Okay, so today we're going to do a change of variables. The theorem, OK. What I'll do is I will first just redo the example we did last time. I just went through it very quickly with this polar coordinate thing. Alright? Then I will motivate what is the change of variables in R N. Ok. And then we'll do two more examples. Ok? So the change of variables theorem, which stated, and then we'll do two more examples and then I'll give you homework problems later on for that. Okay. So here was the question we were working on. Coming to redo this. So can go offline. Ok. So let d be this disc. Ok. Compute double integral cosign of x squared plus y squared da. Da means dx dy over the region t, Right? So we draw this sketch here. This is the origin. Sorry, let me not draw these lights. Okay, that's the region D. And then, so we do polar coordinates. X is r cosine theta. By the way, if you have any problem with the voice was something let me know, okay? Alright. What do you have trouble seeing something? Just interrupt me, please. Okay. And so then I wrote this last time, but there's no harm in doing it. So da is r d r d theta. And the way you figure out the limit is at least this is how I teach it when IT due to 43, I say your fix an angle theta and you draw is strip. Right? So then you move on this trip from the origin to sorry, and this circle is x squared plus y squared equals one, which in polar coordinates becomes r equals one. So your integral, right, it becomes cosine of r-squared. Ok? And then yard in this, r goes from 0 to d one. So d r goes 0 to one. And then you've covered this whole strip. So now you rotate this trip, which means theta goes 0 to two pi. Okay, so before I write that, that is the extra are also ok. So our went from 0 to one and theta 0 to two pi. Have you all okay with that? Right, you've seen this before. And then you can integrate. So you get sine r square over two. R goes from 0 to one. Okay? I equal to 0. So you get sine one divided by two times two pi, which is Craig. Okay? So we discussed this yesterday also. So what I want to extract from this example is what really is a change of variables? Ok, if you're going to make up a change of variables, we need to understand what do we mean by a change of variables. Right? So if you've all written this down, I'm just going to go on. Ok. So what, what change of variables was that there was an integral on a region D. Alright? And we introduce new variables, r and theta. And they were related to the x and y. This was the relation, right? And we reduce the integration two instead of X and Y. And we knew why we didn't want to do int x and y. We showed it yesterday, right? There were difficulties with limits with x or y. So what we did is the following. Right here is I'm going to write it. So what is a change of variable? Okay, so D was this region in the XY plane, right? Okay? And then what we had was the, in the r theta plane, right? We found that we had was between 01, okay? And theta was between 02. You can put it equal to or not. It's not going to matter. Right? That was the limit for r and theta. So you had a region U in the r theta plane. Okay? Where r was between 01 and theta was between 0 and pi. So there was one integration in the xy plane over the region D. And then there is another integration. We converted it to an integration on completely different region. It was integration in the r theta variables. Think of this as r and theta as some other coordinate system. And this, right? So you can show it like this. This is the picture I want you to think about. And this is what really a change of variables is. Here is the r theta plane. R goes from 0 to one, sorry, two pi. R goes from 0 to one, theta goes from 0 to two pi. Okay? There is a generic R theta and this is what I'm calling the region U. And then you have a region in the xy plane. That was your d, right? And this was the point XY. So what you did was he created a map from this rectangle in the r theta plane to your desk. And I'm going to call that map phi. So you created a map phi from this rectangle 01 cross 0 to pi two D, which then r theta to x and y, where x was our cosine theta and y was r sine theta. That's what you did. Instead of computing an integral over the region D by this new change of variables, which sent our theta two r, r cosine theta y r sin theta. From this rectangle, you'd move the integration to a rectangle. That's what you did, right? We didn't think we never thought of it like that, but that's really what we've done. We are integration on this desk, which I could not know with x and y. I found another region which looked rectangular. So it's very easy to do integration on rectangles. And I had a change of variables from here to there, right? And you can see that this rectangle maps exactly onto the disk. Okay? So that's what a change of variables is. A change of variable. This, you want to integrate over some region D. You find another region, you maybe it's simpler. And you create this map. That's a change of variables. Then the question is, what is the impact of this change of variables on an integral? Okay? And that I have to tell you, so you do it. So you can just basically construct any change of variables you make any map, okay, more or less, that's, that's, you can make up any map and that's a change of variables. There are some restrictions, I'll tell you in a second, right? And then you convert integration on one