PSE Lectures Ep 31 - Counting n-torsion points

Transcript

okay so uh today um we're going to have probably a a shorter session than usual ER I want to finish the classification and then H next time we we start with well pairing you will see that basically um after today we did all the preparations for we pairing and it's not going to be very long to to finish the wh pairing um but nevertheless I think it it it deserves a a a fresh start

so so today we're just going to tie the Loose Ends and I remind you what we proved last time um the divisor of GM minus GN and this is for um M bigger than

n MN m + n and M minus n Prime to the characteristic um the divisor of of this GM minus GN is um the divisor generated by all the M plus n torsion

points plus the divisor generated by all the M minus n torsion points umus 2 the divisor generated by the Mion points minus to the divisor generated by n to and if there is an immediate coroller so um taking uh

degrees of these divisors we get a the corer that I mean on the left hand side the divisor of GM minus GN is a principal divisor so we we we know that the degree is zero um and if I denote H DK to be the number of K torsion points

um then I I know that d m+ n this is the the degree of this divisor the degree of this I mean this divisor E M plus n is generated by it's it's this divisor just to illustrate is simply the sum of all points with coefficient one where the the points run over all e m + n so taking the degree is simply summing

um once H how many times the the number of times uh the number of points in M plus n so this is d m + n and then we have d m minus n - 2 D M - 2 D okay is this is this clear and we have it for any for any M bigger than n

now um this kind of a thing is is a what is called a recurrence relation but it's a it's a bit in an unusual form because we have a a two indices instead of one um so the currence relation ER the the typical example to keep in mind is the Fibonacci the Fibonacci sequence

is um a n + 2 is equal to a n + 1 plus a n and um we have starting conditions a um A1 is equal to one and A2 is also equal to one you see um in a recurrence relation like this Fibonacci Sequence if um if I have ER if I if the recurrence

relation like if the recurse recursing relation of the sequence is a two steps then in order to determine a solution I need um I need to know the first first two elements but if the currence relation would be um three steps then then I would need H three initial elements to determine a solution for the sequence if if it exists sometime there is no solution and as we will see there is a there is a way to to solve ER recurrence

relations ER quite generally but but first I I want to say I mean in this corollary star ER we have something a bit different than ER than the the Fibonacci sequence relation because we have M and N but it's not a problem I I can simply take um in Star take Nal 1

um note that D1 which is simply the number of points in E1 E1 is is simply the collection of all all points in E over K Bar such that such that 1 * p is equal to O but 1 * P equal to o means means P equal to O So there is only um there is only one point

okay so D1 is equal to one um and we get er the recurrence relation which I will call RM which is a d m +1 1 plus d m -1 minus 2 DN minus 2 equal to zero so you see the diff so so now we

are in the form that is similar to the Fibonacci sequence but the the difference is that we have a free coefficient so this this is this is called a homogeneous um relation and this is called uh I think in or n non homogeneous in Curren

relation but um we can reduce so okay so so what what is the solution to a recurrence relation a solution to uh recurrence relation is a sequence a m a m let's

say m bigger than one which is just to say um it's like a infinite order tle so A1 A2 and so on up to a n i a m n continues indefinitely um so so a solution is a sequence such that for any um well

to let's say to to RM um is a an infinite sequence such that for any m a m + 1 + a m -1 - 2 a m - 2 is equal to zero one solution

is a m equal m² this is very easy to to check because ER well you can even check it on the on the original a a relation star that has M mnn because if you put m m i mean you put K Square DK equals K square is a is a sequence then you get m +

n² plus M - n² - 2 m² minus 2 n² and this is indeed equal to zero this is um very easy algebraic identity so we we want to show that um um

AER than one equal m² is a unique solution to um either to star or um it's equivalent to say that it's a unique solution to um RM right because question yes go on when you say unique you mean like the

