will be sufficiently small. But somehow this is something which is fighting against that. Because here for a given prime, the degree has to be sufficiently large so that you get this equity distribution. So here I need to introduce a definition in the theory of Markov chains. It's called the mixing time or delta mixing time of the chain. So delta is a parameter between 0 and 1. And this quantity is denoted by d of delta. So this is the smallest j, the smallest time, such that the distribution of the random walk, which I denote by mu j after j steps, is its distance to the uniform distribution is less than or equal to delta. So here I'm using total variation distance, but for the purposes of this talk, it doesn't really matter which norm you are using. OK. So what is known about this? So OK. So here is a result which is based on some work of Konyagin later. So it emerged in the collaboration of these four people. They didn't publish it, but now the details are available in a paper of Emma, Brea, and myself. So here is what you can say. So first of all, you need to make some assumption on A, this element of the finite field. So obviously you don't want it to be 0 because then somehow this random walk will not be very interesting because every time you multiply by 0, you get back to 0. So then this random walk will just move between 0 and 1 and you're surely not actually distributed. You also don't want A to be equal to 1 because then you just have a random walk on Fp when you are either staying where you are or maybe you move to the right by 1. And so you will need to take at least p steps just to have any chance of going around the circle. And in fact, to get to actually distribution, you will need to take p square steps and that's just way too much. And similar problems will occur if the multiplicative order of A is small. So you need to make an assumption that the multiplicative order of A is not too small. So it has to be at least log p to the power 1 sub-silon. This is actually not too bad. So this will exclude some elements of the finite field from this consideration. And you will have to deal with them separately. But there are very few of these guys and it's not going to make you a big problem. So if the multiplicative order of A is not too small, then the mixing time will be actually quite good. So what you should have in mind is that maybe for the delta you take some negative power of p. So what you get, the estimate for the mixing time will be log p to the 2 plus epsilon. So what this says is that what is the relationship between the degree d and the size of the primes p where you are doing the averaging. So what this constraint tells you is that the degree should be at least as big as log p to the 2 plus epsilon. And the other constraint in the Chebotarev was that you didn't want the prime had to be at least maybe the degree to the 10, something like that. So these two things are clearly very much compatible with each other. And then basically we are in business. And so you plug in this 1 over p inside this formula, then you will get precisely 1. And as I explained, the error terms will be controlled. And then the proof just works. And if you work out everything, then you will get the statement that I stated in the beginning. OK, so so far this was kind of old news. And so now I want to tell you about some more recent developments, which will be about other families of random polynomials. So what other families of random polynomials are interesting. So I have to talk first about the classical situation, even though in this setting, this method that I'm talking about hasn't been applied. And probably other methods will do better. So but nevertheless, let me just tell you what this is. So the classical setup is that you fix the degree d and you take a large integer h. And now you take a random polynomial with independent coefficients that are uniformly distributed between minus h and h. So this is kind of dual to the oldest component situation. So in the oldest component situation, you fix the height of the polynomial and you let the degree to infinity. Here you fix the degree and you let the height go to infinity. So this was actually originally studied by Van der Werden, who proved that the probability that the polynomial is irreducible will be roughly 1 minus ordo 1 over h. And this has been made more precise by cellar who worked out the precise value of the constant. So basically the result is that the probability that the polynomial is irreducible is 1 minus some constant over h. And the value of this constant is explicitly determined. But it is not very interesting for the purposes of talk what it is. And it's not very, it would take half a slide to write it down, so I didn't do it. And then there was a big conjecture of Van der Werden, which asked about, OK, so can you say instead of irreducible here that the Galois group is the symmetric group and can you still have the same probability? And so this has been open for quite a while and very recently Manjubar Gava proved this. And so actually he talked about this on this seminar series. It's not too long ago, so I just wanted to mention this, but I will not tell more about this now. So here is another setting which people are interested in. And that's the characteristic polynomials of random matrices. So here is the details of the model. So you take the random matrix M with independent identically distributed entries with respect to some fixed distribution in the integers. So if you like, it can be 0 and 1. We probably have half or something else. And now you take the polynomial to be the characteristic polynomial of this matrix. And then you are interested in various properties of this polynomial. This has a massive literature. And in particular, one question that was asked is is it true that this polynomial is irreducible in zx with high probability? And recently this was proved conditionally on the same hypothesis as above by Sean Eberhard. So it proved that under the extended dream on hypothesis, the probability that the characteristic polynomial is irreducible is 1 minus an exponentially small error in the size of the matrix. So again, this n is supposed to be the same as the d. Apologies for that. And the constants that you have here are just depending on the common law of the entries. And he also obtained the unconditional results which are based on the method of very sort of current cosma. And there he