Thus far in this mini-course, I think it's fair to say all the results I've been talking about are pretty well considered standard, and the proofs I've described are also using more or less standard methods. The title of the mini-course has the word classical, classical transversality results. What I mean by that is any result that you can express in the form for generic domain independent, almost complex structures, J, we have transversality for some particular class of J-halomorphic curves. So that's allowed to include some methods which cannot be called standard, and that's what I'm going to talk about today in particular, because everything I talked about so far with one minor exception when I talked about automatic transversality, everything else makes the restriction that we're only talking about somewhere injective curves, which of course is a big problem if you want to define big invariance such as Gromov-Witton-Thierier or SFT, and to a large extent you cannot really hope for classical transversality results to be true in the generality that you would need to define those theories, and I do already illustrated this briefly and I'll reiterate that a bit in a moment, but there are situations when those results can hold, when you don't need to go to much more general frameworks such as the polyfold theory in order to define everything. If you do get transversality for your honest holomorphic curves, it can make your life easier because the Cauchy-Riemann equation is much easier to handle and carry some sort of natural geometric information that you don't necessarily have in whatever perturbed equation you're going to solve in a more general framework, so sometimes it just requires much more originality in your way of thinking to get what you need. Let me give you an illustration, so for today I'm going to depart a little bit from the title of the course and not really talk about SFT, so I'm not going to talk about punctured holomorphic curves, but just closed holomorphic curves, let's say sigma and sigma prime will be closed surfaces, almost all of what I'm going to say can very likely be generalized to punctured holomorphic curves, but that's work in progress, so I'm not really going to touch upon it besides a few general ideas. I will just tell you about things that I know. So a multiple cover looks like this, I have a k to one holomorphic branched cover from one Riemann surface to another, I assume k is greater than one, sigma prime is the domain of some j-holomorphic curve into an almost complex manifold Wj, capital J, and then the composition of these two holomorphic maps gives me the multiple cover, so let's assume V is somewhere injective, U is now a k-fold covered holomorphic curve, and it'll be helpful to note that there's a relationship between the Euler characteristics of these two domains as they both figure into the index formula, so there's the Riemann-Herwitz formula that says minus Euler characteristic of sigma plus the degree of the cover times Euler characteristic of sigma prime equals this quantity that I like to call Z of d phi because that's literally an algebraic count of the zeros of d phi, or in other words it's the number of branch points counted with the orders of branching, number of critical points of phi, and since phi is holomorphic all of those count positively, so this is an integer that's always greater than or equal to zero. So we have this as a constraint relating the domains, if you've never seen the formula before you can just think of it this way, d phi is a section of a certain complex line bundle which you can easily write down, and then you can compute C1 of that line bundle, the answer is the left-hand side of this formula. That's so hard to understand, but about me, please. You can also draw some picture with, you can triangulate things and draw a picture of what branch points looks like, maybe you prefer that, but it's very quick thinking of it as C1, if I ever forget the formula that's how I remember. So now let's write down the index formulas that we get from Riemann-Roch. So the index of V, so W I assume is 2n dimensional, real 2n dimensional, the index of V by which I mean the virtual dimension of the modularized space of unparameterized holomorphic curves that V lives in. So that's not literally the front home index of the linearized Cauchy-Riemann operator, but it is by basically what I explained before, it is that plus the dimension of the relevant type mirror space. So it's the actual dimension of the modularized space, if transversality holds. And the formula is n minus 3 times Euler characteristic of the domain plus 2 first turn class evaluated on the homology class of V. How does that relate to the index of U? So same formula with a different domain. Now Riemann-Hurwitz tells me I can rewrite Euler characteristic of sigma as k times Euler characteristic of sigma prime minus this count of branch points. And of course C1 of U is just k times C1 of V, so I have 2 times that. And now I see sitting inside this formula, k times the index of V, because there's n minus 3 times Euler characteristic of sigma prime plus C1 of V over here. I just have an extra term. So I get k times the index of V minus n minus 3 times this count of branch points. So in particular, to be a little bit more concrete on what can go wrong with transversality for multiple covers, suppose that the underlying somewhere injective curve had its index 0. So that's generically a rigid isolated object in its modular space. And for generic J, it's also going to be stable under small perturbations of J, which means there's no way you can get rid of the multiply covered curve since there will always be a perturbed, multiple cover of your perturbed somewhere injective curve. But the index of U is then going to be minus n minus 3 times this non-negative count of branch points, which can easily be negative. At least if we're in dimension 8 or upward, unless the cover is unbranched, that's going to be a negative number. Dimension 8 and upwards, that's precisely where we stop being able to assume that our symplectic manifold is semi-positive. So there we run into trouble. This is likely to be a negative number in general. But we see that we cannot perturb that curve away. The situation is actually even worse. If you think about what kind of curves we already know must exist in a neighborhood of U, there's not just U itself. There's the other nearby branched covers of the same V. And those come in a non-trivial modular space in general. So the actual dimension of the modularized space of holomorphic curves near U, and when I say actual dimension, I'm not making the assumption that it's a smooth manifold or anything. It might not be. It might not even be a smooth orbifold. But it does contain a smooth orbifold, which I can identify very clearly, namely the space of branched covers of sigma over sigma prime of the same degree. So we get at least the dimension of the space of k to 1 branched covers modulo reparameterization. And that's a very well understood modularized space. That's just another modularized space of holomorphic curves living in dimension 2, if you like. And you can compute the dimension from the Riemann-Roch formula as usual. The answer is 2 times the number of branch points. And there's sort of a geometric interpretation. You might want to follow up on that. But you can just see that. This is my next sentence that you just interrupted. There's a geometric interpretation of this, because given a branched cover, you can find other branched covers nearby by moving around the positions of the branch points in the image. And those are not equivalent up to reparameterization. They give you different branched covers. And this is precisely the number of parameters you see by doing that. So this is basically classical. So we see that number is never going to match the actual index of U, unless possibly the cover is unbranched, or what, or if we're in dimension 2, but so what. We don't really care about dimension 2. So here it's fair to say transversality is generally not plausible for the multiple covers. I do want to make another observation about this, though. So again, the index of V is 0. So generically, by which I mean after possibly perturbing J, I'm allowed to assume that somewhere injective index 0 curves are immersed. Why is that? So I've discussed this with a couple of people since my last talk or discussion session or whatever it was. There was an exercise I posed about using automatic transversality to prove that in a four-dimensional symplectic co-boardism, generically an index 1 holomorphic cylinder will always be regular. And part of what you have to do to prove that is observe that whatever kind of holomorphic cylinder that one covers, if it's a multiple cover, is also going to be a sufficiently low index so that you can assume it's immersed. And then you can apply the automatic transversality criterion after you know that. So there's this general fact, which I don't have time to talk about in earnest. But one can show that if you take your usual moduli space, add a marked point to it. So that increases the dimension of the moduli space by 2. But now constrain that marked point by asking for the first derivative of your map to vanish at that marked point. So in general, that decreases the dimension of the moduli space by 2n. So the upshot of that is for generic j, the space of a holomorphic somewhere injective curves that are not immersed may be considered to be of co-dimension 2 minus 2n compared to the larger moduli space. So that means if my index is 0, I can assume that that space of non-immersed curves is empty. Therefore, this one's immersed. So I'm not going to say more about that. Let's just accept it for now. It's a j-holomorphic fact analogous to the standard differential topological fact that you can perturb smooth maps to be immersed, given the right dimensional conditions at least. So we have this. Now that means let's look at the normal bundle. I've got. Can you at least say why there's more to say than what you just said about why this statement is true? Yeah, because you have to work out the details. Go ahead and try it. Let me know how it goes. So the normal bundle is not the generalized, but the usual definition of normal bundle here. That's what it's C1 is. The generalized normal bundle of U is very easy to describe. U is not immersed because there can be branch points. There are critical points of U. But of course, the generalized normal bundle of U is just going to be the pullback of this normal bundle in V. So that's 2k times that number over there, C1 of V minus or the characteristic of sigma prime. So remember, I talked last time about restricting the linearized Cauchy-Riemann operator to the generalized normal bundle. That also gives you a Cauchy-Riemann-type operator on a bundle of rank 1 lower. Its threshold index is given directly by the Riemann-Roch formula. Let's see what it is. So d urn is the restriction to the normal bundle. And Riemann-Roch tells me, well, the rank of this bundle is n minus 1 as a complex bundle times the early characteristic of the domain plus 2c1, the generalized normal bundle. So Riemann-Hurwitz tells me what that first term is. We have this extra term counting the number of critical points. And then 2c1 and U, I've just written up here. It's am I missing something? People are saying there's a k missing. But where? No, there shouldn't be. It shouldn't be a k. Yeah. The last one. And the a is missing. Well. From the first line, you have two times the restriction. That's V, and this is U, right? That's why there's the k. No, I think this is OK. I'm doing this a slightly more roundabout way than I planned, accidentally. I've got k times n minus 1 or the characteristic minus another 2 or the characteristic