Thank you very much. And my talk will be kind of my mathematical birthday present to Professor Koshivara. You will see that actually several ideas which you introduce in various subjects will appear in quite surprising way. So I will speak about quantum difference equations. So the basic algebra is the following. Fix a complex number. And that's a Q is between 0 and 1. And denoted by algebra AQ will be 3-algebra generated by two symbols, invertible symbols x hat and y hat. Modular the relations y hat x hat is equal to Q x hat y hat. So as a vector space, it can be written as Laurent polynomials. So consider Laurent polynomials, let's say, in y first, or maybe in x first, and then in y second, you can put a canonical in this form. So first, and then one can go maybe to high dimension. One can speak about AQ to power n for large n. So for such algebra like for differential operations, one can speak about holonomic modules. Maybe you can tell in general what is a holonomic module. M or n module is called holonomic. If there exists a filtration, 0 and 1 and so on, such that compatible with natural filtration on algebra in x and x inverse of y and inverse of degree 1, compatible with natural filtration on algebra, and such that dimension of mn is capital O of n to power n. The model is finally generated here. It's automatic here. OK. And is there an inequality or that it can be that if the dimension is non-zero, that the dimension is for a non-zero module, the dimension is at least something? Yeah. No, I think it's a good definition. So in the equation, both x and x inverse have degree 1? Yeah, yeah. Degree of x, x plus minus i, y plus minus i degrees. OK, now I will first describe the case of n equal to 1. And I'll do it in parallel with differential equations. So we have q difference equations. It is algebra of q, aq. And in usual differential equations, you do something like this. You do polynomial differential equations in one variable. And also this notion of filtration and a similar definition of a polynomial module. So when you study differential equations, first you study it's kind of our formal puncture disk. So it means that you replace your ring by this guy. And automatically, holonomic modules after reduction to the things are finite dimensional spaces over Lorentz series, in doubt with the action differentiation, with the connection. No torsion. No torsion, yeah, because it's a field, yeah, plus connection. And here one can also do kind of formal quantum puncture disk. One considers c of x and then take the similar ring here. And modules, again, will be finite dimensional Lorentz polynomials. And the action of y hat means that instead of connection, you get isomorphism of a bundle and pullback of the bundle where psi is dilation. So it's a completely similar story. And among holonomic modules for a puncture disk, you get a very special case. It's called regular holonomic. And the definition of regular holonomic for usual equation means that there exists a E0 in E, which is 0 is free c of x module and such that E is equal to E0 obtained by localization and invariant under operation x over dx. Because, in fact, here one can write the same ring in this way. E0 is invariant. E0 is invariant under x dx. Sorry? E0 is free, finally generated. But it's finally generated. I think it's automatically free. It's called lattice. So one can repeat the situation here. You get E0 in E, which will be c of x hat modules, and invariant under y hat. So it will be a notion of regular singularity. And first, we want to classify regular modules. And the regular modules in the usual situation are the following. It's all irreducible in the composable modules are correspond to Jordan blocks. Where lambda belongs to c module of z. So it belongs to the quotient set. One can say that one can see the coherent shifts with zero-dimensional support on the kind of bed quotient stack. You divide a fine line by z. So it's kind of pure algebraic description. And here's a story that's similar in the composable modules correspond to coherent shifts. No, there's all regular modules as the same coherent shifts with zero-dimensional support on c star divided by q to power z. But essentially, the same reason. Yeah, you have a frangable basic model. You want to write equations something like y hat of psi is equal to lambda of some kind of z psi. You write this equation as it belongs to c star. But when I replace psi by x to power n psi, you multiply z to some integer power of q. And now it's kind of a transcendental map. This thing is isomorphic to elliptic curve. So we change the algebraic structure. And also we can make a transcendental map. It's isomorphic to c star, c mod z is isomorphic to c star by exponential map. OK, that's all very simple. Now there is also a level to retain the composition for d-modules of a formal puncture disk, kind of a holonomic d-modules of this formal puncture disk. This is a billion category. It's an infinite direct sum over blocks, kind of. And each block is equivalent to this coherent shifts with zero-dimensional support on c mod z. And blocks are labeled by some set which is an uncountable set. And the set correspond to kind of irregular types, or expressions like this. You can see the set of all possible formal pieces of polynomials. Some more ck z to power kn. Consider a set of finite sums, such that for certain maybe m, maybe put, such that great common divisors of k such that ck is non-zero is equal to 1, or just second, what do I mean? We cannot, and m is equal to 1. We cannot divide by a common factor. And you're modelled by cyclic group action. ck goes to exponent 2 pi i k over m ck. You can see the cyclic modulable by free group action. So different branches with the same functions. And if you have such expression, you imagine this kind of, you consider all possible branches. You get this demodule, a typical example of demodule in a block. And here's the same story. I find that amplification. You get it, sorry. Then it contains it by a kind of regular singularity in m-sroot. And here's a very similar theorem in, sorry? x1 over m. Yeah, x1 over m, yes, you're right, yeah. x1 over m, yeah, then it will be right. And then you can multiply by a regular model in x1 over m and get this category. And here's the similar story. It was described by several people, I think by Bronovsky and Ginsburg, but not only. Then consider a halonomic Q difference modules over c, no, put it as x hat. It's again the sum of blocks. Each block is the same coherent shifts with zero dimension on my elliptic curve. Oh, c star for the museum. But a label set is countable. And in fact, it's just rational numbers. Yeah, what is geometric meaning of this? Yeah, so you have to see some kind of big difference here. It's countable sets here. You get continuous parameters. What is the origin of these two sets? In fact, it has a very clean geometric meaning. Again, I can do the following. In differential. Before when you did this thing, when you went to your first thing, you ended up with coherent sheaves. Just one of them, there were no blocks. How come it? No, no, it's regular sum, it corresponds to zero regular. It's one special block. It's one special block, kind of the most basic block. Now, what is the origin of these two sets labeling these blocks? The origin is actually the following one. In differential equation, one can see the spectrum of, let's say, c of x, multiplied by projective line. It's called lambda. Like this. Now, so you get a kind of product of very short interval, and it will be x coordinate, and lambda will be here to hit lambda is equal to infinity, in fact. Now, you can see the product. You can see this formal scheme. And you can see the Poisson bracket, dx. Or you can see maybe two form, dx over x times the lambda. This form has d lambda, sorry, what do I mean? dx over x times the lambda, yeah. Just a second. Yeah, this form has poles of order 1 and 2. It lambda equal to infinity has pole of order 1. And now, what you start to do, you start to make some blow-ups at various points. And sometimes, you see devices when the form has pole of order 1. And it turns out that these devices, which you see in all these things, are exactly corresponding to all possible blocks. Roughly speaking, if you have this Poisson series, you can see the graph of differential. You get certain curve, which maybe goes to intersection point. And you make many, many blow-ups. And eventually, you end to have devices all for the 1. But how do you get something to count? Yeah. Sorry? How do you get something to count? Because when you make a blow-up, you can blow up, and maybe you make some trees. You blow up in some real points inside. Not, yeah. And here, what we do, we consider spec over the, kind of again, spec of c of x hat. And maybe I put also times p1 with coordinate y, say. Or maybe I don't put hat. And consider form, which is the x over x, which dy over y. And this, my form, will have, I get two very short devices of poles, when y equal to 0 infinity and x close to 0. And everywhere, I have pole of 4 to 1. But now, if I make a blow-up, also exactly