Calculus: Taylor Polynomials

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Sequences and Series
Taylor and Maclaurin Polynomials
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Okay, so the name of the game is to try to figure out and understand the nuance of a complicated function by approximating it by simpler functions. And in this case, polynomial functions, because if you want to evaluate a polynomial, it just requires multiplication, addition and subtraction, and that's it. Whereas when you're looking at sines and e^xs and the natural log and all of those other complicated functions requires things that we just don't know how to do by hand. So the polynomial approximation is a great of doing that.
And what's the strategy? The strategy is, you want a polynomial let's say of degree K. So K degree polynomial to approximate a particular function at a particular point, what you want to do is make sure that not only does the point agree at the point C, the function and the parabola or the cubic or whatever, the polynomial agrees, but that all the derivatives agree up to K. If all the derivatives agree, then not only will the slope be okay, and the curvature okay, but even the change in the curvature will be okay. And in fact, you begin to hug better and better the curve at the point.
And this polynomial approximation technique is actually called Taylor Polynomials. So these polynomials that we get are called Taylor Polynomials. So Taylor Polynomial is just the polynomial approximation that we're talking about. And what would the actual formula be for this thing? Well the formula is exactly what we've been discovering here. So if you want to find the K^th Taylor Polynomial centered at C, so the K^th Taylor Polynomial centered about x = C, then what would it equal? Well I just have to make sure it's a polynomial of degree K so that when I evaluate it and all its derivatives up to the K^th derivative at C, they coincide with the derivative of the function f(x). And what I get is the polynomial y equals, I'm going to use the sigma sum notation first, it will equal the sum n going from zero up to K of the following. First I put in the derivative term to make sure the derivative works out okay. So I put in the nth derivative of F evaluated at C divided by n factorial times x minus C^n. Now what in the world does that equal? Well let me write that out for you in long-term. So I write n = 0, n = 1 and so forth. So when n = 0, I see F, the 0^th derivative, so just the function, evaluated at C divided by 0 factorial. By the way, 0 factorial, this is just a convention. So there was a big math convention. And the convention is that 0 factorial is defined to be the number 1, because it doesn't make sense. What does it mean to take 0 and multiply it by all the people that come before it until you get to 1? It doesn't make sense. So the convention is you see 0 factorial, it's just a silly way of saying number 1. Okay, so now on with that. We see that this is just F of C + F prime of C times (x - C) + F double prime, or the second derivative of C divided by 2, (x - C)^2. And you keep doing that until you get to the last term, which is going to be the K^th derivative of F evaluated at C divided by K factorial (x - C)^K. That's the polynomial of degree most K. It turns out that if you take derivatives, derivative after derivatives, what you'll see is every derivative will match up with the derivative of F evaluated at C. So the first derivatives agree. The second derivatives will agree, all the way up to the K^th derivative, they will all agree. So this a polynomial. In fact it's called the Taylor Polynomial, the K^th Taylor Polynomial, because it's a polynomial of degree at most K. We go up to K. And in fact, it estimates the curve really, really, really well.
And why the factorial down here? Well remember the factorial is here because if you don't take K derivatives, just think about it. I have to keep pulling down these exponents. First time I pull it down I get a K. I've got to cancel that. The next time I take a derivative, I have a K - 1. Got to cancel with that. The next time I take a derivative, K - 2, K - 3, all the way out to 3 x 2 x 1. So they all have to be cancelled in order for me just to have the K^th derivative evaluated at C by itself. So in that case, what I need is K(K - 1)(K - 2) all the way out to 3 x 2 x 1. So that's why there's a K factorial here. A great little exercise, and it would be sort of fun to try is just look at like the formula when K = 4, and just take four derivatives. And verify that when you take those derivatives, and plug in x = C, and you evaluate it at x = C, the derivatives will coincide perfectly with the derivative of the function f(x). So that's the point of that factorial and where that comes from.
