Okay, let's continue now. You recall that at the end of the last segment we've reached as far as observing that in order to proceed with our with, with computing the terms needed in our weak form, we needed to figure out a way to write out uh,x, which is sum over A. NA,xi, xi,x, dAe, and the same for wh,x. Okay? And, and in particular the question is how do we do this term, right? Simply because this one is easy, right? This this one is easy. Right, you already know how to compute the derivatives with respect to xi of the basis functions, 'kay? So let's look at how we compute xi,x. And in order to do that, let's just recall that what we've set up for ourselves is a description where omega e, right, and some arbitrary point x in omega e, where xe and xe plus 1 are the nodal positions, right? Some arbitrary point in x is always thought of as being obtained as a mapping from this by unit domain. Okay, now if we could figure out this mapping we may have a chance of figuring out what xi,x is. And what we're going to do now is do exactly that. We're going to define this mapping, okay? So the, we are going to define the mapping. To omega e from omega xi. It's an extremely straightforward manner in which we are going to do it, 'kay? What we are going to do here is the following. We're going to say that any point, x paramaterized by xi over element e is obtained as a, is obtained by representing it in a basis, okay? And I'm not going to complete the limits on that summation sign just yet, 'kay? We're going to write it as some basis function. Right, here, and I'm going to leave myself a blank spot for it times the nodal values, okay? So we're going to write the nodal values here as a let me think, okay, so I'm going to write these nodal values as xeA, okay? Now, we're going to have the sum running over the number of nodes in each element, right? So that is A equals 1 to Nne. When, when I write xAe, right, what I mean here is for A equals 1 I'll get x1e and I get x2e, okay? These are equal to my global nodes, xe and xe plus 1, okay? The coordinate x1e is the same as the coordinate xe. Right, but when I write xe, I'm viewing that as a global node, right, the coordinate is the same. Likewise x2e simply is the second node of element e. Well, globally that is just global node number e plus 1, okay? All right, because of the fact that we need to write this as a summation over the nodes in each element and there are, in this case, two nodes in each element which we're writing generally as N, sub n sub e, 'kay, we need to go to this other sort of numbering for local nodes, okay? So what I have in the brace brackets here are local nodes, and here, the same things appear as global nodes, okay? Those coordinates are the same. It was just that in one case, when we're just viewing a single element, we are regarding those nodes as local nodes, nodes local to that element and numbered 1 and 2 in that element, okay? When we look at them globally, we recognize that they're one of the many nodes in the global domain, and we number them as xe and xe plus 1, okay? So just below this, let me also write the local nodes. So that would be x1 for element e, that would x2 for element e, okay? So the idea there, is that we just want to take our nodes, our two local nodes, describe to them the actual coordinate values, right, that they each possess and essentially interpolate. Now, this blank spot that I've left here in the parenthesis, right, the, the, the parenthesis that I've left unfilled is where we're going to put in our basis function in order to do this interpolation of the geometry. Now, we're free choose whatever basis functions we would want here. It would just mean that we're interpolating the geometry in some way. What is the most straightforward way to do it, the simplest way to do it, one that would probably make our lives the most easy? Yeah, it is to choose the same basis functions as we chose to inter, to, to represent our finite dimensional functions, right? So we'll just use NA, a, A being 1 and 2, right, the corresponding two in fact the same linear Lagrange polynomials that we already know, okay? So, what we are doing here, using the same basis functions. As for representing. And wh, okay? When we do this, when we use the same basis functions to represent our finite dimensional functions as well as to interpolate our geometry, we have what is called an isoparametric formulation, 'kay, isoparametric meaning same sort of parameterization. Okay, so we have here an what we have here is an isoparametric formulation, okay? Isoparametic, but because we're using the same parameterization for our finite dimensional functions as well as for the geometry, okay? Again, it's not necessary, it is just a convenient way to do it. There may be special cases where we don't actually do the same thing, okay, where we don't have an isoparam, parametric formulation. One can design methods of this kind also, right? Finite element methods. Okay, but we will be taking simple approach here, which is to go isoparametric, okay? All right, what that lets us do then, is the following, it lets us say that x,xi is easy to compute, right? x,xi for element e is, now, sum over A, Na,xi xAe, all right? And explicitly, this is just N1,xi, x1 for element e, where x1e is the global node number, right, it's sorry, it's, it's, it's so, I'm sorry, x1e is the local node, right? But the coordinate x1e is the same as xe viewed globally, okay, plus N2,xi x2e, 'kay? And just to make that clear let me state that, that that is just equal to xe viewed globally. That is just equal to xe plus 1, viewed globally. Right, so the coordinates are the same. Okay again, that's easy, easy enough to do. In this particular case, we know exactly what N1 and N2 are. That's easy to compute. So that is just d/d xi of N1, which is 1 minus xi divided by 2 times x1e plus d/d xi of 1 plus xi divided by 2 x to e, okay? Carrying it out, we basically get x 2e minus x1e divided by 2, which is also xe plus 1 minus xe divided by 2, okay? Now, what we will do is observe that xe plus 1 minus xe. For element e is simply the length of that element, right? So we will write this as he over 2, okay? Where 'kay, this is to be viewed as the element link. Importantly what this allows us to do from the ground up, from the, even from the very simplest of finite element formulations is to have elements of unequal lengths, right? So in our partition of the domain omega into subdomains omega e, there was no requirement and indeed there is no requirement that the elements all have to be of the same length, okay? So he can be different for each omega e, 'kay? We don't need a uniform discretization, 'kay? So this is, allows us to have a non-uniform discretization. Okay? And the straight at, at the very outset, it's an important property of finite element methods. Okay let's see how all of this works now. Now so, so we've done very well. We have gone ahead and observed that well where did we start this. Remember we started this calculation all the way up here, okay? So what we've seen is that the derivative of x with respect to xi, which can be thought of as the tangent of this mapping, right, of a tangent of the map that gives us the physical element from this by-unit domain. [INAUDIBLE] Right, the essentially what we've figured out is that we found for this mapping, we know that you know, every little well, well, we know that the scaling of the, the element of the physical domain, relative to the scaling in the parent domain is essentially he over 2, okay? It's a constant because, why, why is it a constant? Because for our geometry we're using linear interpolations, right? We're using linear basis functions, okay? That's what makes it a constant, that's what makes this, this tangent map a cons, a constant.