region two or integration or another region, which is hopefully simpler. Okay? Except when you do a change of variables, the integral from one to the integral to the other, you integrand changes. There is a correction factor that our type of thing, RD, RD theta, right? I have to tell you what that is when you do a change of variables. Okay? So let me define now a change of variables. Okay? I'm going to define something called an open set and a boundary and the closure, okay, it's nothing, right? When you do integration, when you're doing integration on a plane. Are we doing integration on just some 1010 points, 20 points? Or what are we integrating over? Always. When you are integrating in the plane, are we integrating over just a finite collection of points or something? No, right? When you do a double integral, are you just doing things on lines like this? No, right. It's like a region, right? It's some region, maybe it's inside an ellipse and bounded by a parabola in lines or things like that, or you're doing in 3D, right? Is bounded by cones and things. So it's, it's, it's what is called, what you'd think, what's called a region. So I want to define that, that is what is called an open set. Okay? So when we are integrating, very integrating on open sets, okay? It's nothing very complicated, just something very elementary, but I need that term to define that term. Alright? So let me just define what an open sentence. And then changes of variables aren't open sets. Okay, before I even define it, let me just talk of it. Ok, let us do example open set R2. Okay, what is an open set, right? It's going to look like region. Okay, so here is my region D. All right? So when do we mean, what do we mean by saying it's a region which means that if you give me any point in it, and if you draw a little disk around it, there is some disk around that point. So that disc is also inside the regions. That's what an open set is. So let me reset, I'm going to write it down. But Tesla three to this, right? Open set means you give me any point P. Let's see what word I use, P or S or equality. Let's make this bigger alsos inequality. Okay? So it's open means you take any point, this should be true for every point. Then there should be some disk around it, which is also completely contained in U. So such a set is called open. Alright? Okay? All right. Now, what is the boundary of this set? Right? I think we all agree that the boundary of the set is this curve here, this arrow thing. Why do we call that the boundary? So is this a boundary point for you? Now? Is that a boundary point for, you know, right? But this is a boundary point for you. So what, what is it that a boundary point has which these two points tone tab. Right? What is special about this a lot. There's not a boundary point, but this is a boundary point. What's special about that point? Other than tickets on the boundary? You can know all disk on it just because when they go outside of your region. Well, you mean this point is a boundary point. Now by lakes, the boundary points, some of this technically wouldn't be contain in the open, but some of it would not exactly. So a boundary point is one. So that if you draw any disk around it, some part of it will be u and some part will not be in U. Ok? So I've defined two concepts. I have defined what an open set is, and I've defined a boundary point. Alright? So let me just write this down and then we'll do some examples for a second. I mean, it just so that we have a language to talk about, but I'm not really going to use this any, in any great way. Okay, other than saying change of variables is an open set. So Definitions, subset U of R n. So I am defining it in R n. So in two dimensions, our talking of disks, in three dimension, the disk would be a ball, right? N dimensional, there'll be the n-dimensional ball, all the points, but I haven't yet defined a ball, but I think, I think I'm not going to belabor this point. Boyle is what do you think it is? All the points within a certain distance of Eric, Okay, so a subset of R n is said to be an open subset of r n. If for each a in you, there is an n-dimensional ball centered at a with B contained in you. That's what I'm saying. Right here is you've, if it is open for every a, there is some ball B around it. We're just still inside you. For every not just for one day, but every year. That's number one open set. Okay, what is on the boundary? So a point P on the boundary. If every ball around P contains points from you and points not in you. If every n-dimensional ball B around p centered at P points in N points. And then the last thing is the boundary of u is the set of all boundary points of view and is denoted by delta u decimal notation. Right? Let us backtrack for a second. Let's go over it. This is not going to play a very big role in it. It has to be a little more careful and precise. We can't define change of variables on the collection of points are in the plane, you know, just a crisscross of lines. It's always over open sets. That's why I'm doing this. Let's go back. So what is an open set in R n? You take any point a in R n, then you should be able to find a ball around it. So if you're not three, it's a three-dimensional ball, 2D. It's a disk. You should be able to find a ball around it, which is contained in U. And this is true. This should hold for every point a new. And we'll do an example in a second. What is a point