at the section point, again, get pole of 4 to 1, I make a blow-ups iteratively. And what I get, exactly all rational numbers, corresponding to, this explains the difference between situations, so I get countable set, because only countable many such devices. OK, so that's essentially all formal story. You see this big analogy, kind of one can try to describe in some more invariant way, what is this direct sum of copies of blocks. And in fact, it's direct sum of all rational numbers. You can see there's a following thing. You can see the, let's see, something like semi-stable bundles on Eq with slope A over B. Slope is rank over degree, maybe I'll put the same. Yeah, this category are all equivalent to category where Fourier-Muqai transform of shifts with zero-dimensional support by T-A theorem. And that's this kind of more invariant way to see it. OK. Yeah, now the next step, we should go to Stokes' phenomena, yeah? So we get things over formal power series. Now, what would we do the same things over convergent series? So it works also when Q is not a root of unity. It should be enterprise. Yes, yes, the same story works over any field of characteristic zero and Q is not a root of unity. But you don't use elliptic curve as a language, because there's no elliptic curve. And wouldn't you write the slope A over B equal? Slope, it's rank over degree, maybe I should put slope here. Sorry? Is there anything in the right column that is similar to E to the power of Poisson polynomial? Yeah, no, it's the claim these rational numbers are analogues of E to the power of Poisson polynomial. It's some kind of object corresponding to this exponent of Poisson polynomial. There's no exponential, yeah, okay. Now, if you consider over another field, germs of meromorphic functions, then there's this kind of DeLine-Malgrange theory of Stokes' filtrations. Go to it, yeah, so it's deep result, which also Masaki reproved with Piersche-Pirain in different way, in a sense. But what goes on here, one can ask the same question in QD from the equation case. So I can see the modules over this ring. Then there was a very nice result by recent result by Ramesh and Zhang, which can be formulated as a following. This holonomic demodule over this thing is equivalent to the following. It's a billion category. And what is the billion category? It's a vector bundles on elliptic curve and out this anti-hardener-simmert filtration. What does anti-hardener-simmert filtration mean? First, I should explain what is hardener-simmert filtration. It's in general considered bundles on the curve. Then for bundles, one can define the notion of a slope from a zero bundle. And then this notion of semi-stable bundles and this canonical filtration so that the slope increases. And associated gradients will be semi-stable bundles. But now in those bundles with the filtration labeled again by rational numbers, but such that this associated gradients are semi-stable, but the order is opposite. So this thing is not unique. But it turns out that this guy is again a billion category. And this corresponds to this model of an analytic puncture disk. What is the idea? It's again very similar to Deline-Marquardt filtration. Deline-Marquardt filtration, if you have differential equation, then you study asymptotic of solutions along various rays and you get filtration by order of growth and it satisfies some properties. And here, what is going on? You have differential kind of difference equation on puncture disk. So it means that you have your meromorphic bundle, you have some maybe vector space at each point with x, and then you identify with vector space at point with qx, q square x, and so on. And then I approach zero. Then if you have a vector in this bundle, suppose you write some equation, something like c of qx is equal to x to power lambda sub x. You write some equation, let me admit it. You write this equation with lambda, say integer number or rational number. Then what you see is that if you approach many, many times, you get the following. You get qnx to power lambda times qn minus 1x to power lambda times x to power lambda c of x. And then you get something like q to power n square over 2 times lambda times capital of n. So you see that if you transfer to zero, you can have different growth labeled by rational numbers. You get to filter. So it means that you get certain filtration. Before doing filtration, first you can make a vector bundle electric curve because it's an equivariate vector bundle.  And then for each point of this electric curve, you can see the filtration by growth. It turns out to be a holomorphic filtration. A social gradient will be semi-stable, and it will be one-to-one correspondence. Non-trivial result. And the final thing is just associated gradient with vector. Yes, in fact, the formula is associated gradient. Yeah, so it's all completely parallel to the usual story. And finally, there's some kind of little formulation how to go to global result. It's like in Delemargar's theory. If you know things locally, you just add local system and complement you. There is some following theorem, which is essentially corollary of Ramius-Junk result, which is the following. It's that's holonomic AQ modules is canonical equivalent to the following category. You can see the objects. I say what are objects in the morphine, it's obvious. You can see the coherent shifts on electric curve with two anti-hardener oxygen filtration. Now, so coherent shifts are a little bit more than vector bounds to get torsion shifts. What is the geometry of all the story? If you have these things, if you don't do this, consider this essential completion neighborhood over zero, you will have, again, vector bundle on a coherent shifts on the electric curve, or if you mode out ignore torsion. Suppose my bundles equations without torsion. And then you can approach zero and infinity. You can consider positive and negative orbit. And both gives filtration. I don't talk to each other. And this is essentially two anti-filtration. It's zero in stock, phenomenon. Infinity, yeah. That's nothing else, yeah. And this is a little bit of torsion because my module will be not really free as a module over Lorentz polynomial in X. And altogether, one can say that and consider like p1 cross p1 with this form dx over x dy over y. It has first of the pole at four corners. And this terms of filtration will correspond to all possible devices on the blow-up. So the filtration is on the bundle obtained by modding out by the torsion? Yeah, it's a bit, no, it's a bit delicate story how we get coherent shifts and two-filtration. But if my equation was a bundle over cx over x minus 1, then you get a bundle with two-filtration. And just a little story with torsion which I don't want to dwell. So it's not obvious a torsion in X corresponds to torsion on an elliptic curve, right? Yes, yeah, yeah, something like this, yeah. So you get this small story. Now, how to understand this two anti-hard-on-sink filtration? This is a kind of very general remark. Suppose we have a series of triangulated category and we have a bridging stability condition on it. So bridging stability condition means the following. You get a map from k0 of category 2c, which is essentially R plus IR, which is essentially oriented plane, not the complex numbers. It's additive map. And we have a class of semi-stable object with some extract semantics, which, for example, one of the way to say what is it, you say that you have a bounded t structure. And if you consider the heart of this t structure and apply z and remove 0 object, you lend to something set of complex numbers, which is imaginary part is positive or z is negative real number. And then you can define the slope as argument of this thing. And you want to ask that every object has filtration by semi-stable one with kind of canonical slopes. So that's a notion. If you have a category filtration, then I can associate canonical Artinian-Ibelian category, namely consisting of object in the heart of t structure plus 2 anti-hard-on-sink filtration. Object in the heart plus 2 anti-hink infiltrations. And when you consider associated gradients with this hard-on-sink filtration, you get some semi-stable object and apply center charge. You get some numbers. And what you get?  