La regle des signes de Descartes

0:00 The problem is, you mentioned amongst the rules, relatively, reduction. Reduction. So, something has to be reduced to a simpler form, and then you go back in order to get to more complex structures. And what I lack here, in order to understand, is memory. So, there must be an interplay between imagination and intellect and memory, otherwise the whole thing will not work. So, I would like you to talk a little bit on that. And then, just a very short remark, to state it as a presupposition for the whole project, that equations can be understood with full certainty. Your interpretation at the end was equations cannot be understood with full certainty, and I wonder if that is your conclusion, or if there is anything explicit in the regularity of the class. We mentioned memory, and there is another one, general senses, and they didn't survive my structuring of my talk yesterday, just because I was already too long. The original was that this model consists of dreams, silly imagination, and the infinite, that's four. And so the external senses provide information which is kept, which is stored in the imagination and is kept in the memory, as you can see here, which is also kind of screened, but it's built in the memory.

2:30 I mean, it's all like the same. Yes, here it is. Here. This is a confusion, eh? And so the answer is, it is there. It has been created. As we described, it's like taking the material way. Then, and the other one, you say, that is more the understanding of, that's a bit more complicated than I must think. I think, I interpret it that way. I say, yes, you can do that. But I can only support this if I assume that he means the equation, but not the solution of the equation. And if you only see the equation, then that is really built up with addition, multiplication, and division, etc., but not root extraction, because in the equation itself, there are only directional equations. Also, his whole model of understanding allows that you understand the situation where still some of the elements are unknown. So you can't understand an equation, you can't understand what relation is described in the equation, it can't solve it. When I say the process failed, it was that it couldn't solve in the same, within the same criteria, because then you need good expression. So, when the problem was answered for 42 structures you would expect, it would start solving.

5:00 But thank you because that was surprising. In what you said this morning, we realize that it is by the introduction of the symbolic literature that there is a transition between the manipulation of objects or mathematical beings and the way in which they would like to structure their thoughts. That is to say, abandon common sense and look for a method to coordinate, to shape their thinking. And there is, in the French language, a verb that I still don't know, which is the verb mathématiser, as we would say in English, and I... Personally, I can imagine that the verb ''mathematiser'' becomes a verb that sums up this research of the mathematicians of the 17th century, if you want to bring it to the context, for the reflections that are not a form of physics, but rather of the research of the mathematicians. In some ways, I don't understand what you're talking about. It's in the direction of the model that I'm currently working on. I'm working on this model. I'm working on this model. I'm working on this model.

7:30 I'm working on this model. I'm working on this model. I'm working on this model. I'm working on this model. I'm working on this model. I'm working on this model. This is truly a test of France, and we are very happy to know that the most essential of these subjects is the study of mathematics, of mathematics, of physics, of geometry, of mathematics, of mathematics, of mathematics, of mathematics, of mathematics, of mathematics, of mathematics, of mathematics, of mathematics, of mathematics, of mathematics, of mathematics, of mathematics, of mathematics, of mathematics, of mathematics, of mathematics, of mathematics, of mathematics, of mathematics, of mathematics, The subject of today's lecture is the subject of today's lecture is the subject of today's lecture is the subject of It is the alchemist who has misheard, who has misheard the theory that we are doing. The alchemist, in Camden, in London, we call him the alchemist of stability. It is the alchemist who has misheard. The alchemist has taken a system of impurity. If we look back at the past, the scientific evolution of science was in the case of mathematics and the evolution of science is in the case of mathematics and the evolution of science is in the case of mathematics and the evolution of science is in the case of mathematics and the evolution of science is in I think that there are other questions to be answered. It is a question of knowing the future of the French language. It is a model for the French language. There is an emphasis on it. I think it is a little early in the morning to ask this question.

10:00 It is a question of organization. It is a question of organization. I apologize, I will not ask it in French. I understood your question. We have already considered two points of the expression on imagination by Descartes, but I will give you an example. On the one hand, apparently there is no incompatibility between the imagination and the implementation of algebraic procedures in the regulars, or the implementation of a theory of equations, a theory of proportions. Normally, we consider that algebra is a mathematical system that Descartes made up. To exclude, to minimize the role of the imagination within mathematics. You have just shown us that not only is there no incompatibility, but that the imagination plays an important role in the implementation of these processes within the rules. In this sense, what you are saying is not exactly orthodox. It is shared by mathematicians. There are two questions that are more historical than the other two, but I'll wait for my turn to answer them. Well, gives the secret something else to concentrate on, because the secret is not all false. And I think that's certainly what he says. In essence, when he's talking about his memory and about finding the simple parts and the complicated parts, he is very much in favor that when you are exploring your problem, you have seen that when you go this line in the scheme, You can then get from there to there, but it's complicated, but it can be done.

