Calculus: Higher-Degree Approximations

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Sequences and Series
Polynomial Approximations of Elementary Functions
Higher Degree Approximations Page [1 of 3]
What I really want to try to do is understand the nuance of function. How do you do that if you're just given a function and you want to understand its nuance? One way of doing that is to approximate it - inch up to it - by looking at a polynomial that somehow approximates its nuance. But what does it mean to approximate the nuance of a function by polynomial? What that would mean is that at the point that you're looking at, a lot of derivatives agree - they have the same derivatives there. So if the zero^th derivatives agree - what's the zero^th derivative? That's just the function itself. So if the zero^th derivatives agree to the point; that means that they both pass through the same point. If the first derivatives agree, that means that the slope of the curve is the same as the slope of the approximating thing. If the second derivatives agree, that means that the curvature agrees, and so forth. So the idea is to have a lot of coinciding derivatives at a particular point.
As I was trying to think about how we could understand sin(x), which is this very well-known function but actually pretty complicated, right? Sin(4) is a hard thing to compute by hand, even though the curve is so periodic. The question is: Can we approximate this complicated object by simpler objects - by polynomials? The answer is yes, and the way to do it is exactly as I just advertised: Write down the polynomial and make sure all the derivatives agree.
Now, to sort of inspire what the pattern would look like, let's just think for a second about approximating y = f(x) at the point (x = c). So there's the point c I'm looking at. In fact, maybe just a little picture here just to make sure that we're in good shape. I have some exotic curves - very exotic curves, like this - and I'm looking at c. So what I want to do is look at the tangent line approximations to that. How do I find a tangent line approximation? Well, the formula for that - or you can just do it directly, which is probably the best way - is y equals the value of the function, which is f at c, plus the slope at prime at c, times (x - c). And you can actually check and see that in fact if you plug in c into this line - so c for x - so put a c right in here, look what happens. I see (c - c), that's zero. So that's zero right there, so that drops out, and so I'm just left with y = f(c). So that means that they agree at that point. In terms of their height, they agree.
What about in terms of the derivatives? If you take the derivative of this, what do you see? This is a constant, so it's zero. Here I see take the derivative, just f^1(c). So the derivative of this line is equal to the derivative of the curve at that point. So the derivatives agree, which means this is a tangent line.
One really, really peculiar way of looking at this - I'm must going to now write this in a very funny way - so here's something funny. I'm not going to do anything except write this thing over in a very funny way: y equals - and here instead of writing f(c), I'm going to pretend or act as though f(c) is a derivative. It's a zero^th derivative. It means don't take any derivatives; just take the function and no derivatives. So it's the zero^th derivative. So I'd write this as f - and I'll put little parentheses - zero, to mean take the zero^th derivative, which just means f(c). So the zero^th derivative is just the function. I admit, it's silly, but it's supposed to be funny. Then (x - c), I'm going to write that in here, but I've got to make sure that it doesn't really appear, so I'll raise that to the zero power because anything the zero is just one, so that's good. Then I add this term - f1(c)(x - c)^1. Actually, I could write the prime as just a one and put little parentheses around that so it looks like that. This is the exact same thing as this but in a funny way, and the reason why I sort of like this is because it inspires and naturally leads us to what the quadratic approximation would look like. That's where, in fact, not only will the function agree - the zero^th derivative degree at the point and the first derivative agree at the point - but now the curvature - the second derivative - will agree at the point. So what would that look like? Well, it turns out that if you want a quadratic approximation, it's going to look...okay, here it is. y = ....well, first of all I want to agree at the point, so I'll keep all the stuff we had before, then I want the slopes to agree, so I'd better put in this term, just to make sure the slopes agree. The first derivative evaluated at (c) times (x - c)^1. Now I want the second derivatives to agree, so a good guess would be the second derivative evaluated at (c) times (x - c), but I want it to be a quadratic, so I'll put at c^2, and I'll add that term.
Now, does that actually work? Well, it turns out it does not work. Sorry. Because what's wrong? I have to make sure that when I take two derivatives of this, I'm actually going to get the second derivative of this, evaluated as c, and I don't because look what happens. When you take the first derivative, this drops out. This just gives me f^1 at (c), plus - and this two now comes out front - and so I see a two times the second derivative times x. When I take a derivative again, I end up seeing 2f^'' at (c).
So I'm twice as large as I should be. I want this to be the second derivative equal to the second derivative of this; but when I take the second derivative of this, I see two times the second derivative because that two comes out in front. So I'm going to compensate for that, so I'd better put in a "divided by two" here. So that divided by two didn't appear here, although really maybe it did and it was invisible. Here, it would be divided by one. So technically when I bring down the one, it cancels with the one, but we don't see that kind of thing. But with the two, you'd see it. By carrying this two down when I take the derivative and bring this two out on front, it will be killed by this two down below. So when I take a second derivative of this, I get precisely the second derivative of f evaluated at c. So actually the second derivatives now match up. So this is now the second derivative approximation.
Let's see this now in action with the sine function, which was the one we were looking at before. So with the sine function, we're looking at y = sin(x). So here's the curve. What was the straight-line approximation at ? So I'm looking at right here. We saw that was y = 1. So, in fact, the straight-line approximation is not that exciting, it's just the horizontal line at height 1. So not too thrilling, I'd say, but it approximates the curve pretty well right around there, as we saw earlier. It approximates the curve pretty well right around there.