Only Solution yeah unique is is the only the only one thanks yes um because if we know that this is a unique solution I mean look if if I know that there is a unique solution to RM then obviously there would be a unique solution to star to the the orange one right because because the the orange one I can it's it's supposed to be valid for any m and n i mean for any M bigger than n so I can always take n equals one and then I would I would fall

back to to the green relation so if I know that there is a unique solution to the green relation then I immediately can conclude that there is a unique solution to the orange relation do you see this I'm not sure because like the green relation is when you fix n to be one yes so okay I mean I I agree with the the green Sol that okay you can show that the I mean I'm saying like it's just like a this is in math it's called

a a reduction I mean if I show that there is a unique solution to the green relation it would automatically follow that there is a unique solution to the orange relation right okay okay but doesn't that will not will not be the same but not vice versa yeah um er I was a bit um inaccurate so we want to show that this is there is a unique solution to to Star um and it is enough to

show that um am equals m² is a unique solution to a RM RM I remind you is is this um uh D M plus one + d m -1 - 2 DN M - 2 = to

Z okay so this is this okay or no yes I think that now makes more sense thank you very much yeah yeah okay so we did a reduction we it's enough for us to show that there is a unique solution to this a a nonhomogeneous recurrence relation but in fact um so we know one solution um and if if um BN

BM is another solution to this relation RM I guess I can I can just take H take out the the index M because this this relation means that it's for uh for every M for every positive integer m um then so so if BM is another solution to the relation R then I can take the

sequence let's call it CM or maybe H hm um hm is is the sequence of the difference between the two sequences so I take it's like formally I just write this but what it really means I mean what is the difference of two infinite

sequences you take the difference in every spot okay so uh if a a n and BN are two solutions to the relation are then their difference is going to be a solution

to um let me call it our AG um it's another recurrence relation but um in this case it's a homogeneous recurrence relation now to see why why if why would hm be a solution to um to this recurrence relation um I I mean I I

know that a m + one it's basically for all n a m + one plus a n minus1 minus 2 D sorry 2 a m is equal to zero this is because a m is a solution to the recurrence relation R and I also know that b m +1 plus B minus one sorry I should have um minus two right that's the the true relation and this is minus 2 b m

minus 2 is equal to zero right that's um I mean if if I know that A and B are sequences that solve the relation R then I have these two equations but then I can substract the equations and I get a m + 1 minus b m + 1 plus a m -1 - b m -1 - 2 a m - b m M and that's it equals equals zero

because the the free coefficient vanishes or cancels out so so you see um that um hm equals a m minus b m is a solution to our age the the homogeneous recurrence relation that is derived from the original non-homogeneous Rel is this clear or

no yes thank thank you I I just have one question regarding notation this H like kind of poing R say again this is so I have a question regarding notation this are uh with the H like on top yes how do you read this like in in like in in in language yeah um you can say R to the H but it's not the same age it's not the um here this AG m is like H1 H2 and so on sequence yes it's like a sequence of

of integers and here the age is is just like a the first letter of uh of the word homogeneous ah okay okay that's the I mean that's that's my kind of workflow for for notations okay cool thanks um and okay so so let's do another reduction it it is

enough to show that um zero is the unique solution to um to our age

because then if I have two solutions to the original relation are then their difference is a solution to to our age and if we know that zero is the only solution to our age then I know that their difference is is is the zero sequence which means that they as sequences they are equal or in other words for any m a m is

equal to BM okay is this reduction clear okay so how do we show that we want I I claim that a our age has a has a unique solution I mean clearly the the sequence z00 Z all the way is is a solution to our age and how do I show that it is it is unique so our