12:30 Then what you do is you put it in mathematics and then you need it to pick it up. Whether that is symbolism or whatever, it doesn't matter very much. But the idea that you can help yourself by getting a lot of things down in a place where you can go and understand this very well. That doesn't mean that it's so wrong, that the imagination is so wrong. You still need it to have a certain deal of the basics. So there is no contradiction, it's true. He knows very well that you can't have everything in your conscious mind, and so it's a very good thing to have in this case, in the pilot. I go on with the metaphor, but I put it in the pilot, we call it up. But that wouldn't solve any sort of question. So I suppose he's very aware of the economics of thinking, where post-civilism, so that would be my, I don't see the contradiction also, I think, when the historical picture is pictured like that. I say algebra is nice because it allows for consulting, right? That is a little bit peculiar. Do you think this is important? This coexistence between algebra and objective or imaginative dimensions? If it goes back to regularity, we know your work in which you point out this co-existence within geometry. It's the form of a tension, the algebraic force, the susceptibility and the geometric force. Now, if you reverse this co-existence to regularity... I think that this forces us to question the idea that there would have been a César of the Carthaginian French in the mid-30s that geometry would have been a witness, that it would have changed its position towards mathematics by eliminating imagination.

15:00 However, this is already shown in the regulars in the 1920s. Imagination does not exist with the process of generalization. We have to reconsider this chronology. The imagination is no longer present. I wouldn't know where I could find something so extremely formulated. There was simply this issue, because you could see that the physicist had the understanding of solving equations. And he couldn't go further than that on the basis of his view of the origination which he had developed in the ring room. And what you can see in surgery is that, as it were, they're quite split in, they're quite mathematicians and they're quite mythologists. And the rings are less extreme positions about the possibility of a risk. To my mind, this is an understandable academic experience in trying to formulate the book. But a lot of it remains, and what remains certainly is that he wants to have his mathematics certain, and that his certainty is not logical certainty, and it reverses that. And it's actually all of the geometry which says that the combination of motions,

17:30 The essential thing for it is that these curves are so connected that you can see how the one motion guides the other. There is no other way to say that when you look at these G's. You can see them. That's also how you explain it. No, no, I would like to add something to the subject of the seminar. We haven't talked about it yet, but we will talk about it later. The subject of the seminar is the subject of the seminar. We are not going to talk about the subject of the seminar. With the second lecture with Wegman in 1628 and 1629, the general construction and the submission of the equations of the 3rd and 4th centuries, this construction of the 3rd century indicates a conflict between the 3rd century and the 4th century and a conflict between the 3rd century and the 4th century and the 4th century and the 4th century and the 4th century and In which, in the article, it is clear that at the time of the order of the equations, it was not possible to find a problem with the multiplication of the cube. It is to say that the equivalent of the equation 3 to the value of 3 to the value of 0, which is not the equation L, is complicated. It is complicated to consider that it is dived into the equation 3.

20:00 Thank you for your attention. The lectures have given rise to many discussions and conversations that have interested both those who preceded Descartes on the subject and especially those who follow him. We will see in particular that Descartes is still an important classifier. We will see, of course, the work he does. It is essential to know how to respond to such things. Speakers include, on page 373 of the edition of the possessor in the head of Geometry, here it is. So, we also know how many real signs there are and how many false signs there are. We know that there are as many false signs as there are more or less signs. Of course, the terms may have positive and negative roots. And we see that this rule follows an example, as in the last one. The last one is the equation that has served him so far as to expose him to different results and to eliminate them. The equation is the following, which is the following equation. Plus 4, minus 4, minus 3, minus 19x squared, plus 106x, minus 120, equals 0. So how did he get it? Well, he got it by progressive multiplication of x-2, x-3, x-4. So obviously this equation has three true roots. They are 2, 3 and 4, and a false root is 25. And he explains it as follows, that after 0.64, there is 0.63, which is a sign, after 0.10, 0.19, 0.14, there is a plus.

