CHAPTER 07.04: ROMBERG INTEGRATION: Theory: Part 2 of 2
How do we get this 15 here? Let's go ahead and look at it. So basically what you have is that you're going to get true value is approximately equal to Jn, plus c divided by n squared, let's suppose, and the true value will be approximately equal to J2n, plus c divided by now 4 n, squared here, that's what you're going to get, because now what's happening is that as you are doubling the number of segments, that the true error which you have is getting approximately proportional to c h to the power 4, which means the same as c by n squared or c by 2 n, squared, because as you double the number of segments, what's going to happen is that the true error is going to get, not quartered, but one-sixteenth. And if you solve these two equations and two unknowns, this is what you're going to get, you're going to get that the true value is approximately proportional to Jn . . . J2n, plus J2n minus Jn, divided by 15, because the reason why you get 15 is because you have 4, 4 squared is 16, and then there's a 1 here, so 16 minus 1 is 15, and that's how you get the 15 here. And in fact, this 15 can be looked at as 2 to the power 4, minus 1, you can look at it as 2 to the power 4, minus 1, just like when you had true value is approximately proportional to I2n, plus I2n minus In, divided by 3, that can be looked at as 2 to the power 2, minus 1. So you can very well see that how this whole process is working, that once you go from one approximation to another, what the denominator should be, so once you are . . . once you are at the first step, which is the Richardson's extrapolation formula, you're getting 2 to the power 2, minus 1, and then you are getting 2 to the power 4, minus 1, and the next one will be 2 to the power 6, minus 1, and so on and so forth. So that's how this whole approximation stuff is going to work. So if somebody was interested in figuring out that, hey, can you tell me . . . let me do this . . . I think this one should be 4 raised to the power . . . yeah, this should be 4 raised to the power 2 minus . . . 4 raised to the power 2 minus 1, minus 1, and I can look at this as 4 raised to the power 3 minus 1, minus 1. So that's a better way of putting it, although what I put before is not wrong, but the reason why I am putting it like this is because then you have this 2, and then you have this 3, and then 4, like that, because 4 raised to the power 2 minus 1 is 4, 4 minus 1 is 3, then 4 raised to the power 3 minus 1 is 16, minus 1 is 15, and the reason why I rewrote it like this is because you have 4 raised to the power 2, then 4 raised to the power 3 here, and that's how the general formula is going to looks like So what that means is that the Romberg integration is simply going to look like this, it's going to look like this, I, k, comma, j, so you need to follow this a little bit more carefully than other things which we have talked about, k minus 1, j plus 1, plus the I calculated at k minus 1, comma, j plus 1, minus I, k minus 1, comma, j, divided by 4 raised to the power k minus 1 . . . 4 raised to the power k minus 1, minus 1, where k is greater than equal to 2. Now what are these k and j? k basically represents the order of extrapolation. k represents the order of extrapolation, and what we mean by that, let's suppose Richardson's extrapolation formula, the order of extrapolation there is simply 1. So . . . so that's how we're going to . . . so k is equal to 1 corresponds to . . . let me write it down, k equal to 1 corresponds to the trapezoidal rule, so k equal to 1 corresponds to the trapezoidal rule. k is equal to 2 corresponds to the Richardson's . . . Richardson's extrapolation. k is equal to 2 corresponds to the Richardson's extrapolation formula, and so on and so forth, so that is how you are able to figure out what this k represents. Now what does this j represent? j represents, for example, if somebody says j, that represents the less accurate estimate, the less accurate estimate, and j plus 1, which can be at any point, is the next accurate estimate. So what we mean by that is that, when you have, let's suppose, k, comma, 1, that means that that might be the estimate which you obtained by using . . . by using, what's it called? The first . . . the first, with n and 2 n, so let's suppose we look at here, this is the one which we need to look at. You have j and j plus 1, so j might mean that, hey, you are looking at with using n segments, and j plus 1 will mean that what you got by using 2 n segments. So this could be at any point, so this could be j could be 2 n, then in that case, j plus 1 will be 4 n, so that's why we don't talk about what j corresponds to in terms of n, j corresponds to whatever is the less accurate estimate which you are using, and j plus 1 will be the next accurate estimate which you're using, with double the number of segments which you're using here, so that's how this formula is going to work. What I'm going to do is, when I'm going to look at an example, I'm not going to use this particular formula, because it'll be quite complicated to . . . I shouldn't say complicated, but it might be a little bit work to follow the indices as such, so I'm going to write down the indices in the tree, and then I will do the example in the tree itself, so that would be the way you would solve the problem by hand, but if you were going to write an algorithm for this particular program, then you would have to use this particular formula so that you follow the algorithm efficiently and effectively. And that is the end of this segment.