AG which is this m d m + one let me just write it slightly differently M - 2 D M plus d m minus one equals to zero we associate what is called the characteristic polom and the characteristic

polinomial you can I mean basically it's a Lambda squar minus 2 Lambda plus one hello qu qu um I'm going to remove you from the

call if H if you don't speak up what um Can can you speak up yet yet so uh ER what brought you here you you were never h in our session as far as I remember uh I'm from uh orig network but ER why did you join here uh to learn um okay so I I don't think you can

you can catch up with the with what we're doing we are pretty uh pretty Advanced um but what do you think you you want to stay yeah yeah I want to stay I want to learn more about zkp yes this is the I think you are um you are too late it's like we have done lectures for a a more than a year now and you haven't been in any of almost any of them so I don't see how you can um you can

fit uh okay so I suggest let me just give you a link ER and you can uh you can read about it yeah yeah thank you thank you okay um anyway H let's

continue uh so I associate the polom Lambda sare - 2 Lambda + 1 = 0 um and of course this is a this is the same as saying that Lambda minus one square is equal to zero so we know we know the the roots of this polinomial there is one root namely Lambda equals 1 and it's a it's a repeated root or a a root of Multiplicity

to now consider the set of all sequences um let's call the set v v equals to the set of all sequences a m and bigger than one such that the

am are real numbers all the set V is a vector space under the the following two operations um I mean if I tell you that something is a vector space I need to tell you how to add two vectors and how to multiply a vector by

scalar so um if I want to to do a plus BM then this is this is simply a adding the the the M entry of the first sequence to the m entry of the second sequence in each entry and if I want to multiply by a scalar so if Alpha is a a real number

Alpha multiplied by the sequence a is simply the sequence in which I multiply every entry of the original sequence by Alpha and um and you it's it's very easy to check if the the the usual axom of a vector space are satisfied what is the zero Vector can anybody say the zero SE sequence yeah the zero

sequence and what is the the a additive inverse of um of a vector I mean just whatting each element of the sequence say again uh negating each element of the sequence yes exactly okay so it's very easy and now on this sequence sorry on this so this is a um an example of of a so V over

R is an infinite dimensional bit vectory space okay um and it's an interesting Vector space for us because we we want to to look at sequences um Define a function yes uh can you comment a bit maybe with example or something what what it will

be like infinite dimensional Vector space um let's just say do you know you know what is a basis for a a vector space honestly I don't remember anything about Vector spaces okay so so forget about this remark you can you can look at up later um okay but it just says that you you cannot express um all the vectors in v as a a linear combination of some finite set I got it I got it

okay yes um so now I Define um um let's call it Big Lambda which is a fun function from V to V and this function simply takes a sequence a and shifts it by by one so it gives me the sequence a m + one where m is bigger than one so

EG if I take the Fibonacci sequence one 1 2 3 five and so on and I apply Lambda on this Fibonacci Sequence I get the sequence one 2 three five and so okay and I claim so from I didn't call the the the

characteristic polinomial by a name um let's call it f Lambda and F Lambda is is the F Lambda is this Lambda minus one squ but I will use this um this description and now imagine that in F Lambda instead of Lambda I substitute the the linear ER

operation big Lambda I mean I didn't say it but um big Lambda from V to V is a linear map I mean it's very easy to see that if you you add two sequences I mean if you do big Lambda of a m plus b m then this is exactly the same as big Lambda of

a plus big Lambda of BM and also if you multiply by a scalar so if I do Alp time if I do L big Lambda of alpha times a sequence a then it's the same as doing Alpha time Lambda over sequence okay so so it's a linear oper

operator um and I I can I can substitute in instead I mean in the characteristic polinomial f i can substitute a um big Lambda instead of small Lambda so consider F of big Lambda F of big Lambda I need to make sense of what is Lambda Square so Lambda Square I'm going to do Lambda composed with Lambda um and then I have two small

Lambda in F then I just write two big Lambda and finally I have I have to add one but one is is equ I mean is is analogous to the identity map on V so I write identity of V this is a um as a a map from V to V you see it if I have linear maps I mean 2