22:30 So, after 106x, there are 120x, which are two other terms in these numbers that are not really enough, and a break due to the fact that the two signs less than or equal to the x3 and the x2 square are not enough. That's all Descartes left in his book on the rule of sign. There will be a little comment later in a letter, in a correspondence with Pierre de Cartel on this subject, where he will give some more precision. But there will be no demonstration, no explanation, we don't know how he established it. The only thing, apparently, is the 2 of these at the beginning, which could lead us to think that at least he deduced the result that was passing. What is the result that was passing? It is the decomposition of the polynomials for a factor of the first object. It is the result that consists of saying that if a polynomial has a as a root, it is divisible by x minus 1. So we can think that he deleted this rule, but he left no other indication on this subject. So, simply, on this rule, we have to make two or three comments that seem important. First, we notice that the rule is made up of two sentences that are quite symmetrical, one compared to the other. That is, the first sentence concerns the root, namely the fact that there are not as many true signs as there are negative ones, and the second, which concerns the negative root, so many causes that there are as many signs as there are negative ones. The fact of having two statements that are at the same time generous, that is to say that they do not depend at all on a single equation, and that present in a perfectly symmetrical way the positive and negative roots is something that is quite remarkable at a time when the negative numbers are widely accepted by the authors within the group. So, this symmetrical formulation will give you the opportunity to see everything. So, we're going to take a look at what's going on in the media, because among the different authors written on the subject, we find a few localizers of these cases, which are Cargan and Harriot. Cargan and Harriot. So we're going to look at what's going on at Cargan.

25:00 Cardan accepts the negative roots, but the coefficients of the equations are all positive, which forces him to distinguish, for a given degree, as many forms as possible from the terms in the two members of the equation. This means that for the equation of a third degree, Cardan distinguishes 13 forms of terms. He will give some rules for the higher degree equations, but these equations are not really useful. It is essentially that... well, he will not give the four, he will give the resolution, the resolution, etc. But he focuses, in all his magnum, on the two degree equations at most equal to three. So, if we continue the... The fabrication of the forms for a superior equation of 2D2, we have 3 forms for the equation of 2D2, 13 forms for the equation of 2D3, and if we count for the equation of 2D4, we get 45 forms. And for the equation of 2D5, 145 forms. The growth of the length of the exponential form, which makes it extremely difficult to establish a general equation for the equation of 2D2. Cardans are categorized and decided because they give, for a given form, the number of positive and negative roots. Here is an equation of cardans. I took Descartes's equation and rewrote it in the same way as different authors. I tried to respect the symbols used by Descartes. The only thing I didn't manage to do was the equal, which is in the other direction. I put it in modern language. We can see that the symbol of Descartes is very close to what we see today. And in the way of Cartesian, it would give this. 1, square root, square root. Plus, 106 positions of the boost, so 106x, around, so this is equal to 4 boosts, plus 19 boosts, plus 120, that's it.

27:30 So when we look at modern language, it looks like this. This is the symbol of the 4. I'm not going to go back to the 2-carbons, sorry, I'm not going to go back to the 2-carbons, but indeed the cultures are almost symbolic with the abbreviations. And what we notice is that they only take positive conditions, that is to say that the terms that are negative, that are negative, are rejected in the following way. Because they would obtain, for the negative equation of z to the power of 6, they would obtain 45 different conditions. The equation x3 is equal to x squared plus a certain number. He gives the number of roots of this equation. He says how many are positive and how many are negative. And he explicitly says that the negative roots of this equation are... So, we haven't taken into account... We haven't taken into account the people here. I think it's rather this one. The negative roots of this equation are the positive roots of the following equation. Which is true, indeed. So the transition from A to A corresponds to the transformation x gives point x. But nowhere in Cardan will we find a statement that will allow us to say, for a given degree equation, that there are so many positive and so many negative roots, by looking at the sequence of coefficients. The comparison of term-to-term equations, i.e. comparing the coefficient of degree 4 with the coefficient of degree 3, the coefficient of degree 3 with the coefficient of degree 2, does not appear anywhere. This comparison does not go without saying that the equivalent of a sign change or a sign permanence from one term to the next would be the fact that when we move from one term to the next, we change members or we return to the same member. This sign appears later. The only thing that appears is that. So, indeed, we can say that in terms of generality, the cardinal results are specific to a given form and can hardly be generalized to one of any kind and a given form, which is quite the case, however, for the given forms.