on the boundary of you? Appoint pees on the boundary of you. If every disk, every ball around it centered at bag contains points from you and points not in U. Ok. So particle, sorry. So some part of those things, some part of this ball should be new in, some should not be in years. Alright? And then the boundary is the collection of all the boundary points. Okay? Alright, so let's do an OK. There's one more term, okay? You get the closure of u is defined as u bar. This is called the closure and it consists of U and the union, a union, so I don't want to mix it up. Okay. That is u bar contest. You and the boundary points of view. Okay, I'm going to do an example will become very clear. It looks confusing at the moment. Alright. So three things I have defined. What is an open set? What is a boundary point, and what's the closure? Okay? It's just so that I have language. Let's look an example in 2D. Ok. So let you be all the points, x comma y, right? And that region, what does that mean? What region is that? Can you describe this circle centered at the origin? So that again, it's a circle centered at the origin. You notice I make a distinction between a semi-colon or disk. Essay says tell is only the circle, right? So what is this unit disk centered at the origin? Is it the unit disk? Look carefully. It doesn't quite get to one, so okay. So it's the open unit disc, if you like? Correct. Okay. But there's still one more thing. What else? Is it? The open unit disc, meaning it doesn't include this Hegel, that's what it is. It just that orange and it doesn't have the origin either? Correct? Right. So let's make it dashed line so that you can see that, oops. Is it open? Is an open set. If I took any point anywhere, you can always draw. Right? You take any point a, there is always a disk around it contained a new, right? Okay. So U is open. Okay, what's, what are the boundary points for you? Some things are obvious. Remember the boundary does not have to be new. So what are the boundary points of view? Okay, now 00 on the plane. I didn't see different setting. Active speaker video. Okay. That's good. Sorry, someone said something. I didn't know if it is healthier picture. So what's on the boundary of you? The origin and the unit circle, okay? Is a unit circle on the boundary, right? If it took any point p on there, every disk around it will be pointing you in points not correct. That's clear. As the origin and is the boundary point? Is the origin a boundary point? If I took any disk around it, does it contain points from you? Is origin a boundary point of view or not? Right. So if I took any disk around the origin, does it contain points from you? Yeah. Okay. Does this contain points not in, you know, again, what's near? The origin itself is not a new exactly, right? Is a boundary point because every disk around 0 contains a point from you. And a point not in New. Namely, it just has such thing. It's not going to play a critical role for us. Okay. I'm just telling you. Okay. So the boundary of this punctured unit disc is this circle and the earth, our origin. Okay? What is the closure? Closure consists of IU union. This. Alright. And so how do you describe the closure of this? What are all the points in the closure? So what's your Ladies and Gentlemen? That's the set of x, y such that 0 is less than or equal to x squared plus y squared less than or equal one. The whole desecrating the closed desk. I don't need to do does require this fall with greater than or equal to 0 anyway. All right, so that's an example. Let's do another example. Ok. So let's, I'm not going to quality and this time, let's call it k. Open, right? Every point MK, You should be able to find a small disk around that point which is still n k. Does that work? Yeah, those are the true every point MK is. Why dig this point, Tim kay? A disk around it which is still inside k, does not apply to every point in K. Now, which point does it fail? So everything that's on the parabola itself doesn't contain all of you, correct? Right. So it fails because of the red. Remember I have equal to here. So a here, points in k on y equals x square, failed the open set test. Right? Ok. So you understand opens, that means the edge is not there. Basically that's what it means. What is the boundary of k? What are the boundary points, okay. Y equals x squared, the parabola, correct. So delta k is equal to all the points. Get comfortable with these notions. It's what your intuition tells you exactly correct. Alright. So this is the boundary. Okay, let's go back to the figure. What's the closure of K? Remember, closure consists of points in k and the boundary of k. So what's the closure of K? Any point where y is greater than or equal to x squared. So is it different from K? No. So closure of K is actually K itself in this case. Alright? Okay? So actually these sets like here called closed sets. Okay, just one last thing that you understand the distinction. So let's see you is all these things. And y is strictly greater than x square. Is you open is an open set. Yes. Right? Because you take any point here, there is a disk around it, still inside you, and you can't get to the edge because y is greater than x, right? This, this greater than KU is open. Or does the boundary of what is the boundary of u? In this case, y equals x squared, right, is still the same, so its tail. Okay, and what's the closure? Is you combine with delta u. So