Lambda is is just a linear map from V to V that takes a sequence 2 Lambda is is a linear M that takes a sequence a m and sends it to m 2 * a m + 1 for M bigger or equal to one okay so obviously if I have a a linear map and I multiply it by by some

some element in the the ground field then I get another linear map and if I have two linear Maps like a well obviously I can do [Music] minus and the identity map on V is also a linear map and I can take um let's say minus 2 Lambda plus identity V this is another linear map from from V to V because um I mean if you want to see

what it does to a sequence this is going to be minus 2 a m + [Music] 1 plus a m which is just to say um the sequence minus 2 a m + 1 plus a

n it's I can put it as one sequence and now if you want to do Lambda composed with Lambda as a linear map then it's very easy it's like a you take a sequence and you shift it not by one but by two two places so this is a n + 2 um M bigger than one and all in

all if I do F of big Lambda which I just defined it as um Lambda composed Lambda - 2 Lambda plus the identity and I apply F of Lambda on a sequence a then what do I get I get the sequence a m + 2

minus 2 a m + 1 plus a m and this is exactly the relation you see you see now the the reason that we we introduced this um polinomial I mean the relation are um our age that we have is exactly this it's like DM + one minus 2 DM plus

H or yeah plus DM minus one so and in our case okay and I'll just write it here our H is M D I write it like this d m + 2 um minus 2 d m and + one plus

DM okay this is just a a rewriting of um um of the relation RH I just shifted our age by uh by one is this clear uh now Lisa yeah all good okay so what I'm saying th the solutions to our age are precisely the

kernel of this linear map F Lambda the kernel I remind you it's the set of all sequences a m such that F Lambda of a m is equal to zero to zero I mean the zero sequence okay so now we have we have translated the the question about a solution of a a homogeneous

linear recurrence relation into a question of a linear algebra finding the kernel of some s now obviously Kel of of any any linear math I mean f of Lambda is a linear map from v2v so Kel F of Lambda is

some linear Subspace of V so it is it is a linear it is a vector space in its own right I mean you can add if you have two h two elements in the kernel you add them up you you get another element in the kernel if you multiply an element in the kernel by a scalar you get another element in the kernel which is equivalent to say that if I have a solution to R

Ag and I multiply the solution by some scalar Alpha then I get another solution to to r or and if I add two solutions I get another solution and I claim that the kernel of f Lambda is in fact a finite dimensional Vector space the map let's call it um I from Kel F of

Lambda to our square and given by I I stands for initial I take I take a sequence am that I it's in the kernel so it is a it is a solution to our age um and I m the sequence I need to to give you two a pair of elements I mean

in R square so I just give you the two initial elements of the sequence um this map is well first linear this is a a map between a linear Vector spaces H it's it's again very easy to see because h i of am plus BM if am and BM are two solutions this is just um A1 A2 sorry this is just A1 Plus

B1 a2+ B2 which is by definition I of a m that's I of BM and this map is a objective because any solution I mean any um solution a

n our age is completely determined by the first two elements A1 and A2 and if you give me just a a a two initial initial elements A1 and A2 I can cook up a solution there is a unique solution then I that I can cook up from the the initial elements A1 and A2 what

do I do I look at the relation r i put here here A A A1 and A2 I mean whatever values you give me and then I just recursively figure out what is A3 what is A4 and so on there is a unique way to do it so this map is is a linear a linear map between Vector spaces which is bjective so it is an isomorphism of vector spaces the caral of f of

Lambda is isomorphic as Vector spaces as a vector space a vector space over R to r² okay so we started with an infinite dimensional Vector space that was V the space of all possible sequences of real numbers and we um we focused our attention to the space of sequences that are actual solutions to our homogeneous equation our homogeneous recurrence

relation and this turned out to be a finite dimensional Vector space um any questions up to here maybe my only question is that um last session we stopped at case two h of the theorem or the proof of the theorem ah ah and this time we load to something very different no okay I thought that I I don't think that we I

thought that we finished no I I I think that we finished all the cases last time oh no my bad my bad my bad my bad my bad we finished I forgot yeah I I I remember we finished so we had I mean the starting point is this uh the the theorem about the the divisor of GM minus GN yeah but you're okay with all this all this stuff about the the relations uh in in in general yes maybe if you can repeat like the very high of a logic that we had that theorem about