Equipment to one filtration, another filtration. Artinian, you mean object of finite length. Obra object of finite length, yeah. And particularly, if I apply this construction to the coherency of scientific curve, we get this category, which we obtain in this way. There is something nice about this category, namely the following. On the set of stability structures, there is action of a universal cover of p-jail plus 2R, or drill plus 2R. Universal cover of drill matrices is positive determinant. It's a kind of essentially rotate your central charge, and you really find heart of destruction appropriate way. And this action identifies corresponding kabilian categories. How to see it, why it's identified corresponding kabilian categories? If you fix some number of rays, you get maybe N plus and N minus will be number of rays going up and down. You can see it as a representation of function from the quiver to your triangulated category. You consider quiver A and quiver N as a total number of 6 minus 1. And on this quiver up to shift to get N plus and plus and minus specific objects. And you want this object to go to semi-stable objects, some condition on a factor from the quiver to your category. And then it's things in variant under rotation group. And the corollary of this thing, it's the following one. On this algebra, on Aq, x, sl2z, maybe semi-steric product pc star squared, just by monomial transformation, replace x hat by y hat by monomials multiplied by powers. So it gets action of some group. The question, why this group acts on this kabilian category? Yeah, it's not clear at all. But when you have elliptic curve, then the space of stability structures, it's essentially one orbit of gel to R. That is well known. And hence, you get essentially canonical kabilian category associated to derived category of elliptic curve. That's the corollary of all this picture. So it explains this symmetry in a nice way. SL2z, acting by automapism of the algebra, it goes from one halonomic model to another. And I want to, in this description, to see. The man in the pacing x and y by monomial. Yeah, you get Fourier transform. x goes to y, y goes to x inverse. So when you consider those things, there's two artichokes and the same confectrations. Is it always true that it's a nabilian category? Yeah, it's a nabilian category, yeah. It's not obvious. Yeah, but it's not difficult here. OK, yeah, so that's a nice story. And you can essentially no conductors here. Now I want to go to high dimensions. But surprisingly, in the meta-predictive version, you have the double cover. Yeah, no, no, it hits universal cover because it will give you shift functions. It's not too periodic. Now my goal will be to formulate Riemann-Hilbert correspondence for a q tends to power n for any n. Yeah, I will not prove it. Yeah, but I don't know how to prove it, I have to say. But the formulation is already to get exact formulation. What is the right-hand side? It's already on a trivial task, and you'll see it's something much, much simpler than a usual differential equation when you get various, this again, stocks phenomena in high dimensions. And here are the stories and the following. First of all, when you consider holonomic aq2n modules, it is inductive limit when you get of certain subcategory and get fully-faithful embeddings of smaller categories. And indexing set will be something which can be called analog of singular support. Consider bound support by some larger, larger set in a certain sense. What is support? In this case, the definition of support will be closed Lagrangian polyhedral subset, rational polyhedral subset, polyhedral cone, even called L in R2n, which contains rational lattice, q2n, with standard symplectic form. What means it's just closed subset, which is finite union of a simplicial Lagrangian polyhedral cones. So you have maybe n vectors in q2n. So it's a linear independent. And the symplectic pairing is 0. You can see the sum of what it will be. Where is the subrational network? With rational, yeah, it will be non-zero vectors, which are linearly independent. And to your general basis, you take all possible subsets. In case of n equal 1, what you will do? In case of n equal 1, you get the same picture which you have before. You get just finite collection of rational rays, rays in n equal to 1. So what's the answer to the similar single support? Yeah, I will explain. Yeah, it's some better notion of singular support. In a sense. First of all, if you have a model, why you get such a gadget? It's essentially you follow various filtrations. Maybe I'll take the following analogy. Suppose we have an algebraic Lagrangian subr... Algebraic Lagrangian subvariety, L. City can see start to power n. Instead of polynomial demodule, they take Lagrangian variety. Then, this is start to power n is log Calabi-Yau. Oh, and in fact, it's log symplectic. What is the definition of d x, d y, d y over, d x over h? Yes, yes, yes, yeah, it's a standard form. We have this form. It's a log symplectic manifold. In particular, it's log Calabi-Yau. It has a volume element. And then, one can see the various toric compactifications of this guy. And then, if you have toric compactification, you see what intersection of device that's infinity intersect. And you get subset, conical subset, just for any algebraic variety. And for Lagrangian, you get conical Lagrangian variety. I think it's maybe called Tevelev construction or something like this, yeah. It's similar to a projection of Berkowitz spectrum, which yesterday was heard from Tony Youtok, yeah. Yeah, so you got kind of, you consider maybe curves going to infinity. You consider logarithm of norms and you get various thing, yeah. In the same story, it works for holonomic demodals. One can work with appropriate filtration. Holonomic demodals says that it's supported at infinity, in a sense, on this union of Lagrangian con. Yeah, by the way, the story, I have a suggestion what to do for usual differential equations. What will be the same look of better single of support. In general, if m omega is algebraic symplectic variety, you can call an analog of this possible support as a following sets. First, we can consider some huge set, divisorial valuations on field of rational functions on m, puts complex numbers, but it's kind of irrelevant. Is that differential equation or differential equation? No, in general, I want to say something which works both for differential, differential equations and even more. You consider divisorial valuations, so it means that you kind of glue device at infinity to m, set at near generic point. My form is, look like this, it's a logarithmic form and it has the pole. Yeah, so you get this set of divisorial relations. It's in a kind of trigonometric case, it will be rational race. Rational race, and for the case of cotangent bundle, if m is to start to part in, d is to be a rational race in q to n, and if m is cotangent bundle, it's uncountable continuous parameters. It's exactly correspond to this irregular terms for singular points in the case of dimension one. And then one considers a following analog of polyhedral cone will be the following object. Suppose you get some compactification, your top form has the following shape. You get intersection of n divisors and form has local coordinates looks like this. So you get n divisors, normally crossing divisors. And then you can start to all make, you get n various n points in these things, but also you can start make blow-ups and you feel rational points in a simplex. Get rational, so you get analog of this n dimensional simplices and you consider maybe n minus one dimensional simplices, it will be impracticization of the story. And then you consider finite unions of such subsets will be analog of support, yeah. Okay, yeah, but it's- So that discovers support in both cases. Yeah, but for differential equation, it's kind of supported some more delicate ways. You treat all irregular paths and it's kind of, yeah, not stock's phenomenon, formal classification, yeah. Court is it related to the support of the associated graded of a good filtration? No, usual conical support obtained by some, you remove a lot of information from this, I think it's, it's kind of better notion. Yeah, but I have to, sorry? Sorry? If you want to, you say it applies to the AQ to answer n. Yeah. What's the simplex? She start to power to n, yeah. Yeah, yeah, yeah. But now one can do the following. If you have this L, which is kind of polyhedral, whatever, whatever stuff in polyhedral Lagrangian cone, it's a singular Lagrangian variety, kind of singular Lagrangian variety. It's not smooth. It's kind of maybe, here's a bit less linear. And there was a general idea that if you have such variety, you get canonical kind of combinatorial type of categories of Foucaille type category associated to L. In fact, L will be automatically, L will be automatically exact Lagrangian cone because it's conical. If you have form, PDQ, you can integrate along race. Yes, it's even better than just Lagrangian variety. But anyhow, we have canonical Foucaille type category which is described. You take some two-wheeler neighborhood, make some flows. But thanks to kind of progress with Guillemot, Shapira and Kishavara, we know that it can be described as category of constructible shifts or maybe on some domain on a C infinity manifold. So that singular support in microlocal support is belongs to some conical, Legendre and subset. And so there's really no good way to go through this conical in symplectic case, the conical in cotangent space case. One can, for example, add one variable because Lagrangian manifold is exact one variable and then produce eventually something in n plus one dimension, certain conical in cotangent direction, direction subspace. And then you get something very, very concrete. And in the case of this collection of race, you get A n categories, A n number of spikes minus one. And the description will be