 K is introduced by 40 and eventually by 30M, which is a real number, defined by the whole of 8. K is in its top. Umbrella, Umbra, Umbra, Umbra, Umbra, Umbra, Umbra, Umbra, Umbra, Umbra, Umbra, Umbra, Umbra, Umbra, Umbra, Umbra, Umbra, Umbra, Umbra. Where M1 and M1 are the integer satisfies this inequality, which is a real number. Where K is in some sequence, finite sequence, integers, which are non-native and positive, and first one is a straight-to-large design. K is called an index. A straight-to-large design is called an index. And basically, each of them is called just an index. And we denote by the absolute value of K. The sum is the depth of the people. If we define the depth of the index to the n, we know these multiplicative values. And this is by corresponding the key, the key in the prime number. The multiplicative values are denoted by the basis of P of K, which is in the multiplicative values are introduced. A professional asked the following question. It belongs to the rate of integers of Kp. It belongs to the definition that V of Cp generated by the divisor divided by P. We have to explain some application to them to be able to find the multiplicative values. We need to introduce some more motivation to explain the sum. We have to introduce a multiplicative value here. It was studied by the second formula of multiplicative values. The integer is a Q, that is, this is the Q algebra. So we can consider the Q variance actually generated by these finite multiplicative values of weight K. This is the respect to K of this space. In section 2, I will give the proof of the first theorem. What is the details of the proof? We first recall the definition. And here, I would like to see the numbers. The index of our series, the summation, the regular summation is the same as the multiplicative value. This is the power series of this coefficient. It is called the mass of the function. K1 is strictly larger than 1. Then, the p-value is function. The multiplicative value is strictly smaller than 1. However, we can find the p-value of multiplicative values. The sense that it is allergen on p is strictly larger than 1. In my explanation, we can take the function to be... But anyway, for any such set, the weight of K is strictly larger than that of k5. you need element that belongs to this ring. Multiplicable logarithm. At the substitute, there is a variable e by width, e to the p over e to the p y, that's e by that one, e to the p. Into the way that k is the original multiplicable logarithms. Some other terms, where in the sum, k prime runs over all units, whose weight is strictly smaller than that of k, h, k prime. Where can compute this function, h, so k, k prime, but explicitly by using some partial integration. Is it a little bit larger? It's almost smaller. Yes, it's bigger. And what can you find? So I have a question on your formulas. So one question is whether the empty index is allowed in some formulas, and another question is... Empty is allowed. Yes, so then... And so there are conventions to define things. And so the empty index is considered as admissible as... Yes, admissible. So then h, k, k prime lies in which... Is what kind of thing? Is it in the ring of dagger or...? Yes. Ah, okay, okay. Okay. Okay. And then we can define... the p i is about to consider the value of the formula. So the value of the empty can go to the extreme, it's not going to be an assumption, but which is... Nice, nice comparison, proper insertion, which can be known as t is equal to 1, to define p i is about to consider the value. In this section, we use alternative definition by using a threshold line. In our case, we have straight sections, 0, 0, 0, 1, infinity, which is as long as this value is p. And every value... And when we got the sum of this straight point as a normal cross-indicator, we got this as a low-resimist scheme associated with the normal cross-indicator. We have an obvious code, d, and x is a union of x minus p, which is an open sub-skin, where p runs over the subset of straight points, which means that the subset of t don't have any subset of straight... of the set of straight i events in which... p i is equal to this as a closed sub-skin of each cross-indicator to p. In any way, this defines an affine of covering of the underlying scheme of this, such as a direct flow of the covering, so that each is an affine of covering. That's a product, a final product. Moment 1 has a canonical bit, which is compatible with the log structure. At the base symmetry there, for any other i, we have seen that we've got these problems. For any non-indicator, the subset of these three points is by calling that... i is different. ...some straight transform or some... ...devirages. I know it's a trivial one. Maybe if it's... ...except cross-indicator. Similarly, ...except