Lesson 12: Taylor and Maclaurin Polynomials (part 2)

Welcome back! Today is the second part of a two-part lesson on Taylor polynomials. You can see part 1 in Lesson 11 (link here). We’ll revisit Taylor polynomials when we see Taylor series in Lesson 20.

Learning goals:

Describe the procedure for finding a Taylor polynomial of a given order for a function.
Explain the relationship between the family of Taylor polynomials and the function.
Explain the meaning and significance of Taylor’s theorem with remainder.

Topic:

Volume 2, Section 6.3 Taylor and Maclaurin Series (link to textbook section)

WeBWorK:

Series – Taylor and Maclaurin Polynomials

Motivating question

We saw in Lesson 11 that Taylor polynomials approximate a function near the center. Today we ask: how good is that approximation? We also begin to ask: where is the approximation good? We’ll make this second question more precise in Lesson 20.

Reminders

Don’t forget that a Taylor polynomial centered at $a = 0$ is sometimes called a Maclaurin polynomial. We will sometimes use the notation $p_{N} (x)$ and sometimes use the notation $T_{N} (x)$ to represent a degree $N$ Taylor polynomial and we will sometimes use $c$ and sometimes use $a$ to represent the center.

Warmup exercise 1

Find the Taylor polynomial of degree $4$ for $f (x) = \cos (x)$ centered at $a = 0$ .

Show answer 1

Jump to the 4-minute mark of Video 1 below. Notice that this solution shows the degree 5 Taylor polynomial, while we were asked for the degree 4 Taylor polynomial.

Video 1

The video shows us that $T_{5} (x) = 1 - \frac{1}{2} x^{2} + \frac{1}{24} x^{4}$ . In particular, there is no degree $5$ term because the Taylor coefficient $\frac{f^{(5)} (a)}{5!}$ is zero. This means that the degree 4 Taylor polynomial is equal to the degree $5$ Taylor polynomial, so $T_{4} (x) = 1 - \frac{1}{2} x^{2} + \frac{1}{24} x^{4}$ as well.

Warmup exercise 2

Find the Taylor polynomial of degree $3$ for $f (x) = e^{x}$ centered at $a = 0$ .

Show answer 2

We’re lucky because $f (x) = e^{x}$ is one of the easiest functions to differentiate. Since we’re looking for the degree $3$ Taylor polynomial, we’ll need the first three derivatives: $f (x) = e^{x}$ , $f^{'} (x) = e^{x}$ , $f ” (x) = e^{x}$ , and $f^{(3)} (x) = e^{x}$ . Since the Taylor polynomial is centered at $a = 0$ , we need to calculate $f (0) = e^{0} = 1$ , $f^{'} (0) = e^{0} = 1$ , $f ” (0) = e^{0} = 1$ , and $f^{(3)} (0) = e^{0} = 1$ . Now we can plug our values into the formula for the degree $3$ Taylor polynomial:

$T_{3} (x) = f (a) + f^{'} (a) (x - a) + \frac{f ” (a)}{2} (x - a)^{2} + \frac{f^{(3)} (a)}{3!} (x - a)^{3}$

$= 1 + x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3} .$

We know that Taylor polynomials approximate the functions they’re built from, We know that the nearer to the center, the better the approximation. And we know that the higher the degree, the better the approximation. Now we want to make this more precise.

Taylor’s Theorem with Remainder: Let $f (x)$ be a function that can be differentiated at least $n + 1$ times on an interval $I$ containing $a$ . Let $f (x) = T_{n} (x) + R_{n} (x)$ (so $R_{n} (x)$ is the remainder or error function). Then for each $x$ -value in $I$ , there exists a $c$ between $a$ and $x$ such that $R_{n} (x) = \frac{f^{(n + 1)} (c)}{(n + 1)!} (x - a)^{n + 1} .$

Let $M$ be an upper bound for $f (x)$ on the interval $I$ . Then for all $x$ -values in the interval $I$ ,

$R_{n} (x) \leq \frac{M}{(n + 1)!} | x - a |^{n + 1} .$

Video 2 below takes us through what this means and how to use it in two examples based on the warmup exercises above.

Video 2

Jump to minute 3:42 in Video 3 below (you saw the first part of this video in Lesson 11 but you can re-watch for a reminder) to see the application of the remainder theorem to the example we discussed in Lesson 11: $f (x) = \ln (1 + x)$ , centered at $a = 0$ .

Video 3

Looking ahead…

Take a look back at this interactive Desmos graph from Lesson 11 (link to graph here). Click the grey circles on the left to display graphs of many Taylor polynomials of $f (x) = \ln (1 + x)$ centered at $a = 0$ up to degree 80. Notice how the behavior of the family of graphs is different for different $x$ -values. Ask yourself two questions:

If finite-degree Taylor polynomials approximate $f (x)$ near the center, and the approximation gets better with higher-degree Taylor polynomials, what would “the best approximation” be and how might we build it? What would its graph look like?
Notice how the family of graphs of Taylor polynomials behave differently for different values of $x$ as the degree increases. In particular, what is happening to the graphs between $x = - 1$ and $x = 1$ ? What is happening outside this interval? How can you think of this behavior in light of your answer to question 1?

Additional Video Resources

Taylor and Maclaurin Polynomials

Exit ticket

Let $f (x) = x^{2} e^{x} + x$ . Find the Taylor polynomial of degree 4 for $f (x)$ centered at $a = 1$ .

Lesson 12: Taylor and Maclaurin Polynomials (part 2)

Learning goals:

Motivating question

Reminders

Looking ahead…

Additional Video Resources

MAT 1575 Calculus II Course Hub

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The OpenLab at City Tech:A place to learn, work, and share

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