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MIT 18 01 - Lecture Notes

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MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus, Fall 2007 Please use the following citation format: David Jerison, 18.01 Single Variable Calculus, Fall 2007. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/termsMIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus, Fall 2007 Transcript – Lecture 25 The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation, or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Now, today I need to get started by finishing up what I did last time. Namely, talking about numerical methods. And I want to just carry out one example. And then I want to fill in one loose end. And then we'll talk about the unit overall. We were talking, last time, about numerical integration. I'm going to illustrate this just with the simplest example that I can. We're going to look at the integral from 1 to 2 of dx / x. Which we know perfectly well already is the log of x evaluated between 1 and 2, which is ln 2 - ln 1. Which is just ln 2. Now, if you punch that into your calculator, you're going to get something like this. I hope I saved it here. Yeah. It's about 0.693147. That's more digits than we're going to get in our discussion here. Anyway, that's about how big this number is. And the numerical integration methods will give you about as much accuracy as you can get on the function itself. And, of course, some functions we may have more trouble approximating. But the function 1 / x, we know pretty well how to do, because we know how to divide. So since the function that we're integrating here is 1 / x, it's going to be not too difficult to get some arithmetic. Nevertheless, I'm going to do this in the simplest possible case. Namely, just with two intervals. Now, you really can't expect things to work so well with two intervals. That's a pretty ridiculous approximation to your function. When you have two intervals, that means you're looking at the graph of this hyperbola. And you have 1 here, and you have 2 here and you have 3/2. And you're really only keeping track of the values at these three spots. So the idea that you can approximate the area just by knowing the values of three places is a little bit of a stretch of the imagination. But we're going to try it anyway. Now, the trapezoidal rule is the following formula. It's delta x ( 1/2 the first value + the second value + 1/2 the third value). In this case, the pattern is 1/2, 1, 1, 1, 1, 1, 1/2. And in this case, delta x = 1/2 because this interval's of length 1. The b - a, right. Let's just point that out here. Here, b = 2. a = 1. b - a = 1. And the number n = 2. And so, delta x, which is b - a / n = 1/2. So here's what we get. And let's just see what this number is. It's 1/2 of the value at here. Well, so let's just check what these values are. This value is 1, this value over here is 2/3, and the last value is 1/2. Because the function, of course, was y = 1 / x. And those were the three values that we have. So y0, this one is y0, this one is y1, and this one is y2. Now, here we have (1/2* 1 + 2/3 + 1/2 * 1/2). Now, on an exam, I don't expect you to add up long messes of numbers like this. When you have two numbers, I expect you to add them up if they're reasonable, or subtract them. Just as we do when we take antiderivatives. Like, for example, I don't want you to leave the answer to an integration like this in this form. I want you to simplify it at least down to here. And I of course don't expect you to know the numericalapproximation. But I certainly expect you to be able to do that. On the other hand, when the arithmetic gets a little bit long, you can relax a little bit. But I did carry this out on my calculator. Unless I'm mistaken, it's about 0.96. It's pretty far off. So remember what it was. It's what you get when you get these straight lines. And there are these little extra pieces of junk there. Now, don't trust that too much, but the point is that it's far off. So now, let's take a look at Simpson's Rule. And I claim that Simpson's Rule is surprisingly accurate. In this case, really, even a little more than it deserves to be. The formula is (delta x / 3) ( y0 + 4 y1 + y2). So the pattern is 1, 4, 1, or 1, 4 and then it alternates 2's and 4's until 4, 1 at the very end. And if I just plug in the numbers now, what I get is 1/6, because delta x = 1/2 again. And the value for y0 = 1. And the value for y1 = 2/3. And the value for y2 = 1/2. So here's the estimate in this case. And this one I did carry out carefully. And it came out to 0.69444. Which is actually pretty impressive, if you think about it. Given what the logarithm is. Now, what's going on with Simpson's Rule in general is this. If you -- Simpson's minus the exact answer. In absolute value, is approximately of the size of (delta x)^ 4. That's really the way it behaves. Which means that if delta x is about 1/10, so if we had divided this up into 10, intervals which we didn't, but if we'd divided it up into 10 intervals, then you could expect that delta x, the error would be about 10 ^ -4. In other words, four digits of accuracy here for this thing. But the exact analysis of this, a more careful analysis of this, is in your textbook. And I'm not going to do. But I just want to point out that it is an effective method. It really does give you nice four-digit with manageable, you could even really do it by hand. It's so convenient. The Simpson's Rule. Whereas the other rules aren't really that impressive as far as giving fairly accurate answers. The last little remark to make is that the reason is that Simpson's Rule is matching a parabola. And somehow the parabola follows this curve better. It's giving the exact answer. So I'll mention this. Simpson's Rule is derived using the exact answer for all degree 2 polynomials. In other words, parabolas. All parabolas. But even all the ones of lower degree. So straight lines …


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MIT 18 01 - Lecture Notes

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