What is a Maclaurin Series? — BUders Math Concepts

Simple Definition

A Maclaurin series is a special case of the Taylor series where the center point is $a = 0$. Instead of expanding around an arbitrary point, you use everything the function tells you at the origin — its value, its slope, its curvature, and every higher derivative — to build a polynomial that matches it perfectly nearby.

Because $a = 0$, the $(x - a)^n$ terms simplify to just $x^n$, making the formula as clean as it gets.

Maclaurin Series

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

Expanded form

$$f(x) = f(0) + f'(0)\,x + \frac{f''(0)}{2!}\,x^2 + \frac{f'''(0)}{3!}\,x^3 + \frac{f^{(4)}(0)}{4!}\,x^4 + \cdots$$

Every coefficient $\dfrac{f^{(n)}(0)}{n!}$ is fixed once you evaluate the $n$-th derivative at zero. The $n!$ in the denominator is not optional — it cancels exactly the factorial that appears when you differentiate $x^n$ repeatedly, keeping the coefficients honest.

Intuition

🔬

A microscope at the origin

Imagine zooming into a function at $x = 0$ with increasingly powerful lenses. The first lens shows you the value $f(0)$ — a flat constant. The second shows the slope $f'(0)$ — now you see a line. Each new lens reveals another layer of curvature, encoded by a higher derivative.

The Maclaurin series is the result of looking through all infinitely many lenses at once. Near $x = 0$, the polynomial built from these observations is indistinguishable from the original function.

⚡

Why $a = 0$ specifically?

Most standard functions — $e^x$, $\sin x$, $\cos x$ — are perfectly well-defined at the origin and their derivatives there turn out to be simple integers or zero. This makes the coefficients clean and the series easy to memorize and apply. Centering elsewhere (Taylor series) only becomes necessary when $f$ is singular at 0, like $\ln x$.

How to Derive One

The method is always the same three steps:

Step 1. Compute $f(0),\, f'(0),\, f''(0),\, \ldots$ until a pattern appears.
Step 2. Write each coefficient as $\dfrac{f^{(n)}(0)}{n!}$.
Step 3. Express as a sum $\sum_{n=0}^\infty c_n x^n$ and identify the pattern in closed form.

Example walkthrough — $f(x) = e^x$:

Step

Derivative $f^{(n)}(x)$

At $x=0$

$n = 0$

$e^x$

$1$

$n = 1$

$e^x$

$1$

$n = 2$

$e^x$

$1$

$n = k$

$e^x$ (always)

$1$ (always)

Result

$$e^x = \sum_{n=0}^{\infty} \frac{1}{n!}\,x^n = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \qquad \text{for all } x$$

Example walkthrough — $f(x) = \sin x$:

Step

Derivative $f^{(n)}(x)$

At $x=0$

$n = 0$

$\sin x$

$0$

$n = 1$

$\cos x$

$1$

$n = 2$

$-\sin x$

$0$

$n = 3$

$-\cos x$

$-1$

$n = 4$

$\sin x$ (cycle repeats)

$0$

Even derivatives vanish at zero; only odd powers survive, alternating in sign.

Result

$$\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\,x^{2n+1} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \qquad \text{for all } x$$

Standard Series — Reference

These are the six Maclaurin series worth knowing cold. Every other standard expansion can be derived from them by substitution, differentiation, or integration.

Function	Maclaurin Series	Converges
$e^x$	$\displaystyle 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \sum_{n=0}^{\infty} \frac{x^n}{n!}$	all $x$
$\sin x$	$\displaystyle x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$	all $x$
$\cos x$	$\displaystyle 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$	all $x$
$\ln(1+x)$	$\displaystyle x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}$	$-1 < x \le 1$
$\dfrac{1}{1-x}$	$\displaystyle 1 + x + x^2 + x^3 + \cdots = \sum_{n=0}^{\infty} x^n$	$\|x\| < 1$
$(1+x)^k$	$\displaystyle 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + \cdots$	$\|x\| < 1$

Deriving New Series from Old

You rarely need to start from scratch. Most series you'll encounter can be obtained by applying one of these operations to a known series.

Substitution

Replace $x$ with an expression

Substitute directly into a known series. No differentiation needed.

Known: $e^x = \sum \dfrac{x^n}{n!}$

Replace $x \to -x^2$:
$e^{-x^2} = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}$

Differentiation

Differentiate term by term

If you know the series for $f$, differentiate each term to get the series for $f'$.

From $\dfrac{1}{1-x} = \sum x^n$:

Differentiate:
$\dfrac{1}{(1-x)^2} = \displaystyle\sum_{n=1}^{\infty} n\,x^{n-1}$

Integration

Integrate term by term

Integrate a known series to get one that's harder to derive directly.