the following. The conjecture is the following. This category of holonomic modules, holonomic Aq to power n modules in support is contained in given L. So it will be category which is much like we specify irregularity and gratis should be equivalent to the following things. You can see it will be full subcategory in dB of n's power of elliptic curve, multiply it, make tensor product, it makes sense for this category of this constructible shifts are sorted to L. Satisfying some property. And the property will be completely analogous to what I said that for each ray, I have a semi-stable object over the certain slope. And the property is the following. When you consider objects on this guy, you get, just forgetting about dB of elliptic curve of a billion variety, you get a smooth locus of this singular grand set will be just union of some cones. And then on each interior of this, each simplicial cone, you will have a complex and then we'll be some related in one way to some very intersection points. In particular, you get a function from this category to dB of vector space for each smooth point. Strictly speaking, also it's fun to define up to shift. You should choose argument of restriction of volume element. And also if you have a rational Lagrangian variety, a rational Lagrangian subspace. It's kind of Lagrangian subspace in Q2N. You have a canonical subcategory in dB of a billion variety. It's defined the following way. On this gadget, there's section of some, again, universal, some central extension of Cp2Nz. But for real, okay, transform. Yeah, yeah, so a Cartesian product, yeah. You can see the n-dimensional of a billion variety and then free MOC transform gives the action of simplecative group of extension of, central extension of simplecative group here. And the same group acts on the space of Lagrangian subspace, it's transitively, for example, you have kind of horizontal subspace. And for horizontal subspace, QN, we are sourced to category dB, consider category of coherent shifts with zero-dimensional support on e to power n. And then, then they put the constraint that if you apply each factor, lent to the corresponding category, it's everything's well-defined. It's completely mutation of the story which I explained for n-cal-1. And the conjecture, that's its right-hand side answer for the deli, Riemann-Hilbert correspondence. Yeah, so there are many reasons to believe in it. I don't have a pro, but I think it should be really easy. What sounds is the analogous to what you said for n-cal-1, because there had some... For n-cal-1, sorry? Yeah, you have coherent shifts on elliptic current in here. Yes, yes, no, I said that this thing that I put on each array, it's a copy of coherent shifts, but they apply some element of fsl to z, yeah. And what does it mean, support is contained in L? Support contained of d-modules contained in L. It means that if you, it has certain, it should put maybe certain filtration, new filtration on algebra, and you can imitate notion if you have coherent shift on Lagrange variety, when Lagrange variety goes in a certain direction, you imitate the definition, and you do sense it was quantum algebra. Yeah, in fact, the conjecture, it's all works. If you replace C by non-archimedean field, and you're still between less than one, you can just, you can do the same story, but for non-archimedean field, or any characteristic, yeah, it will be, again, this kind of Riemann-Hilbert correspondence, and it's actually a very nice example, because like in usual Riemann-Hilbert correspondence, we get kind of two algebraic variety, two complex varieties of the same algebraic structures, but it doesn't work for Q difference equation. No, for differential equation, for non-archimedean field, because they don't have monodromy, but here for Q difference, I think it's really analytic problems are kind of the same, both in complex numbers and non-archimedean. Field, and maybe the last very small remark, it's about another SL2Z, which goes here. When you write Q is equal to exponent of two pi I tau, and you take maybe tau prime is equal to one one tau, get Q prime, you get the same category, how it acts on, let's say, let's say in one variable, how it acts from one equation to another, and here I get the following suggestion, it's similar through the following story. Like for differential equations, one can consider space of Schwarz functions, dual space to Schwarz functions of C. It's a module over Cx over dx, but also it's a model of a complex conjugate. It's a bimodule. And it was Kashiwara conjecture proven by Michizuki that if you have a holonomic module, consider Holmes through this guy, you get a holonomic module over dual guy, and it's anti-equivalence of categories. And here it seems to be the same story, but without holomorphic complex conjugation goes on. How go from tau to one over tau? Namely, you can see there's a following thing, you can see that Schwarz space is dual to, distributions are dual to Schwarz space. And analog of Schwarz space will be the following. It will be a holomorphic function on C, some coordinate Z, with the following property. You get one tau and consider basic trees. Yeah, so now a tau will be not a negative real number. It's actually works even for real numbers. And you can see the functions said that along these directions have exponential decay. Yeah, so you get this set of vector space of functions. And there is kind of a natural, at least some class of dual expressions. Namely, dual space will have functions which has poles in the neighborhood of two sectors. And you integrate over this, over the path. And now you have exponential growth along these lines. Yeah, so it's some kind of hyper function story. And also you get analog of S prime. And there are two commuting algebras by shift, Z goes to Z plus one. And function multiplied by exponent of two pi i tau Z. You get one algebra, or also Z goes to Z plus one over tau. And F goes to exponent of two pi i Z. It's kind of Fadeev proposed two commuting quantum torus acting here and contractures it the same story holds. That if you go home, you get space of distributions. And if you have harmonic demos, you want to consider solutions. So you get dual harmonic demos on dual torus. And this will be eventually be exactly this model transformation on logarithm of tau, okay. Thank you. Thank you very much. Mike seems to have a question, so, three more, yeah? Is there an analog of regular? No, no, no, nothing, yeah, it's all irregular. In one dimensional case, there is, but. Yeah, so that's what we're talking about. Yeah, because there are only two singular points.  Nothing, yeah, yeah? Yeah, for example, you're supposed to be informed in terms of the queue. Yeah. And if you change the queue, what do you do? I have separate queues, yeah. Then the category will be much, much smaller, yeah. If the logarithm is linear, it will be no object, I think it's interesting. Oh, yeah, yeah. And similarly, in algebraic situation, you'll have no logarithm variety if you take generic, simplecative form, yeah. Oh, yeah, yeah. Yeah. Was that a question? Yeah. So in the conjecture, do you start from autonomic modules or DB? No, autonomic ones, abelian category. Abelian category, actually. And then you have a DB of. Yeah, but consider object in this category with the property, when you have restriction, you go to appropriate t-structure in, yeah. Yeah, one can formulate it pretty exactly, yeah. More questions? At the beginning, when you went through your, I think, different equations and different equations, you had C star on one side and the other on the other side. Was that C star really interior of a nodal curve, or? C star, I mean, I was wondering if it was like a limit of elliptic curves, or a limit. Yeah, no, no, there is something called elliptic algebras. I have to say, you can see the CP2 minus smooth cubic, you can remove, kind of, instead of triangle, which we can see the C star square, you can see the smooth cubic, and there's some algebra, but here's the whole story, it's kind of really trivial, get only one divisor, you can do, and I think this whole series have only one block, but still get Riemann-Hilbert's correspondence by Eric Reins, I think, make some Hilbert, Riemann-Hilbert's correspondence for this sync with itself, with different parameters. Any other questions? Cluster, mutations? Yeah, no, no, no, in principle, all the C star to part two, and it's only approximation to cluster variety. Yeah, it should be some bigger story, I would tell you. But you don't know what would be the other side? Yeah, I don't know, yeah. That's what I meant, yeah. Instead of this AQ, the Ns take these deformations of the cluster out, but you expect some. No, in fact, I expect the right-hand set will be Foucaille's category of the same variety. With the symplectic form, it will be real part of symplectic form, B-field imaginary part of symplectic form. That's a general Riemann-Hilbert's correspondence which you have as Yann for arbitrary symplectic manifolds. Okay, if there are no other questions, we start in five minutes and we thank Maxime again.


