Oggi vorrei parlare del sistema quantum, in particolare dei modelli fermi, che sono introdurati da Matteo già l'altro giorno, e vorrei discutere delle proprietà per equilibrio dei gas fermi e della quantità, che è la energia di correlazione, che sarà introdurata in un momento. Lo che vedete è basato in un lavoro con Nils Benedicter, in Milano, Pham Tham Nam, in Munich, Benjamin Schlein, in Zurich, e Robert Salinger, in Vienna. Ok. So, let me tell you what is the setting that I would like to consider. So, the setting is about an interacting and confined... Oh, sorry, I should probably turn this off. Sorry for that. Sorry. Right. So, N interacting and confined thermionic particles, where N is going to be a large number, eventually it will be taken to infinity, but the system will stay confined. Ok, so it's a mean field regime. So, we're talking about a quantum system, so the state of the system is specified by the wave function. Ok, and this is going to be an anti-symmetric wave function for a system confined in a box lambda, which I will take with periodic boundary conditions. Ok, so my lambda is going to be a three-dimensional torus. Right. And the side of my torus is going to be 2 pi for convenience. Ok, so you should think of that as being populated by N particles and the typical distance between two particles is N to the minus one third. So, it's a very high density regime. Ok, so anti-symmetric means, let me just remind you that if you look at the wave function and you permute, well, if you permute the arguments, if you permute the labels of the particles, what you do, what you get, well, you basically just pick up a sign, the sign of the permutation. Right, so that's my setting, so it's a very high density situation and I will consider a mean field Hamiltonian. It's going to be a self-adjoint operator on L2 of lambda N. Ok, of the type that Matteo already introduced, but now I will focus on the mean field regime. So, there's going to be a first part which encodes the kinetic energy of the particles which comes with a sort of semi-classical parameter epsilon squared. This is going to be part, I mean I will motivate that. For the moment let me just write it and then on top of it I will introduce a many bad interaction of mean field type. Ok, so it's a very interaction with the two-body potential which later I will give condition, I will specify what kind of potential I will focus on for the moment just think of B to be a bound potential. So, what about this parameter epsilon? Well, this parameter epsilon is an independent parameter it's n to the minus one-third, or order n to the minus one-third. Ok, so you see that with respect to the bosonic, I mean classical mean field or bosonic mean field, the difference is that we have to, we are taking into account this semi-classical parameter which makes the problem harder because it makes the spacing between the eigenvalues of the Laplacian smaller and smaller as I am accustomed to. Ok, so what we are focused, what we are interested in for the purpose of this thought is the ground state energy of course on the fermionic sector of the Hilbert space so en is going to be denoted, is going to denote the infimum overall wave function in L2 anti-symmetric lambda n which are normalized of the expectation value of hn over epsilon. Ok, so that's the quantity that I would like to discuss. I mean the point is not really, I mean the number per say is not that interesting physically but what's interesting is to get some amount to compute this number with high precision which is in fact what we are going to do, you have to develop a precise understanding of how the ground state looks like and that's actually the content of this talk. Alright, so that's about the model now in the remaining in the next minutes I would like to motivate why we have this epsilon here so the reason is that I mean what you want to do really is to be in a setting in which there is a balance between kinetic and potential energy so as usual I mean if you think of V to be a bounded potential if you look at the interaction energy over a reasonable class of states then you would expect this guy here of order n and so basically the role of this parameter epsilon is to make the kinetic energy also for the n ok so I want to kinetic energy and potential energy to have a chance to compute. Ok, so to understand that let me consider as an example the free Fermigas ok so or if you want a non-interacting problem and I would like to discuss the ground state of this non-interactive problem so the Eigen states of my Laplacian the Fermionic Eigen states of the Laplacian they are just anti-symmetric tensor product as we have seen from Mathieu's talk these are called later determinants ok so you have some of our permutations and you got the sign of the permutation and then you have the product of Eigen states of the Laplacian single particle ok so this fk denotes a plane wave so fkx is 1 over 2 pi to the power 3 over 2 that's correct l2 normalization e to the ikx and I want my wave function to be compatible with the periodic boundary conditions so I will take k to be a point in z3 and then another feature of this wave function is poly principle so the k's have to be all different otherwise you get identically 0 so k i has to be different from k j for i different from j and that's due to the anti-symmetry of the wave function if you want poly principle alright so computing the ground state in the absence of interaction amounts to find the choice of k that minimizes the energy of the state ok so let me do that so the energy of any such state so it's just kinetic energy let me forget for a moment about the parameter epsilon squared that's just simple computation somewhere k of or if you want some over i from 1 to n of k i squared where those k i's are those that appear in this later determinant ok so you want to minimize this quantity enforcing the constraints that k has to be a point in z3 and the k's have to be different so you can feel only one state every particle I mean you cannot have w occupied states so how do you do this well that's clear so let's look at ok so you should think of that as being z3 no plane because otherwise I would make a mess so what do you do you start to occupy states keeping somehow making sure that the energy is as small as possible right so you start from k equal to 0 and then you go on and you start to populate the states that have the smallest norm once you measure it from the origin ok so you get this 4 and then you keep going you get those guys you start to feel a lattice and you see that this thing here as a radical symmetry ok so you start to populate a ball which has radius kf this is called the fermi radius or fermi momentum which grows as n to the one third ok so clearly if you just give me any n particles then you might have that the ball is not completely filled there might be boundary terms and from now on I will always assume that n is chosen in such a way that the fermi ball is completely filled so the number of points in the fermi ball which is this ball here is equal to my number of particles and that's of course in general there might be boundary terms ok so in this setting then you can ok so if you just look at this quantity here for for k in the fermi ball then it's an exercise I mean it's not difficult to see that if you just sum those numbers what you get is something which is asymptotic to n to the 5 over 3 ok so probably principle it's making kinetic energy higher with respect to the bosonic case so that's of course something you can just somehow it's an insight that you get started from a special class of states but there are general inequalities that give you a lower bounds for the kinetic energy that are compatible with this this is called the leap theory in equality so this epsilon squared is there then to to turn this kinetic energy into something of order n alright so that's a non interacting problem now I would like to go to the interacting problem ok v non 0 and that's of course much more difficult because while in general the ground state of the system is not going to be uncorrelated but you will have to be able to take into account correlations among the particles ok so even in the mean field regime computing the ground state energy of this problem is a difficult task because you have to understand correlations among the n particles alright so that's what we want to do and so I mean of course that's a problem which is somehow it's been studied in physics and a popular approximation for this for this problem is the Hartree-Fock approximation which tells you basically that while it's based on the assumption or on the simplification that you replace the full fermionic the space of the fermionic wave functions by the space of uncorrelated fermionic states so you replace the two anti-symmetric by the space of the set of slater determinants ok where those f i's they are orthonormal functions in the two of single particles ok so these are the slater determinants that we have seen in Mathieu's talk these are fermionic uncorrelated states and the hope is that in this replacement the problem simplifies because if you don't have correlations among the particle then in this mean field regime you would expect every particle to experience a mean field due to everybody else that's the same mechanism that we have seen in Peter's talk yesterday so it's a large number type of thing ok so from a mathematical point of view an advantage of slater determinants is that they are sort of Gaussian states in the sense that they are k particle density matrices that have been introduced by Mathieu completely determined by the one particle density matrix completely determined by the one particle density matrix exactly as for a Gaussian field ok so in that case it would be the covariance so the one particle density matrix the general definition is just a trace over all particles except the first of the project over the state of the matrix of the project over the state psi n you can compute that for a slater determinant and what you get is a rank n projector over the sub space of the Hilbert space which is spent by the n orthonormal wave functions that enter in the slater determinant ok so it's a rank n orthogonal projection ok so that's a an interesting class of states where they're simpler than usual fermionic states because there are no correlations and you can compute the energy of such states ok so psi n here is a slater determinant you can compute the energy of those states and as Mathieu told us I mean in general the energy the interaction energy of a many body problem depends on the two particle density matrix but the two particle density matrix can be expressed in terms of the one particle density matrix for slater determinants so what you get is the final expression is this expression here so there is a first part which is exact is the kinetic energy plus a term which encodes the many body interaction and which is quadratic in the density matrix ok so there is a first part which is of density density type exactly as for a classical mean field or right so there would be just density density type of interaction and then you have to correct the fact that particles are not really uncorrelated but there is the anti-symmetry requirement and that's why you have to take into account these other extra terms so this is called the direct term and this is called the exchange term ok so this row by the way is row of x is the density of particle at x is just the diagonal of our density matrix ok so this is just you see it's a non-linear functional of the density matrix omega it's also called the Hartree-Fock energy function so that's a Hartree-Fock energy functional which plays a central role in Hartree-Fock theory ok so that's of course an easier quantity to consider than the full many body problem but still it's non trivial task to understand