 So, yes, the title sounds a bit strange. And I truly will lecture on linear forms and logs. So Bakel's theory and focus on applications. And maybe an emblematic application is a question of rational approximation to as a right number. So, for instance, we call that if you have a real number sigh. Say, irrational. Then, you know, by the theory of continued traction that there are infinitely many rational X of a Y. You're a bit too far up. So maybe move the piece of your paper down by one or two centimeters. I'm a bit too far up. Nice. Maybe one more line. One line, one extra line feet. I still can't see the top line. I can see half of it now about. It's very determined because. It's good. And my screen is fine. Okay. So, okay, maybe it's just an issue with my screen. So, which I stole says it's okay for him to so please continue. Sorry. So. Then there are infinitely many rational X of a Y. So, what happens is that. Close to side at other tools, such that side, minus X of a Y is at most one, one of a square. So, what happens with. There is a famous theorem of you will think that if XI is algebraic of degree B, then there exists C of XI such that. Simon X of a Y is at least several C of XI divided by Y to the D. So, it's. It's a Y, so C of XI is positive. So, you rush an algebraic number of degree D cannot be approximable. At an order greater than D, then it's the degree value rationality. So, we know more from words of two or dies and Z, etc. communicating with the result of rough, which says that for every epsilon positive, there exists C side epsilon positive such that Simon X of a Y is at least two times epsilon of a Y to the epsilon. So, but the disadvantage in rough result is that this C side epsilon is ineffective, the sense that we cannot compute it. So, no formula in terms of sign it's silent for it. Why do you feel the result, we have a clean value explicit value for several times. So the question important questions when these that is three. Is it possible to improve the result. Or say, maybe to get something smaller than D in the denominator here in the exponent, but still with some effectively computable number at the numerator. And this is question. I mean, under this form was answered. The right man is 71. And proof that there exists. How positive effectively computable to XI C of XI such that Simon, if it's a Y is at least C of XI, why does a D minus tau for all these other ones. So tau is positive. So you have an effective improvement upon Louverry's inequality. And that we are very very far from rough result power is very small. So the ingredient of the proof is Baker theory of linear something. So I will now give a crash course on this. And this theory. So the goal is to get a lower bound for the distance between one on a product of powers of algebraic numbers. So alpha one alpha and algebraic numbers and the experiment be I be I integers. So many problem in different approximation differential equation. Can be reduced to expression of this form. That we can bound from above. And it's then we can say something. So if we can prove that this expression is non zero, and then you apply Baker theory to get a lower bound and combining the information with the upper bound, then we did use, for instance, finiteness of the number of solutions, sometimes I'm going to quit. Okay. So. Fine stands. In the case where the alpha rational number. Is trivial estimate, which is one of them. It's one. And to the be. I put, I put one of the denominator. I put one of the denominator. X I X I is the maximum of. I Y I on capital B is the maximum of the. So this is really the trivial estimates will give nothing interesting in the applications I have in mind. And you see it can be rewritten as log of lambda. I put some of log. Times. Well, I will put to be if all the via one. And. Sorry. And what gives a better theory. It's precisely the improvement I wrote. So at least minus your fan source. Effective. So explicit. The product not to some product of the log of. And times. So what is crucial for applications is to get here. Something which is. So in most of the applications, it's enough when you get something. If you get something. Which is little off capital B. Okay. So method. Give some. That's possible. So I will. Write this more generally with. As you write numbers. And we use. The very high. And. It's convenient to use modified. Version like this. So for instance, you get that age of a rational number. Is not max. It is written in under its reduced form. For what. Make a theory. And I mean classical form. Is an inequality of this type. So. H star. Times. Where. Is the degree of the number field. Generated by. And C and D is an explicit constant. This is not my point. My point is. This capital B here. And. Try now to explain to you that. It's, it's not enough. So if you. If you look. If you are interested in. Improvement upon. So. With. Simonus. Why. What. The plan is to. Bound from above the size of solutions of two. Equations. So. I have. I have a. Side one. Side to. Gala conjugates. Consider this. Equation. So for your homogeneous. For you on your. I want to. You take. An integral. And you consider this equation. And. For the maximum of x and y. In terms of, for this theory, the better theory is extremely suitable. And there's a connection between the first problem on this one is that can check that psi minus x over y is essentially m over x to z d. Okay, because say be very brief. So if so, and it's fixed. And it's a wire large. So here you have a product of D terms. One will be very small so psi will be very close to each other why. And the X over why we cannot be close to the to any of the other Galois conjugates of sign. So it's minus side to why we'll be roughly speaking capital X to some constant, depending on. On the distance between sign and it's Galois conjugates. So you end up with this. And if you manage to prove that it is at most, and to some power C, then you get that psi minus X, otherwise, is at least one of x d minus one of our C. Some constant. So the problem is to bound x from above in terms of. So when