Welcome back to emergency medicine video series. In this segment, we will discuss the various rewarming techniques and the potential complications in the hypothermic patient. These rewarming techniques range from the least to the most invasive. As you can imagine, they also range from the slowest to the fastest way to rewarm the patient. For a patient who is in mild hypothermia, so 32 to 35 degrees Celsius, we will use passive external rewarming technique. That means dry, warm blankets, that is after the wet clothing has been removed in your primary survey, and p.o. hydration with dextrose containing drinks. This is a slow technique, and the patient needs to be able to produce heat to rewarm themselves. For moderate hypothermia, we will add active external rewarming technique. We will use forced air blankets that has warm air coming out from the little holes in the blanket themselves. We will give warmed IV fluids. We also would give warmed humidified oxygen. Since, as we discussed from part two, that 30% of the heat lost in the body comes from the lungs. Warm humidified oxygen decreases that amount of heat loss. For severe hypothermia, we will use all the above mentioned methods and then active internal rewarming. This method warms the different body cavities. There are four main body cavities that rewarm actively. First, the pleural space. Two chest tubes are inserted in each lung. Like so. Warm saline is pushed through two of the chest tubes and drain out through the remaining two. This is called pleural lavage. Next, the bladder. A three-way catheter is put in and warm fluid is used to do a continuous bladder irrigation. Third space is the peritoneal space. Two catheters are put in the peritoneal space. Similar to pleural lavage, warm fluid is put in through the first tube and drain through the second catheter. This is known as peritoneal lavage. The last system that will be warmed is the vascular system. We shove the blood into a warming device and then give it back to the patient. It can be done by dialysis or we can do this through bypass. It can be done by traditional cardiopulmonary bypass or through ECMO, which stands for extra corporeal membrane oxygenation. These are done in conjunction with specialists. The basic premise is that the blood is shunted through something that warms them and returned back to the patient warmed. What complications can occur with rewarming? When we start to warm up the central core of the patient, heat can still conduct from the relatively warmer core to the peripheral tissues. That will cause the core temperature to drop. This is called afterdrop. When we apply heat, we will cause peripheral vasodilation. That can cause hypotension, not so known as rewarming shock. Lastly, lactic acid from peripheral tissue can decrease the pH of the blood once it goes back into central circulation. Once the peripheral tissues start to rewarm, it can cause increased metabolic demand on the hypothermic heart. That might also cause more acidosis. In terms of disposition, patients with moderate to severe hypothermia, especially those who have had active internal rewarming, need to stay in the hospital. To summarize, we've discussed the causes and effects of hypothermia, the approach and the rewarming techniques you might want to use in the emergency department for these patients. Thank you for watching.


Hello in this lecture we're going to work a problem using a process cost system. So remember when we use a process cost system that means that we are producing inventory. We're manufacturing and when we manufacture we either generally talk about a job cost system or a process cost system. Job cost system usually has inventory that is different in nature and difference in size and whatnot. So we have to allocate information per job. Process cost means that we usually have very similar type of inventory and therefore we're going to allocate the cost in a process through the process like if we're refining oil or something. Support accounting instruction by clicking the link below giving you a free month membership to all of the content on our website broken out by category further broken out by course. Each course then organized in a logical reasonable fashion making it much more easy to find what you need then can be done on a YouTube page. We also include added resources such as Excel practice problems PDF files and more like QuickBooks backup files when applicable. So once again click the link below for a free month membership to our website and all the content on it something like that. So process cost system here we're going to calculate the equivalent units at this time. So here's our data on the left hand side and we are going to use this data in order to calculate the equivalent units in an Excel type worksheet and just thinking about kind of the format of the worksheet whether you do it by hand or whether it's in Excel is half the challenge like it is for many type of accounting problems just setting up the way this thing should look. So we're going to start off calculating the costs that we need to account for during this time period meaning where we allocated the cost obviously in the prior time period we're looking at the current cost that we're going to have to allocate we're going to use equivalent units in order to do so. So let's think about just dollar amounts first if we have beginning work in process we're going to start and they're going to give us these numbers these are going to be the dollar amounts we're not talking units here so we're going to say the direct materials are nine nine zero zero and the conversion are one one zero nine seven zero and therefore the total if we use our trusty sum function equals sum brackets of the nine thousand nine hundred plus the one hundred and ten seven nine seventy we come up to the one twenty eight seven we're going to add to that the dollar amount that was incurred during the month for direct materials and conversion remember that conversion means that it's the stuff that changed the direct material to the end product so that's going to be things like direct labor and overhead things that converted the direct material so that's going to be the two forty eight four and the one oh eight two nine seven zero for conversion once again we can sum those up summing these up using the sum function one two forty eight four plus the one million eighty two nine seventy gives us the one million three thirty one three seventy we can then sum them up this way as well having a total column allocating the total cost whoops need an equal sign there equals the sum of the nine thousand nine plus the two forty four tab equals the sum of the one hundred and ten nine seventy plus the one million eighty two nine seventy tab equals the sum of the one twenty eight seventy and the one million three thirty one three seventy and enter and of course it should also add up this way as well one million four uh four fifty two two forty is our total now that's pretty straightforward we know what the costs are in terms of the dollar amount what we don't know really is how to uh allocate those costs between these two departments we're assuming that department a happens before department b so we have to do something like packaging or you know a production before the packaging department and something like that so we have to allocate the items to a department uh and then it goes through a department and then we're going to take that inventory that working process apply it to the second process that process being b similar to a production if we're if we're making uh something we're going to do the do the production of the item and then maybe the packaging of the item next we're going to move from talking about dollar amounts at this point to talking about unit amounts and we're going to use the two of these of course to calculate the cost for equivalent units so we're moving to uh units and note we're going to basically have a calculation like this that we got to just kind of memorize and think through whenever we do this type of problem where we want to break this out into the format of beginning units that we have plus those units started and completed and then minus the ending uh units and i mean plus the total ending units gives us the total units available now i'll explain why we need to break that out in this way as we go and that's what we will be putting in here so we're going to say that the 3000 beginning inventory in units is given to us here so we have units 3000 those are in process so again if uh if they're if they're in the processing part they're already in there in in this month they're already in the working process in the processing what has not happened yet is the conversion the labor and the overhead in order to process those raw materials to finished goods so in terms of the raw material when we think about a FIFO method we're generally going to say that we're we're not going to have any uh conversion or any cost related to the material that's already in the process at the beginning that's going to be a usual assumption under FIFO because uh all the material went in there last time the only thing that's lacking in the process is going to be the direct labor and the overhead therefore the amount of cost that we're going to allocate of this additional cost is not going to be included because um all that material cost went in there prior before so we're going to say of course then the 3000 units times zero is going to be nothing for this month this period whatever this period may be and then on the conversion side it says that the beginning working process from last month that's already in process has 40 complete so how complete is it I mean what are we going to have to do to it this time in terms of labor and overhead we're going to have to say one minus 40 percent and it's going to have to be 60 percent so if it was 40 percent complete last time and it's a first in first out we're going to finish whatever's still in there this time and therefore we're going to have to do the other 60 percent so of these 3 000 total units we're accounting for equivalent units we're going to take times 60 percent is going to be that 1800 equivalent units so then we're going to have to calculate the units that are started and completed now the reason we want to do this is because of course if they started it and completed it then the equivalent units are going to be 100 percent both for materials and conversion that's why we want to break this out in this kind of funny