plus twice c1 of V, then minus n minus 1 times count of Bradford. This is fine, because this number in the brackets here is 0, because I assume that V has index 0. That's the index formula. So I'm just left with minus n minus 1 Z of d phi. So what I notice about this is unlike the index of U that I wrote down, just looking at the normal operator, there's some predictable pattern. This number is always non-positive, which means it's conceivable that this normal operator might actually always be injective. And that's something geometrically meaningful, if it's true. So I'm going to actually state this as a conjecture. About a year ago at this time, I was calling it a theorem, but then an error was discovered in that proof. So this is a conjecture that says, so for generic J, all multiple covers U of somewhere injective index 0 curves V have normal Cauchy-Riemann operator injective, which has a nice consequence if it's true. What that actually implies is all of the other curves in the neighborhood of U are precisely the ones you already know about. They're just the other multiple covers of V. So I'm saying this result would give you a precise description of the modularized space of curves in near U as having exactly this dimension of the space of branched colors. Is that strong enough to get PPS integrality from a number of curves? That would be the intention, right? So the intention of a result like this would be to prove something like the Gopakumar-Vaffa conjecture, right? So the reason why this kind of thing is supposed to be interesting is that it means, in certain settings, if you want to compute Gromov-Witton invariance, you really only have to understand the somewhere injective curves. And the rest of it, of course, the multiple covers are not regular in the usual sense. So you have to do some kind of perturbation if you want to actually count them. But there's a standard way of doing this within homogeneous perturbations to the Cauchy-Riemann equation. And you can predict the count that you'll get because you see the entire modularized space in terms of the space of branched covers. It has an obstruction bundle. And you can compute the Euler class of that obstruction bundle. That will give you the answer, right? So this has nice corollaries in Gromov-Witton theory. What was the answer to that question? I think the answer to your question was yes. I'm not really sure how to interpret the index of this kind of view. Right. The main thing I want to say about it right now is that since it is generally negative, the operator can be injective. And what I really want to explain is how you interpret the fact that that operator is injective. So if the kernel of this operator is trivial, think about it like this. The following is a scenario you don't want. You don't want to have a sequence of curves that have different images from you converging to you. My claim is that all the sequences of curves that can converge to you are of the form v composed with some other branched cover. So they all have the same image. Now, if you have a sequence of curves with different images converging towards you, you can do this trick where you look at that as curves living in the normal bundle of you and rescale that normal bundle. So that as the curves approach, you rescale so that you see them not actually approaching but staying in some bounded subset of the normal bundle, apply Gromov compactness to that. That sequence is going to converge to some generally nodal holomorphic curve, which will have some component that you can interpret as something either in the kernel of this operator or in the kernel of some related operator corresponding to a branched cover of lesser degree. So if you know that these kernels are all trivial, that precludes this scenario. So it says really that the space of branched covers or the space of use that are branched covers of v is an open subset of the whole modular space. This is all speculation. I mean, the argument that I just described can be made fully rigorous. But of course, the conjecture is only a conjecture. We don't know if this is true. So I do want to talk about a special case of it, which we do know. Yes? Can you give an example when the hypotheses of the conjecture aren't satisfied where the desired result isn't true? You know, hopefully covered curves are approximated by curves with different images? I'd have to think about that a little bit. I mean, I certainly could come up with examples given enough time where, I mean, in particular, there's hardly any conditions here. The main condition is just that the simple curve has indexed zero. So in situations where that's not true, the possibility of a sequence of simple curves converging to something multiply covered is always something you have to worry about. You usually have to make some effort to avoid it. Doesn't Tyvers have some examples when he attacks Tori, and he has multiply-covered Tori that's a generically branched off double-covered in most kind of times? And I'm going to talk about that a little bit. But I don't think it's an example of this phenomenon. It's much nicer than that. Richard Heim has some nice example where you need generalistic analysis. OK, I'm not aware of that. OK. So here's an actual theorem, which is one of the cases of this conjecture. And this is in a joint paper that I wrote with Chris Garrick last year. I'm going to make the statement a little bit unnecessarily more complicated than I need just to illustrate how different it is from the results we talked about so far. So let's say fix an open subset u in our closed manifold m, closed symplectic manifold of dimension 2n, and fix also a tame, almost complex structure, J fix. Then I will say there exists a commieger subset J reg living, as you'd expect, in the space of all j's that are omega tame and match J fix outside of this subset, such that for all j's of this class, all unbranched covers, u, of somewhere injective index 0 curves, considered, call it v, contained in the perturbation domain are regular. So remember, if I'm talking about unbranched covers, then the disaster scenario I described with this index relation doesn't happen. That's the case where index v equals u implies index v equals 0 implies index u is also 0. So in that case, regularity in the usual sense is plausible numerically. And I'm saying for generic j, it actually happens, at least if we're looking at curves contained entirely inside this perturbation domain. That's one major difference with the theorems I explained earlier. I'm not saying we get regularity for all the curves that intersect or have any point mapping into the perturbation domain. But they have to be contained in it entirely. So we'll see why that seems to be necessary. I mean, I don't know if that condition can be dropped. I certainly don't know a way of dropping it. Is there anything you can't just say to you to be the whole manifold? No, no, you can certainly take u to be the whole manifold, right? But if you wanted to just do perturbations in some subset, then you have to restrict yourself a bit. Do you ask that u has to have compact closure or something like that? Well, u is closed. Yeah. u is closed. Or sorry, u, m is closed. Therefore, u has compact closure. Yes. I mean, I could also allow m to be not compact, and then I would indeed have to require that u has compact closure. That's a good question, in fact. OK? So I'm not going to explain the proof in quite this level of generality, but I'm going to explain a case of it, which is somewhat older than our result, but somehow very badly known or badly understood. So in the case n equals 2, so it's just dimension 4, and where both domains are the torus, and the underlying simple curve is actually embedded. So this is what Dusa alluded to with her question a moment ago. This was done by Taubes in 1996, a paper in Journal of the AMS. The proof there is kind of hidden. In fact, when I went back to it last night to figure out where it was, it took me a while. And it's very sketchy, but somehow the ideas in it are extremely potent. So of course, this means we're getting regularity for a multiply-covered holomorphic tori that are covering embedded tori. Taubes needed this because his definition of the Gromov invariant actually counted those things, and it did it without doing abstract perturbations. It did it for generic j. So in this situation on the torus, there are some convenient things. The fact that v has indexed 0 means that its normal bundle has to be trivial. So I can write normal bundle of v is going to be identified with trivial bundle over t2, complex line bundle, and use similarly. And a Cauchy-Riemann operator, if I write it in those terms, just looks like the usual d bar operator plus some 0th order term. So here, by d bar operator, I mean literally just partial by s plus i times partial by t acting on complex valued functions. And a is, say, a c infinity map from the torus to the space of real linear n-demorphisms of c. That's what a 0th order term looks like. And then the pulled-back normal operator is literally the pulled-back of that. So the normal operator for u is going to be the standard d bar plus this 0th order term a composed with the cover of phi. So just a few general points before I really get into the argument. It suffices to show the following. Suffices to show that for all j, of course, of the tame class I'm thinking about, v is a j-holomorphic embedding of the torus into this perturbation domain.  Phi is necessarily unbranched, since it's a torus covering a torus. Then I want to show that I can perturb j to j prime such that v is still j prime holomorphic. But the normal operator for u becomes an isomorphism, defined with respect to the perturbed almost complex structure, j, j prime. Can you explain, like, one more time how to think about the normal operator geometry? Definition is very simple. You take the ordinary Cauchy-Riemann operator, which is defined in the pulled-back tangent bundle. You restrict it to sections of the normal bundle. And now you get some section of a larger bundle that you don't want, but also has a projection to a corresponding normal part. It takes you to some section of the bundle hom bar of t sigma to u star tw, tm, whatever it's called. There's a normal projection in that as well. It takes you to hom bar of t sigma to normal bundle. So you just compose the Cauchy-Riemann operator with that normal projection. This is the terms of deformation, sort of, you know, match the tangent bundle, and I think arbitrary definitions by now. Ah, yeah. OK, yeah, there is another nice way to think about this. So there's an alternative way of describing a neighborhood of a curve in its modularized space that I didn't talk about. If the curve is immersed, specifically. So if the curve is immersed, one way of describing all the other curves nearby is actually write this a little bit. So u is immersed. All the other curves nearby can be assumed to be of the form x u h for some h section of the normal bundle. OK, now, in fact, that's going to hit a unique parameterization of every nearby curve in the modularized space, but you don't get to choose what complex structure you have in the domain at all. So what you have to do then is not actually look for nearby maps of this form that solve the Cauchy-Riemann equation with respect to some specific j, but these are all going to be immersed also. So just look for nearby maps like this whose tangent spaces are j invariant, so that automatically you can pull that back to some complex structure in the domain. You don't get to prescribe it. So that's another way of seeing all the holomorphic curves in the vicinity of this one immersed curve. And well, you can write down some nonlinear operator that does that for you. Its linearization is essentially this