needs to assume something about the common law of the entries. And it is that there is an integer capital n such that when you take the distribution of the entries, modulo capital n, then it will be uniform. And capital n has to be divisible by at least four distinct primes. So it is somewhat more restricted. And also the error term, so in that case, he just proves that the probability is converging to 1. On the other hand, that's an unconditional result. So the proof of this result is based on the same scheme that I told you about. And somehow if you remember what I said, what this scheme of proof really cared about, about the random polynomial is what is the probability that a given element of a given finite field is 0 of this polynomial. And so somehow if that can be estimated well, then you can basically run the same scheme and prove that the polynomial is irreducible. So the main input in this proof is this result, which says that if you take a prime and you take a fixed element lambda in Z mod pz, then the probability that m minus lambda times the identity is not singular will be equal to this thing that you see here. And so what you should have in mind is that the prime p is kind of large. So actually the first factor will be the most relevant one, which is 1 minus 1 over p. So what you have here is roughly 1 minus 1 over p. And so basically what this tells you is that the probability that lambda is a root of the characteristic polynomial is roughly 1 over p. And this is precisely the input that we need for that scheme of proof to work. And so what is important here is that the constant that you have here, it depends only on the common law of the entries of m. And it doesn't depend on lambda over the prime. So actually that's the main point that it doesn't depend on lambda. And so in fact, Sean proved a much more general statement. So the crucial assumption is that the entries of the matrix are independent, but they don't really have to be in this special form. So in fact, the distributions can be kind of arbitrary distributions in Fp, modulo some very mild non-degenerative conditions. So I mean, one thing that you certainly don't want to see is that a particular entry is equal to some fixed element of the finite field with probability 99%. So those are the sort of things that you need to exclude here. But it's actually quite general. And then very recently, there was another paper by Ferber, Jain, Sach and Sony, who considered the variation of this problem, where the matrix is symmetric matrix. So you take a random symmetric matrix with independent plus minus 1 entries. So they are independent on and above the diagonal. And below the diagonal, they are just what they have to be to make the matrix symmetric. And so again, this is conditional on the extended Riemann hypothesis. And they conclude that the probability that the characteristic polynomial is irreducible is equal to 1 minus some error. And here the error is a little bit worse than what was before. And so the main point of this work is that you need to get a version of this result where the entries of the matrix are not independent. But you have a symmetric matrix. So only the entries on the diagonal and above are independent. And the entries below the diagonal are forced to be what they have to be to make the matrix symmetric. And that was a quite non-trivial step. So I stated the theorem in a form that appears in the paper. So I stated it with plus minus 1 entries. But they do say in the paper that they can handle more general distributions. And they leave the details to the reader. So you can work it out yourself if you want. Yeah, so I guess it's probably there is a superfluous minus also here. Sorry about that. Excuse me, I have a question. So the first theorem is actually unconditional. But the second one, you need the extended Riemann hypothesis. So for the theorems about irreducibility of the other paper. Yeah, they are both conditional on the extended Riemann hypothesis. This theorem about the random matrices, these are unconditional. How about the last one? The last one is about the same statement, but assuming symmetry. They need the extended Riemann hypothesis. The extended Riemann hypothesis is only coming in in this scheme of proof that I explained. So to show that the characteristic polynomial is irreducible with high probability, for that you need to assume the Riemann hypothesis. For these various statements about the distributions of the characteristic polynomials modulo p, these are unconditional statements. I see, thank you. OK, thank you for the question. OK, so in the last 10 minutes or so, I want to tell you about the other setting that we consider now. And here now we move from polynomials in one variables to polynomials in several variables. So let me describe the setting. So we take a connected, simply connected, semi-simple linear algebraic group G defined over the rationals. So you can just think about SL2 if you want. I actually, I also prefer to think about just SL2. And I deal by this simple G, the complex points of this algebraic group. OK, then I also need a finitely presented group. So this is generated by k generators, S1 up to SK. And these generators are subject to some equations, which is that you have different relations between them. So these are all words in the letters S1 up to SK. And the negative powers are also allowed. And so the condition is that these must give the, when you evaluate these words on the generators, you must get the unit element in the group. So now if you have this finitely presented group gamma and you have your algebraic group G, then you can talk about the representation of a variety of gamma. And so it is denoted by home gamma WG. And it is defined as follows. So you take k two poles in your group. And you take those k two poles, which satisfies the defining relation of your finitely presented group. OK, so what is this? So if you want to write down a homomorphism from this finitely presented group to G, then what you need to do is that you need to decide on where you map the generator. So for S1 up to SK, you need to choose elements G1 up to Gk in the group G, which will be the images of your generators. Now you need to test that these elements G1 up to Gk satisfies the relations of the group. Because if they don't satisfy the relation of the group, then this will not be a homomorphism from gamma W to G. And on the other hand, if these conditions