What's the difference between a Taylor and a Maclaurin series?

A Maclaurin series is a Taylor series centered at x = 0. Every Maclaurin series is a Taylor series, but Taylor series can be centered anywhere.

Which Maclaurin series should I memorize for AP Calculus BC?

Know the series for e^x, sin x, cos x, and 1/(1 − x). The series for ln(1 + x) is also worth knowing, and it comes from integrating a geometric series.

Why do I have to test the endpoints separately?

At the endpoints the ratio test gives a limit of exactly 1, which is inconclusive. A series can converge at both endpoints, one, or neither.

Taylor and Maclaurin series for AP Calculus BC

Updated November 9, 2026

A Taylor polynomial is a simple polynomial built to match a function at one point. Taylor series come up on nearly every AP Calculus BC exam, often as a full free-response question.

This guide covers the formula, the series worth memorizing, the shortcuts that save you from taking derivatives, and how to find an interval of convergence.

The Taylor polynomial formula

The Taylor series for $f$ centered at $x = a$ is $\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n$ . A series centered at $0$ is called a Maclaurin series. The polynomial of degree $n$ stops at the $(x - a)^n$ term.

From derivative values

A function

f

has

f(2) = 3

f'(2) = -1

f''(2) = 4

, and

f'''(2) = 18

. Write the third-degree Taylor polynomial for

f

about

x = 2

Divide each derivative by the matching factorial: $\frac{4}{2!} = 2$ and $\frac{18}{3!} = 3$ .

Answer: $P_3(x) = 3 - (x - 2) + 2(x - 2)^2 + 3(x - 2)^3$

The series worth memorizing

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$ , for all $x$
$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$ , for all $x$
$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$ , for all $x$
$\frac{1}{1 - x} = 1 + x + x^2 + x^3 + \cdots$ , for $|x| < 1$
$\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$ , for $-1 < x \le 1$

See how close a few terms get. Near the center, even a third-degree polynomial is almost on top of sine.

y = \sin x

y = x - \frac{x^3}{6}

Make new series without derivatives

It's usually faster to change a series you know. Substitute, multiply by a power of $x$ , differentiate, or integrate, term by term.

Substitute

Write the first four nonzero terms of the Maclaurin series for

e^{-x^2}

Replace $x$ with $-x^2$ in the series for $e^x$ : $1 + (-x^2) + \frac{(-x^2)^2}{2!} + \frac{(-x^2)^3}{3!}$ .

Answer: $1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6}$

Integrate

Write the first four nonzero terms of the Maclaurin series for

F(x) = \int_0^x e^{-t^2}\,dt

Integrate the series above term by term from $0$ to $x$ . There's no elementary antiderivative for $e^{-t^2}$ , so a series is the natural tool.

Answer: $x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{x^7}{42}$

Read a derivative from a series

Let

f(x) = x^2\cos x

. Find

f^{(6)}(0)

Multiply the series for cosine by $x^2$ : $x^2 - \frac{x^4}{2} + \frac{x^6}{24} - \cdots$ . The coefficient of $x^6$ equals $\frac{f^{(6)}(0)}{6!}$ , so $f^{(6)}(0) = \frac{720}{24} = 30$ .

Answer: $30$

Finding the interval of convergence

1Use the ratio test and find the values of $x$ that make the limit less than $1$ .
2That gives the radius of convergence and an open interval.
3Test each endpoint separately, by putting it into the series. The ratio test can't decide endpoints.

Check both endpoints

Find the interval of convergence of

\sum_{n=1}^{\infty} \frac{(x - 2)^n}{n \cdot 3^n}

The ratio test gives $\lim_{n \to \infty} \frac{n}{n+1} \cdot \frac{|x - 2|}{3} = \frac{|x - 2|}{3}$ . This is less than $1$ when $|x - 2| < 3$ , so $-1 < x < 5$ .

At $x = 5$ , the series is $\sum \frac{1}{n}$ , the harmonic series, which diverges.
At $x = -1$ , the series is $\sum \frac{(-1)^n}{n}$ , which converges by the alternating series test.

Answer: $-1 \le x < 5$

How close is the approximation?

A Taylor polynomial is only an estimate, and the exam often asks how far off it could be. See Lagrange and alternating series error bounds. To check whether a series converges in the first place, see which series test to use.

Keep going

Practice on exam-style questions

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Questions

What's the difference between a Taylor and a Maclaurin series?: A Maclaurin series is a Taylor series centered at x = 0. Every Maclaurin series is a Taylor series, but Taylor series can be centered anywhere.
Which Maclaurin series should I memorize for AP Calculus BC?: Know the series for e^x, sin x, cos x, and 1/(1 − x). The series for ln(1 + x) is also worth knowing, and it comes from integrating a geometric series.
Why do I have to test the endpoints separately?: At the endpoints the ratio test gives a limit of exactly 1, which is inconclusive. A series can converge at both endpoints, one, or neither.