Direct Vandermonde Method of Interpolation – Linear Interpolation

In this segment, we'll talk about the direct method of interpolation which is also called the Vandermonde method, and we will limit our discussion to linear interpolation.

So, if you look at the general order direct method of interpolation, you are basically given n plus one data points, let's suppose, or what we should say is that you are taking n plus minus one data points to do the interpolation. Because many times you'll be given more points, but you will choose only a certain specific number of points to do the interpolation. So, if you have n plus one data points what you can do is you can make a nth order polynomial to pass through these n plus 1 data points, and that's what is being denoted here in this in this picture here, and what we'll do is we will take it easy, we will only do the first order polynomial, or what we call as linear interpolation in this segment.

Let's talk about linear interpolation. So, linear interpolation would also be called as straight-line interpolation, or first order polynomial interpolation so, they are the same names. So, how linear interpolation works in order to be able to make sense of the data, is that hey we are given x and y values only at two points. So, let's suppose x naught comma y naught and this is x1 comma y1, and what you are interested in is to find out the value of y at some value x, which is somewhere between x naught and x1, let's suppose. So, one of the things which you might have already done in other classes that you say, let me go ahead and draw a straight line which goes between those two data points. So, it does go pass through those two data points, and that's why it's called interpolation. So, the bottom line becomes that hey, if I’m able to find a naught and a1, then I can find the value of y at any point between x naught and y naught, x naught and x1. So, how do I find a naught and a1 is by saying that hey, this particular straight is going through this point, and it's going to this point. That sets up two equations, two unknowns. Hence, you'll be able to find out what the value of a naught and a1 is. Let's go and do this through an example, we better illustrate it that way.

So, in this example, we're asked to find the value of the velocity at t equal to 16 seconds, when the upward velocity for rocket is given at these six data points. And you can very well see that this velocity at 16 seconds is not one of the data points, so we have to somehow figure out how we can use interpolation to do that. We're also asked to use the direct method of interpolation, and also, we're asked to use the first order polynomial. So, as we have gathered that for the first order polynomial, we need two data points. So, the question arises that which of these six data points will I choose? Which two of the data points will I choose out of these six to find the value of the velocity at 16? So, it makes perfect sense that we choose the two closest points which are to 16, but they also should, should bracket the point of t equal to 16. So, in this case, we have t equal to 15 and 20 which are the closest point to 16, but at the same time they also bracket the value at t equal to 16. So, basically, what I’ll be doing is I’ll be choosing these two points in order to be able to find the value of the velocity at 16. So, let's go ahead and do that and see how it goes.

So, we have v sub t equal to a naught plus a1 d, and since we are using the values of the time at 15 and 20 to find what the velocity will be between 15 and 20, that would be the domain in which this particular first order polynomial will be valid. We already, let's rewrite v sub 15 is already given to us as 362.78, and the value of the velocity at 20 is also given to us as 517.35. So, let's go and use this information to be able to find a naught and a1, because the velocity at 15 is a naught plus a1 times 15, and that’ll be equal to 362.78. Similarly, the value of the velocity at 20 will be a naught plus a1 times 20, and that will be 517.35. So, you can very well see these are two equations two unknowns. That's how it gets set up, so let's write it in a cleaner fashion. We get this with coefficients in the front, and keep in mind this coefficient is 1, not 0. And then the next equation will be this one, and now we can set up this as two equations two unknowns in the matrix form, because we should be using numerical methods to solve our problems. So, it's 1 15 120 by using our matrix multiplication and you get 362.78 here and 517.35. So, all we have to do is, to solve these two equations two unknowns to find a naught and a1.

Let's also graphically look at this this particular problem. We have time here, and velocity here. So, what we have basically done is that we have two data points 15 comma 362.78 and then we have another data point coordinate as follows. And all we're doing is we're drawing a straight line between these two data points, which is given by a naught plus a1 t. So, let's go and find out what n naught and a1 turn out to be.

So, we find a naught to be equal to minus 100.93 and a1 turns out to be 30.941. So, our original expression, which was v sub t is equal to a0 plus a1 t, and keep in mind this expression is valid only between 15 and 20; so, we get this. So, if I want to find the value of the velocity at 16, all I have to do is to substitute the value of 16 into the straight-line formula, or the first order polynomial formula, and this is what I get. I get 393.7 meters per second, and that is the end of this segment.