cross-indicator from here... ...is the... ...here, grow-ups and... ...some... ...yes... ...it sounds like a... ...this grow-up is necessary only when... ...the computability of p is equal to 2. It sounds like, ...this and grow-up... ...with this point of code I need to do. If we are in a canonical... ...except cross-immersion F, ...this scheme has a... ...you... ...you will have a long-round outbreak. Yet, the UN denotes a... ...close-immersion F. The put-up is defined as... ...we know that this scheme is an defined scheme, ...so we can... ...regard these complex... ...just as the complex of... ...the key modules. And then by passing... ...a new balance... ...key one... ...looks straight at three points... ...also... ...also... ...the front... ...the canonical... ...the problem is defined as... ...natural action... ...on front of these... ...action and components... ...so we can... ...regard with the component... ...and... ...properties in the system... ...is given as complex... ...mine... ...uh... ...it's in... ...special devices... ...but... ...and then we can... ...define... ...the English language... ...it's defined through the... ...the playing sound... ...like... ...another... ...inverted by the sub-I... ...where... ...this sub-I is... ...defined... ...to the put-up... ...this is the coordinate... ...of the RS component... ...I structure... ...of the... ...x to the I... ...on one... ...for the H... ...where I run over... ...properties and... ...rotors... ...so that the server is... ...to the... ...one stack... ...the... ...k is... ...to that... ...to... ...disabletting... ...it does not... ...induce itself... ...from components... ...or... ...it's related also... ...we define... ...with... ...each... ...we have to define... ...the... ...multi-ism... ...of components... ...specifically by... ...all things that... ...eat... ...for example... ...if you want to... ...learn... ...from... ...the building... ...in a W-compact... ...part of... ...its place at... ...a grade zero... ...and we develop it... ...show... ...by using... ...the... ...potion frame as a... ...cursive and compromise... ...in this... ...this space... ...that... ...is the... ...potion frame... ...and it's a canonical base... ...which we do not find... ...at a level... ...where a W...is... ...at a... ...world... ...the frame of noise... ...to have the additional structure... ...of the whole space... ...hand-spanning... ...and... ...mix state... ...of noise... So this is infinite dimension... ...is it? Yes. And... ...each of the C... ...dot... ...is... ...as far as dimensional... ...comology? Yes. And so when you take the... ...total complex... ...is it... ...a total complex... ...where... ...where it's infinite... ...that is that... ...you have to take infinite direct sum... ...or infinite direct product or...? Yeah. Direct sum. Infinite direct sum. Yes. Okay, I understand what you are doing. Also that... ...um... ...the gradient... ...is represented by infiltration... ...is equal to zero... ...where... ...the... ...inertis... ...out... ...and when it is even... ...that probably is... ...after this gradient... ...finished by... ...this product. In fact that... ...then it is 60 unique... ...also the dimension of... ...one-dimensional... ...and this is... ...the... ...Bruze series one-dimensional... ...Bruze series... ...generated by... ...the image of the... ...E-double-M... ...is the fact that... ...then... ...it is 60 unique... ...mini about... ...from H2QP... ...sets that... ...satisfying the following properties. The first one... ...is now... ...is likely to be... ...not quite secure... ...so... ...the theta is a problem in this element... ...and the second part of this element... ...is that... ...the image of... ...the return of the... ...element... ...eat up... ...south... ...in-stream... ...is needed to... ...one... ...to find that... ...one minus one would be... ...the return of the element... ...e to south... ...and the k... ...is equal to the... ...to the... ...next of k times... ...e at... ...one minus one theta thousand... ...and the second observation is that... ...the... ...the image... ...of phi... ...of... ...the... ...qk... ...we can define... ...the... ...concentration... ...of the... ...conference level... ...in... ...in terms of... ...wing... ...I should be able to see... ...of the... ...k's... ...concentration... ...is contained... ...in the... ...k's divided by the times... ...and the other times... ...the rest of the complex. But using these two observations... ...the... ...concentration... ...of... ...outside of the waveform... ...produced another version of P... ...and both will see the values... ...in the list by doing... ...the... ...geographic values... ...the form of completion... ...the natural structure... ...of the vision... ...and the home-artification... ...is given