are satisfied, then you do get a homomorphism. So these are precisely the homomorphisms from gamma W to G. And if you think about it this way, then this is an algebraic variety. It's a sub-variety of the group. Because it is defined by these equations. And these are polynomial equations. OK. So we got some nice sub-variety of the group. But we don't care about all of it. So inside this, there is something called the degenerate locus, denoted by V sub-deck. And this is the collection of two poles, G1 up to Gk, that do not generate a Zarisky then sub-group in G. So it is something that is contained in some proper algebraic sub-group of G. So for instance, in the SL2 case, what you should think about is that maybe all of these guys can be upper triangular matrices. And then somehow that part of this variety is not so nice, and you want to exclude it. So the object that I really care about is the representation variety minus the degenerate locus. So this will be some other variety. And I'm interested in the properties of this. And I'm interested in the properties of this. So actually, this is something that is studied a lot in group theory. You can, it is useful for some purposes. And so what I'm going to study is that I will consider a random group. So what is a random group? So I take W1 up to WR randomly. So I fix some numbers L and K and R. K will be the number of generators. R will be the number of relators. And the relators are chosen randomly, and they will be independent. And there are two possible distributions that you can consider. So one is that you can choose a uniform, the chosen element in reduced words or flanked L. Or the other option that you can consider is that you take a simple symmetric random walk on the finite group, and you choose your relators according to the distribution of this. So the proofs will work for Eber model. And then, so this is what we can prove. So this is a joint work with Oren Becker, Emanuel Breher, and Emanuel Breher. And so I say work in progress. So everything that I'm going to state in this theorem is actually written up. But we want to, we are still working on to make the result nicer. So we haven't yet released the paper. So again, you need to assume the extended Riemann hypothesis. So this is the Riemann hypothesis for some data kinzata functions that can come up. So if you want, you can assume it for all data kinzata functions. And then the statement is that there are some constants, epsilon and c, such that for all L, the following statements hold with very high probability. So the probability of failure is exponential in L to the epsilon. And this time I managed to write this down without the minus. OK, so here are the things that we can say. So the first is that if the number of relators is at least as many as the number of generators, then this variety Zw will be empty. If the number of relators is strictly smaller than the number of generators, then we can tell you what the dimension of this variety will be. It will be k minus R times the dimension of the group. OK, so before I give you the rest of the statements, let me tell you why this is what you would expect. So the ambient space for this variety is g to the k. So that has a dimension k times the dimension of g. And then you have R equations of this form. But in fact, these equations are as many equations as the dimension of the group. So for each relator, you lose R times the dimension of g dimension. So this is very much what is expected. So perhaps that is just one thing, which might be a little bit surprising. And that's the case when R is equal to k. So then the natural guess would be that the dimension of this variety is 0. That would mean that you maybe have some finite number of isolated points. But this turns out to be not the case. And the reason for this is that if you look at this representation variety, then this is invariant under conjugation by elements of g. So if you conjugate each coordinate by an element of g, then it will preserve this variety. So if this is not empty, then it has to be at least of dimension g. So this is the reason why it's empty when R is equal to k. So if R is equal to k minus 1, then on top of determining the dimension of this variety, we can also show that it is irreducible over the rationales. And if R is even smaller, so it is at most k minus 1, then we can also show that the variety will be absolutely irreducible. OK, so I am roughly out of time, so I'm not going to say. I wouldn't have said too much about the proof anyway. So the final comment I want to make is about this error term. So one aspect of this theorem that I'm personally not so satisfied with is that we don't have the exponent. I would really like to have here an exponential error rate. And so I would want to have this statement with epsilon equals to 1. And so that basically comes from a statement about random walks, about mixing times of random walks. And so the point is that if you do, so there are some available results in special cases due to health got. So he did the SL2 and the SL3 case, and maybe a little bit more. And then in full generality by Brea, Gintow, and Piver and Sabo, who proved that if you take the simple symmetric random walk on the FP points of these semi-simple groups, then the mixing time will be at most polylogarithmic in P. And instead of this polylogarithmic, we would like to have a logarithmic to have the exponential rate. And so that's available in the SL2 case, but with a caveat. And the caveat is, so this is a result of Brea and Gumball. So what they showed is that if you don't consider all primes, but you allow for a small exceptional set of primes, then outside of this exceptional set of primes, you will have the logarithmic mixing time. And so now basically the point is that it turns out that this method that I described to you to show irreducibility. And it is also, I adjusted to work in this setting of this theorem. So you don't have to know this result for all primes. So you don't have to know the mixing time results for all primes. It is fine to exclude a small set of primes. And it will still work. So it is actually enough for us to plug in this result. But the problem is that currently this is only for SL2. And my co-authors are working on extending this to the general case. And if that is done, then we will have the exponential rate here with which I will be much happier. So I stop here and I thank you for your attention.


