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### Course: AP®︎/College Calculus BC > Unit 10

Lesson 11: Finding Taylor polynomial approximations of functions- Taylor & Maclaurin polynomials intro (part 1)
- Taylor & Maclaurin polynomials intro (part 2)
- Worked example: Maclaurin polynomial
- Worked example: coefficient in Maclaurin polynomial
- Worked example: coefficient in Taylor polynomial
- Taylor & Maclaurin polynomials
- Visualizing Taylor polynomial approximations

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# Taylor & Maclaurin polynomials intro (part 1)

A Taylor series is a clever way to approximate any function as a polynomial with an infinite number of terms. Each term of the Taylor polynomial comes from the function's derivatives at a single point. Created by Sal Khan.

## Want to join the conversation?

- How could we apply this to a real world case?(143 votes)
- Have you ever wondered how a calculator is able to give you a value for sine at any value, while you're stuck memorizing just the important ones? You can use this concept to do things like approximate trig functions like sine or cosine for any value. Computers can find the sum for a polynomial series with, say, 1000 terms in a snap and give an accurate approximation of that function. I find Taylor series to be very clever.(509 votes)

- what dose a 3rd derivative represent? the first derivative is the slope of the tangent line. the second derivative is the degree that the tangent line of one point differs from the tangent line of a point next to it. so is there any basis for having a third derivative other then using it in a Maclauren series?(12 votes)
- The way I think of it is;

1st derivative: rate of change of the function. (which is equal to the slope in any given point on the graph)

2nd derivative: rate of change of the 1st derivative. (which, when focused on a specific point, is equal to the changing rate of the slope in that point which is what you said)

3rd derivative: rate of change of the 2nd derivative (I guess this would be the acceleration of the slope change in a point)

4th derivative: rate of change of the 3rd derivative (and here I just give up trying to relate the derivatives to the original function)

5th derivative: you guessed it - rate of change of the 4th derivative.

And so on... To me it's just one of those things you simply have to accept that you can't rely on your visualisation skills to understand it. Like a 5-dimensional space, try imagining what that looks like. Still, it's not difficult to calculate the distance between two points in it for example, by just accepting the space as "a space where a point requires 5 coordinates to specify it's location".(33 votes)

- Why at9:45do we use n! in the denominator instead of just the n. I understood up to the second derivative why we introduce 1/2 but then on the third it I do not understand why we have 1/3! as this is 1/6 and not 1/3 which means that it will not cancel out. I hope you understand my questions.

Thank you very much for your help :)(13 votes)- For the second derivative, it's not 1 / 2 it's really 1 / 2!

It's just that 2! happens to be equal to 2.

The factorial works well for repeated differentiation:

d/dx x^5 / 5! = x^4 / 4!

d/dx x^4 / 4! = x^3 / 3!

d/dx x^3 / 3! = x^2 / 2!

d/dx x^2 / 2! = x / 1!

d/dx x / 1! = 1 / 0!

d/dx 1 / 0! = 0(10 votes)

- at about3:09, Sal adds an "x" after f '(0). Why?(9 votes)
- It's so p'(0) still equals f'(0) after he takes the derivative of his expression f(0)+f'(0)x. If the x wasn't there, then the f'(0) would end up equaling zero when he took the derivative, just as the f(0) did. Reviewing power rule may be helpful in clarifying this.(16 votes)

- At10:04why is it (1/2) multiplied by (1/3)? According to Sal's logic from before, shouldn't it just have been (1/3)? I know there was an answer to this question later on, but I didn't understand at all from that... ><(5 votes)
- We want p´´´(c) to be equal to f´´´(c).