normal Cauchy-Riemann operator. That also tells you why it happens that the front home index of the normal operator is the same as the dimension of the modularized space in the immersed case. That's basically because if you preserve in the tangent direction, then you just re-parameterize in the case of this. Right, perturbations in the tangent direction are sort of not meaningful for studying this modularized space because this is giving you re-parameterizations of the same curve. OK, so I don't know if everyone had a chance in the meantime to think about why what I just said here is true. Suffice is to show given a curve and a multiple cover, you can perturb J to one that makes that specific multiple cover regular in the sense of the normal operator being an isomorphism. So this is an exercise using the Taub strict that I explained last time. You can exhaust the space you're interested in with a countable union of compact subsets. In this case, finite subsets, even. So as long as you're able to achieve transversality for each of those subsets, then you can find some set of Js that's a countable intersection of open dense sets that does everything you want. So that's all I'm going to say. This is a version of the Taub strict to get you from there to the result we really want. The other thing I'm going to say is we can reduce this problem to something that's really only involving linear Cauchy-Riemann type operators on an abstract vector bundle. Because I can say for all 0th order terms a prime, so 0th order term is just some c infinity function valued in the space of real linear maps on c, we can find the J prime such that J prime equals J in the tangent directions on the curve v. So T sub v is what I was calling the generalized tangent bundle before. Literally that just means the image of the differential of v. And we can also say J prime equals J outside some neighborhood of the image of v. But the normal operator for v expressed in this trivialization and expressed with respect to the perturbed almost complex structure is d bar plus a prime. So I'm only saying here, give me any perturbed Cauchy-Riemann operator you want in the space of all real linear Cauchy-Riemann operators, I can find a perturbed J that realizes that operator for you. And this is not terribly deep. Actually in the embedded case, when v is embedded, this is fairly easy to prove. It's just a matter of choosing the normal first derivative of your perturbed J in the right way to produce the right 0th order term. One can also do it in the immersed case, and that's a bit more painful. I'm not going to talk about that. OK. So let's also take this as given. And just look at a problem involving Cauchy-Riemann operators on line bundles. Sorry, I have a question about this. I don't understand in the middle of the first line, such that j prime is equal to j on t sub v is equal to t. The image of dv, in other words, the tangent space is to the curve. Oh, I thought that was the index of v, I see. Yeah. Any other questions? All right, so here's a claim. And I'm even going to label this one an improbable claim. Because when I first saw this in Chaubes' 96th paper, I had no idea why I should believe this is true. And I'm still not sure I can explain to you why you should believe this is true, but I can prove it. So here we go. Let's suppose d of the form d bar plus a is a Cauchy-Riemann operator on the trivial bundle over t2, trivial line bundle. And b is a bundle map on that trivial bundle, which I'm going to assume is a complex anti-linear bundle isomorphism. So the two key properties are it's complex anti-linear and it's a bundle isomorphism. So given that the bundle is trivial, it's obvious that you can do this. This is more or less equivalent to saying b of z acts on a vector eta in a certain fiber by some complex valued function beta of z times complex conjugate of eta, where this complex valued function beta is assumed to be nowhere 0. So it's mapping into c star. So that's the assumption. I want to just mention quickly, if I were not working on the torus, but with an unbranched cover with more general domains, sigma and sigma prime, the assumption that the index of the simple curve is 0 allows me again to do this, then I would be able to find a complex anti-linear bundle isomorphism between the relevant bundles, even though they're not trivial. That's actually equivalent to the fact that the index of the simple curve is 0. So that's something also that is easy to check. Now, the statement will be that I can define a perturbed operator d tau as tau plus a real parameter times this extra bundle map b treated as a 0th order term. That is an isomorphism for all tau real numbers outside some discrete subset. So this is one way of perturbing a Cauchy-Riemann operator that might not be an isomorphism and making it into an isomorphism. In particular, what you need to notice about this perturbation is it doesn't care at all about symmetry. I require this extra term b to be a complex anti-linear isomorphism. I don't require it to be anything else. If my bundle is a pullback of some bundle that's defined by a simple curve, but I have been looking actually at the normal operator on a multiply-covered curve, I can do a perturbation like this just by changing j along the simple curve. Now, I pull that back to the multiple cover. My perturbation of my operator is going to be invariant now under deck transformations of the cover. This claim does not care, where symmetry messes up the usual argument and this is the reason why I need somewhere injective in the usual arguments. Symmetry ruins this sard-smale argument. This is impervious to that. So I need to convince you this is true. Sorry, like before, an isomorphism, would it still predict the index of the original b? Well, it's a compact perturbation of the operator. So yeah, this doesn't change the index. So basically, all the Fredholm operators in this talk have index zero. So they're either isomorphisms or they have both kernel and co-crow. Yeah? Can you remind me or tell me how we're going to let's say uproar? So would we have this or are we going to use it? Right, so if we have this, that means I can find, sorry if I have my given j and my given cover, which is maybe not regular, I can find a perturbed j that perturbs the normal operator along the cover in this way and therefore makes that curve regular. So first step, this one parameter family of operators, d tau, is injective, which of course, since its index zero, implies it's an isomorphism for all tau sufficiently large. So this step is interesting because I'm quite convinced that this argument couldn't have come from somebody who was mainly a symplectic topologist. It had to come from a gauge theorist. In particular, if you're familiar with Tauves's work relating Gromov invariance and Cybergwitten invariance, you'll notice a parallel here. There's something that Tauves does in Cybergwitten theory where you write down the Cybergwitten equations with this perturbation term depending on a real parameter. And you can prove that for topological reasons, if you make that parameter very large, you don't have any solutions. Or you do have solutions, but they're converging in the sense of currents to holomorphic curves, something like that. So this is going to be a much easier version of that. And it's an argument that I've never seen anywhere else in symplectic topology. So let's say, well, think about, well, I haven't really specified at all what Bonnock space is I'm working with. But just to make my life easy, let's choose Hilbert spaces and say the operators are going from h1 to l2. So let's suppose I have some non-zero element eta and h1 on the torus. And the idea is to operate on that with dTau and look at the l2 norm squared. So dTau is a sum of two terms, which means I can expand this l2 pairing. And I'll get three terms. I have d eta l2 norm squared plus tau squared times l2 norm of the perturbation, B, which I'm writing here as beta times eta bar. So let's write it beta eta bar l2 norm squared. And then there's a cross term. I'm assuming that my inner product is real valued because the linear operator is real linear, not complex linear. So I can use a Hermitian inner product on the trivial complex line bundle, but then I have to take the real part. So this is going to look like two tau times the real part of the l2 inner product of beta eta prime with d of eta. Let's write that out as d bar eta plus a eta. So a few observations about this. Of course, this first term is non-negative, obviously. The second term, since I assumed that B is a bundle isomorphism, which means this map beta is nowhere zero, I can bound this term from below by the l2 norm of eta. So this is greater than or equal to some constant c1 times the l2 norm of eta squared. And that's all I need to care about that term. Over here, I really have two terms in the cross term. So let's look at the more harmless one first. The pairing of beta eta bar with eta, that also is clearly its absolute value is bounded above by another constant times the l2 norm of eta squared. And I have to worry a little bit about the other part of the cross term. So I need to work a little bit to estimate that properly. So I'd be much happier with that term if it didn't involve a derivative of eta. So you have an l2 pairing of something with something else that's a derivative. What do you do? You can integrate by parts. Let's see. Real part of inner product beta eta bar with d bar eta is literally the real part of the integral of the conjugate beta bar times eta times d bar eta integrated over t2.  Integrating d bar of something over a closed manifold would give me 0 due to Stokes theorem, or probably even reduce that to the fundamental theorem of calculus. That's 0. Over here, I can expand a little bit further and say minus real part d bar beta bar times eta times eta, still minus. This is one of those arguments where if you get one sign wrong, you're really dead, beta bar d bar eta. OK, hopefully I got the signs right. And I am missing an eta. Yeah, here. Thank you. That's important because I want this term to be the same as that term that I had on the left-hand side. So I can put this last term on the left-hand side. So I have twice that equals this. So the thing I'm trying to estimate actually equals minus 1 half real part integral of d bar beta bar eta times eta. And I don't really have to care about the details of this anymore either. I just want to say the absolute value of all this is now less than or equal to some other constant times the L2 norm of eta squared. So I put that all together. Now this whole thing. Sorry, Chris, are you somehow saying that d bar operated as self-projector? No. I am saying that if you take a complex valued function on the torus and integrate d bar of it over the whole torus, you will always get 0. And that you can prove that using Stokes theorem. But that means you're just throwing d bar from one term onto the other. Yeah. Isn't that some L2 in her product? No, no, no. I mean, it was an L2 product to start with, but I wrote it as a product of complex numbers over here. I took conjugates of the terms on the left. Yeah? So this is an integral of a complex valued function that's expressed as a product of complex valued functions. It equals that L2 in a product. It's just the real part of it. Right. So I'm defining my L2 product to be the real part of this integral. So when you write eta times eta, what do you mean? You've got eta's a complex number. Eta's a complex number. So it's just a square of that. Yes. It's a function. Yes. And you've got some measure on T2 that you're integrating. The usual one. Yes. So these are good questions, and they kind of allude to the fact that one can do all this in a much more general setting, but it causes a bit more of