If we look at this graph here, which is an illustration of actually a linear basis function, this is the dashed line in a linear finite element type of context, we would have three elements, and this would be one basis function basically linking one edge of an element to the other edge of an element. That was the classic linear finite elements scheme. We will now, and you see this here in the graph, this is the solid line which is one of the basis functions we will encounter later. This is a Lagrange basis function, we want to go down to the element level, and we basically do that by simply defining even if it would be in the entire physical space, the basis functions just to basically have support inside one element. Now, in order to achieve that, we can also, and that makes life simpler, go down to a local coordinate system, and because we use Lagrange polynomials and they are defined between minus one and plus one, we make a coordinate transform to this interval. So, this is defined as, the entire domain is capital D, and we want to get to a local elemental level which we denote capital D sub e, e for each element and it will be different for each elements, so we have to find a mapping. Now, the mapping we denote as capital F, so x will actually be capital F of xi, xi is as usual our local coordinate, and xi will then be given basically as the inverse function of capital F of x. So, we're going to have a look at the specific form of this coordinate transform. So, an example graphically is given here for three elements. So, we want to basically replace the physical coordinates here with coordinates minus one to one for each element in a general way. So, the way to do is to get the mapping that xi of x is actually xe, which is the left side of the element plus he, which is the element size, multiplying xi plus one over two. Now, it's very easy to see that if you put xi equals minus one, you get the left boundary xe, and if you put xi equals one you get the right boundary which is then actually xe plus he, which would be in that case then in physical space. So, this is forward mapping that we have to transform the coordinates, but we also need the inverse mapping. So, the inverse mapping xi by x is twice x minus xe divided by he, which is the element size minus one. It's very easy to see if you put the element boundaries xe or xe plus he into that equation, you will actually recover xi equals minus one xi equals one, and that's the demonstration that it actually works. The next question is what happens if we have to take integrals with the changed coordinates? So, remember we have to calculate a lot of integrals for the elements of our system matrices. So, we have to figure out what to do when integrating with our local coordinates. So, when we move from the x coordinates to the xi coordinates, our integral in our elemental space de over x. So, integral over de, f of x dx will actually turn into an integral from minus one to one, of f of xi. We integrate over xi, so it's dxi, but then we also needed dx divided by dxi. Now, this dx divided by dxi, that's classic. I'm sure you know this. This is called the Jacobian. So, basically from the mapping, we have seen in the last section we can now calculate this Jacobian. By the way, Carl Jacobi was a child prodigy of mathematics in the 19th century, is very famous also for his work on elliptic functions. So, what does the Jacobian look like? Well, we have our mapping x of xi. So, we actually need the derivative dx by dxi, which is he the element width divided by two. We also need the inverse Jacobian, which is just the inverse which is two over he. Now, all this is extremely simple in one dimensions. Actually that calculation of this Jacobian will be a matrix in 2D and 3D can become very complicated, and also actually the problem for curved elements. You can see here an example of curved elements in 3-dimensions that you always want to map on such a regular Cartesian shape, and we're not going to go into this, but that's an actual real world problems, this has to be solved.