the minimization problem associated to the Hartree-Fock energy functional ok so let me define the Hartree-Fock runs at energy as the smallest energy that is later determinant can reach ok so since there is a one to one correspondence between those rank and orthogonal projections and this later determinants you can just minimize over rank and projections your Hartree-Fock energy functional ok so that's that defines the Hartree-Fock grand state energy ok so that's the same thing as minimizing over later determinants so that's the definition and clearly since we are minimizing over a smaller class of states with respect to the original problem that gives an upper bound on the many body energy ok so trivially the true many body grand state energy is less or equal than the Hartree-Fock grand state energy ok so that's so if you're able to compute this grand state energy at least you're able to come up with an upper bound for the original many body problem so of course much harder is to understand the lower bound of the many body energy in terms of Hartree-Fock and that's the thing I would like to focus on and in particular I would like to discuss the so-called correlation energy which as the name suggests is the energetic contribution due to the fact that fermionic states are not exactly the later determinants so the definition of the correlation energy is just the many body grand state energy minus the Hartree-Fock grand state energy ok so what I would like to discuss today so our goal is to is to understand the asymptotics of this guy as angostin field ok so that's what we would like to focus on so let me make a few remarks no sorry I have to find out oops let me make a few remarks about this Hartree-Fock problem so as I said in general it's a difficult minimization problem I mean it's finding out the exact grand state of the Hartree-Fock problem by itself a very difficult problem however in the present confined case it's actually not so difficult due to the fact that after all the Laplacian has a gap so you can exploit that so let me just give it to you as a remark in our setting so confined system with bounded potential and then later I will make more precise assumption on the potential the Hartree-Fock grand state is actually equal to the plane wave state in general this is not too so let me call it plane wave to say that this is the free Fermigas in general that's not true so when I say in general I mean in the thermodynamic limit where the spectral gap of the Laplacian is just not there anymore you might have symmetry breaking in the infinite volume limit and in fact there is a very nice paper of so in general not true there is a very nice paper of mathematically rigorous paper by Gontier Heinzel and Levine from 2018 where what they proved is that in general what I mean what they prove is that this plane wave state in general is not the grand state but is exponentially close energetically to the true grand state of the system exponentially close I mean exponentially close in the density ok let me not write down the precise estimate but that's just what I want to do to emphasize here ok now what about rigorous work on in fact not really computing the Hartree-Fock grand state energy but understanding the validity of Hartree-Fock theory starting from the many body problem ok so let me make here another remark rigorous work on Hartree-Fock on on validity of Hartree-Fock equilibrium ok so I will mention two words one is work by Falker Bach in 92 where what he did is to prove well he didn't really consider this setting I mean he considered the setting of large atoms which however can be recasted in terms of mean field problem by large atoms I mean that the nucleus has a large charge of the order of N so what Bach proved is that the correlation energy in this setting here would be a little o of N to the one third where N to the one third is the smallest contribution entering the Hartree-Fock energy which unfortunately I raised the exchange term so N to the one third is actually the size of the exchange term and what he proved is that this correlation energy is smaller than that so actually Hartree-Fock gives the leading approximation one question may I ask a bit late sorry so you made this remark that the Hartree-Fock energy is actually the so the minimizer do it's claim that what you say there is that there is a minimizer yeah and it's a plain way state yes in the confined case yes so the other work I'm mentioning here is a work by Graf and Solovey where in 92 94 where they extended this work to the case of Jellium so Jellium means Coulomb system in the thermodynamic limit ok so that's ok so that's what I wanted to say and now I would like to present it here ok let me take it here so this theorem is about the asymptotics of the correlation energy and it goes as follows so let me take v such that the Fourier transform of v is non-negative that's important and such that sum over k of 1 plus k v hat k is finite so we need v to be bounded and then to have a bounded derivative ok then the correlation energy is equal to an explicit thing in which I call e rpa rpa stands for random phase approximation up to some sub bleeding term so alpha is some explicit rate ok where this e rpa is a rather complicated expression that I will write once here ok so this e rpa is a highly non trivial function of the many body interaction ok, so let me write it here there's a 1 over pi it's an integral and there's a log I mean the explicit expression is not enlightening at all maybe let me put here there's a k naught then there is a function of lambda here which is 1 minus lambda arc tan of 1 over lambda and then here I close this parenthesis so this is integrated in d lambda minus a term which is linear in v which is this one, so this cancels the first order term in this expansion of the log ok so it's this complicated expression here and this k naught is just 3 over 4 pi to the one third ok, so this is complicated but explicit and in fact it corresponds to a I mean it has interpretation of an infinite resumption of perturbation theory certain class of diagrams so just a few remarks so the first proof of this result was given through matching upper and lower bounds in a work with benedicternam Schlein and Seieringer 2020 and 2021 this was actually under the assumption that v was small for small potentials the version representing here is actually joint work with Niels benedicternam Schlein and Seieringer in 2021 that's for this version in particular there is no restriction on the size of the potential and so that was not the first result about the correlation energy, it was a previous work that I had with Christian Heinzel and a master student Felix Rix in 2018 this was a sort of a rigorous second order result ok, so we are only able to compute the correlation energy at second order and let me also say that the same result has been obtained well the same time by Christian Heinzel and Nam in 21 with a different method so same result but different slightly different method so that's about some of the history of that now that's of course these are rigorous works as I said this number here captures an infinite resumation of Feynman graphs in the expansion for the ground energy which is called the random phase resumation or random phase approximation so the result agrees with the random phase approximation which was an approximation scheme introduced by introduced in physics by Bohm and Pines in the 50s and the last comment I would like to make is about the fact that so this besides agreeing with the random phase approximation this number, this complicated expression that you see there also agrees with the ground state energy of a quasi free bose gas non interacting bose gas so and that's not an accident in fact but that's because the method of the proof which in fact it's the interesting thing about this result is allows essentially to describe in a precise way excitations around this fermi ball in terms of an emergent non interacting bose gas so in a sense this simplification arises from the fact that those fluctuations around the fermi surface can be understood in terms of a non interacting problem that you can diagonalize and that's the ground state energy ok so that's a rigorous version of what in physics is called bosonization and that's applied to a three dimensional fermi gas ok so sorry say it again do you really mean non interacting verbally non interacting what is the role of the interaction potential so the interaction potential is there to essentially define the non interacting of the boson so this interaction potential for the fermions enters in the bosonic theory in defining the non interacting Hamiltonian so the non interacting Hamiltonian is not just a Laplacian it's something more complicated sort of can think of a strange external potential being there which is written down in terms of Fourier transform of what we call it so not real space but yeah all right ok so this approach of course this kind of methods were developed in the condensed matter physics community and in this mean field regime one can actually use them to prove theorems rigorous version of them let me also say that this bosonization techniques really started from a mathematical physics paper namely from the exact solution of the Laftinger model by Mattis and Lieb ok, so then maybe in the remaining part of the talk I would like to tell you something about the proof which goes through a fox space description of the problem and it's based on a matching upper and lower bounds for the ground serenade so let me introduce the fermionic fox space exactly as Serena did but this time the you start from L2 antisymmetric and not from L2 symmetric ok, so that means that the general element in the fox space is an infinite sequence of fermionic wave function which have arbitrary number of particles sort of a grand canonical picture ok and in this fox space the Hamiltonian also can also be lifted so you can introduce a fox space Hamiltonian as the direct sum of Hamiltonian's acting on states with a given little n number of particles and this fox space Hamiltonian has a form very similar to what we've seen in Serena's talk so there is a kinetic part and there is a many body part which is quartic in this creation and annihilation operators so there is a sum of 3 momenta here the full transform of the potential and then you have a quartic operator in the fermions that describes the scattering among our fermions ok, so Serena gave a nice interpretation of this many body interaction let me just say that now this fox space operators a, k, a, q they don't commute but they rather anti commute ok, so this curly brackets is the anti commutator same for the star version and the only non trivial anti commutator is a, k, a, k ok so you can use those guys to build slater determinants so if you apply n of them over what is called the vacuum so the vector in fox space describes a case in which you have no particle at all you get a slater determinant with plane waves f, k, 1, f, k ok, so this vacuum vector has a 1 in the first component it's a vector that describes a case in which you have no particles ok so ok, so what do we do next while we introduce a, no maybe, well I will just continue with those two so I would like to efficiently compare my ground state energy with the energy of the of the free Fermi gas which is my Hartree-Fock ground state and to do that I will introduce the notion of fermionic Bogolubov transformation so these are unitary operators on the fox space so Bogolubov transformation these are unitary operators in the fox space that preserve the anti commutation relations ok, which so it has the following feature so the existence of this guy can be proved so it has two important properties the first one is that if you act with this fermionic Bogolubov transformation on the vacuum you get the free Fermi gas so you can understand