Worked example: recognizing function from Taylor series | Series | AP Calculus BC | Khan Academy

So we're given this expression: Is the Taylor series about zero for which of the following functions? They give us some choices here, so let's just think a little bit about this series that they gave us.

If we were to expand it out, let's see. When n is equal to 0, it would be ((-1)^0) power, which is 1, times (X^0), which is 1, over 0 factorial, which is 1. So it'll be 1 plus...

When n is 1, well then this is going to be negative. So it's going to be minus, and then (x^1) over 1 factorial. Well, I could just write that as (X) right over here.

Then when n is 2, the negative ((-1)^2) is going to be positive, so plus (x^2) over 2. Then it's going to be minus (x^3) over 3 factorial, and then it’s going to be plus, and I can keep going.

You've seen this before: (X^4) over 4 factorial. It's going to keep going, minus plus; it's going to keep alternating on and on and on.

Now, our general form for a Taylor series about zero, which we could also call a Maclaurin series, would be, our general form would be (F(0)) plus (F'(0) \cdot X) plus (F''(0) \cdot \frac{X^2}{2}) plus the third derivative at 0 (\cdot \frac{X^3}{3!}) plus the fourth derivative, you get the idea, evaluated at 0 (\cdot \frac{X^4}{4!}), and we would just go on and on and on.

Now to figure out which function in order for this, in order what I wrote in blue, to be the Maclaurin series, that means that (F(0)) needs to be equal to 1. It means that (F'(0)), actually let me write this down, it means that (F(0)) needs to be equal to 1. (F(0) = 1).

It means that (F''(0)) needs to be the coefficient on the (X) here, which is negative 1. And we could keep going. It means that the second derivative at zero, well that's going to be the coefficient on this (\frac{X^2}{2}), so that's got to be equal to 1.

And you see the general idea that the third derivative at 0 is equal to 1. It's the coefficient on (\frac{X^3}{3!}), which is negative 1 right over here.

So just using this information, can we figure out which of these it is? You could do a little bit of deductive reasoning here.

Let's evaluate all of these functions at zero and see which of these are 1. So (s(0)), well that's 0 just by looking at this first constraint. (s(0)) isn't 1; we can rule that out.

Cosine of 0 is 1, so that's still in the running. (e^0 = 1), and then the natural log of 1 + 0, that's the natural log of 1, which is 0, so that's out of the running.

So just from that first constraint, knowing that (F(0) = 1), we're able to rule out two of the choices.

Then knowing that the first derivative evaluated at zero is going to be negative 1, well what's the first derivative of cosine of (x)? What's negative sine of (x)? If we evaluate that at zero, we're not going to get negative 1; we're going to get 0, so we can rule this out.

Now, the first derivative of (e^x) is (e^x). If we evaluate that at zero, we're going to get 1, not negative 1, so we can rule that out.

Not even looking at anything else, we have a pretty good sense that D is probably our answer. But we could check the first derivative here: (F'(x)) is going to be negative (e^x), so (F'(0)) is going to be (e^0) or negative 1, so it meets that one.

And if you were curious, you could keep going and see that it meets all the other constraints. But Choice D is the only one that meets even the first two constraints for the function at zero and the first derivative at zero.