Let's try an example and see this sort of in practice, and then finally actually begin to understand what these more exotic functions look like. So here is an example. Let's take a look at the natural log function. And what I want to do is I want to try to find an approximation for this, a polynomial approximation for this, the Taylor Polynomial around, let's find, in fact, the fourth Taylor Polynomial around centered at C = 1. Well to do this I've got to take a lot of derivatives. In fact I actually have to sort of find all the Taylor Polynomials up to four. So I've got to find the first one, the second one, so on and so forth. So the formula requires me to take a whole bunch of derivatives. So let's do that right now. So let's take a whole bunch of derivatives. So this is f(x). So the first thing I need is the 0^th derivative, which is just a function itself. The 0^th derivative is just a silly way of saying the function itself. Take no derivatives. Then the first derivative will be 1 over x. Then the second derivative is going to be what? Well this is x^-1, so the derivative is -1x^-2, which is -1 over x^2. Now what's the third derivative? The third derivative, well this is equal to -x^-2, so the derivative is going to be what? Well I bring down the -2, and it becomes a positive 2 over x^3. And then if I take the fourth derivative, well this is actually 2 times x^-3. The derivative will be -6x^-4, which is -6 over x^4. So there's the long laundry list of derivatives. And I'm going to evaluate them all at C, because remember the formula that we've derived requires the derivative to be evaluated at the point that we're centering this around. That's the point that we're going to be expanding. So that's the point that's sort of the center of attention. So we evaluate all those things at that number. So these should all be numbers here. The only variable, x, should appear over here. If I do that, I evaluate all these things, watch this one fell swoop. I evaluate all of them at once at x = 1. Here we go. This equals natural log of 1 is 0. And I plug in a 1 here. That's 1 over 1. That equals 1. Plug in a 1 here it equals -1. Plug in a 1 here it equals 2. Plug in 1 here it equals -6. That's what we get. Those are the derivative. All we have to do now is literally just insert them into this formula. We insert them in this formula and write out what the fourth degree of everything looks like. If I do that, what I would see is the following. How would you do this if you were me? I guess I'll start here, and I think I'm going to draw a line here so we don't get confused. We've got a lot of writing. And I think maybe a change in font color might be appropriate. So let's start here. And we say that y--this is going to be the fourth degree of Taylor Polynomial, it's going to be equal to--what's the first thing we do? We just evaluate the function at the point. And we get 0. So in fact this is going to be 0. So I put a 0 there, which I won't write. And then, plus the derivative, so the derivative is just 1. Put in 1 times (x - C), which is x - 1, to the first power divided by 1 factorial. All those things are 1, so I'm just going to keep it like that. And then, plus the next derivative. So that's going to be a -1 divided by 2 factorial. So that's going to be just divided by (2x - 1)^2. Then plus the third derivative, which is 2 divided by 3 factorial, which is 6 times x minus the point of the center point, (x - 1)^3. And then plus the fourth derivative, which is -6 divided by 4 factorial, which is 24. Now of course these numbers can be canceled. This is just one third and so forth. But I want to keep them all in so you can see it in all its glory to the fourth power. So this is the fourth degree Taylor Polynomial for the natural log function expanded around the center point C = 1. And in fact, for free, you actually get all the lower polynomials. This is the linear polynomial or the tangent line. This is the parabola. This is the cubic. And this is the quartic.
So in fact, this is a major approximation. Let's take a look at the pictures. So visually, so here's the linear approximation. And here's a great picture of this. So the black is the natural log function. And here's 1. And then the green represents the straight-line approximation. Notice it's a really good approximation for quite a while, but then if you wait long enough they diverge and the approximation is not perfect. Of course, after all, it is an approximation.
What about for the quadratic? Well if you stop at the quadratic, then it should have a little bend to it. It should have a little parabola look to it. And sure enough, you can see that the blue, you can't even really see it here, because it hugs so closely to the black, but then afterwards it's a little divergent, because of course the parabola wants to come and go down, whereas the natural log continues to go up. So there's a little bit of splitting. But it's certainly better than the green, which only agreed for a while. This agrees for a lot longer. It's a better approximation. The higher degree polynomial you look at, the better the approximation.
What about for the cube? So, we look at the third degree Taylor Polynomial. That should look a little bit better because it should allow us the ability to sort of head back up. You see remember the quadratic will come up and then go down. But the cubic now has the ability to sort of take a point of deflection and head up a little bit. So you can see that the red cubic actually agrees even longer than the quadratic. It goes out a little further. So it's a really good approximation.
And the quartic, which of course is really long, does even better still. So it's this purple one. I don't know if you can really see that or not, but the purple one is really, really close in, hugs that black curve beautifully, comes out, and of course there's error. But you'll notice here the error is a lot smaller than all the previous error. So in fact, we are getting closer and closer.
Now what's the point of all this? Well the point of all this is to actually try to understand this function in terms of plugging in and evaluating. We can now do that pretty easily. For example, suppose that what we wanted to do was estimate the natural log of 1.2. What would I do? Well what I would is, one way of doing it is get one of the Taylor Polynomials. In this case I'll use the quartic one since we worked it out. And say well that purple polynomial is really, really close to the black natural log function nearby 1. Especially 1.2 is really nearby. So it should be almost the same, almost, right? So in fact, instead of figuring this out, which I have no idea how to do by hand, the natural log of 1.2, why don't I just plug in 1.2 for x here. That's just a whole bunch of multiplications and additions. And I can actually do that. If you do that, what do you see? So if we approximate this, that's approximately equal to what I get when I plug in here. If you plug in 1.2 here, 1.2 - 1 is just 0.2. So in fact, this is just 0.2 - (0.2^2 over 2) + (2 x 0.2^3 over 6) - (6 x 0.2^4) all over 24. And you can actually multiply all this out by hand. 0.2 after all is just 2 over 10 or one fifth. And so you can actually evaluate all this literally by hand. And if you go on to divide, you would see that this equals something like 0.182266 and so forth.
So that's what we can do by hand. What's the actual retail value? Well if you use a computer, and by the way, how do they do it? Well I think that basically these things are similar to this, but somehow we trust the computer. The computer estimates this to be--I guess we consider the computer to be basically God-like. So this equals 0.18232155 and stuff. So look at the amazing precision we got just for the fourth degree polynomial. We can literally do this by hand. We get certainly three significant digits, and almost the fourth one. So now we can finally understand elaborate complicated looking functions by approximating by these K^th Taylor Polynomials. Neat. I'll see you the next lecture.