the free Fermi gas as a unitary conjugation of the vacuum so this K i's that really form the the Fermi ball and the other property is that those operators they introduce a so-called particle all transformation so if you take the operator that annihilates a plane wave with momentum K and you conjugate it with R what you get is again a K if K does not belong to the Fermi ball or a star K if K belongs to the Fermi ball ok, so these are useful algebraic properties why are they useful well as a consequence of the first property the first property implies that you can compute the Hartree-Fock energy as the expectation value over this unitary conjugation of the vacuum of your Fox space Hamiltonian ok, that's clear because this guy is precisely the free Fermi gas and that's an n particle vector in the Fox space so that's just rewriting of the energy of the free Fermi gas the second property is tells you something about the fact that if you take any vector in the Fox space not necessary are omega by using the unitarity of R what you can do is to rewrite the energy of Psi as R star Psi R star H R R star so that's just one because it's unitary Psi so now what you do next is to use the algebra that I gave you I mean this simple relation to transform this operator it's just an algebraic computation there is nothing deep about it and you can bring the result into normal order meaning that you first write the you first start it from the less you first write the creation operator and then the annihilation operator so you have to anti-commute things around if you do that you get a constant term which is precisely the vacuum expectation value which is the energy of the free Fermi gas plus something let me call this vector here Psi plus the average of what I call the correlation Hamiltonian over this vector Psi I'm sorry I'm sorry in your first line you say that the first concept then the Hartree-Fock energy is the matrix element of interacting Hamiltonian with the ground state with the non-interacting ground state yes so as I said in this special confined case the Hartree-Fock ground state is the free Fermi gas so I'm just rewriting the average of my Hamiltonian over the free Fermi gas in this fancy Fox space language but then I'm saying ok in general I might have any state on the Fox space and I want to compare it with that so how do I do that while I just use unitarity to extract what I want and then I have to show that this is sort of small ok so then the task reduces to a study of this correlation Hamiltonian which I'm not going to write down because it really involves many terms I'm just going to tell you what are the two most important terms la rappresentazione Hamiltonian è, come ho detto, una parte di molte termi è quello di conjugare l'original Fox space Hamiltonian con questo operatore r quindi in termi di questo tipo puoi ripetire l'energia di correlazione come l'enfimum di Rolks psi che dimenticano la conjugazione della restriczione in particolare del Fox space con R star perché poi è tutto quello che stiamo facendo di l'energia di Rolks psi quindi questo correlation Hamiltonian è l'esterno di qualcosa e questo qualcosa si può scrivere, è una parte di molte termi i due mostri importanti, sono un tipo di termi di energia e un termo quartico e poi ci sono i termi più piccoli che non posso scrivere quindi questi due ragazzi sono uno termo che arriva dalla conjugazione di l'energia che è una energia relativata e questo ek è una quantità positiva, è solo il valore di epsilon squared k squared quindi è l'energia kinetica di l'energia con momentum k con rispetto all'esterno di mu quindi mu è epsilon squared e l'esterno di momentum squared è una quantità in ordano 1 quindi è un operatore positivo e poi hai questo q che è quartico che arriva dalla conjugazione della interazione di molte termi che ha la seguina in termini di la seguina di b operatori quindi è il sum per p hai la interazione di molte termi e poi hai qualcosa che sembra quadratico in termini di quei operatori bp b-p plus la misione di conjugazione più 2 b star p quindi cosa sono questi b ragazzi in termini di quadratico perché tutt'ora si deve essere quartico quindi per una buona risposta quindi questi b operatori sono molto importanti creano le excitazioni particole quindi b star è un creatore particolo è diventato da un po' di punti in termini in termini di p che k plus p non è una termina di b a k star a star k plus p ok, quindi cosa sta succedendo pensiamo di avere un stato in il foc spazio che ha un po' di articoli ok, quindi quello che avete aspettato è che dopo conjugare con r star, la maggior parte del stato è trascogliata il fatto che si trascoglie e se avete un b star su questo quello che hai è il sum per p ok exactly the same thing where now you add the two particles ok, one is here it has momentum k plus p outside of the fermi ball and one is inside ok so that is this important constraint which means that if p is fixed these two points have to be closed because the radius of this guy is n to the one third so these are excitations around the fermi surface and they behave as almost bosonic operator told us last time it's a very important thing that you could think pairs of fermions as behaving as a boson and that's precisely what we want to do here so this b operator behave as a boson in the sense that well if you look at the commutation relation for this guy they are bosonic so they commute ok and if you look at the anti commutative of b star p with b q what you get let me be very informal here you get a constant term which is of order there is a Kronecker delta in p and q and then there is epsilon n so that's n to the two thirds then there is p plus something which is small but small means on states with few particles I'm not going to be more precise let me just say that if the state here where you are acting has few particles so much less than n to the two thirds then this guy wins and you can think of that as being an error term so if you normalize these operators by this n to the two thirds then those guys really behave as bosons ok these are almost bosonic operators so if everything was quadratic in this almost bosonic operator then you would be left with almost quadratic, almost bosonic, Hamiltonian and quadratic operators can be diagonalized and you would be done essentially just one detail is that actually your commutator is the other way around it could be yes so maybe there is a minus sign here yeah so you say almost bosonic but epsilon n is not all the one it's n to the two thirds when I say almost bosonic I mean that you have to normalize them ok ok the error is much less than the two thirds when you normalize that's a bleeding with respect to one ok so I would like to get to a point right so if everything was quadratic in those guys then we would be done we would essentially use the Bogle-Luboff transformation that Serena told us and this problem could be exactly diagonalized you could find a spectrum and you could compute this correlation Hamiltonian as the ground set energy of this bosonic problem the problem is that it's not quadratic in the B's ok so this is quadratic in the fermions not in the bosons so there's no in this try here seems a little hopeless but what you could do is to compute just to probe how far this is from being a quadratic operator to compute the commutators with the B and what you get let me just tell you what the result is so this Ip ok, let me just write the things out k plus p does not belong to the fermi-bol and k belongs to the fermi-bol and what you have is k plus p a k and then here it is an retrieval function of k which is e k plus p plus e k that's just twice epsilon squared k dot p plus epsilon squared p squared so forget about this term, that's of order epsilon because k is of order epsilon a is of order epsilon to the minus 1 and so you see that if this guy was a constant that somehow this would look good because this is precisely a B operator and then this would suggest that on certain class of states H not behave as a bosonic operator however that's not a constant ok so then the trick is to localize so we would like to replace k dot p by a constant omega dot p and this would drop out of the sum if you can do that this goes out of the sum this replaces a B and then you have sort of a bosonic problem so just in a nutshell how does this go well this goes by a suitable localization technique on the fermi surface so you take your fermi surface your fermi-bol and you take composite in patches so these patches they are away from each other in a precise construction that I don't have time to review so there is also north and south pole and so on every patch is called B alpha so let me say that this B alpha it has an extension N to the 130 divided by N to the 1 alpha this is B alpha and it has a center which I call omega alpha so M is the number of patches ok so this introduces a localization on the fermi surface and this omega you have to think of it as the center of a patch ok so you introduce a new localized operators they are basically like the B except that now you only look at momenta which are inside a patch so there is a normalization forget about that then you have to sum over K let me call IP this this set here so K is in IP and then you also want this to intersect a patch so one thing that you can tell is that these patches are actually 3 dimensional objects so they extend a little bit inside and a little bit outside the fermi ball by order 1 ok so you have a K plus p a K in terms of those guys then the moral is that in terms of those guys this commutator looks indeed like a bosonic I mean the H naught essentially satisfies a bosonic relation in terms of those guys the conclusion and I will soon stop is that the commutation the commutator of H naught minus a certain bosonic kinetic energy with let's say c alpha p is essentially 0 on states with few particles so up to error terms that's really 0 and this db is a quadratic operator in the bosons sum over alpha and p of some bosonic dispersion relation which I'm going to write here so the bosonic dispersion relation would be epsilon times the length of p and then you have the scalar product of the normalized center of the patch with the normalized vector p times c star alpha p c alpha so that's the idea so this is the localization technique that allows to think of this correlation Hamiltonian in terms of a almost bosonic problem which is now quadratic in this operator c so what we do at the end is to study this almost bosonic problem and of course we have a lot of insight on this almost bosonic problem by the exact bosonic problem which is exactly solvable ok so in particular and I will stop here for the upper bound so let the exact bosonic problem so that would be db plus qb so this qb would be q after replacing the b by the sum over alpha of n alpha c alpha ok so if you look at this operator here and you treat those guys as being exactly bosonic then the ground state would be given by something very explicit would be a Bogolubov transformation of the vacuum in the ground state of the exact bosonic problem would be a vector that may have me call it phi that would just be, or maybe let me call it psi which would be a Bogolubov transformation of the vacuum where this Bogolubov transformation is just a unitary operator generated by something quadratic in the bosons so there is a certain Bogolubov kernel it's an m times i matrix then you have to make everything unitary so you remove the mission conjugate anyway it's something explicit forget about the explicit form and that would be the exact bosonic ground state so what you can do for the fermionic problem is to get inspired by that and take this as a trial state by taking the c as being fermionic as built by a fermionic operator ok so for the upper bound you have to take a trial state which in the end I mean which would be just an almost bosonic