Worked example: coefficient in Maclaurin polynomial | Series | AP Calculus BC | Khan Academy

Nth derivative of g at x equals 0 is given by. So the nth derivative of G evaluated at x equal 0 is equal to n + 7 over n 3r for n is greater than or equal to 1. What is the coefficient for the term containing x^2 in the McLaurin series of G?

So let's just think about the McLaurin series for G. If I were to have my function G of x, the McLaurin series could say approximately equal to, especially if I'm not going to list out all of the terms, is going to be equal to. Well, it's going to be equal to G of 0 plus G Prime of 0 * x plus G Prime Prime of 0 divided by. I could say 2 factorial, but that's just 2 * x^2.

And that's about as far as we go because we just have to think about what is the coefficient for the term containing x. If they said what's the coefficient for the term containing x to the 3, I would keep going. I'd go G Prime. I would take the third derivative evaluated at 0 over 3 factorial. I could view this as a factorial too, but that just evaluates to 2. I could view this as 1 factorial. I could view this as 0 factorial, just so you see it's a consistent idea here.

And I could, of course, keep on going, but we just care about—they're just asking us what is the coefficient for the term containing x squared. So they just want us to figure out this. What is this thing right over here?

So now that we need to figure out what is the second derivative of G evaluated at x equals 0, well, they tell us that over here. It's a little bit unconventional where they give us a formula, a general formula for any derivative evaluated at x equals 0, but that's what they're telling us here.

In this case, the N isn't zero; the N is the derivative we're taking, and that's going to be our second derivative. So this is... if I wanted to figure out G, if I am figuring out the second derivative, I could write it like that evaluated at 0 or I could write it like this just so the notation is consistent.

I could write it like that. The second derivative evaluated at x equals 0 is going to be equal to—well, our n is 2, so this is going to be the square root of 2 + 7 over 2 to the 3 power. So 2 + 7 is 9. Take the principal root of that; it's going to give us pos3 over 2 cubed, which is 8.

So this part right over here is 38. The whole coefficient is going to be 38, that’s this numerator divided by 2, which of course is equal to 3 over 16, and we're done. They didn't want us to figure out, you know, a couple of terms of this which we could call the McLaurin polynomial, an nth degree McLaurin polynomial. They just wanted to find one coefficient right here, the coefficient on the second-degree term, which we just figured out is 3 over 16.