Alright guys welcome back to the channel. It's a special kind of video I guess everyone's been asking for more DIY stuff So I'm out here today with one of San Diego's finest mr. Zack Doughty Been meeting to meet up for a minute How are you sir? You've been like just put it on the record favorite youtuber you Dan Dell. Oh, man.  I feel honored Thank you dude But yeah, we're out at this slab spot out here in San Diego that Zach has brought me to and There used to be a DIY here back. I don't know six seven eight years ago now But we're gonna test the waters. We're gonna build a couple things today and see how it goes Where the spot is it's in like a big nature reserve here This house is that are far enough away keep out of the view of the DIY spot So they can't really see what's going on unless they walk up here But yeah, this slide has been sitting like this for God knows how long so we're gonna Check this place out and then head down to either Home Depot or Lowe's pick up some stuff and then get to building So yeah, I think it's a DIY diaries ever so nine maybe or ten. I don't know. It's been a while. Yeah Yeah, we forgot our mask. I got you more massive if they have them inside. Yeah, thank you Thank you Man with the mask. Thank you, sir. Thank you So we're probably gonna need like between 18 to 20 bricks to be able to build this in the block ledge It's gonna be coping on both sides. So I need the bricks angle iron Forking guns and some liquid nails as usual Teamwork makes the dream work Smart guy by prepping up the spot at the hardware store You don't want to get to the spot and then have to go back to the hardware store to pick up supplies You're better off getting more than less You want to be and then you'll better to have it not need it Come back here and that's be a Back said this a lot was poking sir. I'm trying to find where they're poking maybe over there You want to not these bend and break like you can Is that the right size Okay, all right, we got the angle iron you've got a coughing gun We've got a box now. We've got all the bricks Seeing as they don't cut metal here at home deeper. We had to buy a stainless in the vault Metal hacksaw to get it done ourselves We just got hooked up big time right now Zach was just explaining to the lady on check out what we were building we're building a DIY skate spot for the community And then she was like whereabouts is it whereabouts is the area? Kanto, and she's like oh cool. That's what's up, and she gave us a discount of $50. Yeah, she looked it up Yeah, 50 bucks. That's a massive discount. So well shouts out to that lady at home Depot here Shouts out to Jesse for coming out today bringing the truck For all our stuff loaded up here. Oh I would have been I would have been having some trouble trying to put this in the back of my car. Have a look at this bush here guys bunch of gum trees you wouldn't be able to tell if I'm in Western Sydney or San Diego Thankfully, I'm in San Diego today. We got all the cinder blocks up to the slab here We just got to figure out where to place this ledge and we're gonna get the angle iron cut down to size Get it all glued down. Doughty wants to put in a pole jean here today So he's got a bag of quick crete which you can get in this country. Thankfully, we don't have quick crete in Australia He's got this choice Bar here. It looks like it might have been a Street sign or something in its past life. There's a tag on it here. Someone's done with it Solid marker or something, but yeah Pretty keen to rip into this DIY spot. This is DIY from Campbelltown to California. This is DIY building I don't even know what I'm saying. I'm just super excited to be able to build this spot out here It's been a minute and I'm out here in California life is good All right, we got the angle cut shot glued down got some rocks Some older rubble actually from the old DIY build that was here. It's a bunch over there perfect use for A half pipe. Sorry a quarter pipe when the time is right. But yeah, we're gonna let this thing dry keep these Rocks on it and hold it all down in the California Sun Should dry and set pretty quickly. I'd imagine it's quite hot out here today So this spot right here. It's like a nature reserve, but it looks like the ruins of like a navy kind of barracks and There's like a little slab down there. It looks like it could have been a gas station or something But yeah, check out all the rubble from the last DIY build that was here If I can find photos I'll chuck them in but there was like a DIY build here And they went massive straight away like it was big as you can see they are the city knocked it Because I got too crazy and here the here the leftover is the remnants you can tell because It's like super smooth right there and then like all nasty So this would have been the top of the quarter maybe and that would have been the inside It's like a big piece of quarter pipe concrete cake right here Can serve this to someone at their wedding a bit of wedding cake But like honestly look at it out here Like I said, you don't know if I mean Western Sydney this could be like walking through a nature reserve in Penrith or if it's San Diego, you know You just don't know Some more rubble right here next to the green slab here They just knocked it and they didn't want to clean up so they just left it But yeah, check it out like this could have been like a refuelling station here It's got this little island in the middle here some mental pipes and stuff Could come here and fill up the truck And go to wherever you're going If you're in the Navy Another little concrete slab down here Not as smooth as the main one up there But apparently there's a bunch of scattered concrete slabs like this around this whole bit of land here We're building on the main one all right guys. We've got everything Situated we're about to dip right now get some food building dare-wise you work up a big appetite So I think we're gonna get some Mexican food down here in San Diego so keen to skate this thing check it out Honestly a kicker with a pole jam It's gonna be rad All right guys, so this is a few days later had to work all weekend So just walking back up to the spot right now to meet up with Doughty But uh, I believe I've got a bit of sad news So the following day after we built Doughty went down to check out the spot everything seemed all right Couple days after that he went back to check out the spot again And it was completely gone like everything was flattened like there wasn't even like one piece of like rubble from a Cinderblock or anything so the city definitely ripped it out, but yeah kind of sad but Check it out Gone Gone with the wind It's like completely like it's like you're watching a movie and then the movie just ended. Yeah, I had no ending Yeah, it's just flattened just took straight to the titles. Yeah My god, this is where the ledge was you can see the Liquid nails kind of ripped up the pain a little bit Put a ton of liquid nails on this thing too, you can see It really looks like the