Bogolubov transformation and the vacuum and if you take this trial state you evaluate the energy of this guide and you see that leading order this gives a random phase approximation energy plus error terms of course you have to prove a matching lower bound and I'm not going to be able to tell you anything about it but to do that essentially well you can imagine what you really have to do is to use the fact that I mean instead of having the vacuum here can be written in this way where here you don't have the vacuum you have something which instead has few particles inside and so essentially what you have to do is to prove that actually this number of particles is very small and you have to keep track of course you have to use a lot positivity of the interaction to bound quite a few error terms from below ok so that part I just I won't have time to tell you anything about it so I think I will just stop here and I thank you for your attention you're a little bit late so you are perfectly in time and we have time for questions is it possible to tell in short time which is the new ingredient that I love you to go from small potential to large potential yeah so I mean it's at the end so ok it's really technically in the sense that it boils down to more precise control of so there are several ingredients but one important one is to understand come controllare questi ragazzi da sotto. Quindi, l'unico bambino è già certo per un grande potenziale. Non abbiamo questo potenziale di potenziale, per l'unico bambino, il modo in cui le cose erano raccontate, erano usando molto il potenziale di potenziale di potenziale per controllare alcuni di questi error termini con questa energetica kinetica. E essenzialmente, cosa che dobbiamo fare ora, beh, essenzialmente, siamo capaci di avvenire con più dei migliori estimati per questi error termini qui e per esplorare in un modo più efficiente il modo di positività del potenziale. In un modo in cui non abbiamo costruito completamente questa energetica kinetica e, in realtà, il lavoro di questa energetica kinetica, cioè, abbiamo bisogno di controllare alcuni termini di positività, ma non usiamo più questa energetica kinetica, ma vi piaceremo di usare la positività del spectrum di spazio bosonico. Quindi, abbiamo sacrificato il spectrum di spazio bosonico in cui, nei primi paperi, abbiamo solo abbandonato da sotto per zero. Ora, abbiamo usato la positività di questo ragazzo per controllare, in realtà, alcuni dei error termini non bosonici. Questo è, ovviamente, quello che abbiamo fatto. Vedi che potete provare il potenziale? Quindi, ovviamente, ci sono due questioni naturali. All'anno, siamo solo abbandoni per considerare il potenziale con uno derivativo. Ci sono due problemi. Uno è di quanto fare questo per Coulomb. E' un problema molto interessante e molto difficile, e questo è molto difficile, e non sappiamo come fare questo. Il bambino d'acqua potrebbe essere... Ovviamente, il bambino d'acqua potrebbe essere più semplice, ma già che è abbastanza difficile. Ma, da mio punto di vista, io credo che cosa che trovo davvero più interessante è di poter considerare il limito tramodernamico, anche per un potenziale abbandonato. Quindi, tutto questo è fatto per un sistema confinato. E molti di nostri estimativi, quando si intende a un caso in cui un volume scega all'infinità, non scelano l'infinità con il volume, quindi bisogna riaggiare tutto e fare cose localmente e provare a metterne cose insieme. Nel momento, non sappiamo come fare questo. Con le necessità tecniche, c'è l'integrazione fisica di positività della tramodernamica? Beh, è relato con la repulsione dei particoli, cioè che è una notionale. Se cambiate l'area reale, il reale è un po' di tramodernamico, che non è così... Sì, sull'esercito di... C'è un potenziale che soddisfa... Potenzialmente non cercherà se è positiva o definitiva. No, no, in realtà, no. C'è un potenziale che vuoi, è che quando si inclusiva questo potenziale, l'energia sceglie. E, ovviamente, con la positività della tramodernamica, puoi assurre che questo succede, ma magari la storia è tutt'altro un'altra classe di tramodernamica. L'esercito di positività nella tramodernamica può essere generalizzato, ma, a l'altro, è vero per il potenziale di tramodernamica, che è un potenziale reale e fisicamente relevante. Quindi, in un senso, sì, è vero, è un po' di tramodernamica, ma è anche quello che il potenziale fisico e relevante soddisfa. Quindi, sì. Poi potete anche dire qualcosa di questo? O potrebbe essere l'esercito della tramodernamica? È una buona questione, infatti. Quindi, come è, ora abbiamo... Sangue, le tramodernamica, la tramodernamica finirà in basso? E' un'altra questione. Quindi, la quale ti bisogna direCome è l'esercito del potenziale di tramodernamica? Il potenziale di tramodernamica è, ma, se non ci sono due di queste piacute, che facciamo così? Si, e' un'altra questione. non riesco a raccogliere come questa metrizia densa sembra. Quindi se mi chiedi cosa sono le feature di questa metrizia densa comparedo uno dei gas frittimi, non lo conosco davvero. Lo che posso dire è che in una storia dimensionale, che è più o meno diverso da questo, il rilascio di questa exetazione bosonistica è per smutare il profilo in spazio momentumale. Quindi è una cosa cruciale che Mattis ha scoperto che con queste exetazioni bosonistiche stavano scoperto la discontinuità dell'occupazione. Quindi la superficie di fermi era desaparece, in un senso avevano qualcosa che, a un livello fermi, aveva l'infinito derivativo, ma tutto l'altro era smutato. Quindi, ovviamente, la cosa naturale è per provare a vedere se algo simile a 3D succede. Credo che ci sono argomenti che credono che questo non succede, ma non lo so, ma è una questione molto interessante. Ok, no other questions. Vi salte via il macchello. Thanks.


Hello, Earth citizens. In this video, I will show you a 10 minute routine to do for your finger joints. This routine is particularly good for those of you who work in an office and do this all day. Type type type on the computer, reports, emails, or maybe you're on the phone all the time and you have a lot of tension in your finger joints. So the great thing about this routine is that you can do it sitting down in your office chair. I'm currently sitting in my office right now too. So you can do it sitting down in your office chair. When you have some time, take 10 minutes to exercise your fingers with this routine to help energy, blood, and all the good stuff that should be moving around in your hands to get going again. So first, I recommend sitting with your spine straight. When you're doing any kind of energy work, it's better to be sitting straight rather than slouching over like this. So sit with your spine straight. If you can, feet flat on the floor.  Shake your wrists. But when you shake your wrists, feel this all the way to your fingers. So meaning, don't just shake like this because this is just shaking your wrist. Really shake all the way from your wrist to your fingers. Rest to your fingertips. Let's shake for 30 seconds. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4. Can you feel all the way to your fingertips? 6, 7, 8, 9, 20, 1, 2, 3, 4, 5, 6, 7, 8, 9, 30. Stop. Now clasp your hands together like this, except clasp on the first knuckle. So this is your first knuckle up here. This is your second bottom knuckle. Hold on the first knuckle. So you're not all the way down. You're on the first knuckle like this. Bring your thumbs together. Pretend like you're holding an egg and rotate.  Rotate. This is really good for your wrist. So we want to open the wrist first and then move to the fingers. So rotate like you have an egg in the middle. So don't squish the egg. Don't drop the egg. Keep a nice rounded space on the first knuckle and move. 30 seconds one way, 30 seconds the other way. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Now sometimes when you do this, you can feel tension in your shoulders. If you do, just exhale like that. 1, 2, 3, 4, 5, 6, 7, 8, 9, 20. Last 10, 2, 3, 4, 5, 6, 7, 8, 9, 30. Now the other way. 30 seconds. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Make sure your thumbs are staying put here. Don't let your thumbs float. 1, 2, 3, 4, 5, 6, 7, 8, 9, 20. 10, 2, 3, 4, 5, 6, 7, 8, 9, 30. Great job. Shake it out again. 10 seconds. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Stop. Now we're going to stretch out our joints. So how you're going to do that is put your palms like this facing forward. Close them into a fist like you're squeezing. So really squeeze from your fingers. Okay, you see I'm squeezing. So not like this. Squeeze. Squeeze and then open. So open like this. Not like this. Open. Squeeze. Open. Squeeze. Open like that. For both hands, not weak, weak, weak, weak. Squeeze. Open. Squeeze. Open. We're going to do this 30 times, okay? Count with me. 1, 2, 3. Stretch all the way to your fingertips. 4, 5, 6, 7, 8, 9. Are you squeezing and opening? 10. Squeeze. Open. 11, 12, 13, 14, 15, 16. Squeeze. Open. 17, 18, 19, 20. 1, 2, last 10. 3, can you feel energy opening? 4, 5, 6, 7, 8, 9, and 10. Shake it out again for 10 seconds. Really shake to your fingertips. Not like this. Shake. Get energy going to your fingertips. Breathe. Okay, stop. Now we're going to go double time, okay? So same thing. Squeeze and open, except you're going to... like this. Double time. 30 seconds. Are you ready? 1, 2, 3, 4, 5, 6. Really squeeze and open. 8, 9. If you feel pain, if you have a lot of tension, you're going to feel pain. If you feel pain, that means there's a lot of blockage there. 1, 2, 3, 4, 5, 6, 7, 8, 9, 20. You're almost there. 10 more. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Great job. Shake, shake, shake. Shake. I bet the last one was really challenging. Yeah, we carry a lot more tension in our fingers than we think. All those emails, all that typing takes a toll on your joints. Let's shake here for 10 seconds. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Okay, now clap. Clap so you can feel vibration on your palms. So not like this, okay? Not weak claps. Clap. 10 seconds. 1, 2. Okay, now tap the top of your left hand. Tap, tap, tap. Like you're really tapping up to break up blockage there. 10 seconds. 1, 3, 4, switch. Left hand on your right hand. Top of your right hand. 10 seconds. Tap to break up blockage. 1, 2, 3, and great job. Shake it out again. All the way to your fingertips. Shake, shake, shake. Great job. Stop. Rest your hands on your knees. And even with your eyes open, feel your hands. Feel your palms. When you feel your palms, can you feel some blood circulating? Maybe you feel like a pulsing feeling. Wham, wham, wham like that.  Yes. Maybe you felt a lot of pain as you did the opening and closing or shaking or any part of this routine. If you feel pain, that means you have blockage. So don't be discouraged. It's okay. Just practice this a lot to get rid of the blockage. To get energy circulating again, especially open, really big, close, really tight. Really exaggerate, and you can get energy circulating in your finger joints once again. You did a great job. Experience a full one-hour brain education-based class. Please visit a local body and brain yoga studio near you.


My first day on campus was an interesting one. Immediately walking on the yard, I was approached by several friends, different fraternity people wanting to see if I was interested on what year I was with GPII transferred in, if I was eligible to even attend interest meetings and things like that. I met a lot of football players, basketball players, and cheerleaders as well, and just mingled with a lot of people. It was very interesting to know that I'll be living in the same area with these people opposed to having to commute back and forth to school when I was at the community college. It was a very interesting experience, something different to be living on campus.