the visors and then we had a corer um and we we wanted to arrive to this like last corollary that Kel is isomorphic to R square right this was the end call this would be I mean we we aim to to show that R A so we we want to show that our age has a a a unique solution which is zero if if we know that then we can H work out backwards into a coroller that says that the original recurrence relation has a

unique solution because yes okay yes right ER so we we now um concluded that the space of all the the vector space of all possible solutions to the the homogeneous recurrence relation RH um here RH is a two-dimensional Vector space so we will soon so see that this this Vector space characterization is very

helpful but is this clear now yes you are able to follow the the whole thing yeah I I mean I'm surprised because the last time previous week when you started to go into the sequence stuff I was uh like not following pretty much but today things look very clear so I don't know okay then um let's take a break and uh we come back to H to finish this uh this linear algebra argument and then finish the whole thing maybe if we wanted to make it shorter H

maybe we can do without a break if it work um I'm not sure it's going to be shorter eventually I think it's going to be it will take us a while because after this linear algebra argument we then need to do a a an argument about with the classification of the finite AB bilan groups so let's take a break I think eventually we we will end up with a almost normal time session I feel like I will need to leave in half an hour because you thought it would be shorter and I put a working meeting on that ah okay so I'm sorry for that ER okay okay so then are now you

okay with the that we continue sure yeah okay so um what does it mean that something has a a something is a two dimensional Vector space um it means that you can express any Vector in the space as as a linear combination of h of two basis Vector um so let's consider the

following LMA let C1 um C2 be re numbers suppose the polom Lambda sare - C1 Lambda - C2 is equal to zero um has only one root let's call this root

r a sequence um am is a solution to the recurrence relation um A and N a a nus C1 a n -1 minus C2 A nus 2 =

Z if and only if there are constants alpha 1 and Alpha 2 such that a n or a is of the form alpha 1 R to the M plus Alpha 2 m r to the N for all m in other

words I'm saying that the vectors I mean the the sequences of a r to the m and Mr R to the m are two basis vectors of a [Music] um kernel of f Lambda you see I mean if I say that every every solution to this relation is some kind of linear combinations alpha 1 plus Alpha 2 I mean alpha 1 times some Vector plus Alpha

2 times some other factor I mean I'm just saying here you can you can rewrite it as saying that the sequence a is equal to Alpha 1 times the sequence rpn plus Alpha 2 times the sequence m r to the m so I'm saying that these two vectors are a basis for the space of all solutions for the recurrence relation here ER and I suggest that ER you just

read about the the proof of dilemma in the notes it's um it's like a very easy High School level algebra algebra manipulation okay so you see the corer from the Lemma is that um for the relation r h that that is H let's write it like

this D m+ 2 or sorry d m - 2 d m- one h plus d m- 2al zero now I I just shifted it um so that it would fit to the uh the green box in the LMA so so for the relation here with characteristic polinomial I look at the characteristic

polinomial F of Lambda which is just Lambda minus one squ R the only solution to F of Lambda is R equals 1 so is is the only root and hence all solutions to our age are of the form a

equals um Alpha 1 time R to the m is always one so it's it's one plus Alpha 2 um m r to the m is just m is it okay are now are you following okay yes thanks so now um

now let um a am am be the the solution M Square to the original in BM the solutions to the original relation the the non-homogeneous relation um r r is a d m + 1 plus m d m Min -1 - 2

DM -2 um and then um m² minus BM is equal to m alpha 1 plus Alpha 2 m for all M for every every M bigger than one in particular

um for m equals 1 we know that BM has to be equal to one because it's a solution to ER to the recurrence relation right the the solution to the recurrence relation has to start I mean we are only interested in solutions to the recurrence relation that that are compatible with the information we know about torsion points and the information we know about torsion points is that e E1 is equal to one right I mean the number of points in in E1 is equal to one so BM B sorry