But now let's try to do better. So I wanted a quadratic approximation, so how would I get the quadratic approximation? What I would do is I would use this quadratic fact that we have, which is that if you want a quadratic approximation that would be what? Well, y = f^zeroth evaluated at times (x - )^0. I'm writing this in a very funny way. All I had to really do was write f(). But let me write it this funny way so we can try to detect a pattern. Now I'll put the first derivative, evaluated times (x - )^1 - that's to make sure that the tangent - the slopes - agree. And now to make sure that the second derivatives agree, I take the second derivative, evaluate it at . I have to divide that by two because I have an ^(x - ^^^^^^) all squared, and that squared, when I take the second derivative, that's going to come out in front, and I'm going to kill it away with this.
Now what happens when you plug in here? Well, that requires us to take a couple of derivatives. So what's this piece right here? This piece right here is just f(). What's sin()? Well, sin() is one, so this is just a one. Then this is just (x - )^0, so that's just a one, too, so that's just one. Plus - and now f^1 - so the derivative of sine is cosine. What's cos()? Cos() is zero. So, in fact, that whole term just drops out. That drops out. Now I have a second derivative. So what's the derivative of the derivative? Or what's the derivative of now cosine? The derivative of cosine is negative sine. And what's negative sine evaluated at ? That would be negative one. So what I see here is a negative one, then divided by this two, times (x - ), all squared. So, in fact, that should be the approximation - the quadratic approximation - to the sine at .
Let's just think about that for a second. If you untangle this and square it out, with this negative (sign? sine?) there we see this is going to be a sad-faced parabola. And is that consistent with our picture? Of course, the parabola should be sad faced. It should be pointing down like this. Now what's the graph of this going to look like? Well the graph of this, in fact, looks like this. And you can see, it really does sort of look like the sine curve there. And, in fact, if you graph them both together, here's where we are so far. So the straight-line approximation approximated only very, very close; but now what we see is that the parabola - the quadratic approximation given by this quadratic - approximates really well, and for actually quite a while. Look at how it sort of hugs. It hugs that sine curve for a long time. Now, of course, if you wait long enough, of course they diverge and, in fact, there is a lot of error. But very, very close for a while so you can't even tell them apart. They're almost indistinguishable. So it's a fantastic approximation at this point, and we got that by making sure all the derivatives agree.
Well, now there is an idea here, which we can exploit. What if we wanted an even better approximation? Well, then I want more derivatives to agree. So instead of a quadratic polynomial, why don't we think about a cubic polynomial? If you think about a cubic polynomial, we can think about a fourth-degree polynomial and so forth, and what would these things look like? Well, this inspires a cubic approximation, and the cubic approximation would look like this: y = f^0 at a point (c), (x - c)0 - so that's to make sure their function dot is at the right spot. I want to make sure their slopes are at the right spot, so I put the first derivative evaluated at (c) - (x - c)^1 - and then plus now I want the second derivative to coincide, so I have the second derivative evaluated at (c), divided by two, times (x - c)^2, but now I want the third derivatives to coincide, so I'll need a third derivative. This is going to be a cubic, so (x - c)^3, but what do I have to divide by? Well, you might say, "Gee, I should divide by three." But let's think about it. If we're going to take three derivatives, what kind of coefficients are we going to get? When I take the first derivative, it will bring a three out in front, so I certainly have to cancel away that three. But then I have a two here, and when I take the second derivative, I get another two out in front, so I'm going to actually cancel away the two as well. So, in fact, I've got to actually divide by two times three, or six.
We're beginning to see a pattern. It seems as though the thing I have to divide by is just all of the numbers starting with three and then decreasing by one until I get to one, and take their products, because if I take the first derivative, I get a three, if I take another derivative, I get another two, if I take a derivative again, I get the one in front. It looks like three times two times one, so it's six, which is (3 x 2 x 1), which is, by the way, is three factorial. Now we're heading toward the pattern. You can imagine what the fourth one would look like. If I wanted a fourth one, I would put in the fourth derivative, evaluate at that number, (x - c)^4, and then what would the denominator be? It would be...well, I've got to bring that four for the first derivative, that's a four, then a three, then a two, then a one. That's four factorial, which is 24. So I should divide by 24 in order to cancel out all those coefficients so that the fourth derivative of this approximate polynomial will actually coincide with the fourth derivative of the actual function at hand.
Just to show you visually what this looks like if you do this, if you actually take all of the derivatives and plug in, it's not at all hard with the sine functions to actually do this, and you'd see that the fourth-degree approximation turns out to be this. And what's the graph of that look like? Well, it's this beautiful red curve as the fourth-degree approximation. Notice that the first-degree approximation was only good for a short period of time. The second-degree approximation actually was a good approximation for a longer time. The third-degree approximation actually equals this one again. And the fourth degree turns out to be this red on, and notice that now not even am I really hugging the sine curve close by, but now it's picking up the nuance as you go further out. Whereas the parabola just went down to negative infinity and you never came back, this person now - the cordex - now comes down and starts to capture the going-up part. So it's even agreeing further than the quadratic. And if you want better approximation, you would just look at a higher degree polynomial.
So this idea of polynomial approximations where all of the derivatives individually agree with all of the derivatives of the function at hand provide the really effective approximation for a complicated function such as sine.
All right, we'll take a look at more of these ideas up at the next lecture. I'll see you.