you do some algebraic number theory. You end up with the linear forms, which is of this form. This quantity where the alpha one up to alpha and are well, well defined is essentially a unit suitable number of it. I'll find plus one is so control so it's eight is expressed in terms of M. And we can prove that this expression is very small, not zero. And when you use the theory, they have three to do to bound it from from below. If you do this with the estimate I gave you here. So then you get, and one gets that X is bounded by M log log M, which is good, but it gives obviously an improvement of you will see the quality. So you can then deduce that Simon speaks of a Y is at least one of a Y d minus C by log log Y. You have to trust me. But this is not as clean as I have asked a clean result. So one of them I gave. So we are not very far but we need to remove this exponent log log M to replace it by C. There is some point is that there's one information that we haven't used that here. I said that this number alpha one and two alpha N are known control on this one as a large eight depending on M. It's it's exponent is very small it's one. And in this particular case, we have an improvement of the theory. This has been also noticed by Baker 72 that if BN is one, then get the logarithm of the same quantity. And here you have what I like to call be prime is be divided by H star of alpha. So here you have a game. So be. The application B is large. H of alpha is large true so this be prime is much smaller than B. So this lower one is much better than this one. So as a result, as a result, we get an appointment like this, instead of X. Sorry, it's me again. So yeah, great thanks. I moved sorry. So, if you say differently if you, if you consider irrationality exponent. So what you will give is this result rough if this result. But I call it a rationality exponent. It's just the experiment of why. So which measures the order of approximation by rationality. And with this upper bound cannot deduce anything better than this. But with this one can deduce the point in the, in the secret of the tool. And we're trying to convince you that that this situation occurs extremely frequently for a variety of problems. And that we have essentially three results. The trivial one obtained by you will see an equality or whatever. Trivial or easy one. And the best possible one using rough there and also Schmidt substrate. So, so the best possible one is ineffective. And the easy one is effective. And by the use of a bigger theory, one can effectively improve the easy one by some small amount, small quantity, and the key for our work to get this improvement. So in terms of the exponent, some exponent of approximation is precisely this. We find meant here. Be called the crime. So I will now describe one, another problem with some approximation will see that it works essentially in a similar way. And it will be on the functional parts of power of, of algebraic numbers. So here you, I will, I will take three hours to see. So double bombing distance to the nearest integer. So here, you have some easy lower bound, one of the other denominator. And this has been improved by by Mahler 57. And large enough, this is, this is at least two to the power minus epsilon M. And for any epsilon body, you have this result, and it's a consequence of write out theorem, so p. The question is, can you get some effective improvement. Upon the classical, so to your result here. And the answer is yes, it was done by Baker and course. Yeah, yeah, there's a, there was a question in the chat by Dimitri Vazian. Dimitri maybe you can just ask directly was about the dependency of the constant in terms of the height and the degree of the algebraic number. How small is C of algebraic number in terms of the height and the degree. Guess it's in there for which for which question from for. It wasn't in the previous slide, you said that you get the original exponent of D minus C of zeta. And I'm wondering how small she was it is very, very small. Right. It's very small. So that point. I could upset that too. So, you mean this slide. Yes, yes. Yes. This is very small. But this is a generic result valid for all. So parties in certain cases. There are other methods which give something effective and effective in between two and D. But for specific number for, for instance, if you take integral roots of rational numbers. So these root of A over B, then you can, you can improve this effectively. And even you can get as close as possible to two. But what about the C at the top of the page. Sorry. What about the C at X is less than M to the C. It's very big. It's one of our town. Okay, so it's always bigger than log log M in practice. In practice. So better on codes. Improving effectively this and the proof that there is this town such that three over two to the end module of one is at least one month. And what I put this like this, so tau is positive. And it's effective. It's also very small. In this case you can put 0.99999999999999999. That's been proof. Really? Small. Using theory of linear functions. And this is the Archimedean, the non-Archimedean theory. So what I gave previously was a lower bound for the distance between one and a product of powers of algebraic numbers for the Archimedean absolute value. But you can also consider a periodic absolute value. And I will try to explain you rapidly, very rapidly this. So I write a n being the nearest integral to 3 over 2 to the n. And then you define mn, which is 3n minus a n times 2 to the n. And you consider the 20th variation of 3n minus mn, which is clearly at least n, because 3n minus mn is divided by 2 to the n. So the 20th v2 of something is the highest power of 2 dividing by number. And here, with Baker theory, we have also an upper bound, I mean for the 20th variation, which is a lower bound for the 20th absolute value, of a similar thing. So I take h, which should be h star of 3, h star of mn, and log of n, the exponent, divided by h star of mn. So mn and 3 are integers, so it's simply the log. So if you do this, then you end up