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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC>Unit 11

Lesson 3: AP Calculus BC 2011

# 2011 Calculus BC free response #6b

Taylor Series for cos x at x=0. Created by Sal Khan.

## Video transcript

We're on part b. Write the first four non-zero terms of the Taylor series for cosine of x about x equals 0. Use this series and the series for sine of x squared, found in part A, to write the first four non-zero terms of the Taylor series for f-- so, for this f right over here-- about x equals 0. So let's just do the first part. Let's find the first four non-zero terms of the Taylor series for cosine of x about x equals 0. So let's just say-- well, we'll put our formula for the Taylor series up here around x is equal to 0. So we're centering it around x equals 0 right over here. And so if we say that h of x is equal to cosine of x, then h of 0, is going to be equal to cosine of 0, which is equal to 1. And then, if you have h prime of x, that's going to be equal to negative sine of x. And you could look over here. This is essentially h of x. And this is going to be h prime of x. So it's essentially one step ahead of where we were with g of x. Well, we'll just redo it over here just so it's clear. So the first derivative at 0 is negative sine of 0, which is just 0. And then you have the second derivative of h of x. So the derivative of sine of x is cosine of x. But we have that negative out front, so it's negative cosine of x. So h prime prime of 0 is going to be equal to cosine of 0 is 1. You have that negative out front. So it's negative 1. And then you have the third derivative of h is equal to derivative of negative cosine of x is positive sine of x. So the third derivative at 0 is going to be equal to 0. And then you take the derivative of this, you're going to get cosine of x again. So then the cycle starts again. So I could write, this is also equal to the fourth derivative at 0. Right, you take-- let me just make it clear what I'm saying. If I take the derivative of this, I get cosine of x. The fourth derivative of h is cosine of x again. So the fourth derivative at 0 is going to be the same thing as the function evaluated at 0, which is 1. And so the fifth derivative at 0 is going to be 0. The sixth derivative at 0 is going to be negative 1. And then the seventh the derivative at 0 is going to be 0. So that's enough, now. I think we definitely have four non-zero terms. We have this term-- let me pick a nice color here-- we have this term, because we have a 1 coefficient. This is going to be a 0 because we have a 0 coefficient. This is a non-zero term. Then this is a non-zero term because we have the 1 again. And then this is a non-zero term because we have the negative 1. So we could say that cosine of x is approximately equal to-- because we're just approximating it with the first four non-zero terms. So f of 0 is 1. So that's this term right over here. Instead of f of x, we're calling it h of x right here, so that we don't get confused with that original f defined in the problem, and that we're going to work on in the second part of this section of the problem. So cosine of x is equal to 1. h prime of 0 is 0, so there's no term there. Then f prime prime of 0, second derivative, it's negative 1. So it's negative x squared over 2 factorial. And then you get another 0 for the third degree term. Then you get a 1 for the fourth degree term. So plus 1. But we don't have to write the 1, it's implicitly there. Plus 1 times x to the 4th over 4 factorial. And then the next non-zero term is this negative 1 over here. So minus 1-- we can just write minus-- and that's the sixth degree term. So x to the sixth over 6 factorial. And we're done. Those are the first four non-zero terms of cosine of x. Now let's do the second part of the problem. So we did this first part. Use this series and the series for sine of x squared, which we found right over here, sine of x squared-- actually, I think we wrote it, this is the series for sine of x squared when we simplified it. Use those-- that was found in part A-- to write the first four non-zero terms of the Taylor series for f about x equals 0. So where's our function again? So our function is sine of x squared plus cosine of x. Let me rewrite that. So f of x is equal to sine of x squared plus cosine of x. So let's write the first four non-zero terms. And you want to start with the lowest degree. So you could say that f of x is going to be approximately equal to-- because we're just approximating it with this polynomial right over here. So, from cosine of x-- so you have this right over here is cosine of x. And sine of x is over there. So between both of these, the lowest degree term right over here is this 1. So I'll write this 1 over here. And then after that, the next lowest-- in either of these, we have no first degree terms. We have no x terms. But then, in both of these, we have an x squared term. Actually, let me write it this way to simplify what I'm doing. Let me just write them out. So sine of x squared-- let me do this in a new color. Sine of x squared is this thing. This is what we figured out in part A. So that is x squared minus x to the sixth over 3 factorial-- I'm going to just write the whole thing, and then later we'll pick out the first four terms-- plus x to the 10th over 5 factorial minus x to the 14th over 7 factorial. So that's the first four terms of that. So this isn't exactly that. That's approximately equal to that. So approximation. And then cosine of x, we just figured out, is plus 1 minus x squared over 2 factorial plus x to the 4th over 4 factorial minus x to the 6th over 6 factorial. Now we can pick out the first four non-zero terms. And since we're assuming this is a Taylor series, the first term is going to be the lowest degree, then we're going to keep going at higher degree terms. So the lowest degree term right over here. So once again, f of x is going to be approximately equal to. The lowest degree term over here is this 1. So let's write the 1 over here. And then, we don't have any first degree terms. We don't how many just x to the firsts over here. But we have a couple of x squareds. So that will actually be the next term, because we can add those x squareds. So this right over here, you have x squared-- so let me just write it this way. So plus this x squared minus x squared over 2 factorial. And x squared over 2 factorial is just x squared over 2. So minus, I could write 1/2 x squared. And I'll simplify this in the next step. That's just going to be one term, because this is going to end up being 1/2 x squared. All right. Do we have any third degree terms here? We do not have any third degree terms. Do we have any fourth degree terms here? We do. We have this character right over here. It's a fourth degree term. So plus x to the 4th over 4 factorial. And then finally, do we have any 5th degree terms? We don't have any 5th degree terms, no x to the 5ths over here. Do we have any 6th degree terms? We do. We have this one and this one. So this is going to be minus-- there's a negative sign out in front of both of them. So it's going to be minus x to the 6th over 3 factorial plus x to the 6th over 6 factorial. You could say it's minus x to the 6th over 3 factorial, and then you could distribute the negative sign if you want. I just factored it out. And now we can just simplify this thing right over here. So this is going to be equal to-- so f of x is approximately equal to-- 1 plus-- and then, if you have an x squared minus half of an x squared, you're going to have 1/2 x squared. Or I could just write x squared over 2, plus x squared over 2. And then you have, in that blue color, plus x to the 4th over 4 factorial. And then, to simplify this, 3 factorial 6 factorial. So this 6 factorial is the same thing, just to make everything clear to us. 6 factorial is the same thing as 3 factorial times 4 times 5 times 6, or 3 factorial times 120. And so what we can do is multiply the numerator and the denominator here by 4 times 5 times 6, or by 120. So let me just rewrite this. So I'll do this over here because I want this to be my final answer. So if I have x to the sixth over 3 factorial, I can multiply the denominator by 4 times 5 times 6, and then multiply the numerator times 4 times 5 times 6, which is the same thing as multiplying it times 120. So you have 120 up there. 4 times 5 times 6 is also 120. And now we have the same denominator, because this is the same thing as 6 factorial. So you have 120 x to the 6 over 6 factorial plus x to the 6th over 6 factorial gives you 121 x to the sixth over 6 factorial. And we have this negative sign out here. So it becomes negative 121 x to the sixth over 6 factorial. And we're done. We've figured out the first four non-zero terms of our f of x.