by... ...one theta thousand... ...fifth branch... ...where... ...the value is... ...in the original... ...this is what we're doing. We're going to find... ...phi of kz... ...and... ...this is the sum... ...from theta... ...so say the... ...fragment of the P... ...and... ...the image of the vapor... ...and the theta of... ...the eta... ...salt... ...gallium... ...gallium... ...your shaft... ...and your shaft... ...then it is... ...for what? ...to see the values... ...the... ...home-computer... ...for what? ...for what? ...for vision. Ah, yes... ...home-computer... ...for two rounds. We define the... ...gallium's version... ...of... ...so say the... ...we substitute... ...the... ...derivative of the A and B... ...with... ...A over P and... ...B over P... ...and... ...take in context... ...for example... ...the number... ...of... ...the data... ...is... ...less... ...understand. ...for our version... ...the original version... ...of the... ...10 times plus 1... ...equal to the... ...a key change... ...which is... ...the final sum... ...is the same... ...expressing at the... ...more to be... ...the amount of data that is... ...not really valid... ...not under... ...down to the final feedback... ...of the version of that... ...right? And this is for the final... ...half... ...half of the final sum... ...this is the conjecture... ...second to... ...one to be called... ...conjected to... ...that's here... ...but... ...it is already... ...to... ...by... ...a key... ...that's the same... ...and you can... ...and... ...you go precisely to the spiritual face... ...who gives them... ...the self-containing... ...this... ...conviction... ...or one of the seven... ...so for first... ...just... ...just the... ...the convention of the indices... ...that... ...so you had the... ...in the beginning... ...because of the... ...the condition that K1 is big... ...at least two... ...at least two... ...at least two... ...but then... ...here... ...here we talked about this condition... ...or... ...or what the... ...this... ...this assumption is unnecessary... ...so all the definitions of the... ...peadic things are without this assumption... ...or only that this... ...all the definitions are without this? ...OK... ...OK... ...OK... ...in the whole... ...peadic... ...story... ...you don't assume any admissibility... ...of the index... ...or just here? ...Yes... ...Yes... ...No... ...over... ...we don't assume... ...OK... ...here... ...in true... ...yes... ...yes... ...I'd like to give a sketch... ...of this proposition... ...No... ...no... ...we need to introduce some... ...more... ...in the case... ...with... ...the zero... ...in the solution... ...of Kc... ...this assumption... ...we need to introduce... ...against percentages... ...like for the show... ...OK... ...and we see that... ...some form of... ...power series... ...of the form of power series... ...for the patient in... ...subring... ...and common... ...a bit called... ...common function... ...using... ...power series expressions... ...so we... ...all... ...we get it as a... ...subring of... ...power series... ...and... ...a... ...power series... ...and polynomial... ...over... ...the... ...power series... ...of the index... ...is equal to the... ...so look at... ...the level of error... ...in... ...still at like five... ...these properties... ...and... ...and there's a coefficient... ...of... ...and some... ...some... ...some weak conditions... ...but... ...and also... ...there is some... ...more... ...period... ...just some more... ...rotation... ...and it is... ...also very... ...like... ...also... ...so we can... ...evaluate it... ...as it is equal to one... ...but it is equal to the... ...one... ...one plus the value... ...but the important is the following properties... ...in this rightful... ...the... ...the... ...so this is a... ...a flow-through... ...for a composition... ...for the... ...for the... ...for the... ...for the... ...of W... ...which is equal to... ...the things to work for... ...survive... ...which ends with E... ...so... ...this current is... ...the... ...for the... ...of B.S.M.A. ...of... ...cor Maker... ...to us which is not belongs... ...to... ...Gamor buttons... ... Which... ... vectorization to us... ...the ... ... AG again to the problem. Take this calculator... ...and then... ...contain theingtinger. Then... ...it ends with the... ...ah... ...the worst is it kind of... ...for a little time... In the top of the lower polynomial log, taking the considered coefficient on spice, it stands at 8W or 8W, the other one's from the center. In the center, which pole is more likely to be something like this? It's a ring of something like this. In some integrals, some sub-ring of I1, bit to the round, towards the top. It's a capillary, it's a pole. It marks a bit around the orientation of the surface. When the