city did it because like there's no evidence I was looking around for the pool or the ledges the angle iron. There's nothing. Yeah, I look at all the bushes It's completely gone. Yeah, they did a good job like whoever Did it a good job? They did a good job of destroying it though because like whoever did it cleaned up everything Yeah, they would have even picked up the nails. Yeah nails from the ground There would have been way more red you reckon they would have just chipped it and cleaned it It's like if like you just knocked it out like there would have been some all would have been Yeah, I've had people knock out spots. I've built before when you know, it's like like a personal person or a resident It's not like the city. Yeah leave some stuff. They'll leave but there's like it's just flat They would have for sure gotten a truck in here with a bunch of guys with the crew and just knocked it out Yeah so I like how positive you were though like when I facetime you're like well Like I feel like your third thing that you said was like well, do you have any other spots to build out? Yeah I think that's like the right that's what I was trying to do Yeah, cuz I didn't want to be like this isn't my first rodeo of a DIY being destroyed I'm sure it's not yours either. So I'm just like where else can we build that's like the next question I think like anybody.  You're like there's a good chance.  So yeah, they just stop there never try Yeah, imagine if like you didn't land your trick right away. You just stopped. Yeah, that's pretty much All right, so Doughty and I are at this other little DIY spot not too far from where we were some cool stuff here It's like a really nice manual pad Really steep fun little quick quarter with some beer can coping check it out But yeah, it's an abandoned slab, I guess There might have been a restaurant or business here because there's a car park right here But yeah, we've got a bunch of parking curbs everywhere Whoo for this cool little manual pad There's a tiny little kicker over there. That's pretty fun But there's a lot of open space here, but yeah The difference being this is private property. So the city can't legally go in and tear it out. So um, yeah I'm just gonna skate here for a bit get some skating clips try to make the most of the day Before the Sun goes down. I think we want to hit a couple other spots too. So yeah, that's where we're at right now


So, the next example that we will look at is a compressor. So, in the case of a compressor as you can imagine air is taken in at a steady rate, it is compressed. So, the pressure increases across the compressor. So, the input work is utilized to increase the enthalpy of the fluid ok, whereas in a turbine the enthalpy of the fluid is converted to output work whereas here the work that is put in is converted into to increase the enthalpy of the fluid and most importantly to increase the pressure of the fluid ok. So, the compressor usually works with the prescribed pressure ratio between the outlet and the inlet ok. Once again what we have illustrated what I have illustrated here is an axial flow compressor with similar moving blades and fixed blades and in fact, let us just take a look at the picture of a compressor here. So, here you can see the moving blades and you can see that the moving blades are more numerous than the turbine blades. There is a reason for this which will be discussed later on in a in a turbomissionary course. So, you can see the moving blades and you can see the height of the blades decreasing as we go along. So, the cross sectional area of the passage decreases as the fluid is compressed and its density increases. So, m dot is equal to rho times A times V at any at any cross section. Since the density increases and we wish to keep velocity almost constant because the idea here is to convert the input work into increase in density not increase in kinetic energy or not a change in kinetic energy. So, we want to keep this approximately constant which means we allow the area to decrease so that m dot is a constant at any cross section that is the basic idea in compressor design. Let us look at couple of worked examples involving compressor. So, air at 100 kPa 27 degree Celsius enters an insulated compressor. So, q dot is equal to 0 steadily at a rate of 1 meter cube per minute where it is compressed to a pressure of 1 mPa. So, in compression process to obey PV raise to gamma equal to constant determine the required compressor power k e and P e changes may be neglected. So, our control volume for analyzing the compressor is shown here. So, this is our control volume and if you go ahead and apply SFE and then simplify it for this problem with the q dot equal to 0 k e and P e changes neglected this is what we end up with. And since air is an ideal gas we may write this as m dot Cp T1 minus T2. Now, m dot itself has to be evaluated, we may evaluate m dot at the inlet. So, m dot is equal to rho times the volume flow rate. So, what we are saying here is m dot is equal to the density at the inlet times the volume flow rate at the inlet. This is the value of mass flow rate at inlet it will also be equal to the mass flow rate at the exit because it is the compressor is operating at steady state. However, this expression if you apply it at the outlet the density and the volume flow rate may be different because the thermodynamic state is different. So, we calculate m1 dot using this expression and we say that that is equal to m dot through the compressor which is correct. So, we are given the volume flow rate to be 1 meter cube per minute and we convert that to meter cube per second divided by 60 and we can evaluate the mass flow rate through the compressor as 0.01924. Since the process obeys PV raise to gamma equal to constant we may evaluate T2 as 579 Kelvin. So, if you substitute these values in the expression that we had before m dot cp times T1 minus T2 we get this to be 5423 watts. Notice that this expression technically will yield a negative number, but we say that the compressor power is 5423 watts the understanding is compressor requires input power. So, the negative sign is dropped here with that understanding. The next example involves a refrigerant R134A. So, refrigerant is compressed steadily in a compressor from 140 kPa minus 10 degree Celsius to 700 kPa 120 degree Celsius. Volumetric flow rate at the inlet is 0.06 meter cube per minute. The heat loss to the ambient is 7 percent of the input power determined the power required ke and pe changes may be neglected. Again when volumetric flow rate is given you must bear in mind that mass flow rate at inlet and exit are the same because it is operating at steady state. But volumetric flow rate at inlet and exit will not be the same because the thermodynamic state is different. Densities are different. So, volumetric flow rate will be different. So, the volumetric flow rate at the inlet is given. So, we can calculate the mass flow rate at inlet by saying that it is equal to density times volumetric