B1 has to be equal to one and B2 has to be equal to four so I get um 1 minus 1 is equal to Alpha 1 + Alpha 2 and when I put m equals 2 I I get four - 4 equals to Alpha 1

+ 2 Alpha 2 which is this is a a at least two equation I mean this is two equations that has to be satisfied but already from these two equations I can see that alpha 1 has to be equal to Alpha 2 I mean both both of the alphas the only solution for this equation is alpha 1al alpha

2al Z right because the first equation means that alpha 1 has to be equal to us Alpha 2 and then you plug it into the second equation it cannot be equal to minus Alpha 2 and to minus 2 Alpha 2 unless both of them are zero so we prove that our AG has a unique solution namely zero I mean and and this means that

BM I mean BM has to be equal to m² for all M IE the relation R has a unique solution and the the corer for all from all this is that the the number of points in e m is simply m s we know that the number of points in EM um this is D

DM we know that it has to satisfy some recurrence relation and and the only solution for this recurrence relation is the m² is it clear so far it's okay oh now yes um so and this is what we wanted to prove s some sessions ago right like the yeah this is of this yes this is a this has been a a a

serious goal for the last few sessions I mean how do we how do we count the number of M torsion points we we first had a a LMA that it is finite and then we analyzed the divisor of this GM minus D GN and then we got a recurrence relation and we showed that this recurrence relation has a unique solution okay yeah cool I I will need to go over like the alen of these past sessions to

to see how dots connect because I think I feel like okay I've been able to follow each of the steps but uh it will be helpful to just go over like in a high level like the thing that you just describe but like following myself but but yeah it makes sense I think so thank you so let let me let me just revise the proof quickly okay I think we we still have time um I'm saying we have a LMA that says um this maybe this part of the difficulty is that I didn't prove the LMA because I thought it it may take

some time we don't have um but I I was saying um dilemma says that if you have a a characteristic polinomial that has only one root I mean if you have a a a recurrence relation as in green with characteristic polom that has only one root this is exactly our case right then all all the solutions to the recurrence relation are some linear combination of a a the geometric

sequence basically I mean you can think of this this way I mean for this this homogeneous recurrence relation a a a typical guess for for a solution is the the geometric sequence R to the m or r r to the N because if you if you substitute R to the n i mean you have R to the N minus c h C1 R let me put it in another Colum you have a

um R to the nus C1 R to the nus1 minus C2 R to the nus 2 equal to zero it's the same as saying um that F of R I just I just cancel out R to the N minus 2 and this is the same as saying that F of R is equal to zero you see or no yes yes thank you okay so so

intuitively when I when I have a a a require a homogeneous recurrence relation the geometric sequence are I mean r to the m as a sequence is is always going to be a solution right because of the what is written in pink ER I mean if I I take a solution to the characteristic polom and I take powers of this of this root of the characteristic polinomial I get a sequence that is automatically a solution to the

recurrence relation you you with me yes yes yes yes yes um okay maybe so when I okay yeah and what this LMA says is that um you can of course you can also take the the sequence R to the m and multiply it by m this is also going to be a solution it's very easy to see that these are solutions because you multiply it by m you okay this you can check but it's a it's very easy to see and dma says that the all the all the possible solutions

to this recurrence relation have to be a linear combination of these two solutions H but basically in order to to show The Dilemma all we need to show is that these two solutions RM and M M RM are a a a linearly independent vectors because we know that the space of solutions has Dimension two right so it is enough to to find some um some pair of a of linearly independent

vectors and Via this LMA oh okay ER so so you see the pointer now it's like this LMA says yes that that any solution is is a linear combination of these two solutions and from this LMA I can deduce very easily a what happens in our case because in our case we have

a um we have this R AG which is exactly a relation exactly like in dma it just C C1 is equal to two and C2 is equal to one right yes okay and so I I I can use the corer of D the the corer of dilemma um and I look at the characteristic polom of this um currence relation it has a unique solution R equals 1 so I know that all the solutions to to our age are of the form

