CHAPTER 05.03: NEWTON DIVIDED DIFFERENCE METHOD: Newtons Divided Difference Polynomial Interpolation: Quadratic Interpolation: Example Part 1 of 2

 

In this segment, we're going to take an example for the Newton's divided difference polynomial method, that's what the NDDP stands for, Newton's divided difference polynomial, and we're going to take an example for quadratic interpolation. We're going to take an example. So again, this is the same example as we have taken for linear interpolation, but if you've not seen it, we're going to rewrite it here. So we're given the upward velocity of a rocket, times is given in seconds, velocity is given in meters per second, and it is given at these points, 0 it's 0, at 10 it is 227.04, at 15 it is 362.78, at 20 it is given as 517.35, at 22.5 it is given as 602.97, and 30 it's given as 901.67. So you're given six data points, but we are only doing quadratic interpolation, so we need to figure out what are the three points which we need.  So we are asked to find out what the value of the velocity at 16 is, which is some point at which we don't know the velocity.  So first we have to figure out, because we have quadratic interpolation, what are the three points which we're going to choose? We already know that we've got to choose 15 and 20, because those are the immediate points which do surround 16, so we've got to have 15 and 20, and then the question arises do we 10 or do we choose 22.5? And the reason why I'm going to say that, hey, we will choose 10 is because of the fact that the difference between 16 and 10 is only 6, and the difference between 16 and 22.5 is 6.5.  So the difference between 16 and 10 is 6, so far as the magnitude of the difference is concerned, it is 6 here, and the magnitude of the difference between 16 and 22.5, it is 6.5. So the three data points which Iím going to choose for my interpolation are basically these three, 10, 15, and 20.  So let's go ahead and see that how we're going to go about solving this particular problem by using those three data points.  So this becomes my t0, t1, and t2, v of t0, v of t1, v of t2.  So I'm going to write it down here.  So we're going to choose t0 equal to 10, v of t0 is equal to 227.04, t1 is 15, v of t1 will be equal to 362.78, and t2 is 20, and v of t2 is 517.35, so those are the three data points which are given to us. Now, let's go ahead and find out what b0 b1, and b2 are in this expression for the second-order polynomial for the velocity, which will be b0, plus b1 times t minus t0, plus b2 times t minus t0, times t minus t1. So we want to find b0, b0 is nothing but the value of the function at t0, which is nothing but the value of the velocity at 10, which we get as 227.04.  So we already found out one of the constants, b0.  How do we find b1?  b1 is given as the divided difference between t1 and t0, the first divided difference between t1 and t0. So we need to be . . . we need to figure out what that means. What that means is simply the velocity at t1 minus the velocity at t0, divided by t1 minus t0, so it's the finite difference approximation of the first derivative of velocity based on this expression here. So we know that this is the velocity at t1, which is 15, minus the velocity at 10, divided by 15 minus 10, and the velocity at 15 is how much?  362.78, the velocity at 10 is 227.04, we divide it by 15 minus 10, and this number here turns out to be 27.148, so that's what I get for b1. Now, I need to find what b2 is. So b2 as we said that is the second divided difference based on the values which we have for the velocity at t2, t1, and t0.  So as we said previously that this is nothing but the first divided difference at these two data points minus the first divided difference at these two data points, divided by the difference between the extreme data points here. We just found out what this number is, so this number turned out to be 27.148.  So all we have to do is to calculate this number here, we already know what these two numbers are.  So I'm going to write this down separately, what is v2?  What is the first divided difference between the value at t2 and t1?  So that turns out to be the value of the velocity at t2 minus the velocity at t1, divided by t2 minus t1, and this one here turns out to be the value of the velocity at 20 minus the value of the velocity at 15, divided by 20 minus 15, and this one here turns out to be, what is the value of the velocity at 20? It's 517.35, the value of the velocity at 15 is 362.78, and divided by 20 minus 15, and this value here turns out to be 30.914. So that is the divided difference value which you are getting at t2, first divided difference approximation, or value of this expression here.  So I put it back in here, and so I get b2, which is the value of the first divided difference here minus the first divided difference here, divided by this minus this, so I just calculated the first number here to be 30.914, I just calculated this number previously, it is 27.148, divided by what, t2, which is 20, t0 is 10, and this value here turns out to be 0.3766. So I have been able to find out the values of b0, b1, and b2 by these divided difference definitions. So what that tells me is that the second-order polynomial approximation, which I had written as b0, plus b1 times t minus t0, plus b2 times t minus t0, times t minus t1 is basically b0, which is calculated as 227.04, b1, which was just calculated as 27.148, t minus t0, the first value is 10, plus b2, which was calculated as 0.3766, times t minus t0, which is 10, times t minus t1, which is 15.  So that is the second-order polynomial approximation which we are getting by using Newton's divided difference polynomial method.  And again, this particular polynomial, or this particular interpolant, is valid between the minimum value of t which you and the maximum value of t which you have, which is 10 less than t less than equal to 20.