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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 7

Lesson 13: Taylor & Maclaurin polynomials intro- 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
- Taylor polynomial remainder (part 1)
- Taylor polynomial remainder (part 2)
- Worked example: estimating sin(0.4) using Lagrange error bound
- Worked example: estimating eˣ using Lagrange error bound
- Lagrange error bound
- Visualizing Taylor polynomial approximations
- Worked example: Taylor polynomial of derivative function

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# Taylor polynomial remainder (part 1)

The more terms we have in a Taylor polynomial approximation of a function, the closer we get to the function. But

*HOW*close? Let's embark on a journey to find a bound for the error of a Taylor polynomial approximation. Created by Sal Khan.## Want to join the conversation?

- Why wouldn't the n+1'th derivative of the function be 0 as well?(56 votes)
- Remember also that the function f(x) isn't necessarily a polynomial. p(x) is its polynomial representation, but f(x) can be any function - sin(x), e^x, sqrt(x), etc. Those are examples of functions that will never go to 0 no matter how many times you take the derivative.(112 votes)

- when he derives p(a), and says it is equal to f'(a), how is that possible when the term that has f'(x) is multiplied by (x-a), so at (a) this term should also be zero?(10 votes)
- I know this was posted four years ago, but I may as well write this in case it helps anyone. The reason p'(a) = f'(a) (and p''(a) = f''(a), etc) is because of the following:

We are given:

p(x)=f(a)+f'(a)(x-a)+f''(a)((x-a)^2)/2!+...

To find p'(x), we have to take the derivative of each term in p(x). Since f(a) is a constant (since a is just a number that the function is centered around), the derivative of that would be 0. When taking the derivative of the second term, ie f'(a)(x-a), using the product rule (u'v +uv') gives you f''(a)(x-a)+f'(a)(1), which, when evaluated at a, is f''(a)(a-a)+f'(a)(1) which simplifies to f'(a) since f''(a)(a-a)=0. All of the terms after that are 0 since the product rule doesn't lead to the (x-a)^n part of each term 'cancelling out' as it did here. You can verify this last bit by trying the product rule on them yourself. This whole concept applies to every derivative of the polynomial up through the degree of the polynomial, as far as I know. Hope this helps someone!(19 votes)

- What do you mean by bound toward the end of the video?(7 votes)
- Being "bound" means that you know that a value is definitely between two limits. For instance, you might be interested in knowing that your approximation is good to 1% or to 0.01% or to one part in a million.

With calculators, it's often easy enough to add another term and get your error to be bound to any level you want. However, when math meets the physical world, you could get a scenario where you have 3000 cables supporting a load on a bridge. The math of exactly what's happening on each cable is too hard so you turn to approximations. If you can show your approximation is valid (or bound) to within 2% and you know that all materials have a 25% safety tolerance, you don't need to spend any more time or money on figuring out the details of your cable loads more precisely.(21 votes)

- Is this related to
**Big O Notation**?

And, by the way, what*is*"Big O Notation"?(4 votes)- Big O notation is computer science term, It is used to describe the processing complexity of the algorithm.

For example, If you have n random numbers and you want to find minimum number out of it then Big O for it will be n. Which means that we will have to check every number once to find minimum number out of it.(10 votes)

- Well, there's the obvious problem that the degree of functions such as the sine or e^x is infinite. Does this mess up the principle of bounding the error function? Why or why not?(6 votes)
- Although there might be no end to the expansion of the Taylor series of f(x), we have already seen how an infinite series like this will converge to a fixed number. The goal of this error function is to see how close P(x) is to f(x) with just the first n terms.(7 votes)

- At8:43, When we want to find the N+1 dervivative of the error (E(X)), why is the N+1 derivative of theTaylor Polynomial (P(x)) equals to zero, and not the N+1 derivative of the function, Why is the last on equal to the N+1 derivative of the error?(4 votes)
- Remember that P(x) is an nth polynomial if you try to figure out the 3rd derivative of x^2 you will get zero, In fact if you have a polynomial function with highest degree n and you get the (n+1)th derivative you get zero that is because every time you take the derivative you apply the power rule where you decrease the power by one until it becomes 0 in which case you have a constant function ex: f(x) = 5 x^0 = 5 and the derivative of a constant function will equal zero, point is that take the derivative of a polynomial enough times you will have a constant function take the derivative one more time you will get 0.

one more thing is that we are assuming we don't know what kind of function is f(x) it could be a polynomial in which case f'^(n+1)(x) will equal zero or it could be an exponential function like 2e^x in which case f'^(n+1)(x) will equal 2e^a at x=a so we actually don't know which one could it be so we leave it as it is and we write it f'^(n+1)(x).

