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

And going back to the thing that we want to find out what the value of the velocity at 16 is, we're simply plugging the value of t equal to 16 in here. So you don't have to expand it, so that's something which you've got to understand is that when you are plugging in the value of 16 in these expressions, you don't have to expand it to be able to find out what the velocity at a particular point is, you simply can put it into the expression itself without expansion, because that will take time, and if you're doing by computer, that will take computational time, so there's no need to do that and this one here turns out to be equal to 392.19 meters per second. So that's what you get as the value of the velocity at 16. What I want to show to you is that how the values which we just calculated for b0, b1, and b2, that how we can write them in a tree form.  What we mean by a tree is that we had t0, t1, and t2, which was 10, 15, and 20.  Then we had the corresponding values of the . . . of the velocity, which were given to us, 227.04, and then 362.78, and then 517.35, so those are the numbers which were given to us. And what we're going to do is we're going to calculate our first divided differences.  So we're going to calculate the first divided difference by using these two numbers. So that'll be the first divided difference between t0 and t1, or t1 and t0, it doesn't matter, and this one will be the first divided difference between t2 and t1.  By just taking these two numbers, we will be able to calculate this first divided difference, by taking these two numbers, we'll be able to calculate the divided difference there, which we just did, and we calculated this number to be equal to what?   We calculated this number to be 27.148, and this number we calculated to be 30.914.  And then what we can do is we can take these two numbers now, so those two divided . . . the first divided differences to calculate our second divided difference between t2, t1, and t0, and that turns out to be simply this minus this, which is 30.914, that's the way we calculated it, so I'm just re-showing what we just calculated, divided by t2 minus t0, which is 20 minus 10,  and again, that number we already calculated, which is 0.3766. So what I'm trying to show you here is, graphically, that how you can write down what the data values are, calculate all the divided differences which you are going to calculate, then again calculate . . . keep on calculating the divided differences until you cannot calculate anymore, and then what happens is that this becomes your b0, this becomes your b1, and this becomes your b2. So if you were graphically trying to write down this . . . write down to find out what the coefficients are, that's how you would go about doing this, and this is also going to form a basis for how you would write your computer program also.  Now, keep in mind that somebody might say, hey, you're finding b0, b1, and b2, why do you have to find this?  The reason why you have to find this is because that's what's required to find b2.  So all of these divided differences have to be calculated.  So your b0 turns out to be 227.04, which is the top number here, and the top number here is b1, and the top number here, which is the only number there is turns out to be 0.3766. So it's very important to understand how this tree works for these forward divided differences, and then that you are going from the zeroth divided differences to the first divided differences to the second divided differences, and you continue to do this, and that's how you're going to find out the coefficients of the second-order polynomial.  We will illustrate this again when we do the general order polynomial interpolation by using Newton's divided difference, and, but I wanted to show you to here, because by showing you to here, it will be simpler for you to understand when we do the general order polynomial using Newton's divided difference polynomial method.  And that's the end of this segment.