Maclaurin Series: Common Series & Examples

A Maclaurin series is a Taylor series centered at 0: $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$ . The ones to know are the series for $e^x$ , sin x, cos x, $\frac{1}{1 - x}$ , ln(1 + x) and arctan x. To get a new series, substitute into one of these, such as −x² for x in the series for $e^x$ .

Before this lesson: Taylor Series, Power Series and Taylor Polynomials

Why Maclaurin series are so useful

Centering at 0 makes the derivatives easy to evaluate, and the series use plain powers of x. A handful of these series can be memorized, and then hundreds of other series can be built from them without taking a single derivative.

Calculators use the same idea: to find sin 0.3, they add a few terms of the sine series. The terms shrink so quickly that a few are enough.

Graph of y = sin x (thick navy curve) for −2π ≤ x ≤ 2π with Maclaurin polynomials P₁(x) = x (gray dashed line), P₃(x) = x − x³/3! (solid green) and P₇(x), the sum through x⁷/7! (dotted orange). Each added term keeps the partial sum on the sine curve over a wider interval around 0. — More terms of the sine series hug the curve over a wider stretch around x = 0.

Common Maclaurin series

Function	Maclaurin series	Valid for
$e^x$	$\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$	every x
$\sin x$	$\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$	every x
$\cos x$	$\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$	every x
$\frac{1}{1 - x}$	$\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + \cdots$	−1 < x < 1
$\ln(1 + x)$	$\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$	−1 < x ≤ 1
$\arctan x$	$\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n + 1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots$	−1 ≤ x ≤ 1

Patterns help you remember them. Sine is odd, so it has only odd powers; cosine is even, so it has only even powers. The ln(1 + x) and arctan x series look like the sine and cosine pattern but divide by n instead of n!.

How to build a new Maclaurin series

Find the known series that matches the outer function.
Replace every x in that series with the new inside expression, such as 2x or −x².
Multiply or divide every term by any outside factor, such as x or x².
Change the interval to match: for example, |−x²| < 1 instead of |x| < 1.

Maclaurin series examples

Example 1: The series for eˣ

Find the Maclaurin series of $e^x$ and show that it converges for every x.

f^{(n)}(0) = e^0 = 1

Every derivative of eˣ is eˣ (see derivatives of exponential functions).

c_n = \dfrac{1}{n!}

Divide by n!.

e^x = \sum_{n=0}^{\infty} \dfrac{x^n}{n!}

Write the series in powers of x.

\left|\dfrac{x^{n+1}/(n + 1)!}{x^n/n!}\right| = \dfrac{|x|}{n + 1} \to 0

Ratio test: the limit is 0 for every x, so R = ∞.

Answer: $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$ for every x. At x = 1 this gives $e = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \cdots \approx 2.718$ .

Example 2: Substitute into the series for eˣ

Find the Maclaurin series of $e^{-x^2}$ .

e^u = \sum_{n=0}^{\infty} \dfrac{u^n}{n!}

Start from the known series, with u in place of x.

e^{-x^2} = \sum_{n=0}^{\infty} \dfrac{(-x^2)^n}{n!}

Replace u with −x².

\sum_{n=0}^{\infty} \dfrac{(-x^2)^n}{n!} = \sum_{n=0}^{\infty} \dfrac{(-1)^n x^{2n}}{n!}

(−x²)ⁿ = (−1)ⁿx²ⁿ (see exponent rules).

Answer: $e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots$ for every x.

Example 3: Substitute, then multiply

Find the Maclaurin series of $x\sin(2x)$ and its first three nonzero terms.

\sin(2x) = \sum_{n=0}^{\infty} \dfrac{(-1)^n (2x)^{2n+1}}{(2n+1)!}

Replace x with 2x in the sine series.

\sum_{n=0}^{\infty} \dfrac{(-1)^n (2x)^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} \dfrac{(-1)^n 2^{2n+1} x^{2n+1}}{(2n+1)!}

(2x)²ⁿ⁺¹ = 2²ⁿ⁺¹x²ⁿ⁺¹.

x\sin(2x) = \sum_{n=0}^{\infty} \dfrac{(-1)^n 2^{2n+1} x^{2n+2}}{(2n+1)!}

Multiply every term by x: add 1 to each exponent.

n = 0, 1, 2: \quad \dfrac{2x^2}{1!}, \; -\dfrac{8x^4}{3!}, \; \dfrac{32x^6}{5!}

Put in n = 0, 1 and 2.

2x^2 - \dfrac{4}{3}x^4 + \dfrac{4}{15}x^6

Simplify: 8/6 = 4/3 and 32/120 = 4/15.

Answer: $x\sin(2x) = 2x^2 - \frac{4}{3}x^4 + \frac{4}{15}x^6 - \cdots$ for every x.

Example 4: Use a series to find a limit

Find $\lim_{x \to 0} \dfrac{\sin x - x}{x^3}$ .