30:00 Then, Ariot. I would like to point out that what I just said concerns only the work that was published in a posthumous manner, which was published in 1631, ten years after the death of Ariot, after the manuscripts that had been left to Ariot, and we know that the manuscripts were transcribed in an extremely... Partially by those in charge of the unit, which means that the artist is a bit of a traitor of Ariote's writing. Recently, in the United States, the reality was to treat Ariote from manuscripts. It happened that there was a fairly important difference between the artist and the artist, and then this trait was restored. So what I'm going to say here is that the artist... So, Arion would write an equation that interests us in the following way. It is an unknown multiplication. a a a a, minus 4 a a a, minus 9 a a, plus 106 a, equal, he uses the equation of Robert de Korn. It's all the same. Ariadne, at the beginning of the treatise, presents the equation with all the terms of a member, from zero to zero. But then, when in the treatise we see the resolution of the equation, we find ourselves systematically with this form. Everything that includes an unknown term on one side equals the known, the completely known of the other. Indeed, on an example of this kind, we will have a sequence of coefficients, which, by the way, is not entirely acceptable, but we still have a problem here, because here we have a sequence of coefficients, and here we end up with, instead of a sequence of coefficients, a random number. So here too, to be able to see the line of signs in all its generality, there is a problem. In the different texts that talk about the sign rule, this rule will be attributed either to Descartes or to Harriot. It seems that the first to attribute the sign rule to Harriot is the Hellenists.

32:30 We discussed this with the Hellenist, so there are many questions about Harriot. She told me that in her opinion, it seems to her that this bad attribution comes from a bad reading of Harriot by the Hellenists. More precisely, a good reading of Wannis by Leibniz, Wannis by Andrade. This attribution will then be taken up by many authors, and in all the demonstrations that we will find on mathematics, we will find either the demonstration of Descartes's theorem, or the demonstration of Harriot's theorem, or Harriot's theorem and Descartes's theorem according to the authors. Mrs. Stedal was very clear. She said that neither in the artist, nor in the literature, nor in the manuscripts, is there any mention of something that looks like mathematics. And Wallace, who, however, has indicated a lot of things for Harriot, because this text is quite polemic, he said, to recover a certain degree of discovery for a certain identity, even Wallace's text does not reflect at all this rule for Harriot. Wally is simply content to say that we can deduce from the example of Harriot the reign of Descartes, but he takes it for himself, he does not take it for himself. So now I'm going to go back to my reign and then we're going to start following the path of the famous mathematician Descartes. We also know how much there is to know about the first phase. What can be said about it? Well, it will indeed make you want a lot of ink, because from the publication of a paper, several authors will say whether it is a general rule, whether it is particular, in which case it applies. What does the IPE mean? Is it a certainty? Is it a possibility? Descartes will clearly say in his paper that the IPE means that I consider it as a possibility and not as a certainty. That said, if we look at this line and if we take it straight to the point, we will see that this line is false.

35:00 What does the sequence of signs have? Well, it has two signs, one sign plus and one sign minus. In other words, if I count like Descartes, I count a change of sign and a change of sign and a permanent zero on both sides. So what does that mean? It means that this equation has, according to Descartes' rule, either a positive root or a negative root. There is a positive and a negative ratio, which are plus 1 and minus 1. So, obviously, this equation presents a particularity, it presents an absence of terms, of real terms. And that is indeed what is going to be an imprecision of Descartes, and I do not rule him out at any time in any way if we have to make terms. The only rule that Descartes gives concerning missing terms is that he says that in an equation, we can always get rid of missing terms by making a variable change, a small translation. He does it on the example x5-b equals 0. He says that here, there are missing several terms, and by making y equals x-a or x plus a... First, he transforms it into x6-bx equal to 0 and by changing here x by y plus a, I obtain an equation where all the places are not allowed to be entered. As small as the quantity a is, all the places are not allowed to be entered. By doing this, he does not justify it either. We realize that it is true, indeed, locally, provided that h is sufficiently small. We will obtain an equation in which all the cards are present, but it does not justify more than that. There is a continuity argument. Does this mean that the problem of missing terms was not a problem because we could solve it with a variable change of this kind? So, at the beginning of this case study, there was a critique that started to come in saying that this was not a general idea.