alpha 1 * 1 plus Alpha 2 * m i mean this is like a equals this alpha 1 plus Alpha 2 m for some alpha 1 Alpha 2 that I I have no idea what they are at the moment okay and now I go back to the original problem of finding a solution to the the recurrence the the nonhomogeneous recurrence

relation and I know that a m equals m² is one solution to this H relation right yes okay and and the the the other thing I mean so let's say I have another solution BM then I subtract the two solutions m² minus BM and I know that it has to be a solution to the the homogeneous relation R sorry RH

right but we we concluded that solutions to the the homogeneous Rel relation have to be of the form alpha 1 plus Alpha 2 m now yes sorry I think I had internet problems but yeah I I think I follow this part um okay so you prove that it's it's it's Unique because like the difference is yeah so now I I have this equation let's call it double star and this equation double star I I I mean I know that b B1 has to be equal to

one and B2 has to be equal to four because B1 needs to be the number of points of [Music] E1 and B2 needs to be the number of points of E2 so they have to be one and four it's like I have a recurrence relation and I have two initial conditions I mean I have it's like if you give me just the relation of the Fibonacci sequence there could be many solutions but if you tell me that the first two elements have to be one and one then I know that there is a unique

solution so anyway so this B B1 I know that B1 and B2 have to be equal to to one and to four so I plug it in into this equation double star and I get two equations these two equations it's very easy to check that I mean these these are two equations in the the the unknowns alpha 1 and Alpha 2 but alpha 1 and Alpha 2 have um um I mean so alpha 1 and Alpha two have to be zero these are the only solution for for these two equations okay so so all in all I get I

get that um m² minus BM needs to be equal to to zero for all M it needs to be equal to this is the equation double star if I know that this is zero and this is zero then M Square minus BM is equal to zero so so BM is equal to M Square for all M okay ER yes and that completes the proof that this means that that the only solution to I mean the the number of points in EM is M

squ and I think it's good to stop now and we next time we ER we do the the final argument next time we show that em is in fact isomorphic as a group to integers mod M times integer mod okay nice thanks yeah so I I will try to go um back uh I mean so this this last

part was clear thank you very much I think I think that I it will be helpful for me to to to spend some time on trying to connect um this last part to the previous sessions uh stuff uh but but yeah overall uh I have a good feeling on it I mean today's session was was very clear so um okay thank you very much yeah I mean the connection should be fairly easy I mean the the you see in the start of this session we said we have a theorem about a divisors and in particular I can take the degree of the the divisors I get

some a a relation this relation right yes and I I know the two initial condition for the relations I mean D1 has to be one and D D2 has to be four yeah and this condition in in Orange here it has to be satisfied for all M bigger than n um yes and and then um in particular I can substitute if it has to be true for all

M bigger than n then in particular um I can substitute a nals 1 and then um I get a relation DM this this RH sorry R what we called r i just substitute Nal 1 I get d m + 1 plus dm-1 - 2 DM - 2 is equal to Z you see that's the relation to to what and then I I start soling I mean yeah so from from this point yeah okay so this is the this is

the connection I mean I I want to I want to classify I mean I want to to know what are the number of points in EK and the the the theorem about the divisors give me a a a relation between the decays so now I just solve solve use techniques to solve recurrence relations and I get that there is a unique solution a unique sequence Decay that satisfy this relation I don't know triangle you see and that's so that's that's what we do

today cool yeah thank you thank you very much ER now I see that asked for a I don't have a a uh a full link AR now you have the wait let me I I don't have it and I'm from a tablet so I cannot uh find it now but I think if you go on YouTube and you just search like uh um yeah like privacy scaling Explorations math seminar something like this yeah there will be like a bunch of videos and

and a list yeah okay you yeah I see here [Music] yeah yeah okay so um let's finish today and um qu ER I think it's better for you to just watch the videos there are many of them and um I mean I don't think you can get much from coming to the next session it's but up to you so uh yeah that's it for

today yeah thank you thank you very much yeah thank you very much bye have a nice day bye