CHAPTER 05.03: NEWTON DIVIDED DIFFERENCE METHOD: General Order: Newton's Divided Difference Polynomial: Example: Part 2 of 2
So I'm going to calculate the first divided difference based on these two points, then the first divided difference between these two points, and the first divided difference between these two points. And the first divided difference here will be simply 362.78 minus 227.04, so this minus this, divided by this minus this, which is 15 minus 10. Same thing here, the first divided difference will be this minus this, which will be 517.35 minus 362.78, divided by 20 minus 15. And again, the first divided difference would be this minus this, divided by this minus this, which is 602.97 minus 517.35, divided by 22.5 minus 20. So this number here turns out to be 27.148, and this number here turns out to be 30.914, and this number here turns out to be 34.248. Now my whole idea of keeping everything on one board has fallen apart here, but that's okay. Now, what this number is, v of t0, okay? So that's v of t0, and that's my same thing as b0. Now this one here, this number here which I have here, is the first divided difference between t1 and t0, and that is nothing but my b1. So I have been basically able to calculate the two constants which I need for the third-order polynomial. Now I need to calculate the other two divided differences, so I'm going to go to the next board here to show it. So what I'm going to do is I'm going to put 27.148 here, so these are the first divided differences, and then I have 30.914 here, and 34.248, so these are the three divided differences, first divided differences which I just calculated, so I'm going to put those. Now I'm going to calculate the second difference based on these two numbers. So based on these two numbers, what is the second divided difference? It is this minus this, 30.914 minus 27.148, divided by t2 minus t0, which means that it is 20 minus 10, and same thing here, it'll be this minus this, which is 34.248 minus 30.914, divided by t3 minus t1, which is 22.5 minus 15. And those numbers now, this one turns out to be 0.37660, and this number here turns out to be 0.44453. So what I have just calculated is the second divided difference, so this is my second divided difference, which I have calculated. And this one here is nothing but my second divided difference at the points of t2, t1, and t0, which is nothing but my b2. So that's the third constant which I need. Now what I'm going to do is I'm going to take the second divided difference, and use those to calculate what my third divided difference is. So I'm going to write down here 0.37660, so these are the third differences which I have, and then I have 0.44453, which I just calculated, so that's this number and this number, I'm writing them down here. I'm going to calculate the . . . sorry, this is the second divided difference . . . this is the second divided difference, and now I want to calculate the third divided difference right here. I want to calculate my third divided difference . . . difference here, and what that will turn out to be is this minus this, which is 0.44453 minus 0.37660 in the numerator, and then in the denominator I will have t3 minus t0, which is 22.5 minus 20, that's what I will have there, and this number here turns out to be 5.4347 times 10 to the power -3. So that's what I get for the third divided difference, and this third divided difference is nothing but the divided difference calculated at the data points of t3, t2, t1, and t0, and that will be nothing but b3. So what I have basically done is that I have finished calculating the values of b0, b1, b2, and b3 for a third-order polynomial, so that gives me the third-order polynomial now. So I'm going to write it down here, that the third-order polynomial which I get based on the Newton's divided difference polynomial is of the form b0, plus b1 times t minus t0, plus b2 times t minus t0, times t minus t1, plus b3, t minus t0, times t minus t1, t minus t2. And I'm going to substitute the values of b0, b1, b2, and b3, which I just calculated, and I'm also going to substitute the values of the t0, t1, and t2 here. So b0 which I obtained turns out to be 227.04, that's what I calculated as b0, and then b1 I calculated as 27.148, times t minus 10, t0 is 10, so that's why I'm substituting 10 in there, plus b2, which I calculated just now, which is 0.37660, times t minus t0, which is 10, times t minus t1, which is 15, plus b3, which I just calculated as 5.4347 times 10 to the power -3, that's t minus 10, t minus 15, and t2, which is 20, t minus 20. So that is the third-order polynomial calculated by the Newton's divided difference polynomial, which is going to be valid, so again, the domain in which this particular interpolant is valid also has to be written, it's part of the interpolant, that this interpolant is valid between 10 and 20, because that's the minimum value of t and the maximum value of t which we have . . . not 20, 22.5 . . . 22.5, because if you look at the four data points, the minimum value is 10, and the maximum value is 22.5, so it's going to be valid between those two data points. We are interested in calculating the value of the velocity at 16, so all I'm going to do is to substitute 16 in here, plus 27.148 times 16 minus 15, plus 0.37660 times 16 minus 10, times 16 minus 15, plus 5.4347 times 10 to the power -3 times 16 minus 10, 16 minus 15, times 16 minus 20. So just substituting the value of t equal to 16 into the formula, and the number which I get here is 392.06 meters per second. So that's how you are able to use the Newton's third- order divided difference polynomial to be able to calculate the value of the velocity at a certain point. And that is the end of this segment. |