37:30 The first questions came from Robert Valle, and by the way, in the correspondence with Pierre de Carcabie that I showed you earlier, the cases will be as follows. I once again criticized Rohr-Werbal by saying that his second objection was a manifest falsehood. The second objection of Rohr-Werbal was that the rule was not general. And I said that when it could not work, it was when there were imaginary roots. So, indeed, the imaginary martians have posed you problems in this rule, as well as the absences of terms, and we will see how the two will be linked, and it will be this that will constitute the rest of the questions. So, the first to be interested in the problems of the missing terms, it will be Claude Ravel, in the commentary of 1730. He, too, will ask himself how to get rid of the children's terms. The situation that interests me is especially this one. When a term of the equation is evaluated, i.e. the terms of x squared, minus y, plus x squared, equals 0, one can assume more or less in the evaluative term as if the equation was more or less 0y. And in relation to this only term, we have to imagine a true and a false ratio when we compare a math 3 with a facet, and a true and a false ratio when we compare it to a math that follows it. So here is an example of a good idea but with an erroneous conclusion. That is to say that if we read the text, it means that if we look at the following sequence, we must have both a positive and a negative ratio, and if we look at the same sequence, we must have a positive and a negative ratio. These are two root terms, since we have two here and two there. So here, the interpretation is clearly wrong. However, the idea is actually quite good, because if I apply it to my example,

40:00 what happens if I assume that there is a coefficient that is written as more or less than zero? x squared minus one, I can write it as x squared plus or minus zero x minus one equals zero. And if I consider that this sign is fixed at plus, for example, I obtain a permanence, that is to say a negative root, and a variation, that is to say a positive root. So we find our two roots. If on the contrary I consider that it is a minus, I obtain here a change of sign, that is to say a positive root, and here a permanence, that is to say a negative root. So we find our two roots. This is the first experiment to generate the idea that we can fill the gaps by adding terms that are more or less zero. According to the authors, it will be more or less zero or, on the contrary, more or less infinitely small. It will come later, but it will be an infinitely small term of fixed signs. Since then, the time for demonstrations has come. I'm not going to go into the details of these demonstrations, because it's going to be more than 30 minutes long. I've just listed some of the works I've done in the last few years, all the demonstrations I've done here, which you can apparently find in the... In the period 1740-1830, the first is taken from the memory of the Academy of Sciences and is the work of Jean-Paul de Noël de Mal, and dates from 1740 and is published in the memory of the Academy of Sciences in 1740. The last one will be the work of Carl Friedrich Gauss. It dates from 1828 and was published in the journal of Kreml. Between the two, there will be a good ten of them, which I have sorted into two parts. I will quickly present the principles of these demonstrations and then I will focus more precisely on what... These authors are the ones I've mentioned, the first and the last, because they are the two least important ones that I've mentioned. These two authors have made missing terms in their demonstrations, and these missing terms will bother them until the end, which will result in a complete failure.

42:30 I have therefore arranged this into two demonstrations. What I have called algebraic demonstrations are demonstrations that will be based on the fact of examining What happens when, given a polynomial P of x, I multiply it by a binomial x plus or minus a. And the basis of these demonstrations will be to show that when I multiply P by x plus a, well, in x plus a, P, I get... An additional sign permanence in relation to P, and when I multiply by x-a P, I obtain an additional sign change in relation to the polybe of P. And as such, by recurrence, we will arrive at the fact that there is as much sign change than the positive ratio, and as much sign permanence than the negative ratio. This being, of course, in the case where the positive ratio is 100 VL. And we make sure that even in this, it can happen that this multiplication disappears one of the terms of the product and the authors will find themselves blocked by that. So what they will do with this missing term, either they will completely ignore it, they will not load it. This is the case of many others, for example. Kessner. There are a few names that completely ignore the problem of the term in which we are not talking about. And then there are others that will indeed attack it, but they will attack it in an imperfect way so far. What I call analytical demonstrations are demonstrations that, on the contrary, use the results that appeared at the end of the 17th century. They are based on the theorem that today says that between two roots of Px, there is of course a root of Tx. In other words, it is going to be another recurrence that is going to be established, it is a recurrence that is going to be established by successive derivations. So the derivations are not presented like that, the derivations are presented from the polynomial P by multiplying each of the coefficients. We can also use a arithmetic progression. For example, if we have here x1 plus 1x4 plus x3 plus 5x2 plus x plus 2 plus 0, we are interested in the polynomial we obtain term by term, this polynomial here, by an arithmetic progression. So it can be, for example, minus 3, minus 2, minus 1, 0, 1.

