Main content

### 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

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 2015 AP Calculus BC 6c

Multiplying two Taylor polynomials.

## Want to join the conversation?

- You said g^x when meant g OF x.(4 votes)
- Can we not multiply the general term of e^x with the general term of f(x) to get the general term for e^x*f(x) and expand it?(2 votes)
- no you cannot this would give you an inaccurate answer because the terms from each series are not multiplied properly with one another.(1 vote)

- Why can we not combine e^x and f(x) to make g(x) = e^(2x) and then substitute 2x for x in the original e^x series?(1 vote)
- If we assume g(x) = e^(2x) and we also know that g(x) = f(x)*e^x, then that would imply that f(x) is also equal to e^x. Since e^x * e^x = e^(2x).

But, in this case, f(x) is NOT equal to e^x. As stated in part a, f(x) is an infinite polynomial, namely f(x) = sum from n=1 to infinity [ ((-3)^(n-1))/n * x^n ], so you have to multiply it out term by term.(1 vote)

## Video transcript

- [Voiceover] Write
the first nonzero terms of the Maclaurin series for e to the x. Use the Maclaurin series for e to the x to write the third-degree
Taylor polynomial for g to the x is equal to e to the x times f of x about x equals zero. So Maclaurin series, if
that looks familiar to you, watch the videos on Khan
Academy on Maclaurin series. I'll give a little bit of primary here. For taking the Maclaurin series of f of x, that's going to be equal to f of zero. We could view this f of
zero times x to the zero over zero factorial, if you assume zero
factorial is equal to one, and then plus f prime of zero times x to the first over one factorial plus f prime prime, the second
derivative evaluated at zero, times x squared over two factorial. I think you see the pattern here. Plus f prime prime prime,
the third derivative evaluated at zero of x to the third power over three factorial. I think you see where this is going. And of course, this
first one, x to the zero over zero factorial, that's just one. So oftentimes it will
just be written f of zero. And this term one factorial is just one. So oftentimes it's just
a zero, and this f prime of zero times x, so on and so forth. So, actually, let me just write approximately right over there. And so let's do it for e to the x. So e to the x is approximately equal to, well it's going to be e
to the zero which is one. Plus, you might already know that f of x. If f of x is equal to e to the x, then f prime of x is
also equal to e to the x. That's one of the magical
things about e to the x. The slope of the tangent
line at any point is, well, equal to the value is
equal to the x value there. And if you take the third
derivative or second derivative, it's also, you're gonna get as
many derivatives as you want. You'll still get e to the x. That's one of the special things about e. So the first derivative evaluated zero, well that's still e to the zero
power times x to the first. So plus x. And then we have the second
derivative evaluated at zero. Well, that's still one. So it's going to be times,
or so it's one times x squared over two factorial. So we just say plus x squared over two, plus x squared over two. Two factorials is two times one and then plus. Once again, the third
derivative evaluated at zero, that's just e to the x evaluated at zero which is e to the zero which is one. So x to the third over three factorial. So plus x to the third
over three factorial. We could write it as three factorial or that's three times two times one. That's equal to six. And then we keep going. So we just wrote the
first four nonzero terms of the Maclaurin series for e to the x. That's one, two, three,
four nonzero terms. Now we wanna use, so let
me just underline that. That's part of what
they're asking us to do. Then they say use the
Maclaurin series for e to the x to write the third-degree
Taylor polynomial for g of x which is equal to the product of e to the x and f of x. So what I'm gonna do is I'm gonna write our original f of x. Let's write down the
first few terms of it. And then what we could
do this thing about, well, how can we multiply
those two polynomials? And we just have to know
enough about the multiplication of those two polynomials to just get us our terms that are no higher
than third degree. So f of x is approximately equal to, let's see, it's x, I have a bad memory, it is x minus 3/2 x squared. X minus 3/2 x squared and then, plus three x to the third power, plus three x to the third power. Plus three x to the third power. Plus, and actually you
could say minus if you like because that is gonna be, you are gonna have, it's
plus minus plus minus, however you want us to do it. And that's enough. And, why do I feel confident
that that's enough? Well, we only want we only want to write the third degree Taylor polynomial. So if we multiply this
and we involve terms that are higher than third degree, well, we're gonna, that's going to give us the terms that are polynomial that are higher than third degree. So let's just think about it. What's gonna be the product? So e to the x times f of x. That's going to be approximately equal to, well let's see. We are going to multiply
this infinite polynomial times this infinite polynomial and that might seem
intimidating to you at first. What you could do is you could
go for each of these terms. Must start multiplying it times each of these terms at here. Essentially, you know, when
you're multiplying polynomials, you're just repeatedly doing
the distributive property so you take this and
distribute it on that. But we should only worry about the terms up to third degree because
anything beyond that, well, that's going to add up to a higher than third degree term. So x times one is x. X times x squared, or sorry, x times x is x squared so plus x squared. X times x squared over two is x to the third over two. So I'll just write plus 1/2 x to the third power. And I'm gonna stop there because if I do x times that, that's gonna be fourth degree term and I don't wanna worry about that. We are writing the
third-degree Taylor polynomial. So that's x, x squared,
x to the third over two. So now let's distribute
the negative 3/2 x squared. I'll use another color here just to help explain it a little bit. So if you multiply that times one, that's going to be minus 3/2 x squared. I'm just going to a second line here so I can line things
up and add them nicely. And then this times this is negative 3/2 x to the third. Negative 3/2 x to the third. And once again, I'm gonna start there. If I multiply these two, I'm
gonna get a fourth degree term. I don't care about the
fourth degree terms. And then, and then, let's do, let's worry about this guy. And so let's start distributing. So if I multiply it times this one, I'm gonna get three x to the third power. Three x to the third power. And I'm gonna stop there. Because then if I start
multiplying it times that guy, it's getting me a fourth degree term and then a fifth degree term and then a sixth degree term which I don't need to worry about. So these are all the pieces
that are going to make up that third degree polynomial. And so, what is that going to be equal to? What is that going to be equal to? This is a little bit of,
you know, getting your math intuition for multiplying
infinite polynomials. Let's see, you're gonna have x. And then you're going to have, this is one x squared
minus 1 1/2 x squared. So that's going to be
negative 1/2 x squared. Now let's see. Here, you have plus 1/2 minus 1 1/2 which would give you negative two and then plus three, plus, no, sorry, 1/2 minus 1 1/2 which would be negative one plus three is positive two plus two x to the third power. So we could say e to the x times f of x, the third degree Taylor
polynomial for this is x minus 1/2 x squared plus two x to the third power. And we are done. So this was a little bit tricky. You have to appreciate how
to, and besides calculus, it's a little bit of just algebra. I just appreciate, okay, I
only need the third degree. I don't have to distribute this times the infinite number of terms. Because, at first you might
say, "That's super hard. "How do I multiply two
infinite polynomials?" The key is we only worry
about the third degree, up to the third degree.