Let's look at that one term.

p(c) = terms + f´´´(c) * 1/(2*3) * x^3

Let's look at its first derivative.

p´(c) = terms + f´´´(c) * 1/(2*3) * 3 * x^2

p´(c) = terms + f´´´(c) * 1/2 * x^2

Let's take the second derivative.

p´´(c) = term + f´´´(c) * 1/2 * 2 * x^1

p´´(c) = term + f´´´(c) * x

Then let's take the third derivative.

p´´´(c) = f´´´(c) * 1 * x^0

p´´´(c) = f´´´(c)

You have to have both 1/2 and 1/3 so that the coefficients eventually cancel each other out.(10 votes)

- what is the difference between a polynomial and a function ? dont we write the function of x as a polynomial?(6 votes)
- If it is a polynomial, then it is a function. This is not true the other way around (If its a function, then it is a polynomial). However, the distinction Sal is trying to make in this video is that we can approximate a function using a talyor series which looks like a polynomial. The key here is that we can approximate functions such as sin(x) or cos(x), or ln(x). Clearly, these functions are not polynomials, but they can be written as a talyor series which when you "expand it" looks like a polynomial with x raised to some power, thus functions can be approximated to look like polynomial.

I hope this distinction makes a little more sense.(9 votes)

- i can't visualize why when we add new term of new polynomial then the new curve nearer the old curve !(3 votes)
- Well, by analogy, think of trying to identify a specific person without saying their name. Instead, you will just be listing off facts about that person. Each fact you list off will eliminate a certain number of persons, until you eventually have enough descriptions that only one human on earth meets all the qualifications.

Really, that is all we are doing here. We are adding more and more facts about the original function. And, we're doing it in such a way that it does not interfere with previously established facts. While with this method we might need infinitely many facts added together to get a perfect match, very often we are not interested in the entire domain of the function, but just a particular region. Given that, we only need enough facts for the estimate to be pretty good for the region we're interested in.(12 votes)

- I don't really understand the intuition of these series, I mean, the mechanisms seem perfect and comprehensible, but why can we be confident that P(x) which is expanded at the point zero, would fit into all x when n approaches infinity? we are just evaluating it at zero!(7 votes)
- Why are Taylor series so important and how can they help with representing series?(4 votes)
- Amongst other things, they are very good at modeling functions that are exceedingly difficult to work with or calculate. Suppose you were an engineer and you had to add 50 functions that were rather complicated. If it were the case that they all met the conditions for Maclaurin or Taylor, then you might elect to use Maclaurin or Taylor to be able to add them more practically.(6 votes)

- As with many such "special cases" there is a simple answer and a more complex one. The simple answer is that it JUST IS - it is a convention agreed upon by the mathematical community at some point, which is perfectly legitimate since the entire field of mathematics is a series of agreed-upon conventions. Other similar examples include 0^0=1 and the very concept of zero itself. A more involved answer might be that IT NEEDS TO BE to make some other mathematical things work - Sal has a combinatorics video that explains why here: https://www.khanacademy.org/math/precalculus/prob_comb/combinatorics_precalc/v/zero-factorial-or-0(4 votes)