calculation we haven't really worked with too much before and break out the amount that are started and completed for this equivalent unit calculation we can do that in this case by saying okay here's the amount that was completed uh the units completed and transferred out 22 200 and we're going to assume minus the the ones that were that those 22 200 includes the beginning inventory because we're assuming FIFO so if it was in there at the beginning that that's part of the unit that were transferred out therefore that minus the 3000 means that the amount that was started and completed this time is the 19 200 and if it was started and completed then 100 of the materials went in this time so we're going to say the 19 200 times 100 equivalent units are the same same with conversion we completely can finish the conversion process so 19 200 times 100 percent that's why we do this started and completed and then we're going to have the stuff that's still in there it's still in the processing department at the end of the time period so we're going to say that that's going to be this uh given to us in this 2400 still in working process 2400 and they're going to have to tell us well how converted are those we started them we put them into process now and we haven't yet finished them how finished are they problems going to have to give us a number on that and they're going to say okay it's 40 percent complete but that 40 percent complete is going to be allocated to the conversion part of it because remember that we don't really need to have any number to tell us how complete it was in terms of material because the usual assumption for these when we talk about first in first out is that all the materials went into process so 100 percent of the materials are kind of assumed for most problems that went into process the only thing that hasn't happened is we haven't finished converting those materials using direct labor using the overhead to produce the finished inventory so we're going to say that for materials the 2004 times 100 percent they are uh total the same as far as conversion the conversion of those direct materials they're only 80 percent complete so for equivalent units for them for the conversion we've got the 2004 times the 80 percent and that's going to be only uh 1920 so if we sum these up then we're going to say sum of we've got the 3000 units plus the 19 started and completed plus the ending those are the units that we're gonna have to allocate out in terms of the cost that we had during this time period and in terms of equivalent units for materials we're summing up none of the beginning because those were all in there last time all of the ones that are started and completed and all of the ones that that were started and not completed are going to be in there and then in terms of the conversion equivalent units equals the sum of the uh 1008 the stuff we finished in the beginning plus of course all of that was started and completed and this and the portion of the stuff that's not yet done in the end so note what we have here we've got the total units in terms of total units we need to account for when we talk about equivalent units we're talking about equivalent and units in terms of either materials or conversion that's why we have two of them it's not like we have added these up and we have 44 520 equivalent units no we have equivalent units related to materials of 21 6 always going to be equal to or less than the total units and we have equivalent units in terms of conversion of 22 9 20 always of course being equal to or less than the total units that we're going to account for now that we have the total dollar amounts and the equivalent units we can calculate the cost per equivalent unit here and before I do that just want to point out that in order to do any of these problems again they're going to be very similar in nature just like many accounting problems and if you memorize just kind of this table whether you write it by hand in a test format or do it in excel or something like that so you want to basically be able to put this table into place then what we're going to do is we're just going to take the cost incurred this period in terms of dollar amounts divided by the equivalent units so the cost of this period just remember you're taking the cost that was incurred this period not the total cost because we're allocating the cost that were incurred during this time so we're going to say that's going to be the materials 248 4 and for the conversion we're talking about the dollar amount cost it's going to be the 1,082,970 for conversion and then we're going to divide that by equivalent units so for materials we had the 21,6 equivalent units and for the conversion we had the 22,9 so we may even want to put the dollar signs here so if this is going to be dollars this is going to be dollars i'm going to format right-click and go to format cells and make it maybe currency and add the add the dollar and take off the decimals so so this is units these are dollars so this is the current dollar amount divided by the units is going to give us the cost per unit the cost per equivalent unit we should say that's going to be the 248 4 divided by the 21,006 gives us 12 if i go to the home tab numbers and add decimals uh not that many decimals we got to 1150 let's do the same thing here we're going to say we're going to take the 1,082,970 divided by the 22,920 and once again go to the home tab numbers add decimals and we get the 4,725 those are rounded numbers so uh you know obviously we could be different by penny you could have some rounding differences okay so now that we have that then we can go to our uh this type of calculation out here the total cost accounted for and try to see okay how much of the cost this period are we going to allocate to the current production and remember what we're doing we're in production a and how much got transferred out to the next cycle so kind of like if we're producing something and then what goes from one department to like the packaging department we made something and then we're going to package something or something in a process system like that so now that we have our equivalent units we're going to come down here we're going to say let's take our beginning number in terms of total cost in working process that's going to equal total cost given here so we've got the total cost and then we're going to be allocating the cost to beginning inventory again you want to learn basically this format how to set up this format of table when you when you're working these problems now note that this table is going to be similar to this table we're going to be drawing data from this table down here and putting it into here so we're trying to calculate out the cost per equivalent unit and we're going to need the units and we're going to need the cost per equivalent units so we're talking about the cost at the beginning inventory we've got the direct materials and we're going to use the equivalent units i'm going to say where does this number come from it's going to be up here and and we're talking about material so it's this number here remember equivalent units was zero and therefore if we pull out the rest of the calculation zero equivalent units times whatever the cost per equivalent unit was which is of course 1150 will result in zero times the 1150 or of course zero when we talk about the conversion for the beginning inventory see we're going to go back this table i'm going to say this equals the equivalent unit table for the materials and i'm sorry for the conversion in this case the beginning for the conversion and that's going to be this 1008 and then we're going to get the cost per equivalent unit from our cost per equivalent unit table which is of course is 4725 tab and then if we multiply that out we're going to say okay we have 1008 times the 4725 that gives us the 8550 so therefore the total cost to complete the beginning inventory in terms of both the direct materials and conversion is going to equal the sum of the zero and the 85 150 now we're going to do the same thing for for the next one on this table so we just basically did this piece of it and i'll make that green we'll say we did that and now we're going to move down we're going to do this piece similar fashion down here and do our similar calculations so once again we've got the direct materials and conversion for um costs of units started and completed so i'm going to say this equals and we're going to go back up to our equivalent units table the 192002 tab sorry i'm going to go back up here for the direct materials it's going to be the same but it's going to be the materials here that's this number here tab and then we're going to get the numbered for the materials for the uh cost per equivalent unit 1150 tab multiply that out so this is the number of units for materials that we started and completed and it's 11.5 1150 per unit conversion same thing we're going to say this equals we're going to go to our units table so we're on this one we're on the conversion here 19200 because they all started and completed same number because they were started and completed and then the cost per unit the 47 25 if we multiply that out then we get the 19200 times the conversion per equivalent unit and that gives us the 907 200 let's sum that up summing up this the cost of units started and completed then being the sum of these here so then i'm going to i'm going to go back up here and say okay we found a home for this one and then we're going to do a similar calculation of course for the last one for the ending working process here and before we do that let's clean up some subtotals that we have so we have a subtotal here that's going to say the total cost of units at the beginning inventory so remember we're talking about the working process that's still in department a and the units that we're in there at the beginning allocating the cost that cost that's already in there is the stuff that happened last time like last month the 128 70 in terms of dollars that was given to us way up here at the beginning that's what's in the working process at the beginning then we allocate kated another 85 50 to those units that were in process at the beginning therefore if we add these two up we're going to say the sum of what was already in the working process plus the stuff that we allocated the dollar amount we allocated to those units that were in process at the beginning means that we have 205 920 there then we have this subtotal down here total