This lecture is part of a course on elementary number theory. More precisely, it's the Berkeley course Math 1.15, introduction to the introduction to number theory. What it will cover is things like primes, congruences, quadratic forms, quadratic reciprocity, and Dirichlet's theorem. The course will be very loosely following the textbook by Niven, Zuckerman, and Montgomery on an introduction to the theory of numbers. So number theory is, well, quite a lot of it is about primes. So you recall that a prime number is a number bigger than one that's only divisible by one and itself. And the first few primes look like this. And we can ask some basic questions about prime numbers. So the first question is, how do you find prime numbers? And one way is to use the sieve of Eratosthenes, where what you do is you write out all numbers. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, so on, 20, 30, 40. And we keep going. And you cross off all the ones that aren't prime. Well, first of all, 0 and 1 don't count because 0 and 1 is a unit. And the first number you haven't crossed out is 2. So we write down 2 as a prime. And we then cross off all the multiples of 2 because these can't be prime. And the first one, number not on this list, is 3. So 3 is our next prime. And we now cross off all multiples of 3. So we go 3, 6, 9, 12, 13, 14, 15, 18. So we cross them off like this. And I'm going to miss out a few that should be on that one. And next, we look at the first number we haven't crossed out. And this number is now 5. So 5 is our next prime. And we cross off all multiples of 5, which is easy because those are just those ones there. And then we get to 7, which is the first number not crossed out. And we cross off all multiples of 7. And actually, all of these have already been crossed out, except for that one there. And we can actually stop now because any bigger prime has a square bigger than 49. So it can't be the smallest prime divisible by a number. So what we've done is we've crossed off all numbers less than 50. Those are not divisible by 2, 3, 5, and 7. Well, these are the primes less than the square root of 50. And if we look and see what numbers we've got left, we've got 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. And if you're doing number 3, you very soon learn all the primes up to about 100. So the next question is, how many primes are there? Well, if we look at this table, primes seem to be fairly common. There's no sign of them suddenly stopping. And so this suggests that the number of primes should actually be infinite. And this is the first theorem of number theory. Due to Euclid, the number of primes is infinite. Well, actually, Euclid didn't say the number of primes is infinite because the Greeks didn't really like the concept of infinity. What Euclid actually said is something equivalent. He said that if you take any finite set of primes p1 up to pn, you can find another prime not in this set. And what he did was he took the product p1 times p2, and so up to times pn, and you add one to it. And what we do is we take a prime factor pn plus 1 of this. And we notice the prime factor pn plus 1 can't be any of these primes p1 up to pn because if it was p1 up to pn, then it would divide this number here, and it would also divide this number here, so it would have to divide 1. So pn plus 1 is a new prime. So let's just see how this works. We start off with all the primes we can think of, which is absolutely no primes at all because we haven't yet started yet. So we take the product of all the primes we thought of, which is actually 1 because this is the product of no primes. And then we add 1 to this and we get our first prime, which is 2. You notice that if you take the product of an empty set of numbers, you get 1 for the same reason that if you take the sum of an empty set of numbers, you just get 0. So this is our first prime, and now we take the product of all the primes we thought of so far, which is just 2, and we add 1 to this and we get 3. And then we take the product of all the primes we thought of and add 1 to this and we get 7. And that's prime. So we now take the product of all the primes we thought of and add 1 to this and this gives us 43. And now we take 2 times 3 times 7 times 43 and add 1 to this. And this time it's not a prime. It's 1807 and this factors as 13 times 139. So our next new prime is actually 13. So it seems to be a common misunderstanding that if you take the product of all the primes you thought of and add 1, then you get another prime. But this isn't actually true. I mean, it breaks down for the first time here. So this shows there are infinitely many primes. The next question is how to find large primes. So there's a sort of game in Play by Number Theist where they sort of compete to see who can find the biggest prime. So how can you do this? Well, one good way is to look at things called Mercena primes. These are named after Mercena who was the first person to discuss them. These are primed to the form 2 to the n minus 1. So let's have a look at a few. Then where we have 2 to the 2 minus 1, which is 3, and we can 2 to the 3 minus 1, which is 7, or we can 2 to the 4 minus 1. Well, this is 15 and it's not prime, which is 3 times 5. Well, you notice the problem here is that 4 isn't a prime because 2 to the a b minus 1 is divisible by 2 to the a minus 1. This is 2 to the a minus 1 times 2 to the a b minus a plus various things up to plus 1. So if we want 2 to the n minus 1 to be prime, we should take n to be a prime. So we can speed it up by just looking at 2 to the primes minus 1. So we have 2 to the 5 minus 1, which is 31, and 2 to the 7 minus 1 is 127. And this seems to be a rather good way of generating primes because you notice whenever we take a prime and take 2 to the p minus 1, we always seem to get a prime. Well, this doesn't actually quite work because if we go to 2 to the 11 minus 1, this is actually equal to 2047, which is just 23 times 199. So this doesn't always work. However, it's still a reasonably good way to generate primes. So 2 to the p minus 1 is quite often prime. We have an open problem. Are there infinitely many Mercerna primes? So no one really knows, although the answer is probably yes for a reason I'll discuss later. So people have used computers to find large numbers of Mercerna primes. And in fact, there are very good tests to see if a very large number is a Mercerna prime. And you can actually test numbers with millions of digits to see if they're Mercerna primes. In fact, most of the time, the largest known prime is actually a Mercerna prime because they're particularly easy to check. And so another famous sort of prime are Fermat primes. Well, if we're going to look at primes to the form 2 to the n minus 1, we should also look at primes to the form 2 to the n plus 1. So let's look at the first few cases just to see what's going on. So 2 to the 0 plus 1 is 2. 2 to the 1 plus 1 is 3. 2 to the 2 plus 1 is 5. 2 to the 3 plus 1 is, well, 2 to the 3 plus 1 is now 9. Now that's not prime, it's 3 times 3. 2 to the 4 plus 1 is 17, that's prime. 2 to the 5 plus 1 is no good because that's 33. And if you think about it a bit, you notice that if we've got a number 2 to the power of AB with A odd, and we add one to this, and this is divisible by 2 to the A plus 1. So the exponent can't be divisible by any odd numbers if we want it to be a prime set. So we should only take n to be power of 2. In other words, we should look at numbers of the form 2 to the 2 to the m plus 1. So let's restart. We have 2 to the 2 to the 0 plus 1 is now 3.  And all of these numbers are prime. These are called the Fermat primes. So 2 doesn't really count as a Fermat prime because 0 is kind of a bit of a special case. And Fermat sort of suggested that in fact, maybe 2 to the 2 to the m plus 1 was always prime. But in fact, this is false because 2 to the 2 to the 5 plus 1 turns out to be divisible by 641. Actually a bit of a historical puzzle about why Fermat didn't notice this because he was perfectly capable of checking that 2 to the 2 to the 5 was divisible by 641. And the consensus seems to be that he probably did check but just made some sort of numerical error and got it wrong. So Fermat primes turn up in a rather unexpected place. There's a famous problem about which regular polygons can you construct using a ruler and compass? And it's easy to figure out how to construct polygons of side four and eight and 16 and some because you can bisect angles. And the Greeks figured out how you can construct polygons of sides three or five or 15. And what Gauss discovered is that if a number is a Fermat prime, then you can construct a polygon with that number of sides using ruler and compass which is really rather bizarre. And in fact, it turns out the only numbers you can construct with ruler and compass are ones that are products of distinct Fermat primes times the power of two. So we have an almost complete list of polygons you can construct by ruler and compass except that nobody knows whether there are any further Fermat primes. So people have checked the next 30 or so cases and haven't found any further Fermat primes. And it's a completely open question about whether there are an infinite number of these or not. Well, so Fermat suggested this might be an easy way to find primes, but it doesn't quite work. So we can ask, is there any easy way to generate large primes? And this is a sort of open question. I mean, it depends a bit on what we mean by an easy way to generate large primes since we have some reasonably fast methods that computers can use to generate large primes. But it'd be really nice if we could find say a polynomial f of n such that f of n is always prime. So are there any polynomials f such that f of n is always prime? Well, the answer is yes, it's quite easy to generate these. We can just write f of n equals three for all n. It's the constant polynomial. So what we really want is f to be non-constant. Otherwise we've got a rather trivial answer. So we want to find a non-constant polynomial such that f of n is prime for all positive integers. And you can get pretty close to this. Suppose I take the polynomial n squared plus n plus 41, just found by Euler. And if you try n equals zero, one, two, three. Well, you'll probably find this is always prime for as long as your patience holds up. In fact, it's prime for all numbers n up to 40. However, it's not prime for n equals 41. And it's obviously not prime for n equals 41 because if you put n equals 41 here, this is divisible by 41, of course. So we can get polynomials that try very hard to be prime, but they don't quite make it. So you can ask, is there any polynomial such that f of n is always prime such that f isn't constant? And the answer is no. And we can see this as follows. Let's look at f of n equals n to the k plus, so something times n to the k plus something times n plus some constant term. And now suppose this constant term is bigger than one, let's call it k. Well, then if we put n equals k, this will be, f of n will be divisible by k because all the terms are divisible by k, so it's not prime. So if the constant term of f is bigger than one, then it can't always be prime. So what happens if the constant term is equal to one? Well, we can just change f of n to f of n plus k to some number k such that the constant term is not equal to one and repeat our arguments. So there's no polynomial that always generates primes apart from the trivial case of a polynomial that's constant. And no one has ever found anything better. I mean, there just doesn't seem to be a really easy, simple way that will always generate primes. Well, we've shown there are infinite number of primes, but we can ask another question. I mean, instead of asking how many primes there are, we can ask how many primes are there less than or equal to x, where x is some number. Now, the number of primes less than or equal to x is often denoted by pi of x, and you've got to be a little bit careful not to muddle this up with the number three, quote, one, four, one, and so on. So pi is the Greek letter for p, which stands for prime. And it's not very easy to find an exact formula for pi of x, and you can see primes are a little bit chaotic, and it would be a bit unreasonable to expect an explicit formula for them, although rather surprisingly there's seen a moment, there is one, but we can ask, what is pi of x roughly equal to? So we can ask roughly how many primes are there. And Gauss came up with a pretty good answer to this. One version of Gauss's suggestion is that pi of x should