in this case, the closure is actually already witnessed. The Clojure includes the edge. Alright. Alright. Okay, why am I talking off all disclosure and stuff? There was a reason here. Suppose you did a double integral on this region, the region above the parabola. Okay? And you did this some function, let's say sine of x squared plus y squared. You integrate that over the region you. And you integrate over the region k. What's the difference between UN k, just the edge, right? Let's extra, or the two integrals different, right? They're not. And I'm going to just make use of that, right? So if you have an integral over a region and then you add the edge of that region. It's not going to change the integral. That's the point I want to make here. Okay, anyway, I think you've talked enough. Let me now define what is the change of variables. Okay? So change of variables is exactly what I said before in the beginning, I just had to do this diversion, right? Change of variables is a region in D and a region you, and it's a map. But it's not just the arbitrary map, right? There has to be a one-to-one correspondence is not as if you know, three points in a, you go to 1 in D. You don't allow that, right? Every point-of-view goes to some point in D, and every point in D corresponds to some point of view. There has to be a one-to-one correspondence between them. That's what a change of variables as against what that parameter, right? Okay. Definition. If u, d are open in red, then a map fee from you to the right with you going to phi of u, which I'm just going to quit. So I'm going to call the variables and you will literally when the variables in D X is said to be a change of variables. If three conditions hold, one, phi is a math pages have seen this terms. Others may not have. Bijection means there is a one-to-one correspondence between the points of view and points of D. Okay? I'm going to come back to this again. It just basically means there's no duplication and everything is covered. That's really all it means. Two, I'm going to come back to this. So if he has a derivative at every point, okay, and then some continuity, I'm just going to ignore it. Third thing is this matrix, ok? So the derivative is an n by n matrix. Which means I can talk of its determinant is not equal to 0 for every, OK. Basically, all these things will follow. You just have to set up a one-to-one corresponded like it did from a rectangle to a disc. For the polar coordinates, you just have to set up a one-to-one correspondence. That's a change of variable, that's nothing more than that. Right? Let's, let's just go back and see how, how the polar coordinates fit into this. Okay, let us go back and look at what we did for the polar coordinates. So D was x squared plus y squared. Okay, sorry, let me just backtrack for a second. Okay? So let me draw a figure here. So here is u, here is d. This is the map. Phi sends you to X. Okay? Okay, let us be clear about what is the meaning of bidirectional, right? So some of you have seen this before. Why Jackson a one-to-one correspondence if you've taken to 45 or courses like that, Can anyone explain what that means? Anyone want to give it a shot? What does it mean to say a one-to-one correspondence between u and d? Or a bio reactions. Every point in the set u gets mapped to exactly 1, the set D. Okay? So firstly, what is a mapping? Mapping means every point to new gets mapped to something. And exactly one thing, right? That's what a map, if you take points here, it's centered somewhere in d. That's what a map is. Right? But that's not a bi-directional, that's not a one-to-one corresponded that just any map. You could take ten points here and map them to the same point. Every point of view would just go to the same point. That's taylor map. That's not a bijection, not someone said something. Could you repeat that again? So someone said, give me a definition. So can you reap? So we'll just go through the nuances and make sure we understand this. That every point in you gets mapped to a unique point in D, correct? Right. Every point in you gets mapped to a unique point in D. Value is not getting mapped to five different points anyway. That's just the definition of a map. So I think we have to change the language a little, is not that every point in you gets mapped to a unique point. Well, of course it only goes to 1 in D. What is the correct way to say this? What you're trying to say, every point in D has only one y from here. Correct. Ok. But let us come back to that. You said something a little stronger and that is really where we're headed. But let's come back to what the person said, what the person was really trying to say. It was two different points and you will go to different points in D. Right? That's what I think he's trying to say. Different points and you will go to different points in D. That is not true for a general map, but for a bisection that is required, you cannot send two points to the same point. All right, for a bijection, that's number one. But that's not a bijection, right? Different points and you go to different points in D. That's part a by dx and that's not a one-to-one correspondence. There is one more thing needed, which young lady said something, which was what really it was. What am I missing here from by ejection? Any 2 suffers. The every point of view goes to some point in D, that's a map to different points in you will go to two