cost of units transferred out meaning they're going out from the department we are looking at department a going to department b kind of like if we were producing something and then it goes out of the production department to the packaging department so that means that under a five-fold method everything that was in there at the beginning this whole 205 920 we're talking dollars now of the units that were in there at the beginning the dollar amount applied to those units plus everything that of course was started and completed that's the dollar amount that we need to take out of department a and move to department b so this is going to be the sum of we're just going to add up the 205 920 dollars allocated to the beginning units plus the 1 million 128 thousand dollars allocated to the unit started and completed this is the amount going out from the department a we're looking at to the next department department b now we're going to figure out what is still left in department a that has not yet gone out to department b in terms of dollar amounts we're going to use of course a similar calculation direct materials is going to be here we're going to look at the equivalent units i'm going to say this equals and scroll up to our equivalent units in terms of materials so here's the ending materials uh two four of course all of those are in there because we started it and that's the first thing we do when we start the new thing and we're going to say cost per equivalent units for materials we're going to equals that's this 1150 we calculated tab and then we're going to multiply that out so this equals the 2004 times the 11 dollars and 50 cents gives us the 2760 i'm sorry 27600 and then we're going to say this equals the conversion side of it same type of thing we're going to say the equivalent units for conversion end ending inventory is going to be this 80 complete or the 109 20 and the equivalent the cost per equivalent unit is this 4725 we calculated therefore we're going to say that this equals the equivalent units 109 20 times the 4725 gives us the 902 70 if we sum those up equals the sum of the 27600 plus the 90700 that gives us the 118 uh 118 320 now this last number is just basically a check figure so what we're doing is we're just going to say this equals the sum of the the total amount that's going to be transferred out plus the amount that's still in the working process for department a that gives us the 1,045 240 why is that a check figure because that should match what we calculated up top in terms of total costs the 1,042 240 the 1,042 240 so we allocated this total cost including what was in working process and that amount that was added during the process instead of breaking it out this way we have now breaking it out to the amount that needs to be uh transferred from the department a to the amount that's still in there in the working process for a so if we did that we could take a look at our journal entry over here and see what would the journal entry look like if we had a trial balance it might look something like this we just put I just made up some numbers here so we got the assets here we have this amount in the work in uh working process for a because that of course is the beginning plus all the uh costs for that time period and the idea of course being that we have to take it out of department a and put it into department b for the amount that was completed through the production process that's now going to like that second process possibly to a packaging department or something like that of course we're in balance here and we can make this journal entry so all it's doing is going from one uh asset account to another asset account that asset account being of course inventory type of counts of working process so it's going to go into b so i'm going to copy b that's going to be the debit and paste it 123 and the amount then is going to be equal to you know if we got to go scroll back over here i'm scrolling down the month table and we're going to take this number here why that number because that's the amount the cost that we're transferred out and once we take that out we'll be left with this number the 118 320 so we're going to take this and that's going to be the debit and the credit and we're going to take it out of depart working process for a so i'm going to right click and paste it 123 then if we were to post this to see what this does to an actual kind of trial balance here we would say that uh here we're going to say this equals the uh this debit and now of course working process goes up for b like the packaging department we're going to say and then working process for a what's still in there we're going to take out this amount with the credit and that leaves us that ending balance on our worksheet the 118 320 of course there's no effect on the income statement for this process because we're only uh allocating the asset through our production process at this point it will affect the income statement in the form of cost of good sold when we finally sell our inventory but of course it's going to have to go from b to finish goods and then finish goods to cost of good so


Okay, how many of you guys are using Instagram right now? How many of you guys are using Instagram for business? How many of you guys are satisfied with the amount of business you're getting from Instagram? Okay, so hopefully we can help you improve that today. I'm going to share with you some tips really quick about how you can generate leads utilizing Instagram. A little bit about me. I've been in corporate America over 10 years. Work with some of the larger corporations, AT&T, YP, Web.com. And I developed a passion for online. I still do it today. I help businesses create a 24-7 online sales force, also known as a website. And then from that I show them how to generate new clients and leads utilizing Instagram, Facebook, and PPC. So that's just a little bit about me. Let's dig in right now. This is going to be hands-on to get your phones out, if you like, and get on Instagram because I'm going to take you through some steps that can improve what to do. Why Instagram? Instagram has over 800 million users. I hear this from so many people. My customers on on Instagram, it's for teens, right? No, it's not. 33% of all Internet users are on Instagram and they're between the ages, and I think about 24 and 49. Your customers are using Instagram. It's got more engagement than any other social network out there right now. I say it all the time to people. It's white, hot. It's owned by Facebook. If you know anything about doing business on Facebook, Facebook has turned into a pay-to-play atmosphere. I feel like Instagram is going to go that way, but it's not right now. Ten times more engagement on Instagram. Over 70% of businesses are on Instagram, and they're not just playing around. They're doing business on Instagram. One of the things you can do is every business has a visual story. That's what the platform is all about, creating a visual story. This is a room of creatives, so everyone has a visual story to tell. So look for your visual story. So let's get into it. How to attract your ideal client with your bio. I don't have a lot of time today, so I'm going to give you some tips and tricks with your bio. Number one, use a close-up profile photo. I was going through Instagram yesterday and saw someone, they've got a full shot of them. When that's on mobile, you can't see it.  If it's your logo, use your logo. If it's an avatar, use an avatar. I use my avatar everywhere. That's just my brand. The other thing on Instagram, a lot of people don't know, you've got two names. You've got a username. Now the username, that's the name you utilize when you first set up your account. Okay? So that's usually your company name or your brand name. The other name you've got is an Instagram name. And a lot of people, they miss this because Instagram is searchable. People are searching on hashtags for things that you do and services that you offer. A lot of times people say, no, that's my brand. I'm leaving it just the way it is. But you know what, if no one knows you, they're never going to find you. That's one of the only searchable areas in your bio. So I suggest using keywords in that. If you do web design, make sure you include web design or Atlanta web design. So that when people type in, for instance, with me, social media marketing tips, my goal is to either come up under top people or hashtags. So make your utilize keywords in that bio. That's huge because people want to be like, oh, I'm a guru or this, or I'm a king of this or this or that. No one's searching for that. They're searching for products and services or things that they can get help with a lot of time. And they're also searching for recreational things, which also people offer those services as well. Other thing, a website link, step back real quick because if you don't have your Instagram account set up as a business page, you need to do that first. You get analytics with that. On that website link, make sure that you use a website shortener, like a Bitly or a Google, so that you make it trackable so that you can get analytics. You'll know how many people are clicking on my link. There are also other tools out there that will allow you to utilize multiple links. One is called Link Tree, where you can utilize multiple links to send them to different calls of to action. As the gentleman before me mentioned having a landing page, and that brings me to another thing. You want to have a call to action before they even click that link. Give them a freebie, give them a tip, give them a reason to click your website link. Also a bonus tip on this, use your notes app. You can use Evernote. You can use one that I like a lot is called WordCounter, because you've got 150 characters in that bio to tell who you are, what you do, and how you can help someone. Utilize those tools. The Character Counter or Word Counter, I love that because it'll tell you you've got 150 characters and you're good to go. Also utilize emojis in there. It's a great atmosphere.  