45:00 It could be any arithmetic progression of any kind. We can see that the polynomial we have by multiplying the terms of this polynomial is a linear combination of the initial polynomial and the derivative polynomial. So, indeed, on this principle, we have not only the appearance of the results that say the roots of P and the roots of P, but we also have the beginning of the obligate division of P by P' which will then be used by the successors of the law of sign of Degas. I will say a word at the end if I have time. So, we will focus on the demonstration of the law and mainly on what makes these terms important. So, De Bois' demonstration. So, when I say the demonstration, he gives two demonstrations in his memory plus a historical part, so De Bois' work is extremely important, on the one hand because he works on the history of the rule, and on the other hand, indeed, he took two demonstrations, one which is essentially algebraic and the other which uses the results, so the relations between this famous polynomial B and the polynomial multiplied by the mathematical progression. So, if one of the terms of the combination is missing, then we could suppose a positive or negative infinitive coefficient for this missing term, and look indifferently at the combination as a variation, as a permanent one. So, indeed, he approaches a problem and says, well, I'm going to consider that my missing term is a plus epsilon xk or minus epsilon xk, and I'm going to see what happens. So, everything goes very well, except that at some point, indeed, in his demonstration, He multiplies the polynomial p by x plus a by x plus p, where p is an additional number. At this point, he gets to the point of three quarters. What he says happens. In his multiplication, he gets to a coefficient of this kind.

47:30 f and g are coefficients of the polynomial. In this case, I have to look at the signs of these terms in relation to the signs of the preceding terms and the terms that follow them. There are three different cases. The first one is PF-g0, the second one is PF-g0, and the third one is the term. It's the same. And here, indeed, there is no continuity argument. That is to say, I will demonstrate that it works, that what I tell works in the case of this positive coefficient, that it also works in the case of the negative coefficient, so I could deduce, but I absolutely do not justify that, that it is the same when it is equal to 0. There is no reference to a result. If these voices arrive when they are strictly positive and when they are strictly negative, it must be the same in these two. We obviously know the mathematical results that are true for a strictly positive quantity and strictly negative quantities, but not true for a specific quantity. So this is to say that even someone like De Bois who is extremely scrupulous and who actually attacks the problems of terms and terms will be stuck at some point. He will be even more stuck in the second demonstration that he does. In the second demonstration, he uses this. He uses multiplication by an arithmetic progression. And so, obviously, when we multiply this progression to a zero in the middle between the positive and negative terms. So, in the multiplication, he will obviously obtain a term that will disappear. But he is not going to distinguish two cases, and in this case, he is obliged to assume that here, the term in x2 becomes plus zero in x2, he is obliged to assume that it is positive, so that his demonstration is conscious. It must be, indeed... He says that this additional hypothesis poses a big problem. It was made in the theorem to suppose that the term was thus determined, but here he says, I am entitled to assume that this coefficient that disappears, this term that disappears, is positive. Indeed, if he considers that it is negative, this demonstration does not work. So why does it not work? Simply because he wants to show that ...

50:00 There is an additional permanence of signs compared to what we had before. However, in his demonstration, we find here, in which he has written, a coefficient of quantization, a coefficient in risk, or both. So, if we show that there is an additional permanence, it is necessary to add here... If he assumes that it is negative, he obtains two variations and his demonstration fails. So there he is even more embarrassed in the second demonstration because the problem of the term is heard in the first one. And all the people who will succeed must conduct exactly the same problem. No one will give a satisfactory solution to this problem. We will have to wait for Carl Friedrich Gauss in 1828. So, I'm going to give you the situation. It stops because it's a bit more interesting. It's already starting to read. I put it in the original version. He begins by saying that we do not see in the different authors of these theoretical demonstrations the clarity, the precision and the generality for the work to be done on such an elementary subject. And that's why the new treatment of this one does not seem superfluous. Of course, from the point of view of Diogo, it may seem very pretentious, but the fact is that it is exactly the same in his case. While the demonstrations of his predecessors will sometimes do 10, 20, 30 pages, his one does 4 pages. There is not an equation in it, there is only text and it is clear and everything is regulated in the text of X. So he begins by saying, indeed, a polynomial that we call an anti-algebraic function of x to the power of m, ordered according to the power of 360 of x. We assume, without prejudice for generality, that the highest term is n and that the lowest term without x is not missing. There is sufficient power, and we will even get to the problem of Venus. Only the truly present terms must be represented. Therefore, the truly missing terms will not be represented in the equation.