From $\dfrac{1}{1+x} = \sum (-1)^n x^n$:

Integrate (with $C=0$):
$\ln(1+x) = \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}$

Multiplication

Multiply two series (Cauchy product)

Multiply series when the function is a product. Truncate to needed degree.

$e^x \cdot \sin x$ to degree 4:

$\approx \!\left(1+x+\tfrac{x^2}{2}+\cdots\right)\!\left(x - \tfrac{x^3}{6}+\cdots\right)$
$= x + x^2 + \tfrac{x^3}{3} - \tfrac{x^5}{30} + \cdots$

Worked Examples

Example 1 — By substitution

Find the Maclaurin series for $f(x) = e^{-x^2}$.

Start from $e^u = \displaystyle\sum_{n=0}^{\infty} \dfrac{u^n}{n!}$

Set $u = -x^2$:

$e^{-x^2} = \displaystyle\sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}$

$= 1 - x^2 + \dfrac{x^4}{2} - \dfrac{x^6}{6} + \cdots$

Converges for all $x$. Used in the Gaussian integral $\int_{-\infty}^\infty e^{-x^2}dx = \sqrt{\pi}$.

Example 2 — By differentiation

Find the Maclaurin series for $f(x) = \cos x$ using $\sin x$.

Start from $\sin x = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$

Differentiate term by term $\bigl(\cos x = (\sin x)'\bigr)$:

$\cos x = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n (2n+1)\, x^{2n}}{(2n+1)!}$

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

$= 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots$ — only even powers, as expected.

Example 3 — By integration

Find the Maclaurin series for $\arctan x$.

Note $\dfrac{d}{dx}\arctan x = \dfrac{1}{1+x^2}$

From geometric series, set $u = -x^2$:
$\dfrac{1}{1+x^2} = \displaystyle\sum_{n=0}^{\infty}(-1)^n x^{2n}$

Integrate (constant $= 0$ since $\arctan 0 = 0$):

$\arctan x = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}$

$= x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \cdots$ Converges for $|x| \le 1$.

Example 4 — Limit via series

Evaluate $\displaystyle\lim_{x \to 0} \frac{\sin x - x}{x^3}$ using Maclaurin.

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

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

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

As $x \to 0$: limit $= \boldsymbol{-\dfrac{1}{6}}$. Faster and cleaner than L'Hôpital's rule.

Common Mistakes

✗ Evaluating derivatives at $x$ instead of $x = 0$

The coefficient of $x^n$ is $\dfrac{f^{(n)}(0)}{n!}$ — the derivative must be evaluated at zero. Writing $f^{(n)}(x)$ as the coefficient gives a function, not a number, and the whole construction breaks.

✗ Dropping the $n!$ denominator

The factorial is essential. Without it, differentiating $\sum c_n x^n$ would not recover $f^{(n)}(0)$. The series for $e^x$ would incorrectly become $1 + x + x^2 + x^3 + \cdots = \frac{1}{1-x}$ — a completely different function.

✗ Using Maclaurin when the function isn't defined at 0

If $f(0)$ is undefined — like $\ln x$ or $\dfrac{1}{x}$ — a Maclaurin series doesn't exist. You need a Taylor series centered at some $a \ne 0$ where the function is defined and smooth.

✗ Ignoring the radius of convergence on substitution

When you substitute $x \to g(x)$ into a series with radius $R$, the new series converges where $|g(x)| < R$, not where $|x| < R$. For example, $\ln(1+x^2)$ converges where $|x^2| \le 1$, i.e. $-1 \le x \le 1$.

✗ Confusing $\sin x$ and $\cos x$ parity

$\sin x$ is an odd function — only odd powers of $x$ appear. $\cos x$ is an even function — only even powers appear. If your derivation gives even powers for $\sin x$, something went wrong with the sign pattern.

Related Concepts

Function	Maclaurin Series	Converges
\(e^x\)	\(\displaystyle 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \sum_{n=0}^{\infty} \frac{x^n}{n!}\)	all \(x\)
\(\sin x\)	\(\displaystyle x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}\)	all \(x\)
\(\cos x\)	\(\displaystyle 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}\)	all \(x\)
\(\ln(1+x)\)	\(\displaystyle x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}\)	\(-1 < x \le 1\)
\(\dfrac{1}{1-x}\)	\(\displaystyle 1 + x + x^2 + x^3 + \cdots = \sum_{n=0}^{\infty} x^n\)	\(\|x\| < 1\)
\((1+x)^k\)	\(\displaystyle 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + \cdots\)	\(\|x\| < 1\)

a Maclaurin Series?

Ready to test your knowledge?