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### Course: Calculus, all content (2017 edition) > Unit 8

Lesson 2: AP Calculus BC questions- 2015 AP Calculus BC 2a
- 2015 AP Calculus BC 2b
- 2015 AP Calculus BC 2c
- 2015 AP Calculus BC 2d
- 2015 AP Calculus BC 5a
- 2015 AP Calculus BC 5b
- 2015 AP Calculus BC 5c
- 2015 AP Calculus BC 5d
- 2015 AP Calculus BC 6a
- 2015 AP Calculus BC 6b
- 2015 AP Calculus BC 6c
- AP Calculus BC exams: 2008 1 a
- AP Calculus BC exams: 2008 1 b&c
- AP Calculus BC exams: 2008 1 c&d
- AP Calculus BC exams: 2008 1 d
- Calculus BC 2008 2 a
- Calculus BC 2008 2 b &c
- Calculus BC 2008 2d
- 2011 Calculus BC free response #1a
- 2011 Calculus BC free response #1 (b & c)
- 2011 Calculus BC free response #1d
- 2011 Calculus BC free response #3a
- 2011 Calculus BC free response #3 (b & c)
- 2011 Calculus BC free response #6a
- 2011 Calculus BC free response #6b
- 2011 Calculus BC free response #6c
- 2011 Calculus BC free response #6d

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# 2011 Calculus BC free response #6b

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

## Want to join the conversation?

- Didn't Sal get the Taylor Series and the Maclauren Series mixed up?(0 votes)
- No, a Maclauren Series is a Taylor Series centered on x=0 or the a is 0.(19 votes)

## 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.