When do I use Maclaurin instead of Taylor Series?

Use Maclaurin when expanding around x = 0; use Taylor Series to expand around any other point. Maclaurin is a special case of Taylor with center at zero.

Why do some derivatives vanish at zero, like for \sin(x)?

Sine is an odd function: \sin(-x) = -\sin(x). Odd functions have all even derivatives equal to zero at the origin, which forces even powers out of the series and leaves only odd powers.

How many terms of the series should I include?

It depends on your accuracy goal and how close x is to zero. For most practical problems, 3–5 terms suffice when x is small. Use the Ratio Test or an error bound formula to determine how many terms you need for your desired precision.

Can every function have a Maclaurin series?

No. The function must be infinitely differentiable at x = 0 with all derivatives finite there. For example, |x| has a corner at zero and is not differentiable, so it has no Maclaurin series. Additionally, even if the series exists, it may not converge for all x.

Maclaurin Series

A Maclaurin Series expresses a function as an infinite polynomial using its derivatives at zero, letting you approximate functions without a calculator.

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n

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What each symbol means

What Maclaurin Series takes

Symbol	Meaning
$f$	The original function being approximated; $f^{(n)}(0)$ means the $n$ -th derivative of $f$ evaluated at zero, so if $f$ is not differentiable at zero, the series does not exist.
$x$	The input variable; the series is a polynomial in $x$ , and it converges only for values within a certain distance of zero (the radius of convergence).
$n$	The index running from zero to infinity, counting which derivative term appears; $n!$ (n factorial) grows rapidly, which is why later terms contribute less to the sum.

When to use it

Reach for a Maclaurin Series when you need to approximate a transcendental function or evaluate it near $x = 0$ without a calculator.

Level

Usually taught in: Calculus II

Worked examples

1. Find the first four terms of the Maclaurin series for $e^x$

Problem

Find the first four non-zero terms of the Maclaurin series for

f(x) = e^x

$f(x) = e^x \Rightarrow f(0) = 1, \quad f'(x) = e^x \Rightarrow f'(0) = 1$
Identify the function and compute its derivative at $x = 0$ .
$f''(x) = e^x \Rightarrow f''(0) = 1, \quad f'''(x) = e^x \Rightarrow f'''(0) = 1$
Continue computing derivatives; all derivatives of $e^x$ equal $e^x$ , so they all equal 1 at $x = 0$ .
$\sum_{n=0}^{3} \frac{f^{(n)}(0)}{n!}x^n = \frac{1}{0!}x^0 + \frac{1}{1!}x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3$
Substitute the derivatives into the Maclaurin formula.
$= 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$
Evaluate the factorials: $0! = 1$ , $1! = 1$ , $2! = 2$ , $3! = 6$ .

Answer: $e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$

Because all derivatives of $e^x$ equal $e^x$ , and $e^0 = 1$ , every term has numerator 1. The factorial in the denominator prevents divergence and ensures each term's contribution decreases as $n$ increases.

2. Find the Maclaurin series for $\sin(x)$ with alternating pattern

Problem

Find the first five non-zero terms of the Maclaurin series for

f(x) = \sin(x)

$f(x) = \sin(x), \quad f'(x) = \cos(x), \quad f''(x) = -\sin(x), \quad f'''(x) = -\cos(x), \quad f^{(4)}(x) = \sin(x)$
Compute successive derivatives, which cycle with period 4.
$f(0) = 0, \quad f'(0) = 1, \quad f''(0) = 0, \quad f'''(0) = -1, \quad f^{(4)}(0) = 0, \quad f^{(5)}(0) = 1$
Evaluate each derivative at $x = 0$ ; even derivatives are zero and odd derivatives alternate between 1 and $-1$ .
$\sum_{n=0}^{5} \frac{f^{(n)}(0)}{n!}x^n = 0 + x + 0 - \frac{x^3}{3!} + 0 + \frac{x^5}{5!}$
Substitute into the Maclaurin formula; even-indexed terms vanish because their numerators are zero.
$= x - \frac{x^3}{6} + \frac{x^5}{120}$
Evaluate the factorials: $3! = 6$ and $5! = 120$ .

Answer: $\sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots$

Sine is an odd function, so its Maclaurin series contains only odd powers of $x$ . The alternating signs come from the pattern of odd derivatives: at $x = 0$ , they cycle as $1, -1, 1, -1, \ldots$

3. Approximate exponential cooling with Maclaurin series

Problem

A hot cup cools according to

T(t) = 100e^{-t}

, where

T

is temperature in Celsius and

t

is time in seconds. Use the first four terms of the Maclaurin series to approximate the temperature at

t = 0.2

seconds.