flow rate at inlet just like what we did in the previous problem. So, the control volume remains the same. So, we simplify the steady flow energy equation for this with ke and pe changes neglected q dot is actually given in this problem. So, we cannot neglect it. It is obvious from the given information 140 kPa minus 10 degree Celsius since pressure and temperature are given it is going to be superheated or compressed liquid. In this case we can easily show that it is a superheated state and get h2 from the superheated refrigerant table both at the inlet and the exit. The mass flow rate m dot may be evaluated as so, this is equal to rho 1 times v1 dot or if I write it in terms of a specific volume I may write this as v1 dot over v1. So, the specific volume at inlet has been retrieved from the refrigerant table. So, we can evaluate the mass flow rate to be 0.4107 kilogram per minute. Now, you have to be a little bit careful with the numbers and the signs for q dot and w x dot in this case remember w x dot is negative because it is a compressor we have to supply power to the compressor q dot is also negative in this case because there is heat lost to the ambient. So, we have explicitly taken this into account by writing in the q dot term as minus 0.07 times w x dot because that is 7 percent of the input power and this has also been taken into account by changing this to a plus sign and using an absolute sign around w x dot. So, once I do all this I get the absolute value of w x dot to be 49.68 kilo joule per minute. So, the compressor input power is 49.68 kilo joule per minute. So, you have to be careful with this equation when you actually plug in the numbers. Saturated liquid water at 45 degree Celsius is pumped to 20 bar pressure steadily by a pump. So, we neglect heat losses so, q dot equal to 0 K e and P e changes determine the power required per unit mass flow rate. So, for this particular problem so, here is the diagram of a pump. So, our control volume in this case would look like this. So, this is our control volume for analyzing just the pump. Power is supplied water comes in at a certain thermodynamic state it leaves at a higher pressure. So, that is and we are asked to calculate the power required per unit mass flow rate. So, if you apply S of e to this control volume, q dot may be taken to be 0, K e and P e changes are neglected. So, the equation reduces to something like this.  Since there is no substantial change in the temperature of the liquid as it goes through the pump u 1 is approximately taken to be equal to u 2. Furthermore, since it is a liquid it is practically incompressible. So, we may also take v 2 equal to v 1. So, with this simplifications the above equation reduces to W x dot equal to m dot v 1 times p 1 minus p 2. And v 1 is v f at 45 degree Celsius because it is given that the liquid water is saturated at 45 degree Celsius. So, v 1 is equal to v f at 45 degree Celsius and p 1 is equal to p sat at 45 degree Celsius because it is a saturated liquid that is also given. So, we now substitute these values into this expression and we get the pump power to be 2010.11 watt per unit mass flow rate. What is how we have used the fact that it is an incompressible liquid to simplify this expression and write it in terms of pressure rise while keeping the density constant. The next example is this. Here we are actually using a pump to lift water from sump to an overhead tank like this. So, basically the power that we are putting in is being converted to potential energy of the water as it is lifted from the sump at the ground level to an overhead tank. So, the control volume for this would look something like this. So, this is the control volume for this problem. All the other information that is required for the analysis is given and we are asked to determine the power required to run the pump. So, for the control volume shown, we take q dot equal to 0 and v 2 equal to v 1 since the pipe diameters are the same. We also assume that there is no significant change in the temperature of the liquid as it goes through the pump. Hence, u 2 is equal to u 1. So, with this simplifications, the steady flow energy equation assumes this form. And if you look at the illustration, since this water flows out like this, it is at atmospheric pressure. So, p 2 is equal to p atmosphere. But p 1 is equal to p atmosphere plus the height of the liquid column above the pump inlet. So, p 1 is equal to p atmosphere plus rho g times 2 meters which is the height of the water column above the pipe inlet, I am sorry, above the pipe inlet. So, with m dot equal to rho times v dot, where v dot is the volume flow rate, we may evaluate the pump power to be 2.06 kilowatts. So, one of the most interesting things that this example and the previous example illustrate is how the same equation takes on different forms depending upon the control volume that we have. So, for the control volume in blue, we ended up with something like this that there was no potential energy change. Whereas, for the control volume in red, there is a potential energy change and that is because of the power that we are putting in. In the case of the control volume with blue, the power that was being put in is converted to increase in enthalpy of the fluid. Whereas, with the red control volume, the input power is being used to increase the potential energy of the water as it is lifted from state point 1 to state point 2. The last example that we are going to look at under the category of steady flow devices is the heat exchanger. Now, functionally, heat exchanger is almost the same as the mixing chamber that we saw earlier in the sense that both of them accomplish transfer of enthalpy from one stream to another. However, in the case of a heat exchanger, the transfer takes place without physical mixing. Now, in the case of the mixing chamber, because the transfer takes place by virtue of mixing, the device operates at a constant pressure. So, the incoming streams and the outgoing stream or streams all have to be at the same pressure. Now, in the case of the heat exchanger, because there is no physical mixing of the streams, the streams can be at different pressure. So, basically the stream to be heated or cooled flows in a separate line and the stream that is heating or cooling flows in a separate line. So, they can be at different pressures. And the fluids are brought into contact indirectly and heat transfer takes place from one stream to the another or enthalpy transfer takes place from one stream to another. Now, heat exchangers are used extensively either for heating a stream or for cooling a stream. So, the example that we are going to look at next actually involves cooling a stream, but it can be used for heating a stream as well as cooling a stream. Now, many different types of heat exchangers or in use in practical applications probably three of the most commonly seen would be the parallel flow heat exchanger, where the two streams flow in the same direction. So, this is the heated stream