\sin x - x = \left(x - \dfrac{x^3}{6} + \dfrac{x^5}{120} - \cdots\right) - x

Replace sin x with its series; 3! = 6 and 5! = 120.

\sin x - x = -\dfrac{x^3}{6} + \dfrac{x^5}{120} - \cdots

The x terms cancel.

\dfrac{\sin x - x}{x^3} = -\dfrac{1}{6} + \dfrac{x^2}{120} - \cdots

Divide every term by x³.

\lim_{x \to 0} \left(-\dfrac{1}{6} + \dfrac{x^2}{120} - \cdots\right) = -\dfrac{1}{6}

Every term with a power of x goes to 0.

Answer: the limit is −1/6. This is faster than using L’Hôpital’s rule three times.

Example 5: Continuous interest (word problem)

A bank pays 4% interest compounded continuously, so $5,000 grows to $5000e^{0.04}$ dollars in one year. Use the first three terms of the series for $e^x$ to estimate the balance.

e^{0.04} \approx 1 + 0.04 + \dfrac{(0.04)^2}{2}

Use 1 + x + x²/2 with x = 0.04.

1 + 0.04 + \dfrac{0.0016}{2} = 1 + 0.04 + 0.0008 = 1.0408

Square, divide, then add.

5000(1.0408) = 5204

Multiply by the starting amount.

5000 \cdot \dfrac{(0.04)^3}{6} \approx 0.05

The next term is the size of the error, roughly.

Answer: about $5,204.00. The next term adds only about 5 cents; the exact balance is $5,204.05.

Common mistakes

Mistake	Example of the error	Fix
Substituting into only part of the term	Writing $e^{2x} = \sum \frac{2x^n}{n!}$	Replace all of x with 2x: $\sum \frac{(2x)^n}{n!} = \sum \frac{2^n x^n}{n!}$ .
Mixing up the sine and cosine powers	Writing $\sin x = 1 - \frac{x^2}{2!} + \cdots$	Sine is odd: x, x³, x⁵. Cosine is even: 1, x², x⁴.
Confusing (2n)! with 2n!	Writing (2n)! = 2 · n!	(2n)! is the factorial of 2n: for n = 2 it is 4! = 24, not 2 · 2! = 4.
Forgetting where a series is valid	Using the ln(1 + x) series at x = 3	It works only for −1 < x ≤ 1.
Not changing the interval after substituting	Saying $\frac{1}{1 + x^2} = \sum (-1)^n x^{2n}$ for all x	The geometric series needs \|−x²\| < 1, so \|x\| < 1.
Dropping the sign pattern	Writing $\cos x = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots$	The signs alternate: $1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$ .

Related methods

Where ln(1 + x) and arctan x come from

These two series come from the geometric series. Integrating $\frac{1}{1 + x} = \sum (-1)^n x^n$ term by term gives ln(1 + x), and integrating $\frac{1}{1 + x^2}$ gives arctan x. The next lesson shows how to differentiate and integrate power series this way.

Series for any center

When you need a series centered somewhere other than 0, use the general Taylor series formula with powers of (x − a).

Practice problems

Find the Maclaurin series of $e^{3x}$ .
Find the Maclaurin series of $\cos(x^2)$ .
Find the Maclaurin series of $x^2 e^x$ .
Find the Maclaurin series of $\dfrac{\sin x}{x}$ (for x ≠ 0).
Find the Maclaurin series of ln(1 + 3x) and where it is valid.
Find the first three nonzero terms of the Maclaurin series of arctan(2x).
Use a series to find $\lim_{x \to 0} \dfrac{1 - \cos x}{x^2}$ .
Use the Maclaurin series of sin(2x) to find the fifth derivative of sin(2x) at x = 0.
What function has the Maclaurin series $\sum_{n=0}^{\infty} \dfrac{(-1)^n 9^n x^{2n}}{(2n)!}$ ?
A radioactive sample keeps the fraction $e^{-0.06}$ of its mass after 3 years. Use three terms of the series for $e^x$ to estimate this fraction.

Show answers

e^{3x} = \sum_{n=0}^{\infty} \dfrac{(3x)^n}{n!}

Replace x with 3x.

\sum_{n=0}^{\infty} \dfrac{(3x)^n}{n!} = \sum_{n=0}^{\infty} \dfrac{3^n x^n}{n!}

(3x)ⁿ = 3ⁿxⁿ.

Answer: $\sum \dfrac{3^n x^n}{n!} = 1 + 3x + \dfrac{9x^2}{2} + \cdots$ for every x

\cos(x^2) = \sum_{n=0}^{\infty} \dfrac{(-1)^n (x^2)^{2n}}{(2n)!}

Replace x with x² in the cosine series.

\sum_{n=0}^{\infty} \dfrac{(-1)^n (x^2)^{2n}}{(2n)!} = \sum_{n=0}^{\infty} \dfrac{(-1)^n x^{4n}}{(2n)!}

(x²)²ⁿ = x⁴ⁿ.