## Video transcript

I've draw an arbitrary
function here. And what we're
going to try to do is approximate this
arbitrary function-- we don't know what it
is-- using a polynomial. We'll keep adding terms
to that polynomial. But to do this,
we're going to assume that we can evaluate
the function at 0, that it gives us some
value, and that we can keep taking the
derivative of the function and evaluating the
first, the second, and the third derivative, so
on and so forth, at 0 as well. So we're assuming that
we know what f of 0 is. We're assuming that we
know what f prime of 0 is. We're assuming that we know
the second derivative at 0. We're assuming that we know
the third derivative at 0. So maybe I'll write
it-- third derivative. I'll just write f prime prime
at 0, and so forth and so on. So let's think about how we
can approximate this using polynomials of ever
increasing length. So we could have a
polynomial of just one term. And it would just
be a constant term. So this would be a
polynomial of degree 0. And if we have a constant
term, we at least might want to make that constant
polynomial-- it really is just a constant function-- equal
the function at f of 0. So at first, maybe
we just want p of 0, where p is the polynomial
that we're going to construct, we want p of 0 to
be equal to f of 0. So if we want to do that
using a polynomial of only one term, of only one
constant term, we can just set p of x
is equal to f of 0. So if I were to graph it,
it would look like this. It would just be a
horizontal line at f of 0. And you could say, Sal, that's
a horrible approximation. It only approximates the
function at this point. Looks like we got lucky at
a couple of other points, but it's really bad
everywhere else. And now I would
tell you, well, try to do any better using
a horizontal line. At least we got it
right at f of 0. So this is about as good as we
can do with just a constant. And even though-- I just
want to remind you-- this might not look
like a constant, but we're assuming that
given the function, we could evaluate
it at 0 and that will just give us a number. So whatever number that was, we
would put it right over here. We'd say p of x is
equal to that number. It would just be a horizontal
line right there at f of 0. But that obviously
is not so great. So let's add some
more constraints. Beyond the fact that we want
p of 0 to be equal to f of 0, let's say that we
also want p prime at 0 to be the same thing
as f prime at 0. Let me do this in a new color. So we also want,
in the new color, we also want-- that's
not a new color. We also want p prime. We want the first derivative
of our polynomial, when evaluated at 0,
to be the same thing as the first derivative of the
function when evaluated at 0. And we don't want to lose
this right over here. So what if we set p of x
as being equal to f of 0? So we're taking our old
p of x, but now we're going to add another term so
that the derivatives match up. Plus f prime of times x. So let's think about
this a little bit. If we use this as our new
polynomial, what happens? What is p is 0? p of 0 is going to
be equal to-- you're going to have f of 0 plus
whatever this f prime of 0 is times 0. If you put a 0 in for x, this
term is just going to be 0. So you're going to be left
with p of 0 is equal to f of 0. That's cool. That's just as good
as our first version. Now what's the
derivative over here? So the derivative is p
prime of x is equal to-- you take the derivative of this. This is just a constant,
so its derivative is 0. The derivative of a
coefficient times x is just going to
be the coefficient. So it's going to
be f prime of 0. So if you evaluate it
at 0-- so p prime of 0. Or the derivative of
our polynomial evaluated at 0-- I know it's a little
weird because we're not using-- we're doing a p prime of x
of f of 0 and all of this. But just remember, what's the
variable, what's the constant, and hopefully, it'll make sense. So this is just obviously
going to be f prime of 0. Its derivative is
a constant value. This is a constant
value right here. We're assuming that we can take
the derivative of our function and evaluate that thing at
0 to give a constant value. So if p prime of x is equal
to this constant value, obviously, p prime
of x evaluated at 0 is going to be that value. But what's cool about
this right here, this polynomial that has a 0
degree term and a first degree term, is now this polynomial
is equal to our function at x is equal to 0. And it also has the
same first derivative. It also has the same
slope at x is equal to 0. So this thing will look,
this new polynomial with two terms-- getting
a little bit better-- it will look
something like that. It will essentially have--
it'll look like a tangent line at f of 0, at x is equal to 0. So we're doing better, but still
not a super good approximation. It kind of is going in
the same general direction as our function around 0. But maybe we can do
better by making sure that they have the
same second derivative. And to try to have the same
second derivative while still having the same first derivative
and the same value at 0, let's try to do
something interesting. Let's define p of x. So let's make it clear. This was our first try. This is our second
try right over here. And I'm about to embark
on our third try. So in our third try, my goal is
that the value of my polynomial is the same as the value
of the function at 0. They have the same
derivative at 0. And they also have the same
second derivative at 0. So let's define my
polynomial to be equal to-- so I'm going
to do the first two terms of these guys
right over here. So it's going to be
f of 0 plus f prime of 0 times x, so exactly
what we did here. But now let me add another term. I'll do the other
term in a new color. And I'm going to
put a 1/2 out here. And hopefully it might make