substrate, this substrate needs to be filtered out of the form, but I think it is. That's the instruction of the gradient of the waterproof. If you take something in its surface room, and be in the dilute sound, and if there is an intersection of the progeny, decrease, taking a quality of the heat to the heat part. So we put heat in the map. But this is not the gradient of the walls. You can see the map that means part of the user. The idea generated by the element usually denoted by T. This doesn't depend on the choice of the beginnings, and we can take the potential to be just the heat part of the map. By using this, you obtain a graph C of final level of each part. There is some value for them to be, which stands an element in the VK, through the collection of information that is seen in the graph, divided by some power of the consequence of I1 and some result. The graph is finished. The C area is contained in the sub-regional final level, and if you add it back, it has an extra structure, the top of zero has more than enough output. The version of the motif is multi-layered. If you take the V in its boundary, taking the V that has a multi-layered level. Multi-layered extension. But we can approach me raising the value usually. If it is sufficiently large for fixed index, then the result must be centered. It is merely the configuration of the untyped version of I1 or the divided part of I2. Also, finally, I would like to address this. In some argument, we can show that this equality point if you obtain the graph from the expression of the value of I1 and I2, we obtain the two subjections, and then it is a non-professional term, and also it is subjected by some sense, by an end of the term. The question is, do you expect these two subjections to be an answer? Thank you very much. Thank you very much. Are there any questions from Paris? Okay, so any questions from Paris? So the relation, so can you explain the relation of the, you define certain expressions using truncations in the definition of the, you define the periodic multiples of the values, but also you define those in the connectors, so there were finite sums in F, I don't find it in my, ah, this one. Yes, so can you explain the relation between the finite sum and the multiples that are, so, do you? This way? This is the finite sum, so we can consider this is the untyped version of this, then we are obtained from certain ideas. Consider what? This is the finite sum, this is the finite sum, and if most of P is equal to the both P, of which we see the value, and it is called, and it is the member of finite, of which we see the values, and I'm sorry, this is the P, but this is the delineated P, of which we see the values, but it is related, it goes from definition to the untyped version of the, and P, of which we see the values, so by using this formula, we can relate the finite sum, and some reduction of the delineated sum within the values. So, as a combination of, as a combination of, it's more than 2 by 1, then, that appearing in this, finite sum is a P and integer, in this case, and it holds P, and this we can relate in this case from other, other. I have to explain it, but also I have to not explain it, but also the image of what we think, how to consider values, and how to consider values. So, you also saw in some map, was not a graded map, in the blackboard there, and H was graded, I suppose. So, the map for H to be, is not graded. So, what is it? So, what is it? What, what, what? What is it doing? I have this one. Yes. Yes. This is not graded, yeah. And does it send? Yes. This has some gradient, this has gradient for, which follows the definition, and this also has a natural gradient. Yeah, I don't, so every element goes to a finite summation of things, so every element goes to a finite sum, or? Yeah, as far as sum. Okay. So, other questions? Okay. That's all for us. Okay, any questions or comments from Beijing? Questions. So, I wonder if there's any relationship between the real and the value of the real and the value of the value? Oh, yes. There is another comparison of visual comparison map from H, just a map from H to R by using the whole Beijing drawing from, by using a Beijing drawing from this point, and this map to the left to the right to the left to the right to see the values. Oh, yes, but in terms of by using these moves, it is proved that the space of the value of fixed space is bounded by the dimension D to K. Oh, no, no, no, no. The map itself is only here for a finite object. Oh. Okay. There are no more questions from Beijing. So, at the end of the program, we have one. So, you have mentioned two key of observation. So, the second of observation says that key of control of the program is. So, is this some sort of sample? Yes, yes. Okay. Okay. Okay. Thank you. Okay. Let's have a second.