a headache. So this is not a uniquely low-dimensional phenomenon I'm describing. Even though I'm describing a low-dimensional proof, you can do it in higher dimensions. You can also do it on more general domains. It just requires several extra steps that I don't have time for. So summarize this. This d tau of eta L2 norm squared is now greater than or equal to some constant, which I'm going to change the name of the constant, c1 prime times t squared minus another constant, c2 prime times tau, oops, tau squared tau, times L2 norm of eta. And that's the proof. If eta is non-trivial, then this cannot be zero. As long as tau is sufficiently large. Why did you do it? You felt like there's less than 1, but you immediately say let's take tau to the big, rich stuff? That the first one goes with tau squared and the second one? First one, yes, goes like tau squared. I mean, the main difficulty in this argument was that I had to get control over this term that has the derivative of eta in it and relate that to just the plain L2 norm of eta. That's what the integration by parts is for. And that's also where I used. I mean, you may not have noticed it so explicitly, but that's where I used the assumption that my perturbation term is a complex anti-linear perturbation. So it's specifically because it appears in there as beta times the conjugate of eta. If I didn't have eta bar on the upper left, this argument wouldn't have worked. As close as I can come to giving you an intuitive reason to believe this. I'm sorry. The constants there, like C3 and C2, but C3 is like the norm of del bar beta bar. It's whatever is convenient. There is one. Do you believe that there is some constant that makes that true? You stare at it a little bit longer and you'll be fine. All right, so I'm almost done. I have one more bit of magic to pull off to finish this proof. And that's a bit of analytic perturbation theory. So I won't have too much time to. How much time do I actually have? That's always the answer, isn't it? Analytic perturbation theory. So what I'm about to say can be done in the real analytic category, but I don't want to because I'm a little bit allergic to the real analytic category. So instead, I'm going to work in the complex analytic category and complexify my operator first. My operator is only real linear, even though it's acting in a complex vector bundle. So what I can do at the expense of having two complex structures in the picture instead of one is I can complexify the domain and the target of the operator and consider the canonical complex linear operator, the extension of that, to the complexification. So, oops, more precisely. So we've got d tau is a real linear operator from some space of complex-valued h1 functions to complex-valued l2 functions. Let's take, actually to be more precise, we can say this is h1 functions from s1 to r2. And calling it r2 instead of c will help avoid some confusion coming up, as I'm about to complexify it. Sorry, yeah, t2, not s1. I'm a nasty feeling it even says s1 in my notes. That's not so good. So if we complexify, this becomes an operator I'll call d tau c from, call it h1c, which you can think of as, well, the tensor product of this Hilbert space with c, which is equivalent to the space of h1 functions from t2 to c2. And it's going to map that to corresponding space l2c. And it's complex linear. And what I can do now is allow my parameter tau to be complex instead of just real. So now the map taking tau to the operator d tau is a holomorphic map from c to the space of Fredholm index 0 operators, complex linear, from h1 complexified to l2 complexified, which, of course, is an open subset of the space of all bounded linear operators, complex linear between those spaces. So this is just a complex Bonnock space. I have a map from c into that Bonnock space. It is easily seen to be differentiable with respect to the complex variable tau. It's, in fact, it's an affine map. So that's a holomorphic map into this open subset of Fredholm index 0 operators. So the last step is then to observe the following. The space of non-invertible operators, so operators that are not isomorphisms, sitting inside this space of Fredholm index 0 operators, is what we call an analytic sub-variety, a complex analytic sub-variety, which means locally you can express it as the zero set of some holomorphic function on that infinite dimensional, so open subset of an infinite dimensional complex Bonnock space. But that function is valued in something finite dimensions, valued in c, in fact. So what we end up with is the set of all parameters tau in c with the property that the complexified operator d tau c is not an isomorphism looks locally like the zero set of a holomorphic function from c to c. That's why that set is discrete. We already showed that it's not everything, because for very large tau, the operator is an isomorphism. So once you know then that the zero set must be discrete, you're done. And then, of course, you have to think a little bit about relating this statement about the complexification to the original real linear operator. That's fine. Basically, you can convince yourself that if tau is real, then the complexified operator being an isomorphism implies that the real linear operator is an isomorphism. So I was going to prove this lemma. In fact, I was going to say other stuff after that, but there's no time. So if you want to know the proof of the lemma, I can tell you whenever, if my voice holds out. But not now. But not now. So let's have lunch. Are there any other questions? Why don't you say it's a discrete set? Do you really mean it's a finite set? I mean, it can't accumulate on zero on count. So I don't know what happens in, for other, large, complex parameters that are not necessarily in positive real line, that's probably how it works. I mean, you were talking of, I mean, your main interest was in the zeroes when tau was real, and near zero, because we were making a little count of it. This was why I haven't thought about the answer to your question very much. So it's probably true. But if you're saying a zero is a homomorphic function, they're isolated, aren't they? They're isolated. They're isolated. So in other words, if it's zero, it's zero, and it's a whole neighborhood where it's not zero. Correct. There's a whole neighborhood where it's discrete. Well, oh, I see, a discrete including zero. OK. Please repeat what I mean up for just the picture itself. It's always d tau, I mean, if this is what you do with this, d is d, and d can't be tau. How are we going to kind of show you the picture? So, OK, my goal, I had a specific normal Cauchy V1 operator for this multiply-covered curve. So this is what in this abstract discussion at the end, I'm calling just plain d. And then d tau becomes just some perturbation of this normal operator, which corresponds to some specific choice of perturbation of the almost complex structure in the neighborhood of the curve. I'm going to reset the clock and restart at 2. If you're new, let's thank Chris again.


Section 0 of Persian Self-Tought in Roman Characters with English Phenetic Pronunciation by Shaykh Hassan. This is a LibriVox recording. All LibriVox recordings are in the public domain. For more information or to volunteer, please visit LibriVox.org. Recording by Nicholas James Bridgewater. Persian Self-Tought in Roman Characters with English Phenetic Pronunciation. Containing the alphabet, transliteration and pronunciation, outline of grammar, classified vocabularies and conversations, travel talk, trade and commerce, Persian handwriting, the numerals, money, weights and measures, etc. etc. by Shaykh Hassan. Preface. This volume is primarily intended to supply a working and practical knowledge of the Persian language for the benefit of those who have not the time or the inclination to master the grammar and yet require to use the spoken tongue for purposes of business or pleasure. With this object in view, it supplies many vocabularies of words carefully selected to suit the needs of those holding communication with Persia and the Persians, classified according to subject and a large number of colloquial phrases and sentences of a practical character similarly classified. Throughout these sections of the volume, the Persian words are given in the Romanized form and the English Phenetic pronunciation is added in the third column in accordance with Marlborough System of Phenetics, for which, of course, the publishers alone are responsible, a system which has grown out of long experience and is of so simple a nature that, by means of it, a stranger to the language who knows English only can read off the words at a glance and thus easily make himself understood amongst Persian-speaking people. An outline of the grammar is added for the use of students of the language. The work thus forms a useful and valuable handbook of Persian for tourists and travellers, commercial men, naval and military officers, students and others. We may add that the Persian language is very sweet and compared with other languages is easy to master. It has a complete literature of its own. It is not spoken in Persia alone but in Afghanistan and many parts of India and is the official language of some of the courts of that country. It is also used in Turkish Arabia, Baghdad and by educated people in Russian Turkestan. The study of Persian is almost essential to one who wishes to learn Hindustani and modern Turkish as the former of these contains about 40% and the latter about 25% of Persian words. Sheikh Hassan, Cambridge, 1909 End of Section 0, Recording by Nicholas James Bridgewater Recorded in Aman, Jordan Section 1 of Persian self-taught in Roman characters with English phonetic pronunciation by Sheikh Hassan This is the LibriVox recording, all LibriVox recordings on the public domain. For more information or to volunteer, please visit LibriVox.org Recording by Nicholas James Bridgewater Persian self-taught in Roman characters with English phonetic pronunciation by Sheikh Hassan Section 1 The alphabet The Percy Arabic letters at first sight appear very bewildering to Europeans and are a frequent cause of stumbling to many who would find the knowledge of the language of immense importance to them. It is to meet this difficulty that the Romanized characters have been adopted in this work. Students who desire to study the wonderful literature of the East at first hand must however overcome this initial difficulty and to do this the best plan is to copy each letter repeatedly and with care until it is fixed in the memory only two or three being dealt with at a time and not the whole alphabet In this way one is surprised how soon the characters in their various positions become familiarized to the mind Having mastered the alphabet the words given in the vocabularies should next be learned pronouncing the phonetic aloud and writing the word from memory in the Percy Arabic letters which can then be corrected by reference to the following tables In this way the rudiments of the language will be easily acquired The letters and sentences are written and read from right to left as in all the Semitic languages, the Percy Arabic being one of these although it should always be borne in mind that the Persian language proper is not of Semitic origin but is akin to the old Sanskrit The alphabet in Percy Arabic with Roman characters letters marked with asterisks are Arabic characters and the words in which they occur are Arabic Name Alif Roman character A Name Be Roman character B Name Pe Roman character H Name Te Roman character T Name Se Roman character S Name Jean Roman character J as J in Jewel Name Chim Roman character C-H as C-H in Church Name He Roman character H Name Che Roman character K-H a guttural aspirate like the C-H in the Scottish Loch Name Dol