I hope I was helpful to you.

Good luck and never give up :)(5 votes)

- since this approximation completely matched f(x), why is the (n+1)th derivative of P(x) zero since it goes on and on to infinity? Which means, why is P(x) confined to the nth term?(3 votes)
- Or another way of saying it is that you assume you've created an approximation of order n. When you take the (n+1)th derivative you get 0 and I assume the video explains that well enough.

Why, then, would you not do an infinite number of terms? Practical considerations. Calculators used to use Taylor series expansions to calculate sin, cos, and e^x. (I think most use lookup tables and interpolation now.) However, your calculator can't do an infinite number of terms. Instead, they know they're going to show the answer to, say, 8 decimal places. How many terms do you need for your sin(x) approximation in order to know that your Taylor's series expansion is good to at least 8 decimal places? That's an example of bounding your error AND of why you'd stop at some point.(2 votes)

- At5:51, is there any proof that the derivative of E is also equal to the difference of derivatives of f and P? I mean, we only defined E(x) as the difference of a function's value and the value of its taylor polynomial at some x. But how can we be sure about the derivatives too?(3 votes)
- Hi, Tushar!

The Error function is just the difference of two functions.

E(x) = f(x) - P(x)

Take the derivative from both sides:

d/dx[E(x)] = d/dx[f(x) - P(x)]

By derivative properties, the derivative of a difference of functions is the difference of the derivatives of the functions. In other words, the right hand-side can be rewritten as:

d/dx[E(x)] = d/dx[f(x)] - d/dx[P(x)]

Changing from Leibniz's to Lagrange's notation:

E'(x) = f'(x) - P'(x)

I hope it's clear now.

Nice question!

Edit:

Now, for a formal proof of the abovementioned derivative property, here's the link to it here on Khan Academy:

https://www.khanacademy.org/math/differential-calculus/dc-diff-intro/dc-basic-diff-rules/a/justifying-the-basic-derivative-rules(1 vote)

- Mr. Sal says that Taylor series is useful because the derivative at a is equal to the derivative of the function at a... But I don't understand why we need to use Taylor series in the first place. Why not just use the function?

In other words.. what's the point of approximating a function when you actually have the"real" function?(2 votes)- Because the real function sometimes might be very annoying and approximating a function helps us to simplify it.(1 vote)

- at6:00Sal says, "if we assume that this is higher than degree one, we know that these derivatives are going to be the same at a". Is the "this" that he refers to the degree of the function? If so, wouldn't the equivalence hold even if the function (or the polynomial for that matter) were of zero degree? Why does he qualify the equivalence?(2 votes)
- Good question! At first, I thought you were 100% correct, but now I think I know what he was trying to say. The "this" is P(x), and he should really have said "if we assume P(x) has degree of AT LEAST one", or "if we assume P(x) is higher than degree ZERO". He is writing down the fact that f'(a) and P'(a) are equal (i.e., f'(a) - P'(a) =0), but that will only be true if our Taylor polynomial is of degree at least one (i.e., we are at least making a tangent line, so that the slope matches the slope of f(x) ). It would be pretty silly to work with a zeroth degree Taylor polynomial (i.e. a constant horizontal line at y = f(a) ), but it would not be technically forbidden.