Strategies for Instagram success, again use Instagram as a business page. Uh oh, make your profile public. I've seen a lot of businesses. They have a business page, but then it's private. No one's going to find you. Also link your Instagram business account to your Facebook page as well. Uh oh, hold on. I'm getting clicker happy here. Let me make sure I got those. Tips for engagement. You want to follow, like, and comment on related accounts, especially like competition or accounts that you admire that do similar things to what you do. Engage with their followers because they're obviously interested in what you do as well. Make your feed visually appealing. Again, we're in a room of creatives. That shouldn't be hard to do at all. Look at what some other people are doing. I've seen some great things from graphic designers out there, people in fashion. Kim Kardashian, look at her feed. It's visually appealing. Another thing, use relevant hashtags. That's a whole other subject matter in its own because, again, Instagram is people are using hashtags to search for entertaining things, services, things that interest them. I've picked up web design projects on Instagram. I've picked up social media accounts on Instagram. I've hired a writer on Instagram. Business is being done on Instagram, and it's white hot right now. Also use Instagram stories. 300 million people are using Instagram stories. That's a way for your business to have an online television show every single day. And you get a lot of engagement with that as well. Also, again, get inspiration for other Instagram accounts. Five common Instagram mistakes, how to avoid them. There's no clear message to what you do or what you offer. People come to your page. They don't know if it's your personal page or your business page. Keep them separate. If I'm coming for Instagram or social media tips, I don't want to see what you ate last night. Save that for your personal page. Putting links in your caption. You've got that one clickable link, and that's in your bio. I've seen people put links in their caption or description. It doesn't work because it's not clickable. Buying followers, trying to be like the Joneses.  And the other thing is that, it reminds me of a quote I read one time. Once, everyone starts at ground zero on Instagram. And then the other, don't compare your chapter one to somebody else's chapter 40. Took a lot of hard work to get there. Be patient with the process and it'll happen. Also using Instagram as a billboard. Me, me, me, me. No one wants to hear that. Help people. Give advice. Show some behind the scenes stuff. Show you working on your last website project. There's a ton of different things you could do. And then occasionally throw in the promo. And people will appreciate that more. The other one, no hashtag strategy. Because again, no one's going to find you. I've had people say, no, I don't want you to want a hashtag. Okay, what you're doing, you're limiting the amount of people who can actually find you for the types of services that you offer. Because that's a key strategy to Instagram. And that's a whole nother can of worms right there. What I'd like to do is invite anyone who's interested here to Instagram hashtag strategy workshop for businesses. It's absolutely free. It's this Thursday from 12 p.m. to 1.30 p.m. I'm going to cover things like how to use hashtags in 2018 to promote your brand. Why hashtags are a must on Instagram. Which hashtags to use? How many? That's a question I get a lot. How many should I be using? How to do proper hashtag research? Because you can't just say, hey, I'm going to use hashtag I love. Hashtag happy and expect to be found. It doesn't work like that. There's a strategy to it and I'm going to show you that. And again, this is going to be an absolutely free workshop. Also, how to use hashtags to spy on your competition. How to organize your hashtags as well. And then some of the best hashtag tools as well. So if you're interested, all you have to do is text IG hashtag to 38470. The other thing, if you guys are interested in a deeper dive into how you set up your bio, I've got a step by step video on YouTube. All you have to do is go to YouTube, type in your marketing lady, and you'll see that video. It's probably about five, seven minutes long. Maybe a little longer because it goes into detail. I'm going to show you step by step what I just covered for you in terms of how to optimize your bio. And you're also welcome to follow me there as well, because I'll be sharing weekly social media tips. Questions? Right there. Every day if you can. If you're doing it for business, I maybe use it three times a day at most. Sometimes I'll use mine only once a day, but I'll add to that story throughout the day. That's a whole other subject as well. And in fact, I'm going to have a workshop on that coming up soon, because I was in your same position. I was hearing all this Instagram stories, and it's not that difficult. Again, it's like your own television show. You've got 15 second clips to basically tell little tidbits about your company things you're doing that day. And you can add on to those clips as well. So that's like, I can't even, I don't have enough time to go into that. I wish I could. Effective both ways, but the strategy behind putting the hashtags in the comments is so that it doesn't look spammy. Because you get up to 30 hashtags, and a lot of people will tell you, well, you need to use three hashtags to start, then use seven hashtags. No, use between 15 and 18 hashtags. My theory behind that is I've got 30 lottery tickets I want to give to you. How many of those do you want? And that's what I'm saying. There's no difference in that. The biggest difference is that one, you just got a lot of hashtags after the other. If you put them in the comment, it hides it so it doesn't look as spammy. So both of them are going to get found. If you're using the right hashtag. And there's again a strategy behind that because hashtags that gets millions and millions of searches, it's going to be, you're going to get buried. Your goal is to be discovered. They send you reminders, most of them. Some of them do auto posting, but they won't include hashtags. I usually, I do my posting manually still. I'll use Buffer to remind me to post. Even for my clients, I still post manually just because of the hashtag strategy. I use Planoly just to visually look at mine before I'm posting. Yeah, this class is on hashtag strategies, but there's going to be some tools for hashtags. I also do workshops and things all the time. We're talking about different tools to you. So that those will be coming up. And I always take input from what people are wanting. I couldn't hear you. The best what? Best verify to get it verified on Instagram. For one, you need a huge account. So that's a whole nother topic in itself. And then there's a lot of misconceptions about how to get verified. You've got a lot of people put these hacks and things on there where they show you how to get verified, but only you can see it. So the bigger thing now is worrying about building up your community. And then you can take that step next. Any other questions? Great question. How do hashtags work and why do people use them? When you go to Google, you're looking for something, right? So you're typing in a search query, right? That's the same concept that people utilize on Instagram, except they're using hashtags. So even if I type in social media tips, it's going to give me top. It's going to give me people and it's going to give me hashtags.  There's going to be top results and recent results. So I hope I answered your question in the sense that it works similar to Google, the way that people are actually searching. All right. So folks, Tanya is local here in Atlanta. She gives workshops like this one that she mentioned all the time. So


So let's consider a few more cases of the derivative. So it's helpful to remember that on the graph of y equals f of x, the difference quotient is going to be the slope of the line between x equals a, and x equals a plus h on the graph. So let's see if we can find the limit as h goes to 0 from above of the difference quotient, as h goes to 0 from below of the difference quotient, and then the limit as h goes to 0. So in this case, the difference quotient itself is going to be the slope of the secant line between x equals 3 and x equals 3 plus h. Now since we want the limit as h goes to 0 from above, then x equals 3 plus h will be to the right of x equals 3. And so here we have our point at x equals 3, and someplace off to the right we have our point at x equals 3 plus h. And we'll draw the secant line. And now we're going to let h go to 0, and that means this point on the right is going to come closer to the point at x equals 3. And if we do that, it appears that as h goes to 0 from above, the slope of the secant line will be positive. Again, the difference quotient itself will be the slope of the secant line between two points, but since we want the limit as h goes to 0 from below, then x equals 3 plus h will be to the left of x equals 3. So we have our point at x equals 3, a point somewhere to the left, the secant line, and as the point gets close, we see that it appears that as h goes to 0 from below, the slope of the secant line will be negative. Now because the two limits disagree, a positive number can't be equal to a negative number. This means that the limit itself doesn't exist. Now remember the derivative is an algebraic statement, but it also has to do with the geometry of the graph. And note that the graph has a corner at x equal to 3. This is called a cusp. What about a graph like this? So notice that the point is actually at x equal to 3. So we can try to find the left and right limits. So again the difference quotient will be the slope of the secant line between x equals 3 and x equals 3 plus h, and since we want the limit as h goes to 0 from above, then x equals 3 plus h will be to the right of x equals 3. And as we let h go to 0, it appears that as h goes to 0 from above, the secant line becomes more and more horizontal, and so the slope of the secant line tends to 0. And so we might say that our limit as h goes to 0 from above is going to be 0. Again the difference quotient is the slope of the secant line between x equals 3 and x equals 3 plus h. Since we want the limit as h goes to 0 from below, then x equal to 3 plus h will actually be to the left of x equals 3. And we see that as h goes to 0 from below, the secant line will be more and more vertical, and so the slope of the secant line tends to infinity. And again this disagreement between the limit from the left and from the right means that the limit itself doesn't exist. And this time notice the graph is discontinuous at x equal to 3. And this leads to the following result, which is an important consequence of the derivative. The derivative is undefined wherever the graph has a cusp or a discontinuity.