different points that's required for a bijection. What else does one more thing are? Correct? And every point in d has to come from some very new. It's not as if you just gets mapped and covers only this much and leave that much out. Okay? So bijection means there is a one-to-one correspondence from points in you, two points in D, you cannot miss anything out. In D, you cannot miss anything anew and there is no doublings. Ok? So let me redraw the sketch here. Nothing in, you know, doubling. Get mapped to different points in D. Okay? You cover all of you, you cover all of D with the 5x. And no doubling. You can send two different things. Two, One thing. That's what abide textualis and ascertained, forget ever out everything else. 2n three. That's really what a bijection is, is a change of variables as you re representing d by some other collection of points. That is, that's all there is to a change of variables. Right? And you do it for open set. Okay? I'm going to do an example in a second. So here does the main theorem. Well, you do the change of variables. Where does that extra x by rho squared sine phi, where does that come from? What is it? And this is the statement. So suppose you and be open subsets of fee from you to dy is a change of variables. Then, okay, so here is it. Okay, I'm going to write it as a double integral, but there is not really double is depends on the dimensional, Okay? So you're trying to integrate f x dx. Okay? So you're doing an integral on D. I'm going to switch it to an integral on you. Okay? The function doesn't change. You just have to write it like you it was an x. I have to write it in polar coordinates, so you write it in terms of you. The important thing is the correction factor. Here is the correction factor. It's the absolute value. Okay? I didn't leave myself enough space. Let me just rewrite this. Integral on d becomes an integral on you. The integrand, this part does not change, it, just put it in the u variables. Here is the correction factor. Correction factor is the absolute value of the determinant of a matrix. Where what is dy dx t? His is the n by n matrix. Phi prime mu. That is delta x by delta u is the n by n matrix. Delta x by delta huge L. That's the correction factor. That is the crucial thing. Alright? You can make up your own change of variables. Just make sure it's a bijection. Then you can switch from one region to the integral on the other. Excuse me. You just have to do this correction. Notice the determinant of a matrix. So where is this coming from? And this is exactly question for from your homework. Exactly. This says, if you do question from, from your homework, you will begin to understand where this comes from. Okay, so question for the homework or determinant is going to play a very important role. So go back and look in your math to 43 stuff. Somewhere there was a determinant. Correct? That's where this is, this theorem is actually motivated by. Okay? So I've talked a lot. Let me how much time we have. We have 15 minutes, right? Enough time to do one example. Anyone have any questions? I know it's been very theoretical, but letting them, letting everything becomes clear. Where B is this equation. In the first quadrant. Bounded by, excuse, excuse me, mu sub y equals x squared, y equals x squared plus one plus x square equals two. Okay, I'm going to draw a sketch of the region. So this is y equals x square. Y equals x squared plus one, y plus x square over two. This is an upside down parabola. Okay? This is one. Okay? You can think about it, but this is an upside down parabola and the other one. That's he'll region D. Alright. Now you can try to do this as we were discussing the example on Tuesday. So I wanted to compute this integral here, x, y d a. Okay? You can try to do this by drawing vertical lines and splitting it into reader. It's going to be complicated, right? You don't want to do it like that, right? We are trying to compute. That's what we're trying to compute. And you can try to do this and you'll find limits will be complicated, integration will be hard all there'll be square roots and things, right? So, but you can try to introduce new variables so that this complicated region goes into a rectangle. Just like for polar coordinates. You just make up your own variables. Right? So what's a good choice for variables? So I need two variables, right? Instead of x, y, I need to, I'm going to call them u and v. What do you suggest? So that this region goes into something nice. You can u equal x squared, okay? U equals x squared, v equals y. I could just like that, right? So if been to u equals x squared, this will become y. So n v is y. So v, v plus u equals to V plus U equals four, v equals u. So these all become lines. That's already makes it easier. Correct? That's one choice. So if you replace u equals x squared, remember you are all in the first quadrant, so it should work out. There all become lines, lines. It becomes some sort of parallelogram, maybe, maybe a rectangle, Right? But it's not going to be parallel to the x and y. So we can actually do slightly better if you do a different change of variables. What you said is correct, it'll work to take a little bit of effort, more effort than what I, what I am going to show you. So u equals x squared and y equals v equals y would have done it. But it'll be a little more complicated than what we will do next. Any other change of variables? Anyone