One other quick note is to remember that the FUNCTION, f(x), very likely DOES NOT HAVE ANY DEGREE AT ALL. f(x) is often something like sin(x), e^x, ln(x), which is not a polynomial, and that is generally the reason we are trying to find a Taylor polynomial approximation. Of course, we can (and sometimes do) find Taylor polynomials to approximate other (higher-degree) polynomials, but that is fairly rare.(1 vote)

## Video transcript

- [Voiceover] Let's say
that we have some function f of x right over here. And let me graph an arbitrary f of x. So, that's my y-axis, that is my x-axis and maybe f of x looks
something like that. And what I wanna do is I
wanna approximate f of x with a Taylor polynomial centered around x is equal to a. So this is the x-axis, this is the y-axis. So I want a Taylor polynomial
centered around there. And we've seen how this works. The Taylor polynomial
comes out of the idea that for all of the derivatives up to and including the degree
of the polynomial, those derivatives of that
polynomial evaluated at a should be equal to the derivatives of our function evaluated at a. And that polynomial evaluated at a should also be equal to that
function evaluated at a. So our polynomial, our Taylor polynomial approximation would look
something like this. So, I'll call it P of x. And sometimes you might see
a subscript, a big N there to say it's an Nth degree approximation and sometimes you'll
see something like this. Sometimes you'll see
something like N comma a to say it's an Nth degree
approximation centered at a. Actually, I'll write that right now. Maybe we might lose it if
we have to keep writing it over and over but you
should assume that it is an Nth degree polynomial centered at a. And it's going to look like this. It is going to be f of a, plus f prime of a, times x minus a, plus f prime prime of a, times x minus a squared over-- Either you could write
two or two factorial, they're the same value. I'll write two factorial. You could write a divided by one factorial over here, if you like. And then plus, you go to the
third derivative of f at a times x minus a to the third power, I think you see where this is going, over three factorial. And you keep going, I'll
go to this line right here, all the way to your Nth degree term which is the Nth derivative of f evaluated at a times x minus a to the N over N factorial. And this polynomial right over here, this Nth degree polynomial centered at a, f or P of a is going to be
the same thing as f of a. And you can verify that
because all of these other terms have an x minus a here. So if you put an a in the polynomial, all of these other terms
are going to be zero. And you'll have P of a is equal to f of a. Let me write that down. P of a is equal to f of a. And so it might look something like this. And it's going to fit the curve better the more of these terms
that we actually have. So it might look something like this. I'll try my best to show what it might look like. So this is all review,
I have this polynomial that's approximating this function. The more terms I have, the
higher degree of this polynomial, the better that it will fit this curve the further that I get away from a. But what I wanna do in
this video is think about if we can bound how good it's fitting this function as we move away from a. So what I wanna do is
define a remainder function. Or sometimes, I've seen some text books call it an error function. And I'm going to call this-- I'll just call it an error-- Just so you're consistent with all the different notations you
might see in a book, some people will call
this a remainder function and sometimes they'll
write a remainder function for an Nth degree
polynomial centered at a. Sometimes you'll see this
as an error function. The error function is sometimes avoided because it looks like expected
value from probability. But you'll see this often,
this is E for error. E for error, R for remainder. And sometimes they'll also have the subscripts over there like that. And what we'll do is, we'll just define this function to be the difference between f of x and our approximation
of f of x for any given x. So it's really just going to be, I'll do it in the same colors,
it's going to be f of x minus P of x. Where this is an Nth degree polynomial centered at a. So for example, if
someone were to ask you, or if you wanted to visualize. What are they talking
about if they're saying the error of this Nth degree
polynomial centered at a when we are at x is equal to b. What is thing equal to or how
should you think about this. Well, if b is right over here. So the error of b is going to be f of b minus the polynomial at b. So f of b there, the
polynomial's right over there. So it'll be this distance right over here. So if you measure the error at a, it would actually be zero. Because the polynomial and the
function are the same there. F of a is equal to P of a, so the error at a is equal to zero. And let me actually write that down because that's an interesting property. It'll help us bound it
eventually so let me write that. The error function at a. And for the rest of this
video you can assume that I could write a subscript. This is for the Nth degree
polynomial centered at a. I'm just gonna not write that everytime just to save ourselves a
little bit of time in writing, to keep my hand fresh. So the error at a is equal to f of a minus P of a. And once again, I won't
write the sub-N, sub-a. You can assume it, this is an Nth degree polynomial centered at a. And these two things