sense why I'm about to do this. Plus 1/2 times the
second derivative of our function
evaluated at 0 x squared. And when we evaluate
the derivative of this, I think you'll see
why this 1/2 is there. Because now let's evaluate
this and its derivatives at 0. So if we evaluate p of 0, p of
0 is going to be equal to what? Well, you have
this constant term. If you evaluate it at 0,
this x and this x squared are both going to be 0. So those terms are
going to go away. So p of 0 is still
equal to f of 0. If you take the derivative
of p of x-- so let me take the
derivative right here. I'll do it in yellow. So the derivative
of my new p of x is going to be equal to-- so
this term is going to go away. It's a constant term. It's going to be
equal to f prime of 0. That's the coefficient on this. Plus-- this is the power
rule right here-- 2 times 1/2 is just 1, plus f prime
prime of 0 times x. Take the 2, multiply
it times 1/2, and decrement that
2 right there. I think you now have a sense
of why we put the 1/2 there. It's making it so that we don't
end up with the 2 coefficient out front. Now what is p prime of 0? So let me write it right
here. p prime of 0 is what? Well, this term right here
is just going to be 0, so you're left with this
constant value right over here. So it's going to
be f prime of 0. So so far, our third
generation polynomial has all the properties
of the first two. And let's see how it does
on its third derivative, or I should say the
second derivative. So p prime prime of
x is equal to-- this is a constant, so
its derivative is 0. So you just take the
coefficient on the second term is equal to f prime prime of 0. So what's the second
derivative of p evaluated at 0? Well, it's just going to
be this constant value. It's going to be f
prime prime of 0. So notice, by adding
this term, now, not only is our polynomial value
the same thing as our function value at 0, its derivative
at 0 is the same thing as the derivative of
the function at 0. And its second derivative
at 0 is the same thing as the second derivative
of the function at 0. So we're getting
pretty good at this. And you might guess that
there's a pattern here. Every term we add, it'll allow
us to set up the situation so that the n-th derivative
of our approximation at 0 will be the same thing
as the n-th derivative of our function at 0. So in general, if we
wanted to keep doing this, if we had a lot of
time on our hands and we wanted to just keep
adding terms to our polynomial, we could-- and let me
do this in a new color. Maybe I'll do it in a
color I already used. We could make our
polynomial approximation. So the first term, the constant
term, will just be f of 0. Then the next term will
be f prime of 0 times x. Then the next term
will be f prime prime of 0 times
1/2 times x squared. I just rewrote that in a
slightly different order. Then the next term, if we want
to make their third derivative the same at 0, would be
f prime prime prime of 0. The third derivative
of the function at 0, times 1/2 times 1/3,
so 1 over 2 times 3 times x to the third. And we can keep going. Maybe you you'll start
to see a pattern here. Plus, if we want to make
their fourth derivatives at 0 coincide, it would
be the fourth derivative of the function. I could put a 4 up
there, but this is really emphasizing-- it's the fourth
derivative at 0 times 1 over-- and I'll change the order. Instead of writing it
in increasing order, I'll write it as 4 times 3
times 2 times x to the fourth. And you can verify
it for yourself. If we just had this
only, and if you were to take the fourth
derivative of this, evaluate it at 0,
it'll be the same thing as the fourth derivative of
the function evaluated at 0. And in general, you
can keep adding terms where the n-th term
will look like this. The n-th derivative of your
function evaluated at 0 times x to the n over n factorial. Notice this is the same
thing as 4 factorial. 4 factorial is equal to 4
times 3 times 2 times 1. You don't have to
write the 1 there, but you could put it there. This right here is the same
thing as 3 factorial-- 3 times 2 times 1. I didn't put the 1 there. This right here is the same
thing as 2 factorial, 2 times 1. This is the same thing. We didn't write anything,
but you could divide this by 1 factorial, which
is the same thing as 1. And you can divide
this by 0 factorial, which also happens to be 1. We won't have to study
it too much over here. But this general series that
I've kind of set up right here is called the Maclaurin series. And you can approximate
a polynomial. And we'll see it leads to
some pretty powerful results later on. But what happens-- and I
don't have the computing power in my brain to draw
the graph properly-- is that when only
the functions equal, you get that horizontal line. When you make the
function equal 0 and their first
derivatives equal at 0, then you have something that
looks like the tangent line. When you add another
degree, it might approximate the polynomial
something like this. When you add another degree, it
might look something like that. And as you keep adding
more and more degrees, when you keep adding
more and more terms, it gets closer
and closer around, especially as you get
close to x is equal to 0. But in theory, if you add
an infinite number of terms, you shouldn't be able to do--
I haven't proven this to you, so that's why I'm saying that. I haven't proved it yet to you. But if you add an
infinite number of terms, all of the derivatives
should be the same. And then the function
should pretty much look like each other. In the next video, I'll do
this with some actual functions just so it makes a
little bit more sense. And just so you know,
the Maclaurin series is a special case
of the Taylor series because we're centering it at 0. And when you're doing
a Taylor series, you can pick any center point. We'll focus on the
Maclaurin right now.