be approximately x over log of x. And I just warn everybody, since this is a mathematics course, log means log to base e, not to base 10. There's actually no point in taking logs to base 10. So in mathematics, log always means log to base e. And this was finally proved by Hadamard and Dela Valet Poussin, in about 1896, and at the time it was proved this was probably one of the deepest and most difficult theorems proved in the 19th century. So informally, this means, suppose you take all the numbers up to x, it means about one over log x of them are prime. So we can think of this as saying the chance of n being prime is about one over log of x. And you've got to be a little bit careful about what you mean by this, because the chance of n being prime isn't one over log of n. It's either the chance of n being prime is one if n is prime and zero if n is not prime. So it's kind of a little bit meaningless to say what is the chance of n being prime. But whatever, it's sort of reasonably obvious what's meant. So let's have an application of this. Let's ask, is the number of Mersenne primes infinite? So you remember a Mersenne prime is the number of the form two to the n minus one. So what's the chance of two to the n minus one being prime? Well, it's about one over log of two to the n minus one. And this is about some constant evolving log of two over n. So in order to find out the total number of Mersenne primes, what we could do is we could just take the sum over all n of the chance of two to the n minus one being prime. So we take the sum over n of k over n. And this will hopefully give us a rough estimate of the number of Mersenne primes. Well, this is just equal to k times one plus a half plus a third and so on. And this is the famous harmonic series which you remember from calculus is infinite. So this sort of suggests the number of Mersenne primes is infinite. Well, fine, but there's a bit of a problem with this argument. First of all, it assumes these numbers are random and they're obviously not random because they're two to the n minus one. There's another very obvious problem. Suppose we try and ask how many numbers will form two to the n minus two are prime? Well, we can apply exactly the same argument and it suggests there should be an infinite number of primes in the form two to the n minus two.  So these probability arguments are a useful guideline to what is true or not but you should always be really cautious when using them because they can give horribly misleading answers. A rule of thumb is that in number theory, arguments using these sort of probabilistic arguments are generally regarded as worth almost nothing because they're very easy to do but they're extremely hard to make rigorous. And they sometimes give the wrong answer and they sometimes give extremely ambiguous answers. Suppose, for example, we try to count how many firma primes are there? Well, what we should do is we should take the sum over one over log of two to the two to the n plus one. Well, the logarithm of this is going to be some constant times one over two to the n if we ignore the plus one. And this is going to be finite so this suggests there should be a finite number of firma primes. And indeed, numerical evidence sort of suggests that maybe the one's firma and you're about to the only ones they have. But there's another argument you could say. You could ask how many primes are there of the form two to the n plus one? Well, here we would take sum over one over log of two to the n plus one and this is going to be about k times one over n and this is going to be infinite. So this argument suggests there should be infinite number of primes in the form two to the n plus one all of which have to be firma primes. So we've got two probabilistic arguments. One of which says there should be an infinite number of firma primes and one of which says there should be a finite number in which is correct. I have no idea and nobody else really has any idea either. So Riemann actually found a rather astonishing explicit formula for the number of primes. What you do is you define the logarithmic integral of x to be the integral from zero to x of one over log of x dx. Well, actually there's a little bit of a problem with this integral because if x is equal to one the integral actually becomes infinite and sort of diverges. So when I say it's equal to that it's not actually equal to that. You need to fit around with this a little bit but I'm not going to worry about this. The logarithmic integral is going to be a given by something slightly modified version of this. And Riemann found the following astonishing explicit formula for primes. Well, first of all, when he wasn't counting primes he was counting prime powers. So Riemann discovered that p to the n should count as one over n of a prime. So in some sense four is really half is sort of 50% of a prime and eight is one third of a prime in some sense. And he found that instead of counting primes it's easier to count prime powers in this funny sense. So pi prime of x is the number of prime powers less than x where p to the n counts as one over n of a prime power. And he found a completely explicit formula for this number here. So this is just equal to the logarithmic integral of x minus sum over some funny numbers row of the logarithmic integral of x to the row. So here the sum is over all these numbers row which are some so called zeros of the zeta function. Shall I explain in a moment? And now this is a really astonishing formula because these are all nice continuous real valued functions. However, this function here is certainly not continuous because it sort of jumps by one whenever x goes beyond a prime power. So we've got this absolutely astonishing formula of a discontinuous function being given as a natural sum of all these nice continuous functions. And here the logarithmic integral of x turns out to be approximately x over log of x. So this explains why the number of primes less than x is approximately given by this number here. So what are these mysterious numbers row? Well, the numbers row that zeros of the zeta function which is given by this number here, one over one to the s plus one over two to the s plus one over three to the s and so on. So for example, zeta of two is equal to one over one squared plus one over two squared plus one over three squared. And this is a rather difficult sum to evaluate. Euler became famous by discovering that it was actually equal to pi squared over six. And the numbers row are the numbers row such that the zeta of a row is equal to zero. Well, there's a bit of a problem here because the known zeros of row tend to be complex numbers and they're complex numbers for which this series doesn't actually converge. So we need to do quite a lot of work in order to make sense of the Riemann zeta function but other complex numbers and then try and work out where the zeros of the Riemann zeta function are. At this point, we come to the probably the most, one of the most famous open problems in mathematics which is the Riemann hypothesis which just says the real part of any zero row is most a half. So, there's a million dollar prize for solving this if you ever managed to solve it. In fact, solving this Riemann hypothesis isn't the hard part. The really hard problem is not in solving the Riemann hypothesis but trying to get someone else to read your solution when you've solved it because crackpots produce several proofs of the Riemann hypothesis every month. So, if you show a proof of the Riemann hypothesis to a mathematician, the mathematician will just throw it in the bin unread because all mathematicians are sick of getting sent proofs of this. Anyway, Riemann's work gives an astonishingly accurate formula for the number of primes less than the number. For example, the number of primes less than a hundred million is exactly equal to 5761455. Riemann's estimate not counting the zeros over row turns out to be 5761522. As you can see, it's sort of already accurate to four significant figures. So, this is using l i of x as an approximation not to pi of x but to the number of prime powers. So, you have to subtract out the squares and cubes and so on of primes. And if you do that, you get this astonishingly accurate formula for the number of numbers less than number. So, one of the basic facts about primes we're going to prove is the fundamental theorem for arithmetic. So, the fundamental theorem of arithmetic says that every integer greater than or equal to naught, sorry, greater than or equal to one, is a product of primes in a unique way. That means unique up to order. And there's a very nice formula for this involving the Riemann's data function, which is one over one to the s plus one over two to the s plus one over three to the s and so on. And Euler discovered that you could write this as follows. It's the product of one over one minus two to the minus s times one over one minus three to the minus s times one over one minus five to the minus s and so on where this is a product over all primes. And this looks rather astonishing if you see it for the first time but it's actually quite easy to prove. If you just write each of these factors as a geometric series, we get one over one plus one over two to the s plus one over four to the s plus one over eight to the s and so on times one plus one over three to the s plus one over nine to the s and so on. And now if you multiply out this product of an infinite number of factors, each of the terms you get by taking one term from each of these factors and multiplying them together. So for instance, if we take this term, this term and this term and ones everywhere else, we get one over two to the s times one over three to the s squared times one over five to the s which is equal one over two times three squared times five which corresponds to one over 90. And now here we have a term one over 90 to the s. So you see every term in this series here comes from a unique term in this infinite product because every integer can be written as a product of prime powers in the unique way. So this is Euler's version of the fundamental theorem of arithmetic. It's this amazing factorization for the Riemann zeta function. Euler also used this to give another proof of Euclid's theorem that there are infinitely many primes. What you do is you just take s equals one. And now you notice this term here is infinite because it's just one over one plus one over half plus a third and sum which is infinite. So it means this product over primes must be, also must be infinite. But if a number of primes were finite, then this would just be a finite non-zero product. So it couldn't be infinite. So the number of primes has to be infinite because this series here diverges for s equals one. So this is a completely bizarre proof. You're saying the number of primes is infinite because the harmonic series diverges. So next thing you can discuss in number theory are diefantine equations. So diefantine equations are named after diefantis of Alexandria and not very much is known about him. I mean, he lived a few centuries in the common era but no one even knows the exact date when he lives. In fact, pretty much all that anybody knows about him is he wrote a book about diefantine equations. So diefantine equations are equations where you want solutions in integers. Actually, diefantis himself didn't discuss solutions in integers, he mostly discussed solutions in rational numbers but whatever. So let's have some examples. Famous example that everyone knows about is Pythagoras's equation, x squared plus y squared equals z squared. And this is solutions in integers that everyone knows. We all know three squared plus four squared equals five squared or five squared plus 12 squared equals 13 squared and so on. Another example you might have linear equations. So can you solve 27x plus 11y equals one? And we'll see fairly soon how to solve all linear diefantine equations using Euclid's algorithm. So in this particular case, we can, for example, take x equals minus two y equals five, gives a simple solution. Another example is the equation two x squared equals y squared. And this is an obvious solution, x equals y equals zero, but it has no other solutions. So this is like asking, is y over x equal to the square root of two? I mean, in other words, this is the square root of two rational and the answer is it can't be in the, when the Greeks discovered this, they got a bit upset about it because there was a feeling that all numbers should be rational and having irrational numbers was a little bit disturbing. So