52:30 That is to say, he uses only the missing terms, I don't care about them. He is going to make a demonstration where he doesn't look at the missing terms at all. So, he makes his demonstration. In fact, how is he going to do his demonstration? P, the first integer following M, whose coefficient is not zero. Q, the first integer following P, whose coefficient is not zero, etc. And with that, he will build his demonstration on the principal GP by saying, having described my polynomial in the following way, what happens when I do exclusives? Very clearly and without the need for any subdivision of case, the fact that the equation y equals 0, in fact it is the equation x equals 0, which is the polynomial of the two terms, cannot have more than a positive root because there is a change of sign in y. So I stop there for the first part because we see that in the first place it does not demonstrate. There is only one way to calculate the root of the signs, and he says, how do I do it to calculate the negative roots? Well, he is the first to give up the calculation of the previous pairs. He says, I count the variations for the positive roots, and for the negative roots, what do I do? Well, he makes the variable change, y equals 10. In this case, it's not y, because it's his equation. That is to say, he says, I have my polynomial. I have my polynomial Px equal to zero, I count the variations of signs, so it means O plus, and then I make the change x gives minus x, so I get a polynomial T1, whose lines are alternately changed, and the lines of the coefficients are alternately changed from 1 to 2. And what do I do? I also count the variations, because the positive roots of this polynomial are the negative roots of this one. There is L-variation, which means that there is, in addition, a negative L-root in Descartes' polynomial. In other words, contrary to all his predecessors and contrary to what a large part of Descartes' theory had done,

55:00 Descartes' theory is dissymmetric. What really struck me was the fact that Descartes had a perfect symmetry between the positive root of the rule and the negative root of the rule. Gauss no longer needed this. He said, to establish a certain rule, I will establish a rule on positive roots and I will use another symmetry, which is the transformation of x to get the number of negative roots. So he did say, we can also count negative roots on the initial polydome, but that gives a rule that is much less clear than this one, because this one has no more negative roots. It is better to hear the change of variables than to hear the equation. If we can learn the linear root of the equation, I can also establish it without making a change, but it gives a rule that is much, much less clear. It may be greater than the sum of the number of successive immediate signs interrupted by an odd number of missing terms and the number of interruptions of signs interrupted by an odd number of missing terms in the equation of state. Obviously, we can, with reason, prefer this answer to this one. If none of the terms are still correct, then the negative number is not greater than the number of successive signs, which we find here in the equation. What he calls immediate succession of signs interrupted by a number of long-term terms means that, for example, if I actually have two consecutive terms, for example here, if I delete these two terms, I obtain an equation in which there is one. A succession of these is interrupted by a missing term. When it counts the number of missing terms, it means that one, two, three, four, five, six are missing. This is the result of the interruption of the terms. This is the rule that we find in the few manuals of modern mathematics. We have had the opportunity to study this rule. These are the key terms used to define the rule as it was written by Descartes. It is from the work of Neidei-Glotsky or Weber that we begin to find, at the end of the 20th century, the rule as it was written, that is to say, with this desymitation of logic.

57:30 So, Gauss, with Gauss, takes the end of the enginement of the class. At the moment, there is a little bit of silence. I didn't do the end of all the demonstrations, but it would have taken a lot more time to finish the part. So, I just want to come back to finish with Descartes' rule and what follows Descartes' rule. Just after the paragraph of the class, the next paragraph is Descartes' paragraph. By changing the signs of the order terms 1, 3, 5, 7 and all the odd rows from the term of the second equation, the positive roots of this second equation are the negative roots of the first equation and vice versa. In other words, did Descartes at the beginning... We could also manipulate the negative roots as being the opposite of the positive roots and so, in the end, its denunciation for the negative roots could be enforced, could be removed, so it was superfluous. That too, there is nothing more to it. But it is striking to see that the denunciation of the subsistence, this denunciation there in the text of Descartes, is precisely the one that will be used in Gauss to completely prove it in the first place. So then, one last word of conclusion. From 1828, the Signe's rule will lose its quality, on the one hand because it has been fully demonstrated, with Gauss it is over, and on the other hand because from then on, other rules will intervene much more precisely, these are the rules of the tooth. Fourier and Sturm. In this case, there are two versions, 1807 and 1811. Fourier in 1820 and Sturm in 1835. But it should be noted that all these rules will be based on variations of symbols. The rule of Newton, what does it do? It makes the translation, it replaces x by x minus p, and it shows that there is no more than, if we change, if in the polynome t of x I transform it into q of x minus p, well, in q we lose p, sign change.