Roman character D Name Zoll Roman character Z Name Ray Roman character R Name Z Roman character Z Name J Roman character J as S in Pleasure or J in the French Jardin Name Seen Roman character S Name Sheen Roman character S-H Name Sword Roman character S Name Zord Roman character Z Name Te Roman character T Name Z Roman character Z Name Ain Roman character A Name Bain Roman character G-H as G in Good Very guttural Name Fe Roman character F Name Gough Roman character Q Very guttural Name Cough Roman character K Name Gough Roman character G as G in Ambiguity Name Long Roman character L Name Mime Roman character M Nune Roman character N Name Roman character V Name Hey Roman character H Name Yay Roman character Y Remarks on the alphabet The Persian alphabet contains no vowels properly so called The short vowel sounds are indicated by Zabar Zir Peesh The first is a short A as in At The second Zir Short E as in Effort Third Peesh Gives the short O sound Something like the sound of O in Should Or of You in the French Plum Pronounced short For example Mime with a Zabar Mime with a Zir Mime with a Peesh Pronounced respectively Ma Me Mo When the long vowel sound O as in Art Occurs at the beginning of a word It is indicated by the sign Mad Placed over Alif as in Or Dam Or Mad But elsewhere in a word It is indicated simply by Alif As in Shard Glad There is no sign in Persian For the vowel O as in Note But Ah With Zabar Followed by Vov Is pronounced O Example O der Eternal uncle O rang Throne Ye Following any letter with Zir Is pronounced as long E IE as double E in Sheep Example Iran Persia Di Shab Last night There are two other signs Namely Jasm And Tash Deed The first of these is placed over any letter Which is followed by no vowel sound As over Fe In Raftan To go The second doubles the letter over which it is placed As over Ghorf In Ghorsh Artist These vowel signs are in fact never used in Persian writing But their omission causes no ambiguity As any doubtful word is determined by the context The Romanised characters with phonetic pronunciation A Like the English A in At A with a macron as A long in Art Or AI Like AI in Rain A B As B in But B CH as CH in Church CH D as in English D E as E in Egg E F as F in Friend F G like the G in Ambiguity G GH like the English G in Good Very guttural G H as Pyrated as in English H I as I in Bit I I with a Macron as E in Deed E J as J in Jam G J as Z in Azure G K as K in Keep K KH a guttural aspirate like the CH in the Scottish Lough CH L as in English L M as in English M N as in English N O as O in Hope O P as in English P R as in English Very Always to be pronounced R S as Sibilant as in Sister S SH as in Shine SH T as in Tutor T U short as U in Should or U in French Lomme O U with a Macron as the double O in Moon O V as in English V Y as in U Y as in Zodiac Z Preliminary Notes Persian Equivalence It is necessary to point out that in the third part of the book conversational phrases and sentences the Persian equivalents of the English phrases are given in the second column and not merely a translation of the English words. How different the equivalent and the literal translation may be is shown by the few instances that are over-given. See page 65 Pronunciation The student will do well to master the pronunciation at the outset. The use and value of each phonetic sign adopted are shown on page 11 and should be carefully noted. Marlborough's system of phonetics is based on the principle of one sign, one sound and conversely one sound, one sign. The particular use of each English letter in representing the sounds of the characters should be learnt and each must always be pronounced in the same manner, one sign one sound. To facilitate the correct pronunciation of the short vowels the words in the vocabularies have to a great extent been divided up by hyphens, but these divisions do not always correspond with the true syllables e.g. Sadom Pronunciation written S-A-D-E-U-M properly S-A-D-E-U-M negah pronunciation written n-e-g-a with a macron h for n-e-g-a with a macron h. In such words as das demann where the letter e stands for esor fat it will be noticed that it is represented in the third column das demann to remind the reader that it is esor fat where e does not stand for the esor fat, it is denoted by e in the third column also. The conjunction va and in such words as das dva pa hand and foot ruz vah shab day and night is pronounced o in conversation like the French u in plume das dva pa ruz vah shab In all questions the interrogative oya is inserted in conversation however it is often omitted the question being inferred from the inflection of the voice.  Fishes moheon, moheon, fishes moheon. Kod moheikod, moheikod, kod moheikod. Crab khachang, khachang, crab khachang, eel marmahi, eel marmahi, herring shahmahi, herring shahmahi, lobster khachang, lobster khachang. Oyster gushe maahi, gushe maahi, oyster gushe maahi, salmon azad maahi, salmon azad maahi, sea fish moheyeh darya, sea fish moheyeh darya, trout moheyeh qezelala, moheyeh qezelala, trout moheyeh qezelala, turbot jaris, jaris, turbot jaris, game shakar, shakar, game shakar.  Insects and reptiles, John Avar and Kerm, ant, moor, moor, ant, moor, bee zanbore, zanbore, bee zanbore, beetle chazook, chazook, beetle chazook. Butterfly shah-parak, shah-parak, butterfly shah-parak, caterpillar sad-paw, sad-paw, caterpillar sad-paw, flea cake, cake, flea cake. Fly magas, magas, fly magas, frog gook, gook, frog gook, nat pashe, pashe, nat pashe, grasshopper malach, malach, grasshopper malach. Insect parande, parande, insect parande, or John Avar, John Avar, insect John Avar. Lizard sus-se-mor, sus-se-mor, lizard sus-se-mor. Mosquito pas-she, pas-she, mosquito pas-she, silkworm kerm-pe-le, kerm-pe-le, silkworm kerm-pe-le, snail hal-zoon, hal-zoon. Snail hal-zoon, snake mor, mor, snake mor. Spider an-caboot, an-caboot, spider an-caboot, or cartane, cartane, spider cartane. Or cartane, cartane, spider cartane, toad gook, gook, toad gook, or gor-ba-ge, gor-ba-ge, toad gor-ba-ge. Wasp zan-boor, zan-boor, wasp zan-boor. Worm kerm, kerm, worm kerm. Flowers, fruits, trees, and vegetables. Gol-mi-ve-deracht pas-sab-zi-or-lot. For shopping, see page 86. Flowers gol-ho, gol-ho, flowers gol-ho. Lily sus-san, sus-san, Lily sus-san, or zan-bag, zan-bag. Rose gol, gol, rose gol, or gole sor-ch, gole sor-ch, rose gole sor-ch. Tulip lo-le, lo-le, tulip lo-le, violet bana-f-she, bana-f-she, violet bana-f-she. Fruits-mi-ve-deracht, mi-ve-deracht, fruits-mi-ve-deracht. Apple-see-b, see-b, apple-see-b. Apricot, zar-do-loo, zar-do-loo, apricot, zar-do-loo, or be-see, be-see, apricot, be-see. Cherry-g-loss, g-loss, cherry-g-loss. Currants-ze-res-k, ze-res-k, currants-ze-res-k. Fig-an-jeer, an-jeer, fig-an-jeer. Grapes-an-gour, an-gour, grapes-an-gour. Lemon-li-mu, li-mu, lemon-li-mu. Lime-deracht-li-mu, deracht-li-mu, lime-deracht-li-mu. Melon-xar-be-ze, xar-be-ze, melon-xar-be-ze. Nut-fond-og, fond-og, nut-fond-og. Orange-tor-onge, tor-onge, orange-tor-onge. Peach-shaft-or-loo, shaft-or-loo, peach-shaft-or-loo. Or-ho-loo, ho-loo, peach-ho-loo. Pear-gul-a-bee, gul-a-bee, pear-gul-a-bee. Pineapple-an-an-os, an-an-os, pineapple-an-an-os. Plum-a-loo, a-loo, plum-a-loo. Or-a-loo-che, a-loo-che, plum-a-loo-che. Strawberry-toot-de-far-ang-ee, toot-de-far-ang-ee. Strawberry-toot-de-far-ang-ee. Walnut-ger-doo, ger-doo, walnut-ger-doo. Trees-der-ach-dan, der-ach-dan, trees-der-ach-dan. Ash-zaban-gon-jeshk, zaban-gon-jeshk. Ash-zaban-gon-jeshk. Peach-al-ash, al-ash, beach-al-ash. Birch-goo-she, goo-she, birch-goo-she. Leaf-barg, barg, leaf-barg. Oak-der-ach-de-baloot, der-ach-de-baloot. Palm-der-ach-de-chur-mo, der-ach-de-chur-mo, palm-der-ach-de-chur-mo. Pine-der-ach-de-codge, der-ach-de-codge, pine-der-ach-de-codge. Vine-talk, talk, vine-talk, or-mayve, mayve, vine-mayve. Vegetables-sab-zi-a-lot, sab-zi-a-lot, vegetables-sab-zi-a-lot. Asparagus-mar-chube, mar-chube, asparagus-mar-chube. Beans-lubia, lubia, beans-lubia, or-bar-li, bar-li, beans-bar-li. Beetroot-chog-on-dor, chog-on-dor, beetroot-chog-on-dor. Cabbage-kalam, kalam, cabbage-kalam. Carrot-have-ee-j, have-ee-j, carrot-have-ee-j. Cauliflower-kalame-ru-si, kalame-ru-si, cauliflower-kalame-ru-si. Celery-caraffs, caraffs, celery-caraffs. Cucumber-khiaar, khiaar, cucumber-khiaar. Lettuce-khaw-hoo, khaw-hoo, lettuce-khaw-hoo. Onion-pears, pears, onion-pears. Peas-green-marsh, marsh. Peas-green-marsh. Potatoes-sibe-zamini, sibe-zamini. Potatoes-sibe-zamini. Spinach-espinage, espinage. Spinach-espinage. Tomato-bod-in-john-ferang-ee, bod-in-john-ferang-ee. Tomato-bod-in-john-ferang-ee. Turnip-shal-gam, shal-gam, turnip-shal-gam. End of section 2, recording by Nicholas James Bridgewater, recorded in Amman Jordan. Section 3 of Persian self-taught in Roman characters with English phonetic pronunciation by Sheikh Hassan. This is a LibriVox recording. All LibriVox recordings are in the public domain. For more information or to volunteer, please visit LibriVox.org, recording by Nicholas James Bridgewater. Persian self-taught in Roman characters with English phonetic pronunciation by Sheikh Hassan. Section 3. Colours-rang-ha, for shopping, see page 86. Black-si-ya, si-ya, black-si-ya. Or-mesh-gi, mesh-gi, black-mesh-gi. Blue-o-b, o-b, blue-o-b. Brown-nim-rang, nim-rang, brown-nim-rang. Crimson-soor-ch, soor-ch, crimson-soor-ch. Soor-ch, o-lo-ki, lo-ki, crimson-lo-ki. Dark-deep, ti-re, ti-re, dark-ti-re. Or-tond, tond, dark-deep, tond. Or-seer, seer, dark-deep, seer. Green-sabs, sabs, green-sabs. Gray-chokestari, chokestari, gray-chokestari. Light-pale, roshan, roshan, light-pale, roshan. Pink-goli, go-li, pink-goli. Purple-argavoni, argavoni, purple-argavoni. Red-ghermez, ghermez, red-ghermez. Or-soor-ch, soor-ch, red-soor-ch. Scarlet-soor-ch, soor-ch, scarlet-soor-ch. Violet-banafsh, banafsh, violet-banafsh. White-sefid, sefid, white-sefid. Yellow-zard, zard, yellow-zard. Times and seasons, ogat-vamosem-ha, for conversations, see page 81. Time-vaqt, vaqt, time-vaqt. Or-hen-galm, hen-galm, time-hen-galm. Day-ruze, ruze, day-ruze. Fortnite-do-haf-de, do-haf-de, Fortnite-do-haf-de. Or-ni-maw, ni-maw, Fortnite-ni-maw. Hour-soor-at, soor-at, hour-soor-at. Minute-da-bi-ghe, da-bi-ghe, minute-da-bi-ghe. Month-maw, maw, month-maw. Quarter-three-months, se-maw-he, se-maw-he. Quarter-three-months, se-maw-he. Second-dam, dam, second-dam. Or-son-ye, son-ye, second-son-ye. Week-haf-de, haf-de, week-haf-de. Year-sol, sol, year-sol. Day-time-vaqt-de-ruze, vaqt-de-ruze, day-time-vaqt-de-ruze. Afternoon-asr, asr. Afternoon-asr. Afternoon-asr. Dawn-tolu-e-af-daab, tolu-e-af-daab. Dawn-tolu-e-af-daab. Or-soob, soob. Dawn-soob. Evening-vaqt-de-ruze, vaqt-de-ruze. Evening-vaqt-de-ruze. Or-masa-u, masa-u. Evening-masa-u. Or-shawm, shawm. Evening-shawm. Midday-noon-zor, zor. Midday-noon-zor. Midnight-nes-feshab, nes-feshab. Midnight-nes-feshab. Morning-soob, soob. Morning-soob. Or-bomodod, bomodod. Morning-bomodod. Night-shab, shab. Night-shab. Sunrise-tolu-e-af-daab, tolu-e-af-daab. Sunrise-tolu-e-af-daab. Sunset-gorube-af-daab. Gorube-af-daab. Sunset-gorube-af-daab. Holidays-taqtil-haaw. Taqtil-haaw. Holidays-taqtil-haaw. Christmas-molude-hazrat-e-saaw. Molude-hazrat-e-saaw. Christmas-molude-hazrat-e-saaw. Easter-eed-eed. Easter-eed. Good Friday-jom-e-e-a-ziz. Jom-e-e-e-a-ziz. Good Friday-jom-e-e-a-ziz. Days of the week-gruze-ha-e-haf-dae. Gruze-ha-e-haf-dae. Days of the week-gruze-ha-e-haf-dae. Sunday-yeksham-dae. Monday-yeksham-dae. Sunday-yeksham-dae. Monday-do-sham-dae. Do-sham-dae. Monday-do-sham-dae. Tuesday-seh-sham-dae. Seh-sham-dae. Tuesday-seh-sham-dae. Wednesday-cha-har-sham-dae. Wednesday-cha-har-sham-dae. Thursday Pan Shambay Friday Jom E Or Odeine Saturday Shambay The Months January Jean-Vier January Jean-Vier February February March Mars April Avril May Me June Juan July Jouillet August and August September September October October November November December December The Four Seasons The Four Seasons Spring Bahar Summer Taubestan Autumn Poise Winter Zemestan Today Amrouz Tomorrow Faridah Tomorrow Morning Faridah Sub Tonight Mshab Twilight Shafag Yearly Solyane Yesterday Dirous The Persian Months Moharramul Haram Moharramul Haram Moharram Sephar al-Muzaffar Sephar al-Muzaffar Sephar Rabio Laval Rabio Soni Jamodeo Laval Jamodeo Soni Rajabul Marojab Kloquil Rajab Sha'ban al-Muazzam Kloquil Sha'ban Ramazan al-Mubarak Kloquil Ramazan Shavvalo Mukarram Kloquil Shavval Zelqadatul Haram Kloquil Zelqadat Zelhadjatul Haram Kloquil Zelhadjat The first day of Moharramul Haram is the beginning of the year Hedjri, A.H. Anohegira, i.e. the Muhammadan era. Town and Country Bridge Building Edifice Building Edifice Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building Building


