Math Core

A tangent line matches a function's value and slope at one point, and it gives good approximations nearby. Taylor polynomials push that idea further: by also matching the second derivative, the third derivative, and so on, you get polynomials that hug the curve much more closely and over a wider range.

From tangent lines to better approximations

The linearization of $f$ at $x = a$ is $P_1(x) = f(a) + f'(a)(x - a)$ . It agrees with $f$ in value and first derivative at $a$ . To capture how the curve bends, add a quadratic term chosen so the second derivatives also agree:

P_2(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^2.

Check: $P_2''(x) = f''(a)$ , as intended. The $\frac{1}{2}$ is needed because differentiating $(x-a)^2$ twice produces a factor of $2$ . In the same way, differentiating $(x - a)^k$ $k$ times produces $k!$ , which is why factorials appear in the general formula.

Definition

Taylor polynomial

The $n$ th-degree Taylor polynomial for $f$ centered at $x = a$ is

P_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k.

It is the unique polynomial of degree at most $n$ whose value and first $n$ derivatives at $a$ match those of $f$ . When $a = 0$ , it is also called a Maclaurin polynomial.

The coefficient formula

The coefficient of $(x - a)^k$ in a Taylor polynomial is $\dfrac{f^{(k)}(a)}{k!}$ . Read in reverse:

f^{(k)}(a) = k! \times \big(\text{coefficient of } (x - a)^k\big).

That reverse direction is a favorite AP question: given a Taylor polynomial, find a derivative value.

Watching the approximations improve

For $f(x) = e^x$ at $a = 0$ , every derivative equals $e^0 = 1$ , so

P_1(x) = 1 + x, \qquad P_2(x) = 1 + x + \frac{x^2}{2}, \qquad P_3(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}.

Each new term makes the polynomial follow $e^x$ over a wider interval around $0$ .

For example, to estimate $e^{0.5} \approx 1.6487$ : $P_1(0.5) = 1.5$ , $P_2(0.5) = 1.625$ , and $P_3(0.5) \approx 1.6458$ . Each added term cuts the error sharply, because the new term involves a higher power of the small number $0.5$ divided by a larger factorial. How small the error is guaranteed to be is the subject of the next lesson.

y = eˣ with its Taylor polynomials of degree 1, 2 and 3 centered at 0. Higher degree means a closer fit over a wider interval.Open in grapher →

A recipe

Make a table of $f, f', f'', \dots, f^{(n)}$ .
Evaluate each at $x = a$ .
Divide the $k$ th value by $k!$ and attach $(x - a)^k$ .

Keep the terms in powers of $(x - a)$ ; don't expand them. The factored form is easier to read and is what the AP exam expects.

Common mistake

Two classic errors: forgetting the factorials (writing $f''(a)(x-a)^2$ instead of $\frac{f''(a)}{2}(x-a)^2$ ), and writing powers of $x$ instead of $(x - a)$ when the center is not $0$ . Also remember that "the third-degree Taylor polynomial" includes every term up through $(x-a)^3$ , even if some coefficients are zero.

Worked examples

Worked example: A Maclaurin polynomial for sine

Find the fifth-degree Maclaurin polynomial for $f(x) = \sin x$ .

Solution. The derivatives cycle: $\sin x, \cos x, -\sin x, -\cos x, \sin x, \cos x$ . At $x = 0$ they are $0, 1, 0, -1, 0, 1$ . So

P_5(x) = 0 + x + 0 - \frac{x^3}{3!} + 0 + \frac{x^5}{5!} = x - \frac{x^3}{6} + \frac{x^5}{120}.

Worked example: Centered away from zero

Find the third-degree Taylor polynomial for $f(x) = \ln x$ centered at $x = 1$ .

Solution.

$k$	$f^{(k)}(x)$	$f^{(k)}(1)$	$\frac{f^{(k)}(1)}{k!}$
0	$\ln x$	$0$	$0$
1	$x^{-1}$	$1$	$1$
2	$-x^{-2}$	$-1$	$-\frac12$
3	$2x^{-3}$	$2$	$\frac{2}{6} = \frac13$

P_3(x) = (x - 1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3}.

Worked example: Working from given derivative values

A function $f$ has $f(2) = 3$ , $f'(2) = -1$ , $f''(2) = 4$ and $f'''(2) = 12$ . Write the third-degree Taylor polynomial for $f$ about $x = 2$ and use it to approximate $f(2.1)$ .

Solution.

P_3(x) = 3 - (x - 2) + \frac{4}{2}(x-2)^2 + \frac{12}{6}(x-2)^3 = 3 - (x-2) + 2(x-2)^2 + 2(x-2)^3.

With $x - 2 = 0.1$ : $\; P_3(2.1) = 3 - 0.1 + 2(0.01) + 2(0.001) = 2.922$ .

Worked example: Reading derivatives off a polynomial

The third-degree Taylor polynomial for $g$ about $x = -1$ is $P_3(x) = 4 - 5(x+1) + 3(x+1)^2 - \dfrac{2}{3}(x+1)^3$ . Find $g''(-1)$ and $g'''(-1)$ .

Solution. Multiply each coefficient by the matching factorial:

g''(-1) = 2! \cdot 3 = 6, \qquad g'''(-1) = 3! \cdot \left(-\frac23\right) = -4.

Tip

A Taylor polynomial is most accurate near its center. If you need to approximate $f(3.9)$ , center the polynomial at a nearby point where you know the derivatives, such as $a = 4$ , not at $a = 0$ .

Practice

Practice 1

In the third-degree Maclaurin polynomial for $f(x) = e^x$ , what is the coefficient of $x^3$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Find the second-degree Taylor polynomial for $f(x) = \ln x$ centered at $x = 1$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 3

Find the second-degree Taylor polynomial for $f(x) = \sqrt{x}$ centered at $x = 4$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 4

A function $f$ has $f(2) = 3$ , $f'(2) = -1$ , $f''(2) = 4$ and $f'''(2) = 12$ . Use the third-degree Taylor polynomial about $x = 2$ to approximate $f(2.1)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

The third-degree Taylor polynomial for $h$ about $x = 1$ is $P_3(x) = 5 + 2(x - 1) - 3(x - 1)^2 + \dfrac{4}{3}(x-1)^3$ . Find $h''(1)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

For the same polynomial, $P_3(x) = 5 + 2(x - 1) - 3(x - 1)^2 + \dfrac{4}{3}(x-1)^3$ , find $h'''(1)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Use the fourth-degree Maclaurin polynomial for $\cos x$ to approximate $\cos(0.2)$ . Round to five decimal places.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

Which is the third-degree Taylor polynomial for $f(x) = \dfrac{1}{x}$ centered at $x = 1$ ?