or cooled stream and this is the heating stream or cooled stream and they both flow in the same direction. Now, in the case of counter flow heat exchangers, the streams move in opposite directions. So, for instance, in a parallel flow heat exchanger, one stream moves in this direction and the other stream also moves in the same direction. So, this is a parallel flow heat exchanger. Now, in the case of a counter flow heat exchanger, one stream moves in this direction and the other stream moves in this direction. So, they move in opposite direction. So, this is the counter flow heat exchanger. Now, the cross flow heat exchanger as the name suggests is a device where the streams flow in perpendicular direction feature. So, one stream flows like this and the other stream flows like this, but without mixing. So, they move in perpendicular direction. So, this is the cross flow heat exchanger. The example that we are going to look at next is a counter flow heat exchanger. So, let us see what this does. So, basically what we are going to do here is we are going to accomplish the compression that we saw in one of the earlier examples. Let us go back and look at that. So, you may recall that we looked at a compressor where we had air that comes in. So, air at 100 kPa 27 degree Celsius enters the compressor and it leaves at the pressure of 1 MPa and the compressor is insulated. So, we looked at the steady flow operation of this compressor. So, what we are going to do in this example is to accomplish this compression in two stages. So, rather than going from 100 kPa to 1 MPa in one go, we will first compress the air from 100 kPa to 316.22 kPa in the first stage and then compress from 316.22 to 1000 kPa in the second stage. Now, in between the two stages, the air is taken to a heat exchanger where it is cooled down. So, it enters at a higher temperature after being compressed to 316.22 kPa. It is then cooled in the heat exchanger back to its original temperature, initial temperature of 27 degree Celsius before being sent to the next stage compressor. So, basically the arrangement would look something like this. So, this is the first stage compressor. So, this air is then taken into this. So, here the air comes in at 27 degree Celsius. So, it is at a higher temperature when it comes out, it is then taken to this one, the heat exchanger where it is cooled down to 27 degree Celsius. 27 degree Celsius, the air is then taken to the next stage compressor which looks like this. So, this we label as state 3 and the air leaves at 1 MPa as state 4. Now, water enters at 30 degree Celsius and after cooling the air, the water leaves at 60 degree Celsius. We are asked to calculate the total compression work and compare with the single stage compression work and also determine the required mass flow rate of water. So, the mass flow rate of water that is required to accomplish the set cooling which is to cool the air from whatever temperature it comes in to 27 degree Celsius which is the same as the inlet temperature here. So, we apply SFEE to the first stage compressor, we take u dot to be 0 because it is insulated, KEP changes are neglected. So, we end up with an expression like this and since air is an ideal gas, we can replace H as CP times T. So, we do this and the mass flow rate is the same as before. I remember that the volume flow rate at the inlet to the compressor was given and we calculated the mass flow rate from that. So, the mass flow rate remains the same as before at 0.01924 kg per second and we also know that the compression process obeys PV raise to gamma equal to constant. So, the temperature at the end of the first stage compression is 417 Kelvin. And we can compute the compressor power for the first stage by substituting the values. So, we know the exit, we know the exit temperature, we know the inlet temperature, this is 300 Kelvin, inlet temperature is 300 Kelvin. So, we can calculate the power for the first stage to be 227.45. Now, at the second stage, if you see the air enters at 27 degree Celsius, same as in the first stage and then it is compressed to a pressure of 1000 k part. Now, if you look at the pressure ratios p4 over p3 for the second stage and p2 over p1 for the first stage, you get the pressure ratio to be the same. You see that the pressure ratio is the same and the inlet temperature to the compressor is also the same, which means that the power required in the second stage compressor is also the same. So, the total power required is 4548.9 watts and this represents a 16 percent reduction in power when compared with single stage compression. Obviously, when you take the air out, cool it down before compressing it further, the power required will definitely go down and that is what we are seeing here. There is a 16 percent reduction in power when compared to single stage compression. This is why in many practical applications, this strategy is almost always employed. So, such a device is usually called, although we are calling it a heat exchanger in our example, such a device is usually called an intercooler to denote the fact that cooling is employed in between stages of compression. So, normally there is a two-stage compressor and we use intercooling in between the stages to reduce the compression work. So, we now apply SFE to the heat exchanger. So, this is the control volume that we are using. We assume that there is no heat loss from the heat exchanger as a whole. So, we have two streams, so two inlets and two outlets. So, we apply SFE to the heat exchanger, q dot is 0, wx dot is 0, no significant ke and pe changes. So, we get 0 equal to m dot water times this, hx and hy are the enthalpies of the water at the inlet and exit through the heat exchanger. Since pressure, since information relating to the pressure of the water is not given, we simply assume hx equal to hf at 30 degree Celsius and that are the saturated liquid and hy equal to hf at 60 degree Celsius. If the pressure is given, we can probably fix the thermodynamic state more precisely and evaluate hx and hy but probably not required for this problem. We will simply take hx and hy to be the enthalpy of the saturated liquid at the corresponding temperature and if you substitute the numerical values, we get the mass flow rate of water to be 0.0181 kilogram per second. So, one important aspect that you should notice is that we are saying q dot from the heat exchanger is 0. However, there is enthalpy transfer that is taking place between the streams. So, there is enthalpy transfer that is taking place between the streams. So, that is always there but our control volume encloses both the streams. So, the q dot that we say is the q dot that we are saying is equal to 0 is actually heat loss from the heat exchanger to the surroundings. So, whatever happens inside the control volume is accounted for through the enthalpy change in the two streams. So, this completes the examples that we have been looking at on steady flow analysis. What we will do next is look at unsteady flow analysis.


