So, let's suppose that, observe that u h at x equals x e. All right? Is basically u h at x evaluated with C equals minus 1, right? It's the left-hand of the element, right? It's the left-hand side node, which in the omega of C space is represented by C equals minus 1, right? So, this then, is Sum over A. N A evaluated at C uh,1. Right? Because remember, C 1 in the omega of C space is what we identified as being equal to minus 1. Right? So, it's any evaluated at C 1, d A e, okay? But then, we recall that this node, that these basic functions reconstructed had this Kronecker delta property, right? So, this is essentially, the Kronecker delta, delta A 1. Okay? Okay. And therefore, applied to d A e, this becomes equal to simply d 1 E, okay? All right? Essentially, what we've done here is to observe that if we want to evaluate the trial solution at the left node, right? At, it's effectively the first node of this element what we get, is that the trial solution is indeed, equal to the degree of freedom, corresponding to that node. And the same thing happens, if you want to evaluate u h at x e plus 1. This is equal to u h at x with C equal to 1, okay? This is equal to Sum over A, and A C 2, recognizing that C2 is the point C equals 1 times d A e, okay? But then, this again becomes delta, A 2, by the Kronecker delta property of the basis functions. [COUGH] And therefore, this can apply to d A e, gives us d 2 e. Okay? All right. Essentially, the Kronecker delta property ensures, for us that the trial solution evaluated at the corresponding node is indeed, the degree of freedom value at the node. Okay? Stated differently, the node of degrees of freedom are indeed, the trial solutions themselves. Using these particular basis functions, okay? So and and, the same thing, by the way, happens, of course, with W h, as well. I won't go through the details with W h evaluated at x e is equal to C 1 e, and W h evaluated at x e plus1 equals C 2 E. Okay? So, so the remark here is that, the Kronecker delta property. Then the Kronecker delta property of The basis functions. Right? And by this, we mean the specific basis functions that we're working with, right? The Kronecker delta property of the basis functions ensures That The node of Degrees of freedom, right? Which are these ones, right? Or these for the function, right? This ensures that the node of degrees of freedom of the solution field. Are indeed its values. At the nodes. Okay? This thing, this is sometimes called this, this is sometimes referred to as the interpolatory property of these spaces functions. It is not a universal property of, of arbitrary basis functions. We specifically chosen basis functions, that give us this property, all right? Okay, why did I need to say all of this? I needed to say all of this because it is what we are using here. Okay? We're using it in particular in this to make this identification of W h at L being equal to C 2, for element N here. Okay? All right. So, we have this, okay? And this is the reason, we are able to do that. Okay? Fine. What we want to do here now is go back to what we had when we put together the finite dimensional weak form. When we made all of these observations about, how it can be written out in terms of the matrix-vector products. Okay? You have it on your notes and thanks to the technology here, I have it in front of me, and you can't quite see it. But I can see, that I have it here. Okay. So, what I want you to do is go back to your notes to the point, just before we made the slight digression. To talk about this this Kronecker delta property of the basis functions, okay? Go back to, how we'd written out the weak form, okay? And we are going to work off there. Okay? All right. So, if you look at the way I've written out the, the finite dimensional weak form. Just preceding this this remark, you will see that we are now, able to pull things together from there. Okay? And in order to pull things together, what we need to do, is to observe that there is a mapping between the the degrees of freedom as expressed using local node numbers, and global node numbers. Okay? So, note That The following map Holds Between Local, and global, global node numbers, or, or local, and global degrees of freedom. Okay? I will express this map first in terms of the, degrees of freedom that are used to interpolate the trial solution. Okay? So If we have this particular local degree of freedom, and recall that this refers to local degree of freedom A, in element e. Okay? This can be written as global degree freedom d e plus 1, okay? Let me see. So, d e plus 1. Sorry, it's not d e plus 1 mm, it should be d e plus A minus 1. Okay? So, this let, let me write out in words, what this means. On the left-hand side, I have local degree of freedom A in element Number e, okay? That's what this refers to. On the right-hand side, I have global, Global degree of freedom. Number e, e plus A, minus 1. Okay? Let's check that it works. Always a useful thing to do. All right. So, let's just check, right here. Okay? So, let's suppose we are in element 1, and we are talking of the first degree of freedom. Okay? That would be d 1 1 that is the same on the right-hand side for the global degrees of freedom. d is equal to 1, A equals 1, so we get d1, okay? Likewise, d 2 1 is equal to d e equals 1, A equals 2 minus 1 d 2. Okay? And this goes on. Let's check what happens in the very last element, right? So, when we have d 1 N e L. This is equal to, now, e equals N e L, A equals 1, so we get d N e L, and d 2 N e L is equal to global degree of freedom equals N e L equals 2. We get d N e L plus 1, okay? This works ou,t because if we look at our domain, all right? And this is the very first element. Right? This is omega 1. Right? I will write up here, the local degrees of freedom Right? The global degrees of freedom for this are d 1, and d 2. Okay? For the very last element, which is this 1. Right? This is omega N e L. Local degrees of freedom here are d 1 N e L, and d 2 N e L. And here we have d N e L, d N e L plus 1, right? Makes sense. We have one node more. Than the numbers of elements. Right? The number of the total number of nodes is one more than the number of elements. And that is born out by the way things work out here. Okay? All right. And of course, the same thing holds for the other ones, also. All right? Similarly, we have C e, e equals C e plus A minus 1. All right? So, we've made this the, made the, made the observation about this mapping between local, and global degrees of freedom. [COUGH] we will use this to go back to that weak form that we had written out. In terms of matrix and vector products over each element. And carry out an important process of finite element assembly. We should probably, do this in the next segment. So, we'll come back to this in just a little bit.