So we're at lecture 43. Today's goal is to kind of gather things up, concepts, problems, expectations, all those things for a test that we have in this classroom tomorrow. It covers the supplement, at least the part of it that is appropriate for this class, 777, 7879. Hang on to that supplement because there are other things in there that you'll need in a future course, which I printed one out for you. I did not print out the entire supplement. I just printed out what was part of this course. Back to the book, the first three sections of chapter 8, even though on the syllabus it looks like we have still some things in 8.3 yet to do. We have finished 8.3, so it will be on this test. That'll actually, I think, you'll be glad that we have it on this test because it's one fewer or actually two fewer convergence tests that are in the rest of chapter 8. So we've got a lot of things going on in the rest of chapter 8, which that'll help the next test. So any question pending, any web assign question, or any concept before we kind of review? Yes, Jacob? I have just a general test layout question. Is it going to be enough to know if it converges or diverges? And to what value it converges? Or do you have to write all the, which test you used? You do have to write the test. And I'll mention this again. But some of them I hope will be pretty obvious that I won't need to tell you what test to use. Like if it looks like one that, let's say, could be integrated and you use the integral test, pretty obviously, you're going to show your work. You're not going to use the integral test on a piece of scrap paper and then just put divergent. So if you have your work for the integral and you determine that it doesn't have a solution, does not exist, therefore, the series diverges, then you could just put in parentheses by the integral test. I mean, it's pretty obvious that's how you got there. But that just lets me know that you know if the integral is divergent, so is the series associated with it. Some of them I will actually tell you to use the comparison test or the limit comparison test. So that, I hope, helps because then you don't have to go through this search process in your mind. Is it a geometric? Is it a p-series? Integral test, comparison test, limit comparison test.  But I'm not going to do that on all of them. So some of them you will decide, hopefully, those will be the ones that are real clear. Others I'll ask you to do the problem a certain way. Any other kind of questions like that or clarifications of second-order differential equations or sequence and series convergence and divergence? So we're not going to have one all the way through of the non-homogeneous code or whatever. Well, let's discuss that. If it is, it'll be a kind of a special, shorter version than the so-called typical one. So from the supplement, these were second-order linear non-homogeneous differential equations. We had three cases. I'll just go ahead and tell you because you don't like mysteries on tests. I don't like to give mystery-type questions on tests. All three cases will be used in some way, shape, or form in at least one of the problems. So it's not like you can say, well, I know the first two and I don't think he'll ask the third one. He will ask the third one. So these will all be covered. Now, not necessarily in the non-homogeneous, one of them might be into the homogeneous case, but all three of these cases. So if you have two distinct real roots to the characteristic equation, now, how much of that you show in going from the original second-order differential equation to the characteristic equation? If you want to just make that leap, that's fine. Or if you want to say, what are these things like that solve this, y equals e to the rx? And then go through y and y prime and y double prime and sub n and get the characteristic equation. That's up to you. But you don't have to show every step in getting to the characteristic equation. So if we have y in terms of x, then what would our solution look like that had two distinct real roots to the characteristic equation? Does that look familiar? I hope. So that's to the homogeneous part. If the left side is all the second-order stuff and the right side is 0, then you're done, unless there are some initial conditions. And if there are initial conditions, the y of 0 equals 7 and the y prime of 0 equals 4, then those will help you find c1 and c2. If you have a double root, it's real. So r1 is equal to r2. And we'll just say that it's r, or just r. So in fact, if you want to generate new solutions, and this is one of them, you can continue to modify. I think they called it a modification factor in the supplement. You can continue to multiply by x until you get something that actually does cause the left side when you put in second derivative, first derivative, original function to end up being 0. Again, you could have some initial conditions that would allow you then to solve for c1 and c2. Remember, if you do have to take the first derivative, that you do have a product here. So you have to use the product rule. If you're given y prime of 0 equals 5, then this term is a product rule, so don't forget that. Two complex roots. I guess the others are complex also, but I mean complex in a sense that they have an imaginary term. So the roots look something like that. So we end up with a negative under the radical and not real numbers. So we've got a completely different category. They start out looking like this, but they don't stay that way very long because of the fact that we have e to the i theta in there, and that then converts these exponential-looking things to trigonometric-looking things. So that's something that we used in converting this to this form. You're not going to be tested over that, but that's why it ends up looking very different. What do they look like when we have two complex roots that have the imaginary part as part of the answer? So I would, if I were you, know all three of these, and you will have a question on the test that kind of puts you in a position where you have to use all three of these. Yes, Nicole? I don't know if it was on homogeneous or the homogeneous equations, but actually I think it was the other one. When you have the first, I don't know what I'm trying to say. Two parts to the solution, the homogeneous part and the particular part? Right. When the particular part, the first part, when you take the derivative and it's the same as, I don't know, the homogeneous solution, don't you have to change the particular? Are we going to have any of those? That's possible. Can we do one of those? Yeah. Any questions on this? So the only thing that could kind of change the basics of these problems would be if you had some initial conditions. y of 0 equals something y prime of 0. And you can solve for c1 and c2. And you should plug those back in and then move on to the next problem. All right. These are the second order linear. What did I say? Yeah. The other one non-genius. Yeah, here we go. Just caught an error. Bring that up. These are homogeneous, right? Scrap that. So homogeneous means we've got all this stuff over here equal to 0. So if you were thinking that was incorrect, you were correct in thinking that was incorrect. So we want here non-homogeneous. So we've got all this stuff on the left side equals. And we have a subset of things. It's not wide open on the right side. But it's non-zero. And it comes from three different categories. Polynomial. So we get a generic version of that polynomial. If it's linear, our generic linear is what? What's a generic linear equation? Ax plus b. If it's a constant, then our generic constant is a or b or some letter. If it's quadratic, then our generic quadratic is x squared plus bx plus c. So we have to solve for those values potentially. So polynomial, exponential. So we need to, and this is what Nicole brought up a second ago, we need to make sure that the exponential on the right side is not also one of the homogeneous solutions. If it is, then we don't want to start with those because we know what that'll give us. That'll give us 0. And we just keep modifying it until we get one that is not. I'll make up some examples here or look one up of that type. And either sine of something, cosine of something, or sine plus cosine. So it's not wide open in terms of trigonometry. You will see some different problems. Excuse me. And as you're looking through the supplement, there is another technique in 7.8 that we did not do, variation of parameters. And you'll see some problems in here that have some things on the right side of the equal sign that we did not do, deal with, encounter. Don't be concerned if you're looking through the problem sets, if you see those, you're not responsible for those. So our technique is to put in the generic version of what we see on the right side, take first derivative, second derivative, plug them in, and equate coefficients. The coefficient of e to the 2x on the left side better be equal to the coefficient of e to the 2x on the right side, and so on. Term by term, equating coefficients of like terms on both sides of the equal sign. All right, so the first part of these that are non-homogeneous is to get the homogeneous solution. We just covered that. There are three categories, so the same three categories apply. That becomes part of your solution. Here's where the, what do we have on the right side question comes in. Do we have a polynomial, an exponential, or sine, or cosine, or both? Because we have to establish a generic version of what we see on the right side and get these three quantities, the original y sub p, plug it in where y is, first derivative of y sub p, plug it in where y prime is, second derivative of y sub p, plug it in where y double prime is, and then we start distributing, gathering like terms together, and equating like terms left side and right side of the equation. And then our final solution is y sub h plus y sub p. Then there is the possibility, again, it'd have to be a special problem that's shorter than most, where we could then put in some initial conditions and solve for some letters that still remain, the C1s and the C2s that still remain. But once we get to this stage, we'll have, we shouldn't have a's and b's and c's, right? We'll have those values established. So there won't be any doubt as to the values of the particular solution. There will be some C1s and C2s in the homogeneous solution that we could potentially solve for with some initial conditions or boundary conditions. All right, Nicole asked for a certain type of problem. Let me kind of build the problem as we go, OK? So I'll make sure it's one of that type. We're going to have something on the right side, but I'm going to wait and see what we get for the characteristic equation and then put this in so we get that situation that Nicole asked about. So the characteristic equation is what? And it looks like