it's quite easy to prove there are no non-zero solutions of this, for instance. This equation implies y must be equal to even because if it were odd, then this side would be odd and this side would be even. So y is even and you can then divide out a factor of two from both sides and you then find x squared is equal to two y squared. Now you found a smaller solution of two x squared equals y squared. So given any non-zero solution, you can find a smaller non-zero solution. So there can't be any non-zero solutions. Another particularly notorious example is actually the n plus y to the n equals z to the n. So this is Fermat's last theorem and we want n to be at least three and we want x, y and z to be greater than zero because if you allow x to be zero, for example, it has some rather obvious trivial solutions. And for many years, this was the most famous open problem in mathematics and there are thousands of incorrect solutions of it until it was finally solved by Andrew Wiles in the 1990s. There are also some variations of this. For example, we can ask what about x to the four plus y to the four plus c to the four equals t to the four. Can we find a non-zero solution of this? Well, you can, but it's really rather difficult. So one solution would be 20615673 to the four plus 26, sorry, equals 2682440 to the four plus 15365639 to the four plus 18796760 to the four. Obviously people had to use a computer to find this because it'd be a bit difficult to find it by hand using trial and error. So as you see from this, it can be really hard to tell whether or not a diaphanetine equation has a solution or not. Sometimes there may be no solutions but it's really difficult to prove. Sometimes they might be solutions but they're really difficult to find because they're rather big. So you can ask the following problem. This is Hilbert's 10th problem. Is there an algorithm to solve all diaphanetine equations? What this means is given a diaphanetine equation, f, x, y, z, and so on, the algorithm should either give you the output tell you there are no solutions or it should find a solution. Now having an algorithm to find a solution if there is a solution is actually really easy. All you have to do is to check all possible solutions in order until you find a solution. The problem is that if the diaphanetine equation has no solutions, your algorithm might just go on forever without ever stopping. So we want the algorithm to actually stop and tell you there are no solutions if there are no solutions. And the answer is that this is not possible in general. It doesn't mean it's impossible for any particular diaphanetine equation. I mean for particular diaphanetine equations we can quite often solve this. And the problem is there's no algorithm that will work for all diaphanetine equations. So this was originally shown by Robinson, Davies, Putnam, and the final step was done by Matje Sevic. Actually he did the last step in showing that this was impossible when he was about 22 or 23 or something. So he was pretty young. So another example of a diaphanetine equation is variations of Pell's equation. So we can ask, can you solve the equation x squared plus equals dy squared plus or minus what? So let me give some examples of this. So we can ask, can you solve x squared equals 94 y squared plus one? Well it turns out the smallest solution is x equals 2143295 and y equals 221064. So again, this is not something that would be easy to do by trial and error. But as we'll see later on in the course that there's a sort of reasonably efficient way of finding these solutions using continued fractions. This is another historical example of this problem. It's just done by Dudney who had some puzzle books. These are actually puzzles that I think he gave in a newspaper column or something. So here he has a puzzle about the Battle of Hastings where he says, well, the men of Harold formed 61 squares, all of the same size. And when Harold joined, they all formed one big square. So what you're trying to do is to solve the equation x squared equals 61y squared plus one. Oops, let me make a fight down a bit so you can see that. And again, the smallest solution of this turned out to be ridiculously big. So here's Dudney's solution. If you look here, he has the total number of men is this rather ridiculously large number here. So even harmless looking diaphanetine equations with rather small coefficients can have absurdly large smallest solutions. Another variation of this is the equation x cubed plus y cubed equals z cubed plus t cubed. And obviously there are trivial solutions because you can take x equals z and y equals t and so on. But we would like a non-trivial solution. And the smallest non-trivial solution would be 1729 is equal to 12 cubed plus one cubed, which is equal to 10 cubed plus nine cubed. There's a famous legend about this number that the mathematician Hardy claimed that he was once visiting Ramanujan when he was ill and mentioned to Ramanujan that his taxicab number was 1729 and said this is a very boring number. And Ramanujan said, no, it's a very interesting number. It's the smallest number that's the sum of two cubes in two different ways. And while this is a charming story, I'm rather skeptical of it because even among mathematicians you don't start conversations by telling the other person your taxicab number. I suspect what might have happened is that Hardy knew perfectly well that this number was the sum of two cubes in two different ways and was just giving it to Ramanujan in a sort of attempt to cheer him up because Ramanujan was feeling a bit miserable or something like that. But anyway, this is now notorious or famous or something as being Ramanujan's taxicab number. So the next part of the lecture I'll record separately because this is getting a little bit long and I'll be continuing examples of some problems in number theory.


Good morning, my beautiful internet friends. I am so very excited to bring this video to you guys because it has been a long time coming and because you are a part of it. When I had 100,000 subscribers on this channel a couple weeks ago, someone suggested that I make a community video of all of us together. So you set in submissions talking to me and I'm going to watch them right now and put them together and react to them. Before we start, please know that I will watch every bit of every one of these videos. You're only gonna see little clips of the videos because I don't want this video to be 40 minutes long but if you only see your face for a second please know that I did listen to your whole message. I haven't watched any of these at all yet. I wanted to save it all for right now. And I think that's all I have to say before we actually dive in. So without further ado, footless show community. Let's do this. Hello my beautiful internet friends. Now I can say to you, here I am in my bathroom in the Netherlands to tell you how much you inspire me every day because you are really honest and true to yourself and I really admire that. Ugh, I don't know how I'm gonna get through these without just bawling my eyes out. It's beyond amazing to actually see faces. You guys have helped me in some of my darkest times probably without even realizing it. Back in September of 2018, I ended up falling and pretty much destroying my left ankle. I've had a couple surgeries. I'm about to have a third. And in the beginning of this journey, I was in a very, very dark place and then I found you guys. And Joe, I've been able to hear your stories and see how you keep it real with everybody. And it's helped me realize I'm not alone and know that it's gonna be okay. I'm so incredibly glad that this community could help you and I really hope that your surgery went well. Surgeries are definitely not my favorite either. My name is Natalie. I'm from, I'm living in Edmonton, Alberta, Canada. Hi Natalie. And over the past few months, your videos both on Trauma Talk and Phyllis Joe have comforted me and made me feel not as alone in my struggles. While I don't deal with chronic illness or pain and I don't have a mental health diagnosis, I do believe in my early 20s, I went through a traumatic experience and all my emotional distress and tolerance and struggles have kind of come from that. My partner doesn't watch YouTube. And now when I talk to him about topics or reflections that come from watching your videos, he knows you as the Footless Joe girl. That's how much I'm worried about coming. Also, reading the comments of your videos, I know that I'm not alone in my struggles with things like leaving a faith and religion and that's really comforting. It can be a really terrifying thing. And so your channel and this community and the other channels that you've introduced me to, have been an amazing resource in my own journey. So thank you and congrats on 100,000 subscribers. Thank you so much. Hey Joe, my name is Elizabeth. Hello. When I was growing up, I had a lot of physical challenges that I had to deal with and I didn't have anything like this.  Information access was very difficult. It would have been very different. Having that information access is huge to settle some of your fears. I think that's so true. Having some idea of what you're looking at can make things so much simpler and it's so hard when you don't have that. Channels like yours also really provide support in a sense that you can connect with other people who have experienced similar things and you can feel like you're not alone, I guess is really what it comes down to. I also really think it's great that you're willing to show all of the ups and the downs because life isn't just a straight line. There's a lot of zigzagging and that's normal and I think it's important for people to see that that's normal. Thank you so much Elizabeth. That's incredibly encouraging for me to hear. Your words are beautiful. Hi Joe from my giant beanbag and Mr. Link says hi. Link's the sausage, I follow him on Instagram. I just wanted to point out how you go through a lot and you're so genuinely positive about it, you actively take the steps to be positive about it and you're genuine when you can't be. You've definitely been vulnerable. Like you said you had a hard time doing and I do the same thing and actually this past year has been a wild ride for me. I mean, yeah, I still have all four of my limbs but I have gotten a new diagnosis for my mental health and things have been really crazy surrounding that so it's really refreshing to see. Dealing with a mental health diagnosis is huge. I don't think there's any comparison in all of this. It's not like I lost a leg and anyone else is dealing with here or there up here and I think it's just, we're all dealing with stuff and it's sometimes really heavy and really hard. I'm sorry that this year has been so much for you.  Say goodbye, Link's. Okay. Bye Link's. My name is John to Jarnit. Oh John. When I happened upon your videos, I never expected for them and for you to reach me so deeply. When you present topics and events that you've been dealing with, I find myself drawn into your stories. Often they evoke both happy and traumatic memories from my life but you provide a safe and comfortable place for me to do that. Thank you for that. Since I started following you, I feel like I'm hearing from a friend now when I see you post a new video. I find myself cheering for your successes and shedding tears when something has gone south. Thank you for letting me be a part of your video life. I knew there's like no actual way that I could get through this without crying. Thank you, John. Your words mean a lot to me. My name is Kayla. Hi, Kayla. I wanted to say that your channel is one that I feel like I've been waiting for since I lost my leg.  