Answer: $1 - \dfrac{x^4}{2!} + \dfrac{x^8}{4!} - \cdots$ for every x

x^2 e^x = x^2 \sum_{n=0}^{\infty} \dfrac{x^n}{n!}

Start from the series for eˣ.

x^2 \sum_{n=0}^{\infty} \dfrac{x^n}{n!} = \sum_{n=0}^{\infty} \dfrac{x^{n+2}}{n!}

Multiply every term by x².

Answer: $x^2 + x^3 + \dfrac{x^4}{2!} + \dfrac{x^5}{3!} + \cdots$ for every x

\dfrac{\sin x}{x} = \dfrac{1}{x}\sum_{n=0}^{\infty} \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}

Start from the sine series.

\dfrac{1}{x}\sum_{n=0}^{\infty} \dfrac{(-1)^n x^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} \dfrac{(-1)^n x^{2n}}{(2n+1)!}

Divide every term by x: subtract 1 from each exponent.

Answer: $1 - \dfrac{x^2}{3!} + \dfrac{x^4}{5!} - \cdots$

\ln(1 + 3x) = \sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}(3x)^n}{n} = \sum_{n=1}^{\infty} \dfrac{(-1)^{n-1} 3^n x^n}{n}

Replace x with 3x.

-1 \lt 3x \le 1 \quad \Leftrightarrow \quad -\dfrac{1}{3} \lt x \le \dfrac{1}{3}

Divide the interval for ln(1 + x) by 3.

Answer: $3x - \dfrac{9x^2}{2} + 9x^3 - \cdots$ for −1/3 < x ≤ 1/3

\arctan(2x) = 2x - \dfrac{(2x)^3}{3} + \dfrac{(2x)^5}{5} - \cdots

Replace x with 2x in the arctan series.

2x - \dfrac{8x^3}{3} + \dfrac{32x^5}{5}

(2x)³ = 8x³ and (2x)⁵ = 32x⁵.

Answer: $2x - \dfrac{8}{3}x^3 + \dfrac{32}{5}x^5$

1 - \cos x = \dfrac{x^2}{2} - \dfrac{x^4}{24} + \cdots

Subtract the cosine series from 1; the 1s cancel.

\dfrac{1 - \cos x}{x^2} = \dfrac{1}{2} - \dfrac{x^2}{24} + \cdots

Divide by x².

Answer: the limit is 1/2

\sin(2x) = 2x - \dfrac{8x^3}{6} + \dfrac{32x^5}{120} - \cdots

Replace x with 2x in the sine series.

\dfrac{f^{(5)}(0)}{5!} = \dfrac{32}{120}

The x⁵ coefficient of a Maclaurin series is f⁽⁵⁾(0)/5!.

f^{(5)}(0) = 32

Multiply both sides by 5! = 120.

Answer: 32

\dfrac{(-1)^n 9^n x^{2n}}{(2n)!} = \dfrac{(-1)^n (3x)^{2n}}{(2n)!}

9ⁿx²ⁿ = (3x)²ⁿ.

\sum_{n=0}^{\infty} \dfrac{(-1)^n (3x)^{2n}}{(2n)!} = \cos(3x)

This is the cosine series with 3x in place of x.

Answer: cos(3x)

10.

e^{-0.06} \approx 1 - 0.06 + \dfrac{(0.06)^2}{2}

Use 1 + x + x²/2 with x = −0.06.

1 - 0.06 + \dfrac{0.0036}{2} = 1 - 0.06 + 0.0018 = 0.9418

Square, divide, then combine.

Answer: about 0.9418, or 94.18% of the mass (true value 0.94176)

Maclaurin series FAQ

What is a Maclaurin series?

It is a Taylor series centered at x = 0: $\sum \frac{f^{(n)}(0)}{n!}x^n$ . It writes a function as an infinite polynomial in powers of x.

Which Maclaurin series should I memorize?

Learn the series for $e^x$ , sin x, cos x and $\frac{1}{1 - x}$ , with their intervals. The series for ln(1 + x) and arctan x are also worth knowing, since they come up often.

How do you find a Maclaurin series by substitution?

Start from a known series, replace every x with the new expression, then simplify. For example, $e^{2x} = \sum \frac{(2x)^n}{n!} = \sum \frac{2^n x^n}{n!}$ .

What is the Maclaurin series for eˣ?

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ , and it is valid for every real number x.

Summary

A Maclaurin series is a Taylor series centered at 0.
Know the series for eˣ, sin x, cos x and 1/(1 − x), plus ln(1 + x) and arctan x.
Build new series by substituting for x and multiplying by powers of x.
Update the interval after substituting.
Series give quick estimates and make some limits easy.

Previous lesson: Taylor Series

Next lesson: Differentiating and Integrating Power Series

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