Calculus: Maclaurin Polynomials

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Sequences and Series
Taylor and Maclaurin Polynomials
Maclaurin Polynomials Page [1 of 2]
So the Taylor Polynomial is a great way to approximate a complicated function by just producing a polynomial for which all the derivatives up to the K^th derivative will agree with the original function. And the formula for it actually looks a little bit complicated, but really is not that big of a deal. If you remember that I just want to, as long as I want to be centered around a particular point C, and I want to have a K^th degree polynomial, all I've got to do is write these terms out, where the coefficient makes sure that the derivatives will agree after I take the appropriate number of derivatives. And that these denominators are placed that as I take derivatives and I start to accumulate some factors on top, they will be cancelled away by these placed factors on the bottom.
So this is the K^th Taylor Polynomial of a function f(x) centered around a point x = C, where C can be any number. An easy example of this, by the way, would be the example when C = 0. So what happens when C = 0? Well when C = 0, it sort of makes the formula a little bit easier, because first of all C = 0, I guess I could put a 0 in here. Make this C a 0. So I'm going to make this a 0. So that becomes a 0. So actually it just goes away, right. And I can just erase it, in fact, let's just erase it really fast. So there it goes away. So it's not there. So it's just x^n[. ]Here this C could become a 0. Here I subtract C. It just doesn't do anything, because C is 0. So I can just get rid of all of that. So it's just that. How easy. We can do bookkeeping right here. This would be a 0 here. So I evaluate all the derivatives at 0, and then I just have x^n. And I sum all those up. So that's all that's required to look at the very special case. The K^th Taylor Polynomial when I'm centering around the point 0, it looks like that. Pretty easy, just the n^th derivative evaluated at 0 divided by n factorial times x^n.
Now in fact, this has a name believe it or not. This special case of the Taylor Polynomial is actually called the Maclaurin Polynomial, so the Maclaurin approximation. So basically the Maclaurin Polynomial is just the Taylor Polynomial, but we're evaluating it at the point C = 0, so not a big deal at all. So good old Maclaurin got his name in there, sneaked it in. Let's look at an example. Let's actually find the fourth Maclaurin Polynomial, so therefore centered around 0. If I just say Maclaurin Polynomial, that means the center point is 0. That's what is the distinguishing feature about the Maclaurin Polynomial. And let's find the fourth degree polynomial for f(x) equals e^x, the exponential function, another exotic function that's difficult to understand without a calculator. So let's give this a whirl.
What do I want to do? Well I need to make a whole bunch of derivatives. In fact, I need to take a total of four derivatives. So let's just keep a little chart here. So the 0^th derivative just means do nothing. So that's e^x. The first derivative means take one derivative. That's e^x. The second derivative is the derivative of the derivative. That's just e^x. The third derivative is the derivative of the derivative of the derivative, which is e^x. And the fourth derivative equals e^x. The exponential function is a fantastic function, because you can differentiate. In fact, without even doing any work, right now, I want you to make a guess. What's the 37^th derivative? I heard you. So it's really, really easy. Now what do I want to do? I want to evaluate this. So I'm trying to find the Maclaurin Polynomial. That just means that all I've got to do is find the Taylor Polynomial around the point C = 0. So I'm going to evaluate all these things at 0. So I'm going to evaluate all the derivatives in one fell swoop at x = 0. And what happens when you plug in 0 everywhere? You get even 0 which is 1. So everybody equals 1. Everyone equals 1. So in fact, all the derivatives evaluated at 0 give 1. Very easy, very easy to find those coefficients.
So what do I do? Well the formula tells me that all I've got to do now, all of these things we're seeing are just 1. So the formula is just going to be these things, but don't forget the factorials on the bottom. So we put all of this together, x^n, what I see is--what we need to use a green font for this. I see that y will equal--well first of all just 1 times, well what is the value at 1. It's 1, so just 1. And then plus 1 divided by 1 factorial times x. So that's x + 1 times x^2 divided by 2 factorial. So that's going to be 1 over 2 factorial times x^2 + 1 divided by 3 factorial. See how those factorials are coming in there x^3 + 1 divided by 4 factorial x^4. In fact you can now find a fifth Maclaurin Polynomial and a sixth and so forth. I think you see the pattern, 1 over 5 factorial times x^5 and so forth, because all the derivatives are so easy. They'll all just work out to be 1. When you center around C = 0. Okay, so there's the answer.
Great, and what can you do with that? Well what you can do with that--oh there's a little typo here by the way. But no problem, we'll make it up right on the fly. This is a plus sign here. See having many equal signs, it makes you see in stereo. You know if you can sort of see those things. And it's like you see it here and you see it there. No, no, no, but if this was a plus sign, and there's no stereo, it's just mono. There it is.
Now what can you do with that? Well since this polynomial is supposed to approximate this function really well near the point C = 0, that means that if you have a question about the exponential function around the origin, you can actually convert it to an approximate question around this polynomial. Why bother? Well what is, for example, e^.1? How would you figure that out without a calculator or computer? Answer, I don't know. We can't do it. We can't wrap our minds around this exponential function. However, we can approximate it by this fourth degree Maclaurin Polynomial. And now it's a piece of cake. If I plug in 0.1 here, we can literally do this by hand. Multiplication and addition and that's it. You can figure out what this is. If you plug it in here, 0.1, what you would see is this would be approximately equal to . And you can actually do this. In fact, this won't even take you that long to do it literally by hand. And I did this. And if you do it what you get is 1.105170 and so forth. And what's the actual retail value? Let's see what the computer claims. The computer claims this is 1.1051709 and so forth. In fact, if you want some more accuracy here, I'll even tell you this is 8 something.
So look at the amazing precision we're getting, the absolutely incredible precision. One, two, three, four, five, six, six places to the right of the decimal point we're getting accuracy with just this simple calculation, this simple approximation. So now we're really getting a sense of what's going on here. And by the way, just in terms of a picture, to see what this looks like if you really were to plot it, let me show this to you. Here's the exponential function. And if you want to now take a look at the graph of this, what you would see is it's a quartic. If you line it up, you can see that near x = 0, it's almost a perfect fit. And actually to the right it hugs pretty well. And not surprising, though, if you go off too far to the left, we do see a rise, because after all, this is a quartic polynomial. So it should head up north as it goes off to the left horizon. And we're seeing that here. So there's a little bit of error here. But nearby 0, we get amazing approximation. And in fact, you can see that numerically here.
So the Maclaurin Polynomial, nothing more than the Taylor Polynomial evaluated at C = 0, but again we see evidence that when you have an exotic function that we can't understand, we can make some sense of it in a real way by looking at these approximate polynomials. Enjoy these approximations and I'll see you at the next lecture.