1:00:00 In other words, it is a line that is more precise than the one of Descartes because it allows to locate the roots in a precise interval between between zero and B, for example, and neither, this time, between infinity and infinity. Fourier will do almost the same thing, if a rule is then considered equivalent. He will see the succession of signs P , P' , P' , and P , P' , P' , etc., successive derivatives between the two values a and b. And by counting, indeed, the variations lost between these two signs, he will deduce the number of roots that there are between these two values a and b. And finally, we will give a much more complete rule where he will mix the techniques of Poulan and Fourier and he will apply the grid algorithm in fact to the PP' couple and he will count the sequences, the number of lost variations between the sequence applied to the value of A and the sequence applied to the value of B. Last but not least, this rule of Fourier was envisaged by Debois in his second memoir, Calibri and Sciences, published in 1741, where he writes that to find the number of imaginary rational variables, one must report to the sum of the values of Fourier. Thank you very much. This morning, it seems to me, we went from the most general to the most pointed, but without ever losing sight of the most pointed in the general, or the general in general. We have therefore worked very well and participated in a very happy way in this session. All of these terms are related to mathematics and mathematics of the past, of the present, of the future, of the future, of the future, of the future, of the future, of the future, of the future, of the future, of the future, of the future, of the future, of the future, of the future, of the future, of the future, of the future, of the future, of the future, of the future, of the future, of the future, of the future, of the future, of the future, of the future, I'm not going to go into all the details, but let's take a look at the names of Lagrange.

1:02:30 Lagrange, in his book, The Digital Resolution of Equations of the Two Worlds , there is a note, which is the number 8, in which he comes back to the equation of the equations. He demonstrates it. He demonstrates it, here too, by being led by the problem of non-sense terms. On the other hand, there is one... I don't know if you can see it, but there is one... One thing that is interesting, he pays a great tribute to Debois, we can see that he has many fiches of this field, and he justifies one of Debois' results which, at the end of his demonstration of the sign language, began to try to characterize the presence of imaginary roots in an equation from missing terms. That's why Debois' work is so interesting. What Lagrange will actually specify by saying that... So he decomposes the polynomial, the formula of function development, so the formula of Taylor, into the polynomial, and he says that we can make a random term disappear containing, for example, the power of Zn by determining a in such a way that we are at the end of a and z. F of x equals zero. The values of f derived... So, these are the derivatives, not the decomposed or the powers. The derivative of n-th in a, n-1 in a, and the derivative of n-1 in a will necessarily be the opposite for all the values of a resulting from the equation f of a equals zero, that is, for all the roots. In the case of a polynomial of the derivative n, if f has all its real ratios, it must be the opposite of this one and that one. He shows this and finds the very beautiful result of Witten-Connes-Hawking. I think that the principle at the edge of the equation is a completely marginal part of his theory. It is essential to justify the result of the law with this and to begin to specify the characterization of the imaginary roots. Consequently, we will have imaginary roots if the terms are close to those that disappear in energy. And from there, we can conclude that any equation that lacks terms will necessarily have imaginary roots if the terms are close to those that disappear in energy.

1:05:00 Any other questions? I just have a remark. It's a good time to make sure that you don't only have a concept, you have a theory, you have a science, you have a positive theory, but you always have the opposite of the positive theory. You have to find a way to adapt it to the science, to the terminology, to the theory. Yes, there are other rules, by the way, that are there, of course, yes, yes, yes. It's not exactly this result. It's a result that is after, where he says, by making a sufficient translation, we can get to the literature of the future. Yes, well, he decided that he assembled for Descartes, and in relation to his time, it also happened in the winter, the positive numbers and the negative numbers. They should be placed on a level of equality, both in the collections and in mathematics. In Cardan, we see that coefficients are important. In Ariodt, he proposed to consider any kind of root, any kind of coefficient. He even proposed to put all the terms in the first member. From what I've seen, it's the first one who did that. All the terms, the equation Px equals 11, the first one who did that. Do you want me to go now? Thank you for your attention.

1:07:30 So, if you give me your answer, I'm going to ask you a couple of questions. I'll see you tomorrow. I would love to see you again. Thank you for your attention.