that n is bounded from above by some constant, depending on log mn. And if you do things correctly after that, then you get a result like this. It's just to illustrate on an example. So the fact that here in this number is a constant, so you move h star of mn, so you get that n divided by h star of mn is less than some constant times the logarithm of n divided by h star of mn. So you get something, it implies this. So if you don't have this denominator here, you get only, you have an extra log log mn, and not a clean result like this. It's exactly the same thing, but the application here is much simpler. So when you see this result with 3 half, the natural question is what happens if an algebraic number of psi, if an algebraic number of psi, can you get a similar result? And this is with a nu. So what about, so you take psi and algebraic number. What can be said on the distance of the nearest integer of psi to the n? So the problem that, for instance, can you prove the same thing, so is it true that psi to the n is always at least 2 to the power minus epsilon n when n is sufficiently large and the number of psi is true for 3 to the n. Well, of course, you have to assume that psi is not an integer, but that's not the only restriction. You have to assume that psi is not pizzo. We call that pizzo number as an algebraic integer, which is real greater than 1. So if whose conjugates are in absolute value strictly less than 1, so you have to assume this. But you have psi to the n, so you have also to assume that psi to the h is not pizzo for an arbitrary integer h. So if psi to the h is not pizzo for every integer, integer h is 1, then psi to the n is greater than this, private, and this was proved by Kovaya and Tzanyi in 2004 by means of the Schmitt's substrate theorem. So the question then is whether you get some, you can get some effective result, trigger one first and whether you can improve it in some way. And actually it's not too difficult to prove that you will type a flow of bound, which is this one. And for some C of psi, so for some, I will not write it explicitly. It depends on the conjugates of psi and the leading coefficient of the minimum of defining polynomial of psi. But this is really, this is really elementary, rather an expression of C of psi. On the point is that also as long as psi is not a quadratic pizzo unit, then you can improve this that there exists tau, possibly such that psi to the n is at least C psi minus 1 minus tau and for every n large enough. And this is explicit. So this is another example where you have best possible result with Schmitt's substrate theorem. You will type easy result and some small effective improvement using Baker theory. Here you need to use as an actually an entry and also a non actually get to get this. It's not, it's not really difficult. And so again, the key for this is this B prime. So there are other, I will not give all the examples of the experiment of approximation. It works for all of them works like this. And now I will give a last application to another question, which is a question as part of it. So the problem is a, is a following. So you have a set N which is an infinite subset of N. Yeah, so the top line again, great. Thank you. I don't not care for enough. So you take some infinite subset of N, it has its final set of crimes. And as is the largest divisor of N. Composed. From us. And the question is. So. And. So for instance, say, consider the prime students free. And so said. And composed of integers of the firm and times N plus one. Can I pound. So the greatest divisor of N and plus one composed only of two three five. So, of course, here also you have an easy bound. Which is N and plus one. Good. It's unlikely that N and N plus one have only the divisors 235, but why not. It's not. Obvious. Yeah. So what is interesting is that. That. More generally input with us. So. Right. Enter this form. One half. So it's for every positive. Yes. For every N large enough. And then you have to use rough terrain to produce. You want to have you cannot do better because as you can, you can always take a power of two for. So if you have two in your set S, you take four S for N, a power of two. So here the experiment here is one of the degree of the polynomial that you get here. So the question is to have some effective improvement upon this obvious bound. Yes, you can. So there is this tau positive such that N and plus one. Yes. Yes. And then plus one. For every N large enough. And this is. Think. I can schedule a proof. Very quickly. So it means that. And so this is much more general. So you can replace this by. You have to be careful. So the point is that to prove this, you consider the set. And one S of crime members. And you write. And is some. And one new one. And plus one is some. Say not a one. Yes. And then what you do is. And plus one minus N. And plus one over N. Is very close to one. So this is, you will recognize your linear form. You have your linear form here. So it's going to be is a V. I, or minus UI. So you apply. So the result. I gave to. To bound lambda here you don't know. The eight of this element L one L S are fixed. So you're. Here's a smallness of your lambda. So the logarithms of log N. Here numerical constant depending of this. Then you have the eight of a one. A not. The log of the maximum of the exponent. So I would say max UI. So this maximum is nothing more than essentially log N. So again, you get that. Log N. Bonded from above. That's this. So he is. If you prefer it takes a log max. And not a one. So that ice. One of them. Is at least N to some power. So I saw a not or a one or both. Is N to some power. So you have a saving because the S part of N. Plus one. Is at least N. To the bar one minus. Oh. It's N divided by a not. It's at least N minus. N to the power of one minus. To have this. This. Is one minus. It's again. Because of. Okay. As I will stop here because my time is over. And I've just put on the archive. This morning. Survey paper on this subject. So someone's interested. It should appear tomorrow. On the. Thank you. Thank you.