Hi, and welcome to the session. Let's work out the following question. The question says solve the differential equation dy by dx plus 1 upon x into y is equal to y cube Let us start with the solution to this question here dy by dx Plus 1 upon x into y is equal to y q is a differential equation We divide both the sides by y cube and we get 1 upon y cube into dy by dx Plus 1 upon x into 1 upon y square is equal to 1 we put 1 upon y square to be equal to t this implies Minus 2 by y cube into dy by dx is Equal to dt by dx this we get by differentiating both the sides with respect to x this implies minus 1 upon 2 dt by dx Plus 1 upon x Into t is equal to 1 this we get from this equation because we see that 1 upon y cube into dy by dx is equal to minus 1 by 2 dt by dx Plus 1 upon x into t is equal to 1 this implies that dt by dx minus 2 upon x Into t is equal to minus 2 this we get by multiplying throughout by minus 2 This is a linear differential equation Here where p is minus 2 by x q is minus 2 therefore integrating factor is e raised to power integral p dx That is equal to e raised to power integral minus 2 upon x dx which is equal to e raised to power minus 2 log x Which is equal to e raised to power? log x raised to power minus 2 that is equal to x raised to power minus 2 or 1 upon x square so the required solution will be t into integrating factor is equal to integral minus 2 into integrating factor dx plus the constant This implies that t into 1 upon x square is equal to integral minus 2 into 1 upon x square dx Plus the constant C which is equal to 2 upon x plus C Now we put back the value of t is 1 by y square so we have 1 upon x square y square is equal to Now we see that this is same as integral Sorry, here we have This to be equal to 2 upon x plus the constant C hence 2 x square or 2 y square plus C x square y square is equal to 1 This we get by multiplying throughout by x square y square So this is our answer to this question. I hope that you understood the solution and enjoyed the session. Have a good day


this fuzzy logic flask application. I did provide the HTML JavaScript and Python code above. I'm still getting errors. This seemed to be a mismatch between what the backend is provided and what JavaScript is expecting in terms of the names of the different components. This potentially can be untraced by co-pilot suggesting to use traceback. It's probably not a bad idea. Traceback and Analyze. Try assisting code. When we have the exception, give us the traceback. I have to select all code so it's actually sends it to the co-pilot. It's almost 300 lines of code. Fasty-fast, no specific suggestions suggesting to normalize. Wouldn't this normalize the y-axis, whereas the problem is the x-axis, the length of the array. The membership function should be defined over the same range as the input signal to print and for aggregated it's just aggregated. Actually co-pilot. Okay, while they float, once you shift C, select the whole code for the right test problem. Okay, that is just one. Obviously just resizing would not work. It's better than what we currently have. Sexually working. That must be fake. This is changing at all if this Android is not moving. No, this is not right. It's just not right. Otherwise, I'm sure the zero should be all orange. This is the updated Python code. It doesn't work as expected and it doesn't produce any errors. There are a couple of things. First of all, we would like to limit the number of digits after the decimal point 2, 3. Also, we'd like to know why the rule square array has all the same numbers. Round 2, 3 digits. F low, wrong. That shouldn't take a signal, noisy signal. The name signal is potentially incorrect. Yes, that should be a local variable. A variable called signal. Is it potentially conflicting throughout the code in terms of its name? Not matter, but we just want to change the names for consistency sake. Stop something up. It's really easy to blame the port for it, but the noise of the shoot doesn't go to zero. What we would like for right to go to zero is that used to. Interestingly, the detection error is 99%. How is it 99%? So that the gpt has the updated code. First of all, how to make sure all the variables in the code are rounded to three decimal places. Round and empty round. Can I just round everything by default instead of rounding each variable individually? Round all variables to three decimal points. It's actually doing it changed 10 lines. It keeps counting the number of lines that changed that 9, 10, 9. What? Oh, CPU really going. Okay, made a lot of changes. No, no. Did it add a function? I'm a bit confused. And when they are generated, generated, initiated, could be different things. I just need to delete all that. Yeah, it's going places. It's not right as aggregated. Symmetry universe was in waveform and MF low. Converting it to least. Where are we printing? Posit dot. This values a control I limit to three decimal points. Why doing it differently? Does it matter how many lines I select or what? Right. Yeah, obviously, that build is wrong because it's just probably padding with zeros or something. Yeah, this is a mystery why sometimes generating. Okay, I don't mind which one isn't just adding extra lines. That's okay. Here at first. Okay, this is better. Where are we printing noisy, wafer? We're rounding it down. Okay, might have to continue next time. See you in a bit. Bye.