CHAPTER 06.04: NONLINEAR REGRESSION: Polynomial Model: Example: Part 2 of 2
So let's go ahead and review what kind of right-hand side summations do we need. So if you go back here, you will be able to see that what kind of right-hand sides you need. You need to add all the alpha values, you need to add all the alpha times T values, and the alpha times Ti squared values. Those are the summations which you need for the right-hand side. So I'm going to show you a couple of those, and the third one which you can do as homework. So what is the summation of alpha-i values, which is simply adding all the alpha values which are given to you, which are 6.47, plus 6.24, plus 5.72, plus 5.09, plus 4.30, plus 3.33, and this value here turns out to be 31.15. So similarly you can calculate summation of alpha-i Ti, for example, it will turn out to be -1.865 times 10 to the power 3, that's the summation of the, when you multiply each alpha value by its corresponding temperature value, and add them up. And then summation of alpha-i Ti squared, so let's go through this process here. So alpha-i is 6.47, and the corresponding temperature is 80, so that's, you'll have 80 squared, the next one will be 6.24 times 40 squared, and then you will add all of them, and the last one will be simply 3.33 times -280 squared. So there are several other expressions right here which you have, which you are supposed to fill in, and you get 5.669 times 10 to the power 5. I would strongly advise you to calculate all these summations yourself, just to . . . the ones which I skipped and the ones which I did, just to convince yourself that all of these numbers are right, and that how to be able to do that. So since I have calculated all the numbers, all I have to do is now fill in my coefficient matrix and fill in my right-hand side so that I can set up my three equations and three unknowns. So I'll get something like this. So if you look at this, I'll get 5 . . . 6 here, because I have six data points, summation of Ti is -520, and this one is minus . . . no, 1.424 times 10 to the power 5, then I have -520 here, then 1.424 times 10 to the power 5, then -3.1 . . . -3.117 times 10 to the power 7, and then I have 1.424 times 10 to the power 5 here, then -3.117 times 10 to the power 7 here, and then 8 times 10 to the power 9 here, so those are the numbers which you are getting here. One of the things which I would like to point out here is that you can see that the difference of the order of the numbers, you go from 10 to the power 0 order here to 10 to the power 9 numbers, and imagine that if you had a higher order, higher polynomial order for the polynomial regression, you will get still bigger numbers in your coefficient matrix, and that's why many times you find out that you are going to get some kind of ill conditioning in the coefficient matrix once . . . when you are doing polynomial regression. The right-hand sides are 31.15, -1.865 times 10 to the power 3, 5.669 times 10 to the power 5, so those are the values which you obtain for the right-hand side. And now, all I have to do is to solve these three equations, three unknowns, which I can do my using any of the methods which I have learned in my simultaneous linear equations, such as Gauss elimination, and this is what I get for the value of a0, a1, and a2, so a0 turns out to be 6.013, a1 turns out to be 6.424 times 10 to the power-3, and a2 turns out to be -1.13 . . . 113 . . . -1.113 times 10 to the power -5. So what that means is that the relationship, the polynomial, second-order polynomial relationship between alpha, which is the thermal expansion coefficient, and temperature, is 6.013, plus 6.424 10 to the power -3 times the temperature, minus 1.113 times 10 to the power -5 times temperature squared. So that's the second-order polynomial regression model which we have just found out. And we were interested in finding out what the thermal expansion coefficient, coefficient of thermal expansion is at temperature, T equal to 70. So I'm just going to put 70 in there, I get 6.013, 6.424 times 10 to the power -3 times 70, minus 1.113 times 10 to the power -5 times 70 squared and this value here turns out to be 6.408, and the units are the same as which we used for alpha, which are micro- inch per inch per degree Fahrenheit. And that's the end of this segment. |