$T(0) = 100e^0 = 100, \quad T'(t) = -100e^{-t}, \quad T'(0) = -100, \quad T''(0) = 100, \quad T'''(0) = -100$
Evaluate $T$ and its derivatives at $t = 0$ using the chain rule repeatedly.
$T(t) = \frac{100}{0!} + \frac{-100}{1!}t + \frac{100}{2!}t^2 + \frac{-100}{3!}t^3 + \cdots = 100 - 100t + 50t^2 - \frac{100}{6}t^3 + \cdots$
Substitute the derivatives into the Maclaurin formula; simplify factorials: $\frac{100}{2} = 50$ and $\frac{100}{6} \approx 16.67$ .
$T(0.2) \approx 100 - 100(0.2) + 50(0.04) - \frac{100}{6}(0.008) = 100 - 20 + 2 - \frac{0.8}{6} = 82 - 0.133 \approx 81.87$
Substitute $t = 0.2$ : $(0.2)^2 = 0.04$ , $(0.2)^3 = 0.008$ , and $\frac{0.8}{6} \approx 0.133$ .

Answer: $T(0.2) \approx 81.87 \text{ °C}$

The Maclaurin series converts the transcendental function into a polynomial, avoiding the need for a calculator to evaluate $e^{-0.2}$ . The approximation is accurate near $t = 0$ because higher-order terms are negligible for small $t$ .

Common mistakes

Where Maclaurin Series usually goes wrong

The mistake	Why it is wrong	The fix
Writing the series starting at $n = 1$ and forgetting the constant term $f(0)$ .	Misreading the formula's index: $\sum_{n=0}^{\infty}$ starts at $n = 0$ , not $n = 1$ .	Always include the $n = 0$ term, which equals $f(0)$ ; this is the constant term.
Using the derivative function itself instead of evaluating it at zero, e.g., writing $\sin(x) = \cos(x) + \cdots$ .	Confusing the notation $f^{(n)}(0)$ with $f^{(n)}(x)$ : the zero in parentheses means evaluate at $x = 0$ , not leave the derivative as a function.	Always compute $f^{(n)}(0)$ as a number (the value of the $n$ -th derivative at $x = 0$ ) before placing it in the series.
Attempting to write a Maclaurin series for $f(x) = \ln(x)$ centered at $x = 0$ .	The function $\ln(x)$ is undefined at $x = 0$ , so its derivatives do not exist there; a Maclaurin series requires the function and all its derivatives to be defined at the center.	Before constructing a Maclaurin series, check that $f(0)$ is defined and all relevant derivatives exist at $x = 0$ .

Tips and when to use something else

The Maclaurin series converges only within a finite interval around $x = 0$ , called the radius of convergence; for $e^x$ it is all real numbers, but for $\frac{1}{1-x}$ it is only $|x| < 1$ .
If you need to expand a function around a point other than zero (say, near $x = 2$ ), use the Taylor Series instead—Maclaurin is the special case where the center is zero.
Maclaurin series are most powerful when evaluating transcendental functions (like $\sin(x)$ or $e^x$ ) at small values without a calculator.
To verify your work, check that the constant term equals $f(0)$ and that the derivative of your series at $x = 0$ matches $f'(0)$ .

Related formulas

Taylor SeriesA Taylor Series expresses a function as an infinite polynomial centered at a point, approximating smooth functions using just derivatives.FactorialFactorial multiplies all positive integers up to n to find permutations. Use it to count the number of ways to arrange distinct objects in order.Sigma NotationSigma notation compresses long sums into a compact formula, letting you write and calculate the sum of many terms efficiently and accurately.Infinite Geometric SeriesInfinite Geometric Series gives you the sum of all terms in a geometric sequence when the common ratio lies strictly between negative one and one.

Frequently asked questions

When do I use Maclaurin instead of Taylor Series?: Use Maclaurin when expanding around $x = 0$ ; use Taylor Series to expand around any other point. Maclaurin is a special case of Taylor with center at zero.
Why do some derivatives vanish at zero, like for $\sin(x)$ ?: Sine is an odd function: $\sin(-x) = -\sin(x)$ . Odd functions have all even derivatives equal to zero at the origin, which forces even powers out of the series and leaves only odd powers.
How many terms of the series should I include?: It depends on your accuracy goal and how close $x$ is to zero. For most practical problems, 3–5 terms suffice when $x$ is small. Use the Ratio Test or an error bound formula to determine how many terms you need for your desired precision.
Can every function have a Maclaurin series?: No. The function must be infinitely differentiable at $x = 0$ with all derivatives finite there. For example, $|x|$ has a corner at zero and is not differentiable, so it has no Maclaurin series. Additionally, even if the series exists, it may not converge for all $x$ .

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Reviewed 2026-09-18