want to suggest something? Okay, I'm going to pick someone. Mr. snyder Howard here. K suggest something to way we change the variables to use actuary was why worrying. We're really on things, but that's one choice. But I'm saying we can do better than that because it'll give you a region with lines, but it'll give you a, maybe a parallelogram. But the pilot may not be parallel to the UV access, so you'll have to split it into three regions or something. Can anybody suggest another change of variables? Okay, login. How about you? Come on, I'm not going to hold you responsible for giving a wrong I'm not answering, so don't don't feel pencil out. We're saying that using u equals x squared and v goes, yeah, that's, I mean, you said we want something that's more square-shaped than a parallelogram is what we're aiming for. That one wouldn't already do it. Now I just said u equals x squared, v equals y will already give you a parallelogram, or in fact probably a rectangle. But the rectangle maybe like I'm just guessing. I'm not sure if it did that. I'm just drawing a sketch here. It might look like this. Right? This line will be x i, u plus v equals to u plus v equals four. And like that, something like that, right? That it'll still maybe split it into three parts or correct? It is correct. You can do it. It's not a problem. There are no square roots and everything will be fine. But is there a better one? So the region is even simpler than this. It's easier to do integration over that. Couldn't be u equals x squared plus y and v equals u squared or x squared minus y. How about that one? Right, let me write that citation. So u is y plus x square. Let me stick to the way I wrote it so I don't let my calculations will not be messed up. Okay, I got it other way around. So he said take y minus x square u, and v to be y plus x squared. Do you see why you'd want to do that? Then what does this line become? This curve is equation would be in uv coordinates. What will be the equation of this? Y plus X squared equals to b, to be equal to this one would be vehicles for this would be, I guess you can all see it. U equal to 0. This will be u equals one. So what's the region in the uv plane? V between 24, between 01. Isn't that an easier region to integrate over? All right, so you've choices. There is no algorithm here for choosing a change of variables. It's just convenient. Alright? So, okay. Let's just think about one or two other things. Alright, so this is, I'm going to write this more carefully in a second, right? This, this, okay. So this is going to be, okay. So now, so that's going to change. All of this is by the way, in the notes. Okay. I typed it up. It's there. Let me just ask another question. Right? So you're going to go from the x, y, which is this change of variables. I need this quantity here now, the correction factor, correct? That's what I need. And then I'll be able to switch to the UV variables. U is here is uv, right? Two components. So I need the correction factor is the absolute value of the determinant of right? It was delta x by delta u. Just shorthand notation for, you know, put all the x's on the top and the bottom. Ok, that's a correction factor. So it's delta x, y delta uv. So you take that matrix and then you take its determinant. Okay? You take the determinant of that and you ever take the absolute value of that. Alright? Okay. The reason I'm showing you is there's another issue here. So I have to compute the determinant of this matrix. Write x sub u, x sub v, y is y sub v. That's what I have to do. That's a correction factor according to change of variables. Right? Is nice, nice region to integrate the very good. But I now need to compute this correction factor to switch to the UV variable. I'm going to, as I said, I'm going to write this carefully and it's all written in the notes. I'm just thinking aloud right now. So there's one issue here. So to compute delta x by delta u, this is my change of variable. So I need to write firstly, what is x in terms of u and v? What is Y in terms of u and v? Correct? I need to solve. That's one way. But there is actually a very clever way, which is you compute the opposite. Because notice u and v are there, right? It's easier to compute this. This is u x uy, vx, vy, right? You, and we're already given like that. This is far easier. Is there a connection between them? Then I can skip solving for x and y in terms of u and v. Do you understand my point? Right? Sometimes to compute this directly, the correction factor is complicated because you wrote your variables like that. Whereas given the variables, I can compute this more easily. Which is that, is there a connection between them? Right? Can there is, Can you, can you just guess? Just as a wild guess, what is the connection? And they inversions there exactly as matrices. They are inverses of each other. And if you have determinants of pain versus their reciprocals of each other. Okay. And why is that true? That is from the chain rule, which we did in the last lecture, which I'll show you later on. Why that is true. So this is a trick I'm going to show you next time. My time is up. But this is a trick. It's all explained in the no time were deposited. Alright, give me one minute. Let's do a poll. I'm going to stop firstly, stop the recording.
Change of variables - the theorem
From Rakesh Rakesh September 10, 2020
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