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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 #6a

Taylor Series approximation of sin(x). Created by Sal Khan.

## Want to join the conversation?

- When the problem says "about x=0" why would you not evaluate the terms at 0?

(beside the fact that all the terms would then be 0)(7 votes)- The phrasing used is pretty standard for describing Taylor approximations. Nonetheless, it is a little deceptive. The question is not asking for a numerical answer, but a new function. It's more like saying, "What does sin(x) look like near ("about") x=0?"(7 votes)

- Does anyone know a proof for the property Sal uses at6:40where he just replaces x with x^2?(6 votes)
- I get it now. My trouble was I was thinking chain rule with d/dx (x^2) would mess things up. But I realize now that the Taylor series is now created by taking the derivative with respect to x^2, not wrt. x.(8 votes)

- i dont get it

f'x=dy/dx sin(x^2)=2x*cos(x^2)

if you replace 0 you get 2*0*1=0

thus all the cosines terms will zero out, where is it stated that in taylor we only do the first part of the chain rule?(1 vote)- If we try to find dy/dx[sin(x^2)], d^2y/dx^2[sin(x^2)], etc., we're gonna get something messy. We can just find dy/dx[sin(x)], d^2y/dx^2[sin(x)], etc. for the Taylor series of sin(x) about x=0, and, for the Taylor series of sin(x^2) about x=0, substitute x^2 for x for the derivatives with respect to x^2 for a good approximation.(1 vote)

## Video transcript

Problem number six. Let f of x is equal to sine
of x squared plus cosine of x. The graph of y is equal
to the absolute value of the fifth derivative
of f at x is shown above. And I haven't shown it here
just so we have some space. I'll show it when
we need to show it. I think we have to
show it in part D. So first, let's do
part A right over here. Write the first non-zero
terms for the Taylor series for sine of x about
x is equal to 0. And write the first for non-zero
terms of the Taylor series for sine of x squared
about x equals 0. So let's do this first part. And just a reminder,
a Taylor Series is a polynomial
approximation of a function. So just to give us
a quick reminder. We go into much
more detail in this on the videos about
Taylor series. But if you have a function
that looks like this, and if you wanted to approximate
it with a Taylor series around 0, if you only have one
term in your Taylor series, you would just
literally have it-- it would just be a
constant, just like that. If you have two terms
in your Taylor series it would be a line
that looks like that. If you have three terms
in your Taylor series, you'd get to a
second degree term. And you'd start approximating
it something like that. If you get to the
third degree term it might start
looking like that. And as you add more
and more terms, you get a better, better
approximation of your function. And if you add an
infinite number, it might actually
converge to your function. So let's just remind
ourselves, if I have a function-- if I have f
of x, this Taylor series-- I can approximate it with
this Taylor series. And if we want to center
that approximation around 0, it'll be equal to f of
0 plus f prime of 0. The derivative at 0 times x
plus the second derivative at 0 times x squared
over 2 factorial. You could have divided
this term right here by one factorial,
which is just 1. Two factorials is just 2. And then plus the third
derivative of f at 0 times x to the third
over 3 factorial. Plus the fourth derivative-- I
think you get the idea here-- at the fourth
derivative at 0 times x to the fourth
divided by 4 factorial, and so on and so forth. So what they want us to
do is find the first four non-zero terms of the
Taylor series for sine of x. And some of you might
already know this. We actually cover
this in the video where we show Euler's identity. But we'll do it again
right over here. So if we just
take-- I'll call it g of x, because they already
defined f of x up here. So let's say that g of x--
let me do this in a new color just to ease the monotony. Let's say that g of x
is equal to sine of x. Then we know that g of 0
is going to be equal to 0. And then if we take
the derivative, g prime of x is going to be
equal to negative-- no, it's going to be positive
cosine of x. g prime of 0 is now
going to be equal to 1. Cosine of 0 is 1. Then if you take the
second derivative, the derivative of cosine
of x is negative sine of x. Negative sine of x and
the second derivative at 0 once again is going to
equal zero, sine of 0, 0. And now let's take
the third derivative. Third derivative of
our function of g. The derivative of negative sine
of x is negative cosine of x. And the third
derivative at 0 is now going to be equal to negative 1. And we could keep going. You can already guess
where this might lead to, but I'll just keep
going just in case. The fourth derivative
is once again going to be equal to sine of x. It is now going to be
equal to sine of x. And then the fourth
derivative at 0, because it's the same
thing as the function, is now going to be 0 as well. So this is going to be equal
to g to the fourth at 0. And then we could keep going. This is going to be the same
thing as g to the fifth, because we start cycling as we
take more and more derivatives. So this is going to
be the same thing as the fifth derivative at 0. This is going to
be the same thing as the sixth derivative at 0. And this is going
to be the same. This is an equal sign
right here, equals 1. And this is going
to be the same thing as the seventh derivative at 0. So if you want the first
four non-zero terms, let's just work through it. So first of all, f of 0--
I'll do this in a new color. If we want to
approximate g of x, if we want to
approximate sine of x. So we'll say sine of x is going
to be approximately equal to. So this first term right
over here, that is sine of 0. That's g of 0. That's going to be 0, so we
don't even have to write it. Then we go to this
term right over here. Well, the first derivative, f
prime of 0-- or in this case, g prime of 0-- is
going to be equal to 1. So it's going to be 1 times x. So you're going to
have an x there. Then the next term
is going to be 0. We see that right over there. Because the second derivative
of our function evaluated at 0 is 0. Our third derivative of our
function evaluated at 0 is 1. So that term is going
to show up again. Actually, that
term is negative 1. I don't want to
make a mistake here. That is negative 1, negative 1. It was negative cosine of x. You evaluate negative cosine
of x and 0, you get negative 1 right over here. So the third derivative,
this thing right over here, is negative 1. So then we have minus x to
the third over 3 factorial. The fourth derivative,
once again, is 0. The fifth derivative
evaluated at 0 is 1. So then you're going
to have plus 1 times x to the fifth over 5 factorial. Then the six derivative is 0. So that term is
going to disappear. I didn't even write it up here. And then the seventh term,
the coefficient is negative 1. Or the seventh derivative
evaluated at 0 is negative 1. So you have negative 1 times
x to the seventh over 7 factorial. And we have to go all the
way to the seven degree term to find the first four non-zero
terms for our Taylor Series. And so we're done with
at least the first part. We found the first four
non-zero terms of sine of x Now, what about sine of x squared? We have to be careful here. Because you might just say, OK,
let me just apply this formula. And what you're going
to find very quickly is when you start taking the
second and third derivatives of this thing right over here,
it's going to get really messy. But what you can say
is, look, sine of x is approximately this
thing right over here. What happens if I just replace
the x with the x squared? Then I get sine of x
squared is approximately equal to-- instead of an x here,
I'm going to put an x squared. Instead of an x
to the third here, I'm going to put an x
squared to the third over here, over 3 factorial. Instead of an x to the fifth. I can put an x squared to the
fifth power over 5 factorial. And instead of an
x to the seventh I could write x squared to the
seventh power over 7 factorial. So this is a very
important thing to realize. Because if you had
started directly take the Taylor Series around
0 of this thing right here, you would have taken
up all your time trying to take the derivatives. And you probably wouldn't have
been able to do it anyway, because it would have
gotten really messy. And the key here is just to
realize that you substitute x squared for x,
you're then going to get the approximation
for sine of x squared. We could simplify
this a little bit. This is going to
be approximately equal to x squared minus
x to the sixth over 3 factorial plus x
to the 10th over 5 factorial minus x to the
14th over 7 factorial. So that's our second
part of the problem.