Main content

### Course: Integral Calculus > Unit 5

Lesson 11: Taylor and 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
- Visualizing Taylor polynomial approximations

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Worked example: Maclaurin polynomial

Finding the second degree Maclaurin polynomial of 1/√(x+1).

## Want to join the conversation?

- I still confuse about the function of X in Taylor formula.

This formula uses for estimate at point A, so what is the X use for?(7 votes)- a is the point where you base the approximation, but you can vary x in order to get an approximation of the function itself using the polynomial.

For example, f(a) = P(a) (because you know the value of a), but f(x) ~ P(x) (because P(x) gets you an APPROXIMATION of f(x))(10 votes)

- What kinds of functions are exactly the same as their taylor series and what kinds of functions are not? In other words, what are the conditions for a function to equal its taylor infinite sum?(8 votes)
- Greetings!

If we could take the Taylor Series to an infinite number of terms, it would be exactly equal to the original function.

Since we aren't able to get an infinite number of terms, it is only an increasingly better approximation.

I hope this helps!(4 votes)

- Why would one want to approximate the function with a polynomial at a specific x if you could just fill in x in the original function?(7 votes)
- If you don't have the original function, and are, instead, using a graph, then a polynomial would be necessary. (Am I understanding your question right?)(2 votes)

- When writing out the Taylor/Maclaurin series, what is the purpose of dividing each term by a factorial? How does that make the polynomial a better approximation as opposed to not dividing by the factorial? He probably answered this in the intro videos but I may have missed it or simply didn't understand.(5 votes)
- It comes from the fact that you differentiate several times to get eg the second or third derivative. If you know the value you want eg. the third derivative to have, you need to work out what original term would give you that value when differentiated 3 times.

Say you know at the point you are centering you the third derivative is a, then the original coefficient for the term in the polynomial to give that would be a/(3*2*1). Try for a Maclaurin series: a/(3*2*1) * x^3.

differentiate once: a/(2 * 1) * x^2

differentiate second time: ax

differentiate third time: a(5 votes)

- we take the principle root for f(0), because the function is defined for positive root , isnt it ?(2 votes)
- How can you find
`limit`

toand to**0**using Taylor's polynomial formula?**positive infinity**

like:`lim(x->0)`

of [`(a^x-6)/(x^3)`

] ?(2 votes)- Limit to 0 can be found by replacing functions with their Taylor polynomials to high enough degree to evaluate directly plugging in x=0. So (cos(x)-1) / x^2 would be approximated by (-x^2/2 +x^4/4! -...) / x^2 = - 1/2 + x^2-stuff = -1/2 as x->0.(1 vote)

- what is [1-(-1)^n] as n goes to infinity(1 vote)
- It doesn't converge, it oscillates between 2 and 0.(2 votes)

- Why create a Maclaurin polynomial here? We already know the function, and therefor all its derivatives. What does the Maclaurin (or Taylor) polynomials give us that the function itself does not?(1 vote)
- They can be useful for approximating functions when computing the actual function value would be too time-consuming. This comes up in computer science when trying to optimize a program.

Taylor series are also good for approximating antiderivatives that have no closed form. For example, the integral of e^x^2 cannot be expressed in terms of the standard set of functions (polynomials, trig functions, etc). But we know the first derivative of this function is e^x^2, and we can compute its Taylor series from there in order to approximate this function and understand it better.

Taylor series can also be applied to certain limit problems, since polynomials tend to be easier to understand.(2 votes)

- Does it make sense to have a Taylor polynomial for a function at it's vertical asymptote line?

For Example: A Maclaurin polynomial for f(x)=1/x(1 vote)- No, the Taylor series is a series of successive polynomial approximations at a point. You can't approximate a function at point where it doesn't exist.(2 votes)

- Is the variable "x" affected by the derivative of the function?(1 vote)
- What exactly do you mean by "affected"? x itself won't change as it's the same variable throughout, but with each derivative, x gains an extra power (x to x^2 to x^3 and so on)(2 votes)

## Video transcript

- [Instructor] We're
told that f of x is equal to one over the square root of x plus one, and what we want to figure out is what is the second degree
Maclaurin polynomial of f? And like always, pause this video and see if you could have a go at it. so let's remind ourselves what
a Maclaurin polynomial is, a Maclaurin polynomial is
just a Taylor polynomial centered at zero, so the
form of this second degree Maclaurin polynomial,
and we just have to find this Maclaurin expansion
until our second degree term, it's going to look like this. So p of x, I'm using p for polynomial, it's going to be our f of zero plus, and we could do
that as f of zero times x to the zero power, well,
that's just f of zero. F of zero plus f prime of zero x, plus f prime prime of zero, divided by, we could think of it as two factorial, but
it's really just two, we could think of this as
dividing by one factorial, which is just one, this
dividing by zero factorial, well that's just one again,
so we have f prime prime of zero, the second derivative evaluated zero divided by two, x squared. Now if we wanted a higher
degree we could keep on going, but remember, they're just
asking us for the second degree. So this is the form that we're gonna need, we're gonna have these three terms. So let's see if we can evaluate the function and its derivatives at zero. So f of zero, f of zero
is equal to one over the square root of zero plus one, well that's one over
the square root of one, the principle root of one,
which is positive one, so that's just going to be equal to one. So that right over there is equal to one. Now let's evaluate f prime of x, and then I'll evaluate f prime of zero. F prime of x is equal to, well, one over the square
root of x plus one, this is the same thing as x plus one, let me write it this way, this is the same thing as, actually let me write it down this way. Another way of writing f of
x is this is the same thing as x plus one to the negative one half. And so if I'm thinking
the first derivative of f, well, I could use the chain rule here, the derivative of x plus
one with respect to x, well, that's just going to be one, and then I'll take the
derivative of this whole thing with respect to x plus
one and I'll just use the power rule there, this is
gonna be negative one half, times x plus one, to the, and I decrement the exponent,
negative three halves. And so the first derivative
evaluated of zero is just negative one half, times, if this is zero, zero
plus one is just one, one to the negative three halves, one to the negative three half power, well, that's just going to be one. So this whole thing, f prime of zero is just negative one half. So that is this right over
here is negative one half. And now let's figure out
the second derivative. All right, let me do
this in this green color. So the second derivative
with respect to x, well I did the same thing again, the derivative of x plus
one with respect to x, that's just one, so I just have to take the derivative of the whole thing with respect to x plus one. So I take my exponent, bring it out front, negative three halves times one half, times negative one half, is going to be positive three fourths times x plus one, and then I decrement the
exponent here by one, or by two halves, so that's
gonna be negative five halves, and so the second derivative
evaluated at zero, well if this is equal to
zero you're gonna have one to the negative five
halves which is just one, times three fourths, is
gonna be three fourths. So this part right over
here is three fourths. And so you're gonna have
three fourths divided by two, three fourths divided
by two is three eighths. So our Taylor, I should say,
our Maclaurin polynomial, our second degree Maclaurin polynomial, p of x is going to be equal to, and I'll do it in the same colors, going to be equal to one, plus, maybe I'll just
write it as minus one half, minus one half x, plus three eighths x squared, plus three eighths x squared, and we are done, there you have it. We have our second degree
Maclaurin polynomial of f, which could be used to
provide an approximation for our function, especially
if our x is near zero.