Yes, yes, exactly. Thing to see, like especially like your stretches and yoga for amputees or for that you do. Literally, I've said before like I would love to try yoga but it's difficult with not being able to bend my knee all the way and stuff. I saw that I'm like no way. Like that's perfect. And I can't wait to see you walk and hike and run and do all that stuff. Me too, thank you. I hope yoga goes well for you. I hope you get a chance to try it. That's really cool. That's so cool. Beautiful. Hi there, Joe. It's Dominique from New Zealand. Hi. I wanna say thank you so much for your videos. You've helped so many people. You're so positive. I just wish you the best on your recovery and hopefully, hopefully, fingers crossed that you can be running in no time. Love you so much. Bye. Thank you, fingers crossed. I really appreciate that. Hi, Joe. It's my land. Oh, hi. Congratulations and 100,000 subscribers. You have taught me so much about amputees and disability and I am so thrilled to be able to follow your journey and I can't wait to see more. My face literally is like hurting right here from smiling so much from seeing all your faces. Hi, Joe. I think your channel is amazing and I think you're so sweet and I love that on your channel you talk about whatever you wanna talk about and think is important. I like the little kitty in the corner there. Here's my, one of my cats. You guys are so cute together. I am literally so very proud of you. I feel like all of us have been on this journey along with you. We've cried with you. I actually found your channel, both of your channels. I was actually sitting alone in a theater. It was right before a musical rehearsal and I was literally feeling really down. I felt so horrible inside. I was like really, really, really depressed and to the point where I was thinking I'm not gonna see my 18th birthday. I'm down just ending it right now. I had no idea what I was really feeling or processing in that moment but I remember that I found one of your videos. I think it was the one where you were saying goodbye to your foot and I sat there and I don't know why it just really hit and impacted me in that moment. And I remember going, wow, like even though I'm going through this somebody's going through something as well. Somebody out there and then I kept binge watching your videos right there in that theater. You being vulnerable and saying, wow, I'm feeling this at this moment and I feel like quitting but I'm not gonna quit. And for me that was like, it made me realize that I hadn't processed emotions since the age of six. And I realized that I needed to learn how to love myself. So I just wanna say thank you so much for allowing me to feel my emotions with you. And now I reached my 18th birthday. It's crazy, I graduated. I don't think I'm gonna include the part of a video where I just started sobbing. I'm so happy that you made it to your 18th birthday and you're so strong and so brave for doing so. And I think you are so many of us. You're not alone. You're really, really not alone in what you're feeling and what you're going through and congratulations and happy birthday. Your channel always makes me feel very positive and you are positive always about everything.  Oh, thank you. I never think of myself as someone who really makes other people laugh so that means a lot to me. Hey Joe, congratulations on 100K, woo! I am currently studying to become a prosthetist. That's right. I found your channel very helpful because you're very vulnerable. As you mentioned in your video, although I'll never be able to experience what you are experiencing because no two people walk through the same journey and I also have no experience wearing a prosthesis myself. I'm gonna do the best I can to see it from the other side because I know from my experience wearing orthoses, it was very hard not having people treating me who understood what it was like to go to school in a back brace. A medical device is part of your everyday life and I wanna give that back. Thank you for all you do. I really appreciate it. Thank you for what you're doing. That's incredible. I think so few people would put the time and effort in to really understand. Thank you so much for being brave and sharing many life journeys with the world. I found you when I felt extremely alone in 2016 and your stories made me feel a validation that I wasn't getting anywhere else because I found strength in the stories of others like you. I now work with trauma survivors and strive to make the world a little bit better. That's amazing. That's really incredible. Hi, Joe. This would really need to be long when I'm gonna try and make a shot. I'm autistic and I had my diagnosis really late, which meant that I didn't really have support growing up and it would have been good to have it because I ended up finding myself in really traumatic situations and having to cope with a number of mental illnesses since I was 10. I started following your channel at the beginning and it matched it with the time when I was going back inpatient for in disorder treatment. Once again, to try to get around a few things that I had left untouched before. And I was terrified. I really was because I felt that there was no place for me in a happy and fulfilling life. I felt like I had no access to that. Something really weird happened. Watching your videos throughout the year made me feel understood and made me feel that I had a chance and a happy, loving life. Despite the amount of trauma and damage done, that sometimes makes me feel like I cannot live life and not just survive it. What you decided to share and the way you decided to do it made me see a possibility and it felt healing. If you feel like, please keep doing that because it does save lives. Thank you, thank you. And congratulations on getting the help that you need and deserve. I really understand the feeling that you're talking about. Like not feeling like there's a place for you in a happy and fulfilling life and it's a fight to feel like there is. We're both fighting for it and it is there. I'm Eva. Hey, Eva. When my dad just passed away last year, I found a channel right after I got jobless. And it helped me see that there's more, that I'm not that alone and that there's something I can watch forward to every day, every couple of days. And I get actually really excited when I get a notification that you uploaded a video or something to Instagram. So it brightens up my day a little bit. So for that, thank you. That means so much to me. Thank you for telling me and I'm so sorry for the year that you've had and I'm happy to be a small part of it. Hi, Joe, it's Beth. Oh, hi, Beth. Super uncomfortable making a video, but thank you so much for doing it for me. Love your channel. Look forward to every time you're on. Your channel has helped me to recognize the mental health struggles that I continue to struggle with, as well as chronic pain. It helps me to get through the day and just wanted to say thank you so much for everything you do. Thank you, Beth. My name is Heather. Hey, Heather. Joe, I've been watching your videos since you first started making them on this channel. I'm not sure why. I just stumbled onto them via YouTube algorithms. And I've really appreciated watching you grow and learning and growing with you and supporting you through your struggles while I'm going through my struggles in life. And I remember watching you on your island trip when I was writing a really stressful exam and going through a hard time in my life. And it was really nice watching people around the world support each other and work together to persevere. Your videos impact me a lot and I love seeing everyone's faces every day. Oh, thank you, Heather. I hope that exam went well for you. Hey, Joe. It's Charmaine here. Charmaine, hello. So basically, I think it's on a few levels really. Emotionally, you help us because you still have the time to show us your raw emotions, even when you have so much going on in your life and that kind of thing you could leave behind and not discuss. But I think you are amazing in the way that you handle all of that kind of thing. Also for me, it hits home when you talk and you show when you are vulnerable, when you are feeling pain mentally and physically. Like for me, it definitely hits home because I struggle to tell people how I'm feeling all the time. I just pretend and say, yes, I'm okay when I'm really not. Yeah, I get that. Your show has helped me practically because I had not even heard about the eye walk before I saw it on your channel. So I'd just like to say thank you so much for everything you do. Every single video is true and honest and caring. And also you take the time to read every single one of our comments, which I find absolutely amazing. I really do too. I don't have the time anymore to respond to all of them. I used to try to respond to like every single comment. It's a gift, but I feel bad that I can't. There's too many to respond to now, but I do read every single one. Thank you so much for putting your life on this YouTube channel, Footless Joe, that you have created. I came across your channel probably a month and a half ago and I binge watched all of your videos. That's crazy, thank you. And I just want to say thank you so much for putting your story out there because it's helped me come to some conclusions with doctors and I've been able to talk to my doctors and not feel afraid. I've been able to speak up for myself. I guess, I don't know. I feel like I have more courage after watching your videos and I really needed that. So thank you so much for that. Sending love from Texas. Thank you, you rock. Well, as a one amputee to another, my name is Chris. Hey, Chris. It's just been so different seeing someone dealing with the same things I've had to deal with. I mean, it's a case of, you know, it gives me strength and hope and courage and just also helps me to keep reminding me where I've come from. I've only had my prosthetic for about a year so I'm still relatively new to this. I had my amputation in March of 2018. So yeah, I'm very new to this. It's good to see how far I've been able to come. I'm really glad. Nice to see somebody on YouTube putting a face to this because it doesn't always happen. I think a lot of people don't understand and appreciate the struggles and then successes. It hasn't been easy. I was wheelchair bound for about five months following my amputation. But thank you for your site. Thank you for your page. Thank you. Thank you for putting a face to this. Thank you for putting a reality to this. Thank you. Hey guys, I'm Austin Patrick from Virginia. My house is way over here, somewhere in this area, about 15 miles away. It was probably at an abortion point in early December of 2018. Then, when my pain was at its peak, I happened upon a footless Jo's channel. I think there was only a few thousand subscribers back then but she reached... I think that's true. My heart quickly. Her videos became medicine. And I needed my view of other people or the way I treat other people to be adjusted. Jo showed me by example, what sort of person I should try to be. When I watch your videos or message with you or talk to you in the comments, I regain perspective in my life. I see what changes I should make. And I know and feel that I'm not alone. With personal direct help from you, I was convinced I'm not a lost cause and I finally saw help. That means so much to me.  Now I have God back on my side. Thank you for being so good with me. I can honestly say I might not be alive today without you. Thank you. It was so important that you made the decision to create this channel. And if there's ever a time where you think that you are not as important as other people because of your disability, which I know we couldn't sometimes think, then believe me, you are even more important than other people who have no disability because of the things you do for this community. I hope you know what you mean to us.  Because watching you grow like that made me really happy because there's no one who would deserve it more than you. Thank you. This is Sophie, the artist who did the artwork on the fundraiser shirt that we did. There are so many reasons why you're actually as special to me. I couldn't thank whoever is responsible for that, if it's destiny or someone else. And that you helped me so much and helped me grow and helped me get through stuff that was hard for me. And I think you do this for so many people and that's why I think that you deserve this. Thank you so much, Sophie.  I feel like I'm gonna be absorbing your stories and your videos and your faces for a long while to come. Thank you so much for sharing yourselves with me. Thank you for taking the time to make videos, especially for people who aren't usually video people and audio files. I feel like I could just keep repeating thank you for a long time. Thank you for this community. Thanks for being a part of it. Thank you to everybody who wanted to send in a video and couldn't or didn't just thanks for being here. Thanks for fighting through. Thanks for holding on. Thanks for watching my videos. Thanks. Thanks for being who you are. I am so overwhelmed in a very, very good way by watching all of this and so grateful. I think I'm gonna leave it there. I love you guys. I am thinking of you. I will be thinking of you and I will see you in the next video. Bye guys. Bye guys. Thank you. Have her from the sky.