are equal to each other. So this is going to be equal to zero. And we see that right over here. The distance between the
two functions is zero there. Now let's think about something else. Let's think about what the derivative of the error function evaluated at a is. Well that's going to be the
derivative of our function at a minus the first derivative of our polynomial at a. And if we assume that this is higher than degree one, we
know that these derivates are going to be the same at a. You can try to take the
first derivative here. If you take the first
derivative of this whole mess-- And this is actually why Taylor
polynomials are so useful, is that up to and including
the degree of the polynomial when you evaluate the derivatives
of your polynomial at a they're going to be the same as the derivatives of the function at a. And that's what starts to
make it a good approximation. But if you took a derivative here, this term right here will
disappear, it'll go to zero. I'll cross it out for now. This term right over here
will just be f prime of a and then all of these other
terms are going to be left with some type of an x minus a in them. And so when you evaluate it at a, all the terms with an x minus a disappear, because you have an a minus a on them. This one already disappeared and you're literally just left with P prime of a will equal f prime of a. And we've seen that before.
So let me write that. So because we know that P prime of a is equal to f prime of a, when you evaluate the error function, the derivative of the error function at a, that also is going to be equal to zero. And this general property right over here, is true up to an including N. So let me write this down. So we already know that P
of a is equal to f of a. We already know that P prime
of a is equal to f prime of a. This really comes straight
out of the definition of the Taylor polynomials. And this is going to be true all the way until the Nth derivative
of our polynomial is going, evaluated at a, not everywhere,
just evaluated at a, is going to be equal to the Nth derivative of our function evaluated at a. So what that tells us is
that we can keep doing this with the error function all
the way to the Nth derivative of the error function evaluated at a is going to be equal to, well that's just going
to be the Nth derivative of f evaluated at a,
minus the Nth derivative of our polynomial evaluated at a. And we already said that
these are going to be equal to each other up
to the Nth derivative when we evaluate them at a. So these are all going
to be equal to zero. So this is an interesting property and it's also going to
be useful when we start to try to bound this error function. And that's the whole
point of where I'm going with this video and
probably the next video, is we're gonna try to bound it so we know how good of an estimate we have. Especially as we go further and further from where we are centered. >From where are approximation is centered. Now let's think about when we
take a derivative beyond that. So let's think about what
happens when we take the N plus oneth derivative. What's a good place to write? Well I have some screen
real estate right over here. What is the N plus oneth derivative of our error function? And not even if I'm just evaluating at a. If I just say generally,
the error function E of x, what's the N plus oneth derivative of it? Well it's going to be the
N plus oneth derivative of our function minus the N plus oneth derivative of our-- We're not just evaluating
at a here either. Let me write a x there. I'm literally just taking
the N plus oneth derivative of both sides of this
equation right over here. So it's literally the
N plus oneth derivative of our function minus the
N plus oneth derivative of our Nth degree polynomial. The N plus oneth derivative
of our Nth degree polynomial. I could write a N here,
I could write an a here to show it's an Nth degree centered at a. Now, what is the N plus onethe derivative of an Nth degree polynomial? And if you want some hints, take the second derivative
of y is equal to x. It's a first degree polynomial,
take the second derivative, you're gonna get zero. Take the third derivative
of y is equal to x squared. The first derivative is 2x,
the second derivative is 2, the third derivative is zero. In general, if you take
an N plus oneth derivative of an Nth degree polynomial, and you could prove it for yourself, you could even prove it generally but I think it might make
a little sense to you, it's going to be equal to zero. It is going to be equal to zero. So this thing right here, this
is an N plus oneth derivative of an Nth degree polynomial. This is going to be equal to zero. Let me write this over here. The N plus oneth derivative
of our error function or our remainder function,
we could call it, is equal to the N plus oneth derivative of our function. And so, what we could do now and we'll probably have to
continue this in the next video, is figure out, at least can we bound this? Can we bound this and if
we are able to bound this, if we're able to figure out an upper bound on its magnitude-- So actually, what we want to do is, we wanna bound its overall magnitude. We wanna bound its absolute value. If we can determine that it is less than or equal to some value M, so if we can actually bound it, maybe we can do a little bit of calculus, we could keep integrating it and maybe we can go back
to the original function and bound that in some way. If we do know some type of
bound like this over here. So I'll take that up in the next video.