what? Minus 1, minus 1. So it looks like we have a double root. So the homogeneous part, and this is going to be non-homogeneous. I'm just waiting to fill this in here. The homogeneous part is what? C2x, e to the x. Does that work? And let's say over here we have 4 e to the x. So we've got an exponential. That's one of our three things on the right side. So here's our first piece, our second piece. We would go to the particular solution. So if we want to generate some e to the x's, don't we need to start with some e to the x's? We just don't know how many to start with. But what about that? What about a, e to the x? Can that be part of the particular solution? No, because that is part of the homogeneous solution. So if we put in some a, e to the x's, we're not going to be able to kick out e to the x's. We're going to kick out 0 because it's part of the homogeneous solution. So we scrap that. So we use our little modification factor. We continue to keep multiplying by x. How about that? That's not going to work either because that is also part of the homogeneous solution. So there is the first one that really has a chance of generating some e to the x terms. The other two won't generate anything because they're part of the homogeneous solution. They're going to kick out the 0. They're going to disappear. And there's a chance that this one too. We don't know that, but there's a chance that this one could not work. If, in fact, it doesn't work, then we continue to modify it by multiplying by x. But we would go ahead anyway with this, use the product rule to get the first derivative. And we're going to have the product rule again involved in finding the second derivative. But that's when you look at the term on the right side, if it is present in any way, shape, or form in the homogeneous solution, you better change it because it's not going to work.  Take it, it's first derivative. It's second derivative. You plug it in, you're going to get 0 on the left side. Yes? Your right side is like x sine of x. How do you set up a particular equation? So we have something over here. And on the right side, we have x sine of x. So we would find the homogeneous solution. We don't have that. So you go to the particular solution. What do we really have on the right side? A linear times a sine or cosine, or both. So we need to kind of generalize x. Well, we know our generic linear term is going to be ax plus b. We also need a, well, when we have sine of x, what do we have to allow for in the particular solution? Sines and cosines, right? Because we're using the original function, which has sines and cosines. It's derivative, which has cosines and sines. And the second derivative, which is back to sines and cosines. So we would want the cosine terms to all disappear. That can happen. But we would want the sine terms to remain. So in generic form, we would say sine of x would be what? A. Now, I'm sorry. I probably should use a c. I was thinking a step ahead where we're not going to need those anyway. So let's scrap that. Now, if we were to distribute this thing, we'd have ax times c times sine of x. Well, we don't know a, and we don't know c. But a times c is some unknown constant. So can we get rid of c? We can kind of get rid of c, because we're going to have unknown arbitrary coefficients anyway. We don't need to have product of a and c or a and b or b and c. It's a little redundant. Same thing with b times d times cosine. There's too many letters there. Or ax times d times cosine. So we don't need that one either. So that'll take care of it. So when you distribute stuff and you have two letters, you don't need two letters. There's just one coefficient. So that's the coefficient we would be searching for. Somebody might say, well, let's suppose that we do all this work. And there is a, I don't know if I like what I just said, to be honest with you. I'm always trying to think a couple steps ahead. Sometimes that, when I'm thinking ahead and writing behind, let's see if we can accommodate this. And this is what kind of troubles me right now. So let's say that we want there to be a 2 there. And let's say some other number here. And we want there to be a 4 here and a minus 3 here. I'm saying that the final answer looks something like this. That we're kind of allowing for a generic linear. And we know how to generate signs. We can do that with signs plus cosines. So I'm saying, get rid of the c and the d. But I may come back on that. Could we develop an answer that looks like this by kind of eliminating the c and the d? So we'd have 8x sine of x. We'd have 20x cosine of x. We're not going to be able to come up with something like this from this, are we? Isn't there only going to be one cosine or sine in the final? Not necessarily. We saw that happen when we had one of the terms missing. Like when we had the original y thing and we skipped the first derivative and we went then to second derivative. We saw that when that happened, we had either the sine or the cosine disappear. If we have both, I'm sorry, all present here, if we have y double prime, y prime, and y present, we're probably going to have something that has a sine and a cosine in it. I guess we are getting rid of the a's and b's because the example we did that happened. I don't think we did an example where there were both sines and cosines in them. Yeah, I may recant on getting rid of the c and the d because although there's one letter here, and if you track that back, that would be ax times c times sine of x. That's really a times c. I don't know that c and d are the same. That's where I think this falls apart, that I could factor them out and put the number that I factored out from here and here into here and have it absorbed in this unknown amount of x and this unknown constant. That makes me think that I want to go back to this, that we better allow for a different coefficient for sine of x and cosine of x. If we knew they were both the same, we could get rid of both of them and absorb them in the linear term, but we don't know that. I don't think we could generate something like this without having an unknown value here and here. So I'm going to come back to that. Might be a little redundant, but I'm just afraid that I'm not going to be able to generate something like this, which came from this, if I get rid of the c and the d. That's probably a little too healthy of a chunk for a test because if you start with this, what would be your next step? Find a derivative of that, and probably a little too lengthy for a test. And then find the second derivative, which is going to have probably more terms than this one, so probably a little too healthy. But I do think we need the a and the b and the c and the d. OK, in 7.9, we had applications. These could be applications of homogeneous or applications of non-homogeneous. So once we translate the problem, circuit problem, spring problem, whatever it is, into the form that it becomes a second order differential equation, it's the same stuff we've already looked at today. And these have similar things in their equation. So once we get everything translated, then we just use our solving techniques that we've accumulated to this point. So spring problem, what would be the coefficient of the second derivative term? Comes from force, one way of looking at force, not the Hooke's law, but just kind of force in general. Mass times acceleration is force. So there's mass times acceleration. We can kind of have that on one side of the equal sign. On the other side of the equal sign, we have other things, the restoring force from Hooke's law, which is, if it were on the right side, it'd be negative kx. If we bring it over the left side, it's kx, right? What other things could be on the right side of the equation for a spring problem that once we bring them over to the left side, they're going to appear over here? The damping constant, or damping coefficient, that term. And it says, in theory, we're supposed to assume that the damping term, the c value, is directly proportional to the velocity, right? So on the right side, it's what? Negative c dv over dt. And we bring that to this side, and it's plus, not dv over dt dx over dt, which is velocity. So there's our velocity, our damping coefficient, which is directly proportional to velocity. Here's our restoring force, negative kx. On the right side of the equation, we bring it to the left side. It becomes positive kx. And if we have something on the right side, what is it? Is it some coefficient or something? Some external force, right? So I would say for most of the cases that we really examined, we're going to have a homogeneous equation here. All right, similar terms on the circuit equation. And more than likely, we'll have something non-zero on the right side, because what goes on the right side? The electromotive force, so voltage or it'll tell you what belongs on the right side. And how do these terms go on the circuit? OK, it's 1 over c, right? Q. What is Q in this charge? Right? And what is Q prime? Sometimes they ask us to find, once we're done solving this second order differential equation for Q, they ask us to find something else, current. And that is first derivative, right? So what is L? Where are we going to pick up L? What is the unit? Or what is L representing? Inductance, and that is in entries, right? What is R? Resistance, and it is in ohms. And 1 over c, c is it is capacitance, and it is in ferrets. Right, so we've got Henry's ohms and ferrets here, and electromotive force over here. So they're the same. Once we get them set up, they're really the same thing. Their second order, either homogeneous differential equations or non-homogeneous. In fact, there is a kind of a nice little parallel on page 1178 in the supplement. It's got kind of a nice little parallel going from the spring system to the electric circuit system. But once you get it set up, it's stuff we've already reviewed. All right, back to the book. Really the purpose of sequences at the start of a unit of study on sequences in series, once you're beyond that, it has kind of outlived its purpose. When does a sequence converge? Remember, sequences are lists. Nothing is added. So if we want to see what happens to the sequence as we work our way out to the right to remember their commas, so we're just concerned with that nth term. Is the value of that nth term way out to the right? Is it getting closer and closer and closer to any certain number? If so, the sequence is convergent. Why is that important? Because in order for a series to be convergent, the sequence of partial sums is convergent. So we define convergence of series in terms of convergence of a sequence. So if we generate our own sequence, the sequence being the sum of the first one terms, the sum of the first two terms, the sum of the first three terms, and so on, we're not adding anything together in terms of this list. But as we work our way out here to the right, if we can develop a pattern in this particular sequence of partial sums converges, then the series associated with it converges. Yes? If it's n squared over 3n squared minus 1, we just know it's 1 third. It converges to 1 third because of the coefficient of one third. But if it's a series, then it's not convergent because it doesn't get a zero. Yeah, that's actually pretty good. Let's take a look at that. So let's say we have a sequence, and the nth term of the sequence is, give me what you just gave me. n squared over 3n squared minus 1. So we've got this sequence, a sub 1, when n is 1, what do we get? 1 1⁄2, a sub 2, 4 1⁄11. Let's get one more. A sub 3, OK? I don't know if it's clear if we have anything going on. Are we getting closer and closer to any one number? Well, we can look at the description of the nth term. We're not adding anything together here. We're not adding these terms together, but we're looking at the value of the nth term. It looks like it converges to 1 third, right? 