Finding Taylor polynomial approximations of functions
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Taylor & Maclaurin polynomials intro (part 2)
In the last several videos, we learned how we can approximate an arbitrary function, but a function that is differentiable and twice and thrice differentiable and all of the rest. How we can approximate a function around x is equal to 0 using a polynomial. If we just have a zero-degree polynomial, which is just a constant, you can approximate it with a horizontal line that just goes through that point. Not a great approximation. If you have a first-degree polynomial, you can at least get the slope right at that point. If you get to a second-degree polynomial, you can get something that hugs the function a little bit longer. If you go to a third-degree polynomial, maybe something that hugs the function even a little bit longer than that. But all of that was focused on approximating the function around x is equal to 0. And that's why we call it the Maclaurin series or the Taylor series at x is equal to 0. What I want to do now is expand it a little bit, generalize it a little bit, and focus on the Taylor expansion at x equals anything. So let's say we want to approximate this function when x-- so this is our x-axis-- when x is equal to c. So we can do the exact same thing. We could say, look, our first approximation is that our polynomial at c should be equal to-- or actually, even let me better it-- our polynomial could just be-- if it's just going to be a constant, it should at least equal to function whatever the function equals at c. So it should just equal f of c. f of c is a constant. It's that value right over there. We're assuming that c is given. And then you would have-- this would just be a horizontal line that goes through f of c. That's p of x is equal to f of c. Not a great approximation, but then we could try to go for having this constraint matched, plus having the derivative matched. So what this constraint gave us-- just as a reminder-- this gave us the fact that at least p of c, the approximation at c, our polynomial at c, at least is going to be equal to f of c, right? If you put c over here, it doesn't change what's on the right-hand side, because this is just going to be a constant. Now, let's get the constraint one more step. What if we want a situation where this is true, and we want the derivative of our polynomial to be the same thing as the derivative of our function when either of them are at c. So for this situation, what if we set up our polynomial-- and you'll see a complete parallel to what we did in earlier videos. We're just going to shift it a little bit for the fact that we're not at 0. So now, let's define p of x to be equal to f of c plus f prime of c. So whatever the slope is at this point of the function, whatever the function slope is, times-- and you're going to see something slightly different over here-- x minus c. Now, let's think about what this minus c is doing. So let's test, first of all, that we didn't mess up our previous constraint. So let's evaluate this at c. So now, we know that p of c-- and I'm using this exact example-- so p of c-- let me do this in a new color. Let me try it out. So p-- that's not a new color. p of c is going to be equal to f of c plus f prime of c times c minus c. Wherever you see an x, you put a c in there. c minus c. Well, this term right over here is going to be 0. And so this whole term right over here is going to be 0. And so you're just left with p of c is equal to f of c. You're just left with that constraint right over there. And the only reason why we were able to blank out this second term right over here is because we had f prime of c times x minus c. The x minus c makes all of the terms after this irrelevant. We can go now verify that this is now true. So let's try-- p prime of x is going to be the derivative of this, which is just 0, because this is going to be a constant, plus the derivative of this right over here. And what's that going to be? Well, that's going to be-- you can expand this out to be f prime of c times x minus f prime of c times c, which would just be constant. So if you take the derivative of this thing right here, you're just going to be left with an f prime of c. So the derivative of our polynomial is now constant. So obviously, if you were to evaluate this at c, p prime at c, you're going to get f prime of c. So once again, it meets the second constraint. And now when you have both of these terms, maybe our approximation will look something like this. It will at least have the right slope as f of x. Our approximation is getting a little bit better. And if we keep doing this-- and we're using the exact same logic that we used when we did it around 0, when we did the Maclaurin expansion-- you get the general Taylor expansion for the approximation of f of x around c to be the polynomial. So the polynomial p of x is going to be equal to-- and I'll just expand it out. And this is very similar to what we saw before. f of c plus f prime of c times x minus c. You might even guess what the next few terms are going to be. It's the exact same logic. Watch the videos on Maclaurin series where I go for a few more terms into it. It becomes a little bit more complicated taking the second and third derivatives, and all of the rest just because you have to expand out these binomials, but it's the exact same logic. So then you have plus your second-degree term, f prime prime of c, divided by 2 factorial. And this is just like what we saw in the Maclaurin expansion. And just to be clear, you could say that there's a 1 factorial down here. I didn't take the trouble to write it because it doesn't change the value. And then that times x minus c squared plus the third derivative of the function evaluated at c over 3 factorial times x minus c to the third power. And I think you get the general idea. You can keep adding more and more and more terms like this. Unfortunately, it makes it a little bit harder, especially if you're willing to do the work. It's not so bad, but instead of having just x here and instead of just having an x squared here, and having an (x-c) squared having an (x-c) to the third, this makes the analytical math a little bit hairier, a little bit more difficult. But this will approximate your function better as you add more and more terms or on an arbitrary value as opposed to just around x is equal to 0. And I'll show you that using WolframAlpha in the next video.
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