1n squared over 3n squared. You could divide through by n squared, and you could validate this, but we've looked at that shortcut numerous times. So this is a convergent sequence. The value of the nth term gets closer and closer and closer to 1 third. Now we're switching horses to the series, so we're not talking about just values of individual terms, which is the sequence. Now we're adding terms together. So we want the first term added to the second term, added to the third term, and so on. So although the sequence is convergent, what about the series? This could be, I think, an easy test question. What's the first test that you have on this sheet, which I think this will be a handy study, is the nth term. If the limit of the nth term is not 0, haven't we already proved that? The limit of the nth term is clearly not 0. Therefore, this particular series diverges. Because when we add those terms way out to the right, aren't we adding in like a third, practically a third, to another third, to another third? We keep adding a third. Well, there's no way it has a chance of converging if that term that we're adding doesn't eventually disappear. In fact, even all those don't converge. So because, in this case, the limit of n squared over 3n squared minus 1, way out to the right, is equal to 1 third, which is not 0. Therefore, the series is divergent. So that's that first test. The limit of the nth term, as n approaches infinity, is non-zero. It does not have a chance to converge. Therefore, it diverges. Chandler? I was looking at a homework problem last night, and I mean, I've finally got it. But I don't know how. It was the sum of 3 over n. And that is divergent, and I just wanted to know why. OK. 1, 2, infinity? I guess. That'd be an easy test question. OK, it's p-series. If you want to call it that, you could factor the 3 out in front and then just look at the part that has to do with the summation. The 3 is not going to change the decision. 3 times a convergent thing is going to converge 3 times a divergent thing is going to diverge. So what about 1 over n? There could be 3, what I think are 3, really, really good, safe, correct answers. This is harmonic. We went through that whole process of harmonic series, kind of the harmonic, the basic harmonic, diverging. Somebody said p-series. It is a p-series, 1 over n to the p, and what? p is equal to 1. And in fact, when p is what? Less than or equal to 1, it's divergent. That'll work. I like the first one myself. Write down a word and you're done. There is another one. You could take the integral. It's 1 over that thing that's being scrolled, 1 over n, in this case, n's being scrolled from 1 to infinity. And it's not to any power other than 1. So in this case, 1 over n to the first is harmonic. And we could have variations of that. In this case, just a 3. And let me add to that. I don't think we've had this in an example. But if it happened to be an n over an n squared, that eventually is also harmonic in nature. So you had 2n plus 1 over 5n squared minus n minus 11. Basically, you have an n over an n squared. That behaves as a harmonic as well. n over n squared. Can't you simplify it to 1 over n? Yes. Well, you can as long as there's no baggage. So what I'm talking about is let's say you had 3n minus 1. You have baggage. And over here, you have 5n squared plus 2n minus 3. I mean, the dominant terms in the numerators 3n, the dominant term in the denominators 5n squared, that's harmonic right there based on the dominant terms. Clearly, out to the right, minus 1 doesn't have much meaning. And this gets overtaken, the plus 2n minus 3, those become secondary to this term, which becomes dominant. So that's also going to be harmonic in nature. But this is the type we're talking about. And any of those will work. This is a divert. This is an integral, which the limit does not exist. Therefore, the conclusion is the same. Let's finish with this one. Now, lots of things can go through your mind. I'm just trying to kind of hit a couple of these all at the same time. If it were just 7 over 3 to the n, you could bring the 7 out in front. And if we're left with a 1 over 3 to the n, that's a pretty good, that's infinite geometric, right? That's the same thing as 1 third to the n. So that'd be geometric. We don't have that, but I'm just trying those. There's our thought process that might get us there on another problem. What's the minus 2 due to the denominator, which makes the value of the fraction itself do what? If the fraction were getting smaller, wouldn't we have to make the denominator larger, right? We're making the denominator smaller, which makes the value of the fraction bigger. So yes, that's important to know that this is an infinite geometric series. The ratio is 1 third. The absolute value of the ratio is less than 1. Converges. If this one were even smaller than this one, then we would just use the what? Comparison test. It's not smaller. It's a little bit larger. So if you kind of think the comparison test is going to work, but it fails right at the end, what's the next bigger net that's going to capture convergence or divergence, even though comparison test? So this one fails. It'd be different if that were plus 2. Then we could use the comparison test. It's minus 2. We think it's a lot like this. We think it's going to converge. But we would use the limit comparison test. And how does that go? We want to see way out to the right. It doesn't matter which one goes in the numerator. Let's say the one in question. Didn't we determine that that one, which is a whole lot like this one, converges? You could use 1 over 3 to the end. It's fine. I'm just trying to make it as close to this one as I can. So I know this is convergent. If I get an answer, the one in question also converges, right? If the one that we know something about does something, converges or diverges, we get an answer to the limit problem. The one in question also does the same thing. So how's that limit go on this problem? Actually, we don't need these, right? So we get rid of the 7s. Divide everything by 3 to the end. That'll work. So that divided by 3 to the end. Now what happens as n approaches infinity? What happens to this term? That's 1. This is also 1. And what happens to this guy right here? 2 over 3 to the end as n approaches infinity. Zero?  If this one converges, which it does, this one must also converge. So we would just write a concluding statement. Well, now there's one that we haven't come to yet. So if you know some other tests like the ratio test. I remember you saying one of them because equal to 1 it fails. Yeah, and I think I may have mentioned the fact that we're going to encounter one later in this chapter that when the limit is 1, we don't get a conclusion. This is not it. It will happen later in the ratio test. So if we get an answer, a finite limit, the one in question does the same thing that the other one does that we know something about. Limit comparison test. And that's really the only one on the back of this sheet that we have done. So draw the line there. We have not done alternating series for this test. All right, see you tomorrow. So you don't have a sample test for this? I think there's one posted online. On your website? Yeah, for the distance add classes.


Okay, so I've just received the pine phone. This is the box and I have to say that the box is absolutely beautiful I love it. I would buy the phone just for the box and then we have the keyboard which is a rat and Again, I love the fact that it's right like super pretty Then we have an hub which we can use to connect the pine phone to like an external monitor or USB C Keyboard or mouse super useful even ethernet is in and then finally we have the actual phone that I have already turned on So we can see first of all that the UI is super small, but that is a bug I've already worked around it. So don't worry, but what you will see now is a bug It's not supposed to be this small. Anyway, let's try the pin, which is wrong. Let's try again this time It will be right and And then we can see the phone with all the applications in it we can see index to manage files and it's pretty snappy and Actually, if I go and close it again animations are just fine in the tax manager as well and It's really snappy like you would say it's it was a much more laggy, but it isn't if we open the camera There's this camera up, but I forgot to turn on the camera. So forget about it and We also have settings settings and We can use it to like connect to Wi-Fi and stuff But also change the theme if I set it to breeze dark We can see that the status bar at the top got dark and Then we can see Calendary which is used to as a calendar and Again the calendar that you will see at the center is much smaller than it's supposed to be so don't worry about it it will be much bigger and we've got like the gestures to open the drawers and In theory, I should also be able to like create a new task on the calendar But I actually forgot how it's done. So I'm pressing random buttons, but I Could not find the right one. Sorry about that And then if we go back again if the UI was a bit bigger, I would not miss click stuff We can go back and see the calculator a couple of seconds to loot and then I've got to say that the calculator is Very pretty like you also have a slider from the right to use more advanced stuff It works really well And then what else do we have? we have Nota which is the application that you can use as a text editor and Again, this takes a bit of time to load But that's mostly because it's a my way up my way up needs a couple of seconds more to load Everything, but after it's loaded. It's absolutely beautiful as an application We can change the brightness from the status bar and then we can open the virtual keyboard and start typing It works really well again. I think that this design of the virtual keyboard is just on spot Really well done. And then we can also create new files again. The UI is a bit small, but When we have created a new file, we can also see that tab cells are working. So this one is really a power device and Then we can also save the file. We've got like File pop-up to choose where to save it. We can change the name. I'll just go with like test.txt Again UI a bit small, but that's a bug. Don't worry about it And then by the way, the wallpaper is beautiful. We also have key runner So if we type something as a test within a couple of seconds, we'll see all the results show up And I think this is super cool because key runner on a phone is like a killer feature You can search between apps and files and stuff super cool And then what else do we have? We have console for like power users. We had a bit of a frame drop there, but It happens So again console we also have a small bar to like go around control delete and the arrow keys And here I actually forgot how to close the keyboard So I like try to swipe from the sides and from the bottom But it did not work because in order to close the keyboard you have to swipe down from the keyboard itself So I had to had some troubles here All I had to do was to swipe down from the keyboard. So ignore my stupidity here I actually opened up settings from the status bar And I think this one is a bug because it takes much longer to load compared that To the settings in the list of apps. So I think this is a bug And if we go past it again the settings in the list of apps is much quicker So it's not supposed to be this low. It's not a problem of the machine that will be fixed And then we have the task manager. We can close the active apps we have and everything And then we can also Look at the box again because it's beautiful, isn't it?